Engineering Mechanics: Statics and Dynamics, 3rd Edition [3 ed.] 9781265255411, 1265255415

Table of contents :
Cover
Title Page
Bookmarks
About the Authors
Dedications
Brief Contents
Statics Table of Contents
Dynamics Table of Contents
Preface
Acknowledgments
1 Introduction to Statics
1.1 Engineering and Statics
1.2 Topics That Will Be Studied in Statics
1.3 ABriefHistory of Statics
Galileo Galilei
Isaac Newton
1.4 Fundamental Principles
Newton’s laws of motion
1.5 Force
1.6 Units and Unit Conversions
Dimensional homogeneity and unit conversions
Prefixes
Angular measure
Small angle approximations
Accuracy of calculations
1.7 Newton’s Law of Gravitation
Relationship between specific weight and
density
1.8 Failure
1.9 Chapter Review
2 Vectors: Force and Position
2.1 Basic Concepts
Introduction—force, position, vectors, and tides
Denoting vectors in figures
Basic vector operations
Performing vector operations
Resolution of a vector into vector components
2.2 Cartesian Representation of Vectors in Two
Dimensions
Introduction—Cartesian representation and a walk
to work
Unit vectors
Cartesian coordinate system
Cartesian vector representation
Addition of vectors using Cartesian components
Position vectors
2.3 Cartesian Representation of Vectors in Three
Dimensions
Right-hand Cartesian coordinate system
Cartesian vector representation
Direction angles and direction cosines
Position vectors
Use of position vectors to write expressions for
force vectors
Some simple structural members
2.4 Vector Dot Product
Dot product using Cartesian components
Determination of the angle between two vectors
Determination of the component of a vector in a particular direction
Determination of the component of a vector perpendicular to a direction
2.5 Vector Cross Product
Cross product using Cartesian components
Evaluation of cross product using determinants
Determination of the normal direction to a
plane
Determination of the area of a parallelogram
Scalar triple product
2.6 Chapter Review
3 Equilibrium of Particles
3.1 Equilibrium of Particles in Two
Dimensions
Free body diagram (FBD)
Modeling and problem solving
Cables and bars
Pulleys
Reactions
3.2 Behavior of Cables, Bars, and Springs
Equilibrium geometry of a structure
Cables
Bars
Modeling idealizations and solution of Σ 𝐹⃗ = 0 ⃗
Springs
3.3 Equilibrium of Particles in Three
Dimensions
Reactions
Solution of algebraic equations
Summing forces in directions other than 𝒙, 𝒚, or 𝒛
3.4 Engineering Design
Objectives of design
Particle equilibrium design problems
3.5 Chapter Review
4 Moment of a Force and Equivalent Force
Systems
4.1 Moment of a Force
Scalar approach
Vector approach
Varignon’s theorem
Which approach should I use: scalar or vector?
4.2 Moment of a Force About a Line
Vector approach
Scalar approach
4.3 Moment of a Couple
Vector approach
Scalar approach
Comments on the moment of a couple
Equivalent couples
Equivalent force systems
Resultant couple moment
Moments as free vectors
4.4 Equivalent Force Systems
Transmissibility of a force
Equivalent force systems
Some special force systems
Wrench equivalent force systems
Why are equivalent force systems called
equivalent
4.5 Chapter Review
5 Equilibrium of Bodies
5.1 Equations of Equilibrium
5.2 Equilibrium of Rigid Bodies in Two
Dimensions
Reactions
Free body diagram (FBD
Alternative equilibrium equations
Gears
Examples of correct FBDs
Examples of incorrect and/or incomplete FBDs
5.3 Equilibrium of Bodies in Two Dimensions—Additional Topics
Why are bodies assumed to be rigid
Treatment of cables and pulleys
Springs
Superposition
Supports and fixity
Static determinacy and indeterminacy
Two-force and three-force members
5.4 Equilibrium of Bodies in Three Dimensions?
Reactions
More on bearings
Scalar approach or vector approach
Solution of algebraic equations
Examples of correct FBDs
Examples of incorrect and/or incomplete FBDs
5.5 Engineering Design
Codes and standards
Design problems
5.6 ChapterReview
6 Structural Analysis and Machines
Structures and machines
6.1 Truss Structures and the Method
of Joints
When may a structure be idealized as a
truss
Method of joints
Zero-force members
Typical truss members
6.2 Truss Structures and the Method of
Sections
Treatment of forces that are not at joints
Static determinacy and indeterminacy
Design considerations
6.3 Trusses in ThreeDimensions
Stability of space trusses and design
considerations
6.4 Frames andMachines
Analysis procedure and free body diagrams (FBDs)
Examples of correct FBDs
Examples of incorrect and/or incomplete FBDs
6.5 Chapter Review
7 Centroids and Distributed Force Systems
7.1 Centroid
Introduction—center of gravity
Centroid of an area
Centroid of a line
Centroid of a volume
Unification of concepts
Which approach should I use: composite shapes or integration?
Finer points: surfaces and lines in three
dimensions
7.2 Center of Mass and Center of Gravity
Center of mass
Center of gravity
7.3 Theorems of Pappus and Guldinus
Area of a surface of revolution
Volume of a solid of revolution
Proof of the Pappus–Guldinus theorems
7.4 Distributed Forces, Fluid and Gas Pressure
Loading
Distributed forces
Distributed forces applied to beams
Fluid and gas pressure
Forces produced by fluids
Forces produced by gases
7.5 Chapter Review
8 Internal Forces
8.1 Internal Forces in Structural Members
Why are internal forces important
Internal forces for slender members in two
dimensions
Internal forces for slender members in three
dimensions
Determination of internal forces
8.2 Internal Forces in Straight Beams
Determination of 𝑉 and 𝑀 using equilibrium
Shear and moment diagrams
8.3 Relations Among Shear, Moment, and
Distributed Force
Relations among 𝑉, 𝑀, and 𝑤
Determination of 𝑉 and 𝑀 using integration
Which approach should I use?
Tips and shortcuts for drawing shear and moment
diagrams
Design considerations
8.4 Chapter Review
9 Friction
9.1 Basic Concepts
A brief history of tribology
A simple experiment
Coulomb’s law of friction
Coefficients of friction
Dry contact versus liquid lubrication
Angle of friction
Problems with multiple contact surfaces
Wedges
Coulomb’s law of friction in three
dimensions
Design considerations
9.2 Problems with Multiple Contact Surfaces
Determination of sliding directions
9.3 Belts and Cables Contacting Cylindrical
Surfaces
Equilibrium analysis
9.4 Chapter Review
10 Moments of Inertia
10.1 AreaMoments of Inertia
An example—test scores
An example—beam loading
Definition of area moments of inertia
What are area moments of inertia used for?
Radius of gyration
Evaluation of moments of inertia using
integration
10.2 Parallel Axis Theorem
Use of parallel axis theorem in integration
Use of parallel axis theorem for composite
shapes
10.3 MassMoments of Inertia
An example—figure skating
Definition of mass moments of inertia
What are mass moments of inertia used for?
Radius of gyration
Parallel axis theorem
Evaluation of moments of inertia using
integration
Evaluation of moments of inertia using composite
shapes
10.4 Chapter Review
Preface
11 Introduction to Dynamics
11.1 The Newtonian Equations
11.2 Fundamental Concepts in Dynamics
Space and time
Force, mass, and inertia
Particle and rigid body
Vectors and their Cartesian representation
Useful vector “tips and tricks”
Units
11.3 Dynamics and Engineering Design
System modeling
12 Particle Kinematics
12.1 Position, Velocity, Acceleration, and Cartesian Coordinates
Position vector
Trajectory
Velocity vector and speed
Acceleration vector
Cartesian coordinates
12.2 One-Dimensional Motion
Rectilinear motion relations
Circular motion and angular velocity
12.3 ProjectileMotion
12.4 Planar Motion: Normal-Tangential
Components
Normal-tangential components
12.5 Planar Motion: Polar Coordinates
The time derivative of a vector
Polar coordinates and position, velocity, and
acceleration
12.6 Relative Motion Analysis and Differentiation of Geometrical Constraints
Relative motion
Differentiation of geometrical constraints
12.7 Motion in Three Dimensions
Cartesian coordinates
Tangent-normal-binormal components
Cylindrical coordinates
Spherical coordinates
12.8 Chapter Review
13 Force and Acceleration Methods for
Particles
13.1 Rectilinear Motion
Applying Newton’s second law
Force laws
Equation(s) of motion
Inertial reference frames
Degrees of freedom
13.2 Curvilinear Motion
Newton’s second law in 2D and 3D component systems
13.3 Systems of Particles
Engineering materials one atom at a time
Newton’s second law for systems of particles
13.4 Chapter Review
14 Energy Methods for Particles
14.1 Work-Energy Principle for a Particle
Work-energy principle and its relation
with
Work of a force
14.2 Conservative Forces and Potential Energy
Work done by the constant force of gravity
Work of a central force
Conservative forces and potential energy
Work-energy principle for any type of force
When is a force conservative
14.3 Work-Energy Principle for Systems of
Particles
Internal work and work-energy principle for a
system
Kinetic energy for a system of particles
14.4 Power and Efficiency
Power developed by a force
Efficiency
14.5 Chapter Review
15 Momentum Methods for Particles
15.1 Momentumand Impulse
Impulse-momentum principle
Conservation of linear momentum
15.2 Impact
Impacts are short, dramatic events
Definition of impact and notation
Line of impact and contact force between impacting
objects
Impulsive forces and impact-relevant FBDs
Coefficient of restitution
Unconstrained direct central impact
Unconstrained oblique central impact
Impact and energy
15.3 Angular Momentum
Moment-angular momentum relation for a
particle
Angular impulse-momentum for a system of
particles
Euler’s first and second laws of motion
15.4 Orbital Mechanics
Determination of the orbit
Energy considerations
15.5 Mass Flows
Steady flows
Variable mass flows and propulsion
15.6 Chapter Review
16 Planar Rigid Body Kinematics
16.1 Fundamental Equations, Translation, and
Rotation About a Fixed Axis
Crank, connecting rod, and piston motion
Qualitative description of rigid body motion
General motion of a rigid body
Elementary rigid body motions: translations
Elementary rigid body motions: rotation about a
fixed axis
Planar motion in practice
16.2 Planar Motion: Velocity Analysis
Vector approach
Differentiation of constraints
Instantaneous center of rotation
16.3 Planar Motion: Acceleration Analysis
Vector approach
Differentiation of constraints
Rolling without slip: acceleration analysis
16.4 Rotating Reference Frames
The general kinematic equations for the motion of a
point relative to a rotating reference frame
Coriolis component of acceleration
16.5 Chapter Review
17 Newton-Euler Equations for Planar Rigid
BodyMotion
17.1 Newton-Euler Equations: Bodies
Symmetric with Respect to the Plane
ofMotion
Linear momentum: translational equations
Angular momentum: rotational equations
Graphical interpretation of the equations of
motion
17.2 Newton-Euler Equations: Translation
17.3 Newton-Euler Equations: Rotation About a FixedAxis
17.4 Newton-Euler Equations: General Plane
Motion
Newton-Euler equations for general plane
motion
17.5 Chapter Review
18 Energy and Momentum Methods for
Rigid Bodies
18.1 Work-Energy Principle for Rigid Bodies
Kinetic energy of rigid bodies in planar motion
Work-energy principle for a rigid body
Work done on rigid bodies
Potential energy and conservation of energy
Work-energy principle for systems
Power
18.2 Momentum Methods for Rigid Bodies
Impulse-momentum principle for a rigid
body
Angular impulse-momentum principle for a rigid body
18.3 Impact of Rigid Bodies
Rigid body impact: basic nomenclature and
assumptions
Classification of impacts
Central impact
Eccentric impact
Constrained eccentric impact
18.4 Chapter Review
19 Mechanical Vibrations
19.1 Undamped Free Vibration
Oscillation of a railcar after coupling
Standard form of the harmonic oscillator
Linearizing nonlinear systems
Energy method
19.2 Undamped Forced Vibration
Standard form of the forced harmonic oscillator
19.3 Viscously Damped Vibration
Viscously damped free vibration
Viscously damped forced vibration
19.4 ChapterReview
20 Three-Dimensional Dynamics of
Rigid Bodies
20.1 Three-Dimensional Kinematics of
Rigid Bodies
Computation of angular accelerations
Summing angular velocities
20.2 Three-Dimensional Kinetics of
Rigid Bodies
Newton-Euler equations for three-dimensional motion
Kinetic energy of a rigid body in three-dimensional motion
20.3 Chapter Review
A Technical Writing
B Answers to Even-Numbered Problems
C Mass Moments of Inertia
Definition of mass moments and products of inertia
How are mass moments of inertia used
Radius of gyration
Parallel axis theorem
Principal moments of inertia
Moment of inertia about an arbitrary axis
Evaluation of moments of inertia using composite
shapes
D Angular Momentum of a Rigid Body
Angular momentum of a rigid body undergoing three-dimensional motion
Angular momentum of a rigid body in planar
motion
Index

Citation preview

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This International Student Edition is for use outside of the U.S.

Statics and

Engineering Mechanics:

Dynamics Third Edition

Gary L. Gray Francesco Costanzo Robert J. Witt Michael E. Plesha

ISTUDY

Engineering Mechanics

STATICS & DYNAMICS THIRD EDITION

Michael E. Plesha Professor Emeritus, Department of Engineering Physics University of Wisconsin–Madison

Gary L. Gray Department of Engineering Science and Mechanics Penn State University

Robert J. Witt Department of Engineering Physics University of Wisconsin–Madison

Francesco Costanzo Department of Engineering Science and Mechanics Penn State University

ISTUDY

ENGINEERING MECHANICS: STATICS & DYNAMICS Published by McGraw Hill LLC, 1325 Avenue of the Americas, New York, NY 10019. Copyright © 2023 by McGraw Hill LLC. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw Hill LLC, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 LWI 26 25 24 23 22 ISBN 978-1-265-25541-1 MHID 1-265-25541-5 Cover Image: NASA/MSFC; NASA All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.

The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw Hill LLC, and McGraw Hill LLC does not guarantee the accuracy of the information presented at these sites. mheducation.com/highered

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ABOUT THE AUTHORS Michael E. Plesha is a Professor Emeritus of Engineering Mechanics in the Department of Engineering Physics at the University of Wisconsin–Madison. Professor Plesha received his B.S. from the University of Illinois–Chicago in structural engineering and materials, and his M.S. and Ph.D. from Northwestern University in structural engineering and applied mechanics. His primary research areas are computational mechanics, focusing on the development of finite element and discrete element methods for solving static and dynamic nonlinear problems, and the development of constitutive models for characterizing the behavior of materials. Much of his work focuses on problems featuring contact, friction, and material interfaces. Applications include nanotribology, high-temperature rheology of ceramic composite materials, modeling geomaterials including rock and soil, penetration mechanics, and modeling crack growth in structures. He is coauthor of the book Concepts and Applications of Finite Element Analysis (with R. D. Cook, D. S. Malkus, and R. J. Witt). In addition to teaching statics and dynamics, he also has extensive experience teaching courses in basic and advanced mechanics of materials, mechanical vibrations, and finite element methods. Gary L. Gray is an Associate Professor of Engineering Science and Mechanics in the Department of Engineering Science and Mechanics at Penn State in University Park, PA. He received a B.S. in mechanical engineering (cum laude) from Washington University in St. Louis, MO, an S.M. in engineering science from Harvard University, and M.S. and Ph.D. degrees in engineering mechanics from the University of Wisconsin–Madison. His primary research interests are in dynamical systems, dynamics of mechanical systems, and mechanics education. For his contributions to mechanics education, he has been awarded the Outstanding and Premier Teaching Awards from the Penn State Engineering Society, the Outstanding New Mechanics Educator Award from the American Society for Engineering Education, the Learning Excellence Award from General Electric, and the Collaborative and Curricular Innovations Special Recognition Award from the Provost of Penn State. In addition to dynamics, he also teaches mechanics of materials, mechanical vibrations, numerical methods, advanced dynamics, and engineering mathematics. Robert J. Witt retired from the University of Wisconsin–Madison, Department of Engineering Physics, in 2020 after a 33-year teaching career. He received his B.S. in mechanical engineering from the University of California–Davis and his M.S. and Ph.D. in nuclear engineering from MIT. His research interests were in computational methods in fluid and solid mechanics, with particular applications to nuclear systems. He taught 20 different courses over the span of his career, including statics, dynamics, mechanics of materials, and a variety of other classes in applied mechanics, computational methods, and nuclear engineering. He is coauthor of the book Concepts and Applications of Finite Element Analysis (with R. D. Cook, D. S. Malkus, and M. E. Plesha).

iii

Francesco Costanzo is a Professor of Engineering Science and Mechanics in the Engineering Science and Mechanics Department at Penn State. He received the Laurea in Ingegneria Aeronautica from the Politecnico di Milano, Milan, Italy. After coming to the United States as a Fulbright scholar, he received his Ph.D. in aerospace engineering from Texas A&M University. His primary research interest is the mathematical and numerical modeling of material behavior. His specific research interests include the theoretical and numerical characterization of dynamic fracture in materials subject to thermo-mechanical loading, the development of multi-scale methods for predicting continuum-level material properties from molecular calculations, and modeling and computational problems in bio-medical applications. In addition to scientific research, he has contributed to various projects for the advancement of mechanics education under the sponsorship of several organizations, including the National Science Foundation. For his contributions, he has received various awards, including the 1998 and the 2003 GE Learning Excellence Awards and the 1999 ASEE Outstanding New Mechanics Educator Award.

iv

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The authors thank their families for their patience, understanding, and, most importantly, encouragement during the long years it took to bring these books to completion. Without their support, none of this would have been possible.

BRIEF CONTENTS Statics 1

Introduction to Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2

Vectors: Force and Position . . . . . . . . . . . . . . . . . . . . . . . . .

29

3

Equilibrium of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

4

Moment of a Force and Equivalent Force Systems . . . . . . 203

5

Equilibrium of Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

Structural Analysis and Machines . . . . . . . . . . . . . . . . . . . . 363

7

Centroids and Distributed Force Systems . . . . . . . . . . . . .

8

Internal Forces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493

9

Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533

10 Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

271

431

573

Dynamics 11 Introduction to Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 619 12 Particle Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 649 13 Force and Acceleration Methods for Particles . . . . . . . . . . 789 14 Energy Methods for Particles . . . . . . . . . . . . . . . . . . . . . . . . 861 15 Momentum Methods for Particles . . . . . . . . . . . . . . . . . . . . 933 16 Planar Rigid Body Kinematics . . . . . . . . . . . . . . . . . . . . . . . 1055 17 Newton-Euler Equations for Planar Rigid Body Motion . . . 1145 18 Energy and Momentum Methods for Rigid Bodies . . . . . . 1207 19 Mechanical Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1289 20 Three-Dimensional Dynamics of Rigid Bodies . . . . . . . . . . 1349 Appendices

A

Technical Writing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

A-1

B

Answers to Even-Numbered Problems . . . . . . . . . . . .

A-5

C

Mass Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . . . A-65

D

Angular Momentum of a Rigid Body . . . . . . . . . . . . . . A-75

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U.S. Customary and SI Unit Systems Properties of Lines and Area Moments of Inertia Properties of Solids and Mass Moments of Inertia

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I-1

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STATICS TABLE OF CONTENTS

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2.3

xiv

Introduction to Statics . . . . . . . . . . . . . . .

1

1.1

Engineering and Statics . . . . . . . . . . . . . . .

1

1.2

Topics That Will Be Studied in Statics . . . .

3

1.3

A Brief History of Statics . . . . . . . . . . . . . Galileo Galilei 4

3

Cartesian vector representation 66 Direction angles and direction cosines 66 Position vectors 67 Use of position vectors to write expressions for force vectors 68 Some simple structural members 68

Isaac Newton 5 1.4

Fundamental Principles . . . . . . . . . . . . . .

2.4

5

1.5

Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.6

Units and Unit Conversions . . . . . . . . . . . . 9 Dimensional homogeneity and unit conversions 10

Vector Dot Product . . . . . . . . . . . . . . . . . .

84

Dot product using Cartesian components 85 Determination of the angle between two vectors 85

Newton’s laws of motion 7 8

Determination of the component of a vector in a particular direction 86 Determination of the component of a vector perpendicular to a direction 87

Prefixes 10 Angular measure 11 Small angle approximations 12 Accuracy of calculations 13

2.5

Vector Cross Product . . . . . . . . . . . . . . . . 101 Cross product using Cartesian components 102

1.7

Newton’s Law of Gravitation . . . . . . . . . . . Relationship between specific weight and density 17

16

1.8

Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

Evaluation of cross product using determinants 103 Determination of the normal direction to a plane 105 Determination of the area of a parallelogram 105

1.9

Chapter Review . . . . . . . . . . . . . . . . . . . . .

23

Scalar triple product 105 2.6

2

Cartesian Representation of Vectors in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 66 Right-hand Cartesian coordinate system 66

Vectors: Force and Position . . . . . . . . . . . 2.1

Basic Concepts . . . . . . . . . . . . . . . . . . . . .

29 29

Introduction—force, position, vectors, and tides 29 Denoting vectors in figures 31

2.2

Chapter Review . . . . . . . . . . . . . . . . . . . . . 116

3

Equilibrium of Particles . . . . . . . . . . . . . . . 125 3.1

Equilibrium of Particles in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 125 Free body diagram (FBD) 126

Basic vector operations 32 Performing vector operations 34 Resolution of a vector into vector components 34

Modeling and problem solving 130 Cables and bars 131

Cartesian Representation of Vectors in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . .

Pulleys 133 Reactions 134

48

Introduction—Cartesian representation and a walk to work 48 Unit vectors 48 Cartesian coordinate system 49 Cartesian vector representation 49 Addition of vectors using Cartesian components 51 Position vectors 52

3.2

Behavior of Cables, Bars, and Springs . . . . 151 Equilibrium geometry of a structure 151 Cables 151 Bars 152 Modeling idealizations and solution ∑ of 𝐹⃗ = 0⃗ 152 Springs 152

vii

3.3

Equilibrium of Particles in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 166 Reactions 166

Reactions 272 Free body diagram (FBD) 274 Alternative equilibrium equations 276 Gears 277 Examples of correct FBDs 277

Solution of algebraic equations 166 Summing forces in directions other than 𝑥, 𝑦, or 𝑧 167 3.4

Examples of incorrect and/or incomplete FBDs 279

Engineering Design . . . . . . . . . . . . . . . . . . 181 Objectives of design 181

5.3

Particle equilibrium design problems 182 3.5

4

Treatment of cables and pulleys 298 Springs 299

Chapter Review . . . . . . . . . . . . . . . . . . . . . 195

Superposition 300 Supports and fixity 300 Static determinacy and indeterminacy 301

Moment of a Force and Equivalent Force Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 4.1

Moment of a Force . . . . . . . . . . . . . . . . . .

203

Two-force and three-force members 303

Scalar approach 204 Vector approach 205

5.4

Varignon’s theorem 207 Which approach should I use: scalar or vector? 208 4.2

Examples of correct FBDs 325 Examples of incorrect and/or incomplete FBDs 327

Moment of a Couple . . . . . . . . . . . . . . . . . 232 Vector approach 233 Scalar approach 233

5.5

Comments on the moment of a couple 233 Equivalent couples 234 Equivalent force systems 234

5.6

6

Equivalent Force Systems . . . . . . . . . . . . . 246 Transmissibility of a force 246

Structures and machines 363 6.1

6.2

viii

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364

Zero-force members 367 Typical truss members 369

Chapter Review . . . . . . . . . . . . . . . . . . . . . 263

Equilibrium of Bodies . . . . . . . . . . . . . . . .

Truss Structures and the Method of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . When may a structure be idealized as a truss? 365 Method of joints 365

Wrench equivalent force systems 250 Why are equivalent force systems called equivalent? 250

5

Chapter Review . . . . . . . . . . . . . . . . . . . . . 353

Structural Analysis and Machines . . . . . . 363

Equivalent force systems 247 Some special force systems 248

4.5

Engineering Design . . . . . . . . . . . . . . . . . . 344 Codes and standards 346 Design problems 347

Resultant couple moment 235 Moments as free vectors 235 4.4

Equilibrium of Bodies in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 322 Reactions 322 More on bearings 322 Scalar approach or vector approach? 324 Solution of algebraic equations 324

Moment of a Force About a Line . . . . . . . . 220 Vector approach 220 Scalar approach 221

4.3

Equilibrium of Bodies in Two Dimensions— Additional Topics . . . . . . . . . . . . . . . . . . . 298 Why are bodies assumed to be rigid? 298

271

5.1

Equations of Equilibrium . . . . . . . . . . . . . 271

5.2

Equilibrium of Rigid Bodies in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 272

Truss Structures and the Method of Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 Treatment of forces that are not at joints 382 Static determinacy and indeterminacy 383 Design considerations 384

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6.3

6.4

6.5

7

Internal forces for slender members in three dimensions 495

Trusses in Three Dimensions . . . . . . . . . . . 399 Stability of space trusses and design considerations 400

Determination of internal forces 495

Frames and Machines . . . . . . . . . . . . . . . . 408 Analysis procedure and free body diagrams (FBDs) 408

8.2

Examples of correct FBDs 409 Examples of incorrect and/or incomplete FBDs 411

8.3

Determination of 𝑉 and 𝑀 using equilibrium 503 Shear and moment diagrams 503

Chapter Review . . . . . . . . . . . . . . . . . . . . . 426

Centroid of an area 433 Centroid of a line 434 Centroid of a volume 435

8.4

9

7.3

7.4

9.1

Basic Concepts . . . . . . . . . . . . . . . . . . . . . A brief history of tribology 533

Center of Mass and Center of Gravity . . . . 449

A simple experiment 534 Coulomb’s law of friction 535 Coefficients of friction 535

Center of mass 449 Center of gravity 450

Dry contact versus liquid lubrication 537 Angle of friction 537

Theorems of Pappus and Guldinus . . . . . . 461 Area of a surface of revolution 461

Problems with multiple contact surfaces 537 Wedges 538

Volume of a solid of revolution 462 Proof of the Pappus–Guldinus theorems 462

Coulomb’s law of friction in three dimensions 538 Design considerations 538

Distributed Forces, Fluid and Gas Pressure Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 Distributed forces 468 Distributed forces applied to beams 468 Fluid and gas pressure 469 Forces produced by fluids 471 Forces produced by gases 473

7.5

Chapter Review . . . . . . . . . . . . . . . . . . . . . 527

Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533

Finer points: surfaces and lines in three dimensions 436 7.2

Problems with Multiple Contact Surfaces . 551 Determination of sliding directions 551

9.3

Belts and Cables Contacting Cylindrical Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . Equilibrium analysis 560

9.4

560

Chapter Review . . . . . . . . . . . . . . . . . . . . . 567

Chapter Review . . . . . . . . . . . . . . . . . . . . . 486

Internal Forces . . . . . . . . . . . . . . . . . . . . . . 493 8.1

533

9.2

10 Moments of Inertia . . . . . . . . . . . . . . . . . . 8

514

Tips and shortcuts for drawing shear and moment diagrams 517 Design considerations 518

Centroid . . . . . . . . . . . . . . . . . . . . . . . . . . 431 Introduction—center of gravity 431

Unification of concepts 435 Which approach should I use: composite shapes or integration? 435

Relations Among Shear, Moment, and Distributed Force . . . . . . . . . . . . . . . . . . . Relations among 𝑉 , 𝑀, and 𝑤 514

Determination of 𝑉 and 𝑀 using integration 515 Which approach should I use? 516

Centroids and Distributed Force Systems 431 7.1

Internal Forces in Straight Beams . . . . . . . 503

Internal Forces in Structural Members . . . 493 Why are internal forces important? 493 Internal forces for slender members in two dimensions 494

573

10.1 Area Moments of Inertia . . . . . . . . . . . . . . 573 An example—test scores 573 An example—beam loading 574 Definition of area moments of inertia 574 What are area moments of inertia used for? 575

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Radius of gyration 576 Evaluation of moments of inertia using integration 577 10.2 Parallel Axis Theorem . . . . . . . . . . . . . . . . 585 Use of parallel axis theorem in integration 586 Use of parallel axis theorem for composite shapes 586 10.3 Mass Moments of Inertia . . . . . . . . . . . . . . 593 An example—figure skating 593

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Definition of mass moments of inertia 593 What are mass moments of inertia used for? 594 Radius of gyration 595 Parallel axis theorem 595 Evaluation of moments of inertia using integration 596 Evaluation of moments of inertia using composite shapes 597 10.4 Chapter Review . . . . . . . . . . . . . . . . . . . . . 612

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DYNAMICS TABLE OF CONTENTS

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11 Introduction to Dynamics . . . . . . . . . . . . . 619

Cylindrical coordinates 761 Spherical coordinates 762 12.8 Chapter Review . . . . . . . . . . . . . . . . . . . . . 775

11.1 The Newtonian Equations . . . . . . . . . . . . . 619 11.2 Fundamental Concepts in Dynamics . . . . . 625 Space and time 625 Force, mass, and inertia 625

13.1 Rectilinear Motion . . . . . . . . . . . . . . . . . . Applying Newton’s second law 789

Particle and rigid body 626 Vectors and their Cartesian representation 626

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12 Particle Kinematics . . . . . . . . . . . . . . . . . . 649 12.1 Position, Velocity, Acceleration, and Cartesian Coordinates . . . . . . . . . . . . . . . . 649 Position vector 650 Trajectory 650 Velocity vector and speed 650 Acceleration vector 652 668

Rectilinear motion relations 668 Circular motion and angular velocity 670 12.3 Projectile Motion . . . . . . . . . . . . . . . . . . . . 689 12.4 Planar Motion: Normal-Tangential Components . . . . . . . . . . . . . . . . . . . . . . . Normal-tangential components 703

703

12.5 Planar Motion: Polar Coordinates . . . . . . . 717 The time derivative of a vector 717 Polar coordinates and position, velocity, and acceleration 719 12.6 Relative Motion Analysis and Differentiation of Geometrical Constraints . . . . . . . . . . . . 738 Relative motion 738 Differentiation of geometrical constraints 739 12.7 Motion in Three Dimensions . . . . . . . . . . . 759 Cartesian coordinates 759 Tangent-normal-binormal components 759

Inertial reference frames 794 Degrees of freedom 794 13.2 Curvilinear Motion . . . . . . . . . . . . . . . . . . 812 Newton’s second law in 2D and 3D component systems 812 13.3 Systems of Particles . . . . . . . . . . . . . . . . . . 837 Engineering materials one atom at a time 837 Newton’s second law for systems of particles 837 13.4 Chapter Review . . . . . . . . . . . . . . . . . . . . . 854

14 Energy Methods for Particles . . . . . . . . . . 861

Cartesian coordinates 652 12.2 One-Dimensional Motion . . . . . . . . . . . . .

789

Force laws 791 Equation(s) of motion 793

Useful vector “tips and tricks” 629 Units 630 11.3 Dynamics and Engineering Design . . . . . . System modeling 647

13 Force and Acceleration Methods for Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 789

14.1 Work-Energy Principle for a Particle . . . . Work-energy principle and its relation with 𝐹⃗ = 𝑚𝑎⃗ 861 Work of a force 863

861

14.2 Conservative Forces and Potential Energy . 877 Work done by the constant force of gravity 877 Work of a central force 877 Conservative forces and potential energy 878 Work-energy principle for any type of force 880 When is a force conservative? 880 14.3 Work-Energy Principle for Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . Internal work and work-energy principle for a system 901 Kinetic energy for a system of particles 902

901

14.4 Power and Efficiency . . . . . . . . . . . . . . . . . 919 Power developed by a force 919 Efficiency 919 14.5 Chapter Review . . . . . . . . . . . . . . . . . . . . . 926

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15 Momentum Methods for Particles . . . . . . 933 15.1 Momentum and Impulse . . . . . . . . . . . . . . 933 Impulse-momentum principle 933 Conservation of linear momentum 936 15.2 Impact . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955 Impacts are short, dramatic events 955 Definition of impact and notation 955 Line of impact and contact force between impacting objects 955 Impulsive forces and impact-relevant FBDs 956 Coefficient of restitution 956 Unconstrained direct central impact 958 Unconstrained oblique central impact 958 Impact and energy 959 15.3 Angular Momentum . . . . . . . . . . . . . . . . . 982 Moment-angular momentum relation for a particle 982 Angular impulse-momentum for a system of particles 983 Euler’s first and second laws of motion 986 15.4 Orbital Mechanics . . . . . . . . . . . . . . . . . . . 1007 Determination of the orbit 1007 Energy considerations 1013 15.5 Mass Flows . . . . . . . . . . . . . . . . . . . . . . . . 1023 Steady flows 1023 Variable mass flows and propulsion 1026

16.3 Planar Motion: Acceleration Analysis . . . . 1097 Vector approach 1097 Differentiation of constraints 1097 Rolling without slip: acceleration analysis 1098 16.4 Rotating Reference Frames . . . . . . . . . . . . 1118 The general kinematic equations for the motion of a point relative to a rotating reference frame 1118 Coriolis component of acceleration 1122 16.5 Chapter Review . . . . . . . . . . . . . . . . . . . . . 1136

17 Newton-Euler Equations for Planar Rigid Body Motion . . . . . . . . . . . . . . . . . . . . . . . 1145 17.1 Newton-Euler Equations: Bodies Symmetric with Respect to the Plane of Motion . . . . . . . . . . . . . . . . . . . . . . . . . 1145 Linear momentum: translational equations 1145 Angular momentum: rotational equations 1146 Graphical interpretation of the equations of motion 1150 17.2 Newton-Euler Equations: Translation . . . . 1153 17.3 Newton-Euler Equations: Rotation About a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . 1163 17.4 Newton-Euler Equations: General Plane Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 1177 Newton-Euler equations for general plane motion 1177 17.5 Chapter Review . . . . . . . . . . . . . . . . . . . . . 1199

15.6 Chapter Review . . . . . . . . . . . . . . . . . . . . . 1044

16 Planar Rigid Body Kinematics . . . . . . . . . 1055

18 Energy and Momentum Methods for Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . 1207

16.1 Fundamental Equations, Translation, and Rotation About a Fixed Axis . . . . . . . . . . . 1055 Crank, connecting rod, and piston motion 1055

18.1 Work-Energy Principle for Rigid Bodies . . 1207 Kinetic energy of rigid bodies in planar motion 1207

Qualitative description of rigid body motion 1056 General motion of a rigid body 1057 Elementary rigid body motions: translations 1059 Elementary rigid body motions: rotation about a fixed axis 1060 Planar motion in practice 1062 16.2 Planar Motion: Velocity Analysis . . . . . . . 1075 Vector approach 1075 Differentiation of constraints 1076 Instantaneous center of rotation 1077

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Work-energy principle for a rigid body 1209 Work done on rigid bodies 1209 Potential energy and conservation of energy 1210 Work-energy principle for systems 1212 Power 1212 18.2 Momentum Methods for Rigid Bodies . . . . 1243 Impulse-momentum principle for a rigid body 1243 Angular impulse-momentum principle for a rigid body 1244

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18.3 Impact of Rigid Bodies . . . . . . . . . . . . . . . 1265 Rigid body impact: basic nomenclature and assumptions 1266 Classification of impacts 1266 Central impact 1266 Eccentric impact 1268 Constrained eccentric impact 1269

Newton-Euler equations for three-dimensional motion 1366 Kinetic energy of a rigid body in three-dimensional motion 1371 20.3 Chapter Review . . . . . . . . . . . . . . . . . . . . . 1388

A

Technical Writing . . . . . . . . . . . . . . . . . . . .

A-1

B

Answers to Even-Numbered Problems . .

A-5

C

Mass Moments of Inertia . . . . . . . . . . . . . A-65

18.4 Chapter Review . . . . . . . . . . . . . . . . . . . . . 1282

19 Mechanical Vibrations . . . . . . . . . . . . . . . 1289 19.1 Undamped Free Vibration . . . . . . . . . . . . . 1289

Definition of mass moments and products of inertia A-65 How are mass moments of inertia used? A-67 Radius of gyration A-68 Parallel axis theorem A-68 Principal moments of inertia A-70

Oscillation of a railcar after coupling 1289 Standard form of the harmonic oscillator 1291 Linearizing nonlinear systems 1292 Energy method 1293 19.2 Undamped Forced Vibration . . . . . . . . . . . 1307 Standard form of the forced harmonic oscillator 1307

Moment of inertia about an arbitrary axis A-73 Evaluation of moments of inertia using composite shapes A-74

19.3 Viscously Damped Vibration . . . . . . . . . . . 1321 Viscously damped free vibration 1321 Viscously damped forced vibration 1324 19.4 Chapter Review . . . . . . . . . . . . . . . . . . . . . 1340

20 Three-Dimensional Dynamics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . 1349 20.1 Three-Dimensional Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . 1349 Computation of angular accelerations 1350 Summing angular velocities 1350 20.2 Three-Dimensional Kinetics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . 1366

D

Angular Momentum of a Rigid Body . . . . A-75 Angular momentum of a rigid body undergoing three-dimensional motion A-75 Angular momentum of a rigid body in planar motion A-78

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

I-1

U.S. Customary and SI Unit Systems Properties of Lines and Area Moments of Inertia Properties of Solids and Mass Moments of Inertia

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PREFACE

Welcome to statics and dynamics! With this book we hope to provide a teaching and learning experience that is not only effective but also motivates your study and application of statics and dynamics. The major objectives of this book are to help you 1. Learn the fundamental principles of statics and dynamics; and 2. Gain the skills needed to apply these principles in the modeling of real-life problems and for carrying out engineering design. To achieve the objectives cited above, we have developed this book as follows. First, we provide a rigorous introduction to the fundamental principles. In a constantly changing technological landscape, it is by understanding fundamentals that you will be able to adapt to new technologies. This book places great emphasis on developing problem-solving skills. The need for an engineer to be able to accurately model real-life problems and solve them is obvious. Beyond this, it is only by mastering problem-solving skills that a true, deep understanding of fundamentals can be achieved. Second, we incorporate pedagogical principles that research in math, science, and engineering education has identified as essential for improving learning. These include teaching concepts in parallel with problem-solving skills, identifying and addressing misconceptions, and assessing conceptual understanding via specific problems. Third, we have made modeling the underlying theme of our approach to problem solving. We believe that modeling, understood as the making of reasonable assumptions to reduce a real-life problem to a simpler but tractable problem, is also something that must be taught and discussed alongside the basic principles. Fourth, we emphasize a systematic approach to solving every problem, an integral part of which is creating the aforementioned model. We believe these features make this book new and unique, and we hope that they will help you learn and master statics and dynamics. Several design problems are presented where appropriate throughout the book. These problems can be tackled with the knowledge and skills that are typical of statics and dynamics, although the use of mathematical software is sometimes helpful. These problems are open ended, they allow you to show creativity in developing solutions that solve important and realistic engineering problems, and their solution requires the definition of a parameter space in which the statics and dynamics of the system must be analyzed. Design topics include methods of design, parametric analysis, issues of professional responsibility, ethics, communication, and more. After studying statics and dynamics, you should have a thorough understanding of the fundamental principles, and, at a minimum, key points should remain in your memory for the rest of your life. We say this with a full appreciation that some of you will have careers with new and unexpected directions. Regardless of your eventual professional responsibilities, knowledge of the fundamentals of statics and dynamics will help you to be technically literate. If you are actively engaged in the practice of engineering and/or the sciences, then your needs go well beyond mere technical literacy, and you must also be accomplished at applying these fundamentals so that you can study more advanced subjects that build on statics and dynamics, and because you will apply concepts of statics and dynamics on a daily basis in your career.

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Why Another Statics and Dynamics Series? These books provide thorough coverage of all the pertinent topics traditionally associated with statics and dynamics. Indeed, many of the currently available texts also provide this. However, these texts offer several major innovations that enhance the learning objectives and outcomes in these subjects. What Then Are the Major Differences Between These Books and Other Engineering Mechanics Texts? A Consistent and Systematic Approach to Problem Solving One of the main objectives of these texts is to foster the habit of solving problems using a systematic approach. Therefore, the example problems in these texts follow a structured problemsolving methodology that will help you develop your problem-solving skills not only in statics and dynamics, but also in all other mechanics subjects that follow. This structured problem-solving approach consists of the following steps: Road Map, Modeling, Governing Equations, Computation, and Discussion & Verification. The Road Map provides some of the general objectives of the problem and develops a strategy for how the solution will be developed. Modeling is next, where a real-life problem is idealized by a model. This step results in the creation of a free body diagram and the selection of the balance laws needed to solve the problem. In dynamics, it is difficult to separate these steps, so they are usually done in combination. The Governing Equations step is devoted to writing all the equations needed to solve the problem. These equations typically include the equilibrium equations or balance laws and, depending upon the particular problem, force laws (e.g., spring law, failure criteria, frictional sliding criteria and/or friction laws) and kinematic equations. In the Computation step, the governing equations are solved. In the final step, Discussion & Verification, the solution is interrogated to ensure that it is meaningful and accurate. This problemsolving methodology is followed for all examples that involve equilibrium concepts or balance principles. Some problems (e.g., determination of the center of mass for an object or kinematic problems) do not involve equilibrium concepts, and for these the Modeling step is not needed. Contemporary Examples, Problems, and Applications The examples, homework problems, and design problems were carefully constructed to help show you how the various topics of statics and dynamics are used in engineering practice. Statics and dynamics are immensely important subjects in modern engineering and science, and one of our goals is to excite you about these subjects and the career that lies ahead of you. A Focus on Design A major difference between these texts and other books is the systematic incorporation of design and modeling of real-life problems throughout. In statics, topics include important discussions on design, ethics, and professional responsibility. In Dynamics, the emphasis is on parametric analysis and motion over ranges of time and space. These books show you that meaningful engineering design is possible using the concepts of statics and dynamics. Not only is the ability to develop a design very satisfying, but it also helps you develop a greater understanding of basic concepts and helps sharpen your ability to apply these concepts. Because the main focus of statics and dynamics textbooks should be the establishment of a firm understanding of basic concepts and correct problem-solving techniques, design topics do not have an overbearing presence in the books. Rather, design topics are

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included where they are most appropriate. While some of the discussions on design could be described as “common sense,” such a characterization trivializes the importance and necessity for discussing pertinent issues such as safety, uncertainty in determining loads, the designer’s responsibility to anticipate uses, even unintended uses, communications, ethics, and uncertainty in workmanship. Perhaps the most important feature of our inclusion of design and modeling topics is that you get a glimpse of what engineering is about and where your career in engineering is headed. The book is structured so that design topics and design problems are offered in a variety of places, and it is possible to pick when and where the coverage of design is most effective. Computational Tools Some examples and problems are appropriate for solution using computer software. The use of computers extends the types of problems that can be solved while alleviating the burden of solving equations. Such examples and problems give you insight into the power of computer tools and further insight into how statics and dynamics are used in engineering practice. Modern Pedagogy Numerous modern pedagogical elements have been included. These elements are designed to reinforce concepts, and they provide additional information to help you make meaningful connections with real-world applications. Marginal notes (i.e., Helpful Information, Common Pitfalls, Interesting Facts, and Concept Alerts) help you place topics, ideas, and examples in a larger context. These notes will help you study (e.g., Helpful Information and Common Pitfalls), will provide real-world examples of how different aspects of statics and dynamics are used (e.g., Interesting Facts), and will drive home important concepts or help dispel misconceptions (e.g., Concept Alerts and Common Pitfalls). Mini-Examples are used throughout the text to immediately and quickly illustrate a point or concept without making readers wait for the worked-out examples at the end of the section. Answers to Problems The answers to most even-numbered problems have been included in the back matter for ease of use as Appendix B. Providing answers in this manner allows for the inclusion of more complex information than would otherwise be possible. In addition to final numerical and/or symbolic answers, shear and moment diagrams and/or plots for Computer Problems are included.

Changes to the Third Edition The third editions of Engineering Mechanics: Statics and Engineering Mechanics: Dynamics retain all of the major pedagogical features of the previous editions, including a structured problem-solving methodology for all example problems, contemporary engineering applications in the example problems and homework exercises, the inclusion of engineering design and its implications for problem solving and applications, and use of computational tools where applicable. In addition, as a result of the author-based typesetting process, the outstanding accuracy of the earlier editions has been preserved, leading to books whose accuracy is unrivaled among textbooks. The third editions contain revised and enhanced textual discussions and example problems, additional figures where effective, and new homework exercises. In Connect, the online homework system, there are significant updates, including an autograded FBD tool and interactive learning tools. These interactive assignments help reinforce what is being covered in the text and show students how to tie the material to real-world situations. These tools complement the hundreds of auto-graded, algorithmic-exercises that are included in Connect from the text.

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The following individuals have been instrumental in ensuring the highest standard of content and accuracy. We are deeply indebted to them for their tireless efforts.

Third Edition Reviewers David Ancalle-Reyes Kennesaw State University Hossein Ataei University of Illinois at Chicago Sarah Baxter University of St. Thomas John Burkhardt U.S. Naval Academy Amiya Chatterjee University of California at Los Angeles Mona Eskandari University of California at Riverside

Serhan Guner University of Toledo Anant Honkan Perimeter College, Georgia State University Yufeng Hu Western Michigan University Hesam Moghaddam Northern Arizona University Karim Nohra University of South Florida Anurag Purwar Stony Brook University

Adrian Rodriguez University of Texas at Austin Scott D. Schif Kansas State University Frank Sup University of Massachusetts at Amherst Jekan Thangavelautham University of Arizona Vimal Viswanathan San Jose State University

Michael W. Keller The University of Tulsa Yohannes Ketema University of Minnesota Ahmed Khalafallah University of Central Florida Raymond P. LeBeau University of Kentucky Kevin Lyons North Carolina State University Kevin Mackie University of Central Florida Stephanie Magleby Brigham Young University Mohammad Mahinfalah Milwaukee School of Engineering Dragomir C. Marinkovich Milwaukee School of Engineering Tom Mase California Polytechnic State University–San Luis Obispo Richard McNitt Penn State University William R. Murray California Polytechnic State University

Chris Passerello Michigan Technological University Gordon R. Pennock Purdue University Gary A. Pertmer University of Maryland W. Tad Pfeffer University of Colorado at Boulder Vincent C. Prantil Milwaukee School of Engineering Carisa H. Ramming Oklahoma State University Robert Rizza Milwaukee School of Engineering Bidhan C. Roy University of Wisconsin– Platteville David A. Rubenstein Oklahoma State University Jill Schmidt University of Wisconsin– Milwaukee John Schmitt Oregon State University Oleg Shiryayev Wright State University

Second Edition Combined Reviewers George G. Adams Northeastern University Manohar Arora Colorado School of Mines Stephen Bechtel The Ohio State University J. A. M. Boulet University of Tennessee– Knoxville Mark C. Bourland Lamar University Janet Brelin-Fornari Kettering University Louis Brock University of Kentucky Suren Chen Colorado State University Pasquale Cinnella Mississippi State University Asad Esmaeily Kansas State University Nicola Ferrier University of Wisconsin– Madison Christopher G. Gilbertson Michigan Technological University

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Larry M. Silverberg North Carolina State University Joseph C. Slater Wright State University Richard E. Stanley Kettering University

Candace S. Sulzbach Colorado School of Mines Carl Vilmann Michigan Technological University Thomas H. Wenzel Marquette University

T. W. Wu University of Kentucky Jack Xin Kansas State University Mohammed Zikry North Carolina State University

Stephanie Magleby Brigham Young University Tom Mase California Polytechnic State University–San Luis Obispo

Gregory Miller University of Washington Carisa H. Ramming Oklahoma State University

Mark Nagurka Marquette University David W. Parish North Carolina State University Gordon R. Pennock Purdue University Michael T. Shelton California State Polytechnic University–Pomona

Joseph C. Slater Wright State University Arun R. Srinivasa Texas A&M University Carl R. Vilmann Michigan Technological University Ronald W. Welch The University of Texas at Tyler

Riad Alakkad University of Dayton Farid Amirouche University of Illinois at Chicago Manohar L. Arora Colorado School of Mines Stephen Bechtel Ohio State University Kenneth Belanus Oklahoma State University Glenn Beltz University of California–Santa Barbara

Haym Benaroya Rutgers University Sherrill B. Biggers Clemson University James Blanchard University of Wisconsin– Madison Janet Brelin-Fornari Kettering University Neeraj Buch Michigan State University Manoj Chopra University of Central Florida

Focus Group Participants Brock E. Barry United States Military Academy Daniel Dickrell, III University of Florida Ali Gordon University of Central Florida– Orlando

First Edition Board of Advisors Janet Brelin-Fornari Kettering University Manoj Chopra University of Central Florida Pasquale Cinnella Mississippi State University Ralph E. Flori Missouri University of Science and Technology Christine B. Masters Penn State University

Reviewers Shaaban Abdallah University of Cincinnati Makola M. Abdullah Florida Agricultural and Mechanical University Mohamed Aboul-Seoud Rensselaer Polytechnic Institute Murad Abu-Farsakh Louisiana State University George G. Adams Northeastern University Shahid Ahmad SUNY–Buffalo

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Pasquale Cinnella Mississippi State University Ted A. Conway University of Central Florida Joseph Cusumano Penn State University K. L. (Larry) DeVries University of Utah Roger M. Di Julio, Jr. California State University– Northridge Bogdan I. Epureanu University of Michigan Ralph E. Flori Missouri University of Science and Technology Frank J. Fronczak University of Wisconsin– Madison Barry Goodno Georgia Institute of Technology Kurt Gramoll University of Oklahoma Hartley T. Grandin, Jr. Professor Emeritus, Worcester Polytechnic Institute Walter Haisler Texas A&M University Roy J. Hartfield, Jr. Auburn University G.A. Hartley Carleton University Scott L. Hendricks Virginia Tech University Paul R. Heyliger Colorado State University Lih-Min Hsia California State University– Los Angeles Nancy Hubing University of Missouri–Rolla David A. Jenkins University of Florida Erik A. Johnson University of Southern California James D. Jones Purdue University Yohannes Ketema University of Minnesota Carl R. Knospe University of Virginia

Jeff Kuo California State University– Fullerton Sang-Joon John Lee San Jose State University Shaofan Li University of California– Berkeley Jia Lu The University of Iowa Mohammad Mahinfalah Milwaukee School of Engineering Francisco Manzo-Robledo Washington State University Christine B. Masters Penn State University Ron McClendon University of Georgia Paul Mitiguy Consulting Professor, Stanford University William R. Murray California Polytechnic State University–San Luis Obispo Mark Nagurka Marquette University Karim Nohra University of South Florida Robert G. Oakberg Montana State University James J. Olsen Wright State University David W. Parish North Carolina State University Chris Passerello Michigan Technological University Gordon R. Pennock Purdue University Gary A. Pertmer University of Maryland W. Tad Pfeffer University of Colorado at Boulder John J. Phillips Oklahoma State University David Richardson University of Cincinnati

Scott D. Schiff Clemson University William C. Schneider Texas A&M University Michael T. Shelton California State Polytechnic University–Pomona Sorin Siegler Drexel University Joseph C. Slater Wright State University Ahmad Sleiti University of Central Florida Arun R. Srinivasa Texas A&M University Ganesh Thiagarajan Louisiana State University Jeffrey Thomas Northwestern University Josef S. Torok Rochester Institute of Technology John J. Uicker Professor Emeritus, University of Wisconsin–Madison David G. Ullman Professor Emeritus, Oregon State University Steven A. Velinsky University of California–Davis Carl R. Vilmann Michigan Technological University Ronald W. Welch The University of Texas at Tyler Claudia M. D. Wilson Florida State University C. Ray Wimberly University of Texas at Arlington T. W. Wu University of Kentucky X. J. Xin Kansas State University Joseph R. Zaworski Oregon State University Xiangwu (David) Zeng Case Western Reserve University M. A. Zikry North Carolina State University

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Symposium Attendees Farid Amirouche University of Illinois at Chicago Subhash C. Anand Clemson University Manohar L. Arora Colorado School of Mines Stephen Bechtel Ohio State University Sherrill B. Biggers Clemson University J. A. M. Boulet University of Tennessee Janet Brelin-Fornari Kettering University Louis M. Brock University of Kentucky Amir Chaghajerdi Colorado School of Mines Manoj Chopra University of Central Florida Pasquale Cinnella Mississippi State University Adel ElSafty University of North Florida Ralph E. Flori Missouri University of Science and Technology Walter Haisler Texas A&M University Kimberly Hill University of Minnesota James D. Jones Purdue University

Yohannes Ketema University of Minnesota Charles Krousgrill Purdue University Jia Lu The University of Iowa Mohammad Mahinfalah Milwaukee School of Engineering Tom Mase California Polytechnic State University–San Luis Obispo Christine B. Masters Penn State University Daniel A. Mendelsohn The Ohio State University Faissal A. Moslehy University of Central Florida LTC Mark Orwat United States Military Academy at West Point David W. Parish North Carolina State University Arthur E. Peterson Professor Emeritus, University of Alberta W. Tad Pfeffer University of Colorado at Boulder David G. Pollock Washington State University Robert L. Rankin Professor Emeritus, Arizona State University Mario Rivera-Borrero University of Puerto Rico at Mayaguez

Hani Salim University of Missouri Michael T. Shelton California State Polytechnic University–Pomona Lorenz Sigurdson University of Alberta Larry Silverberg North Carolina State University Joseph C. Slater Wright State University Arun R. Srinivasa Texas A&M University David G. Ullman Professor Emeritus, Oregon State University Carl R. Vilmann Michigan Technological University Anthony J. Vizzini Mississippi State University Andrew J. Walters Mississippi State University Ronald W. Welch The University of Texas at Tyler T. W. Wu University of Kentucky Musharraf Zaman University of Oklahoma–Norman Joseph R. Zaworski Oregon State University

Manoj Chopra University of Central Florida Yohannes Ketema University of Minnesota Christine B. Masters Penn State University Mark Nagurka Marquette University

Michael T. Shelton California State Polytechnic University–Pomona C. Ray Wimberly University of Texas at Arlington M. A. Zikry North Carolina State University

Focus Group Attendees Shaaban Abdallah University of Cincinnati Manohar L. Arora Colorado School of Mines Janet Brelin-Fornari Kettering University

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Accuracy Checkers Walter Haisler Texas A&M University Richard McNitt Penn State University

Mark Nagurka Marquette University

Carl R. Vilmann Michigan Technological University

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ACKNOWLEDGMENTS The authors are grateful to the many instructors who have provided valuable feedback on the previous editions of this book. We are especially thankful to Professor Christine B. Masters of Penn State University, Professor Carl R. Vilmann of Michigan Technological University, Professor Mark Nagurka of Marquette University, and Dr. Antonio Hernandez, University of Wisconsin–Madison for the guidance they have provided over the years these books have been in development and use. The authors are also grateful for helpful feedback from Professor Rani El-Hajjar of the University of Wisconsin–Milwaukee, Professor Richard Keles of the Center for Technology, Innovation and Community Engagement, New York, Professor Charles E. Bakis of Penn State University, and Dr. Suzannah Sandrik of the University of Wisconsin–Madison. Special thanks go to Andrew Miller for infrastructure he created to keep the authors, manuscript, and solutions manual in sync. His knowledge of programming, scripting, subversion, and many other computer technologies made a gargantuan task feel just a little more manageable.

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GUIDED TOUR Mini-Examples

force of attraction between two bodies. The gravitational force on a mass 𝑚1 due to a mass 𝑚2 a distance 𝑟 away from 𝑚1 is

Mini-examples are used throughout the text to immediately and quickly illustrate a point or concept without having to wait for the worked-out examples at the end of the section.

𝐹⃗12 =

𝐺𝑚1 𝑚2 𝑟2

𝑁 𝑟

𝑢, ̂

(11.5) 𝐹⃗𝑁𝐽

where 𝑢̂ is a unit vector pointing from 𝑚1 to 𝑚2 and 𝐺 is the universal gravitational constant∗ (sometimes called the constant of gravitation or constant of universal gravitation). The following example demonstrates the application of this law. Mini-Example Using the planets Jupiter and Neptune as an example, the force on Jupiter due to the gravitational attraction of Neptune, 𝐹⃗𝐽𝑁 , is given by (see Fig. 11.2) 𝐹⃗𝐽𝑁 =

𝐺𝑚𝐽 𝑚𝑁 𝑟2

𝑢, ̂

(11.6)

where 𝑟 is the distance between the two bodies, 𝑚𝐽 is the mass of Jupiter, 𝑚𝑁 is the mass of Neptune, and 𝑢̂ is a unit vector pointing from the center of Jupiter to the center of Neptune. The mass of Jupiter is 1.9 × 1027 kg, and that of Neptune is 1.02×1026 kg. Since the mean radius of Jupiter’s orbit is 778,300,000 km and that of Neptune is 4,505,000,000 km, we assume that their closest approach to one another is approximately 3,727,000,000 km. Thus, at their closest approach, the magnitude of the force between these two planets is 𝐹⃗𝐽𝑁 =

6.674×10−11

m3 kg⋅s2

𝐹⃗𝐽 𝑁

𝑢̂

𝐽 (left) JPL/University of Arizona/NASA; (right) NASA/JPL

Figure The gravitational force between the planets Jupiter 𝐽 and Neptune 𝑁. The relative sizes of the planets are accurate, but their separation distance is not.

1.9×1027 kg 1.02×1026 kg 3.727×1012 m

2

(11.7)

= 9.312×1017 N. We can compare this force with the force of gravitation between Jupiter and the Sun. The Sun’s mass is 1.989×1030 kg, and we have already stated that the mean radius of Jupiter’s orbit is 778,300,000 km. Applying Eq. (11.5) between Jupiter and the Sun gives 4.164×1023 N, which is almost 450,000 times larger.

Acceleration due to gravity. Equation (11.5) allows us to determine the force of Earth’s gravity on an object of mass 𝑚 on the surface of the Earth. This is done by noting that the radius of the Earth is 6371.0 km (see the marginal note) and the mass of the Earth is 5.9736×1024 kg and then applying Eq. (11.5): 3

EXAMPLE

5 9736×1024 kg 𝑚

Interesting Fact The radius of the Earth. The Earth is not a perfect sphere. Therefore, there are different notions of “radius of the Earth.” The given value of km is the volumetric

Tension in a Wrecking Ball Cable

The wrecking ball 𝐴 shown in Fig. 1 is released from rest when 𝜃 = 𝜃0 = 30◦ , and it swings freely about the fixed point at 𝑂. Assuming that the weight of the ball is 𝑊 = 2500 lb and 𝐿 = 30 f t, determine the tension in the cable to which the ball is attached when the ball reaches 𝜃 = 0◦ .

𝑂

𝜃

SOLUTION

𝐿

Road Map & Modeling Modeling the wrecking ball as a particle and neglecting all forces except the weight force 𝑊 and the cable tension 𝑇 , the FBD is as shown in Fig. 2. Applying Newton’s second law in the polar component system shown should allow us to find the tension in the cable as a function of its swing angle and thus, find its tension when 𝜃 = 0◦ .

Consistent Problem-Solving Methodology

Figure 𝑢̂ 𝜃

Governing Equations Balance Principles

Examples

𝐴

Referring to the FBD in Fig. 2 and applying Newton’s second law,

𝑂 𝑢̂ 𝑟

we obtain 𝐹𝜃 ∶

−𝑊 sin 𝜃 = 𝑚𝑎𝜃 ,

(1)

𝐹𝑟 ∶

𝑊 cos 𝜃 − 𝑇 = 𝑚𝑎𝑟 ,

(2)

𝐿 𝑇

where 𝑚 = 𝑊 ∕𝑔. Force Laws

𝐴

All forces are accounted for on the FBD. Writing 𝑎𝜃 and 𝑎𝑟 in polar components gives

Kinematic Equations

𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 𝐿𝜃̈

and

𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 = −𝐿𝜃̇ 2 ,

𝑊

(3)

where we have replaced 𝑟 with the constant length 𝐿. Computation

𝜃

Substituting Eqs. (3) into Eqs. (1) and (2), we obtain

− 𝑊 sin 𝜃 = 𝑚𝐿𝜃̈

and

𝑊 cos 𝜃 − 𝑇 = −𝑚𝐿𝜃̇ 2 𝑔 𝜃̈ = − sin 𝜃 and 𝑇 = 𝑊 cos 𝜃 + 𝑚𝐿𝜃̇ 2 . 𝐿

(4)

̈ ̇ Notice that the tension is a function of 𝜃,̇ so we need to integrate 𝜃(𝜃) to find 𝜃(𝜃) using the chain rule, that is, 𝑔 𝑑 𝜃̇ 𝜃̈ = 𝜃̇ = − sin 𝜃 𝑑𝜃 𝐿

𝜃̇ 0

𝑔 𝜃̇ 𝑑 𝜃̇ = − 𝐿

𝜃

sin 𝜃 𝑑𝜃 𝜃0

𝑔 𝜃̇ 2 = 2 cos 𝜃 − cos 𝜃0 . 𝐿

(5)

Substituting Eq. (5) into the expression for 𝑇 in Eq. (4) gives 𝑇 (𝜃) as 𝑇 = 𝑊 (3 cos 𝜃 − 2 cos 𝜃0 )

𝑇 (𝜃 = 0) = 𝑊 3 − 2 cos 𝜃0 = 3170 lb,

(6)

Figure FBD of the wrecking ball as it swings downward.

Every problem in the text employs a carefully defined problem-solving methodology to encourage systematic problem formulation while reinforcing the steps needed to arrive at correct and realistic solutions. Each example problem contains these steps: • Road Map • Modeling • Governing Equations • Computation • Discussion & Verification Some examples include a Closer Look (noted ) that offers with a magnifying glass icon additional insight into the example.

where we have used 𝑊 = 2500 lb and 𝜃0 = 30◦ to obtain the final numerical result. Discussion & Verification

The final result in Eq. (6) is dimensionally correct, and the magnitude of the tension seems reasonable. Interestingly, the tension does not depend on the length of the supporting cable. That is, if the initial angle is 30◦ and the wrecking ball is released from rest, the tension in the cable will always be 3170 lb, regardless of the length of the suspending cable.

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Concept Alerts and Concept Problems

Concept Alert

Two additional features are the Concept Alerts and the Concept Problems. These have been included because research has shown (and it has been our experience) that even though you may do quite well in a science or engineering course, your conceptual understanding may be lacking. Concept Alerts are marginal notes and are used to drive home important concepts (or help dispel misconceptions) that are related to the material being developed at that point in the text. Concept Problems are mixed in with the problems that appear at the end of each section. These are questions designed to get you thinking about the application of a concept or idea presented within that section. They should never require calculation and should require answers of no more than a few sentences.

Direction of velocity vectors. One of the most important concepts in kinematics is that the velocity of a particle is always tangent to the particle’s path. Problem An object is lowered very slowly owly onto a conveyor belt that is moving to the right. What is the direction of the friction force f acting on the object at the instant the object touches the belt? 𝑣

Figure

Problem A person is trying to move a heavy crate by pushing on it. While the person is pushing, what is the resultant force acting on the crate if the crate does not move?

Common Pitfall

Marginal Notes

Cyclic loading and fatigue. The fact that, C Newton’s second law and inertial frames. u under the given conditions, the rotor Since the application of Newton’s second b bearing experiences a cyclic load law requires the use of an inertial referttimes per second means that it will quickly ence frame, the component system shown experience a large number of load cycles. e in Fig. must be understood as originatIIt turns out that even a rather low stress ing from an 𝑥𝑦 coordinate nate system fixed x with after millions ccan cause an object to break b the ground—this is the inertial reference of load cycles. The high o higher the stress, the Helpful Information istake to choose a coframe. It would be a mistake ssmaller the number of cyc cycles required. This ng with the truck beordinate system moving The right-hand rule. In three dimensions, mechanism m of failure f is ca called fatigue. Since celerating with respect cause the truck is decelerating a Cartesian coordinate system uses three tthe number of load cycles cyc on the rotor refore, is not an inertial to the ground and, therefore, orthogonal reference directions. are bearingThese b of a centrifuge gr grows quickly, even frame of reference. the 𝑥, 𝑦, and 𝑧 directions shown a smallbelow. imbalance can ca cause failure due to ffatigue. To T llearn earn more a about fatigue, see 𝑧 W. D. Callister,r Jr., W Jr., Materials Mater Science and Engineering: An Introduct Introduction th ed., John Wiley & Sons, in a

Marginal notes have been implemented that will help place topics, ideas, and examples larger context. This feature will help students study (using Helpful Information and Common Pitfalls) and will provide real-world examples of how different aspects of statics are used (using Interesting Facts).

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Interesting Fact

𝑥

𝑦

Proper interpretation of many vector operations, such as the cross product, requires that the 𝑥, 𝑦, and 𝑧 directions be arranged in a consistent manner. The convention in mechanics and vector mathematics in general is that if the axes are arranged as shown, then, according to the right-hand rule, rotating the 𝑥 direction into the 𝑦 direction yields the 𝑧 direction. The result is called a righthanded coordinate system.

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Sections and End of Section Summary Each chapter is organized into several sections. There is a wealth of information and features within each section, including examples, problems, marginal notes, and other pedagogical aids. Each section concludes with an end of section summary that succinctly summarizes that section. In many cases, cross-referenced important equations are presented again for review and reinforcement before the student proceeds to the examples and homework problems.

Undamped Forced Vibration Many systems are forced to vibrate by an external excitation. This section is devoted to the forced vibration of mechanical systems.

Standard form of the forced harmonic oscillator

𝑘

A standard forced harmonic oscillator is shown in Fig. 19.11, in which the block of mass 𝑚 is attached to a fixed support by a linear spring of constant 𝑘 and is also being driven by the time-dependent force 𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡. Modeling the block as a particle, its FBD is as shown in Fig. 19.12, where 𝐹𝑠 is the spring force acting on the block. Summing forces in the 𝑥 direction, we obtain 𝐹𝑥 ∶

𝑃 (𝑡) − 𝐹𝑠 = 𝑚𝑎𝑥 ,

(19.29)

Figure A forced harmonic oscillator whose equation √ of motion is given by Eq. (19.31) with 𝜔𝑛 = 𝑘∕𝑚. The position 𝑥 is measured from the equilibrium position of the system when 𝐹0 = 0.

̈ where the force law is given by 𝐹𝑠 = 𝑘𝑥 and the kinematic equation is 𝑎𝑥 = 𝑥. Substituting these relations, as well as 𝑃 (𝑡) into Eq. (19.29), we obtain 𝐹0 sin 𝜔0 𝑡 − 𝑘𝑥 = 𝑚𝑥̈

𝑥̈ +

𝐹 𝑘 𝑥 = 0 sin 𝜔0 𝑡. 𝑚 𝑚

𝐹0 𝑚

sin 𝜔0 𝑡,

𝚤̂

𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡

𝐹𝑠 𝑁

Noting that 𝜔2𝑛 = 𝑘∕𝑚, this last equation becomes 𝑥̈ + 𝜔2𝑛 𝑥 =

𝑚𝑔 𝚥̂

(19.30)

𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡

𝑚

(19.31)

Figure FBD of the forced harmonic oscillator in Fig. 19.11.

which is the standard form of the forced harmonic oscillator equation. It is a nonhomogeneous version of Eq. (19.12) on p. 1291 as a result of the term (𝐹0 ∕𝑚) sin 𝜔0 𝑡. The term on the right-hand side of Eq. (19.31) is a function of only the independent variable 𝑡. It is often called a forcing function because it forces the system to vibrate. Interesting Fact This particular type of forcing is harmonic because it is a harmonic function of time. The theory of differential equations tells us that the general solution of Eq. (19.31) How practical is harmonic forcing? The is the sum of the complementary solution 𝑥𝑐 (𝑡) and a particular solution 𝑥𝑝 (𝑡). The answer to this question lies in an amazing ∗ result due to Jean Baptiste Joseph Fourier complementary solution is the solution of the associated homogeneous equation and later cont contributors. It (i.e., Eq. (19.12)) given in Eq. (19.3) (or in Eq. (19.13)). The par particular partic ticula i ular l r soluti sol solution lution ion is says that any periodic piecew piecewise smooth of Eq. One way to obtain a particular solution is to guess its form E n d o f S e c t i oany nn solution Summ a r (19.31). y function can be represented by an infinite and then verify whether or not the guess is correct. Since it seems reasonable that the series of sines and cosines (ca (called Fourier When a harmonic oscillator is subject forcing, the should standard form ofthe thefforcing, we conjecture response of a forced f to harmonic harmonic oscillator resemble series in honor of Fourier). This means T equation of motion is that the particular solution 𝑥 is of the ffor form m that any periodic forcing can be regarded 𝑝 as the sum of harmonic functi functions! In adEq. (19.31), p. 1307 𝑥𝑝 = 𝐷 sin 𝜔0 𝑡, (19.32) dition, given the nature of the left side of 𝑘 𝑃 (𝑡) = 𝐹 sin 𝜔 𝑡 0 it is0 linear), it tur 𝑚 Eq. (i.e., turns out that 𝐹0 2 + a𝜔constant sinbe𝜔determined. where 𝐷𝑥̈ is We can verify whether our guess is correct the overall particular solution for fo a sum of 𝑛 𝑥 = 𝑚 to 0 𝑡, harmonic forcing terms is simply the sum of by substituting Eq. (19.32) into Eq. (19.31). Doing so yields the particular solutions for each individual Figure where 𝐹0 is the amplitude of the forcing and 𝜔0 is its frequency (see Fig. 𝐹019.16). term. These results taken harmonic re 2 2 A forced harmonic oscillator whoseforcing equation sin 𝜔 −𝐷𝜔 sin 𝜔 𝑡 + 𝜔 𝐷 sin 𝜔 𝑡 = 𝑡. (19.33) The general solution to this equation consists of the sum of the complementary solu√ of 0 𝑛 0 0 0 easily obtain together to ea 𝑚 motion is given by Eq. (19.31) with allow 𝜔𝑛 = engineers 𝑘∕𝑚. tion and a particular solution. The complementary solution 𝑥𝑐 is the solution of the any periodic thefrom solutions to problems with a The position 𝑥 is measured the equilibrium Canceling 𝜔0 𝑡 and solving we obtain forcing as a sum of simple force forced harmonic associated homogeneous equation,sinwhich is given by,for for𝐷,example, Eq. (19.13). For position of the block. periodic forcoscillator solutions. Because pe 𝜔0 ≠ 𝜔𝑛 , a particular solution was found to be 𝐹0 ∕𝑘 𝐹0 ∕𝑚 engineering systems, ing is ubiquitous in engineerin 𝐷= = , (19.34) 2 2 2 this is one of the most importan important results in Eq. (19.35), p. 1308 𝜔𝑛 − 𝜔0 1 − 𝜔0 ∕𝜔𝑛 applied mathematics.

∗ The

complementary 𝐹0solution ∕𝑘 is sometimes called the homogeneous solution. 𝑥𝑝 = sin 𝜔0 𝑡, 2 1 − 𝜔0 ∕𝜔𝑛

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Problems Problems 𝐵

𝜔1

If 𝜔1 and 𝜃̇ are both constant, determine the velocity and acceleration of the subreflector 𝐵 in terms of the elevation angle 𝜃.

𝜃

Problem 𝑂

̇ are known functions of time, determine the velocity If 𝜔1 (𝑡) and 𝜃(𝑡) and acceleration of the subreflector 𝐵 in terms of the elevation angle 𝜃.

Problem

𝑥

Problems

Figure

and

The radar dish can rotate about the vertical 𝑧 axis at rate 𝜔1 and about the horizontal 𝑦 ̇ The distance between the center of rotation at 𝑂 axis (not shown in the figure) at rate 𝜃. and the subreflector at 𝐵 is 𝓁.

and

The truncated cone rolls without slipping on the 𝑥𝑦 plane. At the instant shown, the angular speed about the 𝑧 axis is 𝜔1 , and it is changing at 𝜔̇ 1 .

and

𝜔1 Problems

and

𝜃

𝑂 plate A micro spiral pump∗ consists of a spiral𝑥channel attached to a stationary plate. This has two ports, one for fluid inlet and another for outlet, the outlet being farther from the center of the plate than the inlet. The system is capped by a rotating disk. The fluid trapped between the rotating disk and the stationary plateFigure is put in motionand by the rotation of the top disk, which pulls the fluid through the spiral channel. Problem

rotating disk

spin axis 𝜔

𝑢̂ 𝜃 Determine expressions for the angular velocity and angular acceleration

𝜃 Consider a spiral channel with the geometry given by the equation Problem of the cone in terms of 𝓁, 𝑑, 𝜃, 𝜔 , and 𝜔. ̇ Express your answers in the rotating component𝑢̂ 𝑟 𝑟 = 𝜂𝜃 + 𝑟0 , where 𝑟0 = 146 𝜇m is the starting 1radius, 𝑟 is the distance from the spin axis, and 𝜃, system measuredshown. in radians, is the angular position of a point in the spiral channel. Assume that the radius at the outlet is 𝑟out = 190 𝜇m, that the top disk rotates with a pin joint constant angular speed 𝜔, and that the fluid particles in contact with the rotating disk are stationary plate essentially stuck to it. Determine the constant 𝜂 and the value of 𝜔 (in rpm) such that after 1.25 rev of the top disk, the speed of the particles in contact with this disk is 𝑣 = 0.5 m∕s Figure and at the outlet.

Modern Problems Problems of varying difficulty follow each section. These problems allow students to develop their ability to apply concepts of statics and dynamics on their own. The most common question asked by students is “How do I set this problem up?” What is really meant by this question is “How do I develop a good mathematical model for this problem?” The only way to develop this ability is by practicing numerous problems. Answers to most even-numbered problems appear in Appendix B. Providing answers in this manner allows for more complex information than would otherwise be possible.

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inlet

𝑟

outlet spiral groove

Furthermore, the free body diagrams that are provided for some problems as part of the model for that problem will give students ample opportunity to practice constructing FBDs on their own for extra problems. Appendix B gives examples of the additional information provided for particular problems. Each problem in the book is accompanied by a thermometer icon that indicates the approximate level of difficulty. Those considered to be “introductory” are indicated with the symbol . Problems considered to be “representative” are indicated with the symbol , and problems that are considered to be “challenging” are indicated with the symbol .

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Engineering Design and Design Problems Several design problems are presented where appropriate throughout the book. These problems can be tackled with the knowledge and skill set that are typical of introductory-level courses, although the use of mathematical software is strongly recommended. These problems are open ended, and their solution requires the definition of a parameter space in which the statics or dynamics of the system must be analyzed. In this textbook we have chosen to emphasize the role played by parametric analyses in the overall design process, as opposed to cost-benefit analyses or the choice of specific materials and/or components.

Design Problems Design Problem Revisit the calculations done at the beginning of the chapter concerning the determination of the maximum acceleration that can be achieved by a motorcycle without causing the front wheel to lift off the ground. Specifically, construct a new model of the motorcycle by selecting a real-life motorcycle and researching its geometry and inertia properties, including the inertia properties of the wheels. Then analyze your model to determine how the maximum acceleration in question depends on the horizontal and vertical positions of the center of mass with respect to the points of contact between the ground and the wheels. Include in your analysis a comparison of results that account for the inertia of the front wheel with results that neglect the inertia of the front wheel.

Guitar Studio/Shutterstock

Figure

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Introduction to Statics

1

In statics we study the equilibrium of bodies under the action of forces that are applied to them. Our goal is to provide an introduction to the science, skill, and art involved in modeling and designing real life mechanical systems. We begin the study of statics with an overview of the relevant history of the subject. In subsequent sections and chapters, we cover those elements of physics and mathematics (especially vectors) needed to analyze the equilibrium of particles and rigid bodies. Throughout the book are discussions and applications of engineering design.

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The Infinity Bridge crossing the River Tees in England is a dual arch bridge with steel arches and concrete decking. Concepts from statics formed the basis for the analysis and design of this bridge.

1.1 Engineering and Statics Engineers design structures, machines, processes, and much more for the benefit of humankind. In the process of doing this, an engineer must answer questions such as “Is it strong enough?” “Will it last long enough?” and “Is it safe enough?” To answer these questions requires the ability to quantify important phenomena in the design or system at hand, and to compare these measures with known criteria for what is acceptable and what is not. To do this requires an engineer to have thorough knowledge of science, mathematics, and computational tools, and the creativity to exploit the laws of nature to develop new designs. Central to all of this is the ability to idealize real life problems with mathematical models that capture the essential science of the problem, yet are tractable enough to be analyzed. Proficiency in doing this is a characteristic that sets engineering apart from the pure sciences. In most engineering disciplines, understanding the response of materials or objects subjected to forces is important, and the fundamental science concepts

1

2

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Chapter 1

Introduction to Statics

Statics

Dynamics Advanced Dynamics

Mechanics of Materials Advanced Mechanics of Materials

Vibrations Controls

Finite Element (Computer) Analysis

Robotics

Structural Analysis

Astrodynamics

Aerospace Structures

Satellite Mechanics

Machine Design

Machine Dynamics

Steel and Concrete Design Soil and Rock Mechanics Biomechanics

Figure 1.1. Hierarchy of subject matter and courses studied by many engineering students. Courses in statics, dynamics, and mechanics of materials provide fundamental concepts and a basis for more advanced study. Many subjects, such as vibrations and finite element analysis, draw heavily on concepts from both dynamics and mechanics of materials.

governing such response are known as Newtonian physics.∗ This book examines applications of this topic to engineering problems under the special circumstances in which a system is in force equilibrium, and this body of knowledge is called statics. Statics is usually the first engineering course that students take. Statics is an important subject in its own right, and it develops essential groundwork for more advanced study. If you have read this far, then we presume you are embarking on a study of statics, using this book as an aid. Figure 1.1 shows a hierarchy of subjects, many of which you are likely to study en route to an education in engineering. Following a course in statics are introductory courses in dynamics and mechanics of materials. Dynamics studies the motion of particles and bodies subjected to forces that are not in equilibrium. Mechanics of materials introduces models for material behavior and methods for determining stresses and deformations in structures. The concepts learned in these three basic courses are used daily by almost all engineers who are concerned with the mechanical response of structures and materials! This book will provide you with a solid and comprehensive education in statics. Often, when engineering problems are boiled down to their essential elements, they are remarkably simple to analyze. In fact, throughout most of this book, the mathe∗ When

the velocity of an object is close to the speed of light, relativistic physics is required.

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Section 1.3

A Brief History of Statics

matics needed to analyze problems is straightforward. The bigger challenge usually lies in the idealization of a real life problem by a model, and we hope this book helps you cultivate your ability to do this. Regarding mathematics, this book assumes you have knowledge of algebra and basic trigonometry. Later in this book, beginning in Chapter 7, basic calculus involving differentiation and integration of simple functions is used. Vectors is an important topic, and this book assumes that you have no prior knowledge of this; everything you need to know about vectors for statics will be covered in this book.

1.2

Topics That Will Be Studied in Statics

The remainder of Chapter 1 is devoted to a discussion of the physical entities and governing equations that form the basis of statics and dynamics. An important related topic is the choice of the unit system to be used. Chapter 2 introduces vectors and how they are used to represent entities such as force and position. In Chapters 3 through 6, we use statics to solve problems involving particles and systems of particles that are in equilibrium, and bodies and systems of bodies (i.e., frames and machines) that are in equilibrium. Each of these topics builds upon previous topics to enable you to model engineering problems of increasingly greater sophistication. Chapters 1 through 6 constitute the core of topics in statics. Beginning in Chapter 7, we treat systems that have continuous distributions of properties such as mass, weight, and pressure; basic calculus is effective and is used beginning here. Chapter 8 addresses internal forces that develop within structures due to loads that are applied to them; knowledge of internal forces is essential to create designs and to address questions such as “Is it strong enough?” and “Is it safe enough?” Chapter 9 is devoted to friction, which is a type of force between contacting bodies. Friction presents some challenges to engineers to model and account for in engineering problems. Finally, Chapter 10 is devoted to moments of inertia, which characterize how area and mass are distributed; this topic is essential in dynamics and mechanics of materials, and it marks the transition from statics to these subjects.

1.3

A Brief History of Statics∗

Interesting Fact Early structural design codes. While most of our discussion focuses on accomplishments of philosophers, there were also significant accomplishments in the development of structural design codes over a period of thousands of years. Some of these include the ancient books of Ezekiel and Vitruvius and the secret books of the medieval masonic lodges. Additional history is given in J. Heyman, “Truesdell and the History of the Theory of Structures,” a chapter in Essays on the History of Mechanics, edited by A. Becchi, M. Corradi, F. Foce, and O. Pedemonte, Birkhauser, Boston, 2003. These codes were largely empirical rules of proportion that provided for efficient design and construction of masonry structures. The great Greek temples, Roman aqueducts, and Gothic cathedrals are a testament to their effectiveness. While the writers of these codes were not philosophers, their engineering accomplishments were impressive.

The history of statics is not a distinct subject, as it is closely intertwined with the development of dynamics and mechanics of materials. Early scientists and engineers were commonly called philosophers, and their noble undertaking was to use thoughtful reasoning to provide explanations for natural phenomena. Much of their focus was on understanding and describing the equilibrium of objects and the motion of celestial bodies. With few exceptions, their studies had to yield results that were intrinsically beautiful and/or compatible with the dominant religion of the time and place. What follows is a short historical survey of the major figures who profoundly influenced the development of key aspects of mechanics that are especially significant to statics. Bruno Cossa/SOPA/Corbis ∗ This history is based on the excellent works of C. Truesdell, Essays in the History of Mechanics, Springer-

Verlag, Berlin, 1968; I. Bernard Cohen, The Birth of a New Physics, revised and updated edition, W. W. Norton & Company, New York, 1985; R. Dugas, A History of Mechanics, Dover, Mineola, NY, 1988; and James H. Williams, Jr., Fundamentals of Applied Dynamics, John Wiley & Sons, New York, 1996.

The Parthenon in Athens, Greece, was completed in 438 B.C. and is an example of early column and beam masonry construction.

3

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Chapter 1

Introduction to Statics

For centuries, philosophers studied the equilibrium and motion of bodies with less than full understanding, and sometimes incorrect understanding. Notable early contributors include: • Aristotle (384–322 B.C.) wrote about science, politics, economics, and biology, and he proposed what is often called a “physics of common sense.” He studied levers and although he attributed their efficiency to the “magical” properties of the circle, he understood some basic concepts of the moment of a force and its effect on equilibrium. He classified objects as being either light or heavy, and he believed that light objects fall more slowly than heavy objects. He recognized that objects can move in directions other than up or down; he said that such motion is contrary to the natural motion of the body and that some force must continuously act on the body for it to move this way. Most importantly, he said that the natural state of objects is for them to be at rest. • Archimedes (287–212 B.C.) postulated several axioms based on experimental observations of the equilibrium of levers, and using these, he proved several propositions. His work shows further understanding of the effects of the moment of a force on equilibrium. Archimedes is perhaps best known for his pioneering work on hydrostatic fluid mechanics, where one of his discoveries was that a body that floats in fluid will displace a volume of fluid whose weight is equal to that of the body. Recently, evidence has been found that he discovered some elementary concepts of calculus. • Leonardo da Vinci (1452–1519) had bold imagination and tackled a wide variety of problems. He correctly understood the moment of a force and used the terminology arm of the potential lever to describe what we today call the moment arm. While his conclusions were wrong, he studied the equilibrium of a body supported by two strings. He also conducted experiments on the strength of structural materials. Following the progress of these and many other early philosophers came the work of Galileo and Newton. With their work came rapid progress in achieving the essential elements of a theory for the motion of bodies, and their accomplishments represent the most important milestone in the history of mechanics until the work of Einstein. The contributions of Galileo and Newton are discussed in some detail in the remainder of this section.

Galileo Galilei

Fine Art Images/Heritage Image Partnership Ltd/ Alamy Stock Photo

Figure 1.2 A portrait of Galileo painted in 1636 by Justus Sustermans. ISTUDY

Galileo Galilei (1564–1642) had a strong interest in mathematics, mechanics, astronomy, heat, and magnetism. He made important contributions throughout his life, despite persecution from the church for his support of the Copernican theory that the Earth was not the center of the universe. One of his most important contributions was his thought experiment in which he concluded that a body in its natural state of motion has constant velocity. Galileo discovered the correct law for freely falling bodies; that is, the distance traveled by a body is proportional to the square of time. He also concluded that two bodies of different weight would fall at the same rate and that any differences are due to air resistance. Galileo developed a theory (with some minor errors) for the strength of beams, such as that shown in Fig. 1.3. He was the first to use the concept of stress as a fundamental measure of the loading a material supports, and he is viewed as the father of mechanics of materials. He also discovered that the strength of structures does not scale linearly; that is, if the dimensions

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Section 1.4

Fundamental Principles

5

of a beam are doubled, the load the beam can support does not double. He speculated that it is for this reason that trees, animals, and so on have natural limits to the size they could reach before they would fail under their own weight. More importantly, his work showed that newer, larger structures could not necessarily be built by simply scaling the dimensions of smaller structures that were successfully built.

Isaac Newton Newton (1643∗ –1727) was one of the greatest scientists of all time. He made important contributions to optics, astronomy, mathematics, and mechanics, and his collection of three books entitled Philosophiæ Naturalis Principia Mathematica, or Principia as they are generally known, which were published in 1687, is considered by many to be the greatest collection of scientific books ever written. In the Principia, Newton analyzed the motion of bodies in “resisting” and “nonresisting media.” He applied his results to orbiting bodies, projectiles, pendula, and free fall near the Earth. By comparing his “law of centrifugal force” with Kepler’s third law of planetary motion, Newton further demonstrated that the planets were attracted to the Sun by a force varying as the inverse square of the distance, and he generalized that all heavenly bodies mutually attract one another in the same way. In the first book of the Principia, Newton develops his three laws of motion; in the second book he develops some concepts in fluid mechanics, waves, and other areas of physics; and in the third book he presents his law of universal gravitation. His contributions in the first and third books are especially significant to statics and dynamics. Newton’s Principia was the final brick in the foundation of the laws that govern the motion of bodies. We say foundation because it took the work of Daniel Bernoulli (1700–1782), Johann Bernoulli (1667–1748), Jean le Rond d’Alembert (1717–1783), Joseph-Louis Lagrange (1736–1813), and Leonhard Euler (1707–1783) to clarify, refine, and advance the theory of dynamics into the form used today. Euler’s contributions are especially notable since he used Newton’s work to develop the theory for rigid body dynamics. Newton’s work, along with Galileo’s, also provided the foundation for the theory of mechanical behavior of deformable bodies, which is more commonly called mechanics of materials. However, it took the work of CharlesAugustin Coulomb (1736–1806), Claude Louis Marie Henri Navier (1785–1857), and Augustin Cauchy (1789–1857) to further refine the concept of stress into the form used today; the work of Robert Hooke (1635–1703) and Thomas Young (1773–1829) to develop a theory for elastic deformation of materials; and the work of Leonhard Euler (1707–1783) to consider the deformations of a structure (an elastic strip in particular).†

The Picture Art Collection/Alamy Stock Photo

Figure 1.3 A sketch from Galileo’s last book Discourses on Two New Sciences, published in 1638, where he studies the strength of beams, among several other topics.

Imagno/Hulton Fine Art Collection/Getty Images

1.4

Fundamental Principles

Space and time. Most likely you already have a good intuitive understanding of the concepts of space and time. In fact, to refine concepts of space and time is not easy and may not provide the clarification we would like. Space is the collection of all positions in our universe that a point may occupy. The location of a point is usually described using a coordinate system where measurements are made from some ∗ This

birth date is according to the Gregorian, or “modern,” calendar. According to the older Julian calendar, which was used in England at that time, Newton’s birth was in 1642. † Additional comments on the history of mechanics as it pertains to mechanics of materials are given in M. Vable, Mechanics of Materials, Oxford University Press, New York, 2002.

Figure 1.4 A portrait of Newton painted in 1689 by Sir Godfrey Kneller, which is owned by the 10th Earl of Portsmouth. It shows Newton before he went to London to take charge of the Royal Mint and when he was at his scientific peak.

6

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Introduction to Statics

Chapter 1

reference position using the coordinate system’s reference directions. While selection of a reference position and directions is arbitrary, it is usually based on convenience. Because space is three-dimensional, three pieces of information, called coordinates, are required to locate a point in space. Most often we will use a rectangular Cartesian coordinate system where the distances to a point are measured in three orthogonal directions from a reference location. Other coordinate systems, such as spherical and cylindrical coordinates (and polar coordinates in two dimensions), are sometimes more convenient. All engineering problems are three-dimensional, but often we will be able to idealize a problem as being two-dimensional or one-dimensional. Time provides a measure of when an event, or sequence of events, occurs. Mass and force. Mass is the amount of matter, or material, in an object. Force is an agency that is capable of producing motion of an object. Forces can arise from contact or interaction between objects, from gravitational attraction, from magnetic attraction, and so on. As discussed in Section 1.6, interpretation and quantification of mass and force should be viewed as being related by Newton’s second law of motion. Force is discussed further in Section 1.5. Particle. A particle is an object whose mass is concentrated at a point. For this reason, a particle is also called a point mass, and it is said to have zero volume. An important consequence of this definition is that the notion of rotational motion of a particle is meaningless. Clearly there are no true particles in nature, but under the proper circumstances it is possible to idealize real life objects as particles. Objects that are small compared to other objects and/or dimensions in a problem can often be idealized as particles. For example, to determine the orbit of a satellite around the Earth, it is probably reasonable to idealize the satellite as a particle. Objects do not necessarily need to be small to be accurately idealized as particles. For example, for the satellite orbiting Earth, the Earth is not small, but for many purposes the Earth can also be idealized as a particle. Body and rigid body. A body has mass and occupies a volume of space. In nature, all bodies are deformable. That is, when a body is subjected to forces, the distances between points in the body may change. A rigid body is a body that is not deformable, and hence the distance between any two points in the body never changes. There are no true rigid bodies in nature, but very often we may idealize an object to be a rigid body, and this provides considerable simplification because the intricate details of how the body deforms do not need to be accounted for in an analysis. Furthermore, in statics we will be able to make precise statements about the behavior of rigid bodies, and we will establish methods of analysis that are exact.

Concept Alert Vectors. A vector is an entity that has both size and direction. Vectors are immensely useful in mechanics, and the ability to use vectors to represent force, position, and other entities is essential.

Scalars and vectors. A scalar is a quantity that is completely characterized by a single number. For example, temperature, length, and density are scalars. In this book, scalars are denoted by italic symbols, such as 𝑠. A vector is an entity that has both size (or magnitude) and direction. Much will be said about vectors in Chapter 2, but basic notions of vectors will be useful immediately. Statements such as “my apartment is 1 mile northeast of Engineering Hall” and “I’m walking north at 3 km/h” are statements of vector quantities. In the first example, the position of one location relative to another is stated, while in the second example, the velocity is stated. In both examples, commonly used reference directions of north and east are employed. Vectors are immensely useful for describing many entities in mechanics. Vectors offer compact representation and easy manipulation, and they can be transformed. That

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Section 1.4

Fundamental Principles

is, if a vector is known in reference to one set of coordinate directions, then using established rules for transformation, the vector is known in any other set of coordinate directions. In this book, vectors are denoted by placing an arrow above the symbol for the vector, such as 𝑣. ⃗ Position, velocity, and acceleration. Position, velocity, and acceleration are all examples of vectors. If we consider a particle that has position 𝑟⃗ relative to some location, then the velocity of the particle is the time rate of change of its position 𝑣⃗ = 𝑑 𝑟⃗∕𝑑𝑡,

(1.1)

where 𝑑∕𝑑𝑡 denotes the derivative with respect to time.∗ Similarly, the acceleration is the time rate of change of velocity 𝑎⃗ = 𝑑 𝑣∕𝑑𝑡. ⃗

(1.2)

⃗ our discussion of Eqs. (1.1) Since statics is concerned with situations where 𝑎⃗ = 0, and (1.2) will be brief. If a particle’s acceleration is zero, then integration of Eq. (1.2) shows the particle has constant velocity, which may be zero or nonzero. If the velocity is zero, then Eq. (1.1) shows the particle’s position does not change, while if the velocity is nonzero but is constant, integration of Eq. (1.1) shows the particle’s position changes as a linear function of time. If the acceleration is not zero, then the particle will move with velocity and position that change with time.

Newton’s laws of motion Inspired by the work of Galileo and others before him, Newton postulated his three laws of motion in 1687: First Law. A particle remains at rest, or continues to move in a straight line with uniform velocity, if there is no unbalanced force acting on it. Second Law. The acceleration of a particle is proportional to the resultant force acting on the particle and is in the direction of this force. The mathematical statement of this law† is 𝐹⃗ = 𝑚𝑎, ⃗

(1.3)

where 𝐹⃗ is the resultant force acting on the particle, 𝑎⃗ is the acceleration of the particle, and the constant of proportionality is the mass of the particle 𝑚. In Eq. (1.3), 𝐹⃗ and 𝑎⃗ are vectors, meaning they have both size (or magnitude) and direction. Vectors are discussed in detail in Chapter 2. Third Law. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear. Newton’s laws of motion, especially Eq. (1.3), are the basis of mechanics. They are postulates whose validity and accuracy have been borne out by countless experiments ∗ Equations

(1.1) and (1.2) are valid regardless of how a vector might be represented. However, the details of how the time derivative is evaluated depend on the particular vector representation (e.g., Cartesian, spherical, etc.) that is used. Dynamics explores these details further. † Actually, Newton stated his second law in a more general form as 𝐹⃗ = 𝑑(𝑚𝑣)∕𝑑𝑡, ⃗ where 𝑣⃗ is the velocity of the particle and 𝑑(𝑚𝑣)∕𝑑𝑡 ⃗ denotes the time rate of change of the product 𝑚𝑣, ⃗ which is called the momentum of the particle. When mass is constant, this equation specializes to Eq. (1.3). For problems in which mass is not constant, such as in the motion of a rocket that burns a substantial mass of fuel, the more general form of Newton’s second law is required.

Concept Alert Newton’s second law. Newton’s second law, 𝐹⃗ = 𝑚𝑎, ⃗ is the most important fundamental principle upon which statics, dynamics, and mechanics in general are based.

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Chapter 1

Introduction to Statics

Interesting Fact Measuring force. A force can cause an unsupported body to accelerate and also can cause a body (both unsupported and supported) to deform, or change shape. This suggests two ways to measure force. First, for an accelerating body with known mass 𝑚, by measuring the acceleration 𝑎, ⃗ we may then determine the force 𝐹⃗ applied to the body, using Newton’s law 𝐹⃗ = 𝑚 𝑎. ⃗ This approach is common in celestial mechanics and projectile motion, but it cannot be used for objects that are in static equilibrium. A second approach that is more common for both static and dynamic applications is by measuring the deformation (i.e., shape change) that a force produces in an object whose behavior is known. An example is the handheld spring scale shown, which is being used to weigh bananas.

pounds 0 1 2 3 4 5 6

The weight of the bananas causes the spring’s length to change, and because the spring’s stiffness is known, the force the bananas apply to the scale can be determined. A brief historical discussion of mass and force measurements is given in J. C. Maxwell’s notes on dynamics entitled Matter and Motion, Dover Publications, Inc., New York, 1991, the preface of which is dated 1877. A more contemporary discussion of force measurements (and measurements in general) is available from the National Institute of Standards and Technology (NIST) (see http://www.nist.gov/).

and applications for more than three centuries. Unfortunately, there is no fundamental proof of their validity, and we must accept these as rules that nature follows. The first law was originally stated by Galileo. Of the three laws, only the second two are independent. In Eq. (1.3), we see that if the resultant force 𝐹⃗ acting on a particle is zero, then the acceleration of the particle is zero, and hence the particle may move with uniform velocity, which may be zero or nonzero in value. Hence, when there ⃗ the particle is said to be in static equilibrium, or is no acceleration (i.e., 𝑎⃗ = 0), simply equilibrium. The third law will play an important role when drawing free body diagrams, which we will see are an essential aid for applying 𝐹⃗ = 𝑚𝑎. ⃗

1.5

Force

Forces are of obvious importance to us. In statics, we are usually interested in how structures support the forces that are applied to them, and how to design structures so they can accomplish the goal of supporting forces. In dynamics, we are usually interested in the motions of objects that are caused by forces that are applied to them. In this section, we discuss force in some detail, examine some different types of forces, and discuss how forces are produced. Simply stated, a force is any agency that is capable of producing an acceleration of an unsupported body.∗ While this definition may seem vague, it is comprehensive. All forces are produced from the interaction of two or more bodies (or collections of matter), and the interaction between the bodies can take several forms, which gives rise to different ways that forces can be produced. For many purposes, a force can be categorized as being either a contact force or a field force: • Contact force. When two bodies touch, contact forces develop between them. In general, the contact forces are distributed over a finite area of contact, and hence, they are distributed forces with dimensions of force/area. If the bodies touch over only a small region, or if we replace the distributed force by an equivalent concentrated force as discussed in Chapter 7, then the contact forces are concentrated at a point. Contact forces are made up of two parts: a normal-direction force and a tangential-direction force, which is also called the friction force. Examples of contact forces include the forces between your feet and ground when you are standing, and the force applied by air to a building during a blowing wind. • Field force. A force between bodies that acts through space is called a field force. Field forces act throughout the volume of an object and thus have dimensions of force/volume. Field forces are often called body forces. For many applications, we can represent a field force by a concentrated force that acts at a point. Examples of field forces include the weight of an object, the attractive force between the Earth and Moon, and the force of attraction between a magnet and an iron object. Some examples of contact and field forces are shown in Fig. 1.5. Although the preceding definition of contact forces is useful, more careful consideration of contact at an atomic length scale shows that a contact force is a special ∗ Whether

or not a particular body does accelerate depends upon the combined action of all forces that are applied to the body.

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Section 1.6

Units and Unit Conversions

𝑊

𝑊 𝑃 𝐹 𝑁

𝑁 (a)

(b)

Figure 1.5. Examples of contact forces and field forces. (a) A basketball rests on a hard level surface. (b) A book is pushed across a table with your finger. In both examples, the field force is the weight 𝑊 of the object, and the contact forces are the normal force 𝑁, the friction force 𝐹 , and the force 𝑃 applied by your finger to the book. For the basketball, contact occurs over a very small region, and it is reasonable to idealize this as a point. For the book, contact occurs over the entire surface of the book cover, but it is nonetheless possible to model the contact forces by concentrated forces acting at a point.

case of a field force. As an atom from one surface comes very close to an atom on the opposite surface, the atoms never touch one another, but rather they develop a repulsive field force that increases rapidly as the two atoms come closer. However, the range of distances over which these forces act is very small (on the order of atomic dimensions), and for macroscopic applications, our definition of contact forces is useful.

1.6

Units and Unit Conversions

Units are an essential part of any quantifiable measure. Newton’s law 𝐹 = 𝑚𝑎, written here in scalar form, provides for the formulation of a consistent and unambiguous system of units. We will employ both U.S. Customary units and SI units (International System∗ ) as shown in Table 1.1. Table 1.1. U.S. Customary and SI unit systems.

System of Units Base Dimension

U.S. Customary

SI

force

pound (lb)

newtona (N) ≡ kg⋅m∕s2

mass

sluga ≡ lb⋅s2 ∕ft

kilogram (kg)

length

foot (ft)

meter (m)

time

second (s)

second (s)

a Derived

unit.

Each system has three base units and a fourth derived unit. In the U.S. Customary system, the base units measure force, length, and time, using lb, ft, and s, respectively, and the derived unit is obtained from the equation 𝑚 = 𝐹 ∕𝑎, which gives the mass unit as lb⋅s2 ∕f t, which is defined as 1 slug. In the SI system, the base units measure mass, length, and time, using kg, m, and s, respectively, and the derived unit is obtained from the equation 𝐹 = 𝑚𝑎, which gives the force unit as kg⋅m∕s2 , which ∗ SI

has been adopted as the abbreviation for the French Le Syst`eme International d’Unit´es.

Helpful Information Dimensions versus units. Dimensions and units are different. A dimension is a measurable extent of some kind, while units are used to measure a dimension. For example, length and time are both dimensions, and meter and second, respectively, are units used to measure these dimensions.

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Introduction to Statics

Common Pitfall Weight and mass are different. It is unfortunately common for people, especially laypeople, to refer to weight using mass units. For example, when a person says, “I weigh 70 kg,” the person really means “My mass is 70 kg.” In this book, as well as throughout engineering, we must be precise with our nomenclature. Weights and forces will always be reported using appropriate force units, and masses will always be reported using appropriate mass units.

is defined as 1 newton, N. For both systems, we may occasionally use different, but consistent, measures for some units. For example, we may use minutes rather than seconds, inches instead of feet, grams instead of kilograms, and so on. Nonetheless, the definitions of 1 newton and 1 slug are always as shown in Table 1.1.

Dimensional homogeneity and unit conversions Of course, the symbol “=” means that what is on the left-hand side of the symbol is the same as what is on the right-hand side. Hence, for an expression to be correct, it must be numerically correct and dimensionally correct. Normally this means that the left- and right-hand sides have the same numerical value and the same units.∗ All too often units are not carried along during a calculation, only to be incorrectly assumed at the end. Our strong recommendation is that you always use appropriate units in all equations. Such practice helps avoid catastrophic blunders and provides a useful check on a solution, for if an equation is found to be dimensionally inconsistent, then an error has certainly been made. Unit conversions are frequently needed, and are easily accomplished using conversion factors such as those found in Table 1.2 and rules of algebra. The basic idea is to multiply either or both sides of an equation by dimensionless factors of unity, where each factor of unity embodies an appropriate unit conversion. This description perhaps sounds vague, and the procedure is better illustrated by the examples that follow. Table 1.2. Conversion factors between U.S. Customary and SI unit systems. U.S. Customary

Interesting Fact

length

Abbreviation for inch. Notice in Table 1.2 that the abbreviation for inch is “in.”, which contains a period. This is unusual, but is done because without the period, the abbreviation would also be the same as a word in the English language, and this might lead to confusion.

force

mass

SI

1 in.

=

0.0254 m (2.54 cm, 25.4 mm)a

1 f t (12 in.)

=

0.3048 ma

1 mi (5280 f t)

=

1.609 km

1 lb

=

4.448 N

1 kip (1000 lb)

=

4.448 kN

1 slug (1 lb⋅s2 ∕f t)

=

14.59 kg

a Exact.

Prefixes Prefixes are a useful alternative to scientific notation for representing numbers that are very large or very small. Common prefixes and a summary of rules for use are given in Table 1.3. Rules for Prefix Use 1. With few exceptions, use prefixes only in the numerator of unit combinations. One common exception is kg, which may appear in numerator or denominator. 2. Use a dot or dash to denote multiplication of units. For example, use N⋅m or N-m. ∗A

simple example of an exception to this is the equation 12 in. = 1 ft. Such equations play a key role in performing unit conversions.

ISTUDY

Section 1.6

11

Units and Unit Conversions

3. Exponentiation applies to both the unit and prefix. For example, mm2 = (mm)2 . 4. When the number of digits on either side of a decimal point exceeds 4, it is common to group the digits into groups of 3, with the groups separated by commas or thin spaces. Since many countries use a comma to represent a decimal point, the thin space is sometimes preferable. For example, 1234.0 could be written as is, and 12345.0 should be written as 12,345.0 or as 12 345.0. Table 1.3. Common prefixes used in the SI unit systems. Multiplication Factor 1 000 000 000 000 000 000 000 000 1 000 000 000 000 000 000 000 1 000 000 000 000 000 000 1 000 000 000 000 000 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100 10 0.1 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001 0.000 000 000 000 001 0.000 000 000 000 000 001 0.000 000 000 000 000 000 001 0.000 000 000 000 000 000 000 001

1024 1021 1018 1015 1012 109 106 103 102 101 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24

Prefix

Symbol

yotta zetta exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto zepto yocto

Y Z E P T G M k h da d c m 𝜇 n p f a z y

While prefixes can often be incorporated in an expression by inspection, the rules for accomplishing this are identical to those for performing unit transformations, as shown in the examples of this section.

𝐵

Angular measure

𝜃

𝑂

Angles are usually measured using either radians (rad) or degrees The radian measure of the angle 𝜃 shown in Fig. 1.6 is defined to be the ratio of the circumference 𝑐 of a circular arc to the radius 𝑟 of the arc. Thus, as seen in the examples of Fig. 1.7, the angle for one-quarter of a circular arc is 𝜃 = 𝜋∕2 rad (or 1.571 rad), and for a full circular arc the angle is 𝜃 = 2𝜋 rad (or 6.283 rad). Degree measure arbitrarily chooses the angle for a full circular arc to be 360◦ , in which case 1◦ is the angle of an arc that is 1/360 parts of a full circle. Thus, the transformation between radian and degree measure is 2𝜋 rad = 360◦ . (1.4)

𝑐 𝑟

definition of radian measure 𝐴

𝑟

(◦ ).

Transformations are carried out using the procedures described in this section. For example, to convert the angle 𝜃 = 12◦ to radian measure, we use Eq. (1.4) to

𝜃=

𝑐

Figure 1.6 Definition of radian measure for angles. 𝐵 𝜃=

𝜋 rad 2

𝜃 = 2𝜋 rad

𝐴 𝑂

𝑂

Figure 1.7 Examples of angles measured in radians.

𝐴 𝐵

12

Chapter 1

Introduction to Statics

write

2𝜋 rad = 0.209 rad. (1.5) 360◦ Radians are a measure of angle that naturally arises throughout mathematics and science, and most equations derived from fundamental principles use radian measure. Nonetheless, degree measure has intuitive appeal and is used widely. When writing angles, we will always label these as radians or degrees. However, radians and degrees are not units in the same way as those discussed earlier and, while this may be puzzling, both of these measures are dimensionless. This can be seen by examining the definition of radian measure shown in Fig. 1.6, namely 𝜃 = 𝑐∕𝑟. With 𝑐 and 𝑟 having the same units of length, angle 𝜃 is clearly dimensionless. Thus, radians and degrees are not really units, but rather are statements of the convention used for measuring an angle. Nonetheless, for practical purposes we may consider these to be units, and we will transform them using our usual procedures. Further, if we derive an expression that we expect to be dimensionless and we discover it has units of radians or degrees, then we should not necessarily be alarmed. 𝜃 = (12◦ )

Small angle approximations

𝐴

𝐶

The small angle approximations discussed below are frequently used in statics and subjects that follow. Consider the right triangle shown in Fig. 1.8. The sine, cosine, and tangent of angle 𝜃 are defined as

𝜃 𝐵 Figure 1.8 A right triangle. If 𝜃 is measured in radians and is small (𝜃 ≪ 1 rad), then the small angle approximations are sin 𝜃 ≈ 𝜃, cos 𝜃 ≈ 1, and tan 𝜃 ≈ 𝜃.

ISTUDY

sin 𝜃 =

𝐶 , 𝐴

cos 𝜃 =

𝐵 , 𝐴

tan 𝜃 =

and

𝐶 . 𝐵

(1.6)

If 𝜃 is measured in radians, then sin 𝜃 and cos 𝜃 may be expressed using Taylor series expansions as sin 𝜃 = 𝜃 −

𝜃3 𝜃5 + −… 6 120

and

cos 𝜃 = 1 −

𝜃2 𝜃4 + − …. 2 24

(1.7)

When 𝜃 is small (≪ 1 rad), Eq. (1.7) can be truncated after the first terms to obtain the small angle approximations as sin 𝜃 ≈ 𝜃

and

cos 𝜃 ≈ 1,

if 𝜃 ≪ 1 rad.

(1.8)

Thus, if 𝜃 is small, then Eq. (1.6) becomes 𝜃≈

𝐶 , 𝐴

𝐵 ≈ 𝐴,

and

𝜃≈

𝐶 , 𝐵

if 𝜃 ≪ 1 rad.

(1.9)

Mini-Example Use the small angle approximations to determine the sine and cosine of 5◦ , 10◦ , and 15◦ , and compare these results to the exact values. Solution The angles expressed in radians are 𝜃 = 5◦ (𝜋 rad∕180◦ ) = 0.08727 rad, and similarly 𝜃 = 10◦ = 0.17453 rad, and 𝜃 = 15◦ = 0.261799 rad. We then use Eq. (1.8) to obtain the results listed in Table 1.4. Notice that the small angle approximation for cosine is not as accurate as that for sine. Thus, for some applications, an additional term from Eq. (1.7) is retained to yield the small angle approximation cos 𝜃 ≈ 1 − 𝜃 2 ∕2.

ISTUDY

Section 1.6

Units and Unit Conversions

13

Table 1.4. Small angle approximations for 5◦ , 10◦ , and 15◦ angles.

𝜃 5◦ = 0.08727 rad

10◦ = 0.17453 rad

15◦ = 0.261799 rad

Small Angle Approx.

Exact Value

Error

sin 𝜃 ≈ 0.08727

0.087156

0.1%

cos 𝜃 ≈ 1

0.996195

0.4%

sin 𝜃 ≈ 0.17453

0.173648

0.5%

cos 𝜃 ≈ 1

0.984808

1.5%

sin 𝜃 ≈ 0.261799

0.258819

1.2%

cos 𝜃 ≈ 1

0.965926

3.5%

Accuracy of calculations The accuracy of answers obtained for a particular problem is only as precise as the least accurate information used in the analysis. For example, consider the numbers 1.23 and 45.67. By writing these numbers using three and four digits, respectively, the implication is that they are known to three and four significant digits of accuracy. The exact product of these numbers is 56.1741. But it is wrong to imply that the product is known to six-digit accuracy. Rather, it is appropriate to interpret the product as being accurate to the same number of significant digits as the least accurate piece of information used. Hence, we would round the exact product to three significant digits and interpret the answer as being 56.2. The use of number of digits to imply precision, however, can be ambiguous. Consider the number 6000; it is not clear if this number is known to one, two, three, or four significant digits. To embody accuracy information in numbers, it is probably best to use scientific notation. Thus, for example, if the number 6000 were known to three significant digits, we could write 6.00 × 103 with the convention that the number of digits used indicates the accuracy of the number. In this book, we will use a more pragmatic approach and will generally assume that data is known to three significant digits. When you perform computations, it is good practice to carry a few extra digits of accuracy for intermediate computations, and if an electronic device such as a calculator or computer is used, then you certainly want to use the full precision that is available. Nonetheless, final answers should be interpreted as having precision that is commensurate with the precision of the data used. The margin note on this page describes the convention for accuracy of numbers that is used for the calculations carried out in this book.

Helpful Information Throughout this book, we will generally assume that given data is known to three significant digits of accuracy. When we present calculations, all intermediate results are stored in the memory of a calculator or computer using the full precision these machines offer. When intermediate results are reported, they will usually be rounded to four significant digits. Final answers are also usually reported with four significant digits, although they should generally be interpreted as being accurate to only three significant digits. If an intermediate or final result can be exactly represented using fewer than four digits, then we will usually do so (e.g., if a number is exactly 1∕5, we may write this as 0.2). When verifying the calculations described in this book, you may occasionally calculate results that are slightly different from those shown if you do not store intermediate results as we describe.

14

E X A M P L E 1.1

ISTUDY

Chapter 1

Introduction to Statics

Unit Conversion Convert the speed 𝑠 = 5.12 f t∕s to the SI units m/s and km/h.

SOLUTION Starting with 𝑠 = 5.12 f t∕s, we will multiply the right-hand side of this expression by appropriate conversion factors to achieve the desired unit conversion.

Road Map

Governing Equations & Computation

Referring to Table 1.2, we find

1 f t = 0.3048 m.

(1)

Dividing both sides of Eq. (1) by 1 ft provides the middle term of the following equation

Common Pitfall 1= Omitting units in equations. The most serious mistake made when performing unit conversions (as well as when writing equations in general) is to omit units in equations. Although writing units in equations takes a few moments longer, doing so will help avoid the errors that are sure to result if you do not make this a practice.

0.3048 m 1 ft = , 1 ft 0.3048 m

(2)

whereas dividing both sides of Eq. (1) by 0.3048 m provides the last term of Eq. (2). Regardless of which form of Eq. (2) is used, the left-hand side is the number 1, with no units. The form of Eq. (2) that is used in a particular unit transformation will depend on what units need to be replaced, or canceled. To accomplish the unit conversion needed for 𝑠 = 5.12 f t∕s, we write 𝑠 = 5.12

f∕t 0.3048 m m ft = 1.561 . (1) = 5.12 s s s 1 f∕t ⏟⏞⏞⏟⏞⏞⏟

(3)

=1

In writing Eq. (3), we first multiply 5.12 f t∕s by the dimensionless number 1; this changes neither the value nor the units of 𝑠. Since we want to eliminate the foot unit, we substitute for the dimensionless number 1 using the first form of transformation in Eq. (2), namely 1 = 0.3048 m∕1 f t. Finally, we cancel the foot unit in the numerator and denominator to obtain the speed 𝑠 = 1.561 m∕s in the desired SI units. To obtain 𝑠 in units of km/h, we continue with Eq. (3) and perform the following transformations: 𝑠 = 1.561

60 ∕s 60 min m ∕ km ∕ km . = 5.618 3 ∕ h h ∕s 10 m ∕ min ⏟ ⏟ ⏟ ⏟⏟⏟ ⏟⏟⏟ =1

Discussion & Verification

=1

(4)

=1

When possible, answers should be checked to verify that they are reasonable. For example, starting with 𝑠 = 5.12 f t∕s, the result in Eq. (3) is reasonable since a meter is about 3 feet.

ISTUDY

Section 1.6

Units and Unit Conversions

E X A M P L E 1.2

Unit Conversion

The universal gravitational constant, whose physical significance we discuss later in this chapter, is 𝐺 = 66.74 × 10−12 m3 ∕(kg⋅s2 ). Express 𝐺 in base U.S. Customary units.

SOLUTION Road Map

Perhaps the most straightforward solution strategy is to first convert mass in kilograms to mass in slugs and then replace the unit of slug with its fundamental definition. Governing Equations & Computation Beginning our calculation with 𝐺 = 66.74 × 10−12 m3 ∕(kg ⋅ s2 ), we multiply the right-hand side by appropriate conversion factors to achieve the desired unit conversion. Thus, )3 ( slug ∕ ∕ ft m∕3 14.59 kg 𝐺 = 66.74 × 10−12 slug 0.3048 m ∕ ∕ kg⋅s lb⋅s2 ∕ft ∕ 2 ⏟⏞⏟⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏟⏞⏟ =1

= 34.39 × 10−9

=1

=(1)3

ft4 . lb⋅s4

(1)

Alternatively, we could also perform the unit transformation by first introducing the SI force measure newton, followed by conversion to force measure in pounds, followed

by conversion of length. Thus, 𝐺 = 66.74 × 10−12

( )4 ∕ 2 4.448 N ∕ m∕s ∕ m∕3 kg⋅ ft ∕ lb 0.3048 m ∕ kg⋅s ∕ 2 N ⏟⏞⏟⏞⏟ ⏟⏞⏟⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ =1

= 34.39 × 10−9

ft4 . lb⋅s4

=1

=(1)4

(2)

Because of the complexity of the unit combinations for 𝐺, it is not possible to use inspection to verify that Eqs. (1) and (2) are reasonable. Rather, the accuracy of our results relies solely on the use of appropriate conversion factors and accurate cancellation of units in Eqs. (1) and (2). For this reason, it is essential that you carry units throughout all equations.

Discussion & Verification

15

16

𝑚2

𝑟 𝐹 𝑚1

𝐹

Figure 1.9 The force 𝐹 that attracts two particles toward one another is provided by Newton’s law of universal gravitational attraction.

ISTUDY

Chapter 1

Introduction to Statics

1.7

Newton’s Law of Gravitation

Because weight produced by gravity is so omnipresent, it is worthwhile to examine the source of such forces closely, and to understand the limitations of common expressions such as 𝑊 = 𝑚𝑔 where 𝑚 is an object’s mass, 𝑔 is acceleration due to gravity, and 𝑊 is the object’s weight. Consider the force 𝐹 of mutual attraction between two particles, as shown in Fig. 1.9. In 1666, Newton developed his law of universal gravitational attraction as 𝐹 =𝐺

𝑚1 𝑚2 𝑟2

(1.10)

where

Concept Alert Force due to gravity. Force due to gravitational attraction between two objects is a vector, hence it has both magnitude and direction. Equation (1.10) gives the magnitude, and the direction is along a line connecting the centers of gravity of the two objects.

𝑚1 , 𝑚2 = masses of particles 1 and 2; 𝑟 = distance between the particles; 𝐺 = universal gravitational constant, found to be approximately 66.74×10−12 m3 ∕(kg⋅s2 ); 𝐹 = force of attraction between two particles. It has been widely reported that Newton’s inspiration for this law was the motion of an apple falling from a tree, but he also recognized that the same law should apply to the attraction of celestial bodies to one another. Although Newton postulated the law in 1666, it was not until 1687 that he published his ideas in the Principia. This delay was due in part to the need to prove that an object such as the Earth (if assumed to be spherical and uniform) could be treated as a point mass for gravitational effects on neighboring particles, and in the course of proving this he developed calculus.∗ The first accurate measurement of 𝐺 was by Lord Cavendish in 1798, and this value has been refined by more careful experiments over the last two centuries, leading to the value reported here. The law of universal gravitational attraction is a postulate, and as with Newton’s three laws of motion, we must accept this as a rule that nature follows without a fundamental proof of its validity. For the vast majority of applications on Earth, Eq. (1.10) takes the simple and convenient form 𝑊 = 𝑚𝑔, as follows. Let 𝑚1 in Eq. (1.10) denote the mass 𝑚 of an object, and let 𝑚2 denote the mass of the Earth (with an approximate value 𝑚Earth = 5.9736×1024 kg). If the object is on or near the surface of the Earth, then its position 𝑟 is about the same as the mean radius of the Earth (with an approximate value 6.371 × 106 m). The force 𝐹 in Eq. (1.10) is then called the weight 𝑊 of the object, and Eq. (1.10) can be rewritten as 𝑊 = 𝑚𝑔

where

𝑔 ≡ 𝐺𝑚Earth ∕𝑟2 .

(1.11)

From Eq. (1.11), we see that 𝑔 is not a constant because it depends on the value of 𝑟. However, for the vast majority of applications where objects are near the surface of ∗ Calculus was also developed independently by Gottfried Wilhelm Leibniz (1646–1716), and he and New-

ton had a long-standing dispute over who was the true originator. The historical records show that while Newton was the first to discover calculus (about 10 years before Leibniz), Leibniz was the first to publish his discovery (about 15 years before Newton). In some respects, Leibniz won since it is his superior notation that we use in calculus today.

ISTUDY

Section 1.7

Newton’s Law of Gravitation

17

the Earth, effects of small changes in 𝑟 are negligible, and the commonly used values for acceleration due to gravity are 𝑔 = 9.81 m/s2 = 32.2 ft/s2 .

(1.12)

Note that if the values reported above for 𝐺, Earth’s mass, and Earth’s mean radius are used in Eq. (1.11), the value of 𝑔 produced is slightly different than 9.81 m/s2 . The difference between the accepted value of 𝑔 and the theoretically computed value provided by Eq. (1.11) has several sources, including that the Earth is not perfectly spherical and does not have uniform mass distribution, and the effects of centripetal acceleration due to the Earth’s rotation are not accounted for. Because of these sources, the actual acceleration due to gravity is about 0.3% lower at the equator and 0.3% higher at the poles, relative to the numbers given in Eq. (1.12) which are for a north or south latitude of 45◦ at sea level. In addition, there may be small local variations in acceleration due to gravity due to the effects of geology. Nonetheless, throughout this book we will use the standard values of 𝑔 given in Eq. (1.12).

Relationship between specific weight and density The specific weights and densities of some common materials are given in Table 1.5. When using U.S. Customary units, it is common to characterize the density of Table 1.5. Specific weight and density for some common materials. Except for water and ice, numbers reported are generally at 20 ◦ C. Data may vary depending on composition, alloying, temperature, moisture content for wood, etc.

Specific Weight 𝛾

Density 𝜌

Material

(lb/ft3 )

(kg/m3 )

iron (pure)

491

7860

iron (cast)

450 ± 15

7210 ± 240

aluminum (pure)

169

2710

aluminum (alloy)

170 ± 10

2710 ± 160

steel

490

7850

stainless steel

500

8010

537 ± 8

8610 ± 130

280

4480

rubber

70 ± 10

1120 ± 160

nylon

70

1120

concrete

150

2400

rock (dry granite)

165

2640

cortical bone (adult)

119

1900

32 ± 2

510 ± 30

brass titanium

wood (dry Douglas fir) water (fresh, 4

◦ C,

62.4

1000

ice

1 atm)

57

920

JP–4 jet fuel

48

770

Helpful Information Center of gravity. The center of gravity is the point through which the weight of a body, or a collection of bodies, may be considered to act. In figures, we will often denote the center of gravity by using the symbol . To illustrate, imagine a server at a restaurant brings you wine and pasta on a tray. Obviously, the server must position his hand so that the combined weight of the tray and everything on it is located over his hand. 12 N 8N

10 N

30 N

center of gravity The weight of the wine (12 N), pasta (10 N), and tray (8 N) can be thought of as a single 30 N force acting through the center of gravity for the collection of objects. Center of gravity and how it is determined are discussed thoroughly in Chapter 7, where it is seen that the two force systems shown above are equivalent force systems. In the meantime, a working knowledge of this definition will be useful.

18

ISTUDY

Chapter 1

Introduction to Statics

materials using specific weight (sometimes also called weight density, or unit weight), which is defined to be the weight on Earth of a unit volume of material. For example, the specific weight of steel is 𝛾 = 490 lb∕f t 3 (= 0.284 lb∕in.3 ). However, specific weight is not the same as density, although they are related. Density is defined to be the mass of a unit volume of material, and when SI units are used, it is most common to directly report a material’s density. Thus, for steel, the density is 𝜌 = 7850 kg/m3 . These measures are related by Eq. (1.11) as follows. Imagine a certain volume 𝑉 of material has weight (on Earth) 𝑊 and mass 𝑚. Dividing Eq. (1.11) by volume 𝑉 provides 𝑚 𝑊 = 𝑔. (1.13) 𝑉 𝑉 In this expression, 𝑊 ∕𝑉 is the definition of specific weight 𝛾, and 𝑚∕𝑉 is the definition of density 𝜌. Thus, Eq. (1.13) becomes 𝛾 = 𝜌𝑔

or

𝜌=

𝛾 . 𝑔

(1.14)

ISTUDY

Section 1.7

Newton’s Law of Gravitation

E X A M P L E 1.3

19

Weight and Force of Mutual Attraction

Two bowling balls resting on a shelf touch one another. The balls have 220 mm diameter and are made of plastic with density 𝜌𝐴 = 1170 kg/m3 for ball 𝐴 and 𝜌𝐵 = 980 kg/m3 for ball 𝐵. Determine the weight of each ball and the force of mutual attraction, expressing both in SI units and U.S. Customary units.

SOLUTION Road Map The forces to be determined are shown in Fig. 2. The weights of balls 𝐴 and 𝐵 are forces (vectors) with magnitudes 𝑊𝐴 and 𝑊𝐵 , respectively, and these forces act in the downward vertical direction. The force of mutual attraction between the two balls has magnitude 𝐹 , with directions as shown in Fig. 2. Note that Newton’s third law requires the force of mutual attraction between the two balls to have equal magnitude and opposite direction. We will assume both balls are uniform (i.e., the density is the same throughout each ball), and we will neglect the presence of the finger holes. We will first determine the mass of each ball. We will then determine the weight of each ball, using 𝑊𝐴 = 𝑚𝐴 𝑔 and 𝑊𝐵 = 𝑚𝐵 𝑔, and then the force of mutual attraction, using Newton’s law of gravitational attraction.

Lucinda Dowell

220 mm

𝐴

220 mm

𝐵

Figure 1

The mass 𝑚𝐴 of ball 𝐴 is the product of the material’s density 𝜌𝐴 and the ball’s volume 𝑉𝐴 , and similarly for ball 𝐵. Thus,

Governing Equations & Computation

) ( )3 ( ∕ kg 4 0.220 m = 6.523 kg, 𝜋 𝑚𝐴 = 𝜌𝐴 𝑉𝐴 = 1170 2 m∕3 3 )3 ( ) ( ∕ kg 4 0.220 m 𝜋 𝑚𝐵 = 𝜌𝐵 𝑉𝐵 = 980 = 5.464 kg. 2 m∕3 3

(1)

𝑊𝐵

𝐹

𝐹 𝐴

𝐵

(2)

The weight of each ball is ( ) kg⋅m m 𝑊𝐴 = 𝑚𝐴 𝑔 = (6.523 kg) 9.81 2 = 63.99 2 = 63.99 N, s s ( ) kg⋅m m 𝑊𝐵 = 𝑚𝐵 𝑔 = (5.464 kg) 9.81 2 = 53.60 2 = 53.60 N. s s

𝑊𝐴

(3) (4)

Figure 2 The weight of each ball and the force of mutual attraction are vectors with the directions shown. Important Note: The bowling balls are also subjected to other forces that are not shown (see the Helpful Information margin note below).

∕ )(1 lb∕4.448 N ∕ ) = 14.39 lb and 𝑊𝐵 = In U.S. Customary units, 𝑊𝐴 = (63.99 N ∕ )(1 lb∕4.448 N ∕ ) = 12.05 lb. (53.60 N The force of mutual attraction is given by Eq. (1.10) (with subscripts 1 and 2 replaced by 𝐴 and 𝐵) as m∕ ⋅m ⎞ ⎛ ⏞ ⏞ ⏞⎟ ⎜ 𝑚𝐴 𝑚𝐵 ⎜ kg) ∕ m3 ⎟ (6.523 kg)(5.464 −12 𝐹 =𝐺 = ⎜66.74 × 10 2 ⎟ 2 𝑟2 kg⋅s (0.220 m) ∕ ∕ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ kg⋅m = 4.915 × 10−8 N. = 4.91 × 10−8 s2

Helpful Information

2

(5)

In Eq. (5), 𝑟 = 0.220 m is the distance between the center of each ball. In U.S. Customary ∕ )(1 lb∕4.448 N ∕ ) = 1.105 × 10−8 lb. units, 𝐹 = (4.915 × 10−8 N Discussion & Verification

As you might have expected, the force of mutual attraction between the two balls is very small compared to the weight of the balls (9 orders of magnitude smaller). In developing models for engineering problems, the force of mutual attraction will usually be small compared to other forces, and when this is the case, it will be neglected.

Additional forces. The balls shown in Fig. 2 are subjected to additional forces that are not shown. For example, the shelf applies a force to each ball, and there are probably contact forces between the two balls where they touch. Clearly, without these additional forces, the bowling balls could not be in static equilibrium. Chapter 3 will thoroughly discuss these additional forces and how they may be determined.

20

E X A M P L E 1.4

ISTUDY

Chapter 1

Introduction to Statics

Specific Weight and Density The specific weight of a particular aluminum alloy is 𝛾 = 0.099 lb∕in.3 . Determine the density of this alloy, and report this in U.S. Customary units.

SOLUTION Road Map

Beginning with weight per unit volume for an aluminum alloy, we will determine its mass per unit volume. Governing Equations & Computation

We use Eq. (1.14), with appropriate unit trans-

formations 𝜌=

2 0.099 lb∕in.3 f∕t 𝛾 −4 lb⋅s = = 2.562 × 10 𝑔 32.2 f∕t∕s2 12 in. in.4 ∕ ∕ s∕2 slug 12 in. slug lb⋅ = 3.075 × 10−3 3 . = 2.562 × 10−4 4 2 ∕ s∕ ∕f∕t f∕t lb⋅ in. in. ⏟⏟⏟

(1)

∕ in.3 in.⋅

The first expression in Eq. (1), 𝜌 = 2.562 × 10−4 lb⋅s2 ∕in.4 , does not use the conventional U.S. Customary unit for mass, but is otherwise an acceptable and useful answer for the density of this aluminum alloy. The second expression in Eq. (1), 𝜌 = 3.075 × 10−3 slug∕in.3 , incorporates the mass unit slug and provides the density in the expected form of mass per unit volume.

Discussion & Verification

ISTUDY

Section 1.8

1.8

Failure

21

Failure

Among all of the goals confronting engineers when they design structures and machines, the most crucial goal is to develop designs that are as safe as possible. Unfortunately, despite all human efforts to meet this goal, sometimes we do not, and for reasons that are almost always unexpected, failure occurs. When failure occurs, we must learn from it so that our mistakes and/or lack of foresight are not repeated in the future.∗ In this section, some examples of engineering failures are highlighted. • Tacoma Narrows bridge. Only four months after its opening in 1940, the Tacoma Narrows suspension bridge in Washington collapsed violently due to severe vibrations produced by aerodynamic forces that were not fully anticipated and accounted for in its design (see Fig. 1.10). Interestingly, the Deer Isle bridge along the coast of Maine, while smaller, was of similar construction. It opened one year earlier and also experienced severe wind-induced vibrations. However, the designer of this bridge had the foresight and perhaps sufficient time to add wind fairings along the bridge’s length to give it better aerodynamic properties, and additional diagonal cable bracing to provide greater stiffness. This bridge is still in service today.†

Library of Congress Prints and Photographs Division Washington, D.C. 20540 USA

Figure 1.10 Failure of the Tacoma Narrows bridge in Tacoma, Washington, in 1940, due to severe vibrations produced by a 42 mph wind.

• Escambia Bay bridge. Fifty-six sections of the Interstate 10 bridge crossing Escambia Bay in Pensacola, Florida, were dislodged by Hurricane Ivan in September 2004, including numerous sections that were completely washed into the bay (see Fig. 1.11). Each of these sections weighed about 220 tons. The National Weather Service categorizes the intensity of hurricanes using a scale of 1 to 5. When Ivan struck the Escambia Bay bridge, it was a category 3 hurricane with sustained winds of 111 to 130 mph. While Ivan was not an extreme hurricane according to this scale, the damage caused to the Escambia Bay bridge was extreme. • Airbus A300 failure. On November 12, 2001, only minutes after takeoff, American Airlines flight 587, an Airbus A300, crashed into a residential area of Belle Harbor, New York, because the airplane’s vertical stabilizer separated in flight due to failure of the attachment lugs between the stabilizer and fuselage (see Fig. 1.12). All 260 people on board and five people on the ground were killed. The National Transportation Safety Board‡ (NTSB) investigated the accident and attributed the cause to high aerodynamic loads resulting from unnecessary and excessive rudder pedal inputs as the first officer reacted to turbulence caused by another aircraft. The airline’s pilot training program and the airplane’s rudder design were also cited as contributing factors. Among the recommendations made by the NTSB were to modify the rudder control systems to increase protection from high forces due to hazardous rudder pedal inputs at high speeds. ∗ Interesting

case studies of failures and how we can learn from these are given in H. Petroski, Design Paradigms: Case Histories of Error and Judgment in Engineering, Cambridge University Press, New York, 1994. † For additional reading, see B. Moran (1999), “A Bridge That Didn’t Collapse,” Invention and Technology, 15(2), pp. 10–18. ‡ The National Transportation Safety Board (NTSB) is an independent federal agency charged by Congress with investigating every civil aviation accident in the United States and significant accidents in other modes of transportation including railroad, highway, marine and pipeline, and issuing safety recommendations aimed at preventing future accidents. Although implementation of the NTSB’s recommendations is not mandatory, over 80% of their recommendations have been adopted.

ZUMA Press Inc/Alamy Stock Photo

Figure 1.11 Failure of the Escambia Bay bridge in Pensacola, Florida, during Hurricane Ivan in September 2004.

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Chapter 1

Introduction to Statics

Anthony Correia/Getty Images

• Kansas City Hyatt Regency Hotel. On July 17, 1981, two suspended walkways at the Kansas City Hyatt Regency Hotel collapsed during a dance party, killing 114 people and seriously injuring many more. The collapse was caused by connections that failed, as shown in Fig. 1.13(a). The original connection design, shown in Fig. 1.13(b), was changed during construction to the design shown in Fig. 1.13(c), with the agreement of all parties involved. While the original design had satisfactory strength, the revised design was easier to fabricate, featured shorter bars that were more readily available, and was more straightforward than the potentially confusing original design. However, the revised design was never analyzed to determine its adequacy.∗

Figure 1.12 The vertical stabilizer of an Airbus A300 airplane separated in flight and was recovered from Jamaica Bay, about 1 mile from the crash site.

Lee Lowery/Texas A&M University

(a) (b) original design (c) as constructed Figure 1.13. (a) Failure of a connection supporting a walkway at the Kansas City Hyatt Regency Hotel, where a support rod has pulled through a box beam, allowing the walkways to collapse. (b) The original design, which had satisfactory strength. (c) The revised design, which was easier to fabricate.

Donald Kravitz/Getty Images News

Figure 1.14 Inspectors survey a five-story collapsed section of a parking garage under construction at the Tropicana Casino and Resort in Atlantic City, New Jersey, October 30, 2003.

ISTUDY

• Tropicana Casino parking garage. On October 30, 2003, a 10-story parking garage under construction at the Tropicana Casino and Resort in Atlantic City, New Jersey, collapsed, killing four workers and injuring 21 others (see Fig. 1.14). The failure occurred as concrete was being poured for one of the upper floor decks. The Occupational Safety and Health Administration† (OSHA) investigated the failure and fined the concrete contractor for intentional disregard of safety standards for failing to erect, support, brace, and maintain framework that would be capable of supporting all vertical and lateral loads that may reasonably be anticipated during construction. The design of the building itself was adequate, but the design of structures needed for fabrication was not. Note that concrete requires time after pouring (28 days is common) to reach its full design strength.

∗ Additional aspects of this failure are discussed in H. Petroski, Design Paradigms: Case Histories of Error

and Judgment in Engineering, Cambridge University Press, New York, 1994. mission and regulatory powers of the Occupational Safety and Health Administration are described on p. 347.

† The

ISTUDY

Section 1.9

Chapter Review

1.9 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation numbers cited for additional details.

Scalars and vectors A scalar is a quantity that is completely characterized by a single number. A vector has both size (or magnitude) and direction. In this book, scalars are denoted by italic symbols such as 𝑠, and vectors are denoted by placing an arrow above the symbol for the vector, such as 𝑣. ⃗

Position, velocity, and acceleration Position, velocity, and acceleration are all vector quantities. If 𝑟⃗ denotes the position of a particle relative to some location, then the velocity and acceleration of the particle are defined by Eq. (1.1), p. 7 𝑣⃗ = 𝑑⃗𝑟∕𝑑𝑡, Eq. (1.2), p. 7 𝑎⃗ = 𝑑 𝑣∕𝑑𝑡. ⃗ ⃗ the particle is said to be in static equilibrium, or simply equilibrium, and When 𝑎⃗ = 0, ⃗ then the it either moves with constant velocity or remains stationary in space. If 𝑎⃗ ≠ 0, particle will move with velocity and position that change with time.

Laws of motion Newton’s three laws of motion are as follows: First Law. A particle remains at rest, or continues to move in a straight line with uniform velocity, if there is no unbalanced force acting on it. Second Law. The acceleration of a particle is proportional to the resultant force acting on the particle and is in the direction of this force. Eq. (1.3), p. 7 𝐹⃗ = 𝑚𝑎. ⃗ Third Law. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.

Static equilibrium In Eq. (1.3), if the resultant force 𝐹⃗ acting on a particle is zero, then the acceleration of the particle is zero, and hence the particle will have uniform velocity which may be zero or nonzero in value; if nonzero value then the particle will move in a straight line. Hence, ⃗ the particle is said to be in static equilibrium, when there is no acceleration (i.e., 𝑎⃗ = 0), or simply equilibrium.

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Chapter 1

Introduction to Statics

Small angle approximations Consider the right triangle shown in Fig. 1.15. If angle 𝜃 is measured in radians and is small (𝜃 ≪ 1 rad), then the small angle approximations are 𝐴

𝐶

Eq. (1.8), p. 12

𝜃 𝐵

sin 𝜃 ≈ 𝜃

Figure 1.15 If 𝜃 is measured in radians and is small (𝜃 ≪ 1 rad), then the small angle approximations are sin 𝜃 ≈ 𝜃, cos 𝜃 ≈ 1, and tan 𝜃 ≈ 𝜃.

cos 𝜃 ≈ 1,

and

if 𝜃 ≪ 1 rad,

Eq. (1.9), p. 12 𝜃≈

𝐶 , 𝐴

𝐵 ≈ 𝐴,

and 𝜃 ≈

𝐶 , 𝐵

if 𝜃 ≪ 1 rad.

Newton’s law of gravitation Newton’s law of universal gravitational attraction, as shown in Fig. 1.16, is 𝑚2

𝑟

Eq. (1.10), p. 16

𝐹 𝑚1

𝐹

Figure 1.16 The force 𝐹 that attracts two particles toward one another is provided by Newton’s law of universal gravitational attraction.

ISTUDY

𝐹 =𝐺

𝑚 1 𝑚2 𝑟2

where 𝑚1 , 𝑚2 = masses of particles 1 and 2; 𝑟 = distance between the particles; 𝐺 = universal gravitational constant, found to be approximately 66.74×10−12 m3 ∕(kg⋅s2 ); 𝐹 = force of attraction between two particles. When written for objects resting on or near the surface of Earth, this law takes the simple and useful form Eq. (1.11), p. 16 𝑊 = 𝑚𝑔 where 𝑚 is an object’s mass, 𝑔 is acceleration due to gravity (𝑔 = 9.81 m∕s2 = 32.2 f t∕s2 ), and 𝑊 is the object’s weight.

Relationship between specific weight and density. The density 𝜌 of a material is defined to be the material’s mass per unit volume. The specific weight 𝛾 of a material (sometimes also called weight density, or unit weight) is defined to be the material’s weight on Earth per unit volume. The relation between these is Eq. (1.14), p. 18 𝛾 = 𝜌𝑔

or

𝜌=

𝛾 . 𝑔

Attention to units It is strongly recommended that you always use appropriate units in all equations. Such practice helps avoid catastrophic blunders and provides a useful check on a solution, because if an equation is found to be dimensionally inconsistent, then an error has certainly been made.

ISTUDY

Section 1.9

Chapter Review

Review Problems Problem 1.1 (a) Consider a situation in which the force 𝐹 applied to a particle of mass 𝑚 is zero. Multiply the scalar form of Eq. (1.2) on page 7 (i.e., 𝑎 = 𝑑𝑣∕𝑑𝑡) by 𝑑𝑡, and integrate both sides to show that the velocity 𝑣 (also a scalar) is constant. Then use the scalar form of Eq. (1.1) to show that the (scalar) position 𝑟 is a linear function of time. (b) Repeat Part (a) when the force applied to the particle is a nonzero constant, to show that the velocity and position are linear and quadratic functions of time, respectively.

Problem 1.2 Using the length and force conversion factors in Table 1.2 on p. 10, verify that 1 slug = 14.59 kg.

Problems 1.3 through 1.7 Convert the numbers given in U.S. Customary units to the corresponding SI units indicated. Problem 1.3

(a) Length: Convert 𝑙 = 123 in. to m. (b) Mass: Convert 𝑚 = 2.87 slug to kg. (c) Force (weight): Convert 𝐹 = 18.9 lb to N. (d) Moment (torque): Convert 𝑀 = 433 ft⋅lb to N⋅m. Problem 1.4

(a) Length: Convert 𝑙 = 45.6 ft to m. (b) Mass: Convert 𝑚 = 6.36×104 slug to kg. (c) Force (weight): Convert 𝐹 = 22.1 kip to kN. (d) Moment (torque): Convert 𝑀 = 7660 ft⋅lb to kN⋅m. Problem 1.5

(a) Length: Convert 𝑙 = 2.35 in. to m. (b) Mass: Convert 𝑚 = 0.156 slug to kg. (c) Force (weight): Convert 𝐹 = 100 lb to N. (d) Moment (torque): Convert 𝑀 = 32.9 ft⋅lb to N⋅m. Problem 1.6

(a) Length: Convert 𝑙 = 0.001 in. to 𝜇m. (b) Mass: Convert 𝑚 = 0.305 lb⋅s2 ∕in. to kg. (c) Force (weight): Convert 𝐹 = 2.56 kip to kN.

(Recall: 1 kip = 1000 lb.)

(d) Mass moment of inertia: Convert 𝐼mass = 23.0 in.⋅lb⋅s2 to N⋅m⋅s2 . Problem 1.7

(a) Pressure: Convert 𝑝 = 25 lb∕ft2 to N∕m2 . (b) Elastic modulus: Convert 𝐸 = 30 × 106 lb∕in.2 to GN∕m2 . (c) Area moment of inertia: Convert 𝐼area = 63.2 in.4 to mm4 . (d) Mass moment of inertia: Convert 𝐼mass = 15.4 in.⋅lb⋅s2 to kg⋅m2 .

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26

ISTUDY

Chapter 1

Introduction to Statics Problems 1.8 through 1.12

Convert the numbers given in SI units to the corresponding U.S. Customary units indicated. Problem 1.8

(a) Length: Convert 𝑙 = 18.4 m to ft. (b) Mass: Convert 𝑚 = 4.32 kg to slug. (c) Force (weight): Convert 𝐹 = 2120 N to lb. (d) Moment (torque): Convert 𝑀 = 865 N⋅m to in.⋅lb. Problem 1.9

(a) Length: Convert 𝑙 = 312 mm to in. (b) Mass: Convert 𝑚 = 2100 kg to slug. (c) Force (weight): Convert 𝐹 = 25 kN to lb. (d) Moment (torque): Convert 𝑀 = 1.86 kN⋅m to ft⋅lb. Problem 1.10

(a) Length: Convert 𝑙 = 1.53 m to in. (b) Mass: Convert 𝑚 = 65 kg to slug. (c) Force (weight): Convert 𝐹 = 89.2 N to lb. (d) Moment (torque): Convert 𝑀 = 32.9 N⋅m to in.⋅lb. Problem 1.11

(a) Length: Convert 𝑙 = 122 nm to in. (b) Mass: Convert 𝑚 = 3.21 kg to lb⋅s2 ∕in. (c) Force (weight): Convert 𝐹 = 13.2 kN to lb. (d) Mass moment of inertia: Convert 𝐼mass = 93.2 kg⋅m2 to slug⋅in.2 . Problem 1.12

(a) Pressure: Convert 𝑝 = 25 kN∕m2 to lb∕in.2 . (b) Elastic modulus: Convert 𝐸 = 200 GN∕m2 to lb∕in.2 . (c) Area moment of inertia: Convert 𝐼area = 23.5 × 105 mm4 to in.4 . (d) Mass moment of inertia: Convert 𝐼mass = 12.3 kg⋅m2 to in.⋅lb⋅s2 .

Problem 1.13 (a) Convert the kinetic energy 𝑇 = 0.379 kg⋅m2 ∕s2 to slug⋅in.2 ∕s2 . (b) Convert the kinetic energy 𝑇 = 10.1 slug⋅in.2 ∕s2 to kg⋅m2 ∕s2 .

Problem 1.14 If the weight of a certain object on the surface of the Earth is 0.254 lb, determine its mass in kilograms.

Problem 1.15 If the mass of a certain object is 69.1 kg, determine its weight on the surface of the Earth in pounds.

ISTUDY

Section 1.9

Chapter Review

Problem 1.16 Use Eq. (1.11) on p. 16 to compute a theoretical value of acceleration due to gravity 𝑔, and compare this value with the actual acceleration due to gravity at the Earth’s poles, which is about 0.3% higher than the value reported in Eq. (1.12). Comment on the agreement.

Problem 1.17 Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron and are spherical, what diameter in feet must they have for their mutual gravitational attraction to be 1 lb?

Problem 1.18 The mass of the Moon is approximately 7.35 × 1022 kg, and its mean distance from the Earth is about 3.80 × 105 km. Determine the force of mutual gravitational attraction in newtons between the Earth and Moon. In view of your answer, discuss why the Moon does not crash into the Earth.

Problem 1.19 Consider a spacecraft that is positioned directly between the Earth and Moon. The mass of the Moon is approximately 7.35 × 1022 kg, and at the instant under consideration, the Moon is 3.80 × 105 km from Earth. Determine the distances the spacecraft must be from the Earth and Moon for the gravitational force of the Earth on the spacecraft to be the same as the gravitational force of the Moon on the spacecraft.

Problem 1.20 The gravity tractor, as shown in the artist’s rendition, is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to strike the Earth. Due to the gravitational attraction between the two objects and a prolonged period of time over which it acts (several years), the asteroid’s trajectory is changed slightly, thus hopefully diverting it from striking the Earth. If the gravity tractor’s weight on Earth is 20,000 lb and it flies with its center of gravity 160 f t from the surface of the asteroid, and the asteroid is homogeneous pure iron with 1290 f t diameter spherical shape, determine the force of mutual attraction. Idealize the gravity tractor to be a particle.

Problem 1.21

Stockbyte/Getty Images

Figure P1.20

If a person standing at the first-floor entrance to Willis Tower (formerly named Sears Tower) in Chicago weighs exactly 150 lb, determine the weight while he or she is standing on top of the building, which is 1450 ft above the first-floor entrance. How high would the top of the building need to be for the person’s weight to be 99% of its value at the first-floor entrance?

Problem 1.22 The specific weights of several materials are given in U.S. Customary units. Convert these to specific weights in SI units (kN/m3 ), and also compute the densities of these materials in SI units (kg/m3 ). (a) Zinc die casting alloy, 𝛾 = 0.242 lb/in.3 . (b) Oil shale (30 gal/ton rock), 𝛾 = 133 lb/ft3 . (c) Styrofoam (medium density), 𝛾 = 2.0 lb/ft3 . (d) Silica glass, 𝛾 = 0.079 lb/in.3 .

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Chapter 1

Introduction to Statics

𝑡𝑓 = 0.360 in.

𝑑 = 10.17 in.

𝑡𝑤 = 0.240 in.

Problem 1.23 The cross-sectional dimensions for a steel W10 × 22 wide-flange I beam are shown (the top and bottom flanges have the same thickness). The beam is 18 f t long (into the plane of the figure), and the steel has 490 lb∕f t 3 specific weight. Determine the cross-sectional dimensions in millimeters (show these on a sketch), the length in meters, and the mass of the beam in kilograms.

𝑏 = 5.75 in.

Problem 1.24

Figure P1.23

ℎ2 = 50 cm

The cross-sectional dimensions of a concrete traffic barrier are shown. The barrier is 2 m long (into the plane of the figure), and the concrete has 2400 kg∕m3 density. Determine the cross-sectional dimensions in inches (show these on a sketch), the length in feet, and the weight of the barrier in pounds.

Problem 1.25 ℎ1 = 60 cm

𝑏 𝑏 𝑏 𝑏 = 18 cm Figure P1.24

ISTUDY

The densities of several materials are given in SI units. Convert these to densities in U.S. Customary units (slug/ft3 ), and also compute the specific weights of these materials in U.S. Customary units (lb/ft3 ). (a) Lead (pure), 𝜌 = 11.34 g/cm3 . (b) Ceramic (alumina Al2 O3 ), 𝜌 = 3.90 Mg/m3 . (c) Polyethylene (high density), 𝜌 = 960 kg/m3 . (d) Balsa wood, 𝜌 = 0.2 Mg/m3 .

Problem 1.26 A Super Ball is a toy ball made of hard synthetic rubber called Zectron. This material has a high coefficient of restitution so that if it is dropped from a certain height onto a hard fixed surface, it rebounds to a substantial portion of its original height. If the Super Ball has 5 cm diameter and the density of Zectron is about 1.5 Mg/m3 , determine the weight of the Super Ball on the surface of the Earth in U.S. Customary units.

Problem 1.27 An ice hockey puck is a short circular cylinder, or disk, of vulcanized rubber with 3.00 in. diameter and 1.00 in. thickness, with weight between 5.5 and 6.0 oz (16 oz = 1 lb). Compute the range of densities for the rubber, in conventional SI units, that will provide for a puck that meets these specifications.

Problem 1.28 Convert the angles given to the units indicated. (a) Convert 𝜃 = 35.6◦ to rad. (b) Convert 𝜃 = (1.08 × 10−3 )◦ to mrad. (c) Convert 𝜃 = 4.65 rad to degrees. (d) Convert 𝜃 = 0.254 mrad to degrees.

Problem 1.29 Many of the examples of failure discussed in Section 1.8 have common causes, such as loads that were not anticipated, overestimation of the strength of materials, unanticipated use, etc. Using several paragraphs, identify those examples that have common causes of failure and discuss what these causes were. Note: Concept problems are about explanations, not computations.

ISTUDY

2

Vectors: Force and Position

Vectors are immensely useful for describing many entities in mechanics. Vectors can be compactly represented, are easy to manipulate, and can be transformed from one component description to another using established rules. In this chapter, vectors are used to describe force and position. In later chapters, vectors will be used to represent other entities. Early sections of this chapter focus on vectors in two dimensions, and later sections treat vectors in three dimensions. Vector dot product and cross product operations are also presented.

David Sailors/Corbis/Getty Images

A skyward view of the cables supporting the Leonard P. Zakim Bunker Hill Bridge, in Boston, Massachusetts. Each of the cables supports a tensile force whose size and direction can be described using a vector.

2.1

Basic Concepts

Introduction—force, position, vectors, and tides One phenomenon in which vectors play an important role is ocean tides. High and low tides each occur approximately twice per day at most locations on the Earth due to the rotation of the Earth and the Moon’s orbit around the Earth—let’s see how vectors help explain this. Figure 2.1 shows the Earth and Moon in a coordinate system that has its origin at point 𝑂 (for measuring the motion of objects in our solar system, the origin is usually located at the center of the Sun). Your intuitive notion of position is probably close to its mathematical definition, but an important point needs to be made. A description of position, or location, is always relative to the coordinate system that is used. For example, if the coordinates of the Moon (𝑥𝑀 , 𝑦𝑀 ) in Fig. 2.1 are known, along with a coordinate system whose origin is at a known point (perhaps the center of the Sun), then the absolute location of the Moon is known. However, if the location of the origin is unknown, then obviously the coordinates of the Moon have little meaning. Rather than using coordinates, we can describe the position of a point 𝑃 by specifying the distance from a reference point to 𝑃 , and by specifying the direction from

(𝑥𝐸 , 𝑦𝐸 )

𝑦 (𝑥𝑀 , 𝑦𝑀 ) 𝑟⃗𝑀

Moon

Earth 𝑟⃗𝐸

𝑂

𝑥

Figure 2.1 A coordinate system showing the positions of the Earth and Moon as well as their position vectors. The relative sizes of the Earth and Moon are correct, but the distance between them is not drawn to scale.

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30

head 𝑣⃗ tail

line of action

𝑣⃗ = magnitude (size)

Figure 2.2 Vector nomenclature.

ISTUDY

Chapter 2

Vectors: Force and Position

Interesting Fact Why don’t the Earth and Moon “crash” into one another? If the Earth and Moon were not moving, then the forces 𝐹⃗𝑀𝐸 and 𝐹⃗𝐸𝑀 shown in Fig. 2.3 would cause them to accelerate toward one another, and they would eventually collide. However, the Earth and Moon are not stationary, and as you well know, they have not collided. When you study dynamics, you will learn that if objects have appropriate motion, the acceleration 𝑎⃗ of each of them is such that, according to Newton’s second law 𝐹⃗ = 𝑚𝑎, ⃗ the objects will orbit their common center of mass (such as the Earth and Moon do) even though there is a force attracting them.

the reference point to 𝑃 . The reference point could be the origin of the coordinate system or any other point we choose. This description combines statements of both size and direction, and it has the same characteristics as an invention of mathematics called a vector. A vector has both size and direction, and to emphasize this content, vectors are represented in this book∗ by placing an arrow above the symbol used for the vector, such as 𝑟⃗. Figure 2.1 uses arrows to represent the position vectors of the Earth and Moon, 𝑟⃗𝐸 and 𝑟⃗𝑀 , respectively, where these vectors are measured from the origin of our coordinate system. Figure 2.2 defines some useful nomenclature. A vector has a head and a tail, as well as a line of action, which is the line of infinite extent along which the vector is positioned. For force vectors especially, the line of action is important, and it is most often within this context that we refer to it. The magnitude of a vector 𝑟⃗ is the measure of its size, or length, including appropriate units, and is denoted by |⃗𝑟|. The magnitude of a vector is a positive scalar for any vector that has nonzero size, and is zero only for a vector of zero size. The magnitude can never be negative. Now, let’s look at how the Earth and Moon attract one another as a result of Newton’s law of universal gravitation [Eq. (1.10)], which states that the gravitational attraction between two bodies is proportional to the product of their masses and is inversely proportional to the square of the distance between them. Unfortunately, this equation doesn’t tell the whole story, the remainder of which is found in Fig. 2.3(a). 𝑦 Moon

𝑦 Moon

𝐹⃗𝑀𝐸

𝑟⃗𝑀

𝐹⃗𝐸𝑀

𝑥 (a)

𝐹⃗𝐸𝑀

once/day

𝑟⃗𝐸

𝑟⃗𝐸 𝑂

tidal bulge

tidal bulge

Earth

𝑟⃗𝑀

once/month 𝐹⃗𝑀𝐸

𝑥

𝑂 (b)

Figure 2.3. (a) The forces of gravitational attraction between the Earth and Moon. The force 𝐹⃗𝐸𝑀 is the force on 𝐸 due to 𝑀, and similarly 𝐹⃗𝑀𝐸 is the force on 𝑀 due to 𝐸. (b) Orbit of the Moon about the Earth and the rotation of the Earth as seen from above their orbital planes. We also show the positions of the Earth and Moon, the gravitational force between the Earth and the Moon, and the tidal bulge (greatly exaggerated) of the Earth due to the Moon.

While Eq. (1.10) tells us how much two bodies attract one another, it doesn’t tell us the direction of the attraction. Figure 2.3(a) shows that these bodies attract one another toward their centers (actually, toward their centers of gravity). Thus, the gravitational attraction between two bodies is a force that has magnitude [given by Eq. (1.10)] and direction (the center of gravity of one body is pulled toward the center of gravity of the other body), and therefore it possesses the characteristics of a vector. In addition, Newton’s third law (see p. 7) tells us that the forces of attraction between these bodies must be equal, opposite, and collinear. With all of this as background, ∗ The notation used throughout the engineering literature to denote vectors is very diverse. Hence, in other

books you may see vectors denoted by the use of a bar placed above or below a given symbol, such as 𝑟 or 𝑟 or by bold type such as 𝐫.

ISTUDY

Section 2.1

let’s see how the positions of the Earth and Moon work in concert with the forces between them to cause the ocean tides on Earth. The Moon orbits the Earth about once per month, and the Earth rotates on its axis once per day, as shown in Fig. 2.3(b). The position of the Moon as seen by the Earth∗ determines the direction and magnitude of the gravitational force exerted on the Earth by the Moon 𝐹⃗𝐸𝑀 . As you might expect, the side of the Earth closest to the Moon experiences a tidal bulge, but you might not expect that the side of the Earth farthest from the Moon does too! Why is this? The tides occur because the Earth’s shape elongates in the direction of 𝐹⃗𝐸𝑀 due to the gravitational effect of the Moon. This elongation is the result of differential forces exerted by the Moon on the Earth. These differential forces arise because the materials of the Earth (i.e., soil, rock, magma, etc.) that are closer to the Moon are pulled by the Moon more strongly than the materials of the Earth that are farther from the Moon. The net effect of these differential forces is that the Earth is stretched in both directions along the line of action of the force 𝐹⃗𝐸𝑀 and is compressed in the directions perpendicular to 𝐹⃗𝐸𝑀 . This causes a tidal bulge on the side of the Earth facing the Moon and another tidal bulge on the side of the Earth opposite the Moon. Now, since the Earth rotates about its axis once per day, there are approximately two tides per day in most locations on the surface of the Earth.† We say approximately because the Moon orbits the Earth as the Earth simultaneously rotates [see Fig. 2.3(b)], and this results in tides occurring about 50 minutes later each day.‡ We hope this helps to explain why there are two high tides and two low tides at most locations on the Earth every day and why force and position vectors play such an important role in determining these tides.

Denoting vectors in figures In figures, we represent vectors by using arrows. Furthermore, the arrows will follow a consistent color scheme to indicate what the vector physically represents. For example, force vectors will be shown in red , position vectors will be shown in blue , and when vectors with other physical significance are introduced, they will use other colors. Vectors with no particular physical significance will be black , magenta , or gray . This practice makes comprehension of a figure quicker, but otherwise is not needed. Next to each arrow shown in a figure, we will provide a symbol to identify the vector; or if the numerical value of a vector’s magnitude is known, we will often simply show this value with the appropriate units. Consider the example shown in Fig. 2.4(a) where a crate is held in equilibrium on a ramp by two people who apply forces to it. We will assume the person on the left pulls on the rope with a force having 25 lb magnitude oriented at 25◦ from the horizontal, and the person on the right pushes the crate with a horizontal force whose magnitude is unknown. In Fig. 2.4(b), the crate is sketched showing the forces applied by the two people, using the following forms for labeling vectors.§ ∗ Later

in this section we discuss subtraction of vectors, after which you may show that the position of the Moon as seen by the Earth is given by 𝑟⃗𝑀 − 𝑟⃗𝐸 . † Since the orbital plane of the Moon is tilted with respect to the spin axis of the Earth, some locations (at mainly higher latitudes) only experience one significant tide per day. ‡ The Moon orbits around the Earth every 29.5 days, so it takes 24∕29.5 hours = 48.8 minutes longer each day for the Moon to reach the same position above the surface of the Earth. § Note that in addition to the forces shown in Fig. 2.4(b), the crate is subjected to other forces that are not shown, such as the weight of the crate and the contact forces applied by the ramp.

Basic Concepts

31

Interesting Fact More tidal tidbits. Tides on Earth are caused mostly by the Moon, but the Sun also has an effect. Although the Sun’s gravitational force on the Earth is about 180 times that of the Moon, because the Moon is much closer to the Earth, the differential force of the Moon on the Earth is about twice that of the Sun. At full Moon (when the Earth is between the Sun and Moon) and new Moon (when the Moon is between the Earth and Sun), the Sun, Earth, and Moon line up, producing higher than normal tides (called spring tides). When the Moon is at first or last quarter, smaller neap tides form. Since the Moon’s 29.5 day orbit around Earth is not quite circular, when the Moon is closest to Earth (called perigee), spring tides are even higher (called perigean spring tides). Interestingly, because the Moon is continually deforming the Earth (changing its shape), the Earth’s rotational energy is continually being lost, causing the Earth’s rate of rotation to decrease so that the length of a day increases by approximately 1.5 milliseconds per century. Finally, some books and websites claim that the “centrifugal force” on the water due to the rotation of the Earth and/or the rotation of the Earth-Moon system are factors in the creation of tides. This is not true, and when you study dynamics, you will learn that these effects are negligible compared with gravitational effects, and that there is no such thing as a “centrifugal force.”

32

(a)

Chapter 2

Vectors: Force and Position

• In the case of the force applied by the rope to the crate, the magnitude is known to be 25 lb, and in such cases we will simply list this magnitude next to its arrow in the figure, and the direction of the arrow gives the direction of the force. 25◦

• When a vector symbol is used in a figure (e.g., 𝐹⃗ ), the symbol represents an expression that fully describes the magnitude and direction for the vector. This form of labeling will often be used in a figure when a vector has unknown magnitude and direction. 25 lb (b)

25◦

𝐹⃗ or 𝐹

Figure 2.4 Vector representation used in figures. (a) Two people apply forces to a crate to keep it in equilibrium on a ramp, where we assume the person pulling the rope applies a 25 lb force oriented at 25◦ from the horizontal and the person pushing the crate applies a horizontal force with unknown magnitude. (b) Forces applied by the two people to the crate using various forms of representation. For further examples contrasting these forms of representation and the implications of Newton’s third law, you should compare Fig. 2 on p. 40 with Fig. 3 on p. 71.

ISTUDY

• When a scalar symbol is used (e.g., 𝐹 ), it refers to the component of the vector in the direction of the arrow shown in the figure. We will provide a full definition for the word component later, but for the present, this is the amount of a vector that acts in the direction of the arrow. With this definition, a symbol such as 𝐹 can be positive, zero, or negative, as follows. If 𝐹 is positive, then indeed the direction of the vector is the same as the direction of the arrow shown in the figure. If 𝐹 is negative, then the direction of the vector is actually opposite to the direction of the arrow shown in the figure. A vector of zero size will have 𝐹 = 0. Observe that the notations 𝐹 and |𝐹⃗ | are different, but they are closely related. As described above, 𝐹 shown in Fig. 2.4 can be positive, zero, or negative, but in any event, the absolute value of 𝐹 is the same as |𝐹⃗ |. Labeling vectors by using a symbol such as 𝐹 will be done often, and it is especially useful when the line of action for a vector is known but we are unsure of its direction along this line of action. To illustrate, consider again the crate in Fig. 2.4(b). We have assigned the direction of 𝐹⃗ to correspond to the person pushing the crate. However, it is possible that this person may actually need to pull the crate to keep it in equilibrium. Only after the equilibrium equations are solved, taking into account the weight of the crate, friction between the crate and ramp, steepness of the ramp, and so on, will we know the value (or range of values) 𝐹 must have for equilibrium. For Fig. 2.4(b), if 𝐹 > 0, the person is pushing the crate while if 𝐹 < 0, the person is pulling the crate.

Basic vector operations Helpful Information Jargon. Consider the following statement: “The widget is subjected to a force with 123 N magnitude.” This statement is precise in its nomenclature, but a practicing engineer would probably have said, “The widget is subjected to a 123 N force.” The second statement, if taken literally, is imprecise because it states that a force (which is a vector) is equal to 123 N (a scalar), and furthermore it does not explicitly say that the force has 123 N magnitude. Nonetheless, the second expression is widely used among engineers, and it is understood that the statement means the magnitude of the force is 123 N.

The following remarks on equivalent vectors and vector addition, subtraction, and multiplication by a scalar are true regardless of the type of vector representation that is used. However, the details of how these operations are carried out will depend on the vector representation that is employed, and we will see that some forms of vector representation allow for easier manipulation than others. Equivalent vectors Two vectors are said to be equivalent, or equal, if they have the same magnitude and orientation. Note that two equivalent vectors may have different lines of action, provided the lines of action are parallel. Vector addition ⃗ and this operation is Addition of two vectors 𝐴⃗ and 𝐵⃗ produces a new vector 𝑅, ⃗ Figure 2.5 illustrates two methods for performing vector denoted by 𝑅⃗ = 𝐴⃗ + 𝐵. addition.

ISTUDY

Section 2.1

𝐴⃗

Basic Concepts

𝐴⃗

𝑅⃗ = 𝐴⃗ + 𝐵⃗

𝐴⃗

𝐵⃗

Helpful Information

𝑅⃗ = 𝐵⃗ + 𝐴⃗

𝑅⃗ = 𝐴⃗ + 𝐵⃗ 𝐵⃗

𝐵⃗

𝐵⃗

parallelogram addition

𝐴⃗

head-to-tail addition

Figure 2.5. Addition of two vectors using the parallelogram method and the head-to-tail method.

With the parallelogram method of addition, the outcome of 𝐴⃗ + 𝐵⃗ is determined by arranging the vectors 𝐴⃗ and 𝐵⃗ tail-to-tail to form a parallelogram. The vector 𝑅⃗ = 𝐴⃗ + 𝐵⃗ is then the vector whose tail coincides with the tails of 𝐴⃗ and 𝐵⃗ and whose head coincides with the parallelogram’s opposite vertex. Alternatively, 𝑅⃗ can also be determined using the head-to-tail method by which we slide the tail of 𝐵⃗ to ⃗ and the resulting triangle provides 𝑅. ⃗ Alternatively, the tail of 𝐴⃗ can the head of 𝐴, ⃗ ⃗ be slid to the head of 𝐵, and the resulting triangle provides the same 𝑅. Vector addition has the following properties: 𝐴⃗ + 𝐵⃗ = 𝐵⃗ + 𝐴⃗

commutative property,

⃗ + 𝐶⃗ = 𝐴⃗ + (𝐵⃗ + 𝐶) ⃗ associative property, (𝐴⃗ + 𝐵)

(2.1) (2.2)

⃗ 𝐵, ⃗ and 𝐶⃗ are three arbitrary vectors. Equations (2.1) and (2.2) imply that the where 𝐴, result of adding an arbitrary number of vectors is independent of the order in which the addition is carried out. Also, the head-to-tail method of addition generalizes so that an arbitrary number of vectors can be added simultaneously simply by arranging them head to tail, one after another. Doing this gives rise to a vector polygon, which is very useful for visualizing the arrangement of vectors. These comments are explored in greater detail in Example 2.1. Multiplication of a vector by a scalar

Polar vector representation. A simple representation we will sometimes use for writing vectors in two dimensions is called polar vector representation. This representation consists of a statement of the vector’s magnitude and direction referred to a right-hand horizontal reference direction, with a positive angle being measured counterclockwise. For example: 100 lb

⃗ = 𝑠𝐴⃗ + 𝑠𝐵⃗ 𝑠(𝐴⃗ + 𝐵)

distributive property with respect to vector addition,

(2.3)

(𝑠 + 𝑡)𝐴⃗ = 𝑠𝐴⃗ + 𝑡𝐴⃗

distributive property with respect to addition of scalars,

(2.4)

associative property with respect to multiplication by a scalar.

(2.5)

Equation (2.3) states 𝐴⃗ and 𝐵⃗ may be first added and then multiplied by 𝑠; or each vector may be multiplied by 𝑠 first, and then the vector addition is performed. Similar comments apply to Eqs. (2.4) and (2.5).

6m 150◦

30◦ 𝐹⃗ = 100 lb @ 30◦

𝑟⃗ = 6 m @ 150◦

Common Pitfall Addition of vectors versus addition of scalars. Addition of vectors is an operation that is very different, and more complex, than addition of scalars. A common error is to attempt to add vectors as if they were scalars.

𝐹⃗

Multiplication of a vector 𝐴⃗ by a scalar 𝑠 produces a new vector 𝑅⃗ where 𝑅⃗ = 𝑠𝐴⃗ = ⃗ The magnitude of 𝑅⃗ is equal to the magnitude of 𝐴⃗ multiplied by |𝑠|. If 𝑠 is pos𝐴𝑠. itive, then 𝑅⃗ and 𝐴⃗ have the same direction; if 𝑠 is negative, then 𝑅⃗ has direction ⃗ Multiplication of a vector by a scalar is common, and this operation opposite to 𝐴. allows us to change a vector’s size. Multiplication of a vector by a scalar does not change the vector’s line of action; but if the scalar is negative, then the vector’s direction along its line of action is reversed. Figure 2.6 shows some examples. Let 𝐴⃗ and 𝐵⃗ be vectors and 𝑠 and 𝑡 be scalars. Multiplication of a vector by a scalar has the following properties:

⃗ (𝑠𝑡)𝐴⃗ = 𝑠(𝑡𝐴)

33

(a)

2𝐹⃗

(b)

−2𝐹⃗

(c)

Figure 2.6 Examples of multiplication of a vector by a scalar. (a) A force 𝐹⃗ is applied to the handle of a wrench. (b) 𝐹⃗ is multiplied by 2, with the resulting vector drawn to scale. (c) 𝐹⃗ is multiplied by −2, with the resulting vector drawn to scale and with reversed direction.

34

Chapter 2

Vectors: Force and Position

𝐴⃗

𝐴⃗

𝐵⃗

𝐴⃗ + 𝐵⃗

𝐵⃗

Vector subtraction Subtracting a vector 𝐵⃗ from a vector 𝐴⃗ is denoted by 𝐴⃗ − 𝐵⃗ and is defined as ⃗ 𝐴⃗ − 𝐵⃗ = 𝐴⃗ + (−1) 𝐵.

(b)

(a) 𝐴⃗

−𝐵⃗

𝐴⃗ − 𝐵⃗ −𝐵⃗

𝐴⃗ (d)

(c)

Figure 2.7 Comparison of addition and subtraction of two ⃗ 𝐵⃗ vectors. (a) Vectors 𝐴⃗ and 𝐵⃗ are defined. (b) 𝐴+ is evaluated using the head-to-tail method. (c) The direction of 𝐵⃗ is reversed by multiplying it by −1. (d) 𝐴⃗ − 𝐵⃗ is evaluated using the head-totail method.

In this definition, first the direction of 𝐵⃗ is reversed by multiplying it by −1, and then ⃗ In Fig. 2.7, addition and subtraction of vectors are contrasted. the result is added to 𝐴.

Performing vector operations To use parallelogram addition or head-to-tail addition, we could add vectors graphically using a ruler and protractor with very careful drawings, but it is more precise and appropriate to use analytical methods. Since the addition of two vectors will generally involve a triangle or parallelogram of complex geometry, the laws of sines and cosines will be useful. With reference to Fig. 2.8, for a general triangle the law of sines and the law of cosines are sin 𝜃𝑎

𝐴

𝐶

𝜃𝑎

𝜃𝑐

𝐴

𝜃𝑏

𝐵

𝜃𝑏

𝐶

𝜃𝑐

sin 𝜃𝑏 𝐵

=

sin 𝜃𝑐 𝐶

√ 𝐵 2 + 𝐶 2 − 2𝐵𝐶 cos 𝜃𝑎 √ 𝐵 = 𝐴2 + 𝐶 2 − 2𝐴𝐶 cos 𝜃𝑏 √ 𝐶 = 𝐴2 + 𝐵 2 − 2𝐴𝐵 cos 𝜃𝑐

law of sines,

(2.7)

𝐵 = 𝐴 cos 𝜃𝑐 = 𝐴 sin 𝜃𝑏

Figure 2.9 A right triangle.

Common Pitfall Law of sines and obtuse angles. You will avoid errors if you avoid using the law of sines to determine obtuse angles! The pitfall is that the inverse sine function provided by an electronic calculator yields angles between −90◦ and +90◦ only (right or acute angles only). For example, in Fig. 2.8, let 𝐴 = 9 mm, 𝐵 = 6 mm, and 𝜃𝑏 = 30◦ (with these values, Fig. 2.8 is drawn roughly to scale). Using the law of sines, Eq. (2.7), we obtain sin 𝜃𝑎 = (𝐴∕𝐵) sin 𝜃𝑏 = 0.7500. Using an electronic calculator to evaluate 𝜃𝑎 = sin−1 (0.7500) provides 48.59◦ , which is not the correct value of 𝜃𝑎 for this triangle! Of course, the equation 𝜃𝑎 = sin−1 (0.7500) has an infinite number of solutions, and the correct solution for this triangle is 𝜃𝑎 = 131.4◦ , which is an obtuse angle.

law of cosines.

(2.8)

In the special case that the triangle has a right angle, as in Fig. 2.9 where 𝜃𝑎 = 90◦ , the laws of sines and cosines simplify to the familiar expressions

𝐵

ISTUDY

=

𝐴=

Figure 2.8 A general triangle. 𝐴

(2.6)

𝐶 = 𝐴 sin 𝜃𝑐 = 𝐴 cos 𝜃𝑏 𝐴=

√ 𝐵2 + 𝐶 2

Pythagorean theorem.

(2.9)

(2.10)

If it has been some time since you have used trigonometry, you should thoroughly refamiliarize yourself with Eqs. (2.7)–(2.10).

Resolution of a vector into vector components Consider a vector 𝐹⃗ that might represent a force, position, or some other entity. When 𝐹⃗ is expressed as a sum of a set of vectors, then each vector of this set is called a vector component of 𝐹⃗ , and the process of representing 𝐹⃗ as a sum of other vectors is often called resolution of 𝐹⃗ into vector components. For example, we will often ⃗ when added together, will yield 𝐹⃗ ?” In answering ask, “What two vectors 𝐴⃗ and 𝐵, this question, the vectors 𝐴⃗ and 𝐵⃗ that we find are called the vector components of 𝐹⃗ . Of course, as shown in Fig. 2.10, there is not a unique answer to this problem, as there are an infinite number of vectors 𝐴⃗ and 𝐵⃗ such that 𝐴⃗ + 𝐵⃗ = 𝐹⃗ . However, if we ⃗ such as they must have certain place certain constraints or restrictions on 𝐴⃗ and/or 𝐵, prescribed directions and/or magnitudes, then the vectors 𝐴⃗ and 𝐵⃗ whose addition yields 𝐹⃗ may be unique. In statics, we will usually want to find vector components

ISTUDY

Section 2.1

that have directions we specify, such as the directions of structural members or the directions of coordinate axes. Referring to Fig. 2.10, we see that the vector triangles that arise when resolving a vector into vector components may have a general triangular shape or a right triangular shape. General triangular shapes occur for cases 𝐴⃗1 and 𝐵⃗1 , and 𝐴⃗2 and 𝐵⃗2 , and for such vector triangles the laws of sines and cosines are usually needed to determine the vector components. If the vector components are taken to be orthogonal (i.e., perpendicular), then the vector triangles are right triangles as shown for cases 𝐴⃗3 and 𝐵⃗3 , and 𝐴⃗4 and 𝐵⃗4 . For right triangles, elementary trigonometry is sufficient, and resolution of the vector into vector components is usually straightforward. In fact, it is for this reason that Cartesian vector representation, to be discussed in Section 2.2, is so convenient and effective.

End of Section Summary In this section, basic properties of vectors and some operations using vectors have been described, including • Polar vector representation is a simple convention for stating the magnitude and direction of vectors in two dimensions. Polar vector representation is defined in the margin note on p. 33, and an example is 𝐹⃗ = 100 lb @ 30◦ . • Vector addition, subtraction, and multiplication by a scalar. • Resolution of a vector into vector components. • As an aid to help manipulate vectors, the laws of sines and cosines were reviewed.

Basic Concepts

35

𝐵⃗2 𝐵⃗1

𝐴⃗2 𝐴⃗1

𝐵⃗4

𝐹⃗

𝐴⃗4 𝐵⃗3 𝐴⃗3 Figure 2.10 Four examples of the resolution of a vector 𝐹⃗ into two vectors 𝐴⃗ and 𝐵⃗ such that 𝐴⃗ + 𝐵⃗ = 𝐹⃗ .

36

Chapter 2

Vectors: Force and Position

E X A M P L E 2.1

Addition of Vectors

𝐹2 = 60 N

𝐹3 = 100 N 60◦

30◦ 𝐹1 = 40 N

A D ring is sewn on a backpack for use in securing miscellaneous items to the outside of the backpack. If the D ring has three cords tied to it and the cords support the forces shown, determine the resultant force applied to the D ring by the cords, expressing the result as a vector.

SOLUTION Road Map

We first note that the force supported by a cord has the same direction as the cord, and thus, in Fig. 2 we redraw the D ring, showing the forces that are applied to it by the cords. The resultant force vector is the sum of the three force vectors applied to the D ring. Thus, we will add the vectors, choosing the head-to-tail method, and apply the laws of sines and cosines to determine the results of the vector addition.

Figure 1 Governing Equations & Computation

The resultant force vector is

𝑅⃗ = 𝐹⃗1 + 𝐹⃗2 + 𝐹⃗3 . 𝐹3 = 100 N

𝐹2 = 60 N

60◦

30◦ 𝐹1 = 40 N

(1)

The addition is illustrated graphically in Fig. 3, where the first illustration shows forces added in the order 𝐹⃗1 + 𝐹⃗2 + 𝐹⃗3 , while the second illustration shows addition in the order 𝐹⃗3 + 𝐹⃗2 + 𝐹⃗1 , and the third illustration shows addition in the order 𝐹⃗1 + 𝐹⃗3 + 𝐹⃗2 ; according ⃗ The polygons shown in to Eq. (2.2), all three sums provide the same resultant force 𝑅. Fig. 3 are often called force polygons (or more generally vector polygons), and they are useful in statics as well as many areas of mechanics, not just for computation, but also for visualizing the spatial relationship among forces. 𝐹⃗1

Figure 2 D ring showing the forces applied by the cords to the D ring.

𝐹⃗2

𝐹⃗3

𝐹⃗2 𝑅⃗

𝑅⃗ 𝐹⃗2

𝐹⃗3

𝑅⃗ 𝐹⃗3 𝐹⃗1

𝐹⃗1

Figure 3. Vector polygons illustrating head-to-tail addition of 𝐹⃗1 , 𝐹⃗2 , and 𝐹⃗3 in different orders. In all cases, the same resultant vector 𝑅⃗ is obtained.

𝛽 𝐹1 = 40 N

𝑃 𝛼

𝐹2 = 60 N 30◦

Figure 4 Addition of 𝐹⃗1 and 𝐹⃗2 to obtain an intermediate result 𝑃⃗ (i.e., 𝑃⃗ = 𝐹⃗1 + 𝐹⃗2 ).

ISTUDY

⃗ we will use the first force polygon shown in Fig. 3. Unfortunately, the To compute 𝑅, ⃗ geometry of a general polygon is usually too complex to allow a direct evaluation of 𝑅. Thus, we usually must break the polygon into smaller elements, namely, triangles, each of which may be analyzed using the laws of sines and cosines as given in Eqs. (2.7) and (2.8), or in the case of right triangles as given in Eqs. (2.9) and (2.10). Thus, we will first define an intermediate vector sum to be 𝑃⃗ = 𝐹⃗1 + 𝐹⃗2 , as shown in Fig. 4. Using the law of cosines, we write √ 𝑃 = 𝐹 12 + 𝐹 22 − 2𝐹1 𝐹2 cos 𝛼 √ (2) = (40 N) 2 + (60 N) 2 − 2(40 N)(60 N) cos(150◦ ) = 96.73 N, where angle 𝛼 was easily found from Fig. 4 as 𝛼 = 180◦ −30◦ = 150◦ . Now that magnitude 𝑃 is known, we use the law of sines, Eq. (2.7), to find angle 𝛽 shown in Fig. 4 as 𝐹 𝑃 = 2 , sin 𝛼 sin 𝛽

(3)

ISTUDY

Section 2.1

Basic Concepts

hence, sin 𝛽 =

𝐹2

sin(150◦ ) =

𝑃 Solving Eq. (4) for 𝛽 provides

60 N sin(150◦ ) = 0.3101. 96.73 N

−1

𝛽 = sin (0.3101) = 18.07 . ◦

𝑅

⃗ Defining the angle 𝛾 to be Next, as shown in Fig. 5, we evaluate 𝑃⃗ + 𝐹⃗3 to obtain 𝑅. the angle between force vectors 𝑃⃗ and 𝐹⃗3 , we use the geometry shown in Fig. 2 and the value of 𝛽 previously found to write 𝛾 = 90◦ − 60◦ + 18.07◦ = 48.07◦ . The law of cosines provides √ 𝑅= =

√

𝑃 2 + 𝐹32 − 2𝑃 𝐹3 cos 𝛾

(96.73 N) 2 + (100 N) 2 − 2(96.73 N)(100 N) cos(48.07◦ )

= 80.18 N.

(6)

The law of sines is then used to write 𝐹3 sin 𝜃 Hence, sin 𝜃 =

𝐹3 𝑅

sin 𝛾 =

=

𝑅 . sin 𝛾

(7)

100 N sin 48.07◦ = 0.9278, 80.18 N

(8)

which provides 𝜃 = sin −1 (0.9278) = 68.10◦ . (9) ⃗ To complete our solution, we report 𝑅 in vector form using polar vector representation. The magnitude of 𝑅⃗ is 𝑅 = 80.18 N, and its orientation is 𝜃 + 𝛽 = 68.10◦ + 18.07◦ = 86.17◦ counterclockwise from the right-hand horizontal direction. Thus, 𝑅⃗ = 80.18 N @ 86.17◦

.

𝐹3 = 100 N

(4)

(5)

(10)

Discussion & Verification

• Addition of vectors by the method illustrated here is tedious. In fact, as the number of vectors to be added increases, the number of sides of the vector polygon also increases. The Cartesian vector representation, introduced in Section 2.2, will provide for considerably more straightforward addition of vectors. Nonetheless, it is often very helpful to be able to sketch vector polygons, as this provides a visual understanding of the spatial relationship among vectors, which in turn may offer significant simplifications in analysis. • This problem focused on the resultant force applied by the three cords to the D ring. However, the fabric that attaches the D ring to the backpack also applies forces to the D ring. In Chapter 3, we consider the combined effect of all forces applied to objects such as the D ring in this example.

37

𝜃

60◦ 𝛾 18.07◦ 𝑃 = 96.73 N 18.07◦

Figure 5 Addition of 𝑃⃗ and 𝐹⃗3 to obtain the final result 𝑅⃗ (i.e., 𝑅⃗ = 𝑃⃗ + 𝐹⃗3 ).

38

Chapter 2

Vectors: Force and Position

E X A M P L E 2.2

Resolution of a Vector into Components Two people apply forces to push a child’s play structure resting on a patio. The woman at 𝐴 applies a force in the negative 𝑎 direction and the man at 𝐵 applies a force in the 𝑏 direction, with the goal of producing a resultant force of 250 N in the 𝑐 direction. Determine the forces the two people must apply, expressing the results as vectors.

SOLUTION The force applied by the woman at 𝐴 will be called 𝐹⃗𝑎 , and the force applied by the man at 𝐵 will be called 𝐹⃗𝑏 , such that the sum of these is the 250 N force vector 𝐹⃗ shown in Fig. 2 (1) 𝐹⃗ = 𝐹⃗𝑎 + 𝐹⃗𝑏 . Road Map

Lucinda Dowell

Thus, our goal is to resolve 𝐹⃗ into the component vectors 𝐹⃗𝑎 and 𝐹⃗𝑏 . Since the 𝑎 and 𝑏 directions are orthogonal, determining the component vectors will be straightforward using basic trigonometry.

𝑏

𝑐 250 N 30◦

𝑎

Governing Equations & Computation We first note that both vectors 𝐹⃗𝑎 and 𝐹⃗𝑏 have known direction and unknown magnitude. To determine the components, you may find the process shown in Fig. 3 to be helpful. In Fig. 3(a), we begin by sketching a dashed line parallel to the 𝑎 direction that passes through the tail of the vector, followed by sketching a dashed line parallel to the 𝑏 direction that passes through the head of the vector. These dashed lines form a triangle, which leads to the sketch shown in Fig. 3(b). Using elementary trigonometry, the absolute values of the components are

𝐴 𝐵 Figure 1 𝐹 = 250 N

𝑏

|𝐹𝑎 | = (250 N) cos 30◦ = 216.5 N,

(2)

|𝐹𝑏 | = (250 N) sin 30 = 125 N.

(3)

◦

30◦

𝑎

Figure 2 The resultant force applied to the play structure by the two people has 250 N magnitude.

The absolute value signs are used in Eqs. (2) and (3) and in Fig. 3 because, by definition, the components of the vector are positive in the positive 𝑎 and 𝑏 directions. The vector components may now be written as 𝐹⃗𝑎 = −216.5 N

in the 𝑎 direction,

𝐹⃗𝑏 = 125 N in the 𝑏 direction, 𝑏 𝐹 = 250 N (a) 30◦

𝑎 𝑏

(4) (5)

where the negative sign is needed in Eq. (4) because this component, as seen in Fig. 3(b), acts in the negative 𝑎 direction. Further, we state that the scalar components (or simply components) of the 250 N force in the 𝑎 and 𝑏 directions are −216.5 N and 125 N, respectively. Equations (4) and (5) state both magnitude and direction and hence are vector expressions. These results may alternatively be stated using polar vector representation as

𝐹 = 250 N (b)

𝐹𝑏

𝐹⃗𝑎 = 216.5 N @ 180◦ 30◦

𝐹⃗𝑏 = 125 N @ 90

𝑎

𝐹𝑎 Figure 3 Resolution of the 250 N force vector into components in the 𝑎 and 𝑏 directions.

ISTUDY

◦

, .

(6) (7)

Discussion & Verification By examining Fig. 3(b), you should verify that 𝐹𝑎 and 𝐹𝑏 have reasonable values. That is, we see in Fig. 3(b) that the “length” of 𝐹⃗𝑎 is somewhat smaller than the “length” of 𝐹⃗ ; thus we expect 𝐹𝑎 (found above to be 216.5 N) to be somewhat smaller than 250 N. Similarly, we see in Fig. 3(b) that the “length” of 𝐹⃗𝑏 is somewhat smaller than the “length” of 𝐹⃗𝑎 , thus we expect 𝐹𝑏 (found above to be 125 N) to be somewhat smaller than 216.5 N. Furthermore, since the forces in Fig. 3(b) form √ a right triangle, we can use the Pythagorean theorem, Eq. (2.10), to verify that

𝐹𝑎2 + 𝐹𝑏 2 = 250 N.

ISTUDY

Section 2.1

39

Basic Concepts

E X A M P L E 2.3

Resolution of a Vector into Components

Repeat Example 2.2 to determine the forces the two people must apply in the new 𝑎 and 𝑏 directions shown such that a resultant force of 250 N in the 𝑐 direction is produced.

𝑏

𝑐 250 N

SOLUTION

45◦

30◦

𝑎

The people at 𝐴 and 𝐵 apply forces 𝐹⃗𝑎 and 𝐹⃗𝑏 , respectively, to the child’s play structure such that the sum of these is the 250 N force vector 𝐹⃗ shown in Fig. 2, Road Map

𝐹⃗ = 𝐹⃗𝑎 + 𝐹⃗𝑏 .

(1)

𝐴 𝐵 Figure 1

Thus, our goal is to resolve 𝐹⃗ into the component vectors 𝐹⃗𝑎 and 𝐹⃗𝑏 . Since the 𝑎 and 𝑏 directions are not orthogonal, determining the component vectors will be more work than in Example 2.2, and the law of sines and/or cosines will be needed. As in Example 2.2, both vectors 𝐹⃗𝑎 and 𝐹⃗𝑏 have known direction and unknown magnitude. To determine the components, you may find the process shown in Fig. 3 to be helpful. In Fig. 3(a), we begin by sketching a dashed line parallel to the 𝑎 direction that passes through the tail of the vector, followed by sketching a dashed line parallel to the 𝑏 direction that passes through the head of the vector. These dashed lines form a triangle, which leads to the sketch shown in Fig. 3(b). Angle 𝛼 shown in Fig. 3(b) is easily found to be 𝛼 = 180◦ − 30◦ − 45◦ = 105◦ . Using the law of sines, the absolute values of the components |𝐹𝑎 | and |𝐹𝑏 | are related by

𝑏

𝐹 = 250 N

Governing Equations & Computation

|𝐹𝑏 | |𝐹 | 250 N = 𝑎 = , sin 45◦ sin 𝛼 sin 30◦

(2)

|𝐹𝑎 | = 341.5 N and |𝐹𝑏 | = 176.8 N.

(3)

30◦

45◦

𝑎

Figure 2 The resultant force applied to the play structure by the two people has 250 N magnitude.

hence, The absolute value signs are used in Eqs. (2) and (3) and in Fig. 3 because, by definition, the components of a vector are positive in the positive 𝑎 and 𝑏 directions. The vector components may now be written as 𝐹⃗𝑎 = −341.5 N 𝐹⃗𝑏 = 176.8 N

𝑏

𝑏

in the 𝑎 direction, in the 𝑏 direction,

(4) (5)

𝐹 = 250 N (a)

45◦

where the negative sign is needed in Eq. (4) because this component, as seen in Fig. 3(b), acts in the negative 𝑎 direction. Further, the scalar components (or simply components) of the 250 N force in the 𝑎 and 𝑏 directions are −342 N and 177 N, respectively. Equations (4) and (5) can alternatively be rewritten using polar vector representation as 𝐹⃗𝑎 = 341.5 N @ 180◦ 𝐹⃗𝑏 = 176.8 N @ 45◦

, .

(6) (7)

Discussion & Verification

• By examining Fig. 3(b), you should verify that 𝐹𝑎 and 𝐹𝑏 have reasonable values. That is, we expect 𝐹𝑎 to be somewhat larger than 250 N and 𝐹𝑏 to somewhat smaller than 250 N. • To achieve the objective of producing a 250 N resultant force in the 𝑐 direction shown in Fig. 1, the woman at 𝐴 in this example problem must apply a substantially higher force than in the previous example problem (341.5 N versus 216.5 N). Obviously, the man at 𝐵 in this example is applying force in a direction that is not as effective as in Example 2.2.

30◦

45◦

𝑎

𝑏 (b)

𝐹𝑏

𝐹 = 250 N

𝛼

30◦

45◦ 𝐹𝑎

45◦

𝑎

Figure 3 Resolution of the 250 N force vector into components in the 𝑎 and 𝑏 directions.

40

Chapter 2

Vectors: Force and Position

E X A M P L E 2.4

Parallel and Perpendicular Vector Components A tractor becomes stuck in mud, and the operator plans to use a truck positioned at point 𝐴 to pull it free.

𝐵 trees 𝐴

𝑂 30◦

Figure 1 Use of a truck to free a tractor stuck in mud.

(a) If a 400 lb force in the direction of the tractor’s chassis (i.e., in the direction from 𝑂 to 𝐵) is sufficient to free it, determine the force the truck must produce in the cable. Also, determine the portion of the cable force that is perpendicular to the tractor’s direction, since in the tractor’s precarious condition this force will tend to tip it. (b) If the trees were not present, determine the ideal location on the road where the truck should be positioned.

SOLUTION The force applied by the truck to the tractor will be called 𝐹𝑂𝐴 , and this force is shown in Fig. 2. Note that according to Newton’s third law, the force applied by the tractor to the truck has equal magnitude and opposite direction, and this force is also shown in Fig. 2. We will assume that points 𝑂, 𝐴, and 𝐵 lie in a horizontal plane so that this is a two-dimensional problem; if this is not the case, then vectors in three dimensions are needed. With the truck positioned at 𝐴, we ask for the cable force 𝐹𝑂𝐴 needed so that the component of this force in the tractor’s direction (line 𝑂𝐵) is 400 lb. We denote by 𝐹⃗‖ and 𝐹⃗ the vector components of 𝐹⃗ ; i.e., 𝐹⃗ + 𝐹⃗ = 𝐹⃗ , where 𝐹⃗ is the force in the direction Road Map

𝐵 trees 𝑂 𝐹𝑂𝐴

𝐹𝑂𝐴 𝐴

30◦ Force applied by the truck to the tractor.

According to Newton’s third law, the force applied by the tractor to the truck is equal to 𝐹𝑂𝐴 with opposite direction.

Figure 2 Force applied by the truck to the tractor is 𝐹𝑂𝐴 . Although not needed for this problem, the force applied by the tractor to the truck is also shown.

⊥

𝑂𝐴

‖

⊥

‖

Part (a)

The force polygon for the addition 𝐹⃗‖ + 𝐹⃗⊥ = 𝐹⃗𝑂𝐴 is shown in Fig. 3 (alternatively, the force polygon for the addition 𝐹⃗⊥ + 𝐹⃗‖ = 𝐹⃗𝑂𝐴 could be used). Because this force polygon is a right triangle, the basic trigonometric relations in Eq. (2.9) can be used to provide Governing Equations & Computation

𝐹𝑂𝐴 cos 30◦ = 400 lb

⇒

𝐹𝑂𝐴 = 461.9 lb,

(1)

and

𝐹‖ = 400 lb

𝐹⊥ = 𝐹𝑂𝐴 sin 30◦ = 230.9 lb.

𝐹⊥ 30◦ 𝐹𝑂𝐴 Figure 3 Force polygon for the force applied by the truck to the tractor.

ISTUDY

𝑂𝐴

𝑂𝐵, and 𝐹⃗⊥ is the force in the direction perpendicular to 𝑂𝐵. In this problem, 𝐹⃗‖ is known and has 400 lb magnitude in the direction of 𝑂𝐵. Vector 𝐹⃗⊥ has unknown magnitude but known direction that is perpendicular to 𝑂𝐵, and 𝐹⃗𝑂𝐴 has unknown magnitude but known direction 𝑂𝐴. For Part (b), we will determine the location of the truck so that the magnitude of 𝐹⃗⊥ will be zero, and hence there will be no tendency for the tractor to tip while it is being pulled free.

(2)

Discussion & Verification

By examining the force polygon in Fig. 3, you should verify that 𝐹𝑂𝐴 and 𝐹⊥ have reasonable values. That is, we expect 𝐹𝑂𝐴 to be somewhat larger than 400 lb, and 𝐹⊥ to be somewhat smaller than 400 lb. Furthermore, since the forces in Fig. 3 form a right triangle, we can use the Pythagorean theorem to verify that √ (400 lb)2 + (𝐹⊥ )2 = 𝐹𝑂𝐴 . Part (b)

Discussion & Verification

If the trees were not present, then the optimal location for the truck would be point 𝐵. To see that this is true, we consult Fig. 3 and observe that if the truck is positioned at point 𝐵, then 𝐹⊥ = 0 and the entire cable force 𝐹𝑂𝐴 is in the direction of the tractor. Thus, to free the tractor requires 𝐹𝑂𝐴 = 𝐹‖ = 400 lb, which is smaller than the value found in Eq. (1). Furthermore, 𝐹⊥ = 0 so there is no force tending to tip the tractor.

ISTUDY

Section 2.1

Basic Concepts

E X A M P L E 2.5

41

Vector Components and Optimization

Repeat Example 2.4, now using two trucks to free a tractor that is stuck in mud. By using two trucks, it is possible to avoid the tendency to tip the tractor. To free the tractor, the two trucks must produce a combined force of 400 lb in the direction of the tractor’s chassis (i.e., in the direction from 𝑂 to 𝐵). (a) One truck is positioned at point 𝐴 and a second truck is positioned at point 𝐶. Determine the cable forces so that there is no force perpendicular to the tractor’s direction (i.e., there is no tipping force). (b) Determine the point 𝐶 ′ where the truck originally at point 𝐶 should be relocated so that there is no force perpendicular to the tractor’s direction and the force in this truck’s cable is as small as possible.

𝐶

𝐵

45◦ trees

𝐴

𝑂 30◦

Figure 1 Use of two trucks to free a tractor stuck in mud.

SOLUTION Road Map

The main benefit of using two trucks to free the tractor is that it is possible to simultaneously produce a 400 lb force in the direction of the tractor and zero force perpendicular to the tractor. The forces applied by the trucks to the tractor are called 𝐹𝑂𝐴 and 𝐹𝑂𝐶 , as shown in Fig. 2. Although not needed for this problem, Newton’s third law has been used to also show the forces applied by the tractor to the trucks in Fig. 2. We will assume that points 𝑂, 𝐴, 𝐵, and 𝐶 lie in a horizontal plane so that this is a two-dimensional problem; if this is not the case, then vectors in three dimensions are needed. In both parts of this problem, we ask for two force vectors, one in direction 𝑂𝐴 and the other in direction 𝑂𝐶, such that 𝐹⃗𝑂𝐴 + 𝐹⃗𝑂𝐶 = 𝐹⃗‖ . In Part (a), both 𝐹⃗𝑂𝐴 and 𝐹⃗𝑂𝐶 have known direction but unknown magnitude, and 𝐹⃗ is fully known. In Part (b), the only ‖

known information is the direction of 𝐹⃗𝑂𝐴 and the direction and magnitude of 𝐹⃗‖ . Part (a) Governing Equations & Computation

The force polygon is shown in Fig. 3, and it has general triangular shape (i.e., it does not have a right angle). First, angle 𝛼 is easily found as 𝛼 = 180◦ − 30◦ − 45◦ = 105◦ . Then the law of sines provides 𝐹𝑂𝐶 𝐹 400 lb = 𝑂𝐴 = , ◦ sin 45 sin 𝛼 sin 30◦

(1)

hence, 𝐹𝑂𝐴 = 546.4 lb and 𝐹𝑂𝐶 = 282.8 lb.

(2)

Discussion & Verification By examining the force polygon in Fig. 3, you should verify that 𝐹𝑂𝐴 and 𝐹𝑂𝐶 have reasonable values. That is, we expect 𝐹𝑂𝐴 to be somewhat larger than 400 lb, and 𝐹𝑂𝐶 to be somewhat smaller than 400 lb. Part (b) Governing Equations & Computation

In this part, we ask where the truck originally at 𝐶 should be relocated so that the force in its cable is as small as possible, while still producing the necessary 400 lb in the direction of the tractor (from points 𝑂 to 𝐵) and zero force perpendicular to the tractor’s direction; the new location for this truck is denoted by 𝐶 ′ . As in Part (a), we seek two vectors 𝐹⃗ and 𝐹⃗ ′ such that 𝐹⃗ + 𝐹⃗ ′ = 𝐹⃗ 𝑂𝐴

𝑂𝐶

𝑂𝐴

𝑂𝐶

‖

and as before, 𝐹⃗‖ is known and 𝐹⃗𝑂𝐴 has known direction but unknown magnitude. However, this problem is more complicated because 𝐹⃗𝑂𝐶 ′ has unknown magnitude and direction. It is possible to use calculus to determine the optimal position for the truck at 𝐶 ′ (Problem 2.28 explores this solution). However, consideration of force polygons corresponding to possible choices for 𝐹⃗𝑂𝐴 and 𝐹⃗𝑂𝐶 ′ will offer a more straightforward solution as follows.

𝐹𝑂𝐶

𝐶 45◦

𝐵

𝐹𝑂𝐶

trees 𝑂 𝐹 𝑂𝐴

𝐹𝑂𝐴 𝐴

30◦ Figure 2 Forces applied by the trucks to the tractor are 𝐹𝑂𝐴 and 𝐹𝑂𝐶 . Although not needed for this problem, the forces applied by the tractor to the trucks are also shown.

𝐹‖ = 400 lb 30◦

𝛼 45◦

𝐹𝑂𝐶

𝐹𝑂𝐴 Figure 3 Force polygon for the forces applied by the trucks to the tractor.

42

Chapter 2

Vectors: Force and Position 𝐹‖ = 400 lb 30◦

𝑂

In Fig. 4 we first sketch vector 𝐹⃗‖ since it has known magnitude and direction. Since

𝐹𝑂𝐶 𝐴

𝐹𝑂𝐴

Figure 4 Use of a force polygon to optimize the direction of a force. Three possible choices for 𝐹⃗𝑂𝐴 are shown, along with the corresponding vectors 𝐹⃗𝑂𝐶 ′ that complete the force polygon.

𝐹⃗𝑂𝐴 has known direction, we sketch some possible choices for it that have different magnitude, and three such choices are shown in Fig. 4. For each 𝐹⃗𝑂𝐴 we sketch the required vector 𝐹⃗𝑂𝐶 ′ to complete the force polygon, and these three vectors are also shown in Fig. 4. Examination of the vectors 𝐹⃗𝑂𝐶 ′ in Fig. 4 clearly shows that the smallest magnitude of 𝐹⃗𝑂𝐶 ′ is achieved when 𝐹⃗𝑂𝐶 ′ is perpendicular to 𝐹⃗𝑂𝐴 . Hence, to minimize the magnitude of the force in cable 𝑂𝐶 ′ , the truck originally at 𝐶 should be repositioned to point 𝐶 ′ shown in Fig. 5. Now that the direction of 𝐹⃗𝑂𝐶 ′ is known, the magnitudes of 𝐹⃗𝑂𝐴 and 𝐹⃗𝑂𝐶 ′ can be found, as was done in Part (a), except now Eqs. (2.9) and (2.10) for a right triangle may be used. Doing so, we find 𝐹𝑂𝐴 = 346.4 lb and 𝐹𝑂𝐶 ′ = 200 lb. Discussion & Verification

𝐶 𝐵

90◦ trees 𝑂

𝐴

30◦ Figure 5 New position 𝐶 ′ for the truck that was originally at point 𝐶, such that there is no force perpendicular to the tractor’s direction and the force in cable 𝑂𝐶 ′ is as small as possible.

ISTUDY

• Usefulness of force polygons. The solution to Part (b) requires a vector with unknown magnitude and direction be determined (i.e., 𝐹⃗𝑂𝐶 ′ ). This may be determined using calculus, as described in Prob. 2.28, but nonlinear equations must be solved. By drawing a force polygon, the solution for the optimal orientation of 𝐹⃗𝑂𝐶 ′ was found by inspection, with the result that it was easier to solve for the remaining unknown, namely, the magnitude of 𝐹⃗𝑂𝐶 ′ . • Tip. Although in principle vectors can be added in any order, from a practical standpoint, when constructing vector polygons, you should begin with those vectors whose magnitude and direction are known. Vectors that have unknown magnitude and/or direction should be added last. For example, in Part (b), the force polygon resulting from the addition of forces in the order 𝐹⃗𝑂𝐶 ′ , and then 𝐹⃗𝑂𝐴 , followed by 𝐹⃗‖ would be considerably less straightforward to visualize than Fig. 4.

ISTUDY

Section 2.1

43

Basic Concepts

Problems Problem 2.1 For each vector, write two expressions using polar vector representations, one using a positive value of 𝜃 and the other a negative value, where 𝜃 is measured counterclockwise from the right-hand horizontal direction.

12 in. (a)

Problems 2.2 through 2.7

Figure P2.1

⃗ and report your result using polar Add the two vectors shown to form a resultant vector 𝑅, vector representation. 1.8 m

1.23 kip

183 mm 55◦

20◦ 8.2 kN

65◦

45◦

2.3 m 6 kN

1.55 kip

101 mm

(a)

(b)

(a) Figure P2.2

(b) Figure P2.3

20◦ 30◦

100 mm

48 N

30◦

80 mm

54 N

300 lb

30◦ 3 ft

20◦

10◦

(a)

(a)

(b) Figure P2.4

(b) Figure P2.5

6 in.

200 lb

35 kN

1.89 f t 45◦

20◦ 139 lb

400 lb

4 ft

18 kN

8 in.

60◦

60◦ 1.23 f t

(a)

(b) Figure P2.6

(a)

(b) Figure P2.7

Problem 2.8 Let 𝐴⃗ = 2 m @ 0◦ and 𝐵⃗ = 6 m @ 90◦ . Sketch the vector polygons and evaluate 𝑅⃗ for the following, reporting your answer using polar vector representation. ⃗ (a) 𝑅⃗ = 𝐴⃗ + 𝐵, ⃗ (b) 𝑅⃗ = 2𝐴⃗ − 𝐵, ⃗ 𝐵⃗ + |𝐵| ⃗ 𝐴, ⃗ (c) 𝑅⃗ = |𝐴| 𝐴⃗ 𝐵⃗ (d) 𝑅⃗ = + . ⃗ ⃗ |𝐴| |𝐵|

23 N 45◦ (b)

60◦ 15 m/s (c)

44

Chapter 2

Vectors: Force and Position

Problem 2.9 A tow truck applies forces 𝐹⃗1 and 𝐹⃗2 to the bumper of an automobile where 𝐹⃗1 is horizontal. Determine the magnitude of 𝐹⃗2 that will provide a vertical resultant force, and determine the magnitude of this resultant. 𝐹2 60◦ 𝐹1 = 400 lb

Figure P2.9

15◦

𝐹2

Problem 2.10

𝐹1 = 1000 N

One of the support brackets for the lawn mowing deck of a garden tractor is shown where 𝐹⃗1 is horizontal. Determine the magnitude of 𝐹⃗2 so that the resultant of these two forces is vertical, and determine the magnitude of this resultant.

Figure P2.10

𝐶

4 km

𝐵

8

4 km

𝐴

km

Problem 2.11 N E

A buoy at point 𝐵 is located 3 km east and 4 km north of boat 𝐴. Boat 𝐶 is located 4 km from the buoy and 8 km from boat 𝐴. Determine the possible position vectors 𝑟⃗𝐴𝐶 that give the position from boat 𝐴 to boat 𝐶. State your answers using polar vector representation.

3 km

Figure P2.11

ISTUDY

Problem 2.12 Arm 𝑂𝐴 of a robot is positioned as shown. Determine the value for angle 𝛼 of arm 𝐴𝐵 so that the distance from point 𝑂 to the actuator at 𝐵 is 650 mm. 𝐵

300 mm 𝛼 𝐴 400 mm 60◦ 𝑂

Figure P2.12

ISTUDY

Section 2.1

45

Basic Concepts

Problems 2.13 through 2.15 ⃗ and report your result using Add the three vectors shown to form a resultant vector 𝑅, polar vector representation. 60 lb

40 lb

6 kN

15 mm 8 mm

4 kN

5 in.

30◦ 45◦

30◦ 8 mm

80 lb (a)

8 kN (b)

Figure P2.13

(a)

40◦

2m

200 lb 4 in. 3 in.

150 lb 100 lb

30◦

(b)

Figure P2.14

A ship is towed through a narrow channel by applying forces to three ropes attached to its bow. Determine the magnitude and orientation 𝜃 of the force 𝐹⃗ so that the resultant force is in the direction of line 𝑎 and the magnitude of 𝐹⃗ is as small as possible.

Problem 2.17

4m 30◦

(a)

(b)

Problem 2.16

3m

Figure P2.15 𝐹

2 kN

𝜃

30◦

𝑎

60◦ 3 kN Figure P2.16

A surveyor needs to plant a marker directly northeast from where she is standing. Because of obstacles, she walks a route in the horizontal plane consisting of 200 m east, followed by 400 m north, followed by 300 m northeast. From this position, she would like to take the shortest-distance route back to the line that is directly northeast of her starting position. What direction should she travel and how far, and what will be her final distance from her starting point?

Problem 2.18 A utility pole supports a bundle of wires that apply the 400 N and 650 N forces shown, and a guy wire applies the force 𝑃⃗ .

400 N 650 N

30◦ 𝛼

(a) If 𝑃 = 0, determine the resultant force applied by the wires to the pole and report your result using polar vector representation.

𝑃

(b) Repeat Part (a) if 𝑃 = 500 N and 𝛼 = 60◦ . (c) With 𝛼 = 60◦ , what value of 𝑃 will produce a resultant force that is vertical? (d) If the resultant force is to be vertical and 𝑃 is to be as small as possible, determine the value 𝛼 should have and the corresponding value of 𝑃 .

Figure P2.18

Problem 2.19 The end of a cantilever I beam supports forces from three cables.

𝑃 𝛼

(a) If 𝑃 = 0, determine the resultant force applied by the two cables to the I beam and report your result using polar vector representation.

𝑎

(b) Repeat Part (a) if 𝑃 = 1.5 kip and 𝛼 = 30◦ . (c) With 𝛼 = 30◦ , what value of 𝑃 will produce a resultant force that is horizontal? (d) If the resultant force is to be horizontal and 𝑃 is to be as small as possible, determine the value 𝛼 should have and the corresponding value of 𝑃 .

2 kip Figure P2.19

60◦ 1 kip

46

Chapter 2

Vectors: Force and Position Problems 2.20 and 2.21

Determine the smallest force 𝐹1 such that the resultant of the three forces 𝐹1 , 𝐹2 , and 𝐹3 is vertical, and the angle 𝛼 at which 𝐹1 should be applied. 𝐹2 = 20 kN

𝐹2 = 200 lb

𝐹1

𝐹1 𝐹3 = 100 lb

𝛼

𝛼 45◦

30◦ 40◦

Figure P2.20

𝐹2 = 200 N

𝐹1 𝛼

Figure P2.21

𝑎

soil

Problem 2.22

𝐹3 = 400 N

Forces 𝐹1 , 𝐹2 , and 𝐹3 are applied to a soil nail to pull it out of a slope. If 𝐹2 and 𝐹3 are vertical and horizontal, respectively, with the magnitudes shown, determine the magnitude of the smallest force 𝐹1 that can be applied and the angle 𝛼 so that the resultant force applied to the nail is directed along the axis of the nail (direction 𝑎).

30◦ Figure P2.22

Problem 2.23

𝑏 2 km 30◦

2 km 𝑏

30◦

30◦ 𝑎

𝑎 (a)

(b)

Determine the magnitudes of vectors 𝐹⃗𝑎 and 𝐹⃗𝑏 in the 𝑎 and 𝑏 directions, respectively, such that their sum is the 100 lb force vector shown.

𝑏 15◦ 100 lb

100 lb 𝑎 30◦

(a) Figure P2.24

Determine the magnitudes of vectors 𝑟⃗𝑎 and 𝑟⃗𝑏 in the 𝑎 and 𝑏 directions, respectively, such that their sum is the 2 km position vector shown.

Problem 2.24

Figure P2.23

ISTUDY

𝐹3 = 30 kN

𝑎

𝑏

30◦ 30◦

45◦ (b)

Problems 2.25 and 2.26 The child’s play structure from Examples 2.2 and 2.3 on pp. 38 and 39 is shown again. The woman at 𝐴 applies a force in the 𝑎 direction and the man at 𝐵 applies a force in the 𝑏 direction, with the goal of producing a resultant force of 250 N in the 𝑐 direction. Determine the forces the two people must apply, expressing the results as vectors. 𝑐

𝑐

𝑏

𝑏

250 N

250 N

𝐴 65◦

65◦

𝑎

50◦

𝐴 𝐵 Figure P2.25

𝐵

𝑎

Figure P2.26

ISTUDY

Section 2.1

Basic Concepts

Problem 2.27

𝑎

While canoes are normally propelled by paddle, if there is a favorable wind from the stern, adventurous users will sometimes employ a small sail. If a canoe is sailing northwest and the wind applies a 40 lb force perpendicular to the sail in the direction shown, determine the components of the wind force parallel and perpendicular to the keel of the canoe (direction 𝑎).

45◦

N

40 lb

Problem 2.28

20◦

Repeat Part (b) of Example 2.5, using the optimization methods of calculus. Hint: Redraw the force polygon of Fig. 3 and rewrite Eq. (1) on p. 41 with the 45◦ angle shown there replaced by 𝛽, where 𝛽 is defined in Fig. P2.28. Rearrange this equation to obtain an expression for 𝐹𝑂𝐶 ′ as a function of 𝛽, and then determine the value of 𝛽 that makes 𝑑𝐹𝑂𝐶 ′ ∕𝑑𝛽 = 0. While the approach described here is straightforward to carry out “by hand,” you might also consider using symbolic algebra software such as Mathematica or Maple.

𝐶 𝐵

𝛽 trees 𝑂 30◦

Figure P2.28

𝐴

Figure P2.27

47

48

Chapter 2

Vectors: Force and Position

2.2

Cartesian Representation of Vectors in Two Dimensions

There are a variety of ways, or representations, that may be used to embed magnitude and direction information in a vector expression. Cartesian vector representation is straightforward and is the most widely used form for expressing vectors. For special classes of problems, such as when geometry is circular or spherical, other forms of vector representation may be more convenient. Some of these representations are discussed in dynamics.

Introduction—Cartesian representation and a walk to work

work N

(a)

1 km

W

E S

2 km

home north

𝑊 , work

1 km 𝑟⃗𝐻𝑊

(b) 𝚥̂

east

0

𝚤̂ 0 𝐻, home

1 km

2 km

Figure 2.11 (a) A map shows the locations of home and work. (b) A Cartesian coordinate system is used to describe positions and vectors.

ISTUDY

Imagine that to get from your home to where you work, you walk 2 km east and 1 km north, as shown on the map of Fig. 2.11(a). This simple statement implies a Cartesian coordinate system, unit vectors, and a position vector that describes the location of where you work relative to your home. Figure 2.11(b) shows these ingredients, with the following explanations. We will use east and north to describe directions (we could just as well use west and/or south, respectively), and since these directions are orthogonal (that is, their intersection forms a right angle), they constitute a Cartesian coordinate system. We may select the origin of this coordinate system to be any point we choose, and once this is done, the location of other points can be quantified using coordinates. For example, if we take the origin of our coordinate system to be our home, as shown in Fig. 2.11(b), then the coordinates of where we work are (2 km, 1 km), where this pair of numbers is ordered such that the first number is always the east coordinate and the second number is always the north coordinate. Note that positions that are south or west of our origin will have some negative coordinates. The directions “east” and “north” define unit vectors. That is, when we say east, we describe only a direction, and no statement of distance is made. We can describe the position vector from home to work by 𝑟⃗𝐻𝑊 = 2 km east + 1 km north, or in more compact notation by 𝑟⃗𝐻𝑊 = 2 km 𝚤̂ + 1 km 𝚥̂, where 𝚤̂ means east and 𝚥̂ means north. The concepts introduced in this example are more thoroughly described in the rest of this section.

Unit vectors The concept of a unit vector is useful. A unit vector is defined to be a dimensionless vector that has unit magnitude. Given any arbitrary vector 𝑣⃗ having nonzero magnitude, we may construct a unit vector 𝑢̂ that has the same direction, using

𝑢̂ =

𝑣⃗ . |𝑣| ⃗

(2.11)

We will use a “hat” symbol (̂) over unit vectors, whereas all other vectors will use an arrow symbol ( ⃗ ). Note that vector 𝑣⃗ and its magnitude |𝑣| ⃗ have the same units, thus unit vector 𝑢̂ is dimensionless. In this book, we will reserve the symbol 𝑢̂ to represent unit vectors, although we will also use the more primitive form 𝑣∕| ⃗ 𝑣|. ⃗ In figures, unit vectors will be orange .

ISTUDY

Section 2.2

49

Cartesian Representation of Vectors in Two Dimensions 𝑦

Cartesian coordinate system

𝚤̂ = 𝚥̂ = 1

You probably have considerable experience already using Cartesian coordinate systems, but to be complete, we will state exactly what ingredients are needed. In two dimensions a Cartesian coordinate system uses two orthogonal reference directions, which we will usually call the 𝑥 and 𝑦 directions, as shown in Fig. 2.12(a). The intersection of the reference directions is the origin of the coordinate system, and any point in the 𝑥𝑦 plane is identified by its 𝑥 and 𝑦 coordinates. We denote the coordinates of a point 𝑃 using an ordered pair of numbers (𝑥𝑃 , 𝑦𝑃 ) where 𝑥𝑃 and 𝑦𝑃 are the 𝑥 and 𝑦 coordinates of the point, respectively, measured from the origin. Although we will often take the 𝑥 and 𝑦 directions to be horizontal and vertical, respectively, other orientations, provided they are orthogonal, are acceptable and may be more convenient.

𝚥̂

𝑣⃗

𝜃

𝑥

𝚤̂ (a) 𝑦 𝑣⃗ ( 𝑣⃗ sin 𝜃) 𝚥̂ = 𝑣⃗𝑦

𝜃

𝑥

( 𝑣⃗ cos 𝜃) 𝚤̂ = 𝑣⃗𝑥 (b)

Cartesian vector representation We begin by defining unit vectors 𝚤̂ and 𝚥̂ that point in the positive 𝑥 and 𝑦 directions, respectively, as shown in Fig. 2.12(a). Both 𝚤̂ and 𝚥̂ are dimensionless and are used to describe direction in exactly the same way you would use east and north. Now consider a vector 𝑣⃗ in the 𝑥𝑦 plane. Following the ideas used in Section 2.1, we ask for two vectors, one parallel to the 𝑥 axis and one parallel to the 𝑦 axis, such that their vector sum yields 𝑣. ⃗ This can be stated in equation form as 𝑣⃗ = 𝑣⃗𝑥 + 𝑣⃗𝑦 ,

(2.12)

where 𝑣⃗𝑥 and 𝑣⃗𝑦 are vectors parallel to the 𝑥 and 𝑦 directions, respectively, as shown ⃗ Noting that in Fig. 2.12(b). Vectors 𝑣⃗𝑥 and 𝑣⃗𝑦 are called the vector components of 𝑣. vectors 𝑣⃗𝑥 and 𝚤̂ are parallel (although possibly with opposite direction), we may write 𝑣⃗𝑥 = 𝑣𝑥 𝚤̂, and similarly, we may write 𝑣⃗𝑦 = 𝑣𝑦 𝚥̂. Thus, Eq. (2.12) becomes 𝑣⃗ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂.

(2.13)

An expression in the form of Eq. (2.13) is called a Cartesian representation of the vector 𝑣. ⃗ The scalars 𝑣𝑥 and 𝑣𝑦 are called the scalar components (or simply the components) of 𝑣, ⃗ and these may be positive, zero, or negative. Because the vector components 𝑣⃗𝑥 and 𝑣⃗𝑦 are always orthogonal, basic trigonometric relations and the Pythagorean theorem, Eqs. (2.9) and (2.10), are all that is needed to resolve a vector into its Cartesian components. Magnitude and orientation of a vector. For a vector in two dimensions, the magnitude |𝑣| ⃗ and orientation 𝜃 from the horizontal (±𝑥 direction) are related to its Cartesian components by |𝑣| ⃗ =

√ 𝑣2𝑥 + 𝑣2𝑦

and

𝜃 = tan−1

(𝑣 ) 𝑦 . 𝑣𝑥

(2.14)

The magnitude of a vector is a scalar that is positive, or is zero in the case of a zero vector (i.e., a vector having zero values for all its components). As discussed in Fig. 2.4, rather than using the magnitude, we will often represent a vector’s “size” by using a scalar symbol such as 𝑣, where 𝑣 is the component of the vector in an assumed direction for the vector. Because 𝑣 is a component, it can be positive, zero, or negative, as follows. If 𝑣 is positive or zero, then 𝑣 equals the magnitude of the

Figure 2.12 (a) Cartesian coordinate system with unit vectors 𝚤̂ and 𝚥̂ in the 𝑥 and 𝑦 directions, respectively. (b) Resolution of a vector 𝑣⃗ into vector components 𝑣⃗𝑥 and 𝑣⃗𝑦 in the 𝑥 and 𝑦 directions, respectively.

50

Chapter 2

Vectors: Force and Position

vector, and the direction of the vector is the same as the direction we assumed it had. If 𝑣 is negative, then −𝑣 is equal to the magnitude of the vector, and the direction of the vector is opposite the direction we assumed it had. Thus, under all circumstances, the absolute value of 𝑣 gives the magnitude of the vector. The main benefit of this notation is that in many problems, we will know the line of action for a vector, but at the outset of a problem we may not know its direction along that line of action. Thus, for a vector 𝑣⃗ we will assume a direction along its line of action and will often find after our analysis is complete that 𝑣 < 0, which means that in fact the vector’s direction is opposite to the direction we assumed it had. Note that 𝑣 and |𝑣| ⃗ are defined differently, although they both measure size and, under the circumstances described here, yield the same value. Throughout this book, we will be careful to call 𝑣 the magnitude of a vector only when we are sure it is nonnegative. Remark. The angle 𝜃 defined in Eq. (2.14) gives the orientation of a vector measured from the positive 𝑥 direction if the vector’s 𝑥 component is positive (i.e., 𝑣𝑥 > 0), and gives the orientation measured from the negative 𝑥 direction if the vector’s 𝑥 component is negative (i.e., 𝑣𝑥 < 0). If 𝑣𝑥 = 0, then 𝜃 = ±90◦ . Mini-Example Express the vector shown in Fig. 2.13(a), using Cartesian representation.

𝑦 120 N 30◦

𝑥

(a) 𝑦 120 N (120 N) sin 30◦ = +𝐹𝑦

30◦

𝑥 (120 N) cos 30◦ = +𝐹𝑥 (b)

Figure 2.13 Determination of the Cartesian components of a 120 N force vector.

ISTUDY

Solution Let 𝐹⃗ denote the 120 N magnitude force vector shown in Fig. 2.13(a) and let 𝐹𝑥 and 𝐹𝑦 denote the components of 𝐹⃗ in the 𝑥 and 𝑦 directions, respectively. Components 𝐹 and 𝐹 are obtained by constructing projections of 𝐹⃗ that are parallel to 𝑥

𝑦

the 𝑥 and 𝑦 axes, respectively, with the resulting magnitudes (120 N) cos 30◦ and (120 N) sin 30◦ , as shown in Fig. 2.13(b). We next use these projections to assign vector components in the 𝑥 and 𝑦 directions, being careful to note that 𝐹𝑥 and 𝐹𝑦 are positive when acting in the positive 𝑥 and 𝑦 directions, respectively, and are negative when acting in the negative 𝑥 and 𝑦 directions. Thus, 𝐹𝑥 = (120 N) cos 30◦ = 103.9 N, 𝐹𝑦 = (120 N) sin 30◦ = 60 N.

(2.15)

Thus, the Cartesian representation for 𝐹⃗ is 𝐹⃗ = (103.9 𝚤̂ + 60 𝚥̂) N.

(2.16)

Some useful checks. There are a few useful checks that may help you avoid errors when resolving a vector into Cartesian components. Foremost is to verify that the components have reasonable size and are in proper directions. The next check is to verify that the vector’s components give the correct magnitude. Thus, √ for Eq. (2.16), we evaluate (103.9 N)2 + (60 N)2 = 120 N to find that indeed it has the correct magnitude. While these checks are reassuring, they do not guarantee the components are correct. Nonetheless, if we find an incorrect magnitude, then certainly an error has been made.

ISTUDY

Section 2.2

Cartesian Representation of Vectors in Two Dimensions

51

𝑦 12 m

Mini-Example Express the vector shown in Fig. 2.14(a), using Cartesian representation.

4

Solution Geometric information for the 12 m position vector shown in Fig. 2.14(a) is in the form of a 3–4–5 triangle. While it is possible to evaluate the angle 𝜃 and then proceed as in the previous mini-example, this is not necessary, and the geometry of the 3–4–5 triangle (or any other right triangle where the edge lengths are known) can be exploited to directly obtain the vector’s components as follows. From Fig. 2.14(a), the vector’s orientation is such that a “run” of 3 units is always accompanied by a “rise” of 4 units. If we imagine traversing 3 units√of horizontal motion followed by 4 units of vertical motion, the total motion is 32 + 42 = 5 units. Of this 5 units total, 3 parts have been horizontal in the negative 𝑥 direction, and 4 parts have been vertical in the positive 𝑦 direction, and this constitutes a statement of the vector’s components, as shown in Fig. 2.14(b). Hence, we may write ) ( 4 −3 𝚤̂ + 𝚥̂ . (2.17) 𝑟⃗ = (12 m) 5 5 The negative sign in Eq. (2.17) and Fig. 2.14(b) has been included because the horizontal component of the vector is in the negative 𝑥 direction. Note that the vector within the parentheses on the right-hand side of Eq. (2.17) is a unit vector, and hence, 𝑟⃗ has 12 m magnitude, as expected.

3 𝜃

𝑥 (a) 𝑦

12 m (12 m)(4∕5) = +𝑟𝑦

4 3 𝜃

𝑥

(12 m)(3∕5) = −𝑟𝑥 (b) Figure 2.14 Determination of the Cartesian components of a 20 m position vector.

Remarks. In this example, the geometry was especially nice (i.e., 3–4–5 triangle). Nonetheless, even if the rise, run, and/or hypotenuse is noninteger, this method of determining the Cartesian components of a vector is still very straightforward. If you find the heuristic argument used to construct Eq. (2.17) unconvincing, then we can use a more formal construction to show its validity. With 𝜃 as defined in Fig. 2.14(a), we write 𝑟⃗ = −(12 m) (cos 𝜃) 𝚤̂ + (12 m) (sin 𝜃) 𝚥̂ = (12 m) (− cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂).

Next using the geometry for the vector’s orientation (3–4–5 triangle), we note cos 𝜃 = 3∕5 and sin 𝜃 = 4∕5, and combining these with Eq. (2.18) provides the same result as Eq. (2.17).

𝑣⃗2

𝑦

(2.18)

𝜃2

𝑣⃗1

(a) 𝜃1 𝚥̂

Consider the addition of the two vectors shown in Fig. 2.15(a). We first write the vectors using Cartesian representation. Thus, 𝑣⃗1 = 𝑣1𝑥 𝚤̂ + 𝑣1𝑦 𝚥̂, 𝑣⃗2 = 𝑣2𝑥 𝚤̂ + 𝑣2𝑦 𝚥̂.

(2.19)

The resultant vector 𝑅⃗ is the sum of 𝑣⃗1 and 𝑣⃗2 , hence,

= (𝑣1𝑥 + 𝑣2𝑥 ) 𝚤̂ + (𝑣1𝑦 + 𝑣2𝑦 ) 𝚥̂.

𝑦 𝑅⃗ = 𝑣⃗1 + 𝑣⃗2 (b) 𝜃1 𝑣⃗1 𝑣⃗1 cos 𝜃1 = 𝑣1𝑥

𝚥̂ 𝚤̂

𝑅⃗ = 𝑣⃗1 + 𝑣⃗2 = (𝑣1𝑥 𝚤̂ + 𝑣1𝑦 𝚥̂) + (𝑣2𝑥 𝚤̂ + 𝑣2𝑦 𝚥̂)

𝑥

𝚤̂

Addition of vectors using Cartesian components

(2.20)

𝜃2

𝑣⃗2

𝑣⃗2 sin 𝜃2 = 𝑣2𝑦

𝑣⃗2 cos 𝜃2 = 𝑣2𝑥 𝑣⃗1 sin 𝜃1 = 𝑣1𝑦 𝑥

Figure 2.15 (a) Two vectors 𝑣⃗1 and 𝑣⃗2 are to be added. (b) The ⃗ result of the vector addition is 𝑅.

52

Chapter 2

Vectors: Force and Position

The addition is illustrated graphically in Fig. 2.15(b), where we slide the vectors 𝑣⃗1 ⃗ Notice the 𝑥 and 𝑦 components of and 𝑣⃗2 head to tail to yield a resultant vector 𝑅. 𝑅⃗ are, respectively, the sums of the 𝑥 and 𝑦 components of the vectors being added. To complete the addition, we use the magnitudes |𝑣⃗1 | and |𝑣⃗2 | and orientations 𝜃1 and 𝜃2 provided in Fig. 2.15(b) to determine the components 𝑣1𝑥 , 𝑣1𝑦 , 𝑣2𝑥 , and 𝑣2𝑦 . Addition of an arbitrary number 𝑛 of vectors is similarly straightforward, and the last of Eqs. (2.20) generalizes to ( 𝑅⃗ = 𝑣⃗1 + 𝑣⃗2 + ⋯ + 𝑣⃗𝑛 =

𝑛 ∑ 𝑖=1

) 𝑣𝑖𝑥

( 𝚤̂ +

𝑛 ∑

) 𝑣𝑖𝑦

𝚥̂.

(2.21)

𝑖=1

In words, Eq. (2.21) states that the 𝑥 component of the resultant is the sum of the 𝑥 components of all vectors being added, and the 𝑦 component of the resultant is the sum of the 𝑦 components of all vectors being added.

Position vectors 𝑦

head, 𝐻(𝑥𝐻 , 𝑦𝐻 ) 𝑟⃗𝑇 𝐻

𝑦𝐻 − 𝑦𝑇

tail, 𝑇 (𝑥𝑇 , 𝑦𝑇 ) 𝑥 𝐻 − 𝑥𝑇

𝚥̂ 𝚤̂

𝑥

Figure 2.16 Construction of a position vector using Cartesian coordinates of the vector’s head and tail.

ISTUDY

The spatial position of one point (head of vector) relative to another point (tail of vector) is provided by a position vector. For the example of Fig. 2.16, the two points 𝑇 and 𝐻 denote the tail and head of a position vector and have coordinates (𝑥𝑇 , 𝑦𝑇 ) and (𝑥𝐻 , 𝑦𝐻 ), respectively. The position vector from 𝑇 to 𝐻 is denoted by 𝑟⃗𝑇 𝐻 , and it has the 𝑥 and 𝑦 components shown in Fig. 2.16; hence we may write 𝑟⃗𝑇 𝐻 = (𝑥𝐻 − 𝑥𝑇 ) 𝚤̂ + (𝑦𝐻 − 𝑦𝑇 ) 𝚥̂.

(2.22)

In words, Eq. (2.22) states that the components of the position vector are given by “coordinates of the head minus coordinates of the tail,” and this phrase is worth committing to memory. Another useful method for constructing a position vector is to imagine being positioned at point 𝑇 and then asking what distances in the 𝑥 and 𝑦 directions must be traversed to arrive at point 𝐻; this also leads to Eq. (2.22). What does a position vector really tell us? A position vector provides information on relative position only. That is, it tells us where the head of a vector is located relative to its tail. It does not tell us the absolute position of the head and tail. That is, it does not tell us where the head and tail are located. For example, walking 1 mile west and 2 miles north is a statement of a position vector. With this position vector, we know only where our traveling ends relative to where it started, and exactly where our initial or final positions are located is information that is not contained in the position vector. On the other hand, if we state that we begin our traveling at Willis Tower in Chicago (formerly named Sears Tower), and we walk 1 mile west and 2 miles north, then the position vector together with the coordinates of the starting point provides the absolute position of our final location.

ISTUDY

Section 2.2

Cartesian Representation of Vectors in Two Dimensions

End of Section Summary In this section, Cartesian coordinate systems and Cartesian representation for vectors in two dimensions have been defined. Some of the key points are: • The 𝑥 and 𝑦 directions you choose for a Cartesian coordinate system are arbitrary, but these directions must be orthogonal. • The Cartesian representation for a vector 𝑣⃗ is 𝑣⃗ = 𝑣⃗𝑥 + 𝑣⃗𝑦 = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂. In these ⃗ and 𝑣𝑥 and 𝑣𝑦 are expressions, 𝑣⃗𝑥 and 𝑣⃗𝑦 are called the vector components of 𝑣, ⃗ called the scalar components (or simply the components) of 𝑣. • Addition of vectors with Cartesian representation is accomplished by simply summing the components in the 𝑥 direction of all vectors being added to obtain the 𝑥 component of the resultant, and then summing the components in the 𝑦 direction of all vectors being added to obtain the 𝑦 component of the resultant. Addition of even a large number of vectors is straightforward. • A position vector provides the location of one point relative to another point. Given the coordinates of the two endpoints of the vector, Eq. (2.22) can be used to obtain the position vector. An easy way to remember Eq. (2.22) is that the components of the position vector are given by “coordinates of the head minus coordinates of the tail.” Another useful method for constructing a position vector is to imagine being positioned at the tail of a vector and then asking what distances in the 𝑥 and 𝑦 directions must be traversed to arrive at the head of the vector.

53

54

Chapter 2

Vectors: Force and Position

E X A M P L E 2.6

Addition of Vectors and Working Loads A short post 𝐴𝐵 has a commercially manufactured eyebolt screwed into its end. Three cables attached to the eyebolt apply the forces shown.

𝑦 𝐹2 = 500 lb

𝐹1 = 200 lb

(a) Determine the resultant force applied to the eyebolt by the three cables, using a Cartesian vector representation.

3 1 𝐶 𝐹3 = 800 lb

60◦ 𝑥 𝐵

70◦

50◦

𝐴

(b) The manufacturer of the eyebolt specifies a maximum working load∗ of 2100 lb in the direction of the eyebolt’s axis. When loads are not in the direction of the eyebolt’s axis, the manufacturer specifies reduction of the working load using the multipliers given in Fig. 1. Determine if this size eyebolt is satisfactory.

SOLUTION Road Map

𝜃 0◦ 15◦ 30◦ 45◦ Figure 1

ISTUDY

working load multipliers

𝜃 ≤𝜃 ≤𝜃 ≤𝜃 ≤𝜃

< 15◦ < 30◦ < 45◦ ≤ 90◦

100% 60% 33% 20%

The resultant force vector is the sum of the three force vectors applied to the eyebolt. First we will express each of the three force vectors using Cartesian representation, and then we will add these to obtain the resultant force. For Part (b), we will compare this resultant force to the working load criteria for the eyebolt to determine if it is acceptable. Part (a) Governing Equations & Computation

We first express the three forces using

Cartesian representations as 𝐹⃗1 = (200 lb)(cos 60◦ 𝚤̂ + sin 60◦ 𝚥̂), √ √ 𝐹⃗2 = (500 lb)(−1∕ 10 𝚤̂ + 3∕ 10 𝚥̂),

(2)

𝐹⃗3 = (800 lb)(− sin 70◦ 𝚤̂ − cos 70◦ 𝚥̂).

(3)

(1)

The construction of 𝐹⃗2 and 𝐹⃗3 in Eqs. (2) and (3) is accomplished as described in earlier examples, with the only point of notice being that the 70◦ orientation of 𝐹⃗3 is measured from the vertical, which leads to sine and cosine functions providing the 𝑥 and 𝑦 components, respectively. While 𝐹⃗1 can be constructed in an identical manner to 𝐹⃗3 , the following slightly different construction is useful. We first write 𝐹⃗1 in the form 𝐹⃗1 = (200 lb) 𝑢̂ 1 where 𝑢̂ 1 is a unit vector in the direction of 𝐹⃗1 . We next note that 𝑢̂ 1 is described by taking a step of cos 60◦ in the positive 𝑥 direction followed by a step of sin 60◦ in the positive 𝑦 direction. Hence 𝑢̂ 1 = cos 60◦ 𝚤̂ + sin 60◦ 𝚥̂ and, because (cos 𝜃)2 + (sin 𝜃)2 = 1 for any angle 𝜃, 𝑢̂ 1 is clearly a unit vector and the expression for 𝐹⃗1 in Eq. (1) follows. Adding the force vectors yields the resultant force 𝑅⃗ as 𝑅⃗ = 𝐹⃗1 + 𝐹⃗2 + 𝐹⃗3 ] [ √ = 200 cos 60◦ + 500(−1∕ 10) + 800(− sin 70◦ ) lb 𝚤̂ ] [ √ + 200 sin 60◦ + 500(3∕ 10) + 800(− cos 70◦ ) lb 𝚥̂ = (−809.9 𝚤̂ + 373.9 𝚥̂) lb. ∗ Working

(4)

loads (also called allowable loads) are loads a certain piece of hardware is designed to be subjected to on a repeated basis. These loads are obtained by reducing the actual failure strength by an appropriate factor, called the factor of safety. For the particular model of eyebolt cited here, the actual breaking strength is about 9800 lb for loads in the direction of the bolt’s axis. While a working load of 2100 lb provides a reasonable margin of safety, the margin is not quite as generous as you might believe because of fatigue: materials and structures that must withstand many load cycles can be subjected to only a fraction of their one-time load strength. While working load specifications for eyebolts always have a margin of safety, it is dangerous and irresponsible to use any structural component in an application where the manufacturer’s working loads are exceeded.

ISTUDY

Section 2.2

55

Cartesian Representation of Vectors in Two Dimensions

As an alternative to reporting our answer using Cartesian vector representation, we may use Eq. (2.14) on p. 49 to report the magnitude and orientation of the resultant force as √ 𝑅 = (−809.9 lb)2 + (373.9 lb)2 = 892.0 lb, and (5) ) ( 373.9 lb 𝜃 = tan−1 (6) = −24.78◦ , −809.9 lb

𝑦

𝑅 = 892.0 lb 24.78◦

𝑥 𝐵

and this force is shown in Fig. 2. Part (b) Governing Equations & Computation

To determine if the eyebolt is adequate to support the 892.0 lb resultant force, the orientation of 𝑅⃗ with respect to the bolt’s axis is needed. Defining a new Cartesian coordinate system 𝑡𝑛 as shown in Fig. 3, where 𝑛 is in the direction 𝐴𝐵 and 𝑡 is perpendicular to direction 𝐴𝐵, we find angle 𝜃 = 50◦ − 24.78◦ = 25.22◦ . Consulting Fig. 1, we find the eyebolt’s load multiplier for loading angles up to 30◦ is 60%, which yields a net working load of (2100 lb)(60%) = 1260 lb. This working load is larger than our resultant force 𝑅 = 892.0 lb, and thus we conclude that this size eyebolt is satisfactory for supporting the system of three forces considered here.

𝐴

50◦

Figure 2 Resultant 𝑅⃗ of forces 𝐹⃗1 , 𝐹⃗2 , and 𝐹⃗3 . 𝑦 𝑛 𝑡 𝑅 = 892.0 lb

𝜃=

25.22◦

Discussion & Verification

Our analysis thus far indicates the eyebolt is acceptable when all three forces 𝐹⃗1 , 𝐹⃗2 , and 𝐹⃗3 are present simultaneously. However, we should also consider other possible loading scenarios in which forces are applied sequentially, and in various orders, to make sure that at no time are the manufacturer’s working load specifications exceeded. In addition, the manufacturer’s specifications refer specifically to an eyebolt subjected to just a single force. We have applied these criteria to the resultant of three forces, which is reasonable, provided good judgment is used. To clarify further, consider an eyebolt subjected to two equal forces with opposite direction: 𝐹⃗1 = 𝑃 𝚤̂ and 𝐹⃗2 = −𝑃 𝚤̂. The resultant force is zero, regardless of how large 𝑃 is, and thus the manufacturer’s working load specifications are always met. However, if 𝑃 is too large, the eyebolt will clearly fail. Thus, we should consult the manufacturer to see if additional criteria are available for multiple loads, and if these are not, then we must perform more detailed analysis of the strength properties of the eyebolt to ensure that it is acceptable for our application.

SergeyMarina/Shutterstock

Figure 4. An example of a commercially manufactured eyebolt, made of forged steel.

40◦

24.78◦ 50◦

𝑥

𝐵

50◦

𝐴

Figure 3 Orientation of resultant force 𝑅⃗ with respect to 𝑡 and 𝑛 directions.

56

E X A M P L E 2.7

Resolution of a Vector into Components For the structure of Example 2.6, we eventually need to determine if post 𝐴𝐵 is strong enough to support the forces applied to it. Such problems are discussed in later chapters of this book where it is necessary to resolve the forces applied to the post into components parallel and perpendicular to the post’s axis. Thus, resolve the resultant of forces 𝐹⃗1 , 𝐹⃗2 , and 𝐹⃗3 into components parallel and perpendicular to the post’s axis 𝐴𝐵.

𝑦 𝐹2 = 500 lb

𝐹1 = 200 lb

3 1 𝐶 𝐹3 = 800 lb

70◦

60◦ 𝑥

SOLUTION Road Map

𝐵

50◦

𝐴

Figure 1 Post and loading of Example 2.6 repeated.

ISTUDY

Chapter 2

Vectors: Force and Position

Several strategies may be used, and we will consider two solutions here. Our first solution will simply repeat the procedure of Example 2.6, except with vectors written in terms of a new 𝑡𝑛 coordinate system where 𝑡 and 𝑛 are perpendicular and parallel to the post’s axis, respectively. The second solution will use transformation of the results of Example 2.6. Solution 1 Governing Equations & Computation Using the geometry shown in Fig. 1, we first

compute various angles between the 𝑥 and 𝑦 axes and the 𝑡 and 𝑛 axes, as shown in Fig. 2(a). We then determine angles for the three forces with respect to the 𝑡 and 𝑛 directions with the results shown in Fig. 2(b). Computation of the 20◦ angles for 𝐹1 and 𝐹3 𝑦

𝑦 𝑛

𝐹2 = 500 lb 𝑛

𝑡

𝑡 20◦

40◦ 50◦ 40◦ ◦ 50◦ 40

21.57◦

50◦

𝑥

𝑥

𝐹3 = 800 lb 20◦

50◦ 40◦ 50◦

𝐹1 = 200 lb

𝐶

𝐴

𝐵

50◦

𝐴

(b)

(a)

Figure 2. Forces of Fig. 1 redrawn with orientations with respect to 𝑡 and 𝑛 directions.

are straightforward. For the orientation of 𝐹⃗2 , we first evaluate tan−1 (3∕1) = 71.57◦ to obtain the orientation of 𝐹⃗2 with respect to the negative 𝑥 direction, followed by evaluation of 71.57◦ − 50◦ = 21.57◦ to obtain the orientation of 𝐹⃗2 with respect to the 𝑛 direction. We then write the force vectors using Cartesian representation as ̂ 𝐹⃗1 = (200 lb)(cos 20◦ 𝑡̂ + sin 20◦ 𝑛),

(1)

̂ 𝐹⃗2 = (500 lb)(sin 21.57◦ 𝑡̂ + cos 21.57◦ 𝑛),

(2)

̂ 𝐹⃗3 = (800 lb)(− cos 20◦ 𝑡̂ + sin 20◦ 𝑛),

(3)

where 𝑡̂ and 𝑛̂ are defined to be unit vectors in the 𝑡 and 𝑛 directions, respectively. The resultant force is then 𝑅⃗ = 𝐹⃗1 + 𝐹⃗2 + 𝐹⃗3 [ ] = 200 cos 20◦ + 500 sin 21.57◦ + 800(− cos 20◦ ) 𝑡̂ lb [ ] + 200 sin 20◦ + 500 cos 21.57◦ + 800 sin 20◦ 𝑛̂ lb ̂ lb. = (−380.0 𝑡̂ + 807.0 𝑛)

(4)

ISTUDY

Section 2.2

57

Cartesian Representation of Vectors in Two Dimensions

Thus, the resultant force has the 𝑡 and 𝑛 components 𝑅𝑡 = −380.0 lb and 𝑅𝑛 = 807.0 lb, respectively, which √ are shown in Fig. 3. As a partial check on our solution, we evaluate the magnitude 𝑅 = (−380.0 lb)2 + (807.0 lb)2 = 892.0 lb, which agrees with the results found in Example 2.6. Solution 2 Governing Equations & Computation This solution takes advantage of the results of

Example 2.6 by resolving 𝑅⃗ in Eq. (4) on p. 54 into 𝑡 and 𝑛 components. Referring to Fig. 3 on p. 55, angle 𝜃 was easily found to be 𝜃 = 50◦ − 24.78◦ = 25.22◦ . Thus, the components of 𝑅⃗ in the 𝑡 and 𝑛 directions are 𝑅𝑡 = −(892.0 lb)(sin 25.22◦ ) = −380.0 lb,

(5)

𝑅𝑛 = (892.0 lb)(cos 25.22◦ ) = 807.0 lb,

(6)

which agrees with the results reported in Eq. (4) and shown in Fig. 3. Discussion & Verification

• When you are resolving a vector into orthogonal components, such as 𝑅𝑡 and 𝑅𝑛 in this √ example, a useful and quick check of accuracy is to evaluate 𝑅2𝑡 + 𝑅2𝑛 and verify that it agrees with the magnitude of the original vector.

• A disadvantage of Solution 1 is that if the resultant force must be determined for several different coordinate systems, the work for each of these is fully repeated. The advantage of Solution 2 is that once the resultant force is known for one coordinate system, it is straightforward to transform the result to other coordinate systems. • A third solution that is not presented here is discussed in Prob. 2.55. This approach is similar to Solution 2, but is more elegant and effective. It uses vector transformation to take the 𝑥 and 𝑦 components of 𝑅⃗ found in Example 2.6 and transform these into 𝑡 and 𝑛 components.

𝑦 𝑛 𝑡 380.0 lb 40◦

𝑥

𝐶 𝐵 807.0 lb 50◦

𝐴

Figure 3 Components of the resultant force in 𝑡 and 𝑛 directions.

58

Chapter 2

Vectors: Force and Position

E X A M P L E 2.8

Position Vectors

Lake Superior 𝐵

N

10 km

𝑃

𝚥̂ 𝚤̂

E

𝑟⃗𝑂𝐴 = (44 𝚤̂ − 9 𝚥̂) km, 𝑂

𝑟⃗𝑂𝑃 = (61 𝚤̂ + 31 𝚥̂) km,

𝑟⃗𝑂𝐵 = (−12 𝚤̂ + 53 𝚥̂) km.

(1)

(a) If the rescue craft is a helicopter, determine the shortest distance it must fly from 𝐴 to reach the boat at 𝐵.

𝐴

Keweenaw Peninsula

(b) If the rescue craft is a ship, determine the shortest distance it must cruise from 𝐴 to reach the boat at 𝐵.

Figure 1

(c) In Part (b), if the ship cruises 20 km from 𝐴 toward 𝑃 , determine the ship’s coordinates and position vector relative to the base at 𝑂.

SOLUTION

𝐵

𝐵 𝑟⃗𝐴𝐵

𝑟⃗𝑂𝐵 𝑂

𝑟⃗𝐴𝑂 = −⃗𝑟𝑂𝐴

𝑟⃗𝑃 𝐵

𝑟⃗𝑂𝐵

Road Map 𝑃

𝑟⃗𝑃 𝑂 𝑟⃗𝑂𝑃

𝐴

𝑂

𝑟⃗𝐴𝑂

𝑟⃗𝐴𝑃 𝐴

A strategy that could be used for all three parts of this example is to first determine the coordinates of points 𝐴, 𝐵, and 𝑃 and then use Eq. (2.22) on p. 52 to write position vectors 𝑟⃗𝐴𝐵 and so on, where the magnitudes of these vectors give the distances asked for in each question. A better strategy is to use vector addition to obtain the desired position vectors, and this is the approach we will follow. Part (a)

(b)

(a)

Governing Equations & Computation

The shortest route for the helicopter is the straight line from 𝐴 to 𝐵. Thus we seek vector 𝑟⃗𝐴𝐵 , which may be written as

𝑃 𝐶

𝑟⃗𝐴𝐶 = (20 km)

𝑂 𝐴

𝑟⃗𝐴𝑃 = 𝑟⃗𝐴𝑃

𝑟⃗𝐴𝑃 𝑟⃗𝐴𝑃

unit vector

(c) Figure 2 Addition of vectors. Figure (a) is used to obtain the answer to Part (a) of this example, wherein vectors 𝑟⃗𝐴𝑂 and 𝑟⃗𝑂𝐵 are added to obtain 𝑟⃗𝐴𝐵 . Figure (b) shows various vectors that are added to obtain the answer to Part (b). Figure (c) is used to obtain the answer to Part (c), wherein a position vector 𝑟⃗𝐴𝐶 is constructed using the unit vector 𝑟⃗𝐴𝑃 ∕|⃗𝑟𝐴𝑃 |.

ISTUDY

The Keweenaw Peninsula on the south shore of Lake Superior is shown. Point 𝑂 is the location of a U.S. Coast Guard base, point 𝐴 is the location of a Coast Guard rescue craft (either a helicopter or a ship), point 𝐵 is the location of a boat in need of help, and point 𝑃 is the tip of the Keweenaw Peninsula. Letting 𝚤̂ and 𝚥̂ be unit vectors in the east and north directions, respectively, the position vectors are

𝑟⃗𝐴𝐵 = 𝑟⃗𝐴𝑂 + 𝑟⃗𝑂𝐵 .

(2)

A useful way to think of Eq. (2) is shown in Fig. 2(a). Rather than travel directly from 𝐴 to 𝐵, we may first travel from 𝐴 to 𝑂, followed by travel from 𝑂 to 𝐵. Of course, this is not the path to be taken by the helicopter; but if this path were followed, the same final position would result. The merit in writing Eq. (2) is that both vectors on the right-hand side are known: 𝑟⃗𝑂𝐵 is specified in Eq. (1), and 𝑟⃗𝐴𝑂 = −⃗𝑟𝑂𝐴 where 𝑟⃗𝑂𝐴 is also specified in Eq. (1). Thus, Eq. (2) becomes 𝑟⃗𝐴𝐵 = −⃗𝑟𝑂𝐴 + 𝑟⃗𝑂𝐵 = −(44 𝚤̂ − 9 𝚥̂) km + (−12 𝚤̂ + 53 𝚥̂) km = (−56 𝚤̂ + 62 𝚥̂) km,

(3)

and the distance traveled is the magnitude 𝑟𝐴𝐵 =

√

(−56 km)2 + (62 km)2 = 83.55 km.

(4)

Part (b) Governing Equations & Computation

The shortest route for a rescue ship consists of the straight-line distances from 𝐴 to 𝑃 and from 𝑃 to 𝐵. Thus we seek vectors 𝑟⃗𝐴𝑃 and 𝑟⃗𝑃 𝐵 . Using the addition shown in Fig. 2(b), we write 𝑟⃗𝐴𝑃 = 𝑟⃗𝐴𝑂 + 𝑟⃗𝑂𝑃 , 𝑟⃗𝑃 𝐵 = 𝑟⃗𝑃 𝑂 + 𝑟⃗𝑂𝐵 .

(5) (6)

ISTUDY

Section 2.2

Cartesian Representation of Vectors in Two Dimensions

Vectors 𝑟⃗𝑂𝑃 and 𝑟⃗𝑂𝐵 are specified in the problem statement, 𝑟⃗𝐴𝑂 = −⃗𝑟𝑂𝐴 , and 𝑟⃗𝑃 𝑂 = −⃗𝑟𝑂𝑃 . Thus, 𝑟⃗𝐴𝑃 = −⃗𝑟𝑂𝐴 + 𝑟⃗𝑂𝑃 = −(44 𝚤̂ − 9 𝚥̂) km + (61 𝚤̂ + 31 𝚥̂) km = (17 𝚤̂ + 40 𝚥̂) km,

(7)

𝑟⃗𝑃 𝐵 = −⃗𝑟𝑂𝑃 + 𝑟⃗𝑂𝐵 = −(61 𝚤̂ + 31 𝚥̂) km + (−12 𝚤̂ + 53 𝚥̂) km = (−73 𝚤̂ + 22 𝚥̂) km.

(8)

The magnitudes of the vectors in Eqs. (7) and (8) are √ 𝑟𝐴𝑃 = (17 km)2 + (40 km)2 = 43.46 km, √ 𝑟𝑃 𝐵 = (−73 km)2 + (22 km)2 = 76.24 km.

(9) (10)

Thus, the total distance cruised by the rescue ship is 𝑟𝐴𝑃 + 𝑟𝑃 𝐵 = 119.7 km. As a check on the accuracy of our solution, you may want to verify that 𝑟⃗𝐴𝑃 + 𝑟⃗𝑃 𝐵 = 𝑟⃗𝐴𝐵 . Part (c) Governing Equations & Computation Let point 𝐶 shown in Fig. 2(c) denote the position of the rescue ship after it cruises 20 km from 𝐴 toward 𝑃 . The position vector from 𝐴 to 𝐶 is 𝑟⃗𝐴𝐶 , and it is collinear with 𝑟⃗𝐴𝑃 and can be constructed as shown in Fig. 2(c) as follows:

𝑟⃗𝐴𝐶 = (20 km)

𝑟⃗𝐴𝑃 𝑟𝐴𝑃

= (20 km) √

(17 𝚤̂ + 40 𝚥̂) km

(11)

(17 km)2 + (40 km)2

= (7.823 𝚤̂ + 18.41 𝚥̂) km.

(12)

Note that 𝑟⃗𝐴𝑃 ∕𝑟𝐴𝑃 in Eq. (11) is a unit vector that provides the proper direction for the 20 km distance the ship cruises. To find the position vector 𝑟⃗𝑂𝐶 , we use vector addition to write 𝑟⃗𝑂𝐶 = 𝑟⃗𝑂𝐴 + 𝑟⃗𝐴𝐶 . (13) Vectors on the right-hand side are known, and hence, Eq. (13) provides 𝑟⃗𝑂𝐶 = (44 𝚤̂ − 9 𝚥̂) km + (7.823 𝚤̂ + 18.41 𝚥̂) km = (51.82 𝚤̂ + 9.407 𝚥̂) km.

(14)

Finally, to obtain the coordinates of 𝐶, we use Eq. (2.22) to write 𝑟⃗𝑂𝐶 = (𝑥𝐶 − 𝑥𝑂 ) 𝚤̂ + (𝑦𝐶 − 𝑦𝑂 ) 𝚥̂.

(15)

Selecting the coordinates of 𝑂 to be (0, 0) and equating Eqs. (14) and (15) provides (51.82 𝚤̂ + 9.407 𝚥̂) km = 𝑥𝐶 𝚤̂ + 𝑦𝐶 𝚥̂.

(16)

Equation (16) is a vector equation. For it to be satisfied, terms on both sides of the equation that multiply 𝚤̂ must be equal, and similarly terms on both sides of the equation that multiply 𝚥̂ must be equal. Hence, 𝑥𝐶 = 51.82 km and 𝑦𝐶 = 9.407 km.

(17)

Discussion & Verification Vector addition, as illustrated in this example, is a powerful application of vectors for determining positions, and these techniques will be used throughout this book and subjects that follow statics.

59

60

Chapter 2

Vectors: Force and Position

Problems

ISTUDY

Problems 2.29 through 2.36 For the following problems, use an 𝑥𝑦 Cartesian coordinate system where 𝑥 is horizontal, positive to the right, and 𝑦 is vertical, positive upward. For problems where the answers require vector expressions, report the vectors using Cartesian representations. Problem 2.29

Repeat Prob. 2.2 on p. 43.

Problem 2.30

Repeat Prob. 2.3 on p. 43.

Problem 2.31

Repeat Prob. 2.13 on p. 45.

Problem 2.32

Repeat Prob. 2.16 on p. 45.

Problem 2.33

Repeat Prob. 2.17 on p. 45.

Problem 2.34

Repeat Prob. 2.18 on p. 45.

Problem 2.35

Repeat Prob. 2.21 on p. 46.

Problem 2.36

Repeat Prob. 2.22 on p. 46.

Problem 2.37 Let 𝐴⃗ = (150 𝚤̂ − 200 𝚥̂) lb and 𝐵⃗ = (200 𝚤̂ + 480 𝚥̂) lb. Evaluate the following, and for Parts ⃗ (a) and (b) state the magnitude of 𝑅. ⃗ (a) 𝑅⃗ = 𝐴⃗ + 𝐵. ⃗ (b) 𝑅⃗ = 2𝐴⃗ − (1∕2)𝐵. (c) Find a scalar 𝑠 such that 𝑅⃗ = 𝑠𝐴⃗ + 𝐵⃗ has an 𝑥 component only. ⃗ (d) Determine a dimensionless unit vector in the direction 𝐵⃗ − 𝐴.

Problem 2.38 Let 𝐴⃗ = (−6 𝚤̂ + 8 𝚥̂) kN and 𝐵⃗ = (−9 𝚤̂ − 12 𝚥̂) kN. Evaluate the following, and for Parts ⃗ (a) and (b) state the magnitude of 𝑅. ⃗ (a) 𝑅⃗ = 𝐴⃗ + 𝐵. ⃗ (b) 𝑅⃗ = −2𝐴⃗ + 𝐵. (c) Find a scalar 𝑠 such that 𝑅⃗ = −𝐴⃗ + 𝑠𝐵⃗ has a 𝑦 component only. ⃗ (d) Determine a dimensionless unit vector in the direction 𝐴⃗ − 𝐵.

Problem 2.39 Let 𝐴⃗ = (150 𝚤̂ + 200 𝚥̂) mm, 𝐵⃗ = (300 𝚤̂ − 450 𝚥̂) mm, and 𝐶⃗ = (−100 𝚤̂ − 250 𝚥̂) mm. ⃗ Evaluate the following, and for Parts (a) and (b) state the magnitude of 𝑅. ⃗ (a) 𝑅⃗ = 𝐴⃗ + 𝐵⃗ + 𝐶. ⃗ (b) 𝑅⃗ = 3𝐴⃗ − 2𝐵⃗ + 𝐶. (c) Find scalars 𝑟 and 𝑠, if possible, such that 𝑅⃗ = 𝑟𝐴⃗ + 𝑠𝐵⃗ + 𝐶⃗ has zero 𝑥 and 𝑦 components. ⃗ (d) Determine a dimensionless unit vector in the direction 𝐴⃗ + 𝐵⃗ + 𝐶.

ISTUDY

Section 2.2

Cartesian Representation of Vectors in Two Dimensions

61

Problems 2.40 and 2.41 A rope connecting points 𝐴 and 𝐵 supports the force 𝐹 shown in the figure. Write expressions using Cartesian vector representation for the following: (a) 𝑟⃗𝐴𝐵 : the position vector from 𝐴 to 𝐵. (b) 𝑟⃗𝐵𝐴 : the position vector from 𝐵 to 𝐴. (c) 𝑢̂ 𝐴𝐵 : the unit vector in the direction from 𝐴 to 𝐵. (d) 𝑢̂ 𝐵𝐴 : the unit vector in the direction from 𝐵 to 𝐴. (e) 𝐹⃗𝐴𝐵 : the force vector the rope applies to 𝐴. (f) 𝐹⃗𝐵𝐴 : the force vector the rope applies to 𝐵. 𝑦

4m 𝐹 = 12 kN

𝐴

4 ft

𝑥 3m

𝑦

𝐵 30◦

𝐹 = 8 lb 𝑥

𝐵

Figure P2.40

Figure P2.41

Problem 2.42

𝑦 𝐹2 = 200 N

𝐹1 = 400 N 40◦

A cleat on a boat is used to support forces from three ropes as shown. Determine the ⃗ using Cartesian representation, and determine the magnitude 𝑅. resultant force vector 𝑅, ⃗ Also express 𝑅 in polar vector representation.

Problem 2.43

12 5

𝐹3 = 100 N

30◦

𝑥

Figure P2.42

A control arm in a machine supports the three forces shown. Determine the resultant force ⃗ using Cartesian representation, and determine the magnitude 𝑅. Also express vector 𝑅, ⃗ 𝑅 in polar vector representation.

𝑦

𝐹2 = 1600 lb

𝐹1 = 2000 lb 15

1000 mm

50◦ 6 15◦

70◦ 𝑦 18 in.

𝐹3 = 1100 lb 𝑥

𝐹4 = 900 N

𝑥 350 mm

9 20 13 15

55◦

Figure P2.43 10◦

𝐹3 = 800 N

Problem 2.44 A cantilevered bracket supports forces from two bars and two cables, as shown. Determine ⃗ using Cartesian representation, and determine the magnitude the resultant force vector 𝑅, ⃗ 𝑅. Also express 𝑅 in polar vector representation.

𝐹1 = 600 N Figure P2.44

𝐹2 = 700 N

62

Chapter 2

Vectors: Force and Position

Problem 2.45 A model of a person’s arm is used for ergonomics studies. Distances are 𝑟𝐴𝐵 = 35 cm, 𝑟𝐵𝐶 = 28 cm, and 𝑟𝐶𝐷 = 19 cm. 𝑦

𝐴

4

30◦ 𝑥

(a) Determine the position vector 𝑟⃗𝐴𝐷 and its magnitude 𝑟𝐴𝐷 .

𝐷 𝐶

(b) Express 𝑟⃗𝐴𝐷 using polar vector representation, measuring orientation positive counterclockwise from the right-hand horizontal direction.

25◦

3

(c) Determine a unit vector 𝑢̂ 𝐴𝐷 in the direction from 𝐴 to 𝐷.

𝐵

Figure P2.45

Problem 2.46 15 𝐵

36 𝐶 2 55◦ 𝐷

75◦

𝑦

1

A Caterpillar Ultra High Demolition machine is shown. The distances between points 𝐴 and 𝐵 is 12.5 m, points 𝐵 and 𝐶 is 2.8 m, 𝐶 and 𝐷 is 7 m, and 𝐷 and 𝐸 is 2.5 m. Determine the position vectors 𝑟⃗𝐴𝐵 , 𝑟⃗𝐵𝐶 , 𝑟⃗𝐶𝐷 , and 𝑟⃗𝐷𝐸 , where 𝑟⃗𝐴𝐵 is the position vector from point 𝐴 to point 𝐵, and so on. Add these vectors to determine the position vector 𝑟⃗𝐴𝐸 .

𝐸

Problem 2.47 𝑥

The actuator at point 𝐶 of the robotic arm is positioned by specifying angles 𝛼 and 𝛽, where 𝛼 is measured positive counterclockwise from the positive 𝑥 axis and 𝛽 is measured positive clockwise from the 𝐴𝐵 direction to the 𝐵𝐶 direction. Determine the position vector from point 𝑂 to point 𝐶 in terms of angles 𝛼 and 𝛽.

𝐴

Figure P2.46

𝑦

𝐶

300 mm 𝛽

40◦

𝐵

𝐹2

𝐹1

400 mm 𝛼

𝑦

𝐴

15◦

100 mm 𝑥

𝑂 Figure P2.47

𝑥 Figure P2.48

Problem 2.48 𝑃

12

Two ropes are used to lift a pipe in a congested region. Determine the ratio 𝐹2 ∕𝐹1 so that the resultant of 𝐹⃗1 and 𝐹⃗2 is vertical.

5

𝑥 𝐹 = 1000 lb

3

Problem 2.49

4 𝑦

Figure P2.49

ISTUDY

A welded steel tab is subjected to forces 𝐹 and 𝑃 . Determine the largest value 𝑃 may have if 𝐹 = 1000 lb and the magnitude of the resultant force cannot exceed 1500 lb.

ISTUDY

Section 2.2

63

Cartesian Representation of Vectors in Two Dimensions

Problem 2.50

800 lb

The mast of a ship supports forces from three cables as shown. If 𝐹 = 400 lb, determine the value of 𝛼 that minimizes the magnitude of the resultant of the three forces. Also, determine the magnitude of that resultant.

3 4

600 lb 𝛼 𝐹

Problem 2.51 The mast of a ship supports forces from three cables as shown. If 𝛼 = 0◦ , determine the value of 𝐹 that will make the magnitude of the resultant of the three forces smallest.

Figure P2.50 and P2.51

Problem 2.52

𝑦

𝐹

A short cantilever beam is subjected to three forces. If 𝐹 = 8 kN, determine the value of 𝛼 that minimizes the magnitude of the resultant of the three forces. Also, determine the magnitude of that resultant.

𝛼 𝑥 12 kN

Problem 2.53 A short cantilever beam is subjected to three forces. If 𝛼 = 45◦ , determine the value of 𝐹 that will make the magnitude of the resultant of the three forces smallest.

8 kN Figure P2.52 and P2.53

Problem 2.54

𝑦

An eyebolt is loaded by forces 𝐹1 and 𝐹2 . If the eyebolt has a maximum working load of 1200 lb, determine if the working load multipliers given in Fig. 1 of Example 2.6 are met for the following loading scenarios:

𝐹2 = 800 lb 𝐹1 = 200 lb

12 5

(a) Only 𝐹1 is applied.

8

𝑥

15

(b) Only 𝐹2 is applied. (c) Both 𝐹1 and 𝐹2 are applied simultaneously. Figure P2.54

Problem 2.55 An important and useful property of vectors is they may be easily transformed from one Cartesian coordinate system to another. That is, if the 𝑥 and 𝑦 components of a vector are known, then the 𝑡 and 𝑛 components can be found (or vice versa) by applying the formulas

where

or

𝑣⃗ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ = 𝑣𝑡 𝑡̂ + 𝑣𝑛 𝑛, ̂

(1)

𝑣𝑡 = 𝑣𝑥 cos 𝜙 + 𝑣𝑦 sin 𝜙,

(2)

𝑣𝑛 = −𝑣𝑥 sin 𝜙 + 𝑣𝑦 cos 𝜙,

(3)

𝑣𝑥 = 𝑣𝑡 cos 𝜙 − 𝑣𝑛 sin 𝜙,

(4)

𝑣𝑦 = 𝑣𝑡 sin 𝜙 + 𝑣𝑛 cos 𝜙.

(5)

In these equations, 𝑡̂ and 𝑛̂ are unit vectors in the 𝑡 and 𝑛 directions, respectively; 𝜙 is measured positive counterclockwise from the positive 𝑥 direction to the positive 𝑡 direction; and the 𝑦 and 𝑛 directions must be oriented 90◦ counterclockwise from the positive 𝑥 and 𝑡 directions, respectively. (a) Derive the above transformation that gives 𝑣𝑡 and 𝑣𝑛 in terms of 𝑣𝑥 and 𝑣𝑦 . Hint: First consider a vector 𝑣⃗𝑥 that acts in the 𝑥 direction, and resolve this into components in 𝑡 and 𝑛 directions. Then consider a vector 𝑣⃗𝑦 that acts in the 𝑦 direction, and resolve this into components in 𝑡 and 𝑛 directions. Vectorially adding these results yields the transformation.

𝑛

𝑦 𝑣⃗ 𝑡 𝜙

Figure P2.55

𝑥

64

Chapter 2

Vectors: Force and Position

(b) For the eyebolt and post of Example 2.7, the 𝑥 and 𝑦 components of the resultant force are given by Eq. (4) of Example 2.6. Use these 𝑥 and 𝑦 components with the preceding transformation equations to obtain the 𝑡 and 𝑛 components of the resultant force, and verify these are the same as those in Eq. (4) of Example 2.7.

𝑊 = 200 N 𝑛

𝑦

𝑡 𝐹 = 100 N

Problem 2.56 30◦

𝑥

A box weighing 200 N rests on an inclined surface. A worker applies a horizontal force 𝐹 to help position the box. Determine the 𝑥 and 𝑦 components of the resultant of forces 𝑊 and 𝐹 . Also determine the 𝑡 and 𝑛 components of the resultant force vector. Comment on why the 𝑡 and 𝑛 components might be useful to know.

Figure P2.56 𝑦

𝑃2 = 50 lb

𝑃1 = 200 lb 30◦

Problem 2.57

𝐶

𝐵

A motor-driven gear is used to produce forces 𝑃1 and 𝑃2 in members 𝐴𝐵 and 𝐴𝐶 of a machine. Member 𝐴𝐶 is parallel to the 𝑦 axis.

𝑛 30◦

𝐴

𝑥

𝑡

(b) Determine the 𝑡 and 𝑛 components of the resultant force vector. Comment on why the 𝑡 and 𝑛 components might be useful to know.

Figure P2.57 𝑛

40 N

60 N

35◦ 𝐴 𝑥

Problem 2.58 Traction is a medical procedure used to treat muscle and skeletal disorders by strategically applying one or more forces to a person’s body for a specific length of time.

25◦

𝑦

(a) Determine the 𝑥 and 𝑦 components of the resultant force vector at 𝐴 due to forces 𝑃1 and 𝑃2 .

𝑡 15◦

(a) Determine the 𝑥 and 𝑦 components of the resultant force vector at 𝐴 due to the 40 N and 60 N forces. (b) Determine the 𝑡 and 𝑛 components of the resultant force vector.

Figure P2.58

Problem 2.59

𝑛

𝑦 𝑥

𝑡

𝐴

20◦

70◦ 40◦ 6 kip Figure P2.59

ISTUDY

A structure supports forces from a bar and cable as shown. (a) Determine the 𝑥 and 𝑦 components of the resultant force vector at 𝐴 due to the 6 kip and 10 kip forces. (b) Determine the 𝑡 and 𝑛 components of the resultant force vector.

10 kip

Problem 2.60 Two people apply forces 𝑃1 and 𝑃2 to the handle of a wrench as shown. (a) Determine the 𝑥 and 𝑦 components of the resultant force vector applied to the handle of the wrench. (b) Determine the 𝑡 and 𝑛 components of the resultant force vector. Comment on why the 𝑡 and 𝑛 components might be useful to know. 𝑥 𝑃2 = 140 N

𝑡

45◦ 20◦ 𝑦

10◦ 𝑛

𝑃1 = 80 N

Figure P2.60

ISTUDY

Section 2.2

65

Cartesian Representation of Vectors in Two Dimensions

Problem 2.61 Bar 𝐴𝐶 is straight and has 106 in. length, 𝐵 is a pulley that supports forces 𝑊 and 𝐹 , 𝑊 is vertical, and the 𝑡 direction is parallel to bar 𝐴𝐶.

𝑛

(a) If 𝐹 = 150 lb and 𝛼 = 30◦ , determine the coordinates of point 𝐶 so that the 𝑡 component of the resultant of 𝐹 and 𝑊 is zero.

𝐵

𝐶 𝐹 𝛼

𝑦

(b) If 𝐹 = 150 lb and 𝐶 is located at (56, 90) in., determine angle 𝛼 so that the 𝑡 component of the resultant of 𝐹 and 𝑊 is zero. (c) If 𝐶 is located at (56, 90) in. and 𝛼 = 30◦ , determine 𝐹 so that the 𝑡 component of the resultant of 𝐹 and 𝑊 is zero.

𝑡

𝐴

𝜃

𝑊 = 100 lb 𝑥

Figure P2.61

Hint: For each of these questions, first find the 𝑥 and 𝑦 components of the resultant force and then use the transformation given in Prob. 2.55 to obtain the 𝑡 component.

𝐶

Problem 2.62 Screw 𝐴𝐶 is used to position point 𝐷 of a machine. Points 𝐴 and 𝐶 have coordinates (185, 0) mm and (125, 144) mm, respectively, and are fixed in space by the bearings that support the screw. If point 𝐵 is 52 mm from point 𝐴, determine the position vector 𝑟⃗𝐴𝐵 and the coordinates of point 𝐵.

𝐷 𝐵 𝑦

Problem 2.63 Repeat Prob. 2.62 if point 𝐵 is 39 mm from point 𝐶.

Figure P2.62 and P2.63 𝐶

Problem 2.64 Screw 𝐴𝐶 is used to position point 𝐷 of a machine. Point 𝐴 has coordinates (185, 0) mm and is fixed in space by a bearing that supports the screw. The screw nut at point 𝐵 is supported by lever 𝐸𝐷 and at the instant shown has coordinates (160, 60) mm. If the length of the screw from 𝐴 to 𝐶 is 130 mm, determine the position vector 𝑟⃗𝐴𝐶 and the coordinates of point 𝐶.

𝐷 𝐸

𝐵

𝑦

Problem 2.65 In Prob. 2.64 if point 𝐵 is equidistant from points 𝐴 and 𝐶 and if 𝑟⃗𝐸𝐴 (185 𝚤̂ − 50 𝚥̂) mm and 𝑟⃗𝐸𝐵 = (155 𝚤̂ + 22 𝚥̂) mm, determine the position vector 𝑟⃗𝐴𝐶 .

𝐴

𝑥

𝐴

𝑥

=

Figure P2.64 and P2.65

Problem 2.66 A computer numerical control (CNC) milling machine is used to cut a slot in the component shown. The cutting tool starts at point 𝐴 and advances to point 𝐵 where it pauses while the depth of cut is increased, and then the tool continues on to point 𝐶. If the coordinates of points 𝐴 and 𝐶 are (95, 56) mm and (17, 160) mm, respectively, and 𝐵 is located 45 mm from 𝐴, determine the position vectors 𝑟⃗𝑂𝐴 , 𝑟⃗𝑂𝐵 , and 𝑟⃗𝑂𝐶 and the coordinates of 𝐵.

𝑦

𝑂 𝐶 𝐵 𝐴 𝑥 Figure P2.66

66

Chapter 2

Vectors: Force and Position 𝑧

𝑧

2.3 𝑦

𝑦 𝑥

𝑥

(a) 𝑦

𝑦

(b)

𝑦 𝑥

𝑧

𝑧

𝑥

For problems in three dimensions, vectors are especially powerful, and without them many problems would be intractable. Concepts of Section 2.2 apply, with some additional enhancements needed for three dimensions. These include definitions of a right-handed coordinate system, direction angles, and direction cosines.

𝑦

𝑧

𝑥

Cartesian Representation of Vectors in Three Dimensions

Right-hand Cartesian coordinate system

𝑧 𝑥

(c) Figure 2.17 (a) Cartesian coordinate system in three dimensions. (b) A scheme for constructing a right-hand coordinate system. Position your right hand so the positive 𝑥 direction passes into your palm and the positive 𝑦 direction passes through your fingertips. Your thumb then indicates the positive 𝑧 direction. (c) More examples of right-hand coordinate systems.

In three dimensions a Cartesian coordinate system uses three orthogonal reference directions. These will consist of 𝑥, 𝑦, and 𝑧 directions as shown in Fig. 2.17(a). Proper interpretation of many vector operations, such as the cross product to be discussed in Section 2.5, requires the 𝑥, 𝑦, and 𝑧 directions be arranged in a consistent manner. For example, when you are constructing the coordinate system shown in Fig. 2.17(a), imagine the 𝑥 and 𝑦 directions are chosen first. Then, should 𝑧 be taken in the direction shown, or can it be in the opposite direction? The universal convention in mechanics and vector mathematics in general is 𝑧 must be taken in the direction shown, and the result is called a right-hand coordinate system. Figure 2.17(b) describes a scheme for constructing a right-hand coordinate system. You should study this scheme and become comfortable with its use.

Cartesian vector representation We define vectors 𝚤̂, 𝚥̂, and 𝑘̂ to be unit vectors that point in the positive 𝑥, 𝑦, and 𝑧 directions, respectively. A vector 𝑣⃗ can then be written as

𝑧 𝑣⃗𝑧

𝑣⃗

𝑣⃗ = 𝑣⃗𝑥 + 𝑣⃗𝑦 + 𝑣⃗𝑧

𝑘̂ 𝚥̂

𝚤̂

𝑣⃗𝑥

𝑣⃗𝑦

𝑦

𝑥 Figure 2.18 Right-hand Cartesian coordinate system with unit vectors 𝚤̂, 𝚥̂, and 𝑘̂ in the 𝑥, 𝑦, and 𝑧 directions, respectively, and resolution of a vector 𝑣⃗ into vector components 𝑣⃗𝑥 , 𝑣⃗𝑦 , and 𝑣⃗𝑧 . 𝑧 𝑣⃗ 𝑣⃗𝑥 𝑥

𝑣⃗𝑎 𝑣⃗𝑦

𝑣⃗𝑧

𝑦

𝑎

Figure 2.19 Evaluation of a vector’s magnitude in terms of its components.

ISTUDY

̂ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ + 𝑣𝑧 𝑘.

(2.23)

Resolution of 𝑣⃗ into 𝑥, 𝑦, and 𝑧 components is shown in Fig. 2.18. The magnitude of 𝑣⃗ is given by √ |𝑣| ⃗ = 𝑣2𝑥 + 𝑣2𝑦 + 𝑣2𝑧 . (2.24) This equation is obtained using the construction shown in Fig. 2.19 as follows. First, a vector 𝑣⃗𝑎 that lies in the 𝑥𝑦 plane is defined. Because 𝑣𝑥 , 𝑣𝑦 , and 𝑣𝑎 form a right triangle, the Pythagorean theorem provides 𝑣2𝑎 = 𝑣2𝑥 +𝑣2𝑦 . Then 𝑣𝑎 , 𝑣𝑧 , and 𝑣 also form a right triangle, and the Pythagorean theorem provides 𝑣2 = 𝑣2𝑎 + 𝑣2𝑧 . Substituting for 𝑣2𝑎 in this latter expression yields 𝑣2 = 𝑣2𝑥 + 𝑣2𝑦 + 𝑣2𝑧 , and thus Eq. (2.24) follows.

Direction angles and direction cosines An effective way to characterize a vector’s orientation is to use direction angles. Direction angles 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 are shown in Fig. 2.20 and are defined to be the angles measured from the positive 𝑥, 𝑦, and 𝑧 directions, respectively, to the direction of the vector. Direction angles have values between 0◦ and 180◦ . Direction angles can be used to obtain a vector’s components, and vice versa, as follows. Consider the vector polygon shown in Fig. 2.21. This polygon is a right triangle that consists of the vector’s magnitude |𝑣|, ⃗ the 𝑦 component 𝑣𝑦 , and another

ISTUDY

Section 2.3

Cartesian Representation of Vectors in Three Dimensions

vector component that we are not especially interested in, but which is nonetheless present. Although this triangle has rather complicated orientation in space, since it is a right triangle, the relationship between |𝑣|, ⃗ 𝑣𝑦 , and 𝜃𝑦 is given by elemen⃗ cos 𝜃𝑦 . Now imagine sketching a new vector triangle tary trigonometry as 𝑣𝑦 = |𝑣| that contains |𝑣|, ⃗ 𝑣𝑥 , and angle 𝜃𝑥 ; again, elementary trigonometry provides 𝑣𝑥 = ⃗ 𝑣𝑧 , and angle |𝑣| ⃗ cos 𝜃𝑥 . Finally, consideration of the vector triangle containing |𝑣|, ⃗ cos 𝜃𝑧 . Collectively, these may be written as 𝜃𝑧 provides 𝑣𝑧 = |𝑣|

67

𝑧

𝑣⃗ 𝜃𝑧 𝜃𝑥

𝑣⃗ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ + 𝑣𝑧 𝑘̂

𝜃𝑦 𝑦

⃗ cos 𝜃𝑦 𝚥̂ + |𝑣| ⃗ cos 𝜃𝑧 𝑘̂ = |𝑣| ⃗ cos 𝜃𝑥 𝚤̂ + |𝑣| ̂ = |𝑣| ⃗ (cos 𝜃𝑥 𝚤̂ + cos 𝜃𝑦 𝚥̂ + cos 𝜃𝑧 𝑘).

(2.25)

̂ is a unit vector that points in In Eq. (2.25), note that (cos 𝜃𝑥 𝚤̂ + cos 𝜃𝑦 𝚥̂ + cos 𝜃𝑧 𝑘) the direction of 𝑣. ⃗ Because cosines of the direction angles play such an important role in writing Eq. (2.25), the quantities cos 𝜃𝑥 , cos 𝜃𝑦 , and cos 𝜃𝑧 are called direction cosines. Examination of Eq. (2.25) shows that the direction cosines constitute components of a unit vector. That is, substituting the expression for 𝑣𝑥 , 𝑣𝑦 , and 𝑣𝑧 from Eq. (2.25) into Eq. (2.24) provides cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1,

𝑥 Figure 2.20 Definition of direction angles 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 . Direction angles have values between 0◦ and 180◦ . 𝑧

𝑣⃗

(2.26)

where cos2 𝜃 ≡ (cos 𝜃)2 . A subtle point about direction angles is that although there are three of them, only two are independent. For example, if 𝜃𝑥 and 𝜃𝑦 are known, then 𝜃𝑧 may be determined from Eq. (2.26), although care is usually needed to ensure that the desired solution is obtained from among the multiple solutions this equation has. These subtleties are illustrated in some of the example problems of this section. To summarize, important facts about direction angles and direction cosines for a vector 𝑣⃗ are as follows: Summary Box 𝜃𝑥 = angle between positive 𝑥 direction and vector,

𝜃𝑦 𝑦 𝑥

𝑣𝑦 = 𝑣⃗ cos 𝜃𝑦

Figure 2.21 Relation between direction angle 𝜃𝑦 and vector component 𝑣𝑦 . Similar sketches of triangles using direction angles 𝜃𝑥 and 𝜃𝑧 provide 𝑣𝑥 = |𝑣| ⃗ cos 𝜃𝑥 and 𝑣𝑧 = |𝑣| ⃗ cos 𝜃𝑧 .

𝜃𝑦 = angle between positive 𝑦 direction and vector, 𝜃𝑧 = angle between positive 𝑧 direction and vector,

(2.27)

̂ 𝑣⃗ = |𝑣| ⃗ (cos 𝜃𝑥 𝚤̂ + cos 𝜃𝑦 𝚥̂ + cos 𝜃𝑧 𝑘), cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1, ⃗ cos 𝜃𝑥 = 𝑣𝑥 ∕|𝑣|,

cos 𝜃𝑦 = 𝑣𝑦 ∕|𝑣|, ⃗

cos 𝜃𝑧 = 𝑣𝑧 ∕|𝑣|. ⃗ 𝑧

Position vectors

𝐻(𝑥𝐻 , 𝑦𝐻 , 𝑧𝐻 )

Construction of a position vector in three dimensions is analogous to the procedure described in the previous section. Given points 𝑇 and 𝐻 having coordinates (𝑥𝑇 , 𝑦𝑇 , 𝑧𝑇 ) and (𝑥𝐻 , 𝑦𝐻 , 𝑧𝐻 ), respectively, as shown in Fig. 2.22, the position vector from tail 𝑇 to head 𝐻 is denoted by 𝑟⃗𝑇 𝐻 and is written as ̂ 𝑟⃗𝑇 𝐻 = (𝑥𝐻 − 𝑥𝑇 ) 𝚤̂ + (𝑦𝐻 − 𝑦𝑇 ) 𝚥̂ + (𝑧𝐻 − 𝑧𝑇 ) 𝑘.

(2.28)

The components of the position vector are given by “coordinates of the head minus coordinates of the tail,” and this is a useful phrase to remember.

𝑇 (𝑥𝑇 , 𝑦𝑇 , 𝑧𝑇 )

𝑟⃗𝑇 𝐻 head

𝑥𝐻 − 𝑥𝑇

tail

𝑧 𝐻 − 𝑧𝑇 𝑦

𝑦𝐻 − 𝑦𝑇 𝑥 Figure 2.22 Construction of a position vector using Cartesian coordinates of the vector’s head and tail.

68

Chapter 2

Vectors: Force and Position

Use of position vectors to write expressions for force vectors A position vector can often be a useful aid for writing an expression for a force vector. The basic idea is if a force vector 𝐹⃗ lies along the line of action of a position vector 𝑟⃗, or is known to be parallel to the position vector, then a vector expression for the force may be written as 𝐹⃗ = 𝐹 𝑟⃗∕|⃗𝑟|. In this expression, 𝑟⃗∕|⃗𝑟| is a unit vector that gives 𝐹⃗ the proper direction, and 𝐹 is the component of the force 𝐹⃗ along the direction 𝑟⃗. Often, the location of two points on the line of action of a force will be known, and the components of 𝑟⃗ can be obtained by taking the difference between coordinates of the position vector’s head and tail, as given by Eq. (2.28). Note that in an expression such as 𝐹⃗ = 𝐹 𝑟⃗∕|⃗𝑟|, if 𝐹 > 0, then the direction of the force is the same as the direction of 𝑟⃗, whereas if 𝐹 < 0, then the direction of the force is opposite the direction of 𝑟⃗. In statics, most of our applications of this idea will be for writing force vectors and, starting in Chapter 4, moment vectors also. However, this idea easily generalizes so that if any two vectors have the same line of action, or are parallel, then one of them may be used to construct an expression for the other, regardless of the physical interpretations each vector might have. For example, imagine a particle initially at rest is subjected to a known force 𝐹⃗ that has constant orientation, but perhaps magnitude that varies with time. Due to this force, the particle moves with acceleration 𝑎⃗ and velocity 𝑣⃗ where all three of these vectors share the same line of action. Since 𝐹⃗ is known, vector expressions for 𝑎⃗ and 𝑣⃗ may be written as 𝑎⃗ = 𝑎𝐹⃗ ∕|𝐹⃗ | and 𝑣⃗ = 𝑣𝐹⃗ ∕|𝐹⃗ |, where 𝑎 and 𝑣, which are functions of time, need to be determined.

Shutterstock/SARIN KUNTHONG

Figure 2.23 The cross section of a steel cable is shown where the many individual strands that make up the cable can be seen.

Michael Sohn/AP Photo

Figure 2.24 A worker stands next to one of the steel cables that support a bridge.

ISTUDY

Some simple structural members One category of forces that are important in mechanics is forces that develop within structural members. These are called internal forces, and much will be said about these in later chapters of this book. Simply stated, it is because structural members have the capability to develop internal forces that they are used to make structures, and these structures are used to support the external forces that are applied to them. In the following paragraph, we briefly discuss several structural members for which the forces they support (internal forces) are collinear with their orientation. Shown in Fig. 2.25 are structural members consisting of a cable (or rope, string, or cord), a wire, and a straight bar.∗ Consider a simple experiment with a piece of light string. When tensile forces are applied to the two ends of the string, the string obviously assumes the shape of a straight-line segment. Thus we may infer that the forces it supports are directed from one endpoint of the string to the other. Another experiment would be to attach an object to one end of a string and hold the other end of the string with your hand. Again, the string takes the shape of a straight-line segment that is vertical. Newton’s law of gravity tells us that the weight of the object, which is a vector quantity, is in the vertical direction, pointing from the object toward the center of the Earth. The vertical direction of the weight is identical to the orientation of the string, so once again we may infer that the internal force in the string is collinear with the line connecting the string’s endpoints. Cables, ropes, wires, and the like are idealized to behave the same way. Further, all of these members are assumed to be incapable of supporting compressive forces as this would cause them to immediately bend or buckle. A straight bar is a member that is loaded ∗ Bars

can also be curved or have other geometry, but the remarks made here apply to straight bars only.

ISTUDY

Section 2.3

Cartesian Representation of Vectors in Three Dimensions

cable, rope, string, ... wire bar

Figure 2.25. Some common structural members whose internal forces are collinear with their geometry.

by forces at its two ends, as shown in Fig. 2.25. Its behavior is similar to that for cables and ropes, but a bar can also support compressive forces. To summarize, for the simple but important and common structural members shown in Fig. 2.25, the forces they support may be characterized by expressions such as 𝐹⃗ = 𝐹 𝑟⃗∕|⃗𝑟|, where 𝑟⃗ is a position vector that describes the orientation of the member. These structural members belong to a category of members called two-force members, which are examined in detail in Chapter 5.

End of Section Summary In this section, Cartesian coordinate systems and Cartesian representation for vectors in three dimensions have been defined. Some of the key points are: • The 𝑥𝑦𝑧 coordinate system you use must be a right-hand coordinate system. Proper interpretation of some vector operations requires this. • Direction angles provide a useful way to specify a vector’s orientation in three dimensions. A vector has three direction angles 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 , but only two of these are independent. Direction angles satisfy the equation cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1, so if two direction angles are known, the third may be determined. Direction angles have values between 0◦ and 180◦ . • Structural members such as cables, ropes, and bars support forces whose lines of action have the same orientation as the member’s geometry. Thus, if 𝑟⃗ describes a member’s geometry, a vector expression for the force supported by the member may be written as 𝐹⃗ = 𝐹 (⃗𝑟∕|⃗𝑟|).

69

70

Chapter 2

Vectors: Force and Position

E X A M P L E 2.9

Direction Angles Determine the direction angles for boom 𝐴𝐵 of the crane.

𝑦

SOLUTION

35◦

𝐵

𝐴

𝑥

73◦ 𝑧 Figure 1

ISTUDY

Road Map We denote the position vector of the crane’s boom from point 𝐴 to point 𝐵 by 𝑟⃗𝐴𝐵 . The 35◦ and 73◦ angles reported in the figure are indeed direction angles: the 35◦ angle is measured from the positive 𝑦 direction to the direction of 𝑟⃗𝐴𝐵 , and the 73◦ angle is measured from the positive 𝑧 direction to the direction of 𝑟⃗𝐴𝐵 . Thus, 𝜃𝑦 = 35◦ and 𝜃𝑧 = 73◦ . The remaining direction angle 𝜃𝑥 must satisfy Eq. (2.26) on p. 67. Governing Equations & Computation

Knowing that 𝜃𝑦 = 35◦ and 𝜃𝑧 = 73◦ , we solve

Eq. (2.26) to obtain

Common Pitfall Square roots. When we take a square root, it is common to overlook the possibility of a negative solution. For example, in solving for 2 𝑠 where √ 𝑠 = 9, the solution for 𝑠 is always 𝑠 = ± 9 = ±3. Depending on the physics of the problem, both solutions for 𝑠, or perhaps only one solution, may be meaningful. You will avoid serious blunders if you always check to be sure your direction angles and/or direction cosines are reasonable.

cos2 𝜃𝑥 = 1 − cos2 𝜃𝑦 − cos2 𝜃𝑧 , √ cos 𝜃𝑥 = ± 1 − cos2 𝜃𝑦 − cos2 𝜃𝑧 , ) ( √ 𝜃𝑥 = cos−1 ± 1 − cos2 𝜃𝑦 − cos2 𝜃𝑧 ( √ ) = cos−1 ± 1 − cos2 35◦ − cos2 73◦ = cos−1 (±0.4935).

(1) (2)

(3)

Using the inverse cosine function on an electronic calculator provides the solutions 𝜃𝑥 = cos−1 (+0.4935) = 60.43◦ and 𝜃𝑥 = cos−1 (−0.4935) = 119.6◦ . In fact, there are an infinite number of additional solutions, but the only two solutions in the interval 0◦ ≤ 𝜃𝑥 ≤ 180◦ are 60.43◦ and 119.6◦ . Inspection will usually indicate which of the solutions for the direction angle is correct for our physical problem, and you should examine Fig. 1 now to see if you can make this determination. If you are uncertain, then compute the unit vectors corresponding to the two sets of direction angles to see which of the two vectors points in the proper direction. To show this, we calculate these unit vectors as for 𝜃𝑥 = 60.43◦∶ for 𝜃𝑥 = 119.6◦∶

𝑢̂ 1 = cos(60.43◦ ) 𝚤̂ + cos(35◦ ) 𝚥̂ + cos(73◦ ) 𝑘̂ ̂ = 0.4935 𝚤̂ + 0.8192 𝚥̂ + 0.2924 𝑘,

(4)

𝑢̂ 2 = cos(119.6◦ ) 𝚤̂ + cos(35◦ ) 𝚥̂ + cos(73◦ ) 𝑘̂ ̂ = −0.4935 𝚤̂ + 0.8192 𝚥̂ + 0.2924 𝑘.

(5)

Notice that the only difference between 𝑢̂ 1 and 𝑢̂ 2 in Eqs. (4) and (5) is the sign of the 𝑥 component. Figure 1 clearly shows 𝑟⃗𝐴𝐵 has a negative 𝑥 component. So the solution for 𝜃𝑥 and 𝑢̂ 1 in Eq. (4) is physically meaningless and must be discarded. Thus, the direction angles for boom 𝐴𝐵 are 𝜃𝑥 = 119.6◦ , Discussion & Verification

𝜃𝑦 = 35◦ ,

and 𝜃𝑧 = 73◦ .

(6)

As a quick partial check of solution accuracy, you should verify that the values for 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 in Eq. (6) satisfy the equation cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1; if this is not the case, then one or more of the angles is not a direction angle and an error has been made.

ISTUDY

Section 2.3

71

Cartesian Representation of Vectors in Three Dimensions

E X A M P L E 2.10

Position Vectors and Force Vectors

Two cables apply forces to the cantilever I beam. Cable 𝐴𝐵 applies a tensile force of magnitude 𝑃 = 2 kN, and cable 𝐶𝐷 applies a tensile force of magnitude 𝐹 = 1 kN. ⃗ , 𝐹⃗ , and 𝑃⃗ applied to the I beam. Write expressions for the forces 𝑊

𝑦 𝐷 𝐹 = 1 kN

SOLUTION

4 15 m

Road Map

The directions of the forces applied to the I beam by the weight and two cables are shown in Fig. 2. Because the 𝑦 axis is vertical and weight due to gravity always ⃗ will be acts in the downward vertical direction, writing an expression for the weight 𝑊 straightforward. For the cables at points 𝐴 and 𝐶, first we will write position vectors that describe the orientation of each cable, and from these we will construct expressions for the forces each cable applies to the I beam.

𝐴 7m

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

= (2 kN)

𝑃 = 2 kN

(2)

(3)

√

where the magnitude of 𝑟⃗𝐴𝐵 is |⃗𝑟𝐴𝐵 | = (18)2 + (−14)2 + (−3)2 m = 23 m. The 1 kN tensile force that cable 𝐶𝐷 applies to the I beam is directed from point 𝐶 to 𝐷. In other words, the force in cable 𝐶𝐷 tends to draw point 𝐶 closer to point 𝐷. This vector will be called 𝐹⃗ , and Fig. 2 shows it has orientation that is the same as a vector consisting of 2 units in the positive 𝑧 direction, plus 3 units in the negative 𝑥 direction, plus 4 units in the positive 𝑦 direction. Thus, we may immediately write

𝑥

𝑦 𝐹⃗ 4 15 m

3 2

𝐶

𝐴

33 m

8m 7m

⃗ 𝑊

4m 6m 𝐵

𝑃⃗

Helpful Information Comments on Newton’s third law. In our solution, 𝑃⃗ is the force vector that cable 𝐴𝐵 applies to the I beam. This cable also applies force to the support at 𝐵, and Newton’s third law tells us this force vector is −𝑃⃗ , as illustrated below. Similar comments apply to 𝐹⃗ . 𝑦 𝐷 −𝐹⃗

𝐹⃗ 𝐶 𝐴 ⃗ 𝑊

𝑧

As a quick check, you should verify that each force vector points in the proper direction, and that the expressions for 𝑃⃗ and 𝐹⃗ have 2 kN and 1 kN magnitudes, respectively.

𝑥

Figure 2 Directions of forces applied by the weight and cables to the I beam.

̂ (−3 𝚤̂ + 4 𝚥̂ + 2 𝑘) ̂ N. = (−557.1 𝚤̂ + 742.8 𝚥̂ + 371.4 𝑘) 𝐹⃗ = (1 kN) (4) √ 29 √ ̂ Equation (4) uses the unit vector (−3 𝚤̂ + 4 𝚥̂ + 2 𝑘)∕ 29 to give 𝐹⃗ the proper direction, and the 1 kN term gives 𝐹⃗ the proper magnitude and units. Discussion & Verification

4m 6m 𝐵

Figure 1

𝑧

̂ m (18 𝚤̂ − 14 𝚥̂ − 3 𝑘) 23 m

̂ kN, = (1.565 𝚤̂ − 1.217 𝚥̂ − 0.2609 𝑘)

𝑊 = 3 kN

𝑧

(1)

Another way to construct 𝑟⃗𝐴𝐵 is to imagine being positioned at point 𝐴, the tail of the vector, and then ask what distances in the 𝑥, 𝑦, and 𝑧 directions must be traversed to arrive at point 𝐵, the head of the vector; Eq. (2) is again obtained. Finally, we use 𝑟⃗𝐴𝐵 to write the vector expression for 𝑃⃗ as 𝑃⃗ = (2 kN)

33 m

𝐷

The 2 kN tensile force cable 𝐴𝐵 applies to the I beam is directed from point 𝐴 to point 𝐵. In other words, the force in cable 𝐴𝐵 tends to draw point 𝐴 closer to point 𝐵. This vector will be called 𝑃⃗ , and to construct a vector expression for it, we note that its direction is the same as the position vector from points 𝐴 to 𝐵, 𝑟⃗𝐴𝐵 . Figure 1 shows the coordinates of these points to be 𝐴(15, 8, 7) m and 𝐵(33, −6, 4) m, and Eq. (2.28) can be applied to write ̂ m 𝑟⃗𝐴𝐵 = [(33 − 15) 𝚤̂ + (−6 − 8) 𝚥̂ + (4 − 7) 𝑘] ̂ m. = (18 𝚤̂ − 14 𝚥̂ − 3 𝑘)

𝐶

8m

Governing Equations & Computation The weight is a force with 3 kN magnitude that acts in the downward vertical direction, which is the −𝑦 direction in this example. Thus,

⃗ = −3 𝚥̂ kN. 𝑊

3 2

Figure 3

𝑃⃗

𝑥 −𝑃⃗ 𝐵

72

Chapter 2

Vectors: Force and Position

E X A M P L E 2.11

Direction Angles and Position Vectors Observers on Earth at points 𝐴 and 𝐵 measure direction cosines of position vectors to the SpaceX Dragon spacecraft 𝐶 as for 𝑟⃗𝐴𝐶 ∶

cos 𝜃𝑥 = 0.360,

cos 𝜃𝑦 = 0.480,

cos 𝜃𝑧 = 0.800,

(1)

for 𝑟⃗𝐵𝐶 ∶

cos 𝜃𝑥 = −0.515, cos 𝜃𝑦 = −0.606, cos 𝜃𝑧 = 0.606.

(2)

Determine the 𝑥𝑦𝑧 coordinates of the spacecraft. NASA

SOLUTION Road Map

𝑧

𝐶

𝐴

𝑦 𝐵 (520, 640, 0) km

𝑥 Figure 1

𝑧

𝐶

Some careful thought before putting pencil to paper will help you create a successful strategy for this problem. The orientations of position vectors 𝑟⃗𝐴𝐶 and 𝑟⃗𝐵𝐶 are known, but their magnitudes are not. Given the direction cosines cited above, if the distance between the observers at points 𝐴 and 𝐵 (this will be called 𝑟𝐴𝐵 ) were to increase, then the magnitudes of 𝑟⃗𝐴𝐶 and 𝑟⃗𝐵𝐶 would clearly also increase. Thus, our solution must incorporate the relative position of points 𝐴 and 𝐵. A straightforward way of doing this is to write the vector equation 𝑟⃗𝐴𝐶 = 𝑟⃗𝐴𝐵 + 𝑟⃗𝐵𝐶 . The addition 𝑟⃗𝐴𝐶 = 𝑟⃗𝐴𝐵 + 𝑟⃗𝐵𝐶 is shown in Fig. 2. Note that this vector equation contains three scalar equations; that is, equality must be achieved independently for each of the 𝑥, 𝑦 and 𝑧 components, and hopefully this will provide enough equations so the unknown coordinates of point 𝐶 may be determined. Using the specified direction cosines, and letting 𝑟𝐴𝐶 and 𝑟𝐵𝐶 denote the magnitudes of vectors 𝑟⃗𝐴𝐶 and 𝑟⃗𝐵𝐶 , respectively, we find the vector expressions for 𝑟⃗𝐴𝐶 and 𝑟⃗𝐵𝐶 are Governing Equations & Computation

𝑟⃗𝐴𝐶

̂ 𝑟⃗𝐴𝐶 = 𝑟𝐴𝐶 (0.360 𝚤̂ + 0.480 𝚥̂ + 0.800 𝑘), ̂ 𝑟⃗ = 𝑟 (−0.515 𝚤̂ − 0.606 𝚥̂ + 0.606 𝑘).

𝑟⃗𝐵𝐶 𝐴 𝑟⃗𝐴𝐵 𝑥

𝐵𝐶

𝑦

ISTUDY

(4)

The coordinates of points 𝐴 and 𝐵 are given in Fig. 1, so we may write

𝐵 (520, 640, 0) km

Figure 2 Vector addition that provides the position of 𝐶.

𝐵𝐶

(3)

𝑟⃗𝐴𝐵 = (520 𝚤̂ + 640 𝚥̂) km.

(5)

Then,

𝑟𝐴𝐶

𝑟⃗𝐴𝐶 = 𝑟⃗𝐴𝐵 + 𝑟⃗𝐵𝐶 , ̂ = (0.360 𝚤̂ + 0.480 𝚥̂ + 0.800 𝑘) ̂ (520 𝚤̂ + 640 𝚥̂) km + 𝑟𝐵𝐶 (−0.515 𝚤̂ − 0.606 𝚥̂ + 0.606 𝑘).

(6)

As described earlier, Eq. (6) contains three scalar equations as follows: 𝑟𝐴𝐶 (0.360) = 520 km + 𝑟𝐵𝐶 (−0.515),

(7)

𝑟𝐴𝐶 (0.480) = 640 km + 𝑟𝐵𝐶 (−0.606),

(8)

𝑟𝐴𝐶 (0.800) = 𝑟𝐵𝐶 (0.606).

(9)

Solving Eq. (9) for 𝑟𝐴𝐶 provides 𝑟𝐴𝐶 = 𝑟𝐵𝐶 (0.606∕0.800), and substituting this into Eq. (7) [alternatively, Eq. (8) could be used] and rearranging provides the solution for 𝑟𝐵𝐶 as 520 km 𝑟𝐵𝐶 = = 660.1 km. (10) (0.606∕0.800)(0.360) + 0.515 Then, using Eq. (9) [alternatively, Eq. (7) or (8) could be used] provides the solution for 𝑟𝐴𝐶 as 𝑟𝐴𝐶 = 𝑟𝐵𝐶 (0.606∕0.800) = 500.1 km. (11)

ISTUDY

Section 2.3

Cartesian Representation of Vectors in Three Dimensions

To determine the coordinates of the spacecraft at point 𝐶, we use Eq. (2.28) to write an expression for 𝑟⃗𝐴𝐶 (alternatively, an expression for 𝑟⃗𝐵𝐶 could be used) ̂ 𝑟⃗𝐴𝐶 = 𝑥𝐶 𝚤̂ + 𝑦𝐶 𝚥̂ + 𝑧𝐶 𝑘.

(12)

Equating our two expressions for 𝑟⃗𝐴𝐶 , Eq. (3) with 𝑟𝐴𝐶 = 500.1 km and Eq. (11), provides ̂ = 𝑥 𝚤̂ + 𝑦 𝚥̂ + 𝑧 𝑘. ̂ 500.1 km (0.360 𝚤̂ + 0.480 𝚥̂ + 0.800 𝑘) 𝐶 𝐶 𝐶

(13)

Equating 𝑥, 𝑦, and 𝑧 terms provides the solution for the coordinates of point 𝐶 as 𝑥𝐶 = (500.1 km)(0.360) = 180.0 km,

(14)

𝑦𝐶 = (500.1 km)(0.480) = 240.0 km,

(15)

𝑧𝐶 = (500.1 km)(0.800) = 400.1 km.

(16)

Discussion & Verification

In view of the direction cosines given in Eqs. (1) and (2), and the coordinates of points 𝐴 and 𝐵 given in Fig. 2, the coordinates found in Eqs. (14)–(16) are reasonable. Furthermore, 𝑧𝐶 is the altitude of the spacecraft (neglecting the effects of the Earth’s curvature, which are minor for the dimensions of this problem), and the result 𝑧𝐶 = 400.1 km is also reasonable for real life missions that the SpaceX Dragon spacecraft flies. A Closer Look You might wonder why this example has more equations than unknowns. That is, there are three equations available, Eqs. (7)–(9), but they contain only two unknowns, 𝑟𝐴𝐶 and 𝑟𝐵𝐶 . Normally, such a situation indicates an overdetermined problem, where more information is available than is needed to determine the unknowns. Generally, overdetermined problems do not have a single, unique answer for each unknown. Such situations will rarely arise in statics, except when we are working with direction angles (recall that, as discussed on p. 67, of the three direction angles for a vector, only two are independent). In this example, Eqs. (7)–(9) are not overdetermined. Rather, only two of Eqs. (7)– (9) are independent. To see this, if Eqs. (7) and (8) are multiplied by −640 and 520, respectively, and added, the result is an equation that has exactly the same proportions as Eq. (9). In mathematical nomenclature, we say that only two of the three equations are linearly independent. The take-away message of this short discussion is that you should not be surprised or alarmed when subtleties such as these arise when working with direction angles and direction cosines.

73

74

Chapter 2

Vectors: Force and Position

E X A M P L E 2.12

Force Vectors and Vector Addition ⃗ are given using a variety of different forms of orientation information Forces 𝐹⃗ , 𝑃⃗ , and 𝑄 that are commonly encountered.

𝑧

(a) Write a vector expression for each force.

𝑄 = 300 N

𝐹 = 100 N

(b) Determine the resultant of the three forces.

60◦

(c) Determine the direction angles for the resultant force.

30◦ 45◦ 30◦ 8 17 15

45◦ 60◦ 𝑥

𝑦

SOLUTION Road Map

The strategy to write a vector expression depends on the details of the geometry information provided. Once expressions for the three force vectors are obtained, we will add these to obtain the resultant force vector. Once the resultant force vector is known, it will be straightforward to determine its direction angles.

𝑃 = 200 N Figure 1

Part (a) Governing Equations & Computation

𝑧

We carefully examine Fig. 1 to see if the angles reported for 𝐹⃗ are direction angles. Indeed, the two 60◦ angles are measured from the positive 𝑥 and 𝑧 directions to the direction of 𝐹⃗ , so both of these are direction angles and thus 𝜃𝑥 = 60◦ and 𝜃𝑧 = 60◦ . The remaining 45◦ angle is measured from the negative 𝑦 direction to the vector, so it is not a direction angle. However, the appropriate direction angle, as measured from the positive 𝑦 direction, is easily found as 𝜃𝑦 = 180◦ − 45◦ = 135◦ . As a partial check, we evaluate Eq. (2.26) to find cos2 60◦ + cos2 135◦ + cos2 60◦ = 1, and thus, these angles are indeed direction angles. Note, however, that if we incorrectly took 𝜃𝑦 to be 45◦ , Eq. (2.26) would still have been satisfied, but the resulting vector 𝐹⃗ would have a positive 𝑦 component, rather than the correct value, which is negative. Given the direction angles, a vector expression for 𝐹⃗ may be written as Force 𝐹⃗ :

𝑃𝑎

(a)

𝑦

30◦ 8

17 15

𝑥

𝑃 = 200 N

𝑎 −𝑃𝑧

𝑧

𝑃𝑦

(b)

𝑃𝑎

30◦ 8 𝑥

17 15

𝑃 = 200 N

𝑎 −𝑃𝑧

Figure 2 (a) Resolution of force 𝑃 into 𝑧 and 𝑎 direction components. (b) Resolution of the 𝑎 component into 𝑥 and 𝑦 components.

ISTUDY

̂ 𝐹⃗ = (100 N) (cos 60◦ 𝚤̂ + cos 135◦ 𝚥̂ + cos 60◦ 𝑘)

𝑦 𝑃𝑥

̂ N. = (50 𝚤̂ − 70.71 𝚥̂ + 50 𝑘)

(1)

Force 𝑃⃗ :

For this force, we observe that the 30◦ angle shown in Fig. 1 is not a direction angle; while it is measured from the positive 𝑦 direction, it is not measured to the vector’s direction. When geometry is given in this fashion, we use a two-step process to determine the vector’s components. We first use the 8-15 rise-run geometry to compute the components 𝑃𝑧 and 𝑃𝑎 shown in Fig. 2(a): 𝑃𝑧 = −(200 N)

8 = −94.12 N and 17

𝑃𝑎 = (200 N)

15 = 176.5 N. 17

(2)

The negative sign is included in the expression for 𝑃𝑧 because this force component acts in the negative 𝑧 direction. For the second step of this procedure, we use the 30◦ angle to resolve 𝑃𝑎 into 𝑥 and 𝑦 components, as shown in Fig. 2(b). Thus, 𝑃𝑥 = 𝑃𝑎 sin 30◦ = 88.24 N

and 𝑃𝑦 = 𝑃𝑎 cos 30◦ = 152.8 N.

(3)

Collecting results for 𝑃𝑥 , 𝑃𝑦 , and 𝑃𝑧 allows us to write ̂ N. 𝑃⃗ = (88.24 𝚤̂ + 152.8 𝚥̂ − 94.12 𝑘)

(4)

As a partial√check on our work, we compute the magnitude of the above expression to find 𝑃 = (88.24 N)2 + (152.8 N)2 + (−94.12 N)2 = 200.0 N, which is the expected result.

ISTUDY

Section 2.3

75

Cartesian Representation of Vectors in Three Dimensions 𝑧

⃗ Force 𝑄:

For this force, we observe that the 30◦ and 45◦ angles are not direction angles; the 45 angle is not measured to the vector’s direction, and the 30◦ angle is not measured from a coordinate direction. As with force 𝑃⃗ , we again use a two-step process to obtain ⃗ the vector’s components. As shown in Fig. 3(a), we first use the 30◦ angle to resolve 𝑄 into 𝑧 and 𝑏 direction components ◦

𝑏 30◦ 45◦

(a)

𝑄𝑧 = −(300 N) sin 30◦ = −150 N

and

𝑄𝑏 = (300 N) cos 30◦ = 259.8 N.

𝑄𝑦 = −𝑄𝑏 cos 45◦ = −183.7 N.

𝑄𝑏

𝑦

(5) (6)

𝑥

The negative sign is included in the expressions for 𝑄𝑧 because this component acts in the negative 𝑧 direction. We could similarly include a negative sign for 𝑄𝑏 because it acts in the negative 𝑏 direction, but subsequent resolution of 𝑄𝑏 will take into account the appropriate directions in a straightforward manner. As shown in Fig. 3(b), we next use the 45◦ angle to resolve 𝑄𝑏 into the 𝑥 and 𝑦 components 𝑄𝑥 = 𝑄𝑏 sin 45◦ = 183.7 N and

−𝑄𝑧

𝑄 = 300 N

(7)

𝑧 −𝑄𝑧

𝑄 = 300 N

𝑏 30◦ 45◦

(b)

−𝑄𝑦

Collecting results for 𝑄𝑥 , 𝑄𝑦 , and 𝑄𝑧 allows us to write

𝑄𝑏

𝑦

𝑄𝑥

𝑥

⃗ = (183.7 𝚤̂ − 183.7 𝚥̂ − 150 𝑘) ̂ N. 𝑄

(8)

As a partial check of correctness, you should evaluate the magnitude of the above expression to verify it is 300 N. Part (b) Governing Equations & Computation

The resultant of the three forces is

⃗ = (50 𝚤̂ − 70.71 𝚥̂ + 50 𝑘) ̂ N 𝑅⃗ = 𝐹⃗ + 𝑃⃗ + 𝑄 ̂ N + (88.24 𝚤̂ + 152.8 𝚥̂ − 94.12 𝑘) ̂ N + (183.7 𝚤̂ − 183.7 𝚥̂ − 150 𝑘) ̂ N, = (321.9 𝚤̂ − 101.6 𝚥̂ − 194.1 𝑘) and the magnitude of the resultant is √ ⃗ = (321.9 N)2 + (−101.6 N)2 + (−194.1 N)2 = 389.4 N. |𝑅|

(9)

(10)

Part (c)

Once 𝑅⃗ is known, obtaining its direction an⃗ cos 𝜃 = 𝑅 ∕|𝑅|, ⃗ and gles is straightforward. Applying the equations cos 𝜃𝑥 = 𝑅𝑥 ∕|𝑅|, 𝑦 𝑦 ⃗ gives cos 𝜃 = 𝑅 ∕|𝑅|

Governing Equations & Computation

𝑧

𝑧

cos 𝜃𝑥 = 321.9 N∕389.4 N ⇒ 𝜃𝑥 = cos−1 (321.9∕389.4) = 34.24◦ ,

(11)

cos 𝜃𝑦 = −101.6 N∕389.4 N ⇒ 𝜃𝑦 = cos−1 (−101.6∕389.4) = 105.1◦ ,

(12)

cos 𝜃𝑧 = −194.1 N∕389.4 N ⇒ 𝜃𝑧 = cos−1 (−194.1∕389.4) = 119.9◦ .

(13)

As simple checks of accuracy, you should verify that 𝐹⃗ , 𝑃⃗ , ⃗ each point in the proper direction and have the correct magnitude. Depending on and 𝑄 the number of forces being added and their relative sizes and magnitudes, it may also be possible to verify that some (or all) of the components of 𝑅⃗ have reasonable direction and size. Discussion & Verification

Figure 3 (a) Resolution of force 𝑄 into 𝑧 and 𝑏 direction components. (b) Resolution of the 𝑏 component into 𝑥 and 𝑦 components.

76

E X A M P L E 2.13

Position Vectors and Force Vectors The collar 𝐶 slides on a straight bar 𝐴𝐵 and is supported by a cable attached to point 𝐷.

𝑧 120 mm

(a) Determine the coordinates of point 𝐶.

60 mm 𝐴

𝐷 240 mm

280 mm 𝑂

𝑥 Figure 1

ISTUDY

Chapter 2

Vectors: Force and Position

240 mm

SOLUTION 𝐶 𝑦 120 mm 𝐵

300 mm

(b) If the tensile force in the cable is known to be 150 N, write vector expressions using Cartesian representations for the force the cable exerts on point 𝐶 and for the force the cable exerts on point 𝐷.

To determine the coordinates of point 𝐶, our strategy will be to write the position vector from points 𝐴 to 𝐵, and then using this we will write the position vector from 𝐴 to 𝐶. Then the coordinates of point 𝐶 may be determined. With the coordinates of points 𝐶 and 𝐷 known, we will write the position vector 𝑟⃗𝐶𝐷 , and from this the desired force vectors 𝐹⃗𝐶𝐷 and 𝐹⃗𝐷𝐶 can be written. Road Map

Part (a)

To construct the position vector from 𝐴 to 𝐵, we imagine being positioned at point 𝐴 and asking what distances in the 𝑥, 𝑦, and 𝑧 directions must be traversed to arrive at point 𝐵. Referring to Fig. 1, we find these distances to be 120 mm, 300 mm − 60 mm = 240 mm, and −240 mm, respectively. Thus, the position vector and its magnitude are

Governing Equations & Computation

̂ mm, 𝑟⃗𝐴𝐵 = (120 𝚤̂ + 240 𝚥̂ − 240 𝑘) √ |⃗𝑟𝐴𝐵 | = (120)2 + (240)2 + (−240)2 mm = 360 mm.

(1) (2)

Alternatively, we could tabulate the coordinates of points 𝐵 and 𝐴 and apply Eq. (2.28) with 𝐵 and 𝐴 being the head and tail of the position vector, respectively, to arrive at the same result. Noting that position vectors 𝑟⃗𝐴𝐶 and 𝑟⃗𝐴𝐵 have the same direction, we may now write 𝑟⃗𝐴𝐶 = (240 mm)

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

= (240 mm)

120 𝚤̂ + 240 𝚥̂ − 240 𝑘̂ 360

̂ mm. = (80 𝚤̂ + 160 𝚥̂ − 160 𝑘)

(3)

Since 𝑟⃗𝐴𝐶 is known, we may use its components to evaluate the difference between the coordinates of the head and tail of 𝑟⃗𝐴𝐶 , where 𝐶 is the head and 𝐴 is the tail. The coordinates of point 𝐶 are (𝑥𝐶 , 𝑦𝐶 , 𝑧𝐶 ), and from Fig. 1 the coordinates of point 𝐴 are (0, 60, 240) mm, so we may write ̂ mm 𝑟⃗𝐴𝐶 = (80 𝚤̂ + 160 𝚥̂ − 160 𝑘) ̂ = (𝑥𝐶 − 0 mm) 𝚤̂ + (𝑦𝐶 − 60 mm) 𝚥̂ + (𝑧𝐶 − 240 mm) 𝑘.

(4)

Equality must be achieved independently for each of the 𝑥, 𝑦, and 𝑧 components; thus our solutions are 𝑥𝐶 = 80 mm,

(5)

𝑦𝐶 = 160 mm + 60 mm = 220 mm,

(6)

𝑧𝐶 = −160 mm + 240 mm = 80 mm.

(7)

Part (b)

Cable 𝐶𝐷 exerts a force 𝐹⃗𝐶𝐷 of magnitude 150 N on point 𝐶, in the direction from 𝐶 to 𝐷, as shown in Fig. 2. Using 𝐷 and 𝐶

Governing Equations & Computation

ISTUDY

Section 2.3

Cartesian Representation of Vectors in Three Dimensions 𝑧

as the head and tail, respectively, of a position vector where the coordinates of these are 𝐷(120, 0, 280) mm and 𝐶(80, 220, 80) mm, we write ̂ mm 𝑟⃗𝐶𝐷 = [(120 − 80) 𝚤̂ + (0 − 220) 𝚥̂ + (280 − 80) 𝑘] ̂ mm, = (40 𝚤̂ − 220 𝚥̂ + 200 𝑘) √ |⃗𝑟𝐶𝐷 | = (40)2 + (−220)2 + (200)2 mm = 300 mm.

𝐴

𝐷

(8) 𝐹⃗𝐷𝐶

(9)

𝐹⃗𝐶𝐷

The cable force exerted on 𝐶 is 𝐹⃗𝐶𝐷

𝑦 𝐵

(10)

The force that cable 𝐶𝐷 exerts on the support at 𝐷 is 𝐹⃗𝐷𝐶 and, according to Newton’s third law, it is of equal magnitude and opposite direction to the force the cable exerts on 𝐶. Thus, 𝐹⃗𝐷𝐶 = −𝐹⃗𝐶𝐷 and we obtain ̂ N. 𝐹⃗𝐷𝐶 = (−20 𝚤̂ + 110 𝚥̂ − 100 𝑘)

𝐶

𝑂

40 𝚤̂ − 220 𝚥̂ + 200 𝑘̂ = (150 N) = (150 N) 300 |⃗𝑟𝐶𝐷 | 𝑟⃗𝐶𝐷

̂ N. = (20 𝚤̂ − 110 𝚥̂ + 100 𝑘)

77

(11)

Discussion & Verification As quick, partial checks of solution accuracy, you should verify that 𝑟⃗𝐴𝐶 and 𝐹⃗𝐶𝐷 point in proper directions and have correct magnitudes.

𝑥 Figure 2 The force applied by the cable to the collar 𝐶 is 𝐹⃗𝐶𝐷 , and the force applied by the cable to the support at 𝐷 is 𝐹⃗𝐷𝐶 . Newton’s third law requires 𝐹⃗𝐷𝐶 = −𝐹⃗𝐶𝐷 .

78

Chapter 2

Vectors: Force and Position

Problems

ISTUDY

Problems 2.67 through 2.69 A vector 𝑟⃗ has the two direction angles given below. Determine the possible values for the third direction angle, and describe the differences in the orientation of 𝑟⃗. Provide a sketch showing these different vectors. Problem 2.67

𝜃𝑥 = 36◦ and 𝜃𝑦 = 72◦ .

Problem 2.68

𝜃𝑦 = 60◦ and 𝜃𝑧 = 108◦ .

Problem 2.69

𝜃𝑥 = 150◦ and 𝜃𝑧 = 100◦ .

Problem 2.70 Consider a vector 𝑛⃗ that makes equal angles with the 𝑥, 𝑦, and 𝑧 axes (i.e., 𝜃𝑥 = 𝜃𝑦 = 𝜃𝑧 ). Determine the value of these angles and the value of the direction cosines. Note: In the study of stresses in mechanics of materials, this direction 𝑛⃗ plays an important role and is the normal direction to a special surface called the octahedral plane.

Problem 2.71 A person of dubious technical ability tells you that two of the direction angles for a particular vector are 30◦ and 40◦ . If these direction angles are feasible, then determine the possible values for the remaining direction angle. If these direction angles are not feasible, explain why.

Problems 2.72 through 2.77 Write an expression for each force using Cartesian representation, and evaluate the resultant force vector. Sketch the resultant force in the 𝑥𝑦𝑧 coordinate system. 𝑧 𝐹 = 170 lb

𝐹 = 200 N

𝑄 = 60 lb 𝑃 = 200 lb

𝐹 = 30 lb

12

30◦ 20◦

5

𝑦 16 8 11

𝑥

72◦

𝑄 = 300 N

𝑃 = 20 N

20◦

𝑄 = 300 lb

𝑥

3 2

1

45◦

45◦ 60◦

𝑧

𝐹 = 3 kN

𝑦 𝑧

𝑦 Figure P2.75

𝐵 𝑥

𝑄 = 3 lb

Figure P2.76

𝐹 = 5 lb

9 in. 𝑂

3 in. 𝐴

𝑦

45◦

2 in.

60◦

𝑃 = 100 lb

Figure P2.74

50◦ 30◦

2

60◦

45◦

𝑥

𝑧

𝑃 = 2 kN 𝑄 = 4 kN

1 60◦

Figure P2.73

𝐹 = 10 N 6

𝑦

36◦

𝑧

30◦ 40◦ ◦ 60

2

◦ 60◦ 20

Figure P2.72

𝑥

𝑦

𝑧 9 8 12

6 in. 𝑃 = 2 lb

30◦

Figure P2.77

𝑥

ISTUDY

Section 2.3

79

Cartesian Representation of Vectors in Three Dimensions

Problems 2.78 and 2.79 For each vector, find the direction angles and write an expression for the vector using Cartesian representation. Evaluate the sum of the two vectors, and report the direction angles for the resultant. Also, sketch the resultant in the 𝑥𝑦𝑧 coordinate system. 𝑧

𝑦 𝑃 = 75 N

𝐹 = 25 N

𝑟

◦ 50◦ 75 45◦

𝑠 = 25 in.

15 in.

12 in.

𝑦

60◦

6 in.

𝑥

𝑥

9 in.

10 in. 𝑧

Figure P2.78

Figure P2.79

Problem 2.80

𝑧

A spacecraft at 𝑂 uses radar to determine the magnitudes and direction cosines of position vectors to satellites 𝐴 and 𝐵 as

𝐴

𝑂

𝑦

for 𝑟⃗𝑂𝐴∶ |⃗𝑟𝑂𝐴 | = 2 km, cos 𝜃𝑥 = 0.768, cos 𝜃𝑦 = −0.384, cos 𝜃𝑧 = 0.512, for 𝑟⃗𝑂𝐵∶ |⃗𝑟𝑂𝐵 | = 4 km, cos 𝜃𝑥 = 0.743, cos 𝜃𝑦 = 0.557,

𝐵

cos 𝜃𝑧 = −0.371.

Determine the distance between the satellites.

𝑥 Figure P2.80

Problem 2.81

𝑧

A cube of material with 1 mm edge lengths is examined by a scanning electron microscope, and a small inclusion (i.e., a cavity) is found at point 𝑃 . It is determined that the direction cosines for a vector from points 𝐴 to 𝑃 are cos 𝜃𝑥 = −0.485, and cos 𝜃𝑦 = 0.485, and cos 𝜃𝑧 = −0.728; and the direction cosines for a vector from points 𝐵 to 𝑃 are cos 𝜃𝑥 = −0.667, cos 𝜃𝑦 = −0.667, and cos 𝜃𝑧 = 0.333. Determine the coordinates of point 𝑃 .

1 mm 1 mm 𝐴 𝑃 1 mm

𝑥

A theodolite is an instrument that measures horizontal and vertical angular orientations of a line of sight. The line of sight may be established optically, or by laser, and many forms of theodolite are used in surveying, construction, astronomy, and manufacturing. A simple optical theodolite is shown. After the instrument is aligned so that its base is in a horizontal plane and a desired reference direction is selected (such as perhaps north in the case of a surveying instrument), the telescopic sight is used to establish a line of sight, and then horizontal and vertical angles 𝜃ℎ and 𝜃𝑣 are measured.

Figure P2.81 𝑧

and 𝜃𝑣 =

60◦ ,

use your answer to Part (a) to determine 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 .

line of sight 𝑦

𝜃𝑣

(a) If 𝜃ℎ and 𝜃𝑣 are known, derive an expression that gives the direction angles 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 for the line of sight. (b) If 𝜃ℎ =

𝑦 𝐵

Problem 2.82

30◦

𝑂

𝜃ℎ 𝑥

Figure P2.82

80

Chapter 2

Vectors: Force and Position

Problem 2.83 With direction angles, general practice is to use values of 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 between 0◦ and 180◦ , and this is sufficient to characterize the orientation of any vector in three dimensions. When direction angles are measured in this way, the sum of any two direction angles is always 90◦ or greater. Offer an argument that shows this statement is true.

𝑧

𝜃𝑣

Problem 2.84 If the direction cosines for the satellite antenna shown are to be cos 𝜃𝑥 = 0.286, cos 𝜃𝑦 = 0.429, and cos 𝜃𝑧 = 0.857, determine the values of angles 𝜃ℎ and 𝜃𝑣 .

𝜃ℎ

𝑥

𝑦

Problem 2.85

Figure P2.84

𝑧 𝑦 (north) 𝐵

𝐶

𝐴 Figure P2.85

ISTUDY

𝑥 (east)

Using a theodolite with a laser rangefinder, a researcher at point 𝐴 locates a pair of rare birds nesting at point 𝐵 and determines the direction angles 𝜃𝑥 = 53◦ , 𝜃𝑦 = 38◦ , and 𝜃𝑧 = 81◦ and distance 242 m for a position vector from point 𝐴 to 𝐵. For subsequent observations, the researcher would like to use position 𝐶, and thus while at point 𝐴, she also measures the direction angles 𝜃𝑥 = 142◦ , 𝜃𝑦 = 63◦ , and 𝜃𝑧 = 65◦ and distance 309 m for a position vector from point 𝐴 to 𝐶. For observation from point 𝐶, determine the direction angles and distance to the nest at 𝐵.

Problem 2.86 A robot maneuvers itself by using a laser guidance system where the positions of reflective targets, such as at points 𝐵 and 𝐶, are known. By scanning the room with a laser that emanates from point 𝐴, and sensing the reflections from targets 𝐵 and 𝐶, the robot may compute the direction cosines of position vectors as given below. The locations of other points, such as point 𝐷 at the base of the ramp, are also known, but such locations do not have reflective targets. Determine the position vector from points 𝐴 to 𝐷 so that the robot may move from its current position to the base of the ramp at point 𝐷.

for 𝑟⃗𝐴𝐵∶

cos 𝜃𝑥 = −0.4212, cos 𝜃𝑦 = −0.9025, cos 𝜃𝑧 = 0.0902,

for 𝑟⃗𝐴𝐶 ∶

cos 𝜃𝑥 = 0.2676,

cos 𝜃𝑦 = −0.9635, cos 𝜃𝑧 = 0.

𝑧 𝐵 (0, 0, 120) cm

𝐶 (240, −60, 90) cm 𝐴

𝑥

𝐷 (60, 150, 0) cm

Figure P2.86

𝑦

ISTUDY

Section 2.3

81

Cartesian Representation of Vectors in Three Dimensions

Problem 2.87 A mountain climbing team establishes base camps at points 𝐴 and 𝐵, and the next camp is to be at point 𝐶. Camps 𝐴 and 𝐵 have elevations of 11,500 f t and 16,000 f t above sea level, respectively. With an optical telescope, the direction angles for position vectors from 𝐴 to 𝐵, and 𝐵 to 𝐶, and 𝐴 to 𝐶 are determined as follows: for 𝑟⃗𝐴𝐵∶

𝜃𝑥 = 34.3◦ ,

𝜃𝑦 = 63.2◦ ,

𝜃𝑧 = 70.2◦ ,

for 𝑟⃗𝐵𝐶 ∶ 𝜃𝑥 = 104.7 , 𝜃𝑦 = 27.3 ,

𝜃𝑧 = 67.6 ,

◦

for 𝑟⃗𝐴𝐶 ∶

◦

𝜃𝑥 = 59.1 ,

𝜃𝑦 = 42.0 ,

◦

◦

◦

𝑧 𝐶 𝐵

𝑦

𝑥

𝜃𝑧 = 64.6 . ◦

Determine the elevation above sea level of camp 𝐶.

𝐴 Figure P2.87

Problems 2.88 and 2.89 Bars 𝐴𝐶 and 𝐷𝐺 are straight and parallel to the 𝑥 and 𝑦 axes, respectively. 𝐵𝐸 is a cable whose tensile force is 100 lb. For the dimensions given, determine expressions for the force the cable exerts on 𝐵 and the force the cable exerts on 𝐸 using Cartesian vector representation. Problem 2.88

𝑥 = 4 in. and 𝑦 = 7 in.

Problem 2.89

𝑥 = 6 in. and 𝑦 = 12 in. 𝑧 𝑥 𝐴 𝐶

4 in.

𝐵 𝐷

𝑦 𝐸

𝑥

𝐺 𝑦

Figure P2.88 and P2.89

Problems 2.90 through 2.93

𝑦

A crane consists of a quarter-circular bar that lies in a plane parallel to the 𝑥𝑧 plane with a low-friction collar at point 𝐵 that may slide on the bar. The forces supported by cables 𝐴𝐵 and 𝐵𝐶 are 30 lb and 1500 lb, respectively, and the coordinates of point 𝐴 are given in the figure. For the value of angle 𝛼 given below, determine expressions for the forces the two cables apply to the collar at point 𝐵. Problem 2.90

𝛼 = 0◦ .

Problem 2.91

𝛼 = 30◦ .

Problem 2.92

𝛼 = 60◦ .

𝛼 𝐵

10 f t

𝐶

𝐴

10 f t

𝐴 (8, 4.5, 2) f t 𝑧

Problem 2.93

5 ft

General values of 𝛼 where 0◦ ≤ 𝛼 ≤ 90◦ .

Figure P2.90–P2.93

𝑥

82

Chapter 2

Vectors: Force and Position 𝑦

Problems 2.94 and 2.95 𝐵

𝑧

150 mm 𝑏 𝑂 ℎ

𝐷 𝑥 𝐴

𝐺 𝐸 𝑑

360 mm 𝐶 𝑧

The structure consists of a quarter-circular rod 𝐴𝐵 with radius 150 mm that is fixed in the 𝑥𝑦 plane. 𝐶𝐷 is a straight rod where 𝐷 may be positioned at different locations on the circular rod. 𝐺𝐸 is an elastic cord whose support at 𝐺 lies in the 𝑦𝑧 plane, and the bead at 𝐸 may have different positions 𝑑. For the values of 𝑏, ℎ, 𝑑, and 𝑧 provided, determine the coordinates of 𝐸, and if the elastic cord supports a tensile force of 100 N, write a vector expression using Cartesian vector representation for the force 𝐹⃗𝐸𝐺 the cord exerts on bead 𝐸. Problem 2.94

𝑏 = 4, ℎ = 3, 𝑑 = 260 mm, and 𝑧 = 240 mm.

Problem 2.95

𝑏 = 3, ℎ = 4, 𝑑 = 195 mm, and 𝑧 = 270 mm.

Problems 2.96 through 2.99

Figure P2.94 and P2.95

A wall-mounted jib crane consists of an I beam that is supported by a pin at point 𝐴 and a cable at point 𝐶, where 𝐴 and 𝐶 lie in the 𝑥𝑦 plane. A crate at 𝐸 is supported by a cable that is attached to a trolley at point 𝐵 where the trolley may move along the length of the I beam. The forces supported by cables 𝐶𝐷 and 𝐵𝐸 are 3 kN and 5 kN, respectively. For the value of angle 𝛼 given below, determine expressions for the forces the two cables apply to the I beam.

𝑧 𝐷

6m 𝐴 𝛼

Problem 2.96

𝛼 = 0◦ .

Problem 2.97

𝛼 = 40◦ .

Problem 2.98

𝛼 = 70◦ .

Problem 2.99

General values of 𝛼 where −90◦ ≤ 𝛼 ≤ 90◦ .

𝑥 𝐶

𝐵

5m

3m 𝐸

𝑦

Problem 2.100 A coordinate system that is often used for problems in mechanics is the spherical coordinate system, as shown. With this coordinate system, the location of a point 𝑃 in three dimensions is specified using a radial distance 𝑟 where 𝑟 ≥ 0, an angle 𝜃 (sometimes called the azimuthal angle) where 0◦ ≤ 𝜃 ≤ 360◦ , and an angle 𝜙 (sometimes called the polar angle) where 0◦ ≤ 𝜙 ≤ 180◦ .∗ If values for 𝑟, 𝜃, and 𝜙 are known, determine the direction angles for the position vector from the origin of the coordinate system to point 𝑃 .

Figure P2.96–P2.99

ISTUDY

𝑧 𝑃 𝜙

𝑟

𝑂 𝜃 𝑥 Figure P2.100 and P2.101

∗ Other

ranges of values are also commonly used for angles 𝜃 and 𝜙.

𝑦

ISTUDY

Section 2.3

Cartesian Representation of Vectors in Three Dimensions

Problem 2.101 For the spherical coordinate system described in Prob. 2.100, if the direction angles for the position vector from the origin of the coordinate system to point 𝑃 are known, determine the values for 𝑟, 𝜃, and 𝜙. Hint: Several approaches may be used to solve this problem, but a straightforward solution is to first solve Prob. 2.100, and then use those results to solve this problem.

Problem 2.102 The rear wheel of a multispeed bicycle is shown. The wheel has 32 spokes, with one-half being on either side (spokes on sides 𝐴 and 𝐵 are shown in the figure in black and red, respectively). For the tire to be properly centered on the frame of the bicycle, points 𝐴 and 𝐵 of the hub must be positioned at the same distance 𝑑 from the centerline of the tire. To make room for the sprocket cluster, bicycle manufacturers give spokes on side 𝐵 of the wheel a different orientation than spokes on side 𝐴. For the following questions, assume the tire is in the process of being manufactured, so that all spokes on side 𝐴 have the same force 𝐹𝐴 and all spokes on side 𝐵 have the same force 𝐹𝐵 . (a) Determine the ratio of spoke forces 𝐹𝐵 ∕𝐹𝐴 so that the resultant force in the 𝑥 direction applied to the hub by all 32 spokes is zero. Hint: Although each spoke has a different orientation, all spokes on side 𝐴 have the same length, and similarly all spokes on side 𝐵 have the same length. Furthermore, all spokes on side 𝐴 have the same 𝑥 component of force, and all spokes on side 𝐵 have the same 𝑥 component of force. (b) On which side of the wheel are the spokes more severely loaded? (c) Briefly describe a new design in which spokes on both sides of the wheel are equally loaded and points 𝐴 and 𝐵 are at the same distance 𝑑 from the centerline of the tire.

𝑧

𝑧 tire wheel rim

spoke

26.5

sprocket cluster

27

hub hub

Erik Isakson/Rubberball/Getty Images

𝑦 2.5 geometry of typical spoke all dimensions in cm side 𝐴 of wheel Figure P2.102

𝐵

𝐴 spokes 3.8 2.5 𝑑 𝑑

𝑥

83

84

Chapter 2

Vectors: Force and Position

2.4

Vector Dot Product

The dot product between two vectors may be used to (1) determine the angle between the lines of action for the two vectors and (2) determine the component of one vector that acts in the direction of the other. Throughout statics and subjects that follow we will often need to find such information, and the dot product provides a straightforward way to do this, especially for vectors in three dimensions. The dot product between two vectors 𝐴⃗ and 𝐵⃗ is an operation defined as 𝑦

⃗ 𝐵| ⃗ cos 𝜃 𝐴⃗ ⋅ 𝐵⃗ = |𝐴||

𝑦 𝐵⃗

𝐵⃗

(2.29)

where: 𝜃

𝜃 𝑥

𝑥 (b)

(a)

Figure 2.26 (a) Vectors 𝐴⃗ and 𝐵⃗ in two dimensions. The intersection of the lines of action of the two vectors defines angle 𝜃 for evaluation of the dot product. (b) Often we imagine sliding the vectors tail to tail for evaluation of the dot product. 𝑦

𝑦 𝐵⃗ 𝐵⃗ 𝐴⃗

𝑧

(a)

𝜃 𝐴⃗ 𝑥

𝑥 𝑧

(b)

Figure 2.27 (a) Vectors 𝐴⃗ and 𝐵⃗ in three dimensions. (b) To evaluate the dot product, the vectors are arranged tail to tail to define angle 𝜃, which is measured in the plane containing the two vectors.

ISTUDY

𝜃 is the angle between the lines of action of 𝐴⃗ and 𝐵⃗ where∗ 0 ≤ 𝜃 ≤ 180◦ .

𝐴⃗

𝐴⃗

⃗ In words, Eq. (2.29) states “𝐴⃗ dot 𝐵⃗ ” equals the product of the magnitude of 𝐴, ⃗ and cosine of angle 𝜃 between the lines of action of 𝐴⃗ and 𝐵. ⃗ The magnitude of 𝐵, dot product yields a result that is a scalar, and thus the dot product is sometimes called the scalar product. The units for the dot product are equal to the product of ⃗ Figure 2.26 illustrates the dot product between vectors 𝐴⃗ and the units for 𝐴⃗ and 𝐵. 𝐵⃗ in two dimensions. The intersection of the vectors’ lines of action defines angle 𝜃. Regardless of where the two vectors are located in space, and regardless of whether their lines of action intersect, to perform (or interpret) the dot product, it is helpful if we imagine sliding the vectors tail to tail, which again identifies angle 𝜃. Figure 2.27 illustrates the dot product between vectors 𝐴⃗ and 𝐵⃗ in three dimensions. Note that 𝐴⃗ and 𝐵⃗ do not need to lie in the same plane to compute the dot product between them. After we arrange the vectors tail to tail, they define a plane, and the intersection of their lines of action defines angle 𝜃, which is measured in this plane. Depending on the value of 𝜃, the scalar produced by the dot product may be positive, zero, or negative. For example, 0 ≤ 𝜃 < 90◦ (acute angle) gives a dot product that is positive, 𝜃 = 90◦ (orthogonal or perpendicular angle) gives a dot product that is zero, and 90◦ < 𝜃 ≤ 180◦ (obtuse angle) gives a dot product that is negative. The dot product has the following properties: 𝐴⃗ ⋅ 𝐵⃗ = 𝐵⃗ ⋅ 𝐴⃗ ⃗ = (𝑠𝐴) ⃗ ⋅ 𝐵⃗ = 𝐴⃗ ⋅ (𝑠𝐵) ⃗ 𝑠(𝐴⃗ ⋅ 𝐵)

⃗ ⋅ 𝐶⃗ = (𝐴⃗ ⋅ 𝐶) ⃗ + (𝐵⃗ ⋅ 𝐶) ⃗ (𝐴⃗ + 𝐵)

commutative property,

(2.30)

associative property with respect to multiplication by a scalar,

(2.31)

distributive property with respect to vector addition.

(2.32)

The foregoing remarks are true regardless of the type of vector representation that is used. For example, consider the two position vectors shown in Fig. 2.28(a). We first slide the vectors tail to tail as shown in Fig. 2.28(b) and then identify the angle ∗ The

dot product, and cross product described in Section 2.5, can be defined to allow values of 𝜃 outside this range. However, the description of direction for the cross product becomes more intricate. Thus, with little loss of generality it is common to use 0 ≤ 𝜃 ≤ 180◦ for both definitions.

ISTUDY

Section 2.4

Vector Dot Product

between 𝐴⃗ and 𝐵⃗ as 60◦ . Applying Eq. (2.29) provides the dot product between 𝐴⃗ and 𝐵⃗ as 𝐴⃗ ⋅ 𝐵⃗ = (2 m)(3 m) cos 60◦ = 3 m2 . (2.33) Observe that the result is a scalar, with units of m2 . Although these are appropriate units for area, that does not necessarily mean that this result has the interpretation of being an area. The physical significance of the result of the dot product will be discussed in detail soon.

𝑦

85

𝑦

𝐵 = 3m

𝐵 = 3m 75◦ 𝐴 = 2m 15◦

60◦ 𝐴 = 2m 𝑥

𝑥

(a)

(b)

Figure 2.28 Dot product between two vectors.

Dot product using Cartesian components When vectors 𝐴⃗ and 𝐵⃗ have Cartesian representations, the dot product between them may be evaluated using the following convenient computation: 𝐴⃗ ⋅ 𝐵⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 .

(2.34)

Simply stated, for two vectors with Cartesian representation, the dot product between them is the sum of the products of their components. Equation (2.34) can be obtained ⃗ using Cartesian representation: using the following derivation. We first write 𝐴⃗ and 𝐵, ̂ 𝐴⃗ = 𝐴𝑥 𝚤̂ + 𝐴𝑦 𝚥̂ + 𝐴𝑧 𝑘,

(2.35)

̂ 𝐵⃗ = 𝐵𝑥 𝚤̂ + 𝐵𝑦 𝚥̂ + 𝐵𝑧 𝑘.

(2.36)

⃗ using the distributive law of Eq. (2.32) We then take the dot product between 𝐴⃗ and 𝐵, to expand the product term by term, as follows: ̂ ⋅ (𝐵 𝚤̂ + 𝐵 𝚥̂ + 𝐵 𝑘) ̂ 𝐴⃗ ⋅ 𝐵⃗ = (𝐴𝑥 𝚤̂ + 𝐴𝑦 𝚥̂ + 𝐴𝑧 𝑘) 𝑥 𝑦 𝑧 ̂ = (𝐴𝑥 𝚤̂ ⋅ 𝐵𝑥 𝚤̂) + (𝐴𝑥 𝚤̂ ⋅ 𝐵𝑦 𝚥̂) + (𝐴𝑥 𝚤̂ ⋅ 𝐵𝑧 𝑘) ̂ + (𝐴𝑦 𝚥̂ ⋅ 𝐵𝑥 𝚤̂) + (𝐴𝑦 𝚥̂ ⋅ 𝐵𝑦 𝚥̂) + (𝐴𝑦 𝚥̂ ⋅ 𝐵𝑧 𝑘) ̂ + (𝐴𝑧 𝑘̂ ⋅ 𝐵𝑥 𝚤̂) + (𝐴𝑧 𝑘̂ ⋅ 𝐵𝑦 𝚥̂) + (𝐴𝑧 𝑘̂ ⋅ 𝐵𝑧 𝑘).

(2.37)

To continue we must address dot products between combinations of unit vectors. Equation (2.29) shows 𝚤̂ ⋅ 𝚤̂ = (1)(1) cos 0◦ = 1, and similarly 𝚥̂ ⋅ 𝚥̂ = 1 and 𝑘̂ ⋅ 𝑘̂ = 1. Further, the dot product between combinations of different unit vectors is always zero because the vectors are orthogonal. For example, 𝚤̂ ⋅ 𝚥̂ = (1)(1) cos 90◦ = 0. In summary, dot products between combinations of unit vectors are 𝚤̂ ⋅ 𝚤̂ = 1

𝚤̂ ⋅ 𝚥̂ = 0

𝚥̂ ⋅ 𝚤̂ = 0 𝑘̂ ⋅ 𝚤̂ = 0

𝚥̂ ⋅ 𝚥̂ = 1 𝑘̂ ⋅ 𝚥̂ = 0

𝚤̂ ⋅ 𝑘̂ = 0 𝚥̂ ⋅ 𝑘̂ = 0

(2.38)

𝑘̂ ⋅ 𝑘̂ = 1.

Substituting Eqs. (2.38) into Eq. (2.37) then provides Eq. (2.34).

Determination of the angle between two vectors Given two vectors 𝐴⃗ and 𝐵⃗ with Cartesian representation, we use the dot product given by Eq. (2.34) to provide the left-hand side of Eq. (2.29), from which the angle 𝜃 between the two vectors is then solved for. This provides 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 𝐴⃗ ⋅ 𝐵⃗ = and ⃗ 𝐵| ⃗ ⃗ 𝐵| ⃗ |𝐴|| |𝐴|| 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 𝐴⃗ ⋅ 𝐵⃗ 𝜃 = cos−1 = cos−1 . ⃗ 𝐵| ⃗ ⃗ 𝐵| ⃗ |𝐴|| |𝐴|| cos 𝜃 =

(2.39)

Helpful Information Alternatives for evaluation of the dot product. Regardless of the representation used to express vectors, Eq. (2.29) can always be used to evaluate the dot product. For vectors with Cartesian representation, either Eq. (2.29) or Eq. (2.34) can be used, with the latter usually being more convenient. In fact, sometimes both methods are used in combination to help determine useful information, such as the angle between two vectors, as discussed later in this section in connection with Eq. (2.39).

86

Chapter 2

Vectors: Force and Position

Determination of the component of a vector in a particular direction Concept Alert Dot product. In statics, the most important and frequent use of the dot product is for determining the component, or amount, of a vector that acts in a particular direction. It is essential that the direction of interest be described by a unit vector [i.e., in Eq. (2.40), the dot product of 𝐹⃗ must be taken with a unit vector].

Often, we must determine the component of a vector 𝐴⃗ that acts in the direction of another vector. In other words, we must determine how much, or what portion, of 𝐴⃗ acts in a particular direction. If that direction happens to be the same as the positive 𝑥, 𝑦, or 𝑧 direction, then the answer is easy: the portion of 𝐴⃗ in the direction we ⃗ respectively. However, we are interested in is simply the 𝑥, 𝑦, or 𝑧 component of 𝐴, are often not so fortunate, and the direction in which we must find the component of a vector has complicated orientation, and this leads to perhaps the most useful application of the dot product. A common situation is the determination of the component, or the amount, of a force vector that acts in a particular direction. Thus, consider a force vector 𝐹⃗ and a direction 𝑟⃗. The parallel component of 𝐹⃗ , or in other words, the amount of 𝐹⃗ that acts in direction 𝑟⃗, is denoted by 𝐹‖ , where the subscript ‖ means “parallel to 𝑟⃗ ,” and is given by∗ 𝑟⃗ . 𝐹‖ = 𝐹⃗ ⋅ (2.40) |⃗𝑟| In the above expression, 𝐹⃗ has the usual physical significance: it is a force and has appropriate force units, such as lb or N. The physical significance of 𝑟⃗ is not important, other than it defines a direction of interest. Often 𝑟⃗ will be a position vector, but it could have another physical meaning. Notice in Eq. (2.40) that regardless of the physical significance of 𝑟⃗, it is converted into a dimensionless unit vector. Thus, the only purpose it serves is to specify direction. 𝐹‖ is a scalar and may be positive, zero, or negative: a positive value indicates 𝐹‖ is in the same direction as 𝑟⃗, a zero value indicates 𝐹⃗ is orthogonal (or perpendicular) to 𝑟⃗, and a negative value indicates 𝐹 ‖

𝑦

𝑦 𝐹⃗

𝐹⃗ 𝜃

𝑟

𝐹‖ = 𝐹⃗ cos 𝜃

𝑟⃗ 𝑥 (a)

𝑥 (b)

Figure 2.29 Use of basic trigonometry to determine the component of 𝐹⃗ acting in the direction of 𝑟⃗.

ISTUDY

is in the opposite direction to 𝑟⃗. Although most of our applications of Eq. (2.40) will be for finding the component of a force in a particular direction, the idea is directly applicable to vectors with other physical interpretation. To show why Eq. (2.40) accomplishes the task of finding how much of a vector acts in a particular direction, we use the following argument, the beginning of which has nothing to do with the dot product. Consider the two-dimensional example shown in Fig. 2.29(a), where we wish to determine the component of 𝐹⃗ that acts in the direction 𝑟⃗. We arrange 𝐹⃗ and 𝑟⃗ tail to tail, which leads to Fig. 2.29(b). Given angle 𝜃 between the two vectors, basic trigonometry gives us the component of 𝐹⃗ in the direction of 𝑟⃗ as 𝐹‖ = |𝐹⃗ | cos 𝜃. (2.41) Now consider the scalar 𝑠 produced by taking the dot product between 𝐹⃗ and 𝑟⃗ according to Eq. (2.29) 𝑠 = 𝐹⃗ ⋅ 𝑟⃗ = |𝐹⃗ ||⃗𝑟| cos 𝜃. (2.42) For 𝑠 in Eq. (2.42) to be the same as 𝐹‖ in Eq. (2.41), and hence for the dot product to be capable of telling us the portion of 𝐹⃗ that acts in the direction 𝑟⃗, it is necessary that 𝑟⃗ be a unit vector so that |⃗𝑟| = 1 in Eq. (2.42). To elaborate further, if a unit vector is not used in Eq. (2.40), the value 𝑠 produced by the dot product has inappropriate than use subscript ‖ to denote “parallel to 𝑟⃗,” we will occasionally use subscript 𝑡, where this letter means “tangent to 𝑟⃗.” Similarly, rather than using the subscript ⊥ to denote “perpendicular to 𝑟⃗” (discussed later in this section), we will occasionally use subscript 𝑛, where this letter means “normal to 𝑟⃗.”

∗ Rather

ISTUDY

Section 2.4

Vector Dot Product

units and uncertain physical interpretation. To summarize, if we want the dot product to tell us how much of a vector acts in a particular direction, then the direction must be described by a unit vector.

Determination of the component of a vector perpendicular to a direction Once we have determined the component of a vector parallel to a particular direction 𝑟⃗, it is straightforward to determine the component of the vector that is perpendicular, or orthogonal, to 𝑟⃗, and we call this the perpendicular component of the vector. Consider the resolution of 𝐹⃗ into the parallel and perpendicular components shown in Fig. 2.30(a) and (b). The Pythagorean theorem may be used to write 𝑦

𝑦 𝐹⃗

𝑦 𝐹

𝑥 𝑧

(a)

(b)

𝑧

(c)

Figure 2.30. (a) Vectors 𝐹⃗ and 𝑟⃗ in three dimensions. (b) Resolution of 𝐹⃗ into scalar components in directions parallel and perpendicular to 𝑟⃗. (c) Resolution of 𝐹⃗ into vector components in directions parallel and perpendicular to 𝑟⃗.

𝐹 2 = 𝐹‖2 + 𝐹⊥2 .

(2.43)

Once 𝐹‖ is known from Eq. (2.40), the perpendicular component may be obtained from the above equation as √ 𝐹⊥ = 𝐹 2 − 𝐹‖2 . (2.44) Thus, the magnitude of the perpendicular component is easily determined, but its direction is not yet known. For many problems, however, this direction might not be needed. If the direction for the perpendicular component is needed, then consider the resolution of 𝐹⃗ into the parallel and perpendicular vector components shown in Fig. 2.30(c). Vector addition provides 𝐹⃗ = 𝐹⃗‖ + 𝐹⃗⊥ .

(2.45)

If 𝐹‖ is known, then 𝐹⃗‖ = 𝐹‖ (⃗𝑟∕|⃗𝑟|), and Eq. (2.45) may be rearranged to obtain 𝐹⃗⊥ = 𝐹⃗ − 𝐹⃗‖ .

Not using a unit vector. When we use the dot product to determine the component, or amount, of a vector that acts in a particular direction, the most common error is to not use a unit vector. For example, in Eq. (2.40), an incorrect result is produced if 𝐹⃗ ⋅ 𝑟⃗ is evaluated.

Line of action for a force vector. When we analyze equilibrium of objects, the position of the line of action for each force vector is important. While we may think of repositioning the lines of action of vectors for purposes of evaluating the dot product, equilibrium of objects depends on the actual line of action that each force has.

𝑥

𝑥 𝑧

𝐹⃗⊥

𝐹⃗‖

𝐹‖

𝑟⃗

Common Pitfall

Helpful Information 𝐹⃗

𝐹⊥

87

(2.46)

Important equations for using the dot product to resolve a vector 𝐹⃗ into components parallel and perpendicular to a direction 𝑟⃗ are collected in the following summary box:

88

ISTUDY

Chapter 2

Vectors: Force and Position

Summary Box

(See Fig. 2.30.)

𝐹‖ = component of 𝐹⃗ acting in direction 𝑟⃗, 𝐹⃗‖ = vector component of 𝐹⃗ acting in direction 𝑟⃗, 𝐹⊥ = component of 𝐹⃗ acting perpendicular to 𝑟⃗, 𝐹⃗⊥ = vector component of 𝐹⃗ acting perpendicular to 𝑟⃗, 𝑟⃗ 𝐹‖ = 𝐹⃗ ⋅ , |⃗𝑟| √ 𝐹⊥ = 𝐹 2 − 𝐹‖2 ,

(2.47)

𝑟⃗ 𝐹⃗‖ = 𝐹‖ , |⃗𝑟| 𝐹⃗⊥ = 𝐹⃗ − 𝐹⃗‖ .

End of Section Summary In this section, the dot product between two vectors has been defined. Some of the key points are as follows: • The dot product between two vectors 𝐴⃗ and 𝐵⃗ is a scalar 𝑠. The units of 𝑠 are ⃗ and 𝑠 may be positive, zero, or negative. the product of the units of 𝐴⃗ and 𝐵, • The dot product can be used to find the angle between the lines of action of two vectors. • The dot product can be used to determine the component (or amount) of one vector that acts in the direction of another vector 𝑟⃗. This component is often called the parallel component. In addition, the component of a vector perpendicular to the direction 𝑟⃗ , which we call the perpendicular component, can be determined.

ISTUDY

Section 2.4

Vector Dot Product

E X A M P L E 2.14

89

Angle Between Two Vectors

Bar 𝐴𝐵 supports two cables at its end. The position vector from 𝐴 to 𝐵 is 𝑟⃗𝐴𝐵 , the position vector from 𝐵 to some point along the length of cable 1 is 𝑟⃗1 , and the position vector from 𝐵 to some point along the length of cable 2 is 𝑟⃗2 . Determine the angles between bar 𝐴𝐵 and the cables.

𝑦 𝑟⃗1

𝐵

cable 1 𝑟⃗𝐴𝐵

SOLUTION

cable 2

Road Map

We will use the dot product, as given by Eq. (2.39), to evaluate the angles between the bar and cables.

𝑟⃗2 𝐴

Governing Equations & Computation

Evaluating the magnitudes of the vectors shown in Fig. 1 provides 𝑟𝐴𝐵 = 11 m, 𝑟1 = 7 m, and 𝑟2 = 13 m. The desired angles are then obtained from Eq. (2.39) as 𝜃1 = cos−1 = cos−1 = cos−1

𝜃2 = cos−1 = cos−1 = cos−1

𝑟⃗𝐴𝐵 ⋅ 𝑟⃗1 |⃗𝑟𝐴𝐵 | |⃗𝑟1 | (6 m)(6 m) + (9 m)(3 m) + (−2 m)(2 m) (11 m)(7 m) 59 ◦ = 39.98 , 77

𝑧 Figure 1

(1)

𝑟⃗𝐴𝐵 ⋅ 𝑟⃗2 |⃗𝑟𝐴𝐵 | |⃗𝑟2 | (6 m)(−4 m) + (9 m)(−12 m) + (−2 m)(3 m) (11 m)(13 m) −138 = 164.8◦ . 143

𝑦

𝑟⃗𝐴𝐵 𝜃1

𝑟⃗1

(2)

Discussion & Verification

𝑥 𝑧 𝑟⃗𝐴𝐵

• As shown in Fig. 2, and as expected, 𝜃1 is an acute angle and 𝜃2 is an obtuse angle.

𝑦

• This problem asks for the angles between the cables and the direction from 𝐴 to 𝐵, and these angles are given in Eqs. (1) and (2). The angles between the cables and the direction from 𝐵 to 𝐴 are different than these; and to see how they differ, and how they can be evaluated, we consider the angle between the direction from 𝐵 to 𝐴 and cable 2, using one of the following approaches: – Use the value of 𝜃2 found in Eq. (2) to write 180◦ − 164.8◦ = 15.2◦ . – Write an expression for 𝑟⃗𝐵𝐴 and evaluate 𝑟⃗𝐵𝐴 ⋅ 𝑟⃗2 when solving for the angle. – Because 𝑟⃗𝐵𝐴 = −⃗𝑟𝐴𝐵 , we could evaluate −⃗𝑟𝐴𝐵 ⋅ 𝑟⃗2 when solving for the angle. You may wish to verify one of the last two computations for yourself.

𝑥 ̂ m 𝑟⃗𝐴𝐵 = (6 𝚤̂ + 9 𝚥̂ − 2 𝑘) ̂ m 𝑟⃗1 = (6 𝚤̂ + 3 𝚥̂ + 2 𝑘) ̂ m 𝑟⃗2 = (−4 𝚤̂ − 12 𝚥̂ + 3 𝑘)

𝜃2 𝑟⃗2

𝑥

𝑧 Figure 2 The angle 𝜃1 between 𝑟⃗𝐴𝐵 and cable 1 (⃗𝑟1 ) is expected to be acute, and the angle 𝜃2 between 𝑟⃗𝐴𝐵 and cable 2 (⃗𝑟2 ) is expected to be obtuse.

90

Chapter 2

Vectors: Force and Position

E X A M P L E 2.15 𝑦

Angle Between Two Vectors A sheet metal channel is to be fabricated as follows. Part 1, after being cut to the dimensions shown in Fig. 1(b), is bent to a 90◦ angle along the fold line. Then part 2 is welded to the end of part 1. Determine the dimensions 𝑎, 𝑏, and 𝜃 of part 2 so that it fits the end of the channel.

part 1 after folding

𝑥

SOLUTION

part 2 𝑧

Road Map Using basic trigonometry, determination of edge lengths 𝑎 and 𝑏 for part 2 will be straightforward. The angle 𝜃 will be determined using the dot product once the position vectors for the two edges of part 1 (after folding) have been obtained.

(a) 2 in.

𝑎 70◦

Governing Equations & Computation Edge lengths 𝑎 and 𝑏 for part 2 can be determined using the geometry shown in Fig. 2(a):

𝜃

50◦ fold line 3 in.

𝑏

𝑎 sin 70◦ = 2 in. 𝑏 sin 50 = 3 in. ◦

part 1

part 2

(b)

(c)

𝑦 𝑎 70◦

𝐵 70◦

50◦

50◦

3 in.

𝑏

𝐶 (b)

Figure 2 Part 1 of the sheet metal channel shown before bending in (a) with edge lengths 𝑎 and 𝑏 defined, and after bending in (b).

ISTUDY

(1)

⇒

𝑏 = 3 in.∕ sin 50 = 3.916 in.

(2)

◦

𝑢̂ 𝐴𝐵 = − cos 70◦ 𝚤̂ + sin 70◦ 𝚥̂.

𝐴 𝑥

𝑧 (a)

𝑎 = 2 in.∕ sin 70◦ = 2.128 in.

Determination of 𝜃 is more challenging. To aid our discussion, points 𝐴, 𝐵, and 𝐶 on the end of the channel are labeled in Fig. 2(b). If we can determine vectors in the direction 𝐴 to 𝐵, and 𝐴 to 𝐶, then the dot product between these vectors will provide the angle 𝜃 needed for part 2. We may construct a position vector for edge 𝐴𝐵 as follows: beginning at point 𝐴, taking a step of cos 70◦ in the negative 𝑥 direction, followed by a step of sin 70◦ in the positive 𝑦 direction, gives a point that lies on the line connecting points 𝐴 and 𝐵. Calling this vector 𝑢̂ 𝐴𝐵 , we write

Figure 1

2 in.

⇒

(3)

Similarly, beginning at point 𝐴, a step of cos 50◦ in the negative 𝑥 direction, followed by a step of sin 50◦ in the positive 𝑧 direction, gives a point that lies on the line connecting points 𝐴 and 𝐶. Calling this vector 𝑢̂ 𝐴𝐶 , we write ̂ 𝑢̂ 𝐴𝐶 = − cos 50◦ 𝚤̂ + sin 50◦ 𝑘.

(4)

Note that 𝑢̂ 𝐴𝐵 and 𝑢̂ 𝐴𝐶 are unit vectors. Instead of these unit vectors, you could use the position vectors from point 𝐴 to 𝐵, 𝑟⃗𝐴𝐵 = (2.128 in.) 𝑢̂ 𝐴𝐵 , and from point 𝐴 to 𝐶, 𝑟⃗𝐴𝐶 = (3.916 in.) 𝑢̂ 𝐴𝐶 . Finally, we determine the angle 𝜃 needed for part 2 using Eq. (2.39) 𝜃 = cos−1 = cos−1

𝑢̂ 𝐴𝐵 ⋅ 𝑢̂ 𝐴𝐶 |𝑢̂ 𝐴𝐵 ||𝑢̂ 𝐴𝐶 | (− cos 70◦ )(− cos 50◦ ) + (sin 70◦ )(0) + (0)(sin 50◦ ) (1)(1)

= cos−1 0.2198 = 77.30◦ .

(5)

The values for 𝑎, 𝑏, and 𝜃 in Eqs. (1), (2), and (5) appear to be reasonable. That is, we expect 𝑎 to be somewhat larger than 2 in., 𝑏 to be somewhat larger than 3 in., and 𝜃 to be somewhat smaller than 90◦ .

Discussion & Verification

ISTUDY

Section 2.4

91

Vector Dot Product

E X A M P L E 2.16

Component of a Force in a Particular Direction

A tractor is stuck in mud, and to free it, a cable applying a force with magnitude 𝐹 and direction 𝑟⃗1 is attached to the front of the tractor as shown. The operator estimates that a 400 lb force applied in the direction of the tractor’s chassis, which is 𝑟⃗2 , will be sufficient to free it. Determine the cable force 𝐹 that should be applied and the component of this force perpendicular to the direction of the tractor’s chassis. Note: This problem is similar to Example 2.4, except here the force and position vectors are not in a horizontal plane.

𝑧

𝑥

1

𝑟⃗2

SOLUTION Road Map The direction of the cable force is known, but its magnitude is unknown. Thus, 𝐹⃗ = 𝐹 (⃗𝑟1 ∕|⃗𝑟1 |) where 𝐹 is to be determined. We will use the dot product to determine the component of the cable force that acts in the direction of the tractor, and according to the problem statement, this must equal 400 lb.

The dot product between the cable force 𝐹⃗ and a unit vector in the direction of the tractor’s chassis must equal 400 lb:

Governing Equations & Computation

𝐹‖ = 𝐹⃗ ⋅

𝑟⃗2 |⃗𝑟2 |

=𝐹

𝑟⃗1

𝑟⃗2 ⋅

|⃗𝑟1 | |⃗𝑟2 |

̂ ft (−2 𝚤̂ + 4 𝚥̂ − 𝑘) ̂ ft (200 𝚤̂ + 600 𝚥̂ + 150 𝑘) ⋅ √ 650 ft 21 ft 1850 =𝐹 √ 650 21 = 𝐹 (0.6211).

400 lb = 𝐹

(1)

Solving the preceding equation provides the cable force 𝐹 = 644.0 lb.

(2)

Determination of only the magnitude of the perpendicular component of force is straightforward using the Pythagorean theorem, Eq. (2.44): √ √ (3) 𝐹⊥ = 𝐹 2 − 𝐹‖2 = (644.0 lb)2 − (400 lb)2 = 504.8 lb. If the direction of the perpendicular component is also desired, then we may use vector addition to obtain it, as expressed by Eq. (2.46): 𝐹⃗⊥ = 𝐹⃗ − 𝐹⃗‖ = 𝐹 = (644.0 lb)

𝑟⃗1 |⃗𝑟1 |

− 𝐹‖

𝑟⃗2 |⃗𝑟2 |

̂ ft ̂ ft (−2 𝚤̂ + 4 𝚥̂ − 𝑘) (200 𝚤̂ + 600 𝚥̂ + 150 𝑘) − (400 lb) √ 650 f t 21 ft

̂ lb. = (372.7 𝚤̂ + 245.3 𝚥̂ + 235.9 𝑘) Discussion √ & Verification

(4)

As a partial check of our solution, the magnitude of Eq. (4) is |𝐹⃗⊥ | = (372.7)2 + (245.3)2 + (235.9)2 lb = 505 lb, which agrees with the value found in Eq. (3).

̂ ft 𝑟⃗1 = (200 𝚤̂ + 600 𝚥̂ + 150 𝑘) ̂ ft 𝑟⃗2 = (−2 𝚤̂ + 4 𝚥̂ − 𝑘) 𝑦 cable 𝑟⃗

Figure 1

92

Chapter 2

Vectors: Force and Position

E X A M P L E 2.17

Component of a Force in a Particular Direction Rod 𝐴𝐵 is straight and has a bead at 𝐶. An elastic cord having a 3 lb tensile force is attached between the bead and a support at 𝐷.

𝑦 12 in. 𝐷

11 in.

18 in. 8 in.

(a) Determine the components of the cord force in directions parallel and perpendicular to rod 𝐴𝐵.

𝐴

(b) Determine the vector components of the cord force in directions parallel and perpendicular to rod 𝐴𝐵.

8 in.

𝐶 𝑂

𝑥

𝐵

(c) If the bead at 𝐶 is free to slide on rod 𝐴𝐵 and is released from rest, will the cord force tend to make the bead slide toward 𝐴 or 𝐵?

SOLUTION

4 in.

Road Map

After writing expressions for position and force vectors, we will use the dot product to determine the component of the cord force that acts in the direction of the rod. It will then be straightforward to use the Pythagorean theorem to determine the component of the cord force perpendicular to the rod. Both of these results are scalars. In Part (b), we will write vector expressions for the forces parallel and perpendicular to the rod. Consideration of the sign of the parallel component of the force, as provided by the dot product, will determine the direction along the bar in which the bead will tend to slide.

𝑧 Figure 1

Part (a)

By imagining being positioned at 𝐴, the distances to be traversed in the 𝑥, 𝑦, and 𝑧 directions to arrive at 𝐵 may be determined by inspection of Fig. 1, allowing us to write the position vector from 𝐴 to 𝐵, and its magnitude, as

Governing Equations & Computation

̂ in., 𝑟⃗𝐴𝐵 = (−12 𝚤̂ − 4 𝚥̂ + 18 𝑘) √ |⃗𝑟𝐴𝐵 | = (−12)2 + (−4)2 + (18)2 in. = 22 in.

(1) (2)

The position vector from 𝐴 to 𝐶 has 11 in. magnitude and has the same direction as 𝑟⃗𝐴𝐵 , thus, 𝑦

𝑟⃗𝐴𝐶 = (11 in.)

𝐷 𝐹𝐶𝐷 𝐹⊥

𝐴

𝐶 𝑂

𝐹‖ 𝑥

𝐵

𝑧 Figure 2 The force supported by the cord has component 𝐹𝐶𝐷 in the direction shown. The component of the cord force in the direction from 𝐴 to 𝐵 is denoted by 𝐹‖ , and the component of the cord force perpendicular to the direction from 𝐴 to 𝐵 is denoted by 𝐹⊥ .

ISTUDY

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

= (11 in.)

−12 𝚤̂ − 4 𝚥̂ + 18 𝑘̂ 22

̂ in. = (−6 𝚤̂ − 2 𝚥̂ + 9 𝑘)

(3)

To write a vector expression for the cord force from 𝐶 to 𝐷, we will first obtain a position vector from 𝐶 to 𝐷, 𝑟⃗𝐶𝐷 . This can be easily accomplished by computing the coordinates of 𝐶 and then taking the difference between the coordinates of 𝐷 and 𝐶 as the head and tail, respectively, of 𝑟⃗𝐶𝐷 . Alternatively, we will evaluate 𝑟⃗𝐶𝐷 using the vector addition 𝑟⃗𝐶𝐷 = 𝑟⃗𝐶𝐴 + 𝑟⃗𝐴𝐷 = −⃗𝑟𝐴𝐶 + 𝑟⃗𝐴𝐷 ̂ in. + (−12 𝚤̂) in. = −(−6 𝚤̂ − 2 𝚥̂ + 9 𝑘) ̂ in., = (−6 𝚤̂ + 2 𝚥̂ − 9 𝑘) √ |⃗𝑟𝐶𝐷 | = (−6)2 + (2)2 + (−9)2 in. = 11 in.

(4) (5)

Thus, the cord force, which is shown in Fig. 2, is 𝐹⃗𝐶𝐷 = (3 lb)

𝑟⃗𝐶𝐷 |⃗𝑟𝐶𝐷 |

= (3 lb)

−6 𝚤̂ + 2 𝚥̂ − 9 𝑘̂ 11

̂ lb. = (−1.636 𝚤̂ + 0.5455 𝚥̂ − 2.455 𝑘)

(6)

ISTUDY

Section 2.4

Vector Dot Product

93

We denote the component of the cord force 𝐹⃗𝐶𝐷 in direction 𝐴 to 𝐵 as 𝐹‖ , and this is given by the dot product of 𝐹⃗𝐶𝐷 with a unit vector in direction 𝐴 to 𝐵 as follows: 𝐹‖ = 𝐹⃗𝐶𝐷 ⋅ = (3 lb)

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

−6 𝚤̂ + 2 𝚥̂ − 9 𝑘̂ −12 𝚤̂ − 4 𝚥̂ + 18 𝑘̂ ⋅ = −1.215 lb. 11 22

(7)

Note that instead of 𝑟⃗𝐴𝐵 ∕|⃗𝑟𝐴𝐵 | in Eq. (7), we could have used 𝑟⃗𝐴𝐶 ∕|⃗𝑟𝐴𝐶 |. To determine the magnitude of the perpendicular component of the cord force, the Pythagorean theorem, Eq. (2.44), may be applied to write √ √ 2 𝐹⊥ = 𝐹𝐶𝐷 − 𝐹‖2 = (3 lb)2 − (−1.215 lb)2 = 2.743 lb. (8) Part (b) Governing Equations & Computation

Once the value of the parallel component of the force 𝐹‖ has been determined from Eq. (7), we may write the vector quantity 𝐹⃗‖ and then proceed to determine 𝐹⃗ as follows: ⊥

𝐹⃗‖ = 𝐹‖

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

= (−1.215 lb)

−12 𝚤̂ − 4 𝚥̂ + 18 𝑘̂ 22

̂ lb, = (0.6627 𝚤̂ + 0.2209 𝚥̂ − 0.9940 𝑘)

(9)

𝐹⃗⊥ = 𝐹⃗𝐶𝐷 − 𝐹⃗‖ −6 𝚤̂ + 2 𝚥̂ − 9 𝑘̂ ̂ lb − (0.6627 𝚤̂ + 0.2209 𝚥̂ − 0.9940 𝑘) 11 ̂ lb. = (−2.299 𝚤̂ + 0.3246 𝚥̂ − 1.461 𝑘) = (3 lb)

(10)

Part (c)

The sign of 𝐹‖ determines the direction in which the bead will tend to slide due to the cord force. A positive value of 𝐹‖ indicates the parallel component of the force is in the same direction as the position vector used in the dot product, namely, from 𝐴 to 𝐵, as shown in Fig. 2. A negative value of 𝐹‖ indicates the parallel component of the force is in the opposite direction, while if 𝐹‖ is zero, then the cord force is perpendicular to 𝐴𝐵. According to Eq. (7), 𝐹‖ = −1.215 lb, and because this value is negative, the parallel component of the cord force is in the direction from 𝐵 to 𝐴, and the bead will tend to slide toward 𝐴. Governing Equations & Computation

Discussion & Verification

As a partial check on our solutions in Parts (a) and (b), we should compute the magnitudes of 𝐹⃗‖ and 𝐹⃗⊥ in Eqs. (9) and (10) to verify they are the same as those found in Eqs. (7) and (8). We may also examine the components of 𝐹⃗ ‖

and 𝐹⃗⊥ to see if they have reasonable directions and proportions. For 𝐹⃗‖ this check is uncertain, but for 𝐹⃗⊥ the components appear to be in proper directions. In view of our answers in this example, Fig. 2 may be redrawn to more accurately show the direction of 𝐹⃗‖ ; that is, the actual direction of 𝐹⃗‖ is from point 𝐶 toward point 𝐴.

Helpful Information Actual sliding direction of the bead. In Part (c), we considered the direction that the bead would slide due to the force provided by the cord only. Note that other forces are also applied to the bead, such as the bead’s weight, reaction forces, and possible friction between the bead and rod 𝐴𝐵—the motion of the bead is also influenced by these. The treatment of these other forces is considered in later chapters of this book. Nonetheless, if the weight of the bead is negligible, the bead can slide without friction on the rod, and the bead has zero initial velocity, then as determined in Part (c), the bead will slide toward point 𝐴.

94

Chapter 2

Vectors: Force and Position

E X A M P L E 2.18

Smallest Distance Between a Point and Line In the rod and bead of Example 2.17, determine the smallest distance between 𝐷 and rod 𝐴𝐵.

𝑦 12 in. 𝐷 18 in.

11 in.

SOLUTION

𝐴

Road Map 8 in.

𝐶

8 in.

𝑂

𝑥

𝐵 4 in. 𝑧 Figure 1

There are several strategies that can be used to determine the smallest distance between point 𝐷 and rod 𝐴𝐵. The technique we will use is identical in concept to the solution used in Part (b) of Example 2.17, except only position vectors are used. The idea is to write a position vector from any convenient point on rod 𝐴𝐵 to point 𝐷. Then, using the dot product, we resolve this position vector into components parallel and perpendicular to 𝐴𝐵, as shown in Fig. 2 (where point 𝐵 has been selected as a convenient location on rod 𝐴𝐵). While the magnitude of the parallel component will depend on the particular point on 𝐴𝐵 we select, the perpendicular component is the same and is the shortest distance between 𝐷 and rod 𝐴𝐵.

Governing Equations & Computation

We select point 𝐵 as a convenient point on rod

𝐴𝐵 and write ̂ in., 𝑟⃗𝐵𝐷 = (4 𝚥̂ − 18 𝑘) √ |⃗𝑟𝐵𝐷 | = (4)2 + (−18)2 in. = 18.44 in.

𝑦 𝐷 𝑟⃗𝐵𝐷 𝑂

𝑟⊥

𝐶 𝑟‖

𝐴

(1) (2)

We now take the dot product of the above with 𝑟⃗𝐵𝐴 , where 𝑟⃗𝐵𝐴 is easily written by examining Fig. 1 [or we may use Eq. (1) of Example 2.17 to write 𝑟⃗𝐵𝐴 = −⃗𝑟𝐴𝐵 ], to obtain the parallel component of 𝑟⃗𝐵𝐷 as 𝑥

𝐵

𝑟‖ = 𝑟⃗𝐵𝐷 ⋅

̂ ̂ in.] ⋅ 12 𝚤̂ + 4 𝚥̂ − 18 𝑘 = 15.45 in. = [(4 𝚥̂ − 18 𝑘) 22 |⃗𝑟𝐵𝐴 | 𝑟⃗𝐵𝐴

Then using the Pythagorean theorem, we find √ √ 𝑟⊥ = 𝑟2𝐵𝐷 − 𝑟2‖ = (18.44)2 − (15.45)2 in. = 10.06 in.

(3)

(4)

𝑧 Figure 2 Resolution of 𝑟⃗𝐵𝐷 into parallel and perpendicular components to determine the smallest distance between point 𝐷 and rod 𝐴𝐵.

ISTUDY

Thus, the smallest distance between point 𝐷 and rod 𝐴𝐵 is 10.06 in. You should repeat this problem starting with a different position vector, say 𝑟⃗𝐴𝐷 , to verify that the same 10.06 in. distance is obtained. Discussion & Verification

The solution presented here actually determines the smallest distance from point 𝐷 to the infinite line passing through points 𝐴 and 𝐵, which is not necessarily the same as the smallest distance to line segment 𝐴𝐵. To determine if the smallest distance found here is to a point that is actually on rod 𝐴𝐵, we compare the 15.45 in. magnitude 𝑟‖ found in Eq. (3) to the length of member 𝐴𝐵, which is 22 in. Hence, the distance found is indeed to a point that lies on rod 𝐴𝐵.

ISTUDY

Section 2.4

95

Vector Dot Product

Problems Problems 2.103 through 2.106 ⃗ (a) Determine the angle between vectors 𝐴⃗ and 𝐵. ⃗ (b) Determine the components of 𝐴⃗ parallel and perpendicular to 𝐵. ⃗ (c) Determine the vector components of 𝐴⃗ parallel and perpendicular to 𝐵. 𝑦

𝑧

𝑦 𝐵⃗

𝐴⃗

𝐵⃗

𝐴⃗

𝑧 𝐴⃗

𝐵⃗

𝐴⃗

𝐵⃗

𝑥

𝑦

𝑧

𝑧

𝑥

𝑦

𝑥 𝑥

̂ N 𝐴⃗ = (−̂𝚤 + 8 𝚥̂ + 4 𝑘) ̂ mm 𝐵⃗ = (̂𝚤 + 18 𝚥̂ − 6 𝑘)

̂ lb 𝐴⃗ = (6 𝚤̂ − 2 𝚥̂ + 3 𝑘) ̂ in. 𝐵⃗ = (−14 𝚤̂ − 2 𝚥̂ + 5 𝑘)

̂ lb 𝐴⃗ = (−12 𝚤̂ + 16 𝚥̂ − 15 𝑘) ̂ in. 𝐵⃗ = (18 𝚤̂ + 6 𝚥̂ − 13 𝑘)

̂ N 𝐴⃗ = (4 𝚤̂ + 3 𝚥̂ + 12 𝑘) ̂ m 𝐵⃗ = (−2 𝚤̂ + 5 𝚥̂ − 14 𝑘)

Figure P2.103

Figure P2.104

Figure P2.105

Figure P2.106

Problem 2.107 A slide on a child’s play structure is to be supported in part by strut 𝐶𝐷 (railings are omitted from the sketch for clarity). End 𝐶 of the strut is to be positioned along the outside edge of the slide, halfway between ends 𝐴 and 𝐵. End 𝐷 of the strut is to be positioned on the 𝑦 axis so that the angle ∠𝐴𝐶𝐷 between the slide and the strut is a right angle. Determine the distance ℎ that point 𝐷 should be positioned. 𝑧 12 f t

𝐴

𝜃

𝑑 𝐶 𝐵 ℎ

𝐷

𝑑

5 ft 5 cm

𝑦 2 ft

5 cm

60◦ 𝑦

𝑥

60◦ 𝑧

Figure P2.107 𝑥

Problem 2.108 Figure P2.108

A whistle is made of a square tube with a notch cut in its edge, into which a baffle is brazed. Determine the dimensions 𝑑 and 𝜃 for the baffle. 𝑦

Problem 2.109 A flat, triangular-shaped window for the cockpit of an airplane is to have the corner coordinates shown. Specify the angles 𝜃𝐴 , 𝜃𝐵 , and 𝜃𝐶 and dimensions 𝑑𝐴𝐵 , 𝑑𝐵𝐶 , and 𝑑𝐴𝐶 for the window.

𝐴(35, 25, 90) cm 𝐵(−25, 30, 105) cm 𝐶(15, 60, 80) cm 𝐶

𝐵

𝑑𝐵𝐶

𝐶 𝐵

𝑧

𝐴

Figure P2.109

𝑥

𝜃𝐵

𝜃𝐶 𝑑𝐴𝐵

𝑑𝐴𝐶 𝜃𝐴

𝐴

96

Chapter 2

Vectors: Force and Position

Problem 2.110 𝛼

𝑎

𝛼

side piece

12 in. 𝑏 1 in.

end piece

𝛽 𝛽

The corner of an infant’s bassinet is shown. Determine angles 𝛼 and 𝛽, and dimensions 𝑎 and 𝑏 of the side and end pieces so the corners of the bassinet will properly meet when assembled.

2 in.

Problem 2.111

Figure P2.110

The roof of an 8 f t diameter grain silo is to be made using 12 identical triangular panels. Determine the value of angle 𝜃 needed and the smallest value of 𝑑 that can be used.

2 ft 𝑑 𝜃 8 ft

𝑦

𝑧

Figure P2.111 𝐵

Problems 2.112 and 2.113

150 mm 𝑏 𝑂 ℎ

𝐷 𝑥 𝐴

𝐺 𝐸 𝑑

360 mm 𝐶

𝑑

𝑧

Let’s reconsider Probs. 2.94 and 2.95 on p. 82, as described here. The structure consists of a quarter-circular rod 𝐴𝐵 with radius 150 mm that is fixed in the 𝑥𝑦 plane. 𝐶𝐷 is a straight rod where 𝐷 may be positioned at different locations on the circular rod. 𝐺𝐸 is an elastic cord that supports a tensile force of 100 N, and whose support at 𝐺 lies in the 𝑦𝑧 plane, and the bead at 𝐸 may have different positions 𝑑. For the values of 𝑏, ℎ, 𝑑, and 𝑧 provided, determine the components of the 100 N cord force in directions parallel and perpendicular to rod 𝐶𝐷. If released from rest, will the cord force tend to make bead 𝐸 slide toward 𝐶 or 𝐷? Problem 2.112

𝑏 = 4, ℎ = 3, 𝑑 = 260 mm, and 𝑧 = 240 mm.

Problem 2.113

𝑏 = 3, ℎ = 4, 𝑑 = 195 mm, and 𝑧 = 270 mm.

Figure P2.112 and P2.113

Problems 2.114 through 2.117

𝑦 𝑟⃗2

̂ in. 𝑟⃗𝐴𝐵 = (160 𝚤̂ − 20 𝚥̂ + 80 𝑘) ̂ in. 𝑟⃗1 = (−49 𝚤̂ − 4 𝚥̂ + 97 𝑘) ̂ in. 𝑟⃗2 = (10 𝚤̂ + 120 𝚥̂ + 10 𝑘) ̂ kip 𝐹⃗ = (0.3 𝚤̂ + 1.4 𝚥̂ − 1.8 𝑘) 𝐹⃗

𝐴 𝑟⃗1 𝑧

cable 𝑟⃗𝐴𝐵

Figure P2.114–P2.117

ISTUDY

𝐵

A cantilever I beam has a cable at end 𝐵 that supports a force 𝐹⃗ , and 𝑟⃗𝐴𝐵 is the position vector from end 𝐴 of the beam to end 𝐵. Position vectors 𝑟⃗1 and 𝑟⃗2 are parallel to the flanges and web of the I beam, respectively. For determination of the internal forces in the beam (discussed in Chapter 8), and for mechanics of materials analysis, it is necessary to know the components of the force in the axial direction of the beam (𝐴𝐵) and in directions parallel to the web and flanges. Problem 2.114

𝑥

Using the dot product, show that 𝑟⃗1 , 𝑟⃗2 , and 𝑟⃗𝐴𝐵 are orthogonal to one

another. Problem 2.115

Determine the scalar and vector components of 𝐹⃗ in direction 𝑟⃗𝐴𝐵 .

Problem 2.116

Determine the scalar and vector components of 𝐹⃗ in direction 𝑟⃗1 .

Problem 2.117

Determine the scalar and vector components of 𝐹⃗ in direction 𝑟⃗2 .

ISTUDY

Section 2.4

97

Vector Dot Product

Problem 2.118 The gearshift lever 𝐴𝐵 for the transmission of a sports car has position vector 𝑟⃗ whose line of action passes through points 𝐴 and 𝐵. The driver applies a force 𝐹⃗ to the knob of the lever to shift gears. If the component of 𝐹⃗ perpendicular to the gearshift lever must be 5 N to shift gears, determine the magnitude 𝐹 of the force. 𝑟⃗

𝑧 𝐵

𝐹⃗

𝐴

𝑦 ̂ mm 𝑟⃗ = (−5 𝚤̂ + 2 𝚥̂ + 14 𝑘) 1 ̂ 𝐹⃗ = 𝐹 ( )(9 𝚤̂ − 6 𝚥̂ − 2 𝑘)

𝑧 30 in. 𝐴

11

𝑥

8 lb 40 in.

Figure P2.118 𝑂

𝑦 60 in.

Problem 2.119 A force of magnitude 8 lb is applied to line 𝐴𝐵 of a fishing rod 𝑂𝐴. Determine the vector components of the 8 lb force in directions parallel and perpendicular to the rod.

𝑥

𝐵

150 in.

Figure P2.119

Problem 2.120 Rod 𝐴𝐵 is the firing pin for the right-hand barrel of a side-by-side shotgun. During firing, end 𝐴 is struck by the hammer, which imparts a force 𝐹 to the firing pin. The force 𝐹 lies in a plane parallel to the 𝑥𝑦 plane, but the firing pin has a three-dimensional orientation as shown. Determine the components of 𝐹 in directions parallel and perpendicular to the firing pin. 𝑦 0.600 in.

𝐵 𝐴 0.600 in.

0.200 in. 0.200 in. 1.100 in.

𝐹 𝑥 28◦

𝑧

𝑧 𝐴 Figure P2.120

Problem 2.121 The collar 𝐶 is fixed to rod 𝐴𝐵 and supports a weight 𝑊 = 15 N, acting in the negative 𝑧 direction. Determine the vector components of the weight that are parallel and perpendicular to rod 𝐴𝐵.

𝐶

𝑥 𝑊

40 cm

Problem 2.122 The collar 𝐶 is fixed to rod 𝐴𝐵 using a glued bond that allows a maximum force of 35 N parallel to the axis of the rod. The collar has weight 𝑊 acting in the negative 𝑧 direction. Determine the weight 𝑊 of the collar that will cause the glued bond to break.

70 cm

𝐵

40 cm

Figure P2.121 and P2.122

𝑦

98

Chapter 2

Vectors: Force and Position Problems 2.123 through 2.125

Components of a machine (not shown in the figure) are actuated by motion of a threaded collar 𝐶 that slides on a fixed bar 𝐷𝐸𝐹 𝐺𝐻𝐼. Segment 𝐷𝐸 of the bar is straight and vertical, segment 𝐹 𝐺 is straight where the coordinates of points 𝐹 and 𝐺 are given in the figure, and segment 𝐻𝐼 is straight and horizontal. The collar’s motion is controlled by screw 𝐴𝐵, which is turned by the motor at point 𝐴. If the force in the screw is 50 N compression, use the dot product to determine the component of the screw force in the direction of the bar for the positions of the collar given below. Also determine the component of the screw force perpendicular to the bar.

𝑦 𝐻

75 mm

screw 𝐹 𝐸

motor 𝐴

𝐼

𝐺

𝐹 (65, 25) mm 𝐺 (89, 70) mm 𝐶

𝐵

threaded collar 15 mm

𝑥

(a) Collar 𝐶 is 2 mm below point 𝐸.

𝐷

(b) Collar 𝐶 is 1/3 the distance from point 𝐹 to point 𝐺.

40 mm

60 mm

Problem 2.123

20 mm Problem 2.124

Figure P2.123–P2.125

(a) Collar 𝐶 is 2/3 the distance from point 𝐹 to point 𝐺. (b) Collar 𝐶 is 1/2 the distance from point 𝐻 to point 𝐼. Problem 2.125 Let the collar be positioned on the straight segment of the bar between points 𝐹 and 𝐺. Determine the point along this segment of the bar where the force from the screw in the direction of the bar is largest, and determine the value of this force. Also determine the component of the screw force perpendicular to the bar.

𝑧

𝐵 (−45, 0, 130) in.

A winch at point 𝐵 is used to slide a box up a chute 𝐶𝐷 in a warehouse. If the component of the force from cable 𝐴𝐵 in the direction of the chute must be 400 lb, determine the force in the cable for the position of the box given below. The chute is parallel to the 𝑥𝑧 plane. Neglect the size of the box and the winch.

𝐷 50 in. 𝐴 𝑥

120 in.

100 in. 𝐶

Figure P2.126–P2.128

ISTUDY

Problems 2.126 through 2.128

𝑦

Problem 2.126

The box is at the base of the chute, point 𝐶.

Problem 2.127

The box is 1/3 the distance up the chute from point 𝐶.

Problem 2.128

The box is 3/4 the distance up the chute from point 𝐶.

Problem 2.129 The structure consists of a quarter-circular rod 𝐴𝐵 with 150 mm radius that is fixed in the 𝑥𝑦 plane. An elastic cord supporting a force of 100 N is attached to a support at 𝐷 and a bead at 𝐶. Determine the components of the cord force in directions tangent and normal to the curved rod 𝐴𝐵 at point 𝐶. 𝑦 𝐵

𝐶

150 mm

4 3 𝑂

𝐴 120 mm

𝑧

150 mm

𝐷

Figure P2.129

𝑥

ISTUDY

Section 2.4

99

Vector Dot Product

Problems 2.130 and 2.131

𝑦

Bead 𝐴 has negligible weight and slides without friction on rigid fixed bar 𝐵𝐶. An elastic cord 𝐴𝐷, which supports a 10 lb tensile force, is attached to the bead. At the instant shown, the bead has zero velocity. For the position of the bead given below, determine the vector components of the cord tension that act parallel and perpendicular to direction 𝐵𝐶 of the bar (your answers should be vectors). Due to the cord tension, will the bead slide toward point 𝐵 or 𝐶? Problem 2.130

Bead 𝐴 is positioned halfway between points 𝐵 and 𝐶.

Problem 2.131

Bead 𝐴 is 1/4 the distance from point 𝐵 to point 𝐶.

Problems 2.132 and 2.133

8 in.

10 in. 𝑂

𝐵

19 in.

Problem 2.133

Bead 𝐵 is 2/5 the distance from point 𝐴 to point 𝐶.

𝐶

12 in.

𝑧

Figure P2.130 and P2.131 𝑧 60 mm

60 mm 𝐷

𝑂

60 mm

Bead 𝐵 is positioned halfway between points 𝐴 and 𝐶.

𝑥

𝐴

4 in.

Bead 𝐵 has negligible weight and slides without friction on rigid fixed bar 𝐴𝐶. An elastic cord 𝐵𝐷, which supports a 60 N tensile force, is attached to the bead. At the instant shown, the bead has zero velocity. For the position of the bead given below, determine the components of the cord tension that act parallel and perpendicular to direction 𝐴𝐶 of the bar. Due to the cord tension, will the bead slide toward point 𝐴 or 𝐶? Problem 2.132

𝐷

13 in.

𝐶 40 mm 𝑦

𝐵 120 mm

𝐴 60 mm

Problems 2.134 and 2.135 A sports car at point 𝐴 drives down a straight stretch of road 𝐶𝐷. A police car at point 𝐵 uses radar to measure the speed of the car and obtains a reading of 80 km∕h. For the position of the car given below, determine the speed of the car. Note that the 80 km∕h speed measured by the radar gun is the rate of change of the distance between the car and the radar gun. Problem 2.134

The car is 30% of the distance from point 𝐶 to point 𝐷.

Problem 2.135

The car is 60% of the distance from point 𝐶 to point 𝐷.

𝑥 Figure P2.132 and P2.133

𝐵 (100, 600, 40) m 𝐶 (−150, −50, 30) m 𝑧 𝐷 (200, 400, −10) m 𝐵

𝐶

𝐴 𝑦

Problem 2.136 Let the road shown be straight between points 𝐶 and 𝐷 and beyond. Determine the shortest distance between the police car at point 𝐵 and the road. Determine if the point where the road is closest to the police car falls within segment 𝐶𝐷, or outside of this.

𝐷

𝑥 Figure P2.134–P2.136

Problem 2.137 𝑧

The manager of a baseball team plans to use a radar gun positioned at point 𝐴 to measure the speed of pitches for a right-handed pitcher. If the person operating the radar gun measures a speed 𝑠 when the baseball is one-third the distance from the release point at 𝐵 to the catcher’s glove at 𝐶, what is the actual speed of the pitch? Assume the pitch follows a straight-line path, and express your answer in terms of 𝑠. Note that the value 𝑠 measured by the radar gun is the rate of change of the distance between point 𝐴 and the ball.

𝑦 𝐵 𝑥

Problem 2.138 Repeat Prob. 2.137 for a left-handed pitcher whose release point is 𝐷.

𝐴

𝐶

𝐴 (−15, 15, 4) f t 𝐵 (40, 42, 5) f t 𝐶 (−3, −3, 3) f t 𝐷 (42, 40, 5) f t

𝐷

Figure P2.137 and P2.138

100

Chapter 2

Vectors: Force and Position

𝑧

Problem 2.139 90 mm

60 mm

𝐵

𝐷

𝑦 𝐶

20 mm 𝐴

𝐸

Structural member 𝐴𝐵 is to be supported by a strut 𝐶𝐷. Determine the smallest length 𝐶𝐷 may have, and specify where 𝐷 must be located for a strut of this length to be used.

Problem 2.140 Determine the smallest distance between member 𝐴𝐵 and point 𝐸.

𝑥 Figure P2.139 and P2.140

ISTUDY

Problem 2.141 In Example 2.13 on p. 76, determine the smallest distance between point 𝐷 and the infinite line passing through points 𝐴 and 𝐵. Is this distance the same as the smallest distance to rod 𝐴𝐵? Explain.

Problem 2.142 In Example 2.18 on p. 94, determine the smallest distance between point 𝑂 and the infinite line passing through points 𝐴 and 𝐵. Is this distance the same as the smallest distance to rod 𝐴𝐵? Explain.

Problem 2.143 A building’s roof has “6 in 12” slope in the front and back and “8 in 12” slope on the sides. Determine the angles 𝛼 and 𝛽 that should be used for cutting sheets of plywood so they properly meet along edge 𝐴𝐵 of the roof. Hint: Write the position vector 𝑟⃗𝐴𝐵 (where 𝐵 is some point along the edge of the roof) two ways: 𝑟⃗𝐴𝐵 = 𝑟⃗𝐴𝐶 + 𝑟⃗𝐶𝐵 and 𝑟⃗𝐴𝐵 = 𝑟⃗𝐴𝐷 + 𝑟⃗𝐷𝐵 . Then use the roof slopes to help write 𝑟⃗𝐶𝐵 and 𝑟⃗𝐷𝐵 such that the magnitudes of the two expressions for 𝑟⃗𝐴𝐵 are the same. 12 12 𝑦

8 𝐷

𝐵

𝛽 𝑧

𝛼

6 𝐶

𝐴

Figure P2.143

𝑥

ISTUDY

Section 2.5

2.5

Vector Cross Product

101

Vector Cross Product

The cross product between two vectors may be used to determine (1) the direction normal to the plane containing two vectors, (2) the area of a parallelogram formed by two vectors, and (3) the moment produced by a force. The last use is especially important in statics and mechanics and will be discussed extensively in Chapter 4. Compared to the dot product, the cross product is more intricate to evaluate. This section studies techniques for evaluating the cross product and interpreting its results. The cross product between two vectors 𝐴⃗ and 𝐵⃗ is an operation defined as ⃗ 𝐵| ⃗ sin 𝜃) 𝑢̂ 𝐴⃗ × 𝐵⃗ = (|𝐴||

(2.48)

where 𝜃 = angle between the lines of action of 𝐴⃗ and 𝐵⃗ where 0 ≤ 𝜃 ≤ 180◦ ; 𝑢̂ = unit vector normal to the plane containing 𝐴⃗ and 𝐵⃗ according to the right-hand rule. 𝑦

𝑦

𝑦

𝐴⃗ × 𝐵⃗

𝐵⃗

𝐴⃗ × 𝐵⃗

𝐵⃗

𝐵⃗

𝜃 𝐴⃗

𝐴⃗

𝐴⃗ 𝑥

𝑥 𝑧

(a)

𝑧

(b)

𝑧

𝑥 (c)

Figure 2.31. (a) Vectors 𝐴⃗ and 𝐵⃗ in three dimensions. (b) To evaluate the cross product ⃗ the vectors can be arranged tail to tail to define angle 𝜃, which is between vectors 𝐴⃗ and 𝐵, measured in the plane containing the two vectors. (c) The result of 𝐴⃗ × 𝐵⃗ is a vector whose direction is governed by the right-hand rule.

In words, Eq. (2.48) states “𝐴⃗ cross 𝐵⃗ ” equals a vector whose magnitude is the ⃗ magnitude of 𝐵, ⃗ and sine of angle 𝜃 between the lines product of the magnitude of 𝐴, ⃗ ⃗ of action of 𝐴 and 𝐵, and whose direction is perpendicular to the plane containing 𝐴⃗ and 𝐵⃗ as governed by the right-hand rule. The definition of 𝜃 in Eq. (2.48) is the same as that used for the dot product. Since the cross product yields a result that is a vector, the cross product is sometimes called the vector product. The units for the ⃗ cross product are equal to the product of the units for 𝐴⃗ and 𝐵. ⃗ Although most Figure 2.31 illustrates the cross product between vectors 𝐴⃗ and 𝐵. often they will, 𝐴⃗ and 𝐵⃗ do not need to lie in the same plane to compute the cross product between them. After the vectors are arranged tail to tail, they define a plane, and the intersection of their lines of action defines angle 𝜃, which is measured in this plane. The direction of the cross product is determined by applying the righthand rule, where 𝐴⃗ is the first vector and 𝐵⃗ is the second vector. That is, if you let 𝐴⃗ pass into the palm of your right hand and 𝐵⃗ pass through your fingertips as shown in ⃗ Immediately Fig. 2.31, then your thumb will point in the proper direction for 𝐴⃗ × 𝐵. ⃗ ⃗ we see that if 𝐵 is taken as the first vector in the cross product and 𝐴 the second, then ⃗ Hence, the right-hand rule gives a direction for 𝐵⃗ × 𝐴⃗ that is opposite that for 𝐴⃗ × 𝐵. the order in which vectors are taken in the cross product is important.

Concept Alert Applications of the cross product. The cross product between two vectors produces a result that is a vector. The cross product is often used to determine the normal direction to a surface, the area of a parallelogram formed by two vectors, and (as discussed in Chapter 4) the moment produced by a force. The last application is especially important in statics and mechanics.

102 𝑦

𝐵 = 3m

The cross product has the following properties:

𝑦 𝐵 = 3m

75◦

60◦

𝐴 = 2m 15◦ 𝑧

(a)

𝑥

𝐴 = 2m 𝑥 5.20 m2

𝑧

(b)

Figure 2.32 Cross product between two vectors 𝐴⃗ and 𝐵⃗ that lie in the 𝑥𝑦 plane. The result of 𝐴⃗ × 𝐵⃗ is a vector in the 𝑧 direction having magnitude 5.20 m2 .

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Chapter 2

Vectors: Force and Position

𝐴⃗ × 𝐵⃗ = −𝐵⃗ × 𝐴⃗ ⃗ = (𝑠𝐴) ⃗ × 𝐵⃗ = 𝐴⃗ × (𝑠𝐵) ⃗ 𝑠(𝐴⃗ × 𝐵)

⃗ × 𝐶⃗ = (𝐴⃗ × 𝐶) ⃗ + (𝐵⃗ × 𝐶) ⃗ (𝐴⃗ + 𝐵)

anticommutative property,

(2.49)

associative property with respect to multiplication by a scalar,

(2.50)

distributive property with respect to vector addition.

(2.51)

The foregoing remarks are true regardless of the type of vector representation that is used. For example, consider the two position vectors 𝐴⃗ and 𝐵⃗ shown in Fig. 2.32(a), where both vectors lie in the 𝑥𝑦 plane. Sliding the vectors tail to tail as shown in Fig. 2.32(b) provides the angle between 𝐴⃗ and 𝐵⃗ as 60◦ , and applying Eq. (2.48) provides the cross product between 𝐴⃗ and 𝐵⃗ as 𝐴⃗ × 𝐵⃗ = (2 m)(3 m) sin 60◦ 𝑢̂ = 5.20 m2 𝑢, ̂

(2.52)

where 𝑢̂ is a unit vector normal to the plane containing 𝐴⃗ and 𝐵⃗ and whose direction is given by the right-hand rule, namely, the 𝑧 direction shown in Fig. 2.32(b). Observe that the cross product has units of m2 . Although not obvious, the magnitude of this result has the physical interpretation of being the area of the parallelogram formed ⃗ This feature of the cross product is discussed in greater detail soon. by 𝐴⃗ and 𝐵.

Cross product using Cartesian components For two vectors 𝐴⃗ and 𝐵⃗ with Cartesian representations, the cross product between them is given by ̂ 𝐴⃗ × 𝐵⃗ = (𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 ) 𝚤̂ + (𝐴𝑧 𝐵𝑥 − 𝐴𝑥 𝐵𝑧 ) 𝚥̂ + (𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 ) 𝑘.

Helpful Information Alternatives for evaluation of the cross product. Regardless of the representation used to express vectors, Eq. (2.48) can always be used to evaluate the cross product. For vectors with Cartesian representation, either Eq. (2.48) or Eq. (2.53) can be used, with the latter usually being more convenient. In fact, sometimes both methods are used in combination to help determine useful information.

(2.53)

Equation (2.53) can be obtained using the following derivation. We begin by writing 𝐴⃗ and 𝐵⃗ using Cartesian vector representation ̂ 𝐴⃗ = 𝐴𝑥 𝚤̂ + 𝐴𝑦 𝚥̂ + 𝐴𝑧 𝑘,

(2.54)

̂ 𝐵⃗ = 𝐵𝑥 𝚤̂ + 𝐵𝑦 𝚥̂ + 𝐵𝑧 𝑘.

(2.55)

⃗ using the distributive law of We then take the cross product between 𝐴⃗ and 𝐵, Eq. (2.51) to expand the product term by term: ̂ × (𝐵 𝚤̂ + 𝐵 𝚥̂ + 𝐵 𝑘) ̂ 𝐴⃗ × 𝐵⃗ = (𝐴𝑥 𝚤̂ + 𝐴𝑦 𝚥̂ + 𝐴𝑧 𝑘) 𝑥 𝑦 𝑧 ̂ = (𝐴𝑥 𝚤̂ × 𝐵𝑥 𝚤̂) + (𝐴𝑥 𝚤̂ × 𝐵𝑦 𝚥̂) + (𝐴𝑥 𝚤̂ × 𝐵𝑧 𝑘) ̂ + (𝐴𝑦 𝚥̂ × 𝐵𝑥 𝚤̂) + (𝐴𝑦 𝚥̂ × 𝐵𝑦 𝚥̂) + (𝐴𝑦 𝚥̂ × 𝐵𝑧 𝑘) ̂ + (𝐴𝑧 𝑘̂ × 𝐵𝑥 𝚤̂) + (𝐴𝑧 𝑘̂ × 𝐵𝑦 𝚥̂) + (𝐴𝑧 𝑘̂ × 𝐵𝑧 𝑘).

(2.56)

To proceed further we must address cross products between combinations of unit vectors. Equation (2.48) shows 𝚤̂ × 𝚤̂ has magnitude (1)(1) sin 0◦ = 0, and similarly 𝚥̂ × 𝚥̂ and 𝑘̂ × 𝑘̂ have zero magnitude. The cross product between combinations of different unit vectors always provides the magnitude (1)(1) sin 90◦ = 1 and a direction that depends on the order in which the vectors are taken. For example,

ISTUDY

Section 2.5

Vector Cross Product

̂ while 𝚥̂ × 𝚤̂ = −𝑘. ̂ In summary, cross products between combinations 𝚤̂ × 𝚥̂ = 𝑘, of unit vectors are 𝚤̂ × 𝚤̂ = 0⃗

𝚤̂ × 𝚥̂ = 𝑘̂

𝚤̂ × 𝑘̂ = −̂𝚥

𝚥̂ × 𝚤̂ = −𝑘̂

𝚥̂ × 𝚥̂ = 0⃗

𝚥̂ × 𝑘̂ = 𝚤̂

𝑘̂ × 𝚤̂ = 𝚥̂

𝑘̂ × 𝚥̂ = −̂𝚤

⃗ 𝑘̂ × 𝑘̂ = 0,

(2.57)

where 0⃗ = 0 𝚤̂ + 0 𝚥̂ + 0 𝑘̂ is called the zero vector. Substituting Eq. (2.57) into Eq. (2.56) then provides Eq. (2.53). While Eq. (2.53) can always be employed to evaluate the cross product between vectors with Cartesian representation, it is awkward to remember and use. However, if you want to program a digital machine such as your pocket calculator or a computer to evaluate cross products, then Eq. (2.53) is ideal. But for manual evaluation the following procedure is easier.

Evaluation of cross product using determinants By arranging vectors 𝐴⃗ and 𝐵⃗ in a matrix, the determinant of the matrix can be eval⃗ A matrix is an arrangement of elements into a rectangular array uated to yield 𝐴⃗ × 𝐵. of rows and columns. There is a precise and extensive science underlying matrices, and because matrices are of fundamental importance in engineering, you are sure to study these as you advance in your education. Here we present only those details that are needed for evaluation of the cross product. ⃗ we evaluate the determiTo evaluate the cross product between vectors 𝐴⃗ and 𝐵, nant of the matrix ⎡ 𝚤̂ 𝐴⃗ × 𝐵⃗ = det ⎢ 𝐴𝑥 ⎢ ⎣ 𝐵𝑥

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

⎤ || 𝚤̂ ⎥ = || 𝐴 ⎥ || 𝑥 ⎦ | 𝐵𝑥

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

| | | | | | |

(2.58)

where det is an abbreviation indicating the determinant of the matrix that follows is to be evaluated. In the rightmost form of Eq. (2.58), a more abbreviated notation is used where the vertical bars indicate the determinant of the enclosed matrix. In this book we will primarily use vertical bars to denote determinant. The first row of the matrix in Eq. (2.58) consists of the basis vectors for the coordinate system, namely, 𝚤̂, ̂ The second row consists of the components of 𝐴, ⃗ and the third row consists 𝚥̂, and 𝑘. ⃗ Note that if 𝐵⃗ × 𝐴⃗ were to be evaluated, then the second row of the components of 𝐵. of the matrix would contain the components of 𝐵⃗ and the third row would contain the ⃗ While there are many ways to evaluate the determinant of a matrix, components of 𝐴. we discuss two methods that are effective for Eq. (2.58). Determinant by Method 1. To help in the computation of the determinant, we start by repeating the first and second columns after writing the original matrix. We then take the sums and differences of products of the diagonal elements as follows: take the product of the three elements along diagonal 1 [i.e., 𝚤̂ (𝐴𝑦 )(𝐵𝑧 )], add the product of the three elements along diagonal 2, add the product of the three elements along diagonal 3, subtract the product of the three elements along diagonal 4, subtract the product of the three elements along diagonal 5, and finally subtract the product of

103

104

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Chapter 2

Vectors: Force and Position

the three elements along diagonal 6:

𝐴⃗ × 𝐵⃗ =

𝑘̂ 𝐴𝑧 𝐵𝑧

𝚥̂ 𝐴𝑦 𝐵𝑦

𝚤̂ 𝐴𝑥 𝐵𝑥

=

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

𝚥̂ 𝐴𝑦 𝐵𝑦

𝚤̂ 𝐴𝑥 𝐵𝑥

4 5 6 1 2 3 + 𝑘̂ 𝐴𝑥 𝐵𝑦 − 𝑘̂ 𝐴𝑦 𝐵𝑥 − 𝚤̂ 𝐴𝑧 𝐵𝑦 − 𝚥̂ 𝐴𝑥 𝐵𝑧

= 𝚤̂ 𝐴𝑦 𝐵𝑧 + 𝚥̂ 𝐴𝑧 𝐵𝑥 diagonal 2

diagonal 1

𝚤̂ 𝐴𝑥 𝐵𝑥

diagonal 4

diagonal 3

diagonal 5

diagonal 6

̂ = (𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 ) 𝚤̂ + (𝐴𝑧 𝐵𝑥 − 𝐴𝑥 𝐵𝑧 ) 𝚥̂ + (𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 ) 𝑘.

(2.59)

Note that Eq. (2.59) is identical to Eq. (2.53). Determinant by Method 2. Here we expand the determinant of the matrix by using minors. A minor is the determinant of a subset of the original matrix that is obtained as shown in Eq. (2.60) by eliminating the row and column corresponding to a particular element; the determinant of the elements remaining in the matrix constitutes the minor for that element. Each of the minors is found using the procedure described in method 1 for computing a determinant. That is, the determinant of the matrix multiplying 𝚤̂ in Eq. (2.60) is the product of the two elements along diagonal 1, minus the product of the two elements along diagonal 2. The determinant of the matrix multiplying 𝚥̂ is the product of the two elements along diagonal 3, minus the product of the two elements along diagonal 4. Similarly, the determinant of the matrix multiplying 𝑘̂ is the product of the two elements along diagonal 5, minus the product of the two elements along diagonal 6. 𝚤̂ 𝐴𝑥 𝐵𝑥

Common Pitfall ̂ When Don’t forget the negative sign on 𝒋. you use Method 2 to evaluate the determinant, a common error is to forget the negative sign for 𝚥̂.

𝐴⃗ × 𝐵⃗ =

𝚤̂ 𝐴𝑥 𝐵𝑥

𝚥̂ 𝐴𝑦 𝐵𝑦

𝚥̂ 𝐴𝑦 𝐵𝑦 𝑘̂ 𝐴𝑧 𝐵𝑧

𝑘̂ 𝐴𝑧 𝐵𝑧

𝚤̂ 𝐴𝑥 𝐵𝑥

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

𝚤̂ 𝐴𝑥 𝐵𝑥

= 𝚤̂

𝐴𝑦 𝐴𝑧 𝐵𝑦 𝐵𝑧

− 𝚥̂

𝐴𝑥 𝐴𝑧 𝐵𝑥 𝐵𝑧

+ 𝑘̂

2

1

4

3

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

𝐴𝑥 𝐴𝑦 𝐵𝑥 𝐵𝑦

5

6

= 𝚤̂ ( 𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 ) − 𝚥̂ ( 𝐴𝑥 𝐵𝑧 − 𝐴𝑧 𝐵𝑥 ) + 𝑘̂ ( 𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 ) diagonal 1

diagonal 2

diagonal 3

diagonal 4

diagonal 5

diagonal 6

(2.60)

We have not discussed all of the details of why this procedure works, and some of these details explain why the negative sign associated with 𝚥̂ appears in Eq. (2.60). A common error is to forget this negative sign. Remarks. You should try both methods discussed here and adopt the method you prefer. An advantage of the second method is that terms are automatically grouped ̂ In more advanced mathematics and mechanics subjects, with vectors 𝚤̂, 𝚥̂, and 𝑘. you may need to evaluate the determinant of larger matrices than those treated here. Method 1 is limited to matrices that have at most three rows and columns. With some additional details, Method 2 can be applied to larger matrices, but this may involve the use of many minors, and other methods (not discussed here) are usually more effective.

ISTUDY

Section 2.5

Vector Cross Product

105

Determination of the normal direction to a plane The cross product can be used to determine the normal direction 𝐶⃗ to the plane con⃗ 𝐵. ⃗ If we want the normal direction taining two vectors 𝐴⃗ and 𝐵⃗ by evaluating 𝐶⃗ = 𝐴× ⃗ ⃗ 𝐶|. ⃗ Note that even to be a unit vector, then we normalize 𝐶 by evaluating 𝑢̂ = 𝐶∕| ⃗ ⃗ if 𝐴 and 𝐵 are unit vectors, the cross product between these is not a unit vector unless 𝐴⃗ and 𝐵⃗ are orthogonal. A normal direction can also be obtained by evaluating ⃗ and of course, the vector that is produced has opposite direction to 𝐴⃗ × 𝐵. ⃗ 𝐵⃗ × 𝐴, In many problems that call for a normal direction to be computed, either direction can be used, while in some problems it may be desirable or necessary to distinguish between these. 5.20 m2

𝑦

𝑦

Determination of the area of a parallelogram The magnitude of the cross product gives the area of the parallelogram formed by vectors 𝐴⃗ and 𝐵⃗ arranged tail to tail. We illustrate this using the example of Fig. 2.32, where the magnitude of the cross product was found in Eq. (2.52) to be 5.20 m2 , and this area is shown in Fig. 2.33(a). To show that this statement is true, consider the shaded triangle shown in Fig. 2.33(b): it has base 𝐴 and height 𝐵 sin 𝜃. Hence, its area is (base)(height)∕2 = 𝐴𝐵(sin 𝜃)∕2. The remaining triangle shown in Fig. 2.33(b) has the same base and height and thus the same area. Hence, the area of the parallelogram is the same as the magnitude of the cross product given in Eq. (2.48), namely, 𝐴𝐵 sin 𝜃 = 5.20 m2 .

𝐵 = 3m

𝑧

𝐵

60◦

𝜃

𝐵 sin 𝜃

𝐴 = 2m 𝑥

𝐴

𝑧

(b)

(a)

𝑥

Figure 2.33 The magnitude 5.20 m2 of 𝐴⃗ × 𝐵⃗ corresponds to the area of the parallelogram formed by 𝐴⃗ and 𝐵⃗ arranged tail to tail.

Scalar triple product Occasionally the cross product between two vectors 𝐴⃗ and 𝐵⃗ is to be immediately fol⃗ and we call the expression lowed by a dot product with a third vector 𝐶, ⃗ 𝐵) ⃗ ⋅ 𝐶⃗ a scalar triple product, or a mixed triple product.∗ Of course, this product (𝐴× can be evaluated by first computing the cross product, which produces a vector, and then taking the dot product of this with 𝐶⃗ to provide the final result, which is a scalar. Alternatively, the scalar triple product may be computed by evaluating | 𝐶 | 𝑥 | ⃗ ⃗ ⃗ (𝐴 × 𝐵) ⋅ 𝐶 = | 𝐴𝑥 | | 𝐵 | 𝑥

𝐶𝑦 𝐴𝑦 𝐵𝑦

𝐶𝑧 𝐴𝑧 𝐵𝑧

| | | | | | |

𝐶 𝐵

(2.61)

⃗ ⋅ 𝐶⃗ is the volume of Although we do not prove this, the value produced by (𝐴⃗ × 𝐵) ⃗ ⃗ ⃗ the parallelepiped defined by 𝐴, 𝐵, and 𝐶 arranged tail to tail as shown in Fig. 2.34. In Prob. 2.167, the implications of changing the ordering of vectors in the scalar triple product are explored.

⃗ is preferred to distinguish this from the triple product 𝐴⃗ × 𝐵⃗ × 𝐶, which is more accurately called a vector triple product.

∗ The nomenclature scalar triple product

𝑦

𝑧

𝑥 𝐴

Figure 2.34 ⃗ 𝐵, ⃗ and Volume of the parallelepiped formed by 𝐴, 𝐶⃗ arranged tail to tail is given by the scalar triple ⃗ ⋅ 𝐶. ⃗ product (𝐴⃗ × 𝐵)

106

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End of Section Summary In this section, the cross product between two vectors has been defined. Some of the key points are as follows: ⃗ where 𝐶⃗ = 𝐴× ⃗ 𝐵. ⃗ • The cross product between two vectors 𝐴⃗ and 𝐵⃗ is a vector 𝐶, ⃗ and the direction of 𝐶⃗ is The units of 𝐶⃗ is the product of the units of 𝐴⃗ and 𝐵, perpendicular to the plane containing 𝐴⃗ and 𝐵⃗ as governed by the right-hand rule. • The cross product can be used to find the normal direction to a plane. • The cross product can be used to determine the area of a parallelogram and the volume of a parallelepiped. • One of the most important uses for the cross product is for computing the moment of a force, and this application is discussed extensively in Chapter 4.

ISTUDY

Section 2.5

107

Vector Cross Product

E X A M P L E 2.19

Evaluation of the Cross Product

Evaluate the cross product between vectors 𝐴⃗ and 𝐵⃗ where

𝑧

̂ mm, 𝐴⃗ = (3 𝚤̂ + 5 𝚥̂ + 𝑘) ̂ mm. 𝐵⃗ = (−4 𝚤̂ + 6 𝚥̂ + 2 𝑘)

SOLUTION

𝐵⃗

Road Map

Although we could use Eq. (2.53) to evaluate the cross product, it will generally be more effective to use your choice of Method 1 or 2 as shown here. Method 1 Governing Equations & Computation Using Method 1 described in Eq. (2.59), the

cross product between 𝐴⃗ and 𝐵⃗ is

𝑦 𝐴⃗ 𝑥 Figure 1

𝚤̂ 3 −4

𝐶⃗ = 𝐴⃗ × 𝐵⃗ =

𝑘̂ 1 2

𝚥̂ 5 6

𝚤̂ mm2 = 3 −4

𝚥̂ 5 6

𝑘̂ 𝚤̂ 1 3 2 −4

𝚥̂ 5 mm2 6

(1)

𝐶⃗ = 𝐴⃗ × 𝐵⃗

𝑧

= [̂𝚤 (5)(2) + 𝚥̂ (1)(−4) + 𝑘̂ (3)(6) − 𝑘̂ (5)(−4) − 𝚤̂ (1)(6) − 𝚥̂ (3)(2)] mm2 ̂ mm2 . = (4 𝚤̂ − 10 𝚥̂ + 38 𝑘)

(2)

𝐵⃗

⃗ The mm2 term in Eq. (1) comes from factoring the mm dimensions out of both 𝐴⃗ and 𝐵. ⃗ The result of the cross product, vector 𝐶, is shown in Fig. 2. Method 2 Governing Equations & Computation Using Method 2 described in Eq. (2.60), the

cross product between 𝐴⃗ and 𝐵⃗ is 𝐶⃗ = 𝐴⃗ × 𝐵⃗ =

= 𝚤̂

5 6

1 2

𝚤̂ 3 −4

𝚥̂ 5 6

mm2 − 𝚥̂

𝑘̂ 1 2

1 2

𝐴⃗ 𝑥 Figure 2 Evaluation of the cross product between vectors ⃗ 𝐴⃗ and 𝐵.

mm2

3 −4

𝑦

mm2 + 𝑘̂

3 −4

5 6

mm2

(3)

= 𝚤̂ [(5)(2) − (1)(6)] mm2 − 𝚥̂ [(3)(2) − (1)(−4)] mm2 + 𝑘̂ [(3)(6) − (5)(−4)] mm2 ̂ mm2 , = (4 𝚤̂ − 10 𝚥̂ + 38 𝑘)

(4)

which is the same result as obtained by Method 1. Discussion & Verification To verify that 𝐶⃗ is indeed perpendicular to the plane con⃗ you should evaluate the dot products 𝐶⃗ ⋅ 𝐴⃗ and 𝐶⃗ ⋅ 𝐵⃗ to find they are taining 𝐴⃗ and 𝐵, both zero. You may also wish to evaluate 𝐵⃗ × 𝐴⃗ to show that −𝐶⃗ is produced.

108

Chapter 2

Vectors: Force and Position

E X A M P L E 2.20

Components of Force in Directions Normal and Tangent to a Plane A house with 95 Mg mass is built on a steep slope defined by points 𝐴, 𝐵, and 𝐶. To help assess the possibility of slope failure (mud slide), it is necessary to (a) Determine the components of the weight in directions normal and parallel (tangent) to the slope. (b) Determine the component vectors of the weight in directions normal and parallel (tangent) to the slope. (c) Determine the smallest distance from point 𝑂 to the slope.

RIC FRANCIS/ASSOCIATED PRESS/AP Photo

SOLUTION

𝑧

The weight of the house is (95 Mg)(9.81 m∕s2 ) = 932.0 kN, as shown in Fig. 2, and the vector expression for this is

Road Map 𝐵

𝑥

𝐶

𝑂

60 m

⃗ = (−932.0 kN) 𝑘. ̂ 𝑊 180 m 𝐴

130 m 𝑦 Figure 1

Common Pitfall Don’t confuse mass and weight. A common error is to mistake mass for weight. Mass must be multiplied by acceleration due to gravity to obtain weight. In Eq. (1) and Fig. 2, if you incorrectly gave the weight of the house a value of 95 , then all of your answers are about an order of magnitude too small, and they have the wrong units!

(1)

Our strategy for Part (a) will be to use the cross product to determine the normal vector to the slope. Then the dot product between the weight vector and the normal direction vector will yield the component of weight in the normal direction. Once this is obtained, the component of the weight that is parallel, or tangent, to the slope can be determined using the Pythagorean theorem. For Part (b), we will express the weights normal and parallel to the slope as vectors. For Part (c), we write a position vector from point 𝑂 to any convenient point on the slope, and then we use the dot product to determine the component of this vector that is normal to the slope. This result is the shortest distance from point 𝑂 to the slope. Part (a) Governing Equations & Computation

The normal direction can be computed using the cross product between a variety of different position vectors. We will use position vectors from 𝐴 to 𝐵 and from 𝐴 to 𝐶, ̂ m, 𝑟⃗𝐴𝐵 = (−180 𝚥̂ + 60 𝑘)

(2)

𝑟⃗𝐴𝐶 = (130 𝚤̂ − 180 𝚥̂) m.

(3)

The normal direction 𝑛⃗ to the surface and its magnitude |⃗ 𝑛| are then | 𝚤̂ 𝚥̂ 𝑘̂ || | | | 𝑛⃗ = 𝑟⃗𝐴𝐵 × 𝑟⃗𝐴𝐶 = | 0 −180 60 | m2 | | | 130 −180 0 | | | 2 ̂ = (10, 800 𝚤̂ + 7800 𝚥̂ + 23, 400 𝑘) m , √ |⃗ 𝑛| = (10, 800)2 + (7800)2 + (23, 400)2 m2 = 26, 930 m2 .

𝑧 𝑊 = 932.0 kN 𝐵

𝑥

𝐶

𝑂

60 m

180 m 𝐴 130 m 𝑦 Figure 2 The weight of the house is a vertical force that is applied to the slope.

ISTUDY

(4) (5)

Observe that the units for both 𝑟⃗𝐴𝐵 and 𝑟⃗𝐴𝐶 have been factored out of the matrix expression in Eq. (4), and this is helpful to reduce the repetitious writing of units when expanding the determinant. Also, we selected 𝑟⃗𝐴𝐵 as the first vector and 𝑟⃗𝐴𝐶 as the second vector in the cross product so that the normal vector produced would be in the “outward” direction to the slope (i.e., toward the sky). The same result would have been obtained using 𝑟⃗𝐵𝐶 × 𝑟⃗𝐵𝐴 , or 𝑟⃗𝐶𝐴 × 𝑟⃗𝐶𝐵 . Alternatively, we could use the “inward” normal direction for this problem, as given by 𝑟⃗𝐴𝐶 × 𝑟⃗𝐴𝐵 , or 𝑟⃗𝐵𝐴 × 𝑟⃗𝐵𝐶 , or 𝑟⃗𝐶𝐵 × 𝑟⃗𝐶𝐴 ; this vector’s direction is into the Earth.

ISTUDY

Section 2.5

109

Vector Cross Product

⃗ acting in the direction 𝑛⃗ will be called Next, the component of the weight vector 𝑊 𝑊𝑛 , and it is given by ⃗ ⋅ 𝑊𝑛 = 𝑊

(0)(10, 800 m2 ) + (0)(7800 m2 ) + (−932.0 kN)(23, 400 m2 ) 𝑛⃗ = 26, 930 m2 |⃗ 𝑛| = −809.9 kN.

(6)

Since 𝑛⃗ is the outward normal direction, the negative sign for 𝑊𝑛 indicates that the normal ⃗ acts into the surface, which is the expected result. Note that if an inward component of 𝑊 normal vector were used instead, then the sign of 𝑊𝑛 would be positive. The Pythagorean theorem is now used to obtain the component of the weight that is parallel, or tangent to the surface, 𝑊𝑡 , as √ √ 𝑊𝑡 = 𝑊 2 − 𝑊𝑛2 = (−932.0 kN)2 − (−809.9 kN)2 = 461.1 kN. (7) 𝑧

While the magnitude of the tangential component of the force 𝑊𝑡 is known, we do not know its direction other than that it lies in the plane defined by points 𝐴, 𝐵, and 𝐶. Part (b)

Once 𝑊𝑛 is known from Eq. (6), we may then ⃗ ⃗ find the vector components 𝑊𝑛 and 𝑊𝑡 as follows: Governing Equations & Computation

⃗ = 𝑊 𝑛⃗ = (−324.8 𝚤̂ − 234.6 𝚥̂ − 703.8 𝑘) ̂ kN. 𝑊 𝑛 𝑛 |⃗ 𝑛|

𝑥

𝐶

𝑊𝑡 = 461.1 kN

̂ kN. ⃗ =𝑊 ⃗ −𝑊 ⃗ = (324.8 𝚤̂ + 234.6 𝚥̂ − 228.1 𝑘) 𝑊 𝑡 𝑛

(9)

The components 𝑊𝑛 and 𝑊𝑡 are shown in their proper orientations in Fig. 3. As a partial check on our solution, you should evaluate the magnitudes of Eqs. (8) and (9) to verify that they agree with Eqs. (6) and (7), respectively. Part (c) Governing Equations & Computation

The procedure for finding the smallest distance from point 𝑂 to the surface defined by points 𝐴, 𝐵, and 𝐶 is very similar to that outlined in Section 2.4. We first write a position vector from point 𝑂 to any convenient point on the surface. We then take the dot product of this vector with a unit vector that is normal to the surface, and the scalar that is produced is the smallest distance between point 𝑂 and the surface. The position vector from 𝑂 to 𝐶 is 𝑟⃗𝑂𝐶 = (130 m) 𝚤̂.

(10)

The portion of 𝑟⃗𝑂𝐶 in the normal direction is (130 m)(10, 800) + (0)(7800) + (0)(23, 400) 𝑛⃗ = 52.14 m, = 26, 930 |⃗ 𝑛|

(11)

and thus, the smallest distance between point 𝑂 and the surface is 52.14 m. Discussion & Verification

𝑊𝑛 = 809.9 kN 𝐵 𝑂

(8)

⃗ , we evaluate ⃗ =𝑊 ⃗ +𝑊 Since 𝑊 𝑛 𝑡

𝑟𝑛 = 𝑟⃗𝑂𝐶 ⋅

𝑊 = 932.0 kN

The solution in Part (c) finds the smallest distance from point 𝑂 to the plane of infinite extent that contains points 𝐴, 𝐵, and 𝐶. If, however, the surface in question were finite in size, then the next step would be to determine the coordinates of the head of the vector 𝑟⃗𝑛 = 𝑟𝑛 𝑛⃗∕|⃗ 𝑛| to determine if this point lies on or off of the surface in question.

𝐴 𝑦

Figure 3 The weight of the house is resolved into components 𝑊𝑛 and 𝑊𝑡 that are normal and tangential to the slope, respectively.

110

E X A M P L E 2.21 𝑦

Determination of the Normal Direction to a Plane The finite element method is a computer technique that has revolutionized the way structural engineering is performed. With this method, a structure usually having complex geometry is subdivided into small regions called finite elements, each of which has simple geometry and whose behavior can thus be more easily characterized. Shown in Fig. 1 is a flat three-node plate finite element. Its geometry is fully described by the coordinates of the corners, which are called nodes. Determine the unit vector in the direction normal to the surface of the element, and the area of the element.

node 3

node 2 node 1 𝑥 node 1 (12, 15, 10) mm node 2 (32, 24, −2) mm node 3 (22, 35, −10) mm

𝑧 Figure 1

ISTUDY

Chapter 2

Vectors: Force and Position

SOLUTION Road Map

We will use the cross product between two vectors oriented along the edges of the triangle to determine the normal direction 𝑛⃗. Because three points define a unique plane, the same normal direction 𝑛⃗ (or possibly −⃗ 𝑛, depending on the order in which vectors are taken in the cross product) should be obtained regardless of which two vectors are used. Governing Equations & Computation

Interesting Fact The finite element method. Some of the characteristics of this computer method of analysis are described in the problem statement for this example. Shown here is a finite element model with thousands of elements of a crash test dummy used to assess trauma in crash simulations.

Position vectors from points (nodes) 1 to 2 and 1 to 3 will be used for computing the cross product. Thus, taking the differences between the coordinates of the head and tail, we write ̂ mm, 𝑟⃗12 = (20 𝚤̂ + 9 𝚥̂ − 12 𝑘) ̂ mm, 𝑟⃗ = (10 𝚤̂ + 20 𝚥̂ − 20 𝑘) 13

|⃗𝑟12 | = 25 mm,

(1)

|⃗𝑟13 | = 30 mm.

(2)

The vector in the normal direction and its magnitude are | 𝚤̂ | | 𝑛⃗ = 𝑟⃗12 × 𝑟⃗13 = | 20 | | 10 | |⃗ 𝑛| = 422.0 mm2 .

𝚥̂ 9 20

𝑘̂ −12 −20

| | | ̂ mm2 , | mm2 = (60 𝚤̂ + 280 𝚥̂ + 310 𝑘) | | |

(3) (4)

The unit vector 𝑢̂ in the normal direction is obtained by normalizing 𝑛⃗ in Eq. (3), which provides 𝑛⃗ ̂ = 0.1422 𝚤̂ + 0.6635 𝚥̂ + 0.7346 𝑘. (5) 𝑢̂ = |⃗ 𝑛| To obtain the area of the element, we note the magnitude of the cross product in Eq. (3) is the area of the parallelogram formed by 𝑟⃗12 and 𝑟⃗13 , which is twice the area of the triangle formed by these same vectors. Hence, the area 𝐴 of the triangular region is Department of Transportation

𝐴= Discussion & Verification

422.0 mm2 = 211.0 mm2 . 2

(6)

As a casual check, you should use inspection to verify that the direction of the normal vector is reasonable. However, in this example the geometry is complex enough that this is probably difficult or inconclusive. As a rigorous check, you could verify that 𝑢̂ is perpendicular to both 𝑟⃗12 and 𝑟⃗13 by showing that 𝑢̂ ⋅ 𝑟⃗12 = 0 and 𝑢̂ ⋅ 𝑟⃗13 = 0. In closing, we note that finite element computer programs typically perform these very same calculations for every element.

ISTUDY

Section 2.5

111

Vector Cross Product

Problems Problems 2.144 and 2.145 Vectors 𝐴⃗ and 𝐵⃗ lie in the 𝑥𝑦 plane. ⃗ expressing the resulting vector using (a) Use Eq. (2.48) on p. 101 to evaluate 𝐴⃗ × 𝐵, Cartesian representation. (b) Evaluate 𝐴⃗ × 𝐵⃗ by computing the determinant of a matrix, using either Method 1 or Method 2. 𝑧

𝑦 𝐴 = 6 in.

30◦

60◦

𝑦

𝑥

30◦

45◦

𝐵 = 20 N

𝐵 = 15 lb

𝑧

𝐴 = 35 mm

𝑥

Figure P2.144

Figure P2.145

Problems 2.146 through 2.149 ⃗ (a) Evaluate 𝐴⃗ × 𝐵. ⃗ (b) Evaluate 𝐵⃗ × 𝐴. (c) Comment on any differences between the results of Parts (a) and (b). ⃗ (d) Use the dot product to show the result of Part (a) is orthogonal to vectors 𝐴⃗ and 𝐵. 𝑦

𝑧

𝑦

𝐵⃗

𝐵⃗

𝐴⃗

𝐵⃗

𝐴⃗

𝑧 𝐵⃗

𝐴⃗

𝐴⃗

𝑧

𝑧

𝑥

𝑥

𝑥

𝑦

𝑦 𝑥

̂ mm 𝐴⃗ = (−̂𝚤 + 8 𝚥̂ + 4 𝑘) ̂ mm 𝐵⃗ = (̂𝚤 + 18 𝚥̂ − 6 𝑘)

̂ in. 𝐴⃗ = (6 𝚤̂ − 2 𝚥̂ + 3 𝑘) ̂ in. 𝐵⃗ = (−14 𝚤̂ − 2 𝚥̂ + 5 𝑘)

̂ lb 𝐴⃗ = (−12 𝚤̂ + 16 𝚥̂ − 15 𝑘) ̂ in. 𝐵⃗ = (18 𝚤̂ + 6 𝚥̂ − 13 𝑘)

̂ N 𝐴⃗ = (4 𝚤̂ + 3 𝚥̂ + 12 𝑘) ̂ m 𝐵⃗ = (−2 𝚤̂ + 5 𝚥̂ − 14 𝑘)

Figure P2.146

Figure P2.147

Figure P2.148

Figure P2.149

Problem 2.150 Describe how the cross product operation can be used to determine (or “test”) whether two vectors 𝐴⃗ and 𝐵⃗ are orthogonal. Is this test as easy to use as the test based on the dot product? Explain, perhaps using an example to support your remarks. Note: Concept problems are about explanations, not computations.

112

Chapter 2

Vectors: Force and Position

Problem 2.151 Imagine a left-hand coordinate system has inadvertently been used for a problem. That is, if the 𝑥 and 𝑦 directions have been selected first, the 𝑧 direction has been taken in the wrong direction for a right-hand coordinate system. What consequences will this have for dot products and cross products? Perhaps use an example to support your remarks. Note: Concept problems are about explanations, not computations.

𝑧

𝐴

Problem 2.152

150 lb 250 lb

100 lb 𝑥 𝐵

96 in.

𝑂 72 in.

𝐷 72 in.

𝐶

𝑦

The corner of a tent is supported using three ropes having the forces shown. We wish to ⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ where compute the sum of the cross products 𝑀 𝑂 𝑂𝐴 𝐴𝐵 𝑂𝐴 𝐴𝐶 𝑂𝐴 𝐴𝐷 𝑟⃗𝑂𝐴 is the position vector from points 𝑂 to 𝐴, 𝐹⃗𝐴𝐵 is the force directed from points 𝐴 to 𝐵, and so on. ⃗ , do the properties of (a) Rather than compute three separate cross products to find 𝑀 𝑂 ⃗ the cross product permit 𝑀 to be found using just one cross product? Explain. 𝑂

⃗ . (b) Determine 𝑀 𝑂

Figure P2.152

Problem 2.153 For the triangular-shaped window of Prob. 2.109 on p. 95, use the cross product to determine the outward normal unit vector (i.e., pointing away from the origin) and the area of the window.

Problem 2.154

𝑧 node 1 (20, 20, 12) cm node 2 (25, 40, 7) cm node 3 (0, 35, 13) cm node 4 (5, 20, 15) cm 𝑥 Figure P2.154

ISTUDY

node 4

A flat quadrilateral plate finite element is shown (see Example 2.21 on p. 110 for a description of what a finite element is).

node 3 𝑦

node 1 node 2

(a) Describe and perform a test, using two cross products that will verify if all four nodes (corners) lie in the same plane. (b) Determine the unit outward normal direction to the surface (i.e., pointing away from the origin). (c) Determine the surface area of the element.

Problem 2.155 A contractor is required to prepare a soil surface so that it is planar and the normal direction to the surface is within 0.5◦ of vertical. If the elevations at points 𝐴 and 𝐵, relative to the elevation at point 𝐶, are −3.3 f t and −4.2 f t, respectively, determine the angle between the soil surface’s normal direction and the vertical, and if the surface is sufficiently level. 𝑧 500 f t

400 f t 𝐴

𝐶

𝑥

𝐵

𝑦

Figure P2.155 and P2.156

Problem 2.156 Repeat Prob. 2.155 if the elevations at points 𝐴 and 𝐵, relative to the elevation at point 𝐶, are 2.6 f t and −3.1 f t, respectively.

ISTUDY

Section 2.5

Vector Cross Product

113

Problem 2.157 An ergonomically designed key for a computer keyboard has an approximately flat surface defined by points 𝐴, 𝐵, and 𝐶 and is subjected to a 1 N force in the direction normal to the key’s surface. (a) In Fig. P2.157(a), motion of the key is in the 𝑧 direction. Determine the components of the force in directions normal and tangent to the key’s motion. Comment on why it might be important to know these components. (b) In Fig. P2.157(b), the switch mechanism is repositioned so that the motion of the key is in the direction of the line connecting points 𝐷 and 𝐸, where this line has 𝑥, 𝑦, and 𝑧 direction cosines of 0.123, 0.123, and 0.985, respectively. Determine the components of the force in directions normal and tangent to the key’s motion. Comment on why this design might be more effective than that in Part (a).

𝐴 (15, 0, 10) mm 𝑧 𝐵 (0, 0, 15) mm 𝐸 1N 1N 𝐶 (0, 15, 12) mm 𝐵 𝐵 𝐶 𝐶 𝐴 𝑦 𝑦

𝑧

𝐴

𝑥

𝑥

𝐷 (b)

(a) Figure P2.157

Problem 2.158

𝑧

Impact of debris, both natural and artificial, is a significant hazard for spacecraft. When the Space Shuttle was in service (it is now retired), windows were routinely replaced because of damage due to impact with small objects, and flights employed evasive maneuvers to avoid impact with larger objects, whose orbits NASA constantly monitors. For the triangular-shaped window and the relative velocity of approach 𝑣⃗ shown, determine the components and vector components of the velocity in directions normal and tangent to the window. Note that this information is needed before an analysis of damage due to impact can be performed.

𝐶 (0, 0, 50) cm

𝐵 (0, 40, 0) cm 𝑣⃗ 𝑥

𝑦

space debris

𝐴 (70, 10, 5) cm

̂ 𝑣⃗ = (12 km∕s)(−0.545 𝚤̂ − 0.818 𝚥̂ − 0.182 𝑘)

Problem 2.159 The velocity of air approaching the rudder of an aircraft has magnitude 900 ft/s in the 𝑦 direction. The rudder rotates about line 𝑂𝐴. (a) The position vector from 𝐵 to 𝐶 has the 𝑥, 𝑦, and 𝑧 direction cosines sin 𝛼, (cos 𝛼)(cos 20◦ ), and (− cos 𝛼)(sin 20◦ ), respectively. If the rudder is rotated so that 𝛼 = 10◦ , determine the components, and vector components, of the air velocity in directions normal and tangent to the surface of the rudder. (b) Using the geometry shown in Fig. P2.159, verify the direction cosines stated in Part (a). 𝑧 𝐴 𝐵 𝑣 20◦ 𝑂

𝑥 Figure P2.159

𝐶

𝛼 𝑦

Figure P2.158

114

Chapter 2

Vectors: Force and Position

𝑦

Problem 2.160

𝑟⃗1

𝑧

𝑃⃗

Beam 𝐴𝐵 has rectangular cross section where point 𝐴 is at the origin of the coordinate system and point 𝐵 has the coordinates 𝐵 (1.2, −0.3, 2.4) m. The vector 𝑟⃗1 = (1 𝚤̂ + ̂ m is perpendicular to the axis 𝐴𝐵 of the beam (⃗𝑟 ) and is parallel to the 12 𝚥̂ + 1 𝑘) 𝐴𝐵 thin dimension of the cross section.

𝐴 𝑟̂2 𝑥 𝑟⃗𝐴𝐵

𝐵

𝑦

𝑃1

(b) Determine the unit vector 𝑟̂2 that is perpendicular to both 𝑟⃗1 and 𝑟⃗𝐴𝐵 and has the orientation shown in the figure. (c) The force 𝑃⃗ applied to point 𝐵 of the beam has 1000 N magnitude and direction angles 𝜃𝑥 = 144◦ , 𝜃𝑦 = 72◦ , and 𝜃𝑧 = 60◦ . Determine the components of the 1000 N force, namely 𝑃𝐴𝐵 , 𝑃1 , and 𝑃2 , in the directions 𝑟⃗𝐴𝐵 , 𝑟⃗1 , and 𝑟̂2 , respectively, as shown in the figure. Note: For future work in mechanics of materials, it will be necessary to know these forces.

𝑟⃗1

𝑧

(a) Verify that 𝑟⃗1 is indeed perpendicular to 𝑟⃗𝐴𝐵 .

𝐴 𝑟̂2 𝑥 𝑟⃗𝐴𝐵

𝐵

Repeat Prob. 2.160 if the coordinates of point 𝐵 are 𝐵 (130, −60, 180) in., 𝑟⃗1 = (6 𝚤̂ + ̂ in., and the force 𝑃⃗ has 500 lb magnitude and direction angles 𝜃 = 108◦ , 31 𝚥̂ + 6 𝑘) 𝑥 ◦ 𝜃𝑦 = 60 , and 𝜃𝑧 = 36◦ .

𝑃2

𝑃𝐴𝐵

Problem 2.161

Figure P2.160 and P2.161

Problem 2.162 𝑧

Structure 𝐴𝐵𝐶𝐷 is rigid and is built-in at point 𝐴. Point 𝐴 is at the origin of the coordinate system, points 𝐵 and 𝐷 have the coordinates 𝐵 (16, 24, 48) in. and 𝐷 (26, 52, −2) in., and point 𝐶 is at the midpoint of the straight bar 𝐴𝐵. Portion 𝐶𝐷 of the structure is perpendicular to bar 𝐴𝐵.

𝐵

𝐶

𝑃⃗

(a) Verify that portion 𝐶𝐷 of the structure is indeed perpendicular to bar 𝐴𝐵. 𝑦

(b) Determine the unit vector 𝑟̂ that is perpendicular to both 𝐴𝐵 and 𝐶𝐷 such that this vector has a positive 𝑥 component.

𝐴 𝐷

(c) The force 𝑃⃗ applied to point 𝐷 of the structure has 600 lb magnitude and direction angles 𝜃𝑥 = 108◦ , 𝜃𝑦 = 36◦ , and 𝜃𝑧 = 60◦ . Determine the components of the 600 lb force, namely 𝑃𝐴𝐵 , 𝑃𝐶𝐷 , and 𝑃𝑟 , in the directions 𝐴𝐵, 𝐶𝐷, and 𝑟̂, respectively.

𝑥 Figure P2.162 and P2.163

Problem 2.163 Repeat Prob. 2.162 if the coordinates of points 𝐵 and 𝐷 are 𝐵 (20, 60, 90) mm and 𝐷 (40, 110, −15) mm, and the force 𝑃⃗ has 50 N magnitude and direction angles 𝜃𝑥 = 135◦ , 𝜃𝑦 = 60◦ , and 𝜃𝑧 = 60◦ .

𝑧 𝐶

𝑃

Problem 2.164 9 mm 16 mm 𝑂 12 mm

𝐵

𝑦

Determine the smallest distance between point 𝑂 and the infinite plane containing points 𝐴, 𝐵, and 𝐶.

Problem 2.165 𝐴 𝑥 Figure P2.164 and P2.165

ISTUDY

The vector from point 𝑂 to point 𝑃 has magnitude 40 mm and has equal direction angles with the 𝑥, 𝑦, and 𝑧 axes. Determine the smallest distance from point 𝑃 to the infinite plane containing points 𝐴, 𝐵, and 𝐶.

ISTUDY

Section 2.5

Vector Cross Product

Problem 2.166 ⃗ The product 𝑀 ⃗ ⋅ 𝑟⃗ ∕|⃗𝑟 | produces a scalar 𝑀 , The product 𝑟⃗1 × 𝐹⃗ produces a vector 𝑀. 2 2 ‖ ⃗ in the direction of 𝑟⃗ . which is the component of 𝑀 2

⃗ first, followed by the dot product. (a) Evaluate 𝑀‖ by finding 𝑀 (b) Evaluate 𝑀‖ using the scalar triple product. 𝑧 𝐹⃗ 𝑟⃗2

𝑟⃗1 𝑦

𝑥

̂ N 𝐹⃗ = (2 𝚤̂ − 6 𝚥̂ + 9 𝑘) ̂ m 𝑟⃗1 = (3 𝚤̂ + 12 𝚥̂ + 4 𝑘) ̂ m 𝑟⃗2 = (4 𝚤̂ − 𝚥̂ + 8 𝑘) Figure P2.166

Problem 2.167 ⃗ ⋅ 𝐶⃗ As described in connection with Fig. 2.34 on p. 105, the scalar triple product (𝐴⃗ × 𝐵) ⃗ 𝐵, ⃗ and 𝐶. ⃗ Comment on how the provides the volume of the parallelepiped formed by 𝐴, ⃗ ⋅ 𝐶: ⃗ results of the following triple products compare to the value provided by (𝐴⃗ × 𝐵) ⃗ ⋅ 𝐵. ⃗ (a) (𝐴⃗ × 𝐶) ⃗ ⋅ 𝐴. ⃗ (b) (𝐵⃗ × 𝐶) ⃗ ⋅ 𝐴. ⃗ (c) (𝐶⃗ × 𝐵) ⃗ ⋅ 𝐵. ⃗ (d) (𝐶⃗ × 𝐴) Note: Concept problems are about explanations, not computations.

115

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2.6 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

Basic concepts Laws of sines and cosines. See Figs. 2.35 and 2.36. Eqs. (2.7) and (2.8), p. 34

𝐴

𝜃𝑏 𝐶

𝜃𝑎

𝜃𝑐

For a general triangle (Fig. 2.35) . . . sin 𝜃𝑏 sin 𝜃𝑐 sin 𝜃𝑎 = = law of sines, 𝐴 𝐵 𝐶 √ 𝐴 = 𝐵 2 + 𝐶 2 − 2𝐵𝐶 cos 𝜃𝑎 √ law of cosines. 𝐵 = 𝐴2 + 𝐶 2 − 2𝐴𝐶 cos 𝜃𝑏 √ 𝐶 = 𝐴2 + 𝐵 2 − 2𝐴𝐵 cos 𝜃𝑐

𝐵 Figure 2.35 A general triangle.

Eqs. (2.9) and (2.10), p. 34 𝐴

𝜃𝑏

For a right triangle (Fig. 2.36) . . . 𝐵 = 𝐴 cos 𝜃𝑐

𝐶

𝜃𝑐

𝐶 = 𝐴 sin 𝜃𝑐 √ 𝐴 = 𝐵2 + 𝐶 2

𝐵 Figure 2.36 A right triangle.

𝐵 = 𝐴 sin 𝜃𝑏 𝐶 = 𝐴 cos 𝜃𝑏

Pythagorean theorem.

Unit vectors. A unit vector has unit magnitude and is dimensionless. Given any vector 𝑣⃗ having nonzero magnitude, a unit vector 𝑢̂ in the direction of 𝑣⃗ can be written as 𝑦

𝑖̂ = 𝑗̂ = 1

Eq. (2.11), p. 48 𝑣⃗

𝑢̂ = 𝚥̂

𝜃 𝚤̂

𝑥

(a)

𝑦 𝑣⃗ ( 𝑣⃗ sin 𝜃) 𝚥̂ = 𝑣⃗𝑦

𝑣⃗ |𝑣| ⃗

unit vector.

Polar vector representation in two dimensions. Polar vector representation is a simple convention for stating the magnitude and direction of vectors in two dimensions. Polar vector representation is defined in the margin note on p. 33, and an example is 𝐹⃗ = 100 lb @ 30◦ .

Cartesian vector representation

𝜃 𝑥 ( 𝑣⃗ cos 𝜃) 𝚤̂ = 𝑣⃗𝑥 (b)

Figure 2.37 (a) Cartesian coordinate system with unit vectors 𝚤̂ and 𝚥̂ in the 𝑥 and 𝑦 directions, respectively. (b) Resolution of a vector 𝑣⃗ into vector components in 𝑥 and 𝑦 directions.

ISTUDY

. . . or . . .

Cartesian vector representation in two dimensions. A Cartesian coordinate system in two dimensions uses 𝑥 and 𝑦 axes that are orthogonal. A vector 𝑣⃗ can be written in terms of its Cartesian components as shown in Fig. 2.37 as Eqs. (2.12) and (2.13), p. 49 𝑣⃗ = 𝑣⃗𝑥 + 𝑣⃗𝑦 = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂. 𝑣⃗𝑥 and 𝑣⃗𝑦 are called the vector components of 𝑣, ⃗ and 𝑣𝑥 and 𝑣𝑦 are called the scalar components (or simply the components) of 𝑣. ⃗ The magnitude and orientation from the

ISTUDY

Section 2.6

117

Chapter Review

±𝑥 direction are Eq. (2.14), p. 49 (𝑣 ) √ 𝑦 −1 2 2 |𝑣| ⃗ = 𝑣𝑥 + 𝑣𝑦 , 𝜃 = tan magnitude and orientation. 𝑣𝑥

Cartesian vector representation in three dimensions and direction angles. Expressions for vectors in three dimensions are reported here. Vectors in two dimensions are a special case of the following equations with 𝑣𝑧 = 0 and 𝜃𝑧 = 90◦ .

𝑧

Eq. (2.24), p. 66 |𝑣| ⃗ =

𝑣⃗𝑧

√ 𝑣2𝑥 + 𝑣2𝑦 + 𝑣2𝑧

magnitude.

Eq. (2.25), p. 67

𝚥̂

𝚤̂

𝑣⃗𝑥

𝑣⃗ = 𝑣⃗𝑥 + 𝑣⃗𝑦 + 𝑣⃗𝑧

𝑣⃗

𝑘̂ 𝑣⃗𝑦

𝑦

𝑥

= 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ + 𝑣𝑧 𝑘̂ = |𝑣| ⃗ cos 𝜃𝑥 𝚤̂ + |𝑣| ⃗ cos 𝜃𝑦 𝚥̂ + |𝑣| ⃗ cos 𝜃𝑧 𝑘̂ ̂ = |𝑣| ⃗ (cos 𝜃𝑥 𝚤̂ + cos 𝜃𝑦 𝚥̂ + cos 𝜃𝑧 𝑘).

Figure 2.38 Right-hand Cartesian coordinate system with unit vectors 𝚤̂, 𝚥̂, and 𝑘̂ in the 𝑥, 𝑦, and 𝑧 directions, respectively, and resolution of a vector 𝑣⃗ into vector components 𝑣⃗𝑥 , 𝑣⃗𝑦 , and 𝑣⃗𝑧 .

Eq. (2.27), p. 67 𝑧

Summary Box 𝜃𝑥 = angle between positive 𝑥 direction and vector, 𝜃𝑦 = angle between positive 𝑦 direction and vector,

𝑣⃗

𝜃𝑧 = angle between positive 𝑧 direction and vector, ̂ 𝑣⃗ = |𝑣| ⃗ (cos 𝜃𝑥 𝚤̂ + cos 𝜃𝑦 𝚥̂ + cos 𝜃𝑧 𝑘),

𝜃𝑧 𝜃𝑥

cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1,

𝜃𝑦 𝑦

cos 𝜃𝑥 = 𝑣𝑥 ∕|𝑣|, ⃗

cos 𝜃𝑦 = 𝑣𝑦 ∕|𝑣|, ⃗

cos 𝜃𝑧 = 𝑣𝑧 ∕|𝑣|. ⃗

Direction angles have values between 0◦ and 180◦ . If two direction angles (or direction cosines) are known, the third may be determined using cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1.

𝑥 Figure 2.39 Definition of direction angles 𝜃𝑥 , 𝜃𝑦 , and 𝜃𝑧 . Direction angles have values between 0◦ and 180◦ .

Position vectors. The position vector from point 𝑇 (tail) to point 𝐻 (head) is shown in Fig. 2.40 and is written as Eq. (2.28), p. 67 ̂ 𝑟⃗𝑇 𝐻 = (𝑥𝐻 − 𝑥𝑇 ) 𝚤̂ + (𝑦𝐻 − 𝑦𝑇 ) 𝚥̂ + (𝑧𝐻 − 𝑧𝑇 ) 𝑘, where 𝑥𝑇 , 𝑦𝑇 , and 𝑧𝑇 are coordinates of the tail and 𝑥𝐻 , 𝑦𝐻 , and 𝑧𝐻 are coordinates of the head. For applications in two dimensions, Eq. (2.28) also applies with 𝑧𝐻 − 𝑧𝑇 = 0 [this expression is given explicitly by Eq. (2.22)].

𝑧 𝐻(𝑥𝐻 , 𝑦𝐻 , 𝑧𝐻 ) 𝑇 (𝑥𝑇 , 𝑦𝑇 , 𝑧𝑇 ) 𝑥 𝐻 − 𝑥𝑇

𝑟⃗𝑇 𝐻

tail

head

𝑧𝐻 − 𝑧 𝑇 𝑦

𝑦𝐻 − 𝑦𝑇 𝑥

Dot product The dot product between two vectors 𝐴⃗ and 𝐵⃗ produces a scalar 𝑠 and is defined as Eq. (2.29), p. 84, and Eq. (2.34), p. 85 ⃗ 𝐵| ⃗ cos 𝜃 𝑠 = 𝐴⃗ ⋅ 𝐵⃗ = |𝐴|| = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 .

Figure 2.40 Construction of a position vector using Cartesian coordinates of the vector’s head and tail.

118

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The dot product can be used to determine the angle 𝜃 between two vectors using Eq. (2.39), p. 85 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 𝐴⃗ ⋅ 𝐵⃗ = , ⃗ 𝐵| ⃗ ⃗ 𝐵| ⃗ |𝐴|| |𝐴|| 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 𝐴⃗ ⋅ 𝐵⃗ 𝜃 = cos−1 = cos−1 . ⃗ 𝐵| ⃗ ⃗ 𝐵| ⃗ |𝐴|| |𝐴||

cos 𝜃 =

Resolution of a vector into components. Important equations for using the dot product to resolve a vector 𝐹⃗ into parallel and perpendicular components to a direction 𝑟⃗ are collected in the following summary box: Eq. (2.47), p. 88 Summary Box

(See Fig. 2.41.)

𝐹‖ = component of 𝐹⃗ acting in direction 𝑟⃗, 𝐹⃗‖ = vector component of 𝐹⃗ acting in direction 𝑟⃗, 𝐹⊥ = component of 𝐹⃗ acting perpendicular to 𝑟⃗, 𝐹⃗⊥ = vector component of 𝐹⃗ acting perpendicular to 𝑟⃗, 𝑟⃗ , 𝐹‖ = 𝐹⃗ ⋅ |⃗𝑟| √ 𝐹⊥ = 𝐹 2 − 𝐹‖2 , 𝑦

𝑟⃗ 𝐹⃗‖ = 𝐹‖ , |⃗𝑟| 𝐹⃗⊥ = 𝐹⃗ − 𝐹⃗‖ . 𝑦

𝑦 𝐹⃗

𝐹

𝑥

(a)

𝐹⃗⊥

𝐹⃗‖

𝐹‖

𝑟⃗

𝑧

𝐹⃗

𝐹⊥

𝑥 𝑧

(b)

𝑥 𝑧

(c)

Figure 2.41. (a) Vectors 𝐹⃗ and 𝑟⃗ in three dimensions. (b) Resolution of 𝐹⃗ into scalar components in directions parallel and perpendicular to 𝑟⃗. (c) Resolution of 𝐹⃗ into vector components in directions parallel and perpendicular to 𝑟⃗.

Cross product The cross product between two vectors 𝐴⃗ and 𝐵⃗ produces a vector 𝐶⃗ as shown in Fig. 2.42 and is defined as Eq. (2.48), p. 101, and Eq. (2.53), p. 102 𝐶⃗ = 𝐴⃗ × 𝐵⃗ ⃗ 𝐵| ⃗ sin 𝜃) 𝑢̂ = (|𝐴|| ̂ = (𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 ) 𝚤̂ + (𝐴𝑧 𝐵𝑥 − 𝐴𝑥 𝐵𝑧 ) 𝚥̂ + (𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 ) 𝑘.

ISTUDY

Section 2.6

Chapter Review

𝑦

𝑦 𝐵⃗

𝑦

𝐴⃗ × 𝐵⃗

𝐴⃗ × 𝐵⃗

𝐵⃗

𝐵⃗

𝜃 𝐴⃗

𝐴⃗

𝐴⃗

𝑥 𝑧

(a)

𝑥

𝑧

𝑥

𝑧

(b)

(c)

Figure 2.42. (a) Vectors 𝐴⃗ and 𝐵⃗ in three dimensions. (b) To evaluate the cross product ⃗ the vectors can be arranged tail to tail to define angle 𝜃, which is between vectors 𝐴⃗ and 𝐵, measured in the plane containing the two vectors. (c) The result of 𝐴⃗ × 𝐵⃗ is a vector whose direction is governed by the right-hand rule.

For vectors with Cartesian representations, the cross product may conveniently be evaluated by expanding the determinant Eq. (2.58), p. 103 ⎡ 𝚤̂ 𝐴⃗ × 𝐵⃗ = det ⎢ 𝐴𝑥 ⎢ ⎣ 𝐵𝑥

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

⎤ || 𝚤̂ ⎥ = || 𝐴 ⎥ || 𝑥 ⎦ | 𝐵𝑥

𝚥̂ 𝐴𝑦 𝐵𝑦

𝑘̂ 𝐴𝑧 𝐵𝑧

| | | |, | | |

using Method 1 or 2 described in Section 2.5. Also, the magnitude of 𝐴⃗ × 𝐵⃗ is the area of the parallelogram formed by 𝐴⃗ and 𝐵⃗ arranged tail to tail (see Fig. 2.33).

Scalar triple product. A cross product that is followed by a dot product can be simultaneously evaluated using the scalar triple product Eq. (2.61), p. 105 | 𝐶 | 𝑥 ⃗ ⋅ 𝐶⃗ = || 𝐴 (𝐴⃗ × 𝐵) | 𝑥 | 𝐵𝑥 |

𝐶𝑦 𝐴𝑦 𝐵𝑦

𝐶𝑧 𝐴𝑧 𝐵𝑧

| | | |. | | |

⃗ ⋅ 𝐶⃗ is the The scalar triple product produces a scalar. Also, the value produced by (𝐴⃗ × 𝐵) ⃗ ⃗ ⃗ volume of the parallelepiped defined by 𝐴, 𝐵, and 𝐶 arranged tail to tail (see Fig. 2.34).

119

120

Chapter 2

Vectors: Force and Position

Review Problems Problem 2.168 The manufacturer of a welded steel bracket specifies the working loads depicted in the figure of 𝑅𝑛 versus 𝑅𝑡 . Values of 𝑅𝑛 and 𝑅𝑡 that lie within the shaded region are allowable, while values that lie outside of the region are unsafe. Such a diagram is often called an interaction diagram because it characterizes the combined effect that multiple loads have on the strength of a component. For the loading and geometry shown, determine the range of values load 𝑃 may have and still satisfy the manufacturer’s combined loading criterion. 𝑅𝑛 𝑦, 𝑛 5 kN

𝑅𝑛

𝑃

𝑅𝑡 𝑅𝑡

4 10 kN

3

𝐹 = 2 kN

𝑥, 𝑡

5 kN 5 kN Figure P2.168 𝑦 𝑛

Problem 2.169 𝑃 = 4 kN

𝑄 = 2 kN 10◦

𝑡 30◦

For the loading and geometry shown, use the interaction diagram of Prob. 2.168 to determine if the manufacturer’s combined loading criterion is satisfied.

𝑥

Problems 2.170 and 2.171 Figure P2.169

ISTUDY

⃗ (a) Determine the resultant 𝑅⃗ of the three forces 𝐹⃗ + 𝑃⃗ + 𝑄. (b) If an additional force 𝑇⃗ in the ±𝑥 direction is to be added, determine the magnitude it should have so that the magnitude of the resultant is as small as possible. 𝑦

𝑧 𝑄 = 2 lb

𝑃 = 2 kN

𝐹 = 3 lb

𝑄 = 3 kN 50◦ 40◦ 30◦

8 15 60◦

36◦

72◦ 𝑦

40◦

𝑧

30◦ 𝛼 𝛼

36◦ 𝛼

𝐹 = 1 kN

𝑥

𝑃 = 1 lb

𝑥 Figure P2.170

Figure P2.171

ISTUDY

Section 2.6

121

Chapter Review 𝑧

Problems 2.172 and 2.173

new power transmission line

An electrical power transmission line runs between points 𝐴, 𝐵, and 𝐶, as shown, and a new power transmission line between points 𝐵 and 𝐷 is to be constructed. Using the coordinates of points 𝐴, 𝐶, and 𝐷 given below, determine the distance from point 𝐴 to point 𝐵 where the new power line should connect to the existing power line such that the length of the new power line is as small as possible. Also determine the length of the new power line. Assume that power lines 𝐴𝐵𝐶 and 𝐵𝐷 are straight.

𝐷 𝐶 𝑦

𝐵 𝐴

The coordinates of points 𝐴, 𝐶, and 𝐷 are 𝐴(6000, 1000, 500) ft, 𝐶(700, 4500, 900) ft, and 𝐷(2000, 400, 1400) ft.

Problem 2.172

The coordinates of points 𝐴, 𝐶, and 𝐷 are 𝐴(3000, 600, 300) m, 𝐶(600, 2200, 500) m, and 𝐷(1000, 300, 800) m. Problem 2.173

existing power transmission line 𝑥 Figure P2.172 and P2.173 𝑦

Problems 2.174 and 2.175

20 m 𝑄

𝐺

An architect specifies the roof geometry shown for a building. Each of lines 𝐸𝐴𝐵, 𝐵𝐷, 𝐷𝐶𝐺, 𝐸𝐺, and 𝐴𝐶 are straight. Two forces of magnitude 𝑃 and 𝑄 acting in the −𝑦 direction are applied at the positions shown.

𝐶

20 m 40 m

Determine the components of the force 𝑃 in directions normal and parallel to the roof at point 𝐴. Express your answer in terms of 𝑃 .

𝐷 𝑃

Problem 2.174

Determine the components of the force 𝑄 in directions normal and parallel to the roof at point 𝐶. Express your answer in terms of 𝑄. Problem 2.175

Problem 2.176 A concrete surface for a parking lot is to be prepared such that it is planar and the normal direction to the surface is 2◦ ± 0.5◦ of vertical so that rainwater will have adequate drainage. If the coordinates of points 𝐴, 𝐵, and 𝐶 are 𝐴 (300, 50, 2) ft, 𝐵 (20, 150, −5) ft, and 𝐶 (280, 450, −8) ft, determine the angle between the concrete surface’s normal direction and the vertical, and if the concrete surface meets the drainage specification cited above. 𝑧

𝑥

𝐴

𝐵 𝐶

𝑦

Figure P2.176 and P2.177

Problem 2.177 Repeat Prob. 2.176 if the coordinates of points 𝐴, 𝐵, and 𝐶 are 𝐴 (100, 20, 1) m, 𝐵 (10, 50, −3) m, and 𝐶 (110, 160, 2) m.

𝑧

𝐸 20 m

𝐴 20 m

𝐵

Figure P2.174 and P2.175

𝑥

122

Chapter 2

Vectors: Force and Position

Problem 2.178 A specimen of composite material consisting of ceramic matrix and unidirectional ceramic fiber reinforcing is tested in a laboratory under compressive loading. If a 10 kN force is applied in the −𝑧 direction, determine the components, and vector components, of this force in directions parallel and perpendicular to the fiber direction 𝑓 , where this direction has direction angle 𝜃𝑧 = 40◦ and remaining direction angles that are equal (i.e., 𝜃𝑥 = 𝜃𝑦 ). 𝑧

𝑃 = 10 kN 𝑓 unidirectional fiber reinforcement 𝑦

𝑥

𝑧

Figure P2.178

̂ N 𝐹⃗ = (40 𝚤̂ − 80 𝚥̂ − 80 𝑘) 𝐶

𝐹⃗

Problem 2.179 𝐵

𝐴 𝑥

An automobile body panel is subjected to a force 𝐹⃗ from a stiffening strut. Assuming that region 𝐴𝐵𝐶 of the panel is planar, determine the components, and vector components, of 𝐹⃗ in the directions normal and parallel to the panel.

𝑦

𝐴 (130, 0, 60) mm 𝐵 (0, 180, 0) mm 𝐶 (0, 0, 120) mm

Problems 2.180 and 2.181

Figure P2.179

The steering wheel and gearshift lever of an automobile are shown. Points 𝐴, 𝐶, and 𝐵 lie on the axis of the steering column, where point 𝐴 is at the origin of the coordinate system, point 𝐵 has the coordinates 𝐵 (−120, 0, −50) mm, and point 𝐶 is 60 mm from point 𝐴. The gearshift lever from 𝐶 to 𝐷 has 240 mm length and the direction angles given below. 𝑧

𝑃⃗ Problem 2.180

(a) Two of the direction angles for the position vector from point 𝐶 to 𝐷 are 𝜃𝑦 = 50◦ and 𝜃𝑧 = 60◦ . Knowing that this position vector has a negative 𝑥 component, determine 𝜃𝑥 .

𝐷 𝐴

𝐶

(b) Determine the unit vector 𝑟̂ that is perpendicular to directions 𝐴𝐵 and 𝐶𝐷 such that this vector has a positive 𝑧 component.

𝐵 𝑥 𝑦 Figure P2.180 and P2.181

ISTUDY

̂ N, determine the (c) If the force applied to the gearshift lever is 𝑃⃗ = (−6 𝚤̂ − 4 𝚥̂ + 12 𝑘) ⃗ component of 𝑃 , namely 𝑃𝑟 , in the direction of 𝑟̂. Problem 2.181

(a) Two of the direction angles for the position vector from point 𝐶 to 𝐷 are 𝜃𝑦 = 45◦ and 𝜃𝑧 = 70◦ . Knowing that this position vector has a negative 𝑥 component, determine 𝜃𝑥 . (b) Determine the unit vector 𝑟̂ that is perpendicular to directions 𝐴𝐵 and 𝐶𝐷 such that this vector has a positive 𝑧 component. ̂ N, determine (c) If the force applied to the gear shift lever is 𝑃⃗ = (−9 𝚤̂ − 6 𝚥̂ + 18 𝑘) the component of 𝑃⃗ , namely 𝑃𝑟 , in the direction of 𝑟̂.

ISTUDY

Section 2.6

123

Chapter Review 𝑧

Problem 2.182

flanges

An I beam is positioned from points 𝐴 to 𝐵. Because its strength and deformation properties for bending about an axis through the web of the cross section are different than those for bending about an axis parallel to the flanges, it is necessary to also characterize these directions. This can be accomplished by specifying just one of the direction angles for the direction of the web from 𝐴 to 𝐶, which is perpendicular to line 𝐴𝐵, plus the octant of the coordinate system in which line 𝐴𝐶 lies. (a) If direction angle 𝜃𝑧 = this line.

30◦

𝐶 𝐵 web 𝐴 𝐴 (8, 0, 3) f t 𝐵 (0, 6, 3) f t

for line 𝐴𝐶, determine the remaining direction angles for

(b) Determine the unit vector in the direction perpendicular to the web of the beam (i.e., perpendicular to lines 𝐴𝐵 and 𝐴𝐶).

𝑦

𝑥 Figure P2.182

Problem 2.183 Determine the smallest distance between the infinite lines passing through bars 𝐴𝐵 and 𝐶𝐷. 𝑦 11 f t

𝐷

10 f t 8 ft

𝐶 2 ft 𝐵 6 ft

𝑥

2 ft

𝑧

8 ft 9 ft

𝐴

Figure P2.183

Problems 2.184 and 2.185 A portion of a downhill ski run between points 𝐶 and 𝐷 is to be constructed on a mountainside, as shown. Let the portion of the mountainside defined by points 𝐴, 𝐵, and 𝐶 be idealized to be planar. For the values of 𝑥𝐴 , 𝑦𝐵 , and 𝑧𝐶 given below, determine the distance from point 𝐴 to point 𝐷 where the ski run should intersect line 𝐴𝐵 so that the run will be as steep as possible. Hint: Consider a force acting on the slope in the −𝑧 direction, such as perhaps a skier’s weight – or better yet, a 1 lb weight – and follow the approach used in Example 2.20 on p. 108 to resolve this weight into normal and tangential components; the direction of the tangential component will give the direction of steepest descent. Problem 2.184

𝑥𝐴 = 1500 f t, 𝑦𝐵 = 2000 f t, and 𝑧𝐶 = 800 f t.

Problem 2.185

𝑥𝐴 = 1200 f t, 𝑦𝐵 = 1600 f t, and 𝑧𝐶 = 900 f t.

𝑧 𝑥𝐴

𝑧𝐶

𝐶

𝑦𝐵

𝐴

𝑥

ski run 𝐷 𝐵 𝑦

Figure P2.184 and P2.185

124

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Chapter 2

Vectors: Force and Position

Problem 2.186 The tetrahedron shown arises in advanced mechanics, and it is necessary to relate the areas of the four surfaces. Show that the surface areas are related by 𝐴𝑥 = 𝐴 cos 𝜃𝑥 , 𝐴𝑦 = 𝐴 cos 𝜃𝑦 , and 𝐴𝑧 = 𝐴 cos 𝜃𝑧 where 𝐴 is the area of surface 𝐴𝐵𝐶, and cos 𝜃𝑥 , cos 𝜃𝑦 , and cos 𝜃𝑧 are the direction cosines for normal direction 𝑛⃗. Hint: Find 𝑛⃗ by taking the cross product of vectors along edges 𝐴𝐵, 𝐴𝐶, and/or 𝐵𝐶, and note that the magnitude of this vector is 2𝐴. Then, by inspection, write expressions for 𝐴𝑥 , 𝐴𝑦 , and 𝐴𝑧 (e.g., 𝐴𝑥 = 𝑦𝑧∕2 and so on). 𝑦 𝑛⃗ 𝐵 𝑦

𝐴𝑥

𝑥 𝐴𝑧

𝑂

𝐴

𝑧 𝐶

𝐴𝑦

𝑧 Figure P2.186

𝑥

ISTUDY

3

Equilibrium of Particles

Many engineering problems can be accurately idealized as a single particle or system of particles in equilibrium. This chapter discusses criteria for when a particle, or system of particles, is in equilibrium, and it presents a systematic analysis procedure that can be applied to general problems. Early sections of this chapter focus on twodimensional problems, and later sections treat three-dimensional problems. Engineering design applications are also discussed.

Oleg Totskyi/Shutterstock

Tower cranes at a construction site are used for lifting items such as beams, concrete, steel reinforcing bar, and other items. Many of the components in this photo can be idealized as particles in equilibrium.

3.1 Equilibrium of Particles in Two Dimensions In mechanics, a particle is defined to have zero volume, but it may have mass. While there are no true particles in nature, under the proper circumstances we can idealize a real life object as a particle. An object that is small compared to other objects and/or dimensions in a problem may often be idealized as a particle. A large object may often be idealized as a particle, and whether or not this is possible depends on the forces that are applied to the object. In the example shown in Fig. 3.1, the lines of action of all forces intersect at a common point, and this is called a concurrent force system. An object subjected to a concurrent force system may be idealized as a particle as shown in the figure. There are other circumstances in which an object, even if large, may be idealized as a particle. In this chapter, you will begin to develop the ability to recognize when an object may be idealized as a particle, and as you progress through subsequent chapters of this book, this ability will be further sharpened.

𝑦

𝑦 𝐹⃗

𝑃⃗

𝑃⃗

⇒ 𝑅⃗

𝑅⃗ ⃗ 𝑊

𝑥

𝐹⃗ ⃗ 𝑊 𝑥

Figure 3.1 Example of a two-dimensional concurrent force system where the lines of action of all forces intersect at a common point. When subjected to a concurrent force system, an object—even if it is very large—can be idealized as a particle for many purposes.

125

126

ISTUDY

Chapter 3

Equilibrium of Particles

Newton’s laws, discussed in Chapter 1, provide the conditions under which a particle subjected to forces is in static equilibrium. In particular, Newton’s second law ∑ states 𝐹⃗ = 𝑚𝑎, ⃗ where we have included the summation sign to emphasize that all ⃗ For forces applied to the particle must be included. Static equilibrium means 𝑎⃗ = 0. brevity, throughout the rest of this book we will use the word equilibrium to mean static equilibrium. Hence, the conditions for equilibrium of a particle are

or or

Helpful Information More on static equilibrium. A particle is in static equilibrium if its acceleration is zero ⃗ Thus, a particle is in static equilibrium (⃗ 𝑎 = 0). if it has no motion (i.e., is at rest) or if it moves with constant velocity (i.e., has uniform speed and uniform direction). A particle that has changing velocity (i.e., has nonuniform speed and/or nonuniform direction) is not in static equilibrium, and dynamics is needed to determine its response.

∑

⃗ 𝐹⃗ = 0, (∑ ) ⃗ 𝐹𝑦 𝚥̂ = 0, 𝐹𝑥 𝚤̂ + ∑ ∑ 𝐹𝑥 = 0, and 𝐹𝑦 = 0. (∑

)

(3.1)

∑ In Eq. (3.1), the expression 𝐹⃗ = 0⃗ is valid regardless of the type of vector representation used, while the remaining expressions result if Cartesian vector representation ∑ ∑ ∑ is used. The expression 𝐹⃗ = ( 𝐹𝑥 ) 𝚤̂ + ( 𝐹𝑦 ) 𝚥̂ = 0⃗ states conditions for equilib∑ ∑ rium in vector form, while the expressions 𝐹𝑥 = 0 and 𝐹𝑦 = 0 state conditions for equilibrium in scalar form. Both the vector and scalar forms are completely equivalent, and the choice of which of these to use for a particular problem is a matter of convenience. The vector form provides a compact and concise description of equilibrium. In complex problems, especially in three dimensions, and especially when rigid bodies are involved (these are addressed later in this book), it will often be advantageous to use this form. For many problems, especially in two dimensions where the geometry of forces is straightforward, the scalar form will be effective. When using the scalar form, we compute components of forces in 𝑥 and 𝑦 directions and sum forces in each of these directions. Particles and forces. In particle equilibrium problems, the particle under consideration may represent an individual particle of a real body or structure, or the particle may represent a portion of the real body or structure, or the particle may represent the ∑ entire body or structure (as in the example of Fig. 3.1). When applying 𝐹⃗ = 0⃗ to the particle, all forces that are applied to the particle must be included. These forces have a number of sources, as follows. • Some of the forces may be due to interaction of the particle with its environment, such as weight due to gravity, force of wind blowing against a structure, forces from magnetic attraction of nearby objects, and so on. • Some of the forces may be due to structural members that are attached to (or contain) the particle. For example, if a particle has a cable attached to it, the cable will usually apply a force to the particle. • Some of the forces may be due to supports. For example, if a particle (or the body the particle represents) is glued to a surface, the glue will usually apply forces to the particle. We call forces such as these reaction forces, and more is said about these later in this section.

Free body diagram (FBD) A free body diagram (FBD) is a sketch of a body or a portion of a body that is separated, or made free, from its environment and/or other parts of the structural system,

ISTUDY

Section 3.1

Equilibrium of Particles in Two Dimensions

127

and all forces that act on the body must be shown in the sketch. We often use the word cut to describe the path along which the free body is removed from its environment. In this chapter, the FBD will always result in a particle that is in equilibrium. In subsequent chapters, the FBD may result in an object of finite size in equilibrium. An FBD is an essential aid for applying Newton’s law of motion. It is a tool that helps ensure that all forces that are applied to the FBD are accounted for, including ∑ their proper directions. Once an FBD is drawn, application of 𝐹⃗ = 𝑚𝑎⃗ may proceed. ⃗ whereas in subjects that follow statics, In statics we seek conditions so that 𝑎⃗ = 0, such as dynamics and vibrations, we seek to determine 𝑎⃗ and how the motion of a body or structure evolves with time. Among all of the concepts you will learn in mechanics, regardless of how advanced a level at which you eventually study, the ability to draw accurate FBDs is one of the most important skills you will need. For students and practicing engineers alike, FBDs are used on a daily basis.

Procedure for Drawing FBDs 1. Decide on the particle whose equilibrium is to be analyzed. 2. Imagine this particle is “cut” completely free (separated) from the body, structure, and/or its environment. That is: • In two dimensions, think of a closed line that completely encircles the particle. • In three dimensions, think of a closed surface that completely surrounds the particle. 3. Sketch the particle (i.e., draw a point). 4. Sketch the forces: (a) Sketch the forces that are applied to the particle by the environment (e.g., weight). (b) Wherever the cut passes through a structural member, sketch the forces that occur at that location. (c) Wherever the cut passes through a support, sketch the reaction forces that occur at that location. 5. Sketch the coordinate system to be used. Add pertinent dimensions and angles to the FBD to fully define the locations and orientations of all forces.

Because this chapter deals with equilibrium of particles, Step 3 in the above procedure is trivial. But, with very small modification the same procedure will be used for rigid body equilibrium, where this step will require a bit more artistry. The order in which the forces are sketched in Step 4 is irrelevant. For complicated FBDs, it may be difficult to include all of the dimensions and/or angles in Step 5. When this is the case, some of this information may be obtained from a different sketch. In many problems it will be relatively clear which particle should be used for drawing an FBD. In complex problems this selection may require some thought and perhaps some trial and error. For a single particle in two dimensions, Eq. (3.1)

Helpful Information Free body diagram (FBD). An FBD is an essential aid, or tool, for applying Newton’s ∑ law of motion 𝐹⃗ = 𝑚𝑎. ⃗ Among the many skills you will need to be successful in statics, and in the subjects that follow, and as a practicing engineer, the ability to draw accurate FBDs is essential. An incorrect FBD is the most common source of errors in an analysis.

128

Chapter 3

Equilibrium of Particles

∑ ∑ provides two scalar equations 𝐹𝑥 = 0 and 𝐹𝑦 = 0, and if the FBD involves two unknowns, then these equilibrium equations are sufficient to yield the solution.∗ In more complex problems, the FBD will often involve more than two unknowns, in which case FBDs must be drawn and equilibrium equations written for additional particles so that the final system of equations has as many equations as unknowns. Mini-Example A skier uses a tow rope as shown in Fig. 3.2(a) to reach the top of a ski hill. If the skier weighs 150 lb, if the snow-covered slope can be considered frictionless, and if the portion of the tow rope behind the skier is slack,† determine the force required in the tow rope to pull the skier at constant velocity.

cut

𝑊 = 150 lb 15◦

15◦

𝑊 = 150 lb

𝑇

15◦ 𝑇

15◦ 𝐴

𝐴 𝑦

20◦

20◦

20◦ 𝑅

𝑥

20◦

𝑅

Bridget Clyde/StockShot/Alamy Stock Photo

(a)

(b)

(c)

(d)

Figure 3.2. (a) Photograph of a skier being towed up a snow-covered slope at constant velocity. (b) Sketch of the skier showing the orientation of the slope and tow rope, and the cut used for drawing the FBD. (c) Free body diagram of the skier showing the lines of action of all forces intersecting at a common point. (d) Free body diagram where the skier is idealized as a particle. All FBDs assume the portion of the tow rope behind the skier is slack.

ISTUDY

Solution Draw FBD. To draw the FBD and to determine if the skier can be idealized as a particle, we use the cut shown in Fig. 3.2(b) to separate the skier from the environment. The forces applied to the skier’s body by the environment outside of the cut are shown in Fig. 3.2(c), as follows. The 150 lb force represents the skier’s weight. Where the cut passes through the tow rope in front of the skier, the tow rope applies a force 𝑇 to the skier’s hands. Where the cut passes through the tow rope behind the skier, there is no force since we have assumed that portion of the rope is slack. Where the cut passes between the skis and slope, there is a reaction force 𝑅. Note that between the skis and the slope, the actual forces are distributed over the full contact area of the skis, and this distribution is probably complicated. In Fig. 3.2(c), we are modeling this distribution by a single force 𝑅. As seen in Fig. 3.2(c), all of the forces applied to the skier by the environment outside of the cut intersect at a common point (point 𝐴). Thus, the skier’s body may be idealized as a particle. That this is true will be more thoroughly explored when rigid bodies are discussed in Chapter 5. Furthermore, since the skier is to move with constant velocity, the skier’s acceleration is zero, and hence this is a ∗ There

are occasional subtle exceptions to this statement, such as when a structure is simultaneously statically indeterminate and a mechanism. Such exceptions are discussed later in this book. † In Prob. 3.10, the portion of the tow rope behind the skier will be taut.

ISTUDY

Section 3.1

Equilibrium of Particles in Two Dimensions

problem of static equilibrium of a particle. To complete the FBD, we select a coordinate system. Since the 15◦ and 20◦ orientations are given with respect to the horizontal direction, a coordinate system where 𝑥 and 𝑦 are horizontal and vertical is convenient. Once the appropriate particle has been identified and its FBD has been drawn, we may then write the equations of equilibrium, followed by solving these to determine the unknowns (i.e., the force 𝑇 in the tow rope and the reaction 𝑅 between the skis and slope). In what follows, we demonstrate a vector solution, followed by a scalar solution, followed by comments on the merits of using an alternate coordinate system. Vector solution of equilibrium equations. To carry out a vector solution, we begin by writing vector expressions for all forces as follows: ⃗ = −150 lb 𝚥̂, 𝑊 ( ) 𝑇⃗ = 𝑇 cos 15◦ 𝚤̂ + sin 15◦ 𝚥̂ , ( ) 𝑅⃗ = 𝑅 − sin 20◦ 𝚤̂ + cos 20◦ 𝚥̂ .

(3.2) (3.3) (3.4)

129

Helpful Information What forces should be included in the FBD? When drawing an FBD, only forces that are external to the FBD (i.e., external to the cut used to construct the FBD) are included on the FBD. A common source of error is to include forces that are internal to the FBD. For example, the skier’s boots are attached to the skis, and there are forces between the boots and skis that keep them connected to one another. However, these forces are internal to the FBD, because the cut that was used to construct the FBD did not separate the boots from the skis. To take this idea a step further, every atom of material within the skier’s body and clothing exerts forces on neighboring atoms. However, all such forces are internal to the FBD and hence do not appear on the FBD.

Next we apply Newton’s law with 𝑎⃗ = 0⃗ to write ∑

⃗ ⃗ + 𝑇⃗ + 𝑅⃗ = 0, 𝐹⃗ = 0⃗ ∶ 𝑊 ( ) 𝑇 cos 15◦ − 𝑅 sin 20◦ 𝚤̂ ( ) ⃗ + −150 lb + 𝑇 sin 15◦ + 𝑅 cos 20◦ 𝚥̂ = 0.

(3.5)

(3.6)

For the above vector equation to be satisfied, both the 𝑥 and 𝑦 components must be zero independently, which provides two scalar equations. Also, we observe that there are two unknowns, 𝑇 and 𝑅. Thus, the system of equations is determinate, meaning there are as many equations as unknowns, and a solution for the unknowns is obtainable. Writing the two scalar equations contained in Eq. (3.6) provides 𝑇 cos 15◦ − 𝑅 sin 20◦ = 0,

(3.7)

𝑇 sin 15 + 𝑅 cos 20 = 150 lb.

(3.8)

◦

◦

Basic algebra is used to solve these equations. For example, if Eq. (3.7) is multiplied by cos 20◦ and Eq. (3.8) is multiplied by sin 20◦ and the results are added, the terms containing 𝑅 sum to zero and we obtain ) ( 𝑇 cos 15◦ cos 20◦ + sin 15◦ sin 20◦ = (150 lb) sin 20◦ ⇒

𝑇 = 51.50 lb.

(3.9) (3.10)

Once one of the unknowns has been determined, 𝑇 in this case, either of Eqs. (3.7) and (3.8) may be used to determine the other unknown. Using Eq. (3.7), we write 𝑅= which completes the solution.

𝑇 cos 15◦ = 145.4 lb, sin 20◦

(3.11)

Helpful Information Some useful checks of our solution. Intuitively, we know that the portion of the tow rope in front of the skier will be in tension. From Fig. 3.2(c), we observe that the direction of 𝑇 in the FBD has been assigned so that a positive value of 𝑇 corresponds to tension in the rope. Thus, we expect the solution to this problem to display a positive value for 𝑇 , and indeed this is the case. Further, the reaction 𝑅 is positive in compression. Thus, we expect the solution to this problem to display a positive value for 𝑅, and indeed this is also the case. Had we obtained a negative value for either 𝑇 or 𝑅, then we would suspect an error in our solution.

130

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Equilibrium of Particles

Scalar solution of equilibrium equations. We examine Fig. 3.2(d) and sum the 𝑥 and 𝑦 components of forces to write ∑ 𝑇 cos 15◦ − 𝑅 sin 20◦ = 0, (3.12) 𝐹𝑥 = 0 ∶ ∑ 𝐹𝑦 = 0 ∶ 𝑇 sin 15◦ + 𝑅 cos 20◦ − 150 lb = 0. (3.13) These equations are identical to Eqs. (3.7) and (3.8), and thus, the solution is the same as that obtained using the vector approach. Solution with alternate coordinate system. In the foregoing solutions we elected to use a horizontal-vertical Cartesian coordinate system, but this selection was arbitrary, and in some problems a coordinate system with different orientation may be more convenient. For example, consider the tn coordinate system shown in Fig. 3.3. Using a scalar solution approach, we sum forces in 𝑡 and 𝑛 directions to obtain ∑ 𝐹𝑡 = 0 ∶ −(150 lb) sin 20◦ + 𝑇 cos 5◦ = 0, (3.14) ∑ 𝐹𝑛 = 0 ∶ −(150 lb) cos 20◦ − 𝑇 sin 5◦ + 𝑅 = 0. (3.15)

𝑊 = 150 lb 5◦ 𝑛

20◦

5◦

𝑇

𝑡

𝑅 20◦ (a)

Notice that because of the choice of coordinate system, Eq. (3.14) has only one unknown, which is easily found to be

(b)

Figure 3.3 (a) Skier being towed up a slope at constant velocity. (b) Free body diagram with orientation of forces given with respect to a 𝑡𝑛 coordinate system where 𝑡 and 𝑛 are parallel and perpendicular to the slope, respectively. This FBD assumes the portion of the tow rope behind the skier is slack.

ISTUDY

𝑇 =

(150 lb) sin 20◦ = 51.50 lb. cos 5◦

(3.16)

Then, using Eq. (3.15), the remaining unknown 𝑅 is found to be 𝑅 = (150 lb) cos 20◦ + 𝑇 sin 5◦ = 145.4 lb.

(3.17)

As expected, both solutions are the same as those obtained earlier.

Modeling and problem solving The process of drawing an FBD involves modeling, wherein a real life problem is replaced by an idealization. For example, in Fig. 3.2, the actual distribution of forces between the skis and slope was modeled by a single reaction force, friction between the skis and slope was neglected, the portion of the tow rope behind the skier was assumed to be slack, and so on. In modeling, reasonable assumptions are made about what is physically important in a system and what is not, with the goal of developing a model that contains the essential physics while hopefully being simple enough to allow for a tractable mathematical solution. Effective modeling is both an art and a science. Sometimes it may take several iterations to develop a good model for a problem. Once a model has been established, the next step of an analysis is largely an exercise in mathematics. In the example problems in this book that deal with equilibrium concepts,∗ the following structured problem-solving approach will be used:

∗ In

example problems that do not deal with equilibrium concepts, such as those in Chapters 1 and 2, the modeling step may not be required.

ISTUDY

Section 3.1

Equilibrium of Particles in Two Dimensions

131

Structured Problem-Solving Approach • Road Map: Identify data and unknowns and create a solution strategy. • Modeling: Identify the system and make reasonable assumptions leading to the development of an FBD (or possibly multiple FBDs). • Governing Equations: Write the equations of equilibrium, and as needed, additional equations for other physical phenomena pertinent to the problem. • Computation: Solve the system of governing equations. • Discussion and Verification: The results are critically examined and interpreted. Road Map & Modeling: In some cases, the development of the solution strategy is dependent on the model created. In these cases, the first two steps are carried out simultaneously. Governing Equations & Computation: Occasionally, governing equations will be written and immediately solved, and then additional equations will be written and solved, and so on. When this is the case, these steps will be listed together.

Cables and bars In the foregoing example of Fig. 3.2, the FBD was constructed so that a positive value of 𝑇 corresponded to tension in the tow rope that pulls the skier up the slope. It may not be obvious that this statement is true, and it is important that you fully understand why, because construction of FBDs and interpretation of the results of an analysis require this understanding. Consider the structure shown in Fig. 3.4(a), consisting of a bar and two cables that support a bucket weighing 100 N. As discussed in Section 2.3, 30◦ 𝐵 𝐴

30◦ 𝐹𝐴𝐷 𝑦

cut 𝐷

cut 40◦ (a)

𝐶

𝐹𝐴𝐶

(b)

𝐹𝐴𝐷 = 100 N

30◦ 𝐹𝐴𝐵

𝑦

𝐴 𝑥

100 N 𝐷

𝐹𝐴𝐵

40◦ (c)

𝑦

𝐴 𝐹𝐴𝐶

𝑥 𝐹𝐴𝐷 = 100 N

𝑥

40◦ (d)

Figure 3.4. (a) A structure consisting of a bar and two cables supports a bucket weighing 100 N. Two cuts are taken to draw two FBDs. (b) FBD of the bucket. (c) FBD of point 𝐴 where 𝐹𝐴𝐶 is defined to be positive in tension. (d) FBD of point 𝐴 where 𝐹𝐴𝐶 is defined to be positive in compression.

the forces supported by the bar and cables are collinear with their respective axes. To model the structure in Fig. 3.4(a), you might begin by considering the bucket, to arrive at the intuitively obvious conclusion that if the bucket weighs 100 N, then the force

Helpful Information More on Governing Equations. Some of the examples in this book will involve failure criteria, deformable members, friction, or other criteria related to the physical behavior of a structural system. When this is the case, the Governing Equations category in our structured problem solving approach will be subdivided to include, as appropriate, • Equilibrium equations. • Force laws. • Kinematic equations. The specific meanings of force laws and kinematic equations are discussed later in this and subsequent chapters.

132

Chapter 3

Equilibrium of Particles

Jeffrey Markowitz/Getty Images

in cable 𝐴𝐷 must be 100 N. This conclusion is reached more formally by examining Fig. 3.4(a) to identify point 𝐷 as the location of a concurrent force system. A cut is then taken to separate point 𝐷 from its environment, and this cut passes through ∑ cable 𝐴𝐷, leading to the FBD shown in Fig. 3.4(b). With this FBD, writing 𝐹𝑦 = 0 provides 𝐹𝐴𝐷 − 100 N = 0, whose solution is 𝐹𝐴𝐷 = 100 N. Thus, one of the cable forces has been determined. To determine the forces supported by the bar and the remaining cable, we again examine Fig. 3.4(a) to identify point 𝐴 as the location of another concurrent force system. A cut is then taken to separate point 𝐴 from its environment, and this cut passes through one bar and two cables. Two possible FBDs resulting from this cut are shown in Fig. 3.4(c) and (d), where we have taken advantage of the result 𝐹𝐴𝐷 = 100 N. In both Fig. 3.4(c) and Fig. 3.4(d), the direction for the cable force 𝐹𝐴𝐵 is such that a positive value corresponds to a tensile force in the cable. For the force in the bar, the FBD in Fig. 3.4(c) defines the direction of 𝐹𝐴𝐶 such that a positive value corresponds to a tensile force in the bar, while the FBD in Fig. 3.4(d) defines the direction such that a positive value corresponds to a compressive force in the bar. If it is not clear that these statements are true, then drawing additional FBDs will fully clarify the situation. In fact if you are ever in doubt about whether a positive value for a particular cable or bar force means tension or compression, the construction of an additional FBD will always provide clarification. To illustrate, in Fig. 3.5(b) the FBD for point 𝐴 is shown, along with FBDs for portions of the cable and bar. Note that in drawing these FBDs, we have invoked Newton’s third law, which states that forces of action and reaction are equal in magnitude, opposite in direction, and collinear. Examining the FBD of cable 𝐴𝐵 in Fig. 3.5(b) unquestionably confirms that a positive value of 𝐹𝐴𝐵 corresponds to tension in the cable. Examining the FBD of bar 𝐴𝐶 in Fig. 3.5(b) also confirms that a positive value of 𝐹𝐴𝐶 corresponds to tension in the bar. Examining the FBD of bar 𝐴𝐶 in Fig. 3.5(c) shows that in this case a positive value of 𝐹𝐴𝐶 corresponds to compression in the bar. 30◦ 30◦

𝐵 𝐹𝐴𝐵

𝐴

𝐴

cuts

30◦ 𝐹𝐴𝐵

𝐹𝐴𝐵

𝐴 𝐹𝐴𝐶

bar 𝐴𝐶

40◦

𝐶 40◦

(a)

𝑦 𝐹𝐴𝐶

𝑥 𝐹𝐴𝐶

100 N

100 N 𝐷

𝐹𝐴𝐵

cable 𝐴𝐵 𝐹𝐴𝐶

𝐹𝐴𝐵

𝐹𝐴𝐵

(b)

𝐹𝐴𝐶

40◦

𝐹𝐴𝐶

(c)

Figure 3.5. (a) The structure of Fig. 3.4 shown again with more extensive FBDs to clarify different sign conventions for the force supported by a bar. (b) A positive value of 𝐹𝐴𝐶 corresponds to tension in bar 𝐴𝐶. (c) A positive value of 𝐹𝐴𝐶 corresponds to compression in bar 𝐴𝐶.

Comstock/Corbis

Figure 3.6 Examples of pulleys in use.

ISTUDY

Since cables buckle when subjected to very low compressive force, it is common to assume they can support only tensile forces. Thus, we will always assign cable forces in FBDs so that positive values correspond to tension. If, after the analysis of a particular problem, you find that a cable supports a compressive load, then either you have made an error or the structure has a serious flaw; namely, equilibrium of the

ISTUDY

Section 3.1

133

Equilibrium of Particles in Two Dimensions

structure is relying on a cable to support a compressive force, whereas in reality the cable cannot do this. In contrast to cables, bars can support both tensile and compressive forces, and thus, we always confront the question of what direction should be used to represent the forces they support when drawing FBDs. The answer is that it does not matter what convention you follow, provided your FBDs are consistent within a particular problem, and that you understand the proper interpretation for the forces you compute. In a problem as simple as that of Fig. 3.4(a), where there are a small number of structural members and the loading is simple, it is probably obvious that the cables will be in tension and the bar in compression. Thus, many people would draw the FBD shown in Fig. 3.4(d) where the bar force is taken positive in compression. In more complicated problems, however, it is often not possible to identify beforehand which bars will be in tension and which will be in compression. Thus, a good practice when drawing FBDs is to assign the forces supported by all bars such that positive values correspond to tension. Mixed sign conventions, where some bars have forces that are positive in tension and the others are positive in compression, are manageable for simple problems, but they are cumbersome for larger problems since you must always consult the original FBDs to discern which are in tension and which are in compression—the sign of the member’s force alone does not give enough information.

If the pulley has friction, then in general 𝑇1 ≠ 𝑇2 . 𝑇2

If the pulley is frictionless, then 𝑇1 = 𝑇2 always.

𝑇1

Figure 3.7 A cable with negligible weight wrapped around a pulley.

Pulleys Cables are often used in conjunction with pulleys. A pulley, as shown in Figs. 3.6 and 3.7, is a simple device that changes the orientation of a cable and hence, changes the direction of the force a cable supports. If a pulley is idealized as being frictionless (that is, the pulley can rotate on its bearing without friction) and the cable has negligible weight, then the magnitude of the force supported by the cable is unchanged as it wraps around the pulley. For the present, we must accept this statement without proof. In Chapter 5, when rigid bodies are discussed, we will show this statement is true. To carry this concept further, if a cable with negligible weight is wrapped around several frictionless pulleys, such as shown in Fig. 3.8(a), then the magnitude of the force is the same throughout the entire cable, as shown in Fig. 3.8(b). Note that the sketches of pulleys in Fig. 3.8(b) are not FBDs because the pulleys have not been completely removed from their environment, and thus there are additional forces acting on the pulleys that are not shown. If a cable has significant weight, then the force supported by the cable will vary throughout its length, even if frictionless pulleys are used. Throughout most of this book, cables are assumed to have negligible weight and pulleys are assumed to be frictionless. To idealize a pulley as a particle, such as when drawing an FBD, you can look for the point of intersection of all forces applied to the pulley, or better yet, you can simply “shift” the cable forces to the bearing of the pulley, as follows. Figure 3.9(a) shows a pulley that is supported by a shackle. To see how the pulley and shackle may be idealized as a particle, first we remove the pulley from the shackle to begin drawing the FBDs in Fig. 3.9(b) and (c). In Fig. 3.9(b) we introduce two forces, equal in magnitude to the cable force 𝑇 but in opposite directions, on the bearing 𝐴 of the pulley. Then on the FBD of the shackle in Fig. 3.9(c) we use Newton’s third law to include these same two forces on point 𝐴, but in opposite directions. The shackle shown in Fig. 3.9(c) is then easily modeled as a particle since it is a concurrent force system where all forces intersect at the bearing 𝐴 of the pulley. Note that the FBDs shown in Fig. 3.9 are valid regardless of the radius of the pulley, provided the pulley

𝑇 𝑇

𝑇

𝑇 𝑇 𝑇 𝑇

𝑇

Frictionless pulleys (a)

(b)

Figure 3.8 (a) A single cable wrapped around several pulleys and subjected to a force 𝑇 at its end. (b) If the pulleys are frictionless and the cable has negligible weight, then the magnitude of the force throughout the entire cable is the same.

𝐹

𝐹 𝑇

𝑇 𝐴

𝑇 𝐴

𝑇 𝐴 𝑇

(a)

𝑇

𝑇

𝑇 (b)

(c)

Figure 3.9 Demonstration of how a pulley may be idealized as a particle, where the weight of the cable, pulley, and shackle are negligible. (a) A frictionless pulley is supported by a shackle. (b) Free body diagram of the pulley removed from the shackle. (c) Free body diagram of the shackle removed from the pulley, showing how the pulley’s cable forces have been “shifted” to the bearing of the pulley at point 𝐴. These FBDs are valid regardless of the radius of the pulley.

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Equilibrium of Particles

is circular in shape and the bearing is frictionless, and the cable and pulley have negligible weight. 𝑊

Reactions

𝑅 𝑅 (a)

(b)

Figure 3.10 (a) Person standing on a horizontal surface. (b) Free body diagram of the person, showing the reaction the floor exerts on the person’s body.

𝑊

𝐹 𝑅 𝑅 𝐹 (a)

(b)

Figure 3.11 (a) Person standing on a rough slope. (b) Free body diagram of the person, showing the reactions the surface exerts on the person’s body.

ISTUDY

As previously stated, a reaction force, or simply reaction, is a force exerted by a support on a body or a structure. We will study support reactions extensively later in this book, but some remarks in addition to those already made are needed now. We begin with an example that is easy to relate to. Consider yourself as you stand on a horizontal floor, as shown in Fig. 3.10(a). It is because the floor exerts a force on your body that you remain in equilibrium. Thus, the floor exerts on your body a vertical force of magnitude 𝑅, and your body in turn exerts on the floor a force of magnitude 𝑅 in the opposite direction. The principal feature of the floor is that it prevents vertical motion of your body, and it will generate whatever force is needed to accomplish this. Thus, if you also wear a backpack while standing on the floor, the floor will produce an even larger reaction on your body so that you have no vertical motion. In the example shown in Fig. 3.11, imagine you are now standing on a slope, such as a sidewalk on a gentle hill. Assuming the sidewalk is sufficiently rough so that your feet do not slip, there are now two components of reaction forces in the directions shown. The reaction 𝑅 arises because the sidewalk does not let your body move in the direction normal to the surface, and the tangential component 𝐹 arises because the sidewalk does not let your feet slide. The sidewalk provides constraint of motion in two directions, and hence, there are two reactions in those directions. If the sidewalk is covered with ice so that it is frictionless, then the sidewalk no longer prevents slip, 𝐹 = 0, and an unpleasant experience results! When ice-covered, the sidewalk constrains motion in just one direction, and hence, there is only one reaction, 𝑅. The foregoing examples illustrate a thought process that will always allow us to identify the number and direction of reaction forces associated with a particular support. Namely, if a support prevents motion in a certain direction, it can do so only by producing a reaction force in that direction. When we are solving particle equilibrium problems, the supports and associated reactions shown in Fig. 3.12 occur often. It is not necessary to memorize these reactions; rather, you should reconstruct these as needed. For example, consider the particle pinned to a surface. The particle may be subjected to external forces, or may be a connection point between several cables and/or bars, which give rise to the forces 𝐹1 , 𝐹2 , and 𝐹3 shown. The surface to which the particle is fixed prevents motion of the particle in the 𝑦 direction, so there must be a reaction 𝑅𝑦 in this direction; and the surface also prevents motion of the particle in the 𝑥 direction, so there must also be a reaction 𝑅𝑥 in this direction. For the slider on a frictionless bar shown in Fig. 3.12, the fixed bar prevents motion of the slider in the direction normal to the bar, so there must be a reaction 𝑅𝑦 in this direction. The slider is free to move along the frictionless bar, hence there is no reaction in that direction.

End of Section Summary In this section, the equations governing static equilibrium of a particle were discussed, and analysis procedures were described. Some of the key points are as follows:

ISTUDY

Section 3.1

Equilibrium of Particles in Two Dimensions

Reactions

Support 𝐹2 𝑦

𝐹1

𝐹3

𝐹2

𝐹3

𝐹1

𝑥 particle pinned to surface

𝑅𝑥

𝐹2 𝑦

𝐹1

𝑅𝑦

𝐹3 𝑥

For a particle on a frictionless surface, 𝑅𝑥 = 0.

particle on rough surface 𝐹2 𝐹1

𝑦

𝐹3 𝐹2 𝑥

𝐹3

𝐹1

slider on fixed frictionless bar 𝐹2 𝐹3 𝑦 𝐹1

𝑅𝑦 𝑥

pin in frictionless slot Figure 3.12. Some common supports and reactions in two-dimensional particle equilibrium problems. Forces 𝐹1 , 𝐹2 , and 𝐹3 are hypothetical forces applied to a particle by cables and/or bars that might be attached to the particle, and forces 𝑅𝑥 and 𝑅𝑦 are reactions.

• For a particle in two dimensions, the equations of equilibrium written in vector ∑ ∑ ∑ form are 𝐹⃗ = 0⃗ and in scalar form are 𝐹𝑥 = 0 and 𝐹𝑦 = 0. In these summations, all forces applied to the particle must be included. • A free body diagram (FBD) is a sketch of a particle and all forces applied to the particle. The FBD is an essential tool to help ensure that all forces are accounted for when you are writing the equilibrium equations. • Complex problems may require more than one FBD. In two dimensions, each ∑ ⃗ ∑ FBD allows two equilibrium equations to be written [i.e., 𝐹 = ( 𝐹𝑥 ) 𝚤̂ + ∑ ⃗ or ∑ 𝐹 = 0 and ∑ 𝐹 = 0], thus allowing for an increased ( 𝐹𝑦 ) 𝚥̂ = 0, 𝑥 𝑦 number of unknowns to be determined. • Cables and straight bars are structural members supporting forces having the same orientation and line of action as the member’s geometry. • A pulley is used to change the direction of a cable and hence, change the direction of the force supported by a cable. If a single cable with negligible weight is wrapped around any number of frictionless pulleys, then the magnitude of the force throughout the cable is the same everywhere.

135

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E X A M P L E 3.1

Cables, Bars, and Failure Criteria Consider the structure discussed earlier in this section (Figs. 3.4 and 3.5) shown again here in Fig. 1 where the loading now consists of a vertical force 𝑃 applied to point 𝐴.

30◦ 𝐵

(a) If 𝑃 = 100 N, determine the forces supported by cable 𝐴𝐵 and bar 𝐴𝐶.

𝐴

𝑃

40◦

(b) If cable 𝐴𝐵 can support a tensile force of 1200 N before breaking, and bar 𝐴𝐶 can support a compressive force of 1600 N before buckling, determine the largest force 𝑃 that can be applied.

𝐶

Figure 1

SOLUTION Part (a)

Examining Fig. 1, we observe that an FBD of point 𝐴 will involve the force 𝑃 = 100 N, and the unknown cable and bar forces. Thus, we draw the FBD shown in Fig. 2, where positive values of 𝐹𝐴𝐵 and 𝐹𝐴𝐶 correspond to tensile forces in the cable and bar, respectively. We also select the 𝑥𝑦 coordinate system shown. Since the FBD contains two unknown forces, and since there are two equilibrium equations available, we have as many equations as unknowns, and we expect to be able to determine the forces supported by the cable and bar.

Road Map & Modeling 30◦ 𝐹𝐴𝐵

𝑦

𝐴 𝐹𝐴𝐶

𝑥

100 N 40◦ Figure 2 Free body diagram of point 𝐴 where 𝑃 = 100 N.

Governing Equations Using the FBD in Fig. 2, the equilibrium equations are written by summing forces in the 𝑥 and 𝑦 directions

∑

∑

Computation

𝐹𝑥 = 0 ∶

𝐹𝐴𝐵 cos 30◦ + 𝐹𝐴𝐶 cos 40◦ = 0,

(1)

𝐹𝑦 = 0 ∶

𝐹𝐴𝐵 sin 30 − 𝐹𝐴𝐶 sin 40 − 100 N = 0.

(2)

◦

◦

Equations (1) and (2) are easily solved to obtain 𝐹𝐴𝐵 = 81.52 N,

(3)

𝐹𝐴𝐶 = −92.16 N.

(4)

Part (b) Road Map

30◦ 𝐹𝐴𝐵

𝑦

𝐴 𝐹𝐴𝐶 𝑃

𝑥

40◦

Figure 3 Free body diagram of point 𝐴 where 𝑃 is unknown.

ISTUDY

We are given the maximum loads the cable and bar can support. It is important to understand that it is unlikely that both the cable and bar will simultaneously be at their failure loads, and the margin note at the end of this example provides further discussion of this. One solution strategy is to use the results of Part (a) to determine the largest load the structure can support by exploiting linearity, as follows. In Part (a), the solutions for 𝐹𝐴𝐵 and 𝐹𝐴𝐶 were obtained using 𝑃 = 100 N. If 𝑃 is doubled, then 𝐹𝐴𝐵 and 𝐹𝐴𝐶 are doubled, and so on. This assumes the angles shown in Fig. 2 remain the same as the load changes. Thus, we may scale the load 𝑃 until 𝐹𝐴𝐵 = 1200 N or 𝐹𝐴𝐶 = −1600 N. A generally preferable way to solve problems with multiple failure criteria is to determine the forces supported by each member in terms of 𝑃 , where 𝑃 is yet to be determined, and this is the approach we will use. A feature of this approach is that you do not need to guess which of several members will fail first. Modeling

unknown.

The FBD shown in Fig. 3 is the same as that for Part (a) except that 𝑃 is

ISTUDY

Section 3.1

Equilibrium of Particles in Two Dimensions

137

Governing Equations & Computation Equilibrium Equations The equilibrium equations are the same as Eqs. (1) and (2), except with the 100 N force replaced by 𝑃 . ∑ 𝐹𝑥 = 0 ∶ 𝐹𝐴𝐵 cos 30◦ + 𝐹𝐴𝐶 cos 40◦ = 0, (5) ∑ ◦ ◦ 𝐹𝑦 = 0 ∶ 𝐹𝐴𝐵 sin 30 − 𝐹𝐴𝐶 sin 40 − 𝑃 = 0. (6)

Solving Eqs. (5) and (6) provides

Force Laws

𝐹𝐴𝐵 = (0.8152) 𝑃 ,

(7)

𝐹𝐴𝐶 = (−0.9216) 𝑃 .

(8)

Now the various failure criteria can be applied using Eqs. (7) and (8): If 𝐹𝐴𝐵 = 1200 N, If 𝐹𝐴𝐶 = −1600 N,

then 𝑃 = 1472 N. then 𝑃 = 1736 N.

(9) (10)

Only the smaller value of 𝑃 in Eqs. (9) and (10) will simultaneously satisfy the failure criteria for both the cable and bar. Thus, the maximum value 𝑃 may have is 𝑃max = 1472 N,

(11)

and cable 𝐴𝐵 will fail first. Discussion & Verification

As expected from intuition, solutions to both parts of this example show 𝐹𝐴𝐵 is positive, meaning the cable is in tension, and 𝐹𝐴𝐶 is negative, meaning the bar is in compression. After this check, we should substitute the solutions for 𝐹𝐴𝐵 and 𝐹𝐴𝐶 into the original equilibrium equations; in Part (a) both Eqs. (1) and (2) must be satisfied, and in Part (b) both Eqs. (5) and (6) must be satisfied. This is a useful check to avoid algebra errors, but unfortunately it does not catch errors in drawing FBDs and writing equilibrium equations.

Common Pitfall Failure loads. A common error in solving problems with failure criteria, such as Part (b) of this example, is to assume that all members are at their failure loads at the same time. With reference to the FBD of Fig. 3, you will be making this error if you take 𝐹𝐴𝐵 = 1200 N and 𝐹𝐴𝐶 = −1600 N; in fact, if you ∑ do this, you will find that 𝐹𝑥 = 0 (Eq. (5)) cannot be satisfied! Another way to describe this problem is to consider slowly increasing force 𝑃 from zero. Eventually, one of the members will reach its failure load first, while the other will be below its failure load.

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E X A M P L E 3.2

A Problem Requiring Multiple FBDs 𝐷

𝐵

20◦

10◦ 𝐸

𝐶

A small cable car is used to transport passengers across a river. If the cable car and its contents have a mass of 400 kg, determine the forces in cables 𝐴𝐶 and 𝐸𝐷 and the force in bar 𝐴𝐵. Point 𝐴 is a pin.

20◦

𝐴

SOLUTION Road Map & Modeling

We begin by searching for an appropriate particle, or particles, whose equilibrium should be analyzed, and points 𝐴 and 𝐵 are likely choices. To get started, we draw the FBD for point 𝐴 in Fig. 2 where the weight of the cable car and its contents is (400 kg)(9.81 m∕s2 ) = 3924 N = 3.924 kN. This FBD has three unknowns (𝑇𝐴𝐵 , 𝑇𝐴𝐶 , and 𝛼), and since there are only two equilibrium equations available for this FBD, we must draw another FBD so that additional equilibrium equations may be written, with the goal of having as many equations as unknowns. Thus, we draw the FBD for pulley 𝐵 in Fig. 2 where we assume pulley 𝐵 is frictionless so that the tensile force throughout cable 𝐸𝐷 is uniform, and as discussed in Fig. 3.9 on p. 133, the cable forces have been shifted to the bearing of the pulley. There are now four unknowns, and four equilibrium equations may be written.

Figure 1

𝑇𝐸𝐷 𝐵

10◦

20◦

𝑇𝐸𝐷

Governing Equations

𝛼

∑

𝑇𝐴𝐵 𝑦 𝑥

𝑇𝐴𝐵 𝑇𝐴𝐶

𝛼 𝐴

20◦

3.924 kN Figure 2 Free body diagrams of points 𝐴 and 𝐵.

ISTUDY

∑

Common Pitfall Don’t confuse mass and weight. A common error, especially when you are using SI units, is to mistake mass for weight when drawing FBDs, writing equilibrium equations, and so on. Mass must be multiplied by acceleration due to gravity to obtain weight. With reference to Fig. 2, if you incorrectly give the weight of the cable car and its contents a value of 400, then the values you obtain for 𝑇𝐴𝐵 , 𝑇𝐴𝐶 , and 𝑇𝐸𝐷 are all about an order of magnitude too small!

The equilibrium equations for point 𝐴 are

𝐹𝑥 = 0 ∶

−𝑇𝐴𝐵 sin 𝛼 + 𝑇𝐴𝐶 cos 20◦ = 0,

(1)

𝐹𝑦 = 0 ∶

𝑇𝐴𝐵 cos 𝛼 + 𝑇𝐴𝐶 sin 20◦ − 3.924 kN = 0.

(2)

Equations (1) and (2) contain three unknowns. Thus, additional equilibrium equations are needed, and to produce these we use the FBD of point 𝐵 shown in Fig. 2 to write ∑ 𝐹𝑥 = 0 ∶ 𝑇𝐴𝐵 sin 𝛼 − 𝑇𝐸𝐷 cos 10◦ + 𝑇𝐸𝐷 cos 20◦ = 0, (3) ∑ 𝐹𝑦 = 0 ∶ −𝑇𝐴𝐵 cos 𝛼 − 𝑇𝐸𝐷 sin 10◦ + 𝑇𝐸𝐷 sin 20◦ = 0. (4) There are now four equations and four unknowns, 𝑇𝐴𝐵 , 𝑇𝐴𝐶 , 𝑇𝐸𝐷 , and 𝛼. Solving a system of four equations is rarely fun and is often tedious, especially when trigonometric functions of the unknowns are involved. While the use of software such as Mathematica or Maple can considerably ease this burden, in many problems some careful study of the equation system will offer a simple solution strategy, and before reading further, you should examine these equations to see if you can identify such a strategy. Solving for the term 𝑇𝐴𝐵 sin 𝛼 in Eq. (1) and substituting this into Eq. (3), and solving for the term 𝑇𝐴𝐵 cos 𝛼 from Eq. (2) and substituting this into Eq. (4), provide a system of two equations in two unknowns, 𝑇𝐴𝐶 and 𝑇𝐸𝐷 , which can be solved to obtain Computation

𝑇𝐴𝐶 = 1.019 kN and 𝑇𝐸𝐷 = 21.23 kN.

(5)

Now, using Eqs. (1) and (2) to compute the ratio 𝑇𝐴𝐵 sin 𝛼∕(𝑇𝐴𝐵 cos 𝛼), we obtain an expression for tan 𝛼 which then provides 𝛼 = 15.00◦ .

(6)

Finally, 𝑇𝐴𝐵 may be found from any of Eqs. (1)–(4) as 𝑇𝐴𝐵 = 3.701 kN. Discussion & Verification

(7)

As expected, the solution shows all cables are in tension. Furthermore, by intuition we also expect the force supported by bar 𝐴𝐵 to be tensile. After these checks, we should substitute all our solutions into the equilibrium equations to verify that each of them is satisfied.

ISTUDY

Section 3.1

139

Equilibrium of Particles in Two Dimensions

E X A M P L E 3.3

Cables, Pulleys, and Failure Criteria

The cable and pulley system shown is used by a camper to hoist a backpack into a tree to keep it out of the reach of bears. If cables 𝐴𝐵 and 𝐴𝐶 have breaking strengths of 200 lb and cable 𝐷𝐴𝐸 has a breaking strength of 100 lb, determine the largest weight 𝑊 that may be lifted.

SOLUTION

𝐶 50◦ 𝐴

𝐵

Road Map Our strategy is to find the forces in each cable in terms of weight 𝑊 and then to apply the failure criteria to determine the largest value 𝑊 may have.

The force the camper applies to the cable at point 𝐷 will be called 𝑇 , and assuming the cable’s weight is negligible, the orientation of this force is the same as the 30◦ orientation of cable segment 𝐴𝐷. Cable 𝐷𝐴𝐸 is a single continuous cable, and assuming the pulleys are frictionless, in addition to the assumption of negligible weight of the cable, the tensile force throughout this cable is the same with value 𝑇 . The FBDs for points 𝐴 and 𝐸 are shown in Fig. 2,∗ where as discussed in Fig. 3.9 on p. 133, the cable forces applied to pulley 𝐴 have been shifted to the bearing of that pulley, and similarly for pulley 𝐸. Modeling

∑

∑

𝐹𝑥 = 0 ∶

−𝑇𝐴𝐵 + 𝑇𝐴𝐶 cos 50◦ + 𝑇 sin 30◦ = 0,

(1)

𝐹𝑦 = 0 ∶

𝑇𝐴𝐶 sin 50◦ − 𝑇 − 𝑇 − 𝑇 cos 30◦ = 0.

(2)

𝑇𝐴𝐵 = (1.452)𝑊 ,

(4)

𝑇𝐴𝐶 = (1.871)𝑊 .

(5)

𝑊

𝑇𝐴𝐶 𝐴

𝑇𝐴𝐵

50◦ 30◦

𝑦 𝑇 𝑇

𝑇

𝑥 𝑇 𝑇

𝐸

The various failure criteria can now be applied using Eqs. (3)–(5): If 𝑇 = 100 lb,

then 𝑊 = 200 lb.

(6)

If 𝑇𝐴𝐵 = 200 lb,

then 𝑊 = 137.7 lb.

(7)

If 𝑇𝐴𝐶 = 200 lb,

then 𝑊 = 106.9 lb.

(8)

Only the smallest value of 𝑊 in Eqs. (6)–(8) will simultaneously satisfy all three failure criteria. Thus the largest value 𝑊 may have is 𝑊max = 106.9 lb,

𝑊 Figure 2 Free body diagrams of points 𝐴 and 𝐸. The force systems for points 𝐴 and 𝐸 are both concurrent, although these figures show some of the forces to be slightly separated for clarity.

(9)

and cable 𝐴𝐶 will be the first to fail. Discussion & Verification

As expected, Eqs. (3)–(5) show that all cable forces are tensile. After this check, we should substitute our solutions into all of the equilibrium equations to verify that each is satisfied. A common error in problems such as this is to assume that all cables are at their failure loads simultaneously. ∗ The

𝐸

The equilibrium equations for point 𝐴 are

Equations (1) and (2) are a system of two equations with three unknowns (𝑇 , 𝑇𝐴𝐵 , and 𝑇𝐴𝐶 ), and hence, an additional equation is needed. Thus, we also need the FBD for point 𝐸 in Fig. 2, and we write the equilibrium equation ∑ 𝑊 𝐹𝑦 = 0 ∶ 𝑇 + 𝑇 − 𝑊 = 0 ⇒ 𝑇 = . (3) 2 Using the solution 𝑇 = 𝑊 ∕2 obtained in Eq. (3), Eqs. (1) and (2) may be solved for

Force Laws

𝐷

Figure 1

Governing Equations & Computation Equilibrium Equations

30◦

need for the FBD at 𝐸 may not be apparent until the equilibrium equations for point 𝐴 are written and we find there are too many unknowns.

Helpful Information Notation for forces. In problems where we take the force supported by a structural member to be positive in tension, such as 𝑇 , 𝑇𝐴𝐵 , and 𝑇𝐴𝐶 in this example, we will often use the symbol 𝑇 to emphasize that a positive value means tension.

140

Chapter 3

Equilibrium of Particles

E X A M P L E 3.4

Reactions and Force Polygons The structure consists of a collar at 𝐵 that is free to slide along a straight fixed bar 𝐴𝐶. Mounted on the collar is a frictionless pulley, around which a cable supporting a 5 lb weight is wrapped. The collar is further supported by a bar 𝐵𝐷.

𝐶 𝐸

𝐵

𝐴

(a) If 𝛼 = 0◦ , determine the force in bar 𝐵𝐷 needed to keep the system in equilibrium.

𝛼 𝐷

30◦

(b) Determine the value of 𝛼 that will provide for the smallest force in bar 𝐵𝐷, and determine the value of this force.

SOLUTION

5 lb

Part (a)

Figure 1

Road Map & Modeling With 𝛼 = the force supported by bar 𝐵𝐷 is horizontal. To draw the FBD for the collar at 𝐵, we cut through the cable twice and bar 𝐵𝐷 once, and

0◦ ,

𝑅 30◦ 𝐵

𝑦 5 lb 𝑥

𝐹𝐵𝐷

5 lb Figure 2 Free body diagram for point 𝐵 when 𝛼 = 0◦ .

we separate the collar from bar 𝐴𝐶. Assuming the pulley is frictionless and the cable is weightless, the tensile force is the same throughout the cable. Further, there is a reaction between the collar and bar 𝐴𝐶. We could consult Fig. 3.12 on p. 135 to determine the direction for this reaction, but it is easier to construct this as follows. The collar may not move perpendicular to bar 𝐴𝐶, so there must be a reaction in this direction. The collar is free to slide along bar 𝐴𝐶, so there is no reaction in this direction. Hence, the FBD is as shown in Fig. 2. Observe that this FBD contains two unknowns, and since there are two equilibrium equations that may be written, we expect the solution to be straightforward. Governing Equations

∑

∑

Computation

The equilibrium equations for point 𝐵 are

𝐹𝑥 = 0 ∶

−𝑅 sin 30◦ + 𝐹𝐵𝐷 − 5 lb = 0,

(1)

𝐹𝑦 = 0 ∶

𝑅 cos 30◦ − 5 lb = 0.

(2)

Solving Eqs. (1) and (2) provides 𝑅 = 5.774 lb and

𝐹𝐵𝐷 = 7.887 lb.

(3)

Part (b)

With the orientation 𝛼 of member 𝐵𝐷 unknown, the FBD for point 𝐵 will contain three unknowns, and since there are only two equilibrium equations, some additional information is needed. This information can take the form of an equation that states the force in bar 𝐵𝐷 is a minimum (Problem 3.50 guides you to use calculus to do this), or as we do below, by using physical insight gained from a force polygon to determine the solution for 𝛼.

Road Map

𝑅 𝑦 5 lb

𝐹𝐵𝐷

𝑥 5 lb

Figure 3 Free body diagram for point 𝐵 when 𝛼 ≠ 0◦ .

ISTUDY

The FBD for point 𝐵 is shown in Fig. 3 where 𝐹𝐵𝐷 , 𝑅, and 𝛼 are unknown. The easiest and most insightful way to determine the orientation 𝛼 for which the force in member 𝐵𝐷 is smallest is to use the force polygon concepts of Section 2.1. Newton’s ∑ law 𝐹⃗ = 0⃗ can be evaluated by constructing a closed force polygon. Thus, all the forces that are applied to point 𝐵 in the FBD of Fig. 3 are added in head-to-tail fashion to form the closed force polygon shown in Fig. 4. In principle, the order in which forces are added does not matter, but visualization and computation are more straightforward if forces having known magnitude and direction are added first, followed by forces having unknown magnitude and/or unknown direction. Thus, in Fig. 4(a) we assemble the two 5 lb cable forces first. Of the two remaining forces, the reaction 𝑅 has known direction but unknown magnitude, whereas 𝐹𝐵𝐷 has unknown direction and magnitude. Thus, it is best to assemble 𝑅 next and 𝐹𝐵𝐷 last. Note that if 𝛼 = 0◦ , then the polygon corresponds to the forces found in Part (a). Modeling

30◦ 𝐵 𝛼

ISTUDY

Section 3.1

141

Equilibrium of Particles in Two Dimensions

𝛼 𝐹 𝐵𝐷

Governing Equations & Computation

We now study the force polygon in Fig. 4(a) to see what orientation 𝛼 will give the smallest value of 𝐹𝐵𝐷 . Clearly that orientation is when 𝐹𝐵𝐷 is perpendicular to the direction of 𝑅, and the force polygon for this case is shown in Fig. 4(b). Hence, 𝛼 = −30◦ , where the negative sign∗ is used because the angle found here is opposite to the direction defined in Fig. 1. To determine 𝑅 and 𝐹𝐵𝐷 , we could use the geometry of the force polygon shown in Fig. 4(b), but it is easier to use the FBD in Fig. 3 to write the equilibrium equations ∑ 𝐹𝑥 = 0 ∶ −𝑅 sin 30◦ + 𝐹𝐵𝐷 cos 𝛼 − 5 lb = 0, (4) ∑ ◦ 𝐹𝑦 = 0 ∶ 𝑅 cos 30 − 𝐹𝐵𝐷 sin 𝛼 − 5 lb = 0. (5)

𝑅

30◦

𝐵

𝐵

30◦ 𝐹𝐵𝐷

5 lb

𝛼 = −30◦

5 lb

𝑅 5 lb

5 lb

(a)

(b)

Figure 4 Force polygons for equilibrium of point 𝐵.

Substituting 𝛼 = −30◦ into the above expressions, we may solve them to obtain 𝑅 = 1.830 lb

and 𝐹𝐵𝐷 = 6.830 lb.

(6)

Alternate solution. Using a 𝑡𝑛 coordinate system where 𝑡 and 𝑛 are tangent and normal, respectively, to bar 𝐴𝐶 provides for an alternate solution that, for this problem, is effective. Using these 𝑡𝑛 directions, the FBD for point 𝐵 is shown in Fig. 5, where for convenience an angle 𝛽 is defined such that 𝛽 = 30◦ + 𝛼. The equilibrium equations are ∑ 𝐹𝑡 = 0 ∶ −(5 lb) cos 30◦ − (5 lb) sin 30◦ + 𝐹𝐵𝐷 cos 𝛽 = 0, (7) ∑ ◦ ◦ 𝐹𝑛 = 0 ∶ 𝑅 + (5 lb) sin 30 − (5 lb) cos 30 − 𝐹𝐵𝐷 sin 𝛽 = 0. (8)

The merit of using this 𝑡𝑛 coordinate system is that Eq. (7) may be immediately solved to obtain 1 𝐹𝐵𝐷 = (5 lb)(cos 30◦ + sin 30◦ ). (9) cos 𝛽 Inspection of Eq. (9) shows that the smallest value of 𝐹𝐵𝐷 occurs when cos 𝛽 = 1. Hence, 𝛽 = 0◦ and since 𝛽 = 30◦ + 𝛼, we determine that 𝛼 = −30◦ , which is the same conclusion obtained by using the force polygons in Fig. 4. To complete this solution, we use 𝛽 = 0◦ in Eq. (9) to obtain 𝐹𝐵𝐷 = 6.830 lb, and then we solve Eq. (8) to obtain 𝑅 = 1.830 lb, both of which agree with the solutions found in Eq. (6). Discussion & Verification

Part (a) was reasonably straightforward, while Part (b) was more difficult because we were asked to minimize a particular force. In Part (b), three equations were needed, and if you re-solve this problem using the calculus approach described in Prob. 3.50, you will see that the additional equation is the optimization condition 𝑑𝐹𝐵𝐷 ∕𝑑𝛼 = 0. The advantage of using the force polygon approach is that the solution for 𝛼 can be found by inspection, which then leaves the two equilibrium equations that were easily solved.

∗ Rather

than using 𝛼 = −30◦ , we could use 𝛼 = 330◦ instead.

𝑅

𝑛

5 lb 30◦

𝑡 𝐴

𝐶 30◦

𝐵

𝛼 30◦

𝛽 = 30◦ + 𝛼 𝐹𝐵𝐷

5 lb

Figure 5 Free body diagram for point 𝐵 using a 𝑡𝑛 coordinate system.

142

Chapter 3

Equilibrium of Particles

Problems

ISTUDY

General Instructions. In all problems, draw FBDs and label all unknowns. It is recommended that you state forces in cables and bars using positive values for tension. In problems where forces due to gravity are present, unless otherwise stated, these forces are in the downward vertical direction in the illustration provided. Tip: To practice extra problems quickly, draw FBDs and write equilibrium equations, but do not solve for unknowns. Problem 3.1 Consider an airplane whose motion is described below. For each case, state whether or not the airplane is in static equilibrium, with a brief explanation. (a) The airplane flies in a straight line at a constant speed and at a constant altitude. (b) The airplane flies in a straight line at a constant speed while climbing in altitude. (c) The airplane flies at a constant speed and at a constant altitude while making a circular turn. (d) After touching down on the runway during landing, the airplane rolls in a straight line at a constant speed. (e) After touching down on the runway during landing, the airplane rolls in a straight line while its brakes are applied to reduce its speed. Note: Concept problems are about explanations, not computations.

Problems 3.2 through 3.5 The structure shown uses members 𝐴𝐵, 𝐵𝐶, and 𝐵𝐷 to support an object with the weight or mass given in the figure. Determine the force supported by each of these members, indicating whether they are in tension or compression. 𝐴 𝐴

𝐴 𝐶

40◦

90 cm

𝐵

𝐶

3 𝛼=

20◦

𝐵 𝐶

5 𝐷

𝐷

12

400 N Figure P3.3

𝛽 = 65◦ 𝐶

𝐴

60 cm 80 cm

200 lb

Figure P3.2

𝐵 15◦

4

5 kg Figure P3.4

𝐵 𝐷

𝐷 50 lb Figure P3.5

Problem 3.6 In Fig. P3.2, determine what the 20◦ orientation of member 𝐵𝐶 should be changed to (i.e., determine angle 𝛼) so that the force this member supports is as small as possible, and determine the force supported by members 𝐴𝐵, 𝐵𝐶, and 𝐵𝐷.

Problem 3.7 In Fig. P3.5, determine what the 65◦ orientation of member 𝐵𝐶 should be changed to (i.e., determine angle 𝛽) so that the force this member supports is as small as possible, and determine the force supported by members 𝐴𝐵, 𝐵𝐶, and 𝐵𝐷.

ISTUDY

Section 3.1

Equilibrium of Particles in Two Dimensions

Problems 3.8 and 3.9 In a machining setup, workpiece 𝐵, which weighs 20 lb, is supported by a fixed V block 𝐸 and a clamp at 𝐴. All contact surfaces are frictionless, and the clamp applies a vertical force of 35 lb to the workpiece. Determine the reactions at points 𝐶 and 𝐷 between the V block and the workpiece.

𝐴

𝐴

𝐵

𝐵 30◦

𝐶

𝐷

30◦

𝐸

𝐶 30◦

𝐷

◦ 30◦ 𝐸 10

𝑇1 Figure P3.8

20◦

Figure P3.9

Problem 3.10

𝑇2 20◦

A skier uses a taut tow rope to reach the top of a ski hill. The skier weighs 150 lb, the snow-covered slope is frictionless, and the tow rope is parallel to the slope. (a) If 𝑇2 = 200 lb, determine the value of 𝑇1 to move the skier up the slope at constant velocity and the reaction between the skier and the slope.

20◦

Figure P3.10

(b) If 𝑇1 = 200 lb, determine the value of 𝑇2 to move the skier up the slope at constant velocity and the reaction between the skier and the slope. 100 lb

Problem 3.11 The dimension ℎ is to be determined so that a worker can comfortably slide boxes weighing up to 100 lb up and down a frictionless incline. If the worker can apply to the box a 50 lb horizontal force (parallel to the ground), determine the largest value ℎ may have.

ℎ

3 ft

Figure P3.11

Problem 3.12 Blocks 𝐴 and 𝐵 each have 5 kg mass, and all contact surfaces are frictionless.

𝐴

(a) Determine the force 𝐹 needed to keep the blocks in static equilibrium and the forces on all contact surfaces.

20◦ 𝐵

(b) If the value of 𝐹 determined in Part (a) is applied, will the blocks move? Explain. (c) If 𝐹 is smaller than the value determined in Part (a), describe what happens.

𝐹

Figure P3.12

Problem 3.13

𝐶

Bead 𝐴 has 2 kg mass and slides without friction on bar 𝐵𝐶. (a) Determine the force 𝐹 needed to keep the bead in static equilibrium and the reaction force between the bead and bar.

2 kg 𝐵

(b) If the value of 𝐹 determined in Part (a) is applied, will the bead move? Explain. (c) If 𝐹 is larger than the value determined in Part (a), describe what happens.

𝐴

𝐹

Figure P3.13

30◦

143

144

Chapter 3

Equilibrium of Particles

𝐵

Problem 3.14 A crane is used to lift a concrete pipe weighing 5 kN into place. If 𝑑 = 0.25 m, determine the tension in cables 𝐴𝐵 and 𝐴𝐶.

4m 𝐴 𝐶

Problem 3.15

5 kN

20◦

A crane is used to lift a concrete pipe weighing 5 kN into place. For precision positioning, the worker at 𝐶 can apply up to a 400 N force to cable 𝐴𝐶. Determine the largest distance 𝑑 the concrete pipe may be moved.

𝑑 Figure P3.14 and P3.15

Problem 3.16 𝑃

10◦ 𝐴

Guy wire 𝐴𝐵 is used to help support the utility pole 𝐴𝐶. Assuming the force supported by the utility pole is directed along the line 𝐴𝐶,∗ determine the force in wire 𝐴𝐵 and pole 𝐴𝐶 if 𝑃 = 800 N.

60◦

𝐵

𝐶

Problem 3.17 Guy wire 𝐴𝐵 is used to help support the utility pole 𝐴𝐶. If the guy wire 𝐴𝐵 can support a maximum tensile force of 500 lb, and if the pole 𝐴𝐶 can support a maximum compressive force of 800 lb before buckling, determine the largest force 𝑃 that can be supported.

Figure P3.16 and P3.17

Problems 3.18 and 3.19 A component of a machine has a frictionless roller 𝐴 that rests in a slot. The roller is supported by bar 𝐴𝐵 where point 𝐵 is a pin. For the values of 𝑃 and 𝜃 given below, determine the force supported by bar 𝐴𝐵. Problem 3.18

𝑃 = 200 N and 𝜃 = 15◦ .

Problem 3.19

𝑃 = 400 N and 𝜃 = 35◦ .

𝐴

𝑃 3

𝜃

𝐵

4 𝐶

Figure P3.18 and P3.19

Problems 3.20 and 3.21

2 lb 𝑟

𝐴 𝐵

𝜃 𝑟

Figure P3.20 and P3.21

ISTUDY

A bead 𝐴 weighing 2 lb slides on a quarter-circular frictionless bar 𝐵𝐶 with radius 𝑟. The bead is supported by a cable 𝐴𝐶. For the value of 𝜃 given below, determine the force supported by cable 𝐴𝐶. Problem 3.20

𝜃 = 20◦ .

Problem 3.21

General values of 𝜃 where 0◦ < 𝜃 < 90◦ .

∗ Whether

or not this statement is true depends on details of how end 𝐶 of the utility pole is supported. Such issues are explored in Chapter 5.

ISTUDY

Section 3.1

145

Equilibrium of Particles in Two Dimensions

Problem 3.22 Two schemes are shown for hanging a large number of flowerpots side by side on an outdoor porch. The flowerpots are to have 60 cm spacing. Each flowerpot weighs 175 N.

30 cm 30 cm

(a) Determine the force in wire 𝐴𝐵.

𝐵

𝐷

30 cm 𝐸

(b) Determine the forces in wires 𝐶𝐷 and 𝐶𝐸. (c) Compared to the scheme using one wire, the scheme using two wires may be more resistant to adjacent flowerpots hitting one another in high winds. Do you believe this statement is valid? Explain. Hint: Consider the application of a horizontal wind force 𝑃 to points 𝐴 and 𝐶 and speculate on the ability of each system to resist horizontal motion. To do this, compare the values of 𝑃 needed to produce the same horizontal displacement of points 𝐴 and 𝐶 of, say, 10 cm.

40 cm 𝐶

𝐴

Figure P3.22

Problem 3.23 A hydraulic cylinder 𝐴𝐵 on a backhoe produces a 4500 lb compressive force. Determine the forces in members 𝐵𝐷 and 𝐵𝐶.

𝐴 25◦

Problems 3.24 and 3.25 A worker inside a truck at point 𝐴 applies a force to rope 𝐴𝐵𝐶 to slowly lower a box 𝐶 down a ramp 𝐷𝐸. The box has 400 N weight and slides without friction on the ramp, and the pulley at 𝐵 is frictionless. For the position of the box given below, determine the force the worker must apply to the rope and the reaction between the box and ramp. Neglect the size of the box and pulley. Problem 3.24

The box is 1/4 the distance from point 𝐷 to point 𝐸.

Problem 3.25

The box is 3/4 the distance from point 𝐷 to point 𝐸. 0.8 m

𝐵

70◦ 𝐷

𝐸

20◦ 𝐶

Figure P3.23

3.5 m

20◦ 𝐵 𝐴

2.5 m

𝐷

𝐶 1.2 m 𝐸

Figure P3.24 and P3.25

Problem 3.26 The pulley shown is frictionless and all weights are negligible. 𝐹

(a) Show that 𝜃 = 𝛼. (b) By drawing an FBD of the pulley and writing and solving equilibrium equations, determine the force 𝐹 in terms of the force 𝑇 and angle 𝜃. Plot the ratio 𝐹 ∕𝑇 versus 𝜃 for 0 ≤ 𝜃 ≤ 90◦ . (c) Imagine a structure has this pulley and cable arrangement, and you carefully measure 𝜃 and 𝛼 and find they are not equal. Explain possible circumstances that might cause this to occur.

𝑇

𝜃 𝛼 𝑇

Figure P3.26

146

Chapter 3

Equilibrium of Particles

Problem 3.27 Due to settlement of soil, a recently planted tree has started to lean. To straighten it, the cable system shown is used, where a turnbuckle on cable 𝐴𝐵 is periodically tightened to keep the cable taut as the tree gradually straightens. If the force in cable 𝐴𝐵 is 450 N, determine the force in cable CBD.

𝐷

Problems 3.28 and 3.29 𝐶 10◦

The symmetric cable and pulley arrangement shown is used to lift a fragile architectural stone beam. If the only significant mass in the system is the 800 kg mass 𝑚 of the beam, determine the forces in cables ACB and CDE.

𝐵 𝐴

30◦

𝐹

𝐹

Figure P3.27 𝐷

𝐶

60◦ 𝐴

𝐷

𝐵

𝐶

𝐸 𝑚

60◦ 𝐹

𝐺

Figure P3.28

60◦

60◦

30◦

𝐸 45◦

𝐴

𝐵

30◦

45◦

𝑚

𝐹

𝐺

Figure P3.29

Problem 3.30 In Fig. P3.29, cables ACB and FEG each can support a maximum force of 3 kN. Cable CDE can support a maximum force of 9 kN. The spreader bar CE can support a maximum compressive force of 5 kN. Determine the largest mass 𝑚 of the stone beam that may be lifted, assuming this is the only significant mass in the system.

Problem 3.31 10◦

𝐵

𝑃

Bars 𝐴𝐵 and 𝐵𝐶 have the tensile and compressive strengths listed below. If 𝑄 = 0 and 𝑃 ≥ 0, determine the largest value of 𝑃 that may be supported.

20◦

Member

𝑄

𝐴𝐵 𝐵𝐶 𝐴

55◦ 80◦

Strength 4000 lb tension & 2500 lb compression 4500 lb tension & 3000 lb compression

𝐶

Problem 3.32

Figure P3.31 and P3.32

Bars 𝐴𝐵 and 𝐵𝐶 have the tensile and compressive strengths listed in Prob. 3.31. If 𝑃 = 0 and 𝑄 ≥ 0, determine the largest value of 𝑄 that may be supported.

30◦ 𝐴 45◦ Figure P3.33

ISTUDY

𝐵

𝑃

Problem 3.33 Blocks 𝐴 and 𝐵 each weigh 100 lb and rest on frictionless surfaces. They are connected to one another by cable 𝐴𝐵. Determine the force 𝑃 required to hold the blocks in the equilibrium position shown and the reactions between the blocks and surfaces.

ISTUDY

Section 3.1

147

Equilibrium of Particles in Two Dimensions

Problem 3.34 Two weights are supported by cable 𝐴𝐵𝐶𝐷. With the geometry shown, if one of the weights is 2000 N, determine the other weight 𝑊 and the cable tensions. 9m

16 m

12 m

𝐷 12 m 𝐶 12 m 2000 N

𝐵 5m 𝐴 𝑊

Figure P3.34

Problem 3.35 40 in.

Determine the largest weight 𝑊 that may be supported if the failure strengths of the cables and bars are Member 𝐴𝐶 𝐴𝐸 𝐶𝐷 𝐴𝐵 𝐵𝐶

Strength 600 lb 250 lb 300 lb 2000 lb tension & 700 lb compression 1000 lb tension & 300 lb compression

𝐶

12 in. 𝐷

𝐴

9 in.

𝐵

𝐸 𝑊 Figure P3.35

Problem 3.36

𝐴

Determine the largest weight 𝑊 that may be supported if the failure strengths of the cables and bars are Member 𝐴𝐵 𝐵𝐶 𝐶𝐷 𝐵𝐸 𝐶𝐸

45 cm

Strength

𝐵

80 cm

3.5 kN 3 kN 2.5 kN 3.7 kN tension & 1.8 kN compression 4.7 kN tension & 2.8 kN compression

𝐸

𝐶 60 cm

60 cm

𝐷 𝑊

Figure P3.36

Problem 3.37 The system shown consists of cables 𝐴𝐵 and 𝐴𝐶, horizontal cable 𝐶𝐸, and vertical bar 𝐶𝐷. If the cables and bar have the failure strengths shown, determine the largest load 𝑃 that can be supported by the system. Member 𝐴𝐵 𝐴𝐶 𝐶𝐸 𝐶𝐷

𝐵

𝐸

𝐶

Strength 300 lb 300 lb 200 lb 600 lb tension & 200 lb compression

45◦

𝐴 𝑃

Figure P3.37

30◦

𝐷

148

Chapter 3

Equilibrium of Particles

Problem 3.38 The cable-pulley systems shown are used to support a weight 𝑊 . (a) Assuming the segments of cable between the pulleys are vertical, determine the cable tension 𝑇 in terms of 𝑊 . (b) If the segments of cable between the pulleys are not precisely vertical, will your answer to Part (a) be affected? Explain.

𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

𝑇

𝑊

𝑊

𝑊

𝑊

𝑊

𝑊

(a)

(b)

(c)

(d)

(e)

(f)

Figure P3.38 𝑊

𝑊

(a)

(b)

Problem 3.39 The cable-pulley systems shown are used to support a weight 𝑊 . Determine the cable tension 𝑇 in terms of 𝑊 .

Figure P3.39

Problem 3.40 𝐶

12

𝐵

5

𝐷

𝐸

7

The hoist shown is used in a machine shop to position heavy workpieces in a lathe. If the cable between pulleys 𝐴 and 𝐵 can support a force of 300 lb, all other cables can support a force of 500 lb, and bar 𝐶𝐸 can support a compressive force of 600 lb, determine the largest weight 𝑊 that may be lifted.

24

Problem 3.41 𝐹

𝐴 𝑊

The cable system shown is used to help support a water pipe that crosses a creek. The water pipe applies forces 𝑊𝐴 and 𝑊𝐷 to the ends of cables 𝐴𝐵 and 𝐷𝐸, respectively, where it is known that 𝑊𝐴 + 𝑊𝐷 = 800 lb. If cable 𝐵𝐸 is horizontal and cables 𝐴𝐵 and 𝐷𝐸 are vertical, determine the tension in each cable and the forces 𝑊𝐴 and 𝑊𝐷 . 𝐶

Figure P3.40

ISTUDY

45◦

60◦

𝐵

𝐸

𝐴

𝐷 𝑊𝐴 Figure P3.41

𝑊𝐷

𝐹

ISTUDY

Section 3.1

149

Equilibrium of Particles in Two Dimensions

Problem 3.42 Compared to the cable system shown in Prob. 3.41, the system shown here may provide for more uniform support of a water pipe that crosses a creek. The water pipe applies forces 𝑊𝐴 , 𝑊𝐷 , and 𝑊𝐺 to the ends of cables 𝐴𝐵, 𝐷𝐸, and 𝐺𝐻, respectively, where it is known that 𝑊𝐴 + 𝑊𝐷 + 𝑊𝐺 = 800 lb. If cables 𝐴𝐵, 𝐷𝐸, and 𝐺𝐻 are vertical, determine the tension in each cable and the forces 𝑊𝐴 , 𝑊𝐷 , and 𝑊𝐺 . 𝐶 𝐹 45◦

60◦ 70◦ 70◦

𝐵

𝐸

𝐻 𝐴 𝑊𝐴

𝐺

𝐷

𝑊𝐺

𝑊𝐷

Figure P3.42

Problem 3.43 30◦

The frictionless slider 𝐵 has a frictionless pulley mounted to it, around which is wrapped a cable that supports weight 𝑊 = 30 lb as shown. The pulley at 𝐷 is also frictionless, and member 𝐵𝐸 is a bar.

𝐶

𝐴

(a) If 𝛼 = 20◦ , determine the force in bar 𝐵𝐸 and the reaction between the slider and bar 𝐴𝐶. (b) Determine the value of 𝛼 that will provide for the smallest force in bar 𝐵𝐸, and determine the value of this force.

𝐷

𝐵

𝐸 𝑊

𝛼 = 20◦

Figure P3.43

Problem 3.44 If 𝑊 = 225 N, determine the tension in cable 𝐵𝐶𝐴 and angle 𝜃. 𝐵

𝐶 70◦ 𝜃

𝐴

𝑊 Figure P3.44 and P3.45

Problem 3.45 Consider a configuration where 𝜃 =

𝐵

30◦ .

Is this configuration possible? Explain.

Problem 3.46

𝐶 50◦

𝐴 30◦ 𝑊2

𝑊1 50 N

Determine the weights 𝑊1 and 𝑊2 needed for the pulley-cable structure to have the equilibrium configuration shown.

Figure P3.46

150

Chapter 3

Equilibrium of Particles

Problems 3.47 and 3.48 75 in. 𝐴 𝐵

𝐶

ℎ

An object with 100 lb weight is supported by cable 𝐴𝐵𝐶 where the pulley at 𝐵 is frictionless. For the cable length and value of ℎ given below, determine the force supported by cable 𝐴𝐵𝐶 and the value of 𝑑 when the system is in equilibrium. Neglect the size of the pulley. Use of a computer mathematics program such as Mathematica or Maple is necessary. Hint: The solution for 𝑑 is an integer number.

𝑑 100 lb

Problem 3.47

Cable 𝐴𝐵𝐶 has 85 in. length and ℎ = 8 in.

Problem 3.48

Cable 𝐴𝐵𝐶 has 195 in. length and ℎ = 36 in.

Figure P3.47 and P3.48

ISTUDY

Problem 3.49 Load 𝑃 is supported by cables 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷. Cable 𝐴𝐶 is vertical. When you solve for these cable tensions, a “problem” arises. Describe this problem and, if possible, its physical significance. 𝐵

𝐷

𝐶

60◦

30◦ 𝐴 𝑃 Figure P3.49

Problem 3.50 Repeat Part (b) of Example 3.4 on p. 140, using optimization methods of calculus. Hint: Solve Eqs. (4) and (5) in Example 3.4 for 𝐹𝐵𝐷 as a function of 𝛼. Then solve for the value of 𝛼 that makes 𝑑𝐹𝐵𝐷 ∕𝑑𝛼 = 0 (this equation is difficult to solve analytically, and you may need to solve it graphically or by other approximate means). This problem can also be effectively solved using computer mathematics programs such as Mathematica or Maple.

ISTUDY

Section 3.2

3.2

Behavior of Cables, Bars, and Springs

Behavior of Cables, Bars, and Springs

We begin this section by more thoroughly examining the theoretical underpinnings of the equilibrium analyses performed in the previous section. We then introduce springs and the inclusion of deformable members in equilibrium problems. Finally, we present a very brief introduction to statically indeterminate problems in an example problem.

Equilibrium geometry of a structure ∑ ⃗ the oriWhen analyzing equilibrium by summing forces in Newton’s law 𝐹⃗ = 0, entation of forces in the equilibrium configuration must be used, and this generally means the orientation or shape of the structure after loads have been applied must be known, or must be determined. To explore the ramifications of this statement, reconsider the example shown in Fig. 3.13, which was previously analyzed in Example 3.1 on p. 136. The crucial question is, Do the 30◦ and 40◦ angles shown correspond to the geometry of the structure before force 𝑃 is applied, after 𝑃 is applied, or both? To answer this question, we first note that all materials are deformable, meaning if a material is subjected to a force, the material will change shape. After force 𝑃 is applied in Fig. 3.13, the cable 𝐴𝐵 will support a tensile force and hence, will lengthen, or elongate, and the bar 𝐴𝐶 will support a compressive force and hence, will shorten, or contract. These statements are true regardless of the material the cable and bar are made of, or the size of their cross sections.∗ Thus, if the 30◦ and 40◦ angles correspond to the geometry of the structure before 𝑃 is applied, then the geometry of the structure after 𝑃 is applied is different, or vice versa. Typically, the geometry of a structure before loads are applied is known. For example, when constructing the structure shown in Fig. 3.13, we first cut the cable and bar to prescribed lengths and then assemble them, producing a structure with known initial geometry. After loads are applied, the structure’s geometry changes ∑ and is unknown. So we immediately confront a problem when writing 𝐹𝑥 = 0 ∑ and 𝐹𝑦 = 0, namely, What angles should be used to resolve forces into 𝑥 and 𝑦 components? Fortunately, for many problems, the members that constitute a structure are sufficiently stiff and appropriately arranged so that the structure’s geometry changes very little after loads are applied. When this is the case, accurate results may be obtained by taking the geometries of the structure before loads are applied and after loads are applied to be the same. Words such as sufficiently stiff and adequately arranged are imprecise and difficult to quantify, so in most of statics, we idealize materials to be undeformable, as described in the following subsections.

Cables We generally assume cables to be inextensible and perfectly flexible. In other words, we assume a cable’s length does not change, regardless of how large its tensile load is. Also, we assume a cable is perfectly flexible, so that it may be freely bent, such as when it is wrapped around a pulley. A perfectly flexible cable will immediately buckle if it is subjected to a compressive force. In addition, the cable may have a maximum tensile load limit, beyond which the cable will be unsafe or will fail. ∗ Some

151

materials are inherently more resistant to shape change than others, and increasing a cable’s or bar’s cross section size will also help increase its resistance to deformation.

30◦ 𝐵 𝑦

𝐴

𝑥 𝑃

40◦

𝐶

Figure 3.13 A structure consisting of a cable and bar subjected to a force 𝑃 . This structure was analyzed in Example 3.1 on p. 136.

152

ISTUDY

Chapter 3

Equilibrium of Particles

Bars We generally assume bars to be rigid, as defined on p. 6, so that they are inextensible in both tension and compression. In other words, we assume a bar’s length does not change, regardless of how large its tensile or compressive load is. In Chapter 5, some additional features of the behavior of bars are discussed within the context of rigid bodies. In addition, a bar may have maximum tensile load and compressive load limits, beyond which the bar will be unsafe or will fail.

Modeling idealizations and solution of

∑ ⃗ ⃗ 𝐹 =0

The idealizations of behavior discussed above allow us to approximate a real life ∑ problem by a mathematical model where only the equilibrium equations 𝐹⃗ = 0⃗ are needed to obtain the solution. Because of these assumptions, the geometry of a structure after loads are applied is assumed to be known, and the methods of analysis discussed in this book give exact results for the forces in structural members and the support reactions.

Springs A spring is a mechanical device that produces a force when it undergoes a deformation. Springs come in a myriad of shapes, sizes, and materials. A coil spring is a particular type of spring that is constructed of wire (or other material) wound in the shape of a helix (usually). Figure 3.14 shows several examples of coil springs. A few examples of other types of springs include an elastic band, an elastic cord, and a gas spring, which consists of a column of gas in a sealed cylinder that is compressed by forces applied to it.

Courtesy of Peninsula Spring

Figure 3.14. An assortment of small coil springs made of wire.

Springs such as those shown in Fig. 3.14 are important and common structural members. However, cables and bars are also springs. In problems where the deformation of a structure is large, it is usually necessary to model the members of the structure, especially those that are very flexible, using springs. Even if the deformations are not large, it may be necessary to determine the deformation of the structure, in which case its members may be modeled as springs. The behavior of a linear elastic spring is characterized in Fig. 3.15. Linear means there is a linear relation between the force 𝐹𝑠 supported by the spring and the change in its length 𝛿. Elastic means the spring returns to its original length, or shape, after the force 𝐹𝑠 is removed. The equation governing the spring’s behavior is often called

ISTUDY

Section 3.2

Behavior of Cables, Bars, and Springs

𝐹𝑠

𝐿 = final length 𝐿0 = initial length

tension

initial

𝑘 1

𝛿

𝐹𝑠 = 𝑘𝛿

compression

153

𝛿 𝐹𝑠

final 𝑘

contraction

elongation

𝐹𝑠 = 𝑘𝛿 = 𝑘(𝐿 − 𝐿0 )

Figure 3.15. Spring law for a linear elastic spring.

Helpful Information the spring law and is 𝐹𝑠 = 𝑘𝛿 ) ( = 𝑘 𝐿 − 𝐿0 ,

(3.18)

where

𝐹𝑠 > 0 tension, 𝐹𝑠 < 0 compression, 𝛿 = 0 unstretched position,

𝐹𝑠 is the force supported by the spring; 𝛿 is the elongation of the spring from its unstretched or undeformed length; 𝑘 is the spring stiffness (units: force/length); 𝐿0 is the initial (unstretched) spring length; and 𝐿 is the final spring length. As stated in the Helpful Information margin note in Eq. (3.19), the sign convention in the spring law is that positive values of force 𝐹𝑠 correspond to tension and negative values to compression, and positive values of 𝛿 correspond to elongation and negative values to contraction. The constant 𝑘 is called the spring stiffness: it is always positive, and a large value means the spring is stiff whereas a low value means the spring is flexible. It is possible to use other conventions for positive force and deformation. For example, if 𝐹𝑠 and 𝛿 are positive in compression and contraction, respectively, the spring law is still given by Eq. (3.18). If 𝐹𝑠 is positive in compression and 𝛿 is positive in elongation, or vice versa, then the spring law becomes 𝐹𝑠 = −𝑘𝛿. Springs are used to represent a variety of deformable members in engineering problems. When the deformations of cables and bars must be accounted for, they are usually represented by springs where their stiffness depends on the material, length, and cross-sectional area of the bar or cable.∗ Other deformable members or structures, even if very complex, may also be characterized by a simple spring, such as the examples shown in Fig. 3.16. The nail clipper and lock washer each have a metal part that deforms as loads are applied. The teacup rests on a cushion of foam packaging that deflects due to forces from vibrations during handling. The tire of the wheelbarrow supports loads by changing its shape and compressing the air within it. The multistory building deflects when a force, perhaps due to wind, is applied as shown. ∗ For

Spring law sign conventions. The sign conventions for the spring law given in Eq. (3.18) are

both cables and bars, the spring stiffness is 𝑘 = 𝐴𝐸∕𝐿, where 𝐴 and 𝐿 are the cross-sectional area and length, respectively, of the cable or bar, and 𝐸, called the elastic modulus with units of force/area, is a property of the material that characterizes its inherent stiffness. For example, nylon has 𝐸 = 4×105 lb∕in.2 , while steel, which is much stiffer, has 𝐸 = 30×106 lb∕in.2 .

(3.19)

𝛿 > 0 extension, 𝛿 < 0 contraction.

Interesting Fact Springs. Springs are important structural members in their own right, but they are also important for laying the groundwork for characterizing more general engineering materials and members, which you will study in subjects that follow statics. Simply stated, almost all materials are idealized as springs, albeit more complex than that shown in Fig. 3.15, over at least some range of forces.

154

Chapter 3

Equilibrium of Particles

𝐹𝑠 𝛿 fingernail clipper 𝐹𝑠 𝛿 lock washer 𝐹𝑠

𝐹𝑠 𝛿 𝑘

𝛿

china teacup in box with foam packaging

𝐹𝑠 𝛿 wheelbarrow with pneumatic tire 𝐹𝑠

𝛿

𝑘

𝛿

𝐹𝑠

multistory building Figure 3.16. Examples of structures or devices that may be idealized as springs.

𝐹𝑠 𝐹𝑠

𝛿

𝑘 𝛿 𝐹𝑠

1

𝑘 = 𝑘(𝛿) 𝛿

Figure 3.17 Truck axle with multileaf spring suspension. Such springs are elastic, but nonlinear.

ISTUDY

In all cases cited here, the linear elastic spring law 𝐹𝑠 = 𝑘𝛿 may be used over at least a portion of the full range of response that these devices display, although determining the value of 𝑘 usually requires further analysis. Note that in all of these problems, as the deflection 𝛿 increases beyond some point, the response probably becomes nonlinear. For example, the nail clipper becomes very stiff once the cutting edges come into contact. The multistory building, on the other hand, will likely become less stiff as 𝛿 increases beyond some limit because the structural members within the building will sustain damage if they are loaded too severely. Occasionally springs are designed to be nonlinear, such as the leaf spring suspension for a truck shown in Fig. 3.17. When the load supported by the axle is small, only a few of the leaves in the spring will engage and the stiffness is low. As the load increases, more leaves engage and the stiffness becomes progressively higher.

ISTUDY

Section 3.2

Behavior of Cables, Bars, and Springs

Although nonlinear, a leaf spring is still elastic since it returns to its original shape when the load is removed.

End of Section Summary In this section, some of the finer points regarding static equilibrium were reviewed, and springs were discussed in detail. Some of the key points are as follows: ∑ ⃗ the geometry of the • When you are writing equilibrium equations 𝐹⃗ = 0, structure in the equilibrium position must be used. • A spring is a deformable member that undergoes a change of length when subjected to a force. The spring law is 𝐹𝑠 = 𝑘𝛿, where 𝑘 is called the spring stiffness and 𝑘 ≥ 0. In writing this equation it is assumed that force 𝐹𝑠 and displacement 𝛿 are positive in the same direction. The force 𝐹𝑠 and displacement 𝛿 may be measured positive in opposite directions, but it may be necessary to introduce a negative sign in the spring law (i.e., 𝐹𝑠 = −𝑘𝛿) as discussed on p. 153.

155

156

Chapter 3

Equilibrium of Particles

E X A M P L E 3.5

Springs A model for the latch of a briefcase is shown. Spring 𝐴𝐵 has stiffness 𝑘 = 3 N∕cm and 6 cm unstretched length. (a) Determine the force 𝐹 needed to begin closing the latch from the open position shown. (b) Determine the force 𝐹 needed to begin opening the latch from the closed position where member 𝐵𝐶 is horizontal.

SOLUTION Road Map

This problem involves a spring, and to determine the force it supports requires that we use the spring law, Eq. (3.18) on p. 153. This equation is in addition to the equilibrium equations.

Michael Plesha

2 cm 4 cm

6 cm

Part (a)

𝐹

Free body diagrams for points 𝐵 and 𝐶 are shown in Fig. 2. Note in the FBD for 𝐵 there are two reactions 𝑅𝑥 and 𝑅𝑦 between the roller at 𝐵 and the track it rolls in. However, at the instant the latch begins to close, contact between the roller and the vertical surface to the right of the roller is broken and hence, 𝑅𝑥 = 0. Modeling

𝐶

𝐷 1 cm

𝐴 𝐵

Figure 1 Briefcase latch in the open position.

Governing Equations & Computation

𝐹

𝑦 𝑥 𝑄

𝑇𝐵𝐶 𝐵

𝑇𝐵𝐶 𝐶 √ 5 1 𝑅𝑥 2

𝑇𝐶𝐷

𝑅𝑦 Figure 2 Free body diagrams of points 𝐵 and 𝐶 for closing the latch from the open position. 𝑦 𝐹𝑠

𝑥

𝐹𝑠

Newton’s third law requires 𝑄 = −𝐹𝑠

𝑇𝐵𝐶 𝑄

𝐵

Recall that in developing Eq. (3.18), and hence in writing Eq. (1), the spring force 𝐹𝑠 is positive in tension. Thus, the negative sign that appears in Eq. (1) indicates the spring is in compression. However, in the FBD of point 𝐵, the force 𝑄 applied by the spring to point 𝐵 is taken to be positive in compression. Thus, 𝑄 = −𝐹𝑠 = 6 N. This mental accounting of sign is simple to perform; but if there is any confusion, then adding the FBD of the spring to that of point 𝐵, as shown in Fig. 3, provides full clarification, where it is seen that Newton’s third law requires 𝑄 = −𝐹𝑠 . Writing equilibrium equations for point 𝐵 provides √ 𝐹𝑥 = 0 ∶ 𝑄 − 𝑅𝑥 + 𝑇𝐵𝐶 (2∕ 5) = 0, √ ∑ 𝐹𝑦 = 0 ∶ 𝑅𝑦 + 𝑇𝐵𝐶 (1∕ 5) = 0.

Equilibrium Equations 𝑅𝑥

𝑅𝑦

Figure 3 Free body diagrams of point 𝐵 and the spring showing that force 𝑄 and spring force 𝐹𝑠 are related by Newton’s third law with the result 𝑄 = −𝐹𝑠 .

ISTUDY

Force Laws To determine the spring force, one of the forms of Eq. (3.18) will always be applicable, with the choice depending on the form of data available in the problem. Here the initial length of the spring is 𝐿0 = 6 cm, and the final length of the spring, from Fig. 1, is 𝐿 = 4 cm. Hence, in this problem the second form of Eq. (3.18) is more useful. In the geometry shown, when the latch begins to close, the spring force 𝐹𝑠 is ( ) N 𝐹𝑠 = 3 (4 cm − 6 cm) = −6 N. (1) cm

∑

(2) (3)

With 𝑅𝑥 = 0 and 𝑄 = 6 N, Eqs. (2) and (3) may be solved to obtain 𝑇𝐵𝐶 = −6.708 N and 𝑅𝑦 = 3 N.

Writing equilibrium equations for point 𝐶 provides √ ∑ 𝐹𝑥 = 0 ∶ −𝑇𝐵𝐶 (2∕ 5) + 𝑇𝐶𝐷 = 0, √ ∑ 𝐹𝑦 = 0 ∶ −𝑇𝐵𝐶 (1∕ 5) − 𝐹 = 0.

(4)

(5) (6)

By using the result for 𝑇𝐵𝐶 in Eq. (4), Eqs. (5) and (6) may be solved for: 𝑇𝐶𝐷 = −6 N

and 𝐹 = 3 N.

Hence, a downward force 𝐹 = 3 N will cause the latch to begin closing.

(7)

ISTUDY

Section 3.2

157

Behavior of Cables, Bars, and Springs

Part (b) Modeling In the closed position, member 𝐵𝐶 is horizontal as shown in Fig. 4. The FBDs of points 𝐵 and 𝐶 are shown in Fig. 5 where, for convenience, we have reassigned

𝐹 to be positive upward since clearly a force in this direction is needed to open the latch. Note in the FBD for 𝐶 that there is a reaction 𝑅𝐶 between point 𝐶 and the surface below it that it contacts. However, at the instant the latch begins to open, this contact is broken and hence, 𝑅𝐶 = 0. Governing Equations & Computation

To obtain 𝑇𝐵𝐶 , the force 𝑄 exerted by the spring on point 𝐵 is needed. The initial length of spring 𝐴𝐵 is known, and the final length 𝐿 of the spring can be evaluated, using the geometry in Figs. 1 and 4, as √ √ 𝐿 = 4 cm + 2 cm + 6 cm − ( 5 cm + 35 cm) = 3.848 cm. (8)

𝐹

𝐿

𝐷

𝐶

𝐴 𝐵 √ 5 cm

1 cm

√ 35 cm

Figure 4 Briefcase latch in the closed position.

Force Laws

In Eq. (8), 4 cm + 2 cm is obtained from Fig. 1 as the horizontal distance between √ + 6 cm √ points 𝐴 and 𝐷, and 5 cm + 35 cm is obtained from Fig. 4 as the horizontal distance between points 𝐵 and 𝐷. The second form of Eq. (3.18) then provides ( ) N (9) 𝐹𝑠 = 3 (3.848 cm − 6 cm) = −6.456 N. cm Equilibrium Equations From the FBD of point 𝐵 in Fig. 5, the equilibrium equations for point 𝐵 are as follows: ∑ 𝐹𝑥 = 0 ∶ 𝑄 + 𝑇𝐵𝐶 = 0, (10) ∑ 𝐹𝑦 = 0 ∶ 𝑅𝑦 = 0. (11)

As argued earlier in Fig. 3, 𝑄 = −𝐹𝑠 = 6.456 N, and the solutions of Eqs. (10) and (11) are, respectively, 𝑇𝐵𝐶 = −6.456 N and 𝑅𝑦 = 0 . (12) Writing equilibrium equations for point 𝐶 provides ∑

∑

𝐹𝑥 = 0 ∶ 𝐹𝑦 = 0 ∶

(√

) 35 −𝑇𝐵𝐶 + 𝑇𝐶𝐷 = 0, 6 ( ) 1 + 𝑅𝐶 = 0. 𝐹 + 𝑇𝐶𝐷 6

(13) (14)

By using the results for 𝑇𝐵𝐶 from Eq. (12) and noting that 𝑅𝐶 = 0, Eqs. (13) and (14) may be solved for: 𝑇𝐶𝐷 = −6.548 N and 𝐹 = 1.091 N. (15) Hence, an upward force 𝐹 = 1.091 N will cause the latch to begin opening. Discussion & Verification

In this example, it was possible to evaluate the material models (i.e., spring law) independently of solving the equilibrium equations. In some problems, such as Example 3.6, the material models and the equilibrium equations are coupled and must be solved simultaneously.

𝑦 𝑄

𝐹 𝑥

𝐵

𝑇𝐵𝐶 𝑅𝑦

𝐶 𝑅𝐶

𝑇𝐶𝐷 6 1 √ 35

Figure 5 Free body diagrams for points 𝐵 and 𝐶 for opening the latch from the closed position.

158

Chapter 3

Equilibrium of Particles

E X A M P L E 3.6

Introduction to Statically Indeterminate Problems 𝐽

𝐻 𝐵 𝑘2 𝐷

𝐺

𝐴

𝑘2 𝐹

𝐼

𝑘1

𝛿 𝐶

𝐸

Figure 1

A drop hammer is used in some sheet metal forming and forging operations. With this machine, die 𝐵 is dropped onto die 𝐴, and due to impact forces, the workpiece between 𝐴 and 𝐵 is hammered into the shape of the cavity or die between them. Both dies 𝐴 and 𝐵 slide on low-friction fixed vertical bars so they maintain proper alignment. Special coil springs 𝐶𝐷 and 𝐸𝐹 , called die springs, are used to help isolate the severe vibrations of the dies from being transmitted to the floor that supports the machine. The user of the machine determines that additional stiffness is needed for vibration isolation. Although the die springs could be replaced with heavier springs, this requires disassembly of the machine. Instead, additional vertical springs such as 𝐺𝐻 and IJ are more easily added and removed when they are no longer needed. The springs have stiffnesses 𝑘1 = 5000 lb∕in. and 𝑘2 = 1500 lb∕in. If all springs are unstretched when 𝛿 = 0, determine the deflection 𝛿 if die 𝐴 is subjected to a downward vertical force 𝑃 = 1 kip.

𝑃 𝑦

𝑄2

Road Map

𝐴

𝑥 𝑄1

SOLUTION

𝑄2 𝛿 𝑄1

Figure 2 Free body diagram of die 𝐴 modeled as a particle.

ISTUDY

Helpful Information Statically indeterminate structures. A characteristic of a statically indeterminate structure is that there are extra “load paths” available to support the applied loads. In this example, the springs at the base, by themselves, are adequate to support the applied loads, and equations of equilibrium will determine exactly what value of force they must support. Conversely, the springs on the sides, by themselves, are also adequate to support the applied loads, and equations of equilibrium will determine exactly what value of force they must support. With both sets of springs present, it is unclear from the equations of equilibrium alone what portions of the applied force are supported by each set of springs. Either set of springs could be viewed as providing an extra load path. Static indeterminacy makes analysis more difficult, but it is often a desirable feature to have in structures. For example, if one load path in a statically indeterminate structure is lost because of an accident or failure, the remaining load paths may be sufficient for the structure to remain in static equilibrium. Static indeterminacy is studied in greater detail in later chapters of this book.

A solution to this problem will require as many equations as there are unknowns. As described in the margin note, the equations of equilibrium alone are not sufficient in number, and additional equations must be added that describe the behavior of the springs.

Modeling Assuming the vertical guides allow die 𝐴 to undergo only vertical motion with no rotation, both springs at the bottom of 𝐴 are compressed by the same amount, and hence, develop the same force 𝑄1 . Similarly, both springs on the sides of 𝐴 develop the same force 𝑄2 . Idealizing die 𝐴 as a particle allows the FBD shown in Fig. 2 to be drawn, where 𝑄1 and 𝑄2 are taken to be positive in compression and tension, respectively. Equilibrium Equations

The equilibrium equation for 𝐴 is ∑

𝐹𝑦 = 0 ∶ 2𝑄1 + 2𝑄2 − 𝑃 = 0.

(1)

∑ Since no forces act in the 𝑥 direction, the equation 𝐹𝑥 = 0 is automatically satisfied regardless of the values for 𝑄1 , 𝑄2 , and 𝑃 , and hence, this equation provides no useful information. Equation (1) is unlike any we have encountered thus far in this book. Even though 𝑃 = 1 kip, there remain two unknowns and only one equation with which to determine them. Such a problem has no unique solution. Rather, there are an infinite number of solutions for 𝑄1 and 𝑄2 that will satisfy Eq. (1). We call such problems statically indeterminate, because the equations of static equilibrium alone are not sufficient to determine the unknowns. To obtain the appropriate solution among the infinite number of solutions that are possible, additional information is needed. In this problem, this information corresponds to descriptions of the deformability of the springs, which we address next. Force Laws Since all the springs are uncompressed when 𝛿 = 0, the first form of Eq. (3.18) on p. 153 allows us to write for each set of springs

𝑄1 = 𝑘1 𝛿 = (5000 lb∕in.) 𝛿,

(2)

𝑄2 = 𝑘2 𝛿 = (1500 lb∕in.) 𝛿.

(3)

In writing Eq. (2), 𝑄1 is positive in compression, and positive 𝛿 gives contraction of the springs at the base. Hence, the negative signs associated with each of these quantities, compared to our normal convention of spring force being positive in tension and spring deformation being positive in extension, compensate for one another.

ISTUDY

Section 3.2

Computation

Behavior of Cables, Bars, and Springs

Combining Eqs. (1)–(3) provides ) ( lb lb 𝛿 − 𝑃 = 0. 2 5000 + 1500 in in

(4)

With 𝑃 = 1000 lb, we now have one equation and one unknown, and Eq. (4) may be solved for 𝛿 = 0.07692 in. (5) Once 𝛿 is known, if desired, we may return to Eqs. (2) and (3) to determine the portion of the force 𝑃 that is supported by each spring to obtain 𝑄1 = 384.6 lb and 𝑄2 = 115.4 lb. Discussion & Verification

(6)

We expect the die to move downward, and hence, our solution should show 𝛿 > 0, which it does. We expect springs 𝐶𝐷 and 𝐸𝐹 to be in compression, and since 𝑄1 was defined to be positive in compression, our solution should show 𝑄1 > 0, which it does. Similarly, we expect springs 𝐺𝐻 and IJ to be in tension, and since 𝑄2 was defined to be positive in tension, our solution should show 𝑄2 > 0, which it does. After these simple checks, we should verify that our solutions for 𝛿, 𝑄1 , and 𝑄2 satisfy all governing equations, which consist of the equilibrium equation, Eq. (1), and the force laws, Eqs. (2) and (3).

Vershinin89/Shutterstock

Figure 3. A drop hammer used for forging an automobile part.

159

160

Chapter 3

Equilibrium of Particles

Problems Problem 3.51

𝑇

𝐴

𝑇

𝐶

𝑘

ℎ 𝐵

(a) ℎ = 18 mm.

24 mm 24 mm

(b) ℎ = 10 mm.

Figure P3.51

𝛿 𝐵

A device for tensioning recording tape in a video cassette recorder is shown. The tape wraps around small pins at 𝐴, 𝐵, and 𝐶. The pins at 𝐴 and 𝐶 are fixed, and the pin at 𝐵 is supported by a spring and can undergo vertical motion in the frictionless slot. Friction between the tape and pins is negligible. The spring has stiffness 𝑘 = 0.5 N∕mm and is unstretched when ℎ = 25 mm. Neglecting the size of the pins, determine the tension in the tape when

30◦

𝐴

𝐶

500 lb∕in.

𝐸

𝐷 1000 lb∕in. 𝐹

45◦ 𝐺

500 lb∕in. 𝐻

Problem 3.52 The brake linkage for a vehicle is actuated by pneumatic cylinder 𝐴𝐵. Cylinder 𝐴𝐵, springs 𝐶𝐷, 𝐸𝐹 , and 𝐺𝐻, and the slotted tracks at 𝐴, 𝐶, 𝐸, and 𝐺 are all parallel. If the slotted tracks are frictionless, and cylinder 𝐴𝐵 produces a tensile force of 12 kip, determine the deflection 𝛿 and forces in members 𝐴𝐶, 𝐴𝐸, and 𝐴𝐺. All springs are unstretched when 𝛿 = 0.

Figure P3.52

Problems 3.53 and 3.54 𝑘1

The system shown may undergo vertical motion only. For the values of 𝑚𝐴 , 𝑚𝐵 , 𝑘1 , and 𝑘2 given below, determine the displacements 𝛿𝐴 and 𝛿𝐵 .

𝑚𝐴 𝑘2

𝛿𝐴 𝑚𝐵

Problem 3.53

𝑚𝐴 = 100 kg, 𝑚𝐵 = 80 kg, 𝑘1 = 200 N∕mm, and 𝑘2 = 300 N∕mm.

Problem 3.54

𝑚𝐴 = 5 kg, 𝑚𝐵 = 4 kg, 𝑘1 = 6 N∕mm, and 𝑘2 = 3 N∕mm.

𝛿𝐵

Problem 3.55

Figure P3.53 and P3.54 each crate weighs 2000 lb

𝐴

(a)

𝑘

𝛿𝐴

𝐵

𝛿𝐵

𝐶

𝛿𝐶

𝐷

𝛿𝐷

𝐹𝐴 = 1000 lb 𝐹𝐵 = 2000 lb 𝑘 𝐹𝐶 = 2000 lb 𝑘 𝐹𝐷 = 2000 lb 𝑘

𝐸 (b)

In Fig. P3.55(a), four identical crates weighing 2000 lb each are stacked one on top of another, and in Fig. P3.55(b), a simple model for determining the deformation of the stack of crates is shown. In this model, each spring has the same stiffness 𝑘 = 4500 lb∕in. and the forces are determined using the following idealization.∗ Half the weight of the top crate is applied to point 𝐴 and half to point 𝐵. Similarly, half the weight of the next crate is applied to point 𝐵 and half to point 𝐶, and so on. Determine the deflections 𝛿𝐴 , 𝛿𝐵 , 𝛿𝐶 , and 𝛿𝐷 .

Problem 3.56 In Prob. 3.55, let the top two crates weigh 3000 lb each, and the bottom two crates weigh 4000 lb each, and all springs have the same stiffness 𝑘 = 5000 lb∕in. Use the scheme described in Prob. 3.55 to determine the forces 𝐹𝐴 –𝐹𝐷 and determine the deflections 𝛿𝐴 , 𝛿𝐵 , 𝛿𝐶 , and 𝛿𝐷 .

Figure P3.55

ISTUDY

∗ Other idealizations for distributing the weight of the crates are possible, and a better scheme will consider

details of how the contents of the crates are supported within the crates.

ISTUDY

Section 3.2

161

Behavior of Cables, Bars, and Springs

Problem 3.57 𝐷

3 ft

Collar 𝐴 has negligible weight and slides without friction on the vertical bar 𝐶𝐷. Determine the vertical force 𝐹 that will produce 𝜃 = 30◦ if

𝐵 𝜃

(a) Spring 𝐴𝐵 is unstretched when 𝜃 = 0◦ . (b) Spring 𝐴𝐵 has an unstretched length of 2 f t.

𝑘 = 20 lb∕f t

𝐴

(c) Spring 𝐴𝐵 has an unstretched length of 4 f t. 𝐹 𝐶

Problem 3.58

Figure P3.57

Point 𝐴 is supported by springs 𝐵𝐶 and 𝐷𝐸 and cable segments 𝐴𝐵 and 𝐴𝐷. The springs and cables have negligible weight, and the springs have identical stiffness 𝑘 = 10 N∕cm and 20 cm unstretched length. The structure has the geometry shown when 𝐹 = 225 N.

𝐶 𝐵

𝑥

(a) Determine the lengths of cables 𝐴𝐵 and 𝐴𝐷.

𝐸

𝑦 𝐷

120 cm

(b) Determine the coordinates of point 𝐴 when 𝐹 = 0. 𝐴 50 cm

Problem 3.59

90 cm 𝐹 = 225 N

Spring 𝐴𝐵 is supported by a frictionless roller at 𝐵 so that it is always vertical. If the spring is unstretched when 𝜃 = 0◦ , determine 𝜃 and the forces in spring 𝐴𝐵 and bar 𝐴𝐶 when 𝐹 has the value indicated. Hint: The force supported by bar 𝐴𝐶 is zero when 𝜃 < 90◦ , and it may be nonzero when 𝜃 = 90◦ . (a) 𝐹 = 25 lb.

Figure P3.58

𝐵 𝐶

𝑘 = 1 lb∕in. 𝜃

(b) 𝐹 = 50 lb.

38 in.

Problems 3.60 and 3.61

𝐹

A collar with a pulley slides on a frictionless vertical bar 𝐺𝐻. A string 𝐴𝐵𝐶𝐷 is wrapped around the pulley, where portion 𝐴𝐵 of the string is horizontal. A spring with 2 lb∕in. stiffness is placed between the collar and point 𝐻. The spring has 8 in. unstretched length and 6 in. final length. The collar and pulley have a combined weight of 5 lb, and the weights of all other components are negligible. Determine the value of 𝑃 required for equilibrium, and the reaction between the collar and bar 𝐺𝐻. 𝐻

𝐻 spring

𝐷 60◦

spring

𝐵

𝐴

5 lb weight 𝐺 Figure P3.60

3

𝐷

4

pulley

pulley 𝐶

𝐴

𝐵

𝑃

𝐶 5 lb weight 𝐺 Figure P3.61

𝐴

𝑃

Figure P3.59

162

Chapter 3

Equilibrium of Particles

𝐵

Problem 3.62

4N

The machine shown is used for compacting powder. Collar 𝐶 slides on plunger 𝐴𝐵 and is driven by a motor (not shown in the sketch) so that it has the oscillatory vertical motion 𝛿 = (10 mm)(sin 𝜋𝑡) where 𝑡 is time in seconds. Plunger 𝐴𝐵 weighs 4 N and is pressed against the powder by the spring whose end is driven by the motion of collar 𝐶. The spring has stiffness 𝑘 = 0.1 N∕mm and 100 mm unstretched length. The 50 mm dimension shown is the spring length when the machine is started (i.e., 𝛿 = 0 and the powder column is at its initial height ℎ = 110 mm); this dimension changes with the oscillation of collar 𝐶 and as the plunger 𝐴 moves down. Assume there is no friction in the system (other than friction between individual grains of powder), and assume the motion of 𝐶 is slow enough that there are no dynamic effects. Determine the largest and smallest forces the plunger applies to the powder over a full cycle of motion of 𝐶 if

𝐶 𝑘

50 mm

𝛿(𝑡)

𝐴 powder ℎ

(a) The powder column is at its initial height ℎ = 110 mm. (b) The powder column is at its compacted height ℎ = 80 mm.

Figure P3.62

Problem 3.63 A fuel pump is driven by a motor-powered flywheel. The pump behaves as a spring with stiffness 2 N∕mm that is unstretched when 𝛼 = ±60◦ . Neglect any possible dynamic effects. (a) Determine the largest tensile and compressive forces spring 𝐴𝐵 experiences during one revolution of the flywheel, and state the positions 𝛼 where these occur. (b) Without further analysis, is it certain that the largest tensile and compressive forces in crank arm 𝐵𝐶 occur at the same positions 𝛼 determined in Part (a)? Explain.

60 mm fuel pump 𝐴 𝐴

Figure P3.64

ISTUDY

𝛼

𝐷

𝐵 crank arm Figure P3.63

ℎ

𝐹

20 mm

flywheel

𝐶

𝐵

𝐶

Problem 3.64 The suspension for the landing gear of an aircraft is shown. The wheel is attached to bar 𝐵𝐶, which slides vertically without friction in housing 𝐴, which is fixed to the frame of the aircraft. The spring is precompressed so that it does not undergo additional deflection until the force supported by the landing gear is sufficiently large. Further, if the force supported by the landing gear exceeds a limit, the suspension “bottoms out” and deflects no more. Specify the spring stiffness and initial length if ℎ = 18 in. for 𝐹 ≤ 500 lb, and ℎ = 12 in. for 𝐹 ≥ 1500 lb.

ISTUDY

Section 3.2

Behavior of Cables, Bars, and Springs

̃ 3 in.

Problem 3.65 A hanger is made of cord-reinforced rubber. It is used as a spring support with limited travel for a wide variety of applications. If prototype samples are available, an effective means to characterize its nonlinear stiffness is by testing in a laboratory, where forces of known magnitude are applied and the deflections that result are measured. Imagine this produces the load-deflection data provided in the table of Fig. P3.65. (a) Determine the constants 𝑎, 𝑏, and 𝑐 that will fit the general quadratic equation 𝐹 = 𝑎 + 𝑏𝛿 + 𝑐𝛿 2 to the load-deflection data for this hanger. (b) Plot the load-deflection relation determined in Part (a). (c) Speculate on the range of values for 𝐹 for which the relation obtained in Part (a) will be reasonably accurate.

163

𝐹 𝛿 0 lb 0 in. 100 lb 0.50 in. 300 lb 1.00 in.

̃ 4 in.

rubber hanger steel link 𝛿

𝐹

Figure P3.65

Problem 3.66 In Prob. 3.65, imagine that the load-deflection data for the hanger is such that 𝐹 = (100 lb∕in.) 𝛿 + (200 lb∕in.2 ) 𝛿 2 and that three hangers are used to support a straight rigid pipe from an uneven ceiling as shown. Assume the pipe may undergo vertical motion only. Hanger 2 displaces by 0.50 in. more than hangers 1 and 3. Because of symmetry, equal forces are supported by hangers 1 and 3. Further, the sum of the forces supported by all three hangers equals the weight of the pipe and its contents, which is 600 lb. Determine the deflection of the pipe and the forces supported by each of the three hangers.

hanger 2

hanger 1

hanger 3

0.50 in. 𝛿 Figure P3.66

Problem 3.67 A model for the suspension of a vehicle is shown where the spring has stiffness 𝑘 = 200 N∕mm and an unstretched length of 360 mm. (a) Determine the value of 𝑃 and the force supported by member 𝐴𝐵 so that the suspension has the equilibrium position shown where member 𝐴𝐵 is horizontal. (b) If 𝑃 = 0, determine the 𝑥 and 𝑦 coordinates of point 𝐴, namely 𝑥𝐴 and 𝑦𝐴 .

𝑦 tire and wheel

𝐶 𝑘 180 mm 𝐵

𝐴 240 mm

𝑃

Problem 3.68 Repeat Prob. 3.67 if the spring has stiffness 𝑘 = 250 N∕mm and an unstretched length of 340 mm.

Figure P3.67 and P3.68

Problems 3.69 and 3.70 If 𝑘 = 5 N∕mm and 𝑊 = 100 N, determine 𝛿. Springs are unstretched when 𝛿 = 0. 2𝑘

𝑘

𝛿 𝑊

𝛿 𝑊

𝑘

20◦ Figure P3.69

20◦ Figure P3.70

𝑘1

Problem 3.71 The system shown can undergo vertical motion only, and both springs are unstretched when 𝑊 = 0. Determine the deflection 𝛿 in terms of 𝑊 , 𝑘1 , and 𝑘2 . Show that your result is correct for the following five special cases: when 𝑊 = 0; when 𝑘1 = 0; when 𝑘2 = 0; when 𝑘1 → ∞; and when 𝑘2 → ∞.

𝑊 𝛿

Figure P3.71

𝑘2

𝑥

164

Chapter 3

Equilibrium of Particles

Problem 3.72

𝑊 𝑑 𝑘1

The system shown can undergo vertical motion only. The springs have stiffnesses 𝑘1 = 240 lb∕in. and 𝑘2 = 310 lb∕in., and both springs have the same unstretched length of 100 in. The springs are installed such that the foundation where the springs are built-in has ℎ = 8 in.

𝑘2 ℎ

(a) If 𝑊 = 0, determine the distance 𝑑 and the force supported by each spring.

Figure P3.72

(b) If 𝑊 = 5000 lb, determine the distance 𝑑 and the force supported by each spring.

Problems 3.73 and 3.74 The system shown can undergo vertical motion only, and the springs are unstretched when 𝛿𝐴 = 𝛿𝐵 = 0. Determine 𝛿𝐴 , 𝛿𝐵 , and the force supported by each spring if 𝑚𝐴 = 50 kg, 𝑚𝐵 = 80 kg, 𝑘1 = 100 N∕mm, 𝑘2 = 120 N∕mm, and 𝑘3 = 140 N∕mm.

𝑘1 𝑚𝐴 𝛿𝐴

𝑚𝐴 𝑘2

𝛿𝐴

𝑚𝐵 𝛿𝐵

𝑘1 𝑘3

𝑚𝐵 𝑘3

Figure P3.73

𝛿𝐵

𝑘2 Figure P3.74

Problem 3.75

spring 1

𝐵

𝑘1

𝐴

A fragile item 𝐴, with weight 𝑊 , is to be shipped within a box 𝐵 using the vertical spring suspension shown. Springs 1 and 2 have stiffnesses 𝑘1 and 𝑘2 , respectively, and unstretched lengths 𝐿1 and 𝐿2 , respectively. The springs are installed by stretching them to the same length ℎ and then attaching them to 𝐴. The vertical deflection 𝛿 is measured such that 𝛿 = 0 is the position where 𝐴 is in the middle of box 𝐵. 𝑊 − 𝑘1 (ℎ − 𝐿1 ) + 𝑘2 (ℎ − 𝐿2 ) . (a) Show that 𝑊 and 𝛿 are related by 𝛿 = 𝑘1 + 𝑘2

ℎ

𝑊

𝛿 𝑘2

spring 2

(b) Explain why 𝛿 ≠ 0 when 𝑊 = 0.

ℎ

(c) Suggest some mathematical tests you can perform to verify the accuracy of the expression in Part (a). For example, if 𝛿 = 0 and 𝑘2 = 0, show that 𝑊 has the expected value.

Figure P3.75

Problem 3.76

𝐹 𝐴 ℎ 𝛿 𝑘1

𝑘2

A model for the suspension of a truck is shown. Block 𝐴 represents the chassis of the truck, and it may undergo vertical motion only with no rotation. Force 𝐹 represents the portion of the truck’s weight and payload that is supported by this suspension. Due to 𝐹 , block 𝐴 deflects by amount 𝛿. When 𝐹 = 0, both springs are undeformed, and there is a gap ℎ between the second spring and the chassis. When 𝛿 is less than ℎ, only one spring supports 𝐹 ; and when 𝛿 exceeds ℎ, the second spring engages and helps support 𝐹 . If 𝑘1 = 1200 lb∕in., 𝑘2 = 600 lb∕in., and ℎ = 1.3 in., (a) Determine 𝛿 when 𝐹 = 1400 lb.

Figure P3.76

ISTUDY

(b) Determine 𝛿 when 𝐹 = 2800 lb.

ISTUDY

Section 3.2

Behavior of Cables, Bars, and Springs

Problems 3.77 through 3.80

𝐶 𝑘

𝛼

The structure shown is a retractable tool holder that is used in a factory to support a tool at point 𝐷. When the tool is to be used, the worker will grasp the tool and apply a downward force to lower it to the position that is needed. When the tool is not in use, the spring causes the tool to retract so that it is out of the way. The spring has 2 lb∕in. stiffness, and assume 𝑊 is vertical.

𝐴

30 in. 𝜃

𝑊

30 in.

𝐵

If 𝑊 = 10 lb and the structure is to have an equilibrium configuration where 𝜃 = 65◦ , determine the unstretched length of the spring.

Problem 3.77

If the unstretched length of the spring is 30 in., and 𝜃 = 𝛼 = 60◦ when 𝑊 = 0, determine the value of 𝑊 needed so that 𝜃 = 80◦ .

165

𝐷

Problem 3.78

Figure P3.77–P3.80

If the unstretched length of the spring is 30 in., and 𝜃 = 𝛼 = 60◦ when 𝑊 = 0, determine the value of 𝑊 needed so that 𝜃 = 100◦ .

Problem 3.79

If the unstretched length of the spring is 30 in., and 𝜃 = 𝛼 = 60◦ when 𝑊 = 0, determine the value of 𝜃 that occurs when 𝑊 = 15 lb. Note: Unlike Probs. 3.77–3.79, the system of equations for this problem is difficult to solve by hand, and software such as Mathematica or Maple is very helpful. Problem 3.80

Problem 3.81 A model for a push button or key, such as on a calculator or computer keyboard, is shown. The model has the feature that if 𝐹 is sufficiently large, point 𝐴 snaps through to give the user positive tactile feedback that the keystroke was properly entered. Both springs have stiffness 2 N∕mm, and when 𝐹 = 0, the structure has the geometry shown and all springs are unstretched.

𝐹

(a) Derive an expression that gives 𝐹 as a function of 𝛿.

𝐴

(b) Plot 𝐹 versus 𝛿 for 0 ≤ 𝛿 ≤ 5 mm.

𝐵

(c) Determine the approximate largest value that 𝐹 has for 0 ≤ 𝛿 ≤ 2 mm. (d) What deficiency does this model display for representing a push button?

𝛿 5 mm

𝐶

2 mm

𝐶

2 mm

5 mm

Figure P3.81

Problem 3.82 The model of Prob. 3.81 is revised to include a third spring. Springs 𝐴𝐵 and 𝐴𝐶 have stiffness 2 N∕mm, and spring 𝐴𝐷 has stiffness 0.3 N∕mm. When 𝐹 = 0, the structure has the geometry shown and all springs are unstretched. (a) Derive an expression that gives 𝐹 as a function of 𝛿.

𝐹 𝐴 𝐵

𝛿

(b) Plot 𝐹 versus 𝛿 for 0 ≤ 𝛿 ≤ 5 mm. (c) Determine the approximate largest value that 𝐹 has for 0 ≤ 𝛿 ≤ 2 mm. (d) Does spring 𝐴𝐷 serve a useful purpose in this model? Explain.

𝐷 5 mm Figure P3.82

5 mm

166

Chapter 3

Equilibrium of Particles

3.3

Equilibrium of Particles in Three Dimensions

The conditions for static equilibrium of a particle in three dimensions are

or or Geoffrey Taunton/Alamy Stock Photo

(a)

David Cheskin - PA Images/PA Images/Getty Images

(b) Figure 3.18 (a) Temporary grandstand erected at Edinburgh Castle, Scotland. (b) Close-up view showing numerous bar members that intersect at joints. Each of these joints may be modeled as a particle in equilibrium.

ISTUDY

∑

⃗ 𝐹⃗ = 0, (∑ ) (∑ ) ⃗ 𝐹𝑦 𝚥̂ + 𝐹𝑧 𝑘̂ = 0, 𝐹𝑥 𝚤̂ + ∑ ∑ ∑ 𝐹𝑥 = 0 and 𝐹𝑦 = 0 and 𝐹𝑧 = 0. (∑

)

(3.20)

The first two expressions in Eq. (3.20) state conditions for equilibrium in vector form, while the last expression states conditions in scalar form. In three-dimensional problems, it will very often be advantageous to use the vector form. Fundamentally, everything stated in earlier sections for equilibrium of particles in two dimensions also applies to equilibrium in three dimensions, except now all vectors have, in general, three components. The procedure for drawing an FBD of a particle is as described in Section 3.1, except now it is useful to think of the cut as being a closed surface that completely surrounds the particle. The procedure for analysis is also as described in Section 3.1, except now forces in the 𝑧 direction must also sum to zero. All remarks made in Sections 3.1 and 3.2 on pulleys and the forces supported by bars, cables, and springs are still applicable. Some additional remarks on reactions and the solution of algebraic equations are helpful.

Reactions We use the same thought process described in Section 3.1 to construct reactions for a particle in three dimensions. That is, if a support prevents motion of a particle in a certain direction, it can do so only by producing a reaction force in that direction. When you are solving particle equilibrium problems, the supports and associated reactions shown in Fig. 3.19 occur often. It is not necessary to memorize these reactions; rather, you should reconstruct these as needed. For example, consider the slider on a fixed frictionless bar. The bar prevents motion of the slider in both the 𝑦 and 𝑧 directions, so there must be reactions in both of these directions. The slider is free to move in the 𝑥 direction; hence, there is no reaction in this direction.

Solution of algebraic equations Analysis of equilibrium of one particle in three dimensions will typically produce three simultaneous linear algebraic equations (equilibrium equations) with three unknowns. More complicated problems may require several FBDs, and there will generally be three equations and three unknowns for each. If the problem has convenient geometry and/or symmetry, then solving these equations may be straightforward. In general, however, the solution of three or more equations is tedious. Fortunately, efficient techniques for solving systems of simultaneous algebraic equations have been developed. Most scientific pocket calculators include programs for solving such equations, and you may wish to become familiar with this feature. In addition, programs such as Mathematica, Maple, Mathcad, EES (Engineering Equation Solver), MATLAB, and many others are available for use on personal computers and workstations. You are strongly encouraged to become familiar with some of these

ISTUDY

Section 3.3

Equilibrium of Particles in Three Dimensions

Support

Reactions

𝐹2 𝑦

𝐹1

𝐹3

𝐹2

𝑅𝑥

𝑧 particle pinned to surface 𝐹2 𝑦

𝐹1

𝐹3

𝐹1

𝑅𝑦

𝑧

For a particle on a frictionless surface, 𝑅𝑥 = 𝑅𝑧 = 0.

𝑧 particle on rough surface

𝐹2 𝑦

𝑥

𝑅𝑧

𝑥

𝐹2

𝐹3

𝑦

𝐹1

𝑥

𝐹3

𝐹1

𝑥

𝑥 𝑧 slider on fixed frictionless bar

𝐹3

𝑦

𝑅𝑧 𝑧

𝑅𝑦

Figure 3.19. Some common supports and reactions in three-dimensional particle equilibrium problems. Forces 𝐹1 , 𝐹2 , and 𝐹3 are hypothetical forces applied to a particle by cables and/or bars that might be attached to the particle, and forces 𝑅𝑥 , 𝑅𝑦 , and 𝑅𝑧 are reactions.

tools, as they will help you throughout statics and subjects that follow and in your professional practice. While computing tools are indispensable in engineering, it is your responsibility to be sure the answers produced by a calculator or computer are correct. Errors can arise from a variety of sources, including poor modeling, errors in input data and/or program options that are selected, occasional software errors (these are rare, but do occur), outright blunders, and so on. Any solution you obtain, whether by computer or hand calculation, must be thoroughly examined for correctness. This is accomplished by substituting your solutions into all of the governing equations for the problem (i.e., the equilibrium equations plus other equations pertinent to the problem) and verifying that all of these equations are satisfied. This check does not verify that FBDs have been accurately drawn and equations have been accurately written. Thus, it is also necessary to scrutinize the solution to verify that it is physically reasonable. These comments hold for both two-dimensional and three-dimensional problems. However, it is generally more challenging to verify the validity of a solution to a three-dimensional problem.

Summing forces in directions other than 𝒙, 𝒚, or 𝒛 Most often, we choose coordinate directions 𝑥, 𝑦, and 𝑧 for a particular problem because they provide a convenient description of geometry. As a result, writing vector expressions for forces and summing forces in these directions is usually straightforward.

167

168

ISTUDY

Chapter 3

Equilibrium of Particles

Occasionally, it will be useful to sum forces in a direction other than 𝑥, 𝑦, or 𝑧. While we could use a new coordinate system for this purpose where one of the new 𝑥, 𝑦, or 𝑧 directions is coincident with the direction we are interested in, this is often not convenient because the necessary geometric information for writing vector expressions for forces may not be available or may be difficult to determine. An easier way is to use the dot product, as follows. Newton’s law in vector form is ∑ ⃗ (3.21) 𝐹⃗ = 0⃗ ∶ 𝐹⃗1 + 𝐹⃗2 + ⋯ + 𝐹⃗𝑛 = 0.

Helpful Information Vector equation versus scalar equation. You should contrast Eqs. (3.21) and (3.22) to see that the first of these is a vector equation, whereas the second is a scalar equation.

Of course, because the force vectors in Eq. (3.21) are written in terms of 𝑥, 𝑦, and 𝑧 components, Eq. (3.21) amounts to summation of forces in the 𝑥, 𝑦, and 𝑧 directions. Let 𝑟⃗ be the direction in which we would like to sum forces. Often 𝑟⃗ will be a position vector. However, its physical significance is irrelevant, other than that it describes a direction that is useful for summing forces. To sum forces in the direction of 𝑟⃗, we take the dot product of both sides of Eq. (3.21) with the unit vector∗ 𝑟⃗∕|⃗𝑟| to obtain the scalar equilibrium equation Summation of forces in the 𝑟⃗ direction: ∑

𝐹𝑟 = 0 ∶

𝑟⃗ 𝑟⃗ 𝑟⃗ 𝐹⃗1 ⋅ + 𝐹⃗2 ⋅ + ⋯ + 𝐹⃗𝑛 ⋅ = 0. |⃗𝑟| |⃗𝑟| |⃗𝑟|

(3.22)

In the above expression, 𝐹⃗1 ⋅ 𝑟⃗∕|⃗𝑟| is the portion (or component) of force 𝐹⃗1 that acts in the direction 𝑟⃗, 𝐹⃗2 ⋅ 𝑟⃗∕|⃗𝑟| is the portion (or component) of 𝐹⃗2 that acts in the direction 𝑟⃗, and so on. On the right-hand side is the scalar 0, which of course is the portion (or component) of the vector 0⃗ that acts in the direction 𝑟⃗. Use of Eq. (3.22) is especially convenient in problems where we might know that some of the forces are perpendicular to 𝑟⃗, in which case the dot product of these forces with 𝑟⃗ is zero. This technique is illustrated in Example 3.8.

∗ Because

the right-hand side of Eq. (3.22) is zero, a unit vector in the direction of 𝑟⃗ is not needed, and we could just as well evaluate 𝐹⃗1 ⋅ 𝑟⃗ + 𝐹⃗2 ⋅ 𝑟⃗ + ⋯ + 𝐹⃗𝑛 ⋅ 𝑟⃗ = 0.

ISTUDY

Section 3.3

Equilibrium of Particles in Three Dimensions

End of Section Summary In this section, the equations governing static equilibrium of a particle in three dimensions were discussed. The main differences compared to equilibrium in two dimensions are that reactions are more complex, FBDs depict forces in three dimensions, and the equation systems that must be solved are larger. Some of the key points are as follows: • Problems in three dimensions with simple geometry can often be effectively solved using a scalar approach. But very often a vector approach will be more tractable. • The selection of an 𝑥𝑦𝑧 coordinate system is dictated largely by the geometry that is provided for a particular problem. Often it will be effective to sum forces in directions other than the 𝑥, 𝑦, or 𝑧 direction, and this is easily done using the dot product, as given by Eq. (3.22). • Software for solving systems of simultaneous algebraic equations can be very helpful. Regardless of how you obtain your solution for a particular problem, whether it be analytically or by using software, it is your responsibility to verify that the solution is correct, both mathematically and physically.

169

170

Chapter 3

Equilibrium of Particles

E X A M P L E 3.7

Cables, Bars, and Failure Criteria

𝑧 18

𝐶

9 5

6

12

𝐷

𝑂

𝐵

𝐶

𝐴

𝐵 4

2

𝑦

𝑊

8

SOLUTION Dimensions in feet

𝑥

Road Map

In both parts of this problem, the solution approach is the same: make appropriate modeling decisions to draw an FBD, write vector expressions for forces, write equilibrium equations, and then solve these equations. In Part (a), we then apply failure criteria to each member to determine the largest load 𝑊 that can be supported. In Part (b), we are given 𝑊 = 1000 lb, and then we determine the force supported by each member. By inspection, the force in cable 𝐴𝐷 is equal to the weight 𝑊 .∗

Figure 1 𝑧 20 9

𝐴

𝐵 𝑂 12

Part (a)

𝐶 𝐹𝐴𝐵

𝑥

𝐹𝐴𝑂

𝐹𝐴𝐶

𝑊

The FBD for point 𝐴 is shown in Fig. 2, where the cable forces and boom force are shown such that positive values correspond to tension. Vector expressions for the cable forces and weight may be written by inspection, while the force supported by the boom is written using position vector 𝑟⃗𝐴𝑂 , with the following results: Modeling

12 𝑦

⃗ = 𝑊 (−𝑘), ̂ 𝑊

𝐹⃗𝐴𝐵 = 𝐹𝐴𝐵 (−̂𝚥),

𝐹⃗𝐴𝑂 = 𝐹𝐴𝑂

Dimensions in feet

Figure 2 Free body diagram of point 𝐴 where, by inspection, the force in cable 𝐴𝐷 is equal to the weight 𝑊.

ISTUDY

The weight 𝑊 is supported by boom 𝐴𝑂 and cables 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷, which are parallel to the 𝑦, 𝑥, and 𝑧 axes, respectively. (a) If cables 𝐴𝐵 and 𝐴𝐶 can support maximum forces of 5000 lb each, and boom 𝐴𝑂 can support a maximum compressive force of 8000 lb before buckling, determine the largest weight 𝑊 that can be supported. Assume that cable 𝐴𝐷 is sufficiently strong to support 𝑊 . (b) If the supports at points 𝐵 and 𝐶 are relocated to points 𝐵 ′ and 𝐶 ′ , respectively, and 𝑊 = 1000 lb, determine the forces supported by boom 𝐴𝑂 and cables 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷.

𝑟⃗𝐴𝑂 |⃗𝑟𝐴𝑂 |

= 𝐹𝐴𝑂

𝐹⃗𝐴𝐶 = 𝐹𝐴𝐶 (−̂𝚤),

−9 𝚤̂ − 20 𝚥̂ − 12 𝑘̂ . 25

(1)

Governing Equations & Computation Equilibrium Equations

Applying Newton’s law with terms grouped by direction pro-

vides

) ( ) −20 −9 − 𝐹𝐴𝐶 𝚤̂ + 𝐹𝐴𝑂 − 𝐹𝐴𝐵 𝚥̂ 25 25 ) ( −12 ⃗ ̂ + 𝐹𝐴𝑂 (2) − 𝑊 𝑘 = 0. 25 For Eq. (2) to be satisfied, each term must be zero independently. Thus, the following system of equations results: ∑

𝐹⃗ = 0⃗ ∶

(

𝐹𝐴𝑂

−9 − 𝐹𝐴𝐶 = 0, 25 −20 − 𝐹𝐴𝐵 = 0, 𝐹𝐴𝑂 25 −12 𝐹𝐴𝑂 − 𝑊 = 0. 25 The solution to Eqs. (3)–(5) is easily obtained and is 𝐹𝐴𝑂

(3) (4) (5)

5 3 25 𝑊 , 𝐹𝐴𝐵 = 𝑊 , and 𝐹𝐴𝐶 = 𝑊 . (6) 12 3 4 With weight 𝑊 being a positive number that still needs to be determined, Eq. (6) shows the forces supported by both cables are positive and hence, they are in tension, and the force supported by the boom is negative and hence, it is in compression. 𝐹𝐴𝑂 = −

∗ If

this statement is not clear, you may formally arrive at this conclusion by drawing an FBD of point 𝐷 ∑ [as in Fig. 3.4(b) on p. 131] and then use 𝐹𝑧 = 0.

ISTUDY

Section 3.3

Force Laws

Equilibrium of Particles in Three Dimensions

Using Eq. (6), the various failure criteria can be applied as follows: If 𝐹𝐴𝑂 = −8000 lb,

then 𝑊 = 3840 lb.

(7)

If 𝐹𝐴𝐵 = 5000 lb,

then 𝑊 = 3000 lb.

(8)

If 𝐹𝐴𝐶 = 5000 lb,

then 𝑊 = 6670 lb.

(9)

Only the smallest value of 𝑊 in Eqs. (7)–(9) will simultaneously satisfy all three failure criteria, and thus the largest value 𝑊 may have is 3000 lb. Part (b) Modeling The FBD of point 𝐴 is the same as that shown in Fig. 2, except the orientations of 𝐹𝐴𝐵 and 𝐹𝐴𝐶 are slightly different. Vector expressions for the weight, boom force, and cable force are

⃗ = 1000 lb(−𝑘), ̂ 𝑊 𝐹⃗𝐴𝐵 𝐹⃗𝐴𝐶

𝐹⃗𝐴𝑂 = 𝐹𝐴𝑂 Governing Equations

(10)

𝑟⃗ 5 𝚤̂ − 20 𝚥̂ − 4 𝑘̂ = 𝐹𝐴𝐵 𝐴𝐵 = 𝐹𝐴𝐵 , 21 |⃗𝑟𝐴𝐵 | 𝑟⃗ −9 𝚤̂ − 2 𝚥̂ + 6 𝑘̂ , = 𝐹𝐴𝐶 𝐴𝐶 = 𝐹𝐴𝐶 11 |⃗𝑟𝐴𝐶 |

Writing

𝑟⃗𝐴𝑂

|⃗𝑟𝐴𝑂 |

= 𝐹𝐴𝑂

−9 𝚤̂ − 20 𝚥̂ − 12 𝑘̂ . 25

(11) (12) (13)

∑ ⃗ ⃗ 𝐹 = 0 provides the system of equations

5 9 9 − 𝐹𝐴𝐶 − 𝐹𝐴𝑂 = 0, 21 11 25 2 20 20 − 𝐹𝐴𝐶 − 𝐹𝐴𝑂 = 0, −𝐹𝐴𝐵 21 11 25 6 12 4 + 𝐹𝐴𝐶 − 𝐹𝐴𝑂 = 1000 lb. −𝐹𝐴𝐵 21 11 25 𝐹𝐴𝐵

(14) (15) (16)

Computation

This system of equations is not as easy to solve as that in Part (a). Systems of equations that are difficult to solve are a fact of life in engineering, and you must be proficient in solving them. The basic strategy for hand computation (where one of the equations is solved for one of the unknowns in terms of the others, and then this result is substituted into the remaining equations, and so on) is workable for systems of three ⇒ equations, but it rapidly becomes very tedious for larger systems of equations. You should take this opportunity to use a programmable calculator or one of the software packages mentioned earlier to find the solutions to Eqs. (14)–(16), which are 𝐹𝐴𝐵 = 1027 lb,

𝐹𝐴𝐶 = 930 lb,

and 𝐹𝐴𝑂 = −1434 lb.

(17)

⇐ In addition to the results in Eq. (17), the force supported by cable 𝐴𝐷 is equal to 𝑊 , therefore 𝐹𝐴𝐷 = 1000 lb. Discussion & Verification For both parts of this example, we expect the cables to be in tension, and the boom to be in compression, and indeed our solutions show this. A common error made in problems such as Part (a), where each member has its own failure criterion, is to assume that each member is at its failure load at the same time.

171

172

Chapter 3

Equilibrium of Particles

Summing Forces in a Direction Other Than 𝑥, 𝑦, or 𝑧

E X A M P L E 3.8

Bar 𝐴𝐵 is straight and is fixed in space. Spring CD has 3 N∕mm stiffness and 200 mm unstretched length. If there is no friction between collar 𝐶 and bar 𝐴𝐵, determine (a) The weight 𝑊 of the collar that produces the equilibrium configuration shown. (b) The reaction between the collar and bar 𝐴𝐵. 𝑧 120 mm

SOLUTION 60 mm

Road Map

The initial length of the spring is given, and sufficient information is provided to determine the final length of the spring. Thus, the spring force will be determined first. Then we will use the dot product to sum forces applied to the collar in the direction of bar 𝐴𝐵. For Part (b), we will use the equilibrium equation to determine the reaction between the collar and bar.

𝐴

𝐷 240 mm

Part (a) 240 mm

280 mm

𝐶

𝑂 𝑊

𝐵

𝑦 120 mm

300 mm

𝑥 Figure 1

𝑧

Governing Equations 60 mm

Force Laws The spring’s final length is the magnitude of the position vector from 𝐶 to 𝐷 (or vice versa). We could determine the coordinates of point 𝐶, and this along with the coordinates of point 𝐷 would allow us write 𝑟⃗𝐶𝐷 . However, we will use the following approach:

120 mm 𝐴 𝐷 𝑅⃗ 1

𝐹⃗𝐶𝐷

𝑦 120 mm

⃗ 𝑊 300 mm

𝐵

Figure 2 Free body diagram for the collar 𝐶.

ISTUDY

𝑟⃗𝐵𝐴 |⃗𝑟𝐵𝐴 |

+ 𝑟⃗𝐴𝐷

−120 𝚤̂ − 240 𝚥̂ + 240 𝑘̂ ̂ mm + (120 𝚤̂ − 60 𝚥̂ + 40 𝑘) 360 ̂ mm. = (40 𝚤̂ − 220 𝚥̂ + 200 𝑘) = (240 mm)

𝐶

𝑂

𝑥

𝑟⃗𝐶𝐷 = 𝑟⃗𝐶𝐴 + 𝑟⃗𝐴𝐷 = (240 mm)

𝑅⃗ 2

240 mm

280 mm

The FBD for collar 𝐶 is shown in Fig. 2. In accord with Fig. 3.19, there are two reaction components between the collar and bar 𝐴𝐵. If we select the direction for 𝑅⃗ 1 first, so that it is perpendicular to bar 𝐴𝐵, whose direction is 𝑟⃗𝐴𝐵 , then the remaining reaction 𝑅⃗ 2 must be perpendicular to both 𝑅⃗ 1 and 𝑟⃗𝐴𝐵 . Observe that there are an infinite number of choices for the direction of 𝑅⃗ 1 , with the only requirement being that its direction must be perpendicular to 𝑟⃗𝐴𝐵 . Once 𝑅⃗ 1 is selected, the requirement that 𝑅⃗ 2 be perpendicular to both 𝑅⃗ 1 and 𝑟⃗𝐴𝐵 dictates the only direction it may have. With this approach, 𝑅⃗ 1 and 𝑅⃗ 2 have known directions but unknown magnitudes. Alternatively, we may call the reaction ⃗ where 𝑅⃗ is perpendicular to 𝑟⃗ , but otherwise has unknown direction and simply 𝑅, 𝐴𝐵 magnitude; this approach is employed later in this problem. Modeling

(1)

Hence, the final length of the spring is 𝐿 = |⃗𝑟𝐶𝐷 | = 300 mm, and using Eq. (3.18) on p. 153, the force supported by the spring is ) ( N (300 mm − 200 mm) = 300 N. (2) 𝐹𝐶𝐷 = 𝑘(𝐿 − 𝐿0 ) = 3 mm Equilibrium Equations

Using the FBD in Fig. 2, Newton’s law may be written for the

collar as ∑

𝐹⃗ = 0⃗ ∶

⃗ ⃗ = 0. 𝐹⃗𝐶𝐷 + 𝑅⃗ 1 + 𝑅⃗ 2 + 𝑊

(3)

Vector expressions for the weight and spring force are ⃗ = 𝑊 (−𝑘), ̂ 𝑊

𝐹⃗𝐶𝐷 = 𝐹𝐶𝐷

𝑟⃗𝐶𝐷 |⃗𝑟𝐶𝐷 |

= (300 N)

40 𝚤̂ − 220 𝚥̂ + 200 𝑘̂ . 300

(4)

ISTUDY

Section 3.3

Equilibrium of Particles in Three Dimensions

173

Vector expressions for 𝑅⃗ 1 and 𝑅⃗ 2 can be developed using a variety of strategies.∗ However, some careful thought will allow us to avoid the use of these vectors altogether for Part (a) of this example! Referring to the FBD shown in Fig. 2, observe that 𝑅⃗ 1 and 𝑅⃗ 2 are both perpendicular to direction 𝐵𝐴. Thus, if we sum forces in this direction, we know 𝑅⃗ 1 and 𝑅⃗ 2 will not appear in this equation since they have no component in this direction. This can be accomplished by using the dot product on each force to determine how much of them, or what component, acts in the 𝐵𝐴 direction. Thus, as discussed in connection with Eq. (3.22) on p. 168, we take the dot product of both sides of Eq. (3) with the unit vector 𝑟⃗𝐵𝐴 ∕|⃗𝑟𝐵𝐴 | to write 𝐹⃗𝐶𝐷 ⋅

𝑟⃗𝐵𝐴

|⃗𝑟𝐵𝐴 |

+ 𝑅⃗ 1 ⋅

𝑟⃗𝐵𝐴

|⃗𝑟𝐵𝐴 | ⏟⏞⏞⏞⏟⏞⏞⏞⏟ =0

+ 𝑅⃗ 2 ⋅

𝑟⃗𝐵𝐴

|⃗𝑟𝐵𝐴 | ⏟⏞⏞⏞⏟⏞⏞⏞⏟

⃗ ⋅ +𝑊

=0

𝑟⃗𝐵𝐴

|⃗𝑟𝐵𝐴 |

= 0⃗ ⋅

𝑟⃗𝐵𝐴

|⃗𝑟𝐵𝐴 | ⏟⏞⏟⏞⏟

.

(5)

=0

In Eq. (5), the dot product with the zero vector on the right-hand side is of course zero. Because 𝑅⃗ 1 and 𝑅⃗ 2 are orthogonal to 𝑟⃗𝐵𝐴 , their dot products must also be zero, and because of this knowledge we may avoid evaluating those dot products in Eq. (5). Evaluating the remaining two dot products in Eq. (5) results in 40 𝚤̂ − 220 𝚥̂ + 200 𝑘̂ −120 𝚤̂ − 240 𝚥̂ + 240 𝑘̂ ⋅ 300 360 ̂ −120 𝚤 ̂ − 240 𝚥 ̂ + 240 𝑘 ̂ ⋅ + 𝑊 (−𝑘) 360 = (300 N)(0.8889) − 𝑊 (0.6667).

0 = (300 N)

Computation

(6)

(7)

Equation (7) is a scalar equation, and its solution is 𝑊 = 400 N.

(8)

Note that as promised, this clever solution approach avoided having to compute the reactions between the collar and bar. Part (b) Computation

To determine the reactions between the collar and bar, we return to Eq. (3) and solve for the reaction 𝑅⃗ = 𝑅⃗ 1 + 𝑅⃗ 2 as ⃗ 𝑅⃗ = −𝐹⃗𝐶𝐷 − 𝑊 40 𝚤̂ − 220 𝚥̂ + 200 𝑘̂ ̂ − (400 N)(−𝑘) 300 ̂ N, = (−40 𝚤̂ + 220 𝚥̂ + 200 𝑘)

= (−300 N)

(9)

and the magnitude of this reaction is ⃗ = 300 N. |𝑅|

(10)

⃗ given Discussion & Verification As a partial check of our solution, we can verify that 𝑅 by Eq. (9) is perpendicular to bar 𝐵𝐴 by showing 𝑅⃗ ⋅ 𝑟⃗𝐵𝐴 = 0. While 𝑅⃗ must be given by the expression determined in Eq. (9) for equilibrium to prevail, there are an infinite number of vectors 𝑅⃗ 1 and 𝑅⃗ 2 that are perpendicular to bar 𝐴𝐵 and perpendicular to one ⃗ If you construct 𝑅⃗ and 𝑅⃗ using the suggestions provided another and will still sum to 𝑅. 1 2 in the footnote on this page, then you will obtain one of these possible combinations. Problem 3.106 asks you to do this. ∗ One

strategy is to evaluate the cross product 𝑟⃗𝐵𝐴 × 𝑟⃗𝐶𝐷 , then make this result a unit vector called 𝑢̂ 1 , and then 𝑅⃗ 1 = 𝑅1 𝑢̂ 1 . Next evaluate 𝑟⃗𝐵𝐴 × 𝑅⃗ 1 , make this result a unit vector called 𝑢̂ 2 , and then 𝑅⃗ 2 = 𝑅2 𝑢̂ 2 .

Helpful Information Use of dot product to sum forces. Use of the dot product to sum forces in a direction other than a coordinate direction can be very helpful, as illustrated in this example. This example can also be solved by directly ∑ ⃗ ⃗ but this solving Eq. (3) (i.e., 𝐹 = 0), requires writing expressions for 𝑅⃗ 1 and 𝑅⃗ 2 and some possibly tedious algebra.

174

Chapter 3

Equilibrium of Particles

Problems Problem 3.83

𝑦 𝐹

The structure shown consists of three bars pinned together at point 𝐴, and with socket supports at points 𝐵, 𝐶, and 𝐷. Force 𝐹 acts in the −𝑦 direction. If 𝐹 = 3 kN, determine the force supported by each member.

9m 𝐵

𝐴

Problem 3.84

𝐷 2m

𝑧

2m

The structure shown consists of three bars pinned together at point 𝐴, and with socket supports at points 𝐵, 𝐶, and 𝐷. Force 𝐹 acts in the −𝑦 direction. If each of the bars has 5 kN strength in tension and 4 kN strength in compression, determine the largest force 𝐹 that may be supported.

𝐶 6m

Problem 3.85

𝑥

The object at 𝐷 is supported by boom 𝐴𝑂 and by cables 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷, which are parallel to the 𝑥, 𝑦, and 𝑧 axes, respectively. If the mass of object 𝐷 is 𝑚𝐷 = 100 kg, determine the forces supported by the boom and three cables.

Figure P3.83 and P3.84

Problem 3.86

𝑧

In Prob. 3.85, if the cables and boom have the failure strengths given below, determine the largest mass 𝑚𝐷 that can be supported.

2m 𝐶

0.9 m

Member 1.2 m 𝑥

𝐴

𝐵

𝐷 𝑚𝐷

𝐴𝑂 𝐴𝐵 𝐴𝐶 𝐴𝐷

𝑂 1.2 m

Strength 4000 N tension & 2000 N compression 1500 N 1000 N 2000 N

𝑦 Figure P3.85 and P3.86

ISTUDY

Problem 3.87 A weight 𝑊 is supported at 𝐴 by bars 𝐴𝐵 and 𝐴𝐷 and by cable 𝐴𝐶. Bars 𝐴𝐵 and 𝐴𝐷 are parallel to the 𝑦 and 𝑥 axes, respectively. If 𝑊 = 1000 N, determine the forces in each bar and the cable. 𝑧 𝐶 3m 𝐷 2m 𝐵 𝑥

𝑦

𝐴 6m

𝑊

Figure P3.87 and P3.88

Problem 3.88 In Prob. 3.87, if the cable can support a maximum force of 2 kN, and bars 𝐴𝐵 and 𝐴𝐷 can support maximum compressive forces of 1 kN and 3 kN, respectively, determine the largest weight 𝑊 that may be supported.

ISTUDY

Section 3.3

Equilibrium of Particles in Three Dimensions

Problem 3.89

𝑃

When in equilibrium, plate 𝐵𝐶𝐷 is horizontal. If the plate weighs 1.6 kN, determine the forces in cables 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷.

𝐴

175

Problem 3.90 A weight 𝑊 is supported by bar 𝐴𝑂 and cables 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷. Point 𝐴 lies in the 𝑥𝑦 plane. If 𝑊 = 1000 lb, determine the forces in the bar and the three cables.

1.2 m

𝑧 𝐷 0.8 m

𝐵

𝑧

0.8 m

48 in. 15 in.

0.9 m

𝐶

0.9 m

𝐶

𝑥 𝐵

Figure P3.89 36 in.

36 in.

𝑂

𝑦 𝐴 𝐷

𝑥

𝑊 Figure P3.90 and P3.91

Problem 3.91 In Prob. 3.90, if the bar and cables have the strengths listed below, determine the largest value 𝑊 may have. Member 𝐴𝑂 𝐴𝐵 𝐴𝐶 𝐴𝐷

Strength 3000 lb tension & 2000 lb compression 2000 lb 1500 lb 2500 lb

𝐹 𝑟

Problem 3.92 A circular ring with weight 𝑊 and inside radius 𝑟 is supported by three identical springs having stiffness 𝑘 and that are unstretched when 𝑑 = 0. When in equilibrium, the ring is horizontal.

𝑑 120◦ 120◦

(a) Derive an expression that relates the weight 𝑊 to 𝑑 and 𝑟. (b) If 𝑊 = 300 lb, 𝑘 = 25 lb∕in., and 𝑟 = 20 in., determine 𝑑 (an accurate approximate solution is acceptable).

Figure P3.92

120◦

𝑦

176

Chapter 3

Equilibrium of Particles

Problem 3.93 In the cable system shown, point 𝐷 lies in the 𝑦𝑧 plane and force 𝑃 is parallel to the 𝑧 axis. If 𝑃 = 500 lb, determine the force in cables 𝐴𝐵, 𝐴𝐶, 𝐴𝐷, and 𝐴𝐸. 𝐷 𝑧 12 f t 8 ft 𝐵

𝑦 𝐴 9 ft

6 ft 6 ft

𝐶

𝐸 𝑥

𝑃

Figure P3.93 and P3.94

Problem 3.94 In the cable system shown, point 𝐷 lies in the 𝑦𝑧 plane and force 𝑃 is parallel to the 𝑧 axis. If cables 𝐴𝐵 and 𝐴𝐶 have 500 lb breaking strength, and cables 𝐴𝐷 and 𝐴𝐸 have 1000 lb breaking strength, and if all cables are to have a factor of safety∗ of 1.5 against failure, determine the largest force 𝑃 that may be supported.

Problem 3.95 Repeat Prob. 3.93 with point 𝐴 having coordinates 𝐴(1, 8, 0) ft.

Problem 3.96 Repeat Prob. 3.94 with point 𝐴 having coordinates 𝐴(1, 8, 0) ft.

Problem 3.97

𝑧

Force 𝑃 is supported by two cables and a bar. Point 𝐴 lies in the 𝑦𝑧 plane, and points 𝐵 and 𝐶 lie in the 𝑥𝑧 plane. If 𝑃 = 3 kip, determine the forces supported by the cables and bar.

4 ft 𝐶

𝐴

4 ft

Problem 3.98 𝑃 𝐵 6 ft 𝑥

𝑂

𝐷 5 ft

12 f t

12 f t 𝑦

Force 𝑃 is supported by two cables and a bar. Point 𝐴 lies in the 𝑦𝑧 plane, and points 𝐵 and 𝐶 lie in the 𝑥𝑧 plane. The compressive load that causes the bar to buckle and the breaking strength of each cable are specified below. If factors of safety against failure (see the footnote of Prob. 3.94) of 1.7 and 2.0 are to be used for cables and bars, respectively, determine the allowable force 𝑃 that can be supported. Member

Figure P3.97 and P3.98

ISTUDY

𝐴𝑂 𝐴𝐵 𝐴𝐶 ∗ The

Strength 3000 lb compression 6000 lb 5000 lb

factor of safety against failure is defined to be the failure load for a member divided by the allowable load for the member. Thus, the largest load the member may be subjected to is its failure load divided by the factor of safety.

ISTUDY

Section 3.3

177

Equilibrium of Particles in Three Dimensions

Problem 3.99 A photograph of the NASA Apollo 16 Lunar Module (abbreviated by NASA as the LM) is shown on the surface of the Moon. Such spacecraft made six Moon landings during 1969– 72.∗ A simplified model for one of the four landing gear assemblies of the LM is shown. If the LM has 15,000 kg mass, and rests on the surface of the Moon where acceleration due to gravity is 1.62 m∕s2 , determine the force supported by members 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷. Assume the weight of the LM is uniformly supported by all four landing gear assemblies, and neglect friction between the landing gear and the surface of the Moon. 𝑧

𝑦

𝐵 𝑥

𝐻

𝐵

𝐴 (2.5, 2.5, −2.2) m 𝐵 (1.5, 1.5, 0) m 𝐶 (2, 1, −1.2) m 𝐷 (1, 2, −1.2) m

NASA

𝐻

𝐷

𝐶

𝐵 𝐴

𝐷

𝐴 𝐶

Figure P3.99

𝑧

𝐸 𝑥 𝐶

Problem 3.100

𝐴

When in equilibrium, plate 𝐴𝐶𝐷𝐸 is horizontal. The plate weighs 1200 lb and is supported by cables 𝐴𝐵𝐶, 𝐵𝐷, 𝐵𝐸, and 𝐵𝐻. As shown, cable 𝐴𝐵𝐶 wraps around ring 𝐵, which has negligible friction. If the coordinates of point 𝐵 are 𝐵 (40, 20, 80) in., determine the forces in the four cables. Remark: The system of equations can be solved manually, but solution by calculator or computer is recommended.

30 in.

Problem 3.101 Member 𝑂𝐴 buckles when the compressive force it supports reaches 400 N. Cables 𝐴𝐶 and 𝐴𝐷 each have 300 N breaking strength. Assuming the cabling between 𝐴 and 𝐵 is sufficiently strong, determine the force 𝑇 that will cause the structure to fail. Assume the pulleys are frictionless with diameters that are small enough so that all cables between 𝐴 and 𝐵 are parallel to line 𝐴𝐵. 𝑧 𝐴 6m

𝑇

2m

𝐵

𝐶

𝑂 8m

3m

𝑦 𝐷

2m

𝑥 Figure P3.101 ∗ The Apollo 13 mission and the oxygen tank failure that occurred is a spectacular story of human courage

and resourcefulness and shows the LM far exceeding its design requirements.

𝐷 40 in.

60 in. 𝑦 Figure P3.100

𝐸

178

Chapter 3

Equilibrium of Particles

Problem 3.102 A portable tripod hoist for moving objects into and out of a manhole is shown. The hoist consists of identical-length bars 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷 that are connected by a socket at 𝐴 and are supported by equal 8 f t length cables 𝐵𝐶, 𝐵𝐷, and 𝐶𝐷 to prevent ends 𝐵, 𝐶, and 𝐷 of the bars from slipping. Cable FAE passes around a frictionless pulley at 𝐴 and terminates at winch 𝐸, which is fixed to bar 𝐴𝐵. Idealize points 𝐴 and 𝐸 to be particles where all bar and cable forces pass through these points. If the tripod is erected on level ground and is 7 f t high, and the object being lifted in the manhole weighs 300 lb,

𝐴 7 ft 𝐸

𝐷

8 ft

(a) Determine the forces in bars 𝐴𝐶 and 𝐴𝐷 and in portion 𝐴𝐸 of bar 𝐴𝐵. 𝐹

𝐵

8 ft

(b) Determine the force in portion 𝐸𝐵 of bar 𝐴𝐵.

𝐶

8 ft

Hint: Define an 𝑥𝑦𝑧 coordinate system where the 𝑥 and 𝑦 directions lie in the plane defined by points B, C, and 𝐷, and where the 𝑥 or 𝑦 direction coincides with one of cables 𝐵𝐶, 𝐵𝐷, or 𝐶𝐷. Then determine the coordinates of points 𝐵, 𝐶, and 𝐷. The 𝑥 and 𝑦 coordinates of point 𝐴 are then the averages of the coordinates of points 𝐵, 𝐶, and 𝐷.

Figure P3.102

Problem 3.103

𝑦

A worker standing on a truck uses rope 𝐴𝐵 to slowly lower object 𝐵 down a chute. The object weighs 100 lb and fits loosely against the walls of the chute and slides with no friction. In the position shown, the center of object 𝐵 is halfway down the chute. The person’s hand 𝐴 lies in the 𝑥𝑦 plane, and the chute lies in a plane parallel to the 𝑦𝑧 plane. 4 ft 𝐴

(a) Determine the tension in rope 𝐴𝐵 and the reactions between the object and the chute. (b) If the worker wishes to slowly pull the object up the chute, explain how your answers to Part (a) change.

3 ft 𝐶

4 ft 𝑧 Figure P3.103

ISTUDY

𝐷

3 ft 𝑥

𝐵

Problem 3.104 Due to a poorly designed foundation, the statue at point 𝐴 slowly slides down a grasscovered slope. To prevent further slip, a cable is attached from the statue to point 𝐵, and another cable is attached from the statue to point 𝐶. The statue weighs 1000 lb, and idealize the surface on which the statue rests to be frictionless. Determine the minimum tensile strength each cable must have, and the magnitude of the reaction between the statue and the slope. 𝑧

𝐶

grass slope

12 f t 𝐴

𝑦

𝐵 3 ft

12 f t

3 ft 4 ft

4 ft

𝑥 Figure P3.104

ISTUDY

Section 3.3

179

Equilibrium of Particles in Three Dimensions

𝑥

Problem 3.105 Channel 𝐴𝐵 is fixed in space, and its centerline lies in the 𝑥𝑦 plane. The plane containing edges 𝐴𝐶 and 𝐴𝐷 of the channel is parallel to the 𝑥𝑧 plane. If the surfaces of the channel are frictionless and the sphere 𝐸 has 2 kg mass, determine the force supported by cord 𝐸𝐹 , and the reactions 𝑅𝐶 and 𝑅𝐷 between the sphere and sides 𝐶 and 𝐷, respectively, of the channel.

𝐹 20◦ 𝐵

Problem 3.106 𝐸

Follow the suggestions made in the footnote of Example 3.8 on p. 173 to write vector ⃗ , by expressions for 𝑅⃗ 1 and 𝑅⃗ 2 . Then determine the magnitude of these, and weight 𝑊 ∑ ⃗ ⃗ ⃗ ⃗ ⃗ applying 𝐹 = 0. Show that the magnitude of 𝑊 and the vector sum 𝑅1 + 𝑅2 agree with the results reported in Example 3.8.

𝐷

𝐶

30◦

30◦

𝐴

𝑧

30◦

Problem 3.107 Rod 𝐴𝐵 is fixed in space. Spring 𝐶𝐷 has stiffness 1.5 N∕mm and an unstretched length of 400 mm. If there is no friction between the collar and rod, determine the weight of the collar 𝑊 that produces the equilibrium configuration shown, and the reaction between the collar and rod 𝐴𝐵.

𝑦 Figure P3.105

𝑧 𝐴

400 mm

600 mm 𝐷 𝑦 𝐵 𝑥 400 mm 𝐵

𝐶

700 mm

𝐶

𝑃

𝑊

150 mm 400 mm

4

𝑦

3 𝑂

Figure P3.107

Problem 3.108 The structure consists of a quarter-circular rod 𝐴𝐵 with 150 mm radius that is fixed in the 𝑥𝑦 plane. An elastic cord 𝐶𝐷 has stiffness 𝑘 = 2 N∕mm and 100 mm unstretched length. Bead 𝐶 has negligible weight, slides without friction on the rod, and is subjected to a force of magnitude 𝑃 that lies in the 𝑥𝑦 plane and is tangent to the curved rod 𝐴𝐵 in the position shown. Determine the value of 𝑃 needed for equilibrium in the position shown and the reaction between the bead and curved rod 𝐴𝐵.

𝑧

𝐷

150 mm

Figure P3.108

𝑧 60 mm

60 mm 𝐷

Problem 3.109 Bead 𝐵 has negligible weight and slides without friction on rigid fixed bar 𝐴𝐶. An elastic cord 𝐵𝐷 has spring constant 𝑘 = 3 N∕mm and 20 mm unstretched length, and bead 𝐵 has a force of magnitude 𝑃 in direction 𝐵𝐶. If bead 𝐵 is positioned halfway between points 𝐴 and 𝐶, determine the value of 𝑃 needed for equilibrium, and the reaction between bead 𝐵 and rod 𝐴𝐶.

𝑥

𝐴 120 mm

𝑂

60 mm 𝑃 𝐴 60 mm 𝑥 Figure P3.109

𝐵 120 mm

𝐶 40 mm 𝑦

180

Chapter 3

Equilibrium of Particles Problems 3.110 and 3.111

The collar at 𝐶 slides on a frictionless bar 𝐴𝐵, which is fixed in space. Cable 𝐶𝐷𝐸 passes around a frictionless ring at 𝐷. For the values of 𝑥𝐷 and 𝑦𝐷 given below, determine the weight collar 𝐶 must have if the system is in equilibrium. Problem 3.110

𝑥𝐷 = 4 in. and 𝑦𝐷 = 10 in.

Problem 3.111

𝑥𝐷 = 6 in. and 𝑦𝐷 = 15 in. 𝑦 𝑥𝐷 𝐷 𝐵 𝐸 𝑦 𝐷 6 in. 50 lb 𝑥

𝐶 5 in. 𝑦 𝑧

1.2 m

𝐴 Figure P3.110 and P3.111

𝐷

𝐸

8 in.

Problems 3.112 and 3.113 𝐵

The bead 𝐶 weighs 200 N and slides without friction on the quarter-circular bar 𝐴𝐵 with 1.4 m radius. The position of the bead is controlled by rope 𝐶𝐷𝐸, which passes through a frictionless ring at point 𝐷.

𝐶

Problem 3.112

If the bead is in equilibrium with 𝛼 = 60◦ , determine the force in the

rope. 2.5 m

𝑂

𝛼 1.4 m

𝑧 Figure P3.112 and P3.113

ISTUDY

𝐴

𝑥

Problem 3.113 Rope 𝐶𝐷𝐸 has 180 N breaking strength, and the collar 𝐶 starts at the position 𝛼 = 90◦ . If the collar is slowly lowered, determine the value of 𝛼 at which the rope will break.

Problem 3.114 Two identical traffic lights each weighing 120 lb are to be suspended by cables over an intersection. For points 𝐸 and 𝐹 to have the coordinates shown, it is necessary to add an additional weight to one of the lights. Determine the additional weight needed, the light to which it should be added, and forces in each cable. Remark: The system of equations can be solved manually, but solution by calculator or computer is recommended. 𝑧

𝐷 𝑦

𝐴

𝐸

𝐶

𝐹 𝐵

𝐴 (0, 0, 24) f t 𝐵 (30, 2, 24) f t 𝐶 (30, 14, 24) f t 𝐷 (0, 27, 24) f t 𝐸 (8, 8, 20) f t 𝐹 (18, 8, 20) f t 𝑥

Figure P3.114

ISTUDY

Section 3.4

3.4

Engineering Design

181

Engineering Design

While there are many different ways to define engineering design, the description that follows encompasses the key points. Engineering design is a process that culminates in the description and specifications of how a structure, machine, device, procedure, or other entity is to be constructed so that needs and requirements that were identified in the design process are achieved. The word engineering implies that laws of nature, mathematics, and computing are at the core of the design. Design is a creative and exciting activity, and broadly speaking, proficiency in doing this is what sets engineering apart from the sciences. Indeed, the ability to design is a distinct point of pride among all engineers! The word process in the above definition is important. Design should not be conducted solely by iterations of trial and error, but rather by iterations of identifying and prioritizing needs, making value decisions, and exploiting the laws of nature to develop a sound solution that optimizes all of the objectives while satisfying constraints that have been identified. As you continue your education, you will learn about structured procedures and algorithms for design, including exposure to pertinent performance standards, safety standards, and design codes. In the meantime, your knowledge of particle equilibrium allows you to begin some meaningful engineering design.

Objectives of design Our discussion of the objectives of design and the design process will be brief in this section. Chapter 5 contains additional discussion. At a minimum, structures, machines, mechanical devices, and so on must be designed to achieve the following criteria: 1. They will not fail during normal use. 2. Performance objectives are adequately met. 3. Hazards are adequately addressed. 4. All foreseeable uses must be anticipated and accounted for to the extent possible or reasonable. 5. Design work must be thoroughly documented and archived. In addition, a good design will also consider 6. Cost. 7. Ease of manufacture and assembly. 8. Energy efficiency in both manufacture and use. 9. Impact on the environment in manufacture, use, and retirement. The need for item 1 is obvious. Further, your design should not contain unnecessary hazards. Through good design, it may be possible to eliminate or reduce a hazard. Clearly some devices, due to their very nature, contain hazards that are unavoidable. For example, a machine to cut paper, whether it be a simple scissors, a tabletop cutter, or a device for cutting large rolls of paper, must contain sharp cutting

Gorodenkoff/Shutterstock

Gorodenkoff/Shutterstock

Figure 3.20 The upper photo shows an engineer working on the design of an electric automobile using computer aided design (CAD) tools. The lower photo shows engineers working on a prototype of the chassis of this vehicle including frame, suspension, wheels, motor, and batteries. While CAD tools help alleviate the burden of performing complicated calculations, the engineers who use these tools must be experts in modeling, interpretation of results, ability to perform hand calculations on simpler models to help verify the results of more sophisticated models, and in the methodology of design.

182

ISTUDY

Chapter 3

Equilibrium of Particles

cutting wheel (b)

(a)

Figure 3.21. Two tabletop paper cutters.

edges. Consider the two tabletop paper cutters shown in Fig. 3.21. The traditional cutter shown in Fig. 3.21(a) uses a long, heavy steel blade that is unavoidably exposed to a user’s fingers and hands, whereas the more contemporary cutter shown in Fig. 3.21(b) employs a small circular cutting wheel that is less exposed and easier to guard and, in fact, performs better. When hazards are unavoidably present, they should be guarded to the extent feasible, and warnings should be included to pictorially show users what the hazards are. Thus, the hierarchy in managing hazards is as follows: eliminate or reduce the hazard through design, guard users from the hazard, and finally warn users of the hazard. An extremely important aspect of design, and one that is unfortunately sometimes overlooked, is anticipation of and accounting for how a structure, machine, or device might be used, even if it was not intended for such use. While we may occasionally need to consider malicious misuse, here we are more concerned with a well-intentioned user who pushes the limits of a product in an effort to complete some task. For example, imagine you design a jack for use in lifting an automobile so a tire may be changed. If the jack is designed to have 2000 lb lifting capacity, most certainly someone will try to lift more. While your design does not need to be capable of lifting more than 2000 lb plus some reasonable margin for error, the main question is, What is the consequence of attempting to lift more? Will the jack collapse, allowing the vehicle to fall and potentially hurt someone, or can it be designed so the result is less catastrophic? Regardless of how fertile and creative a mind you have, it will be difficult for your ideas to have an impact if you cannot communicate effectively, both orally and in writing, with other engineers, scientists, professionals, and lay people. It is necessary that your design and analysis work be documented and archived. This information may be needed in the future by you or others to show how your design was developed, to aid in modifications, to help support patent rights, and so on. Expectations of engineers are high, and it is necessary that you perform your work precisely and conduct your behavior to the highest standards of ethics and professionalism.

Particle equilibrium design problems For the problems in this section, imagine you are employed as an intern working under the supervision of an engineer who asks you to conduct a design study, or to add details to a design that is started. You will be presented with a problem that is suitable to your level of knowledge along with some pertinent data. But, unlike most textbook homework problems, the information provided may not be complete, and you may not be instructed on everything that needs to be done. This is not an attempt to be artificially vague, but rather is a reflection of how design and modeling of real life problems are conducted. Typically, a design begins with an idea or the

ISTUDY

Section 3.4

183

Engineering Design

desire to accomplish something. What information is needed and the analysis tasks to be performed are often not fully known at the outset, but rather are discovered as the design and analysis process evolves. It is in this spirit that the problems of this section are presented, and you may need to make reasonable assumptions or seek out additional information on your own. Your work should culminate in a short written technical report that is appropriate for an engineer to read, where you present your design, state assumptions made, and so on. Appendix A of this book gives a brief discussion of technical writing that may be helpful.

Table 3.1 Allowable loads for steel cable. Includes suitable factor of safety for repeated loads. Steel Cablea U.S. Customary and SI units

Strength of steel cable, bar, and pipe

Nominal diameter

Many of the problems in this section involve steel cable, bar, and pipe. For these problems, use the allowable loads given in Tables 3.1–3.3, which are generally applicable for situations in which loads may be cyclically applied and removed many times. Allowable loads, or working loads, are the forces to which components can be safely subjected, and these are obtained by dividing the failure strength of a member

1∕4 in. 3∕8 in. 1∕2 in. 5∕8 in. 3∕4 in. 7∕8 in. 1 in. a 1∕4

Table 3.2. Allowable loads for round steel bar. Includes suitable factors of safety for repeated loads.

Steel Bara U.S. Customary units

Nominal diameter

Tensile load not to exceed

1∕8 in.

200 lb

200 lb

or

1700 lb⋅in.2 ∕𝐿2

1∕4 in.

850 lb

850 lb

or

28(10)3 lb⋅in.2 ∕𝐿2

1∕2 in.

3500 lb

3500 lb

or

450(10)3 lb⋅in.2 ∕𝐿2

3∕4 in.

7500 lb

7500 lb

or

2.2(10)6 lb⋅in.2 ∕𝐿2

1 in.

14,000 lb

14,000 lb

or

7.2(10)6 lb⋅in.2 ∕𝐿2

Compressive load: not to exceed the smaller of b

SI units

Nominal diameter

Tensile load not to exceed

3 mm

0.85 kN

0.85 kN

or

3.9(10)6 N⋅mm2 ∕𝐿2

6 mm

3.5 kN

3.5 kN

or

62(10)6 N⋅mm2 ∕𝐿2

Compressive load: not to exceed the smaller of c

12 mm

14 kN

14 kN

or

1(10)9 N⋅mm2 ∕𝐿2

18 mm

30 kN

30 kN

or

5(10)9 N⋅mm2 ∕𝐿2

24 mm

55 kN

55 kN

or

16(10)9 N⋅mm2 ∕𝐿2

a Allowable

tension load and first column of allowable compressive load are obtained using a factor of safety of 2 against yielding for 36 ksi (250 MPa) steel. Second column of allowable compressive load is obtained using a factor of safety of 2 against the theoretical buckling load for a centrically loaded pin-supported bar. b 𝐿 represents the length of the member in inches. c 𝐿 represents the length of the member in mm.

6 mm 10 mm 13 mm 16 mm 19 mm 22 mm 25 mm

Allowable load 1300 lb 3000 lb 5000 lb 8000 lb 11,000 lb 15,000 lb 20,000 lb

6.0 kN 13 kN 20 kN 35 kN 50 kN 70 kN 90 kN

and 3∕8 in. (6 and 10 mm) diameter cable is 6 × 19 construction; others are 6 × 26 construction. Allowable load is obtained using a factor of safety of 5 against breaking loads published by manufacturer: Performance Series 620 Rope, Wire Rope Industries, Ltd. If pulleys are used, pulley diameter is recommended to be at least 34 times the nominal cable diameter, and 51 times is suggested.

184

ISTUDY

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Table 3.3. Allowable loads for standard weight steel pipe. Includes suitable factors of safety for repeated loads.

Steel Pipea U.S. Customary units

Nominal inside diameter

Tensile load not to exceed

0.5 in.

4500 lb

4500 lb

or

2.5(10)6 lb⋅in.2 ∕𝐿2

1 in.

8500 lb

8500 lb

or

12(10)6 lb⋅in.2 ∕𝐿2

1.5 in.

14,000 lb

14,000 lb

or

45(10)6 lb⋅in.2 ∕𝐿2

2 in.

19,000 lb

19,000 lb

or

98(10)6 lb⋅in.2 ∕𝐿2

Compressive load: not to exceed the smaller of b

SI units

Nominal inside diameter

Tensile load not to exceed

13 mm

20 kN

20 kN

or

6.9(10)9 N⋅mm2 ∕𝐿2

25 mm

40 kN

40 kN

or

36(10)9 N⋅mm2 ∕𝐿2

38 mm

60 kN

60 kN

or

120(10)9 N⋅mm2 ∕𝐿2

51 mm

85 kN

85 kN

or

270(10)9 N⋅mm2 ∕𝐿2

Compressive load: not to exceed the smaller of c

a Allowable

tension load and first column of allowable compressive load are obtained using a factor of safety of 2 against yielding for 36 ksi (250 MPa) steel. Second column of allowable compressive load is obtained using a factor of safety of 2 against the theoretical buckling load for a centrically loaded pin-supported bar. Geometric data used to compute these forces is obtained from the Manual of Steel Construction—Load and Resistance Factor Design, vol. I, American Institute of Steel Construction, 1998, and the Metric Conversion Volume of the same title, 1999. b 𝐿 represents the length of the member in inches. c 𝐿 represents the length of the member in mm.

by a factor of safety. Thus, failure load or factor of safety failure load Factor of safety = . allowable load Allowable load =

(3.23)

The failure load can be defined in different ways. In some applications it may be the load at which a member breaks or ruptures, while in other applications it may be the load at which a member begins to yield (i.e., take on permanent deformation) or otherwise fails to perform adequately. The factor of safety helps account for uncertainties in loads and strength, consequences of failure, and many other concerns. Determining a suitable factor of safety requires many value decisions to be made. Sometimes an appropriate factor of safety is recommended by the manufacturer of a component and is reflected in the allowable load data the manufacturer provides, or it may be specified by industry standards or design codes. Often you will need to make such decisions. Determining an appropriate factor of safety may be problem-dependent,

ISTUDY

Section 3.4

and if industry or government standards are available, you may be required to use these. Even if use of these is voluntary, it is obviously prudent to adhere to these as minimum standards. While the data of Tables 3.1–3.3 is safe and reasonable for many real life applications, more stringent factors of safety may be required for specific applications. However, you should not indiscriminately use excessive factors of safety, as this generally implies more massive components, which increases cost and may decrease performance. Cables, bars, and pipes can support tensile forces, and the allowable tensile loads listed in Tables 3.1–3.3 should not be exceeded. Pipes and bars can also support compressive forces, as follows. If the member is short, then its allowable compressive load is similar in magnitude to, or even the same as, its allowable tensile load. However, if the member is long, then it may fail by buckling, and the susceptibility to buckling increases rapidly with increasing length. Thus, the allowable compressive load for bar and pipe is the smaller of the two values listed in Tables 3.2 and 3.3, where it is seen that the buckling load depends on the inverse of the member’s length squared. For most applications of bars and pipes loaded in compression, buckling considerations usually govern allowable loads. Buckling is discussed in greater detail in Chapter 6. A cable, sometimes called wire rope, is a complex composite structural member consisting of an arrangement of wires twisted into strands, which are then twisted into a cable, often having a core of different material that is intended to improve flexibility or performance. Cables are manufactured in an enormous variety of materials, sizes, and constructions for different purposes. Cables that are used in conjunction with pulleys are more severely loaded than cables used solely as straight, static structural tension members. When a cable is wrapped around a pulley, it is subjected to high bending loads, and if the pulley’s diameter is too small, the cable’s life will be substantially reduced. The allowable loads reported in Table 3.1 are for a very common construction of steel cable and use a factor of safety of 5 against breakage of a straight segment of cable. These loads can also be employed for cables used with pulleys, provided the pulleys have a diameter that is at least about 34 times the nominal diameter of the cable (51 is suggested). However, the factor of safety is then not as generous. For many applications, the factor of safety should be taken higher. For example, in passenger elevators a factor of safety of at least 10 is generally used.

Engineering Design

185

186

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E X A M P L E 3.9

Engineering Design 𝐷

𝐴 𝐵

𝐶 22 f t

ℎ1

103 lb 195 lb

ℎ2

Cabling is to be designed to support two traffic lights such that both lights are at the same 19 f t height above the road, with the horizontal spacing shown in Fig. 1. Each of the supporting poles has 22 f t height. Specify dimensions ℎ1 and ℎ2 where ends of the steel cable should be attached to the support poles. Also specify the appropriate diameter of a single steel cable, from the selection in Table 3.1, that will safely support the weight of the two traffic lights.

19 f t

SOLUTION 9 ft

11 f t

15 f t

Road Map

Figure 1

𝑇𝐵𝐴

𝑦

𝑇𝐶𝐷 𝐵

𝑥

𝛼 103 lb

𝑇𝐵𝐶

𝐶 𝛽 195 lb

Possible values for ℎ1 and ℎ2 are between 19 and 22 f t. Further, these values are not independent. That is, if specific values for both ℎ1 and ℎ2 are assumed, it is unlikely the equilibrium configuration will have both traffic lights at the same 19 f t height above the roadway. Rather, if a value for ℎ1 is assumed, equations of equilibrium will determine the value ℎ2 must have, and vice versa. In this problem, the cost of the cabling is small and thus is not a major concern. Rather, our design priorities are driven primarily by performance objectives. The merit of having ℎ1 and ℎ2 as large as possible is that the forces in the cables will be smaller. Some thought about the geometry shown in Fig. 1 suggests that the lights may experience vibratory motion in the vertical direction, most likely due to aerodynamically produced forces from wind, where as one light moves up, the other moves down, in oscillatory fashion. Selecting small values of ℎ1 and ℎ2 will provide for shorter cabling, which may help minimize the possible amplitude of such motion. Thus, for our initial design we will select ℎ2 = 20 f t. Modeling

Governing Equations & Computation

Figure 2 Free body diagrams of points 𝐵 and 𝐶.

ISTUDY

Neglecting the weight of the cables, FBDs are shown in Fig. 2. The equilibrium equations for point 𝐶 are

∑

𝐹𝑦 = 0 ∶

𝑇𝐶𝐷 sin 𝛽 − 195 lb = 0,

(1)

∑

𝐹𝑥 = 0 ∶

−𝑇𝐵𝐶 + 𝑇𝐶𝐷 cos 𝛽 = 0,

(2)

where 15 f t

cos 𝛽 = √

(15 f t)2

sin 𝛽 = √

(15 f t)2 + (ℎ2 − 19 f t)2

+ (ℎ2 − 19 f t)2 ℎ2 − 19 f t

= 0.9978,

(3)

= 0.0665.

(4)

Solving Eqs. (1) and (2) provides 𝑇𝐶𝐷 = 2931 lb,

(5)

𝑇𝐵𝐶 = 2925 lb.

(6)

The equilibrium equations for point 𝐵 are ∑ 𝐹𝑦 = 0 ∶ ∑ 𝐹𝑥 = 0 ∶

𝑇𝐵𝐴 sin 𝛼 − 103 lb = 0,

(7)

−𝑇𝐵𝐴 cos 𝛼 + 𝑇𝐵𝐶 = 0.

(8)

Solving Eqs. (7) and (8) provides tan 𝛼 =

103 lb 𝑇𝐵𝐶

𝑇𝐵𝐴 = 2927 lb.

⇒

𝛼 = 2.02◦ ,

(9) (10)

ISTUDY

Section 3.4

Engineering Design

From the geometry of Fig. 1, ℎ1 − 19 f t

, (11) 9 ft which provides ℎ1 = 19.32 f t, which is less than the maximum permitted value of 22 f t. To specify an appropriate cable, note that the largest tensile force to be supported is 2931 lb in portion 𝐶𝐷 of the cable. Thus, consulting Table 3.1, we see that any cable with 3∕8 in. nominal diameter or larger will be acceptable. However, because the 2931 lb cable force is very close to the 3000 lb allowable load for the 3/8 in. cable, we will select the 1/2 in. diameter size. To summarize, our initial design calls for tan 𝛼 =

𝑇𝐵𝐴 = 2927 lb, ℎ1 = 19.32 f t,

𝑇𝐵𝐶 = 2925 lb,

ℎ2 = 20.00 f t,

𝑇𝐶𝐷 = 2931 lb,

and cable diameter = 1∕2 in.

(12) (13)

Discussion & Verification

The cable forces in this initial design are very large. While they may be acceptable, we might question the ability of the support poles to be subjected to these. Note that even if the support poles themselves were deemed to be strong enough, the soil that supports the bases of the poles may not be. These considerations go beyond information that is available and our analysis ability at this time. Nonetheless, we reanalyze this design, starting with ℎ2 = 21 f t, with the results 𝑇𝐵𝐴 = 1466 lb, ℎ1 = 19.63 f t,

𝑇𝐵𝐶 = 1463 lb,

ℎ2 = 21.00 f t,

𝑇𝐶𝐷 = 1475 lb,

and cable diameter = 3∕8 in.

(14) (15)

Although a subjective decision, the design reported in Eq. (15) would seem preferable to that in Eq. (13), due to the substantially lower forces the support poles are subjected to. To complete the design, we should write a brief technical report, following the guidelines of Appendix A, where we summarize all pertinent information. It should be written using proper, simple English that is easy to read. A sketch along with critical dimensions should be included. We should discuss the objectives and constraints considered in the design and the process used to arrive at our final design. Detailed calculations such as Eqs. (1)–(11) should not be included in the main body of the report, but should be included in an appendix. In addition to the immediate use of the report in helping to erect the traffic lights, it will be archived for possible future reference.

187

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E X A M P L E 3.10

Engineering Design A large construction company plans to add plastic 60 L water coolers to the back of most of its trucks, as shown in Fig. 1. Your supervisor asks you to use the model shown in Fig. 2 to specify the dimension ℎ and the diameter of round steel bars to be used for members 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷. Although a more refined model is possible using concepts discussed in Chapter 5, the model shown in Fig. 2 is useful for this design problem. In this model, load 𝑃 is vertical and member 𝐴𝐵 is parallel to the 𝑦 axis. Your supervisor also tells you to allow for loads up to twice the weight of the cooler in a crude attempt to allow for dynamic forces, and to use the allowable loads given in Table 3.2.

60 L water cooler

600 mm

SOLUTION ℎ

Road Map

The most poorly defined part of this problem is the loading the water cooler support will be subjected to. Thus, assumptions on the sizes of loads will need to be made. Furthermore, different loading conditions may need to be considered, and our design should be safe for all of these. Finally, this design problem does not have a unique solution. That is, there is not a single value of ℎ that will work. Rather, we may need to assume reasonable values for some parameters and then calculate the values of others based on this.

Figure 1 𝑧

𝑃

Modeling

600 mm 𝐵

When full, the weight of water in the cooler is∗ kg (0.1 m)3 m 9.81 2 = 589 N. 𝑊 = 1000 3 60 L L m s ⏟⏞⏟⏞⏟ ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏟⏟

𝐴

𝐷

𝜌

300 mm 𝐶

𝑦

300 mm

ℎ

𝑥 Figure 2 Model for the water cooler support shown in Fig. 1. 𝑧

𝑃⃗ 600 mm 𝐵

𝐹⃗𝐴𝐵 𝐹⃗

𝐴

𝐴𝐷

𝐷 𝐶

𝐹⃗𝐴𝐶

300 mm

300 mm

ℎ

𝑦

ISTUDY

𝑔

We next use our judgment to add to the above value a nominal amount of 60 N to account for the weight of the container itself. Thus, the total weight to be supported is 589 N + 60 N = 649 N. Doubling the weights (or use of other multiplicative factors) is an imprecise but common approach to account for inertial forces produced when the truck hits bumps in the road. Thus, we obtain the approximate load 𝑃 = 1300 N. Before continuing with this load, we should consider other possible load scenarios. For example, if the cooler is removed, the support might make a convenient step for a worker to use while climbing on or off the truck. Consider a person weighing 890 N (200 lb). Since it is extremely unlikely a person would be standing on the support when the truck is moving and hitting bumps, it is not necessary to further increase this value as was done earlier. Since the weight of this person is lower than the force determined earlier, we will proceed with the design, using 𝑃 = 1300 N. Small values of dimension ℎ will lead to large forces in all members, while larger values of ℎ will decrease these forces. However, the allowable load data in Table 3.2 shows that allowable compressive load decreases rapidly as the length of members increases. Thus, we will select for an initial design the value ℎ = 250 mm, in which case the length of members 𝐴𝐶 and 𝐴𝐷 is 716 mm. The FBD for point 𝐴 is shown in Fig. 3. Governing Equations

Vector expressions for forces are ̂ 𝑃⃗ = 1300 N (−𝑘),

𝑥 Figure 3 Free body diagram for point 𝐴.

𝑉

(1)

𝐹⃗𝐴𝐵 = 𝐹𝐴𝐵 (−̂𝚥), ∗ One liter (1

(2) (3)

L) is defined to be the volume of 1 kg of pure water at 4◦ C and pressure of 76 cm of mercury. For practical purposes, however, the transformation 1 L = 1000 cm3 = 0.001 m3 may be used with less than 0.003% error.

ISTUDY

Section 3.4

Engineering Design 300 𝚤̂ − 600 𝚥̂ − 250 𝑘̂ 𝐹⃗𝐴𝐶 = 𝐹𝐴𝐶 , 716 −300 𝚤̂ − 600 𝚥̂ − 250 𝑘̂ . 𝐹⃗𝐴𝐷 = 𝐹𝐴𝐷 716

Writing

(4) (5)

∑ ⃗ ⃗ 𝐹 = 0 and grouping 𝑥, 𝑦, and 𝑧 direction terms provide the system of equations 300 −300 + 𝐹𝐴𝐷 = 0, 716 716 −600 −600 + 𝐹𝐴𝐷 = 0, 𝐹𝐴𝐵 (−1) + 𝐹𝐴𝐶 716 716 −250 −250 + 𝐹𝐴𝐷 = 1300 N. 𝐹𝐴𝐶 716 716 𝐹𝐴𝐶

Computation

(6) (7) (8)

The solution of Eqs. (6)–(8) is

𝐹𝐴𝐵 = 3120 N,

𝐹𝐴𝐶 = −1860 N,

and 𝐹𝐴𝐷 = −1860 N.

(9)

To determine appropriate members to support these loads, we consult Table 3.2 on p. 183. For the tensile load in member 𝐴𝐵, any bar with diameter 6 mm or larger is acceptable. For the compressive load in members 𝐴𝐶 and 𝐴𝐷, we first check the suitability of the 6 mm bar, which can support a maximum compressive force that is the smaller of 3500 N or 62(10)6 N ⋅ mm2 ∕𝐿2 = 62(10)6 N ⋅ mm2 ∕(716 mm)2 = 120.9 N; hence, this bar is not acceptable. The next larger-sized bar has 12 mm diameter, which allows a maximum compressive load that is the smaller of 14 kN and 1.95 kN; hence, this bar is acceptable for members 𝐴𝐶 and 𝐴𝐷. Discussion & Verification

Our design thus far requires ℎ = 250 mm,

member 𝐴𝐵:

use 6 mm diameter steel bar or larger,

members 𝐴𝐶, and 𝐴𝐷:

(10)

use 12 mm diameter steel bar or larger.

For convenience in fabrication, we will recommend that the same size bar be used for all three components, in which case 12 mm diameter bar or larger is to be used. Finally, we consider whether the use of a diameter greater than 12 mm is warranted. This is analogous to incorporating an even greater factor of safety in the design, which does not seem necessary, since the factors of safety incorporated in the allowable loads and forces applied to the structure are already generous. Further, we have applied the entire weight of the water cooler to point 𝐴, whereas in reality, a portion of this weight will be supported by point 𝐵. To summarize, our final design is ℎ = 250 mm, members 𝐴𝐵, 𝐴𝐶, and 𝐴𝐷:

use 12 mm diameter steel bar.

(11)

To complete our work, we should prepare a short report following the guidelines described in Appendix A.

189

190

Chapter 3

Equilibrium of Particles

Design Problems General Instructions. In problems requiring the specification of sizes for steel cable, bar, or pipe, selections should be made from Tables 3.1–3.3. In all problems, write a brief technical report following the guidelines of Appendix A, where you summarize all pertinent information in a well-organized fashion. It should be written using proper, simple English that is easy for another engineer to read. Where appropriate, sketches, along with critical dimensions, should be included. Discuss the objectives and constraints considered in your design, the process used to arrive at your final design, safety issues if appropriate, and so on. The main discussion should be typed, and figures, if needed, can be computer-drawn or neatly hand-drawn. Include a neat copy of all supporting calculations in an appendix that you can refer to in the main discussion of your report. A length of a few pages, plus appendix, should be sufficient. Design Problem 3.1 0.5 in. 𝐴 𝑑 𝐵 𝑘

ℎ

Figure DP3.1

A scale for rapidly weighing ingredients in a commercial bakery operation is shown. An empty bowl is first placed on the scale. Electrical contact is made at point 𝐴, which illuminates a light indicating the bowl’s contents are underweight. A bakery ingredient, such as flour, is slowly poured into the bowl. When a sufficient amount is added, the contact at 𝐴 is broken. If too much is added, contact is made at 𝐵, thus indicating an overweight condition. If the contents of the bowl are to weigh 18 lb ± 0.25 lb, specify dimensions ℎ and 𝑑, spring stiffness 𝑘, and the unstretched length of the spring 𝐿0 . The bowl and the platform on which it rests have a combined weight of 5 lb. Assume the scale has guides or other mechanisms so that the platform on which the bowl rests is always horizontal.

Design Problem 3.2 60 mm 𝐵

𝐴 15 mm

20 mm

𝐶

600 mm

𝑘 𝐷

Design Problem 3.3 15 mm 30 mm

Figure DP3.2 and DP3.3

ISTUDY

A plate storage system for a self-serve salad bar in a restaurant is shown. As plates are added to or withdrawn from the stack, the spring force and stiffness are such that the plates always protrude above the tabletop by about 60 mm. If each plate has 0.509 kg mass, and if the support 𝐴 also has 0.509 kg mass, determine the stiffness 𝑘 and unstretched length 𝐿0 of the spring. Assume the spring can be compressed by a maximum of 40% of its initial unstretched length before its coils begin to touch. Also specify the number of plates that can be stored. Assume the system has guides or other mechanisms so the support 𝐴 is always horizontal.

In Design Prob. 3.2, the spring occupies valuable space that could be used to store additional plates. Repeat Design Prob. 3.2, employing cable(s) and pulley(s) in conjunction with one or more springs to design a different system that will allow more plates to be stored. Pulleys, cables, and springs can be attached to surfaces 𝐴, 𝐵, 𝐶, and 𝐷. For springs in compression, assume they may not contract by more than 40% of their initial unstretched length before their coils begin to touch.

ISTUDY

Section 3.4

191

Engineering Design

Design Problem 3.4 A push button for a pen is shown in cross section. Pressing and releasing the push button advances the ink cartridge so that the pen may be used for writing. Pressing and releasing it again retracts the ink cartridge so that the pen may be stored. The mechanism that keeps the ink cartridge in the advanced or retracted position is not shown. Design the spring for the pen by specifying the spring’s stiffness 𝑘 (units: N/mm), the spring’s unstretched length 𝐿0 , and the dimensions 𝑑 and 𝑡 (units for 𝐿0 , 𝑑, and 𝑡: mm). To prevent the coils of the spring from making contact with one another, the spring should not be compressed by more than 50% of its unstretched length.

push button

𝑡 𝑑 housing

ink cartridge

Design Problem 3.5

Figure DP3.4

A pogo stick is a toy used by a person for jumping up and down, and a model for a pogo stick is shown. Design the spring for the pogo stick assuming static behavior.∗ The design of the spring includes specification of: • The spring stiffness 𝑘 (units: lb/in.). • Unstretched length of the spring 𝐿0 . • The dimension ℎ of the spring housing. ℎ

• The travel 𝑡 of the pogo stick before the spring is fully compressed. Also, specify the force 𝐹1 at which the pogo stick begins to travel and the force 𝐹2 at which the pogo stick undergoes full travel.

𝑡

Design Problem 3.6 You are working for the Peace Corps in an impoverished country that needs your help. Your job is to design a simple but accurate scale for weighing bulk materials such as food and construction materials. A possible idea consists of a loading platform 𝐷 that is supported by cable 𝐴𝐵, where end 𝐴 is attached to a large tree limb. After loading platform 𝐷, weight 𝑊3 is applied, which causes point 𝐵 to move horizontally by a small distance 𝛿 that is to be measured with a yardstick. Assume: 𝑊2 = weight of the loading platform = 100 lb,

Rubberball/Erik Isakson/Getty Images

Figure DP3.5

𝐴

𝑊1 = weight of materials to be weighed ≤ 500 lb, Cable segment 𝐵𝐶 is horizontal, Pulley 𝐶 is frictionless.

𝐿

You are to specify 𝐿 = length of cable 𝐴𝐵, where 10 f t ≤ 𝐿 ≤ 15 f t, 𝑊3 = weight of the counterweight, where 𝑊3 ≤ 40 lb. For the specific values of 𝐿 and 𝑊3 you choose, produce a graph with 𝑊1 and 𝛿 as the vertical and horizontal axes, respectively. Thus, by measuring the deflection 𝛿, your graph will tell the user the weight 𝑊1 .

𝛿

a pogo stick is in use, inertia of the person is an important source of forces that are applied to the pogo stick. Thus, it may be possible to develop a better design if concepts of dynamics are used. Nonetheless, the assumption of static behavior is satisfactory for the design purposes discussed here, provided reasonable values are chosen for the force at which the pogo stick begins to travel, and the force at which the pogo stick undergoes full travel.

𝐶 𝑊1

𝐷

Figure DP3.6 ∗ When

𝐵

𝑊2

𝑊3

192

Chapter 3

Equilibrium of Particles

60 m

𝑙

Design Problem 3.7

𝑙

𝑑

𝑑

𝐵

𝐶 ℎ2

35 m 𝐴

ℎ1

permanent position region for point 𝐴 Figure DP3.7

An aviation museum has exhibits in a room that is 60 m wide by 35 m high. A cable system is to be designed to suspend a new display, namely, a Mercury space capsule. Using this cabling system, point 𝐴 (where the two cables attach to the capsule) will be slowly maneuvered to a final position in the ”permanent position region” indicated in Fig. DP3.7. The cabling system must be sufficiently strong that the factor of safety against breaking is at least 5 while the capsule is maneuvered, and is at least 10 when the capsule is anywhere in the permanent position region. Note that during maneuvering, the display will be closed to the public so that the factor of safety during maneuvering can be lower than the final factor of safety. Your analysis may neglect the size of the pulleys at points 𝐵 and 𝐶. You are to select the smallest cable diameter possible and to specify the dimensions 𝑑, 𝑙, ℎ1 , and ℎ2 .

Design Problem 3.8 A utility pole supports a wire that exerts a maximum horizontal force of 500 lb at the top of the pole as shown. Although the pole is to be buried in soil at end 𝐴, for in-plane behavior it is conservative to idealize this support to be a pin, and thus, the net force supported by the pole is directed along the axis of the pole. Design a support system for the pole, considering that it is located in a congested region. Your design can utilize steel cable and/or steel pipe. Whatever support members you choose, only one support member may be attached to the utility pole, and this must be at the eyebolt 𝐶. Specify the size eyebolt to be used (if needed, see Section 2.2 for a description of these load multipliers). The support system you design cannot be attached to the adjacent building or to any point on the sidewalk or roadway. Safe pedestrian passage on the sidewalk should be considered. eyebolt

Eyebolt working loads Working load parallel to shank Nominal size

500 lb 𝐵

𝐶

3∕4 in. 1 1.5 2 2.5

15 f t

𝜃 street 𝐴

4 ft

𝜃

Working load multipliers

0◦ ≤ 𝜃 < 15◦ 15◦ ≤ 𝜃 < 30◦ 30◦ ≤ 𝜃 < 45◦ 45◦ ≤ 𝜃 ≤ 90◦

sidewalk

1.5 f t

river flow

480 lb 1,200 4,800 8,500 17,800

100% 60% 33% 20%

3 ft Figure DP3.8

𝐶

𝐵

Design Problem 3.9

40 m

𝐴

𝑥 98 m 20 m Figure DP3.9

ISTUDY

20 m

A ferry is being designed to cross a small river having a strong current. The ferry has no engine and is supported by a steel cable ABCA. At 𝐵 the cable is wrapped around a winch so that it does not slip, and at 𝐶 the cable passes around a freely rotating pulley. Both ends of the cable are fixed to the ferry at 𝐴. The drag force on the ferry is highest at midstream and lowest at the banks, and it is given by 𝐹 (𝑥) = (4 kN)[1 + sin(𝜋𝑥∕98 m)]. The winch moves the ferry slowly, so there are no other in-plane forces applied to the ferry beyond 𝐹 and the cable forces. Specify the necessary diameter for cable ABCA. The owner of the ferry also wants to know the position 𝑥 (within ± a few meters) where the load on the cable is the greatest.

ISTUDY

Section 3.4

193

Engineering Design

Design Problem 3.10 A company is designing an aftermarket hoist to be installed on a pickup truck. Thus, the dimensions of the hoist have largely been determined by the available space inside the bed of the truck. The hoist consists of two pipes, 𝐴𝐶 and 𝐴𝐷, and two cables, 𝐴𝐸 and FABA. The hoist is to be capable of lifting a maximum of 4000 lb. You are to specify The diameter of the steel wire rope FABA, The diameter of steel wire rope 𝐴𝐸, The diameter of steel pipes 𝐴𝐶 and 𝐴𝐷, The torque rating for the winch 𝐹 (equal to the force cable FABA supports multiplied by 6 in.). In your calculations, you may assume cable segments 𝐴𝐵 are vertical, and the bed of the truck is horizontal. Although the hoist is to be designed for a maximum load of 4000 lb, surely some user will attempt to lift more. Discuss in your report the consequences of doing this and how your specifications account for this possibility. 28 in.

48 in. 𝐴

28 in. 𝐴

12 in. 48 in.

25◦

12 in.

30◦ 𝐹

𝐸

𝐶, 𝐷

𝐵

𝐶

𝐵

𝐷

𝐷 28 in. 𝐹 𝐸

𝐴

12 in.

𝑧

28 in.

𝑑

𝐶

𝐶 𝑑

Figure DP3.10 𝐵

Design Problem 3.11 𝑂

A bar and cable system is to be designed to support a weight 𝑊 suspended from point 𝐴. Due to handling and positioning of the weight, a small force equal to 10% of 𝑊 lying in a horizontal plane is also applied at point 𝐷. If 𝑊 = 5 kN, specify • Dimensions 𝑑 and ℎ, where 𝑑 ≤ 1.0 m and ℎ ≤ 1.4 m.

2m 𝑥

• The diameter of the steel wire ropes 𝐴𝐵 and 𝐴𝐶. • The diameter of the steel pipe 𝐴𝑂.

𝐴

ℎ

Figure DP3.11

𝑦

𝐷 𝑊

(0.10)𝑊

194

Chapter 3

Equilibrium of Particles Design Problem 3.12

Design the cable support system for a radio tower of height 25 m. This includes specifying the radius 𝑅 of the supports at points 𝐵, 𝐶, and 𝐷 (as measured from the base of the tower) and the diameter of the steel cable. Remarks: • All supports will be erected on a horizontal surface. 𝐴

𝐵

𝐷 120◦

𝐹

120◦

𝑂, 𝐴

𝐵

25 m 𝑅

𝐷 𝑂

Figure DP3.12

ISTUDY

𝐹

𝐶

𝐶

• Supports 𝐵, 𝐶, and 𝐷 will each be located at the same radial distance 𝑅 from the base of the tower (point 𝑂) such that ∠𝐵𝑂𝐶 = ∠𝐵𝑂𝐷 = ∠𝐶𝑂𝐷 = 120◦ . • The cables must be able to support a maximum horizontal storm force 𝐹 = 18 kN from any direction in the horizontal plane. • Neglect the weight of the cables in your calculations. • Assume one of the cables is always slack and supports no load. For example, when force 𝐹 lies in sector 𝐶𝐴𝐷 (as shown), cable 𝐴𝐵 is slack. In your report describe the merits of large versus small values of 𝑅 and why you selected the value you did. State which orientation(s) of the storm force 𝐹 lead to the most severe cable loads and how you determined these orientations.

Design Problem 3.13 An overhead television camera is to be installed in a football stadium. The camera is supported by cables 𝐴𝐵 and 𝐴𝐶. At points 𝐵 and 𝐶 are pulleys that are driven by computercontrolled motors that allow the camera to be maneuvered into various positions, with the maximum altitude of point 𝐴 being limited to ℎ. Design the cabling system to be used by specifying the following: • The material and diameter for cables 𝐴𝐵 and 𝐴𝐶. • The weight 𝑊𝐴 of the camera at 𝐴 that your system uses. • The maximum altitude ℎ that point 𝐴 may have. Some factors to consider in your design include: • The cables should have a high strength-to-weight ratio so that they may be as thin as possible so as to minimize the distraction to spectators. • Assume the camera moves slowly enough so that static equilibrium may be assumed. • Neglect the weight of the cables in your design. • Neglect the diameter of pulleys 𝐵 and 𝐶 in your design. 600 f t 𝐵

180 f t

𝐶

𝐴 𝑊𝐴

Figure DP3.13

maximum altitude of point 𝐴

ℎ

ISTUDY

Section 3.5

Chapter Review

195

3.5 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation numbers cited for additional details.

Equilibrium of a particle A particle is in static equilibrium if

or or

Eq. (3.20), p. 166 ∑ ⃗ 𝐹⃗ = 0, (∑ ) (∑ ) (∑ ) ⃗ 𝐹𝑦 𝚥̂ + 𝐹𝑧 𝑘̂ = 0, 𝐹𝑥 𝚤̂ + ∑ ∑ ∑ 𝐹𝑦 = 0 and 𝐹𝑧 = 0. 𝐹𝑥 = 0 and

The above summations must include all forces applied to the particle. For equilibrium of a system of particles, the above equilibrium equations are written for each particle, and then the resulting system of simultaneous equations is solved.

Free body diagram. The free body diagram (FBD) is an essential aid for helping ensure that all forces applied to a particle are accounted for when you write equations of equilibrium. When you draw an FBD, it is helpful to imagine enclosing the particle by a closed line in two dimensions, or a closed surface in three dimensions. Wherever the cut passes through a structural member, the forces supported by that member must be introduced in the FBD. Wherever the cut passes through a support, the reaction forces that the support applies to the particle must be introduced in the FBD.

Cables, bars, pulleys, and springs Cables and bars. Cables and straight bars are structural members that support forces that are collinear with their axis. We assume cables may support tensile forces only and may be freely bent, such as when wrapped around a pulley. We usually assume cables have negligible weight. Bars may support both tensile and compressive forces. Pulleys. A pulley is a device that changes the direction of a cable, and hence, changes the direction of the force that is supported by the cable. If the pulley is frictionless and the cable is weightless, then the magnitude of the force throughout the cable is uniform. Springs. Behavior of a linear elastic spring is shown in Fig. 3.22 and is described by the spring law Eq. (3.18), p. 153 𝐹𝑠 = 𝑘𝛿 ( ) = 𝑘 𝐿 − 𝐿0 , where 𝑘 is the spring stiffness (units: force/length), 𝛿 is the elongation of the spring from its unstretched length, 𝐿0 is the initial (unstretched) spring length, and 𝐿 is the final spring length. In solving problems with springs, one of the forms of Eq. (3.18) will usually be more convenient than the other, depending on what data is provided. These equations are written with the convention that positive values of force 𝐹𝑠 correspond to tension and positive values of 𝛿 correspond to elongation. Other conventions are possible, such as

Helpful Information Spring law sign conventions. The sign conventions for the spring law given in Eq. (3.18) are as follows: 𝐹𝑠 > 0 tension, 𝐹𝑠 < 0 compression, 𝛿 = 0 unstretched position, 𝛿 > 0 extension, 𝛿 < 0 contraction.

196

ISTUDY

Chapter 3

Equilibrium of Particles

𝐹𝑠

𝐿 = final length 𝐿0 = initial length

tension

initial

𝑘 1

𝛿

𝐹𝑠 = 𝑘𝛿

compression

𝛿 𝐹𝑠

final 𝑘

contraction

elongation

𝐹𝑠 = 𝑘𝛿 = 𝑘(𝐿 − 𝐿0 )

Figure 3.22. Spring law for a linear elastic spring.

force being positive in compression and/or deformation being positive in contraction, but it may be necessary to introduce a negative sign on the right-hand side of Eq. (3.18), as described in Section 3.2.

Summing forces in directions other than 𝒙, 𝒚, or 𝒛 To sum forces in a direction 𝑟⃗, we take the dot product of all vectors in the expression ∑ ⃗ ⃗ 𝐹 = 0 with the unit vector∗ 𝑟⃗∕|⃗𝑟| to obtain Eq. (3.22), p. 168

Summation of forces in the 𝑟⃗ direction: ∑

𝐹𝑟 = 0 ∶

𝑟⃗ 𝑟⃗ 𝑟⃗ + 𝐹⃗2 ⋅ + ⋯ + 𝐹⃗𝑛 ⋅ = 0. 𝐹⃗1 ⋅ |⃗𝑟| |⃗𝑟| |⃗𝑟|

∑ Use of 𝐹𝑟 = 0 is especially convenient when it is known that some forces are perpendicular to 𝑟⃗.

Allowable load Allowable loads, sometimes called working loads, are forces structural components can safely be subjected to, and they are obtained by dividing the failure strength of a member by a factor of safety. Thus, Eq. (3.23), p. 184 failure load or factor of safety failure load . Factor of safety = allowable load Allowable load =

∗ Because

the right-hand side of Eq. (3.22) is zero, a unit vector in the direction of 𝑟⃗ is not needed, and we could just as well evaluate 𝐹⃗1 ⋅ 𝑟⃗ + 𝐹⃗2 ⋅ 𝑟⃗ + ⋯ + 𝐹⃗𝑛 ⋅ 𝑟⃗ = 0.

ISTUDY

Section 3.5

197

Chapter Review

Review Problems Problem 3.115 Consider a problem involving cables and bars only. For the conditions listed below, is the ∑ solution obtained from 𝐹⃗ = 0⃗ using geometry of the structure before loads are applied approximate or exact? Explain. (a) Cables are modeled as inextensible, and bars are modeled as rigid. (b) Cables and bars are modeled as linear elastic springs. Note: Concept problems are about explanations, not computations.

𝑦

Problem 3.116

10 in.

𝐶

The structure consists of two bars and a vertical force 𝑊 . When 𝑊 = 0, point 𝐴 has the position shown (i.e., its coordinates are 𝐴 (10, 0) in.). When 𝑊 = 50 lb, point 𝐴 moves to point 𝐴 ′ , whose coordinates are 𝐴 ′ (9.6, −1.2) in.

5 in.

𝐵

𝐴

(a) Determine the force each bar supports when 𝑊 = 50 lb. (b) If bars 𝐴𝐵 and 𝐴𝐶 are modeled as linear elastic springs, determine the stiffness of each of these.

Problem 3.117

𝑥

𝐴 𝑊 Figure P3.116 𝑦

The structure consists of a bar 𝐴𝐵 and cables 𝐴𝐶 and 𝐴𝐷. When 𝑊 = 20 N, point 𝐴 has the position shown (i.e., its coordinates are 𝐴 (100, 0) mm). When the weight at point 𝐷 is increased to 𝑊 = 40 N, point 𝐴 moves to point 𝐴 ′ , whose coordinates are 𝐴 ′ (99.8, −8) mm.

100 mm

60 mm 𝐶 50 mm

𝐴

𝐵

𝑥

(a) Determine the force in bar 𝐴𝐵 and cable 𝐴𝐶 when 𝑊 = 40 N. (b) If bar 𝐴𝐵 and cable 𝐴𝐶 are modeled as linear elastic springs, determine the stiffness of each of these.

𝐴 𝐷 𝑊

Problem 3.118

Figure P3.117

The frictionless pulley 𝐴 weighs 20 N and supports a box 𝐵 weighing 60 N. When you solve for the force in cable 𝐶𝐷, a “problem” arises. Describe this problem and its physical significance. 𝐶

𝐷 20 N 4 3

3

𝐴

4 𝐵

60 N Figure P3.118

Problem 3.119 To produce a force 𝑃 = 40 N in the horizontal member 𝐶𝐷 of a machine, a worker applies a force 𝐹 to the handle 𝐵. Determine the smallest value of 𝐹 that can be used and the angle 𝛼 it should be applied at.

𝐵 𝐷 𝑃 = 40 N Figure P3.119

𝐶

30◦

𝛼 60◦

𝐴

𝐹

198

Chapter 3

Equilibrium of Particles Problems 3.120 and 3.121

𝐴

𝐴 10◦ 𝐵

𝐶

40◦

𝐷

𝐵

40◦

(a)

(b)

4

Problem 3.120 If a 4000 lb force must be applied to the car, for each scenario determine the force the tow truck must apply to its cable. Problem 3.121 If the tow truck can produce a 6000 lb force in the cable attached to it, for each scenario determine the force that is applied to the car.

Figure P3.120 and P3.121 𝐺

Two scenarios are shown in Figs. P3.120(a) and (b) for using a tow truck 𝐴 to free a car 𝐵 that is stuck on the side of a road. In (a), a cable 𝐴𝐵 is attached directly from the tow truck to the car and in (b), three cables are used where point 𝐶 is a fixed support, such as a concrete signpost footing or a large tree stump. Assume all cables lie in the same plane and cable 𝐶𝐷 is parallel to the road.

Problem 3.122 3 𝐷

𝐸

The structure shown consists of five cables. Cable ABCD supports a drum having weight 𝑊 = 200 lb. Cable 𝐷𝐹 is horizontal, and cable segments 𝐴𝐵 and 𝐶𝐷 are vertical. If contact between the drum and cable ABCD is frictionless, determine the force in each cable.

4

Problem 3.123

𝐹 5 12 3

𝐴 𝐶

𝑊

In Prob. 3.122, if cable ABCD has 600 lb breaking strength and all other cables have 200 lb breaking strength, determine the largest value 𝑊 may have.

𝐵

Problem 3.124

Figure P3.122 and P3.123

The structure shown consists of two bars and three cables. If 𝑊 = 1000 lb, determine the force supported by each bar and cable. 𝐷

𝐶

16 in. 𝐵

𝐴 30 in.

12 in. 𝐸 𝑊

Figure P3.124 and P3.125

Problem 3.125

200 N

300 N

𝐴 30◦ Figure P3.126

ISTUDY

𝜃

𝐵 30◦

The structure shown consists of two bars and three cables. If each bar has 3000 lb strength in tension and 2000 lb strength in compression, and each cable has 1500 lb strength, determine the largest value of 𝑊 that may be supported.

Problem 3.126 Two frictionless pulleys connected by a weightless bar 𝐴𝐵 support the 200 N and 300 N forces shown. The pulleys rest on a wedge that is fixed in space. Determine the angle 𝜃 when the system is in equilibrium and the force in bar 𝐴𝐵.

ISTUDY

Section 3.5

199

Chapter Review

Problems 3.127 and 3.128

𝐶

The hoist shown is used to lift an engine during manufacturing. The hoist consists of a horizontal bar 𝐴𝐵 and a cable 𝐴𝐶, and the engine at 𝐷 is supported by a cable 𝐷𝐴𝐸 that wraps around a frictionless pulley at point 𝐴 and is driven by an electric motor at 𝐸. The engine at 𝐷 weighs 2500 N and other weights in the system are negligible. Determine the force supported by bar 𝐴𝐵 and cables 𝐴𝐶 and 𝐷𝐴𝐸.

Problem 3.127 Problem 3.128

If the bar and cables have the strengths listed below, determine the factor of safety for the hoist [factor of safety is defined in Eq. (3.23) on p. 184]. Member

Strength

𝐴𝐵 𝐴𝐶 𝐷𝐴𝐸

1m 0.1 m

𝐸

𝐴 0.1 m

𝐵

1.4 m 𝐷 2500 N

Figure P3.127 and P3.128

12,000 N tension & 9000 N compression 8000 N 7000 N

turnbuckle

𝑇

Problem 3.129 To glue a strip of laminate to the edge of a circular table, 𝑛 clamps are evenly spaced around the perimeter of the table (the figure shows 𝑛 = 8). Each clamp has a small pulley, and around all pulleys a cable is wrapped. A turnbuckle is used to tighten the cable, producing a force 𝑇 . Assuming the diameter of each pulley is small, show that the force 𝑅 each clamp applies to the edge of the table is given by 𝑅 = 2𝑇 cos[(𝑛 − 2)180◦ ∕(2𝑛)]. Does this expression give expected values when 𝑛 = 2 and 𝑛 = ∞? Explain. Hint: The sum of the interior angles of a polygon with 𝑚 corners is (𝑚 − 2)180◦ .

laminate

𝑇 Figure P3.129

Problem 3.130 Spring 𝐴𝐶 is unstretched when 𝜃 = 0. Force 𝐹 is always perpendicular to bar 𝐴𝐵. Determine the value of 𝐹 needed for equilibrium when 𝜃 = 45◦ .

𝑅

table

𝐵

𝜃

100 mm 𝐹

80 mm 𝐴 𝐶

Problems 3.131 and 3.132 If 𝑊 = 100 N and 𝑘 = 5 N∕mm, determine 𝛿1 and 𝛿2 . Springs are unstretched when 𝛿1 = 𝛿2 = 0. 𝑘

𝑘

𝛿2

𝑘

𝛿2

𝑊

𝛿1

𝑘

𝑊

𝛿1

𝑊

30◦

𝑘

𝑊

30◦

Figure P3.131

Figure P3.132

Figure P3.130

𝑘 = 2 N∕mm

200

Chapter 3

Equilibrium of Particles

Problem 3.133

𝐸 𝛿 𝐴

Frictionless sliders 𝐵 and 𝐶 have frictionless pulleys mounted to them, and are connected by a spring with stiffness 𝑘 = 12 N∕mm. Around the pulleys is wrapped a cable that supports a weight 𝑊 = 100 N. Member 𝐶𝐸 is a bar.

30◦

𝐵

𝐶

𝐷 𝛼

𝑘

(a) If 𝛼 = 30◦ , determine the force in bar 𝐶𝐸 and the motion 𝛿 of slider 𝐵. 𝐹

𝑊

(b) Determine the value of 𝛼 that will provide for the smallest force in bar 𝐶𝐸, and determine this force.

Figure P3.133

Problem 3.134 𝑦

A weight 𝑊 is supported by four cables. Points 𝐵 and 𝐶 lie in the 𝑦𝑧 plane. If 𝑊 = 8 kN, determine the force supported by each cable.

2m 𝐶

2m

Problem 3.135 A weight 𝑊 is suspended by four cables. Points 𝐵 and 𝐶 lie in the 𝑦𝑧 plane. If the allowable strength of each cable is as specified below, determine the largest allowable weight 𝑊 that can be supported.

3m

𝐵

𝐴

𝐷

𝑥

6m

3m

Cable

𝐸

𝐴𝐵 𝐴𝐶 𝐴𝐷 𝐴𝐸

𝑊

𝑧

Figure P3.134 and P3.135

Strength 12 kN 10 kN 5 kN 15 kN

Problem 3.136 Repeat Prob. 3.134 with points 𝐵 and 𝐶 having coordinates 𝐵(0, 6, 3) m and 𝐶(0, 6, −7) m.

Problem 3.137 𝑧

𝐵 (−45, 0, 130) in.

Repeat Prob. 3.135 with points 𝐵 and 𝐶 having coordinates 𝐵(0, 6, 3) m and 𝐶(0, 6, −7) m.

Problems 3.138 and 3.139 𝐷 50 in. 𝐴 𝑥

120 in.

100 in. 𝐶

Figure P3.138 and P3.139

ISTUDY

𝑦

A winch at point 𝐵 is used to slowly slide a box up a chute 𝐶𝐷 in a warehouse. The box weighs 600 lb and fits loosely against the walls of the chute and slides with no friction. For the position of the box given below, determine the force in the cable and the reactions between the box and chute. The chute is parallel to the 𝑥𝑧 plane. Neglect the size of the box and winch. Problem 3.138

The box is 1/5 the distance up the chute from point 𝐶.

Problem 3.139

The box is 3/5 the distance up the chute from point 𝐶.

ISTUDY

Section 3.5

𝑧

Problems 3.140 and 3.141

Problem 3.141

𝑦𝐸

18 in.

The collars 𝐵 and 𝐸 each weigh 20 lb and slide without friction on fixed bars 𝐴𝐶 and 𝐷𝐺, respectively. Bar 𝐴𝐶 is parallel to the 𝑧 axis and bar 𝐷𝐺 is parallel to the 𝑦 axis. For the value of 𝑦𝐸 given below, determine the force 𝑃 needed for equilibrium, the force in cable 𝐵𝐸, and the reactions between each collar and rod. Problem 3.140

201

Chapter Review

𝐸

𝐷

𝐶

𝑃

𝐺

𝐵

𝑦𝐸 = 9 in.

20 in.

14 in.

𝑦𝐸 = 13 in.

𝐴

𝑥

Problem 3.142

𝑦

Figure P3.140 and P3.141

Bead 𝐶 has 2 lb weight and slides without friction on straight bar 𝐴𝐵. The tensile forces in elastic cords 𝐶𝐷 and 𝐶𝐸 are 0.5 and 1.5 lb, respectively. If the bead is released from the position shown with no initial velocity, will it slide toward point 𝐴 or 𝐵 or will it remain stationary?

𝑦 𝐷 𝐴 16 in.

Problem 3.143

12

𝐸

Bead 𝐶 has 2 lb weight and slides without friction on straight bar 𝐴𝐵, and the tensile force in elastic cord 𝐶𝐸 is 0.9 lb. Determine the force needed in cord 𝐶𝐷 for the bead to be in equilibrium, and the magnitude of the reaction between the bead and bar 𝐴𝐵.

3

12

4

9 8

𝑥

𝐶

Problem 3.144

𝑊

A hoist for lifting objects onto and off a truck is shown. All cables lie in the 𝑥𝑦 plane, cable segment 𝐴𝐵 is horizontal, and the plane formed by bars 𝐶𝐸 and 𝐶𝐹 is parallel to the 𝑦𝑧 plane. Pulleys are frictionless and force 𝑃 is vertical. If the object is being slowly lifted and weighs 𝑊 = 600 lb, determine the force 𝑃 and forces supported by all cables and bars. 𝑦 𝐶 𝐵

𝐴

3 20◦ 4 3

𝐺 𝑊

4

𝐹

45◦

𝑃

𝑧

𝐷

𝑥

𝐸

Figure P3.144 and P3.145

Problem 3.145 For Prob. 3.144, determine the largest weight 𝑊 that may be lifted if forces supported by the cables may not exceed 2000 lb and compressive forces supported by the bars may not exceed 2800 lb.

𝑧

12 in.

Figure P3.142 and P3.143

15 in. 𝐵

202

ISTUDY

Chapter 3

Equilibrium of Particles

Problem 3.146 The structure shown in Fig. P3.146(a) consists of a steel I beam 𝐴𝐵 and a steel cable 𝐴𝐶, and load 𝑃 is horizontal. For the geometry and loading shown, the cable will likely be considerably more flexible than the I beam, and the structure shown in Fig. P3.146(b) is an appropriate model where the I beam is rigid and the cable is a spring with stiffness 𝑘 = 2000 lb∕in. (a) Let 𝑃 = 5000 lb, and assume the displacement of the structure is small. Thus, using the geometry shown in Fig. P3.146(b) where the I beam is vertical and the cable has 45◦ orientation, determine the forces supported by the I beam and cable. Using these forces, determine the horizontal displacement 𝑑 of point 𝐴, assuming the spring is unstretched when the I beam is vertical. (b) Let 𝑑 = 4 in.; for this case the I beam is no longer vertical and the cable’s orientation is no longer 45◦ . Determine the force 𝑃 needed to produce this value of 𝑑, assuming the spring is unstretched when the I beam is vertical. (c) Comparing the analyses in Parts (a) and (b), which of these provides the more accurate results (explain)? 𝑑

𝐴

𝑑

𝐴

𝑃

𝑃

𝑘 100 in.

100 in.

𝐶

45◦

𝐵

𝐶

45◦

rigid 𝐵

100 in.

100 in.

(b)

(a) Figure P3.146

ISTUDY

Moment of a Force and Equivalent Force Systems

4

Additional concepts of forces and systems of forces are discussed in this chapter. These concepts are used extensively in the analysis of equilibrium and motion of bodies and throughout more advanced mechanics subjects.

Ed Freeman/Stone/Getty Images

A hovering rescue helicopter lifting a man to safety. Concepts of the moment of a force and equivalent force systems are needed to determine the forces developed by the main rotor and tail rotor of the helicopter.

4.1

Moment of a Force

To help demonstrate some of the features of the moment of a force, we will consider an example of a steering wheel in a car. Figure 4.1 shows a classic Ferrari sports car, and Fig. 4.2 shows the steering wheel in this car. The wheel lies in a plane that is perpendicular to the steering column 𝐴𝐵 (it would not be very comfortable to use if this were not the case), and the steering wheel offers “resistance” to being turned (for most vehicles, this resistance increases for slower speeds and as a turn becomes sharper). Imagine you are driving this car, and you wish to execute a righthand turn. Figure 4.2 shows two possible locations where you could position your left hand to turn the steering wheel, and the directions of forces 𝐹1 and 𝐹2 that you would probably apply, where both of these forces lie in the plane of the steering wheel. For a given speed and sharpness of turn, clearly position 𝐶 will require a

Bob Masters Classic Car Images/Alamy Stock Photo

Figure 4.1 Ferrari 250 GTO sports car, circa 1962–1964.

203

204

Chapter 4

Moment of a Force and Equivalent Force Systems

lower force to turn the wheel than position 𝐷 (i.e., 𝐹1 < 𝐹2 ). Both forces 𝐹1 and 𝐹2 produce a moment (i.e., twisting action) about line 𝐴𝐵 of the steering column, and the size of this moment increases as the force becomes larger and/or as the distance from the force’s line of action to line 𝐴𝐵 increases. If the line of action of 𝐹1 is perpendicular to line 𝐴𝐵 (as we have assumed here), and if we let 𝑑 be the shortest distance between these two lines, then the moment produced by 𝐹1 has size 𝐹1 𝑑 and the direction of the twisting action is about line 𝐴𝐵. Thus, we observe that the

𝐹1

𝐹2

𝐶

𝐷 𝐴

𝐵

(a)

𝐴

𝐵

(b)

Figure 4.2. An accurate sketch of the Nardi Anni ’60 steering wheel in the Ferrari 250 GTO, showing two possible locations where your hand could be positioned to turn the wheel.

𝐹

Figure 4.3 Force applied to a wrench to twist a pipe.

moment of a force has the properties of a vector, namely, magnitude (or size) and direction. In the remainder of this section, we provide a more precise definition and describe methods of evaluation. The moment of a force, or simply moment, is a measure of a force’s ability to produce twisting, or rotation about a point. Moment has both magnitude (or size) and direction and is a vector quantity. For example, consider using a wrench to twist a pipe, as shown in Fig. 4.3. When you apply a force to the handle of the wrench, the force tends to make the pipe twist. Whether or not the pipe actually does twist depends on details of how the pipe is supported. To create a greater tendency to twist the pipe, we could apply a larger force to the wrench, or we could use a wrench with a longer handle, or we could do both. Moment can be evaluated using both scalar and vector approaches, as follows.

Scalar approach 𝑀𝑂 = 𝐹 𝑑 moment arm

𝑂

𝑑

The moment of a force is a vector and can be evaluated using the scalar approach described here. Consider a force 𝐹⃗ with magnitude 𝐹 . As shown in Fig. 4.4, this force produces a moment vector about point 𝑂 (the twisting action shown) where the magnitude of this moment is 𝑀𝑂 , which is given by

𝐹

𝑀𝑂 = 𝐹 𝑑, where

Figure 4.4 Scalar approach to evaluate the moment of a force.

ISTUDY

𝐹 is the magnitude of the force; 𝑑 is the perpendicular distance from point 𝑂 to the line of action of 𝐹⃗ and is called the moment arm; and 𝑀𝑂 has dimensions of force times length.

(4.1)

ISTUDY

Section 4.1

Moment of a Force

The moment of a force is a vector, and it has both magnitude and direction. In the scalar evaluation of the moment, Eq. (4.1) conveniently gives the magnitude. The direction of the moment is not provided by Eq. (4.1), but is understood to be as follows. The line about which the twisting action occurs (this is called the line of action for the moment) is parallel to the axis through point 𝑂 that is perpendicular to the plane containing 𝐹⃗ and the moment arm. The direction of the moment along its line of action is given by the direction of the thumb of your right hand when your fingers curl in the twisting direction of the moment. For summing multiple moments, Eq. (4.1) must be supplemented with the proper directions for each moment, as illustrated in the mini-example that follows on the next page and in Examples 4.1 and 4.2.

Vector approach The magnitude and direction of the moment of a force can be obtained using the cross product as described here. As shown in Fig. 4.5, the moment of a force 𝐹⃗ about ⃗ and is given as a point 𝑂 is denoted by 𝑀 𝑂 ⃗ = 𝑟⃗ × 𝐹⃗ , 𝑀 𝑂

Helpful Information Units for moment. The dimensions for moment are force times length, and the following table gives the units we will typically use. Units for moment of a force U.S. Customary

SI

ft⋅lb or in.⋅lb

N⋅m

From a fundamental point of view, the order in which the units are written (e.g., f t ⋅lb or lb⋅f t) is irrelevant. However, some people follow the convention that moments, whether in U.S. Customary units or SI units, have the force unit first, followed by the length unit.

(4.2)

where 𝐹⃗ is the force vector; and 𝑟⃗ is a position vector from point 𝑂 to any point∗ on the line of action of 𝐹⃗ . In contrast to Eq. (4.1) for the scalar approach, Eq. (4.2) automatically provides both the magnitude and direction of the moment. Remarks • The order in which the vectors are taken when computing the cross product is important, and hence, vectors 𝑟⃗ and 𝐹⃗ in Eq. (4.2) may not be interchanged. • In Fig. 4.5, we show the moment vector using a double-headed arrow to emphasize its physical differences compared to vectors that are represented using a single-headed arrow. That is, vectors describing force and position represent physical phenomena that are directed along a line, and we represent these using a single-headed arrow. Moment vectors represent physical phenomena that are directed about (or twisting around) a line, and we represent these using a double-headed arrow. The convention between the direction of the doubleheaded arrow and the twisting direction is given by the right-hand rule as described in Fig. 4.6. • Observe that the moment of a force depends on the location of the point about which the moment is computed. Hence, the moments of a force 𝐹⃗ computed about two different points are usually different, in both magnitude and direction. ∗ Because the head of the position vector 𝑟 ⃗ can be any point on the line of action of 𝐹⃗ , there is considerable

flexibility in selecting 𝑟⃗. Problem 4.31 asks you to show this is true.

205

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝑂 𝑂

𝑟⃗

𝐹⃗

Figure 4.5 Vector approach to evaluate the moment of a force.

Figure 4.6 Use of the right-hand rule to determine the twisting direction for vectors with double-headed arrows. With your right hand, position your thumb in the direction of the vector’s arrows; then your fingers define the direction of the twisting action. If the direction of the twisting action is known, wrap your fingers in this direction; then your thumb defines the direction of the vector.

206

line of action for the 8 kN force

4 kN 8 kN

𝑦 3m

5 kN 2m

Mini-Example For the balance mechanism shown in Fig. 4.7(a), determine the value of 𝐹 so that the resultant moment of all forces about point 𝑂 is zero.

𝐹 1m

𝑂 2m

𝑥

3m

(a) 4 kN 8 kN 𝑟⃗1 3m

5 kN 2m

𝑦 𝑟⃗2 2m (b)

𝐹 𝑟⃗3

𝑂

1m

𝑥

Solution Using a scalar approach first, we note that each force has a tendency to twist the structure about point 𝑂 in either a clockwise or counterclockwise fashion. In other words, the moment of each force about point 𝑂 is a vector that points either into or out of the plane of Fig. 4.7. Furthermore, we may distinguish between these by taking the counterclockwise direction to be positive (see additional discussion in the margin note). Thus, 𝑀𝑂 = −(8 kN)(3 m) + (4 kN)(4 m) + (5 kN)(2 m) − 𝐹 (3 m) = 0,

(4.3)

𝐹 = 0.667 kN.

(4.4)

⇒

3m

Figure 4.7 A balance mechanism subjected to several forces.

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Chapter 4

Moment of a Force and Equivalent Force Systems

Each term of Eq. (4.3) consists of the product of the magnitude of a force with its corresponding moment arm, where the moment arm is the shortest (perpendicular) distance from point 𝑂 to the line of action of the force. For example, in Fig. 4.7(a), the line of action for the 8 kN force is shown, and the shortest distance from point 𝑂 to this line is 3 m. The sign of each term is positive for counterclockwise twisting about point 𝑂 and negative for clockwise twisting. A vector approach can also be used for this problem with the following force and position vectors [see Fig. 4.7(b)]:

Helpful Information Direction for moments in two-dimensional problems. In two-dimensional problems, such as Fig. 4.7, the moment of a force about a given point will be either clockwise or counterclockwise. That is, the moment of a force about a given point is a vector in either the −𝑧 direction or the +𝑧 direction, respectively. In two-dimensional problems, when using a scalar approach, we will always take counterclockwise to be the positive direction for moments. 𝑦 +𝑀 𝑥 Figure 4.8 This choice is consistent with the right-hand rule for the 𝑥𝑦 coordinate system shown, where the 𝑧 direction, and hence the direction for positive moment vectors, is out of the plane of the figure.

𝐹⃗1 = (8 𝚤̂ − 4 𝚥̂) kN,

𝑟⃗1 = (−4 𝚤̂ + 3 𝚥̂) m,

(4.5)

𝐹⃗2 = −5 𝚥̂ kN,

𝑟⃗2 = −2 𝚤̂ m,

(4.6)

𝐹⃗3 = −𝐹 𝚥̂ kN,

𝑟⃗3 = 3 𝚤̂ m,

(4.7)

⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ , 𝑀 𝑂 1 1 2 2 3 3 ⃗ = [2 kN⋅m − 𝐹 (3 m)] 𝑘̂ = 0, ⇒ 𝐹 = 0.667 kN.

(4.8) (4.9)

Remarks • Since all forces and point 𝑂 lie in the 𝑥𝑦 plane, the forces produce moment about the 𝑧 axis only, and hence a scalar approach is very straightforward. In a fully three-dimensional problem the scalar approach may be difficult to use, and the vector approach is often more effective. • In the vector approach, since the 8 kN and 4 kN forces share the same point of application, they are combined to yield a single force vector 𝐹⃗1 in Eq. (4.5). Thus, their moment can be obtained by evaluating just one cross product. • In the vector approach, the direction of positive moment is the positive 𝑧 direction, which in this example is the direction out of the plane of the figure as provided by the right-hand rule. To see that this is true, use your right hand such that your thumb points in the positive 𝑧 direction; then the direction in which your fingers curl provides the twisting sense of a positive moment.

ISTUDY

Section 4.1

207

Moment of a Force

• In the vector approach there is considerable flexibility in selecting the position vectors. In this example, 𝑟⃗1 and 𝑟⃗2 are taken from point 𝑂 to the actual points of application of 𝐹⃗1 and 𝐹⃗2 , while 𝑟⃗3 is taken from point 𝑂 to a convenient ⃗ is obtained using, for point on the line of action of 𝐹⃗3 . The same result for 𝑀 𝑂 example, 𝑟⃗3 = (3 𝚤̂ + 𝚥̂) m or 𝑟⃗3 = (3 𝚤̂ − 𝚥̂) m. • In general, there is no reason why the resultant moment of all forces about point 𝑂 should be zero. In this example, if 𝐹 = 1.0 kN, then the scalar approach ⃗ = −1 kN⋅m 𝑘. ̂ The yields 𝑀𝑂 = −1 kN⋅m and the vector approach yields 𝑀 𝑂 result of having a nonzero resultant moment about point 𝑂 is that the balance mechanism will tend to rotate about point 𝑂. This topic is explored in detail in Chapter 5.

Varignon’s theorem Varignon’s theorem, also known as the principle of moments, states that the moment of a force is equal to the sum of the moments of the vector components of the force. Thus, if 𝐹⃗ has vector components 𝐹⃗ , 𝐹⃗ , and so on, then the moment of 𝐹⃗ about a 1

2

point 𝐴 is given by ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴 = 𝑟⃗ × (𝐹⃗1 + 𝐹⃗2 + ⋯ ) = 𝑟⃗ × 𝐹⃗1 + 𝑟⃗ × 𝐹⃗2 + ⋯ ,

(4.10)

where 𝑟⃗ is a position vector from point 𝐴 to any point on the line of action of 𝐹⃗ . This principle is simply a restatement of the distributive property of the cross product, but in fairness to Varignon (1654–1722), he discovered the concepts underlying Eq. (4.10) well before vector mathematics and the cross product were invented. Varignon’s theorem remains very useful for evaluating moments, especially for scalar evaluations. While the component forces will often be Cartesian components, in which case 𝐹⃗1 , 𝐹⃗2 , and 𝐹⃗3 would usually be called 𝐹⃗𝑥 , 𝐹⃗𝑦 , and 𝐹⃗𝑧 , in general the components do not need to be orthogonal, and there may be an arbitrary number of them.

𝑦

𝑦 𝐹

𝑑2

Solution Force 𝐹⃗ has magnitude 𝐹 and components 𝐹𝑥 and 𝐹𝑦 . The magnitude of the moment of 𝐹⃗ about a point 𝐴 can be evaluated by the two scalar approaches shown in Fig. 4.9. In the first of these, the magnitude of the moment is computed using the definition given in Eq. (4.1), where the moment arm 𝑑 is the perpendicular distance from point 𝐴 to the line of action of the force and positive values of 𝑀𝐴 are taken to be counterclockwise. In the second evaluation in Fig. 4.9, the 𝑥 and 𝑦 components of the force are used, and the moment of each of these is evaluated and summed, again with counterclockwise being positive. In practice, you should use the first approach if it is easy to find the moment arm 𝑑, and the

𝐹𝑥 𝑑1

𝑑 𝐴

𝑥

𝐴

𝑥

𝑀𝐴 = −𝐹𝑥 𝑑1 + 𝐹𝑦 𝑑2

𝑀𝐴 = −𝐹 𝑑

Mini-Example With reference to Figs. 4.9 and 4.10, use scalar and vector approaches, respectively, to evaluate the moment of force 𝐹⃗ about point 𝐴.

𝐹𝑦

Figure 4.9 Scalar description of Varignon’s theorem where, according to our sign convention, positive moment is counterclockwise. 𝑦

𝑦 𝐹⃗

𝐹⃗𝑦 𝐹⃗𝑥

𝑟⃗ 𝐴

𝑟⃗ 𝑥

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴

𝐴

𝑥

⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ 𝑀 𝐴 𝑥 𝑦

Figure 4.10 Vector description of Varignon’s theorem.

208

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Moment of a Force and Equivalent Force Systems

Chapter 4

second approach if it is easier to obtain the moment arms for the components of the force. The vector approaches shown in Fig. 4.10 can also be used. In the first of these, the position vector is taken from point 𝐴 to the tail of 𝐹⃗ , although a position vector from 𝐴 to any point on the line of action of 𝐹⃗ is acceptable. In the second evaluation in Fig. 4.10, the force is resolved into vector components 𝐹⃗𝑥 and 𝐹⃗𝑦 , and the moment of each of these is computed and summed. In practice, the first of these vector evaluations will almost always be more convenient.

Which approach should I use: scalar or vector?

Common Pitfall What does the moment depend on? A common misconception is that the moment of a force is an inherent property of only the force. The moment of a force depends on both the force and the location of the point about which the moment is evaluated. For example, consider the moment of force 𝐹⃗ about point 𝐴 and about point 𝐵. 𝑧

𝐹⃗ 𝐵

𝑥

𝑦

𝐴

In general, the moment of 𝐹⃗ about these two ⃗ . ⃗ ≠𝑀 points is different, and 𝑀 𝐴 𝐵

In simpler problems, such as two-dimensional problems when all forces are coplanar (i.e., all forces lie in the same plane and hence produce moments about the same axis, which is perpendicular to that plane), the scalar approach for evaluating moments will usually be easier and faster, and the use of clockwise and counterclockwise to distinguish moment direction is effective. For more complicated problems, such as in three dimensions, a scalar approach can sometimes be used effectively, but generally the vector approach is better. You should contrast the scalar and vector solutions for the example problems that follow to help refine your ability to select the easier approach for a particular problem.

End of Section Summary In this section, the moment of a force, or simply moment, is defined, and scalar and vector approaches for evaluation are discussed. Some of the key points are as follows: • The moment is a vector quantity. • When we evaluate the moment using the scalar approach, the magnitude of the moment of a force 𝐹⃗ about a point 𝑃 is given by 𝑀𝑃 = 𝐹 𝑑 where 𝑑 is the moment arm, which is the perpendicular distance from point 𝑃 to the line of action of the force. The scalar approach does not automatically provide the direction of the moment; this must be assigned manually. • When we evaluate the moment using the vector approach, the moment of a ⃗ = 𝑟⃗ × 𝐹⃗ , where 𝑟⃗ is a position vector force 𝐹⃗ about a point 𝑃 is given by 𝑀 𝑃 measured from point 𝑃 to anywhere on the line of action of 𝐹⃗ . The direction ⃗ is automatically provided by this approach. of 𝑀 𝑃

• The moment of a force depends on both the force and the location of the point about which the moment is evaluated. The important characteristics of the force are its magnitude, its orientation or direction, and the position in space of its line of action. • Varignon’s theorem, also known as the principle of moments, states that the moment of a force is equal to the sum of the moments of the components of the force.

ISTUDY

Section 4.1

209

Moment of a Force

E X A M P L E 4.1

Moment: Vector and Scalar Evaluations for a Two-Dimensional Problem

A portion of a structure is acted upon by the 10 lb and 20 lb forces shown. Determine the resultant moment of these forces about point 𝐴.

20 lb

𝑦

10 lb

SOLUTION

𝐵 5 in.

Road Map

Both scalar and vector approaches are effective for this problem. In the scalar approach, we must find the moment arm for each force, and after evaluating the moment of each force, we must be careful to sum the results according to a sign convention for positive moment. With the vector approach, we must develop vector expressions for forces and positions and then carry out cross products to determine the moment. For two-dimensional problems, the vector approach will usually require more computations, but it does not require the careful visualization of the scalar approach.

𝑥 𝐴 10 in. Figure 1

Scalar solution Governing Equations & Computation In Fig. 2, the moment arms for the 10 lb and

20 lb forces are determined. By applying Eq. (4.1) on p. 204 to each force and summing the results, using the convention that positive moment is counterclockwise, the resultant moment about point 𝐴 is 𝑀𝐴 = (10 lb)(5 in.) − (20 lb)(10 in.) = −150 in.⋅lb.

(1)

When writing Eq. (1), we used Eq. (4.1) to evaluate the magnitude of the moment for the 10 lb force as (10 lb)(5 in.), followed by examination of Fig. 1 to determine that this force produces counterclockwise moment about point 𝐴, and hence, it is positive in accordance with our sign convention. We then repeated this process for the 20 lb force to determine the magnitude of its moment as −(20 lb)(10 in.), where the negative sign is used because this force produces clockwise moment about point 𝐴. Carrying out the algebra in Eq. (1) yields the final answer: the moment of the two forces about point 𝐴 is −150 in.⋅lb counterclockwise, or alternatively is +150 in.⋅lb clockwise. If desired, the results of Eq. (1) can be used to write a vector expression for the resultant moment about point 𝐴 as ⃗ = −150 in.⋅lb + 𝑀 𝐴

20 lb 10 lb 𝐵 𝑑1 = 5 in. 𝐴

𝑑2 = 10 in.

Figure 2 A sketch showing the lines of action of each force. The moment arm 𝑑1 for the 10 lb force is the shortest distance from point 𝐴 to the line of action of this force, hence 𝑑1 = 5 in. The moment arm 𝑑2 for the 20 lb force is the shortest distance from point 𝐴 to the line of action of this force, hence 𝑑2 = 10 in.

(2)

where the + symbol specifies the direction and is analogous to the ∡ symbol used in polar vector representation (described in the margin note on p. 33). 𝐹⃗

Vector solution Governing Equations & Computation A vector approach can also be employed using

𝑦

the vectors 𝐹⃗ and 𝑟⃗𝐴𝐵 shown in Fig. 3, which are 𝐹⃗ = (−10 𝚤̂ − 20 𝚥̂) lb,

𝑟⃗𝐴𝐵 = (10 𝚤̂ + 5 𝚥̂) in.

(3)

𝑟⃗𝐴𝐵

𝐵 𝑥

Using Eq. (4.2) on p. 205, the moment vector of the two forces about point 𝐴 is ⃗ = 𝑟⃗ × 𝐹⃗ = −150 𝑘̂ in.⋅lb. 𝑀 𝐴 𝐴𝐵

5 in. 𝐴

(4)

Discussion & Verification While you should be proficient using both scalar and vector approaches, you should contrast these two solutions to help you recognize which of the two will be more efficient for a particular problem. You can also use both approaches for a particular problem to help verify the accuracy of your answers.

10 in. Figure 3 Force and position vectors used to determine the moment of 𝐹⃗ about point 𝐴.

210

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.2

Moment: Different Points Along the Force’s Line of Action May Be Used A machine handle is connected to a shaft at 𝐵. Determine the moment produced by the 15 N force about point 𝐵.

𝑦 200 mm

𝐹 = 15 N 3

4

SOLUTION

𝐴

Road Map

Both scalar and vector approaches are effective for this problem, and several solutions will be carried out and compared. If a scalar solution is used, there are several options for treatment of the force and determination of moment arms. If a vector approach is used, there are options on selection of the position vector. Regardless of your approach, the results should be the same, but some solutions may be easier to perform than others.

350 mm

𝑥

𝐵

Solution 1 Governing Equations & Computation

With the geometry provided, similar triangles and trigonometry can be used (detailed calculations are given in Fig. 2) to determine that the moment arm is 𝑑 = 160 mm, as shown in Fig. 3(a). With our convention that counterclockwise moments are positive, the moment about point 𝐵 is

Figure 1

𝑦 𝐹 = 15 N 200 mm 5 3 4 𝐴

𝑀𝐵 = −(15 N)(160 mm) = −2400 N⋅mm. Solution 2

350 mm

Governing Equations & Computation We resolve the 15 N force into its 𝑥 and 𝑦 components as shown in Fig. 3(b) and use Varignon’s theorem (the principle of moments) to sum the moments produced by each of these components. With our convention that positive moment is counterclockwise,

𝐺 𝑑 𝜃 𝐸

𝐵

𝐷

𝑏

𝑥

𝑀𝐵 = −(12 N)(350 mm) + (9 N)(200 mm) = −2400 N⋅mm.

𝑎 350 mm 𝑎

tan 𝜃 =

⇒

(2)

Solution 3

Triangle 𝐴𝐸𝐷: = 43 ⇒ 𝑎 = 466.7 mm 350 mm 𝑎

(1)

𝜃 = 36.9◦

Triangle 𝐵𝐷 𝐺: 𝑏 = 𝑎 − 200 mm = 266.7 mm

Governing Equations & Computation

The moment of a force does not depend on the specific location where the force is applied, but rather depends on the position and orientation of the force’s line of action. Thus, we may consider the 15 N force as being positioned anywhere along its line of action, and we select point 𝐶 shown in Fig. 3(c) as a convenient location (you should verify for yourself the coordinates of point 𝐶, where Fig. 2 may be helpful). With our convention that positive moment is counterclockwise,

𝑑 = 𝑏 sin 𝜃 = 160 mm Figure 2 Use of similar triangles and trigonometry to determine the moment arm 𝑑 needed for Solution 1. Other strategies for determining 𝑑 are possible.

ISTUDY

𝑀𝐵 = −(12 N)(200 mm) + (9 N)(0 mm) = −2400 N⋅mm.

(3)

Solution 4 Governing Equations & Computation

We use the same approach employed in the previous solution, except here we “move” the force to point 𝐷 shown in Fig. 3(d). With our convention that positive moment is counterclockwise, 𝑀𝐵 = (12 N)(0 mm) − (9 N)(266.7 mm) = −2400 N⋅mm.

(4)

Solution 5 Governing Equations & Computation

A vector approach can also be employed using the force and position vectors shown in Fig. 3(e), which are 𝐹⃗ = (12 𝚤̂ − 9 𝚥̂) N,

𝑟⃗𝐵𝐴 = (−200 𝚤̂ + 350 𝚥̂) mm.

(5)

ISTUDY

Section 4.1

Moment of a Force

𝑦

𝑦 200 mm

𝐹 = 15 N 200 mm 3

4

211

𝐴

𝐴

4 5 3 5

350 mm

350 mm

15 N = 12 N

15 N = 9 N

𝑑 = 160 mm Solution 1

𝐵

𝑥 Solution 2

(a)

𝑥

𝐵 (b)

𝑦

𝑦

𝑦

𝐹⃗ 266.7 mm

𝐶 200 mm

Solution 3

𝐴

12 N

𝑟⃗𝐵𝐴

9N

𝐵

𝑥 Solution 4

(c)

𝐵 (d)

𝑥

𝐷 12 N 9N

Solution 5

𝐵

𝑥

(e)

Figure 3. (a)–(d) Positioning of forces and/or resolution of forces into components so that a scalar approach can be used to determine the moment of the 15 N force about point 𝐵. (e) Force and position vectors that can be used in a vector approach.

Rather than using 𝑟⃗𝐵𝐴 , the position vectors 𝑟⃗𝐵𝐶 and 𝑟⃗𝐵𝐷 are convenient choices that could have been used instead. Evaluating the moment provides ⃗ = 𝑟⃗ × 𝐹⃗ = −2400 𝑘̂ N⋅mm. 𝑀 𝐵 𝐵𝐴

Discussion & Verification

(6)

As expected, all solutions produce the same result. For this example, Solution 1 is not especially convenient because the determination of the moment arm is tedious. Solutions 2–5 are all effective, with Solutions 2 and 5 probably being the most straightforward.

Helpful Information Distance between a point and line. Sometimes, the distance between a point and the line of action of a force (or other vector) is desired, such as 𝑑 in Fig. 2. For situations where a direct evaluation of 𝑑 is tedious, such as in this example, you may use one of Solutions 2–5 to determine the moment of the force about point 𝐵 (namely −2400 N⋅mm), then write Eq. (1) as 𝑀𝐵 = −(15 N) 𝑑 = −2400 N⋅mm, and then solve for 𝑑 = 160 mm.

212

E X A M P L E 4.3

Moment: Vector and Scalar Evaluations for a Three-Dimensional Problem The structure supports vertical forces 𝐹 = 200 lb and 𝑃 = 50 lb. Pipe segments 𝐵𝐶 and 𝐶𝐷 are parallel to the 𝑦 and 𝑥 axes, respectively. Determine the resultant moment of both forces about point 𝑂.

𝑦 16 in.

12 in.

20 in.

𝐴

𝐵

𝑧

𝐶

𝑃 = 50 lb

𝑥 𝐷 𝐹 = 200 lb

Figure 1

SOLUTION

𝑂 18 in.

ISTUDY

Chapter 4

Moment of a Force and Equivalent Force Systems

Road Map

Both vector and scalar solutions are possible, and both approaches will be illustrated and compared. The vector approach is straightforward: we first write expressions for force vectors and position vectors and then carry out the cross products to obtain the desired moment. A scalar solution requires good visualization to identify the appropriate moment arms and a consistent sign convention for moments. Vector solution Governing Equations & Computation We first write expressions for force vectors and

position vectors as follows: 𝐹⃗ = −200 𝚥̂ lb,

̂ in., 𝑟⃗𝑂𝐷 = (18 𝚤̂ − 12 𝚥̂ + 36 𝑘)

(1)

𝑃⃗ = 50 𝚥̂ lb,

̂ in. 𝑟⃗𝑂𝐶 = (−12 𝚥̂ + 36 𝑘)

(2)

Using the vectors in Eqs. (1) and (2), we find the moment about point 𝑂 is ⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝑃⃗ 𝑀 𝑂 𝑂𝐷 𝑂𝐶 ̂ in.⋅lb. = (5400 𝚤̂ − 3600 𝑘)

(3)

Remarks ̂ in. • Rather than using 𝑟⃗𝑂𝐷 and 𝑟⃗𝑂𝐶 in Eq. (3), the position vectors (18 𝚤̂ + 36 𝑘) and 𝑟⃗𝑂𝐵 , respectively, could have been used. These are slightly better choices since they each have fewer components, which will reduce the number of computations needed to evaluate the cross products. ⃗ in Eq. (3) has 𝑥 and 𝑧 components, meaning that 𝐹⃗ and 𝑃⃗ • The result for 𝑀 𝑂 combine to produce twisting action about both the 𝑥 and 𝑧 axes through point 𝑂 as shown in Fig. 2(a). Alternatively, the 𝑥 and 𝑧 components can be added to give a single moment vector with magnitude 𝑀𝑂 , as shown in Fig. 2(b). 𝑦

𝑦 𝑀𝑂𝑧 = −3600 in.⋅lb 𝑧

𝐴

𝐴

𝑧 𝑂

𝑂 𝑀𝑂𝑥 = 5400 in.⋅lb (a)

𝑥

𝑀𝑂 = 6490 in.⋅lb

𝑥

(b)

⃗ of forces 𝐹 and 𝑃 about point 𝑂 is shown. (a) The Figure 2. The resultant moment 𝑀 𝑂 ⃗ are shown. (b) The vector 𝑀 ⃗ is shown in its proper orientation. components of 𝑀 𝑂 𝑂 Scalar solution Governing Equations & Computation A scalar solution is possible, but it requires good

visualization to identify the appropriate moment arms and a consistent sign convention for moments. In three-dimensional problems, it is very difficult to unambiguously categorize

ISTUDY

Section 4.1

Moment of a Force

moments with words like clockwise and counterclockwise. Thus, we will usually take the positive coordinate directions to define the directions of positive moments (i.e., the righthand rule governs the direction of positive moment). As shown in Fig. 3, we extend the lines of action for each force to help identify the appropriate moment arms. Force 𝐹 produces positive moment about the 𝑥 axis where the 𝑦 36 in. 𝐴

𝐵

𝑧

18 in. 𝐶

𝑃 = 50 lb

36 in. 𝑂

𝑥

𝐷 𝐹 = 200 lb

Figure 3. Moment arms to be used in a scalar approach for determining the moment of forces 𝐹 and 𝑃 about point 𝑂.

moment arm is 36 in., and negative moment about the 𝑧 axis where the moment arm is 18 in. Since 𝐹 is parallel to the 𝑦 axis, it produces no moment about this axis. Force 𝑃 produces negative moment about the 𝑥 axis where the moment arm is 36 in. and produces no moment about the other two axes. Thus, 𝑀𝑂𝑥 = (200 lb)(36 in.) − (50 lb)(36 in.) = 5400 in.⋅lb,

(4)

𝑀𝑂𝑦 = 0,

(5)

𝑀𝑂𝑧 = −(200 lb)(18 in.) = −3600 in.⋅lb.

(6)

Discussion & Verification As expected, the 𝑥, 𝑦, and 𝑧 components of the moment about point 𝑂 found by the scalar approach agree with those found using the vector approach. Note that in the scalar approach, if you do not take positive moments to be in the positive coordinate directions, then the signs of moment components may not agree with those in the vector approach. Our recommendation is that you always take moments to be positive in the positive coordinate directions.

213

Common Pitfall Clockwise or counterclockwise? Earlier example problems in this section all have moments that are either into or out of the plane containing the forces for each problem, and when we used the scalar approach, use of clockwise and counterclockwise for describing moment directions was effective. For moments in three dimensions, clockwise and counterclockwise are ambiguous and have little meaning. When using the scalar approach to find moments in three dimensions, you should use the positive coordinate directions and the right-hand rule to define positive moments, as illustrated in this example.

214

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.4

Maximizing the Moment of a Force The belt tensioner 𝐴𝐵𝐶 is attached to an engine using a bearing at 𝐴 having a torsional spring. To release the belt tension, a ratchet wrench 𝐶𝐷 is applied to the tensioner at point 𝐶. If a moment about point 𝐴 of 50 N⋅m is required to release the belt tension, determine the smallest force 𝐹 required and the angle 𝛼 at which the wrench should be positioned. Consider the following two cases:

𝐷

𝐹 𝑦

𝛼 𝑥

300 mm

(a) The force 𝐹 is always perpendicular to the handle of the wrench.

250 mm 𝐶 𝐸

(b) The force 𝐹 is always horizontal (parallel to the 𝑥 axis).

30◦ 𝐵

SOLUTION

𝐴

Road Map

While both scalar and vector solutions are effective, only a scalar solution will be used here. For both parts of this problem, our goal is to produce a 50 N⋅m moment about point 𝐴 using the smallest force possible. This will be accomplished by positioning the wrench handle so that the moment arm is as large as possible.

Figure 1

𝐹

𝐷1

Part (a) Governing Equations & Computation

300 mm 𝑑1

𝐷2

𝑀𝐴 = −𝐹 𝑑 = −50 N⋅m

𝑑2 𝐶

𝐷3

30◦ 𝐴

𝑑3

Figure 2 Depending on the value of 𝛼, there are an infinite number of possible positions for point 𝐷 and force 𝐹 , and three possible positions, denoted by 𝐷1 , 𝐷2 , and 𝐷3 , are shown with their corresponding moment arms 𝑑1 , 𝑑2 , and 𝑑3 , respectively. 𝐷1

The moment of force 𝐹 about point 𝐴 is required

to be 50 N⋅m clockwise. Thus, (1)

where the moment arm 𝑑 is the distance from point 𝐴 to the line of action of 𝐹 , and because 𝐹 produces clockwise moment about point 𝐴, the negative signs are included. To minimize 𝐹 , we seek the position of the wrench handle 𝐷 so that moment arm 𝑑 is maximized. Shown in Fig. 2 is the circular locus of possible locations of point 𝐷, with three example locations shown (points 𝐷1 , 𝐷2 , and 𝐷3 ) and their associated moment arms (𝑑1 , 𝑑2 , and 𝑑3 ). Examination of this figure shows that the largest moment arm is achieved when the handle of the wrench is at point 𝐷2 ; thus 𝛼 = −60◦ and 𝑑 = 𝑑2 = 550 mm. Equation (1) can now be solved to obtain 𝐹 =

−𝑀𝐴 𝑑

=

50 N⋅m = 90.91 N. 0.550 m

(2)

Part (b)

𝐹 Governing Equations & Computation

300 mm 𝑑1

𝐷2

𝑑2 𝐷3

𝐶 30◦

𝑑3

𝐴

Figure 3 Among the infinite number of possible positions for point 𝐷, three possible positions, denoted by 𝐷1 , 𝐷2 , and 𝐷3 , are shown.

ISTUDY

As with Part (a), we wish to maximize the moment arm 𝑑 in Eq. (1) where now force 𝐹 is always horizontal. The locus of possible locations for point 𝐷 is shown in Fig. 3, with three example locations illustrated (points 𝐷1 , 𝐷2 , and 𝐷3 ). Clearly, 𝑑 is largest when the handle of the wrench is vertical. For this position 𝛼 = 0◦ , the moment arm is 𝑑 = (0.250 m) sin 30◦ +0.300 m = 0.425 m. Solving Eq. (1) provides −𝑀𝐴 50 N⋅m = = 117.6 N. 𝐹 = (3) 𝑑 0.425 m

Discussion & Verification

Sketches such as Figs. 2 and 3 can be very helpful in problems where you must maximize or minimize the moment of a force about a particular point. Without the insight these figures offer, this problem would otherwise be more difficult.

ISTUDY

Section 4.1

215

Moment of a Force

Problems 𝑦

General instructions. Unless otherwise stated, in the following problems you may use a scalar approach, a vector approach, or a combination of these. 4

Problems 4.1 and 4.2

𝐴

Compute the moment of force 𝐹 about point 𝐵, using the following procedures.

3 𝐶 𝐹 = 25 lb 12 in.

𝑑

(a) Determine the moment arm 𝑑 and then evaluate 𝑀𝐵 = 𝐹 𝑑.

𝐷

60◦ 𝐵

(b) Resolve force 𝐹 into 𝑥 and 𝑦 components at point 𝐴 and use the principle of moments.

𝑥

Figure P4.1

(c) Use the principle of moments with 𝐹 positioned at point 𝐶. (d) Use the principle of moments with 𝐹 positioned at point 𝐷.

30◦

(e) Use the vector approach.

𝑦 𝐴

𝐹 = 25 lb

𝐶

Problems 4.3 and 4.4

12 in. 60◦

The cover of a computer mouse is hinged at point 𝐵 so that it may be clicked. Repeat Prob. 4.1 to determine the moment about point 𝐵.

𝐷

𝑥

Figure P4.2

𝑦

𝑦

𝑑 𝐵

𝐶 45◦

45◦

𝐹 = 3N

40 mm

𝐴

𝑑

𝐷 𝐵

40 mm 60 mm Figure P4.3

𝑥

𝐴

𝐹 = 3N 20 mm

𝐵 𝑑

𝐷

𝑥

𝐶 Figure P4.4 𝑦

Problem 4.5 An atomic force microscope (AFM) is a state-of-the-art device used to study the mechanical and topological properties of surfaces on length scales as small as the size of individual atoms. The device uses a flexible cantilever beam 𝐴𝐵 with a very sharp, stiff tip 𝐵𝐶 that is brought into contact with the surface to be studied. Due to contact forces at 𝐶, the cantilever beam deflects. If the tip of the AFM is subjected to the forces shown, determine the resultant moment of both forces about point 𝐴. Use both scalar and vector approaches.

12 𝜇m 𝐵

𝐴

5 nN

𝑥

𝐶

1.5 𝜇m

13 nN Figure P4.5

Problem 4.6

3m

A large lever 𝐴𝐵 has a 5 kN force applied to it where the line of action of this force passes through points 𝐵 and 𝐶. Determine the moment of the force about point 𝐴 for any position of the lever where 0◦ ≤ 𝛼 ≤ 180◦ , and plot the moment versus 𝛼.

𝐹 = 5 kN 𝐴 𝛼 1m 𝐵 Figure P4.6

𝐶 2m

216

Moment of a Force and Equivalent Force Systems 3 in.

𝐵 𝜃

Problem 4.7

𝐶

𝐴 4 in. 8 in.

Chapter 4

𝐷 𝑊

A circular disk rotates on a bearing at point 𝐴, and has a string 𝐵𝐶𝐷 attached to it. The pulley at 𝐶 is frictionless, and the string supports a weight 𝑊 = 10 lb. Determine the moment of the force in segment 𝐵𝐶 of the string about point 𝐴 for any position of the disk where 0◦ ≤ 𝜃 ≤ 360◦ , and plot the moment versus 𝜃. Neglect the size of the pulley.

Problem 4.8

Figure P4.7

The weight of a 22 lb oven door acts vertically through point 𝐷. The door is supported by a hinge at point 𝐴 and two springs that are symmetrically located on each side of the door. For the positions specified below, determine the force needed in each of the two springs if the resultant moment about point 𝐴 of the weight and spring forces is to be zero. 22 lb

(a) 𝛼 = 45◦ . (b) 𝛼 = 90◦ .

𝛼

𝐵

𝐷

(c) Based on your answers to Parts (a) and (b), determine the stiffness for the springs.

𝐶

8 in.

𝐴

12 in.

5 in.

10 in.

The ball of a trailer hitch is subjected to a force 𝐹 . If failure occurs when the moment of 𝐹 about point 𝐴 reaches 10,000 in. ⋅ lb, determine the largest value 𝐹 may have and the value of 𝛼 for which the moment about 𝐴 is the largest. Note that the value of 𝐹 you determine must not produce a moment about 𝐴 that exceeds 10,000 in.⋅lb for any possible value of 𝛼.

Figure P4.8

5 in.

𝐵

Problem 4.10

4 in. 𝐴

Repeat Prob. 4.9 if the moment at point 𝐵 may not exceed 5000 in.⋅lb.

6 in.

𝐶

𝛼

Problem 4.9

Problem 4.11

𝐹

The port hull of a catamaran (top view shown) has cleats at points 𝐴, 𝐵, and 𝐶. A rope having 100 N force is to be attached to one of these cleats. If the force is to produce the largest possible counterclockwise moment about point 𝑂, determine the cleat to which the rope should be attached and the direction the rope should be pulled (measured positive counterclockwise from the positive 𝑥 direction). Also, determine the value of 𝑀𝑂 produced. Assume all cleats, point 𝑂, and the rope lie in the same plane.

Figure P4.9 and P4.10 𝑦 2m

3m 𝐵

𝐴

𝐶 1.5 m 𝑥

𝑂

Problem 4.12 Frame 𝐴𝐵𝐶 has a frictionless pulley at 𝐶 around which a cable is wrapped. Determine the resultant moment about point 𝐴 produced by the cable forces if 𝑊 = 5 kN and

Figure P4.11

0.5 m

(a) 𝛼 = 0◦ . 3m

𝐵

𝐶

(b) 𝛼 = 90◦ . (c) 𝛼 = 30◦ .

𝛼

𝑊

𝑇

Figure P4.12 and P4.13

ISTUDY

4m

𝐴

Problem 4.13 Frame 𝐴𝐵𝐶 has a frictionless pulley at 𝐶 around which a cable is wrapped. If the resultant moment about point 𝐴 produced by the cable forces is not to exceed 20 kN⋅m, determine the largest weight 𝑊 that may be supported and the value of 𝛼 for which the moment about 𝐴 is the largest. Note that the value of 𝑊 you determine must not produce a moment about 𝐴 that exceeds 20 kN⋅m for any possible value of 𝛼.

ISTUDY

Section 4.1

217

Moment of a Force

0.5 m

Problem 4.14

3m

𝐵

𝐶

The load-carrying capacity of the frame of Prob. 4.13 can be increased by placing a counterweight 𝑄 at point 𝐷 as shown. The resultant moment about point 𝐴 due to the cable forces and 𝑄 is not to exceed 20 kN⋅m for any possible value of 𝛼.

𝛼 𝑊

(a) Determine the largest value of counterweight 𝑄. Hint: Let 𝑊 = 0 and determine 𝑄 so that 𝑀𝐴 = 20 kN⋅m.

2m

4m

𝑇

𝐷 𝑄

𝐴

(b) With the value of 𝑄 determined in Part (a), determine the largest weight 𝑊 that may be supported and the corresponding value of 𝛼. Figure P4.14

Problem 4.15

𝑦

A bolt whose shank lies on the 𝑦 axis is to be loosened. In Fig. P4.15(a), an open-end wrench is used where the handle lies along the 𝑥 axis. In Fig. P4.15(b), a ratchet wrench is used where the handle is parallel to the 𝑥 axis. If the force applied to the handle of the wrench acts in the −𝑧 direction, determine the moment of this force about point 𝑂 for each wrench. Speculate on which of the wrenches will be more effective in loosening the bolt. Explain.

(a)

Problem 4.16

(b)

8 in. 𝐴

𝑧

𝑥 𝑦 8 in.

𝐵

𝐹 = 30 lb 𝐴

Repeat Prob. 4.15 if the force applied to the handle of the wrench has direction angles 𝜃𝑥 = 72◦ , 𝜃𝑦 = 60◦ , and 𝜃𝑧 = 144◦ .

5 in.

𝑂 𝑧

Problem 4.17 A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the 𝑥 or 𝑦 axis. If 𝐹 = 200 N and 𝑃 = 300 N, determine the resultant moment of all forces about the

𝐹 = 30 lb

𝑂

𝑥

Figure P4.15 and P4.16 𝑎 𝑧

(a) 𝑧 axis.

𝑦 𝐹

(b) 𝑎 axis, which is parallel to the 𝑧 axis. 500 mm

𝑃

600 mm 700 mm

Problem 4.18 A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the 𝑥 or 𝑦 axis. If 𝑃 = 300 N, determine 𝐹 when the resultant moment of all forces about the 𝑧 axis is 𝑀𝑧 = −100 N⋅m.

300 mm

𝐹

Figure P4.17–P4.19

Problem 4.19

𝑧

Repeat Prob. 4.18 if the resultant moment of all forces about the 𝑎 axis is 𝑀𝑎 = −100 N⋅m.

7m 12 m

𝐸

Problem 4.20 10 m

𝐶

4m

𝐹

The structure shown supports a force 𝐹 = 20 kN. Use the vector approach with the position vectors cited below to determine the moment of the force about point 𝐴.

𝐷

6m

𝐴

(a) Use the position vector from point 𝐴 to point 𝐷, namely 𝑟⃗𝐴𝐷 . (b) Use the position vector from point 𝐴 to point 𝐸, namely 𝑟⃗𝐴𝐸 .

𝑃 200 mm 𝑥

𝑥 Figure P4.20

𝐵

𝑦

218

Moment of a Force and Equivalent Force Systems

𝑧

Chapter 4

Problem 4.21

𝑂

𝐹⃗𝐵

𝐵

𝑥 𝐹⃗𝐴

𝐴 𝑦

Figure P4.21

The aircraft shown has three engines (the engine on the right-hand wing is hidden in the figure). During a test, the right-hand engine is shut down. If the thrust produced by ̂ lb, and points 𝐴 and engines 𝐴 and 𝐵 are 𝐹⃗𝐴 = 30,000 𝚤̂ lb and 𝐹⃗𝐵 = (29,700 𝚤̂ + 4000 𝑘) ⃗ ⃗ 𝐵 lie on the lines of action of 𝐹𝐴 and 𝐹𝐵 , respectively, with coordinates 𝐴(6, 25, −7) ft and 𝐵(−80, 0, 20) ft, determine the resultant moment of the two engine thrusts about the origin of the coordinate system, point 𝑂.

𝑦

Problem 4.22

12 in. 20 in.

16 in.

𝑇 = 50 lb 𝑧

18 in.

𝐵

𝐶

(a) point 𝐵. (b) point 𝑂.

𝑂 𝑃 = 100 lb

𝐷 15 in.

Structure 𝑂𝐵𝐶𝐷 is built in at point 𝑂 and supports a 50 lb cable force at point 𝐶 and 100 lb and 200 lb vertical forces at points 𝐵 and 𝐷, respectively. Using a vector approach, determine the moment of these forces about

𝐴

𝑥

Problem 4.23

𝐹 = 200 lb

Repeat Prob. 4.22, using a scalar approach. Figure P4.22 and P4.23

Problem 4.24

𝑧

Structure 𝑂𝐴𝐵 is built in at point 𝑂 and supports forces from two cables. Cable 𝐶𝐴𝐷 passes through a frictionless ring at point 𝐴, and cable 𝐷𝐵𝐸 passes through a frictionless ring at point 𝐵. If the force in cable 𝐶𝐴𝐷 is 250 N and the force in cable 𝐷𝐵𝐸 is 100 N, use a vector approach to determine

𝐵

120 mm

(a) the moment of all cable forces about point 𝐴. 𝐴

(b) the moment of all cable forces about point 𝑂.

𝑦

𝐸

80 mm

Problem 4.25

150 mm 𝑂 𝐶

150 mm

60 mm

Repeat Prob. 4.24, using a scalar approach.

𝐷 𝑥

Problem 4.26 Figure P4.24 and P4.25 𝑧

9 ft

Structure 𝑂𝐴𝐵𝐶 is built in at point 𝑂 and supports forces from two cables. Cable 𝐸𝐴𝐷 passes through a frictionless ring at point 𝐴, and cable 𝑂𝐶𝐺 passes through a frictionless ring at point 𝐶. If the force in cable 𝐸𝐴𝐷 is 800 lb and the force in cable 𝑂𝐶𝐺 is 400 lb, determine

𝐶

𝐵

(a) the moment of forces from cable 𝑂𝐶𝐺 about point 𝐵. (b) the moment of all cable forces about point 𝐴.

6 ft

𝐴

(c) the moment of all cable forces about point 𝑂.

𝑦 𝐷

6 ft 𝑂

Problem 4.27

8 ft 𝐸 8 ft 12 f t

Figure P4.26 and P4.27

ISTUDY

Repeat Prob. 4.26, using a scalar approach. 𝐺 𝑥

ISTUDY

Section 4.1 𝑦

Problem 4.28

400 mm

Structure 𝑂𝐴𝐵 is built in at point 𝑂 and supports two forces of magnitude 𝐹 parallel to the 𝑦 and 𝑧 axes. If the magnitude of the moment about point 𝑂 cannot exceed 1.0 kN⋅m, determine the largest value 𝐹 may have.

𝐹 𝐴 𝑂

Problem 4.29

𝐵

𝐹 Figure P4.28 and P4.29

𝑀𝑦2 + 𝑀𝑧2 ≤ 0.8 kN⋅m).

Problem 4.30

𝑧 3 kN

Forces of 3 kN and 200 N are exerted at points 𝐵 and 𝐶 of the main rotor of a helicopter, and force 𝐹 is exerted at point 𝐷 on the tail rotor. The 3 kN forces are parallel to the 𝑧 axis, the 200 N forces are perpendicular to the main rotor and are parallel to the 𝑥𝑦 plane, 𝐹 is parallel to the 𝑦 axis, and 𝛼 = 45◦ . (a) Determine the value of 𝐹 so that the 𝑧 component of the moment about point 𝑂 of all rotor forces is zero.

200 N

(c) If 𝛼 is different than 45◦ , do your answers to Parts (a) and (b) change? Explain.

𝐵 𝛼

𝐴

1.8 m 𝑥

𝐷 𝐹

𝑂

1.2 m 5m 𝑦

Figure P4.30 𝑧

⃗ The moment of force 𝐹⃗ about point 𝐴 can be computed using 𝑀 ⃗𝐴𝐵 × 𝐹⃗ or using 𝐴,1 = 𝑟 ⃗ = 𝑟⃗ × 𝐹⃗ , where 𝐵 and 𝐶 are points on the line of action of 𝐹⃗ . Noting that 𝑀

𝐵

⃗ ⃗ 𝑟⃗𝐴𝐶 = 𝑟⃗𝐴𝐵 + 𝑟⃗𝐵𝐶 , show that 𝑀 𝐴,1 = 𝑀𝐴,2 .

𝑟⃗𝐴𝐶

𝑟⃗𝐴𝐵

Problem 4.32

𝑦 𝐴

𝑥 Figure P4.31

𝐹⃗ 𝐵

𝑥

𝐶

𝐹⃗

𝐴𝐶

𝑧

3m

200 N 𝐶

Problem 4.31

⃗ and 𝑀 ⃗ . If 𝑀 ⃗ =𝑀 ⃗ , Consider the moment of force 𝐹⃗ about points 𝐴 and 𝐵, namely 𝑀 𝐴 𝐵 𝐴 𝐵 what can be said about the locations of points 𝐴 and 𝐵?

3m

3 kN

(b) Using the value of 𝐹 found in Part (a), determine the resultant moment of all rotor forces about point 𝑂.

𝐴,2

𝑥 90 mm

𝑧

Repeat Prob. 4.28 if the 𝑥 component of the moment (torsional component) at point 𝑂 may not exceed 0.5 kN⋅m and the resultant of the 𝑦 and 𝑧 components (bending components) √ may not exceed 0.8 kN⋅m (i.e.,

219

Moment of a Force

𝑦

𝑧 new power transmission line

𝐴 Figure P4.32

𝐷 𝐶

Problem 4.33 𝐵

An electrical power transmission line runs between points 𝐴, 𝐵, and 𝐶, as shown, and a new power transmission line between points 𝐵 and 𝐷 is to be constructed. The coordinates of points 𝐴, 𝐶, and 𝐷 are 𝐴(6000, 1000, 500) ft, 𝐶(700, 4500, 900) ft, and 𝐷(2000, 400, 1400) ft. Use the tip given in the margin note on p. 211 to determine the smallest length possible for the new power line. Assume that power lines 𝐴𝐵𝐶 and 𝐵𝐷 are straight.

𝐴 existing power transmission line 𝑥 Figure P4.33

𝑦

220

Chapter 4

Moment of a Force and Equivalent Force Systems

4.2 𝑧

𝐶 𝐹⃗ 𝐴 𝑥

𝐵 𝑦

Figure 4.11 The steering wheel in the Ferrari 250 GTO (see the discussion beginning on p. 203), where the driver applies a force 𝐹⃗ at point 𝐶 to turn the steering wheel.

Moment of a Force About a Line

We begin this section by summarizing the main result of the previous section: the moment of a force about a point (usually called simply the moment) is the tendency of the force to cause twisting of the point about which the moment is evaluated. As such, the moment of a force about a point is a vector quantity. In contrast to this, very often it is useful or necessary to determine the moment of a force about a line or about a specific direction. The moment of a force about a line is the tendency of the force to cause twisting about the line. To illustrate, we reconsider the steering wheel from the Ferrari sports car, shown again in Fig. 4.11, except that now the direction of the force 𝐹⃗ applied by the driver’s hand is arbitrary. Clearly, the effectiveness of 𝐹⃗ to turn the steering wheel depends greatly on the orientation of 𝐹⃗ . In fact, if the driver pushes on the steering wheel so that 𝐹⃗ is in the same direction as the steering column 𝐴𝐵 (i.e., 𝐹⃗ = 𝐹 𝑟⃗𝐴𝐵 ∕|⃗𝑟𝐴𝐵 | where 𝑟⃗𝐴𝐵 is the position vector from point 𝐴 to 𝐵), the steering wheel will have no tendency to turn! Problem 4.34 explores this steering wheel further. The moment of a force about a line is defined to be the component of the moment that is in the direction of the line. If the line happens to be parallel to a coordinate axis, then the answer is easily obtained, as described below. If the line has a more general direction, then obtaining the answer is more involved, and both vector and scalar approaches can be used as follows. 𝑧

𝑧

𝑧

𝐷

𝐷 𝐴 𝐵

𝑥

⃗ 𝑂 𝐹

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝑂 𝑂𝐶 𝑂

𝑦

𝐴 𝐵

𝐶

𝑎

𝑥

𝑎

𝐹⃗

𝑂

𝑦

𝑃

𝐵

𝑥

𝑢̂ 𝑎

(a)

𝐹⃗

⃗ 𝑀𝑎 , 𝑀 𝑎 𝐴

𝑟⃗𝑂𝐶 𝐶

Figure 4.12 Hinged door supported by a rope.

ISTUDY

𝐷 ⃗ 𝑀 𝑂

𝑦

𝑟⃗𝑂𝐶 𝐶 𝑀𝑎 = 𝑟⃗𝑂𝐶 × 𝐹⃗ ⋅ 𝑢̂ ⃗ = 𝑀 𝑢̂ 𝑀 𝑎 𝑎 (b)

Figure 4.13. Vector approach to determine the moment of a force about a line 𝑎. (a) The moment of force 𝐹⃗ about some convenient point on line 𝑎 (point 𝑂 is selected here) is evaluated. (b) The dot product is then used to determine the component of the moment vector in the 𝑎 direction, 𝑀𝑎 .

Vector approach Consider the example shown in Fig. 4.12, where a door with rectangular shape is hinged along an axis with direction 𝑎 lying in the 𝑥𝑧 plane. The door is supported by a cable between points 𝐶 and 𝐷, where the tensile force in the cable has magnitude 𝐹 . It will often be necessary to determine the moment of 𝐹 about a direction such as the line passing through the hinges. To use a vector approach, first we find the moment at some convenient point along line 𝑎: points 𝑂, 𝐴, and 𝐵 are likely choices, but any point 𝑃 on line 𝑎, as shown in Fig. 4.13(a), could be used (this statement is ⃗ = 𝑟⃗ × 𝐹⃗ . Then, proved in Prob. 4.43). If we use point 𝑂, we then evaluate 𝑀 𝑂 𝑂𝐶

ISTUDY

⃗ acting in the direction of line 𝑎, we use the dot to determine the component of 𝑀 𝑂 ⃗ ⋅ 𝑢, product to evaluate 𝑀𝑎 = 𝑀 𝑂 ̂ where 𝑢̂ is a unit vector in the direction of line 𝑎, as shown in Fig. 4.13(b). To express the moment about line 𝑎 as a vector, we simply ⃗ = 𝑀 𝑢. write 𝑀 𝑎 𝑎 ̂ If you prefer, you may use the scalar triple product so that the cross product and dot product are evaluated simultaneously to yield 𝑀𝑎 .

𝑧 𝐷

Scalar approach

𝑂

𝐴

𝑥

𝐹‖

𝐶

𝑎

Figure 4.14 Scalar approach to determine the moment of a force about line 𝑎.

𝑦 𝐹

𝐹⊥ ,1

𝑧 𝑎

𝑑2 𝐹⊥ ,2

e1

𝑥 𝑑1

𝐹‖ ,1 𝐹‖ ,2

pl

• When you evaluate the moment of a force about a line using the scalar approach, values for the perpendicular component of the force and the moment arm are not unique, and these will depend on the particular plane you select. The only requirement is that the plane you use must contain the line about which you want to determine the moment. For example, to determine the moment of force 𝐹 about line 𝑎 in Fig. 4.15, two possible planes are shown. Clearly the perpendicular components of 𝐹 are different for the two planes (𝐹⊥,1 ≠ 𝐹⊥,2 ), and the moment arms are also different (𝑑1 ≠ 𝑑2 ). Nonetheless, the moment of 𝐹 about line 𝑎 is the same, so that 𝑀𝑎 = 𝐹⊥,1 𝑑1 = 𝐹⊥,2 𝑑2 .

𝑑

𝐵

• If line 𝑎 is parallel to the 𝑥 axis, and the positive 𝑎 and 𝑥 directions are the same, then the moment about line 𝑎 is simply the 𝑥 component of the moment at any point on line 𝑎. If line 𝑎 is parallel to the 𝑥 axis, and the positive 𝑎 and 𝑥 directions are opposite, then the moment about line 𝑎 is the negative of the 𝑥 component of the moment at any point on line 𝑎. Similar remarks apply if 𝑎 is parallel to the 𝑦 or 𝑧 axis. • The moment about line 𝑎 in the example of Fig. 4.12 can be obtained by evalu⃗ ⋅ 𝑢̂ where the moment 𝑀 ⃗ at any point 𝑃 along line 𝑎 is used. ating 𝑀𝑎 = 𝑀 𝑃 𝑃 ⃗ is usuWhile the same value 𝑀𝑎 results for any point 𝑃 , the moment 𝑀 𝑃 ally different for different points along line 𝑎. Thus, for Fig. 4.13(a), in general ⃗ ≠𝑀 ⃗ ≠𝑀 ⃗ ≠𝑀 ⃗ , however 𝑀 ⃗ ⋅ 𝑢̂ = 𝑀 ⃗ ⋅ 𝑢̂ = 𝑀 ⃗ ⋅ 𝑢̂ = 𝑀 ⃗ ⋅ 𝑢. 𝑀 𝑂 𝐴 𝐵 𝑃 𝑂 𝐴 𝐵 𝑃 ̂

𝑦

𝐹⊥

pl an

Remarks

𝐹

e2

To use a scalar approach, we first resolve 𝐹 into perpendicular and parallel components 𝐹⊥ and 𝐹‖ , respectively, to a plane containing line 𝑎, as shown in Fig. 4.14. Next we evaluate the moment about line 𝑎 produced by the perpendicular component (the parallel component produces no moment about line 𝑎), namely, 𝑀𝑎 = 𝐹⊥ 𝑑, where the moment arm 𝑑 is the perpendicular (shortest) distance from line 𝑎 to the line of action of 𝐹⊥ . For problems with complex geometry, determining 𝐹⊥ and/or 𝑑 may be tedious.

221

Moment of a Force About a Line

an

Section 4.2

Figure 4.15 Two different planes, among an infinite number of possibilities, that may be used to determine the moment of force 𝐹 about line 𝑎. Observe that quantities such as the moment arm and the perpendicular component of 𝐹 depend on the particular plane that is chosen. However, regardless of what plane is chosen, the moment of 𝐹 about line 𝑎 is the same.

222

Moment of a Force and Equivalent Force Systems

End of Section Summary

𝑧 𝐹⃗ ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝑃 ⃗ ⋅ 𝑢̂ 𝑀 =𝑀

𝑟⃗

𝑎

𝑦 𝑃 𝑢̂

𝑃

= 𝑟⃗ × 𝐹⃗ ⋅ 𝑢̂

(a) vector approach

Vector approach:

1. Select a point 𝑃 at a convenient location on line 𝑎. Determine the moment of 𝐹⃗ ⃗ = 𝑟⃗ × 𝐹⃗ , where 𝑟⃗ is a position vector from 𝑃 to any point about 𝑃 , using 𝑀 𝑃 on the line of action of 𝐹⃗ .∗

𝑧 𝐹 𝐹⊥

𝐹‖

𝑀𝑎 = 𝐹 ⊥ 𝑑 𝑦

𝑑 𝑎

(b) scalar approach Figure 4.16 Vector and scalar approaches for determining the moment of a force about a line.

ISTUDY

In this section, the moment of a force about a line is defined to be the component of the moment that is in the direction of the line. The moment of a force about a line can be evaluated using vector and scalar approaches, as follows. To determine the moment 𝑀𝑎 of a force 𝐹 about a line (or direction) 𝑎 as shown in Fig. 4.16:

𝑎

𝑥

𝑥

Chapter 4

⃗ ⋅ 𝑢, 2. 𝑀𝑎 = 𝑀 𝑃 ̂ where 𝑢̂ is a unit vector in the direction of 𝑎. To express this ⃗ = 𝑀 𝑢. moment as a vector quantity, evaluate 𝑀 𝑎 𝑎 ̂ Note: Steps 1 and 2 may be combined to yield 𝑀𝑎 directly using the scalar triple product. Scalar approach:

1. Resolve 𝐹 into components 𝐹⊥ and 𝐹‖ that are perpendicular and parallel, respectively, to a plane containing line 𝑎. 2. 𝑀𝑎 = 𝐹⊥ 𝑑 where 𝑑 is the moment arm (shortest distance) between line 𝑎 and the line of action of 𝐹 . Note: 𝐹‖ produces no moment about 𝑎, so you may skip its evaluation altogether.

⃗ , especially for problems with simple an alternative to using the cross product to determine 𝑀 𝑃 ⃗ , to be followed geometry, you could use a scalar approach to determine the vector expression for 𝑀 𝑃 by taking the dot product as described in Step 2.

∗ As

ISTUDY

Section 4.2

223

Moment of a Force About a Line

E X A M P L E 4.5

Moment of a Force About a Line—Vector and Scalar Solutions

A door with rectangular shape is hinged along an axis having direction 𝑎 lying in the xz plane. The door is supported by a cable that has a tensile force 𝐹 = 100 N. Determine the moment of 𝐹 about line 𝑎.

𝑧 𝐷 dimensions in meters

SOLUTION

𝑂

𝐴

Road Map

Determining the moment of a force about a line is inherently a threedimensional problem, and while both vector and scalar approaches can always be used, the vector approach will usually be more straightforward and methodical. Nonetheless, both solution approaches are carried out.

𝐵

𝑥

We first select point 𝑂 as a convenient location on line 𝑎 to compute the moment of 𝐹 . Needed vectors are

Figure 1

−2 𝚤̂ + 𝚥̂ + 2 𝑘̂ 𝐹⃗ = (100 N) , 3 𝑟⃗𝑂𝐶 = (0.8 m)(cos 15◦ ) 𝚤̂ + (0.4 m) 𝚥̂ − (0.8 m)(sin 15◦ ) 𝑘̂ ̂ m. = (0.7727 𝚤̂ + 0.4 𝚥̂ − 0.2071 𝑘)

𝐶

0.8

0.8

15◦

Vector solution Governing Equations & Computation

2 12

𝐹 = 100 N 𝑦

𝑎

0.4

0.4

(1)

(2)

The moment of 𝐹⃗ about point 𝑂 is ̂ N⋅m. ⃗ = 𝑟⃗ × 𝐹⃗ = (33.57 𝚤̂ − 37.71 𝚥̂ + 52.42 𝑘) 𝑀 𝑂 𝑂𝐶

(3)

⃗ in Eq. (3) with a unit vector in the 𝑎 direction, Finally, we take the dot product of 𝑀 𝑂 ̂ to obtain 𝑢̂ 𝑎 = cos 15◦ 𝚤̂ − sin 15◦ 𝑘, ⃗ ⋅ 𝑢̂ = 18.86 N⋅m. 𝑀𝑎 = 𝑀 𝑂 𝑎

(4)

⃗ that acts in the 𝑎 direction, and hence the moment of 𝐹 about Thus, the portion of 𝑀 𝑂 the 𝑎 axis, is 18.86 N⋅m. The fact that this result is positive means the twisting action of 𝐹 about line 𝑎 is in the positive 𝑎 direction, as governed by the right-hand rule. Scalar solution

𝑧

Governing Equations & Computation

The geometry of this problem is complex enough that a scalar evaluation of the perpendicular component of 𝐹⃗ using trigonometry is tedious. Rather we will start with a vector approach to obtain the perpendicular component of 𝐹⃗ (the magnitude of this vector is 𝐹⊥ shown in Fig. 2) by taking the dot product of 𝐹⃗ ̂ This in Eq. (1) with a unit vector 𝑢̂ 𝑛 normal to the door, where 𝑢̂ 𝑛 = sin 15◦ 𝚤̂ + cos 15◦ 𝑘. provides 𝐹⊥ = 𝐹⃗ ⋅ 𝑢̂ 𝑛 = 47.14 N. (5) To calculate the moment of 𝐹⊥ about line 𝑎, the moment arm needed is the perpendicular distance from line 𝑎 to the line of action of 𝐹⊥ , namely, 𝑑 = 0.4 m. Thus, 𝑀𝑎 = 𝐹⊥ 𝑑 = (47.14 N)(0.4 m) = 18.86 N⋅m.

(6)

In the scalar approach, the sign for 𝑀𝑎 must be assigned manually. Examination of Fig. 2 shows that 𝐹⊥ produces twisting action about line 𝑎, according to the right-hand rule, that is in the positive 𝑎 direction. Thus, we conclude that 𝑀𝑎 = +18.86 N⋅m. Discussion & Verification

As expected, both solutions produce the same result.

𝐷 𝐹 𝐹⊥ 𝐵

𝑥

𝑂

𝐴 𝑑

𝐶

𝑦 𝐹‖

𝑎

Figure 2 Resolution of 𝐹⃗ into perpendicular and parallel components 𝐹⊥ and 𝐹‖ , respectively.

224

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.6

Moment of a Force About a Line—Vector and Scalar Solutions A piece of cardboard is bent along the edge of a table. The 𝑎 axis lies in the 𝑥𝑦 plane and force 𝐹 lies in the 𝑦𝑧 plane. Determine the moment of 𝐹 about the 𝑎 axis.

𝑧 𝐹 = 5 lb 2 ft

SOLUTION

65◦

𝑂

𝑦

𝐵

30◦

𝑥

Road Map

Both vector and scalar solutions are possible, and both are illustrated.

Vector solution Governing Equations & Computation With point 𝑂 selected as a convenient location,

𝑎

the necessary vectors, as shown in Fig. 2, are

Figure 1

𝑟⃗𝑂𝐵 = 2 𝚥̂ ft, ̂ 𝐹⃗ = (5 lb)(− cos 65◦ 𝚥̂ − sin 65◦ 𝑘). 𝑧

𝑂

𝐹⃗ 𝑟⃗𝑂𝐵

⃗ = 𝑟⃗ × 𝐹⃗ = (−9.063) 𝚤̂ ft⋅lb. 𝑀 𝑂 𝑂𝐵 𝑦

30◦ 𝑢̂

𝑎

𝑥

(2)

The moment of 𝐹⃗ about point 𝑂 is

65◦

𝐵

(1)

Figure 2 Force and position vectors to be used in a vector solution.

(3)

Finally, using the unit vector 𝑢̂ = sin 30◦ 𝚤̂ + cos 30◦ 𝚥̂ to describe the 𝑎 direction, the moment about line 𝑎 is ⃗ ⋅ 𝑢̂ = −4.532 ft⋅lb. 𝑀𝑎 = 𝑀 (4) 𝑂 The negative sign for 𝑀𝑎 means the twisting direction of 𝐹 about line 𝑎, according to the right-hand rule, is the negative 𝑎 direction. If desired, the moment of 𝐹 about line 𝑎 can be stated as a vector by evaluating ⃗ = 𝑀 𝑢̂ 𝑀 𝑎 𝑎 = (−4.532 ft⋅lb)(sin 30◦ 𝚤̂ + cos 30◦ 𝚥̂) = (−2.266 𝚤̂ − 3.924 𝚥̂) ft⋅lb.

𝑧

𝐹⊥ = (5 lb)(sin 65◦ )

Scalar solution Governing Equations & Computation We resolve 𝐹 into perpendicular and parallel

2 ft 𝐵

𝑂

65◦

30◦ 𝑥

𝑑 = (2 ft)(sin 30◦ )

𝑦

𝐹‖ = (5 lb)(cos 65◦ )

components, as shown in Fig. 3, and determine the moment arm 𝑑 as the perpendicular distance from the 𝑎 axis to the line of action of 𝐹⊥ . Then 𝑀𝑎 = 𝐹⊥ 𝑑 = (5 lb)(sin 65◦ )(2 ft)(sin 30◦ ) = 4.532 ft⋅lb.

𝑎

Figure 3 Components of force and moment arms to be used in a scalar solution.

ISTUDY

(5)

(6)

In the scalar approach, we must manually provide the sign for our result. Examining Fig. 3, we observe that the twisting action of 𝐹⊥ about line 𝑎 is in the negative 𝑎 direction according to the right-hand rule. Hence, we conclude 𝑀𝑎 = −4.532 ft⋅lb. Discussion & Verification

As expected, both solutions produce the same result.

ISTUDY

Section 4.2

225

Moment of a Force About a Line

E X A M P L E 4.7

Moment of a Force About a Line—Vector Solution

The axle and steering linkage for the front wheel of an off-road vehicle is shown. Force 𝐹 causes the assembly to rotate about line 𝑎 so that the vehicle can be steered. Point 𝐴 is located at (160, −20, 100) mm. Determine the force 𝐹 needed to produce a moment about line 𝑎 of 10 N⋅m if (a) line 𝑎 lies in the 𝑦𝑧 plane and has direction angle 𝜃𝑧 = 10◦ .

𝑧 𝑎 𝐹

(b) line 𝑎 coincides with the 𝑧 axis. 8

𝐴

9

𝑦

12

SOLUTION

𝑂

Road Map

A vector solution is straightforward and will be used here. To determine the moment of 𝐹 about line 𝑎, we will first determine the moment of 𝐹 about a convenient point on line 𝑎 (we will select point 𝑂, and the result of this is a vector). We will then use the dot product to determine the portion of this moment vector that acts in the 𝑎 direction, and according to the problem statement, this must be 10 N⋅m. From this the value of 𝐹 may be determined.

𝑥 Figure 1 𝑧 𝑎

Part (a)

To find the moment of 𝐹 about line 𝑎, a position vector from some point along 𝑎 to some point along the line of action of 𝐹 is needed, and 𝑟⃗𝑂𝐴 , as shown in Fig. 2, is an obvious choice. The necessary vectors are

⃗ 𝑀 𝑂

Governing Equations & Computation

̂ mm, 𝑟⃗𝑂𝐴 = (160 𝚤̂ − 20 𝚥̂ + 100 𝑘)

−8 𝚤̂ − 12 𝚥̂ + 9 𝑘̂ . 𝐹⃗ = 𝐹 17

𝐹⃗ 𝐴

(1)

𝑟⃗𝑂𝐴

The moment of 𝐹⃗ about point 𝑂 is then ̂ mm. ⃗ = 𝑟⃗ × 𝐹⃗ = 𝐹 (60.00 𝚤̂ − 131.8 𝚥̂ − 122.4 𝑘) 𝑀 𝑂 𝑂𝐴

⃗ ⋅ 𝑢̂ = −𝐹 (143.4 mm). 𝑀𝑎 = 𝑀 𝑂 𝑎

(3)

The negative sign in the above result indicates a positive force 𝐹 produces a moment about line 𝑎 that is in the negative 𝑎 direction. To finish this problem, the moment about line 𝑎 is required to have magnitude 10 N⋅m = 10, 000 N⋅mm. Thus, using the absolute value of Eq. (3), we obtain 𝐹 (143.4 mm) = 10, 000 N⋅mm

⇒

𝐹 = 69.75 N.

(4)

Part (b)

Assuming the orientation of 𝐹 and the location of point 𝐴 are unchanged when line 𝑎 coincides with the 𝑧 axis, the moment of 𝐹 about point 𝑂 is unchanged and is given by Eq. (2). The unit vector 𝑢̂ used in Eq. (3) becomes ̂ and reevaluation of Eq. (3) simply provides the 𝑧 component of 𝑀 ⃗ , which is 𝑢̂ = 𝑘, 𝑂 Governing Equations & Computation

𝑀𝑎 = 𝑀𝑂𝑧 = −𝐹 (122.4 mm).

(5)

Requiring the absolute value of the above result to be equal to 10,000 N⋅mm provides 𝐹 = 81.73 N.

𝑦 𝑂

(2)

⃗ in the direction 𝑎 is given by the dot product between 𝑀 ⃗ and a The component of 𝑀 𝑂 𝑂 ̂ unit vector in the direction 𝑎, 𝑢̂ 𝑎 = sin 10◦ 𝚥̂ + cos 10◦ 𝑘:

Discussion & Verification

⃗ = 𝑀 𝑢̂ 𝑀 𝑎 𝑎

(6)

The main difference between the two parts of this example is the orientation of line 𝑎. In Part (b), where line 𝑎 is parallel to the 𝑧 axis, the dot product is a trivial operation that produces the 𝑧 component of the moment vector. The first remark on p. 221 gives additional comments.

𝑥 Figure 2 Force and position vectors 𝐹⃗ and 𝑟⃗𝑂𝐴 , respectively, used to determine the moment of 𝐹⃗ about point 𝑂 and the moment of 𝐹⃗ about line 𝑎.

226

Chapter 4

Moment of a Force and Equivalent Force Systems

Problems General instructions. Unless otherwise stated, in the following problems you may use a scalar approach, a vector approach, or a combination of these. Problem 4.34 The steering wheel of a Ferrari sports car has circular shape with 190 mm radius, and it lies in a plane that is perpendicular to the steering column 𝐴𝐵. Point 𝐶, where the driver’s hand applies force 𝐹⃗ to the steering wheel, lies on the 𝑦 axis with 𝑦 coordinate 𝑦𝐶 = −190 mm. Point 𝐴 is at the origin of the coordinate system, and point 𝐵 has the coordinates 𝐵 (−120, 0, −50) mm. Determine the moment of 𝐹⃗ about line 𝐴𝐵 of the steering column if (a) 𝐹⃗ has 10 N magnitude and lies in the plane of the steering wheel and has orientation such that its moment arm to line 𝐴𝐵 is 190 mm. Also determine the vector expression for this force. 18 𝚤̂ − 3 𝚥̂ + 14 𝑘̂ (b) 𝐹⃗ = (10 N) . 23 −12 𝚤̂ − 5 𝑘̂ . (c) 𝐹⃗ = (10 N) 13

𝑧

𝐶 𝐹⃗ 𝐵

𝐴

190 mm 𝑥

𝑦

Figure P4.34

Problem 4.35 A rectangular piece of sheet metal is clamped along edge 𝐴𝐵 in a machine called a brake. The sheet is to be bent along line 𝐴𝐵 by applying a 𝑦 direction force 𝐹 . Determine the moment of this force about line 𝐴𝐵 if 𝐹 = 200 lb. Use both vector and scalar approaches. 𝑦 20 in. 𝑂

𝐹 𝐴

16 in.

𝐵

𝑥 15 in.

𝐶

𝑧 Figure P4.35 and P4.36

Problem 4.36 In Prob. 4.35 determine 𝐹 if the moment about line 𝐴𝐵 is to be 2000 in.⋅lb. 𝑧

Problem 4.37

dimensions in mm 350 𝑂 𝑥

𝑦

30◦ 350 𝐶 𝐹

Figure P4.37 and P4.38

ISTUDY

𝐴

In the pipe assembly shown, points 𝐵 and 𝐶 lie in the 𝑥𝑦 plane, and force 𝐹 is parallel to the 𝑧 axis. If 𝐹 = 150 N, determine the moment of 𝐹 about lines 𝑂𝐴 and 𝐴𝐵. Use both vector and scalar approaches.

𝐵 300

Problem 4.38 In the pipe assembly shown, points 𝐵 and 𝐶 lie in the 𝑥𝑦 plane and force 𝐹 is parallel to the 𝑧 axis. If a twisting moment (torque) of 50 N⋅m will cause a pipe to begin twisting in the flange fitting at 𝑂 or at either end of the elbow fitting at 𝐴, determine the first fitting that twists and the value of 𝐹 that causes it.

ISTUDY

Section 4.2

Moment of a Force About a Line

Problem 4.39 A chocolate candy bar is molded in the shape of a thin rectangular slab with deep grooves. The deep grooves are represented in the figure by the dashed lines 𝐴𝐵, 𝐴𝐷, and 𝐵𝐶, and these grooves allow small bite-size pieces of candy to be easily broken off. Groove 𝐴𝐵 breaks when the moment about this line is 5 in.⋅lb, and grooves 𝐴𝐷 and 𝐵𝐶 break when the moment about these lines is 4 in.⋅lb. The candy bar is held along the edge of a table, as shown, and a force 𝐹 acting in the −𝑧 direction is applied at point 𝐷. Determine the force 𝐹 required to break off a piece of candy and which of the grooves 𝐴𝐵, 𝐴𝐷, or 𝐵𝐶 will break. Neglect the thickness of the candy bar.

𝑧 𝐵

𝑦

𝐶

𝐴 𝑥

𝐷

𝐹

2 in. 3 in.

Problem 4.40 For the candy bar described in Prob. 4.39, the manufacturer wishes to design the strength of the grooves so that:

Figure P4.39 and P4.40

• Groove 𝐵𝐶 will break when a force 𝐹 = 3 lb is applied in the −𝑧 direction at point 𝐷, as shown in Fig. P4.40, and • Groove 𝐴𝐵 will break when 𝐹 = 0 and a force 𝑄 = 4 lb is applied in the −𝑧 direction at the midpoint of line segment 𝐶𝐷 (this force is not shown in Fig. P4.40). strength

If this design is possible, determine the strengths of grooves 𝐴𝐵 and 𝐵𝐶, namely 𝑀𝐴𝐵 strength and 𝑀𝐵𝐶 , where these have units of in.⋅lb.

Problems 4.41 and 4.42 Determine the moment of force 𝐹 about line 𝐴𝐵 as follows. ⃗ , and then determine the component (a) Determine the moment of 𝐹 about point 𝐴, 𝑀 𝐴 of this moment in the direction of line 𝐴𝐵. ⃗ , and then determine the component (b) Determine the moment of 𝐹 about point 𝐵, 𝑀 𝐵 of this moment in the direction of line 𝐴𝐵. ⃗ ,𝑀 ⃗ , and the moment about (c) Comment on differences and/or agreement between 𝑀 𝐴 𝐵 line 𝐴𝐵 found in Parts (a) and (b). Also comment on the meaning of the sign (positive or negative) found for the moment about line 𝐴𝐵. 𝑦 3

𝐶

̂ lb 𝐹⃗ = (10 𝚤̂ + 80 𝚥̂ − 40 𝑘)

4 mm

𝐶

2 𝐴

𝐹 = 2N

18 in. 20 in. 𝐴

15 in. 𝑥

4 mm

6

𝑧

𝑦

𝑧

𝐵

7 mm

𝐵

𝑥 Figure P4.41

Figure P4.42

𝑧

Problem 4.43

𝐵

The moment of force 𝐹⃗ about line 𝐴𝐵 can be computed using 𝑀𝐴𝐵,1 or 𝑀𝐴𝐵,2 where 𝑀𝐴𝐵,1 = (⃗𝑟𝐴𝐶 × 𝐹⃗ ) ⋅

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

,

𝑀𝐴𝐵,2 = (⃗𝑟𝐵𝐶 × 𝐹⃗ ) ⋅

𝑟⃗𝐴𝐵 |⃗𝑟𝐴𝐵 |

𝐴

𝐹⃗ 𝑦

,

where 𝐶 is a point on the line of action of 𝐹⃗ . Noting that 𝑟⃗𝐴𝐶 = 𝑟⃗𝐴𝐵 + 𝑟⃗𝐵𝐶 , show that 𝑀𝐴𝐵,1 = 𝑀𝐴𝐵,2 .

𝑥 Figure P4.43

𝐶

227

228

Chapter 4

Moment of a Force and Equivalent Force Systems Problems 4.44 and 4.45 𝑦

The system shown is commonly used for supporting a rural mailbox. It has the feature that the mailbox can swing about line 𝑎 if it is subjected to a 𝑧 direction force, preventing possible damage to the mailbox and supporting post. Line 𝑎 lies in the 𝑥𝑦 plane. The 150 N force acts in the −𝑦 direction, the 200 N force has the direction specified below, and the lines of action of each of these pass through point 𝐶. Determine the moment of the 150 N and 200 N forces about line 𝑎.

𝑎 12 𝐵 5

𝑧

Ike Plesha 𝐶

200 N

𝑥

𝐴

150 N

𝐴 (5, 0, 0) cm 𝐵 (10, 12, 0) cm 𝐶 (120, 18, 0) cm

Problem 4.44

The 200 N force acts in the 𝑧 direction.

Problem 4.45

The 200 N force has the direction angles 𝜃𝑥 = 108◦ , 𝜃𝑦 = 60◦ , and

𝜃𝑧 =

36◦ .

Figure P4.44 and P4.45 𝑦

Problems 4.46 and 4.47

𝐶

A flat-screen television is mounted on the wall of a room using bracket 𝐴𝐵𝐶, which is fixed in position. For comfortable viewing, the television may be rotated about line 𝐴𝐵. For the direction given below for the 20 N force at point 𝐷, determine the moment of this force about line 𝐴𝐵. Note: For the coordinates of point 𝐷 given, the top and bottom edges of the television are not parallel to the 𝑥 axis.

𝐵

𝐷 Problem 4.46 20 N

The 20 N force has the direction angles 𝜃𝑥 = 60◦ , 𝜃𝑦 = 72◦ , and 𝜃𝑧 =

144◦ . 𝐴 Michael Plesha

Problem 4.47 The 20 N force has negative 𝑧 component and acts perpendicular to the television screen (i.e., plane containing points 𝐴, 𝐵, and 𝐷).

Problems 4.48 and 4.49 𝑧

𝐴 (0, 150, 30) cm 𝐵 (0, 200, 60) cm 𝐶 (0, 200, 0) cm 𝐷 (−60, 160, 60) cm

Figure P4.46 and P4.47

ISTUDY

𝑥

The door shown provides outdoors access to the basement of a building. The door has hinges at points 𝐵, 𝐶, and 𝐷, and it weighs 80 lb with center of gravity at point 𝐺. The force 𝐹 acts perpendicular to the door. Problem 4.48

If 𝐹 = 50 lb, determine the resultant moment of 𝐹 and the 80 lb weight

about line 𝐵𝐶𝐷. Determine the value of 𝐹 if the resultant moment of 𝐹 and the 80 lb weight about line 𝐵𝐶𝐷 is to be −60 in.⋅lb.

Problem 4.49

𝐴 (34, 15, 5) in. 𝐵 (58, 5, 28) in. 𝐶 (34, 15, 28) in. 𝐷 (10, 25, 28) in. 𝐺 (34, 15, 16) in.

𝑦

𝐹 𝐷 𝐶

𝐺 𝐴 𝐵

𝑧 Figure P4.48 and P4.49

𝑥

ISTUDY

Section 4.2

229

Moment of a Force About a Line

Problem 4.50

𝑦

An automobile windshield wiper is actuated by a force 𝐹⃗ = (120 𝚤̂ − 50 𝚥̂) N. Determine the moment of this force about shaft 𝑂𝐵.

Problem 4.51 𝐵

In the windshield wiper of Prob. 4.50, if 𝐹⃗ has 130 N magnitude, determine the direction in which it should be applied so that its moment about shaft 𝑂𝐵 is as large as possible, and determine the value of this moment.

Problem 4.52

𝑂

𝑧

Wrench 𝐴𝐵 is used to twist a nut on a threaded shaft. Point 𝐴 is located 120 mm from point 𝑂, and line 𝑎 has 𝑥, 𝑦, and 𝑧 direction angles of 36◦ , 60◦ , and 72◦ , respectively. Point ̂ N. 𝐵 is located at (0, 300, 100) mm, and the force is 𝐹⃗ = (−80 𝚥̂ + 20 𝑘) ⃗ , and then find the component of this (a) Determine the moment of 𝐹⃗ about point 𝐴, 𝑀 𝐴 moment about line 𝑎, 𝑀𝑎 .

𝑟⃗𝑂𝐵

̂ mm 𝑟⃗𝑂𝐴 = (−60 𝚥̂ + 80 𝑘) ̂ = (20 𝚤̂ + 90 𝚥̂ + 60 𝑘) mm 𝑥 𝐹⃗

𝐴

Figure P4.50 and P4.51

⃗ , and then find the component of this (b) Determine the moment of 𝐹⃗ about point 𝑂, 𝑀 𝑂 moment about line 𝑎, 𝑀𝑎 . 𝑥

𝑎

𝐴 𝑂 𝑦 𝐵

𝐹⃗ 𝑧 Figure P4.52 𝑧

Problem 4.53 A trailer has a rectangular door 𝐷𝐸𝐺𝐻 hinged about edge 𝐺𝐻. If 𝐹⃗ = (−2 𝚤̂ + 5 𝚥̂ + ̂ lb, determine the moment of 𝐹 about edge 𝐺𝐻. 14 𝑘)

𝐴 (0, 28, 12) in. 𝐵 (24, 4, 12) in. 𝐶 (0, 4, 60) in.

Problem 4.54

𝐷 (−40, 24, 20) in. 𝐸 (−10, 24, 20) in. 𝐺 (−10, 8, 52) in. 𝐻 (−40, 8, 52) in.

In the trailer of Prob. 4.53, if 𝐹 = 15 lb, determine the direction in which it should be applied so that its moment about edge 𝐺𝐻 of the rectangular door is as large as possible, and determine the value of this moment.

𝑥

Problem 4.55 ⃗ = ( 𝚤̂ + 4 𝚥̂ + 8 𝑘) ̂ lb, A trailer has a triangular door 𝐴𝐵𝐶 hinged about edge 𝐵𝐶. If 𝑄 determine the moment of 𝑄 about edge 𝐵𝐶.

Problem 4.56 In the trailer of Prob. 4.55, if 𝑄 = 9 lb, determine the direction in which it should be applied so that its moment about edge 𝐵𝐶 of the triangular door is as large as possible, and determine the value of this moment.

Figure P4.53–P4.56

𝐻

𝐶 𝐺 𝑄 𝐵

𝐹 𝐷

𝐸

𝐴

𝑦

230

Chapter 4

Moment of a Force and Equivalent Force Systems 𝑦

Problem 4.57

𝑎

A poorly leveled crane rotates about line 𝑎, which lies in the 𝑥𝑦 plane and has a 𝑦 direction angle of 5◦ . Line 𝐴𝐵 lies in the 𝑥𝑧 plane. If the crane supports a weight 𝑊 = 5000 lb and 𝛼 = 45◦ , determine the moment of 𝑊 about line 𝑎. 𝐴 𝑊

𝑥

𝛼 ℎ

𝐵

ft = 40

Problem 4.58 If the poorly leveled crane of Prob. 4.57 can have any position 𝛼, where 0 ≤ 𝛼 ≤ 135◦ , determine the largest moment of the weight 𝑊 = 5000 lb about line 𝑎 and the position 𝛼 that produces this moment. Assume ℎ = 40 ft for any position 𝛼 (this requires the operator to slightly change the boom’s height and/or length as the crane rotates about line 𝑎).

𝑧

Figure P4.57 and P4.58

𝑧

Problems 4.59 and 4.60 𝐴

The rudder of an aircraft is shown where the rudder rotates about line 𝑂𝐵𝐴. The force 𝐹 at point 𝐶 acts in the −𝑥 direction. The distance between points 𝑂 and 𝐵 is 1.8 m and the distance between points 𝐵 and 𝐶 is 0.6 m.

𝐵 20◦

𝐶

𝛼

𝐹

When 𝛼 = 15◦ , the position vector from point 𝐵 to 𝐶 has the 𝑥, 𝑦, and 𝑧 direction cosines 0.2588, 0.9077, and −0.3304, respectively, and 𝐹 = 0.8 kN. Determine the moment of 𝐹 about line 𝑂𝐵𝐴. Problem 4.59

𝑦

𝑂

Problem 4.60 The position vector from 𝐵 to 𝐶 has the 𝑥, 𝑦, and 𝑧 direction cosines sin 𝛼, (cos 𝛼)(cos 20◦ ), and (− cos 𝛼)(sin 20◦ ), respectively, and for 0◦ ≤ 𝛼 ≤ 40◦ the force is 𝐹 = 3 kN sin 𝛼. Determine the moment of 𝐹 about line 𝑂𝐵𝐴. Express your answer in terms of 𝛼. Plot this moment versus 𝛼 over the range 0◦ ≤ 𝛼 ≤ 40◦ .

𝑥 Figure P4.59 and P4.60

Problems 4.61 and 4.62 𝑦

𝑃⃗

Beam 𝐴𝐵 has rectangular cross section where point 𝐴 is at the origin of the coordinate system and point 𝐵 has the coordinates 𝐵 (1.2, −0.3, 2.4) m. The vector 𝑟⃗1 = (1 𝚤̂ + 12 𝚥̂ + ̂ m is perpendicular to the axis 𝐴𝐵 of the beam, and is parallel to the thin dimension 1 𝑘) ̂ m is perpendicular to both of the cross section. The vector 𝑟⃗2 = (29.1 𝚤̂ − 1.2 𝚥̂ − 14.7 𝑘) the axis 𝐴𝐵 of the beam and to 𝑟⃗1 . The force 𝑃⃗ applied to point 𝐵 of the beam has the direction angles 𝜃𝑥 = 144◦ , 𝜃𝑦 = 72◦ , and 𝜃𝑧 = 60◦ .

𝑟⃗1

𝐴

𝑧

𝑟⃗2 𝑥 𝐵

Problem 4.61

If the beam will fail when the absolute value of the moment of 𝑃⃗ about 𝑟⃗1 is 4000 N⋅m, or when the absolute value of the moment of 𝑃⃗ about 𝑟⃗2 is 6000 N⋅m, determine the magnitude of 𝑃⃗ that will cause the beam to fail. Problem 4.62

Figure P4.61 and P4.62

ISTUDY

If the magnitude of 𝑃⃗ is 1000 N, determine the moment of this force

about 𝑟⃗1 and 𝑟⃗2 .

ISTUDY

Section 4.2

Moment of a Force About a Line

Problem 4.63 Three-wheel and four-wheel all-terrain vehicles (ATVs) are shown. For both ATVs, the combined weight of the driver and vehicle is 𝑊 = 2600 N. During a hard turn, the tendency for the ATV to tip is modeled∗ by the force 𝐹 , which acts in the negative 𝑦 direction. Determine the value of 𝐹 such that the resultant moment of 𝐹 and the weight about line 𝐴𝐵 is zero. Comment on which ATV is more prone to tip during hard turns. Note: One of these models of ATV is no longer manufactured because of its poor safety record.

sirtravelalot/Shutterstock

0.6 m 0.6 m

0.6 m 0.6 m

𝐵

𝐵 0.7 m

𝐹 𝑊

𝑦

0.9 m

𝑊

𝐴

0.7 m

𝐹 𝐴

𝑥

𝑥

𝑧 0.6 m 0.6 m

𝑧 0.6 m 0.6 m

𝑊

𝑊

𝐹

𝑦

𝐹

𝑦

0.8 m 𝐵

𝐴

𝑦

0.9 m

0.8 m 𝐴

Figure P4.63

∗ When

an ATV moves on a curved path, it is a dynamics problem. We can model this system as a statics problem by treating inertial effects as external forces, as we do in this problem. While this is a crude approximation, it is nevertheless a common modeling approach. You will see in dynamics how to more accurately formulate these problems.

231

232

4.3 𝐹 , 𝐹⃗ 𝐹 , −𝐹⃗

𝐶

𝐸 𝐴

𝐵

Figure 4.17 The steering wheel in the Ferrari 250 GTO (see the discussion beginning on p. 203), where the driver uses two hands to apply two forces of equal magnitude and opposite direction to the steering wheel. Such a force system is defined to be a couple.

ISTUDY

Chapter 4

Moment of a Force and Equivalent Force Systems

Helpful Information Vector notation. Notice in Fig. 4.17 that two types of vector notation are shown (these were described in Chapter 2 on p. 32). When forces are labeled using a scalar such as 𝐹 , then the direction of the arrow in the figure provides the direction of the force. When forces are labeled using a vector such as 𝐹⃗ , then the symbol 𝐹⃗ embodies both the magnitude and direction for the force. Hence, in this figure if the force at point 𝐶 is labeled as 𝐹⃗ , and if the force system is a couple, then the force at 𝐸 must be labeled as −𝐹⃗ .

Moment of a Couple

A couple is defined to be a system of two forces of equal magnitude and opposite direction and whose lines of action are separated by a distance. To understand better what a couple is, reconsider the steering wheel from the Ferrari sports car, shown again in Fig. 4.17, except that now the driver uses two hands to apply forces to turn the steering wheel. Of course, the driver could apply forces of different magnitudes and in any two different directions that he or she pleases. However, if the driver applies forces of equal magnitude and with opposite direction, as shown in Fig. 4.17, then the system of two forces is called a couple. The moment of a couple (sometimes also called a couple moment) is the moment produced by the couple; that is, the moment produced by two forces of equal magnitude and opposite direction. The moment of a couple can always be evaluated using the procedures of Section 4.1. However, the moment of a couple has some special features, and in addition to the methods of Section 4.1, this section discusses other approaches that may be used for evaluation. A couple that is applied to a body produces a moment, but does not apply any net force to the body. To help illustrate some of the implications of this, consider the air hockey table shown in Fig. 4.18(a), and imagine that a rectangular plate is resting motionless on the surface of the table.∗ If the two forces shown in Fig. 4.18(b) are applied to the plate, because the two forces have equal magnitude and opposite direction, there is no net force applied in the 𝑥 direction, and hence, there is no net motion of the plate in the 𝑥 direction. However, the two forces produce a moment, and due to this moment the plate will begin to spin. Thus, the essential feature of a couple is that it produces only a moment. Furthermore, when the moment has the proper magnitude and direction (as described below), then the two force systems shown in Fig. 4.18(b) and (c) are said to be equivalent, and if the plate can be idealized as being rigid, then the motion of the plate in Fig. 4.18(b) and (c) is identical. Section 4.4 discusses equivalent force systems in detail. 𝑧

𝑧 𝑑

𝑀 𝑦

𝑦 =

𝑥

𝐹

𝐹

𝑥

Great American Recreation Equipment

(a)

(b)

(c)

Figure 4.18. (a) An air hockey table. (b) A rectangular plate resting on the surface of the air hockey table is subjected to a couple; that is, two forces with equal magnitude and opposite direction, separated by a distance 𝑑. (c) The moment produced by the couple. The two force systems shown in (b) and (c) are equivalent when 𝑀 has the proper value and direction, as described in this section.

The moment of a couple can be evaluated using both vector and scalar approaches as follows. ∗ If

you are unfamiliar with an air hockey table, you can instead imagine the rectangular plate resting on any horizontal frictionless surface, such as a sheet of ice.

ISTUDY

Section 4.3

Moment of a Couple

233

Vector approach Referring to Fig. 4.19, consider a couple consisting of two parallel forces 𝐹⃗ and −𝐹⃗ . ⃗ of this couple is The moment 𝑀

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴𝐵

𝑧

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴𝐵 = 𝑟⃗𝐵𝐴 × (−𝐹⃗ ),

(4.11)

−𝐹⃗

𝐶

𝐴 𝑟⃗𝐴𝐵

where 𝑟⃗𝐴𝐵 and 𝑟⃗𝐵𝐴 are position vectors; 𝐴 is any point on the line of action of −𝐹⃗ ; and 𝐵 is any point on the line of action of 𝐹⃗ .

𝐹⃗

𝐵

𝑦 𝑥

To see that the two expressions in Eq. (4.11) are equal, note that 𝑟⃗𝐵𝐴 = −⃗𝑟𝐴𝐵 and ⃗ = 𝑟⃗ × (−𝐹⃗ ) = −⃗𝑟 × (−𝐹⃗ ) = 𝑟⃗ × 𝐹⃗ . thus, 𝑀 𝐵𝐴 𝐴𝐵 𝐴𝐵 When evaluating the moment of a couple using the vector approach, you will select one of the forces of the couple to be called 𝐹⃗ . Then the position vector 𝑟⃗ must start somewhere on the line of action of the other force and terminate somewhere on the line of action of 𝐹⃗ . It does not matter which of the two forces of the couple you choose to call 𝐹⃗ ; using the proper position vector as described here will result in the ⃗ same moment 𝑀.

Figure 4.19 Moment of a couple: vector description. 𝑀 = 𝐹𝑑

𝑧 𝐹

𝐶 𝐹

𝑑

Scalar approach

𝑦

Referring to Fig. 4.20, consider a couple consisting of two parallel forces having the same magnitude 𝐹 . The magnitude of the moment of this couple is 𝑀 = 𝐹 𝑑,

(4.12)

𝑥 Figure 4.20 Moment of a couple: scalar description.

where 𝑑 is the perpendicular (shortest) distance between the forces’ lines of action; and the direction of the moment is perpendicular to the plane containing the forces.

𝑧

Comments on the moment of a couple

−𝐹⃗

To see that the vector and scalar descriptions of the moment of a couple are valid and to understand some subtle features of a couple and the moment it produces, consider the forces shown in Fig. 4.21. The moment produced by these forces about point 𝐶 having arbitrary location is ⃗ = 𝑟⃗ × (−𝐹⃗ ) + 𝑟⃗ × 𝐹⃗ . 𝑀 𝐶 𝐶𝐴 𝐶𝐵

(4.13)

Noting that 𝑟⃗𝐶𝐵 = 𝑟⃗𝐶𝐴 + 𝑟⃗𝐴𝐵 , we see Eq. (4.13) becomes

𝐵 𝑟⃗𝐶𝐴 𝑥

𝐹⃗ 𝑦

𝑟⃗𝐶𝐵 𝐶

Figure 4.21 Vectors for computing the moment of forces 𝐹⃗ and −𝐹⃗ about point 𝐶. Because 𝐹⃗ and −𝐹⃗ are a couple, the moment they produce is seen to be the same regardless of where point 𝐶 is located.

⃗ = 𝑟⃗ × (−𝐹⃗ ) + (⃗𝑟 + 𝑟⃗ ) × 𝐹⃗ 𝑀 𝐶 𝐶𝐴 𝐶𝐴 𝐴𝐵 = 𝑟⃗𝐶𝐴 × (−𝐹⃗ ) + 𝑟⃗𝐶𝐴 × 𝐹⃗ + 𝑟⃗𝐴𝐵 × 𝐹⃗ = 𝑟⃗𝐴𝐵 × 𝐹⃗ .

𝐴

(4.14)

Notice from Eq. (4.14) that the moment about point 𝐶 depends only on the position vector between the two forces’ lines of action and does not depend on the location

234

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Chapter 4

Moment of a Force and Equivalent Force Systems

of 𝐶. Thus, regardless of where 𝐶 is located, the moment produced by the couple is the same. In other words, a couple produces a moment vector that has a specific magnitude and orientation, but the vector’s position in space (i.e., the location of its line of action) is arbitrary, because you can let point 𝐶 be located anywhere. Thus, we may consider the point at which a couple moment acts as being anywhere we choose. For the examples in Figs. 4.19 and 4.20, point 𝐶 is shown as being positioned between the two forces, although any other location for point 𝐶 is equally valid. For the foregoing reasons, the moment of a couple is often called a free vector, and this is an important difference compared to all other vectors we have encountered in this book. A free vector is a vector that has a specific magnitude and direction, but its position in space is arbitrary. For example, a force vector has a specific magnitude and direction and a specific location in space for its line of action. The moment of a force about a point also has the same characteristics. In contrast, the moment of a couple has the first two characteristics, namely, a specific magnitude and direction, but its position, or where its line of action is located in space, is arbitrary.

Equivalent couples Two couples are said to be equivalent if the moment vectors they produce are identical. Thus, the couples shown in the examples of Fig. 4.22(a)–(c) are all equivalent since they produce moments having the same magnitude and direction. 𝑧 𝐹 ∕2

𝑎 2𝑎

𝑦

𝑥

𝐹 ∕2 𝐹

𝐹

𝐹

𝐹 (a)

(b)

𝑀 = 𝐹𝑎

(d)

(c)

𝑀 = 𝐹𝑎 𝑀 = 𝐹𝑎

(e)

(f)

Figure 4.22. Equivalent couples and the moments they produce.

Equivalent force systems A couple and the moment it produces are examples of equivalent force systems. This important concept is explored in greater detail in the next section, but for the present we will simply state that equivalent force systems applied to a body or structure produce the same effects for many purposes. For example, consider the identical prisms of material shown in Fig. 4.22. The couple shown in Fig. 4.22(a) produces a moment with magnitude 𝑀 = 𝐹 𝑎 that acts in the 𝑧 direction. Since this moment is a free

ISTUDY

Section 4.3

Moment of a Couple

235

vector, it can be positioned anywhere on, or even off, of the prism of material, and three examples are shown in Fig. 4.22(d)–(f). Since the couples shown in Fig. 4.22(b) and (c) are equivalent to that in Fig. 4.22(a), the moments they produce can also be positioned as shown in Fig. 4.22(d)–(f). In summary, all six force systems shown in Fig. 4.22 are equivalent. If we imagine these prisms rest on a smooth frictionless horizontal surface, such as the air hockey table discussed earlier, and if these prisms are initially motionless, then all six prisms of material shown in Fig. 4.22 will undergo the same motion, namely, spin about an axis parallel to the 𝑧 direction that passes through the prism’s center of mass, and no net translation of the center of mass.

Resultant couple moment Because the moment produced by a couple is a free vector, if an object has more than one couple applied to it, the moment vector produced by each couple can simply be added to yield a resultant couple moment, which is also a free vector. This process is illustrated in Fig. 4.23. ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 2 2 2 𝑧 𝐹⃗2

−𝐹⃗2

⃗ 𝑀 1

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 1 1 1

⃗ 𝑀 2

⃗ =𝑀 ⃗ +𝑀 ⃗ 𝑀 1 2

𝑦 𝑥 −𝐹⃗1

𝐹⃗1 (c)

(b)

(a)

(d)

Figure 4.23. Addition of moments of couples to form a resultant couple moment. (a) Object ⃗ and 𝑀 ⃗ are produced by the couples where 𝑟⃗ is a position with two couples. (b) Moments 𝑀 1 2 1 vector from somewhere on the line of action of −𝐹⃗1 to somewhere on the line of action of 𝐹⃗1 , ⃗ and 𝑀 ⃗ are free vectors, they may be positioned tail to tail and similarly for 𝑟⃗2 . (c) Since 𝑀 1 2 ⃗ and 𝑀 ⃗ to form the resultant couple moment 𝑀. ⃗ at any location. (d) Addition of 𝑀 1

2

𝑦

𝑦

𝑦

𝑀 𝑀

Moments as free vectors Consider the moment applied to the structure shown in Fig. 4.24(a). Because a moment can always be thought of as being produced by a couple, even if the actual agency that produced it is different, all moments are free vectors. For many purposes, moments can be thought of as being positioned anywhere on the structure, or even off of it. Thus the moment 𝑀 in Fig. 4.24 can be positioned at any of the locations shown with “equivalent” effects, where the exact meaning of the word equivalent is discussed in Section 4.4.

𝑀 𝑥 (a)

𝑥 (b)

𝑥 (c)

Figure 4.24 Moments are free vectors and can be positioned anywhere on a rigid structure with “equivalent” results, where the full meaning of this is discussed in Section 4.4.

236

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Moment of a Force and Equivalent Force Systems

Chapter 4

End of Section Summary In this section, the moment of a couple is described. Some of the key points are as follows: • A couple is defined to be a system of two forces of equal magnitude and opposite direction and whose lines of action are separated by a distance. • The moment of a couple, or couple moment, is a vector and can be evaluated using a vector or scalar approach. • The moment of a couple is a free vector, meaning the line of action of the couple can be positioned anywhere on (or even off) an object. • A resultant couple moment is the sum of all couple moments applied to an object.

ISTUDY

Section 4.3

237

Moment of a Couple

E X A M P L E 4.8

Moment of a Couple for a Two-Dimensional Problem

A cantilever beam is subjected to a couple. Determine the moment of the couple.

𝑦

10 N

SOLUTION Road Map

The two forces have equal magnitude and opposite direction, so indeed they are a couple. Both scalar and vector approaches can be used to determine the moment of the couple, as follows. Solution 1 Governing Equations & Computation The moment of the couple can be determined

(1)

Solution 3 Governing Equations & Computation Although the use of Eq. (4.12) as demonstrated

in the preceding solutions will usually be convenient for two-dimensional problems, we may also evaluate the moment of the couple by summing moments about any convenient point we choose. Using, for example, point 𝑂 shown in Fig. 3 with counterclockwise being positive, we obtain

100 mm

= −440 N⋅mm.

4 10 N

𝑦

10 N 𝐴

𝑂

𝑥

𝑑

𝐵

10 N Figure 2 Forces and moment arm used to determine the moment of a couple. 𝑦

8N 6N

𝑂

𝐴

30 mm

100 mm

𝑥

30 mm

𝐵

6N 8N

(2)

Since we know 𝑀𝑂 is the resultant moment from two couples, we know that the same moment would be obtained if any other summation point had been used. On the other hand, if this approach is used for forces that are not couples, then the resultant moment will usually differ from point to point. Solution 4 Governing Equations & Computation A vector solution can also be employed. Using

Eq. (4.11) on p. 233 with 𝑟⃗𝐴𝐵 = (100 𝚤̂ − 60 𝚥̂) mm and 𝐹⃗ = (6 𝚤̂ − 8 𝚥̂) N, we obtain ⃗ = 𝑟⃗ × 𝐹⃗ = −440 𝑘̂ N⋅mm. 𝑀 𝐴𝐵

3

100 mm

𝑥

30 mm

Figure 1

100 mm

𝑀𝑂 = (6 N)(30 mm) + (8 N)(100 mm) + (6 N)(30 mm) − (8 N)(200 mm)

30 mm 𝐵

Solution 2 Governing Equations & Computation The 10 N forces are resolved into the 𝑥 and 𝑦

𝑀 = −(8 N)(100 mm) + (6 N)(60 mm) = −440 N⋅mm.

4 𝐴

𝑂

using the scalar approach in Eq. (4.12) on p. 233 as 𝑀 = −(10 N) 𝑑, where 𝑑 is the perpendicular distance between the lines of action of the forces as shown in Fig. 2, and the negative sign is included because the couple shown in Fig. 1 produces a clockwise moment. However, for this problem finding 𝑑 is tedious, so we will select a different solution approach.

components as shown in Fig. 3 to yield two couples. With 𝐹𝑥 = (10 N)(3∕5) = 6 N and 𝐹𝑦 = (10 N)(4∕5) = 8 N, we then use Eq. (4.12) to sum the moment from each couple, taking counterclockwise to be positive as usual, to obtain

3

(3)

The negative sign in the above result indicates the moment acts in the negative 𝑧 direction according to the right-hand rule, which corresponds to clockwise in Figs. 1–3. You may ⃗ is obtained if expressions for the other force of wish to verify that the same result for 𝑀 the couple and the associated position vector are used. Discussion & Verification As expected, all solutions agree. For this problem, solutions 2 and 3 were the quickest, and solution 4 was only slightly longer.

Figure 3 Resolution of the 10 N forces into components so that the resultant moment of two couples can be determined.

238

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.9

Moment of a Couple for a Three-Dimensional Problem

𝑧 ℎ 𝐹 𝐴

ℎ 𝐸 𝐵 𝐷

ℎ 𝐶

𝑦

A cube with edge lengths ℎ is subjected to a couple whose forces have magnitude 𝐹 . Edges of the cube are parallel to their respective coordinate directions, and the forces are parallel to the 𝑦 axis. Determine the moment of the couple. Express your answers in terms of parameters such as ℎ and 𝐹 .

SOLUTION

𝐹

Road Map

The two forces have equal magnitude and opposite direction, so indeed they are a couple. Although a vector solution will usually be the most straightforward for threedimensional problems, the geometry here is simple enough that a scalar solution can be used.

𝑥 Figure 1

Solution 1 Governing Equations & Computation The perpendicular distance between√ the lines of

𝑧

⃗ = 𝐹 ℎ(−̂𝚤 + 𝑘) ̂ 𝑀

𝑦

𝑥 Figure 2 Moment produced by the couple shown in Fig. 1.

ISTUDY

action of the forces is the distance between points 𝐵 and 𝐶, which is 𝑑 = ℎ 2. Hence, Eq. (4.12) on p. 233 provides √ 𝑀 = 𝐹 (ℎ 2), (1)

where the direction of the moment is perpendicular to the plane containing the two forces, as shown in Fig. 2. Solution 2 Governing Equations & Computation To use a vector solution we will use the force at

point 𝐶 and choose a position vector from points 𝐴 to 𝐶. Equation (4.11) on p. 233 then provides ⃗ = 𝑟⃗ × 𝐹⃗ = ℎ(−̂𝚤 + 𝚥̂ − 𝑘) ̂ × (−𝐹 𝚥̂) = 𝐹 ℎ(−̂𝚤 + 𝑘). ̂ 𝑀 (2) 𝐴𝐶

𝐶

This moment vector is sketched in its proper orientation in Fig. 2. Solution 3 Governing Equations & Computation In this alternative solution we use the force at

point 𝐴, where 𝐹⃗𝐴 = −𝐹⃗𝐶 , and choose a position vector from points 𝐶 to 𝐵, where we note that 𝐵 is on the line of action of the force at 𝐴. Thus, ̂ ̂ × (𝐹 𝚥̂) = 𝐹 ℎ(−̂𝚤 + 𝑘). ⃗ = 𝑟⃗ × 𝐹⃗ = ℎ(̂𝚤 + 𝑘) 𝑀 𝐶𝐵 𝐴

(3)

As expected, the result is the same as that obtained in Eq. (2). Solution 4 Governing Equations & Computation In this solution we sum moments about a point

that is not on either force’s line of action. Using point 𝐷, we write ⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ 𝑀 𝐷𝐶 𝐶 𝐷𝐴 𝐴 ̂ × (𝐹 𝚥̂) = (−ℎ 𝚤̂) × (−𝐹 𝚥̂) + ℎ(−̂𝚥 + 𝑘) ̂ = 𝐹 ℎ(−̂𝚤 + 𝑘),

(4)

which is the same result as those obtained earlier. You may wish to repeat this solution, summing moments about point 𝐸, or any other point, to find the same result. Solution 5

In this solution we add two new forces at point 𝐷, as shown in Fig. 3(a). The argument for why we may do this is that the two forces added at 𝐷 will have no net effect on the object if they have equal magnitude and opposite direction and they have the same line of action. Examining Fig. 3(a) shows that the object Governing Equations & Computation

ISTUDY

Section 4.3

Moment of a Couple 𝑧

𝑧

𝐹 ℎ 𝑘̂ 𝐸

𝐹⃗ 𝐴

𝐵

𝐶

−𝐹⃗

𝑦

𝑦 =

ℎ

ℎ

𝐷 𝑥

−𝐹⃗

𝐹⃗

𝑥

−𝐹 ℎ 𝚤̂ (b)

(a)

Figure 3. (a) Two forces of equal magnitude and opposite direction, and sharing the same line of action are introduced at point 𝐷, so that the object is now subjected to two couples. (b) The moment for each couple may be written by inspection.

is now subjected to two couples, where the forces 𝐹⃗ and −𝐹⃗ at points 𝐴 and 𝐷, respectively, are one couple and the forces −𝐹⃗ and 𝐹⃗ at points 𝐶 and 𝐷, respectively, are the other couple. The merit of this solution strategy is that the moment for each of these couples may easily be evaluated by inspection, yielding the results shown in Fig. 3(b). Summing the two moments in Fig. 3(b) yields ⃗ = −𝐹 ℎ 𝚤̂ + 𝐹 ℎ 𝑘̂ 𝑀 ̂ = 𝐹 ℎ(−̂𝚤 + 𝑘).

Discussion & Verification

(5)

As expected, all solutions agree, except that Eq. (1) gives only the magnitude of the moment, whereas the other solutions are vectors and hence also provide the direction of the moment. Solution 4 does not require that the two forces involved be a couple. If you did not recognize that the two forces were a couple, then you would expect Eq. (4) to be the moment at point 𝐷 only. Because the two forces are a couple, the result in Eq. (4) is obtained for any other moment point that might be used.

239

240

E X A M P L E 4.10

Resultant Couple Moment 𝑃

𝑧 𝛼

𝑃

𝐵

𝛼 𝐴 𝐴 (10, 20, 36) in. 𝐵 (−10, 20, 36) in. 𝑦

𝐹 𝐹

𝐸

𝐷

𝐺

𝐶

𝐹 Figure 1

ISTUDY

Chapter 4

Moment of a Force and Equivalent Force Systems

𝐹 12 in. 𝑥

A gasoline-powered machine for smoothing the surface of wet concrete slabs is shown (the guard for the paddles is removed). If the concrete applies forces 𝐹 = 8 lb to each paddle, where all forces 𝐹 lie in the 𝑥𝑦 plane (the 𝑧 direction forces applied by the concrete to the paddles, and the weight of the machine, are not shown and are not needed for this problem), determine the force 𝑃 and orientation 𝛼 needed so the resultant couple moment of the forces shown is zero. Forces 𝑃 are parallel to the 𝑦𝑧 plane.

SOLUTION Road Map

Because this problem is three-dimensional, a vector solution will probably be the most straightforward. However, with some forethought, a scalar solution can also be used effectively. Scalar solution Governing Equations & Computation If the resultant couple moment is to be zero,

then by inspection of Fig. 1 we see that we must have 𝛼 = 0 so that the moment of couple forces 𝑃 has the same direction as the moments due to the paddle force couples. With this observation, a scalar solution can be used to sum the magnitudes of the couple moments in the 𝑧 direction as 𝑀 = (8 lb)(24 in.) + (8 lb)(24 in.) − 𝑃 (20 in.) = 0

⇒

𝑃 = 19.2 lb,

(1)

where moment is taken to be positive in the positive 𝑧 direction and the distance between points 𝐴 and 𝐵 is 20 in. from the coordinate information provided. Warning: Equation (1) can be used to sum moment magnitudes only if we know that all moments share the same direction. Vector solution Governing Equations & Computation If you did not recognize that 𝛼 = 0, then a vector

solution is better. The resultant couple moment is ) ( ⃗ = 2 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝑃⃗ 𝑀 𝐺𝐷 𝐷 𝐴𝐵 𝐵 [ ] ̂ = 2 (24 in. 𝚥̂) × (−8 lb 𝚤̂) + (−20 in. 𝚤̂) × 𝑃 (cos 𝛼 𝚥̂ + sin 𝛼 𝑘) [ ] ̂ = 𝑃 (20 in.) sin 𝛼 𝚥̂ + 384 in.⋅lb − 𝑃 (20 in.) cos 𝛼 𝑘.

(2)

In the above expression, we have simply doubled the moment of the couple forces at points 𝐷 and 𝐺, since by inspection, the moment from the couple forces at points 𝐶 and 𝐸 has ⃗ each component of Eq. (2) must be ⃗ = 0, the same magnitude and direction. To have 𝑀 zero. Thus, 𝑦 component: 𝑃 (20 in.) sin 𝛼 = 0,

(3)

𝑧 component: 384 in.⋅lb − 𝑃 (20 in.) cos 𝛼 = 0.

(4)

Since 𝑃 ≠ 0, Eq. (3) is satisfied only if 𝛼 = 0. Equation (4) can then be solved for 𝑃 = 19.2 lb. Discussion & Verification

Both solutions provide the same results. If we did not recognize by inspection that 𝛼 = 0, then solution 1 would have been a little more tedious, but could still be carried out.

ISTUDY

Section 4.3

241

Moment of a Couple

Problems Problem 4.64 An open-end wrench applies the forces 𝐹 to the head of a bolt to produce the moment 𝑀, where each force 𝐹 is normal to the surface on which it acts. Determine 𝑀 if 𝐹 = 400 lb. 𝐹

𝐹 𝐹

𝐹

𝑀 5∕8 in.

𝑀 5∕8 in.

𝐹 𝐹

𝐹

open-end wrench Figure P4.64

𝐹

box-end wrench

Figure P4.65

Problem 4.65

50 mm

A box-end wrench applies the forces 𝐹 to the head of a bolt to produce the moment 𝑀, where each force 𝐹 is normal to the surface on which it acts. Determine 𝐹 if 𝑀 = 20 f t ⋅lb.

𝑄 15 mm 𝐹 15 mm

30 mm 𝛼

Problem 4.66 The top view of a workpiece that fits loosely in a fixture for drilling is shown. The drill bit has two edges that apply in-plane cutting forces 𝐹 to the workpiece. (a) If 𝐹 = 600 N, determine the forces 𝑄 between the workpiece and fixture so that the resultant couple moment is zero when 𝛼 = 30◦ . (b) Does your answer for 𝑄 from Part (a) change if 𝛼 has a different value? If yes, then repeat Part (a) with 𝛼 = 60◦ .

Problem 4.67 To increase the volume of a stereo, a person applies the forces 𝑁𝐴 , 𝐹𝐴 , 𝑁𝐵 , and 𝐹𝐵 to the volume control knob. All of these forces lie in the same plane, which is parallel to the 𝑦𝑧 plane, and 𝑁𝐴 and 𝑁𝐵 have the same line of action, which passes through the 𝑥 axis, 𝐹𝐴 is perpendicular to 𝑁𝐴 , and 𝐹𝐵 is perpendicular to 𝑁𝐵 . To prevent the knob from slipping, the person applies the forces so that 𝐹𝐴 = 0.2𝑁𝐴 and 𝐹𝐵 = 0.2𝑁𝐵 . (a) If 𝑁𝐴 = 6 N and the forces applied by the person are to be a couple or a system of couples, determine the values of 𝐹𝐴 , 𝑁𝐵 , and 𝐹𝐵 . (b) Using the results of Part (a), determine the resultant couple moment vector applied to the knob. 𝑦 12 mm 25◦ 𝐹𝐴 𝑧

25◦ Figure P4.67

𝐵 𝐴

𝑁𝐴 Michael Plesha

30 mm

20 mm

𝐹𝐵 𝑥

𝑁𝐵

fixture

𝐹

fixture 120 mm

workpiece 𝑄 Figure P4.66

242

Chapter 4

Moment of a Force and Equivalent Force Systems

36 in.

Problem 4.68

𝐴

𝐶

𝑥 𝛼

24 in.

𝐹𝐴

𝛽

The large wrench shown is shaped so that a worker can use both hands to apply forces to the wrench. If 𝐹𝐴 = 80 lb, 𝛼 = 20◦ , and the forces at 𝐴 and 𝐵 are to be a couple, determine 𝐹𝐵 , 𝛽, and the moment of the couple.

Problem 4.69 𝐵

𝐹𝐵

Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point 𝐴 applies a 400 lb force.

𝑦 Figure P4.68

(a) Determine 𝐹𝐵 and 𝐹𝐶 so that only couples are applied. (b) Using your answers to Part (a), determine the resultant couple moment that is produced. (c) Resolve the forces at 𝐴 and 𝐵 into 𝑥 and 𝑦 components, and identify the pairs of forces that constitute couples. 40 f t

50 f t

𝑦 𝑂

(a)

44 f t

𝑥

𝐵

𝑥 𝐹𝐵

𝐵 𝑟

45◦

45◦

45◦ Figure P4.69

𝑟 = 10 mm

𝑇

𝐶

𝑇

𝐴 𝑟

20 mm

𝑦 𝐴

20 mm

50 mm

𝐹𝐶

400 lb

30◦

Problem 4.70 𝑦

20 mm

50 mm 𝑂

(b)

𝐴

𝑇

𝑇 45◦

20 mm

𝑇

𝐵

𝑇 Figure P4.70

Problem 4.71

𝑧 𝐷

𝐴

𝑦

𝐶 𝐵 𝑥

(a) Use the two tape forces in Fig. P4.70(a) to compute the resultant couple moment. (b) Replace the forces on the two pulleys with point forces on the pinions as shown in Fig. P4.70(b), and compute the resultant couple moment (replacement of pulley forces by bearing forces is discussed in Chapter 3 in connection with Fig. 3.9).

45◦

𝐹⃗𝐴

A bracket with pulleys for positioning magnetic tape in an electronic data storage device is shown. The pulleys are frictionless, the tape supports a force 𝑇 = 2 N, and the thickness of the tape can be neglected.

Consider an object with forces 𝐹⃗𝐴 and 𝐹⃗𝐵 applied. The first column of the following table lists resultant moments at various points due to both of these forces. For each of the following statements, select True or False.

𝐹⃗𝐵

Figure P4.71

ISTUDY

⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ 𝑀 𝐶 𝐶𝐴 𝐴 𝐶𝐵 𝐵 ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐶 𝐴𝐵 𝐵 ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐶 𝐵𝐴 𝐴 ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴 𝐴𝐵 𝐵 ⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ 𝑀 𝐷 𝐶𝐴 𝐴 𝐶𝐵 𝐵

If 𝐹⃗𝐴 and 𝐹⃗𝐵 are not a couple

If 𝐹⃗𝐴 and 𝐹⃗𝐵 are a couple

T or F? T or F? T or F? T or F? T or F?

T or F? T or F? T or F? T or F? T or F?

Note: Concept problems are about explanations, not computations.

ISTUDY

Section 4.3

243

Moment of a Couple

Problem 4.72 (a) For loading (a) shown below, determine 𝐹𝐴 and 𝛼 so that the forces applied at points 𝐴 and 𝐵 are a couple. Determine the moment of the couple. (b) For loading (b), determine 𝐹1 and 𝐹2 so that this loading is equivalent to the couple in Part (a). (c) For loading (c), determine 𝑄1 and 𝑄2 so that this loading is equivalent to the couple in Part (a). (d) For loading (d), determine 𝑀𝐴 so that this loading is equivalent to the couple in Part (a). 5 cm

𝑦

𝑦

7 cm

𝑄1

𝛼

13 cm

200 N

𝐵

𝐵 𝐹1

18 cm

12

30◦

𝐴 5

5 cm

30◦

𝑄2 𝐴

𝐴

𝐴

10 cm

𝑀𝐴 𝑥

𝑂 (a)

𝐵

𝐵

𝐹2 𝐹𝐴

𝑦

𝑦 10◦

𝑥

𝑂 (b)

𝑥

𝑂 (c)

𝑥

𝑂 (d)

Figure P4.72 𝑦

Problem 4.73 Consider the object shown in Fig. P4.73(a) with forces 𝐹⃗𝐴 , 𝐹⃗𝐵 , and 𝐹⃗𝐶 applied, where points 𝐴, 𝐵, and 𝐶 lie on the lines of action of these forces, respectively. If 𝐹⃗𝐴 + 𝐹⃗𝐵 + ⃗ show that these forces constitute a system of couples. Hint: Consider a point 𝐷 𝐹⃗𝐶 = 0, whose location is arbitrary. Evaluate the moment of the three forces about point 𝐷 and show that the result is independent of where 𝐷 is located, as was done in Eqs. (4.13) and (4.14) on p. 233. Using Fig. P4.73(b), sketch a system of forces at points 𝐴, 𝐵, and 𝐶, including magnitudes, that will illustrate an example of the force system described in this problem.

To loosen the cap of a peanut butter jar requires a moment about the 𝑧 axis of 50 in.⋅lb. If 𝐹⃗𝐵 has the 𝑥, 𝑦, and 𝑧 direction cosines 6∕11, −9∕11, and 2∕11, respectively, and the forces 𝐹⃗ and 𝐹⃗ are a couple, determine the magnitude of 𝐹⃗ needed to loosen the cap. 𝐴

𝐵

𝐵

𝑧 15◦ 𝐹⃗𝐴

𝐴

𝐵 1.8 in.

𝑥

Figure P4.74

𝐹⃗𝐵 𝑦

𝐶

𝐶

𝐹⃗𝐵

𝐵 𝐴

𝑥 𝐹⃗𝐴

4 in. 𝑦

𝐵 2 in.

𝐴 3 in.

(a) Figure P4.73

Problem 4.74

1 in.

𝐹⃗𝐶

𝑧

(b)

𝑥

244

Chapter 4

Moment of a Force and Equivalent Force Systems

Problem 4.75 The manufacturer of an automobile is considering the two designs shown for the vehicle’s lug nut wrench. Forces 𝐹𝐴 , 𝐹𝐶 , 𝑄𝐴 , and 𝑄𝐶 are applied by the user to the wrench; these forces are perpendicular to plane 𝐴𝐵𝐶𝐷 and are to be couples. (a) If 𝐹𝐴 = 𝑄𝐴 = 120 N, determine 𝐹𝐶 and 𝑄𝐶 , and for each wrench, determine the moment of the couple. (b) If a moment about the 𝑥 axis at point 𝐷 of 8000 N⋅mm is required to loosen the lug nut, determine the values of 𝐹𝐴 , 𝐹𝐶 , 𝑄𝐴 , and 𝑄𝐶 . 𝑧

𝑧 𝑄𝐴 𝐹𝐴 40 mm

𝐴

40 mm

𝐴

𝐵

𝐵

𝐷

𝑥

𝐶

𝑥

𝑦

25 mm

25 mm 𝑦

𝑄𝐶

𝐹𝐶

𝑦

25 mm

𝐷

(a)

(b) Figure P4.75

𝐶 300 N

300 N 400 N 2m 𝑧

𝐵

Problem 4.76 4m

400 N 𝐴 1.5 m 𝑂

A structure built in at point 𝑂 supports 300 N and 400 N couples. Determine the resultant couple moment vector, using both scalar and vector approaches.

Problem 4.77 𝑥

A structure built in at point 𝑂 supports 70 N and 85 N couples and a tip moment. Determine the resultant couple moment, using both scalar and vector approaches.

Figure P4.76

ISTUDY

𝑦

85 N 70 N

0.8 m

𝑂

1.2 m

2

6

3 𝐵

2

𝐴 6

1.2 m

𝐷 1.2 m 𝐸

70 N

9 8 12 85 N

𝑥

200 N⋅m

0.8 m

3

𝑧

12 9 8

𝐶

1.5 m

Figure P4.77–P4.79

Problem 4.78 Determine the distance between the lines of action of the 70 N forces. Hint: Use the vector approach to determine the moment of this couple. The magnitude of this result is equal to (70 N)𝑑, where 𝑑 is the distance between the lines of action for the two forces.

Problem 4.79 Determine the distance between the lines of action of the 85 N forces. See the hint in Prob. 4.78.

ISTUDY

Section 4.3

245

Moment of a Couple 𝑧

Problem 4.80

500 N⋅m

The input shaft of a speed reducer supports a 200 N ⋅ m moment, and the output shafts support 300 N⋅m and 500 N⋅m moments.

200 N⋅m

(a) Determine the resultant moment applied by the shafts to the speed reducer. (b) If the speed reducer is bolted to a surface that lies in the 𝑥𝑦 plane, speculate on the characteristics of the forces these bolts apply to the speed reducer. In other words, do you expect these forces to constitute couples only, or may they be more general? Remark: Problems such as this are discussed in detail in Chapter 5.

𝑦 300 N⋅m 𝑥 Figure P4.80

Problem 4.81 Satellites and other spacecraft perform attitude positioning using thrusters that are fired in pairs so as to produce couples. If thrusters at points 𝐴, 𝐵, 𝐶, and 𝐷 each produce 3 N forces, determine the resultant moment, and hence, the axis through the center of mass about which an initially nonrotating satellite will begin to rotate. The forces at 𝐴 and 𝐶 are parallel to the 𝑧 direction, and the forces at 𝐵 and 𝐷 are parallel to the 𝑥 direction. 𝑦 0.8 m 𝐵

3N

0.8 m 𝐶

3N

0.6 m 𝑂

𝑧

𝑦 0.6 m 𝑥 𝐴

3N

𝐷

3N

𝐹

Figure P4.81

𝑇 120◦

120◦

Problem 4.82

𝐹

Forces 𝐹 and 𝑇 exerted by air on a rotating airplane propeller are shown. Forces 𝐹 lie in the 𝑥𝑦 plane and are normal to the axis of each propeller blade, and thrust forces 𝑇 act in the 𝑧 direction. Show that the forces 𝐹 can be represented as a couple or system of couples.

Problem 4.83 The structure shown is subjected to couple forces applied at points 𝐴 and 𝐵 where the ̂ lb. If line 𝑎 has direction angles 𝜃 = 72◦ , force applied at 𝐴 is 𝐹⃗ = (8 𝚤̂ + 10 𝚥̂ − 40 𝑘) 𝑥 𝜃𝑦 = 36◦ , and 𝜃𝑧 = 60◦ , determine the moment of the couple about line 𝑎. 𝐹 𝑧

𝐵

𝑎

𝐶 𝑂

𝐴 𝐹

𝑥

𝐴 (16, 36, 10) in. 𝐵 (−8, 24, 18) in. Figure P4.83

𝑦

𝑇

𝐹 𝑧

Figure P4.82

𝑇

𝑥

246

Chapter 4

Moment of a Force and Equivalent Force Systems

4.4

(a)

(b)

Equivalent Force Systems

Equivalent force systems are extremely important in mechanics and are used by engineers daily. Consider, for example, the pliers shown in Fig. 4.25(a), and imagine we wish to determine the force developed at the jaws as we squeeze the handles. Each finger and the palm of our hand apply a complex pressure distribution to the handles, as shown in Fig. 4.25(b). It would be very unfortunate if the complexities of these distributions needed to be included in all analyses. For many purposes, such as for determining the force developed at the jaws, these pressure distributions may be replaced by equivalent forces as shown in Fig. 4.25(c) and (d) with no loss of accuracy. This section defines conditions under which two force systems are equivalent and describes techniques for replacing one force system with another. Concepts are first presented as definitions for rigid bodies, using physical arguments as justification. Implications of these concepts for deformable bodies are discussed. The section closes with a short discussion of Newton’s laws, which provides a theoretical justification and interpretation of concepts.

Transmissibility of a force (c)

(d)

Figure 4.25 (a) A person uses a hand to grip a pair of pliers. (b)–(d) Examples of equivalent force systems.

ISTUDY

Before discussing the principle of transmissibility of a force, we first define the terms external effects of a force applied to an object and internal effects of a force applied to an object. External effects refer to the response of the object as a whole. For example, if the object’s position is fixed in space by supports, external effects include the support reactions. If the object is unsupported so that it can move in space, external effects refer to the object’s displacement, velocity, and acceleration. Internal effects refer to the internal forces supported by (or developed within) a body, and if the body is deformable, the deformations that the body experiences. The principle of transmissibility states that the external effects of a force applied to a rigid body are the same, regardless of the point of application of the force along its line of action. This principle is developed in Fig. 4.26 as follows. In Fig. 4.26(a),

−𝐹⃗ 𝑧

𝑧

𝐹⃗

𝑦 𝐴

𝑥 𝐹⃗

𝑧

𝐵

𝐴

𝑥

𝐵 𝐹⃗

𝑦

𝑦

𝐴

𝑥

𝐹⃗ (a)

(b)

(c)

Figure 4.26. Transmissibility of force.

a force vector 𝐹⃗ is applied at point 𝐴 on an object or structure. In Fig. 4.26(b), two additional forces 𝐹⃗ and −𝐹⃗ are applied at point 𝐵 where 𝐵 lies on the line of action of the force at 𝐴; since the forces at 𝐵 have equal magnitude and opposite direction and are applied at the same point, they have no effect on the object. In Fig. 4.26(c) the forces 𝐹⃗ at 𝐴 and −𝐹⃗ at 𝐵 have been canceled, leaving only the force 𝐹⃗ at 𝐵.

ISTUDY

Section 4.4

Equivalent Force Systems

247

We now conjecture that if the body is rigid, the external effects of 𝐹⃗ on the body are the same whether 𝐹⃗ is applied to point 𝐴 or to point 𝐵. Using definitions to follow, we will see that the principle of transmissibility of a force states conditions that give special cases of equivalent force systems, and that all of the force systems shown in Fig. 4.26 are equivalent. Furthermore, at the end of this section, the reason why equivalent force systems are called equivalent is discussed, and this provides justification for why the principle of transmissibility of a force is valid. Rigid versus deformable objects. To be precise, the principle of transmissibility only holds for forces applied to rigid bodies. If an object is rigid, the distance between any two points, such as points 𝐴 and 𝐵 in Fig. 4.26, is constant regardless of where forces are applied to the object. Because the geometry of the object does not change, the position of lines of action of forces does not change, and hence, the external effects on the rigid body, which are determined by summing forces and moments, are the same. Note that the principle of transmissibility makes no claims about internal forces. In fact, the internal forces depend very much on exactly where a particular force is applied along its line of action. But if the body is rigid, then regardless of the internal forces, the body does not change shape, and hence the external effects of the force are unchanged. If a body is deformable, which is always the situation in nature, then the body will generally change shape when forces are applied. Furthermore, the deformed shape of the body depends on the distribution of internal forces within the body. The body’s change of shape causes lines of action of forces to be repositioned, and hence, the external effects of forces are different. Even though all bodies in nature are deformable, for many purposes, objects and structures may be idealized as being rigid, and the principle of transmissibility may be applied. To illustrate further, consider a block of material resting under its own weight on a rough surface as shown in Fig. 4.27. If the material is stiff, then there is very little change of geometry for the loadings shown in Fig. 4.27(a) and (b), and for many practical purposes the object may be idealized as being rigid. The principle of transmissibility may be applied, and the response of the object, such as the force 𝐹 that will cause the block to slip, is the same regardless of where 𝐹 is applied along its line of action. If the material is flexible, the object may undergo significant changes of shape when subjected to the forces shown in Fig. 4.27(c) and (d). Such objects often cannot be idealized as being rigid since the deformation that occurs may have a strong effect on their response.

Equivalent force systems The concept of an equivalent force system is developed in Fig. 4.28. In Fig. 4.28(a), a rigid object or structure has a force vector 𝐹⃗ applied at point 𝐴. In Fig. 4.28(b), two additional forces 𝐹⃗ and −𝐹⃗ are applied at point 𝐵; since the forces at 𝐵 have equal magnitude and opposite direction and are applied at the same point, they have no effect on the object. Forces 𝐹⃗ at 𝐴 and −𝐹⃗ at 𝐵 are a couple and can be re⃗ = 𝑟⃗ × 𝐹⃗ , where 𝑟⃗ is a position vector from anywhere on the placed by moment 𝑀 𝐵 line of action of the force at 𝐵 to anywhere on the line of action of the force at 𝐴 (Fig. 4.28(b) shows a vector 𝑟⃗ from points 𝐵 to 𝐴 in particular). Thus, in Fig. 4.28(c) a new force system at point 𝐵 has been developed that is equivalent to the original

𝑧

𝑧 stiff material 𝐹

𝐹 𝑦 𝑥

(a)

𝑦

𝑥

𝑧

(b) 𝑧

flexible material 𝐹

𝐹 𝑦

𝑦 𝑥

(c)

𝑥

(d)

Figure 4.27 A block of material resting on a rough surface and subjected to a force. (a) and (b) The material is stiff, and the difference in the block’s shape for the two loadings is very small. (c) and (d) The material is flexible, and the block’s shape is significantly different for the two loadings.

248

Chapter 4

Moment of a Force and Equivalent Force Systems

force system shown in Fig. 4.28(a). Further, since the moment of a couple is a free ⃗ may be applied at any point, such as point 𝐶 shown in Fig. 4.28(d). vector, 𝑀 𝐵 ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐵

−𝐹⃗ 𝑧

𝑧 𝐴 𝐹⃗

𝑥

𝑧

𝐵 𝐹⃗

𝐹⃗

𝑟⃗ 𝐴

𝑦 𝐹⃗

𝑥

(a)

𝑧

𝐵

𝐹⃗

𝑦

𝑦

𝑥

𝑥

(b)

𝐵 𝑦

𝐶 ⃗ 𝑀 𝐵 (d)

(c)

Figure 4.28. Construction of an equivalent force system at point 𝐵.

⃗ 𝑀 𝑖

To generalize, if an object or structure has an arbitrary number of forces and/or moments applied as shown in Fig. 4.29, an equivalent force system at a point 𝐴 con⃗ where sists of a resultant force 𝐹⃗𝑅 and a resultant moment 𝑀 𝑅

𝐹⃗𝑅

⃗ 𝑀 𝑅 𝐹⃗𝑖

𝑧

𝑧 ⃗ 𝑀 2

⃗ 𝑀 1

𝐴

𝑦 𝑥

𝐹⃗2

𝐹⃗1

𝑦 𝑥

⃗ = 𝑀 𝑅

𝑛 ∑ 𝑖=1 𝑛 ∑

𝐹⃗𝑖 , 𝑟⃗𝑖 × 𝐹⃗𝑖 +

𝑖=1

(a)

𝑚 ∑

(4.15) ⃗ , 𝑀 𝑖

𝑖=1

(b)

Figure 4.29 Construction of an equivalent force system at point 𝐴.

ISTUDY

𝐹⃗𝑅 =

where 𝑛 is the number of forces that are applied, 𝑚 is the number of moments that are applied, and 𝑟⃗𝑖 is a position vector from point 𝐴 to anywhere on the line of action of ⃗ is a free vector and may be positioned anywhere. 𝐹⃗𝑖 . As discussed previously, 𝑀 𝑅 Equation (4.15) can be used as a test to determine if two force systems are equivalent. That is, two force systems are equivalent if (

(

𝐹⃗𝑅

⃗ 𝑀 𝑅

)

system 1

)

system 1

( ) = 𝐹⃗𝑅 system 2 and ( ) ⃗ = 𝑀 𝑅 system 2 ,

(4.16)

⃗ ⃗ ) where the moment summation point used for determining (𝑀 𝑅 system 1 and (𝑀𝑅 )system 2 must be the same. Using Eq. (4.15), we may state Eq. (4.16) more explicitly as two force systems are equivalent if 𝑛1 (∑ 𝑛1 (∑ 𝑖=1

𝐹⃗𝑖

𝑖=1 𝑚1

𝑟⃗𝑖 × 𝐹⃗𝑖 +

∑ 𝑖=1

⃗ 𝑀 𝑖

)

system 1

)

system 1

=

=

𝑛2 (∑ 𝑖=1 ( 𝑛2

∑ 𝑖=1

𝐹⃗𝑖

)

system 2

𝑟⃗𝑖 × 𝐹⃗𝑖 +

𝑚2 ∑ 𝑖=1

and ⃗ 𝑀 𝑖

)

system 2

(4.17) ,

where 𝑛1 and 𝑚1 are the number of forces and moments in system 1, respectively, 𝑛2 and 𝑚2 are the number of forces and moments in system 2, respectively, and the moment summation points used for both force systems must be the same.

Some special force systems Various special force systems, categorized as concurrent, coplanar, and parallel, are shown in Figs. 4.30–4.32. Observe that for each of these force systems, there exists

ISTUDY

Section 4.4

Equivalent Force Systems

249

a point (i.e., point 𝐵 in Figs. 4.30–4.32) where an equivalent force system exists that consists of a single force only (no moment).

𝐹⃗2 𝑧

𝐹⃗𝑅

𝑧 𝐹⃗𝑖 𝐵

𝐵

𝑦

𝑥

𝑦

𝑥 𝐹⃗1 (b)

(a)

Figure 4.30. Concurrent force system: the lines of action of all forces intersect at a common point.

𝑧

⃗ 𝑀 1

⃗ 𝑀 𝑖

⃗ 𝑀 𝑅

𝑧

𝐹⃗2

Helpful Information 𝑦

𝑥

𝑧

𝐹⃗𝑖

𝐹⃗1

𝑥

𝐹⃗𝑅

𝑦

𝑑 𝐴

𝑦

𝐴

𝐵

𝑥

𝐹⃗𝑅 𝑑 = 𝑀𝑅 ∕𝐹𝑅

(a)

(b)

(c)

Figure 4.31. Coplanar force system: all forces lie in the same plane, and all moments are perpendicular to that plane.

𝑧

𝑧

𝐹⃗𝑖 𝐹⃗1 𝐹⃗2

𝑧

𝑦

𝑦 𝑥

⃗ 𝑀 1

⃗ 𝑀 𝑖 (a)

𝐹⃗𝑅

𝐹⃗𝑅

𝑥

𝐴

𝑑 𝐵 𝑥

⃗ 𝑀 𝑅 (b)

𝑦

𝐴 𝑑 = 𝑀𝑅 ∕𝐹𝑅 (c)

Figure 4.32. Parallel force system: all forces are parallel, and all moments are perpendicular to the direction of the forces.

Remarks • Concurrent force system. In these force systems, such as shown in Fig. 4.30, no resultant moment is produced by the forces about the point where the lines of action of the forces intersect. When an object is subjected to a concurrent force system, it may be idealized as a particle for purposes of equilibrium analysis, and this was studied extensively in Chapter 3. • Coplanar force system. In these force systems, such as shown in Fig. 4.31, all forces lie in the same plane, and all moments are perpendicular to that plane.

Positioning a force system to eliminate a moment. As discussed in connection with coplanar and parallel force systems, by suitable repositioning of the line of action of the resultant force 𝐹⃗𝑅 , we can construct an equivalent force system that has no moment, as shown in Figs. 4.31(c) and 4.32(c). However, on occasion you may find the distance 𝑑 in these figures to be large enough that the force is repositioned off the object or structure. While this is theoretically acceptable, it presents a practical problem on how such a force can be applied to the object or structure. Similar comments apply to the construction of a wrench force system, as shown in Fig. 4.34.

250

Chapter 4

Moment of a Force and Equivalent Force Systems

The use of Eq. (4.15) allows the determination of an equivalent force system ⃗ . At point 𝐵, as at point 𝐴, as shown in Fig. 4.31(b), consisting of 𝐹⃗𝑅 and 𝑀 𝑅 shown in Fig. 4.31(c), the equivalent force system consists of 𝐹⃗𝑅 only; the location of 𝐵 relative to 𝐴 is determined so that 𝐹⃗𝑅 produces the proper moment (i.e., 𝑀𝑅 = 𝐹𝑅 𝑑 ⇒ 𝑑 = 𝑀𝑅 ∕𝐹𝑅 ). • Parallel force system. In these force systems, such as shown in Fig. 4.32, all forces are parallel, and all moments are perpendicular to the direction of the forces. The use of Eq. (4.15) allows the determination of an equivalent force ⃗ . At point system at point 𝐴, as shown in Fig. 4.32(b), consisting of 𝐹⃗𝑅 and 𝑀 𝑅 𝐵, as shown in Fig. 4.32(c), the equivalent force system consists of 𝐹⃗𝑅 only; the location of 𝐵 relative to 𝐴 is determined so that 𝐹⃗𝑅 produces the proper moment (i.e., 𝑀𝑅 = 𝐹𝑅 𝑑 ⇒ 𝑑 = 𝑀𝑅 ∕𝐹𝑅 ).

Wrench equivalent force systems ⃗ 𝑀 𝑅

𝑧 𝐹⃗𝑅 𝐴 𝑥

⃗ 𝑀 𝑅

𝑧

𝐹⃗𝑅

𝑦

𝐴

𝑦

𝑥 positive wrench (a)

negative wrench (b)

Figure 4.33 Wrench force systems. A positive wrench has 𝐹⃗𝑅 ⃗ positive in the same direction. A negaand 𝑀 𝑅 ⃗ positive in opposite tive wrench has 𝐹⃗𝑅 and 𝑀 𝑅 directions.

ISTUDY

A feature of concurrent, coplanar, and parallel force systems is that for these systems you can always find an equivalent force system consisting of a single force only. More general force systems usually cannot be simplified to this extent, but it is useful for us to address the following question: What is the simplest force system to which any general force system can always be reduced? The answer is a force system called a ⃗ that is wrench, which consists of a resultant force 𝐹⃗𝑅 and a resultant moment 𝑀 𝑅 parallel to 𝐹⃗𝑅 , as shown in Fig. 4.33. A wrench force system is constructed as follows. Starting with force 𝐹⃗𝑅 and mo⃗ into components parallel ⃗ at point 𝐴 as shown in Fig. 4.34(a), we resolve 𝑀 ment 𝑀 𝑅 𝑅 ⃗ ⃗ constitute and perpendicular to 𝐹𝑅 as shown in Fig. 4.34(b). Noting that 𝐹⃗𝑅 and 𝑀 𝑅⊥ ⃗ ⃗ a parallel force system, 𝑀 𝑅⊥ may be eliminated by relocating 𝐹𝑅 to point 𝐵, where its location relative to 𝐴 is determined so that 𝐹⃗𝑅 produces the proper moment (i.e., ⃗ is a free vector, it can also 𝑀𝑅⊥ = 𝐹𝑅 𝑑 ⇒ 𝑑 = 𝑀𝑅⊥ ∕𝐹𝑅 ). Note that because 𝑀 𝑅‖ be relocated from 𝐴 to 𝐵. Example 4.15 illustrates further details of determining a wrench equivalent force system. 𝑧

𝐹⃗𝑅

𝑧 𝐹⃗𝑅

⃗ 𝑀 𝑅

⃗ 𝑀 𝑅

𝐹⃗𝑅

𝑧

⃗ 𝑀 𝑅‖

Helpful Information 𝐴 Replacing a general force system with a wrench. Example 4.15 gives a two-step process for determining a wrench equivalent force system.

⃗ 𝑀 𝑅‖

𝑑 ⃗ 𝑀 𝑅⊥

𝑦 𝑥

𝐴

𝐴

𝐵

𝑦

𝑦 𝑥

𝑥

𝑑 = 𝑀𝑅⊥ ∕𝐹𝑅 (a)

(b)

(c)

Figure 4.34. Construction of a wrench force system.

Why are equivalent force systems called equivalent? If you like, you may consider the concept of an equivalent force system as being simply a definition. However, there is good reason behind the definition and why

ISTUDY

Section 4.4

it is meaningful, and this is seen from Newton’s law of motion. In Chapter 5, we ∑ will see that the conditions for static equilibrium of a rigid body are 𝐹⃗ = 0⃗ and ∑ ⃗ ⃗ If the first of these equations is not satisfied, the body will have a trans𝑀 = 0. lational acceleration; and if the second equation is not satisfied, the body will have a rotational acceleration. Regardless of whether the body is in static equilibrium or is undergoing accelerations, the response in terms of external effects (e.g., reaction forces, accelerations, etc.) is the same for all systems of external forces for which ∑ ⃗ ∑ ⃗ 𝐹 and 𝑀 are the same, and the definition of equivalent force systems given in Eq. (4.15) is a statement of these conditions. Since the principle of transmissibility of a force is a special case of equivalent force systems, this explanation also shows why it is valid.

End of Section Summary In this section, the concepts of equivalent force systems were described. Some of the key points are as follows: • The principle of transmissibility states that the effects of a force applied to a rigid body are the same regardless of the point of application of the force along its line of action. • Two force systems are said to be equivalent if the sum of forces for the two systems is the same and if the sum of moments (about any common point) for the two force systems is the same. If two force systems are equivalent, their external effects on a rigid body or structure, such as support reactions, are the same. • Some special force systems that often occur were defined, including concurrent, coplanar, and parallel force systems. Each of these force systems has the feature that they have an equivalent force system that consists of a single force only (no moment). • A wrench force system consists of a force and moment, where these have the same line of action. Any force system can be represented by an equivalent force system that is a wrench.

Equivalent Force Systems

251

252

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.11 𝑦

Determine an equivalent force system

4N

𝑂

Determination of an Equivalent Force System 2N

(a) at point 𝐴,

3N 𝐶

𝐵

𝐴 2 mm

2 mm

𝑥

(b) at point 𝐵, (c) consisting of a single force only, and specify the 𝑥 coordinate of the point where the force’s line of action intersects the 𝑥 axis.

2 mm

Figure 1

SOLUTION Helpful Information

Road Map

Reaction forces and moments applied to the beam. The structure shown in Fig. 1 is supported at point 𝑂, and this support applies reaction forces and moments to the structure so that it is in equilibrium. However, our focus throughout this section has been on determining equivalent systems of external forces. Thus, whether an object is supported or unsupported is irrelevant to the task of determining equivalent systems of external forces.

For all three parts of this problem, an equivalent system of external forces will be developed using Eq. (4.15) on p. 248. Part (a)

Governing Equations & Computation

We apply Eq. (4.15) with a scalar approach, where positive moment is counterclockwise, to obtain ∑ (1) 𝐹𝑥 = 0, 𝐹𝑅𝑥 = ∑ (2) 𝐹𝑅𝑦 = 𝐹𝑦 = −4 N + 3 N − 2N = −3 N, ∑ (3) 𝑀𝑅𝐴 = 𝑀𝐴 = (4 N)(4 mm) − (3 N)(2 mm) = 10 N⋅mm. This force system is illustrated in Fig. 2. Part (b)

3N

𝑦

𝐴

𝑂

𝑥

10 N⋅mm Figure 2 This force system is equivalent to that shown in Fig. 1. 3N

𝑦 𝑂

𝐵

Figure 3 This force system is equivalent to those shown in Figs. 1 and 2.

Governing Equations & Computation

A force system that consists of a single force only is shown in Fig. 4, where the distance 𝑑 is determined so that this force system is equivalent to the force systems shown in Figs. 1 through 3. Thus, selecting point 𝑂 as a convenient location to sum moments for the force systems shown in Figs. 2 and 4 provides (∑

3N 𝑂

This force system is illustrated in Fig. 3, where 𝑀𝑅𝐵 is shown as 2 N⋅mm clockwise rather than −2 N ⋅ mm counterclockwise. As an alternative to using the force system shown in Fig. 1 to write Eq. (4), we could have used the force system shown in Fig. 2 to write ∑ 𝑀𝑅𝐵 = 𝑀𝐵 = 10 N⋅mm − (3 N)(4 mm) = −2 N⋅mm which, as expected, agrees with Eq. (4). Part (c)

𝑥

2 N⋅mm

𝑦

For an equivalent force system at point 𝐵, the resultant forces 𝐹𝑅𝑥 and 𝐹𝑅𝑦 are unchanged from Eqs. (1) and (2), and referring to the forces shown in Fig. 1, the resultant moment about 𝐵 is ∑ (4) 𝑀𝑅𝐵 = 𝑀𝐵 = (3 N)(2 mm) − (2 N)(4 mm) = −2 N⋅mm.

Governing Equations & Computation

𝑀𝑅𝑂

)

Fig. 2

=

(∑

𝑀𝑅𝑂

)

−(3 N)(6 mm) + 10 N⋅mm = −(3 N)𝑑 ⇒

𝑥

𝑑 = 2.667 mm.

Fig. 4

(5) (6)

𝑑 Figure 4 With the appropriate value of 𝑑, this force system is equivalent to those shown in Figs. 1–3.

ISTUDY

Discussion & Verification

The force systems shown in Figs. 1 through 3 are all equivalent, and the force system shown in Fig. 4 is also equivalent to these provided 𝑑 = 2.667 mm. When developing an equivalent force system that consists of a single force only, as in Part (c), sometimes you may find that the line of action of the force does not intersect the structure.

ISTUDY

Section 4.4

253

Equivalent Force Systems

E X A M P L E 4.12

Determining if Force Systems Are Equivalent

Determine which of the force systems shown are equivalent.

10 lb 𝐴

SOLUTION

3 in.

For two force systems to be equivalent, both the resultant force and the resultant moment, taken about any convenient point, must be the same. Thus, for each of the force systems shown, we will evaluate the resultant forces in the 𝑥 and 𝑦 directions and the resultant moment, and those force systems for which all of these are the same are equivalent. Using point 𝐵 as a convenient location for moment summation, with positive moments being counterclockwise, we use Eq. (4.15) on p. 248 to find

Governing Equations & Computation

10 lb 𝐶

𝐵

Road Map

30 lb

6 in.

𝐹𝑅𝑥 = 𝐹𝑅𝑦 = 𝑀𝑅𝐵 =

∑

∑

∑

𝐹𝑥 = −10 lb,

(1)

𝐹𝑦 = 10 lb − 10 lb = 0,

(2)

𝑀𝐵 = −(10 lb)(6 in.) = −60 in.⋅lb.

(3)

Force system (b)

𝐹𝑅𝑥 = 𝐹𝑅𝑦 = 𝑀𝑅𝐵 =

∑

∑

∑

𝐹𝑥 = −40 lb + 30 lb = −10 lb,

(4)

𝐹𝑦 = −10 lb + 10 lb = 0,

(5)

𝑀𝐵 = −(30 lb)(3 in.) + (10 lb)(6 in.) = −30 in.⋅lb.

(6)

∑

𝐹𝑥 = 10 lb − 20 lb = −10 lb,

(7)

𝐹𝑦 = 0,

(8)

Force system (c)

𝐹𝑅𝑥 = 𝐹𝑅𝑦 = 𝑀𝑅𝐵 =

∑

∑

𝑀𝐵 = 20 in.⋅lb − 20 in.⋅lb − (10 lb)(3 in.)

= −30 in.⋅lb.

(9)

Force system (d)

𝐹𝑅𝑥 = 𝐹𝑅𝑦 = 𝑀𝑅𝐵 =

∑

∑

∑

𝐹𝑥 = −10 lb,

(10)

𝐹𝑦 = 10 lb + 10 lb − 20 lb = 0,

(11)

𝑀𝐵 = 30 in.⋅lb + (10 lb)(3 in.) − (20 lb)(6 in.)

= −60 in.⋅lb.

(12)

Discussion & Verification Force systems (a) and (d) have the same resultant force and moment about point 𝐵, and hence, they are equivalent to one another. Also, force systems (b) and (c) have the same resultant force and moment about point 𝐵, and hence, they are equivalent to one another. In summary:

Force systems (a) and (d) are equivalent, and force systems (b) and (c) are equivalent.

10 lb

10 lb

(a)

(b) 𝑦 𝑥

20 in.⋅lb 10 lb

10 lb 10 lb 20 lb

20 lb

20 in.⋅lb

Force system (a)

10 lb

40 lb

(c) Figure 1

10 lb

30 in.⋅lb (d)

254

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.13

A table supports the vertical forces shown.

𝑧

100 N

(b) Determine an equivalent force system consisting of a single force, and specify the 𝑥 and 𝑦 coordinates of the point where the force’s line of action intersects the table.

𝐶 𝑂 0.3 m 𝑥

𝐵

𝐴

(a) Determine an equivalent force system at the center of the table, point 𝑂.

120 N

260 N 0.4 m

Determination of an Equivalent Force System

𝑦

0.5 m

SOLUTION Road Map

Both scalar and vector approaches are effective for Part (a) of this problem, and we will use a vector approach. For Part (b), a straightforward scalar evaluation will provide the necessary location of an equivalent force system that consists of a single force only.

0.2 m 0.2 m

Figure 1

Part (a) Governing Equations & Computation

𝑧 480 N 114 N⋅m 𝑂

𝑦

With the following force and position vectors

𝐹⃗𝐴 = −260 𝑘̂ N,

𝑟⃗𝑂𝐴 = (0.5 𝚤̂ + 0.2 𝚥̂) m,

(1)

𝐹⃗𝐵 = −120 𝑘̂ N,

𝑟⃗𝑂𝐵 = 0.4 𝚥̂ m,

(2)

𝐹⃗𝐶 = −100 𝑘̂ N,

𝑟⃗𝑂𝐶 = (−0.4 𝚤̂ − 0.3 𝚥̂) m,

(3)

we use Eq. (4.15) on p. 248 to evaluate the resultant force and moment at point 𝑂 as 𝐹⃗𝑅 = 𝐹⃗𝐴 + 𝐹⃗𝐵 + 𝐹⃗𝐶 𝑥 Figure 2 This force system is equivalent to that shown in Fig. 1. 𝑧

= −480 𝑘̂ N, ⃗ 𝑀 ⃗𝑂𝐴 × 𝐹⃗𝐴 + 𝑟⃗𝑂𝐵 × 𝐹⃗𝐵 + 𝑟⃗𝑂𝐶 × 𝐹⃗𝐶 𝑅𝑂 = 𝑟

𝑀𝑅𝑂 480 N

= (−70 𝚤̂ + 90 𝚥̂) N⋅m, √ = (−70)2 + (90)2 N⋅m = 114.0 N⋅m.

𝐷

Governing Equations & Computation

As shown in Fig. 4.32 on p. 249, an equivalent force system consisting of a single force is obtained by moving 𝐹⃗𝑅 to a new position, point ⃗ 𝐷, where 𝐷 is located a distance 𝑑 perpendicular to the plane containing 𝐹⃗𝑅 and 𝑀 𝑅𝑂

𝑥 Figure 3 This force system is equivalent to those shown in Figs. 1 and 2. 90 114.0

𝑑=

𝑀𝑅𝑂 𝐹𝑅

=

114.0 N⋅m = 0.2375 m. 480 N

𝑦

𝑥 Figure 4 Use of similar triangles to locate point 𝐷.

(7)

This force system is shown in Fig. 3. To determine the coordinates of 𝐷, similar triangles with the geometry shown in Fig. 4 may be used to write 90 = 0.1875 m, 114.0 70 𝑦 = (0.2375 m) = 0.1458 m. 114.0

𝑥 = (0.2375 m)

𝑥 90 𝑑 = 0.2375 m 70 𝐷 𝑦

ISTUDY

(6)

Part (b) 𝑦

𝑑 = 0.2375 m

𝑂

(5)

This force system is illustrated in Fig. 2.

𝑂

70

(4)

Discussion & Verification

(8) (9)

The force systems shown in Figs. 1 through 3 are all equivalent. When developing an equivalent force system that consists of a single force only, as in Part (b), sometimes you may find that the line of action of the force does not intersect the structure.

ISTUDY

Section 4.4

255

Equivalent Force Systems

E X A M P L E 4.14

Determination of an Equivalent Force System

A casting supports the forces and moment shown where the 100 N forces are parallel to the 𝑥𝑦 plane and the moment at 𝐶 has direction angles 𝜃𝑥 = 60◦ , 𝜃𝑦 = 60◦ , and 𝜃𝑧 = 135◦ . Determine the equivalent force system acting at point 𝑂.

𝑦 10 N⋅m 400 N

SOLUTION

𝐶 𝑂

𝑧 50 mm

Road Map

Because of the complexity of the geometry, a vector solution is preferable to a scalar solution. Note that the two 100 N forces have equal magnitude and opposite direction, hence they are a couple. Governing Equations & Computation

100 N 30 mm

100 N

𝐷 70 mm

𝐴

𝐵

60◦

We first write expressions for force, position,

60◦

𝑥

and moment vectors: 60 mm 40 mm

𝐹⃗𝐵 = (100 N)(cos 60◦ 𝚤̂ + sin 60◦ 𝚥̂),

𝑟⃗𝐴𝐵 = −60 𝑘̂ mm,

(1)

𝐹⃗𝐷 = −400 𝚥̂ N,

̂ mm, 𝑟⃗𝑂𝐷 = (50 𝚤̂ + 70 𝑘)

(2)

̂ ⃗ = (10 N⋅m)(cos 60◦ 𝚤̂ + cos 60◦ 𝚥̂ + cos 135◦ 𝑘). 𝑀 𝐶

Figure 1

(3)

𝑦

Taking care to convert millimeter dimensions to meter dimensions, we use Eq. (4.15) on p. 248 to obtain the resultant force and moment at point 𝑂 as 𝐹⃗𝑅 = 𝐹⃗𝐷 = −400 𝚥̂ N,

(4)

𝐹𝑅 = 400 N 𝑂

𝑧

⃗ ⃗ 𝑀 ⃗𝐴𝐵 × 𝐹⃗𝐵 + 𝑟⃗𝑂𝐷 × 𝐹⃗𝐷 + 𝑀 𝑅𝑂 = 𝑟 𝐶

𝑀𝑅𝑂

̂ N⋅m, = (38.20 𝚤̂ + 2 𝚥̂ − 27.07 𝑘) √ = (38.20)2 + (2)2 + (−27.07)2 N⋅m = 46.86 N⋅m.

(5) (6)

In writing Eqs. (4) and (5), we have noted that the two forces at 𝐴 and 𝐵 are a couple. Hence, they produce no net force and are not included in the expression for 𝐹⃗𝑅 in Eq. (4). The equivalent force system at point 𝑂 is illustrated in Fig. 2. Discussion & Verification

The force systems shown in Figs. 1 and 2 are equivalent. As discussed above, we took advantage of the properties of a couple when writing the ex⃗ pressions for 𝐹⃗𝑅 and 𝑀 𝑅𝑂 in Eqs. (4) and (5). If you did not recognize that these forces are a couple, then you would need to include the two 100 N forces in the expression for ⃗ you would replace 𝑟⃗ × 𝐹⃗ with 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ . 𝐹⃗𝑅 , and in the expression for 𝑀 𝑅𝑂 𝐴𝐵 𝐵 𝑂𝐴 𝐴 𝑂𝐵 𝐵 ⃗ in Eqs. (4) Nonetheless, you would have obtained the same results for 𝐹⃗𝑅 and 𝑀 𝑅𝑂 and (5).

𝑥

𝑀𝑅𝑂 = 46.86 N⋅m

Figure 2 This force system is equivalent to that shown in Fig. 1.

256

Chapter 4

Moment of a Force and Equivalent Force Systems

E X A M P L E 4.15

Determination of a Wrench Equivalent Force System Replace the three forces with a wrench force system, specifying the force and moment of the wrench and the 𝑥 and 𝑦 coordinates of point 𝐷 where the wrench’s line of action intersects the 𝑥𝑦 plane.

𝑧 200 lb

7 in. 𝐶

𝐴

𝑦

𝐷(𝑥, 𝑦) 100 lb

Road Map

The construction of a wrench equivalent force system can always be accomplished using a two-step procedure, as illustrated here.

12 in. 𝐵

3 in. 8 in.

SOLUTION

Governing Equations & Computation

50 lb

Step 1: Determine an equivalent force system at point 𝐷.

𝑥

By using the following force

and position vectors

Figure 1

Helpful Information Determining a wrench equivalent force system. This example presents a twostep process that can always be used to determine a wrench equivalent force system.

𝐹⃗𝐴 = −200 𝑘̂ lb,

𝑟⃗𝐷𝐴 = −𝑥 𝚤̂ + (8 in. − 𝑦) 𝚥̂,

(1)

𝐹⃗𝐵 = 50 𝚥̂ lb,

𝑟⃗𝐷𝐵 = (12 in. − 𝑥) 𝚤̂ + (8 in. − 𝑦) 𝚥̂,

(2)

𝐹⃗𝐶 = 100 𝚤̂ lb,

𝑟⃗𝐷𝐶 = (7 in. − 𝑥) 𝚤̂ − 𝑦 𝚥̂ + 3 𝑘̂ in.,

(3)

the resultant force and moment at point 𝐷 are 𝐹⃗𝑅 = 𝐹⃗𝐴 + 𝐹⃗𝐵 + 𝐹⃗𝐶 ̂ lb, = (100 𝚤̂ + 50 𝚥̂ − 200 𝑘) √ 𝐹𝑅 = (100)2 + (50)2 + (−200)2 lb = 229.1 lb,

(4) (5)

⃗ ⃗𝐷𝐴 × 𝐹⃗𝐴 + 𝑟⃗𝐷𝐵 × 𝐹⃗𝐵 + 𝑟⃗𝐷𝐶 × 𝐹⃗𝐶 𝑀 𝑅𝐷 = 𝑟 = [−1600 in.⋅lb + (200 lb)𝑦] 𝚤̂ + [300 in.⋅lb − (200 lb)𝑥] 𝚥̂ ̂ + [600 in.⋅lb − (50 lb)𝑥 + (100 lb)𝑦] 𝑘.

(6)

If we specify values for 𝑥 and 𝑦, Eqs. (4) and (6) provide the resultant force and moment at that location. However, these are not likely to have the same direction and hence will not be a wrench force system. In the next step of this solution, we determine the coordinates ⃗ ⃗ of point 𝐷 so that 𝑀 𝑅𝐷 is parallel to 𝐹𝑅 . The requirement that 𝐹⃗𝑅 and ⃗ ⃗ ∕𝐹 = 𝑀 ⃗ ∕𝑀 . Each of the 𝑥, 𝑦, and 𝑧 components 𝑀 be parallel is stated by 𝐹 𝑅𝐷 𝑅 𝑅 𝑅𝐷 𝑅𝐷 of this equation can be written separately.

Step 2: Make the equivalent force system a wrench.

𝑥 component: 𝑀𝑅𝐷 = 1157 in.⋅lb

𝑦 component:

𝑧

𝑧 component: 𝐹𝑅 = 229.1 lb 𝑦 𝐷

2.762 in.

5.476 in.

−1600 in.⋅lb + (200 lb)𝑦 100 lb = , 229.1 lb 𝑀𝑅𝐷 300 in.⋅lb − (200 lb)𝑥 50 lb = , 229.1 lb 𝑀𝑅𝐷 600 in.⋅lb − (50 lb)𝑥 + (100 lb)𝑦 −200 lb = . 229.1 lb 𝑀𝑅𝐷

Figure 2 A wrench equivalent force system.

ISTUDY

(8) (9)

Solving Eqs. (7)–(9) for unknowns 𝑥, 𝑦, and 𝑀𝑅𝐷 provides 𝑥 = 2.762 in.,

𝑦 = 5.476 in., 𝑀𝑅𝐷 = −1157 in.⋅lb,

(10)

and, from Eq. (5), the magnitude of the force in the wrench force system is 𝐹𝑅 = 229.1 lb.

𝑥

(7)

(11)

⃗ ⃗ The negative value for 𝑀𝑅𝐷 indicates 𝑀 𝑅𝐷 is in the direction opposite 𝐹𝑅 , and hence, this is a negative wrench as defined in Fig. 4.33 on p. 250. This wrench force system is shown in Fig. 2.

ISTUDY

Section 4.4

Discussion & Verification

The force systems shown in Figs. 1 and 2 are equivalent. When determining a wrench force system, we generally identify the location of the wrench by finding one point on its line of action (point 𝐷 in this example). Often, we will find where the wrench’s line of action intersects a plane, such as the 𝑥𝑦, 𝑦𝑧, or 𝑧𝑥 plane. This way one of the coordinates of a point on the wrench’s line of action is known (in this example the 𝑧 coordinate of point 𝐷 is known to be zero). A common mistake is to let all three coordinates of a point on the wrench’s line of action be unknowns; then there are four unknowns but still only three equations available to determine them, and hence a unique solution is not possible.

Equivalent Force Systems

257

258

Chapter 4

Moment of a Force and Equivalent Force Systems

Problems

ISTUDY

Problems 4.84 through 4.86 Determine which of the force systems shown, if any, are equivalent.

3 in.

20 lb

80 in.⋅lb 30 lb

5 in.

2 in.

20 lb

20 lb

10 lb

(a)

5 lb

10 lb

5 lb

6 in. 4 in.

7 in.

5 lb

15 lb

5 lb

40 in.⋅lb

(c)

(b)

(d)

Figure P4.84 100 N

200 N

100 N

200 N

100 N

100 N

500 N⋅m

300 N⋅m

1m

4m

200 N⋅m 3m (b)

(a)

500 N⋅m 2m 2m

4m (c)

(d)

Figure P4.85 100 lb

100 lb 100 lb

100 lb

2 ft

300 lb 4 ft

100 lb

100 lb

100 lb

200 ft⋅lb

100 lb

100 lb

100 lb

2 ft (a)

100 lb (b)

(c)

(d)

Figure P4.86

Problem 4.87 Determine values for forces 𝐹 and 𝑃 and moment 𝑀, if possible, so that the force systems shown in Fig. P4.87(b)–(d) are equivalent to the force system shown in Fig. P4.87(a). 𝑀

100 N

𝐹

𝑃

2m 𝐹

100 N

𝐹

2m

(a)

(b)

(c) Figure P4.87

(d)

ISTUDY

Section 4.4

259

Equivalent Force Systems

Problem 4.88 Determine values for forces 𝐹 and 𝑃 , if possible, so that the force systems shown in Fig. P4.88(b)–(d) are equivalent to the force system shown in Fig. P4.88(a). 𝐹

2 kip

8 ft 4 kip

𝑃

4 ft 4 ft 4 ft 4 ft 4 ft 4 ft (a)

(b)

𝐹 2 kip

𝑃

2 kip

2 kip 𝐹

𝑃

1200 N

(d)

(c)

600 N 𝑦

Figure P4.88

𝐷

800 N

Problem 4.89 The floor of an airplane cargo bay is shown in a horizontal position.

𝑂

(a) Determine an equivalent force system at point 𝑂.

𝐵

(b) Determine an equivalent force system consisting of a single force, and specify the 𝑥 and 𝑧 coordinates of the point where the force’s line of action intersects the floor. (c) Keeping points 𝐴–𝐷 at the locations shown, suggest a repositioning of the forces so that the location of the force system described in Part (b) is closer to point 𝑂.

Problem 4.90 In a vehicle collision reconstruction analysis, an engineer estimates the tire forces shown. Determine an equivalent force system at point 𝑂. 𝑦 200 lb 300 lb 34 in.

𝐷

34 in.

𝐶

𝑂

400 lb 𝐴 200 lb

300 lb

𝐵

70 in.

50 in.

𝐶

300 N

20◦ 𝑥 20◦

400 lb

Figure P4.90 and P4.91

Problem 4.91 For Prob. 4.90, determine an equivalent force system consisting of a single force, specifying the 𝑥 coordinate of where the force’s line of action intersects the 𝑥 axis.

𝐴 𝑧 Figure P4.89

2m

2m 1m 1m

𝑥

260

Chapter 4

Moment of a Force and Equivalent Force Systems

Problem 4.92 A Boeing CH-47 Chinook helicopter is shown. This helicopter was first produced in 1962, is still in production, and it sees extensive military and civilian use worldwide. The rotors counter-rotate so that a vertical stabilizing rotor is not needed. The rotor forces 𝐹𝐴 and 𝐹𝐵 lie in the 𝑥𝑦 plane where 𝛼𝐴 and 𝛼𝐵 are angles about the −𝑧 axis. Forces 𝐹𝐶 and 𝐹𝐷 are due to the thrust from the engines, and they act in the 𝑦 direction. Forces 𝑄𝐴 and 𝑄𝐵 are lift forces from the rotors, and they act in the 𝑧 direction. Moments 𝑀𝐴 and 𝑀𝐵 are due to rotation of the rotors, and they act in the ±𝑧 direction. Points 𝐴–𝐷 lie on the lines of action of the forces and moments (point 𝐷 is on the opposite side of the helicopter and is not seen in this figure), and these points have the coordinates given below. If the forces, moments, and angles have the values given below, determine an equivalent force system acting at the origin of the coordinate system, point 𝑂. 𝑧

𝑀𝐵 𝛼𝐵 𝐹 𝐵

𝑀𝐴

𝑄𝐵

𝐹𝐶

𝐶 𝑥

𝐹𝐴 = 700 lb,

𝛼𝐴 = 5◦

𝛼𝐴

𝑄𝐴 = 9000 lb, 𝐹𝐵 = 600 lb,

𝑀𝐴 = 16,000 ft⋅lb

𝑦

𝑄𝐵 = 8000 lb,

𝛼𝐵 = 8◦ 𝑀𝐵 = 14,000 ft⋅lb

𝐹𝐶 = 400 lb,

𝐹𝐷 = 400 lb.

𝑄𝐴

𝐵

𝐴 𝐹𝐷

𝐴 (0, 35, 12) ft , 𝐵 (0, −40, 20) ft 𝐶 (9, −40, 8) ft , 𝐷 (−9, −40, 8) ft

𝑂

𝐹𝐴

Figure P4.92

Problem 4.93 dani3315/Shutterstock

20 m

𝐹1 𝐹2

𝑦

140 m 𝐹3

𝛼1

150 m 𝐹5

𝑂 17 m

𝛼2 9 m

𝑥

𝐹4

Figure P4.93

The RMS Queen Mary II is a luxury ocean liner that has a propulsion system consisting of four pods mounted at its stern. Each pod contains an electric motor that is powered by a combination of diesel and gas turbine generators. The frontmost two pods are fixed in position parallel to the 𝑥 axis, while the rearmost two pods can rotate 360◦ , thus eliminating the need for a rudder. The ship also has three bow thrusters whose net thrust is shown as 𝐹5 ; these thrusters are used only at low speeds to provide greater maneuverability. Assuming all forces lie in the same plane, determine an equivalent force system at point 𝑂 that is valid for 0◦ ≤ 𝛼1 ≤ 360◦ and 0◦ ≤ 𝛼2 ≤ 360◦ . Express your answer in terms of 𝐹1 –𝐹5 , 𝛼1 , and 𝛼2 .

Problem 4.94 If 𝛼 = 30◦ , 𝑊 = 1 kN, and 𝑄 = 2 kN in Fig. P4.14 on p. 217, determine an equivalent force system consisting of a single force, and specify the distance from point 𝐶 where the force’s line of action intersects member 𝐶𝐵𝐷. 𝑦 20 in.

5 in.

Problem 4.95 If 𝐹 = 200 N and 𝑃 = 300 N in Fig. P4.17 on p. 217, determine an equivalent force system consisting of a single force, and specify the 𝑥 coordinate of the point where the force’s line of action intersects the 𝑥 axis.

10◦ 𝐵 𝑟

𝑟 = 2 in. 𝐴 𝑂

𝑊 Figure P4.96

ISTUDY

𝑟

4 in. 𝑥

Problem 4.96 The structure has two frictionless pulleys around which a string is wrapped. If 𝑊 = 50 lb, determine a force system that is equivalent to the forces applied by the string, where this force system consists of a single force, and specify the 𝑥 coordinate of the point where the force’s line of action intersects the 𝑥 axis.

ISTUDY

Section 4.4

261

Equivalent Force Systems 𝑧

Problem 4.97 30 cm

Two bottles with the weights shown rest on a table.

40 cm

12 N

(a) Determine an equivalent force system at point 𝑂.

8N

𝐴

(b) Determine an equivalent force system consisting of a single force, specifying the 𝑥 and 𝑦 coordinates of the point where the force’s line of action intersects the top of the table.

Problem 4.98

𝑂

40◦

𝐵

𝑦

𝑥 Figure P4.97

A thick circular plate is used in a machine called a gyratory compactor to determine the mechanical properties of hot asphalt concrete. The plate is instrumented with load cells at points 𝐴, 𝐵, and 𝐶 where each load cell is located at the same 80 mm radial distance from point 𝑂. If the load cells measure forces 𝐹𝐴 = 500 N, 𝐹𝐵 = 600 N, and 𝐹𝐶 = 700 N, determine

𝐹𝐶

𝑧

𝐹𝐵

𝐹𝐴

(a) an equivalent force system at point 𝑂,

30◦

(b) an equivalent force system consisting of a single force, specifying the 𝑥 and 𝑦 coordinates of the point where the force’s line of action intersects the 𝑥𝑦 plane.

𝐴

𝐶

𝑂 80 mm

𝐵 30◦ 𝑦

𝑥

Problem 4.99 A boat trailer is subjected to the forces shown where the forces at points 𝐴–𝐸 are vertical and the forces at points 𝐹 and 𝐺 lie in the 𝑦𝑧 plane. Determine

Figure P4.98

(a) an equivalent force system at point 𝑂, (b) an equivalent force system consisting of a single force, specifying the 𝑥 and 𝑦 coordinates of the point where the force’s line of action intersects the 𝑥𝑦 plane. 𝑧 400 lb

350 lb

300 lb 900 lb 500 lb

400 lb 𝐵 18 in. 18 in. 𝐴

300 lb

30◦ 𝐺 𝐷

𝐶

45◦ 𝐹

𝐸

𝑦

𝑂 18 in.

24 in.

36 in.

24 in.

24 in. 𝑦

𝑥 Figure P4.99 𝑂

Problem 4.100 The tip of an atomic force microscope (AFM) is subjected to the forces shown (if you are curious, see the description in Prob. 4.5 on p. 215 of what an AFM is used for). Use a vector approach to determine (a) an equivalent force system at point 𝐵, (b) an equivalent force system at point 𝑂.

Problem 4.101 Repeat Prob. 4.100, using a scalar approach.

𝑥

80 𝜇m

10 𝜇m 𝐵

𝑧

12 𝜇m

𝐴 5 nN 10 nN

Figure P4.100 and P4.101

50 nN

262

Chapter 4

Moment of a Force and Equivalent Force Systems 𝑦

Problem 4.102

2 kN

𝐷

𝐴 𝑧

(a) a vector approach,

𝐸

𝐶

Determine an equivalent force system at point 𝐴, using

10 kN

𝐵

2 kN

5 kN⋅m 2m

(b) a scalar approach.

2m

Problem 4.103 Repeat Prob. 4.102 to determine an equivalent force system at point 𝐵.

𝑥

2m

Problems 4.104 and 4.105

Figure P4.102 and P4.103

Using inspection, if possible, determine a wrench equivalent force system and specify the 𝑥 and 𝑦 coordinates of the point where the wrench’s line of action intersects the 𝑥𝑦 plane. Express your answers in terms of parameters such as 𝐹 , 𝑃 , and 𝑟. 𝑧

𝑧

𝐹

𝐹

𝐹 𝐹

𝑃

3𝑟

𝑃

𝑦

𝑃 𝑟 𝑦

𝑦

𝑟 3m

𝑥

𝑥 Figure P4.104

3m 𝑂

𝑃 2𝑟

𝑥

Figure P4.105

Problem 4.106

3m 𝐴

̂ N and 𝑀 ⃗ = The object shown is subjected to a wrench having 𝐹⃗𝑅 = (2 𝚤̂ + 3 𝚥̂ − 4 𝑘) 𝑅 ̂ N ⋅ m and whose line of action intersects the 𝑥𝑦 plane at 𝑥 = 2 m and (4 𝚤̂ + 6 𝚥̂ − 8 𝑘) 𝑦 = 1 m. Determine an equivalent force system at point 𝐴, stating this in vector form.

𝑧 Figure P4.106

Problem 4.107 Determine a wrench equivalent force system and specify the 𝑥 and 𝑦 coordinates of the point where the wrench’s line of action intersects the 𝑥𝑦 plane. 𝑦

4 N⋅m 1m

0.8 m 𝑥 6 N⋅m

𝑧 𝐹 𝐹 𝑏 𝑥 Figure P4.108

ISTUDY

10 N

𝑧 𝑦

Figure P4.107

𝑎

Problem 4.108 Determine a wrench equivalent force system and specify the 𝑥 and 𝑦 coordinates of the point where the wrench’s line of action intersects the 𝑥𝑦 plane. Express your answers in terms of parameters such as 𝐹 , 𝑎, and 𝑏.

ISTUDY

Section 4.5

Chapter Review

4.5 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the section numbers cited for additional details. 𝑀𝑂 = 𝐹 𝑑

Moment of a force Moment of a force — scalar approach. As shown in Fig. 4.35, the moment of a force 𝐹⃗ about a point 𝑂 is a vector (represented by the twisting action shown), and the magnitude of this moment is 𝑀𝑂 , which is given by

moment arm

𝑂

𝑑

𝐹

Eq. (4.1), p. 204 𝑀𝑂 = 𝐹 𝑑, where

Figure 4.35 Scalar definition of the moment of a force.

𝐹 is the magnitude of the force; 𝑑 is the perpendicular distance from point 𝑂 to the line of action of 𝐹⃗ and is called the moment arm; and 𝑀𝑂 has units of force times length. The direction of the moment is not provided by Eq. (4.1), but is understood to be as follows. The line of action of the moment is parallel to the axis through point 𝑂 that is perpendicular to the plane containing 𝐹⃗ and the moment arm. The direction of the moment along the line of action is given by the direction of the thumb of your right hand when your fingers curl in the twisting direction of the moment.

Moment of a force — vector approach. As shown in Fig. 4.36, the moment of a ⃗ and is defined as force 𝐹⃗ about a point 𝑂 is denoted by 𝑀 𝑂

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝑂 𝑂

𝑟⃗

𝐹⃗

Eq. (4.2), p. 205 ⃗ = 𝑟⃗ × 𝐹⃗ , 𝑀 𝑂 where 𝐹⃗ is the force vector; and 𝑟⃗ is a position vector from point 𝑂 to any point on the line of action of 𝐹⃗ .

Varignon’s theorem (principle of moments) Varignon’s theorem, also known as the principle of moments, is a restatement of the distributive property of the cross product. The principle states that the moment of a force is equal to the sum of the moments of the vector components of the force. Thus, if 𝐹⃗ has vector components 𝐹⃗1 , 𝐹⃗2 , and so on, then the moment of 𝐹⃗ about a point 𝐴 is given by Eq. (4.10), p. 207 ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴 = 𝑟⃗ × (𝐹⃗1 + 𝐹⃗2 + ⋯) = 𝑟⃗ × 𝐹⃗1 + 𝑟⃗ × 𝐹⃗2 + ⋯ . The principle of moments is most commonly used for scalar evaluations of the moment of a force. Often, the vector components will be orthogonal, but the principle is also valid for nonorthogonal vector components.

Figure 4.36 Vector definition of the moment of a force.

263

264

Chapter 4

Moment of a Force and Equivalent Force Systems

Moment of a force about a line

𝑧 𝐹⃗ ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝑃 ⃗ ⋅ 𝑢̂ 𝑀𝑎 = 𝑀 𝑃 = 𝑟⃗ × 𝐹⃗ ⋅ 𝑢̂ 𝑦

𝑟⃗ 𝑃 𝑢̂

Vector approach:

𝑎

𝑥

1. Select a point 𝑃 at a convenient location on line 𝑎. Determine the moment of 𝐹⃗ ⃗ = 𝑟⃗ × 𝐹⃗ , where 𝑟⃗ is a position vector from 𝑃 to any point on about 𝑃 , using 𝑀 𝑃 the line of action of 𝐹⃗ .∗ ⃗ ⋅ 𝑢, 2. 𝑀𝑎 = 𝑀 ̂ is a unit vector in the direction of 𝑎. To express this moment 𝑃 ̂ where 𝑢 ⃗ = 𝑀 𝑢. as a vector quantity, evaluate 𝑀 ̂

(a) vector approach 𝑧 𝐹

𝑎

𝐹⊥

𝐹‖

The moment of a force about a line is defined to be the component of the moment that is in the direction of the line. The moment of a force about a line is discussed in Section 4.2 and can be evaluated by using vector and scalar approaches, as follows. To determine the moment 𝑀𝑎 of a force 𝐹 about a line (or direction) 𝑎 as shown in Fig. 4.37:

𝑀𝑎 = 𝐹 ⊥ 𝑑 𝑦

𝑑

𝑎

Note: Steps 1 and 2 may be combined to yield 𝑀𝑎 directly by using the scalar triple ̂ product, 𝑀𝑎 = (⃗𝑟 × 𝐹⃗ ) ⋅ 𝑢. Scalar approach:

1. Resolve 𝐹 into components 𝐹⊥ and 𝐹‖ that are perpendicular and parallel, respectively, to a plane containing line 𝑎.

𝑎

𝑥

(b) scalar approach Figure 4.37 Vector and scalar approaches for determining the moment of a force about a line.

2. 𝑀𝑎 = 𝐹⊥ 𝑑, where 𝑑 is the moment arm (shortest distance) between line 𝑎 and the line of action of 𝐹 . (Note: 𝐹‖ produces no moment about 𝑎, so you may skip its evaluation altogether.)

Couple and the moment produced by a couple ⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴𝐵

𝑧 −𝐹⃗ 𝐴 𝑟⃗𝐴𝐵

𝐶 𝐵

A couple is defined to be a system of two forces of equal magnitude and opposite direction and whose lines of action are separated by a distance. The moment of a couple (sometimes also called a couple moment) is the moment produced by the couple. A couple that is applied to a body produces a moment, but does not apply any net force to the body. The moment of a couple can be evaluated using both vector and scalar approaches as follows.

Moment of a couple — vector approach. Consider a couple consisting of two par⃗ of this couple is allel forces 𝐹⃗ and −𝐹⃗ as shown in Fig. 4.38. The moment 𝑀

𝐹⃗

Eq. (4.11), p. 233

𝑦

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐴𝐵

𝑥 Figure 4.38 Moment of a couple: vector description.

𝐹 𝑑

Moment of a couple — scalar approach. Consider a couple consisting of two parallel forces having the same magnitude 𝐹 as shown in Fig. 4.39. The magnitude of the moment of this couple is Eq. (4.12), p. 233

𝐶 𝐹 𝑦

𝑥 Figure 4.39 Moment of a couple: scalar description. ISTUDY

where 𝑟⃗𝐴𝐵 and 𝑟⃗𝐵𝐴 are position vectors; 𝐴 is any point on the line of action of −𝐹⃗ ; and 𝐵 is any point on the line of action of 𝐹⃗ .

𝑀 = 𝐹𝑑

𝑧

= 𝑟⃗𝐵𝐴 × (−𝐹⃗ ),

𝑀 = 𝐹 𝑑, ⃗ . As an alternative, especially for probdetermine 𝑀 𝑃 ⃗ , lems with simple geometry, you could use a scalar approach to determine the vector expression for 𝑀 𝑃 to be followed by taking the dot product as described in Step 2.

∗ This procedure suggests using the cross product to

ISTUDY

Section 4.5

Chapter Review

265

where 𝑑 is the perpendicular (shortest) distance between the forces’ lines of action; and the direction of the moment is perpendicular to the plane containing the forces. Two couples are said to be equivalent if the moment vectors they produce are identical (both the magnitude and direction of the moments must be the same). If an object or structure has multiple couples applied, a resultant couple moment may be determined by summing the individual couple moment vectors. The moment of a couple is a free vector, meaning the moment may be positioned anywhere on an object or structure. Further explanation of this subtle feature is given throughout Section 4.3.

Transmissibility of a force The principle of transmissibility of a force, described in Section 4.4, states that the external effects of a force applied to a rigid body are the same, regardless of the point of application of the force along its line of action.

Equivalent force systems If an object or structure has an arbitrary number of forces and/or moments applied as shown in Fig. 4.40, an equivalent force system at a point 𝐴 consists of a resultant force ⃗ where 𝐹⃗𝑅 and a resultant moment 𝑀 𝑅 Eq. (4.15), p. 248

𝑧

𝐹⃗𝑅 =

∑

𝐹⃗𝑖 ,

∑

𝑟⃗𝑖 × 𝐹⃗𝑖 +

𝑖=1

𝑚 ∑

𝐹⃗2

𝐹⃗1

𝑖=1

(a)

Eq. (4.16), p. 248 𝐹⃗𝑅

)

system 1

) ( ⃗ 𝑀 𝑅 system 1

( ) = 𝐹⃗𝑅 system 2 and ) ( ⃗ = 𝑀 𝑅 system 2 ,

⃗ ) ⃗ where the moment summation points used for determining (𝑀 𝑅 system 1 and (𝑀𝑅 )system 2 must be the same. Equation (4.16) may be stated more explicitly as two force systems are equivalent if Eq. (4.17), p. 248 𝑛1 (∑ 𝑛1 (∑ 𝑖=1

𝐹⃗𝑖

𝑖=1 𝑚1

𝑟⃗𝑖 × 𝐹⃗𝑖 +

∑ 𝑖=1

⃗ 𝑀 𝑖

)

=

)

=

system 1

system 1

𝑧

⃗ 𝑀 2

𝑥

⃗ , 𝑀 𝑖

where 𝑛 is the number of forces that are applied, 𝑚 is the number of moments that are applied, and 𝑟⃗𝑖 is a position vector from point 𝐴 to anywhere on the line of action of 𝐹⃗𝑖 . ⃗ is a free vector, it may be positioned anywhere on the object or structure. Because 𝑀 𝑅 Two force systems are equivalent if they have the same resultant force and produce the same resultant moment about any common point. Thus, force system 1 and force system 2 are equivalent if (

𝐹⃗𝑖

𝐴

𝑦

𝑖=1 𝑛

⃗ = 𝑀 𝑅

⃗ 𝑀 𝑖

⃗ 𝑀 1

𝑛

𝑛2 (∑

𝑖=1 ( 𝑛2

∑ 𝑖=1

𝐹⃗𝑖

)

and

system 2

𝑟⃗𝑖 × 𝐹⃗𝑖 +

𝑚2 ∑ 𝑖=1

⃗ 𝑀 𝑖

)

, system 2

𝐹⃗𝑅

⃗ 𝑀 𝑅

𝑦 𝑥 (b)

Figure 4.40 Construction of an equivalent force system at point 𝐴.

266

ISTUDY

Moment of a Force and Equivalent Force Systems

Chapter 4

where 𝑛1 and 𝑚1 are the number of forces and moments in system 1, respectively, 𝑛2 and 𝑚2 are the number of forces and moments in system 2, respectively, and the moment summation point used for both force systems must be the same.

Some special force systems. Several categories of force systems, including concurrent, coplanar, parallel, and wrench force systems, are defined and studied in Section 4.4. If a particular force system is concurrent, coplanar, or parallel, then an equivalent force system that consists of a single resultant force (no moment) with appropriate position can always be found. For more general force systems, an equivalent force system called a wrench, which consists of a single resultant force with appropriate position plus a moment that is parallel to the resultant force, can always be found.

ISTUDY

Section 4.5

267

Chapter Review

Review Problems Problem 4.109

𝑧 𝐹⃗2

In orthodontics, teeth are repositioned by applying forces to them for prolonged periods ̂ N, 𝐹⃗ = of time. If point 𝐴 has coordinates (2, 7, 8) mm, 𝐹⃗1 = (0.8 𝚤̂ − 0.3 𝚥̂ − 0.1 𝑘) 2 ̂ ⃗ ̂ (−0.7 𝚤̂ − 0.1 𝚥̂ + 0.05 𝑘) N, and 𝑀 = (0.1 𝚤̂ + 1.4 𝚥̂ − 0.3 𝑘) N⋅mm, determine an equivalent force system at point 𝑂.

𝐹⃗1

Problem 4.110

⃗ 𝑀

𝐴 𝑂

𝑦

𝑥

The 60 N forces lie in planes parallel to the 𝑦𝑧 plane. Determine the resultant couple moment vector for the force system shown using a

Figure P4.109

(a) vector approach, (b) scalar approach.

60 N

𝑧 60 N

Problem 4.111

60◦

50 N

Two force systems are applied to a right circular cylinder. Points 𝐴 to 𝐷 lie on the 𝑥𝑧 plane, and points 𝐸 and 𝐹 lie on the 𝑦𝑧 plane. Determine if these force systems are equivalent. 𝑧 150 N 1m 𝐶

3m

100 N

𝐸

𝐵

𝐵

3m

50 N

𝐴

𝑦

𝑥 150 N

50 N

50 N 𝐹

𝑥

(a)

𝑦

𝐶

40 N

𝐺

40 N

𝐹 𝐴

300 N⋅m 100 N 𝐶 150 N

60◦

𝐸 𝐷

𝑧

𝐷

𝐵

2m

50 N

6m

𝑥 Figure P4.110

100 N 𝑦

𝑧 0.1 m

(b)

𝐶

𝐴

0.2 m

scratch

𝐺

Figure P4.111 𝐷

𝐵

Problems 4.112 and 4.113 A thin cantilever-supported rectangular plate is made of a brittle material. The plate will crack along line 𝐴𝐵 when the moment about this line reaches 200 N⋅m. The plate also has a deep scratch along line 𝐶𝐷 and will crack when the moment about this line reaches 100 N⋅m. The forces 𝐹 and 𝑄 act in the −𝑧 direction. Problem 4.112

If 𝑄 = 0, determine the value of 𝐹 that will cause the plate to crack.

Problem 4.113

If 𝐹 = 0, determine the value of 𝑄 that will cause the plate to crack.

0.3 m

0.15 m

𝑥

𝑄

𝐸 𝐹

Figure P4.112 and P4.113

𝑦 𝑎 40◦ 𝑂

𝐹⃗ 𝑥

Problem 4.114 The landing gear for a fighter jet rotates about line 𝑎 so that it retracts into the fuselage. Point 𝐴 has the coordinates given, line 𝑎 lies in the 𝑦𝑧 plane, and 𝐹⃗ = (300 𝚤̂ + 400 𝚥̂ − ̂ N. 200 𝑘)

𝑧

(b) Determine a new direction for line 𝑎, which probably no longer lies in the 𝑦𝑧 plane, so that the moment of 𝐹⃗ about this line is as large as possible.

𝐴 𝐴 (0.1, −0.4, 0.6) m

(a) Determine the moment of 𝐹⃗ about line 𝑎. Figure P4.114

𝑦

268

𝑦

Problem 4.115

𝑎

6 ft

𝑊

15◦

𝐶

𝐵 𝑧

Chapter 4

Moment of a Force and Equivalent Force Systems

𝐸

𝑥 2 ft

𝐴

𝐷 𝐹

Figure P4.115

Problem 4.116

1500 lb 𝐶

A jack stand for supporting an automobile or truck during servicing is shown. The stand has three legs with equal radial positioning. The height ℎ of the stand can be adjusted between 9 in. and 15 in. If the stand supports a 1500 lb vertical force, determine the smallest horizontal force 𝐹 that will cause the moment about line 𝐴𝐵 to be zero and hence, cause the stand to tip.

𝐹

ℎ

The device shown is a pointer that mounts on the front of a tractor to help its operator position the tractor relative to other rows of seeds that have been planted. Bracket 𝐸 is bolted to the front of the tractor, which drives in the 𝑧 direction. The bracket supports the bent boom 𝐴𝐵𝐶, the end 𝐶 of which has a weight 𝑊 and a pointer 𝐶𝐷. The boom is allowed to rotate about line 𝑎, which lies in the 𝑦𝑧 plane, so that if the boom strikes an obstruction, the boom will rotate backward and upward to help avoid damage to it. If 𝑊 = 50 lb, determine the force 𝐹 , which acts in the −𝑧 direction, that will cause the boom to begin rotating about line 𝑎.

6 in. 𝑂 120◦

𝐴

Problem 4.117 Rather than the traditional horizontal and vertical stabilizers, some aircraft such as the Bonanza 35 single-engine airplane (pictured) and the F-117 Stealth fighter feature a V tail. Points 𝐴 and 𝐵 are located at 𝐴 (112, 7, 0) mm and 𝐵 (62, −93, 0) mm, and direction 𝑎 has direction angles 𝜃𝑥 = 150◦ , 𝜃𝑦 = 60◦ , and 𝜃𝑧 = 90◦ . If the forces are 𝐹⃗1 = (40 𝚤̂ − ̂ N and 𝐹⃗ = (50 𝚥̂ + 120 𝑘) ̂ N, determine the resultant moment of 𝐹⃗ and 𝐹⃗ 100 𝚥̂ + 280 𝑘) 2 1 2 about line 𝑎.

𝐵

Figure P4.116

𝑦 𝑎 𝑎

𝑧 𝑂

𝐷

𝐹⃗2

𝐶

𝐵 𝐴

𝐴 (40, 0, 0) in. 𝐵 (−4, 0, 0) in. 𝐶 (0, 0, 24) in. 𝐷 (20, 0, 54) in. 𝐸 (5, −30, 28) in. 𝑦 (west)

Figure P4.118

ISTUDY

𝑥

𝐵

𝐸

𝑥 (north)

𝐴

𝑧

𝐹⃗1 Figure P4.117

Problem 4.118 A solar panel is supported by a fixed structure 𝐴𝐵𝐶𝐷. The screw-driven actuator 𝐴𝐸 causes the solar panel to rotate on portion 𝐶𝐷 of the structure so that the panel tracks the motion of the Sun during the day. If the actuator applies a 12 lb force to point 𝐸, where the 𝑧 component of this force is positive, determine the moment of this force about line 𝑎.

ISTUDY

Section 4.5

269

Chapter Review

Problem 4.119

𝐹⃗𝐸

𝑧

The front of a bicycle is shown. For all parts of this problem, the force applied to point 𝐷 ̂ N. is 𝐹⃗ = (2 𝚤̂ + 3̂𝚥 − 6 𝑘)

𝐸 𝐶

𝐷

(a) Determine the moment of 𝐹⃗𝐷 about line 𝐴𝐵. ⃗ be the moment of the couple. For each of the (b) Let 𝐹⃗𝐷 and 𝐹⃗𝐸 be a couple, and let 𝑀 following statements, select True or False.

𝑦

𝐹⃗𝐷

𝐵 𝐴

𝐷

⃗ = 𝑟⃗ × 𝐹⃗ 𝑀 𝐷𝐸 𝐷 ⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ 𝑀 𝐵𝐷 𝐷 𝐵𝐸 𝐸 ⃗ = 𝑟⃗ × 𝐹⃗ + 𝑟⃗ × 𝐹⃗ 𝑀 𝐶𝐷 𝐷 𝐶𝐸 𝐸

𝑥

T or F? T or F? T or F?

(c) If 𝐹⃗𝐷 and 𝐹⃗𝐸 are a couple, determine the moment of the couple.

𝐴 (0, 150, 50) mm 𝐵 (0, 100, 180) mm 𝐷 (300, −100, 200) mm 𝐶 (0, 40, 290) mm 𝐸 (−300, −100, 200) mm

Problems 4.120 and 4.121 Frictionless pulleys at 𝐵 and 𝐶 are attached to member 𝐵𝐶𝐸𝐺, which is horizontal when 𝛼 = 0◦ . Wire 𝐴𝐵𝐶𝐷 wraps around the pulleys. End 𝐴 of the wire supports a weight 𝑊 = 200 N, and end 𝐷 of the wire is supported by a frictionless pulley so that segment 𝐶𝐷 of the wire is always vertical.

Figure P4.119

𝐷

Determine the moment that wire 𝐴𝐵𝐶𝐷 applies to member 𝐵𝐶𝐸𝐺

Problem 4.120

if 𝛼 = 20◦ .

30 mm 𝐺 𝛼

Problem 4.121

(a) Determine the moment that wire 𝐴𝐵𝐶𝐷 applies to member 𝐵𝐶𝐸𝐺 if 0◦ ≤ 𝛼 ≤ 90◦ . Plot this moment versus 𝛼. (b) If 𝛼 may exceed 90◦ , determine the largest value of 𝛼, namely 𝛼max , for which your answer to Part (a) is valid.

𝐸

𝑟

𝑟 𝐵 𝑟 = 10 mm 𝐴 𝑊 𝐶

Figure P4.120 and P4.121

Problem 4.122 Determine values for 𝑃 , 𝑄, 𝑅, 𝑆, and 𝑀𝐷 so that the two force systems shown are equivalent. In these figures, the 5 lb force, 10 lb force, 𝑃 , and 𝑄 are parallel to the 𝑥 direction; the 20 lb force, 50 in.⋅lb moment, 𝑆 and 𝑀𝐷 are parallel to the 𝑦 direction; and 𝑅 acts in the −𝑧 direction. 𝑧 𝐸 50 in.⋅lb 𝐴

𝐶

𝐵 𝐷

𝑥

5 in.

5 in. 10 lb

𝑧

5 lb

𝐹

𝑃 𝑅

4 in. 6 in.

20 lb

𝑆

𝑦 𝑄

𝑦

𝑀𝐷

𝑥 system 2

system 1 Figure P4.122

Problem 4.123

𝑦

4 kN 2 kN

3 kN

A beam is subjected to the three forces shown. (a) Determine an equivalent force system at the midspan of the beam 𝑥 = 3 m. (b) Determine an equivalent force system consisting of a single force, and specify the 𝑥 coordinate of the point where its line of action intersects the 𝑥 axis.

𝑥 2m Figure P4.123

1m

2m

1m

270

Chapter 4

Moment of a Force and Equivalent Force Systems 𝑦 0.4 lb

Problem 4.124 𝐶

4

3

A speed control mechanism for a small gasoline engine is shown. 0.3 in.

(a) Determine an equivalent force system at point 𝑂.

0.9 in.

(b) Determine an equivalent force system consisting of a single force, and specify the 𝑦 coordinate of the point where its line of action intersects the 𝑦 axis.

𝐵

0.6 lb

𝑂

𝑥 0.8 in.

𝐴

Problem 4.125 A weldment is a structure or component built by welding an assembly of pieces together. The weldment shown is subjected to three cable forces. Determine an equivalent force system consisting of a single force, and specify the 𝑥 coordinate of the point where its line of action intersects the 𝑥 axis.

1.2 in. 0.3 lb Figure P4.124

Problem 4.126 800 N

600 N

12

𝑦

5 3

The wing of a jet supports a 900 lb force due to the weight of an engine and 300 lb and 600 lb forces due to the weight of fuel.

4

(a) Determine an equivalent force system at point 𝐴.

15 mm

(b) Determine an equivalent force system consisting of a single force, and specify the 𝑥 and 𝑧 coordinates of the point where its line of action intersects the 𝑥𝑧 plane.

400 N

𝐶

𝐵

55 mm

12

5

𝑧

𝐴

50 mm

𝑦

𝑂 50 mm 55 mm 15 mm

𝑥

600 lb 7 ft 3 ft

Figure P4.125

5 ft

𝐶

7 ft

60 mm

300 lb 15 ft

900 lb 𝐵

𝐴

150 N 3 ft

𝑥

100 N

150 N

𝑂 200 N

𝑧

𝑦

30 mm 𝑦

𝑥 Figure P4.126

Figure P4.127

Problem 4.127 𝐶

A thin rectangular flat plate is loaded by the forces shown.

40 lb

(a) Determine an equivalent force system at point 𝑂.

18 in.

(b) Determine a wrench equivalent force system, and specify the 𝑦 and 𝑧 coordinates of the point where the wrench’s line of action intersects the 𝑦𝑧 plane.

90 lb

30 lb 𝐵

10 in.

𝐴

Problem 4.128 25 lb 10 lb

36 in. 30◦

26 in.

30◦ 𝐷

𝐸 𝑂

18 in. 15 in.

8 in.

8 in.

𝑧 Figure P4.128 and P4.129

ISTUDY

𝑥

A seat of a roller coaster is subjected to the forces shown during a turn. The force at 𝐴 is vertical, the forces at 𝐵 and 𝐶 are parallel to the 𝑥 and 𝑧 directions, respectively, and the forces at 𝐷 and 𝐸 lie in planes parallel to the 𝑦𝑧 plane. Determine an equivalent force system at point 𝑂, where the seat is attached to the car.

Problem 4.129 For the roller coaster seat described in Prob. 4.128, determine a wrench equivalent force system and specify the 𝑥 and 𝑧 coordinates of the point where the wrench’s line of action intersects the 𝑥𝑧 plane.

ISTUDY

5

Equilibrium of Bodies

This chapter discusses equilibrium of a single rigid body. The rigid body may be a single object, or it may be an assemblage of numerous members whose arrangement is such that the assemblage as a whole can be treated as a single rigid body. Because all materials are deformable, there are no true rigid bodies in nature. Nonetheless, it is remarkable how often a rigid body idealization can be used.

NASA/Kennedy Space Center

The Space Shuttle Columbia sits on a launch pad at Kennedy Space Center, Florida, before its maiden flight in 1981. For many purposes, the Space Shuttle with its attached external fuel tank and two solid fuel rocket boosters may be modeled as a single body.

5.1 Equations of Equilibrium We begin this chapter with a general discussion of the equations of equilibrium for bodies in three dimensions. Subsequent sections consider applications to both twoand three-dimensional problems. Shown in Fig. 5.1 are objects whose equilibrium we will analyze. The object may be a single body of solid material as shown in Fig. 5.1(a), or it may consist of an arrangement of numerous members as shown in Fig. 5.1(b). In either case, we idealize the object to be a rigid body so that the distance between any two points within the object remains constant under all circumstances of loading and/or motion. The object will have forces and/or moments applied to it, and it will often also have supports that fix its position in space. Note that there are many objects whose equilibrium we

𝑧

𝑧 𝑦

𝑥

𝑦 𝑥

(a)

(b)

Figure 5.1 Examples of objects that may be idealized as rigid bodies.

271

272

Chapter 5

Equilibrium of Bodies

are interested in that are unsupported or are only partially supported against motion. Examples include aerospace vehicles, ships, automobiles rolling on a highway, moving components in machines, and so on. Our objective in this chapter is to determine conditions under which equilibrium of a single rigid body is obtained. In the case of objects that consist of an assemblage of members, we may also be interested in the forces supported by the individual members, and this topic is taken up in Chapter 6. Similarly, for objects that consist of solid material only, we may also be interested in the internal forces that develop within the material, and this topic is taken up in Chapter 8. The equations governing the static equilibrium of a body are ∑

𝐹⃗ = 0⃗

and

∑

⃗ ⃗ = 0, 𝑀 𝑃

(5.1)

where the summations include all forces and moments that are applied to the body, and the moments are evaluated about any convenient point 𝑃 that we may select. In scalar form, Eq. (5.1) is ∑

∑

∑

𝐹𝑥 = 0 𝐹𝑦 = 0

and

𝐹𝑧 = 0

∑

𝑀𝑃 𝑥 = 0

∑

𝑀𝑃 𝑧 = 0.

∑

𝑀𝑃 𝑦 = 0

(5.2)

∑ The force equilibrium equations 𝐹⃗ = 0⃗ are identical to those used for equilibrium ∑ ⃗ ⃗ of particles. The moment equilibrium equations 𝑀 𝑃 = 0 are new, and the rationale for these is that if they are not satisfied, the body will undergo angular accelerations due to the twisting action of moments. Both equations are postulates, and no fundamental proof of their validity exists. Rather, we must accept these as laws that nature follows.

5.2

Equilibrium of Rigid Bodies in Two Dimensions

The object shown in Fig. 5.2 can be modeled using two dimensions if all forces, including the forces applied by the supports, lie in the same plane and all moments have direction perpendicular to this plane. In two dimensions, with 𝑥 and 𝑦 being the ∑ ∑ in-plane coordinates as shown in Fig. 5.2, the equations 𝐹𝑧 = 0, 𝑀𝑃 𝑥 = 0, and ∑ 𝑀𝑃 𝑦 = 0 in Eq. (5.2) are always satisfied, leaving

𝑦

∑

𝑥 Figure 5.2 An object in two dimensions in equilibrium.

ISTUDY

𝐹𝑥 = 0,

∑

𝐹𝑦 = 0,

and

∑

𝑀𝑃 𝑧 = 0.

(5.3)

For brevity, in the rest of this section, as well as for two-dimensional problems in general, we will drop the 𝑧 subscript in the moment equilibrium equation. As stated above, the summations in Eq. (5.3) include all forces and moments applied to the body, including the reaction forces and moments.

Reactions A reaction is a force or moment exerted by a support on a structure. Supports and the reactions they produce take many forms, as shown in Fig. 5.3. In all cases, the

ISTUDY

Section 5.2

Equilibrium of Rigid Bodies in Two Dimensions

reactions for a particular support may be determined by considering the motion the support prevents. That is, if a support prevents translation (or rotation) in a certain direction, it can do so only by producing a reaction force (or moment) in that direction. For example, in the case of a pin support, the pin prevents translation of the bar in both the 𝑥 and 𝑦 directions and thus must produce reaction forces in each of these directions. The pin does not prevent rotation of the bar, and hence, there is no moment reaction. It is not necessary to memorize the reactions shown in Fig. 5.3. Rather, you should reconstruct these as needed.

Support

Reactions

𝑦 𝑥 roller

rocker

roller 𝑅𝑦

𝑅𝑦 frictionless surface

roller in loosefitting track

collar on frictionless bar

𝜃 link or bar

𝜃 cable

𝑅 𝜃

𝜃

𝜃 cable with self-weight

curved link or bar

𝑦 or

𝑅𝑥 𝑥 pin

𝑅𝑦

rough surface

𝛼

𝑦 𝑅𝑥

𝑀

𝑀 or

𝑥 built-in or clamped

𝑅

𝑅𝑦

𝑅 𝛼

𝑦

𝑥 moment-resisting collar

𝑀 𝑅𝑦

Figure 5.3. Common supports for bodies in two dimensions and their associated reactions. Cable, link, and bar support members are assumed to be weightless except where otherwise noted.

273

274

ISTUDY

Chapter 5

Equilibrium of Bodies

Free body diagram (FBD) A free body diagram (FBD) is a sketch of a body or structure that includes all forces and moments that are applied to the body or structure. The FBD is an essential aid for accurate application of the equations of equilibrium, and all of the remarks made on FBDs in Chapter 3 are applicable here.

Procedure for Drawing FBDs 1. Decide on the body (or portion of a body) whose equilibrium is to be analyzed. 2. Imagine this body is “cut” completely free (separated) from its environment. That is: • In two dimensions, think of a closed line that completely encircles the body. • In three dimensions, think of a closed surface that completely surrounds the body. 3. Sketch the body. 4. Sketch the forces and moments: (a) Sketch the forces and moments that are applied to the body by the environment (e.g., weight). (b) Wherever the cut passes through a structural member, sketch the forces and moments that occur at that location. (c) Wherever the cut passes through a support (i.e., where a support is removed from the body), sketch the reaction forces and/or moments that occur at that location. 5. Sketch the coordinate system to be used. Add pertinent dimensions and angles to the FBD to fully define the locations and orientations of all forces and moments.

The order in which the forces are sketched in Step 4 is irrelevant. For complicated FBDs, it may be difficult to include all of the dimensions and/or angles in Step 5. When

this is the case, some of this information may be obtained from a different sketch. In this chapter we consider equilibrium of an entire body or structure only. Hence, when we construct the FBD, the cut taken in Step 2 of the procedure will encompass the entire object. Thus, in this chapter the forces cited in Step 4(b) do not arise. Occasionally, objects are supported by cables and/or bars, and when cutting through these supports to draw an FBD, we categorize their forces as reactions. Once the FBD is drawn, equilibrium equations may be written and solved to obtain the solution to the problem. In this process, drawing the FBD is the most important step, since once a proper FBD is available, writing the equilibrium equations and solving these are routine. The solution of a typical problem is outlined in the following mini-example.

ISTUDY

Section 5.2

275

Equilibrium of Rigid Bodies in Two Dimensions

200 mm

Mini-Example Determine the support reactions for the structure shown in Fig. 5.4.

300 mm 80 N

𝐴 50 mm

Solution The completed FBD is shown in Fig. 5.5, and it is constructed as follows. We first sketch the structure and then choose an 𝑥𝑦 coordinate system. At each support that is cut through (or removed from the structure) we introduce the appropriate reactions. Thus, for the pin support at 𝐴 we introduce reactions 𝐴𝑥 and 𝐴𝑦 , and at the roller support at 𝐵 we introduce reaction 𝐵𝑦 . Next, we use Eq. (5.3) to write the equilibrium equations

60 N

𝐵

Figure 5.4 A structure with supports at points 𝐴 and 𝐵. 𝑦 𝑥 𝐴𝑥

200 mm

300 mm 80 N

𝐴

𝐶

50 mm

∑

∑

∑

𝐹𝑥 = 0 ∶

𝐴𝑥 + 60 N = 0,

(5.4)

𝐹𝑦 = 0 ∶

𝐴𝑦 + 𝐵𝑦 − 80 N = 0,

(5.5)

𝑀𝐴 = 0 ∶

𝐵𝑦 (500 mm) − (80 N)(200 mm) + (60 N)(50 mm) = 0.

(5.6)

𝐴𝑦

𝐵

60 N

𝐵𝑦 Figure 5.5 Free body diagram showing all forces applied to the object.

Solving Eqs. (5.4)–(5.6) provides 𝐴𝑥 = −60 N,

𝐴𝑦 = 54 N,

and

𝐵𝑦 = 26 N.

(5.7)

Helpful Information

Remarks. The following remarks pertain to the foregoing mini-example. • Coordinate system. While we will most often use a coordinate system whose directions are horizontal and vertical, occasionally other choices may be more convenient and will be used. • Direction of reactions in FBD. When putting reaction forces and moments in the FBD, we often do not know the actual directions these forces will have until after the equilibrium equations are solved. In Fig. 5.5 we have elected to take 𝐴𝑥 , 𝐴𝑦 , and 𝐵𝑦 to be positive in their positive coordinate directions. After solving the equilibrium equations, we may find some of these are negative (such as 𝐴𝑥 above), meaning they actually act in the directions opposite those shown in Fig. 5.5. When you encounter such a situation, you should resist the temptation to revise the direction of those reactions in the FBD, because this often leads to errors since all equilibrium equations also need to be revised. Instead, just allow those reactions to have negative values. • Number of unknowns. After we draw the FBD, it is a good idea to count the number of unknowns. In Fig. 5.5, there are three unknowns, namely, 𝐴𝑥 , 𝐴𝑦 , and 𝐵𝑦 ; since there are three equilibrium equations, we expect to have a determinate system of algebraic equations that can be solved to obtain a unique solution for the unknowns. If more or fewer than three unknowns are present in the FBD and you rule out any errors in your FBD, then the object you are modeling is statically indeterminate or is partially fixed, as discussed in the next section.

Direction for moment summation. When writing moment equilibrium equations in two-dimensional problems, such as Eq. (5.6), we will always take counterclockwise to be the positive direction for moments. 𝑦 +𝑀 𝑥 Figure 5.6 This choice is consistent with the right-hand rule for the 𝑥𝑦 coordinate system shown.

276

ISTUDY

Chapter 5

Equilibrium of Bodies

• Selection of moment summation point. In Eq. (5.6), moments were evaluated about point 𝐴. While any point can be used, the merit of point 𝐴 is that the unknown reactions at 𝐴 produce no moment about point 𝐴, and hence, they do not enter the moment equilibrium equation. This leaves 𝐵𝑦 as the only unknown in that expression, which is easily solved for. When you sum moments, it is helpful to use a moment summation point that omits as many unknowns as possible. ∑ • Alternative equilibrium equations. Rather than using 𝐹𝑦 = 0, Eq. (5.5), we could use in its place the moment equilibrium equation ∑ 𝑀𝐶 = 0 ∶ −𝐴𝑦 (500 mm) + (80 N)(300 mm) + (60 N)(50 mm) = 0,

(5.8)

where point 𝐶 is located at the intersection of the lines of action of 𝐴𝑥 and 𝐵𝑦 , as shown in Fig. 5.5. Observe that the moment summation point does not need to lie on the object. The merit of using point 𝐶 for moment summation is that it will provide an equation that contains 𝐴𝑦 as the only unknown. Solving Eqs. (5.4), (5.6), and (5.8) provides the same results given in Eq. (5.7). Additional comments on the use of alternative equilibrium equations follow.

Alternative equilibrium equations As demonstrated in the foregoing mini-example, we can replace either or both of the ∑ ∑ 𝐹𝑥 = 0 and 𝐹𝑦 = 0 equations with moment equilibrium equations where the moment summation points are different. The merit of doing this is that, with appropriate selection of moment summation points, we can often obtain an equation system that ∑ ∑ is easier to solve. It might seem as if “trading” a 𝐹𝑥 = 0 and/or 𝐹𝑦 = 0 equation for an additional moment equilibrium equation violates the fundamental principle of equilibrium stated in Eq. (5.3). In reality, writing multiple moment equilibrium equa∑ ∑ tions, if properly done, still ensures 𝐹𝑥 = 0 and 𝐹𝑦 = 0. To explain why this is true, we use concepts of linear algebra (Prob. 5.123 on p. 343 guides you to develop a different explanation for why these alternative equilibrium equations may be used). ∑ ∑ ∑ Consider the three equilibrium equations 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀𝑃 = 0. Each of these equations may be multiplied by any nonzero number, and the resulting equations may be added or subtracted. None of these manipulations change the basic fact that there are still three independent equations whose solutions are the ∑ ∑ same as those of the original equations. Replacing the 𝐹𝑥 = 0 (or 𝐹𝑦 = 0) ∑ equation with the moment equilibrium equation 𝑀𝐴 = 0, subject to the minor restrictions discussed below on where point 𝐴 may be located, is identical to mul∑ ∑ tiplying the 𝐹𝑥 = 0 (or 𝐹𝑦 = 0) equation by a suitable nonzero number and ∑ adding the result to the 𝑀𝑃 = 0 equation. To illustrate, in the preceding mini∑ example, the 𝑀𝐶 = 0 expression, Eq. (5.8), is identical to Eq. (5.6) after the product of Eq. (5.5) and 500 mm is subtracted from it. Summary. The various alternative equilibrium equations that may be used in place of Eq. (5.3) are ∑ ∑ ∑ 𝑀𝐴 = 0, and 𝑀𝑃 = 0. (5.9) 𝐹𝑥 = 0, Points 𝐴 and 𝑃 must have different 𝑥 coordinates.

ISTUDY

Section 5.2

277

Equilibrium of Rigid Bodies in Two Dimensions

∑

𝑀𝐴 = 0,

∑

𝐹𝑦 = 0,

and

∑

𝑀𝑃 = 0.

(5.10) 𝑟𝐵

Points 𝐴 and 𝑃 must have different 𝑦 coordinates. ∑

𝑀𝐴 = 0,

∑

𝑀𝐵 = 0,

and

∑

𝑟𝐴 𝐵

𝑀𝑃 = 0.

(5.11)

𝐴

Points 𝐴, 𝐵, and 𝑃 may not lie on the same line.

(a)

The restrictions on the locations of moment summation points given in Eqs. (5.9) through (5.11) ensure that the three equilibrium equations that are written are independent and hence, have a unique solution. While it might seem important to memorize the rules for locating moment summation points, you do not need to. The strategy you are likely to use to write your equilibrium equations to find your unknowns will almost always be sufficient to ensure that these rules are obeyed.

𝛼

(b)

Gears Gears are manufactured in an enormous variety of shapes and forms, and methods of analyzing and designing gears are a specialized topic. In this book we consider the simplest and most common type of gear, called a spur gear, as shown in Fig. 5.7(a). Radii 𝑟𝐴 and 𝑟𝐵 are the effective radii of the gears and are called the pitch radii. In general, two meshing gears have multiple teeth in contact at the same time. If, for example, gear 𝐴 shown in Fig. 5.7(a) rotates counterclockwise and applies power to gear 𝐵, then the forces between the contacting teeth are as shown in Fig. 5.7(b). The shape of gear teeth is such that the force supported by an individual tooth is not perpendicular to the gear’s radius at the point of contact [i.e., angle 𝛼 shown in Fig. 5.7(b) is not 90◦ ]. All of the forces acting on the teeth of a particular gear can be vectorially summed to obtain an equivalent force system that represents all the tooth forces as shown in Fig. 5.7(c). This force system is positioned on the line connecting the gears’ centers and is at the pitch radius of each gear. Note that in addition to the tangential force 𝐺, there is a normal force 𝑁 that tends to push the gears apart. While 𝑁 is smaller than 𝐺, it is not insignificant, and its size as a fraction of 𝐺 depends on the shape of the gears.∗ When dealing with gears in this book, we will usually neglect 𝑁 when this assumption is reasonable. While not perfect, this is a common simplification that is useful and adequate for many purposes. Examples 5.5 and 5.6 at the end of this section elaborate further on this issue; in Example 5.5, it is reasonable to neglect 𝑁, while in Example 5.6, 𝑁 must be included.

Examples of correct FBDs Figure 5.8 shows several examples of properly constructed FBDs. Comments on the construction of these FBDs follow. Bulletin board. After sketching the bulletin board, we apply the 40 N weight through its center of gravity at point 𝐶. At point 𝐴, the support does not permit horizontal and vertical motion, and therefore there are two reactions 𝐴𝑥 and 𝐴𝑦 . At point 𝐵, the board cannot move perpendicular to the interface between the screw and wire hanger, and therefore there is a reaction 𝐵 in the direction shown. ∗ One of the primary measures of gear shape is the pressure angle. The pressure angle is selected by a gear

designer (among several industry-standard values) so that the gear has acceptable force transmission ability, noise, life, and so on. The pressure angle also determines how large 𝑁 is as a fraction of 𝐺.

𝑟𝐴

𝐺 𝑁

𝑁

𝑟𝐵

𝐺 (c) Figure 5.7 (a) Meshing spur gears with pitch radii 𝑟𝐴 and 𝑟𝐵 . (b) Multiple teeth on each gear are usually in contact at the same time, and this example shows two teeth on each gear in simultaneous contact. The force on a particular tooth is not perpendicular to the gear’s radius (𝛼 ≠ 90◦ ). (c) Regardless of the number of teeth in contact, all tooth forces on a gear may be represented by a single tangential force 𝐺 and a normal force 𝑁, positioned on the line connecting the gears’ centers and at the pitch radius of each gear. For many applications, 𝑁 is neglected.

Interesting Fact Standards for gears. The American Gear Manufacturers Association (AGMA) is an organization of gear manufacturers from around the world that develops standards for the manufacture, performance, and use of gears. When possible, an engineer will select gears from the standard shapes and sizes that are commercially available. Sometimes special-purpose gears are needed, and these must be custom designed and manufactured.

278

ISTUDY

Chapter 5

Equilibrium of Bodies

Examples of correct FBDs Bulletin board. A bulletin board with 40 N weight and center of gravity at point 𝐶 hangs on a wall by screws at points 𝐴 and 𝐵. Draw the FBD for the bulletin board. 30◦

30◦

30◦

𝐴𝑦

𝐴𝑥

30◦

30◦

𝐵 40 N

𝐴

𝐵

bulletin board

𝑦

𝐶

𝑥

Lifting machine. Member 𝐴𝐵 of a machine weighs 40 lb with center of gravity at point 𝐶. Draw the FBD for member 𝐴𝐵. 200 lb

200 lb 𝐵𝑥

𝑀𝐵

𝐵 40 lb

𝐶

𝑦

frictionless 𝑥

𝐴𝑦

𝐴 𝐹

Folding desk. The folding desk is used in a lecture hall of a university. Assume a person may apply up to a 75 lb force to the desk. Draw the FBD, neglecting the weights of members.

𝑦 𝑥 75 lb

𝐷 𝐴

𝐹𝐶𝐷

𝐸

𝐶 𝐵

Figure 5.8. Examples of properly constructed FBDs.

𝐸 𝐹𝐴𝐵

ISTUDY

Section 5.2

Equilibrium of Rigid Bodies in Two Dimensions

Lifting machine. After sketching the frame of the lifting machine, we apply the 200 lb force and the 40 lb weight through its center of gravity at point 𝐶. At point 𝐴, the gear applies a tangential force 𝐴𝑦 to the frame, and the normal force between the gear and frame is neglected. At point 𝐵, the support does not permit horizontal translation and rotation, and therefore there are two reactions consisting of force 𝐵𝑥 and moment 𝑀𝐵 . Folding desk. We are instructed to neglect the weights of members. Member 𝐴𝐵 prevents motion of point 𝐴 in the direction along line 𝐴𝐵, and hence, there must be a reaction force 𝐹𝐴𝐵 whose line of action is line 𝐴𝐵. Similarly, member 𝐶𝐷 produces a reaction 𝐹𝐶𝐷 whose line of action is line 𝐶𝐷. At point 𝐸, we assume the contact is smooth, in which case there is one reaction force 𝐸 that is perpendicular to the interface between the desk and the seat in front of it. While we are instructed that a person can apply a 75 lb force, no information is given on the position or orientation of this force. Thus, we have chosen to give this force a position and orientation so that it maximizes the severity of the loading on the desk (i.e., it is placed at the edge of the desk, perpendicular to the desktop).

Examples of incorrect and/or incomplete FBDs Figure 5.9 shows several examples of incorrect and/or incomplete FBDs. Comments on how these FBDs must be revised follow, but before reading these, you should study Fig. 5.9 to find as many of the needed corrections and/or additions on your own as possible. Pickup truck 1. The 15 kN weight should be vertical. 2. The rear wheels should have a reaction force 𝐴𝑥 in the 𝑥 direction. Hand truck 1. The reaction force at 𝐵 should be directed from point 𝐵 through the bearing of the wheel. We also note that if the bearing of the wheel is frictionless, then there will be no friction force at 𝐵. 2. The force 𝐹 applied by the operator’s hand is properly shown, but it should be noted that both its magnitude and direction are unknown. Machine control 1. Although not really a deficiency with the FBD, a coordinate system should be selected and shown. 2. The reaction force at 𝐵 should be perpendicular to the slot (i.e., vertical). 3. The pin at 𝐷 should also have a horizontal force 𝐷𝑥 .

279

280

ISTUDY

Chapter 5

Equilibrium of Bodies

Examples of incorrect and/or incomplete FBDs Pickup truck. A rear-wheel-drive pickup truck drives up an incline with steady speed. The truck weighs 15 kN with center of gravity at point 𝐶. Draw the FBD for the truck.

𝑦

15 kN 𝑥

𝐶

𝐵𝑦 𝐵

15◦

𝐴𝑦

𝐴

Hand truck. A hand truck with 75 N weight and center of gravity at point 𝐶 is used to move a keg weighing 400 N with center of gravity at point 𝐷. Draw the FBD if the truck is on the verge of rolling over the bump at point 𝐵.

𝐹 75 N 400 N 𝑦 𝑥

𝐶

𝐷

𝐵𝑦 𝐵 𝐴 Machine control. Slotted bracket 𝐸𝐵𝐷 is used to control the height of a machine at point 𝐷. Draw the FBD of member 𝐸𝐵𝐷, neglecting its weight. 𝐶

𝐷𝑦 𝐹 𝐸𝑥 𝐵 𝐷

𝐸

𝐵

𝐸𝑦

20◦

20◦ 𝐴

Figure 5.9. Examples of incorrect and/or incomplete FBDs.

ISTUDY

Section 5.2

Equilibrium of Rigid Bodies in Two Dimensions

End of Section Summary In Sections 5.1 and 5.2, the equations governing static equilibrium of a rigid body were presented, and analysis procedures for bodies in two dimensions were discussed. Some of the key points are as follows: • Both vector and scalar approaches can be used for problems in two dimensions. However, unless the geometry is complicated, a scalar approach will often be more straightforward. • The equilibrium equations for a body in two dimensions, with 𝑥 and 𝑦 being ∑ ∑ ∑ the in-plane directions, are 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀𝑃 = 0 where 𝑃 is the moment summation point you select. ∑ • Alternative sets of equilibrium equations are available where the 𝐹𝑥 = 0 ∑ and/or 𝐹𝑦 = 0 equations are replaced by additional moment equilibrium equations, subject to certain restrictions on where the moment summation points may be located. In some problems, use of these alternative equilibrium equations will reduce the amount of algebra required to solve for the unknowns. It is important to remember that a body in two dimensions has only three independent equilibrium equations available. • A free body diagram (FBD) is a sketch of a body and all of the forces and moments applied to it. The FBD is an essential tool to help ensure that all forces are accounted for when writing the equilibrium equations. • Gears are common and are often used in connection with shafts. Typically, when two gears mesh, multiple teeth from each gear are in simultaneous contact. However, for many purposes these forces may be replaced with an equivalent force system consisting of a tangential force, and possibly a normal force, as shown in Fig. 5.7.

281

282

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.1 𝐹

Equilibrium Analysis

𝐵

𝐴

𝐶

𝐷

8 mm 20 mm 4 mm

35 mm 45 mm 50 mm workpiece

A device for clamping a flat workpiece in a machine tool is shown. If a 200 N clamping force is to be generated at 𝐶, and if the contact at 𝐶 is smooth (no friction), determine the force 𝐹 required and the reaction at 𝐵.

SOLUTION Road Map Determining the force 𝐹 required to produce a 200 N clamping force will be accomplished by analyzing the equilibrium of the clamp 𝐴𝐵𝐶. Member 𝐵𝐷 is viewed as a support for the clamp (as shown in Fig. 5.3), and thus it is not necessary to explicitly address its equilibrium.

Figure 1

Modeling 𝐴𝑦 𝐹

𝐴

𝑦 𝑥

𝐵

𝐹𝐵𝐷 𝐷

𝐶

35 mm 45 mm 50 mm

8 mm 24 mm

𝐶𝑦 = 200 N

Figure 2 Free body diagram. 𝑦

𝑥

𝐵 𝐷 Figure 3 The link 𝐵𝐷 prevents translation of point 𝐵 in the 𝑥 ′ direction, and therefore the link must apply a reaction force to the clamp in this direction. Translation of point 𝐵 in the 𝑦 ′ direction is possible, therefore there is no reaction force in this direction. Further, rotation of the clamp about point 𝐵 is possible, therefore there is no moment reaction.

The completed FBD for the clamp is shown in Fig. 2, and it is constructed as follows. We first sketch clamp 𝐴𝐵𝐶 and then choose an 𝑥𝑦 coordinate system. The roller support at 𝐴 prevents translation of the clamp in the 𝑦 direction, and thus, there must be a 𝑦 direction reaction 𝐴𝑦 that enforces this constraint. Similarly, at the smooth support at 𝐶 the only reaction is 𝐶𝑦 , which we want to be 200 N. The link 𝐵𝐷 supports the clamp at point 𝐵, and the reaction forces and/or moments at point 𝐵 are determined by consulting Fig. 5.3 on p. 273, or better yet by the discussion given in Fig. 3. Hence, the reaction force due to the link is 𝐹𝐵𝐷 with direction from point 𝐵 to point 𝐷.

Governing Equations & Computation Prior to writing equilibrium equations, we resolve the reaction at 𝐵 into horizontal and vertical components so that moment arms are more easily obtained. The revised FBD is shown in Fig. 4. When writing equilibrium equations, we usually try to write them so that each equation contains only one unknown, which may then be found immediately. When this is possible, simultaneous algebraic equations do not need to be solved. In Fig. 4, we observe ∑ ∑ that both the 𝐹𝑥 = 0 and 𝐹𝑦 = 0 equations involve two unknowns each, and thus, these equations are not especially convenient to start with, although no harm is done if ∑ you do this. We next consider 𝑀 = 0, and we look for a point where the lines of action of two (or more) unknown forces intersect; in Fig. 4 point 𝐴 is such a point.∗ Thus, writing ∑ 𝑀𝐴 = 0, with positive moments taken counterclockwise, we immediately find 𝐹𝐵𝐷 as follows: ( ) ( ) ∑ 24 45 (8 mm) − 𝐹𝐵𝐷 (80 mm) 𝑀𝐴 = 0 ∶ − 𝐹𝐵𝐷 51 51

+ (200 N)(130 mm) = 0 𝐴𝑦 𝐹

𝐹𝐵𝐷

𝐹𝐵𝐷

24 51

𝐵

𝐴

𝑦

45 51

𝐶

8 mm 24 mm

𝐷 𝑥

35 mm 45 mm 50 mm

𝐶𝑦 = 200 N

Figure 4 Revised FBD with 𝐹𝐵𝐷 resolved into horizontal and vertical components (where √ (45)2 + (24)2 = 51) so that moment arms are more easily obtained.

ISTUDY

⇒

𝐹𝐵𝐷 = 581.6 N.

(1)

∑ ∑ Now that 𝐹𝐵𝐷 is known, the equations 𝐹𝑥 = 0 and 𝐹𝑦 = 0 may be written in any order to obtain the remaining unknowns ( ) ∑ 45 =0 ⇒ 𝐹𝑥 = 0 ∶ 𝐹 − 𝐹𝐵𝐷 𝐹 = 513.2 N, (2) 51 ( ) ∑ 24 + 200 N = 0 ⇒ 𝐴𝑦 = 73.68 N. (3) 𝐹𝑦 = 0 ∶ 𝐴𝑦 − 𝐹𝐵𝐷 51 Discussion & Verification

• Intuitively, we expect the solution of our equilibrium equations to show that 𝐹 > 0, and this is what we obtain. If we had obtained a negative value for 𝐹 , we would ∗ The

intersection of the lines of action of 𝐴𝑦 and 𝐹𝐵𝐷 shown in Fig. 2 is another such point, as is the intersection of the lines of action of 𝐹 and 𝐹𝐵𝐷 . However, determining the locations of these intersection points requires some work, and for this reason, point 𝐴 is a better choice.

ISTUDY

Section 5.2

Equilibrium of Rigid Bodies in Two Dimensions

immediately suspect an error and would correct this before continuing with our solution. • Once we verify that our solution passes the preceding casual but essential check, we should then verify that all unknowns (𝐹 , 𝐹𝐵𝐷 , and 𝐴𝑦 ) satisfy all of the equilibrium equations. If any of the equilibrium equations is not satisfied, then a math error has been made. However, this check does not guarantee that the equilibrium equations have been accurately written. Thus, it is essential that the FBD be accurately drawn and that the equilibrium equations accurately include all forces and moments from the FBD. • An additional check on our solution, which is usually easy to carry out, is to use the FBD of Fig. 2 or 4 to evaluate the moment of all forces about additional points to verify that moment equilibrium is satisfied. For example, if we evaluate moments ∑ about point 𝐶 in Fig. 4, a correct solution must show that 𝑀𝐶 = 0.

283

284

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.2

Two-Dimensional Idealization of a Three-Dimensional Problem A large number of identical uniform cantilever beams, each with 40 in. length and 10 lb∕ft weight, are used to support 4 ft sections of pipes 𝐵 and 𝐶. A typical beam 𝐴𝐵𝐶𝐷 is shown. If pipes 𝐵 and 𝐶 plus their contents weigh 25 lb∕ft and 35 lb∕ft, respectively, determine the reactions at the built-in end of the cantilever beams.

4 ft 4 ft

SOLUTION 4 ft

Road Map We will idealize this problem to be two-dimensional and will draw an FBD of one cantilever beam, assuming it supports the weights of a 4 f t length of each pipe plus its own weight. Our goal is to determine the reactions at the built-in support.

The FBD is shown in Fig. 2 and is constructed as follows. Beam 𝐴𝐵𝐶𝐷 is sketched first, and an 𝑥𝑦 coordinate system is chosen. The reactions at the built-in support at point 𝐴 may be determined by consulting Fig. 5.3, but it easier to simply construct these by considering the motion that the support prevents. That is, the support at 𝐴 prevents horizontal translation of the beam, and hence there must be a reaction 𝐴𝑥 . The support also prevents vertical translation of the beam, and hence there must be a reaction 𝐴𝑦 . Finally, the support prevents rotation of the beam, and hence there must be a moment reaction 𝑀𝐴 . The beam’s total weight is ( ) ( ) lb 1 ft 𝑊𝐸 = 10 (40 in.) = 33.33 lb, (1) ft 12 in. Modeling

𝐵

𝐷 𝐶 10 in.

16 in.

𝐴 14 in.

Figure 1

𝑊𝐶 = 140 lb

𝑦

𝑊𝐵 = 100 lb 𝑀𝐴

𝑥 𝐷 𝐶 𝐸 10 in. 10 in.

𝐵

𝐴 14 in.

6 in. 𝑊𝐸 = 33.33 lb Figure 2 Free body diagram.

ISTUDY

𝐴𝑦

𝐴𝑥

and since the beam is uniform, meaning its cross-sectional dimensions and material are the same over its entire length, this weight is placed at the center of gravity of the beam, which is its midpoint 𝐸. The weights of 4 ft lengths of pipes 𝐵 and 𝐶 are ( ) ( ) lb lb 𝑊𝐵 = 25 (4 ft) = 100 lb, 𝑊𝐶 = 35 (4 ft) = 140 lb. (2) ft ft Governing Equations & Computation

actions at 𝐴 as ∑ ∑ ∑

Writing equilibrium equations provides the re-

𝐹 𝑥 = 0 ∶ 𝐴𝑥 = 0

(3) ⇒ 𝐴𝑥 = 0,

𝐹𝑦 = 0 ∶ 𝐴𝑦 − 140 lb − 33.33 lb − 100 lb = 0 ⇒ 𝐴𝑦 = 273.3 lb,

(4) (5) (6)

𝑀𝐴 = 0 ∶ 𝑀𝐴 + (140 lb)(30 in.) + (33.33 lb)(20 in.) + (100 lb)(14 in.) = 0 ⇒ 𝑀𝐴 = −6267 in.⋅lb = −522.2 ft⋅lb.

(7) (8)

Discussion & Verification

• Because of the loading and geometry, we expect 𝐴𝑦 > 0 and 𝑀𝐴 < 0, which are obtained. We should also verify that 𝐴𝑥 , 𝐴𝑦 , and 𝑀𝐴 satisfy all of the equilibrium equations. • The biggest assumption made is that an individual beam is responsible for supporting the weight of a 4 ft length of each pipe only. Whether or not this assumption is valid cannot be verified with the limited data given here. However, the intent of using multiple beams to support the pipes is that each beam shares equally in supporting the weight. This assumption is reasonable if the two pipes are sufficiently flexible and/or all of the support beams are accurately aligned.

ISTUDY

Section 5.2

E X A M P L E 5.3

285

Equilibrium of Rigid Bodies in Two Dimensions

Two-Dimensional Idealization of a Three-Dimensional Problem

The rear door of a minivan is hinged at point 𝐴 and is supported by two struts; one strut is between points 𝐵 and 𝐶, and the second strut is immediately behind this on the opposite side of the door. If the door weighs 350 N with center of gravity at point 𝐷 and it is desired that a 40 N vertical force applied by a person’s hand at point 𝐸 will begin closing the door, determine the force that each of the two struts must support and the reactions at the hinge.

500 mm

𝐸

100 mm

650 mm 150 mm 𝐷

80 mm

𝐵 250 mm

𝐴

200 mm

SOLUTION

𝐶

Road Map

Although the problem is really three-dimensional, a two-dimensional idealization is sufficient and will be used here. We will neglect the weights of the two struts since they are probably very small compared to the weight of the door. Modeling The FBD is shown in Fig. 2 and is constructed as follows. The door is sketched first and an 𝑥𝑦 coordinate system is chosen. The person’s hand at 𝐸 applies a 40 N downward vertical force, and the 350 N weight of the door is a vertical force that acts through point 𝐷. The hinge (or pin) at 𝐴 has horizontal and vertical reactions 𝐴𝑥 and 𝐴𝑦 . The force in one strut is 𝐹𝐵𝐶 , with a positive value corresponding to compression. Thus, the total force applied by the two struts is 2𝐹𝐵𝐶 . In Fig. 2, the horizontal and vertical components of the strut force are determined using similar triangles with the geometry shown. Figure 1

Summing moments about point 𝐴 is convenient because it will produce an equilibrium equation where 𝐹𝐵𝐶 is the only unknown: ∑ 𝑀𝐴 = 0 ∶ (40 N)(1.150 m) + (350 N)(0.800 m) ) ) ( ( 450 570 − 2𝐹𝐵𝐶 (0.650 m) + 2𝐹𝐵𝐶 (0.250 m) = 0 (1) 726.2 726.2 𝐹𝐵𝐶 = 789.2 N. (2) ⇒

Governing Equations & Computation

Thus, the force supported by each of the two struts is 𝐹𝐵𝐶 = 789.2 N. The reactions at point 𝐴 are found by writing the remaining two equilibrium equations ) ( ∑ 570 𝐹𝑥 = 0 ∶ − 2𝐹𝐵𝐶 + 𝐴𝑥 = 0 (3) 726.2 (4) ⇒ 𝐴𝑥 = 1239 N, ) ( ∑ 450 + 𝐴𝑦 = 0 (5) 𝐹𝑦 = 0 ∶ − 40 N − 350 N + 2𝐹𝐵𝐶 726.2 (6) ⇒ 𝐴𝑦 = −588.0 N. Discussion & Verification

• Because of the geometry and loading for this problem, we intuitively expect the struts to be in compression. Since the strut force 𝐹𝐵𝐶 was defined to be positive in compression, we expect the solution to Eq. (1) to give 𝐹𝐵𝐶 > 0, which it does. • You should verify that the solutions are mathematically correct by substituting 𝐹𝐵𝐶 , 𝐴𝑥 , and 𝐴𝑦 into all equilibrium equations to check that each of them is satisfied. However, this check does not verify the accuracy of the equilibrium equations themselves, so it is essential that you draw accurate FBDs and check that your solution is reasonable.

𝐵 450 mm

726.2 mm 570 mm

𝐶

𝐹𝐵𝐶

350 N 𝐸

570 726.2

𝐹𝐵𝐶

650 mm 150 mm

500 mm

100 mm

𝐹𝐵𝐶

450 726.2

𝐷

2𝐹𝐵𝐶

570 726.2

𝑦 𝑥

250 mm 40 N

2𝐹𝐵𝐶

Figure 2 Free body diagram.

450 726.2

𝐴𝑥 𝐴𝑦

286

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.4

Numerous Roller Supports

40◦ 6 in. 6 in.

𝑇 15◦

𝐵 𝐷

𝐴 𝐶

𝐸

SOLUTION 8 in. 9 in.

500 lb Figure 1

𝑇

A trolley rolls on the flange of a fixed I beam to move a 500 lb vertical force. The trolley has a total of eight rollers; four of these are shown in Fig. 1 at points 𝐴, 𝐵, 𝐶, and 𝐷, and the remaining four are located immediately behind these points on the opposite side of the trolley. Thus, there is one pair of rollers located at 𝐴, another pair at 𝐵, and so on. The rollers at 𝐴 and 𝐵 are a loose fit so that only one pair will make contact with the flange of the I beam, and similarly for the rollers at 𝐶 and 𝐷, as well as for those on the opposite side of the trolley. Determine the cable force 𝑇 and the reactions at each of the four pairs of rollers.

At the outset of this problem it is not known whether the pair of rollers at 𝐴 or the pair of rollers at 𝐵 will make contact. Similarly, it is not known whether the pair of rollers at 𝐶 or the pair of rollers at 𝐷 will make contact. To proceed, we will assume the rollers at 𝐵 and 𝐷 make contact (the rollers at 𝐴 and 𝐶 could just as well have been chosen), and after the analysis is complete, we will interpret our results to determine which rollers are actually in contact.

Road Map

Modeling The FBD is shown in Fig. 2. Because the geometry is given with respect to the I beam’s axis of orientation, use of an 𝑥𝑦 coordinate system with this same orientation will allow for easy determination of moment arms. In drawing the FBD, we have assumed that the upper pairs of rollers at 𝐵 and 𝐷 make contact, and we label their reactions as 𝑅1 and 𝑅2 in the FBD. After analysis is complete, we must examine the signs of 𝑅1 and 𝑅2 ; a positive value indicates the assumption was correct, while a negative value means the assumption was incorrect and that the adjacent pair of rollers actually makes contact. 15◦

𝐵 𝐴 𝐸

𝑅1

6 in. 6 in.

𝑦

𝐷 𝑅2

𝐶

𝑥

8 in. 9 in.

40◦

Governing Equations & Computation The following sequence of equilibrium equations provides for easy determination of the unknowns: ∑ 𝐹𝑥 = 0 ∶ −𝑇 cos 15◦ + (500 lb) sin 40◦ = 0 (1)

⇒ 𝑇 = 332.7 lb,

∑

𝑀𝐴 = 0 ∶

ISTUDY

−(500 lb) cos 40◦ (6 in.) + (500 lb) sin 40◦ (9 in.) + 𝑅2 (12 in.) = 0

500 lb Figure 2 Free body diagram assuming contact is made at roller pairs at 𝐵 and 𝐷.

(2)

(3)

⇒ 𝑅2 = −49.53 lb, ∑

𝐹𝑦 = 0 ∶

(4)

𝑅1 + 𝑅2 − 𝑇 sin 15 − (500 lb) cos 40 = 0 ◦

◦

⇒ 𝑅1 = 518.7 lb.

(5) (6)

Discussion & Verification Since 𝑅1 is positive, the assumption of contact at the pair of rollers at 𝐵 was correct; thus, we write 𝐴𝑦 = 0 and 𝐵𝑦 = 𝑅1 = 518.7 lb where positive 𝐵𝑦 is measured in the positive 𝑦 direction. Since 𝑅2 is negative, it is the pair of rollers at 𝐶 rather than the pair at 𝐷 that actually makes contact; hence, we write 𝐶𝑦 = 𝑅2 = −49.53 lb and 𝐷𝑦 = 0 where positive 𝐶𝑦 is measured in the positive 𝑦 direction. Thus, our solutions for the cable tension and the reactions at each pair of rollers may be summarized as

𝐴𝑦 = 0,

𝐵𝑦 = 518.7 lb,

𝐶𝑦 = −49.53 lb,

𝐷𝑦 = 0, and 𝑇 = 332.7 lb.

(7)

ISTUDY

Section 5.2

Equilibrium of Rigid Bodies in Two Dimensions

E X A M P L E 5.5

Loose-Fitting Gears

A drum for mixing material rotates clockwise under the power of a geared motor at 𝐴. The drum weighs 320 lb and is supported by a bearing at point 𝐵, and the material being mixed weighs 140 lb with center of gravity at point 𝐷. If gear 𝐴 is a loose fit with the gear on the drum, determine the reactions at point 𝐵 and the gear tooth force required to operate the machine. Assume the machine operates at steady speed and the material being mixed maintains the same shape and position as the drum rotates.

8 in. 𝐵

SOLUTION

30◦

𝐷

Road Map Because the drum rotates at constant speed and the material being mixed maintains the same shape and position, this is a problem of static equilibrium. To determine the support reactions at bearing 𝐵 and the gear tooth force, we will analyze the equilibrium of the drum and the material being mixed. Modeling The FBD is shown in Fig. 2, where we simplify the gear tooth force (as discussed in Fig. 5.7) to be a single force 𝐺 tangent to the gear and positioned at the pitch radius of the gear, which is 19 in. from point 𝐵. The 320 lb weight of the drum is positioned at the drum’s center, point 𝐵, and the weight of the material being mixed is placed at its center of gravity, point 𝐷. Governing Equations & Computation

⇒ 𝐺 = 58.95 lb,

∑

𝐹𝑥 = 0 ∶

∑

𝐹𝑦 = 0 ∶

𝐴

Figure 1

8 in.

Using the FBD in Fig. 2, the equilibrium equa-

tions and solutions are as follows: ∑ 𝑀𝐵 = 0 ∶ −𝐺(19 in.) + (140 lb)(8 in.) = 0 𝐵𝑥 − 𝐺 cos 30 = 0 ◦

⇒ 𝐵𝑥 = 51.05 lb, 𝐵𝑦 − 320 lb − 140 lb + 𝐺 sin 30◦ = 0 ⇒ 𝐵𝑦 = 430.5 lb.

19 in.

320 lb 140 lb

(1)

𝐵𝑥

𝐵

19 in.

(2) (3) (4) (5)

30◦ 𝐺 30◦

(6)

Discussion & Verification

• If the drum does not rotate at uniform speed, then accelerations are not zero and concepts of dynamics must be used to determine the gear tooth force and the reactions at 𝐵. If the material being mixed does not maintain the same shape and position as the drum rotates, then, at a minimum, the moment arm in Eq. (1) changes with time, and depending on how the shape and/or position changes with time, the problem may also be dynamic. • The idealization of the gear tooth force consisting of only a tangential force is appropriate, provided the gears are a loose fit, because both gears rotate on bearings (or axes) that are fixed in space. You should contrast this situation with that in Example 5.6.

𝐵𝑦

𝐷

𝑦 𝑥

Figure 2 Free body diagram.

287

288

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.6

Gears Actively Pressed Together The mixing machine described in Example 5.5 is modified so that the drum is supported by a motor-powered gear at point 𝐴 and a frictionless roller at point 𝐶. Determine the reactions at 𝐶 and the gear tooth force required to operate the machine.

8 in. 18 in.

𝐵

19 in. 𝐷

30◦ 30◦

𝐶

𝐴

SOLUTION Road Map

Because the drum rotates at constant speed and the material being mixed maintains the same shape and position, this is a problem of static equilibrium. To determine the support reactions at the roller 𝐶 and the gear tooth forces, we will analyze the equilibrium of the drum and the material being mixed. In contrast to Example 5.5, here the gears are actively pressed together, and thus, the gear tooth forces consist of both tangential and normal forces. Modeling

The FBD is shown in Fig. 2, where the gear tooth forces consist of a tangential force 𝐺 and a normal force 𝑁. The 320 lb weight of the drum is positioned at its center, point 𝐵, and the weight of the material being mixed is placed at its center of gravity, point 𝐷.

Figure 1

Governing Equations & Computation

tions are ∑

8 in. 320 lb 140 lb 𝐵 𝐷

19 in.

30◦

30◦ 30◦

30◦ Figure 2 Free body diagram.

ISTUDY

𝐶 𝑦

∑

∑

𝐺 𝑁

𝑀𝐵 = 0 ∶

30◦

Using the FBD in Fig. 2, the equilibrium equa-

−𝐺(19 in.) + (140 lb)(8 in.) = 0

(1)

⇒ 𝐺 = 58.95 lb,

(2)

𝐹𝑥 = 0 ∶

𝑁 sin 30◦ − 𝐺 cos 30◦ − 𝐶 sin 30◦ = 0,

(3)

𝐹𝑦 = 0 ∶

𝑁 cos 30 + 𝐺 sin 30 + 𝐶 cos 30 − 320 lb − 140 lb = 0.

(4)

◦

◦

◦

Equation (1) was solved immediately for 𝐺, and the result is the same as that obtained in Example 5.5. With the solution for 𝐺 known, Eqs. (3) and (4) can be solved simultaneously to obtain ⇒ 𝑁 = 299.6 lb and 𝐶 = 197.5 lb. (5)

𝑥 Discussion & Verification

The main difference between this example problem and Example 5.5 is the method by which the drum is supported, and this is reflected in the way the gear tooth forces are idealized. In Example 5.5 both gears rotated on fixed bearings and it was therefore possible to model the gear tooth force as consisting of a single tangential force only. In this example, only the gear at 𝐴 rotates on a fixed bearing, and the gear on the drum is actively pressed into the other gear under the action of the 320 lb and 140 lb forces. Thus, the normal component of the gear tooth force 𝑁 plays an important role in supporting the drum, and it cannot be neglected.

ISTUDY

Section 5.2

289

Equilibrium of Rigid Bodies in Two Dimensions

Problems General instructions. In problems involving gears, you may assume the gears are loose fitting if this assumption is reasonable.

4 in. 16 in.

18 in.

Problem 5.1 A freezer rests on a pair of supports at 𝐴 and a pair of supports at 𝐵. Weights for various parts of the freezer are shown. Neglecting friction, determine the reactions at each pair of supports.

45 lb

150 lb

8 in.

Problem 5.2 Weights of various parts of a computer are shown. Neglecting friction, determine the reactions at the supports at points 𝐴 and 𝐵.

9 in. 60 lb

36 in. 𝐵 12 in.

𝐴

120 mm 150 mm

Figure P5.1 30 N

55 N 45 N

𝐴

𝐵

6 in.

20 lb

10 in.

200 mm 130 mm

𝐶

Figure P5.2 40◦

Problem 5.3 The tool shown is used in a gluing operation to press a thin laminate to a thicker substrate. If the wheels at points 𝐴 and 𝐵 both have 2 in. diameter and have frictionless bearings, and a 20 lb vertical force is applied to the handle of the tool, determine the forces applied to the top of the laminate and the bottom of the substrate.

Problem 5.4

substrate 𝐴 Figure P5.3 12 in. 4 in.

𝐹

An escalator is driven by a chain connected to point 𝐴 that supports a force 𝐹 . The rollers at points 𝐴 and 𝐵 have frictionless bearings and ride in a loose-fitting track. If a person weighing 200 lb is being lifted at a steady speed and other weights may be neglected, determine the required chain force 𝐹 and the reactions at the wheels at 𝐴 and 𝐵. If the person is being lifted at a variable speed, will the force 𝐹 be different than that calculated earlier? Explain.

Problem 5.5

laminate

𝐵

200 lb 𝐴

9 in. 𝐵

Figure P5.4

Bar 𝐴𝐵𝐶𝐷 is supported by a pin at point 𝐶 and by a cable from points 𝐷 to 𝐸. Determine the reactions at 𝐶 and the force supported by the cable.

1m

1.5 m

2 kN

3 kN

𝐷 4

𝐶 3 𝐸

Figure P5.5

𝐵

1.5 m 3 kN

𝐴

1 kN

290

Chapter 5

Equilibrium of Bodies

𝐸

Problem 5.6 15 in.

𝐷

80 in.

A steel plate weighing 200 lb with center of gravity at point 𝐺 is supported by a roller at point 𝐴, a bar 𝐷𝐸, and a horizontal hydraulic cylinder 𝐵𝐶. Neglecting the weights of all members except for the plate, determine the force supported by the hydraulic cylinder and by the bar, and the reaction at the roller.

𝐺 𝐴

Problem 5.7

𝐶 𝐵

20◦ 40 in.

In Example 5.1 on p. 282, let the straight link 𝐵𝐷 be replaced by a curved link, as shown here. Discuss how the FBDs and solution to Example 5.1 will change. Note: Concept problems are about explanations, not computations.

15 in. 20 in. 12 in.

Figure P5.6 𝐹

𝐵

𝐴 𝐷 60 mm

20 N

180 mm

𝐶

8 mm 20 mm 4 mm

35 mm 45 mm 50 mm workpiece Figure P5.7

𝐷 𝐶 70 mm

𝐴

Problem 5.8 𝐵

A foldable tray for the paper supply of a photocopy machine is shown. The tray is supported by a single hinge at 𝐴 and two slotted links (one on each side of the tray). If the stack of paper weighs 20 N and other weights may be neglected, determine the reactions at the hinge and at point 𝐵 for one of the links.

25 mm

Figure P5.8

Problem 5.9 𝑊 = 6000 N 0.8 m

𝐴

𝑇

0.5 m 𝐷

𝛽

𝐶

𝛼 𝐵

To wrap a length of wire onto a large spool, the setup shown may be used, where a force 𝑇 is applied to end 𝐷 of the wire, such as by attaching it to a truck that drives slowly to the left. If the spool weighs 𝑊 = 6000 N with center of gravity at point 𝐴, 𝛼 = 3◦ , and 𝛽 = 8◦ , determine the cable tension 𝑇 needed to slowly wrap wire onto the spool, and determine the reactions at 𝐵. Assume the spool rolls without slipping.

Problem 5.10 In Prob. 5.9, let the spool be on a level surface (i.e., 𝛼 = 0◦ ). If 𝛽 is small enough, the spool will roll to the left, and if 𝛽 is large enough, the spool will roll to the right. Considering 0◦ ≤ 𝛽 ≤ 180◦ , determine the range of values for 𝛽 for which the spool will roll to the left and the range of values for which the spool will roll to the right.

Figure P5.9 and P5.10

Problem 5.11 7 ft 𝑊𝑚

9 ft 𝑊𝑏 𝑊𝑡

𝐵 3 ft Figure P5.11 and P5.12

ISTUDY

𝑑

𝐴 11 ft

A boat rests on a trailer with two wheels, where the boat, motor, and trailer have weights 𝑊𝑏 = 600 lb, 𝑊𝑚 = 125 lb, and 𝑊𝑡 = 300 lb, respectively. If the distance from the front of the boat to the hitch is 𝑑 = 4 ft, determine the vertical reaction at point 𝐴 and the reaction on each of the wheels.

Problem 5.12 For the boat trailer of Prob. 5.11, determine the distance 𝑑 so that the vertical reaction at point 𝐴 is 100 lb.

ISTUDY

Section 5.2

291

Equilibrium of Rigid Bodies in Two Dimensions

𝑑

Problem 5.13 A hand cart weighing 800 N is used for moving heavy loads in a warehouse. If each axle (pair of wheels) can support a maximum of 10 kN, determine the largest weight 𝑊 that may be supported and the position 𝑑 where it should be placed, assuming both axles are loaded to their capacity.

𝑊 𝐴

𝐵

𝐶

800 N

𝐷

𝐹 𝐸 400 mm 400 mm 300 mm 200 mm Figure P5.13

Problems 5.14 and 5.15 A variety of structures with pin, roller, and built-in supports are shown. In Figs. P5.14(c) and P5.15(c), the rollers at point 𝐷 allow vertical translation and constrain horizontal translation and rotation. Determine all reactions. Express your answers in terms of parameters such as 𝐿, 𝐹 , 𝑃 , and/or 𝑀.

𝐴

𝐹

𝑃

𝐵

𝐶

𝐿 3

𝐷

𝐿 3 (a)

𝐴

𝐴

𝐹

𝑃

𝐵

𝐶

𝐿 3

𝐷 30◦

(b)

𝐹

𝑃

𝐵

𝐶

𝐷

𝐴

(c)

𝐹

𝑃

𝐵

𝐶

𝐷

(d) Figure P5.14 𝐹

𝑀

𝐴

𝐴 𝐶

𝐵

𝐹

𝑀

𝐶

𝐵

𝐿 30◦

𝐷 𝐿

𝐷 2 kN

𝐵

𝐿 (a)

𝐴

3 kN

𝐷 (b)

𝐹

𝑀

𝐴 𝐶

𝐵

𝐹

𝑀

𝐴

𝐶

𝐵

𝐸 𝐶 1 kN

𝐷

𝐷

(c)

Figure P5.16 𝐵

(d)

𝐷

𝐹

Figure P5.15

Problems 5.16 and 5.17 In the structures shown, all members have the same 2 m length. Determine all support reactions.

𝐴

1 kN 𝐶 2 kN

Figure P5.17

𝐺 𝐸 3 kN

292

Chapter 5

Equilibrium of Bodies

Problem 5.18 The top chord of the truss is subjected to a uniform distributed load that gives rise to the forces 𝑄 shown. If 𝑄 = 1 kip, determine all support reactions. 𝑄∕2 𝐴 7 in.

11 in. 120 lb

𝑄

𝑄

𝑄

𝐵

𝐷

𝐹

𝐶

𝐸

𝐺

𝑄∕2 𝐻

𝐽

𝐿

𝑁

𝑃

4 ft 𝐼

𝐼

𝐾 𝑀

𝑂

3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 𝐹

Figure P5.18

𝐻 𝐺

80 lb

Problem 5.19 𝐶

𝐸

𝐷

Determine the support reactions for the short flight of stairs due to the forces shown on the two steps.

45◦ 𝐴

Problem 5.20

𝐵 8 in.

8 in.

8 in.

Figure P5.19

The forklift has a vehicle weight of 𝑊𝑉 = 15,000 lb, fuel weight of 𝑊𝐹 = 300 lb, and operator weight of 𝑊𝑂 = 160 lb. If 𝑃 = 2000 lb, determine the reactions on each pair of wheels. 1.5 ft

6.5 ft

3 ft 2 ft 𝑊𝐹

𝑊𝑂 𝑊𝑉 𝑃 𝐴

𝐵 𝐷

𝐶 7 ft 𝑊

2 ft

Figure P5.20

40 cm

Problem 5.21 50 cm

𝐴

6 ft

𝐵 𝜃

Figure P5.21

𝐸

A fixture for positioning 40 cm by 50 cm cardboard boxes with weight 𝑊 = 150 N in a packaging company is shown. If the weight of the box and its contents acts through the center of the box, determine the largest value of 𝜃 if the magnitude of the reactions at points 𝐴 and 𝐵 may not exceed 800 N, at which point the box begins to crush. The notch at point 𝐴 prevents horizontal and vertical translation of the box and assume the contact at point 𝐵 is frictionless.

𝑃 𝐵

𝐴 𝑎

𝐷 Figure P5.22 ISTUDY

𝐶 𝑎

Problem 5.22 Bar 𝐴𝐵𝐶 is supported by a roller at 𝐵 and a frictionless collar at 𝐴 that slides on a fixed bar 𝐷𝐸. Determine the support reactions at points 𝐴 and 𝐵. Express your answers in terms of parameters such as 𝑃 , 𝑎, etc.

ISTUDY

Section 5.2

Equilibrium of Rigid Bodies in Two Dimensions

293

Problem 5.23 A steel I beam is supported by a pin at point 𝐴 and a rocker at 𝐵, and a cable applies a 2000 lb force. The beam is uniform and weighs 26 lb∕f t. Determine the support reactions.

10 ft

0.5 ft 𝐴

3 ft 𝐵

10◦

10 ft

0.5 ft 0.5 ft 𝐶

𝐴

3 ft 10◦

𝐵

0.5 ft 𝐶

𝐷

10◦ 25◦ 2000 lb Figure P5.23

2000 lb Figure P5.24 𝑎

Problem 5.24

𝑃

𝐺

A steel I beam is supported by a pin at point 𝐴 and by a vertical link 𝐶𝐷, and a cable applies a 2000 lb force. The beam is uniform and weighs 26 lb∕f t. Determine the support reactions.

𝐵

𝐴

𝐶 𝐷 𝑏 𝑐

𝐸

30◦

Problem 5.25 Structure 𝐴𝐵𝐶𝐷𝐸 is supported by a roller at 𝐷 and a frictionless collar at 𝐴 that slides on a fixed bar 𝐹 𝐺. Determine the support reactions at points 𝐴 and 𝐷. Express your answers in terms of parameters such as 𝑃 , 𝑄, 𝑎, 𝑏, 𝑐, etc.

𝑎

𝐹 𝑄

Figure P5.25

Problem 5.26 A hand-operated winch for winding and unwinding rope on a spool is shown. The mean radius of the spool is 8 in., and the mean radius of the sprocket is 5 in. The spool, sprocket, and handle 𝐴𝐵 are one piece and rotate on a frictionless bearing at 𝐵. The pawl 𝐶𝐷 engages the sprocket at point 𝐷 and prevents the drum from rotating when 𝑃 = 0. The weights of all members are negligible.

3 in. 𝐶 15◦

6 in.

𝑇

(a) To wind rope onto the spool, if the operator can apply a maximum force 𝑃 = 50 lb, and if this force is perpendicular to the handle 𝐴𝐵 of the winch, determine the largest value of 𝑇 the rope may have. Also determine the reactions at 𝐵. (b) If 𝑃 = 0 and 𝑇 = 75 lb, determine the reactions at 𝐵 and the force supported by the pawl 𝐶𝐷.

𝐷

8 in. 5 in.

4 in.

𝐵

4 3 𝑃

12 in. 𝐴 Figure P5.26

Problem 5.27 The front view of one of the wings of a Cessna 172 airplane is shown. The wing is supported by a pin-connected strut 𝐵𝐸 and by a connection at 𝐴 that may be idealized as a pin. The plane’s mass, including fuel, passengers, and cargo, is 1000 kg, and it flies so that it is in static equilibrium. The strut 𝐵𝐸 has negligible mass, and the wing 𝐴𝐵𝐶𝐷 has 200 kg mass with center of gravity at point 𝐵. Model the wing as a two-dimensional structure where all forces lie in the same plane. (a) Determine the resultant lift force 𝑃 that this wing supports, assuming this force is vertical and that only the wings provide lift for the plane. (b) Determine the forces supported by the strut 𝐵𝐸 and the pin at 𝐴.

𝑃 𝐴

𝐵 𝐷

𝐶 1.4 m 𝐸 2m

Figure P5.27

0.6 m

2.4 m

294

Chapter 5

Equilibrium of Bodies

Problem 5.28 The automobile shown has 1200 kg mass with center of gravity at point 𝐺, and all wheels have the same 0.34 m radius. If the automobile has rear wheel drive, determine the reactions at each pair of wheels when the automobile just begins to slowly drive over the curb in front of it at point 𝐶. 𝐶 𝐺 0.7 m 0.2 m

𝐴

1.2 m

1.4 m

0.34 m 𝐵

Figure P5.28 and P5.29 0.6 m

0.6 m

𝐵

𝐷 𝐹t

Repeat Prob. 5.28 if the automobile has front wheel drive.

𝐸

𝑃

Problem 5.29

1.1 m

Problem 5.30

𝐺 1.8 m 1.3 m

𝐴

𝐹b

𝐶 0.6 m

0.6 m

A sliding screen door for a house or apartment is shown. The door has rollers at points 𝐴, 𝐵, 𝐶, and 𝐷; the rollers 𝐴 and 𝐵 are a loose fit, and the rollers 𝐶 and 𝐷 are a loose fit. The door weighs 90 N with center of gravity at point 𝐺. The rollers are frictionless, but when the door slides to the right, horizontal forces 𝐹t = 10 N and 𝐹b = 40 N arise due to friction from the weather sealing along the top and bottom of the door. To slide the door to the right, a person applies the horizontal force 𝑃 at point 𝐸. (a) Determine the value of 𝑃 so that the door slides to the right with constant speed. (b) For the value of 𝑃 obtained in Part (a), determine which rollers make contact and the reactions at these rollers.

Figure P5.30

Problem 5.31 8◦

𝐸

𝑇

𝐺 0.7 m 1.2 m

𝐶

𝐴

0.4 m

𝐷

(a) Determine the cable force 𝑇 needed to move the bucket up or down the rail at constant speed.

0.4 m

𝐵

1.4 m

𝐻

(b) For the value of 𝑇 obtained in Part (a), determine which rollers make contact and the reactions at these rollers.

20◦

Figure P5.31 12 in. 𝐴

6 in. 𝐶

𝐵

𝐷

rail

𝑃

𝑊 8 in. 2 in.

Figure P5.32–P5.34

ISTUDY

𝐼

A bucket used in a mining operation for moving ore has frictionless rollers at points 𝐴, 𝐵, 𝐶, and 𝐷. The bucket moves up and down a fixed rail 𝐻𝐼. The rollers at 𝐴 and 𝐵 are a loose fit on the rail, and the rollers at 𝐶 and 𝐷 are a loose fit. The mass of the bucket and its contents is 1600 kg with center of mass at point 𝐺.

Problems 5.32 through 5.34 A motor and mounting hardware with weight 𝑊 = 65 lb are supported by rollers 𝐴, 𝐵, 𝐶, and 𝐷. Each pair of rollers is loose-fitting so that only one roller of the pair will make contact with the fixed rail. This problem may be idealized as two-dimensional if the torque the motor applies to the pulleys is neglected. Determine which rollers make contact and the reactions at these rollers if: 𝑄

Problem 5.32

𝑃 = 0 and 𝑄 = 125 lb.

Problem 5.33

𝑃 = 200 lb and 𝑄 = 0.

Problem 5.34

𝑃 = 200 lb and 𝑄 = 125 lb.

ISTUDY

Section 5.2

295

Equilibrium of Rigid Bodies in Two Dimensions

𝑊

Problem 5.35 During assembly in a factory, a compressor with weight 𝑊 = 120 lb rests on a bench. A cover plate is to be attached using four bolts. To speed production, a special machine that simultaneously tightens all four bolts is used. Determine the largest torque 𝑀 that may be simultaneously applied to each bolt before the compressor begins to tip. 𝑀

Problem 5.36

3 3 4 𝐴 𝑀

A tractor is fitted with a hole-drilling attachment. The tractor has weight 𝑊𝑉 = 2000 lb, the operator has weight 𝑊𝑂 = 180 lb, and the supplemental weights at the front of the tractor weigh 𝑊𝐸 = 300 lb.

3 ft 𝑊𝑂

𝑊𝑉

𝑊𝐸

𝐸 𝐶

𝐵 7 ft

Figure P5.35

5 ft

𝐷 𝐴

𝑀 𝐵

5 3 3 5 dimensions in inches

(a) Determine the largest thrust that may be produced by the drilling attachment at point 𝐴. (b) Describe some simple ways the drilling thrust determined in Part (a) can be increased (e.g., addition of more weight at 𝐸, removal of weight at 𝐸, repositioning of weight from 𝐸 to 𝐷, and so on).

𝑀

6 ft

2 ft

Figure P5.36 𝑃

Problem 5.37 A spanner wrench is used to apply torque to circular shafts and other similar shapes. Such wrenches are routinely used in the setup of tools, such as milling machines, lathes, and so on. The wrench makes contact with the shaft at point 𝐴, which may be assumed to be frictionless, and at 𝐵, where a small pin fits into a hole in the shaft. If 𝑃 = 80 N, 𝐿 = 120 mm, 𝑟 = 25 mm, and 𝛼 = 120◦ , determine the reactions at points 𝐴 and 𝐵.

𝐵

𝐶

𝛼

𝑟

𝐴 𝐿 Figure P5.37 and P5.38

Problem 5.38 In the spanner wrench of Prob. 5.37, determine the range of values for angle 𝛼 so that the pin at 𝐵 will not slip out of its hole when force 𝑃 is applied for (a) 𝐿∕𝑟 = 4.

ℎ 𝐷

(b) Any value of 𝐿∕𝑟. 𝛼

Problem 5.39 An arbor press is used to apply force to the workpiece at point 𝐶. The length ℎ of handle 𝐷𝐸 can be adjusted by sliding it through a hole in shaft 𝐹 such that 50 mm ≤ ℎ ≤ 250 mm. The press has weight 𝑊 = 350 N and simply rests on the rough surface of a table. Force 𝑃 is applied perpendicular to the handle. (a) If the handle is horizontal (𝛼 = 0◦ ), determine the smallest force 𝑃 that will cause the press to tip about point 𝐴. (b) For any possible position of the handle, determine the smallest force 𝑃 and the corresponding value of 𝛼 that will cause the press to tip about point 𝐴.

𝐸

𝐹

𝑃

𝑊 300 𝐶 𝐴

𝐵 100 30 120 170 dimensions in mm

Figure P5.39

296

Chapter 5

Equilibrium of Bodies

Problem 5.40 A portion of a structure is supported by a frictionless V-shaped notch as shown. Which are the proper support reactions: those shown in Fig. P5.40(b) or those in Fig. P5.40(c), or are both of these acceptable? Explain. Note: Concept problems are about explanations, not computations.

𝐶 100 lb

𝑦 𝜃 24 ft

𝑅2

7 in.

𝑥

𝜃

𝑅𝑦

(b)

(a) 300 lb

5 in.

𝑅𝑥

𝑅1

(c)

Figure P5.40

Problem 5.41

120 lb 𝐴 18 in.

An antenna used for research at a university is shown. It is easily raised and lowered so that it can be outfitted with different equipment. The antenna is pinned to a supporting frame at point 𝐴, and it is raised and lowered using the gear at 𝐵 to which a hand crank is attached. The antenna and its attached sector gear weigh 300 and 120 lb, respectively, and the 100 lb horizontal force at point 𝐶 models the effect of wind loads during a storm. If the gear at 𝐵 is locked so it does not rotate, determine the gear tooth force and the reactions at 𝐴.

20◦

𝐵

Problem 5.42 Figure P5.41

Frame 𝐴𝐵𝐶𝐷 is rigidly attached to a gear 𝐸, which engages two parallel geared tracks 𝐹 and 𝐺 that are fixed in space. The gear has 4 in. pitch radius. If the gear has weight 𝑊 = 8 lb and 𝑃 = 10 lb, determine the tooth forces between gear 𝐸 and each track. Hint: A normal force is present at either 𝐹 or 𝐺, but not both.

𝐶

6 in.

8 in. 𝐵

8 in. 𝐹

𝐷 𝐸

𝐷

4 in. 𝐹

15◦ 𝐸

400 mm 𝐶

70◦

𝑊 Figure P5.43

ISTUDY

𝑊𝐺

𝐵

𝑊

𝐺

5 in. 𝑃

𝐴

Figure P5.42

Problem 5.43

𝐴 120 mm

The gear shown slides in a frictionless vertical slot. The gear weighs 𝑊𝐺 = 300 N with center of gravity at point 𝐴. A chain with negligible weight is wrapped around the gear. Member 𝐵𝐶 has negligible weight and is bolted to the gear. (a) If 𝑊 = 50 N, determine the forces supported by the chain and the reactions at the bearing 𝐴 of the gear. (b) Determine the value 𝑊 (where 𝑊 > 0) when the chain is no longer in tension throughout its entire length.

ISTUDY

Section 5.2

297

Equilibrium of Rigid Bodies in Two Dimensions

Problem 5.44 The main sprocket, chain, and pedals of a bicycle are shown. If 𝑃𝐴 = 400 N, 𝑃𝐶 = 30 N, and 𝑇b = 60 N, determine the tensile force in the top portion of the chain and the reactions at the bearing 𝐵 of the sprocket. Assume all forces shown lie in the same plane. Hint: It may be helpful to use the vector approach to evaluate the moments of some of the forces. 7◦

𝑃𝐴

𝐴 𝑇t

50◦ 80 mm

𝐵

𝑇b

2◦

10◦

15◦ 𝑃𝐶 200 mm

𝐶

Figure P5.44

𝑊 30 mm

Problem 5.45 110 mm

Frame 𝐵𝐶𝐷 is rigidly attached to a gear 𝐴, which engages a geared track 𝐸 that is fixed in space. The gear has 30 mm pitch radius, and portions 𝐵𝐶 and 𝐶𝐷 of the frame are perpendicular. If the gear has weight 𝑊 = 6 N and 𝑃 = 4 N, determine the value of 𝑄 needed so that angle 𝜃 = 30◦ , and also determine the reactions (tooth forces) between the gear and track.

Problem 5.46 In the frame and gear of Prob. 5.45, if 𝑊 = 6 N, 𝑃 = 4 N, and 𝑄 = 8 N, determine the value of 𝜃 when the assembly is in static equilibrium, and determine the reactions (tooth forces) between the gear and track.

15◦

𝐶

𝜃

𝐴 120 mm

𝐵 𝐸 𝑃 𝐷 𝑄 Figure P5.45 and P5.46

298

ISTUDY

Chapter 5

Equilibrium of Bodies

5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

We begin this section by more thoroughly examining the theoretical underpinnings of the equilibrium analyses performed in the previous section. We then introduce springs and the inclusion of deformable members in equilibrium problems. Finally, we present a very brief introduction to statically indeterminate problems in an example.

Why are bodies assumed to be rigid? Throughout statics, when analyzing the equilibrium of bodies, objects, and structures, we almost always assume they are rigid (springs are a notable exception). This assumption is made out of necessity for the same reasons as those discussed in Section 3.2 in connection with cables and bars being idealized as inextensible. When ∑ ∑ ∑ we write equilibrium equations such as 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀𝑃 = 0, the geometry of the object is needed so that forces can be resolved into components and moment arms can be determined. If the object is deformable, as real objects are, then this geometry is usually unknown, and the equations of static equilibrium alone are too few to solve for the unknown reactions and the new geometry. Structures or objects that are stiff typically undergo little change of geometry when forces are applied, and hence they can often be idealized as rigid. For structures that are flexible, we usually have no alternative but to incorporate deformability into our analysis, and this topic is addressed in subjects that follow statics and dynamics, such as mechanics of materials.

Helpful Information Cable Geometry. A cable wrapped around two pulleys with the same radius is common.

𝑟

𝐴

𝐿 𝛼

𝐵

𝑟

ℎ Figure 5.10 With 𝑟 being the radius of the pulleys and ℎ being the distance between the pulleys’ bearings, the orientation 𝛼 and the length 𝐿 of the cable segment between the pulleys are 2𝑟 𝛼 = sin−1 , ℎ (5.12) 𝐿 = ℎ cos 𝛼. Derivation of these expressions is given as an exercise in Prob. 5.47.

Treatment of cables and pulleys Cables and pulleys are common components in structures, and there are several options for how these may be treated in an analysis. Consider the example shown in Fig. 5.11(a), where force 𝑊 is specified and the support reactions are to be determined. Since the pulleys are idealized as frictionless and the cable is continuous, perfectly flexible, and weightless, all portions of the cable support the same tensile force, which is equal to 𝑊 . Three possible FBDs are shown in Fig. 5.11(b)–(d). In the first of these, Fig. 5.11(b), the pulleys are left on the structure, and when we take the cut to draw the FBD, the forces shown are exposed. In the next two FBDs, Fig. 5.11(c) and (d), the pulleys are removed from the structure. That is, when taking the cut to produce these FBDs, we remove the pulleys, and this exposes the forces that the pulleys apply to the structure. In Fig. 5.11(c), the pulley forces are simply “shifted” to the bearings of the pulleys. Justification for this is seen in the FBDs of the pulleys themselves shown on the right-hand side of Fig. 5.11(c), where the bearing forces have been constructed so that each pulley is in equilibrium. The FBD shown in Fig. 5.11(d) is very similar to that shown above it, except the bearing forces are further resolved into 𝑥 and 𝑦 components. For most purposes, you will likely find the FBD approaches in Fig. 5.11(b) and (c) to be the most useful.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

𝑦

4m

3m 𝑥

(a)

299

𝐵

𝐴

𝐶

1m

𝛼 1m

1m 𝑊

𝐴𝑦 (b)

𝐴𝑥

4m

3m 𝐵

𝐴

𝑀𝐴

𝐶 1m

1m

𝑊

𝑊

𝐴𝑦 (c)

𝐴𝑥

3m

(d)

𝐴𝑥 𝑀𝐴

𝛼

𝑊

𝑀𝐴

𝐴𝑦

𝐵

𝐴

4m 𝑊 𝛼

𝐶

𝑊

𝑊

𝐴

𝐵𝑦

𝐶 𝐵𝑥

𝑊

𝛼 𝑊

𝐶𝑦

𝐵𝑥 𝐵

𝐶𝑥

𝐶 𝑊

𝑊

𝐶𝑦

𝐵𝑦

4m 𝐵

𝑊

𝑊

𝑊

3m

𝐵

𝑊

𝛼

𝐶

𝐶𝑥

𝑊 𝑊

Figure 5.11. (a) Cantilever beam structure with frictionless pulleys at points 𝐵 and 𝐶. (b)–(d) Free body diagrams that may be used to determine the reactions at point 𝐴. Because the pulleys have the same radius, angle 𝛼 may be easily computed using Eq. (5.12).

Springs Springs are among the few deformable members we consider in statics. Some examples of springs are shown in Fig. 5.12, and a schematic representation is shown in Fig. 5.13. All of the remarks made about springs in Section 3.2 are applicable here, and for linear elastic behavior, the spring law is 𝐹𝑠 = 𝑘𝛿 ) ( = 𝑘 𝐿 − 𝐿0 ,

Courtesy of Peninsula Spring

Figure 5.12 An assortment of small coil springs made of wire.

(5.13)

where 𝐹𝑠 is the force supported by the spring; 𝛿 is the elongation of the spring from its unstretched or undeformed length; 𝑘 is the spring stiffness (units: force/length); 𝐿0 is the initial (unstretched) spring length; and 𝐿 is the final spring length. The spring stiffness 𝑘 is always positive. Because the force 𝐹𝑠 and the elongation 𝛿 in Eq. (5.13) are directed along an axis, or line, these springs are sometimes called

𝐿 = final length 𝐿0 = initial length

initial 𝑘 final

𝛿 𝐹𝑠

Figure 5.13 A spring produces a force 𝐹𝑠 that is proportional to its elongation 𝛿. Such springs are sometimes called axial springs.

300

Chapter 5

Equilibrium of Bodies

Figure 5.14 Example of a torsional spring used in a clothespin.

axial springs to differentiate them from torsional springs, which are discussed next. Equation (5.13) can be written using other sign conventions for the positive directions of 𝐹𝑠 and 𝛿, but this may require introducing a negative sign in Eq. (5.13) as discussed in Section 3.2. Torsional springs are also common, and a simple example of a torsional spring is shown in the clothespin of Fig. 5.14. Figure 5.15 shows some small torsional springs made of wire, and you should contrast the construction of these with the axial springs shown in Fig. 5.12. A torsional spring produces a moment that is proportional to the relative rotation, or twist, of the spring. A torsional spring is shown schematically in Fig. 5.16, and for linear elastic behavior, the spring law is 𝑀𝑡 = 𝑘𝑡 𝜃,

(5.14)

where 𝑀𝑡 is the moment produced by the spring; 𝜃 is the rotation of the spring measured in radians from the unloaded geometry; and 𝑘𝑡 is the spring stiffness (units: moment/radian).∗

Courtesy of Peninsula Spring

Figure 5.15 An assortment of small torsional springs made of wire. 𝜃

𝑘𝑡 𝑀𝑡

Figure 5.16 A torsional spring produces a moment 𝑀𝑡 that is proportional to its rotation 𝜃.

ISTUDY

By definition, the spring stiffness 𝑘𝑡 is always positive. Note that units for the stiffness of a torsional spring are different than units for the stiffness of an axial spring. Equation (5.14) assumes 𝑀𝑡 and 𝜃 are positive in the same direction. If you choose to take 𝑀𝑡 and 𝜃 to be positive in opposite directions, then a negative sign must be introduced in Eq. (5.14).

Superposition Superposition is a concept that can be used to replace a problem having complex loading with a sum (or superposition) of problems having simpler loading. Often, each of the problems with the simpler loading is easier to analyze than the original problem. Even when the problems with the simpler loading are no easier to analyze than the original problem, superposition may still be useful. Example 5.10 illustrates superposition.

Supports and fixity The way an object is supported determines its fixity and whether it is statically determinate or indeterminate. These concepts are defined as follows: Complete fixity. A body with complete fixity has supports that are sufficient in number and arrangement so that the body is completely fixed in space and will undergo no motion (either translation or rotation) in any direction under the action of any possible set of forces. Partial fixity. A body with partial fixity† has supports that will allow the body to undergo motion (translation and/or rotation) in one or more directions. Whether or not such motion occurs depends on the forces and/or moments that are applied and whether the body is initially in motion. No fixity. A body with no fixity has no supports and is completely free to translate and rotate in space. ∗ The

stiffness for torsional springs is sometimes given using degrees to measure rotation. However, for analytical work it is best to use radians. † Some statics textbooks describe partial fixity as “improper supports.” However, such nomenclature is undesirable since it implies that all objects should be fully fixed when, of course, this is not the case.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

All objects fall into one of the above three categories. Rather than the word fixity, it is common to refer to a body’s ability to undergo rigid body motion. A body in two dimensions can undergo three types of rigid body motion: translation in two orthogonal directions and rotation, or any combination of these. Thus, equivalent nomenclature for complete fixity, partial fixity, and no fixity is: no rigid body motion capability, partial rigid body motion capability, and full rigid body motion capability, respectively. For a body in two dimensions, complete fixity requires supports having a total of three or more reaction forces and/or moments with suitable arrangement. Partial fixity results when there are only one or two reaction forces, or it may also result if there are three or more reaction forces and/or moments that do not have sufficient arrangement to fully prevent motion.

Static determinacy and indeterminacy Static determinacy and indeterminacy were introduced in Example 3.6 on p. 158. These are defined as follows: Statically determinate body. For a statically determinate body, the equilibrium equations of statics are sufficient to determine all unknown forces and/or other unknowns that appear in the equilibrium equations. Statically indeterminate body. For a statically indeterminate body, the equilibrium equations of statics are not sufficient to determine all unknown forces and/or other unknowns. All objects that are in static equilibrium fall into one of these two categories. For a statically indeterminate body, there are more unknowns than equilibrium equations, and in general none of the unknowns can be found. On occasion, however, the equilibrium equations may be sufficient so that some (but not all) of the unknowns in a statically indeterminate problem can be found. A simple rule of thumb to help ascertain whether an object is statically determinate or indeterminate is to compare the number of unknowns to the number of equilibrium equations, and we call this equation counting. With 𝑛 being the number of unknowns, the rule of thumb for a single body in two dimensions is as follows: If 𝑛 < 3

The body is statically determinate; it has partial fixity if 𝑛 = 1 or 2, and it has no fixity if 𝑛 = 0.

If 𝑛 = 3

The body is statically determinate if it has full fixity, or the body is statically indeterminate if it has partial fixity.

If 𝑛 > 3

(5.15)

The body is statically indeterminate, and it can have full fixity or partial fixity.

Successful use of equation counting, Eq. (5.15), requires some judgment on your part, which is why it is called a rule of thumb rather than a rule. Nonetheless, it is very useful and it will be employed throughout this book with enhancements as needed. Thorough understanding of equation counting here will make these subsequent enhancements self-evident. The basis for this rule of thumb and subtleties in its application are explored in the following mini-example.

301

302

ISTUDY

Chapter 5

Equilibrium of Bodies

Mini-Example A body with a variety of support schemes is shown in Fig. 5.17. For each support scheme, specify whether the body has partial or full fixity and whether it is statically determinate or indeterminate.

(a)

(b)

(c)

Helpful Information 𝑛 = 3 and visual inspection shows complete fixity ⇒ statically determinate.

Cables and rollers. For each object in Fig. 5.17, the number of unknowns 𝑛 is determined assuming that cables are in tension and rollers are in compression. If the loading is such that, for example, a particular cable is not in tension, then the force it supports is known to be zero, and it does not count as an unknown when determining 𝑛. Similarly, if a particular roller is not in compression, then the force it supports is known to be zero, and it does not count as an unknown when determining 𝑛.

(d)

(e)

𝑛 = 2 ⇒ statically determinate with partial fixity.

(g)

(f)

(h)

𝑛 = 4 ⇒ statically indeterminate.

(i)

(j)

𝑛 = 3 and visual inspection shows partial fixity ⇒ statically indeterminate. Figure 5.17. Examples of equation counting to determine if a body is statically determinate or indeterminate. For bodies that are not fully fixed, a possible displaced position is shown by the dashed outlines.

Solution In Fig. 5.17(a), the roller has one unknown reaction force and the pin has two, for a total of 𝑛 = 3 (if it is not clear that 𝑛 = 3, then draw the FBD and count the number of reactions to obtain 𝑛 = 3). Inspection of the body shows that the supports are sufficient in number and arrangement so that it has complete fixity. That is, the pin prevents horizontal translation, the pin or the roller prevents vertical translation, and the combined effect of the pin and roller prevents rotation; in making these observations we assume the roller is in compression.∗ Hence, Eq. (5.15) indicates the body is statically determinate. Similar remarks apply to Fig. 5.17(b) and (c). The bodies shown in Fig. 5.17(d) and (e) each have two unknown reactions, and Eq. (5.15) indicates both bodies are statically determinate with partial fixity. Inspection of Fig. 5.17(d) shows this body can undergo horizontal translation, while the body of Fig. 5.17(e) can undergo horizontal translation and rotation.

∗ If

the roller is not in compression and the left-hand side of the object lifts off its support, then the object is essentially supported by only a pin, in which case 𝑛 = 2 and the object has only partial fixity.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

When stating that these bodies are statically determinate, we presume the loading is such that the problems are indeed static rather than dynamic. For example, if the body of Fig. 5.17(d) has a net horizontal force applied to it, then it will undergo horizontal acceleration and Newton’s law 𝐹 = 𝑚𝑎 must be used. The bodies shown in Fig. 5.17(f), (g), and (h) each have four unknown reactions, and Eq. (5.15) indicates that all three bodies are statically indeterminate. Inspection of Fig. 5.17(f) and (g) shows that both of these have complete fixity. Inspection of Fig. 5.17(h) shows that it can undergo horizontal translation, and hence, it has only partial fixity. The bodies shown in Fig. 5.17(i) and (j) each have three unknown reactions, but examination shows that each has partial fixity. The motion capability in Fig. 5.17(i) is obvious; it can undergo horizontal translation. The motion capability in Fig. 5.17(j) is more subtle; the body can undergo small rotation about the pin before the link’s orientation changes enough to restrain further motion. Hence, Eq. (5.15) indicates that both bodies are statically indeterminate. 𝑃

Remarks. The following remarks pertain to the examples shown in Fig. 5.17. • The supports for Fig. 5.17(c) and (j) are very similar, and it is perhaps subtle why the latter of these has only partial fixity. Another perspective that might help show that it has partial fixity is to consider the response of the body of Fig. 5.17(j) when it is subjected to a force 𝑃 , as shown in Fig. 5.18. After we draw the FBD and sum moments about point 𝐴, it is very clear that the only ∑ possible value of 𝑃 that satisfies 𝑀𝐴 = 0 is 𝑃 = 0. If a nonzero value of 𝑃 were applied, the supports would be incapable of developing reactions that could equilibrate 𝑃 , and dynamic motion would occur, at least until such time that the geometry of link 𝐶𝐷 changes. Use of a force to determine fixity is helpful, but is not foolproof. A force with the wrong position or orientation may not reveal that a body has partial fixity. • Whether or not a body has rigid body motion capability has nothing to do with the forces that are applied to it. However, if a body does have rigid body motion capability, whether or not motion occurs depends strongly on the forces and/or moments that are applied. • Equation counting is easy, but successful use relies on your ability to examine a body to determine its fixity. Essentially, equation counting provides only a scalar characterization of a body’s supports. A more complete characterization would need to include the vector aspect of the supports, namely, the positions and directions of the reactions. But such a method of analysis is considerably more complex.

Two-force and three-force members If a body or structural member is subjected to forces at two points or three points only, as described next, then when the body is in equilibrium, the orientation of the forces supported by the body has special properties. These situations are defined as follows:

𝐵

𝐴

(a)

𝐶 𝑦

(b)

𝐴𝑦

𝑃 𝑥

𝐴 𝐴𝑥

𝐵 𝐶 𝐹 𝐶𝐷

Figure 5.18 Use of a force to determine a body’s fixity.

303

304

ISTUDY

Chapter 5

Equilibrium of Bodies

Two-force member. A body subjected to forces at two points (no moment loading and no distributed forces such as weight) is called a two-force member. The special feature of a two-force member is that, when in equilibrium, the two forces have the same line of action and opposite direction. Examples are shown in Fig. 5.19. Three-force member. A body subjected to forces at three points (no moment loading and no distributed forces such as weight) is called a three-force member. The special feature of a three-force member is that, when in equilibrium, the lines of action of all three forces intersect at a common point. If the three forces are parallel (this is called a parallel force system), then their point of intersection can be thought of as being at infinity. Examples are shown in Fig. 5.20. If a body is not a two-force or three-force member, then we will refer to it as either a zero-force member, if it is subjected to no force at all, or a general multiforce member, if it is subjected to forces at more than three points and/or has moment loading and/or has distributed loading. Zero-force members will be routinely encountered in truss structures, discussed in Chapter 6, and we will see they can play an important role in improving the strength of a truss. general body 𝐵

𝐴

Common Pitfall Geometry of a two-force member. A common misconception is that a two-force member is always straight. While two-force members are often straight, they can be curved or have other geometry. Chapter 8 discusses the merits of two-force members that are straight versus two-force members that are not straight (see Prob. 8.22 on p. 502).

cable, rope, string, ... wire straight bar

curved bar

Figure 5.19. Examples of two-force members. In all cases, the two forces supported by the member have equal magnitude, opposite direction, and have the same line of action.

Remarks on two-force members • The characteristics of a two-force member can be proved using the following argument. Consider the general body shown in Fig. 5.19, and imagine the forces at 𝐴 and 𝐵 may not have the orientations shown. Moment equilibrium about 𝐴 is satisfied only if the line of action of the force at 𝐵 passes through 𝐴. Similarly, moment equilibrium at 𝐵 is satisfied only if the force at 𝐴 passes through 𝐵. Hence, the two forces must have the same line of action, which is the line containing 𝐴 and 𝐵. Then summing forces in the direction from 𝐴 to 𝐵 shows that the two forces must have equal magnitude and opposite direction.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

𝐹1

𝐹2

𝐹1

𝐹2

𝐹3

𝐹3

three-force member subjected to a concurrent force system

three-force member subjected to a parallel force system

(a)

(b) 𝑃

𝑃 𝐴 𝐵

𝐴

𝐷

𝐹𝐶𝐷

𝐶 (c)

(d)

𝐵

𝐹𝐵𝐶 𝑊

𝐶

𝑊

𝐴 𝐴

(e)

𝑊

frictionless 𝐴

(f)

𝑊

𝐵

𝐵 𝜃

𝜃 𝐴 (g)

(h)

Figure 5.20. Examples of three-force members. (a) Three-force member with a concurrent force system: all forces intersect at a common point. (b) Three-force member with a parallel force system. (c)–(h) Examples of three-force members with concurrent force systems. In (g) and (h), the bar’s equilibrium position has a value of 𝜃 so that all forces intersect at a common point.

305

306

𝜃 cable

𝜃 link or bar 𝑅

𝜃 curved link or bar

𝜃 reaction

Figure 5.21 Support reactions for cable, link, and bar supports repeated from Fig. 5.3 on p. 273. Because all of these supports are two-force members, the reaction force 𝑅 is directed between the endpoints of these members.

ISTUDY

Chapter 5

Equilibrium of Bodies

• The support reactions for cable, link, and bar supports were shown in Fig. 5.3 on p. 273, and these are repeated in Fig. 5.21.∗ In Section 5.2, the support reactions for these members were obtained by considering the motion that the support prevents. Alternatively, the reaction forces for these members can be obtained by recognizing that the cable, link, and bar supports are all two-force members; hence, the reaction force 𝑅 is directed between the endpoints of these members. • The ability to identify two-force members is important and can help simplify the analysis of complex problems, such as with frames and machines, discussed in Chapter 6. Remarks on three-force members • The characteristics of a three-force member with a concurrent force system can be proved using the following argument. Consider the body shown in Fig. 5.20(a), and imagine force 𝐹3 may not have the orientation shown. Moment equilibrium about the intersection point of the lines of action of 𝐹1 and 𝐹2 is satisfied only if the line of action of 𝐹3 passes through this point. The same argument can be made by considering the orientations of 𝐹1 and 𝐹2 to be different. Hence, moment equilibrium for a three-force member with a concurrent force system requires the lines of action of the three forces to intersect at a common point. • Although the forces applied to a three-force member can be parallel, such as shown in Fig. 5.20(b), there are no especially significant remarks that can be made about this situation. Rather, analysis proceeds in the usual fashion, where ∑ ∑ ∑ we write equations such as 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀𝑃 = 0.

End of Section Summary In this section, some of the finer points regarding static equilibrium of a rigid body were reviewed. Some of the key points are as follows: • When equilibrium equations are written, the geometry of the structure in the equilibrium position must be used. By assuming a structure or object is rigid, the geometry before and after forces are applied is the same, and hence, the solution to the equilibrium equations provides an exact solution for the problem. • When a structure or object includes pulleys and cables, there are several options for how these may be treated when FBDs are drawn. Sometimes it is most effective if a particular pulley is left on the structure and the cut that is used to draw the FBD is taken through that pulley’s cables. In other cases it may be most effective to take a cut that removes a pulley from the structure, in which case the forces that the pulley applies to the structure must be included in the FBD; this is often called shifting the cable forces to the bearing of a pulley. A more detailed discussion of these issues is given in connection with Fig. 5.11. • Springs were described in detail in Section 3.2 and were reviewed again in this section.

∗ The

cable with self-weight shown in Fig. 5.3 is not a two-force member, and hence it is not shown here.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

• A torsional spring is a deformable member that produces a moment 𝑀𝑡 proportional to the amount of twist 𝜃 of the spring. The spring law is 𝑀𝑡 = 𝑘𝑡 𝜃 where 𝑘𝑡 is called the spring stiffness, 𝑘𝑡 ≥ 0, and 𝑘𝑡 has units of moment/radian. In writing this equation it is assumed that 𝑀𝑡 and 𝜃 are positive in the same direction. • Superposition was described. In simple terms, superposition is most often used to replace a problem having complex loading by two or more problems with simpler loading that are easier to analyze. The solution to the original problem is obtained by superposing the solutions to the simpler problems. • An object’s fixity is determined by the type, number, and arrangement of its supports. All objects have either full fixity, partial fixity, or no fixity. Throughout statics and dynamics, we rely on visual inspection of a structure and its supports to determine its fixity. Rather than the above nomenclature, it is also common to refer to an object’s ability to undergo rigid body motion. • Statically determinate and statically indeterminate structures and objects were described. For a statically determinate structure or object, the equations of equilibrium are sufficient to determine all unknowns. For a statically indeterminate structure or object, the equations of equilibrium are not sufficient to determine all unknowns. Equation counting was described as an effective way to determine whether a structure was statically determinate or indeterminate. • Two-force and three-force members were described. If a member is a two-force or three-force member, then the forces it supports have special properties. The ability to identify two-forces members is especially important as this will allow for simplifications that make equilibrium analysis easier.

307

308

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.7

Springs

0.6 m 𝐿 𝐷

𝜃 𝑘 𝛿

airflow

𝐵 0.4 m

𝐴 𝑘𝑡

A wind tunnel is used to experimentally determine the lift force 𝐿 and drag force 𝐷 on a scale model of an aircraft. The bracket supporting the aircraft is fitted with an axial spring with stiffness 𝑘 = 0.125 N∕mm and a torsional spring with stiffness 𝑘𝑡 = 50 N⋅m∕rad. By measuring the deflections 𝛿 and 𝜃 of these springs during a test, the forces 𝐿 and 𝐷 may be determined. If the geometry shown in Fig. 1 occurs when there is no airflow, and if the springs are calibrated so that 𝛿 = 0 and 𝜃 = 0◦ when there is no airflow, determine 𝐿 and 𝐷 if 𝛿 = 2.51 mm and 𝜃 = 1.06◦ are measured.

SOLUTION

Figure 1

Road Map

This problem involves spring elements, and these have equations that govern their load–deformation response. Thus, the problem-solving methodology used here will be enhanced to emphasize that in addition to the need for equilibrium equations, force laws that describe the behavior of the springs are needed.

The FBD is shown in Fig. 2, where we assume that bar 𝐴𝐵, which supports the aircraft, is slender enough that it does not develop any lift or drag forces.

Modeling

Governing Equations

0.6 m

𝑦

𝐿

Equilibrium Equations

𝐷

𝐵

∑

0.4 m

𝐴 𝐴𝑥 = 𝑘𝛿

Summing forces in the 𝑥 direction and summing moments about

point 𝐴 provide

𝑥

𝑀𝐴 = 𝑘𝑡 𝜃 𝐴𝑦

Figure 2 Free body diagram.

∑

𝐹𝑥 = 0 ∶

𝑀𝐴 = 0 ∶

𝐴𝑥 − 𝐷 = 0,

(1)

−𝑀𝐴 + 𝐷(400 mm) + 𝐿(600 mm) = 0.

(2)

In writing Eq. (2), we assume the deformation of the torsional spring is small so that the geometry of the support bracket 𝐴𝐵 is essentially unchanged from its original geometry. Notice that Eqs. (1) and (2) have five unknowns, so clearly more equations must be written to obtain a determinate system of equations. Force Laws The force supported by the axial spring is related to its deformation 𝛿, and the moment supported by the torsional spring is related to its rotation 𝜃, by

𝐴𝑥 = 𝑘𝛿,

(3)

𝑀𝐴 = 𝑘𝑡 𝜃.

(4)

Computation

Using Eq. (3), Eq. (1) may be solved for ) ( N 𝐷 = 𝑘𝛿 = 0.125 (2.51 mm) = 0.3138 N. mm Using Eq. (4) and the solution for 𝐷 just obtained, Eq. (2) may be solved for 𝐿=

−𝐷(400 mm) + 𝑘𝑡 𝜃 600 mm −(0.3138 N)(400 mm) + (50

=

Matt Cardy/Getty Images News/Getty Images

Figure 3 A model of an Airbus airplane is prepared for testing in a large wind tunnel.

ISTUDY

(5)

N⋅ m 103 mm 𝜋 rad )( m )( 180◦ )1.06◦ rad

600 mm

= 1.333 N.

(6)

Discussion & Verification If the springs are not sufficiently stiff, then 𝛿 and/or 𝜃 may be substantially larger, and the original geometry cannot be used when writing Eq. (2). An additional disadvantage of a soft torsional spring is that if 𝜃 is large, then the angle of attack of the aircraft also changes appreciably, which is undesirable. Since 𝛿 and 𝜃 are known in this problem, it is easy to verify that the difference between the original geometry and the deformed geometry is small. You can explore this issue in greater detail in Probs. 5.79 through 5.82.

ISTUDY

Section 5.3

309

Equilibrium of Bodies in Two Dimensions—Additional Topics

E X A M P L E 5.8

Free Body Diagram Choices for Pulleys

The stationary crane is supported by a pin at point 𝐵 and a bar between points 𝐴 and 𝐸. A winch at point 𝐶 is used to raise and lower loads. The pulleys at points 𝐷, 𝐹 , and 𝐺 have 1 ft radius. Determine the support reactions due to the vertical force 𝑃 if 𝑃 = 3 kip.

3 ft

𝐹 3 ft

The structure has a single cable wrapped around pulleys at 𝐷, 𝐹 , and 𝐺. Assuming these pulleys are frictionless and neglecting the weight of the cable, the force supported by the cable has the same value 𝑃 throughout its length. When treating the cable and pulleys in FBDs, we may choose to leave these on the FBD, or to remove them, or to use a combination of these. Our choice of approach will be determined by the ease with which moment arms for the cable forces can be determined. Road Map

3 ft

3 ft

𝐺

𝐷

𝐸

SOLUTION

3 ft

1 ft 𝑃

3 ft

3 ft 𝐴

𝐶

𝐵

Figure 1

Modeling

In considering the FBD options as discussed in connection with Fig. 5.11 on p. 299, we first consider leaving the cable and all the pulleys on the structure. The resulting FBD is shown in Fig. 2, where we observe that all moment arms are easily obtained, and thus it is not necessary to consider other options for drawing the FBD. Governing Equations & Computation

tions and solutions are ∑ 𝐹𝑥 = 0 ∶ ∑

𝑀𝐵 = 0 ∶

∑

Using the FBD in Fig. 2, the equilibrium equa-

3ft

𝐵𝑥 = 0

(1) 𝐵𝑥 = 0,

(2)

𝑇𝐸𝐴 (3 ft) − (3 kip)(1 ft) − (3 kip)(10 ft) = 0

(3)

⇒

𝑇𝐸𝐴 = 11 kip,

(4)

−𝑇𝐸𝐴 + 𝐵𝑦 − 3 kip − 3 kip = 0

(5)

𝐵𝑦 = 17 kip.

(6)

⇒

Discussion & Verification

• In the FBD of Fig. 2, a positive value of 𝑇𝐸𝐴 corresponds to tension in bar 𝐴𝐸, and a positive value of 𝐵𝑦 corresponds to vertical compression in the pin support at 𝐵. Intuitively, we expect bar 𝐴𝐸 to be in tension and the pin support at 𝐵 to be in compression, and indeed our solutions show 𝑇𝐸𝐴 > 0 and 𝐵𝑦 > 0. • In this example, the weight of the crane was neglected, and thus, the reactions we computed are those due to supporting the force 𝑃 only. The weight of the crane is probably not small. Problem 5.84 asks you to determine the reactions due to only the weight of the crane and then to use superposition with the results of this example to determine the total support reactions.

3ft

3ft

𝐷

𝐸

⇒ 𝐹𝑦 = 0 ∶

3ft

𝐹

1ft

3ft

3ft

𝑦 𝑥

𝑇𝐸𝐴

𝐺

𝑃 = 3 kip

𝑃 = 3 kip

3ft 𝐵

𝐵𝑥 𝐵𝑦

Figure 2 Free body diagram leaving all pulleys on the structure.

310

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.9 3ft

𝐹 3ft

3ft

Free Body Diagram Choices for Pulleys

3ft

3ft

𝜃

𝐷

𝐸

𝐺

SOLUTION

1ft 𝑃

3ft

3ft 𝐴

𝐶

𝐵

Figure 1

3ft

𝐹 3ft

3ft

3ft

𝑦

1ft

3ft

𝑥 𝑇𝐸𝐴

𝐺

20◦

𝐷

𝐸

3ft

𝑃 = 3 kip

𝑃 = 3 kip

3ft 𝐵

Figure 2 Free body diagram leaving all pulleys on the structure.

𝐹

This problem is identical to that in Example 5.8, except that the cable segment at pulley 𝐺 is oriented 20◦ from the vertical. The structure has a single cable wrapped around pulleys at 𝐷, 𝐹 , and 𝐺. Assuming these pulleys are frictionless and neglecting the weight of the cable, the force supported by the cable has the same value 𝑃 throughout its length. We will begin by using an FBD where all of the pulleys are left on the structure and will consider if the moment arms needed are easy to obtain. If the moment arms are not easy to obtain, we will then consider if it is more convenient to use an FBD where some or all of the pulleys are removed from the structure. Modeling In considering the FBD options as discussed in connection with Fig. 5.11 on p. 299, we first consider leaving the cable and all the pulleys on the structure, and the resulting FBD is shown in Fig. 2. Unfortunately, determining the moment arm for the cable force at pulley 𝐺 in this FBD is tedious. Thus, we draw a new FBD, as shown in Fig. 3, where the pulley at 𝐺 is removed, and we observe that all moment arms are now easily obtained. Note that the horizontal forces at pulley 𝐹 and point 𝐺 do not have the same line of action. Governing Equations & Computation

𝐵𝑥

3ft 𝑃

Road Map

tions and solutions are ∑ 𝐹𝑥 = 0 ∶

𝐵𝑦

3ft

The stationary crane is supported by a pin at point 𝐵 and a bar between points 𝐴 and 𝐸. A winch at point 𝐶 is used to raise and lower loads. The pulleys at points 𝐷, 𝐹 , and 𝐺 have 1 ft radius. Determine the support reactions due to the force 𝑃 if 𝑃 = 3 kip and 𝜃 = 20◦ .

3ft

3ft 𝑃

∑

𝑀𝐵 = 0 ∶

𝐵𝑥 + 3 kip − 3 kip + (3 kip)(sin 20◦ ) = 0 ⇒

pulley 𝐷

20◦

𝐷

𝐸

1ft

3ft

𝑃 = 3 kip

𝑦 𝑥

𝑇𝐸𝐴

𝑃 = 3 kip

3ft 𝐵

𝐵𝑥 𝐵𝑦

Figure 3 Free body diagram where the pulley at 𝐺 has been removed. Observe that the horizontal forces at pulley 𝐹 and point 𝐺 do not have the same line of action.

ISTUDY

𝐵𝑥 = −1.026 kip,

pulley 𝐹

− (3 kip)(cos 20◦ )(9 ft) − (3 kip)(sin 20◦ )(9 ft) = 0 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

𝐺

𝐹𝑦 = 0 ∶

(2)

pulley 𝐺

(3)

pulley 𝐺

⇒ ∑

(1)

𝑇𝐸𝐴 (3 ft) − (3 kip)(1 ft) − (3 kip)(10 ft) + (3 kip)(9 ft) ⏟⏞⏞⏞⏟⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏟⏞⏞⏞⏟

pulley 𝐺

3ft

Using the FBD in Fig. 3, the equilibrium equa-

𝑇𝐸𝐴 = 13.54 kip,

−𝑇𝐸𝐴 + 𝐵𝑦 − 3 kip − (3 kip)(cos 20◦ ) = 0 ⇒

𝐵𝑦 = 19.35 kip.

(4) (5) (6)

Discussion & Verification

• The comments made in the Discussion & Verification in Example 5.8 also apply here, and indeed we find that bar 𝐴𝐸 is in tension (𝑇𝐸𝐴 > 0) and the pin at 𝐵 is in vertical compression (𝐵𝑦 > 0). • Comparing our answers in Eqs. (2), (4), and (6) with those in Example 5.8 shows that the reactions in this example are slightly larger in magnitude. This can be explained by comparing the FBD shown in Fig. 2 of Example 5.8 with the FBD shown in Fig. 2 of this example, where it is seen that the moment arm for the force 𝑃 at pulley 𝐺 is larger in this example, hence, the larger reaction forces found in this example are expected.

ISTUDY

Section 5.3

E X A M P L E 5.10

Superposition

The cantilever beam shown in Fig. 1 is loaded by two 1 kN forces. Use superposition to determine the reactions at point 𝐴.

1 kN

2m

Road Map

The loading for this problem can be broken into the two simpler loadings shown in Fig. 2(a), namely load case 1 and load case 2, which we will simply call case 1 and case 2.

𝐶 2m

Figure 1

The FBDs for the original loading and for each load case are shown in Fig. 2(b). load case 1 1 kN

1 kN 𝐵

𝐴

(a)

2m

𝑦 𝑥 (b)

1 kN 𝐵

𝐴

SOLUTION

Modeling

311

Equilibrium of Bodies in Two Dimensions—Additional Topics

𝐶

1 kN 𝐵

(𝐴𝑥 )total

1 kN 𝐵

𝐴

=

1 kN 𝐴

+

𝐶

2m

1 kN

(𝑀𝐴 )total

load case 2

2m (𝐴𝑦 )total

(𝑀𝐴 )case 1

1 kN

𝐶 = (𝐴𝑥 )case 1

1 kN

(𝑀𝐴 )case 2

𝐵 +

(𝐴𝑥 )case 2

𝐶

2m (𝐴𝑦 )case 1

(𝐴𝑦 )case 2

Figure 2. (a) Two load cases to be used for superposition. (b) Free body diagrams corresponding to the original loading and each load case. Governing Equations & Computation

solve the equilibrium equations reactions for load case 1 as (𝐴𝑥 )case 1 = 0,

∑

Using the FBD for load case 1, we write and ∑ ∑ 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀 = 0 to obtain the

(𝐴𝑦 )case 1 = 1 kN,

and (𝑀𝐴 )case 1 = 2 kN⋅m.

(1)

Similarly, using the FBD for load case 2, we write and solve the equilibrium equations ∑ ∑ ∑ 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀 = 0 to obtain the reactions for load case 2 as (𝐴𝑥 )case 2 = 0,

(𝐴𝑦 )case 2 = 1 kN,

and (𝑀𝐴 )case 2 = 4 kN⋅m.

(2)

To obtain the reactions for the total loading, we add (or superpose) the results for each load case. Thus, (𝐴𝑥 )total = (𝐴𝑥 )case 1 + (𝐴𝑥 )case 2 = 0 kN + 0 kN = 0 kN,

(3)

(𝐴𝑦 )total = (𝐴𝑦 )case 1 + (𝐴𝑦 )case 2 = 1 kN + 1 kN = 2 kN,

(4)

(𝑀𝐴 )total = (𝑀𝐴 )case 1 + (𝑀𝐴 )case 2 = 2 kN⋅m + 4 kN⋅m = 6 kN⋅m.

(5)

Discussion & Verification

To further demonstrate the utility of superposition, consider the loading shown in Fig. 3. Rather than reanalyze this problem from the beginning, we may use our previous superposition analysis where we simply scale the results for each of the constituent load cases. Thus, for Fig. 3, the total moment at point 𝐴 is

= 3(2 kN⋅m) + 5(4 kN⋅m) = 26 kN⋅m. Similarly, the remaining reactions are found to be (𝐴𝑥 )total = 0 and (𝐴𝑦 )total = 8 kN.

𝐵

𝐴 2m

(𝑀𝐴 )total = 3(𝑀𝐴 )case 1 + 5(𝑀𝐴 )case 2 (6)

5 kN

3 kN

𝐶 2m

Figure 3 Using the results of the superposition solution for Fig. 1, this example is very easy to analyze.

312

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.11

Equilibrium of a Three-Force Member

folded position

A handle for pushing a cart is shown, where the handle may be easily folded against the side of the cart when it is not needed. Determine the forces supporting handle 𝐴𝐵𝐶 when the 50 N force is applied.

𝐸 50 N 40◦

𝐷

𝐵 𝐶

SOLUTION

90 mm

15 mm

Road Map

This problem is readily solvable by the methods and approaches discussed in Section 5.2. That is, an FBD is drawn, and then equilibrium equations are written and solved to determine the support reactions. However, once the FBD is drawn, a close examination of this shows that member 𝐴𝐵𝐶 is a three-force member, and this offers an alternative solution approach. Both solutions are carried out and compared.

30 mm

𝐴 70 mm 90 mm

Solution 1 Figure 1

Members 𝐴𝐷 and 𝐵𝐸 are both two-force members. Thus, the force supported by member 𝐴𝐷 acts along the line connecting points 𝐴 and 𝐷, and similarly, the force supported by member 𝐵𝐸 acts along the line connecting points 𝐵 and 𝐸. Taking advantage of these features of two-force members gives the FBD shown in Fig. 2, where we have assumed that the contact between the handle and cart at point 𝐴 is frictionless. Modeling

150 120 𝐸 90

𝑦 𝑥 50 N

𝐹𝐵𝐸 𝐹𝐴𝐷

15 mm

40◦

𝐵 𝐶 70 mm

Governing Equations & Computation Using the FBD in Fig. 2, equilibrium equations can be written and immediately solved as follows: ∑ 𝑀𝐵 = 0 ∶ −(50 N)(cos 40◦ )(15 mm) + (50 N)(sin 40◦ )(70 mm)

𝐴𝑥

𝐴 90 mm

− 𝐹𝐴𝐷 (90 mm) = 0

Figure 2 Free body diagram.

∑ ∑

150 120 90 𝐹𝐵𝐸

50 N

𝑅𝐴

40◦ (a) 𝐺 𝑅𝐴

120 40◦ 𝑅𝐴

(c)

𝛼

𝐹𝐵𝐸

50 N

𝛼 𝐴𝑥

90 𝐹𝐴𝐷

Figure 3 Alternate solution for a three-force member. (a) Free body diagram. (b) A closed force triangle ∑ ⃗ (c) Once 𝑅 is determined, 𝐴 enforces 𝐹⃗ = 0. 𝐴 𝑥 and 𝐹𝐴𝐷 may be determined.

ISTUDY

𝐹𝑥 = 0 ∶

(2) (3) (4) (5) (6)

Solution 2

𝛼

(b)

𝐹𝑦 = 0 ∶

⇒ 𝐹𝐴𝐷 = 18.61 N, ) ( 120 − 𝐹𝐴𝐷 = 0 (−50 N)(sin 40◦ ) + 𝐹𝐵𝐸 150 ⇒ 𝐹𝐵𝐸 = 63.44 N, ) ( 90 − 𝐴𝑥 = 0 (50 N)(cos 40◦ ) + 𝐹𝐵𝐸 150 ⇒ 𝐴𝑥 = 76.37 N.

(1)

Modeling

This solution begins with all of the modeling considerations from Solution 1, resulting in the FBD shown in Fig. 2. Examination of member 𝐴𝐵𝐶 in Fig. 2 shows it is a three-force member, because it has forces applied at three points only (points 𝐴, 𝐵, and 𝐶). To carry out a solution that exploits the properties of a three-force member, we combine the two forces 𝐴𝑥 and 𝐹𝐴𝐷 at point 𝐴 in Fig. 2 into a single force, which we will √ 2 call 𝑅𝐴 and whose magnitude is 𝑅𝐴 = 𝐴2𝑥 + 𝐹𝐴𝐷 . Both the magnitude and orientation of 𝑅𝐴 are unknown. The FBD in Fig. 2 is revised to give the FBD shown in Fig. 3(a), where we use the fact that the lines of action of the three forces applied to a three-force member must intersect at a common point, which is point 𝐺 in Fig. 3(a). Governing Equations & Computation Using the FBD in Fig. 3(a), the equilibrium ∑ ∑ equation 𝑀𝐺 = 0 is automatically satisfied, and the equilibrium equation 𝐹⃗ = 0⃗ is satisfied using the force polygon shown in Fig. 3(b) where the three forces are added head to tail to form a closed polygon. The calculations needed to determine the unknowns 𝐹𝐵𝐸 , 𝑅𝐴 , and 𝛼 from Fig. 3(b) are tedious, so we will only summarize the procedure. The 50 N force and 𝐹𝐵𝐸 both have known lines of action, and in the FBD shown in Fig. 3(a), we must determine the location where these lines of action intersect, point 𝐺. This calculation involves applications of the laws of sines and/or cosines. Because member 𝐴𝐵𝐶

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

is a three-force member, the line of action of 𝑅𝐴 must also intersect point 𝐺, and this provides the orientation 𝛼 that this force must have. To determine the value of 𝑅𝐴 , we construct the force polygon (for a three-force member this will always be a triangle) as ∑ ∑ shown in Fig. 3(b) (this enforces 𝐹𝑥 = 0 and 𝐹𝑦 = 0). Application of the laws of sines and/or cosines provides the value of 𝑅𝐴 . Finally, 𝐴𝑥 and 𝐹𝐴𝐷 are determined by resolving 𝑅𝐴 into components as shown in Fig. 3(c). Discussion & Verification

Occasionally, the second solution approach may provide a clever solution for a particular three-force member problem. However, the first solution approach is more methodical and is very robust (that is, it can always be used, regardless of whether a member is a three-force member or not). Furthermore, even for three-force member problems, the first solution approach will usually be more straightforward than the second.

313

314

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.12 𝐸

𝐹

𝐵

𝐴

Introduction to a Statically Indeterminate Problem

𝐶

𝐷

𝐺 2m

2m

1m 1m

A balcony in a motel consists of an I beam 𝐴𝐵𝐶𝐷 supported by a pin at point 𝐴 and two vertical cables 𝐵𝐸 and 𝐶𝐹 . The weight of the beam and the materials it supports (this is often called the dead load), plus the weight from the maximum number of people the balcony can accommodate (this is called the live load), is 24 kN with center of gravity at point 𝐺. Determine the support reactions at 𝐴 and the forces supported by the cables.

SOLUTION Examining Fig. 1, there are two unknown reactions at pin 𝐴, and each cable supports an unknown force. Hence, 𝑛 = 4 in Eq. (5.15) on p. 301, therefore the beam is statically indeterminate, and it will not be possible to determine all of the unknowns by using only the three equations of static equilibrium. While the analysis of statically indeterminate problems is normally beyond the limits of statics, we can occasionally perform accurate analysis by using simple springs to introduce deformability into the model for the problem.

Road Map

Figure 1

rigid

𝐸

𝐹

𝑘

𝑘 Modeling

𝐴

𝐵

𝐶

𝐷

𝐺 2m

2m

1m 1m

Figure 2 Beam 𝐴𝐵𝐶𝐷 is modeled as being rigid, and cables 𝐵𝐸 and 𝐶𝐹 are modeled as springs having stiffness 𝑘.

𝑇𝐵𝐸

𝐴𝑦 𝐴𝑥

24 kN

𝐵

𝑇𝐶𝐹

𝑥

𝐶

2m

𝛿𝐷 2m

1m 1m

𝐷

Figure 3 Free body diagram for the beam, and the displaced position of the beam where 𝛿𝐷 is the vertical displacement of point 𝐷.

𝐴 2m

𝛿𝐵𝐸

𝛿𝐶𝐹

𝐵

𝐶 2m

Using the FBD to sum forces in the 𝑥 direction provides one of

Equilibrium Equations

the reactions as

𝐷 2m

𝐹𝑥 = 0 ∶

𝐴𝑥 = 0

⇒

𝐴𝑥 = 0.

Summing forces in the 𝑦 direction and summing moments about point 𝐴 provides ∑ 𝐹𝑦 = 0 ∶ 𝐴𝑦 + 𝑇𝐵𝐸 + 𝑇𝐶𝐹 − 24 kN = 0, ∑ 𝑀𝐴 = 0 ∶ 𝑇𝐵𝐸 (2 m) + 𝑇𝐶𝐹 (4 m) − (24 kN)(3 m) = 0.

(1)

(2) (3)

Notice that Eqs. (2) and (3) contain three unknowns, thus additional equations must be written to obtain as many equations as unknowns.

Force Laws The spring forces 𝑇𝐵𝐸 and 𝑇𝐶𝐹 are related to the deformations of those springs, namely 𝛿𝐵𝐸 and 𝛿𝐶𝐹 , respectively, by

𝑇𝐵𝐸 = 𝑘 𝛿𝐵𝐸 Kinematic Equations

𝛿𝐷

Figure 4 Displaced geometry of the beam that may be used to determine the deformations of springs 𝐵𝐸 and 𝐶𝐹 in terms of the displacement 𝛿𝐷 . That is, the triangle containing points 𝐴 and 𝐵 is similar to the triangle containing points 𝐴 and 𝐶, and these are similar to the triangle containing points 𝐴 and 𝐷.

ISTUDY

Governing Equations

∑

𝑦

𝐺

The cables in Fig. 1 are probably considerably more deformable than the beam, and thus it is reasonable to model the beam as being rigid and the cables as springs with stiffness 𝑘, as shown in Fig. 2. We will take 𝑘 = 800 kN∕m for both springs, although more generally, these stiffnesses can be precisely determined given the cable’s cross-sectional area, length, and material. The FBD for the beam is shown in Fig. 3.

and

𝑇𝐶𝐹 = 𝑘 𝛿𝐶𝐹 .

By inspection, you might be able to conclude that

2 1 𝛿 and 𝛿𝐶𝐹 = 𝛿𝐷 . 3 𝐷 3 To see this more formally, we use Fig. 4 with similar triangles. 𝛿𝐵𝐸 =

Computation

(4)

(5)

Combining Eqs. (2) through (5) and solving provides the solutions 𝛿𝐵𝐸 = 0.009 m, 𝐴𝑦 = 2.4 kN,

Discussion & Verification

𝛿𝐶𝐹 = 0.018 m, 𝑇𝐵𝐸 = 7.2 kN,

𝛿𝐷 = 0.027 m, 𝑇𝐶𝐹 = 14.4 kN.

(6) (7)

This analysis assumes that the displacement of the beam is small enough so that the original geometry of the structure may be used when writing the equilibrium equations and for determining the spring deformations in Eq. (5). Notice that the deformations obtained in Eq. (6) are indeed very small compared to the 6 m length of the beam, so the assumption of small deformations appears to be reasonable.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

Problems Problem 5.47 Derive Eq. (5.12) on p. 298.

Problems 5.48 through 5.50 All pulleys are frictionless. Determine the reactions at point 𝐴 and the force supported by the cable.

0.5 f t 𝐷

𝐶

100

𝐵

𝐴

𝐴

𝐺

100 50 N

𝐶

𝐵

𝑟 = 2 in. 𝑟 𝐸 40◦

𝐷

𝐸

100 200 lb

0.5 f t

2 ft

2 ft

Figure P5.48

50 lb

𝑟

𝐴

𝑟 𝐷

𝐵

400 300 300 200 dimensions in mm

2 ft

𝐻

𝐶

10 in.

Figure P5.49

8 in.

12 in.

Figure P5.50

Problems 5.51 through 5.53 The horizontal bar 𝐴𝐵𝐶𝐷 is supported by a link 𝐵𝐸 with the shape shown. All pins are frictionless, and slots (if present) are loose fitting. Determine the reactions that support bar 𝐴𝐵𝐶𝐷.

400 in.⋅lb

𝐴 𝐵

12 in.

3 kN

8 kN⋅m

𝐴 0.7 m 𝐵 𝐶

𝐷

𝐶

4 in. 𝐸 4 in.

30 lb

0.5 m 8 in.

3m

Figure P5.51

𝐸 1m

80 N 𝐶

𝐷

20◦ 𝐵

𝐷 400 N⋅mm

0.4 m 2m

8 mm

𝐸

20 mm 8 mm 20 mm

Figure P5.52

Figure P5.53

Problems 5.54 and 5.55 The crane is supported by a pin at point 𝐴 and a roller at point 𝐵. A winch at point 𝐶 is used to raise and lower loads. The pulleys all have 350 mm radius, and cable segment 𝐸𝐷 is horizontal. Determine the support reactions if 𝑃 = 10 kN and (a) 𝜃 = 0◦ . (b) 𝜃 = 30◦ . 2m

2m

2m

2m

350 mm 𝐸

2m

2m

𝐷

𝜃

15◦

2m

20◦

𝑃

2m 𝐵

𝑃

2m 𝐶

𝐴 Figure P5.54

2m 2m

𝐸

2m

𝐶

2m

350 mm 𝐹

𝐷 𝜃

2m

𝐵 𝐴

Figure P5.55

𝐴

315

316

Chapter 5

Equilibrium of Bodies

12 in. 𝐹 𝐴

24 in.

𝐵

12 in. 5 in. 10 in. 5 in. 13 in. 𝑟

𝐶 𝐺

13 in. 𝐸

Problem 5.56 𝐷

𝑟

𝑊

𝑟 = 4 in.

Problem 5.57

Figure P5.56 5 in.

3 in. 4 in.

8 in.

3 in.

2 in. soda

soda

𝐺 milk

𝐸

𝐴

2 in.

𝐵

20◦

𝐹

Problem 5.58

𝐶 𝐻

30◦ Figure P5.57

ISTUDY

The structure 𝐴𝐵𝐶𝐷𝐸 shown is used to support two bottles of soda and one bottle of milk. Each bottle of soda weighs 2 lb, the bottle of milk weighs 4 lb, and the dimensioning lines shown locate the center of gravity for each bottle. The weight of all other components is negligible. The structure is supported by a roller at 𝐵, a collar at 𝐶 that slides on a fixed frictionless vertical bar, and by a cable 𝐷𝐹 𝐺 that is subjected to a 6 lb force. The portion of the cable attached to point 𝐷 is horizontal, and the roller and pulleys are frictionless. Determine the support reactions.

6 lb

40◦

𝐷 3 in.

Bar 𝐴𝐵𝐶𝐷 is supported by a link 𝐵𝐸 with circular shape and a cable wrapped around two frictionless pulleys that have 4 in. radius. Cable segment 𝐶𝐹 is horizontal, and segments 𝐶𝐺 and 𝐷𝐺 are vertical. If the weight supported by pulley 𝐺 is 𝑊 = 30 lb, determine the reactions at 𝐴 and 𝐵.

A walkway for loading and unloading ships at a wharf is shown. The elevation of the walkway is controlled by cable 𝐵𝐶𝐷, which is attached to a drum on a geared motor at 𝐵. If the 1 kip force is vertical and is positioned halfway between points 𝐴 and 𝐶, determine the forces supported by cables 𝐵𝐶𝐷 and 𝐷𝐸, the reactions at 𝐴, and the force supported by bar 𝐷𝐹 . 8 in. radius

1 kip

5◦ 𝐷

𝐶 10◦ 10◦ 6 ft

ship

6 in. radius

𝐴

𝐵 3 ft

3 4

𝐹

𝐸

wharf

Figure P5.58

Problem 5.59 Repeat Prob. 5.58 if the 1 kip vertical force is positioned at the bearing of pulley 𝐶.

Problems 5.60 through 5.62 In the structure shown, member 𝐴𝐵𝐶𝐷 is supported by a pin at 𝐶 and a cable that wraps around pulley 𝐸, which is frictionless. (a) Specify if member 𝐴𝐵𝐶𝐷 has complete fixity or partial fixity and whether it is statically determinate or statically indeterminate. (b) Draw the FBD for member 𝐴𝐵𝐶𝐷, and determine the cable tension in terms of force 𝐹 and length 𝐿. Comment on any difficulties that might arise in your analysis. 𝐸

𝐸 2

𝐴 𝐵

𝐹 𝐿

1

1 1 𝐶

𝐿

Figure P5.60

𝐷

2

𝐴 𝐵

𝐹 𝐿

𝐿

1

𝐸

1 1 𝐶

𝐿

Figure P5.61

𝐷

1

𝐴 𝐵

𝐹 𝐿

𝐿

1

1

1

𝐶 𝐿

Figure P5.62

𝐿

𝐷

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

317

Problem 5.63 An office chair has a compressed spring that allows the chair to tilt backward when a sufficiently large force 𝐹 is applied. When 𝐹 is small, the stop at point 𝐴 prevents the chair from tilting forward. If the spring has 10 lb∕in. stiffness and 15 in. unstretched length, determine the value of 𝐹 that will cause the chair to begin tilting backward.

2 in.

2 in.

𝐹

120 lb 𝐴

1 in. 2 in.

120 lb

14 in.

𝐶

6 in.

14 in.

𝐴

1 in. 2 in.

8 in.

𝐵

𝐹

𝐵

𝐷

Figure P5.63

Figure P5.64

Problem 5.64 An office chair has a prewound torsional spring that allows the chair to tilt backward when a sufficiently large force 𝐹 is applied. When 𝐹 is small, the stop at point 𝐴 prevents the chair from tilting forward. If the spring has 100 in.⋅lb∕rad stiffness and is prewound by 3/4 of a turn, determine the value of 𝐹 that will cause the chair to begin tilting backward.

Problem 5.65 The brake pedal 𝐴𝐵𝐶𝐷 for an automobile has a torsional spring at pin 𝐶. The spring has 4000 N⋅mm∕rad stiffness and is prewound by 1/2 of a turn so that contact is made at point 𝐸 when 𝑃 = 𝑄 = 0. If 𝑃 = 100 N and the pedal has negligible weight, determine the value of 𝑄 that will cause the pedal to begin to move and the reactions at 𝐶. 80

200

10◦

160 𝐶 𝐴

𝐸

𝐵

90 Figure P5.65

𝐷

10◦

3

3

4 𝑄

80

200 𝐷

𝑃 dimensions in mm

4 𝑄

160 𝐶 𝐴 90

𝐵

𝐸

𝐹

𝑃 dimensions in mm

80

Figure P5.66

Problem 5.66 The brake pedal 𝐴𝐵𝐶𝐷 for an automobile has a horizontal spring 𝐵𝐹 . The spring has 5 N∕mm stiffness and 110 mm unstretched length. If 𝑃 = 100 N and the pedal has negligible weight, determine the value of 𝑄 that will cause the pedal to begin to move and the reactions at 𝐶.

𝐷

Problem 5.67 The geometry of a structure before the 500 N force is applied is shown, where member 𝐴𝐵𝐶 is horizontal and spring 𝐴𝐷 is unstretched. The pulley at 𝐶 is frictionless. Determine the vertical displacement 𝑑 of point 𝐴, the forces supported by cable 𝐵𝐶𝐷 and the spring, and the reaction between the roller 𝐴 and the wall to the right of it, due to the 500 N force. Assume 𝑑 is small enough so that the geometry of the displaced structure is essentially the same as the geometry before loads are applied.

12 𝐶

5

𝐴

0.4 m 2m Figure P5.67

30 kN∕m

500 N 𝐵 0.4 m 𝑑 2m

318

Chapter 5

Equilibrium of Bodies

2 in. radius

10 in. 𝐷

𝐹

𝐶

4

8 in. 𝐸

3 𝐷

𝐵 8 in.

𝑃 =0

𝐵

𝐴 7 in..

10 in.

𝑃

𝐶 8 in.

8 in.

Figure P5.68

Problem 5.68 When force 𝑃 = 0, portion 𝐴𝐵𝐶 of the structure is vertical, portion 𝐶𝐷 is horizontal, and spring 𝐸𝐵 is horizontal and unstretched. When a sufficiently large value of 𝑃 is applied, points 𝐵, 𝐶, and 𝐷 move to locations 𝐵 ′ , 𝐶 ′ , and 𝐷 ′ , respectively. The force 𝑃 is always vertical, the spring has 20 lb∕in. stiffness, the pulley at 𝐷 is frictionless, and weight of members is negligible. Determine the value of 𝑃 so that portion 𝐴𝐵 ′ 𝐶 ′ of the structure is horizontal and portion 𝐶 ′ 𝐷 ′ is vertical. Also determine the force supported by the spring and the reactions at point 𝐴.

Problem 5.69 A truck pulling a trailer with a tractor on it is shown in Fig. P5.69(a). The trailer’s mass is 500 kg and the tractor’s mass is 900 kg, with centers of gravity at points 𝐸 and 𝐺, respectively. A model for the trailer is shown in Fig. P5.69(b) where point 𝐴 is idealized as a pin and the tires are modeled by a vertical spring with stiffness 𝑘 = 600 kN∕m. Assuming the trailer is in static equilibrium, determine the vertical deflection of point 𝐷 due to the weight of the trailer and tractor. Also determine the reactions at 𝐴 and the force supported by the spring. 1.1 m 1 m 0.6 m 𝐺

𝐺 𝐷

𝐴 𝐸

𝐷

𝐴

𝛿𝐷

𝐸

𝐵 2.8 m

0.7 m

𝑘

𝛿𝐷

𝐵 1.8 m

(a)

(b) Figure P5.69

Problem 5.70 A clipboard that is used for holding and writing on paper is shown. The clamp 𝐴𝐵𝐶 has a torsional spring at pin 𝐶 with stiffness 𝑘𝑡 = 2 in.⋅lb∕rad. To release the clamp so that paper may be inserted or removed, the user applies a vertical force 𝑃 . Determine the pretwist of the torsional spring (in terms of the number of turns) so that the clamp provides a clamping force of 4 lb at point 𝐴 when 𝑃 = 0, and determine the force 𝑃 needed to release the clamp. Assume the contact at 𝐴 is frictionless. 𝑃 𝐵

𝐷 𝑄

1.5 in. 25 mm

15◦

30 mm

ISTUDY

𝐶 2.2 in.

𝐶

0.5 in. 0.8 in.

15 mm Figure P5.70

𝐵 𝐸 𝐴

Figure P5.71

𝐴

𝑘

Problem 5.71 A model for a 110 V electrical wall switch is shown where force 𝑄 that operates the switch is perpendicular to line 𝐵𝐶𝐷. If the spring has stiffness 𝑘 = 1.5 N∕mm, determine the unstretched length of the spring 𝐿0 so that the switch begins to move when 𝑄 = 2 N.

ISTUDY

Section 5.3

319

Equilibrium of Bodies in Two Dimensions—Additional Topics

Problem 5.72

𝐹 𝐵

A stiff fiberglass antenna is supported by a coiled spring at point 𝐴 that has torsional stiffness 𝑘𝑡 = 50 N⋅m∕rad. Force 𝐹 at point 𝐵 is always horizontal and it models the wind forces on the antenna. Determine the rotation 𝜃 of the antenna if (a) 𝐹 = 5 N.

800 mm

(b) 𝐹 = 50 N.

𝜃

(c) Discuss why the answer for Part (b) is not 10 times greater than the answer for Part (a). Hint: When appropriate in this problem, you should use the original geometry when writing equilibrium equations. When this simplification is employed, you should discuss its validity.

𝐴

𝑘𝑡

Figure P5.72

Problems 5.73 and 5.74 ∑ Draw the FBD for the structure shown. Then write the four equilibrium equations 𝐹𝑥 = ∑ ∑ ∑ 0, 𝐹𝑦 = 0, 𝑀𝐶 = 0, and 𝑀𝐷 = 0. If possible, solve these equations to determine the support reactions. Discuss the difficulties that arise. 600 lb

12 mm 120 N 40 N 5 mm

𝐵

𝐴

4 mm

𝐴

𝑦

3 ft 𝑥

𝐷

10 mm

200 lb

𝑦

𝑥

𝐶

𝐵

𝐶

𝐷

4 ft

20 mm

Figure P5.73

5 ft

Figure P5.74 1000 lb

Problem 5.75 The I beam shown is statically indeterminate. Under certain circumstances, it may be appropriate to use a model where the I beam is rigid and the roller supports at points 𝐵 and 𝐶 are replaced by vertical springs of equal stiffness so that the support reactions may be determined. Do this and find the reactions at points 𝐴, 𝐵, and 𝐶.

Problem 5.76 The I beam shown in Fig. P5.76(a) is statically indeterminate. Under certain circumstances, it may be appropriate to use the model in Fig. P5.76(b) where the I beam is rigid, the built-in support at point 𝐴 is replaced by a pin and torsional spring with stiffness 𝑘𝑡 , and the roller support at point 𝐶 is replaced by a vertical spring with stiffness 𝑘. Use this model to determine the reactions at points 𝐴 and 𝐶. Express your answers in terms of parameters such as 𝐹 , 𝐿, 𝑘, 𝑘𝑡 , etc. 𝐹 𝐴

𝐹 𝐶

𝐵

𝑘𝑡

𝐴

𝐿 2

𝐿 2 (a)

𝐿 2 (b)

Figure P5.76

𝐶

𝐵 𝐿 2

𝑘

𝐴 2 ft Figure P5.75

𝐶

𝐵 2 ft

320

ISTUDY

Chapter 5

Equilibrium of Bodies Problems 5.77 and 5.78

For each object shown, specify whether it has partial fixity or full fixity and whether it is statically determinate or statically indeterminate. Assume that cables are in tension and rollers are in compression.

ℎ

45◦

ℎ (a)

(b)

(c)

(a) 45◦

(b)

(c)

45◦ ℎ

45◦ 45◦ (d)

(e)

ℎ (d)

(f)

Figure P5.77

(e)

(f)

Figure P5.78

Problem 5.79 Without solving, speculate on the difficulty of each of Probs. 5.80 through 5.82. Note: Concept problems are about explanations, not computations.

Problem 5.80 Repeat Example 5.7 on p. 308, using the actual geometry when writing the equilibrium equations to determine the lift and drag forces 𝐿 and 𝐷. Assume 𝐿 is vertical and 𝐷 is horizontal. In your opinion, are the differences between your answers here and those in Example 5.7 acceptable? Describe some ways the design of the wind tunnel of Example 5.7 could be changed so that these differences are reduced (e.g., change of dimensions, spring stiffnesses, etc.).

Problem 5.81 In Example 5.7 on p. 308, if 𝐿 = 2 N and 𝐷 = 0.3 N, determine 𝛿 and 𝜃, using the original geometry when writing the equilibrium equations.

Problem 5.82 In Example 5.7 on p. 308, if 𝐿 = 2 N and 𝐷 = 0.3 N, determine 𝛿 and 𝜃, using the actual geometry when writing the equilibrium equations. Assume 𝐿 is vertical and 𝐷 is horizontal.

Problem 5.83 Can the solution to Prob. 5.34 on p. 294 be obtained by superposing the solutions to Probs. 5.32 and 5.33? Explain. Note: Concept problems are about explanations, not computations.

ISTUDY

Section 5.3

Equilibrium of Bodies in Two Dimensions—Additional Topics

Problem 5.84

3 ft

𝐹

Consider the structure from Example 5.8 on p. 309, shown again here where 𝑊 is the weight of the structure with center of gravity at point 𝐻. (a) If 𝑊 = 2 kip, determine the support reactions due to the weight of the structure only (i.e., 𝑃 = 0). (b) Use superposition of the results from Part (a) and Example 5.8 to determine the total values of the support reactions when 𝑊 = 2 kip and 𝑃 = 3 kip. (c) Use superposition of the results from Part (a) and Example 5.8 to determine the total values of the support reactions when 𝑊 = 1.8 kip and 𝑃 = 4 kip.

3 ft

For each of the support schemes shown in Fig. 5.17 on p. 302, apply a vertical downward force 𝑃 at location 𝐵, as shown in Fig. 5.18, and specify if the object is a two-force, three-force, or general multiforce member.

Problems 5.86 through 5.98 Identify each of the members cited below as a zero-force, two-force, three-force, or multiforce member. Problem 5.86

Members 𝐴𝐵𝐶 and 𝐵𝐷 in Example 5.1 on p. 282.

Problem 5.87

Member 𝐴𝐵𝐶𝐷 in Example 5.2 on p. 284.

Problem 5.88

Door 𝐴𝐵𝐷𝐸 and strut 𝐵𝐶 in Example 5.3 on p. 285.

Problem 5.89

Plate 𝐴𝐵𝐶𝐷𝐸 in Example 5.4 on p. 286.

Problem 5.90

Drum and contents in Example 5.5 on p. 287.

Problem 5.91

Drum and contents in Example 5.6 on p. 288.

Problem 5.92

Member 𝐴𝐵𝐶 in Prob. 5.3 on p. 289.

Problem 5.93

Step 𝐴𝐵 in Prob. 5.4 on p. 289.

Problem 5.94

Members 𝐴𝐵𝐶𝐷 and 𝐷𝐸 in Prob. 5.5 on p. 289.

Problem 5.95

Tray 𝐴𝐵 and link 𝐵𝐶 in Prob. 5.8 on p. 290.

Problem 5.96

Member 𝐴𝐵𝐶 in Prob. 5.22 on p. 292.

Problem 5.97

Member 𝐴𝐵𝐶𝐷𝐸 in Prob. 5.25 on p. 293.

Problem 5.98

Wrench 𝐴𝐵𝐶 in Prob. 5.37 on p. 295.

3ft

1 ft 𝐻 𝑊

3 ft

2 ft

3 ft 𝐴

3 ft

𝐺

𝐷

𝐸

𝐵

Figure P5.84

Problem 5.85

3ft

321

𝐶

𝑃

322

ISTUDY

Chapter 5

Equilibrium of Bodies

Helpful Information

5.4

Moment directions in the scalar approach. In two-dimensional problems all moment vectors are parallel to one another because they are all perpendicular to the plane of the problem. As a consequence, the moment directions may be fully described using words like clockwise and counterclockwise. In three-dimensional problems, moment vectors generally have components in all three coordinate directions, and if a scalar approach is used, you must decide on a sign convention for each of these directions. Thus, words like clockwise and counterclockwise have little meaning, and a more robust scheme must be used, as illustrated in the examples of this section.

Helpful Information Self-aligning bearings. Moment reactions in bearings are sometimes undesirable, and there are some ways these may be eliminated. A special type of bearing, called a self-aligning bearing, is common, and an example is shown below in Fig. 5.22. 𝑧

𝑦 𝑥

The equations governing static equilibrium of a body in three dimensions were stated in Section 5.1 and are repeated here: ∑

𝐹⃗ = 0⃗

and

∑

⃗ ⃗ = 0, 𝑀 𝑃

(5.16)

where subscript 𝑃 denotes the moment summation point you select. In scalar form, Eq. (5.16) is ∑ ∑ 𝐹𝑥 = 0 𝑀𝑃 𝑥 = 0 ∑ ∑ 𝐹𝑦 = 0 and 𝑀𝑃 𝑦 = 0 (5.17) ∑ ∑ 𝐹𝑧 = 0 𝑀𝑃 𝑧 = 0. As illustrated in the examples of this section, we can use different points to write each of the moment summation expressions in Eq. (5.17). Fundamentally, the analysis of rigid body equilibrium in three dimensions is the same as analysis in two dimensions. The major differences in three dimensions are that FBDs are usually more intricate, reactions are more complex, and the number of unknowns to be determined and the number of equations to be solved are greater. Comments made in Section 5.2 on alternative equilibrium equations also apply here (Prob. 5.123 on p. 343 guides you to explore further details on alternative equilibrium equations).

Reactions Common supports and their associated reactions are shown in Fig. 5.23. In all cases, the reaction forces and moments for a particular support may be determined by considering the motion the support prevents. For example, in the case of a bar supported by a pin, the pin prevents motion of the bar in the 𝑥, 𝑦, and 𝑧 directions, and thus the pin must produce reaction forces in each of these directions. The pin also prevents rotation of the bar about the 𝑦 and 𝑧 axes, and hence there must also be moment reactions about these axes. The pin does not prevent rotation about the 𝑥 axis; therefore, there is no moment reaction about this axis. It is not necessary to memorize the reactions shown in Fig. 5.23. Rather, you should reconstruct these as needed.

More on bearings

Michael Plesha

𝑧

𝑥

Equilibrium of Bodies in Three Dimensions

𝑧

𝑦

𝑥

𝑦

Figure 5.22 A self-aligning bearing allows the shaft to undergo small rotations about the 𝑦 and 𝑧 axes so that reactions 𝑀𝑦 and 𝑀𝑧 are zero.

As shown in Fig. 5.23, a bearing nominally has two moment reactions. Moment reactions in bearings are sometimes undesirable, especially when two or more bearings are used to support a rotating shaft. The two most common ways to eliminate moment reactions in bearings are by the use of a special type of bearing called a self-aligning bearing or by the use of two or more perfectly aligned bearings. Self-aligning bearings are described in the Helpful Information margin note on this page. Perfectly aligned bearings are conventional bearings that are required to be perfectly aligned so that it is justified to assume the moment reactions are zero.

ISTUDY

Section 5.4

Equilibrium of Bodies in Three Dimensions

Support

Reactions

𝑧 𝑦 𝑥 frictionless surface

𝑅𝑧

𝑧

𝑧

socket

𝑥 rough surface

𝑧

𝑧

𝑥

𝑥

𝑅𝑧

𝑅𝑥 𝑦

𝑦

𝑅𝑦

𝑅𝑥

𝑦

𝑦

𝑅𝑦

𝑀𝑦

𝑅𝑦 𝑅𝑥

𝑅𝑧

𝑅𝑧

𝑥 hinge

pin

𝑀𝑦

𝑀𝑧

𝑀𝑧 𝑅𝑦

𝑧

𝑀𝑦

𝑅𝑧

𝑦

𝑀𝑧

𝑥 bearing

If also a thrust bearing, add 𝑅𝑥 . If self-aligning, 𝑀𝑦 = 𝑀𝑧 = 0.

𝑧

𝑅𝑥 𝑦 𝑥 built-in or clamped

𝑀𝑥

𝑅𝑦

𝑀𝑦

𝑅𝑧 𝑀𝑧

Figure 5.23. Common supports for bodies in three dimensions and the associated reactions.

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Equilibrium of Bodies

For example, consider a shaft supported by two bearings as shown in Fig. 5.24(a). Normally, bearings will have moment reactions, and the FBD for this is shown in Fig. 5.24(b). This FBD has eight unknown reactions and is statically indeterminate because the six available equilibrium equations are not sufficient to determine all of the reactions. If the bearings at 𝐴 and 𝐶 are self-aligning or are perfectly aligned (subject to the warning in the margin note), then the moment reactions are zero and the FBD is shown in Fig. 5.24(c), which is statically determinate. 𝑧

Common Pitfall Warning: perfectly aligned bearings. You must be very cautious when assuming the moment reactions for a bearing are zero under the premise that the bearings are perfectly aligned. Despite your assumption, there are many reasons for which the moment reactions may not be zero, including poor construction (i.e., the bearings are not as well aligned as you anticipated), misalignment that develops due to wear or vibration, or large deflection of the shaft or object being supported that will cause moment reactions to develop, even if the bearings are perfectly aligned.

𝐹

𝑦 𝑥

𝐶

𝐹

𝐹

𝐴𝑦 𝐴𝑧

𝑀𝐶𝑧 𝑀𝐴𝑦

𝐶𝑦 𝐶𝑧

𝐶𝑧

𝐵 𝐴

𝐶𝑦 𝑀𝐶𝑦

𝐴𝑦 𝐴𝑧

𝑀𝐴𝑧 (a)

(b) FBD if bearings have moment reactions

(c) FBD if bearings are self-aligning or are perfectly aligned

Figure 5.24. (a) A shaft supported by two bearings. (b) The FBD if the bearings have moment reactions. (c) The FBD if the bearings are self-aligning or are perfectly aligned—observe that the moment reactions are zero.

Bearings are occasionally designed to prevent axial motion of the object they support, and such bearings are called thrust bearings. For example, if the bearing shown in Fig. 5.23 is a thrust bearing, the bearing will not allow translation of the shaft in the 𝑥 direction, and thus the bearing will have a reaction force 𝑅𝑥 in addition to those shown in Fig. 5.23. The remarks made here for perfectly aligned bearings also apply to objects that are supported by two or more hinges. In particular, it is possible to assume that the moment reactions for the hinge shown in Fig. 5.23 are zero if two or more hinges are perfectly aligned.

Scalar approach or vector approach? The scalar approach requires good visualization ability, skill is needed to correctly identify moment arms, and it is necessary to be consistent with positive and negative directions for moments. Problems with simple geometry can often be effectively solved using a scalar approach. The vector approach can be used for both simple and complex problems and does not require careful visualization, and positive and negative moment directions are automatically accounted for. While the selection of an analysis approach for a particular problem is your choice, you should still be comfortable with both approaches. Most of the example problems of this section use both approaches, and you should contrast the merits of these approaches to help you learn which is more effective for a particular problem.

Solution of algebraic equations If a scalar approach is used, a thoughtful strategy on the order in which equilibrium equations are written, and the selection of moment summation points and axes, may

ISTUDY

Section 5.4

Equilibrium of Bodies in Three Dimensions

provide equilibrium equations whose unknowns are uncoupled or are weakly coupled. Uncoupled or weakly coupled equations can be solved with a minimum of algebra. If a vector approach is used and an effective moment summation point can be identified, the resulting equilibrium equations may also be uncoupled or weakly coupled. For both scalar and vector approaches, more complex problems will usually result in equilibrium equations that are highly coupled and are tedious to solve by hand. For such situations, you are encouraged to use computer software or a programmable calculator to solve the equilibrium equations.

Examples of correct FBDs Figure 5.25 shows several examples of properly constructed FBDs. Comments on the construction of these FBDs follow. Cable-supported cantilever. After sketching the structure, we apply the 200 lb force at 𝐷, and the 50 lb, 30 lb, and 20 lb weights of portions 𝑂𝐵, 𝐵𝐶, and 𝐶𝐷 of the structure, respectively; because each of these portions is uniform, these weights are applied at the center of each portion. Because the cable passes over a frictionless ring at 𝐴, and because we are neglecting the weight of the cable, the force 𝑇 supported throughout the cable is the same and is taken to be positive in tension. The support at 𝑂 prevents the structure from translating in the 𝑥, 𝑦, and 𝑧 directions, and hence there are reaction forces 𝑂𝑥 , 𝑂𝑦 , and 𝑂𝑧 . Similarly, the support at 𝑂 prevents the structure from rotating about the 𝑥, 𝑦, and 𝑧 axes, and hence there are reaction moments 𝑀𝑂𝑥 , 𝑀𝑂𝑦 , and 𝑀𝑂𝑧 . Observe that this FBD is statically indeterminate: there are seven unknown forces and moments and only six equilibrium equations. Storage chest. After sketching the lid, we apply its 15 lb weight. This weight is vertical, and because the lid is uniform, it acts through the center of the lid, point 𝐸. The cord force is 𝑇𝐶𝐷 , taken to be positive in tension. Each hinge prevents translation in all three coordinate directions and prevents rotation about the 𝑦 and 𝑧 axes; hence there are three force reactions and two moment reactions at each hinge. Rather than use reactions 𝐴𝑦 , 𝐴𝑧 , 𝑀𝐴𝑦 , and 𝑀𝐴𝑧 , we could have used reactions 𝐴𝑛 , 𝐴𝑡 , 𝑀𝐴𝑛 , and 𝑀𝐴𝑡 , where 𝑛 and 𝑡 are normal and tangent directions, respectively, to the lid (similar remarks apply to the reactions at 𝐵). The FBD as shown in Fig. 5.25 is statically indeterminate: there are 11 unknown forces and moments and only six equilibrium equations. If the hinges are perfectly aligned, we may then assume the four moment reactions are zero. If we further assume that 𝐴𝑥 or 𝐵𝑥 is zero (this assumption is not warranted unless further information is given in the problem statement), then the problem becomes statically determinate. Aircraft landing gear. After sketching the landing gear, we apply the 200 N and 300 N weights. Member 𝐴𝐵 is a two-force member, so the force it applies to the landing gear is directed along line 𝐴𝐵, and we have selected 𝐹𝐴𝐵 to be positive in tension. The bearing at 𝑂 prevents translation in all directions, which gives rise to reaction forces 𝑂𝑥 , 𝑂𝑎 , and 𝑂𝑏 , where 𝑎 is the direction about which the landing gear pivots and 𝑏 is perpendicular to the 𝑥 and 𝑎 directions. The bearing also prevents rotation about the 𝑥 and 𝑏 directions, so there are reaction moments in these directions. This problem has six reactions and six equilibrium equations, so we expect it to be statically determinate.

325

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Equilibrium of Bodies

Examples of correct FBDs Cable-supported cantilever. The structure is built-in at 𝑂 and is also supported by a single cable that passes through a frictionless ring at 𝐴. Portions 𝑂𝐵, 𝐵𝐶 , and 𝐶𝐷 of the structure are each uniform and 𝑦 weigh 50 lb, 30 lb, and 20 lb, respectively, and the weight 𝐴 supported at 𝐷 is 200 lb. Draw the FBD for member 𝑂𝐵𝐶𝐷. 𝐵 𝑧 𝐶 𝑂 𝐷

𝑦 𝑀𝑂𝑦

𝑂𝑦 𝑂𝑧 𝑀𝑂𝑧

𝑇 𝑧 𝑇

20 lb

𝑂𝑥 50 lb

𝑥

𝑀𝑂𝑥

𝑥

200 lb

30 lb

Storage chest. The lid of a storage chest is uniform with 15 lb weight. It is supported by hinges at points 𝐴 and 𝐵, and a cord between points 𝐶 and 𝐷. Draw the FBD 𝐶 for the lid.

𝑦

15 lb

𝐴

𝐷

𝐴

𝑀𝐴𝑦

𝑇𝐶𝐷

𝐴𝑦

𝐵 𝑧 𝑀𝐴𝑧

𝐴𝑧

𝐸

𝑀𝐵𝑦

𝐴𝑥 𝑀𝐵𝑧

𝐵𝑦 𝐵𝑧

𝐵𝑥 𝑥

Aircraft landing gear. The landing gear of a jet pivots on a bearing at 𝑂 and is actuated by two-force member 𝐴𝐵. Member 𝑂𝐶𝐴 has 300 N weight whose line 𝑦 of action passes through 𝑎 point 𝐶, and the tire and 40◦ wheel weigh 200 N. Draw the FBD of 𝐵 𝑂 member 𝑂𝐶𝐴 𝑥 and the wheel. 𝐶 𝑧

𝐴

𝑏

𝑦 𝑎

𝑀𝑂𝑥

40◦

𝑂𝑎

𝑂𝑥

𝑀𝑂𝑏 𝑂𝑏 𝑥 𝐹𝐴𝐵

𝑧 300 N

200 N

Figure 5.25. Examples of properly constructed FBDs.

ISTUDY

Section 5.4

Equilibrium of Bodies in Three Dimensions

Examples of incorrect and/or incomplete FBDs Figure 5.26 shows several examples of incorrect and/or incomplete FBDs. Comments on how these FBDs must be revised follow, but before reading these, you should study Fig. 5.26 to find as many of the needed corrections and/or additions on your own as possible. Beam 1. The forces shown at points 𝐵 and 𝐶 have the proper directions and positions where 𝐶 is midway between 𝐴 and 𝐵, but their magnitudes should be (50 kg)(9.81 m∕s2 ) = 490.5 N at 𝐵 and (25 kg)(9.81 m∕s2 ) = 245.3 N at 𝐶. 2. The beam is free to translate in the 𝑥 and 𝑦 directions, so 𝐴𝑥 and 𝐴𝑦 are not reactions and should not be shown. 3. The beam is free to rotate about the 𝑧 axis, so 𝑀𝐴𝑧 is not a reaction and should not be shown. 4. The support prevents the beam from rotating about the 𝑥 and 𝑦 axes, so moment reactions 𝑀𝐴𝑥 and 𝑀𝐴𝑦 must be included. In summary, the reactions at the support consist of 𝐴𝑧 , 𝑀𝐴𝑥 , and 𝑀𝐴𝑦 only. The beam has partial fixity since the number of reactions is less than the number of equilibrium equations. Shifting fork 1. The 5 N force applied by the gear at 𝐺 to the fork is properly shown, but there is also a 𝑧 direction force 𝐺𝑧 at that location. If the fit between the fork and the shaft at 𝐺 is not loose, then additional reaction forces and moments are possible. 2. Because of the thrust collar at 𝐻, there is a reaction 𝐹𝑥 . 3. The shifting fork may freely rotate on shaft 𝐴𝐵; therefore 𝑀𝐹𝑥 is not a reaction and should not be shown. 4. Shaft 𝐴𝐵 prevents the shifting fork from rotating about the 𝑦 axis; therefore a moment reaction 𝑀𝐹𝑦 must be included. In summary, the reactions on the shifting fork consist of 𝐹𝑥 , 𝐹𝑦 , 𝐹𝑧 , 𝑀𝐹𝑦 , 𝑀𝐹 𝑧 , and 𝐺𝑧 only. This problem has six reactions and six equilibrium equations, so we expect it to be statically determinate. This problem is revisited again in Example 8.3 on p. 499, and the correct FBD is shown there in Fig. 2. Crop planting positioner 1. The support at 𝐸 allows the boom to rotate about the 𝑎 direction; therefore 𝑀𝐵𝑎 is not a reaction and should not be shown. 2. The support at 𝐸 prevents the boom from rotating about the 𝑏 direction; therefore a moment reaction 𝑀𝐵𝑏 must be included. In summary, the reactions at 𝐵 consist of 𝐵𝑥 , 𝐵𝑎 , 𝐵𝑏 , 𝑀𝐵𝑥 , and 𝑀𝐵𝑏 only. This problem has five reactions and six equilibrium equations, so it has partial fixity. Although the problem description does not provide these details, most likely the device has a stop at 𝐸 so that if 𝐹 = 0, rotation about the 𝑎 direction is prevented.

327

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Equilibrium of Bodies

Examples of incorrect and/or incomplete FBDs Beam. A uniform beam with 25 kg mass is supported by a slotted block at 𝐴 that is built-in. The slot is frictionless and allows the beam to slide freely. End 𝐵 of the beam supports a 50 kg mass. Draw the FBD for the beam. 𝐴

𝑧 𝑀𝐴𝑧 𝐴𝑧

𝐴𝑥

𝐶 𝐵

𝐴𝑦

𝐵

𝑦 25 kg

𝑥 50 kg

Shifting fork. Member 𝐸𝐹 𝐺 is a shifting fork used in a transmission to move the gear at 𝐺 along shaft 𝐶𝐷 . It is actuated by the 10 N and 20 N forces at 𝐸, and the gear at 𝐺 applies a 5 N force to the fork in the direction from 10 N 20 N 𝐷 to 𝐶. The fork is 𝐸 supported by a 𝐵 fixed shaft 𝐴𝐵 that has a thrust collar at 𝐻. Draw 𝐷 𝐻 𝐹 the FBD for the shifting fork, assuming the 𝐺 fork is a loose fit 𝐴 on the shaft at 𝐺. 𝐶

Crop planting positioner. The device shown is a pointer that mounts on the front of a tractor to help its operator position the tractor relative to other rows of seeds that have been planted. Bracket 𝐸 is bolted to the front of the tractor, which drives in the 𝑧 direction. The bracket supports the bent boom 𝐴𝐵𝐶, the end 𝐶 of which has a weight 𝑊 and a pointer 𝐶𝐷. The boom is allowed to rotate about line 𝑎, which lies in the 𝑦𝑧 plane, so that if the boom strikes an obstruction, the boom will rotate backward and upward to help avoid damage to it. Draw the FBD of the boom and pointer, lumping 𝑦 all of the reaction forces at 𝐴 and 𝐵 𝑎 into a single system 𝑊 ◦ 15 of forces and/or 𝐶 𝐵 𝑥 moments at point 𝐵. 𝑧 𝐸 𝐴 𝐷 𝐹

𝑧

10 N

20 N

𝐸 𝑀𝐹 𝑥 𝑦

𝐹𝑦

5N 𝑥

𝐹𝑧 𝑀𝐹 𝑧

𝑎 𝑀𝐵𝑎

𝑦

𝐵𝑎 𝐵𝑥

𝑀𝐵𝑥 𝑧 𝑏

𝑊

15◦ 𝐶 𝐵𝑏

𝑥 𝐷

𝐹

Figure 5.26. Examples of incorrect and/or incomplete FBDs.

ISTUDY

Section 5.4

Equilibrium of Bodies in Three Dimensions

End of Section Summary In this section, static equilibrium of a body in three dimensions was discussed. Some of the key points are as follows: • Problems in three dimensions with simple geometry can often be effectively solved using a scalar approach. But very often a vector approach will be more tractable. • If a scalar approach is used, a sign convention for positive moments about the three Cartesian coordinate directions must be adopted (the right-hand rule is customary), and great care must be exercised when summing moments. • In general, a bearing has moment reactions, and often these play an important role in the equilibrium of a structure. In structures with multiple bearing supports, moment reactions at the bearings may be undesirable and may be eliminated by using special bearings called self-aligning bearings or by using conventional bearings, but requiring their axes to be perfectly aligned. If you design or analyze a structure and assume reaction moments at bearings are zero by virtue of perfect alignment, you must be very careful to ensure this assumption is valid. • In general, equilibrium of a single rigid body requires the solution of a system of six algebraic equations, which can be tedious to solve by hand. Software for solving such systems of equations can be very helpful.

329

330

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Equilibrium of Bodies

E X A M P L E 5.13

Socket and Cable Supports

𝑧 1.1 m

Boom 𝑂𝐴𝐵 is supported by a socket at point 𝑂 and two cables. Determine the support reactions at 𝑂 and the forces supported by the two cables. 𝐷

SOLUTION

1.6 m

Road Map 𝐶

Vector solution 𝐴

0.8 m 1.6 m

𝑥

𝐵 𝑦

2m ̂ kN 𝐹⃗ = (5 𝚤̂ − 12 𝑘) Figure 1

𝑧 1.1 m

𝐷

1.6 m

𝐶 0.8 m 𝑥

𝑂𝑦

𝑂𝑧

Modeling The FBD is shown in Fig. 2 where 𝑇𝐴𝐶 and 𝑇𝐴𝐷 are defined to be positive in tension. The reactions at the socket support may be determined by consulting Fig. 5.23, but it is easier to simply construct these by considering the motion the support prevents. That is, the support at 𝑂 prevents translation of the boom in each of the 𝑥, 𝑦, and 𝑧 directions, and hence, there must be reaction forces 𝑂𝑥 , 𝑂𝑦 , and 𝑂𝑧 in these directions. The socket allows rotation of the boom about all three coordinate axes, and thus, there are no moment reactions. ∑ Governing Equations & Computation The equilibrium equations to be used are 𝐹⃗ = ∑ ⃗ ⃗ 0⃗ and 𝑀 𝑃 = 0. Careful consideration of which of these equations to write first and where to locate point 𝑃 may help reduce the algebra required to solve for the unknowns. ∑ ⃗ ⃗ In this problem, 𝑀 𝑂 = 0 is an effective choice: unknowns 𝑂𝑥 , 𝑂𝑦 , and 𝑂𝑧 pass through the moment summation point and hence, do not enter the expression, leaving 𝑇𝐴𝐶 and 𝑇𝐴𝐷 as the only unknowns.∗ With the following vector expressions,

𝑇𝐴𝐷 𝑂𝑥 1.6 m

1.6 𝚤̂ − 1.6 𝚥̂ + 0.8 𝑘̂ , 𝑇⃗𝐴𝐶 = 𝑇𝐴𝐶 2.4 −1.1 𝚤̂ − 1.6 𝚥̂ + 0.8 𝑘̂ , 𝑇⃗𝐴𝐷 = 𝑇𝐴𝐷 2.1

𝑟⃗𝑂𝐴 = 1.6 𝚥̂ m,

𝐴

𝑇𝐴𝐶

𝑟⃗𝑂𝐵 = 3.6 𝚥̂ m, 𝐵 2m

(1) (2)

we obtain 𝑦

⃗ 𝑟⃗𝑂𝐴 × 𝑇⃗𝐴𝐶 + 𝑟⃗𝑂𝐴 × 𝑇⃗𝐴𝐷 + 𝑟⃗𝑂𝐵 × 𝐹⃗ = 0, [ ]𝑇 [ ]𝑇 𝐴𝐶 𝐴𝐷 + (1.28 m) 𝚤̂ + (1.76 m) 𝑘̂ (1.28 m) 𝚤̂ + (−2.56 m) 𝑘̂ 2.4 2.1 [ ] ⃗ + (−43.2) 𝚤̂ + (−18) 𝑘̂ kN⋅m = 0. ∑

̂ kN 𝐹⃗ = (5 𝚤̂ − 12 𝑘) Figure 2 Free body diagram for the vector approach.

ISTUDY

Both vector and scalar solutions are effective, and both are used.

𝑂

Interesting Fact Partial fixity or full fixity? You might have noticed there are only five unknowns in this example, while there are six equilibrium equations. Generalizing the equation counting idea described in Section 5.3 to three dimensions, we see that the boom has only partial fixity. Examining the FBD in Fig. 2 reveals that none of the forces prevent rotation of the boom about the 𝑦 axis. However, this partial fixity may be an artifact of our modeling idealization, and the real structure may have some minor support details that render it fully fixed.

⃗ = 0⃗ ∶ 𝑀 𝑂

(3)

(4)

Grouping all terms that multiply 𝚤̂ in Eq. (4) provides the first of the following equations, and grouping all terms that multiply 𝑘̂ provides the second: ( ) ( ) 1.28 m 1.28 m 𝑇𝐴𝐶 + 𝑇𝐴𝐷 − 43.2 kN⋅m = 0, (5) 2.4 2.1 ) ( ) ( 1.76 m −2.56 m 𝑇𝐴𝐶 + 𝑇𝐴𝐷 − 18.0 kN⋅m = 0. (6) 2.4 2.1 Solving Eqs. (5) and (6) provides the solutions ⇒

𝑇𝐴𝐶 = 23 kN

and 𝑇𝐴𝐷 = 50.75 kN.

The reactions at point 𝑂 can now be determined by writing ∑ ⃗ 𝐹⃗ = 0⃗ ∶ 𝑂𝑥 𝚤̂ + 𝑂𝑦 𝚥̂ + 𝑂𝑧 𝑘̂ + 𝑇⃗𝐴𝐶 + 𝑇⃗𝐴𝐷 + 𝐹⃗ = 0,

(7)

(8)

which is easily solved for ⇒

𝑂𝑥 = 6.25 kN,

𝑂𝑦 = 54 kN,

and 𝑂𝑧 = −15 kN.

(9)

⃗ 𝐴 = 0 is another effective choice: unknowns 𝑇𝐴𝐶 , 𝑇𝐴𝐷 , and 𝑂𝑦 pass through the moment summation point, leaving 𝑂𝑥 and 𝑂𝑧 as the only unknowns.

∗∑𝑀 ⃗

ISTUDY

Section 5.4

331

Equilibrium of Bodies in Three Dimensions

Remark.

A clever solution for this problem is obtained by examining the FBD in Fig. 2 and recognizing that the sum of moments about line 𝑂𝐷, which of course must be zero for equilibrium, involves only one unknown,∗ namely, 𝑇𝐴𝐶 . To carry out this solution, we first evaluate the moment about some point on line 𝑂𝐷, such as point 𝑂, in which case Eq. (3) is written, and then we take the dot product of this with a unit vector in the 𝑂𝐷 direction to write ∑ ( ) 𝑀𝑂𝐷 = 0 ∶ 𝑟⃗𝑂𝐴 × 𝑇⃗𝐴𝐶 + 𝑟⃗𝑂𝐴 × 𝑇⃗𝐴𝐷 + 𝑟⃗𝑂𝐵 × 𝐹⃗ ⋅ 𝑢̂ 𝑂𝐷 = 0, (10) ( ) 𝑟⃗𝑂𝐴 × 𝑇⃗𝐴𝐶 + 𝑟⃗𝑂𝐵 × 𝐹⃗ ⋅ 𝑢̂ 𝑂𝐷 = 0, (11)

𝑇𝐴𝐷 𝑇𝐴𝐷

1.6 2.1

1.6 2.4

𝑇𝐴𝐶

𝑇𝐴𝐶 𝑇𝐴𝐶

𝑂𝑥

Modeling

The FBD is shown in Fig. 3, where for convenience the cable forces have been resolved into 𝑥, 𝑦, and 𝑧 components.

0.8 2.4

1.6 2.4

𝑂𝑧

𝑂𝑦

Scalar solution

1.1 2.1

𝑇𝐴𝐷

𝑧

where the latter expression can be written because we know 𝑇⃗𝐴𝐷 produces no moment about line 𝑂𝐷 [if you are uncertain about this statement, you should evaluate (⃗𝑟𝑂𝐴 × 𝑇⃗𝐴𝐷 ) ⋅ 𝑢̂ 𝑂𝐷 to verify that it is zero]. Problem 5.107 explores this solution in greater detail.

0.8 2.1

𝐴

1.6 m

𝑥

𝐵 2m

5 kN

𝑦

Governing Equations & Computation

In this solution we will elect to sum moments about axes passing through point 𝐴. With a scalar solution it is necessary to be absolutely consistent with a sign convention for positive and negative moments, and we will generally take moments to be positive in the positive coordinate directions according to the righthand rule. Thus, we write† ∑ 𝑀𝐴𝑥 = 0 ∶ −𝑂𝑧 (1.6 m) − (12 kN)(2 m) = 0 ⇒ 𝑂𝑧 = −15 kN, (12) ∑ (13) 𝑀𝐴𝑧 = 0 ∶ 𝑂𝑥 (1.6 m) − (5 kN)(2 m) = 0 ⇒ 𝑂𝑥 = 6.25 kN, ( ) ) ( ∑ 1.6 1.1 𝐹𝑥 = 0 ∶ 𝑂𝑥 + 𝑇𝐴𝐶 − 𝑇𝐴𝐷 + 5 kN = 0, (14) 2.4 2.1 ( ) ) ( ∑ 0.8 0.8 + 𝑇𝐴𝐷 − 12 kN = 0. (15) 𝐹𝑧 = 0 ∶ 𝑂𝑧 + 𝑇𝐴𝐶 2.4 2.1 The first two equations were immediately solved for 𝑂𝑧 and 𝑂𝑥 , and the last two equations can be solved simultaneously to obtain ⇒

𝑇𝐴𝐶 = 23 kN

and 𝑇𝐴𝐷 = 50.75 kN.

(16)

12 kN Figure 3 Free body diagram for the scalar approach.

Helpful Information More on +∕− moment directions. If the convention for positive and negative moment directions in the scalar approach is not clear, then let’s examine Eq. (13) more closely. Consider a 𝑧 ′ direction that is parallel to the 𝑧 axis and that passes through point 𝐴. 𝑧

Finally, the last unknown is obtained from ( ( ) ) ∑ 1.6 1.6 𝐹𝑦 = 0 ∶ 𝑂𝑦 − 𝑇𝐴𝐶 − 𝑇𝐴𝐷 =0 2.4 2.1

⇒

𝑂𝑦 = 54 kN.

(17)

𝑧

𝑂

As expected, all of these solutions agree with those obtained using a vector approach. 𝑥

𝑂𝑥

𝐴 𝐵

Discussion & Verification

We expect both cables to be in tension, and indeed our solution shows 𝑇𝐴𝐶 > 0 and 𝑇𝐴𝐷 > 0. After these simple checks, we should verify that our solutions satisfy all of the equilibrium equations.

∗ Similarly, † In

words,

the sum of moments about line 𝑂𝐶 will involve 𝑇𝐴𝐷 as the only unknown. ∑ 𝑀𝐴𝑥 means “sum of moments about the 𝑥 axis passing through point 𝐴.”

5 kN

𝑦

If you point the thumb of your right hand in the 𝑧 ′ direction, then your fingers define the direction for positive moment about the 𝑧 axis through point 𝐴. When writing Eq. (13), observe that the moment produced by 𝑂𝑥 about the 𝑧 ′ axis is in the same direction as your fingers curl, hence it is positive, while the moment produced by the 5 kN force is in the opposite direction, hence it is negative.

332

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.14

Pin and Cable Supports

𝑧

Boom 𝑂𝐴𝐵 is supported by a pin at point 𝑂 and a cable. Determine the support reactions at 𝑂 and the force supported by the cable.

1.1 m 𝐷

SOLUTION

0.8 m

Road Map

This problem is similar to Example 5.13, with the difference being in the details of how the boom is supported. Both vector and scalar solutions are effective, and both are illustrated.

𝑂 𝐴 𝑥

Vector solution

1.6 m 𝐵 𝑦

2m ̂ kN 𝐹⃗ = (5 𝚤̂ − 12 𝑘)

Governing Equations & Computation

An effective choice for the first equilibrium ∑ ⃗ ⃗ because reaction forces 𝑂 , 𝑂 , and 𝑂 do not enter this = 0 equation to write is 𝑀 𝑂 𝑥 𝑦 𝑧 equation. However, reaction moments 𝑀𝑂𝑦 and 𝑀𝑂𝑧 do enter this equation, as does 𝑇𝐴𝐷 . ∑ ⃗ Nonetheless, 𝑀 = 0⃗ provides a system of three scalar equations that can be solved to

Figure 1 𝑧 1.1 m 𝑀𝑂𝑧 𝑀𝑂𝑦 𝑂𝑦

𝑂

𝐷 0.8 m

find the three unknowns it contains. With the following vector expressions 𝑟⃗𝑂𝐴 = 1.6 𝚥̂ m,

𝑂𝑧

𝑂𝑥

𝐴

∑

1.6 m 𝐵 2m

𝑦

̂ kN 𝐹⃗ = (5 𝚤̂ − 12 𝑘) Figure 2 Free body diagram for the vector solution.

ISTUDY

𝑟⃗𝑂𝐵 = 3.6 𝚥̂ m,

−1.1 𝚤̂ − 1.6 𝚥̂ + 0.8 𝑘̂ 𝑇⃗𝐴𝐷 = 𝑇𝐴𝐷 , 2.1

(1)

we obtain

𝑇𝐴𝐷

𝑥

The FBD is shown in Fig. 2. The support at 𝑂 prevents translation of the boom in each of the 𝑥, 𝑦, and 𝑧 directions, hence there must be reaction forces 𝑂𝑥 , 𝑂𝑦 , and 𝑂𝑧 in these directions. The pin prevents rotation of the boom about the 𝑦 and 𝑧 axes, hence there must be reaction moments 𝑀𝑂𝑦 and 𝑀𝑂𝑧 about these axes. The cable force 𝑇𝐴𝐷 is taken to be positive in tension.

Modeling

Helpful Information Directions for reactions. In most problems the actual directions of reactions are unknown until after the equilibrium equations are solved. Thus, when drawing an FBD, we will routinely take reactions (both forces and moments) to be positive in the positive coordinate directions, such as shown at point 𝑂 in Fig. 2. Doing so makes it easy to write a vector expression for the reaction, which in ̂ Other this example is 𝑂⃗ = 𝑂𝑥 𝚤̂ + 𝑂𝑦 𝚥̂ + 𝑂𝑧 𝑘. conventions can be used, provided you are careful when writing vector expressions. For instance, if 𝑂𝑥 is taken to be positive in the opposite direction from that shown in Fig. 2, ̂ then you write 𝑂⃗ = −𝑂𝑥 𝚤̂ + 𝑂𝑦 𝚥̂ + 𝑂𝑧 𝑘.

⃗ = 0⃗ ∶ 𝑀 𝑂

𝑀𝑂𝑦 𝚥̂ + 𝑀𝑂𝑧 𝑘̂ + 𝑟⃗𝑂𝐴 × 𝑇⃗𝐴𝐷 + 𝑟⃗𝑂𝐵 × 𝐹⃗ = 0⃗ [ ]𝑇 𝐴𝐷 𝑀𝑂𝑦 𝚥̂ + 𝑀𝑂𝑧 𝑘̂ + (1.28 m) 𝚤̂ + (1.76 m) 𝑘̂ 2.1 [ ] ⃗ + (−43.2) 𝚤̂ + (−18) 𝑘̂ kN⋅m = 0.

(2)

(3)

Grouping all terms that multiply 𝚤̂ in Eq. (3) provides the first of the following equations, grouping terms that multiply 𝚥̂ provides the second, and grouping terms that multiply 𝑘̂ provides the third: ( ) 1.28 m (4) 𝑇𝐴𝐷 − 43.2 kN⋅m = 0, 2.1 (5) 𝑀𝑂𝑦 = 0, ) ( 1.76 m 𝑇𝐴𝐷 − 18 kN⋅m = 0. (6) 𝑀𝑂𝑧 + 2.1 Equations (4)–(6) are easily solved for: 𝑇𝐴𝐷 = 70.88 kN,

⇒

𝑀𝑂𝑦 = 0,

and 𝑀𝑂𝑧 = −41.4 kN⋅m.

The reactions at point 𝑂 can now be determined by writing ∑ ⃗ 𝐹⃗ = 0⃗ ∶ 𝑂𝑥 𝚤̂ + 𝑂𝑦 𝚥̂ + 𝑂𝑧 𝑘̂ + 𝑇⃗𝐴𝐷 + 𝐹⃗ = 0,

(7)

(8)

which is easily solved for: ⇒

𝑂𝑥 = 32.13 kN,

𝑂𝑦 = 54 kN,

and 𝑂𝑧 = −15 kN.

(9)

ISTUDY

Section 5.4

333

Equilibrium of Bodies in Three Dimensions Scalar solution 𝑧

Modeling

The FBD is shown in Fig. 3, where for convenience the cable force has been resolved into 𝑥, 𝑦, and 𝑧 components.

𝑀𝑂𝑧

Governing Equations & Computation

A variety of different strategies for the order in which the equilibrium equations are written and solved are effective for this problem. In the following equations, we take forces and moments to be positive in the positive coordinate directions. ) ( ∑ 0.8 𝑀𝑂𝑥 = 0 ∶ 𝑇𝐴𝐷 (1.6 m) − (12 kN)(3.6 m) = 0, (10) 2.1 (11) ⇒ 𝑇𝐴𝐷 = 70.88 kN, ∑ 𝑀𝑂𝑦 = 0 ∶ 𝑀𝑂𝑦 = 0, (12) ∑

∑

𝑀𝑂𝑧 = 0 ∶

𝐹𝑥 = 0 ∶

∑

𝐹𝑦 = 0 ∶

∑

𝐹𝑧 = 0 ∶

⇒ 𝑀𝑂𝑦 = 0, ) ( 1.1 (1.6 m) − (5 kN)(3.6 m) = 0, 𝑀𝑂𝑧 + 𝑇𝐴𝐷 2.1 ⇒ 𝑀𝑂𝑧 = −41.4 kN⋅m,

) 1.1 + 5 kN = 0, 2.1 ) ( 1.6 = 0, 𝑂𝑦 − 𝑇𝐴𝐷 2.1 ) ( 0.8 − 12 kN = 0, 𝑂𝑧 + 𝑇𝐴𝐷 2.1

𝑂𝑥 − 𝑇𝐴𝐷

(

𝑂𝑦

𝑂𝑧

𝑇𝐴𝐷

1.6 2.1

𝑇𝐴𝐷

𝑇𝐴𝐷

𝑂𝑥 𝑥

0.8 2.1 1.1 2.1

1.6 m 𝐵 2m

5 kN

𝑦

(13)

(14) (15)

⇒ 𝑂𝑥 = 32.13 kN,

(16)

⇒ 𝑂𝑦 = 54 kN,

(17)

⇒ 𝑂𝑧 = −15 kN.

(18)

As expected, all of these solutions agree with those obtained using a vector solution. Discussion & Verification

𝑀𝑂𝑦

We expect the cable to be in tension, and indeed our solution shows 𝑇𝐴𝐷 > 0. We should also verify that our solutions satisfy all of the equilibrium equations.

12 kN Figure 3 Free body diagram for the scalar solution.

Common Pitfall Summing moments. In both the vector and scalar approaches, a common error when summing moments, such as about point 𝑂 in Eqs. (2), (12), and (14) of this example, is to forget the moment reactions, the misconception being that they make no contribution because their lines of action pass through the moment summation point.

334

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.15

Bearing Supports A heavy door seals a furnace used to heat-treat metal parts. The door’s weight is 𝑊 = 200 lb, which acts through point 𝐺 located at the center of the 18 in. by 18 in. door. Determine the force supported by cable 𝐴𝐵 and the reactions at the bearings at points 𝐶 and 𝐷.

𝑦 𝐵

SOLUTION

𝐴

4 in. 24 in.

𝑊 𝐺

10 in.

18 in.

4 in. 8 in.

𝐶

𝐷 10 in.

6 in. 𝐸

𝑧

𝑥

Road Map

Our objective is to analyze equilibrium of the furnace door. The door is supported by a cable and two bearings, and appropriate modeling idealizations regarding the bearings will be needed to obtain a statically determinate problem. Both vector and scalar solutions are effective, and both are illustrated. Vector solution Modeling

The FBD of the furnace door is shown in Fig. 2, where we have assumed both bearings are perfectly aligned (or are self-aligning) so that the bearings have no moment reactions. If this assumption is not valid, then moment reactions are present at the bearings, and this problem is statically indeterminate and cannot be solved using only the equations of static equilibrium. Governing Equations In anticipation of summing moments about point 𝐷, the required position vectors∗ and force vectors are

Figure 1

̂ in., 𝑟⃗𝐷𝐺 = (−6 𝚤̂ + 5 𝚥̂ + 17 𝑘) 𝑟⃗𝐷𝐶 = 10 𝚥̂ in.,

𝑦 𝐵 𝑇 𝐴

4 in.

24 in.

𝑊

𝐶𝑧

𝐺

10 in.

18 in.

4 in. 8 in.

10 in.

6 in. 𝑧

Figure 2 Free body diagram for the vector solution.

ISTUDY

(1) (2)

̂ in., 𝑟⃗𝐷𝐴 = (−6 𝚤̂ + 14 𝚥̂ + 26 𝑘) 6 𝚤̂ + 20 𝚥̂ − 26 𝑘̂ ̂ = 𝑇 (0.1799 𝚤̂ + 0.5998 𝚥̂ − 0.7797 𝑘). 𝑇⃗ = 𝑇 √ 1112

(4)

(3400 in.⋅lb) 𝚤̂ + (1200 in.⋅lb) 𝑘̂ + 𝐶𝑧 (10 in.) 𝚤̂ [ ] ⃗ − 𝐶𝑥 (10 in.) 𝑘̂ + 𝑇 (−26.51 in.) 𝚤̂ + (−6.117 in.) 𝑘̂ = 0.

(6)

(3)

Because there are no 𝚥̂ terms in Eq. (6), moment equilibrium about the 𝑦 axis is automatically satisfied, regardless of the values of 𝑇 and the reactions. Collecting 𝚤̂ and 𝑘̂ terms in Eq. (6) provides the following two equations, respectively,

𝐸

𝑥

̂ 𝐶⃗ = 𝐶𝑥 𝚤̂ + 𝐶𝑧 𝑘,

The remainder of this solution uses the four-digit numbers in Eq. (4). Summing moments† about point 𝐷 gives ∑ ⃗ ⃗ = 0⃗ ∶ 𝑟⃗ × 𝑊 ⃗ + 𝑟⃗ × 𝐶⃗ + 𝑟⃗ × 𝑇⃗ = 0, 𝑀 (5) 𝐷 𝐷𝐺 𝐷𝐶 𝐷𝐴

𝐶𝑥 𝐷𝑧 𝐷𝑥

⃗ = −200 𝚥̂ lb, 𝑊

𝑇 (−26.51 in.) + 𝐶𝑧 (10 in.) + 3400 in.⋅lb = 0,

(7)

𝑇 (−6.117 in.) − 𝐶𝑥 (10 in.) + 1200 in.⋅lb = 0.

(8)

Equations (7) and (8) contain three unknowns, thus additional equilibrium equations must be written before these unknowns can be determined. Equilibrium of forces requires ∑ ⃗ ⃗ + 𝐶⃗ + 𝐷 ⃗ + 𝑇⃗ = 0. 𝐹⃗ = 0⃗ ∶ 𝑊 (9) ⃗ = 𝐷 𝚤̂ + 𝐷 𝑘, ̂ and collecting terms Substituting for the force vectors in Eq. (9), where 𝐷 𝑥 𝑧 ̂ multiplying 𝚤̂, 𝚥̂, and 𝑘 provides

∗ Rather

than 𝑟⃗𝐷𝐴 , the vector 𝑟⃗𝐷𝐵 = 34 𝚥̂ in. could be used, and this would provide for easier evaluation of the cross product in Eq. (5). † In Eq. (5), you may find it convenient to evaluate 𝑟 ⃗𝐷𝐶 × 𝐶⃗ by inspection (i.e., scalar approach) rather than by formal evaluation of the cross product.

ISTUDY

Section 5.4

335

Equilibrium of Bodies in Three Dimensions (0.1799)𝑇 + 𝐶𝑥 + 𝐷𝑥 = 0,

(10)

(0.5998)𝑇 − 200 lb = 0,

(11)

(−0.7797)𝑇 + 𝐶𝑧 + 𝐷𝑧 = 0.

(12)

Equation (11) may immediately be solved for 𝑇 . Then Eqs. (7) and (8) may be solved for 𝐶𝑥 and 𝐶𝑧 , followed by solution of Eqs. (10) and (12) for 𝐷𝑥 and 𝐷𝑧 . These solutions are Computation

𝑇 = 333.4 lb,

⇒ 𝐶𝑥 = −83.98 lb,

𝐶𝑧 = 544.0 lb,

𝐷𝑥 = 23.99 lb,

𝐷𝑧 = −284.0 lb.

(13)

Scalar solution

𝑦

Modeling

The FBD is shown in Fig. 3, where we have assumed both bearings are perfectly aligned (or are self-aligning) so that the bearings have no moment reactions. For convenience, the cable force has been resolved into 𝑥, 𝑦, and 𝑧 components.

Governing Equations & Computation

The following strategy for writing and solving equilibrium equations is one of several that will allow for determination of the unknowns with minimal algebra: ∑ 𝐹𝑦 = 0 ∶ (0.5998)𝑇 − 200 lb = 0 ⇒ 𝑇 = 333.4 lb, (14) ∑ 𝑀𝐷𝑥 = 0 ∶ −(0.7797)𝑇 (14 in.) − (0.5998)𝑇 (26 in.) + (200 lb)(17 in.) + 𝐶𝑧 (10 in.) = 0

𝑀𝐷𝑧 = 0 ∶

∑

(0.7797)𝑇

(16)

+ (200 lb)(6 in.) − 𝐶𝑥 (10 in.) = 0

𝐹𝑥 = 0 ∶

(0.1799)𝑇 + 𝐶𝑥 + 𝐷𝑥 = 0

𝐹𝑧 = 0 ∶

−(0.7797)𝑇 + 𝐶𝑧 + 𝐷𝑧 = 0

⇒ ⇒

𝐷𝑧 = −284.0 lb.

𝐶𝑧

𝐺

10 in.

18 in.

4 in. 8 in.

𝐶𝑥 𝐷𝑧 𝐷𝑥

10 in.

𝑧

𝐸

(18) 𝑥

(19) (20)

As expected, all solutions agree with those obtained using the vector approach. Discussion & Verification

24 in.

6 in.

(17)

𝐷𝑥 = 23.99 lb,

4 in.

𝑊

(0.1799)𝑇

−(0.5998)𝑇 (6 in.) − (0.1799)𝑇 (14 in.)

⇒ 𝐶𝑥 = −83.98 lb, ∑

(0.5998)𝑇

(15)

⇒ 𝐶𝑧 = 544.0 lb, ∑

𝐵

We expect the cable to be in tension, and indeed our solution shows 𝑇 > 0. We should also verify that our solutions satisfy all of the equilibrium equations.

Figure 3 Free body diagram for the scalar solution.

336

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.16 front view

Equilibrium Analysis Over a Range of Motion 𝑦

𝐵 𝑥

𝑧 𝐴

Michael Plesha

The dump of a heavy-duty articulated dump truck is actuated by two symmetrically positioned hydraulic cylinders. One of the cylinders is visible in Fig. 1 between points 𝐴 and 𝐵, while the cylinder on the opposite side of the truck is not visible. The dump’s tilt angle is denoted by 𝜃, where the dump is fully lowered when 𝜃 = 0◦ and is fully raised when 𝜃 = 80◦ . If the dump and its contents weigh 400 kN with center of gravity at point 𝐺, determine the largest force the hydraulic cylinders must produce to raise the dump through its full range of motion. Coordinates of points 𝐴, 𝐵, and 𝐺 are [ ] 𝐴 0.3 m, −0.5 m, 3.2 m , [ ] (1) 𝐵 1.6 m, (1.1 m) sin(50◦ + 𝜃), (1.1 m) cos(50◦ + 𝜃) , [ ] ◦ ◦ 𝐺 0, (1.8 m) sin(30 + 𝜃), (1.8 m) cos(30 + 𝜃) .

rear view 𝑦

SOLUTION Road Map

We will neglect the weights of the two hydraulic cylinders, since they are likely very small compared to the weight of the dump and its contents. Thus, the hydraulic cylinders are two-force members. We will also assume the contents of the dump remain fixed in shape and position within the dump as it is being raised. With this assumption, the weight of the dump and its contents is constant, and its line of action always passes through point 𝐺, whose position is known from Eq. (1) as a function of tilt angle 𝜃. We also assume the truck is on level ground.

𝑧

𝑥 𝑂

𝐶

𝐶

Michael Plesha

Helpful Information Symmetry. For a problem to be symmetric, it must have symmetric geometry, loading, and supports. In other words, the geometry, loading, and supports that are present on one side of the problem’s plane of symmetry must be a perfect mirror image of those on the other side. Taking advantage of symmetry makes a problem easier to analyze because the number of unknowns is reduced. 𝑦 𝑊 𝐺 𝐹𝐴𝐵 , 𝐹𝐴 𝐵 𝑧 𝐴, 𝐴

𝐵, 𝐵 𝑂 𝜃 2𝐶𝑧 𝑥 2𝐶𝑦

Figure 2 Free body diagram when the tilt angle is 𝜃 = 0◦ . The 𝑥 direction is perpendicular to the figure. Forces 𝐹𝐴𝐵 and 𝐹𝐴 ′ 𝐵 ′ have components in the 𝑥 direction which are not visible in this twodimensional drawing.

ISTUDY

Inspection of Fig. 1 shows the dump is supported by bearings at points 𝐶 and 𝐶 ′ , and we assume these bearings are perfectly aligned so there are no moment reactions. Assuming the forces produced by the two hydraulic cylinders are the same, this problem is symmetric about the 𝑦𝑧 plane; and when drawing the FBD shown in Fig. 2, we may take the reactions at points 𝐶 and 𝐶 ′ to be the same. In Fig. 2, the hydraulic cylinder forces are taken to be positive in compression, and points 𝐴′ and 𝐵 ′ are defined to be symmetrically positioned on the opposite side of the dump from points 𝐴 and 𝐵, respectively.

Modeling

Figure 1

Governing Equations

Using the following position vectors

[ ] [ ] 𝑟⃗𝐴𝐵 = 1.3 m 𝚤̂ + (1.1 m) sin(50◦ + 𝜃) + 0.5 m 𝚥̂ [ ] ̂ + (1.1 m) cos(50◦ + 𝜃) − 3.2 m 𝑘, {[ ]2 [ ]2 𝑟𝐴𝐵 = 1.3 m + (1.1 m) sin(50◦ + 𝜃) + 0.5 m [ ]2 }1∕2 + (1.1 m) cos(50◦ + 𝜃) − 3.2 m , [ ] [ ] 𝑟⃗𝐴 ′ 𝐵 ′ = −1.3 m 𝚤̂ + (1.1 m) sin(50◦ + 𝜃) + 0.5 m 𝚥̂ [ ] ̂ + (1.1 m) cos(50◦ + 𝜃) − 3.2 m 𝑘, 𝑟𝐴 ′ 𝐵 ′ = 𝑟𝐴𝐵 ,

(2)

(3)

(4) (5)

vector expressions for the hydraulic cylinder forces and the 400 kN weight can be written as 𝑟⃗ 𝑟⃗ ′ ′ ⃗ = −400 kN 𝚥̂. 𝐹⃗𝐴𝐵 = 𝐹𝐴𝐵 𝐴𝐵 , 𝐹⃗𝐴 ′ 𝐵 ′ = 𝐹𝐴 ′ 𝐵 ′ 𝐴 𝐵 , 𝑊 (6) 𝑟𝐴𝐵 𝑟𝐴 ′ 𝐵 ′ Summation of moments about point 𝑂 is given by ∑ ⃗ ⃗ = 0⃗ ∶ 𝑟⃗ × 𝐹⃗ + 𝑟⃗ ′ × 𝐹⃗ ′ ′ + 𝑟⃗ × 𝑊 ⃗ = 0, 𝑀 𝑂 𝑂𝐴 𝐴𝐵 𝑂𝐴 𝐴 𝐵 𝑂𝐺

(7)

ISTUDY

Section 5.4

Equilibrium of Bodies in Three Dimensions

337

where the position vectors are ̂ 𝑟⃗𝑂𝐴 = (0.3 m) 𝚤̂ − (0.5 m) 𝚥̂ + (3.2 m) 𝑘, ̂ 𝑟⃗𝑂𝐴 ′ = (−0.3 m) 𝚤̂ − (0.5 m) 𝚥̂ + (3.2 m) 𝑘, [ ] [ ] ◦ ̂ 𝑟⃗𝑂𝐺 = (1.8 m) sin(30 + 𝜃) 𝚥̂ + (1.8 m) cos(30◦ + 𝜃) 𝑘.

(8) (9) (10)

2𝐹𝐴𝐵 𝑟𝐴𝐵

[

] kN ) cos(30◦ + 𝜃) = 0. (11) (−0.55) cos(50◦ + 𝜃) − (3.52) sin(50◦ + 𝜃) + (720 m

Computation

Solving Eq. (11) for 𝐹𝐴𝐵 provides 𝐹𝐴𝐵 =

𝑟𝐴𝐵

(720

kN ) cos(30◦ m

+ 𝜃)

2 (0.55) cos(50◦ + 𝜃) + (3.52) sin(50◦ + 𝜃)

,

(12)

where 𝑟𝐴𝐵 is given by Eq. (3). To complete this problem, we plot 𝐹𝐴𝐵 from Eq. (12) for values of 𝜃 over the range 0◦ to 80◦ to obtain the results shown in Fig. 3. Thus, we see the largest force the hydraulic cylinders must produce is about 320 kN, and this occurs when the dump first starts to open, when 𝜃 = 0◦ . A more accurate value of 𝐹𝐴𝐵 is obtained by evaluating Eq. (12) with 𝜃 = 0◦ , which produces 𝐹𝐴𝐵 = 318.5 kN. Discussion & Verification

• This solution assumes the contents of the dump remain fixed in shape and position within the dump as it is being raised. For purposes of determining the largest hydraulic cylinder force needed, this assumption is useful. A more accurate model would allow the weight of the dump’s contents to change with tilt angle 𝜃. Exactly how the weight varies with 𝜃 is a difficult problem, but a simple yet useful model might let this weight vary in some fashion (perhaps linear) from full value when 𝜃 = 0◦ , to zero when 𝜃 = 80◦ . Additionally, we may also consider allowing the center of gravity for the dump’s contents to change as a function of 𝜃. • If desired, the reactions at bearings 𝐶 and 𝐶 ′ as functions of 𝜃 can be determined by writing ∑ ⃗ ⃗ + 2𝐶⃗ = 0, (13) 𝐹⃗ = 0⃗ ∶ 𝐹⃗𝐴𝐵 + 𝐹⃗𝐴 ′ 𝐵 ′ + 𝑊 ̂ where 𝐶⃗ = 𝐶𝑦 𝚥̂ + 𝐶𝑧 𝑘.

• Although not illustrated here, a scalar solution for this problem is also effective. • Problem 5.100 gives some additional suggestions for verifying the accuracy of the solution to this example.

𝐹𝐴𝐵 = 𝐹𝐴 𝐵 (kN)

Despite the perhaps formidable appearance of the vector expressions in this problem, the cross products in Eq. (7) are easy to evaluate. Using 𝐹𝐴𝐵 = 𝐹𝐴 ′ 𝐵 ′ and 𝑟𝐴𝐵 = 𝑟𝐴 ′ 𝐵 ′ , Eq. (7) results in only an 𝚤̂ term whose coefficient must be zero for equilibrium, thus,

400 300 200 100 0 −100 −200 −300

0

20 40 60 tilt angle 𝜃 (degrees)

80

Figure 3 Force in hydraulic cylinders 𝐴𝐵 and 𝐴′ 𝐵 ′ as a function of tilt angle.

338

Chapter 5

Equilibrium of Bodies

Problems 𝑧

Problem 5.99 A handheld mixer for blending cooking ingredients is shown. To minimize operator fatigue, the reaction forces on the operator’s hand should be as small as possible. With this goal in mind, should the beaters rotate in the same direction or opposite directions? Assume each beater produces a moment about the 𝑧 axis, and if needed, assume reasonable values for dimensions. Explain your reasoning. Note: Concept problems are about explanations, not computations.

𝑦 𝑥

Problem 5.100 When you evaluate the solution to a problem, you should always verify the accuracy of the solution, and when possible, performing simple checks can help with verification. In Example 5.16 on p. 336, consider the specific position 𝜃 = 0, analyze the problem afresh to determine the hydraulic cylinder force required to begin opening the dump, and verify that the results of Example 5.16 are in agreement. As another check, Fig. 3 in Example 5.16 shows that the hydraulic cylinder force is zero for a particular tilt angle. Explain why this occurs and, if possible, perform a simple analysis that confirms the value of 𝜃 at which this force is zero.

Figure P5.99 𝑧 8 in.

𝐷

Problem 5.101 𝐶

10 in. 34 in.

(a) Both sanding drums rotate about the positive 𝑧 direction.

𝐴 𝐵 15 in. 𝑥 Figure P5.101

ISTUDY

A machine for sanding wood floors is shown. The machine weighs 80 lb with center of gravity along the 𝑧 axis. At each sanding drum a moment of 60 in. ⋅ lb is applied to the machine in the direction opposite the rotation of the drum; in addition to this moment, each drum has a 𝑧-direction reaction force, and all other reactions are zero. Assume the operator’s hands, positioned at points 𝐶 and 𝐷, can apply forces in the positive or negative 𝑥 direction. Determine the forces on the operator’s hands if (b) The sanding drums at 𝐴 and 𝐵 rotate about the positive and negative 𝑧 directions, respectively.

𝑦

Problem 5.102 Bar 𝐸𝐹 has a square cross section and is fixed in space. The structure 𝐴𝐵𝐶 has negligible weight and has a collar at 𝐶 that has a square hole that slides freely on bar 𝐸𝐹 . The structure 𝐴𝐵𝐶 supports a uniform rectangular sign with weight 1 kN (the two vertical edges of the sign align with points 𝐴 and 𝐵). Determine the magnitude of the tension in cable 𝐴𝐷 and all of the reaction components at 𝐶 referred to the 𝑥, 𝑦, and 𝑧 directions provided. 𝑧

5m

𝐸 2m 𝐵

𝐶

3m 𝐷

𝐹 𝑥

𝐴

ng eeri n i g y En ar r

Lib

0.5 m Figure P5.102

𝑦 4m

ISTUDY

Section 5.4

339

Equilibrium of Bodies in Three Dimensions 𝑧

Problem 5.103

𝐺

Structure 𝐴𝐵𝐶𝐷𝐸 is supported by a frictionless collar at 𝐵 and by cable 𝐴𝐺. Bar 𝐹 𝐺 has circular cross section and is built-in at 𝐹 . Portion 𝐷𝐸 of the structure is parallel to the 𝑥 axis. 150 mm

(a) Is structure 𝐴𝐵𝐶𝐷𝐸 statically determinate or statically indeterminate, and is it fully fixed or partially fixed? Explain. 𝑥

(b) Determine the force supported by the cable and all of the reactions at 𝐵.

80 mm 𝐸

𝐵 𝐷 90 mm

𝐴

500 N

80 mm

Problem 5.104 Structure 𝐴𝐵𝐶𝐷𝐸𝐹 is supported by a pin at point 𝐴, and point 𝐹 rests on a frictionless horizontal surface. Points 𝐴 and 𝐹 lie in the 𝑥𝑦 plane, and portion 𝐵𝐶𝐷𝐸 of the structure has the same 8 in. 𝑧 coordinate. The 10 lb and 20 lb forces act in the −𝑧 direction. Determine all of the reactions at points 𝐴 and 𝐹 .

120 mm 𝐶 𝐹 𝑦 Figure P5.103

𝑧 𝑧 4 in. 12 in.

4 in. 𝐸

𝐵 4

𝐷

8 in.

𝐹

𝐶

𝐴

10 lb 8 in. 𝐴

𝐹 20 lb 𝑥

𝑥 Figure P5.104

𝑦

𝐷 𝐸

4 𝐵

𝑦

dimensions in ft

𝐶 4

2

4 𝑃 = 100 lb Figure P5.105 𝑦

Problem 5.105

𝐷

Bar 𝐴𝐵𝐶𝐷𝐸 is supported by cable 𝐵𝐹 , a ball and socket at 𝐴, and a self-aligning bearing at 𝐸. Determine the tension in cable 𝐵𝐹 and the reactions at points 𝐴 and 𝐸.

Problem 5.106

6 in.

Vertical bar 𝐸𝐷 has circular cross section and is built in at 𝐸. Member 𝐴𝐵𝐶 is a single member that lies in a horizontal plane, with portion 𝐵𝐶 parallel to the 𝑧 axis and with cable 𝐶𝐷 attached to point 𝐶. The collar at 𝐴 can freely slide in the 𝑦 direction and can freely rotate about the 𝑦 axis. (a) Does the structure 𝐴𝐵𝐶 have complete fixity or partial fixity, and is it statically determinate or statically indeterminate? Explain. (b) When point 𝐵 is subjected to a downward vertical force of 18 lb, determine the force supported by the cable and all support reactions at 𝐴.

Problem 5.107 Follow the suggestion made in Eqs. (10) and (11) of Example 5.13 on p. 331 to find the tension in cable 𝐴𝐶 by summing moments about line 𝑂𝐷.

Problem 5.108 In Prob. 5.107, find the tension in cable 𝐴𝐷 by summing moments about line 𝑂𝐶.

18 lb

6 in.

𝐵

𝐴

3 in. 𝐶 𝑧 𝐸

Figure P5.106

𝑥

340

Chapter 5

Equilibrium of Bodies

Problem 5.109 Member 𝐴𝐺𝐷𝐵 is supported by a cable 𝐷𝐸, a self-aligning bearing at 𝐴, and a selfaligning thrust bearing at 𝐵. (a) Draw the FBD for 𝐴𝐺𝐷𝐵, labeling all forces and moments. (b) Rate the solution strategies listed below for ease of obtaining the magnitude of the tension 𝑇𝐷𝐸 in cable 𝐷𝐸. Rate the best choice as number 1, second-best choice number 2, and so on. If a solution strategy does not work, then label it with zero.

𝑦 𝐶

𝐸

Rating 𝐵 𝐴

𝑂

⃗ solve for 𝑇 . ⃗ + 𝑟⃗ × 𝑇⃗ = 0, Write 𝑟⃗𝐴𝐺 × 𝑊 𝐴𝐷 𝐷𝐸 𝐷𝐸 ∑ ∑ ⃗ solve for 𝑇 . ⃗ = 0, Write 𝐹⃗ = 0⃗ and 𝑀 𝑂 𝐷𝐸

𝐷

𝐺

Solution strategy ∑ ∑ ⃗ solve for 𝑇 . ⃗ = 0, Write 𝐹⃗ = 0⃗ and 𝑀 𝐵 𝐷𝐸

𝑥

⃗ + 𝑟⃗ × 𝑇⃗ ) ⋅ 𝑟⃗ = 0, solve for 𝑇 . Write (⃗𝑟𝐴𝐺 × 𝑊 𝐴𝐷 𝐷𝐸 𝑂𝐶 𝐷𝐸 ∑ ∑ ⃗ solve for 𝑇 . ⃗ = 0, Write 𝐹⃗ = 0⃗ and 𝑀

𝑊 𝑧

𝐴

Figure P5.109

𝐷𝐸

Note: Concept problems are about explanations, not computations.

Problem 5.110 ⃗ = −200 𝑘̂ lb acting at its center, point 𝐹 , is supported by A circular plate with weight 𝑊 cord 𝐷𝐸 and a thrust bearing at 𝐵. Shaft 𝐴𝐶 is fixed and frictionless. (a) Draw the FBD for the plate, labeling all forces and moments. 𝑧 12 in.

(b) Rate the solution strategies listed below for ease of obtaining the magnitude of the tension 𝑇𝐷𝐸 in cable 𝐷𝐸. Rate the best choice as number 1, second-best choice number 2, and so on. If a solution strategy does not work, then label it with zero.

12 in.

𝐸

Rating 11 in.

10 in.

𝐴 𝐵 𝐶

𝐹

5 in.

𝐷

𝑂

𝑦

⃗ + 𝑟⃗ × 𝑇⃗ ) ⋅ 𝑟⃗ = 0, solve for 𝑇 . Write (⃗𝑟𝐵𝐹 × 𝑊 𝐵𝐷 𝐷𝐸 𝐴𝐶 𝐷𝐸 ∑ ∑ ⃗ ⃗ ⃗ = 0, solve for 𝑇 . Write 𝐹⃗ = 0 and 𝑀 𝐷

Figure P5.110 and P5.111 𝑧

⃗ solve for 𝑇 . ⃗ + 𝑟⃗ × 𝑇⃗ = 0, Write 𝑟⃗𝐵𝐹 × 𝑊 𝐵𝐷 𝐷𝐸 𝐷𝐸 ∑ ∑ ⃗ solve for 𝑇 . ⃗ = 0, Write 𝐹⃗ = 0⃗ and 𝑀 𝑂 𝐷𝐸

𝐷 (9, 18, 3) in. 𝐹 (5, 13, 5) in.

200 lb

𝑥

Solution strategy ∑ ∑ ⃗ solve for 𝑇 . ⃗ = 0, Write 𝐹⃗ = 0⃗ and 𝑀 𝐵 𝐷𝐸

𝐷𝐸

Note: Concept problems are about explanations, not computations.

Problem 5.111

𝐹

Determine the cable tension for the circular plate of Prob. 5.110. 𝐹 𝑃

𝑃 𝑟 𝑦

𝑥 Figure P5.112

ISTUDY

𝐴

ℎ

Problem 5.112 A circular plate of radius 𝑟 is welded to a post with length ℎ that is built in at point 𝐴. Determine the reactions at point 𝐴. Express your answers in terms of parameters such as 𝑟, ℎ, 𝐹 , and 𝑃 .

ISTUDY

Section 5.4

341

Equilibrium of Bodies in Three Dimensions

𝑧

Problem 5.113

𝑦

Object 𝐴𝐵𝐶𝐷𝐸𝐹 is a sliding door that is supported by a frictionless bearing at 𝐴 and a wheel at 𝐹 that rests on a frictionless horizontal surface. The object has weight 𝑊 = 800 N, which acts at the midpoint of the rectangular region 𝐵𝐶𝐷𝐸. Determine all support reactions.

𝐻 𝐶 𝐴

Problem 5.114

0.8 m

𝐵

𝑥

𝐺

The tonearm 𝐴𝐵𝐶𝐷𝐸 of a phonograph (i.e., record player) is shown. The support at point 𝐵 prevents all motions of the tonearm except for rotations about the 𝑦 and 𝑧 axes. At point 𝐸, the tip of the tonearm rests on the lightly grooved (i.e., rough) surface of a vinyl record. If the tonearm weighs 0.2 lb with center of gravity at point 𝐺, and the force 𝑃 = 0.01 lb acts in the 𝑥 direction, determine the reactions at points 𝐵 and 𝐸. Hint: In addition to the force 𝑃 , there are also reaction forces in the 𝑦 and 𝑧 directions at point 𝐸.

𝑊 0.3 m

0.5 m 𝐸

𝐷

1m 𝐹

frictionless surface

Figure P5.113

𝑧 𝐴

𝑦 𝐵

3 in. 7 in.

0.5 in.

1 in.

𝐺 tonearm

𝐶

𝐷 𝐸

0.5 in.

𝐶 0.1 m

𝑥

𝐷 0.1 m 𝑦

𝐵 𝐹

𝑃 vinyl record

𝑃

𝑄

𝑧

𝐴

0.6 m

0.3 m

2m

Figure P5.114 𝑥

Problem 5.115 One of the wings of a Cessna 172 airplane is shown. The wing is supported by a strut 𝐵𝐸 with ball-and-socket joints at 𝐵 and 𝐸, and by bearings at 𝐴 and 𝐹 . The bearings are perfectly aligned and are supported by a straight shaft which is fixed to the fuselage, and bearing 𝐴 is a thrust bearing. Bearings 𝐴 and 𝐹 lie on the 𝑥 axis, point 𝐵 lies on the 𝑦 axis, point 𝐸 lies on the 𝑧 axis, and points 𝐶 and 𝐷 lie in the 𝑥𝑦 plane. The airplane’s mass, including fuel, passengers, and cargo, is 1000 kg, and it flies so that it is in static equilibrium. The strut 𝐵𝐸 has negligible mass and the wing 𝐴𝐵𝐶𝐷𝐹 has 200 kg mass with center of gravity at point 𝐵. The force 𝑄 acts in the −𝑥 direction and is due to air drag. Remark: A two-dimensional model of this wing was treated in Prob. 5.27 on p. 293.

1.4 m

𝐸

Figure P5.115

(a) Determine the resultant lift force 𝑃 that this wing supports, assuming this force is vertical and that only the wings provide lift for the plane.

0.6 m 0.2 m

(b) If the drag force 𝑄 is 10% of the lift force 𝑃 , determine the force supported by the strut 𝐵𝐸 and the reactions at bearings 𝐴 and 𝐹 .

𝑦

0.3 m

𝑧

1 kN 𝐷

𝐶 𝑥

Problem 5.116 The control surface of an aircraft is supported by a thrust bearing at point 𝐶 and is actuated by a bar connected to point 𝐴. The 1 kN force acts in the negative 𝑧 direction, and the line connecting points 𝐴 and 𝐵 is parallel to the 𝑧 axis. Determine the value of force 𝐹 needed for equilibrium and all support reactions.

𝐵

𝐹 5

0.1 m 14 2 𝐴

Figure P5.116

342

Chapter 5

Equilibrium of Bodies 𝑦

Problem 5.117 2m

1.5 m

An L-shaped bar is supported by a bearing at 𝐴 and a smooth horizontal surface at 𝐵. Determine the reactions at 𝐴 and 𝐵.

1.5 m

𝐴

𝑂 𝐶 𝐵

𝑥

Problem 5.118

𝐹 = 800 N

Structure 𝐴𝐵𝐶𝐷 is supported by a collar at 𝐷 that can rotate and slide along bar 𝐸𝐹 , which is fixed and is frictionless. Structure 𝐴𝐵𝐶𝐷 makes contact with smooth surfaces at 𝐴 and 𝐶 where the normal direction 𝑛⃗ to the surface at 𝐴 lies in a plane that is parallel to the 𝑥𝑦 plane. Force 𝑃 is parallel to the 𝑦 axis. If 𝑃 = 10 kN, determine the reactions at 𝐴, 𝐶, and 𝐷.

𝑧 Figure P5.117

𝑦 2m 𝑃

𝐸

𝐹

𝐷

𝑥 4m 3m 𝐶

𝐵 𝑛⃗

𝑧

2m

𝐴 30◦

Figure P5.118

Problem 5.119 𝑧

𝐹 = 5000 lb

4 in. gear radius = 3 in.

𝐹 𝐶

𝐷

𝐵 𝐴 𝑦

𝐺 18 in. 𝑥 𝐸 𝑄

24 in. 𝑊 = 1000 lb

Figure P5.119 and P5.120

ISTUDY

The rear wheel, axle, and final drive gear assembly of a John Deere 2720 tractor is shown. These components form a rigid object where the axle rotates on bearings at points 𝐴 and 𝐶 that are attached to the frame of the tractor. The engine and transmission apply a force 𝐹 = 5000 lb in the −𝑧 direction at point 𝐺 of the gear, which has 3 in. radius. The force 𝑊 = 1000 lb acts in the 𝑧 direction and is due to the weight of the tractor. The force 𝑄 acts in the 𝑥 direction and is due to the wheel applying tractive force to the ground. Points 𝐴, 𝐶, 𝐷, and 𝐸 lie in the 𝑦𝑧 plane, and point 𝐺 lies in the 𝑥𝑦 plane. Neglect the weight of the wheel, axle, and gear. (a) Draw the FBD for the wheel, axle, and gear assembly if the bearings at 𝐴 and 𝐵 resist moments and the bearing at 𝐴 is a thrust bearing. Label all unknowns. Is the assembly statically determinate or statically indeterminate? Explain. (b) Repeat Part (a) if the bearings at 𝐴 and 𝐵 are self-aligning and the bearing at 𝐴 is a thrust bearing. (c) Use the FBD from Part (b) and assume the assembly is in equilibrium. Determine force 𝑄 and the reactions at the bearings.

Problem 5.120 Repeat Prob. 5.119, Parts (b) and (c), if the wheel, axle, and gear assembly weigh 150 lb with center of gravity at point 𝐷, which lies on the 𝑦 axis with coordinate 𝑦𝐷 = −18 in.

343

Equilibrium of Bodies in Three Dimensions 𝑧

Problem 5.121 A model for the control arm 𝐴𝐵𝐶𝐷 for the suspension of one of the wheels of an automobile is shown. The control arm is supported by ball-and-socket joints at points 𝐴 and 𝐶, and by a spring 𝐵𝐸. Points 𝐴, 𝐶, and 𝐸 lie in the 𝑦𝑧 plane and are attached to the frame of the automobile. The vertical force 𝑃 = 4 kN is due to the weight of the automobile, and the control arm has negligible weight.

𝐸

240 mm 𝐶

𝑦

70 mm

390 mm

(a) Is the control arm statically determinate or statically indeterminate? Explain. 𝐴

(b) Determine the force in spring 𝐵𝐸.

𝐵 𝐷

Problem 5.122 Frame 𝐴𝐵𝐶𝐷𝐸 is supported by ball-and-socket joints at point 𝐴 and 𝐸, and by cable 𝐶𝐹 . The frame has negligible mass compared to the 100 kg object supported at point 𝐷.

𝑃

𝐵 (180, 120, 30) mm 𝐷 (300, 120, 10) mm

𝑥

(a) Is this frame statically determinate or indeterminate? Explain. Figure P5.121

(b) Determine the force in cable 𝐶𝐹 . 𝑦 2m 𝐴

𝐵 3m

𝐹

𝐶

𝑥

6m 𝐷 𝐸 𝑧

100 kg Figure P5.122

Problem 5.123 Alternative equilibrium equations: Consider the FBD shown where 𝐹1 , 𝐹2 , etc., are the forces acting on the body (including reaction forces from supports), and 𝑀1 , 𝑀2 , etc., are the moments acting on the body (including reaction moments from supports, if any). Let the total number of forces and moments be 𝑛 and 𝑚, respectively. Consider the equilibrium equations

∑

⃗ = 0⃗ ∶ 𝑀 𝐴

∑

⃗ = 0⃗ ∶ 𝑀 𝐵

⃗ 𝐹⃗𝑖 = 0,

𝑖=1 𝑛

∑

𝑟⃗𝐴𝑖 × 𝐹⃗𝑖 +

∑

𝑟⃗𝐵𝑖 × 𝐹⃗𝑖 +

∑

𝑟⃗𝐴𝑖

(2)

𝑖=1 𝑚

𝑖=1 𝑛

𝑖=1

𝑟⃗𝐵𝑖

⃗ ⃗ = 0, 𝑀 𝑖

𝑦 𝐴

𝑥

⃗ ⃗ = 0, 𝑀 𝑖

(3)

𝐹⃗ 𝑖

𝑖=1

where 𝑟⃗𝐴𝑖 is a position vector from point 𝐴 to anywhere on the line of action of 𝐹⃗𝑖 , and 𝑟⃗𝐵𝑖 is a position vector from point 𝐵 to the same point on the line of action of 𝐹⃗𝑖 . Equations (1) and (2) are the “standard” equilibrium equations as given in Eq. (5.16) on p. 322, while Eqs. (2) and (3) are alternative equilibrium equations that may be used instead. Show that satisfying Eqs. (2) and (3) means that Eq. (1) is also satisfied. Hint: Note that 𝑟⃗𝐵𝑖 = 𝑟⃗𝐵𝐴 + 𝑟⃗𝐴𝑖 where 𝑟⃗𝐵𝐴 is the position vector from point 𝐵 to point 𝐴, combine Eqs. (2) and (3), and after simplification and imposition of necessary restrictions on the locations of points 𝐴 and 𝐵, show that Eq. (1) results.

𝐹⃗ 2

𝐵

(1) 𝑚 ∑

⃗ 𝑀 1

t

𝐹⃗ = 0⃗ ∶

𝑛 ∑

𝑧

Ni t

∑

𝐹⃗ 1 Mc

ISTUDY

Section 5.4

Figure P5.123

⃗ 𝑀 2

344

ISTUDY

Chapter 5

Equilibrium of Bodies

5.5

Engineering Design

Engineering design is a subject in its own right, and you will study this subject in detail as you advance in your education. The objective of engineering design is to develop the description and specifications for how a structure, machine, device, or procedure can be produced so that needs and requirements that have been identified are met. The objective of engineering design theory is to establish structured procedures for design that help you develop the most effective and optimal design possible in a timely manner. Good references are available on this subject, such as Ullman (2003), Dominick et al. (2001), and Middendorf and Engelmann (1998).∗ Design is an iterative process. Whether the design is of a simple paper clip or a fuel pump for a rocket engine, a process is certain to be followed. This process may vary between individual engineers and may depend on the specific nature of what is being designed, the magnitude of the project, and other factors, but the process shown in Fig. 5.27 is common. While each of the elements in Fig. 5.27 is present in almost all design activities, there may be reordering of tasks and/or additional components needed. For example, in civil construction a project will often begin with a request for proposals (RFP) being issued by a government unit where the specifications for a project are stated. Companies that respond to the RFP will conduct preliminary design work and perform a thorough cost analysis, and only after their proposal has been accepted will detailed design work begin. In concurrent design, the process shown in Fig. 5.27 is interjected with input from interested parties, such as other professionals, customers, and so on, so that the final design is more certain to satisfy the needs of all affected parties. In concurrent design, tasks such as developing the manufacturing systems needed to fabricate the design may take place while the design is still being developed. In the following paragraphs, we discuss some of the items of Fig. 5.27 in greater detail. • Problem identification. Most designs begin with the identification of a need that is not met. When such a need is identified, you must thoroughly review the state of the art to confirm the need, why it exists, why this need has not been recognized and/or addressed by others, and so on. Very simply, you must establish that there is indeed an opportunity for you to make a contribution before investing substantial time and resources in design. • Problem evaluation. Here you must identify all of the needs your design must satisfy, not just those that might be new or novel. The needs consist of all the requirements, specifications, goals, and so on that your design must satisfy. Some needs may be achievable only at the expense of other needs (e.g., strength and light weight), so value decisions may be required to develop needs that are realistic and achievable. In the course of doing this, you must review applicable standards, codes, patents, industry practices, and so on. Depending on what you are designing, you may be required by law to conform to certain standards. Other standards may be voluntary, but acceptance of your design by customers may depend on compliance with these. Additional comments on standards and ∗ D. G. Ullman, The Mechanical Design Process, 3rd ed., McGraw-Hill, New York, 2003. P. G. Dominick,

J. T. Demel, W. M. Lawbaugh, R. J. Freuler, G. L. Kinzel and E. Fromm, Tools and Tactics of Design, John Wiley & Sons, New York, 2001. W. H. Middendorf and R. H. Engelmann, Design of Devices and Systems, 3rd ed., Marcel Dekker, New York, 1998.

ISTUDY

Section 5.5

Should the problem identification or problem evaluation be revised in view of knowledge gained?

Engineering Design

Problem identification. • Recognition of need. • Study state of the art.

Problem evaluation. • Comprehensive needs analysis. • Establish requirements, specifications, goals. • Study applicable standards, codes, patents, industry practices, etc. • Feasibility study.

Formulate solutions. • Develop alternative designs. • Preliminary analysis. • Tentative selection of most promising design.

Develop detailed design. • Selection of modeling approaches. – Analytical. – Computer. – Prototype and/or physical testing. • Cost analysis. • Manufacturing considerations.

O p t i m i z a t i o n

Finalize design. • Documentation. • Communication. • Production. • Follow-up.

Figure 5.27. A process for engineering design.

codes follow later in this section. Once all of your needs and requirements have been established, you may be able to conduct a feasibility study to determine if your ideas are possible. In a feasibility study, you might consider if it is physically possible to satisfy your needs and objectives, although this may be difficult to fully assess until more careful design work is done. Other issues commonly considered in a feasibility study include a company’s manufacturing capability, compatibility with existing product line, marketing, and economics. • Formulate solutions. You will likely want to develop several preliminary solutions that are capable of satisfying your needs. Based on your preliminary analyses, you will select the most promising of these for deeper investigation. Because of the iterative nature of design, a solution that is initially discarded here may be revived later for further study.

345

346

ISTUDY

Chapter 5

Equilibrium of Bodies

• Detailed design analysis. Here you will use all of the engineering tools you have at your disposal, including analytical, computer, and/or experimental methods of analysis to thoroughly evaluate your design, including safety for foreseeable uses and misuses. If a serious flaw in your design is discovered, you may need to revise your design or return to one of the earlier steps in the design process. • Finalize design. Your design must be documented and communicated to others. This documentation may be used by you or others for some future design revision, may help support patent rights, or may be needed for litigation. Follow-up refers to design enhancements that may result from feedback from users.

Codes and standards A code is a comprehensive set of instructions and procedures for specific applications that help you develop a successful design. It is developed by engineers usually under the sponsorship of professional engineering societies, and it embodies an enormous amount of knowledge and wisdom gained over years. A sampling of organizations that sponsor design codes follows: • American Institute of Steel Construction (AISC). This organization represents the structural steel design community and the construction industry and publishes a code for applications to framed steel structures. • American Concrete Institute (ACI). This organization is dedicated to improving the design, construction, and maintenance of concrete structures and publishes a code for these applications. • American Society of Mechanical Engineers (ASME). This organization provides comprehensive representation of a wide range of the mechanical engineering specialties and publishes numerous codes, one of which governs the design of boilers, pressure vessels, and piping, including applications to nuclear reactor components. Use of these design codes is often voluntary (codes for nuclear components are mandatory), but if they are suitable for your design work, then they should be followed. A standard is a minimum performance measure that a design should meet, but typically it will not tell you how to develop the design. In addition to the organizations stated earlier, the following organizations develop and publish numerous standards: • Society of Automotive Engineers (SAE). This organization represents a wide range of interests pertaining to transportation and develops standards that apply to land, sea, air, and space vehicles. • American Institute of Aeronautics and Astronautics (AIAA). This organization is the principal society representing aerospace engineers and scientists, and it publishes numerous standards governing the performance of aircraft and spacecraft. • American National Standards Institute (ANSI). This organization develops safety standards based on the consensus of affected parties, including

ISTUDY

Section 5.5

manufacturers and users. Use of ANSI standards is voluntary, but if applicable standards are available, these should be used. • Consumer Product Safety Commission (CPSC). This is a U.S. federal agency that sets voluntary and mandatory safety standards for consumer products. They have the power to issue product recalls and can ban hazardous products. • Occupational Safety and Health Administration (OSHA). This is a U.S. federal agency that encourages employers and employees to cooperate to reduce workplace hazards. They develop voluntary and mandatory workplace safety and health standards and enforce these. The U.S. federal agencies cited here set standards for the United States only. The other organizations cited have participation and impact throughout other parts of North America. Most other countries, especially in other parts of the world, have similar organizations of their own. There are efforts to develop uniform standards that are applicable worldwide, and a notable organization that promotes this is the International Organization for Standardization (ISO). This organization has membership drawn from the standards organizations of approximately 150 countries. Sometimes the difference between codes and standards is not distinct. You must be familiar with all codes and standards that could affect your design. Your design must conform to mandatory standards, and if voluntary standards are available, these should be followed if at all possible. If it is not possible to follow voluntary standards or if standards do not exist, then suitable performance and safety measures must be established. Whether or not you use a code and/or standards to develop a design, the responsibility for the safety and performance of your design is yours.

Design problems For the design problems in this section and throughout this book, imagine you are employed as an intern working under the supervision of an engineer who asks you to conduct a design study, or to add details to a design that is started. You will be presented with a problem that is suitable to your level of knowledge along with some pertinent data. Sometimes, the information provided may not be complete, and you may not be instructed on everything that needs to be done. As shown in Fig. 5.27, this is not an attempt to be artificially vague, but rather is a reflection of how design and modeling of real life problems are carried out. It is in this spirit that the design problems throughout this book are presented, and you may need to make reasonable assumptions or seek out additional information on your own. Furthermore, our work in this book will focus on design based on first principles. That is, we will use the laws of physics plus good judgment to establish a design. We generally will not survey and apply standards that may be available, because this is not central to the theme of this book. Your work should culminate in a short written technical report that is appropriate for an engineer to read, where you present your design, state assumptions made, and so on. Appendix A of this book gives a brief discussion of technical writing that may be helpful.

Engineering Design

347

348

Chapter 5

Equilibrium of Bodies

E X A M P L E 5.17

Engineering Design The forklift has a vehicle weight of 𝑊𝑉 = 12,500 lb, fuel weight 𝑊𝐹 , which is 300 lb when fully fueled, and operator weight 𝑊𝑂 . The drive wheels at 𝐶 are supported by a fixed axle, and the wheels at 𝐷 steer the vehicle. Determine the capacity rating for this forklift, assuming it is operated at slow speed on a smooth, level surface. The capacity rating is the maximum load this forklift is designed to lift.

SOLUTION Road Map Lucinda Dowell

The following needs must be satisfied by the capacity rating we determine:

• The capacity rating should be as large as possible while still being safe. 10 in. 11 in. 9 in.

• The forklift should not tip forward while supporting a load, regardless of whether the operator is on or off the vehicle. 𝑊𝑂

fork face

𝑊𝐹 𝑊𝑉

𝑃 𝐵

𝐴

𝐷

𝐶 48 in. 13 in.

64 in.

14 in.

Figure 1

• The rear tires should support enough force so they are capable of steering the forklift. This requirement applies when the forklift is being driven, and thus the operator may be assumed to be seated on the forklift. All our analyses will take the fuel weight to be zero, as the forklift is likely to see occasional use when its fuel supply is low, and this leads to a more conservative capacity rating. While somewhat arbitrary, but reasonable, we will assume the steering requirement is satisfied if the forces supported by the wheels at 𝐷 are at least 10% of the forces supported by the wheels at 𝐶.∗ When we analyze this requirement, an operator’s weight may be included. However, to obtain a conservative capacity rating, we must use the minimum possible operator weight, and this is difficult to determine with any precision. Obviously, this weight is small compared to the forklift’s weight, and it is conservative to take it as zero. If this is the case, then the no tipping requirement will automatically be satisfied if the steering requirement is met.

58 in. 11 in. 9 in.

𝑊𝐹 = 0

𝑥 𝑑

𝑊𝑉 = 12,500 lb

𝑃 𝐴

The FBD is shown in Fig. 2, where we have imposed the constraint 𝐷𝑦 = (0.1) 𝐶𝑦 to enforce the need for steering ability, and the fuel and operator weights are zero. Clearly, as distance 𝑑 from the load to the fork face increases (𝑑 is called the load center), the value of 𝑃 that can be lifted decreases. The most conservative idea is to place 𝑃 at the worst possible position, which is at 𝑑 = 48 in. However, a quick investigation of industry practice shows that for purposes of specifying capacity rating, manufacturers customarily position 𝑃 at the midpoint of the forks, which is 𝑑 = 24 in. In the following calculation we will consider both values of 𝑑, although the “official” capacity rating will be based on 𝑑 = 24 in. Modeling

𝑦

𝐵 𝐶 48 in. 13 in. 𝐶𝑦

𝐷 64 in.

14 in.

𝐷𝑦 = (0.1) 𝐶𝑦

Governing Equations

∑

𝑀𝐶 = 0 ∶ ∑ 𝐹𝑦 = 0 ∶

Figure 2 Free body diagram.

ISTUDY

Computation

Using the FBD in Fig. 2, the equilibrium equations are 𝑃 (𝑑 + 13 in.) − (12,500 lb)(58 in.) + (0.1) 𝐶𝑦 (64 in.) = 0,

(1)

𝐶𝑦 + (0.1) 𝐶𝑦 − 𝑃 − 12,500 lb = 0.

(2)

Solving Eqs. (1) and (2) for 𝑃 provides ( 0.1 ) 64 in. 1.1 . 𝑃 = 12,500 lb ( 0.1 ) 𝑑 + 13 in. + 1.1 64 in. 58 in. −

∗ Some

research of industry practices may yield a better criterion for ensuring steerability.

(3)

ISTUDY

Section 5.5

Engineering Design

Discussion & Verification When 𝑑 = 24 in., we obtain 𝑃 = 15,230 lb from Eq. (3), which we will round down slightly to obtain the capacity rating for the forklift as

Capacity rating = 15,000 lb (load positioned 24 in. from the fork face).

(4)

Remarks • The capacity rating in Eq. (4) will be used for advertising and marketing the forklift to customers. It will also be prominently displayed on the forklift, along with other important information, so that the operator is sure to see it and understand its meaning. • If the forklift is used with 𝑑 = 48 in., the load that can safely be lifted, as given by Eq. (3), reduces to 9762 lb. This number will be rounded down and will be displayed on the machine so that the operator knows that the maximum load that can safely be lifted is about 9000 lb when the load is positioned at 48 in. from the fork face. • To complete our work, we will write a short report that details our assumptions, calculations, and conclusions.

349

350

Chapter 5

Equilibrium of Bodies

Design Problems

𝑑 𝑊 𝐴

𝐵

𝐶

800 N

𝐷

General Instructions. In all problems, write a brief technical report following the guidelines of Appendix A, where you summarize all pertinent information in a wellorganized fashion. It should be written using proper, simple English that is easy to read by another engineer. Where appropriate, sketches along with critical dimensions should be included. Discuss the objectives and constraints considered in your design, the process used to arrive at your final design, safety issues if appropriate, and so on. The main discussion should be typed, and figures, if needed, can be computer-drawn or neatly handdrawn. Include a neat copy of all supporting calculations in an appendix that you can refer to in the main discussion of your report. A length of a few pages, plus appendix, should be sufficient. Design Problem 5.1

𝐹 𝐸 400 mm 400 mm 300 mm 200 mm Figure DP5.1

ISTUDY

A hand cart for moving heavy loads in a warehouse is shown. If each axle (pair of wheels) can support a maximum of 10 kN and if the wheels are not allowed to lift off the pavement, determine the largest weight 𝑊 that may be supported for any position 0 ≤ 𝑑 ≤ 1.3 m.

Design Problem 5.2 Specify the weight 𝑊 , width 𝑤, and depth 𝑑 for the base of a fluorescent desk lamp. For this lamp, the shade and lightbulb assembly 𝐶 weighs 1.8 lb, and the movable arm 𝐴𝐵, including the attachments at 𝐴 and 𝐵, weighs 1.5 lb with center of gravity approximately midway between points 𝐴 and 𝐵. For the weight of the base 𝑊 that you select, the dimensions 𝑑 and 𝑤 should be such that the lamp is as tip-resistant as possible.

𝐵 1.5 lb 1.8 lb

𝐴

𝐶 𝑊

𝑤

𝑑 𝐵

1.5 in.

18 in.

𝛼

𝐶

𝑤 𝐴

𝛽

𝐴

𝑑

30◦ ≤ 𝛽 ≤ 60◦ 𝑊

𝐵

𝛼

−30◦ ≤ 𝛼 ≤ 30◦

𝑑 Figure DP5.2

𝐶

19 in.

ISTUDY

Section 5.5

351

Engineering Design

Design Problem 5.3 𝑇 = 6 lb

The electric fan shown has the following specifications: • The entire unit weighs 𝑊 = 15 lb with the center of gravity shown. • The fan produces a maximum thrust of 𝑇 = 6 lb.

𝑊 = 15 lb

• The height ℎ can be adjusted by the user between 24 in. and 48 in.

ℎ

• The fan can be rotated to any horizontal position desired by the user. Specify the dimension 𝑏 for a base having three equally spaced legs. In your work, assume that the weight of the legs is already included in the 15 lb weight of the unit. Your design should consider a reasonable degree of safety against overturning. Also make at least two suggestions for how the stability of the fan can be improved without increasing the weight 𝑊 or increasing the dimension 𝑏.

Design Problem 5.4 The chute of a concrete truck for delivering wet concrete to a construction site is shown. The length of the chute may be changed by adding or removing segments 𝐵𝐶 and 𝐶𝐷. Chute segments 𝐴𝐵, 𝐵𝐶, and 𝐶𝐷 each weigh 50 lb, and the maximum length of the chute is 144 in. The chute has semicircular shape with 8 in. inside radius, and the hydraulic cylinder 𝐺𝐻 is used to raise and lower the chute such that 0◦ ≤ 𝜃 ≤ 50◦ . Specify the force capacity of the hydraulic cylinder 𝐺𝐻.

concrete Peter Righteous/Alamy Stock Photo

48 in. 30 in.

8 in. 48 in.

𝐴 7 in.

48 in.

𝐵 chute 𝜃

𝐶

0◦ ≤ 𝜃 ≤ 50◦

𝐷 Figure DP5.4

24 in.

𝐻 𝐺

𝑏

Figure DP5.3

352

ISTUDY

Chapter 5

Equilibrium of Bodies Design Problem 5.5

A telescopic boom aerial lift is shown. The lift is designed to support one worker, plus miscellaneous tools and supplies that are likely to be used, and is to be operated on a hard, level surface. Portion 𝐵𝐶 of the boom is retractable and supports a work platform whose floor is always horizontal. Turret 𝐷 can rotate 360◦ on base 𝐸. The design of the lift is essentially complete, and weights of the various components are shown. The masses corresponding to these weights are Boom 𝐴𝐵:

𝑚𝐴𝐵 = 450 kg,

Boom 𝐵𝐶 plus platform at 𝐶:

𝑚𝐵𝐶 = 500 kg,

Turret 𝐷:

𝑚𝐷 = 1800 kg,

Base 𝐸:

𝑚𝐸 = 1400 kg.

In addition to these are the mass of the worker, tools, and supplies 𝑚𝐶 and the mass of a counterweight 𝑚𝐹 whose center of gravity is to be placed below point 𝐴. You are asked to do the following: • Specify the lifting mass rating 𝑚𝐶 . • Specify the mass of the counterweight 𝑚𝐹 . • Specify the force capacity of the hydraulic cylinder between points 𝐺 and 𝐻. 𝑊𝐶

𝑊𝐵𝐶

𝐶

8

1

𝑊𝐹

𝑑

𝑊𝐷

𝜃

𝐴

1.4

𝐴 0.8

1

𝐷 0.3

4

𝐵

4

𝐸

1.3 1.6

𝑊𝐴𝐵 𝑊𝐸

−15◦ ≤ 𝜃 ≤ 75◦ 0 ≤ 𝑑 ≤ 8m dimensions in meters

𝐻 𝐺

𝐷

1.5

1.3 Figure DP5.5

𝑊𝐸

𝐸

1.5

ISTUDY

Section 5.6

Chapter Review

353

5.6 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

Equations of equilibrium In three dimensions, the equations governing the static equilibrium of a body are Eq. (5.1), p. 272 ∑

𝐹⃗ = 0⃗ and

∑

⃗ ⃗ = 0, 𝑀 𝑃

where the summations include all forces and moments that are applied to the body. In scalar form, Eq. (5.1) is Eq. (5.2), p. 272 ∑ ∑ ∑

𝐹𝑥 = 0

∑

𝑀𝑃 𝑥 = 0

𝐹𝑦 = 0

∑

𝑀𝑃 𝑦 = 0

∑

𝑀𝑃 𝑧 = 0.

and

𝐹𝑧 = 0

In two dimensions, with 𝑥 and 𝑦 being the in-plane coordinates, the equations ∑ ∑ 𝐹𝑧 = 0, 𝑀𝑃 𝑥 = 0, and 𝑀𝑃 𝑦 = 0 in Eq. (5.2) are always satisfied, and the remaining equilibrium equations are (with subscript 𝑧 dropped from the moment equation)

∑

Eq. (5.3), p. 272 ∑

𝐹𝑥 = 0,

∑

𝐹𝑦 = 0,

and

∑

𝑀𝑃 = 0.

∑ ∑ If desired, the 𝐹𝑥 = 0 (and/or 𝐹𝑦 = 0) equation in Eq. (5.3) may be replaced by an additional moment equilibrium equation provided a suitable moment summation point is selected, as discussed in Section 5.2. Similar remarks apply to alternative equilibrium equations in three dimensions.

Springs

𝐿 = final length

A linear elastic spring, shown schematically in Fig. 5.28, produces an axial force 𝐹𝑠 that is proportional to its change of length 𝛿 according to

𝐿0 = initial length

initial

Eq. (5.13), p. 299 𝐹𝑠 = 𝑘𝛿 = 𝑘(𝐿 − 𝐿0 ), where 𝑘 is the spring stiffness (units: force/length), 𝛿 is the elongation of the spring from its unstretched length, 𝐿0 is the initial (unstretched) spring length, and 𝐿 is the final spring length. Because the force 𝐹𝑠 and the elongation 𝛿 in Eq. (5.13) are directed along an axis, or line, these springs are sometimes called axial springs to differentiate them from torsional springs. Springs were discussed extensively in Chapter 3, and key points were repeated in Section 5.3.

𝑘 final

𝛿 𝐹𝑠

Figure 5.28 A spring produces a force 𝐹𝑠 that is proportional to its elongation 𝛿. Such springs are sometimes called axial springs.

354

Chapter 5

Equilibrium of Bodies

Torsional springs 𝜃

𝑘𝑡

A linear elastic torsional spring, shown schematically in Fig. 5.29, produces a moment 𝑀𝑡 that is proportional to the relative rotation, or twist, 𝜃 according to Eq. (5.14), p. 300

𝑀𝑡 Figure 5.29 A torsional spring produces a moment 𝑀𝑡 that is proportional to its rotation 𝜃.

ISTUDY

𝑀𝑡 = 𝑘𝑡 𝜃, where 𝑘𝑡 is the spring stiffness (units: moment/radian).

Supports and fixity Fixity refers to an object’s ability to move in space as a rigid body. All objects fall into one of the following three categories: Complete fixity. A body with complete fixity has supports that are sufficient in number and arrangement so that the body is completely fixed in space and will undergo no motion (either translation or rotation) in any direction under the action of any possible set of forces. Partial fixity. A body with partial fixity has supports that will allow the body to undergo motion (translation and/or rotation) in one or more directions. Whether or not such motion occurs depends on the forces and/or moments that are applied and whether the body is initially in motion. No fixity. A body with no fixity has no supports and is completely free to move in space under the action of forces that are applied to it.

Static determinacy and indeterminacy An object is either statically determinate or statically indeterminate, as follows: Statically determinate body. For a statically determinate body, the equilibrium equations of statics are sufficient to determine all unknown forces and/or other unknowns that appear in the equilibrium equations. Statically indeterminate body. For a statically indeterminate body, the equilibrium equations of statics are not sufficient to determine all unknown forces and/or other unknowns. A simple rule of thumb to help ascertain whether an object is statically determinate or indeterminate is to compare the number of unknowns to the number of equilibrium equations, and we call this equation counting. With 𝑛 being the number of unknowns, the rule of thumb for a single body in two dimensions is Eq. (5.15), p. 301 If 𝑛 < 3

The body is statically determinate; it has partial fixity if 𝑛 = 1 or 2, and it has no fixity if 𝑛 = 0.

If 𝑛 = 3

The body is statically determinate if it has full fixity, or the body is statically indeterminate if it has partial fixity.

If 𝑛 > 3

The body is statically indeterminate, and it can have full fixity or partial fixity.

Successful use of equation counting requires good judgment on your part, which is why it is called a rule of thumb rather than a rule.

ISTUDY

Section 5.6

Two-force and three-force members If a body or structural member is subjected to forces at two points or three points only, as described below, then when the body is in equilibrium, the orientation of the forces supported by the body has special properties. These situations are defined as follows: Two-force member. A body subjected to forces at two points (no moment loading and no distributed forces such as weight) is called a two-force member. The special feature of a two-force member is that, when in equilibrium, the two forces have the same line of action and opposite directions. Three-force member. A body subjected to forces at three points (no moment loading and no distributed forces such as weight) is called a three-force member. The special feature of a three-force member is that, when in equilibrium, the lines of action of all three forces intersect at a common point. If the three forces are parallel (this is called a parallel force system), then their point of intersection can be thought of as being at infinity. If a body is not a two-force or three-force member, then we refer to it as either a zeroforce member, if it is subjected to no force at all (these are often encountered in trusses, discussed in Chapter 6), or a general multiforce member, if it is subjected to forces at more than three points and/or has moment loading and/or has distributed loading.

Engineering design Engineering design was discussed in Section 5.5, and a process for developing a design was described (see Fig. 5.27 on p. 345). Codes and standards were described, and an overview of some professional and government organizations that sponsor codes and standards and/or have regulatory power over performance and safety were reviewed.

Chapter Review

355

356

Chapter 5

Equilibrium of Bodies

Review Problems 10 kN

8 kN

Problem 5.124

𝐶

15 kN 𝐷

10 m

20 kN 7m 15 kN 𝐸

𝐵 optional guy wire

20 kN

The transmission tower 𝐴𝐵𝐶 shown supports forces from three electrical wires at points 𝐶, 𝐷, and 𝐸. The tower is supported at point 𝐴 by a concrete foundation, and this foundation is normally adequate to support the tower without the guy wire 𝐵𝐹 . For towers that are sited in poor soil, or when the horizontal forces from the electrical wires are high due to a sharp turn, an optional guy wire 𝐵𝐹 may also be used to help support the tower. Neglect the weight of the tower. (a) If guy wire 𝐵𝐹 is not used, determine the support reactions at 𝐴.

8m 22 m

20 m

(c) If guy wire 𝐵𝐹 is used, determine the smallest force it must support so that the moment reaction at 𝐴 is no larger than 500 kN⋅m.

𝐴

𝐹 9m

Problem 5.125

Figure P5.124

The propellers of the twin engine airplane shown rotate in the same direction, and each propeller exerts a moment 𝑀 = 1.3 kN⋅m on the wings of the plane. To equilibrate this moment, trim tabs on the vertical stabilizer are used to produce trim forces 𝑇 . Determine the value of 𝑇 , assuming the trim forces are vertical.

0.8 m 𝑊 𝑇

𝑇

𝑀

Problem 5.126 𝑀

In Prob. 5.125, in what direction do the propellers rotate? Specify clockwise or counterclockwise with respect to the view shown in Fig. P5.125, and explain your reasoning. Note: Concept problems are about explanations, not computations.

1m 3.2 m

(b) If guy wire 𝐵𝐹 is used, determine the force it must support so that the moment reaction at 𝐴 is zero.

𝐿 3.2 m

Problem 5.127

Figure P5.125 and P5.126

A bracket is supported by a loose-fitting pair of rollers at points 𝐴 and 𝐵, and another loose-fitting pair at 𝐶 and 𝐷, and a frictionless pin at 𝐹 . The forces at 𝐸 and 𝐺 are horizontal and vertical, respectively. Determine the reactions at the pin and each of the four rollers. 14 in.

80 N

12 in.

15 in. 𝐶

200

𝑂

11 in.

𝐷 𝐴 𝐶 30◦

70

𝐺

𝐹 8 in.

20 lb

80

30◦ 𝐵 30◦

Figure P5.127 dimensions in mm

𝐴 35 𝐷 𝐸 Figure P5.128

ISTUDY

15 lb

30◦

20

16 in.

𝐵 𝐸

Problem 5.128 A semicircular geared bracket is subjected to a vertical 80 N force at point 𝐶. The bracket is supported by frictionless pins at 𝐴 and 𝐵 and a gear at 𝐷. The pins and gear are fixed to plate 𝐸, and the gear at 𝐷 is not allowed to rotate. Determine the tangential force supported by the gear at 𝐷 and the reactions at pins 𝐴 and 𝐵.

ISTUDY

Section 5.6

5 cm

Problem 5.129 A frame supports three frictionless pulleys that guide a tape that runs at constant speed. Determine the reactions at support 𝐴 if 𝛼 = 𝛽 = 0◦ .

357

Chapter Review

𝐸

6 cm 2 cm 𝐷

6 cm 5 cm 2 cm 𝐵

𝛼

𝛽 45◦

Problem 5.130

45◦ 𝐶 2 cm

14 N

Repeat Prob. 5.129 if 𝛼 = 30◦ and 𝛽 = 0◦ .

𝐴

14 N

Figure P5.129–P5.131

Problem 5.131 Repeat Prob. 5.129 if 𝛼 = 30◦ and 𝛽 = 45◦ .

Problem 5.132 A horizontal wooden beam is used to display iron pots and pans in a store. The wooden beam is suspended from the ceiling by one cable whose ends are pinned at points 𝐴 and 𝐼, and which wraps around seven frictionless pulleys. The beam is also supported by a vertical bar 𝐽 𝐾. Points 𝐵, 𝐷, 𝐹 , 𝐻, and 𝐼 have equal horizontal spacing, and each of these has a pot or pan hanging immediately below it. The straight cable segments are oriented at 30◦ from the vertical (as shown for segment 𝐴𝐵). The wooden beam weighs 60 lb with center of gravity immediately below the bearing of pulley 𝐹 . Determine the forces supported by the cable and bar 𝐽 𝐾.

Problem 5.133 The ramp shown is used on the back of a trailer for loading and unloading equipment. The lowered position is shown by points 𝐴, 𝐵, 𝐶, and 𝐷, and the raised position is shown by points 𝐴, 𝐵 ′ , 𝐶 ′ , and 𝐷 ′ . The ramp has 800 N weight with center of gravity at point 𝐶, and in the raised position, point 𝐶 ′ is immediately above point 𝐴. To raise the ramp, the user applies a vertical force 𝑃 at handle 𝐷. To lower the ramp, the user applies a horizontal force 𝑄 to handle 𝐵 ′ . To make raising and lowering the ramp easier, there is a torsional spring at pin 𝐴. Determine the stiffness 𝑘𝑡 of the torsional spring and the pretwist 𝜃0 it must have when in the raised position so that 𝑃 = 100 N and 𝑄 = 60 N.

𝐴

𝐶

𝐸

𝐺

𝐼 𝐾

30◦

𝐵

20 lb

𝐹

𝐷

20 lb

𝐻

10 lb

10 lb

10 lb

Figure P5.132

𝐷

𝐶 𝑄 𝐸

𝐵 0.8 m 𝐴 25◦ 𝐵 1.4 m

𝑃

𝐶

𝐷

Problem 5.134 A momentary-on electrical toggle switch is shown. To operate the switch, the user applies a horizontal force 𝑄. The switch has a torsional spring that causes lever 𝐴𝐵 to make contact at point 𝐶 when 𝑄 = 0 (this is the off position). When 𝑄 is sufficiently large, lever 𝐴𝐵 moves to position 𝐴′ 𝐵 and contact is made at point 𝐷 (this is the on position). Determine the stiffness 𝑘𝑡 of the spring and the pretwist 𝜃0 it must have so that 𝑄 = 3 lb will cause contact to be broken at point 𝐶, and 𝑄 = 4 lb will cause contact to be made at point 𝐷. Note: The pretwist 𝜃0 is the amount the torsional spring is twisted when the lever is on the off position with contact at 𝐶.

1.2 m Figure P5.133

𝑄

𝐴

𝛼

Problem 5.135

𝐴 30◦

1.25 in.

For the toggle switch described in Prob. 5.134, let the stiffness and pretwist of the torsional spring be 𝑘𝑡 = 0.6 in.⋅lb∕rad and 𝜃0 = 10 rad. Determine the force 𝑄 as a function of angle 𝛼 over the full range of motion for lever 𝐴𝐵 (i.e., 0◦ ≤ 𝛼 ≤ 60◦ ). Plot 𝑄 versus 𝛼, and determine the value of 𝑄 needed to break contact at point 𝐶, and the value of 𝑄 needed to make contact at point 𝐷.

𝐽

𝐶

𝐷 𝐵

Figure P5.134 and P5.135

358

Chapter 5

Equilibrium of Bodies

Problem 5.136 𝑘𝑡

𝐴

𝐺 35 mm

𝜃

𝐹 𝐵 𝐶

55 mm 𝑘𝑎 𝐸

𝐷

30 mm

The trigger of a high-pressure washer gun is shown. The torsional spring at point 𝐴 has stiffness 𝑘𝑡 = 1100 N⋅mm∕rad and is prewound by 6◦ when it is installed (i.e., when the trigger makes contact at point 𝐺). The axial spring 𝐷𝐸 has stiffness 𝑘𝑎 = 0.1 N∕mm and has 40 mm unstretched length. The trigger operates the washer on low pressure when 0◦ < 𝜃 ≤ 15◦ , and when 15◦ < 𝜃 ≤ 25◦ , the trigger operates the washer on high pressure. Assume force 𝐹 remains horizontal with the same line of action for all trigger positions. (a) Determine the force 𝐹 that causes the trigger to begin movement. (b) Determine the force 𝐹 that causes the trigger to first make contact with the plunger at 𝐶 (𝜃 = 15◦ ). (c) Determine the force 𝐹 required to fully pull the trigger (𝜃 = 25◦ ). Assume the plunger at 𝐶 contacts the back of the trigger at a right angle when 𝜃 = 25◦ .

Figure P5.136

Problem 5.137 A truck pulling a trailer with a tractor on it is shown in Fig. P5.137(a). The trailer’s mass is 500 kg and the tractor’s mass is 900 kg, with centers of gravity at points 𝐸 and 𝐺, respectively. A model for the trailer is shown in Fig. P5.137(b) where point 𝐴 is idealized as a pin and the tires are modeled by vertical springs with stiffness 𝑘 = 300 kN∕m. Assuming the trailer is in static equilibrium, determine the vertical deflection of point 𝐷 due to the weight of the trailer and tractor. Also determine the reactions at 𝐴 and the force supported by each spring. 1.1 m 1 m 0.6 m 𝐺

𝐺 𝐷

𝐴 𝐸 2.8 m

𝛿𝐷

𝐷

𝐴 𝐸

𝐵 𝐶 1.2 m 0.6 m 0.7 m

𝑘

𝑘

𝐵

𝐶

𝛿𝐷

(b)

(a) Figure P5.137

Problem 5.138 𝐺

Beam 𝐴𝐵𝐶𝐷 is supported by a vertical bar 𝐴𝐸, a pin at point 𝐵, and a vertical cable 𝐶𝐺. The weights of all members are negligible.

𝐶

(a) If the value of 𝑃 is known, is beam 𝐴𝐵𝐶𝐷 statically determinate or statically indeterminate? Explain.

𝐸 𝐵

𝐴 24 in. Figure P5.138

ISTUDY

24 in.

24 in.

𝐷

30◦ 𝑃

(b) Let bar 𝐴𝐸 and cable 𝐶𝐺 be modeled by springs with stiffnesses 𝑘𝐴𝐸 = 10,000 lb∕in. and 𝑘𝐶𝐺 = 14,000 lb∕in., respectively. If 𝑃 = 2000 lb, determine the forces supported by the bar and cable, and the reactions at point 𝐵.

ISTUDY

Section 5.6

359

Chapter Review

𝐹𝐷

Problem 5.139

𝐷

Member 𝐴𝐵𝐶𝐷 has negligible weight. (a) If member 𝐴𝐵𝐶𝐷 is to be a two-force member, which (if any) of 𝐹𝐵 , 𝐹𝐷 , and 𝑀𝐶 must be zero? (b) If member 𝐴𝐵𝐶𝐷 is to be a three-force member, which (if any) of 𝐹𝐵 , 𝐹𝐷 , and 𝑀𝐶 must be zero? (c) If 𝐹𝐷 = 0, 𝑀𝐶 = 0, and 𝐹𝐵 > 0, draw the FBD for member 𝐴𝐵𝐶𝐷 and sketch the ∑ ⃗ force polygon corresponding to 𝐹⃗ = 0.

Problems 5.140 and 5.141 Draw the FBD for each object shown, and specify whether it has partial fixity or full fixity and whether it is statically determinate or statically indeterminate. Assume that cables are in tension.

+𝐶

+

+

(a)

(b)

(c)

+

+

+

(d)

(e)

(f)

Figure P5.140

(a)

(b)

(c)

(d)

(e)

(f)

Figure P5.141

𝐸

30◦ 𝑀𝐶

ℎ 𝐵

𝐶

𝐴 𝐹𝐵

ℎ Figure P5.139

ℎ

360

Chapter 5

Equilibrium of Bodies

Problem 5.142 Boom 𝑂𝐴𝐵𝐶 is supported by a socket at 𝑂, cable 𝐸𝐴𝐵𝐹 that passes through small frictionless loops at 𝐴 and 𝐵, and a cable at 𝐶 that supports a force 𝑇1 and whose line of action is directed toward 𝐷. The distances between points 𝑂 and 𝐴, 𝐴 and 𝐵, and 𝐵 and 𝐶 are equal. (a) If 𝑇1 = 0, qualitatively describe the equilibrium position of the boom. (b) For the static equilibrium position shown, determine the value of 𝑇1 , the force 𝑇2 supported by cable 𝐸𝐴𝐵𝐹 , and the reactions at 𝑂. Hint: A numerical solution of the equilibrium equations is recommended. 𝑧 12 in. 𝐸

10 in.

𝐹

𝐷 12 in. 𝑂 𝐴

12 in.

𝑧

𝑇1

𝑦 𝐵

𝐹 𝑥

𝐶 (9, 12, 0) in.

8m

𝑊 = 30 lb 𝐴

Figure P5.142 𝐵

8m 𝑥

Problem 5.143 𝐸 𝐷

6m

𝐶 12 m

𝐺

𝑦

𝑚𝐺

Figure P5.143

Frame 𝐴𝐵𝐶𝐷𝐸 consists of four bars welded together to form a single object. Bar 𝐶𝐷 is parallel to the 𝑥 axis, bars 𝐴𝐵 and 𝐷𝐸 are parallel to the 𝑦 axis, and bar 𝐵𝐶 is parallel to the 𝑧 axis. The frame is supported by a ball and socket at point 𝐴, a frictionless surface at 𝐸 that lies in the 𝑥𝑧 plane, cable 𝐵𝐹 , and cable 𝐸𝐹 . The frame has negligible weight and supports an object 𝐺 with mass 𝑚𝐺 . If the cables have the allowable loads listed below, and assuming the other members are sufficiently strong, determine the largest value 𝑚𝐺 may have.

5 in. 𝑦

22 in. 15 in.

𝐸

𝐴

𝑧 20 in.

𝑃 𝐵

12 in. 𝑥

𝐷

150 lb Figure P5.144

Member

Allowable load

𝐵𝐹 𝐸𝐹

8 kN 12 kN

𝐶

Problem 5.144

250 lb

ISTUDY

𝐶

The landing gear for an airplane is shown just prior to landing. Frame 𝐴𝐵𝐶𝐷𝐸 lies in the 𝑦𝑧 plane. The 150 lb weight of the wheel and frame acts at point 𝐷 in the −𝑦 direction, and the 250 lb force due to drag acts at point 𝐷 in the −𝑥 direction. The frame is supported by a self-aligning thrust bearing at 𝐴, and a self-aligning bearing at 𝐵. A torsional spring is installed at 𝐶; it has 20 in.⋅lb∕degree stiffness, and the moment it produces is zero when the landing gear is fully raised (i.e., frame 𝐴𝐵𝐶𝐷𝐸 rotates about the positive 𝑧 axis until it lies in the 𝑥𝑧 plane). The force 𝑃 applied at point 𝐸 has 𝑥, 𝑦, and 𝑧 direction cosines of −3∕5, 0, and 4∕5, respectively. Assuming the landing gear is in static equilibrium, determine the value of 𝑃 and all support reactions.

ISTUDY

Section 5.6

361

Chapter Review

Problem 5.145 A machine for sawing concrete is shown. It is supported by a cutting disk at point 𝐶 and two wheels at points 𝐴 and 𝐵 (the wheel at 𝐵 is not shown). The wheels at 𝐴 and 𝐵 are separated by a 0.8 m distance along the 𝑥 axis. Determine the dimension 𝑑 where the cutting disk should be located so that the force supported by wheel 𝐴 is 20% of the force supported by wheel 𝐵. 𝑦 0.35 m

0.4 m

𝑊

𝐴 𝑥 𝑑

0.8 m 𝐶

𝑧

𝑧

Figure P5.145

Problem 5.146

100 lb 8 in.

A structure consists of a thin flat plate and two short bars with bearing supports at 𝐴 and 𝐵, where the bearing at 𝐵 is self-aligning. The plate is loaded at its center by a 100 lb vertical force and by a 40 lb force in the 𝑥 direction at one of the corners. (a) Does the plate have complete fixity or partial fixity, and is it statically determinate or statically indeterminate? Explain. (b) Determine all reactions at 𝐴 and 𝐵.

𝑥 Figure P5.146

Problem 5.147 Bar 𝐴𝐵𝐶𝐷 is supported by a cable 𝐴𝐸𝐷, which passes over a frictionless pulley at point 𝐸, and a collar 𝐵 that slides without friction on a vertical shaft with square cross section. If the tip 𝐴 is subjected to a 5 kN vertical force, determine the tension in the cable and all support reactions at collar 𝐵. 𝑦 𝐸

6m

𝐶

3m 𝐷

𝐵

2m

𝐴 8m

𝑧

𝐴

5 kN Figure P5.147

𝑥

𝑦 𝐵

8 in. 2 in.

2 in.

40 lb

ISTUDY

Chapter 5

Equilibrium of Bodies

Problems 5.148 through 5.151 The FBDs for two objects are shown. The forces and moments act at the points indicated, in the directions indicated, and have positive values; i.e., 𝐹1 > 0, 𝐹2 > 0, ... , 𝑀1 > 0, and so on. Assuming the forces and moments have the proper values, select one of the choices below to indicate whether these bodies could be or could never be in equilibrium. If a body could never be in equilibrium, explain why. (a) Object 1 could be in equilibrium; Object 2 could be in equilibrium. (b) Object 1 could never be in equilibrium; Object 2 could never be in equilibrium. (c) Object 1 could be in equilibrium; Object 2 could never be in equilibrium. (d) Object 1 could never be in equilibrium; Object 2 could be in equilibrium. (e) Cannot determine without additional information. Note: Concept problems are about explanations, not computations. 𝐹3

𝐹3

𝐹4

𝐹2

𝑀1

𝐹6 𝐹4

𝐹1

.S t e if

362

𝑀1

P. S

𝐹1

𝐹2

𝐹5 Object 2

Object 1

Figure P5.148

Figure P5.149 𝐹5

𝐹3

𝐹1

𝐹4

𝑀2 𝐹3

𝑀1 𝐹5

𝐹2

Object 1

𝐹1 𝐹2

𝐹6

𝑀1

𝐹5 Object 2

Object 1

𝐹4

𝐹7 Object 2

Figure P5.150

Object 1

Object 2

Figure P5.151

ISTUDY

Structural Analysis and Machines

6

Structures and machines often consist of an assemblage of many members. In this chapter, we analyze the equilibrium of these. One of the goals is to determine the forces supported by the individual members of a structure or machine. Design considerations, especially for trusses, are also discussed.

David Dawson Image/Getty Images

The Prince Felipe Science Museum at the City of Arts and Sciences, in Valencia, Spain. The building is supported by an arrangement of trusses that surround its exterior.

Structures and machines Before we go into details, we should take a moment to consider the meaning of structure and machine. We use the word structure to describe an arrangement of material and/or individual members that, as a whole, is intended to support forces that are applied to it. The word machine describes an arrangement of material and/or individual members where the goal is usually to transmit motion and/or force. These definitions are broad and have considerable overlap, and often a particular device could appropriately be called either a structure or a machine. Although there are frequent exceptions, a structure will often use a stationary arrangement of members (that is, the individual members will have little or no motion relative to one another), while a machine will have members with significant relative motion. Structures are further categorized as trusses or frames. The differences between these, and methods of analysis, are discussed in detail.

363

364

ISTUDY

Chapter 6

Structural Analysis and Machines

6.1

Truss Structures and the Method of Joints

A truss is a structure that consists of two-force members only, where members are organized so that the assemblage as a whole behaves as a single object. Some examples of truss structures are shown in Fig. 6.1. Observe that trusses use very little material, yet they are very strong.

Michael Plesha

David R. Frazier Photolibrary, Inc./Alamy Stock Photo

(a)

(b)

Figure 6.1. Examples of truss structures. (a) The roof of this highway maintenance building under construction uses 124 identical wooden roof trusses with 16 in. spacing. (b) Mores Creek bridge over Lucky Peak Reservoir near Boise, Idaho.

Throughout this section and the next, we discuss truss structures in two dimensions, and these are called plane trusses. For a two-dimensional structure to be a plane truss, it must have the following characteristics: • All members must be connected to one another by frictionless pins, and the locations of these pins are called joints. • Each member may have no more than two joints. • Forces may be applied at joints only. • The weight of individual members must be negligible. If all of the above characteristics are satisfied, then it is guaranteed that all members of the structure are two-force members. The important consequence of having only two-force members is that equilibrium analysis of the structure reduces to equilibrium analysis of a system of particles where the number of particles equals the number of joints in the truss. Some common types of truss structures are shown in Fig. 6.2.

Pratt

Pratt

Warren

Howe

Howe

K

typical roof trusses

typical bridge trusses

Figure 6.2. Some common types of truss structures.

ISTUDY

Section 6.1

Truss Structures and the Method of Joints

365

Recall from Chapter 5 that a two-force member does not need to be straight. While trusses are most often constructed using straight members, they may contain members with curved or other complex shapes, and Fig. 6.3 shows such an example.

When may a structure be idealized as a truss? Relatively few real life structures fully conform to the definition of a truss, the main reason usually being that connections between members are not pins. Figure 6.4 shows some examples of typical connections in structures. These connections are capable of supporting moments, and thus the members that emanate from such connections usually are not two-force members. For precise analysis, structures with connections such as these should be modeled as frame structures, which are discussed in Section 6.4. If a structure is sufficiently stiff, then the moments at connections such as those shown in Fig. 6.4 may be small, and it may be reasonable to assume they are negligible, in which case the structure could be modeled as a truss. To summarize, if a particular structure would qualify as a truss except for the connection details, then in engineering practice these structures are often modeled as trusses anyway, and the methods of analysis discussed in this section and the next are used. This idealization is not perfect, but it is widely used, and with good judgment and generous factors of safety, it is very successful.

𝐴 𝐵 Michael Plesha

Michael Plesha

(a)

(b)

Figure 6.4. Typical connections between members of a structure. (a) In this wooden roof truss, a rectangular nail plate is used to connect members. (b) In this steel truss, connection 𝐴 is welded while connection 𝐵 is pinned.

Another common departure from the definition of a truss is that structures often have forces applied at locations other than joints. In Section 6.2, a method is discussed where such loads are replaced by equivalent loads positioned at joints so that truss analysis can be performed.

Method of joints In general, the objectives in analyzing a truss are to determine the support reactions for the truss and the forces supported by the individual members of the truss. In the method of joints, a truss is analyzed by treating each joint as a particle. If the entire truss is in equilibrium, then every joint within the truss is also in equilibrium (indeed, all material within the truss is in equilibrium). Analysis proceeds by drawing FBDs of the joints throughout the truss, writing equilibrium equations for each joint, and solving the equilibrium equations for the unknowns. Prior to this, it may be necessary or desirable to determine the reactions for the truss as a whole, and the methods of

NI QIN/Getty Images

Figure 6.3 While most truss structures are composed entirely of straight members, this truss uses a curved upper chord.

366

Chapter 6

Structural Analysis and Machines

Chapter 5 can be used for this. Analysis by the method of joints is illustrated in the following mini-example. 𝐵

𝐷 3 kN 4m 𝐶

𝐴

3m

3m

𝐸

10 kN 3m

Solution We begin by obtaining the support reactions for the truss by drawing the FBD of the structure as a whole as shown in Fig. 6.6. Writing equilibrium equations provides ∑ 𝑀𝐴 = 0 ∶ −(10 kN)(6 m) + 𝐸𝑦 (12 m)

3m

Figure 6.5 A plane truss.

3 kN

𝑦 𝑥

4m 𝐴𝑥

𝐴𝑦

𝐸𝑦

10 kN 6m

6m

Figure 6.6 Free body diagram to determine the support reactions.

𝑦

𝐵 𝛼

𝑥

𝛼

𝛼

3 kN 6 kN

𝐷 𝛼

𝐹𝐴𝐶

3 kN 𝛼 𝐹𝐷𝐸

𝐹𝐶𝐷 𝐶 𝛼 𝛼 𝛼 𝐹𝐶𝐸

𝐹𝐵𝐶

𝐹𝐴𝐵 𝐴

𝐹𝐵𝐷

10 kN

𝐸 4 kN

Figure 6.7 Free body diagrams of joints (pins) in the truss. All member forces are defined to be positive in tension. For example, a positive value of 𝐹𝐴𝐵 means that member 𝐴𝐵 is in tension.

ISTUDY

Mini-Example Use the method of joints to determine the force supported by each member of the truss shown in Fig. 6.5.

∑ ∑

+ (3 kN)(4 m) = 0

⇒

𝐸𝑦 = 4 kN,

(6.1)

𝐹𝑦 = 0 ∶

𝐴𝑦 + 𝐸𝑦 − 10 kN = 0

⇒

𝐴𝑦 = 6 kN,

(6.2)

𝐹𝑥 = 0 ∶

𝐴𝑥 − 3 kN = 0

⇒

𝐴𝑥 = 3 kN.

(6.3)

Next, we draw FBDs of each joint (pin) as shown in Fig. 6.7. Since each joint has a concurrent force system, each joint is treated as a particle in equilibrium, and hence each joint permits two equilibrium equations to be written. Among the many joints that the truss has, a good strategy is to always (if possible) select a joint that has no more than two unknowns, since then the unknowns may be immediately solved for. Among the five possible joints shown in Fig. 6.7, joints 𝐴 and 𝐸 are the only two having only two unknowns. Noting that 𝛼 = tan−1 (4∕3) = 53.1◦ , we select joint 𝐴 and write∗ Joint A: ∑ ⇒ 𝐹𝐴𝐵 = −7.5 kN, (6.4) 𝐹𝑦 = 0 ∶ 6 kN + 𝐹𝐴𝐵 sin 𝛼 = 0 ∑ 𝐹𝑥 = 0 ∶ 3 kN + 𝐹𝐴𝐵 cos 𝛼 + 𝐹𝐴𝐶 = 0 ⇒ 𝐹𝐴𝐶 = 1.5 kN. (6.5) We may now repeat this procedure for joint 𝐸, but we will continue with joint 𝐵 instead, as now it has only two unknowns. Using the FBD for joint 𝐵 gives Joint B: ∑ (6.6) 𝐹𝑦 = 0 ∶ −𝐹𝐴𝐵 sin 𝛼 − 𝐹𝐵𝐶 sin 𝛼 = 0 ⇒ 𝐹𝐵𝐶 = 7.5 kN, ∑ 𝐹𝑥 = 0 ∶ −𝐹𝐴𝐵 cos 𝛼 + 𝐹𝐵𝐶 cos 𝛼 + 𝐹𝐵𝐷 = 0 ⇒

𝐹𝐵𝐷 = −9 kN.

(6.7)

Joints 𝐶, 𝐷, and 𝐸 remain to be analyzed, and of these, joints 𝐷 and 𝐸 both have two unknowns. Furthermore, examination of the FBDs for joints 𝐷 and 𝐸 shows that 𝐸 will entail slightly less algebra to solve for its unknowns, and thus, we elect to write equilibrium equations for it Joint E: ∑ ⇒ 𝐹𝐷𝐸 = −5 kN, (6.8) 𝐹𝑦 = 0 ∶ 4 kN + 𝐹𝐷𝐸 sin 𝛼 = 0 ∑ 𝐹𝑥 = 0 ∶ −𝐹𝐶𝐸 − 𝐹𝐷𝐸 cos 𝛼 = 0 ⇒ 𝐹𝐶𝐸 = 3 kN. (6.9) ∗ Because

of the nice geometry of this truss, we could avoid computing 𝛼 and simply use cos 𝛼 = 3∕5 and sin 𝛼 = 4∕5.

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Section 6.1

Truss Structures and the Method of Joints

367

There is now only one unknown remaining, namely, 𝐹𝐶𝐷 , and either joint 𝐶 or 𝐷 may be used to determine it. We select 𝐷 because its FBD has one less force than the FBD for joint 𝐶. Thus, Joint D: ∑ 𝐹𝑦 = 0 ∶ −𝐹𝐶𝐷 sin 𝛼 − 𝐹𝐷𝐸 sin 𝛼 = 0 ⇒ 𝐹𝐶𝐷 = 5 kN. (6.10) Remarks • This solution began with finding the support reactions for the entire structure. Depending on the geometry of the structure and how it is supported, finding the reactions at the outset may not be necessary. In this example, it was necessary because otherwise all of the FBDs shown in Fig. 6.7 have three or more unknowns. Example 6.3 considers a structure where the support reactions are not needed prior to analysis by the method of joints.

Helpful Information Method of joints. The method of joints is really nothing more than the method of analysis discussed in Chapter 3, namely, equilibrium of a system of particles.

• In the FBD shown in Fig. 6.6, the truss is represented as a solid object to emphasize that for the analysis of support reactions, the arrangement of members within the structure is irrelevant provided it is sufficient to support the forces and reactions that are applied to it. When this assumption is not true, this phase of the solution must be revised, as discussed later. • In the FBDs of Fig. 6.7, all of the member forces are defined so that positive values correspond to tension. In the solutions obtained here, members 𝐴𝐶, 𝐵𝐶, 𝐶𝐸, and 𝐶𝐷 support positive forces, hence they are in tension; and members 𝐴𝐵, 𝐵𝐷, and 𝐷𝐸 support negative forces, hence they are in compression. • To obtain the solutions for all unknowns, only one equilibrium equation for joint 𝐷 was used, and neither of the two equilibrium equations for joint 𝐶 was used. To help check the accuracy of your solution, you should write these three equilibrium equations and verify that they are satisfied. If they are not, then an error has been made. • In view of the preceding comment, you might wonder why all of the unknowns could be determined without using all of the available equilibrium equations. Recall from Section 5.3 that having more equilibrium equations than unknowns usually indicates a mechanism. However, the truss of Fig. 6.5 is clearly not a mechanism: it is fully fixed and is statically determinate. Indeed, because the support reactions were found at the outset, the three equilibrium equations in question were exhausted in writing Eqs. (6.1)–(6.3), and in fact the number of unknowns and the number of available equilibrium equations are the same. For this reason, the three “extra” equilibrium equations for joints 𝐷 and 𝐸 were not needed to find the unknowns.

Zero-force members A truss member (or any member) that supports no force is called a zero-force member. Trusses often contain many zero-force members. While it might seem that such members play no role in strengthening a truss, in fact they may be very important, as discussed in Section 6.2. Here we discuss a method to identify zero-force members by inspection. Consider the situation shown in Fig. 6.8; joint 𝐴 connects three

Interesting Fact Computer analysis. Although not discussed in this book, the method of joints can be automated for solution using a computer. With knowledge of statics, mechanics of materials, and a basic understanding of matrices, developing such software is surprisingly straightforward. Computer programs for truss analysis see widespread use.

368

truss members, two of the members are collinear (𝐴𝐵 and 𝐴𝐷), and no external force is applied to joint 𝐴. With the 𝑥𝑦 coordinate system shown, summing forces in the 𝑦 direction provides ∑ 𝐹𝑦 = 0 ∶ 𝐹𝐴𝐶 sin 𝛼 = 0. (6.11)

𝐶 𝐹𝐴𝐶 𝐵

Chapter 6

Structural Analysis and Machines

𝐹𝐴𝐵

𝑦

𝛼

𝑥 𝐹𝐴𝐷

𝐴

𝐷

Figure 6.8 Geometry of members in a truss allowing a zeroforce member to be recognized by inspection.

Unless 𝛼 = 0◦ or 180◦ , in which case member 𝐴𝐶 is collinear with the other two members, 𝐹𝐴𝐶 = 0 and member 𝐴𝐶 is a zero-force member. Note that if joint 𝐴 has an external force applied to it, then if that force has any component in the 𝑦 direction shown in Fig. 6.8, then 𝐹𝐴𝐶 ≠ 0. Similarly, if joint 𝐴 has more than three members connected to it, then in general 𝐹𝐴𝐶 ≠ 0. Identification of zero-force members in a truss. Figure 6.8 defines a pattern that allows zero-force members in a truss to be easily identified by inspection. To use this, we examine the joints in a truss, and if a particular joint • Has three members connected to it, • Two of these members are collinear, and • The joint has no external force applied to it, then the noncollinear member is a zero-force member. Rather than memorize this list, it is easier to simply understand the rationale behind a zero-force member; namely, Fig. 6.8 and Eq. (6.11). Furthermore, understanding these features allows you to recognize other situations in which a zero-force member occurs. For example, if member 𝐴𝐷 in Fig. 6.8 is not present, then summing forces in the 𝑦 direction shows that member 𝐴𝐶 is zero-force, and with this result, summing forces in the 𝑥 direction shows that member 𝐴𝐵 is also a zero-force member.

𝐵

𝐷

Mini-Example Identify the zero-force members in the truss shown in Fig. 6.9.

𝐹 𝐼

𝐴

Solution By inspection of each joint, we identify the following zero-force members: 𝐻

𝐶

𝐸

𝐺 10 kN

Figure 6.9 Example of a truss containing several zero-force members.

ISTUDY

Examination of joint 𝐶 shows that member 𝐵𝐶 is zero-force. Examination of joint 𝐷 shows that member 𝐷𝐸 is zero-force. Examination of joint 𝐼 shows that member 𝐺𝐼 is zero-force. Because member 𝐺𝐼 is zero-force, examination of joint 𝐺 then shows that member 𝐹 𝐺 is zero-force. Remark. For a member such as 𝐷𝐸, the presence of the 10 kN force applied to joint 𝐸 is a common source of confusion, as intuition may suggest (wrongly) that member 𝐷𝐸 must participate in supporting the 10 kN force. Note that our conclusion that member 𝐷𝐸 is a zero-force member is based entirely on the conditions at joint 𝐷. Before closing, we note that most of our discussion focused on sufficient conditions for a member to be zero-force. That is, if an equilibrium equation for a particular joint can be written in the form of Eq. (6.11), and 𝛼 ≠ 0◦ or 180◦ , then this is a sufficient condition to conclude that a member is zero-force, and furthermore, it is possible to easily identify such members by inspection. However, it is not a necessary condition. That is, a truss may have other zero-force members. For example,

ISTUDY

Section 6.1

Truss Structures and the Method of Joints

369

in the truss shown in Fig. 6.9, if other forces were applied at joints 𝐵 and/or 𝐹 and they had proper magnitudes and directions, then it is possible that by chance other members, in addition to the four members already cited, could be zero-force members. However, it is typically not possible to identify such members by inspection; rather if any such members are present, they are seen to be zero-force only after the equilibrium equations have been solved. This situation arises in Example 6.3.

Typical truss members The vast majority of trusses are constructed entirely of straight, slender members. The reason is that straight members are very efficient at supporting axial forces. Because of this, straight members can usually be slender, meaning the length of a member is substantially greater than the size of its cross section. While a truss may contain members that are not straight, such as the curved upper chord shown in Fig. 6.3, such members are considerably less efficient at supporting forces because they also experience bending, and therefore substantially more material is required so that the member is strong enough. When curved members are used in a truss, it is almost always for nonstructural reasons such as aesthetics, ease of manufacture, and so on. Efficiency of straight members versus curved members for supporting loads is discussed in detail in Chapter 8, where internal forces and bending are treated (see Prob. 8.22 on p. 502). An additional, important factor controlling the strength of straight truss members is the possibility of buckling due to compressive forces. Buckling is briefly discussed in Section 6.2.

Helpful Information An experiment: straight versus nonstraight truss members. Take a metal or plastic clothes hanger and cut it at the two locations shown. Then, use your hands to apply forces to the non-straight portion. Clearly it is very flexible and will undergo significant bending. In contrast, use your hands to apply tensile forces to the straight portion and observe that no visible deformation is produced, even if you apply as much force as possible. The straight member is very efficient at supporting the forces applied to it, whereas the non-straight member is not as efficient because of bending.

End of Section Summary In this section, truss structures are defined and the method of joints is developed for determining the forces supported by individual members of a truss. Some of the key points are as follows: • A truss is a structure that consists of two-force members only, where members are organized so that the assemblage as a whole behaves as a single object. For all members of a structure to be two-force members, the structure must have the following characteristics: – All members must be connected to one another by frictionless pins, and the locations of these pins are called joints. – Each member may have no more than two joints. – Forces may be applied at joints only. – The weight of individual members must be negligible. Real structures usually do not satisfy all of these requirements. Nonetheless, many real structures are idealized as trusses. • In the method of joints, the force supported by each member of a truss is determined by drawing FBDs for each joint and then requiring that all joints be in ∑ ∑ equilibrium by writing the equations 𝐹𝑥 = 0 and 𝐹𝑦 = 0 for each joint. • A truss member (or any member) that supports no force is called a zero-force member. Trusses often contain many zero-force members, and the role played by these in strengthening a truss is discussed in Section 6.2.

non-straight two-force member

straight two-force member

370

E X A M P L E 6.1

Truss Analysis by the Method of Joints The structure shown consists of 13 members, where the hollow circles indicate joints. Determine the force supported by each member of the structure.

𝐷 𝐵 𝐴

30◦

2m

𝐹 𝐸

𝐶 4 kN 2m

𝐻

𝐺

𝑦

In this structure, we will neglect the weights of individual members under the assumption they are small compared to the 4 kN and 6 kN forces. Because forces are applied at joints only and all members are connected to one another by frictionless pins, all members of the structure are two-force members and hence, the structure is a truss. To analyze this truss, we begin by determining the support reactions, followed by use of the method of joints to determine the force supported by each member.

2m

𝐷 𝑥

𝐴

30◦ 𝐴𝑦 2m

𝐵

Modeling To determine the reactions for the truss as a whole, we draw the FBD for the entire structure as shown in Fig. 2.

𝐹

𝐶 4 kN 2m

𝐸 6 kN 2m

𝐻

𝐺 𝐻𝑦 2m

Figure 2 Free body diagram for determining the support reactions.

ISTUDY

SOLUTION Road Map

6 kN 2m

Figure 1

𝐴𝑥

Chapter 6

Structural Analysis and Machines

Governing Equations & Computation

Using the FBD shown in Fig. 2, we write solve the following equilibrium equations: ∑ 𝑀𝐴 = 0 ∶ −(4 kN)(2 m) − (6 kN)(4 m) + 𝐻𝑦 (8 m) = 0 ⇒ 𝐻𝑦 = 4 kN, ∑ 𝐹𝑦 = 0 ∶ 𝐴𝑦 + 𝐻𝑦 − 4 kN − 6 kN = 0 ⇒ 𝐴𝑦 = 6 kN, ∑ 𝐹𝑥 = 0 ∶ 𝐴𝑥 = 0 ⇒ 𝐴𝑥 = 0.

and

(1) (2) (3)

Modeling

Now that the support reactions are known, we proceed with the method of joints. Examination of joint 𝐺 shows that member 𝐹 𝐺 is zero-force, and with this knowledge, examination of joint 𝐹 then shows that member 𝐸𝐹 is zero-force. Recognizing zero-force members at the outset of an analysis may save you some work, but in the following analysis we will use this knowledge as a partial check of our solution. To use the method of joints, we draw FBDs of all joints in the truss, as shown in Fig. 3. In these FBDs, all member forces are taken to be positive in tension. For example, if we find that 𝐹𝐴𝐵 is positive, then member 𝐴𝐵 is in tension, whereas if we find that 𝐹𝐴𝐵 is negative, then member 𝐴𝐵 is in compression. Governing Equations & Computation

We begin writing equilibrium equations at joint 𝐴, because there are only two unknowns there (joint 𝐻 is an equally good choice). Following joint 𝐴, we proceed to joint 𝐶, as there are then only two unknowns there, and so on. 𝐷

𝑦

𝛼 𝐵

𝛼 𝛼

𝛼 𝐴

𝛼 6 kN

𝐹𝐴𝐶

𝐹𝐷𝐹

𝐹𝐵𝐷

𝑥

𝐹𝐴𝐵

𝛼

𝐹𝐵𝐶 𝐶 4 kN

𝛼 𝛼

𝐹𝐷𝐸 𝐹𝐸𝐹

𝐹𝐵𝐸 𝐹𝐶𝐸

𝛼

𝐸

𝛼 = 30◦

𝛼

𝐹𝐸𝐺

𝐹 𝛼 𝐹𝐹 𝐺 𝐺

𝐹𝐹 𝐻 𝐹𝐺𝐻

6 kN

Figure 3. Free body diagrams of joints where 𝛼 = 30◦ .

𝛼 4 kN

𝐻

ISTUDY

Section 6.1

Truss Structures and the Method of Joints

Joint A: ∑

𝐹𝑦 = 0 ∶

𝐹𝐴𝐵 sin 30◦ + 6 kN = 0

⇒

𝐹𝐴𝐵 = −12 kN,

(4)

∑

𝐹𝑥 = 0 ∶

𝐹𝐴𝐶 + 𝐹𝐴𝐵 cos 30◦ = 0

⇒

𝐹𝐴𝐶 = 10.39 kN.

(5)

Joint C: ∑

𝐹𝑥 = 0 ∶

−𝐹𝐴𝐶 + 𝐹𝐶𝐸 = 0

⇒

𝐹𝐶𝐸 = 10.39 kN,

(6)

∑

𝐹𝑦 = 0 ∶

𝐹𝐵𝐶 − 4 kN = 0

⇒

𝐹𝐵𝐶 = 4 kN.

(7)

Joint B: ∑

𝐹𝑥 = 0 ∶

−𝐹𝐴𝐵 cos 30◦ + 𝐹𝐵𝐷 cos 30◦ + 𝐹𝐵𝐸 cos 30◦ = 0,

(8)

∑

𝐹𝑦 = 0 ∶

−𝐹𝐴𝐵 sin 30 + 𝐹𝐵𝐷 sin 30 − 𝐹𝐵𝐸 sin 30 − 𝐹𝐵𝐶 = 0

(9)

◦

⇒ Joint D: ∑ 𝐹𝑥 = 0 ∶ ∑ 𝐹𝑦 = 0 ∶

◦

◦

𝐹𝐵𝐷 = −8 kN and 𝐹𝐵𝐸 = −4 kN.

−𝐹𝐵𝐷 cos 30◦ + 𝐹𝐷𝐹 cos 30◦ = 0 ⇒ 𝐹𝐷𝐹 = −8 kN, −𝐹𝐵𝐷 sin 30◦ − 𝐹𝐷𝐸 − 𝐹𝐷𝐹 sin 30◦ = 0 ⇒ 𝐹𝐷𝐸 = 8 kN.

(10)

(11) (12)

Joint H: ∑

𝐹𝑦 = 0 ∶

𝐹𝐹 𝐻 sin 30◦ + 4 kN = 0

⇒

𝐹𝐹 𝐻 = −8 kN,

(13)

∑

𝐹𝑥 = 0 ∶

−𝐹𝐹 𝐻 cos 30◦ − 𝐹𝐺𝐻 = 0

⇒

𝐹𝐺𝐻 = 6.928 kN.

(14)

Joint G: ∑

𝐹𝑦 = 0 ∶

𝐹𝐹 𝐺 = 0

⇒

𝐹𝐹 𝐺 = 0,

(15)

∑

𝐹𝑥 = 0 ∶

−𝐹𝐸𝐺 + 𝐹𝐺𝐻 = 0

⇒

𝐹𝐸𝐺 = 6.928 kN.

(16)

Joint F: ∑

𝐹𝑥 = 0 ∶

−𝐹𝐷𝐹 cos 30◦ − 𝐹𝐸𝐹 cos 30◦ + 𝐹𝐹 𝐻 cos 30◦ = 0 ⇒

𝐹𝐸𝐹 = 0.

(17) (18)

Discussion & Verification

• Notice that we were able to determine all of the member forces without using one of the equilibrium equations for joint 𝐹 and the two equilibrium equations for joint 𝐸. As a check on our solution, you should write these three equations and verify that the member forces we determined satisfy all of them. • In Eqs. (15) and (18), two members were found to have zero force, and these are the same members that were identified by inspection at the outset of our analysis as being zero-force members. This is a useful partial check of solution accuracy. • This solution neglected the weight of the truss. An approximate way to include this weight is illustrated in Example 6.2.

371

372

Chapter 6

Structural Analysis and Machines

E X A M P L E 6.2

A Truss with Curved Members and Weights of Members The structure shown consists of five pin-connected members (members 𝐴𝐶 and 𝐵𝐷 do not intersect). In addition to the 4 kip force shown, the individual members of the structure have the weights listed in Table 1. Idealize the structure to be a truss and determine the forces supported by each member.

4 kip 𝐶

𝐵

𝐴

30◦

30◦

SOLUTION

𝐷

Road Map

Because the weights listed in Table 1 are distributed forces, the members of this structure are not two-force members, and hence this structure is not a truss. Nonetheless, because these weights are small compared to the 4 kip force applied at joint 𝐶, it is possible to approximate this structure as a truss, and to do so under these circumstances is common.

10 ft Figure 1 Table 1 Weights of individual members. Member

Weight

𝐴𝐵 𝐴𝐶 𝐵𝐶 𝐵𝐷 𝐶𝐷

𝑊𝐴𝐵 = 400 lb 𝑊𝐴𝐶 = 100 lb 𝑊𝐵𝐶 = 200 lb 𝑊𝐵𝐷 = 100 lb 𝑊𝐶𝐷 = 400 lb

𝑃𝐵

𝑦 𝑥

𝐵

Modeling

The weights (distributed forces) listed in Table 1 will be replaced with forces that are applied to joints by using the following procedure. For each member, that member’s weight is proportioned equally between the two joints it connects to. Thus, for member 𝐴𝐵, one-half of its weight will be applied to joint 𝐴 and one-half will be applied to joint 𝐵. Similarly, for member 𝐴𝐶, one-half of its weight will be applied to joint 𝐴 and one-half will be applied to joint 𝐶. Doing this for all members results in the following forces at joints 𝑃𝐴 = (𝑊𝐴𝐵 + 𝑊𝐴𝐶 )∕2 = 250 lb,

(1)

4 kip + 𝑃𝐶

𝑃𝐵 = (𝑊𝐴𝐵 + 𝑊𝐵𝐶 + 𝑊𝐵𝐷 )∕2 = 350 lb,

(2)

𝐶

𝑃𝐶 = (𝑊𝐴𝐶 + 𝑊𝐵𝐶 + 𝑊𝐶𝐷 )∕2 = 350 lb,

(3)

𝑃𝐷 = (𝑊𝐵𝐷 + 𝑊𝐶𝐷 )∕2 = 250 lb.

(4)

𝑃𝐷

𝑃𝐴

10 ft

Treatment of distributed forces in this fashion is sometimes called load lumping because a distributed force is being replaced by concentrated forces.∗ To determine the reactions for the truss as a whole, we draw the FBD shown in Fig. 2. Then, to use the method of joints to determine the force supported by each member, we draw the FBDs shown in Fig. 4, where member forces are taken to be positive in tension.

Figure 2 Free body diagram for determining the support reactions.

Governing Equations & Computation To find the reactions for the truss as a whole, we use the FBD shown in Fig. 2 and write the following equilibrium equations (with the

𝐴

30◦

𝐷

30◦

𝐴𝑦

𝐷𝑦

𝐷𝑥

4 kip + 𝑃𝐶 (10 f t) cos 30◦ = 8.660 f t

𝐴

𝐶

aid of Fig. 3 to determine moment arms): ∑ 𝑀𝐴 = 0 ∶ −𝑃𝐵 (2.5 f t) − (4000 lb + 𝑃𝐶 )(7.5 f t) + (𝐷𝑦 − 𝑃𝐷 )(10 f t) = 0 ∑

𝛼 = 60◦ 𝐷

30◦

∑

10 ft Figure 3 Geometry of member 𝐴𝐶 and joints 𝐴, 𝐶, and 𝐷 to determine moment arms for writing equilibrium equations.

ISTUDY

𝐹𝑥 = 0 ∶

𝐷𝑦 = 3600 lb,

(5)

⇒

𝐴𝑦 = 1600 lb,

(6)

𝐴𝑦 + 𝐷𝑦 − 𝑃𝐴 − 𝑃𝐵 − 4000 lb − 𝑃𝐶 − 𝑃𝐷 = 0

𝑑 𝑑 = (8.660 f t) cos 30◦ = 7.5 f t

𝐹𝑦 = 0 ∶

⇒

𝐴𝑥 = 0

⇒

𝐴𝑥 = 0.

(7)

To apply the method of joints, we draw the FBDs of all the joints in the truss, as shown in Fig. 4, where member forces are taken to be positive in tension. We then write and ∗ Problem

7.103 shows that the load lumping procedure outlined here provides an equivalent force system for straight members with uniform weight distributions. For members that are not straight (as in this example) or if the weight distribution is not uniform, this load lumping procedure is an approximation, although it is commonly used and provides good results provided the shape and weight distribution of members does not deviate greatly from being straight and uniform, and the weights of members are small compared to other forces the structure supports.

ISTUDY

Section 6.1

Truss Structures and the Method of Joints

4 kip + 𝑃𝐶

𝑃𝐵 30◦

𝐵

𝑦 60◦

𝑥

𝐹𝐵𝐷

𝐹𝐴𝐵 𝐹𝐴𝐶

𝑃𝐴 𝐴

𝐹𝐵𝐶

30◦

𝐶 60◦

𝐹𝐴𝐶

𝐹𝐶𝐷 𝑃𝐷

𝐹𝐵𝐷

30◦

30◦

60◦

60◦

1600 lb

3600 lb

𝐷

Figure 4. Free body diagrams of joints.

solve the equilibrium equations for the joints as follows: Joint A: ∑

𝐹𝑥 = 0 ∶

𝐹𝐴𝐵 cos 60◦ + 𝐹𝐴𝐶 cos 30◦ = 0,

(8)

∑

𝐹𝑦 = 0 ∶

𝐹𝐴𝐵 sin 60◦ + 𝐹𝐴𝐶 sin 30◦ + 1600 lb − 𝑃𝐴 = 0

(9)

⇒ Joint B: ∑ 𝐹𝑦 = 0 ∶ ∑ 𝐹𝑥 = 0 ∶

𝐹𝐴𝐵 = −2338 lb and 𝐹𝐴𝐶 = 1350 lb.

(10)

−𝐹𝐴𝐵 sin 60◦ − 𝐹𝐵𝐷 sin 30◦ − 𝑃𝐵 = 0 ⇒ 𝐹𝐵𝐷 = 3350 lb,

(11)

−𝐹𝐴𝐵 cos 60◦ + 𝐹𝐵𝐷 cos 30◦ + 𝐹𝐵𝐶 = 0 ⇒ 𝐹𝐵𝐶 = −4070 lb.

(12)

Joint C: ∑

𝐹𝑦 = 0 ∶

−𝐹𝐴𝐶 sin 30◦ − 𝐹𝐶𝐷 sin 60◦ − 4000 lb − 𝑃𝐶 = 0 ⇒ 𝐹𝐶𝐷 = −5802 lb.

Discussion & Verification

(13) (14)

We were able to determine all of the member forces without using one of the equilibrium equations for joint 𝐶 and the two equilibrium equations for joint 𝐷. As a check on our solution, you should write these three equations and verify that the member forces we determined satisfy all of them.

373

374

Chapter 6

Structural Analysis and Machines

E X A M P L E 6.3

A Truss with a Cable and Pulleys In the exercise machine shown, the stack of weights at 𝐻 weighs 50 lb. If cable segment 𝐴𝐵 is vertical, determine the force supported by each member of the machine.

7 in. 𝐵

36 in.

20 in. 7 in.

SOLUTION

𝐷 15 in. 𝐶

𝐸

𝐴

20◦ 𝐹 Figure 1

ISTUDY

𝐻 𝐺

Road Map

In this structure, we will neglect the weights of individual members, including the weight of the cable, and will compute the member forces due to the person lifting the 50 lb weight only. In contrast to Examples 6.1 and 6.2, in this problem the geometry of the structure is such that the support reactions are not needed before we can proceed with the method of joints. Further comments on how to recognize such situations are given at the end of this example. Modeling To use the method of joints, we draw FBDs of all the joints in the truss, as shown in Fig. 2. In these FBDs, all member forces are taken to be positive in tension. Also, we have elected to leave the pulleys on the joints. Alternatively, we could have removed the pulleys by shifting the cable forces to the bearings, as discussed in connection with Fig. 5.11 on p. 299. Indeed, if either cable segment 𝐴𝐵 or 𝐸𝐻 were not vertical, this would be the preferred approach because the moment arms would be easier to obtain. Finally, we will assume the pulleys at 𝐵 and 𝐸 are frictionless, and we will neglect the weight of the cable itself so that the tensile force supported by the cable is the same throughout its entire length and is equal to the 50 lb weight at 𝐻. Since the pulleys at 𝐵 and 𝐸 have the same radius, the orientation of cable segment 𝐵𝐸 is given by 𝛼 = tan−1 (15 in.∕56 in.) = 15.00◦ . 𝛼

50 lb

𝐵

𝐷

𝑇𝐵𝐷

𝑇𝐵𝐷 36

50 lb

25

15

𝑇𝐵𝐶 39

𝑇𝐶𝐷

15

𝑇𝐷𝐸

20 𝑇𝐷𝐸

𝑇𝐶𝐷

𝑇𝐵𝐶 𝑇𝐶𝐸

𝐶

𝛼 = 15◦

50 lb

𝛼 𝑇𝐶𝐸

𝐸 50 lb

𝑇𝐸𝐹

𝑦

𝑇𝐸𝐺

𝑇𝐶𝐹 𝑥

20◦ 𝑇𝐸𝐺

𝑇𝐶𝐹 𝑇𝐸𝐹 20◦ 𝐹

𝐺 𝐹𝑥 𝐹𝑦

𝐺𝑦

Figure 2. Free body diagrams of joints. Governing Equations & Computation

We begin writing equilibrium equations at joint 𝐵, because there are only two unknowns there. Furthermore, examination of the FBD for

ISTUDY

Section 6.1

Truss Structures and the Method of Joints

375

∑ joint 𝐵 shows that the 𝐹𝑦 = 0 equation will contain only one unknown, whereas the ∑ ∑ 𝐹𝑥 = 0 will contain two, thus we will write the 𝐹𝑦 = 0 expression first. Then we will proceed to joint 𝐷, as there are only two unknowns there (once 𝑇𝐵𝐷 is found), and so on. Joint B: ∑

∑

) 15 =0 39 𝑇𝐵𝐶 = −163.6 lb, ⇒ ( ) 36 + 𝑇𝐵𝐷 = 0 (50 lb) cos 𝛼 + 𝑇𝐵𝐶 39 𝑇𝐵𝐷 = 102.8 lb. ⇒

𝐹𝑦 = 0 ∶

−50 lb − (50 lb) sin 𝛼 − 𝑇𝐵𝐶

𝐹𝑥 = 0 ∶

Joint D: ∑

𝐹𝑥 = 0 ∶

∑

𝐹𝑦 = 0 ∶

Joint C: ∑

𝐹𝑥 = 0 ∶

∑

𝐹𝑦 = 0 ∶

) 20 =0 25 ( ) 15 =0 −𝑇𝐶𝐷 − 𝑇𝐷𝐸 25 −𝑇𝐵𝐷 + 𝑇𝐷𝐸

( ) 36 + 𝑇𝐶𝐸 = 0 −𝑇𝐵𝐶 39 ( ) 15 + 𝑇𝐶𝐷 − 𝑇𝐶𝐹 = 0 𝑇𝐵𝐶 39

Joint E: ∑ 𝐹𝑥 = 0 ∶

∑

𝐹𝑦 = 0 ∶

(

−𝑇𝐶𝐸 − 𝑇𝐷𝐸

𝑇𝐷𝐸

(

(

(

(2) (3) (4)

⇒

𝑇𝐷𝐸 = 128.4 lb,

(5)

⇒

𝑇𝐶𝐷 = −77.06 lb.

(6)

⇒

𝑇𝐶𝐸 = −151.0 lb,

(7)

⇒

𝑇𝐶𝐹 = −140.0 lb.

(8)

) 20 − (50 lb) cos 𝛼 − 𝑇𝐸𝐹 sin 20◦ = 0 25 𝑇𝐸𝐹 = 0, ⇒

15 + (50 lb) sin 𝛼 − 50 lb − 𝑇𝐸𝐹 cos 20◦ − 𝑇𝐸𝐺 = 0 25 ⇒ 𝑇𝐸𝐺 = 40.00 lb. )

(1)

(9) (10) (11) (12)

Discussion & Verification

• The forces supported by all members were determined without using the equilibrium equations for joints 𝐹 and 𝐺. If desired, the equilibrium equations for these joints may also be written to obtain the support reactions, which are found to be 𝐹𝑥 = 0, 𝐹𝑦 = 140.0 lb, and 𝐺𝑦 = 40.0 lb. • After Eq. (9) is solved, member 𝐸𝐹 is found to be a zero-force member. Because neither joint 𝐸 nor 𝐹 fits the pattern shown in Fig. 6.8 on p. 368, it is not possible to use inspection to determine that this member is zero-force. • This problem was convenient to solve without first determining the support reactions because joint 𝐵 had only two unknowns, and thus, it was possible to begin the analysis at this location. In contrast, examination of the FBDs for Examples 6.1 and 6.2 shows that if the reactions are not obtained first, then the FBDs for all joints in those examples contain three or more unknowns.

Jill Braaten

Figure 3 An exercise machine similar to that shown in Fig. 1 in use.

376

Chapter 6

Structural Analysis and Machines

Problems Problems 6.1 and 6.2 Determine the force supported by each member of the truss if 𝑃 = 1000 lb. 𝐵 4 ft 𝐴

𝐵

𝐷

4 ft 𝐶

𝐴

𝐷

𝐸 𝐶

𝑃 3 ft

3 ft

𝑃 3 ft

Figure P6.1

3 ft

Figure P6.2

Problems 6.3 and 6.4 For the truss shown in Fig. P6.1, each member can support a maximum tensile force of 4000 lb and a maximum compressive force of 2000 lb. Problem 6.3

If 𝑃 = 1000 lb, determine the factor of safety for the truss.

Problem 6.4

Determine the largest positive value of 𝑃 that can be applied.

Problems 6.5 and 6.6 For the truss shown in Fig. P6.2, each member can support a maximum tensile force of 6000 lb and a maximum compressive force of 4000 lb.

𝑅 𝐵

𝐴

𝑄

𝐷

𝐷

Problem 6.8

𝐹

All members of the truss have the same length. Determine the force supported by members 𝐶𝐸 and 𝐷𝐹 if 𝑃 = 1 kN, 𝑄 = 2 kN, and 𝑅 = 3 kN.

𝑃

Figure P6.8

ISTUDY

𝐺

𝐸 𝑄

Determine the largest positive value of 𝑃 that can be applied.

All members of the truss have the same length. Determine the force supported by each member if 𝑃 = 1 kN, 𝑄 = 2 kN, and 𝑅 = 3 kN.

Figure P6.7

𝐶

Problem 6.6

Problem 6.7 𝑃

𝐴

If 𝑃 = 1000 lb, determine the factor of safety for the truss.

𝐸

𝐶

𝐵

Problem 6.5

𝑅

Problems 6.9 and 6.10 Determine the force supported by each member of the truss if 𝑃 = 4 kN and 𝑄 = 1 kN. 𝑄 𝑃 𝐴 1m 1m

𝐵

𝐶

𝐹

𝐵 3m

𝑄

2m

𝐷

𝐴

𝐸 𝐶

𝐸 𝐷

3m

Figure P6.9

𝑃 3m

3m

Figure P6.10

ISTUDY

Section 6.1

377

Truss Structures and the Method of Joints

Problems 6.11 and 6.12 For the truss shown in Fig. P6.9, each member can support a maximum tensile force of 15 kN and a maximum compressive force of 10 kN. Problem 6.11

If 𝑃 = 4 kN and 𝑄 = 1 kN, determine the factor of safety for the truss.

Problem 6.12

If 𝑄 = 𝑃 ∕4, determine the largest positive value of 𝑃 that can be

applied.

Problems 6.13 and 6.14 For the truss shown in Fig. P6.10, each member can support a maximum tensile force of 20 kN and a maximum compressive force of 12 kN. Problem 6.13

If 𝑃 = 4 kN and 𝑄 = 1 kN, determine the factor of safety for the truss.

Problem 6.14

If 𝑄 = 𝑃 ∕4, determine the largest positive value of 𝑃 that can be

applied. 𝑃

Problem 6.15

𝑄 𝐶

𝐴

Determine the force supported by each member of the truss. Express your answers in terms of 𝑃 and 𝑄.

50◦ 𝐵

Problem 6.16 Each member can support a maximum tensile force of 15 kip and a maximum compressive force of 10 kip. If 0 ≤ 𝑃 ≤ 4 kip and 0 ≤ 𝑄 ≤ 4 kip, determine the lowest factor of safety for the truss. Hint: The answer is the lowest factor of safety among the three load cases 𝑃 = 4 kip and 𝑄 = 0, 𝑃 = 0 and 𝑄 = 4 kip, and 𝑃 = 𝑄 = 4 kip.

𝐸

𝐹 50◦

50◦

𝐷

Figure P6.15 and P6.16

𝑄

Problem 6.17

𝐴 𝑃

Determine the force supported by each member of the truss. Express your answers in terms of 𝑃 and 𝑄.

𝐵

Problem 6.18

𝐶

Each member can support a maximum tensile force of 3 kN and a maximum compressive force of 2 kN. If 0 ≤ 𝑃 ≤ 400 N and 0 ≤ 𝑄 ≤ 400 N, determine the lowest factor of safety for the truss. Hint: The answer is the lowest factor of safety among the three load cases 𝑃 = 400 N and 𝑄 = 0, 𝑃 = 0 and 𝑄 = 400 N, and 𝑃 = 𝑄 = 400 N.

𝐷 45◦ 45◦ 𝐸

𝐹

Figure P6.17 and P6.18

Problems 6.19 through 6.21 The structure consists of seven pin-connected members. Determine the force supported by all members. Express your answers in terms of parameters such as 𝐹 and 𝐿. 𝐹 𝐸

𝐶

𝐹

𝐿 𝐷

Figure P6.19

𝐸

𝐿

𝐴

𝐿

𝐶

𝐸

𝐿 𝐶

𝐵 𝐿

𝐹 𝐷

𝐴

𝐵 𝐿 Figure P6.20

𝐴 𝐿

𝐵

𝐷 𝐿 Figure P6.21

𝐿

378

Chapter 6

Structural Analysis and Machines Problems 6.22 through 6.24

The structure consists of pin-connected members. Determine the force supported by all members. 𝐵

𝐵

6 ft 𝐶

𝐴 8 ft

𝐵

2m

𝐸

2m 𝐸

𝐴

6 ft

𝐷

𝐴

𝐶 2m

2m

2000 lb

1m

9 kN Figure P6.22

10 ft

10 ft 5 kip 𝐵

𝐴

1m

𝐶

1m

Figure P6.24

Problem 6.25 A Bollman truss is shown. Determine the force supported by each of the five bars and four cables.

𝐹 10 ft

𝐸

𝐶

1m

𝐹

2 kN 3 kN 4 kN

Figure P6.23

10 ft 5 kip 𝐷

𝐸

Problems 6.26 through 6.31 By inspection, identify the zero-force members in the truss.

Figure P6.25 𝐸

𝐶

𝐺

𝐸

𝐶

𝐵

𝐺

𝐵

𝐵

𝐴

𝐼 𝐷

𝐴

𝐻

𝐹

𝐼 𝐷

Figure P6.26

𝐹

𝐶

𝐻

𝐷 𝐸

𝐹 𝐺 𝐻

𝐴

Figure P6.27

Figure P6.28

𝐵 𝐸 𝐸

D

B

𝐶 30◦

𝐷

𝐵 4 kN

2m

𝐹

6 kN 2m

2m

𝐻

Figure P6.29

𝐹

𝐴

𝐹 𝐷 Figure P6.30

𝐺

𝐴

𝐻

Figure P6.31

2m

12 kN

(a) By inspection, identify the zero-force members in the truss.

𝐻

(b) Determine the force supported by all members of the truss.

𝐵

𝐽

6 kN

Problem 6.33 𝐴

𝐿 𝐶 𝐸

𝐺

𝐼 𝐾

6@3 m Figure P6.33

ISTUDY

H

𝐸

Problem 6.32 𝐷

4m

E C

𝐹

𝐶

𝐺

𝐵

Figure P6.32

4m

𝐶

G

𝐺

A 𝐴

F

𝐷

(a) By inspection, identify the zero-force members in the truss. (b) Determine the force supported by member 𝐹 𝐺 and all of the members to the left of it. (c) Determine the force supported by all of the members to the right of member 𝐹 𝐺.

ISTUDY

Section 6.1

379

Truss Structures and the Method of Joints

Problems 6.34 and 6.35 A model for a two-story building is shown. Determine the force supported by each member of the truss. 4 kip

4 kip 𝐴 30 ft

𝐵

5 kip

10 kN 3 kip

8 kN

5 kip

𝐶

12 kN 2 kip

𝐷

10 kN 𝐵

𝐴

6 kN

6m

12 kN 𝐷

𝐶

30 ft

6m 𝐸

𝐹

𝐹

𝐸

40 ft

8m

Figure P6.34

6 kip

Figure P6.35

60◦

𝐴

𝐶

𝐵

𝐷

𝐸

Problem 6.36 4 ft

(a) By inspection, identify the zero-force members in the truss. (b) Determine the force supported by member 𝐶𝐻 and all of the members to the left of it. (c) Determine the force supported by all of the members to the right of member 𝐶𝐻.

𝐽

𝐹

𝐺

𝐻

4 kip 3 ft

3 ft

𝐼 8 kip

3 ft

3 ft

Figure P6.36

Problem 6.37 The truss has frictionless pulleys at points 𝐵, 𝐽 , and 𝐾. Cable segment 𝐵𝐿 is vertical and 𝑊 = 1 kip. (a) By inspection, identify the zero-force members in the truss. (b) Determine the force supported by member 𝐶𝐸. (c) Determine the force supported by member 𝐺𝐼. 1 ft

𝐵

𝐷

𝐹

𝐻

𝐽

1 ft

4 ft 𝐴

𝐶

𝐸

𝐺

𝐼

1 ft

𝐿 3 ft

20 lb

𝐾

𝐴 𝑊

3 ft

3 ft

3 ft

3 ft

3 ft

3 ft

3 ft

3 ft

3 ft

25 in.

20 lb

50 lb

𝐶 𝐵 50 lb

Figure P6.37 𝐷

Problem 6.38 A shelf used in a store for displaying shoes and handbags is modeled as a truss having 13 members, as shown. Determine the force supported by all members of the truss.

25 in.

In Fig. P6.38, let the 20 lb, 50 lb, and 30 lb forces acting at joints 𝐴, 𝐵, and 𝐶 be due to the weight of shelves 𝐴𝐵 and 𝐵𝐶, and the items these shelves support. Assuming the weights are uniformly distributed, follow the procedure described in Example 6.2 on p. 372 to determine the weight of shelves 𝐴𝐵 and 𝐵𝐶 (including the items they support).

𝐹

40 lb 𝐻

𝐺 15 in. 15 in. Figure P6.38

30 lb

𝐸

40 lb

Problem 6.39

30 lb

380

Chapter 6

Structural Analysis and Machines

6.2

𝐵

𝐷

𝐹 4m

𝐴

𝐻 𝐶 3m

𝐸 3m

𝐺 10 kN 3m

3m

Truss Structures and the Method of Sections

The method of sections is an effective and popular approach for determining the forces supported by individual members of a truss, especially when forces in only a portion of the members must be found. In this method, we select a member whose force we want to determine. Then a cut is taken that passes through this member, subdividing the truss into two parts. If the entire truss is in equilibrium, then both parts of the truss are also in equilibrium. Analysis proceeds by drawing an FBD for one of the parts of the truss, writing equilibrium equations, and solving these for the unknowns. Prior to this, we may want or need to determine the reactions for the truss as a whole. Analysis by the method of sections is illustrated in the following mini-example.

Figure 6.10 A plane truss.

Mini-Example Use the method of sections to determine the force supported by member 𝐵𝐷 of the truss shown in Fig. 6.10.

𝑦

Solution We begin by obtaining the support reactions for the truss by drawing the FBD of the structure as a whole, as shown in Fig. 6.11. Writing equilibrium equations provides ∑ 𝑀𝐴 = 0 ∶ −(10 kN)(6 m) + 𝐻𝑦 (12 m) = 0 ⇒ 𝐻𝑦 = 5 kN, (6.12) ∑ 𝐹𝑦 = 0 ∶ 𝐴𝑦 + 𝐻𝑦 − 10 kN = 0 ⇒ 𝐴𝑦 = 5 kN, (6.13) ∑ 𝐹𝑥 = 0 ∶ 𝐴𝑥 = 0 ⇒ 𝐴𝑥 = 0. (6.14)

𝑥

𝐴𝑥

𝐴𝑦

𝐻𝑦

10 kN 6m

6m

Figure 6.11 Free body diagram to determine the support reactions. 𝐵

Because the geometry and loading for this problem are symmetric, we could have determined these reactions by inspection. In order to find the force in member 𝐵𝐷, it is necessary to use an FBD where the cut passes through member 𝐵𝐷, and Fig. 6.12(a) shows the start of such a cut. In Fig. 6.12(b), two possible paths 𝑎𝑎 and 𝑏𝑏 (among many) for continuing the cut are shown. Figure 6.12(c) shows completed cuts where in each case, the cut is a closed line that fully encompasses the left-hand or right-hand portion of the structure. Once a cut has been taken and an FBD of a portion of the structure has been drawn, there are three equilibrium equations that may be written. Thus, a good strategy is to select a cut (when possible) so that no more than three members are cut, since then the unknown forces for these members may be immediately solved for by using the three equilibrium equations that are available. Compare the two cuts shown in Fig. 6.12(b); observe that cut 𝑎𝑎 passes through three members and hence its FBD will have three unknowns, while cut 𝑏𝑏 passes through four members and hence its FBD will have four unknowns.∗ Thus, cut 𝑎𝑎 is the better choice. Using cut 𝑎𝑎, we draw the FBD for the left-hand and right-hand portions of the structure as shown in Fig. 6.13, where member forces 𝐹𝐵𝐷 , 𝐹𝐵𝐸 , and 𝐹𝐶𝐸 are positive in tension. Normally, we would draw only one of these FBDs, and we

𝐷

(a)

𝑏

𝑎 (b) 𝑎

𝑏

𝑎

𝑎

𝑎

𝑎

𝑏

𝑏

(c) 𝑏

𝑏

Figure 6.12 Possible cuts that will allow the force in member 𝐵𝐷 to be determined by the method of sections. (a) The cut starts by passing through member 𝐵𝐷. (b) Two possible paths (among many) for continuing the cut started in (a). (c) Four possible paths for completing the cut so that an FBD can be drawn.

ISTUDY

∗ If

you recognized that member 𝐷𝐸 is a zero-force member, then cut 𝑏𝑏 has only three unknowns and is therefore as good a choice as cut 𝑎𝑎.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

𝑦

𝐵 𝑥

𝐹𝐵𝐸 𝐹𝐶𝐸

𝐴 𝐶 5 kN 3m

𝐹𝐵𝐷

𝐹𝐵𝐷

𝐸 3m

4m

𝐷

𝐹

𝐹𝐵𝐸 𝐹𝐶𝐸

4m 𝐻 𝐸

𝐺 10 kN 3m

5 kN 3m

Figure 6.13. Free body diagrams.

usually select the side of the truss that has fewer forces and/or more straightforward geometry. Using the FBD for the left-hand portion of the structure, we notice that point 𝐸 is a convenient location for summing moments, because two of the unknown forces pass through this point, leaving only 𝐹𝐵𝐷 . Thus, taking positive moment counterclockwise, we write ∑ 𝑀𝐸 = 0 ∶ −(5 kN)(6 m) − 𝐹𝐵𝐷 (4 m) = 0 ⇒ 𝐹𝐵𝐷 = −7.5 kN. (6.15) Although this problem does not ask for the forces in members 𝐵𝐸 and 𝐶𝐸, we will nonetheless determine these by writing and solving the following equilibrium equations ∑ −(5 kN)(3 m) + 𝐹𝐶𝐸 (4 m) = 0 ⇒ 𝐹𝐶𝐸 = 3.75 kN, (6.16) 𝑀𝐵 = 0 ∶ ( ) ∑ 3 + 𝐹𝐵𝐷 = 0 ⇒ 𝐹𝐵𝐸 = 6.25 kN. 𝐹𝑥 = 0 ∶ 𝐹𝐶𝐸 + 𝐹𝐵𝐸 (6.17) 5 Remarks • Compared to the method of joints, the method of sections is often easier to use when the forces in only a portion of a structure’s members are desired. If the method of joints were used for this example, FBDs would need to be drawn and equilibrium equations written for joints 𝐴, 𝐶, and 𝐵 (in that order) before 𝐹𝐵𝐷 could be determined. • When a cut is taken, the truss is subdivided into two (or more) parts. Regardless of which part you choose to draw an FBD for, there are only three equilibrium equations that may be used to determine the unknown forces. For example, if the FBD shown in Fig. 6.13 for the right-hand portion of the structure is used, ∑ ∑ ∑ the 𝑀𝐸 = 0, 𝑀𝐵 = 0, and 𝐹𝑥 = 0 expressions that are written are identical to Eqs. (6.15)–(6.17). If different moment summation points are used, then the resulting equilibrium equations will differ from Eqs. (6.15) and (6.16). Nonetheless they will be a linear combination of these, so there are only three independent equations available for determining the unknown forces. • To further elaborate on the consequence of the last remark, imagine that cut 𝑏𝑏 shown in Fig. 6.12 was used. The FBD for this cut has four unknown member forces. To determine the four unknowns, you might consider writing three equilibrium equations for the left-hand portion of the truss and then writing an additional equilibrium equation for the right-hand portion. However, in view of the foregoing remark, this strategy will fail because of the four equilibrium equations, only three are independent, and thus we cannot determine all four unknowns in this fashion.

381

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• If the problem had asked for the forces in members 𝐵𝐷 and 𝐷𝐸, then the method of sections with cut 𝑏𝑏 shown in Fig. 6.12 (or a similar cut) would be needed, because it passes through both of these members. However, as explained in the preceding remarks, the method of sections with cut 𝑏𝑏 will not be sufficient to determine these forces. If the truss is statically determinate (discussed later in this section), then multiple applications of the method of sections and/or the method of joints will be needed. In this example, we could first use the method of sections with cut 𝑎𝑎 to find 𝐹𝐵𝐷 , followed by the method of sections with cut 𝑏𝑏 to find the remaining unknowns. Alternatively, we could first use the method of joints for joint 𝐷 to find 𝐹𝐷𝐸 . Then we could use the method of sections with cut 𝑏𝑏; because 𝐹𝐷𝐸 is known, the three remaining unknowns could then be determined.

𝑦 𝑥 𝐵 equivalent 𝐴

𝐴 𝐹 𝑑 𝐿

𝐵 𝐹2 𝐹1 force systems 𝐹1 = (1 − 𝑑∕𝐿)𝐹 𝐹2 = (𝑑∕𝐿)𝐹

Figure 6.14 For many purposes, a force that is not located at a joint may be replaced by an equivalent force system where forces are positioned at joints.

ISTUDY

Treatment of forces that are not at joints In real life structures, forces often occur at locations other than joints. For many purposes such forces may be replaced by equivalent force systems so that truss analysis may be used. Consider, for example, the vertical force 𝐹 shown in Fig. 6.14. To see that 𝐹1 and 𝐹2 have the values reported in Fig. 6.14, we construct an equivalent force system using the procedure given by Eq. (4.16): (∑

and

(∑

𝐹𝑦

𝑀𝐴

)

system 1

=

(∑

𝐹𝑦

)

system 2

𝐹 = 𝐹1 + 𝐹2 , ) (∑ = 𝑀𝐴

)

system 1

(6.18)

system 2

−𝐹 𝑑 = −𝐹2 𝐿.

(6.19)

Solving Eqs. (6.18) and (6.19) provides

𝐹1 = (1 −

𝑑 )𝐹 𝐿

and

𝐹2 =

𝑑 𝐹. 𝐿

(6.20)

You may wish to verify that Eq. (6.20) provides the proper results for 𝐹1 and 𝐹2 for the cases 𝑑 = 0 and 𝑑 = 𝐿. Although we say the two force systems shown in Fig. 6.14 are equivalent, exactly how equivalent they really are depends on what we are interested in. In the left-hand case shown in Fig. 6.14, where member 𝐴𝐵 is subjected to 𝐹 with 0 < 𝑑 < 𝐿, the member experiences bending. In the right-hand case shown in Fig. 6.14, where member 𝐴𝐵 is subjected to 𝐹1 and 𝐹2 , the member does not experience bending. Nonetheless, the two force systems are equivalent in the sense that effects that are external to member 𝐴𝐵, such as reactions and the forces supported by all other members, are the same. The issue of when two “equivalent” force systems really are equivalent is sometimes subtle, and you may not be able to fully understand these subtleties until you study internal forces in Chapter 8. Nonetheless, the discussion of why equivalent force systems are called equivalent in Section 4.4 always applies, and it is the source for resolving any questions of interpretation.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

Static determinacy and indeterminacy In a statically determinate truss, the equations of equilibrium are sufficient to determine the forces supported by all members of the truss and the support reactions. In a statically indeterminate truss, the equations of equilibrium are not sufficient to determine all of these. Compared to a statically determinate truss, a statically indeterminate truss has extra members and/or supports. As a result, the forces supported by each member, and possibly the support reactions, also depend on the material the members are made of and the sizes of their cross sections. Methods of analysis for statically indeterminate trusses are covered in more advanced subjects. A simple rule of thumb, called equation counting, can be used to help determine whether a truss is statically determinate or statically indeterminate. The rule is developed by comparing the number of unknowns for a truss to the number of equilibrium equations. Consider a plane truss having 𝑚 = number of members, 𝑟 = number of support reactions,

(6.21)

𝑗 = number of joints. Each member has one unknown force and each support reaction has one unknown force, so 𝑚 + 𝑟 is the total number of unknowns. For a plane truss, each joint has two equilibrium equations, so 2𝑗 is the total number of equations. Thus, the rule of thumb is as follows:

If 𝑚 + 𝑟 < 2𝑗

The truss is a mechanism and/or has partial fixity.

If 𝑚 + 𝑟 = 2𝑗

The truss is statically determinate if it has full fixity. The truss is statically indeterminate if it has partial fixity.

If 𝑚 + 𝑟 > 2𝑗

(6.22)

The truss is statically indeterminate, and it can have full fixity or partial fixity.

• If 𝑚 + 𝑟 < 2𝑗, not all of the equilibrium equations can be satisfied, implying that some of them may not be satisfied, in which case motion will occur. If the truss has only partial fixity, then the truss as a whole may undergo a rigid body motion. If the truss has full fixity, then it is a mechanism, meaning it may collapse. The presence of a mechanism in a truss is disastrous, unless the truss is intended to be part of a machine. Trusses are almost always supported so that they are fully fixed. Thus, the situation 𝑚 + 𝑟 < 2𝑗 is rarely permitted. • If 𝑚 + 𝑟 = 2𝑗, there are as many unknowns as equilibrium equations. Barring the possibility of having members and/or supports with insufficient arrangement, all unknowns can be determined and all equilibrium equations can be satisfied. • If 𝑚 + 𝑟 > 2𝑗, not all of the unknowns can be determined, and additional equations, in the form of models that characterize the deformability of all members of the truss, must be introduced.

383

384

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Structural Analysis and Machines

Several examples of equation counting are shown in Fig. 6.15.

(a) 7 + 2 < 2(5) 𝑚 + 𝑟 < 2𝑗 ⇒

(b) 6 + 3 < 2(5) partial fixity or a mechanism

(c) 7 + 3 = 2(5)

(d) 6 + 4 = 2(5)

𝑚 + 𝑟 = 2𝑗 and visual inspection shows complete fixity ⇒ statically determinate

(e) 8 + 2 = 2(5)

(f) 6 + 4 = 2(5)

𝑚 + 𝑟 = 2𝑗 and visual inspection shows partial fixity or a mechanism ⇒ statically indeterminate

(g) 8 + 3 > 2(5) 𝑚 + 𝑟 > 2𝑗

(h) 7 + 4 > 2(5) ⇒

(i) 8 + 3 > 2(5)

statically indeterminate

Figure 6.15. Examples of equation counting to determine if a truss is statically determinate or statically indeterminate. For trusses that are not fully fixed or are mechanisms, a possible displaced position is shown by the dashed outlines.

Design considerations Simple, compound, and complex trusses

Figure 6.16 In a simple truss the arrangement of members begins with three members that form a triangle, and adding to this two new members (shown by dashed lines) for each new joint that is added.

ISTUDY

In a simple truss,∗ members are arranged beginning with three members joined to one another at their ends in the shape of a triangle, as shown in Fig. 6.16, and adding to this two new noncollinear members for each new joint that is added. All of the trusses shown in Fig. 6.2 on p. 364 are simple trusses. A compound truss is formed by interconnecting two or more simple trusses to form a stable structure, where stability is provided by a sufficient number and arrangement of members that connect the simple trusses. A complex truss is formed by interconnecting two or more simple ∗ Additional

details and more advanced methods of analysis are contained in T. Au and P. Christiano, Fundamentals of Structural Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1993; and E. C. Rossow, Analysis and Behavior of Structures, Prentice-Hall, Upper Saddle River, NJ, 1996.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

385

trusses, along with a sufficient number and arrangement of supports to make the structure stable. These three truss categories are compared in Fig. 6.17.

simple truss

compound truss (Fink)

complex truss

Figure 6.17. Examples of simple, compound, and complex trusses. The compound truss shown is called a Fink truss.

A truss does not need to be a simple truss to be a good design. For example, the compound and complex trusses shown in Fig. 6.17 and the Baltimore truss shown in Fig. 6.18 are not simple trusses, but they are nonetheless popular and effective designs. However, one of the main features of a simple truss is guaranteed stability: if the pattern of construction described in Fig. 6.16 is used, the resulting truss will have no mechanisms, and if it is fully supported, it will be stable. The worst blunder you can make in designing a truss is to produce a structure that is unstable. At a minimum, you will be very embarrassed, and at worst, if the structure is built, people may be injured or killed. Note that as the arrangement of members in a truss becomes more complex, it becomes more difficult to ensure there are no mechanisms. The use of equation counting in Eq. (6.22) is helpful, but it is not foolproof. Buckling of truss members A typical truss uses remarkably little material, yet provides impressive strength. Because truss members are usually slender, they are susceptible to buckling under compressive forces. To understand this phenomenon better, perform the experiment shown in Fig. 6.19, using a wood or metal yardstick (or meterstick), ideally one that is very straight. Under low compressive force, the yardstick remains straight. But under sufficiently high compressive force, you should observe that the yardstick suddenly buckles. When it buckles, it begins to bend and it loses strength. In fact, the loss of strength is very dramatic, and for this reason, buckling failures in structures are often catastrophic. Buckling is a fascinating subject and is crucially important. The following mini-example explores buckling for a simple problem, but its results are characteristic of the responses of more complicated structures, such as columns and truss members.

Mini-Example A simple model for buckling analysis of an initially straight truss member is shown in Fig. 6.20(a), where bars 𝐴𝐵 and 𝐵𝐶 are rigid, and the ability of the structure to deform is modeled by the torsional spring. Determine the value of 𝑃 for which the structure buckles. Solution Although this is a crude model for a straight truss member, it is easy to analyze and its results are revealing. The FBD for the buckled structure is shown in Fig. 6.20(b). Note that because of the torsional spring, members 𝐴𝐵 and 𝐵𝐶 are not two-force members, although the structure 𝐴𝐵𝐶 as a whole is. For this reason ∑ ∑ (or by writing 𝑀𝐴 = 0 and 𝑀𝐶 = 0), we determine 𝐴𝑥 = 𝐶𝑥 = 0, and

Baltimore Figure 6.18 The Baltimore truss is not a simple truss, but it is a popular and effective design.

low force

Under low force, the yardstick remains straight.

high force

Under high force, the yardstick buckles.

Figure 6.19 An experiment showing how a yardstick (or meterstick) buckles under high compressive force.

386

Chapter 6

Structural Analysis and Machines

𝑃

𝑃

𝐴 torsional spring

𝐿∕2

𝐴𝑥 = 0

𝐴𝑥

𝐿∕2

2𝜃

𝜃 𝑀𝑡

𝑦

𝐵

𝜃

𝐿∕2

𝑥

𝐶

𝐵𝑥 𝐵𝑦

𝐶𝑥 𝐶𝑦

(a)

𝐹𝑦 = 0 gives 𝐶𝑦 = 𝑃 . Next, we remove∗ member 𝐴𝐵 from the structure, giving the FBD shown in Fig. 6.20(c). Then ∑ 𝐿 𝑀𝐵 = 0 ∶ −𝑀𝑡 + 𝑃 sin 𝜃 = 0. (6.23) 2

∑

𝑃

(c)

(b)

Noting that the torsional spring is twisted by 2𝜃, the spring law, Eq. (5.14) on p. 300, gives 𝑀𝑡 = 𝑘(2𝜃). Assuming small angles with 𝜃 measured in radians, and with sin 𝜃 ≈ 𝜃, Eq. (6.23) becomes ) ( 𝐿 𝜃 = 0. (6.24) 2𝑘 − 𝑃 2 Equation (6.24) has two solutions:

Figure 6.20 A simple model to study buckling.

Solution 1: Solution 2:

𝜃=0 𝐿 2𝑘 − 𝑃 = 0 2

trivial solution,

(6.25)

buckling solution.

(6.26)

Solution 1 is called the trivial solution because it says that structure 𝐴𝐵𝐶 remains straight. Solution 2 applies when 𝑃critical = 4𝑘∕𝐿,

𝑃 bifurcation 𝑃critical = 4𝑘∕𝐿 𝜃 Figure 6.21 Load-deformation results for buckling of the model shown in Fig. 6.20.

𝐵 −

𝐴

+

0

𝐶

3m

𝐷

−

0

+ +

𝐸

3m

− + +

10 kN 3m

𝐹 −

0

𝐺

+

4m 𝐻

3m

Figure 6.22 A truss subjected to a 10 kN force. Members that are in tension and compression are identified by + and − symbols, respectively, and zero-force members are identified by 0.

ISTUDY

(6.27)

where the subscript critical is added to emphasize that this value of 𝑃 causes the structure to buckle. When 𝑃 = 𝑃critical , according to Eq. (6.24), 𝜃 can have arbitrary value. Observe that if the spring stiffness is increased, then 𝑃critical increases, whereas if the length of the structure increases, then 𝑃critical decreases. This solution and its ramifications are summarized in Fig. 6.21. Let force 𝑃 slowly increase from zero value. As long as 𝑃 < 𝑃critical = 4𝑘∕𝐿, equilibrium [Eq. (6.24)] requires 𝜃 = 0. When 𝑃 = 𝑃critical = 4𝑘∕𝐿, equilibrium is satisfied regardless of the value of 𝜃, which can have arbitrary positive or negative values. Thus, as 𝑃 reaches 𝑃critical , the structure undergoes an abrupt change from 𝜃 = 0 to 𝜃 ≠ 0. Further, there is no particular preference for 𝜃 to have a positive or negative value, and this branching of the response is often called a bifurcation. Any attempt to apply a force greater than 𝑃critical will cause a dynamic event in which the structure rapidly collapses. The buckling analysis of a truss member is more complicated than the foregoing example, and this is a topic studied in mechanics of materials. The result for the buckling load is 𝑃critical = 𝑐∕𝐿2 where 𝑐 is a constant that depends on the member’s material and cross-sectional geometry and 𝐿 is the length of the member. The most important observation is the strong dependence of the buckling load on length: if the length of a member is doubled, its buckling load decreases by a factor of 4! For this reason, long compression members are generally undesirable in trusses. Use of zero-force members Consider the example analyzed at the start of this section, repeated again in Fig. 6.22. By inspection, members 𝐵𝐶, 𝐷𝐸, and 𝐹 𝐺 are zero-force members. You are probably wondering why zero-force members are present in the truss, and if it is possible to eliminate them without reducing the strength of the truss. ∗ Section

6.4 discusses procedures for taking a structure such as this apart for purposes of determining the forces supported by multiforce members.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

To answer these questions requires knowledge of which members are in tension and which are in compression, and these are designated in Fig. 6.22 using + and − symbols, respectively. In Fig. 6.22, members 𝐵𝐷 and 𝐷𝐹 both have 3 m length, and as such, the compressive load at which they will buckle is proportional to 1∕(3 m)2 . If member 𝐷𝐸 were not present, then joint 𝐷 would not be present, member 𝐵𝐹 would have 6 m length, and its buckling load would be 4 times lower than that for members 𝐵𝐷 and 𝐷𝐹 , presuming the material and cross-sectional geometry of the members are the same. Thus, even though member 𝐷𝐸 is a zero-force member, its effect on increasing strength is remarkable. Members 𝐴𝐶 and 𝐶𝐸 are in tension, and buckling is not an issue for them. Thus, it may be possible to eliminate member 𝐵𝐶 without reducing the strength of the truss. Similarly, it may be possible to eliminate member 𝐹 𝐺. It is important to note that there may be reasons for which members 𝐵𝐶 and 𝐹 𝐺 are important. For instance, the loading in this example is very simple, and in real life applications, joints 𝐶 and 𝐺 may be subjected to loads, either continuously or periodically, in which case members 𝐵𝐶 and/or 𝐹 𝐺 would not be zero-force members.

Statically determinate versus statically indeterminate trusses Both statically determinate and statically indeterminate trusses are popular, and both have advantages and disadvantages. Statically determinate trusses are usually straightforward to design and analyze using the methods discussed in this chapter. Statically indeterminate trusses are usually more difficult to design and analyze. One of the primary features of statically indeterminate trusses is the potential for greater safety because failure of an individual structural member or support does not necessarily mean the entire structure will fail. Figure 6.23 discusses an example of failure in a statically determinate structure.

John Weeks III/AP Images

MANDEL NGAN/AFP/Getty Images

Figure 6.23. Failure of the Interstate 35W bridge in Minneapolis, Minnesota. This bridge opened to traffic in 1967 and collapsed without warning during the evening rush hour on August 1, 2007, killing 13 people and injuring over 100 more. The National Transportation Safety Board’s (NTSB’s) investigation determined that some of the gusset plates were undersized and that failure of some of these initiated collapse of the bridge. One of the criticisms of this bridge’s design is that it was statically determinate, and hence, failure of one member or connection was sufficient to cause collapse of the entire structure.

387

Concept Alert Buckling of truss members. Because of buckling, long compression members are undesirable in trusses. Zero-force members are effective for reducing the length of compression members, and this can substantially improve the strength of a truss.

388

ISTUDY

Chapter 6

Structural Analysis and Machines

End of Section Summary In this section, the method of sections is developed for determining the forces supported by individual members of a truss, and several characteristics affecting the design and performance of trusses are discussed. Some of the key points are as follows: • In the method of sections, a cut is passed through the truss, and the FBD that ∑ results is required to be in equilibrium by writing the equations 𝐹𝑥 = 0, ∑ ∑ 𝐹𝑦 = 0, and 𝑀 = 0. • In a statically determinate truss, the equations of equilibrium are sufficient to determine the forces supported by all members of the truss and the support reactions. In a statically indeterminate truss, there are more unknowns than the number of equilibrium equations, and hence, not all of the unknowns (perhaps none of them) can be determined. Equation counting is an effective way to determine whether a truss is statically determinate or statically indeterminate. • A simple truss has members arranged in a triangular pattern, as shown in Fig. 6.16 on p. 384. Compound and complex trusses consist of two or more simple trusses connected to one another. Simple, compound, and complex trusses are popular, but other designs for trusses are also common and can perform very well. One of the main features of simple, compound, and complex trusses is that they are always stable, presuming that no members or connections fail. • Because truss members are usually straight and slender, compression members are susceptible to buckling. Furthermore, the force at which buckling occurs in a straight member decreases very rapidly as a member becomes longer (the buckling load is proportional to 1∕𝐿2 , where 𝐿 is the member’s length). Thus, it is undesirable to have long compression members in a truss. Zero-force members can be very effective in reducing the length of compression members and thus, increasing the overall strength of a truss.

ISTUDY

Section 6.2

389

Truss Structures and the Method of Sections

E X A M P L E 6.4

Truss Analysis by the Method of Sections

The top chord of the steel truss is subjected to a combined dead load and live load of 6 kN∕m. Determine the force supported by member 𝐹 𝐻. All angles are 60◦ .

1m 1m 1m 1m 1m 1m 𝐴

𝐶

𝐸

𝐼

𝐺

𝐾

𝑀

SOLUTION Because of the 6 kN∕m distributed load, the members of the top chord of this structure are not two-force members, and hence this structure is not a truss. Nonetheless, the structure will be approximated as a truss by lumping the distributed load into forces at joints. After this, support reactions will be determined followed by use of the method of sections. Before proceeding, we determine if the truss is statically determinate or indeterminate. Examination of Fig. 1 shows that the numbers of members, support reactions, and joints are 𝑚 = 23, 𝑟 = 3, and 𝑗 = 13, respectively. Application of Eq. (6.22) on p. 383 shows 𝑚 + 𝑟 = 2𝑗, and since the truss is fully fixed, it is statically determinate. Member 𝐴𝐶 is subjected to a vertical force of (6 kN∕m)(1 m) = 6 kN. Of this, half will be applied to joint 𝐴 and half to joint 𝐶. Similarly, member 𝐶𝐸 is subjected to a 6 kN force, and half will be applied to joint 𝐶 and half to joint 𝐸, and so on for all members of the top chord of the truss. The final forces are shown in Fig. 2. To determine the support reactions, the FBD of the whole structure is drawn in Fig. 2.

𝐷 𝐹 𝐻 𝐽 All angles are 60◦

𝐵

Road Map

Figure 1

1m 1m 1m 1m 1m 1m

Modeling

6 kN each 3 kN

+ (6 kN)(3 m) + (6 kN)(2 m) + (6 kN)(1 m) = 0 ⇒ 𝐴𝑦 = 18 kN.

3 kN

𝑦 𝐴

𝐶

𝐸

𝐼

𝐺

𝐾

𝑀

𝑀𝑥

𝑥

Governing Equations & Computation

In anticipation of using the method of sections, we will need the reactions for only one side of the truss, and we will use the left-hand side (for this problem, the right-hand side is an equally good choice). Using the FBD of Fig. 2, the only equilibrium equation needed is ∑ 𝑀𝑀 = 0 ∶ −𝐴𝑦 (6 m) + (3 kN)(6 m) + (6 kN)(5 m) + (6 kN)(4 m)

𝐿

𝐷

𝐵

𝐹

𝐻

𝐽

𝐿

𝐴𝑦

𝑀𝑦

Figure 2 Free body diagram for determination of the support reactions.

(1) (2)

Modeling

To use the method of sections to determine the force supported by member 𝐹 𝐻, we must take a cut that passes through this member. As usual, once the cut through 𝐹 𝐻 is started, it must be closed. In Fig. 2, we elect to cut through the three members shown (𝐸𝐺, 𝐹 𝐺, and 𝐹 𝐻), and then we complete the cut so that it encompasses the lefthand portion of the structure. The resulting FBD is shown in Fig. 3, where member forces are taken to be positive in tension. The unknown force 𝐹𝐹 𝐻 can quickly be determined by applying moment equilibrium about point 𝐺: ∑ 𝑀𝐺 = 0 ∶ − (18 kN)(3 m) + (3 kN)(3 m) + (6 kN)(2 m) + (6 kN)(1 m) + 𝐹𝐹 𝐻 (0.8660 m) = 0

(3)

⇒ 𝐹𝐹 𝐻 = 31.18 kN.

(4)

Although not asked for, forces supported by members 𝐸𝐺 and 𝐹 𝐺 are easily found: ∑ 𝐹𝑦 = 0 ∶ 18 kN − 3 kN − 6 kN − 6 kN + 𝐹𝐹 𝐺 sin 60◦ = 0 (5) ∑

𝐹𝑥 = 0 ∶

Discussion & Verification

⇒ 𝐹𝐹 𝐺 = −3.464 kN,

(6)

𝐹𝐸𝐺 + 𝐹𝐹 𝐺 cos 60 + 𝐹𝐹 𝐻 = 0

(7)

⇒ 𝐹𝐸𝐺 = −29.44 kN.

(8)

◦

You should also try to estimate the effort required to use the method of joints to solve this problem. Doing so will help you judge which of these methods, or combination of them, is most efficient for a particular problem.

1m

1m

1m

6 kN

6 kN

𝐶

𝐸

3 kN 𝑦

𝐴

𝐹𝐸𝐺

𝐺 𝐹𝐹 𝐺

𝑥 𝐵

𝐷

𝐹 60◦

0.8660 m 𝐹𝐹 𝐻

𝐴𝑦 = 18 kN Figure 3 Free body diagram using the method of sections.

390

Chapter 6

Structural Analysis and Machines

E X A M P L E 6.5

1500 lb 𝑃 𝑀 𝐼

Truss Analysis by the Method of Sections

𝑄

𝑁

𝐽

𝐵

𝐴

20

𝑂 𝐾

𝐶

𝐷

𝐿 20 20

SOLUTION

20

Road Map

𝐸 𝐹

dimensions in inches

1000 lb

20 𝐻

𝐺

15

30

15

Figure 1

1500 lb

𝐼

𝐵

𝐴 𝐷

𝐸

𝑏

𝑐 𝑎 𝑏

1000 lb

𝐹

𝐺

A quick inspection of the members of interest in Fig. 1 indicates that because of the number of unknowns involved with any candidate cut (some possible cuts are shown in Fig. 2), use of the method of sections will not be as straightforward as in Example 6.4. Although we will use the method of sections in this example, more generally you should also consider the method of joints, or a combination of these, and select the approach that is most efficient.† Before proceeding, by inspection we observe that there are eight zero-force members present in Fig. 1: 𝐽𝑀, 𝐽𝑃 , 𝐾𝑂, 𝐾𝑄, 𝐼𝐽 , 𝐴𝐽 , 𝐼𝑀, and 𝑀𝑃 . Given the complexity of this truss, we may also suspect that it is statically indeterminate. While we do not have details of the entire structure and thus cannot make a definitive determination, if we imagine that joints 𝐺 and 𝐻 are supported by a pin and roller, respectively, then the truss has the number of members 𝑚 = 31, number of support reactions 𝑟 = 3, and number of joints 𝑗 = 17, and the truss is fully fixed. Application of Eq. (6.22) on p. 383 shows 𝑚 + 𝑟 = 2𝑗; and since the truss is fully fixed, it is statically determinate, and we should be able to determine the desired member forces.

𝐿

𝐶

𝑐 𝑎

A transmission tower for supporting electric wires is shown. During construction,∗ the tower will, at certain times, support the load of only one wire (the 1000 lb vertical force at 𝐿), and may also be subjected to the force of a storm (crudely modeled by the 1500 lb horizontal force at 𝑃 ). To help determine if the tower has a sufficient factor of safety for this loading scenario, determine the forces supported by members 𝐴𝐷, 𝐵𝐷, 𝐵𝐸, and 𝐶𝐸.

𝐻

Figure 2 Various cuts for use by the method of sections.

Modeling Cut 𝑎𝑎 shown in Fig. 2 passes through all of the members we are interested in. But, because the resulting FBD will contain four unknowns, additional FBDs will be needed so that additional equilibrium equations can be written. Cut 𝑏𝑏 could be used, and while this also results in an FBD with four unknowns, two of these are the same as those from cut 𝑎𝑎. Thus, this solution strategy will work: draw two FBDs, one for cut 𝑎𝑎 and one for cut 𝑏𝑏, and write equilibrium expressions to obtain six equations with six unknowns. While the foregoing strategy is straightforward, some additional thought gives a clever solution using cut 𝑐𝑐, which results in the FBD shown in Fig. 3: while it has four unknowns, it is nonetheless possible to determine two of these. Then cut 𝑎𝑎, with the FBD shown in Fig. 4, can be used to obtain the remaining unknowns.

Using cut 𝑐𝑐, the FBD is shown in Fig. 3. We cannot determine 𝐹𝐴𝐵 and 𝐹𝐵𝐶 , but we can determine the remaining unknowns as follows:

Governing Equations & Computation 1500 lb

𝑦

20

𝑥 𝐼

𝐴

𝐹𝐴𝐵 𝐹𝐵𝐶

𝐹𝐴𝐷 dimensions in inches

𝐿 20

𝐶

∑

1000 lb 15

15

𝑀𝐴 = 0 ∶ −(1500 lb)(40 in.) − (1000 lb)(60 in.) − 𝐹𝐶𝐸 (30 in.) = 0 ⇒

𝐹𝐶𝐸

Figure 3 Free body diagram for cut 𝑐𝑐.

ISTUDY

∑

𝐹𝑦 = 0 ∶

−𝐹𝐴𝐷 − 𝐹𝐶𝐸 − 1000 lb = 0 ⇒

30

∗ Safety

𝐹𝐶𝐸 = −4000 lb,

𝐹𝐴𝐷 = 3000 lb.

(1) (2) (3) (4)

during construction is a difficult and sometimes overlooked aspect of the overall design of structures. † After examining Fig. 1, hopefully you determined that for this problem, the method of sections is likely more efficient than the method of joints.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

Then consider the FBD for cut 𝑎𝑎, shown in Fig. 4. Since there are now only two unknowns, only two equilibrium equations are needed, and we elect to use the following: ( ) ( ) ∑ 15 15 𝐹𝑥 = 0 ∶ 1500 lb − 𝐹𝐵𝐷 + 𝐹𝐵𝐸 = 0, (5) 25 25 ( ) ( ) ∑ 20 20 − 𝐹𝐵𝐸 − 𝐹𝐶𝐸 − 1000 lb = 0. (6) 𝐹𝑦 = 0 ∶ −𝐹𝐴𝐷 − 𝐹𝐵𝐷 25 25

1500 lb

𝑦

20

𝑥 𝐼

𝐹𝐵𝐷 = 1250 lb and 𝐹𝐵𝐸 = −1250 lb.

𝐹𝐴𝐷

(7) dimensions in inches

Discussion & Verification

As an alternative to the solution followed here, the use of cuts 𝑎𝑎 and 𝑏𝑏 is also a good solution strategy, although it entails more algebra. Nonetheless, you may wish to use this strategy, and you should obtain the same solution. If you are using a computer to solve the equilibrium equations, then the extra algebra is irrelevant.

𝐵

𝐴

Solving Eqs. (5) and (6) provides ⇒

391

𝐹𝐵𝐷 𝐹𝐵𝐸

25 20 15

15

𝐿 20

𝐶 𝐹𝐶𝐸

1000 lb

20 25 15

15

Figure 4 Free body diagram for cut 𝑎𝑎.

30

392

Chapter 6

Structural Analysis and Machines

Problems Problems 6.40 and 6.41 All members of the truss have the same length. Determine the force supported by members 𝐵𝐷, 𝐶𝐷, and 𝐶𝐸. 𝐷

𝐵

𝐵

𝐸

𝐴

𝐷

𝐹

𝐴

𝐶

𝐺 𝐶

𝑃

𝐸 𝑃

Figure P6.40

𝑃

Figure P6.41

Problems 6.42 and 6.43 The structure shown below is supported by a pin at joint 𝐴 and a horizontal link at joint 𝐵. Determine the force supported by the members indicated below. Problem 6.42

Members 𝐶𝐸, 𝐷𝐸, and 𝐷𝐹 .

Problem 6.43

Members 𝐸𝐻, 𝐹 𝐻, and 𝐹 𝐺.

60cm 60cm 60cm 60cm 1 kN 𝐴

4 kN 𝐸

2 kN 𝐶

𝐵

𝐷

𝐹 𝐺

90 cm 𝑃 3 kN 12 in. 𝐻 90 cm 𝐹 12 in. 𝐼 3 kN

Figure P6.42 and P6.43

18 in. 𝑃

18 in. 𝑃 𝐸

𝐻

18 in. 𝑃

18 in. 𝑃

𝐷

𝐼

𝐶

𝐽

𝐵 𝐴

𝐺 Figure P6.44 and P6.45

Problems 6.44 and 6.45 The structure shown above is supported by a pin and roller on an inclined surface that is parallel to the line connecting joints 𝐴 and 𝐺. If 𝑃 = 500 lb, determine the force supported by the members indicated below.

𝑄∕2 𝐴

𝑄

𝑄

𝑄

𝑄∕2

𝐵

𝐷

𝐹

𝐻

𝐶

𝐸

𝐺

𝐼

𝐽

𝐿

𝑁

4 ft

Members 𝐶𝐷, 𝐶𝐼, and 𝐼𝐽 .

Problem 6.45

Members 𝐷𝐸, 𝐸𝐼, and 𝐻𝐼.

Problem 6.46 (a) By inspection, identify the zero-force members in the truss.

𝐾

𝑀

𝑂

3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft Figure P6.46

ISTUDY

𝑃

Problem 6.44

(b) Of the zero-force members identified in Part (a), which could possibly be eliminated without reducing the strength of the truss? Explain. (c) Find the force supported by member 𝐺𝐻.

ISTUDY

Section 6.2

393

Truss Structures and the Method of Sections

Problem 6.47 𝐷

𝐵

(a) By inspection, identify the zero-force members in the truss. (b) Of the zero-force members identified in Part (a), which could possibly be eliminated without reducing the strength of the truss? Explain.

𝐹

𝐻

𝐽

2m 𝐴

𝐿

(c) Find the force supported by member 𝐹 𝐻.

𝐸

𝐶

𝐺

𝐾

𝐼

1 kN

Problem 6.48

1m 1m 1m 1m 1m 1m

(a) By inspection, identify the zero-force members in the truss. (b) Of the zero-force members identified in Part (a), which could possibly be eliminated without reducing the strength of the truss? Explain.

Figure P6.47

(c) Find the force supported by member 𝐹 𝐺. 1m

1m 𝐶

1m

1m

2m

𝐸

𝐺

1m

1 kN

1m 𝐴

𝐼 𝐵

𝐷

1m

𝐾

𝑀

𝑁

𝐹

1m

1 kN

𝐻

𝐽

𝐿

Figure P6.48

Problem 6.49 The structure has a frictionless pulley at 𝐺. (a) By inspection, identify the zero-force members in the truss. (b) Of the zero-force members identified in Part (a), which could possibly be eliminated without reducing the strength of the truss? Explain. (c) Find the force supported by member 𝐺𝐻. (d) Find the force supported by member 𝐽 𝐾. 𝐻 dimensions in inches 𝐿

𝐽

𝐹

5 5

𝐷

8 ft

20 ft

𝐿

8 ft

𝑀

4 ft 𝐺

𝐵

32

𝐶

𝑀 4

𝐸

𝐼 𝐾

𝐴

10 𝑁

100 lb 24

12

12

12

12

5 5

24

𝐻

4 kip 12 ft 8 kip

𝐺

12 ft 𝐶

𝐷

𝐸

12 ft

The electric power transmission tower supports two wires that apply the 4 kip vertical forces, and the wind force during a storm is crudely modeled by the 8 kip horizontal force.

𝐴

(a) Determine if the truss is statically determinate or indeterminate.

5 ft 5 ft

(b) Determine the force supported by the four members that emanate from joint 𝐷.

Figure P6.50

𝐾 4 kip

𝐹

Figure P6.49

Problem 6.50

𝐽

𝐼

𝐵

394

ISTUDY

Chapter 6

Structural Analysis and Machines

Problem 6.51 (a) Determine if the truss is statically determinate or statically indeterminate. (b) Determine the force supported by the five members that emanate from joint 𝐺. 𝐻 1m 𝐷

𝐵

𝐽

𝐹

𝐿

𝑁 1m

𝐴

𝐶

𝐺

𝐸

𝑀

𝐾

𝐼 3 kN

1m

1m

1m

1m

1m

1m

Figure P6.51

Problems 6.52 through 6.55 A truss used for supporting signs over an expressway is shown. The truss is supported by a pin at joint 𝐴 and a horizontal link at joint 𝐵. Let 𝑊 = 500 lb. 4 ft

4 ft

4 ft

4 ft

𝐵

𝐷

𝐹

𝐻

𝐴

𝐶

𝐸

𝐺

𝑊

𝑊

𝑊

Los Angeles next exit

4 ft

4 ft

𝐽 𝐾 𝐼

Pasadena 2 mi

𝑊

4 ft

𝐿 𝑀 𝑊

4 ft

𝑁

𝑃 𝑄

𝑂 𝑊

Rose Bowl 3 mi

𝑅

𝑊

3 ft 𝑊

rest area 4 mi

Figure P6.52–P6.55 Problem 6.52

Determine the force supported by the four members that emanate from

joint 𝐸. Problem 6.53

Determine the force supported by the four members that emanate from

joint 𝑁. Problem 6.54

Determine the force supported by the five members that emanate from

joint 𝐼. Problem 6.55

Without carrying out detailed calculations, answer the following ques-

tions. (a) The top chord of the truss has eight members (i.e., members 𝐵𝐷, 𝐷𝐹 , ..., 𝑃 𝑅). Are these members in tension or compression? Which member of the top chord will have the absolute largest force? Explain. (b) The bottom chord of the truss has eight members (i.e., members 𝐴𝐶, 𝐶𝐸, ..., 𝑂𝑄). Are these members in tension or compression? Which member of the bottom chord will have the absolute largest force? Explain. (c) There are eight diagonal members of the truss (i.e., members 𝐵𝐶, 𝐷𝐸, ..., 𝑂𝑅). Which of these members are in tension and which are in compression? Which diagonal member will have the absolute largest force? Explain.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

Problem 6.56 The boom of a tower crane is shown. The pulleys at 𝐴, 𝐵, and 𝑄 are frictionless, and 𝑊 = 10 kN. (a) Determine the force supported by cable 𝐽 𝑇 . (b) Determine the force supported by the five members that emanate from joint 𝐽 . 𝑇

15◦ 0.2 m 𝐵

𝐹

𝐽

𝐷

𝐻

𝑁

𝐿

𝑄 0.2 m

1m

𝑅

𝐴 0.2 m 𝐶

𝐸

𝐺

𝐾

𝐼

𝑀

𝑂

𝑆 (a)

(b)

(c)

(d)

𝑃

𝑊 1m

1m

1m

1m

1m

1m

1m

1m

Figure P6.57 Figure P6.56

Problems 6.57 and 6.58 (a)

(b)

(c)

(d)

Determine if each truss is statically determinate, statically indeterminate, or a mechanism.

Problem 6.59 For the K truss shown, determine the force supported by members 𝐸𝐻, 𝐹 𝐻, 𝐹 𝐼, and 𝐺𝐼. Hint: Use the method of sections twice with cuts such as those described in Example 6.5 on p. 390. 𝐸

𝐵

𝐻

𝐽

𝑀

6 ft 6 ft𝐴

𝐹

𝐶

𝐾

𝑁 𝑃

𝐷

𝐺

𝐼

𝐿

𝑂

2 kip 2 kip 4 kip 6 kip 6 kip 8 ft

8 ft

8 ft

8 ft

8 ft

8 ft

Figure P6.59 and P6.60

Problem 6.60 For the K truss shown, determine the force supported by members 𝐻𝐽 , 𝐻𝐾, 𝐼𝐾, and 𝐼𝐿. Hint: Use the method of sections twice with cuts such as those described in Example 6.5 on p. 390.

Figure P6.58

395

396

Chapter 6

Structural Analysis and Machines

Problem 6.61 A gambrel room-in-attic truss is shown. (a) Use equation counting, Eq. (6.22) on p. 383, to show that this truss is a mechanism.

𝐸

(b) This truss type is popular for many applications. Discuss why this truss design does not collapse in view of the fact that it is a mechanism according to truss theory, as shown in Part (a).

𝐺

𝐷 𝐹 𝐵

𝐼 𝐽

𝐴 𝐶

𝐻

Figure P6.61

(c) If the roller support at 𝐽 is replaced by a pin support, equation counting indicates that the number of equations and the number of unknowns is the same. With a pin support at 𝐽 , is the structure no longer a mechanism according to truss theory? Explain. Note: Concept problems are about explanations, not computations.

Problem 6.62 The two truss designs shown differ only in their depths. (a) Determine the forces supported by all members of both trusses, and compare values for corresponding members in each truss. (b) Offer reasons why truss (b) is the better design. (c) Offer reasons why truss (a) is the better design. (d) Overall, which of the two trusses do you believe is stronger, assuming all members are of the same material and have the same cross-sectional shape? 𝐵 𝐷

𝐵 1m

𝐹

𝐴

𝐷

2m 𝐴

𝐹

𝐶 𝐸 𝑊 𝑊

𝐶 𝐸 𝑊 𝑊

1m 1m 1m

1m 1m 1m

(a)

(b)

𝑃 𝐵 𝑘 𝐿

Figure P6.63

Problem 6.64

𝑃 𝐵

𝐿

𝑘𝑡 𝐴

ISTUDY

Problem 6.63 A bar 𝐴𝐵 is supported by a pin at 𝐴 and a horizontal spring 𝐵𝐶 with stiffness 𝑘. When the bar is vertical, the force in the spring is zero. The force 𝑃 applied to the bar is vertical. Determine the value of 𝑃 at which the structure buckles (i.e., becomes unstable). Express your answer in terms of parameters like 𝑘, 𝐿, etc. Hint: For small angles, with 𝜃 measured in radians, sin 𝜃 ≈ 𝜃 and cos 𝜃 ≈ 1.

𝐴

Figure P6.64

Figure P6.62

𝐶

A bar 𝐴𝐵 is supported by a pin at 𝐴 and a torsional spring at 𝐵 with stiffness 𝑘𝑡 . When the bar is vertical, the moment in the spring is zero. The force 𝑃 applied to the bar is vertical. Determine the value of 𝑃 at which the structure buckles (i.e., becomes unstable). Express your answer in terms of parameters like 𝑘𝑡 , 𝐿, etc. Hint: For small angles, with 𝜃 measured in radians, sin 𝜃 ≈ 𝜃 and cos 𝜃 ≈ 1.

ISTUDY

Section 6.2

Truss Structures and the Method of Sections

Design Problems General Instructions. In all problems, write a brief technical report, following the guidelines of Appendix A, where you summarize all pertinent information in a wellorganized fashion. It should be written using proper, simple English that is easy to read by another engineer. Where appropriate, sketches along with critical dimensions should be included. Discuss the objectives and constraints considered in your design, the process used to arrive at your final design, safety issues if appropriate, and so on. The main discussion should be typed, and figures, if needed, can be computer-drawn or neatly handdrawn. Include a neat copy of all supporting calculations in an appendix that you can refer to in the main discussion of your report. A length of a few pages, plus appendix, should be sufficient. Design Problem 6.1 Design a footbridge for crossing a small stream. The bridge is intended for residential use only. The bridge consists of two identical trusses, spaced 3 f t apart. For ease of fabrication, each truss is to be constructed of one size of welded steel pipe from Table 3.3 on p. 184, with all members having the same 3 f t length. The 100 lb forces represent the dead loads. You are to specify the maximum safe live load for the bridge and the diameter of pipe to be used.

𝐵

3 ft 𝐴

𝐷

𝐹

𝐸 100 lb 100 lb 100 lb 100 lb 3 ft 3 ft 3 ft 𝐶

𝐺

Figure DP6.1

Design Problem 6.2 Design a truss to be made of welded steel pipe to support the loads that are given. The truss is to be supported at joints 𝐴 and 𝐸 and should have joints at points 𝐵, 𝐶, and 𝐷 (joints at other locations are also permissible). For ease of fabrication, the truss should be constructed from only one size of pipe from Table 3.3 on p. 184. You may consider the weight of the truss as already being approximately accounted for through the loads that are given, and no additional factors of safety beyond those already incorporated in Table 3.3 are needed. While it is not necessary to determine the truss design with minimum weight, it is desirable to use a small-diameter pipe so that the truss is reasonably economical.

2000 lb 2000 lb 2000 lb 1000 lb 1000 lb 𝐶 𝐵 𝐷 𝐸 4 ft 𝐴 6 ft Figure DP6.2

6 ft

6 ft

6 ft

397

398

Chapter 6

Structural Analysis and Machines Design Problem 6.3

𝐶 5 ft 𝐵

𝐴

Design a simply supported wooden roof truss for residential construction. The truss must have the span and height shown, and it will support the following loads: • The bottom chord of the truss is unloaded.

10 ft

10 ft

• The top chord of the truss is subjected to a dead load of 40 lb∕f t of chord length and a maximum live load of 80 lb∕f t of chord length (all loads are vertical).

Figure DP6.3

• All other loads can be neglected.

Table 6.1 Allowable loads for dimensioned lumber where 𝐿 is the length of the member.

ISTUDY

Dimensioned Lumber Allowable load 2 × 4 in.

2 × 6 in.

Tension:

8000 lb

14,000 lb

Compression: 0 < 𝐿 ≤ 2 ft 2 ft < 𝐿 ≤ 4 ft 4 ft < 𝐿 ≤ 6 ft 6 ft < 𝐿 ≤ 8 ft

8000 lb 2000 lb 1000 lb 0

14,000 lb 3000 lb 1500 lb 1000 lb

The members of the truss are to be wood with 2 × 4 in. and/or 2 × 6 in. cross section (nominal dimensions). For ease of fabrication, the top chord should be entirely one dimension of member, and similarly for the bottom chord. The wood members have the allowable strengths given in Table 6.1. You are to specify the member size for both the top and bottom chords, and specify the size and placement of additional truss members between the top and bottom chords that you determine are needed. Your design does not need to be fully optimized in terms of minimizing the amount of wood used, but your design should be reasonably economical (e.g., a truss constructed entirely of 2 × 6 in. wood is probably not economical). If you determine that it is not possible to achieve the strength needed using 2 × 4 in. and/or 2 × 6 in. members, then you should specify the number of identical trusses that should be attached to one another so that the specified loads can be supported.

ISTUDY

Section 6.3

6.3

Trusses in Three Dimensions

Trusses in Three Dimensions

A space truss is a three-dimensional structure that consists of two-force members only, where members are organized so that the assemblage as a whole behaves as a single object. Some examples of space trusses are shown in Fig. 6.24.

Glow Images

jiawangkun/Shutterstock

(a)

Perry Mastrovito/Digital Stock/CORBIS

(b)

Michael Plesha

(c)

(d)

Figure 6.24. Examples of space trusses. (a) The base of the Eiffel Tower in Paris, France, showing an intricate arrangement of individual truss members that form larger structural members, which serve as truss members for the entire structure. (b) The Expo Center, constructed in Vancouver, Canada, for the 1986 World Exposition, uses a spherical truss structure. (c) A view from the base of a tower that supports electric wires. (d) A large crane in a lowered position in anticipation of an approaching storm.

For a three-dimensional structure to be a space truss, it must have the following characteristics: • All members must be connected to one another by frictionless ball-and-socket joints. • Each member may have no more than two joints. • Forces may be applied at joints only. • The weight of individual members must be negligible. If all of the those characteristics are satisfied, then it is guaranteed that all members of the structure are two-force members. Note that while trusses are most often

399

400

Chapter 6

Structural Analysis and Machines

Interesting Fact Computer modeling. Because of the complexity of most space trusses, computer methods of analysis such as the finite element method are usually used. These computer programs automate the handling of data, generation of equilibrium equations, and solving for unknowns. However, you are responsible for modeling, determination of loads, interpretation of results, and—hugely important—assessing the accuracy of results. The methods of analysis discussed for trusses in this book give you many of the tools you need to interpret the results of computer models and to help ensure their accuracy.

constructed using straight members, they may contain members with curved or other complex shapes, provided they are two-force members. For practical applications, the definition of a space truss is restrictive, because in real life, members are usually attached to one another using connections that are not ball-and-socket joints. Similar to the discussion in Fig. 6.4 on p. 365, the connections in real structures are usually capable of supporting moments, and thus, the members that emanate from such connections usually are not two-force members. For precise analysis, structures with moment-resisting connections should be modeled as frame structures, which are discussed in Section 6.4. Nonetheless, if a particular structure would qualify as a truss except for the connection details, in engineering practice these structures are often modeled as trusses anyway. Fundamentally, the methods for equilibrium analysis of space trusses are the same as those for plane trusses. That is, both the method of joints and method of sections can be used. The major differences for space trusses are that FBDs are usually more intricate, reactions are more complex, and the number of unknowns to be determined and the number of equations to be solved are greater.

Stability of space trusses and design considerations

Figure 6.25 In a simple space truss the arrangement of members begins with six members that form a tetrahedron, and adding to this three new members (shown by dashed lines) for each new joint that is added.

ISTUDY

A significant difference between space trusses and plane trusses is the difficulty in assessing stability. If a space truss is unstable, it is often difficult, even for experienced engineers, to identify this defect based on inspection only. An effective approach to help avoid instability is to design space trusses starting with the tetrahedral arrangement of members shown in Fig. 6.25; such structures are called simple space trusses. While there are advanced analytical methods to help determine the stability of a truss, the simple rule of thumb called equation counting, which was introduced in Section 6.2 for plane trusses, is very useful with the modifications described here. To help determine whether a truss is statically determinate or statically indeterminate, and whether it is stable or unstable, we compare the number of unknown forces to the number of equilibrium equations. Consider a space truss having 𝑚 = number of members, 𝑟 = number of support reactions,

(6.28)

𝑗 = number of joints. Each member has one unknown force, and each support reaction has one unknown force, so 𝑚 + 𝑟 is the total number of unknowns. For a space truss, each joint has three equilibrium equations, so 3𝑗 is the total number of equations. Thus, the rule of thumb∗ is as follows: If 𝑚 + 𝑟 < 3𝑗

The truss is a mechanism and/or has partial fixity.

If 𝑚 + 𝑟 = 3𝑗

The truss is statically determinate if it has full fixity. The truss is statically indeterminate if it has partial fixity.

If 𝑚 + 𝑟 > 3𝑗 ∗ The

The truss is statically indeterminate, and it can have full fixity or partial fixity.

comments following Eq. (6.22) on p. 383 also apply here (with 2𝑗 replaced by 3𝑗).

(6.29)

ISTUDY

Section 6.3

Use of this rule of thumb requires that we inspect a truss structure to determine its fixity. A truss (or any object) in three dimensions has the potential for six types of rigid body motion: translation in each of the 𝑥, 𝑦, and 𝑧 directions and rotation about each of the 𝑥, 𝑦, and 𝑧 axes, or a combination of these. Thus, a structure that is fully supported against rigid body motion (i.e., fully fixed) requires a minimum of six support reactions that are properly arranged. If a structure in three dimensions has less than six support reactions, then it will have only partial fixity. Several examples are shown in Fig. 6.26.

Trusses in Three Dimensions

401

(a) 11 + 6 < 3(6) (b) 8 + 6 < 3(5) 𝑚 + 𝑟 < 3𝑗 ⇒ mechanism

End of Section Summary In this section, methods of analysis for space trusses were described. Some of the key points are as follows: • The methods for equilibrium analysis of space trusses are the same as those for plane trusses. That is, both the method of joints and the method of sections can be used. • A simple space truss has members arranged in a tetrahedral pattern, as shown in Fig. 6.25. One of the main features of simple space trusses is that they are always stable.

(c) 6 + 6 = 3(4) (d) 12 + 6 = 3(6) 𝑚 + 𝑟 = 3𝑗 ⇒ statically determinate

• A rule of thumb called equation counting, when used with good judgment, can be effective for determining the stability and static determinacy of a space truss.

(e) 10 + 6 > 3(5) (f) 13 + 6 > 3(6) 𝑚 + 𝑟 > 3𝑗 ⇒ statically indeterminate Figure 6.26 Examples of equation counting to determine if a truss is statically determinate or indeterminate and whether it is stable. All trusses shown here have six support reactions that provide full fixity.

402

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Structural Analysis and Machines

E X A M P L E 6.6

Space Truss Analysis by the Method of Joints A boom for a crane is shown in a horizontal position. If 𝑊 = 1 kN, determine the force supported by member 𝐷𝐺.

2m 2m 2m

1m

𝐽

1m 𝑁 𝑀

𝐺

𝑧 𝑥

𝐷

𝑦 𝑇

𝐿 𝐼 𝐹

2m

𝐶

𝐴 2m

𝐵 2m

𝐸 2m

𝐻 2m

𝐾 2m

𝑊 Figure 1

SOLUTION Road Map

2m ⃗ 𝑇𝐷𝐺 𝐺

1m 𝐷

𝑇⃗𝐷𝐴 ⃗ ⃗ 𝑇𝐷𝐶 𝑇𝐷𝐵

2m

𝑇⃗𝐴𝐷 𝐴

2m

𝑇⃗𝐴𝐶 𝑇⃗

𝐶

𝑧 𝑥

𝑦 𝐵

𝐴𝐵

1 kN

2m 1 kN Figure 2 Free body diagrams of joints 𝐴 and 𝐷.

ISTUDY

Before beginning our analysis, we determine if the truss is statically determinate or indeterminate. Examination of Fig. 1 shows that the number of members is 𝑚 = 33, number of support reactions is 𝑟 = 9, number of joints is 𝑗 = 14, and the truss is fully fixed. Application of Eq. (6.29) on p. 400 provides 𝑚 + 𝑟 = 3𝑗, and since the truss is fully fixed, it is statically determinate. To use the method of joints to determine the force supported by member 𝐷𝐺, we examine joint 𝐷 in Fig. 1 to see that there are four members that emanate from it, and thus its FBD will involve four unknowns, while only three equilibrium equations are available. Examination of joint 𝐴 shows that its FBD will have three unknowns, and therefore this is an ideal joint to begin with. The FBDs for joints 𝐴 and 𝐷 are shown in Fig. 2, where member forces are positive in tension, and the weight of the members is neglected. Assuming the pulley is frictionless and the cable is weightless, the force supported throughout the entire cable is the same, with value 𝑇 = 1 kN. Either we can leave the pulley attached to the truss,∗ or as shown in Fig. 2, we can remove the pulley by shifting the pulley forces to the bearing of the pulley.

Modeling

Governing Equations & Computation Referring to the FBD for joint 𝐴 shown in Fig. 2, ∑ we observe that the equilibrium equation 𝐹𝑧 = 0 will have only one unknown, namely, 𝑇𝐴𝐷 . The vector expression for the force that member 𝐴𝐷 exerts on joint 𝐴 is

( ) −̂𝚤 + 2𝑘̂ 𝑇⃗𝐴𝐷 = 𝑇𝐴𝐷 . √ 5

(1)

∗ If we elect to leave the pulley on the truss, then we will need to assume a value for the radius of the pulley.

The radius we select, however, is arbitrary, since all member forces, reactions, etc., will be the same as those found in this solution regardless of the pulley’s radius. If you are uncertain that these comments are true, or how to proceed if the pulley is left on the truss, then you should review Fig. 5.11 on p. 299.

ISTUDY

Section 6.3

Trusses in Three Dimensions

Using a scalar approach to sum forces in the 𝑧 direction provides Joint A: ( ) ∑ 2 𝐹𝑧 = 0 ∶ −1 kN + 𝑇𝐴𝐷 √ = 0 ⇒ 𝑇𝐴𝐷 = 1.118 kN. 5

(2)

We omit finding the remaining unknowns for joint 𝐴 since they are not needed for the equilibrium of joint 𝐷. Referring to the FBD for joint 𝐷 shown in Fig. 2, we see that it will probably be easier to write the equilibrium equations using a vector approach. Vector expressions for the forces that emanate from joint 𝐷 are ( ) 𝚤̂ − 2𝑘̂ , 𝑇⃗𝐷𝐴 = −𝑇⃗𝐴𝐷 = (1.118 kN) √ 5 ( −̂𝚤 + 𝚥̂ − 2𝑘̂ ) , 𝑇⃗𝐷𝐵 = 𝑇𝐷𝐵 √ 6 ( −̂𝚤 − 𝚥̂ − 2𝑘̂ ) , 𝑇⃗𝐷𝐶 = 𝑇𝐷𝐶 √ 6 𝑇⃗ = 𝑇 (−̂𝚤 ). 𝐷𝐺

𝐷𝐺

The equilibrium equations for joint 𝐷 are Joint D: ∑ ⃗ 𝐹⃗ = 0⃗ ∶ 𝑇⃗𝐷𝐴 + 𝑇⃗𝐷𝐵 + 𝑇⃗𝐷𝐶 + 𝑇⃗𝐷𝐺 = 0.

(3) (4) (5) (6)

(7)

Substituting Eqs. (3) through (6) into the above along with 𝑇𝐴𝐷 = 1.118 kN, and grouping terms in the 𝑥, 𝑦, and 𝑧 directions, provide the following three equations: ) ) ) ( ( ( −1 1 −1 (8) 𝑇𝐷𝐵 √ + 𝑇𝐷𝐶 √ + 𝑇𝐷𝐺 (−1) = (−1.118 kN) √ , 6 6 5 ( ) ) ( 1 −1 = 0, (9) 𝑇𝐷𝐵 √ + 𝑇𝐷𝐶 √ 6 6 ( ( ) ) ( ) −2 2 −2 𝑇𝐷𝐵 √ + 𝑇𝐷𝐶 √ = (1.118 kN) √ . (10) 6 6 5 Solving these equations provides ⇒

𝑇𝐷𝐵 = 𝑇𝐷𝐶 = −0.6124 kN and

𝑇𝐷𝐺 = 1 kN.

(11)

Discussion & Verification Because the loading for this truss is simple, based on inspection we expect members 𝐴𝐷 and 𝐷𝐺 to be in tension, and members 𝐷𝐵 and 𝐷𝐶 to be in compression, and indeed our solution shows this. Further, because of the symmetry of the problem, we expect the force supported by members 𝐷𝐵 and 𝐷𝐶 to be the same, and our solution also shows this. Other checks include substituting the solutions into the original equilibrium equations to verify that all of these are satisfied. However, this does not provide a check that the equilibrium equations are correct. Therefore, it is essential that FBDs be accurately drawn and equilibrium equations be correctly written.

403

404

Chapter 6

Structural Analysis and Machines

E X A M P L E 6.7

Space Truss Analysis by the Method of Sections For the structure in Example 6.6 with 𝑊 = 1 kN, determine the force supported by all members that emanate from the supports at points 𝐾, 𝐿, and 𝑁.

SOLUTION

𝐷

𝐺

𝐽

𝐴 𝐵

𝑁

𝐸

𝑇

𝐿

𝐼 𝐻

Road Map

𝑧 𝑥

𝐹

𝐶

𝑀

𝑦

𝐾

𝑊

Figure 1 To use the method of sections, a cut is taken that passes through the structure, intersecting the members of interest.

ISTUDY

We will use the method of sections with a cut that passes through the members of interest, as shown in Fig. 1. Observe that the cut passes through six members. Since six equilibrium equations are available, we expect to be able to determine all of the unknown member forces using the FBD that results from this cut. Modeling Using the cut shown in Fig. 1, the FBD for the left-hand portion of the structure is shown in Fig. 2, where member forces are positive in tension, and the weight of the members is neglected. Assuming the pulley is frictionless and the cable is weightless, the force supported throughout the entire cable is the same, with value 𝑇 = 1 kN. We can either leave the pulley attached to the truss or, as shown in Fig. 2, remove the pulley by shifting the pulley forces to the bearing of the pulley. In this problem, removing the pulley from the truss will be slightly easier since we will be able to combine the two 1 kN forces at point 𝐴 in Fig. 2 into a single force vector, and thus we will be able to evaluate its moment using one cross product. 1m

2m 2m 2m

1m

𝑇⃗𝑀𝐿

𝐺

𝐷 𝐼

𝑇⃗𝐼𝐿

𝐹 𝐶

2m 𝐴 2m

𝐵 2m

1 kN

𝑁

𝑀

𝐽

𝐸 2m

𝐻 2m

𝑇⃗𝑀𝑁 𝑧

𝑇⃗𝑀𝐾 𝑥 𝐿 𝑇⃗𝐻𝐿

𝑦

𝐾 𝑇⃗𝐻𝐾 2m

1 kN Figure 2. Free body diagram. Governing Equations & Computation

Before we write equilibrium equations, vector expressions for truss member forces and the cable forces at 𝐴, which we will call 𝐹⃗ , are needed: ( −2̂𝚤 − 2̂𝚥 ) 𝑇⃗𝐻𝐿 = 𝑇𝐻𝐿 𝑇⃗𝐻𝐾 = 𝑇𝐻𝐾 (−̂𝚤 ), , (1) √ 8 ( −̂𝚤 + 𝚥̂ − 2𝑘̂ ) 𝑇⃗𝑀𝐾 = 𝑇𝑀𝐾 , (2) 𝑇⃗𝐼𝐿 = 𝑇𝐼𝐿 (−̂𝚤 ), √ 6 ( −̂𝚤 − 𝚥̂ − 2𝑘̂ ) , 𝑇⃗𝑀𝑁 = 𝑇𝑀𝑁 (−̂𝚤 ), (3) 𝑇⃗𝑀𝐿 = 𝑇𝑀𝐿 √ 6 𝐹⃗ = (1 kN)(−̂𝚤 − 𝑘̂ ).

(4)

Equilibrium of forces will provide three scalar equations with six unknowns, while equilibrium of moments about point 𝑀 will provide three scalar equations with three unknowns; thus, we begin by summing moments. Using the following position vectors ̂ m, 𝑟⃗𝑀𝐴 = (7̂𝚤 − 2𝑘)

̂ m, 𝑟⃗𝑀𝐻 = (̂𝚤 + 𝚥̂ − 2𝑘)

̂ m, 𝑟⃗𝑀𝐼 = (̂𝚤 − 𝚥̂ − 2𝑘)

(5)

ISTUDY

Section 6.3

Trusses in Three Dimensions

we sum moments about point 𝑀, ∑ ⃗ ⃗ = 0⃗ ∶ 𝑟⃗ × 𝐹⃗ + 𝑟⃗ ⃗ ⃗ 𝑀 ⃗𝑀𝐼 × 𝑇⃗𝐼𝐿 = 0. 𝑀 𝑀𝐴 𝑀𝐻 × (𝑇𝐻𝐾 + 𝑇𝐻𝐿 ) + 𝑟

(6)

Carrying out the cross products in Eq. (6), canceling the meter unit, and grouping terms in the 𝑥, 𝑦, and 𝑧 directions give the following equations ( ) −4 (7) √ 𝑇𝐻𝐿 = 0, 8 ( ) 4 2 𝑇𝐻𝐾 + √ 𝑇𝐻𝐿 + 2 𝑇𝐼𝐿 = −9 kN, (8) 8 𝑇𝐻𝐾 − 𝑇𝐼𝐿 = 0.

(9)

Equations (7)–(9) are easily solved to obtain 𝑇𝐻𝐿 = 0 and 𝑇𝐻𝐾 = 𝑇𝐼𝐿 = −2.25 kN.

⇒

Equilibrium of forces provides ∑ ⃗ 𝐹⃗ = 0⃗ ∶ 𝐹⃗ + 𝑇⃗𝐻𝐾 + 𝑇⃗𝐻𝐿 + 𝑇⃗𝐼𝐿 + 𝑇⃗𝑀𝐾 + 𝑇⃗𝑀𝐿 + 𝑇⃗𝑀𝑁 = 0.

(10)

(11)

By using the solutions given in Eq. (10) and grouping terms in the 𝑥, 𝑦, and 𝑧 directions, Eq. (11) gives the following equations: ) ) ( ( −1 −1 (12) √ 𝑇𝑀𝐾 + √ 𝑇𝑀𝐿 − 𝑇𝑀𝑁 = −3.5 kN, 6 6 ) ( ) ( −1 1 (13) √ 𝑇𝑀𝐾 + √ 𝑇𝑀𝐿 = 0, 6 6 ) ) ( ( −2 −2 (14) √ 𝑇𝑀𝐾 + √ 𝑇𝑀𝐿 = 1 kN. 6 6 Equations (12)–(14) are easily solved to obtain ⇒

𝑇𝑀𝑁 = 4 kN and 𝑇𝑀𝐾 = 𝑇𝑀𝐿 = −0.6124 kN.

(15)

Discussion & Verification

• Because of the complexity of most space trusses, verification of the solution can be challenging. In this problem, based on inspection, we expect member 𝑀𝑁 to be in tension and members 𝐻𝐾 and 𝐼𝐿 to be in compression, and indeed our solution shows this. Further, because of the symmetry of the problem, we expect the forces supported by members 𝐻𝐾 and 𝐼𝐿 to be the same and the forces supported by members 𝑀𝐾 and 𝑀𝐿 to be the same, and our solution also shows this. • A simple partial check of our solution’s quantitative accuracy can be obtained by using a scalar approach to evaluate moment equilibrium about line 𝐾𝐿 in the FBD of Fig. 2. Taking moments to be positive in the 𝑦 direction, we obtain ∑ 𝑀𝐾𝐿 = 0 ∶ (1 kN)(8 m) − 𝑇𝑀𝑁 (2 m) = 0 ⇒ 𝑇𝑀𝑁 = 4 kN, (16) which agrees with the result found earlier. You should study Fig. 2 to see if other scalar equilibrium equations can be written to help you to check the remaining solutions.

405

406

Chapter 6

Structural Analysis and Machines

Problems Problem 6.65

𝑧

The truss shown has a socket support at point 𝐷, rollers at points 𝐸 and 𝐹 , and a link from 𝐸 to 𝐺 that prevents motion in the 𝑥 direction.

𝑅 𝑄

(a) Does the truss have partial fixity or full fixity?

𝐵

(b) Determine if the truss is statically determinate or statically indeterminate.

𝐶 𝑃

(c) Determine the force supported by all members of the truss if 𝑃 = 1 kip, 𝑄 = 2 kip, and 𝑅 = 3 kip.

𝐴 20 ft 𝐺

𝐹

𝐸 12 ft

𝐷

Problem 6.66 𝑦

The truss shown has a socket support at point 𝐴 and rollers at points 𝐵 and 𝐶. Points 𝐴, 𝐵, and 𝐶 lie in the 𝑥𝑦 plane, and point 𝐸 lies in the 𝑦𝑧 plane. (a) Does the truss have partial fixity or full fixity?

9 ft

𝑥

(b) Determine if the truss is statically determinate or statically indeterminate.

Figure P6.65

(c) Determine the force supported by all members of the truss if 𝑃 = 1 kN. Force 𝑃 is vertical.

𝑧 5m

𝐷

𝐸 𝑃

3m 𝐶 𝐴 1m 𝑥

0.5 m 0.5 m

𝑦

Problem 6.67 The box shown has cardboard sides and wood strips along the edges and from corner to corner. The strength of the box is provided primarily by the wood strips, and a truss model with 18 members for the wood strips is also shown. Let the truss be supported by rollers at points 𝐴 and 𝐵, a socket at point 𝐶, and a short link at 𝐸 that is parallel to the 𝑦 direction. For the 100 N and 150 N forces shown, determine the force supported by the six members that emanate from joint 𝐴.

𝐵

𝑧 150 N

Figure P6.66

ISTUDY

𝐻

𝐸

𝐺 252 mm

𝐹

100 N 𝐷

𝐴 𝑥 240 mm

𝐶

𝐵 275 mm

𝑦

Figure P6.67 and P6.68

Problem 6.68 Repeat Prob. 6.67 to determine the force supported by the six members that emanate from joint 𝐶.

ISTUDY

Section 6.3

Trusses in Three Dimensions

Problem 6.69 Imagine you have been retained by an attorney to serve as an expert witness for possible litigation regarding an accident that occurred on the playground structure shown. You have been given only preliminary information on this structure, including the following: • It was designed as a truss structure. • It had been in service for several years, without incident. • It collapsed while an adult was climbing on it, causing serious injury to the person. • It was thought to be in good condition at the time of the accident. • Detailed information on the design, fabrication, installation, and so forth has not yet been made available to you. Until you have fully studied all of the information available, it is not possible for you to render an informed opinion on this structure and/or the accident that occurred. Nonetheless, the attorney retaining you is anxious to informally hear your initial impressions. With the information provided, what would you tell the attorney? In answering this, consider responses such as: the structure may have been defective in its design, it may have been defective in its fabrication or installation, the adult probably should not have been on the structure and thus, he or she is probably responsible for its failure, or I really don’t know until I have further information.

Problems 6.70 through 6.74 The truss shown has a socket support at point 𝐿, a roller at point 𝐾, and a cylindrical roller at point 𝐴 that prevents motion in the 𝑥 and 𝑧 directions. Problem 6.70

Determine if the truss is statically determinate or statically indetermi-

nate. Problem 6.71 Use the method of joints to determine the force supported by member 𝐷𝐺 if 𝑃 = 40 kN and 𝑄 = 0 . Problem 6.72 Use the method of joints to determine the force supported by member 𝑀𝐽 if 𝑃 = 𝑄 = 10 kN. Problem 6.73 Use the method of sections to determine the force supported by member 𝐽 𝐺 if 𝑃 = 40 kN and 𝑄 = 0 . Problem 6.74 Use the method of sections to determine the force supported by member 𝐽 𝐺 if 𝑃 = 𝑄 = 10 kN. 1m 𝑧 𝑥

𝑄

2m

𝑃

2m

𝑀

2m

𝐽

1m

𝐺

𝑦

𝐷

𝐿 𝐼 𝐹

𝐾

2m

𝐶 2m

𝐻 2m

𝐸 2m

𝐴 𝐵 2m

Figure P6.70–P6.74

2m

Figure P6.69

407

408

Chapter 6

Structural Analysis and Machines

6.4

Frames and Machines

Analysis of frames and machines is discussed in this section, and these are defined as follows: Frame. A frame is a structure that contains one or more three-force and/or multiforce members. Typically, a frame is fully fixed in space and uses a stationary arrangement of members with the goal of supporting forces that are applied to it.

3DMI/Shutterstock

Figure 6.27 The mountain bike frame shown is an example of a frame structure where multiforce members are used and the connections between members are often moment-resisting. Indeed, this structure is also a machine because the spring suspension mechanism allows for some relative motion of components as the bike travels over rough terrain.

Machine. A machine is an arrangement of members where typically the members can have significant motion relative to one another. The usual goal of a machine is transmission of motion and/or force. While a machine can be composed entirely of two-force members, in which case it may be analyzed as a truss, in this section we are interested in machines that contain one or more multiforce members. These definitions are broad and have considerable overlap. Nonetheless, when multiforce members are involved, the methods of static analysis for frames and machines are the same. Throughout statics, we assume that if motion is possible, it is slow ∑ ⃗ ∑𝑀 ⃗ ⃗ ≠ 0, enough that inertia can be neglected. If this is not the case, then 𝐹⃗ ≠ 0, and methods of analysis from dynamics must be used.

Analysis procedure and free body diagrams (FBDs)

Sandia National Laboratories

Figure 6.28 This microelectromechanical system (MEMS) machine was fabricated at Sandia National Laboratory. The machine features a mirror whose tilt angle can be adjusted by a motor (not shown) that drives gears and a rack and pinion. The largest gear is about 150 𝜇m in diameter (a human hair is about 100 𝜇m in diameter). The methods discussed in this section can be used to analyze such systems. A large challenge is controlling contact and friction forces in MEMS.

ISTUDY

The main objective in structural analysis of frames and machines is to determine the forces supported by all members. Depending on the arrangement and loading of members, a particular problem may involve a combination of particles in equilibrium (discussed in Chapter 3), two-force members in equilibrium (truss analysis, discussed in Sections 6.1 and 6.2), and multiforce members in equilibrium (discussed in Chapter 5). Thus, there are no new technical tools needed to analyze frames and machines. Rather, the challenge is to combine the many skills you have already developed. Determining the forces supported by the individual members of a frame or machine requires that each member be separated from its neighboring members, with an FBD being drawn for each. The forces of action and reaction between the various members must be shown in the FBDs so that Newton’s third law is satisfied. Using these FBDs, equilibrium equations are then written for each member, and presuming the frame or machine is statically determinate, these equations are solved to determine the forces supported by all members. Because frames and machines usually involve many members and because many of these will typically be three-force members and/or multiforce members,∗ many FBDs are usually needed, and sometimes these are complex. You will likely find the FBDs for frames and machines to be among the most challenging you will encounter in statics, and all of the skills you have cultivated for drawing FBDs will be needed. To draw the FBDs needed for the analysis of a frame or machine, the procedure for drawing the FBD for a single body, discussed in Section 5.2, is repeatedly applied to each of the members of the frame or machine. This procedure is as follows: ∗ If only two-force members were involved, then the structure or machine would be a truss, and the methods

of Chapter 3 and Sections 6.1 and 6.2 would be sufficient.

ISTUDY

Section 6.4

409

Frames and Machines

Procedure for Drawing FBDs 1. Disassemble the frame or machine into individual members. For each member: 2. Imagine this member is “cut” completely free (separated) from its environment. That is:

Helpful Information Two-force members. When drawing FBDs for frames and machines, it is helpful to first identify two-force members and label their unknowns. Doing this provides for less complicated FBDs, fewer equilibrium equations that need to be written, and fewer unknowns to be determined.

• In 2D, think of a closed line that completely encircles the member. • In 3D, think of a closed surface that completely surrounds the member. 3. Sketch the member. 4. Sketch the forces and moments. (a) Sketch the forces and moments that are applied to the member by the environment (e.g., weight). (b) Wherever the cut passes through a structural member, sketch the forces and moments that occur at that location. (c) Wherever the cut passes through a support (i.e., where a support is removed from the member), sketch the reaction forces and/or moments that occur at that location. 5. Sketch the coordinate system to be used. Add pertinent dimensions and angles to the FBD to fully define the locations and orientations of all forces and moments.

Common Pitfall Statics versus dynamics. Consider the structure shown, where member 𝐴𝐵 is pinned to member 𝐵𝐶 and force 𝐹 is known. 𝐹 𝐵

The order in which the forces are sketched in Step 4 is irrelevant. For complicated FBDs, it may be difficult to include all of the dimensions and/or angles in Step 5. When this is the case, some of this information may be obtained from a different sketch.

𝐶

𝐴 The following FBD

Examples of correct FBDs Figure 6.29 shows several examples of properly constructed FBDs for frames and machines. Comments on the construction of these FBDs follow.

Traffic signal pole. After sketching each member, we apply external forces, which consist of the weights 𝑊𝐶 and 𝑊𝐻 . If you want to consider the weights of other members (e.g., the weight of member 𝐴𝐵𝐹 may be relatively large), then their weights can also be included in the FBDs. Next, we apply the appropriate reactions for the built-in support at point 𝐴. Then we search for two-force members and identify cables 𝐷𝐸 and 𝐺𝐹 , and we assign their forces as 𝑇𝐹 𝐺 and 𝑇𝐷𝐸 , respectively. Finally, at locations where members are joined to one another, we include the appropriate forces, ensuring that forces of action and reaction between interacting bodies are equal in magnitude and opposite in direction (i.e., Newton’s third law is satisfied).

𝐹

𝑦

𝐴𝑥

𝑥 𝐴𝑦

𝐶𝑦

is properly drawn. However, a common pitfall for problems like this is to write equilib∑ rium equations such as 𝑀𝐴 = 0 to determine the reaction 𝐶𝑦 . The error in writing such equilibrium equations is that this is not a problem of static equilibrium. Clearly, in this structure with this loading, members 𝐴𝐵 and 𝐵𝐶 will undergo accelerations, and ∑ hence 𝑀𝐴 ≠ 0, and concepts of dynamics are needed to determine reactions, velocities, accelerations, etc.

410

ISTUDY

Chapter 6

Structural Analysis and Machines

Examples of correct FBDs Traffic signal pole. A pole supporting traffic and pedestrian signals is shown. Neglecting the weights of individual members, except for the two signals, draw FBDs of each component.

𝑇𝐹 𝐺 𝑇𝐹 𝐺 𝑇𝐹 𝐺

𝐹

𝐺

𝑇𝐹 𝐺 𝐸𝑦 𝐸𝑥 𝑇𝐷𝐸

𝑇𝐷𝐸

𝐸𝑦

𝐻

𝐸 𝐷

𝐶

𝐸𝑥 𝑊𝐻

𝐵

𝑇𝐷𝐸

𝐵𝑥 𝑇𝐷𝐸

𝐵𝑦 𝐵𝑥

WALK

𝑊𝐻

𝐵𝑦 𝑦

𝑊𝐶

𝑥

𝑀𝐴

𝐴𝑥

𝐴

𝐴𝑦

𝑊𝐶 𝑊𝐽

Aircraft service vehicle. A vehicle for servicing aircraft is shown. Draw FBDs so that the forces supported by members of the lift mechanism due to the weights of containers 𝐽 and 𝐾 may be determined.

𝐹𝑥 𝐸𝑦

𝐹𝑥 𝐽 𝐼

𝐹 𝐸

𝐶

𝐺 𝐶𝑦

𝐶𝑥

𝑦 𝑥

𝐻 𝐵𝑦 𝐹𝑦 𝐹𝑥 𝑦 𝑥

𝑃 𝐹

𝐻 𝐺

𝐶

𝐸𝑦

𝐷𝑦

𝐵

𝑃

𝐴

𝐶𝑦

𝐻𝑦

Pickup tool. The tool shown is used for picking up items that otherwise would be out of a person’s reach. When 𝑃 = 0, the tool is held open by a torsional spring at point 𝐷. Neglecting the weights of individual members, draw FBDs of each component.

𝐷

𝐻𝑦

𝐻𝑦

𝐴𝑦

𝐵

𝐸𝑥 𝐶𝑥

𝐷𝑦

𝐷

𝐺𝑦

𝐻𝑦

𝐸𝑥

𝐾

𝐻 𝐴

𝐺𝑦

𝐹𝑦

𝐹𝑦

𝑊𝐾

𝐸

𝐷𝑦 𝑀𝐷

𝐷𝑥 𝐶𝑥

𝑃

𝑀𝐷

𝐺𝑦

𝐶𝑥

𝐺𝑦 𝐺𝑦 𝐹𝑥

𝐷𝑥

𝐷𝑦 𝐶𝑥

𝐺𝑦

𝐹𝑦 𝐶𝑥

𝑃 Figure 6.29. Examples of properly constructed FBDs.

ISTUDY

Section 6.4

Aircraft service vehicle. We are instructed to consider the weights of the crates at 𝐽 and 𝐾 only, although clearly the weights of several other components are large. When sketching each member’s shape, we elect to leave the wheels on the vehicle. Assuming the vehicle is parked on a level surface, the support reactions consist of 𝐴𝑦 and 𝐵𝑦 . The only two-force member is the hydraulic cylinder 𝐻𝐼, and the force it supports is labeled 𝐻𝑦 . Finally, at locations where members are joined to one another, we include the appropriate forces, ensuring that forces of action and reaction between interacting bodies are equal in magnitude and opposite in direction. Pickup tool. We are instructed to neglect the weights of individual members. Rod 𝐶𝐸 and the item 𝐺𝐻 being gripped are the only two-force members, and the forces they support are labeled 𝐶𝑥 and 𝐺𝑦 , respectively. At pin 𝐷, in addition to the usual two forces 𝐷𝑥 and 𝐷𝑦 , we have a moment 𝑀𝐷 due to the torsional spring.

Examples of incorrect and/or incomplete FBDs Figure 6.30 shows several examples of incorrect and/or incomplete FBDs for frames and machines. Comments on how these FBDs must be revised follow, but before reading these, you should study Fig. 6.30 to find as many of the needed corrections and/or additions on your own as possible. Frame with pulley 1. The forces on the pulley must be revised in one of the following ways: • Either the force 𝑊 applied to the bearing of the pulley must be placed on the right-hand rim of the pulley, or • The force 𝑊 oriented at 30◦ on the left-hand rim of the pulley must be shifted to the bearing. The first of the above choices corresponds to including the pulley in the FBD, while the second choice corresponds to removing the pulley from the FBD by shifting the cable forces to the bearing of the pulley. 2. The connection forces at pin 𝐵 must also include vertical forces 𝐵𝑦 . 3. The support reactions at 𝐴 must also include a moment reaction 𝑀𝐴 . Scraper 1. The connection at 𝐸 allows no rotation about the 𝑧 axis (direction out of the plane of the figure); therefore, a moment 𝑀𝐸 must be applied to the scoop and tractor such that they are in opposite directions. 2. The front tire exerts a horizontal force on the ground due to the axle torque 𝑀𝐷 , and therefore, a horizontal force 𝐵𝑥 is needed between the tire and ground. Ratchet pruner 1. A coordinate system should be selected and shown. 2. With the information given, the positions of the two 10 lb forces on the handles are uncertain. Either we must do more detailed work to precisely locate these (this may be difficult), or we must make a reasonable decision for these positions. In these FBDs, the latter approach is used, and while the locations

Frames and Machines

411

412

ISTUDY

Chapter 6

Structural Analysis and Machines

Examples of incorrect and/or incomplete FBDs Frame with pulley. The pulley at 𝐶 is frictionless and points 𝐵–𝐸 are frictionless pins. Neglecting the weights of individual members, draw FBDs of each component. 𝐸

𝐶

𝐵 𝑊

30◦

𝑇𝐷𝐸

𝑊

𝐵

𝐵 𝑇𝐷𝐸

𝑇𝐷𝐸

30◦ 𝑦

𝐷 𝑊

𝑥

𝐴𝑥

𝑇𝐷𝐸

𝐴𝑦

𝑊 𝐴

Scraper. A scraper for moving earth is shown. The center of gravity for the scoop and its contents is at point 𝐺, and the center of gravity for the tractor is at point 𝐻. The rear wheels at 𝐶 roll freely while the front wheels at 𝐷 are powered by the tractor’s engine. The tractor and scoop are connected by a vertical pin at 𝐸. Draw FBDs of the wheels, scoop, and tractor.

𝑊𝐺

𝐺 𝐶𝑥

𝐸𝑦

𝐶𝑦 𝐹𝑦

𝐶𝑦

𝐻 𝐷

𝐴

𝐹

𝐷𝑦

𝑥

𝐴𝑦

𝑃

𝑃

𝐴𝑦 𝐵𝑦

𝑃

𝐹𝐶𝐷

𝐹𝐶𝐷

𝐷 𝐶

𝐵𝑦

10 lb

𝐴𝑦

𝐴 𝐵

𝐵𝑦

𝐵

𝐵𝑦

𝐸

𝐷𝑥

𝑀𝐷

The sketch of the ground is not an FBD.

𝐴𝑦

Ratchet pruner. A ratchet pruner for gardening is shown. All connections are pins, and assume that the branch being cut is a two-force member. If 10 lb forces are applied to the handles of the pruner, draw FBDs of each component.

𝐹

𝐷𝑦

𝑦

𝐶𝑥

𝐺

𝑀𝐷

𝐷𝑥

𝐹𝑥

𝐸 𝐶

𝐸𝑥

𝐸𝑥

𝑊𝐻

𝐸𝑦

𝐹𝐶𝐷

𝐹𝐶𝐷

𝑃 10 lb

Figure 6.30. Examples of incorrect and/or incomplete FBDs.

ISTUDY

Section 6.4

of the 10 lb forces are approximate, they will probably be acceptable for purposes of analysis and design. The main deficiency shown by the 10 lb forces in these FBDs is that they have different lines of action; if this is the case, then the pruner will tend to rotate due to the moment they produce, it will not be in moment equilibrium, and it will undergo angular acceleration. Hence, the FBDs must be revised so that the two 10 lb forces have the same line of action. 3. The connection forces at pin 𝐴 must also include horizontal forces 𝐴𝑥 . 4. The connection forces at pin 𝐵 must also include horizontal forces 𝐵𝑥 .

End of Section Summary In this section, methods of analysis for frames and machines are discussed. Some of the key points are as follows: • A frame is a structure that contains one or more three-force and/or multiforce members. Typically, a frame is fully fixed in space and uses a stationary arrangement of members with the goal of supporting forces that are applied to it. • A machine is an arrangement of members where typically the members can have significant motion relative to one another. The usual goal of a machine is transmission of motion and/or force. Throughout statics, when motion is possible, we assume it is slow enough that inertia can be neglected. When this is not the case, methods of analysis from dynamics must be used. • When drawing FBDs for frames and machines, it is helpful to first identify twoforce members and label their unknowns. Doing this provides for less complicated FBDs, fewer equilibrium equations that need to be written, and fewer unknowns to be determined. If you do not identify the two-force members, or if you miss some, you should still obtain the correct solution to a problem, although it may require more work.

Frames and Machines

413

414

E X A M P L E 6.8

𝐶

Analysis of a Bridge The design shown is common in multispan highway bridges. If the bridge supports a uniform vertical load of 8 kN∕m, determine the support reactions and the force supported by the link 𝐶𝐷.

𝐷 𝐶

𝐴 25 m

𝐵

𝐷

𝐸 20 m

5m Figure 1

ISTUDY

Chapter 6

Structural Analysis and Machines

SOLUTION Road Map This bridge has a beam from 𝐴 to 𝐶 and a beam from 𝐷 to 𝐸. Because the two beams, when connected by link 𝐶𝐷, do not behave as a single body, multiple FBDs will be required. Modeling

The FBDs are shown in Fig. 2 and are constructed as follows. We start by sketching beams 𝐴𝐶 and 𝐷𝐸, and link 𝐶𝐷 which we assume is vertical. We examine the structure for two-force members and find that link 𝐶𝐷 is the only two-force member present (disregarding the vertical support column at 𝐵). We label the force supported by link 𝐶𝐷 in the FBDs as 𝑇𝐶𝐷 , where positive corresponds to tension. Next, we add the support reactions to the FBDs. Finally, beam 𝐴𝐶 has 30 m length, and therefore the force it supports due to the distributed load is (8 kN∕m)(30 m) = 240 kN, which we place at the center of beam 𝐴𝐶. Similarly, beam 𝐷𝐸 has 20 m length, and therefore the force it supports is (8 kN∕m)(20 m) = 160 kN, which we place at the center of beam 𝐷𝐸. 15 m

Helpful Information Two-force members. When we draw FBDs, identifying and treating two-force members first will simplify the analysis because all the equilibrium equations for the two-force members are satisfied. For example, the FBD for link 𝐶𝐷 shown in Fig. 2 satisfies ∑ ∑ ∑ 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀 = 0, leaving only the equilibrium equations for the remaining FBDs to be written and solved.

𝐴𝑥

10 m

15 m 240 kN

𝐴

𝑇𝐶𝐷

𝐶 𝐵𝑦

𝐴𝑦

𝑇𝐶𝐷

𝑇𝐶𝐷

10 m 160 kN

𝑇𝐶𝐷

𝑦 𝐸𝑥

𝑥

𝐸

𝐷

𝐸𝑦

25 m

20 m 5m Figure 2. Free body diagrams.

Governing Equations & Computation Examining Fig. 2 shows that beam 𝐴𝐶 has four unknowns, while beam 𝐷𝐸 has only three. Thus, we will begin writing equilibrium equations for beam 𝐷𝐸 first, followed by equations for beam 𝐴𝐶. Member DE: ∑ 𝐸𝑥 = 0, (1) 𝐹𝑥 = 0 ∶ 𝐸𝑥 = 0 ⇒ ∑ (2) 𝑀𝐸 = 0 ∶ −𝑇𝐶𝐷 (20 m) + (160 kN)(10 m) = 0 ⇒ 𝑇𝐶𝐷 = 80 kN, ∑ 𝐸𝑦 = 80 kN. (3) 𝐹𝑦 = 0 ∶ 𝑇𝐶𝐷 − 160 kN + 𝐸𝑦 = 0 ⇒

Member AC: ∑ 𝐹𝑥 = 0 ∶ ∑ 𝑀𝐴 = 0 ∶ ∑

𝐹𝑦 = 0 ∶

𝐴𝑥 = 0

𝐴𝑥 = 0,

⇒

−(240 kN)(15 m) + 𝐵𝑦 (25 m) − 𝑇𝐶𝐷 (30 m) = 0

𝐴𝑦 − 240 kN + 𝐵𝑦 − 𝑇𝐶𝐷 = 0

⇒

(4) (5)

⇒ 𝐵𝑦 = 240 kN,

(6)

𝐴𝑦 = 80 kN.

(7)

ISTUDY

Section 6.4

Frames and Machines

Discussion & Verification

• Does the solution appear to be reasonable? To the extent possible, you should verify that your solution is reasonable. For example, beam 𝐷𝐸 is simply supported, and with a 160 kN load at its midspan, the solutions 𝑇𝐶𝐷 = 80 kN and 𝐸𝑦 = 80 kN given by Eqs. (1) and (2) are clearly correct. • Verification of the solution. You should verify that the solution is mathematically correct by substituting forces into all equilibrium equations to check that each of them is satisfied. However, this check does not verify the accuracy of the equilibrium equations themselves, so it is essential that you draw accurate FBDs and check that your solution is reasonable. • Why didn’t we draw an FBD of the entire structure? You probably noticed that, as the first step in our solution process, we did not draw an FBD of the entire structure for purposes of determining the support reactions. To see why this would not have been fruitful for this particular problem, consider the FBD for the entire structure, as shown in Fig. 3 (this FBD is easily constructed by putting together all three FBDs shown in Fig. 2, in which case the four 𝑇𝐶𝐷 forces no longer appear). This 15 m

25 m 240 kN

𝐴𝑥

𝐴

10 m 160 kN

𝐶 𝐵𝑦 25 m

𝑥

𝐸

𝐷 𝐴𝑦

𝑦 𝐸𝑥 𝐸𝑦

25 m

Figure 3. Free body diagram of the entire structure. FBD has five unknown reactions, and there are only three equilibrium equations

that can be written, so clearly it will not be possible to determine all of these unknowns using only this FBD. Thus, in this example, our solution began with the FBDs shown in Fig. 2. In Example 6.9 it will be effective to draw an FBD of the entire structure at the outset of the solution, while in Example 6.10 it will not.

415

416

Chapter 6

Structural Analysis and Machines

E X A M P L E 6.9

Analysis of a Basketball Hoop

0.4 f t 𝐸 𝐺 𝐻

𝐹

𝐴

1.6 f t

0.5 f t

1.6 f t

𝐵 2 ft

0.4 f t 𝐼

2.5 f t

𝐶

4 ft

SOLUTION 10 f t

5 ft

𝐽

A basketball hoop whose rim height at point 𝐻 is adjustable is shown. The supporting post 𝐴𝐵𝐶𝐷 weighs 90 lb with center of gravity at point 𝐶, and the backboard-hoop assembly weighs 50 lb with center of gravity at point 𝐺. The height of the rim is adjustable by means of the screw and hand crank 𝐼𝐽 , where the screw is vertical. If a person with 180 lb weight hangs on the rim, determine the support reactions at 𝐷 and the forces supported by all members.

Road Map To determine the support reactions at the base of the structure, we first draw an FBD of the entire structure, using a cut that passes through point 𝐷 only, and we then write and solve the equilibrium equations for this FBD. To determine the forces supported by the individual members, we will draw FBDs of each member followed by writing and

solving equilibrium equations for these.

4 ft 𝐷

Modeling To determine the support reactions for the structure, we draw an FBD, using a cut that passes through the structure at point 𝐷 only, as shown in Fig. 2. The 90 lb and 50 lb weights are vertical forces placed at their respective centers of gravity, and we assume the weights of the other components are negligible. We assume the person hanging on the rim is in static equilibrium and thus applies a 180 lb vertical force at point 𝐻.

Figure 1

Helpful Information Hanging on the rim. If the person hanging on the rim is in static equilibrium, then the force applied to the rim is his or her weight. If the person is in motion (e.g., swinging or jumping), then because of inertia, the force applied is usually different than his or her weight, with magnitude and direction that are time-dependent.

Governing Equations & Computation

Using the FBD in Fig. 2, we write and solve the

following equilibrium equations: ∑

𝐹𝑥 = 0 ∶

𝐷𝑥 = 0

∑

𝐹𝑦 = 0 ∶

𝐷𝑦 − 90 lb − 50 lb − 180 lb = 0

∑

𝑀𝐷 = 0 ∶

𝐷𝑥 = 0,

(1)

𝐷𝑦 = 320 lb,

(2)

⇒ ⇒

𝑀𝐷 − (50 lb)(2.4 f t) − (180 lb)(4.5 f t) = 0, ⇒ 𝑀𝐷 = 930 ft⋅lb.

𝐺

180 lb 𝐻

0.5 f t

2.5 f t

2 ft 90 lb 𝑦

𝑥

Modeling To determine the forces supported by the individual members, we draw FBDs for each member, as shown in Fig. 3. These FBDs are constructed by first sketching the members. We then identify any two-force members and label their forces first. In Fig. 3, members 𝐴𝐸 and 𝐼𝐽 are the only two-force members, and we label their forces as 𝑇𝐴𝐸 and 𝑇𝐼𝐽 , respectively. After the FBDs for the two-force members are completed, we add to the remaining FBDs the forces from the two-force members such that Newton’s third law is satisfied. That is, we apply force 𝑇𝐴𝐸 to the post 𝐴𝐵𝐶𝐷, and to the backboardhoop, making sure they are applied in directions opposite to the forces on member 𝐴𝐸. Similarly, we apply force 𝑇𝐼𝐽 to the post 𝐴𝐵𝐶𝐷 and to member 𝐼𝐵𝐹 . The FBDs are completed by adding the support reactions at 𝐷 and adding horizontal and vertical forces at pins where structural members are attached. Governing Equations & Computation

𝐷𝑥 𝑀𝐷 𝐷𝑦 Figure 2 Free body diagram of the entire structure for determining the support reactions.

ISTUDY

(4)

50 lb

0.4 f t

𝐶

(3)

By treating two-force members as described ∑ ∑ ∑ here, members 𝐴𝐸 and 𝐼𝐽 automatically satisfy 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀 = 0. Of the three members remaining, the backboard-hoop assembly has the fewest unknowns (three), and thus, we begin writing and solving equilibrium equations for it. Using the geometry shown in Fig. √ 1, the horizontal component of the force supported by member 𝐴𝐸 is 𝑇𝐴𝐸 (2 f t)∕ (2 f t)2 + (1.6 f t)2 = 𝑇𝐴𝐸 (2∕2.561), and the vertical component is 𝑇𝐴𝐸 (1.6∕2.561). Thus, equilibrium equations can be written as

ISTUDY

Section 6.4

Frames and Machines

𝑇𝐴𝐸

𝐴 𝑇𝐼𝐽

𝐵𝑦

1.6 f t

4.4 f t 𝑇𝐼𝐽 𝐽 𝑇𝐼𝐽

𝑇𝐴𝐸

𝑇𝐴𝐸

𝑀𝐷 𝐷𝑦

𝐹𝑦 𝐹𝑥

𝐵𝑥

0.4 f t 𝑇𝐼𝐽

𝐷𝑥

𝐹𝑦

𝑇𝐴𝐸 1.6 f t

4 ft

180 lb 1.6 f t 𝐹

𝐹𝑥

90 lb 𝐶

5 ft

50 lb

0.4 f t

𝐵𝑥

𝐵

417

2.5 f t

2.561

𝐵𝑦

1.6

𝑦

2

𝑥

2 ft

0.5 f t

Figure 3. Free body diagrams for determining the forces supported by each member.

Member EFH: ( ∑ 𝑀𝐹 = 0 ∶ 𝑇𝐴𝐸

) 2 (1.6 f t) − (50 lb)(0.4 f t) − (180 lb)(2.5 f t) = 0 2.561 ⇒ 𝑇𝐴𝐸 = 376.2 lb, ) ( ∑ 2 = 0 ⇒ 𝐹𝑥 = 293.8 lb, 𝐹𝑥 = 0 ∶ 𝐹𝑥 − 𝑇𝐴𝐸 2.561 ( ) ∑ 1.6 𝐹𝑦 = 0 ∶ 𝐹𝑦 − 𝑇𝐴𝐸 − 50 lb − 180 lb = 0 ⇒ 𝐹𝑦 = 465.0 lb. 2.561

(5) (6) (7) (8)

The FBDs remaining for members 𝐴𝐵𝐶𝐷 and 𝐼𝐵𝐹 now have three unknowns each, and either could be considered next. We will use member 𝐼𝐵𝐹 , and we write Member IBF: ∑ 𝑀𝐵 = 0 ∶ 𝑇𝐼𝐽 (0.5 f t) + 𝐹𝑥 (1.6 f t) − 𝐹𝑦 (2 f t) = 0 (9) ∑

𝐹𝑥 = 0 ∶

∑

𝐹𝑦 = 0 ∶

⇒

𝑇𝐼𝐽 = 920.0 lb,

(10)

𝐵 𝑥 − 𝐹𝑥 = 0

⇒

𝐵𝑥 = 293.8 lb,

(11)

−𝑇𝐼𝐽 + 𝐵𝑦 − 𝐹𝑦 = 0

⇒

𝐵𝑦 = 1385 lb.

(12)

With Eq. (12), all of the unknowns have been determined, and it is not necessary to write the equilibrium equations for the post 𝐴𝐵𝐶𝐷. Discussion & Verification

• You should check that the solutions appear to be reasonable. In this problem the loading is simple enough that we expect members 𝐴𝐸 and 𝐼𝐽 to be in tension, and indeed our solution shows this. By inspection, the reactions at the base of the post, point 𝐷, also have the proper signs. • In this solution we found the support reactions first. Doing this used three equilibrium equations, and thus, it was possible to find all of the remaining unknowns without writing the equilibrium equations for the post 𝐴𝐵𝐶𝐷. To help verify the solution’s accuracy, you could write and solve these equations, and you should find that 𝐷𝑥 , 𝐷𝑦 , and 𝑀𝐷 agree with the results in Eqs. (1), (2), and (4).

Helpful Information Strategy for finding reactions. Rather than begin our solution by finding the support reactions, we could have proceeded directly to the FBDs in Fig. 3, and this eventually would have produced the same reactions. This was the procedure used in Example 6.8 because in that problem it was not possible to determine the support reactions at the outset.

418

E X A M P L E 6.10

Analysis of a Ladder A ladder used by swimmers to climb in and out of a pool is shown. The ladder has rungs at points 𝐸 and 𝐹 . Neglecting forces applied by the water to the ladder, determine the forces supported by each member.

20 lb 𝐺

dimensions in inches

18

9

𝐶

30 lb

9

𝐹 𝐴

40 lb 9

𝐵 𝐸 4

4

4

SOLUTION Road Map

𝐷

4

Figure 1

ISTUDY

Chapter 6

Structural Analysis and Machines

Since the ladder contains at least one multiforce member, this structure must be analyzed as a frame. In fact, all three members of the ladder are multiforce members (three-force members, to be precise). What is especially challenging in this problem is that point 𝐴 has three members joined by a single pin. While we could disassemble the structure and draw FBDs for the three members only, for problems such as this where three or more members are supported by a single pin, it is especially helpful to draw a separate FBD for the pin. Hence, four FBDs will be used in our solution. We disassemble the structure into its three members plus the pin at 𝐴, which leads to the FBDs shown in Fig. 2. These FBDs are constructed as follows. Beginning with member 𝐴𝐺𝐷, point 𝐷 is pin-supported so it has the usual two orthogonal reactions 𝐷𝑥 and 𝐷𝑦 . In addition, the member is supported by a pin at 𝐴; the pin applies forces 𝐴1𝑥 and 𝐴1𝑦 to the member, and the member applies these same forces to the pin, but in opposite directions in accord with Newton’s third law. Next, member 𝐴𝐹 𝐶 is supported by a roller at 𝐶, so there is one reaction that is perpendicular to the wall of the pool. In addition, the member is supported by a pin at 𝐴; the pin applies forces 𝐴2𝑥 and 𝐴2𝑦 to the member, and the member applies these same forces to the pin, but in opposite directions. The FBD for member 𝐴𝐸𝐵 is similarly drawn. Modeling

Helpful Information 20 lb

𝑥 18

9 9

𝐴1𝑥

𝐴

𝐴2𝑦 𝐴1𝑥 𝐴2𝑥 𝐴3𝑥

4

𝐴3𝑥

dimensions in inches

𝐴1𝑦 4

4

4

𝐴3𝑦 40 lb 𝐵𝑥 𝐸

4

9

𝐹

𝐴2𝑥

𝐴1𝑦 𝐴2𝑦 𝐴3𝑦

𝐷𝑥 𝐷𝑦

𝐶𝑥

30 lb

𝑦

𝐺

Three or more members supported by a pin. When we draw FBDs of frames and machines where three or more members are supported by a single pin, such as the pin at point 𝐴 in this example, it is helpful to also draw an FBD of the pin.

9

𝐵𝑦

4 4 4

4

4

Figure 2. Free body diagrams.

Governing Equations & Computation

Examination of the FBDs shows that member 𝐴𝐹 𝐶 has only three unknown forces, and hence, we choose this to begin writing equilibrium equations. Thus, Member AFC: ∑ 𝑀𝐴 = 0 ∶ −(30 lb)(8 in.) + 𝐶𝑥 (9 in.) = 0 ⇒ 𝐶𝑥 = 26.67 lb, (1)

ISTUDY

Section 6.4

Frames and Machines

∑

𝐹𝑦 = 0 ∶

∑

𝐹𝑥 = 0 ∶

𝐴2𝑦 − 30 lb = 0 𝐴2𝑥 − 𝐶𝑥 = 0

𝐴2𝑦 = 30 lb,

(2)

𝐴2𝑥 = 26.67 lb.

(3)

⇒

⇒

419

Examination of the remaining FBDs shows they each contain four unknowns, so there is no preferred order for treating them since a system of coupled algebraic equations will need to be solved. Thus, for the remaining three FBDs, the equilibrium equations are as follows: Member AGD: ∑ 𝑀𝐴 = 0 ∶ −(20 lb)(8 in.) − 𝐷𝑥 (18 in.) + 𝐷𝑦 (16 in.) = 0, (4) ∑ 𝐹𝑥 = 0 ∶ 𝐴1𝑥 + 𝐷𝑥 = 0, (5) ∑ 𝐹𝑦 = 0 ∶ 𝐴1𝑦 + 𝐷𝑦 − 20 lb = 0. (6) Member AEB: ∑ ∑ ∑

𝑀𝐴 = 0 ∶

𝐵𝑥 (9 in.) + 𝐵𝑦 (12 in.) − (40 lb)(4 in.) = 0,

(7)

𝐹𝑥 = 0 ∶

𝐴3𝑥 + 𝐵𝑥 = 0,

(8)

𝐹𝑦 = 0 ∶

𝐴3𝑦 + 𝐵𝑦 − 40 lb = 0.

(9)

20 lb 𝐺

Pin A:

𝑦

∑

𝐹𝑥 = 0 ∶ −𝐴1𝑥 − 𝐴2𝑥 − 𝐴3𝑥 = 0,

(10)

∑

𝐹𝑦 = 0 ∶

(11)

−𝐴1𝑦 − 𝐴2𝑦 − 𝐴3𝑦 = 0.

𝑥

𝐷𝑦

Equations (4)–(11) are a system of eight equations with eight unknowns. While these may is more effective, and provides be solved manually, the use of a computer 𝐵𝑥 = −19.56 lb

𝐵𝑦 = 28 lb

𝐷𝑥 = 46.22 lb

𝐷𝑦 = 62 lb

𝐴1𝑥 = −46.22 lb

𝐴1𝑦 = −42 lb

𝐴3𝑥 = 19.56 lb

𝐴3𝑦 = 12 lb

𝐷𝑥

𝐴2𝑥 𝐴3𝑥

(12)

Discussion & Verification

• Verification of solution. You should verify that the solution is mathematically correct by substituting forces into all equilibrium equations to check that each of them is satisfied. However, this check does not verify the accuracy of the equilibrium equations themselves, so it is essential that you draw accurate FBDs and check that your solution is reasonable. • Alternate FBDs. The use of a separate FBD for the pin at 𝐴 was helpful in this problem. However, other FBDs, provided they are properly drawn, can also be used. For example, if pin 𝐴 is left on member 𝐴𝐺𝐷 so that the remaining two members directly apply forces to 𝐴𝐺𝐷, then the FBDs are as shown in Fig. 3. You should sketch for yourself the FBDs that result if pin 𝐴 is left on member 𝐴𝐹 𝐶, or member 𝐴𝐸𝐵. • The bigger picture. The solution outlined here is the first step (a major first step) in fully designing and/or analyzing this pool ladder. In mechanics of materials, a subject that follows statics, you will use the forces determined here to investigate if the ladder is strong enough, if its deformations are acceptable, and so on.

𝐴2𝑦 𝐴3𝑦 𝐶𝑥

30 lb 𝐹

𝐴2𝑥 𝐴2𝑦

𝐴3𝑥

𝐴3𝑦 40 lb 𝐵𝑥 𝐸

𝐵𝑦

Figure 3 Alternate FBDs where pin 𝐴 and its forces are left on member 𝐴𝐺𝐷.

420

Chapter 6

Structural Analysis and Machines

Problems Problem 6.75 15 N

15 N

𝑀

𝐴

𝐶

𝐸 𝐵

𝐷 𝐹

𝐻

Problem 6.76 The keyboard mechanism for a piano consists of a large number of parts, but for purposes of modeling the transmission of force and motion from the key to the piano wire, the model shown can be used. Draw FBDs for all of the members (i.e., members 𝐴𝐵𝐶, 𝐷𝐶𝐸, 𝐸𝐹 , and 𝐺𝐹 𝐻), labeling all unknowns. Neglect the weights of members.

𝐺 𝐼

𝐿

A cork puller for removing the cork from a bottle is shown. Draw FBDs for all of the members of the cork puller and the cork (i.e., members 𝐴𝐵, 𝐶𝐷, 𝐿𝑀, 𝐻𝐵𝐸𝐷𝐼, and 𝐽 𝐾𝐿), labeling all unknowns. Neglect the weights of members.

𝐽 𝐾

𝐹 𝐺 𝐻 Figure P6.75

piano wire jack

𝐷

hammer

𝐶

5N

𝐸

𝐴

whippen

key

𝐵 Figure P6.76

Problem 6.77 A frame used for a chair is shown. Determine the force supported by members 𝐴𝐵𝐶𝐷, 𝐸𝐶𝐹 𝐺, and 𝐵𝐹 . Neglect friction at points 𝐴 and 𝐺.

60 lb

0.3 m 0.4 m 0.2 m 𝐷

40 lb 0.3 m

𝐷

𝐸

8 in. 𝐶 𝐵

𝐹

5 in. 3 in.

𝐴

𝐺

𝐶 𝐸

𝐹

𝐺

0.4 m 𝐵 0.6 m 𝐻

𝑊

20◦ 𝐴

4 in. 4 in. Figure P6.77

Figure P6.78

Problem 6.78 A frame used for supporting a weight 𝑊 = 800 N is shown. Determine the force supported by cable 𝐷𝐻 and members 𝐴𝐵𝐶𝐷, 𝐶𝐸𝐹 𝐺, and 𝐵𝐸. 12 in.

4 in. 5 lb 𝐵 2 in.

𝐶

𝐴 5 lb Figure P6.79

ISTUDY

𝐷

2 in.

3 in. 𝐹 𝐸

Problem 6.79 𝐻 𝐺

The tool shown is used for picking up items that otherwise would be out of a person’s reach. In the position shown, the torsional spring at 𝐶 supports a moment of 10 in.⋅lb, which tends to open the jaws of the tool. Neglect the weights of the individual members and neglect friction at points 𝐻 and 𝐺. Determine the forces supported by all members of the tool and the force applied by the tool to the item being gripped at 𝐻 and 𝐺.

ISTUDY

Section 6.4

Frames and Machines

Problem 6.80 Compare the designs for the pickup tools shown in Fig. P6.79 and in Fig. 6.29 on p. 410. If the dimensions of the members in these tools are approximately the same, speculate on which of these tools will be capable of applying the greater force to the item being gripped at 𝐻 and 𝐺 before one of the members of the tool fails. Note: Concept problems are about explanations, not computations. 18 in.

Problems 6.81 and 6.82 The elevation and tilt angle of table 𝐴𝐵 are controlled by an operator’s hand at point 𝐷. The table supports a box weighing 10 lb with center of gravity at point 𝐺. Determine the force the operator must apply to keep the table in equilibrium, and determine the force supported by all members of the machine. Problem 6.81

The force applied by the operator is vertical.

Problem 6.82

The force applied by the operator is horizontal.

𝐺 30◦ 𝐴 𝐸

𝐵

𝐶

30◦

𝐷 10 in.

12 in.

Figure P6.81 and P6.82

Problems 6.83 and 6.84

30 cm

The elevation and tilt angle of table 𝐴𝐵 are controlled by an operator’s hand at point 𝐷. The table supports a box weighing 40 N with center of gravity at point 𝐺. Determine the force the operator must apply to keep the table in equilibrium, and determine the force supported by all members of the machine. Problem 6.83

The force applied by the operator is vertical.

Problem 6.84

The force applied by the operator is horizontal.

Problem 6.85 For the ladder shown, if 𝑊 = 800 N and ℎ = 0.3 m, determine the force supported by each of the spreader bars 𝐵𝐺 and 𝐷𝐺. Neglect the weight of the ladder and friction at points 𝐴 and 𝐸. 0.4 m 0.5 m 0.3 m 0.4 m 𝐶 0.9 m

ℎ 𝑊

0.4 m 𝐵

𝐺

𝐷 1.2 m

𝐴

𝐸

Figure P6.85 and P6.86

Problem 6.86 For the ladder shown, if 𝑊 = 800 N and ℎ can have any value such that 0 ≤ ℎ ≤ 0.9 m, determine the largest force supported by each of the spreader bars 𝐵𝐺 and 𝐷𝐺, and the value(s) of ℎ for which this occurs. Neglect the weight of the ladder and friction at points 𝐴 and 𝐸.

𝐴

𝐺 𝐵 𝐶

20 cm 𝐸

𝐷 8 cm

30 cm

Figure P6.83 and P6.84

25 cm

421

422

Chapter 6

Structural Analysis and Machines 400 lb 𝐵

dimensions in inches 𝐹

Problem 6.87 𝐴

𝐶 40◦

Stairs leading to a balcony are shown. For the 400 lb and 600 lb forces, and if 𝐹 = 200 lb and 𝑃 = 0, determine the reaction at point 𝐷 and the force supported by bar 𝐵𝐸.

𝐸

Problem 6.88 𝑃

144

Repeat Prob. 6.87 if 𝐹 = 0 and 𝑃 = 200 lb.

600 lb

Problem 6.89

𝐷 32 22 16 38

30

Stairs leading to a balcony are shown. For the 400 lb and 600 lb forces, and if 𝐹 = 200 lb and 𝑃 = 0, determine the reaction at point 𝐷 and the force supported by bar 𝐵𝐸.

30

Figure P6.87 and P6.88

Problem 6.90 400 lb

Repeat Prob. 6.89 if 𝐹 = 0 and 𝑃 = 200 lb. 𝐵

dimensions in inches 𝐹

𝐴

𝐶

Problem 6.91

56 𝑃

𝐸

144

In the structure shown, cable segment 𝐸𝐷 and member 𝐸𝐶𝐹 are horizontal. If 𝑊 = 600 lb, determine the force supported by members 𝐴𝐵𝐶𝐷, 𝐸𝐶𝐹 , and 𝐺𝐵𝐹 .

600 lb

84 in. 8 in. radius

𝐷 32 22 16 38

30

30

21 in. 𝐷 𝐶 𝐵

𝐸

𝑊

Figure P6.89 and P6.90

𝐹 28 in.

80 in. 40 lb 𝐴

𝐵

60 lb

80 lb

𝐶

𝐷

𝐺

𝐴 60 in.

𝐸

Figure P6.91

6 in. 𝐺 8 in.

8 in.

8 in.

Problem 6.92

8 in.

Figure P6.92 𝐺

The structure shown consists of five members with pins at points 𝐵, 𝐶, 𝐷, 𝐸, and 𝐺. Determine the force supported by members 𝐵𝐺, 𝐶𝐺, and 𝐷𝐺 due to the 40 lb, 60 lb, and 80 lb loads.

𝐻

𝐹 2 ft

Problem 6.93

2 ft 𝐸

𝐷 𝐵

𝐴 4 ft Figure P6.93

ISTUDY

3 ft

𝐶 2 ft

2 ft

A hydraulically powered lift supports the two boxes shown. Each box weighs 200 lb with center of gravity in the middle of each box. Neglect the weight of individual members. (a) Draw four FBDs, one each for members 𝐴𝐸𝐹 , 𝐴𝐵𝐶, 𝐸𝐷, and 𝐵𝐸, labeling all forces. (b) Determine the force in the hydraulic actuator 𝐵𝐸 required to keep the lift in equilibrium.

ISTUDY

Section 6.4

423

Frames and Machines

Problem 6.94 𝐶

The shovel of an end loader has pins at points 𝐴, 𝐵, 𝐶, and 𝐷. The scoop supports a downward vertical load of 2000 lb, which is not shown in the figure, at point 𝐸.

𝐴 45◦

(a) Draw five FBDs, one each for parts 𝐴𝐵 and 𝐴𝐶, hydraulic cylinder 𝐴𝐷, shovel 𝐶𝐷𝐸, and member 𝐵𝐷, labeling all forces.

20 in. 𝐸

𝐵

(b) Determine the force the hydraulic cylinder 𝐴𝐷 must generate to keep the shovel in equilibrium.

𝐷 20 in. 12 in. 12 in.

Problem 6.95 A backhoe is shown. Hydraulic cylinders 𝐺𝐹 and 𝐵𝐶 are horizontal and vertical, respectively. Neglect the weights of members.

Figure P6.94 1

𝐹 1

𝐸

1.5 45◦ 45◦

𝐷

0.5 𝐶

dimensions in feet

Problem 6.96 The hand brake for a bicycle is shown. Portions 𝐷𝐸 and 𝐹 𝐺 are free to rotate on bolt 𝐴, which is screwed into the frame 𝐵𝐶 of the bicycle. The brake is actuated by a shielded cable where 𝑇1 is applied to point 𝐸 and 𝑇2 is applied to point 𝐺. A spring having 50 N compressive force is placed between points 𝐸 and 𝐺 so that the brake stays open when it is not being used. Assume the change in the spring’s force is negligible when the brake is actuated to produce the 200 N forces at points 𝐷 and 𝐹 .

1 0.8

𝐺

(a) Draw six FBDs, one each for members 𝐷𝐸𝐹 and 𝐹 𝐶𝐴, shovel 𝐴𝐵, and hydraulic cylinders 𝐵𝐶, 𝐶𝐸, and 𝐹 𝐺. (b) If each of the three hydraulic cylinders is capable of producing a 1000 lb force (in both tension and compression), determine the largest weight 𝑊 that can be supported in the position shown.

1

1.5

2 𝑊 𝐵

𝐴

Figure P6.95 𝑇1 cable 𝑇2

(a) Draw three FBDs, one each for 𝐷𝐸 and 𝐹 𝐺 and bolt 𝐴, labeling all forces. spring

(b) Determine the necessary cable forces 𝑇1 and 𝑇2 .

𝐺

(c) Determine the forces exerted by 𝐷𝐸 and 𝐹 𝐺 on bolt 𝐴.

15◦ 60 mm

𝐸

Problem 6.97 A highchair for seating a child at a table is shown. The chair slips onto the edge of the table and makes contact at points 𝐶 and 𝐺, where friction at 𝐺 is negligible. To prevent the back of the chair from rotating counterclockwise, there is a stop that makes frictionless contact at point 𝐻. Members 𝐵𝐹 and 𝐸𝐽 are vertical. Hint: With negligible loss of accuracy, you may assume that point 𝐶 lies on the line connecting points 𝐴 and 𝐵.

𝐴

𝐵 70 mm 𝐷

(a) Determine the reactions at points 𝐶 and 𝐺. (b) Draw six FBDs, one each for members 𝐴𝐵𝐶, 𝐷𝐸𝐹 𝐺, 𝐴𝐷𝐼, 𝐻𝐼𝐽 , 𝐵𝐹 , and 𝐸𝐽 .

𝐶

200 N

𝐹

Figure P6.96

(c) Determine the forces supported by members 𝐵𝐹 and 𝐸𝐽 . 16 N

Problem 6.98 The highchair in Prob. 6.97 can be collapsed (i.e., folded flat) so that it may be easily transported and stored. To collapse the chair, the seatback 𝐴𝐷𝐼 is rotated clockwise about hinge 𝐼. When designing the chair, it is essential that it does not tend to collapse when a child is seated in it (as modeled by the 16 N and 100 N loads shown). With the 16 N and 100 N loads shown, determine if the chair is safe from collapsing. Hint: If the reaction at point 𝐻 is compressive, then the chair will be safe, and if this reaction is tensile, then the chair will begin to collapse.

8 cm 18 cm 8 cm 𝐻

2 cm

𝐴

table 𝐶 𝐺

𝐵 100 N 𝐸 𝐹

𝐷 𝐼

12 cm

𝐽

6 cm 14 cm 6 cm 8 cm

Figure P6.97 and P6.98

34 cm

424

Chapter 6

Structural Analysis and Machines

Problem 6.99 20 in.

16 in. 𝐶 radius = 1.5 in.

14 in.

28 in.

𝐷

𝐸 𝐴

Figure P6.99

ISTUDY

𝐹 6 in.

𝐵 5◦ radius = 3 in.

A bicycle is pedaled up a gentle incline. The rider and bicycle weigh 120 lb with center of gravity at point 𝐶. For the position shown, determine the force the rider must apply to the pedal at 𝐹 for the bicycle to move at constant speed. Assume the rider applies a force to the pedals using her right foot only, the force applied by the rider to pedal 𝐹 is perpendicular to the crank 𝐸𝐹 , and the lower portion of the chain between sprockets 𝐷 and 𝐸 is slack.

14 in.

Problem 6.100 The linkage shown is used on a garbage truck to lift a 2000 lb dumpster. Points 𝐴–𝐺 are pins, and member 𝐴𝐵𝐶 is horizontal. This linkage has the feature that as the dumpster is lifted and rotated, the front edge 𝐴 of the dumpster is simultaneously lowered, placing it at a more convenient height for emptying. (a) Draw five FBDs, one each for members 𝐴𝐵𝐶, 𝐶𝐷, and 𝐸𝐷𝐺, dumpster 𝐻, and hydraulic cylinder 𝐶𝐹 , labeling all forces. (b) For the position shown, when the dumpster just fully lifts off the ground, determine the force in hydraulic cylinder 𝐶𝐹 .

𝐹

12 in.

𝐸

24 in.

𝐷 45◦

24 in.

𝐴 𝐶

2000 lb

𝐵 45◦

𝐻

𝐺 45◦

18 in. 18 in.

60 in.

Figure P6.100

Problem 6.101 An end loader for a small garden tractor is shown. All connections are pinned. The only significant weight is 𝑊 . (a) Draw six FBDs, one each for members 𝐵𝐶, 𝐷𝐺, and 𝐴𝐸𝐹 , plate 𝐶𝐷𝐸, hydraulic cylinder 𝐸𝐺, and shovel 𝐴𝐵, labeling all forces. (b) If the hydraulic cylinder 𝐸𝐺 is capable of producing a 3000 lb force, determine the largest weight 𝑊 that may be lifted in the position shown. 10 in. 18 in. 14 in.

48 in. 𝐷

𝐶

𝐹 14 in.

24 in.

𝐸

𝑊 𝐵

6 in. 𝐺

20 in. 𝐴 Figure P6.101

ISTUDY

Section 6.4

425

Frames and Machines

Problem 6.102 A trash can weighing 12 lb has a lid that weighs 4 lb; both weights have the same line of action, and the weights of all other components are negligible. A foot-operated lever 𝐴𝐵𝐶𝐷 is used to simultaneously tilt the can and open the lid, as shown in the inset of the figure. Pin 𝐵 is attached to the trash can, and it rides in a frictionless slot in lever 𝐴𝐵𝐶𝐷. Determine the force 𝑃 required to begin tilting and opening the trash can.

7 in.

3 in.

8 in. 4 lb

𝐺

trash can in the opened position

𝐻 12 lb

Problem 6.103 The hoist shown is powered by a motor at point 𝐼. Cable segment 𝐵𝐸 is vertical, and segments 𝐸𝐺 and 𝐼𝐵 are horizontal. If 𝑊 = 10 kN, determine the reactions at point 𝐴, and the force supported by members 𝐶𝐹 𝐺 and 𝐷𝐹 𝐻.

𝑃

𝐴 𝐵

𝐶

𝐷

4

3 𝐸

9 in.

𝐹

0.6 m 0.6 m 0.6 m 𝐸

8 in. 7 in. 5 in. 9 in. 4 in.

𝐺

0.8 m

𝐷

0.2 m

Figure P6.102

𝐻 𝐹

0.8 m 1m

𝐶

𝑊

𝐵 𝐼

2m

0.4 m 𝐴

all pulleys have 0.2 m radius

Figure P6.103 𝐴

Problem 6.104 A prosthetic arm and hand assembly is shown. Points 𝐵 and 𝐻 are fixed to the arm. The hand is actuated by a pneumatic cylinder 𝐵𝐹 that opens and closes the hand. The spring 𝐹 𝐻 helps keep the hand in alignment with the arm. If the person holds a bag of groceries that weighs 30 N and grips the bag at 𝐴 and 𝐷 with a 10 N force, determine the forces supported by the pneumatic cylinder and the spring. Assume that while holding the bag of groceries, the arm and hand have the geometry shown where the pneumatic cylinder is horizontal.

𝐸

𝐵

𝐹

𝐷

𝐶 10 cm

8 cm

𝐻 2 cm

Figure P6.104

𝐶

𝐷

Problem 6.105 Three common designs for a bridge to span a multilane highway are shown. Design 1 is the same as that considered in Example 6.8 on p. 414, where link 𝐶𝐷 is used to connect the two spans of the bridge. Design 2 uses two simply supported I beams. Design 3 uses a single I beam from points 𝐴 to 𝐸 with a roller support at midspan. Comment on the pros and cons of each of these designs, considering issues such as the following:

8 cm

3 cm 3 cm

𝐶 𝐴

𝐵

𝐷

𝐸

Design 1 𝐵 𝐶 𝐸

𝐴

• Which are straightforward to analyze for determining the support reactions, and which may be difficult to analyze?

Design 2

• Which may be easier to construct? • Which may be more sensitive to unexpected support motion, such as if the support at 𝐵 settles with time? • Which may be safer, such as if the support at 𝐵 were accidentally struck by a vehicle passing under the bridge? Note: Concept problems are about explanations, not computations.

𝐴

𝐵 Design 3

Figure P6.105

𝐸

426

Chapter 6

Structural Analysis and Machines

6.5 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

Trusses A truss is a structure that consists of two-force members only, where members are organized so that the assemblage as a whole behaves as a single object. For a two-dimensional structure, for all members to be two-force members, the structure must have the following characteristics: • All members must be connected to one another by frictionless pins, and the locations of these pins are called joints. • Each member may have no more than two joints. • Forces may be applied at joints only. • The weight of individual members must be negligible. A structure having the above characteristics is called a plane truss.

Method of joints. In the method of joints, the force supported by each member of a truss is determined by drawing FBDs for each joint, and then requiring that each joint be ∑ ∑ in equilibrium by writing the equations 𝐹𝑥 = 0 and 𝐹𝑦 = 0 for each joint. Method of sections. In the method of sections, the force supported by members of a truss is determined by taking a cut through the structure, and the FBD that results is ∑ ∑ ∑ required to be in equilibrium by writing the equations 𝐹𝑥 = 0, 𝐹𝑦 = 0, and 𝑀 = 0. 𝐶 𝐹𝐴𝐶 𝐵

𝐹𝐴𝐵

Zero-force members. A truss member (or any member) that supports no force is called a zero-force member. Zero-force members in a truss can be identified by finding joints that match the pattern shown in Fig. 6.31. If

𝑦

𝛼

• A particular joint has three members connected to it,

𝑥 𝐴

• Two of these members are collinear, and

𝐹𝐴𝐷 𝐷

Figure 6.31 Geometry of members in a truss allowing a zeroforce member to be recognized by inspection.

ISTUDY

• The joint has no external force applied to it, then the noncollinear member at that joint is a zero-force member. Zero-force members can be very effective in improving the overall strength of a truss by reducing the length of compression members and thus improving their resistance to buckling.

Statically determinate and indeterminate trusses. In a statically determinate truss, the equations of equilibrium are sufficient to determine the forces supported by all members of the truss and the support reactions. In a statically indeterminate truss, the equations of equilibrium are not sufficient to determine all of these. A simple rule of thumb to help you determine whether a truss is statically determinate or statically indeterminate is to compare the number of unknowns to the number of equilibrium equations. We call this equation counting. For a plane truss having Eq. (6.21), p. 383 𝑚 = number of members, 𝑟 = number of support reactions, 𝑗 = number of joints,

ISTUDY

Section 6.5

Chapter Review

the rule of thumb is as follows: Eq. (6.22), p. 383 If 𝑚 + 𝑟 < 2𝑗

The truss is a mechanism and/or has partial fixity.

If 𝑚 + 𝑟 = 2𝑗

The truss is statically determinate if it has full fixity. The truss is statically indeterminate if it has partial fixity.

If 𝑚 + 𝑟 > 2𝑗

The truss is statically indeterminate, and it can have full fixity or partial fixity.

Successful use of equation counting requires good judgment on your part, which is why it is called a rule of thumb rather than a rule.

Simple, compound, and complex trusses. A simple truss has members arranged in a triangular pattern, as shown in Fig. 6.16 on p. 384. Compound and complex trusses consist of two or more simple trusses connected to one another. While simple trusses (and compound and complex trusses) are popular, other designs for trusses are also common and can perform very well. One of the main features of simple trusses (and compound and complex trusses) is that they are always stable. Buckling. Because truss members are usually straight and slender, compression members are susceptible to buckling. Furthermore, the force at which buckling occurs in a straight member decreases very rapidly as a member becomes longer (the buckling load is proportional to 1∕𝐿2 where 𝐿 is the length of the member). Thus, it is undesirable to have long compression members in a truss.

Space trusses A space truss is a three-dimensional structure that consists of two-force members only, where members are organized so that the assemblage as a whole behaves as a single object. The comments made earlier regarding the requirements for two-force members in a plane truss also apply, but in a space truss members must be connected to one another by balland-socket joints. Space trusses can be analyzed using the method of joints and the method of sections, and the equation-counting rule of thumb can also be used, except each joint has three equilibrium equations.

Frames and machines Frame. A frame is a structure that contains one or more three-force and/or multiforce members. Typically, a frame is fully fixed in space and uses a stationary arrangement of members with the goal of supporting forces that are applied to it. Machine. A machine is an arrangement of members where typically the members can have significant motion relative to one another. The usual goal of a machine is transmission of motion and/or force. Throughout statics, when motion is possible, we assume it is slow enough that inertia can be neglected. When this is not the case, methods of analysis from dynamics must be used. FBDs. When we draw FBDs for frames and machines, it is helpful to first identify twoforce members and label their unknowns. Doing this provides for less complicated FBDs, fewer equilibrium equations to write, and fewer unknowns to be determined.

427

428

Chapter 6

Structural Analysis and Machines

Review Problems Problem 6.106 Member 𝐴𝐺𝐵 is a single member with pin connections at points 𝐴, 𝐺, and 𝐵. Similarly, member 𝐵𝐻𝐶 is a single member with pin connections at points 𝐵, 𝐻, and 𝐶. All other connections are also pins.

𝐵 𝐺

𝐻

(a) Is this structure a truss? Explain.

2m

(b) Determine the forces supported by all truss members. 𝐴

𝐷 1m

𝐸

𝐹

1m

1m

𝐶 1m

2 kN 3 kN 4 kN Figure P6.106

Problem 6.107 (a) Determine if the truss is statically determinate or statically indeterminate. (b) By inspection, identify the zero-force members in the truss. (c) Regardless of your answer to Part (a), determine the force supported by member 𝑄𝐼. 2m

2m 2m 2m 2m 𝐶 𝐺 𝐸 𝐼

2m

2m 2m 2m 𝑂 𝐾 𝑀

𝑄

𝐴 𝐵

𝑅 𝐻

𝐷 𝐹 4 kN

1m

𝐽

𝑃 𝐿

𝑁

Figure P6.107

Problem 6.108 (a) Determine if the truss is statically determinate or statically indeterminate. (b) By inspection, identify the zero-force members in the truss. (c) Regardless of your answer to Part (a), determine the force supported by member 𝐷𝐺. 2m

2m 2m 2m 2m 𝐶 𝐸 𝐺 𝐼

2m

2m 2m 2m 𝑂 𝐾 𝑀

𝑄 𝑃

𝐴 𝐵

𝐴

𝐵

𝐷 𝑃 (a)

𝐶

𝐷 𝐹 4 kN

𝐻

1m

𝐽

𝐿

𝑁

Figure P6.108 𝑃 (b)

Problem 6.109 Compared to the truss shown in Fig. P6.109(a): • The truss in (b) subdivides members 𝐴𝐵 and 𝐵𝐶 in half. • The truss in (c) subdivides members 𝐴𝐵, 𝐵𝐶, 𝐴𝐷, and 𝐷𝐶 in half.

𝑃 (c) Figure P6.109

ISTUDY

𝑃 (d)

• The truss in (d) subdivides members 𝐴𝐵 and 𝐵𝐶 into thirds. Consider failure due to in-plane buckling only. If the truss in figure (a) fails when 𝑃 = 1440 lb, specify (if possible) the values of 𝑃 at which the trusses in figures (b), (c), and (d) will fail.

ISTUDY

Section 6.5

429

Chapter Review

Problem 6.110 The boom hangs in the vertical 𝑥𝑧 plane and is supported by a socket at point 𝑁 and cables 𝑂𝐵 and 𝐵𝑃 . The bottom flange of the boom is equipped with a roller system so that a load can be moved along its length. Even though the top and bottom flanges of the boom are continuous members, idealize the structure as a truss, and determine the force supported by the two cables and member 𝐶𝐷 when the 600 lb load is positioned at point 𝐶. 𝑧 3 ft 𝑃

3 ft

4 ft 𝑂

3m 𝐴

𝑥

1m

𝐵

𝐶

𝐷

𝐵 𝐷

𝐹

𝐻

𝐿

𝐽

4 ft

𝑁

3m

4 ft 𝐸

𝐶

𝐴

𝐺

𝐸

𝐾

𝐼

𝐹

𝑀

𝑦

3m 200 N

2 ft

2 ft

𝐻

𝐺

600 lb 2 ft

2 ft

2 ft

pulleys have 300 mm radius

2 ft

Figure P6.110

3m 𝐽

𝐼

Problem 6.111 𝑀

A hoist for lifting building materials is attached to a scaffold. The hoist has frictionless pulleys at points 𝐴 and 𝐵, and both pulleys have 300 mm radius. Figure P6.111

(c) Determine the force supported by member 𝐸𝐻.

Problem 6.112 If 𝑊 = 400 lb, determine the force supported by members 𝐵𝐺 and 𝐶𝐹 . 4 ft 4 ft 4 ft 4 ft 𝐻 𝐵 𝐸 𝐾

5 in. 35 in. 30 in. 25 in. 𝐶

𝐷

𝐵

𝐴

𝐿 3 ft

𝑀 40 in. 𝑊

𝐴 𝐻

𝐺

𝐹

𝐸

Figure P6.112

𝐶

𝐽

𝐺

𝐷 𝐹

𝐼

radius = 0.8 f t

3 ft 𝑊

Figure P6.113

Problem 6.113 In the truss shown, cable segment 𝑀𝐿 is horizontal. If 𝑊 = 6000 lb, determine the force supported by members 𝐶𝐹 , 𝐶𝐺, 𝐸𝐺, and 𝐸𝐻. Hint: Use the method of sections twice with cuts such as those described in Example 6.5 on p. 390.

𝐿

𝐾 4m

(a) By inspection, identify the zero-force members in the truss. (b) Determine the force supported by member 𝐷𝐸.

3m

430

Chapter 6

Structural Analysis and Machines Problems 6.114 and 6.115 Member 𝐴𝐵𝐶𝐷 is a single member, and points 𝐴–𝐻 are pins. (a) Is this structure a truss? Explain.

(b) Is this structure statically determinate or statically indeterminate? Explain. 7 in.

(c) Determine as many reactions and the forces supported by as many members as possible.

11 in. 𝐼

120 lb

6m 2 kN 𝐴

𝐹

𝐵

𝐸

𝐵 8 in.

2 kN

8 in.

40 40 𝐽

80

80

80

8 ft 𝐺

𝐴 𝐸

6m

5 ft 𝐻 5 ft

𝐹

Figure P6.115

Problem 6.117 A lifting platform is supported by members 𝐽 𝐾𝐹 and 𝐿𝐻 and hydraulic cylinder 𝐾𝐿. All connections are frictionless pins.

80 𝐷

𝐶

𝐹

𝐼 𝐺

𝐾

𝐻

(a) Determine the force the hydraulic cylinder must support for equilibrium. (b) Use the method of joints to determine the force supported by member 𝐺𝐻. (c) Use the method of sections to determine the force supported by member 𝐵𝐶.

𝐿 dimensions in cm

Figure P6.117

ISTUDY

𝐻

5 ft

𝐶

Steps 𝐶𝐷𝐸 and 𝐹 𝐺𝐻 are supported by a truss structure having nine members. Determine the forces supported by all nine members of the truss.

2 kN

𝐵

𝐸

𝐺

𝐷

Problem 6.116 8 in.

Figure P6.116

80

𝐹 3m 3m

4 kip

8 ft

𝐵

Figure P6.114

45◦

8 ft 6 kip

𝐷 4m

𝐷

𝐴

𝐶

8 ft

8 kip 𝐸

80

3m 1 kN

𝐺

𝐶

80

3m

𝐻

80 lb

𝐴

6m

Problem 6.118 A small trailer-mounted dumper is shown. All connections are pinned, and member 𝐸𝐹 is horizontal. The hydraulic cylinder simultaneously tilts the dump and opens the gate. (a) Draw five FBDs, one each for hydraulic cylinder 𝐴𝐸, members 𝐵𝐷𝐸 and 𝐸𝐹 , gate 𝐹 𝐺, and the dump, labeling all forces. (b) Determine the force supported by hydraulic cylinder 𝐴𝐸. 6 in. 6 in. 600 lb

5

7 in. 12

𝐷

𝐺

𝐸

𝐹

12 in. 𝐴

𝐵

50 lb

𝐶 35 in.

14 in. Figure P6.118

24 in.

5 in.

ISTUDY

Centroids and Distributed Force Systems

7

This chapter begins with discussions of centroid, center of mass, and center of gravity. The common element among these is that centroid, center of mass, and center of gravity represent average positions of distributions. Applications of these concepts to distributed force systems, fluid pressure loading, and gas pressure loading are then considered.

Emily Exon/National Geographic Creative

Hoover Dam, in Boulder City, Nevada. Concepts of centroid, center of mass, center of gravity, and fluid pressure loading were needed to design this structure. The Mike O’Callaghan–Pat Tillman Memorial Bridge, shown under construction in the foreground of Hoover Dam, was opened to traffic in 2010; concepts of centroid, center of mass, and center of gravity were also used to design this structure.

7.1 Centroid Introduction—center of gravity While centroids are the focus of this section, it is helpful to begin with a short discussion about center of gravity. Consider the photo on this page of Hoover Dam, located on the Colorado River on the border between Arizona and Nevada. The dam is made primarily of concrete, but it also includes steel piping, turbine generators, and so on. Every atom of material in the dam has a weight due to gravity, and in the aggregate, these yield the total weight of the dam, which of course is quite substantial. The dam retains water in the reservoir, and this water applies pressure to the dam whose direction and intensity vary with every point on the face of the dam. In the aggregate, this water pressure gives rise to a substantial force. For many analysis and design purposes, it is sufficient to know only the total weight of the dam and the total force exerted by the water on the dam. Indeed, each of these constitutes an equivalent force system—that is, part of an equivalent force system—the remaining information needed is the location of

431

432

ISTUDY

Chapter 7

Centroids and Distributed Force Systems

the line of action for the weight, called the center of gravity, and the location of the line of action for the fluid force, sometimes called the center of pressure. Among the topics discussed in this chapter are methods for determining the center of gravity and the center of pressure. Thus, the concepts of this chapter build on and enhance the concepts of equivalent force systems discussed in Chapter 4. However, the concepts of this chapter go beyond forces and equivalent force systems. For example, the centroid of an area is important in governing the strength of a beam, and while the centroid is mathematically similar to the center of gravity, it is nonetheless very different. The underlying concept of this chapter is determination of the average position of a distribution, where the distribution may be shape, mass, weight, fluid pressure, gas pressure, and so on. We may quantify some of these ideas by considering a simple example, as shown in Fig. 7.1(a), where a server at a restaurant brings you wine and pasta on a tray. The center of gravity is defined to be the average position of the weight distribution of the objects the server carries. If the server is any good at his job, he will be sure that the center of gravity of the tray, wine, and pasta is positioned directly over the hand he uses to carry these. If this is not the case, then the tray will fall as he tries to carry it. In Fig. 7.1(a), the weights of the wine, tray, and pasta are 12 N, 8 N, and 12 N 8N

𝐹

10 N

𝐴

= 𝑥

12 18 16 dimensions in cm

𝑦

𝐴 𝑥̄ center of gravity

(a)

(b)

Figure 7.1. (a) A tray with wine and pasta. (b) Construction of an equivalent force system provides the weight of the collection of objects and the location of this weight, which is called the center of gravity.

10 N, respectively, with the positions shown. We would like to replace these weights with a single force, which is to be located at the center of gravity for the collection of objects. We can determine the force and its position (the center of gravity) by constructing an equivalent force system, as shown in Fig. 7.1(b). Using Eq. (4.16) on p. 248 for constructing an equivalent force system provides ∑ ∑ (𝐹𝑦 )system 2 (𝐹𝑦 )system 1 = 12 N + 8 N + 10 N = 𝐹 ⇒

𝐹 = 30 N,

∑ ∑ (𝑀𝐴 )system 2 (𝑀𝐴 )system 1 = (12 N)(12 cm) + (8 N)(30 cm) + (10 N)(46 cm) = 𝐹 𝑥̄ ⇒

𝑥̄ = 28.13 cm.

(7.1) (7.2) (7.3) (7.4) (7.5) (7.6)

The force 𝐹 = 30 N represents the entire weight of the collection of objects, and the center of gravity 𝑥̄ = 28.13 cm is the location of the line of action for this force. The

ISTUDY

Section 7.1

Centroid

433

important idea here is that the center of gravity is the average position of the weight distribution. Equations (7.1)–(7.6) can be generalized to give the center of gravity for a collection of an arbitrary number of objects as 𝑛 ∑

𝑥̄ =

𝑥̃ 𝑖 𝑊𝑖

𝑖=1 𝑛 ∑

𝑖=1

(7.7)

, 𝑊𝑖

Concept Alert

where 𝑛 is the number of objects, 𝑊𝑖 is the weight of object 𝑖, and 𝑥̃ 𝑖 is the location of 𝑊𝑖 (i.e., the moment arm) measured from the origin of the coordinate system. In Eq. (7.7), we use the superposed tilde ̃ to emphasize that 𝑥̃ 𝑖 is the position of the center of gravity of object 𝑖. In Eq. (7.7), the products 𝑥̃ 𝑖 𝑊𝑖 are called the first moments of the weights 𝑊𝑖 . The concept underlying Eq. (7.7) can be generalized so that by evaluating the first moment of any distribution, the average position of that distribution may be determined. For example, if we replace 𝑊𝑖 in Eq. (7.7) with the mass 𝑚𝑖 of each object, 𝑥̄ then measures the average position of the mass for a collection of objects, and this is called the center of mass. Similarly, if we replace 𝑊𝑖 in Eq. (7.7) with the volume 𝑉𝑖 of each object, 𝑥̄ then measures the average position for the volume of a collection of objects, and this is called the centroid.

Centroid, etc. Centroid, center of mass, and center of gravity all measure the average position of distributions. Furthermore, the calculation procedures for determining these are almost identical. • The centroid is the average position of a distribution of shapes, such as lines, areas, or volumes. • The center of mass is the average position of a distribution of mass. • The center of gravity is the average position of a distribution of weight.

Centroid of an area The centroid is defined to be the average position of a distribution of shapes. If the distribution consists of a single shape, the shape can be a line (straight or curved), an area, or a volume. For a distribution (or collection) of shapes, the distribution can include multiple lines, or multiple areas, or multiple volumes—but not combinations of these. For the present, we consider the area shown in Fig. 7.2. By considering the area to be a collection of composite areas 𝐴𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 and 𝑦̃𝑖 , the position of the centroid for the entire area, 𝑥̄ and 𝑦, ̄ can be determined by generalizing Eq. (7.7) to obtain 𝑛 ∑

𝑥̄ =

𝑛 ∑

𝑥̃ 𝑖 𝐴𝑖

𝑖=1 𝑛 ∑

𝑖=1

and

𝑦̄ =

𝐴𝑖

𝑦̃𝑖 𝐴𝑖

𝑖=1 𝑛 ∑

𝑖=1

,

∫ 𝑥̃ 𝑑𝐴 ∫ 𝑑𝐴

and

𝑦̄ =

∫ 𝑦̃ 𝑑𝐴 ∫ 𝑑𝐴

𝑦 𝐴

𝑦̄

𝐶

𝑥̄

𝐴𝑖

𝑦̃𝑖 𝑥

𝑥̃ 𝑖

𝑥

Figure 7.2 The centroid 𝐶 of area 𝐴 is located at 𝑥̄ and 𝑦. ̄ Area 𝐴 can be considered to consist of composite areas 𝐴𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 and 𝑦̃𝑖 .

(7.8)

Helpful Information

𝐴𝑖

where 𝑛 is the number of composite shapes that constitute the entire area, 𝐴𝑖 is the area of composite shape 𝑖, and 𝑥̃ 𝑖 and 𝑦̃𝑖 are the locations of the centroid of 𝐴𝑖 measured ∑ from the origin of the coordinate system. In Eq. (7.8), the numerator 𝑥̃ 𝑖 𝐴𝑖 is called ∑ the first moment of the area about the 𝑦 axis, and 𝑦̃𝑖 𝐴𝑖 is called the first moment of the area about the 𝑥 axis. By taking the limits of Eq. (7.8) as 𝐴𝑖 → 0, the summations become integrals, and Eq. (7.8) becomes

𝑥̄ =

𝑦

,

(7.9)

̃ The expressions Don’t confuse 𝒙̄ and 𝒙. for 𝑥̄ in Eqs. (7.8) and (7.9) can be written together as 𝑛 ∑

𝑥̄ =

𝑖=1 𝑛

𝑥̃ 𝑖 𝐴𝑖

∑

𝑖=1

= 𝐴𝑖

∫ 𝑥̃ 𝑑𝐴 ∫ 𝑑𝐴

.

In these expressions, 𝑥̄ is the centroid of the entire shape, whereas 𝑥̃ 𝑖 is the centroid of composite shape 𝑖 whose area is 𝐴𝑖 , and 𝑥̃ is the centroid of area element 𝑑𝐴.

434

Chapter 7

Centroids and Distributed Force Systems

where 𝑥̃ and 𝑦̃ are the locations of the centroid of area element 𝑑𝐴 measured from the origin of the coordinate system. In Eq. (7.9), the numerator ∫ 𝑥̃ 𝑑𝐴 is called the first moment of the area about the 𝑦 axis, and ∫ 𝑦̃ 𝑑𝐴 is called the first moment of the area about the 𝑥 axis.

Centroid of a line 𝑦

Consider the line shown in Fig. 7.3. By considering the line to be a collection of composite lines with lengths 𝐿𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 and 𝑦̃𝑖 , the position of the centroid for the entire line, 𝑥̄ and 𝑦, ̄ is given by

𝑦 𝐿

𝑦̄

𝐶 𝑦̃𝑖 𝑥

𝑥̄

𝐿𝑖 𝑥̃ 𝑖

𝑛 ∑

𝑥

Figure 7.3 The centroid 𝐶 of a line with length 𝐿 is located at 𝑥̄ and 𝑦. ̄ Length 𝐿 can be considered to consist of composite lengths 𝐿𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 and 𝑦̃𝑖 .

𝑥̄ =

𝑛 ∑

𝑥̃ 𝑖 𝐿𝑖

𝑖=1 𝑛 ∑

𝑖=1

and

𝑦̄ =

𝐿𝑖

𝑦̃𝑖 𝐿𝑖

𝑖=1 𝑛 ∑

𝑖=1

,

(7.10)

𝐿𝑖

where 𝑛 is the number of composite lines that constitute the entire length, 𝐿𝑖 is the length of composite line 𝑖, and 𝑥̃ 𝑖 and 𝑦̃𝑖 are the locations of the centroid of 𝐿𝑖 measured from the origin of the coordinate system. By taking the limits of Eq. (7.10) as 𝐿𝑖 → 0, this equation becomes 𝑥̄ =

∫ 𝑥̃ 𝑑𝐿 ∫ 𝑑𝐿

and

𝑦̄ =

∫ 𝑦̃ 𝑑𝐿 ∫ 𝑑𝐿

,

(7.11)

where 𝑥̃ and 𝑦̃ are the locations of the centroid of length element 𝑑𝐿 measured from the origin of the coordinate system.

𝑦

𝑑𝐿 𝑥

𝑑𝑦

𝑑𝑥

Figure 7.4 Length increment 𝑑𝐿 is related to increments 𝑑𝑥 and 𝑑𝑦 by the Pythagorean theorem.

ISTUDY

Integration along a path. In Eq. (7.11) [and Eq. (7.18) later in this section], integrations are to be carried out along the path of a line. Such integrals are often called line integrals (or path integrals, or curve integrals). Rather than integrate along the path of the line, it will often be more convenient to integrate with respect to 𝑥 or 𝑦, and to do this, a transformation between 𝑑𝐿 and 𝑑𝑥 or 𝑑𝑦 is needed. As shown in Fig. 7.4, length increment 𝑑𝐿 is related to increments 𝑑𝑥 and 𝑑𝑦 by the Pythagorean theorem: √ 𝑑𝐿 = (𝑑𝑥)2 + (𝑑𝑦)2 . (7.12) Multiplying the right-hand side of Eq. (7.12) by 𝑑𝑥∕𝑑𝑥 and bringing 1∕𝑑𝑥 within the radical give √ ( 𝑑𝑦 )2 𝑑𝐿 = 1 + 𝑑𝑥. (7.13) 𝑑𝑥 Similarly, multiplying the right-hand side of Eq. (7.12) by 𝑑𝑦∕𝑑𝑦 and bringing 1∕𝑑𝑦 within the radical give √ ( )2 𝑑𝑥 𝑑𝐿 = + 1 𝑑𝑦. (7.14) 𝑑𝑦 Rather than memorize Eqs. (7.13) and (7.14), it is perhaps easier to simply remember Eq. (7.12) along with multiplication by 𝑑𝑥∕𝑑𝑥 or 𝑑𝑦∕𝑑𝑦. Example 7.5 on p. 443 illustrates the use of Eq. (7.13).

ISTUDY

Section 7.1

Centroid

Centroid of a volume

𝑦

Consider the volume shown in Fig. 7.5. By considering the volume to be a collection of composite volumes 𝑉𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 , 𝑦̃𝑖 , and 𝑧̃ 𝑖 , the position of the centroid for the entire volume, 𝑥, ̄ 𝑦, ̄ and 𝑧, ̄ is given by 𝑛 ∑

𝑥̄ =

𝑛 ∑

𝑥̃ 𝑖 𝑉𝑖

𝑖=1 𝑛 ∑

𝑖=1

,

𝑦̄ =

𝑉𝑖

𝑛 ∑

𝑦̃𝑖 𝑉𝑖

𝑖=1 𝑛 ∑

𝑖=1

,

and

𝑧̄ =

𝑉𝑖

𝑖=1

∫ 𝑥̃ 𝑑𝑉 ∫ 𝑑𝑉

,

𝑦̄ =

∫ 𝑦̃ 𝑑𝑉 ∫ 𝑑𝑉

,

and

𝑧̄ =

(7.15)

𝑉𝑖

∫ 𝑧̃ 𝑑𝑉 ∫ 𝑑𝑉

𝐶(𝑥, ̄ 𝑦, ̄ 𝑧) ̄

𝑧

,

where 𝑛 is the number of composite volumes that constitute the entire volume, 𝑉𝑖 is the volume of composite volume 𝑖, and 𝑥̃ 𝑖 , 𝑦̃𝑖 , and 𝑧̃ 𝑖 are the locations of the centroid of 𝑉𝑖 measured from the origin of the coordinate system. By taking the limits of Eq. (7.15) as 𝑉𝑖 → 0, this equation becomes 𝑥̄ =

𝑦

𝑉𝑖 𝑥̃ 𝑖 , 𝑦̃𝑖 , 𝑧̃𝑖 𝑥

𝑥

𝑧̃ 𝑖 𝑉𝑖

𝑖=1 𝑛 ∑

435

𝑧

Figure 7.5 The centroid 𝐶 of a volume 𝑉 is located at 𝑥, ̄ 𝑦, ̄ and 𝑧. ̄ Volume 𝑉 can be considered to consist of composite volumes 𝑉𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 , 𝑦̃𝑖 , and 𝑧̃ 𝑖 .

Helpful Information ,

(7.16)

where 𝑥, ̃ 𝑦, ̃ and 𝑧̃ are the locations of the centroid of volume element 𝑑𝑉 measured from the origin of the coordinate system.

Centroid of symmetric areas. If an area has an axis of symmetry, such as the 𝑦 axis in the figure below, the centroid 𝐶 of that area lies on the axis of symmetry. 𝑦

Unification of concepts In this section, expressions using composite shapes and integration for the centroid of lines, areas, and volumes in one, two, and three dimensions have been given. The unifying idea behind all of these is that the centroid is the average position of a distribution of shape. While the number of equations presented is rather large, Eq. (7.19), discussed in the End of Section Summary on p. 437, boils all of these down into one compact expression. If you remember and understand this expression, then it is easy to extrapolate it for other applications. In addition to this, Eq. (7.12) [and/or Eqs. (7.13) and (7.14)] is also needed to determine the centroid of lines by using integration.

Which approach should I use: composite shapes or integration? Determining the centroid using composite shapes is usually very straightforward provided the centroid position for each composite shape is readily available, either by inspection or from tabulated data for common geometric shapes. To this end, the tables given at the end of this book for Properties of Lines and Areas and for Properties of Solids will be useful. Use of integration to determine the centroid is simple in concept, but not always simple in evaluation. Nonetheless, integration can always be used. When we evaluate integrals, many choices are possible for coordinate systems (Cartesian coordinates, polar coordinates, and cylindrical coordinates are common), and area and volume integrations may require double or triple integrals. These various choices are extensively discussed in calculus. In this book, our discussion of integration for determining the centroid is more focused. We will evaluate all area and volume integrations by using single integrals. Computer mathematics software such as Mathematica and Maple can make evaluation of integrals easy, but you must still properly set up the integrals, including the limits of integration.

axis of symmetry 𝑥̃ 𝐶

ℎ

ℎ 𝑥

𝑑𝑥

𝑑𝑥

To understand why this is true, consider the two symmetrically located area elements shown. For every area element 𝑑𝐴 = ℎ 𝑑𝑥 to the right of the axis of symmetry, there is an equal area element to the left, with equal but negative 𝑥̃ /value. Thus, when we evaluate 𝑥̄ = ∫ 𝑥̃ 𝑑𝐴 ∫ 𝑑𝐴, the integral in the numerator is zero and hence 𝑥̄ = 0. If an area has two axes of symmetry, then the centroid is located at the intersection of the two axes of symmetry. Centroid of symmetric volumes. If a volume has a plane of symmetry, then the centroid of that volume lies somewhere on the plane of symmetry. If the volume has multiple planes of symmetry, then the centroid lies somewhere on the intersection of the planes of symmetry.

436 𝑦

Finer points: surfaces and lines in three dimensions

𝑦 𝐴𝑖 𝐶(𝑥, ̄ 𝑦, ̄ 𝑧) ̄ 𝑥̃ 𝑖 , 𝑦̃𝑖 , 𝑧̃𝑖 𝑥

𝑥 𝑧

𝑧

Figure 7.6 The centroid 𝐶 of a surface area 𝐴 in three dimensions is located at 𝑥, ̄ 𝑦, ̄ and 𝑧. ̄ Area 𝐴 can be considered to consist of composite areas 𝐴𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 , 𝑦̃𝑖 , and 𝑧̃ 𝑖 . 𝑦

𝑦 𝐿 𝐶(𝑥, ̄ 𝑦, ̄ 𝑧) ̄

𝐿𝑖 𝑥̃ 𝑖 , 𝑦̃𝑖 , 𝑧̃𝑖

𝑥 𝑧

𝑥

𝑧

Figure 7.7 The centroid 𝐶 of a line of length 𝐿 in three dimensions is located at 𝑥, ̄ 𝑦, ̄ and 𝑧. ̄ Length 𝐿 can be considered to consist of composite lengths 𝐿𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 , 𝑦̃𝑖 , and 𝑧̃ 𝑖 .

ISTUDY

Chapter 7

Centroids and Distributed Force Systems

So far in this section, we have considered the centroid of lines and areas in two dimensions. However, Eqs. (7.8) through (7.11) are also applicable to lines and areas in three dimensions, provided an additional expression is written for the 𝑧 location of the centroid. For example, for a surface area 𝐴 in three dimensions, as shown in Fig. 7.6, the centroid positions 𝑥̄ and 𝑦̄ are as given in Eqs. (7.8) and (7.9), and the 𝑧 position is given by 𝑛 ∑

𝑧̄ =

𝑧̃ 𝑖 𝐴𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝐴𝑖

∫ 𝑧̃ 𝑑𝐴 ∫ 𝑑𝐴

.

(7.17)

Similarly, for a line of length 𝐿 in three dimensions, as shown in Fig. 7.7, the centroid positions 𝑥̄ and 𝑦̄ are as given in Eqs. (7.10) and (7.11), and the 𝑧 position is given by 𝑛 ∑ 𝑧̃ 𝑖 𝐿𝑖 ∫ 𝑧̃ 𝑑𝐿 𝑖=1 𝑧̄ = 𝑛 = . (7.18) ∑ ∫ 𝑑𝐿 𝐿𝑖 𝑖=1

ISTUDY

Section 7.1

Centroid

End of Section Summary The centroid is defined to be the average position of a distribution of shapes. If the distribution consists of a single shape, the shape can be a line (straight or curved), an area, or a volume. For a collection of shapes, the distribution can include multiple lines, or multiple areas, or multiple volumes—but not combinations of these. The centroid can be determined by the use of composite shapes or by integration. The many formulas presented in this section can be compactly summarized. Consider the case of the centroid of an area or a distribution of areas, as shown in Fig. 7.2 on p. 433. The expressions for 𝑥̄ in Eqs. (7.8) and (7.9) can be written together as 𝑛 ∑

𝑥̄ =

𝑥̃ 𝑖 𝐴𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝐴𝑖

∫ 𝑥̃ 𝑑𝐴 ∫ 𝑑𝐴

,

(7.19)

with the following remarks: • The summation form is used if the centroid is to be determined using composite shapes. This approach is straightforward, provided the centroid position 𝑥̃ 𝑖 for each of the composite shapes is readily available. • The integral form can always be used to determine the centroid. Many of the example problems of this section illustrate this approach. • To determine the 𝑦 position of the centroid, an equation for 𝑦̄ is written by replacing all of the 𝑥’s that appear in Eq. (7.19) with 𝑦’s. • If the 𝑧 position of the centroid is needed, such as for an object in three dimensions (e.g., the area in three dimensions shown in Fig. 7.6), then an equation for 𝑧̄ is written by replacing all of the 𝑥’s that appear in Eq. (7.19) with 𝑧’s. • To determine the centroid of a line or a distribution of lines (straight and/or curved), replace all of the 𝐴’s that appear in Eq. (7.19) with 𝐿’s. • To determine the centroid of a volume or a distribution of volumes, replace all of the 𝐴’s that appear in Eq. (7.19) with 𝑉 ’s. The final point to summarize is that if an area has an axis of symmetry, then the centroid lies on the axis of symmetry. Furthermore, if an area has multiple axes of symmetry, then the centroid is located at the intersection of the axes of symmetry. Similarly, if a volume has a plane of symmetry, then the centroid lies on the plane of symmetry. If a volume has multiple planes of symmetry, then the centroid lies on the intersection of the planes of symmetry.

437

438

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.1

Centroid of an Area Using Composite Shapes The cross section of an extruded aluminum channel is shown. Determine the 𝑥 and 𝑦 positions of the cross section’s centroid.

𝑦 40 mm 8 mm

SOLUTION Road Map

The shape of the channel’s cross section is an arrangement of simple geometric shapes, namely rectangles. Thus, it will be straightforward to determine the position of the centroid by using composite shapes with Eq. (7.8) on p. 433. There are several possible arrangements of composite shapes that may be used, and we will consider solutions for two of these.

50 mm 10 mm 8 mm

𝑥

Figure 1

Solution 1 Governing Equations & Computation

In Fig. 2, the shape is subdivided into three rectangles, where 𝐴1 is the area of rectangle 1, 𝐶1 denotes the centroid of rectangle 1, and so on. Because the cross-sectional shape is symmetric about the line 𝑦 = 25 mm, the centroid must lie on this line, and therefore the 𝑦 position of the centroid is 𝑦̄ = 25 mm. The areas and centroid positions of each composite shape are collected in Table 1 followed by evaluation of Eq. (7.8).

𝑦 10 mm 30 mm 𝐴2 8 mm 𝐶2 𝐴1 𝐶 1 50 mm 10 mm 𝐴3 8 mm 𝐶3

Table 1. Areas and centroid positions for composite shapes in Fig. 2. Shape no. 𝑥

1 2 3

Figure 2 A selection of three composite shapes to describe the channel’s cross section.

3 ∑ 𝑖=1

𝑥̄ =

𝑥̃ 𝑖 𝐴𝑖 =

3

∑ 𝑖=1

𝐴𝑖

𝐴𝑖

𝑥̃ 𝑖 2

(10 mm)(50 mm) = 500 mm (30 mm)(8 mm) = 240 mm2 (30 mm)(8 mm) = 240 mm2

5 mm 25 mm 25 mm

(5 mm)(500 mm2 ) + (25 mm)(240 mm2 ) + (25 mm)(240 mm2 ) 500 mm2 + 240 mm2 + 240 mm2

= 14.80 mm. 𝑦

Solution 2

10 mm 30 mm

40 mm

Governing Equations & Computation

8 mm 𝐴1

𝐴2

50 mm

34 mm

Shown in Fig. 3 is a selection of two composite shapes for the cross section of the channel, where the second of these has negative area. The areas and centroid positions of each composite shape are collected in Table 2 followed by evaluation of Eq. (7.8).

𝐶2

𝐶1 𝑥

Table 2. Areas and centroid positions for composite shapes in Fig. 3. 8 mm

𝑥

Shape no.

Figure 3 A selection of two composite shapes to describe the channel’s cross section.

ISTUDY

(1)

𝑦

1 2 2 ∑

𝑥̄ =

𝑖=1

𝑥̃ 𝑖 𝐴𝑖 =

2

∑ 𝑖=1

𝐴𝑖

𝐴𝑖

𝑥̃ 𝑖 2

(40 mm)(50 mm) = 2000 mm −(30 mm)(34 mm) = −1020 mm2

20 mm 25 mm

(20 mm)(2000 mm2 ) + (25 mm)(−1020 mm2 ) = 14.80 mm. 2000 mm2 − 1020 mm2

Discussion & Verification

(2)

As expected, Eqs. (1) and (2) agree. As an exercise, you should add a column for 𝑦̃𝑖 to Tables 1 and 2 and use Eq. (7.8) to show 𝑦̄ = 25 mm.

ISTUDY

Section 7.1

Centroid

E X A M P L E 7.2

Centroid of a Volume Using Composite Shapes

A solid has the shape of a cylinder with a truncated cone. Determine the location of the centroid.

𝑧 2 in.

SOLUTION

2 in.

Road Map

This object consists of an arrangement of simple geometric shapes. Thus, compared to using integration, it will be easier to determine the position of the centroid by using composite shapes with Eq. (7.15) on p. 435. This object is symmetric about the 𝑥𝑧 plane, and therefore its centroid must lie in the 𝑥𝑧 plane. The object is also symmetric about the 𝑦𝑧 plane, and therefore its centroid must also lie in the 𝑦𝑧 plane. The intersection of these planes is the 𝑧 axis. Therefore, we can conclude that the 𝑥 and 𝑦 positions of the centroid are 𝑥̄ = 𝑦̄ = 0. Governing Equations & Computation In Fig. 2 the object is subdivided into three shapes. The volume and centroid position of each of the composite shapes are given in the Table of Properties of Solids at the end of this book, and these values are collected in Table 1. For example, for shape 2, the volume is 𝑉2 = 𝜋𝑟2 ℎ∕3, where 𝑟 = 2 in. is the radius

3 in. 𝑦 𝑥

2 in.

Figure 1

𝑉3

4 in.

𝑉2 Shape no.

𝑉𝑖 2

𝑧̃ 𝑖 3

𝜋(2 in.) (3 in.) = 37.70 in. 𝜋 (2 in.)2 (4 in.) = 16.76 in.3 3 𝜋 − 3 (1 in.)2 (2 in.) = −2.094 in.3

2 in. 𝑧

1.5 in. 3 in. + 14 (4 in.) = 4 in. 5 in. + 14 (2 in.) = 5.5 in.

𝑉1

of the cone’s base, and ℎ = 4 in. is its height. The centroid for the cone measured from its base is ℎ∕4, and to this value, the distance from the origin of the coordinate system to the cone’s base (i.e., 3 in.) must be added. Evaluating the expression for 𝑧̄ in Eq. (7.15) provides 3 ∑

𝑧̄ =

𝑖=1

𝑖=1

=

(1.5 in.)(37.70 in.3 ) + (4 in.)(16.76 in.3 ) + (5.5 in.)(−2.094 in.3 ) 37.70 in.3 + 16.76 in.3 − 2.094 in.3

Discussion & Verification

𝑦 𝑥

2 in.

Helpful Information

𝑉𝑖

= 2.140 in.

3 in.

Figure 2 A selection of three composite shapes to describe the solid.

𝑧̃ 𝑖 𝑉𝑖

3 ∑

2 in.

1 in.

Table 1. Volumes and centroid positions for composite shapes in Fig. 2.

1 2 3

439

(1)

Short of re-solving this problem by using a different selection of composite shapes, or using a different method (e.g., integration), there are no definitive checks of accuracy for our solution. Nonetheless, we expect 𝑧̄ to be somewhat larger than 1.5 in., which is the centroid of volume 𝑉1 only, and in view of this, the value found in Eq. (1) is reasonable.

Solid of revolution. The shape shown in Fig. 1 is called a solid of revolution because it can be produced by revolving a planar shape about a straight line, called the axis of revolution. For example, rotating the area shown below by 360◦ about the 𝑧 axis produces the shape shown in Fig. 1: 𝑧

𝑦 𝑥

440

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.3 𝑦 (mm) 15

Centroid of an Area Using Integration The cross section of a turbine blade in a pump is shown. Determine the 𝑥 and 𝑦 positions of the centroid.

√ 𝑦𝑡 = 3 𝑥

10 𝑦𝑏 =

5

SOLUTION

3 𝑥 5

Road Map

𝑥 (mm)

0

0

10

5

15

20

25

Figure 1

The cross section shape of the turbine blade is an area, and thus Eq. (7.9) on p. 433 will be used to determine the 𝑥 and 𝑦 positions of its centroid. In Eq. (7.9), three ingredients are needed: expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦. ̃ These expressions can be developed using either a vertical area element or a horizontal area element, and solutions using both of these are demonstrated. Solution 1 – Vertical area element

𝑦 (mm) 15

Governing Equations & Computation

To evaluate Eq. (7.9), expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦̃ are needed, and these can be developed using the vertical area element shown in Fig. 2 as follows. ( √ ) 𝑑𝐴 = (𝑦𝑡 − 𝑦𝑏 ) 𝑑𝑥 = 3 𝑥 − 53 𝑥 𝑑𝑥, (1)

√ 𝑦𝑡 = 3 𝑥

𝑑𝑥

10

𝑦𝑏 =

𝑦̃

3 𝑥 5 𝑥 (mm)

0

0

𝑥̃ 10

15

20

25

Figure 2 A vertical area element is used to develop expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦. ̃

Helpful Information Notation. Throughout this problem, subscripts are used to help distinguish top and bottom curves, and left and right curves, as follows:

∫

25 mm

(3)

) ( √ 𝑥 3 𝑥 − 35 𝑥 𝑑𝑥

) 6𝑥5∕2 𝑥3 |25 mm − | ∫ 𝑥̃ 𝑑𝐴 0 5 |0 = = ( 5 𝑥̄ = ) 25 mm ( √ ∫ 𝑑𝐴 3𝑥2 |25 mm ) 2𝑥3∕2 − | ∫ 3 𝑥 − 35 𝑥 𝑑𝑥 10 |0 (

0

625 mm3 = 10 mm, = 62.5 mm2

• Subscript 𝑏 denotes “bottom.”

𝑦̄ =

• Subscript 𝑙 denotes “left.”

(2)

( √ ) 𝑦̃ = 12 (𝑦𝑡 + 𝑦𝑏 ) = 12 3 𝑥 + 53 𝑥 .

Substituting Eqs. (1) through (3) into Eq. (7.9), the position of the centroid is given by

• Subscript 𝑡 denotes “top.” • Subscript 𝑟 denotes “right.”

𝑥̃ = 𝑥,

∫ 𝑦̃ 𝑑𝐴 ∫ 𝑑𝐴

∫

25 mm

=

0

1( 3 2

(4)

√ )( √ ) 𝑥 + 35 𝑥 3 𝑥 − 35 𝑥 𝑑𝑥

∫

25 mm ( 0

√ ) 3 𝑥 − 35 𝑥 𝑑𝑥

) 9𝑥2 3𝑥3 |25 mm − | 50 |0 = ( 4 ) 3𝑥2 |25 mm 2𝑥3∕2 − | 10 |0 (

468.75 mm3 = = 7.5 mm. 62.5 mm2

(5)

Solution 2 – Horizontal area element 𝑦 (mm) 15

Governing Equations & Computation

𝑦2 𝑥𝑙 = 9

10 𝑦̃ 0

𝑑𝑦

𝑥𝑟 =

5 𝑦 3 𝑥 (mm)

0

𝑥̃ 10

15

20

25

Figure 3 A horizontal area element is used to develop expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦. ̃

ISTUDY

The centroid can also be determined using the horizontal area element shown in Fig. 3. Since integrations will be over 𝑦, it is necessary to express the shape of the object as functions of 𝑦. Thus, the equation for the bottom curve 𝑦𝑏 = (3∕5)𝑥 is rearranged to obtain 𝑥𝑟 = (5∕3)𝑦. Similarly, the equation for the top √ curve 𝑦𝑡 = 3 𝑥 is rearranged to obtain 𝑥𝑙 = 𝑦2 ∕9. Then expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦̃ can be written as follows. 𝑦2 ) 𝑑𝑦, 9 ( 𝑦2 ) 𝑥̃ = 12 (𝑥𝑟 + 𝑥𝑙 ) = 21 35 𝑦 + , 9 𝑦̃ = 𝑦.

𝑑𝐴 = (𝑥𝑟 − 𝑥𝑙 ) 𝑑𝑦 =

(5

3

𝑦−

(6) (7) (8)

ISTUDY

Section 7.1

Centroid

441

Substituting Eqs. (6) through (8) into Eq. (7.9), the position of the centroid is given by

𝑥̄ =

∫ 𝑥̃ 𝑑𝐴 ∫ 𝑑𝐴

∫

15 mm

=

=

0

1 2

(

( 25𝑦3 𝑦2 )( 5 𝑦2 ) 𝑦5 )|15 mm 𝑦 − 𝑑𝑦 − | 3 9 9 |0 = ( 54 2 810 ) 3 15 mm ( ) 15 2 𝑦 5𝑦 | mm 𝑦 5 − | ∫ 𝑑𝑦 𝑦 − 3 6 27 |0 9 0 5 𝑦 3

+

625 mm3 = 10 mm, 62.5 mm2 ∫

(9)

( 5𝑦3 𝑦4 ) 15 mm 𝑦2 ) | 𝑑𝑦 − | 9 ∫ 𝑦̃ 𝑑𝐴 0 9 36 |0 𝑦̄ = = ( 2 = 15 mm ( ∫ 𝑑𝐴 5𝑦 𝑦3 )|15 mm 𝑦2 ) 5 − | ∫ 𝑦 − 𝑑𝑦 3 6 27 |0 9 0 (

15 mm

=

𝑦

5 𝑦 3

−

468.75 mm3 = 7.5 mm. 62.5 mm2

(10) 𝑦 (mm)

As expected, the solutions for both 𝑥̄ and 𝑦̄ agree with those obtained earlier in Eqs. (4) and (5).

15 10

Discussion & Verification

• The location of the centroid is shown as point 𝐶 in Fig. 4, and this location appears to be reasonable. • The solution procedure outlined here can be used for a wide variety of area centroid problems by simply using the appropriate expressions for the bounding curves 𝑦𝑡 and 𝑦𝑏 , or 𝑥𝑟 and 𝑥𝑙 . Specifically:

𝑦̄

𝐶

5 0

𝑥̄ = 10 mm 𝑦̄ = 7.5 mm 𝑥 (mm)

0

5 𝑥̄

10

15

20

25

Figure 4 The centroid 𝐶 of the area is located at 𝑥̄ and 𝑦. ̄

– If a vertical area element is used, then expressions for the top and bottom curves 𝑦𝑡 and 𝑦𝑏 , respectively, as functions of 𝑥 are needed, and Eqs. (1) through (3) can be used to obtain expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦. ̃ – If a horizontal area element is used, then expressions for the right and left curves 𝑥𝑟 and 𝑥𝑙 , respectively, as functions of 𝑦 are needed, and Eqs. (6) through (8) can be used to obtain expressions for 𝑑𝐴, 𝑥, ̃ and 𝑦. ̃ • For some problems, a vertical area element may be more convenient than a horizontal area element, or vice versa. For example, for the area shown in Fig. 5, a vertical area element will be more convenient because both 𝑦𝑡 and 𝑦𝑏 are given by single equations. To use a horizontal area element, two different equations are needed for 𝑥𝑟 , depending on where the area element is located. • Software such as Mathematica and Maple can make evaluation of integrals easy, but obtaining correct results with such software requires that you provide the correct integrand and limits of integration.

𝑦

𝑥 Figure 5 Example of a shape for which a vertical area element is more convenient than a horizontal area element.

442

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.4

Centroid of a Line Using Integration Determine the 𝑥 and 𝑦 positions of the centroid for the circular arc shown.

𝑦

SOLUTION 𝑟 Road Map 𝛼 𝛼

𝑥

Because the arc is symmetric about the 𝑥 axis, the centroid must lie on the 𝑥 axis. Therefore, by inspection, the 𝑦 position of the centroid is 𝑦̄ = 0. To determine the 𝑥 position of the centroid, we will evaluate the first expression in Eq. (7.11) on p. 434, and for developing expressions for 𝑑𝐿 and 𝑥, ̃ polar coordinates will be very convenient.

Governing Equations & Computation

Expressions for 𝑑𝐿 and 𝑥̃ will be written using the line element shown in Fig. 2 as follows:

Figure 1

𝑑𝐿 = 𝑟 𝑑𝜃, 𝑥̃ = 𝑟 cos 𝜃.

𝑦 𝑑𝐿 = 𝑟 𝑑𝜃 𝜃 𝑟

𝑑𝜃 𝑥̃

(2)

Substituting Eqs. (1) and (2) into the first expression of Eq. (7.11), the 𝑥 position of the centroid is given by

𝑥

𝑥̄ = Figure 2 A line element is used to develop expressions for 𝑑𝐿 and 𝑥. ̃

ISTUDY

(1)

∫ 𝑥̃ 𝑑𝐿 ∫ 𝑑𝐿

∫ 𝑟 cos 𝜃 𝑟 𝑑𝜃 𝛼

=

−𝛼

∫ 𝑟 𝑑𝜃 𝛼

−𝛼

=

( )|𝛼 𝑟2 sin 𝜃 | 2𝑟2 sin 𝛼 |−𝛼 = = ( )|𝛼 2𝑟𝛼 𝑟 𝜃 | |−𝛼

𝑟 sin 𝛼 . 𝛼

(3)

(4)

Discussion & Verification

• For a circular arc, 𝑟 = constant, and the integrals in Eq. (3) are simplified, as follows. Because 𝑟 is not a function of 𝜃, the 𝑟2 term in the numerator and the 𝑟 term in the denominator of Eq. (3) may be brought outside their integrals, leaving simple functions of 𝜃 to be integrated. For a line that is not a circular arc, 𝑟 is a function of 𝜃, giving rise to generally more difficult integrals to evaluate. For such cases, the approach used in Example 7.5, where integrations are carried out with respect to 𝑥 (or 𝑦), may be more convenient. • To evaluate Eq. (4) for particular values of 𝑟 and 𝛼, it is necessary to measure 𝛼 in radians. • The validity of our results can be partially checked by considering the location of 𝑥̄ for particular values of 𝛼. When 𝛼 → 0, the arc approaches a point, and the centroid should approach 𝑥̄ = 𝑟, which Eq. (4) does.∗ When 𝛼 = 𝜋, the arc becomes a full circle, and the centroid should become 𝑥̄ = 0, which Eq. (4) does. • As an exercise, you can evaluate the second expression of Eq. (7.11) on p. 434 by using 𝑦̃ = 𝑟 sin 𝜃 to confirm that 𝑦̄ = 0.

∗ Recall

the small-angle approximation sin 𝛼 ≈ 𝛼 when 𝛼 ≪ 1 (𝛼 measured in radians).

ISTUDY

Section 7.1

443

Centroid

E X A M P L E 7.5

Centroid of a Line Using Integration

Determine the 𝑥 and 𝑦 positions of the centroid for the uniform curved bar shown. 𝑦

SOLUTION

0.25 m

𝑦 = 𝑥2

Road Map

To determine the position of the centroid, we will evaluate Eq. (7.11) on p. 434. Because the equation for the line is expressed in terms of 𝑥 and 𝑦, it will be more convenient to perform integrations with respect to 𝑥 or 𝑦, rather than with respect to length along the path of the line. To this end, we will use Eqs. (7.13) and (7.14) on p. 434.

𝑥 0.5 m Figure 1

Governing Equations

In Eq. (7.11), 𝑥̃ and 𝑦̃ locate the centroid of length element 𝑑𝐿, as shown in Fig. 2. An expression for 𝑑𝐿 will be written in terms of 𝑑𝑥 by using Eq. (7.13) on p. 434.∗ Noting that 𝑑𝑦∕𝑑𝑥 = 2𝑥, √ ( 𝑑𝑦 )2 √ (1) 𝑑𝐿 = 1 + 𝑑𝑥 = 1 + 4𝑥2 𝑑𝑥, 𝑑𝑥 𝑥̃ = 𝑥, (2) 𝑦̃ = 𝑦 = 𝑥2 .

(3)

Substituting Eqs. (1)–(3) into Eq. (7.11), the 𝑥 and 𝑦 positions of the centroid are given by

𝑥̄ =

∫ 𝑥̃ 𝑑𝐿 ∫ 𝑑𝐿

√ ∫ 𝑥 1 + 4𝑥2 𝑑𝑥

0.5 m

=

0

∫

0.5 m √

(4)

, 1+

4𝑥2

𝑑𝑥

0

𝑦̄ =

∫ 𝑦̃ 𝑑𝐿 ∫ 𝑑𝐿

√ ∫ 𝑥2 1 + 4𝑥2 𝑑𝑥

0.5 m

=

0

∫

0.5 m √

.

(5)

1 + 4𝑥2 𝑑𝑥

0

Computation

The integrals in Eqs. (4) and (5) are perhaps tedious to evaluate. Using software to evaluate these integrals provides 0.1524 m2 = 0.2655 m, 0.5739 m 0.05252 m2 = 0.09151 m. 𝑦̄ = 0.5739 m

𝑥̄ =

(6) (7)

Discussion & Verification

• When using software, we must be especially cautious to check the accuracy of the answers it produces. In the intermediate results in Eqs. (6) and (7), the denominator (i.e., 0.5739 m) is√the total length of the line. Note that if the line were straight, its length would be (0.5)2 + (0.25)2 m = 0.5590 m, and as expected, the length of the curved line is slightly greater than this. Also, if the line were straight, by inspection its centroid would be at the 𝑥 and 𝑦 positions 0.25 m and 0.125 m, respectively. Thus, the 𝑥̄ and 𝑦̄ values in Eqs. (6) and (7) are reasonable. • As an exercise, you should repeat this example, using Eq. (7.14) on p. 434 to relate 𝑑𝐿 to 𝑑𝑦.

∗ Alternatively,

we could use Eq. (7.14) to write an expression for 𝑑𝐿 in terms of 𝑑𝑦.

𝑦 0.25 m 𝑑𝐿 𝑦̃ 𝑥̃

𝑥 0.5 m

Figure 2 A line element is used to develop expressions for 𝑑𝐿 and 𝑥. ̃

444

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.6

Centroid of a Volume Using Integration

𝑦

Determine the position of the centroid for the solid hemisphere of radius 𝑟 shown. 𝑟

SOLUTION Road Map

𝑥 𝑧

Because the object is symmetric about both the 𝑥𝑦 and 𝑥𝑧 planes, the centroid lies on the 𝑥 axis, and we can conclude that 𝑦̄ = 𝑧̄ = 0. To determine the 𝑥 position of the centroid, we will evaluate the expression for 𝑥̄ in Eq. (7.16) on p. 435. Expressions for 𝑑𝑉 and 𝑥̃ are needed, and these can be developed using either a thin disk volume element or a thin shell volume element, and both of these approaches are shown. Solution 1 – Thin disk volume element Governing Equations & Computation Expressions for 𝑑𝑉 and 𝑥̃ can be written using

𝑥2 + 𝑦2 + 𝑧2 = 𝑟2

the thin disk volume element shown in Fig. 2 as follows:

Figure 1

𝑑𝑉 = 𝜋𝑦2 𝑑𝑥 = 𝜋(𝑟2 − 𝑥2 ) 𝑑𝑥,

𝑦

𝑦=

√ 𝑟2 − 𝑥2

In Eq. (1), 𝜋𝑦2 is the area of a disk where 𝑦 is the radius, and 𝑑𝑥 is the thickness. The entire disk element is at the same 𝑥 position, so that 𝑥̃ = 𝑥. Substituting Eqs. (1) and (2) into the first expression of Eq. (7.16) provides the 𝑥 position of the centroid as 𝑟

) ( 2 𝑥4 |𝑟 𝜋𝑟4 2 𝑥 − 𝜋 𝑟 | ∫ 𝑥̃ 𝑑𝑉 0 4 |0 = 4 = 3𝑟 . = 𝑟 𝑥̄ = = ( 2 3) 𝑟 8 ∫ 𝑑𝑉 𝑥 2𝜋𝑟3 | 𝜋 𝑟2 𝑥 − | ∫ 𝜋(𝑟2 − 𝑥2 ) 𝑑𝑥 | 0 3 3 0

𝑥̃

(3)

Solution 2 – Thin shell volume element Governing Equations & Computation Expressions for 𝑑𝑉 and 𝑥̃ can be written using

𝑑𝑥 Figure 2 A thin disk volume element is used to develop expressions for 𝑑𝑉 and 𝑥. ̃ The expression for the radius 𝑦 of the disk is obtained by rearranging 𝑥2 + 𝑦2 + 𝑧2 = 𝑟2 , with 𝑧 = 0. 𝑦 √ 𝑥 = 𝑟2 − 𝑦2 𝑑𝑦 𝑥

the thin shell volume element shown in Fig. 3 as follows: √ 𝑑𝑉 = 2𝜋𝑦𝑥 𝑑𝑦 = 2𝜋𝑦 𝑑𝑦 𝑟2 − 𝑦2 , √ 𝑥 𝑥̃ = = 12 𝑟2 − 𝑦2 . 2

Figure 3 A thin shell volume element is used to develop expressions for 𝑑𝑉 and 𝑥. ̃ The expression for the length 𝑥 of the cylinder is obtained by rearranging 𝑥2 + 𝑦2 + 𝑧2 = 𝑟2 , with 𝑧 = 0.

(4) (5)

In Eq. (4), 2𝜋𝑦 is the circumference of the shell where 𝑦 is the radius, 𝑥 is the length of the shell, and 𝑑𝑦 is the thickness. In Eq. (5), the 𝑥 position of the centroid of the shell element 𝑥̃ is given by 𝑥∕2. Substituting Eqs. (4) and (5) into the first expression of Eq. (7.16) provides the 𝑥 position of the centroid as

𝑥̄ =

∫ 𝑥̃ 𝑑𝑉 ∫ 𝑑𝑉

𝑧

ISTUDY

(2)

∫ 𝑥𝜋(𝑟2 − 𝑥2 ) 𝑑𝑥

𝑥

𝑧

(1)

𝑥̃ = 𝑥.

∫

𝑟

=

0

1 2

√ √ 𝑟2 − 𝑦2 2𝜋𝑦 𝑟2 − 𝑦2 𝑑𝑦 √ ∫ 2𝜋𝑦 𝑟2 − 𝑦2 𝑑𝑦 𝑟

=

3𝑟 . 8

(6)

0

Discussion & Verification

As expected, the solutions for 𝑥̄ obtained by using the two different volume elements agree. Also, the result for the denominator of Eq. (3), namely, 2𝜋𝑟3 ∕3, is the correct volume of a hemisphere (as reported in the Table of Properties of Solids at the end of this book).

ISTUDY

Section 7.1

445

Centroid

Problems General instructions. For shapes that have one or more axes or planes of symmetry, you may use inspection to determine some of the coordinates of the centroid. When composite shapes is used, the tables at the end of this book may be helpful. Problems 7.1 through 7.6 For the area shown, use composite shapes to determine the 𝑥 and 𝑦 positions of the centroid. 𝑦 0.5 in.

0.5 in.

𝑦 20 mm 20 mm

1 in.

0.25 in. 𝑥

1.5 in. 1.5 in.

1 in.

20 mm

0.75 in.

𝑦

100 mm

0.5 in.

1 in. 4 in. 10 mm

0.125 in.

0.125 in.

𝑥

Figure P7.1

𝑥

Figure P7.2

𝑦

Figure P7.3

𝑦

𝑦 20 mm

0.5 in. 4 in.

12 mm

30 mm

0.5 in.

60 mm

20 mm 𝑥

12 mm

18 mm

12 mm

30 mm 𝑥

𝑥 10 mm

4 in. Figure P7.4

100 mm Figure P7.5

Figure P7.6 𝑦

Problem 7.7 The solid shown consists of a circular cylinder and a hemisphere. Use composite shapes to determine the 𝑥, 𝑦, and 𝑧 locations of the centroid.

3 in.

5 in. 3 in.

𝑥

Problem 7.8 The rectangular block is a solid with rectangular cutouts. Use composite shapes to determine the 𝑥, 𝑦, and 𝑧 locations of the centroid.

𝑧 Figure P7.7

𝑧 2.5 in.

2 in. 1.5 in. 2 in. 𝑦 2 in. 1 in. 2 in.

1.5 in. 2 in. 𝑥 Figure P7.8

446

Chapter 7

Centroids and Distributed Force Systems

𝑦

Problem 7.9 𝑥2 + 𝑦2 + 𝑧2 = 𝑅2

The solid shown consists of a hemisphere with a conical cavity. Use composite shapes to determine the 𝑥, 𝑦, and 𝑧 locations of the centroid. Express your answers in terms of 𝑅.

𝑅 𝑅∕2 𝑥

Problems 7.10 through 7.13 𝑧

For the area shown, use integration to determine the 𝑥 and 𝑦 positions of the centroid. (a) Use a vertical area element.

Figure P7.9

ISTUDY

𝑦

(b) Use a horizontal area element.

𝑦=

1 in.

√ 𝑥 𝑦

𝑦=

√ 𝑥

𝑦

2 mm

𝑦

𝑦 = 𝑥2

𝑦 = 𝑥2 ∕8

𝑦=

√ 𝑥

𝑥

𝑥 1 in.

𝑥

4 mm

Figure P7.10

𝑦 = 2𝑥 − 𝑥2

1 mm

2m

𝑥 2 mm

4m

Figure P7.11

Figure P7.12

Figure P7.13

Problems 7.14 through 7.17 For the area shown, use integration to determine the 𝑥 and 𝑦 positions of the centroid. 𝑦 1.5 in.

𝑦

𝑦 = 1 + 𝑥∕2

𝑦

𝑦 = 8 − 5𝑥∕12

1 in. 𝑦

𝑦 = 𝑥 − 𝑥2 ∕40

10 in. 8 in.

10 mm

𝑥2 + 𝑦2 = 𝑟2

𝑦 = 𝑥2 𝑦 = 5𝑥∕6

𝑟 1 in.

𝑥

Figure P7.14

𝑥

12 in.

Figure P7.15

Figure P7.16

𝑥

𝑦 = −30 + 2𝑥 15 mm 20 mm

𝑥

Figure P7.17

Problems 7.18 through 7.20 Determine expressions for lines 𝑦1 and 𝑦2 , and then use integration to determine the 𝑥 and 𝑦 positions of the centroid. 𝑦

𝑦 14 cm 16 cm

𝑦

60 in. 𝑦2

𝑦2 5m

20 in.

𝑥

−20 in.

𝑦1 Figure P7.18

28 cm

1m

80 in.

0

𝑦1

𝑦1

𝑥 8m

Figure P7.19

1m

𝑦2 23 cm Figure P7.20

𝑥

ISTUDY

Section 7.1

Centroid

Problems 7.21 through 7.23 For the area in the figure cited below, use composite shapes to determine the 𝑥 and 𝑦 positions of the centroid. Problem 7.21

Figure P7.18.

Problem 7.22

Figure P7.19.

Problem 7.23

Figure P7.20.

𝑦 𝑦2 ℎ 𝑦1

Problem 7.24 For the triangle shown, having base 𝑏 and height ℎ, use integration to show that the 𝑦 position of the centroid is 𝑦̄ = ℎ∕3. Hint: Begin by writing an expression for the width of a horizontal area element as a function of 𝑦.

𝑥

𝑏 𝑐 Figure P7.24

Problem 7.25

𝑦

Use integration to determine the 𝑥 and 𝑦 positions of the centroid. Express your answers in terms of 𝑟 and 𝛼.

𝑟 𝛼 𝛼

Problem 7.26 Determine constants 𝑐1 and 𝑐2 so that the curves intersect at 𝑥 = 𝑎 and 𝑦 = 𝑏. Use integration to determine the 𝑥 and 𝑦 positions of the centroid. Express your answers in terms of 𝑎 and 𝑏.

𝑥

Figure P7.25

Problems 7.27 and 7.28 For the straight line shown, set up the integrals, including the limits of integration, that will yield the length of the line, and the 𝑥 and 𝑦 positions of the centroid. Evaluate these integrals. 𝑦

𝑦 𝑏

𝑦

8 in.

𝑦 = 𝑐2 𝑥3

6 cm

𝑦 = 8 − 𝑥∕2

𝑎

3 cm

𝑦 = 2 + 𝑥∕3

𝑥

𝑥 3 cm

16 in. Figure P7.27

12 cm

Figure P7.28

Problems 7.29 and 7.30 For the line shown: (a) Set up the integrals for integration with respect to 𝑥, including the limits of integration, that will yield the 𝑥 and 𝑦 positions of the centroid. (b) Repeat Part (a) for integrations with respect to 𝑦. (c) Evaluate the integrals in Parts (a) and/or (b) by using computer software such as Mathematica or Maple. 𝑦 𝑦

1 in.

2m

𝑦 = 𝑥3 √ 𝑦= 𝑥 𝑥 4m Figure P7.29

√ 𝑦 = 𝑐1 𝑥

1 in. Figure P7.30

𝑥

Figure P7.26

𝑥

447

448

Chapter 7

Centroids and Distributed Force Systems Problems 7.31 and 7.32 For the line shown:

(a) Set up the integrals for integration with respect to 𝑥, including the limits of integration, that will yield the 𝑥 and 𝑦 positions of the centroid. (b) Evaluate the integrals in Part (a) by using computer software such as Mathematica or Maple. 𝑦

𝑦 = 2𝑥 − 𝑥2

1 mm

𝑧 𝑥2 + 𝑦2 = 𝑟2 1 −

𝑧 ℎ

2 mm

2

𝑦

𝑥2 + 𝑦2 = 𝑟2

𝑟

𝑥

𝑥

Figure P7.31

Figure P7.32

Problem 7.33

ℎ

A solid cone is shown. Use integration to determine the position of the centroid. Express your answers in terms of 𝑟 and ℎ. 𝑦

Problem 7.34

𝑟

𝑥

The solid shown has a cylindrical hole with 1 in. radius. Use integration to determine the volume of the solid and the coordinates of the centroid.

Figure P7.33

𝑦

𝑦 4 in.

8 cm 𝑦 = 3 − 𝑥2 ∕8 6 cm

3 in. 1 in.

4 cm

2 cm

𝑥

𝑥

𝑧

𝑧 Figure P7.34

Figure P7.35

Problem 7.35 𝑦 2 in. 𝑦 = 1∕𝑥

Problem 7.36

1 in.

For the hemisphere with a conical cavity shown in Fig. P7.9 on p. 446, use integration to determine the 𝑥 location of the centroid. Express your answers in terms of 𝑅.

0.5 in.

𝑥 1 in.

𝑧 Figure P7.37

ISTUDY

The punch shown is used for cutting holes in fabric. It has the shape of a truncated cone with a conical hole. Use integration to determine the volume of the punch and the coordinates of the centroid.

Problem 7.37 A solid of revolution is produced by revolving the shaded area shown 360◦ around the 𝑦 axis. Use integration to determine the coordinates of the centroid.

ISTUDY

Section 7.2

7.2

Center of Mass and Center of Gravity

449

Center of Mass and Center of Gravity

Centroid, center of mass, and center of gravity are different, but the methods used to determine them are essentially the same, as discussed in this section. In fact, under certain circumstances, which occur often, all of these have the same values. To understand the differences between these and when their values differ or are the same, we expand on the definitions that were given in the previous section: Centroid. The centroid is defined to be the average position of a shape or a distribution of shapes. The shape or distribution of shapes can consist of lines (straight and/or curved), areas, or volumes, but not combinations of these. The centroid depends on geometry only and is independent of the material an object might be made of and the presence of gravity. Center of mass. The center of mass is defined to be the average position of a distribution of mass. The center of mass depends on the geometry (shape) of an object and the density of the material it is made of and is independent of the presence of gravity. Center of gravity. The center of gravity is defined to be the average position of a distribution of weight. The center of gravity depends on the geometry (shape) of an object, the density of the material, and the presence of gravity.

Center of mass Using the concepts of the previous section, the center of mass for an object, or a collection of objects, in three dimensions is given by 𝑛 ∑

𝑥̄ =

𝑥̃ 𝑖 𝑚𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝑚𝑖

∫ 𝑥̃ 𝑑𝑚 ∫ 𝑑𝑚

𝑛 ∑

𝑦̄ =

,

𝑦̃𝑖 𝑚𝑖

𝑖=1 𝑛 ∑

𝑖=1 𝑛 ∑

𝑧̄ =

𝑧̃ 𝑖 𝑚𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝑚𝑖

= 𝑚𝑖

∫ 𝑧̃ 𝑑𝑚 ∫ 𝑑𝑚

∫ 𝑦̃ 𝑑𝑚 ∫ 𝑑𝑚

, (7.20)

.

The above equations are also applicable to objects in two dimensions, in which case only the 𝑥̄ and 𝑦̄ expressions are needed, and for objects in one dimension in which case only the 𝑥̄ equation is needed. Furthermore, Eq. (7.20) is applicable to solids, surfaces, and wires, and combinations of these, where these various terms are defined as follows:

A solid is formed when mass is distributed throughout a volume. A surface is formed when mass is distributed over an area. Usually, the thickness of the surface is small compared to the other dimensions of the surface. Surfaces are often called plates or shells. A wire is formed when mass is distributed along a line (either straight or curved). Usually, the dimensions of the wire’s cross section are small compared to the length of the wire.

Helpful Information Centroid, center of mass and center of gravity. Because the centroid, center of mass, and center of gravity often have the same values, some people use these words synonymously. However, these are fundamentally different—even when their values are the same—and thus we will always use proper nomenclature.

450

Solid volume = 𝑉 density = 𝜌 (mass∕volume)

𝑥 𝑧 Surface area = 𝐴 density = 𝜌𝐴

𝑛 ∑

𝑥̄ =

𝑥̃ 𝑖 𝑚𝑖

𝑖=1 𝑛 ∑

𝑖=1

(mass∕area)

Figure 7.8 Example of an object that consists of a solid, a surface, and a wire, with some of the information needed to determine its center of mass.

Gravity gradient satellites. For some applications, such as satellite mechanics and orbital mechanics, a more accurate treatment of gravity, and hence a more appropriate definition of the center of gravity, is needed. Consider a satellite orbiting Earth. center of mass center of gravity

= 𝑚𝑖

𝑥̃ 1 𝜌𝑉 + 𝑥̃ 2 𝜌𝐴 𝐴 + 𝑥̃ 3 𝜌𝐿 𝐿 𝜌𝑉 + 𝜌𝐴 𝐴 + 𝜌𝐿 𝐿

(7.21)

,

with similar expressions for 𝑦̄ and 𝑧. ̄ In Eq. (7.21), 𝑥̃ 1 , 𝑥̃ 2 , and 𝑥̃ 3 are the 𝑥 positions of the centers of mass for the solid, surface, and wire, respectively, for the object in Fig. 7.8. Using integration, the 𝑥 position of the center of mass for the object in Fig. 7.8 is given by the first expression in Eq. (7.20) as 𝑥̄ =

∫ 𝑥̃ 𝑑𝑚 ∫ 𝑑𝑚

=

∫ 𝑥𝜌 ̃ 𝐿 𝑑𝐿 ̃ 𝑑𝑉 + ∫ 𝑥𝜌 ̃ 𝐴 𝑑𝐴 + ∫ 𝑥𝜌 ∫ 𝜌 𝑑𝑉 + ∫ 𝜌𝐴 𝑑𝐴 + ∫ 𝜌𝐿 𝑑𝐿

,

(7.22)

with similar expressions for 𝑦̄ and 𝑧. ̄ In Eqs. (7.21) and (7.22), the densities 𝜌, 𝜌𝐴 , and 𝜌𝐿 have different definitions, and these should not be confused with one another. 𝜌 is the density of a material, according to the usual definition, with dimensions of mass/volume. In contrast, 𝜌𝐴 is the density of a surface, with dimensions of mass/area. Similarly, 𝜌𝐿 is the density of a wire, with dimensions of mass/length. Summaries of these definitions and relationships between them are as follows:

Interesting Fact

Earth

For example, consider the object shown in Fig. 7.8, which consists of a solid, a surface, and a wire. Using composite shapes, the 𝑥 position of the center of mass is given by the first expression in Eq. (7.20) as

Wire length = 𝐿 density = 𝜌𝐿 (mass∕length)

𝑦

ISTUDY

Chapter 7

Centroids and Distributed Force Systems

𝜌 = density of a material (dimensions: mass/volume), 𝜌𝐴 = density of a surface (dimensions: mass/area)

As indicated by Eq. (1.10) on p. 16, portions of the satellite’s mass that are closer to Earth experience slightly higher forces of attraction (weight) than portions that are farther away. Hence, for the satellite shown here, the center of gravity is closer to Earth than the center of mass. In fact, assuming the satellite has an overall length of, say, 20 m and is in a low Earth orbit with an altitude of, say, 800 km, and depending on details of its mass distribution, the center of mass and center of gravity may differ by only about 10−5 m. While this difference may seem insignificant, it is often sufficient to stabilize the satellite’s orientation, and such satellites are called gravity gradient satellites.

(7.24) 𝜌𝐴 = 𝜌 𝑡,

𝜌𝐿 = density of a wire (dimensions: mass/length)

satellite

flight path

(7.23)

(7.25) 𝜌𝐿 = 𝜌𝐴,

where in Eq. (7.24), 𝑡 is the thickness of the surface, and in Eq. (7.25), 𝐴 is the crosssectional area of the wire.

Center of gravity As we have done throughout this book, and for the vast majority of applications in statics, we assume the special case of a uniform gravity field, where the acceleration due to gravity 𝑔, as given by Eq. (1.11) on p. 16, is assumed to be constant, and the direction of the attractive forces between Earth and every element of material in a body is assumed to be parallel. With this idealization, the magnitude of the weight 𝑑𝑤 of a volume 𝑑𝑉 of material is 𝑑𝑤 = 𝛾 𝑑𝑉 = 𝜌𝑔 𝑑𝑉 , where 𝛾 is the material’s specific weight (dimensions: weight/volume) and 𝜌 is the material’s density (dimensions: mass/volume). Furthermore, since the direction for the weight is the same for each volume element 𝑑𝑉 , it is customary to define the center of gravity to be the average position of a weight distribution in the same fashion as the center of mass is the average position of a mass distribution. Hence, for a uniform gravity field, formulas for the center of gravity are identical to Eq. (7.20), with appearances of 𝑚𝑖 replaced by the weight of an object 𝑤𝑖 and appearances of 𝑑𝑚 replaced by the weight element

ISTUDY

Section 7.2

Center of Mass and Center of Gravity

451

𝑑𝑤. Hence, 𝑛 ∑

𝑥̄ =

𝑥̃ 𝑖 𝑤𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝑤𝑖

∫ 𝑥̃ 𝑑𝑤 ∫ 𝑑𝑤

𝑛 ∑

𝑦̄ =

,

𝑦̃𝑖 𝑤𝑖

𝑖=1 𝑛 ∑

𝑖=1 𝑛 ∑

𝑧̄ =

𝑧̃ 𝑖 𝑤𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝑤𝑖

= 𝑤𝑖

∫ 𝑧̃ 𝑑𝑤 ∫ 𝑑𝑤

∫ 𝑦̃ 𝑑𝑤 ∫ 𝑑𝑤

, (7.26)

.

For applications to objects in two dimensions, only the 𝑥̄ and 𝑦̄ expressions are needed, and for objects in one dimension only the 𝑥̄ equation is needed. Furthermore, Eq. (7.26) is applicable to solids, surfaces, and wires, and combinations of these. Substituting 𝑤𝑖 = 𝑔𝑚𝑖 and 𝑑𝑤 = 𝑔 𝑑𝑚 into Eq. (7.26), and noting that 𝑔 is a constant which can be canceled from the numerator and denominator, we see that Eqs. (7.20) and (7.26) are identical, and thus the center of mass and the center of gravity are identical when the gravity field is uniform. In Fig. 7.9 we reconsider our earlier example of an object that consists of a solid, a surface, and a wire. Using composite shapes, the 𝑥 position of the center of gravity is given by the first expression in Eq. (7.26) as 𝑛 ∑

𝑥̄ =

𝑥̃ 𝑖 𝑤𝑖

𝑖=1 𝑛 ∑

𝑖=1

= 𝑤𝑖

𝑥̃ 1 𝛾𝑉 + 𝑥̃ 2 𝛾𝐴 𝐴 + 𝑥̃ 3 𝛾𝐿 𝐿 𝛾𝑉 + 𝛾𝐴 𝐴 + 𝛾𝐿 𝐿

(7.27)

,

with similar expressions for 𝑦̄ and 𝑧. ̄ In Eq. (7.27), 𝑥̃ 1 , 𝑥̃ 2 , and 𝑥̃ 3 are the 𝑥 positions of the centers of gravity for the solid, surface, and wire, respectively. Using integration, the 𝑥 position of the center of gravity for the object in Fig. 7.9 is given by the first expression in Eq. (7.26) as 𝑥̄ =

∫ 𝑥̃ 𝑑𝑤 ∫ 𝑑𝑤

=

∫ 𝑥𝛾 ̃ 𝑑𝑉 + ∫ 𝑥𝛾 ̃ 𝐴 𝑑𝐴 + ∫ 𝑥𝛾 ̃ 𝐿 𝑑𝐿 ∫ 𝛾 𝑑𝑉 + ∫ 𝛾𝐴 𝑑𝐴 + ∫ 𝛾𝐿 𝑑𝐿

,

(7.28)

with similar expressions for 𝑦̄ and 𝑧. ̄ In Eqs. (7.27) and (7.28), the specific weights 𝛾, 𝛾𝐴 , and 𝛾𝐿 have different definitions. 𝛾 is the specific weight (or unit weight) of a material, according to the usual definition, with dimensions of weight/volume. In contrast, 𝛾𝐴 is the specific weight of a surface, with dimensions of weight/area. Similarly, 𝛾𝐿 is the specific weight of a wire, with dimensions of weight/length. Summaries of these definitions and relationships between these are as follows: 𝛾 = specific weight of a material

(7.29)

(dimensions: weight/volume), 𝛾𝐴 = specific weight of a surface (dimensions: weight/area)

(7.30) 𝛾𝐴 = 𝛾 𝑡,

𝛾𝐿 = specific weight of a wire (dimensions: weight/length)

(7.31) 𝛾𝐿 = 𝛾𝐴,

where in Eq. (7.30), 𝑡 is the thickness of the surface, and in Eq. (7.31), 𝐴 is the crosssectional area of the wire.

Solid volume = 𝑉 specific weight = 𝛾 𝑦 𝑥 𝑧 Surface area = 𝐴 specific weight = 𝛾𝐴

(weight∕volume) Wire length = 𝐿 specific weight = 𝛾𝐿 (weight∕length)

(weight∕area)

Figure 7.9 Example of an object that consists of a solid, a surface, and a wire, with some of the information needed to determine its center of gravity.

452

ISTUDY

Centroids and Distributed Force Systems

Chapter 7

End of Section Summary The center of mass is defined to be the average position of a distribution of mass. The center of mass for an object can be determined using composite shapes and/or integration, and the object can be a solid, surface, or wire, or a combination of these. The center of mass is affected by an object’s shape and mass distribution. The center of gravity is defined to be the average position of a distribution of weight. For the vast majority of applications in statics, we assume the special case of a uniform gravity field, where the acceleration due to gravity 𝑔 is assumed to be constant, and the direction of the attractive forces between Earth and every element of material in a body is assumed to be parallel. With this idealization, the center of gravity can be defined in similar fashion as the center of mass, and indeed, for a uniform gravity field, the center of mass and center of gravity of an object are always the same. The final point to summarize is that if an object has an axis of symmetry, which means that both the object’s shape and its mass distribution are symmetric, then the center of mass lies on the axis of symmetry. Furthermore, if the object has two axes of symmetry, then the center of mass is located at the intersection of the axes of symmetry.

ISTUDY

Section 7.2

453

Center of Mass and Center of Gravity

E X A M P L E 7.7

Center of Mass Using Composite Shapes

A kite consists of wooden members, paper, and string. The wooden members 𝐴𝐵 and 𝐶𝐷 have density 𝜌w = 510 kg∕m3 and 5 mm by 5 mm square cross section. The paper 𝐴𝐷𝐵𝐶 has density 𝜌p = 0.039 kg∕m2 . Around the perimeter of the kite (between points 𝐴, 𝐷, 𝐵, 𝐶, and 𝐴) is a taut string with density 𝜌s = 2.6(10)−5 kg∕m. Determine the weight of the kite and the center of mass.

𝑦 𝐷 0.3 m 𝐴

SOLUTION

𝐵

𝑥

0.3 m

Road Map

The kite consists of an arrangement of objects having simple shapes. Thus, it will be easiest to determine the center of mass by using composite shapes with Eq. (7.21) on p. 450. The kite is symmetric about the 𝑥 axis, thus its center of mass must lie on the 𝑥 axis, and hence 𝑦̄ = 0.

Figure 1

Governing Equations & Computation In Fig. 2 the kite is subdivided into eight composite shapes, or objects. The mass of each composite shape and the position of its center of mass are collected in Table 1.

𝑚5

𝐶

0.2 m

0.5 m

𝐷

𝑚7

𝐷

Table 1. Masses and center of mass positions for the composite objects in Fig. 2. Object no.

𝑚𝑖 (kg) 510 mkg3 (0.005 m)(0.005 m)(0.7 m) = 0.008925

1 (wood 𝐴𝐵) 2 (wood 𝐶𝐷) 3 (paper 𝐴𝐷𝐶) 4 (paper 𝐵𝐶𝐷) 5 (string 𝐴𝐷)

𝐴

𝑥̃ 𝑖 (m)

510 mkg3 (0.005 m)(0.005 m)(0.6 m) 0.039 mkg2 21 (0.6 m)(0.2 m) 0.039 mkg2 21 (0.6 m)(0.5 m) √ (0.2 m)2 + (0.3 m)2 = 2.6(10)−5 kg m

= 0.007650

2.6(10)−5 kg m

𝑚2

𝑚6

0

𝐶

𝑚8

𝐶

= 0.002340 −0.06667 = 0.005850

𝑚1

𝐴

0.1667

𝐵

Figure 2 A selection of composite objects to describe the kite.

9.374(10)−6 −0.1

(. . . ''. . . ) = 9.374(10) −0.1 √ (0.5 m)2 + (0.3 m)2 = 1.516(10)−5 0.25 (. . . ''. . . ) = 1.516(10)−5

8 (string 𝐵𝐶)

𝐵

0.15

−6

6 (string 𝐴𝐶) 7 (string 𝐵𝐷)

𝑚4

𝑚3

0.25

Using Eq. (7.21) provides the 𝑥 position of the center of mass as 𝑛 ∑

𝑥̄ =

𝑖=1 𝑛

𝑥̃ 𝑖 𝑚𝑖

∑

𝑖=1

= 𝑚𝑖

(0.15 m)(0.008925 kg) + ⋯ + (0.25 m)(1.516(10)−5 kg) 0.008925 kg + ⋯ + 1.516(10)−5 kg

(1)

𝑦 𝐷

0.002163 kg⋅m = 0.08719 m. = 0.02481 kg

(2)

For brevity, numbers for only the first and eighth objects are shown in Eq. (1), and you should verify the results that are given. The location of the center of mass is shown in Fig. 3. In Eq. (2), the denominator is the total mass of the kite, 𝑚 = 0.02481 kg. Hence, the weight of the kite is

𝑥̄ = 0.08719 m

0.3 m 𝐴

𝐵 center of mass

0.3 m 0.2 m

2

𝑤 = 𝑚𝑔 = (0.02481 kg)(9.81 m∕s ) = 0.2434 N. Discussion & Verification

(3)

We do not have a definitive check of our solution’s accuracy. Nonetheless, the center of mass has a reasonable location, and the weight of the kite is reasonable (it may be helpful to express the weight in U.S. Customary units—you should find that the kite weighs slightly less than 1 ounce).

𝐶

0.5 m

Figure 3 Location of the kite’s center of mass.

𝑥

454

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.8

Center of Gravity Using Composite Shapes A sign for a restaurant is constructed of rectangular sheets of plywood 𝐴𝐵𝐶𝐷 and 𝐹 𝐺𝐻𝐼, steel bar 𝐶𝐸𝐷, and chains 𝐷𝐹 and 𝐸𝐺. The plywood weighs 2 lb∕f t 2 , the steel bar weighs 5 lb∕f t, and the weight of the chains is negligible.

𝑦 𝐵

6 ft

3 ft

𝐶

(a) Determine the center of gravity. 3 ft 𝐸

𝐷 𝐴

0.5 ft

2 ft

𝑥

𝐹

𝐺

𝐼

𝐻

SOLUTION Road Map

Because the sign consists of an arrangement of objects having simple shapes, it will be easiest to determine the center of gravity by using composite shapes with Eq. (7.27) on p. 451. Once the center of gravity is determined, it will be straightforward to determine the support reactions.

Figure 1

𝐶 𝐵

𝑤1

Part (a)

𝑤3

𝐶

Governing Equations & Computation

The sign is subdivided into four composite shapes, or objects, as shown in Fig. 2. The weight of each composite shape and the position of its center of gravity are collected in Table 1. For object 3 in Table 1 (the quarter circular

𝐸 𝐷 𝐴

(b) Determine the support reactions at 𝐴 and 𝐵.

𝐸 𝑤4

𝐷 𝐹 𝐼

𝐺 𝑤2

Table 1. Weights and center of gravity positions for composite objects.

𝐻

Object no.

Figure 2 A selection of composite objects to describe the sign.

ISTUDY

𝑤𝑖 (lb)

1 (wood 𝐴𝐵𝐶𝐷)

2 ftlb2 (3 f t)(6 f t) = 36

2 (wood 𝐹 𝐺𝐻𝐼)

2 ftlb2 (2 f t)(3 f t) = 12 lb 5 ft 𝜋(3 f t)∕2 = 23.56 5 lbft (3 f t) = 15

3 (steel 𝐶𝐸) 4 (steel 𝐷𝐸)

𝑥̃ 𝑖 (ft)

𝑦̃𝑖 (ft)

3

1.5

7.5

−1.5

7.910

1.910

7.5

0

pipe), 𝜋(3 f t)∕2 is the length of the pipe, and we consult the Table of Properties of Solids at the end of this book to determine 𝑥̃ = 6 f t + 2(3 f t)∕𝜋 = 7.910 f t and 𝑦̃ = 2(3 f t)∕𝜋 = 1.910 f t. Using Eq. (7.27) provides the 𝑥 and 𝑦 positions of the center of gravity as 𝑛 ∑

𝑥̄ =

𝑖=1 𝑛

𝑥̃ 𝑖 𝑤𝑖 (1)

∑ 𝑖=1

𝑤𝑖

(3 f t)(36 lb) + (7.5 f t)(12 lb) + (7.910 f t)(23.56 lb) + (7.5 f t)(15 lb) 36 lb + 12 lb + 23.56 lb + 15 lb 496.9 f t ⋅lb = = 5.740 f t, 86.56 lb =

𝑛 ∑

𝑦̄ =

𝑖=1 𝑛

(3)

𝑦̃𝑖 𝑤𝑖

∑ 𝑖=1

(4) 𝑤𝑖

(1.5 f t)(36 lb) + (−1.5 f t)(12 lb) + (1.910 f t)(23.56 lb) + (0 f t)(15 lb) 86.56 lb 81.00 f t ⋅lb = = 0.9357 f t. 86.56 lb

=

(2)

(5) (6)

ISTUDY

Section 7.2

Center of Mass and Center of Gravity

The denominators in Eqs. (1) through (6) are the total weight of the sign, hence 𝑤 = 86.56 lb.

455

𝑦 𝐵𝑦

(7)

𝐵

𝐵𝑥

Part (b) Modeling

The FBD for the sign is shown in Fig. 3, where the sign’s 86.56 lb weight is placed at the center of gravity.

5.740 ft

𝑤 = 86.56 lb 0.9357 ft

3 ft 𝐴𝑥

𝑥

𝐴

Governing Equations & Computation

With the FBD in Fig. 3, equilibrium equations are written and easily solved as follows: ∑ 𝑀𝐵 = 0 ∶ 𝐴𝑥 (3 f t) − (86.56 lb)(5.740 f t) = 0 (8) ⇒ 𝐴𝑥 = 165.6 lb, ∑

𝐹𝑥 = 0 ∶

𝐴𝑥 + 𝐵𝑥 = 0 ⇒ 𝐵𝑥 = −165.6 lb,

∑

𝐹𝑦 = 0 ∶

𝐵𝑦 − 86.56 lb = 0 ⇒ 𝐵𝑦 = 86.56 lb.

Figure 3 Free body diagram where the sign’s weight is placed at the center of gravity.

(9) (10) (11) (12) 𝑦

(13)

𝐵𝑦 𝐵𝑥

𝐵

Discussion & Verification

• The center of gravity for the sign, given by Eqs. (3) and (6), appears to be in a reasonable position. The reactions have proper directions and appear to have reasonable values. • If we required the answer to only Part (b) of this example, we could use the solution as carried out above (note that the FBD shown in Fig. 3 requires determination of the center of gravity); or alternatively, it may be quicker to use the FBD shown in Fig. 4, where the center of gravity of the sign is not needed. In Fig. 4, the weight of each component object is placed at its center of gravity.

36 lb

23.56 lb 15 lb

𝐴𝑥

𝑥

𝐴 12 lb

Figure 4 Free body diagram where the weight of each component object is placed at that object’s center of gravity.

456

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.9

Center of Gravity Using Integration A solid cone is constructed of a functionally graded material that is designed to be hard at the tip and softer farther away. As a consequence, the specific weight of the material varies linearly from 𝛾0 at the tip to 𝛾0 ∕2 at 𝑥 = 𝐿. Determine the location of the center of gravity and the weight of the cone.

𝑦 𝑅 𝑥

SOLUTION 𝐿

𝑧

Road Map

Because the cone has specific weight that changes with position, the center of gravity will most likely not coincide with the centroid. To determine the center of gravity, we have little choice but to use integration, with the integral expressions in Eq. (7.26) on p. 451, or equivalently, the volume integral expressions in Eq. (7.28). Noting that the specific weight changes with the 𝑥 coordinate only, the cone is symmetric about the 𝑥𝑦 and 𝑥𝑧 planes, and thus, its center of gravity must lie on the 𝑥 axis, and 𝑦̄ = 𝑧̄ = 0.

Figure 1 𝑦 𝑑𝑥 𝑟

𝑅

Governing Equations & Computation

𝑥

𝑧

Eq. (7.26) with 𝑑𝑤 = 𝛾 𝑑𝑉 gives

𝑥̃ 𝐿

𝑥̄ =

Figure 2 A thin disk volume element is used to develop expressions for 𝑑𝑉 and 𝑥. ̃

ISTUDY

Linear functions. You will often need to develop expressions for linear functions, such as for 𝑟 and 𝛾 in this problem. While expressions such as Eq. (3) can be written by inspection or by using a little trial and error, the following more rigorous procedure can always be used. Consider the expression for the specific weight 𝛾. The problem description states that 𝛾 varies linearly, hence, 𝛾 = 𝑎 + 𝑏𝑥,

(4)

where 𝑎 and 𝑏 are constants that are determined by noting that 𝛾 = 𝛾0 at 𝑥 = 0 and 𝛾 = 𝛾0 ∕2 at 𝑥 = 𝐿. Hence, 𝛾 = 𝑎 + 𝑏(0) = 𝛾0 ,

(5)

𝛾 = 𝑎 + 𝑏(𝐿) = 𝛾0 ∕2.

(6)

and 𝑏 = −

𝛾0 . 2𝐿

(7)

Substituting 𝑎 and 𝑏 into Eq. (4) results in Eq. (3). Note that regardless of the approach you use to obtain a linear expression, you can always check its validity.

∫ 𝛾 𝑑𝑉

(1)

.

and

𝑥̃ = 𝑥.

(2)

In Eq. (2), 𝜋𝑟2 is the area of a disk where, as shown in Fig. 2, 𝑟 is the radius, and 𝑑𝑥 is the thickness. The center of gravity of the disk element is located at 𝑥̃ = 𝑥. Before we evaluate Eq. (1), it is necessary to express both the radius 𝑟 and specific weight 𝛾 of the cone as functions of 𝑥 position. These expressions are ( ) 𝑥 𝑥 𝑟=𝑅 . (3) and 𝛾 = 𝛾0 1 − 𝐿 2𝐿 The validity of Eqs. (3) is easily verified by evaluating each expression at 𝑥 = 0 and 𝑥 = 𝐿 to see that they produce the proper results. The margin note on this page provides some tips for developing expressions such as these. Substituting Eqs. (2) and (3) into Eq. (1) provides the center of gravity 𝐿 ) ( ) ( 𝑥 2 𝑥 3𝜋𝐿2 𝑅2 ∫ 𝑥𝛾0 1 − 𝑑𝑥 𝜋 𝑅 𝛾0 2𝐿 𝐿 ∫ 𝑥𝛾 ̃ 𝑑𝑉 0 18𝐿 20 = . = = 𝑥̄ = 2 𝐿 ( ) ( ) 25 ∫ 𝛾 𝑑𝑉 5𝜋𝐿𝑅 𝑥 𝑥 2 𝛾 ∫ 𝛾0 1 − 𝑑𝑥 𝜋 𝑅 0 24 2𝐿 𝐿 0

(8)

The denominator in Eq. (8) is the total weight of the cone, hence, 𝑤=

Solving Eqs. (5) and (6) for 𝑎 and 𝑏 provides 𝑎 = 𝛾0

∫ 𝑥𝛾 ̃ 𝑑𝑉

In Eq. (1), 𝑥̃ is the location of the center of gravity for the weight element 𝛾 𝑑𝑉 . Because of the variable specific weight, a thin disk weight (volume) element, as shown in Fig. 2, will be more convenient than a thin shell weight (volume) element (see Prob. 7.51 for further discussion). Expressions for 𝑑𝑉 and 𝑥̃ are 𝑑𝑉 = 𝜋𝑟2 𝑑𝑥

Helpful Information

Using the expression 𝑥̄ = ∫ 𝑥̃ 𝑑𝑤 in

Discussion & Verification

5𝜋𝐿𝑅2 𝛾0 . 24

(9)

The centroid of a solid cone, as given in the Table of Properties of Solids at the end of this book, is 3𝐿∕4 measured from the tip, while the center of gravity in this example, from Eq. (8), is 𝑥̄ = 18 𝐿∕25 = (0.72)𝐿. As expected, because of the specific weight distribution in this problem, we see that the center of gravity is closer to the tip than is the centroid. Of course, if the specific weight were uniform, the center of mass and the centroid would have the same location.

ISTUDY

Section 7.2

457

Center of Mass and Center of Gravity

Problems General instructions. For shapes that have one or more axes or planes of symmetry, you may use inspection to determine some of the coordinates of the center of mass or center of gravity. When composite shapes is used, the tables at the end of this book may be helpful. Problem 7.38

𝑑𝐴

Two planets 𝐴 and 𝐵 have circular orbits, and each rotates in the same plane about the center of mass 𝐺 for the two-planet system. The distance between the planets’ centers of mass (points 𝐴 and 𝐵) is always 4×105 km. Planet 𝐴 has spherical shape with radius 𝑟𝐴 = 6000 km and uniform density 𝜌𝐴 = 5000 kg∕m3 . Planet 𝐵 has spherical shape with radius 𝑟𝐵 = 1500 km and uniform density 𝜌𝐵 = 4000 kg∕m3 . Determine the distances 𝑑𝐴 and 𝑑𝐵 from each planet’s center of mass to the system’s center of mass 𝐺.

𝑑𝐵

𝐴 𝐺

𝐵

Figure P7.38

Problem 7.39 An instrument for measuring the direction of wind is shown. It consists of a hemispherical aluminum cap, a cylindrical plastic body, and a triangular sheet metal fin. The instrument is symmetric about the 𝑥𝑦 plane, and the aluminum, plastic, and sheet metal have specific weights of 𝛾a = 0.1 lb∕in.3 , 𝛾p = 0.04 lb∕in.3 , and 𝛾sm = 0.09 lb∕in.2 , respectively. Determine the 𝑥, 𝑦, and 𝑧 positions of the center of gravity.

sheet metal

dimensions in inches

𝑦 6 in.

9 in.

5 in.

50 30

20

2 in.

circuit box

40

𝐴

2 in. 𝑧

𝑧 𝑥

4 in. diameter plastic Figure P7.39

aluminum

aluminum 15 15 15 15

60

2 in.

15 15 30 15 2 in.

Figure P7.40

Problem 7.40 The radio antenna shown is symmetric about its horizontal member. The horizontal member weighs 0.6 lb∕f t, the vertical members weigh 0.1 lb∕f t, and the circuit box weighs 8 lb with center of gravity at point 𝐴. Determine the location of the antenna’s center of gravity, measured from the left-hand end.

Problem 7.41 In the cylinder and cone from Example 7.2 on p. 439 (shown again here), let the cylindrical portion be steel with specific weight 490 lb∕f t 3 and the truncated cone portion be aluminum alloy with specific weight 170 lb∕f t 3 . Determine the coordinates of the center of gravity.

3 in. steel 𝑥

2 in.

Figure P7.41 𝑦 3 in.

5 in. 3 in.

Problem 7.42 In the cylinder and hemisphere from Prob. 7.7 on p. 445 (shown again here), let the cylindrical portion be cast iron with specific weight 450 lb∕f t 3 and the hemispherical portion be aluminum alloy with specific weight 170 lb∕f t 3 . Determine the coordinates of the center of gravity.

𝑦

cast iron

𝑧 Figure P7.42

𝑥 aluminum

458

Chapter 7

Centroids and Distributed Force Systems

Problem 7.43 A square plate having 1.25 lb∕in.2 weight has a circular hole. Along the right-hand edge of the plate, a circular cross section rod having 0.75 lb∕in. weight is welded to it. Determine the 𝑥 position of the center of gravity. 𝑦 1.1 in. 0.25 in. 𝑧

0.5 in. 𝑥

6 cm

6 cm

wire 12 cm 9 cm

solid

0.5 in.

0.5 in. 𝑧

10 cm

1 in. Figure P7.43 𝑦

plate

Problem 7.44

20 cm

An object consists of a rectangular solid with 𝜌s = 5 g∕cm3 , a rectangular thin plate with 𝜌p = 10 g∕cm2 , and a thin wire with 𝜌w = 60 g∕cm. The object is symmetric about the 𝑦𝑧 plane. Determine the 𝑧 position of the center of mass.

𝑥 Figure P7.44

Problem 7.45 The frame shown consists of a semicircular bar 𝐴𝐵𝐶 and straight bars 𝐴𝐷𝐶 and 𝐷𝐸, all with 𝛾b = 0.5 lb∕in. weight. A circular counterweight with 𝛾c = 0.1 lb∕in.2 is to be welded to the frame at point 𝐸. Determine the radius 𝑟 of the counterweight so that the center of gravity for the assembly is located at point 𝐷. 𝐴

40 mm

flywheel counterweight

𝐷 𝑟

𝑂

15 in.

𝐸

counterweight and bolt 𝑚𝐶 , 𝑚𝐵

𝐵

12 in.

𝐶 Figure P7.45

100 mm

45 mm 50 mm

Figure P7.46

𝑦 𝐵

Problem 7.46

15 in. 𝐶

60 in. 𝐴

10 in.

Problem 7.47 𝑥

𝑧 Figure P7.47

ISTUDY

The assembly shown consists of a flywheel, a counterweight, and a bolt that attaches the counterweight to the flywheel. The flywheel is uniform with mass per area of 2 g∕mm2 and has a 40 mm diameter hole as shown. Determine the mass 𝑚𝐶 of the counterweight so that the center of mass of the assembly is at point 𝑂. The counterweight is attached to the flywheel by a bolt with mass 𝑚𝐵 = 800 g that is screwed into a 10 mm diameter hole that passes completely through the flywheel.

A floor lamp consists of a half-circular base with weight per area of 0.06 lb∕in.2 , tubes 𝐴𝐵 and 𝐵𝐶, each having weight 0.05 lb∕in., and the lamp shade at 𝐶 with weight 2 lb. Tube 𝐵𝐶 is parallel to the 𝑥 axis. Determine the coordinates of the center of gravity.

ISTUDY

Section 7.2

Center of Mass and Center of Gravity

Problem 7.48 The assembly shown is made of a semicircular plate having weight per area of 0.125 lb∕in.2 and uniform rods having weight per length of 0.05 lb∕in. Determine the coordinates of the center of gravity. 𝑧 𝑧 3 in. 6 mm 𝑦 5 in.

𝑦

8 mm 𝑥

𝑥 Figure P7.48

10 mm

6 mm Figure P7.49

Problem 7.49 The object shown consists of a rectangular solid, a plate, and a quarter-circular wire, with densities as follows: solid: 0.0005 g∕mm3 , plate: 0.01 g∕mm2 , wire: 0.05 g∕mm. Determine the coordinates of the center of mass.

𝐹 30◦ 𝐵

Problem 7.50

2m

A sign frame is constructed of metal pipe having a mass per length of 4 kg∕m. Portion 𝐶𝐷𝐸 is semicircular. The frame is supported by a pin at 𝐵 and a weightless cable 𝐶𝐹 . Determine the coordinates of the center of mass for the frame, the reactions at 𝐵, and the force in the cable.

Problem 7.51 In Example 7.9 on p. 456, the center of gravity of a cone with variable specific weight is found using integration with a thin disk mass element. Discuss why, for this example, a thin disk mass element is considerably more convenient than a thin shell mass element. Note: Concept problems are about explanations, not computations.

Problems 7.52 and 7.53 The object shown consists of a metal outside housing with density 𝜌0 and a hole filled with plastic having density 𝜌0 ∕2. Use integration to determine the mass of the solid and the 𝑥 position of the center of mass. 𝑦

𝑦

8 cm

4 in. 𝑦 = 3 − 𝑥2 ∕8 6 cm

3 in. 1 in.

𝑧

𝐶

4 cm

plastic, 𝜌0 ∕2 metal, 𝜌0 Figure P7.52

2 cm

𝑥 𝑧

𝑥

plastic, 𝜌0 ∕2 metal, 𝜌0 Figure P7.53

𝐷 𝐸

𝐴 2m Figure P7.50

459

460

Chapter 7

Centroids and Distributed Force Systems Problems 7.54 through 7.57 For the object indicated in the subproblem below:

(a) Fully set up the integral, including the limits of integration, that will yield the center of mass of the object.

𝑦 𝑟 2

𝑟 2

(b) Evaluate the integral determined in Part (a), using computer software such as Mathematica or Maple.

𝜌0 2

Problem 7.54 The solid hemisphere of radius 𝑟 shown has density 𝜌0 ∕2 for 0 ≤ 𝑥 ≤ 𝑟∕2 and 𝜌0 for 𝑟∕2 ≤ 𝑥 ≤ 𝑟.

𝜌0 𝑥

Problem 7.55 The hollow cone shown is constructed of a material with uniform density 𝜌0 and has wall thickness that varies linearly from 2 𝑡0 at 𝑥 = 0 to 𝑡0 at 𝑥 = 𝐿. Assume 𝑡0 is much smaller than 𝐿 and 𝑅.

𝑧 𝑥2 + 𝑦2 + 𝑧2 = 𝑟2

Problem 7.56

A solid of revolution is formed by revolving the area shown 360◦ about the 𝑥 axis. The material has uniform density 𝜌0 .

Figure P7.54

Problem 7.57

A solid of revolution is formed by revolving the area shown 360◦ about the 𝑦 axis. The material has uniform density 𝜌0 . 𝑦

𝑦 𝑅

𝑦=ℎ 1−

2 𝑡0 𝑥

𝑧

𝐿

𝑡0

Figure P7.55

𝑦

𝑥 𝐿

𝑦=𝑅

𝑥2 𝑎2

ℎ 𝑎

𝑥

Figure P7.56 and P7.57

Problems 7.58 through 7.60

1∕3

For the object indicated in the subproblem below:

𝑅 𝑥 𝑅 2

𝐿

(a) Fully set up the integral, including the limits of integration, that will yield the center of gravity of the object. (b) Evaluate the integral determined in Part (a), using computer software such as Mathematica or Maple.

𝑧 Problem 7.58 Figure P7.58

The solid shown has a cone-shaped cavity. The material has uniform

specific weight 𝛾0 . 𝑦

𝑦=

1 + 𝑥2

Problem 7.59 A solid of revolution is formed by revolving the area shown 360◦ about the 𝑥 axis. The material has uniform specific weight 𝛾0 .

2 cm Problem 7.60 A solid of revolution is formed by revolving the area shown 360◦ about the 𝑦 axis. The material has uniform specific weight 𝛾0 . 1 cm 𝑦=𝑥 𝑥

0 0 Figure P7.59 and P7.60

ISTUDY

1 cm

ISTUDY

Section 7.3

7.3

461

Theorems of Pappus and Guldinus

Theorems of Pappus and Guldinus

The two theorems discussed here are jointly attributed to Pappus of Alexandria, a mathematician who was active at the end of the 3rd century A.D. (the years of his birth, his death, and publication of his works are uncertain), and Paul Guldin, or Guldinus (original name Habakkuk Guldin) (1577–1643), who was a mathematician and astronomer. These theorems are useful and straightforward to apply for determining the area of a surface of revolution and the volume of a solid of revolution. We first present these theorems and then prove their validity.

CL

CL

generating curve

𝜃

Area of a surface of revolution

𝑟̄

A surface of revolution is produced by rotating a generating curve, as shown in Fig. 7.10, by an angle 𝜃 (in radians) about an axis of revolution. The axis of revolution and the generating curve lie in the same plane, and the generating curve must not intersect the axis of revolution, although portions of the generating curve may lie on the axis of revolution. Based on physical considerations, the angle 𝜃 through which the generating curve is rotated must be positive. If 𝜃 = 2𝜋, a full surface of revolution is produced, whereas if 0 < 𝜃 < 2𝜋, then a partial surface of revolution is produced. Values of 𝜃 > 2𝜋 are normally not possible, but may be considered for unusual applications (e.g., a very thin surface wrapped upon itself multiple times). If the generating curve has arc length 𝐿 and the position of its centroid relative to the axis of revolution is 𝑟̄, as shown in Fig. 7.10, then the area of the surface of revolution is 𝐴 = 𝜃 𝑟̄𝐿.

(7.32)

𝐶

axis of revolution

𝐴=𝜃

𝑛 ∑

𝑟̃𝑖 𝐿𝑖 .

(7.33)

𝑖=1

For the example in Fig. 7.11, Eq. (7.33) becomes

𝐴=𝜃

2 ∑ 𝑖=1

𝑟̃𝑖 𝐿𝑖 = 𝜃 (̃𝑟1 𝐿1 + 𝑟̃2 𝐿2 ).

(7.34)

area of surface = 𝐴

Figure 7.10 A surface of revolution is produced by rotating a generating curve by an angle 𝜃 about an axis of revolution. The generating curve has arc length 𝐿 and centroid 𝐶 located a distance 𝑟̄ from an axis of revolution.

CL

CL

𝜃

𝐶2

Equation (7.32) gives the area of only one side of the surface. The generating curve does not need to be smooth for Eq. (7.32) to apply. For example, the generating curve shown in Fig. 7.11 consists of two straight-line segments. Equation (7.32) may be applied directly if we first determine the centroid location for the entire generating curve. An approach that is usually simpler for generating curves that consist of simple composite shapes, such as in Fig. 7.11, is to recognize that 𝑟̄𝐿 in Eq. (7.32) is the first moment of the generating curve’s length with respect to the axis ∑ of revolution, in which case 𝑟̄𝐿 = 𝑛𝑖=1 𝑟̃𝑖 𝐿𝑖 , where 𝐿𝑖 is the arc length of composite shape 𝑖 (which is a line segment, either straight or curved, but usually with simple shape), 𝑟̃𝑖 is the position of the centroid of 𝐿𝑖 measured from the axis of revolution, and 𝑛 is the number of composite shapes that constitute the generating curve. Hence, Eq. (7.32) can be expressed as

𝐿

𝑟̃2 𝐶1 𝑟̃1

𝐿2 𝐿1

axis of generating revolution curve

area of surface = 𝐴

Figure 7.11 Example of a surface of revolution produced by a generating curve consisting of simple composite shapes. The two straight-line segments have lengths 𝐿1 and 𝐿2 and centroids located at 𝐶1 and 𝐶2 .

462

Chapter 7

Centroids and Distributed Force Systems

CL

Volume of a solid of revolution

CL

generating area 𝐴

𝜃 𝑟̄ 𝐶

axis of revolution

volume of solid = 𝑉

Figure 7.12 A solid of revolution is produced by rotating a generating area by an angle 𝜃 about an axis of revolution. The generating area 𝐴 has its centroid 𝐶 located a distance 𝑟̄ from the axis of revolution.

CL

CL

𝑟̃2 𝑟̃1

𝐴2 𝐶 2

𝜃

𝐴1 𝐶1

A solid of revolution is produced by rotating a generating area, as shown in Fig. 7.12, by an angle 𝜃 (in radians) about an axis of revolution. The axis of revolution and the generating area lie in the same plane, and the generating area must not intersect the axis of revolution, although portions of the generating area may lie on the axis of revolution. Based on physical considerations, the angle 𝜃 through which the generating area is rotated must be positive. If 𝜃 = 2𝜋, a full solid of revolution is produced, whereas if 0 < 𝜃 < 2𝜋, then a partial solid of revolution is produced. Values of 𝜃 > 2𝜋 are not possible. If the centroid 𝐶 of the generating area 𝐴 has position 𝑟̄ from the axis of revolution, as shown in Fig. 7.12, then the volume of the solid of revolution is 𝑉 = 𝜃 𝑟̄𝐴.

For applications when the generating area consists of composite shapes, such as shown in Fig. 7.13, Eq. (7.35) may be applied directly if we first determine the centroid location for the entire generating area. However, it is usually easier to replace ∑ 𝑟̄𝐴 in Eq. (7.35) with 𝑟̄𝐴 = 𝑛𝑖=1 𝑟̃𝑖 𝐴𝑖 , where 𝐴𝑖 is the area of composite shape 𝑖, 𝑟̃𝑖 is the position of the centroid of 𝐴𝑖 measured from the axis of revolution, and 𝑛 is the number of composite shapes that constitute the generating area. Hence, Eq. (7.35) can be expressed as 𝑉 =𝜃

axis of revolution

volume of solid = 𝑉

generating area

Figure 7.13 Example of a solid of revolution produced by a generating area consisting of simple composite shapes. The triangle and rectangle have areas 𝐴1 and 𝐴2 and centroids located at 𝐶1 and 𝐶2 , respectively.

(7.35)

𝑛 ∑

𝑟̃𝑖 𝐴𝑖 .

(7.36)

𝑖=1

For the example in Fig. 7.13, Eq. (7.36) becomes

𝑉 =𝜃

2 ∑

𝑟̃𝑖 𝐴𝑖 = 𝜃 (̃𝑟1 𝐴1 + 𝑟̃2 𝐴2 ).

(7.37)

𝑖=1

Proof of the Pappus–Guldinus theorems

CL

generating area 𝐴 𝑟̃ 𝜃

The theorems of Pappus and Guldinus are easy to prove using calculus and the concepts of centroid defined in Section 7.1. In recognition of their contribution, we note that Pappus and Guldinus developed these theorems long before calculus was invented. Because of the similarity between the proofs for a solid of revolution and a surface of revolution, we discuss only the solid of revolution, using the generating area shown in Fig. 7.14. The area element 𝑑𝐴 is located a distance 𝑟̃ from the axis of revolution, and this area element gives rise to a volume 𝑑𝑉 = 𝜃 𝑟̃ 𝑑𝐴. The total volume of the solid of revolution is then

𝑑𝐴

𝑉 = axis of revolution Figure 7.14 The area element 𝑑𝐴 is located a distance 𝑟̃ from the axis of revolution, and it gives rise to a volume 𝑑𝑉 = 𝜃 𝑟̃ 𝑑𝐴.

ISTUDY

∫

𝑑𝑉 = 𝜃

∫

𝑟̃ 𝑑𝐴.

(7.38)

Using Eq. (7.9) from Section 7.1 (p. 433), with position measured from the axis of revolution, the centroid of the generating area is 𝑟̄ = ∫ 𝑟̃ 𝑑𝐴∕𝐴, and hence Eq. (7.38) becomes 𝑉 = 𝜃 𝑟̄𝐴, which is identical to Eq. (7.35).

(7.39)

ISTUDY

Section 7.3

Theorems of Pappus and Guldinus

End of Section Summary The theorems of Pappus and Guldinus provide a straightforward means to determine the surface area and volume of objects of revolution. A surface of revolution is produced by rotating a generating curve, as shown in Fig. 7.10, by an angle 𝜃 (in radians) about an axis of revolution. A solid of revolution is produced by rotating a generating area, as shown in Fig. 7.12, by an angle 𝜃 about an axis of revolution. A full surface or volume of revolution will have 𝜃 = 2𝜋, whereas a partial surface or volume of revolution will have 0 < 𝜃 < 2𝜋.

463

464

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.10 200 cm 35 cm

Volume and Surface Area of an Object of Revolution 30 cm 15 cm

The cross section of a solid-fuel rocket motor is shown, where the fuel is a solid of revolution. (a) Determine the volume of fuel. (b) Determine the total surface area of the fuel.

CL

5 cm solid rocket fuel motor casing

SOLUTION

Figure 1

Road Map The fuel for the rocket is a solid of revolution, as shown in Fig. 2. We will use the theorems of Pappus and Guldinus to determine the volume of the solid and the surface area of the solid.

CL

Part (a) Governing Equations & Computation

Figure 2 The fuel for the rocket is a solid of revolution. 𝐴1 CL

𝐴2 𝑟̃1

𝐴3 𝑟̃3

𝑟̃2

Figure 3 Three composite areas (shown with a slight horizontal separation for clarity) that describe the generating area for a solid of revolution.

In Fig. 3, three simple composite areas are selected to describe the generating area for a solid of revolution (i.e., the volume of the rocket’s fuel). Before evaluating Eq. (7.36) on p. 462, we collect in Table 1 the area of each composite shape and the distance from the axis of revolution to the centroid of each shape. Table 1. Areas and centroid positions relative to the axis of revolution for the composite shapes in Fig. 3. Shape no.

𝐴𝑖

𝑟̃𝑖

2

4 (35 cm) 3𝜋

2

𝜋(35 cm) ∕4 = 962.1 cm (230 cm)(30 cm) = 6900 cm2 − 12 (30 cm)(15 cm) = −225 cm2

1 2 3

= 14.85 cm 20 cm 20 cm + 23 (15 cm) = 30 cm

Since the fuel is a full solid of revolution, 𝜃 = 2𝜋, and Eq. (7.36) provides 𝑉 = 2𝜋

3 ∑

𝑟̃𝑖 𝐴𝑖

𝑖=1

[ ] = 2𝜋 (14.85 cm)(962.1 cm2 ) + (20 cm)(6900 cm2 ) + (30 cm)(−225 cm2 ) = 9.145×105 cm3 .

(1)

Converting Eq. (1) from cm3 to m3 provides

𝐿1 CL

𝐿6

𝐿2

𝑉 = (9.145×105 cm3 )

𝐿3

Figure 4 Six composite lines that describe the generating curve for a surface of revolution. The arc lengths of the lines are given by 𝐿1 through 𝐿6 , and the centroid positions of the lines are indicated by the solid dots (the centroid position of 𝐿6 is not shown).

ISTUDY

m 100 cm

)3

= 0.9145 m3 .

(2)

Part (b)

𝐿4 𝐿5

(

Governing Equations & Computation

In Fig. 4, six simple composite lines are identified to describe the generating curve for a surface of revolution (i.e., the surface area of the rocket’s fuel). Before evaluating Eq. (7.33) on p. 461, we collect in Table 2 the arc length of each composite line and the distance from the axis of revolution to the centroid of each line. Since the surface of the fuel is a full surface of revolution, 𝜃 = 2𝜋, and Eq. (7.33) provides

ISTUDY

Section 7.3

Theorems of Pappus and Guldinus

Table 2. Lengths and centroid positions relative to the axis of revolution for the composite shapes in Fig. 4. Shape no.

𝐿𝑖

𝑟̃𝑖

1 𝜋(35 cm)∕2 = 54.98 cm 2 √ 200 cm 3 (30 cm)2 + (15 cm)2 = 33.54 cm 4 15 cm 5 230 cm 6 5 cm

𝐴 = 2𝜋

6 ∑

2 (35 cm) 𝜋

= 22.28 cm 35 cm 20 cm + 12 (15 cm) = 27.50 cm 5 cm + 12 (15 cm) = 12.50 cm 5 cm 2.5 cm

𝑟̃𝑖 𝐿𝑖

𝑖=1

[ ] = 2𝜋 (22.28 cm)(54.98 cm) + ⋯ + (2.5 cm)(5 cm) = 65,960 cm2 .

(3)

For brevity, numbers for only the first and sixth line elements are shown in Eq. (3), and you should verify the results that are given. Converting Eq. (3) from cm2 to m2 provides 𝐴 = (65,960 cm2 )

(

m 100 cm

)2

= 6.596 m2 .

(4)

Discussion & Verification To help judge if our solutions are reasonable, we will compare the volume and surface area of the fuel, as given by Eqs. (1) and (3), with those for a circular cylinder with radius 𝑟 and length 𝐿. Using 𝑟 = 35 cm and 𝐿 = 265 cm, the volume of this cylinder is 𝑉 = 𝜋𝑟2 𝐿 = 1.02×106 cm3 , and the surface area including the ends is 𝐴 = 2𝜋𝑟𝐿 + 2𝜋𝑟2 = 66,000 cm2 . Observe that with the values of 𝑟 and 𝐿 cited, the volume of the cylinder is larger than the exact volume of the fuel, while it is unclear if the surface area of the cylinder should be larger or smaller than the exact surface area of the fuel. The volume estimate of 1.02×106 cm3 is in reasonable agreement with Eq. (1) and, as expected, is larger than Eq. (1). The surface area estimate of 66,000 cm2 is in good agreement with Eq. (3).

465

466

Chapter 7

Centroids and Distributed Force Systems

Problems General instructions. Use the theorems of Pappus and Guldinus to determine the volume and/or surface area for the following problems. Problem 7.61 0.5

The cross section of a rubber V belt is shown. If the belt has circular shape about the axis of revolution with an inside radius of 6 in., determine the volume of material in the belt and the surface area of the belt. 0.4

0.25

6

0.25

Problem 7.62

dimensions in inches CL Figure P7.61

A pharmaceutical company’s design for a medicine capsule consists of hemispherical ends and a cylindrical body. Determine the volume and outside surface area of the capsule.

Problem 7.63 A solid is produced by rotating a triangular area 360◦ about the vertical axis of revolution shown. Determine the volume and surface area of the solid. CL

3 mm

6 mm

3 mm

𝑦 9 mm 2ℎ

Figure P7.62

10 mm ℎ 5 mm

ℎ

8 mm

𝑥 CL

Figure P7.63 and P7.64

ℎ Figure P7.65 and P7.66

Problem 7.64 Repeat Prob. 7.63 if the solid is produced by rotating the triangular area about the horizontal axis shown.

Problem 7.65 A solid is generated by rotating the shaded area shown 360◦ about the 𝑦 axis. Determine the volume and surface area of this solid in terms of the dimension ℎ.

𝑧

Problem 7.66

2 in.

Repeat Prob. 7.65 if the solid is generated by rotating the shaded area about the 𝑥 axis. 2 in.

Problem 7.67 Determine the volume and surface area for the solid of revolution in Example 7.2 on p. 439, shown again here.

3 in. 𝑦 𝑥 Figure P7.67

ISTUDY

2 in.

ISTUDY

Section 7.3

467

Theorems of Pappus and Guldinus

CL

Problem 7.68 A metal Sierra cup is used by campers as a multipurpose utensil for drinking, holding food, and so on. It is an object of revolution that has the generating curve shown. If the cup is made of titanium sheet that weighs 0.004 lb∕in.2 , determine the volume of fluid the cup is capable of holding and the total weight of the cup.

2 in.

Problem 7.69

1 in.

1.25 in.

A funnel is to be made of thin sheet metal using the generating curve shown. If the funnel is to be plated with 0.005 mm thick chrome on all surfaces, determine the volume of chrome required.

Figure P7.68 CL

Problem 7.70

60 mm 50 mm

A pressure vessel is to be constructed as a solid of revolution using the generating area shown. 70 mm

(a) Determine the volume of material needed to construct the pressure vessel.

20 mm

(b) Determine the outside surface area. (c) Determine the inside surface area.

30 mm 10 mm

Problem 7.71 The penstock shown is retrofitted to an existing dam to deliver water to a turbine generator so that electricity may be produced. The penstock has circular cross section with 24 in. diameter throughout its length, including section 𝐵𝐶, which is a 90◦ elbow. Determine the total weight of portion 𝐴𝐵𝐶𝐷 of the penstock, including the water that fills it. The penstock is made of thin-walled steel with a weight of 0.07 lb∕in.2 , and the water weighs 0.036 lb∕in.3 . 𝐴

48 in.

Figure P7.69

CL

𝐵

20 mm

5 mm

24 in.

36 in.

55 mm

45 mm

8 mm

Figure P7.70 𝑦

12 in.

𝑥2 + 𝑦2 + 𝑧2 = 𝑅2

𝐶

20 in. 𝐷

𝑅

24 in. 𝑅∕2 𝑥

Figure P7.71 𝑧

Problem 7.72 Determine the volume and surface area for the hemispherical solid with a conical cavity from Prob. 7.9, shown again here.

Figure P7.72 𝑦

Problem 7.73 The area of Prob. 7.12, shown again here, is revolved 360◦ about the line 𝑥 = −1 m to create a solid of revolution. Determine the volume and surface area of the solid. Hint: This problem is straightforward if you first solve Prob. 7.12 on p. 446 and Prob. 7.29 on p. 447.

2m 𝑦=

√ 𝑥 𝑥 4m

Figure P7.73

468 𝑦

𝑤 𝑤

𝑧

(b)

(a)

Figure 7.15 (a) A line force 𝑤 is a distributed force that acts along a line; 𝑤 has dimensions of force/length. (b) An example of a line force applied to an I beam. 𝑦

𝑝

(a)

𝑝 (b)

Figure 7.16 (a) A surface force 𝑝 is a distributed force that acts over a surface; 𝑝 has dimensions of force/ area. (b) An example of a surface force is the air pressure applied to the bottom surface of an airplane wing. 𝑦

𝑏 𝑏

𝑥 (a)

(b)

Figure 7.17 (a) A volume force 𝑏 is a distributed force that acts throughout a volume; 𝑏 has dimensions of force/volume. (b) An example of a volume force is the weight of material in a dumbbell used for weightlifting exercises.

ISTUDY

Distributed Forces, Fluid and Gas Pressure Loading

Distributed forces Distributed forces are forces that are distributed along a line, over a surface, or throughout a volume. Specific types of distributed forces are: Line force. A force distributed along a line is called a line force, or a line load. Examples of a line force 𝑤 are shown in Fig. 7.15, where 𝑤 has dimensions of force/length. Line forces are often used in connection with beam-type structures.

𝑥 𝑧

7.4

Throughout this book, we have idealized all forces to be point forces. A point force is a force that is concentrated at a single point. Point forces do not exist in nature. Rather, all forces are distributed as discussed below. In this section, we discuss procedures for how distributed forces can be idealized, or modeled, by point forces.

𝑥

𝑧

Chapter 7

Centroids and Distributed Force Systems

Surface force. A force distributed over a surface is called a surface force, or a traction. Examples of a surface force 𝑝 are shown in Fig. 7.16, where 𝑝 has dimensions of force/area. Surface forces result from fluid and gas pressure loadings, contact (with or without friction), and a variety of other situations. Volume force. A force distributed throughout a volume is called a volume force, or a body force. Examples of a volume force 𝑏 are shown in Fig. 7.17, where 𝑏 has dimensions of force/volume. Common examples of volume forces are weight due to gravity and attractive forces within an iron object due to a magnetic field. Line forces, surface forces, and volume forces are all vectors. Line forces and surface forces do not need to be perpendicular to the objects or structures to which they are applied, although often they will be. For example, consider standing on a gently sloped sidewalk on a cold winter day. Under one foot there is dry pavement, while under your other foot is ice. For the foot touching dry pavement, there is a surface force distribution that consists of a normal direction component and a tangential direction component. In fact, the tangential direction component keeps you from slipping! For the foot touching ice, the surface force distribution has only a normal direction component. In nature, all forces are either surface forces or volume forces. Line forces and point forces are useful idealizations for surface forces and volume forces. In this section, we use Chapter 4 concepts on equivalent force systems and the concepts of centroid discussed earlier in this chapter to replace distributed forces with point forces.

Helpful Information Units for surface forces. For convenience, special units for surface forces have been defined. In the U.S. Customary system, 1 psi = 1 lb∕in.2 , and in the SI system, 1 Pa = 1 N∕m2 . These units are read as follows: “psi” means pound per square inch, and “Pa” means pascal, which is a unit named in honor of the French scientist Blaise Pascal (1623–1662).

Distributed forces applied to beams Consider the example shown in Fig. 7.18(a), where a 30 m long cantilever beam is subjected to a distributed force. The distributed force 𝑤 varies linearly from 12 kN∕m at the left-hand end to 6 kN∕m at the right-hand end. Using the coordinate system shown in Fig. 7.18(a), the distributed force 𝑤 as a function of position 𝑥 is (for tips on developing expressions for linear functions, see the Helpful Information margin note on p. 456) (7.40) 𝑤 = 12 kN∕m − (0.2 kN∕m2 )𝑥.

ISTUDY

Section 7.4

Note that in writing Eq. (7.40), we consider the distributed force 𝑤 to be positive when acting in the −𝑦 direction, and we will follow this convention throughout this book. As shown in Fig. 7.18(b), over a small length of beam 𝑑𝑥, the distributed force 𝑤 produces a force 𝑑𝐹 = 𝑤 𝑑𝑥. Hence, the total force 𝐹 produced by the distributed loading is 𝐹 =

∫

𝑑𝐹 =

=

∫

[

𝑦

𝑦 12 kN∕m 6 kN∕m

𝐴

∫

𝑤 𝑑𝑥

𝑥

2

12 kN∕m − (0.2 kN∕m )𝑥 𝑑𝑥 = 270 kN.

∫ 𝑑𝐹

=

∫ 𝑤 𝑑𝑥

[ ] ∫ 𝑥 12 kN∕m − (0.2 kN∕m2 )𝑥 𝑑𝑥

(7.43)

30 m 0

270 kN

=

3600 kN⋅m = 13.33 m. 270 kN

Fluid and gas pressure Pressures from fluids and gases are surface forces and are always compressive. We consider an incompressible stationary (nonflowing) fluid, as shown in Fig. 7.20. The surface of the fluid lies in the 𝑥𝑦 plane and is subjected to a gas pressure 𝑝0 . We consider an infinitesimally small cube of fluid located at a depth 𝑑 below the surface. Regardless of the cube’s orientation, such as the two different orientations shown in Fig. 7.20, all six faces of the cube are subjected to the same compressive pressure 𝑝 as follows = 𝑝0 + 𝛾𝑑,

𝑁𝐴

𝑥̄ = 13.33 m (c)

𝐹 = 270 kN

𝑉𝐴 𝑥 13.33 m (d)

Figure 7.18 (a) A distributed force applied to a cantilever beam. (b) A distributed force 𝑤 acting over a small length 𝑑𝑥 produces a force 𝑑𝐹 = 𝑤 𝑑𝑥. (c) An equivalent force system consisting of a force 𝐹 at position 𝑥. ̄ (d) An FBD that can be used to determine the support reactions.

(7.44)

Using the results of Eqs. (7.42) and (7.44), the equivalent force system shown in Fig. 7.18(c) can be constructed. Then an FBD may be drawn as shown in Fig. 7.18(d), and reactions may be determined (you should verify for yourself that the reactions are 𝑁𝐴 = 0, 𝑉𝐴 = 270 kN, and 𝑀𝐴 = −3600 kN⋅m). The forgoing treatment demonstrates the use of integration to determine the force and centroid of a distributed load. For distributed loads that have simple shape, such as uniform and linear distributions, it may be more convenient to use composite shape concepts, as was done in earlier sections of this chapter. Returning to the cantilever beam shown in Fig. 7.18(a), either of the composite shapes shown in Fig. 7.19 may be used. In the case of Fig. 7.19(a), the “area” and centroid of the triangular distribution are 𝐹1 = (1∕2)(base)(height) = (1∕2)(30 m)(6 kN∕m) = 90 kN and 𝑥̃ 1 = 10 m. The “area” and centroid of the rectangular distribution are 𝐹2 = (base)(height) = (30 m)(6 kN∕m) = 180 kN and 𝑥̃ 2 = 15 m. Similar comments apply for determining the values of 𝐹1 , 𝑥̃ 1 , 𝐹2 , and 𝑥̃ 2 for Fig. 7.19(b). For Fig. 7.19(a) and (b), you should draw FBDs and verify that the reactions are the same as those reported earlier.

𝑝 = 𝑝0 + 𝜌𝑔𝑑

𝑦 𝑀𝐴

𝑥

𝑥̄ =

𝑑𝑥 (b)

(7.42)

Using Eq. (7.19) on p. 437, with 𝑑𝐴 replaced by the force 𝑑𝐹 , the centroid of the distributed load, and hence the position of the line of action of the force 𝐹 , is ∫ 𝑥̃ 𝑤 𝑑𝑥

𝑥

𝐹 = 270 kN

0

∫ 𝑥̃ 𝑑𝐹

𝑤

30 m (a) 𝑦

]

𝑑𝐹 = 𝑤 𝑑𝑥

𝐵

(7.41)

30 m

=

469

Distributed Forces, Fluid and Gas Pressure Loading

(7.45)

𝑦

𝑦 𝑥̃ 2

𝑥̃ 1 𝐹1

𝐴

𝑥̃ 1

𝐹1

𝐹2

𝑥

𝑥̃ 2

𝐵 30 m

𝐴

𝐹2

𝑥

𝐵 30 m

𝐹1 = 90 kN 𝑥̃ 1 = 10 m

𝐹1 = 180 kN 𝑥̃ 1 = 10 m

𝐹2 = 180 kN 𝑥̃ 2 = 15 m

𝐹2 = 90 kN 𝑥̃ 2 = 20 m

(a)

(b)

Figure 7.19 Two examples of using composite shapes to determine the forces due to the distributed load shown in Fig. 7.18(a). (a) Triangular and rectangular distributions. (b) Two triangular distributions.

470

Chapter 7

Centroids and Distributed Force Systems

surface of fluid

where

𝑝0

𝑦

𝑝0 is the pressure at the surface of the fluid; 𝑑 is the depth below the fluid’s surface; 𝑔 is acceleration due to gravity; 𝜌 is the density of the fluid; 𝛾 is the specific weight of the fluid (𝛾 = 𝜌𝑔); and 𝑝 is the pressure at a depth 𝑑 due to the fluid and gas, with dimensions of force/area.

𝑥

𝑑

𝑝 𝑝

𝑝 𝑝 𝑝

𝑝

𝑧

fluid

Remarks

Figure 7.20 A volume of incompressible stationary fluid with density 𝜌. The surface of the fluid lies in the 𝑥𝑦 plane and is subjected to a pressure 𝑝0 . Infinitesimally small cubes of fluid are subjected to hydrostatic compressive pressure 𝑝.

• The state of pressure due to gas and/or fluid loading is called hydrostatic because the pressure at a particular point has the same intensity in all directions. • Observe that Eq. (7.45) is a linear function of depth 𝑑. • Equation (7.45) gives the absolute pressure at a point, that is, the total pressure due to both gas and fluid loading. For many applications, only the portion of the pressure due to the fluid loading is needed, and this is often called the gage pressure, and Eq. (7.45) with 𝑝0 omitted becomes 𝑝 = 𝜌𝑔𝑑 = 𝛾𝑑.

(7.46)

• If there is no gas pressure acting on the surface of the fluid, then 𝑝0 = 0, and the absolute pressure and the gage pressure are identical. However, even when the gas pressure is significant, exact analyses can often be carried out using only the gage pressure, and this is the approach normally used. The margin note on p. 473 and Example 7.14 on p. 478 discuss this issue in greater detail.

𝐴

𝐵

𝑝0

𝑦

𝐴 𝑝0

𝑦

𝑝0

𝐵 𝑝0

𝑊

𝑥 𝑑

𝑑 𝐶

𝐷

𝑎

𝑝

𝑎

𝐷 𝑧 (a)

𝑝 𝐶

𝑝 𝑧 (b)

Figure 7.21 (a) A rectangular prism of fluid with height 𝑑, width 𝑎, and thickness 𝑎, whose surface is subjected to a pressure 𝑝0 . (b) A two-dimensional view of the FBD for the prism of fluid.

ISTUDY

• If there is no fluid pressure loading, then Eq. (7.45) shows that all points, regardless of position 𝑑, are subjected to the same gas pressure 𝑝0 . While this model is suitable for some applications, you should note that in contrast to fluids where the assumption of incompressibility is reasonable over a very large range of conditions, gases are highly compressible, and pressures are also strongly dependent on temperature. To understand more deeply why Eq. (7.45) describes gas and fluid pressure, consider the rectangular prism of fluid shown in Fig. 7.21(a). The FBD for this prism of fluid is three-dimensional, but only a two-dimensional view is needed, and this is shown in Fig. 7.21(b). Note that at point 𝐴, the pressure acting on the vertical surface is the same as the pressure acting on the horizontal surface. Identical remarks apply for points 𝐵, 𝐶, and 𝐷. Summing forces in the 𝑧 direction provides ∑ (7.47) 𝐹𝑧 = 0 ∶ 𝑝0 𝑎2 + 𝑊 − 𝑝𝑎2 = 0, where 𝑊 is the weight of the prism of fluid, which is equal to the product of the density of the fluid 𝜌, acceleration due to gravity 𝑔, and the volume of the prism 𝑎2 𝑑; thus 𝑊 = 𝜌𝑔𝑎2 𝑑. In writing Eq. (7.47), the force from the gas pressure on surface 𝐴𝐵 is given by the product of pressure at that location 𝑝0 and the area over which it acts, which is 𝑎2 . Similarly, the force on surface 𝐶𝐷 is the product of pressure at that location 𝑝 and the area over which it acts, which is also 𝑎2 . Substituting 𝑊 = 𝜌𝑔𝑎2 𝑑 into Eq. (7.47) and solving for 𝑝 yields Eq. (7.45).

ISTUDY

Section 7.4

Distributed Forces, Fluid and Gas Pressure Loading

471

Forces produced by fluids When a fluid makes contact with the surface of a body, a surface force that is compressive is exerted on the body, and this can be a significant source of loading. In the following discussion, we consider approaches for determining the forces that fluids apply to surfaces of bodies and structures. Fluid forces on flat surfaces In Fig. 7.22, we consider the situation of a rectangular plate submerged in an incompressible stationary fluid. surface of fluid, with 𝑝0 = 0 𝑦 𝑧

𝑝𝐴 𝑠

𝑝𝐴

𝑥 𝐴

𝑃 𝐿

𝑝𝐵 𝐵 (a)

surface of fluid

surface of fluid

𝑦

𝑃1 𝑃2 𝐿∕2 𝐿∕3

𝑧

𝑑𝐴

𝑧

𝑝𝐴 𝑝𝐴

𝐴

𝐿 𝑝𝐵

𝑦

𝐴 𝑊

𝑑𝐵

𝐴 𝑃

𝑃 𝐵

𝑝𝐵

𝐵

𝐵

𝑃 = 𝑃 1 + 𝑃2 (b)

(c)

Figure 7.22. (a) Fluid pressure acting on one surface of a flat rectangular plate and a force 𝑃 that is equivalent to the fluid pressure distribution. The other surfaces of the plate also have fluid pressures acting on them, but these pressures are not shown. (b) A view of the plate looking down the 𝑥 axis showing the pressure distribution and an approach for determining the force the fluid applies to the surface of the plate. (c) An alternative approach using an FBD of a volume of fluid for determining the force the fluid applies to the surface of the plate.

The upper and lower edges of the plate, denoted by 𝐴 and 𝐵, are parallel to the surface of the fluid and have depths 𝑑𝐴 and 𝑑𝐵 , respectively. We focus on the gage pressure, which is the pressure due to the fluid only. The pressure on the upper edge is given by Eq. (7.46) and has value 𝑝𝐴 = 𝛾𝑑𝐴 , and the pressure on the lower edge has value 𝑝𝐵 = 𝛾𝑑𝐵 . Between edges 𝐴 and 𝐵 the pressure distribution is linear, and, very importantly, the direction of the pressure is perpendicular to the surface of the plate over the entire surface of the plate. Two approaches for determining the force 𝑃 and/or 𝑃1 and 𝑃2 that the fluid applies to the surface of the plate are shown in Fig. 7.22(b) and (c). In Fig. 7.22(b), we use concepts of equivalent force systems from Chapter 4 and the concepts of centroid discussed earlier in this chapter to break the pressure distribution into two simple composite shapes. The first shape is a rectangular “volume” with dimensions 𝑝𝐴 by 𝐿 by 𝑠; the force corresponding to this is 𝑃1 = 𝑝𝐴 𝐿𝑠; and the position of the line of action of 𝑃1 is through the centroid of a rectangular volume, namely, 𝐿∕2 from edge 𝐵 as shown. The second shape is a triangular “volume” with height 𝑝𝐵 − 𝑝𝐴 , length 𝐿, and width 𝑠; the force corresponding to this is 𝑃2 = (1∕2)(𝑝𝐵 − 𝑝𝐴 )𝐿𝑠; and the position of the line of action of 𝑃2 is through the centroid of a triangular volume, namely, 𝐿∕3 from edge 𝐵 as shown. After 𝑃1 and 𝑃2 are known and their

Helpful Information Don’t confuse 𝒑 and 𝑷 . In Fig. 7.22, and elsewhere in this section, lowercase 𝑝 represents pressure (dimensions: force/area), and uppercase 𝑃 represents force. Thus, 𝑝 and 𝑃 are different—but they are related. Integrating pressure 𝑝 over the area it acts upon yields the force 𝑃 .

Concept Alert Fluid and gas pressure. The direction of the pressure that a fluid or gas applies to the surface of a body is always perpendicular to the surface over which the pressure acts.

472

ISTUDY

Centroids and Distributed Force Systems

Common Pitfall Errors in FBDs. A common error when drawing FBDs is to have fluid and gas pressures that are not perpendicular to the surfaces over which they act. For example, consider a gravity dam that retains water. correct

incorrect 𝑊

𝑊

𝐴 𝑅 (a)

𝐹 (b)

𝑝𝐴 𝑅

𝐹

𝑝𝐴

(c)

Both of the FBDs in (b) and (c) correctly have linear water pressure distributions, but the FBD in (b) is incorrect because the water pressure is not perpendicular to the surface over which it acts.

Chapter 7

lines of action have been located, we proceed to determine similar quantities for other surfaces of the body that may be in contact with fluid. We then draw an FBD where all forces acting on the body are included, followed by writing equilibrium equations and so on. In Fig. 7.22(c), we use a different approach wherein equilibrium concepts are used to determine the force the fluid applies to the surface of the structure. We begin by drawing an FBD of a triangular wedge of fluid; although the FBD is three-dimensional, a two-dimensional view as shown in Fig. 7.22(c) is adequate for our purposes. The wedge of fluid has weight 𝑊 , the upper edge of the fluid is subjected to a uniform pressure 𝑝𝐴 (= 𝛾𝑑𝐴 ), the vertical surface is subjected to a linear pressure distribution that changes from 𝑝𝐴 at the top to 𝑝𝐵 (= 𝛾𝑑𝐵 ) at the bottom, and rather than show the pressure distribution on surface 𝐴𝐵, we instead show the force 𝑃 that this pressure ∑ ∑ distribution produces. We then write equilibrium equations 𝐹𝑦 = 0 and 𝐹𝑧 = 0 ∑ to determine the 𝑦 and 𝑧 components of 𝑃 , and we write 𝑀 = 0 to locate the line of action of 𝑃 . Note that by determining the force 𝑃 that acts on edge 𝐴𝐵 of the fluid, we have, according to Newton’s third law of motion, determined the force that the fluid applies to the surface of the structure, also shown in Fig. 7.22(c).

Fluid forces on nonflat surfaces In Fig. 7.23, we consider the situation of a curved plate with uniform width 𝑠 submerged in an incompressible stationary fluid. The upper and lower edges of the plate are parallel to the surface of the fluid and have depths 𝑑𝐴 and 𝑑𝐵 , respectively. The pressures on the upper and lower edges have values 𝑝𝐴 and 𝑝𝐵 , respectively. Between edges 𝐴 and 𝐵 the value of the pressure has a linear distribution; for example, if edge 𝐵 is twice as deep as edge 𝐴, then 𝑝𝐵 is twice as large as 𝑝𝐴 . Despite the linearity in the value of the pressure, because the direction of the pressure is always perpendicular to the surface of the plate, the pressure distribution is complex. To determine the force 𝑃 that the fluid applies to the plate by using the approach shown in Fig. 7.23(b), it is necessary to develop expressions for the 𝑦 and 𝑧 components of the pressure 𝑝𝑦 and 𝑝𝑧 and to integrate these over the area of the plate. We obtain the force 𝑃𝑦 = ∫ 𝑝𝑦 𝑠 𝑑𝐿, where 𝐿 is the arc length of the curved plate and 𝑠 is its width, and similarly 𝑃𝑧 = ∫ 𝑝𝑧 𝑠 𝑑𝐿. Once 𝑃𝑦 and 𝑃𝑧 are known, the force the fluid applies √ to the plate is 𝑃 = 𝑃𝑦2 + 𝑃𝑧2 , and then we must determine the location of its line of action. Carrying out these calculations can be tedious. Fortunately, for many problems the alternative procedure shown in Fig. 7.23(c) will be much easier to use. If the shape of the volume of fluid in the FBD is sufficiently simple that its volume and centroid can be readily determined, then this approach is straightforward. Observe in Fig. 7.23(b) and (c) that while the fluid pressures are always perpendicular to the surfaces they act on, the line of action of force 𝑃 is not necessarily perpendicular to the surface.

Remarks We have presented two solution approaches for determining the force that a fluid applies to the surface of a structure. The examples of this section give further details on how these two solution approaches are carried out, and we compare them to help you decide which approach is more convenient for a particular problem. For surfaces with more complicated shape and/or orientation than those considered here, we may

ISTUDY

Section 7.4

473

Distributed Forces, Fluid and Gas Pressure Loading

surface of fluid, with 𝑝0 = 0 𝑦 𝑧

surface of fluid

𝑦 𝑝𝐴

surface of fluid

𝑦

𝑑𝐴 𝑠

𝑃

𝑝𝐴 𝐴

𝑥 𝐴

𝐿

𝑝𝐵

𝑝𝐴

𝐴

𝐴

𝑃 𝐿

𝑧

𝑝𝐴

𝑧

𝑊

𝑑𝐵

𝑝𝐵

𝑃

𝑃 𝐵

𝑝𝐵

𝐵

𝐵

𝐵 (a)

(b)

(c)

Figure 7.23. (a) Fluid pressure acting on one surface of a curved plate, and a force 𝑃 that is equivalent to the fluid pressure distribution. The other surfaces of the plate also have fluid pressures acting on them, but these pressures are not shown. (b) A view of the plate looking down the 𝑥 axis showing the pressure distribution. (c) An approach using an FBD of a volume of fluid for determining the force the fluid applies to the surface of the plate.

have no choice but to carry out the integrations previously described to determine the forces due to the fluid loading.

Forces produced by gases Loads applied to structures from a pressurized gas are often important, and the preceding discussion of fluid pressure loading is generally applicable with the following simplifications. In a pressurized gas, the density of the gas is typically negligible,∗ so all surfaces of a structure over which the gas makes contact are subjected to the same pressure 𝑝0 . As with fluids, gas pressure is compressive and always acts perpendicular to the surface the gas makes contact with. Thus, Figs. 7.22 and 7.23 are applicable with 𝑝𝐴 = 𝑝𝐵 = 𝑝0 , and in Figs. 7.22(c) and 7.23(c), the weight of the volume of gas in the FBDs is 𝑊 = 0. Example 7.14 provides further details on the treatment of forces due to a pressurized gas.

Helpful Information Can gas pressure be neglected in FBDs? Provided all surfaces of an object are subjected to the same gas pressure, then for most purposes we can omit this pressure in FBDs with no error or loss of accuracy. For example, consider a glass full of your favorite beverage resting on a table, as shown in (a) below. Figure (b) shows a two-dimensional view of the FBD including air pressure, and (c) shows the FBD omitting air pressure, where 𝑊 is the weight of the glass and contents and 𝑅 is the reaction between the glass and table. Although we do not prove this, when summing forces in the vertical direction, the downward component of all forces due to the air pressure in (b) is equilibrated by the upward component of all forces due to the air pressure. Hence, the same reaction 𝑅 is obtained for both FBDs, namely, 𝑅 = 𝑊. 𝑝0 𝑊

𝑊

𝑝0

𝑝0 𝑝0

(a)

∗ If

the density of a gas is large enough that it cannot be neglected, then the gas is essentially a fluid and is treated as such. Further, if the gas is stationary and can be assumed to have constant density, then the methods for treatment of fluids discussed in this section can be applied. However, because gases are highly compressible, the assumption of constant density is more restrictive for gases than for fluids.

𝑅 (b)

𝑅 (c)

474

ISTUDY

Centroids and Distributed Force Systems

Chapter 7

End of Section Summary In this section, distributed forces and their treatment were discussed. Distributed forces are forces that are distributed along a line, over a surface, or throughout a volume. A force distributed along a line is called a line force, or a line load, and this has dimensions of force/length. A force distributed over a surface is called a surface force, or a traction, and this has dimensions of force/area. A force distributed throughout a volume is called a volume force, or a body force, and this has dimensions of force/volume. Fluid pressure is a surface force. If the fluid is incompressible and at rest, then the pressure within the fluid is given by Eq. (7.45) as 𝑝 = 𝑝0 + 𝜌𝑔𝑑, where 𝑝0 is the gas pressure (constant) at the surface of the fluid. For many purposes, only the portion of the pressure due to fluid loading is important, and this pressure (𝑝 = 𝜌𝑔𝑑) is sometimes called the gage pressure. In either case, fluid pressure acts normal to the surface the fluid makes contact with. Gas pressure is a surface force. If the gas has negligible density and is at rest, then the pressure throughout the gas has constant value 𝑝0 , and all surfaces that make contact with the gas are subjected to this pressure, which acts normal to the surfaces. Also discussed in this section were methods for determining the forces that fluid and gas pressure distributions apply to structures.

ISTUDY

Section 7.4

E X A M P L E 7.11

475

Distributed Forces, Fluid and Gas Pressure Loading

Distributed Forces Applied to a Beam

A bookshelf holding 11 books is supported by brackets at points 𝐴 and 𝐵. The weights and thicknesses of the books are given in Table 1. Determine the support reactions.

8 9 10 11 1

2

3

4

SOLUTION

5

7

6

Road Map

We could determine the support reactions for the shelf by applying 11 individual loads according to the weights given in Table 1, but because of the large number of forces, it will be better to model the weight of the books by using distributed forces. Once this is accomplished, an FBD is drawn and equilibrium equations are written and solved.

𝐴 4 in.

Modeling

Books 1–4 have identical weight (4 lb each) and thickness (2 in. each), and thus over the portion of the shelf where these books rest, they apply a uniform distributed force of 4 lb∕2 in. = 2 lb∕in. Similarly, books 5–7 have identical weight and thickness, and thus over the portion of the shelf where these books rest, they apply a uniform distributed force of 3 lb∕3 in. = 1 lb∕in. Books 8–11 have the same thickness and show a linear variation in their weights, and thus the force they apply will be modeled using a linear distributed force that varies from 4 lb∕in. at book 8 to 1 lb∕in. at book 11. The model for books 8–11 is approximate, although reasonably accurate; Prob. 7.84 gives suggestions for developing a more accurate model. The distributed forces are shown in Fig. 2. The problem statement does not provide details of the supports at 𝐴 and 𝐵, and thus we will idealize the support reactions to consist of vertical forces only, and the FBD is shown in Fig. 2. At point 𝐴 and/or 𝐵, there is probably also a horizontal force reaction, but its presence is not important to this problem. Governing Equations & Computation

Shown in Figs. 2 and 3 are FBDs; the first of these shows the distributed loads, while the second has replaced these with equivalent concentrated loads as follows: 𝐹1 = (2 lb∕in.)(8 in.) = 16 lb,

𝐹2 = (1 lb∕in.)(9 in.) = 9 lb,

𝐹3 = (1 lb∕in.)(4 in.) = 4 lb,

𝐹4 = 12 (3 lb∕in.)(4 in.) = 6 lb,

⇒ 𝐵𝑦 = 17.12 lb, ∑

𝐹𝑦 = 0 ∶

𝐴𝑦 + 𝐵𝑦 − 𝐹1 − 𝐹2 − 𝐹3 − 𝐹4 = 0 ⇒ 𝐴𝑦 = 17.88 lb.

4 in.

13 in.

Figure 1 Table 1 Weight and thickness of each book. Books

Weight

Thickness

1–4 5–7 8 9 10 11

4 lb 3 lb 4 lb 3 lb 2 lb 1 lb

2 in. 3 in. 1 in. 1 in. 1 in. 1 in.

𝑦 𝑥 2 lb∕in.

4 lb∕in.

1 lb∕in.

1 lb∕in.

(1)

and the locations of these forces are shown in Fig. 3. Each of the forces given in Eq. (1) is determined by evaluating the “area” of the distributed load and placing this at the centroid of the distributed load. For example, the distributed load for books 1–4 is a rectangle with height 2 lb∕in. and base 8 in.; the “area” of this rectangle is 𝐹1 = (2 lb∕in.)(8 in.) = 16 lb, and the centroid is at the interface between books 2 and 3. The distributed load for books 8–11 is broken into two simpler distributions consisting of a rectangle and a triangle, as shown by the dashed lines in Fig. 3. The support reactions are found by writing and solving the following equilibrium equations: ∑ 𝑀𝐴 = 0 ∶ 𝐵𝑦 (13 in.) − 𝐹2 (8.5 in.) − 𝐹3 (15 in.) − 𝐹4 (14.33 in.) = 0 (2)

𝐵

𝐴𝑦 4 in.

4 in.

𝐵𝑦 9 in.

4 in.

Figure 2 Free body diagram showing distributed loads. 2 (4 in.) 3

1 (4 in.) 3

4 in.

4 in. 4.5 in. 4.5 in.

𝑦 𝑥

𝐹1

(3)

𝐹2

𝐹4

2 in.

𝐹3

(4)

Discussion & Verification To help verify that 𝐹1 through 𝐹4 are evaluated accurately, note that their sum is equal to the total weight of the books from Table 1, which is 35 lb. Also note that the sum of the reactions 𝐴𝑦 and 𝐵𝑦 is also 35 lb.

𝐵𝑦

𝐴𝑦

(5) 4 in.

4 in.

9 in.

4 in.

Figure 3 Free body diagram showing concentrated loads.

476

E X A M P L E 7.12 0.7 m

Fluid Pressure Loading and FBD Choices Freshwater in a channel is retained by a flat rectangular gate with 0.6 m width (into the plane of the figure) that is supported by a pin at 𝐵. The vertical wall 𝐵𝐷 is fixed in position. If the weight of the gate is negligible, determine the force 𝐹 required to begin opening the gate.

𝐷

𝐹 0.4 m

𝐵 𝐶

SOLUTION

1m 𝐴

Road Map When drawing an FBD, we can use either of the approaches shown in Fig. 7.22 on p. 471 for treating the fluid. That is, we may draw an FBD of the gate alone, including the pressures the water applies to it, as shown in Fig. 7.22(b). Or we may draw an FBD of the gate and a judiciously selected volume of water, as shown in Fig. 7.22(c). Both approaches are effective for this problem, although the first approach is slightly easier and will be used here.

1m Figure 1

0.7 m 𝐵𝑦 𝐹 𝐵𝑥 𝑝𝐵 1m

Modeling √

While the gate is a three-dimensional structure with length from 𝐴 to 𝐵 of 𝐿 = (1 m)2 + (1 m)2 = 1.414 m and a width of 0.6 m, an FBD in two dimensions is suitable, as shown in Fig. 2. When the gate begins to open, contact at 𝐴 is broken and 𝐴𝑦 = 0.

𝑦 𝑥

𝐿

1m

𝐴𝑦 = 0

In Fig. 3, the FBD of Fig. 2 is redrawn with the pressure distribution replaced by the forces the water applies to the structure. These forces are determined by breaking the pressure distribution into simpler composite shapes consisting of a rectangular “volume” with height 𝑝𝐵 , length 𝐿 = 1.414 m, and width 0.6 m, and a triangular “volume” with height 𝑝𝐴 − 𝑝𝐵 and the same length and width. Thus, the forces corresponding to these shapes are

𝐿∕2

0.7 m 𝐵𝑦 𝐹 𝐵𝑥

2𝐿∕3 𝑃1 𝑃2

𝑦 𝑥

𝐿

1m

𝐴𝑦 = 0

Figure 3 Free body diagram showing the forces the water applies to the gate.

ISTUDY

The density of freshwater is 𝜌 = 103 kg∕m3 , and from Eq. (7.46) on p. 470, the pressures at points 𝐴 and 𝐵 are )( ) ( kg m N kN 9.81 2 (1.4 m) = 13,730 2 = 13.73 2 , (1) 𝑝𝐴 = 𝜌 𝑔𝑑𝐴 = 103 3 m s m m ( )( ) kg m N kN 𝑝𝐵 = 𝜌 𝑔𝑑𝐵 = 103 3 9.81 2 (0.4 m) = 3924 2 = 3.924 2 . (2) m s m m Governing Equations & Computation

𝑝𝐴

Figure 2 Free body diagram showing the pressure distribution the water applies to the gate.

1m

Chapter 7

Centroids and Distributed Force Systems

𝑃1 = 𝑝𝐵 𝐿(0.6 m) = 3.330 kN,

𝑃2 = 21 (𝑝𝐴 − 𝑝𝐵 )𝐿(0.6 m) = 4.162 kN,

(3)

and the locations of the lines of action of these forces are shown in Fig. 3. To determine the force 𝐹 required to begin opening the gate, we sum moments about point 𝐵 as follows: ( ) ( ) ∑ 𝐿 2𝐿 𝑀𝐵 = 0 ∶ 𝐹 (0.7 m) − 𝑃1 − 𝑃2 =0 (4) 2 3 ⇒ 𝐹 = 8.969 kN. (5) If desired, the reactions 𝐵𝑥 and 𝐵𝑦 can be determined by writing the equilibrium equations ∑ ∑ 𝐹𝑥 = 0 and 𝐹𝑦 = 0. Discussion & Verification

The force 𝐹 determined above is large, but it is reasonable considering the depth of the water and the size of the gate. To contrast the solution approaches cited in the Road Map discussion, you should consider re-solving this problem using an FBD that includes a volume of fluid.

ISTUDY

Section 7.4

477

Distributed Forces, Fluid and Gas Pressure Loading

E X A M P L E 7.13

Fluid Pressure Loading and FBD Choices

Freshwater in a channel is retained by a cylindrical gate with 0.6 m width (into the plane of the figure) that is supported by a pin at 𝐵. The vertical wall 𝐵𝐷 is fixed in position. If the weight of the gate is negligible, determine the force 𝐹 required to begin opening the gate.

0.7 m

𝐷

𝐹 0.4 m

𝐵

SOLUTION

𝐶

1m

1m Road Map

This example differs from Example 7.12 only in the geometry of the gate. Fundamentally, we may develop an FBD using either of the approaches shown in Fig. 7.23(b) or (c) on p. 473 for treating the fluid. However, for this problem it will be considerably easier to use the latter approach where the FBD includes an appropriate volume of fluid.

𝐴 1m Figure 1

Modeling

An FBD of the gate and a cylindrical volume of fluid is shown in Fig. 2. When the gate begins to open, contact at 𝐴 is broken and 𝐴𝑦 = 0. The density of freshwater is 𝜌 = 103 kg∕m3 , and from Eq. (7.46) on p. 470, the pressures at points 𝐴 and 𝐵 are ( )( ) m N kN 3 kg 𝑝𝐴 = 𝜌 𝑔𝑑𝐴 = 10 3 9.81 2 (1.4 m) = 13,730 2 = 13.73 2 , (1) m s m m ( )( ) kg m N kN 𝑝𝐵 = 𝜌 𝑔𝑑𝐵 = 103 3 9.81 2 (0.4 m) = 3924 2 = 3.924 2 . (2) m s m m

0.7 m 𝐵𝑦 𝐹 𝑝𝐵 𝐵𝑥

Governing Equations & Computation

Note that these pressures are identical to those found in Example 7.12. In Fig. 3, the FBD of Fig. 2 is redrawn with the pressure distributions replaced by the forces 𝑃1 = 𝑃2 = 𝑝𝐵 (1 m)(0.6 m) = 2.354 kN, 𝑃3 =

1 (𝑝 2 𝐴

− 𝑝𝐵 )(1 m)(0.6 m) = 2.943 kN,

(3)

The line of action of 𝑊 is obtained by using the centroid position of a quarter-circular area from the Table of Properties of Lines and Areas at the end of this book and then evaluating 1 m − 4(1 m)∕3𝜋 = 0.5756 m. To determine the force 𝐹 required to begin opening the gate, we sum moments about point 𝐵 as follows: ∑ 𝑀𝐵 = 0 ∶ 𝐹 (0.7 m) − 𝑃1 (0.5 m) − 𝑃2 (0.5 m) − 𝑃3 (0.667 m) ⇒ 𝐹 = 9.968 kN.

𝑦

1m

𝑥

𝑝𝐴 1m

𝐴𝑦 = 0

Figure 2 Free body diagram showing the gate, a cylindrical volume of fluid, and the pressure distribution acting on the volume of fluid.

(4)

where locations of the lines of action of these forces are shown in Fig. 3. The weight 𝑊 of the cylindrical volume of water is the product of the water’s density 𝜌, acceleration due to gravity 𝑔, and the cylindrical volume 𝑉 : )( ) ( m 𝜋(1 m)2 3 kg 9.81 2 (0.6 m) = 4623 N = 4.623 kN. (5) 𝑊 = 𝜌𝑔𝑉 = 10 3 4 m s

−𝑊 (0.5756 m) = 0

𝑝𝐵

𝑊

0.7 m 𝐵𝑦 𝐹 𝐵𝑥

𝑃2

𝑊

𝑦

𝑃3 0.5756 m 1m

𝑥

𝐴𝑦 = 0

Figure 3 Free body diagram showing the forces produced by the pressure distributions. 𝐵𝑦 𝐹

Discussion & Verification The force 𝐹 determined here is slightly larger than the result for Example 7.12. The ease of carrying out the solution strategy in this example depends strongly on how easy it is to obtain the volume and centroid position of the fluid volume used in the FBD. To help you contrast this solution approach with that used in Example 7.12, the FBD of only the gate is shown in Fig. 4, where the complexity of the pressure distribution is apparent.

0.6667 m

1m

(6)

If desired, the reactions 𝐵𝑥 and 𝐵𝑦 can be determined by writing the equilibrium equations ∑ ∑ 𝐹𝑥 = 0 and 𝐹𝑦 = 0.

0.5 m 𝑃1

𝐵𝑥

𝑝𝐵 𝑝𝐴

𝑦 𝑥

𝐴𝑦 = 0 Figure 4 Free body diagram showing the pressure distribution the water applies to the gate.

478

Chapter 7

Centroids and Distributed Force Systems

E X A M P L E 7.14

Gas Pressure Loading A pressure cooker is a large pot used for cooking and canning∗ foods under high pressure and temperature. The pressure cooker shown is cylindrical, operates at an internal pressure of 30 psi (psi = lb∕in.2 ), and has a lid that is clamped to the base using six hand-tightened screws symmetrically located around the pot’s circumference. Determine the force each screw must support due to the pressure within the cooker.

SOLUTION Road Map

To determine the force supported by the screws, we can use an FBD of the lid alone, or we can use an FBD of the lid and a judiciously selected volume of air. Because the surface of the lid over which the air makes contact has complex shape, the latter approach will be considerably easier. The air will be treated as a gas, so it is assumed to have negligible weight.

16 in. Figure 1

Modeling

We will assume that all six screws support the same force 𝐹 . An FBD of the lid and a volume of air is shown in Fig. 2, where we have neglected the weight of the lid and air. To help provide clarification, the volume of air included in this FBD is shown in Fig. 3, and the uniform pressure distribution shown in the FBD acts on the flat bottom surface of this volume of air.

6𝐹 total 𝑦 air

𝑥

Governing Equations & Computation In Fig. 4, the FBD of Fig. 2 is redrawn with the pressure distribution replaced by the force

𝑃 = 𝑝0 𝜋(8 in.)2 = 6032 lb,

𝑝0 Figure 2 Free body diagram of the lid and a volume of air.

𝑦 16 in.

(1)

where the pressure is 𝑝0 = 30 lb∕in.2 , 𝜋(8 in.)2 is the area of the circular surface over which the pressure acts, and the line of action of 𝑃 passes through the centroid of the pressure distribution. To determine the force 𝐹 supported by each of the six screws, we use the FBD of Fig. 4 to sum forces in the 𝑦 direction to obtain ∑ 𝐹𝑦 = 0 ∶ −6𝐹 + 𝑃 = 0 ⇒ 𝐹 = 1005 lb. (2) Discussion & Verification

𝑧

𝑥

Figure 3 A sketch of the volume of air included in the FBD of Fig. 2. 6𝐹 total 𝑦 air

𝑥 𝑝0

𝑃

• The force 𝑃 determined in Eq. (1) and the bolt force 𝐹 determined in Eq. (2) are impressively large, and this underscores the need for safety when designing and operating a pressure cooker (and pressure vessels in general). • The screw force 𝐹 determined here is due to the pressurization only. In addition to this force, each of the screws supports a force due to the hand tightening, but this value is probably small compared to Eq. (2). Also, the assumption to neglect the weight of the lid is clearly acceptable in view of the results of Eq. (1). • The pressure 𝑝0 = 30 lb∕in.2 is the gage pressure. In the FBDs, atmospheric pressure (approximate value at sea level is 14.7 lb∕in.2 ) is not included, because it has no net effect on equilibrium. That is, if atmospheric pressure were included, it would completely surround the objects shown in the FBDs, and when we wrote equilib∑ rium equations such as 𝐹𝑦 = 0, its effects would fully cancel in these expressions.

Figure 4 Free body diagram showing the force produced by the pressure distribution.

ISTUDY

∗ Canning

is a process whereby food in sealed jars is placed in a pressure cooker with a small amount of water that is raised to high pressure and temperature to kill harmful bacteria. After heat treatment, the jars may be stored at room temperature without spoilage of the food.

ISTUDY

Section 7.4

479

Distributed Forces, Fluid and Gas Pressure Loading

Problems General instructions. Unless otherwise stated, in the following problems you may use integration, composite shapes, or a combination of these. The specific weight and density of water are 0.0361 lb∕in.3 and 103 kg∕m3 , respectively. Problem 7.74 A shelf in a grocery store supports 100 bags of rice, each bag weighing 1 lb. Consider the arrangements shown: (a) The bags are stacked at a uniform height, (b) the bags are stacked twice as high on the right-hand side as on the left-hand side, (c) the bags are stacked twice as high on the right-hand side as on the left-hand side with a linear variation, and (d) the bags are stacked twice as high in the middle as at the two ends with linear variations. For each of the arrangements, develop an expression (or multiple expressions if needed) for the distributed force 𝑤 as a function of position 𝑥. 𝑦

𝑦 𝑦

𝐴

𝐶

𝑥 27 in. (a)

𝐴

𝑥

𝐵 18 in. 9 in. 9 in. (b)

𝐵 9 in.

𝑦

𝑦

𝐶 𝐴

𝐵

𝑥

𝐴

4m (a)

𝑦

2m

𝐶

𝑥 27 in. (c)

𝐴

𝐵 9 in.

𝑥

𝐵 18 in. 9 in. 9 in. (d)

𝐶

𝐴

𝑦 𝐵

𝑥

Problem 7.75 A cantilever beam supports a wall built of 1000 bricks, each brick having 5 kg mass. Consider the arrangements shown: (a) The wall has uniform height, (b) the wall is twice as high on the left-hand side as on the right-hand side, (c) the wall is twice as high on the left-hand side as on the right-hand side with a linear variation, and (d) the wall is twice as high in the middle as at the two ends with linear variations. For each of the arrangements, develop an expression (or multiple expressions if needed) for the distributed force 𝑤 as a function of position 𝑥.

Problems 7.76 through 7.79 Determine the support reactions, using composite shapes to represent the distributed load. Loading (a), shown above, in Prob. 7.74.

Problem 7.77

Loading (d), shown above, in Prob. 7.74.

Problem 7.78

Loading (a), shown above, in Prob. 7.75.

Problem 7.79

Loading (d), shown above, in Prob. 7.75.

𝐴

4m (c)

Figure P7.74

Problem 7.76

2m (b)

𝑦 𝐴

𝐵

𝑥

𝐵

𝑥 2m

2m (d)

Figure P7.75

480

Chapter 7

Centroids and Distributed Force Systems Problems 7.80 through 7.83 For the loading indicated in the subproblem below:

(a) Use integration [Eqs. (7.41) and (7.43) on p. 469] with the appropriate expression(s) for the distributed load 𝑤 to determine the 𝑥 position of the line of action for the resultant force produced by the distributed load. (b) Determine the support reactions using the results of Part (a). (c) Determine the support reactions using composite shapes to represent the distributed load. Problem 7.80

Loading (b) in Prob. 7.74 on p. 479.

Problem 7.81

Loading (c) in Prob. 7.74 on p. 479.

Problem 7.82

Loading (b) in Prob. 7.75 on p. 479.

Problem 7.83

Loading (c) in Prob. 7.75 on p. 479.

Problem 7.84 𝑦

𝑦

Consider modeling the weights from books 8–11 from Example 7.11 on p. 475, using the linear force distribution 𝑤 = 𝑎 + 𝑏𝑥, as shown here. The constants 𝑎 and 𝑏 can be determined by requiring the two force systems shown to be equivalent. That is,

8 9 10 11

𝑤 = 𝑎 + 𝑏𝑥

4 in.

∫ 0

𝑥 𝐵

𝑥 𝐵

4 in.

𝑤 𝑑𝑥 =

11 ∑ 𝑖=8

4 in.

𝑊𝑖

and

∫ 0

𝑥𝑤 𝑑𝑥 =

11 ∑

𝑀𝐵𝑖 ,

𝑖=8

where 𝑀𝐵𝑖 is the moment of weight 𝑊𝑖 about point 𝐵 (the origin). (a) Using the weights and geometry of books 8–11 given in Table 1 of Example 7.11, lb evaluate the above expressions and solve for 𝑎 and 𝑏 to show that 𝑤 = 35 − 8 in. ( 15 lb ) 𝑥. 16 in.2

4 in.

Figure P7.84

(b) Evaluate the distributed force from Part (a) at 𝑥 = 0 and 𝑥 = 4 in., and compare these to the values used in Example 7.11, namely, 4 lb∕in. and 1 lb∕in., respectively. (c) Discuss why the distributed force from Part (a) is better than that used in Example 7.11. (d) Using the distributed force from Part (a), determine the support reactions for the bookshelf and compare to those found in Example 7.11.

Problem 7.85 𝑦

(a) For the distributed loading shown, develop an expression (or multiple expressions if needed) for the distributed force 𝑤 as a function of position 𝑥.

4 kip∕ft 𝐴

𝑥 6 ft

Figure P7.85

ISTUDY

𝐶

𝐵 6 ft

(b) Use integration [Eqs. (7.41) and (7.43) on p. 469] with the results of Part (a) to determine the total force produced by the distributed load and the 𝑥 position of its line of action. (c) Determine the support reactions using the results of Part (b). (d) Determine the support reactions using composite shapes for the distributed load.

ISTUDY

Section 7.4

481

Distributed Forces, Fluid and Gas Pressure Loading

Problem 7.86

𝑦

A beam is loaded by a distributed force that begins at the left-hand end as an 800 N∕m uniform load with 𝑑𝑤∕𝑑𝑥 = 0 and decreases to zero at the right-hand end. (a) Determine the constants 𝑎, 𝑏, and 𝑐 so that the quadratic polynomial 𝑤 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 describes this loading. (b) Determine the constants 𝑑 and 𝑓 so that the trigonometric function 𝑤 = 𝑑 cos(𝑓 𝑥) describes this loading.

800 N∕m 𝐴

𝐵

𝑥 10 m

Figure P7.86 and P7.87

Problem 7.87 (a) Use integration [Eqs. (7.41) and (7.43) on p. 469] with the results of Part (a) of Prob. 7.86 to determine the total force produced by the distributed load, the 𝑥 position of its line of action, and the support reactions. (b) Use integration [Eqs. (7.41) and (7.43) on p. 469] with the results of Part (b) of Prob. 7.86 to determine the total force produced by the distributed load, the 𝑥 position of its line of action, and the support reactions. (c) Compare the results of Parts (a) and (b), and discuss if these should be the same or if differences are expected.

Problems 7.88 through 7.93 Determine the support reactions for the loading shown.

4 kN∕m 𝐴

𝐶

𝐵 3m

2 kN∕m 𝐵

𝐴 2 kN∕m

3m

Figure P7.89

6 kN∕m 𝐴

4 kN∕m 3m

20 f t

6m

𝐵 2m

Figure P7.91

20 f t

Figure P7.90

400 lb∕ft

200 lb∕ft

3m

𝐴 6 kip∕ft

6m

Figure P7.88

6 kip∕ft 𝐵

50 kip⋅ft 𝐵

𝐴 1 kip 20 f t

20 f t

Figure P7.92

Problem 7.94 In Fig. P7.88, replace the pin and roller supports with a built-in support at 𝐶, and determine the support reactions.

Problem 7.95 In Fig. P7.89, replace the pin and roller supports with a built-in support at 𝐴, and determine the support reactions.

Problem 7.96 In Fig. P7.90, reposition the roller support to the midspan of the beam, and determine the support reactions.

15 kN

4 kN∕m 𝐴

𝐵 6m

6m

Figure P7.93

1 kN∕m

482

Centroids and Distributed Force Systems

Chapter 7

Problem 7.97 𝑤1

In Fig. P7.91 on p. 481, replace the pin and roller supports with a built-in support at 𝐴, and determine the support reactions.

𝑤2 𝐵

𝐴

Problem 7.98

𝑎 𝐿

Determine the support reactions for the cantilever beam. Express your answers in terms of parameters such as 𝑤1 , 𝑤2 , 𝑎, and 𝐿.

Figure P7.98

Problem 7.99

𝑤0 𝐴

Determine the support reactions for the simply supported beam. Express your answers in terms of parameters such as 𝑤0 , 𝑎, and 𝐿.

𝐵 𝑎 𝐿

Problem 7.100

Figure P7.99 𝑤𝐶 = 6 kN∕m

𝑦 𝑤𝐵 = 4 kN∕m

𝐶

𝑤𝐴 = 1 kN∕m

𝐵

𝐴

𝑥

3m

A blade of the main rotor of a hovering helicopter is subjected to the 𝑦 direction distributed forces shown, where values are known at points 𝐴, 𝐵, and 𝐶. Determine the constants 𝑎, 𝑏, and 𝑐 so that the quadratic polynomial 𝑤 = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 describes this loading. Using this polynomial, determine the total force produced by this distribution and the 𝑥 position of its line of action.

Problem 7.101

3m 0.4 m 𝑧

The distributed forces on the main rotor of the helicopter described in Prob. 7.100 may be approximated by using a linear distribution where values of the distributed force are known at points 𝐴 and 𝐶. Determine the constants 𝑎 and 𝑏 so that the linear polynomial 𝑤 = 𝑎 + 𝑏𝑥 describes this loading. Using integration with this polynomial, determine the total force produced by this distribution and the 𝑥 position of its line of action.

Figure P7.100 and P7.101

Problem 7.102 𝑦 1m 1m 1m torsional 5 kN spring, 𝑘𝑡

3 kN 𝐵 𝑥

3m 4 kN

𝐶

𝐷 𝐸

𝐴

The wing of a jet is modeled as a two-dimensional structure that is supported by a pin and torsional spring at point 𝐴. The distributed load on the bottom of the wing is due to the lift forces, and the 5 kN, 3 kN, and 4 kN forces are due to the weights of the fuel, engine, and wing, respectively. When there are no forces applied to the wing, the wing is horizontal and the torsional spring produces no moment. Due to the forces shown, the tip of the wing (point 𝐸) deflects upward by 8 cm. Assuming the wing is rigid, determine the stiffness 𝑘𝑡 of the torsional spring. Report your answer using degrees.

Problem 7.103 10 kN (1 − m

𝑥2 36 m2

Consider a straight uniform member with length 𝐿 and weight 𝑊 . Two force systems for representing the weight of this member are shown. In system 1, the distributed force is uniform with value 𝑤0 = 𝑊 ∕𝐿.

)

Figure P7.102 𝐵𝑦 𝐿 𝜃

𝐴𝑦

𝐵 𝑦

𝐴

𝑤0 system 1

Figure P7.103

ISTUDY

𝜃 𝐿

𝑥 system 2

(a) Use Eq. (4.16) on p. 248 to determine 𝐴𝑦 and 𝐵𝑦 in terms of 𝑊 so that the two force systems are equivalent. (b) Are the results for Part (a) in agreement with the load lumping scheme described for trusses in Example 6.2 on p. 372? (c) Without calculations, but perhaps by use of an appropriate sketch, show that the results of Part (a) do not apply for a member that is not straight and/or has nonuniform weight distribution. Hint: Redraw Fig. P7.103 using a member that is not straight or using a straight member with nonuniform weight distribution. Then argue that the results of Part (a) do not constitute an equivalent force system for this situation.

ISTUDY

Section 7.4

Distributed Forces, Fluid and Gas Pressure Loading

Problem 7.104

𝑦

A uniform curved beam with circular shape and weight 𝑊 has a built-in support at 𝐴. Determine the support reactions. Express your answers in terms of parameters such as 𝑊 and 𝑅.

𝑧 𝐴 𝑅

Problem 7.105

𝑥

A uniform curved beam with circular shape and weight 𝑊 is supported by a frictionless bearing at 𝐴 and a roller at 𝐵. Determine the support reactions. Express your answers in terms of parameters such as 𝑊 and 𝑅. 𝑧

𝐵 Figure P7.104

𝑦

𝐷 𝑅 𝐵

𝑥

𝐴

𝑇

𝐶 Figure P7.105

𝛾 𝑑

Problem 7.106 A cube of material with edge lengths 𝑑 and specific weight 2𝛾 is suspended by a cable and is submerged to a depth 𝑑 in a fluid having specific weight 𝛾. Determine the force 𝑇 in the cable. Express your answer in terms of parameters such as 𝑑, 𝛾, etc.

𝑑

2𝛾 𝑑

Figure P7.106

Problem 7.107 In Fig. P7.107(a), the concrete wall of a building has a small water-filled gap between it and the adjacent soil. In Fig. P7.107(b), a concrete wall is used to retain water. If the depth 𝑑 of water is the same for both walls, which wall will be subjected to the larger forces due to water pressure? Explain. Note: Concept problems are about explanations, not computations.

small gap filled with water

𝑑

𝑑

Problems 7.108 and 7.109 (a)

Water in a channel is retained by a gate with 5 in. width (into the plane of the figure). The gate is supported by a pin at 𝐵 and a vertical cable at 𝐴, and the contact between the gate and the bottom of the channel at 𝐴 is frictionless. The vertical wall 𝐵𝐶 is fixed in position. If the gate’s 50 lb weight is uniformly distributed, determine the cable force 𝑇 required to begin opening the gate. 𝑇

𝑇 𝐶

𝐶 4 in.

4 in. 𝐵

𝐵

𝑊𝐴𝐵 = 50 lb

𝑊𝐴𝐵 = 50 lb

8 in.

8 in. 𝐴 6 in. Figure P7.108

𝐴 8 in. Figure P7.109

Figure P7.107

(b)

483

484

Chapter 7

Centroids and Distributed Force Systems Problems 7.110 and 7.111

Water in a channel is retained by a gate with 10 cm width (into the plane of the figure). The gate is supported by a pin at 𝐵 and by frictionless contact with the bottom of the channel at 𝐴. The gate is outfitted with a prewound torsional spring at 𝐵 with stiffness 𝑘𝑡 = 75 N⋅m∕rad. The vertical wall 𝐵𝐶 is fixed in position. If the gate has negligible weight, determine the amount 𝜃0 the spring must be prewound so that the gate will begin to open when 𝑑 = 50 cm. 𝐶

𝐶 𝑑

𝑑

15 cm

𝐵

𝐵 30 cm

30 cm 𝐴

𝐴

Figure P7.110

Figure P7.111

Problems 7.112 and 7.113 A breakwater along an oceanfront is to be constructed of concrete, and the formwork for retaining the concrete while it is poured is shown. For every 6 f t of width (into the plane of the figure), the formwork has a horizontal support 𝐵𝐶, a support 𝐷𝐹 , and stakes at points 𝐴, 𝐸, and 𝐹 . The stakes may be modeled as pins and the weights of all members except the concrete may be neglected. If the concrete is modeled as a fluid with 150 lb∕f t 3 specific weight, determine the force supported by members 𝐵𝐶 and 𝐷𝐹 . 𝐶

𝐶

𝐵 3 ft

3 ft

𝐷

𝐷 6 ft

4.5 f t

6 ft

8 ft 𝐸

𝐹

𝐵

concrete

4 ft

6 ft

8 ft 𝐸

𝐴

Figure P7.112

𝐹

4.5 f t

4 ft

concrete 8 ft

𝐴

Figure P7.113

gas

Problem 7.114

fluid

36 in.

The cross section through a spherical tank is shown. The upper and lower portions of the tank are attached using 72 bolts that are uniformly spaced around the perimeter of the tank. The upper portion of the tank has a small dome that contains a gas. The fluid in the tank has approximately spherical shape and specific weight of 0.06 lb∕in.3 . Assuming all the bolts support the same force, determine the force each bolt supports due to the fluid and gas pressures if: (a) The gas is not pressurized (i.e., it is at atmospheric pressure).

Figure P7.114

ISTUDY

(b) The gas is pressurized to 5 psi.

ISTUDY

Section 7.4

Distributed Forces, Fluid and Gas Pressure Loading

Problem 7.115 The tank shown has a cylindrical midsection and hemispherical ends. Each of the hemispherical ends is attached to the cylindrical midsection using 60 bolts that are uniformly spaced around the perimeter of the tank. At 𝐷, the tank has a circular access plate that is attached using 12 bolts that are uniformly spaced. The tank is fully filled with a fluid having density 900 kg∕m3 . Assume that each of the bolts at 𝐵 supports the same force, each of the bolts at 𝐶 supports the same force, and each of the bolts at 𝐷 supports the same force. However, the forces supported by the bolts at 𝐵, 𝐶, and 𝐷 are probably different. Assume the piping that enters the tank at 𝐴 is flexible and has negligible weight. Determine the force each bolt supports due to the fluid pressure if the fluid at 𝐴 is at atmospheric pressure.

𝐴 0.8 m 𝐵

2m

𝐶 0.8 m

𝐷

Problem 7.116

0.5 m

Repeat Prob. 7.115 if the fluid at 𝐴 is at

10 kN∕m2

pressure.

Figure P7.115 and P7.116

Problem 7.117 The cross section through the valve of a fuel injector for an engine is shown, where the tip of the valve has conical shape. If the fuel is at 500 kN∕m2 pressure, determine the force 𝐹 that must be applied to keep the valve closed. Hint: The pressure due to the weight of the fuel is negligible compared to 500 kN∕m2 . fuel

valve 2 mm

𝐹

24 in.

𝐹 3 mm Figure P7.117

grain

18 in.

𝐸

32 in.

𝐺 𝐷

𝐶

Problem 7.118 Grain is contained in a silo. The walls of the silo are fixed, and the door 𝐴𝐵𝐶𝐷 can be opened to allow the grain to pour out. Door 𝐴𝐵𝐶𝐷 is flat, with 8 in. depth (into the plane of the figure). Idealize the grain to be a fluid with 0.025 lb∕in.3 specific weight. In the position shown, the hydraulic cylinder 𝐸𝐺 is horizontal. Neglect the weights of the individual members. Determine the force the hydraulic cylinder 𝐸𝐺 must support to keep the door in equilibrium. Report your answer, using a positive value for tension in the hydraulic cylinder and a negative value for compression.

22 in.

22 in. 𝐵

𝐴

12 in. Figure P7.118

vent

Problem 7.119 A tank with two compartments contains pressurized gas on the left and oil on the right that are separated by a gate 𝐴𝐵 having 20 in. width into the plane of the figure. The gate is supported by a hinge at 𝐴 and a stop at 𝐵. The oil compartment is vented so that the surface of the oil is subjected to atmospheric pressure. Determine the value of the gas pressure that will cause the gate to begin to open. The specific weight of the oil is 0.02 lb∕in.3 , and the specific weight of the gas is negligible.

10 in.

𝐴 gas

5 in. 15 in.

𝐵 7 in. Figure P7.119

oil

485

486

Chapter 7

Centroids and Distributed Force Systems

7.5 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

Centroid, center of mass, and center of gravity 𝑦

𝑦 𝐴

𝑦̄

𝐶

𝐴𝑖

𝑦̃𝑖

Centroid. The centroid is defined to be the average position of a distribution of shape. The equations for determining the centroid of lines, areas, and volumes can be compactly summarized as follows. Consider the example of the centroid of an area or a distribution of areas, as shown in Fig. 7.24. The 𝑥 position of the centroid of this area can be determined by using composite shapes or integration as follows: Eq. (7.19), p. 437

𝑥̄

𝑥

𝑥̃ 𝑖

𝑥

Figure 7.24 The centroid 𝐶 of area 𝐴 is located at 𝑥̄ and 𝑦. ̄ Area 𝐴 can be considered to consist of composite areas 𝐴𝑖 , where the centroid of each of these is located at 𝑥̃ 𝑖 and 𝑦̃𝑖 .

ISTUDY

𝑛 ∑

𝑥̄ =

𝑖=1 𝑛

𝑥̃ 𝑖 𝐴𝑖

∑

𝑖=1

= 𝐴𝑖

∫ 𝑥̃ 𝑑𝐴 ∫ 𝑑𝐴

.

The expressions for the 𝑦 position of the centroid are identical to Eq. (7.19) with all of the 𝑥’s replaced by 𝑦’s. Similarly, the expressions for the 𝑧 position of the centroid are identical to Eq. (7.19) with all of the 𝑥’s replaced by 𝑧’s. To determine the centroid of a line or a distribution of lines (straight and/or curved), all of the 𝐴’s that appear in Eq. (7.19) are replaced by 𝐿’s. To determine the centroid of a volume or a distribution of volumes, all of the 𝐴’s that appear in Eq. (7.19) are replaced by 𝑉 ’s.

Center of Mass. The center of mass is defined to be the average position of a distribution of mass. The equations for determining the center of mass are identical to Eq. (7.19) with all of the 𝐴’s replaced by 𝑚’s; hence, Eq. (7.20), p. 449 𝑛 ∑

𝑥̄ =

𝑖=1 𝑛

𝑥̃ 𝑖 𝑚𝑖

∑

𝑖=1

= 𝑚𝑖

∫ 𝑥̃ 𝑑𝑚 ∫ 𝑑𝑚

,

with similar expressions for 𝑦̄ and 𝑧. ̄ Furthermore, the summations and integrations in Eq. (7.20) may be over a combination of volumes, surfaces, and/or wires, as shown in Eqs. (7.21) and (7.22) on p. 450.

Center of Gravity. The center of gravity is defined to be the average position of a distribution of weight. For objects in a uniform gravity field, the equations for determining the center of gravity are identical to Eq. (7.19) with all of the 𝐴’s replaced by 𝑤’s, hence, Eq. (7.26), p. 451 𝑛 ∑

𝑥̄ =

𝑖=1 𝑛

𝑥̃ 𝑖 𝑤𝑖

∑

𝑖=1

= 𝑤𝑖

∫ 𝑥̃ 𝑑𝑤 ∫ 𝑑𝑤

,

with similar expressions for 𝑦̄ and 𝑧. ̄ Furthermore, the summations and integrations in Eq. (7.26) may be over a combination of volumes, surfaces, and/or wires, as shown in Eqs. (7.27) and (7.28) on p. 451.

ISTUDY

Section 7.5

487

Chapter Review

Theorems of Pappus and Guldinus

CL

CL

generating curve

A surface of revolution is produced by rotating a generating curve, as shown in Fig. 7.25, by an angle 𝜃 (in radians) about an axis of revolution. From the theorems of Pappus and Guldinus, the area of one side of the surface of revolution is

𝜃 𝑟̄

Eqs. (7.32) and (7.33), p. 461 𝐴 = 𝜃 𝑟̄𝐿 = 𝜃

𝑛 ∑

𝑟̃𝑖 𝐿𝑖 .

axis of revolution

𝑖=1

The latter expression, Eq. (7.33), is useful when the generating curve consists of simple composite shapes, as shown in Fig. 7.11 on p. 461. A solid of revolution is produced by rotating a generating area, as shown in Fig. 7.26, by an angle 𝜃 about an axis of revolution. From the theorems of Pappus and Guldinus, the volume of the solid of revolution shown in Fig. 7.26 is Eqs. (7.35) and (7.36), p. 462 𝑉 = 𝜃 𝑟̄𝐴 = 𝜃

𝑛 ∑

𝐿

𝐶

area of surface = 𝐴

Figure 7.25 A surface of revolution is produced by rotating a generating curve by an angle 𝜃 about an axis of revolution. The generating curve has arc length 𝐿 and centroid 𝐶 located a distance 𝑟̄ from an axis of revolution. CL

CL

generating area, 𝐴

𝑟̃𝑖 𝐴𝑖 .

𝑖=1

𝜃

The latter expression, Eq. (7.36), is useful when the generating area consists of simple composite shapes, as shown in Fig. 7.13 on p. 462.

𝑟̄ 𝐶

Distributed Forces Distributed forces are forces that are distributed along a line, over a surface, or throughout a volume. A force distributed along a line is called a line force, or a line load, and this has dimensions of force/length. A force distributed over a surface is called a surface force, or a traction, and this has dimensions of force/area. A force distributed throughout a volume is called a volume force, or a body force, and this has dimensions of force/volume. Line forces, surface forces, and volume forces are all vectors. Line forces and surface forces do not need to be perpendicular to the objects or structures they are applied to, although often they will be.

Distributed Forces Applied to Beams. Beams are often subjected to distributed forces that are line loads. With 𝑤 being the line load, with dimensions of force/length, and 𝑥 being a coordinate along the length of the beam, the total force 𝐹 produced by the distributed load and the position 𝑥̄ of its line of action are

∫

𝑑𝐹 =

∫

𝑤 𝑑𝑥,

𝑥̄ =

∫ 𝑥̃ 𝑑𝐹 ∫ 𝑑𝐹

volume of solid = 𝑉

Figure 7.26 A solid of revolution is produced by rotating a generating area by an angle 𝜃 about an axis of revolution. The generating area 𝐴 has its centroid 𝐶 located a distance 𝑟̄ from the axis of revolution.

surface of fluid

𝑝0

𝑦

Eqs. (7.41) and (7.43), p. 469 𝐹 =

axis of revolution

=

∫ 𝑥̃ 𝑤 𝑑𝑥 ∫ 𝑤 𝑑𝑥

.

Fluid and Gas Pressure. If a fluid is incompressible and at rest, as shown in Fig. 7.27, then the pressure within the fluid is given by Eq. (7.45), p. 469 𝑝 = 𝑝0 + 𝜌𝑔𝑑, = 𝑝0 + 𝛾𝑑, where 𝑝0 is the gas pressure (constant) at the surface of the fluid. For many purposes, only the portion of the pressure due to fluid loading is important, and this pressure (𝑝 = 𝜌𝑔𝑑 = 𝛾𝑑) is sometimes called the gage pressure.

𝑥

𝑑

𝑝 𝑝

𝑝 𝑝 𝑝

𝑝

fluid

𝑧

Figure 7.27 A volume of incompressible stationary fluid with density 𝜌. The surface of the fluid lies in the 𝑥𝑦 plane and is subjected to a pressure 𝑝0 . Infinitesimally small cubes of fluid are subjected to hydrostatic compressive pressure 𝑝.

488

ISTUDY

Centroids and Distributed Force Systems

Chapter 7

Forces produced by fluids and gases. The forces that a fluid and/or gas applies to a structure can, in principle, always be determined by integration. However, the use of composite shapes and/or a judicious FBD, as discussed in Section 7.4, will often be more straightforward. An important point is that fluid and gas pressures are always perpendicular to the structure or surface over which they act.

ISTUDY

Section 7.5

Chapter Review

489

Review Problems General instructions. For problems involving shapes or objects having one or more axes or planes of symmetry, you may use inspection to determine some of the coordinates of the centroid, center of mass, and/or center of gravity. Problems 7.120 through 7.123 For the area shown, use composite shapes to determine the 𝑥 and 𝑦 positions of the centroid. 40 mm

𝑦

10 mm 2 cm

4 cm

15 mm 𝑦

4 cm

𝑦

6 cm

3 in. 5 cm

5 cm

Figure P7.120

𝑥

0.5 in.

0.5 in.

60 mm

𝑦

𝑥

30 mm

𝑟𝑜

20 mm

𝑥

𝑟𝑖

6 in. Figure P7.121

Figure P7.122

𝑥

Figure P7.123

Problem 7.124

𝑦

𝑦=

√

𝑥

2m

For the area shown, use integration to determine the 𝑥 and 𝑦 positions of the centroid.

Problem 7.125 For the truncated circular cone shown, use composite shapes to determine the location of the centroid.

𝑥 𝑦 = 𝑥∕4

4m

Figure P7.124

Problem 7.126

𝑦

For the truncated circular cone shown, use integration to determine the location of the centroid.

𝑦 = 60 mm − 𝑥∕4 60 mm

Problem 7.127

40 mm 𝑥

The truncated circular cone shown has a truncated conical hole. (a) Fully set up the integral, including the limits of integration, that will yield the centroid of the object. (b) Evaluate the integral determined in Part (a) using computer software such as Mathematica or Maple.

𝑧

80 mm

Figure P7.125 and P7.126

𝑦

Problem 7.128

6 mm

The truncated circular cone shown has a truncated conical hole and is made of a material with density 0.002 g∕mm3 . Let the conical hole be filled with a material with density 0.003 g∕mm3 . (a) Fully set up the integral, including the limits of integration, that will yield the center of mass of the object. (b) Evaluate the integral determined in Part (a) using computer software such as Mathematica or Maple.

4 mm

3 mm 2 mm

𝑥 1 mm

𝑧 Figure P7.127 and P7.128

490

𝑦

𝑦=

𝜌2

𝜌1

√

Problem 7.129

𝑥 𝑥

𝑧

Chapter 7

Centroids and Distributed Force Systems

ℎ

The bullet-shaped object is a solid of revolution that is composed of materials with densities 𝜌1 and 𝜌2 . Set up the integral, including the limits of integration, that will yield the 𝑥 position of the center of mass.

ℎ

Problem 7.130

Figure P7.129

A solid of revolution is produced by revolving the area shown 360◦ about the 𝑥 axis. Use integration to determine the 𝑥 coordinate of the centroid.

𝑦 2m

Problem 7.131 𝑦=

√

𝑥 4m

𝑥

Figure P7.130 and P7.131

A solid of revolution is produced by revolving the area shown 360◦ about the 𝑦 axis. Use integration to determine the 𝑦 coordinate of the centroid.

Problems 7.132 and 7.133 For the solid of revolution described in the subproblem below: (a) Fully set up the integral, including the limits of integration, that will yield the centroid of the object.

𝑦 2 cm

(b) Evaluate the integral determined in Part (a) using computer software such as Mathematica or Maple.

𝑦 = 1 + 𝑥2

Problem 7.132 1 cm

The solid of revolution is produced by revolving the area shown about

the 𝑥 axis. Problem 7.133

0 0

The solid of revolution is produced by revolving the area shown about

the 𝑦 axis.

𝑥 1 cm

Problem 7.134

Figure P7.132 and P7.133

For the line shown:

𝑦

(a) Set up the integrals for integration with respect to 𝑥, including the limits of integration, that will yield the 𝑥 and 𝑦 positions of the centroid.

4 km 𝑦 = 𝑥2

(b) Evaluate the integrals in Part (a) using computer software such as Mathematica or Maple.

Problem 7.135 𝑥 −1 km 0 Figure P7.134

ISTUDY

1 km 2 km

A truss-supported solar panel for a satellite is shown. The truss members are hollow tubes with mass per unit length of 20 kg∕m, and the solar panel has mass per unit area of 12 kg∕m2 . Determine the mass of the structure and the 𝑥 position of the center of mass. If the structure is on the surface of Earth, with acceleration due to gravity acting in the −𝑦 direction, determine the support reactions at points 𝐴 and 𝐵. 𝑦 𝐴 𝐵 0.6 m

𝐸 1.5 m solar panel

𝐶 𝐷 2m

𝐹

2m

1.5 m 5m

Figure P7.135

𝑥

ISTUDY

Section 7.5

491

Chapter Review 𝑦

Problem 7.136

30 in.

The structure shown is used to support a sign for a restaurant. The structure has five truss members, and cables 𝐴𝐶, 𝐷𝐹 , and 𝐸𝐺. The sign weighs 0.2 lb∕in.2 , the truss members weigh 0.5 lb∕in., and the cables have negligible weight. Determine the 𝑥 and 𝑦 positions of the center of gravity, and determine the support reactions at points 𝐴 and 𝐵.

𝐴

30 in. 𝐶

18 in.

𝐸 𝐵

10 in.

𝐷 𝐺

𝐹

20 in.

Problem 7.137 The area of Prob. 7.15 on p. 446 is revolved 360◦ about the line 𝑥 = −𝑟 to create a solid of revolution. Determine the volume and surface area of the solid.

30 in. 12 in. Figure P7.136

Problem 7.138

𝑦

The area of Prob. 7.15 on p. 446 is revolved 360◦ about the 𝑥 axis to create a solid of revolution. Determine the volume and surface area of the solid.

6 ft

18 ft

Problem 7.139

9 ft

A solar panel has the shape of a 90◦ sector of a truncated right circular cone, and hence it is a surface of revolution. Use the Pappus-Guldinus theorem to determine the outside surface area of the panel.

𝑧

𝑥

Figure P7.139

Problem 7.140 A scoop for handling animal food is shown. The scoop’s shape is one-half of a truncated circular cone. Use the Pappus-Guldinus theorem to determine the volume of food the scoop will hold, assuming the food is “level.” Also, disregarding the handle, determine the area of sheet metal, in cm2 , required to fabricate the scoop.

12 cm 20 cm 8 cm

Problems 7.141 through 7.143

Figure P7.140

Determine the support reactions for the loading shown.

2 kN 1 kN∕m

600 lb∕ft 𝐴

400 lb∕ft 𝐵

6 ft

10 ft

Figure P7.141

200 lb∕ft 𝐵

𝐴 6 ft

6 ft

Figure P7.142

Problem 7.144 In Fig. P7.141, reposition the pin support 6 f t to the right of point 𝐴, and determine the support reactions.

Problem 7.145 In Fig. P7.142, replace the pin and roller supports with a built-in support at 𝐴, and determine the support reactions.

Problem 7.146 In Fig. P7.143, reposition the roller support 2 m to the left of point 𝐵, and determine the support reactions.

𝐴

𝐵 3 kN ⋅m 2m 2m 2m 2m Figure P7.143

𝑥

492

Chapter 7

Centroids and Distributed Force Systems 𝑦

Problem 7.147

𝑦 25

lb∕in.2

10 lb∕in.2

𝑥

A circular plate with 21 in. radius is subjected to the pressure distribution shown. By treating the pressure distribution as a solid of revolution, use the theorems of Pappus and Guldinus to determine the total force applied to the plate.

𝑥

𝑧

9 in. 12 in.

Figure P7.147

Problem 7.148 Water in a channel is retained by a gate with 0.5 f t width (into the plane of the figure). The gate is supported by a pin at 𝐴 and a roller at 𝐶. The vertical wall 𝐴𝐷 is built into the bottom of the channel. If the gate has negligible weight, determine the support reactions. 𝐶

𝐴 3

1.5 m

3 ft 𝐵 1.5 ft

𝐴 3 ft

4

1.5 m 𝐵

𝐷

Figure P7.148

𝐶

Figure P7.149

Problem 7.149 Water in a channel is retained by a cylindrical gate with 2 m width. The gate is supported by a pin at 𝐵 and a cable between 𝐴 and 𝐶. If the gate has negligible weight, determine the force supported by the cable and the reactions at 𝐵.

Problem 7.150 40 mm 40 mm 𝐷

𝐵 20 mm 20 mm

Figure P7.150

ISTUDY

100 mm

𝐶

𝑑

𝐴

A uniform right circular cone 𝐶 with 40 mm radius at its base and 0.1 N weight is attached to a beam 𝐴𝐵 with negligible weight. The cone is partially submerged in water. A block 𝐷 with 0.2 N weight is placed a distance 𝑑 from the support at 𝐴. Determine the value of 𝑑 so that the system is in equilibrium in the position shown. Report 𝑑 such that a positive value means block 𝐷 is to the left of 𝐴, and a negative value means block 𝐷 is to the right of 𝐴. Hint: The theorems of Pappus and Guldinus may be useful.

Problem 7.151 A model for the gate of a dump truck is shown. Gate 𝐴𝐵𝐶 is hinged at 𝐴, is actuated by a horizontal hydraulic cylinder 𝐵𝐷, and has 48 in. width into the plane of the figure. The pins at 𝐴 and 𝐷 are fixed in space. The gate retains a 72 in. depth of sand that may be modeled as a fluid with 0.06 lb∕in.3 specific weight. If gate 𝐴𝐵𝐶 is uniform with 400 lb weight that acts midway between points 𝐴 and 𝐶, determine the force in the hydraulic cylinder that will allow the gate to begin to open. 15 in.

21 in. 𝐴 𝐵

𝐷 weight = 400 lb

72 in. sand

𝐶 Figure P7.151

ISTUDY

Internal Forces

8

When a structure is subjected to external forces, the various members of the structure develop internal forces within them. In fact, it’s because of these internal forces that the structure can support the external forces that are applied to it. In earlier chapters, we routinely determined the internal forces that cables and bars support. In this chapter, we discuss internal forces for more complex structural members, such as beams.

Carl & Ann Purcell/The Image Bank Unreleased/Getty Images

The glass roof at Canada Place, Vancouver, British Columbia, Canada. The design of the beams that support the weight of the roof and the forces from wind, rain, and snow requires determination of the internal forces throughout all of the beams in the roof system.

8.1

Internal Forces in Structural Members

Why are internal forces important? Internal forces are forces and moments that develop within structural members and/or materials due to the external forces that are applied. Knowledge of the internal forces that a particular member must support is essential before the member can be designed. The design of a member includes: specification of the material(s) it is constructed of, its shape and dimensions, methods of support and/or attachment to other members, and so on. Obviously, as the internal forces that a member must support become larger, the member must be larger and/or be constructed of stronger material(s) so that it has sufficient strength. A structural member is said to be slender if the dimensions of its cross section are small compared to its length. Slender members are very common, and the methods

493

494

Chapter 8

Internal Forces

of analysis of statics and mechanics of materials are very effective for these. Many structural members are not slender, and determining internal forces and design are usually more difficult and require more advanced theory and methods of analysis. Thus, the main focus of this chapter is determination of internal forces in slender members.

Internal forces for slender members in two dimensions (a)

𝐴

𝑀

𝑦 (b)

𝑀 𝑁

𝑥

𝑁 𝑉

𝑉

Figure 8.1 Internal forces that develop on a particular cross section of a slender member in two dimensions.

Consider the structure in two dimensions shown in Fig. 8.1(a), and imagine we are interested in the internal forces that act on cross section 𝐴. We take a cut through cross section 𝐴, thus separating the left-hand portion of the structure from the righthand portion. The forces that develop on this cross section are called internal forces, as shown in Fig. 8.1(b), and these forces and moment must be present if the material immediately to the left of the cut is to remain bonded or attached to the material immediately to the right of the cut. Another way to understand the nature of these forces is by analogy with the reaction forces for a built-in support: the material to the left of cut 𝐴 may not translate in the horizontal direction relative to the material to the right of the cut, thus a force 𝑁 must exist. Similarly, the material to the left may not translate vertically relative to material to the right, thus, a force 𝑉 must exist; and the material to the left may not rotate relative to material to the right, thus, a moment 𝑀 must exist. Observe that the internal forces in Fig. 8.1(b) satisfy Newton’s third law. Remarks • For several reasons, most of which are not clear until we study mechanics of materials, it is most useful to determine internal forces referenced to directions that are along and transverse to the axis of the member. Thus, we will routinely select an 𝑥𝑦 system (or 𝑡𝑛 system for members that are curved or have shape that is not straight), such as in Fig. 8.1(b), where 𝑥 is in the axial direction and 𝑦 is in the transverse direction. Furthermore, we will usually take the origin of the coordinate system (the 𝑦 = 0 position) to coincide with the centroid of the cross section.

(a)

𝑁

𝑁 axial deformation

𝑦 (b)

𝑥

𝑉

𝑉

𝑦 𝑥

𝑀

𝑀

bending deformation Figure 8.2 In a structure made of deformable material, internal forces 𝑁, 𝑉 , and 𝑀 produce the types of deformation shown by the dashed outlines.

ISTUDY

– 𝑁 is called the normal force or axial force. We will usually follow the sign convention shown in Figs. 8.1(b) and 8.2(a), wherein a positive value of 𝑁 corresponds to tension, and a negative value to compression. Normal force gives rise to the axial deformation shown in Fig. 8.2(a). – 𝑉 is called the shear force. Although it’s an arbitrary choice, we will usually follow the sign convention shown in Figs. 8.1(b) and 8.2(b). Shear force gives rise to the shear deformation shown in Fig. 8.2(b).

shear deformation

(c)

• The internal forces shown in Fig. 8.1(b) are often categorized as follows:

– 𝑀 is called the bending moment. Although it is an arbitrary choice, we will usually follow the sign convention shown in Figs. 8.1(b) and 8.2(c). Bending moment gives rise to the bending deformation shown in Fig. 8.2(c). • Later sections of this chapter focus on structures consisting of a single straight member, and it will be straightforward (and important) to follow the sign conventions shown in Figs. 8.1 and 8.2. In this section, our main focus is determination of the absolute values of internal forces at various locations in a structure, and many of the structures we consider have multiple members and/or

ISTUDY

Section 8.1

Internal Forces in Structural Members

495

members that are not straight. Thus, we will sometimes adopt different sign conventions that are more convenient (e.g., Example 8.3).

Internal forces for slender members in three dimensions The internal forces for a slender member in three dimensions are shown in Fig. 8.3 where it is seen that on every cross section, there exist six internal forces, as follows: 𝑁 is called the normal force or axial force, and it gives rise to axial deformation; 𝑉𝑦 and 𝑉𝑧 are called shear forces, and these give rise to shear deformations; 𝑀𝑦 and 𝑀𝑧 are called bending moments, and these give rise to bending deformations; and 𝑀𝑥 is called a torque, and this gives rise to twisting deformation, or torsional deformation. For convenience, the forces and moments acting on the left-hand side of the cut are taken to be positive in the positive coordinate directions, while the forces and moments on the right-hand side of the cut are positive in the negative coordinate directions, as required by Newton’s third law. Other sign conventions may be used provided Newton’s third law is respected.

𝑀𝑦

𝑀𝑦

𝑦

𝑉𝑦 𝑁 𝑀𝑥

𝑥

𝑉𝑦 𝑀𝑥 𝑁

𝑧 𝑀𝑧

𝑉𝑧 𝑉𝑧

𝑀𝑧

Figure 8.3 Internal forces that develop on a particular cross section of a slender member in three dimensions.

Interesting Fact Effects of internal forces on strength. All of the internal forces that a member supports are important, and one of the main topics of mechanics of materials is how to design a member so that it can safely support all of these. Nonetheless, for slender members, the following generalizations are often true: Alisa Ch/Shutterstock

National Geographic Image Collection/Alamy Stock Photo

Figure 8.4. The Grand Canyon Skywalk is an observation platform on the grounds of the Hualapai Indian Nation in Arizona. It extends 70 f t over the edge of the Grand Canyon and is 4000 f t above the canyon’s floor. The platform has steel framing, its deck is 3 in. thick glass, and it is designed to accommodate 120 people. Knowledge of the internal forces due to loads from the materials, spectators, and other sources is essential for the design of this structure.

●

●

●

Determination of internal forces First, we must decide where in a structure we wish to determine internal forces. Note that, in general, the internal forces on every cross section of every member of a structure are different, and usually one of our goals is to determine all of these. Sometimes it is sufficient to determine only the maximum values of the internal forces for each member. With some experience, you will often be able to identify, by inspection, the cross section or sections that experience the largest internal forces. For example, in a cantilever beam the internal forces are usually highest at the built-in support, and in a simply supported beam the internal moment is usually highest near the midspan of the beam. Once a cross section of interest is identified, we proceed by taking a cut through that cross section and drawing an FBD of a portion of the structure, followed by writing and solving equilibrium equations. This process is then repeated for other cross sections of interest. Other approaches are possible (e.g., solution of differential equations), and some of these are discussed later in this chapter.

●

●

Straight members are extremely effective at supporting axial tensile internal forces. Straight members are susceptible to buckling, and often they can support only low axial compressive internal forces. Compared to the effects of other internal forces, shear internal forces are often not of great concern. An important exception is for materials that are weak in shear, such as wood and some fiber-reinforced composite materials. Members are flexible in bending, and thus, bending moments are of great importance. Members are flexible in torsion, and thus, twisting moments (torques) are of great importance.

496

ISTUDY

Chapter 8

Internal Forces

End of Section Summary In this section, internal forces in slender structural members were discussed. For every cross section where the internal forces are to be determined, we select a coordinate system in which one of the coordinates (often the 𝑥 coordinate) is in the axial direction of the member and the other coordinates are perpendicular to the axis of the member. Furthermore, the origin of the coordinate system is usually taken to coincide with the centroid of the cross section. At each cross section of a member in two dimensions, there are three internal forces consisting of an axial force 𝑁, a shear force 𝑉 , and a bending moment 𝑀. At each cross section of a member in three dimensions (where 𝑥 is in the axial direction of the member), there are six internal forces consisting of an axial force 𝑁, two shear forces 𝑉𝑦 and 𝑉𝑧 , two bending moments 𝑀𝑦 and 𝑀𝑧 , and a torque 𝑀𝑥 . In this section, internal forces are determined by taking a cut through the cross section of interest, drawing an FBD of a portion of the structure, followed by writing and solving equilibrium equations.

ISTUDY

Section 8.1

E X A M P L E 8.1

497

Internal Forces in Structural Members

Internal Forces for a Two-Dimensional Problem

A metal bottle cap opener is shown. If a vertical force of 10 N at point 𝐴 is required to remove a bottle cap, determine the internal forces that develop on cross section 𝐷.

SOLUTION Road Map We are especially interested in the internal forces on cross section 𝐷 because, among all the locations in the bottle opener, this location will have very high internal forces. To explain further, imagine the bottle opener is made of metal that is too thin or is used to open a particularly stubborn bottle. Intuitively, we expect the bottle opener to severely bend and deform at cross section 𝐷. As engineers, our ultimate objective is to design the bottle opener (i.e., specify the material it is made of and its thickness and width) so that it is sufficiently strong for its intended use. Before this can be done, the maximum internal forces in the bottle opener must be determined. We will neglect the weight of the bottle opener since it is clearly small compared to the other forces in the system. The forces acting on the bottle opener are obtained first. Then, to determine the internal forces on cross section 𝐷, a cut will be taken through that cross section, an FBD will be drawn, and equilibrium equations will be written and solved.

Michael Plesha

12 mm 14 mm 𝐶

58 mm

𝐷 𝐵

30◦

10 N

𝐴

Modeling

The FBD for the bottle opener is shown in Fig. 2. Although there may be horizontal force components at 𝐵 and 𝐶, we will assume they are negligible compared to the vertical components 𝐵𝑦 and 𝐶𝑦 .

Figure 1 12 mm 14 mm

Governing Equations & Computation

Using the FBD shown in Fig. 2, we write and solve the following equilibrium equations: ∑ −𝐵𝑦 (12 mm) + (10 N)(84 mm) = 0 ⇒ 𝐵𝑦 = 70 N, (1) 𝑀𝐶 = 0 ∶ ∑ 𝐹𝑦 = 0 ∶ 𝐶𝑦 − 𝐵𝑦 + 10 N = 0 ⇒ 𝐶𝑦 = 60 N. (2)

Modeling To determine the internal forces on cross section 𝐷, we take a cut through that cross section on the FBD of Fig. 2 to draw the two FBDs shown in Fig. 3. Observe that the assignment of directions for the internal forces matches those shown in Fig. 8.1(b).

58 mm

10 N 𝐶𝑦

𝑦

𝐵𝑦

𝑥 𝐴 Figure 2 Free body diagram for the bottle opener.

Governing Equations & Computation

Either of the FBDs shown in Fig. 3 may be used to determine the internal forces 𝑁𝐷 , 𝑉𝐷 , and 𝑀𝐷 , and we will select the right-hand FBD, since it has simpler geometry and contains fewer forces. Using the 𝑡𝑛 coordinate system shown, where 𝑡 is oriented along the axis of the member and 𝑛 is in the transverse direction, we write and solve the following equations: ∑ 𝐹𝑡 = 0 ∶ −𝑁𝐷 − (10 N)(sin 30◦ ) = 0 ⇒ 𝑁𝐷 = −5 N, (3) ∑ 𝑉𝐷 = −8.660 N, (4) 𝐹𝑛 = 0 ∶ 𝑉𝐷 + (10 N)(cos 30◦ ) = 0 ⇒ ∑ 𝑀𝐷 = 0 ∶ −𝑀𝐷 + (10 N)(58 mm) = 0 ⇒ 𝑀𝐷 = 580 N⋅mm. (5)

12 mm 14 mm 𝑀𝐷 𝐶𝑦

𝑉𝐷 𝐵𝑦

𝑁𝐷

58 mm 𝑁𝐷 𝑀𝐷

𝑉𝐷 𝑛 10 N

𝑡

30◦

𝑦 𝑥

Discussion & Verification With some foresight, we might have anticipated using the right-hand FBD in Fig. 3, in which case the reaction forces 𝐵𝑦 and 𝐶𝑦 are not needed and we could have avoided writing Eqs. (1) and (2). Using the left-hand FBD in Fig. 3, you

should verify that the same internal forces 𝑁𝐷 , 𝑉𝐷 , and 𝑀𝐷 are obtained.

𝐴

Figure 3 Free body diagrams with a cut taken through cross section 𝐷 to determine the internal forces on that cross section.

498

Chapter 8

Internal Forces

E X A M P L E 8.2 10◦

4 in.

SOLUTION

𝐸

𝐵 7 in.

A hand saw for cutting firewood is shown. The blade of the saw is tensioned to 200 lb using the wing nut at 𝐷. Neglecting the forces due to cutting, determine the internal forces acting on cross section 𝐺.

𝐷

𝐶 5 in. 𝐺

Internal Forces for a Two-Dimensional Problem

10◦

Road Map

We will neglect the weights of individual members of the saw under the assumption they are small. We also neglect the forces due to cutting, although these forces may not be small, and it may be warranted to consider an analysis where these are included. The forces acting on member 𝐴𝐵𝐶𝐸 are obtained first. To determine the internal forces on cross section 𝐺, a cut will be taken through that cross section, an FBD will be drawn, and equilibrium equations will be written and solved.

𝐹

𝐴 Figure 1 𝐶

𝑇𝐶𝐷

Modeling

𝐸𝑦

5 in.

𝐸𝑥 𝐵

𝑦

7 in. 𝑥 200 lb 𝐴 Figure 2 Free body diagram for member 𝐴𝐵𝐶𝐸. 10◦

𝐶 𝑛

𝑡

𝑇𝐶𝐷 = 280 lb

𝑉𝐺

4 in. 𝑀𝐺

𝑀𝐺

𝑁𝐺 𝑁𝐺

𝐸𝑦

𝑉𝐺 𝐵

𝐸𝑥 𝑦 𝑥

200 lb Figure 3 Free body diagrams with a cut taken through cross section 𝐺 to determine the internal forces on that cross section. 𝐽

Figure 4 Additional cross sections in the vicinity of 𝐵 where internal forces should be determined.

ISTUDY

Governing Equations & Computation

Using the FBD shown in Fig. 2, we write solve the following equilibrium equations: ∑ 𝑀𝐸 = 0 ∶ −𝑇𝐶𝐷 (5 in.) + (200 lb)(7 in.) = 0 ⇒ 𝑇𝐶𝐷 = 280 lb, ∑ 𝐹𝑦 = 0 ∶ 𝐸𝑦 = 0 ⇒ 𝐸𝑦 = 0, ∑ 𝐹𝑥 = 0 ∶ 𝑇𝐶𝐷 + 200 lb + 𝐸𝑥 = 0 ⇒ 𝐸𝑥 = −480 lb.

and

(1) (2) (3)

Modeling

To determine the internal forces on cross section 𝐺, we take a cut through that cross section on the FBD of Fig. 2 to draw the two FBDs shown in Fig. 3. Observe that the assignment of directions for the internal forces matches those shown in Fig. 8.1(b). Governing Equations & Computation

Either of the FBDs shown in Fig. 3 may be used to determine the internal forces 𝑁𝐺 , 𝑉𝐺 , and 𝑀𝐺 , and we will select the upper FBD, since it has simpler geometry and contains fewer forces. Using the 𝑡𝑛 coordinate system shown, where 𝑡 is oriented along the axis of the member and 𝑛 is in the transverse direction, we write and solve the following equations: ∑ 𝐹𝑡 = 0 ∶ −𝑁𝐺 + 𝑇𝐶𝐷 (sin 10◦ ) = 0 ⇒ 𝑁𝐺 = 48.62 lb, (4) ∑ (5) 𝐹𝑛 = 0 ∶ 𝑉𝐺 − 𝑇𝐶𝐷 (cos 10◦ ) = 0 ⇒ 𝑉𝐺 = 275.7 lb, ∑ (6) 𝑀𝐺 = 0 ∶ −𝑀𝐺 − 𝑇𝐶𝐷 (cos 10◦ )(4 in.) = 0 ⇒ 𝑀𝐺 = −1103 in.⋅lb. Discussion & Verification

𝐴

𝐵 𝐻

Separating member 𝐴𝐵𝐶𝐸 from the saw gives the FBD shown in Fig. 2.

The internal forces on cross section 𝐺 are high. However, there are other locations in member 𝐴𝐵𝐶𝐸, as well as member 𝐷𝐸𝐹 , which also will likely support high internal forces, and a complete analysis requires that these also be considered. Regarding junction 𝐵, Fig. 4 identifies some additional cross sections we should consider (see Prob. 8.2).

ISTUDY

Section 8.1

E X A M P L E 8.3

Internal Forces for a Three-Dimensional Problem

Member 𝐸𝐹 𝐺 is a shifting fork used in a transmission to move gear 𝐺 along shaft 𝐶𝐷. It is actuated by the 10 N and 20 N forces at 𝐸, which act in the −𝑧 and −𝑥 directions, respectively, and gear 𝐺 applies a 5 N force to the fork in the 𝑥 direction. The fork is supported by a fixed shaft 𝐴𝐵 that has a thrust collar at 𝐻, and the fork is a loose fit on the shaft at 𝐺 so that there is only a 𝑧 direction reaction. Determine the internal forces acting on cross section 𝐽 .

45 mm 20 N

10 N

𝑧

𝐸 25 mm

𝑥

𝐾

𝐷

𝐻

𝐹 𝐽

Road Map

Neglecting weight and friction, an FBD of the entire shifting fork will be drawn, followed by writing and solving equilibrium equations to determine the necessary support reactions. Observe in Fig. 1 that portion 𝐹 𝐺 is cantilever-supported by the remainder of the structure, and hence the internal forces at cross section 𝐽 are expected to be large. To determine the internal forces on cross section 𝐽 , a cut will be taken through this cross section, an FBD will be drawn, and equilibrium equations will be written and solved.

𝑦

𝐵

SOLUTION

Modeling

499

Internal Forces in Structural Members

𝐺 𝐴

15 mm

𝐶 40 mm

10 mm Figure 1 𝑧

10 N

20 N

𝐸

The FBD for the shifting fork 𝐸𝐹 𝐺 is shown in Fig. 2.

𝐹𝑥

Governing Equations & Computation

Using the FBD shown in Fig. 2, we write and solve the following equilibrium equation for the reaction 𝐺𝑧 : ∑ 𝑀𝑥 = 0 ∶ (10 N)(45 mm) − 𝐺𝑧 (50 mm) = 0 ⇒ 𝐺𝑧 = 9 N. (1)

𝑀𝐹 𝑦

To determine the internal forces on cross section 𝐽 , we take a cut through that cross section on the FBD of Fig. 2 to draw the FBD shown in Fig. 3, where we have placed the origin of the coordinate system at the centroid of cross section 𝐽 . For simplicity, we have assigned the directions for the internal forces so that all forces and moments are positive in the positive coordinate directions. Rather than the FBD of Fig. 3, we could have drawn an FBD of the left-hand portion of the shifting fork, although this will clearly have many more forces and moments, and hence is not as judicious a choice as Fig. 3. Governing Equations & Computation

following equations: ∑ 𝐹𝑥 = 0 ∶ ∑ 𝐹𝑦 = 0 ∶ ∑ 𝐹𝑧 = 0 ∶ ∑ 𝑀𝑥 = 0 ∶ ∑ 𝑀𝑦 = 0 ∶ ∑ 𝑀𝑧 = 0 ∶ Discussion & Verification

Using the FBD of Fig. 3, we write and solve the

𝑥

𝐹𝑧 𝐺𝑧 𝑀𝐹 𝑧

Figure 2 Free body diagram for the shifting fork. 𝑧 𝑀𝐽 𝑧 𝑀𝐽 𝑦

𝑁𝐽 𝑦

𝑉𝐽 𝑧 𝑦 5N

𝐽

𝑉𝐽 𝑥 + 5 N = 0

⇒

𝑉𝐽 𝑥 = −5 N,

(2)

𝑁𝐽 𝑦 = 0

⇒

𝑁𝐽 𝑦 = 0,

(3)

𝑉𝐽 𝑧 − 9 N = 0

⇒

𝑉𝐽 𝑧 = 9 N,

(4)

𝑀𝐽 𝑥 − (9 N)(40 mm) = 0

⇒

𝑀𝐽 𝑥 = 360 N⋅mm,

(5)

𝑀𝐽 𝑦 − (5 N)(15 mm) = 0

⇒

𝑀𝐽 𝑦 = 75 N⋅mm,

(6)

𝑀𝐽 𝑧 − (5 N)(40 mm) = 0

⇒

𝑀𝐽 𝑧 = 200 N⋅mm.

(7)

Cross section 𝐾 shown in Fig. 1, where portion 𝐸𝐹 is built into the shifting fork, is also a likely location of high internal forces, and Prob. 8.23 asks you to determine these.

5N

𝐽

By writing additional equilibrium equations, the remaining reactions can be determined (some additional dimensions in Fig. 1 may be needed). However, some foresight into the FBD that will be used to determine the internal forces on cross section 𝐽 shows that the remaining reactions are not needed, and hence we will avoid determining them. Modeling

𝑦

𝐹𝑦

𝑉𝐽 𝑥 𝑥

15 mm

𝑀𝐽 𝑥 40 mm

𝐺𝑧 = 9 N

Figure 3 Free body diagrams with a cut taken through the cross section at 𝐽 to determine the internal forces on that cross section.

500

Chapter 8

Internal Forces

Problems General instructions Report the axial internal force using positive and negative values for tension and compression, respectively. For shear and moment internal forces, the absolute values are acceptable. 6 mm cover

150 mm

90 mm 𝐹

𝐶𝐵

𝐴

Problem 8.1 A screwdriver is used to pry open the cover of a can containing paint. If a force 𝐹 = 75 N is required to open the cover, and assuming the contact forces between the screwdriver and can are vertical, determine the internal forces acting on cross sections 𝐴, 𝐵, and 𝐶 of the screwdriver. Cross sections 𝐵 and 𝐶 are immediately to the right and left, respectively, of where the screwdriver makes contact with the can.

can paint Figure P8.1

Problem 8.2 500 N

Determine the internal forces acting on cross sections 𝐻 and 𝐽 in Fig. 4 of Example 8.2 on p. 498. Cross section 𝐻 is located 6 in. from point 𝐴, measured along line 𝐴𝐵.

15 cm

Problem 8.3 𝐴

Determine the internal forces acting on cross sections 𝐴 and 𝐵 of the bicycle seat stem.

10 cm 𝐵

Problems 8.4 and 8.5 15◦

Member 𝐴𝐵𝐶 and brace 𝐵𝐷 are used to suspend an electric transmission line from a utility pole. Determine the internal forces acting on:

Figure P8.3

Problem 8.4 Cross sections 𝐸 and 𝐹 , which are located immediately to the left of point 𝐴 and to the right of point 𝐵, respectively. 𝐶 𝐻 24 in. 𝐺

Problem 8.5 𝐹 24 in. 𝐸

𝐵

𝐷 30◦

30◦ 400 lb

Figure P8.4 and P8.5

ISTUDY

Cross sections 𝐺 and 𝐻, which are located immediately to the left of point 𝐵 and to the right of point 𝐶, respectively.

Problems 8.6 and 8.7 𝐴

The structure shown consists of a single member 𝐴𝐵𝐶𝐷𝐸 with a pin support at 𝐴 and a roller support at 𝐸. Points 𝐵 and 𝐷 are at the midpoints of their respective segments. Determine the internal forces acting on: Problem 8.6

Cross sections 𝐹 , 𝐺, 𝐻, and 𝐼, which are located immediately to the right of 𝐴, the left of 𝐵, the right of 𝐵, and the left of 𝐶, respectively.

Problem 8.7 Cross sections 𝐽 , 𝐾, 𝐿, and 𝑀, which are located immediately to the right of 𝐶, the left of 𝐷, the right of 𝐷, and the left of 𝐸, respectively. 3 kN 4 kN

𝐻

𝐽 𝐶

𝐺 𝐹

𝐼

𝐾

2 kN 𝐿 𝑀 𝐸 𝐷 3m

𝐵

𝐴 4m

5m

Figure P8.6 and P8.7

ISTUDY

Section 8.1

Internal Forces in Structural Members

501

Problems 8.8 and 8.9 The structure of Probs. 8.6 and 8.7 is revised to consist of two members, 𝐴𝐵𝐶 and 𝐶𝐷𝐸, that are pinned to one another at 𝐶 and have pin supports at 𝐴 and 𝐸. Points 𝐵 and 𝐷 are at the midpoints of their respective members. Determine the internal forces acting on: Problem 8.8 Cross sections 𝐹 , 𝐺, 𝐻, and 𝐼, which are located immediately to the right of 𝐴, the left of 𝐵, the right of 𝐵, and the left of 𝐶, respectively. Problem 8.9 Cross sections 𝐽 , 𝐾, 𝐿, and 𝑀, which are located immediately to the right of 𝐶, the left of 𝐷, the right of 𝐷, and the left of 𝐸, respectively. 3 kN 4 kN

𝐻

𝐶

𝐺 𝐹

𝐼

𝐽

𝐾

2 kN 𝐿 𝑀 𝐸 𝐷 3m

𝐵

𝐴 4m

5m

Figure P8.8 and P8.9

Problems 8.10 and 8.11 A hacksaw for cutting metal is shown. Assume contact between the frame 𝐴𝐵𝐶 and the handle assembly occurs at points 𝐵 and 𝐶 only, neglect friction, and neglect the size of the pin and notch at point 𝐵. If the blade is tensioned to 100 lb, determine the internal forces acting on:

𝐴 1 in.

1.5 in. 0.5 in.

Cross sections 𝐷, 𝐸, and 𝐹 , where cross section 𝐷 is located immediately above point 𝐴.

3 in.

1.5 in. 𝐹 𝐺 45◦ 𝐸

𝐼 𝐻

𝐽

Problem 8.10

Problem 8.11 Cross sections 𝐺, 𝐻, 𝐼, and 𝐽 , where cross sections 𝐻 and 𝐼 are located immediately to the right and left of point 𝐵, respectively, and cross section 𝐽 is immediately to the right of point 𝐶.

𝐵

𝐶

𝐷

1.5 in. 1.5 in. 0.5 in.

Figure P8.10 and P8.11

Problems 8.12 through 8.16 A weight 𝑊 = 2 kN is supported by a cable that passes over frictionless pulleys at 𝐷 and 𝐹 . The cable is attached to a winch at 𝐺, and cable segment 𝐷𝐺 is vertical. Member 𝐴𝐵𝐶 is built in at 𝐴, and members 𝐴𝐵𝐶, 𝐷𝐶𝐸𝐹 , and 𝐵𝐸 are attached using pins at points 𝐵, 𝐶, and 𝐸. Neglecting the weights of individual members, determine the internal forces acting on:

dimensions in meters 0.2

Problem 8.12

Cross sections 𝐻 and 𝐼, located immediately to the left of point 𝐹 and the right of point 𝐸, respectively. Problem 8.13

Cross sections 𝐽 and 𝐾, located immediately to the left of point 𝐸 and the right of point 𝐶, respectively.

1.6 𝑂

𝐻

𝐷

𝐹 𝐶 𝑃 𝑄

Problem 8.14

Cross sections 𝐿 and 𝑂, located immediately to the left of point 𝐶 and the right of point 𝐷, respectively.

𝐸

𝑅

1.2

𝐵 2.6

Problem 8.15

Cross sections 𝑄 and 𝑃 , located immediately below and above point

𝐵, respectively. Problem 8.16 Cross section 𝑅 located at the midpoint of member 𝐵𝐸. Also determine the internal forces (i.e., the reactions) at the built-in support at 𝐴.

𝐺

0.2

2.8

0.9 𝐿 𝐾 𝐽 𝐼

𝐴

Figure P8.12–P8.16

𝑊

502

Chapter 8

Internal Forces Problems 8.17 through 8.20 𝐸

𝐷 𝐿

𝐾

𝐽

A machine for lifting heavy objects on an assembly line is shown. It consists of a straight member 𝐴𝐵𝐶, a quarter-circular member 𝐶𝐷𝐸, and a hydraulic cylinder 𝐵𝐷. In the position shown, the hydraulic cylinder is vertical. If 𝑊 = 800 lb, and neglecting the weight of the components of the machine, determine the internal forces acting on:

𝑀

𝐼

Problem 8.17

𝐶

Cross sections 𝐹 and 𝐺, located immediately to the left of point 𝐴 and to the right of point 𝐵, respectively.

𝐺 𝐻

𝐹

𝐵

2 ft

30◦

Problem 8.18 Cross sections 𝐻 and 𝐼, located immediately to the left of point 𝐵 and to the right of point 𝐶, respectively.

𝐴

2 ft Problem 8.19 Cross sections 𝐽 and 𝐾, located immediately above point 𝐶 and below and to the left of point 𝐷, respectively.

𝑊 Figure P8.17–P8.20

Problem 8.20 Cross sections 𝐿 and 𝑀, located immediately to the right and above point 𝐷 and to the left of point 𝐸, respectively.

𝑃

𝑃

𝐴

Problem 8.21

𝑦 𝐴

𝐶

𝑥

𝜃

𝐶

2𝑟

Two structural members are shown; one is straight and the other is semicircular. Neglect the weight of the members. (a) For the straight member, show that the internal forces acting on cross section 𝐶 are 𝑉 = 0, 𝑁 = −𝑃 , and 𝑀 = 0.

𝑟 𝐵

(b) For the semicircular member, show that the internal forces acting on cross section 𝐶 are 𝑉 = −𝑃 cos 𝜃, 𝑁 = −𝑃 sin 𝜃, and 𝑀 = −𝑃 𝑟 sin 𝜃.

𝐵

Figure P8.21 and P8.22

Problem 8.22 𝑦

20 in.

16 in. 𝑧

𝐴

𝐵 12 in.

𝑂 𝐶

𝑥 18 in.

𝑃 = 50 lb

Both the straight and semicircular members shown in Fig. P8.22 are two-force members, and hence either can be used as members of a truss. However, it is most common that truss members are straight. Compare the answers to Prob. 8.21 (which are given in the problem description) to argue why straight truss members are preferable to curved truss members. Note: Concept problems are about explanations, not computations.

Problems 8.23 and 8.24

𝐷

Determine the internal forces acting on: 𝐹 = 200 lb

Figure P8.24 𝑧

𝐶 𝐵

120 N 𝑥

Cross section 𝐾 in Fig. 1 of Example 8.3 on p. 499.

Problem 8.24

The cross section at 𝐴 in Example 4.3 on p. 212, shown again here.

Problems 8.25 through 8.28

𝐷

25 mm 45 mm 𝐴

Problem 8.23

30 mm

One of the cranks of a child’s bicycle is shown. The entire crank lies in the 𝑥𝑦 plane. Determine the internal forces acting on:

15 mm Problem 8.25

Cross section 𝐴.

Problem 8.26

Cross section 𝐵.

Problem 8.27

Cross section 𝐶.

Problem 8.28

Cross section 𝐷, which is located immediately next to the sprocket.

𝑦 120 mm

70 mm Figure P8.25–P8.28

ISTUDY

ISTUDY

Section 8.2

8.2

Internal Forces in Straight Beams

503

Internal Forces in Straight Beams

In this section and the next, we focus on determination of the internal forces in straight beams in two dimensions. In this section, we use equilibrium concepts to accomplish this, while in Section 8.3 we use differential equations.

𝑤

𝑦 (a)

𝑥

Determination of 𝑉 and 𝑀 using equilibrium Our goal in this section is the determination of the internal forces everywhere throughout a straight beam, and we will accomplish this using the equilibrium approach. We begin by assigning an 𝑥𝑦 coordinate system, as shown in Fig. 8.5(a), where 𝑥 is along the axis of the beam. Usually, the 𝑥 = 0 position will be at one of the ends of the beam, and the 𝑦 = 0 position is taken to coincide with the centroid of the beam’s cross section. With this coordinate system, we define positive distributed force 𝑤 to act in the −𝑦 direction. The cross section shown in Fig. 8.5(a) is located at position 𝑥, and the directions for positive internal forces on this cross section are defined in Fig. 8.5(b). In contrast to the previous section, where we found the internal forces at only selected cross sections, here we are interested in determining the internal forces everywhere. Hence, in this section, we will draw multiple FBDs as needed by taking cuts at arbitrary positions 𝑥, and we will use equilibrium equations to determine the internal forces 𝑁, 𝑉 , and 𝑀, which in general will be functions of position 𝑥. There is considerable advantage to knowing the internal forces as functions of position. For one, it allows for easy plotting of internal forces, as discussed below. Furthermore, when you study mechanics of materials, you will see that all aspects of the behavior of a beam are governed by differential equations (some aspects of this are discussed in Section 8.3). Thus, if the shear and/or moment distribution is known, then it will be possible to determine behavior such as deflections of a deformable beam, reactions (including reactions for statically indeterminate beams), and more. All of these applications require a sign convention that must be rigorously followed, and we will use the sign convention defined in Fig. 8.5. However, other sign conventions are common, and if you consult other references such as textbooks, technical papers, or handbooks, you may see other sign conventions used.

Shear and moment diagrams Shear and moment diagrams are plots of the shear 𝑉 and moment 𝑀 as functions of position 𝑥. These diagrams help us understand how the internal forces change throughout a beam and show locations where these have large values. This knowledge helps us to design a beam that has sufficient strength, and drawing shear and moment diagrams is a routine part of the design process. For some applications, plots of the axial force are also important, although in this book our main focus is on the shear and moment.

End of Section Summary In this section the shear and moment at all locations in a straight beam are determined. The equilibrium approach is used where cuts are taken at arbitrary (variable) positions as needed, FBDs are drawn, and equilibrium equations are written. Finally, the usefulness of shear and moment diagrams is discussed.

𝑤

𝑦 (b)

𝑀

𝑀 𝑁

𝑥

𝑁 𝑉

𝑉

Figure 8.5 Internal forces that develop on a particular cross section of a straight beam in two dimensions. (b) Sign convention for positive 𝑁, 𝑉 , and 𝑀.

Helpful Information Forces in the 𝒙 direction. In addition to the forces shown in Fig. 8.5(a), it is possible to have point forces and distributed forces that act in the 𝑥 direction. If this is the case, then these forces must also be included in all FBDs that are drawn (Prob. 8.57 explores this situation). Nonetheless, beams are very often used in a horizontal position with most of the forces being due to gravity, and hence the situation depicted in Fig. 8.5, while a special case, is very common and is the focus in this chapter. Further, since distributed loads usually arise due to gravity, it is convenient to take positive distributed load 𝑤 to be downward (−𝑦 direction).

504

Chapter 8

Internal Forces

E X A M P L E 8.4

Internal Forces in a Straight Beam with a Distributed Load A simply supported beam with an overhang is subjected to a uniformly distributed load. Determine the shear and moment throughout the beam, and plot these versus position.

𝑦 100 lb∕f t 𝐴

𝑥 4 ft

SOLUTION

𝐶

𝐵

Road Map

We begin by drawing an FBD of the entire beam and writing equilibrium equations to determine the reactions at 𝐴 and 𝐵. To determine the internal forces throughout the beam, we will take a cut through a cross section (subsequently called cross section 𝐷), where this cross section has general (or variable) position in the beam. We will draw FBDs as needed and apply equilibrium equations to determine the shear and moment as functions of position.

2 ft

Figure 1 𝑦

(100 lb∕f t) (6 f t) = 600 lb 3 ft

𝐴𝑥

Modeling The FBD for the entire beam is shown in Fig. 2, where the distributed force has been replaced by an equivalent force consisting of a single 600 lb force placed at the centroid of the distributed force’s shape (i.e., centroid of a rectangle).

𝑥 4 ft

2 ft

𝐴𝑦

𝐵𝑦

Figure 2 Free body diagram for the beam to determine the support reactions.

Common Pitfall

𝐴 0

𝑉 𝐴𝑦 = 150 lb 𝑥

(100 lb∕f t) (6 f t − 𝑥) 𝐷

𝑁

𝐶

𝑥

𝑉 𝐵𝑦 = 450 lb (6 f t − 𝑥)

Figure 3 Free body diagrams when cross section 𝐷 is between points 𝐴 and 𝐵; i.e., 0 ≤ 𝑥 ≤ 4 f t.

ISTUDY

Fig. 2, we write and 𝐵𝑦 = 450 lb,

(1)

𝐴𝑦 = 150 lb,

(2)

𝐴𝑥 = 0.

(3)

To determine the internal forces throughout the beam, we consider cross section 𝐷, whose location is variable throughout the beam. As was done in Section 8.1, to determine the internal forces acting on cross section 𝐷, we take a cut that passes through this cross section, followed by construction of FBDs. Thus, when cross section 𝐷 is located between points 𝐴 and 𝐵, the two FBDs that result are shown in Fig. 3. A few important points regarding Fig. 3 are as follows: • The length of the beam in the left-hand FBD is 𝑥, hence the force on this portion of the beam due to the distributed load is (100 lb∕f t) 𝑥. Similarly, the length of the beam in the right-hand FBD is 6 f t − 𝑥, hence the force on this portion of the beam due to the distributed load is (100 lb∕f t)(6 f t − 𝑥). • The directions for the internal forces on both the left-hand and right-hand FBDs match the sign convention shown in Fig. 8.5(b) on p. 503. Governing Equations & Computation

𝑀

𝑀 𝐷

Using the FBD shown in solve the following equilibrium equations to obtain the reactions. ∑ 𝑀𝐴 = 0 ∶ −(600 lb)(3 f t) + 𝐵𝑦 (4 f t) = 0 ⇒ ∑ 𝐹𝑦 = 0 ∶ −600 lb + 𝐴𝑦 + 𝐵𝑦 = 0 ⇒ ∑ 𝐹𝑥 = 0 ∶ 𝐴𝑥 = 0 ⇒ Modeling

Equivalent forces. The 600 lb equivalent force shown in Fig. 2 may be used only for determination of the support reactions 𝐴𝑥 , 𝐴𝑦 , and 𝐵𝑦 . For determinations of the internal forces in Figs. 3 and 4, the original 100 lb∕f t distributed loading must be used. Example 8.5 offers additional explanation.

𝑦 (100 lb∕f t) 𝑥

Governing Equations & Computation

Either of the FBDs shown in Fig. 3 may be used to determine the internal forces acting on cross section 𝐷, and we will use the left-hand FBD. With some experience, we could have anticipated using this FBD, and we would have omitted drawing the FBD of the right-hand portion of the structure. We write and solve the following equations. ∑ 𝐹𝑦 = 0 ∶ −𝑉 − (100 lb∕f t)𝑥 + 150 lb = 0 (4) 𝑉 = 150 lb − (100 lb∕f t)𝑥, 𝑥 𝑀 + (100 lb∕f t)𝑥 − (150 lb)𝑥 = 0 2 ⇒

∑

𝑀𝐷 = 0 ∶

⇒ ∑

𝐹𝑥 = 0 ∶

𝑁 =0

𝑀 = (150 lb)𝑥 − (50 lb∕f t)𝑥2 , ⇒

𝑁 = 0.

(5) (6) (7) (8)

The results obtained in Eqs. (5), (7), and (8) are valid as long as the FBD is valid, which is all cross sections whose 𝑥 coordinate lies in the region 0 ≤ 𝑥 ≤ 4 f t.

ISTUDY

Section 8.2

𝑦 (100 lb∕f t) 𝑥

Modeling

When cross section 𝐷 is located between points 𝐵 and 𝐶, the FBDs that result are shown in Fig. 4. Notice that reaction 𝐵𝑦 now appears on the left-hand FBD. Either of the FBDs shown in Fig. 4 may be used to determine the internal forces acting on cross section 𝐷, and we will use the right-hand FBD. We write and solve the following equations. ∑ 𝐹𝑦 = 0 ∶ 𝑉 − (100 lb∕f t)(6 f t − 𝑥) = 0 (9) ⇒ 𝑀𝐷 = 0 ∶ ⇒ ∑

𝐹𝑥 = 0 ∶

𝑉 = 600 lb − (100 lb∕f t)𝑥, −𝑀 − (100 lb∕f t)(6 f t − 𝑥)

6 ft − 𝑥 =0 2

𝑀 = −1800 f t ⋅lb + (600 lb)𝑥 − (50 lb∕f t)𝑥2 , 𝑁 =0

⇒

𝑁 = 0.

(10) (11)

𝑀 0

𝑀 𝐶

𝑁 𝑉

𝑉 𝐴𝑦 = 150 lb 𝐵𝑦 = 450 lb

𝑥

𝐷

6 ft − 𝑥

𝑥

Figure 4 Free body diagrams when cross section 𝐷 is between points 𝐵 and 𝐶; i.e., 4 f t ≤ 𝑥 ≤ 6 f t.

(12)

Helpful Information

(13)

The results obtained in Eqs. (10), (12), and (13) are valid as long as the FBD is valid, which is for all cross sections whose 𝑥 coordinate lies in the region 4 f t ≤ 𝑥 ≤ 6 f t. Discussion & Verification The results for shear 𝑉 and moment 𝑀 are plotted in Fig. 5, where Eqs. (5) and (7) are used for 0 ≤ 𝑥 ≤ 4 f t, and Eqs. (10) and (12) are used for 4 f t ≤ 𝑥 ≤ 6 f t. These plots display some interesting features of internal forces:

• The shear at 𝑥 = 0 is equal to the support reaction 𝐴𝑦 . • The change in shear at 𝑥 = 4 f t is equal to the support reaction 𝐵𝑦 . That is, the shear just to the right of 𝐵 (200 lb) minus the shear just to the left of 𝐵 (−250 lb) is equal to 𝐵𝑦 (450 lb).

𝑽 at 𝒙 = 𝟒 𝐟 𝐭. The careful reader will notice that Eq. (5) is valid only for 0 ≤ 𝑥 < 4 f t and Eq. (10) is valid only for 4 f t < 𝑥 ≤ 6 f t, because at 𝑥 = 4 f t, 𝑉 is undefined (it undergoes a discontinuity as shown in Fig. 5). Nonetheless, as a matter of practice, in this chapter we will use ≤ signs for the regions of validity for 𝑉 and 𝑀 expressions with the understanding that there will often be a discontinuity in 𝑉 and/or 𝑀 at interval endpoints.

• The moment at the right-hand end is zero, because this end is unsupported and has no concentrated moment applied. • The largest moment throughout the beam is −200 f t ⋅lb, which occurs at 𝑥 = 4 f t. In mechanics of materials, this value will play an extremely important role in the design of the beam (i.e., determination of the material and cross-sectional shape for the beam). The values of the shear force are sometimes also important, especially for materials that are weak in shear, such as wood and some composite materials. • The largest value of the moment between 𝐴 and 𝐵 may be of interest. This value can be determined by evaluating Eq. (7), using trial and error with different values of 𝑥 in the neighborhood of about 1.5 f t. However, a local maximum value of moment occurs where 𝑉 = 0 (this is proved in Section 8.3).∗ Thus, we could solve Eq. (5) for the value of 𝑥 that makes 𝑉 = 0 and then substitute this value into Eq. (7) to obtain the moment. If you verify this for yourself, you should find that 𝑉 = 0 at 𝑥 = 1.5 f t and 𝑀 = 112.5 f t ⋅lb at this location.

local maximum value of the moment can also occur at an interval endpoint where 𝑉 ≠ 0, such as at 𝑥 = 4 f t in this example.

moment 𝑀 (f t⋅lb)

• The moment at 𝑥 = 0 is zero, because the support at that location has no moment reaction.

shear 𝑉 (lb)

• The shear at the right-hand end is zero, because this end is unsupported and has no concentrated force applied.

∗A

(100 lb∕f t) (6 f t − 𝑥)

𝐷

𝐴

Governing Equations & Computation

∑

505

Internal Forces in Straight Beams

300 200 100 0 −100 −200 −300

200 lb 150 lb

−250 lb 0

2

300 200 100 0 −100 −200 −300

𝑥 ( f t)

4

6

−200 f t⋅lb 0

2

𝑥 ( f t)

Figure 5 Shear and moment diagrams.

4

6

506

Chapter 8

Internal Forces

E X A M P L E 8.5 𝑦

A Simply Supported Beam with Distributed Load and Point Load Two simply supported beams with different loadings are shown.

𝑦 𝑤0

𝐴

(a) Determine the shear and moment as functions of position for the uniformly distributed loading 𝑤0 shown in Fig. 1(a).

𝑃 𝐵

𝑥

𝐴

𝐵

𝑥 𝐿 2

𝐿 2

𝐿

(c) Let 𝑃 = 𝑤0 𝐿. Plot the shear and moment determined in Parts (a) and (b) and comment on the differences, if any, between these.

(b)

(a)

(b) Determine the shear and moment as functions of position for the point load 𝑃 shown in Fig. 1(b).

Figure 1

SOLUTION Road Map 𝑦

𝑦

𝑤0 𝐿

𝑥 𝐴𝑥

For both Parts (a) and (b), we first determine the support reactions. We then proceed to determine the shear and moment by taking cuts through cross sections of the beam, drawing FBDs as needed, and applying equilibrium equations. Finally, for Part (c) we let 𝑃 = 𝑤0 𝐿 and we compare the results for the shear and moment.

𝑃

𝑥 𝐴𝑥

𝐵𝑦

𝐴𝑦

𝐵𝑦

𝐴𝑦 (b)

(a)

Figure 2 Free body diagrams to determine the support reactions for each beam.

𝑦

∑ ∑ By writing the equations 𝑀 = 0, 𝐹𝑦 = 0, and 𝐹𝑥 = 0, you should verify that the support reactions for the uniformly loaded beam in Fig. 2(a) are 𝐴𝑥 = 0 and 𝐴𝑦 = 𝐵𝑦 = 𝑤0 𝐿∕2. Similarly, you should verify that the support reactions for the beam with a point load in Fig. 2(b) are 𝐴𝑥 = 0 and 𝐴𝑦 = 𝐵𝑦 = 𝑃 ∕2. In anticipation of Part (c), note that if 𝑃 = 𝑤0 𝐿, the two FBDs in Fig. 2 are identical and the support reactions are also identical. Governing Equations & Computation

∑

Modeling 𝑁

𝑥

𝐷 𝑉 𝐴𝑦 = 𝑤0 𝐿∕2 𝑥

Figure 3 Free body diagram for the left-hand portion of the uniformly loaded beam of Fig. 1(a).

ISTUDY

of 𝑤0 𝐿 has been used in place of the original distributed loading.

Part (a)

(𝑤0 ) 𝑥 𝑀

0

Modeling For purposes of determining the support reactions for the beams in Parts (a) and (b), FBDs for each beam are drawn in Fig. 2, where in Fig. 2(a), an equivalent force

For the beam with the distributed load in Fig. 1(a), we determine the internal forces by taking a cut through a cross section to draw the FBD shown in Fig. 3. Governing Equations & Computation Using the FBD shown in Fig. 3, we write and solve the following equations. ∑ ∑

𝐹𝑦 = 0 ∶

𝑀𝐷 = 0 ∶

∑

𝑀 + 𝑤0 𝑥

(

𝐿 =0 2

⇒

𝑉 = 𝑤0

𝑥 𝑤0 𝐿 − 𝑥=0 2 2

⇒

𝑀=

𝑁 =0

⇒

𝑁 = 0.

−𝑉 − 𝑤0 𝑥 + 𝑤0

𝐹𝑥 = 0 ∶

) 𝐿 −𝑥 , 2

𝑤0 ( ) 𝐿𝑥 − 𝑥2 , 2

(1) (2) (3)

The results obtained in Eqs. (1) through (3) are valid as long as the FBD is valid, which is for any cross section between ends 𝐴 and 𝐵 (i.e., 0 ≤ 𝑥 ≤ 𝐿). Part (b) Modeling For the beam with a point load 𝑃 in Fig. 1(b), we determine the internal forces by taking a cut through a cross section to draw the FBDs shown in Fig. 4.

Using the FBD shown in Fig. 4(a) for the region 0 ≤ 𝑥 ≤ 𝐿∕2, we write and solve the following equations.

Governing Equations & Computation

∑ ∑

𝐹𝑦 = 0 ∶

𝑀𝐷 = 0 ∶

𝑃 =0 2

⇒

𝑉 =

𝑃 , 2

(4)

𝑃 𝑥=0 2

⇒

𝑀=

𝑃 𝑥, 2

(5)

−𝑉 + 𝑀−

ISTUDY

Section 8.2 ∑

𝐹𝑥 = 0 ∶

𝑁 =0

𝑁 = 0.

⇒

∑

𝑃 𝑉 =− , 2

(7)

) ( 𝑃 𝐿 − 𝑥=0 𝑀 +𝑃 𝑥− 2 2

⇒

𝑃 𝑀 = (𝐿 − 𝑥), 2

(8)

𝑁 =0

⇒

𝑁 = 0.

(9)

−𝑉 − 𝑃 +

𝑀𝐷 = 0 ∶

∑

⇒

𝐹𝑦 = 0 ∶

𝐹𝑥 = 0 ∶

𝑃 =0 2

𝑦

(6)

Using the FBD shown in Fig. 4(b) for the region 𝐿∕2 ≤ 𝑥 ≤ 𝐿, we write and solve the following equations. ∑

507

Internal Forces in Straight Beams

𝑀 (a)

𝑥

𝑁 𝐷

0

𝑉 𝐴𝑦 = 𝑃 ∕2 𝑥 𝑦

𝑃 𝑀

Part (c) Discussion & Verification

For the beam with uniformly distributed load, the shear and moment given by Eqs. (1) and (2) are plotted in Fig. 5(a), and for the beam with a point load, the shear and moment given by Eqs. (4), (5), (7), and (8) are plotted in Fig. 5(b). due to distributed load 𝑤0

𝑉 𝑃 2

2

𝐿

0 −

𝑤0 𝐿

𝐿

0

𝑥

𝑥

− 𝑃2

2

𝑁 𝐷

0

𝑥

𝑉

𝐴𝑦 = 𝑃 ∕2 𝑥

due to point load 𝑃

𝑉

𝑤0 𝐿

(b)

Figure 4 Free body diagrams for the beam with a point load shown in Fig. 1(b). The first FBD is valid for the region 0 ≤ 𝑥 ≤ 𝐿∕2, and the second is valid for 𝐿∕2 ≤ 𝑥 ≤ 𝐿.

𝑀

𝑀 𝑃𝐿 4

𝑤0 𝐿2 8

𝑥

0 (a)

𝐿

𝑥

0 (b)

𝐿

Figure 5. (a) Shear and moment diagrams for a simply supported beam with uniformly distributed load. (b) Shear and moment diagrams for a simply supported beam with a point load at midspan.

If 𝑃 = 𝑤0 𝐿, then the loadings shown in Fig. 1 are equivalent according to the definition given in Chapter 4 by Eq. (4.16) on p. 248, and as mentioned earlier in this example, the FBDs in Fig. 2 are identical and the support reactions are identical. However, the internal forces shown in Fig. 5 are clearly very different. Observe that with 𝑃 = 𝑤0 𝐿, all of the plots in Fig. 5 are drawn to scale, and among the many differences, the maximum value of the moment for the beam with a concentrated load is twice as large as that for the beam with uniform distributed loading. Part (c) of this example demonstrates some of the subtleties and important limitations of the definition of equivalent force systems given in Chapter 4. To summarize, two force systems that are equivalent, according to the definition given by Eq. (4.16) on p. 248, will produce the same external effects on a rigid structure (e.g., the support reactions will be the same), but internal effects (e.g., internal forces such as the shear and moment) are not necessarily the same.

Common Pitfall When are equivalent force systems really equivalent? A common error when determining internal forces is to replace a distributed force, such as 𝑤0 shown in Fig. 1(a), with an equivalent force, such as 𝑃 = 𝑤0 𝐿 shown in Fig 1(b). This replacement is valid only for purposes of determining forces that are external to the FBD, such as support reactions. For determination of internal forces, the original loading must be used.

508

Chapter 8

Internal Forces

E X A M P L E 8.6

Superposition For the cantilever beam shown, determine the shear and moment as functions of position, and draw the shear and moment diagrams.

0.3 kN∕m

𝑦

1 kN

𝐴

𝑥

𝐵

2m

𝐶 2 kN⋅m 2m 2m

SOLUTION

𝐷

Road Map

We could follow the procedure used in earlier examples in this section, wherein we would take a cut in Fig. 1 between points 𝐴 and 𝐵, draw an FBD and apply equations of equilibrium to determine the internal forces in that region, and then repeat this for cuts between 𝐵 and 𝐶, and between 𝐶 and 𝐷. However, we will employ an alternative solution using superposition where we break the original loading into simpler load cases, as shown in Fig. 2. Each of the load cases is analyzed independently, and then the total shear and moment are obtained by adding those for each of the load cases.

Figure 1

Load case 1

Load case 2

Load case 3

Total Loads 0.3 kN∕m

0.3 kN∕m

𝑦

1 kN

1 kN

𝐴

𝑥 2m

𝐵

𝐷 2m

𝐴

𝐶 2 kN⋅m

2m

shear 𝑉

shear 𝑉 𝑥

0 −1 kN

𝐷

𝐴

𝐷

𝐴

0 −1 kN

𝐶 2 kN⋅m

𝐷

shear 𝑉 (total)

shear 𝑉 𝑥

𝐵

𝑥

0 −1 kN

𝑥

0 −1 kN

−0.9 kN

−1.9 kN

−2 kN 2m 4m 6m

0

moment 𝑀 𝑥

0

0

2m 4m 6m

moment 𝑀 2 kN⋅m 0

0

2m 4m 6m

0

0

moment 𝑀 𝑥

2m 4m 6m

−1.8 kN⋅m 0

2m 4m 6m

moment 𝑀 (total) 𝑥

0 −2 kN⋅m

−4 kN⋅m 0

2m 4m 6m

0 −2 kN⋅m −4 kN⋅m

2m 4m 6m

𝑥 −3.8 kN⋅m 0 2m 4m 6m

Figure 2. Selection of three load cases to be used for superposition. The shear and moment for each load case as found in Eqs. (1)–(3) are shown. The total shear and moment as given by Eqs. (4) and (5) are also shown.

Load case 1 Modeling 𝑦

For the 1 kN force applied at 𝐵, the shear and moment between 𝐴 and 𝐵 are clearly zero. For the region between 𝐵 and 𝐷, the FBD is shown in Fig. 3.

1 kN 𝑀

𝐴

𝐵 𝑥

Governing Equations & Computation 𝑁

solve the equilibrium equations {

𝑉

Figure 3 Free body diagram for load case 1.

ISTUDY

𝑥

Load case 1:

𝑉 = 0,

∑

Using the FBD in Fig. 3, you should write and ∑ 𝐹𝑦 = 0 and 𝑀 = 0 to obtain 𝑀 = 0,

𝑉 = −1 kN, 𝑀 = 2 kN⋅m − (1 kN)𝑥,

Plots of 𝑉 and 𝑀 in Eq. (1) are shown in Fig. 2.

for 0 ≤ 𝑥 ≤ 2 m,

for 2 m ≤ 𝑥 ≤ 6 m.

(1)

ISTUDY

Section 8.2

509

Internal Forces in Straight Beams Load case 2

𝑦

Modeling

For the 2 kN⋅m moment applied at 𝐶, the shear and moment between 𝐴 and 𝐶 are clearly zero. For the region between 𝐶 and 𝐷, the FBD is shown in Fig. 4.

𝑀 𝐴

Governing Equations & Computation

Using the FBD in Fig. 4, you should write and ∑ ∑ solve the equilibrium equations 𝐹𝑦 = 0 and 𝑀 = 0 to obtain { Load case 2:

𝑉 = 0, 𝑀 = 0,

for 0 ≤ 𝑥 ≤ 4 m,

𝑉 = 0, 𝑀 = 2 kN⋅m, for 4 m ≤ 𝑥 ≤ 6 m.

(2)

𝑥

𝑦

Load case 3

𝑃

Modeling

Governing Equations & Computation

To determine the force 𝑃 in the FBD of Fig. 5, first we write an expression for the distributed load as 𝑤 = (0.05 kN∕m2 )𝑥. This expression is easily verified by noting that 𝑤 = 0 when 𝑥 = 0 and 𝑤 = 0.3 kN∕m when 𝑥 = 6 m (for tips on developing expressions for linear functions, see the Helpful Information margin note on p. 456). The force 𝑃 due to the distributed load is the “area” of the triangular distributed force (i.e., 1/2 the “base” 𝑥 multiplied by the “height” (0.05 kN∕m2 )𝑥), hence 𝑃 = (0.025 kN∕m2 )𝑥2 . ∑ Using the FBD in Fig. 5, you should write and solve the equilibrium equations 𝐹𝑦 = ∑ 0 and 𝑀 = 0 to obtain { ( ) ) ( kN 1 kN Load 0.025 2 𝑥3 , for 0 ≤ 𝑥 ≤ 6 m. (3) 𝑉 = − 0.025 2 𝑥2 , 𝑀 = − case 3: 3 m m Plots of 𝑉 and 𝑀 in Eq. (3) are shown in Fig. 2. Superposition for total 𝑉 and 𝑀 Discussion & Verification The total shear 𝑉 and moment 𝑀 are obtained by adding the results given in Eqs. (1), (2), and (3), paying careful attention to use the appropriate expressions for the various regions for 𝑥. Hence,

Total shear:

( ) ⎧𝑉 = − 0.025 kN 𝑥2 , ⎪ m2 ( ) ⎨ ⎪𝑉 = −1 kN − 0.025 kN 𝑥2 , ⎩ m2

for 0 ≤ 𝑥 ≤ 2 m, for 2 m ≤ 𝑥 ≤ 6 m,

) ( ⎧𝑀 = − 1 0.025 kN 𝑥3 , for 0 ≤ 𝑥 ≤ 2 m, 3 m2 ⎪ ( ) 1 kN Total ⎪ 0.025 2 𝑥3 , for 2 m ≤ 𝑥 ≤ 4 m, ⎨𝑀 = 2 kN⋅m − (1 kN)𝑥 − moment: ⎪ 3 m ( ) ⎪𝑀 = 4 kN⋅m − (1 kN)𝑥 − 1 0.025 kN 𝑥3 , for 4 m ≤ 𝑥 ≤ 6 m. ⎩ 3 m2

(4)

(5)

The shear and moment diagrams for the total loading are obtained by summing the shear and moment diagrams for each load case, as shown in Fig. 2.

𝑥

Figure 4 Free body diagram for load case 2.

Plots of 𝑉 and 𝑀 in Eq. (2) are shown in Fig. 2. For the linearly increasing distributed load, we take a cut between points 𝐴 and 𝐷 to draw the FBD shown in Fig. 5.

𝑁

𝐶 2 kN⋅m 𝑉

𝑥 3 𝑀 𝑁

𝐴 𝑉 𝑥

Figure 5 Free body diagram for load case 3.

𝑥

510

Chapter 8

Internal Forces

Problems Problem 8.29 For the simply supported beam shown, let 𝑎 = 6 f t, 𝑏 = 4 f t, and 𝑃 = 1000 lb. Determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝑦

𝑃

𝐴

Problem 8.30

𝐵

𝑥

𝐶

Repeat Prob. 8.29, using 𝑎 = 3 m, 𝑏 = 1 m, and 𝑃 = 10 kN. Determine the shear and moment as functions of position, and draw the shear and moment diagrams.

𝑏

𝑎 Figure P8.29–P8.31

Problem 8.31 For arbitrary values of 𝑃 , 𝑎, 𝑏, determine the shear and moment as functions of position. Express your answers in terms of parameters such as 𝑃 , 𝑎, 𝑏, etc. Draw the shear and moment diagrams. Hint: The answers to this problem are given in the statement of Prob. 8.54.

Problem 8.32 A simply supported beam with two equal forces applied equidistant from the supports is called four-point bending. This loading arrangement is commonly used for experimentally determining the strength of beams in a laboratory. (a) Determine the shear and moment as functions of position. Express your answers in terms of parameters such as 𝑃 , 𝐿, etc. Draw the shear and moment diagrams. (b) Comment on any interesting features the shear and moment display in the region between points 𝐵 and 𝐶. 𝑦 𝐴

𝑥 𝐿 3

𝑃

𝑃

𝐵

𝐶

𝐿 3

𝐷

𝐿 3

Figure P8.32

Problems 8.33 and 8.34 A simply supported beam with a 2000 ft⋅lb moment is shown. Determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝑦 𝐴

𝑦 2000 f t⋅lb 𝐶

𝑥 16 f t

𝑦

Figure P8.33

1m Figure P8.35

ISTUDY

𝐶

𝐵

𝐴

3m

𝑥

𝐴

2000 f t⋅lb 𝑥 𝐵 8 ft

𝐶

8 ft

Figure P8.34

Problem 8.35 A diver stands on the end of a diving board. If the diver’s mass is 70 kg, determine the shear and moment as functions of position, and draw the shear and moment diagrams.

ISTUDY

Section 8.2

Internal Forces in Straight Beams

511

Problems 8.36 through 8.39 For the cantilever beam shown, determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝑦

𝑦

1000 lb

𝐴

𝑥

𝐴

𝐶

𝑦 500 lb 𝑥 10 f t

20 f t Figure P8.36

8 kN∕m

500 lb 𝐶

𝐵

𝑦

𝐴

10 kN∕m 6 kN∕m 𝑥

𝐴

𝐶

𝑥

3m

6m

10 f t

Figure P8.37

Figure P8.38

𝐶

𝐵 3m

Figure P8.39 𝑦

Problem 8.40

600 N∕m

A simply supported beam has the linear distributed load shown. Determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝐴

𝐵

𝑥 6m

Problem 8.41 A beam with an overhang is subjected to the uniformly distributed load shown. Determine the shear and moment as functions of position, and draw the shear and moment diagrams.

Figure P8.40

𝑦 500 lb∕f t

Problem 8.42

𝐴

Replace the distributed load of Prob. 8.41 with a single force so that the two force systems are equivalent. Determine the shear and moment as functions of position (it should be possible to do this by inspection). Even though the loadings are equivalent, do you expect the results for this problem to be the same as those for Prob. 8.41? Note: Concept problems are about explanations, not computations.

𝑥 9 ft

𝑦

One of the beams that supports a balcony is shown. To design a beam for this purpose, it is common to use a uniformly distributed load that includes the dead loads (e.g., weight of materials) and live loads (e.g., weight of a large, but reasonable number of people distributed over the balcony). If the uniformly distributed load is 2500 N∕m, determine the shear and moment as functions of position, and draw the shear and moment diagrams. Idealize the supports at 𝐵 and 𝐶 to be a roller and pin, respectively.

𝐴

𝑥 1m

Figure P8.43

Problems 8.44 and 8.45 The wooden beam shown can support an allowable absolute maximum shear of 3000 lb and an allowable absolute maximum moment of 6000 f t ⋅lb. Determine the largest value of 𝑤0 that is allowable. 𝑦

𝑦

2𝑤0

𝑤0 𝑥 3 ft

𝐶

Figure P8.41

Problem 8.43

𝐴

𝐵 9 ft

𝑤0 𝐶

𝐵 10 f t

Figure P8.44

𝐴

𝐵

𝑥 12 f t

6 ft

Figure P8.45

𝐶

𝐵 2m

512

Chapter 8

Internal Forces Problems 8.46 through 8.49

In Example 4.11 on p. 252, numerous equivalent force systems for a cantilever beam were developed. For each of the following force systems, determine the shear and moment as functions of position, and draw the shear and moment diagrams: Problem 8.46

Figure 1 on p. 252.

Problem 8.47

Figure 2 on p. 252.

Problem 8.48

Figure 3 on p. 252.

Problem 8.49

Figure 4 on p. 252 with 𝑑 = 2.67 mm.

Problem 8.50 Without solving Probs. 8.46 through 8.49, comment on the agreement you expect between the shear and moment distributions for each of these load cases. Are there particular points in the beam where you know the shear and moment must be the same for all of these loadings? Note: Concept problems are about explanations, not computations.

Problems 8.51 and 8.52 Consider the simply supported beam shown with a uniformly distributed load and a force at midspan. Use superposition of the results from Example 8.5 on p. 506 to determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝑦

𝑦

6N

2 kip 100 lb∕f t

0.5 N∕mm 𝐴

10 mm

𝐴

𝐵

𝑥

6 ft

10 mm

Figure P8.51

𝐵

𝑥 6 ft

Figure P8.52

15 kN∕m

𝑦

Problem 8.53 𝐴

𝑥

2m

𝐷 𝐵 3 kN 4m

A wing of a jet is crudely modeled as a beam with the loadings shown. Use superposition of the results from Example 8.6 on p. 508 to determine the shear and moment in the wing as functions of position, and draw the shear and moment diagrams.

Figure P8.53

Problem 8.54

𝑦 𝐴

900 N 𝑥 1m

Figure P8.54

ISTUDY

𝐵 1m

1200 N 𝐶 1m

𝐷

A simply supported beam is subjected to the 900 N and 1200 N forces shown. Use superposition to determine the shear and moment as functions of position, and draw the shear and moment diagrams. Hint: The answers to Prob. 8.31 on p. 510, given below, are helpful for this problem. 𝑃𝑏 , 𝑎+𝑏 𝑃𝑎 , 𝑉 =− 𝑎+𝑏 𝑉 =

𝑃𝑏 𝑥, 𝑎+𝑏 ( 𝑀 = 𝑃𝑎 1 −

𝑀=

for

0 ≤ 𝑥 ≤ 𝑎,

for

𝑎 ≤ 𝑥 ≤ 𝑎 + 𝑏.

)

𝑥 , 𝑎+𝑏

ISTUDY

Section 8.2

513

Internal Forces in Straight Beams

Problem 8.55 𝑦

A person uses a wrench to apply a force 𝐹𝐴 and a moment 𝑀𝐴 to the end of a cantilever beam. The weight of the beam is represented by the uniform distributed load 𝑤0 .

(a) For 𝐹𝐴 ≠ 0, 𝑀𝐴 = 0, and 𝑤0 = 0, determine the shear and moment as functions of position. Express your answers in terms of 𝐹𝐴 .

(b) For 𝐹𝐴 = 0, 𝑀𝐴 ≠ 0, and 𝑤0 = 0, determine the shear and moment as functions of position. Express your answers in terms of 𝑀𝐴 . (c) For 𝐹𝐴 = 0, 𝑀𝐴 = 0, and 𝑤0 ≠ 0, determine the shear and moment as functions of position. Express your answers in terms of 𝑤0 .

𝑤0

𝐹𝐴 𝑥 𝐴

𝐵

𝑀𝐴 𝐿

Figure P8.55

(d) If 𝐹𝐴 = 20 lb, 𝑀𝐴 = 200 in.⋅lb, 𝑤0 = 0.5 lb∕in., and 𝐿 = 30 in., use superposition of the results of Parts (a) through (c) to determine the shear and moment as functions of position, and draw the shear and moment diagrams.

Problem 8.56 A cross section through a railroad bed is shown. The rails are supported by ties that are made of wood or sometimes concrete, and the ties rest on ballast, which is usually crushed stone. Assuming the ballast applies a uniformly distributed load to the ties, determine the shear and moment in a tie due to the 10 kip forces and the distributed load from the ballast as functions of position. Draw the shear and moment diagrams. 10 kip

10 kip

rails 20 in.

59 in.

tie 20 in.

ballast

Figure P8.56

𝑦 𝐴

Problem 8.57

𝑥

One of the beams of a staircase is to support a 200 lb∕f t uniformly distributed vertical force. Determine the axial force, shear, and moment as functions of position and draw the normal force, shear, and moment diagrams.

15 f t

Figure P8.57

200 lb∕f t

3

4

𝐵

514

Chapter 8

Internal Forces

8.3 𝑦 𝑤

(a)

𝑥

𝑉

(b)

𝑀 + 𝛥𝑀

𝑀 𝑥

𝑁

𝐴 𝑉 + 𝛥𝑉 𝛥𝑥

Figure 8.6 (a) A straight beam subjected to a distributed force 𝑤. (b) An FBD of a small length Δ𝑥 of the beam. The directions for positive 𝑉 , 𝑀, and 𝑤 follow the sign conventions given in Fig. 8.5 on p. 503.

ISTUDY

In this section we develop differential equations that relate the distributed force, shear, and moment for a straight beam with transverse loads. These relations are useful in statics and are also useful in mechanics of materials, where they are supplemented with additional differential equations that describe other aspects of the behavior of beams.

Relations among 𝑉 , 𝑀, and 𝑤

𝑤 𝛥𝑥

𝑦

Relations Among Shear, Moment, and Distributed Force

Consider the straight beam shown in Fig. 8.6(a); the supports for the beam and/or other forces that might be applied to it are not important for our purposes here, other than we assume the beam is in equilibrium. We take cuts through two cross sections of this beam, one at position 𝑥 and the other at 𝑥 + Δ𝑥, and the FBD that results is shown in Fig. 8.6(b). The internal forces on the cross section at 𝑥 are the normal (axial) force 𝑁, shear force 𝑉 , and moment 𝑀. The shear and moment on the cross section at 𝑥 + Δ𝑥 may be different, and these are 𝑉 + Δ𝑉 and 𝑀 + Δ𝑀. If Δ𝑥 is small, then the distributed force 𝑤 in Fig. 8.6(b) is approximately uniform, and the total force is therefore 𝑤 Δ𝑥. The small piece of beam in the FBD of Fig. 8.6(b) is in equilibrium, hence we may write ∑ 𝐹𝑦 = 0 ∶ 𝑉 − (𝑉 + Δ𝑉 ) − 𝑤 Δ𝑥 = 0. (8.1) Rearranging the above expression and dividing by the length Δ𝑥 provide Δ𝑉 = −𝑤. Δ𝑥

(8.2)

Taking the limit of Eq. (8.2) as Δ𝑥 → 0 provides 𝑑𝑉 = −𝑤. 𝑑𝑥

Helpful Information Drawing shear and moment diagrams. Equations (8.3) and (8.6) are useful for drawing shear and moment diagrams. In words, Eq. (8.3) says, “The slope of the shear diagram is equal to the negative of the distributed force’s value,” and Eq. (8.6) says, “The slope of the moment diagram is equal to the value of the shear.” Example 8.9 on p. 522 illustrates the usefulness of this.

(8.3)

In words, Eq. (8.3) says, “The change in shear divided by the change in length of the beam at position 𝑥 is equal to the negative of the distributed force at that location.” Summing moments about point 𝐴 in Fig. 8.6(b) provides ∑

𝑀𝐴 = 0 ∶

−𝑀 + 𝑀 + Δ𝑀 − 𝑉 Δ𝑥 + 𝑤 Δ𝑥

Δ𝑥 = 0. 2

(8.4)

Rearranging the above expression and dividing by the length Δ𝑥 provides Δ𝑥 Δ𝑀 =𝑉 −𝑤 . Δ𝑥 2

(8.5)

Taking the limit of Eq. (8.5) as Δ𝑥 → 0 provides 𝑑𝑀 =𝑉. 𝑑𝑥

(8.6)

In words, Eq. (8.6) says, “The change in moment divided by the change in length of the beam at position 𝑥 is equal to the shear at that location.”

ISTUDY

Section 8.3

Relations Among Shear, Moment, and Distributed Force

Determination of 𝑉 and 𝑀 using integration Equations (8.3) and (8.6) are differential equations that can be solved to obtain the shear 𝑉 and moment 𝑀, given the distributed force 𝑤. We call this the integration approach, and it is developed by multiplying both sides of Eq. (8.3) by 𝑑𝑥 to write 𝑑𝑉 = −𝑤 𝑑𝑥.

(8.7)

Integrating both sides of the above expression provides 𝑉

∫

𝑥

𝑑𝑉 = −

∫

𝑤 𝑑𝑥,

(8.8)

𝑥𝑃

𝑉𝑃

where the limits of integration∗ for the right-hand integral are chosen to be 𝑥𝑃 to 𝑥. Thus, the limits of integration for the left-hand integral are 𝑉𝑃 to 𝑉 , where 𝑉𝑃 is the shear at position 𝑥𝑃 and 𝑉 is the shear at position 𝑥. Evaluating the integral on the left-hand side of Eq. (8.8) and adding 𝑉𝑃 to both sides provides 𝑥

𝑉 = 𝑉𝑃 −

∫

𝑤 𝑑𝑥.

(8.9)

𝑥𝑃

Similarly, Eq. (8.6) may be written as 𝑑𝑀 = 𝑉 𝑑𝑥,

(8.10)

which upon integration gives 𝑥

𝑀 = 𝑀𝑃 +

∫

𝑉 𝑑𝑥.

(8.11)

𝑥𝑃

Thus, if the distributed force 𝑤 is known as a function of position, then by integration using Eq. (8.9), the shear as a function of position is obtained. Once the shear is known, then by integration using Eq. (8.11) the moment as a function of position is obtained. Remarks • When using Eqs. (8.9) and (8.11), we select the location of point 𝑃 , whose coordinate is 𝑥𝑃 . Point 𝑃 will be a convenient location where the shear 𝑉𝑃 and moment 𝑀𝑃 are known or can be easily determined. • Often, point 𝑃 in Eqs. (8.9) and (8.11) will be taken to be at one of the ends of the beam, because these are locations where it will often be straightforward to determine the shear 𝑉𝑃 and moment 𝑀𝑃 . For example, consider the common situation of a horizontal beam of length 𝐿 with the origin of the coordinate system at the left-hand end. Taking point 𝑃 at the left-hand end means 𝑥𝑃 = 0, while taking point 𝑃 at the right-hand end means 𝑥𝑃 = 𝐿. Example 8.8 on p. 520 illustrates the use of both of these choices. we may use indefinite integrals, in which case Eq. (8.8) would be written as ∫ dV = − ∫ 𝑤 𝑑𝑥. Evaluation of the integral on the left-hand side provides 𝑉 = − ∫ 𝑤 𝑑𝑥. Note that when the integral on the right-hand side is evaluated, a constant of integration is produced, which by comparison with Eq. (8.9) is seen to have the physical interpretation of being the shear at some location in the beam.

∗ Alternatively,

515

516

ISTUDY

Chapter 8

Internal Forces

• For some problems, it may be effective to take point 𝑃 to be somewhere between the ends of the beam. • In all of the integrals written in this section, we could replace the limit of integration 𝑥 with the coordinate of a point 𝑄, which is 𝑥𝑄 . Then Eqs. (8.9) and (8.11) become 𝑥𝑄

𝑉𝑄 = 𝑉𝑃 −

∫

𝑥𝑄

and

𝑤 𝑑𝑥 ,

𝑀𝑄 = 𝑀𝑃 +

𝑥𝑃

∫

𝑉 𝑑𝑥 .

(8.12)

𝑥𝑃

⏟⏞⏟⏞⏟

⏟⏞⏟⏞⏟

area under 𝑤 vs. 𝑥 plot between 𝑥𝑃 and 𝑥𝑄

area under 𝑉 vs. 𝑥 plot between 𝑥𝑃 and 𝑥𝑄

The integrals in the above expressions represent, respectively, the area under the 𝑤 vs. 𝑥 plot and the area under the 𝑉 vs. 𝑥 plot, between 𝑥𝑃 and 𝑥𝑄 . These expressions provide a graphical method for drawing shear and moment diagrams, as illustrated in Example 8.9 on p. 522. • Another way to interpret Eqs. (8.9) and (8.11) is that they describe the change of shear Δ𝑉 and the change of moment Δ𝑀 between positions 𝑥𝑃 and 𝑥. Hence, Eqs. (8.9) and (8.11) can be rewritten as 𝑥

Δ𝑉 = −

∫

𝑥

𝑤 𝑑𝑥

𝑥𝑃

and

Δ𝑀 =

∫

𝑉 𝑑𝑥.

(8.13)

𝑥𝑃

As discussed in connection with Eq. (8.12), you may also elect to take 𝑥 = 𝑥𝑄 , in which case Δ𝑉 and Δ𝑀 are the change in shear and moment, respectively, between points 𝑃 and 𝑄. These expressions can be useful for drawing shear and moment diagrams.

Helpful Information Combining approaches. You may choose to use a combination of equilibrium and integration approaches to determine the shear and moment in beams. For example, consider the following beam and loading:

𝑃 𝐵

𝑥 𝐿 2

In this chapter we have discussed two methods for determining shear and moment as functions of position: 1. Equilibrium approach, discussed in Section 8.2, where cuts are taken as needed, FBDs are drawn, and equilibrium equations are written and solved. 2. Integration approach, where the expression for the distributed force 𝑤 is integrated to obtain the shear, which is then integrated to obtain the moment.

𝑦

𝐴

Which approach should I use?

𝐿 2

Rather than use the equilibrium approach exclusively or the integration approach exclusively (both of which are effective for this problem), you could choose to use the equilibrium approach to determine the shear, followed by use of integration to obtain the moment.

The equilibrium approach is very robust. It is straightforward for uniform distributed forces, is a bit more tedious for linear distributed forces, and becomes unwieldy when the distributed force is more complicated. The integration approach is elegant, but sometimes has subtleties. It is straightforward when there is a single function that describes a distributed force that acts over the full length of the beam and when the supports and/or point forces and moments are at the ends of the beam. Subtleties arise when there are multiple functions that describe the distributed force, when a beam has point forces and moments between its ends, and when a beam has supports between its ends; the examples in this section point out some of these subtleties. Very often, a combination of the equilibrium and integration approaches will be effective.

ISTUDY

Section 8.3

Relations Among Shear, Moment, and Distributed Force

517

Sometimes, we will only require the shear and moment diagrams, and we will not need to know the shear and moment as functions of position. For such situations we may elect to use the equilibrium approach of Section 8.1 to determine the shear and moment at specific points within a beam, and then we use Eq. (8.3) to relate the slopes of the shear diagram to the values of the distributed force, and Eq. (8.6) to relate the slopes of the moment diagram to the values of the shear.

Tips and shortcuts for drawing shear and moment diagrams In the remainder of this section, we discuss some of the characteristics of shear and moment distributions. Many of these characteristics you have probably noticed in the examples and problems of Section 8.2, while the expressions derived in this section provide further insights. Knowledge of these characteristics will provide you with additional tools that will be helpful for drawing shear and moment diagrams and for detecting errors. Equations (8.9) and (8.11) on p. 515 show the following: • In regions of a beam where the distributed force is zero, the shear is constant and the moment is linear.

𝐹𝐶 𝑦

• In regions of a beam where the distributed force is constant (i.e., uniform), the shear is linear and the moment is quadratic. • In regions of a beam where the distributed force is linear, the shear is quadratic and the moment is cubic. The following remarks pertain to Fig. 8.7: 𝑤

𝑦

∑

𝐹𝐶 𝑀𝐸

𝑥 𝐴

𝑥

∑

𝐵

𝐶

𝐷

𝐸

𝐹

Figure 8.7. A straight beam with a variety of loadings and supports.

• Point 𝐴 is an unsupported end of a beam with no concentrated force and no moment applied. At 𝐴, the shear and moment are zero. This is true regardless of the presence of a distributed force 𝑤. • At point 𝐵, a distributed force ends. The shear and moment just to the right of 𝐵 are the same as those just to the left of 𝐵. The same comments apply to points where a distributed force begins. • A concentrated force 𝐹𝐶 acting in the negative 𝑦 direction is applied at point 𝐶. The shear just to the right of 𝐶 is lower than the shear just to the left of 𝐶 by amount 𝐹𝐶 . The moment just to the right of 𝐶 is the same as that just to the left of 𝐶. The FBD and equilibrium equations shown in Fig. 8.8 justify the validity of these remarks. • A roller support is positioned at point 𝐷. The shear just to the right of 𝐷 is higher than the shear just to the left of 𝐷 by amount 𝐷𝑦 , where 𝐷𝑦 is the reaction the roller applies to the beam, with positive 𝐷𝑦 acting in the positive

𝑉𝐶 −

𝑀𝐶 −

𝑀𝐶 +

𝑁

𝑁 ∼0

𝑉𝐶 +

𝐹𝑦 = 0 ∶ (𝑉𝐶 − ) − 𝐹𝐶 − (𝑉𝐶 + ) = 0 ⇒ (𝑉𝐶 + ) = (𝑉𝐶 − ) − 𝐹𝐶 𝑀𝐶 = 0 ∶ −(𝑀𝐶 − ) + (𝑀𝐶 + ) = 0 ⇒ (𝑀𝐶 + ) = (𝑀𝐶 − )

Figure 8.8 An FBD for an infinitesimally small length of beam at point 𝐶 in Fig. 8.7. 𝐹𝐶 is the concentrated force. 𝑉𝐶 − and 𝑀𝐶 − are the shear and moment just to the left of 𝐶. 𝑉𝐶 + and 𝑀𝐶 + are the shear and moment just to the right of 𝐶.

𝑉𝐷 −

𝑦

∼0 𝑀𝐷+

𝑀𝐷− 𝑥

𝑁

𝑁 𝑉𝐷 +

∑ ∑

𝐷𝑦 𝐹𝑦 = 0 ∶ (𝑉𝐷− ) + 𝐷𝑦 − (𝑉𝐷+ ) = 0 ⇒ (𝑉𝐷+ ) = (𝑉𝐷− ) + 𝐷𝑦 𝑀𝐶 = 0 ∶ −(𝑀𝐷− ) + (𝑀𝐷+ ) = 0 ⇒ (𝑀𝐷+ ) = (𝑀𝐷− )

Figure 8.9 An FBD for an infinitesimally small length of beam at point 𝐷 in Fig. 8.7. 𝐷𝑦 is the support reaction. 𝑉𝐷− and 𝑀𝐷− are the shear and moment just to the left of 𝐷. 𝑉𝐷+ and 𝑀𝐷+ are the shear and moment just to the right of 𝐷.

518

𝑦

𝑀𝐸 − 𝑥

∑ ∑

Chapter 8

Internal Forces

𝑉𝐸 −

∼0 𝑀𝐸 +

𝑀𝐸

𝑁

𝑁 𝑉𝐸 +

𝐹𝑦 = 0 ∶ (𝑉𝐸 − ) − (𝑉𝐸 + ) = 0 ⇒ (𝑉𝐸 + ) = (𝑉𝐸 − ) 𝑀𝐶 = 0 ∶ − (𝑀𝐸 − ) + 𝑀𝐸 + (𝑀𝐸 + ) = 0 ⇒ (𝑀𝐸 + ) = (𝑀𝐸 − ) − 𝑀𝐸

Figure 8.10 An FBD for an infinitesimally small length of beam at point 𝐸 in Fig. 8.7. 𝑀𝐸 is the moment applied at 𝐸. 𝑉𝐸 − and 𝑀𝐸 − are the shear and moment just to the left of 𝐸. 𝑉𝐸 + and 𝑀𝐸 + are the shear and moment just to the right of 𝐸.

saweang.w/Shutterstock

𝑦 direction. The moment just to the right of 𝐷 is the same as that just to the left of 𝐷. The FBD and equilibrium equations shown in Fig. 8.9 justify the validity of these remarks. • A concentrated moment 𝑀𝐸 acting counterclockwise is applied at point 𝐸. The shear just to the right of 𝐸 is the same as that just to the left of 𝐸. The moment just to the right of 𝐸 is lower than the moment just to the left of 𝐸 by amount 𝑀𝐸 . The FBD and equilibrium equations shown in Fig. 8.10 justify the validity of these remarks.

Design considerations The design of a beam includes: specification of the material(s) it is constructed of, its shape and dimensions, methods of support and/or attachment to other members, and so on. When possible or convenient, beams that are commercially manufactured are used, such as the examples shown in Fig. 8.11. Some of the benefits of commercially manufactured beams are economy and rapid availability. When we use commercially manufactured beams, our task is to determine the material, size, and shape of beams that are needed. Commercially manufactured beams are available in a variety of materials, such as various grades of steel and aluminum, and a variety of shapes, such as I beam, L channel, U channel, and so on. Prefabricated reinforced concrete beams are also available, as are beams made of wood, including high-strength beams made of laminated wood. Often a beam will need to be constructed to our specifications. While cost may be higher and it may take longer to fabricate, performance may be better. Beams are usually designed to satisfy a variety of criteria, such as these: • Acceptable strength. A beam must be strong enough to support the forces applied to it without failing. • Acceptable deformations. A beam should not deflect excessively. • Acceptable fatigue life. A beam should be able to withstand the number of load cycles that will be applied to it over its life span.

Javier Larrea/Pixtal/Superstock

Figure 8.11 A steel yard showing commercially manufactured beams with an assortment of shapes.

ISTUDY

There are often additional criteria such as cost, manufacturability, corrosion resistance, high-temperature resistance, and so on.

End of Section Summary In this section, differential equations that relate the shear, moment, and distributed force were derived for straight beams. Equation (8.3) relates the shear to the distributed force by 𝑑𝑉 ∕𝑑𝑥 = −𝑤, and Eq. (8.6) relates the moment to the shear by 𝑑𝑀∕𝑑𝑥 = 𝑉 . These equations follow the sign conventions for positive 𝑉 , 𝑀, and 𝑤 given in Fig. 8.5 on p. 503. These relations are useful for drawing shear and moment diagrams. Also, the solutions to these differential equations were given in Eq. (8.9) where the shear is obtained by integrating the distributed force, and in Eq. (8.11) where the moment is obtained by integrating the shear.

ISTUDY

Section 8.3

Relations Among Shear, Moment, and Distributed Force

E X A M P L E 8.7

Integration to Determine Shear and Moment

A simply supported beam of length 𝐿 with a uniformly distributed force 𝑤0 is shown. Use integration to determine the shear and moment as functions of position.

𝑦 𝑤0

SOLUTION

𝐴

𝐵

𝑥

Road Map

This problem was previously solved using the equilibrium approach in Part (a) of Example 8.5 on p. 506. Here we develop the solution using integration. To determine the shear and moment as functions of position using Eqs. (8.9) and (8.11) on p. 515, the shear 𝑉𝑃 and moment 𝑀𝑃 at some convenient point 𝑃 must be known, and for this example, either of points 𝐴 or 𝐵 is such a location. We will use 𝐴, and comments are provided at the end of the solution regarding the use of 𝐵.

𝐿 Figure 1 𝑦

To determine the shear and moment at 𝐴, the reactions at that location are needed. Thus, we draw the FBD for the entire beam as shown in Fig. 2.

Governing Equations & Computation By writing and solving the equilibrium equations for the FBD in Fig. 2, the vertical support reaction at 𝐴 is 𝐴𝑦 = 𝑤0 𝐿∕2. Modeling To determine the shear 𝑉𝐴 and moment 𝑀𝐴 at point 𝐴 of the beam, we take a cut just to the right of 𝐴 to draw the FBD shown in Fig. 3. Governing Equations & Computation

𝑤0 𝐿

𝑥

Modeling

𝐴𝑥

𝐴𝑦

𝐵𝑦

Figure 2 Free body diagram for the entire beam. By writing equilibrium equations, the support reactions are found to be 𝐴𝑥 = 0 and 𝐴𝑦 = 𝐵𝑦 = 𝑤0 𝐿∕2. 𝑦

Using the FBD in Fig. 3, we write and solve the

0

equilibrium equations, ∑ ∑

𝑤0 𝐿

𝐹𝑦 = 0 ∶

2

𝑀𝐴 = 0 ∶

− 𝑉𝐴 = 0

⇒ 𝑉𝐴 =

𝑤0 𝐿

2 ⇒ 𝑀𝐴 = 0.

𝑀𝐴 = 0

,

(1)

𝑀𝐴 ∼0

(2)

Noting that 𝑉𝐴 = 𝑤0 𝐿∕2 and 𝑤 = 𝑤0 (a constant), we see Eq. (8.9) provides 𝑉 = 𝑉𝐴 −

∫

𝑤0 𝑑𝑥

⇒

𝑉 =

𝑤0 𝐿 2

− (𝑤0 )𝑥.

(3)

Noting that 𝑀𝐴 = 0 and 𝑉 = 𝑤0 𝐿∕2 − (𝑤0 )𝑥, we see Eq. (8.11) provides ∫

[

𝑤0 𝐿 2

] − (𝑤0 )𝑥 𝑑𝑥

⇒

𝑥

𝑉𝐴 𝐴𝑦 = 𝑤0 𝐿∕2

𝑥𝐴 =0

𝑥

𝑁𝐴

𝐴𝑥 = 0

𝑥

𝑀 = 𝑀𝐴 +

519

𝑀=

𝑤0 𝐿 2

𝑥 − 𝑤0

𝑥2 . 2

Figure 3 Free body diagram obtained by taking a cut just to the right of point 𝐴. Because the length of this portion of beam is zero, the force due to the distributed load is zero.

(4)

𝑥𝐴 =0

Discussion & Verification

• As expected, since the distributed force is uniform, the shear is linear and the moment is quadratic. Also, Eqs. (3) and (4) agree with the results of Part (a) in Example 8.5. • Rather than take point 𝑃 to be the left-hand end of the beam in Eqs. (8.9) and (8.11), we could use the right-hand end. To do this, we draw the FBD shown in Fig. 4, and we write equilibrium equations to determine 𝑉𝐵 = −𝑤𝑜 𝐿∕2 and 𝑀𝐵 = 0. In Eqs. (3) and (4), the lower limits become 𝑥𝐵 = 𝐿, and 𝑉𝐴 and 𝑀𝐴 are replaced by 𝑉𝐵 and 𝑀𝐵 . The same results for 𝑉 and 𝑀 are obtained. • As a partial check of our solutions, you should evaluate Eqs. (3) and (4) at 𝑥 = 0 to verify that they yield the proper results for 𝑉𝐴 and 𝑀𝐴 . Another check is to determine the shear and moment at another point on the beam, such as point 𝐵 in Fig. 4, to verify that the correct results are obtained there.

0

𝑦 𝑀𝐵 𝑥

∼0

𝑁𝐵 𝑉𝐵

𝐵𝑦 = 𝑤0 𝐿∕2 Figure 4 Free body diagram obtained by taking a cut just to the left of point 𝐵. Because the length of this portion of beam is zero, the force due to the distributed load is zero.

520

Chapter 8

Internal Forces

E X A M P L E 8.8

Integration to Determine Shear and Moment—a Subtlety Reconsider the beam and loading of Example 8.4 on p. 504, where a simply supported beam with an overhang is subjected to a uniformly distributed load. Use integration to determine the shear and moment as functions of position.

𝑦 100 lb∕f t 𝐴

𝑥

𝐵

4 ft

𝐶

SOLUTION

2 ft

Figure 1

Road Map

This problem was previously solved using the equilibrium approach in Example 8.4 on p. 504. Here we develop the solution using integration, and we point out a subtlety that arises when a support is between the ends of the beam. In Fig. 1 the distributed loading is uniform throughout the beam with value 𝑤 = 100 lb∕f t. We will determine the shear 𝑉 and moment 𝑀, using Eqs. (8.9) and (8.11) with point 𝑃 taken to be at end 𝐴 of the beam. The results for 𝑉 and 𝑀 will be valid only for 0 ≤ 𝑥 ≤ 4 f t, as we know that the shear undergoes a discontinuity (i.e., jump in value) at the roller at 𝐵, and clearly this discontinuity is not contained in the results from Eq. (8.9). To determine 𝑉 and 𝑀 in the remainder of the beam, we will reapply Eqs. (8.9) and (8.11) with point 𝑃 taken to be at end 𝐶 of the beam, and the results will be valid for 4 f t ≤ 𝑥 ≤ 6 f t.

𝑦 0

0

𝑀𝐴 ∼0

𝑁𝐴

𝑀𝐶 𝑥

𝑉𝐴

0

∼0

𝑁𝐶 𝑉𝐶

𝐴𝑦 = 150 lb (a)

(b)

Figure 2 Free body diagrams obtained by taking cuts immediately adjacent to the ends of the beam. (a) A cut just to the right of point 𝐴. (b) A cut just to the left of 𝐶. Because the lengths of these portions of beam are infinitesimally small, the force due to the distributed load is zero.

ISTUDY

Modeling The FBD for the entire beam was shown in Fig. 2 of Example 8.4 on p. 504, and the support reactions were determined to be 𝐴𝑦 = 150 lb and 𝐵𝑦 = 450 lb. In this solution, we will need the shear and moment at points 𝐴 and 𝐶, and to this end we draw the FBDs shown in Fig. 2. Governing Equations & Computation Using the FBD of end 𝐴 of the beam shown in Fig. 2(a), we write the equilibrium equations.

∑ ∑

𝐹𝑦 = 0 ∶

150 lb − 𝑉𝐴 = 0

⇒

𝑉𝐴 = 150 lb,

(1)

𝑀𝐴 = 0 ∶

𝑀𝐴 = 0

⇒

𝑀𝐴 = 0.

(2)

Similarly, using the FBD of end 𝐶 of the beam shown in Fig. 2(b), we write the equilibrium equations. ∑ ∑

𝐹𝑦 = 0 ∶

𝑉𝐶 = 0

⇒

𝑉𝐶 = 0,

(3)

𝑀𝐶 = 0 ∶

−𝑀𝐶 = 0

⇒

𝑀𝐶 = 0.

(4)

To obtain the shear and moment for the left-hand portion of the beam, we take point 𝑃 to be at end 𝐴. Noting that 𝑉𝐴 = 150 lb and 𝑤 = 100 lb∕f t, Eq. (8.9) provides 𝑥

𝑉 = 𝑉𝐴 −

∫

( 100

) lb 𝑑𝑥 ft

⇒

( ) lb 𝑉 = 150 lb − 100 𝑥. ft

(5)

𝑥𝐴 =0

Common Pitfall Noting that 𝑀𝐴 = 0 and 𝑉 is given in Eq. (5), Eq. (8.11) provides Determining 𝑽 and 𝑴 by integration. A common error in problems, such as this example, is to develop the solution up to Eq. (6) and then mistakenly believe that the results for 𝑉 and 𝑀 are valid throughout the entire beam.

𝑥

𝑀 = 𝑀𝐴 +

∫

[

) ] ( ) ( lb 2 lb 𝑥 𝑑𝑥 ⇒ 𝑀 = (150 lb)𝑥 − 50 𝑥 . 150 lb − 100 ft ft

𝑥𝐴 =0

The shear and moment determined in Eqs. (5) and (6) are valid for 0 ≤ 𝑥 ≤ 4 f t.

(6)

ISTUDY

Section 8.3

521

Relations Among Shear, Moment, and Distributed Force

To obtain the shear and moment for the right-hand portion of the beam, we take point 𝑃 to be at end 𝐶. Noting that 𝑉𝐶 = 0 and 𝑤 = 100 lb∕f t, Eq. (8.9) provides∗ 𝑥

𝑉 = 𝑉𝐶 −

) ( lb 𝑑𝑥 100 ft

∫

⇒

) ( lb 𝑥. 𝑉 = 600 lb − 100 ft

(7)

𝑥𝐶 =6 f t

Noting that 𝑀𝐶 = 0 and 𝑉 is given in Eq. (7), Eq. (8.11) provides 𝑥

𝑀 = 𝑀𝐶 +

∫

[

( ) ] lb 𝑥 𝑑𝑥 600 lb − 100 ft

(8)

𝑥𝐶 =6 f t

⇒

) ( lb 2 𝑥 . 𝑀 = −1800 f t ⋅lb + (600 lb)𝑥 − 50 ft

(9)

The shear and moment determined in Eqs. (7) and (9) are valid for 4 f t ≤ 𝑥 ≤ 6 f t.

• Because the distributed force is uniform over the full length of the beam, the shear is linear, as expected. However, Eqs. (5) and (7) are different linear functions, although they share the same slope. Because 𝑑𝑉 ∕𝑑𝑥 = −𝑤 (i.e., the slope of the shear is equal to the negative of the distributed load), the slopes of Eqs. (5) and (7) are both −100 lb∕f t. • Since the shear is linear, the moment is quadratic. However, Eqs. (6) and (9) are different quadratic functions. • Observe in the moment diagram in Fig. 3 that the moment displays a local maximum at a position between 1 f t and 2 f t. To determine exactly where this occurs, and the value of the moment at that location, we use the expression 𝑑𝑀∕𝑑𝑥 = 𝑉 as follows. At this local maximum, 𝑑𝑀∕𝑑𝑥 = 0, which means that 𝑉 = 0 at this location. Thus, we set Eq. (5) equal to zero and solve for 𝑥 to obtain 1.5 f t. We then substitute 𝑥 = 1.5 f t into Eq. (6) to obtain the moment 112.5 f t ⋅lb. • The shear 𝑉 is undefined when 𝑥 is exactly equal to 4 f t (point 𝐵). However, we know that the support reaction 𝐵𝑦 at 𝑥 = 4 f t is responsible for the jump that 𝑉 undergoes from just to the left of 𝐵 to just to the right of 𝐵. Because there is no discrete moment applied at 𝑥 = 4 f t, there is no discontinuity in the moment at that location. However, because the slope of 𝑀 is equal to 𝑉 and 𝑉 has a discontinuity at 𝑥 = 4 f t, the slope of 𝑀 changes at 𝑥 = 4 f t; hence, the moment curve shows a kink at that location.

∗ You should carry out the integrations in Eqs. (7) and (8) and carefully evaluate the limits of integration, to

verify the results that are reported. Incorrect evaluation of such limits of integration is a common source of error.

moment 𝑀 (f t⋅lb)

• The results for shear and moment agree with those obtained in Example 8.4. Of course, the shear and moment diagrams will also be the same, and these are shown again in Fig. 3 for purposes of the following discussion.

shear 𝑉 (lb)

Discussion & Verification

300 200 100 0 −100 −200 −300

300 200 100 0 −100 −200 −300

200 lb 150 lb

−250 lb 0

2

0

2

𝑥 ( f t)

4

6

−200 f t⋅lb 𝑥 ( f t)

Figure 3 Shear and moment diagrams.

4

6

522

Chapter 8

Internal Forces

E X A M P L E 8.9

Constructing Shear and Moment Diagrams Beam 𝐴𝐵𝐶𝐷 supports a wall built of concrete block. The concrete blocks have a total weight of 26 kN, and the wall is three times as high on the right-hand side as on the left. Draw the shear and moment diagrams.

𝑦

𝑥

𝐴

3m

𝐵 4m

𝐶 3m

𝐷

Figure 1

3𝑤0 = 6 kN∕m

𝑦

𝑥 3m

𝐷 4m

3m

𝐵𝑦 = 7 kN 𝐶𝑦 = 19 kN Figure 2 Free body diagram of the entire beam. Equation (1) is used to determine that 𝑤0 = 2 kN∕m, ∑ and then equilibrium equations 𝑀 = 0 and ∑ 𝐹𝑦 = 0 are written and solved to determine that the support reactions are 𝐵𝑦 = 7 kN and 𝐶𝑦 = 19 kN.

ISTUDY

Road Map We will begin by using the description of the wall’s geometry to determine the distributed force the concrete blocks apply to the beam. We will then use Eq. (8.12) on p. 516 to directly construct the shear and moment diagrams without explicitly determining the shear and moment as functions of position. Modeling

𝑤0 = 2 kN∕m 𝐴

SOLUTION

Given the description of the wall’s geometry, we may conclude that the distributed force from the concrete block is as shown in Fig. 2, where we let the value of the distributed force at end 𝐴 be 𝑤0 , and then the value at end 𝐷 is 3𝑤0 . Since the total force from the distributed loading must equal the 26 kN weight of the concrete blocks, we may solve for 𝑤0 by using∗ 𝑤0 (10 m) + 21 (3 m)(2𝑤0 ) = 26 kN

⇒

𝑤0 = 2 kN∕m.

(1)

With this result, the support reactions may then be determined, and you should verify that these are 𝐵𝑥 = 0, 𝐵𝑦 = 7 kN, and 𝐶𝑦 = 19 kN. Governing Equations & Computation

We will directly construct the shear and moment diagrams without finding 𝑉 and 𝑀 as functions of position. Equation (8.12) is repeated here as 𝑥𝑄

𝑉𝑄 = 𝑉𝑃 −

∫

𝑥𝑄

and 𝑀𝑄 = 𝑀𝑃 +

𝑤 𝑑𝑥

𝑥𝑃

∫

𝑉 𝑑𝑥,

(2)

𝑥𝑃

⏟⏞⏟⏞⏟

⏟⏞⏟⏞⏟

area under 𝑤 vs. 𝑥 plot between 𝑥𝑃 and 𝑥𝑄

area under 𝑉 vs. 𝑥 plot between 𝑥𝑃 and 𝑥𝑄

where we select points 𝑃 and 𝑄. Our strategy will be to begin by taking point 𝑃 to be at the left-hand end of the beam (point 𝐴), where the shear and moment are easily found. We will then sequentially work toward the right-hand end of the beam, evaluating areas for the 𝑤 vs. 𝑥 diagram and constructing the shear diagram in the process. Once the shear diagram is complete, we will repeat this process to construct the moment diagram. Shear diagram

By inspection, the shear at 𝐴 is zero, because end 𝐴 is unsupported and has no concentrated force applied. Hence, 𝑉𝐴 = 0,

(3)

𝑉𝐵 −

(4)

𝑉𝐵 + 𝑉𝐶 − 𝑉𝐶 +

) ( kN (3 m) = −6 kN, = 𝑉𝐴 − 2 m = (𝑉𝐵− ) + 𝐵𝑦 = −6 kN + 7 kN = 1 kN, ) ( kN = (𝑉𝐵+ ) − 2 (4 m) = −7 kN, m = (𝑉𝐶 − ) + 𝐶𝑦 = −7 kN + 19 kN = 12 kN,

𝑉𝐷 = 0.

(5) (6) (7) (8)

∗ While it is obvious that Eq. (1) is valid, we are in fact applying the first expression in Eq. (4.16) on p. 248

for construction of an equivalent force system.

ISTUDY

Section 8.3

523

The shear diagram is shown in Fig. 3 and is constructed as follows. We first plot the values of the shear from Eqs. (3) through (8). Since 𝑤 is constant from 𝐴 to 𝐵 and from 𝐵 to 𝐶, the shear is linear in those regions; so two straight lines are drawn. As a check on our solutions, you should compute the slopes for the shear between 𝐴 and 𝐵, and 𝐵 and 𝐶, to verify that both are −2 kN∕m. Since 𝑤 is linear between 𝐶 and 𝐷, the shear is quadratic in that region. We use the expression 𝑑𝑉 ∕𝑑𝑥 = −𝑤 to determine that the slope just to the right of 𝐶 is −2 kN∕m, and the slope at 𝐷 is −6 kN∕m. Moment diagram

By inspection, the moment at 𝐴 is zero, because end 𝐴 is unsupported and has no concentrated moment applied. Also, values of the moment will be the same on each side of the supports at 𝐵 and 𝐶, so there is no need to distinguish between these. Hence, 𝑀𝐴 = 0,

[

] 1 𝑀𝐵 = 𝑀𝐴 + (3 m)(−6 kN) = −9 kN⋅m, [2 ] 1 kN + (−7 kN) 𝑀𝐶 = 𝑀𝐵 + (4 m) = −21 kN⋅m, 2 𝑀𝐷 = 0.

(10) (11) (12) (13)

Equations (10) and (13) were written by inspection, while Eq. (2) was used to write Eqs. (11) and (12), and you should verify the areas that are computed using the 𝑉 vs. 𝑥 diagram shown in Fig. 3. The moment diagram is shown in Fig. 3 and is constructed as follows. We first plot the values of the moment from Eqs. (10) through (13). Since 𝑉 is linear from 𝐴 to 𝐵 and from 𝐵 to 𝐶, the shear is quadratic in these regions, and between 𝐶 and 𝐷 the shear is quadratic so the moment is cubic. We use the expression 𝑑𝑀∕𝑑𝑥 = 𝑉 to determine the slopes of the moment diagram, given the values of the shear, and this allows us to draw the moment diagram with the correct slopes and curvatures. Discussion & Verification

Numerous checks have been used and suggested in the course of developing this solution. Observe that the moment diagram has a local maximum at about 𝑥 ≈ 3.5 m, and this location corresponds to 𝑉 = 0, as expected.

15 10 5 0 −5 −10 −15

12 kN slope = −2 kN∕m

1 kN slope = 6 k N∕m

−6 kN 0

moment 𝑀 (kN⋅m)

In this example, the shear at 𝐴 was determined by inspection. If you are unsure that 𝑉𝐴 = 0 (or if end 𝐴 had a concentrated force or a support), you draw an appropriate FBD of end 𝐴, similar to Fig. 3 on p. 519, to determine the shear at this location. Equation (2) is used to write Eqs. (4) and (6), and you should verify the areas that are computed using the 𝑤 vs. 𝑥 diagram shown in Fig. 2. Equations (5) and (7) are applications of the results shown in Fig. 8.9 on p. 517. Equation (8) is written by inspection, although a useful check of accuracy would be to use Eq. (2) for [ ] 2 kN∕m + 6 kN∕m 𝑉𝐷 = (𝑉𝐶 + ) − (3 m) = 0. (9) 2

shear 𝑉 (kN)

Relations Among Shear, Moment, and Distributed Force

5 0 −5 −10 −15 −20 −25

3

−7 kN 𝑥 (m)

10

7

slope = 0 slope = −6 kN slope = 1 kN

−9 kN⋅m

slope = 12 kN

slope = −7 k N

−21 kN⋅m 0

3

𝑥 (m)

Figure 3 Shear and moment diagrams.

7

10

524

Chapter 8

Internal Forces

Problems General instructions Use the integration approach for the following problems. Shear and moment diagrams should show the maximum values of the shear and moment and the locations where these occur.

Problems 8.58 through 8.63 Determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝑦

𝐵

𝑥

𝐴

𝐴

𝐵

𝑥

−𝑤0 𝐿 Figure P8.64

𝑦

𝐴

Figure P8.59

𝐵 10 m

𝐵

𝐴

4 kN∕m 𝐵

𝑥 6m Figure P8.60 𝑦

𝑦 600 lb∕f t

600 lb∕f t

𝑥

𝐴

𝐵

Figure P8.58

8 kN∕m

300 lb∕f t 𝐴

𝐵

𝑥

600 lb∕f t

𝐵

𝑥

20 f t

20 f t

20 f t

Figure P8.61

Figure P8.62

Figure P8.63

Problem 8.64 A simply supported beam has a distributed force that varies linearly from 𝑤0 at the lefthand end to −𝑤0 at the right-hand end. Determine the shear and moment as functions of position, and draw the shear and moment diagrams.

Problem 8.65

𝑤

𝑥

𝑥 6m

𝑦 𝑤0

𝐴

𝑦

8 kN∕m

6m

𝑦

𝐴

𝑦

8 kN∕m

The beam is shaped so that its cross section is deeper near midspan than near the ends. As a consequence, its weight distribution is 𝑤 = (0.8 kN∕m)[1 + sin(𝜋𝑥∕10 m)]. Determine the shear and moment as functions of position due to the beam’s weight distribution, and draw the shear and moment diagrams.

Figure P8.65

ISTUDY

Problems 8.66 through 8.68 For the beam and loading shown, determine the shear and moment as functions of position, and draw the shear and moment diagrams. 𝑦 𝑦

𝑃

𝑃

𝑦

600 lb∕f t

400 lb 1500 f t⋅lb

𝐴

𝑥 𝐿 2

𝐵

𝐶

𝐿 2

Figure P8.66

𝐴

𝑥 5 ft

𝐶

𝐵 5 ft

Figure P8.67

𝐴

𝑥 4 ft

𝐶

𝐵 8 ft

Figure P8.68

ISTUDY

Section 8.3

525

Relations Among Shear, Moment, and Distributed Force

Problem 8.69 Beam 𝐴𝐵𝐶𝐷 is used to support a machine tool. The beam weighs 60 lb∕f t of length, and the machine weighs 1200 lb with center of gravity at point 𝐸. Assuming the machine applies only vertical forces to the beam at points 𝐵 and 𝐶, determine the shear and moment within beam 𝐴𝐵𝐶𝐷 as functions of position, and draw the shear and moment diagrams. 𝑦 6 ft 𝐸 𝐴

𝑥 4 ft

𝐵 6 ft

𝐶 4 ft

𝐷

Figure P8.69

𝑦

Problem 8.70

2.5 m

Beam 𝐴𝐵𝐶𝐷 is used to support an automobile so that it may be serviced. The beam weighs 1 kN∕m of length, and the automobile weighs 10 kN with center of gravity at point 𝐸. Assuming the automobile’s tires apply only vertical forces to the beam at points 𝐵 and 𝐶, determine the shear and moment as functions of position, and draw the shear and moment diagrams.

𝐸 𝐴

𝑥

𝐵

𝐷

𝐶

1.3 m

2.7 m

1.5 m

Figure P8.70

Problems 8.71 through 8.75 Draw the shear and moment diagrams for the beam and loading shown. Determination of the shear and moment as functions of position is not required. 𝑦

𝑃

𝐴 𝑑

𝑃

𝑃

𝐵 𝑥

𝐶

𝑑

𝑑

𝑦

𝑃

𝐷 𝐸 𝑑

𝐹

𝑃

𝐵 𝑥

𝐴

𝑑

𝑑

Figure P8.71

𝑑

𝑃

𝑃

𝐶 𝐷 𝐸 𝑑

𝑑

Figure P8.72 𝑦

𝑦 6 lb

2 lb 𝐴

𝑃

4 lb

𝐶

𝑥

𝐵

𝐷

5 in. 5 in. 5 in. 5 in. Figure P8.73

𝐸

𝑦

30 N 𝐴

𝑥 30 mm

3N

6N

0.5 N∕mm 𝐶

𝐵 50 mm

Figure P8.74

Problem 8.76 Draw the shear and moment diagrams for the bookshelf shown in Fig. 1 of Example 7.11 on p. 475. Determination of the shear and moment as functions of position is not required.

1 N∕mm 𝐴

𝑥

𝐵

𝐶

6 mm 6 mm 6 mm Figure P8.75

𝐷

526

ISTUDY

Chapter 8

Internal Forces

Problem 8.77 Three beams are shown, along with four possible shear diagrams and four possible moment diagrams. All forces and moments act in the directions shown. Without calculation, complete the table provided by selecting the appropriate shear and moment diagrams that correspond to each beam. Your answers may use a shear and/or moment diagram more than once. As an example, the answers for Beam 1 are provided in the table. 𝑦

𝑦

𝑦

𝐹1

𝐹1

𝐹1

𝐹2

Shear Moment Beam diagram diagram 𝑀1

𝑥

𝑥

𝑥

Beam 1

Beam 2

Beam 3

𝑉

𝑉

𝑉 𝑥

0

𝑥

0

(a) 𝑀 𝑥

0

𝑥

𝑥

(e)

(d)

𝑀

0

𝑥

0

(c)

𝑀

𝑀 𝑥

0

(f)

h

𝑉

0

(b)

c

1 2 3

𝑥

0

(g)

(h)

Figure P8.77

Problem 8.78 Three beams are shown, along with four possible shear diagrams and four possible moment diagrams. The distributed force acts in the direction shown. Without calculation, complete the table provided by selecting the appropriate shear and moment diagrams that correspond to each beam. Your answers may use a shear and/or moment diagram more than once. 𝑦

𝑦

𝑦

𝑤 𝑥

𝑥

𝑥

Beam 1

Beam 2

Beam 3

𝑉

𝑉

𝑉 𝑥

0

𝑥

0

(a)

𝑥 (e)

𝑥

(d)

𝑀 𝑥

0

𝑀 𝑥

0

(f)

(g) Figure P8.78

𝑥

0

(c)

𝑀

0

𝑉

0

(b)

𝑀

Shear Moment Beam diagram diagram 1 2 3

𝑤

𝑤

𝑥

0 (h)

ISTUDY

Section 8.4

Chapter Review

527

8.4 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

𝑤

𝑦 (a)

Internal forces Internal forces are forces and moments that develop within structural members and/or materials due to the external forces that are applied. A structural member is said to be slender if the dimensions of its cross section are small compared to its length. The methods of analysis discussed in this chapter are appropriate and effective for slender members. In two dimensions, the internal forces consist of two forces and one moment, as shown in Fig. 8.12, where 𝑁 is called the normal force or axial force, 𝑉 is called the shear force, and 𝑀 is called the bending moment. In three dimensions, the internal forces consist of three forces and three moments, as shown in Fig. 8.13, where 𝑁 is the normal force or axial force, 𝑉𝑦 and 𝑉𝑧 are shear forces, 𝑀𝑦 and 𝑀𝑧 are bending moments, and 𝑀𝑥 is called the torque. When we find the internal forces in straight beams in two dimensions, it is necessary to follow a consistent sign convention as shown in Fig. 8.12.

Methods for determining internal forces Two methods for determining internal forces were discussed in this chapter: the equilibrium approach (used in Sections 8.1 and 8.2) and the integration approach (used in Section 8.3). In the equilibrium approach, cuts are taken as needed, FBDs are drawn, and equilibrium equations are written to obtain the internal forces either at specific locations in the structure or as functions of position. In the integration approach, differential equations (as summarized below) are solved to obtain the internal forces as functions of position. Shear and moment diagrams are plots of the shear and moment as functions of position.

Relations among 𝑉 , 𝑀, and 𝑤. By drawing an FBD of a small portion of a straight beam (as shown in Fig. 8.6 on p. 514) using the sign convention shown in Fig. 8.12, the shear, moment, and distributed force are related by Eqs. (8.3) and (8.6), p. 514 𝑑𝑉 = −𝑤 𝑑𝑥

and

𝑑𝑀 =𝑉. 𝑑𝑥

In words, Eq. (8.3) says, “The change in shear divided by the change in length of the beam at position 𝑥 is equal to the negative of the distributed force at that location,” and Eq. (8.6) says, “The change in moment divided by the change in length of the beam at position 𝑥 is equal to the shear at that location.”

Determination of 𝑉 and 𝑀 using integration. Equations (8.3) and (8.6) are differential equations that can be rearranged and integrated to obtain Eqs. (8.9) and (8.11), p. 515 𝑥

𝑉 = 𝑉𝑃 −

∫ 𝑥𝑃

𝑥

𝑤 𝑑𝑥

and

𝑀 = 𝑀𝑃 +

∫

𝑉 𝑑𝑥.

𝑥𝑃

If the distributed force 𝑤 is known as a function of position, then by integration using Eq. (8.9), the shear as a function of position is obtained. Once the shear is known, then by integration using Eq. (8.11), the moment as a function of position is obtained.

𝑥

𝑤

(b)

𝑀

𝑀

𝑦

𝑁

𝑥

𝑁 𝑉

𝑉

Figure 8.12 Internal forces that develop on a particular cross section of a slender member in two dimensions. (b) Sign convention for positive 𝑁, 𝑉 , and 𝑀.

𝑀𝑦

𝑦

𝑉𝑦 𝑁 𝑀𝑥

𝑥

𝑀𝑦 𝑉𝑦 𝑀𝑥 𝑁

𝑧 𝑀𝑧

𝑉𝑧 𝑉𝑧

𝑀𝑧

Figure 8.13 Internal forces that develop on a particular cross section of a slender member in three dimensions.

528

ISTUDY

Chapter 8

Internal Forces

Useful forms of Eqs. (8.9) and (8.11), especially for purposes of drawing shear and moment diagrams, are Eq. (8.12), p. 516 𝑥𝑄

𝑉𝑄 = 𝑉𝑃 −

∫

𝑥𝑄

𝑤 𝑑𝑥

𝑥𝑃

⏟⏞⏟⏞⏟

area under 𝑤 vs. 𝑥 plot between 𝑥𝑃 and 𝑥𝑄

and

𝑀𝑄 = 𝑀𝑃 +

∫

𝑉 𝑑𝑥 .

𝑥𝑃

⏟⏞⏟⏞⏟

area under 𝑉 vs. 𝑥 plot between 𝑥𝑃 and 𝑥𝑄

Design considerations Beams are usually designed to satisfy a variety of criteria, such as acceptable strength, acceptable deformations, acceptable fatigue life, low cost, manufacturability, and so on.

ISTUDY

Section 8.4

529

Chapter Review

Review Problems Problem 8.79

𝑦

A beam is supported by a roller at 𝐴 and a device at 𝐶 that allows vertical motion of the beam while preventing horizontal motion and rotation. Use the equilibrium approach to determine the shear and moment as functions of position, and draw the shear and moment diagrams.

10 N

𝐴

𝑥

𝐶

𝐵

8 mm 8 mm Figure P8.79

Problems 8.80 and 8.81

4 ft

Members 𝐴𝐵𝐶 and 𝐶𝐷𝐸 are pinned to one another at 𝐶 and have pin supports at 𝐴 and 𝐸. Determine the internal forces acting on: Problem 8.80

Cross sections 𝐹 and 𝐺, which are located immediately to the right of 𝐴 and the left of 𝐵, respectively.

4 ft 2 kip

2 kip 1 kip 𝐹 𝐴

Problem 8.81

Cross sections 𝐻 and 𝐼, which are located immediately to the right of 𝐵 and the left of 𝐶, respectively.

4 ft

𝐻

𝐺

𝐵

2 kip

𝐼

1 kip 𝐶

𝐷

300 mm

200 mm

400 N 𝐷

𝐶 40 mm

𝐵

𝐹

𝐺

Problem 8.83

200 mm

A historically important shipwreck is to be recovered, and a number of fragile wooden beams must be lifted. Specify the dimension 𝑑 (in terms of length 𝐿) where the two lifting slings should be placed so that the maximum absolute value of the moment is as small as possible. Assume the beams are straight with uniform weight distribution, and the cables attached to the slings are vertical. Hint: The optimal value of 𝑑 will give the same absolute value of the moment at the slings as at the midpoint of the beam.

𝐴 Figure P8.82

slings 𝑑

Use the equilibrium approach to determine the shear and moment as functions of position, and draw the shear and moment diagrams.

𝐴

𝑦

6 kN∕m 𝑥 4m

𝐶

𝐵 4m

Figure P8.84

Problem 8.86 Repeat Prob. 8.84 using the integration approach.

Problem 8.87 Repeat Prob. 8.85 using the integration approach.

𝐴

𝑤0 𝐵 𝑥 𝐿 3

𝐶

𝐿 3

𝐷

𝐿 3

Figure P8.85

𝐸

45◦

Problems 8.84 and 8.85

𝑦

𝐸

20◦

20◦

Figure P8.80 and P8.81

Problem 8.82 To provide generous leg room, a quarter-circular member 𝐴𝐵 is used to support a wallmounted desk 𝐶𝐷. Neglecting the weight of the members, determine the internal forces acting on cross sections 𝐸, 𝐹 , and 𝐺, which are located immediately above point 𝐴, at the midpoint of member 𝐴𝐵, and immediately to the right of point 𝐵, respectively.

4 ft

Figure P8.83

𝐿

𝑑

530

Chapter 8

Internal Forces Problems 8.88 through 8.90

200 mm

𝑦

50 mm 𝐷 𝐴

𝑥

𝐵

𝐶

𝑑

The device shown is used in a factory to support a tool that applies a 60 N vertical force at 𝐷. It has the feature that portion 𝐶𝐷 can slide to the right when the tool is needed, and can slide to the left when the tool is to be stored, such that 50 mm ≤ 𝑑 ≤ 200 mm. For the value of 𝑑 indicated in the subproblem below, determine the shear and moment in beam 𝐴𝐵 as functions of position, and draw the shear and moment diagrams. Neglect the size of the rollers at 𝐵 and 𝐶. Problem 8.88

𝑑 = 200 mm.

Problem 8.89

𝑑 = 100 mm.

Problem 8.90

𝑑 = 50 mm.

60 N

200 mm Figure P8.88–P8.90

Problem 8.91 𝑦 (a) 𝐴

𝑃 𝐷

(a) For the cantilever beam shown in Fig. P8.91(a), determine the shear and moment as functions of position. Express your answers in terms of parameters such as 𝑃 , 𝑎, and 𝑏.

200 lb 100 lb 150 lb

(b) For the cantilever beam with the three forces shown in Fig. P8.91(b), use superposition of the results of Part (a) to determine shear and moment as functions of position, and draw the shear and moment diagrams.

𝑥 𝑎 𝑦

(b) 𝐴

𝑥

𝐵

𝑏

𝐶

10 in. 10 in. 10 in. Figure P8.91

ISTUDY

𝐷

Problems 8.92 and 8.93 Draw the shear and moment diagrams for the beam and loading shown. Determination of the shear and moment as functions of position is not required. 𝑦

𝑦 2000 lb

100 lb∕f t 𝐴

𝑥 10 f t

𝐶 𝐵 5 ft 5 ft

2 kN∕m

𝐷

𝐴

Figure P8.92

𝑥

𝐵 2m 2m

𝐶 2m

𝐷

Figure P8.93

Problem 8.94 (a) Use the equilibrium approach to determine the shear as a function of position in the region 0 ≤ 𝑥 ≤ 3 m. (b) Use the integration approach with the results of Part (a) to determine the moment in the region 0 ≤ 𝑥 ≤ 3 m. (c) Values for the shear and moment just to the right of 𝐵 and at 𝐶 are shown in the shear and moment diagrams provided. By inspection, complete these diagrams for the region 3 m ≤ 𝑥 ≤ 6 m. Accurately draw the shapes of the curves and label the slopes at 𝑥 = 3 m and 𝑥 = 6 m. 𝑉 (kN)

4 kN∕m

𝑦

𝑀(kN⋅m)

9 𝐴

𝑥 3m

0

𝐶

𝐵 3m

0

3

Figure P8.94

6

𝑥 (m) −15

3

6

ISTUDY

Section 8.4

Chapter Review

Problem 8.95 (a) Use the equilibrium approach to determine the shear as a function of position in the region 6 f t ≤ 𝑥 ≤ 12 f t. (b) Use the integration approach with the results of Part (a) to determine the moment in the region 6 f t ≤ 𝑥 ≤ 12 f t. (c) Values for the shear and moment at 𝐴 and just to the left of 𝐵 are shown in the shear and moment diagrams provided. By inspection, complete these diagrams for the region 0 ≤ 𝑥 ≤ 6 f t. Accurately draw the shapes of the curves and label the slopes at 𝑥 = 0 and 𝑥 = 6 f t. 𝑦

𝑉 (lb)

400 lb∕f t 200 lb∕f t 𝑥

𝐴

𝐵

6 ft

𝑀 (f t⋅lb)

1700

4200

𝐶

6 −100

6 ft

𝑥 ( f t)

0

6

𝑥 ( f t)

Figure P8.95

Problem 8.96 (a) Use the equilibrium approach to determine the moment as a function of position in the region 0 ≤ 𝑥 ≤ 6 f t.

(b) Use the integration approach to determine the shear in the region 6 f t ≤ 𝑥 ≤ 12 f t.

(c) Values for the shear and moment just to the right of 𝐵 and at 𝐶 are shown in the shear and moment diagrams provided. By inspection, complete these diagrams for the region 6 f t ≤ 𝑥 ≤ 12 f t. Accurately draw the shapes of the curves and label the slopes at 𝑥 = 6 f t and 𝑥 = 12 f t. 𝑦

𝑉 (kip) 4 kip∕f t 𝑥

𝐴

20 𝐶

𝐵

6 ft

𝑀 (f t⋅kip) 6 12 𝑥 ( f t) 0

8 0

6 ft

6

−72

12

Figure P8.96

Problem 8.97 (a) Use the equilibrium approach to determine the moment as a function of position in the region 0 ≤ 𝑥 ≤ 3 m.

(b) Use the integration approach to determine the shear in the region 6 m ≤ 𝑥 ≤ 9 m.

(c) Values for the shear and moment just to the right of 𝐵 and left of 𝐶 are shown in the shear and moment diagrams provided. By inspection, complete these diagrams for the region 3 m ≤ 𝑥 ≤ 6 m. Accurately draw the shapes of the curves and label the slopes at 𝑥 = 3 m and 𝑥 = 6 m. 𝑦

4 kN∕m 0

2 kN∕m 𝐴

𝑥 3m

𝐵 𝐶 3m 3m

𝐷

𝑉 (kN) 3

6

−6

𝑥 (m)

0 −9 −39

−15 Figure P8.97

𝑀 (kN⋅m) 3 6

𝑥 (m)

531

532

ISTUDY

Chapter 8

Internal Forces

Problem 8.98 (a) Use the equilibrium approach to determine the moment as a function of position in the region 6 m ≤ 𝑥 ≤ 12 m.

(b) Use the integration approach to determine the shear in the region 0 ≤ 𝑥 ≤ 6 m.

(c) Values for the shear and moment at 𝐴, 𝐵, and 𝐶 are shown in the shear and moment diagrams provided (the shear just to the left and right of 𝐵 is 4 kN and −11 kN, respectively). By inspection, complete these diagrams. Accurately draw the shapes of the curves, and label the slopes at 𝑥 = 0, 𝑥 = 6 m, and 𝑥 = 12 m. 𝑦

𝑉 (kN) 4 kN∕m 15 kN

𝐴

𝑥 1 kN∕m 6m

𝑀 (kN⋅m)

16

48 12

4 0 𝐶 −5 −11

𝐵 6m

𝑥 (m) 0

6

12

𝑥 (m)

Figure P8.98

Problem 8.99 (a) Use the equilibrium approach to determine the shear as a function of position in the region 0 ≤ 𝑥 ≤ 2 m. (b) Use the integration approach with the results of Part (a) to determine the moment in the region 0 ≤ 𝑥 ≤ 2 m. (c) Values for the shear and moment at 𝐵 and 𝐶 are shown in the shear and moment diagrams provided. By inspection, complete these diagrams. Accurately draw the shapes of the curves, and label the slopes at 𝑥 = 2 m and 𝑥 = 4 m. 𝑉 (N)

𝑦 600 N∕m 𝐴

𝑥

1200

4

𝑥 (m)

𝐶

𝐵

2m

𝑀 (N⋅m) 2 0

0

2m

2

4

𝑥 (m) −1200

Figure P8.99

Problem 8.100 (a) Use the equilibrium approach to determine the shear as a function of position in the region 0 ≤ 𝑥 ≤ 4 f t. (b) Use the integration approach with the results of Part (a) to determine the moment in the region 0 ≤ 𝑥 ≤ 4 f t. (c) Values for shear and moment just to the right of 𝐵 and at 𝐶 are shown in the shear and moment diagrams provided. By inspection, complete these diagrams. Accurately draw the shapes of the curves, and label the slopes at 𝑥 = 4 f t and 𝑥 = 12 f t. 𝑉 (lb) 𝑦

𝑀 (f t⋅lb)

600 lb∕f t 1500 800 f t⋅lb

𝑥 𝐴 4 ft

𝐶

𝐵

𝑥 ( f t)

0 4

0

12

8 ft −1330

−1700 Figure P8.100

4

12

𝑥 ( f t)

ISTUDY

Friction

9

When two objects are in contact, friction forces generally develop between them. This chapter presents models for quantifying friction forces and discusses methods for analyzing problems with friction and sliding.

Ivan Kovbasniuk/Shutterstock

The driver of a motorcycle spins its rear wheel before a race, generating intense heat due to frictional slip between the tire and pavement. The increase in temperature of the tire increases the frictional resistance between the tire and pavement, thus improving performance during the race.

9.1

Basic Concepts

When two objects are in contact, there is often a tendency for them to slide relative to one another. When this is the case, friction forces that resist the sliding motion develop on the contact surfaces between the objects. A number of factors influence how large the friction forces can be, and the field of study that addresses this topic is called tribology, which is derived from the Greek word tribos, which means rubbing.

A brief history of tribology Leonardo da Vinci (1452–1519) used a scientific approach in his studies of friction and recognized that friction force and normal force were proportional. In 1699, Guillaume Amontons (1663–1705), a French architect turned engineer, presented a

Interesting Fact Contact surfaces. “Putting two solids together is rather like turning Switzerland upside down and standing it on Austria—the area of intimate contact will be small.” These words were used by the pioneering tribologist F. P. Bowden in a BBC broadcast in 1950. Virtually all contact surfaces, despite any apparent degree of smoothness, are inherently rough at small length scales.

533

534

Chapter 9

Friction

paper to the French Academy where he reported on friction tests using various combinations of iron, copper, lead, and wood, lubricated with pork fat (suet). It is rather remarkable that he found the friction force 𝐹 during sliding and the normal force 𝑁 were related by 𝐹 ≈ 𝑁∕3, and very importantly, that the friction force was independent of the apparent area of contact. In 1781, Charles Augustin de Coulomb (1736–1806), a French physicist and engineer who also did groundbreaking work in electricity and magnetism, confirmed the findings of Amontons and distinguished between static and kinetic friction. While Coulomb’s work is notable, he did not improve substantially on the findings of Amontons. Nonetheless, the friction model discussed in this chapter is usually called Coulomb’s law, although it is occasionally referred to as Amontons’ law or the Amontons-Coulomb law.∗

A simple experiment Art Collection 3/Alamy Stock Photo

Figure 9.1 A portrait of Coulomb painted by Hippolyte Lecomte.

𝑊 𝑃 𝐺 𝐸

𝑦 𝑥 (a)

(b)

𝐹 𝑁

𝐹 slip

stick 𝐵

𝐹𝑠

𝐷

𝐶

(c) 𝐴 0

1 0

1

𝐹𝑘 𝑃

Figure 9.2 (a) An empty coffee cup with weight 𝑊 and center of gravity at point 𝐺 subjected to a horizontal force 𝑃 . (b) Free body diagram. (c) Relationship between 𝑃 and the friction force 𝐹 that is likely to be observed. Notice that point 𝐸, the location where the contact forces 𝑁 and 𝐹 act, is gener∑ ally not directly below 𝐺; writing 𝑀 = 0 will determine where this point is located.

ISTUDY

The basic features of friction between contacting bodies, and a model to quantify this phenomenon, can be developed by considering the example shown in Fig. 9.2(a), where an empty coffee cup with weight 𝑊 rests on a table. Imagine using your fingers to apply a slowly increasing horizontal force 𝑃 to the cup, starting from a value of zero, so as to slide the cup to the right. Between the cup and table, in addition to the normal force 𝑁, a friction force 𝐹 develops, as shown in the FBD in Fig. 9.2(b). The relationship between 𝑃 and the friction force 𝐹 that you are likely to observe is shown in Fig. 9.2(c). In region 𝐴𝐵 of this figure, the cup undergoes no motion ∑ (this is often called stick), hence static equilibrium prevails, and by writing 𝐹𝑥 = 0 we observe that 𝑃 = 𝐹 . At point 𝐵, we say that motion is impending, meaning that slip is about to occur. For values of 𝑃 beyond point 𝐵, the interface between the cup and table is not capable of supporting friction forces that are high enough to provide equilibrium. In this regime, 𝑃 > 𝐹 , sliding occurs, the cup accelerates, and ∑ Newton’s law 𝐹𝑥 = 𝑚𝑎𝑥 must be used where the acceleration in the 𝑥 direction is nonzero. If we repeat this experiment using a cup filled with enough coffee to double the weight of the original empty cup, we will find that the force at which sliding starts is approximately doubled. Typically, and as shown in Fig. 9.2(c), the friction force 𝐹𝑠 at which sliding starts is somewhat higher than the friction force 𝐹𝑘 for sustained sliding, and these values are called the static friction force and the kinetic friction force. Figure 9.2(c) shows an instantaneous decrease of the friction force from 𝐹𝑠 to 𝐹𝑘 ; in reality, this decrease is rapid but is not instantaneous. Figure 9.3 offers a simple theory for why 𝐹𝑠 > 𝐹𝑘 . Contact surfaces are inherently rough, and actual contact occurs at a relatively small number of prominent asperities, two of which are shown in Fig. 9.3, where the 𝑥 direction is oriented along the mean plane of the interface. Before sliding, the force supported by the two asperities shown is 𝑅⃗ 𝑠 and −𝑅⃗ 𝑠 , and shortly after sliding starts, these forces change to 𝑅⃗ 𝑘 and −𝑅⃗ 𝑘 . Notice that after a small amount of sliding, the asperities of one surface have slid up and over those of the other surface. Hence, the 𝑥 component of 𝑅⃗ 𝑘 relative to its 𝑦 component is lower than that just before sliding. In essence, this model presumes that prior to sliding the asperities of one surface rest in the troughs of the other surface, and that during gross sliding the asperities are more likely to make contact closer to their summits. Reasons why the asperities of one surface settle into ∗ Additional

historical perspectives, as well as an excellent study of the field of tribology, are given by K. C. Ludema, Friction, Wear and Lubrication, CRC Press, Boca Raton, 1996.

ISTUDY

Section 9.1

Basic Concepts

the troughs of the other surface before sliding include microvibrations and viscous deformation (i.e., time-dependent flow) of the asperities.

𝑦 𝑥 𝑅⃗ 𝑠

Coulomb’s law of friction A model that describes the relationship between the normal force and friction force between two contacting surfaces is called Coulomb’s law, as follows: |𝐹 | ≤ 𝜇𝑠 𝑁

|𝐹 | = 𝜇𝑘 𝑁

before sliding ( 6.998 kN, gross sliding motion occurs, with no tipping, and the barrier will accelerate.

𝑃 𝐺 45 cm

60 cm

27 cm 𝐴 𝐹 𝑁 Figure 3 Free body diagram for tipping of the traffic barrier.

542

Chapter 9

Friction

E X A M P L E 9.3

Wedges and Multiple Contact Surfaces Two nylon blocks 𝐴 and 𝐵 are used to level one corner of a heavy refrigerator. The floor 𝐶 and guides 𝐷 and 𝐸 are fixed. Block 𝐴 supports a 200 lb force from the refrigerator and is a loose fit between guides 𝐷 and 𝐸. The coefficient of static friction between all contact surfaces is 0.2.

200 lb 𝐸 𝐴

(a) Determine the force 𝑃 needed to raise block 𝐴.

𝐷 𝑃

(b) When 𝑃 = 0, determine if the system is at rest.

𝐵 10◦ 𝐶

Figure 1

SOLUTION Road Map

We will neglect the weights of the blocks, will assume the system is initially at rest, and will analyze motion due to sliding only (i.e., we assume the blocks will not tip). Because block 𝐴 is a loose fit between guides 𝐷 and 𝐸, it will make contact with 𝐷 or 𝐸, but not both at the same time. In Part (a), block 𝐵 slides to the right while block 𝐴 is raised: there are three surfaces in contact, and the kinematics are such that if one of these surfaces slides, then they all must slide. In Part (b), if this is a good design, there will be no sliding when 𝑃 = 0. If sliding does occur, 𝐵 will move to the left and 𝐴 will be lowered, and the FBDs from Part (a) will be revised accordingly.

200 lb

𝐹3

𝐴

𝑁3

𝐹2 10◦

The FBD for the two blocks is shown in Fig. 2, where the directions of the friction forces oppose the directions of sliding.

𝑦

𝑁2 𝑃

Part (a) Modeling 𝑥 Governing Equations

10◦ 𝐹2 𝐹1

𝐵 𝑁1

Equilibrium Equations

Block A:

∑

Figure 2 Free body diagrams for raising block 𝐴.

ISTUDY

∑

Block B:

∑

∑

Using the FBD from Fig. 2, the equilibrium equations are 𝑁2 (sin 10◦ ) + 𝐹2 (cos 10◦ ) − 𝑁3 = 0,

𝐹𝑥 = 0 ∶

◦

(1)

𝐹𝑦 = 0 ∶

◦

𝑁2 (cos 10 ) − 𝐹2 (sin 10 ) − 𝐹3 − 200 lb = 0.

(2)

𝐹𝑥 = 0 ∶

𝑃 − 𝐹1 − 𝑁2 (sin 10◦ ) − 𝐹2 (cos 10◦ ) = 0,

(3)

𝐹𝑦 = 0 ∶

𝑁1 − 𝑁2 (cos 10◦ ) + 𝐹2 (sin 10◦ ) = 0.

(4)

Note that Eqs. (1)–(4) contain seven unknowns. Force Laws

Assuming that sliding is impending, and noting that if one surface slides then all surfaces must slide, Coulomb’s law, Eq. (9.3) on p. 535, for each sliding surface provides 𝐹1 = 𝜇𝑁1 = (0.2)𝑁1 ,

Helpful Information Mechanical advantage. Notice that the force 𝑃 = 128.0 lb needed to raise block 𝐴, given by Eq. (9), is less than the 200 lb force being lifted. This feature is sometimes called mechanical advantage. The mechanical advantage of a wedge can be increased by decreasing the angle of the wedge (e.g., reducing the 10◦ angle shown in Fig. 1), or by using materials with a lower coefficient of friction.

𝐹2 = 𝜇𝑁2 = (0.2)𝑁2 ,

𝐹3 = 𝜇𝑁3 = (0.2)𝑁3 ,

(5)

where the coefficient of static friction is used and the directions of 𝐹1 , 𝐹2 , and 𝐹3 have been properly accounted for in the FBDs of Fig. 2. Computation

There are now seven equations and seven unknowns. These equations are easily solved by hand, and you should verify that the solutions are ⇒

𝑁1 = 216.9 lb,

𝐹1 = 43.38 lb,

(6)

𝑁2 = 228.3 lb,

𝐹2 = 45.66 lb,

(7)

𝑁3 = 84.62 lb, 𝐹3 = 16.92 lb, and

𝑃 = 128.0 lb.

(8) (9)

ISTUDY

Section 9.1

Basic Concepts

Part (b)

200 lb

Modeling

We assume the system is in equilibrium with no sliding, and at the end of the solution we must verify this assumption. The FBDs are shown in Fig. 3, with the following comments. Because we assume there is no sliding, we may not use 𝐹1 = 𝜇𝑁1 , 𝐹2 = 𝜇𝑁2 , and 𝐹3 = 𝜇𝑁3 ; rather, we are assuming 𝐹1 < 𝜇𝑁1 , and so on. Thus, 𝑁1 , 𝐹1 , 𝑁2 , 𝐹2 , 𝑁3 , and 𝐹3 are six independent unknowns. While we assume there is no sliding, the directions of 𝐹1 , 𝐹2 , and 𝐹3 are nonetheless correctly shown for block 𝐵 sliding to the left and block 𝐴 moving down. Governing Equations & Computation

The FBDs in Fig. 3 contain six unknowns, and there are only four equilibrium equations available; hence, this problem is statically indeterminate.∗ We proceed by assuming that block 𝐴 does not contact either of the guides, which is reasonable if there is no sliding. Thus, 𝑁3 = 𝐹3 = 0. By inspection of Fig. 3, ∑ the equation 𝐹𝑥 = 0 for blocks 𝐴 and 𝐵 taken together (note that the forces 𝑁2 and 𝐹2 become internal to the FBD and hence, they do not appear in this equilibrium equation) ∑ provides 𝐹1 = 0, and writing 𝐹𝑦 = 0 provides 𝑁1 = 200 lb. Finally, the equilibrium equations for block 𝐴 alone in Fig. 3 (with 𝑁3 = 𝐹3 = 0) are ∑ Block A: 𝐹𝑥 = 0 ∶ 𝑁2 (sin 10◦ ) − 𝐹2 (cos 10◦ ) = 0, (10) ∑ ◦ ◦ 𝐹𝑦 = 0 ∶ 𝑁2 (cos 10 ) + 𝐹2 (sin 10 ) − 200 lb = 0, (11) ⇒

𝑁2 = 197.0 lb,

and 𝐹2 = 34.73 lb.

𝑁3 𝐴 𝐹3 10◦

𝑦

𝑁2 𝐹2 10◦

𝑥

𝐵 𝐹1

𝑁1

Figure 3 Free body diagrams with 𝑃 = 0.

(12)

Values for all six unknowns have been obtained (i.e., 𝑁1 = 200 lb, 𝐹1 = 0, 𝑁2 = 197.0 lb, 𝐹2 = 34.73 lb, 𝑁3 = 𝐹3 = 0), and we must now verify that these satisfy Coulomb’s law. Clearly, surface 1 does not slide, and surface 3 is not in contact. For surface 2, Coulomb’s law states that sliding is impending when 𝐹2 = 𝜇𝑁2 = 39.39 lb. Since this value is larger than the result for 𝐹2 obtained in Eq. (12), we conclude that surface 2 does not slide. Hence, the solution we obtained satisfies all equilibrium equations and Coulomb’s law; therefore our initial assumptions were correct, and we may conclude that the system will be at rest when 𝑃 = 0.

200 lb 𝑁4

𝐴 𝐹4

Alternate solution.

As an alternative solution, we can demonstrate that a nonzero force is required to slide block 𝐵 to the left. The solution procedure of Part (a) is repeated, and the FBDs in Fig. 4 are used with the following comments. Force 𝑄 is defined to be positive in the −𝑥 direction. At the outset of this solution, it may not be clear whether block 𝐴 contacts guide 𝐷 or 𝐸, and in the course of this solution you will find that contact is made at 𝐸, whose contact forces are defined to be 𝑁4 and 𝐹4 . Observe that the directions of all friction forces oppose the directions of sliding. Problem 9.9 asks you to carry out this solution, and you should determine that 𝑄 = 44.37 lb will cause block 𝐴 to be lowered. From this we may infer there is no motion when 𝑄 = 0, and hence when 𝑃 = 0. Discussion & Verification Impending upward motion of block 𝐴 occurs when 𝑃 = 128.0 lb, and the system is at rest when 𝑃 = 0. Hence, this design is acceptable from the point of view that the wedge will not slip out of place when the insertion force 𝑃 is removed.

∗ While

𝐹2

one moment equilibrium equation could be written for each of blocks 𝐴 and 𝐵, three additional unknowns in the form of the locations of the contact force systems must also be introduced; hence, the problem would remain statically indeterminate.

10◦

𝑦

𝑁2 𝑄

𝐹2

𝐹2 10◦

𝑥

𝐵 𝐹1

𝑁1

Figure 4 Free body diagrams for lowering block 𝐴.

543

544

Chapter 9

Friction

Problems Problems 9.1 and 9.2 10 in.

A worker applies the force described below to push a box that weighs 40 lb with center of gravity at point 𝐺. The surface between the box and ramp has coefficient of friction 𝜇𝑠 = 𝜇𝑘 = 0.25. Determine the normal force and friction force between the box and ramp, and determine if the box will slide up the ramp, down the ramp, or remain at rest.

5 in. 5 in.

10 in. 𝐺 30◦

Problem 9.1

The worker applies a 30 lb force parallel to the ramp.

Problem 9.2

The worker applies a 35 lb force that is horizontal.

Figure P9.1 and P9.2

Problem 9.3 A person applies a horizontal force 𝑃 to a rectangular box that is resting on a horizontal surface. The box weighs 𝑊 = 20 N with center of gravity at point 𝐺. The surface has coefficients of static and kinetic friction that are equal, with value 𝜇 = 0.4. Assume the box does not tip. 10 cm 10 cm

(a) An FBD is shown; this FBD has an error. Explain the error and draw the correct FBD. 𝑊

8 cm

𝑃

(b) If 𝑃 = 7 N, determine the normal force 𝑁 and friction force 𝐹 . Does the box remain at rest or does it slide? If the box slides, does it do so with constant speed or does it accelerate?

𝐺

𝐺 8 cm

𝐹

𝜇 = 0.4

𝑁

(c) Repeat Part (b) if 𝑃 = 9 N. (d) Determine if the assumption that the box does not tip is correct.

Figure P9.3

𝑃

Problem 9.4

𝑃 𝐴

A cross-sectional view through the wheel of a bicycle is shown, where points 𝐴, 𝐵, and 𝐶 lie in the same plane. The vertical force applied to the axle is 𝑊 = 300 N, and the horizontal forces 𝑃 are applied by the brake to the rim of the wheel (if you are curious, Prob. 6.96 on p. 423 shows the brake mechanism that applies forces 𝑃 ). If the coefficient of friction between the wheel and pavement is 1.1, and the coefficient of friction between the brake material and wheel rim is 0.4, determine the value of 𝑃 that will cause the wheel to skid on the pavement.

28 cm 𝐵 𝑊 32 cm wheel rim tire

Problem 9.5 ∑ For Example 9.1 on p. 540, write 𝑀 = 0 about some convenient point to verify that the distances from the lower left-hand corner of the box (point 𝐷) to points 𝐴 and 𝐵 are 5.18 in. and 11.2 in., respectively. In view of these results, is motion of the box in fact due to sliding (as assumed in Example 9.1) or is it due to tipping? Explain.

𝐶 Figure P9.4

Problem 9.6 𝜃 𝐴

𝐶 𝐵

Figure P9.6

ISTUDY

The apparatus shown can be used to experimentally determine the angle of static friction, and hence the coefficient of static friction, for many combinations of contacting materials. A block of material 𝐶 rests on a beam 𝐴𝐵. Starting with 𝜃 = 0◦ , point 𝐵 is slowly lowered until block 𝐶 begins to slide. Assuming block 𝐶 does not tip, show that the value of 𝜃 when sliding starts is equal to the angle of friction 𝜙, given by Eq. (9.4) on p. 537.

ISTUDY

Section 9.1

Basic Concepts

100 cm

Problem 9.7 A tool chest has 800 N weight that acts through the midpoint of the chest. The chest is supported by feet at 𝐴 and rollers at 𝐵. The surface has a coefficient of friction of 0.3. Determine the value of the horizontal force 𝑃 necessary to cause motion of the chest to the right, and determine if the motion is sliding or tipping.

𝑃 140 cm 100 cm

Problem 9.8

𝐵

In Example 9.2 on p. 541, determine the value of ℎ so that sliding and tipping motion of the traffic barrier are simultaneously impending.

𝐴

70 cm 15 cm

15 cm

Figure P9.7

Problem 9.9 Carry out the alternate solution described in Example 9.3 on p. 543.

Problem 9.10 The photograph shows two U.S. Coast Guard icebreakers in an ice field, and a simple model for an ice breaking operation. If the coefficients of static and kinetic friction for contact between the ship’s bow and ice are 0.08 and 0.06, respectively, and if the ship produces a thrust of 106 lb, determine the normal and friction forces acting on each side of the ship’s bow as it moves through the ice field with constant velocity. Assume the ship makes contact with the ice only on its bow, and neglect all forces between the ship’s hull and water except for the thrust.

United States Coast Guard

Problem 9.11 The structure consists of two uniform members 𝐴𝐵 and 𝐵𝐶, each weighing 2 kN. The members are pinned to each other at 𝐵, and the structure is supported by a pin at 𝐶 and a surface at 𝐴 having coefficient of static friction 1.2. Determine the largest positive value 𝑃 the structure can support.

Figure P9.10

𝑃 𝐴

𝐶 1 ft

𝐵

4

3m

3

1 ft

𝐴 4m

5m Figure P9.11

𝐵

𝐶

Figure P9.12

Problem 9.12 Water (𝛾 = 62.4 lb∕f t 3 ) is retained by a uniform thin semicircular dam having 1 f t radius and 3 f t depth into the plane of the figure. The dam weighs 60 lb and is supported by a cable 𝐴𝐶 and by frictional contact with the bottom of the channel at 𝐵. Determine the minimum coefficient of friction required so that the dam does not slip at 𝐵.

1.2 m 1.2 m 𝐷

𝐶

concrete

Problem 9.13 A long concrete retaining wall with 2.4 × 103 kg∕m3 density is used to support a soil embankment. Assuming the soil behaves as a fluid with 1.5 × 103 kg∕m3 density, determine the required coefficient of friction so the wall does not slip along its base, and determine if the wall is safe from tipping.

3m soil 𝐴 Figure P9.13

𝐵

545

546

Chapter 9

Friction

Problem 9.14 8m

A long concrete gravity dam retains water in a reservoir. The surface between the dam and earth has coefficients of friction 𝜇𝑠 = 0.8 and 𝜇𝑘 = 0.7. When the reservoir is completely full (i.e., ℎ = 8 m), determine if the dam is safe from both overturning (tipping) and sliding along its base. The specific weight of concrete is 𝛾𝑐 = 25 kN∕m3 and the density of water is 𝜌𝑤 = 103 kg∕m3 .

ℎ

6m Figure P9.14

Problem 9.15 1 ft

1 ft

Repeat Prob. 9.14 if the reservoir is on the left-hand side of the dam.

Problem 9.16 3 ft

The owner of a small concrete gravity dam is considering attaching steel plate to the face of the dam so that a greater depth of water can be retained. The surface between the dam and earth has coefficients of friction 𝜇𝑠 = 0.6 and 𝜇𝑘 = 0.55. Determine if the dam is safe from both overturning (tipping) and sliding along its base. The specific weights of concrete and water are 𝛾𝑐 = 150 lb∕f t 3 and 𝛾𝑤 = 62.4 lb∕f t 3 , respectively. Neglect the weight of the steel plate.

2 ft

Figure P9.16 300 mm

300 mm 𝑊

Problem 9.17

𝐴 80 mm

𝐵 15◦

𝐶

𝑃

A wedge is used to level a structure. All contact surfaces have coefficients of static and kinetic friction of 0.3 and 0.25, respectively, and 𝑊 = 500 N. Assume the dimensions of the wedge are small. Determine the value of 𝑃 to cause impending motion of the wedge: (a) To the left.

Figure P9.17

(b) To the right.

Problems 9.18 and 9.19

𝐴 20◦

𝑃

Blocks 𝐴 and 𝐵 each have 2 kg mass. All contact surfaces have the same coefficient of friction. Determine the force 𝑃 needed to cause impending motion of block 𝐵 to the left if the coefficient of static friction is

𝐵 Figure P9.18 and P9.19

ISTUDY

Problem 9.18

0.4.

Problem 9.19

0.3.

Problem 9.20 A roll of paper weighs 5 N with center of gravity at point 𝐴. The roll is supported by a steel bar 𝐴𝐵 that has negligible weight, and the roll rests against a vertical wall with equal coefficients of static and kinetic friction of 0.5. If the paper tears when angle 𝜃 reaches 20◦ , determine the strength of the sheet of paper. 𝐵 80 mm

𝜃 60 mm

𝐴

𝐶

Figure P9.20 and P9.21

ISTUDY

Section 9.1

Basic Concepts

Problem 9.21 In Prob. 9.20, let the strength of the paper be 4 N. Determine the largest value of angle 𝜃 that will permit paper to be pulled off the roll without tearing. Hint: The equilibrium equations are difficult to solve analytically. Thus, a solution using software such as Mathematica or Maple is helpful, or an approximate solution (i.e., graphical or by trial and error) to obtain 𝜃 with two-digit accuracy is also acceptable.

Problem 9.22 The bicycle shown has a brake for each wheel. The rider and bicycle weigh 120 lb with center of gravity at point 𝐶, and the bicycle descends a straight slope with 𝜃 = 15◦ . For each of the braking situations given below, determine the minimum coefficient of kinetic friction between the tires and pavement so that the bicycle will descend the slope at a uniform speed, and determine if the bicycle is safe from tipping. (a) Only the rear brake is applied, causing the rear wheel to skid on the pavement. (b) Only the front brake is applied, causing the front wheel to skid on the pavement. (c) Both the rear and front brakes are applied, causing both wheels to skid on the pavement.

16 in.

20 in.

28 in. 14 in. 𝐴

𝜃

14 in. 𝐵

Figure P9.22

Problem 9.23 The machine shown is used to move boxes. Bar 𝐴𝐵𝐶 slides horizontally in the bearing of the fixed machine housing. Points 𝐵, 𝐶, and 𝐷 are pins, and point 𝐶 has a frictionless roller. The flywheel 𝐸 rotates clockwise under the action of moment 𝑀𝐸 . The horizontal surface on which the box rests has coefficients of friction 𝜇𝑠 = 0.3 and 𝜇𝑘 = 0.25, and all other contact surfaces are frictionless. If the box weighs 900 N, determine the moment 𝑀𝐸 that must be applied to the flywheel to initiate motion of the box, and determine if the motion is sliding or tipping.

𝐵 𝐴 0.2 m 𝐸

𝐶

0.6 m

𝐷 20◦

0.5 m

45◦ 𝑀𝐸 Figure P9.23

0.6 m

547

548

Chapter 9

Friction

Problem 9.24

𝐵

𝜃 = 60◦ 𝐴 Figure P9.24 and P9.25

ISTUDY

An 8 f t long ladder has seven rungs. The rungs are spaced 1 f t apart, and the top and bottom rungs are 1 f t from their respective ends of the ladder. The top of the ladder has a roller. Neglect the weight of the ladder and assume the worker’s hand applies no force to the ladder. (a) If the worker weighs 140 lb and stands on the middle rung, determine the minimum value of the coefficient of friction required so that the ladder does not slide. (b) If the worker weighs 140 lb and stands on a different rung, does your answer to Part (a) change? Explain. (c) If the worker weighs more than 140 lb and stands on the middle rung, does your answer to Part (a) change? Explain.

Problem 9.25 In Prob. 9.24, the roller at 𝐵 is removed and the surfaces at 𝐴 and 𝐵 both have the same coefficient of friction. If the worker weighs 140 lb and stands on the middle rung, determine the minimum value of the coefficient of friction required so that the ladder does not slide. Hint: The use of mathematical software for solving equations is helpful, but is not required.

Problems 9.26 through 9.28 The mechanism for a gumball machine is shown in Figs. P9.26 and P9.27. To begin dispensing a gumball, as shown in Fig. P9.26, a horizontal force 𝑃1 is applied to block 𝐴 to cause it to begin sliding to the right while member 𝐶𝐷 simultaneously begins to lift door 𝐵. After the door 𝐵 is fully open, as shown in Fig. P9.27, the gumball drops out of the machine. Blocks 𝐴 and 𝐵 weigh 0.3 lb and 0.2 lb, respectively, and the weights of member 𝐶𝐷 and the gumballs may be neglected. The pins at points 𝐶 and 𝐷 are frictionless, blocks 𝐴 and 𝐵 are a loose fit in their tracks, and all contact surfaces have a coefficient of friction of 0.3. The motion of the blocks is slow enough so that static equilibrium may be assumed. Problem 9.26

Determine the value of 𝑃1 that will cause block 𝐴 to begin moving to

the right. Problem 9.27

Just before the door 𝐵 is fully opened, as shown in Fig. P9.27, determine

the value of 𝑃2 . In Fig. P9.27, where the door is fully open, replace the force 𝑃2 with a force 𝑄 in the opposite direction. Determine the value of 𝑄 so that block 𝐴 will begin moving to the left.

Problem 9.28

gumballs

door

gumballs

door

𝐷 4.8 cm 𝐷 𝐵

3 cm 𝑃1

𝐴 1 cm

𝐵

𝐶 4 cm

Figure P9.26

𝑃2

𝐴 1 cm

𝐶 1.4 cm

Figure P9.27

ISTUDY

Section 9.1

Basic Concepts

Problem 9.29

549

8 in. 𝐸

The tool shown is used by a person to carry a large sheet of plywood. The tool consists of aluminum clamps 𝐴 and 𝐵, and steel member 𝐶𝐷𝐸, with frictionless pins at points 𝐶 and 𝐷. The weights of all members are negligible, except for the weight of the plywood. If the coefficient of friction between aluminum and plywood is 0.2, determine the largest value of ℎ required so that the plywood does not slip out of the tool. Assume that the person rests the plywood on their hip so that the sheet of plywood is vertical.

𝐷 ℎ 𝐶 𝐵

𝐴 plywood

Problem 9.30 A table saw for cutting wood is shown. The blade rotates counterclockwise, and the operator pushes the wood into the blade using a stick to help keep his or her hand away from the blade. Despite this safety precaution, it is possible for the wood to be propelled by the blade with great force and speed into the operator, causing injury. To help prevent this accident, the saw is outfitted with an antikickback device, which weighs 0.3 lb with center of gravity at point 𝐵. Neglecting friction between the wood and saw table, determine the minimum coefficient of friction between the wood and the antikickback device that will prevent the wood workpiece from kicking back. 1 in. 𝐴 2 in.

𝐵

Figure P9.29

0.5 in. antikickback device 𝐶

𝐷 𝐸 saw table wood workpiece Figure P9.30 𝑧

Problem 9.31 Bar 𝐴𝐵𝐶 has square cross section and can slide in the square hole that is in collar 𝐷. Collar 𝐷 is supported by fixed vertical post 𝐸 that is built in at its base. The collar 𝐷 can translate in the 𝑧 direction and rotate about the 𝑧 axis without friction; other translations and rotations are constrained. End 𝐶 of the square cross section bar rests on a horizontal surface having coefficients of friction 𝜇𝑠 = 0.6 and 𝜇𝑘 = 0.5. If bar 𝐴𝐵𝐶 is initially motionless, determine the positive value of 𝑃 that will cause impending motion. For this value of 𝑃 , also determine the reactions between bar 𝐴𝐵𝐶 and the collar 𝐷.

𝑃

𝐴

100 lb

𝐷

30 lb

3 in.

𝐵

10 in. 𝑥 Figure P9.31

𝐸

𝐶

5 in.

𝑦

550

Chapter 9

Friction

Design Problems General Instructions. In all problems, write a brief technical report following the guidelines of Appendix A, where you summarize all pertinent information in a wellorganized fashion. It should be written using proper, simple English that is easy for another engineer to read. Where appropriate, sketches along with critical dimensions should be included. Discuss the objectives and constraints considered in your design, the process used to arrive at your final design, safety issues if appropriate, and so on. The main discussion should be typed, and figures, if needed, can be computer-drawn or neatly hand-drawn. Include a neat copy of all supporting calculations in an appendix that you can refer to in the main discussion of your report. A length of a few pages, plus appendix, should be sufficient. Design Problem 9.1

𝑃

detent housing

𝑑

detent 2 mm

track

ℎ Figure DP9.1

Design a detent mechanism for precision positioning control. This mechanism consists of a spring-loaded plunger (detent) with a spherical end that is normally positioned at the bottom of the spaced notches on the track. The track is fixed in space, and the detent housing can undergo horizontal translational motion only. It is desired that the detent housing move to the next available notch when the force 𝑃 is approximately 2 N. Specify the spring stiffness (in N/mm), the spring’s unstretched length, and dimensions ℎ and 𝑑 where 𝑑 is the length of the spring cavity when the detent is at the bottom of a notch. The coefficient of friction between the detent and the track is 0.3, and friction between the detent and the housing can be assumed to be negligible because of sufficient lubrication.

Design Problem 9.2 ladder rail plate

𝐸 40 40

dimensions in mm 𝐶

𝐴 30

𝑑

𝐵 cam

𝐷 80 Figure DP9.2

ISTUDY

Design a safety device for use on fixed steel ladders. The device consists of a cam, a plate, and several roller bearings attached to the plate. The device slips onto the rail of the ladder, and it is attached by a short cable to a harness worn by the person using the ladder. It has the feature that if the person were to accidentally fall, the device automatically would lock in position on the ladder and limit the distance the person would fall. A small upward force on the cam allows it to slide to a new position on the ladder. All materials are steel. The pins at 𝐸 and 𝐷 are equipped with roller bearings. The cam has a circular surface that contacts the flange of the ladder at 𝐶 and is hinged to the plate at point 𝐴. Specify the dimension 𝑑. Note that small values of 𝑑 produce large contact forces, but they also lead to more rapid wear of the cam and bearings. If 𝑑 is too large, it may not provide sufficient frictional resistance. Thus, you should specify a value of 𝑑 with these competing factors in mind.

ISTUDY

Section 9.2

9.2

551

Problems with Multiple Contact Surfaces

Problems with Multiple Contact Surfaces

This section addresses friction problems that have multiple contact surfaces where, in general, only some of the surfaces slide while the others do not. Such problems are challenging because we usually do not know at the outset of a solution which of the surfaces will slip. Note that it is only for surfaces that slide or have impending motion that we may write 𝐹 = 𝜇𝑁, and for surfaces that do not slide, 𝐹 < 𝜇𝑁. In Section 9.1, some problems with multiple contact surfaces were considered, but these problems were simpler because if one surface slid, then all surfaces were required to slide. The subject of geometry of motion is called kinematics, and the problems considered in this section have more complicated kinematics than those in Section 9.1. The basic concepts needed to address the problems of this section were all covered in Section 9.1. The new ingredient needed here is an understanding that only a subset of the contact surfaces will slide.

𝑃 𝐷

𝐸 (a) 𝐴

𝑃

𝐷𝑦

𝐸

End of Section Summary This section considers problems with multiple contact surfaces where only a subset of the surfaces slide. Some of the key points are: • At the outset of an analysis, it is often uncertain which surfaces will slide. Note that 𝐹 = 𝜇𝑁 may be used only for those surfaces that slide or have impending slip, while 𝐹 < 𝜇𝑁 applies for those that do not slide. • Simple kinematics help clarify the directions for sliding so that friction forces can be given the correct directions in FBDs.

𝐷𝑦 𝐷𝑥

Determination of sliding directions When you draw FBDs, it is necessary to know the directions in which sliding may occur so that friction forces can be given the proper directions. Determination of sliding directions is sometimes challenging, and simple kinematics usually help clarify this, as demonstrated in the following example. Consider the use of a wrench to twist a pipe as shown in Fig. 9.7(a). In Fig. 9.7(b), we begin drawing FBDs; most of the forces in these FBDs are straightforward to assign, but the directions of the friction forces are, at this point, probably uncertain. As an example of how you might determine these directions, imagine that pipe 𝐶 in Fig. 9.7(a) is fixed, and let there be no slip at 𝐵. If handle 𝐸 moves down, then clearly jaw 𝐴 also moves down, and hence there is sliding at that location. The friction force 𝐹1 the pipe applies to the jaw at 𝐴 must be upward, and by Newton’s third law, the friction force the jaw applies to the pipe is in the opposite direction. We now repeat this thought process by letting there be no slip at 𝐴. If handle 𝐸 moves down, then clearly jaw 𝐵 moves up, and hence there is sliding at that location. The friction force 𝐹2 the pipe applies to the jaw at 𝐵 must be downward, and the friction force the jaw applies to the pipe is in the opposite direction. The final FBDs are shown in Fig. 9.7(c). Once the FBDs are drawn, we proceed with the analysis by writing equilibrium equations, applying Coulomb’s law, etc., to determine whether surface 𝐴 or surface 𝐵 will slide; Prob. 9.48 asks you to do this.

𝐵

𝐶

?

?

𝐴 𝑦

(b)

? 𝑥

? 𝑁1

𝐵

𝑁1 𝑁2 𝑀𝐶

? ? 𝑁2

?

? 𝑃

𝑃

𝐷𝑦

𝐸

𝐴 (c)

𝐷𝑦 𝐹1

𝑦

𝐷𝑥 𝑁1

𝑁2

𝑀𝐶

𝑥 𝑁1

𝐵 𝐹2

𝐹2 𝑁2

𝐹1 𝑃

Figure 9.7 (a) A wrench is used to twist a pipe 𝐶. The wrench consists of jaw 𝐵𝐷 and handle-jaw 𝐴𝐷𝐸, with a pin at 𝐷. (b) Partially complete FBDs where the directions of the friction forces are uncertain, as indicated by the “ ? ” symbols and dashed arrowheads. The force 𝑃 and moment 𝑀𝐶 on the pipe are reactions from the pipe’s support that equilibrate the force 𝑃 applied to handle 𝐸 of the wrench. (c) Complete FBDs. Note: The FBDs can be further simplified by noticing that member 𝐵𝐷 is a two-force member.

552

Chapter 9

Friction

E X A M P L E 9.4

Multiple Contact Surfaces 𝐴

12

5 4

𝐵

3 𝐷

𝐶

A makeshift crane consists of a cable 𝐴𝐵, a steel bar 𝐴𝐶, concrete blocks 𝐵 and 𝐶, and a pulley system. While the crane is primitive, it is inexpensive and portable, and it may be especially useful in impoverished countries for lifting heavy objects on and off trucks and carts. Concrete blocks 𝐵 and 𝐶 weigh 8 kN and 5 kN, respectively, and the coefficient of friction between the blocks and the soil on which they rest is 0.5. Determine the weight 𝑊 of the heaviest object 𝐷 that may be lifted, and specify whether block 𝐵 or 𝐶 slides.

SOLUTION

Figure 1

Road Map

𝑇𝐴𝐵

5 kN

𝑁1

𝐶 𝐹1 𝐹2

𝑊

13 5 5 3 𝑃𝐴𝐶 12 4 𝑦 𝑃𝐴𝐶

8 kN 𝑇 𝐴𝐵 𝐵

𝐴

Modeling The FBDs for point 𝐴 and blocks 𝐵 and 𝐶 are shown in Fig. 2, where the cable force 𝑇𝐴𝐵 is positive in tension and the bar force 𝑃𝐴𝐶 is positive in compression. Note that if block 𝐵 slides, it will move to the right, and if block 𝐶 slides, it will move to the left, and the friction forces in these FBDs oppose these motions.

𝑥 Governing Equations & Computation

𝑁2

Figure 2 Free body diagrams in which the force the person applies to the pulley system is assumed to be small.

ISTUDY

We will neglect the weights of the cable, bar, and pulley system and will assume that blocks 𝐵 and 𝐶 do not tip. Because of the pulley system, the force applied by the person is small relative to the weight being lifted,∗ and we will therefore neglect the force the person applies to the cable. It is not likely that both blocks 𝐵 and 𝐶 will slide at the same time, and it is not clear which of these will reach its sliding threshold first. Our strategy will be to determine the forces supported by the cable and bar in terms of the weight 𝑊 of object 𝐷. We will then proceed to analyze the equilibrium of block 𝐵, and then block 𝐶, to determine which of these slides, and the weight 𝑊 necessary to cause this.

Equilibrium Equations

Using the FBDs in Fig. 2, equilibrium equations are written and

solved as follows: Point A:

∑

∑

𝐹𝑥 = 0 ∶

−𝑇𝐴𝐵 12 + 𝑃𝐴𝐶 45 = 0, 13

(1)

𝐹𝑦 = 0 ∶

5 + 𝑃𝐴𝐶 53 − 𝑊 = 0, −𝑇𝐴𝐵 13 𝑇𝐴𝐵 = 13 𝑊 and 𝑃𝐴𝐶 = 15 𝑊. 4 4

(2)

⇒

Common Pitfall Overuse of 𝑭 = 𝝁𝑵. A common error in problems with friction is to use 𝐹 = 𝜇𝑁 for all contact surfaces. While this is sometimes true (as in the example problems of Section 9.1), in general this is not the case. In Fig. 2 above, taking 𝐹1 = 𝜇𝑁1 and 𝐹2 = 𝜇𝑁2 is incorrect because it implies that both blocks have impending motion at the same time, and further, the equilibrium equations cannot be satisfied. At the end of this solution we see that when 𝑊 reaches its maximum value, that 𝐹1 = 𝜇𝑁1 and 𝐹2 < 𝜇𝑁2 .

Block B:

∑

∑

𝐹𝑥 = 0 ∶

12 𝑇𝐴𝐵 13 − 𝐹1 = 0,

(4)

𝐹𝑦 = 0 ∶

5 + 𝑁1 = 0, −8 kN + 𝑇𝐴𝐵 13

(5)

∑

∑

(6)

𝐹𝑥 = 0 ∶

−𝑃𝐴𝐶 45 + 𝐹2 = 0,

(7)

𝐹𝑦 = 0 ∶

−5 kN − 𝑃𝐴𝐶 35 + 𝑁2 = 0,

(8)

⇒

𝐹1 = 3𝑊

𝐹2 = 3𝑊

and 𝑁1 = 8 kN −

5 𝑊 4

.

⇒

Block C:

(3)

and 𝑁2 = 5 kN +

9 𝑊 4

.

(9)

Note that if point 𝐴 and blocks 𝐵 and 𝐶 are in equilibrium, then all the preceding expressions are valid, regardless of whether block 𝐵 or block 𝐶 has impending slip. ∗ Exactly

what force the person applies relative to the weight being lifted depends on details of the pulley system, which are not described here.

ISTUDY

Section 9.2

Problems with Multiple Contact Surfaces

553

Force Laws We now apply Coulomb’s law to determine the value of 𝑊 needed to cause each of the blocks to slide.

If block 𝐵 slides: using Eq. (6) →

𝐹1 = 𝜇𝑁1

(11)

𝑊 = 1.103 kN.

(12)

⇒ If block 𝐶 slides: using Eq. (9) →

(10)

) ( 3𝑊 = (0.5) 8 kN − 45 𝑊

𝐹2 = 𝜇𝑁2

(13)

( ) 3𝑊 = (0.5) 5 kN + 94 𝑊

(14)

𝑊 = 1.333 kN.

(15)

⇒

When 𝑊 reaches the smaller of the values given by Eqs. (12) and (15), sliding is impending. Hence, the largest value of weight that may be lifted is 𝑊max = 1.103 kN,

(16)

and when this occurs, sliding of block 𝐵 is impending. Discussion & Verification

An analysis that includes the force the person applies to the cable will give more precise results than those obtained here. To carry out this analysis, details of the pulley system must be known, and the angle at which the person applies the force must be known, or more likely, a range of reasonable values should be considered.

Helpful Information Failure criteria. In this example, Eqs. (10) and (13) are failure criteria that govern when each block will have impending slip, and until the solution is complete we do not know which one of these will actually occur. You should contrast the approach used here to the many examples earlier in this book that involve failure criteria for cables and bars, such as Eqs. (9) and (10) in Example 3.1 on p. 137. The approach used throughout this book to apply failure criteria, whether due to frictional sliding or to strength of structural members, is the same.

554

Chapter 9

Friction

E X A M P L E 9.5

Multiple Contact Surfaces A 5◦ steel wedge 𝐶 is used to level a 10 f t long uniform steel I beam 𝐴𝐵 weighing 400 lb. The ends of the beam are supported by concrete walls 𝐷 and 𝐸. Determine the force needed to drive the wedge to the left, and describe the motion of the beam that results (i.e., determine if end 𝐵 of the beam is lifted or if the entire beam slides to the left). The coefficients of static friction are 0.4 for steel-on-steel contact and 0.3 for steel-on-concrete contact.

10 f t 𝐴

𝐵 𝐶

𝐸

𝐷

5◦

Figure 1

SOLUTION Road Map

We will neglect the weight of the wedge. There are three potential sliding surfaces in this problem. While sliding must occur between the bottom of the wedge and the stationary wall at 𝐷, it is unclear which of the remaining two contacts slides. In the first part of our solution we determine the force needed to slide the beam and wedge together to the left. In the second part, we determine the force needed to slide only the wedge. The smaller of these values is the force that causes motion to occur. Sliding of beam and wedge together

Modeling

The FBD for the beam and wedge together is shown in Fig. 2, where the direction of the friction forces opposes sliding of the beam to the left. When writing equilibrium equations, we will assume the lines of action of 𝐹1 , 𝐹2 , and 𝑃 are the same, which is reasonable if the dimensions of the wedge are small. We also assume that the support reaction at 𝐴 is positioned at the end of the beam. 5 ft

𝑦

5 ft 400 lb

𝑥 𝐵 𝐶 𝑃

𝐴 𝐹2

𝑁2

Governing Equations

𝐹1

𝑁1

Figure 2 Free body diagram for the beam and wedge sliding together to the left.

ISTUDY

Equilibrium Equations

∑

𝑀𝐴 = 0 ∶ ∑ 𝐹𝑦 = 0 ∶ ∑ 𝐹𝑥 = 0 ∶

Using the FBD in Fig. 2, the equilibrium equations are 𝑁1 (10 f t) − (400 lb)(5 f t) = 0

⇒

𝑁1 = 200 lb,

(1)

𝑁1 + 𝑁2 − 400 lb = 0

⇒

𝑁2 = 200 lb,

(2)

𝐹1 + 𝐹2 − 𝑃 = 0.

(3)

These three equations contain five unknowns, although we were able to solve for 𝑁1 and 𝑁2 . Force Laws

Assuming that sliding occurs at surfaces 1 and 2, Coulomb’s law becomes 𝐹1 = 𝜇𝑐 𝑁1 = (0.3)(200 lb) = 60 lb,

(4)

𝐹2 = 𝜇𝑐 𝑁2 = (0.3)(200 lb) = 60 lb,

(5)

where 𝜇𝑐 is the coefficient of static friction for steel-on-concrete. Computation

Solving Eq. (3) for 𝑃 provides ⇒

𝑃 = 120 lb.

(6)

Sliding of wedge only Modeling

Here we consider the wedge sliding to the left while end 𝐵 of the beam is lifted. Note that the FBD in Fig. 2 is still valid, and the equilibrium equations [Eqs. (1)– (3)] are still valid, hence 𝑁1 = 200 lb and 𝑁2 = 200 lb. To continue, we may use an FBD of beam 𝐴𝐵 or wedge 𝐶, and we choose the latter, as shown in Fig. 3.

ISTUDY

Section 9.2

Problems with Multiple Contact Surfaces

555

Governing Equations Equilibrium Equations

∑

∑

Using the FBD shown in Fig. 3, the equilibrium equations are

𝐹𝑥 = 0 ∶

𝐹1 + 𝑁3 (sin 5◦ ) + 𝐹3 (cos 5◦ ) − 𝑃 = 0,

(7)

𝐹𝑦 = 0 ∶

𝑁1 − 𝑁3 (cos 5◦ ) + 𝐹3 (sin 5◦ ) = 0.

(8)

Assuming that sliding occurs at surfaces 1 and 3, Coulomb’s law becomes 𝐹1 = 𝜇𝑐 𝑁1 = (0.3)(200 lb) = 60 lb,

(9)

𝐹3 = 𝜇𝑠 𝑁3 = (0.4)𝑁3 ,

(10)

where 𝜇𝑐 and 𝜇𝑠 are the coefficients of static friction for steel-on-concrete and steel-onsteel, respectively, and the directions of 𝐹1 and 𝐹3 have been assigned in the FBD of Fig. 3 to oppose sliding motion of the wedge to the left. Computation

⇒

Solving Eqs. (7)–(10), with 𝑁1 = 200 lb, provides 𝑁3 = 208.0 lb,

𝐹3 = 83.22 lb,

and

𝑃 = 161.0 lb.

5◦

𝐹3

(11)

Discussion & Verification Sliding occurs when 𝑃 reaches the smaller of Eqs. (6) and (11). Thus, when 𝑃 = 120 lb, the beam and wedge slide together to the left. Unfortunately, this is not the desired effect. Some possible remedies that might result in end 𝐵 of the beam being lifted are as follows:

• An additional weight could be temporarily added to the left-hand end of the beam while the wedge is inserted. For example, a worker could stand on end 𝐴. This will increase the normal force at end 𝐴, and hence increase the friction force there. • A lubricant could be applied to the contact between the beam and wedge so that the coefficient of friction there is lowered. • All the results obtained here are for static equilibrium. It may be possible to drive the wedge dynamically, such as by tapping it into position with a hammer. Analysis of this requires dynamics.

𝑁3 𝐶

𝐹1

Because 𝑁1 is known from Eq. (1), these two equations contain four unknowns. Force Laws

5◦

𝑦 𝑃

𝑥

𝑁1

Figure 3 Free body diagram for the wedge sliding to the left.

556

Chapter 9

Friction

Problems Problem 9.32 Three books rest on a table. Books 𝐴, 𝐵, and 𝐶 weigh 2 lb, 3 lb, and 4 lb, respectively. Determine the horizontal force applied to book 𝐴 that causes impending motion of any of the books to the right, and determine which books move. 𝐴 𝐵 𝐶 𝑃

𝜇𝑠 = 0.6 𝜇𝑠 = 0.3 𝜇𝑠 = 0.1

Figure P9.32

Problem 9.33 𝐷 𝜇2 = 0.4

To glue blocks 𝐵 and 𝐶 to one another, the blocks are temporarily held together until after a force 𝑃 = 6 lb is applied. Both blocks weigh 2 lb. During the gluing operation, the glue between blocks 𝐵 and 𝐶 acts as a lubricant; hence, this surface is frictionless. Using the coefficients of friction given in the figure, determine if this gluing operation can be performed without either block sliding, and if sliding does occur, determine which of the blocks will slide.

𝐶

frictionless 15◦ 𝜇1 = 0.2

𝐵 𝐴

Problem 9.34

Figure P9.33

A portion of a machine that is used to lift an egg is shown. The clamps at 𝐴 and 𝐵 apply horizontal forces 𝑃 to the egg, and the egg can have any of the three positions shown. If 𝑃 = 0.5 N and an egg has 60 g mass, determine the smallest value for the coefficient of friction at clamps 𝐴 and 𝐵 that will prevent the egg from slipping. 35 mm 25 mm 40 mm 𝐵 𝐴 𝐶 𝑃 (a)

𝐴 𝑃

𝑃

𝐵

𝐴

𝐵 𝐶

𝑃

𝑃

(b)

𝐶

𝑃

(c)

Figure P9.34

Problem 9.35 𝐴

𝑃

𝐵 1

2

3

𝑃

4

Figure P9.35

A portion of a machine that is used to lift a stack of four identical books in a printing factory is shown. The clamps at 𝐴 and 𝐵 apply horizontal forces 𝑃 to the books. The coefficient of friction between a clamp and book is 0.6, and the coefficient of friction between two books is 0.4. If each book weighs 8 N, determine the smallest value of 𝑃 that will enable the stack of books to be lifted.

Problem 9.36 𝐸 𝐷

Problem 9.37

6 in. 3 in.

𝐵

𝑃 𝐶 Figure P9.37

ISTUDY

In Prob. 9.35, if 𝑃 = 75 N, determine how many books may be lifted.

𝐴

Block 𝐴 and spool 𝐵 weigh 8 lb and 6 lb, respectively. The spool has a string wrapped around it to which a horizontal force 𝑃 is applied. The coefficient of static friction between the spool and the contact surfaces at 𝐴 and 𝐶 is 0.25. Determine the value of 𝑃 that causes impending motion, and determine if slip occurs at 𝐴, or 𝐶, or both locations simultaneously.

ISTUDY

Section 9.2

Problems with Multiple Contact Surfaces

Problems 9.38 through 9.40

6 in.

2 in. 𝐶

Bars 𝐴𝐵 and 𝐶𝐷 are uniform and each weighs 7 lb. The coefficient of static friction at surfaces 𝐴 and 𝐵 is the same.

2 in.

If 𝑃 = 𝑄 = 0, determine the minimum coefficient of static friction that will prevent motion in the system.

𝐷

𝐵 𝑃

2 in.

Problem 9.38

𝑄 2 in.

If 𝑄 = 0 and the coefficient of static friction for all surfaces is 0.6, determine the value of 𝑃 that causes impending motion of bar 𝐴𝐵 to the right.

Problem 9.39

𝐴 2 in.

Figure P9.38–P9.40

If 𝑃 = 0 and the coefficient of static friction for all surfaces is 0.6, determine the value of 𝑄 that causes impending motion of bar 𝐴𝐵 to the left. Problem 9.40

Problems 9.41 and 9.42 Blocks 𝐴 and 𝐵 each have 2 kg mass. All contact surfaces have the same coefficient of friction. Determine the force 𝑃 needed to cause impending motion of block 𝐵 to the right if the coefficient of static friction is: Problem 9.41

0.4.

Problem 9.42

0.3.

𝐴 𝑃

20◦ 𝐵 Figure P9.41 and P9.42

Problems 9.43 and 9.44 A truck with 1500 kg mass and center of gravity at point 𝐶 is used to pull a dumpster with 1700 kg mass and center of gravity at point 𝐺. The coefficient of static friction between the tires and pavement is 1.1, and between the dumpster and pavement is 0.5. Assume the truck’s engine has sufficient power and the dumpster does not tip. Hint: The use of mathematical software is helpful, but not required. Determine if the truck can pull the dumpster if: Problem 9.43

The truck has rear-wheel drive.

Problem 9.44

The truck has four-wheel drive and end 𝐸 of the cable is moved to

point 𝐻. dimensions in meters 1.5 𝐶 𝐴 1.3

𝐻 𝐷 0.8 𝐸 𝐵 0.5 1.5 1.2 1.7

1.6

𝐺 1.6

Figure P9.43 and P9.44

𝑃

𝐷

Problem 9.45

𝐶

Blocks 𝐴 and 𝐵 weigh 10 N and 15 N, respectively. Point 𝐶 is at the midpoint of member BD, and both members AC and BD have negligible weight. The coefficient of static friction for all contact surfaces is 0.3. Determine the positive value of 𝑃 that causes impending motion, and determine which of blocks 𝐴 or 𝐵 slides and the direction of motion.

12 𝐴 Figure P9.45

5

3

4 𝐵

557

558

Chapter 9

Friction

Problem 9.46 The structure consists of two uniform members AB and BC, each weighing 2 kN. The members are pinned to one another at 𝐵, and the structure is supported by surfaces at 𝐴 and 𝐶 having coefficients of static friction of 1.2 and 0.5, respectively. Determine the largest positive value 𝑃 the structure can support.

𝑃 𝐶 𝐵 3m 𝐴

The structure consists of two uniform members AB and BC, each weighing 2 kN and 𝑃 ≥ 0. Determine the minimum coefficient of static friction needed at surface 𝐴 and the minimum coefficient of static friction needed at surface 𝐶 so that neither surface slips for any value of 𝑃 (this is called self-locking).

5m

4m

Problem 9.47

Figure P9.46 and P9.47

Problem 9.48 A wrench is used to twist a pipe 𝐶. The wrench consists of jaw BD and handle-jaw 𝐴𝐷𝐸, with a pin at 𝐷. The coefficient of static friction 𝜇𝑠 for all contact surfaces is the same. Determine the minimum value of 𝜇𝑠 so that there is no slip at 𝐴 or 𝐵 regardless of the value of force 𝑃 (this is called self-locking). 18 cm

𝑃

𝐺

2 cm

4 cm

𝐷

𝐸

𝐴

𝐶

𝐵

5 cm

Figure P9.48 𝐵 𝐷

𝐶 𝐴

2 in. Figure P9.49

ISTUDY

1.5 in.

0.8 in.

Problem 9.49

0.8 in.

A small lamp is clamped on a fixed circular bar 𝐶. The lamp assembly weighs 4 lb with center of gravity at point 𝐺, and the weights of all other members are negligible. Determine the minimum moment the torsional spring at 𝐷 must produce so the lamp does not slip on the bar. The contact surfaces at 𝐴 and 𝐵 have coefficients of static friction of 𝜇𝐴 = 0.25 and 𝜇𝐵 = 0.35.

Problem 9.50 A bracket 𝐴𝐵𝐶𝐷𝐸𝐹 is clamped to a fixed circular bar. The bracket consists of a horizontal member 𝐴𝐵𝐶 that is pinned at 𝐶 to member 𝐶𝐷𝐸𝐹 , and a vertical bolt 𝐵𝐷. The force 𝑃 at point 𝐹 is vertical, and the bracket makes contact with the fixed bar at points 𝐴, 𝐺, and 𝐸. If all contact surfaces have the same coefficient of static friction of 0.4, and if 𝑃 = 50 lb, determine the force that bolt 𝐵𝐷 must support to prevent slip.

𝐷

𝐹

𝐸 1.2 in.

𝐺 𝐶

𝑃 𝐵

𝐴

1.2 in. 1.4 in.

7 in. Figure P9.50

ISTUDY

Section 9.2

Problems with Multiple Contact Surfaces

Design Problems General Instructions. In all problems, write a brief technical report following the guidelines of Appendix A, where you summarize all pertinent information in a wellorganized fashion. It should be written using proper, simple English that is easy to read by another engineer. Where appropriate, sketches along with critical dimensions should be included. Discuss the objectives and constraints considered in your design, the process used to arrive at your final design, safety issues if appropriate, and so on. The main discussion should be typed, and figures, if needed, can be computer-drawn or neatly handdrawn. Include a neat copy of all supporting calculations in an appendix that you can refer to in the main discussion of your report. A length of a few pages, plus appendix, should be sufficient.

40 mm 10 mm

Design Problem 9.3 Design a message holder that is to be mounted on a door or wall for holding paper notes. It is made of a 20 mm diameter circular aluminum bar and an extruded aluminum bracket. Both aluminum pieces have an anodized finish. Specify the angle 𝜃 so the holder is selflocking. That is, the note cannot be pulled out of the holder without manually lifting the circular bar. A large value of 𝜃 is probably desirable so that the circular bar may be easily lifted.

𝐴+𝐵 100 mm

paper note

Design Problem 9.4 A large concrete surface is to be removed by sawing it into rectangular pieces.∗ In each rectangular piece, three holes of depth 𝑑 are drilled perpendicular to the slab’s surface with the positioning shown relative to the slab’s center of gravity 𝐺. In each hole a steel plug is inserted, and the slab is lifted with cables. Design this lifting scheme by specifying the depth 𝑑 the holes should have and the height ℎ so that the plugs will self-lock when the slab is lifted. Assume the concrete slab is at least 4 in. thick.

ℎ 3 ft 3 ft

𝑑 𝐺

3 ft 3 ft Figure DP9.4

∗A

perhaps more conventional method of removal is to pulverize the slab using a tractor-mounted impact hammer. However, this method produces waste that is difficult to recycle. The scheme described here provides for waste that has useful salvage value. For example, such slabs are often used to line waterfronts for erosion control.

Figure DP9.3

𝜃

559

560

Chapter 9

Friction

9.3

Belts and Cables Contacting Cylindrical Surfaces

In this section we use Coulomb’s law to develop a theory for the tensile force in belts and cables sliding on cylindrical surfaces.

Equilibrium analysis Consider the situation of a flexible belt or cable in contact with a cylindrical surface, as shown in Fig. 9.9. At points 𝐴 and 𝐵 where the belt meets the cylindrical surface, the values of the tensile force in the belt are 𝑇1 and 𝑇2 , respectively. We will use the convention that 𝑇2 ≥ 𝑇1 . Thus 𝑇1 is the force on the low-tension side, and 𝑇2 is the force on the high-tension side. Between points 𝐴 and 𝐵 the tensile force 𝑇 is variable such that 𝑇1 ≤ 𝑇 ≤ 𝑇2 . In Fig. 9.9 the cylinder is fixed, and if slip occurs, the belt will move in the direction of 𝑇2 .∗ 𝑦 𝛥𝜃∕2 𝐸

Mar1Art1/Shutterstock

Figure 9.8 Front view of a diesel engine for an automobile. The belt for this engine wraps around seven pulleys and is needed to operate a variety of essential accessories such as the power steering pump and alternator. The belt and pulleys must be designed so there is no slip.

ISTUDY

direction of slip or 𝐵 impending slip

𝐸

𝐷

𝛽

𝜃

𝑇2

𝐶

𝑇2 ≥ 𝑇1

𝑥 𝑇 𝛥𝐹 = 𝜇 𝛥𝑁

𝑇 + 𝛥𝑇

𝐴 𝛥𝜃

𝐷 𝛥𝜃∕2

𝛥𝑁 𝑟

𝑇1

Figure 9.9 A flexible belt or cable wrapped around a fixed cylindrical surface.

𝛥𝜃 𝐶 Figure 9.10 Free body diagram of a small segment of belt.

An angular coordinate 𝜃 is defined in Fig. 9.9, where 𝜃 = 0 corresponds to point 𝐴 on the low-tension side, and 𝜃 = 𝛽 corresponds to point 𝐵 on the high-tension side. Angle 𝛽 is called the angle of wrap. Consider a small segment of the belt obtained by taking cuts through the belt at points 𝐷 and 𝐸 in Fig. 9.9. The FBD for this segment of the belt is shown in Fig. 9.10, and the equilibrium equations are ∑

∑

𝐹𝑥 = 0 ∶

−(𝑇 + Δ𝑇 ) cos(Δ𝜃∕2) + 𝑇 cos(Δ𝜃∕2) + Δ𝐹 = 0,

(9.6)

𝐹𝑦 = 0 ∶

−(𝑇 + Δ𝑇 ) sin(Δ𝜃∕2) − 𝑇 sin(Δ𝜃∕2) + Δ𝑁 = 0.

(9.7)

In these expressions, 𝑇 is the tensile force at position 𝜃 (point 𝐷), and 𝑇 + Δ𝑇 is the tensile force at position 𝜃 + Δ𝜃 (point 𝐸). The normal force between the belt and cylindrical surface is Δ𝑁, and the friction force is Δ𝐹 with direction that opposes the direction of relative slip. Assuming the belt slides on the surface or has impending slip, Coulomb’s law provides Δ𝐹 = 𝜇 Δ𝑁, where 𝜇 is the coefficient of kinetic friction in the case of slip and 𝜇 is the coefficient of static friction in the case of impending slip. Because Δ𝜃 is small and is measured in radians, sin Δ𝜃 ≈ Δ𝜃 and cos Δ𝜃 ≈ 1. Using these expressions, dividing Eqs. (9.6) and (9.7) by Δ𝜃, taking ∗ The FBD

in Fig. 9.10 and all of the expressions derived in this section are valid if Fig. 9.9 is revised so that the belt is fixed on the high-tension side and the cylinder rotates clockwise. Indeed, the FBD and all expressions are valid if both the belt and cylinder have motion provided the direction of slip or impending slip of the belt relative to the cylinder is in the direction of 𝑇2 .

ISTUDY

Section 9.3

Belts and Cables Contacting Cylindrical Surfaces

561

the limit as Δ𝜃 → 0, and simplifying provide 𝑑𝑇 = 𝜇𝑇. 𝑑𝜃

(9.8)

This expression is a differential equation that describes how the tensile force 𝑇 changes with position 𝜃. It can be solved by multiplying both sides by 𝑑𝜃∕𝑇 and integrating to write 𝑇2

𝛽

1 𝑑𝑇 = 𝜇 𝑑𝜃. ∫ 𝑇 ∫ 𝑇1

(9.9)

0

Assuming the coefficient of friction is uniform over the entire surface, the above integrals can be evaluated to obtain ln

𝑇2 𝑇1

= 𝜇𝛽.

(9.10)

Taking the base 𝑒 exponent (𝑒 = 2.71828 …) of both sides and solving for 𝑇2 provide 𝑇2 = 𝑇1 𝑒𝜇𝛽 ,

𝑇2 ≥ 𝑇1 ,

(9.11)

where 𝑇1 is the force in the low-tension side of the belt during slip or impending slip; 𝑇2 is the force in the high-tension side of the belt during slip or impending slip; 𝜇 is the coefficient of static or kinetic friction; and 𝛽 is the angle of wrap (𝛽 ≥ 0, measured in radians). You should be mindful of the assumptions underlying Eq. (9.11): the surface must be cylindrical, friction is governed by Coulomb’s law where the coefficient of friction is uniform, and the belt and drum are in the process of sliding relative to each other or sliding is impending. If sliding is not imminent, the most we can say is (𝑇2 )max > 𝑇2 ≥ 𝑇1 where (𝑇2 )max = 𝑇1 𝑒𝜇𝛽 . Problem solving Solving problems with belt friction can occasionally be challenging. To use Eq. (9.11), you must be able to identify the high-tension and lowtension sides of a belt. In other words, you must be able to determine the direction of relative slip. Example 9.6 discusses some intuitive and fundamental methods to determine the high-tension and low-tension sides of a belt. Problems with multiple contact surfaces can be challenging, and the remarks made earlier in this chapter also apply here. Depending on the kinematics of a problem, only a portion of the contacts may slip, and Eq. (9.11) may be used only for those belt-cylinder contacts that are sliding or have impending slip.

End of Section Summary In this section, Coulomb’s law was used to develop the relationship 𝑇2 = 𝑇1 𝑒𝜇𝛽 between values of the tensile forces on the high-tension side (𝑇2 ) and low-tension side (𝑇1 ) of a belt or cable in contact with a cylindrical surface under conditions of slip or impending slip. Prior to slip and impending slip, such as may occur in problems with multiple contact surfaces, this expression cannot be used.

Common Pitfall Overuse of 𝑻𝟐 = 𝑻𝟏 𝒆𝝁𝜷 . This expression relates belt tensions only when there is slip or impending slip. Prior to slip and impending slip, this expression cannot be used.

562

Chapter 9

Friction

E X A M P L E 9.6

Belt Friction An accessory belt for an engine is shown. Pulley 𝐴 is attached to the engine’s crankshaft and rotates clockwise. The belt tensioner consists of a frictionless idler pulley at 𝐵 that is mounted to a horizontal bar 𝐷 that slides in a frictionless track with a horizontal force 𝑃 . Pulley 𝐶 operates a hydraulic pump that requires 200 N⋅m. The coefficients of static friction for pulleys 𝐴 and 𝐶 are 0.4 and 0.6, respectively, and the radii of pulleys 𝐴 and 𝐶 are 110 and 80 mm, respectively. Determine the minimum value of 𝑃 needed so the belt does not slip.

130◦ 45◦

𝐶 𝐷

𝑃

𝐵 𝐴

60◦

155◦

SOLUTION Road Map

We will first identify the portions of the belt having uniform values of force. We then draw FBDs of selected pulleys and apply equilibrium equations to obtain some of the requirements the belt forces must satisfy. For example, the 200 N⋅m moment required of pulley 𝐶 must be produced by the net moment of the belt forces about this pulley’s bearing. Slip may occur between the belt and pulley 𝐴, or the belt and pulley 𝐶, and both of these possibilities must be considered to determine which occurs.

Figure 1

𝑇𝐴𝐵𝐶

𝐶 𝐵

𝑇𝐴𝐶 𝐴 𝑇𝐴𝐶 > 𝑇𝐴𝐵𝐶

Figure 2 Regions of the belt having uniform values of force 𝑇𝐴𝐵𝐶 and 𝑇𝐴𝐶 . The force is variable in portions of the belt shown by the dashed lines.

Modeling In Fig. 2 we identify portions of the belt that have uniform tensile force. Note that because pulley 𝐵 is frictionless, the force in the entire left-hand portion of the belt from where the belt breaks contact with pulley 𝐴 to where it makes contact with pulley 𝐶 has the same value, 𝑇𝐴𝐵𝐶 . The force in the right-hand portion of the belt is 𝑇𝐴𝐶 . In regions of the belt shown by dashed lines in Fig. 2, the force changes value, although the exact details of this are not needed. The FBDs for the pulleys are shown in Fig. 3. Governing Equations Equilibrium Equations

Pulley B: 𝑀𝐶 = 200 N⋅m 𝑟𝐶 = 80 mm 𝜇𝐶 = 0.6

𝑇𝐴𝐵𝐶 𝐶𝑥

45◦ 𝑇𝐴𝐵𝐶 𝑃

60◦

Pulley C:

𝐹𝑥 = 0 ∶

𝑇𝐴𝐵𝐶 (cos 45◦ ) + 𝑇𝐴𝐵𝐶 (cos 60◦ ) − 𝑃 = 0,

∑

𝑇𝐴𝐵𝐶 = (0.8284)𝑃 ,

𝑀𝐶 = 0 ∶

𝑇𝐴𝐵𝐶 (0.08 m) + 200 N⋅m − 𝑇𝐴𝐶 (0.08 m) = 0,

⇒

𝑇𝐴𝐶

𝑇𝐴𝐶 = 𝑇𝐴𝐵𝐶 + 2500 N.

(1) (2) (3) (4)

Combining Eqs. (2) and (4) provides 𝑇𝐴𝐶 = (0.8284)𝑃 + 2500 N.

(5)

𝑥 𝑇𝐴𝐶

𝐴𝑦

𝑇𝐴𝐵𝐶 𝐴𝑥

𝑟𝐴 = 110 mm 𝜇𝐴 = 0.4 𝑀𝐴

Figure 3 Free body diagrams for the pulleys.

ISTUDY

Using the FBDs from Fig. 3, the equilibrium equations are

⇒

𝑦

𝐵𝑦 𝑇𝐴𝐵𝐶

𝐶𝑦

∑

Equations (2) and (5) contain three unknowns 𝑇𝐴𝐵𝐶 , 𝑇𝐴𝐶 , and 𝑃 , and thus they cannot yet be solved. Force Laws

Before using Eq. (9.11) on p. 561 for belt friction, we must first identify the high-tension and low-tension belt forces. There are a variety of intuitive and fundamental ways to do this. You should study the problem description and Figs. 1 and 2 now to see if you can determine by inspection that 𝑇𝐴𝐶 > 𝑇𝐴𝐵𝐶 . Some explanations for why 𝑇𝐴𝐶 > 𝑇𝐴𝐵𝐶 follow: • Pulley 𝐴 is powered by moment 𝑀𝐴 , which is supplied by the crankshaft of the engine. Hence, due to 𝑀𝐴 , the right-hand portion of the belt (𝑇𝐴𝐶 ) is pulled toward the left-hand side (𝑇𝐴𝐵𝐶 ), hence 𝑇𝐴𝐶 > 𝑇𝐴𝐵𝐶 . • You could examine Eq. (4) to determine 𝑇𝐴𝐶 > 𝑇𝐴𝐵𝐶 . • Starting with the FBD of pulley 𝐴, shown in Fig. 3 and again in Fig. 4(a), we remove the belt from the pulley to draw the two FBDs shown in Fig. 4(b) and (c). In Fig. 4(b), the friction forces, which are distributed over the contact region

ISTUDY

Section 9.3

563

Belts and Cables Contacting Cylindrical Surfaces

between the pulley and belt, must be in a direction so that the pulley is in moment equilibrium. Knowing the direction of the friction forces in Fig. 4(b), the friction forces applied to the belt, shown in Fig. 4(c), must be in the opposite direction. ∑ Consideration of moment equilibrium 𝑀𝐴 = 0 for the FBD in Fig. 4(c) shows that 𝑇𝐴𝐶 > 𝑇𝐴𝐵𝐶 .

𝑇𝐴𝐵𝐶

If pulley A slips:

𝑇𝐴𝐶 = 𝑇𝐴𝐵𝐶 𝑒

(a)

(0.8284)𝑃 + 2500 N = (0.8284)𝑃 𝑒 ⇒

𝑃 = 1547 N.

(7)

𝑦 𝑥 𝑀𝐴

𝐴𝑦

(6) (0.4)(155◦ )(𝜋 rad∕180◦ )

𝑇𝐴𝐶

𝐴𝑥

We now apply Eq. (9.11) for belt friction to determine the value of 𝑃 needed to cause each of the pulleys to slip. 𝜇𝐴 𝛽𝐴

𝐴𝑦

𝑇𝐴𝐶 𝑇𝐴𝐵𝐶

𝐴𝑥

𝐴

(8) 𝑀𝐴

If pulley 𝐶 slips:

𝑇𝐴𝐶 = 𝑇𝐴𝐵𝐶 𝑒𝜇𝐶 𝛽𝐶

(0.8284)𝑃 + 2500 N = (0.8284)𝑃 𝑒 ⇒

(9) (0.6)(130◦ )(𝜋 rad∕180◦ )

𝑃 = 1040 N.

(10) (11)

Comparing Eqs. (8) and (11), we observe that if 𝑃 is larger than 1547 N, then neither of the pulleys will slip. When 𝑃 = 1547 N, slip is impending at pulley 𝐴. Hence, the smallest value of 𝑃 needed to prevent slip is 𝑃min = 1547 N,

(12)

(b)

(c)

Figure 4 Free body diagrams for pulley 𝐴 to help determine the high-tension and low-tension portions of the belt. (a) Free body diagram for pulley 𝐴 and the belt together. (b) Free body diagram for only the pulley, showing the friction and normal forces the belt applies to the pulley. (c) Free body diagram for only the belt, showing the friction and normal forces the pulley applies to the belt.

and when this occurs, slip is impending at pulley 𝐴. Discussion & Verification

• If it is not clear that slip first occurs when 𝑃 is the larger of Eqs. (8) and (11), consider the situation in which 𝑃 is very large so that neither of the pulleys slips. Then as 𝑃 decreases, slip will occur when the larger of Eqs. (8) and (11) is achieved. • To provide a reasonable margin of safety against slip, we will want to use a value of 𝑃 greater than 1547 N. Some of the sources of uncertainty are as follows. Pulley 𝐵 may have nonnegligible friction, in which case slip at pulley 𝐴 may occur at a slightly higher value of 𝑃 . Probably the greatest source of uncertainty is the possibility that the coefficients of friction might be lower at some time during the belt’s life, such as might occur if the belt comes in contact with a small amount of lubricant or other fluid from the engine. • The FBD of pulley 𝐴 was not used in this solution. However, if it is necessary to determine the moment 𝑀𝐴 the crankshaft applies to this pulley, or the reactions 𝐴𝑥 and 𝐴𝑦 , then this FBD would be used to write the appropriate equilibrium equations.

Common Pitfall Measure 𝜷 in radians. A common error when using the expression 𝑇2 = 𝑇1 𝑒𝜇𝛽 is to forget to express the angle of wrap 𝛽 in radians.

564

Chapter 9

Friction

Problems Problem 9.51

5 in.

Two objects are connected by a weightless cord that is wrapped around a fixed cylindrical surface. The coefficient of static friction between the cord and surface is 0.3. One of the objects weighs 100 lb, and the other has weight 𝑊 . If the system is initially at rest, determine the range of values for 𝑊 for which there is no motion.

𝑊

100 lb Figure P9.51

Problem 9.52 𝐴

𝑃

A 10 N weight is supported by a weightless cable where portion 𝐴𝐵 of the cable is horizontal. The coefficients of static and kinetic friction are 0.5. (a) Determine the value of 𝑃 so that downward motion of the 10 N weight is impending. (b) If 𝑃 = 3 N, determine the force in portion 𝐶𝐷 of the cable. Hint: Do not assume, based on static equilibrium of 𝐷, that the answer is 10 N. Is the system in equilibrium or is there motion? Explain.

𝐵 𝜃 𝐶

20 mm

𝐷 10 N Figure P9.52 and P9.53

Problem 9.53 A 10 N weight is supported by a weightless cable where portion 𝐴𝐵 of the cable is horizontal. A test is conducted where it is found that 𝑃 = 4 N will cause impending motion of the 10 N weight at 𝐷. (a) Determine the coefficient of static friction between the cable and cylindrical surface. (b) Determine the tension in portion 𝐵𝐶 of the cable as a function of angular position 𝜃. Plot this function.

Problem 9.54 A hoist for lifting lightweight building materials is shown. The pulley at 𝐵 is frictionless, and the pulley at 𝐶 is driven by an electric motor rotating counterclockwise at a speed high enough to always cause slip between the cable and pulley 𝐶. The cable makes contact with one-fourth of the surface of pulley 𝐶, and the coefficients of static and kinetic friction are 0.25 and 0.2, respectively. If the cable has negligible weight and the building materials and platform at 𝐷 have a combined weight of 100 lb, determine the force the worker must apply to the cable to: (a) Raise the building materials at a uniform speed. (b) Lower the building materials at a uniform speed. (c) Hold the building materials in a fixed, suspended position.

𝐶

𝐵

𝐷 75◦

𝐴

Problem 9.55 Figure P9.54 and P9.55

In Prob. 9.54, if the cable were given an additional full wrap around the pulley at 𝐶 and if the worker can apply a force of 50 lb to the cable, determine the largest weight that may be lifted at 𝐷.

Problem 9.56 𝐴 𝐵 15 in.

9 in.

Figure P9.56 and P9.57

ISTUDY

𝐶

3 in.

𝐷 𝑊

The machine shown is used to slowly lower and raise the object at point 𝐷. Cable 𝐵𝐶𝐷 is wrapped around the drum at 𝐵 a large number of turns, and cable segment 𝐵𝐶 is horizontal. The coefficients of static and kinetic friction are 0.22 and 0.18, respectively. Let 𝑃 be the force the operator applies to point 𝐴 of the machine, where this force is perpendicular to the crank 𝐴𝐵. If 𝑊 = 20 lb, determine the range of values for 𝑃 for which there is no motion of the object at 𝐷.

ISTUDY

Section 9.3

565

Belts and Cables Contacting Cylindrical Surfaces

Problem 9.57 In Prob. 9.56, if the person operating the machine can apply a maximum force of 25 lb to the handle 𝐴 of the machine, where this force is perpendicular to the crank 𝐴𝐵, and if the cable has 50 lb strength, determine the largest value of 𝑊 that may be lowered and the largest value of 𝑊 that may be raised. Hint: Lowering and raising motions will be initiated from at-rest conditions.

Problems 9.58 through 9.61 A brake for reducing the speed of a rotating drum is shown. The braking moment is defined as the resultant moment the belt produces about the drum’s bearing, point 𝐴. Determine the braking moment if the coefficient of kinetic friction is 0.35 and: Problem 9.58

The drum rotates clockwise and 𝑃 = 90 N.

Problem 9.59

The drum rotates counterclockwise and 𝑃 = 90 N.

Problem 9.60

The drum rotates clockwise and 𝑃 = 20 lb.

Problem 9.61

The drum rotates counterclockwise and 𝑃 = 20 lb. 𝐶 𝐵

20 cm vertical cable 𝐷 segment

𝑃

𝑃 20 in.

20 cm

𝐴

𝐶

15 cm

30 cm

Figure P9.58 and P9.59

𝐵 10 in.

45◦ 𝐴 45◦

10 in.

𝐷 Figure P9.60 and P9.61

Problem 9.62

𝐻

𝑃

𝐷

A spring 𝐴𝐵 with 2 N∕mm stiffness is attached to a cable at 𝐵. The cable wraps around two cylindrical surfaces, and cable segments BC, DE, and GH are horizontal. All contact surfaces have coefficients of static and kinetic friction of 0.1 and 0.09, respectively. The spring is initially unstretched. (a) Determine the value of 𝑃 needed to stretch the spring by 4 mm.

𝐺

20 mm

𝐸 𝐴

𝑘

𝐵 𝐶

40 mm

Figure P9.62

(b) Once the spring is stretched by 4 mm, determine the value to which 𝑃 must be reduced so that the spring begins to contract.

𝐴

𝑀𝐴

𝐵

Problem 9.63 Pulley 𝐴 of the treadmill is driven by a motor that can supply a moment 𝑀𝐴 = 200 in.⋅lb. Pulley 𝐵 is frictionless, and its bearing slides in a frictionless slot where force 𝑃 tensions the belt. Both pulleys 𝐴 and 𝐵 have 1.5 in. radius, and the coefficient of static friction between pulley 𝐴 and the belt is 0.3. Determine the smallest value of 𝑃 needed to prevent slip between pulley 𝐴 and the belt.

𝑃

5◦

Figure P9.63

4 in.

8 in. 𝐵

𝐶

Problem 9.64 The winch is driven by an electric motor at 𝐴. The motor weighs 50 lb with center of gravity at the center of pulley 𝐴. The platform supporting the motor has negligible weight and is hinged at 𝐷. The motor turns the pulley at 𝐵 via a belt between 𝐴 and 𝐵. Pulley 𝐵 has a spool upon which the rope BCG is wrapped. The coefficient of static friction at 𝐴 is 0.3 and at 𝐵 is 0.2. Assuming the motor has sufficient power, determine the largest weight 𝑊 that may be lifted if there is no slip between the belt and pulleys 𝐴 and 𝐵. Hint: The use of mathematical software is helpful, but not required.

15◦

15◦ 𝐴

8 in. 𝐸 𝑟 = 3 in. Figure P9.64

𝑊

𝐷 𝐺

12 in.

566

Chapter 9

Friction

𝜇𝐶 = 0.12

𝜇𝐴 = 0.08 20 mm

Problem 9.65

20 mm 𝐴

𝐶 𝐵 30 mm 𝜇𝐵 = 0.1

40 N

60 N

Problem 9.66

Figure P9.65 and P9.66 80◦

In Prob. 9.65, determine the weight of object 𝐵 that causes it to have impending upward motion, and describe which of surfaces 𝐴, 𝐵, and/or 𝐶 will have impending slip.

𝑟𝐶 = 40 mm

Problem 9.67

45◦ 𝐺

𝑃

An accessory belt for an engine is shown. Pulley 𝐴 is attached to the engine’s crankshaft and rotates clockwise. The belt tensioner consists of a frictionless pulley at 𝐵 that is mounted to a horizontal bar 𝐺 that slides in a frictionless track with a horizontal force 𝑃 . Pulley 𝐶 operates an alternator that requires 100 N⋅m, and pulley 𝐷 operates a hydraulic pump that requires 300 N⋅m. Pulley 𝐸 is frictionless. The coefficients of static friction for pulleys 𝐴, 𝐶, and 𝐷 are 0.5, and the radii of the pulleys are shown. Determine the minimum value of 𝑃 required to prevent the belt from slipping. Hint: The use of mathematical software is helpful, but not required.

𝐷 𝐸

𝐵 60◦

Surfaces 𝐴, 𝐵, and 𝐶 are cylindrical with the coefficients of static friction shown. Object 𝐵 can undergo vertical motion only. Determine the weight of object 𝐵 that causes it to have impending downward motion, and describe which of surfaces 𝐴, 𝐵, and/or 𝐶 will have impending slip.

205◦ 𝑟𝐷 = 60 mm

𝐴

180◦ 𝑟𝐴 = 100 mm Figure P9.67

Problem 9.68 The portable winch shown consists of a rotating motor-powered drum that is attached to a skid. Winches such as this were once used in logging operations to move logs and to lift them onto trucks and railcars. To move the winch to a new location, a worker stands on the skid and wraps a rope 𝐴𝐵𝐶𝐷 around the drum. End 𝐴 of the rope is attached to a fixed object, portion 𝐴𝐵 of the rope is horizontal, and the rope makes less than one full turn around the drum. The skid, drum, and worker weigh 800 lb with center of gravity at point 𝐺, and the drum has 10 in. radius.

𝐷 10 in.

𝐴

𝐵 𝐶

rotating drum 𝜇 = 0.3 3 ft

40◦ 𝐺

𝜇 = 0.6

skid

2 ft

(a) Assuming the skid does not tip, determine the smallest force the worker must apply to portion 𝐶𝐷 of the rope so that the skid will slide. (b) Determine if the assumption in Part (a) that the skid does not tip is correct.

2.5 f t

3.5 f t

Figure P9.68 4 cm

Problem 9.69 diamond wire

𝐵 2N 𝐶

𝐴 workpiece table

30◦

3 cm

𝐸

𝐹

𝐷 4 cm

24 cm

24 cm 𝑊

Figure P9.69

ISTUDY

The schematic for a precision diamond wire saw that is used in a laboratory for cutting very hard, brittle materials is shown. It consists of a wire with diamond particles bonded to its entire surface. The wire wraps around a motor-driven pulley at 𝐴 that rotates counterclockwise, and frictionless pulleys at 𝐵 and 𝐷. At 𝐶 is a fixed table where a workpiece is cut. Due to cutting, the workpiece applies a 2 N vertical force to the wire in the direction shown. The wire is tensioned by weight 𝑊 . Portions 𝐴𝐵 and 𝐵𝐶𝐷 of the wire are horizontal and vertical, respectively. The angle of wrap at the motor is 150◦ , and the coefficients of static and kinetic friction between the motor pulley and the wire are both 0.7. Neglecting all weight except 𝑊 , determine the value of 𝑊 so that the wire does not slip on the pulley at 𝐴.

ISTUDY

Section 9.4

Chapter Review

9.4 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

Coulomb’s law of friction Coulomb’s law of friction states Eqs. (9.1) and (9.2), p. 535 |𝐹 | ≤ 𝜇𝑠 𝑁

|𝐹 | = 𝜇𝑘 𝑁

before sliding ( 𝐼𝑥 ′ . In general, the moment of inertia 𝐼𝑥 about any 𝑥 axis parallel to (but not coincident with) the centroidal 𝑥 ′ axis is larger than 𝐼𝑥 ′ . Similar remarks apply to the other moments of inertia. • The physical interpretation of the radius of gyration is shown in Fig. 3 for the case of 𝐼𝑥 . If the area 𝐴 = 𝑏ℎ is redistributed into a thin strip, the strip must be positioned at 𝑘𝑥 = (0.5774)ℎ from the 𝑥 axis if it is to have the same moment of inertia about the 𝑥 axis as the original shape shown in Fig. 1. • To determine the moments of inertia 𝐼𝑦 and 𝐼𝑦 ′ , this solution procedure is repeated using an area element parallel to the 𝑦 axis (see Prob. 10.1).

ℎ

𝑘𝑥 = (0.5774)ℎ 𝑥 𝑏

Figure 3 If the entire cross-sectional area 𝐴 = 𝑏ℎ is redistributed into a thin strip parallel to the 𝑥 axis, the radius of gyration 𝑘𝑥 is the distance it must be positioned from the 𝑥 axis to have the same moment of inertia as the original shape.

580

Chapter 10

Moments of Inertia

E X A M P L E 10.2

Area Moments of Inertia Using Integration

𝑦

A circular area with outside radius 𝑟𝑜 is shown. Determine the area moment of inertia 𝐼𝑦 and the polar moment of inertia 𝐽𝑂 about point 𝑂.

𝑟𝑜

𝑥

𝑂

SOLUTION Road Map

We will use Eq. (10.2) on p. 574 to determine 𝐼𝑦 and Eq. (10.3) to determine

𝐽𝑂 . Figure 1

𝐼𝑦 Governing Equations & Computation

To determine 𝐼𝑦 using a single integration, expressions for 𝑑𝐴 and 𝑥̃ are needed, and these must be developed using an area element parallel to the 𝑦 axis, as shown in Fig. 2: 𝑦

𝑑𝐴 = ℎ 𝑑𝑥

𝑑𝑥 𝑟𝑜

ℎ

𝑥

𝑥̃ Figure 2 To determine 𝐼𝑦 , an area element parallel to the 𝑦 axis is used to develop expressions for 𝑑𝐴 and 𝑥. ̃

and 𝑥̃ = 𝑥.

(1)

The height ℎ of the area element is related to 𝑥 and radius 𝑟𝑜 by the Pythagorean theorem 𝑟2𝑜 = 𝑥2 + (ℎ∕2)2 , hence, √ (2) ℎ = 2 𝑟2𝑜 − 𝑥2 . Substituting Eqs. (1) and (2) into Eq. (10.2), and using a table of integrals or computer software to carry out the integration, the moment of inertia about the 𝑦 axis is √ 𝑟𝑜 𝐼𝑦 = 𝑥̃ 2 𝑑𝐴 = 𝑥2 2 𝑟2𝑜 − 𝑥2 𝑑𝑥 ∫ ∫−𝑟 𝑜 [ ( )] 𝑟 𝑟2𝑜 𝑥 √ 𝑟4𝑜 |𝑜 ) ( −𝑥 2 𝑥 −1 2 3∕2 | + 𝑟𝑜 − 𝑥 = 𝑟2𝑜 − 𝑥2 + sin | 2 4 4 𝑟𝑜 |−𝑟𝑜 =

𝜋𝑟4𝑜 4

(3)

. 𝐽𝑂

Governing Equations & Computation To determine 𝐽𝑂 using a single integration, an area element having a uniform value of 𝑟̃ is needed, and this necessitates the use of a thin circular area element as shown in Fig. 3. With 𝑟 being a radial coordinate, expressions for 𝑑𝐴 and 𝑟̃ are obtained from Fig. 3 as

𝑦

𝑟𝑜 𝑟̃ 𝑂𝑟

𝑑𝑟

𝑑𝐴 = 2𝜋𝑟 𝑑𝑟

and 𝑟̃ = 𝑟.

Substituting Eq. (4) into Eq. (10.3), the polar moment of inertia about point 𝑂 is 𝑟𝑜

Figure 3 To determine 𝐽𝑂 , a circular area element is used to develop expressions for 𝑑𝐴 and 𝑟̃.

ISTUDY

(4)

𝑥

𝐽𝑂 =

∫

2

𝑟̃ 𝑑𝐴 =

∫

𝑟2 2𝜋𝑟 𝑑𝑟 =

𝜋𝑟4𝑜 2

.

(5)

0

Discussion & Verification

• Because of symmetry of the circular shape in Fig. 1, it is obvious that 𝐼𝑥 = 𝐼𝑦 . • Noting from Eq. (10.3) that 𝐽𝑂 = 𝐼𝑥 + 𝐼𝑦 , and with 𝐼𝑥 = 𝐼𝑦 as discussed above, we could have obtained Eq. (5) by inspection from the result in Eq. (3), or vice versa.

ISTUDY

Section 10.1

Area Moments of Inertia

E X A M P L E 10.3

Area Moments of Inertia Using Integration

The cross section of a turbine blade in a pump is shown. Determine the area moments of inertia 𝐼𝑥 and 𝐼𝑦 .

𝑦 (mm) 15

√ 𝑦𝑡 = 3 𝑥

10

SOLUTION Road Map

This geometry was considered in Example 7.3 on p. 440 where the location of the centroid was determined. In this example, the location of the centroid is not needed, although knowledge of its approximate location may help with verification of the area moments of inertia, as discussed below. The expressions developed in Example 7.3 for quantities such as 𝑑𝐴, 𝑥, ̃ and 𝑦̃ will be used again here.

𝑦𝑏 = 3 𝑥

5

5 0

0

=

∫ (

6𝑥7∕2 7

−

∫0

3𝑥4 20

)

√ 𝑥2 (3 𝑥 − 53 𝑥) 𝑑𝑥

|25 mm | |0

15

20

25

𝑥 (mm)

Notation. Throughout this problem, subscripts are used to distinguish top and bottom curves, and left and right curves, as follows:

Substituting Eq. (1) into Eq. (10.2) on p. 574, the moment of inertia about the 𝑦 axis is 25 mm

10

Helpful Information

To determine 𝐼𝑦 using a single integration, expressions for 𝑑𝐴 and 𝑥̃ are needed, and these must be developed using an area element parallel to the 𝑦 axis, as shown in Fig. 2: √ and 𝑥̃ = 𝑥. (1) 𝑑𝐴 = (𝑦𝑡 − 𝑦𝑏 ) 𝑑𝑥 = (3 𝑥 − 35 𝑥) 𝑑𝑥

𝑥̃ 2 𝑑𝐴 =

5

Figure 1

𝐼𝑦 Governing Equations & Computation

𝐼𝑦 =

581

= 8371 mm4 .

●

Subscript 𝑡 denotes “top.”

●

Subscript 𝑏 denotes “bottom.”

●

Subscript 𝑟 denotes “right.”

●

Subscript 𝑙 denotes “left.”

(2)

𝐼𝑥 Governing Equations & Computation To determine 𝐼𝑥 using a single integration, expressions for 𝑑𝐴 and 𝑦̃ are needed, and these must be developed using an area element parallel to the 𝑥 axis, as shown in Fig. 3: ) ( 𝑦2 𝑑𝑦 and 𝑦̃ = 𝑦. (3) 𝑑𝐴 = (𝑥𝑟 − 𝑥𝑙 ) 𝑑𝑦 = 35 𝑦 − 9

Substituting Eq. (3) into Eq. (10.1) on p. 574, the moment of inertia about the 𝑥 axis is ( ) 15 mm 𝑦2 𝐼𝑥 = 𝑦̃2 𝑑𝐴 = 𝑦2 53 𝑦 − 𝑑𝑦 ∫ ∫0 9 ) ( 4 𝑦5 |15 mm 5𝑦 − = 4219 mm4 . (4) = | 12 45 |0 Discussion & Verification

• By inspection of Fig. 1 (or by consulting the results of Example 7.3 on p. 440), the centroid of the area is farther from the 𝑦 axis than from the 𝑥 axis, so we expect 𝐼𝑦 > 𝐼𝑥 , and Eqs. (2) and (4) show this. Despite this observation, 𝐼𝑦 ≠ 𝑥̄ 2 𝐴 and 𝐼𝑥 ≠ 𝑦̄2 𝐴; Prob. 10.2 discusses this further. • Knowing the moments of inertia about the centroidal 𝑥 ′ and 𝑦 ′ axes is also desirable. This may be accomplished by rewriting Eqs. (1)–(4) with integrations carried out with respect to 𝑥 ′ and 𝑦 ′ , and with expressions for 𝑦𝑡 , 𝑦𝑏 , 𝑥𝑟 , and 𝑥𝑙 written in terms of 𝑥 ′ and 𝑦 ′ . However, the parallel axis theorem discussed in the next section will be more convenient for evaluating 𝐼𝑥 ′ and 𝐼𝑦 ′ .

𝑦 (mm) 15

√ 𝑦𝑡 = 3 𝑥

𝑑𝑥

10

𝑦𝑏 = 3 𝑥 5

5 0

𝑥 (mm) 0

𝑥̃ 10

15

20

25

Figure 2 To determine 𝐼𝑦 , an area element parallel to the 𝑦 axis is used to develop expressions for 𝑑𝐴 and 𝑥. ̃ 𝑦 (mm) 𝑥𝑙 =

15

𝑦2 9

10 𝑑𝑦

𝑦̃ 0

𝑥𝑟 = 5 𝑦 3 𝑥 (mm)

0

5

10

15

20

25

Figure 3 To determine 𝐼𝑥 , an area element parallel to the 𝑥 axis is used to develop expressions for 𝑑𝐴 and 𝑦. ̃

582

Chapter 10

Moments of Inertia

Problems Problem 10.1 For the rectangular area with base 𝑏, height ℎ, and centroid at point 𝐶 shown in Fig. 1 of Example 10.1 on p. 579, determine the area moments of inertia 𝐼𝑦 and 𝐼𝑦 ′ and the radii of gyration 𝑘𝑦 and 𝑘𝑦 ′ .

Problem 10.2 In Example 7.3 on p. 440 the centroid 𝑥, ̄ 𝑦, ̄ and cross-sectional area 𝐴 of a turbine blade were determined, and in Example 10.3 on p. 581 the area moments of inertia 𝐼𝑥 and 𝐼𝑦 were determined. Show that 𝐼𝑥 ≠ 𝑦̄2 𝐴 and 𝐼𝑦 ≠ 𝑥̄ 2 𝐴, and discuss why area moments of inertia may not be determined in this way. Note: Concept problems are about explanations, not computations.

𝑦 𝑏

𝑏 ℎ

Problem 10.3

ℎ 𝑥 Figure P10.3

Problems 10.4 and 10.5 The cross section of a hollow tube has rectangular shape with uniform wall thickness.

𝑦 𝑦 8𝑎

𝐶

𝑎 𝑥 𝑥

6𝑎 Figure P10.4 and P10.5

ISTUDY

For the rectangular shape shown, determine the moments of inertia 𝐼𝑥 and 𝐼𝑦 .

Problem 10.4

Determine the moments of inertia 𝐼𝑥 and 𝐼𝑦 .

Problem 10.5

Determine the centroidal moments of inertia 𝐼𝑥 ′ and 𝐼𝑦 ′ .

Problems 10.6 and 10.7 Determine 𝐼𝑥 and 𝐼𝑦 . 𝑦 0.5 in.

0.5 in. 0.25 in.

20 mm 20 mm 𝑥

20 mm

0.75 in.

𝑦

100 mm

10 mm 0.125 in.

0.125 in.

𝑥

Figure P10.6

Figure P10.7

Problems 10.8 and 10.9 Determine 𝐼𝑥 and 𝐼𝑦 . Express your answers in terms of 𝑏 and ℎ. 𝑦

𝑦 2ℎ

ℎ

ℎ

𝑥 𝑏 Figure P10.8

𝑥 𝑏

𝑏

Figure P10.9

ISTUDY

Section 10.1

Area Moments of Inertia

Problem 10.10 A circular area has outside radius 𝑟𝑜 and a concentric hole with inside radius 𝑟𝑖 . Show that the polar moment of inertia about point 𝑂 is 𝐽𝑂 = 𝜋(𝑟4𝑜 − 𝑟4𝑖 )∕2. 𝑦 𝑟𝑜 𝑟𝑖 𝑂

𝑥

Figure P10.10

Problems 10.11 and 10.12 The area shown has outside radius 𝑟𝑜 . Determine 𝐼𝑦 and 𝐽𝑂 . Comment on how your answers compare with the results of Example 10.2 on p. 580. 𝑦 𝑦 𝑟𝑜 𝑂

𝑥 𝑟𝑜 𝑂

Figure P10.11

𝑥

Problems 10.13 and 10.14 A semicircular area has outside radius 𝑟𝑜 . Problem 10.13

𝑦

Figure P10.12

𝑥2 + (𝑦 − 𝑟𝑜 )2 = 𝑟𝑜2

𝑟𝑜 𝐴 𝑟𝑜

Determine 𝐼𝑥 . Hint: The use of mathematical software is helpful, but

𝑥

is not required. Figure P10.13 and P10.14 Problem 10.14

Determine 𝐽𝐴 .

Problem 10.15 Consider two shafts having cross sections with the same area. The cross section for the first shaft is solid, and the cross section for the second shaft has a concentric hole whose inside radius is one-half its outside radius. Determine the ratio of the centroidal polar moments of inertia for the two shafts 𝐽hollow ∕𝐽solid . Hint: Use the polar moment of inertia given in the statement of Prob. 10.10.

Problem 10.16 The strength of long bones, as well as slender structural members in general, is directly related to the area moments of inertia, such that if the moments of inertia increase, then strength increases. From the point of view of the ratio of strength to weight, discuss why most long bones in humans and animals, such as the human femur shown, are hollow. Note: Concept problems are about explanations, not computations.

Figure P10.16

583

584

Chapter 10

Moments of Inertia Problems 10.17 through 10.22 (a) Determine 𝐼𝑥 . (b) Determine 𝐼𝑦 . 𝑦 1 in. 𝑦=

√

𝑥

𝑦

𝑦=

√

𝑥

2 mm 𝑦 = 𝑥2 1 in.

𝑦 = 𝑥2 ∕8

𝑥

𝑥 4 mm

Figure P10.17

Figure P10.18 𝑦

𝑦 2m

𝑦 = 2𝑥 − 𝑥2

1 mm 𝑦=

√

𝑥

𝑥

𝑥 2 mm

4m Figure P10.19

Figure P10.20 𝑦

𝑦 𝑦 = 1 + 𝑥∕2

2 in.

1.5 in.

𝑦 = 1∕𝑥 1 in. 1 in. 𝑦 = 𝑥2 1 in. 𝑦

Figure P10.21 𝑦2

𝑏

𝑥

𝑐

𝑏

For the triangle shown, having base 𝑏 and height ℎ, show that the area moment of inertia about the 𝑥 axis is 𝐼𝑥 = 𝑏ℎ3 ∕12.

Determine constants 𝑐1 and 𝑐2 so that the curves intersect at 𝑥 = 𝑎 and 𝑦 = 𝑏. Express your answers in terms of 𝑎 and 𝑏. Then determine the area moment of inertia indicated in the subproblem below.

√ 𝑦 = 𝑐1 𝑥

𝑦 = 𝑐 2 𝑥3 𝑎 Figure P10.24 and P10.25

ISTUDY

Figure P10.22

Problems 10.24 and 10.25

Figure P10.23 𝑦

𝑥 0.5 in. 1 in.

Problem 10.23

ℎ 𝑦1

𝑥

𝑥

Problem 10.24

Determine 𝐼𝑥 .

Problem 10.25

Determine 𝐼𝑦 .

ISTUDY

Section 10.2

10.2

Parallel Axis Theorem

585

Parallel Axis Theorem

The parallel axis theorem relates area moments of inertia 𝐼𝑥 , 𝐼𝑦 , 𝐽𝑂 , and 𝐼𝑥𝑦 to the centroidal area moments of inertia 𝐼𝑥 ′ , 𝐼𝑦 ′ , 𝐽𝑂 ′ , and 𝐼𝑥 ′ 𝑦 ′ , where the 𝑥 and 𝑥 ′ axes are parallel and the 𝑦 and 𝑦 ′ axes are parallel. The parallel axis theorem is important and is used on a daily basis by most engineers. Consider the area 𝐴 with centroid 𝐶 shown in Fig. 10.8. A centroidal 𝑥 ′ 𝑦 ′ coordinate system is defined, with origin 𝑂 ′ positioned at the centroid of the area. The 𝑥 and 𝑦 axes are parallel to the 𝑥 ′ and 𝑦 ′ axes, respectively, with separation distances 𝑑𝑥 and 𝑑𝑦 . The parallel axis theorem relates the moments of inertia with respect to the 𝑥 and 𝑦 axes to the centroidal moments of inertia as follows: 𝐼𝑥 = 𝐼𝑥 ′ + 𝑑𝑥2 𝐴,

(10.6)

𝐼𝑦 = 𝐼𝑦 ′ + 𝑑𝑦2 𝐴,

(10.7)

2

𝐽𝑂 = 𝐽𝑂 ′ + 𝑑 𝐴,

(10.8)

𝐼𝑥𝑦 = 𝐼𝑥 ′ 𝑦 ′ + 𝑑𝑥 𝑑𝑦 𝐴.

(10.9)

𝑦 𝑦 𝑑𝑦 𝐴

𝐶 𝑂

𝑑

𝑥 𝑑𝑥 𝑥

𝑂

Figure 10.8 An area 𝐴 with centroid at point 𝐶. The 𝑥 and 𝑥 ′ axes are parallel with separation distance 𝑑𝑥 , and the 𝑦 and 𝑦 ′ axes are parallel with separation distance 𝑑𝑦 .

To see how the above expressions are obtained, consider the example of determining 𝐼𝑥 for the area shown in Fig. 10.8. Beginning with the basic definition of 𝐼𝑥 given by Eq. (10.1) on p. 574, we use the area element shown in Fig. 10.9 to write 𝐼𝑥 =

∫

𝑦̃2 𝑑𝐴

(10.10)

=

∫

(𝑦̃ ′ + 𝑑𝑥 )2 𝑑𝐴

(10.11)

=

∫ ⏟⏞⏞⏞⏟⏞⏞⏞⏟

∫ ⏟⏞⏟⏞⏟

∫ ⏟⏟⏟

=𝐼𝑥 ′

=0

=𝐴

(𝑦̃ ′ )2 𝑑𝐴 + 2𝑑𝑥

= 𝐼𝑥 ′ + 𝑑𝑥2 𝐴.

𝑦̃ ′ 𝑑𝐴 + 𝑑𝑥2

𝑑𝐴

𝑦 𝑦

𝑦̃

(10.12) 𝑂

(10.13)

Equation (10.11) is obtained from Eq. (10.10) by noting in Fig. 10.9 that 𝑦̃ = 𝑦̃ ′ + 𝑑𝑥 . Equation (10.12) is obtained by expanding (𝑦̃ ′ + 𝑑𝑥 )2 , writing separate integrals for each term, and bringing the constant 𝑑𝑥 outside the integrals. The first integral in Eq. (10.12) is the same as Eq. (10.1) written in terms of a centroidal coordinate; hence, it yields the centroidal moment of inertia 𝐼𝑥 ′ . The second integral is zero because 𝑦 ′ is measured from the centroid of the shape, and the third integral is the area 𝐴 of the shape. The proof for the other parallel axis theorems is similar to Eqs. (10.10)–(10.13). Remarks

• It is clear from Eqs. (10.6)–(10.8) that 𝐼𝑥 ≥ 𝐼𝑥 ′ , 𝐼𝑦 ≥ 𝐼𝑦 ′ , and 𝐽𝑂 ≥ 𝐽𝑂 ′ . • When transforming to obtain the product of inertia 𝐼𝑥𝑦 using Eq. (10.9), the signs of 𝑑𝑥 and 𝑑𝑦 are important. Transformation of the product of inertia arises in advanced subjects and is not explored further in this book. • The parallel axis theorem may also be used to obtain radii of gyration, and Prob. 10.31 explores this further.

𝑦̃ 𝑂

𝑥

𝑑𝑥 𝑥

Figure 10.9 Selection of an area element for determining the moment of inertia about the 𝑥 axis, 𝐼𝑥 .

586

Chapter 10

Moments of Inertia

Use of parallel axis theorem in integration 𝑦 1 =

2

1 3

=

2

3

𝑥 (a)

(b)

(c)

Figure 10.10 Use of composite shapes with the parallel axis theorem to determine area moments of inertia. (a) An area consisting of a semicircle and two rectangles. (b) A possible set of composite shapes. (c) Another possible set of composite shapes where shape 3 has negative area.

ISTUDY

The parallel axis theorem may be used to remove the restriction on the orientation and shape of the area element that was needed in Section 10.1. That is, with the parallel axis theorem, the moments of inertia 𝐼𝑥 and 𝐼𝑦 may be determined using a single integration with area elements that are perpendicular to the 𝑥 and 𝑦 axes, respectively. Example 10.4 illustrates this approach.

Use of parallel axis theorem for composite shapes The most common use of the parallel axis theorem is for determining the moments of inertia using composite shapes. For example, consider the area shown in Fig. 10.10 where 𝐼𝑥 is desired. The parallel axis theorem, written for composite shapes, becomes 𝐼𝑥 =

𝑛 ∑ ( ) 𝐼𝑥 ′ + 𝑑𝑥2 𝐴 𝑖 ,

(10.14)

𝑖=1

Common Pitfall Parallel axis theorem. Consider the parallel axes 𝑥1 , 𝑥2 , and centroidal axis 𝑥 ′ , such as 𝑦 𝑥2 centroid 𝐶

𝑥 𝑥1

where the centroid is at point 𝐶. If 𝐼𝑥 is 1 known, a common error is to use the parallel axis theorem to directly determine 𝐼𝑥 . 2 In the parallel axis theorem, one of the axes is always a centroidal axis. Thus, with 𝐼𝑥 1 known, the parallel axis theorem must be used twice. The first time 𝐼𝑥 is used to de1 termine 𝐼𝑥 ′ . The second time 𝐼𝑥 ′ is used to determine 𝐼𝑥 . See Prob. 10.40. 2

where 𝑛 is the number of shapes, 𝐼𝑥 ′ is the moment of inertia for shape 𝑖 about its centroidal 𝑥 ′ axis, 𝑑𝑥 is the shift distance for shape 𝑖 (i.e., the distance between the 𝑥 axis and the 𝑥 ′ axis for shape 𝑖), and 𝐴 is the area for shape 𝑖. Similar expressions may be written for 𝐼𝑦 and 𝐽𝑂 . To use Eq. (10.14), it is necessary to know the centroidal moment of inertia for each of the composite shapes, and this generally must be obtained by integration, as was done in Section 10.1. Fortunately, the centroidal moments of inertia for many basic shapes have been tabulated, such as given in the Table of Properties of Lines and Areas at the end of this book. Examples 10.5 and 10.6 illustrate this approach.

End of Section Summary In this section, the parallel axis theorem for area moments of inertia was presented. Some of the key points are as follows: • The parallel axis theorem relates area moments of inertia 𝐼𝑥 , 𝐼𝑦 , 𝐽𝑂 , and 𝐼𝑥𝑦 to the centroidal moments of inertia 𝐼𝑥 ′ , 𝐼𝑦 ′ , 𝐽𝑂 ′ , and 𝐼𝑥 ′ 𝑦 ′ , where the 𝑥 and 𝑥 ′ axes are parallel and the 𝑦 and 𝑦 ′ axes are parallel. A common error is to use the parallel axis theorem to relate moments of inertia between two parallel axes where neither of them is a centroidal axis. • The parallel axis theorem has applications for determining area moments of inertia by integration, where it permits use of an area element that is perpendicular to the axis about which the moment of inertia is being determined. • The most common application of the parallel axis theorem is for determining area moments of inertia using composite shapes. For use of composite shapes, the moment of inertia of each composite shape about its centroidal axis must be known. This can always be determined by using integration, but information for many basic shapes has been tabulated, such as given in the Table of Properties of Lines and Areas at the end of this book.

ISTUDY

Section 10.2

Parallel Axis Theorem

E X A M P L E 10.4

Area Moments of Inertia Using Integration with the Parallel Axis Theorem 𝑦

Use integration with an area element perpendicular to the 𝑥 axis to determine the area moment of inertia about the 𝑥 axis, 𝐼𝑥 .

3 in.

SOLUTION

0

Road Map An area element perpendicular to the 𝑥 axis, in conjunction with the parallel axis theorem, may be used to determine 𝐼𝑥 . The strategy is to write an expression for the moment of inertia of the area element about its centroidal axis, and to use the parallel axis theorem to obtain the moment of inertia of this area element about the 𝑥 axis, followed by integration over all area elements to obtain the total moment of inertia.

An area element perpendicular to the 𝑥 axis is shown in Fig. 2. In this approach, we evaluate the moment of inertia using 𝐼𝑥 =

∫

𝑦 = 21 𝑥 6 in.

0

𝑦

𝑑𝑥

3 in. ℎ

𝑑𝐼𝑥 ,

(1)

𝑦̃ 6 in.

where 𝑑𝐼𝑥 is the moment of inertia about the 𝑥 axis for the area element shown in Fig. 2. The area element is rectangular, and the parallel axis theorem, Eq. (10.6) on p. 585, is used to write its moment of inertia about the 𝑥 axis as 𝑑𝐼𝑥 =

1 3 ℎ 12

𝑑𝑥 + 𝑦̃2 ℎ 𝑑𝑥.

(2)

The first term in Eq. (2) is the moment of inertia of a rectangle about its centroid,∗ namely, (1∕12)(base)(height)3 where the base of the area element in Fig. 2 is 𝑑𝑥 and the height is ℎ. The second term in Eq. (2) is the parallel axis shift, namely, (shift distance)2 (area) where the shift distance is 𝑦̃ and the area is ℎ 𝑑𝑥. Letting 𝑦𝑡 and 𝑦𝑏 be the expressions for the top and bottom curves, respectively, that define the area in Fig. 2, we write ( 𝑥) 𝑑𝑥, (3) 𝑑𝐴 = (𝑦𝑡 − 𝑦𝑏 ) 𝑑𝑥 = 3 − 2 ) ( 𝑥 𝑦̃ = 21 (𝑦𝑡 + 𝑦𝑏 ) = 12 3 + , (4) 2 𝑥 (5) ℎ = 𝑦𝑡 − 𝑦 𝑏 = 3 − . 2 Substituting Eqs. (2)–(5) into Eq. (1) provides

=

∫0

6 in.

1 12

6 in. [ ( ) )] ( ) ( 𝑥 𝑥 3 𝑥 2 1 𝑑𝑥 + 3 − 𝑑𝑥 3− 3 + 2 ∫0 2 2 2

81 4 in. = 40.5 in.4 2

(6)

Discussion & Verification

To help judge if our answer is reasonable, we may compare our result to that for a rectangular area with 𝑥 and 𝑦 dimensions of 6 in. by 3 in. From the results of Example 10.1 on p. 579, or by consulting the Table of Properties of Lines and Areas at the end of this book, the moment of inertia of a rectangular area about the axis through its base is 𝐼𝑥 = (1∕3)(base)(height)3 = (1∕3)(6 in.)(3 in.)3 = 54 in.4 , which, as expected, is somewhat larger than the result in Eq. (6). Hence, our answer appears to be reasonable. moment of inertia of a rectangle about its centroid, namely (1∕12)(base)(height)3 , was obtained in Example 10.1 on p. 579, but more generally expressions such as this are obtained from the Table of Properties of Lines and Areas at the end of this book.

∗ The

𝑥

Figure 1

Governing Equations & Computation

𝐼𝑥 =

587

𝑥

Figure 2 To determine 𝐼𝑥 , an area element perpendicular to the 𝑥 axis may be used in conjunction with the parallel axis theorem. Expressions for 𝑑𝐴, 𝑦, ̃ and ℎ are needed.

588

Chapter 10

Moments of Inertia

E X A M P L E 10.5

Area Moments of Inertia Using Composite Shapes

𝑦

The cross section of an I beam with cover plates welded to its flanges is shown. The crosssectional area is symmetric about the 𝑥 and 𝑦 axes. Determine the moment of inertia about the 𝑥 axis of the beam’s cross-sectional area with and without the cover plates.

240 mm 16 mm

12 mm

SOLUTION

𝑥

150 mm 10 mm

Road Map

12 mm

The cross-sectional area consists of rectangular shapes, so the use of composite shapes for determining 𝐼𝑥 will be convenient. We will first determine 𝐼𝑥 for the I beam without cover plates and will then determine 𝐼𝑥 with cover plates.

16 mm 180 mm

Figure 1

I beam without cover plates Governing Equations & Computation Two combinations of composite shapes are shown

in Fig. 2. Using Eq. (10.14) on p. 586 with the three shapes shown in Fig. 2(b), the moment of inertia about the 𝑥 axis is

Interesting Fact Strengthening beams. Usually, only portions of a beam will support large internal forces. For example, a simply supported beam with uniformly distributed load has maximum moment at midspan and zero moment at the supports (see Example 8.5 on p. 506). If the beam has a uniform cross section, then the cross section must be selected based on the maximum moment (and possibly the shear). Selective reinforcement using cover plates is a simple way to strengthen a beam. I beam

cover plates

=

1

2

(b)

(c)

1 𝑥

2

= 3

(a)

] + (81 mm)2 (180 mm)(12 mm) 2

= 3.121×107 mm4 .

(2) (3)

In Eq. (2), the centroidal axis of shape 1 coincides with the 𝑥 axis, so the parallel axis shift for this shape is zero. Also, the contributions to 𝐼𝑥 for shapes 2 and 3 are identical, and thus the terms within square brackets are simply doubled.

in Fig. 3(b), the moment of inertia about the 𝑥 axis is 𝐼𝑥 =

3 ∑ ) ( 𝐼𝑥 ′ + 𝑑𝑥2 𝐴 𝑖

(5)

𝑖=1

= 3.121×107 mm4 ] [ 1 (240 mm)(16 mm)3 + (95 mm)2 (240 mm)(16 mm) 2 + 12

(6)

= 1.007×108 mm4 .

(7)

(b)

Figure 3 (a) Cross-sectional area of an I beam with cover plates. (b) A combination of composite shapes.

ISTUDY

1 (180 mm)(12 mm)3 12

I beam with cover plates Governing Equations & Computation Using Eq. (10.14) with the three shapes shown

Figure 2 (a) Cross-sectional area of an I beam without cover plates. (b) and (c) Two combinations of composite shapes to describe the area. 𝑦

[

+

(1)

3

3 (a)

𝑖=1 1 (10 mm)(150 mm)3 12

which is identical to Eq. (3). Note that the centroidal axis of each composite shape coincides with the 𝑥 axis, thus the parallel axis shift in Eq. (4) is zero for each shape.

2 1

=

3 ∑ ( ) 𝐼𝑥 ′ + 𝑑𝑥2 𝐴 𝑖

Alternate solution We may also determine 𝐼𝑥 using the three composite shapes shown in Fig. 2(c), where the areas of shapes 2 and 3 are negative. Equation (10.14) provides ] [ 1 1 𝐼𝑥 = 12 (180 mm)(174 mm)3 − 12 (85 mm)(150 mm)3 2 = 3.121×107 mm4 , (4)

𝑦 𝑥 =

𝐼𝑥 =

Discussion & Verification

The addition of the cover plates increases the beam’s crosssectional area by about 130%, yet 𝐼𝑥 increases dramatically by about 220%.

ISTUDY

Section 10.2

589

Parallel Axis Theorem

E X A M P L E 10.6

Area Moments of Inertia Using Composite Shapes

The cross section of a bar is symmetric about the 𝑥 axis. Determine 𝑑 so that the origin of the coordinate system, point 𝑂, is positioned at the centroid of the area, and determine the area moment of inertia about the 𝑦 axis.

𝑦 𝑑

1.2 in.

1.6 in.

SOLUTION

𝑥

𝑂

Road Map

The cross-sectional area consists of rectangular and semicircular shapes, so the use of composite shapes for determining the position of the centroid and 𝐼𝑦 will be convenient. Concepts from Section 7.1 will be used to locate the centroid, and then the parallel axis theorem will be used to determine the centroidal moment of inertia 𝐼𝑦 . Centroid Governing Equations & Computation A set of composite shapes is shown in Fig. 2,

2.4 in. Figure 1

𝑛

where for purposes of locating the centroid, a convenient 𝑡𝑛 coordinate system is defined. The area and centroidal position of each composite shape are collected in Table 1.

1 𝑡

Table 1. Areas and centroidal positions for composite shapes in Fig. 2. Shape no.

𝑡̃𝑖

𝐴𝑖 (1.6 in.)(2.4 in.) = 3.840 in. 𝜋 (1.2 in.)2 = 2.262 in.2 2

1 2

2

2

𝑡̃1

1.2 in. 4 2.4 in. + 3𝜋 (1.2 in.) = 2.909 in.

Figure 2 A combination of composite shapes and a 𝑡𝑛 coordinate system for use in determining the location of the centroid.

Evaluating Eq. (7.8) on p. 433 for the location of the centroid 𝑡̄ provides 2 ∑

𝑡̄ =

𝑖=1

𝑡̃𝑖 𝐴𝑖

2 ∑ 𝑖=1

=

) ( ) ( (1.2 in.) 3.840 in.2 + (2.909 in.) 2.262 in.2 3.840 in.2 + 2.262 in.2

𝐴𝑖

= 1.834 in.

(1)

Thus, the origin of the coordinate system in Fig. 1 should be located at 𝑑 = 1.834 in.

𝑦

Moment of inertia Governing Equations & Computation To determine 𝐼𝑦 , the same composite shapes are

used, and the distance from the centroid of each composite shape to the 𝑦 axis is shown in Fig. 3. From the Table of Properties of Lines and Areas at the end of this book, the centroidal moments of inertia for each shape are 𝐼𝑦 ′ = 1

𝐼𝑦 ′

2

1 (1.6 in.)(2.4 in.)3 12

(

= 1.843 in.4

(2)

)

8 𝜋 (1.2 in.)4 = 0.2276 in.4 − = 8 9𝜋

(3)

Using the parallel axis theorem, Eq. (10.14) on p. 586, with Eqs. (2) and (3), the moment of inertia about the centroidal 𝑦 axis is 𝐼𝑦 =

𝑡̃2

2 ∑ ( ) ) ( 𝐼𝑦 ′ + 𝑑𝑦2 𝐴 𝑖 = 1.843 in.4 + (0.6336 in.)2 3.840 in.2

1.834 in. 𝑦1 1 𝑂

0.6336 in.

𝑦2 𝑥 2

1.076 in.

Figure 3 A combination of composite shapes to determine 𝐼𝑦 , and distances from the centroid of each composite shape to the 𝑦 axis.

Helpful Information

𝑖=1

( ) + 0.2276 in.4 + (1.076 in.)2 2.262 in.2 = 6.230 in.4

Discussion & Verification

(4)

As a rough check to see if our answer in Eq. (4) is reasonable, the centroidal moment of inertia of a rectangle with dimensions that approximate Fig. 1 is (1∕12)(1.6 in.)(3.6 in.)3 = 6.221 in.4 , which agrees surprisingly well with Eq. (4).

Alternate strategy. Rather than evaluate 𝐼𝑦 directly as in Eq. (4), we could evaluate 𝐼𝑛 first (you should find 𝐼𝑛 = 26.75 in.4 ) and then use the parallel axis theorem to determine 𝐼𝑦 = 𝐼𝑛 − 𝑑 2 𝐴 = 26.75 in.4 − (1.834 in.)2 (6.102 in.2 ) = 6.230 in.4 .

590

Chapter 10

Moments of Inertia

Problems Problems 10.26 and 10.27 Use integration with an area element perpendicular to the 𝑥 axis to determine the area moment of inertia 𝐼𝑥 for the shape indicated. Problem 10.26

Figure P10.20 on p. 584.

Problem 10.27

Figure P10.21 on p. 584.

Problems 10.28 and 10.29 The beam cross sections shown are symmetric about the 𝑥 and 𝑦 axes. Determine the area moments of inertia 𝐼𝑥 and 𝐼𝑦 and the radii of gyration 𝑘𝑥 and 𝑘𝑦 . 𝑦

𝑦

10 mm 0.2 in.

0.2 in. 𝑥

250 mm 𝑦

6 mm 0.360 in.

0.2 in. 140 mm Figure P10.28

𝑥 0.240 in.

10.17 in.

𝑥

2 in. 0.2 in. 1.2 in. Figure P10.29

Problem 10.30

5.75 in. Figure P10.30

0.5 in.

The cross-sectional dimensions for a W10 × 22 wide-flange I beam are shown. By idealizing the cross section to consist of rectangular shapes, determine the area moment of inertia 𝐼𝑥 , and compare this to the value of 118 in.4 reported in the AISC Steel Construction Manual. Note: The value reported by AISC accounts for the effects of small fillets where the web and flanges join.

Problem 10.31 3 in. 24 in. 2 in.

Using Eq. (10.6) on p. 585, show that the parallel axis theorem for the radius of gyration about the 𝑥 axis is 𝑘2𝑥 = 𝑘2𝑥 ′ + 𝑑𝑥2 .

2 in. 0.5 in.

Problem 10.32 𝑥

For the T shape shown, determine the radius of gyration about the 𝑥 axis.

Figure P10.32

Problem 10.33 𝑦 0.5 in.

For the channel shown, determine the radii of gyration about the 𝑥 and 𝑦 axes.

0.5 in. 0.25 in.

0.75 in.

0.125 in. Figure P10.33

ISTUDY

0.125 in.

𝑥

Problem 10.34 Let each of the beams shown in Fig. 10.2 on p. 574 be constructed of identical planks of wood with 2 in. by 6 in. cross-sectional dimensions. Determine the area moments of inertia for each beam’s cross-sectional area about the 𝑥 and 𝑦 axes.

ISTUDY

Section 10.2

Parallel Axis Theorem

Problem 10.35 The cross sections of two beams are constructed by arranging the 2 cm by 16 cm strips of wood as shown. Determine the area moments of inertia for each beam about the horizontal and vertical axes passing through the centroid of each area.

Problems 10.36 through 10.39

16 cm

2 cm 2 cm

Determine the area moments of inertia 𝐼𝑥 and 𝐼𝑦 .

Figure P10.35 𝑦

𝑦 20 mm

0.2 in.

0.2 in. 30 mm 20 mm

0.2 in.

0.2 in.

𝑥

𝑥

10 mm

0.2 in. 0.2 in. 0.2 in. Figure P10.36

Figure P10.37

𝑦 0.5 in.

4 in.

𝑦

12 mm

1 in. 60 mm

0.5 in.

12 mm 12 mm

18 mm 𝑥 30 mm

𝑥 4 in.

100 mm

Figure P10.38

Figure P10.39

𝑦 𝑥2

Problem 10.40 9 mm

(a) Determine 𝐼𝑥 . 1

(b) Use the result of Part (a) with the parallel axis theorem to determine 𝐼𝑥 . Hint: See 2 the common pitfall discussed on p. 586.

6 mm Figure P10.40

Problem 10.41 A circular hole is to be drilled through the side of a beam as shown. Describe where the hole should be positioned so that the area moment of inertia about the centroidal 𝑥 ′ axis, at the cross section that passes through the hole, is reduced as little as possible. Note: Concept problems are about explanations, not computations.

Figure P10.41

6 mm

𝑥1

591

592

ISTUDY

Chapter 10

Moments of Inertia Problems 10.42 through 10.47

The cross section of the bar shown is symmetric about either the 𝑥 or 𝑦 axis. (a) Determine 𝑑 so that the origin of the coordinate system, point 𝑂, is positioned at the centroid of the area. (b) Determine the area moment of inertia about the 𝑥 axis. (c) Determine the area moment of inertia about the 𝑦 axis. Hint: If you carried out some of the exercises from Section 7.1, beginning on p. 445, you may have already solved Part (a) of Probs. 10.46 and 10.47. 𝑦 10 mm 𝑦 2 in. 𝑂

4 in. 6 in.

𝑥

𝑂 𝑑

𝑥

6 in.

4 in.

𝑑

10 mm

3 @ 5 mm

Figure P10.42

Figure P10.43 𝑦

𝑦

2 in. 2 in.

𝑑 3 mm

2 in.

𝑥 2 mm

𝑂

𝑥

𝑂

4 in.

𝑑

3 mm 2 mm 4 mm

2 in. 1 in.

Figure P10.44

Figure P10.45

𝑦

𝑦

20 mm 20 mm

1.5 in.

1 in.

20 mm 100 mm 10 mm

𝑥

𝑂 𝑑

Figure P10.46

0.5 in. 4 in.

𝑥

𝑂

1 in. Figure P10.47

𝑑

ISTUDY

Section 10.3

10.3

Mass Moments of Inertia

593

Concept Alert

Mass Moments of Inertia

Mass moments of inertia are measures of how an object’s mass is distributed about particular axes. While mass and area moments of inertia have some similarities, they are different in that mass moments of inertia are inherently volume- and densityrelated, whereas area moments of inertia are area-related. Before delving into the precise definitions of mass moments of inertia, we first discuss an example to help you understand why a measure of mass dispersion is useful and what this measure quantifies.

Mass moments of inertia. Mass moments of inertia are measures of how the mass of an object is distributed about particular axes. Mass moments of inertia depend on the geometry of an object (size and shape), the density of the material(s) it is made of, and the axes you select.

An example—figure skating Consider the figure skater shown in Fig. 10.11(a) as she spins on one skate with her arms and leg extended from her body. If friction between her skate and the ice is negligible, then she will spin at a constant rate. In Fig. 10.11(b) she draws her arms together and her leg closer to her body, thus making her mass distribution more compact and thus, reducing the mass moment of inertia of her body about the axis on which she spins. The reduction in her mass moment of inertia between Fig. 10.11(a) and (b) causes her to spin at a higher rate; exactly how fast she spins and how this relates to the mass moment of inertia are addressed in dynamics.

Definition of mass moments of inertia

Jill Braaten

Jill Braaten

(a)

The mass moments of inertia for the object shown in Fig. 10.12 are defined as 𝐼𝑥 =

∫

𝑟̃2𝑥 𝑑𝑚 =

∫

(𝑦̃2 + 𝑧̃ 2 ) 𝑑𝑚,

(10.15)

𝐼𝑦 =

∫

𝑟̃2𝑦 𝑑𝑚 =

∫

(𝑥̃ 2 + 𝑧̃ 2 ) 𝑑𝑚,

(10.16)

𝐼𝑧 =

∫

𝑟̃2𝑧 𝑑𝑚 =

∫

(𝑥̃ 2 + 𝑦̃2 ) 𝑑𝑚,

(10.17)

𝐼𝑥𝑦 =

∫

𝑥̃ 𝑦̃ 𝑑𝑚,

𝐼𝑦𝑧 =

∫

𝑦̃𝑧̃ 𝑑𝑚,

(10.19)

𝐼𝑥𝑧 =

∫

𝑥̃ 𝑧̃ 𝑑𝑚,

(10.20)

(b)

Figure 10.11 A figure skater spins with (a) her arms and leg extended and (b) her arms drawn together and her leg drawn close to her body. By making her body more compact in (b), she reduces the mass moment of inertia of her body about the axis on which she spins, causing her to spin at a higher rate. 𝑦

(10.18)

𝑑𝑚 = 𝜌𝑑𝑉 𝑟̃𝑦 𝑦̃

where

𝑧

𝑟̃𝑥 , 𝑟̃𝑦 , and 𝑟̃𝑧 are defined in Fig. 10.12 as the radial distances (i.e., moment arms) from the 𝑥, 𝑦, and 𝑧 axes, respectively, to the center of mass for mass element 𝑑𝑚; 𝑥, ̃ 𝑦, ̃ and 𝑧̃ are defined in Fig. 10.12 as the 𝑥, 𝑦, and 𝑧 distances, respectively, to the center of mass for mass element 𝑑𝑚; 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 are the mass moments of inertia about the 𝑥, 𝑦, and 𝑧 axes, respectively; and 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 are the products of inertia of the mass about the 𝑥𝑦, 𝑦𝑧, and 𝑥𝑧 axes, respectively.

𝑟̃𝑥

𝑟̃𝑧

𝑥 𝑥̃

𝑧̃

Figure 10.12 An object with mass 𝑚, density 𝜌, and volume 𝑉 ; 𝑟̃𝑥 , 𝑟̃𝑦 , and 𝑟̃𝑧 are radial distances from the 𝑥, 𝑦, and 𝑧 axes, respectively, to the center of mass for mass element 𝑑𝑚.

594

Chapter 10

Moments of Inertia Remarks

• When referring to mass moments of inertia, we often omit the word mass when it is obvious from the context that we are dealing with mass moments of inertia as opposed to area moments of inertia. • In each of Eqs. (10.15)–(10.17), two integral expressions are provided, and each is useful depending on the geometry of the object under consideration. • The moments of inertia in Eqs. (10.15)–(10.20) measure the second moment of the mass distribution. That is, to determine 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 in Eqs. (10.15)–(10.17), the moment arms 𝑟̃𝑥 , 𝑟̃𝑦 , and 𝑟̃𝑧 are squared. The second integral in each of these expressions is obtained by noting that 𝑟̃2𝑥 = 𝑦̃2 + 𝑧̃ 2 , and similarly for 𝑟̃2𝑦 and 𝑟̃2𝑧 . For the products of inertia 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 in Eqs. (10.18)–(10.20), the product of two different moment arms is used. • In Eqs. (10.15)–(10.20), 𝑥, ̃ 𝑦, ̃ and 𝑧̃ have dimensions of length, and 𝑑𝑚 has dimension of mass. Hence, all mass moments of inertia have dimensions of (mass)(length)2 , such as slug⋅in.2 or kg⋅m2 . • When the 𝑥, 𝑦, and 𝑧 axes pass through the center of mass of an object, we sometimes define the axes as 𝑥 ′ , 𝑦 ′ , and 𝑧 ′ , and we refer to the moments of inertia as mass center moments of inertia with the designations 𝐼𝑥 ′ , 𝐼𝑦 ′ , etc. In dynamics, moments of inertia about mass center axes are especially useful. • For any object with positive mass and finite volume, 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 are always positive.∗ The products of inertia 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 may be positive, zero, or negative, as discussed below. • Evaluation of moments of inertia using composite shapes is possible using the parallel axis theorem discussed later in this section.

What are mass moments of inertia used for? It is useful to discuss why there are six mass moments of inertia, what is different about them, and what they are used for. 𝑦 𝑧

𝑥

Moments of inertia 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 . In Fig. 10.13, the International Space Station with a docked Space Shuttle is shown. Imagine that a moment 𝑀𝑥 about the 𝑥 axis is applied to the Space Station. Assuming the Space Station is rigid,† it will begin to undergo angular acceleration about the 𝑥 axis. The value of the angular acceleration is directly dependent on the mass moment of inertia about the 𝑥 axis 𝐼𝑥 . Furthermore, the larger 𝐼𝑥 is, the lower the angular acceleration will be for a given value of 𝑀𝑥 . Similar remarks apply to moments applied about the 𝑦 and 𝑧 axes, and the influence that moments of inertia 𝐼𝑦 and 𝐼𝑧 have on angular accelerations about these axes.

European Space Agency

Figure 10.13 The International Space Station with a Space Shuttle docked to it.

ISTUDY

∗ For

the thin rod shown in the Table of Properties of Solids at the end of this book, 𝐼𝑥 is in fact positive, but since it is much smaller than 𝐼𝑦 and 𝐼𝑧 , it is usually taken to be zero. † The International Space Station is very flexible, as are most space structures. If a moment were applied about the 𝑥 axis shown in Fig. 10.13, then in addition to the rotations discussed above, the structure would vibrate. Control of vibrations in space structures is very important and receives considerable attention.

ISTUDY

Section 10.3

595

Mass Moments of Inertia

Products of inertia 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 . Products of inertia measure the asymmetry of a mass distribution with respect to the 𝑥, 𝑦, and 𝑧 axes. Products of inertia can have positive, zero, or negative value, depending on the shape and mass distribution of an object, the selection of the 𝑥, 𝑦, and 𝑧 directions, and the location of the origin of the 𝑥𝑦𝑧 coordinate system. An object is said to be symmetric if it has at least one plane about which both the shape and mass distribution are symmetric. For example, if an object is symmetric about the 𝑥𝑦 plane, such as the mallet shown in Fig. 10.14, then 𝐼𝑥𝑧 = 𝐼𝑦𝑧 = 0, and 𝐼𝑥𝑦 may be positive, zero, or negative. If an object is symmetric about at least two of the 𝑥𝑦, 𝑦𝑧, and 𝑥𝑧 planes, then all of the products of inertia are zero. Thus, the products of inertia are zero for a solid of revolution when one of the 𝑥, 𝑦, or 𝑧 axes coincides with the axis of revolution. Objects that have one or more nonzero products of inertia may display complicated dynamics in three-dimensional motions. For example, the Space Station is unsymmetric about the 𝑥𝑦𝑧 axes shown in Fig. 10.13, and it has nonzero products of inertia. As a consequence, if a moment 𝑀𝑥 about the 𝑥 axis is applied, the Space Station, in addition to rotating about the 𝑥 axis, will rotate about the 𝑦 and/or 𝑧 axes. The dynamics of objects with nonzero products of inertia is an advanced subject, and until you study this, you will not need to use products of inertia. Hence, methods for evaluating 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 are not discussed in this book.

𝑦

𝑧

𝑥

Figure 10.14 A mallet with a metal head and wood handle is shown. If the mallet’s geometry and mass distribution are both symmetric about the xy plane, then the mallet is said to be a symmetric object. If the mallet is also symmetric about the yz plane, it may be called a doubly symmetric object.

Radius of gyration Rather than use mass moments of inertia to quantify how mass is distributed, it is sometimes convenient to use the radii of gyration, which are directly related to the mass moments of inertia and are defined as √ 𝑘𝑥 =

√ 𝐼𝑥 𝑚

,

𝑘𝑦 =

𝐼𝑦 𝑚

√ ,

𝑘𝑧 =

𝐼𝑧 𝑚

,

(10.21)

where 𝑘𝑥 , 𝑘𝑦 , and 𝑘𝑧 are the radii of gyration of the mass about the 𝑥, 𝑦, and 𝑧 axes, respectively; and 𝑚 is the mass of the object.

𝑦 𝑦

The radii of gyration have units of length, and have a straightforward physical interpretation, as discussed in Example 10.12.

𝑑𝑦

𝑚

𝐺

𝑥

Parallel axis theorem The parallel axis theorem relates mass moments of inertia 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 to the mass center moments of inertia 𝐼𝑥 ′ , 𝐼𝑦 ′ , and 𝐼𝑧 ′ , where the 𝑥 and 𝑥 ′ axes are parallel, the 𝑦 and 𝑦 ′ axes are parallel, and the 𝑧 and 𝑧 ′ axes are parallel. The parallel axis theorem is important and is used often in dynamics and vibrations. Consider the object with mass 𝑚 and center of mass 𝐺 shown in Fig. 10.15. An 𝑥 ′ 𝑦 ′ 𝑧 ′ coordinate system is defined, with origin at the center of mass of the object. The 𝑥, 𝑦, and 𝑧 axes are parallel to the 𝑥 ′ , 𝑦 ′ , and 𝑧 ′ axes, respectively, with separation distances 𝑑𝑥 , 𝑑𝑦 , and 𝑑𝑧 , respectively. The parallel axis theorem relates the mass moments of inertia with respect to the 𝑥, 𝑦, and 𝑧 axes to the mass center

𝑑𝑥 𝑥

𝑧 𝑧

𝑑𝑧

Figure 10.15 An object with mass 𝑚 and center of mass at point 𝐺. The 𝑥 and 𝑥 ′ axes are parallel with separation distance 𝑑𝑥 , the 𝑦 and 𝑦 ′ axes are parallel with separation distance 𝑑𝑦 , and the 𝑧 and 𝑧 ′ axes are parallel with separation distance 𝑑𝑧 .

596

Chapter 10

Moments of Inertia

moments of inertia as follows: 𝐼𝑥 = 𝐼𝑥 ′ + 𝑑𝑥2 𝑚,

(10.22)

𝐼𝑦 ′ + 𝑑𝑦2 𝑚, 𝐼𝑧 ′ + 𝑑𝑧2 𝑚.

(10.23)

𝐼𝑦 = 𝐼𝑧 =

(10.24)

The derivation of these expressions is similar to that for the area moment of inertia given by Eqs. (10.10)–(10.13) on p. 585.

Evaluation of moments of inertia using integration 𝑦 General object

𝑟̃𝑦

(𝑥̃ 2 + 𝑧̃ 2 )𝜌 𝑑𝑉

𝐼𝑦 =

(a)

𝑑𝑚 = 𝜌𝑑𝑉

𝑑𝑉 = 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑥̃ = 𝑥

𝑧

𝑥̃

𝑧̃

𝑥

𝑧̃ = 𝑧

𝑦 𝑟̃𝑦

Object of revolution 𝑑𝑚 = 𝜌𝑑𝑉

(b)

ℎ

𝐼𝑦 =

𝑟̃𝑦2 𝜌 𝑑𝑉

𝑑𝑉 = ℎ 2𝜋𝑥 𝑑𝑥 𝑟̃𝑦 = 𝑥 𝑥

𝑧 𝑦 Object of revolution 𝑑𝐼𝑦

𝐼𝑦 =

(c)

𝑑𝐼𝑦 = 𝑥

=

𝑑𝐼𝑦 1 (𝑚disk )(𝑟disk )2 2 1 (𝜌𝜋𝑥2 𝑑𝑦)(𝑥2 ) 2

𝑧 𝑦 Symmetric about 𝑥𝑦 and 𝑦𝑧 planes 𝑑𝐼𝑦 (d)

𝐼𝑦 =

𝑑𝐼𝑦

𝑥

Mass moments of inertia are volume- and density-related and hence, are typically more difficult to evaluate by integration than area moments of inertia. Figure 10.16 shows some strategies for determining the mass moments of inertia for an object, using 𝐼𝑦 as an example, with comments as follows. • Object with general shape. The triple integral approach shown in Fig. 10.16(a) has straightforward expressions for 𝑑𝑉 , 𝑥, ̃ 𝑦, ̃ and 𝑧. ̃ This approach is applicable to bodies with general geometry and density, and it can be used to determine 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 . If you are comfortable with determining limits of integration and evaluating triple integrals, then this approach is robust and effective. • Object of revolution—thin shell volume element. An object of revolution where the 𝑦 axis is the axis of revolution is shown in Fig. 10.16(b). To determine 𝐼𝑦 using a single integration, a thin cylindrical shell volume element can be used. The key feature of this volume element is that all material within the element has the same moment arm 𝑟̃𝑦 from the 𝑦 axis. The radius of the shell is taken to be 𝑥, and the thickness is 𝑑𝑥 (these could just as well be taken as 𝑧 and 𝑑𝑧). Example 10.8 uses this approach. The object shown in Fig. 10.16(b) also has mass moments of inertia 𝐼𝑥 and 𝐼𝑧 , and the strategy shown in Fig. 10.16(a) or (c) must be used to determine these. • Object of revolution—thin disk volume element. An object of revolution where the 𝑦 axis is the axis of revolution is shown in Fig. 10.16(c). To determine 𝐼𝑦 using a single integration, a thin circular disk volume element can be used. We consult a table of moments of inertia for common shapes, such as the Table of Properties of Solids at the end of this book, to obtain the mass moment of inertia for a thin circular plate, which is 𝑚𝑟2 ∕2, where 𝑚 is the mass of the plate and 𝑟 is its radius; this expression assumes the density of the plate is uniform. By taking 𝑚 and 𝑟 in this expression to be the mass and radius of the thin disk element, we obtain an expression for the mass moment of inertia for the thin disk element, which we call 𝑑𝐼𝑦 . Once an expression for 𝑑𝐼𝑦 is known, the moment of inertia for the entire object is obtained by 𝐼𝑦 =

∫

𝑑𝐼𝑦 ,

(10.25)

𝑧 Figure 10.16 Strategies for evaluating the mass moments of inertia, using 𝐼𝑦 as an example.

ISTUDY

which requires only a single integration. This volume element may also be used to evaluate the moments of inertia 𝐼𝑥 and 𝐼𝑧 . Examples 10.8 and 10.9 illustrate this.

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Section 10.3

Mass Moments of Inertia

597

• Symmetric object—thin plate volume element. An object that is symmetric about both the 𝑥𝑦 and 𝑦𝑧 planes is shown in Fig. 10.16(d). A thin plate volume element is used, where the geometry of the plate depends on the object’s geometry [Fig. 10.16(d) shows a rectangular plate]. We consult a table of moments of inertia for common shapes, such as the Table of Properties of Solids at the end of this book, to obtain the mass moment of inertia for the volume element 𝑑𝐼𝑦 and then use Eq. (10.25) to obtain the moment of inertia for the entire object. This volume element may also be used to evaluate the moments of inertia 𝐼𝑥 and 𝐼𝑧 . Example 10.10 illustrates this.

Evaluation of moments of inertia using composite shapes The parallel axis theorem, written for composite shapes, becomes 𝐼𝑥 =

𝑛 ∑ ) ( 𝐼𝑥 ′ + 𝑑𝑥2 𝑚 𝑖

Common Pitfall (10.26)

𝑖=1

where 𝑛 is the number of shapes, 𝐼𝑥 ′ is the mass moment of inertia for shape 𝑖 about its mass center 𝑥 ′ axis, 𝑑𝑥 is the shift distance for shape 𝑖 (i.e., the distance between the 𝑥 axis and the 𝑥 ′ axis for shape 𝑖), and 𝑚 is the mass for shape 𝑖. Similar expressions may be written for 𝐼𝑦 and 𝐼𝑧 . To use Eq. (10.26), it is necessary to know the mass moment of inertia for each of the composite shapes about its mass center axis, and this generally must be obtained by integration or, when possible, by consulting a table of moments of inertia for common shapes, such as the Table of Properties of Solids at the end of this book. A common error is to use the parallel axis theorem to relate moments of inertia between two parallel axes where neither of them is a mass center axis. Example 10.12 illustrates this approach.

End of Section Summary In this section, mass moments of inertia 𝐼𝑥 , 𝐼𝑦 , 𝐼𝑧 , 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 were defined to be the second moments of a mass distribution. Some of the key points are as follows: • Mass moments of inertia are measures of how the mass for a particular object is distributed. Moments of inertia have dimensions of (mass)(length)2 . • The products of inertia 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 are measures of the asymmetry of an object’s mass distribution with respect to the 𝑥, 𝑦, and 𝑧 axes. The need for products of inertia typically arises in advanced subjects, so methods for evaluating these quantities are not discussed in this book. • Radii of gyration are alternative measures of the dispersion of mass. The radii of gyration are easily determined if the mass moments of inertia are known, and vice versa.

Parallel axis theorem. Consider the parallel axes 𝑥1 and 𝑥2 , mass center axis 𝑥 ′ , and center of mass at point 𝐺, as shown below. 𝑥1

𝑥

𝑥2

𝐺 If 𝐼𝑥 is known, a common error is to use the 1 parallel axis theorem to directly determine 𝐼𝑥 . In the parallel axis theorem, one of the 2 axes is always a mass center axis. Thus, with 𝐼𝑥 known, the parallel axis theorem must be 1 used twice. The first time 𝐼𝑥 is used to de1 termine 𝐼𝑥 ′ . The second time 𝐼𝑥 ′ is used to determine 𝐼𝑥 . See Example 10.11. 2

598

Chapter 10

Moments of Inertia

E X A M P L E 10.7

Thin, Flat Plate-Type Objects A thin rectangular plate with base 𝑏, height ℎ, thickness 𝑡, and uniform density 𝜌 is shown. Determine the mass moments of inertia 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 about the center of mass for the plate, expressing each in terms of the mass 𝑚 of the plate.

𝑦 𝑡 ℎ

SOLUTION

𝐺 𝑧

𝑥 𝑏

Road Map

We will use Eqs. (10.15)–(10.17) on p. 593 to determine the three moments

of inertia. 𝐼𝑥

Figure 1

Governing Equations & Computation

𝐼𝑥 =

∫

We begin with Eq. (10.15) with 𝑑𝑚 = 𝜌 𝑑𝑉 (𝑦̃2 + 𝑧̃ 2 )𝜌 𝑑𝑉 .

(1)

If the plate is thin, then 𝑧̃ 2 ≪ 𝑦̃2 and hence, 𝑧̃ 2 can be neglected. Using a volume element parallel to the 𝑥 axis as shown in Fig. 2, we obtain 𝑑𝑉 = 𝑏𝑡 𝑑𝑦

𝑑𝑉 = 𝑏𝑡 𝑑𝑦

𝑑𝑦

and 𝑦̃ = 𝑦.

(2)

𝑡 𝑦̃ = 𝑦 𝐺

𝑧

𝑥

Substituting Eq. (2) into Eq. (1), the moment of inertia about the 𝑥 axis is 𝐼𝑥 =

𝑏 Figure 2 To determine 𝐼𝑥 , a volume element parallel to the 𝑥 axis is used to develop expressions for 𝑑𝑉 and 𝑦. ̃

∫

ℎ∕2

𝑦̃2 𝜌 𝑑𝑉 =

∫−ℎ∕2

𝑦2 𝜌𝑏𝑡 𝑑𝑦 = 𝜌𝑡

𝑏𝑦3 |ℎ∕2 𝑏ℎ3 = 𝜌𝑡 . | | −ℎ∕2 3 12

(3)

While the above result is satisfactory, it is customary to report mass moments of inertia in terms of the total mass of an object. The mass of the thin plate is 𝑚 = 𝜌𝑏ℎ𝑡. We express Eq. (3) in terms of mass 𝑚, using the following procedure: 𝐼𝑥 = 𝜌𝑡

𝑏ℎ3 12

(

) 𝑚 𝑚ℎ2 . = 𝜌𝑏ℎ𝑡 12 ⏟⏟⏟

(4)

=1

𝐼𝑦 Governing Equations & Computation 𝑦 𝑡

𝐼𝑦 =

𝑥̃ = 𝑥

𝐺

𝑥

𝑑𝑉 = ℎ𝑡 𝑑𝑥

𝑑𝑉 = ℎ𝑡 𝑑𝑥 𝑑𝑥 Figure 3 To determine 𝐼𝑦 , a volume element parallel to the 𝑦 axis is used to develop expressions for 𝑑𝑉 and 𝑥. ̃

ISTUDY

(𝑥̃ 2 + 𝑧̃ 2 )𝜌 𝑑𝑉 ,

(5)

where 𝑧̃ 2 can be neglected because 𝑧̃ 2 ≪ 𝑥̃ 2 . Using a volume element parallel to the 𝑦 axis as shown in Fig. 3, we obtain

ℎ 𝑧

∫

We begin with Eq. (10.16)

and 𝑥̃ = 𝑥.

(6)

Substituting Eq. (6) into Eq. (5), the moment of inertia about the 𝑦 axis is 𝐼𝑦 =

∫

𝑏∕2

𝑥̃ 2 𝜌 𝑑𝑉 =

∫−𝑏∕2

𝑥2 𝜌ℎ𝑡 𝑑𝑥 = 𝜌𝑡

ℎ𝑥3 |𝑏∕2 ℎ𝑏3 . = 𝜌𝑡 | 3 |−𝑏∕2 12

(7)

Expressing Eq. (7) in terms of mass 𝑚 provides 𝐼𝑦 = 𝜌𝑡

𝑚𝑏2 ℎ𝑏3 𝑚 = . 12 𝜌𝑏ℎ𝑡 12

(8)

ISTUDY

Section 10.3

Mass Moments of Inertia

𝐼𝑧 Governing Equations & Computation

𝐼𝑧 =

∫

We begin with Eq. (10.17) (𝑥̃ 2 + 𝑦̃2 ) 𝜌 𝑑𝑉 .

(9)

For the geometry of the plate in this problem, 𝑥̃ and 𝑦̃ are of comparable size, so neither can be neglected. This presents a difficulty if Eq. (9) is to be evaluated using a single integration. However, some ingenuity provides a straightforward solution, as follows. We break Eq. (9) into two separate integrals 𝐼𝑧 =

∫

𝑥̃ 2 𝜌 𝑑𝑉 +

∫

𝑦̃2 𝜌 𝑑𝑉 .

(10)

The first of the above integrals is identical to Eq. (7), and the second is identical to Eq. (3). Hence, 𝑚𝑏2 𝑚ℎ2 𝑚 2 𝐼𝑧 = (11) + = (𝑏 + ℎ2 ). 12 12 12 In essence, the first and second integrals in Eq. (10) were evaluated using different volume elements. Discussion & Verification

• A thin plate is a common object in structures and machines. Hence, the results obtained here are usually tabulated in handbooks and texts, such as the Table of Properties of Solids at the end of this book. • By letting ℎ or 𝑏 in Fig. 1 become small, the plate considered here becomes a thin rod. For example, if ℎ → 0, then Eqs. (4), (8), and (11) become 𝐼𝑥 = 0 and 𝐼𝑦 = 𝐼𝑧 = 𝑚𝑏2 ∕12. You should verify that these mass moments of inertia agree with those reported for a thin rod in the Table of Properties of Solids at the end of this book (when making this comparison, you must note that some coordinate directions and/or dimensions may be defined differently). A Closer Look While mass moments of inertia and area moments of inertia are generally quite different, they are related for thin flat plates, such as in this example. To see this relationship, rewrite Eq. (3) as

𝐼𝑥 =

∫

𝑦̃2 𝜌 𝑑𝑉 = 𝜌𝑡

∫

𝑦̃2 𝑑𝐴.

(12)

The second integral is obtained by using 𝑑𝑉 = 𝑡 𝑑𝐴 and factoring 𝜌𝑡 outside of the integral because the plate has uniform density and thickness. The second integral is seen to be the definition of the area moment of inertia about the 𝑥 axis [i.e., from Eq. (10.1) on p. 574, (𝐼𝑥 )area = ∫ 𝑦̃2 𝑑𝐴]. Hence, 𝐼𝑥 = 𝜌𝑡(𝐼𝑥 )area = 𝜌𝑡

𝑏ℎ3 , 12

(13)

where we have used the result from Example 10.1 on p. 579 that (𝐼𝑥 )area = 𝑏ℎ3 ∕12 for a rectangular area.

599

600

Chapter 10

Moments of Inertia

E X A M P L E 10.8

Solid of Revolution—Moment of Inertia About Axis of Revolution Determine the mass moment of inertia about the 𝑥 axis for the solid hemisphere of radius 𝑟 and uniform density 𝜌. Express the result in terms of the mass 𝑚 of the object.

𝑦 𝑟

SOLUTION Road Map 𝑥 𝑧

We will show two solutions, the first using a thin shell volume element and the second using a thin disk volume element. Note that the centroid of this object was determined in Example 7.6 on p. 444, and the expressions for 𝑑𝑉 developed in this solution are identical to those used earlier.

Solution 1—Thin shell volume element Governing Equations & Computation We begin with Eq. (10.15) with 𝑑𝑚 = 𝜌 𝑑𝑉 , 𝑥2 + 𝑦 2 + 𝑧2 = 𝑟 2

𝐼𝑥 =

Figure 1

∫

𝑟̃2𝑥 𝜌 𝑑𝑉 .

The thin shell volume element shown in Fig. 2 is used to obtain expressions for 𝑑𝑉 and 𝑟̃𝑥 . The merit of this volume element is that all the material in the element is at the same radial distance 𝑟̃𝑥 from the 𝑥 axis. With 𝑦 being the radius of the shell,

𝑦

𝑑𝑉 = 2𝜋𝑦𝑥 𝑑𝑦 𝑑𝑉

𝑟̃𝑥 = 𝑦 𝑥 𝑥

𝑧

(1)

𝑑𝑦 𝑥=

√ 𝑟2 − 𝑦2

and

𝑟̃𝑥 = 𝑦.

(2)

In Eq. (2), 2𝜋𝑦 is the circumference of the shell, 𝑦 is the radius, 𝑥 is√ the length of the shell, and 𝑑𝑦 is the thickness. Substituting Eq. (2) into Eq. (1), with 𝑥 = 𝑟2 − 𝑦2 , the moment of inertia about the 𝑥 axis is 𝑟 √ 2𝜋𝜌 2 |𝑟 4𝜋𝜌𝑟5 (𝑦 − 𝑟2 )3∕2 (2𝑟2 + 3𝑦2 )| = . (3) 𝐼𝑥 = 𝑦2 𝜌 2𝜋𝑦 𝑟2 − 𝑦2 𝑑𝑦 = |0 ∫0 15 15 Using the expression for 𝑑𝑉 in Eq. (2), the mass 𝑚 of the hemisphere is 𝑟 √ 2𝜋𝜌 2 |𝑟 2𝜋𝜌𝑟3 𝜌 2𝜋𝑦 𝑟2 − 𝑦2 𝑑𝑦 = 𝑚= 𝜌 𝑑𝑉 = (𝑦 − 𝑟2 )3∕2 | = . |0 ∫ ∫0 3 3

(4)

Finally, we express Eq. (3) in terms of mass 𝑚 as follows: Figure 2 A thin shell volume element is used to develop expressions for 𝑑𝑉 and 𝑟̃𝑥 .

4𝜋𝜌𝑟5 ( 𝑚 ) 2𝑚𝑟2 . = 3 15 5 2𝜋𝜌𝑟 ∕3

(5)

Solution 2—Thin disk volume element Governing Equations & Computation We begin with Eq. (10.25) on p. 596,

𝑦 𝑑𝐼𝑥 𝑦=

√

𝐼𝑥 =

𝑟2 − 𝑥 2

𝑥

𝑧

𝑑𝑥 Figure 3 A thin circular disk volume element is used to develop an expression for 𝑑𝐼𝑥 .

ISTUDY

𝐼𝑥 =

∫

𝑑𝐼𝑥 ,

(6)

where 𝑑𝐼𝑥 is the moment of inertia of the thin circular disk mass element shown in Fig. 3 about the 𝑥 axis. From the Table of Properties of Solids, or from Fig. 10.16(c), 𝑑𝐼𝑥 = 21 (𝑚disk )(𝑟disk )2 = 21 (𝜌 𝜋𝑦2 𝑑𝑥)(𝑦)2 ,

(7) √ where 𝑦 is the radius of the disk. Substituting Eq. (7) into Eq. (6), with 𝑦 = 𝑟2 − 𝑥2 , the moment of inertia about the 𝑥 axis is 𝑟 ) ( 4 ( 2 ) 𝑟 𝑥 𝑟2 𝑥3 𝑥5 |𝑟 4𝜋𝜌𝑟5 1 2 2 − + , (8) 𝜌𝜋 𝑟 − 𝑥 𝑑𝑥 = 𝜋𝜌 𝐼𝑥 = | = ∫0 2 2 3 10 |0 15 which is identical to Eq. (3). Hence, we obtain the same result, 𝐼𝑥 = 2𝑚𝑟2 ∕5. Discussion & Verification

Knowing the mass moments of inertia for a cylinder and cone, it is possible to show that the result obtained here is reasonable; see Prob. 10.59.

ISTUDY

Section 10.3

601

Mass Moments of Inertia

E X A M P L E 10.9

Solid of Revolution—Additional Moments of Inertia

For the solid hemisphere of radius 𝑟 and uniform density 𝜌 considered in Example 10.8, shown again in Fig. 1, determine the mass moments of inertia about the 𝑦 and 𝑧 axes. Express the result in terms of the mass 𝑚 of the object.

𝑦 𝑟

SOLUTION Road Map Because of the symmetry of the object, 𝐼𝑦 = 𝐼𝑧 , and hence we will determine only one of these, arbitrarily choosing 𝐼𝑧 . Unlike Example 10.8 where we had the choice of two volume elements, in this example we must use a thin circular disk volume element (that is, if we are to determine 𝐼𝑧 using a single integration). Governing Equations & Computation

𝑥 𝑧

We begin with Eq. (10.25) on p. 596 (written in

terms of 𝑧) 𝑑𝐼𝑧 , (1) ∫ where 𝑑𝐼𝑧 is the moment of inertia of the thin circular disk mass element shown in Fig. 2 about the 𝑧 axis. Using the parallel axis theorem, Eq. (10.24) on p. 596, the moment of inertia of the thin disk element about the 𝑧 axis is 𝐼𝑧 =

𝑑𝐼𝑧 = 𝑑𝐼𝑧 ′ +

𝑑𝑧2

𝑥2 + 𝑦 2 + 𝑧2 = 𝑟 2 Figure 1

𝑦 𝑑𝐼𝑧

(2)

𝑑𝑚,

𝑦=

√

𝑟2 − 𝑥2

𝑧′

where 𝑑𝐼𝑧 ′ is the moment of inertia of the disk element about the axis passing through the disk’s mass center, 𝑑𝑚 is the mass of the disk, and 𝑑𝑧 is the distance between the 𝑧 and 𝑧 ′ axes. From the Table of Properties of Solids at the end of this book, the moment of inertia of a thin disk is 𝑑𝐼𝑧 ′ = 14 (𝑚disk )(𝑟disk )2 = 14 (𝜌𝜋𝑦2 𝑑𝑥)(𝑦)2 ,

𝑥

(3)

𝑧

𝑑𝑚 = 𝜌𝜋𝑦 𝑑𝑥,

(4)

𝑑𝑧 = 𝑥

𝑑𝑧 = 𝑥,

(5)

2

√ where 𝑦 is the radius of the disk. Using Eqs. (2)–(5) and noting that 𝑦 = 𝑟2 − 𝑥2 , Eq. (1) becomes 𝑟

𝑟 𝜋𝜌 2 (𝑟 − 𝑥2 )2 𝑑𝑥 + 𝑥2 𝜋𝜌(𝑟2 − 𝑥2 ) 𝑑𝑥 ∫0 4 ∫0 ) ) ( 2 3 𝜋𝜌 ( 4 2𝑟2 𝑥3 𝑥5 |𝑟 𝑥5 |𝑟 𝑟 𝑥 + − 𝑟 𝑥− = | + 𝜋𝜌 | 4 3 5 |0 3 5 |0 4𝜋𝜌𝑟5 . = 15

𝐼𝑧 =

(6)

Noting from Example 10.8 that the mass of the hemisphere is 𝑚 = 2𝜋𝜌𝑟3 ∕3, Eq. (6) expressed in terms of 𝑚 is 𝐼𝑧 =

4𝜋𝜌𝑟5 ( 𝑚 ) 2𝑚𝑟2 . = 3 15 5 2𝜋𝜌𝑟 ∕3

(7)

Because of symmetry, 𝐼𝑦 = 𝐼𝑧 , hence 𝐼𝑦 = 2𝑚𝑟2 ∕5. Comparing 𝐼𝑦 and 𝐼𝑧 found here with the results for 𝐼𝑥 from Example 10.8 shows the unexpected result that all three of these are equal. Discussion & Verification

𝑧

𝑑𝑥 Figure 2 A thin circular disk volume element is used to develop an expression for 𝑑𝐼𝑧 .

602

Chapter 10

Moments of Inertia

E X A M P L E 10.10

Symmetric Object The tapered prism is made of aluminum and is symmetric about the xy and xz planes. Determine the mass moments of inertia about the 𝑥 and 𝑦 axes.

𝑧 10 mm 10 mm

SOLUTION

50 mm

Road Map

We will use a thin rectangular plate volume element, with the strategy described in Fig. 10.16(d) on p. 596, to evaluate 𝐼𝑥 and 𝐼𝑦 .

20 mm 20 mm

𝑦

𝑥 10 mm 5 mm

We begin with Eq. (10.25) on p. 596 (written in

terms of 𝑥) 𝑑𝐼𝑥 , (1) ∫ where 𝑑𝐼𝑥 is the moment of inertia of the thin rectangular plate mass element shown in Fig. 2 about the 𝑥 axis. Note that the 𝑥 axis and the 𝑥 ′ axis passing through the mass center of the thin plate element coincide, and therefore no parallel axis shift is required. From the Table of Properties of Solids at the end of this book, 𝐼𝑥 =

Figure 1

𝑧

𝑑𝐼𝑥 , 𝑑𝐼𝑦

𝐼𝑥 Governing Equations & Computation

𝑑𝐼𝑥 =

𝑏

1 (𝑚plate )(𝑏2 12

+ ℎ2 ).

(2)

The dimensions 𝑏 and ℎ of the thin plate element are functions of position as follows: 𝑦 ℎ

𝑥

𝑦

𝑥

𝑏 = 20 mm − (0.2)𝑥

(3)

The mass of the plate is the product of density 𝜌 and the volume of the plate, hence, 𝑚plate = 𝜌 𝑏ℎ 𝑑𝑥.

𝑑𝑥 Figure 2 A thin rectangular plate volume element is used to develop expressions for 𝑑𝐼𝑥 and 𝑑𝐼𝑦 .

ISTUDY

and ℎ = 40 mm − (0.4)𝑥.

(4)

Combining Eqs. (1), (2), and (4) provides 𝐼𝑥 =

∫0

50 mm

1 𝜌 𝑏ℎ(𝑏2 12

+ ℎ2 ) 𝑑𝑥.

(5)

From Table 1.5 on p. 17, the density of aluminum is 𝜌 = 2710 kg∕m3 = 2.710 × 10−6 kg∕mm3 . Substituting 𝑏 and ℎ from Eq. (3) and the value for 𝜌 into Eq. (5) and carrying out the integration yield 𝐼𝑥 = 7.001 kg⋅mm2 .

(6)

𝐼𝑦 Governing Equations & Computation

We begin with Eq. (10.25) on p. 596,

𝐼𝑦 =

∫

𝑑𝐼𝑦 ,

(7)

where 𝑑𝐼𝑦 is the moment of inertia of the thin rectangular plate mass element shown in Fig. 2 about the 𝑦 axis. Since the 𝑦 and 𝑦 ′ axes do not coincide, the parallel axis theorem, Eq. (10.23) on p. 596, is used to write the moment of inertia of the thin plate element about the 𝑦 axis as 𝑑𝐼𝑦 = 𝑑𝐼𝑦 ′ + 𝑑𝑦2 𝑑𝑚, (8) where 𝑑𝐼𝑦 ′ is the moment of inertia of the plate element about the 𝑦 ′ axis passing through its mass center, 𝑑𝑚 is the mass of the plate element, and 𝑑𝑦 is the distance between the 𝑦 and 𝑦 ′ axes, namely, 𝑑𝑦 = 𝑥. From the Table of Properties of Solids at the end of this book, the moment of inertia of a thin plate is 𝑑𝐼𝑦 ′ =

1 (𝑚plate )(ℎ)2 , 12

(9)

ISTUDY

Section 10.3

Mass Moments of Inertia

603

where 𝑚plate is given by Eq. (4). Combining Eqs. (4) and (7)–(9) provides 𝐼𝑦 =

∫0

50 mm

1 𝜌 𝑏ℎ ℎ2 12

𝑑𝑥 +

∫0

𝑧

50 mm

𝑥2 𝜌 𝑏ℎ 𝑑𝑥.

(10) 50 mm

Substituting 𝑏 and ℎ from Eq. (3) and the value for 𝜌 into Eq. (10) and carrying out the integration yield 𝐼𝑦 = 41.73 kg⋅mm2 .

(11)

30 mm

𝑦

𝑥 Discussion & Verification

A rough check of accuracy can be obtained by using the Table of Properties of Solids at the end of this book to evaluate the moments of inertia for the uniform rectangular prism shown in Fig. 3, where dimensions have been chosen to approximate the shape and volume of the tapered prism. For the uniform prism shown in Fig. 3, you should verify that 𝐼𝑥 = 5.716 kg⋅mm2 and 𝐼𝑦 = 55.39 kg⋅mm2 . The results for the tapered prism in Eqs. (6) and (11) are in reasonable agreement with these.

15 mm Figure 3 A uniform rectangular prism with dimensions that approximate those of the tapered prism shown in Fig. 1.

604

Chapter 10

Moments of Inertia

E X A M P L E 10.11

Experimental Determination of Mass Moments of Inertia A connecting rod from a gasoline engine weighs 3.28 lb and has its center of gravity at point 𝐺. When the connecting rod is supported at point 𝑂 and is allowed to oscillate as a pendulum, 0.816 s is required for one full cycle of motion. Determine the mass moments of inertia about points 𝐴 and 𝐺.

𝑦 𝑂

𝑥 0.45 f t

SOLUTION 𝐺

Road Map

We will use Eq. (1), described in the margin note on this page, to determine 𝐼𝑂𝑧 . We will then use the parallel axis theorem to determine the moments of inertia about points 𝐴 and 𝐺. Note that, as discussed in the margin note on p. 597, to determine 𝐼𝐴𝑧 it is necessary to first know 𝐼𝐺𝑧 .

0.22 f t 𝐴 Figure 1

ISTUDY

𝐼𝐺𝑧 Governing Equations & Computation

Using Eq. (1), the mass moment of inertia of the connecting rod about the 𝑧 axis through point 𝑂 is 𝐼𝑂𝑧 =

Helpful Information Experimental determination of mass moments of inertia. The object shown has mass 𝑚 and center of gravity at point 𝐺. 𝑦

(2)

To express Eq. (2) in terms of the U.S. Customary mass unit (recall from the definition in Table 1.1 on p. 9, that 1 slug ≡ 1 lb⋅s2 ∕f t), we carry out the unit conversion ( )( 𝐼𝑂𝑧 = 0.02489 f t ⋅lb⋅s2

slug ) = 0.02489 slug⋅f t 2 . lb⋅s2 ∕f t

(3)

The parallel axis theorem, Eq. (10.24) on p. 596, rearranged for the moment of inertia about the 𝑧 axis through the mass center 𝐺, is

𝑥

𝑂

(0.816 s)2 (3.28 lb)(0.45 f t) = 0.02489 f t ⋅lb⋅s2 . 4𝜋 2

𝐿

𝐼𝐺𝑧 = 𝐼𝑂𝑧 − 𝑑𝑧2 𝑚

𝐺

By supporting the object with a smooth pin at point 𝑂 and allowing it to oscillate as a pendulum, the mass moment of inertia about the 𝑧 axis through point 𝑂 (i.e., perpendicular to the figure) is 𝐼𝑂𝑧 =

𝑇 2 𝑚𝑔𝐿 , 4𝜋 2

) 3.28 lb 32.2 f t∕s2 ( slug ) = 0.02489 slug⋅f t 2 − 0.02063 f t ⋅lb⋅s2 lb⋅s2 ∕f t = 0.02489 slug⋅f t 2 − (0.45 f t)2

𝑚

(

= 4.267×10−3 slug⋅f t 2 .

(4)

𝐼𝐴𝑧 Governing Equations & Computation

(1)

where 𝑇 is the time required for one full cycle of motion. Thus, by locating the center of gravity for an object and carefully measuring 𝑇 , 𝑚 (or weight 𝑚𝑔), and 𝐿, the mass moment of inertia is easily determined. Remark: You will be able to derive Eq. (1) when you study dynamics. You may find this surprising, but the time 𝑇 required for one full cycle of motion is independent of the amplitude of motion provided the amplitude is not too large.

With the moment of inertia about the mass center known from Eq. (4), the parallel axis theorem can be used again to determine the moment of inertia about the 𝑧 axis through point 𝐴. Thus, 𝐼𝐴𝑧 = 𝐼𝐺𝑧 + 𝑑𝑧2 𝑚

) 3.28 lb 32.2 f t∕s2 ( slug ) = 4.267×10−3 slug⋅f t 2 + 4.930×10−3 f t ⋅lb⋅s2 lb⋅s2 ∕f t = 4.267×10−3 slug⋅f t 2 + (0.22 f t)2

= 9.198×10−3 slug⋅f t 2 .

Discussion & Verification

(

(5)

A rough check of accuracy can be obtained by approximating the connecting rod by a slender rod with the same mass and with length approximating that of the connecting rod. Problem 10.76 asks you to carry out this calculation.

ISTUDY

Section 10.3

605

Mass Moments of Inertia

E X A M P L E 10.12

Composite Shapes

The hammer consists of a cast iron head and a wood handle. The iron head has 7000 kg∕m3 density, and its shape is a rectangular prism with a circular hole. The wood handle has 500 kg∕m3 density, and its shape is a circular cylinder. Determine the mass moment of inertia about the 𝑧 axis and the corresponding radius of gyration.

𝑦 30 mm

30 mm

80 mm

𝑥

20 mm

SOLUTION

𝑧

Road Map

The hammer consists of a rectangular prism and circular cylinder, so it will be convenient to determine 𝐼𝑧 using composite shapes with the parallel axis theorem. Governing Equations & Computation A combination of three composite shapes is shown in Fig. 2, where the mass for shape 3 is negative. The mass of each composite shape is

𝑚1 = 𝜌wood 𝑉1 = (500 kg∕m3 )𝜋(0.010 m)2 (0.150 m) = 0.02356 kg,

(1)

𝑚2 = 𝜌iron 𝑉2 = (7000 kg∕m3 )(0.040 m)(0.040 m)(0.060 m) = 0.6720 kg,

(2)

𝑚3 = −𝜌iron 𝑉3 = −(7000 kg∕m3 )𝜋(0.010 m)2 (0.040 m) = −0.08796 kg.

(3)

Figure 1 𝑦 30 mm 120 mm 45 mm 𝑥

20 mm 𝑧

3 ∑ ( ) 𝐼𝑧 ′ + 𝑑𝑧2 𝑚 𝑖 .

𝑦

𝑚2

𝑥

(4)

100 mm

𝑧

40 mm

80 mm

𝑥 100 mm

𝑧

(5)

𝑧

𝑚3

Figure 2 A combination of three composite shapes to determine 𝐼𝑧 , and distances from the center of mass of each composite shape to the 𝑧 axis. Note that 𝑚3 < 0.

Noting that the total mass of the hammer is 𝑚 = 𝑚1 + 𝑚2 + 𝑚3 = 0.6076 kg, the radius of gyration is easily determined using Eq. (10.21) on p. 595 as √ √ 𝐼𝑧 6210 kg⋅mm2 𝑘𝑧 = (6) = = 101.1 mm. 𝑚 0.6076 kg

𝑦 𝑚 = 0.6076 kg 𝑥

Discussion & Verification

Imagine that all of the mass 𝑚 of the hammer is concentrated at a point, as shown in Fig. 3. While it is physically impossible to do this, the radius of gyration 𝑘𝑧 has the interpretation of being the distance from the 𝑧 axis to the point so that the point mass has the same moment of inertia 𝐼𝑧 as the original object. Considering that the mass of the wood handle is small compared to the iron head, and that the shift distance 𝑑𝑧 for the wood handle in the parallel axis theorem is small compared to that for the head, we expect the radius of gyration to locate a point close to the center of mass of just the head. Thus, the value determined in Eq. (6) is reasonable.

40 mm

𝑦

𝑚1

[3(10 mm)2 + (150 mm)2 ] + (45 mm)2 𝑚1 12 𝑚 + 2 [(40 mm)2 + (60 mm)2 ] + (100 mm)2 𝑚2 12 𝑚3 [3(10 mm)2 + (40 mm)2 ] + (100 mm)2 𝑚3 + 12

𝑧 40 mm

For each composite shape, the Table of Properties of Solids at the end of this book is used to obtain the mass moment of inertia about the appropriate axis through the composite shape’s mass center. Note that when using such a table, it is often the case that the xyz axes in the table are different from the xyz axes used in your particular problem. Hence, you must be careful to obtain the correct expressions. Equation (4) becomes

= 6210 kg⋅mm2 .

30 mm

80 mm

𝑖=1

𝐼𝑧 =

𝑚1

𝑧

The parallel axis theorem for composite shapes, Eq. (10.26) on p. 597, is 𝐼𝑧 =

40 mm

40 mm

𝑧

𝑘𝑧 = 101.1 mm

Figure 3 The radius of gyration 𝑘𝑧 is the distance from the 𝑧 axis that a point mass 𝑚 should be positioned so that it has the same mass moment of inertia 𝐼𝑧 as the original object.

606

Chapter 10

Moments of Inertia

Problems Problem 10.48 The uniform slender rod has mass 𝑚 and length 𝑙. Use integration to show that the mass moment of inertia about the 𝑦 axis is 𝐼𝑦 = 𝑚𝑙2 ∕12.

𝑦 𝑙∕2 𝑙∕2

𝑧

Problems 10.49 and 10.50

𝑥

Figure P10.48

For the subproblem below, the uniform circular plate has thickness 𝑡, radius 𝑟, and density 𝜌. 𝑦

Problem 10.49 Use integration with a thin cylindrical shell mass element to determine the mass moment of inertia about the 𝑥 axis. Express your answer in terms of the mass 𝑚 of the plate.

𝑡 𝑟 𝑥

𝑧

Problem 10.50

Assuming 𝑡 ≪ 𝑟, use integration with a mass element parallel to the 𝑦 axis (as in Example 10.2 on p. 580) to determine the mass moment of inertia about the 𝑦 axis. Express your answer in terms of the mass 𝑚 of the plate.

Figure P10.49 and P10.50

Problems 10.51 and 10.52 For the subproblem below, the uniform cylinder has length 𝐿, radius 𝑅, and density 𝜌.

𝑦 𝑅 𝑥

Problem 10.52 Use integration with a thin disk mass element to determine the mass moment of inertia about the 𝑦 axis. Express your answer in terms of the mass 𝑚 of the cylinder.

𝐿

𝑧

Problem 10.51 Use integration with a thin cylindrical shell mass element to determine the mass moment of inertia about the 𝑥 axis. Express your answer in terms of the mass 𝑚 of the cylinder.

Figure P10.51 and P10.52

Problems 10.53 and 10.54

𝑦

For the uniform solid cone with length 𝐿 and radius 𝑅, use integration to determine the mass moment of inertia indicated in the subproblem below, expressing your answer in terms of the mass 𝑚 of the cone.

𝑅 𝑥

Problem 10.53

𝐼𝑥 .

Problem 10.54

𝐼𝑧 .

𝐿

𝑧

Figure P10.53 and P10.54

Problems 10.55 through 10.57 𝑦

𝑎 𝑎 𝑥

𝑧

ℎ

The tapered solid prism shown has density 𝜌 and rectangular cross section. Use integration to determine the mass moment of inertia indicated in the subproblem below, expressing your answer in terms of the mass 𝑚 of the prism and parameters such as 𝑎, 𝑏, and ℎ. Problem 10.55

𝐼𝑥 .

Problem 10.56

𝐼𝑦 .

Problem 10.57

𝐼𝑧 .

𝑏 𝑏

Figure P10.55 and P10.57

ISTUDY

ISTUDY

Section 10.3

607

Mass Moments of Inertia

Problem 10.58 The Table of Properties of Solids at the end of this book shows the mass moments of inertia for a uniform sphere and hemisphere are both 𝐼𝑧 = 2𝑚𝑟2 ∕5. If a sphere and hemisphere have the same radius and density, does this mean their mass moments of inertia 𝐼𝑧 are the same? Explain. Note: Concept problems are about explanations, not computations.

Problem 10.59 In Example 10.8 on p. 600, the mass moment of inertia of a hemisphere about its axis of revolution was found to be 2𝑚𝑟2 ∕5. Show that this result is between those for a cylinder and a cone, both having radius 𝑟 and length 𝑟 (the mass moments of inertia for these are given in the Table of Properties of Solids at the end of this book). Discuss why this result is expected. Note: Concept problems are about explanations, not computations.

Problems 10.60 and 10.61 For the subproblem below, a solid hemisphere is constructed of materials with densities 𝜌0 and 𝜌0 ∕2 as shown.

𝑦

𝑟 2

Problem 10.60

𝑟 2

𝜌0 2

(a) Fully set up the integrals, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis.

𝜌0 𝑥

(b) Evaluate the integrals obtained in Part (a) using computer software such as Mathematica or Maple. 𝑧

Problem 10.61

(a) Fully set up the integrals, including limits of integration, that will yield the mass moment of inertia about the 𝑦 axis. (b) Evaluate the integrals obtained in Part (a) using computer software such as Mathematica or Maple.

𝑥2 + 𝑦 2 + 𝑧2 = 𝑟 2 Figure P10.60 and P10.61

Problem 10.62 For the solid of revolution shown, determine the mass moment of inertia about the 𝑥 axis. The material has specific weight 𝛾 = 0.409 lb∕in.3 . Report your answer using slugs and inches.

A solid of revolution is produced by revolving the area shown 360◦ around the 𝑦 axis. Use integration to determine the mass moment of inertia about the axis of revolution assuming the solid has uniform density. Express your answer in terms of the mass 𝑚 of the object. 𝑦 𝑦 = 1∕𝑥

𝑦=ℎ 1−

𝑥2 𝑎2

1 in. ℎ 𝑥 𝑧

1 in.

Figure P10.63

𝑥2 + 𝑦 2 = (0.22 in.)2 𝑦

𝑎

𝑥

Figure P10.64

𝑥

0.44 in. 0.80 in. Figure P10.62

𝑦

0.5 in.

𝑥 𝑧

Problems 10.63 and 10.64

2 in.

𝑦

0.18 in.

608

Chapter 10

Moments of Inertia Problems 10.65 and 10.66

For the subproblem below, a solid has a cone-shaped cavity and uniform density.

𝑦 = 𝑅( 𝐿𝑥 )1∕3

𝑦

𝑅

Problem 10.65 𝑥

𝑅 2

𝐿

𝑧

(a) Fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis. (b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple. Express your answer in terms of the mass 𝑚 of the object. Problem 10.66

Figure P10.65 and P10.66

(a) Fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑦 axis. (b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple. Express your answer in terms of the mass 𝑚 of the object.

Problems 10.67 and 10.68 For the subproblem below, a thin-walled hollow cone has uniform density with thicknesses 𝑡1 at the left-hand end and 𝑡2 at the right-hand end.

𝑦 𝑅

𝑡1

Problem 10.67 𝑥

𝑧

𝐿

𝑡2

Figure P10.67 and P10.68

(a) If the cone’s thickness is uniform with 𝑡1 = 𝑡2 = 𝑡0 , fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis. (b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple. Express your answer in terms of the mass 𝑚 of the object. Problem 10.68

(a) If the cone’s thickness varies linearly from 𝑡1 = 2𝑡0 at the left-hand end to 𝑡2 = 𝑡0 at the right-hand end, fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis. (b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple. Express your answer in terms of the mass 𝑚 of the object.

𝑦 2 cm

𝑦 = 1 + 𝑥2

A plastic part with 1100 kg∕m3 density is produced by revolving the area shown 360◦ around an axis of revolution. Determine the mass moment of inertia about the axis of revolution indicated in the subproblem below. Express your answer in units of kg⋅cm2 .

1 cm

0 0

𝑦=𝑥 𝑥 1 cm

Figure P10.69 and P10.70

ISTUDY

Problems 10.69 and 10.70

Problem 10.69

The axis of revolution is the 𝑥 axis.

Problem 10.70

The axis of revolution is the 𝑦 axis.

ISTUDY

Section 10.3

609

Mass Moments of Inertia

Problems 10.71 through 10.74 For the subproblem below, a solid hemisphere has a cone-shaped cavity and uniform density.

𝑦

Problem 10.71

(a) Fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis.

𝑅 𝑅∕2

(b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple. Express your answer in terms of the mass 𝑚 of the object.

𝑥

Problem 10.72 Determine the mass moment of inertia about the 𝑥 axis using composite shapes. Express your answer in terms of the mass 𝑚 of the object.

𝑧 𝑥2 + 𝑦 2 + 𝑧2 = 𝑅2

Problem 10.73

(a) Fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑧 axis.

Figure P10.71–P10.74

(b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple. Express your answer in terms of the mass 𝑚 of the object. Problem 10.74 Determine the mass moment of inertia about the 𝑧 axis using composite shapes. Express your answer in terms of the mass 𝑚 of the object.

Problem 10.75 A throwing toy is molded of uniform foam. It consists of an oblong-shaped portion, a cylindrical portion, and four rectangular fins. The density of the oblong and cylindrical shapes is 100 kg∕m3 , and each of the fins has 1.8×10−3 kg mass. (a) Fully set up the integral, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis for the oblong-shaped portion. (b) Evaluate the integral obtained in Part (a) using computer software such as Mathematica or Maple.

𝑦

𝑦 = 𝑥(1 − 𝑥∕200 mm) 10 mm 𝑥 30 mm

𝑧

200 mm 40 mm

60 mm

Figure P10.75

(c) You should find the result of Part (b) to be 79.8 kg⋅mm2 . Using this value, determine the total mass moment of inertia about the 𝑥 axis for the toy.

Problem 10.76 Approximate the connecting rod in Example 10.11 on p. 604 with a uniform slender rod. Take the length of this rod to be 0.8 f t and the mass to be the same as that for the connecting rod. Determine the mass moment of inertia of this rod about its end, and compare to the value for 𝐼𝑂𝑧 found for the connecting rod in Example 10.11.

Problems 10.77 through 10.79 𝑦

A uniform rectangular prism with mass 𝑚 is shown. Beginning with the appropriate mass moment of inertia given in the Table of Properties of Solids at the end of this book, use the parallel axis theorem to determine the mass moment of inertia of the prism about the axis indicated in the subproblem below. Problem 10.77

𝑏 𝑥 𝑐

The 𝑥 axis. 𝑧

Problem 10.78

The 𝑦 axis.

Problem 10.79

The 𝑧 axis.

𝑎

Figure P10.77–P10.79

610

Chapter 10

Moments of Inertia

𝑂

Problem 10.80 A handwheel for a machine has 0.8 kg mass and center of gravity at point 𝐺. When the handwheel is supported at point 𝑂 and is allowed to oscillate as a pendulum, 1.12 s is required for one full cycle of motion. Determine the mass moment of inertia about the axis perpendicular to the figure and passing through point 𝐵. Hint: See the helpful information margin note on p. 604.

120 mm 𝐵

40 mm 𝐺

70 mm

𝐴

Problem 10.81 Figure P10.80 𝑂 0.25 in.

𝐴

1.35 in.

A sector gear for a machine has 0.355 lb weight and center of gravity at point 𝐺. When the sector gear is supported at point 𝑂 and is allowed to oscillate as a pendulum, 0.379 s is required for one full cycle of motion. Determine the mass moment of inertia about the axis perpendicular to the figure and passing through point 𝐴. Hint: See the helpful information margin note on p. 604.

Problem 10.82 𝐺

The plate shown has uniform thickness and is made of material with specific weight 0.7 lb∕in.2 . Determine the mass moment of inertia about the axis perpendicular to the plate and passing through point 𝐴. Figure P10.81

𝑦

0.4 in.

0.4 in.

0.2 in. 0.4 in.

1m 0.1 in.

𝐴

0.4 in.

0.2 in.

1m 0.2 in. Figure P10.82

1m

𝑥

1m

𝑧 Figure P10.83

Problem 10.83 The antenna shown is constructed of identical small-diameter uniform rods, each having 0.25 kg∕m mass. Determine the mass moment of inertia of the antenna about the:

𝑦

(a) 𝑥 axis. (b) 𝑦 axis. 4 in.

4 in.

(c) 𝑧 axis. 𝑥

4 in.

Problem 10.84 An object is constructed by welding together three small-diameter uniform identical rods, each having quarter-circular shape and 0.5 lb weight. Determine the mass moment of inertia of the object about the 𝑥 axis.

𝑧 Figure P10.84

60 mm 50 mm 𝑥

𝑧

20 mm Figure P10.85 and P10.86

ISTUDY

Problems 10.85 and 10.86

160 mm aluminum

brass

𝑦

An object is constructed of a brass rod and aluminum cylinder having densities of 8500 kg∕m3 and 2700 kg∕m3 , respectively. The brass rod fills the hole in the aluminum cylinder. Determine the mass moment of inertia of the object about the axis indicated in the subproblem below. Problem 10.85

𝑥 axis.

Problem 10.86

𝑦 axis.

ISTUDY

Section 10.3

611

Mass Moments of Inertia

Problem 10.87 The cam shown consists of a circular cylinder 𝐴 with a circular hole to reduce its moment of inertia and a circular shaft 𝐵𝐶 about which it rotates. If the cam is made of cast iron with 7200 kg∕m3 density, determine the mass moment of inertia about the 𝑧 axis. 15 mm 𝐴 20 mm

20 mm 40 mm

25 mm 𝐵 𝐶

5 mm

50 mm

𝑧 Figure P10.87

Problems 10.88 through 10.90 A bracket is constructed of a thin semicircular plate of uniform thickness having 0.05 lb∕in.2 specific weight and a thick rectangular plate of uniform thickness having 0.25 lb∕in.3 specific weight. The bracket is symmetric about the 𝑥𝑧 plane. Determine the mass moment of inertia of the object about the axis indicated in the subproblem below. Report your answer using slugs and inches. Problem 10.88

𝑥 axis.

Problem 10.89

𝑦 axis.

Problem 10.90

𝑧 axis.

𝑧

0.8 in. 1.5 in.

3 in.

0.6 in. radius

4 in. 𝑦

𝑥 Figure P10.88–P10.90 𝑦

Problems 10.91 and 10.92

5 in.

An object with 6 lb weight consists of a cylinder and hemisphere of the same material. Determine the mass moment of inertia of the object about the axis indicated in the subproblem below. Report your answer using slugs and inches. Problem 10.91

𝑥 axis.

Problem 10.92

𝑦 axis.

3 in.

3 in. 𝑥

𝑧 Figure P10.91 and P10.92

Problems 10.93 through 10.95 𝑧

The object shown is made of uniform plastic and weighs 2.34 lb. Determine the mass moment of inertia of the object about the axis indicated in the subproblem below. Report your answer using slugs and feet. Problem 10.93

𝑥 axis.

Problem 10.94

𝑦 axis.

Problem 10.95

𝑧 axis.

2.5 in.

2 in. 1.5 in. 2 in.

1.5 in. 2 in. 𝑥 Figure P10.93–P10.95

2 in. 1 in. 2 in.

𝑦

612

Chapter 10

Moments of Inertia

10.4 C h a p t e r R e v i e w Important definitions, concepts, and equations of this chapter are summarized. For equations and/or concepts that are not clear, you should refer to the original equation and page numbers cited for additional details.

Area moments of inertia The area moments of inertia for the area shown in Fig. 10.17 are defined to be

𝑦 𝐴

Eqs. (10.1)–(10.4), p. 574 𝐶

𝑦̃ 𝑟̃ 𝑂

𝑑𝐴

𝐼𝑥 =

∫

𝑦̃2 𝑑𝐴,

𝐼𝑦 =

∫

𝑥̃ 2 𝑑𝐴,

𝐽𝑂 =

∫

𝑟̃2 𝑑𝐴

𝐼𝑥𝑦 =

∫

𝑥̃ 𝑦̃ 𝑑𝐴,

𝑥

𝑥̃

Figure 10.17 An area 𝐴 with centroid 𝐶. 𝑥̃ and 𝑦̃ are the 𝑥 and 𝑦 positions, respectively, of the centroid of area element 𝑑𝐴. 𝑟̃ is the distance from point 𝑂 to area element 𝑑𝐴.

=

𝐼𝑥 + 𝐼𝑦 ,

where 𝑥, ̃ 𝑦, ̃ and 𝑟̃ are defined in Fig. 10.17. Note that the definitions of these are identical to those used in Chapter 7. For example, 𝑥̃ is the distance (i.e., moment arm) from the 𝑦 axis to the centroid of area element 𝑑𝐴; 𝐼𝑥 is the area moment of inertia about the 𝑥 axis; 𝐼𝑦 is the area moment of inertia about the 𝑦 axis; 𝐽𝑂 is the polar moment of inertia of the area about point 𝑂; and 𝐼𝑥𝑦 is the product of inertia of the area about point 𝑂. Area moments of inertia have dimensions of (length)4 .

Area radius of gyration. The radii of gyration of an area 𝐴 are defined as

√ 𝑘𝑥 =

𝐼𝑥 𝐴

,

Eq. (10.5), p. 576 √ √ 𝐼𝑦 𝐽𝑂 𝑘𝑦 = , 𝑘𝑂 = , 𝐴 𝐴

where 𝑘𝑥 is the radius of gyration of the area about the 𝑥 axis;

𝑦 𝑦

𝑘𝑦 is the radius of gyration of the area about the 𝑦 axis;

𝑑𝑦 𝐴

𝑂 𝑑 𝑂

𝑘𝑂 is the polar radius of gyration of the area about point 𝑂; and

𝐶 𝑥

Radii of gyration have units of length.

𝑑𝑥 𝑥

Figure 10.18 An area 𝐴 with centroid at point 𝐶. The 𝑥 and 𝑥 ′ axes are parallel with separation distance 𝑑𝑥 , and the 𝑦 and 𝑦 ′ axes are parallel with separation distance 𝑑𝑦 .

ISTUDY

𝐴 is the area of the shape.

Parallel axis theorem for area moments of inertia. Consider the area 𝐴 with centroid 𝐶 shown in Fig. 10.18. A centroidal 𝑥 ′ 𝑦 ′ coordinate system is defined, with origin 𝑂 ′ positioned at the centroid of the area. The 𝑥 and 𝑦 axes are parallel to the 𝑥 ′ and 𝑦 ′ axes, respectively, with separation distances 𝑑𝑥 and 𝑑𝑦 . The parallel axis theorem relates the area moments of inertia with respect to the 𝑥 and 𝑦 axes to the centroidal area moments of inertia as follows:

ISTUDY

Section 10.4

Chapter Review

613

Eqs. (10.6)–(10.9), p. 585 𝐼𝑥 = 𝐼𝑥 ′ + 𝑑𝑥2 𝐴, 𝐼𝑦 = 𝐼𝑦 ′ + 𝑑𝑦2 𝐴, 𝐽𝑂 = 𝐽𝑂 ′ + 𝑑 2 𝐴, 𝐼𝑥𝑦 = 𝐼𝑥 ′ 𝑦 ′ + 𝑑𝑥 𝑑𝑦 𝐴. The parallel axis theorem, written for composite shapes, is Eq. (10.14), p. 586 𝐼𝑥 =

𝑛 ∑ ) ( 𝐼𝑥 ′ + 𝑑𝑥2 𝐴 𝑖 , 𝑖=1

where 𝑛 is the number of shapes, 𝐼𝑥 ′ is the area moment of inertia for shape 𝑖 about its centroidal 𝑥 ′ axis, 𝑑𝑥 is the shift distance for shape 𝑖 (i.e., the distance between the 𝑥 axis and the 𝑥 ′ axis for shape 𝑖), and 𝐴 is the area for shape 𝑖. Similar expressions may be written for 𝐼𝑦 and 𝐽𝑂 . A common error is to use the parallel axis theorem to relate moments of inertia between two parallel axes where neither is a centroidal axis.

Mass moments of inertia The mass moments of inertia for the object shown in Fig. 10.19 are defined as Eqs. (10.15)–(10.20), p. 593 𝑦

𝐼𝑥 =

∫

𝑟̃2𝑥 𝑑𝑚 =

∫

(𝑦̃2 + 𝑧̃ 2 ) 𝑑𝑚,

𝐼𝑦 =

∫

𝑟̃2𝑦 𝑑𝑚 =

∫

(𝑥̃ 2 + 𝑧̃ 2 ) 𝑑𝑚,

𝐼𝑧 =

∫

𝑟̃2𝑧 𝑑𝑚 =

∫

(𝑥̃ 2 + 𝑦̃2 ) 𝑑𝑚,

𝑑𝑚 = 𝜌 𝑑𝑉 𝑟̃𝑦 𝑦̃ 𝑟̃𝑥

𝑟̃𝑧

𝑥 𝑥̃

𝐼𝑥𝑦 =

∫

𝑥̃ 𝑦̃ 𝑑𝑚,

𝐼𝑦𝑧 =

∫

𝑦̃𝑧̃ 𝑑𝑚,

𝐼𝑥𝑧 =

∫

𝑥̃ 𝑧̃ 𝑑𝑚,

where 𝑟̃𝑥 , 𝑟̃𝑦 , and 𝑟̃𝑧 are defined in Fig. 10.19 as the radial distances (i.e., moment arms) from the 𝑥, 𝑦, and 𝑧 axes, respectively, to the center of mass for mass element 𝑑𝑚; 𝑥, ̃ 𝑦, ̃ and 𝑧̃ are defined in Fig. 10.19 as the 𝑥, 𝑦, and 𝑧 distances, respectively, to the center of mass for mass element 𝑑𝑚; 𝐼𝑥 , 𝐼𝑦 , and 𝐼𝑧 are the mass moments of inertia about the 𝑥, 𝑦, and 𝑧 axes, respectively; and 𝐼𝑥𝑦 , 𝐼𝑦𝑧 , and 𝐼𝑥𝑧 are the products of inertia of the mass about the 𝑥𝑦, 𝑦𝑧, and 𝑥𝑧 axes, respectively. Mass moments of inertia have dimensions of (mass)(length)2 .

𝑧̃

𝑧 Figure 10.19 An object with mass 𝑚, density 𝜌, and volume 𝑉 ; 𝑟̃𝑥 , 𝑟̃𝑦 , and 𝑟̃𝑧 are radial distances from the 𝑥, 𝑦, and 𝑧 axes, respectively, to the center of mass for mass element 𝑑𝑚.

614

Chapter 10

Moments of Inertia

Mass radius of gyration. The radii of gyration of an object with mass 𝑚 are defined as Eq. (10.21), p. 595 √ √ √ 𝐼𝑦 𝐼𝑥 𝐼𝑧 , 𝑘𝑦 = , 𝑘𝑧 = , 𝑘𝑥 = 𝑚 𝑚 𝑚 where 𝑘𝑥 is the radius of gyration of the mass about the 𝑥 axis; 𝑘𝑦 is the radius of gyration of the mass about the 𝑦 axis; 𝑘𝑧 is the polar radius of gyration of the mass about the 𝑧 axis; and 𝑚 is the mass of the object. Radii of gyration have units of length.

Parallel axis theorem for mass moments of inertia. Consider the object with mass 𝑚 and center of mass 𝐺 shown in Fig. 10.20. An 𝑥 ′ 𝑦 ′ 𝑧 ′ coordinate system is defined, with origin at the center of mass of the object. The 𝑥, 𝑦, and 𝑧 axes are parallel to the 𝑥 ′ , 𝑦 ′ , and 𝑧 ′ axes, respectively, with separation distances 𝑑𝑥 , 𝑑𝑦 , and 𝑑𝑧 , respectively. The parallel axis theorem relates the mass moments of inertia with respect to the 𝑥, 𝑦, and 𝑧 axes to the mass center moments of inertia as follows:

𝑦 𝑦 𝑑𝑦

𝑚

𝐺

𝑥

Eqs. (10.22)–(10.24), p. 596

𝑑𝑥 𝑥

𝑧 𝑧

𝑑𝑧

Figure 10.20 An object with mass 𝑚 and center of mass at point 𝐺. The 𝑥 and 𝑥 ′ axes are parallel with separation distance 𝑑𝑥 , the 𝑦 and 𝑦 ′ axes are parallel with separation distance 𝑑𝑦 , and the 𝑧 and 𝑧 ′ axes are parallel with separation distance 𝑑𝑧 .

ISTUDY

𝐼𝑥 = 𝐼𝑥 ′ + 𝑑𝑥2 𝑚, 𝐼𝑦 = 𝐼𝑦 ′ + 𝑑𝑦2 𝑚, 𝐼𝑧 = 𝐼𝑧 ′ + 𝑑𝑧2 𝑚. The parallel axis theorem, written for composite shapes, is Eq. (10.26), p. 597 𝐼𝑥 =

𝑛 ∑ ( ) 𝐼𝑥 ′ + 𝑑𝑥2 𝑚 𝑖 , 𝑖=1

where 𝑛 is the number of shapes, 𝐼𝑥 ′ is the mass moment of inertia for shape 𝑖 about its mass center 𝑥 ′ axis, 𝑑𝑥 is the shift distance for shape 𝑖 (i.e., the distance between the 𝑥 axis and the 𝑥 ′ axis for shape 𝑖), and 𝑚 is the mass for shape 𝑖. Similar expressions may be written for 𝐼𝑦 and 𝐼𝑧 . A common error is to use the parallel axis theorem to relate moments of inertia between two parallel axes where neither of them is a mass center axis.

ISTUDY

Section 10.4

615

Chapter Review

Review Problems Problem 10.96 A semicircular area has outside radius 𝑟𝑜 and a concentric hole with inside radius 𝑟𝑖 . Determine the polar moment of inertia of the area about point 𝑂.

𝑦 𝑟𝑜

Problems 10.97 and 10.98 (a) Fully set up the integral, including limits of integration, that will give the area moment of inertia about the 𝑦 axis.

𝑟𝑖 𝑂

𝑥

Figure P10.96

(b) Evaluate the integral in Part (a). 𝑦 2 cm

𝑦 = 1 + 𝑥2 𝑦

1 cm

2m

0 0

𝑥 1 cm

Figure P10.97

𝑦=

√

𝑥

𝑦 = 𝑥∕4

4m

𝑥

Figure P10.98

Problems 10.99 and 10.100 Determine the constants 𝑐1 , 𝑐2 , and 𝑐3 so the curves pass through the points shown. Then determine the moment of inertia indicated in the subproblem below.

𝑦 𝑦 = 𝑐 1 + 𝑐2 𝑥

6 in.

Problem 10.99

(a) Fully set up the integral, including limits of integration, that will give the area moment of inertia about the 𝑥 axis. (b) Evaluate the integral in Part (a). Problem 10.100

3 in. √ 𝑦 = 𝑐3 𝑥

0

𝑥 9 in.

0

Figure P10.99 and P10.100

(a) Fully set up the integral, including limits of integration, that will give the area moment of inertia about the 𝑦 axis. (b) Evaluate the integral in Part (a).

Problem 10.101 (a) Fully set up the integral, including limits of integration, that will give the area moment of inertia about the 𝑥 axis.

𝑦 2 in.

(b) Evaluate the integral in Part (a). 𝑦 = 2(𝑥 − 1)2

Problem 10.102 (a) Fully set up the integral, including limits of integration, that will give the area moment of inertia about the 𝑦 axis. (b) Evaluate the integral in Part (a).

𝑥

0 0

1 in.

Figure P10.101 and P10.102

616

Chapter 10

Moments of Inertia Problems 10.103 and 10.104 Determine the area moments of inertia 𝐼𝑥 and 𝐼𝑦 . 40 mm

10 mm 15 mm 𝑦 𝑦 0.5 in. 3 in.

0.5 in.

60 mm 𝑥

30 mm 20 mm

𝑥 6 in. Figure P10.103

Figure P10.104

Problem 10.105 The cross section of a symmetric W8 × 15 wide-flange I beam has area 𝐴 = 4.44 in.2 and area moment of inertia about the 𝑥1 axis 𝐼𝑥 = 121 in.4 . Determine the moment of inertia 1 of the area about the 𝑥2 axis 𝐼𝑥 .

8.11 in.

2

𝑥1 6 in. 𝑥2 Figure P10.105

ISTUDY

Problem 10.106 (a) For the L-shaped channel shown, determine the dimension 𝑑 so that the origin of the coordinate system, point 𝑂, is positioned at the centroid of the area. (b) Determine the area moment of inertia about the 𝑥 axis. 𝑦 𝑑

0.5 in.

4 in. 𝑥

𝑂

𝑑 4 in.

0.5 in.

Figure P10.106

Problem 10.107 The cross section of the bar shown is symmetric about the 𝑥 axis. (a) Determine 𝑑 so that the origin of the coordinate system, point 𝑂, is positioned at the centroid of the area. (b) Determine the area moment of inertia about the 𝑥 axis. (c) Determine the area moment of inertia about the 𝑦 axis. 𝑦 𝑑

4 mm

𝑥

28 mm 10 mm 16 mm 20 mm Figure P10.107

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Section 10.4

617

Chapter Review

Problems 10.108 and 10.109 𝑥2 + 𝑦 2 = 𝑟 2 1 − 𝑧 ℎ

For the uniform solid cone with height ℎ and radius 𝑟, use integration to determine the mass moment of inertia indicated in the subproblem below, expressing your answer in terms of the mass 𝑚 of the cone. Problem 10.108

𝐼𝑧 .

Problem 10.109

𝐼𝑥 .

ℎ

𝑦

Problems 10.110 through 10.113

𝑟

𝑥

The truncated cone shown has 2000 kg∕m3 density. Report your answers for the subproblems below using kg and mm units. Problem 10.110

Figure P10.108 and P10.109

Use integration to determine the mass moment of inertia about the

𝑦

𝑥 axis. Problem 10.111

𝑦 = 60 mm − 𝑥∕4

Determine the mass moment of inertia about the 𝑥 axis using com60 mm

posite shapes. Problem 10.112

2

40 mm 𝑥

Use integration to determine the mass moment of inertia about the

𝑧 axis. Problem 10.113

Determine the mass moment of inertia about the 𝑧 axis using com-

posite shapes.

𝑧

80 mm

Figure P10.110–P10.113

Problems 10.114 and 10.115 The object shown has 8000 kg∕m3 density and has a conical hole. Use integration to determine the mass moment of inertia about the axis indicated in the subproblem below. Problem 10.114

𝑥 axis.

Problem 10.115

𝑦 axis. 𝑦 6 mm 4 mm

3 mm 2 mm

𝑥 1 mm

𝑧 Figure P10.114 and P10.115

Problems 10.116 and 10.117 For the subproblem below, a solid of revolution consists of materials with densities 𝜌1 and 𝜌2 .

𝑦

𝜌2

𝜌1

𝑦=

Problem 10.116

Fully set up the integrals, including limits of integration, that will yield the mass moment of inertia about the 𝑥 axis. You are not required to evaluate the integrals. Problem 10.117

Fully set up the integrals, including limits of integration, that will yield the mass moment of inertia about the 𝑦 axis. You are not required to evaluate the integrals.

√

𝑥

𝑥

𝑧

ℎ

ℎ

Figure P10.116 and P10.117

618

Chapter 10

Moments of Inertia Problems 10.118 and 10.119

A solid of revolution is produced by revolving the area shown 360◦ around the 𝑦 axis. Use integration to determine the mass moment of inertia about the axis of revolution, assuming the solid has uniform density of 2000 kg∕m3 . 𝑦 2 cm

𝑦 = 1 + 𝑥2

𝑦 1 cm

2m 𝑦=

√

𝑥 4m

0

𝑥

0

Figure P10.118

𝑥 1 cm

Figure P10.119

Problems 10.120 and 10.121

40 mm

For the subproblem below, a beam is constructed of three identical pieces of wood. Each piece has 40 mm by 160 mm by 500 mm dimensions, and 2 kg mass that is uniformly distributed. The cross section of the beam is symmetric about the 𝑥 and 𝑦 axes.

160 mm

Problem 10.120

𝑦

𝑥

40 mm

𝑧 500 mm 160 mm Figure P10.120 and P10.121

ISTUDY

(a) Determine the area moment of inertia for the cross section about the 𝑥 axis. (b) Determine the mass moment of inertia for the beam about the 𝑥 axis. Problem 10.121

(a) Determine the area moment of inertia for the cross section about the 𝑦 axis. (b) Determine the mass moment of inertia for the beam about the 𝑦 axis.

Problems 10.122 through 10.124 The object shown is made of thin plate with 0.02 lb∕in.2 specific weight. Determine the mass moment of inertia about the axis indicated in the subproblem below. Problem 10.122

𝑥 axis.

Problem 10.123

𝑦 axis.

Problem 10.124

𝑧 axis. 𝑧

9 in. 𝑥 12 in. 𝑦

8 in. 10 in. Figure P10.122–P10.124

ISTUDY

Introduction to Dynamics

11

In Section 11.1, we introduce Isaac Newton’s (1643–1727) laws of motion and his universal law of gravitation. In Section 11.2, we review those elements of physics and vector algebra needed to develop the material in the remainder of the book. In Section 11.3, we touch upon the role of dynamics in engineering design.

MShieldsPhotos/Alamy Stock Photo

Raphael’s School of Athens depicts ancient Greek philosophers, such as Aristotle, Plato, Euclid, and Pythagoras. This fresco celebrates the kinship that the renaissance humanists felt with the great minds from antiquity as they explored new ways of thinking about the arts, sciences, and engineering.

11.1

The Newtonian Equations

The dynamics we study in this book is the part of mechanics concerned with the motion of bodies, the forces causing their motion, and/or the forces caused by their motion. Dynamics builds upon statics in that the ability to draw free body diagrams and to write the corresponding balance equations for particles and rigid bodies are fundamental to dynamics. Dynamics also complements mechanics of materials in that it develops your ability to find forces due to the acceleration of objects, forces that can be used to find stresses using mechanics of materials. Since the middle of the 20th century, dynamics has also included the study and analysis of any time-varying process, be it mechanical, electrical, chemical, or biological. While we focus on mechanical processes, much of what we study is also

619

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Introduction to Dynamics

applicable to other time-varying phenomena. Our goal is to provide an introduction to the science, skill, and art involved in modeling mechanical systems to predict their motion. Newton’s laws of motion Newton’s three laws of motion are First Law. A particle remains at rest, or moves in a straight line with a constant speed, as long as the total force acting on the particle is zero. Second Law. The time rate of change of momentum of a particle is equal to the resultant force acting on that particle. Third Law. The forces of action and reaction between interacting particles are equal in magnitude, opposite in direction, and collinear. The second law, stated mathematically, is 𝑑⃗ 𝑝 𝑑(𝑚𝑣) ⃗ 𝐹⃗ = = , 𝑑𝑡 𝑑𝑡

Interesting Fact Newton’s third law in modern mechanics. Modern mechanics generally discards Newton’s third law and replaces it with a much more general result based on the concept of angular momentum. Since the 1950s, it has been proposed that an even more general notion called the principle of material frame indifference could be used to replace Newton’s third law. This latter principle states that the properties of materials and the actions of bodies on one another are the same for all observers. 𝑧 𝑃𝑖 𝑟⃗𝑖

𝐹⃗𝑗𝑖

𝐹⃗𝑖𝑗 𝐹⃗𝑖𝑘

𝑟⃗𝑗

𝑟⃗𝑘

𝐹⃗𝑘𝑖

𝑃𝑗

𝐹⃗ = 𝑚𝑎, ⃗

𝑦

Figure 11.1 A system of particles interacting with one another.

(11.2)

which explicitly accounts for the fact that a particle is generally understood to have constant mass.∗ We will learn in Chapter 13 that the first law is simply a special case of the second. The second and third laws, along with the ideas developed by Leonhard Euler (1707–1783) for rigid body dynamics, are all that is needed to solve a broad spectrum of problems involving particles and rigid bodies. The third law, stated mathematically, is

𝐹⃗𝑘𝑗

𝑥

ISTUDY

where 𝐹⃗ is the net force acting on the particle, 𝑝⃗ is the momentum of the particle, 𝑚 is the mass of the particle, and 𝑣⃗ is the velocity of the particle. We have used the definition of momentum, which is 𝑝⃗ = 𝑚𝑣. ⃗ Throughout this book, we will denote vectors by using either a superposed arrow ( ⃗ ) or a superposed caret or hat ( ̂ ) if the vector is a unit vector. Newton’s second law is often written as

𝐹⃗𝑗𝑘

𝑃𝑘

(11.1)

𝐹⃗𝑖𝑗 = −𝐹⃗𝑗𝑖 ,

(11.3)

⃗ 𝐹⃗𝑖𝑗 × (⃗𝑟𝑖 − 𝑟⃗𝑗 ) = 0,

(11.4)

where, for any interacting particles 𝑖 and 𝑗, 𝐹⃗𝑖𝑗 is the force on particle 𝑖 due to particle 𝑗 and 𝑟⃗𝑖 is the position of the ith particle (Fig. 11.1). Some people refer to Newton’s third law as just Eq. (11.3), while others require both Eqs. (11.3) and (11.4). Requiring both equations is sometimes referred to as the strong form of Newton’s third law. Newton’s universal law of gravitation Newton used his laws of dynamics along with the laws postulated by Johannes Kepler (1571–1630) to deduce the universal law of gravitation, which describes the

∗ The

application of Eq. (11.1) to variable mass systems will be considered in Section 15.5.

ISTUDY

Section 11.1

The Newtonian Equations

force of attraction between two bodies. The gravitational force on a mass 𝑚1 due to a mass 𝑚2 a distance 𝑟 away from 𝑚1 is 𝐹⃗12 =

𝐺𝑚1 𝑚2 𝑟2

𝑁 𝑟

𝑢, ̂

(11.5) 𝐹⃗𝑁𝐽

where 𝑢̂ is a unit vector pointing from 𝑚1 to 𝑚2 and 𝐺 is the universal gravitational constant∗ (sometimes called the constant of gravitation or constant of universal gravitation). The following example demonstrates the application of this law. Mini-Example Using the planets Jupiter and Neptune as an example, the force on Jupiter due to the gravitational attraction of Neptune, 𝐹⃗𝐽𝑁 , is given by (see Fig. 11.2) 𝐹⃗𝐽𝑁 =

𝐺𝑚𝐽 𝑚𝑁 𝑟2

621

𝑢, ̂

(11.6)

where 𝑟 is the distance between the two bodies, 𝑚𝐽 is the mass of Jupiter, 𝑚𝑁 is the mass of Neptune, and 𝑢̂ is a unit vector pointing from the center of Jupiter to the center of Neptune. The mass of Jupiter is 1.9 × 1027 kg, and that of Neptune is 1.02×1026 kg. Since the mean radius of Jupiter’s orbit is 778,300,000 km and that of Neptune is 4,505,000,000 km, we assume that their closest approach to one another is approximately 3,727,000,000 km. Thus, at their closest approach, the magnitude of the force between these two planets is )( ) ( )( 3 1.9×1027 kg 1.02×1026 kg m −11 |𝐹⃗𝐽𝑁 | = 6.674×10 )2 ( kg⋅s2 (11.7) 3.727×1012 m

𝐹⃗𝐽 𝑁

𝑢̂

𝐽 (left) JPL/University of Arizona/NASA; (right) NASA/JPL

Figure 11.2 The gravitational force between the planets Jupiter 𝐽 and Neptune 𝑁. The relative sizes of the planets are accurate, but their separation distance is not.

= 9.312×1017 N.

We can compare this force with the force of gravitation between Jupiter and the Sun. The Sun’s mass is 1.989×1030 kg, and we have already stated that the mean radius of Jupiter’s orbit is 778,300,000 km. Applying Eq. (11.5) between Jupiter and the Sun gives 4.164×1023 N, which is almost 450,000 times larger.

Acceleration due to gravity. Equation (11.5) allows us to determine the force of Earth’s gravity on an object of mass 𝑚 on the surface of the Earth. This is done by noting that the radius of the Earth is 6371.0 km (see the marginal note) and the mass of the Earth is 5.9736×1024 kg and then applying Eq. (11.5): ) ( )( 3 5.9736×1024 kg 𝑚 −11 m 𝐹𝑠 = 6.674×10 ( )2 kg⋅s2 (11.8) 6371.0×103 m ( ) = 9.8222 m∕s2 𝑚.

This result† tells us that the force of gravity (in N) on an object on the Earth’s surface is about 9.8 times the object’s mass (in kg). This factor of 9.8 is so prevalent in engineering that it is given the label 𝑔, and it is called the acceleration due to ∗ Henry

Cavendish (1731–1810) was the first to measure 𝐺 and did so in 1798. The generally accepted value is 𝐺 = 6.674×10−11 m3 ∕(kg⋅s2 ) = 3.439×10−8 f t 3 ∕(slug⋅s2 ). † We will normally round the result of all calculations to 4 significant digits. Here we are using 5 significant figures because the data used in this particular calculation is known to that degree of accuracy.

Interesting Fact The radius of the Earth. The Earth is not a perfect sphere. Therefore, there are different notions of “radius of the Earth.” The given value of 6371.0 km is the volumetric radius when rounded to 5 significant digits. The Earth’s volumetric radius is the radius of a perfect sphere with volume equal to that of the Earth. Other measures of the Earth’s radius, rounded to 5 significant digits, are the quadratic mean radius, the authalic mean radius, and the meridional Earth radius, which are equal to 6372.8, 6371.0, and 6367.4 km, respectively.

622

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Introduction to Dynamics

gravity because it has units of acceleration and its value is the acceleration of objects in free fall near the surface of the Earth. We will take the value of 𝑔 to be 9.81 m∕s2 in SI units and 32.2 f t∕s2 in U.S. Customary units. Notice that the value of 𝑔 obtained in Eq. (11.8) is slightly greater than the 9.81 m∕s2 that we will use in this book. The difference between these values has several causes, including that the Earth is not perfectly spherical, does not have uniform mass distribution, and is rotating. Because of these factors, the actual acceleration due to gravity is about 0.27% lower at the equator, and 0.26% higher at the poles, relative to the standard value of 𝑔 = 9.81 m∕s2 , which is for a north or south latitude of 45◦ at sea level. There may also be small local variations in gravity due to geological formations. Nonetheless, throughout this book we will use the standard value of 𝑔 stated above. Change in acceleration due to altitude. There is a formula that allows us to find how the acceleration due to gravity changes with altitude. To find it, we begin by equating Eqs. (11.2) and (11.5) to determine the acceleration 𝑎 at a height ℎ above the surface of the Earth 𝐺𝑚𝑒 , (11.9) 𝑎= (𝑟𝑒 + ℎ)2 where 𝑟𝑒 is the radius of the Earth, 𝑚𝑒 is the mass of the Earth, and we have canceled the mass of the object on both sides of the equation. Now, at the surface of the Earth, we know that 𝑎 = 𝑔 and ℎ = 0, so Eq. (11.9) becomes 𝑔 = 𝐺𝑚𝑒 ∕𝑟2𝑒

⇒

𝐺𝑚𝑒 = 𝑔𝑟2𝑒 .

(11.10)

Substituting Eq. (11.10) into Eq. (11.9), we see that 𝑎 is given by 𝑎=𝑔

𝑟2𝑒 (𝑟𝑒 + ℎ)2

,

(11.11)

where 𝑔 is the acceleration due to gravity at the surface of the Earth. Equation (11.11) is very handy because it requires knowledge of only the radius of the Earth to get the acceleration due to gravity rather than having to know both the radius of the Earth and the universal gravitational constant 𝐺. Building on statics to develop mastery in dynamics Students undertaking dynamics have often just finished a prior course in statics. It is useful to point out the similarities and differences between the two subjects before outlining the content in dynamics. The approach to solving a large subset of statics problems can be summarized as follows: 1. Draw one or more Free Body Diagrams (FBDs) representing the loads and reactions acting on rigid bodies. 2. Write the equations of static equilibrium informed by those FBDs. 3. Solve the resulting equations for reactions or internal loads. Unless we examine problems involving distributed loads, mathematical proficiency required for such problems is restricted to geometry, trigonometry, and algebra. FBDs are as important in dynamics as they are in statics, but we will use more than Cartesian components to solve problems. Our governing equation, as embodied

ISTUDY

Section 11.1

The Newtonian Equations

in Eq. (11.1), now contains the time rate of change of the linear momentum, so force imbalance implies motion. Because velocity is time rate of change of position, and acceleration is time rate of change in velocity, we will find ourselves in need of calculus from the outset of dynamics. An additional complication arises from our differentiating a vector with respect to time. Because the velocity vector has a direction as well as a magnitude, ⃗ and 𝑢̂ 𝑣 is a unit vector in the direction of 𝑣, ⃗ ∗ the acceleration 𝑣⃗ = 𝑣 𝑢̂ 𝑣 , where 𝑣 = |𝑣| in Eq. (11.2) generally requires the product rule from calculus: 𝑎⃗ =

𝑑 𝑢̂ 𝑑 𝑣⃗ 𝑑𝑣 = 𝑢̂ + 𝑣 𝑣 . 𝑑𝑡 𝑑𝑡 𝑣 𝑑𝑡

(11.12)

This tells us that acceleration can occur not just from the familiar change in speed, as embodied in the first term, but from change in direction, as embodied in 𝑑 𝑢̂ 𝑣 ∕𝑑𝑡. As an example, a satellite in a circular orbit about the Earth has a constant speed but is always accelerating because the direction of the velocity unit vector is constantly changing. There is no contradiction between a satellite at altitude ℎ above the surface of the Earth having the acceleration as described by Eq. (11.11) and stating that the satellite moves at constant speed. The ability to account for changing directions of unit vectors in different component systems is of the utmost importance in describing the motion of particles. In statics, we generally begin from simpler 2D and 3D particle equilibrium problems before moving on to 2D and 3D equilibrium of individual rigid bodies and then, finally, 2D and 3D equilibrium of assemblies of rigid bodies in the form of trusses, frames, and machines. Once we transition from particles to rigid bodies, we need to impose rotational as well as translational equilibrium, and the former requires a rotational constraint in the form that the sum of moments about a point, on or off the rigid body, must equal zero. In dynamics, we likewise begin with an examination of particle behavior. Because particle motion is an entirely new subject for us, we devote an entire chapter (Chapter 12) to descriptions of motion in different component systems. One of the challenges is selecting the appropriate component system for each problem. In some problems, the forces are naturally described in one component system, while the motion is most appropriately described in a different component system. This requires us to map either the forces or the motion into the other component system before writing our equations of motion. Once we understand how to represent particle motion in different component systems, we turn to the examination of force imbalance as embodied in Eq. (11.2). In Chapters 13, 14, and 15 we provide different methods for examining force imbalance on particles. In statics, we found many ways to draw FBDs and solve for internal loads and reactions, but some processes were more efficient than others. In dynamics, we might begin with Eq. (11.2) and follow the evolution of a particle’s motion from some initial state to a final state. This is an approach we use in Chapter 13, and this can be characterized as a path-dependent approach to particle dynamics. In contrast, we will see that some problems are amenable to a path-independent approach, where the final state can be inferred from information about the initial state, and without having to follow all the details from initial to final state. This is the subject of Chapters 14 and 15. In Chapter 14, we examine transitions over distance, while in Chapter 15, we examine transitions through time. All three chapters begin from the ∗ We

will cover this in detail in Section 12.4.

623

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Introduction to Dynamics

Chapter 11

same fundamental equation, Eq. (11.2), and the choice of problem-solving method is often telegraphed by the form of the problem statement. Chapters 16, 17, and 18 follow a similar development for the dynamics of rigid bodies. Chapter 16 is devoted to a description of rigid body motion, and is to rigid bodies what Chapter 12 is to particles. To keep problems relatively simple, we restrict our attention in these chapters to planar motion, where all rotational motion is in the out-of-plane direction. For rigid bodies in general plane motion, we will see that every material point in the body has, in general, a different velocity and acceleration at any one instant in time. We will show that the treatment of a rigid body as an assembly of particles leads to a translational dynamics equation similar to Eq. (11.2), but with the acceleration on the right-hand side being equal to that of the body’s center of mass. We will also be able to formulate an equation describing the rotational dynamics of the body. Unlike statics, the rotational dynamics equation is not equally simple at all points. Rotational dynamics is most simply expressed when taking moments about the center of mass. Chapter 17 is to rigid bodies as Chapter 13 is to particles. Chapter 18 articulates path-independent processes for planar rigid body dynamics, and is to rigid bodies as Chapters 14 and 15 are to particles. Chapter 19 is devoted to an introduction to mechanical vibrations, where we examine the interplay of inertia and compliance. Vibrations is an important subject all its own, providing a foundation for structural dynamics. The subject matter examined in this chapter is confined to single degree-of-freedom problems, where the vibration of a single translational or rotational degree-of-freedom is of interest. Finally, Chapter 20 relaxes the constraint on planar motion and presents 3D dynamics of rigid bodies. Rotational inertia must now be treated as a full tensor, as rotation about any axis is possible. To the uninitiated, it may seem as though the transition from 2D to 3D rigid body motion is relatively straightforward. But there is an order of magnitude increase in complexity associated with this transition, and a quick skim of the equations in Chapter 20 should disabuse the reader of the notion that this is a simple transition. For now, we devote a bit of space to discussing some fundamental concepts in dynamics as well as reviewing important vector mechanics operations. The ability to resolve a vector into orthogonal components in any coordinate system is especially useful when we go back and forth between different coordinate systems in Chapter 12.

ISTUDY

Section 11.2

11.2

Fundamental Concepts in Dynamics

625

𝑧

Fundamental Concepts in Dynamics

Space and time Space Space is the environment in which objects move, and we consider it to be a collection of locations or points. The position of a point is indicated by the point’s coordinates in a chosen coordinate system. Figure 11.3 shows a three-dimensional Cartesian coordinate system with origin at 𝑂 and mutually orthogonal axes 𝑥, 𝑦, and 𝑧. The Cartesian coordinates of the point 𝑃 are 𝑥𝑃 , 𝑦𝑃 , and 𝑧𝑃 , which are scalars obtained by measuring the distance between 𝑂 and the perpendicular projections of the point 𝑃 onto the axes 𝑥, 𝑦, and 𝑧, respectively. Note that 𝑥𝑃 , 𝑦𝑃 , and 𝑧𝑃 have a positive or negative sign depending on whether, in going from 𝑂 to the projections of 𝑃 along each axis, one moves in the positive or negative direction of these axes. In Chapter 12, we will introduce additional coordinate systems.

𝑃

𝑥𝑃

𝑂

𝑦𝑃

𝑥

𝑧𝑃 𝑦

Figure 11.3 A point in a three-dimensional space.

Helpful Information Time Time is a scalar variable that allows us to specify the order of a sequence of events. In classical mechanics and in this book, the most important assumption about time is that it is absolute. We assume that the duration of an event is independent of the motion of the observer making time measurements. Einstein’s theory of relativity rejects this assumption.

The right-hand rule. In three dimensions, a Cartesian coordinate system uses three orthogonal reference directions. These are the 𝑥, 𝑦, and 𝑧 directions shown below. 𝑧

Force, mass, and inertia Force The force acting on an object is the interaction between that object and its environment. A more precise description of this interaction requires that we know something about the interaction in question. For example, if two objects collide or slide against one another, we say that they interact via contact forces. Regardless of the type, the characteristics of a force are its magnitude, its line of action, and its orientation or direction. This is why we use vectors to represent forces. Mass The mass of an object is a measure of the amount of matter in the object. Along with the concept of force, the concept of mass is considered a primitive concept—that is, not explainable via more elementary ideas. Newton’s second law postulates that the force acting on a body is proportional to the body’s acceleration—the constant of proportionality is the mass of the body. Inertia Inertia is commonly understood as a body’s resistance to changing its state of motion in response to the application of a force system. In this book, we use inertia as an umbrella term encompassing both the idea of mass and that of mass distribution over a region of space. The inertia properties of an object are its mass and a quantitative description of the mass distribution.

𝑥

𝑦

Proper interpretation of many vector operations, such as the cross product, requires that the 𝑥, 𝑦, and 𝑧 directions be arranged in a consistent manner. The convention in mechanics and vector mathematics in general is that if the axes are arranged as shown, then, according to the right-hand rule, rotating the 𝑥 direction into the 𝑦 direction yields the 𝑧 direction. The result is called a righthanded coordinate system.

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Particle and rigid body Particle A particle is an object whose mass is concentrated at a point; therefore, it is also called a point mass. The inertia properties of a particle consist only of its mass. A particle is generally understood to have zero volume. It is meaningless to talk about the rotation of a particle whose position is held fixed, although we do say that a particle can “rotate about a point,” meaning that a particle can move along a path around a point. Regardless of its volume, when we choose to model an object as a particle, we neglect the possibility that the object might rotate in the sense of “change its orientation” relative to some chosen reference. Rigid body A rigid body is an object whose mass is (1) distributed over a region of space and (2) such that the distance between any two points on it never changes. Since its mass is not concentrated at a point, the rigid body is the simplest model for the study of motions that include the possibility of rotation—that is, a change of orientation relative to a chosen reference. We model objects as rigid bodies when we want to account for the possibility of rotation while neglecting the effects of deformation. Finally, the mass distribution of a rigid body does not change relative to an observer moving with the body. This fact makes it possible to describe the inertia properties of a threedimensional rigid body with seven pieces of information consisting of the body’s mass and six mass moments of inertia.∗

Vectors and their Cartesian representation Notation 𝜔 ⃗ or 𝜔

𝑦

⃗ or 𝑀 𝑀 𝑣⃗ or 𝑣

𝐹⃗ or 𝐹 𝛼⃗ or 𝛼

𝑟⃗ or 𝑟

𝚥̂ 𝚤̂

𝑎⃗ or 𝑎 If the label is a signed length (scalar), the arrow’s direction will establish the positive direction. 𝑥

Figure 11.4 Notation and colors for commonly used vectors. Position vectors will always be blue (⃗𝑟), velocity (linear and angular) vectors purple (𝑣⃗ and 𝜔), ⃗ and acceleration (linear and angular) vectors green (𝑎⃗ and 𝛼⃗ ). Forces and moments will al⃗ and unit vectors orways be red (𝐹⃗ and 𝑀) ange (̂𝚤 and 𝚥̂). Vectors with no particular physical significance will be black , magenta , or gray .

ISTUDY

Scalars. By scalar we mean a real number. Scalars will be denoted by italic roman characters (e.g., 𝑎, ℎ, or 𝑊 ) or by Greek letters (e.g., 𝛼, 𝜔, or 𝛿). Vectors. We will always denote vectors by placing arrows or carets (in the case of unit vectors) over letters, such as 𝐹⃗ or 𝚤̂. The conventions we use to depict vectors in figures are shown in Fig. 11.4. The color scheme used in the figure is defined in the caption. Depending on what we want or need to emphasize in a figure, a vector will be labeled with a letter that has an arrow placed above it (e.g., 𝑎⃗ or 𝜔 ⃗ ) or with just a letter (e.g., 𝑎 or 𝜔) according to the following conventions: • In figures, a vector will be labeled with arrows over letters when it is important to emphasize the arbitrary directional nature of the vector (e.g., a velocity) or the vectorial nature of the quantity (e.g., a unit vector). • Base vectors in Cartesian components will be designated using the unit vectors ̂ A unit vector is a vector with magnitude equal to 1. In any other 𝚤̂, 𝚥̂, and 𝑘. context, e.g., in other component systems, unit vectors will be designated using a caret over the letter 𝑢, that is, 𝑢, ̂ often accompanied by a subscript indicating the direction of the vector, such as 𝑢̂ 𝑟 . • In figures, the label of a vector with known direction will generally not be a letter with an arrow over it. A vector with known direction will usually be ∗ For a definition of the mass moments of inertia of a rigid body, see Section 10.3 on p. 593 of M. E. Plesha,

G. L. Gray, R. J. Witt, and F. Costanzo, Engineering Mechanics: Statics, McGraw-Hill, Dubuque, IA, 2023.

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Section 11.2

627

Fundamental Concepts in Dynamics

labeled as a signed length (i.e., a scalar component) whose positive direction is that of the arrow in the figure. • Double-headed arrows will designate vectors associated with “rotational” quantities—that is, moments, angular velocities, and angular accelerations (angular velocities and accelerations will be discussed in Chapter 12). Cartesian vector representation We now review those aspects of vectors in two dimensions that are most important for our applications. This presentation is easily extended to three dimensions. Figure 11.5 shows the position of a point 𝑃 with respect to the origin 𝑂 of a 𝑦 𝑃 𝑟⃗𝑃 ∕𝑂 = 𝑟⃗ 𝑟𝑦 𝚥̂

𝜃 𝑢̂ 𝑟

𝑂

𝑥

𝚤̂ 𝑟𝑥

Figure 11.5. Description of the position of a point 𝑃 . The curved arrows indicating an angle with a single arrowhead designate an angle’s positive direction.

rectangular coordinate system. The position of 𝑃 is represented by the arrow that starts at 𝑂 and ends at 𝑃 , which we call the vector 𝑟⃗𝑃 ∕𝑂 . The subscript “𝑃 ∕𝑂” is read “𝑃 relative to 𝑂,” or “𝑃 as seen by an observer at 𝑂,” or “𝑃 with respect to 𝑂.” When only one point is being discussed, we typically drop the “𝑃 ∕𝑂” part of the notation and simply indicate position as 𝑟⃗. The Cartesian representation of 𝑟⃗ is 𝑟⃗ = 𝑟𝑥 𝚤̂ + 𝑟𝑦 𝚥̂,

(11.13)

where 𝚤̂ and 𝚥̂ are unit vectors in the 𝑥 and 𝑦 directions, respectively. The quantities 𝑟𝑥 and 𝑟𝑦 are the (scalar) Cartesian components of 𝑟⃗. Using trigonometry, we have and

𝑟𝑦 = |⃗𝑟| sin 𝜃,

(11.14)

where 𝜃 is the orientation of the segment 𝑂𝑃 (the bar over the letters 𝑂 and 𝑃 designates the line segment connecting the points 𝑂 and 𝑃 ) relative to the 𝑥 axis and |⃗𝑟|, called the magnitude of 𝑟⃗ or length of 𝑟⃗, is the length of 𝑂𝑃 . Equation (11.13) could be written as 𝑟⃗ = 𝑟⃗𝑥 + 𝑟⃗𝑦 , where the vectors 𝑟⃗𝑥 = 𝑟𝑥 𝚤̂ and 𝑟⃗𝑦 = 𝑟𝑦 𝚥̂ are called the 𝑥 and 𝑦 vector components of 𝑟⃗, respectively. In this book, component will always mean scalar component. When talking about vector components, we will explicitly say vector components. Generalizing what we said about 𝑟⃗𝑃 ∕𝑂 , given points 𝐴 and 𝐵 with coordinates (𝑥𝐴 , 𝑦𝐴 ) and (𝑥𝐵 , 𝑦𝐵 ), respectively, the vector ( ) ( ) 𝑟⃗𝐴∕𝐵 = 𝑥𝐴 − 𝑥𝐵 𝚤̂ + 𝑦𝐴 − 𝑦𝐵 𝚥̂ (11.15) will be called the position of 𝐴 with respect to 𝐵, or position of 𝐴 relative to 𝐵 (Fig. 11.6).

𝑦 𝐴 𝑦𝐴 − 𝑦𝐵

𝑟𝑥 = |⃗𝑟| cos 𝜃

𝑟⃗𝐴∕𝐵 𝐵

𝚥̂

𝑂

𝑥 𝐴 − 𝑥𝐵 𝚤̂

𝑥

Figure 11.6 Vector representation of the position of 𝐴 relative to 𝐵.

628

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Introduction to Dynamics Vector operations

𝑦

Here are the vector operations we will use: ⃗ 𝑤

𝑤𝑦 𝑟⃗

1. A vector 𝑟⃗ can be multiplied by a scalar 𝑎 in the following way: 𝑎⃗𝑟 = 𝑎𝑟𝑥 𝚤̂ + 𝑎𝑟𝑦 𝚥̂.

⃗ 𝑟⃗ + 𝑤

(11.16)

This scales the magnitude of 𝑟⃗ by the factor |𝑎|. The object 𝑎⃗𝑟 is a vector (not a scalar) with the same line of action as 𝑟⃗; the direction of 𝑎⃗𝑟 is the same as that of 𝑟⃗ if 𝑎 > 0, whereas it is opposite to 𝑟⃗ if 𝑎 < 0.

𝑟𝑦 𝚥̂ 𝑂

𝑥

𝚤̂ 𝑟𝑥

𝑤𝑥

Figure 11.7 Graphical representation of the vector addition of ⃗ showing the “triangle law.” 𝑟⃗ and 𝑤

2. Two vectors can be summed to obtain another vector as follows: ) ( ) ( ⃗ = 𝑟𝑥 + 𝑤𝑥 𝚤̂ + 𝑟𝑦 + 𝑤𝑦 𝚥̂, 𝑟⃗ + 𝑤

(11.17)

which conforms to the triangle law of vector addition (see Fig. 11.7).

3. The operation of summing a scalar with a vector is not defined. ⃗ is 4. Referring to Fig. 11.8, the dot or scalar product of two vectors 𝑟⃗ and 𝑤 ⃗ and yields the following scalar quantity: denoted by 𝑟⃗ ⋅ 𝑤 ⃗ = |⃗𝑟||𝑤| ⃗ cos 𝜃. 𝑟⃗ ⋅ 𝑤

⃗ 𝑟⃗ × 𝑤

⃗ is the vector 5. Referring to Fig. 11.8, the cross product of two vectors 𝑟⃗ and 𝑤 ⃗ with denoted by 𝑟⃗ × 𝑤

⃗ 𝑤

𝜃

(11.18)

(a) magnitude

𝑟⃗

⃗ = |⃗𝑟||𝑤| ⃗ sin 𝜃, |⃗𝑟 × 𝑤|

(11.19)

⃗ and (b) line of action perpendicular to the plane containing 𝑟⃗ and 𝑤,

Figure 11.8 Graphical representation of the vector cross prod⃗ uct of 𝑟⃗ and 𝑤.

(c) direction determined by the right-hand rule. In Eqs. (11.18) and (11.19), 𝜃 is the smallest angle that will rotate one of the vectors into the other. For the cross product, this choice of 𝜃 ensures that Eq. (11.19) always yields a nonnegative value. For the dot product, since cos 𝜃 = cos(2𝜋 − 𝜃), 𝜃 can be replaced by 2𝜋 − 𝜃. Finally, the definition of cross product implies that the cross product is anticommutative, that is,

𝑦

⃗ × 𝑟⃗. ⃗ = −𝑤 𝑟⃗ × 𝑤 𝑃 𝑟⃗𝑃 ∕𝑂 = 𝑟⃗

𝑟𝑦 𝚥̂

𝜃 𝑢̂ 𝑟

𝑂

𝑥

𝚤̂ 𝑟𝑥

Figure 11.9 Description of the position of a particle 𝑃 .

ISTUDY

(11.20)

Referring to Fig. 11.9, we recall that 𝑟⃗ represents the length and orientation of the segment 𝑂𝑃 . Using the Pythagorean theorem and trigonometry, and expressing angles in radians, we have √ length of 𝑟⃗ = |⃗𝑟| = 𝑟2𝑥 + 𝑟2𝑦 , (11.21) and

direction of 𝑟⃗ = 𝜃 = tan−1

(𝑟 ) 𝑦

𝑟𝑥

± 𝑛𝜋,

𝑛 = 0, 1, 2, …

(11.22)

where 𝑛 is found by identifying the quadrant containing 𝑃 (for the case in Fig. 11.9, 𝑛 = 0). Finally, Eqs. (11.13) and (11.14) allow us to rewrite 𝑟⃗ as 𝑟⃗ = |⃗𝑟| 𝑢̂ 𝑟 ,

where

𝑢̂ 𝑟 = cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂.

(11.23)

Since 𝑢̂ 𝑟 is a unit vector in the direction of 𝑟⃗, Eq. (11.23) implies that The information carried by any vector can be written as the product of its magnitude and a unit vector pointing in the direction of that vector.

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Section 11.2

629

Fundamental Concepts in Dynamics

Useful vector “tips and tricks”

𝐵2

Components of a vector

𝐵

We now review how to find the components of a vector, since this operation occurs often in dynamics. Figure 11.10 shows two perpendicular and oriented lines 𝓁1 and 𝓁2 , where by oriented we mean that they have a positive and a negative direction. The lines are oriented using the unit vector 𝑢̂ 1 for 𝓁1 and 𝑢̂ 2 for 𝓁2 . We also have a vector 𝑞⃗ oriented arbitrarily relative to 𝓁1 and 𝓁2 . Our goal is to find the scalar components of 𝑞⃗ along 𝓁1 and 𝓁2 . ⃗ be 𝑢̂ 1 , and 𝜃 be 𝜃1 , we see If we apply Eq. (11.18) to Fig. 11.10 and let 𝑟⃗ be 𝑞, ⃗ 𝑤 that the dot product gives us 𝑞1 directly, that is, 𝑞1 = 𝑞⃗ ⋅ 𝑢̂ 1 = |𝑞|| ⃗ 𝑢̂ 1 | cos 𝜃1 = |𝑞| ⃗ cos 𝜃1 .

(11.24)

The quantity 𝑞1 in Eq. (11.24) is what we were looking for because, according to the definition of scalar component of a vector and Fig. 11.10,

𝑞2

𝑞⃗ 𝜃2

𝐴2 𝜃2

𝐴

𝜃1

𝐴1 𝑞1

𝑢̂ 2

𝑢̂ 1

𝐵1

𝓁1

𝓁2 Figure 11.10 Diagram showing the components of 𝑞⃗ in the directions of 𝑢̂ 1 and 𝑢̂ 2 .

1. |𝑞1 | is the distance between 𝐴1 and 𝐵1 .

2. The sign of 𝑞⃗ ⋅ 𝑢̂ 1 is determined by the sign of cos 𝜃1 , which is positive if 0◦ ≤ 𝜃1 < 90◦ and negative if 90◦ < 𝜃1 ≤ 180◦ (if 𝜃1 = 90◦ , 𝐴1 and 𝐵1 coincide so that 𝑞1 = 0).

In summary, component of 𝑞⃗ along 𝓁1 = 𝑞1 = 𝑞⃗ ⋅ 𝑢̂ 1 .

𝚥̂

(11.25)

Repeating the foregoing discussion in the case of 𝑞2 we have that the component of 𝑞⃗ along 𝓁2 = 𝑞2 = 𝑞⃗ ⋅ 𝑢̂ 2 = |𝑞| ⃗ cos 𝜃2 = (11.26) ( ) since 𝜃2 = 𝜃2′ + 𝜋 and cos 𝜃2′ + 𝜋 = − cos 𝜃2′ . In dynamics we often face the situation depicted in Fig. 11.11, in which we need to calculate the Cartesian components of two mutually orthogonal vectors 𝑞⃗ and 𝑟⃗. If the angle 𝜃 defining the orientation of 𝑞⃗ relative to the 𝑦 axis is given, to express 𝑞⃗ and 𝑟⃗ in components, we can write 𝑞⃗ as ) ( ⃗ cos 𝜃 𝚥̂ 𝑞⃗ = (𝑞⃗ ⋅ 𝚤̂) 𝚤̂ + (𝑞⃗ ⋅ 𝚥̂) 𝚥̂ = |𝑞| ⃗ cos 𝜃 + 𝜋2 𝚤̂ + |𝑞| = |𝑞| ⃗ (− sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂), ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

(11.27)

unit vector

and then we can write 𝑟⃗ as ( ) 𝑟⃗ = (⃗𝑟 ⋅ 𝚤̂) 𝚤̂ + (⃗𝑟 ⋅ 𝚥̂) 𝚥̂ = |⃗𝑟| cos(𝜋 − 𝜃) 𝚤̂ + |⃗𝑟| cos 𝜃 + 𝜋2 𝚥̂ = −|⃗𝑟| cos 𝜃 𝚤̂ − |⃗𝑟| sin 𝜃 𝚥̂

= |⃗𝑟| (− cos 𝜃 𝚤̂ − sin 𝜃 𝚥̂) . ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

𝑞𝑦 𝑦

−|𝑞| ⃗ cos 𝜃2′ ,

= −|𝑞| ⃗ sin 𝜃 𝚤̂ + |𝑞| ⃗ cos 𝜃 𝚥̂

𝑟𝑦

𝚤̂

(11.28)

unit vector

Equations (11.27) and (11.28) demonstrate that the two equations have the following structure: • Each vector is equal to its magnitude times a unit vector with one sine and one cosine term.

𝑞𝑥 𝑟𝑥 𝑥

𝑟⃗

𝑞⃗

𝜃 𝜃

Figure 11.11 Diagram showing the Cartesian components of the vector 𝑞⃗ as well as the vector 𝑟⃗ that is orthogonal to 𝑞. ⃗

630

Chapter 11

Introduction to Dynamics

• The argument of the sine and cosine terms is the angle orienting one of the vectors with respect to one of the component directions. • In the two equations there will always be three positive terms and one negative term or three negative terms and one positive term. As a final check that the decomposition has been performed correctly, note that since 𝑞⃗ and 𝑟⃗ are orthogonal, their dot product should yield zero. The dot product of the unit vectors in (11.27) and (11.28) give us sin 𝜃 cos 𝜃 − sin 𝜃 cos 𝜃 = 0. Cross products

𝚤̂

𝑘̂

𝚥̂

Figure 11.12 A little “trick” to help remember the cross products between Cartesian unit vectors.

ISTUDY

Since we will often encounter cross products in dynamics, here is a useful procedure to help us evaluate them. Consider three unit vectors 𝚤̂, 𝚥̂, and 𝑘̂ such that, according ̂ Arrange the three vectors as shown in Fig. 11.12. to the right-hand rule, 𝚤̂ × 𝚥̂ = 𝑘. ̂ just move around the circle, starting from 𝚥̂ To calculate the product of, say, 𝚥̂ × 𝑘, ̂ Now notice that (1) the next vector on the circle is 𝚤̂ and (2) in and going toward 𝑘. going from 𝚥̂ to 𝑘̂ we move with the arrow (counterclockwise). Hence, 𝚥̂ × 𝑘̂ = +̂𝚤. Now consider 𝑘̂ × 𝚥̂ and notice that in going from 𝑘̂ toward 𝚥̂, the next vector along the circle is 𝚤̂, and we move opposite to the arrow. Therefore, the result is negative, and we have 𝑘̂ × 𝚥̂ = −̂𝚤.

Helpful Information Cross products using determinants. You may be familiar with the following determinant method of evaluating the cross product of two vectors: | 𝚤̂ | | ⃗ 𝑎⃗ × 𝑏 = |𝑎𝑥 | |𝑏𝑥 |

𝚥̂ 𝑎𝑦 𝑏𝑦

𝑘̂ || 𝑎𝑧 || . | 𝑏𝑧 ||

Units Units are essential to any quantifiable measure. Newton’s second law in scalar form, 𝐹 = 𝑚𝑎, provides for the formulation of a consistent and unambiguous system of units. We will use both U.S. Customary units and SI units (International System∗ ) as shown in Table 11.1. Each system has three base dimensions and a fourth derived Table 11.1. U.S. Customary and SI unit systems.

For vectors in 3D, this method provides a very efficient evaluation. As an alternative, the cross product may be evaluated on a term-by-term basis by expanding the following product: ) ( 𝑎⃗ × 𝑏⃗ = 𝑎𝑥 𝚤̂ + 𝑎𝑦 𝚥̂ + 𝑎𝑧 𝑘̂ ) ( × 𝑏𝑥 𝚤̂ + 𝑏𝑦 𝚥̂ + 𝑏𝑧 𝑘̂ .

When expanded, nine terms such as 𝑎𝑥 𝚤̂ × 𝑏𝑥 𝚤̂ and 𝑎𝑥 𝚤̂ × 𝑏𝑦 𝚥̂ must be evaluated. This is accomplished quickly using Fig. 11.12. In this book, we primarily do cross products of vectors in 2D, and we will likely find the termby-term evaluation to be quicker.

System of units Base dimension

U.S. Customary

SI

force

pound (lb)

newtona (N) ≡ kg⋅m∕s2

mass

sluga ≡ lb⋅s2 ∕ft

kilogram (kg)

length

foot (ft)

meter (m)

time

second (s)

second (s)

a derived

unit

dimension. In the U.S. Customary system, the base dimensions are force, length, and time, whose corresponding base units are lb (pounds), ft (feet), and s (seconds), respectively. The corresponding derived dimension is mass, which is obtained from the equation 𝑚 = 𝐹 ∕𝑎. This gives the mass unit as lb⋅s2 ∕ft. This unit of mass is often called the slug. In the SI system, the base dimensions are mass, length, and time, whose corresponding base units are kg (kilogram), m (meter), and s (second), respectively. The corresponding derived dimension is force, the unit of which is obtained from the ∗ SI

has been adopted as the abbreviation for the French Le Syst`eme International d’Unit´es.

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Section 11.2

631

Fundamental Concepts in Dynamics

equation 𝐹 = 𝑚𝑎, which gives the force unit as kg⋅m∕s2 . This unit of force is referred to as a newton, and its abbreviation is N. Because of the difference in base dimensions between the U.S. Customary system and the SI system, when using the U.S. Customary system, we normally specify the weight of an object (typically in lb) instead of its mass; and, conversely, when using the SI system, we normally specify the mass of an object (typically in kg) instead of its weight. For both systems, we may occasionally use different, but consistent, units for some dimensions. For example, we may use minutes rather than seconds, inches instead of feet, grams instead of kilograms. Plane angles are dimensionless quantities (they are defined as the ratio of two lengths). In both the U.S. Customary and SI systems, angles are expressed in radians, abbreviated rad. Another commonly used unit to express angle measurements is the degree, indicated by the symbol ◦ . Angle measurements in degrees and in radians are related as follows: 180◦ = 𝜋 rad.

Common Pitfall Weight and mass. Unfortunately, it is common to refer to weight using mass units. For example, the person who says, “I weigh 70 𝗄𝗀” really means “My mass is 70 𝗄𝗀.” In science and engineering it is essential that accurate nomenclature be used. Weights and forces must be reported using appropriate force units, and masses must be reported using appropriate mass units.

(11.29)

Dimensional homogeneity and unit conversions Equations must be dimensionally homogeneous. This means that the quantities on the two sides of the equal sign must have the same dimensions. Our strong recommendation is that appropriate units always be used in all equations during a calculation to make sure that the results are dimensionally correct. Such practice helps avoid catastrophic blunders and provides a useful check on a solution, for if an equation is found to be dimensionally inconsistent, then an error has certainly been made. In September 1999, NASA (National Aeronautics and Space Administration) lost a $125 million Mars orbiter because the climate orbiter spacecraft team at the contractor who built the spacecraft used U.S. Customary units when computing rocket thrust, while the mission navigation team at NASA used metric units for this key spacecraft operation. This units error, which came into play when the spacecraft was to be inserted into orbit around Mars, caused the spacecraft to approach Mars at too low an altitude, thus causing it to burn up in Mars’ atmosphere. Unit conversions are often needed and are easily done using conversion factors, such as those shown in Table 11.2, and rules of algebra. The basic idea is to multiply either or both sides of an equation by dimensionless factors of unity, where each factor of unity embodies an appropriate unit conversion. This procedure is illustrated in the examples at the end of this section. Prefixes Prefixes are a useful alternative to scientific notation for representing numbers that are very large or very small. Common prefixes and a summary of rules for their use are given in Table 11.3. Here is a list of common rules for correct prefix use: 1. With few exceptions, prefixes should be used only in the numerator of unit combinations; for example, use the unit km∕s (kilometer per second) and avoid the unit m∕ms (meter per millisecond). One common exception to this rule is kg, which may appear in numerator or denominator; for example, use the unit kW∕kg (kilowatt per kilogram) and avoid the unit W∕g (watt per gram).

Table 11.2 Conversion factors between U.S. Customary and SI unit systems. U.S. Customary

length

1 in.

SI

1 f t (12 in.) 1 mi (5280 f t)

0.0254 m (25.4 mm) 0.3048 m 1.609 km

force

1 lb 1 kip (1000 lb)

4.448 N 4.448 kN

mass

1 slug (1 lb⋅s2 ∕f t)

14.59 kg

632

ISTUDY

Chapter 11

Introduction to Dynamics Table 11.3. Common prefixes used in the SI unit systems. Multiplication factor 1 000 000 000 000 000 000 000 000 1 000 000 000 000 000 000 000 1 000 000 000 000 000 000 1 000 000 000 000 000 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100 10 0.1 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001 0.000 000 000 000 001 0.000 000 000 000 000 001 0.000 000 000 000 000 000 001 0.000 000 000 000 000 000 000 001

1024 1021 1018 1015 1012 109 106 103 102 101 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24

Prefix

Symbol

yotta zetta exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto zepto yocto

Y Z E P T G M k h da d c m 𝜇 n p f a z y

2. Double prefixes must be avoided; for example, use the unit GHz (gigahertz) and avoid the unit kMHz (kilo-megahertz). 3. Use a center dot or dash to denote multiplication of units, for example, N ⋅ m or N-m. In this book, we denote multiplication of units by a dot, for example, N⋅m. 4. Exponentiation applies to both the unit and prefix, for example, mm2 = (mm)2 . 5. If the number of digits on either side of a decimal point exceeds 4, it is common to group the digits into groups of 3, with the groups separated by commas or thin spaces. Since many countries use a comma to represent a decimal point, the thin space is sometimes preferable; for example, 1234.0 could be written as is, but by contrast, 12345.0 should be written as 12,345.0 or as 12 345.0. While prefixes can often be incorporated into an expression by inspection, the rules for doing this are identical to those for performing unit transformations. Accuracy of numbers in calculations Throughout this book, we will generally assume that the data given for problems is accurate to three significant digits. When calculations are performed, such as in example problems, all intermediate results are stored in the memory of a calculator or computer using the full precision these machines offer. However, when these intermediate results are reported in this book, they are rounded to four significant digits. Final answers are usually reported with three significant digits. If you verify the calculations described in this book using the rounded numbers that are reported, you may occasionally calculate results that are slightly different from those shown.

ISTUDY

Section 11.2

633

Fundamental Concepts in Dynamics

End of Section Summary

𝑦

Review of vector operations. Referring to Fig. 11.13, the Cartesian representation of a two-dimensional vector 𝑟⃗ takes the form Eq. (11.13), p. 627 𝑟⃗ = 𝑟𝑥 𝚤̂ + 𝑟𝑦 𝚥̂, where 𝚤̂ and 𝚥̂ are unit vectors in the positive 𝑥 and 𝑦 directions, respectively, and where 𝑟𝑥 and 𝑟𝑦 are the 𝑥 and 𝑦 (scalar) components of 𝑟⃗, respectively. Using trigonometry, 𝑟𝑥 and 𝑟𝑦 are given by Eqs. (11.14), p. 627 𝑟𝑥 = |⃗𝑟| cos 𝜃

and

𝑟𝑦 = |⃗𝑟| sin 𝜃,

where 𝜃 is the orientation of the segment 𝑂𝑃 relative to the 𝑥 axis and |⃗𝑟|, called the magnitude of 𝑟⃗ or length of 𝑟⃗, is the length of 𝑂𝑃 . ⃗ gives the scalar The dot or scalar product of the vectors 𝑟⃗ and 𝑤 Eq. (11.18), p. 628 ⃗ = |⃗𝑟||𝑤| ⃗ cos 𝜃, 𝑟⃗ ⋅ 𝑤

with 𝜃 being the smallest angle that will rotate one of the vectors into the other. ⃗ is denoted by Referring to Fig. 11.14, the cross product of two vectors 𝑟⃗ and 𝑤 ⃗ and gives a vector 𝑟⃗ × 𝑤 1. With magnitude equal to Eq. (11.19), p. 628 ⃗ = |⃗𝑟||𝑤| ⃗ sin 𝜃, |⃗𝑟 × 𝑤|

where 𝜃 is the smallest angle that will rotate one of the vectors into the other ⃗ sin 𝜃 ≥ 0). (to ensure that |⃗𝑟||𝑤|

⃗ and 2. Whose line of action is perpendicular to the plane containing 𝑟⃗ and 𝑤 whose direction is determined by the right-hand rule. ⃗ = −𝑤 ⃗ × 𝑟⃗. 3. For which the cross product is anticommutative; i.e., 𝑟⃗ × 𝑤 Referring to Fig 11.13, given 𝑟𝑥 and 𝑟𝑦 , we compute |⃗𝑟| and 𝜃 as follows: Eqs. (11.21) and (11.22), p. 628 √ length of 𝑟⃗ = |⃗𝑟| = 𝑟2𝑥 + 𝑟2𝑦 , (𝑟 ) 𝑦 direction of 𝑟⃗ = 𝜃 = tan−1 . 𝑟𝑥

Another useful representation of a vector 𝑟⃗ is as follows: Eq. (11.23), p. 628 𝑟⃗ = |⃗𝑟| 𝑢̂ 𝑟 ,

where

𝑢̂ 𝑟 = cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂,

where 𝑢̂ 𝑟 is the unit vector in the direction of 𝑟⃗.

𝑃 𝑟⃗𝑃 ∕𝑂 = 𝑟⃗ 𝑟𝑦 𝚥̂

𝜃 𝑢̂ 𝑟

𝑂

𝑥

𝚤̂ 𝑟𝑥

Figure 11.13 Description of the position of a particle 𝑃 .

⃗ 𝑟⃗ × 𝑤

𝜃

⃗ 𝑤

𝑟⃗ Figure 11.14 Graphical representation of the vector cross prod⃗ uct of 𝑟⃗ and 𝑤.

634

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Introduction to Dynamics

Useful vector “tips and tricks.” Referring to Fig. 11.15, the component of a vector in a given direction can be computed using the dot product. For example,

𝐵2 𝐵 𝑞2

Eq. (11.24), p. 629

𝑞⃗ 𝜃2

𝐴2 𝜃2

𝐴

𝜃1

that is,

Eq. (11.25), p. 629

𝐴1 𝑞1

𝑢̂ 2

𝑢̂ 1

𝐵1

𝓁1

𝓁2 Figure 11.15 Diagram showing the components of 𝑞⃗ in the directions of 𝑢̂ 1 and 𝑢̂ 2 . 𝚥̂

𝑟𝑦

𝚤̂

𝑞𝑦

𝑟⃗

𝑞⃗

𝜃 𝜃

Figure 11.16 Diagram showing the Cartesian components of the vector 𝑞⃗ as well as the vector 𝑟⃗ that is orthogonal to 𝑞. ⃗

ISTUDY

Referring to Fig. 11.16, for two mutually orthogonal vectors 𝑟⃗ and 𝑞⃗ the following relations hold: Eqs. (11.27) and (11.28), p. 629 ( ) 𝑞⃗ = (𝑞⃗ ⋅ 𝚤̂) 𝚤̂ + (𝑞⃗ ⋅ 𝚥̂) 𝚥̂ = |𝑞| ⃗ cos 𝜃 + 𝜋2 𝚤̂ + |𝑞| ⃗ cos 𝜃 𝚥̂ = |𝑞|(− ⃗ sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂),

) ( 𝑟⃗ = (⃗𝑟 ⋅ 𝚤̂) 𝚤̂ + (⃗𝑟 ⋅ 𝚥̂) 𝚥̂ = |⃗𝑟| cos(𝜋 − 𝜃) 𝚤̂ + |⃗𝑟| cos 𝜃 + 𝜋2 𝚥̂ = −|⃗𝑟| cos 𝜃 𝚤̂ − |⃗𝑟| sin 𝜃 𝚥̂

𝑞𝑥

𝑥

component of 𝑞⃗ along 𝓁1 = 𝑞1 = 𝑞⃗ ⋅ 𝑢̂ 1 .

= −|𝑞| ⃗ sin 𝜃 𝚤̂ + |𝑞| ⃗ cos 𝜃 𝚥̂

𝑦

𝑟𝑥

𝑞1 = 𝑞⃗ ⋅ 𝑢̂ 1 = |𝑞|| ⃗ 𝑢̂ 1 | cos 𝜃1 = |𝑞| ⃗ cos 𝜃1 ,

= |⃗𝑟|(− cos 𝜃 𝚤̂ − sin 𝜃 𝚥̂).

ISTUDY

Section 11.2

635

Fundamental Concepts in Dynamics

E X A M P L E 11.1

Components of Vectors

At the instant shown, the acceleration of the airplane in Fig. 1 is the vector 𝑎⃗ = (5.63 𝑢̂ 𝑡 + 37.2 𝑢̂ 𝑛 ) m∕s2 ,

𝑎⃗

(1)

𝑦

𝜙

𝑢̂ 𝑛

where the mutually perpendicular unit vectors 𝑢̂ 𝑡 and 𝑢̂ 𝑛 are tangent and perpendicular to the airplane’s path, respectively. The angle between 𝑢̂ 𝑡 and the horizontal direction is 𝜃 = 26◦ . Determine 𝜙, the angle between 𝑎⃗ and 𝑢̂ 𝑡 , and the expression of 𝑎⃗ relative to the unit vectors 𝚤̂ and 𝚥̂, which are horizontal and vertical, respectively.

𝜃 𝑢̂ 𝑡

𝚥̂

airplane’s path

SOLUTION Road Map Letting 𝑎𝑡 and 𝑎𝑛 be the components of 𝑎⃗ in the (𝑢̂ 𝑡 , 𝑢̂ 𝑛 ) component system, Eq. (1) implies that 𝑎𝑡 = 5.63 m∕s2 and 𝑎𝑛 = 37.2 m∕s2 . Since 𝑎𝑡 and 𝑎𝑛 are known, 𝜙 can be found via Eq. (11.22) on p. 628 after replacing the quantities 𝑟𝑥 and 𝑟𝑦 in that equation with 𝑎𝑡 and 𝑎𝑛 , respectively. To express 𝑎⃗ using 𝚤̂ and 𝚥̂, we need to determine the components of 𝑎⃗ in the 𝚤̂ and 𝚥̂ directions. This can be done via Eq. (11.27) on p. 629.

𝑥

𝚤̂ Figure 1 An airplane performing a maneuver.

Determination of 𝜙 Computation

we have

Replacing 𝑟𝑥 and 𝑟𝑦 in Eq. (11.22) on p. 628 with 𝑎𝑡 and 𝑎𝑛 , respectively, ( ) 𝑎𝑛 −1 ± 𝑛𝜋, 𝑛 = 0, 1, 2, … . (2) 𝜙 = tan 𝑎𝑡

Since the components 𝑎𝑛 and 𝑎𝑡 are both positive, the vector 𝑎⃗ lies in the first quadrant of the (𝑢̂ 𝑡 , 𝑢̂ 𝑛 ) system. Therefore, we can choose 𝑛 = 0 in Eq. (2). Recalling that 𝑎𝑡 = 5.63 m∕s2 and 𝑎𝑛 = 37.2 m∕s2 , Eq. (2) can be evaluated to obtain 𝜙 = 81.39◦ .

(3)

Expression of 𝑎⃗ via 𝚤̂ and 𝚥̂

Replacing the vector 𝑞⃗ with 𝑎⃗ = 𝑎𝑡 𝑢̂ 𝑡 + 𝑎𝑛 𝑢̂ 𝑛 in the first equality in Eq. (11.27) on p. 629, we have [ ] [ ] 𝑎⃗ = (𝑎𝑡 𝑢̂ 𝑡 + 𝑎𝑛 𝑢̂ 𝑛 ) ⋅ 𝚤̂ 𝚤̂ + (𝑎𝑡 𝑢̂ 𝑡 + 𝑎𝑛 𝑢̂ 𝑛 ) ⋅ 𝚥̂ 𝚥̂ Computation

= (𝑎𝑡 𝑢̂ 𝑡 ⋅ 𝚤̂ + 𝑎𝑛 𝑢̂ 𝑛 ⋅ 𝚤̂) 𝚤̂ + (𝑎𝑡 𝑢̂ 𝑡 ⋅ 𝚥̂ + 𝑎𝑛 𝑢̂ 𝑛 ⋅ 𝚥̂) 𝚥̂.

(4)

Referring to Fig. 2, the angles between 𝑢̂ 𝑡 and the unit vectors 𝚤̂ and 𝚥̂ are 𝜃 and 90◦ − 𝜃, respectively. Similarly, the angles between 𝑢̂ 𝑛 and the unit vectors 𝚤̂ and 𝚥̂ are 𝜃 + 90◦ and 𝜃, respectively. Therefore, we have ◦

𝑢̂ 𝑡 ⋅ 𝚤̂ = cos 𝜃,

𝑢̂ 𝑡 ⋅ 𝚥̂ = cos(90 − 𝜃) = sin 𝜃,

(5)

𝑢̂ 𝑛 ⋅ 𝚤̂ = cos(𝜃 + 90◦ ) = − sin 𝜃,

𝑢̂ 𝑛 ⋅ 𝚥̂ = cos 𝜃.

(6)

Using Eqs. (5) and (6), Eq. (4) can be simplified to 𝑎⃗ = (𝑎𝑡 cos 𝜃 − 𝑎𝑛 sin 𝜃) 𝚤̂ + (𝑎𝑡 sin 𝜃 + 𝑎𝑛 cos 𝜃) 𝚥̂.

𝑢̂ 𝑛

𝜃 90◦ − 𝜃 𝑢̂ 𝑡

𝜃 𝚤̂

(7)

Recalling that 𝑎𝑡 = 5.63 m∕s2 , 𝑎𝑛 = 37.2 m∕s2 , and 𝜃 = 26◦ , Eq. (7) gives 𝑎⃗ = (−11.25 𝚤̂ + 35.90 𝚥̂) m∕s2 .

𝜃 + 90◦

𝚥̂

(8)

Discussion & Verification The value of 𝜙 seems reasonable given how much larger 𝑎𝑛 is relative to 𝑎𝑡 . To verify the result in Eq. (8), we can calculate the magnitude of 𝑎⃗ using its components relative to both the (̂𝚤, 𝚥̂) and the (𝑢̂ 𝑡 , 𝑢̂ 𝑛 ) systems and check that we obtain the )1∕2 ( m∕s2 = 37.62 m∕s2 . same value. In the (̂𝚤, 𝚥̂) system, we have |𝑎| ⃗ = −11.252 + 35.902 ) ( 1∕2 m∕s2 = 37.62 m∕s2 . Since the In the (𝑢̂ 𝑡 , 𝑢̂ 𝑛 ) system we have |𝑎| ⃗ = 5.632 + 37.22 result is as expected, we can say that the result in Eq. (8) appears to be correct.

Figure 2 Orientation of the unit vectors 𝑢̂ 𝑡 and 𝑢̂ 𝑛 relative to the unit vectors 𝚤̂ and 𝚥̂.

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E X A M P L E 11.2

Position Vectors, Relative Position Vectors, and Components

𝑦, north

The map of the state of Minnesota in Fig. 1 shows four of its cities and defines a coordinate system whose origin is at 𝑂. The coordinates of the four cities relative to 𝑂 are given in Table 1. Assuming the Earth is flat and ignoring errors due to the map projection used, the 𝑥 and 𝑦 directions can be considered to be east and north, respectively. Using the information in Table 1, determine

International Falls

(a) The position of Duluth (𝐷) relative to Minneapolis/St. Paul (𝑀), 𝑟⃗𝐷∕𝑀 .

Fargo/ Moorhead

(b) The orientation, relative to north, of the position of International Falls (𝐼) relative to Fargo/Moorhead (𝐹 ).

Duluth

(c) The east and north (scalar) components of the position of Fargo/Moorhead relative to Minneapolis/St. Paul. (d) The position of the point 𝐻 halfway between Minneapolis/St. Paul and International Falls.

Minneapolis/ St. Paul

100 mi 𝑥, east

𝑂

Table 1. Coordinates of the four cities in the state of Minnesota shown in Fig. 1. All coordinates are relative to the origin 𝑂. City

Figure 1 A map of the state of Minnesota with the locations of four of its cities. An east-north or 𝑥𝑦 coordinate frame with origin at 𝑂 is also defined.

Minneapolis/St. Paul (𝑀) Duluth (𝐷) Fargo/Moorhead (𝐹 ) International Falls (𝐼)

x, east (mi)

y, north (mi)

216 259 12 195

130 267 278 413

SOLUTION 𝑦

Part (a) Road Map Since the coordinates of points 𝐷 and 𝑀 are available, the components of 𝑟⃗𝐷∕𝑀 can be found by using Eq. (11.15) on p. 627 to compute the difference between the coordinates of 𝐷 and 𝑀.

𝐼 𝜃𝐼∕𝐹 𝑢̂ 𝐼∕𝐹

𝐹

𝑟⃗𝐻∕𝑀 𝐻

𝑟⃗𝐹 ∕𝑀

) ) ( ( 𝑟⃗𝐷∕𝑀 = 𝑟⃗𝐷 − 𝑟⃗𝑀 = 𝑥𝐷 − 𝑥𝑀 𝚤̂ + 𝑦𝐷 − 𝑦𝑀 𝚥̂ [ ] = (259 − 216) 𝚤̂ + (267 − 130) 𝚥̂ mi

𝐷

𝑟⃗𝐻

𝑦𝐹 ∕𝑀

Computation Referring to Fig. 2 and Table 1 and then taking the difference between 𝑟⃗𝐷 and 𝑟⃗𝑀 , we obtain

𝑟⃗𝐷∕𝑀

= (43.00 𝚤̂ + 137.0 𝚥̂) mi = 143.6 mi @ 72.57

𝑥𝐹 ∕𝑀

𝜃𝐷∕𝑀

𝚤̂

𝑥

Figure 2 Vectors, projections, and angles needed to compute the quantities of interest.

ISTUDY

Referring to Fig. 2 and taking advantage of the scale indicated on the map, we can graphically verify that our answers are correct. We note that in this problem, the answer given as a length (143.6 mi) and a direction (𝜃𝐷∕𝑀 = 72.57◦ ) is probably more straightforward to verify than the answer given in terms of Cartesian components, since we could directly measure the distance from Minneapolis/St. Paul to Duluth on the map itself. Discussion & Verification

𝑟⃗𝑀 100 mi

𝑂

,

where 𝜃𝐷∕𝑀 = 72.57◦ .

𝑀 𝚥̂

(1) ◦

Part (b)

To determine the orientation of 𝐼 relative to 𝐹 , finding either the unit vector 𝑢̂ 𝐼∕𝐹 or the angle 𝜃𝐼∕𝐹 will suffice. To find 𝑢̂ 𝐼∕𝐹 , we can find 𝑟⃗𝐼∕𝐹 and then divide by its magnitude [Eq. (11.21) on page 628]. Road Map

ISTUDY

Section 11.2

Computation

Fundamental Concepts in Dynamics

Again referring to Fig. 2 and Table 1 and using Eq. (11.15), the vector

𝑢̂ 𝐼∕𝐹 is given by 𝑢̂ 𝐼∕𝐹 =

𝑟⃗𝐼∕𝐹 |⃗𝑟𝐼∕𝐹 |

(195 − 12) 𝚤̂ + (413 − 278) 𝚥̂ = √ = 0.8047 𝚤̂ + 0.5936 𝚥̂. (195 − 12)2 + (413 − 278)2

(2)

Now, we can find the angle 𝜃𝐼∕𝐹 by applying Eq. (11.22), which gives 𝜃𝐼∕𝐹 = tan−1

(𝑥

𝐼∕𝐹

𝑦𝐼∕𝐹

)

= tan−1

(

183 135

)

= 53.58◦ ,

(3)

which would be approximately northeast. As with Part (a), we have expressed the answer in two different ways, and the version given in terms of an angle [i.e., Eq. (3)] probably allows for an easier verification that the solution is reasonable. Discussion & Verification

Part (c) Road Map

To find the east (𝑥) and north (𝑦) components of 𝑟⃗𝐹 ∕𝑀 , we can use Eq. (11.25)

on p. 629. Computation

Referring to Fig. 2 and using Eq. (11.25), we find that

[ ] 𝑥𝐹 ∕𝑀 = 𝑟⃗𝐹 ∕𝑀 ⋅ 𝚤̂ = (12 − 216) 𝚤̂ + (278 − 130) 𝚥̂ ⋅ 𝚤̂ mi = −204.0 mi, [ ] 𝑦𝐹 ∕𝑀 = 𝑟⃗𝐹 ∕𝑀 ⋅ 𝚥̂ = (12 − 216) 𝚤̂ + (278 − 130) 𝚥̂ ⋅ 𝚥̂ mi = 148.0 mi,

(4) (5)

where we have used the fact that 𝚤̂ ⋅ 𝚤̂ = 𝚥̂ ⋅ 𝚥̂ = 1 and 𝚤̂ ⋅ 𝚥̂ = 𝚥̂ ⋅ 𝚤̂ = 0. Discussion & Verification We calculated that Fargo/Moorhead is 148.0 mi north of Minneapolis/St. Paul, and it is 204.0 mi west (i.e., −204 mi east), which certainly seem reasonable given the figure. Part (d)

As we can see from Fig. 2, the position of the point 𝐻, which is halfway between Minneapolis/St. Paul and International Falls, is given by the vector 𝑟⃗𝐻 . The key to the solution is then seeing that we can write this position as 𝑟⃗𝐻 = 𝑟⃗𝑀 + 𝑟⃗𝐻∕𝑀 . Road Map

Computation

We begin with the decomposition of 𝑟⃗𝐻 , which is given by 𝑟⃗𝐻 = 𝑟⃗𝑀 + 𝑟⃗𝐻∕𝑀 .

(6)

We can write 𝑟⃗𝐻∕𝑀 as 𝑟⃗𝐻∕𝑀 = 21 𝑟⃗𝐼∕𝑀 =

1[ (195 2

] − 216) 𝚤̂ + (413 − 130) 𝚥̂ mi = (−10.50 𝚤̂ + 141.5 𝚥̂) mi,

(7)

which, when substituted into Eq. (6), gives

[ ] 𝑟⃗𝐻 = (216 𝚤̂ + 130 𝚥̂) + (−10.50 𝚤̂ + 141.5 𝚥̂) mi = (205.5 𝚤̂ + 271.5 𝚥̂) mi,

(8)

where we have used 𝑟⃗𝑀 = (216 𝚤̂ + 130 𝚥̂) mi from Table 1. Discussion & Verification Again referring to Fig. 2, the result in Eq. (8) looks reasonable. In addition, the power of vectors starts to come into focus in this part of the example. That is, once we knew the position of International Falls relative to Minneapolis/St. Paul, computing the position at any fraction of the distance in between was trivial. Once that was computed, vector addition allowed us to easily find the position of the halfway point relative to the origin at 𝑂.

637

638

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Introduction to Dynamics

E X A M P L E 11.3

Dimensional Analysis and Unit Usage In a collision, the bumper of a car can undergo both elastic (i.e., reversible) and permanent deformation. A model for the force 𝐹 transmitted by the bumper to the car is given by 𝐹 = 𝑃 + 𝑘(𝑠 − 𝑠0 ) + 𝜂 𝑑𝑠∕𝑑𝑡, where 𝑠 − 𝑠0 is the compression experienced by the bumper and has the dimension of length, 𝑠0 is a constant with dimension of length, and 𝑑𝑠∕𝑑𝑡 is the time rate of change of 𝑠, which has the dimension of length over time. The quantity 𝑃 is the force needed to permanently deform the bumper, 𝑘 is the stiffness of the system, and 𝜂 is a constant that relates the overall force to the speed of deformation. Determine

𝑃, 𝜂

(a) The dimensions of 𝑃 , 𝑘, and 𝜂, and 𝑠 − 𝑠0

(b) The units that these quantities would have in the SI and the U.S. Customary systems.

𝑘 Figure 1 A car about to collide with a concrete block.

ISTUDY

SOLUTION Part (a)

The first step in dimensional analysis is the identification of a basic relation, such as a law of nature, containing the quantities to analyze and for which the dimensions are known. Since we are dealing with the expression of a force, we can use Eq. (11.2) on p. 620 as the basic relation. Notice that the given force law consists of the sum of three terms. For this sum to be meaningful, each of the terms in question must have dimensions of force. We call this property dimensional homogeneity, and it is the key to solving the given problem. Road Map

Let 𝐿, 𝑀, and 𝑇 denote length, mass, and time, respectively. Writing “[something]” to mean the “dimensions of something,” for Eq. (11.2), we have

Computation

[𝐹 ] = [𝑚𝑎] = [𝑚] [𝑎] = 𝑀

Concept Alert Dimensional homogeneity. In writing Eq. (2), we have used a property stating that “[something + something else] = [something] + [something else].” This property expresses the requirement that the sum of two physical quantities makes sense only when these quantities have the same dimensions. Using a more formal language, we say that the quantities in question must satisfy dimensional homogeneity or, equivalently, must be dimensionally homogeneous.

𝐿 . 𝑇2

Considering the given force law, we have ] [ ] [ 𝑑𝑠 𝑑𝑠 = [𝑃 ] + [𝑘(𝑠 − 𝑠0 )] + 𝜂 . [𝐹 ] = 𝑃 + 𝑘(𝑠 − 𝑠0 ) + 𝜂 𝑑𝑡 𝑑𝑡

(1)

(2)

Comparing Eq. (1) with Eq. (2) and enforcing dimensional homogeneity between them, we see that [𝑃 ], [𝑘(𝑠 − 𝑠0 )], and [𝜂 𝑑𝑠∕𝑑𝑡] must each be 𝑀𝐿∕𝑇 2 . Therefore, the quantity 𝑃 must have the dimensions of force: [𝑃 ] = 𝑀

𝐿 . 𝑇2

(3)

For the term 𝑘, we have 𝐿 , (4) 𝑇2 where we have used the fact that the dimension of 𝑠 and 𝑠0 is 𝐿. Simplifying the last equality in Eq. (4), we have 𝑀 [𝑘] = 2 . (5) 𝑇 Next, considering the term with 𝜂 in Eq. (2), we have ] [ ] [ 𝐿 𝐿 𝑑𝑠 𝑑𝑠 (6) = [𝜂] = [𝜂] = 𝑀 2 , 𝜂 𝑑𝑡 𝑑𝑡 𝑇 𝑇 [𝑘(𝑠 − 𝑠0 )] = [𝑘] [𝑠 − 𝑠0 ] = [𝑘]𝐿 = 𝑀

where we have used the fact that the dimensions of 𝑑𝑠∕𝑑𝑡 are 𝐿∕𝑇 . Simplifying the last equality in Eq. (6), we have 𝑀 . [𝜂] = (7) 𝑇

ISTUDY

Section 11.2

Fundamental Concepts in Dynamics

The correctness of our dimensional analysis in the case of 𝑃 is apparent, since 𝑃 must have the dimensions of a force. In the case of 𝑘 and 𝜂, we can verify the correctness of our solution by substituting these quantities back into the expression for 𝐹 . Doing so confirms that the dimensions of 𝑘 and 𝜂 are correct. Discussion & Verification

Part (b)

To solve Part (b) of the problem, we need to match the dimensions obtained in Part (a) with their corresponding units according to the conventions established by the SI and U.S. Customary systems. To do this, we need only look at Table 11.1 on p. 630. Road Map

Computation Since 𝑃 has the same dimensions as a force, its SI units can be simply taken to be N (newtons) or, using the SI system base units, kg⋅m∕s2 . In the U.S. Customary system, 𝑃 is measured in lb (pounds). Since the dimensions of 𝑘 are mass over time squared, the corresponding units are kg∕s2 in the SI system and lb∕ft in the U.S. Customary system. Notice that in the SI system, the units of 𝑘 can also be expressed as N∕m. Recalling that 𝜂 has dimensions of mass over time, the units of 𝜂 are kg∕s in the SI system and slug∕s in the U.S. Customary system. Using the base units of the U.S. Customary system, the units of 𝜂 are lb⋅s∕ft. All of these results are summarized in Table 1. Table 1. Summary of the solution to the second part of the problem. Quantity 𝑃 𝑘 𝜂

SI units

U.S. Customary units 2

N or kg⋅m∕s kg∕s2 or N∕m kg∕s

lb lb∕ft lb⋅s∕ft

The correctness of our results can be verified by replacing the units with the dimensions they correspond to. For example, for 𝑃 we have that the dimensions corresponding to the unit of pound are those of a force, i.e., 𝑀𝐿∕𝑇 2 , which are the dimensions of 𝑃 obtained in Part (a) of the problem solution. Repeating this process for the other quantities and for both the SI system and the U.S. Customary system, we see that our results are correct. Discussion & Verification

639

640

E X A M P L E 11.4

Dimensional Analysis and Unit Conversion

𝑣 (0, 𝑡)

𝑠

Figure 1 A moving string with one end being dragged.

ISTUDY

Chapter 11

Introduction to Dynamics

In studying the motion of a string, it is determined that the speed at various points along the string is given by the function 𝑠 (1) 𝑣(𝑠, 𝑡) = 𝛼 + 𝛽𝑡2 − 𝛾𝑠 + 𝛿 , 𝑡 where 𝑠 is the coordinate of points along the string, 𝑡 is time, and 𝛼, 𝛽, 𝛾, and 𝛿 are constants. (a) What are the dimensions of 𝛼, 𝛽, 𝛾, and 𝛿? (b) If 𝛼, 𝛽, 𝛾, and 𝛿 are all equal to 1 in SI units, what are they in U.S. Customary units?

SOLUTION Part (a)

Interesting Fact Are engineers really interested in the motion of strings? The answer is “Actually, yes!” Highly detailed models of real physical objects tend to be quite complicated, difficult to tackle from a numerical standpoint, and often difficult to interpret. Therefore, before delving into the complexity of highly detailed models, both physicists and engineers tend to “simplify things” and model real systems as simple objects, such as strings (or beams). Simple models can be very effective in capturing the essential physical behavior of real physical objects and, in that way, give us useful insight into the physical world.

Road Map Since the dimensions of speed are 𝐿∕𝑇 (with corresponding units of m∕s in SI and ft∕s in U.S. Customary), the dimensions of every term on the right-hand side of

Eq. (1) must also be 𝐿∕𝑇 . Computation Begin with [𝛼], which must have the same dimensions as those of the speed 𝑣, so they are simply 𝐿∕𝑇 . As for [𝛽], we know that [ ] [𝛽] = 𝐿∕𝑇 3 . (2) 𝛽𝑡2 = [𝛽]𝑇 2 = 𝐿∕𝑇 ⇒

To get [𝛾], we proceed as in Eq. (2) to obtain [𝛾𝑠] = [𝛾]𝐿 = 𝐿∕𝑇 Finally, [𝛿] is obtained similarly as [ ] 𝑠 = [𝛿]𝐿∕𝑇 = 𝐿∕𝑇 𝛿 𝑡

⇒

⇒

[𝛾] = 1∕𝑇 .

[𝛿] is dimensionless.

(3)

(4)

The verification of the results for 𝛼 is immediate since no calculations were performed to obtain them. In the case of 𝛽, 𝛾, and 𝛿, substituting the results in Eqs. (2)–(4), we see that our results are correct. Discussion & Verification

Road Map

Part (b) After expressing 𝛼, 𝛽, and 𝛾 in SI units, we need to convert them into U.S.

Customary units. Note that no conversion is needed for 𝛿 since it is dimensionless. Based on Part (a), the SI units of 𝛼, 𝛽, and 𝛾 are m∕s, m∕s3 , and s−1 , respectively. Converting unit values for 𝛼, 𝛽, and 𝛾 to U.S. Customary units gives ( ) m ∕ ft = 3.281 f t∕s, 𝛼 = 1 m∕s = 1 (5) s 0.3048 m ∕ ( ) m ∕ ft 𝛽 = 1 m∕s3 = 1 3 (6) = 3.281 f t∕s3 , 0.3048 m ∕ s

Computation

𝛾 = 1 s−1 ,

(7)

where no conversion is needed for 𝛾 since 1 s−1 is the same in either SI or U.S. Customary units. The only results to verify are those for 𝛼 and 𝛽, which have dimensions of 𝐿∕𝑇 and 𝐿∕𝑇 3 , respectively. Therefore, since the base unit for time is the same in both the SI and U.S. Customary systems, the unit conversion was expected to affect only the 𝐿 dimension and yield the same value (namely, 3.281) for both 𝛼 and 𝛽. Finally, the value in question was expected to be close to 3 since a meter is a little over 3 f t. Thus, our results appear to be correct. Discussion & Verification

ISTUDY

Section 11.2

641

Fundamental Concepts in Dynamics

Problems Problem 11.1 ( ) ( ) Determine 𝑟𝐵∕𝐴 𝑥 and 𝑟𝐵∕𝐴 𝑦 , the 𝑥 and 𝑦 components of the vector 𝑟⃗𝐵∕𝐴 , so as to be ( ) ( ) able to write 𝑟⃗𝐵∕𝐴 = 𝑟𝐵∕𝐴 𝑥 𝚤̂ + 𝑟𝐵∕𝐴 𝑦 𝚥̂.

6 𝐷

5

𝑞

4 𝑦 (f t)

Problem 11.2 If the positive direction of line 𝓁 is from 𝐷 to 𝐶, find the component of the vector 𝑟⃗𝐵∕𝐴 along 𝓁.

3 2

1 𝚥̂ 𝑢̂ 𝑢̂ 𝑞 𝑝

Problem 11.3

𝑂 𝚤̂ 1

Find the components of 𝑟⃗𝐵∕𝐴 along the 𝑝 and 𝑞 axes.

Problem 11.4

𝐶

𝐴

𝑢̂ 𝓁 𝑝

𝜃 = 22.5◦ 2

3 4 𝑥 (f t)

𝓁 𝐵 5

6

Figure P11.1–P11.5

Determine expressions for the vector 𝑟⃗𝐵∕𝐴 using both the 𝑥𝑦 and the 𝑝𝑞 coordinate systems. Next, determine |⃗𝑟𝐵∕𝐴 |, the magnitude of 𝑟⃗𝐵∕𝐴 , using both the 𝑥𝑦 and the 𝑝𝑞 representations and establish whether or not the two values for |⃗𝑟𝐵∕𝐴 | are equal to each other.

Problem 11.5 Suppose that you were to compute the quantities |⃗𝑟𝐵∕𝐴 |𝑥𝑦 and |⃗𝑟𝐵∕𝐴 |𝑝𝑞 , that is, the magnitude of the vector 𝑟⃗𝐵∕𝐴 computed using the 𝑥𝑦 and 𝑝𝑞 frames, respectively. Do you expect these two scalar values to be the same or different? Why?

Problem 11.6 The magnitude of the velocity vector of the car is |𝑣| ⃗ = 80 f t∕s. If the vector 𝑣⃗ forms an angle 𝜃 = 0.09 rad with the horizontal direction, determine the Cartesian representation of 𝑣⃗ relative to the (̂𝚤, 𝚥̂) component system. 𝑦 𝑣⃗

𝚥̂

𝜃 𝚤̂

𝑥

Figure P11.6 and P11.7

Problem 11.7 The velocity of the car has the following representation: 𝑣⃗ = (8.30 𝚤̂ + 0.726 𝚥̂) m∕s. Determine the magnitude of 𝑣. ⃗ In addition, knowing that the angle 𝜃 describes the orientation of 𝑣, ⃗ determine 𝜃 and express its value in degrees.

Problem 11.8 The acceleration of the car has the following representation: 𝑎⃗ = −(3.53 𝚤̂ + 0.309 𝚥̂) m∕s2 . Knowing that 𝑎⃗ is parallel to the incline, determine the angles 𝜃 and 𝜙 and express their value in radians.

𝑦 𝚥̂

𝜃 𝑎⃗ 𝜙 𝚤̂

Figure P11.8

𝑥

642

Chapter 11

Introduction to Dynamics

Problem 11.9 𝐴

𝑣⃗0

𝑂 𝛽

𝑢̂ 𝑞 𝑢̂ 𝑝

𝜃

𝑔

𝐵

𝚥̂ 𝜃

𝚤̂ Figure P11.9

A jaguar 𝐴 leaps from 𝑂 with a velocity 𝑣⃗0 to try and intercept a panther 𝐵. The unit vectors 𝑢̂ 𝑝 and 𝑢̂ 𝑞 are parallel and perpendicular to the incline, respectively. The unit vectors 𝚤̂ and 𝚥̂ are horizontal and vertical, respectively. While airborne, the jaguar is subject to a constant acceleration with magnitude 𝑔 and direction opposite to 𝚥̂. Denoting the magnitude of 𝑣⃗0 by 𝑣0 and denoting the (vector) acceleration of the jaguar by 𝑎⃗𝐴 , provide the expression of 𝑣⃗0 in the (̂𝚤, 𝚥̂) component system and the expression of 𝑎⃗𝐴 in the (𝑢̂ 𝑝 , 𝑢̂ 𝑞 ) component system. Treat the angles 𝛽 and 𝜃 as known.

Problem 11.10 The velocity vector of the airplane is 𝑣⃗ = 𝑣0 𝚤̂, with 𝑣0 = 420 mph. Determine the components of the vector 𝑣⃗ in the 𝑢̂ 𝑟 and 𝑢̂ 𝜃 directions for 𝜃 = 35◦ . Express the result in feet per second.

𝑢̂ 𝑟

𝑢̂ 𝜃

𝑣⃗ = 𝑣0 𝚤̂

Problem 11.11

𝐵

𝜃

The motion of the telescopic arm is such that the velocity and acceleration vectors of the gear 𝐵 are 𝑣⃗ = −𝑣0 𝚥̂ and 𝑎⃗ = −𝑎0 𝚥̂, respectively, with 𝑣0 = 8 f t∕s and 𝑎0 = 0.5 f t∕s2 . Determine the components of 𝑣⃗ and 𝑎⃗ in the direction of the unit vectors 𝑢̂ 𝑟 and 𝑢̂ 𝜃 for 𝜃 = 32◦ .

𝚥̂

𝐴

𝚤̂

Problem 11.12 At the instant shown, the velocity and acceleration vectors of the airplane have the following expressions:

Figure P11.10

𝑢̂ 𝜃 𝐴

𝑣⃗ = (215 𝚤̂ + 332 𝚥̂) ft∕s

𝑢̂ 𝑟 𝐵

Use Eq. (11.18) on p. 628 to determine the angle 𝜙, the smaller of the two angles formed by 𝑣⃗ and 𝑎. ⃗ Express the result in degrees.

𝑣0

𝑦

𝑎0 𝜃 𝑂

𝜙

𝑦 𝑣⃗ 𝑎⃗

𝚥̂

𝜙

𝑣⃗

𝑢̂ 𝑛

𝑢̂ 𝑡

𝑎⃗

𝚤̂

57◦

𝚥̂

𝚥̂ Figure P11.11

ISTUDY

and 𝑎⃗ = (−190 𝚤̂ + 76.0 𝚥̂) ft∕s2 .

𝑥

𝚤̂ Figure P11.12

𝑥

𝚤̂ Figure P11.13

Problem 11.13 At the instant shown, when expressed via the (𝑢̂ 𝑡 , 𝑢̂ 𝑛 ) component system, the airplane’s velocity and acceleration are 𝑣⃗ = 135 𝑢̂ 𝑡 m∕s

and 𝑎⃗ = (−7.25 𝑢̂ 𝑡 + 182 𝑢̂ 𝑛 ) m∕s2 .

Determine the angle 𝜙 between the velocity and acceleration vectors. In addition, treating the (𝑢̂ 𝑡 , 𝑢̂ 𝑛 ) and (̂𝚤, 𝚥̂) component systems as stationary relative to one another, express the airplane’s velocity and acceleration in the (̂𝚤, 𝚥̂) component system.

ISTUDY

Section 11.2

643

Fundamental Concepts in Dynamics

𝑥1

Problem 11.14 6

The components of the position vector 𝑟⃗ of point 𝑃 relative to the (̂𝚤1 , 𝚥̂1 ) component system are 𝑟𝑥1 = 2 f t and 𝑟𝑦1 = 5 f t. If 𝜃 = 30◦ , determine coordinates of 𝑃 relative to the (𝑥2 , 𝑦2 ) coordinate system.

𝑃 𝑟⃗

4

Problem 11.15

𝑦1

The velocity of point 𝑃 relative to frame 𝐴 is 𝑣⃗𝑃 ∕𝐴 = (−14.9 𝚤̂𝐴 + 19.4 𝚥̂𝐴 ) ft∕s, and the acceleration of 𝑃 relative to frame 𝐵 is 𝑎⃗𝑃 ∕𝐵 = (3.97 𝚤̂𝐵 + 4.79 𝚥̂𝐵 ) ft∕s2 . Frames 𝐴 and 𝐵 do not move relative to one another. Determine the expressions for the velocity of 𝑃 in frame 𝐵 and the acceleration of 𝑃 in frame 𝐴. 𝑦

𝚥̂2

𝚤̂2

45◦

𝑦𝐵

𝚤̂2

𝑦2 𝚤̂1 𝑥2

𝜃

2

𝚥̂𝐵 𝚤̂𝐵

𝚤̂𝐴

𝑣⃗𝐴

𝚥̂ 𝑥𝐴

𝐴

72◦

𝑥𝐵 23◦

𝐵

Figure P11.14

𝑎⃗𝐴

𝑃

𝐴

𝚥̂2

4 𝑎⃗𝑃

𝚥̂𝐴

𝚥̂1 𝑂

𝑃2

𝑣⃗𝑃

𝑦𝐴

2

𝑃1

𝚤̂

Figure P11.15

𝚥̂1 𝚤̂1 𝑥

Figure P11.16

𝐵

Problem 11.16

𝑟

𝑠𝐴𝐵 𝜃

Two Coast Guard patrol boats 𝑃1 and 𝑃2 are stationary while monitoring the motion of a surface vessel 𝐴. The velocity of 𝐴 with respect to 𝑃1 is expressed by

𝐴

𝑣⃗𝐴 = (−23 𝚤̂1 − 6 𝚥̂1 ) ft∕s, whereas the acceleration of 𝐴, expressed relative to 𝑃2 , is given by 𝑎⃗𝐴 = (−2 𝚤̂2 − 4 𝚥̂2 ) ft∕s2 .

Figure P11.17

Determine the velocity and the acceleration of 𝐴 expressed with respect to the land-based component system (̂𝚤, 𝚥̂).

Problem 11.17 The measure of angles in radians is defined according to the following relation: 𝑟𝜃 = 𝑠𝐴𝐵 , where 𝑟 is the radius of the circle and 𝑠𝐴𝐵 denotes the length of the circular arc. Determine the dimensions of the angle 𝜃.

𝑟

Problem 11.18 Letting 𝐶 denote the circumference of a circle, a 1◦ angle is, by definition, an angle that subtends an arc of length 𝓁 such that 𝐶∕𝓁 = 360. Apply the definition of degree and determine the radius of the circle shown knowing that the length 𝑠 of the arc subtended by the 4◦ angle in the figure is 1.84 mm.

𝑠 4◦

Figure P11.18

644

Chapter 11

Introduction to Dynamics

Problem 11.19 A simple oscillator consists of a linear spring fixed at one end and a mass attached at the other end, which is free to move. Suppose that the periodic motion of a simple oscillator is described by the relation 𝑦 = 𝑌0 sin(2𝜋𝜔0 𝑡), where 𝑦 has units of length and denotes the vertical position of the oscillator, 𝑌0 is the oscillation amplitude, 𝜔0 is the oscillation frequency, and 𝑡 is time. Recalling that the argument of a trigonometric function is an angle, determine the dimensions of 𝑌0 and 𝜔0 , as well as their units in both the SI and the U.S. Customary systems.

𝑘 𝑦

𝑚

Figure P11.19

Problem 11.20 𝜔 ⃗ 𝑧 ⃗ 𝑀

𝛼⃗

𝑦 Figure P11.20

ISTUDY

To study the motion of a space station, the station can be modeled as a rigid body and the equations describing its motion can be chosen to be Euler’s equations, which read ( ) 𝑀𝑥 = 𝐼𝑥𝑥 𝛼𝑥 − 𝐼𝑦𝑦 − 𝐼𝑧𝑧 𝜔𝑦 𝜔𝑧 , ( ) 𝑀𝑦 = 𝐼𝑦𝑦 𝛼𝑦 − 𝐼𝑧𝑧 − 𝐼𝑥𝑥 𝜔𝑥 𝜔𝑧 , ( ) 𝑀𝑧 = 𝐼𝑧𝑧 𝛼𝑧 − 𝐼𝑥𝑥 − 𝐼𝑦𝑦 𝜔𝑥 𝜔𝑦 .

In the previous equations, 𝑀𝑥 , 𝑀𝑦 , and 𝑀𝑧 denote the 𝑥, 𝑦, and 𝑧 components of the moment applied to the body;∗ 𝜔𝑥 , 𝜔𝑦 , and 𝜔𝑧 denote the corresponding components of the angular velocity of the body, where angular velocity is defined as the time rate of change of an angle; 𝛼𝑥 , 𝛼𝑦 , and 𝛼𝑧 denote the corresponding components of the angular acceleration of the body, where angular acceleration is defined as the time rate of change of an angular velocity. The quantities 𝐼𝑥𝑥 , 𝐼𝑦𝑦 , and 𝐼𝑧𝑧 are called the principal mass moments of inertia of the body. Determine the dimensions of 𝐼𝑥𝑥 , 𝐼𝑦𝑦 , and 𝐼𝑧𝑧 and determine their units in SI, as well as in the U.S. Customary system.

Problem 11.21 The lift force 𝐹𝐿 generated by the airflow moving over a wing is often expressed as follows: 𝐹𝐿 = 21 𝜌𝑣2 𝐶𝐿 (𝜃)𝐴, where 𝜌, 𝑣, and 𝐴 denote the mass density of air, the airspeed (relative to the wing), and the wing’s nominal surface area, respectively. The quantity 𝐶𝐿 is called the lift coefficient, and it is a function of the wing’s angle of attack 𝜃. Find the dimensions of 𝐶𝐿 and determine its units in the SI system. 𝐹𝐿 𝜃 𝑣 leading edge trailing edge Figure P11.21 ∗ The

𝑥𝑦𝑧 frame used in Euler’s equations is a special frame that moves with the body, but for the purpose of the problem solution, it will suffice to say that it consists of three mutually orthogonal axes.

ISTUDY

Section 11.2

Fundamental Concepts in Dynamics

Problem 11.22 Are the words units and dimensions synonyms?

Problem 11.23 A rock is released from rest into water. The magnitude 𝐹𝑑 of the drag force acting on the rock due to its motion through water can be modeled as 𝐹𝑑 = 𝐶𝑑 𝑣, where 𝑣 is the speed of the rock and 𝐶𝑑 is a constant drag coefficient. Determine the units used to measure 𝐶𝑑 in the U.S. Customary system. 𝑠

Problem 11.24 In elementary beam theory, for a uniform beam supported as shown, the relation between the force 𝑃 applied at the end of a beam and the corresponding end deflection 𝛿 is 𝑃 = (3𝐸𝐼∕𝐿3 )𝛿, where 𝐸 is a constant called the modulus of elasticity, 𝐼 is a constant called the centroidal area moment of inertia, and 𝐿 is the length of the beam. If the dimensions of 𝐼 are length to the power 4, determine the SI units used to measure the constant 𝐸. 𝑂 𝛿

𝑃 𝐿 Figure P11.24

Problem 11.25 In orbital mechanics, we find that the period of an elliptical orbit 𝜏 depends on only its semimajor axis, 𝑎, and the strength of the attractor, 𝜇, as given by [this is Eq. (15.97) on p. 1012; also see Fig. 15.28]: ( 3 )1∕2 𝑎 𝜏 = 2𝜋 . 𝜇 In this expression, 𝜏 has units of seconds and the semimajor axis 𝑎 has units of kilometers. Find the SI units of the attractor 𝜇. In orbital mechanics, km is a more natural length scale than meters, so express your answer in terms of km and s.

Problem 11.26 Docking maneuvers in orbital mechanics can be described using linear orbit theory. This theory provides a simplified description of the required changes in position and velocity of a vehicle as it approaches and seeks to dock with another vehicle in a reference orbit. When the reference orbit is elliptical, a quantity 𝑞 appears in the expression for position change with 𝑞 given by: (𝑝∕ℎ)(𝑝 + 𝑟)𝑟 sin 𝜃 − 3𝑒𝑝𝑡 𝑞= 1 − 𝑒2 In this equation, 𝑒 is a dimensionless measure of the elliptical nature of the orbit and is called the eccentricity (𝑒 < 1). The radial position of the reference orbit, 𝑟, is given in units of km. Time 𝑡 is given in seconds, and 𝜃 is an angular displacement measured from the perigee of the reference orbit. Find the SI units of ℎ in terms of km and s as well as the corresponding units of 𝑞.

Figure P11.23

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11.3

Library of Congress Prints and Photographs Division Washington, D.C. 20540 USA

Figure 11.17 Dynamic loading due to the interaction of the structure with the wind played a crucial role in the collapse of the Tacoma Narrows bridge on November 7, 1940.

ISTUDY

Dynamics and Engineering Design∗

Design is the goal of all engineering endeavors. While there is no single accepted definition of engineering design, we can define it as the process that results in the specification of how a system can be produced to meet the needs and requirements identified in the design process. The “system” to which we refer might be as simple as a bolt or as complex as the Apollo missions. The need might be to hold two steel plates together or to put a human on the moon. The requirements can be technical, environmental, political, or financial, among others. One of the hard things about engineering design is that neither the process used to reach the final design nor the final result of the design is unique (this is in contrast to the typical homework or exam problem you have been given, where there is only one correct answer). There are generally many (sometimes infinitely many) designs that will satisfy the requirements of most design specifications. In the design process, mechanicians† are generally concerned with the determination of a variety of mechanical responses, such as deflection, stress, strain, and temperature in machines or structures due to imposed forces. Mechanicians are also concerned with relating these responses to the prediction of a structure’s or machine’s fatigue life, durability, and safety, as well as to the choice of materials and fabrication procedures. Dynamics plays an important role in this determination since accelerations are an important cause of forces in mechanical systems (via the equations developed by Newton and Euler). There are several methods to structure the design process. Regardless of the method used, the design process should be done iteratively by identifying needs, prioritizing, making value decisions, and exploiting physical laws to develop a sound solution that optimizes the objectives while satisfying the constraints that have been identified. As you continue your education, you will learn about structured and standard procedures for design, including information about performance standards, safety standards, and design codes. For now, we will focus on that part of the design process that can be introduced based on your knowledge of calculus, statics, and dynamics. Objectives of design At a minimum, a mechanical design product must • Accomplish the design goals. • Not fail during normal use. • Minimize hazards. • Attempt to anticipate and account for all foreseeable uses. • Be thoroughly documented and archived. In addition, a design should also take into consideration • Cost of manufacture, purchase, and ownership. • Ease of manufacture and maintenance. ∗ See

D. G. Ullman, The Mechanical Design Process, 3rd edition, McGraw-Hill, 2003, for a thorough treatment of design. † Mechanicians are people who study mechanics. Mechanics are people who fix cars or other machinery.

ISTUDY

Section 11.3

Dynamics and Engineering Design

• Energy efficiency in manufacture and use. • Impact on the environment in its manufacture, use, and retirement. Our goal is not to teach you engineering design, but to introduce you to some aspects of it. The aspects that we will focus on most heavily are system modeling and parametric studies. It can be argued that dynamics is about modeling a mechanical system and then studying its behavior as parameters of the system are varied.

System modeling In dynamics, we will consider modeling to be the process of translating real life into mathematical equations to predict the behavior of the model. Once you learn to model mechanical systems, you will have a very powerful tool at your disposal. Unfortunately, that modeling process does not come easily for many students, and it is only mastered through a lot of practice. When creating models, we have to remember that models are not exact representations of reality, but they should give us enough information to tell us something meaningful about real physical systems. For example, the wings of a commercial airliner undergo significant bending and torsion during flight. However, in building a model of an airplane’s performance, we typically start by assuming that the wings are perfectly rigid. To assume that the wings are rigid allows us to simplify the equations of our model and more directly predict the airplane’s behavior for a range of flight conditions. Based on these predictions, we can then refine the model to study the airplane’s behavior in those circumstances where we need to account for the wings’ deformation. In general, once a model is created, you should compare the predictions of that model with data obtained from the real system. If the behavior predicted by the model agrees with the real behavior of the system, then, within the assumptions used to create the model, you can use that model to make further predictions about the system’s behavior. If the predictions do not agree, then you must determine how your model can be improved. It is not always possible to compare the results of a model with the results of a laboratory test. For example, when engineers are creating a new type of aircraft, they cannot, for financial reasons and/or time constraints, build a prototype for every design they create to test it. They must use experience garnered from previous models and designs and extrapolate, using their knowledge of engineering and physics to create a new system. In fact, when new commercial or military aircraft are designed, the first prototypes built are generally those that are flight-tested—there are no intermediate steps.

647

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ISTUDY

Particle Kinematics

12

technotr/Getty Images

Kinematics studies the geometry of motion without reference to the causes of motion. It is rooted in vector algebra and calculus, and to many “it looks and feels like math.” Kinematics is essential for the application of Newton’s second law, 𝐹⃗ = 𝑚𝑎, ⃗ since it allows us to describe the 𝑎⃗ in 𝐹⃗ = 𝑚𝑎. ⃗ Unfortunately, just writing 𝐹⃗ = 𝑚𝑎⃗ doesn’t tell us how something moves; it only describes the relationship between force and acceleration. In general, we also want to know a particle’s motion, where by motion we mean “all the positions occupied by an object over time.” It is kinematics that allows us to translate 𝑎⃗ into the motion of a point by using calculus. This is why in this chapter we study the concepts of position, velocity, and acceleration, and how these quantities relate to one another. In addition, we will learn how to write position, velocity, and acceleration in the various component systems most commonly used in dynamics.

An athlete performing the long jump. Projectile motion can be used to model the long jump.

12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

Kinematics describes motion without reference to the causes and the effects of motion. Kinematics is a core component of any theory relating motion to forces. It is often a vital aspect of many branches of technology, such as tracking aircraft so as to make air traffic control possible, the design of mechanisms, or relating the shape of a roller coaster track and the speed of a roller coaster car to the accelerations experienced during the ride. In this section, we develop those concepts that allow us to characterize the motion of a body as a function of time. Vectors play a fundamental role in this development. In fact, the three main fundamental concepts we cover, namely, position, velocity, and acceleration, are all vector quantities. A notation for time derivatives In kinematics we write derivatives with respect to time so often that it is convenient to have a shorthand notation. If 𝑓 (𝑡) is a function of time, we place a dot over it to mean 𝑑𝑓 (𝑡)∕𝑑𝑡. Furthermore, the number of dots over a quantity indicates the order

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of the derivative; that is, 𝑑𝑓 (𝑡) 𝑓̇ (𝑡) = ; 𝑑𝑡

𝑑 3𝑓 (𝑡) 𝑓⃛(𝑡) = ; 𝑑𝑡3

etc.

(12.1)

Position vector

𝑦 𝑥𝑃 (𝑡)

path of 𝑃 𝑃

𝑟⃗𝑃 (𝑡) 𝚥̂ 𝑂

𝑑 2𝑓 (𝑡) ; 𝑓̈(𝑡) = 𝑑𝑡2

𝑦𝑃 (𝑡)

Referring to Fig. 12.1, the position vector of the point 𝑃 at time 𝑡 relative to the origin 𝑂 is the vector 𝑟⃗𝑃 (𝑡) going from 𝑂 to 𝑃 at time 𝑡. In two dimensions (Section 12.7 on p. 759 covers the three-dimensional case) and using Cartesian components, 𝑟⃗𝑃 (𝑡) = 𝑥𝑃 (𝑡) 𝚤̂ + 𝑦𝑃 (𝑡) 𝚥̂.

𝑥

𝚤̂

Figure 12.1 A point 𝑃 and its position vector relative to 𝑂.

(12.2)

When studying the motion of a single point, we can drop the subscript indicating the point and simply write 𝑟⃗ = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂ instead of 𝑟⃗𝑃 (𝑡) = 𝑥𝑃 (𝑡) 𝚤̂ + 𝑦𝑃 (𝑡) 𝚥̂. The notion of position is meaningless without the specification of a reference point relative to which position is defined. If the reference point is not the origin of a coordinate system, or if several coordinate systems are used concurrently, then we use the notation introduced in Section 11.2 [see Eq. (11.15) on p. 627] to indicate the position of a point relative to another.

Trajectory trajectory or path of 𝑃 line of action 𝑣⃗avg

𝑦

𝑃 (𝑡) 𝑟⃗(𝑡)

𝚥̂ 𝑂

𝚤̂

Figure 12.2 Displacement of point 𝑃 between times 𝑡 and 𝑡 + Δ𝑡, with Δ𝑡 > 0. The distance traveled by 𝑃 between 𝑡 and 𝑡 + Δ𝑡 is highlighted in yellow.

ISTUDY

The trajectory of a point is the line traced through space by the point during its motion (Fig. 12.2). Synonyms for trajectory are path and space curve. The trajectory of a point can be an open line, a closed loop, or some other kind of self-intersecting line. An object’s trajectory by itself cannot tell us anything about how fast an object moves or how many times a point goes past a specific location.

Velocity vector and speed Displacement vector. Referring to Fig. 12.2, we define the displacement (or position change) of 𝑃 between 𝑡 and 𝑡+Δ𝑡 (with Δ𝑡 > 0) as the vector Δ⃗𝑟 = 𝑟⃗(𝑡+Δ𝑡)−⃗𝑟(𝑡). In general, the length of Δ⃗𝑟 does not measure the distance traveled by 𝑃 between 𝑡 and 𝑡+Δ𝑡 (highlighted in yellow in Fig. 12.2). This is so because the distance traveled between 𝑡 and 𝑡 + Δ𝑡 depends on the geometry of the path of 𝑃 , whereas Δ⃗𝑟 depends only on 𝑟⃗(𝑡) to 𝑟⃗(𝑡 + Δ𝑡) without reference to how 𝑃 moved from one position to the other. Average velocity vector. Referring again to Fig. 12.2, we define the average velocity vector of 𝑃 between 𝑡 and 𝑡 + Δ𝑡 as 𝑣⃗avg =

[ ] Δ⃗𝑟 1 𝑟⃗(𝑡 + Δ𝑡) − 𝑟⃗(𝑡) = . (𝑡 + Δ𝑡) − 𝑡 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ Δ𝑡 ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ vector

(12.3)

scalar

The vectors 𝑣⃗avg and Δ⃗𝑟 have the same direction since the term 1∕Δ𝑡 in Eq. (12.3) is a positive scalar. Velocity vector.

Given 𝑣⃗avg between 𝑡 and 𝑡 + Δ𝑡, the velocity vector at time 𝑡 is 𝑣(𝑡) ⃗ = lim 𝑣⃗avg = lim Δ𝑡→0

Δ𝑡→0

Δ⃗𝑟 . Δ𝑡

(12.4)

ISTUDY

Section 12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

651

The second limit in Eq. (12.4) is the time derivative of 𝑟⃗(𝑡), so that we write 𝑣(𝑡) ⃗ =

𝑑⃗𝑟(𝑡) = 𝑟⃗̇ (𝑡). 𝑑𝑡

(12.5)

The velocity vector is the time rate of change of the position vector. Speed.

The speed of a point is the magnitude of its velocity: 𝑣(𝑡) = ||𝑣(𝑡) ⃗ || .

tangent line at 𝑃 (𝑡)

𝑃 (𝑡)

(12.6)

By definition, the speed is never negative. The velocity vector is always tangent to the path. The velocity has an important property: The velocity vector at a point along the trajectory is tangent to the trajectory at that point! To see this we recall that 𝑣⃗avg between 𝑡 and 𝑡 + Δ𝑡 has the same direction as the displacement Δ⃗𝑟 between the same time instants. Referring to Fig. 12.3, let’s consider the sequence of vectors Δ⃗𝑟𝑛 = 𝑟⃗(𝑡 + Δ𝑡∕𝑛) − 𝑟⃗(𝑡) between 𝑡 and 𝑡 + Δ𝑡∕𝑛 (𝑛 = 1, 2, …). As 𝑛 → ∞, Δ⃗𝑟𝑛 and the associated average velocity become tangent to the path, thus implying that 𝑣(𝑡) ⃗ is tangent to the path at 𝑃 (𝑡). Distance traveled and speed. Let 𝑠(𝑡) be the distance traveled as a function of time. This quantity only increases with time, and calculus tells us that 𝑑𝑠∕𝑑𝑡 ≥ 0. We now show that the quantity 𝑑𝑠∕𝑑𝑡 coincides with the speed. Referring to Fig. 12.4, we consider the infinitesimal displacement 𝑑⃗𝑟 between time instants 𝑡 and 𝑡 + 𝑑𝑡, with 𝑑𝑡 > 0. Recalling that 𝑣⃗ = 𝑑⃗𝑟∕𝑑𝑡, we can write 𝑑⃗𝑟 = 𝑣⃗ 𝑑𝑡.

(12.7)

Figure 12.3 Displacement vectors Δ⃗𝑟𝑛 between time 𝑡 and subsequent times 𝑡 + Δ𝑡∕𝑛 (𝑛 = 1, 2, …).

Concept Alert Direction of velocity vectors. One of the most important concepts in kinematics is that the velocity of a particle is always tangent to the particle’s path.

Since 𝑑⃗𝑟 is infinitesimal, it is tangent to the path and has magnitude equal to 𝑑𝑠: |𝑑⃗𝑟| = 𝑑𝑠 = |𝑣⃗ 𝑑𝑡|

⇒

𝑑𝑠 = |𝑣| ⃗ 𝑑𝑡

𝑑𝑠 = 𝑣 𝑑𝑡,

⇒

(12.8) trajectory or path of 𝑃

where we have used Eq. (12.6). The last of Eqs. (12.8) implies that

𝑃 (𝑡 + 𝑑𝑡)

𝑦

𝑑𝑠 = 𝑣. 𝑑𝑡

(12.9)

The speed is therefore both the magnitude of the velocity and the time rate of change of the distance traveled. Another consequence of the last of Eqs. (12.8) is obtained by integrating this equation between two time instants 𝑡1 and 𝑡2 , with 𝑡1 < 𝑡2 : ∫𝑠

𝑠2

1

𝑑𝑠 =

∫𝑡

𝑡2

𝑣 𝑑𝑡

⇒

𝑠2 − 𝑠1 =

1

∫𝑡

𝑡2

𝑣 𝑑𝑡.

(12.10)

1

) ( 𝑡 Recalling that the average speed between 𝑡1 and 𝑡2 is 𝑣avg = ∫𝑡 2 𝑣 𝑑𝑡 ∕(𝑡2 − 𝑡1 ), 1 dividing both sides of the last of Eqs. (12.10) by 𝑡2 − 𝑡1 , we have 𝑣avg =

𝑠2 − 𝑠1 𝑡2 − 𝑡1

.

(12.11)

The ratio of the distance traveled between two time instants 𝑡1 and 𝑡2 and the time interval size 𝑡2 − 𝑡1 measures the average speed between 𝑡1 and 𝑡2 .

𝑟⃗(𝑡 + 𝑑𝑡)

𝑑𝑠 𝑑⃗𝑟

𝑃 (𝑡)

𝑟⃗(𝑡) 𝚥̂ 𝑂

𝚤̂

𝑥

Figure 12.4 A point 𝑃 and its path. The distance traveled (indicated in yellow) between time instants 𝑡 and 𝑡 + 𝑑𝑡 is 𝑑𝑠, which, as 𝑑𝑡 → 0, represents the magnitude of the vector 𝑑⃗𝑟.

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Chapter 12

Particle Kinematics

𝑦

Acceleration vector

path of 𝑃

The acceleration vector is the time rate of change of the velocity vector, i.e., 𝑣(𝑡 ⃗ 𝐶)

𝑣(𝑡 ⃗ 𝐵)

𝐶

𝐵

𝑃

𝚥̂ 𝑂

𝐴

𝑎(𝑡) ⃗ = 𝑣(𝑡 ⃗ 𝐴) 𝑥

𝚤̂

Figure 12.5 Velocity vectors of a particle 𝑃 at three instants in time 𝑡𝐴 , 𝑡𝐵 , and 𝑡𝐶 . The particle is moving with constant speed.

𝑦

path of 𝑃

𝚥̂

𝑃

𝑂

(approximate) direction of 𝑎(𝑡) ⃗

𝑣(𝑡 ⃗ + Δ𝑡) − 𝑣(𝑡) ⃗ , 𝑎(𝑡) ⃗ = 𝑣(𝑡) ⃗̇ = lim Δ𝑡→0 Δ𝑡

−𝑣(𝑡) ⃗ 𝑥

Figure 12.6 Velocity of 𝑃 at times 𝑡 and 𝑡 + Δ𝑡 showing the approximate direction of 𝑎(𝑡). ⃗

(12.12)

The acceleration is generally not tangent to the path. Contrary to what we discovered about the velocity, the acceleration vector is generally not tangent to the path. To see why, consider a point 𝑃 moving at constant speed as shown in Fig. 12.5. Although the speed is constant, the acceleration 𝑎(𝑡) ⃗ of 𝑃 , which measures the changes in both the magnitude and the direction of 𝑣(𝑡), ⃗ is not zero because 𝑣(𝑡) ⃗ changes direction so as to remain tangent to the curved path. Note that the velocity change, i.e., 𝑎(𝑡), ⃗ is more pronounced between points 𝐴 and 𝐵 than it is between 𝐵 and 𝐶 since the path’s curvature is larger between 𝐴 and 𝐵 than between 𝐵 and 𝐶.∗ Furthermore, even if the speed 𝑣 is constant, 𝑎(𝑡) ⃗ depends on 𝑣 because 𝑣 determines how quickly 𝑃 moves along its path and therefore, how fast 𝑣(𝑡) ⃗ changes direction. Hence, 𝑎(𝑡) ⃗ along a curved path depends on both the speed and the path’s curvature. Finally, to get a sense of the direction of 𝑎(𝑡), ⃗ consider the velocity between times 𝑡 and 𝑡 + Δ𝑡, as shown in Fig. 12.6. Recalling that

𝑣(𝑡) ⃗

𝚤̂

𝑑 𝑣(𝑡) ⃗ 𝑑 2 𝑟⃗(𝑡) = 𝑣(𝑡) ⃗̇ = 𝑟⃗̈(𝑡). = 𝑑𝑡 𝑑𝑡2

(12.13)

the direction of 𝑎(𝑡) ⃗ Δ𝑡, which is the same as that of 𝑎(𝑡), ⃗ can be approximated by the direction of 𝑣(𝑡 ⃗ + Δ𝑡) − 𝑣(𝑡) ⃗ if Δ𝑡 is small enough. Referring to Fig. 12.6, the vector 𝑣(𝑡 ⃗ + Δ𝑡) − 𝑣(𝑡), ⃗ and therefore 𝑎(𝑡), ⃗ points toward the concave side of the trajectory instead of being tangent to it. While the arguments just provided are qualitative, in Section 12.4 we will quantitatively show that, for a curved path, not only does the acceleration have a component pointing toward the concave side of the trajectory, but also this component is proportional to the square of the speed and inversely proportional to the path’s radius of curvature.

Cartesian coordinates path of 𝑃 𝑦

𝑎(𝑡) ⃗

𝑣(𝑡) ⃗ 𝑥 coordinate line through 𝑃 (𝑡) 𝑃 (𝑡)

𝑟⃗(𝑡) 𝑦(𝑡) 𝚥̂ 𝑂

𝚤̂ 𝑥(𝑡)

𝑥 𝑦 coordinate line through 𝑃 (𝑡)

Figure 12.7 Position, velocity, and acceleration of a moving point 𝑃 along with a Cartesian coordinate system.

ISTUDY

We now discuss the relation between the Cartesian coordinates of a point and the components of the point’s position, velocity, and acceleration vectors. Coordinates and position. Referring to Fig. 12.7, the Cartesian coordinates of 𝑃 are 𝑥(𝑡) and 𝑦(𝑡) (for the three-dimensional case, see Section 12.7). Cartesian coordinate systems have the distinctive property that the coordinates of a point are also the components of the position vector of the point relative to the system’s origin. That is, if 𝑟𝑥 and 𝑟𝑦 are the Cartesian components of 𝑟⃗ in the 𝑥 and 𝑦 directions, respectively, then 𝑟𝑥 = 𝑥(𝑡) and 𝑟𝑦 = 𝑦(𝑡). Therefore, we express the position vector 𝑟⃗(𝑡) in a Cartesian component system as 𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂.

(12.14)

Velocity and acceleration components in terms of time derivatives of the coordinates. The velocity 𝑣(𝑡) ⃗ and the acceleration 𝑎(𝑡) ⃗ of 𝑃 are vectors based at 𝑃 (𝑡), as shown in Fig. 12.8. The components of 𝑣(𝑡) ⃗ and 𝑎(𝑡) ⃗ are understood to be ∗ We

will mathematically define curvature in Section 12.4.

ISTUDY

Section 12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

path of 𝑃 𝑦

𝑎(𝑡) ⃗

𝑣(𝑡) ⃗ 𝑥 coordinate line through 𝑃 (𝑡)

𝚥̂ 𝚤̂ 𝑃 (𝑡)

𝑥

𝑂

𝑦 coordinate line through 𝑃 (𝑡)

Figure 12.8. A moving point 𝑃 at various times with its corresponding companion base vectors.

relative to the vectors 𝚤̂ and 𝚥̂ that are also based at 𝑃 (𝑡) and tangent to the coordinate lines. Computing 𝑣⃗ as the time derivative of 𝑟⃗ in Eq. (12.14), we have 𝑣⃗ = 𝑥(𝑡) ̇ 𝚤̂ + 𝑥(𝑡)

𝑑̂𝚥 𝑑̂𝚤 + 𝑦(𝑡) ̇ 𝚥̂ + 𝑦(𝑡) , 𝑑𝑡 𝑑𝑡

(12.15)

where the terms 𝑑̂𝚤∕𝑑𝑡 and 𝑑̂𝚥∕𝑑𝑡 are due to the fact that, since 𝚤̂ and 𝚥̂ are based at the moving point 𝑃 (𝑡), 𝚤̂ and 𝚥̂ need to be regarded as implicit functions of time. However, Cartesian coordinate systems have the property that their base vectors do not change direction from point to point, i.e., 𝚤̂ and 𝚥̂ are constants so that 𝑑̂𝚤 ⃗ = 0 and 𝑑𝑡

𝑑̂𝚥 ⃗ = 0, 𝑑𝑡

(12.16)

⃗ in a Cartesian component system becomes and the velocity vector 𝑣(𝑡) 𝑣⃗ = 𝑥(𝑡) ̇ 𝚤̂ + 𝑦(𝑡) ̇ 𝚥̂ = 𝑣𝑥 (𝑡) 𝚤̂ + 𝑣𝑦 (𝑡) 𝚥̂,

(12.17)

where the velocity components are obtained by direct time differentiation of the coor⃗ in a Cartesian dinates, i.e., 𝑣𝑥 = 𝑥̇ and 𝑣𝑦 = 𝑦.̇ Similarly, the acceleration vector 𝑎(𝑡) component system is 𝑎⃗ = 𝑥(𝑡) ̈ 𝚤̂ + 𝑦(𝑡) ̈ 𝚥̂ = 𝑣̇ 𝑥 (𝑡) 𝚤̂ + 𝑣̇ 𝑦 (𝑡) 𝚥̂ = 𝑎𝑥 (𝑡) 𝚤̂ + 𝑎𝑦 (𝑡) 𝚥̂,

(12.18)

where 𝑎𝑥 = 𝑣̇ 𝑥 = 𝑥̈ and 𝑎𝑦 = 𝑣̇ 𝑦 = 𝑦. ̈ In later sections, we will discover that in non-Cartesian coordinate systems the components of 𝑣⃗ and 𝑎⃗ are also influenced by the time rate of change of their unit vectors.

End of Section Summary Position. The position of a point is a vector going from the origin of the chosen frame of reference to the point in question. Trajectory. The trajectory of a moving point is the line traced by the point during its motion. Another name for trajectory is path. Displacement. The displacement between positions 𝐴 and 𝐵 is the vector going from 𝐴 to 𝐵. In general, the magnitude of the displacement between two positions is not the distance traveled along the path between these positions.

653

654

Chapter 12

Particle Kinematics

Velocity. The velocity vector is the time rate of change of the position vector. The velocity is always tangent to the path. Speed. The speed is the magnitude of the velocity and is a nonnegative scalar quantity. The speed measures the time rate of change of the distance traveled along the path.

path of 𝑃 𝑦

𝑎(𝑡) ⃗

𝑣(𝑡) ⃗ 𝑥 coordinate line through 𝑃 (𝑡)

Acceleration. The acceleration vector is the time rate of change of the velocity vector. Contrary to the velocity vector, the acceleration vector is, in general, not tangent to the trajectory. Cartesian coordinates. The Cartesian coordinates of a point 𝑃 moving along some path are shown in Fig. 12.9. The position vector is given by

𝑃 (𝑡)

Eq. (12.14), p. 652

𝑟⃗(𝑡) 𝑦(𝑡)

𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂.

𝚥̂ 𝑂

𝚤̂ 𝑥(𝑡)

𝑥 𝑦 coordinate line through 𝑃 (𝑡)

In Cartesian components, the velocity and acceleration vectors are given by Eqs. (12.17) and (12.18), p. 653

Figure 12.9 The position vector 𝑟⃗(𝑡) of the point 𝑃 in Cartesian coordinates.

ISTUDY

𝑣(𝑡) ⃗ = 𝑥(𝑡) ̇ 𝚤̂ + 𝑦(𝑡) ̇ 𝚥̂ = 𝑣𝑥 (𝑡) 𝚤̂ + 𝑣𝑦 (𝑡) 𝚥̂, 𝑎(𝑡) ⃗ = 𝑥(𝑡) ̈ 𝚤̂ + 𝑦(𝑡) ̈ 𝚥̂ = 𝑣̇ 𝑥 (𝑡) 𝚤̂ + 𝑣̇ 𝑦 (𝑡) 𝚥̂ = 𝑎𝑥 (𝑡) 𝚤̂ + 𝑎𝑦 (𝑡) 𝚥̂, where 𝑣𝑥 = 𝑥, ̇ 𝑣𝑦 = 𝑦,̇ 𝑎𝑥 = 𝑣̇ 𝑥 = 𝑥, ̈ and 𝑎𝑦 = 𝑣̇ 𝑦 = 𝑦. ̈ An important note regarding example problems. In Chapter 12, all examples will begin to employ part of the problem-solving framework that is formally introduced in Chapter 13. That is, each example problem will include a Road Map step, a Computation step, and a Discussion & Verification step.∗ The Road Map step will lay out the path to the solution by identifying given information and unknowns and then proposing a problem-solving strategy. The Computation step will set up the appropriate equations and will solve them. Finally, the Discussion & Verification step will check whether or not the solution is reasonable.

∗ The

problem-solving framework introduced in Chapter 13 includes additional steps since it is applied to kinetics problems, which are generally more involved than kinematics problems.

ISTUDY

Section 12.1

655

Position, Velocity, Acceleration, and Cartesian Coordinates

How Do You Get to Carnegie Hall? . . . Practice!

Carnegie Hall

SOLUTION

E5 2nd St

Radio City Music Hall

St

E5

Museum Of Modern Art

1st S t

E

500 f t

Figure 1 Cab route. t

𝐶 𝐷

Rink Cen tral Par kS W5 8th St

Carnegie Carne Hall all

(1)

𝑇ℎ𝑒𝑒 𝑃𝑜𝑛𝑑 𝑇 𝑇ℎ 𝑃 𝑛𝑑 𝑃𝑜

𝑥 𝐵

Plaza Hotel Pl

Gra nd A rmy Pl

Δ⃗𝑟𝐴𝐷 = 𝑟⃗𝐷 − 𝑟⃗𝐴 = (1500 𝚤̂ + 906.0 𝚥̂) ft.

3rd

𝐴

𝐵

Plaza Hotel

E5

We s

We select a Cartesian coordinate system with origin at 𝐴 and aligned with the city grid (Fig. 2). Using the given information, the coordinates of 𝐴, 𝐵, 𝐶, and 𝐷 are given in Table 1. The displacement from 𝐴 to 𝐷, which we will denote by Δ⃗𝑟𝐴𝐷 , is the difference between 𝑟⃗𝐷 and 𝑟⃗𝐴 , namely, the position vectors of 𝐴 and 𝐷:

Computation

N

Ave

To solve the problem, we need to set up a coordinate system and identify the coordinates of points 𝐴, 𝐵, 𝐶, and 𝐷 that define the cab’s path. Then the problem’s questions can be answered by applying the definitions of displacement, average velocity, distanced traveled, and speed.

8th

Road Map

The Pond Pl

𝐷

Cen tral Par kS W5 8th St

Arm y

8th Ave

𝐶

Gra nd

t

A cab picks up a passenger outside Radio City Music Hall on the corner of the Avenue of the Americas and E 51st St. (point 𝐴) and drops her off in front of Carnegie Hall on 7th Ave. (point 𝐷) after 5 min, following the route shown. Find the cab’s displacement, average velocity, distance traveled, and average speed in going from 𝐴 to 𝐷. The distance from 𝐴 to 𝐵 is 2200 f t, from 𝐵 to 𝐶 is 906 f t, and from 𝐶 to 𝐷 is 700 f t.

We s

E X A M P L E 12.1

Applying the definition of average velocity in Eq. (12.3) between 𝑡𝐴 and 𝑡𝐷 we have E5 2nd St

𝑦

𝑣⃗avg =

Δ⃗𝑟𝐴𝐷 𝑡𝐷 − 𝑡𝐴

= (5.000 𝚤̂ + 3.020 𝚥̂) ft∕s,

(2)

where 𝑡𝐷 − 𝑡𝐴 = 5 min = 300 s. Next, the distance traveled by the cab, which we will denote by 𝑑, is given by the sum of the lengths of the segments 𝐴𝐵, 𝐵𝐶, and 𝐶𝐷, i.e, 𝑑 = (𝑥𝐵 − 𝑥𝐴 ) + (𝑦𝐶 − 𝑦𝐵 ) + (𝑥𝐶 − 𝑥𝐷 ) = 3806 f t.

(3)

As shown in Eq. (12.11) on p. 651, the average speed between 𝑡𝐴 and 𝑡𝐷 is the ratio between the distance traveled between these time instants and the time interval duration 𝑡𝐷 − 𝑡𝐴 . Hence, we have 𝑣avg =

𝑑 = 12.69 f t∕s. 𝑡𝐷 − 𝑡𝐴

(4)

Discussion & Verification

The results obtained are dimensionally correct. We observe that 𝑣avg in miles per hour is 8.650 mph. Hence, we can consider the result acceptable since such a speed is typical of midtown Manhattan. A Closer Look Observing that |Δ⃗𝑟𝐴𝐷 | = 1752 f t, and that |𝑣⃗avg | = 5.841 f t∕s, we see that this example reinforces the idea that we should never confuse the magnitude of the displacement for the distance traveled or the magnitude of the average velocity for the average speed.

N

𝚥̂

𝐴 E 𝚤̂

Radio adio City C y Music Hall allll

E5 3rd St

Museum um Of Modern dern ern Art

51s t St

500 5 ft

E

Figure 2 Cartesian coordinate system with origin at the pickup point 𝐴. The city grid is such that the line through 𝐵 and 𝐶 is parallel to the 𝑦 axis and the line through 𝐶 and 𝐷 is parallel to the 𝑥 axis. Table 1 Coordinates of the points defining the cab’s route. Point

x (ft)

y (ft)

𝐴 𝐵 𝐶 𝐷

0 2200 2200 1500

0 0 906 906

656

Chapter 12

Particle Kinematics

E X A M P L E 12.2

Trajectory, Velocity, and Acceleration Ship 𝐵 tracks an unmanned aerial vehicle 𝑃 launched from 𝐴 and flying parallel to the water. Relative to the Cartesian frame shown in Fig. 1 and for the first 8 seconds of flight, the recorded motion of 𝑃 is [( ) ( ) ] 𝑟⃗(𝑡) = 225 + 2.13𝑡3 𝚤̂ + 225 + 0.993𝑡3 𝚥̂ m, (1)

𝑦 𝑃

𝐴

where 𝑡 is in seconds. Determine

𝑟⃗(𝑡)

𝚥̂ 𝐵

(a) The trajectory of 𝑃 for 0 ≤ 𝑡 ≤ 5 s.

𝑥

𝚤̂

(b) The velocity and the speed of 𝑃 . (c) The acceleration of 𝑃 and its orientation relative to the path.

Figure 1 Point 𝑃 represents an unmanned aerial vehicle launched from ship 𝐴.

SOLUTION Part (a): Path of 𝑷

We have the motion of 𝑃 as a function of time. Since the trajectory of 𝑃 is the line traced by 𝑃 in space, we can find the path by eliminating time from the motion.

Road Map

Helpful Information Trajectory and time. The trajectory (or path) is what the motion looks like once time is removed from the motion’s description.

Computation Letting 𝑥(𝑡) and 𝑦(𝑡) denote the coordinates of 𝑃 relative to the coordinate system in Fig. 1, 𝑟⃗(𝑡) can be expressed as 𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂, where, by comparison with Eq. (1), ( ) ( ) 𝑥(𝑡) = 225 + 2.13𝑡3 m and 𝑦(𝑡) = 225 + 0.993𝑡3 m. (2)

To eliminate time, we solve the first of Eqs. (2) for 𝑡3 and then substitute the result in the second of Eqs. (2). This yields [ ] ( ) ) 0.993 ( 𝑦 = 225 + (3) 𝑥 − 225 m ⇒ 𝑦 = 120.1 + 0.4662𝑥 m. 2.13 𝑦 trajectory of 𝑃

𝑟⃗(5 s)

𝚥̂ 𝑟⃗(0)

𝑥

𝚤̂ 225 m

491.2 m

Figure 2 Path of 𝑃 as seen by frames 𝐴 and 𝐵.

ISTUDY

𝑟⃗(0) = (225.0 𝚤̂ + 225.0 𝚥̂) m

𝑃 (5 s)

𝑃 (0)

𝐵

ℒ

The last of Eqs. (3) is the equation of the straight line ℒ depicted in Fig. 2. The answer to Part (a) of the problem is the segment of ℒ traveled by 𝑃 for 0 ≤ 𝑡 ≤ 5 s. To find this segment, we first determine the positions of 𝑃 at 𝑡 = 0 and 𝑡 = 5 s: and 𝑟⃗(5 s) = (491.2 𝚤̂ + 349.1 𝚥̂) m.

(4)

We observe that 𝑥(𝑡) and 𝑦(𝑡) in Eq. (2) only increase with time so that, for 0 < 𝑡 < 5 s, 𝑃 is on ℒ between the points identified by 𝑟⃗(0) and 𝑟⃗(5 s). Therefore, referring to Eqs. (4) and Fig. 2, the trajectory of 𝑃 for 0 ≤ 𝑡 ≤ 5 s is the segment of ℒ for 225.0 m ≤ 𝑥 ≤ 491.2 m, corresponding to 225.0 m ≤ 𝑦 ≤ 349.1 m: ( ) 𝑦 = 120.1 + 0.4662𝑥 m for 225.0 m ≤ 𝑥 ≤ 491.2 m. (5) The fact that the trajectory 𝑃 is a straight-line segment is reasonable since both 𝑥 and 𝑦 depend on time in the same way, namely, via the term 𝑡3 . Also, the trajectory was expected to be some finite segment since the time interval indicated by the problem was finite.

Discussion & Verification

Part (b): Velocity and speed of 𝑷 Road Map Since 𝑟⃗(𝑡) is given, we can compute the velocity 𝑣(𝑡) ⃗ by differentiating 𝑟⃗(𝑡) with respect to time. The speed is then found by computing the magnitude of 𝑣(𝑡). ⃗ Computation

Differentiating Eq. (1) with respect to time and simplifying, we have ( ) 𝑣(𝑡) ⃗ = 6.390𝑡2 𝚤̂ + 2.979𝑡2 𝚥̂ m∕s. (6)

Using Eq. (6) and the Pythagorean theorem, the magnitude of 𝑣(𝑡) ⃗ is ( ) 𝑣(𝑡) = 7.050𝑡2 m∕s.

(7)

ISTUDY

Section 12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

657

Discussion & Verification Since the position is a cubic polynomial in 𝑡, we expect the velocity to be a quadratic polynomial in 𝑡, as is in fact the case. As far as 𝑣(𝑡) is concerned, we observe that it is greater or equal to zero for any possible value of time, as it should be. Part (c): Acceleration of 𝑷 & its orientation

Since 𝑣(𝑡) ⃗ is known, we can find the acceleration 𝑎(𝑡) ⃗ by differentiating 𝑣(𝑡) ⃗ with respect to time. To determine the orientation of 𝑎(𝑡) ⃗ relative to the path, we recall that 𝑣(𝑡) ⃗ is tangent to the path. Hence, the orientation of 𝑎(𝑡) ⃗ relative to the path is the same as that of 𝑎(𝑡) ⃗ relative to 𝑣(𝑡). ⃗ In turn, this orientation is described by the angle between 𝑎(𝑡) ⃗ and 𝑣(𝑡). ⃗ Road Map

Computation

Differentiating Eq. (6) with respect to time, we have ( ) 𝑎(𝑡) ⃗ = 12.78𝑡 𝚤̂ + 5.958𝑡 𝚥̂ m∕s2 .

(8)

Denoting by 𝜃𝑣⃗𝑎⃗ the angle between 𝑣(𝑡) ⃗ and 𝑎(𝑡), ⃗ we recall that the dot product of 𝑣(𝑡) ⃗ and 𝑎(𝑡) ⃗ has the form 𝑣(𝑡) ⃗ ⋅ 𝑎(𝑡) ⃗ = |𝑣(𝑡)|| ⃗ 𝑎(𝑡)| ⃗ cos 𝜃𝑣⃗𝑎⃗

⇒

cos 𝜃𝑣⃗𝑎⃗ =

𝑣(𝑡) ⃗ ⋅ 𝑎(𝑡) ⃗ . |𝑣(𝑡)|| ⃗ 𝑎(𝑡)| ⃗

(9)

To determine the numerator of the right-hand side of the last of Eqs. (9), we use the component expressions for 𝑣(𝑡) ⃗ and 𝑎(𝑡) ⃗ in Eqs. (6) and (8), respectively: [( ] ) ] [( ) 𝑣(𝑡) ⃗ ⋅ 𝑎(𝑡) ⃗ = 6.390𝑡2 𝚤̂ + 2.979𝑡2 𝚥̂ m∕s ⋅ 12.78𝑡 𝚤̂ + 5.958𝑡 𝚥̂ m∕s2 ( ) (10) = 99.41𝑡3 m2 ∕s3 . As for the denominator of the right-hand side of the last of Eqs. (9), we observe that |𝑣(𝑡)| ⃗ is given in Eq. (7) and |𝑎(𝑡)| ⃗ can be obtained from Eq. (8) via the Pythagorean theorem: ( ) |𝑎(𝑡)| ⃗ = 14.10𝑡 m∕s2 . (11) Substituting the results in Eqs. (7), (10), and (11) into the last of Eqs. (9) we have cos 𝜃𝑣⃗𝑎⃗ = 1,

(12)

𝜃𝑣⃗𝑎⃗ = 0.

(13)

which can be solved for 𝜃𝑣⃗𝑎⃗ to obtain

Equation (13) implies that 𝑎(𝑡) ⃗ is tangent to the trajectory at every time instant. Discussion & Verification

The result for the acceleration appears to be reasonable since it is a first-order polynomial in 𝑡, which is consistent with the fact that the velocity is a second-order polynomial in 𝑡. Also, the fact that 𝑎(𝑡) ⃗ is tangent to the path is not surprising since the trajectory computed earlier in the problem turned out to be a straight-line segment.

Common Pitfall Does Eq. (13) contradict what we said earlier about the acceleration not being tangent to the trajectory? Earlier in the section we stated that in general the acceleration is not tangent to the path. We also indicated that if the trajectory is curved, then we must expect the acceleration not to be parallel to the path. Therefore, we can conclude that (1) there can be points along a path at which the acceleration is tangent to the path and that (2) at these points the path’s curvature must be equal to zero. This is consistent with what we found in our example since the path in this example is a straight line, that is, a path with no curvature.

658

Chapter 12

Particle Kinematics

E X A M P L E 12.3

Relating the Path and the Speed to Velocity and Acceleration A particle 𝑃 moves with constant speed 𝑣0 along the parabola 𝑦2 = 4𝑎𝑥 with 𝑦̇ > 0. Determine each of the following.

𝑦

(a) The Cartesian components of the velocity vector as a function of 𝑦. (b) The Cartesian components of the acceleration vector as a function of 𝑦.

𝑃

0.5

(c) The angle between the velocity and acceleration vectors as a function of 𝑦. In addition, plot the Cartesian components of the velocity and acceleration vectors as functions of 𝑦 for 𝑎 = 0.2 m and 𝑣0 = 3 m∕s.

𝑥

0

SOLUTION −0.5

Road Map

0

We are given the path of the particle and its speed along the path, but we do not know how 𝑥 and 𝑦 vary with time, so we cannot immediately apply Eqs. (12.17) and (12.18). On the other hand, we can differentiate the equation √ of the curve with respect to time and make use of the fact that the speed is given by 𝑥̇ 2 + 𝑦̇ 2 .

1

Figure 1 Parabolic path of the particle 𝑃 for 𝑎 = 0.2 m.

Helpful Information Why find components as a function of 𝒚 instead of 𝒙? We chose to find the velocity and acceleration components as a function of 𝑦 rather than 𝑥 because of the fact that for any given value of 𝑦, there is one value of 𝑥. The converse is not true; that is, for a given value of 𝑥, there are two values of 𝑦, and this makes the analysis more complicated.

Computation

We start by differentiating the particle’s path with respect to time to ob-

tain

) 𝑑(𝑦2 ) 𝑑𝑦 𝑑(4𝑎𝑥) 𝑑( 2 = ⇒ 2𝑦𝑦̇ = 4𝑎𝑥, ̇ (1) 𝑦 = 4𝑎𝑥 ⇒ 𝑑𝑡 𝑑𝑦 𝑑𝑡 𝑑𝑡 where, in differentiating with respect to time, we have used the chain rule. Solving the last of Eqs. (1) for 𝑦,̇ we obtain 2𝑎𝑥̇ 𝑦̇ = . (2) 𝑦 Now, we also know that the speed is related to the components of the velocity via ( 𝑣20 = 𝑥̇ 2 + 𝑦̇ 2

𝑣20 = 𝑥̇ 2 +

⇒

2𝑎𝑥̇ 𝑦

)2 ,

(3)

where we have substituted in Eq. (2). We can now solve Eq. (3) for 𝑥̇ to obtain

velocity component (m∕s)

𝑥̇ = √ 3 2

𝑥̇

0 −1 −2 −1

−0.5

0 𝑦 (m)

0.5

1

Figure 2 The 𝑥 and 𝑦 components of the velocity as a function of 𝑦.

ISTUDY

𝑦2 + 4𝑎2

,

(4)

where we have chosen the plus sign when taking the square root to be consistent with Eq. (2), which, given that 𝑦̇ > 0, tells us that for 𝑦 < 0 we must have 𝑥̇ < 0 and for 𝑦 > 0 we must have 𝑥̇ > 0. Now we can find 𝑦(𝑦) ̇ by substituting Eq. (4) into Eq. (2), which gives 2𝑣 𝑎 , 𝑦̇ = √ 0 (5) 𝑦2 + 4𝑎2

𝑦̇

1

𝑣0 𝑦

where 𝑦̇ > 0 as expected. Figure 2 shows 𝑥̇ and 𝑦̇ as functions of 𝑦. There are several ways to obtain the 𝑥 and 𝑦 components of the acceleration. We will differentiate the last of Eqs. (1) with respect to time, and after simplifying, we obtain 𝑦̇ 2 + 𝑦𝑦̈ = 2𝑎𝑥̈

⇒

𝑥̈ =

) 1( 2 𝑦̇ + 𝑦𝑦̈ . 2𝑎

(6)

We see that we need 𝑦̈ to know 𝑥. ̈ Differentiating Eq. (5) with respect to time, we obtain ] [ −𝑦𝑦̇ 𝑦̈ = 2𝑣0 𝑎 ( (7) )3∕2 , 𝑦2 + 4𝑎2

ISTUDY

Section 12.1

659

Position, Velocity, Acceleration, and Cartesian Coordinates

−4𝑣20 𝑎2 𝑦 𝑦̈ = ( )2 . 𝑦2 + 4𝑎2

(8)

We can now get the final version of 𝑥̈ by substituting Eqs. (5) and (8) into the second of Eqs. (6): [ ] 4𝑣20 𝑎2 𝑦 4𝑣20 𝑎2 1 −𝑦 ( 𝑥̈ = )2 2𝑎 𝑦2 + 4𝑎2 𝑦2 + 4𝑎2

⇒

8𝑣20 𝑎3 𝑥̈ = ( )2 . 𝑦2 + 4𝑎2

(9)

√ Figure 3 shows 𝑥, ̈ 𝑦, ̈ and 𝑎 = 𝑥̈ 2 + 𝑦̈2 as a function of 𝑦. Finally, to find the angle between the velocity and acceleration vectors, we first need to form those vectors as 𝑣⃗ = 𝑥̇ 𝚤̂ + 𝑦̇ 𝚥̂ and 𝑎⃗ = 𝑥̈ 𝚤̂ + 𝑦̈ 𝚥̂.

acceleration component (m∕s2 )

which, after substituting 𝑦̇ from Eq. (5), becomes

20 15 10 5 0 −5

𝑥̈

𝑦̈

−1

(𝑥̇ 𝚤̂ + 𝑦̇ 𝚥̂) ⋅ (𝑥̈ 𝚤̂ + 𝑦̈ 𝚥̂) 𝑥̇ 𝑥̈ + 𝑦̇ 𝑦̈ 𝑣⃗ ⋅ 𝑎⃗ = √ = √ . √ √ 2 2 2 2 2 |𝑣|| ⃗ 𝑎| ⃗ 𝑥̇ + 𝑦̇ 𝑥̈ + 𝑦̈ 𝑥̇ + 𝑦̇ 2 𝑥̈ 2 + 𝑦̈2

cos 𝜙 = 0

𝜙 = 90◦ .

1

𝑣⃗

(11)

𝜙

0.5 𝑎⃗ 0

Recall that 𝑦̇ > 0. This agrees with the plot in Fig. 2, as it should, since we chose the plus sign when taking the square root to get Eq. (5). We also see that 𝑦̇ is largest at 𝑦 = 0. This also makes sense due to the fact that the particle has constant speed. Recalling that the velocity is always tangent to the path, note that at 𝑦 = 0, the velocity must be completely in the 𝑦 direction. Therefore, at 𝑦 = 0, 𝑥̇ must be zero, and so 𝑦̇ achieves a maximum equal to 𝑣0 = 3 m∕s, as can be seen in Fig. 2. As for the 𝑥 component of the velocity, it is negative for 𝑦 < 0 and positive for 𝑦 > 0. This is what we should expect, given that 𝑥 is decreasing for 𝑦 < 0 and is increasing for 𝑦 > 0 [this is why we chose the + sign in Eq. (4)]. In addition, notice that 𝑥̇ = 0 when 𝑦 = 0, which is also to be expected from the parabola shown in Fig. 1. To verify our acceleration results, recall that the particle moves at a constant speed along a parabola. Since the particle’s path is curved, we expect the components of the acceleration vector to be different from zero, as can be seen in Fig. 3. In addition, we expect 𝑦̈ = 0 at 𝑦 = 0 because, as argued earlier, 𝑦̇ is maximum at 𝑦 = 0. In addition, note that 𝑥̈ is largest at 𝑦 = 0 even though 𝑥̇ = 0 there. While perhaps counterintuitive, this result is consistent with the fact that part of the acceleration is proportional to the curvature of the path on which the particle moves. Since the curvature √ of the parabola is largest at 𝑦 = 0, the magnitude of the acceleration, given by 𝑎 = 𝑥̈ 2 + 𝑦̈2 , achieves a maximum at 𝑦 = 0.∗ As for the angle 𝜙 between 𝑣⃗ and 𝑎, ⃗ we discovered that it is always equal to 90◦ . Since the speed is constant, we know that 𝑣⃗ cannot be changing along its line of action, and so any change in 𝑣⃗ (that is, any 𝑎) ⃗ must be perpendicular to 𝑣. ⃗ We will discuss this topic in detail in Section 12.4.

−0.5

⇒

0.5

𝑦

(12)

⇒

0 𝑦 (m)

(10)

Although it is tedious, we can substitute Eqs. (4), (5), (8), and (9) into the numerator of the last expression in Eq. (11) to find 𝑥̇ 𝑥̈ + 𝑦̇ 𝑦̈ = 0

−0.5

Figure 3 The 𝑥 component (blue), 𝑦 component (red), and the total acceleration (green) of the particle as a function of 𝑦.

Then, using Eq. (11.18) on p. 628, the angle 𝜙 between these two vectors is cos 𝜙 =

𝑎

𝑥

⃗ Since 𝜙 = 90◦ for all values of 𝑦, 𝑣⃗ is always perpendicular to 𝑎. Discussion & Verification

∗ This

topic is discussed in detail in Section 12.4.

0

1

Figure 4 The velocity vector, acceleration vector, and the angle 𝜙 for one position of the particle as it moves along the parabola. 𝜙 150◦ 100◦ 50◦ 0◦ −1.0

−0.5

0.0 𝑦 (m)

0.5

1.0

Figure 5 The heavy red line shows the angle 𝜙(𝑦) between the velocity vector and the acceleration vector.

660

Chapter 12

Particle Kinematics

Problems Problem 12.1 If 𝑣⃗avg is the average velocity of a point 𝑃 over a given time interval, is |𝑣⃗avg |, the magnitude of the average velocity, equal to the average speed of 𝑃 over the time interval in question?

Problem 12.2 A car is seen parked in a given parking space at 8:00 A .M . on a Monday morning and is then seen parked in the same spot the next morning at the same time. What is the displacement of the car between the two observations? What is the distance traveled by the car during the two observations?

Problem 12.3 Figure P12.2

Is it possible for the vector 𝑣⃗ shown to represent the velocity of the point 𝑃 ? 𝑎⃗ 𝑃 𝑣⃗ path of 𝑃 Figure P12.3 and P12.4

Problem 12.4 Is it possible for the vector 𝑎⃗ shown to be the acceleration of the point 𝑃 ?

Problem 12.5 Two points 𝑃 and 𝑄 happen to go by the same location in space (though at different times). (a) What must the paths of 𝑃 and 𝑄 have in common if, at the location in question, 𝑃 and 𝑄 have identical speeds? STOP

𝑟 Figure P12.6

STOP

𝑢̂ 𝑟

0

(b) What must the paths of 𝑃 and 𝑄 have in common if, at the location in question, 𝑃 and 𝑄 have identical velocities?

Problem 12.6 The position of a car traveling between two stop signs along a straight city block is given by 𝑟 = [9𝑡 − (45∕2) sin(2𝑡∕5)] m, where 𝑡 denotes time (in seconds), and where the argument of the sine function is measured in radians. Compute the displacement of the car between 2.1 and 3.7 s, as well as between 11.1 and 12.7 s. For each of these time intervals compute the average velocity.

Problem 12.7

Figure P12.7

ISTUDY

A city bus covers a 15 km route in 45 min. If the initial departure and final arrival points coincide, determine the average velocity and the average speed of the bus over the entire duration of the ride. Express the answers in m∕s.

ISTUDY

Section 12.1

661

Position, Velocity, Acceleration, and Cartesian Coordinates 𝑦

Problem 12.8 An airplane 𝐴 is performing a loop with constant radius 𝜌. When 𝜃 = 120◦ , the speed of the airplane is 𝑣0 = 210 mph. Modeling the airplane as a point, find the velocity of the airplane at this instant using the component system shown. Express your answer in ft∕s.

𝜌 𝐶

𝑣 𝜃

Problem 12.9 An airplane 𝐴 is performing a loop with constant radius 𝜌 = 300 m. We will see shortly that the acceleration of a point in uniform circular motion (i.e., circular motion at constant speed) is directed toward the center of the circle and has magnitude equal to 𝑣2 ∕𝜌, where 𝑣 is the speed and 𝜌 is the radius of the circle. Assuming that 𝐴 can maintain its speed constant and using the component system shown, provide the expressions of the velocity and acceleration of 𝐴 when 𝜃 = 40◦ and |𝑎| ⃗ = 3𝑔, where 𝑎⃗ is the acceleration of 𝐴 and 𝑔 is the acceleration due to gravity.

𝐴

𝚥̂ 𝚤̂

𝑥

𝑂 Figure P12.8 and P12.9

Problem 12.10 An airplane takes off as shown following a trajectory described by equation 𝑦 = 𝜅𝑥2 , where 𝜅 = 2 × 10−4 f t −1 . When 𝑥 = 1200 f t, the speed of the plane is 𝑣0 = 110 mph. Using the component system shown, provide the expression for the velocity of the airplane when 𝑥 = 1200 f t. Express your answer in ft∕s.

𝑦

path of 𝐴

Problem 12.11

𝚥̂

The position of a car as a function of time 𝑡, with 𝑡 > 0 and expressed in seconds, is [( ) ( ) ] 𝑟⃗(𝑡) = 5.98𝑡2 + 0.139𝑡3 − 0.0149𝑡4 𝚤̂ + 0.523𝑡2 + 0.0122𝑡3 − 0.00131𝑡4 𝚥̂ ft.

𝑂

𝐴

𝑣0 𝑥

𝚤̂

Figure P12.10

Determine the velocity, speed, and acceleration of the car for 𝑡 = 15 s. 𝑦 𝚥̂ 𝜃 𝚤̂

𝑥

Figure P12.11 and P12.12

Problem 12.12 The position of a car as a function of time 𝑡, with 𝑡 > 0 and expressed in seconds, is [ ( ) ( ) ] 𝑟⃗(𝑡) = 12.3 𝑡 + 1.54𝑒−0.65𝑡 𝚤̂ + 2.17 𝑡 + 1.54𝑒−0.65𝑡 𝚥̂ m.

Find the difference between the average velocity over the time interval 0 ≤ 𝑡 ≤ 2 s and the true velocity computed at the midpoint of the interval, i.e., at 𝑡 = 1 s. Repeat the calculation for the time interval 8 s ≤ 𝑡 ≤ 10 s. Explain why the difference between the average velocity and the true velocity over the time interval 0 ≤ 𝑡 ≤ 2 s is not equal to that over 8 s ≤ 𝑡 ≤ 10 s. 𝑦

Problem 12.13 The position of a car as a function of time 𝑡, with 𝑡 > 0 and expressed in seconds, is [( ) ( ) ] 𝑟⃗(𝑡) = 66𝑡 − 120 𝚤̂ + 1.2 + 31.7𝑡 − 8.71𝑡2 𝚥̂ ft. If the speed limit is 55 mph, determine the time at which the car will exceed this limit.

𝚥̂ 𝜃 𝑂

𝚤̂

Figure P12.13 and P12.14

𝑥

662

Chapter 12

Particle Kinematics

Problem 12.14 The position of a car as a function of time 𝑡, with 𝑡 > 0 and expressed in seconds, is [( ) ( ) ] 𝑟⃗(𝑡) = 66𝑡 − 120 𝚤̂ + 1.2 + 31.7𝑡 − 8.71𝑡2 𝚥̂ ft. Determine the slope 𝜃 of the trajectory of the car for 𝑡1 = 1 s and 𝑡2 = 3 s. In addition, find the angle 𝜙 between velocity and acceleration for 𝑡1 = 1 s and 𝑡2 = 3 s. Based on the values of 𝜙 at 𝑡1 and 𝑡2 , argue whether the speed of the car is increasing or decreasing at 𝑡1 and 𝑡2 .

Problem 12.15 [ ( ) ] Let 𝑟⃗ = 𝑡 𝚤̂ + 2 + 3𝑡 + 2𝑡2 𝚥̂ m describe the motion of the point 𝑃 relative to the Cartesian frame of reference shown. Determine an analytic expression of the type 𝑦 = 𝑦(𝑥) for the trajectory of 𝑃 for 0 ≤ 𝑡 ≤ 5 s. 𝑦 𝑎⃗

𝑟⃗ 𝚥̂

𝜙

𝑃

path of 𝑃 𝚤̂

𝑥

Figure P12.15 and P12.16

Problem 12.16 [ ( ) ] Let 𝑟⃗ = 𝑡 𝚤̂ + 2 + 3𝑡 + 2𝑡2 𝚥̂ ft describe the motion of a point 𝑃 relative to the Cartesian frame of reference shown. Recalling that for any two vectors 𝑝⃗ and 𝑞⃗ we have that 𝑝⃗ ⋅ 𝑞⃗ = |𝑝| ⃗ |𝑞| ⃗ cos 𝛽, where 𝛽 is the angle formed by 𝑝⃗ and 𝑞, ⃗ and recalling that the velocity vector is always tangent to the trajectory, determine the function 𝜙(𝑥) describing the angle between the acceleration vector and the tangent to the path of 𝑃 . 𝑦

Problems 12.17 and 12.18 The motion [ √ of(a point 𝑃 with respect ) ] to a Cartesian coordinate system is described by 𝑟⃗ = 2 𝑡 𝚤̂ + 4 ln(𝑡 + 1) + 2𝑡2 𝚥̂ ft, where 𝑡 denotes time, 𝑡 > 0, and is expressed in seconds.

𝑃 𝜃

𝚥̂ 𝚤̂

Figure P12.17 and P12.18

ISTUDY

𝑥

Problem 12.17 Determine the angle 𝜃 formed by the tangent to the path and the horizontal direction at 𝑡 = 3 s. Problem 12.18 Determine the average acceleration of 𝑃 between times 𝑡1 = 4 s and 𝑡2 = 6 s, and find the difference between it and the true acceleration of 𝑃 at 𝑡 = 5 s.

Problem 12.19 The motion of a stone thrown into a pond is described by [( ) ( ) ] 𝑟⃗(𝑡) = 1.5 − 0.3𝑒−13.6𝑡 𝚤̂ + 0.094𝑒−13.6𝑡 − 0.094 − 0.72𝑡 𝚥̂ m, where 𝑡 is time expressed in seconds, and 𝑡 = 0 s is the time when the stone first hits the water. Determine the stone’s velocity and acceleration. In addition, find the initial angle

ISTUDY

Section 12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

663

of impact 𝜃 of the stone with the water, i.e., the angle formed by the stone’s trajectory and the horizontal direction at 𝑡 = 0. 𝑦 𝚥̂ 𝚤̂

𝑥 𝜃

Figure P12.19

Problem 12.20 As part of a mechanism, a peg 𝑃 is made to slide within a rectilinear guide with the following prescribed motion: [ ] 𝑟⃗(𝑡) = 𝑥0 sin(2𝜋𝜔𝑡) − 3 sin(𝜋𝜔𝑡) 𝚤̂, where 𝑡 denotes time in seconds, 𝑥0 = 1.2 in., and 𝜔 = 0.5 rad∕s. Determine the displacement and the distance traveled over the time interval 0 ≤ 𝑡 ≤ 4 s. In addition, determine the corresponding average velocity and average speed. Express displacement and distance traveled in ft, and express velocity and speed in ft∕s. You may find useful the following trigonometric identity: cos(2𝛽) = 2 cos2 𝛽 − 1.

Problems 12.21 and 12.22

The position of point 𝑃 as a function of time 𝑡, 𝑡 ≥ 0 and expressed in seconds, is [ ] 𝑟⃗(𝑡) = 2.0 [0.5 + sin(𝜔𝑡)] 𝚤̂ + 9.5 + 10.5 sin(𝜔𝑡) + 4.0 sin2 (𝜔𝑡) 𝚥̂,

𝑦 𝚥̂

𝑃 𝚤̂

𝑥

𝑂 Figure P12.20

𝑦

where 𝜔 = 1.3 rad∕s and the position is measured in meters. Find the trajectory of 𝑃 in Cartesian components and then, using the 𝑥 component of 𝑟⃗(𝑡), find the maximum and minimum values of 𝑥 reached by 𝑃 . The equation for the trajectory is valid for all values of 𝑥, yet the maximum and minimum values of 𝑥 as given by the 𝑥 component of 𝑟⃗(𝑡) are finite. What is the origin of this discrepancy? Problem 12.21

𝑃 𝑥

Problem 12.22

(a) Plot the trajectory of 𝑃 for 0 ≤ 𝑡 ≤ 0.6 s, 0 ≤ 𝑡 ≤ 1.4 s, 0 ≤ 𝑡 ≤ 2.3 s, and 0 ≤ 𝑡 ≤ 5 s.

(b) Plot the 𝑦(𝑥) trajectory for −10 m ≤ 𝑥 ≤ 10 m.

(c) You will notice that the trajectory found in (b) does not agree with any of those found in (a). Explain this discrepancy by analytically determining the minimum and maximum values of 𝑥 reached by 𝑃 . As you look at this sequence of plots, why does the trajectory change between some times and not others?

Problems 12.23 through 12.25 A bicycle is moving to the right at a speed 𝑣0 = 20 mph on a horizontal and straight road. The radius of the bicycle’s wheels is 𝑅 = 1.15 f t. Let 𝑃 be a point on the periphery of

Figure P12.21 and P12.22

664

Chapter 12

Particle Kinematics

the front wheel. One can show that the 𝑥 and 𝑦 coordinates of 𝑃 are described by the following functions of time: [ ] 𝑥(𝑡) = 𝑣0 𝑡 + 𝑅 sin(𝑣0 𝑡∕𝑅) and 𝑦(𝑡) = 𝑅 1 + cos(𝑣0 𝑡∕𝑅) . 𝑦

𝑃 𝑣0 𝐶 𝑥 Figure P12.23–P12.25 Problem 12.23

Determine the expressions for the velocity, speed, and acceleration of 𝑃 as functions of time. Determine the maximum and minimum speed achieved by 𝑃 , as well as the 𝑦 coordinate of 𝑃 when the maximum and minimum speeds are achieved. Finally, compute the acceleration of 𝑃 when 𝑃 achieves its maximum and minimum speeds.

Problem 12.24

Problem 12.25 Plot the trajectory of 𝑃 for 0 ≤ 𝑡 ≤ 1 s. For the same time interval, plot the speed as a function of time, as well as the components of the velocity and acceleration of 𝑃 .

Problem 12.26 Find the 𝑥 and 𝑦 components of the acceleration in Example 12.3 (except for the plots) by simply differentiating Eqs. (4) and (5) with respect to time. Verify that you get the results given in Example 12.3. 𝑦

Problem 12.27 Find the 𝑥 and 𝑦 components of the acceleration in Example 12.3 (except for the plots) by differentiating the first of Eqs. (3) and the last of Eqs. (1) with respect to time and then solving the resulting two equations for 𝑥̈ and 𝑦. ̈ Verify that you get the results given in Example 12.3.

𝜌 𝐶

𝑣 𝜃 𝐴

𝚥̂

Problem 12.28

𝚤̂

𝑥

𝑂 Figure P12.28

where 𝑥𝐶 = 0 and 𝑦𝐶 = 1500 f t are the coordinates of the center of the loop. If the plane were capable of maintaining its speed constant and equal to 𝑣0 = 160 mph, determine the velocity and acceleration of the plane for 𝜃 = 30◦ .

𝑦

path of 𝐴

𝚤̂

Figure P12.29

ISTUDY

Problem 12.29 𝐴

𝚥̂ 𝑂

Airplane 𝐴 is performing a loop with constant radius 𝜌 = 1000 f t. The equation describing the loop is as follows: ( )2 ( )2 𝑥 − 𝑥𝐶 + 𝑦 − 𝑦𝐶 = 𝜌2 ,

𝑣0 𝑥

An airplane 𝐴 takes off as shown with a constant speed equal to 𝑣0 = 160 km∕h. The path of the airplane is described by the equation 𝑦 = 𝜅𝑥2 , where 𝜅 = 6×10−4 m−1 . Using the component system shown, provide the expression for the velocity and acceleration of the airplane when 𝑥 = 400 m. Express the velocity in m∕s and the acceleration in m∕s2 .

ISTUDY

Section 12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

665

Problem 12.30 A test track for automobiles has a portion with a specific profile described by: [ ] 𝑦 = ℎ 1 − sin(𝑥∕𝑤) , where ℎ = 0.5 f t and 𝑤 = 8 f t, and where the argument of the sine function is understood to be in radians. A car travels in the positive 𝑥 direction such that the horizontal component of velocity remains constant and equal to 55 mph. Modeling the car as a point moving along the given profile, determine the maximum speed of the car. Express your answer in ft∕s. 𝑦 𝚥̂ 𝑥

𝚤̂ Figure P12.30–P12.32

Problem 12.31 A test track for automobiles has a portion with a specific profile described by: [ ] 𝑦 = ℎ 1 − cos(𝑥∕𝑤) , where ℎ = 0.20 m and 𝑤 = 2 m, and where the argument of the cosine function is understood to be in radians. A car travels in the positive 𝑥 direction with a constant 𝑥 component of velocity equal to 100 km∕h. Modeling the car as a point moving along the given profile, determine the velocity and acceleration (expressed in m∕s and m∕s2 , respectively) of the car for 𝑥 = 24 m.

Problem 12.32 A test track for automobiles has a portion with a specific profile described by: [ ] 𝑦 = ℎ 1 − cos(𝑥∕𝑤) , where ℎ = 0.75 f t and 𝑤 = 10 f t, and where the argument of the cosine function is understood to be in radians. A car drives at a constant speed 𝑣0 = 35 mph. Modeling the car as a point moving along the given profile, find the velocity and acceleration of the car for 𝑥 = 97 f t. Express velocity in ft∕s and acceleration in ft∕s2 .

Problem 12.33 The orbit of a satellite 𝐴 around planet 𝐵 is the ellipse shown and is described by the equation (𝑥∕𝑎)2 + (𝑦∕𝑏)2 = 1, where 𝑎 and 𝑏 are the semimajor and semiminor axes of the ellipse, respectively. When 𝑥 = 𝑎∕2 and 𝑦 > 0, the satellite is moving with a speed 𝑣0 as shown. Determine the expression for the satellite’s velocity 𝑣⃗ in terms of 𝑣0 , 𝑎, and 𝑏 for 𝑥 = 𝑎∕2 and 𝑦 > 0.

𝑦 𝑣0 𝐴

𝚥̂

𝑏

𝑂

𝐵

𝑏

𝑥

𝚤̂

𝑎

𝑎

Figure P12.33 𝑦

follower 𝐴

Problems 12.34 and 12.35 In the mechanism shown, block 𝐵 is fixed and has a profile described by the following relation: [ ( )2 ( )4 ] 1 𝑥 1 𝑥 − . 𝑦=ℎ 1+ 2 𝑑 4 𝑑 The follower moves with the shuttle 𝐴, and the tip 𝐶 of the follower remains in contact with 𝐵.

𝐶

𝚥̂

𝐵 ℎ 𝑑

𝑂

𝚤̂

𝑥 𝑑

Figure P12.34 and P12.35

666

Chapter 12

Particle Kinematics

Problem 12.34 Assume that ℎ = 0.25 in., 𝑑 = 1 in., and the horizontal position of 𝐶 is 𝑥 = 𝑑 sin(𝜔𝑡), where 𝜔 = 2𝜋 rad∕s, and 𝑡 is time in seconds. Determine an analytical expression for the speed of 𝐶 as a function of 𝑥 and the parameters 𝑑, ℎ, and 𝜔. Then, evaluate the speed of 𝐶 for 𝑥 = 0, 𝑥 = 0.5 in., and 𝑥 = 1 in. Express your answers in ft∕s. Problem 12.35 Assume that ℎ = 2 mm, 𝑑 = 20 mm, and 𝐴 is made to move from 𝑥 = −𝑑 to 𝑥 = 𝑑 with a constant speed 𝑣0 = 0.1 m∕s. Determine the acceleration of 𝐶 for 𝑥 = 15 mm. Express your answer in m∕s2 .

Problem 12.36 The Center for Gravitational Biology Research at NASA’s Ames Research Center runs a large centrifuge capable of 20𝑔 of acceleration, where 𝑔 is the acceleration due to gravity (12.5𝑔 is the maximum for human subjects). The distance from the axis of rotation to the ( )1∕2 cab at either 𝐴 or 𝐵 is 𝑅 = 25 f t. The trajectory of 𝐴 is described by 𝑦𝐴 = 𝑅2 − 𝑥2𝐴 )1∕2 ( for 𝑦𝐴 ≥ 0 and by 𝑦𝐴 = − 𝑅2 − 𝑥2𝐴 for 𝑦𝐴 < 0. If 𝐴 moves at the constant speed 𝑣𝐴 = 120 f t∕s, determine the velocity and acceleration of 𝐴 when 𝑥𝐴 = −20 f t and 𝑦𝐴 > 0. 𝑦 𝑣𝐴

𝐴

𝚥̂ 𝑂

𝐴 𝑥

𝐵

𝚤̂

𝑥

𝑅 𝐵

NASA

Figure P12.36

Problems 12.37 and 12.38 Point 𝐶 is a point on the connecting rod of a mechanism called a slider-crank. The 𝑥 and ( )1∕2 𝑦 coordinates of 𝐶 can be expressed as follows: 𝑥𝐶 = 𝑅 cos 𝜃 + 21 𝐿2 − 𝑅2 sin2 𝜃 and 𝑦𝐶 = (𝑅∕2) sin 𝜃, where 𝜃 describes the position of the crank. The crank rotates at a constant rate such that 𝜃 = 𝜔𝑡, where 𝑡 is time.

𝐿 𝐶 𝜃 𝑦 𝑅

Figure P12.37 and P12.38

ISTUDY

Problem 12.37 Find expressions for the velocity, speed, and acceleration of 𝐶 as functions of the angle 𝜃 and the parameters, 𝑅, 𝐿, and 𝜔. Problem 12.38 Let 𝑡 be expressed in seconds, 𝑅 = 0.1 m, 𝐿 = 0.25 m, and 𝜔 = 250 rad∕s. Plot the trajectory of point 𝐶 for 0 ≤ 𝑡 ≤ 0.025 s. For the same interval of time, plot the speed as a function of time, as well as the components of the velocity and acceleration of 𝐶.

Problem 12.39 A classic problem in vibrations is the single degree-of-freedom damped vibration problem in which the one-dimensional amplitude of an oscillating particle is described by: 𝑥(𝑡) = 𝐴𝑒−𝛼𝑡 cos 𝛽𝑡 where 𝐴 is the initial amplitude and 𝛼 and 𝛽 describe the strength of the decay and the frequency of oscillation, respectively. For 𝐴 = 0.1 m, 𝛼 = 0.1 s−1 and 𝛽 = 0.2 s−1 , at what time does the velocity first go to zero? What is the value of the acceleration at this instant?

ISTUDY

Section 12.1

Position, Velocity, Acceleration, and Cartesian Coordinates

Problem 12.40 A particle is sliding along a frictionless surface with varying elevation, 𝑦(𝑡). In Chapter 14, we will see that the relationship between the particle’s speed, 𝑣(𝑡), and its elevation, 𝑦(𝑡), can be expressed as: 1 2 𝑚𝑣 (𝑡) + 𝑚𝑔𝑦(𝑡) = 𝐶, 2 where 𝑚 is the particle’s mass and 𝐶 is[ a constant. Suppose the particle’s position as it ] navigates a hill can be described by 𝑟⃗ = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂ m, where the elevation varies with 𝑥(𝑡) according to: 𝐴 𝑦(𝑡) = [ ]2 . 1 + 𝑥(𝑡)∕𝑎 In this expression for 𝑦(𝑡), 𝐴 represents the height of the hill and parameter 𝑎 characterizes the severity of the grade. For 𝐴 = 10 m and 𝑎 = 2 m, what is the rate of change of the particle’s speed at the instant when the particle has passed over the hill and fallen to an elevation of 5 m?

Figure P12.40

667

668

Chapter 12

Particle Kinematics

12.2

One-Dimensional Motion

We now look at motions that can be completely described by a single scalar coordinate. These motions can be in a straight line or circular. In all cases, the sign on the position, velocity, and acceleration will tell us the direction of the corresponding quantity, so we are essentially working with one-dimensional vectors. NICOLAS ASFOURI/AFP/Getty Images

Figure 12.10 Usain Bolt winning the gold medal in the 100 m sprint at the 2008 Beijing Olympic Games. In this race, each runner is in rectilinear motion.

𝑠=0

𝑃 𝑠(𝑡)

𝑣(𝑡) 𝑠 𝑎(𝑡)

Figure 12.11 The position coordinate 𝑠(𝑡) for rectilinear motion.

ISTUDY

Rectilinear motion relations Rectilinear motion is motion that occurs along a straight line, though the relations that govern it are useful for describing other types of one-dimensional motions, even when the trajectory is not a straight line (see circular motion on p. 670). Referring to Fig. 12.11, we denote by 𝑠(𝑡) the rectilinear position coordinate of a point 𝑃 . The velocity and acceleration of 𝑃 will be denoted by 𝑣(𝑡) = 𝑠(𝑡) ̇ and 𝑎(𝑡) = 𝑠(𝑡), ̈ respectively. Since physical measurements and Newton’s second law usually provide us with data of the form 𝑎(𝑡), 𝑎(𝑣), or 𝑎(𝑠), we now investigate how to relate acceleration information of these types to velocity and position for rectilinear motions. For each of the three cases, we want to emphasize the process used to obtain the final result and not the final result itself. If 𝒂(𝒕) is known If we know the acceleration as a function of time 𝑎(𝑡), then we can determine 𝑣(𝑡) and 𝑠(𝑡) by integration in time. In this case, our only option is to write 𝑎(𝑡) = 𝑑𝑣∕𝑑𝑡 as 𝑑𝑣 = 𝑎(𝑡) 𝑑𝑡, and letting 𝑣 = 𝑣0 for 𝑡 = 𝑡0 , we have

Interesting Fact Relation to differential equations. If you have had a course in differential equations, you may recognize that the equation 𝑎(𝑡) = 𝑑𝑣∕𝑑𝑡 is actually a first-order differential equation. This equation is separable in that everything involving 𝑣 can be put on one side of the equals sign and everything involving 𝑡 can be put on the other side. This whole section is really about solving separable first-order differential equations.

∫𝑣

𝑣

𝑑𝑣 =

0

∫𝑡

𝑡

𝑎(𝑡) 𝑑𝑡,

(12.19)

0

or 𝑣(𝑡) = 𝑣0 +

∫𝑡

𝑡

𝑎(𝑡) 𝑑𝑡.

(12.20)

0

Rewriting 𝑣(𝑡) = 𝑑𝑠∕𝑑𝑡 as 𝑑𝑠 = 𝑣(𝑡) 𝑑𝑡 and letting 𝑠 = 𝑠0 for 𝑡 = 𝑡0 , we can determine 𝑠(𝑡) from Eq. (12.20) as follows: ∫𝑠

𝑠

𝑑𝑠 =

0

∫𝑡

𝑡

𝑣(𝑡) 𝑑𝑡 = 0

∫𝑡

𝑡

[ 𝑣0 +

0

or 𝑠(𝑡) = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) +

∫𝑡

𝑡 0

[

∫𝑡

∫𝑡

𝑡

]

𝑡

𝑎(𝑡) 𝑑𝑡 𝑑𝑡,

(12.21)

0

] 𝑎(𝑡) 𝑑𝑡 𝑑𝑡.

(12.22)

0

In summary, if acceleration is a function of time, then we integrate 𝑎(𝑡) = 𝑑𝑣∕𝑑𝑡 to get 𝑣(𝑡) and then 𝑣(𝑡) = 𝑑𝑠∕𝑑𝑡 to obtain 𝑠(𝑡). If 𝒂(𝒗) is known If we know the acceleration as a function of velocity 𝑎(𝑣), then finding 𝑣 and 𝑠 by integration is slightly more complicated and involves making one of two choices.

ISTUDY

Section 12.2

One-Dimensional Motion

669

Which choice you make will depend on what you are trying to find. Starting from 𝑎 = 𝑑𝑣∕𝑑𝑡, when 𝑎(𝑣) is given, we can separate the 𝑣 and 𝑡 variables by writing 𝑑𝑡 =

𝑑𝑣 . 𝑎(𝑣)

(12.23)

If 𝑎(𝑣) ≠ 0 during the time interval considered, we can integrate Eq. (12.23) as follows: ∫𝑡

𝑡

𝑑𝑡 = 0

∫𝑣

𝑣

1 𝑑𝑣 𝑎(𝑣)

0

𝑣

⇒

𝑡(𝑣) = 𝑡0 +

1 𝑑𝑣, ∫𝑣 𝑎(𝑣) 0

(12.24)

where 𝑣0 is the value of 𝑣 for 𝑡 = 𝑡0 . While the last of Eqs. (12.24) gives time as a function of velocity, sometimes we can invert 𝑡(𝑣) to find 𝑣(𝑡). To compute the position we can try to solve the last of Eqs. (12.24) for 𝑣(𝑡). Then we can try to integrate 𝑣(𝑡) with respect to 𝑡 to obtain 𝑠(𝑡) as was done in Eq. (12.22). Unfortunately, this is often difficult or impossible to do. Alternatively, we can obtain 𝑠(𝑣) instead of 𝑠(𝑡), using the chain rule of calculus, i.e., 𝑎=

𝑑𝑣 𝑑𝑣 𝑑𝑠 𝑑𝑣 = =𝑣 . 𝑑𝑡 𝑑𝑠 𝑑𝑡 𝑑𝑠

(12.25)

Then, recalling that we have 𝑎(𝑣), we can separate the variables 𝑠 and 𝑣 as follows: 𝑑𝑠 =

𝑣 𝑑𝑣. 𝑎(𝑣)

(12.26)

Setting 𝑠 = 𝑠0 when 𝑣 = 𝑣0 , Eq. (12.26) can be integrated to obtain 𝑠

∫𝑠

𝑑𝑠 =

0

∫𝑣

𝑣

𝑣 𝑑𝑣 𝑎(𝑣)

0

𝑣

⇒

𝑠(𝑣) = 𝑠0 +

𝑣 𝑑𝑣. ∫𝑣 𝑎(𝑣) 0

(12.27)

In summary, if 𝑎(𝑣) is known, we can integrate in time by separating variables in 𝑎 = 𝑑𝑣∕𝑑𝑡 or we can integrate in space by letting 𝑎 = 𝑣 𝑑𝑣∕𝑑𝑠. The former results in 𝑡(𝑣) and the latter in 𝑠(𝑣). If 𝒂(𝒔) is known When the acceleration is known as a function of position, i.e., 𝑎 = 𝑎(𝑠), our only option is to start from 𝑎 = 𝑣 𝑑𝑣∕𝑑𝑠 given in Eq. (12.25). This time the correct separation of variables gives 𝑣 𝑑𝑣 = 𝑎(𝑠) 𝑑𝑠. Letting 𝑣 = 𝑣0 for 𝑠 = 𝑠0 , we can obtain the velocity as a function of position 𝑣(𝑠) as follows: ∫𝑣

𝑣

0

𝑣 𝑑𝑣 =

∫𝑠

𝑠

𝑎(𝑠) 𝑑𝑠

⇒

1 2 𝑣 2

− 12 𝑣20 =

0

∫𝑠

𝑠

𝑎(𝑠) 𝑑𝑠,

(12.28)

0

or 𝑣2 (𝑠) = 𝑣20 + 2

∫𝑠

𝑠

𝑎(𝑠) 𝑑𝑠.

(12.29)

0

Finally, once 𝑣(𝑠) is known through Eq. (12.29), we can obtain time as a function of position 𝑡(𝑠) starting from 𝑣 = 𝑑𝑠∕𝑑𝑡. Then, recalling that we have 𝑣(𝑠), we can separate the variables 𝑡 and 𝑠 as: 𝑑𝑡 =

𝑑𝑠 𝑣(𝑠)

⇒

∫𝑡

𝑡

𝑠

𝑑𝑡 = 0

𝑑𝑠 , ∫𝑠 𝑣(𝑠) 0

(12.30)

Helpful Information The chain rule. Since it is often used, let’s look more closely at the chain rule. Taking a bit of liberty, the chain rule can be presented as follows: 𝑑(Adam) 𝑑(Ryan) 𝑑(Adam) = . 𝑑(Jack) 𝑑(Ryan) 𝑑(Jack) Although Ryan did not appear on the lefthand side of the equation above, we were able to force him to appear on the righthand side. This chain-rule-based technique will come in handy. The reason for using the brothers from the band AJR in this example is that the chain rule works only if all its terms are related to one another (mathematically, they must be functions of one another). Now, the connection with Eq. (12.25) is that we needed to make the variable 𝑠 come into the picture even though, at first, it was not present in 𝑎 = 𝑑𝑣∕𝑑𝑡. Therefore, we made 𝑣 pose as our Adam and 𝑡 as our Jack. Letting 𝑠 take on the role of Ryan, we were able to accomplish what we wanted.

670

Chapter 12

Particle Kinematics

where we have again let 𝑠 = 𝑠0 when 𝑡 = 𝑡0 . Completing the integration of the lefthand side, we obtain 𝑠 𝑑𝑠 . 𝑡(𝑠) = 𝑡0 + (12.31) ∫𝑠 𝑣(𝑠) 0 In summary, if 𝑎(𝑠) is known, we apply the chain rule to obtain 𝑎(𝑠) = 𝑣 𝑑𝑣∕𝑑𝑠. Separating variables and integrating results in 𝑣(𝑠). Then letting 𝑣(𝑠) = 𝑑𝑠∕𝑑𝑡, separating variables, and then integrating results in 𝑡(𝑠). What if 𝒂 is constant? If the acceleration is a constant, then the equations we have derived simplify substantially. The constant acceleration relations are important because there are many problems in dynamics in which the acceleration is constant. For example, in studying the motion of a projectile, we generally assume that the projectile’s acceleration is constant. If the acceleration is a constant 𝑎𝑐 , Eq. (12.20) becomes ( ) 𝑣 = 𝑣0 + 𝑎𝑐 𝑡 − 𝑡0 (constant acceleration),

(12.32)

Eq. (12.22) becomes ( ) ( )2 𝑠 = 𝑠0 + 𝑣0 𝑡 − 𝑡0 + 12 𝑎𝑐 𝑡 − 𝑡0

(constant acceleration),

(12.33)

and Eq. (12.29) becomes ( ) (constant acceleration). 𝑣2 = 𝑣20 + 2𝑎𝑐 𝑠 − 𝑠0

(12.34)

Circular motion and angular velocity 𝑣⃗𝐴 (𝑡)

𝑣⃗𝐴 (𝑡 + 𝛥𝑡)

𝐴 𝛥𝜃 𝑟 𝑂

𝑠

𝜃

fixed line, 𝜃 = 0, 𝑠 = 0

Figure 12.12 Particle 𝐴 with speed 𝑣𝐴 = |𝑣⃗𝐴 | moving in a circle of radius 𝑟 centered at 𝑂.

ISTUDY

The relationships for rectilinear motion are applicable to any one-dimensional motion. To demonstrate this idea, we will now apply them to a common one-dimensional curvilinear motion: circular motion. In Fig. 12.12, a particle 𝐴 is moving in a circle of radius 𝑟 and center 𝑂. Since 𝑟 is constant, the position of 𝐴 can be described with a single coordinate, such as the oriented arc length 𝑠 or the angle 𝜃. If the line 𝑂𝐴 rotates through the angle Δ𝜃 in the time Δ𝑡, then we can define an average time rate of change of the angle 𝜃 as 𝜔avg = Δ𝜃∕Δ𝑡. Letting Δ𝑡 → 0, we obtain the instantaneous time rate of change of ̇ called the angular velocity, as 𝜃, i.e., 𝜃, 𝜔(𝑡) = lim

Δ𝑡→0

𝑑𝜃(𝑡) Δ𝜃 ̇ = = 𝜃(𝑡). Δ𝑡 𝑑𝑡

(12.35)

We can then define angular acceleration 𝛼 by differentiating Eq. (12.35) with respect to time, i.e., 𝑑𝜔(𝑡) ̈ 𝛼(𝑡) = (12.36) = 𝜔(𝑡) ̇ = 𝜃(𝑡). 𝑑𝑡 When using the coordinate 𝑠, since 𝑠 = 𝑟𝜃 and 𝑟 is constant, we can write 𝑠̇ = 𝑣 = 𝑟𝜃̇ = 𝜔𝑟 and where 𝑣 is the speed of the particle.

𝑠̈ = 𝑣̇ = 𝑟𝜃̈ = 𝛼𝑟,

(12.37)

ISTUDY

Section 12.2

One-Dimensional Motion

671

Circular motion relations All of the relationships we developed for rectilinear motion apply equally well to circular motion, except that we need to replace the rectilinear variables with their circular counterparts. For example, Eq. (12.20) becomes 𝜔(𝑡) = 𝜔0 +

∫𝑡

𝑡

𝛼(𝑡) 𝑑𝑡.

(12.38)

0

Table 12.1 lists each kinematic variable in rectilinear motion and the corresponding kinematic variable for circular motion. Replacing each rectilinear motion variable with its circular motion counterpart in Eqs. (12.20)–(12.34), we obtain the corresponding circular motion equations. Finally, if the angular acceleration is constant, we can use the constant acceleration relations with 𝑎𝑐 replaced by 𝛼𝑐 .

End of Section Summary In this section we have developed relationships that link a single coordinate and its time derivatives. Note that, except for the constant acceleration relations, they are rarely applied directly, and the acceleration is integrated as was done in the development of these equations. 1. If the acceleration is provided as a function of time, i.e., 𝑎 = 𝑎(𝑡), for velocity and position, we have Eqs. (12.20) and (12.22), p. 668 𝑣(𝑡) = 𝑣0 +

∫𝑡

𝑡

𝑎(𝑡) 𝑑𝑡, 0

𝑠(𝑡) = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) +

∫𝑡

𝑡 0

[

∫𝑡

]

𝑡

𝑎(𝑡) 𝑑𝑡 𝑑𝑡. 0

2. If the acceleration is provided as a function of velocity, i.e., 𝑎 = 𝑎(𝑣), for time and position, we have Eqs. (12.24) and (12.27), p. 669 𝑣

𝑡(𝑣) = 𝑡0 +

1 𝑑𝑣, ∫𝑣 𝑎(𝑣) 0 𝑣

𝑠(𝑣) = 𝑠0 +

𝑣 𝑑𝑣. ∫𝑣 𝑎(𝑣) 0

3. If the acceleration is provided as a function of position, i.e., 𝑎 = 𝑎(𝑠), for velocity and time, we have Eq. (12.29), p. 669, and Eq. (12.31), p. 670 𝑣2 (𝑠) = 𝑣20 + 2

∫𝑠

𝑠

𝑎(𝑠) 𝑑𝑠,

0

𝑠

𝑡(𝑠) = 𝑡0 +

𝑑𝑠 . ∫𝑠 𝑣(𝑠) 0

Table 12.1 Correspondence of kinematic variables between rectilinear and circular motion. Kinematic variable time position velocity acceleration a Except

Rectilinear motion 𝑡 𝑠 𝑣 𝑎

Circular motiona 𝑡 𝜃 𝜔 𝛼

for time, each of these should have the word angular in front of its kinematic variable name.

672

ISTUDY

Chapter 12

Particle Kinematics

4. If the acceleration is a constant 𝑎𝑐 , for velocity and position, we have Eqs. (12.32)–(12.34), p. 670 𝑣 = 𝑣0 + 𝑎𝑐 (𝑡 − 𝑡0 ), 𝑠 = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) + 12 𝑎𝑐 (𝑡 − 𝑡0 )2 , 𝑣2 = 𝑣20 + 2𝑎𝑐 (𝑠 − 𝑠0 ). Circular motion. For circular motion, the equations summarized in items 1–4 above hold as long as we use the replacement rules 𝑠 → 𝜃,

𝑣 → 𝜔,

𝑎 → 𝛼,

where 𝜔 = 𝜃̇ and 𝛼 = 𝜃̈ are the angular velocity and angular acceleration, respectively.

ISTUDY

Section 12.2

One-Dimensional Motion

E X A M P L E 12.4

Acceleration as a Function of Time

A rocket sled accelerates from rest along a straight rail (Fig. 1). The sled’s acceleration is 𝑎 = 𝛽𝑡2 , where 𝛽 is a constant and 𝑡 is time in seconds. The sled achieves a final speed 𝑣𝑓 = 180 mph after traveling a distance 𝑑 = 300 f t. Determine 𝛽 and the time required to achieve 𝑣𝑓 .

SOLUTION The acceleration 𝑎 is given as a function of time and needs to be related to speed and position. The speed is obtained from the velocity, which is obtained by integrating 𝑎 with respect to time. Once the velocity is known as a function of time, the position is obtained by integrating the velocity with respect to time.

Road Map

Since 𝑎 = 𝑑𝑣∕𝑑𝑡, where 𝑣 denotes the velocity, and since the acceleration is a known function of time, we can separate the variables 𝑣 and 𝑡 as follows: 𝑑𝑣 = 𝑎(𝑡) 𝑑𝑡. Using the given expression for 𝑎(𝑡), we can then write

Computation

𝑑𝑣 = 𝛽𝑡2 𝑑𝑡

∫0

⇒

𝑣

𝑑𝑣 =

∫0

𝑡

𝛽𝑡2 𝑑𝑡

𝑣 = 31 𝛽𝑡3 ,

⇒

(1)

where the limits of integration reflect that 𝑣 = 0 for 𝑡 = 0 (the sled starts from rest). Next we recall that 𝑣 = 𝑑𝑠∕𝑑𝑡, where, referring to Fig. 1, 𝑠 denotes the position. Hence, we can write 𝑑𝑠 = 𝑣 𝑑𝑡, which, using the last of Eqs. (1) gives 𝑑𝑠 = 31 𝛽𝑡3 𝑑𝑡.

(2)

Letting 𝑠0 denote the position at time 𝑡 = 0, we can integrate Eq. (2) as follows: ∫𝑠

𝑠

0

𝑑𝑠 =

∫0

673

𝑡

1 3 𝛽𝑡 3

𝑑𝑡

⇒

𝑠 − 𝑠0 =

1 𝛽𝑡4 . 12

(3)

Let 𝑡𝑓 denote the time at which 𝑣 = 𝑣𝑓 and 𝑠 − 𝑠0 = 𝑑. Then, for 𝑡 = 𝑡𝑓 , the last of Eqs. (1) and the last of Eqs. (3) become 𝑣𝑓 = 13 𝛽𝑡3𝑓

and 𝑑 =

1 𝛽𝑡4 . 12 𝑓

(4)

Equations (4) form a system of two equations in the two unknowns 𝛽 and 𝑡𝑓 , whose solution is 3𝑣4𝑓 𝑑 and 𝑡𝑓 = 4 . (5) 𝛽= 𝑣𝑓 64𝑑 3 Recalling that 𝑑 = 300 f t and 𝑣𝑓 = 180 mph = 264.0 f t∕s, we can evaluate the expressions in Eqs. (5) to obtain 𝛽 = 8.433 f t∕s4

and 𝑡𝑓 = 4.545 s.

(6)

Discussion & Verification Since 𝑎 = 𝛽𝑡2 , 𝛽 has dimensions of length over time to the fourth power. Also, 𝑡𝑓 has dimensions of time. Hence, our results have the correct dimensions and units. To check whether or not the numerical results are reasonable, we can find the time 𝑡𝑓 under constant acceleration conditions and then compare this result to what we have obtained. Under constant acceleration, 𝑡𝑓 = 2𝑑∕𝑣𝑓 , i.e., half of what we obtained. This is consistent with the fact that under constant acceleration the velocity increases linearly with time, whereas in our problem the velocity increases cubically. Therefore, our result can be considered acceptable. In turn, assuming that we have not made any algebra mistakes, we can say that the corresponding result for 𝛽 is also acceptable.

𝑎 𝑠 Figure 1 Rocket sled on rectilinear rail.

674

Chapter 12

Particle Kinematics

E X A M P L E 12.5

Acceleration as a Function of Velocity

𝑠=0

𝑎

A sphere is dropped from rest at 𝑠 = 0 in a thick polymer fluid (say, shampoo). A typical model for the motion of a sphere in such a fluid tells us that the acceleration of the sphere has the form 𝑎 = 𝑔 − 𝜂𝑣, where 𝑔 is the acceleration due to gravity, 𝑣 is the velocity, and 𝜂 is a constant coefficient with dimensions of one over time. If 𝜂 = 30 s−1 , determine the time it takes for the velocity of the sphere to become equal to 1 f t∕s. Also, determine the corresponding position of the sphere.

𝑠

SOLUTION Figure 1 A sphere sinking in a beaker filled with a polymer fluid.

ISTUDY

Road Map

To relate a change in velocity to a corresponding time interval, we will apply the definition of acceleration, since, in such a definition, velocity and time appear directly. To relate a change in position to a corresponding change in velocity, we rewrite the acceleration in such a way that the expression for the acceleration is in terms of velocity and position. This can be done by applying the chain rule. Computation Recalling that 𝑎 = 𝑑𝑣∕𝑑𝑡, since the acceleration is a given function of velocity, we can separate the variable 𝑣 and 𝑡 as follows: 𝑑𝑡 = 𝑑𝑣∕𝑎(𝑣). Then, using the given expression for the acceleration, we have

𝑑𝑡 =

𝑑𝑣 . 𝑔 − 𝜂𝑣

(1)

Letting 𝑡𝑓 be the time at which 𝑣 = 𝑣𝑓 = 1 f t∕s, and recalling that 𝑣 = 0 for 𝑡 = 0, we can integrate Eq. (1) as follows: ∫0

𝑡𝑓

𝑑𝑡 =

∫0

𝑣𝑓

𝑑𝑣 𝑔 − 𝜂𝑣

⇒

𝑡𝑓 = −

] 1[ ln(𝑔 − 𝜂𝑣𝑓 ) − ln 𝑔 . 𝜂

(2)

Recalling that 𝜂 = 30 s−1 , 𝑔 = 32.2 f t∕s2 , and 𝑣𝑓 = 1 f t∕s (the subscript 𝑓 stands for final), we can evaluate the last of Eqs. (2) to obtain 𝑡𝑓 = 0.08945 s.

(3)

To determine the position of the sphere corresponding to 𝑣 = 𝑣𝑓 , we use the chain rule to express the acceleration as 𝑎 = 𝑣𝑑𝑣∕𝑑𝑠, which allows us to write ( ) 𝑔∕𝜂 𝑑𝑣 𝑣 1 𝑔 − 𝜂𝑣 = 𝑣 ⇒ 𝑑𝑠 = 𝑑𝑣 ⇒ 𝑑𝑠 = − + 𝑑𝑣, (4) 𝑑𝑠 𝑔 − 𝜂𝑣 𝜂 𝑔 − 𝜂𝑣 where in the second of Eqs. (4) we have separated the variables 𝑠 and 𝑣, and in the last of Eqs. (4) we have expressed the function 𝑣∕(𝑔 − 𝜂𝑣) in a way that is convenient for integration. Recalling that 𝑣 = 0 for 𝑠 = 0, and letting 𝑠𝑓 denote the position for 𝑣 = 𝑣𝑓 , we can integrate the last of Eqs. (4) as follows: ) 𝑣𝑓 ( 𝑠𝑓 𝑣𝑓 ] 𝑔∕𝜂 𝑔[ 1 − + 𝑑𝑠 = 𝑑𝑣 ⇒ 𝑠𝑓 = − − 2 ln(𝑔 − 𝜂𝑣𝑓 ) − ln 𝑔 . (5) ∫0 ∫0 𝜂 𝑔 − 𝜂𝑣 𝜂 𝜂 Recalling again that 𝜂 = 30 s−1 , 𝑔 = 32.2 f t∕s2 , and 𝑣𝑓 = 1 f t∕s, we can evaluate the last of Eqs. (5) to obtain 𝑠𝑓 = 0.06268 f t. Discussion & Verification

(6)

Our results have the correct dimensions and units. As for their numerical values, we expect that to achieve the same velocity 𝑣𝑓 = 1 f t∕s, a sphere sinking in a thick fluid will take longer and will need to sink deeper than a sphere falling freely due to gravity. In this latter case, the acceleration of the sphere would be equal to 𝑔 and 𝑡𝑓 and 𝑠𝑓 would be 𝑣𝑓 ∕𝑔 = 0.03106 s and 𝑣2𝑓 ∕(2𝑔) = 0.01553 f t, respectively, which are indeed smaller than the results we have obtained in Eqs. (3) and (6), respectively.

ISTUDY

Section 12.2

E X A M P L E 12.6

One-Dimensional Motion

675

Acceleration as a Function of Position

In a braking test, a car traveling at a speed 𝑣0 = 80 km∕h is made to apply the brakes so as to lock the wheels and slide. The pavement has a roughness that increases as the car moves to the right. Letting 𝑠 = 0 characterize the location at which sliding begins, the pavement’s roughness is such that the acceleration of the car is 𝑎 = −𝑔𝜇0 [1 + (𝑠∕𝜆)], where 𝑔 is the acceleration due to gravity, 𝜇0 = 0.6, and 𝜆 is a constant with dimensions of length. Determine 𝜆 knowing that the car stops after a distance 𝑑 = 30 m.

𝑣0

𝑠 𝑠=0 Figure 1

SOLUTION Road Map We can determine 𝜆 by relating the acceleration to the braking distance. Since the acceleration is given as a function of position, we first apply the chain rule to rewrite the acceleration in terms of velocity and position. Then we will be able to relate the change in velocity to the corresponding change in position, thus obtaining an equation we can solve for 𝜆. Computation By definition, 𝑎 = 𝑑𝑣∕𝑑𝑡, where 𝑣 is the velocity. Hence, making use of the chain rule, we have [ ] 𝑠 𝑑𝑣 𝑑𝑣 𝑑𝑣 𝑑𝑣 𝑑𝑠 =𝑣 , = =𝑣 ⇒ −𝑔𝜇0 1 + (1) 𝑎= 𝑑𝑡 𝑑𝑠 𝑑𝑡 𝑑𝑠 𝜆 𝑑𝑠

where we have used the given expression for 𝑎 and the fact that 𝑣 = 𝑑𝑠∕𝑑𝑡. The last of Eqs. (1) can be rewritten to separate the variables 𝑠 and 𝑣, i.e., [ ] 𝑔𝜇 − 𝑔𝜇0 + 0 𝑠 𝑑𝑠 = 𝑣 𝑑𝑣, (2) 𝜆 where we have expanded the product on the left-hand side of the last of Eqs. (1). We now observe that 𝑠 = 0 when 𝑣 = 𝑣0 and that 𝑠 = 𝑑 when 𝑣 = 0. Hence, the expression in Eq. (2) can be integrated as follows: ] 0 𝑑[ 𝑔𝜇 𝑑 2 𝑔𝜇 𝑔𝜇0 + 0 𝑠 𝑑𝑠 = 𝑣 𝑑𝑣 ⇒ −𝑔𝜇0 𝑑 − 0 = − 12 𝑣20 . (3) − ∫𝑣 ∫0 𝜆 2𝜆 0 The last of Eqs. (3) can be solved for 𝜆 to obtain 𝜆=

𝑔𝜇0 𝑑 2 𝑣20 − 2𝑔𝜇0 𝑑

.

(4)

Recalling that 𝑔 = 9.81 m∕s2 , 𝜇0 = 0.6, 𝑑 = 30 m, and 𝑣0 = 80 km∕h = 22.22 m∕s, we can evaluate the expression in Eq. (4) to obtain 𝜆 = 37.66 m.

(5)

Discussion & Verification The expression on the right-hand side of Eq. (4) has dimensions of length, as it should. To check whether or not the result is acceptable, we observe that Eq. (4) implies that 𝜆 → 0 for 𝑑 → 0 and 𝜆 → ∞ for 𝑑 → 𝑣20 ∕(2𝑔𝜇0 ), where 𝑣20 ∕(2𝑔𝜇0 ) = 41.95 m can be shown to be the stopping distance if the acceleration of the car were constant and equal to −𝑔𝜇0 . Recalling that 𝑎 = −𝑔𝜇0 [1 + (𝑠∕𝜆)], this behavior is consistent with the fact that if 𝜆 → 0, 𝑎 tends to negative infinity and for 𝜆 → ∞, 𝑎 becomes equal to −𝑔𝜇0 . Therefore, overall, we can say that the value of 𝜆 we found is certainly within the range of admissible values for the parameter in question.

676

Chapter 12

Particle Kinematics

E X A M P L E 12.7

Measuring the Depth of a Well by Relating Time, Velocity, and Acceleration We can estimate the depth of a well by measuring the time it takes for a rock dropped from the top of the well to reach the water below. Assuming that gravity is the only force acting on the rock, estimate a well’s depth under two different assumptions: the speed of sound is (a) finite and equal to 𝑣𝑠 = 340 m∕s and (b) infinite. Also, compare the two estimates to provide a “rule of thumb” as to when we can assume that the speed of sound is infinite.

SOLUTION 𝑠

Road Map 𝐷

Figure 1 A well of depth 𝐷 showing the positive direction of the coordinate 𝑠.

ISTUDY

In this problem our time measure is the sum of two parts: (1) the time taken by the rock to go from the top to the bottom of the well and (2) the time taken by sound to go from the bottom to the top of the well. The well’s depth can be related to the first time by assuming that the rock travels at a constant acceleration, namely, 𝑔 = 9.81 m∕s2 . The well’s depth can also be related to the second time by assuming that sound travels at a constant speed, which will be assumed to be finite in Part (a) and infinite in Part (b). By requiring that the two depth estimates be identical, we will be able to find a relation between the well’s depth and the overall measured time. Part (a): Finite sound speed Computation Figure 1 shows a well of unknown depth 𝐷. Let 𝑡𝑚 , 𝑡𝑖 , and 𝑡𝑠 be the (total) measured time, the time taken by the rock to fall the distance 𝐷 and hit the water, and the time it takes sound to go back up, respectively, so that

𝑡𝑚 = 𝑡𝑖 + 𝑡𝑠 .

(1)

Helpful Information Solving Eq. (4) for 𝑫. To solve Eq. (4) for 𝐷, we first rewrite it to isolate the square root term, i.e., √ 𝐷 − 𝑣𝑠 𝑡𝑚 = −𝑣𝑠 2𝐷∕𝑔. We then square each side to obtain 𝐷2 − 2𝐷𝑣𝑠 𝑡𝑚 + 𝑣2𝑠 𝑡2m = 𝑣2𝑠

2𝐷 , 𝑔

which can be rearranged to read ) ( 𝑣 𝐷2 − 2𝑣𝑠 𝑡𝑚 + 𝑠 𝐷 + 𝑣2𝑠 𝑡2m = 0. 𝑔 This is a quadratic equation in 𝐷 with the following two roots: √ ) ( 𝑣2𝑠 2𝑡𝑚 𝑔 . 1± 1+ 𝐷 = 𝑣𝑠 𝑡𝑚 + 𝑔 𝑣𝑠 Only one of these roots is physically meaningful. The solution with the plus sign in front of the square root term yields a nonzero value for 𝐷 when 𝑡𝑚 = 0. This result contradicts Eq. (4), so the only acceptable solution is the one with the minus sign.

The motion of the rock falling the distance 𝐷 is a rectilinear motion with constant acceleration 𝑔 = 9.81 m∕s2 . Hence, by choosing a coordinate axis pointing from the top to the bottom of the well, noting that the rock starts at 𝑠0 = 0 m, and assuming that the rock is released with initial velocity 𝑣0 equal to zero, Eq. (12.33) tells us that √ 𝐷 = 21 𝑔𝑡2𝑖

⇒

𝑡𝑖 =

2𝐷 . 𝑔

(2)

As soon as the rock hits the water, a sound wave traveling with a constant velocity 𝑣𝑠 = 340 m∕s, and therefore, with constant acceleration 𝑎𝑠 = 0 m∕s2 , goes from the bottom of the well up to the observer’s ear at 𝑠 = 0. Hence, the time taken by the sound is 𝑡𝑠 =

𝐷 . 𝑣𝑠

(3)

Next, using the expressions for 𝑡𝑖 and 𝑡𝑠 in Eqs. (2) and (3), respectively, Eq. (1) becomes √ 𝑡𝑚 =

2𝐷 𝐷 + . 𝑔 𝑣𝑠

(4)

This equation can be solved for 𝐷 to obtain (see the Helpful Information note in the margin for details) (√ ) 𝑣2𝑠 2𝑡𝑚 𝑔 −1 . 𝐷 = 𝑣𝑠 𝑡𝑚 − 1+ (5) 𝑔 𝑣𝑠

ISTUDY

Section 12.2

One-Dimensional Motion

677

Part (b): Infinite sound speed

If the speed of sound were infinite, Eq. (3) would imply that 𝑡𝑠 = 0. Hence, from Eq. (1), we see that 𝑡𝑚 = 𝑡𝑖 so that the first of Eqs. (2) yields

Computation

𝐷 = 12 𝑔𝑡2𝑚 .

(6)

Discussion & Verification

The result in Eq. (5) is dimensionally correct. Since the given data 𝑡𝑚 , 𝑔, and 𝑣𝑠 have dimensions of time (𝑇 ), length over time squared (𝐿∕𝑇 2 ), and length over time (𝐿∕𝑇 ), respectively, then note that the argument of the square root term in Eq. (5) is nondimensional, i.e., [ ] 2𝑡𝑚 𝑔 1 𝐿 𝑇 = [𝑡𝑚 ][𝑔] = 𝑇 2 = 1. (7) 𝑣𝑠 [𝑣𝑠 ] 𝑇 𝐿 Consequently, the dimensions of 𝐷 in Eq. (5) are [ ] 𝑣2𝑠 𝐿 𝐿2 𝑇 2 1 [𝐷] = 𝑣𝑠 𝑡𝑚 + = 𝑇+ 2 = 𝐿, = [𝑣𝑠 ][𝑡𝑚 ] + [𝑣𝑠 ]2 𝑔 [𝑔] 𝑇 𝑇 𝐿

(8)

as expected. The result in Eq. (6) can be shown to be dimensionally correct in a similar manner. 5

A Closer Look

∗ This

formula for the aerodynamic drag is often discussed in fluid mechanics courses.

“true” 𝑡𝑚 -D curve

4 𝑡𝑚 (s)

We now compare the solutions with finite and infinite sound speed to understand under what conditions it is important to account for the finiteness of the speed of sound. Consider Fig. 2, which presents three curves derived under three different sets of assumptions (the curves have 𝐷 on the horizontal axis to allow us to more easily make comments based on the well’s depth). Figure 2 not only shows the functions in Eqs. (5) and (6), but also the solution we would obtain by taking into account both the finiteness of the speed of sound and air resistance. This latter curve was obtained by assuming that the rock used for the measurement (1) is spherical with a radius 𝑟 = 1 cm and (2) is made out of granite, with a density of 2.75 g∕cm3 ; and it is subject to an aerodynamic drag force given by 𝐹𝐷 = 𝐶𝐷 𝜌𝐴𝑣2 ∕2, where the dimensionless drag coefficient 𝐶𝐷 was chosen to be equal to 0.3, 𝜌 is the density of air at ground level, and 𝐴 is the frontal area of the spherical rock.∗ Treating the curve obtained by this drag model as the “true” relation between 𝐷 and 𝑡𝑚 , observe that the red curve, representing Eq. (5), and the black curve, corresponding to Eq. (6), diverge from the true curve as the depth of the well increases. However, the three curves essentially coincide near the origin of the plot. So estimating the depth of a well while disregarding air resistance and the finiteness of the speed of sound is not that bad for shallow wells, where one could define shallow to mean, say, less than about 30 m. The second conclusion we can draw is that, by accounting for the finiteness of the speed of sound, our formula can now be applied for depths all the way up to 80 m without having to resort to complex theories of rock-air interaction. Finally, notice that the curves provided here allow us to get a quantitative appreciation for the error we would make in estimating the depth of the well depending on the curve used. For example, consider the case in which we measure 𝑡𝑚 = 4 s. How deep is the well? Well, according to the red curve, the depth is roughly 70.5 m, whereas the black line indicates a depth of 78.5 m. Therefore, we could go with the estimate given by the black line (since it is easier to compute) with the knowledge that the error is of the order of 12%.

7.93 m

3 2

𝑣𝑠 = 340 m∕s 𝑣𝑠 → ∞

1 0

0

20

40

60 80 𝐷 (m)

100 120

Figure 2 Measured time vs. well depth curves for various sets of assumptions.

678

Chapter 12

Particle Kinematics

E X A M P L E 12.8

Acceleration Function of Velocity: Descent of a Skydiver The skydiver shown in Figs. 1 and 2 has deployed his parachute after free-falling at 𝑣0 = 44.5 m∕s. As we will learn in Chapter 13, if we model the drag force 𝐹𝑑 due to air resistance as being proportional to the square of the skydiver’s velocity, Newton’s second law tells us that the skydiver’s acceleration is 𝑎 = 𝑔 − 𝐶𝑑 𝑣2 ∕𝑚, where 𝑔 is the acceleration due to gravity, 𝑚 is the skydiver’s mass, and 𝐶𝑑 is a constant drag coefficient.∗ Letting 𝐶𝑑 = 43.2 kg∕m, 𝑚 = 110 kg, and 𝑔 = 9.81 m∕s2 , determine the skydiver’s velocity as a function of time. In addition, determine the skydiver’s terminal velocity.

SOLUTION Part (a): From acceleration to velocity

Since 𝑎 = 𝑑𝑣∕𝑑𝑡 and the acceleration is of the form 𝑎 = 𝑎(𝑣), we obtain an expression of the type 𝑑𝑡 = 𝑑𝑣∕𝑎(𝑣). Therefore, in this problem we first obtain time as a function of velocity, i.e., 𝑡 = 𝑡(𝑣), and then we will try to invert this relationship to obtain 𝑣 = 𝑣(𝑡). This is the strategy followed in developing Eq. (12.24) for the case when 𝑎 = 𝑎(𝑣).

Road Map BORODIN DENIS/Shutterstock

Figure 1 A skydiver descending.

drag force 𝐹𝑑

Computation As discussed in the Road Map, starting with 𝑑𝑡 = 𝑑𝑣∕𝑎(𝑣) and using the given expression for 𝑎(𝑣), we have

𝑑𝑡 =

𝑑𝑣 𝑔−

𝐶𝑑

𝑣2 𝑚

⇒

∫0

𝑡

𝑣

𝑑𝑡 =

∫𝑣 𝑔 − 0

𝑣

𝑑𝑣

⇒

𝐶𝑑

𝑣2 𝑚

𝑡=

𝑑𝑣 , ∫𝑣 𝑔 − 𝐶𝑑 𝑣2 0 𝑚

(1)

where we have used the fact that 𝑣 = 𝑣0 for 𝑡 = 0. To carry out the integral on the righthand side of the last of Eqs. (1), we can factor out the term 𝐶𝑑 ∕𝑚 at the denominator and rewrite the integrand as follows: 1 𝑔− weight force 𝑚𝑔 Figure 2 Skydiver with drag and weight forces depicted.

ISTUDY

𝐶𝑑 𝑚

= 𝑣2

𝑚 𝐶𝑑

𝑚𝑔 𝐶𝑑

1 . − 𝑣2

(2)

We observe that the denominator of the last fraction in Eq. (2) contains the term 𝑚𝑔∕𝐶𝑑 , which has the same dimensions as a velocity squared. Hence, we define the quantity √ 𝑚𝑔 , (3) 𝑣̃ = 𝐶𝑑 with the same dimensions as a velocity, and rewrite Eq. (2) as follows: ) ( 1 1 𝑚 𝑚 1 1 . + = = 𝐶 𝐶𝑑 𝑣̃2 − 𝑣2 2𝑣𝐶 ̃ 𝑑 𝑣̃ − 𝑣 𝑣̃ + 𝑣 𝑔 − 𝑚𝑑 𝑣2

(4)

Substituting the last expression in Eq. (4) into the last of Eqs. (1) and integrating, we have ] [ 𝑚 |𝑣 |𝑣 𝑡=− (5) ln(𝑣̃ − 𝑣)| − ln(𝑣̃ + 𝑣)| . |𝑣0 |𝑣0 2𝑣𝐶 ̃ 𝑑 Recalling that ln 𝑎 − ln 𝑏 = ln(𝑎∕𝑏), Eq. (5) can be simplified to read ) ( 𝑚 𝑣̃ − 𝑣 𝑣̃ + 𝑣0 𝑡=− . ln 2𝑣𝐶 ̃ 𝑑 𝑣̃ + 𝑣 𝑣̃ − 𝑣0

(6)

As anticipated in the Road Map, we now have an expression for time as a function of velocity. To obtain the desired result, namely, velocity as function of time, we need to solve ∗ The

coefficient 𝐶𝑑 is related to the coefficient 𝐶𝐷 in Example 12.7 as follows: 𝐶𝑑 = 12 𝐶𝐷 𝜌𝐴.

ISTUDY

Section 12.2

One-Dimensional Motion

The last of Eqs. (7) can be solved for 𝑣 to obtain 𝑣 = 𝑣̃

35 25 15

̃ 𝑑 ∕𝑚)𝑡 𝑣̃ + 𝑣0 − (𝑣̃ − 𝑣0 )𝑒−(2𝑣𝐶

𝑣̃ + 𝑣0 + (𝑣̃ − 𝑣0

45 𝑣 (m∕s)

Eq. (6) for 𝑣. To do so, we first isolate the logarithmic term and then take the exponential of both sides of the equation, i.e., ) ( 2𝑣𝐶 ̃ 𝑑 𝑣̃ − 𝑣 𝑣̃ + 𝑣0 𝑣̃ − 𝑣 𝑣̃ + 𝑣0 ̃ 𝑑 ∕𝑚)𝑡 − = ⇒ 𝑒−(2𝑣𝐶 𝑡 = ln . (7) 𝑚 𝑣̃ + 𝑣 𝑣̃ − 𝑣0 𝑣̃ + 𝑣 𝑣̃ − 𝑣0

679

̃ 𝑑 ∕𝑚)𝑡 )𝑒−(2𝑣𝐶

.

(8)

The expression in Eq. (8) has been plotted in Fig. 3 for the parameters given. Notice that the skydiver starts out at 44.5 m∕s at 𝑡 = 0 s and quickly (in about one second) slows down to approximately 5 m∕s.

5 0

0.5

1

1.5

𝑡 (s) Figure 3 Velocity of the skydiver as he descends with his parachute deployed.

Part (b): Terminal velocity Road Map

The terminal velocity is defined as the velocity reached after an infinite amount of time. Therefore, we can find the terminal velocity by examining the behavior of Eq. (8) as 𝑡 grows larger and larger.

Computation As 𝑡 becomes larger and larger, the two exponential terms in Eq. (8) go to zero, and the expression ̃ From Eq. (3), √ for 𝑣 simplifies to the following expression: 𝑣 = 𝑣. we recall that 𝑣̃ = 𝑚𝑔∕𝐶𝑑 , and therefore, we can conclude that

√ 𝑣term =

𝑚𝑔 = 4.998 m∕s, 𝐶𝑑

(9)

where we have used the fact that 𝑚 = 110 kg, 𝑔 = 9.81 m∕s2 , and 𝐶𝑑 = 43.2 kg∕m. This agrees with the plot in Fig. 3, which shows that for larger time values the velocity tends to become constant and equal to the value 5 m∕s. In defining 𝑣̃ we had observed that this quantity has the same dimensions as velocity. Using this fact, we can then readily verify that Eq. (8) has the correct dimensions since the fraction on the right-hand side is nondimensional. Also, the fact that a finite terminal velocity exists does match experience, since we observe that bodies free falling in air stop accelerating after some time (provided they can fall for long enough time).

Discussion & Verification

Helpful Information Another commonly accepted and consistent definition of terminal velocity. When a body is in free fall in a medium such as air (or water), the body experiences an aerodynamic (or fluid dynamic) resistance, called drag, that opposes gravity and increases with speed. If the body falls for enough time, the aerodynamic drag will end up equilibrating the force of gravity and the body will stop accelerating. The terminal velocity can be defined as the value of the velocity at which the acceleration becomes equal to zero.

680

Chapter 12

Particle Kinematics

E X A M P L E 12.9

Constant Angular Acceleration: Propeller and Supersonic Effects In Fig. 2, the propeller shown has radius 𝑟𝑝 = 19 f t, and it rotates about its axis while keeping the propeller disk stationary.∗ Suppose that the propeller starts from rest with a constant angular acceleration 𝛼 = 50 rad∕s2 .† Knowing that the speed of sound at sea level under standard conditions is 𝑣𝑠 = 1130 f t∕s, find (a) 𝜔𝑠 , the angular speed of the propeller, expressed in rpm, at which 50% of the area of the propeller disk operates in the supersonic regime. (b) 𝑡𝑠 , the time it takes to achieve 𝜔𝑠 . (c) 𝜃𝑠 , the number of revolutions experienced by the propeller in going from rest to 𝜔𝑠 .

U.S. Navy photo by Seaman Daniel A. Barker

Figure 1 V-22 Osprey aircraft.

SOLUTION Road Map

𝑧

𝛼 𝜔

radial direction

𝑟

As the propeller spins up, different points along a propeller’s blade experience different speeds. This is so because the motion of each point on the propeller is circular, and while the angular velocity and acceleration for this motion are the same for all points, the distance from the axis of rotation is not. The overall motion is a constant angular acceleration motion, and we can use the formulas derived for this case. Part (a): Calculation of 𝝎𝒔 Computation

Referring to Fig. 3 and using Eq. (12.37) to describe the velocity of points in circular motion, we have that the speed |𝑠| ̇ of a point at a distance 𝑟 from the spin axis is

propeller disk

|𝑠| ̇ = 𝑟|𝜔|,

Figure 2 View of one of the engines and its companion propeller of the V-22 Osprey. The radius of the V-22 propellers is 19 ft. 𝑟𝑝 𝑟𝑖

𝑠 𝑟

𝜃

𝑂

(1)

which shows that |𝑠| ̇ is proportional to the distance from the axis of rotation. Hence, if a point interior to the propeller disk moves at supersonic speeds, all of the points between it and the propeller’s periphery will also move at supersonic speeds. Letting 𝑟𝑖 be the radius of the inner disk of the propeller whose area is 50% of the total disk area (see Fig. 3) yields 𝜋𝑟2𝑝

𝑟𝑝 𝑟𝑖 = √ = 13.44 f t. (2) 2 If points on the circle of radius 𝑟𝑖 have achieved the speed of sound, then combining Eq. (1) and the second of Eqs. (2), we have √ 𝑣𝑠 2 = 84.11 rad∕s = 803.2 rpm, 𝜔𝑠 = (3) 𝑣𝑠 = 𝑟𝑖 𝜔𝑠 ⇒ 𝑟𝑝 𝜋𝑟2𝑖 =

2

⇒

where 𝜔𝑠 > 0 since it is an angular speed. Part (b): Calculation of 𝒕𝒔 𝜔 𝛼 Figure 3 Definition of the coordinates 𝑠 of a point a distance 𝑟 from the center of rotation 𝑂. Using the angular coordinate 𝜃, we have that 𝑠 = 𝑟𝜃. The shaded circle has an area equal to 50% of the total area of the propeller disk.

ISTUDY

To establish how long it takes to achieve 𝜔𝑠 , recall that the angular acceleration 𝛼 is the time derivative of the angular velocity 𝜔, i.e., 𝛼 = 𝜔̇ = 𝑑𝜔∕𝑑𝑡, so that we can write 𝑑𝜔 = 𝛼 𝑑𝑡. Therefore, observing that 𝛼 is constant and letting 𝜔0 = 0 be the initial angular velocity (the propeller starts from rest), we have Computation

∫𝜔

𝜔

0

𝑑𝜔 =

∫0

𝑡𝑠

𝛼 𝑑𝑡

⇒

𝜔𝑠 = 𝜔0 +

∫0

𝑡𝑠

Combining the results in Eqs. (3) and (4), we have √ 𝑣𝑠 2 = 1.682 s. 𝑡𝑠 = 𝛼𝑟𝑝 ∗ The

𝛼 𝑑𝑡

⇒

𝜔𝑠 = 𝛼𝑡𝑠 .

(4)

(5)

propeller disk is the disk spanned by the propeller blades as they rotate. is roughly what it takes to go from 0 to 1430 rpm in 3 s and is therefore very easy to achieve even with an average small-car engine.

† This

ISTUDY

Section 12.2

One-Dimensional Motion Part (c): Calculation of 𝜽𝒔

Since a revolution is an angular displacement equal to 2𝜋 rad, we can calculate the number of revolutions experienced by the propeller by computing the difference in the propeller’s angular position 𝜃 between 𝑡 = 0 and 𝑡 = 𝑡𝑠 . Since the angular acceleration is constant, we can use Eq. (12.34) along with Table 12.1 to arrive at the following relationship between 𝜔𝑠 , 𝛼, and 𝜃𝑠 : 𝜔2𝑠 = 𝜔20 + 2𝛼(𝜃𝑠 − 𝜃0 ), (6) where 𝜃0 is the angular position of a point on the propeller disk at 𝑡 = 0. Since all points on the propeller disk experience the same angular displacement, by choosing a point with 𝜃0 = 0 and recalling 𝜔0 = 0, Eq. (12.34) can be solved to obtain 𝜃𝑠 =

𝜔2𝑠 2𝛼

= 70.74 rad = 11.26 rev.

(7)

Discussion & Verification Let 𝐿 and 𝑇 denote dimensions of length and time, respectively. Then, referring to Eq. (3), we see that the dimensions of 𝜔𝑠 are given by [ ] 𝑣 1 𝐿1 1 [𝜔𝑠 ] = 𝑠 = [𝑣𝑠 ] = = , (8) 𝑟𝑝 [𝑟𝑝 ] 𝑇 𝐿 𝑇

as expected. In addition, recalling that a radian is nondimensional, we see that 𝜔𝑠 has the right dimensions and is expressed in appropriate units. Next, referring to Eq. (5), the dimensions of 𝑡𝑠 are given by ] [ 𝑣𝑠 𝐿 1 1 1 1 = = 𝑇, (9) = [𝑣𝑠 ] [𝑡𝑠 ] = 𝛼𝑟𝑝 [𝛼] [𝑟𝑝 ] 𝑇 𝑇 −2 𝐿 as expected. Considering Eq. (5) again, we see that 𝑡𝑠 has been expressed in appropriate units. Finally, referring to Eq. (7), the dimensions of 𝜃𝑠 are given by [ [𝜃𝑠 ] =

𝜔2𝑠 𝛼

] = [𝜔2𝑠 ]

1 1 1 = = 1, [𝛼] 𝑇 2 𝑇 −2

(10)

as expected. Considering Eq. (7) again and recalling that a radian is nondimensional, 𝜃𝑠 has been expressed in appropriate units. As far as the numerical values of our results are concerned, because the propeller is accelerating from rest to the angular speed 𝜔𝑠 during the time interval 0 ≤ 𝑡 ≤ 𝑡𝑠 , the angular displacement 𝜃𝑠 we computed must be smaller than the value 𝑡𝑠 𝜔𝑠 , which represents the angular displacement the propeller would have experienced if it had been rotating at the constant angular speed 𝜔𝑠 for 0 ≤ 𝑡 ≤ 𝑡𝑠 . Since 𝑡𝑠 𝜔𝑠 = 141.5 rad, our expectation is met.

681

Interesting Fact Propellers in high-performance planes. Propeller propulsion was extremely popular until the early 1950s. However, especially for military applications, jet propulsion quickly replaced propeller propulsion starting at the end of World War II. By the end of World War II, the German Luftwaffe had already put in service four different jet planes, while the U.S. Army Air Forces (USAAF), the British RAF, and the Japanese Imperial Navy were testing their first prototypes. Propeller propulsion was abandoned because of an intrinsic aerodynamic problem: propellers do not perform well at supersonic speeds. This same problem contributes to keeping helicopters in the realm of “slow” flying machines. To understand this issue, we need to realize that, aerodynamically, each blade on a propeller is a wing. As a blade rotates, it deflects air from one side to the other side of the propeller disk. Due to Newton’s third law, the backward motion imparted to the air by the propeller results in a corresponding forward motion of the propeller and of the plane to which the propeller is attached. Therefore the air velocity as seen by a propeller’s blade is the sum of the velocity due to rotatory motion and the velocity due to the airplane’s forward motion. Consequently, the propeller blades experience supersonic speeds much sooner than the plane does, and unless special design strategies are implemented, this quickly decreases the propeller’s efficiency.

682

Chapter 12

Particle Kinematics

Problems Problems 12.41 through 12.44 STOP

STOP

The following four problems refer to a car traveling between two stop signs, in which the car’s velocity is assumed to be given by 𝑣(𝑡) = [9 − 9 cos(2𝑡∕5)] m∕s for 0 ≤ 𝑡 ≤ 5𝜋 s.

𝑢̂ 𝑠

𝑠

𝑠=0 20 15

𝑣 (m∕s)

Problem 12.41 Determine 𝑣max , the maximum velocity reached by the car. Furthermore, determine the position 𝑠𝑣 and the time 𝑡𝑣 at which 𝑣max occurs.

𝑣max

max

10

Problem 12.42 5 𝜋 2

0

𝜋

2𝜋

𝑡 (s)

Figure P12.41–P12.44

ISTUDY

Determine the time at which the brakes are applied and the car starts

to slow down.

5 0

max

3𝜋

4𝜋

5𝜋

Problem 12.43

Determine the average velocity of the car between the two stop signs.

Problem 12.44 Determine |𝑎|max , the maximum of the magnitude of the acceleration reached by the car, and determine the position(s) at which |𝑎|max occurs.

Problem 12.45

√ The acceleration of a sled is prescribed to have the following form: 𝑎 = 𝛽 𝑡, where 𝑡 is time expressed in seconds, and 𝛽 is a constant. The sled starts from rest at 𝑡 = 0. Determine 𝛽 in such a way that the distance traveled after 1 s is 25 f t. 𝑎 𝑠 Figure P12.45 and P12.46

Problem 12.46

√ The acceleration of a sled can be prescribed to have one of the following forms: 𝑎 = 𝛽1 𝑡, 𝑎 = 𝛽2 𝑡, and 𝑎 = 𝛽3 𝑡2 , where 𝑡 is time expressed in seconds, 𝛽1 = 1 m∕s5∕2 , 𝛽2 = 1 m∕s3 , and 𝛽3 = 1 m∕s4 . The sled starts from rest at 𝑡 = 0. Determine which of the three cases allows the sled to cover the largest distance in 1 s. In addition, determine the distance covered for the case in question.

Problems 12.47 and 12.48 A peg is constrained to move in a rectilinear guide and is given the following acceleration: 𝑎 = 𝑎0 sin 𝜔𝑡, where 𝑎0 = 20 f t∕s2 , 𝜔 = 250 rad∕s, and 𝑡 is time expressed in seconds. 𝑥

Figure P12.47 and P12.48 Problem 12.47

If 𝑥 = 0 and 𝑣 = 0 for 𝑡 = 0, determine the position of the peg at

𝑡 = 4 s. Problem 12.48

periodic.

Determine the value of the velocity of the peg at 𝑡 = 0 so that 𝑥(𝑡) is

ISTUDY

Section 12.2

683

One-Dimensional Motion

Problem 12.49 A ring is thrown straight upward from a height ℎ = 2.5 m off the ground and with an initial velocity 𝑣0 = 3.45 m∕s. Gravity causes the ring to have a constant downward acceleration 𝑔 = 9.81 m∕s2 . Determine ℎmax , the maximum height reached by the ring.

𝑠

Problem 12.50 A ring is thrown straight upward from a height ℎ = 2.5 m off the ground. Gravity causes the ring to have a constant downward acceleration 𝑔 = 9.81 m∕s2 . Letting 𝑑 = 5.2 m, if the person at the window is to receive the ring in the gentlest possible manner, determine the initial velocity 𝑣0 the ring must be given when first released.

𝑣0

𝑑

ℎ Figure P12.49 and P12.50

Problem 12.51 A hot air balloon is climbing with a velocity of 7 m∕s when a sandbag (used as ballast) is released at an altitude of 305 m. Assuming that the sandbag is subject only to gravity and that therefore its acceleration is given by 𝑦̈ = −𝑔, 𝑔 being the acceleration due to gravity, determine how long the sandbag takes to hit the ground and its impact velocity.

𝑦

Figure P12.51

Problem 12.52 Approximately 1 h 15 min into the movie King Kong (the one directed by Peter Jackson), there is a scene in which Kong is holding Ann Darrow (played by the actress Naomi Watts) in his hand while swinging his arm in anger. A quick analysis of the movie indicates that at a particular moment Kong displaces Ann from rest by roughly 10 f t in a span of four frames. Knowing that the DVD plays at 24 frames per second and assuming that Kong subjects Ann to a constant acceleration, determine the acceleration Ann experiences in the scene in question. Express your answer in terms of the acceleration due to gravity 𝑔. Comment on what would happen to a person really subjected to this acceleration.

Moviestore Collection Ltd/Alamy Stock Photo

Figure P12.52

Problem 12.53 A car travels on a rectilinear stretch of road at a constant speed 𝑣0 = 65 mph. At 𝑠 = 0 the driver applies the brakes hard enough to cause the car to skid. Assume that the car keeps sliding until it stops, and assume that throughout this process the car’s acceleration is given by 𝑠̈ = −𝜇𝑘 𝑔, where 𝜇𝑘 = 0.76 is the kinetic friction coefficient and 𝑔 is the acceleration of gravity. Compute the car’s stopping distance and time.

𝑠 𝑠=0 Figure P12.53

684

Chapter 12

Particle Kinematics Problems 12.54 and 12.55

𝑣0

𝑠 Figure P12.54 and P12.55

If the truck brakes and the crate slides to the right relative to the truck, the horizontal acceleration of the crate is given by 𝑠̈ = −𝑔𝜇𝑘 , where 𝑔 is the acceleration of gravity, 𝜇𝑘 = 0.87 is the kinetic friction coefficient, and 𝑠 is the position of the crate relative to a coordinate system attached to the ground (rather than the truck). Problem 12.54 Assuming that the crate slides without hitting the right end of the truck bed, determine the time it takes to stop if its velocity at the start of the sliding motion is 𝑣0 = 55 mph. Problem 12.55

Assuming that the crate slides without hitting the right end of the truck bed, determine the distance it takes to stop if its velocity at the start of the sliding motion is 𝑣0 = 75 km∕h.

Problems 12.56 and 12.57 A sphere is dropped from rest at the free surface of a thick polymer fluid. The acceleration of the sphere has the form 𝑎 = 𝑔 − 𝜂𝑣, where 𝑔 is the acceleration due to gravity, 𝜂 is a constant, and 𝑣 is the sphere’s velocity. 𝑎

Problem 12.56 The sphere is observed to reach a constant sinking velocity equal to 0.1 m∕s. Determine 𝜂. 𝑠

If 𝜂 = 50 s−1 determine the velocity of the sphere after 0.02 s. Express the result in feet per second.

Problem 12.57 Figure P12.56 and P12.57

Problems 12.58 and 12.59 The motion of a peg sliding within a rectilinear guide is controlled by an actuator in such a way that the peg’s acceleration takes on the form 𝑥̈ = 𝑎0 (2 cos 2𝜔𝑡 − 𝛽 sin 𝜔𝑡), where 𝑡 is time, 𝑎0 = 3.5 m∕s2 , 𝜔 = 0.5 rad∕s, and 𝛽 = 1.5. 𝑥

Figure P12.58 and P12.59 Problem 12.58

Determine the expressions for the velocity and the position of the peg as functions of time if 𝑥(0) ̇ = 0 m∕s and 𝑥(0) = 0 m.

Problem 12.59 Determine the total distance traveled by the peg during the time interval 0 s ≤ 𝑡 ≤ 5 s if 𝑥(0) ̇ = 𝑎0 𝛽∕𝜔.

Problems 12.60 and 12.61 𝑣

𝑥 𝜃 Figure P12.60 and P12.61

ISTUDY

A package is pushed up an incline at 𝑥 = 0 with an initial speed 𝑣0 . The incline is coated with a thin viscous layer so that the acceleration of the package is given by 𝑎 = −(𝑔 sin 𝜃 + 𝜂𝑣), where 𝑔 is the acceleration due to gravity, 𝜂 is a constant, and 𝑣 is the velocity of the package. If 𝜃 = 30◦ , 𝑣0 = 10 f t∕s, and 𝜂 = 8 s−1 , determine the time it takes for the package to come to a stop.

Problem 12.60

Problem 12.61 If 𝜃 = 25◦ , 𝑣0 = 7 m∕s, and 𝜂 = 8 s−1 , determine the distance 𝑑 traveled by the package before it comes to a stop.

ISTUDY

Section 12.2

One-Dimensional Motion

Problem 12.62 Referring to Example 12.8 on p. 678, and defining terminal velocity as the velocity at which a falling object stops accelerating, determine the skydiver’s terminal velocity without performing any integrations.

Problem 12.63 Referring to Example 12.8 on p. 678, determine the distance 𝑑 traveled by the skydiver from the instant the parachute is deployed until the difference between the velocity and the terminal velocity is 10% of the terminal velocity.

Problems 12.64 and 12.65

Figure P12.62 and P12.63

In a physics experiment, a sphere with a given electric charge is constrained to move along a rectilinear guide with the following acceleration: 𝑎 = 𝑎0 sin(2𝜋𝑠∕𝜆), where 𝑎0 = 8 m∕s2 , 𝜋 is measured in radians, 𝑠 is the position of the sphere measured in meters, −𝜆 ≤ 𝑠 ≤ 𝜆, and 𝜆 = 0.25 m. 𝑠

If the sphere is placed at rest at 𝑠 = 0 and then gently nudged away from this position, what is the maximum speed that the sphere could achieve, and where would this maximum occur?

Problem 12.64

Suppose that the velocity of the sphere is equal to zero for 𝑠 = 𝜆∕4. Determine the range of motion of the sphere, that is, the interval along the 𝑠 axis within which the sphere moves. Hint: Determine the speed of the sphere and the interval along the 𝑠 axis within which the speed has admissible values. Problem 12.65

𝑂 Figure P12.64 and P12.65

Problems 12.66 and 12.67 The acceleration of an object in rectilinear free fall while immersed in a linear viscous fluid is 𝑎 = 𝑔 − 𝐶𝑑 𝑣∕𝑚, where 𝑔 is the acceleration of gravity, 𝐶𝑑 is a constant drag coefficient, 𝑣 is the object’s velocity, and 𝑚 is the object’s mass. Problem 12.66 Letting 𝑡0 = 0 and 𝑣0 = 0, find the velocity as a function of time and find the terminal velocity. Problem 12.67

Letting 𝑠0 = 0 and 𝑣0 = 0, find the position as a function of velocity.

Problem 12.68 A 1.5 kg rock is released from rest at the surface of a calm lake. If the resistance offered by the water as the rock falls is directly proportional to the rock’s velocity, the rock’s acceleration is 𝑎 = 𝑔 − 𝐶𝑑 𝑣∕𝑚, where 𝑔 is the acceleration of gravity, 𝐶𝑑 is a constant drag coefficient, 𝑣 is the rock’s velocity, and 𝑚 is the rock’s mass. Letting 𝐶𝑑 = 4.1 kg∕s, determine the rock’s velocity after 1.8 s.

Problems 12.69 and 12.70 A 3.1 lb rock is released from rest at the surface of a calm lake, and its acceleration is 𝑎 = 𝑔 − 𝐶𝑑 𝑣∕𝑚, where 𝑔 is the acceleration of gravity, 𝐶𝑑 = 0.27 lb⋅s∕f t is a constant drag coefficient, 𝑣 is the rock’s velocity, and 𝑚 is the rock’s mass. Problem 12.69

Determine the depth to which the rock will have sunk when the rock achieves 99% of its terminal velocity. Problem 12.70

Determine the rock’s velocity after it drops 5 f t.

𝑠 Figure P12.66–P12.70

685

686

Chapter 12

Particle Kinematics

Problem 12.71 Suppose that the acceleration of an object of mass 𝑚 along a straight line is 𝑎 = 𝑔−𝐶𝑑 𝑣∕𝑚, where the constants 𝑔 and 𝐶𝑑 are given and 𝑣 is the object’s velocity. If 𝑣(𝑡) is unknown and 𝑣(0) is given, can you determine the object’s velocity with the following integral? 𝑣(𝑡) = 𝑣(0) +

∫0

𝑡(

𝑔−

𝐶𝑑 𝑚

) 𝑣 𝑑𝑡.

Problem 12.72

𝑠 𝑠=0 Figure P12.72

Heavy rains cause a particular stretch of road to have a coefficient of friction that changes as a function of location. Specifically, measurements indicate that the friction coefficient has a 3% decrease per meter. Under these conditions the acceleration of a car skidding while trying to stop can be approximated by 𝑠̈ = −(𝜇𝑘 − 𝑐𝑠)𝑔 (the 3% decrease in friction was used in deriving this equation for acceleration), where 𝜇𝑘 is the friction coefficient under dry conditions, 𝑔 is the acceleration of gravity, and 𝑐, with units of m−1 , describes the rate of friction decrement. Let 𝜇𝑘 = 0.5, 𝑐 = 0.015 m−1 , and 𝑣0 = 45 km∕h, where 𝑣0 is the initial velocity of the car. Determine the distance it will take the car to stop and the percentage of increase in stopping distance with respect to dry conditions, i.e., when 𝑐 = 0.

Problem 12.73 A car stops 4 s after the application of the brakes while covering a rectilinear stretch 337 f t long. If the motion occurred with a constant acceleration 𝑎𝑐 , determine the initial speed 𝑣0 of the car and the acceleration 𝑎𝑐 . Express 𝑣0 in mph and 𝑎𝑐 in terms of 𝑔, the acceleration of gravity.

𝑠 Figure P12.73

Problems 12.74 through 12.77 As you will learn in Chapter 13, the angular acceleration of a simple pendulum is given by 𝜃̈ = −(𝑔∕𝐿) sin 𝜃, where 𝑔 is the acceleration of gravity and 𝐿 is the length of the pendulum cord.

𝜃

𝐿

Problem 12.74 Derive the expression of the angular velocity 𝜃̇ as a function of the ̇ angular coordinate 𝜃. The initial conditions are 𝜃(0) = 𝜃0 and 𝜃(0) = 𝜃̇ 0 .

Let the length of the pendulum cord be 𝐿 = 1.5 m. If 𝜃̇ = 3.7 rad∕s determine the maximum value of 𝜃 achieved by the pendulum.

Problem 12.75

when 𝜃 =

Figure P12.74–P12.77

ISTUDY

14◦ ,

Problem 12.76 The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let 𝐿 = 5.3 f t and assume that the pendulum is placed in motion at 𝜃 = 0◦ . What is the minimum angular velocity at this position for the pendulum to swing through a full circle? Problem 12.77 Let 𝐿 = 3.5 f t and suppose that at 𝑡 = 0 s the pendulum’s posi̇ tion is 𝜃(0) = 32◦ with 𝜃(0) = 0 rad∕s. Determine the pendulum’s period of oscillation, i.e., from its initial position back to this position.

ISTUDY

Section 12.2

One-Dimensional Motion

Problems 12.78 through 12.80 As we will see in Chapter 13, the acceleration of a particle of mass 𝑚 suspended by a linear spring with spring constant 𝑘 and unstretched length 𝐿0 (when the spring length is equal to 𝐿0 , the spring exerts no force on the particle) is given by 𝑥̈ = 𝑔 − (𝑘∕𝑚)(𝑥 − 𝐿0 ).

𝑘, 𝐿0

𝑥

Derive the expression for the particle’s velocity 𝑥̇ as a function of position 𝑥. Assume that at 𝑡 = 0, the particle’s velocity is 𝑣0 and its position is 𝑥0 .

Problem 12.78

Problem 12.79 Let 𝑘 = 100 N∕m, 𝑚 = 0.7 kg, and 𝐿0 = 0.75 m. If the particle is released from rest at 𝑥 = 0 m, determine the maximum length achieved by the spring.

𝑚 Figure P12.78–P12.80

Problem 12.80 Let 𝑘 = 8 lb∕f t, 𝑚 = 0.048 slug, and 𝐿0 = 2.5 f t. If the particle is released from rest at 𝑥 = 0 f t, determine how long it takes for the spring to achieve its maximum length. Hint: A good table of integrals will come in handy.

Problem 12.81 A weight 𝐴 with mass 𝑚 = 18 kg is attached to the free end of a nonlinear spring such that the acceleration of 𝐴 is 𝑎 = 𝑔 − (𝛾∕𝑚)(𝑦 − 𝐿0 )3 , where 𝑔 is the acceleration due to gravity, 𝛾 is a constant, and 𝐿0 = 0.5 m. Determine 𝛾 such that 𝐴 does not fall below 𝑦 = 1 m when released from rest at 𝑦 = 𝐿0 . 𝑦

Problems 12.82 and 12.83 Two masses 𝑚𝐴 and 𝑚𝐵 are placed at a distance 𝑟0 from one another. Because of their mutual gravitational attraction, the acceleration of sphere 𝐵 as seen from sphere 𝐴 is given by ) ( 𝑚 𝐴 + 𝑚𝐵 , 𝑟̈ = −𝐺 𝑟2

𝑚

Figure P12.81

where 𝐺 = 6.674×10−11 m3 ∕(kg⋅s2 ) = 3.439×10−8 f t 3 ∕(slug⋅s2 ) is the universal gravitational constant. Problem 12.82

If the spheres are released from rest, determine

(a) The velocity of 𝐵 (as seen by 𝐴) as a function of the distance 𝑟. (b) The velocity of 𝐵 (as seen by 𝐴) at impact if 𝑟0 = 7 f t, the weight of 𝐴 is 2.1 lb, the weight of 𝐵 is 0.7 lb, and (i) The diameters of 𝐴 and 𝐵 are 𝑑𝐴 = 1.5 f t and 𝑑𝐵 = 1.2 f t, respectively. (ii) The diameters of 𝐴 and 𝐵 are infinitesimally small. Problem 12.83

Assume that the particles are released from rest at 𝑟 = 𝑟0 .

(a) Determine the expression relating their relative position 𝑟 and time. Hint: ∫

√ (√ ) √ 𝑥∕(1 − 𝑥) 𝑑𝑥 = sin−1 𝑥 − 𝑥(1 − 𝑥).

(b) Determine the time it takes for the objects to come into contact if 𝑟0 = 3 m, 𝐴 and 𝐵 have masses of 1.1 and 2.3 kg, respectively, and (i) The diameters of 𝐴 and 𝐵 are 𝑑𝐴 = 22 cm and 𝑑𝐵 = 15 cm, respectively. (ii) The diameters of 𝐴 and 𝐵 are infinitesimally small.

𝐴

𝐵 𝑟

Figure P12.82 and P12.83

687

688

Chapter 12

Particle Kinematics

Problem 12.84

𝑣0

If the truck brakes hard enough that the crate slides to the right relative to the truck, the distance 𝑑 between the crate and the front of the trailer changes according to the relation { 𝜇𝑘 𝑔 + 𝑎𝑇 for 𝑡 < 𝑡𝑠 , 𝑑̈ = 𝜇𝑘 𝑔 for 𝑡 > 𝑡𝑠 ,

𝑑

Figure P12.84

𝑣𝐴 𝐴

where 𝑡𝑠 is the time it takes the truck to stop, 𝑎𝑇 is the acceleration of the truck, 𝑔 is the acceleration of gravity, and 𝜇𝑘 is the kinetic friction coefficient between the truck and the crate. Suppose that the truck and the crate are initially traveling to the right at 𝑣0 = 60 mph and the brakes are applied so that 𝑎𝑇 = −10.0 f t∕s2 . Determine the minimum value of 𝜇𝑘 so that the crate does not hit the right end of the truck bed if the initial distance 𝑑 is 12 f t. Hint: The truck stops before the crate stops. 𝑣𝐵

𝑑

Problem 12.85 𝐵 𝑠

Figure P12.85

Cars 𝐴 and 𝐵 are traveling at 𝑣𝐴 = 72 mph and 𝑣𝐵 = 67 mph, respectively, when the driver of car 𝐵 applies the brakes abruptly, causing the car to slide to a stop. The driver of car 𝐴 takes 1.5 s to react to the situation and applies the brakes in turn, causing car 𝐴 to slide as well. If 𝐴 and 𝐵 slide with equal accelerations, i.e., 𝑠̈𝐴 = 𝑠̈𝐵 = −𝜇𝑘 𝑔, where 𝜇𝑘 = 0.83 is the kinetic friction coefficient and 𝑔 is the acceleration of gravity, compute the minimum distance 𝑑 between 𝐴 and 𝐵 at the time 𝐵 starts sliding to avoid a collision.

Problems 12.86 through 12.88 The spool of paper used in a printing process is unrolled with velocity 𝑣𝑝 and acceleration 𝑎𝑝 . The thickness of the paper is ℎ, and the outer radius of the spool at any instant is 𝑟.

𝑟

𝜔𝑠

𝛼𝑠 𝑣𝑝

Pressmaster/Shutterstock

Figure P12.86–P12.88

ISTUDY

𝑎𝑝

Problem 12.86 If the velocity at which the paper is unrolled is constant, determine the angular acceleration 𝛼𝑠 of the spool as a function of 𝑟, ℎ, and 𝑣𝑝 . Evaluate your answer for ℎ = 0.0048 in., for 𝑣𝑝 = 1000 f t∕min, and two values of 𝑟, that is, 𝑟1 = 25 in. and 𝑟2 = 10 in. Problem 12.87 If the velocity at which the paper is unrolled is not constant, determine the angular acceleration 𝛼𝑠 of the spool as a function of 𝑟, ℎ, 𝑣𝑝 , and 𝑎𝑝 . Evaluate your answer for ℎ = 0.0048 in., 𝑣𝑝 = 1000 f t∕min, 𝑎𝑝 = 3 f t∕s2 , and two values of 𝑟, that is, 𝑟1 = 25 in. and 𝑟2 = 10 in. Problem 12.88 If the velocity at which the paper is unrolled is constant, determine the angular acceleration 𝛼𝑠 of the spool as a function of 𝑟, ℎ, and 𝑣𝑝 . Plot your answer for ℎ = 0.0048 in. and 𝑣𝑝 = 1000 f t∕min as a function of 𝑟 for 1 in. ≤ 𝑟 ≤ 25 in. Over what range does 𝛼𝑠 vary?

Problem 12.89 Derive the constant acceleration relation in Eq. (12.32), starting from Eq. (12.24). State what assumption you need to make about the acceleration 𝑎 to complete the derivation. Finally, use Eq. (12.27), along with the result of your derivation, to derive Eq. (12.33). Be careful to do the integral in Eq. (12.27) before substituting your result for 𝑣(𝑡) (try it without doing so, to see what happens). After completing this problem, notice that Eqs. (12.32) and (12.33) are not subject to the same assumption you needed to make to solve both parts of this problem.

ISTUDY

Section 12.3

12.3

Projectile Motion

689

Projectile Motion

In this section we present a simple model to study the motion of projectiles. We define projectile motion to be a motion with constant acceleration given by 𝑎horiz = 0 and

𝑎vert = −𝑔,

(12.39)

where, referring to Fig. 12.14, 𝑎horiz and 𝑎vert are the components of the acceleration vertical direction (positive upward ↑) 𝐵 path

Thomas Barwick/Getty Images

horizontal direction

Figure 12.13 A person shooting a basketball. The trajectory of the basketball is accurately described using the definition of projectile motion given in Eq. (12.39).

𝑔

Figure 12.14. Acceleration of a basketball 𝐵 in projectile motion.

in the horizontal and vertical directions, respectively, and where we have chosen the vertical direction to be positive upward. Since these accelerations are both constant, we can use Eqs. (12.32)–(12.34) to analyze the motion in the horizontal and vertical directions independently of one another. Letting 𝑥 be aligned with the horizontal direction and 𝑦 be aligned with the vertical direction, Eqs. (12.39) can be integrated twice to yield 𝑥(𝑡) = 𝑥(0) + 𝑥(0)𝑡 ̇ and

𝑦(𝑡) = 𝑦(0) + 𝑦(0)𝑡 ̇ − 12 𝑔𝑡2 ,

(12.40)

where 𝑥(0) and 𝑦(0) are the object’s initial position and 𝑥(0) ̇ and 𝑦(0) ̇ are the 𝑥 and 𝑦 components of the object’s initial velocity. Because of the linear relationship between horizontal position and time, we can eliminate time from the equation for vertical position to obtain a parabolic trajectory, 𝑦(𝑥), of the object’s path. This relatively simple trajectory permits us to obtain analytical relationships between a particle’s range, initial velocity and launch angle, and we explore these relationships in the examples to follow. Our definition of projectile motion comes from the assumption that the only force acting on the projectile during flight is the constant force of gravity. As we will formally study in Chapter 13 and as you probably already know, the free body diagram of a projectile in flight looks like that in Fig. 12.15, where 𝑊 = 𝑚𝑔 is the weight force, 𝑚 is the mass of the basketball, and we have assumed no air resistance. There are no forces in the 𝑥 or horizontal direction, which means there is no acceleration in that direction. In the vertical direction, applying Newton’s second law immediately leads to the second of Eqs. (12.39). Our definition of projectile motion is a very simplified description of true projectile motion because it neglects air resistance and changes in gravitational attraction with changes in height. How would our free body diagram change with the inclusion of air resistance, and how can we determine when it’s important to include this effect? Consider the player executing the jump shot in Fig. 12.13. There are two separate drag forces acting on the basketball. The first, which we designate as the primary drag force, results from air navigating around the ball as it moves on its arc toward the basket. In Fig. 12.15, this primary drag force would be represented as a force

𝚥̂ 𝚤̂ path

weight force 𝑊 = 𝑚𝑔

Figure 12.15 Free body diagram of a basketball in flight.

690

ISTUDY

Chapter 12

Particle Kinematics

tangent to the dashed path and opposite to the ball’s velocity. In addition, there is a secondary drag force associated with the ball’s spin. A well-executed jump shot spins backward toward the shooter along an axis parallel to the ground. The superposition of spin along the arc further complicates the air distribution around the ball, creating a bit of extra pressure underneath the ball (and a bit of diminished pressure on the top of the ball), contributing to a lofting force opposing the weight in Fig. 12.15. This secondary drag force creates a “softness” to the well-executed jump shot. In contrast, there is a name for a poorly executed, spinless shot, particularly one of low launch angle—a brick! It was mentioned in Example 12.7 that the primary drag force on an object moving through a medium of density 𝜌 with speed 𝑣 is frequently represented as 1 𝐹𝐷 = 𝐶𝐷 ⋅ 𝐴 ⋅ 𝜌𝑣2 . 2

(12.41)

This primary drag force acts opposite to the object’s velocity. The quantity 𝜌𝑣2 ∕2 is called the dynamic pressure, 𝐴 is a characteristic cross-sectional area, often the silhouette area of the object projected on to a plane normal to its path, and 𝐶𝐷 is a dimensionless drag coefficient. The complicated fluid mechanics responsible for the drag force are frequently buried in this coefficient. We can say that the drag force is negligible as long as 𝐹𝐷 ∕𝑊 ≪ 1. Because 𝐹𝐷 scales with 𝑣2 , the drag force becomes increasingly important as the object’s speed increases. Secondary drag forces become increasingly important in many ball sports where the ball is imparted with significant spin. These secondary drag forces act normal to the plane defined by the ball’s velocity and spin axis. It we want to understand the crude trajectory of a baseball as it leaves a pitcher’s hand and moves toward home plate, our definition of projectile motion will suffice. If we want to make a distinction between the trajectories of a fastball, curveball, and slider, we need to modify our projectile motion to include secondary drag. In some sports, our definition of projectile motion works very well for some shots, not well at all for others. The volleyball setter who passes the ball to a hitter typically launches the pass with minimal spin. The pass is described well with our definition of projectile motion. In contrast, an overhead serve struck with significant spin may dive (if struck with topspin) or move laterally (if struck so the spin axis is more vertically oriented). Our definition of projectile motion can be used to describe a lob shot in tennis, but will be deficient in describing a drop shot or one struck with significant topspin. In Example 12.10, we use our definition of projectile motion to understand the trajectory of a golfer’s gentle chip shot. The trajectory of a drive, however, is typically far more complicated than a simple parabola. Golf ball manufacturers invest significant sums of money designing dimple patterns on golf balls to facilitate lofting forces that extend the range of drives. Similarly, it would be impossible to understand the differences in shaped golf shots (fades and draws) without incorporating secondary drag forces. The advantage of our definition of projectile motion is its relative simplicity. That said, we will want to make a mental note of circumstances where we expect our model to come up short.

ISTUDY

Section 12.3

Projectile Motion

E X A M P L E 12.10

691

Time of Impact and Range of a Projectile

A golfer chips the ball from the rough at the edge of the green. If the ball leaves the rough with a speed 𝑣0 = 15 f t∕s and an angle 𝛽 = 45◦ , determine the flight time of the ball and the corresponding distance 𝑑 covered.

𝑦

SOLUTION

𝑣0

Road Map & Modeling We model the motion of the golf ball as a projectile motion. This will allow us to use constant acceleration equations to determine the 𝑥 and 𝑦 coordinates of the ball as a function of time. The time of flight is the value of time for which the ball comes back to the ground, i.e., to 𝑦 = 0. The distance 𝑑 is then the corresponding value of the coordinate 𝑥.

𝛽 𝑥 𝑑 Figure 1

Computation The positive direction of the 𝑦 axis is opposite to that of gravity. Therefore, the 𝑥 and 𝑦 components of the constant acceleration of the ball are

𝑥̈ = 0

and 𝑦̈ = −𝑔.

(1)

Letting 𝑡 denote time, Eqs. (1), along with Eq. (12.33) on p. 670, tell us that the 𝑥 and 𝑦 coordinates of the ball as functions of time are 𝑥(𝑡) = 𝑥(0) + 𝑥(0)𝑡 ̇

and 𝑦(𝑡) = 𝑦(0) + 𝑦(0)𝑡 ̇ − 21 𝑔𝑡2 ,

(2)

where 𝑥(0) ̇ and 𝑦(0) ̇ are the 𝑥 and 𝑦 components of the golf ball’s initial velocity. Since the ball starts at the origin of the 𝑥𝑦 coordinate system, and using the quantities 𝑣0 and 𝛽, we have 𝑥(0) = 0,

𝑦(0) = 0,

𝑥(0) ̇ = 𝑣0 cos 𝛽,

and 𝑦(0) ̇ = 𝑣0 sin 𝛽.

(3)

Substituting Eqs. (3) into Eqs. (2), 𝑥(𝑡) and 𝑦(𝑡) become: 𝑥(𝑡) = 𝑣0 𝑡 cos 𝛽

and 𝑦(𝑡) = 𝑣0 𝑡 sin 𝛽 − 21 𝑔𝑡2 .

(4)

Letting 𝑡𝑓 be the time at which the ball comes back to the green, we have 𝑦(𝑡𝑓 ) = 0

⇒

𝑣0 𝑡𝑓 sin 𝛽 − 12 𝑔𝑡2𝑓 = 0

⇒

𝑡𝑓 =

2𝑣0 sin 𝛽 𝑔

Helpful Information .

(5)

Recalling that 𝑣0 = 15 f t∕s, 𝛽 = 45◦ , and 𝑔 = 32.2 f t∕s2 , the last of Eqs. (5) yields 𝑡𝑓 = 0.6588 s.

(6)

The distance 𝑑 is equal to 𝑥(𝑡𝑓 ). Substituting the last of Eqs. (5) into the first of Eqs. (4), we have 𝑣2 2𝑣2 sin 𝛽 cos 𝛽 = 0 sin 2𝛽, (7) 𝑑= 0 𝑔 𝑔 which can be evaluated to obtain 𝑑 = 6.988 f t.

(8)

Discussion & Verification The expressions for 𝑡𝑓 and 𝑑 have correct dimensions, and the corresponding numerical values are expressed in appropriate units. To check whether or not the value of 𝑡𝑓 is reasonable, we observe that 𝑡𝑓 ∕2 is the time that the golf ball takes to drop from its maximum height above ground. This indicates that the maximum elevation reached by the golf ball is ℎ = 𝑔(𝑡𝑓 ∕2)2 ∕2 = 1.747 f t, which is quite reasonable given the low initial speed of the ball and the value of the initial angle 𝛽. Note also the second form of Eq. (7) indicates a maximum range in projectile motion beginning and ending at the same elevation is achieved with a launch angle 𝛽 = 45◦ .

Determination of 𝒕𝒇 . The time 𝑡𝑓 is the solution of the second of Eqs. (5). The equation in question is a second-order algebraic equation in 𝑡𝑓 , and therefore we expect two roots. One of these roots is that reported in the last of Eqs. (5). The second is simply 𝑡𝑓 = 0. This latter result simply indicates that 𝑦 = 0 for 𝑡 = 0, which is true since the ball starts out on the ground. However, from the viewpoint of the calculation of the time of flight, 𝑡𝑓 = 0 makes no sense, and it has therefore been disregarded.

692

Chapter 12

Particle Kinematics

E X A M P L E 12.11

Range of Elevation Angles of a Projectile

𝑦 𝐵

𝑣0 𝑂

ℎ ℎ1

𝜃

𝑥

𝑅 Figure 1 A projectile launched from 𝑂 in an attempt to hit the target at 𝐵. Not drawn to scale.

ISTUDY

A projectile is launched from 𝑂 at speed 𝑣0 = 1100 f t∕s to hit a point 𝐵 on a target that is 𝑅 = 1000 f t away. The bottom of the target is ℎ1 = 4 f t above the ground, and the target is ℎ = 3 f t high. Determine the range of angles at which the projectile can be fired in order to hit the target, and compare this with the angle subtended by the target as seen from 𝑂.

SOLUTION Road Map

This is a projectile motion with given initial and final positions, as well as initial speed 𝑣0 . Referring to Fig. 1, the components of the projectile’s acceleration are 𝑎𝑥 = 0 and 𝑎𝑦 = −𝑔. We will relate the projectile’s time of flight 𝑡𝑓 to 𝑦𝐵 , the vertical position of 𝐵. We can then write the launch angle in terms of 𝑦𝐵 and, in turn, infer the range of angles that allows the projectile to hit the target. Because 𝑅 is so much larger than ℎ and ℎ1 , we should expect the range of angles we will find to be small (i.e., our aim will need to be very accurate). Therefore, in carrying out our calculations we will use more significant digits than we normally do. We will discuss the practical implications of this choice in the Discussion & Verification section. The initial and final 𝑥 and 𝑦 positions of 𝐵 can be related to 𝑡𝑓 by applying Eq. (12.33) (on p. 670) in the 𝑥 and 𝑦 directions

Computation

𝑥𝐵 = 𝑥0 + 𝑣0𝑥 𝑡𝑓

and 𝑦𝐵 = 𝑦0 + 𝑣0𝑦 𝑡𝑓 − 21 𝑔𝑡2𝑓 .

(1)

Considering Fig. 1, we have 𝑥0 = 𝑦0 = 0, 𝑣0𝑥 = 𝑣0 cos 𝜃, 𝑣0𝑦 = 𝑣0 sin 𝜃, and 𝑥𝐵 = 𝑅. We will set 𝑦𝐵 = ℎ1 = 4 f t to find 𝜃 = 𝜃min and 𝑦𝐵 = ℎ1 + ℎ = 7 f t to find 𝜃 = 𝜃max . Therefore, treating 𝑦𝐵 as a known quantity, we have 𝑅 = 𝑣0 𝑡𝑓 cos 𝜃

and 𝑦𝐵 = 𝑣0 𝑡𝑓 sin 𝜃 − 21 𝑔𝑡2𝑓 .

(2)

Equations (2) are two equations in the unknowns 𝑡𝑓 and 𝜃. Solving the first of Eqs. (2) for 𝑡𝑓 and substituting the result into the second of Eqs. (2), we obtain ( 𝑦𝐵 = 𝑣0

𝑅 𝑣0 cos 𝜃

)

( sin 𝜃 −

1 𝑔 2

𝑅 𝑣0 cos 𝜃

)2 .

Since sin 𝜃∕ cos 𝜃 = tan 𝜃 and 1∕ cos2 𝜃 = sec2 𝜃, we have ) ( 𝑔𝑅2 sec2 𝜃. 𝑦𝐵 = 𝑅 tan 𝜃 − 2𝑣20 Finally, by noting that sec2 𝜃 = 1 + tan2 𝜃 and then rearranging, Eq. (4) becomes ( ) ( ) 𝑔𝑅2 𝑔𝑅2 2 tan 𝜃 − 𝑅 tan 𝜃 + 𝑦𝐵 + 2 = 0, 2𝑣20 2𝑣0

(3)

(4)

(5)

which is a quadratic equation in tan 𝜃. Dividing through by the coefficient of the tan2 𝜃 term, we obtain ( 2) ) ( 2𝑣0 2𝑦𝐵 𝑣20 2 tan 𝜃 − tan 𝜃 + + 1 = 0. (6) 𝑔𝑅 𝑔𝑅2 Equation (6) can be solved for tan 𝜃 to obtain the two solutions √ ( ) 𝑣20 ± 𝑣40 − 𝑔 𝑔𝑅2 + 2𝑦𝐵 𝑣20 tan 𝜃 = , 𝑔𝑅

(7)

ISTUDY

Section 12.3

Projectile Motion

which means that for each value of 𝑦𝐵 , there are two possible values of the firing angle 𝜃: 𝜃1 and 𝜃2 . Substituting in values for all constants, including the two different values for 𝑦𝐵 , we obtain { 𝜃1 = 0.991678◦ , 𝑦𝐵 = 4 f t ⇒ (8) 𝜃2 = 89.2375041◦ , { 𝜃1 = 1.163590◦ , (9) 𝑦𝐵 = 7 f t ⇒ 𝜃2 = 89.2374736◦ . Equations (8) and (9) give us the values of 𝜃 needed to hit the bottom and top of the target, respectively. Referring to Eq. (8), it is probably intuitive that if we chose 𝜃1 < 𝜃 < 𝜃2 , we would overshoot the bottom of the target, whereas we would undershoot it for 𝜃 < 𝜃1 and 𝜃 > 𝜃2 . For example, substituting 𝜃 = 45◦ [i.e., a value of 𝜃 between those in Eq. (8)] into Eq. (4) gives 𝑦𝐵 = 973.4 f t, which, as expected, is larger than 4 f t. Extending the discussion to Eq. (9), we can then say that if 𝜃1 < 𝜃 < 𝜃2 in Eq. (9), we would overshoot the top of the target, whereas we would undershoot it for 𝜃 < 𝜃1 and 𝜃 > 𝜃2 . We can therefore conclude that there are two ranges of firing angles that will enable our projectile to hit the target and that these ranges are given by 0.991678◦ ≤ 𝜃 ≤ 1.163590◦

(10)

89.23747360◦ ≤ 𝜃 ≤ 89.23750406◦ .

(11)

693

Helpful Information Number of digits in calculations. In the numerical calculations shown in this example, the differences in some of the numbers are so small that we are keeping many more digits than we normally would. If we did not, the differences would not be apparent.

and

Discussion & Verification In obtaining the angle ranges in Eqs. (10) and (11), we went through a simple verification step by computing the answer for 𝜃 = 45◦ , in which we saw that our results were as expected. To extend our discussion, let’s now consider the size of the ranges in Eqs. (10) and (11):

Δ𝜃1 = 1.163590◦ − 0.991678◦ = 0.171912◦ , ◦

(12)

◦

◦

Δ𝜃2 = 89.23747360 − 89.23750406 = −0.00003046 .

(13)

Equations (10) and (12) tell us that to hit our target, we need to elevate our launcher about 1◦ with an accuracy of 0.17◦ . Equations (11) and (13) tell us that we can also hit the target if we elevate our launcher to approximately 89.2◦ , but this time we have an extremely small margin of error. In fact, we need to be accurate to within three one-hundred thousandths of a degree! In addition, we also need to keep in mind that our model does not account for aerodynamic effects, which place an additional accuracy burden on our aim. These considerations tell us that, in practical applications, we need to rely on an active guidance system rather than the accuracy of the launch angles. Let’s complete this example by comparing the angle Δ𝜃1 with the angle subtended by the target at a distance of 1000 f t. Referring to Fig. 2, we can see that the angle subtended is given by 𝛽 = 𝛾 − 𝜙, (14) where ( 𝛾 = tan−1

7 1000

(

) = 0.401064◦

and 𝜙 = tan−1

4 1000

) = 0.229182◦ ,

(15)

so that 𝛽 = 0.171882◦ . Since 𝛽 is very close to the Δ𝜃1 in Eq. (12) and since computing 𝛽 is simpler than computing Δ𝜃1 , we might think that we could have computed 𝛽 to approximate Δ𝜃1 . However, in general, the elevation angle ranges and angle subtended by the target can be substantially different.

3 ft

𝑂

𝛽 𝜙

𝛾

4 ft

1000 f t Figure 2 Not drawn to scale—the vertical dimension has been greatly exaggerated so that the angles can be easily seen.

694

Chapter 12

Particle Kinematics

E X A M P L E 12.12

Initial Speed and Elevation Angle of a Projectile A baseball batter makes contact with a ball about 4 f t above the ground and hits it hard enough that it just clears the center field wall, which is 400 f t away and is 9 f t high. How fast must the ball be moving and at what angle must it be hit so that it just clears the center field wall as shown in Fig. 1? 𝑦

𝑣0

ℎ

𝑤

𝜃

𝑥 𝑑 Figure 1. Side view, drawn to scale, of the given baseball field with all parameters defined. In the trajectory shown, the baseball was hit 4 f t off the ground, at 123.2 f t∕s, and at a 30◦ angle so that it just clears a 9 f t fence that is 400 f t away.

SOLUTION trajectory

Road Map

Referring to Fig. 2, we model the ball as a projectile with acceleration given by 𝑎𝑥 = 0 and 𝑎𝑦 = −𝑔. We know the starting and ending locations of the projectile, and we wish to determine the 𝑣0 and 𝜃 required to get it from start to finish. Therefore, we can proceed as in Example 12.11, i.e., by writing the projectile’s 𝑥 and 𝑦 positions as a function of time and then, eliminating time, we will obtain an expression for 𝑣0 in terms of 𝜃.

𝚥̂ 𝑔

𝚤̂

Figure 2 The only nonzero component of acceleration of the baseball.

Computation

Since both components of acceleration are constant, we can apply the constant acceleration equation, Eq. (12.33) (on p. 670), in both the 𝑥 and 𝑦 directions to obtain 𝑥 = 𝑥0 + 𝑣0𝑥 𝑡 𝑦 = 𝑦0 + 𝑣0𝑦 𝑡 +

⇒ 1 𝑎 𝑡2 2 𝑦

𝑑 = 0 + 𝑣0 𝑡 cos 𝜃,

⇒ 𝑤 = ℎ + 𝑣0 𝑡 sin 𝜃 −

(1) 1 2 𝑔𝑡 . 2

(2)

Equations (1) and (2) are two equations for the three unknowns 𝑣0 , 𝜃, and 𝑡. Since we are interested in 𝑣0 and 𝜃, we can eliminate 𝑡 from these two equations and then solve for 𝑣0 as a function of 𝜃 to obtain √ 𝑣0 = 𝑑

𝑔 . 2 cos 𝜃[(ℎ − 𝑤) cos 𝜃 + 𝑑 sin 𝜃]

(3)

This result tells us that there are infinitely many combinations of 𝑣0 and 𝜃 that will just get the baseball over the center field fence. Discussion & Verification

𝑣0 (f t∕s)

1500 1000 500 0

0

15

30 45 60 𝜃 (degrees)

75

Figure 3 Plot of Eq. (3), that is, 𝑣0 as a function of 𝜃.

ISTUDY

90

The solution to the problem is an expression rather than a specific quantitative answer. To verify that Eq. (3) is correct, we first check that its dimensions are correct. Letting 𝐿 and 𝑇 denote dimensions of length and time, we have [𝑔] = 𝐿∕𝑇 2 and [ℎ] = [𝑤] = [𝑑] = 𝐿. Therefore, the dimensions of the argument of the square root in Eq. (3) are 𝑇 −2 , so that the overall dimensions of the right-hand side of Eq. (3) are 𝐿∕𝑇 , as expected. Further verification requires that we study the behavior of the expression in Eq. (3), as is done next. A Closer Look For 𝑑 = 400 f t, ℎ = 4 f t, 𝑤 = 9 f t, and 𝑔 = 32.2 f t∕s2 , the plot of the required 𝑣0 for 0◦ ≤ 𝜃 ≤ 90◦ is shown in Fig. 3. Inspection of Eq. (3) indicates the speed becomes asymptotically large as 𝜃 approaches a critical shallow angle 𝜃𝑐 given

ISTUDY

Section 12.3

Projectile Motion

by tan 𝜃𝑐 = (𝑤 − ℎ)∕𝑑. For the numbers given in this problem, 𝜃𝑐 = 0.7162◦ . This is so because the point at which the batter makes contact with the ball is lower than the height of the fence, so that even if the batter hits the ball infinitely hard, there is an angle below which the ball will not clear the fence. On the right side of the curve, we see that the required speed again approaches infinity, but now it is as the angle approaches 90◦ . Finally, it is also apparent that the optimal angle to hit the ball is near 45◦ , where by optimal angle we mean the angle corresponding to the smallest possible 𝑣0 . As it turns out, the optimal angle is not exactly at 45◦ since we are trying to get the maximum distance for the minimum speed between two points of unequal height. To find the optimal angle, we could differentiate Eq. (3) with respect to 𝜃, set the result equal to zero, and then solve for the optimal angle. We can also square both sides and differentiate that. Doing so and setting the result equal to zero, we have ( ) } { 2 𝑑 𝑣20 || 𝑔𝑑 2 −𝑑 cos2 𝜃0 + 2(ℎ − 𝑤) cos 𝜃0 sin 𝜃0 + 𝑑 sin 𝜃0 = 0, (4) = | [ ] 2 𝑑𝜃 ||𝜃=𝜃 2 cos2 𝜃 (ℎ − 𝑤) cos 𝜃 + 𝑑 sin 𝜃 0

0

0

0

where we have replaced 𝜃 with 𝜃0 , which is the optimal value of 𝜃. We can solve Eq. (4) by setting the numerator of the fraction within curly braces to zero. Doing this and using some trigonometric identities,∗ we obtain ( ) ( ) −𝑑 cos 2𝜃0 + (ℎ − 𝑤) sin 2𝜃0 = 0, (5) or

( ) tan 2𝜃0 =

𝑑 = −80. ℎ−𝑤 Equation (6) has infinitely many solutions given by 2𝜃0 = 90.72◦ ± 𝑛180◦ ,

𝑛 = 0, 1, … , ∞,

(6)

(7)

but the only meaningful solution in this context is 2𝜃0 = 90.72◦ , or 𝜃0 = 45.36◦ . This means that the optimal angle to hit the ball when trying to hit it over an object that is higher than the initial position of the ball is greater than 45◦ . From Eq. (5) we can also see that the optimal angle is less than 45◦ when we try to hit the ball over a lower object, and it is exactly equal to 45◦ only if ℎ = 𝑤, i.e., if we are trying to hit the ball over an object whose height is the same as the ball’s initial position. (This is the result we obtained in Example 12.10.) This discussion leads to the conclusion that the solution we derived does indeed have a behavior that matches our expectations and physical intuition.

∗ sin2

𝑥 − cos2 𝑥 = − cos(2𝑥) and 2 sin 𝑥 cos 𝑥 = sin(2𝑥).

695

696

Chapter 12

Particle Kinematics

Problems Problem 12.90 The “Green Monster” is the nickname of the left field wall at Fenway Park in Boston. It is 37 f t high and only 310 f t from home plate at the left field foul pole. Assuming a batter makes good contact with a fastball 3 ft off the ground, find the optimal (in the same sense as in Example 12.12) launch angle and corresponding speed to clear the “Monster” just inside the foul pole.

Problem 12.91 Winslow Townson/MLB/Getty Images

A stomp rocket is a toy consisting of a hose connected to a “blast pad” (i.e., an air bladder) at one end and to a short pipe on a tripod at the other end. A rocket with a hollow body is placed on the pipe and is propelled by “stomping” on the blast pad. Some manufacturers claim that one can shoot a rocket over 200 f t in the air. Neglecting air resistance, determine the rocket’s minimum initial speed such that it reaches a maximum flight height of 200 f t.

Figure P12.90

𝜃

Problem 12.92

Figure P12.91

An airplane flying horizontally at elevation ℎ = 150 f t and at a constant speed 𝑣0 = 80 mph drops a package 𝑃 when passing over point 𝑂. Determine the horizontal distance 𝑑 between the drop point and point 𝐵 at which the package hits the ground.

𝑦 𝑣0 𝑃

Problem 12.93 ℎ

𝑑 𝚥̂ 𝚤̂ 𝑂

𝐵

Problem 12.94

Figure P12.92 and P12.93 𝐵

𝐴

30 f t

𝑥

An airplane flying horizontally at elevation ℎ = 60 m and at a constant speed 𝑣0 = 120 km∕h drops a package 𝑃 when passing over point 𝑂. Determine the time it takes for the package to hit the ground at point 𝐵. In addition, determine the velocity of the package at 𝐵.

Stuntmen 𝐴 and 𝐵 are shooting a movie scene in which 𝐴 needs to pass a gun to 𝐵. Stuntman 𝐵 is supposed to start falling vertically precisely when 𝐴 throws the gun to 𝐵. Treating the gun and the stuntman 𝐵 as particles, find the velocity of the gun as it leaves 𝐴’s hand so that 𝐵 will catch it after falling 30 f t.

Problem 12.95 20 f t

The jaguar 𝐴 leaps from 𝑂 at speed 𝑣0 = 6 m∕s and angle 𝛽 = 35◦ relative to the incline to try to intercept the panther 𝐵 at 𝐶. Determine the distance 𝑅 that the jaguar jumps from 𝑂 to 𝐶 (i.e., 𝑅 is the distance between the two points of the trajectory that intersect the incline), given that the angle of the incline is 𝜃 = 25◦ . 𝑣0

Figure P12.94

ISTUDY

𝐴

parabolic trajectory of 𝐴

𝑂 𝛽 ℎmax 𝑔

𝑅

𝐵 𝜃 𝐶

Figure P12.95

ISTUDY

Section 12.3

697

Projectile Motion

Problem 12.96 If the projectile is released at 𝐴 with initial speed 𝑣0 and angle 𝛽, derive the projectile’s trajectory, using the coordinate system shown. Neglect air resistance. 𝑦

𝑣0 𝑤

𝛽 𝐴

ℎ 𝑦

𝑂

𝑚 = 50 kg

𝑣⃗0

𝑥 Figure P12.96

𝑥

𝑂

Problem 12.97 ℎ = 4.5 m

A trebuchet releases a rock with mass 𝑚 = 50 kg at point 𝑂. The initial velocity of the projectile is 𝑣⃗0 = (45 𝚤̂ + 30 𝚥̂) m∕s. Neglecting aerodynamic effects, determine where the rock will land and its time of flight.

Problem 12.98

Figure P12.97

A golfer chips the ball into the hole on the fly from the rough at the edge of the green. Letting 𝛼 = 4◦ and 𝑑 = 2.4 m, verify that the golfer will place the ball within 10 mm of the center of the hole if the ball leaves the rough with a speed 𝑣0 = 5.03 m∕s and an angle 𝛽 = 41◦ . 𝑦

𝑣0 𝛼

𝛽

𝑥

𝑑 Figure P12.98

Problems 12.99 and 12.100 In a movie scene involving a car chase, a car goes over the top of a ramp at 𝐴 and lands at 𝐵 below. If 𝛼 = 20◦ and 𝛽 = 23◦ , determine the distance 𝑑 covered by the car if the car’s speed at 𝐴 is 45 km∕h. Neglect aerodynamic effects.

Problem 12.99

Problem 12.100 Determine the speed of the car at 𝐴 if the car is to cover distance 𝑑 = 150 f t for 𝛼 = 20◦ and 𝛽 = 27◦ . Neglect aerodynamic effects.

𝐴

𝛼 𝛽 𝑑

Figure P12.99 and P12.100

𝐵

698

Chapter 12

Particle Kinematics

Problem 12.101 The M777 lightweight 155 mm howitzer is a piece of artillery whose rounds are ejected from the gun with a speed of 829 m∕s. Assuming that the gun is fired over a flat battlefield and ignoring aerodynamic effects, determine (a) the elevation angle needed to achieve the maximum range, (b) the maximum possible range of the gun, and (c) the time it would take a projectile to cover the maximum range. Express the result for the range as a percentage of the actual maximum range of this weapon, which is 30 km for unassisted ammunition. 𝐵 14 f t 𝑣0 𝑂 Stocktrek Images/Getty Images

Figure P12.101

𝐴 8 ft

𝜃 30 f t Figure P12.102

Problem 12.102 You want to throw a rock from point 𝑂 to hit the vertical advertising sign 𝐴𝐵, which is 𝑅 = 30 f t away. You can throw a rock at the speed 𝑣0 = 45 f t∕s. The bottom of the sign is 8 f t off the ground and the sign is 14 f t tall. Determine the range of angles at which the projectile can be thrown in order to hit the target, and compare this with the angle subtended by the target as seen from an observer at point 𝑂. Compare your results with those found in Example 12.11.

𝑦

Problem 12.103

𝑣0 𝑣0 𝜃2 𝑂

𝜃1

𝐷

𝑅

𝑥

Suppose that you can throw a projectile at a large enough 𝑣0 so that it can hit a target a distance 𝑅 downrange. Given that you know 𝑣0 and 𝑅, determine the general expressions for the two distinct launch angles 𝜃1 and 𝜃2 that will allow the projectile to hit 𝐷. For 𝑣0 = 30 m∕s and 𝑅 = 70 m, determine numerical values for 𝜃1 and 𝜃2 .

Figure P12.103

Problem 12.104

ℎ

𝐴

𝛼 𝛽

Figure P12.104

ISTUDY

𝐵

An alpine ski jumper can fly distances in excess of 100 m∗ by using his or her body and skis as a “wing” and therefore, taking advantage of aerodynamic effects. With this in mind and assuming that a ski jumper could survive the jump, determine the distance the jumper could “fly” without aerodynamic effects, i.e., if the jumper were in free fall after clearing the ramp. For the purpose of your calculation, use the following typical data: 𝛼 = 11◦ (slope of ramp at takeoff point 𝐴), 𝛽 = 36◦ (average slope of the hill),† 𝑣0 = 86 km∕h (speed at 𝐴), ℎ = 3 m (height of takeoff point with respect to the hill). Finally, for simplicity, let the jump distance be the distance between the takeoff point 𝐴 and the landing point 𝐵. ∗ On

March 20, 2005, using the very large ski ramp at Planica, Slovenia, Bjørn Einar Romøren of Norway set the world record by flying a distance of 239 m. † While the given average slope of the landing hill is accurate, you should know that, according to regulations, the landing hill must have a curved profile. Here, we have chosen to use a landing hill with a constant slope of 36◦ to simplify the problem.

ISTUDY

Section 12.3

Projectile Motion

699

Problems 12.105 and 12.106 A soccer player practices kicking a ball from 𝐴 directly into the goal (i.e., the ball does not bounce first) while clearing a 6 f t tall fixed barrier. 80 f t 58 f t 𝐴

6 ft

8 ft

Figure P12.105 and P12.106 Problem 12.105

Determine the minimum speed that the player needs to give the ball to accomplish the task. Hint: Consider the equation for the projectile’s trajectory of the form 𝑦 = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 , with the 𝑦 axis parallel to the direction of gravity, for the case in which the ball reaches the goal at its base. Solve this equation for the initial speed 𝑣0 as a function of the initial angle 𝜃, and finally find (𝑣0 )min as you learned in calculus. Don’t forget to check whether or not the ball clears the barrier. Problem 12.106 Find the initial speed and angle that allow the ball to barely clear the barrier while barely reaching the goal at its base. Hint: A projectile’s trajectory can be given the form 𝑦 = 𝐶1 𝑥 − 𝐶2 𝑥2 , where the coefficients 𝐶1 and 𝐶2 can be found by forcing the parabola to go through two given points.

Problems 12.107 and 12.108

𝐵

In a circus act a tiger is required to jump from point 𝐴 to point 𝐶 so that it goes through the ring of fire at 𝐵. Hint: A projectile’s trajectory can be given the form 𝑦 = 𝐶1 𝑥 − 𝐶2 𝑥2 , where the coefficients 𝐶1 and 𝐶2 can be found by forcing the parabola to go through two given points.

3m 𝐴 𝑑

Problem 12.107

Determine the tiger’s initial velocity if the ring of fire is placed at a distance 𝑑 = 5.5 m from 𝐴. Furthermore, determine the slope of the tiger’s trajectory as the tiger goes through the ring of fire.

𝐶

0.5 m 9m

Figure P12.107 and P12.108

Problem 12.108 Determine the tiger’s initial velocity, as well as the distance 𝑑 so that the slope of the tiger’s trajectory as the tiger goes through the ring of fire is completely horizontal.

Problem 12.109 A jaguar 𝐴 leaps from 𝑂 at speed 𝑣0 and angle 𝛽 relative to the incline to attack a panther 𝐵 at 𝐶. Determine an expression for the maximum perpendicular height ℎmax above the incline achieved by the leaping jaguar, given that the angle of the incline is 𝜃.

𝑣0 𝐴

parabolic trajectory of 𝐴

𝑂 𝛽

Problems 12.110 and 12.111 The jaguar 𝐴 leaps from 𝑂 at speed 𝑣0 and angle 𝛽 relative to the incline to intercept the panther 𝐵 at 𝐶. The distance along the incline from 𝑂 to 𝐶 is 𝑅, and the angle of the incline with respect to the horizontal is 𝜃. Problem 12.110

𝑔

𝑅

Derive 𝑣0 as a function of 𝛽 to leap a given distance 𝑅 along with the optimal value of launch angle 𝛽, i.e., the value of 𝛽 necessary to leap a given distance 𝑅 with the minimum 𝑣0 . Then plot 𝑣0 as a function of 𝛽 for 𝑔 = 9.81 m∕s2 ,

𝐵 𝜃

Determine an expression for 𝑣0 as a function of 𝛽 for 𝐴 to be able to

get from 𝑂 to 𝐶. Problem 12.111

ℎmax

𝐶 Figure P12.109–P12.111

700

Chapter 12

Particle Kinematics

𝑅 = 7 m, and 𝜃 = 25◦ , and find a numerical value of the optimal 𝛽 and the corresponding value of 𝑣0 for the given set of parameters.

Problem 12.112 A ball is projected horizontally off of a 150 f t cliff at 50 f t∕s and it lands at 𝐶. Wind is blowing horizontally to give the ball a constant 𝑥 component of acceleration 𝑎𝑤 . Neglecting air resistance in the vertical direction, determine (a) the time it takes for the ball to hit the ground, (b) the 𝑥 component of the ball’s acceleration, (c) the time for the ball to get to 𝐵, that is, when its velocity vector is at 45◦ .

Problems 12.113 and 12.114 A stomp rocket is a toy consisting of a hose connected to a blast pad (i.e., an air bladder) at one end and to a short pipe mounted on a tripod at the other end. A rocket with a hollow body is mounted onto the pipe and is propelled into the air by stomping on the blast pad.

Figure P12.112

𝐵 ℎ

𝜃

If the rocket can be imparted an initial speed 𝑣0 = 120 f t∕s, and if the Problem 12.113 rocket’s landing spot at 𝐵 is at the same elevation as the launch point, i.e., ℎ = 0 f t, neglect air resistance and determine the rocket’s launch angle 𝜃 such that the rocket achieves the maximum possible range. In addition, compute 𝑅, the rocket’s maximum range, and 𝑡𝑓 , the corresponding flight time. Problem 12.114 Assuming the rocket can be given an initial speed 𝑣0 = 120 f t∕s, the rocket’s landing spot at 𝐵 is 10 f t higher than the launch point, i.e., ℎ = 10 f t, and neglecting air resistance, find the rocket’s launch angle 𝜃 such that the rocket achieves the maximum possible range. In addition, as part of the solution, compute the corresponding maximum range and flight time. To do this:

Figure P12.113 and P12.114

(a) Determine the range 𝑅 as a function of time. (b) Take the expression for 𝑅 found in (a), square it, and then differentiate it with respect to time to find the flight time that corresponds to the maximum range, and then find that maximum range. (c) Use the time found in (b) to then find the angle required to achieve the maximum range.

Problem 12.115 𝑦 𝑚 = 50 kg

𝑣⃗0

𝑥

𝑂

ℎ = 4.5 m

A trebuchet releases a rock with mass 𝑚 = 50 kg at the point 𝑂. The initial velocity of the projectile is 𝑣⃗0 = (45 𝚤̂ + 30 𝚥̂) m∕s. If one were to model the effects of air resistance via a drag force directly proportional to the projectile’s velocity, the resulting accelerations in the 𝑥 and 𝑦 directions would be 𝑥̈ = −(𝜂∕𝑚)𝑥̇ and 𝑦̈ = −𝑔 − (𝜂∕𝑚)𝑦,̇ respectively, where 𝑔 is the acceleration of gravity and 𝜂 = 0.64 kg∕s is a viscous drag coefficient. Find an expression for the trajectory of the projectile.

Problem 12.116 Continue Prob. 12.115 and, for the case where 𝜂 = 0.64 kg∕s, determine the maximum height from the ground reached by the projectile and the time it takes to achieve it. Compare the result with what you would obtain in the absence of air resistance.

Figure P12.115–P12.118

ISTUDY

Problem 12.117 Continue Prob. 12.115 and, for the case where 𝜂 = 0.64 kg∕s, determine 𝑡𝐼 and 𝑥𝐼 , the value of 𝑡, and the 𝑥 position corresponding to the projectile’s impact with the ground.

ISTUDY

Section 12.3

Problem 12.118 With reference to Probs. 12.115 and 12.117, assume that an experiment is conducted so that the measured value of 𝑥𝐼 is 10% smaller than what is predicted in the absence of viscous drag. Find the value of 𝜂 that would be required for the approach in Prob. 12.115 to match the experiment.

Projectile Motion

701

702

Chapter 12

Particle Kinematics

Design Problems

ISTUDY

Design Problem 12.1 In the manufacture of steel balls of the type used for ball bearings, it is important that their material properties be sufficiently uniform. One way to detect gross differences in their material properties is to observe how a ball rebounds when dropped on a hard strike plate. The rebound characteristics of a ball can be assessed via a quantity called the coefficient of restitution (COR).∗ Specifically, if (𝑣𝑛 )strike is the component of a ball’s impact velocity normal to the strike plate, then the COR is given by COR =

|(𝑣𝑛 )rebound | |(𝑣𝑛 )strike |

,

(COR equation)

where (𝑣𝑛 )rebound is the component of the rebound velocity normal to the strike plate. Given that each ball has a radius 𝑅 = 0.2 in., design a sorting device to select the balls with 0.800 < COR < 0.825. The device consists of an incline defined by the angle 𝜃 and length 𝐿. The strike plate is placed at the bottom of a well with depth ℎ and width 𝑤. Finally, at a distance 𝓁 from the incline, there is a thin vertical barrier with a gap of size 𝑑 placed at a height 𝑏 from the bottom of the well. Releasing a ball from rest at the top of the incline and assuming that the ball rolls without slip, we know the ball will reach the bottom of the incline with a speed† √ 𝑔𝐿 sin 𝜃, 𝑣0 = 10 7 where 𝑔 is the acceleration due to gravity. After rolling off the incline, each ball will rebound off the strike plate such that the horizontal component of velocity is unaffected by the impact while the vertical component will behave as described in the COR equation. After rebounding, the balls to be isolated will pass through the vertical gap, whereas the rest of the balls will not go through the gap. In your design, choose appropriate values of 𝐿, 𝜃 < 45◦ , ℎ, 𝑤, 𝑑, and 𝓁 to accomplish the desired task while ensuring that the overall dimensions of the device do not exceed 4 f t in both the horizontal and vertical directions. 𝐿

strike plate 𝓁

𝜃 𝑑 ℎ

𝑏

𝑤 Figure DP12.1

∗ We † We

will study the COR in detail in Section 15.2. will see how to derive this formula in Chapters 17 and 18.

ISTUDY

Section 12.4

12.4

Planar Motion: Normal-Tangential Components

703

Planar Motion: Normal-Tangential Components

It is not always convenient to study motion using a Cartesian coordinate system. In this section and the next, we will learn about two additional ways to describe motion. Here we introduce a way to describe the velocity and acceleration vectors of a particle that is based entirely on the path of the particle.

Normal-tangential components Referring to Fig. 12.16, the particle 𝑃 is moving along an arbitrary path. We denote by 𝑢̂ 𝑡 the unit vector tangent to the path at 𝑃 and pointing in the direction of motion. Since the velocity vector is always tangent to the path, we can write it using 𝑢̂ 𝑡 as 𝑣⃗ = 𝑣 𝑢̂ 𝑡 ,

𝑎⃗ = 𝑣⃗̇ = 𝑣̇ 𝑢̂ 𝑡 + 𝑣 𝑢̂̇ 𝑡 .

(12.43)

To determine a convenient expression for 𝑢̂̇ 𝑡 , we begin by introducing the arc length 𝑠(𝑡), defined as the distance traveled by 𝑃 along its path from 𝑡 = 0 to the current time 𝑡 (Fig. 12.17). Since 𝑠 is the distance traveled, 𝑠 ≥ 0 and, no matter which direction 𝑃 moves along the path, 𝑠 continues to increase. Now, since 𝑣⃗ changes direction as 𝑃 moves along the path, we can view the unit vector 𝑢̂ 𝑡 as a function of 𝑠(𝑡) (unless the path is straight). We can use the rate of change of 𝑢̂ 𝑡 (𝑠) with respect to 𝑠, i.e., the vector 𝑑 𝑢̂ 𝑡 (𝑠)∕𝑑𝑠, to describe how “bendy” the path is, that is, its curvature. The curvature of the path, traditionally denoted by the Greek letter 𝜅 (kappa), is defined as | 𝑑 𝑢̂ (𝑠) | 𝜅(𝑠) = || 𝑡 ||, (12.44) | 𝑑𝑠 | which has dimensions of 1 over length. If the path is straight, 𝑢̂ 𝑡 (𝑠) = constant and 𝜅(𝑠) = 0. Writing 𝑢̂̇ 𝑡 using the chain rule, we see that 𝑑 𝑢̂ 𝑡 𝑑𝑠 𝑑𝑠 𝑑𝑡

⇒

𝑃

path of 𝑃

(12.42)

where 𝑣 is the speed of the particle. To obtain the acceleration of 𝑃 , we differentiate Eq. (12.42) with respect to time to obtain

𝑢̂̇ 𝑡 =

𝑢̂ 𝑡

𝑣⃗ = 𝑣 𝑢̂ 𝑡

𝑎⃗ = 𝑣̇ 𝑢̂ 𝑡 + 𝑣2

𝑑 𝑢̂ 𝑡 𝑑𝑠

,

Figure 12.16 The velocity vector 𝑣⃗ defined in terms of the unit tangent vector 𝑢̂ 𝑡 . path of 𝑃

𝑠

𝑃

𝑠=0 Figure 12.17 A point 𝑃 moving along a path showing the arc length 𝑠.

(12.45)

where we have used Eq. (12.43) and 𝑣 = 𝑑𝑠∕𝑑𝑡 from Eq. (12.9) on p. 651. We now find an expression for 𝑑 𝑢̂ 𝑡 ∕𝑑𝑠 by recalling that 𝑢̂ 𝑡 (𝑠) is a unit vector, which implies that 𝑢̂ 𝑡 (𝑠) ⋅ 𝑢̂ 𝑡 (𝑠) = 1 = constant. Therefore, we have 𝑑 𝑢̂ 𝑑 𝑢̂ ) 𝑑 𝑢̂ 𝑑 ( 𝑢̂ 𝑡 ⋅ 𝑢̂ 𝑡 = 𝑡 ⋅ 𝑢̂ 𝑡 + 𝑢̂ 𝑡 ⋅ 𝑡 = 2𝑢̂ 𝑡 ⋅ 𝑡 = 0, 𝑑𝑠 𝑑𝑠 𝑑𝑠 𝑑𝑠

(12.46)

which means that 𝑢̂ 𝑡 and 𝑑 𝑢̂ 𝑡 ∕𝑑𝑠 are orthogonal to one another. Consequently, when 𝜅 ≠ 0, we can define a curvature-related unit normal to the path as: 𝑢̂ 𝑛 (𝑠) =

𝑑 𝑢̂ 𝑡 (𝑠)∕𝑑𝑠

=

1 𝑑 𝑢̂ 𝑡 (𝑠) , 𝜅(𝑠) 𝑑𝑠

𝑢̂ 𝑡 (𝑠)

𝑃 𝑠

𝑢̂ 𝑛 (𝑠)

(12.47)

𝑠=0

where the subscript 𝑛 stands for normal. Referring to Fig. 12.18, the unit vector 𝑢̂ 𝑛 (𝑠) is called the principal unit normal to the path, and it always points toward the

Figure 12.18 Principal unit normal to the path at 𝑃 .

|𝑑 𝑢̂ 𝑡 (𝑠)∕𝑑𝑠|

704

Chapter 12

Particle Kinematics

𝑎𝑡 = 𝑣̇

𝑢̂ 𝑡

concave side of the curve. Because 𝑢̂ 𝑛 (𝑠) is only defined when 𝜅 ≠ 0, we cannot use 𝑢̂ 𝑛 (𝑠) when dealing with straight lines or at inflection points of curved lines. If at 𝑃 the curvature 𝜅 ≠ 0, then Eq. (12.47) can be rewritten as

𝑃 𝑠 𝑢̂ 𝑛

path of 𝑃

𝑎𝑛 =

𝑑 𝑢̂ 𝑡 𝑑𝑠

𝑣2

=

1 𝑢̂ , 𝜌 𝑛

where

𝜌(𝑠) =

1 𝜅(𝑠)

(12.48)

𝜌

is called the radius of curvature of the path and, in general, it changes along the path. If 𝜌(𝑠) is constant, then the path is a circle. Going back to the expression for the acceleration, substituting the first of Eqs. (12.48) into the second of Eqs. (12.45), we obtain

𝑠=0

𝐶 𝜌

𝑎⃗ = 𝑣̇ 𝑢̂ 𝑡 +

osculating circle Figure 12.19 Acceleration in normal-tangential components. The osculating circle is the circle tangent to the path at 𝑃 with radius 𝜌 and center 𝐶 on the concave side of the path. Even for three-dimensional motions, the acceleration vector lies always in the same plane as the osculating circle.

𝑣2 𝑢̂ = 𝑎𝑡 𝑢̂ 𝑡 + 𝑎𝑛 𝑢̂ 𝑛 , 𝜌 𝑛

(12.49)

where 𝑎𝑡 = 𝑣̇ and 𝑎𝑛 = 𝑣2 ∕𝜌 are the tangential and normal components of the acceleration, respectively (see Fig. 12.19). In a three-dimensional context, the direction perpendicular to the plane defined by 𝑢̂ 𝑡 and 𝑢̂ 𝑛 is identified by the unit vector 𝑢̂ 𝑏 (𝑠) = 𝑢̂ 𝑡 (𝑠) × 𝑢̂ 𝑛 (𝑠),

(12.50)

which is called the binormal unit vector. The three unit vectors 𝑢̂ 𝑡 , 𝑢̂ 𝑛 , and 𝑢̂ 𝑏 are known as the intrinsic triad or the Serret-Frenet triad. 𝑦 𝜌2 = 0.231 .23

3 2

Radius of curvature in Cartesian coordinates. In Cartesian coordinates, the planar path of a particle is usually given the form 𝑦 = 𝑦(𝑥). In this case, geometry tells us that the radius of curvature at any position 𝑥 is given by

𝜌3 = 66.770

1 2 −1

4

6

𝑥

𝜌1 = 0.450

Figure 12.20 A 2D path showing its radius of curvature at three different points.

𝑣⃗ = 𝑣 𝑢̂ 𝑡

𝑢̂ 𝑡

[ ]3∕2 1 + (𝑑𝑦∕𝑑𝑥)2 𝜌(𝑥) = . | 2 | |𝑑 𝑦∕𝑑𝑥2 | | |

(12.51)

As an example, Fig. 12.20 shows a plot of the path 𝑦(𝑥) = (1−𝑥) sin 𝑥 for 0 ≤ 𝑥 ≤ 2𝜋. We have used Eq. (12.51) to compute the radii of curvature of 𝑦(𝑥) at three different points. The first point, the local minimum at 𝑥 = 2.24, has 𝜌1 = 0.450. The second, the local maximum at 𝑥 = 4.96, has 𝜌2 = 0.231. Finally, at 𝑥 = 5.60, 𝜌3 = 6.70. Figure 12.20 demonstrates the intuitive notion that the “tighter the bend” in the path, the smaller its radius of curvature and the “straighter the path,” the larger its radius of curvature.

End of Section Summary

𝑃

path of 𝑃

In this section we have derived expressions for the velocity and acceleration of a point using the normal-tangential component system. Referring to Fig. 12.21, the velocity vector has the form Eq. (12.42), p. 703

Figure 12.21 The velocity vector 𝑣⃗ defined in terms of the unit tangent vector 𝑢̂ 𝑡 .

ISTUDY

𝑣⃗ = 𝑣 𝑢̂ 𝑡 , where 𝑣 is the speed and 𝑢̂ 𝑡 is the tangent unit vector at the point 𝑃 . Referring to Fig. 12.22, the acceleration vector in normal-tangential components

ISTUDY

Section 12.4

Planar Motion: Normal-Tangential Components

𝑎𝑡 = 𝑣̇

𝑢̂ 𝑡

path of 𝑃

𝑃 𝑠 𝑢̂ 𝑛 𝑎𝑛 =

𝑣2 𝜌

𝑠=0

𝐶 𝜌

osculating circle

Figure 12.22. Acceleration in normal-tangential components.

has the form Eq. (12.49), p. 704 𝑎⃗ = 𝑣̇ 𝑢̂ 𝑡 +

𝑣2 𝑢̂ = 𝑎𝑡 𝑢̂ 𝑡 + 𝑎𝑛 𝑢̂ 𝑛 , 𝜌 𝑛

where 𝑣̇ = 𝑎𝑡 is the tangential component of acceleration and 𝑣2 ∕𝜌 = 𝑎𝑛 is the normal component of acceleration. When we are using a Cartesian coordinate system, for a path expressed by a relation, such as 𝑦 = 𝑦(𝑥), the path’s radius of curvature is given by 𝜌: Eq. (12.51), p. 704 [ ]3∕2 1 + (𝑑𝑦∕𝑑𝑥)2 𝜌(𝑥) = . | 2 | |𝑑 𝑦∕𝑑𝑥2 | | |

705

706

Chapter 12

Particle Kinematics

E X A M P L E 12.13 T1 2 112 70 1.2 2.0 89 143 2

4 225 140

Sainte Devote

Beau Rivage

2.2 80 128 2

55 88 2

Anthony Noghes

2 80 50 Mirabeau

2 88 55

6 273 170 Casino

165 265 6

T3

Relating the Shape of the Path to Acceleration

Tabac Nouvelle Chicane 4 204 127 6 282 175 T2 Piscine 3 193 120 1.7

Portier Grand Hotel Hairpin

2 64 55

1 45 30 Tunnel 6 264 165 2.6

2 141 88 1.5 Rascasse

Let’s assume that by lateral 𝐺-force the Federation Internationale de l’Automobile (FiA) that compiled the map in Fig. 1 really meant to provide a measurement of the acceleration normal to the path of the racing cars expressed in “units of 𝑔,” where 𝑔 is the acceleration due to gravity. Use this information along with the reported speed to estimate the radius of curvature of the Monaco Formula 1 track (1) at the stretch preceding the Rascasse and (2) at the end of the Tunnel.

SOLUTION Road Map

2 128 80

Key Saint 5 251 156 2.4 T3 Devote Start Gear Speed Speed Lateral Turn Timing Sector Target Finish Location (Km/h) (mph) G-force Name Sector Time Lap Time Location

Figure 1 Formula 1 track at Monaco. The locations of the Tunnel and the Rascasse are indicated in gold. Additional information is provided including typical speed, acceleration, and gear information at various important points along the track.

If we assume that a Formula 1 car can be modeled as a particle, then the solution to this problem is obtained by using the relation linking the speed to the component of the acceleration normal to the path, that is, 𝑎𝑛 = 𝑣2 ∕𝜌. Computation

As indicated in the problem statement, if one then assumes that the lateral 𝐺-force is the component of the acceleration normal to the path, then at the end of the stretch preceding the Rascasse (see Fig. 2) we have 𝑣 = 141 km∕h = 39.17 m∕s and

𝑎𝑛 = 1.5𝑔 = 14.72 m∕s2 ,

(1)

so that 𝜌=

𝑣2 = 104.2 m. 𝑎𝑛

(2)

Proceeding in a similar way for the Tunnel (see Fig. 3), we have Anthony Noghes

𝑢̂ 𝑛

𝑢̂ 𝑡

𝑎𝑛 = 1.5𝑔

𝑎𝑛 = 2.6𝑔 𝑢̂ 𝑛 𝑢̂ 𝑡

Tunnel 6 264 165 2.6

Figure 3 Normal tangential component system at the end of the Tunnel. The normal component of the acceleration is also shown. Refer to Fig. 1 for the key to the numbers shown in the blue boxes.

𝑎𝑛 = 2.6𝑔 = 25.51 m∕s2 ,

(3)

so that 𝜌=

Rascasse

Figure 2 Normal tangential component system right before the Rascasse. The normal component of the acceleration is also shown. Refer to Fig. 1 for the key to the numbers shown in the blue boxes.

ISTUDY

𝑣 = 264 km∕h = 73.33 m∕s and

2 141 88 1.5

𝑣2 = 210.8 m. 𝑎𝑛

(4)

Discussion & Verification

The solution was obtained as a direct application of the formula for the normal component of the acceleration. It is therefore elementary to verify that the dimensions are correct and proper units have been used. A Closer Look A basic question to ask is, How good are these estimates? After some research, the authors were able to obtain a map of the circuit from which the radii of curvature of most turns could be directly measured, although without taking into account changes in elevation along the curves. From this map, the radius of curvature in the first calculation was measured to be roughly 90 m, whereas the radius of curvature midway through the tunnel was found to be roughly 200 m. Hence, we can conclude that the gross estimates we have derived from the map in Fig. 1 and those that were obtained by direct (although still approximate) measurement are in reasonable agreement.

ISTUDY

Section 12.4

707

Planar Motion: Normal-Tangential Components

E X A M P L E 12.14

Curvature and Projectile Motion

A pumpkin has been launched from a trebuchet at the annual World Championship “Punkin Chunkin” competition in Millsboro, Delaware. The pumpkin, shown in Fig. 1, is assumed to have been released at 𝑂 with an initial speed 𝑣0 and an elevation angle 𝛽. Determine the time rate of change of the speed at the time of release and the radius of curvature of the pumpkin’s trajectory at its highest point.

𝐻

𝑦

𝑣0

𝚥̂ 𝑂

𝛽 𝚤̂

𝑥

SOLUTION Road Map

By modeling the pumpkin’s motion as a projectile motion, the pumpkin’s acceleration is the acceleration of gravity 𝑔 in the negative 𝑦 direction. The key to the solution is then to compute the tangential and normal components of the acceleration, since the rate of change of the speed is 𝑣̇ = 𝑎𝑡 and the radius of curvature 𝜌 is such that 𝑎𝑛 = 𝑣2 ∕𝜌.

Figure 1 A pumpkin launched by a trebuchet.

Computation Referring to Fig. 2, recall that the velocity is always tangent to the path. Therefore, the tangent to the path at 𝑂 is oriented at an angle 𝛽 with respect to the 𝑥 axis. Therefore, we have

𝑣̇ 𝑂 = 𝑎𝑂𝑡 = −𝑔 sin 𝛽,

and 𝑎𝐻𝑛 = 𝑔.

𝑣𝐻 = 𝑣0 cos 𝛽.

⇒

(3)

2

Using the normal component of Eq. (12.49), we have 𝑎𝑛 = 𝑣 ∕𝜌. Therefore, using Eqs. (2) and (3), we have 𝜌𝐻 =

𝑣2𝐻 𝑎𝐻𝑛

=

𝑣20 cos2 𝛽 𝑔

.

𝑢̂ 𝑡

𝛽

𝑂

𝑥 𝑢̂ 𝑛

𝛽 𝑔

Figure 2 Normal-tangential component system at 𝑂.

(2)

Furthermore, the velocity at 𝐻 is completely in the 𝑥 direction. From projectile motion, we know that the horizontal component of the velocity remains constant throughout the motion. Therefore, we have 𝑣⃗𝐻 = 𝑣𝐻𝑥 𝚤̂ = 𝑣0 cos 𝛽 𝚤̂

𝑣0

(1)

where 𝑣𝑂 and 𝑎𝑂𝑡 denote the speed at 𝑂 and the tangential component of the acceleration at 𝑂, respectively. Considering the situation at the highest point on the trajectory, referring to Fig. 3, observe that the tangent to the trajectory at this point is completely horizontal. Therefore, calling 𝐻 the highest point on the trajectory, since the pumpkin’s acceleration is completely in the 𝑦 direction, we have 𝑎𝐻𝑡 = 0

𝑦

(4)

Discussion & Verification Both the results in Eqs. (1) and (4) are dimensionally correct. Referring to Eq. (1), observe that sin 𝛽 is nondimensional, and therefore, 𝑣̇ 𝑂 has the same dimensions of 𝑔, i.e., the dimensions of acceleration, as expected. As far as Eq. (4) is concerned, recall that the dimensions of 𝑣0 are length over time and those of 𝑔 are length over time squared. Therefore, 𝜌 has dimensions of length, as expected. From Eq. (1), observe that 𝑣̇ 𝑂 is negative. This is to be expected since, up until the projectile reaches point 𝐻, the speed of the projectile is expected to decrease. Finally, from Eq. (4), observe that 𝜌 increases with the horizontal component of the initial velocity. This is to be expected since the “flatness” of the overall trajectory is governed by the horizontal component of the velocity. Hence, the solution we have obtained appears to be correct.

𝑣0 cos 𝛽

𝐻 𝑢̂ 𝑡

𝑢̂ 𝑛 𝑔

Figure 3 Normal-tangential component system at 𝐻.

708

Chapter 12

Particle Kinematics

E X A M P L E 12.15

𝐴

Accelerations in Circular Motion The Center for Gravitational Biology Research at NASA’s Ames Research Center runs a large centrifuge capable of simulating 20𝑔 of acceleration (12.5𝑔 is the maximum for human subjects). The radius of the centrifuge is 29 f t, and the distance from the axis of rotation to the cab at either 𝐴 or 𝐵 is about 25 f t. Assuming that the centrifuge accelerates uniformly to reach its final speed and that it takes 12.5 s to do so, determine

𝐵

NASA

Figure 1 The 20𝑔 centrifuge at the Center for Gravitational Biology Research, which is part of NASA’s Ames Research Center in Moffett Field, California.

(a) The angular velocity of the centrifuge 𝜔𝑓 required to maintain a final acceleration of 20𝑔 in cab 𝐴. (b) The magnitude of the acceleration of cab 𝐴 as a function of time from the moment the centrifuge starts spinning until its final speed is achieved.

SOLUTION Road Map Since we know the magnitude of the final acceleration of 𝐴 (20𝑔), we can use the acceleration relationships developed in this section to determine the final value of the angular velocity of the centrifuge. Because the centrifuge accelerates uniformly during spin-up, we can use the constant acceleration relations from Section 12.2 to relate the centrifuge’s final angular velocity to the needed angular acceleration during spin-up. Once the centrifuge’s acceleration is known, we will be able to derive an expression for the acceleration of 𝐴 as a function of time during spin-up.

𝑢̂ 𝑏 𝑢̂ 𝑛

𝜔 𝐶

𝐴

𝐵 Computation

25 f t

𝑢̂ 𝑛

𝐴 𝑢̂ 𝑡

𝐶 simplified top view 𝜌

The key element of the solution of this problem is the relation between the motion of 𝐴 and the motion of the centrifuge as a whole. To establish this relation, consider Fig. 2 and observe that the path of 𝐴 is a circle with center 𝐶 on the spin axis of the centrifuge. Thus, the angular speed of the centrifuge and the speed of 𝐴 can be related using Eq. (12.37) to obtain 𝑣 (1) 𝜔= , 𝜌 where 𝑣 is the speed of 𝐴 and 𝜌 is the radius of the path of 𝐴. When the magnitude of the acceleration of 𝐴 reaches its final value 𝑎𝑓 = 20𝑔, the speed of 𝐴 will become constant, i.e., 𝑣̇ 𝑓 = 0, and 𝑎𝑓 will coincide with just the normal component of Eq. (12.49) so that 𝑎𝑓 =

path of 𝐴 Figure 2 The 20𝑔 centrifuge showing the axis of rotation and the path of 𝐴 as well as the normal-tangential component system at 𝐴.

ISTUDY

𝑣2𝑓 𝜌

= 𝜌𝜔2𝑓

⇒

20𝑔 = 20(32.2 f t∕s2 ) = 𝜌𝜔2𝑓 ,

(2)

where we have used Eq. (1), 𝜔𝑓 is the final value of 𝜔, and we have used 𝑔 = 32.2 f t∕s2 . Since 𝜌 = 25 f t, solving for 𝜔𝑓 , we obtain 𝜔𝑓 = 5.075 rad∕s = 48.47 rpm.

(3)

We can now use Eq. (12.32), along with Table 12.1, to obtain the angular acceleration of the centrifuge during spin-up as 𝜔𝑓 = 𝜔0 + 𝛼𝑡𝑓

⇒

𝛼 = 𝜔𝑓 ∕𝑡𝑓

⇒

𝛼 = 0.4060 rad∕s2 ,

(4)

where we have used 𝜔0 = 0 and 𝑡𝑓 = 12.5 s. To relate 𝛼 to the acceleration of 𝐴, we rewrite Eq. (1) as 𝑣 = 𝜌𝜔, and differentiating with respect to time, we obtain 𝑣̇ = 𝜌𝜔̇ = 𝜌𝛼, (5) where we have used the fact that 𝜌 is constant. The quantity 𝑣̇ is the tangential component of the acceleration of 𝐴. The normal component of the acceleration of 𝐴 is given by 𝑎𝑛 =

𝑣2 = 𝜌𝜔2 = 𝜌𝛼 2 𝑡2 , 𝜌

(6)

ISTUDY

Section 12.4

Planar Motion: Normal-Tangential Components

where we have used 𝜔 = 𝛼𝑡 from Eq. (4), which is the value of 𝜔 at an arbitrary time 𝑡 during spin-up. Therefore, the magnitude of the acceleration during spin-up is given by √ √ √ 𝑎 = 𝑎2𝑡 + 𝑎2𝑛 = 𝜌2 𝛼 2 + 𝜌2 𝛼 4 𝑡4 = 𝜌𝛼 1 + 𝛼 2 𝑡4 . (7) Recalling that 𝜌 = 25 f t and using the result in Eq. (4), we have ( )√ 𝑎 = 10.15 f t∕s2 1 + (0.1649 s−4 )𝑡4 .

(8)

Discussion & Verification Equations (2) and (7) present our results in symbolic form and allow us to easily verify that our results are dimensionally correct. In addition, the numerical form of the results presented in Eqs. (3) and (8) is expressed using appropriate and consistent units. Overall, our solution indicates that the speed of 𝐴 increases uniformly, and this is consistent with the given piece of information that the spin-up of the centrifuge occurs at a constant rate. Hence, our solution appears to be correct. A Closer Look The magnitude of the acceleration of 𝐴, as given in Eq. (8), is an increasing function of time. Hence, 𝑎 is largest at 𝑡 = 12.5 s (the end of the spin-up) and its value is given by 𝑎max = 644.1 f t∕s2 = 20.00𝑔, (9)

which appears to be the same as the value of 𝑎𝑓 . The result in Eq. (9) is somewhat unexpected because the value of 𝑎𝑓 is based only on the normal acceleration of 𝐴 at the end of spin-up, whereas 𝑎max includes the contributions of both the normal and tangential components of the acceleration of 𝐴. This indicates that the tangential component of acceleration contributes an insignificant amount to the total acceleration. When Eq. (7) is evaluated at the end of spin-up (𝑡 = 12.5 s), 𝑎𝑡 = 10.15 f t∕s2 [see the leading term on the right-hand side of Eq. (8)], and 𝑎𝑛 = 644.0 f t∕s2 , so we can see that 𝑎𝑛 is over 63 times larger than 𝑎𝑡 . We conclude by observing that our solution assumes that the angular acceleration is constant until 𝑡 = 12.5 s, at which time it becomes zero, so the angular velocity becomes constant. This assumption is not entirely realistic since such abrupt changes in acceleration are not typically found in applications.

709

710

Chapter 12

Particle Kinematics

Problems Problem 12.119 𝑣⃗

1.0 𝑃 𝑦 (in.)

A particle 𝑃 is moving along the curve 𝐶, whose equation is given by

𝜙

( 2 ) ( )2 𝑦 − 𝑥2 (𝑥 − 1)(2𝑥 − 3) = 4 𝑥2 + 𝑦2 − 2𝑥 ,

𝑎⃗

0.0

at a constant speed 𝑣𝑐 . For any position on the curve 𝐶 for which the radius of curvature is defined (i.e., not equal to infinity), what must be the angle 𝜙 between the velocity vector 𝑣⃗ and the acceleration vector 𝑎? ⃗ 𝐶

−1.0

Problem 12.120 0.0

1.0 𝑥 (in.)

Figure P12.119

2.0

A particle 𝑃 is moving along a path with the velocity shown. Is the sketch of the normaltangential component system at 𝑃 correct? 𝑣⃗

𝑣⃗ 𝑃 𝑢̂ 𝑛

𝑢̂ 𝑡 𝑢̂ 𝑡

path of 𝑃

𝑢̂ 𝑛 𝑃

path of 𝑃

Figure P12.120

Figure P12.121

Problem 12.121 A particle 𝑃 is moving along a path with the velocity shown. Is the sketch of the normaltangential component system at 𝑃 correct?

Problem 12.122 A particle 𝑃 is moving along a straight line with the velocity and acceleration shown. What is wrong with the unit vectors shown in the figure? 𝑢̂ 𝑛

𝑢̂ 𝑛

𝑢̂ 𝑡

𝑣⃗ path of 𝑃

𝑃

Figure P12.122

𝑎⃗

𝑣⃗

𝑢̂ 𝑡

𝑎⃗ 𝑃

Figure P12.123

Problem 12.123 A particle 𝑃 is moving along some path with the velocity and acceleration shown. Can the path of 𝑃 be the straight line shown?

𝑣0 𝛽

Problem 12.124 Figure P12.124

ISTUDY

The water jet of a fountain is let out at a speed 𝑣0 = 80 f t∕s and at an angle 𝛽 = 60◦ . Determine the radius of curvature of the jet at its highest point.

ISTUDY

Section 12.4

711

Planar Motion: Normal-Tangential Components

Problem 12.125 A telecommunications satellite is made to orbit the Earth in such a way as to appear to hover in the same point in the sky as seen by a person standing on the surface of the Earth. Assuming that the satellite’s orbit is circular with radius 𝑟𝑔 = 1.385 × 108 f t and knowing that the speed of the satellite is constant and equal to 𝑣𝑔 = 1.008 × 104 f t∕s, determine the magnitude of the acceleration of the satellite.

𝑟𝑔 Earth 𝑣𝑔

Problem 12.126 A car travels along a city roundabout with radius 𝜌 = 30 m. At the instant shown, the speed of the car is 𝑣 = 35 km∕h and the magnitude of the acceleration of the car is 4.5 m∕s2 . If the car is increasing its speed, determine the time rate of change of the speed of the car at the instant shown.

Figure P12.125

𝑣

Problem 12.127 A car travels along a city roundabout with radius 𝜌 = 100 f t. At the instant shown, the speed of the car is 𝑣 = 25 mph and the speed is decreasing at the rate 8 f t∕s2 . Determine the magnitude of the acceleration of the car at the instant shown.

𝜌

Problem 12.128 Making the same assumptions stated in Example 12.13, consider the map of the Formula 1 circuit at Hockenheim in Germany and estimate the radius of curvature of the curves S¨udkurve and Nordkurve (at the locations indicated in gold). 1 70 43 186 300 6

2 110 62

168 270 6

T2 Spitzkehre

Figure P12.126 and P12.127

2 150 94 3.5 Mobil 1 Kurve Südkurve

4 244 152

2 150 94 3.0

4 174 109 192 310 6

155 250 5

3.0 65 105 1

Hochgeschwindigkeits (Parabolica) - Kurve

T3 Sachs

T1

74 120 2

Pit Lane

6 270 168 3 200 125 3.4

Nordkurve

68 110 2 Einfarhrt Parabolica

Key

𝑦

Mobil 1 T3 Kurve Start Gear Speed Speed Lateral Turn Timing Sector Target Finish Location (Km/h) (mph) G-force Name Sector Time Lap Time Location 5

251

156

2.4

Figure P12.128

𝐶 𝐿

Problem 12.129 The position of the piston 𝐶, as a function of the crank angle 𝜙√and the lengths of the crank 𝐴𝐵 and connecting rod 𝐵𝐶, is given by 𝑦𝐶 = 𝑅 cos 𝜙 + 𝐿 1 − (𝑅 sin 𝜙∕𝐿)2 and 𝑥𝐶 = 0. Using the component system shown, express 𝑢̂ 𝑡 , the unit vector tangent to the trajectory of 𝐶, as a function of the crank angle 𝜙 for 0 ≤ 𝜙 ≤ 2𝜋 rad.

𝐵

𝜙 𝐴

𝑅 Figure P12.129

𝑦𝐶

𝚥̂ 𝚤̂

𝑥

712

Chapter 12

Particle Kinematics

Problem 12.130 An aerobatics plane initiates the basic loop maneuver such that, at the bottom of the loop, the plane is going 140 mph, while subjecting the plane to approximately 4𝑔 of acceleration. Estimate the corresponding radius of the loop.

𝜌

Problems 12.131 through 12.133 Figure P12.130

The portion of a race track between points 𝐴 (corresponding to 𝑥 = 0) and 𝐵 is part of a parabolic curve described by the equation 𝑦 = 𝜅𝑥2 , where 𝜅 is a constant. Let 𝑔 denote the acceleration due to gravity.

𝑦

Determine 𝜅 such that a car driving at constant speed 𝑣0 = 180 mph experiences at 𝐴 an acceleration with magnitude equal to 1.5𝑔.

Problem 12.131 𝐵

If 𝜅 = 0.4 × 10−3 f t −1 , determine 𝑑 such that a car driving at constant speed 𝑣0 = 180 mph experiences at 𝐵 an acceleration with magnitude equal to 𝑔. Problem 12.132

𝑣0

𝐴

𝑥

𝑑

Problem 12.133 Suppose a car travels from 𝐴 to 𝐵 with a constant speed 𝑣0 = 180 mph. Let |𝑎| ⃗ min and |𝑎| ⃗ max denote the minimum and maximum values of the magnitude of the acceleration, respectively. Determine |𝑎| ⃗ min if 𝑑 = 1200 f t and |𝑎| ⃗ max = 1.5𝑔.

Figure P12.131–P12.133

Problems 12.134 through 12.136 An airplane is flying straight and level at a constant speed 𝑣0 when it starts climbing along a path described by the equation 𝑦 = ℎ + 𝛽𝑥3 , where ℎ and 𝛽 are constants. Let 𝑔 denote the acceleration due to gravity. 𝑦 𝑣0 ℎ

𝚥̂ 𝚤̂

𝑥

Figure P12.134–P12.136 Problem 12.134

Determine the acceleration of the airplane at 𝑥 = 0.

If 𝛽 = 0.05 × 10−3 m−2 and 𝑣0 remains constant, find 𝑣0 such that the magnitude of the acceleration of the airplane is equal to 3𝑔 for 𝑥 = 300 m. Problem 12.135

𝜌

𝑣0 𝜌

𝐶

𝐵 𝐴 path of 𝐵

Figure P12.138

ISTUDY

Problem 12.137 A jet is flying at a constant speed 𝑣0 = 750 mph while performing a constant speed circular turn. If the magnitude of the acceleration needs to remain constant and equal to 9𝑔, where 𝑔 is the acceleration due to gravity, determine the radius of curvature of the turn.

Figure P12.137

𝑃

If 𝑣0 = 600 km∕h and 𝛽 = 0.025×10−4 m−2 , determine the acceleration of the airplane for 𝑥 = 350 m and express it in the Cartesian component system shown. Problem 12.136

path of 𝐴

Problem 12.138 Particles 𝐴 and 𝐵 are moving in the plane with the same constant speed 𝑣, and their paths are tangent at 𝑃 . Do these particles have zero acceleration at 𝑃 ? If not, do these particles have the same acceleration at 𝑃 ?

ISTUDY

Section 12.4

Planar Motion: Normal-Tangential Components

Problem 12.139 Uranium is used in light water reactors to produce a controlled nuclear reaction for the generation of power. When first mined, uranium comes out as the oxide U3 O8 , 0.7% of which is the isotope U-235 and 99.3% the isotope U-238.∗ For it to be used in a nuclear reactor, the concentration of U-235 must be in the 3–5% range.† The process of increasing the percentage of U-235 is called enrichment, and it is done in a number of ways. One method uses centrifuges, which spin at very high rates to create artificial gravity. In these centrifuges, the heavy U-238 atoms concentrate on the outside of the cylinder (where the acceleration is largest), and the lighter U-235 atoms concentrate near the spin axis. Before centrifuging, the uranium is processed into gaseous uranium hexafluoride or UF6 , which is then injected into the centrifuge. Assuming that the radius of the centrifuge is 20 cm and that it spins at 70,000 rpm, determine (a) The velocity of the outer surface of the centrifuge. (b) The acceleration in 𝑔 experienced by an atom of uranium that is on the inside of the outer wall of the centrifuge.

20 cm

rotation axis 𝜔 array of enrichment centrifuges

centrifuge cross section

Time Life Pictures/Department Of Energy (DOE)/ The LIFE Picture Collection/Getty Images)

𝜔

Figure P12.139

𝑅

Problem 12.140 Treating the center of the Earth as a fixed point, determine the magnitude of the acceleration of points on the surface of the Earth as a function of the angle 𝜙 shown. Use 𝑅 = 6371 km as the radius of the Earth.

Problems 12.141 through 12.143

Figure P12.140

An airplane is flying straight and level at a speed 𝑣0 = 150 mph and with a constant time rate of increase of speed 𝑣̇ = 20 f t∕s2 , when it starts to climb along a circular path with a radius of curvature 𝜌 = 2000 f t. The airplane maintains 𝑣̇ constant for about 30 s. 𝚥̂ Problem 12.141

Determine the acceleration of the airplane right at the start of the climb and express the result in the Cartesian component system shown.

𝚤̂ 𝜌

Problem 12.142 Determine the acceleration of the airplane 25 s after the start of the climb and express the result in the Cartesian component system shown. Problem 12.143

Determine the acceleration of the airplane after it has traveled 150 f t along the path and express the result in the Cartesian component system shown. ∗ The

U-235 atom has 92 protons and 143 neutrons, giving an atomic mass of 235. The nucleus of U-238 also has 92 protons, but it has 146 neutrons, giving it an atomic mass of 238. † For nuclear weapons, the concentration of U-235 must be about 90%.

Figure P12.141–P12.143

𝜙

713

714

Chapter 12

Particle Kinematics

𝐵

Problem 12.144 Suppose that a highway exit ramp is designed to be a circular segment of radius 𝜌 = 130 f t. A car begins to exit the highway at 𝐴 while traveling at a speed of 65 mph and goes by point 𝐵 with a speed of 25 mph. Compute the acceleration vector of the car as a function of the arc length 𝑠, assuming that the tangential component of the acceleration is constant between points 𝐴 and 𝐵.

𝜌 𝑠

Problem 12.145 Suppose that a highway exit ramp is designed to be a circular segment of radius 𝜌 = 130 f t. A car begins to exit the highway at 𝐴 while traveling at a speed of 65 mph and goes by point 𝐵 with a speed of 25 mph. Compute the acceleration vector of the car as a function of the arc length 𝑠, assuming that between 𝐴 and 𝐵 the speed was controlled so as to maintain constant the rate 𝑑𝑣∕𝑑𝑠.

𝐴 Figure P12.144 and P12.145

Problem 12.146 A water jet is ejected from the nozzle of a fountain with a speed 𝑣0 = 12 m∕s. Letting 𝛽 = 33◦ , determine the rate of change of the speed of the water particles as soon as these are ejected as well as the corresponding radius of curvature of the water path.

𝑣0 𝛽

Problem 12.147 A water jet is ejected from the nozzle of a fountain with a speed 𝑣0 . Letting 𝛽 = 21◦ , determine 𝑣0 so that the radius of curvature at the highest point on the water arch is 10 f t.

Figure P12.146 and P12.147

Problem 12.148 𝑦

A car traveling with a speed 𝑣0 = 65 mph almost loses contact with the ground when it reaches the top of the hill. Determine the radius of curvature of the hill at its top.

𝑣0 𝑥

𝑂 Figure P12.148 and P12.149

Problem 12.149 A car is traveling at a constant speed over a hill. If, using a Cartesian coordinate system with origin 𝑂 at the top of the hill, the hill’s profile is described by the function 𝑦 = −(0.003 m−1 )𝑥2 , where 𝑥 and 𝑦 are in meters, determine the minimum speed at which the car would lose contact with the ground at the top of the hill. Express the answer in km∕h.

Problem 12.150 A race boat is traveling at a constant speed 𝑣0 = 130 mph when it performs a turn with constant radius 𝜌 to change its course by 90◦ as shown. The turn is performed while losing speed uniformly in time so that the boat’s speed at the end of the turn is 𝑣𝑓 = 125 mph. If the maximum allowed normal acceleration is equal to 2𝑔, where 𝑔 is the acceleration due to gravity, determine the tightest radius of curvature possible and the time needed to complete the turn.

𝑣0

𝜌 𝜌 𝑣𝑓

𝐶

Figure P12.150 and P12.151

ISTUDY

Problem 12.151 A race boat is traveling at a constant speed 𝑣0 = 130 mph when it performs a turn with constant radius 𝜌 to change its course by 90◦ as shown. The turn is performed while losing speed uniformly in time so that the boat’s speed at the end of the turn is 𝑣𝑓 = 116 mph. If the magnitude of the acceleration is not allowed to exceed 2𝑔, where 𝑔 is the acceleration due to gravity, determine the tightest radius of curvature possible and the time needed to complete the turn.

ISTUDY

Section 12.4

Planar Motion: Normal-Tangential Components

Problem 12.152 A truck enters an exit ramp with an initial speed 𝑣0 . The ramp is a circular arc with radius 𝜌. Derive an expression for the magnitude of the acceleration of the truck as a function of the path coordinate 𝑠 (and the parameters 𝑣0 and 𝜌) if the truck stops at 𝐵 and travels from 𝐴 to 𝐵 with a constant rate of change of the speed with respect to 𝑠. 𝐵 𝜌 𝑠

𝑣0

𝐴

Figure P12.152 𝑣𝑓

Problem 12.153 A jet is flying straight and level at a speed 𝑣0 = 1100 km∕h when it turns to change its course by 90◦ as shown. In an attempt to progressively tighten the turn, the speed of the plane is uniformly decreased in time while keeping the normal acceleration constant and equal to 8𝑔, where 𝑔 is the acceleration due to gravity. At the end of the turn, the speed of the plane is 𝑣𝑓 = 800 km∕h. Determine the radius of curvature 𝜌𝑓 at the end of the turn and the time 𝑡𝑓 that the plane takes to complete its change in course. Hint: Although the instantaneous center of curvature varies throughout the turn, moving from 𝐶 to 𝑂, ̇ where 𝜃 the speed of the plane can be written in terms of the local curvature as 𝑣 = 𝜌𝜃, represents angular position, beginning at 0 at 𝐶 and ending at 𝜋∕2 at 𝑂.

𝑂

𝜌𝑓

𝑣0 𝜌0 𝐶

Figure P12.153

Problem 12.154 A car is traveling over a hill with a constant speed 𝑣0 = 70 mph. Using the Cartesian coordinate system shown, the hill’s profile is given by the function 𝑦 = −(0.0005 f t −1 )𝑥2 , where 𝑥 and 𝑦 are measured in feet. At 𝑥 = −300 f t, the driver applies the brakes, causing a constant time rate of change of speed 𝑣̇ = −3 f t∕s2 until the car arrives at 𝑂. Determine the distance traveled while applying the brakes along with the time to cover this distance. Hint: To compute √ √ the distance traveled by the car along the car’s path, observe that 𝑑𝑠 = 𝑑𝑥2 + 𝑑𝑦2 = 1 + (𝑑𝑦∕𝑑𝑥)2 𝑑𝑥, and that ( ) √ √ 𝑥√ 1 ln 𝐶𝑥 + 1 + 𝐶 2 𝑥2 . 1 + 𝐶 2 𝑥2 𝑑𝑥 = 1 + 𝐶 2 𝑥2 + ∫ 2 2𝐶 𝑦 𝑣0

𝑑𝑦 𝑑𝑠

𝑂

𝑦

𝑥

𝑥2 + 𝑦2 = 𝑅2

𝑑𝑥 𝑅

Figure P12.154

𝑂

Problem 12.155 Recalling that a circle of radius 𝑅 and center at the origin 𝑂 of a Cartesian coordinate system with axes 𝑥 and 𝑦 can be expressed by the formula 𝑥2 + 𝑦2 = 𝑅2 , use Eq. (12.51) to verify that the radius of curvature of this circle is equal to 𝑅.

Figure P12.155

𝑥

715

716

Chapter 12

Particle Kinematics

Problem 12.156 A particle moves over a hill whose shape is described by: 𝑦(𝑥) =

𝐴 1 + (𝑥∕𝑎)2

In this expression 𝐴 represents the height of the hill and parameter 𝑎 characterizes the severity of the grade. For 𝐴 = 10 m and 𝑎 = 2 m, find the radius of curvature at the top of the hill (𝑥 = 0 m). If the particle’s speed at the top of the hill is 𝑣 = 1 m∕s and its rate of change of speed is 𝑣̇ = 0 m∕s, find the particle’s acceleration at the top of the hill. Figure P12.156

ISTUDY

ISTUDY

Section 12.5

12.5

717

Planar Motion: Polar Coordinates

Planar Motion: Polar Coordinates

In this section we describe the position, velocity, and acceleration of a point moving in a plane when the point’s coordinates are given relative to a polar coordinate system. Before doing that, it will be helpful here and in future developments to know how to take the time derivative of any type of vector.

The time derivative of a vector Time derivatives of vector quantities are ubiquitous in dynamics. We have already seen the time derivative of position and velocity vectors. In coming chapters, we will also see the time derivatives of quantities, such as angular velocity, momentum, and angular momentum, all of which are vectors. It is therefore worthwhile to devote a little time to the time derivative of a vector so as to strengthen our understanding of this operation. Vectors can change with time in two ways: ⃗ 𝐴(𝑡)

1. They can change in magnitude. 2. They can change in direction due to rotation. Consider a vector 𝐴⃗ in the plane of the page at time 𝑡 and time 𝑡 + Δ𝑡, as shown in Fig. 12.23. The vector 𝐴⃗ is changing in length and direction, as well as in the position ⃗ as depicted in Fig. 12.24, in which the of its “tail.” We now consider the change in 𝐴, tail points at 𝑡 and 𝑡+Δ𝑡 are made to coincide. Since we can express 𝐴⃗ as a magnitude ⃗ differentiating it with respect to time, we times a unit vector 𝑢̂ in the direction of 𝐴, 𝐴

⃗ + 𝛥𝑡) 𝐴(𝑡

𝑢̂ 𝐴 (𝑡)

Figure 12.23 ⃗ to 𝐴(𝑡 ⃗ + Δ𝑡). A vector 𝐴⃗ changing from 𝐴(𝑡)

obtain 𝑑 𝑢̂ ] [ ⃗̇ = 𝑑 𝐴(𝑡) 𝑢̂ (𝑡) = 𝑑𝐴 𝑢̂ + 𝐴 𝐴 = 𝐴̇ 𝑢̂ + 𝐴 𝑢̂̇ , 𝐴(𝑡) 𝐴 𝐴 𝐴 𝐴 𝑑𝑡 𝑑𝑡 𝑑𝑡

⃗ + 𝛥𝑡) 𝐴(𝑡

(12.52)

⃗ is the magnitude of 𝐴, ⃗ and we have used the product rule of differenwhere 𝐴 = |𝐴| tiation. Equation (12.52) shows that the time derivative of 𝐴⃗ consists of two parts: 1. The term 𝐴̇ 𝑢̂ 𝐴 , which is a vector in the direction of 𝐴⃗ measuring the time rate ⃗ and of change of the magnitude of 𝐴, 2. The term 𝐴 𝑢̂̇ 𝐴 , which is the magnitude of 𝐴⃗ multiplied by the time derivative ⃗ of the unit vector 𝑢̂ 𝐴 measuring the time rate of change of the direction of 𝐴.

⃗ 𝐴(𝑡)

change in the length of 𝐴⃗ 𝛥𝜃

Figure 12.24 ⃗ and 𝐴(𝑡 ⃗ + Δ𝑡) Figure 12.23 with vectors 𝐴(𝑡) drawn with their tail points coinciding.

⃗̇ we now need to understand the vector 𝑢̂̇ . To understand 𝐴, 𝐴

𝑦 𝑢̂̇ 𝐴

Time derivative of a unit vector

𝚥̂

To understand and obtain a formula for the time derivative of a unit vector, we begin by expressing it as (see Fig. 12.25) 𝑢̂ 𝐴 = cos 𝜃 𝚤̂ + sin 𝜃̂𝚥,

(12.53)

̇ sin 𝜃 𝚤̂ + cos 𝜃̂𝚥). 𝑢̂̇ 𝐴 = 𝜃(−

(12.54)

𝐴

𝑢̂ 𝐴

𝑂

𝚤̂ 1

so that The term − sin 𝜃 𝚤̂ + cos 𝜃̂𝚥 in Eq. (12.54) is a unit vector perpendicular to 𝑢̂ 𝐴 . This implies that 𝑢̂̇ 𝐴 is perpendicular to 𝑢̂ 𝐴 (this is shown in Fig. 12.25). The term 𝜃̇ in

Figure 12.25 A rotating unit vector 𝑢̂ 𝐴 .

𝜃 𝑥

718

Chapter 12

Particle Kinematics

𝜃̇

Eq. (12.54) is the rotation rate of 𝑢̂ 𝐴 since 𝜃 measures the orientation of 𝑢̂ 𝐴 . Equation (12.54) gives us 𝑢̂̇ 𝐴 in this case, but can we find a more general expression? We can if we define the angular velocity of 𝑢̂ 𝐴 as a vector. In particular, we define the following:

𝜃

axis of rotation is the “hinge line” about which an object rotates. For the door in Fig. 12.26, the indicated hinge line 𝐴𝐵 is the axis of rotation of the door.

𝐴

𝐵

hinge line Figure 12.26 The hinge line of a swinging door. direction of rotation axis of rotation

𝛥𝜃 Figure 12.27 Direction of rotation defined using the right-hand rule.

ISTUDY

direction of rotation orients the axis of rotation via the right-hand rule, with the thumb pointing in the direction of rotation (see Fig. 12.27). This direction can be indicated by a unit vector. With these definitions in mind, we can now define the angular velocity vector 𝜔 ⃗ 𝐴 of the unit vector 𝑢̂ 𝐴 in Fig. 12.25 as ̂ 𝜔 ⃗ 𝐴 = 𝜃̇ 𝑘,

(12.55)

where we note that since 𝑢̂ 𝐴 is rotating in the 𝑥𝑦 plane, its angular velocity must ⃗ 𝐴 are be perpendicular to it. Equation (12.55) implies that the vectors 𝑢̂ 𝐴 , 𝑢̂̇ 𝐴 , and 𝜔 mutually perpendicular and that the vector 𝜔 ⃗ 𝐴 × 𝑢̂ 𝐴 is parallel to 𝑢̂̇ 𝐴 , that is, 𝜔 ⃗ 𝐴 × 𝑢̂ 𝐴 = 𝜃̇ 𝑘̂ × (cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂) ̇ = 𝜃(cos 𝜃 𝑘̂ × 𝚤̂ + sin 𝜃 𝑘̂ × 𝚥̂) ̇ = 𝜃(cos 𝜃 𝚥̂ − sin 𝜃 𝚤̂).

(12.56)

Equation (12.56) is a remarkable result because, by comparing it with Eq. (12.54), it tells us that 𝜔 ⃗ × 𝑢̂ 𝐴 is not just parallel to 𝑢̂̇ 𝐴 , it is 𝑢̂̇ 𝐴 ! That is, 𝑢̂̇ 𝐴 = 𝜔 ⃗ 𝐴 × 𝑢̂ 𝐴 .

(12.57)

It turns out that the result in Eq. (12.57) is universal: whether in 2D or 3D, the time rate of change of a unit vector can always be represented as the cross product of the angular velocity of that unit vector with the unit vector in question. We now have a general and physically motivated way to express the time derivative of any unit vector 𝑢⃗ as 𝑢̂̇ = 𝜔 ⃗ × 𝑢, ̂ (12.58) 𝑢

where 𝜔 ⃗ 𝑢 is the angular velocity of 𝑢⃗. Equation (12.58) is best remembered as The time derivative of a unit vector is the angular velocity of the vector crossed with the vector itself. Time derivative of an arbitrary vector Using Eq. (12.58), the interpretation of the term 𝐴 𝑢̂̇ 𝐴 in Eq. (12.52) is now clear: it is the magnitude of 𝐴⃗ times 𝜔 ⃗ 𝐴 × 𝑢̂ 𝐴 , so Eq. (12.52) becomes ⃗̇ = 𝐴̇ 𝑢̂ + 𝐴 𝜔 𝐴(𝑡) ⃗ 𝐴 × 𝑢̂ 𝐴 = 𝐴̇ 𝑢̂ 𝐴 + 𝜔 ⃗ 𝐴 × 𝐴 𝑢̂ 𝐴 , 𝐴

(12.59)

⃗ ⃗̇ = 𝐴̇ 𝑢̂ + 𝜔 ⃗ 𝐴 × 𝐴. 𝐴(𝑡) 𝐴

(12.60)

or

ISTUDY

Section 12.5

719

Planar Motion: Polar Coordinates

We can now interpret the time derivative of any vector as the time rate of change of the magnitude of the vector plus a time rate of change of direction of that vector ⃗̇ = 𝐴̇ 𝑢̂ 𝐴 + 𝜔 ⃗ 𝐴 × 𝐴⃗ , 𝐴(𝑡) ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ change in 𝐴⃗

change in magnitude

(12.61)

change in direction

where the time rate of change of direction is given by the cross product of the vector’s angular velocity and the vector itself. This relationship applies to any vector, and we will use it extensively. Normal-tangential components and the time derivative of a vector We conclude our discussion of the time derivative of a vector by going back to Eq. (12.45) on p. 703 to obtain 𝑎⃗ in normal-tangential components using the ideas we just obtained. Applying Eq. (12.60) gives ⃗ 𝑣 × 𝑣⃗ = 𝑣̇ 𝑢̂ 𝑡 + 𝑣𝜔 ⃗ 𝑣 × 𝑢̂ 𝑡 . 𝑎⃗ = 𝑣⃗̇ = 𝑣̇ 𝑢̂ 𝑡 + 𝜔

(12.62)

Comparing Eq. (12.62) with Eq. (12.49), we have 𝑣𝜔 ⃗ 𝑣 × 𝑢̂ 𝑡 =

𝑣2 𝑢̂ . 𝜌 𝑛

(12.63)

⃗ 𝑣 is Canceling 𝑣, noting from Eq. (12.50) that 𝑢̂ 𝑛 = 𝑢̂ 𝑏 × 𝑢̂ 𝑡 , and assuming that 𝜔 perpendicular to the plane containing 𝑢̂ 𝑡 and 𝑢̂ 𝑛 , we obtain 𝜔 ⃗ 𝑣 × 𝑢̂ 𝑡 =

𝑣 𝑢̂ × 𝑢̂ 𝑡 𝜌 𝑏

⇒

𝜔 ⃗𝑣 =

𝑣 𝑢̂ . 𝜌 𝑏

(12.64)

That is, the rate of rotation of the velocity vector is directly proportional to the speed of the particle, as well as the curvature (i.e., 1∕𝜌) of the path.

Polar coordinates and position, velocity, and acceleration Figure 12.28 shows a particle 𝑃 moving along a path in the 𝑥𝑦 plane. The position of 𝑃 relative to the origin at 𝑂 is given by 𝑟⃗. The distance 𝑟 between 𝑂 and 𝑃 and the angle 𝜃, measured with respect to the 𝑥 axis, identify the position of 𝑃 in the plane of motion. The quantities 𝑟 and 𝜃 are called the polar coordinates∗ of 𝑃 relative to the origin 𝑂 and the reference line coinciding with the 𝑥 axis. The position vector of 𝑃 is then 𝑟⃗ = 𝑟 𝑢̂ 𝑟 , (12.65) where 𝑢̂ 𝑟 is the unit vector pointing from 𝑂 to 𝑃 . Although 𝜃 does not explicitly appear in Eq. (12.65), 𝑟⃗ depends on 𝜃 because 𝜃 defines the direction of 𝑢̂ 𝑟 . Taking the time derivative of Eq. (12.65) to find the velocity in polar coordinates, we obtain (12.66) 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟 𝑢̂̇ 𝑟 . The time derivative of the unit vector 𝑢̂ 𝑟 can be evaluated using Eq. (12.58), which allows us to rewrite Eq. (12.66) as ⃗ 𝑟 × 𝑢̂ 𝑟 , 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟 𝜔 ∗ Polar

(12.67)

coordinates are also called radial-transverse coordinates, in which radial refers to the 𝑟 direction and transverse refers to the 𝜃 direction. In Section 12.7, we will introduce the cylindrical coordinate system, which is the three-dimensional generalization of the polar coordinate system.

𝑦

path 𝑢̂ 𝜃 𝑃

𝑟

𝑢̂ 𝑟

𝑟⃗ 𝜃 𝑂

𝑥

Figure 12.28 The polar coordinate system defining the position of the plane at 𝐵.

720

Chapter 12

Particle Kinematics

where 𝜔 ⃗ 𝑟 is the angular velocity of 𝑢̂ 𝑟 . Since 𝜃 increases in the counterclockwise ̂ Using 𝜔 ̂ the last direction, using the right-hand rule, we have 𝜔 ⃗ 𝑟 = 𝜃̇ 𝑘. ⃗ 𝑟 = 𝜃̇ 𝑘, ̇ ̂ ̇ ̂ term in Eq. (12.67) becomes 𝑟𝜃 𝑘 × 𝑢̂ 𝑟 = 𝑟𝜃 𝑢̂ 𝜃 , where 𝑢̂ 𝜃 is given by 𝑘 × 𝑢̂ 𝑟 = 𝑢̂ 𝜃 . Equation (12.67) then becomes

Concept Alert Direction of 𝒖̂ 𝒓 and 𝒖̂ 𝜽 . The unit vector 𝑢̂ 𝑟 always points away from the origin. The unit vector 𝑢̂ 𝜃 always points in the direction of increasing 𝜃.

𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 = 𝑣𝑟 𝑢̂ 𝑟 + 𝑣𝜃 𝑢̂ 𝜃 ,

(12.68)

where 𝑣𝑟 = 𝑟̇ and

𝑣𝜃 = 𝑟𝜃̇

(12.69)

are the radial and transverse components of the velocity, respectively. The time derivative of Eq. (12.68) yields 𝑎⃗ = 𝑟̈ 𝑢̂ 𝑟 + 𝑟̇ 𝑢̂̇ 𝑟 + 𝑟̇ 𝜃̇ 𝑢̂ 𝜃 + 𝑟𝜃̈ 𝑢̂ 𝜃 + 𝑟𝜃̇ 𝑢̂̇ 𝜃 .

(12.70)

̂ Eq. (12.70) becomes ⃗ 𝑟 × 𝑢̂ 𝑟 , 𝑢̂̇ 𝜃 = 𝜔 ⃗ 𝜃 × 𝑢̂ 𝜃 , and 𝜔 ⃗𝑟 = 𝜔 ⃗ 𝜃 = 𝜃̇ 𝑘, Since 𝑢̂̇ 𝑟 = 𝜔 𝑎⃗ = 𝑟̈ 𝑢̂ 𝑟 + 𝑟̇ 𝜃̇ 𝑘̂ × 𝑢̂ 𝑟 + 𝑟̇ 𝜃̇ 𝑢̂ 𝜃 + 𝑟𝜃̈ 𝑢̂ 𝜃 + 𝑟𝜃̇ 2 𝑘̂ × 𝑢̂ 𝜃 .

(12.71)

Noting that 𝑘̂ × 𝑢̂ 𝜃 = −𝑢̂ 𝑟 and combining the coefficients of 𝑢̂ 𝑟 and 𝑢̂ 𝜃 , we can rewrite Eq. (12.71) as ) ( ( ) 𝑎⃗ = 𝑟̈ − 𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝑢̂ 𝜃 = 𝑎𝑟 𝑢̂ 𝑟 + 𝑎𝜃 𝑢̂ 𝜃 , (12.72) where 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2

and

𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇

(12.73)

are the radial and transverse components of the acceleration, respectively. Normal-tangential and polar components for circular motion Referring to Fig. 12.29, say that a particle 𝑃 is traveling on a circular path centered at 𝑂. Writing its velocity and acceleration in both polar and normal-tangential components, we have that (12.74) 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 = 𝑣 𝑢̂ 𝑡 , where the unit vectors for the two component systems are as shown in Fig. 12.29. We see that, for circular motion, 𝑟 is constant, so 𝑟̇ = 0, and the 𝑢̂ 𝑡 and 𝑢̂ 𝜃 vectors are parallel to one another (in this case, they are in the same direction). Therefore, the coefficients of 𝑢̂ 𝑡 and 𝑢̂ 𝜃 must be equal, which implies Figure 12.29 The position 𝑟⃗ of a particle defined using the polar coordinates 𝑟 and 𝜃.

ISTUDY

̇ 𝑣 = 𝑟𝜃.

(12.75)

This equation is incredibly important for circular motion since it gives us a translation ̇ For acceleration, we have that from speed 𝑣 to angular speed 𝜃. ( ( ) ) 𝑣2 𝑢̂ . (12.76) 𝑎⃗ = 𝑟̈ − 𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝑢̂ 𝜃 = 𝑣̇ 𝑢̂ 𝑡 + 𝜌 𝑛 Since both 𝑟̇ and 𝑟̈ are zero, and 𝜌 = 𝑟, this becomes 𝑣2 𝑎⃗ = −𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ 𝑢̂ 𝜃 = 𝑣̇ 𝑢̂ 𝑡 + 𝑢̂ . 𝑟 𝑛 Now 𝑢̂ 𝑛 = −𝑢̂ 𝑟 , so when we equate coefficients now, we get 𝑣2 𝑟, 𝑛 coefficients ∶ 𝑟𝜃̇ 2 = 𝑟 ̈ 𝜃, 𝑡 coefficients ∶ 𝑟𝜃 = 𝑣̇

(12.77)

⇒ 𝑣2 = 𝑟2 𝜃̇ 2 ⇒

̈ 𝑣̇ = 𝑟𝜃.

We see that the first result is just the square of Eq. (12.75) and the second is the derivative. So, for circular motion, Eq. (12.75) is the one equation that rules them all!

ISTUDY

Section 12.5

721

Planar Motion: Polar Coordinates

End of Section Summary The time derivative of a vector consists of a part due to the vector’s change in length and a part due to the vector’s change in direction. Since a unit vector never changes its length, its time derivative only consists of a contribution from a change in direction: Eq. (12.58), p. 718 𝑢(𝑡) ̂̇ = 𝜔 ⃗ 𝑢 × 𝑢̂ . ⏟⏟⏟ ⏟⏟⏟ change in 𝑢⃗

𝐴̇ 𝑢̂ 𝐴 𝑑𝑡

change in direction

⃗ 𝐴(𝑡)

⃗ the following relationship provides its time derivative: For an arbitrary vector 𝐴, Eq. (12.60), p. 718

𝜔 ⃗ 𝐴 × 𝐴⃗ 𝑑𝑡

⃗̇ = 𝐴̇ 𝑢̂ 𝐴 + 𝜔 ⃗ 𝐴 × 𝐴⃗ , 𝐴(𝑡) ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ change in 𝐴⃗

change in magnitude

change in direction

⃗ 𝐴̇ = 𝑑|𝐴|∕𝑑𝑡, ⃗ where, referring to Fig. 12.30, 𝑢̂ 𝐴 is a unit vector in the direction of 𝐴, ⃗ and 𝜔 ⃗ 𝐴 is the angular velocity of the vector 𝐴. Referring to Fig. 12.31, 𝑟 and 𝜃 are the polar coordinates of point 𝑃 . The coordinate 𝜃 was chosen positive in the counterclockwise direction as viewed down the positive 𝑧 axis. Using polar coordinates, the position vector of 𝑃 is

⃗ + 𝑑𝑡) 𝐴(𝑡 𝜔𝐴 Figure 12.30 Depiction of the vector 𝐴⃗ at time 𝑡 and at time 𝑡+𝑑𝑡 showing its change in magnitude and change in direction.

𝑦

path

Eq. (12.65), p. 719

𝑢̂ 𝜃 𝑟

𝑟⃗ = 𝑟 𝑢̂ 𝑟 .

𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 = 𝑣𝑟 𝑢̂ 𝑟 + 𝑣𝜃 𝑢̂ 𝜃 , where Eqs. (12.69), p. 720 𝑣𝑟 = 𝑟̇

and

𝑣𝜃 = 𝑟𝜃̇

are the radial and transverse components of the velocity, respectively. Differentiating Eq. (12.68) with respect to time, we see that the acceleration vector in polar coordinates takes on the form Eq. (12.72), p. 720 ( ( ) ) 𝑎⃗ = 𝑟̈ − 𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝑢̂ 𝜃 = 𝑎𝑟 𝑢̂ 𝑟 + 𝑎𝜃 𝑢̂ 𝜃 , where Eqs. (12.73), p. 720 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2

and

𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇

are the radial and transverse components of the acceleration, respectively.

𝑢̂ 𝑟

𝑟⃗

Differentiating Eq. (12.65) with respect to time, we obtain the following result for the velocity vector in polar coordinates Eq. (12.68), p. 720

𝑃

𝜃 𝑂

𝑥

Figure 12.31 The position 𝑟⃗ of a particle defined using the polar coordinates 𝑟 and 𝜃.

722

Chapter 12

Particle Kinematics

E X A M P L E 12.16 𝑢̂ 𝜃

Constant Velocity Motion in Polar Coordinates

𝑢̂ 𝑟 𝑣⃗ = 𝑣0 𝚤̂ 𝐵

SOLUTION

𝑟 ℎ

𝑢̂ 𝜃 𝐴

𝜃 𝑢̂ 𝑟 ℎ𝐴

Figure 1 Radar station at 𝐴 tracking a plane at 𝐵.

ISTUDY

What relationships do the radar readings obtained by the station at 𝐴 need to satisfy to conclude that the plane at 𝐵 shown in Fig. 1 is flying straight and level at altitude ℎ and at a constant speed 𝑣0 ?

It is not hard to imagine that a radar station, such as the one at 𝐴, would record discrete readings of 𝑟(𝑡) and 𝜃(𝑡) for any object it is tracking. Given a tabulated list of 𝑟 and 𝜃 as a function of time, the question is, How can we use these readings to determine whether or not a plane is flying (1) straight and level and (2) at a constant speed? The solution strategy used in this example will be used often. We will use expressions for positions, velocities, and accelerations in a chosen coordinate system to give form to specific requirements (e.g., that the altitude be constant or that the speed be maintained). Once a requirement is obtained, it can be manipulated further (e.g., differentiated with respect to time) to obtain new relationships that will be consistent with the requirements’ initial statements. Road Map

Computation

To assess the plane’s altitude, first we need to express the altitude using the given data, and then we need to enforce the requirement that the altitude be maintained constant. Hence, from Fig. 1, we have ℎ = 𝑟 sin 𝜃 + ℎ𝐴 .

(1)

So ℎ will remain constant as long as 𝑟(𝑡) sin 𝜃(𝑡) = constant.

(2)

Since the speed is obtained as the magnitude of the velocity vector, to find what relationship must be satisfied in order for the plane to maintain the given constant speed, let’s look at the expression for the plane’s velocity. The velocity vector in polar coordinates is given by Eq. (12.68), that is, 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 , and so its magnitude is given by the square root of the sum of the squares of its components, i.e., √ |𝑣| ⃗ = 𝑣 = 𝑟̇ 2 + 𝑟2 𝜃̇ 2 . (3) Therefore, for the speed to be a constant equal to 𝑣0 we must have 𝑣0 =

√

𝑟̇ 2 + 𝑟2 𝜃̇ 2 = constant.

(4)

Since 𝑟 has dimensions of length and 𝜃 is nondimensional, then 𝑟̇ and 𝑟𝜃̇ both have dimensions of length over time, i.e., of velocity. Hence, we can conclude that Eq. (4) is dimensionally correct.

Discussion & Verification

A Closer Look While Eq. (4) gives the relationship we wanted, it is often desirable to have relations that require as few mathematical operations as possible. For example, it turns out that there is no need to compute a square root to verify whether or not the speed is a constant—if it is true that 𝑣0 is a constant, then we necessarily must have that 𝑣20 is also a constant. Therefore, as long as

𝑟̇ 2 (𝑡) + 𝑟2 (𝑡)𝜃̇ 2 (𝑡) = constant, we can conclude that the plane is flying at a constant speed.

(5)

ISTUDY

Section 12.5

Planar Motion: Polar Coordinates

E X A M P L E 12.17

723

Acceleration During Orbital Motion

For a satellite orbiting a planet, empirical observations (see Kepler’s laws in Chapter 11) tell us that in the polar coordinate system shown in Fig. 1, the quantity 𝑟2 𝜃̇ remains constant throughout the satellite’s motion. Show that for such a motion, the satellite’s acceleration is purely in the radial direction; i.e., the transverse component of the satellite’s acceleration is equal to zero.

satellite 𝑟 𝜃 planet

orbit center 𝑏

SOLUTION We need to show that 𝑎𝜃 = 0. To do this, we can determine how the expression for 𝑎𝜃 given by Eq. (12.72) relates to the law of Kepler stating that 𝑟2 𝜃̇ is constant. Road Map

Equation (12.72) tells us that in polar coordinates the 𝑟 and 𝜃 components of the satellite’s acceleration are given by

Computation

𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2

̇ and 𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃.

(1)

Kepler’s observations indicate that 𝑟2 𝜃̇ = 𝐾,

(2)

where 𝐾 is a constant whose value depends on the particular orbit followed by the satellite. Now, if we differentiate Eq. (2) with respect to time, we obtain 2𝑟𝑟̇ 𝜃̇ + 𝑟2 𝜃̈ = 0,

(3)

where we have used the fact that 𝐾 is constant so that 𝐾̇ = 0. Equation (3) tells us that ( ) 2𝑟𝑟̇ 𝜃̇ + 𝑟2 𝜃̈ = 𝑟 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 0 ⇒ 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 0, (4) where we have used the fact that 𝑟 is never zero to obtain the final expression. Comparing Eq. (4) with Eq. (1), we see that 𝑎𝜃 = 0, (5) which is what we set out to show. Discussion & Verification The derivation of our result is elementary, and it is correct because we have correctly applied the chain and product rules of calculus in taking the derivative of Eq. (2). A Closer Look The result we have obtained is based on astronomical observations that predate the work of Newton. From a historical viewpoint this is important because, in formulating his second law of motion and his law of universal gravitation, Newton needed to formulate a theory consistent with Kepler’s observations. Therefore, it is not by chance that Newton’s law of gravitation demands that the force of gravity between two particles be directed along the line connecting the particles. This requirement, along with Newton’s second law, 𝐹⃗ = 𝑚𝑎, ⃗ causes the acceleration of the planet in Fig. 1 to be completely along the radial line connecting the satellite and the planet (i.e., 𝑎𝜃 = 0) so that, overall, both the universal law of gravitation and Newton’s second law are consistent with Kepler’s observations. We will come across Eq. (2) again because Eq. (2) is also the mathematical expression of an important physical principle. The principle is that, under the influence of a purely central force, when no moment is applied to a body, its angular momentum is conserved. You may not have heard of the term angular momentum yet, but we will examine it in great detail in Section 15.3.

𝑐 𝑎 Figure 1 A satellite orbiting a planet. The orbit is an ellipse with major and minor semiaxes√equal to 𝑎 and 𝑏, respectively. The length 𝑐 = 𝑎2 − 𝑏2 denotes the distance between either focus (one of which is occupied by the planet) and the orbit center.

724

Chapter 12

Particle Kinematics

E X A M P L E 12.18

Rectilinear Motion and Polar Coordinates As a part of an assembly process, the end effector 𝐴 on the robotic arm in Fig. 2 needs to move the gear 𝐵 along the vertical line shown in a specified fashion. Arm 𝑂𝐴 can vary its length by telescoping via internal actuators. A motor at 𝑂 allows the arm to pivot in the vertical plane. When 𝜃 = 50◦ , 𝐵 is moving downward with a speed 𝑣0 = 8 f t∕s and a downward acceleration with magnitude 𝑎0 = 0.5 f t∕s2 . At this instant, determine the required length of the arm, the rate at which the arm is extending, and its rotation rate ̇ In addition, determine the second time derivatives of both the arm’s length and the 𝜃. angle 𝜃.

Charles O’Rear/The Image Bank Unreleased/ Getty images

Figure 1 A robotic arm. 𝐵

𝐴

SOLUTION Road Map We know the gear’s path, velocity, and acceleration as well as the angle 𝜃 at the instant shown. Therefore, we can determine the length 𝑟 at this instant by simple trigonometry. As far as determining 𝑟̇ and 𝜃̇ is concerned, observe that the known velocity of 𝐵 is easily written in terms of the unit vector 𝚥̂ shown in Fig. 3. Once this is done, we can use trigonometry to rewrite the velocity of 𝐵 via 𝑢̂ 𝑟 and 𝑢̂ 𝜃 . Then 𝑟̇ and 𝜃̇ are found by equating this specific expression for the velocity of 𝐵 to the general expression of the velocity vector in polar coordinates. Finally, 𝑟̈ and 𝜃̈ are found by applying to the acceleration the same strategy just described for the velocity. Computation

𝜃

We know that 𝜃 = 50◦ at this instant, so the geometry in Fig. 3 tells us

that

𝑂

𝑟= 4 ft

4 ft = 6.223 f t. cos 𝜃

(1)

Since 𝐵 moves vertically downward, the velocity of the end effector is

Figure 2 Schematic of robot arm shown in Fig. 1.

𝑣⃗ = −𝑣0 𝚥̂ = −(8 f t∕s) 𝚥̂.

(2)

Equating Eq. (2) with the expression of the velocity in polar coordinates, we have 𝑢̂ 𝜃 𝐴

𝑢̂ 𝑟 𝐵 𝑣0 = 8 f t∕s

𝑟

𝑎0 = 0.5 f t∕s2 𝜃

4 ft Figure 3 Robotic arm showing the polar and Cartesian coordinate systems we will use, as well as the velocity and acceleration of the end effector.

ISTUDY

(3)

Now, we can write 𝚥̂ in terms of the polar unit vectors as 𝚥̂ = sin 𝜃 𝑢̂ 𝑟 + cos 𝜃 𝑢̂ 𝜃 = sin 50◦ 𝑢̂ 𝑟 + cos 50◦ 𝑢̂ 𝜃 .

(4)

Substituting Eq. (4) into Eq. (3) and equating components, we have 𝚥̂ 𝚤̂

𝑂

−(8 f t∕s) 𝚥̂ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 .

𝑟̇ = −8 sin 50◦ ft∕s and 𝑟𝜃̇ = −8 cos 50◦ ft∕s,

(5)

which, by using the result in Eq. (1), can then be solved for 𝑟̇ and 𝜃̇ to obtain 𝑟̇ = −6.128 f t∕s

and 𝜃̇ = −0.8264 rad∕s.

Dealing with the acceleration of 𝐵 as we have done for the velocity, we have ) ( 𝑎⃗ = −𝑎0 𝚥̂ = − 0.5 f t∕s2 𝚥̂. Equating Eq. (7) with the expression for acceleration in polar coordinates, we have ( ) ( ) ( ) − 0.5 f t∕s2 𝚥̂ = 𝑟̈ − 𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝑢̂ 𝜃 .

(6)

(7)

(8)

Using Eq. (4) again and equating components, we have 𝑟̈ − 𝑟𝜃̇ 2 = −0.5 sin 50◦ ft∕s2

and 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = −0.5 cos 50◦ ft∕s2 .

(9)

ISTUDY

Section 12.5

Planar Motion: Polar Coordinates

̈ we have Solving Eqs. (9) for 𝑟̈ and 𝜃, 𝑟̈ = −0.5 sin 50◦ ft∕s2 + 𝑟𝜃̇ 2

) 1( and 𝜃̈ = − 0.5 cos 50◦ ft∕s2 + 2𝑟̇ 𝜃̇ . 𝑟

(10)

Substituting in Eqs. (10) the results we have already obtained for 𝑟, 𝑟,̇ and 𝜃,̇ we have 𝑟̈ = 3.866 f t∕s2

and 𝜃̈ = −1.679 rad∕s2 .

(11)

Discussion & Verification Our results are dimensionally correct, and appropriate units have been used. To understand whether or not the signs of our results are correct, we begin by noticing that in Eqs. (6) 𝑟̇ < 0 and 𝜃̇ < 0, which means that the arm is actually getting shorter while rotating clockwise. Intuition tells us that if the end effector is moving straight down, then the arm does need to get shorter and rotate clockwise. Fortunately, both of these observations are consistent with the results in Eq. (6). As for the sign of 𝑟̈ in Eqs. (11), the fact that 𝑟̈ > 0 indicates that while the arm is getting shorter (i.e., 𝑟̇ < 0), the rate at which this happens is decreasing. That is, since 𝑟̈ has a sign opposite to 𝑟,̇ we should expect that as 𝐵 keeps moving down, the length of the arm will stop shortening. This result is correct because the length of the arm will stop shortening when the arm becomes horizontal, and then it will start increasing for negative values of 𝜃. As far as the sign of 𝜃̈ is concerned, our result indicates that the rate of clockwise rotation is increasing. This result is to be expected even if the acceleration ̇ the sign of 𝜃̈ is opposite to of 𝐵 were equal to zero. In this case, i.e., if 𝑎𝜃 = 0 = 𝑟𝜃̈ + 2𝑟̇ 𝜃, ̇ For us, such a product is positive because we found that both the sign of the product 𝑟̇ 𝜃. 𝑟̇ and 𝜃̇ were negative. In our case 𝑎𝜃 ≠ 0, but 𝐵 is accelerating downward and therefore, provides no contribution to 𝜃̈ in the counterclockwise direction.

725

726

Chapter 12

Particle Kinematics

E X A M P L E 12.19 parabolic path

𝑦 𝑢̂ 𝜃 𝑢̂ 𝑟 (0) 𝐴

Projectile Motion in Polar Coordinates In Section 12.3, we saw that the equations describing the motion of a projectile in Cartesian coordinates took the simple form 𝑥̈ = 0 and 𝑦̈ = −𝑔. Referring to Fig. 1, revisit the projectile problem and derive the equations describing the motion of a projectile released at 𝐴, using polar coordinates. In addition, derive the expressions for the initial conditions of the motion in the same polar coordinate system, given that the projectile is launched from point 𝐴 at speed 𝑣0 in the direction shown in Fig. 1.

𝑢̂ 𝑟

𝑣0 𝜃 𝑎⃗𝑔

𝜙

𝑢̂ 𝜃 (0)

SOLUTION ℎ

Road Map

𝑟

𝜃 𝑂

𝑥

Figure 1 Projectile motion in polar coordinates. The point 𝑂 is the origin of the coordinate system. The point 𝐴 is the point at which the object was released, and it is taken to be the initial position of the projectile.

ISTUDY

What we want to show is that any problem can be formulated using any coordinate system we choose. However, this freedom of choice comes with the consequence that if we do not choose wisely, the mathematical complexity of the problem can be substantial. Since we know the acceleration of the projectile at every point along its path, the idea is to relate that known acceleration to the direction of 𝑎𝑟 and 𝑎𝜃 at every point. A similar idea will apply to finding the initial conditions; that is, we know the initial conditions relative to point 𝐴, and we just need to translate those to the polar coordinate system whose origin is at point 𝑂. Computation

We begin by finding the polar components of the projectile’s acceleration due to gravity, which we denote by 𝑎⃗𝑔 . The components of this vector along the radial and transverse directions are 𝑎𝑔𝑟 = 𝑎⃗𝑔 ⋅ 𝑢̂ 𝑟 = −𝑔 sin 𝜃,

(1)

𝑎𝑔𝜃 = 𝑎⃗𝑔 ⋅ 𝑢̂ 𝜃 = −𝑔 cos 𝜃,

(2)

where 𝑔 is the acceleration due to gravity. Equating the general expressions for the components of acceleration in polar coordinates given by Eq. (12.73) to the corresponding components in Eqs. (1) and (2) gives 𝑟̈ − 𝑟𝜃̇ 2 = −𝑔 sin 𝜃, 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = −𝑔 cos 𝜃.

Concept Alert Unit vectors in path and polar component systems are functions of time. When we express vectors using normal-tangential components or polar components, it is important to keep in mind that the unit vectors of these component systems are themselves functions of time. This is why, in expressing the velocity at time 𝑡 = 0, we had to use the unit vectors 𝑢̂ 𝑟 and 𝑢̂ 𝜃 at 𝑡 = 0.

(3) (4)

Equations (3) and (4) form a system of coupled, nonlinear ordinary differential equations which, mathematically, are far more complex than the equations we obtained in Cartesian coordinates. To integrate Eqs. (3) and (4) so as to determine the motion of the projectile, we would need to complement these differential equations with corresponding initial conditions. These conditions consist of the position and velocity of 𝐴 at time 𝑡 = 0 expressed in polar coordinates. Referring to Fig. 1, we see that at time 𝑡 = 0 the projectile is at 𝐴, and therefore, we have 𝜋 𝑟(0) = ℎ and 𝜃(0) = . (5) 2 For the velocity at 𝑡 = 0 we have 𝑣(0) ⃗ = 𝑣0 sin 𝜙 𝑢̂ 𝑟 (0) − 𝑣0 cos 𝜙 𝑢̂ 𝜃 (0),

(6)

where the quantities 𝑣0 and 𝜙 are known. Equating Eq. (6) to the general expression for the velocity vector in polar coordinates given by Eq. (12.68), we have ̇ 𝑢̂ (0) = 𝑣 sin 𝜙 𝑢̂ (0) − 𝑣 cos 𝜙 𝑢̂ (0), 𝑣(0) ⃗ = 𝑟(0) ̇ 𝑢̂ 𝑟 (0) + 𝑟(0)𝜃(0) 𝜃 0 𝑟 0 𝜃

(7)

that is, 𝑟(0) ̇ = 𝑣0 sin 𝜙

̇ and 𝑟(0)𝜃(0) = −𝑣0 cos 𝜙.

(8)

ISTUDY

Section 12.5

Planar Motion: Polar Coordinates

In summary, the initial conditions for this problem take on the form 𝑟(0) = ℎ, 𝑟(0) ̇ = 𝑣0 sin 𝜙,

𝜋 , 2 𝑣 ̇ 𝜃(0) = − 0 cos 𝜙, ℎ

𝜃(0) =

(9) (10)

where we have used the fact that 𝑟(0) = ℎ from Eq. (5). Recalling that 𝑟 has dimensions of length and that 𝜃 is nondimensional, we can verify that the left-hand sides of Eqs. (3) and (4) have dimensions of length over time squared, as expected since the dimensions of the right-hand sides of these equations are those of the acceleration of gravity 𝑔. We can verify the dimensional correctness of Eqs. (9) and (10) in a similar manner.

Discussion & Verification

A Closer Look Equations (3), (4), (9), and (10) are much more complicated-looking than the corresponding equations in Cartesian coordinates [i.e., 𝑥̈ = 0 and 𝑦̈ = −𝑔, along with the initial conditions 𝑥(0) = 0 and 𝑦(0) = ℎ]. However, the trajectory obtained by solving Eqs. (3) and (4) is identical to that obtained by solving the corresponding equations in Cartesian coordinates (i.e., 𝑥̈ = 0 and 𝑦̈ = −𝑔); that is, we would obtain exactly the same parabola in either case (provided, of course, that the same initial position and velocity are used). A question often asked by students is, Can I solve this problem using this or that coordinate system? The answer to this question is, in general, yes. However, whether you are solving the problem analytically or numerically, the coordinate system chosen does make a difference in how involved the problem’s solution becomes. In this example, we have derived (although not solved) the equations governing the motion of a projectile. Clearly, the choice of coordinate system did not change the underlying physics of the problem. However, the resulting system of equations cannot be easily solved by hand because they are nonlinear and, above all, coupled; that is, the equation containing 𝑟̈ also ̇ and the equation containing 𝜃̈ also contains 𝑟 and 𝑟.̇ Furthermore, even contains 𝜃 and 𝜃, the derivation of the initial conditions required several steps to complete. Another lesson to be learned concerns the polar coordinate system in particular. We should be aware of the fact that this problem becomes ill-posed if point 𝐴, the initial position of the projectile, is chosen to coincide with point 𝑂, i.e., the origin of the chosen coordinate system. In this case, although we could still write Eqs. (3) and (4), we can no longer write meaningful initial conditions because the value of 𝜃 for a point at the origin is arbitrary and (as ℎ would be equal to zero) 𝜃̇ becomes undefined. This implies that when using the polar coordinate system, we should choose the origin of the coordinate system so as not to coincide with points where initial conditions are specified.

727

728

Chapter 12

Particle Kinematics

Problems Problem 12.157 ̂ Compute the following Consider the vectors 𝑎⃗ = 2 𝚤̂ + 1 𝚥̂ + 7 𝑘̂ and 𝑏⃗ = 1 𝚤̂ + 2 𝚥̂ + 3 𝑘. quantities. (a) 𝑎⃗ × 𝑏⃗ (b) 𝑏⃗ × 𝑎⃗ (c) 𝑎⃗ × 𝑏⃗ + 𝑏⃗ × 𝑎⃗ (d) 𝑎⃗ × 𝑎⃗ (e) (𝑎⃗ × 𝑎) ⃗ × 𝑏⃗ ⃗ (f) 𝑎⃗ × (𝑎⃗ × 𝑏) Parts (a)–(d) of this problem are meant to be a reminder that the cross product is an anticommutative operation, while Parts (e) and (f) are meant to be a reminder that the cross product is an operation that is not associative.

Problem 12.158 Consider two vectors 𝑎⃗ = 1 𝚤̂ + 2 𝚥̂ + 3 𝑘̂ and 𝑏⃗ = −6 𝚤̂ + 3 𝚥̂. (a) Verify that 𝑎⃗ and 𝑏⃗ are perpendicular to one another. ⃗ (b) Compute the vector triple product 𝑎⃗ × (𝑎⃗ × 𝑏). ⃗ with the vector −|𝑎| ⃗ (c) Compare the result from calculating 𝑎⃗ × (𝑎⃗ × 𝑏) ⃗ 2 𝑏. The purpose of this exercise is to show that as long as 𝑎⃗ and 𝑏⃗ are perpendicular to one ⃗ = −|𝑎| ⃗ This identity turns out to be very another, you can always write 𝑎⃗ × (𝑎⃗ × 𝑏) ⃗ 2 𝑏. useful in the study of the planar motion of rigid bodies.

𝜔1 𝑘̂

𝚥̂

𝑘̂

Problem 12.159

𝚥̂

𝚤̂

𝚤̂

𝜔2

The three propellers shown are all rotating with the same angular speed of 1000 rpm about different coordinate axes. (a) Provide the proper vector expressions for the angular velocity of each of the three propellers.

𝜔3 𝓁 𝑘̂

𝚥̂

𝑘̂ 𝚤̂

Figure P12.159

ISTUDY

𝚥̂

𝑢̂ 𝓁 𝚤̂

(b) Suppose that an identical propeller rotates at 1000 rpm about the axis 𝓁 oriented by the unit vector 𝑢̂ 𝓁 . Let any point 𝑃 on 𝓁 have coordinates such that 𝑥𝑃 = 𝑦𝑃 = 𝑧𝑃 . Find the vector representation of the angular velocity of this fourth propeller. Express the answers using units of radians per second.

Problem 12.160 A particle 𝑃 is moving along a path with the velocity shown. Discuss in detail whether or not there are incorrect elements in the sketch of the polar component system at 𝑃 . 𝑣⃗ 𝑢̂ 𝜃 𝑟

𝑃

𝑟

𝜃 𝑂

𝜃 path of 𝑃

Figure P12.160

𝑢̂ 𝜃

path of 𝑃

𝑢̂ 𝑟

𝑢̂ 𝑟

𝑃

𝑂 Figure P12.161

𝑣⃗

ISTUDY

Section 12.5

Planar Motion: Polar Coordinates

729

Problem 12.161 A particle 𝑃 is moving along a path with the velocity shown. Discuss in detail whether or not there are incorrect elements in the sketch of the polar component system at 𝑃 .

𝑣⃗

𝑢̂ 𝑟

𝑃 𝑟

Problem 12.162

𝜃

A particle 𝑃 is moving along a path with the velocity shown. Discuss in detail whether or not there are incorrect elements in the sketch of the polar component system at 𝑃 .

𝑢̂ 𝜃

𝑂

path of 𝑃 Figure P12.162

Problem 12.163

𝑢̂ 𝜃

A particle 𝑃 is moving along a circle with center 𝐶 and radius 𝑅 in the direction shown. Letting 𝑂 be the origin of a polar coordinate system with the coordinates 𝑟 and 𝜃 shown, discuss in detail whether or not there are incorrect elements in the sketch of the polar component system at 𝑃 .

𝑢̂ 𝑟 𝑟

𝑃

𝑣

𝜃

path of 𝑃

𝑂

Problem 12.164

𝑅

𝐶

Figure P12.163

A radar station is tracking a plane flying at a constant altitude with a constant speed 𝑣0 = 550 mph. If at a given instant 𝑟 = 7 mi and 𝜃 = 32◦ , determine the corresponding values ̇ 𝑟̈, and 𝜃. ̈ of 𝑟,̇ 𝜃, 𝑣0

𝑟 𝑢̂ 𝜃

𝜃 𝑢̂ 𝑟

path

𝑦 ℎ𝐴 Figure P12.164 𝑟

Problem 12.165 A basketball moves along the trajectory shown. Modeling the motion of the ball as a projectile motion, determine the radial and transverse components of the acceleration when 𝜃 = 65◦ . Express your answer in SI units.

𝜃 𝑂

𝑥

Figure P12.165

Problem 12.166 At a given instant, the merry-go-round is rotating with an angular velocity 𝜔 = 20 rpm while the child is moving radially outward at a constant rate of 0.7 m∕s. Assuming that the angular velocity of the merry-go-round remains constant, i.e., 𝛼 = 0, determine the magnitudes of the speed and of the acceleration of the child when he is 0.8 m away from the spin axis.

𝑧 𝛼 𝜔

𝑘̂

𝑢̂ 𝜃 𝑢̂ 𝑟

𝑃

Problem 12.167 At a given instant, the merry-go-round is rotating with an angular velocity 𝜔 = 18 rpm, and it is slowing down at a rate of 0.4 rad∕s2 . When the child is 2.5 f t away from the spin axis, determine the time rate of change of the child’s distance from the spin axis so that the child experiences no transverse acceleration while moving along a radial line.

Figure P12.166–P12.168

𝑟

730

Chapter 12

Particle Kinematics

Problem 12.168 At a given instant, the merry-go-round is rotating with an angular velocity 𝜔 = 18 rpm. When the child is 0.45 m away from the spin axis, determine the second derivative with respect to time of the child’s distance from the spin axis so that the child experiences no radial acceleration.

𝑦

𝑟

Problem 12.169

ℎ 𝜃

𝑂

𝑥 𝑑

A ball is dropped from rest from a height ℎ = 5 f t. If the distance 𝑑 = 3 f t, determine the radial and transverse components of the ball’s velocity and acceleration when the ball has traveled a distance ℎ∕2 from its release position.

Problem 12.170

Figure P12.169

The polar coordinates of a particle are the following functions of time: 𝑟 = 𝑟0 [1 + sin(𝑡3 ∕𝜏 3 )]

and 𝜃 = 𝜃0 cos(𝑡∕𝜏),

where 𝑟0 and 𝜃0 are constants, 𝜏 = 1 s, and where 𝑡 is time in seconds. Determine 𝑟0 and 𝜃0 such that, at 𝑡 = 𝜋∕4 s, the magnitude of the transverse velocity is half that of the radial velocity and the speed of the particle is 6 m∕s. 𝑦

path

0.22 rad∕s

𝑃

𝑟 𝑟⃗ 𝜃

𝑥

𝑂 Figure P12.170

Figure P12.171 and P12.172

Problem 12.171 A space station is rotating in the direction shown at a constant rate of 0.22 rad∕s. A crew member travels from the periphery to the center of the station through one of the radial shafts at a constant rate of 1.3 m∕s (relative to the shaft) while holding onto a handrail in the shaft. Taking 𝑡 = 0 to be the instant at which travel through the shaft begins and knowing that the radius of the station is 200 m, determine the velocity and acceleration of the crew member as a function of time. Express your answer using a polar coordinate system with origin at the center of the station.

Problem 12.172 Solve Prob. 12.171 and express your answers as a function of position along the shaft traveled by the astronaut.

𝑢̂ 𝜃 (𝑡)

𝑢̂ 𝑟 (𝑡)

𝑟(𝑡) 𝜃(𝑡)

Problem 12.173 During a given time interval, a radar station tracking an airplane records the readings 𝑟(𝑡) ̇ = [449.8 cos 𝜃(𝑡) + 11.78 sin 𝜃(𝑡)] mph,

Figure P12.173

ISTUDY

̇ = [11.78 cos 𝜃(𝑡) − 449.8 sin 𝜃(𝑡)] mph, 𝑟(𝑡)𝜃(𝑡)

ISTUDY

Section 12.5

731

Planar Motion: Polar Coordinates

where 𝑡 denotes time. Determine the speed of the plane. Furthermore, determine whether the plane being tracked is ascending or descending and determine the corresponding climbing rate (i.e., the rate of change of the plane’s altitude) expressed in ft∕s. 𝑦

Problems 12.174 and 12.175 The polar coordinates of a particle are the following functions of time: ) ( 𝑡2 𝑡 and 𝜃 = 𝜃0 2 , 𝑟 = 𝑟0 1 + 𝜏 𝜏

𝑟

where 𝑟0 = 3 f t, 𝜃0 = 1.2 rad, 𝜏 = 20 s, and 𝑡 is time in seconds.

𝜃

𝚥̂

Determine the velocity and the acceleration of the particle for 𝑡 = 35 s and express the result using the polar component system formed by the unit vectors 𝑢̂ 𝑟 and 𝑢̂ 𝜃 at 𝑡 = 35 s.

Problem 12.174

Problem 12.175 Determine the velocity and the acceleration of the particle for 𝑡 = 35 s and express the result using the Cartesian component system formed by the unit vectors 𝚤̂ and 𝚥̂.

𝑢̂ 𝑟

𝑢̂ 𝜃

𝑂

𝑥

𝚤̂

path Figure P12.174 and P12.175

Problems 12.176 and 12.177

𝑦

A particle is moving such that the time rate of change of its polar coordinates are 𝑟̇ = constant = 3 f t∕s

𝑢̂ 𝑟

and 𝜃̇ = constant = 0.25 rad∕s.

Knowing that at time 𝑡 = 0, a particle has polar coordinates 𝑟0 = 0.2 f t and 𝜃0 = determine the position, velocity, and acceleration of the particle for 𝑡 = 10 s. Express your answers in the polar component system formed by the unit vectors 𝑢̂ 𝑟 and 𝑢̂ 𝜃 at 𝑡 = 10 s. Problem 12.176

𝑢̂ 𝜃

𝑟

path

15◦ ,

Knowing that at time 𝑡 = 0, a particle has polar coordinates 𝑟0 = 0.2 f t and 𝜃0 = 15◦ , determine the position, velocity, and acceleration of the particle for 𝑡 = 10 s. Express your answers in the Cartesian component system formed by the unit vectors 𝚤̂ and 𝚥̂.

𝜃 𝚥̂ 𝑂

Problem 12.177

𝚤̂

𝑥

Figure P12.176 and P12.177

Problems 12.178 and 12.179 A micro spiral pump∗ consists of a spiral channel attached to a stationary plate. This plate has two ports, one for fluid inlet and another for outlet, the outlet being farther from the center of the plate than the inlet. The system is capped by a rotating disk. The fluid trapped between the rotating disk and the stationary plate is put in motion by the rotation of the top disk, which pulls the fluid through the spiral channel.

∗ The

spiral pump was originally invented in 1746 by H. A. Wirtz, a Swiss pewterer from Zurich. Recently, the spiral pump concept has seen a comeback in microdevice design. Some of the data used in this problem is taken from M. I. Kilani, P. C. Galambos, Y. S. Haik, and C.-J. Chen, “Design and Analysis of a Surface Micromachined Spiral-Channel Viscous Pump,” Journal of Fluids Engineering, 125, pp. 339–344, 2003.

inlet

𝑟 𝑢̂ 𝜃

Problem 12.178

Consider a spiral channel with the geometry given by the equation 𝑟 = 𝜂𝜃 + 𝑟0 , where 𝑟0 = 146 𝜇m is the starting radius, 𝑟 is the distance from the spin axis, and 𝜃, measured in radians, is the angular position of a point in the spiral channel. Assume that the radius at the outlet is 𝑟out = 190 𝜇m, that the top disk rotates with a constant angular speed 𝜔, and that the fluid particles in contact with the rotating disk are essentially stuck to it. Determine the constant 𝜂 and the value of 𝜔 (in rpm) such that after 1.25 rev of the top disk, the speed of the particles in contact with this disk is 𝑣 = 0.5 m∕s at the outlet.

spin axis 𝜔

rotating disk

𝑢̂ 𝑟

𝜃

pin joint stationary plate Figure P12.178 and P12.179

outlet spiral groove

732

Chapter 12

Particle Kinematics

𝐿

Problem 12.179 Consider a spiral channel with the geometry given by the equation 𝑟 = 𝜂𝜃 + 𝑟0 , where 𝜂 = 12 𝜇m is called the polar slope, 𝑟0 = 146 𝜇m is the starting radius, 𝑟 is the distance from the spin axis, and 𝜃, measured in radians, is the angular position of a point in the spiral channel. If the top disk rotates with a constant angular speed 𝜔 = 30,000 rpm, and assuming that the fluid particles in contact with the rotating disk are essentially stuck to it, use the polar coordinate system shown and determine the velocity and acceleration of one fluid particle when it is at 𝑟 = 170 𝜇m. 𝑣𝑠

Problem 12.180

𝜃

The cutaway of the gun barrel shows a projectile that, upon exit, moves with a speed 𝑣𝑠 = 5490 f t∕s relative to the gun barrel. The length of the gun barrel is 𝐿 = 15 f t. Assuming that the angle 𝜃 is increasing at a constant rate of 0.15 rad∕s, determine the speed of the projectile right when it leaves the barrel. In addition, assuming that the projectile acceleration along the barrel is constant and that the projectile starts from rest, determine the magnitude of the acceleration upon exit.

Figure P12.180

Problem 12.181 𝑦 𝑢̂ 𝑟 𝑟 𝑢̂ 𝜃

𝜃 𝑂

𝑥

A particle moves along a spiral described by the equation 𝑟 = 𝑟0 + 𝛾𝜃, where 𝑟0 and 𝛾 are constants, and where 𝜃 is in radians. Assume that 𝜃̇ = 𝛼𝑡, where 𝛼 = 0.15 rad∕s2 and 𝑡 is time expressed in seconds. If 𝑟 = 0.25 m and 𝜃 = 0 for 𝑡 = 0, determine 𝛾 such that, for 𝑡 = 2 s, the acceleration is completely in the radial direction. In addition, determine the value of the polar coordinates of the point for 𝑡 = 2 s.

Problem 12.182 Figure P12.181 and P12.182

ISTUDY

A point is moving counterclockwise at constant speed 𝑣0 along a spiral described by the equation 𝑟 = 𝑟0 + 𝛾𝜃, where 𝑟0 and 𝛾 are constants with dimensions of length. Determine the expressions of the velocity and the acceleration of the particle as a function of 𝜃 expressed in the polar component system shown.

Problem 12.183 A person driving along a rectilinear stretch of road is fined for speeding, having been clocked at 75 mph when the radar gun was pointing as shown. The driver claims that, because the radar gun is off to the side of the road instead of directly in front of his car, the radar gun overestimates his speed. Is he right or wrong and why? 𝑑∕2 𝐶 𝐴 radar gun Figure P12.183

𝑑∕2 𝐻 𝜃

𝐿

𝐵

𝑂 Figure P12.184

Problem 12.184 A motion tracking camera is placed along a rectilinear stretch of a racetrack (the figure is not to scale). A car 𝐶 enters the stretch at 𝐴 with a speed 𝑣𝐴 = 110 mph and accelerates uniformly in time so that at 𝐵 it has a speed 𝑣𝐵 = 175 mph, where 𝑑 = 1 mi. Letting the distance 𝐿 = 50 f t, if the camera is to track 𝐶, determine the camera’s angular velocity and the time rate of change of the angular velocity when the car is at 𝐴 and at 𝐻.

ISTUDY

Section 12.5

733

Planar Motion: Polar Coordinates

Problem 12.185 The radar station at 𝑂 is tracking a meteor 𝑃 as it moves through the atmosphere. At the instant shown, the station measures the following data for the motion of the meteor: 𝑟 = 21,000 f t, 𝜃 = 40◦ , 𝑟̇ = −22,440 f t∕s, 𝜃̇ = −2.935 rad∕s, 𝑟̈ = 187,500 f t∕s2 , and 𝜃̈ = −5.409 rad∕s2 . (a) Determine the magnitude and direction (relative to the 𝑥𝑦 coordinate system shown) of the velocity vector at this instant. (b) Determine the magnitude and direction (relative to the 𝑥𝑦 coordinate system shown) of the acceleration vector at this instant.

𝑃 𝑟

𝑦

𝜃

𝑥

𝑂 Figure P12.185

Problem 12.186 The time derivative of the acceleration, i.e., 𝑎, ⃗̇ is usually referred to as the jerk.∗ Starting from Eq. (12.72), compute the jerk in polar coordinates.

𝜔

Problem 12.187 A disk rotates about its center, which is the fixed point 𝑂. The disk has a straight channel whose centerline passes by 𝑂 and within which a collar 𝐴 is allowed to slide. If, when 𝐴 passes by 𝑂, the speed of 𝐴 relative to the channel is 𝑣 = 14 m∕s and is increasing in the direction shown with a rate of 5 m∕s2 , determine the acceleration of 𝐴 given that 𝜔 = 4 rad∕s and is constant. Express the answer using the component system shown, which rotates with the disk.

𝑟

The reciprocating rectilinear motion mechanism shown consists of a disk pinned at its center at 𝐴 that rotates with a constant angular velocity 𝜔𝐴𝐵 , a slotted arm 𝐶𝐷 that is pinned at 𝐶, and a bar that can oscillate within the guides at 𝐸 and 𝐹 . As the disk rotates, the peg at 𝐵 moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between 𝐶 and 𝐵. Letting 𝜃 = 30◦ , 𝜔𝐴𝐵 = 50 rpm, 𝑅 = 0.3 f t, and ̈ i.e., the angular velocity and angular acceleration of the ℎ = 0.6 f t, determine 𝜙̇ and 𝜙, slotted arm 𝐶𝐷, respectively. 𝜔𝐴𝐵

𝐷 𝑅

disk

𝐴 𝜃

𝜙 𝐵

ℎ 𝐶 𝑑

bar 𝐸

𝐹 Figure P12.188

∗ For

passengers riding in vehicles, high values of jerk usually make for an uncomfortable ride. Elevator manufacturers are very interested in jerk since they want to move passengers quickly from one floor to another without large changes in acceleration.

𝚥̂ 𝚤̂

𝐴 𝑣

Figure P12.187

Problem 12.188

𝑂

734

Chapter 12

Particle Kinematics Problems 12.189 and 12.190

𝐵

As a part of an assembly process, the end effector at 𝐴 on the robotic arm needs to move the gear at 𝐵 along the vertical line shown with some known velocity 𝑣0 and acceleration 𝑎0 . Arm 𝑂𝐴 can vary its length by telescoping via internal actuators, and a motor at 𝑂 allows it to pivot in the vertical plane.

𝐴 𝑟

When 𝜃 = 50◦ , it is required that 𝑣0 = 8 f t∕s (down) and that it be slowing down at 𝑎0 = 2 f t∕s2 . Using ℎ = 4 f t, determine, at this instant, the values for 𝑟̈ (the extensional acceleration) and 𝜃̈ (the angular acceleration). Problem 12.189

𝜃 𝑂 ℎ

Problem 12.190 Letting 𝑣0 and 𝑎0 be positive if the gear moves and accelerates uṗ and 𝜃̈ that are valid for any value of 𝜃. ward, determine expressions for 𝑟, 𝑟,̇ 𝑟̈, 𝜃,

Figure P12.189 and P12.190

Problems 12.191 and 12.192

𝐶 𝐴 𝜙

𝐷

In the cutting of sheet metal, the robotic arm 𝑂𝐴 needs to move the cutting tool at 𝐶 counterclockwise at a constant speed 𝑣0 along a circular path of radius 𝜌. The center of the circle is located in the position shown relative to the base of the robotic arm at 𝑂.

𝑟 𝜌 cutting path

ℎ

̇ 𝑟̈, and 𝜃̈ as Problem 12.191 When the cutting tool is at 𝐷 (𝜙 = 0), determine 𝑟, 𝑟,̇ 𝜃, functions of the given quantities (i.e., 𝑑, ℎ, 𝜌, 𝑣0 ).

𝜃

For all positions along the circular cut (i.e., for any value of 𝜙), determine 𝑟, 𝑟,̇ 𝜃,̇ 𝑟̈, and 𝜃̈ as functions of the given quantities (i.e., 𝑑, ℎ, 𝜌, 𝑣0 ). These quantities can be found “by hand,” but it is tedious, so you might consider using symbolic algebra software, such as Mathematica or Maple.

Problem 12.192

𝑂 𝑑 Figure P12.191 and P12.192 𝑦

Problem 12.193 The cam is mounted on a shaft that rotates about 𝑂 with constant angular velocity 𝜔cam . The profile of the cam is described by the function 𝓁(𝜙) = 𝑅0 (1 + 0.25 cos3 𝜙), where the angle 𝜙 is measured relative to the segment 𝑂𝐴, which rotates with the cam. Letting 𝜔cam = 3000 rpm and 𝑅0 = 3 cm, determine the velocity and acceleration of the follower when 𝜃 = 33◦ . Express the acceleration of the follower in terms of 𝑔, the acceleration due to gravity. Hint: Since the follower is not part of the cam rigid body, its acceleration is only 𝑟̈.

follower

cam 𝓁(𝜙)

𝐴

𝜙 𝜃

𝑥

𝑂 𝜔cam Figure P12.193

ISTUDY

shaft

Problem 12.194 The collar is mounted on the horizontal arm shown, which is originally rotating with the angular velocity 𝜔0 . Assume that after the cord is cut, the collar slides along the arm in such a way that the collar’s total acceleration is equal to zero. Determine an expression of the radial component of the collar’s velocity as a function of 𝑟, the distance from the spin ̇ axis. Hint: Using polar coordinates, observe that 𝑑(𝑟2 𝜃)∕𝑑𝑡 = 𝑟𝑎𝜃 .

ISTUDY

Section 12.5

𝑧

Planar Motion: Polar Coordinates

𝑟0 𝐴

cord

𝑅 𝑟

𝛼

𝜔0

𝜙

𝜃

𝐶

𝑂

𝜔 Figure P12.194

Figure P12.195

Problem 12.195 Particle 𝐴 slides over the semicylinder while pushed by the arm pinned at 𝐶. The motion of the arm is controlled such that it starts from rest at 𝜃 = 0, 𝜔 increases uniformly as a function of 𝜃, and 𝜔 = 0.5 rad∕s for 𝜃 = 45◦ . Letting 𝑅 = 4 in., determine the speed and the magnitude of the acceleration of 𝐴 when 𝜙 = 32◦ .

Problem 12.196 The mechanism shown is called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door-closing systems. If the disk is rotating with a constant angular velocity 𝜃̇ = 60 rpm, 𝐻 = 4 f t, 𝑅 = 1.5 f t, ̇ 𝑟̈, and 𝜙̈ when 𝜃 = 90◦ . and 𝑟 denotes the distance between 𝐵 and 𝑂, compute 𝑟,̇ 𝜙, 𝑥

swinging block

𝐵 𝑅 𝑦

𝐴

𝜃

𝜙

𝑂 𝑆

𝐻 Figure P12.196

Problem 12.197 A satellite is moving along the elliptical orbit shown. Using the polar coordinate system in the figure, the satellite’s orbit is described by the equation √ 𝑎 + 𝑎2 − 𝑏2 cos 𝜃 2 , 𝑟(𝜃) = 2𝑏 2 𝑎 + 𝑏2 − (𝑎2 − 𝑏2 ) cos(2𝜃)

satellite 𝑟 𝜃 planet

which implies the following identity 𝑟𝑟′′ − 2(𝑟′ )2 − 𝑟2 𝑎 = − 2, 𝑟3 𝑏 where the prime indicates differentiation with respect to 𝜃. Using this identity and knowing that the satellite moves so that 𝐾 = 𝑟2 𝜃̇ with 𝐾 constant (i.e., according to Kepler’s laws), show that the radial component of acceleration is proportional to −1∕𝑟2 , which is in agreement with Newton’s universal law of gravitation. Hint: This is an exercise in using the chain rule. Given 𝑟(𝜃(𝑡)), find expressions for 𝑟̇ and 𝑟̈ in terms of 𝑟′ and 𝑟′′ . This can be accomplished without having to make explicit use of 𝑟(𝜃) in the orbital equation.

orbit center 𝑏

𝑐 𝑎 Figure P12.197

735

736

Chapter 12

Particle Kinematics

Problem 12.198

𝑢̂ 𝜃 (0)

𝑢̂ 𝑟 (0)

𝑟(0) 𝜃(0)

ℎ0

At a given instant, an airplane flying at an altitude ℎ0 = 10,000 f t begins its descent in preparation for landing when it is 𝑟(0) = 20 mi from the radar station at the destination’s airport. At that instant, the airplane’s speed is 𝑣0 = 300 mph, the climb rate is constant and equal to −5 f t∕s, and the horizontal component of velocity is decreasing steadily at a ̇ 𝑟̈, and 𝜃̈ that would be observed by the radar station. rate of 15 f t∕s2 . Determine the 𝑟,̇ 𝜃,

Figure P12.198

ISTUDY

Problem 12.199 Considering the system analyzed in Example 12.19, let ℎ = 15 f t, 𝑣0 = 55 mph, and 𝜙 = 25◦ . Plot the trajectory of the projectile in two different ways: (1) by solving the projectile motion problem using Cartesian coordinates and plotting 𝑦 versus 𝑥 and (2) by using a computer to solve Eqs. (3), (4), (9), and (10) in Example 12.19. You should, of course, get the same trajectory regardless of the coordinate system used.

ISTUDY

Section 12.5

737

Planar Motion: Polar Coordinates

Design Problems Design Problem 12.2 As a part of a robotics competition, a robotic arm is to be designed so as to catch an egg without breaking it. The egg is released at point 𝐴 from rest while the arm is initially also at rest in the position shown. The arm starts moving when the egg is released, and it catches the egg at 𝐵 with the same speed and acceleration that the egg has at that instant (matching the speed avoids impact and matching the acceleration keeps them together after impact). Assume that the arm will start slowing the egg down at 𝐵 and will bring the egg to a complete stop at 𝐶. Also assume that the vertical rate of change of deceleration (the time derivative of the acceleration) felt by the egg remains constant and that the egg arrives at 𝐶 with zero acceleration. For the motion between 𝐵 and 𝐶, determine (a) The rate of change of the deceleration. (b) The function 𝑦(𝑡) of the vertical motion. (c) The time it takes for the arm to bring the egg to a stop. (d) The vertical position 𝐶 at which they come to a stop. (e) Then, using the fact that 𝑎⃗̇ ⋅ 𝚤̂ is 0 and 𝑎⃗̇ ⋅ 𝚥̂ is the time rate of change of the vertical acceleration due to the motion of the arm along a vertical line, plot the functions 𝑟(𝑡) and 𝜃(𝑡) required to achieve the given motion from 𝐵 to 𝐶. (f) Finally, use the geometrical constraints 0.5 tan 𝜃 = 𝑦 and 𝑟2 = 𝑦2 + (0.5)2 to determine analytical expressions for 𝑟(𝑡) and 𝜃(𝑡), and compare the plots of these analytical expressions with the plots found in Part (e). They should, of course, be the same.

Design Problem 12.3 As a part of a robotics competition, a robotic arm is to be designed so as to catch an egg without breaking it. The egg is released at point 𝐴 from rest while the arm is initially also at rest in the position shown. The arm starts moving when the egg is released, and it is to catch the egg at point 𝐵 in such a way as to avoid any impact between the egg and the robot hand. See Design Problem 12.2 for how the robot arm needs to be moving for it to ̈ catch the egg. With this in mind, find an acceleration profile [i.e., 𝑟̈(𝑡) and 𝜃(𝑡)] of the arm as it moves from its initial position to 𝐵 that satisfies this condition.

All dimensions are in meters. 𝚥̂

𝐴 𝚤̂

𝑟 𝐵 𝑦

𝜃 𝑂

0.2

𝐶

0.5 Figure DP12.2 and DP12.3

0.5

0.6

0.8

738

Chapter 12

Particle Kinematics

12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

In this section we discuss relative motion and differentiation of constraints. These concepts are used to solve problems with multiple moving objects and are important in the development of rigid body kinematics. We will study relative motion using frames of reference that only translate relative to one another. The general case, which includes frames that also rotate relative to one another, is presented in Section 16.4.

𝑌

Relative motion

𝑦

path of 𝐴

Consider two points 𝐴 and 𝐵 moving on separate paths in a plane (Fig. 12.32). The 𝑋𝑌 Cartesian frame of reference has base vectors 𝐼̂ and 𝐽̂ and is fixed — this is the stationary frame. The 𝑥𝑦 Cartesian frame of reference has base vectors 𝚤̂ and 𝚥̂, it is attached to 𝐴, and it only translates relative to the 𝑋𝑌 frame — this is the moving frame. Using Eq. (11.15) on p. 627, the position of 𝐵 relative to 𝐴 in the 𝑋𝑌 frame is ( ) ( ) (12.78) 𝑟⃗𝐵∕𝐴 = 𝑋𝐵 − 𝑋𝐴 𝐼̂ + 𝑌𝐵 − 𝑌𝐴 𝐽̂,

𝚥̂ 𝐴

𝑥

𝚤̂ 𝑟⃗𝐵∕𝐴

𝐵

𝑟⃗𝐴

where (𝑋𝐴 , 𝑋𝐵 ) and (𝑋𝐵 , 𝑌𝐵 ) are the coordinates of 𝐴 and 𝐵 relative to the 𝑋𝑌 frame, respectively, and where we recall that the subscript 𝐵∕𝐴 is read “𝐵 relative to 𝐴” or “𝐵 as seen by an observer at 𝐴” [see discussion of Eq. (11.15) on p. 627]. The position of 𝐵 relative to 𝐴 as viewed by the 𝑥𝑦 frame is

𝑟⃗𝐵 path of 𝐵

𝐽̂ 𝑂

𝑋

𝐼̂

Figure 12.32 Two particles 𝐴 and 𝐵 and the definition of their relative position vector 𝑟⃗𝐵∕𝐴 . The 𝑥𝑦 frame only translates relative to the 𝑋𝑌 frame.

ISTUDY

Helpful Information The notation 𝒓⃗𝑩∕𝑨 . The relative position vector 𝑟⃗𝐵∕𝐴 is sometimes written as 𝑟⃗𝐴𝐵 , which is read as “the position vector from 𝐴 to 𝐵.” In fact, this is the notation used in the Statics book. This notation is potentially confusing when we are talking about velocities and accelerations because the vector 𝑣⃗𝐴𝐵 might be read as “the velocity vector from 𝐴 to 𝐵” as opposed to “the velocity vector of 𝐵 relative to 𝐴,” the latter being the correct way to refer to the relative velocity.

𝑟⃗𝐵∕𝐴 = 𝑥𝐵 𝚤̂ + 𝑦𝐵 𝚥̂,

(12.79)

where 𝑥𝐵 and 𝑦𝐵 are the coordinates of 𝐵 relative to the 𝑥𝑦 frame. Since the 𝑥𝑦 frame is Cartesian, when the moving observer differentiates the vector 𝑟⃗𝐵∕𝐴 with respect to time, this observer obtains ) (̇ 𝑟⃗𝐵∕𝐴 𝑥𝑦 frame = 𝑥̇ 𝐵 𝚤̂ + 𝑦̇ 𝐵 𝚥̂.

(12.80)

By contrast, when the stationary observer computes the same time derivative, the result is ) ( ) ( ) (̇ 𝑟⃗𝐵∕𝐴 𝑋𝑌 frame = 𝑋̇ 𝐵 − 𝑋̇ 𝐴 𝐼̂ + 𝑌̇ 𝐵 − 𝑌̇ 𝐴 𝐽̂ = 𝑥̇ 𝐵 𝚤̂ + 𝑥𝐵 𝚤̂̇ + 𝑦̇ 𝐵 𝚥̂ + 𝑦𝐵 𝚥̂̇ = 𝑥̇ 𝐵 𝚤̂ + 𝑦̇ 𝐵 𝚥̂,

(12.81)

where the second line in Eq. (12.81) is obtained by letting the stationary observer compute the time derivative of Eq. (12.79) and we have used the fact that 𝚤̂̇ = 𝚥̂̇ = 0 since 𝑥𝑦 is not rotating. Comparing Eqs. (12.80) and (12.81) we see that ) ( ) (̇ 𝑟⃗𝐵∕𝐴 𝑥𝑦 frame = 𝑟⃗̇ 𝐵∕𝐴 𝑋𝑌

frame

,

(12.82)

which states that the time rate of change of the vector 𝑟⃗𝐵∕𝐴 is the same for the stationary and moving observers when these observers only translate relative to one another. To relate position, velocity, and acceleration between the 𝑥𝑦 and 𝑋𝑌 reference frames, we now consider the vector triangle 𝑂𝐴𝐵, for which we have (see Fig. 12.32) 𝑟⃗𝐵 = 𝑟⃗𝐴 + 𝑟⃗𝐵∕𝐴 ,

(12.83)

ISTUDY

Section 12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

where 𝑟⃗𝐴 and 𝑟⃗𝐵 are the position vectors of 𝐴 and 𝐵 relative to the 𝑋𝑌 reference frame, respectively. The time derivative of Eq. (12.83) gives 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 ,

(12.84)

where 𝑣⃗𝐵∕𝐴 = 𝑑⃗𝑟𝐵∕𝐴 ∕𝑑𝑡 is the relative velocity of 𝐵 with respect to 𝐴. Differentiating Eq. (12.84) with respect to time, we have 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 ,

(12.85)

where 𝑎⃗𝐵∕𝐴 = 𝑑 2 𝑟⃗𝐵∕𝐴 ∕𝑑𝑡2 is the relative acceleration of 𝐵 with respect to 𝐴. Because of Eq. (12.82), 𝑣⃗𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 are identical in the 𝑥𝑦 and 𝑋𝑌 frames. Equation (12.84) says that the velocity of 𝐵, as seen by the stationary observer, is equal to the velocity of 𝐴, as seen by the stationary observer, plus the relative velocity of 𝐵 with respect to 𝐴, which is the velocity of 𝐵 as seen by the moving observer. By replacing velocity with acceleration, Eq. (12.85) can be read in a similar way.

Differentiation of geometrical constraints In addition to the vector-based methods we have studied, sometimes we can find the motion of points by writing geometrical constraint equations involving the position of these points and then differentiating the constraint equations with respect to time. We will show how this works with two short examples. A rigid system sliding down a wall Referring to Fig. 12.33(a), the particles 𝐴 and 𝐵 are connected by a rigid bar of length 𝐿. The system is sliding down the wall at 𝐴 and across the floor at 𝐵. By writing constraint equations involving the positions of 𝐴 and 𝐵 and the angle 𝜃, we can derive equations relating the time rate of change of these quantities. Since the bar is 𝑦 𝐴

𝐴

𝜃

𝑦𝐴

𝐿

𝐿

𝜃

𝐵

𝐵

𝑂

𝑥𝐵

(a)

𝑥

(b)

Figure 12.33. (a) A bar sliding down a wall. (b) Positions of 𝐴 and 𝐵 measured in the Cartesian frame with origin at 𝑂.

rigid, points 𝐴 and 𝐵 move under the constraints that 𝐴 only moves vertically, 𝐵 only moves horizontally, and the distance 𝐿 between 𝐴 and 𝐵 remains constant. Therefore, referring to Fig. 12.33(b), neglecting the size of 𝐴 and 𝐵, and using the Cartesian coordinate system shown, the positions of 𝐴 and 𝐵 are completely characterized in terms of the 𝜃 and 𝐿 as follows: 𝑥𝐵 = 𝐿 sin 𝜃

and

𝑦𝐴 = 𝐿 cos 𝜃.

(12.86)

739

740

Chapter 12

Particle Kinematics

Differentiating Eqs. (12.86) with respect to time, we obtain the velocity and acceleration of 𝐴 and 𝐵: 𝑥̇ 𝐵 = 𝑣𝐵 = 𝐿𝜃̇ cos 𝜃, 𝑥̈ = 𝑎 = 𝐿𝜃̈ cos 𝜃 − 𝐿𝜃̇ 2 sin 𝜃, 𝐵

𝐵

𝑦̇ 𝐴 = 𝑣𝐴 = −𝐿𝜃̇ sin 𝜃, 𝑦̈ = 𝑎 = −𝐿𝜃̈ sin 𝜃 − 𝐿𝜃̇ 2 cos 𝜃. 𝐴

𝐴

(12.87) (12.88)

Equations (12.87) and (12.88) can be used in several ways. For example, if we are given the position of 𝐵 as a function of time, we could solve the six expressions in ̇ ̈ Eqs. (12.86)–(12.88) to find the 𝜃(𝑡), 𝜃(𝑡), 𝜃(𝑡), 𝑦𝐴 (𝑡), 𝑦̇ 𝐴 (𝑡), and 𝑦̈𝐴 (𝑡). A simple pulley system 𝐽

𝐴

+𝑦

𝐷

𝐻

𝐸 𝑦𝑄

𝑦𝑃 𝐵

𝐺

𝐶 𝐹 𝑄

𝐼 𝑃

Figure 12.34 Simple pulley system used to demonstrate the principles behind the differentiation of geometric constraints.

ISTUDY

As an another example of a constrained system, we consider the pulley systems in Fig. 12.34, for which we want to determine how the velocity and acceleration of block 𝑃 are related to the velocity and acceleration of block 𝑄 under the constraint that the cords in the system are inextensible. The key to the analysis of any pulley system is the notion of cord (or cable or rope) length and its first and second time derivatives. In Fig. 12.34, there are three cords. However, cords 𝐺𝐼 and 𝐽 𝐻 simply keep block 𝑃 attached to pulley 𝐺 and pulley 𝐻 attached to the fixed ceiling at 𝐽 , respectively. Therefore, 𝑣𝑃 = 𝑣𝐺

⇒

𝑎𝑃 = 𝑎𝐺 ,

(12.89)

𝑣𝐻 = 𝑣𝐽 = 0

⇒

𝑎𝐻 = 𝑎𝐽 = 0,

(12.90)

where Eqs. (12.90) hold because 𝐽 is fixed. We now look at the cord 𝐴𝐵𝐶𝐷𝐸𝐹 . Let the length of cord 𝐴𝐵𝐶𝐷𝐸𝐹 be 𝐿, which is a constant since all cords are assumed inextensible. We now express 𝐿 in terms of the quantities shown in Fig. 12.34, i.e., ̃ + 𝐶𝐷 + 𝐷𝐸 ̃ + 𝐸𝐹 , 𝐿 = 𝐴𝐵 + 𝐵𝐶

(12.91)

̃ and 𝐷𝐸 ̃ are the lengths of the curved segments of cord that wrap around where 𝐵𝐶 pulleys 𝐺 and 𝐻, respectively, and the letters with overbars represent the lengths ̃ and 𝐷𝐸 ̃ are both of the corresponding straight line segments. The lengths of 𝐵𝐶 constant, the sum of which we will call 𝐾. The lengths of the straight segments can be written in terms of the coordinates of blocks 𝑃 and 𝑄, which we do as follows:

Helpful Information What if the cord length is not constant? In some problems the cord length is not constant, such as when a motor or winch is pulling in or letting out cord. In these cases, we modify the kinematics by setting the time derivative of the appropriate cord length equal to the rate at which the cord is becoming longer or shorter, i.e., ⎧ + rate of increase, ⎪ 𝐿̇ = ⎨ or ⎪ − rate of decrease. ⎩

𝐴𝐵 = 𝑦𝑃 − 𝐺𝐼,

𝐶𝐷 = 𝑦𝑃 − 𝐺𝐼 − 𝐽 𝐻,

𝐸𝐹 = 𝑦𝑄 − 𝐽 𝐻.

(12.92)

Then Eq. (12.91) becomes 𝐿 = 2𝑦𝑃 + 𝑦𝑄 − 2𝐺𝐼 − 2𝐽 𝐻 + 𝐾.

(12.93)

We now differentiate Eq. (12.93) with respect to time and recognize that 𝑑(𝐺𝐼)∕𝑑𝑡 = 0 and 𝑑(𝐽 𝐻)∕𝑑𝑡 = 0, due to Eqs. (12.89) and (12.90), respectively. Therefore, we obtain 𝐿̇ = 2𝑦̇ 𝑃 + 𝑦̇ 𝑄 . (12.94) Recalling that 𝐿 is constant because the cord is inextensible, 𝐿̇ = 0 and Eq. (12.94) yields the relation between the velocities of blocks 𝑃 and 𝑄, i.e., 2𝑦̇ 𝑃 + 𝑦̇ 𝑄 = 0 or

2𝑣𝑃 + 𝑣𝑄 = 0.

(12.95)

We now note the following about Eq. (12.95): • Equation (12.95) says that if 𝑃 is moving 4 m∕s down, then 𝑄 must be moving 8 m∕s up. This is so because 𝑣𝑄 = −2𝑣𝑃 and we have defined both 𝑦𝑃 and 𝑦𝑄 to be positive downward.

ISTUDY

Section 12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

741

• We can differentiate Eq. (12.95) with respect to time to obtain a relationship between the accelerations of blocks 𝑃 and 𝑄 2𝑎𝑃 + 𝑎𝑄 = 0.

(12.96)

• We did not need to know the length of each cord — we only needed to know that each length was constant. This will often be the case.

End of Section Summary Relative motion. Referring to Fig. 12.35, consider the planar motion of points 𝐴 and 𝐵. The positions of 𝐴 and 𝐵 relative to the 𝑋𝑌 frame are 𝑟⃗𝐴 and 𝑟⃗𝐵 , respectively. Attached to 𝐴 there is a frame 𝑥𝑦 that translates but does not rotate relative to frame 𝑋𝑌 . In either frame, the position of 𝐵 relative to 𝐴 is given by 𝑟⃗𝐵∕𝐴 . Using vector addition, the vectors 𝑟⃗𝐴 , 𝑟⃗𝐵 , and 𝑟⃗𝐵∕𝐴 are related as follows:

𝑌

𝑦

𝚥̂

Eq. (12.83), p. 738

𝐴

𝑟⃗𝐵

Eqs. (12.84) and (12.85), p. 739

where 𝑣⃗𝐵∕𝐴 = 𝑟⃗̇ 𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 = 𝑟⃗̈𝐵∕𝐴 are the relative velocity and acceleration of 𝐵 with respect to 𝐴, respectively. In general, the vectors 𝑣⃗𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 computed by the 𝑥𝑦 observer are different from the vectors 𝑣⃗𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 computed by the 𝑋𝑌 observer. However, the 𝑥𝑦 and 𝑋𝑌 observers compute the same 𝑣⃗𝐵∕𝐴 and the same 𝑎⃗𝐵∕𝐴 if these observers do not rotate relative to one another. Constrained motion. There are very few dynamics problems in which the motion is not constrained in some way. In certain classes of systems, constraints are described by geometrical relations between points in the system. We analyzed a pulley system whose motion was constrained by the inextensibility of the cords in the system. The key to constrained motion analysis is the awareness that we can differentiate the equations describing the geometrical constraints to obtain velocities and accelerations of points of interest.

𝐵

𝑟⃗𝐴

The first and second time derivatives of the above equation are, respectively,

𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 ,

𝑥

𝚤̂ 𝑟⃗𝐵∕𝐴

𝑟⃗𝐵 = 𝑟⃗𝐴 + 𝑟⃗𝐵∕𝐴 .

𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 ,

path of 𝐴

path of 𝐵

𝐽̂ 𝑂

𝐼̂

𝑋

Figure 12.35 Two particles 𝐴 and 𝐵 and the definition of their relative position vector 𝑟⃗𝐵∕𝐴 .

742

Chapter 12

Particle Kinematics

E X A M P L E 12.20 𝑣𝐵

𝐵

𝑎𝐵

𝜃 𝑣𝑃 𝑃

Relative Speed and Acceleration The driver of car 𝐵 sees a police car 𝑃 and applies his brakes, causing the car to decelerate at a constant rate of 25 f t∕s2 . At the same time, the police car is traveling at a constant speed 𝑣𝑃 = 35 mph, and using a radar gun, the police officer sees 𝐵 coming toward her at 65 mph when 𝜃 = 22◦ . At the instant that the radar gun measurement was taken, determine the corresponding true speed of 𝐵 and the magnitude of the relative acceleration of 𝐵 with respect to 𝑃 .

SOLUTION Road Map We need to find the speed of 𝐵 with respect to the road, which we choose as our stationary frame. The speed measured by the radar gun is relative to the moving observer 𝑃 and is the component of the velocity of 𝐵 relative to 𝑃 along the line connecting 𝐵 and 𝑃 . Hence, we will find the velocity of 𝐵 relative to 𝑃 and then consider the component of this velocity along the line 𝑃 𝐵. As for the relative acceleration of 𝐵 with respect to 𝑃 , we can calculate it as a direct application of Eq. (12.85).

Figure 1

Computation 𝑣𝐵

𝐵 𝜃

𝚥̂

𝑣𝑃 𝚤̂

𝑢̂ 𝐵∕𝑃 𝑃

Figure 2 Stationary component system.

ISTUDY

𝑎𝐵

to 𝑃 is

Letting 𝑣𝐵 be the speed of 𝐵, referring to Fig. 2, the velocity of 𝐵 relative 𝑣⃗𝐵∕𝑃 = 𝑣⃗𝐵 − 𝑣⃗𝑃 = −𝑣𝐵 𝚤̂ − 𝑣𝑃 𝚥̂.

(1)

We denote the radar gun speed reading by 𝑣𝑟 , and we observe that 𝑣𝑟 is the magnitude of the component of 𝑣⃗𝐵∕𝑃 along the line connecting points 𝑃 and 𝐵. The sign of the component in question is negative because the police officer sees 𝐵 coming toward her. Therefore, denoting by 𝑢̂ 𝐵∕𝑃 the unit vector pointing from 𝑃 to 𝐵 and observing that 𝑢̂ 𝐵∕𝑃 = sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂, we have 𝑣⃗𝐵∕𝑃 ⋅ 𝑢̂ 𝐵∕𝑃 = −𝑣𝑟

⇒

−𝑣𝐵 sin 𝜃 − 𝑣𝑃 cos 𝜃 = −𝑣𝑟 ,

(2)

where we have used the expression for 𝑣⃗𝐵∕𝑃 in Eq. (1). Solving the second of Eqs. (2) for 𝑣𝐵 , we have 𝑣 − 𝑣𝑃 cos 𝜃 = 86.89 mph. 𝑣𝐵 = 𝑟 (3) sin 𝜃 Applying Eq. (12.85), the acceleration of 𝐵 relative to 𝑃 is 𝑎⃗𝐵∕𝑃 = 𝑎⃗𝐵 − 𝑎⃗𝑃 = (25.00 f t∕s2 ) 𝚤̂,

(4)

since the police car is traveling at a constant velocity. Therefore, we have |𝑎⃗𝐵∕𝑃 | = 25.00 f t∕s2 .

(5)

Since the terms cos 𝜃 and sin 𝜃 are nondimensional, the result in Eq. (3) is dimensionally correct. The result in Eq. (5) is also dimensionally correct since it was derived as a direct application of a dimensionally correct formula for the relative acceleration.

Discussion & Verification

ISTUDY

Section 12.6

743

Relative Motion Analysis and Differentiation of Geometrical Constraints

E X A M P L E 12.21

Relative Motion in a Car Shock Absorbing System

Consider a car traveling over an undulating road (Fig. 1) such that points 𝐴 and 𝐵 have the same constant horizontal velocity 𝑣0 . Determine, as functions of time, the relative position, velocity, and acceleration of 𝐴 with respect to 𝐵. Assume that 𝐴 and 𝐵 move such that 𝑦𝐴 = ℎ𝐴 + 𝐷𝐴 sin(2𝜋𝑥𝐴 ∕𝜆𝐴 ) and 𝑦𝐵 = 𝑅𝑊 + 𝐷𝑅 sin(2𝜋𝑥𝐵 ∕𝜆𝑅 ), respectively. Let ℎ𝐴 and 𝑅𝑤 be the height of the top of the suspension and the center of the wheel, respectively, when the car is not moving.

𝜆𝐴

2𝐷𝐴

2𝐷𝑅

𝜆𝑅

𝐴

𝑦

SOLUTION Road Map Once we describe the position of points 𝐴 and 𝐵, we can find their relative position by subtracting them. The relative velocity and acceleration can then be found by differentiating the relative position with respect to time. Computation

Using the coordinate system shown in Fig. 1, we have 𝑟⃗𝐴 = 𝑥𝐴 𝚤̂ + 𝑦𝐴 𝚥̂ and 𝑟⃗𝐵 = 𝑥𝐵 𝚤̂ + 𝑦𝐵 𝚥̂.

(1)

𝑦𝐴

𝐵

𝑦𝐵

𝚥̂

𝑥

𝚤̂

Figure 1 Very simple schematic of a car’s shock absorbing system. Point 𝐴 is attached to the frame of the car.

Using Eq. (12.83), the relative position becomes 𝑟⃗𝐴∕𝐵 = 𝑟⃗𝐴 − 𝑟⃗𝐵 = (𝑥𝐴 − 𝑥𝐵 ) 𝚤̂ + (𝑦𝐴 − 𝑦𝐵 ) 𝚥̂ = 𝑦𝐴∕𝐵 𝚥̂,

(2)

where 𝑥𝐴 = 𝑥𝐵 since 𝐴 and 𝐵 move with the same constant horizontal velocity. Using the given trajectories of 𝐴 and 𝐵, the relative position vector 𝑟⃗𝐴∕𝐵 takes on the form ) ( )] [ ( 2𝜋𝑣0 𝑡 2𝜋𝑣0 𝑡 − 𝐷𝑅 sin 𝚥̂, 𝑟⃗𝐴∕𝐵 =𝑦𝐴∕𝐵 𝚥̂ = ℎ𝐴 − 𝑅𝑊 + 𝐷𝐴 sin 𝜆𝐴 𝜆𝑅

(3)

where we have used 𝑥𝐴 = 𝑥𝐵 = 𝑣0 𝑡 since the vehicle is moving at the constant speed 𝑣0 . We have set, without loss of generality, 𝑥𝐴 (0) = 𝑥𝐵 (0) = 0. The relative velocity and acceleration are obtained by differentiating Eq. (3) with respect to time, using the chain rule, and noting that 𝑥̇ 𝐴 = 𝑣0 and 𝑥̈ 𝐴 = 0, i.e., 𝑑𝑦𝐴∕𝐵 𝑑𝑥𝐴 𝑑𝑥𝐴

𝑦̈𝐴∕𝐵 = 𝑣0

𝑑 𝑑𝑡

𝑑𝑡 ( 𝑑𝑦

=

𝐴∕𝐵

𝑑𝑦𝐴∕𝐵

)

𝑑𝑥𝐴

𝑑𝑥𝐴 = 𝑣0

𝑥̇ 𝐴 = 𝑑 𝑑𝑥𝐴

𝑑𝑦𝐴∕𝐵

10

𝑣0 ,

(4)

𝑑𝑥𝐴 ( 𝑑𝑦 )

𝑑𝑥𝐴

𝑑𝑥𝐴

𝑑𝑡

𝐴∕𝐵

=

𝑑2𝑦

𝐴∕𝐵

𝑑𝑥2𝐴

𝑣20 ,

(5)

𝑣𝐴∕𝐵 (m∕s)

𝑦̇ 𝐴∕𝐵 =

so that 𝑑𝑥𝐴

𝑣0 𝚥̂ and 𝑎⃗𝐴∕𝐵 = 𝑦̈𝐴∕𝐵 𝚥̂ =

𝑑 2 𝑦𝐴∕𝐵 𝑑𝑥2𝐴

Substituting Eq. (3) into Eqs. (6), we obtain ( ( ) )] [ 𝐷 𝐷 2𝜋𝑥𝐴 2𝜋𝑥𝐵 − 𝑅 cos 𝚥̂, 𝑣⃗𝐴∕𝐵 = 2𝜋𝑣0 𝐴 cos 𝜆𝐴 𝜆𝐴 𝜆𝑅 𝜆𝑅 ( ( ) )] [ 𝐷 𝐷 2𝜋𝑥𝐴 2𝜋𝑥𝐵 𝑎⃗𝐴∕𝐵 = −4𝜋 2 𝑣20 2𝐴 sin − 2𝑅 sin 𝚥̂. 𝜆𝐴 𝜆𝑅 𝜆𝐴 𝜆𝑅

𝑣20 𝚥̂.

−5 0.2

0.4 0.6 𝑡(s)

0.8

1.0

0.2

0.4

0.6

0.8

1.0

(6) 100

(7) (8)

Discussion & Verification

The verification of the dimensional correctness of the result is left to the reader as an exercise. A Closer Look

0

−10 0.0

Figure 2 shows a representation of Eqs. (7) and (8) for a specific choice of the relevant geometrical parameters. Notice that the magnitudes of the vertical velocity are of the same order as the horizontal speed of the car. Furthermore, the maximum acceleration exceeds 100𝑔!

𝑎𝐴∕𝐵 ∕g

𝑣⃗𝐴∕𝐵 = 𝑦̇ 𝐴∕𝐵 𝚥̂ =

𝑑𝑦𝐴∕𝐵

5

50 0 −50 −100 0.0

𝑡(s) Figure 2 Plots of relative velocity and acceleration. The plot of acceleration is given in multiples of 𝑔. The following values were used: 𝐷𝐴 = 𝐷𝑅 = 0.1 m, 𝜆𝐴 = 10 m, 𝜆𝑅 = 1 m, 𝑣0 = 60 km∕h.

744

Chapter 12

Particle Kinematics

E X A M P L E 12.22

Analysis of a Moving Target Problem In an action movie scene, railcar 𝐴 and mobile robot 𝐵 are 4 s away from colliding at the crossing 𝐶 (Fig. 1). At this instant the movie’s hero is traveling on the railcar and is set to fire a gun at 𝐵 to destroy 𝐵 before the collision. 𝐴 and 𝐵 move with constant speeds 𝑣𝐴 = 18 m∕s and 𝑣𝐵 = 40 m∕s, respectively, and the muzzle velocity of the gun’s projectile 𝑃 has a magnitude of 300 m∕s. In what direction 𝜃 would you tell the hero on 𝐴 to point the gun in order to hit 𝐵?

𝐶

𝑣⃗𝐴

48.2◦ 𝜃

𝐴

SOLUTION

𝑣⃗𝐵

𝐵

Figure 1 Railcar and robot approaching a rail crossing. The figure is not drawn to scale. 𝑦 𝚥̂ 𝚤̂ 𝐶 𝛽 = 48.2◦

𝑟⃗𝐴 (𝑡)

𝑟⃗𝑃 (𝑡)

𝐴(𝑡)

𝑥

𝐵(𝑡ℎ ) = 𝑃 (𝑡ℎ )

𝑃 (𝑡)

𝑟⃗𝐵 (𝑡) 𝐵(𝑡)

𝐴(𝑡𝑓 ) = 𝑃 (𝑡𝑓 ) path of 𝑃

𝐵(𝑡𝑓 )

Figure 2 Position vectors of points 𝐴, 𝐵, and 𝑃 at a generic time 𝑡. The figure is not drawn to scale. 𝑦 𝚥̂ 𝚤̂ 𝐶

𝑑

𝑥

𝛽 𝑟⃗𝑃 (𝑡𝑓 ) 𝐴(𝑡𝑓 ) = 𝑃 (𝑡𝑓 )

𝓁 𝑟⃗𝐵 (𝑡𝑓 )

𝐵(𝑡𝑓 ) Figure 3 Position vectors of points 𝑃 and 𝐵 at the time of firing. The figure is not drawn to scale.

ISTUDY

Let 𝑣⃗𝑃 ∕𝐴 be the velocity vector of the projectile 𝑃 relative to railcar 𝐴. The magnitude of 𝑣⃗𝑃 ∕𝐴 is 300 m∕s, and we want to solve for 𝑣⃗𝑃 ∕𝐴 , which provides the direction the hero points the gun and can be described by the angle 𝜃 in Fig. 1. To find 𝜃, we will use relative kinematics to relate 𝑣⃗𝑃 ∕𝐴 , which is measured relative to 𝐴, with 𝑣⃗𝐴 and 𝑣⃗𝐵 , which are relative to the ground. If 𝑟⃗𝑃 (𝑡) and 𝑟⃗𝐵 (𝑡) are the positions of the projectile 𝑃 and of the target 𝐵 as a function of time, respectively, then to “hit the target” means that there must be a time 𝑡ℎ such that 𝑃 and 𝐵 occupy the same position, i.e., Road Map

𝑟⃗𝑃 (𝑡ℎ ) = 𝑟⃗𝐵 (𝑡ℎ ).

(1)

Our strategy will be to describe the velocities of 𝑃 and 𝐵, which we will integrate with respect to time to obtain 𝑟⃗𝑃 (𝑡) and 𝑟⃗𝐵 (𝑡) so that we can satisfy Eq. (1). Computation Referring to Fig. 2, we define a reference frame with origin at 𝐶 and a 𝑦 axis coinciding with the path of 𝐵. We will refer to this frame as stationary. We also observe that, once fired, 𝑃 travels with a constant velocity 𝑣⃗𝑃 relative to the stationary frame. Thus, if we let 𝑡𝑓 be the time of firing, 𝑃 moves along the segment joining the position of 𝐴 at 𝑡𝑓 to the position of 𝐵 at 𝑡ℎ . If we Let 𝑣⃗𝑃 and 𝑣⃗𝐵 be the velocities of 𝑃 and 𝐵, respectively, in the stationary frame, they can be written as

𝑣⃗𝑃 = 𝑣𝑃 𝑥 𝚤̂ + 𝑣𝑃 𝑦 𝚥̂ and 𝑣⃗𝐵 = 𝑣𝐵 𝚥̂,

(2)

where 𝑣𝑃 𝑥 and 𝑣𝑃 𝑦 are unknown and 𝑣𝐵 = 40 m∕s. Since 𝑣⃗𝑃 and 𝑣⃗𝐵 are constant, applying constant acceleration equations component by component, Eq. (2) becomes [ ( ) ( )] [ ( ) ( )] 𝑟⃗𝑃 (𝑡) = 𝑟𝑃 𝑥 𝑡𝑓 + 𝑣𝑃 𝑥 𝑡 − 𝑡𝑓 𝚤̂ + 𝑟𝑃 𝑦 𝑡𝑓 + 𝑣𝑃 𝑦 𝑡 − 𝑡𝑓 𝚥̂, (3) [ ( ) ( )] (4) 𝑟⃗𝐵 (𝑡) = 𝑟𝐵𝑦 𝑡𝑓 + 𝑣𝐵 𝑡 − 𝑡𝑓 𝚥̂, where 𝑟⃗𝑃 (𝑡𝑓 ) = 𝑟𝑃 𝑥 (𝑡𝑓 ) 𝚤̂ + 𝑟𝑃 𝑦 (𝑡𝑓 ) 𝚥̂ and 𝑟⃗𝐵 (𝑡𝑓 ) = 𝑟𝐵𝑦 (𝑡𝑓 ) 𝚥̂ are the positions of 𝑃 and 𝐵 at the time of firing, respectively. Since 𝑃 is fired from 𝐴, for 𝑡 = 𝑡𝑓 we must have 𝑟⃗𝑃 (𝑡𝑓 ) = 𝑟⃗𝐴 (𝑡𝑓 ). In addition, at 𝑡 = 𝑡𝑓 , 𝐴 and 𝐵 are 4 s away from 𝐶, and so we have (see Fig. 3) ( ) ( ) 𝑟⃗𝑃 𝑡𝑓 = 𝑟⃗𝐴 𝑡𝑓 = −𝑑 sin 𝛽 𝚤̂ − 𝑑 cos 𝛽 𝚥̂ with 𝑑 = 𝑣𝐴 (4 s) = 72.00 m, (5) and 𝑟⃗𝐵 (𝑡𝑓 ) = −𝓁 𝚥̂ with

𝓁 = 𝑣𝐵 (4 s) = 160.0 m.

(6)

Since 𝑣⃗𝑃 is constant, its value is determined by the conditions at 𝐴 (i.e., at 𝑡 = 𝑡𝑓 ). Referring to Fig. 4, relative kinematics says that at the time of firing we have 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 ∕𝐴 = 𝑣𝐴 (sin 𝛽 𝚤̂ + cos 𝛽 𝚥̂) + 𝑣𝑃 ∕𝐴 (cos 𝜃 𝚤̂ − sin 𝜃 𝚥̂), = (𝑣𝐴 sin 𝛽 + 𝑣𝑃 ∕𝐴 cos 𝜃) 𝚤̂ + (𝑣𝐴 cos 𝛽 − 𝑣𝑃 ∕𝐴 sin 𝜃) 𝚥̂, Using Eqs. (5)–(7), we can then rewrite Eqs. (3) and (4) as [ ( )] 𝑟⃗𝑃 (𝑡) = −𝑑 sin 𝛽 + (𝑣𝐴 sin 𝛽 + 𝑣𝑃 ∕𝐴 cos 𝜃) 𝑡 − 𝑡𝑓 𝚤̂ )] [ ( + −𝑑 cos 𝛽 + (𝑣𝐴 cos 𝛽 − 𝑣𝑃 ∕𝐴 sin 𝜃) 𝑡 − 𝑡𝑓 𝚥̂,

(7)

(8)

ISTUDY

Section 12.6

745

Relative Motion Analysis and Differentiation of Geometrical Constraints

[ ( )] 𝑟⃗𝐵 (𝑡) = −𝓁 + 𝑣𝐵 𝑡 − 𝑡𝑓 𝚥̂.

𝑦

(9)

𝚥̂

Now we are ready to enforce the condition in Eq. (1), i.e., to set 𝑟⃗𝑃 (𝑡ℎ ) = 𝑟⃗𝐵 (𝑡ℎ ). Letting 𝑡 = 𝑡ℎ in Eqs. (8) and (9) and setting 𝑟⃗𝑃 (𝑡ℎ ) = 𝑟⃗𝐵 (𝑡ℎ ) component by component, we have ( ) −𝑑 sin 𝛽 + (𝑣𝐴 sin 𝛽 + 𝑣𝑃 ∕𝐴 cos 𝜃) 𝑡ℎ − 𝑡𝑓 = 0, (10) ( ) ( ) −𝑑 cos 𝛽 + (𝑣𝐴 cos 𝛽 − 𝑣𝑃 ∕𝐴 sin 𝜃) 𝑡ℎ − 𝑡𝑓 = −𝓁 + 𝑣𝐵 𝑡ℎ − 𝑡𝑓 . (11)

𝚤̂ 𝑑 𝑣𝐴 𝑢̂ 𝑣

Solving Eq. (10) for 𝑡ℎ − 𝑡𝑓 and substituting the result in Eq. (11), we get

𝜃 𝑢̂ 𝑃 ∕𝐴

𝑑 sin 𝛽 − 𝑑 cos 𝛽 + (𝑣𝐴 cos 𝛽 − 𝑣𝑃 ∕𝐴 sin 𝜃) 𝑣𝐴 sin 𝛽 + 𝑣𝑃 ∕𝐴 cos 𝜃 = −𝓁 + 𝑣𝐵

𝑥

𝐶 𝛽

𝓁

𝐴

𝑣𝑃 ∕𝐴

𝑣𝐵

𝐴(𝑡𝑓 ) = 𝑃 (𝑡𝑓 )

𝑑 sin 𝛽 . 𝑣𝐴 sin 𝛽 + 𝑣𝑃 ∕𝐴 cos 𝜃

Multiplying through by the denominator of the two fractions and simplifying give −𝑑𝑣𝑃 ∕𝐴 (cos 𝜃 cos 𝛽 + sin 𝜃 sin 𝛽) = −𝓁(𝑣𝐴 sin 𝛽 + 𝑣𝑃 ∕𝐴 cos 𝜃) + 𝑣𝐵 𝑑 sin 𝛽.

𝐵(𝑡𝑓 )

(12)

(13)

Figure 4 Velocity vectors of points 𝐴 and 𝐵, as well as relative velocity vector of 𝑃 with respect to 𝐴 at the time of firing. The figure is not drawn to scale.

Now, recall that we want to solve for 𝜃, and so isolating sin 𝜃 and cos 𝜃 on one side of the equation, we obtain ) sin 𝛽 ( 𝑣𝐵 𝑑 − 𝑣𝐴 𝓁 . 𝑣𝑃 ∕𝐴

𝑦

(14)

𝚥̂

Now recall that, at 𝑡 = 𝑡𝑓 , 𝐴 and 𝐵 are 4 s away from colliding at 𝐶, and therefore, we must have 4 s = 𝑑∕𝑣𝐴 = 𝓁∕𝑣𝐵

⇒

𝑣𝐴 𝓁 = 𝑣𝐵 𝑑

⇒

𝑣𝐵 𝑑 − 𝑣𝐴 𝓁 = 0.

𝚤̂ 𝑑

(15)

𝛽

Therefore, Eq. (14) becomes

𝐴(𝑡𝑓 )

which can be solved for 𝜃 to obtain ( 𝜃 = tan−1

𝑑 sin 𝛽

(16)

𝓁 − 𝑑 cos 𝛽 𝑑 sin 𝛽

) = 64.40◦ .

(17)

𝜃

𝑢̂ 𝑃 ∕𝐴 𝓁 𝑟⃗𝐵∕𝐴 (𝑡𝑓 )

𝓁 − 𝑑 cos 𝛽

(𝓁 − 𝑑 cos 𝛽) cos 𝜃 − 𝑑 sin 𝛽 sin 𝜃 = 0,

Discussion & Verification

The argument of the inverse tangent in Eq. (17) is nondimensional, as it should be. Observe that our result is independent of 𝑣𝑃 ∕𝐴 . This result is unusual in moving target problems, and in general, it suggests that an observer on 𝐵 sees 𝑃 travel along a straight line that does not change its orientation, almost as if 𝐵 were stationary and 𝑃 traveled along the line connecting 𝐴 and 𝐵 at the time of firing. Although unusual, in this particular problem this result is correct, but only because 𝐴 and 𝐵 are on a collision course while traveling at constant velocity. This means that 𝐴 and 𝐵 see each other moving directly toward each other along a straight line whose orientation relative to 𝐴 or 𝐵 remains fixed. In turn, this means that when the hero aims the gun at 𝐵, the hero must be aiming directly at 𝐵. This can be seen by referring to Fig. 5, which also allows us to obtain the result found in Eq. (17) by using simple trigonometry.

𝑥

𝐶 𝑑 cos 𝛽

(𝓁 − 𝑑 cos 𝛽) cos 𝜃 − 𝑑 sin 𝛽 sin 𝜃 =

tan 𝜃 =

𝓁 − 𝑑 cos 𝛽 𝑑 sin 𝛽

𝐵(𝑡𝑓 ) Figure 5 Position of 𝐵 relative to 𝐴 at the time of firing. This figure is drawn to scale.

746

Chapter 12

Particle Kinematics

E X A M P L E 12.23 𝐴

𝐺

Pulley System Analysis Figure 1 shows a pulley system for which we want to determine how the velocity and acceleration of block 𝑃 are related to the velocity and acceleration of block 𝑄 when they are connected by the pulleys and inextensible cords shown. Find the equations relating 𝑣𝑃 to 𝑣𝑄 and 𝑎𝑃 to 𝑎𝑄 .

𝑀 𝐽

𝑁 𝐾 𝑦𝑄

SOLUTION Road Map

𝑦𝑃

𝐷 𝐼

𝐻

𝐿 𝑄 𝐵

𝐸

𝐶

The key to the analysis of any pulley system is the notion of cord (or cable or rope) length and its first and second time derivatives. We note that in Fig. 1, there are a total of four cords, but two of the cords, 𝐸𝐹 and 𝑀𝑁, simply keep block 𝑃 attached to pulley 𝐸 and pulley 𝑁 attached to the fixed ceiling, respectively. Kinematically, we are saying that 𝑦𝐹 ∕𝐸 = constant

⇒ ⇒

𝐹

𝑦𝑁∕𝑀 = constant

Figure 1 A pulley system for which we want to relate the motion of block 𝑃 to that of block 𝑄.

ISTUDY

Concept Alert Does the length of the cord matter? Again, it is not the length of the cord that is important; only its rate of change of length matters.

⇒

𝑣𝑃 = 𝑣𝐹 = 𝑣𝐸

⇒ 𝑣𝑁∕𝑀 = 0 ⇒

𝑃

𝑣𝐹 ∕𝐸 = 0

𝑎𝐹 ∕𝐸 = 0

⇒

𝑎𝑃 = 𝑎𝐹 = 𝑎𝐸 ,

⇒ 𝑎𝑁∕𝑀 = 0

𝑣𝑀 = 𝑣𝑁 = 0

⇒

𝑎𝑀 = 𝑎𝑁 = 0.

(1) (2) (3) (4)

Recognizing this, we now only need to enforce the constraints expressing the fact that the lengths of the two cords 𝐴𝐵𝐶𝐷 and 𝐺𝐻𝐼𝐽 𝐾𝐿 are constant. Let the length of cord 𝐴𝐵𝐶𝐷 be 𝐿1 and the length of cord 𝐺𝐻𝐼𝐽 𝐾𝐿 be 𝐿2 . Notice that in Fig. 1, the cords have constant length since we are told they are inextensible. Let’s express these lengths in terms of the quantities shown in Fig. 1. The lengths 𝐿1 and 𝐿2 can be written as

Computation

̃ + 𝐶𝐷, 𝐿1 = 𝐴𝐵 + 𝐵𝐶

(5)

̃ + 𝐼𝐽 + 𝐽 ̃ 𝐿2 = 𝐺𝐻 + 𝐻𝐼 𝐾 + 𝐾𝐿,

(6)

̃ 𝐻𝐼, ̃ and 𝐽 ̃ where 𝐵𝐶, 𝐾 represent the lengths of the curved segments of cord that wrap around pulleys 𝐸, 𝐷, and 𝑁, respectively, and the letters with overbars represent the lengths of the corresponding straight line segments. Observe that the lengths of each of the cord segments either are constant or can be written in terms of the coordinates describing the positions of the blocks 𝑃 and 𝑄. Hence, we can write 𝐴𝐵 = 𝑦𝑃 − 𝐸𝐹 ,

𝐶𝐷 = 𝑦𝑃 − 𝐸𝐹 − 𝐺𝐻,

(7)

𝐼𝐽 = 𝐺𝐻 − 𝑀𝑁,

𝐾𝐿 = 𝑦𝑄 − 𝑀𝑁,

(8)

and then Eqs. (5) and (6) become, respectively, ̃ 𝐿1 = 2𝑦𝑃 − 𝐺𝐻 − 2𝐸𝐹 + 𝐵𝐶, ̃+𝐽 ̃ 𝐿2 = 𝑦𝑄 + 2𝐺𝐻 + 𝐻𝐼 𝐾 − 2𝑀𝑁.

(9) (10)

If we now differentiate Eqs. (9) and (10) with respect to time and recognize that • 𝑑(𝐸𝐹 )∕𝑑𝑡 = 𝑑𝑦𝐹 ∕𝐸 ∕𝑑𝑡 = 𝑣𝐹 ∕𝐸 = 0 due to Eq. (1), • 𝑑(𝑀𝑁)∕𝑑𝑡 = 𝑑𝑦𝑁∕𝑀 ∕𝑑𝑡 = 𝑣𝑁∕𝑀 = 0 due to Eq. (3), ̃ ̃ ̃ • 𝑑(𝐵𝐶)∕𝑑𝑡, 𝑑(𝐻𝐼)∕𝑑𝑡, and 𝑑(𝐽 𝐾)∕𝑑𝑡 are all zero because the quantity of cord wrapped around any of the pulleys is always the same,

ISTUDY

Section 12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

then we obtain 𝑑(𝐺𝐻) ̇ = 2𝑦̇ 𝑃 − 𝐺𝐻, 𝐿̇ 1 = 2𝑦̇ 𝑃 − 𝑑𝑡 𝑑(𝐺𝐻) ̇ 𝐿̇ 2 = 𝑦̇ 𝑄 + 2 = 𝑦̇ 𝑄 + 2𝐺𝐻. 𝑑𝑡

(11) (12)

We now make the following observations: 1. Since we have stated that the cords are inextensible, their lengths must be constant, and therefore, the time derivative of their lengths must be zero, i.e., 𝐿̇ 1 = 𝐿̇ 2 = 0. ̇ ̇ 2. We still have 𝐺𝐻 in Eqs. (11) and (12). However, 𝐺𝐻 can be eliminated from these two equations, leaving a single equation relating 𝑦̇ 𝑃 and 𝑦̇ 𝑄 . Therefore, adding 2 times Eq. (11) to Eq. (12), we obtain 4𝑦̇ 𝑃 + 𝑦̇ 𝑄 = 0 or

4𝑣𝑃 + 𝑣𝑄 = 0,

(13)

which can then be differentiated to obtain the acceleration relationship as 4𝑎𝑃 + 𝑎𝑄 = 0. Discussion & Verification

(14)

Equations (13) and (14) are dimensionally homogeneous, as they should be. In addition, notice that Eq. (13) says that if 𝑃 is moving at 4 m∕s down, then 𝑄 must be moving at 16 m∕s up. This is so because we have defined both 𝑦𝑃 and 𝑦𝑄 to be positive downward and 𝑣𝑄 = −4𝑣𝑃 . This result is consistent with the geometry of the pulley system in Fig. 1.

747

748

Chapter 12

Particle Kinematics

E X A M P L E 12.24

Constrained Motion of a Ladder A worker on a maintenance crew has decided that it would be easier to let his truck “do the lifting” than to do it himself. With this in mind, he decides to prop his ladder up against the building and stand on the ladder’s end at 𝐴 while his buddy starts the truck from rest and backs it up with constant acceleration 𝑎𝑇 in the direction shown in Fig. 1. Given a sufficiently strong ladder and that there are no windows in the way, this will cause the worker at 𝐴 to rise up the wall of the building. Observing that end 𝐵 of the ladder is a distance ℎ off the ground, the length of the ladder is 𝐿, and 𝜃 starts at 𝜃0 , determine the velocity and acceleration of the worker at 𝐴 as a function of the angle 𝜃 and, as a part of the solution, determine how 𝜃̇ and 𝜃̈ relate to 𝑎𝑇 .

𝑦

𝐴 𝜃

𝑣𝑇 𝑎𝑇

𝐿

𝐵

𝑂

ℎ 𝑥

Figure 1 The quantities 𝑣𝑇 and 𝑎𝑇 denote the components of the velocity and acceleration, respectively, of the truck in the direction of the truck’s motion, which is to the left.

ISTUDY

SOLUTION Road Map

We can do the required kinematic analysis of this system by noticing the following geometric constraints: 1. The length of the ladder is a constant 𝐿. 2. End 𝐴 of the ladder is constrained to move only in the vertical direction by the wall. 3. End 𝐵 of the ladder is constrained to move only in the horizontal direction by the motion of the truck. All of these can be incorporated into one relationship by noting that point 𝐴, point 𝐵, and the wall of the building will always form a right triangle so that (𝑦𝐴 − ℎ)2 + 𝑥2𝐵 = 𝐿2 .

(1)

Equation (1) is the key to our analysis. Since Eq. (1) is true for any value of 𝑦𝐴 and 𝑥𝐵 , we can differentiate it with respect to time to obtain

Computation

(𝑦𝐴 − ℎ)𝑦̇ 𝐴 + 𝑥𝐵 𝑥̇ 𝐵 = 0,

(2)

where we have used the fact that the length of the ladder is constant so that its time derivative is zero. Rearranging Eq. (2), we obtain ) ( 𝑥𝐵 𝑥̇ . (3) 𝑦̇ 𝐴 = − 𝑦𝐴 − ℎ 𝐵 This does not give us the velocity of 𝐴 in the form we seek since we can still write 𝑥𝐵 and 𝑦𝐴 in terms of the angle 𝜃. Noting that

Helpful Information Signs of 𝒙̇ 𝑩 and 𝒚̇ 𝑨 . We must have the minus sign in the relation 𝑥̇ 𝐵 = −𝑣𝑇 since we have defined positive 𝑥𝐵 to be to the right and since we have designated positive 𝑣𝑇 to be to the left. In addition, since positive 𝑦 is upward and 0◦ < 𝜃 < 90◦ , Eq. (3) tells us that when the truck is backing up, i.e., when 𝑣𝑇 > 0, then the worker at 𝐴 must be moving upward.

𝑥𝐵 = 𝐿 sin 𝜃,

𝑦𝐴 − ℎ = 𝐿 cos 𝜃,

(4)

and 𝑥̇ 𝐵 = −𝑣𝑇 , we see that Eq. (3) becomes 𝑦̇ 𝐴 = 𝑣𝐴 = 𝑣𝑇 tan 𝜃.

(5)

We need to take this a step further since we know only 𝑎𝑇 , the constant acceleration of the truck, and not 𝑣𝑇 . Let’s see how. We begin by applying the chain rule to 𝑣𝑇 to get 𝑎𝑇 , which gives 𝑎𝑇 =

𝑑𝑣𝑇 𝑑𝑡

=

𝑑𝑣𝑇 𝑑𝜃 𝑑𝑣𝑇 = 𝜃̇ , 𝑑𝜃 𝑑𝑡 𝑑𝜃

(6)

in which 𝜃̇ can be found by taking the time derivative of 𝑥𝐵 in Eqs. (4), that is, 𝑥̇ 𝐵 = 𝐿𝜃̇ cos 𝜃

⇒

𝜃̇ =

𝑥̇ 𝐵 𝐿 cos 𝜃

=

−𝑣𝑇 𝐿 cos 𝜃

.

(7)

ISTUDY

Section 12.6

749

Relative Motion Analysis and Differentiation of Geometrical Constraints

Substituting 𝜃̇ from Eq. (7) into Eq. (6), we obtain 𝑎𝑇 =

−𝑣𝑇 𝑑𝑣𝑇 𝐿 cos 𝜃 𝑑𝜃

⇒

Helpful Information

𝑎𝑇 𝐿 cos 𝜃 𝑑𝜃 = −𝑣𝑇 𝑑𝑣𝑇 .

(8)

Now, since 𝑎𝑇 is constant, we can integrate this as follows: 𝑎𝑇 𝐿

∫𝜃

𝜃

cos 𝜃 𝑑𝜃 = −

0

∫0

𝑣𝑇

𝑣𝑇 𝑑𝑣𝑇 ,

𝑎𝑇 𝐿(sin 𝜃 − sin 𝜃0 ) = − 21 𝑣2𝑇 ,

(9) (10)

which gives the following for 𝑣𝑇 as a function of 𝜃 𝑣2𝑇

= 2𝑎𝑇 𝐿(sin 𝜃0 − sin 𝜃).

(11)

𝜽̇ and 𝜽̈ as functions of 𝒂𝑻 . We were also asked to determine how 𝜃̇ and 𝜃̈ relate to 𝑎𝑇 . To do so, notice that substituting Eq. (11) into Eq. (7) gives us 𝜃̇ as a function of 𝜃 in the following form: √ − 2𝑎𝑇 𝐿(sin 𝜃0 − sin 𝜃) . 𝜃̇ = 𝐿 cos 𝜃 ̈ let’s differentiate Eq. (7) with reTo get 𝜃, spect to time to obtain 𝜃̈ =

Finally, substituting Eq. (11) into Eq. (5), we obtain the final result for 𝑣𝐴 √ 𝑣𝐴 =

2𝑎𝑇 𝐿(sin 𝜃0 − sin 𝜃) tan 𝜃.

(12)

To obtain the acceleration of the worker at 𝐴, we could differentiate Eq. (2), (3), or (5), but we will work with Eq. (5) since it is the simplest of the three. Doing so, we obtain 𝑣̇ 𝐴 = 𝑎𝐴 = 𝑣̇ 𝑇 tan 𝜃 + 𝑣𝑇 𝜃̇ sec2 𝜃 = 𝑎 tan 𝜃 + 𝑣 𝜃̇ sec2 𝜃, 𝑇

𝜃̈ = −

𝐿 cos 𝜃

= 𝑎𝑇 tan 𝜃 −

𝐿

2𝑎 sin 𝜃 𝑎𝑇 sec 𝜃 + 𝑇 3 (sin 𝜃0 − sin 𝜃). 𝐿 𝐿 cos 𝜃

(14) 4

sec3 𝜃.

(15)

Substituting Eq. (11) into Eq. (15), we obtain [ ] 𝑎𝐴 = 𝑎𝑇 tan 𝜃 − 2 sec3 𝜃(sin 𝜃0 − sin 𝜃) .

3

𝑣𝐴 (f t∕s)

𝑎𝐴 = 𝑎𝑇 tan 𝜃 −

,

(13)

𝑇

𝑣2𝑇

𝐿2 cos2 𝜃

which, upon substituting in Eq. (11), becomes

so that, after we substitute in 𝜃̇ from Eq. (7), Eq. (14) becomes 𝑣2𝑇 sec2 𝜃

−𝑎𝑇 𝐿 cos 𝜃 − 𝑣𝑇 𝐿𝜃̇ sin 𝜃

2 1

(16)

0

Discussion & Verification

0◦

20◦

40◦ 𝜃

60◦

80◦

Figure 2 Plot of 𝑣𝐴 as given by Eq. (12) for 𝐿 = 28 ft, 𝑎𝑇 = 1 ft/s2 , and 𝜃0 = 80◦ .

4 𝑎𝐴 (f t∕s2 )

Our result was obtained in analytical form, and it is dimensionally correct. To facilitate the discussion of our result, plots of the functions in Eqs. (12) and (16) can be found in Figs. 2 and 3, respectively, for specified values of the parameters. Keep in mind that both Figs. 2 and 3 should be read from right to left since 𝜃 decreases as the truck backs up. It should be observed that the velocity of the person at 𝐴 goes to zero as the ladder reaches the vertical position. This is to be expected, since the ladder has a finite length and as the ladder becomes more and more vertical, the amount of vertical displacement possible decreases. It is interesting that the acceleration of the person at 𝐴 is quite large when the truck starts moving, i.e., when 𝜃 = 𝜃0 = 80◦ . We should also mention that Eq. (12) can also be found by using constant acceleration equations from Section 12.2.

2 0

−2

0◦

20◦

40◦ 𝜃

60◦

80◦

Figure 3 Plot of 𝑎𝐴 as given by Eq. (16) for 𝑎𝑇 = 1 ft/s2 and 𝜃0 = 80◦ .

750

Chapter 12

Particle Kinematics

Problems Problem 12.200 Reference frame 𝐴 is translating relative to reference frame 𝐵. Both frames track the motion of a particle 𝐶. If at one instant the velocity of particle 𝐶 is the same in the two frames, what can you infer about the motion of frames 𝐴 and 𝐵 at that instant?

Problem 12.201 Reference frame 𝐴 is translating relative to reference frame 𝐵 with velocity 𝑣⃗𝐴∕𝐵 and acceleration 𝑎⃗𝐴∕𝐵 . A particle 𝐶 appears to be stationary relative to frame 𝐴. What can you say about the velocity and acceleration of particle 𝐶 relative to frame 𝐵?

Problem 12.202 Reference frame 𝐴 is translating relative to reference frame 𝐵 with constant velocity 𝑣⃗𝐴∕𝐵 . A particle 𝐶 appears to be in uniform rectilinear motion relative to frame 𝐴. What can you say about the motion of particle 𝐶 relative to frame 𝐵?

Problem 12.203 A skier is going down a slope with moguls. Let the skis be short enough for us to assume that the skier’s feet are tracking the moguls’ profile. Then, if the skier is skilled enough to maintain her hips on a straight-line trajectory and vertically aligned over her feet, determine the velocity and acceleration of her hips relative to her feet when her speed is equal to 15 km∕h. For the profile of the moguls, use the formula 𝑦(𝑥) = ℎ𝐼 − 0.15𝑥 + 0.125 sin(𝜋𝑥∕2) m, where ℎ𝐼 is the elevation at which the skier starts the descent. 𝑦 𝑣⃗𝐵

8.53◦

𝐵 𝑦(𝑥) 𝑥 Figure P12.203

𝑣𝐵

𝑣𝐴 𝐵 𝐴

𝑢̂ 𝐵∕𝐴

𝐴 𝑣⃗𝐴

Figure P12.204

Problem 12.204 Two particles 𝐴 and 𝐵 are moving in a plane with arbitrary velocity vectors 𝑣⃗𝐴 and 𝑣⃗𝐵 , respectively. Letting the rate of separation (ROS) be defined as the component of the relative velocity vector along the line connecting particles 𝐴 and 𝐵, determine a general expression for ROS. Express your result in terms of 𝑟⃗𝐵∕𝐴 = 𝑟⃗𝐵 − 𝑟⃗𝐴 , where 𝑟⃗𝐴 and 𝑟⃗𝐵 are the position vectors of 𝐴 and 𝐵, respectively, relative to some chosen fixed point in the plane of motion.

𝜃

Problem 12.205

Figure P12.205

ISTUDY

Airplanes 𝐴 and 𝐵 are flying along straight lines at the same altitude and with speeds 𝑣𝐴 = 660 km∕h and 𝑣𝐵 = 550 km∕h, respectively. Determine the speed of 𝐴 as perceived by 𝐵 if 𝜃 = 50◦ .

ISTUDY

Section 12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

751

65

Problem 12.206 Three vehicles 𝐴, 𝐵, and 𝐶 are in the positions shown and are moving with the indicated directions. We define the rate of separation (ROS) of two particles 𝑃1 and 𝑃2 as the component of the relative velocity of, say, 𝑃2 with respect to 𝑃1 in the direction of the relative position vector of 𝑃2 with respect to 𝑃1 , which is along the line that connects the two particles. At the given instant, determine the rates of separation ROS𝐴𝐵 and ROS𝐶𝐵 , that is, the rate of separation between 𝐴 and 𝐵 and between 𝐶 and 𝐵. Let 𝑣𝐴 = 60 mph, 𝑣𝐵 = 55 mph, and 𝑣𝐶 = 35 mph. Furthermore, treat the vehicles as particles and use the dimensions shown in the figure.

𝑣𝐶

𝐴

54◦

29 𝑣𝐴

𝐶

𝑣𝐵 65

Problem 12.207 𝐵

Car 𝐴 is moving at a constant speed 𝑣𝐴 = 75 km∕h, while car 𝐶 is moving at a constant speed 𝑣𝐶 = 42 km∕h on a circular exit ramp with radius 𝜌 = 80 m. Determine the velocity and acceleration of 𝐶 relative to 𝐴.

42

Problems 12.208 and 12.209 During practice, a player 𝑃 punts a ball 𝐵 with a speed 𝑣0 = 25 f t∕s, at an angle 𝜃 = 60◦ , and at a height ℎ from the ground. Then the player sprints along a straight line and catches the ball at the same height from the ground at which the ball was initially kicked. The length 𝑑 denotes the horizontal distance between the player’s position at the start of the sprint and the ball’s position when the ball leaves the player’s foot. Also, let Δ𝑡 denote the time interval between the instant at which the ball leaves the player’s foot and the instant at which the player starts sprinting.

Figure P12.206 and P12.207

𝑣0 𝑑

𝜃

𝑃 𝐵 ℎ

Figure P12.208 and P12.209

Assume that 𝑑 = 0 and Δ𝑡 = 0, and determine the average speed of the player so that he catches the ball.

Problem 12.208

Problem 12.209 Assume that 𝑑 = 3 f t and Δ𝑡 = 0.2 s, and determine the average speed of the player so that he catches the ball.

𝐴

Problem 12.210 A remote controlled boat, capable of a maximum speed of 10 f t∕s in still water, is made to cross a stream with a width 𝑤 = 35 f t that is flowing with a speed 𝑣𝑊 = 7 f t∕s. If the boat starts from point 𝑂 and keeps its orientation parallel to the cross-stream direction, find the location of point 𝐴 at which the boat reaches the other bank while moving at its maximum speed. Furthermore, determine how much time the crossing requires.

𝑣𝑊 𝑂 Figure P12.210

𝑤

752

Chapter 12

Particle Kinematics

Problem 12.211

𝐴

𝑣𝑊

𝑤

𝑂 Figure P12.211

A remote controlled boat, capable of a maximum speed of 10 f t∕s in still water, is made to cross a stream of width 𝑤 = 35 f t that is flowing with a speed 𝑣𝑊 = 7 f t∕s. The boat is placed in the water at 𝑂, and it is intended to arrive at 𝐴 by using a homing device that makes the boat always point toward 𝐴. Determine the time the boat takes to get to 𝐴 and the path it follows. Also, consider a case in which the maximum speed of the boat is equal to the speed of the current. In such a case, does the boat ever make it to point 𝐴? Hint: To solve the problem, write 𝑣⃗𝐵∕𝑊 = 𝑣𝐵∕𝑊 𝑢̂ 𝐴∕𝐵 , where the unit vector 𝑢̂ 𝐴∕𝐵 always points from the boat to point 𝐴 and is therefore, a function of time.

Problem 12.212 An airplane flying horizontally with a speed 𝑣𝑝 = 110 km∕h relative to the water drops a crate onto a carrier when vertically over the back end of the ship, which is traveling at a speed 𝑣𝑠 = 26 km∕h relative to the water. If the airplane drops the crate from a height ℎ = 20 m, at what distance from the back of the ship will the crate first land on the deck of the ship?

𝑣𝑝 ℎ 𝑣𝑠

Figure P12.212 and P12.213

Problem 12.213

𝛽

3

An airplane flying horizontally with a speed 𝑣𝑝 relative to the water drops a crate onto a carrier when vertically over the back end of the ship, which is traveling at a speed 𝑣𝑠 = 32 mph relative to the water. The length of the carrier’s deck is 𝓁 = 1000 f t, and the drop height is ℎ = 50 f t. Determine the maximum value of 𝑣𝑝 so that the crate will first impact within the rear half of the deck.

3

Problem 12.214

𝑣0

An airplane is initially flying north with a speed 𝑣0 = 430 mph relative to the ground, while the wind has a constant speed 𝑣𝑊 = 12 mph, forming an angle 𝜃 = 23◦ with the north-south direction. The airplane performs a course change of 𝛽 = 75◦ eastward while maintaining a constant reading of the airspeed indicator. Letting 𝑣⃗𝑃 ∕𝐴 be the velocity of the airplane relative to the air and assuming that the airspeed indicator measures the magnitude of the component of 𝑣⃗𝑃 ∕𝐴 in the direction of motion of the airplane, determine the speed of the airplane relative to the ground after the course correction.

𝑣𝑊

3

3

𝜃

Figure P12.214

Problem 12.215 𝐴 𝑣⃗𝐴

𝐵 𝜃

𝚥̂ 𝚤̂ Figure P12.215 and P12.216

ISTUDY

𝑣⃗𝐵

At the instant shown, block 𝐵 is sliding over the ground with a velocity 𝑣⃗𝐵 while block 𝐴 is sliding over block 𝐵 and has an absolute velocity 𝑣⃗𝐴 = −(4 𝚤̂ + 4 𝚥̂) ft∕s. Determine 𝑣⃗𝐵 if 𝜃 = 30◦ .

Problem 12.216 At the instant shown, 𝑣⃗𝐵 = 5 𝚤̂ m∕s. If 𝜃 = 25◦ , determine the speed of 𝐴 relative to 𝐵 in order for 𝐴 to travel only in the vertical direction while sliding over 𝐵.

ISTUDY

Section 12.6

753

Relative Motion Analysis and Differentiation of Geometrical Constraints

Problem 12.217 An interesting application of the relative motion equations is the experimental determination of the speed at which rain falls. Say you perform an experiment in your car in which you park your car in the rain and measure the angle the falling rain makes on your side window. Let this angle be 𝜃rest = 20◦ . Next you drive forward at 25 mph and measure the new angle 𝜃motion = 70◦ that the rain makes with the vertical. Determine the speed of the falling rain. direction of motion 𝜃

Figure P12.217 𝐿

𝑎0

Problem 12.218 A woman is sliding down an incline with a constant acceleration of 𝑎0 = 2.3 m∕s2 relative to the incline. At the same time the incline is accelerating to the right at 1.2 m∕s2 relative to the ground. Letting 𝜃 = 34◦ and 𝐿 = 4 m and assuming that both the woman and the incline start from rest, determine the horizontal distance traveled by the woman with respect to the ground when she reaches the bottom of the slide.

𝑎𝑠

𝜃

Figure P12.218

Problem 12.219 The pendulum bob 𝐴 swings about 𝑂, which is a fixed point, while bob 𝐵 swings about 𝐴. Express the components of the acceleration of 𝐵 relative to the Cartesian component system shown with origin at the fixed point 𝑂 in terms of 𝐿1 , 𝐿2 , 𝜃, 𝜙, and the necessary time derivatives of 𝜙 and 𝜃. 𝑦 𝚥̂ 𝚤̂

𝑥

𝑂 𝜃

𝐿1 𝐴 𝐿2 𝜙

𝐶 𝐵

Figure P12.219

Problem 12.220 Revisit Example 12.22 in which the movie’s hero is traveling on train car 𝐴 with constant speed 𝑣𝐴 = 18 m∕s while the mobile robot 𝐵 is moving at a constant speed 𝑣𝐵 = 40 m∕s (so that 𝑎𝐵 = 0). Recall that 4 s before an otherwise inevitable collision between 𝐴 and 𝐵, a projectile 𝑃 traveling at a speed of 300 m∕s relative to 𝐴 is shot toward 𝐵. Take advantage of the solution in Example 12.22 and determine the time it takes the projectile 𝑃 to reach 𝐵 and the projectile’s distance traveled.

48.2◦

𝑣𝐴 𝐴

𝜃 𝑣𝑃 ∕𝐴

𝐵 Figure P12.220

𝑣𝐵

754

Chapter 12

Particle Kinematics

Problem 12.221 Consider the following variation of the problem in Example 12.22 in which a movie hero needs to destroy a mobile robot 𝐵, except this time they are not going to collide at 𝐶. Assume that the hero is traveling on the train car 𝐴 with constant speed 𝑣𝐴 = 18 m∕s, while the robot 𝐵 travels at a constant speed 𝑣𝐵 = 50 m∕s. In addition, assume that at time 𝑡 = 0 s the train car 𝐴 and the robot 𝐵 are 72 and 160 m away from 𝐶, respectively. To prevent 𝐵 from reaching its intended target, at 𝑡 = 0 s the hero fires a projectile 𝑃 at 𝐵. If 𝑃 can travel at a constant speed of 300 m∕s relative to the gun, determine the orientation 𝜃 that must be given to the gun to hit 𝐵. Hint: An equation of the type sin 𝛽 ± 𝐴 cos 𝛽 = 𝐶 has the solution 𝛽 = ∓𝛾 + sin−1 (𝐶 cos 𝛾), if |𝐶 cos 𝛾| ≤ 1, where 𝛾 = tan−1 𝐴.

𝐶

48.2◦

𝑣𝐴 𝜃

𝐴

𝑣𝑃 ∕𝐴

Problem 12.222 𝐵 Figure P12.221 and P12.222

𝑣𝐵

Consider the following variation of the problem in Example 12.22 in which a movie hero needs to destroy a mobile robot 𝐵. As was done in that problem, assume that the movie hero is traveling on the train car 𝐴 with constant speed 𝑣𝐴 = 18 m∕s and that, 4 s before an otherwise inevitable collision at 𝐶, the hero fires a projectile 𝑃 traveling at 300 m∕s relative to 𝐴. Unlike Example 12.22, assume here that the robot 𝐵 travels with a constant acceleration 𝑎𝐵 = 10 m∕s2 and that 𝑣𝐵 (0) = 20 m∕s, where 𝑡 = 0 is the time of firing. Determine the orientation 𝜃 of the gun fired by the hero so that 𝐵 can be destroyed before the collision at 𝐶.

Problem 12.223 A park ranger 𝑅 is aiming a rifle armed with a tranquilizer dart at a bear (the figure is not to scale). The bear is moving in the direction shown at a constant speed 𝑣𝐵 = 25 mph. The ranger fires the rifle when the bear is at 𝐶 at a distance of 150 f t. Knowing that 𝛼 = 10◦ , 𝛽 = 108◦ , the dart travels with a constant speed of 425 f t∕s, and the dart and the bear are moving in a horizontal plane, determine the orientation 𝜃 of the rifle so that the ranger can hit the bear. Hint: An equation of the type sin 𝛽 ± 𝐴 cos 𝛽 = 𝐶 has the solution 𝛽 = ∓𝛾 + sin−1 (𝐶 cos 𝛾), if |𝐶 cos 𝛾| ≤ 1, where 𝛾 = tan−1 𝐴.

𝑣𝐵 𝐵

𝑅 𝛼

𝐶

𝜃

𝛽 Figure P12.223

Problem 12.224

𝐴 𝑣𝐴 𝐵 Figure P12.224

ISTUDY

The object in the figure is called a gun tackle, and it used to be very common on sailboats to help in the operation of front-loaded guns. If the end at 𝐴 is pulled down at a speed of 1.5 m∕s, determine the velocity of 𝐵. Neglect the fact that some portions of the rope are not vertically aligned.

Problem 12.225 The gun tackle shown is operated with the help of a horse. If the horse moves to the right at a constant speed of 7 f t∕s, determine the velocity and acceleration of 𝐵 when the horizontal distance from 𝐵 to 𝐴 is 15 f t. Except for the part of the rope between 𝐶 and 𝐴, neglect the fact that some portions are not vertically aligned. Also neglect the change

ISTUDY

Section 12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

in the amount of rope wrapped around pulley 𝐶 as the horse moves to the right.

𝐶

𝐴

14 f t 𝐵 6 ft

Figure P12.225

Problem 12.226 The figure shows an inverted gun tackle with snatch block, which used to be common on sailboats. If the end at 𝐴 is pulled at a speed of 1.5 m∕s, determine the velocity of 𝐵. Neglect the fact that some portions of the rope are not vertically aligned. 𝐵

𝐶 𝐴 𝐴

𝑣𝐴

𝑣𝐴

𝐷 𝐵 Figure P12.226

Figure P12.227

Problem 12.227 In maritime speak, the system in the figure is often called a whip-upon-whip purchase and is used for controlling certain types of sails on small cutters (by attaching point 𝐵 to the sail to be unfurled). If the end of the rope at 𝐴 is pulled with a speed of 4 m∕s, determine the velocity of 𝐵. Neglect the fact that the segment of the rope between 𝐶 and 𝐷 is not vertically aligned, and assume that the slope of segment 𝐴𝐶 is constant.

Problem 12.228 The pulley system shown is used to store a bicycle in a garage. If the bicycle is hoisted by a winch that winds the rope at a rate 𝑣0 = 5 in.∕s, determine the vertical speed of the bicycle.

𝑣0

Problem 12.229 Letting 𝜃 = 50◦ , determine the vertical component of the velocity of 𝐴 if 𝐵 is moving downward with a speed 𝑣𝐵 = 3 f t∕s.

Figure P12.228

755

756

Chapter 12

Particle Kinematics

𝑣𝐴 𝑣0 𝐴 𝐴 𝐵

𝜃

𝑣𝐴 𝑣𝐵

𝜃

Figure P12.229

𝐵

Figure P12.230

Problem 12.230 Determine the speed of block 𝐵 if block 𝐴 is sliding down the incline with a speed 𝑣𝐴 = 1.5 m∕s while the cord is retracted by a winch at a constant rate 𝑣0 = 2.5 m∕s.

Problem 12.231 Block 𝐴 is released from rest and starts sliding down the incline with an acceleration 𝑎0 = 3.7 m∕s2 . Determine the acceleration of block 𝐵 relative to the incline. Also, determine the time needed for 𝐵 to move a distance 𝑑 = 0.2 m relative to 𝐴.

𝐵 𝐴 𝑎0 𝜃 𝐸

𝐹

Figure P12.231 𝐵

Problems 12.232 and 12.233 𝐶

In the pulley system shown, the segment 𝐴𝐷 and the motion of 𝐴 are not impeded by the load 𝐺. Assume all ropes are vertically aligned.

𝐷 𝐺

𝑣0

𝐴 𝑎0

Figure P12.232 and P12.233

ISTUDY

Problem 12.232

and 𝑎0 = 1 f t∕s2 .

Determine the velocity and acceleration of the load 𝐺 if 𝑣0 = 3 f t∕s

The load 𝐺 is initially at rest when the end 𝐴 of the rope is pulled with the constant acceleration 𝑎0 . Determine 𝑎0 so that 𝐺 is lifted 2 f t in 4.3 s.

Problem 12.233

Problem 12.234 A crate 𝐴 is being pulled up an inclined ramp by a winch retracting the cord at a constant rate 𝑣0 = 2 f t∕s. Letting ℎ = 1.5 f t, determine the speed of the crate when 𝑑 = 4 f t.

ISTUDY

Section 12.6

Relative Motion Analysis and Differentiation of Geometrical Constraints

757

𝑑 ℎ

𝐴

Figure P12.234 and P12.235

Problem 12.235 A crate 𝐴 is being pulled up an inclined ramp by a winch. The rate of winding of the cord is controlled so as to hoist the crate up the incline with a constant speed 𝑣0 . Letting 𝓁̇ denote the length of cord retracted by the winch per unit time, determine an expression for 𝓁̇ in terms of 𝑣0 , ℎ, and 𝑑.

Problem 12.236 The piston head at 𝐶 is constrained to move along the 𝑦 axis. Let the crank 𝐴𝐵 be rotating counterclockwise at a constant angular speed 𝜃̇ = 2000 rpm, 𝑅 = 3.5 in., and 𝐿 = 5.3 in. Determine the velocity of 𝐶 when 𝜃 = 35◦ .

𝑦 𝐶

Problem 12.237

𝐿 𝑦𝐶

Let 𝜔 ⃗ 𝐵𝐶 denote the angular velocity of the relative position vector 𝑟⃗𝐶∕𝐵 . As such, 𝜔 ⃗ 𝐵𝐶 is also the angular velocity of the connecting rod 𝐵𝐶. Using the concept of time derivative of a vector given in Section 12.5 on p. 717, determine the component of the relative velocity of 𝐶 with respect to 𝐵 along the direction of the connecting rod 𝐵𝐶.

𝐵 𝜃

𝑥

𝐴 𝑅

Problem 12.238

Figure P12.236–P12.238

The piston head at 𝐶 is constrained to move along the 𝑦 axis. Let the crank 𝐴𝐵 be rotating counterclockwise at a constant angular speed 𝜃̇ = 2000 rpm, 𝑅 = 3.5 in., and 𝐿 = 5.3 in. Determine expressions for the velocity and acceleration of 𝐶 as a function of 𝜃 and the given parameters. 𝐶

Problems 12.239 and 12.240 𝐴

In the cutting of sheet metal, the robotic arm 𝑂𝐴 needs to move the cutting tool at 𝐶 counterclockwise at a constant speed 𝑣0 along a circular path of radius 𝜌. The center of the circle is located in the position shown relative to the base of the robotic arm at 𝑂.

𝜙

𝜌

For all positions along the circular cut (i.e., for any value of 𝜙), determine 𝑟, 𝑟,̇ and 𝜃̇ as functions of the given quantities (i.e., 𝑑, ℎ, 𝜌, 𝑣0 ). Use one or more geometric constraints and their derivatives to do this. These quantities can be found “by hand,” but it is tedious, so you might consider using symbolic algebra software, such as Mathematica or Maple. Problem 12.239

ℎ 𝜃 𝑂

For all positions along the circular cut (i.e., for any value of 𝜙), determine 𝑟̈ and 𝜃̈ as functions of the given quantities (i.e., 𝑑, ℎ, 𝜌, 𝑣0 ). These quantities can be found by hand, but it is very tedious, so you might consider using symbolic algebra software, such as Mathematica or Maple.

𝐷

𝑟

Problem 12.240

𝑑 Figure P12.239 and P12.240

cutting path

758

Chapter 12

Particle Kinematics

Problem 12.241 In the pulley system shown, the winch at 𝐴 is retracting rope with constant speed 𝑣𝑟 . Determine the velocity of the payload at 𝐷 as a function of 𝑣𝑟 . Also determine the velocity of 𝐷 relative to 𝐶.

Figure P12.241

Figure P12.242

Problem 12.242 If 𝐵 is moving upward with speed 𝑣0 , determine the speed of 𝐴. In addition, determine the speed of 𝐵 relative to 𝐶. 𝑤

Problem 12.243 At the instant shown, block 𝐴 is moving at a constant speed 𝑣0 = 3 m∕s to the left and 𝑤 = 2.3 m. Using ℎ = 2.7 m, determine how much time is needed to lower 𝐵 0.75 m from this position.

ℎ 𝐵 𝐴

Problem 12.244 At the instant shown, ℎ = 10 f t, 𝑤 = 8 f t, and block 𝐵 is moving with a speed 𝑣0 = 5 f t∕s and an acceleration 𝑎0 = 1 f t∕s2 , both downward. Determine the velocity and acceleration of block 𝐴.

Figure P12.243 and P12.244

Problem 12.245 All dimensions are in meters. 𝐴 𝐶

𝑟

𝜃 𝑂

0.2

0.5 Figure P12.245 and P12.246

ISTUDY

0.6

0.8

As a part of a robotics competition, a robotic arm with a rigid open hand at 𝐶 is to be designed so that the hand catches an egg without breaking it. The egg is released from rest at 𝑡 = 0 from point 𝐴. The arm, initially at rest in the position shown, starts moving when the egg is released. The hand must catch the egg without any impact with the egg. This can be done by specifying that the hand and the egg must be at the same position at the same time with identical velocities. A student proposes to do this using a constant value of 𝜃̈ for which (after a fair bit of work) it is found that the arm catches the egg at 𝑡 = 0.4391 s ̈ determine the acceleration of both for 𝜃̈ = −13.27 rad∕s2 . Using these values of 𝑡 and 𝜃, the hand and the egg at the time of catch. Then, explain whether or not using a constant ̈ as has been proposed, is an acceptable strategy. value of 𝜃,

Problem 12.246 Referring to the problem of a robot arm catching an egg (Prob. 12.245), the strategy is that the arm and the egg must have the same velocity and the same position at the same time for the arm to gently catch the egg. In addition, what should be true about the accelerations of the arm and the egg for the catch to be successful after they rendezvous with the same velocity at the same position and time? Describe what happens if the accelerations of the arm and egg do not match.

ISTUDY

Section 12.7

12.7

759

Motion in Three Dimensions

Motion in Three Dimensions

So far in Chapter 12, we have focused on planar motion, which, while occurring in a three-dimensional environment, is such that we can describe the position of moving points using only two coordinates. In this section, we consider the motion of points that are not constrained to move in a plane and therefore, require that we use three coordinates when describing their position.

Cartesian coordinates The Cartesian coordinate system and its unit vectors in three dimensions are defined as shown in Fig. 12.37. Because the directions of its base vectors do not change, we obtain the following relations for position, velocity, and acceleration, respectively, in Cartesian coordinates: ̂ 𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂ + 𝑧(𝑡) 𝑘, ̂ 𝑣(𝑡) ⃗ = 𝑥̇ 𝚤̂ + 𝑦̇ 𝚥̂ + 𝑧̇ 𝑘̂ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ + 𝑣𝑧 𝑘,

(12.97)

̂ 𝑎(𝑡) ⃗ = 𝑥̈ 𝚤̂ + 𝑦̈ 𝚥̂ + 𝑧̈ 𝑘̂ = 𝑎𝑥 𝚤̂ + 𝑎𝑦 𝚥̂ + 𝑎𝑧 𝑘.

(12.99)

𝑎𝑦 = 0

𝑎𝑧 = −𝑔.

(12.100)

Tangent-normal-binormal components Imagine a system of components describing the space curve of a modern rollercoaster that includes 3D loops, twists and turns. Let the space curve be described by a single parameter, 𝜉 ̂ 𝑟⃗(𝜉) = 𝑥(𝜉) 𝚤̂ + 𝑦(𝜉) 𝚥̂ + 𝑧(𝜉) 𝑘. (12.101) The velocity is still (12.102)

and the acceleration is still confined to the osculating plane, as was shown originally in Fig. 12.19 on p. 704. We still have an acceleration described by Eq. (12.49) 𝑎⃗ = 𝑣̇ 𝑢̂ 𝑡 +

𝑣2 𝑢̂ . 𝜌 𝑛

(12.103)

There is no acceleration component in the binormal direction, 𝑎𝑏 = 0, where we recall from Eq. (12.50) that 𝑢̂ 𝑏 = 𝑢̂ 𝑡 × 𝑢̂ 𝑛 . However, when the motion is no longer planar, the osculating plane is no longer stationary. Let’s investigate how the osculating plane is moving. We already know how the unit tangent vector changes with respect to the arc length variable 𝑠 from Eq. (12.51), which is 𝑑 𝑢̂ 𝑡 𝑑𝑠

=

1 𝑢̂ = 𝜅 𝑢̂ 𝑛 , 𝜌 𝑛

𝑃 𝑟⃗𝑃

This is a straightforward extension of the 2D case, and we explore an example of this in Example 12.26.

𝑣⃗ = 𝑣 𝑢̂ 𝑡 ,

𝑧

(12.98)

Cartesian coordinates are most relevant in three dimensions when the acceleration is fixed in one direction, as was the case of projectile motion in Section 12.3. If the 𝑧-direction is aligned with gravity, for instance, Eq. (12.39) becomes 𝑎𝑥 = 0

StockShot/Alamy Stock Photo

Figure 12.36 A skysurfer inverted and spinning. The motion of every point on the surfer and his board is threedimensional.

(12.104)

𝑧𝑃 𝑥

𝑦𝑃

𝑘̂

𝚤̂

𝚥̂ 𝑥𝑃

𝑦

Figure 12.37 The Cartesian coordinate system and its unit vectors in three dimensions.

760

ISTUDY

Chapter 12

Particle Kinematics

where 𝜅 was defined in Eq. (12.44) to be the curvature of the path. Now considering 𝑢̂ 𝑏 , we note that 𝑢̂ 𝑏 ⋅ 𝑢̂ 𝑏 = 1

⇒

𝑑 𝑢̂ 𝑏 𝑑𝑠

⋅ 𝑢̂ 𝑏 + 𝑢̂ 𝑏 ⋅

𝑑 𝑢̂ 𝑏 𝑑𝑠

=0

⇒

𝑑 𝑢̂ 𝑏 𝑑𝑠

⋅ 𝑢̂ 𝑏 = 0.

(12.105)

This tells us that 𝑑 𝑢̂ 𝑏 ∕𝑑𝑠 only has components in the 𝑢̂ 𝑡 and 𝑢̂ 𝑛 directions. Let’s first look at the component of 𝑑 𝑢̂ 𝑏 ∕𝑑𝑠 in the 𝑢̂ 𝑡 direction 𝑢̂ 𝑡 ⋅ 𝑢̂ 𝑏 = 0

⇒

𝑑 𝑢̂ 𝑡

⇒

𝜅 𝑢̂ 𝑛 ⋅ 𝑢̂ 𝑏 + 𝑢̂ 𝑡 ⋅

𝑑𝑠

⋅ 𝑢̂ 𝑏 + 𝑢̂ 𝑡 ⋅

𝑑 𝑢̂ 𝑏 𝑑𝑠 𝑑 𝑢̂ 𝑏 𝑑𝑠

=0 =0

⇒

𝑢̂ 𝑡 ⋅

𝑑 𝑢̂ 𝑏 𝑑𝑠

= 0,

where we have used Eq. (12.104) and the fact that 𝑢̂ 𝑛 ⋅ 𝑢̂ 𝑏 = 0. This means that 𝑑 𝑢̂ 𝑏 ∕𝑑𝑠 does not have any component in the 𝑢̂ 𝑡 direction. This result, taken with Eq. (12.105) means that 𝑑 𝑢̂ 𝑏 ∕𝑑𝑠 must be parallel to 𝑢̂ 𝑛 . Now define 𝑑 𝑢̂ 𝑏 𝑑𝑠

1 = −𝜏 𝑢̂ 𝑛 = − 𝑢̂ 𝑛 , 𝜎

(12.106)

where 𝜏(𝑠) is the torsion of the path at 𝑠 and 𝜎 is the radius of torsion. Just as the curvature 𝜅 measures how “bendy” a path is, the torsion measures how “twisty” it is. As with curvature 𝜅, the torsion has dimensions of inverse length, and, similarly to radius of curvature, the radius of torsion 𝜎 = 1∕𝜏 has length as its dimension. We now have expressions for 𝑑 𝑢̂ 𝑡 ∕𝑑𝑠 and 𝑑 𝑢̂ 𝑏 ∕𝑑𝑠. To obtain an expression for 𝑑 𝑢̂ 𝑛 ∕𝑑𝑠, we simply use the definition 𝑢̂ 𝑛 = 𝑢̂ 𝑏 × 𝑢̂ 𝑡 and then take the derivative with respect to 𝑠 to obtain 𝑑 𝑢̂ 𝑛 𝑑𝑠

=

𝑑 𝑢̂ ) 𝑑 𝑢̂ 𝑑 ( 𝑢̂ × 𝑢̂ 𝑡 = 𝑏 × 𝑢̂ 𝑡 + 𝑢̂ 𝑏 × 𝑡 = −𝜏 𝑢̂ 𝑛 × 𝑢̂ 𝑡 + 𝑢̂ 𝑏 × 𝜅 𝑢̂ 𝑛 𝑑𝑠 𝑏 𝑑𝑠 𝑑𝑠 = 𝜏 𝑢̂ 𝑏 − 𝜅 𝑢̂ 𝑡 , (12.107)

where we have used Eqs. (12.104) and (12.106), as well as Eq. (12.50) to obtain the final result. Taken together Eqs. (12.104), (12.106), and (12.107) are known as the SerretFrenet formulas, which we summarize here as 𝑑 𝑢̂ 𝑡 𝑑𝑠 𝑑 𝑢̂ 𝑛 𝑑𝑠 𝑑 𝑢̂ 𝑏 𝑑𝑠

= 𝜅 𝑢̂ 𝑛 ,

(12.108)

= 𝜏 𝑢̂ 𝑏 − 𝜅 𝑢̂ 𝑡 ,

(12.109)

= −𝜏 𝑢̂ 𝑛 .

(12.110)

They can be more compactly written as 𝑑 𝑢̂ 𝑖 𝑑𝑠

=𝜔 ⃗ 𝐷 × 𝑢̂ 𝑖 ,

𝑖 = 𝑡, 𝑛, or 𝑏,

(12.111)

where 𝜔 ⃗ 𝐷 is the Darboux vector and is defined as 𝜔 ⃗ 𝐷 = 𝜏 𝑢̂ 𝑡 + 𝜅 𝑢̂ 𝑏 .

(12.112)

We now know how the intrinsic triad (𝑢̂ 𝑡 , 𝑢̂ 𝑡 , 𝑢̂ 𝑏 ) changes with 𝑠, but how does it change with time? To determine this, consider an arbitrary unit vector 𝑢̂ that we have

ISTUDY

Section 12.7

761

Motion in Three Dimensions

written in terms of the intrinsic triad as 𝑢̂ = 𝑙 𝑢̂ 𝑡 + 𝑚 𝑢̂ 𝑛 + 𝑛 𝑢̂ 𝑏 , where 𝑙, 𝑚, and 𝑛 are constants since 𝑢̂ is fixed in the intrinsic triad. Now consider the derivative of 𝑢̂ with respect to time ) 𝑑 𝑢̂ 𝑑𝑠 𝑑 𝑢̂ 𝑑 ( 𝑑 𝑢̂ = =𝑣 =𝑣 𝑙 𝑢̂ 𝑡 + 𝑚 𝑢̂ 𝑛 + 𝑛 𝑢̂ 𝑏 . 𝑢̂̇ = 𝑑𝑡 𝑑𝑠 𝑑𝑡 𝑑𝑠 𝑑𝑠

(12.113)

Distributing the derivative in the parentheses and using Eq. (12.112), this becomes ( ) 𝑑 𝑢̂ 𝑑 𝑢̂ 𝑑 𝑢̂ 𝑢̂̇ = 𝑣 𝑙 𝑡 + 𝑚 𝑛 + 𝑛 𝑏 𝑑𝑠 𝑑𝑠 𝑑𝑠 ( ) = 𝑣 𝑙𝜔 ⃗ 𝐷 × 𝑢̂ 𝑡 + 𝑚𝜔 ⃗ 𝐷 × 𝑢̂ 𝑛 + 𝑛𝜔 ⃗ 𝐷 × 𝑢̂ 𝑏 ( ) = 𝑣𝜔 ⃗ 𝐷 × 𝑙 𝑢̂ 𝑡 + 𝑚 𝑢̂ 𝑚 + 𝑛 𝑢̂ 𝑏 = 𝑣𝜔 ⃗ 𝐷 × 𝑢. ̂ (12.114) Now, recall from Eq. (12.58) that 𝑢̂̇ = 𝜔 ⃗ 𝑢 × 𝑢, ̂ where 𝜔 ⃗ 𝑢 is the angular velocity of ⃗ 𝐷 ! Since 𝑢̂ is fixed in the 𝑢. ̂ Comparing this with Eq. (12.114), we see that 𝜔 ⃗ 𝑢 = 𝑣𝜔 intrinsic triad, it must have the same angular velocity as the intrinsic triad, and so we have the result ( ) ( ) 1 1 ⃗ 𝐷 = 𝑣 𝜏 𝑢̂ 𝑡 + 𝜅 𝑢̂ 𝑏 = 𝑣 𝜔 ⃗ 𝐼𝑇 = 𝑣𝜔 𝑢̂ + 𝑢̂ , 𝜎 𝑡 𝜌 𝑏 where 𝜔 ⃗ 𝐼𝑇 is the angular velocity of the intrinsic triad. So, we see that the intrinsic triad rotates about the binormal direction at the rate 𝑣∕𝜌 and about the tangent direction at the rate 𝑣∕𝜎! A considerable amount of algebraic manipulation can be employed to obtain a more general relationship for 𝜌 than that of Eq. (12.51) as well as an expression for 𝜎. The results are (𝑠′ )3 , 𝜌= √ ( ) ( )2 𝑟⃗′′ ⋅ 𝑟⃗′′ (𝑠′ )2 − 𝑟⃗′ ⋅ 𝑟⃗′′ 𝜎=

(𝑠′ )6 [ )] . ( 𝜌2 𝑟⃗′′′ ⋅ 𝑟⃗′ × 𝑟⃗′′

(12.115)

(12.116)

In these expressions, (𝑠′ )2 = 𝑟⃗′ ⋅ 𝑟⃗′ and primes denote differentiation with respect to parameter 𝜉. Note in the limit that we return to a planar problem, 𝑟⃗′ , 𝑟⃗′′ and 𝑟⃗′′′ are all coplanar, so the mixed triple product in the denominator of 𝜎 goes to zero. In this case, 𝜎 becomes infinite, there is no change in the direction of the binormal along the space curve, and the osculating plane experiences no twist. It is left as an exercise to show that when we return to two dimensions, 𝜉 → 𝑥, 𝑟⃗ = 𝑥 𝚤̂ + 𝑦(𝑥) 𝚥̂, and the expression for 𝜌 in Eq. (12.115) reverts to Eq. (12.51).

𝑧

𝑢̂ 𝑧 𝑢̂ 𝜃

𝑃 𝑟⃗

𝑢̂ 𝑅

𝑂

Cylindrical coordinates

𝑅

The three-dimensional extension of the polar coordinate system is called the cylindrical coordinate system. Figure 12.38 shows the coordinate directions for a cylindrical coordinate system with origin at 𝑂 and with the coordinate 𝜃 defined to be positive counterclockwise as viewed down the positive 𝑧 axis. If 𝜃 is defined to be positive in the clockwise direction, the results that follow will change. Referring to Fig. 12.38, the cylindrical coordinates of a point 𝑃 are 𝑅, 𝜃, and 𝑧,

(12.117)

𝑦

𝑢̂ 𝑧 𝑥

𝜃 𝑄

𝑢̂ 𝜃 𝑢̂ 𝑅

Figure 12.38 Coordinate directions in the cylindrical coordinate system. As with polar coordinates, defined in Section 12.5, the 𝜃 coordinate is implicitly contained in 𝑢̂ 𝑅 .

762

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Particle Kinematics

where 𝑅 and 𝜃 are the polar coordinates of the projection of point 𝑃 onto the 𝑅𝜃 plane, namely, point 𝑄, and the coordinate 𝑧 is the distance between 𝑃 and 𝑄 taken positive or negative, depending on whether or not 𝑃 is reached from 𝑄 by moving in the positive 𝑧 direction. To describe vectors, we introduce the following set of three mutually orthogonal base vectors (12.118) 𝑢̂ 𝑅 , 𝑢̂ 𝜃 , and 𝑢̂ 𝑧 with 𝑢̂ 𝑧 = 𝑢̂ 𝑅 × 𝑢̂ 𝜃 . Here, 𝑢̂ 𝑅 and 𝑢̂ 𝜃 are determined as in the case of the polar coordinate system and are therefore parallel to the 𝑅𝜃 plane, whereas 𝑢̂ 𝑧 is a unit vector parallel to the 𝑧 axis and pointing in the positive 𝑧 direction. It follows that the position vector of 𝑃 relative to the origin 𝑂 is given by 𝑟⃗ = 𝑅 𝑢̂ 𝑅 + 𝑧 𝑢̂ 𝑧 .

(12.119)

Equation (12.119) expresses the fact that we can locate a point by tracking the point’s projection on the 𝑅𝜃 plane, given by the term 𝑅 𝑢̂ 𝑅 , and then adding to this the point’s elevation 𝑧 with respect to the 𝑅𝜃 plane. Differentiating Eq. (12.119) with respect to time, we obtain the following result for the velocity in cylindrical coordinates 𝑣⃗ = 𝑅̇ 𝑢̂ 𝑅 + 𝑅𝜃̇ 𝑢̂ 𝜃 + 𝑧̇ 𝑢̂ 𝑧 = 𝑣𝑅 𝑢̂ 𝑅 + 𝑣𝜃 𝑢̂ 𝜃 + 𝑣𝑧 𝑢̂ 𝑧 ,

(12.120)

where ̇ 𝑣𝑅 = 𝑅,

̇ 𝑣𝜃 = 𝑅𝜃,

and

𝑣𝑧 = 𝑧̇

(12.121)

are the cylindrical components of the velocity vector. In writing Eq. (12.120), we have used Eq. (12.68) on p. 720 and the fact that 𝑢̂ 𝑧 is constant because the 𝑧 axis is fixed. We obtain the acceleration in cylindrical coordinates by differentiating Eq. (12.120) with respect to time, i.e., ) ( ( ) 𝑎⃗ = 𝑅̈ − 𝑅𝜃̇ 2 𝑢̂ 𝑅 + 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ 𝑢̂ 𝜃 + 𝑧̈ 𝑢̂ 𝑧 = 𝑎𝑅 𝑢̂ 𝑅 + 𝑎𝜃 𝑢̂ 𝜃 + 𝑎𝑧 𝑢̂ 𝑧 ,

𝑧

(12.122)

where 𝜙

𝑎𝑅 = 𝑅̈ − 𝑅𝜃̇ 2 ,

𝑟 = 𝑟⃗

𝑟⃗

𝑢̂ 𝜃

𝑦 𝜃

Figure 12.39 Definition of the spherical coordinates and corresponding orthogonal base vectors whose length is one, which are often called orthonormal base vectors.

ISTUDY

𝑎𝑧 = 𝑧̈

(12.123)

Spherical coordinates

𝑢̂ 𝜙

𝑂 𝑥

and

are the cylindrical components of the acceleration vector. In writing Eq. (12.122), we have used Eq. (12.72) on p. 720 and the constancy of 𝑢̂ 𝑧 .

𝑢̂ 𝑟 𝑃

̇ 𝑎𝜃 = 𝑅𝜃̈ + 2𝑅̇ 𝜃,

Another coordinate system that is often used in the study of three-dimensional motion, such as in problems involving navigation and motion relative to the Earth, is the spherical coordinate system. The position of a point 𝑃 in a three-dimensional space can be described by 𝑟, 𝜙, and 𝜃, which are called the spherical coordinates of 𝑃 (Fig. 12.39). The angles 𝜙 and 𝜃 give orientation to the line going from the origin 𝑂 to 𝑃 , while 𝑟 is the distance between 𝑃 and 𝑂. In Fig. 12.39, we have also indicated three mutually orthogonal unit vectors 𝑢̂ 𝑟 , 𝑢̂ 𝜙 , and 𝑢̂ 𝜃 . These unit vectors are oriented such that 𝑢̂ 𝑟 points from 𝑂 to 𝑃 , while 𝑢̂ 𝜙 and 𝑢̂ 𝜃 point in the direction of increasing 𝜙 and 𝜃, respectively, with

ISTUDY

Section 12.7

Motion in Three Dimensions

763

𝑢̂ 𝜙 × 𝑢̂ 𝜃 = 𝑢̂ 𝑟 . If 𝜙 and 𝜃, as well as the unit vectors 𝑢̂ 𝑟 , 𝑢̂ 𝜙 , and 𝑢̂ 𝜃 , are not defined as given here, the results that follow may not be applicable. In spherical coordinates, the position vector of 𝑃 relative to the origin 𝑂 is 𝑟⃗ = 𝑟 𝑢̂ 𝑟 .

(12.124)

Differentiating 𝑟⃗ with respect to time, we obtain 𝑣⃗ = 𝑟⃗̇ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟 𝑢̂̇ 𝑟 ,

Helpful Information (12.125)

which implies that to describe the velocity vector we need to express 𝑢̂̇ 𝑟 in terms of ⃗ bv × 𝑢̂ 𝑟 , where 𝜔 ⃗ bv is the base vectors 𝑢̂ 𝑟 , 𝑢̂ 𝜃 , and 𝑢̂ 𝜙 . To find 𝑢̂̇ 𝑟 , we need to find 𝑢̂̇ 𝑟 = 𝜔 the angular velocity of the unit base vectors. To do this, we note that if the coordinates 𝜃 and 𝜙 undergo infinitesimal changes, the angular velocity corresponding to these infinitesimal changes is 𝜔 ⃗ bv = 𝜙̇ 𝑢̂ 𝜃 + 𝜃̇ 𝑢̂ 𝑧 = 𝜃̇ cos 𝜙 𝑢̂ 𝑟 − 𝜃̇ sin 𝜙 𝑢̂ 𝜙 + 𝜙̇ 𝑢̂ 𝜃 ,

(12.126)

where the last equation is obtained by observing that 𝑢̂ 𝑧 = cos 𝜙 𝑢̂ 𝑟 − sin 𝜙 𝑢̂ 𝜙 . Using the result in Eq. (12.126) along with Eq. (12.58) on p. 718, we can find the time derivative of each of the unit vectors 𝑢̂ 𝑟 , 𝑢̂ 𝜃 , and 𝑢̂ 𝜙 as ⃗ bv × 𝑢̂ 𝑟 = 𝜙̇ 𝑢̂ 𝜙 + 𝜃̇ sin 𝜙 𝑢̂ 𝜃 , 𝑢̂̇ 𝑟 = 𝜔

(12.127)

⃗ bv × 𝑢̂ 𝜙 = −𝜙̇ 𝑢̂ 𝑟 + 𝜃̇ cos 𝜙 𝑢̂ 𝜃 , 𝑢̂̇ 𝜙 = 𝜔

(12.128)

𝑢̂̇ 𝜃 = 𝜔 ⃗ bv × 𝑢̂ 𝜃 = −𝜃̇ sin 𝜙 𝑢̂ 𝑟 − 𝜃̇ cos 𝜙 𝑢̂ 𝜙 .

(12.129)

Substituting Eq. (12.127) into Eq. (12.125), we obtain the velocity in spherical coordinates as 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜙̇ 𝑢̂ 𝜙 + 𝑟𝜃̇ sin 𝜙 𝑢̂ 𝜃 = 𝑣𝑟 𝑢̂ 𝑟 + 𝑣𝜙 𝑢̂ 𝜙 + 𝑣𝜃 𝑢̂ 𝜃 ,

(12.130)

where 𝑣𝑟 = 𝑟,̇

̇ 𝑣𝜙 = 𝑟𝜙,

and

𝑣𝜃 = 𝑟𝜃̇ sin 𝜙

(12.131)

are the spherical components of the velocity vector. Differentiating Eq. (12.130) with respect to time and using Eqs. (12.127)–(12.129) give the acceleration as ( ) 𝑎⃗ = 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙 𝑢̂ 𝑟 ( ) + 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙 𝑢̂ 𝜙 ( ) + 𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙 𝑢̂ 𝜃

(12.132)

= 𝑎𝑟 𝑢̂ 𝑟 + 𝑎𝜙 𝑢̂ 𝜙 + 𝑎𝜃 𝑢̂ 𝜃 , where 𝑎𝑟 = 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙, 𝑎 = 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙, 𝜙

𝑎𝜃 = 𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙 are the spherical components of the acceleration vector.

(12.133)

Where are 𝝓 and 𝜽 in the position vector? We have seen this before, that is, the position vector seems not to depend on all of the “coordinates.” In this case, the position vector 𝑟⃗ = 𝑟 𝑢̂ 𝑟 seems not to be a function of either 𝜃 or 𝜙. However, the angles 𝜃 and 𝜙 are implicitly contained in the definition of the direction of 𝑢̂ 𝑟 .

764

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Particle Kinematics

End of Section Summary Cartesian coordinates. Referring to Fig. 12.40, in the three-dimensional Cartesian coordinate system, the position, velocity, and acceleration, respectively, are given in component form as

𝑧 𝑃 𝑟⃗𝑃 𝚤̂

𝑧𝑃 𝑥

𝑘̂

Eqs. (12.97)–(12.99), p. 759 𝚥̂

𝑦𝑃

𝑦

𝑥𝑃

Figure 12.40 The Cartesian coordinate system and its unit vectors in three dimensions.

̂ 𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂ + 𝑧(𝑡) 𝑘, ̂ 𝑣(𝑡) ⃗ = 𝑥̇ 𝚤̂ + 𝑦̇ 𝚥̂ + 𝑧̇ 𝑘̂ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ + 𝑣𝑧 𝑘, ̂ 𝑎(𝑡) ⃗ = 𝑥̈ 𝚤̂ + 𝑦̈ 𝚥̂ + 𝑧̈ 𝑘̂ = 𝑎𝑥 𝚤̂ + 𝑎𝑦 𝚥̂ + 𝑎𝑧 𝑘.

Tangent-normal-binormal components. acceleration are still given by, respectively,

As in Section 12.4, the velocity and

Eqs. (12.102) and (12.103), p. 759 𝑣(𝑡) ⃗ = 𝑣 𝑢̂ 𝑡 , 𝑎(𝑡) ⃗ = 𝑣̇ 𝑢̂ 𝑡 +

𝑣2 𝑢̂ . 𝜌 𝑛

For three-dimensional paths, the osculating plane is no longer stationary, and the intrinsic unit vectors 𝑢̂ 𝑡 , 𝑢̂ 𝑛 , and 𝑢̂ 𝑏 twist and turn as the point in question moves along the path. As a function of the path coordinate 𝑠, their change is described by the Serret-Frenet formulas Eqs. (12.108)–(12.110), p. 760 𝑑 𝑢̂ 𝑡 𝑑𝑠 𝑑 𝑢̂ 𝑛 𝑑𝑠 𝑑 𝑢̂ 𝑏 𝑑𝑠 𝑧

𝑢̂ 𝜃

𝑟⃗

𝑢̂ 𝑅

𝑂 𝑅

𝑦

𝑢̂ 𝑧 𝑥

𝜃 𝑄

𝑢̂ 𝜃 𝑢̂ 𝑅

Figure 12.41 Coordinate directions in the cylindrical coordinate system. As with polar coordinates defined in Section 12.5, the 𝜃 coordinate is implicitly contained in 𝑢̂ 𝑟 .

ISTUDY

= 𝜏 𝑢̂ 𝑏 − 𝜅 𝑢̂ 𝑡 , = −𝜏 𝑢̂ 𝑛 .

Cylindrical coordinates. Referring to Fig. 12.41, the position of a point 𝑃 in three dimensions can be described by the three quantities 𝑅, 𝜃, and 𝑧 that are the point’s cylindrical coordinates. In addition, we see that cylindrical coordinates involve the orthogonal triad of unit vectors 𝑢̂ 𝑅 , 𝑢̂ 𝜃 , and 𝑢̂ 𝑧 . Using this triad, the position in cylindrical coordinates is given by

𝑢̂ 𝑧 𝑃

= 𝜅 𝑢̂ 𝑛 ,

Eq. (12.119), p. 762 𝑟⃗ = 𝑅 𝑢̂ 𝑅 + 𝑧 𝑢̂ 𝑧 . The velocity vector in cylindrical coordinates is given by Eq. (12.120), p. 762 𝑣⃗ = 𝑅̇ 𝑢̂ 𝑅 + 𝑅𝜃̇ 𝑢̂ 𝜃 + 𝑧̇ 𝑢̂ 𝑧 = 𝑣𝑅 𝑢̂ 𝑅 + 𝑣𝜃 𝑢̂ 𝜃 + 𝑣𝑧 𝑢̂ 𝑧 .

ISTUDY

Section 12.7

Motion in Three Dimensions

765

Finally, the acceleration vector in cylindrical coordinates is given by 𝑧

Eq. (12.122), p. 762 ( ) ( ) 𝑎⃗ = 𝑅̈ − 𝑅𝜃̇ 2 𝑢̂ 𝑅 + 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ 𝑢̂ 𝜃 + 𝑧̈ 𝑢̂ 𝑧

𝜙

= 𝑎𝑅 𝑢̂ 𝑅 + 𝑎𝜃 𝑢̂ 𝜃 + 𝑎𝑧 𝑢̂ 𝑧 . Spherical coordinates. Referring to Fig. 12.42, the position of a point 𝑃 in three dimensions can be described by the three quantities 𝑟, 𝜃, and 𝜙 that are the point’s spherical coordinates. In addition, spherical coordinates involve the orthogonal triad of unit vectors 𝑢̂ 𝑟 , 𝑢̂ 𝜙 , and 𝑢̂ 𝜃 , with 𝑢̂ 𝜙 × 𝑢̂ 𝜃 = 𝑢̂ 𝑟 . Using this triad, the position vector in spherical coordinates is given by Eq. (12.124), p. 763

𝑢̂ 𝑟 𝑃 𝑟 = 𝑟⃗

𝑟⃗

𝑢̂ 𝜃 𝑢̂ 𝜙

𝑂 𝑥

𝑦 𝜃

𝑟⃗ = 𝑟 𝑢̂ 𝑟 . The velocity vector in spherical coordinates is given by Eq. (12.130), p. 763 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜙̇ 𝑢̂ 𝜙 + 𝑟𝜃̇ sin 𝜙 𝑢̂ 𝜃 = 𝑣𝑟 𝑢̂ 𝑟 + 𝑣𝜙 𝑢̂ 𝜙 + 𝑣𝜃 𝑢̂ 𝜃 . Finally, the acceleration vector in spherical coordinates is given by Eq. (12.132), p. 763 ( ) 𝑎⃗ = 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙 𝑢̂ 𝑟 ( ) + 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙 𝑢̂ 𝜙 ( ) + 𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙 𝑢̂ 𝜃 = 𝑎𝑟 𝑢̂ 𝑟 + 𝑎𝜙 𝑢̂ 𝜙 + 𝑎𝜃 𝑢̂ 𝜃 .

Figure 12.42 Definition of the unit vectors in spherical coordinates.

766

Chapter 12

Particle Kinematics

E X A M P L E 12.25

Application of Cylindrical Coordinates A top-slewing crane (also called a tower crane) is lifting an object 𝐶 at a constant rate of 1.5 m∕s while rotating at a constant rate of 0.15 rad∕s in the direction shown in Fig. 1. If the distance between the object and the axis of rotation of the crane’s boom is 45 m, find the velocity and acceleration of 𝐶, assuming that the swinging motion of 𝐶 can be neglected.

0.15 rad∕s

SOLUTION Road Map

𝐶

1.5 m∕s

This problem is readily solved using cylindrical coordinates. We are provided with quantities that are naturally defined in a cylindrical coordinate system with its origin placed at the intersection of the 𝑢̂ 𝑅 and 𝑢̂ 𝑧 vectors at the base of the crane (see Fig. 2). Computation

Figure 1 A top-slewing crane. This kind of crane is very common on big construction sites.

𝑣⃗𝐶 = 𝑅̇ 𝑢̂ 𝑅 + 𝑅𝜃̇ 𝑢̂ 𝜃 + 𝑧̇ 𝑢̂ 𝑧 , ( ( ) ) 𝑎⃗𝐶 = 𝑅̈ − 𝑅𝜃̇ 2 𝑢̂ 𝑅 + 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ 𝑢̂ 𝜃 + 𝑧̈ 𝑢̂ 𝑧 .

(1) (2)

Recalling that 𝑅 = 45 m, 𝑅̇ = 0, 𝑅̈ = 0, 𝑧̇ = 1.5 m∕s, 𝑧̈ = 0, 𝜃̇ = 0.15 rad∕s, and 𝜃̈ = 0, Eqs. (1) and (2) become

𝜃̇ 𝑧

𝑅 𝐴

𝐶 𝑢̂ 𝑧

𝑣⃗𝐶 = (6.750 𝑢̂ 𝜃 + 1.500 𝑢̂ 𝑧 ) m∕s,

(3)

𝑎⃗𝐶 = −1.0125 𝑢̂ 𝑅 m∕s2 .

(4)

The solution seems reasonable, since the motion of 𝐶 is the composition of a uniform circular motion with angular velocity equal to 0.15 rad∕s and a uniform rectilinear motion in the positive 𝑧 direction. We were expecting the velocity not to have a radial component, while the acceleration was expected to have only a radial component, which is what Eqs. (3) and (4) reflect.

Discussion & Verification

𝑧̇

𝑢̂ 𝑅

Figure 2 The top-slewing crane with a cylindrical coordinate system defined.

ISTUDY

According to our choice of cylindrical coordinates, the velocity and acceleration vectors of 𝐶 are given by Eqs. (12.120) and (12.122), respectively, which are

A Closer Look This example demonstrates how easy it can be to find velocities and accelerations when the appropriate component system is used. Even if:

• The trolley at 𝐴 were moving inward or outward with any known 𝑅(𝑡) (so that 𝑅, ̇ and 𝑅̈ were known). 𝑅, • The payload 𝐶 were moving up or down at any known 𝑧(𝑡) (so that 𝑧̇ and 𝑧̈ were known). • The tower crane were rotating with any known 𝜃(𝑡) (so that 𝜃̇ and 𝜃̈ were known), we could still easily determine the values of all the quantities in Eqs. (1) and (2) to find the velocity and acceleration of the payload at 𝐶.

ISTUDY

Section 12.7

Motion in Three Dimensions

E X A M P L E 12.26

767

Application of Cartesian Coordinates

Electrostatic precipitators are used to “scrub” or clean the emissions from coal-fired power plants (see Fig. 1). They work by sending the particulate-laden flue gas through a large structure that slows down the particles so that they can be efficiently collected. The gas enters at 50–60 f t∕s and slows down to 3–6 f t∕s as it expands. Inside the main structure are alternating rows of collection and discharge electrodes. Applying a high voltage (about 45,000–70,000 V for large precipitators) to the discharge electrodes located between the collection plates causes them to emit electrons into the gas, thus ionizing the immediate area. When the flue gas passes through this “corona,” the particles become negatively charged and are then attracted to the positively charged collecting plates from which they can be collected and removed. Using the geometry defined in Fig. 2, determine where the particle 𝑃 will land on the collecting plate given that 𝜃1 = 30◦ , 𝜃2 = 55◦ , 𝑑 = 1.5 f t, |𝑣⃗𝑃 | = 𝑣𝑃 = 5 f t∕s, and that gravity acts in the −𝑧 direction. Assume that the collecting plate imposes a constant 𝑦 component of acceleration on the particle of 𝑎ep = 250 f t∕s2 . 𝑧 𝑦 𝑣⃗𝑃

𝑃

𝜃2 𝜃1

charged collecting plate 𝑑

Gary L. Gray

𝑥 Figure 2. Schematic of a particle moving near a positively charged collecting plate in an electrostatic precipitator. Collection efficiencies of modern precipitators range from 99.5–99.9%.

SOLUTION

Gary L. Gray

Road Map

The key to the solution is to first determine the time it takes the particle to get to the plate. We can find that time since we know the acceleration in the 𝑦 direction, and we can find the initial velocity in the 𝑦 direction. Once that time is known, we can use constant acceleration kinematics to find the position at which the particle hits the plate. We begin by finding the components of the initial velocity of 𝑃 in all three Cartesian coordinate directions. Referring to Fig. 3, we can see that the initial component of 𝑣⃗𝑃 in the 𝑥𝑦 plane is 𝑣𝑃 cos 𝜃2 and that the initial 𝑧 component of 𝑣⃗𝑃 is given by

Computation

𝑣𝑃 𝑧 = 𝑣𝑃 sin 𝜃2 .

(1)

Now that we have the component of 𝑣𝑃 in the 𝑥𝑦 plane, it is easier to see that the initial 𝑥 and 𝑦 components are given by 𝑣𝑃 𝑥 = (𝑣𝑃 cos 𝜃2 ) cos 𝜃1 ,

(2)

𝑣𝑃 𝑦 = (𝑣𝑃 cos 𝜃2 ) sin 𝜃1 .

(3)

Next, we are told that the acceleration of the particle is ( ) 𝑎⃗𝑃 = 𝑎ep 𝚥̂ − 𝑔 𝑘̂ = 250 𝚥̂ − 32.2 𝑘̂ ft∕s2 , which is constant in all component directions.

(4)

Figure 1 Photos of an electrostatic precipitator from a manufacturing plant. The top photo shows the outside of a precipitator with the air inlet in the middle. The bottom photo shows the collecting plates on the inside of the precipitator.

768

Chapter 12

Particle Kinematics 𝑧

𝑣𝑃

Now that we have all accelerations and the initial velocities, we can use the 𝑦 component to determine the time it takes for the particle to hit the plate. Using Eq. (12.33), we have (5) 𝑑 = 𝑣𝑃𝑦 𝑡𝑐 + 21 𝑎ep 𝑡2𝑐 ,

(𝑣𝑃 cos 𝜃2 ) sin 𝜃1 𝑦

𝜃2 𝑃

𝑣𝑃 cos 𝜃2

𝜃1 (𝑣𝑃 cos 𝜃2 ) cos 𝜃1

𝑥

Figure 3 Graphical depiction of the initial velocity of the particle in the precipitator.

ISTUDY

where 𝑡𝑐 is the time for 𝑃 to reach the collector plate. Equation (5) can be solved for 𝑡𝑐 to obtain √ 𝑣 1 (6) 2𝑑𝑎ep + 𝑣2𝑃 cos2 𝜃2 sin2 𝜃1 , 𝑡𝑐 = − 𝑃 cos 𝜃2 sin 𝜃1 ± 𝑎ep 𝑎ep where Eq. (3) has been used. Substituting in the given parameters, we find that 𝑡𝑐 equals either 0.1040 s or −0.1154 s, with the physically appropriate answer being the positive one 𝑡𝑐 = 0.1040 s. (7) Knowing the time to reach the collector, we can now determine the distance traveled in the 𝑥 and 𝑧 directions. Again applying Eq. (12.33), but now in the 𝑥 and 𝑧 directions, we obtain 𝑥𝑐 = 𝑣𝑃 𝑥 𝑡𝑐 = 0.2582 f t, (8) 𝑧𝑐 = 𝑣𝑃 𝑧 𝑡𝑐 − 12 𝑔𝑡2𝑐 = 0.2518 f t.

(9)

Figure 4 shows the trajectory of the particle that has been attracted to the collecting plate.

1.5 𝑦

1.0

0.5

Interesting Fact Is 250 𝗳 𝘁∕𝘀𝟮 a reasonable acceleration? We stated that the particle is accelerating at 250 𝖿 𝗍∕𝗌𝟤 toward the collector plate. This is reasonable when we consider that the particles collected on these plates are very small. For example, in some electrostatic precipitators, the particles range from smaller than 1 𝜇m to up to 3 𝜇m in diameter. A 1 𝜇m particle of sodium would have a mass of about 𝟣𝟢−𝟣𝟧 kg, and so to generate an acceleration of 250 𝖿 𝗍∕𝗌𝟤 , which is equal to 76.20 𝗆∕𝗌𝟤 , requires a force of only about 𝟪 × 𝟣𝟢−𝟣𝟦 N.

0.2 𝑧 0.1 0 𝑥 0.1

0.2

Figure 4. Trajectory of the particle 𝑃 . All dimensions are in feet.

Discussion & Verification

Figure 4 is very helpful in assessing whether or not our solution is reasonable. Given that the initial speed of the particle is 5 f t∕s and that it is accelerating toward the collecting plate at 250 f t∕s2 , it seems reasonable that it would not move very far down the precipitator (i.e., in the 𝑥 direction) before hitting the collecting plate.

ISTUDY

Section 12.7

E X A M P L E 12.27

Application of Spherical Coordinates

Revisit Example 12.16 and determine, using spherical coordinates, what relationships the radar readings obtained by the station at 𝐴 need to satisfy for you to conclude that the jet at 𝐵 shown in Fig. 1 is flying

𝑧

(b) in a straight line at constant speed 𝑣0 .

𝜙

ℎ𝐴

SOLUTION Road Map

As with Example 12.16, the idea is to write the imposed constraints in terms of the chosen coordinate system. For this system, that means that the altitude is constant and that the velocity vector is constant.

𝐵 𝑟

𝐴

(a) at a level altitude.

Computation

769

Motion in Three Dimensions

𝜃

Figure 1 Airliner flying straight and level at altitude ℎ.

Given the coordinate system indicated in the figure, the plane’s altitude

ℎ is given by ℎ = 𝑟 cos 𝜙 + ℎ𝐴 ,

(1)

where ℎ𝐴 is the elevation of the radar antenna above ground. Since ℎ𝐴 is constant, the plane can be said to be flying at a constant altitude as long as 𝑟(𝑡) cos 𝜙(𝑡) = constant.

(2)

However, contrary to what we saw in Example 12.16, this relationship is not sufficient to guarantee that the trajectory of the plane is a straight line since Eq. (2) is satisfied by any trajectory lying on a plane ℎ above the ground. If the plane being tracked is flying along a straight line and at a constant speed, then its velocity vector must be constant. While it is tempting to say that constant velocity means that 𝑟̇ = constant, 𝑟𝜙̇ = constant,

(3)

𝑟𝜃̇ sin 𝜙 = constant,

(5)

(4)

that is, that each component of the velocity must be constant, this is not the case since the base vectors of a spherical coordinate system change direction as the plane moves (see marginal note). Therefore, in order for 𝑣⃗ to be constant, the airplane’s acceleration must be equal to zero, i.e., 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙 = 0, 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙 = 0,

(6)

𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙 = 0. Finally, to measure the airplane’s speed 𝑣0 , we need to compute the magnitude of 𝑣, ⃗ which is given by the square root of the sum of the squares of the components of the velocity vector, that is, √ 𝑣0 =

)2 ( )2 ( 𝑟̇ 2 + 𝑟𝜙̇ + 𝑟𝜃̇ sin 𝜙 .

(7)

Discussion & Verification

Our results are correct since they were obtained in symbolic form by a direct application of Eqs. (12.133) and (12.131), respectively, i.e., without additional manipulations that could have introduced errors. A Closer Look Verifying that the speed is constant does not allow us to say that the plane’s trajectory is straight or level.

Helpful Information What does constant velocity mean? We know that mathematically, constant velocity ⃗ We know that this means that 𝑣⃗̇ = 0. means that the velocity vector doesn’t change magnitude or direction. In spherical coordinates, this relationship takes the form ) ( 𝑑 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜙̇ 𝑢̂ 𝜙 + 𝑟𝜃̇ sin 𝜙 𝑢̂ 𝜃 , 𝑣⃗̇ = 𝑑𝑡 When this derivative is expanded, we must differentiate the scalar coefficients of the unit vectors, as well as the unit vectors themselves. So, as we have already seen, constant scalar components of a vector do not imply that the vector is constant since the unit vectors can change direction. This is why Eqs. (3)–(5) are not sufficient to say that 𝑣⃗ is constant. In addition, this is why it is sufficient in Cartesian coordinates to say that the constancy of the scalar coefficients is sufficient to say that a vector is constant.

770

Chapter 12

Particle Kinematics

Problems Problem 12.247 Although point 𝑃 is moving on a sphere, its motion is being studied with the cylindrical coordinate system shown. Discuss in detail whether or not there are incorrect elements in the sketch of the cylindrical component system at 𝑃 . 𝑧

𝑧 𝑢̂ 𝑧

𝑢̂ 𝑅

𝑃

𝑢̂ 𝜃

𝑢̂ 𝑧

𝑢̂ 𝜃

𝑟⃗

𝑂 𝑥

𝑃

𝑂 𝑦

𝜃

𝑥

𝑣⃗

𝑢̂ 𝑅

𝑟⃗

𝑦 𝜃

𝑅

𝑅

Figure P12.247

Figure P12.248

Problem 12.248 Although point 𝑃 is moving on a sphere, its motion is being studied with the cylindrical coordinate system shown. Discuss in detail whether or not there are incorrect elements in the sketch of the cylindrical component system at 𝑃 .

Problem 12.249 Discuss in detail whether or not (a) there are incorrect elements in the sketch of the spherical component system at 𝑃 and (b) the formulas for the velocity and acceleration components derived in the section can be used with the coordinate system shown. 𝑧

𝑧

𝑃 𝑟⃗ 𝑂 𝑥

𝑧 𝑣0

𝜙

𝑟⃗

𝑢̂ 𝜙

𝑂 𝑦

𝜃

𝑢̂ 𝑟

𝑃 𝑢̂ 𝜃

𝑢̂ 𝜃

Figure P12.249

𝐴

𝑢̂ 𝜙

𝜙

𝑢̂ 𝑟

𝑥

𝑦 𝜃

Figure P12.250

Problem 12.250

𝑅

Discuss in detail whether or not (a) there are incorrect elements in the sketch of the spherical component system at 𝑃 and (b) the formulas for the velocity and acceleration components derived in the section can be used with the coordinate system shown.

Problem 12.251

Figure P12.251

ISTUDY

A glider is descending with a constant speed 𝑣0 = 30 m∕s and a constant descent rate of 1 m∕s along a helical path with a constant radius 𝑅 = 400 m. Determine the time the glider takes to complete a full 360◦ turn about the axis of the helix (the 𝑧 axis).

ISTUDY

Section 12.7

Motion in Three Dimensions

Problem 12.252

771

𝑣0

An airplane is flying horizontally at a constant speed 𝑣0 = 320 mph while its propellers rotate at a constant angular speed 𝜔 = 1500 rpm. If the propellers have a diameter 𝑑 = 14 f t, determine the magnitude of the acceleration of a point on the periphery of the propeller blades.

Figure P12.252

Problem 12.253 A top-slewing crane is lifting an object 𝐶 at a constant rate of 𝑧̇ = 5.3 f t∕s while rotating at a constant rate 𝜔 = 0.12 rad∕s about the vertical axis. If the distance between the object and the axis of rotation of the crane’s boom is 𝑟 = 46 f t and it is being reduced at a constant rate of 6.5 f t∕s, find the velocity and acceleration of 𝐶, assuming that the swinging motion of 𝐶 can be neglected. 𝑧

𝜔

𝐶 𝑟

Figure P12.253

𝜃

𝑦 𝑂 𝑢̂ 𝜙

Problem 12.254 The system depicted in the figure is called a spherical pendulum. The fixed end of the pendulum is at 𝑂. Point 𝑂 behaves as a spherical joint; i.e., the location of 𝑂 is fixed while the pendulum’s cord can swing in any direction in the three-dimensional space. Assume that the pendulum’s cord has a constant length 𝐿, and use the coordinate system depicted in the figure to derive the expression for the acceleration of the pendulum.

𝐿

𝑢̂ 𝜃 𝑢̂ 𝑟 𝑟 𝜙

Problem 12.255

𝑧

Revisit Example 12.27, and assuming that the plane is accelerating, determine the relation(s) that the radar readings obtained by the station at 𝐴 need to satisfy for you to conclude that the jet is flying along a straight line whether at constant altitude or not. 𝑧 𝑟

𝜙 𝐴 𝜃

Figure P12.255

Figure P12.254

𝑥

772

Chapter 12

Particle Kinematics

Problem 12.256 A golfer chips the ball on a flat, level part of a golf course as shown. Letting 𝛼 = 23◦ , 𝛽 = 41◦ , and the initial speed be 𝑣0 = 6 m∕s, determine the 𝑥 and 𝑦 coordinates of the place where the ball will land. 𝑧 𝑢̂ 𝑧

𝐶 𝑧

𝑢̂ 𝑅

𝑑

𝐴

𝑂

𝑅

𝛽 𝜔𝑠

𝑣0

𝐺

𝐿 2

𝑦

𝛽 𝛼

𝑥 Figure P12.256

𝐵

𝐿 2

𝐷

Figure P12.257

Problem 12.257 Relative to the cylindrical coordinate system shown, with origin at 𝑂, the radial and 𝑧 coordinates of point 𝐺 are 𝑅 = 𝑑 + (𝐿∕2) cos 𝛽 and 𝑧 = −(𝐿∕2) sin 𝛽, respectively, where 𝑑 = 0.5 m and 𝐿 = 0.6 m. The shaft 𝐶𝐷 rotates as shown with a constant angular velocity 𝜔𝑠 = 10 rad∕s, and the angle 𝛽 varies with time as follows: 𝛽 = 𝛽0 sin(2𝜔𝑡), where 𝛽0 = 0.3 rad, 𝜔 = 2 rad∕s, and 𝑡 is time in seconds. Determine the velocity and the acceleration of 𝐺 for 𝑡 = 3 s [express the result in the cylindrical component system (𝑢̂ 𝑅 , 𝑢̂ 𝜃 , 𝑢̂ 𝑧 ), with 𝑢̂ 𝜃 = 𝑢̂ 𝑧 × 𝑢̂ 𝑅 ].

𝑧 𝑟

𝜙

𝑦 𝜃 𝑥 𝑥 + 𝑦 = 10 mi

Figure P12.258

ISTUDY

Problem 12.258 An airplane is traveling at a constant altitude of 10,000 f t, with a constant speed of 450 mph, within the plane whose equation is given by 𝑥 + 𝑦 = 10 mi and in the direction of increaṡ 𝜙, ̇ 𝑟̈, 𝜃, ̈ and 𝜙̈ that would be measured when the ing 𝑥. Find the expressions for 𝑟,̇ 𝜃, airplane is closest to the radar station.

Problem 12.259 A carnival ride called the octopus consists of eight arms that rotate about the 𝑧 axis at the constant angular velocity 𝜃̇ = 6 rpm. The arms have a length 𝐿 = 22 f t and form an angle 𝜙 with the 𝑧 axis. Assuming that 𝜙 varies with time as 𝜙(𝑡) = 𝜙0 + 𝜙1 sin 𝜔𝑡 with 𝜙0 = 70.5◦ , 𝜙1 = 25.5◦ , and 𝜔 = 1 rad∕s, determine the magnitude of the acceleration of the outer end of an arm when 𝜙 achieves its maximum value.

𝑧 𝐿

Gary L. Gray

Figure P12.259

𝜙

𝜃̇

ISTUDY

Section 12.7

Motion in Three Dimensions

Problem 12.260

𝑧

𝛽

A particle is moving over the surface of a right cone with angle 𝛽 and under the constraint that 𝑅2 𝜃̇ = 𝐾, where 𝐾 is a constant. The equation describing the cone is 𝑅 = 𝑧 tan 𝛽. Determine the expressions for the velocity and the acceleration of the particle in terms of 𝐾, 𝛽, 𝑧, and the time derivatives of 𝑧.

𝑅

Problem 12.261 Solve Prob. 12.260 for general surfaces of revolution; that is, 𝑅 is no longer equal to 𝑧 tan 𝛽 but is now an arbitrary function of 𝑧, so 𝑅 = 𝑓 (𝑧). The expressions you need to find will contain 𝐾, 𝑓 (𝑧), derivatives of 𝑓 (𝑧) with respect to 𝑧, and derivatives of 𝑧 with respect to time. 𝑅

Figure P12.260

𝑧

𝑧

front wall left wall

𝑂 𝑣0

𝑥 Figure P12.261

𝜃

𝑦

𝛽 𝑃

Figure P12.262

Problem 12.262 In a racquetball court, at point 𝑃 with coordinates 𝑥𝑃 = 35 f t, 𝑦𝑃 = 16 f t, and 𝑧𝑃 = 1 f t, a ball is imparted a speed 𝑣0 = 90 mph and a direction defined by the angles 𝜃 = 63◦ and 𝛽 = 8◦ (𝛽 is the angle formed by the initial velocity vector and the 𝑥𝑦 plane). The ball bounces off the left vertical wall to then hit the front wall of the court. Assume that the rebound off the left vertical wall occurs such that (1) the component of the ball’s velocity tangent to the wall before and after rebound is the same and (2) the component of velocity normal to the wall right after impact is equal in magnitude and opposite in direction to the same component of velocity right before impact. Accounting for the effect of gravity, determine the coordinates of the point on the front wall that will be hit by the ball after rebounding off the left wall.

Problem 12.263 An airplane is being tracked by a radar station at 𝐴. At the instant 𝑡 = 0, the following data is recorded: 𝑟 = 15 km, 𝜙 = 80◦ , 𝜃 = 15◦ , 𝑟̇ = 350 km∕h, 𝜙̇ = −0.002 rad∕s, 𝜃̇ = 0.003 rad∕s. If the airplane is flying to keep each of the spherical velocity components constant for a few minutes, determine the spherical components of the airplane’s acceleration when 𝑡 = 30 s.

Problem 12.264 An airplane is being tracked by a radar station at 𝐴. At the instant 𝑡 = 0, the following data is recorded: 𝑟 = 15 km, 𝜙 = 80◦ , 𝜃 = 15◦ , 𝑟̇ = 350 km∕h, 𝜙̇ = −0.002 rad∕s, 𝜃̇ = 0.003 rad∕s. If the airplane is flying to keep each of the spherical velocity components constant, plot the trajectory of the airplane for 0 < 𝑡 < 150 s.

𝑧 𝑟

𝜙 𝐴 𝜃

Figure P12.263 and P12.264

773

774

ISTUDY

Chapter 12

Particle Kinematics

Problem 12.265 In the limit that we transition from 3D to 2D problems, the parametric description of the space curve transitions from 𝜉 → 𝑥 and 𝑟⃗ = 𝑥 𝚤̂ + 𝑦(𝑥) 𝚥̂. Under these circumstances, show that Eq. (12.115) becomes Eq. (12.51).

Problem 12.266 A portion of a roller coaster loop is described parametrically by: 𝑥 = 𝑎𝜉,

2

𝑦 = 𝑏𝑒−𝜉 sin (2𝜋𝜉),

2

𝑧 = 𝑏𝑒−𝜉 cos (2𝜋𝜉).

For 𝑎 = 20 m and 𝑏 = 10 m, plot 𝜌 and 𝜎 over the interval −2 ≤ 𝜉 ≤ 2. What is the largest acceleration in this interval if the speed is constant at 10 m∕s?

Figure P12.266

ISTUDY

Section 12.8

Chapter Review

775

12.8 C h a p t e r R e v i e w In this chapter we presented some basic definitions needed to study the motion of objects. We also developed some basic tools for the analysis of motion in both two and three dimensions.

Position, velocity, acceleration, and Cartesian coordinates Position. The position of a point is a vector going from the origin of the chosen frame of reference to the point in question. Trajectory. The trajectory of a moving point is the line traced by the point during its motion. Another name for trajectory is path. Displacement. The displacement between positions 𝐴 and 𝐵 is the vector going from 𝐴 to 𝐵. In general, the magnitude of the displacement between two positions is not the distance traveled along the path between these positions. Velocity. The velocity vector is the time rate of change of the position vector. The velocity is always tangent to the path. Speed. The speed is the magnitude of the velocity and is a nonnegative scalar quantity. The speed measures the time rate of change of the distance traveled along the path.

path of 𝑃

𝑦 𝑎(𝑡) ⃗

Acceleration. The acceleration vector is the time rate of change of the velocity vector. Contrary to what happens for the velocity, the acceleration vector is, in general, not tangent to the trajectory.

𝑃 (𝑡)

𝑦(𝑡) 𝚥̂

Eq. (12.14), p. 652

𝑂

Eqs. (12.17) and (12.18), p. 653 𝑣(𝑡) ⃗ = 𝑥(𝑡) ̇ 𝚤̂ + 𝑦(𝑡) ̇ 𝚥̂ = 𝑣𝑥 (𝑡) 𝚤̂ + 𝑣𝑦 (𝑡) 𝚥̂, 𝑎(𝑡) ⃗ = 𝑥(𝑡) ̈ 𝚤̂ + 𝑦(𝑡) ̈ 𝚥̂ = 𝑣̇ 𝑥 (𝑡) 𝚤̂ + 𝑣̇ 𝑦 (𝑡) 𝚥̂ = 𝑎𝑥 (𝑡) 𝚤̂ + 𝑎𝑦 (𝑡) 𝚥̂, where 𝑣𝑥 = 𝑥, ̇ 𝑣𝑦 = 𝑦,̇ 𝑎𝑥 = 𝑣̇ 𝑥 = 𝑥, ̈ and 𝑎𝑦 = 𝑣̇ 𝑦 = 𝑦. ̈

One-dimensional motion In applications, the acceleration of a point can be a function of time, position, or sometimes velocity. Except for the constant acceleration relations, these equations are rarely applied directly, and the acceleration is integrated as was done in their development. 1. If the acceleration is provided as a function of time, i.e., 𝑎 = 𝑎(𝑡), for velocity and position, we have Eqs. (12.20) and (12.22), p. 668 𝑣(𝑡) = 𝑣0 +

∫𝑡

𝑡

𝑎(𝑡) 𝑑𝑡,

0

𝑠(𝑡) = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) +

∫𝑡

0

𝑡

𝚤̂ 𝑥(𝑡)

In Cartesian components, the velocity and acceleration vectors are given by

[

∫𝑡

0

𝑡

] 𝑎(𝑡) 𝑑𝑡 𝑑𝑡.

𝑥 coordinate line through 𝑃 (𝑡)

𝑟⃗(𝑡)

Cartesian coordinates. The Cartesian coordinates of a particle 𝑃 moving along some path are shown in Fig. 12.43. The position vector is given by

𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂.

𝑣(𝑡) ⃗

𝑥 𝑦 coordinate line through 𝑃 (𝑡)

Figure 12.43 The position vector 𝑟⃗(𝑡) of the point 𝑃 in Cartesian coordinates.

776

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Particle Kinematics

2. If the acceleration is provided as a function of velocity, i.e., 𝑎 = 𝑎(𝑣), for time and position, we have Eqs. (12.24) and (12.27), p. 669 𝑣

𝑡(𝑣) = 𝑡0 +

1 𝑑𝑣, ∫𝑣 𝑎(𝑣) 0 𝑣

𝑠(𝑣) = 𝑠0 +

𝑣 𝑑𝑣. ∫𝑣 𝑎(𝑣) 0

3. If the acceleration is provided as a function of position, i.e., 𝑎 = 𝑎(𝑠), for velocity and time, we have Eq. (12.29), p. 669, and Eq. (12.31), p. 670 𝑣2 (𝑠) = 𝑣20 + 2

∫𝑠

𝑠

𝑎(𝑠) 𝑑𝑠, 0

𝑠

𝑡(𝑠) = 𝑡0 +

𝑑𝑠 . ∫𝑠 𝑣(𝑠) 0

4. If the acceleration is a constant 𝑎𝑐 , for velocity and position, we have Eqs. (12.32)–(12.34), p. 670 𝑣 = 𝑣0 + 𝑎𝑐 (𝑡 − 𝑡0 ), 𝑠 = 𝑠0 + 𝑣0 (𝑡 − 𝑡0 ) + 21 𝑎𝑐 (𝑡 − 𝑡0 )2 , 𝑣2 = 𝑣20 + 2𝑎𝑐 (𝑠 − 𝑠0 ).

Circular motion. For circular motion, the equations summarized in items (1)–(4) above hold as long as we use the replacement rules 𝑠 → 𝜃,

𝑣 → 𝜔,

𝑎 → 𝛼,

where 𝜔 = 𝜃̇ and 𝛼 = 𝜃̈ are the angular velocity and angular acceleration, respectively. vertical direction (positive upward) horizontal direction 𝑃 path 𝑔

Projectile motion A common application of the constant acceleration equations summarized above is the study of projectile motion. We defined projectile motion as the motion of a particle in free flight, neglecting the forces due to air drag and neglecting changes in gravitational attraction with changes in height. In this case, referring to Fig. 12.44, the only force on the particle is the constant gravitational force, and the equations describing the motion are Eqs. (12.39), p. 689

Figure 12.44 Acceleration of a point 𝑃 in projectile motion.

ISTUDY

𝑎horiz = 0 and 𝑎vert = −𝑔.

Normal-tangential components and polar coordinates While we can always represent vectors using Cartesian components, it is sometimes convenient to represent vector quantities in other component systems.

Normal-tangential component system. Referring to Fig. 12.45, the velocity vector has the form Eq. (12.42), p. 703 𝑣⃗ = 𝑣 𝑢̂ 𝑡 ,

ISTUDY

Section 12.8

777

Chapter Review

𝑣⃗ = 𝑣 𝑢̂ 𝑡

𝑢̂ 𝑡

𝑃

path of 𝑃

𝑎𝑡 = 𝑣̇

𝑢̂ 𝑡

𝑃 𝑠 𝑢̂ 𝑛

path of 𝑃 Figure 12.45. Representation of the velocity in normal-tangential components.

𝑎𝑛 =

where 𝑣 is the speed and 𝑢̂ 𝑡 is the tangent unit vector at the point 𝑃 . Referring to Fig. 12.46, the acceleration vector in normal-tangential components has the form

𝑣2 𝜌

𝑠=0

𝐶 𝜌

Eq. (12.49), p. 704 𝑣2 𝑢̂ = 𝑎𝑡 𝑢̂ 𝑡 + 𝑎𝑛 𝑢̂ 𝑛 , 𝜌 𝑛

𝑎⃗ = 𝑣̇ 𝑢̂ 𝑡 +

where 𝑣̇ = 𝑎𝑡 is the tangential component of acceleration and 𝑣2 ∕𝜌 = 𝑎𝑛 is the normal component of acceleration. When we are using a Cartesian coordinate system, for a path expressed by a relation, such as 𝑦 = 𝑦(𝑥), the path’s radius of curvature is given by 𝜌:

osculating circle Figure 12.46 Acceleration in normal-tangential components.

Eq. (12.51), p. 704 [ ]3∕2 1 + (𝑑𝑦∕𝑑𝑥)2 . 𝜌(𝑥) = | 2 | |𝑑 𝑦∕𝑑𝑥2 | | |

Polar coordinates. To obtain the velocity and acceleration in polar components, and for use later, it was useful to find the time derivative of a unit vector 𝑢, ̂ which we found to be Eq. (12.58), p. 718

𝑦 𝑢̂ 𝜃

𝑢̂̇ = 𝜔 ⃗ 𝑢 × 𝑢, ̂ where 𝜔 ⃗ 𝑢 is the angular velocity of the unit vector 𝑢. ̂ Referring to Fig. 12.47, 𝑟 and 𝜃 are the polar coordinates of point 𝑃 . The coordinate 𝜃 is chosen positive in the counterclockwise direction as viewed down the positive 𝑧 axis. Using polar coordinates, the position vector of 𝑃 is Eq. (12.65), p. 719 𝑟⃗ = 𝑟 𝑢̂ 𝑟 . Differentiating the position with respect to time, we obtain the velocity vector in polar coordinates as Eq. (12.68), p. 720 𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 = 𝑣𝑟 𝑢̂ 𝑟 + 𝑣𝜃 𝑢̂ 𝜃 , where Eqs. (12.69), p. 720 𝑣𝑟 = 𝑟̇

and 𝑣𝜃 = 𝑟𝜃̇

path

𝑃

𝑟

𝑢̂ 𝑟

𝑟⃗ 𝜃 𝑂

𝑥

Figure 12.47 The position 𝑟⃗ of a particle defined using the polar coordinates 𝑟 and 𝜃.

778

Chapter 12

Particle Kinematics

are the radial and transverse components of the velocity, respectively. Differentiating the velocity with respect to time, the acceleration vector in polar coordinates is Eq. (12.72), p. 720 ) ( ( ) 𝑎⃗ = 𝑟̈ − 𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝑢̂ 𝜃 = 𝑎𝑟 𝑢̂ 𝑟 + 𝑎𝜃 𝑢̂ 𝜃 , where Eqs. (12.73), p. 720 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2

and 𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇

are the radial and transverse components of the acceleration, respectively.

Relative motion analysis and differentiation of geometrical constraints 𝑌

𝑦

Relative motion analysis. In general, physical systems consist of several moving parts. To study the motion of these systems, it is important to be able to describe the motion of one object relative to another. Referring to Fig. 12.48, consider the planar motion of points 𝐴 and 𝐵. The positions of 𝐴 and 𝐵 relative to the 𝑋𝑌 frame are 𝑟⃗𝐴 and 𝑟⃗𝐵 , respectively. Attached to 𝐴 there is a frame 𝑥𝑦 that translates, but does not rotate relative to frame 𝑋𝑌 . In either frame, the position of 𝐵 relative to 𝐴 is given by 𝑟⃗𝐵∕𝐴 . Using vector addition, the vectors 𝑟⃗𝐴 , 𝑟⃗𝐵 , and 𝑟⃗𝐵∕𝐴 are related as follows:

path of 𝐴

𝚥̂ 𝐴

𝑥

𝚤̂ 𝑟⃗𝐵∕𝐴

𝐵

𝑟⃗𝐴

Eq. (12.83), p. 738 𝑟⃗𝐵 = 𝑟⃗𝐴 + 𝑟⃗𝐵∕𝐴 .

𝑟⃗𝐵

𝑂

The first and second time derivatives of the above equation are, respectively,

path of 𝐵

𝐽̂ 𝐼̂

𝑋

Figure 12.48 Two particles 𝐴 and 𝐵 and the definition of their relative position vector 𝑟⃗𝐵∕𝐴 .

ISTUDY

Eqs. (12.84) and (12.85), p. 739 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 , 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 , where 𝑣⃗𝐵∕𝐴 = 𝑟⃗̇ 𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 = 𝑟⃗̈𝐵∕𝐴 are the relative velocity and acceleration of 𝐵 with respect to 𝐴, respectively. In general, the vectors 𝑣⃗𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 computed by the 𝑥𝑦 observer are different from the vectors 𝑣⃗𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 computed by the 𝑋𝑌 observer. However, the 𝑥𝑦 and 𝑋𝑌 observers compute the same 𝑣⃗𝐵∕𝐴 and the same 𝑎⃗𝐵∕𝐴 if these observers do not rotate relative to one another.

Differentiation of geometrical constraints. There are very few dynamics problems in which the motion is not constrained in some way. In certain classes of systems, constraints are described by geometrical relations between points in the system. We analyzed a pulley system whose motion was constrained by the inextensibility of the cords in the system. The key to constrained motion analysis is knowing that we can differentiate the equations describing the geometrical constraints to obtain velocities and accelerations of points of interest.

ISTUDY

Section 12.8

779

Chapter Review

Motion in three dimensions When the motion of points in a system is not constrained to be planar, we need to use component systems that are three-dimensional. We studied four such component systems.

Cartesian coordinates. Referring to Fig. 12.49, in the three-dimensional Cartesian coordinate system, the position, velocity, and acceleration, respectively, are given in component form as Eqs. (12.97)–(12.99), p. 759 ̂ 𝑟⃗(𝑡) = 𝑥(𝑡) 𝚤̂ + 𝑦(𝑡) 𝚥̂ + 𝑧(𝑡) 𝑘, ̂ 𝑣(𝑡) ⃗ = 𝑥̇ 𝚤̂ + 𝑦̇ 𝚥̂ + 𝑧̇ 𝑘̂ = 𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂ + 𝑣𝑧 𝑘, ̂ 𝑎(𝑡) ⃗ = 𝑥̈ 𝚤̂ + 𝑦̈ 𝚥̂ + 𝑧̈ 𝑘̂ = 𝑎𝑥 𝚤̂ + 𝑎𝑦 𝚥̂ + 𝑎𝑧 𝑘.

𝑧 𝑃 𝑟⃗𝑃 𝚤̂

𝑧𝑃 𝑥

𝑘̂ 𝚥̂

𝑦𝑃

𝑦

𝑥𝑃

Figure 12.49 The Cartesian coordinate system and its unit vectors in three dimensions.

Tangent-normal-binormal components. Extending normal-tangential components to three dimensions did not change the expressions for velocity and acceleration, which are still given by Eq. (12.42), p. 703, and Eq. (12.49), p. 704 𝑣⃗ = 𝑣 𝑢̂ 𝑡 , 𝑎⃗ = 𝑣̇ 𝑢̂ 𝑡 +

𝑣2 𝑢̂ . 𝜌 𝑛

On the other hand, because the osculating plane is not longer stationary, we needed to introduce the torsion 𝜏 in addition to the curvature 𝜅. We found that the change in the tangent, normal, and binormal vectors with respect to the path variable 𝑠 is Eqs. (12.108)–(12.110), p. 760 𝑑 𝑢̂ 𝑡 𝑑𝑠 𝑑 𝑢̂ 𝑛 𝑑𝑠 𝑑 𝑢̂ 𝑏 𝑑𝑠

= 𝜅 𝑢̂ 𝑛 , = 𝜏 𝑢̂ 𝑏 − 𝜅 𝑢̂ 𝑡 , = −𝜏 𝑢̂ 𝑛 . 𝑧

Cylindrical coordinates. The position of a point 𝑃 in three dimensions can be described by the three quantities 𝑅, 𝜃, and 𝑧 that are the point’s cylindrical coordinates (Fig. 12.50). In addition, we see that cylindrical coordinates involve the orthogonal triad of unit vectors 𝑢̂ 𝑅 , 𝑢̂ 𝜃 , and 𝑢̂ 𝑧 . Using this triad, the position in cylindrical coordinates is given by Eq. (12.119), p. 762

Eq. (12.120), p. 762 𝑣⃗ = 𝑅̇ 𝑢̂ 𝑅 + 𝑅𝜃̇ 𝑢̂ 𝜃 + 𝑧̇ 𝑢̂ 𝑧 = 𝑣𝑅 𝑢̂ 𝑅 + 𝑣𝜃 𝑢̂ 𝜃 + 𝑣𝑧 𝑢̂ 𝑧 , where Eqs. (12.121), p. 762 ̇ 𝑣𝑅 = 𝑅,

𝑣𝜃 = 𝑅𝜃,̇

and 𝑣𝑧 = 𝑧̇

𝑢̂ 𝜃 𝑢̂ 𝑅

𝑟⃗ 𝑂 𝑅

𝑟⃗ = 𝑅 𝑢̂ 𝑅 + 𝑧 𝑢̂ 𝑧 . The velocity vector in cylindrical coordinates is given by

𝑢̂ 𝑧 𝑃

𝑥

𝑦

𝑢̂ 𝑧

𝜃 𝑄

𝑢̂ 𝜃 𝑢̂ 𝑅

Figure 12.50 Coordinate directions in the cylindrical coordinate system. As with polar coordinates defined in Section 12.5, the 𝜃 coordinate is implicitly contained in 𝑢̂ 𝑟 .

780

Chapter 12

Particle Kinematics

are the components of the velocity vector in the 𝑢̂ 𝑅 , 𝑢̂ 𝜃 , and 𝑢̂ 𝑧 directions, respectively. Finally, the acceleration vector in cylindrical coordinates is given by Eq. (12.122), p. 762 ( ) ( ) 𝑎⃗ = 𝑅̈ − 𝑅𝜃̇ 2 𝑢̂ 𝑅 + 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ 𝑢̂ 𝜃 + 𝑧̈ 𝑢̂ 𝑧 = 𝑎𝑅 𝑢̂ 𝑅 + 𝑎𝜃 𝑢̂ 𝜃 + 𝑎𝑧 𝑢̂ 𝑧 , where Eqs. (12.123), p. 762 𝑎𝑅 = 𝑅̈ − 𝑅𝜃̇ 2 ,

̇ 𝑎𝜃 = 𝑅𝜃̈ + 2𝑅̇ 𝜃,

and 𝑎𝑧 = 𝑧̈

are the components of the acceleration vector in the 𝑢̂ 𝑅 , 𝑢̂ 𝜃 , and 𝑢̂ 𝑧 directions, respectively.

Spherical coordinates. The position of a point 𝑃 in three dimensions can be described by the three quantities 𝑟, 𝜃, and 𝜙 that are the point’s spherical coordinates (Fig. 12.51). In addition, spherical coordinates involve the orthogonal triad of unit vectors 𝑢̂ 𝑟 , 𝑢̂ 𝜙 , and 𝑢̂ 𝜃 , with 𝑢̂ 𝜙 × 𝑢̂ 𝜃 = 𝑢̂ 𝑟 . Using this triad, the position vector in spherical coordinates is given by Eq. (12.124), p. 763

𝑧 𝜙

𝑢̂ 𝑟 𝑃 𝑟 = 𝑟⃗

𝑟⃗

𝑟⃗ = 𝑟 𝑢̂ 𝑟 . 𝑢̂ 𝜃

The velocity vector in spherical coordinates is given by

𝑢̂ 𝜙

Eq. (12.130), p. 763

𝑂 𝑥

𝑣⃗ = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜙̇ 𝑢̂ 𝜙 + 𝑟𝜃̇ sin 𝜙 𝑢̂ 𝜃 = 𝑣𝑟 𝑢̂ 𝑟 + 𝑣𝜙 𝑢̂ 𝜙 + 𝑣𝜃 𝑢̂ 𝜃 ,

𝑦 𝜃

where Eqs. (12.131), p. 763

Figure 12.51 Definition of the unit vectors in spherical coordinates.

ISTUDY

𝑣𝑟 = 𝑟,̇

̇ 𝑣𝜙 = 𝑟𝜙,

and 𝑣𝜃 = 𝑟𝜃̇ sin 𝜙

are the components of the velocity vector in the 𝑢̂ 𝑟 , 𝑢̂ 𝜙 , and 𝑢̂ 𝜃 directions, respectively. Finally, the acceleration vector in spherical coordinates is given by Eq. (12.132), p. 763 ( ) 𝑎⃗ = 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙 𝑢̂ 𝑟 ( ) + 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙 𝑢̂ 𝜙 ( ) + 𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙 𝑢̂

𝜃

= 𝑎𝑟 𝑢̂ 𝑟 + 𝑎𝜙 𝑢̂ 𝜙 + 𝑎𝜃 𝑢̂ 𝜃 , where Eqs. (12.133), p. 763 𝑎𝑟 = 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙, 𝑎 = 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙, 𝜙

𝑎𝜃 = 𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙 are the components of the acceleration vector in the 𝑢̂ 𝑟 , 𝑢̂ 𝜙 , and 𝑢̂ 𝜃 directions, respectively.

ISTUDY

Section 12.8

781

Chapter Review

Review Problems Problem 12.267

𝑦𝐴

𝑦𝐵

The velocity and acceleration of point 𝑃 expressed relative to frame 𝐴 at some time 𝑡 are ) ) ( ( 𝑣⃗𝑃 ∕𝐴 = 12.5 𝚤̂𝐴 + 7.34 𝚥̂𝐴 m∕s and 𝑎⃗𝑃 ∕𝐴 = 7.23 𝚤̂𝐴 − 3.24 𝚥̂𝐴 m∕s2 .

𝑣⃗𝑃

Knowing that frame 𝐵 does not move relative to frame 𝐴, determine the expressions for the velocity and acceleration of 𝑃 with respect to frame 𝐵. Verify that the speed of 𝑃 and the magnitude of 𝑃 ’s acceleration are the same in the two frames.

𝚥̂𝐵 𝐵

Problem 12.268

𝚥̂𝐴

The motion √ of a point 𝑃 with respect to a Cartesian coordinate system is described by 𝑟⃗ = {2 𝑡 𝚤̂ + [4 ln(𝑡 + 1) + 2𝑡2 ] 𝚥̂} ft, where 𝑡 is time expressed in seconds. Determine the average velocity between 𝑡1 = 4 s and 𝑡2 = 6 s. Then find the time 𝑡̄ for which the 𝑥 component of 𝑃 ’s velocity is exactly equal to the 𝑥 component of 𝑃 ’s average velocity between times 𝑡1 and 𝑡2 . Is it possible to find a time at which 𝑃 ’s velocity and 𝑃 ’s average velocity are exactly equal? Explain why. Hint: Velocity is a vector.

𝐴

𝑦 𝑃 (𝑡1 ) 𝑃

𝚥̂ 𝚤̂

𝛥⃗𝑟(𝑡1 , 𝑡2 )

𝚤̂𝐵

𝑃

𝑎⃗𝑃

15◦

𝑥𝐵

𝚤̂𝐴

𝑥𝐴

Figure P12.267

𝑃 (𝑡2 )

𝑣⃗avg 𝑥

path of 𝑃

Figure P12.268

Figure P12.269

Problem 12.269 The figure shows the displacement vector of a point 𝑃 between two time instants 𝑡1 and 𝑡2 . Is it possible for the vector 𝑣⃗avg shown to be the average velocity of 𝑃 over the time interval [𝑡1 , 𝑡2 ]?

Problems 12.270 and 12.271 A dynamic fracture model proposed to explain the behavior of cracks propagating at high velocity views the crack path as a wavy path.∗ In this model, a crack tip appearing to travel along a straight path actually travels at roughly the speed of sound along a wavy path. Let the wavy path of the crack tip be described by the function 𝑦 = ℎ sin(2𝜋𝑥∕𝜆), where ℎ is the amplitude of the crack tip fluctuations in the direction perpendicular to the crack plane and 𝜆 is the corresponding period. Assume that the crack tip travels along the wavy path at a constant speed 𝑣𝑠 (e.g., the speed of sound). Problem 12.270 Find the expression for the 𝑥 component of the crack tip velocity as a function of 𝑣𝑠 , 𝜆, ℎ, and 𝑥.

Denote the apparent crack tip velocity by 𝑣𝑎 , and define it as the average value of the 𝑥 component of the crack velocity, that is, Problem 12.271

𝜆

𝑣𝑎 = ∗ In

1 𝑣 𝑑𝑥. 𝜆 ∫0 𝑥

dynamic fracture, the structural failure of a material occurs at speeds close to the speed of sound (in that material). This field of study is very important in the design of impact- and/or blast-resistant structures. The model mentioned in these problems is due to H. Gao, “Surface Roughening and Branching Instabilities in Dynamic Fracture,” Journal of the Mechanics and Physics of Solids, 41(3), pp. 457–486, 1993.

cracked panel 𝑦 wavy crack path ℎ

𝑥 𝜆

apparent crack path crack faces Figure P12.270 and P12.271

782

Chapter 12

Particle Kinematics

In dynamic fracture experiments on polymeric materials, 𝑣𝑎 = 2𝑣𝑠 ∕3, 𝑣𝑠 is found to be close to 800 m∕s, and 𝜆 is of the order of 100 𝜇m. What value of ℎ would you expect to find in the experiments if the wavy crack theory were confirmed to be accurate?

Problem 12.272 The motion of a peg sliding within a rectilinear guide is controlled by an actuator in such a way that the peg’s acceleration takes on the form 𝑥̈ = 𝑎0 (2 cos 2𝜔𝑡 − 𝛽 sin 𝜔𝑡), where 𝑡 is time, 𝑎0 = 3.5 m∕s2 , 𝜔 = 0.5 rad∕s, and 𝛽 = 1.5. Determine the total distance traveled by the peg during the time interval 0 s ≤ 𝑡 ≤ 5 s if 𝑥(0) ̇ = 𝑎0 𝛽∕𝜔 + 0.3 m∕s. When compared with Prob. 12.59, why does the addition of 0.3 m∕s in the initial velocity turn this into a problem that requires a computer to solve? 𝑥

Figure P12.272

Problem 12.273 The acceleration of an object in rectilinear free fall while immersed in a linear viscous fluid is 𝑎 = 𝑔 − 𝐶𝑑 𝑣∕𝑚, where 𝑔 is the acceleration of gravity, 𝐶𝑑 is a constant drag coefficient, 𝑣 is the object’s velocity, and 𝑚 is the object’s mass. Letting 𝑣 = 0 and 𝑠 = 0 for 𝑡 = 0, where 𝑠 is position and 𝑡 is time, determine the position as a function of time.

𝑠 Figure P12.273

Problem 12.274 Heavy rains cause a stretch of road to have a coefficient of friction that changes as a function of location. Under these conditions, the acceleration of a car skidding while trying to stop can be approximated by 𝑠̈ = −(𝜇𝑘 − 𝑐𝑠)𝑔, where 𝜇𝑘 is the friction coefficient under dry conditions, 𝑔 is the acceleration of gravity, and 𝑐, with units of m−1 , describes the rate of friction decrement. Let 𝜇𝑘 = 0.5, 𝑐 = 0.015 m−1 , and 𝑣0 = 45 km∕h, where 𝑣0 is the initial velocity of the car. Determine the time it will take the car to stop and the percent increase in stopping time with respect to dry conditions, i.e., when 𝑐 = 0. Hint: ∫

√

( ) √ 𝑑𝑥 = ln 𝑥 + 1 + 𝑥2 .

1 1 + 𝑥2

𝑠 𝑠=0 Figure P12.274

𝑥

Problem 12.275 𝑘, 𝐿0 𝑚

Figure P12.275

ISTUDY

The acceleration of a particle of mass 𝑚 suspended by a linear spring with spring constant 𝑘 and unstretched length 𝐿0 (when the spring length is equal to 𝐿0 , the spring exerts no force on the particle) is given by 𝑥̈ = 𝑔−(𝑘∕𝑚)(𝑥−𝐿0 ). Assuming that at 𝑡 = 0 the particle is at rest and its position is 𝑥 = 0 m, derive the expression of the particle’s position 𝑥 as a function of time. Hint: A good table of integrals will come in handy.

ISTUDY

Section 12.8

783

Chapter Review

Problem 12.276

𝛼 𝛽

𝐴

In a movie scene involving a car chase, a car goes over the top of a ramp at 𝐴 and lands at 𝐵 below. Letting 𝛼 = 18◦ and 𝛽 = 25◦ , determine the speed of the car at 𝐴 if the car is to be airborne for a full 3 s. Furthermore, determine the distance 𝑑 covered by the car during the stunt, as well as the impact speed and angle at 𝐵. Neglect aerodynamic effects. Express your answer using the U.S . Customary system of units.

𝑑

Figure P12.276

Problem 12.277 Consider the problem of launching a projectile a distance 𝑅 from 𝑂 to 𝐷 with a known launch speed 𝑣0 . It is probably clear to you that you also need to know the launch angle 𝜃 if you want the projectile to land exactly at 𝑅. But it turns out that the condition determining whether or not 𝑣0 is large enough to get to 𝑅 does not depend on 𝜃. Determine this condition on 𝑣0 . Hint: Find 𝑣0 as a function of 𝑅 and 𝜃, and then remember that the sine function is bounded by 1. 𝑦

𝑣0 𝑣0 𝜃2

𝜃1

𝑂

𝐷

𝑅

𝑥

Figure P12.277

Problem 12.278 A skater is spinning with her arms completely stretched out and with an angular velocity 𝜔 = 60 rpm. Letting 𝑟𝑏 = 0.55 f t, and 𝓁 = 2.2 f t and neglecting the change in 𝜔 as the skater lowers her arms, determine the velocity and acceleration of the hand 𝐴 right when 𝛽 = 0◦ if the skater lowers her arms at the constant rate 𝛽̇ = 0.2 rad∕s. Express the answers using the component system shown, which rotates with the skater and for which the unit vector 𝚥̂ (not shown) is such that 𝚥̂ = 𝑘̂ × 𝚤̂. 𝑧

𝜔

𝛽

𝛽

𝐴

𝓁

𝑘̂ 𝚤̂ 𝑟𝑏 𝑂

𝑣 𝑥

𝐴 𝜌

Figure P12.278 𝐶

Problem 12.279 A roller coaster travels over the top 𝐴 of the track section shown with a speed 𝑣 = 60 mph. Compute the largest radius of curvature 𝜌 at 𝐴 such that the passengers on the roller coaster will experience weightlessness at 𝐴.

Figure P12.279

𝐵

784

Chapter 12

Particle Kinematics

Problem 12.280

𝜔𝐸 𝑃 𝑅𝐸 𝜆

equator

Determine, as a function of the latitude 𝜆, the normal acceleration of the point 𝑃 on the surface of the Earth due to the spin 𝜔𝐸 of Earth about its axis. In addition, determine the normal acceleration of the Earth due to its rotation about the Sun. Using these results, determine the latitude above which the acceleration due to the orbital motion of the Earth is more significant than the acceleration due to the spin of the Earth about its axis. Use 𝑅𝐸 = 6371 km for the mean radius of the Earth, and assume the Earth’s orbit about the Sun is circular with radius 𝑅𝑂 = 1.497×108 km.

Problem 12.281 A car is traveling over a hill with a speed 𝑣0 = 160 km∕h. Using the Cartesian coordinate system shown, the hill’s profile is described by the function 𝑦 = −(0.003 m−1 )𝑥2 , where 𝑥 and 𝑦 are measured in meters. At 𝑥 = −100 m, the driver realizes that her speed will cause her to lose contact with the ground once she reaches the top of the hill at 𝑂. Verify that the driver’s intuition is correct, and determine the minimum constant time rate of change of the speed such that the car will not lose contact with the ground at 𝑂. Hint: √ To compute the distance traveled by the car along the car’s path, observe that 𝑑𝑠 = 𝑑𝑥2 + 𝑑𝑦2 = √ 2 1 + (𝑑𝑦∕𝑑𝑥) 𝑑𝑥 and that

Figure P12.280

∫

√

1 + 𝐶 2 𝑥2 𝑑𝑥 =

) ( √ 𝑥√ 1 ln 𝐶𝑥 + 1 + 𝐶 2 𝑥2 . 1 + 𝐶 2 𝑥2 + 2 2𝐶 𝑦

𝑣0

𝑑𝑦 𝑑𝑠

𝑂

𝑥

𝑑𝑥 Figure P12.281 𝑣𝑓 𝑂

Problem 12.282

𝜌𝑓

𝑣0 𝜌0 𝐶

A jet is flying straight and level at a speed 𝑣0 = 1100 km∕h when it turns to change its course by 90◦ as shown. The turn is performed by decreasing the path’s radius of curvature uniformly as a function of the position 𝑠 along the path while keeping the normal acceleration constant and equal to 8𝑔, where 𝑔 is the acceleration due to gravity. At the end of the turn, the speed of the plane is 𝑣𝑓 = 800 km∕h. Determine the radius of curvature 𝜌𝑓 at the end of the turn and the time 𝑡𝑓 that the plane takes to complete its change in course.

Figure P12.282

ISTUDY

Problem 12.283 The mechanism shown is called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door-closing systems. Let 𝐻 = 1.25 m, 𝑅 = 0.45 m, and 𝑟 denote the distance between 𝐵 and 𝑂. Assuṁ 𝑟̈, and 𝜙̈ when ing that the speed of 𝐵 is constant and equal to 5 m∕s, determine 𝑟,̇ 𝜙, ◦ 𝜃 = 180 .

ISTUDY

Section 12.8

785

Chapter Review 𝑥

swinging block

𝐵 𝑅 𝜃

𝐴

𝑦

𝜙

𝑂 𝑆

𝐻

𝑦

Figure P12.283

follower

Problem 12.284

cam

The cam is mounted on a shaft that rotates about 𝑂 with constant angular velocity 𝜔cam . The profile of the cam is described by the function 𝓁(𝜙) = 𝑅0 (1 + 0.25 cos3 𝜙), where the angle 𝜙 is measured relative to the segment 𝑂𝐴, which rotates with the cam. Letting 𝑅0 = 3 cm, determine the maximum value of angular velocity 𝜔max such that the maximum speed of the follower is limited to 2 m∕s. In addition, compute the smallest angle 𝜃min for which the follower achieves its maximum speed.

𝓁(𝜙)

𝜃

A car is traveling at a constant speed 𝑣0 = 210 km∕h along a circular turn with radius 𝑅 = 137 m (the figure is not to scale). The camera at 𝑂 is tracking the motion of the car. Letting 𝐿 = 15 m, determine the camera’s rotation rate, as well as the corresponding time rate of change of the rotation rate when 𝜙 = 30◦ .

𝐿 𝑅 𝑣𝑊

𝐵 𝜙

𝜃

𝑂

𝑣0 𝜌

𝐴

𝐶

𝐶 Figure P12.285

Figure P12.286

Problem 12.286 A plane is initially flying north with a speed 𝑣0 = 430 mph relative to the ground while the wind has a constant speed 𝑣𝑊 = 12 mph in the north-south direction. The plane performs a circular turn with radius of 𝜌 = 0.45 mi. Assume that the airspeed indicator on the plane measures the absolute value of the component of the relative velocity of the plane with respect to the air in the direction of motion. Then determine the value of the tangential component of the airplane’s acceleration when the airplane is halfway through the turn, assuming that the airplane maintains constant the reading of the airspeed indicator.

𝑥

𝑂 𝜔cam Figure P12.284

Problem 12.285

𝐴

𝜙

shaft

786

Chapter 12

Particle Kinematics Problems 12.287 through 12.289

A fountain has a spout that can rotate [ about 𝑂 and ] whose angle 𝛽 is controlled so as to vary with time according to 𝛽 = 𝛽0 1 + sin2 (𝜔𝑡) , with 𝛽0 = 15◦ and 𝜔 = 0.4𝜋 rad∕s. The length of the spout is 𝐿 = 1.5 f t, the water flow through the spout is constant, and the water is ejected at a speed 𝑣0 = 6 f t∕s, measured relative to the spout.

𝑣0 ℎ

𝐿 𝛽 𝑂

Problem 12.287 Determine the largest speed with which the water particles are released from the spout.

Figure P12.287–P12.289 Problem 12.288 Determine the magnitude of the acceleration immediately before release when 𝛽 = 15◦ . Problem 12.289

Determine the highest position reached by the resulting water arc.

𝑦

Problem 12.290

𝐶 𝐿 𝐵 𝜃 𝐴

𝑥

The piston head at 𝐶 is constrained to move along the 𝑦 axis. Let the crank 𝐴𝐵 be rotating counterclockwise at a constant angular speed 𝜃̇ = 2000 rpm, 𝑅 = 3.5 in., and 𝐿 = 5.3 in. Obtain the angular velocity of the connecting rod 𝐵𝐶 by differentiating the relative position vector of 𝐶 with respect to 𝐵 when 𝜃 = 35◦ . Hint: You will also need to determine the velocity of 𝐵 and enforce the constraint that demands that 𝐶 move only along the 𝑦 axis.

𝑅 Figure P12.290

ISTUDY

Problem 12.291 A child 𝐴 is swinging from a swing that is attached to a trolley that is free to move along a fixed rail. Letting 𝐿 = 3 m, if at a given instant 𝑎𝐵 = 47.98 m∕s2 , 𝜃 = 23◦ , 𝜃̇ = −3.512 rad∕s, and 𝜃̈ = −16 rad∕s2 , determine the magnitude of the acceleration of the child relative to the rail at that instant. 𝑑 = 2.5 m 𝐷

𝑎𝐵

𝐵

𝜃

𝑃

𝑙 = 4m

𝐿

𝐵 𝑎0

𝐴

ℎ = 2m

𝐴 0.25 m Figure P12.291

Figure P12.292

Problem 12.292 Block 𝐵 is released from rest at the position shown, and it has a constant acceleration downward 𝑎0 = 5.7 m∕s2 . Determine the velocity and acceleration of block 𝐴 at the instant that 𝐵 touches the floor.

ISTUDY

Section 12.8

Chapter Review

Problem 12.293 At a given instant, an airplane is flying horizontally with speed 𝑣0 = 290 mph and acceleration 𝑎0 = 12 f t∕s2 . At the same time, the airplane’s propellers rotate at an angular speed 𝜔 = 1500 rpm while accelerating at a rate 𝛼 = 0.3 rad∕s2 . Knowing that the propeller diameter is 𝑑 = 14 f t, determine the magnitude of the acceleration of a point on the periphery of the propellers at the given instant. 𝑎0

𝑣0 𝜔 𝛼

Figure P12.293

𝑧

Problem 12.294 A golfer chips the ball as shown. Treating 𝛼, 𝛽, and the initial speed 𝑣0 as given, find an expression for the radius of curvature of the ball’s trajectory as a function of time and the given parameters. Hint: Use the Cartesian coordinate system shown to determine the acceleration and the velocity of the ball. Then reexpress these quantities, using normaltangential components.

𝑣0 𝑦

𝛽 𝛼

𝑥

Problem 12.295

Figure P12.294

A carnival ride called the octopus consists of eight arms that rotate about the 𝑧 axis with a constant angular velocity 𝜃̇ = 6 rpm. The arms have a length 𝐿 = 8 m and form an angle 𝜙 with the 𝑧 axis. Assuming that 𝜙 varies with time as 𝜙(𝑡) = 𝜙0 + 𝜙1 sin 𝜔𝑡 with 𝜙0 = 70.5◦ , 𝜙1 = 25.5◦ , and 𝜔 = 1 rad∕s, determine the magnitude of the acceleration of the outer end of an arm when 𝜙 achieves its minimum value.

𝑧 𝐿

Gary L. Gray

Figure P12.295

𝜙

𝜃̇

787

ISTUDY

ISTUDY

Force and Acceleration Methods for Particles

13 Newton’s second law is an axiom, that is, a statement we accept as true and not derivable from other principles. Therefore, this chapter will emphasize how 𝐹⃗ = 𝑚𝑎⃗ is applied, and it begins the study of kinetics, which is the study of the forces that cause and are caused by motion. By the time we complete this chapter, we will be able to use the kinematics discussed in Chapter 12, along with Newton’s second law, to either (1) predict the motion of a particle system caused by given forces or (2) determine the forces needed for a particle system to move in a prescribed way.

Steve Whittington/Moment/Getty Images

An aerobatic maneuver performed by the Red Arrows, the aerobatic team of the British Royal Air Force. Using Newton’s second law we can relate the acceleration of these airplanes to the forces acting on them.

13.1

Rectilinear Motion

In this chapter we show how Newton’s second law, which we stated in Eq. (11.2) as 𝐹⃗ = 𝑚𝑎, ⃗

(13.1)

is applied to study the motion of bodies that are modeled as particles. We begin by looking at straight line or rectilinear motion (see Fig. 13.1).

Applying Newton’s second law Newton’s second law has associated with it a rich terminology and a well-established sequence of steps to formulate a problem and find a solution. We now present the procedure and terminology used when applying Newton’s second law to mechanical systems.

Figure 13.1 A man pushing a crate over a flat horizontal surface. The crate is in rectilinear motion.

Step 1. Road Map & Modeling. Review the given information, identify the system, state assumptions, sketch the FBD, and identify a problem-solving strategy. The ultimate goal of this first step is the creation of a model and associated free body diagram (FBD) for the system under consideration. An FBD is

789

790

𝑚𝑔 𝑃

𝐹

𝚥̂ 𝚤̂

𝑁 Figure 13.2 FBD of the crate in Fig. 13.1, where 𝑃 is the force the man is applying to the crate and 𝑁 is the normal force between the crate and the flat surface. We have represented the crate as a dot (in blue) to emphasize that we are modeling it as a particle.

ISTUDY

Chapter 13

Force and Acceleration Methods for Particles

a diagram of the object under consideration showing all the external forces (and, in later chapters, moments) acting on that object (see Fig. 13.2). When we depict a force vector on an FBD, we must include its line of action and its direction, along with the label indicating which force is associated with that vector. Since the application of Newton’s second law is based on the FBD, it is important to remember that forces are vectors, and we need to choose a convenient component system to represent them (see the unit vectors in Fig. 13.2). Finally, we recall that a particle cannot be acted upon by moments. The system of forces we apply to a particle must be a concurrent system of forces. We must remember this when sketching the FBD of an object modeled as a particle. Referring to Fig. 13.2, we do so in this text by giving the object in question a faded appearance and applying the forces to a blue dot that is meant to remind us of the particle-model assumption. Step 2. Governing Equations. Write the balance principles, force laws, and kinematic equations. In this step, we consider three sets of equations: balance principles, force laws, and kinematic equations. Taken together, these constitute the governing equations. Newton’s second law in Eq. (13.1) is the first balance principle we encounter in dynamics. A balance principle, or a balance law, is a relation that equates the change in a property of a physical system, such as its velocity, with the cause of that change. In the case of Newton’s second law, the balance is between the total force 𝐹⃗ acting on a particle of mass 𝑚 and the change in the state of motion of the particle as measured by the quantity 𝑚𝑎. ⃗ In Chapters 14 and 15, we will discover other balance principles, such as the balance of energy, linear momentum, and/or angular momentum. In practice, Newton’s second law is almost always written in the component form as ∑

𝐹𝑎 = 𝑚𝑎𝑎 ,

∑

𝐹𝑏 = 𝑚𝑎𝑏 ,

and

∑

𝐹𝑐 = 𝑚𝑎𝑐 ,

(13.2)

where 𝑎, 𝑏, and 𝑐 are the orthogonal directions of the chosen component system and we usually need only two directions for planar problems. The second set of governing equations we write are the force laws, which are the equations that provide the specific forms of forces in terms of material properties and kinematic variables. The force laws we will most often encounter are (1) gravity, (2) friction laws, (3) spring force laws, and (4) resistance laws due to fluids. Sometimes the description of a force is so elementary that we do not write an explicit equation for it. Often one or more forces will be unknowns in the problem. In such cases, we indicate the force in question on the FBD and then use the rest of the governing equations to determine it. The final set of governing equations are the kinematic equations, which represent the kinematic quantities appearing in the balance laws and the force laws. These include equations that describe geometric constraints, as well as equations that express the velocities and accelerations in terms of the chosen component system(s). Step 3. Computation. Solve the assembled system of equations. After the governing equations are written, the third step is to solve them. This is typically done by substituting the force laws and kinematic equations into the balance laws and then solving the resulting equations for the quantities of interest. Step 4. Discussion & Verification. Study the solution and do a “sanity check.” Once a solution is obtained, the fourth and final step is to assess its reasonableness.

ISTUDY

Section 13.1

Rectilinear Motion

We should ask: does it make physical sense? This means checking the correctness of the units and/or dimensions used to express the solution, as well as the order of magnitude of the results. The solution should be reconciled with practical experience. Sometimes this is as easy as confirming that the solution matches our physical intuition. At other times, our physical intuition will differ, and this means either that we have made a mistake or that our physical intuition is incorrect. A problem should not be declared solved until everything fits.

Force laws We now look more closely at the force laws for two types of forces we will see often: friction and spring forces. Friction Friction plays an important role in many engineering problems and is still the object of active research. Here we only review an elementary model of friction∗ called the Coulomb friction model, which was first published by Charles Augustin Coulomb (1736–1806) around 1773. The friction force is generally defined to be the component of the contact force between two objects that is tangent to the contact surface. The Coulomb friction model states that when two objects slide relative to one another, the friction force between the two objects is equal to the normal force between the objects times a parameter called the coefficient of kinetic friction. When the objects do not slide relative to one another, the friction force is bounded by the value of the normal force times a parameter called the coefficient of static friction. The friction force acts to oppose relative motion or tendency for relative motion between the objects that are in contact. The Coulomb friction model specifies that all friction falls within one of the following three conditions: No slip. When two objects do not move (or do not slip or do not slide) relative to one another, then the friction force satisfies the inequality |𝐹 | ≤ 𝜇𝑠 |𝑁|,

(13.3)

where 𝜇𝑠 is the coefficient of static friction and 𝑁 is the normal force between the two bodies. The direction of 𝐹 on one body is equal and opposite to the direction of 𝐹 on the other body (due to Newton’s third law). In addition, the direction of 𝐹 on a given body is opposite to the direction that the velocity vector of that body would have, relative to the other body, in the absence of friction. Note that the absolute value signs around 𝐹 and 𝑁 are important. Often the direction of the friction force is one of the unknowns of the problem, and writing |𝐹 | allows us not to worry about having to guess the “right” direction for 𝐹 . As far as 𝑁 is concerned, usually it is assumed to be positive when “in compression,” and at least for equilibrium problems, the sign of 𝑁 is easy to predict. However, in dynamics there are important examples, such as mechanisms with pegs moving along slotted guides, in which the force 𝑁 can change sign with time. Therefore, we write |𝑁| instead of 𝑁 to make sure that Eq. (13.3) is always evaluated correctly. ∗ A treatment of friction can be found in Chapter 9 of M. E. Plesha, G. L. Gray, R. J. Witt, and F. Costanzo,

Engineering Mechanics: Statics, 3rd ed., McGraw-Hill, Dubuque, 2023.

791

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Chapter 13

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Impending slip. When slip between two bodies is impending, that is, when one body is just about to slip relative to the other, then the friction and normal forces are related according to (13.4) 𝐹 = 𝜇𝑠 𝑁, where the direction of 𝐹 must be consistent with the fact that the friction force opposes the tendency for relative motion of 𝐴 and 𝐵. 𝐴

𝑊𝐴

𝑣𝐴∕𝐵 𝐹

𝐵 (a)

𝑁 (b)

Figure 13.3 (a) Body 𝐴 sliding on body 𝐵; the relative velocity is given by 𝑣𝐴∕𝐵 . (b) FBD of body 𝐴 showing the direction of kinetic friction. Body 𝐴 has been modeled as a particle.

ISTUDY

Slip. When two bodies 𝐴 and 𝐵 slide relative to one another (see Fig. 13.3), then the friction force on body 𝐴 due to body 𝐵 is 𝐹⃗ = −𝜇𝑘 |𝑁|

𝑣⃗𝐴∕𝐵 |𝑣⃗𝐴∕𝐵 |

,

(13.5)

where 𝜇𝑘 is the kinetic friction coefficient and 𝜇𝑘 ≤ 𝜇𝑠 . Notice that the direction of 𝐹⃗ on body 𝐴 is opposite to the direction of the velocity of 𝐴 relative to 𝐵. Equation (13.5) is usually written as (13.6) 𝐹 = 𝜇𝑘 𝑁, where the direction of 𝐹 must be consistent with the fact that the friction force opposes relative motion of 𝐴 and 𝐵. Mini-Example For the system in Fig. 13.1, given a constant pushing force 𝑃 , the coefficient of kinetic friction 𝜇𝑘 between the crate and ground, the mass 𝑚 of the crate, and that the system starts from rest, we will now determine the distance 𝑑 the crate moves until it reaches a speed 𝑣𝑓 . Solution Modeling the crate as a particle, we obtain the FBD shown in Fig. 13.2 (we have chosen to use a Cartesian component system). If we can determine the acceleration of the crate, we should be able to integrate it to determine the distance the crate moves to achieve the desired speed. To write the governing equations for the crate, we start with the balance laws for the crate, which are the components of Newton’s second law in this case. Applying Newton’s second law to the FBD in Fig. 13.2, we obtain ∑ 𝐹𝑥 ∶ 𝑃 − 𝐹 = 𝑚𝑎𝑥 , (13.7) ∑ 𝐹𝑦 ∶ 𝑁 − 𝑚𝑔 = 𝑚𝑎𝑦 . (13.8) Since we know the crate slides on the ground, the force law for the friction force 𝐹 is (13.9) 𝐹 = 𝜇𝑘 𝑁. The kinematic equations come from the constraint that the crate cannot move vertically and from the need to integrate the 𝑥 component of acceleration, that is, 𝑎𝑥 = 𝑥̈ and

𝑎𝑦 = 0.

(13.10)

For the computation step, we substitute Eqs. (13.9) and (13.10) into Eqs. (13.7) and (13.8) to obtain 𝑃 − 𝜇𝑘 𝑁 = 𝑚𝑥̈ and

𝑁 − 𝑚𝑔 = 0.

(13.11)

ISTUDY

Section 13.1

793

Rectilinear Motion

The second equation tells us that 𝑁 = 𝑚𝑔, which can then be substituted into the first to obtain ( ) 𝑚𝑣2𝑓 𝑃 𝑃 − 𝜇𝑘 𝑔 𝑑 ⇒ 𝑑 = , (13.12) 𝑥̈ = − 𝜇𝑘 𝑔 ⇒ 𝑣2𝑓 = 2 𝑚 𝑚 2(𝑃 − 𝜇𝑘 𝑚𝑔) where we have used the fact that 𝑥̈ is constant and Eq. (12.34) on p. 670 to obtain the second of Eqs. (13.12). For the verification step, the final result for 𝑑 has the dimension of length, as it should. In addition, we see that the distance it takes the crate to reach speed 𝑣𝑓 increases as 𝑣𝑓 increases and decreases as 𝑃 increases, both of which agree with our intuition.

Springs Springs, which are often thought of as having the coiled shape shown in Fig. 13.4, come in many shapes, sizes, and materials. These include bungee cords, beams made of metal or other materials, metal plates, and torsional rods. Many objects that can deform generally do so with some elasticity and can, therefore, be treated as springs. Geometrically, we describe a spring in terms of its length (Fig. 13.5). The unstretched length of the spring is the length of the spring when no force is applied to it. Letting 𝐿 and 𝐿0 denote the current and unstretched lengths of a spring, respectively, we define the stretch, denoted by the Greek letter 𝛿, to be the quantity 𝛿 = 𝐿 − 𝐿0 .

(13.13)

𝐿 𝐿0 unstretched 𝛿

When 𝛿 > 0 the spring is stretched, and when 𝛿 < 0 it is compressed. We will always assume that springs are massless, which means that the force internal to a spring is equal to the external force applied to the spring. Linear elastic springs. We use the term elastic to convey reversibility; that is, a spring is said to be elastic if, after being stretched or compressed, it returns to its original geometry upon being relieved of the load. Elastic springs may be linear (like structural beams) or nonlinear (like bungee cords), the former being springs whose deformation is proportional to the load applied. In contrast, inelastic or plastic behavior is associated with irreversible deformation. If a coiled spring is pulled so hard that its coil begins to straighten, releasing the load leaves the spring in a state of permanent deformation. In almost all engineering applications, this is to be avoided. A spring is said to be linear elastic if the internal force in the spring is linearly related to the amount the spring is stretched or compressed. Referring to Fig. 13.6, the force 𝐹𝑠 required to stretch a linear elastic spring by an amount 𝛿 is given by ( ) 𝐹𝑠 = 𝑘𝛿 = 𝑘 𝐿 − 𝐿0 ,

Gary L. Gray

Figure 13.4 Simple coil springs from a pen.

(13.14)

where 𝑘 is the spring constant and has dimensions of force over length. Unless we indicate otherwise, we will always assume that springs are linear elastic.

Equation(s) of motion The equation(s) whose solution allows us to determine the motion of a particle or rigid body as a function of time is (are) the equation(s) of motion for that particle

𝐹𝑠

current Figure 13.5 Stretch 𝛿 of a spring due to force 𝐹𝑠 .

𝐹𝑠

tension

𝑘 𝛿

compression contraction

𝐹𝑠 = 𝑘𝛿 elongation

Figure 13.6 Spring law for a linear elastic spring.

794

Chapter 13

Force and Acceleration Methods for Particles

or rigid body. The first of Eqs. (13.12) is the equation of motion of the crate since it provides the constant acceleration of the crate, which can be solved for the position of the crate 𝑥 as a function of time 𝑡. In general, equation(s) of motion for a system involve the system’s position variables, their time derivatives, and time.

Inertial reference frames The use of Newton’s second law requires that the acceleration 𝑎⃗ be measured with respect to an inertial frame of reference. An inertial reference frame is one in which Newton’s first and second laws are valid, at least to the level of accuracy desired. In addition, if we have found a frame that acceptably satisfies this definition (i.e., it is inertial), then any frame that is not accelerating relative to such an inertial frame is also inertial. For all but a small class of engineering problems (e.g., those that involve relativistic effects or orbital mechanics), a frame attached to the surface of the Earth can be considered inertial. 𝑥 𝑥𝐴 (𝑡)

Degrees of freedom

𝚥̂ 𝚤̂ 𝐴

𝑂

𝑦 𝑦𝐶 𝐶 𝜃(𝑡) 𝑥𝐶

𝐵

Figure 13.7 An illustration of degrees of freedom using a rigid bar with a roller at one end. The length of the bar is 𝐿.

ISTUDY

Helpful Information We will structure the solution of kinetics problems using the following four steps: 1. Road Map & Modeling: Identify data and unknowns, identify the system, state assumptions, sketch the FBD, and identify a problem-solving strategy. 2. Governing Equations: Write the balance principles, force laws, and kinematic equations. 3. Computation: Solve the assembled system of equations. 4. Discussion & Verification: Study the solution and perform a “sanity check.”

A system’s degrees of freedom are the independent coordinates needed to uniquely specify the position of that system. A more intuitive way of thinking about degrees of freedom is to think of the number of degrees of freedom as the number of different coordinates in a system that must be fixed in order to keep the system from moving. For example, the bar shown in Fig. 13.7 can pivot about the hinge at 𝐴 and can move horizontally along the guide bar. This bar has two degrees of freedom since we would need to fix the two coordinates 𝑥𝐴 and 𝜃 to prevent the system from moving. It is useful to identify the degrees of freedom of a system because the number of equations of motion of a system is equal to its number of degrees of freedom. We saw this with the sliding crate — it has one degree of freedom and we ended up with one equation of motion.

End of Section Summary Applying Newton’s second law. We developed a four-step problem-solving procedure for applying Newton’s second law to mechanical systems. The central element of this procedure is the derivation of the governing equations, which originate from the balance principles, force laws, and kinematic equations. In this chapter, the balance principle is Newton’s second law, which we write as Eq. (13.1), p. 789 𝐹⃗ = 𝑚𝑎, ⃗ and which we apply in component form as Eqs. (13.2), p. 790 ∑

𝐹𝑎 = 𝑚𝑎𝑎 ,

∑

𝐹𝑏 = 𝑚𝑎𝑏 ,

and

∑

𝐹𝑐 = 𝑚𝑎𝑐 ,

where 𝑎, 𝑏, and 𝑐 are the orthogonal directions of the chosen component system and where we usually need only two directions for planar problems. The four-step procedure we presented for applying Newton’s second law, outlined in the margin note, is given as a guide to the order in which things should be done, and we will use it consistently in each example we present.

ISTUDY

Section 13.1

Rectilinear Motion

Governing equations and equations of motion. The governing equations for a system consist of the (1) balance principles, (2) force laws, and (3) kinematic equations. The equations of motion are the equations derived from the governing equations that allow for the determination of the motion. Friction. We will include friction via the Coulomb friction model. According to this model, in the absence of slip, the magnitude of the friction force 𝐹 satisfies the following inequality: Eq. (13.3), p. 791 |𝐹 | ≤ 𝜇𝑠 |𝑁|, where 𝜇𝑠 is the coefficient of static friction and 𝑁 is the force normal to the contact surface. The relation Eq. (13.4), p. 792 𝐹 = 𝜇𝑠 𝑁 defines the case of impending slip. If 𝐴 and 𝐵 are two objects sliding with respect to one another, the magnitude of the friction force exerted by 𝐵 onto 𝐴 is given by Eq. (13.6), p. 792 𝐹 = 𝜇𝑘 𝑁, where 𝜇𝑘 is the coefficient of kinetic friction and where the direction of the friction force must be consistent with the fact that friction opposes the relative motion of 𝐴 and 𝐵. Springs. A spring is said to be linear elastic if the internal force in the spring is linearly related to the amount the spring is stretched or compressed. The force 𝐹𝑠 required to stretch a linear elastic spring by an amount 𝛿 is given by Eq. (13.14), p. 793 ( ) 𝐹𝑠 = 𝑘𝛿 = 𝑘 𝐿 − 𝐿0 , where 𝑘 is the spring constant and where 𝐿 and 𝐿0 are the current and unstretched lengths of the spring, respectively. When 𝛿 > 0, the spring is said to be stretched, and when 𝛿 < 0, the spring is said to be compressed.

795

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Force and Acceleration Methods for Particles

E X A M P L E 13.1

A Chameleon Capturing an Insect When a chameleon propels its long tongue to snatch an insect for a meal (see Fig. 1), the process occurs so quickly that a high-speed video camera is needed to capture the event. Video data tells us that it takes 0.15 s for the chameleon to completely retrieve the insect. We will assume that the initial speed of the insect is zero and that the final speed of the insect must be zero since it ends up in the chameleon’s mouth for ingestion. The distance traveled is experimentally found to be 0.3 m along a straight line, so we will assume the velocity profile 𝑣(𝑡) = 2[1 − cos(41.89 𝑡)] m∕s (𝑡 is time in seconds), a plot of which is shown in Fig. 2. Given that the insect is a dragonfly whose mass is 6 g, determine the “stickiness” force of the chameleon’s tongue required to retrieve the insect.

kuritafsheen/Getty Images

SOLUTION

Figure 1 A chameleon capturing an insect.

Road Map & Modeling The FBD of the insect as it is being pulled in by the chameleon is shown in Fig. 3, where the 𝑥 axis is aligned with the direction of motion. Knowing 𝑣(𝑡) = 𝑥(𝑡) ̇ for the insect, we can determine acceleration and then apply Newton’s second law to determine the required force. Governing Equations Balance Principles

Figure 2 Velocity versus time profile for the tip of the chameleon’s tongue. The area under this curve is equal to the 0.3 m distance traveled by the insect.

We are told that the insect moves along a straight line, so 𝑎𝑦 = 0. To determine 𝑎𝑥 , we differentiate 𝑣(𝑡) with respect to time to obtain 𝑎𝑥 = 𝑑𝑣∕𝑑𝑡 = (83.78 m∕s2 ) sin(41.89 𝑡).

(3)

Substituting 𝑎𝑦 = 0 and Eq. (3) into Eqs. (1) and (2), we obtain 𝑅𝑦 = 𝑚𝑔,

(4)

which gives a total required force of

𝚥̂ 𝚤̂

Figure 3 FBD of the insect retrieved by the chameleon.

⃗ = |𝑅|

√ √ [ ]2 2 2 𝑅𝑥 + 𝑅𝑦 = 𝑚 (83.78 m∕s2 ) sin(41.89 𝑡) + 𝑔 2 .

Figure 4 Required force on the insect as it is pulled into the mouth of the chameleon.

(5)

⃗ given by Eq. (5), divided by the insect’s weight, has been plotted in The value of |𝑅| Fig. 4. Discussion & Verification

ISTUDY

(2)

All forces are accounted for on the FBD.

𝑅𝑥 = [(83.78 m∕s2 ) sin(41.89 𝑡)]𝑚 and

𝑅𝑥

(1)

Kinematic Equations

Computation

𝑊 = 𝑚𝑔

𝑅𝑦

Force Laws

Applying Newton’s second law to the FBD in Fig. 3, obtain ∑ 𝐹𝑥 ∶ 𝑅𝑥 = 𝑚𝑎𝑥 , ∑ 𝐹𝑦 ∶ 𝑅𝑦 − 𝑚𝑔 = 𝑚𝑎𝑦 .

Figure 4 shows that the maximum force required is almost 9𝑚𝑔, and this occurs at two different points in time. Referring to Eqs. (1) and (3), the first time the maximum force is achieved corresponds to a positive value of 𝑅𝑥 so that the force is directed toward the chameleon. The second maximum corresponds to a negative value of 𝑅𝑥 , which is needed to slow down the insect before it comes to a stop in ⃗ can the chameleon’s mouth. For a dragonfly with mass 𝑚 = 6 g, the maximum value of |𝑅| be calculated by differentiating Eq. (5) with respect to time and setting the result equal to zero, which tells us that 𝑅max = 0.5061 N. This value occurs at 𝑡1 = 0.03750 s and 𝑡2 = 0.1125 s.

ISTUDY

Section 13.1

797

Rectilinear Motion

E X A M P L E 13.2

Spring Stopping a Moving Crate on an Incline

The crate of mass 𝑚 = 45 kg shown in Fig. 1 is moving down the incline with speed 𝑣0 = 8 m∕s at the instant that the spring with elastic constant 𝑘 = 100 N∕m is unstretched. Modeling the crate as a particle, determine the distance 𝑑 the crate moves from the given position before it momentarily comes to a stop. Assume that friction between the crate and the incline is negligible, that the unstretched length of the spring is 𝐿0 = 1.2 m, and that 𝜃 = 25◦ .

𝑦

𝑘

𝑣0 𝑚

SOLUTION

𝜃

Road Map & Modeling

Our goal is to use Newton’s second law to determine the acceleration of the crate. The acceleration can then be manipulated to obtain the position of the crate. Since the crate is sliding on the incline, the FBD is as shown in Fig. 2, where 𝐹𝑠 is the force exerted on the crate by the spring, 𝑁 is the normal force between the crate and the incline, and 𝑚𝑔 is the force of gravity. The position 𝑥 is measured from the wall to which the spring is attached. As such, 𝑥 measures the current length of the spring.

𝑥 Figure 1 A crate moving with speed 𝑣0 on an incline. 𝑚𝑔 𝜃

Governing Equations Balance Principles

𝚥̂

Referring to the FBD in Fig. 2, Newton’s second law gives ∑ 𝐹𝑥 ∶ −𝐹𝑠 + 𝑚𝑔 sin 𝜃 = 𝑚𝑎𝑥 , ∑ 𝐹𝑦 ∶ 𝑁 − 𝑚𝑔 cos 𝜃 = 𝑚𝑎𝑦 .

𝐹𝑠

(1) (2)

Force Laws Since 𝑥 measures the current length of the spring, the expression for 𝐹𝑠 for any position 𝑥 is ( ) (3) 𝐹𝑠 = 𝑘 𝑥 − 𝐿 0 . Kinematic Equations

Since the motion is constrained to just the 𝑥 direction, the kine-

matic equations are 𝑎𝑥 = 𝑥̈ Computation

and

𝑎𝑦 = 0.

(4)

Substituting Eqs. (3) and (4) into Eqs. (1) and (2), we obtain ) ( −𝑘 𝑥 − 𝐿0 + 𝑚𝑔 sin 𝜃 = 𝑚𝑥̈ and 𝑁 = 𝑚𝑔 cos 𝜃.

(5)

Dividing through by 𝑚, applying the chain rule to the first of Eqs. (5), and then integrating from the initial position to the point where the crate first comes to a stop, we obtain 𝑣

) 𝑑𝑣 𝑘( = 𝑔 sin 𝜃 − 𝑥 − 𝐿0 𝑑𝑥 𝑚 ⇒

∫𝑣

0

0

𝑣 𝑑𝑣 =

∫𝐿

𝐿0 +𝑑

0

[

𝑔 sin 𝜃 +

)] 𝑘( 𝐿0 − 𝑥 𝑑𝑥 𝑚

⇒

− 12 𝑣20 = 𝑔𝑑 sin 𝜃 −

𝑘𝑑 2 , 2𝑚

(6)

which is a quadratic equation for the distance 𝑑. Solving and then substituting in given data, we obtain √ ( ) 𝑚𝑔 1 𝑑= sin 𝜃 ± 𝑚 𝑘𝑣20 + 𝑚𝑔 2 sin2 𝜃 𝑘 𝑘 𝑑 = 7.547 m, (7) ⇒ 𝑑 = −3.816 m and 𝑑 = 7.547 m ⇒ where we have chosen the physically meaningful result as the final answer. Discussion & Verification

The dimensions in Eqs. (6) and (7) are both correct, and the stopping distance seems physically reasonable given the initial speed and lack of friction.

𝚤̂ 𝑁

Figure 2 The FBD of the crate as it slides down the incline.

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Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.3

Friction and Impending Slip The truck shown in Fig. 1 is traveling at 𝑣0 = 100 km∕h when the driver slams on the brakes and comes to a stop as quickly as possible. If the coefficient of static friction between the crate 𝐴 and the bed of the truck is 0.35, determine the minimum stopping distance 𝑑min and the minimum stopping time 𝑡min of the truck for which the crate does not slide forward on the truck.

𝑣0 𝐴

Figure 1 A truck hauling a large crate.

SOLUTION Road Map & Modeling

𝑚𝑔 𝐹

𝚥̂ 𝚤̂ 𝑁

Figure 2 FBD of the crate.

ISTUDY

Common Pitfall Newton’s second law and inertial frames. Since the application of Newton’s second law requires the use of an inertial reference frame, the component system shown in Fig. 2 must be understood as originating from an 𝑥𝑦 coordinate system fixed with the ground—this is the inertial reference frame. It would be a mistake to choose a coordinate system moving with the truck because the truck is decelerating with respect to the ground and, therefore, is not an inertial frame of reference.

The truck must stop as fast as possible without causing the crate to slide. Therefore, the system to analyze is the crate, which we model as a particle. In the FBD shown in Fig. 2, we have assumed that the only relevant forces are the crate’s weight 𝑚𝑔, the normal force 𝑁 between the crate and the truck, and the friction force 𝐹 between the crate and the truck. The force 𝐹 points left because if the crate were to slip, it would slip to the right relative to the truck, and 𝐹 must oppose this motion. Since the motion is rectilinear, we have chosen a Cartesian coordinate system with the 𝑥 axis parallel to the direction of motion. The maximum deceleration without sliding corresponds to the maximum possible friction force acting on the crate, which is the force corresponding to impending slip. Our solution strategy will be to relate the maximum friction force to the acceleration via Newton’s second law. Once we have an expression for the maximum acceleration, we will apply our knowledge of kinematics to compute 𝑑min and 𝑡min . Governing Equations Balance Principles

Force Laws

Referring to Fig. 2, Newton’s second law yields ∑ 𝐹𝑥∶ −𝐹 = 𝑚𝑎𝑥 , ∑ 𝐹𝑦∶ 𝑁 − 𝑚𝑔 = 𝑚𝑎𝑦 .

(1) (2)

The friction law for impending slip is 𝐹 = 𝜇𝑠 𝑁.

Kinematic Equations

(3)

The kinematic equations are 𝑎𝑥 = 𝑎max

and 𝑎𝑦 = 0,

(4)

which express the fact that while 𝑎𝑥 is still unknown, we are seeking the maximum value of acceleration, and also express the fact that the crate is not moving vertically. Computation Equations (1)–(4) are four equations in the unknowns 𝑁, 𝑎𝑦 , 𝑎max , and 𝐹 . Equation (2) and the second of Eqs. (4) tell us that 𝑁 = 𝑚𝑔. Substituting this result into Eq. (3) and, in turn, substituting that into Eq. (1), we obtain

−𝜇𝑠 𝑚𝑔 = 𝑚𝑎max

⇒

𝑎max = −𝜇𝑠 𝑔,

(5)

where the negative sign indicates that the crate (and truck) is decelerating when the driver applies the brakes. Since the maximum deceleration is constant (both 𝜇𝑠 and 𝑔 are constants), we can determine the stopping distance by using Eq. (12.34) on p. 670: 𝑣2 = 𝑣20 + 2𝑎𝑐 (𝑥 − 𝑥0 )

⇒

0 = 𝑣20 − 2𝜇𝑠 𝑔𝑑min ,

(6)

so that 𝑑min =

𝑣20 2𝜇𝑠 𝑔

= 112.4 m,

(7)

ISTUDY

Section 13.1

where we have used the conversion 100 km∕h = 27.78 m∕s. To determine the stopping time, we apply Eq. (12.32) on p. 670:

so that 𝑡min =

⇒ 𝑣0 𝜇𝑠 𝑔

0 = 𝑣0 − 𝜇𝑠 𝑔𝑡min ,

= 8.090 s.

(8)

(9)

Discussion & Verification

The symbolic forms of the results in Eqs. (7) and (9) have the correct dimensions for 𝑑min and 𝑡min . Also, the final numerical results have been expressed using appropriate units. As far as the values we have obtained for 𝑑min and 𝑡min are concerned, these are certainly reasonable since, in absolute value, the maximum deceleration possible under the stated assumptions is roughly one-third of the acceleration of gravity. A Closer Look

The model used here, which is based on the Coulomb friction model, is such that the deceleration we calculated is completely independent of the crate’s mass. Therefore, the crate could have been twice as heavy, and we would have obtained the same answer. This is a basic property of this friction model. Another important observation to make is that, again because of the friction model used, the minimum time required to stop is linearly proportional to the initial speed 𝑣0 , and the minimum distance required to stop is proportional to the initial speed squared 𝑣20 . This is shown in Figs. 3 and 4, respectively.

10 8 6 4 2 0

0

10

30 20 𝑣0 (m∕s)

799

40

Figure 3 The minimum stopping time of the crate as a function of the initial speed of the truck.

200 𝑑min (m)

𝑣 = 𝑣0 + 𝑎𝑐 𝑡

𝑡min (s)

Rectilinear Motion

150 100 50 0

0

10

20 30 𝑣0 (m∕s)

40

Figure 4 The minimum stopping distance of the crate as a function of the initial speed of the truck.

800

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.4

Transition from Static to Kinetic Friction

𝑃 (𝑡) 𝜃 𝑚 𝜇𝑠 , 𝜇𝑘 Figure 1 A person pushing a crate of mass 𝑚 over a rough surface with a time-dependent force 𝑃 (𝑡).

𝜃

𝚥̂

𝑚𝑔

𝑃0 𝑡

𝐹

𝚤̂ 𝑁

A simple problem of practical importance that illustrates the nature of friction is the study of the initiation of motion of an object on a rough surface. Let’s look at what happens when a person pushes a crate of mass 𝑚 on a rough floor, as shown in Fig. 1. Letting 𝜇𝑠 and 𝜇𝑘 , with 𝜇𝑘 < 𝜇𝑠 , be the static and kinetic friction coefficients between the crate and the ground, respectively, consider a case in which the force 𝑃 (𝑡) exerted by the person increases linearly with time; i.e., 𝑃 (𝑡) = 𝑃0 𝑡. For this case, determine the time at which the motion begins and the friction force as a function of time.

SOLUTION Road Map & Modeling

We are given one of the forces applied to the crate as a function of time. To determine the time at which the motion starts, we need to determine when the horizontal component of the applied force 𝑃 overcomes the maximum friction force allowed by the static friction coefficient, which is done by applying Newton’s second law. Referring to Fig. 2, we model the crate as a particle under the action of gravity 𝑚𝑔, the friction force 𝐹 , the normal force 𝑁 between the crate and the ground, and the force 𝑃 (𝑡) applied by the person. The force 𝑃 (𝑡) has been plotted as a function of time in Fig. 3. For the sake of generality, the direction of the applied force 𝑃 (𝑡) is assumed to form an angle 𝜃 with the horizontal direction. Governing Equations

Figure 2 FBD of the crate. 𝑃 (𝑡)

Balance Principles

𝑃0 𝑡 1

𝑃0

ISTUDY

(1) (2)

The force law for friction will depend on the value of 𝐹 . If there were no friction, the crate would move to the right, and so the force 𝐹 must point to the left, as indicated on the FBD in Fig. 2. In addition, 𝐹 ≤ 𝜇𝑠 𝑁 when the crate is not moving, and 𝐹 = 𝜇𝑘 𝑁 after the crate starts moving. As long as the component of 𝑃 (𝑡) propelling the crate forward is less than the maximum possible static friction, the crate will not move, that is, as long as 𝑃 (𝑡) cos 𝜃 ≤ 𝜇𝑠 𝑁. Once 𝑃 (𝑡) cos 𝜃 > 𝜇𝑠 𝑁, then the crate starts to move, kinetic friction comes into play, and the friction force immediately drops to 𝜇𝑘 𝑁 (since 𝜇𝑘 < 𝜇𝑠 ). Therefore, to compute the time at which the motion starts, the force law to use is 𝐹 = 𝜇𝑠 𝑁. (3) Force Laws

𝑡 Figure 3 Plot of 𝑃 (𝑡) versus time.

Based on the FBD in Fig. 2, Newton’s second law gives ∑ 𝐹𝑥∶ 𝑃0 𝑡 cos 𝜃 − 𝐹 = 𝑚𝑎𝑥 , ∑ 𝐹𝑦∶ 𝑁 − 𝑃0 𝑡 sin 𝜃 − 𝑚𝑔 = 𝑚𝑎𝑦 .

Once the crate starts moving, this equation will need to be replaced by 𝐹 = 𝜇𝑘 𝑁.

(4)

The force law describing the weight of the crate is elementary and has been indicated directly on the FBD as 𝑚𝑔. We do not have an explicit force law for 𝑁 since its value is determined by the fact that the crate cannot move in the 𝑦 direction. Kinematic Equations

Since the crate does not move in the 𝑦 direction, we have 𝑎𝑦 = 0.

(5)

In the 𝑥 direction, before the motion starts, we have 𝑎𝑥 = 0. Once the motion starts, 𝑎𝑥 becomes an unknown.

(6)

ISTUDY

Section 13.1

Computation

801

Rectilinear Motion Substituting Eq. (5) into Eq. (2) and solving for 𝑁, we obtain 𝑁 = 𝑃0 𝑡 sin 𝜃 + 𝑚𝑔.

(7)

Recalling that 𝑎𝑥 = 0 until the motion begins, we solve Eq. (1) for 𝐹 and set the result equal to 𝜇𝑠 𝑁, where 𝑁 comes from Eq. (7). This yields ( ) 𝜇𝑠 𝑃0 𝑡𝑠 sin 𝜃 + 𝑚𝑔 = 𝑃0 𝑡𝑠 cos 𝜃, (8) where 𝑡𝑠 is the time at which motion begins. Solving for 𝑡𝑠 in Eq. (8), we obtain 𝑡𝑠 =

) 𝜇𝑠 𝑚𝑔 ( . 𝑃0 cos 𝜃 − 𝜇𝑠 sin 𝜃

(9)

Once 𝑡 = 𝑡𝑠 , the crate starts to move, and as indicated in Eq. (4), 𝐹 immediately drops from 𝜇𝑠 𝑁 to 𝜇𝑘 𝑁. Also, for 𝑡 > 𝑡𝑠 , 𝑎𝑥 is no longer known. Therefore, for 𝑡 > 𝑡𝑠 , we have four equations—Eqs. (1), (2), (4), and (5)—to solve for the four unknowns 𝑎𝑥 , 𝑎𝑦 , 𝐹 , and 𝑁. Solving, we obtain ⎫ 𝑃0 𝑡 ( ) cos 𝜃 − 𝜇𝑘 sin 𝜃 − 𝜇𝑘 𝑔,⎪ 𝑚 ⎪ ⎪ 𝑎𝑦 = 0, ⎬ for 𝑡 > 𝑡𝑠 . ( ) ⎪ 𝐹 = 𝜇𝑘 𝑚𝑔 + 𝑃0 𝑡 sin 𝜃 , ⎪ ⎪ 𝑁 = 𝑚𝑔 + 𝑃0 𝑡 sin 𝜃. ⎭

90◦

𝑎𝑥 =

+∞

60◦

(10)

𝑃0 𝑚𝑔

𝜃 30◦

0◦

Observing that the quantity 𝑃0 has dimensions of force per unit time, we can easily verify that the dimensions of the results in Eqs. (9) and (10) are correct.

Discussion & Verification

As we have just seen, Eq. (10) gives us 𝐹 for 𝑡 > 𝑡𝑠 . A plot of 𝐹 vs. 𝑡 for some particular values of the system parameters is shown in Fig. 5. Notice the discontinuity in the friction force when the crate starts to slip at 𝑡 = 𝑡𝑠 . This is a general feature of Coulomb friction and will always happen as long as 𝜇𝑠 ≠ 𝜇𝑘 . Finally, note that 𝐹 keeps increasing with time after the crate starts moving because the value of 𝑁 increases to balance the corresponding increasing value of the vertical component of 𝑃 (𝑡).

0

0

0.5

1

𝜇𝑠

1.5

2

2.5

Figure 4 A contour plot of the term in parentheses in Eq. (9). Note that combinations of 𝜃 and 𝜇𝑠 lying in the gray area above the yellow curve correspond to values for which the crate will never move.

800 𝐹 (N)

This example demonstrates several features of the Coulomb friction model. First, Eq. (9) tells us that the crate will never move if cos 𝜃 ≤ 𝜇𝑠 sin 𝜃 (mathematically, the denominator becomes negative; physically, the applied force is too vertical to initiate motion). Also, the closer cos 𝜃 is to 𝜇𝑠 sin 𝜃, the closer the denominator in Eq. (9) is to zero and the longer it takes to get the crate moving. An interesting representation of Eq. (9) can be found in Fig. 4, which shows 𝑡𝑠 as a function of 𝜃 and 𝜇𝑠 . The yellow curve at the edge of the red region represents the 𝜃 and 𝜇𝑠 values for which the denominator of Eq. (9) is zero. Since the crate only moves for positive values of 𝑡𝑠 , no matter how long we wait or how hard we push, the crate will never move for any 𝜃 and 𝜇𝑠 values above and to the right of the yellow curve. As we approach the yellow curve from below, it takes longer and longer to get the crate moving. For example, as 𝜃 approaches 90◦ , it becomes impossible to move the crate no matter how small 𝜇𝑠 is, unless 𝜇𝑠 = 0. Since friction is the key ingredient in this problem, it is illustrative to plot the friction force 𝐹 as a function of time. Until the crate starts to move, Eq. (1) gives 𝐹 as 𝐹 = 𝑃0 𝑡 cos 𝜃, for 𝑡 < 𝑡𝑠 . (11) A Closer Look

𝑡𝑠

600 400 𝑡 = 𝑡𝑠

200 0

0

2

4

6

8

10

𝑡(s) Figure 5 Friction 𝐹 vs. time 𝑡. The parameters used were 𝑚 = 50 kg, 𝑔 = 9.81 m∕s2 , 𝑃0 = 200 N∕s, 𝜃 = 40◦ , 𝜇𝑠 = 0.6, and 𝜇𝑘 = 0.45.

802

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.5

Motion Under the Action of Spring Forces

4 mph 𝑥

Figure 1 Railcar running into a large spring.

A 60 ton railcar and its cargo, a 27 ton trailer, are moving to the right at 4 mph, as shown in Fig. 1, when they encounter a large linear spring that has been designed to stop a 60 ton railcar moving at 5 mph in a distance of 3 f t when it is initially uncompressed. If the trailer does not slip relative to the railcar and if the spring is initially uncompressed, determine (a) how much the spring compresses in stopping the 87 ton loaded railcar and (b) how long it takes for the spring to stop the railcar.

SOLUTION 𝚥̂

Road Map & Modeling

𝚤̂ 𝑚𝑔 𝐹𝑠 𝑁 Figure 2 FBD of the railcar shown in Fig. 1, where 𝑚 is either the mass of the railcar or the railcar and trailer.

ISTUDY

The given information consists of (1) the conditions under which the railcar and its cargo impact the spring and (2) the spring’s design criterion. We need to compute the stopping distance and time of the railcar and its cargo, and so “railcar and cargo” is the system to analyze. We will model this system as a particle, and its FBD is shown in Fig. 2. We have assumed that the only relevant forces are gravity, the reaction with the rails, and the spring force. Note that we are not given direct information about the spring constant or its unstretched length. However, using the stated design criteria in conjunction with Newton’s second law, we will be able to first determine the spring’s force law and then use the result to compute the system’s acceleration. Once we have the system’s acceleration, we will need to apply kinematics to obtain the corresponding stopping time and distance information. Governing Equations Balance Principles

Based on the FBD in Fig. 2, Newton’s second law gives ∑ 𝐹𝑥∶ −𝐹𝑠 = 𝑚𝑎𝑥 , ∑ 𝐹𝑦∶ 𝑁 − 𝑚𝑔 = 𝑚𝑎𝑦 .

(1) (2)

While the spring constant 𝑘 is unknown, the spring’s force law can still be given the form 𝐹𝑠 = 𝑘𝑥, (3)

Force Laws

where 𝑥 is measured from the uncompressed end position of the spring. Kinematic Equations

The kinematic relations are 𝑎𝑥 = 𝑥̈

and 𝑎𝑦 = 0,

(4)

where we have chosen to represent 𝑎𝑥 as 𝑥̈ because the problem is asking us to relate acceleration to position, and where the second of Eqs. (4) states that there is no motion in the vertical direction. Since there is no motion in the 𝑦 direction, we will disregard the equations in the vertical direction. Next, substituting the first of Eqs. (4) and Eq. (3) into Eq. (1) and rearranging, we obtain 𝑘 (5) 𝑥̈ + 𝑥 = 0. 𝑚 Rearranging Eq. (5) and making use of the chain rule, we obtain

Computation

𝑥̈ = 𝑥̇

𝑑𝑥̇ 𝑘 =− 𝑥 𝑑𝑥 𝑚

⇒

𝑘 𝑥̇ 𝑑𝑥̇ = − 𝑥 𝑑𝑥, 𝑚

(6)

which can be integrated as ∫𝑣

𝑖

𝑥

𝑣

𝑥̇ 𝑑𝑥̇ = −

𝑘 𝑥 𝑑𝑥 ∫𝑥 𝑚 𝑖

⇒

𝑣2 − 𝑣2𝑖 = −

) 𝑘( 2 𝑥 − 𝑥2𝑖 , 𝑚

(7)

ISTUDY

Section 13.1

Rectilinear Motion

803

where 𝑣 is the speed of the railcar at the location 𝑥 and where the subscript 𝑖 stands for initial. Letting 𝑣𝑓 be the speed corresponding to the final position 𝑥𝑓 , solving Eq. (7) for 𝑘, and substituting in numbers corresponding to the spring’s design criteria, we obtain ( ) 𝑚𝑟 𝑣2𝑓 − 𝑣2𝑖 𝑘= = 22,270 lb∕f t, (8) 𝑥2𝑖 − 𝑥2𝑓 where we have used the unit conversions 5 mph = 7.333 f t∕s and 1 ton = 2000 lb, 𝑚𝑟 is the mass of the railcar, 𝑣𝑓 = 0, 𝑥𝑖 = 0, and 𝑥𝑓 = 3 f t. Part (a)

Now that we have 𝑘, we can determine how far the spring compresses with the loaded railcar by applying Eq. (7) again. Letting 𝑣 = 𝑣𝑓 for 𝑥 = 𝑥𝑓 , solving Eq. (7) for 𝑥𝑓 , and substituting in the numbers for the loaded railcar, we obtain √ 𝑥𝑓 =

𝑥2𝑖 −

𝑚𝑡 ( 2 ) 𝑣𝑓 − 𝑣2𝑖 = 2.890 f t, 𝑘

(9)

where we have used the unit conversion 4 mph = 5.867 f t∕s and 𝑚𝑡 is the total mass of the railcar and the trailer. Part (b)

To determine the stopping time 𝑡𝑓 , we go back to Eq. (7), solve it for 𝑣, and rearrange the result to integrate it with respect to time, i.e., √ 𝑡𝑓 𝑥𝑓 ) 𝑑𝑥 𝑘 ( 2 𝑑𝑥 = 𝑣= 𝑑𝑡. (10) ⇒ = 𝑣2𝑖 − 𝑥 − 𝑥2𝑖 √ ( ) ∫𝑡 ∫𝑥 𝑑𝑡 𝑚𝑡 𝑖 𝑖 𝑣2𝑖 − 𝑚𝑘 𝑥2 − 𝑥2𝑖 𝑡

Using a table of integrals, we obtain √ 𝑡𝑓 − 𝑡𝑖 =

√ 𝑘 ⎛ ⎞ ⎡ 𝑥 𝑚𝑡 ⎢ −1 ⎜ 𝑚𝑡 𝑓 ⎟ sin ⎜√ ⎟ 𝑘 ⎢⎢ ⎜ 𝑣2 + 𝑘𝑥2 ∕𝑚 ⎟ 𝑡 𝑖 𝑖 ⎣ ⎝ ⎠ √ 𝑘 ⎞⎤ ⎛ 𝑥 𝑚𝑡 𝑖 ⎟⎥ −1 ⎜ − sin ⎜ √ ⎟⎥ ⎜ 𝑣2 + 𝑘𝑥2 ∕𝑚 ⎟⎥ 𝑡 𝑖 𝑖 ⎠⎦ ⎝

The integral in Eq. (10). In computing the integral in Eq. (10), it might help to notice that if we let √ √ 𝑘 2 𝑘 2 𝑥, 𝑎 = 𝑣𝑖 + 𝑥𝑖 and 𝑢 = 𝑚𝑡 𝑚𝑡 then Eq. (10), with lower limits set equal to zero, becomes

⇒

𝑡𝑓 = 0.7738 s,

∫0

(11)

where we substituted in 𝑥𝑓 = 2.890 f t, 𝑣𝑖 = 5.867 f t∕s, 𝑘 = 22,270 lb∕f t, 𝑚𝑡 = (174,000∕ 32.2) slug, 𝑡𝑖 = 0 s, and 𝑥𝑖 = 0 f t. Discussion & Verification

Helpful Information

The result in Eq. (9) tells us that a 174,000 lb loaded railcar traveling at 4 mph compresses the spring a little less than 3 f t. This result seems reasonable. Although the loaded car is 45% heavier than when it is empty, the spring compression should have not been expected to necessarily increase because the loaded car is moving with a speed that is only 80% of the design speed for an impact between an unloaded railcar and the bumper. Clearly the increase in weight is larger than the decrease in speed; however, the spring compression depends on the square of the speed, i.e., in this problem, speed is a more significant factor than weight.

√

𝑡𝑓

𝑑𝑡 =

√

𝑚𝑡 𝑘 ∫0

𝑘 𝑚𝑡

𝑥𝑓

𝑑𝑢 , √ 𝑎2 − 𝑢2

or √

√

( )| 𝑚 𝑥𝑓 𝑚𝑡 𝑢 | 𝑡 sin−1 𝑡𝑓 = 𝑘 𝑎 ||0 √ 𝑘 ⎛ ⎞ √ 𝑥 𝑚𝑡 𝑚𝑡 𝑓 ⎜ ⎟ −1 = sin ⎜ √ ⎟. 𝑘 𝑘 ⎜ 𝑣2𝑖 + 𝑥2𝑖 ⎟ 𝑚 ⎝ ⎠ 𝑡 𝑘

804

Chapter 13

Force and Acceleration Methods for Particles

Problems Problem 13.1

Finish

𝑑 𝐵

𝐴 Start 𝐹

𝐹

Two curling stones 𝐴 and 𝐵, with masses 𝑚 and 4𝑚, respectively, and initially at rest on the start line, are pushed by two identical forces 𝐹 over the distance 𝑑. Which stone arrives first to the finish line?

Problem 13.2 An object is lowered very slowly onto a conveyor belt that is moving to the right. What is the direction of the friction force acting on the object at the instant the object touches the belt?

Figure P13.1 𝑣

Figure P13.2

Problem 13.3 A person is trying to move a heavy crate by pushing on it. While the person is pushing, what is the resultant force acting on the crate if the crate does not move?

Figure P13.3

ISTUDY

Problem 13.4 A person is lifting a 75 lb crate 𝐴 by applying a constant force 𝑃 = 40 lb to the pulley system shown. Neglecting friction and the inertia of the pulleys, determine the acceleration of the crate. Treat all rope segments as purely vertical.

motor 𝑀

𝑃

cargo 𝐶

𝐴 Figure P13.4

Figure P13.5

Problem 13.5 The motor 𝑀 is at rest when someone flips a switch and it starts pulling in the rope. The acceleration of the rope is uniform and is such that it takes 1 s to achieve a retraction rate of 4 f t∕s. After 1 s the retraction rate becomes constant. Determine the tension in the rope during and after the initial 1 s interval. The cargo 𝐶 weighs 130 lb, the weight of the ropes and pulleys is negligible, and friction in the pulleys is negligible.

ISTUDY

Section 13.1

Rectilinear Motion

Problems 13.6 and 13.7 A crate of weight 𝑊 = 550 lb has been attached to a pickup truck by a rope whose tensile strength is 𝑇max = 350 lb. If the truck and crate start at rest with 𝜃 = 30◦ , determine the maximum acceleration of the truck such that the rope does not break. 𝑊

𝜃

Problem 13.6

Determine the solution for the case in which friction between the crate and the ground is negligible.

Problem 13.7

Determine the solution for the case in which friction between the crate and the ground is not negligible and 𝜇𝑠 = 0.4 and 𝜇𝑘 = 0.25.

Figure P13.6 and P13.7

Problem 13.8 The crate 𝐴 of mass 𝑚 and the wedge 𝐵 on which it rests are both initially at rest. The wedge, whose face is inclined at 𝜃 = 30◦ with the horizontal, is given an acceleration 𝑎𝐵 to the left as shown. Given that the coefficient of static friction between the crate and the wedge is 𝜇𝑠 = 0.6, determine the maximum value of 𝑎𝐵 such that the crate does not slip on the wedge. 𝜇𝑠

𝐴 𝑎𝐵

𝐴

𝜇𝑠 𝜃

𝐵

Figure P13.8

𝐵

𝑎𝐵

𝜃

Figure P13.9

Problem 13.9 The crate 𝐴 of mass 𝑚 and the wedge 𝐵 on which it rests are both initially at rest. The wedge, whose face is inclined at 𝜃 = 30◦ with the horizontal, is given an acceleration 𝑎𝐵 to the right as shown. Given that the coefficient of static friction between the crate and the wedge is 𝜇𝑠 = 0.6, determine the maximum value of 𝑎𝐵 such that the crate does not slip on the wedge.

Problem 13.10 A hammer hits a mass 𝑚 on the end of a metal bar. In Chapter 15, we will see that this imparts an instantaneous initial velocity 𝑣0 at 𝑥 = 0 to the mass. Treating the bar as a massless spring, determine the equation of motion of the mass 𝑚. The equivalent spring constant of a bar in compression is given by 𝑘eq = 𝐸𝐴∕𝐿, where 𝐸 is Young’s modulus of the bar, 𝐴 is the cross-sectional area of the bar, and 𝐿 is the length of the bar.∗

𝑥 𝑑

Problem 13.11 For the mass described in Prob. 13.10: (a) Integrate the equation of motion to determine the speed of the mass 𝑣(𝑥) as a function of 𝑥. (b) Use the result found in Part (a) to obtain the position of the mass as a function of time 𝑥(𝑡) from the initial time up until the mass stops for the first time. (√ )−1∕2 ) √ ( 𝑑𝑥 = sin−1 𝑏𝑥∕𝑎 ∕ 𝑏. Hint: ∫ 𝑎2 − 𝑏𝑥2 ∗ From

strength of materials, 𝜎 = 𝐹 ∕𝐴 = 𝐸𝜖 = 𝐸Δ𝐿∕𝐿. Therefore, 𝐹 = (𝐸𝐴∕𝐿)Δ𝐿 = (𝐸𝐴∕𝐿)𝑥. Since 𝐹 = 𝑘eq 𝑥 for this system, we see that 𝑘eq = 𝐸𝐴∕𝐿.

𝑚 hammer Figure P13.10 and P13.11

𝐿

805

806

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.12 The crate 𝐴 of mass 𝑚 and the wedge 𝐵 on which it rests are moving together down the incline with the acceleration 𝑎𝐵 as shown. The angle of the incline is 𝜃 = 30◦ with respect to the horizontal. Given that the coefficient of static friction between the crate and the wedge is 𝜇𝑠 = 0.6, determine the maximum value of 𝑎𝐵 before the crate starts to slip on the wedge. 𝐴

𝑎𝐵

𝜇𝑠

𝜃

𝐴

𝜇𝑠

𝜃 𝑎𝐵 𝐵

𝐵

Figure P13.12

Figure P13.13

Problem 13.13 The crate 𝐴 of mass 𝑚 and the wedge 𝐵 on which it rests are moving together up the incline with the acceleration 𝑎𝐵 as shown. The angle of the incline is 𝜃 = 30◦ with respect to the horizontal. Given that the coefficient of static friction between the crate and the wedge is 𝜇𝑠 = 0.6, determine the maximum value of 𝑎𝐵 before the crate starts to slip on the wedge. 𝐴 𝓁 𝜃 𝐵

A suitcase is released from rest at 𝐴 on the 𝜃 = 30◦ ramp. It slides a distance 𝓁 = 25 f t and then goes over the edge at 𝐵 and drops a height ℎ = 5 f t. Determine the horizontal distance 𝑑 to the landing spot at 𝐶. Problem 13.14

Assume that friction on the incline between 𝐴 and 𝐵 is negligible.

ℎ

𝐶 𝑑

Figure P13.14 and P13.15

ISTUDY

Problems 13.14 and 13.15

Problem 13.15 Assume that the coefficient of static friction is insufficient to prevent slipping and that the coefficient of kinetic friction on the incline between 𝐴 and 𝐵 is 𝜇𝑘 = 0.3.

Problems 13.16 and 13.17 A vehicle is stuck on the railroad tracks as a 430,000 lb locomotive is approaching with a speed of 75 mph. As soon as the problem is detected, the locomotive’s emergency brakes are activated, locking the wheels and causing the locomotive to slide. 𝑑 Figure P13.16 and P13.17 Problem 13.16

If the coefficient of kinetic friction between the locomotive and the track is 0.45, what is the minimum distance 𝑑 at which the brakes must be applied to avoid a collision? What would that distance be if instead of a locomotive, there was a 30×106 lb train? Treat the locomotive and the train as particles, assume that the railroad tracks are rectilinear and horizontal, and note that only the locomotive’s brakes are applied. Problem 13.17

locomotive.

Continue Prob. 13.16 and determine the time required to stop the

ISTUDY

Section 13.1

Rectilinear Motion

Problem 13.18 As the skydiver moves downward with a speed 𝑣, the air drag exerted by the parachute on the skydiver has a magnitude 𝐹𝑑 = 𝐶𝑑 𝑣2 (𝐶𝑑 is a drag coefficient) and a direction opposite to the direction of motion. Determine the expression of the skydiver’s acceleration in terms of 𝐶𝑑 , 𝑣, the mass of the skydiver 𝑚, and the acceleration due to gravity.

Problem 13.19 A car is driving down a 23◦ rough incline at 55 km∕h when its brakes are applied. Treating the car as a particle and neglecting all forces except gravity and friction, determine the stopping distance if (a) The tires slide and the coefficient of kinetic friction between the tires and the road is 0.7.

Figure P13.18

(b) The car is equipped with antilock brakes and the tires do not slide. Use 0.9 for the coefficient of static friction between the tires and the road.

23◦ Figure P13.19

Problems 13.20 through 13.22 The truck shown is traveling at 𝑣0 = 60 mph when the driver applies the brakes to come to a stop. The deceleration of the truck is constant, and the truck comes to a complete stop after braking for a distance of 350 f t. Treat the crate as a particle so that tipping can be neglected.

𝑣0 𝑑 𝐴

Problem 13.20 Determine the minimum coefficient of static friction between the crate 𝐴 and the truck so that the crate does not slide relative to the truck. Problem 13.21 If the coefficient of kinetic friction between the crate 𝐴 and the bed of the truck is 0.3 and static friction is not sufficient to prevent slip, determine the minimum distance 𝑑 between the crate and the truck 𝐵 so that the crate never hits the truck at 𝐵.

Figure P13.20–P13.22

If the coefficient of kinetic friction between the crate 𝐴 and the bed of the truck is 0.3, static friction is not sufficient to prevent slip, and the distance 𝑑 from the front of the crate to the truck at 𝐵 is 10 f t, determine the speed relative to the truck with which the crate strikes the truck at 𝐵.

Problem 13.22

Problem 13.23 A metal ball with mass 𝑚 = 0.15 kg is dropped from rest into a fluid. The magnitude of the resistance due to the fluid is given by 𝐶𝑑 𝑣, where 𝐶𝑑 is a drag coefficient and 𝑣 is the ball’s speed. If 𝐶𝑑 = 2.1 kg∕s, determine the ball’s speed 4 s after release.

Problem 13.24 A metal ball weighing 0.35 lb is dropped from rest into a fluid. The magnitude of the resistance due to the fluid is given by 𝐶𝑑 𝑣, where 𝐶𝑑 is a drag coefficient and 𝑣 is the ball’s speed. It is observed that 2 s after release, the speed of the ball is 25 f t∕s. Determine the value of 𝐶𝑑 .

Figure P13.23 and P13.24

𝐵

807

808

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.25 A horse is lifting a 500 lb crate by moving to the right at a constant speed 𝑣0 = 3 f t∕s. Observing that 𝐵 is fixed and letting ℎ = 6 f t and 𝓁 = 14 f t, determine the tension in the rope when the horizontal distance 𝑑 between 𝐵 and point 𝐴 on the horse is 10 f t. Treat all rope segments between 𝐵 and 𝐶 as vertical, and ignore the change in the amount of rope that wraps around the pulley at 𝐵. 𝑑

𝐵

𝑣0

𝐶 𝐴

𝓁 ℎ

Figure P13.25

Problem 13.26 𝐵

𝐴

The centers of two spheres 𝐴 and 𝐵 with masses 𝑚𝐴 = 1 kg and 𝑚𝐵 = 2 kg are a distance 𝑟0 = 1 m apart. 𝐵 is fixed in space, and 𝐴 is initially at rest. Using Eq. (11.5) on p. 621, which is Newton’s universal law of gravitation, determine the speed with which 𝐴 impacts 𝐵 if the radii of the two spheres are 𝑟𝐴 = 0.05 m and 𝑟𝐵 = 0.15 m. Assume that the two masses are infinitely far from any other mass so that they are only influenced by their mutual attraction.

𝑟0 𝑟 Figure P13.26

Problems 13.27 and 13.28 Spring scales work by measuring the displacement of a spring that supports both the platform and the object, of mass 𝑚, whose weight is being measured. Neglect the mass of the platform on which the mass sits and assume that the spring is uncompressed before the mass is placed on the platform. In addition, assume that the spring is linear elastic with spring constant 𝑘.

𝑚 platform

Problem 13.27 If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the maximum reading on the scale after the mass is released. Figure P13.27 and P13.28 Problem 13.28 If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the maximum speed attained by the mass 𝑚 as the spring compresses.

weight

Problem 13.29

felt pad

Figure P13.29

ISTUDY

felt pad

A scale is to be used on a wood countertop made from some very nice Brazilian cherry. To protect the countertop, the owner attaches self-sticking felt pads to the feet of the scale. When the weight was placed on the scale before the felt pads were applied, the scale read a certain value. Will the value be higher, lower, or the same when the same weight is placed on the scale but with the felt pads between the scale and the countertop? Ignore the transient dynamic effects that occur immediately after the weight is placed on the scale.

ISTUDY

Section 13.1

Rectilinear Motion

Problems 13.30 through 13.32 Car bumpers are designed to limit the extent of damage to the car in the case of lowvelocity collisions. Consider a 3300 lb passenger car impacting a concrete barrier while traveling at a speed of 4.0 mph. Model the car as a particle, and consider two types of bumpers: (1) a simple linear spring with constant 𝑘 and (2) a linear spring of constant 𝑘 in parallel with a shock-absorbing unit generating a nearly constant force of 700 lb over 0.25 f t.

𝑘

If the bumper is of type 1 and if 𝑘 = 6500 lb∕f t, find the spring compression necessary to stop the car.

(1)

If the bumper is of type 1, find the value of 𝑘 necessary to stop the car when the bumper is compressed 0.25 f t.

(2)

Problem 13.30

𝐹𝑆

Problem 13.31

If the bumper is of type 2, find the value of 𝑘 necessary to stop the car when the bumper is compressed 0.25 f t.

Problem 13.32

𝑘 Figure P13.30–P13.32

Problem 13.33 What would it mean for the static or kinetic friction coefficients to be negative? Is this possible? Can either the static or kinetic friction coefficients be greater than 1? If yes, explain and give an example.

Problem 13.34 Packages for transporting delicate items (e.g., a laptop or glass) are designed to “absorb” some of the energy of the impact in order to protect the contents. These energy absorbers can get pretty complicated to model (e.g., the mechanics of styrofoam peanuts is not easy), but we can begin to understand how they work by modeling them as a linear elastic spring of constant 𝑘 that is placed between the contents (an expensive vase) of mass 𝑚 and the package 𝑃 . Assuming that 𝑚 = 3 kg and that the box is dropped from a height of 1.5 m, determine the magnitude of the maximum displacement of the vase relative to the box, as well as the magnitude of the maximum force exerted on the vase by the packaging if 𝑘 = 3500 N∕m. Treat the vase as a particle, and neglect all forces except for gravity and the spring force. Hint: Divide the transient into two stages: (1) In the first stage, before impact with the floor, the vase and packaging move as one and there is no restoring force exerted on the vase, and (2) at impact with the floor, the vase moves relative to the packaging and the restoring force is present.

𝑚

𝑘 𝑃 Figure P13.34

Problems 13.35 and 13.36 Car bumpers are designed to limit the extent of damage to the car in the case of lowvelocity collisions. Consider a passenger car impacting a concrete barrier while traveling at a speed of 4.0 mph. Model the car as a particle of mass 𝑚, and assume that the bumper has a spring element in parallel with a shock absorber so that the overall force exerted by the bumper is 𝐹𝐵 = 𝑘𝛿 + 𝜂 𝛿,̇ where 𝑘, 𝛿, and 𝜂 denote the spring constant, the spring compression, and the bumper damping coefficient, respectively. Problem 13.35

𝜂

Derive the equations of motion for the car during the collision. 𝑘

Let the weight of the car be 3300 lb, 𝑘 = 6500 lb∕f t, and 𝜂 = 300 lb⋅s∕f t, and let the car be traveling at 4.0 mph at impact. Determine the maximum compression of the bumper necessary to bring the car to a stop. Also determine the time required to stop the car. Problem 13.36

Figure P13.35 and P13.36

809

810

Chapter 13

Force and Acceleration Methods for Particles Problems 13.37 through 13.39 𝑣𝑖 𝑥

A railcar with an overall mass of 75,000 kg traveling with a speed 𝑣𝑖 is approaching a barrier equipped with a bumper consisting of a nonlinear spring whose force vs. compression law is given by 𝐹𝑠 = 𝛽𝑥3 , where 𝛽 = 640×106 N∕m3 and 𝑥 is the compression of the bumper. Problem 13.37

Treating the system as a particle and assuming that the contact between the railcar and rails is frictionless, determine the maximum value of 𝑣𝑖 so that the compression of the bumper is limited to 20 cm.

Figure P13.37–P13.39

Problem 13.38 Treating the system as a particle, assuming that the contact between railcar and rails is frictionless, and letting 𝑣𝑖 = 6 km∕h, determine the bumper compression necessary to bring the railcar to a stop. Problem 13.39 Treating the system as a particle, assuming that the contact between the railcar and rails is frictionless, and letting 𝑣𝑖 = 6 km∕h, determine how long it takes for the bumper to bring the railcar to a stop.

Problem 13.40 A 6 lb collar is constrained to travel along a rectilinear and frictionless bar of length 𝐿 = 5 f t. The springs attached to the collar are identical and are unstretched when the collar is at 𝐵. Treating the collar as a particle, neglecting air resistance, and knowing that at 𝐴 the collar is moving to the right with a speed of 11 f t∕s, determine the linear spring constant 𝑘 so that the collar reaches 𝐷 with zero speed. Points 𝐸 and 𝐹 are fixed. 𝐿∕4

𝐿∕2 𝐴

𝐷

𝐵

𝑘

𝑘

𝐸 𝐿

𝐿∕2

𝐹

Figure P13.40 and P13.41 𝐿∕2

Problem 13.41 𝐹

𝐷 𝑘

𝐿∕2 𝐵

𝐿

A 10 lb collar is constrained to travel along a rectilinear and frictionless bar of length 𝐿 = 5 f t. The springs attached to the collar are identical, they have a spring constant 𝑘 = 4 lb∕f t, and they are unstretched when the collar is at 𝐵. Treating the collar as a particle, neglecting air resistance, and knowing that at 𝐴 the collar is moving to the right with a speed of 14 f t∕s, determine the speed with which the collar arrives at 𝐷. Points 𝐸 and 𝐹 are fixed.

Problem 13.42 𝐴

𝑘

𝐿∕4 𝐸 Figure P13.42

ISTUDY

An 11 kg collar is constrained to travel along a rectilinear and frictionless bar of length 𝐿 = 2 m. The springs attached to the collar are identical, and they are unstretched when the collar is at 𝐵. Treating the collar as a particle, neglecting air resistance, and knowing that at 𝐴 the collar is moving upward with a speed of 23 m∕s, determine the linear spring constant 𝑘 so that the collar reaches 𝐷 with zero speed. Points 𝐸 and 𝐹 are fixed.

ISTUDY

Section 13.1

811

Rectilinear Motion

Problem 13.43 Derive the equation of motion of the mass 𝑚 released from rest at 𝑥 = 𝑥0 from the slingshot-like device (the mass 𝑚 is attached to the elastic cords). Assume that the cords connecting the mass to the device are linear springs with spring constant 𝑘 and unstretched length 𝐿0 . In addition, assume that the mass is equidistant from the two supports and that the mass and both springs lie in the 𝑥𝑦 plane. Ignore gravity and assume 𝐿 > 𝐿0 .

𝑥0

elastic springs

𝑧

𝑚 𝑦

𝑥

Problem 13.44 Determine the speed of the mass 𝑚 when it reaches 𝑥 = 0 if it is released from rest at 𝑥 = 𝑥0 from the device in Prob. 13.43. Assume that the cords connecting the mass to the device are linear springs with spring constant 𝑘 and unstretched length 𝐿0 . Ignore gravity and assume 𝐿 > 𝐿0 .

2𝐿

Figure P13.43–P13.46

Problem 13.45 Given the approximation (𝑥∕𝐿)2 1 ≈1− , √ 2 (𝑥∕𝐿)2 + 1

(1)

show that the equation of motion for the mass 𝑚 when it is released from rest at 𝑥 = 𝑥0 from the device in Prob. 13.43 can be written as ) ( 𝑥̈ + 𝜔20 𝑥 1 + Λ𝑥2 = 0, (2) (

where 2𝑘 𝜔20 = 𝑚

𝐿 − 𝐿0 𝐿

) and Λ =

𝐿0 ( ). 2𝐿2 𝐿 − 𝐿0

(3)

Assume that the cords connecting the mass to the device are linear springs with spring constant 𝑘, length 𝐿, and unstretched length 𝐿0 . Ignore gravity and assume 𝐿 > 𝐿0 . Equation (2) is a famous equation in mechanics called Duffing’s equation.

Problem 13.46

Force for device with 𝐿 = 2 and 𝐿0 = 1. Force for device with 𝐿 = 1 and 𝐿0 = 2.

1 0.5 𝐹𝑒 ∕𝑘

The force on the mass for the device in Prob. 13.43 and the force on the mass in the Duffing equation obtained from the device equation of motion [and defined by Eqs. (2) and (3)] can be plotted as a function of 𝑥. The nature of that force depends on whether or not the springs are initially stretched (𝐿 > 𝐿0 ) or initially compressed (𝐿 < 𝐿0 ). The figure shows the elastic restoring force on the mass 𝑚 as a function of the displacement 𝑥 for four different cases:

0 −0.5 −1 −1

Force for the Duffing equation with 𝐿 = 2 and 𝐿0 = 1. Force for the Duffing equation with 𝐿 = 1 and 𝐿0 = 2. For small 𝑥, the force given by the Duffing equation is a good approximation to the force in the device. Explain which of the curves corresponds to a “hardening” spring (a spring that gets stiffer as you pull it) and which corresponds to a “softening” spring (a spring that gets less stiff as you pull it), and explain physically why we see this behavior.

Figure P13.46

−0.5

0 𝑥

0.5

1

812

Chapter 13

Force and Acceleration Methods for Particles

13.2

Curvilinear Motion

This section builds on what we learned in Section 13.1 by applying Newton’s second law to systems undergoing curvilinear motion. No new material is introduced, thus allowing us to strengthen our modeling and problem-solving skills. The application of Newton’s second law to curvilinear problems is formally identical to what we saw in Section 13.1 with the same four-step problem-solving procedure.

Newton’s second law in 2D and 3D component systems

Jonathan Ferrey/Getty Images

Figure 13.8 Marco Andretti at the 2007 Indianapolis 500. When going around a turn, the Formula One cars are approximately in planar motion, but are not traveling in a straight line.

When applying 𝐹⃗ = 𝑚𝑎, ⃗ it is done in component form, using Eq. (13.2) on p. 790, after choosing a convenient component system. For systems in curvilinear motion, we will choose one or more component systems from those described in Chapter 12. We now present Newton’s second law as it appears in each component system studied in Chapter 12. Two-dimensional component systems

𝑦

𝚥̂ 𝑂

Cartesian components. Cartesian components is

𝑎⃗

∑

𝐹⃗

Referring to Fig. 13.9, Newton’s second law in planar

𝐹𝑥 = 𝑚𝑎𝑥

∑

𝐹𝑦 = 𝑚𝑎𝑦 .

(13.15)

𝑃 𝚤̂

𝑥

The associated kinematic equations are given by Eqs. (12.18), which imply

Figure 13.9 Force, acceleration, and a Cartesian component system.

ISTUDY

and

𝑎𝑥 = 𝑥̈ Path components.

and

𝑎𝑦 = 𝑦. ̈

(13.16)

Referring to Fig. 13.10, Newton’s second law in planar path 𝑢̂ 𝑡

𝑃

𝐹⃗

𝑢̂ 𝑛

𝑎⃗ 𝜌

path of 𝑃 𝐶 osculating circle

Figure 13.10. Force, acceleration, and a path component system.

components is ∑

𝐹𝑛 = 𝑚𝑎𝑛

and

∑

𝐹𝑡 = 𝑚𝑎𝑡 .

(13.17)

The associated kinematic equations are given by Eqs. (12.49), which imply 𝑎𝑛 =

𝑣2 𝜌

and

𝑎𝑡 = 𝑣.̇

(13.18)

ISTUDY

Section 13.2

Polar components. components is

Curvilinear Motion

Referring to Fig. 13.11, Newton’s second law in planar polar ∑

𝐹𝑟 = 𝑚𝑎𝑟

and

∑

𝐹𝜃 = 𝑚𝑎𝜃 .

𝑎⃗

𝐹⃗

(13.19) 𝜃

𝑥

𝑂

̇ 𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃.

and

(13.20)

Three-dimensional component systems Cartesian components.

𝑃

𝑟 𝑢̂ 𝑟

𝑢̂ 𝜃

The associated kinematic equations are given by Eqs. (12.73), which are 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2

813

Figure 13.11 Force, acceleration, and a polar component system.

Referring to Fig. 13.12, Newton’s second law in three𝑧 𝐹⃗

𝑘̂ 𝑂

𝑦

𝚥̂

𝚤̂

𝑥

𝑎⃗

𝑃

Figure 13.12. Force, acceleration, and a 3D Cartesian component system.

dimensional Cartesian components is ∑

𝐹𝑥 = 𝑚𝑎𝑥 ,

∑

𝐹𝑦 = 𝑚𝑎𝑦 ,

and

∑

𝐹𝑧 = 𝑚𝑎𝑧 .

(13.21) 𝑧

The associated kinematic equations are given by Eq. (12.99), which imply 𝑎𝑥 = 𝑥, ̈ Path components. path components is ∑

𝑎𝑦 = 𝑦, ̈

and

𝑎𝑧 = 𝑧. ̈

binormal direction

(13.22)

𝑢̂ 𝑡

𝑢̂ 𝑏

𝑃

Referring to Fig. 13.13, Newton’s second law in three-dimensional

𝐹𝑛 = 𝑚𝑎𝑛 ,

∑

𝐹𝑡 = 𝑚𝑎𝑡 ,

and

∑

𝐹⃗

𝐹𝑏 = 𝑚𝑎𝑏 .

(13.23)

The associated kinematic equations are given by Eq. (12.103), which imply 𝑎𝑛 =

𝑣2 𝜌

𝑢̂ 𝑛

𝑎⃗ 𝑥

𝜌 𝐶 path of 𝑃

,

𝑎𝑡 = 𝑣,̇

and

𝑎𝑏 = 0,

𝑦

(13.24) osculating circle

where 𝑎𝑏 = 0 because, by definition, 𝑢̂ 𝑏 is perpendicular to the plane containing the velocity and the acceleration vectors (see Section 12.4). Cylindrical components. Referring to Fig. 13.14, Newton’s second law in threedimensional cylindrical components is ∑

𝐹𝑅 = 𝑚𝑎𝑅 ,

∑

𝐹𝜃 = 𝑚𝑎𝜃 ,

and

∑

𝐹𝑧 = 𝑚𝑎𝑧 .

(13.25)

The associated kinematic equations are given by Eqs. (12.123), which are 𝑎𝑅 = 𝑅̈ − 𝑅𝜃̇ 2 ,

𝑎𝜃 = 𝑅𝜃̈ + 2𝑅̇ 𝜃,̇

and

𝑎𝑧 = 𝑧. ̈

(13.26)

Figure 13.13 Force, acceleration, and a 3D path component system.

814

ISTUDY

Chapter 13

Force and Acceleration Methods for Particles

𝑧

𝑢̂ 𝑧 𝑃

𝑢̂ 𝜃 𝑢̂ 𝑅

𝑟⃗ 𝐹⃗

𝑂

𝑎⃗ 𝑅 𝑦 𝜃

𝑥

Figure 13.14. Force, acceleration, and a cylindrical component system. 𝑧 𝜙

𝐹⃗ 𝑎⃗ 𝑟 = 𝑟⃗ 𝑟⃗

𝑢̂ 𝑟 𝑃 𝑢̂ 𝜃 𝑢̂ 𝜙

𝑂 𝑦

𝑥

𝜃

Figure 13.15. Force, acceleration, and a spherical component system.

Spherical components. Referring to Fig. 13.15, Newton’s second law in threedimensional spherical components is ∑

𝐹𝑟 = 𝑚𝑎𝑟 ,

∑

𝐹𝜙 = 𝑚𝑎𝜙 ,

and

∑

𝐹𝜃 = 𝑚𝑎𝜃 .

(13.27)

The associated kinematic equations are given by Eqs. (12.133), which are 𝑎𝑟 = 𝑟̈ − 𝑟𝜙̇ 2 − 𝑟𝜃̇ 2 sin2 𝜙, 𝑎 = 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 2 sin 𝜙 cos 𝜙, 𝜙

𝑎𝜃 = 𝑟𝜃̈ sin 𝜙 + 2𝑟̇ 𝜃̇ sin 𝜙 + 2𝑟𝜙̇ 𝜃̇ cos 𝜙.

(13.28)

ISTUDY

Section 13.2

Curvilinear Motion

E X A M P L E 13.6

Tension in a Wrecking Ball Cable

The wrecking ball 𝐴 shown in Fig. 1 is released from rest when 𝜃 = 𝜃0 = 30◦ , and it swings freely about the fixed point at 𝑂. Assuming that the weight of the ball is 𝑊 = 2500 lb and 𝐿 = 30 f t, determine the tension in the cable to which the ball is attached when the ball reaches 𝜃 = 0◦ .

𝑂

𝜃

SOLUTION

𝐿

Road Map & Modeling

Modeling the wrecking ball as a particle and neglecting all forces except the weight force 𝑊 and the cable tension 𝑇 , the FBD is as shown in Fig. 2. Applying Newton’s second law in the polar component system shown should allow us to find the tension in the cable as a function of its swing angle and thus, find its tension when 𝜃 = 0◦ .

𝐴

Figure 1 𝑢̂ 𝜃

Governing Equations Balance Principles

Referring to the FBD in Fig. 2 and applying Newton’s second law,

𝑂 𝑢̂ 𝑟

we obtain ∑ ∑

𝐹𝜃 ∶

−𝑊 sin 𝜃 = 𝑚𝑎𝜃 ,

(1)

𝐹𝑟 ∶

𝑊 cos 𝜃 − 𝑇 = 𝑚𝑎𝑟 ,

(2)

𝐿

𝐴

All forces are accounted for on the FBD. Writing 𝑎𝜃 and 𝑎𝑟 in polar components gives

Kinematic Equations

𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 𝐿𝜃̈

and 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 = −𝐿𝜃̇ 2 ,

𝑊

(3)

where we have replaced 𝑟 with the constant length 𝐿. Computation

𝜃

𝑇

where 𝑚 = 𝑊 ∕𝑔. Force Laws

815

Substituting Eqs. (3) into Eqs. (1) and (2), we obtain

− 𝑊 sin 𝜃 = 𝑚𝐿𝜃̈

and 𝑊 cos 𝜃 − 𝑇 = −𝑚𝐿𝜃̇ 2 𝑔 ⇒ 𝜃̈ = − sin 𝜃 and 𝑇 = 𝑊 cos 𝜃 + 𝑚𝐿𝜃̇ 2 . 𝐿

(4)

̈ ̇ Notice that the tension is a function of 𝜃,̇ so we need to integrate 𝜃(𝜃) to find 𝜃(𝜃) using the chain rule, that is, 𝑔 𝑑 𝜃̇ 𝜃̈ = 𝜃̇ = − sin 𝜃 𝑑𝜃 𝐿

⇒

∫0

𝜃̇

𝜃

𝑔 sin 𝜃 𝑑𝜃 𝜃̇ 𝑑 𝜃̇ = − 𝐿 ∫𝜃0 ⇒

) 𝑔( 𝜃̇ 2 = 2 cos 𝜃 − cos 𝜃0 . 𝐿

(5)

Substituting Eq. (5) into the expression for 𝑇 in Eq. (4) gives 𝑇 (𝜃) as 𝑇 = 𝑊 (3 cos 𝜃 − 2 cos 𝜃0 )

⇒

( ) 𝑇 (𝜃 = 0) = 𝑊 3 − 2 cos 𝜃0 = 3170 lb,

(6)

where we have used 𝑊 = 2500 lb and 𝜃0 = 30◦ to obtain the final numerical result. Discussion & Verification

The final result in Eq. (6) is dimensionally correct, and the magnitude of the tension seems reasonable. Interestingly, the tension does not depend on the length of the supporting cable. That is, if the initial angle is 30◦ and the wrecking ball is released from rest, the tension in the cable will always be 3170 lb, regardless of the length of the suspending cable.

Figure 2 FBD of the wrecking ball as it swings downward.

816

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.7

Projectile Motion with Drag

Andrew Redington/Getty Images

Figure 1 Rory McIlroy of Northern Ireland hits a shot in the 17th fairway during the second round of the 111th U.S. Open.

The projectile motion model presented in Section 12.3, in which a projectile is subject only to constant gravity and the trajectory is a parabola, is not adequate for studying the trajectory of objects like golf balls. In general, the trajectory of a golf ball is not a parabola, and it is affected by many factors, such as the dimple pattern on the ball, the spin imparted to the ball, and the local density of the air. Here we can consider a simple improvement on the model in Section 12.3 by lumping the effects just mentioned into an aerodynamic drag, which we assume to be proportional to the square of the ball’s speed and directed opposite to the ball’s velocity. Use this model to derive the equations of motion of the golf ball.

SOLUTION Road Map & Modeling

𝚥̂ 𝚤̂ 𝑢̂ 𝑡 𝐹𝑑

𝜃

𝜃

golf ball 𝑚𝑔

We model the golf ball as a particle subject to its own weight 𝑚𝑔 and a drag force 𝐹𝑑 (Fig. 2). We choose a Cartesian component system with the 𝑦 axis parallel to gravity because it simplifies the representation of the weight force and it allows for an easy comparison between the current model and that in Section 12.3. The direction of the drag force is opposite to the ball’s velocity, which is in the 𝑢̂ 𝑡 direction. Although the orientation of 𝑢̂ 𝑡 is currently unknown, for convenience we have oriented it using the timedependent angle 𝜃. As we saw in Section 13.1, the ball’s equations of motion are derived by combining Newton’s second law with the force laws and the kinematic equations. Governing Equations

Balance Principles

path Figure 2 FBD of a golf ball in flight, modeled as a particle and subject to gravity and the aerodynamic drag force 𝐹𝑑 . The unit vector 𝑢̂ 𝑡 is tangent to the ball’s path and indicates the direction of the ball’s velocity.

Using the FBD in Fig. 2, Newton’s second law, in component form,

gives ∑ ∑ Force Laws

𝐹𝑥∶

−𝐹𝑑 cos 𝜃 = 𝑚𝑎𝑥 ,

(1)

𝐹𝑦∶

−𝐹𝑑 sin 𝜃 − 𝑚𝑔 = 𝑚𝑎𝑦 .

(2)

Because 𝐹𝑑 is proportional to the square of the speed, we have 𝐹𝑑 = 𝐶𝑑 𝑣2 ,

(3)

where 𝐶𝑑 is a drag coefficient∗ and 𝑣 is the ball’s speed. Kinematic Equations

In the chosen component system, we have 𝑎𝑥 = 𝑥̈

𝜃

𝑦̇ = 𝑣 sin 𝜃

(4)

Finally, given that 𝜃 is the orientation of 𝑣⃗ relative to the 𝑥 direction, we have

𝑥̇ = 𝑣 cos 𝜃 Figure 3 Components of the velocity vector of the golf ball.

ISTUDY

𝑎𝑦 = 𝑦. ̈

In addition, because the ball’s velocity can be written as 𝑣⃗ = 𝑥̇ 𝚤̂ + 𝑦̇ 𝚥̂ and the speed is the magnitude of 𝑣, ⃗ we have √ (5) 𝑣 = 𝑥̇ 2 + 𝑦̇ 2 .

𝑣 𝑢̂ 𝑡

and

cos 𝜃 =

𝑥̇ 𝑥̇ = √ 𝑣 2 𝑥̇ + 𝑦̇ 2

and

sin 𝜃 =

𝑦̇ 𝑦̇ , = √ 𝑣 2 𝑥̇ + 𝑦̇ 2

(6)

where the components of 𝑣⃗ are depicted in Fig. 3. ∗ The

coefficient 𝐶𝑑 in Eq. (3) has dimensions of mass over length and should not be confused with the nondimensional drag coefficient normally used in aerodynamics.

ISTUDY

Section 13.2

Computation

Curvilinear Motion

Substituting Eqs. (3)–(6) into Eqs. (1) and (2), we obtain √ ̈ −𝐶𝑑 𝑥̇ 𝑥̇ 2 + 𝑦̇ 2 = 𝑚𝑥, √ −𝐶𝑑 𝑦̇ 𝑥̇ 2 + 𝑦̇ 2 − 𝑚𝑔 = 𝑚𝑦. ̈

(7) (8)

Equations (7) and (8) are the equations of motion for this problem. Discussion & Verification Since the dimensions of 𝐶𝑑 are mass over length, Eqs. (7) and (8) are dimensionally correct. In addition, referring to Eq. (7), note that for a positive 𝑥, ̇ the left-hand side is negative, thus indicating that the effect of the drag is that of slowing the particle down, as expected. A similar argument can be made for Eq. (8). Finally, note that if we set 𝐶𝑑 = 0 in Eqs. (7) and (8), we recover the equations of motion of a projectile according to the model in Section 12.3, namely, 𝑥̈ = 0 and 𝑦̈ = −𝑔. Therefore, we can conclude that, overall, the equations of motion we have derived appear to be correct.

A Closer Look When 𝐶𝑑 ≠ 0, Eqs. (7) and (8) cannot be solved analytically. However, they can easily be solved numerically using mathematical software as long as we provide values for 𝑚, 𝑔, and 𝐶𝑑 , as well as the initial position and velocity of the ball. A typical golf ball weighs 1.61 oz, corresponding to a mass of 3.125 × 10−3 slug. To estimate 𝐶𝑑 , the authors have calibrated the current model, using experimental data pertaining to the swing of Tiger Woods who, using a driver, imparted to a ball an initial speed of 186 mph and an upward direction of 11.2◦ with respect to the horizontal, and caused the ball to travel on the fly a distance of roughly 270 yd, or 810 f t. Using this information and with the help of a computer, we estimated that 𝐶𝑑 = 4.71×10−7 lb⋅s2 ∕f t 2 . Figure 4 shows a comparison between the trajectory corresponding to the stated values of 𝐶𝑑 and the initial conditions with a trajectory having the same initial conditions but with 𝐶𝑑 = 0. The reduction of the range of the ball due to drag is about 8%. 40 30 𝑦 (f t) 20 10 0 0

no drag with drag 200

400 𝑥 (f t)

600

800

Figure 4. Trajectories of a golf ball with and without drag computed using the parameters and initial conditions stated in the text. To allow the two trajectories to be easily distinguished, the 𝑥 and 𝑦 axes have been given different scales.

817

818

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.8

Maximum Speed Allowed by Friction 𝑣

𝜌

A race car of mass 𝑚 moves at a constant speed 𝑣 along a banked turn on the track shown. Let the bank angle and turn radius of curvature be those of the Talladega Superspeedway in East Aboga, Alabama, which means that 𝜌 is 1100 f t and the turn bank angle is 33◦ . For this turn, determine the maximum value of 𝑣 such that the car does not slide. Assume that the static friction coefficient between the car and track is 𝜇𝑠 = 0.9.

SOLUTION Road Map & Modeling

Figure 1 Top view of the racetrack with banked turns. 𝑚𝑔

𝑢̂ 𝑏 𝑢̂ 𝑛 𝐹

𝑁 𝜃 Figure 2 FBD of the car on the banked turn. The car’s velocity vector is pointing out of the page. The force 𝐹𝑡 (not shown) driving the car acts perpendicular to and out of the page.

ISTUDY

Modeling the car as a particle and neglecting all forces except the contact force with the track and gravity, the car’s FBD is that shown in Fig. 2. We have chosen to use path components, although cylindrical components would have been just as convenient for this problem because the path of the car is circular. We note that the 𝑡 direction (the direction of motion) is perpendicular to the plane of the drawing in Fig. 2. When dealing with friction, it is important to recall that the friction force is the vector component of the contact force tangent to the contact surface. Here the contact surface is the track and, for convenience, we break the (total) friction force into two components 𝐹𝑡 and 𝐹 . The component 𝐹𝑡 drives (propels or slows) the car in the direction of motion and therefore, cannot be seen in Fig. 2. The component 𝐹 is tangent to the track and perpendicular to the 𝑡 direction. We show 𝐹 pointing toward the inside of the track because if 𝑣max (the maximum value of 𝑣 for the car not to slip) were exceeded, the car would slip toward the outside of the track. To find 𝑣max , we assume that the car is not slipping but slip is impending. Letting 𝐹tot denote the magnitude of the total friction force, the impending slip condition requires that 𝐹tot = 𝜇𝑠 𝑁. Governing Equations Balance Principles

Using the FBD in Fig. 2, Newton’s second law gives ∑ 𝐹𝑛∶ 𝐹 cos 𝜃 + 𝑁 sin 𝜃 = 𝑚𝑎𝑛 , ∑ 𝐹𝑡∶ 𝐹𝑡 = 𝑚𝑎𝑡 , ∑ 𝐹𝑏∶ −𝑚𝑔 + 𝑁 cos 𝜃 − 𝐹 sin 𝜃 = 𝑚𝑎𝑏 ,

(1) (2) (3)

where, as argued earlier, 𝐹𝑡 is the driving force in the tangent direction due to friction between the car’s tires and the track. Force Laws

Using the impending slip assumption stated above, we have √ 𝐹tot = 𝐹 2 + 𝐹𝑡2 = 𝜇𝑠 𝑁.

(4)

Kinematic Equations

In path components, the acceleration has only components in the 𝑛𝑡 plane. Therefore, we have 𝑎𝑛 = 𝑣2max ∕𝜌,

𝑎𝑡 = 𝑣̇ = 0,

and 𝑎𝑏 = 0,

(5)

where 𝑎𝑡 = 0 because the speed of the car is constant. Computation

Equations (1)–(5) are the governing equations for this problem. The quantities 𝑔, 𝜃, and 𝜌 are known, and so we have seven equations in the seven unknowns 𝐹𝑡 , 𝑁, 𝐹 , 𝑎𝑡 , 𝑎𝑏 , 𝑎𝑛 , and 𝑣max . These equations can be solved by hand to obtain 𝑎𝑡 = 0 = 𝑎 𝑏 , ( ) 𝑔 𝜇𝑠 + tan 𝜃 = 120.1 f t∕s2 , 𝑎𝑛 = 1 − 𝜇𝑠 tan 𝜃 𝐹𝑡 = 0,

(6) (7) (8)

ISTUDY

Section 13.2

Curvilinear Motion 𝜇𝑠 𝑚𝑔

𝐹 =

( ) = 83.16 f t∕s2 𝑚,

cos 𝜃 − 𝜇𝑠 sin 𝜃 ( ) 𝑚𝑔 = 92.40 f t∕s2 𝑚, 𝑁= cos 𝜃 − 𝜇𝑠 sin 𝜃 𝑣max

√ = 𝜌𝑔

√

𝜇𝑠 + tan 𝜃 1 − 𝜇𝑠 tan 𝜃

= 363.4 f t∕s = 247.8 mph.

819

(9) (10)

(11)

Discussion & Verification

The results in Eqs. (6)–(11) are dimensionally correct, and are expressed using appropriate units. Hence, we need to answer whether or not the solution is consistent with the impending slip assumption used to derive it. The answer is yes. Notice that the obtained values of 𝐹 and 𝑁 are positive; i.e., 𝐹 and 𝑁 point exactly as shown on the FBD. In addition, the value of 𝑎𝑛 is positive, as it should be, since the normal component of acceleration can only point toward the center of the circular path. Hence, we can conclude that the solution is physically acceptable, that the working assumption was correct, and that the problem is solved.

𝑎𝑡 = 0 = 𝑎𝑏 ,

(12)

2

𝑎𝑛 = 𝑣 ∕𝜌,

(13)

𝐹𝑡 = 0,

(14)

𝑚𝑣2

cos 𝜃 − 𝑚𝑔 sin 𝜃, 𝜌 𝑚𝑣2 sin 𝜃 + 𝑚𝑔 cos 𝜃. 𝑁= 𝜌 𝐹 =

(15) (16)

By using Eqs. (15) and (16), and recalling that this solution assumes 𝜇𝑠 > 1∕ tan 𝜃, it is possible to verify that the no-slip condition is indeed satisfied for any value of 𝑣, i.e., 𝑣max is not restricted by insufficient friction. Going back to Fig. 3, observe that since 𝜇𝑠 is a positive quantity, the portion of the figure shaded in yellow concerns situations that are not physically achievable. However, the plot does indicate that if there were no friction (𝜇𝑠 = 0), the car would not slip as long as it traveled at roughly 150 f t∕s.

1500 𝑣max (f t∕s)

A Closer Look To better understand the verification of the impending slip condition, suppose that the given value of 𝜇𝑠 were 1.6 instead of 0.9. Then 𝜇𝑠 > 1∕ tan 𝜃 = 1.540, and the term 1 − 𝜇𝑠 tan 𝜃 in the expression for 𝑣max would be negative so that 𝑣max would involve the square root of a negative number. This result tells us that if 𝜇𝑠 > 1∕tan 𝜃, then the original impending slip assumption becomes incorrect! The question now is, If 𝜇𝑠 > 1∕ tan 𝜃, how do we solve the problem? To answer this question, we need to analyze the impending slip solution more closely. Consider the plot of 𝑣max vs. 𝜇𝑠 given by Eq. (11) and shown in Fig. 3. As 𝜇𝑠 approaches 1∕ tan 𝜃, 𝑣max becomes infinite. This implies that we would never need a 𝜇𝑠 larger than 1.54 no matter how fast we wanted to go, and it indicates that for 𝜇𝑠 > 1∕ tan 𝜃, the working assumption must change from impending slip to no slip. Under this new assumption, Eq. (4) is replaced by the inequality 𝐹tot ≤ 𝜇𝑠 |𝑁|, and we solve Eqs. (1)–(3) as well as Eqs. (5) for 𝐹𝑡 , 𝑁, 𝐹 , 𝑎𝑡 , 𝑎𝑏 , and 𝑎𝑛 as functions of 𝑣:

1000 500 0

−0.5

0

0.5 𝜇𝑠

1

1.5

Figure 3 Plot of 𝑣max vs. 𝜇𝑠 as given by Eq. (11), i.e., under the impending slip condition assumption. The yellow dot corresponds to 𝜇𝑠 = 0.9 used in the example. The silver dot corresponds to 𝜇𝑠 = 0.

820

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.9

Analysis of Circular Motion A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right so that it slides along the surface. Determine the angle 𝜃 at which the sphere separates from the surface.

𝜃 = 0 at release

SOLUTION 𝑅 𝜃

𝑅

Road Map & Modeling

It may not be immediately obvious that the sphere will separate from the surface before it reaches 𝜃 = 90◦ , though it is easy to set up a simple experiment to verify the claim. Referring to Fig. 2, to analytically determine the separation angle 𝜃 𝑚𝑔

Figure 1 A small sphere sliding down a smooth semicylindrical surface.

ISTUDY

𝑢̂ 𝑟 𝑢̂ 𝜃 𝑁

Figure 2. FBD of the particle when it is at an arbitrary position on the semicylinder.

𝜃𝑠 , we model the sphere as a particle and sketch its FBD, which we have drawn at an arbitrary value of 𝜃 given that 𝜃𝑠 is not yet known. Friction has been neglected because the sliding surface is frictionless. Until the sphere separates, its motion is circular, and we use polar coordinates to study it. To find 𝜃𝑠 , recall that separation implies “lack of contact,” a condition for which 𝑁 = 0. Therefore, we will determine 𝑁 as a function of 𝜃 and find 𝜃𝑠 by finding the value of 𝜃 for which 𝑁 = 0. Governing Equations Balance Principles

Using the FBD in Fig. 2, Newton’s second law yields ∑ 𝐹𝑟∶ 𝑁 − 𝑚𝑔 cos 𝜃 = 𝑚𝑎𝑟 , ∑ 𝐹𝜃 ∶ 𝑚𝑔 sin 𝜃 = 𝑚𝑎𝜃 .

(1) (2)

Force Laws Since the only force law needed in this problem is that describing gravity, we can say that all force laws are accounted for on the FBD. Kinematic Equations

The kinematics relations are 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 = −𝑅𝜃̇ 2 , ̈ 𝑎 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 𝑅𝜃, 𝜃

(3) (4)

where we have let 𝑟 = 𝑅 = constant since the path is circular. Computation

Substituting Eqs. (3) and (4) into Eqs. (1) and (2) and simplifying, we

obtain 𝑁 − 𝑚𝑔 cos 𝜃 = −𝑚𝑅𝜃̇ 2 , ̈ 𝑔 sin 𝜃 = 𝑅𝜃.

(5) (6)

̇ Because we need 𝑁 as a function of Equation (5) gives us 𝑁 as a function of both 𝜃 and 𝜃. only 𝜃, we need to find an expression for 𝜃̇ as a function of 𝜃. We can find this expression by considering Eq. (6), which gives us 𝜃̈ as a function of 𝜃. To turn this relation into an

ISTUDY

Section 13.2

Curvilinear Motion

821

̈ using the chain rule, and then expression of 𝜃̇ as a function of 𝜃, we begin by rewriting 𝜃, using that in Eq. (6), i.e., 𝑑 𝜃̇ 𝑑𝜃 𝑑 𝜃̇ 𝜃̈ = = 𝜃̇ 𝑑𝜃 𝑑𝑡 𝑑𝜃

⇒

𝑔 𝑑 𝜃̇ 𝜃̇ = sin 𝜃. 𝑑𝜃 𝑅

(7)

Multiplying both sides of Eq. (7) by 𝑑𝜃 and then integrating the result from the top of the cylinder to an arbitrary position, we obtain ∫0

𝜃̇

𝜃

𝑔 sin 𝜃 𝑑𝜃, ∫0 𝑅 |𝜃 𝑔 𝑔 = − cos 𝜃 || = (1 − cos 𝜃), 𝑅 |0 𝑅 2𝑔 (1 − cos 𝜃), = 𝑅

𝜃̇ 𝑑 𝜃̇ =

(8)

𝜃̇ 2 2

(9)

𝜃̇ 2

(10) 1

where we used the fact that 𝜃̇ = 0 when 𝜃 = 0 to obtain the lower limits of integration. Substituting Eq. (10) into Eq. (5), we obtain 𝑁 as a function of 𝜃 𝑁 = 𝑚𝑔(3 cos 𝜃 − 2),

(11)

𝑁 𝑚𝑔

0 −1 48.19

which has been plotted in Fig. 3. Force 𝑁 goes to zero when 3 cos 𝜃𝑠 − 2 = 0

⇒ ⇒

2 3 𝜃𝑠 = ±48.19◦ ± 𝑛360◦ , 𝑛 = 0, 1, … , ∞.

cos 𝜃𝑠 =

Since we are only interested in solutions in the range meaningful solution is 𝜃𝑠 = 48.19◦ .

0◦

≤𝜃≤

90◦ ,

(12) (13)

the only physically (14)

Discussion & Verification

The result in Eq. (11) is dimensionally correct since the term 𝑚𝑔 has dimensions of force and the term in parentheses is nondimensional. In addition, the result appears to be reasonable because, as intuition suggests, 𝑁 is equal to 𝑚𝑔 when 𝜃 = 0, and it decreases as 𝜃 increases. Figure 3 shows that for 𝜃 > 48.19◦ , the normal force becomes negative. What does this mean physically? Since the governing equations are written under the assumption that the sphere moves along the circular path with radius equal to 𝑅, the solution yields the value of 𝑁 that is required for the sphere to move as prescribed. This means that for 48.19◦ < 𝜃 ≤ 90◦ , for the sphere to remain in contact with the sliding surface, the surface must actually pull the sphere in. At 𝜃 = 90◦ , the mass must be pulled in at twice its weight. Finally, notice that even though the normal force is a function of the weight of the sphere, the angle at which the sphere leaves the surface is independent of 𝑚 and 𝑔. A Closer Look

−2 0

60

30

90

𝜃 (deg) Figure 3 The normal force 𝑁 (normalized by the weight of the particle 𝑚𝑔) as a function of 𝜃 as the particle slides down the semicylinder.

822

E X A M P L E 13.10

Central Force Motion

𝑟

𝑂

𝑘

𝑚

Figure 1 A disk of mass 𝑚 moving on a smooth horizontal plane. The disk’s motion is constrained only by the linear elastic spring.

ISTUDY

Chapter 13

Force and Acceleration Methods for Particles

Consider a disk of mass 𝑚 at one end of a linear elastic spring, the other end of which is attached to a freely rotating pin. The disk is free to move in the smooth horizontal plane. Determine the equations of motion of the disk for any values of the initial position and velocity, as well as the system parameters, namely, the spring constant 𝑘, the unstretched length of the spring 𝑟𝑢 , and the mass of the disk 𝑚. After doing this, plot the disk’s trȧ jectory using 𝑟(0) = 0.35 m, 𝜃(0) = 0 rad, 𝑟(0) ̇ = 0 m∕s, 𝜃(0) = 0.5 rad∕s, 𝑟𝑢 = 0.25 m, and four different values of 𝑘∕𝑚: 5, 20, 100, and 500 s−2 . Do each of the four plots for 0 ≤ 𝑡 ≤ 10 s.

SOLUTION Road Map & Modeling

The equations of motion result from applying Newton’s second law to the disk and then combining it with the force laws and kinematic equations. Since there are as many equations of motion as there are degrees of freedom, we should expect two equations of motion for the disk. Referring to Fig. 2, given that the motion of the disk 𝑢̂ 𝜃

𝑚

𝑟 𝑢̂ 𝑟

𝐹𝑠 𝜃 𝑂

Figure 2. FBD of the system as viewed from above.

occurs over a smooth horizontal plane, we will model the disk as a particle subject only to the spring force 𝐹𝑠 . The motion is planar, and while we could use Cartesian coordinates to describe the motion, the fact that 𝐹𝑠 always points toward 𝑂 makes it convenient to use a polar coordinate system with origin at 𝑂. Governing Equations Balance Principles

Force Laws

Referring to the FBD in Fig. 2, Newton’s second law yields ∑ 𝐹𝑟∶ −𝐹𝑠 = 𝑚𝑎𝑟 , ∑ 𝐹𝜃 ∶ 0 = 𝑚𝑎𝜃 .

The only force law needed is that of the linear spring, i.e., ) ( 𝐹𝑠 = 𝑘 𝑟 − 𝑟𝑢 .

Kinematic Equations

(1) (2)

(3)

The kinematic equations in polar coordinates are 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 , ̇ 𝑎 = 𝑟𝜃̈ + 2𝑟̇ 𝜃. 𝜃

(4) (5)

Computation

Substituting Eqs. (3)–(5) into Eqs. (1) and (2) and rearranging, we obtain the equations of motion for this system as∗ 𝑟̈ − 𝑟𝜃̇ 2 +

∗ Notice

) 𝑘( 𝑟 − 𝑟𝑢 = 0, 𝑚 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 0,

(6) (7)

that the spring constant 𝑘 and the mass 𝑚 only appear as a ratio in Eq. (6). Therefore, we don’t need to specify 𝑘 and 𝑚 individually, we need only specify their ratio.

ISTUDY

Section 13.2

Curvilinear Motion

where we have divided 𝑚 out of Eq. (7). Solving Eqs. (6) and (7) for 𝑟(𝑡) and 𝜃(𝑡) would tell us the motion of the mass at the end of the spring for all time. Unfortunately, these equations cannot be solved analytically. Therefore, we must resort to computer solutions. ⇒ To solve Eqs. (6) and (7) on a computer, we will need to use the values for all the constants in the system, as well as the initial conditions, that are given in the problem statement. We will compute four different solutions, all of which will have the same initial ̇ conditions, which are given as 𝑟(0) = 0.35 m, 𝜃(0) = 0 rad, 𝑟(0) ̇ = 0 m∕s, and 𝜃(0) = 0.5 rad∕s. In addition, all of the four simulations will have the same value of the spring’s unstretched length, namely, 𝑟𝑢 = 0.25 m. The simulations in question will differ only in the values chosen for the ratio 𝑘∕𝑚, which we set to 5, 20, 100, and 500 s−2 . Using mathematical software and integrating our equations for 0 ≤ 𝑡 ≤ 10 s, we obtain the trajectories shown in Fig. 3. ⇐ 20 s−2

0.3 5 s−2 𝑟

0.15

𝜃 0

𝑦 position (m)

−0.15 −0.3 0.3 100 s−2

500 s−2

0.15 0 −0.15 −0.3 −0.3 −0.15

0

0.15

0.3

−0.3 −0.15

0

0.15

0.3

𝑥 position (m) Figure 3. Trajectories of the disk for four different values of 𝑘∕𝑚, which are shown in the upper left corner of each figure. The values of 𝑥 and 𝑦 are computed as 𝑟 cos 𝜃 and 𝑟 sin 𝜃, respectively. The green dots indicate the start of each trajectory, and the red dots indicate the end after 10 s. Discussion & Verification

The “answer” to this example is not a number, but is a pair of ordinary differential equations given by Eqs. (6) and (7). The number of equations of motion matches the number of degrees of freedom, which is equal to two because the disk is free to move in the horizontal plane, which is two-dimensional. Since the equation of motion we obtained cannot be solved analytically, we had to resort to a computer solution for specific values of the system parameters and over a specific interval of time.

823

824

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.11

Anomalous Acceleration on a Merry-Go-Round

spin axis

Figure 1 shows a typical playground merry-go-round. In walking over the platform while spinning, not only does the child feel like he is being thrown radially outward, but also he feels like he is being thrown sideways. Investigate this motion, and determine the forces required to walk radially at a constant rate 𝑣0 on a platform of radius 𝜌, while spinning at constant angular rate 𝜔0 .

SOLUTION Road Map & Modeling

rotation

We are told how a child moves, and we are asked to find what forces are necessary for the motion to occur as described, which means that we are given the 𝑎⃗ in 𝐹⃗ = 𝑚𝑎. ⃗ Referring to Fig. 2, we model the child as a particle and describe his 𝑚𝑔

Figure 1 A small child walking radially on a spinning merry-go-round.

ISTUDY

𝑢̂ 𝜃

𝐹𝜃 𝜌

𝑁

𝐹𝑅 𝑢̂ 𝑅

Figure 2. FBD of the child walking radially outward on the merry-go-round.

motion using a cylindrical coordinate system with origin on the spin axis of the merry-goround. We assume that the child is subject to gravity, namely, 𝑚𝑔, the normal reaction 𝑁 at the floor, and any other forces that might be required for the motion to be as described. This is why the FBD includes the forces 𝐹𝑅 and 𝐹𝜃 , in the radial and transverse directions, respectively. Generally, 𝐹𝑅 and 𝐹𝜃 will be due to friction between the child and the spinning platform, as well as the fact that the child can hold onto the railings mounted on the merry-go-round. Governing Equations Balance Principles

Using the FBD in Fig. 2, Newton’s second law gives ∑ 𝐹𝑅 ∶ 𝐹𝑅 = 𝑚𝑎𝑅 , ∑ 𝐹𝜃 ∶ 𝐹𝜃 = 𝑚𝑎𝜃 , ∑ 𝐹𝑧 ∶ 𝑁 − 𝑚𝑔 = 𝑚𝑎𝑧 .

(1) (2) (3)

All known forces are accounted for on the FBD. The forces 𝑁, 𝐹𝑅 , and 𝐹𝜃 are unknowns of the problem. Kinematic Equations Using cylindrical components, since 𝜃̇ = 𝜔0 and 𝑟̇ = 𝑣0 are both constant, we have Force Laws

𝑎𝑅 = 𝑅̈ − 𝑅𝜃̇ 2 = −𝑅𝜔20 , 𝑎𝜃 = 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ = 2𝑣0 𝜔0 ,

(5)

𝑎𝑧 = 𝑧̈ = 0.

(6)

(4)

Equations (3) and (6) simply tell us that 𝑁 = 𝑚𝑔. Equations (1), (2), (4), and (5) tell us that

Computation

𝐹𝑅 = −𝑚𝑅𝜔20 ,

(7)

𝐹𝜃 = 2𝑚𝑣0 𝜔0 ,

(8)

ISTUDY

Section 13.2

825

Curvilinear Motion

which are the forces required to keep the child moving in the radial direction at a constant speed as the merry-go-round spins at a constant angular velocity. Discussion & Verification Equation (7) says that 𝐹𝑅 is independent of how fast the child walks and that it always acts inward given that 𝑚, 𝑟, and 𝜔20 are positive. As intuition might suggest, this corresponds to a physical sensation of being “thrown” outward.∗ By contrast, 𝐹𝜃 in Eq. (8) can be positive or negative because 𝑣0 and 𝜔0 can each be positive or negative, thus leading to four possible cases. In each of the cases, when we use right and left to indicate direction, they will be relative to the child, who we assume is facing in the direction of motion.

Case 1 In this case, let 𝑣0 > 0 and 𝜔0 > 0, so that the child is walking outward and the merry-go-round is spinning as shown in Fig. 1, i.e., counterclockwise (ccw) as viewed from above. This situation is shown in Fig. 3(a). For this case, Eq. (8) tells us that 𝐹𝜃 > 0 so that the force applied to the child while walking outward will be to his left. The child will then feel as though he is thrown to the right. Case 2 In this case, 𝑣0 < 0 and 𝜔0 > 0, so that the child is walking inward and the merrygo-round is spinning counterclockwise (ccw). This situation is shown in Fig. 3(b). For this case, Eq. (8) tells us that 𝐹𝜃 < 0 so that the force applied to the child while walking inward will again be to his left. Once again, the child will then feel as though he is thrown to the right. Case 3 In this case, 𝑣0 > 0 and 𝜔0 < 0, so that the child is walking outward and the merry-go-round is now spinning clockwise (cw). This situation is shown in Fig. 3(c). For this case, Eq. (8) tells us that 𝐹𝜃 < 0 so that the force applied to the child while walking outward will now be to his right. This time, the child will then feel as though he is thrown to the left. Case 4 In this case, 𝑣0 < 0 and 𝜔0 < 0, so that the child is walking inward and the merrygo-round is again spinning clockwise (cw). This situation is shown in Fig. 3(d). For this case, Eq. (8) tells us that 𝐹𝜃 > 0 so that the force applied to the child while walking inward will now be to his right, and the child will again feel as though he is thrown to the left. The analysis of these four cases tells us that whether the child walks inward or outward, as long as the merry-go-round spins counterclockwise (ccw), the child is always thrown to the right. Similarly, whether or not the child walks inward or outward, as long as the merry-go-round spins clockwise (cw), the child is always thrown to the left. It turns out that we can generalize these results for a person walking in any direction to state the following: No matter which direction a person walks on a rotating reference frame, if the frame is rotating counterclockwise (ccw), then the person is thrown to his or her right, whereas the person is thrown to her or his left if the reference frame is rotating clockwise (cw). This result will be easy to demonstrate once we learn about the kinematics of motion relative to rotating reference frames, which we will cover in Section 16.4.

∗ For

example, the force on you as you round a curve in a car is in the same direction as your acceleration, i.e., toward the center of the curve. However, the physical sensation you experience is that of being thrown away from the center of the curve.

𝜔0 (a)

𝑣0 𝜔0

(c)

𝑣0

𝜔0

𝑣0

(b)

𝜔0

𝑣0

(d)

Figure 3 (a) Case 1: 𝑣0 > 0, 𝜔0 > 0; (b) Case 2: 𝑣0 < 0, 𝜔0 > 0; (c) Case 3: 𝑣0 > 0, 𝜔0 < 0; (d) Case 4: 𝑣0 < 0, 𝜔0 < 0.

826

Chapter 13

Force and Acceleration Methods for Particles

Problems Problem 13.47 A train is traveling with constant speed 𝑣0 along the level track 𝑂𝐴𝐵, which lies in the horizontal plane. The section between points 𝑂 and 𝐴 is straight, and the section between points 𝐴 and 𝐵 is circular with radius of curvature 𝜌. The train starts at point 𝑂 at time 𝑡 = 0, reaches point 𝐴 at time 𝑡𝐴 , and reaches point 𝐵 at time 𝑡𝐵 . For this motion, sketch the magnitude of the acceleration vector as a function of time. Would you want to design a train track with this shape? 𝑂

𝐴 𝑣0 𝜌

𝑣𝑓 𝑂

𝐵

𝜌𝑓

Figure P13.47

𝑣0

Problem 13.48

𝜌0

A plane is turning along a horizontal path at constant speed. What is the component of force in the direction of the path acting on the plane?

𝐶

Problem 13.49 Body 𝐷 is in equilibrium when cable 𝐴𝐶 suddenly breaks. If 𝐵 is fixed, does the tension in the inextensible cable 𝐵𝐶 increase or decrease at the instant of release?

Figure P13.48

𝐴

𝐵

𝐵

𝐶 𝐷

Figure P13.49 𝑣

𝐴

𝐶 𝐷

Figure P13.50

Problem 13.50

𝑣0

Body 𝐷 is in equilibrium when cable 𝐴𝐶 suddenly breaks. If 𝐵 is fixed, does the tension in the inextensible cable 𝐵𝐶 increase or decrease at the instant of release?

Problem 13.51 𝜌 𝑂 Figure P13.51

ISTUDY

An aircraft carrier is turning with a constant speed 𝑣 along a circular path with radius 𝜌 and center 𝑂. During the maneuver, a forklift is being driven across the deck of the ship at a constant speed 𝑣0 , relative to the deck. Does the friction force between the forklift and the deck have a component perpendicular to the relative velocity of the forklift with respect to the deck?

ISTUDY

Section 13.2

827

Curvilinear Motion

Problem 13.52 A jet is coming out of a dive, and a sensor in the pilot’s seat measures a force of 800 lb for a pilot whose weight is 180 lb. If the jet’s instruments indicate that the plane is traveling at 850 mph, determine the radius of curvature of the plane’s path at this instant.

𝜌 path of jet

𝑣 Figure P13.52

Problems 13.53 and 13.54 A partial cross section of an amusement park ride is shown. While the ride spins up to the angular speed 𝜔𝑐 , there is a small platform at 𝐹 on which the person 𝑃 stands. Once the ride reaches the desired angular speed, the platform falls away and only friction keeps the person from sliding to the floor of the ride. The wall, against which the person lies, is inclined at the angle 𝜃 = 15◦ with respect to the vertical. Model the person as a particle that is a distance 𝑑 = 20 f t from the spin axis 𝐴𝐵 and let the coefficient of static friction between the person and the wall be 𝜇𝑠 = 0.7.

𝑑

𝜃 𝑃 𝐹

𝐵 𝜔𝑐

Problem 13.53

Determine the minimum value of 𝜔𝑐 at which the platform at 𝐹 can

be withdrawn. 𝐴 Problem 13.54 Determine whether or not the person can slide up the wall and out of the ride. If yes, determine the value of 𝜔𝑐 at which the person begins to slide.

Figure P13.53 and P13.54

Problem 13.55 A person is swinging a ball of mass 𝑚 above their head in a horizontal plane. When the ball is moving at a constant speed, the string forms an angle 𝜃 with respect to the horizontal. Assuming that the motion of point 𝑂 is negligible and that the distance between the point 𝑂 and the mass 𝑚 is 𝐿, determine the speed of the ball. 𝑂 𝐿 𝑚

𝑂

𝜃 Figure P13.55 𝜃 𝐿

Problem 13.56 Initially, the wrecking ball 𝐴 of mass 𝑚 is held stationary by the horizontal cable 𝐴𝐵. The cable 𝐴𝐵 is then released so that the wrecking ball 𝐴 starts swinging about the fixed point 𝑂. Determine the tension in the cable 𝑂𝐴 before the cable 𝐴𝐵 is released and immediately after it is released. What is the percent change in cable tension if 𝜃 = 30◦ ?

𝐵

Figure P13.56

𝐴

828

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.57 A car traveling over a hill starts to lose contact with the ground at the top of the hill at 𝑂. If the radius of curvature of the hill is 282 f t, determine the speed of the car at 𝑂.

𝑦

𝜌 𝑣0 𝑥

𝑂 Figure P13.57

Figure P13.58

Problem 13.58 A 950 kg aerobatics plane initiates the basic loop maneuver at the bottom of a loop with radius 𝜌 = 110 m and a constant speed of 225 km∕h. At this instant, determine the magnitude of the plane’s acceleration, expressed in terms of 𝑔, the acceleration due to gravity, and the magnitude of the lift provided by the wings. 𝑚

𝐴

𝑅 𝑂

𝜙

𝜃

𝜔 Figure P13.59

Problem 13.59 The ball of mass 𝑚 is guided along the vertical circular path of radius 𝑅 = 1 m using the arm 𝑂𝐴. If the arm starts from 𝜙 = 90◦ and rotates clockwise with a constant angular velocity 𝜔 = 0.87 rad∕s, determine the angle 𝜙 at which the particle starts to leave the surface of the semicylinder. Neglect all friction forces acting on the ball, neglect the thickness of the arm 𝑂𝐴, and treat the ball as a particle. Hint: A relatively simple relationship can be established between the angles 𝜙 and 𝜃 if we note the distance from 𝑂 to 𝑚 forms the base of an isosceles triangle.

Problem 13.60 Referring to Example 13.10, instead of using polar coordinates as was done in that example, work the problem using a Cartesian coordinate system with origin at 𝑂 (cf. Fig. 1 of Example 13.10) and derive the problem’s equations of motion. 𝑟 𝑚 𝑂

𝑘

Figure P13.60 and P13.61

Problem 13.61

𝑂

𝑘

Problem 13.62

𝜃 𝑚 Figure P13.62 and P13.63

ISTUDY

𝑟

Continue Prob. 13.60 and using mathematical software, numerically solve the equations of motion. Use the same parameters and initial conditions that were used in Example 13.10, and compare your results with those presented in that example.

Derive the equations of motion for the pendulum supported by a linear spring of constant 𝑘 and unstretched length 𝑟𝑢 . Neglect friction at the pivot 𝑂, the mass of the spring, and air resistance. Treat the pendulum bob as a particle of mass 𝑚, and use polar coordinates.

ISTUDY

Section 13.2

Curvilinear Motion

829

Problem 13.63 Using mathematical software, solve the equations of motion for the pendulum supported by a linear spring of constant 𝑘 (derived in Prob. 13.62). Plot the trajectory of the mass 𝑚 in the vertical plane for a number of different values of 𝑘∕𝑚. The unstretched length of the spring is 0.25 m, and the mass is released when the pendulum is vertical, the spring is stretched 0.75 m, and the mass is moving to the right at 1 m∕s.

Problems 13.64 and 13.65 𝐴

The system shown is initially at rest when the bent bar starts to rotate about the vertical axis 𝐴𝐵 with constant angular acceleration 𝛼0 = 3 rad∕s2 . The coefficient of static friction between the collar of mass 𝑚 = 2 kg and the bent bar is 𝜇𝑠 = 0.35, and the collar is initially 𝑑 = 70 cm from the spin axis 𝐴𝐵.

𝑚

𝜇𝑠

𝛼0

Assuming the motion starts at 𝑡 = 0, determine the time at which the collar starts to slip relative to the bent bar.

Problem 13.64

Problem 13.65 Determine the number of rotations undergone by the bent bar when the collar starts to slip relative to it.

𝑑

𝐵 Figure P13.64 and P13.65

Problem 13.66 Revisiting Example 13.7, assume that the drag force in the 𝑥 direction is proportional to the square of the 𝑥 component of velocity, but that the ball’s trajectory is shallow enough to neglect the drag force in the 𝑦 direction. Using this assumption, letting 𝑂 be the initial position of the ball, and letting the ball’s initial velocity have a magnitude 𝑣0 and letting it form an angle 𝜃0 with the 𝑥 axis, determine an expression for the trajectory of the ball of the form 𝑦 = 𝑦(𝑥).

𝑦

𝑂

𝑣0 𝜃0

−𝑥 𝑅

Figure P13.66 and P13.67

Problem 13.67 Revisit Example 13.7 and assume that the trajectory of the ball is shallow enough that the effects of the drag force in the 𝑦 direction can be neglected and that the component of the drag force in the 𝑥 direction is proportional to the square of the 𝑥 component of velocity. Using this assumption and the same drag coefficient discussed in the example (𝐶𝑑 = 4.71 × 10−7 lb⋅s2 ∕f t 2 ), compute the horizontal distance 𝑅 traveled by a 1.61 oz golf ball subject to the same initial conditions given in the example, i.e., the initial velocity has a magnitude 𝑣0 = 187 mph and an initial orientation 𝜃0 = 11.2◦ .

Problem 13.68 Using a cylindrical component system whose origin is at the center of curvature of the path of the car, derive the governing equations for Example 13.8 on p. 818. Solve the equations and verify that you get the same solution. 𝜌

𝜃 Figure P13.68–P13.70

830

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.69 A race car is traveling at a constant speed over a circular banked turn. Oil on the track has caused the static friction coefficient between the tires and the track to be 𝜇𝑠 = 0.2. If the radius of the car’s trajectory is 𝜌 = 320 m and the bank angle is 𝜃 = 33◦ , determine the range of speeds within which the car must travel not to slide sideways.

Problem 13.70 A race car is traveling at a constant speed 𝑣 = 200 mph over a circular banked turn. Let the weight of the car be 𝑊 = 4000 lb, the radius of the car’s trajectory be 𝜌 = 1100 f t, the bank angle be 𝜃 = 33◦ , and the coefficient of static friction between the car and the track be 𝜇𝑠 = 1.7. Determine the component of the friction force perpendicular to the direction of motion.

𝜃 = 0 at release

Problem 13.71 𝑅 𝜃

Revisit Example 13.9 on p. 820 by letting the sphere be released at 𝜃 = 0 with a speed 𝑣0 = 0.5 m∕s. Neglecting friction, compute the angle at which the sphere separates from the cylinder if 𝑅 = 1.35 m.

𝑅

Problem 13.72 Figure P13.71

The small object of mass 𝑚 is placed on the rotating conical surface at the radius shown. If the coefficient of static friction between the object and the rotating surface is 0.8, calculate the maximum angular velocity 𝜔𝑐 of the cone about the vertical axis for which the object will not slip. Assume the 𝜔𝑐 is increased very gradually so that the angular acceleration of the cone can be ignored. 0.2 m

𝐴

𝑚

𝑎𝐵

30◦ 𝜔𝑐

𝜇𝑘

𝚤̂

𝜃

𝐵

Figure P13.72

𝚥̂

Figure P13.73

Problem 13.73 The wedge-shaped cart 𝐵 is moving to the left with acceleration 𝑎𝐵 . The coefficient of static friction between the crate and the cart is insufficient to prevent slipping between the two. If the mass of 𝐴 is 𝑚 and the coefficient of kinetic friction between the crate and the cart is 𝜇𝑘 , determine the acceleration of the crate in the component system shown. 𝐿

𝑣𝑠 𝜃

Problem 13.74 The cutaway of the gun barrel shows a projectile moving through the barrel. If the projectile’s exit speed is 𝑣s = 1675 m∕s (relative to the barrel), the projectile’s mass is 18.5 kg, the length of the barrel is 𝐿 = 4.4 m, the acceleration of the projectile down the gun barrel is constant, and 𝜃 is increasing at a constant rate of 0.18 rad∕s, determine (a) the acceleration of the projectile (b) the pressure force acting on the back of the projectile

Figure P13.74

ISTUDY

(c) the normal force on the gun barrel due to the projectile

ISTUDY

Section 13.2

Curvilinear Motion

as the projectile leaves the gun, but while it is still in the barrel. Assume that the projectile exits the barrel when 𝜃 = 20◦ , and ignore friction between the projectile and the barrel.

Problems 13.75 and 13.76 A simple sling can be built by placing a projectile in a tube and then spinning it. Consider a simple model in which the tube is pinned as shown and is rotated about the pin in the horizontal plane at constant angular velocity 𝜔. Assume that there is no friction between the projectile and the inside of the tube and that the projectile is initially kept fixed at a distance 𝑑 from the open end of the tube.

𝐿

projectile 𝜃

Problem 13.75

After the projectile is released, compute the normal force exerted by the inside of the tube on the projectile as a function of position of the projectile along the tube.

𝑑

𝜔 Figure P13.75 and P13.76

Letting 𝑑 = 3 f t and 𝐿 = 7 f t, determine the value of the tube’s angular velocity 𝜔 if, after release, the projectile exits the tube with a speed of 90 f t∕s. Problem 13.76

Problem 13.77 The wedge-shaped cart 𝐵 is moving up and to the left with acceleration 𝑎𝐵 . The coefficient of static friction between the crate and the cart is insufficient to prevent slipping between the two. If the mass of 𝐴 is 𝑚 and the coefficient of kinetic friction between the crate and the cart is 𝜇𝑘 , determine the acceleration of the crate in the component system shown. 𝑎𝐵

𝐴

𝜇𝑘

𝜃

𝚥̂ 𝚤̂

𝜃

𝐵

Figure P13.77

Figure P13.78

Problem 13.78 The pendulum is released from rest when 𝜃 = 0◦ . If the string holding the pendulum bob breaks when the tension is twice the weight of the bob, at what angle does the string break? Treat the pendulum as a particle, ignore air resistance, and let the string be inextensible and massless.

𝑇 𝐿 𝜃

Problem 13.79

𝑚

The trolley 𝑇 moves along rails on the horizontal truss of the tower crane. The trolley and the load of mass 𝑚 are both initially at rest (with 𝜃 = 0) when the trolley starts moving to the right with constant acceleration 𝑎𝑇 = 𝑔. Determine (a) The maximum angle 𝜃max achieved by the load 𝑚. (b) The tension in the supporting cable as a function of 𝜃. Treat the load 𝑚 as a particle, and ignore the mass of the supporting cable.

Figure P13.79

831

832

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.80 If the particle is constrained to only move back and forth in the plane of the center of the bowl, how many degrees of freedom must it have? Recall that this will also be the number of equations of motion of the particle. 𝑦 cross section 𝑦(𝑥) = 1 + 0.5𝑥2

particle 4 3 𝑦 2 1

−2

−1

𝑥

parabolic bowl

0 𝑥

1

2

Figure P13.80 and P13.81

Problem 13.81 Derive the equation(s) of motion for a particle moving on the inner surface of the smooth parabolic bowl shown. Assume that the particle only moves in the vertical 𝑥𝑦 plane that goes through the center of the bowl, that the equation of the parabola is 𝑦(𝑥) = 1 + 0.5𝑥2 , and that gravity is acting in the −𝑦 direction. The bowl’s cross section is shown on the right side of the figure.

satellite

𝑃 , perigee

𝐴, apogee

Problems 13.82 and 13.83 A satellite orbits the Earth as shown. Model the satellite as a particle, and assume that the center of the Earth can be chosen as the origin of an inertial frame of reference.

𝑅𝑃

𝑅𝐴 Problem 13.82 Using a polar coordinate system and letting 𝑚𝑒 be the mass of the Earth, determine the equations of motion of the satellite. Problem 13.83

The minimum and maximum distances from the center of the Earth are 𝑅𝑃 = 4.5 × 107 m and 𝑅𝐴 = 6.163 × 107 m, respectively, where the subscripts 𝑃 and 𝐴 stand for perigee (the point on the orbit closest to Earth) and apogee (the point on the orbit farthest from Earth), respectively. If the satellite’s speed at 𝑃 is 𝑣𝑃 = 3.2 × 103 m∕s, determine the satellite’s speed at 𝐴. Hint: Since this is a central force problem, 𝑟2 𝜃̇ is a constant of the motion.

Figure P13.82 and P13.83

𝑟

𝐵

Problem 13.84 3 𝑟 2

track 𝑘

𝑚 𝐴

Figure P13.84

ISTUDY

The package handling system is designed to launch the small package of mass 𝑚 from 𝐴, using a compressed linear spring of constant 𝑘. After launch, the package slides along the track until it lands on the conveyor belt at 𝐵. The track has small, well-oiled rollers, making any friction between the packages and the track negligible. Modeling the package as a particle, determine the minimum initial compression of the spring so that the package gets to 𝐵 without separating from the track, and determine the corresponding speed with which the package reaches the conveyor at 𝐵.

ISTUDY

Section 13.2

Curvilinear Motion

Problem 13.85 A particle moves over the inner surface of an inverted cone. Assuming that the cone’s surface is frictionless and using the cylindrical coordinate system shown, show that the particle’s equations of motion are ( ) 𝑅̈ 1 + cot 2 𝜙 − 𝑅𝜃̇ 2 = −𝑔 cot 𝜙, 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ = 0. 𝑧

𝑧

𝜙

𝑅 𝑥

𝑦

Figure P13.85 and P13.86

Problem 13.86 Continue Prob. 13.85 by integrating the particle’s equations of motion and plotting its trajectory for 0 ≤ 𝑡 ≤ 25 s. Use the following parameter values and initial conditions: ̇ ̇ 𝑔 = 9.81 m∕s2 , 𝜙 = 30◦ , 𝜃(0) = 0◦ , 𝜃(0) = 1.00 rad∕s, 𝑅(0) = 5 m, 𝑅(0) = 0 m∕s.

Problem 13.87 The disk shown, weighing 3 lb, rotates about 𝑂 by sliding without friction over the horizontal surface shown. The spring is linear with constant 𝑘 and unstretched length 𝐿0 = 0.75 f t. The maximum distance achieved by the disk from point 𝑂 is 𝑑max = 1.85 f t while traveling at a speed 𝑣0 = 20 f t∕s. Determine the value of 𝑘 such that the minimum distance between the disk and 𝑂 is 𝑑min = 𝑑max ∕2. Hint: Since this is a central force problem, 𝑟2 𝜃̇ is a constant of the motion. Use this constant of the motion to eliminate 𝜃̇ from the radial component of 𝐹⃗ = 𝑚𝑎, ⃗ reducing the radial component equation to 𝑟̈ = 𝑟̈(𝑟). The kinematic expression 𝑟̈𝑑𝑟 = 𝑟𝑑 ̇ 𝑟,̇ a variation of 𝑎𝑑𝑠 = 𝑣𝑑𝑣 first integrated in Eq. (12.29), is then useful in finding 𝑘. 𝑟 𝑚 𝑂

𝑘

𝐿∕2 𝜃

𝐿

Figure P13.87

Problems 13.88 and 13.89 A pendulum with cord length 𝐿 = 6 f t and a bob weighing 3 lb is released from rest at an angle 𝜃𝑖 . Once the pendulum has swung to the vertical position (i.e., 𝜃 = 0), its cord runs into a small fixed obstacle. In solving this problem, neglect the size of the obstacle; model

𝜙

Figure P13.88 and P13.89

833

834

Chapter 13

Force and Acceleration Methods for Particles

the pendulum’s bob as a particle and the pendulum’s cord as massless and inextensible; and let gravity and the tension in the cord be the only relevant forces. Problem 13.88

What is the maximum height reached by the pendulum, measured from its lowest point, if 𝜃𝑖 = 20◦ ?

Problem 13.89

If the bob is released from rest at 𝜃𝑖 = 90◦ , at what angle 𝜙 does the

cord go slack?

Problems 13.90 through 13.93 A spherical pendulum is suspended from point 𝑂 and put in motion.

𝑂

Problem 13.90 𝜙0

Derive the pendulum’s equations of motion using Cartesian coordi-

nates.

𝐿

Problem 13.91

𝑣0

Derive the pendulum’s equations of motion using cylindrical coordi-

nates.

Figure P13.90–P13.93

Problem 13.92

Derive the pendulum’s equations of motion using spherical coordi-

nates. Problem 13.93 At the lowest point on its trajectory, the pendulum in the figure has a speed 𝑣0 = 2.5 f t∕s while 𝜙0 = 15◦ . Letting 𝐿 = 2 f t, plot the trajectory of the pendulum bob for 5 s.

𝑧

Problems 13.94 through 13.97 𝑟0

𝑑

restraint 𝜔0

𝑚

Consider a collar with mass 𝑚 that is free to slide with no friction along a rotating arm, which has negligible mass. The system is initially rotating with an angular velocity 𝜔0 while the collar is kept a distance 𝑟0 away from the 𝑧 axis. At some point, the restraint keeping the collar in place is removed so that the collar is allowed to slide.

arm

collar

Problem 13.94

𝑀𝑧

Derive the collar’s equations of motion when 𝑀𝑧 = 0.

Problem 13.95

If no external forces and moments are applied to the system, what are the radial speed and the total speed of the collar when it reaches the end of the arm? Use 𝑚 = 2 kg, 𝜔0 = 1 rad∕s, 𝑟0 = 0.5 m, and 𝑑 = 1 m.

Figure P13.94–P13.97

Compute the moment 𝑀𝑧 that you would need to apply to the arm, as a function of 𝑟, to keep the arm rotating at a constant angular velocity 𝜔0 . In addition, determine the radial speed, as well as the total speed with which the collar would reach the end of the arm with this moment applied. Hint: The moment 𝑀𝑧 applied to the arm of negligible mass is equivalent to a force 𝑀𝑧 ∕𝑟 applied to the collar in the plane of motion and perpendicular to the arm. Problem 13.96

Problem 13.97 Compute and plot the collar’s trajectory from the moment of release until the collar reaches the end of the arm. Use the parameters and initial conditions given in Prob. 13.95.

𝑃 𝑟 𝜃 𝑂 Figure P13.98

ISTUDY

𝑦

Problem 13.98 𝑥

The radar station at 𝑂 is tracking the meteor 𝑃 as it moves through the atmosphere. The data measured by the radar station indicates that the acceleration vector of 𝑃 is almost exactly in the opposite direction of the velocity vector. Explain why this is what we should expect.

ISTUDY

Section 13.2

Curvilinear Motion

835

Problem 13.99 The particle 𝑃 is placed on the turntable, and both are initially at rest. The turntable is then turned on so that the disk starts spinning. Assuming that there is no friction between the turntable disk and the particle, what will be the motion of the particle after the disk starts spinning? Explain.

𝑃

Problem 13.100

√ A small object enters the top 𝐴 of a circular path with horizontal speed 𝑣0 = 𝑔𝑅∕2. Treating the object as a particle and neglecting friction, determine the angle 𝜃, in degrees, at which the object separates from the path and becomes a projectile. Figure P13.99

Figure P13.100

Figure P13.101

Problem 13.101

√ A small object of mass 𝑚 enters a circular track with speed 𝑣0 = 𝑔𝑅 at 𝐴, which is at 𝜃 = 0. Treating the object as a particle and neglecting friction, determine the angle 𝜃, in degrees, at which the normal force between the object and the track is equal to 4𝑚𝑔.

𝐴 𝜃 𝜌

Problem 13.102 The conveyor belt moves parts, each with mass 𝑚, at a constant speed 𝑣0 . When the parts get to 𝐴, they begin moving over a circular path of radius 𝜌. If the coefficient of static friction between the belt and the parts is 𝜇𝑠 , determine the angle 𝜃 at which the parts will start to slide on the belt. Neglect the size of the parts. After determining 𝜃 in terms of 𝜇𝑠 , 𝑣0 , 𝜌, and 𝑔, evaluate 𝜃 for 𝜇𝑠 = 0.6, 𝑣0 = 3 mph, and 𝜌 = 14 in.

𝑣0 Figure P13.102

Problem 13.103 The drum with inner radius 𝑟 rotates in the vertical plane about the horizontal axis at 𝑂 with constant angular speed 𝜔0 . The small block 𝐴 has no motion relative to the surface of the drum as it passes the bottom position at 𝜃 = 0. If the block is observed to slip at an angle 𝜃𝑠 , determine, as a function of 𝑔, 𝑟, 𝜃𝑠 , and 𝜔0 , the coefficient of static friction 𝜇𝑠 between the surface of the drum and the block. What is the condition on 𝜔0 so that the mass must slip before it reaches 𝜃 = 180◦ ?

Figure P13.103

836

Chapter 13

Force and Acceleration Methods for Particles

Design Problems

ISTUDY

Design Problem 13.1 The mechanism in the figure needs to be designed so as to capture parcels at 𝐴 and to deliver them at 𝐵. The impact speed at 𝐵 must not exceed 1.5 m∕s. The parcels have masses ranging between 5 and 10 kg incoming at speeds between 3 and 6 m∕s. Determine the acceptable ranges of the spring constant, the free length of the spring, and the kinetic friction coefficient between the sliding surface and the cradle to achieve the desired result. 𝑣0

𝜇

𝐴

𝑘

𝑅 = 4m

𝐵 𝑅∕2 Figure DP13.1

Design Problem 13.2 Consider the cam-follower system in the figure. The cam is on a shaft that rotates with speeds up to 3000 rpm. The profile of the cam is described by the function 𝓁(𝜙) = 𝑅0 (1 + 0.25 cos3 𝜙), where the angle 𝜙 is measured relative to the segment 𝑂𝐴, which rotates with the cam, and where 𝑅0 = 3 cm. The follower has a mass of 90 g. Assuming that the contact between cam and follower is frictionless, choose appropriate values of the spring constant 𝑘 and the spring’s unstretched length 𝐿0 such that the follower always remains in contact with the cam while minimizing the value of the contact force between the follower and the cam. 𝑦 𝑘, 𝐿0

follower

cam 𝓁(𝜙)

𝐴

𝜙 𝜃

𝑥

𝑂 𝜔cam

shaft Figure DP13.2

ISTUDY

Section 13.3

13.3

Systems of Particles

837

Systems of Particles

Engineering materials one atom at a time When applied to large collections of atoms, the particle dynamics we are studying plays an important role in the burgeoning field of nanotechnology. The dynamics of systems of particles is a core component of a nanoscale modeling approach called molecular dynamics (MD), in which the force laws describing atomic interactions are used together with Newton’s second law to study the motion of a system of atoms. Using information from these simulations, we can estimate and predict a variety of physical properties (see Fig. 13.16). We will now develop the equations that allow us to apply Newton’s second law to a system of particles.

Newton’s second law for systems of particles In Fig. 13.17, we consider a collection of 𝑛 particles. The quantities 𝑚𝑖 and 𝑟⃗𝑖 denote 𝑧

𝐹⃗𝑖 𝑓⃗𝑖𝑗

𝑚𝑖

𝑓⃗𝑖2

𝑓⃗𝑖1

𝑟⃗𝑖

𝐺

𝑓⃗1𝑖 𝑚1 𝐹⃗1 𝑟⃗1

𝐹⃗𝑗 𝑓⃗𝑗𝑖

𝑓⃗1𝑗

𝑟⃗𝐺

𝑓⃗𝑗2

𝑟⃗𝑗 𝑓⃗2𝑖

𝑓⃗12

𝑓⃗2𝑗

𝑓⃗21 𝑟⃗2

𝑚𝑗

𝑓⃗𝑗1

𝑚2

𝐹⃗2 𝑦

𝑥 Figure 13.17. System of 𝑛 particles showing position vectors 𝑟⃗𝑖 , external forces 𝐹⃗𝑖 , and internal forces 𝑓⃗𝑖𝑗 .

the mass and the position of particle 𝑖, respectively. A force acting on a particle of the system is called internal if it arises from the interaction between that particle and other particles within the system; otherwise the force is called external. The force denoted by 𝐹⃗𝑖 is the sum of all external forces acting on particle 𝑖 (e.g., gravity, air drag, etc.), and it would exist even if the rest of the particles were removed from the system. With 𝑓⃗𝑖𝑗 we denote the force on particle 𝑖 due to particle 𝑗. Therefore, the forces 𝑓⃗ , 𝑓⃗ , … , 𝑓⃗ are the internal forces acting on particle 𝑖. 𝑖1

𝑖2

𝑖𝑛

Applying Newton’s second law to each particle within the system, we have 𝐹⃗1 +

𝑛 ∑ 𝑗=1 𝑗≠1

𝑓⃗1𝑗 = 𝑚1 𝑎⃗1 , 𝐹⃗2 +

𝑛 ∑ 𝑗=1 𝑗≠2

𝑓⃗2𝑗 = 𝑚2 𝑎⃗2 , … 𝐹⃗𝑛 +

𝑛 ∑ 𝑗=1 𝑗≠𝑛

𝑓⃗𝑛𝑗 = 𝑚𝑛 𝑎⃗𝑛 ,

(13.29)

U.S. Navy

Figure 13.16 Snapshot from an MD simulation of fracture in silicon used to estimate the fracture toughness of the material under various conditions.

838

ISTUDY

Chapter 13

Force and Acceleration Methods for Particles

where we have noted that in equation 𝑖, we cannot have 𝑗 = 𝑖 because we assume that a particle does not exert a force on itself. To understand the effect of Newton’s third law on the system as a whole, we sum all 𝑛 equations in Eqs. (13.29) to obtain 𝑛 ∑

𝐹⃗𝑖 +

𝑖=1

𝑛 ∑ 𝑛 ∑

𝑓⃗𝑖𝑗 =

𝑖=1 𝑗=1 𝑗≠𝑖

𝑛 ∑

𝑚𝑖 𝑎⃗𝑖 .

(13.30)

𝑖=1

∑ The first term in Eq. (13.30), 𝑛𝑖=1 𝐹⃗𝑖 , is the sum of all external forces acting on the system of particles, which we will denote by 𝐹⃗ , i.e., 𝑛 ∑

𝐹⃗ =

𝐹⃗𝑖 .

(13.31)

𝑖=1

The second term in Eq. (13.30) is the sum of all internal forces. Referring to Fig. 13.17, observe that for every 𝑓⃗𝑖𝑗 in the system there will always be an 𝑓⃗𝑗𝑖 . By Newton’s third law, these forces are equal and opposite, and they will always cancel one another. For example, if 𝑛 = 3, the second term in Eq. (13.30) yields 3 3 ∑ ∑

⃗ 𝑓⃗𝑖𝑗 = 𝑓⃗12 + 𝑓⃗13 + 𝑓⃗21 + 𝑓⃗23 + 𝑓⃗31 + 𝑓⃗32 = 0,

(13.32)

𝑖=1 𝑗=1 𝑗≠𝑖

given that 𝑓⃗12 = −𝑓⃗21 , 𝑓⃗13 = −𝑓⃗31 , and 𝑓⃗23 = −𝑓⃗32 . Therefore, by Newton’s third law, the second sum on the left-hand side of Eq. (13.30) always vanishes, i.e., 𝑛 𝑛 ∑ ∑

⃗ 𝑓⃗𝑖𝑗 = 0.

(13.33)

𝑖=1 𝑗=1 𝑗≠𝑖

Finally, the last term in Eq. (13.30), 𝑛 ∑

𝑚𝑖 𝑎⃗𝑖 =

𝑖=1

∑𝑛

⃗𝑖 , 𝑖=1 𝑚𝑖 𝑎

𝑛 ∑

𝑚𝑖

𝑑 2 𝑟⃗𝑖

𝑖=1

𝑑𝑡2

can be written as

=

𝑛 𝑑2 ∑ 𝑚 𝑟⃗ , 𝑑𝑡2 𝑖=1 𝑖 𝑖

(13.34)

where, in moving 𝑑 2 ∕𝑑𝑡2 outside the summation, we used the fact that the mass of ∑ each particle is constant. We now recall that the term 𝑛𝑖=1 𝑚𝑖 𝑟⃗𝑖 defines the position of the system’s center of mass through the relation 𝑚⃗𝑟𝐺 =

𝑛 ∑

𝑚𝑖 𝑟⃗𝑖 ,

(13.35)

𝑖=1

∑ where 𝑚 = 𝑛𝑖=1 𝑚𝑖 is the total mass of the system and 𝑟⃗𝐺 is the position of the mass center 𝐺 (see Fig. 13.17). Using Eq. (13.35), Eq. (13.34) becomes 𝑛 ∑ 𝑖=1

𝑚𝑖 𝑎⃗𝑖 =

𝑑 2 𝑟⃗𝐺 𝑑2 (𝑚⃗𝑟𝐺 ) = 𝑚 = 𝑚𝑎⃗𝐺 , 𝑑𝑡2 𝑑𝑡2

(13.36)

where we have used the fact that, because each mass 𝑚𝑖 is constant, the total mass 𝑚 must also be constant. Substituting Eqs. (13.31), (13.33), and (13.36) into Eq. (13.30), we obtain 𝑛 ∑ 𝐹⃗ = 𝑚𝑖 𝑎⃗𝑖 = 𝑚𝑎⃗𝐺 . (13.37) 𝑖=1

ISTUDY

Section 13.3

Systems of Particles

Equation (13.37) says that the mass center of a system of particles moves as if it were a particle of mass 𝑚 subjected to only the total external force 𝐹⃗ . Consequently, ⃗ then the mass center moves with constant velocity. Therefore, if • If 𝐹⃗ = 0, ⃗ ⃗ 𝐹 = 0 and the mass center is initially at rest, the mass center will remain at rest. ⃗ then 𝑚𝑎⃗ = 0, ⃗ and its time integral must be constant, i.e., • If 𝐹⃗ = 0, 𝐺 ∫

𝑚𝑎⃗𝐺 𝑑𝑡 = 𝑚𝑣⃗𝐺 = constant,

(13.38)

which is a statement of the conservation of linear momentum for a system of particles. We will discuss this conservation law in Section 15.1. Finally, note that a body is simply a very large assembly of particles. The equation governing translational dynamics of a body is identical to Eq. (13.37), and it forms one part of the governing equations of rigid body dynamics in Chapter 17.

839

840

E X A M P L E 13.12 𝐸

Chapter 13

Force and Acceleration Methods for Particles

𝐿

Motion with Conservation of Linear Momentum The canoeist 𝑃 has paddled her canoe 𝐷 to the dock, as shown in Fig. 1. After the front end of her canoe reaches the end of the dock, she decides to walk the distance 𝐿 from the back of the canoe to the front and exit onto the dock. Assuming that the canoe can slide in the water with negligible resistance, determine the distance between her and the dock when she reaches the front end of the canoe to determine whether or not she will be able to exit onto the dock without getting wet. Assume she weighs 185 lb, the canoe weighs 70 lb, and 𝐿 = 10 f t.

𝐹

𝑃 𝐷

Figure 1 A canoeist who has just reached the dock and is about to walk the length of the canoe in order to exit onto the dock.

SOLUTION Road Map & Modeling Viewing the person and canoe as a two-particle system, Fig. 2 shows the FBD of the system as a whole, which includes only forces external to the system. 𝚥̂

𝑊𝑃

𝚤̂ 𝑊𝐷 𝐷 𝐹𝐵

Figure 2. FBD of the canoe and canoeist. We have modeled the canoe and the canoeist as particles.

Our model includes the weight of both the person and the canoe, as well as the buoyancy force 𝐹𝐵 . We have neglected any horizontal resistance offered by the water and/or air. Because the total horizontal force is equal to zero and the system is initially at rest, the position of the mass center will not change even as the person moves relative to the canoe. We will write Newton’s second law for the system as a whole and integrate it with respect to time to determine the position of each part of the system. Governing Equations Balance Principles Referring to the FBD in Fig. 2, applying the first of Eqs. (13.37) in the 𝑥 direction, we have ∑ 𝐹𝑥 ∶ 0 = 𝑚𝑃 𝑎𝑃 𝑥 + 𝑚𝐷 𝑎𝐷𝑥 , (1) 𝐿

where 𝑎𝑃 𝑥 and 𝑎𝐷𝑥 are the 𝑥 components of the acceleration of 𝑃 and 𝐷, respectively.

𝑑

Force Laws

𝑥𝑃 ℎ

Kinematic Equations

Because our analysis is limited to the 𝑥 direction, the kinematic

relations we need are 𝑎𝑃 𝑥 = 𝑣̇ 𝑃 𝑥 ,

𝐷 𝑥𝐷 Figure 3 Definitions of distances and coordinate directions for the problem’s kinematics.

ISTUDY

All forces are accounted for on the FBD.

𝑣𝑃 𝑥 = 𝑥̇ 𝑃 ,

𝑎𝐷𝑥 = 𝑣̇ 𝐷𝑥 ,

and

𝑣𝐷𝑥 = 𝑥̇ 𝐷 ,

(2)

where, referring to Fig. 3, the coordinate 𝑥 is measured from the end of the dock. Computation

Integrating Eq. (1) with respect to time yields 𝑚𝑃 𝑣𝑃 𝑥 + 𝑚𝐷 𝑣𝐷𝑥 = 𝐶1 ,

(3)

where 𝐶1 is a constant of integration. This equation says that for any arbitrary position of 𝑃 and 𝐷, the sum of the product of their masses and the 𝑥 component of their velocities is

ISTUDY

Section 13.3

Systems of Particles

constant. Hence, we can evaluate 𝐶1 by recalling that both the canoe 𝐷 and the canoeist 𝑃 have zero velocity after they reach the dock and before the canoeist starts to walk the length of the canoe, that is, 𝑚𝑃 ⋅ 0 + 𝑚 𝐷 ⋅ 0 = 𝐶 1

⇒

𝐶1 = 0.

(4)

Next, substituting 𝐶1 = 0 into Eq. (3) and integrating with respect to time, we have 𝑚𝑃 𝑥𝑃 + 𝑚𝐷 𝑥𝐷 = 𝐶2 ,

(5)

where 𝐶2 is a second constant of integration. Referring to Fig. 3, to find 𝐶2 we evaluate Eq. (5) when ℎ = 0, i.e., when the canoe is touching the dock and the canoeist is at the left end of the canoe. This yields 𝑚𝑃 (𝑑 + 𝐿) + 𝑚𝐷 (𝑑 + 𝐿∕2) = 𝐶2 ,

(6)

where 𝑑 is the constant distance from the end of the canoe to the point in the canoe at which the canoeist would step out. Substituting the expression for 𝐶2 given by Eq. (6) into Eq. (5), we have that the positions of the canoeist and the canoe are related as follows: 𝑚𝑃 𝑥𝑃 + 𝑚𝐷 𝑥𝐷 = 𝑚𝑃 (𝑑 + 𝐿) + 𝑚𝐷 (𝑑 + 𝐿∕2).

(7)

We can now determine where the canoe ends up when the canoeist walks to the front end. When the canoeist is at the front end of the canoe, we must have 𝑥𝐷∕𝑃 = 𝑥𝐷 − 𝑥𝑃 = 𝐿∕2. Therefore, 𝑥𝐷 = 𝑥𝑃 + 𝐿∕2, and Eq. (7) becomes 𝑚𝑃 𝑥𝑃 + 𝑚𝐷 (𝑥𝑃 + 𝐿∕2) = 𝑚𝑃 (𝑑 + 𝐿) + 𝑚𝐷 (𝑑 + 𝐿∕2). Solving for 𝑥𝑃 , we have 𝑥𝑃 =

𝑚𝑃 (𝑑 + 𝐿) + 𝑚𝐷 𝑑 . 𝑚𝑃 + 𝑚𝐷

(8)

(9)

Finally, since ℎ = 𝑥𝑃 − 𝑑 when the canoeist is at the front end of the canoe, for this situation we obtain 𝑚𝑃 ℎ= (10) 𝐿 = 7.255 f t, 𝑚𝑃 + 𝑚𝐷 where we have used the given data to obtain the final numerical answer. The result in Eq. (10) indicates that the poor canoeist will not be able to step from the canoe onto the dock without getting wet! Discussion & Verification

The result in Eq. (10) has the dimension of length, as it should. The distance ℎ is smaller than the length of the canoe and is therefore within reason. A Closer Look One solution to the problem of the canoe moving away from the dock would be for someone on the dock to tie the canoe to the dock after the canoe arrives. An interesting feature of Eq. (10) is that the speed with which the person walks along the canoe from one end to the other does not appear in the solution. In addition, it would be reasonable to expect that if the canoe were tied to the dock with a rope and then the canoeist walked the length of the canoe, there would be a tension in the rope. It turns out that it is not the speed that determines the tension in the rope — it is the acceleration of the person that determines the rope’s tension. We will see an interesting variation of this problem in Chapter 15 when we study momentum methods for systems of particles (see Example 15.4).

841

842

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.13

Pulley System Analysis The pulley system in Fig. 1 is designed to lift a heavy load at 𝐴 by attaching a mass at 𝑃 . Assuming that the payload 𝐴 has mass 𝑚𝐴 and the pulley housing 𝐵 (which includes the pulleys) has mass 𝑚𝐵 , determine the accelerations of 𝐴 and 𝑃 if a mass 𝑚𝑃 is attached at 𝑃 . Neglect friction in, and rotational inertia of, the pulleys.

𝐶

𝐸 cable 𝐺

SOLUTION Road Map & Modeling

We want to determine the acceleration of specific elements of the system, so we isolate these elements and sketch an FBD for each (as opposed to sketching an FBD for the whole system, as was done in Example 13.12). Modeling each moving element as a particle, ignoring the masses of the cables and assuming they are inextensible, we obtain the FBDs shown in Fig. 2. Since the motion of 𝐴, 𝐵, and 𝑃 is only in the

𝐵 cable 𝐻 𝑃 𝐴

𝑇𝐺

𝑇𝐺

𝑇𝐺

𝑇𝐺 𝑇𝐺

𝑇𝐻

Figure 1 A pulley system designed for lifting large loads.

𝚥̂

𝑚𝐵 𝑔

𝐴

𝑃

𝐵 𝑚𝐴 𝑔

𝑚𝑃 𝑔

𝑇𝐻

Figure 2. FBD of mass 𝐴, mass 𝑃 , and pulley housing 𝐵.

vertical direction, the problem is one-dimensional, and our component system consists of only one unit vector (̂𝚥), as shown in Fig. 2. Governing Equations Balance Principles

Using the FBDs in Fig. 2, application of Newton’s second law to 𝐴,

𝐵, and 𝑃 yields (∑

)

∶ 𝑚𝐴 𝑔 − 𝑇𝐻 = 𝑚𝐴 𝑎𝐴 , )𝐴 𝐹𝑦 ∶ 𝑚𝐵 𝑔 + 𝑇𝐻 − 4𝑇𝐺 = 𝑚𝐵 𝑎𝐵 , (∑ )𝐵 𝐹𝑦 ∶ 𝑚𝑃 𝑔 − 𝑇𝐺 = 𝑚𝑃 𝑎𝑃 .

(∑ 𝑦𝐵 𝑦𝐴

𝑃

𝑦𝑃

(1) (2) (3)

Force Laws The weights of 𝐴, 𝐵, and 𝑃 are already accounted for in Fig. 2. As far as the cables are concerned, their inextensibility is enforced by kinematic constraints instead of force laws.

𝐵 𝑃 𝐴 Figure 3 The positions of 𝐴, 𝐵, and 𝑃 .

ISTUDY

𝐹𝑦

Kinematic Equations

Since the cables are inextensible, from Fig. 3 we have 𝐿𝐺 = 4𝑦𝐵 + 𝑦𝑃

⇒

4𝑎𝐵 + 𝑎𝑃 = 0,

(4)

where 𝐿𝐺 denotes the length of cable 𝐺 (less a number of short, fixed-length segments). In addition, the inextensibility of the cable 𝐻 connecting 𝐴 and 𝐵 demands that 𝑎𝐴 = 𝑎𝐵 .

(5)

ISTUDY

Section 13.3

Systems of Particles

843

Computation

Equations (1)–(5) provide a system of five equations in the five unknowns 𝑎𝐴 , 𝑎𝐵 , 𝑎𝑃 , 𝑇𝐺 , and 𝑇𝐻 that can be solved to obtain ) ( 5𝑚𝑃 𝑚𝐴 + 𝑚𝐵 𝑇𝐺 = (6) ( ) 𝑔, 16𝑚𝑃 + 𝑚𝐴 + 𝑚𝐵 20𝑚𝐴 𝑚𝑃 (7) 𝑇𝐻 = ( ) 𝑔. 16𝑚𝑃 + 𝑚𝐴 + 𝑚𝐵 and

) ( 4𝑚𝑃 − 𝑚𝐴 + 𝑚𝐵 𝑎 𝐴 = 𝑎𝐵 = − ( ) 𝑔, 16𝑚𝑃 + 𝑚𝐴 + 𝑚𝐵 [ ( )] 4 4𝑚𝑃 − 𝑚𝐴 + 𝑚𝐵 𝑎𝑃 = ( ) 𝑔. 16𝑚𝑃 + 𝑚𝐴 + 𝑚𝐵

(8) (9)

Discussion & Verification

The fractions in Eqs. (8) and (9) are dimensionless, so the results in these equations have the dimensions of 𝑔, i.e., acceleration, as they should. The fractions in Eqs. (6) and (7) have dimensions of mass, so the results in these equations have dimensions of force, as they should. Equations (8) and (9) tell us that the accelerations of 𝐴 and 𝑃 are in the opposite direction, again as expected. Finally, the cable tension 𝑇𝐺 is always positive, so the cable never goes slack, as it should. Hence, our solution seems to be correct. A Closer Look Focusing on either Eq. (8) or (9), we see that the “bang for the buck” that we get with the mass at 𝑃 is 4 times its mass since the numerator of either of these equations contains the term 4𝑚𝑃 − (𝑚𝐴 + 𝑚𝐵 ). This term is positive if 𝑚𝑃 > 41 (𝑚𝐴 + 𝑚𝐵 ); and furthermore, if 𝑚𝑃 > 41 (𝑚𝐴 + 𝑚𝐵 ), then 𝑎𝑃 > 0 (otherwise, 𝑎𝑃 < 0). Thus, we see that it is not 𝑚𝑃 that must be greater than the combined mass of 𝐴 and 𝐵 for 𝑃 to pull 𝐴 up: 𝑚𝑃 need only be greater than 41 (𝑚𝐴 + 𝑚𝐵 )! The sometimes mysterious and serpentine configuration of pulley cables is used to reduce the effort required to lift loads.

Helpful Information Sign conventions. The positive direction of the 𝑦 axis is downward. Hence, when 𝑎𝐴 and 𝑎𝑃 have opposite signs, it means that they are accelerating in opposite directions. Indeed, 𝐴 and 𝑃 move in opposite directions, as can be seen by differentiating the first of Eqs. (4) with respect to time.

844

Chapter 13

Force and Acceleration Methods for Particles

E X A M P L E 13.14

Sliding with Friction

𝜇𝑘2 , 𝜇𝑠2

𝑚2 𝑚1

𝜇𝑘1

A pair of stacked books with masses 𝑚1 = 1.5 kg and 𝑚2 = 1 kg is thrown on a table (Fig. 1). The books strike the table with essentially zero vertical speed, and their common horizontal speed is 𝑣0 = 0.75 m∕s. Letting 𝜇𝑘1 = 0.45 be the coefficient of kinetic friction between the bottom book and the table, and letting 𝜇𝑠2 = 0.4 and 𝜇𝑘2 = 0.3 be the coefficients of static and kinetic friction between the two books, respectively, determine the books’ final positions relative to where they hit the table and their position relative to one another. Model both books as particles with the same initial horizontal position.

SOLUTION Figure 1 A pair of stacked books thrown horizontally on a rough table.

𝑚2 𝑔

𝐹2

𝐹2

𝑁2

𝑁2

𝚥̂ 𝚤̂

𝑚1 𝑔

Road Map & Modeling

When the books strike the table, the bottom book must slide on the table because it would otherwise experience an infinite deceleration in going from 𝑣0 to zero. Since there is no similar argument telling us that the top book must slide relative to the bottom book, we begin by assuming that 𝑚2 does not slip on 𝑚1 . After we obtain the solution using this assumption, we will check whether or not the results are consistent with it. If not, we will conclude that the books slide relative to one another, and we will have to compute a new solution. Since we need to determine their individual positions, we draw an FBD of each book as shown in Fig. 2. We have chosen the 𝑥 axis to be parallel to the trajectory of the books, with the origin taken to be the point at which the books first impact the table. Governing Equations Balance Principles

𝐹1

Applying Newton’s second law to the FBDs in Fig. 2, we obtain (∑ ) 𝐹𝑥 ∶ −𝐹2 = 𝑚2 𝑎2𝑥 , (1) (∑ ) 2 𝐹𝑦 ∶ 𝑁2 − 𝑚2 𝑔 = 𝑚2 𝑎2𝑦 , (2) (∑ )2 𝐹𝑥 ∶ 𝐹2 − 𝐹1 = 𝑚1 𝑎1𝑥 , (3) (∑ ) 1 𝐹𝑦 ∶ 𝑁1 − 𝑁2 − 𝑚1 𝑔 = 𝑚1 𝑎1𝑦 . (4)

𝑁1

Figure 2 FBDs for the two books modeled as particles.

ISTUDY

1

Force Laws Using the no-slip assumption between books 1 and 2 and the sliding assumption between book 1 and the table, the friction laws are

𝐹1 = 𝜇𝑘1 𝑁1 Kinematic Equations

and |𝐹2 ∕𝑁2 | < 𝜇𝑠2 .

(5)

Letting 𝑎 be the common acceleration of the books, we have

𝑎2𝑥 = 𝑎,

𝑎2𝑦 = 0,

𝑎1𝑥 = 𝑎,

and 𝑎1𝑦 = 0.

(6)

Computation

The first of Eqs. (5), along with the equations resulting from substituting Eqs. (6) into Eqs. (1)–(4), form a system of five equations in the five unknowns 𝐹1 , 𝐹2 , 𝑎, 𝑁1 , and 𝑁2 . Since our first objective is to verify whether or not the no-slip assumption is correct, we begin by focusing 𝐹2 and 𝑁2 . After a bit of algebra, 𝐹2 = 𝜇𝑘1 𝑚2 𝑔 = 4.414 N and 𝑁2 = 𝑚2 𝑔 = 9.810 N. Discussion & Verification

(7)

Substituting the results of Eqs. (7) into the inequality in

Eq. (5), we have |𝐹2 ∕𝑁2 | = 0.45 ≮ 𝜇𝑠2 = 0.4, which means that the no-slip assumption is incorrect and that 𝑚2 does slip over 𝑚1 .

(8)

ISTUDY

Section 13.3

Systems of Particles

845

𝒎𝟐 slides over 𝒎𝟏

We now rework the problem, assuming that the books slide relative to one another. The FBDs in Fig. 2 still apply, so Eqs. (1)–(4) are still valid. However, the force laws and corresponding kinematic equations need to reflect the new working assumption. Force Laws

The force laws are now 𝐹1 = 𝜇𝑘1 𝑁1

Kinematic Equations

and 𝐹2 = 𝜇𝑘2 𝑁2 .

(9)

The 𝑥 components of acceleration of 𝑚1 and 𝑚2 are now different,

so we have 𝑎2𝑥 = 𝑎2 ,

𝑎2𝑦 = 0,

𝑎1𝑥 = 𝑎1 ,

and 𝑎1𝑦 = 0.

(10)

Computation

The relations obtained by substituting Eqs. (10) into Eqs. (1)–(4), along with Eqs. (9), form a system of six equations in the six unknowns 𝑎1 , 𝑎2 , 𝐹1 , 𝐹2 , 𝑁1 , and 𝑁2 . Solving this system, we have ] ) )[ ( ( (11) 𝑎1 = − 𝑔∕𝑚1 𝜇𝑘1 𝑚1 + 𝑚2 − 𝜇𝑘2 𝑚2 = −5.396 m∕s2 , 𝑎2 = −𝜇𝑘2 𝑔 = −2.943 m∕s2 , ( ) 𝐹1 = 𝜇𝑘1 𝑔 𝑚1 + 𝑚2 = 11.04 N,

(12) (13)

𝐹2 = 𝜇𝑘2 𝑚2 𝑔 = 2.943 N, ( ) 𝑁1 = 𝑔 𝑚1 + 𝑚2 = 24.52 N,

(14)

𝑁2 = 𝑔𝑚2 = 9.810 N.

(16)

(15)

Now that we know 𝑎1 and 𝑎2 , and recalling that 𝑣0 is the initial speed of both books, we can compute how far each of them slides, using constant acceleration kinematics from Section 12.2. This yields 0 = 𝑣20 + 2𝑎1 𝑥1 = 𝑣20 −

) ] 2𝑔 [ ( 𝜇 𝑚 + 𝑚2 − 𝜇𝑘2 𝑚2 𝑥1 , 𝑚1 𝑘1 1

0 = 𝑣20 + 2𝑎2 𝑥2 = 𝑣20 − 2𝜇𝑘2 𝑔𝑥2 .

𝑥1 = 𝑥2 =

] = 0.05213 m, ) [ ( 2𝑔 𝜇𝑘1 𝑚1 + 𝑚2 − 𝜇𝑘2 𝑚2 𝑣20 2𝑔𝜇𝑘2

= 0.09557 m.

𝑥2 𝑥1

(17) (18)

Solving for 𝑥1 and 𝑥2 and using the given parameters, we have 𝑚1 𝑣20

𝑥2∕1

Figure 3 Relative position kinematics of the two books.

(19) (20)

Referring to Fig. 3, our solution implies that book 2, as viewed by an inertial observer, i.e., someone sitting on the table, slides 9.6 cm. However, relative to the bottom book, it slides 𝑥top∕bottom = 𝑥2∕1 = 𝑥2 − 𝑥1 = 0.04344 m. (21) Discussion & Verification The signs of the results in Eqs. (19)–(21) are what we expect given that both books move to the right, and book 2 stops to the right of book 1. Also, the distances slid by the books are on the order of centimeters, which seems reasonable. Because their accelerations are constant and the two books start their motion from the same position with identical speeds, the fact that 𝑥1 < 𝑥2 implies that book 1 stops first. This tells us that the friction laws used in the present solution remained valid throughout the motion of both books. Had the situation been reversed, we would have had to compute a new solution, although such a scenario is not achievable using the Coulomb friction model. In conclusion, the current solution appears to be correct.

846

Chapter 13

Force and Acceleration Methods for Particles

Problems Problem 13.104 𝐴

𝐵

The driver of the truck suddenly applies the brakes, and the truck comes to a stop. During braking, either the crate slides or it does not. Considering the forces acting on the truck during braking, will the truck stop in a shorter distance (or time) if the crate slides, or will the distance (or time) be shorter if it does not? Justify your answer. Figure P13.104

Problem 13.105 A car is being pulled to the right in the two ways shown. Neglecting the inertia of the pulleys and rope, as well as any friction in the pulleys, if the car is allowed to roll freely, will the acceleration of the car in (a) be smaller, equal to, or larger than the acceleration of the car in (b)?

(b)

(a) 500 lb

500 lb

Figure P13.105 𝐵

Problem 13.106 Particles 𝐴 and 𝐵, which are connected with a massless linear spring, have been thrown up in the air and are moving under the action of the spring force and their own weight. Assuming that no other forces are affecting the motion of the particles, what will be the acceleration of their center of mass?

𝐴

Problem 13.107

Figure P13.106

Two particles 𝐴 and 𝐵 with masses 𝑚𝐴 and 𝑚𝐵 , respectively, are placed at a distance 𝑟0 from one another. Assuming that the only force acting on the masses is their mutual gravitational attraction, determine the acceleration of particle 𝐵 as seen from particle 𝐴. 𝐴

𝐵 𝑟 Figure P13.107

𝐸

Problem 13.108 𝐷

𝐴

A person lifts the 80 kg load 𝐴 by pulling down on the rope with a constant force 𝐹 as shown. Neglecting any source of friction as well as the inertia of the ropes and the pulleys, determine 𝐹 if 𝐴 accelerates upward at 0.5 m∕s2 .

Problem 13.109 𝐹 Figure P13.108–P13.110

ISTUDY

The load 𝐴 weighs 185 lb. Neglecting any source of friction as well as the inertia of the ropes and the pulleys, determine the acceleration of 𝐴 if a person pulls down on the rope with a constant force 𝐹 = 185 lb as shown.

ISTUDY

Section 13.3

Systems of Particles

Problem 13.110 A person lifts the 80 kg load 𝐴 by pulling down on the rope with a constant force 𝐹 as shown. Neglecting friction, the inertia of the ropes, and the rotational inertia of the pulleys, but accounting for the fact that pulley 𝐷 has a mass 𝑚𝐷 = 8 kg, determine 𝐹 if 𝐴 accelerates upward at 2.5 m∕s2 .

Problem 13.111 Revisit Example 13.13 and determine the expression for the acceleration of 𝐴 if the load at 𝑃 is replaced by a force with magnitude equal to the load’s weight, i.e., 𝐹 = 𝑚𝑃 𝑔.

𝐶

𝐸

motor 𝑀 cable 𝐺

𝐵

𝐵

𝐴

𝑃 cable 𝐻 𝐹 = 𝑚𝑃 𝑔

𝐴 Figure P13.111

cargo 𝐶

Figure P13.112

Problem 13.112 The motor 𝑀 is at rest when someone flips a switch and it starts pulling in the rope. The acceleration of the rope is uniform, and it takes 1 s to achieve a retraction rate of 4 f t∕s. After 1 s the retraction rate becomes constant. Determine the tension in the cable during and after the initial 1 s interval. The cargo 𝐶 weighs 130 lb, pulleys 𝐴 and 𝐵 each weigh 12 lb, and the weight of the ropes is negligible. Neglect friction in the pulleys and the rotational inertia of the pulleys.

Problem 13.113 Revisit Example 13.14 and assume that the static coefficient of friction between the two books is 𝜇𝑠2 = 0.55, while all other parameters stay as specified in the example. Determine the acceleration of each of the books. 𝜇𝑘2 , 𝜇𝑠2

𝑚2 𝑚1

𝜇𝑘1

Figure P13.113

847

848

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.114 As seen in Fig. P13.114(a), a window washing platform is controlled by the two pulley systems at 𝐴𝐵 and 𝐶𝐷. The workers 𝐸 and 𝐹 can raise and lower the platform 𝑃 by pulling on the cables 𝐻 and 𝐼, respectively. The weight of each of the workers is 185 lb, and the platform 𝑃 weighs 200 lb. A schematic representation of the pulley system is shown in Fig. P13.114(b). If the workers start from rest and, in 1.5 s, uniformly start pulling the cable in at 2.5 f t∕s, determine the force each worker must exert on the cables 𝐻 and 𝐼 during that 1.5 s. Neglect the mass of the pulleys, friction in the pulleys, and the mass of the cable. Assume each worker pulls with the same force, and ignore the departure from vertical of segments 𝐻 and 𝐼. 𝐶

𝐵

𝐵 𝐻

𝐼 𝐿 1 𝐿 2 𝐿 3 𝐿4 𝐿5

𝐹

𝐸 𝐴

𝐻 𝐷

𝐴

𝑃 (b) the pulleys

(a) the platform Figure P13.114

Problem 13.115 𝐴

A man 𝐴 is trying to keep his balance while on a metal wedge 𝐵 that is sliding down an icy incline. Letting 𝑚𝐴 = 78 kg and 𝑚𝐵 = 25 kg be the masses of 𝐴 and 𝐵, respectively, and assuming that there is enough friction between 𝐴 and 𝐵 for 𝐴 not to slide with respect to 𝐵, determine the value of the normal reaction force between 𝐴 and 𝐵, as well as the magnitude of their acceleration if 𝜃 = 20◦ . Friction between the wedge and the incline is negligible. 𝐵

ice

𝜃

A man 𝐴 is trying to keep his balance while on a metal wedge 𝐵 that is sliding down an icy incline. The weights of 𝐴 and 𝐵 are 𝑊𝐴 = 181 lb and 𝑊𝐵 = 50 lb, respectively. Determine the minimum coefficient of static friction 𝜇𝑠 between 𝐴 and 𝐵 required for 𝐴 not to slide with respect to 𝐵 if 𝜃 = 23◦ . Friction between the wedge and the incline is negligible.

Figure P13.115 and P13.116

𝐹0

𝐴 𝐵

Problem 13.116

𝜇1 𝜇2

Figure P13.117

Problem 13.117 A force 𝐹0 of 400 lb is applied to block 𝐴. Letting the weights of 𝐴 and 𝐵 be 55 and 73 lb, respectively, and letting the static and kinetic friction coefficients between blocks 𝐴 and 𝐵 be 𝜇1 = 0.25, and the static and kinetic friction coefficients between block 𝐵 and the ground be 𝜇2 = 0.45, determine the accelerations of both blocks.

Problem 13.118 𝐹0

Figure P13.118

ISTUDY

𝐴 𝐵

𝜇1 𝜇2

A force 𝐹0 of 400 lb is applied to block 𝐵. Letting the weights of 𝐴 and 𝐵 be 55 and 73 lb, respectively, and letting the static and kinetic friction coefficients between blocks 𝐴 and 𝐵 be 𝜇1 = 0.25, and the static and kinetic friction coefficients between block 𝐵 and the ground be 𝜇2 = 0.45, determine the accelerations of both blocks.

ISTUDY

Section 13.3

Systems of Particles

Problem 13.119 Blocks 𝐴 and 𝐵 are connected by a pulley system. The coefficient of kinetic friction between the block 𝐴 and the incline is 𝜇𝑘 = 0.25, and static friction is insufficient to prevent slipping. The mass of 𝐴 is 𝑚𝐴 = 7 kg, the mass of 𝐵 is 𝑚𝐵 = 20 kg, and the angle between the incline and the horizontal is 𝜃 = 30◦ . Determine the acceleration of 𝐴, the acceleration of 𝐵, and the tension in the rope after the system is released.

Problem 13.120

𝜃

Two identical balls, each of mass 𝑚, are connected by a string of negligible mass and length 2𝑙. A short string is attached to the middle of the string connecting the two balls and is pulled vertically with a constant force 𝑃 . If the system starts from rest at 𝜃 = 𝜃0 and assuming that the balls only move in the horizontal direction, determine the expression for the speed of the two balls as 𝜃 approaches 90◦ . Neglect the size of the balls as well as friction between the balls and the surface on which they slide. Hint: Taking advantage of symmetry, let 𝑥 be the horizontal distance measured from the center of the system toward one of the balls. After drawing an FBD and writing the equation of motion for a ball, make use of the kinematic relationship 𝑥̈ 𝑑𝑥 = 𝑥̇ 𝑑 𝑥. ̇

𝑙

𝑙

𝜃

𝜃

Figure P13.120

Problem 13.121 The two sliders 𝐴 and 𝐵 of mass 𝑚𝐴 = 4 kg and 𝑚𝐵 = 3 kg, respectively, move with negligible friction in the slots shown, which lie in the vertical plane. They are connected by a rigid bar of negligible mass and length 𝐿 = 0.5 m. If, for 𝑦𝐵 = 0.3 m, the system is initially at rest, determine the acceleration of each slider and the force in the bar immediately after release. Hint: The force that the bar exerts on each slider has the same direction as the bar itself. 𝐵

𝐵

𝐿

𝑦𝐵

𝐴

𝑦𝐵

𝐿

𝐴 𝑥𝐴 Figure P13.121

𝐴

𝑃

𝑥𝐴 Figure P13.122

Problem 13.122 The two sliders 𝐴 and 𝐵 of weight 𝑊𝐴 = 8 lb and 𝑊𝐵 = 6 lb, respectively, move with negligible friction in the slots shown, which lie in the vertical plane. They are connected

𝜇𝑘 𝐵

Figure P13.119

849

850

Force and Acceleration Methods for Particles

Chapter 13

by a rigid bar of negligible weight and length 𝐿 = 1.75 f t. Slider 𝐴 is also subject to the force 𝑃 = 15 lb. If, for 𝑦𝐵 = 1.0 f t, the system is initially at rest, determine the acceleration of each slider and the force in the bar immediately after release. Hint: The force that the bar exerts on each slider has the same direction as the bar itself.

Problems 13.123 and 13.124 Spring scales work by measuring the displacement of a spring that supports both the platform of mass 𝑚𝑝 and the object, of mass 𝑚, whose weight is being measured. Most scales read zero when no mass 𝑚 has been placed on them; that is, they are calibrated so that the weight reading accounts for the mass of the platform 𝑚𝑝 . Assume that the spring is linear elastic with spring constant 𝑘.

𝑚 𝑚𝑝

If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the maximum reading on the scale after the mass is released.

Problem 13.123

Figure P13.123 and P13.124

If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the expression for maximum speed attained by the mass 𝑚 as the spring compresses.

Problem 13.124

Problem 13.125 A simple elevator consists of a 15,000 kg car 𝐴 connected to a 12,000 kg counterweight 𝐵. Suppose that a failure occurs when the car is at rest and 50 m above its buffer, causing the elevator car to fall. Model the car and the counterweight as particles and the cord as massless and inextensible; and model the action of the emergency brakes using a Coulomb friction model with kinetic friction coefficient 𝜇𝑘 = 0.5 and a normal force equal to 35% of the car’s weight. Determine the speed with which the car impacts the buffer.

car 𝐴

Problems 13.126 and 13.127 counterweight 𝐵 car buffer

counterweight buffer

Figure P13.125

𝐴

𝑘

𝐵

Figure P13.126 and P13.127

The linear elastic spring with stiffness 𝑘 and unstretched length 𝓁𝑢 is attached to both the vertical wall and the metal block 𝐴 of mass 𝑚𝐴 . The metal block 𝐵 is pushed into 𝐴 so that the spring compresses a distance 𝑑. The block 𝐵 is then released from rest. Problem 13.126 Assuming that friction between the blocks and the horizontal surface is negligible, determine the distance the blocks slide before they separate and their speed at separation. Hint: The blocks will start to separate when the normal force between them goes to zero. Problem 13.127 Assuming that friction between the blocks and the surface is nonnegligible and that the coefficient of static friction is insufficient to prevent motion, determine the condition on the compression distance 𝑑 for the blocks to separate. The coefficient of kinetic friction between the blocks and the surface is 𝜇𝑘 . Hint: The blocks will start to separate when the normal force between them goes to zero.

Problem 13.128 𝐴

𝐵 𝑟0 𝑟

Figure P13.128

ISTUDY

Two particles 𝐴 and 𝐵 with masses 𝑚𝐴 and 𝑚𝐵 , respectively, are a distance 𝑟0 apart, and both masses are initially at rest. Using Eq. (11.5) on p. 621, determine the amount of time it takes for the two masses to come into contact if 𝑚𝐴 = 1 kg, 𝑚𝐵 = 2 kg, and 𝑟0 = 1 m. Assume that the two masses are only influenced by their mutual gravitational attraction. 𝑟 √ 3∕2 Hint: ∫0 0 𝑟0 𝑟∕(𝑟0 − 𝑟) 𝑑𝑟 = 21 𝜋𝑟0 .

ISTUDY

Section 13.3

851

Systems of Particles

Problem 13.129 Energy storage devices that use spinning flywheels to store energy are becoming available. To maximize energy storage, the flywheel must spin as fast as possible. Unfortunately, if it spins too fast, internal stresses in the flywheel cause it to come apart catastrophically. Therefore, it is important to keep the speed at the periphery of the flywheel below about 1000 m∕s. It is also critical that the flywheel be well balanced to avoid the damaging vibrations that would otherwise result. With this in mind, let the flywheel 𝐷 with diameter 0.3 m rotate at 𝜔 = 60,000 rpm. In addition, assume that the cart 𝐵 is constrained to move rectilinearly along the guide tracks. Given that the flywheel is not perfectly balanced, that the unbalanced weight 𝐴 has mass 𝑚𝐴 , and that the total mass of the flywheel 𝐷, cart 𝐵, and electronics package 𝐸 is 𝑚𝐵 , determine the constraint force between the wheels of the cart and the guide tracks as a function of 𝜃, the masses, the diameter, and the angular speed of the flywheel. What is the maximum constraint force between the wheels of the cart and the guide tracks? Finally, evaluate your answers for 𝑚𝐴 = 1 g (about the mass of a paper clip) and 𝑚𝐵 = 70 kg. Assume that the unbalanced mass is at the periphery of the flywheel.

Problem 13.130

𝜔

𝐸 𝐵 guide tracks

top view

𝜃

Figure P13.129

The double pendulum shown consists of two particles with masses 𝑚1 = 7.5 kg and 𝑚2 = 12 kg connected by two inextensible cords of length 𝐿1 = 1.4 m and 𝐿2 = 2 m and negligible mass. If the system is released from rest when 𝜃 = 10◦ and 𝜙 = 20◦ , determine the tension in the two cords at the instant of release. Hint: The acceleration of 𝑚2 is most naturally written as 𝑎⃗2 = 𝑎⃗1 + 𝑎⃗2∕1 , where the absolute acceleration 𝑎⃗1 and the relative acceleration 𝑎⃗2∕1 are most easily described in the local polar coordinate systems attached to the respective masses.

Problems 13.131 and 13.132

𝐴 𝐷

𝐿1

𝑢̂ 𝜃

𝜃 𝑚1

𝜙

𝐴 𝜃

𝑢̂ 𝜃 𝑢̂ 𝑟

𝛽 𝑑

𝜙

𝑅

𝑢̂ 𝜙

𝐵

Using the angle 𝜃 as the dependent variable, derive the equation of motion for the particle system from the moment of release until particle 𝐵 reaches point 𝐷. Problem 13.131

Determine the expression for the speed of the spheres immediately before 𝐵 reaches point 𝐷.

𝑢̂ 𝜌

𝑚2

Figure P13.130

Two small spheres 𝐴 and 𝐵, each of mass 𝑚, are attached at either end of a rod of length 𝑑. The system is released from rest in the position shown. Neglect friction, treat the spheres as particles (assume that their diameter is negligible), neglect the mass of the rod, assume the rod is rigid, and assume that 𝑑 < 𝑅. Hints: (1) The force that the rod exerts on either ball has the same direction as the rod itself; (2) the kinematics and kinetics of each particle is most easily described in the local polar coordinate system attached to each particle.

Problem 13.132

𝑢̂ 𝜙

𝐿2

𝑢̂ 𝑟

𝑢̂ 𝜌

𝐷

Figure P13.131 and P13.132

Problem 13.133 A 62 kg woman 𝐴 sits atop the 60 kg cart 𝐵, both of which are initially at rest. The cart is rigidly attached to a wall by the rope 𝐶𝐷. If the woman slides down the frictionless incline of length 𝐿 = 3.5 m, determine the tension in the rope 𝐶𝐷 as she slides down the incline. Ignore the mass of the wheels on which the cart can roll. The angle 𝜃 = 26◦ . Hint: The woman’s acceleration is most naturally written as 𝑎⃗𝐴 = 𝑎⃗𝐵 + 𝑎⃗𝐴∕𝐵 , where the direction of the relative acceleration is along the incline.

Problem 13.134 A 62 kg woman 𝐴 sits atop the 60 kg cart 𝐵, both of which are initially at rest. If 𝜃 = 26◦ and the woman slides down the incline of length 𝐿 = 3.5 m, determine the velocity of

𝐿

𝜃

𝐶 𝐷 Figure P13.133

𝐴

𝐵

852

Force and Acceleration Methods for Particles

Chapter 13

both the woman and the cart when she reaches the bottom of the incline. Ignore the mass of the wheels on which the cart rolls and friction in their bearings, and neglect friction between the woman and the incline. Hint: The woman’s acceleration is most naturally written as 𝑎⃗𝐴 = 𝑎⃗𝐵 + 𝑎⃗𝐴∕𝐵 , where the direction of the relative acceleration is along the incline.

𝐴 𝐿

Problems 13.135 through 13.138 𝜃

𝐵

Figure P13.134

Two blocks 𝐴 and 𝐵 weighing 123 and 234 lb, respectively, are released from rest as shown. At the moment of release the spring is unstretched. In solving these problems, model 𝐴 and 𝐵 as particles, neglect air resistance, and assume that the cord is inextensible. Hint: If 𝐵 hits the ground, then its maximum displacement is equal to the distance between the initial position of 𝐵 and the ground. Problem 13.135

𝑘

Determine the maximum displacement and the maximum speed of block 𝐵 if 𝛼 = 0◦ , the contact between 𝐴 and the surface is frictionless, and the spring constant is 𝑘 = 30 lb∕f t.

𝐴

Problem 13.136

Determine the maximum displacement and the maximum speed of block 𝐵 if 𝛼 = 20◦ , the contact between 𝐴 and the incline is frictionless, and the spring constant is 𝑘 = 30 lb∕f t.

𝛼 𝐵

Problem 13.137

Determine the maximum displacement and the maximum speed of block 𝐵 if 𝛼 = 20◦ , the contact between 𝐴 and the incline is frictionless, and the spring constant is 𝑘 = 300 lb∕f t.

2 ft

Figure P13.135–P13.138

Problem 13.138

Determine the maximum displacement and the maximum speed of block 𝐵 if 𝛼 = 35◦ , the static and kinetic friction coefficients are 𝜇𝑠 = 0.25 and 𝜇𝑘 = 0.2, respectively, and the spring constant is 𝑘 = 25 lb∕f t.

𝐵

𝜃

Problems 13.139 and 13.140 In the ride shown, a person 𝐴 sits in a seat that is attached by a cable of length 𝐿 to a freely moving trolley 𝐵 of mass 𝑚𝐵 . The total mass of the person and the seat is 𝑚𝐴 . The trolley is constrained by the beam to move in only the horizontal direction. The system is released from rest at the angle 𝜃 = 𝜃0 and is allowed to swing in the vertical plane. Neglect the mass of the cable, and treat the person and the seat as a single particle.

𝐿 𝐴

Figure P13.139 and P13.140

ISTUDY

Problem 13.139 Derive the system’s equations of motion, using the position of the trolley and the angle 𝜃 as dependent variables. Problem 13.140 Derive the system’s equations of motion, using the position of the trolley and the angle 𝜃 as dependent variables, and then use a computer to solve these equations for one full period/cycle of the motion. Plot the speed of the trolley and the speed of the person vs. the angle 𝜃 for 𝑚𝐴 = 45 kg, 𝑚𝐵 = 10 kg, 𝐿 = 3 m, and 𝜃0 = 70◦ .

ISTUDY

Section 13.3

853

Systems of Particles

Design Problems Design Problem 13.3 In pulley problems, we have assumed that the ropes are inextensible and massless. In some engineering problems, such as in the design of fast elevators, this may not be a reasonable assumption. Here we will confront one aspect of any design process concerning the quality of the information provided by models of different complexity. In this problem, we will consider the effects of the rope’s deformation. However, to avoid excessive complexity, we will model the rope’s elasticity as being “lumped” at one end. Let the system be released from rest and the velocity of 𝐴 be controlled so that it accelerates uniformly downward to a speed of 3 m∕s in 1.2 s. Over this time interval, determine and plot, for different values of 𝑘, the position of 𝐵 as a function of time, as well as the tension in the rope as a function of time. Keep in mind that the inextensible rope model can be viewed as a special case of the deformable rope model with an infinite stiffness 𝑘. Consider various values of 𝑘, starting with extremely large values (to simulate infinity) and gradually decreasing the value of 𝑘 to understand at what point the two models provide significantly different answers (you will need to decide what significant means in this context).

𝑘 𝐴

𝐵

𝑊𝐴 = 54 lb 𝑊𝐵 = 54 lb Figure DP13.3

Design Problem 13.4 In this problem we will explore the assumption that we can neglect the mass of cables or ropes in pulley systems. Consider the simple pendulum in the left part of the figure. Let the pendulum bob be set in motion in the position shown with a speed 𝑣0 = 3 m∕s. Determine the motion of the pendulum bob and, in particular, its maximum displacement from its starting position. Next consider a double pendulum with an equally long cord, but with a cord of mass 𝑚𝐶 . To gain an appreciation for the effect of the mass of the cord on the pendulum’s motion, let the entire mass of the cord be lumped at its midpoint (we then neglect the mass of the two cords connecting 𝑂 to 𝑚𝐶 and 𝑚𝐶 to 𝑚𝐵 ). Next let this double pendulum be set in motion in the same way as in the previous case. Again, determine the motion of the pendulum bob 𝑚𝐵 , and determine the value of 𝑚𝐶 necessary to cause a 10% difference in the maximum displacement from its starting position.

𝑂

𝑚𝐶

2m

𝑚𝐵 = 12 kg

𝑚𝐵 = 12 kg

𝑣0

𝑣0

Figure DP13.4

Design Problem 13.5 In this problem we will explore the assumption that we can neglect the mass of cables or ropes in pulley systems. Consider the simple pendulum in the left part of the figure. Let the pendulum bob be set in motion in the position shown with a speed 𝑣0 = 3 m∕s. Determine the motion of the pendulum bob and, in particular, its maximum displacement from its starting position. Next consider a pendulum with an equally long cord but with a cord of mass 𝑚𝐶 . To gain an appreciation for the effect of the mass of the cord on the pendulum’s motion, let the entire mass of the cord be lumped at its midpoint (we then neglect the mass of the single rigid rod connecting 𝑂 to 𝑚𝐶 to 𝑚𝐵 ). Next let this rigid pendulum be set in motion in the same way as in the previous case. Again, determine the motion of the pendulum bob 𝑚𝐵 , and determine the value of 𝑚𝐶 necessary to cause a 10% difference in the maximum displacement from its starting position. Hint: Assume that the rigid bar can provide only forces that are parallel to the bar itself.

𝑂

𝑚𝐶

2m

𝑚𝐵 = 12 kg

𝑚𝐵 = 12 kg

𝑣0

𝑣0

Figure DP13.5

854

Chapter 13

Force and Acceleration Methods for Particles

13.4 C h a p t e r R e v i e w

ISTUDY

In this chapter, we have presented a general approach to the solution of kinetics problems involving the motion of a single particle or a system of particles. We now present a concise summary.

Applying Newton’s second law. We developed a four-step problem-solving procedure for applying Newton’s second law to mechanical systems. The central element of this procedure is the derivation of the governing equations, which originate from the 1. Balance principles. 2. Force laws. 3. Kinematic equations. In this chapter, the balance principle was Newton’s second law, which we write as Eq. (13.1), p. 789

Helpful Information 𝐹⃗ = 𝑚𝑎. ⃗ We will structure the solution of kinetics problems using the following four steps: 1. Road Map & Modeling: Identify data and unknowns, identify the system, state assumptions, sketch the FBD, and identify a problem-solving strategy. 2. Governing Equations: Write the balance principles, force laws, and kinematic equations. 3. Computation: Solve the assembled system of equations. 4. Discussion & Verification: Study the solution and perform a “sanity check.”

We applied Newton’s second law in component form as Eq. (13.2), p. 790 ∑

𝐹𝑎 = 𝑚𝑎𝑎 ,

∑

𝐹𝑏 = 𝑚𝑎𝑏 ,

and

∑

𝐹𝑐 = 𝑚𝑎𝑐 ,

where 𝑎, 𝑏, and 𝑐 are the orthogonal directions of the chosen component system, and we usually need only two directions for planar problems. The four-step “recipe” we presented for applying Newton’s second law, outlined in the margin note, is given as a guide to the order in which things should be done, and we will use it consistently in each example we present.

Governing equations and equations of motion. The governing equations for a system consist of the (1) balance principles, (2) force laws, and (3) kinematic equations. The equations of motion are the equations derived from the governing equations that allow for the determination of the motion. Degrees of freedom. A system’s degrees of freedom are the independent coordinates needed to completely describe a system’s position. The number of degrees of freedom is equal to the number of different coordinates in a system that must be fixed in order to keep the system from moving. The required number of equations of motion is equal to the number of degrees of freedom for a system. Friction. We will include friction via the Coulomb friction model. According to this model, in the absence of slip, the magnitude of the friction force 𝐹 satisfies the inequality Eq. (13.3), p. 791 |𝐹 | ≤ 𝜇𝑠 |𝑁|, where 𝜇𝑠 is the coefficient of static friction and 𝑁 is the force normal to the contact surface. The relation Eq. (13.4), p. 792 𝐹 = 𝜇𝑠 𝑁 defines the case of impending slip.

ISTUDY

Section 13.4

Chapter Review

If 𝐴 and 𝐵 are two objects sliding with respect to one another, the magnitude of the friction force exerted by 𝐵 onto 𝐴 is given by Eq. (13.6), p. 792 𝐹 = 𝜇𝑘 𝑁, where 𝜇𝑘 is the coefficient of kinetic friction and where the direction of the friction force must be consistent with the fact that friction opposes the relative motion of 𝐴 and 𝐵.

Springs. A spring is said to be linear elastic if the internal force in the spring is linearly related to the amount the spring is stretched or compressed. The force 𝐹𝑠 required to stretch or compress a linear elastic spring by an amount 𝛿 is given by Eq. (13.14), p. 793 ) ( 𝐹𝑠 = 𝑘𝛿 = 𝑘 𝐿 − 𝐿0 , where 𝑘 is the spring constant and where 𝐿 and 𝐿0 are the current and unstretched lengths of the spring, respectively. When 𝛿 > 0, the spring is said to be stretched; and when 𝛿 < 0, the spring is said to be compressed.

Systems of particles In practice, applying Newton’s second law to a system of particles is essentially identical to applying it to a single particle — we write (𝐹⃗𝑖 )tot = 𝑚𝑖 𝑎⃗𝑖 , 𝑖 = 1, … , 𝑛, for each of the 𝑛 particles in the system, where (𝐹⃗𝑖 )tot is the total force acting on particle 𝑖, thus including both the external and internal forces acting on the particle. In addition to Newton’s second law, we need to enforce Newton’s third law, which is crucial to account for the effect of the internal forces.

Motion of the mass center. The motion of the mass center of a system of particles is governed by the relation Eq. (13.37), p. 838 𝐹⃗ =

𝑛 ∑

𝑚𝑖 𝑎⃗𝑖 = 𝑚𝑎⃗𝐺 ,

𝑖=1

where 𝐹⃗ is the total external force on the system, 𝑚 is the total mass of the system, and 𝑎⃗𝐺 is the acceleration of the center of mass of the system. This equation has some important consequences: ⃗ then the mass center moves with constant velocity; and if the mass center • If 𝐹⃗ = 0, is initially at rest, it will remain at rest. ⃗ then we can conclude that 𝑚𝑎⃗ must be zero and its time integral must • If 𝐹⃗ = 0, 𝐺 be a constant.

855

856

Chapter 13

Force and Acceleration Methods for Particles

Review Problems Problem 13.141 𝐹

A constant force 𝑃 is applied at 𝐴 to the rope running behind the load 𝐺, which has a mass of 300 kg. Assuming that any source of friction and the inertia of the pulleys can be neglected, determine 𝑃 such that 𝐺 has an upward acceleration of 1 m∕s2 .

𝐸

𝐵

Problem 13.142 𝐶

A constant force 𝑃 = 300 lb is applied at 𝐴 to the rope running behind the load 𝐺, which weighs 1000 lb. If each of the pulleys weighs 7 lb, and assuming that any source of friction and the rotational inertia of the pulleys can be neglected, determine the acceleration of 𝐺 and the tension in the rope connecting pulleys 𝐵 and 𝐶.

𝐷

𝐺

Problem 13.143 𝐴 𝑃

Figure P13.141 and P13.142

A metal ball weighing 0.2 lb is dropped from rest into a fluid. If the magnitude of the resistance due to the fluid is given by 𝐶𝑑 𝑣, where 𝐶𝑑 = 0.5 lb⋅s∕f t is a drag coefficient and 𝑣 is the ball’s speed, determine the depth at which the ball will have sunk when the ball achieves a speed of 0.3 f t∕s.

Problem 13.144 A metal ball weighing 0.2 lb is dropped from rest into a fluid. After falling 1 f t, the ball has a speed of 2.25 f t∕s. If the magnitude of the resistance due to the fluid is given by 𝐶𝑑 𝑣, where 𝐶𝑑 is a drag coefficient and 𝑣 is the ball’s speed, determine the value of 𝐶𝑑 .

Problem 13.145

Figure P13.143 and P13.144

Two particles 𝐴 and 𝐵, with masses 𝑚𝐴 and 𝑚𝐵 , respectively, are a distance 𝑟0 apart. Particle 𝐵 is fixed in space, and 𝐴 is initially at rest. Using Eq. (11.5) on p. 621 and assuming that the diameters of the masses are negligible, determine the time it takes for the two particles to come into contact if 𝑚𝐴 = 1 kg, 𝑚𝐵 = 2 kg, and 𝑟0 = 1 m. Assume that the 𝑟 √ 3∕2 two masses are infinitely far from any other mass. Hint: ∫0 0 𝑟0 𝑟∕(𝑟0 − 𝑟) 𝑑𝑟 = 21 𝜋𝑟0 . 𝐵

𝐴 𝑟0 𝑟 Figure P13.145

Problems 13.146 and 13.147 𝐴

𝑑

𝑚

𝜇𝑠

𝜃 𝛼0 𝐵

The system shown is initially at rest when the bent bar starts to rotate about the vertical axis 𝐴𝐵 with constant angular acceleration 𝛼0 = 3 rad∕s2 . The coefficient of static friction between the collar of mass 𝑚 = 2 kg and the bent bar is 𝜇𝑠 = 0.6, the angle of the bend in the bar is 𝜃 = 30◦ , and the collar is initially at 𝑑 = 70 cm from the spin axis 𝐴𝐵. Assuming the motion starts at 𝑡 = 0, determine the time at which the collar starts to slip relative to the bent bar.

Problem 13.146

Problem 13.147 Figure P13.146 and P13.147

ISTUDY

Determine the number of rotations undergone by the bent bar when the collar starts to slip relative to it.

ISTUDY

Section 13.4

Chapter Review

Problem 13.148 The centers of two spheres 𝐴 and 𝐵, with weights 𝑊𝐴 = 3 lb and 𝑊𝐵 = 7 lb, respectively, are a distance 𝑟0 = 5 f t apart when they are released from rest. Using Eq. (11.5) on p. 621, determine the speed with which they collide if the diameters of spheres 𝐴 and 𝐵 are 𝑑𝐴 = 2.5 in. and 𝑑𝐵 = 4 in., respectively. Assume that the two masses are only influenced by their mutual gravitational attraction. 𝐵

𝐴 𝑟0 Figure P13.148

𝑣

Problem 13.149

𝐴 𝜌

A roller coaster goes over the top 𝐴 of the track shown with a speed 𝑣 = 135 km∕h. If the radius of curvature at 𝐴 is 𝜌 = 60 m, determine the minimum force that a restraint must apply to a person with a mass of 85 kg to keep the person on his or her seat.

Problem 13.150

𝐶

Figure P13.149

A 50,000 lb aircraft is flying along a rectilinear path at a constant altitude with a speed 𝑣 = 720 mph when the pilot initiates a turn by banking the plane 20◦ to the right. Assuming that the initial rate of change of speed is negligible, determine the components of the acceleration of the aircraft right at the beginning of the turn if the pilot does not adjust the attitude of the aircraft so that the magnitude of the lift remains the same as when the plane is flying straight and the aerodynamic drag remains in the horizontal plane. Also determine the radius of curvature at the beginning of the turn. 𝜃 = 0 at release

𝑣0 𝑅 𝜃

𝜌

𝑅

𝐶

Figure P13.150

Figure P13.151

Problem 13.151 Referring to Example 13.9 on p. 820, let 𝑅 = 1.25 f t and let the angle at which the sphere separates from the cylinder be 𝜃𝑠 = 34◦ . If the sphere were placed in motion at the very top of the cylinder, determine the sphere’s initial speed.

𝜌

Problem 13.152 Referring to Example 13.8 on p. 818, show that, for 𝜃 = 33◦ and under the assumption that 𝜇𝑠 > 1∕ tan 𝜃, the no-slip solution in Eqs. (15) and (16) satisfies the no-slip condition |𝐹 | ≤ 𝜇𝑠 |𝑁| for any value of the car’s speed.

𝜃 Figure P13.152

857

858

Chapter 13

Force and Acceleration Methods for Particles

Problem 13.153 Revisit Example 13.7 and assume that the drag force acting on the ball has the form 𝐹⃗𝑑 = −𝜂 𝑣, ⃗ where 𝑣⃗ is the velocity of the ball and 𝜂 is a drag coefficient. Determine the trajectory of the ball, expressing it in the form 𝑦 = 𝑦(𝑥). 𝑣0 𝛽

𝑂

𝑅 Figure P13.153 and P13.154

Problem 13.154 Revisit Example 13.7 and assume that the drag force acting on the ball has the form 𝐹⃗𝑑 = −𝜂 𝑣, ⃗ where 𝑣⃗ is the velocity of the ball and 𝜂 is a drag coefficient. Determine the value of 𝜂 such that a 1.61 oz ball has a range 𝑅 = 270 yd when put in motion with an initial velocity of magnitude 𝑣0 = 186 mph and initial direction 𝛽 = 11.2◦ .

Problem 13.155 The load 𝐵 has a mass 𝑚𝐵 = 250 kg, and the load 𝐴 has a mass 𝑚𝐴 = 120 kg. Let the system be released from rest, and neglecting any source of friction, as well as the inertia of the ropes and the pulleys, determine the acceleration of 𝐴 and the tension in the cord to which 𝐴 is attached.

𝐸

𝐷

Problem 13.156

𝐴

The load 𝐵 weighs 300 lb. Neglecting any source of friction, as well as the inertia of the ropes and the pulleys, determine the weight of 𝐴 if, after the system is released from rest, 𝐵 moves upward with an acceleration of 0.75 f t∕s2 . 𝐵

Problem 13.157 Figure P13.155 and P13.156

ISTUDY

A crate of mass 𝑚 is gently placed with zero initial velocity on an inextensible conveyor belt that is moving to the right at a constant speed 𝑣0 . Treating the crate as a particle and assuming that the coefficients of static and kinetic friction between the crate and conveyor are 𝜇𝑠 and 𝜇𝑘 , respectively, determine: (a) the distance the crate slides before it stops slipping relative to the belt, and (b) the time it takes for the crate to stop sliding. 𝑣𝑐

𝑚

𝑚

𝑣𝑏

𝑣0 Figure P13.157

Figure P13.158

Problem 13.158 A crate of mass 𝑚 is thrown horizontally with speed 𝑣𝑐 onto an inextensible conveyor belt that is moving to the right at a constant speed 𝑣𝑏 . Treating the crate as a particle, knowing

ISTUDY

Section 13.4

859

Chapter Review

that 𝑣𝑏 > 𝑣𝑐 , and assuming that the coefficients of static and kinetic friction between the crate and conveyor are 𝜇𝑠 and 𝜇𝑘 , respectively, determine: (a) the distance the crate slides before it stops slipping relative to the belt, (b) the time it takes for the crate to stop sliding, and (c) the distance the crate moves relative to the belt. 𝐴

Problem 13.159 A man 𝐴 is trying to keep his balance while on a metal wedge 𝐵 that is sliding down an icy incline. Let 𝑚𝐴 = 78 kg and 𝑚𝐵 = 25 kg be the masses of 𝐴 and 𝐵, respectively. In addition, let the static and kinetic friction coefficients between 𝐴 and 𝐵 be 𝜇𝑠 = 0.4 and 𝜇𝑘 = 0.35, respectively. Determine the acceleration of 𝐴 if 𝜃 = 23◦ . Friction between the wedge and the incline is negligible.

𝐵 𝜃

Problem 13.160 Derive the equations of motion for the double pendulum shown.

𝐿1

Figure P13.159

𝑢̂ 𝜃

𝜃 𝑚1

𝐿2

𝑢̂ 𝑟 𝜙

𝑚2

𝑢̂ 𝜙 𝑢̂ 𝜌

Figure P13.160 and P13.161

Problem 13.161 Derive the equations of motion for the double pendulum shown. After doing so, let 𝐿1 = 1.4 m, 𝐿2 = 2 m, 𝑚1 = 7.5 kg, and 𝑚2 = 12 kg, and release the pendulum from rest with 𝜃(0) = 25◦ and 𝜙(0) = −37◦ . Integrate the equations of motion, and plot the trajectory of each of the particles for at least 5 s.

Problem 13.162 Blocks 𝐴 and 𝐵 are connected by the pulley system shown. Friction between the block 𝐴 and the incline is negligible. The weight of 𝐴 is 𝑊𝐴 = 12 lb, the weight of 𝐵 is 𝑊𝐵 = 30 lb, and the angle between the incline and the horizontal is 𝜃 = 30◦ . Determine the acceleration of 𝐴, the acceleration of 𝐵, and the tension in the rope after the system is released.

𝐴

𝜃 𝐵 Figure P13.162

ice

ISTUDY

ISTUDY

Energy Methods for Particles

14 This chapter presents the concepts of work of a force and kinetic energy of a particle. These two quantities play a crucial role in a balance law called the work-energy principle, which is intimately related to Newton’s second law. We will see that the work-energy principle can be derived from 𝐹⃗ = 𝑚𝑎⃗ by integrating 𝐹⃗ = 𝑚𝑎⃗ with respect to position. We sometimes refer to the work-energy principle as the “pre-integrated” form of Newton’s second law to remind us of the connection between these two fundamental laws of physics. The chapter will conclude with a presentation of the concepts of power and efficiency, which are important in measuring the performance of engines and machines.

Gouhier-Kempinaire/Cameleon/ABACAPRESS/Alamy

Blanka Vlaˇsi´c performing a high jump at the 2008 Indoor World Championship in Valencia, Spain. Using the work-energy principle we can see how speed is converted to height.

14.1

Work-Energy Principle for a Particle

Various problems in Chapter 13 required us to find a change in speed from an acceleration that was a function of position (e.g., Examples 13.2 and 13.5). The solution was based on the application of the chain rule followed by a corresponding integration. In this chapter, we will learn a more direct solution method based on the application of a balance law that pre-integrates Newton’s second law with respect to position.

Work-energy principle and its relation with 𝑭⃗ = 𝒎⃗ 𝒂 Consider a particle of mass 𝑚 moving along a path ℒ1-2 between points 𝑃1 and 𝑃2 under the action of a force 𝐹⃗ (Fig. 14.1). Newton’s second law says that 𝐹⃗ = 𝑚𝑎⃗ for the particle. Dotting both sides of 𝐹⃗ = 𝑚𝑎⃗ with the particle’s infinitesimal displacement 𝑑⃗𝑟, we have 𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝑚𝑎⃗ ⋅ 𝑑⃗𝑟. (14.1) Recalling that 𝑑⃗𝑟 = 𝑣⃗ 𝑑𝑡 and 𝑎⃗ = 𝑑 𝑣∕𝑑𝑡, ⃗ we can rewrite Eq. (14.1) as 𝑑 𝑣⃗ ⋅ 𝑣⃗ 𝑑𝑡 = 𝑚𝑣⃗ ⋅ 𝑑 𝑣, ⃗ 𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝑚 𝑑𝑡

(14.2)

𝑃2 ℒ1-2 𝐹⃗ 𝑃1 𝑚

𝑣⃗ 𝑢̂ 𝑡

Figure 14.1 Particle of mass 𝑚 moving along the path ℒ1-2 while being subjected to the force 𝐹⃗ .

861

862

ISTUDY

Chapter 14

Energy Methods for Particles

where the last expression was obtained by canceling 𝑑𝑡. We now observe that the differential of 12 𝑚𝑣⃗ ⋅ 𝑣⃗ is equal to the last term in Eq. (14.2), that is, 𝑑

(1 2

) ( ) ( ) 𝑚𝑣⃗ ⋅ 𝑣⃗ = 12 𝑚 𝑑 𝑣⃗ ⋅ 𝑣⃗ + 𝑣⃗ ⋅ 𝑑 𝑣⃗ = 12 𝑚 2𝑣⃗ ⋅ 𝑑 𝑣⃗ = 𝑚𝑣⃗ ⋅ 𝑑 𝑣, ⃗

(14.3)

where 𝑑( ) is the differential of the quantity in parentheses. Substituting Eq. (14.3) into Eq. (14.2) and integrating along the path of the particle from the initial point 𝑃1 to the final point 𝑃2 , we obtain ∫ℒ

𝐹⃗ ⋅ 𝑑⃗𝑟 =

1-2

Helpful Information Line integral of an exact differential. All introductory calculus textbooks discuss line integrals (see, e.g., G. B. Thomas, Jr., and R. L. Finney, Calculus and Analytic Geometry, 9th ed., Addison-Wesley, Boston, 1996). A theorem pertaining to line integrals states that if 𝜑 = 𝜑(⃗𝑟 ), then ∫ℒ

𝑑

(1 2

) 𝑚𝑣⃗ ⋅ 𝑣⃗ .

(14.4)

1-2

Both integrals in Eq. (14.4) are path or line integrals, which you have probably studied in your calculus courses. The integral on the right-hand side of Eq. (14.4) is the line integral of an exact differential (see the Helpful Information marginal note) and can be written as ( ) 𝑑 12 𝑚𝑣⃗ ⋅ 𝑣⃗ = 12 𝑚𝑣⃗2 ⋅ 𝑣⃗2 − 12 𝑚𝑣⃗1 ⋅ 𝑣⃗1 , (14.5) ∫ℒ 1-2

where 𝑣⃗1 and 𝑣⃗2 are the velocities of the particle at points 𝑃1 and 𝑃2 , respectively. Since 𝑣⃗ ⋅ 𝑣⃗ = 𝑣2 , substituting Eq. (14.5) into Eq. (14.4) yields

𝑑𝜑 = 𝜑(⃗𝑟2 ) − 𝜑(⃗𝑟1 ),

∫ℒ

1-2

where 𝑟⃗1 is the initial point of ℒ1-2 and 𝑟⃗2 is the endpoint of ℒ1-2 . The only restrictions on this result are that 𝜑(⃗𝑟 ) must be continuous and single-valued in the region containing ℒ1-2 and that ℒ1-2 must be smooth. It is this theorem that we apply in Eq. (14.5).

∫ℒ

𝐹⃗ ⋅ 𝑑⃗𝑟 = 12 𝑚𝑣22 − 12 𝑚𝑣21 .

(14.6)

1-2

We now introduce the following definitions: 𝑈1-2 =

∫ℒ

𝐹⃗ ⋅ 𝑑⃗𝑟 = the work done by 𝐹⃗ on the particle 1-2 in moving from 𝑃1 to 𝑃2 along the path ℒ1-2 ,

𝑇 = 12 𝑚𝑣2 = the kinetic energy of the particle.

Concept Alert Work equals change in kinetic energy. The work done on an object of mass 𝑚 equals the object’s change in kinetic energy regardless of the object’s mass. That is, the same force 𝐹 acting through the same distance always results in the same amount of work done, whether the mass of a particle is, say, 𝑚, 𝟦𝑚, or 𝟣𝟢𝟢𝑚. This is not to say that the speed of particles of different mass will be the same, but they will have exactly the same kinetic energy.

(14.7)

(14.8)

By using these definitions, it can be seen why we refer to Eq. (14.6) as the workenergy principle. In words, we interpret the work-energy principle as saying that the change in kinetic energy of a particle is equal to the work done on that particle. Work and kinetic energy are scalar quantities and, by definition, the kinetic energy is never negative. Using the definitions in Eqs. (14.7) and (14.8), the work-energy principle can be given the form 𝑇1 + 𝑈1-2 = 𝑇2 .

(14.9)

Units for work and kinetic energy The dimension of both work and kinetic energy are (force) × (length) or, equivalently, (mass) × (length)2 ∕(time)2 . Table 14.1 gives the units of energy we will use in both Table 14.1. Units of work, energy, and moment.

Quantity energy or work moment

U.S. Customary

ft⋅lb ft⋅lb

SI

J (joule) ≡ 1 N⋅m N⋅m

the U.S. Customary system and the SI system. The choice for the U.S. Customary

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

863

system is somewhat arbitrary since ft⋅lb is the same as lb⋅ft. We have also included the units of a moment in Table 14.1. From a dimensional viewpoint, energy, work, and moment are equivalent. This is not a problem in the SI system since, when referring to energy, the unit N⋅m is called a joule and is written with the symbol J. By contrast, in the U.S. Customary system the unit ft⋅lb is used for moment, work, and energy. 𝑃2

Work of a force

ℒ1-2

We now want to more closely examine what it means for a force to do work. Referring to Fig. 14.2, the work of a force 𝐹⃗ can be written in all of the following ways:

𝐹⃗ 𝑃1

𝑈1-2 =

∫ℒ

1-2

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑡

𝑡2

𝐹⃗ ⋅ 𝑣⃗ 𝑑𝑡 =

1

=

∫𝑡 ∫𝑡

𝑡2

𝑚

𝐹⃗ ⋅ 𝑣 𝑢̂ 𝑡 𝑑𝑡

1

𝑡2 1

𝑠

2 𝑑𝑠 𝐹⃗ ⋅ 𝐹⃗ ⋅ 𝑢̂ 𝑡 𝑑𝑠, 𝑢̂ 𝑑𝑡 = ∫𝑠 𝑑𝑡 𝑡 1

(14.10)

where 𝑡1 and 𝑡2 are the times at which the particle is at 𝑃1 and 𝑃2 , respectively, 𝑠1 and 𝑠2 are the arc lengths at 𝑃1 and 𝑃2 , respectively, and we have used 𝑑⃗𝑟 = 𝑣⃗ 𝑑𝑡. The last expression in Eqs. (14.10) clearly shows that the work done by the force 𝐹⃗ in moving along the path ℒ1-2 depends only on the component of 𝐹⃗ in the direction of motion, i.e., 𝐹⃗ ⋅ 𝑢̂ 𝑡 . Equations (14.10) also tell us that • The sign of the work is determined by the sign of 𝐹⃗ ⋅ 𝑢̂ 𝑡 : it is positive if 𝐹⃗ promotes the motion, whereas it is negative if 𝐹⃗ hinders the motion (see Fig. 14.3). • Since the second integral in Eqs. (14.10) is with respect to time, 𝐹⃗ ⋅ 𝑣⃗ can be interpreted as the time rate of work done by the force 𝐹⃗ . We will explore this idea in Section 14.4 where we study power and efficiency. • Forces of constraint, such as normal forces, never contribute to 𝑈1-2 because they are always perpendicular to the path.∗ 𝐹⃗𝑝 𝑚 𝐹⃗ℎ

𝑣⃗ 𝜃𝑝 𝑢̂ 𝑡

ℒ1-2

𝜃ℎ

Figure 14.3. The angle 𝜃𝑝 between the force 𝐹⃗𝑝 and the velocity vector 𝑣⃗ is acute. Hence, cos 𝜃 is positive and the work of 𝐹⃗ is positive. By contrast, the angle 𝜃 between the force 𝐹⃗ 𝑝

𝑝

ℎ

ℎ

and the velocity vector 𝑣⃗ is obtuse. Hence, cos 𝜃ℎ is negative and the work of 𝐹⃗ℎ is negative.

Work of a constant force Here we consider a simple two-dimensional example in which we compute the work done by a constant force. The generalization to three dimensions is done similarly. ∗ If

the constraint is moving and it has a component of velocity normal to the constraint surface, then Eqs. (14.10) tell us that the constraint force will do work. We will not consider such cases in this book, as this is a topic for advanced dynamics courses.

𝑢̂ 𝑡

𝑣⃗

Figure 14.2 Particle of mass 𝑚 moving along the path ℒ1-2 while being subjected to the force 𝐹⃗ .

864 𝑦

𝑑⃗𝑟

𝑑𝑦 𝚥̂

𝑚

𝑃2 𝑑𝑥 𝚤̂

𝐹⃗

𝐹𝑦 𝚥̂

𝐹𝑥 𝚤̂ 𝚥̂ 𝑂

Chapter 14

Energy Methods for Particles

𝑃1 ℒ1-2

𝑥

𝚤̂

We will consider more complex examples in Section 14.2, where we will compute the work of forces, such as spring forces and gravitation as given by Newton’s universal law of gravitation. We can express a force 𝐹⃗ in two dimensions in Cartesian components as 𝐹⃗ = 𝐹𝑥 𝚤̂ + 𝐹𝑦 𝚥̂ (Fig. 14.4). We will assume that 𝐹⃗ is constant. Letting 𝑑⃗𝑟 denote the infinitesimal displacement of the point of application of 𝐹⃗ , in Cartesian components we can write 𝑑⃗𝑟 as 𝑑⃗𝑟 = 𝑑𝑥 𝚤̂ + 𝑑𝑦 𝚥̂. We then have the following expression for the work done by 𝐹⃗ : 𝑈1-2 =

Figure 14.4 A constant force 𝐹⃗ acting on a particle.

∫ℒ

𝐹⃗ ⋅ 𝑑⃗𝑟 =

1-2

=

∫𝑥

𝑥2 1

∫ℒ

( ) 𝐹𝑥 𝚤̂ + 𝐹𝑦 𝚥̂ ⋅ (𝑑𝑥 𝚤̂ + 𝑑𝑦 𝚥̂)

1-2

𝐹𝑥 𝑑𝑥 +

∫𝑦

𝑦2

𝐹𝑦 𝑑𝑦 = 𝐹𝑥

1

∫𝑥

𝑥2

𝑑𝑥 + 𝐹𝑦

1

( ) ( ) ( ) = 𝐹𝑥 𝑥2 − 𝑥1 + 𝐹𝑦 𝑦2 − 𝑦1 = 𝐹⃗ ⋅ 𝑟⃗2 − 𝑟⃗1 .

∫𝑦

𝑦2

𝑑𝑦

1

(14.11)

Equation (14.11) tells us that the work of a constant force only depends on the endpoints of the path over which the force acts and not the path itself.

End of Section Summary 𝑃2 ℒ1-2 𝐹⃗ 𝑃1 𝑚

𝑢̂ 𝑡

𝑣⃗

Figure 14.5 Particle of mass 𝑚 moving along the path ℒ1-2 while being subjected to the force 𝐹⃗ .

ISTUDY

The work-energy principle is expressed by the following equation: Eq. (14.9), p. 862 𝑇1 + 𝑈1-2 = 𝑇2 , where 𝑈1-2 is the work and is defined as (see Fig. 14.5) Eq. (14.7), p. 862 𝑈1-2 =

∫ℒ

𝐹⃗ ⋅ 𝑑⃗𝑟

1-2

and 𝑇 is the kinetic energy and is defined as Eq. (14.8), p. 862

Common Pitfall Kinetic energy is never negative. If you compute a kinetic energy using a velocity component that is negative, don’t be tempted to write 𝑇 = − 12 𝑚𝑣2 ; since 𝑣 is squared, regardless of the sign of the velocity component, the kinetic energy can never be negative.

𝑇 = 12 𝑚𝑣2 . Work and kinetic energy have the following basic properties: • Work and kinetic energy are scalar quantities. • The work depends on only the component of 𝐹⃗ in the direction of motion, that is, on 𝐹⃗ ⋅ 𝑢̂ 𝑡 , and its sign is determined by the sign of 𝐹⃗ ⋅ 𝑢̂ 𝑡 . • The kinetic energy is never negative.

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

E X A M P L E 14.1

865

Ball Sliding Down a Cylindrical Surface

As shown in Fig. 1, a small ball is given a slight nudge from rest at the top of a semicylinder and slides downward. If the mass of the ball is 𝑚, the radius of the semicylinder is 𝑅, and friction between the ball and the semicylinder is negligible, determine the angle 𝜃 at which the ball separates from the semicylinder. We solved this problem using Newton’s second law in Example 13.9 on p. 820.

1 2

𝑅 𝑅

𝜃

SOLUTION 𝑂

Road Map & Modeling

The FBD of the ball as it slides down the semicylinder is shown in Fig. 2, where we have used polar components to describe the motion. As in Example 13.9, the key is to find the normal force 𝑁 as a function of the angle 𝜃 and then ̇ determine where 𝑁 becomes zero. We can apply Newton’s second law to find 𝑁(𝜃, 𝜃) ̇ and then the work-energy principle to find 𝜃(𝜃).

Figure 1 A particle sliding on a frictionless semicylinder.

𝑚𝑔

𝜃 𝑢̂ 𝑟

Governing Equations Application of Newton’s Second Law Balance Principles

Force Laws

Applying Newton’s second law in the radial direction, we obtain ∑ 𝐹𝑟∶ 𝑁 − 𝑚𝑔 cos 𝜃 = 𝑚𝑎𝑟 . (1)

All forces are accounted for on the FBD.

Kinematic Equations Computation

Since the motion of the ball is circular, we have that 𝑎𝑟 = −𝑅𝜃̇ 2 .

𝑢̂ 𝜃

𝑁

Figure 2 FBD at an arbitrary 𝜃 of the ball as it slides down the semicylinder.

Substituting the expression for 𝑎𝑟 into Eq. (1), we obtain 𝑁 = 𝑚𝑔 cos 𝜃 − 𝑚𝑅𝜃̇ 2 ,

(2)

̇ where we see that we still need to find 𝜃(𝜃). Application of the Work-Energy Principle Balance Principles

Applying the work-energy principle between 1 and 2, we obtain 𝑇1 + 𝑈1-2 = 𝑇2 ,

(3)

where the kinetic energies at 1 and 2 are, respectively, 𝑇1 = 0 and 𝑇2 = 21 𝑚𝑣22 ,

(4)

𝑢̂ 𝑟

where 𝑣2 is the speed of the ball in 2. 2

Force Laws

Only the weight force does work on the ball, so we compute its work using the last of Eqs. (14.10) as 𝑈1-2 =

∫𝑠

𝑠2

) ( 𝑚𝑔 sin 𝜃 𝑢̂ 𝜃 − cos 𝜃 𝑢̂ 𝑟 ⋅ 𝑢̂ 𝑡 𝑑𝑠 = 𝑚𝑔𝑅

1

∫0

sin 𝜃 𝑑𝜃 = 𝑚𝑔𝑅(1 − cos 𝜃), (5)

2

Substituting Eqs. (4), (5), and 𝑣2 = 𝑅𝜃̇ into Eq. (3), we obtain

̇ 2 𝑚𝑔𝑅(1 − cos 𝜃) = 12 𝑚(𝑅𝜃) ⇒

2𝑔 𝜃̇ 2 = (1 − cos 𝜃) 𝑅 𝑁 = 𝑚𝑔 cos 𝜃 − 2𝑚𝑔(1 − cos 𝜃) = 𝑚𝑔(3 cos 𝜃 − 2),

Discussion & Verification

2 3

or 𝜃 = 48.19◦ .

This is the same angle we obtained in Example 13.9.

𝜃

𝑅

𝑂 Figure 3 Equality of 𝑢̂ 𝜃 and 𝑢̂ 𝑡 for circular motion.

⇒

which, for 𝑁 = 0, implies that cos 𝜃 =

𝑢̂ 𝜃 , 𝑢̂ 𝑡

𝑅

𝜃

where we have used 𝑢̂ 𝜃 = 𝑢̂ 𝑡 (see Fig. 3) and 𝑑𝑠 = 𝑅 𝑑𝜃 to obtain the second equation. ̇ Kinematic Equations Using polar components, we can write 𝑣 = 𝑅𝜃. Computation

𝑢̂ 𝑛

(6)

866

Chapter 14

Energy Methods for Particles

E X A M P L E 14.2

Relating Kinetic Energy to Average Force An F/A-18 Hornet (see Fig. 1) takes off from an aircraft carrier using two separate propulsion systems: its two jet engines and a steam-powered catapult. During launch, a fully loaded Hornet weighing 50,000 lb goes from 0 (relative to the carrier) to 165 mph (relative to the surface of the Earth) in a distance of 300 f t (relative to the carrier) while each of its two engines is at full power, generating about 22,000 lb of thrust, and the catapult is engaged. For a stationary aircraft carrier, neglecting aerodynamic forces, determine: (a) The total work done on the aircraft during launch. (b) The work done by the catapult on the aircraft during launch. (c) The average force exerted by the catapult on the aircraft.

U.S. Navy photo by Mass Communication Specialist 3rd Class Torrey W. Lee

Figure 1 An F/A-18 Hornet taking off from an aircraft carrier. 𝚥̂ 𝚤̂

𝑚𝑔

3

𝐹𝐶

𝐹𝑇

3

N NAVY

𝑁 Figure 2 FBD of the aircraft as it is propelled by the jet engines and by the catapult.

ISTUDY

SOLUTION Road Map & Modeling Referring to the FBD in Fig. 2, we will model the airplane as a particle under the action of gravity, the engines’ thrust 𝐹𝑇 , the force of the catapult 𝐹𝐶 , and the normal reaction between the aircraft and the deck. Based on this model, the work-energy principle tells us that the work done on the airplane is equal to the airplane’s change in kinetic energy. We can calculate the kinetic energy because we are given the airplane’s weight and change in speed. Since the engines’ thrust is constant and we know the takeoff distance, we can compute the work done by the engines by a direct application of the definition of work. Subtracting the work of the engines from the total work will allow us to find the work of the catapult. Governing Equations Balance Principles

The work-energy principle gives 𝑇1 + 𝑈1-2 = 𝑇2 ,

(1)

where 1 is at the start of the launch, 2 is right before takeoff, 𝑈1-2 is the total work done on the aircraft between 1 and 2, and 𝑇1 and 𝑇2 are the aircraft’s kinetic energy at 1 and 2, respectively, which are given by 𝑇1 = 12 𝑚𝑣21

and 𝑇2 = 12 𝑚𝑣22 ,

(2)

where 𝑣1 and 𝑣2 are the speed of the aircraft at 1 and 2, respectively. Force Laws

Both gravity and the engines’ thrust are modeled as constant forces. Because we are only interested in the average value of the catapult force, we will model 𝐹𝐶 as a constant. Since the work is computed using forces and force laws, in work-energy problems, we will always determine expressions for the work in the Force Laws portion of our solution procedure. Hence, the expressions for the work done by the engines and catapult are ( ) 𝑈1-2 engines =

∫ℒ

𝐹⃗𝑇 ⋅ 𝑑⃗𝑟 =

∫ℒ

𝐹⃗𝐶 ⋅ 𝑑⃗𝑟 =

1-2

( ) 𝑈1-2 catapult =

1-2

300 f t

∫0 f t

(44,000 lb) 𝑑𝑥 = 1.320×107 f t ⋅lb,

(3)

𝐹𝐶 𝑑𝑥 = (300.0 f t)𝐹𝐶 ,

(4)

300 f t

∫0 f t

where ℒ1-2 is the 300 f t rectilinear stretch traveled by the plane between 1 and 2. Note that the aircraft’s weight and the normal reaction 𝑁 do no work since the motion of the airplane is assumed to be horizontal and therefore, perpendicular to these forces. Kinematic Equations

Recalling the definition of 1 and 2, we have 𝑣1 = 0 and 𝑣2 = 165 mph = 242.0 f t∕s.

(5)

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

Computation Combining Eqs. (2) and (5) with Eq. (1) and solving for 𝑈1-2 , we find the total work done on the aircraft during launch to be

𝑈1-2 = 𝑇2 − 𝑇1 = 4.547×107 f t ⋅lb.

(6)

Therefore, the work done by the catapult is the total work done minus the work done by the engines, that is, ( ) ) ( 𝑈1-2 catapult = 𝑈1-2 − 𝑈1-2 engines . (7) Substituting Eqs. (3) and (6) into Eq. (7), we have ( ) 𝑈1-2 catapult = 3.227×107 f t ⋅lb.

(8)

Finally, solving Eq. (4) for 𝐹𝐶 and using the result in Eq. (8), we have 𝐹𝐶 =

) ( 𝑈1-2 catapult 300 f t

= 107,600 lb.

(9)

Discussion & Verification

The calculation of the total work done on the aircraft and the work done by the catapult was done by a direct application of the work-energy principle, and therefore, we need only verify that proper units were used to express our result, which is indeed the case. As far as the result in Eq. (9) is concerned, since the dimensions of work are force times length, we know that the dimensions of our result are correct and that proper units were used to express it. A Closer Look This problem emphasizes an important point: no matter how many or what kinds of forces are acting on a particle, the work done by all forces is equal to the change in kinetic energy of the body. The amount of work done on the F/A-18 is equivalent to the work required to push a 70 lb wooden crate over level concrete (𝜇𝑘 ≈ 0.6) at a constant speed for more than 205 miles! Also, notice that the force due to the catapult is more than twice that provided by the aircraft’s engines. This tells us that an airplane such as an F/A-18 could not take off unassisted from an aircraft carrier. This problem was simplified by the fact that the carrier was stationary. Fighter aircraft are often launched while the carrier is moving at significant speed. If the carrier speed is constant, both the Earth and the carrier deck are inertial coordinate systems. We could use either inertial system as a basis for the work-energy principle, but choosing the carrier deck leads to a more efficient solution. This is explored in Problem 14.25.

867

868

Chapter 14

Energy Methods for Particles

E X A M P L E 14.3

Relating Speed to Position

𝐷

ℎ

𝐸

The block of mass 𝑚 = 20 kg is connected by a pulley at 𝐷 to the winch at 𝐸 by an inextensible cord. The winch, which is mounted at a fixed location, can exert a constant force 𝑃 = 130 N on the cord. The friction between the block and the horizontal bar on which the block slides is negligible. Letting ℎ = 0.8 m, if the block starts from rest in the position shown, determine the speed of the block when it has moved the distance 𝑑 = 1.15 m and is directly under the pulley.

𝑚

SOLUTION Road Map & Modeling

Since we are interested in relating speed to position, we will solve this problem by using the work-energy principle. Referring to the FBD in Fig. 2, we will model the block as a particle subject to its own weight, the normal reaction between

𝑑 Figure 1 Winch and block system geometry.

𝑚𝑔

𝑃 𝜃 𝚥̂ 𝚤̂

𝑁

Figure 2. FBD of the block in a generic position between its initial and final positions.

the block and the horizontal guide, and the force in the cord. Notice that neither the weight nor the force 𝑁 does any work, given that the block’s motion is in the horizontal direction.

𝑦

Governing Equations Balance Principles

Referring to Fig. 3 and applying the work-energy principle between

1 and 2, we obtain 𝑇1 + 𝑈1-2 = 𝑇2 .

ℎ

(1)

The kinetic energies are given by 𝑥

𝑇1 = 21 𝑚𝑣21 1

𝑑

2

Figure 3 Winch and block system showing 1 and 2 and the Cartesian coordinate system used.

ISTUDY

and 𝑇2 = 12 𝑚𝑣22 ,

(2)

where 𝑣1 and 𝑣2 are the block’s speed at 1 and 2, respectively. Force Laws The only force doing work is the force 𝑃 , whose magnitude is constant and given. As mentioned in Example 14.2, when using the work-energy principle, we devote the Force Laws step of our solution procedure to the computation of the work done by the forces appearing in the FBD. Hence, we have

𝑈1-2 =

∫ℒ

1-2

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑑

0

−𝑃 cos 𝜃 𝑑𝑥,

(3)

where we have used 𝐹⃗ = −𝑃 cos 𝜃 𝚤̂ + 𝑃 sin 𝜃 𝚥̂ and 𝑑⃗𝑟 = 𝑑𝑥 𝚤̂. Kinematic Equations Recalling that the block is released from rest and that the speed at 2 is the unknown of this problem, we have

𝑣1 = 0.

(4)

Observe also that the integral in Eq. (3) contains the variable 𝜃 in the integrand and uses the variable 𝑥 as the variable of integration. To compute this integral, we need to express

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

869

either the variable 𝜃 as a function of 𝑥 or the variable 𝑥 as a function of 𝜃. We will choose the latter strategy (the final result must be the same no matter which strategy we use) and will use differentiation of constraints (see Section 12.6 on p. 738) to obtain the relation we need. Noticing that 𝑥 tan 𝜃 = ℎ and then taking the differential of this relation, we have tan 𝜃 𝑑𝑥 + 𝑥 sec2 𝜃 𝑑𝜃 = tan 𝜃 𝑑𝑥 +

ℎ sec2 𝜃 𝑑𝜃 = 0, tan 𝜃

(5)

where we have used the fact that 𝑥 = ℎ∕ tan 𝜃 and that 𝑑ℎ = 0 since ℎ is a constant. Solving for 𝑑𝑥, we find that −ℎ 𝑑𝑥 = 𝑑𝜃. (6) sin2 𝜃 Computation

Substituting Eq. (6) into Eq. (3), we have ( ) 𝜋∕2 ℎ cos 𝜃 𝑃 𝑑𝜃, 𝑈1-2 = ∫tan−1 (ℎ∕𝑑) sin2 𝜃

(7)

where we have used the fact that 𝜃 = tan−1 (ℎ∕𝑑) when 𝑥 = 𝑑 and 𝜃 = 𝜋∕2 rad when 𝑥 = 0. Evaluating the integral in Eq. (7), we obtain ) ( 𝜋∕2 √ ℎ || = −𝑃 ℎ − 𝑑 2 + ℎ2 , | sin 𝜃 |tan−1 (ℎ∕𝑑) √ ) ( where we have used the identity sin tan−1 𝑥 = 𝑥∕ 1 + 𝑥2 (see Fig. 4). Combining Eqs. (2), (4), and (8) with Eq. (1), we obtain (√ ) 𝑃 𝑑 2 + ℎ2 − ℎ = 21 𝑚𝑣22 , 𝑈1-2 = −𝑃

(8)

𝜃

(9)

which, upon solving for 𝑣2 , gives √ 𝑣2 =

( ) 2𝑃 √ 2 𝑑 + ℎ2 − ℎ = 2.795 m∕s. 𝑚

(10)

Discussion & Verification Observing that the term 𝑃 ∕𝑚 has dimensions of acceleration √ and that the term 𝑑 2 + ℎ2 − ℎ has dimensions of length, we see that 𝑣2 has dimensions of length over time, as it should. Furthermore, the final numerical result was expressed with appropriate units. Also, observe that the argument of the square root in Eq. (10) ) (√ contains the term 𝑃 𝑑 2 + ℎ2 − ℎ , which can be interpreted as the work of a constant force√ (i.e., constant in both magnitude and direction) with magnitude 𝑃 along the distance Δ = 𝑑 2 + ℎ2 − ℎ. What makes Δ interesting is that it corresponds to the amount of cord that is wound onto the winch as the block goes from 1 to 2. In fact, this is the work done by the tension in the vertical branch of the cord, i.e., the portion of the cord that goes from the pulley to the winch. So overall our result appears to be correct. A Closer Look Can we obtain the result in Eq. (10) without resorting to the rather involved integration in Eqs. (7) and (8)? The answer lies in computing 𝑈1-2 in some more physically based way. We will explore this question in Example 14.4.

Figure 4 Demonstration of the trigonometric identity used in Eq. (8).

870

Chapter 14

Energy Methods for Particles

E X A M P L E 14.4

Choosing a Convenient FBD to Relate Speed to Position Let’s revisit Example 14.3 and again determine the speed of the block when it has moved the distance 𝑑 and is directly under the pulley. However, instead of focusing on the FBD of only the block, solve the problem by using an FBD that includes the pulley at 𝐷.

𝐷

ℎ

𝐸

SOLUTION Road Map & Modeling As in Example 14.3, we will apply the work-energy principle, but we will use the FBD suggested in the problem statement, which is shown in Fig. 2.

𝑚

Again, the only force doing work on the block is 𝑃 because the point of application of the reactions 𝑅𝑥 and 𝑅𝑦 is fixed (hence, 𝑅𝑥 and 𝑅𝑦 do no work) and the forces 𝑚𝑔 and 𝑁 are perpendicular to the direction of motion of the block.

𝑑 Figure 1 Winch and block system geometry.

Governing Equations Balance Principles

𝑅𝑥

Letting 1 be the position of the block at the instant of release and 2 be the position of the block when the block is directly under the pulley, applying the work-energy principle between 1 and 2, we obtain

𝑅𝑦

𝑇1 + 𝑈1-2 = 𝑇2 .

𝚥̂ 𝚤̂

(1)

The block’s kinetic energies are given by

𝑚𝑔

𝑇1 = 21 𝑚𝑣21

𝑃

and 𝑇2 = 12 𝑚𝑣22 ,

(2)

where 𝑣1 and 𝑣2 are the block’s speed at 1 and 2, respectively. Force Laws

This time 𝑈1-2 takes on the following simple form:

𝑁

𝑈1-2 = 𝑃 Δ,

Figure 2 FBD of the block, cord, and pulley as the block moves from 1 to 2.

(3)

where Δ is the amount of cord wound up on the winch as the block moves from 1 to 2 and is therefore given by (see Fig. 3) (√ ) √ Δ = 𝐿1 − 𝐿 2 = 𝑑 2 + ℎ2 + 𝓁 − (ℎ + 𝓁) = 𝑑 2 + ℎ2 − ℎ, (4) where 𝐿1 and 𝐿2 are the length of the cord at 1 and 2, respectively. Kinematic Equations

√ 𝑑 2 + ℎ2

𝓁

𝑣1 = 0.

ℎ Computation

1

𝑑

2

Figure 3 The length of the cord at 1 and 2. The length 𝓁 is the constant vertical distance between the pulley and winch.

ISTUDY

Recalling that 𝑚 starts from rest and that 𝑣2 is the unknown of

this problem, we have

Substituting Eqs. (2)–(5) into Eq. (1), we obtain ) (√ 𝑑 2 + ℎ2 − ℎ = 12 𝑚𝑣22 , 𝑃

(5)

(6)

which, upon solving for 𝑣2 , gives √ 𝑣2 = Discussion & Verification

( ) 2𝑃 √ 2 𝑑 + ℎ2 − ℎ . 𝑚

(7)

As expected, we obtained the same result found in Exam-

ple 14.3. A Closer Look The approach followed in this example is much more straightforward than that in Example 14.3 even though the only difference between the two examples is how we computed 𝑈1-2 . The lesson here is that judiciously selecting the system to analyze can save significant time and effort.

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

Problems Problem 14.1 A rocket lifts off with an acceleration 𝑎. During liftoff, in terms of absolute values, is the work done on an astronaut by gravity larger than, equal to, or smaller than the work done by the normal reaction between the astronaut and her or his seat?

𝑣2

𝑣1 NASA

Figure P14.1

Figure P14.2

Problem 14.2 A soft rubber ball bounces against a wall. Assuming that the wall’s deformation due to the ball’s impact is negligible, does the contact force due to the wall do positive work, no work, or negative work on the ball?

Problem 14.3 Determine the kinetic energy of the bodies listed below, when modeled as particles. Express all answers using both U.S. Customary units and SI units. (a) A .30-06 bullet weighing 150 gr (1 lb = 7000 gr) and traveling at 3000 f t∕s. (b) A 25 kg child traveling in a car at 45 km∕h. (c) A 415,000 lb locomotive traveling at 75 mph. (d) A 20 g metal fragment from a space vehicle traveling at 8000 km∕s. (e) A 3000 lb car traveling at 60 mph. 𝑎𝑒

Problem 14.4 A man whose mass is 𝑚 = 80 kg is in an elevator that accelerates at 𝑎𝑒 = 1.5 m∕s2 over a distance of 𝑑 = 2 m. Determine the work done by the weight force acting on the man and the work done by the normal force the floor of the elevator exerts on him. Explain why the magnitudes of the work of the forces are not the same.

Problems 14.5 and 14.6

Figure P14.4

Consider a 3000 lb car whose speed is increased by 30 mph. Problem 14.5

Modeling the car as a particle and assuming that the car is traveling on a rectilinear and horizontal stretch of road, determine the amount of work done on the car throughout the acceleration process if the car starts from rest. Problem 14.6

Modeling the car as a particle and assuming that the car is traveling on a rectilinear and horizontal stretch of road, determine the amount of work done on the car throughout the acceleration process if the car has an initial speed of 45 mph.

Figure P14.5 and P14.6

871

872

Chapter 14

Energy Methods for Particles

Problem 14.7 A 75 kg skydiver is falling at a speed of 250 km∕h when the parachute is deployed, allowing the skydiver to land at a speed of 4 m∕s. Modeling the skydiver as a particle, determine the total work done on the skydiver from the moment of parachute deployment until landing.

Problems 14.8 and 14.9 The crate of mass 𝑚 is pushed to the left until the linear elastic spring of constant 𝑘 has compressed a distance 𝑑 from its unstretched length 𝓁0 . The crate is released from rest, and friction between the crate and the horizontal surface is negligible. 𝓁0

Figure P14.7 𝑘

𝑑 𝑚

Figure P14.8 and P14.9 Problem 14.8

Using the work-energy principle, determine the speed of the crate at the instant the spring becomes uncompressed. Using the work-energy principle, determine the value of 𝑘 so that the crate is moving at 3 m∕s the instant the spring becomes uncompressed. Use 𝑚 = 20 kg and 𝑑 = 0.75 m.

Problem 14.9

Problems 14.10 and 14.11 Consider a 1500 kg car whose speed is increased by 45 km∕h over a distance of 50 m while traveling up an incline with a 15% grade.

100

15

Figure P14.10 and P14.11 Problem 14.10

Modeling the car as a particle, determine the work done on the car if the car starts from rest. Problem 14.11 Modeling the car as a particle, determine the work done on the car if the car has an initial speed of 60 km∕h.

Problems 14.12 and 14.13 𝑃

𝛽

𝑊 𝑑 𝑘 𝜃

Figure P14.12 and P14.13

ISTUDY

The crate moves up the incline a distance 𝑑 = 4.5 f t due to the action of the constant force 𝑃 = 100 lb. The weight of the crate is 𝑊 = 65 lb, and the spring with constant 𝑘 = 10 lb∕f t is unstretched before the crate starts moving. Friction between the crate and the incline is negligible, and the angle between the force 𝑃 and the surface on which the crate slides is 𝛽 = 30◦ . Letting 𝜃 = 0◦ , that is, the surface is horizontal, determine the total work done by all forces after the crate has moved the distance 𝑑. Problem 14.12

Problem 14.13 Letting 𝜃 = 30◦ , determine the total work done by all forces after the crate has moved the distance 𝑑.

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

Problem 14.14

𝑣

A 350 kg crate is sliding down a rough incline with a constant speed 𝑣 = 7 m∕s. Assuming that the angle of the incline is 𝜃 = 33◦ and that the only forces acting on the crate are gravity, friction, and the normal force between the crate and the incline, determine the work done by friction over every meter slid by the crate.

Problem 14.15

𝜃 Figure P14.14

A vehicle 𝐴 is stuck on the railroad tracks as a train 𝐵 approaches with a speed of 120 km∕h. As soon as the problem is detected, the train’s emergency brakes are activated, locking the wheels and causing the train’s wheels to slide relative to the tracks. If the coefficient of kinetic friction between the wheels and the track is 0.2, use the work-energy principle to determine the minimum distance 𝑑min at which the brakes must be applied to avoid a collision under the following circumstances: (a) The train consists of just a 195,000 kg locomotive. (b) The train consists of a 195,000 kg locomotive and a string of cars whose mass is 10×106 kg, all of which can apply brakes and lock their wheels. (c) The train consists of a 195,000 kg locomotive and a string of cars whose mass is 10×106 kg, but only the locomotive can apply its brakes and lock its wheels. Treat the train as a particle. Assume that the railroad tracks are rectilinear and horizontal. 𝐴

𝑑

𝐵

Figure P14.15

Problems 14.16 and 14.17 A classic car is driving down a 20◦ incline at 45 km∕h when the brakes are applied. Treat the car as a particle, and neglect all forces except gravity and friction. Problem 14.16

Using the work-energy principle, determine the stopping distance if the tires slide and the coefficient of kinetic friction between the tires and the road is 0.7.

Problem 14.17

Using the work-energy principle, determine the minimum stopping distance if the car is retrofitted with antilock brakes. Although tire motion is complicated when antilock brakes are present, you should model the effect as though the tires are perpetually in a state of impending slip. Use 0.9 for the coefficient of static friction between the tires and the road.

20◦ Figure P14.16 and P14.17

Problem 14.18 Two identical cars travel at a speed of 60 mph, one along a newly asphalt-paved straight and horizontal road and the other on a straight and horizontal dirt road. If brakes are applied and if the second car slips during the braking process, what difference will there be in the amount of work done to stop each car?

Problem 14.19 Two identical locomotives 𝐴 and 𝐵 are coupled with one and two passenger cars, respectively. Suppose that each passenger car is identical to the others in all respects. If each train (locomotive plus cars) starts from rest, each locomotive exerts the maximum tractive effort (traction force), and assuming that gravity and the tractive effort are the only relevant forces acting on the trains, which of the two trains will have greater kinetic energy after the locomotives have moved 50 m along a horizontal and rectilinear stretch?

Figure P14.19

𝐴

50 m

𝐵

50 m

873

874

Chapter 14

Energy Methods for Particles Problems 14.20 through 14.22

𝑥

𝐴

𝐵

𝜇𝑘

Cable Force, 𝑃 (lb)

The crate 𝐴 of weight 𝑊 = 30 lb is being pulled to the right by the winch at 𝐵. The crate starts from rest at 𝑥 = 0 and is pulled a total distance of 15 f t over the rough surface for which the coefficient of kinetic friction is 𝜇𝑘 = 0.3. The force 𝑃√in the cable due to the winch varies according to the plot, where 𝑃 is in lb, 𝑏 is in lb∕ ft, and 𝑥 is in ft. The coefficient of static friction is insufficient to prevent slipping. 60 40

√ 𝑃 = 20 + 𝑏 𝑥

20 00

3

9 12 6 Crate Position, 𝑥 (f t)

15

Figure P14.20–P14.22 Problem 14.20 √ Using the work-energy principle, determine the speed of the block

when 𝑏 = 11 lb∕ f t and 𝑥 = 15 f t.

Using the work-energy principle, determine the value of 𝑏 so that the block is moving at 35 f t∕s when 𝑥 = 10 f t.

Problem 14.21

Problem 14.22

√Determine how far the crate slides before its speed becomes 20 f t∕s with 𝑏 = 11 lb∕ f t.

Problem 14.23 Pullout Force (N)

0.20 0.15 0.10 0.05 0.00

0

10 20 30 Displacement (𝜇m)

40

Figure P14.23

Many advanced materials consist of fibers (e.g., made of glass, Kevlar, or carbon) placed within a matrix (such as epoxy, a titanium alloy, etc.). For these materials it is important to assess the bond strength between fibers and matrix, and this is often done with a pullout test, in which the tip of a fiber is exposed, the material sampled is properly clamped, and the fiber is pulled out of the matrix. The data collected often consists of a graph like the one shown, in which the force exerted on the fiber is recorded as a function of pullout displacement. With this in mind, the interface toughness assessment process may require a measure of the energy expended to pull out the fiber. Use the force-displacement graph shown, which is typical for a glass-fiber reinforced epoxy, to measure the total pullout energy. Hint: The work of the pullout force is given by the area under the curve.

Problem 14.24 Indentation Force (10−4 N)

30 Unloading 25 20 Loading 15 10 5 0 0

20 40 60 80 100 120 Indentation Depth (nm)

Figure P14.24

ISTUDY

Components subjected to large contact forces (e.g., brake disks) are often made of highgrade steel coated with a thin film of a very hard material, such as diamond. A common test to assess the mechanical properties (hardness, elastic moduli, etc.) of the coating is the nanoindentation test, which consists of making a controlled dent in the film using a nail-like object, called an indentor. During the test, the force applied to the indentor and the indentation depth are measured. The graph shows the curves interpolating the loading and unloading data for a diamond film on steel. Since the unloading curve does not go back to the origin of the plot, the film is permanently deformed during the indentation process. Determine the energy lost to permanent deformation if the indentation force is given by { 5 2 25 𝑥 + 32 𝑥 during loading, 4 𝐹𝐼 = 145 7 2 300 − 6 𝑥 + 18 𝑥 during unloading, where 𝐹𝐼 is expressed in 𝜇N and the indentation depth 𝑥 is given in nm.

ISTUDY

Section 14.1

Work-Energy Principle for a Particle

875

Problem 14.25 An F/A-18 Hornet takes off from an aircraft carrier, using two separate propulsion systems: its two jet engines and a steam-powered catapult. During launch, a fully loaded Hornet weighing about 50,000 lb goes from 0 mph (relative to the aircraft carrier) to 165 mph (measured relative to surface of the Earth) in a distance of 300 f t (measured relative to the aircraft carrier) while each of its two engines is at full power generating about 22,000 lb of thrust. Assuming that the aircraft carrier is traveling in the same direction as the takeoff direction and at a constant speed of 30 knot (1 knot = 1.6878 ft∕s), determine: (a) The total work done on the aircraft during launch. (b) The work done by the catapult on the aircraft during launch. (c) The force exerted by the catapult on the aircraft. In solving this problem, model the aircraft as a particle; assume that its trajectory is horizontal and that the catapult assists the aircraft the full 300 f t needed for takeoff; and finally, let all forces be constant and neglect air resistance and friction. Use an inertial reference frame attached to the aircraft carrier.

U.S. Navy photo by Mass Communication Specialist 3rd Class Torrey W. Lee

Figure P14.25

Problem 14.26 Rubber bumpers are commonly used in marine applications to keep boats and ships from getting damaged by docks. Treating the boat 𝐶 as a particle, neglecting its vertical motion, and neglecting the drag force between the water and the boat 𝐶, what is the maximum speed of the boat at impact with the bumper 𝐵 so that the deflection of the bumper is limited to 6 in.? The weight of the boat is 70,000 lb, and the force compression profile for the rubber bumper is given by 𝐹𝐵 = 𝛽𝑥3 , where 𝛽 = 3.5 × 106 lb∕f t 3 and 𝑥 is the compression of the bumper. 5 𝐵

𝐹𝐵 (× 105 lb)

𝐶

4 3 2 1 0 0

0.1

0.2 0.3 𝑥 (f t)

0.4

0.5

Figure P14.26

Problems 14.27 and 14.28 Packages for transporting delicate items (e.g., a laptop or glass) are designed to “absorb” some of the energy of the impact in order to protect their contents. These energy absorbers can get pretty complicated (e.g., the mechanics of Styrofoam peanuts is not easy), but we can begin to understand how they work by modeling them as a linear elastic spring of constant 𝑘 that is placed between the contents (an expensive vase) of mass 𝑚 and the package 𝑃 . Assume that the vase weighs 6 lb and that the box is dropped from a height of 5 f t. Treat the vase as a particle, and neglect all forces except for gravity and the spring force.

𝑚

𝑘 𝑃

Problem 14.27

Determine the maximum displacement of the vase relative to the box and the maximum force on the vase due to the spring if 𝑘 = 264 lb∕f t. Problem 14.28 Plot the maximum displacement of the vase relative to the box and the maximum force on the vase due to the spring as a function of the linear elastic spring constant 𝑘. What do these plots tell you about the problem you would encounter in trying to minimize the force on the vase?

Figure P14.27 and P14.28

876

Chapter 14

Energy Methods for Particles

Problem 14.29 A block 𝐴 moves horizontally under the action of a force 𝐹 whose line of action is parallel to the motion. If the kinetic energy of 𝐴 as a function of 𝑥 is that shown (𝑥1 and 𝑥2 are extrema for 𝑇𝐴 ), what can you say about the sign of 𝐹 for 𝑥0 < 𝑥 < 𝑥1 and 𝑥1 < 𝑥 < 𝑥2 ? 𝑦

𝑇𝐴

𝐴

bungee cord 400 f t bungee jumper

𝐹 𝑥

𝑥0

𝑥1

𝑥2

𝑥

Figure P14.29

Problem 14.30 While the stiffness of an elastic cord can be quite constant (i.e., the force versus displacement curve is a straight line), some cords exhibit nonlinear force-displacement behavior. As a bungee cord is stretched, it softens; that is, the cord tends to get less stiff as it gets longer. Assuming a softening force-displacement relation of the form 𝑘𝛿 − 𝛽𝛿 3 , where 𝛿 (measured in ft) is the displacement of the cord from its unstretched length, considering a bungee cord whose unstretched length is 150 f t, and letting 𝑘 = 2.58 lb∕f t, determine the value of the constant 𝛽 such that a bungee jumper weighing 170 lb and starting from rest gets to the bottom of a 400 f t tower with zero speed.

Figure P14.30

ISTUDY

Problems 14.31 through 14.33 Car bumpers are designed to limit the extent of damage to the car in the case of lowvelocity collisions. Consider a 1420 kg passenger car impacting a concrete barrier while traveling at a speed of 5.0 km∕h. Model the car as a particle and consider two bumper models: (1) a simple linear spring with constant 𝑘 and (2) a linear spring of constant 𝑘 in parallel with a shock-absorbing unit generating a nearly constant force 𝐹𝑆 = 2000 N over 10 cm. If the bumper is of type 1 and if 𝑘 = 9 × 104 N∕m, find the spring compression (distance) necessary to stop the car. Problem 14.31

If the bumper is of type 1, find the value of 𝑘 necessary to stop the car when the bumper is compressed 10 cm. Problem 14.32

If the bumper is of type 2, find the value of 𝑘 necessary to stop the car when the bumper is compressed 10 cm. Problem 14.33

𝑘 (1) 𝐹𝑆 (2) 𝑘 Figure P14.31–P14.33

ISTUDY

Section 14.2

14.2

877

Conservative Forces and Potential Energy

Conservative Forces and Potential Energy

In Example 14.1 on p. 865, we saw that the work done by the weight force only depended on the change in height 𝑅(1 − cos 𝜃) of the ball — the fact that the path of the ball was circular did not matter. In this section, we will see that this is a general property of the weight force. We will also see that there are other forces whose work is independent of the path followed by their points of application in going from an initial to a final position.

Work done by the constant force of gravity In Fig. 14.6, the particle of mass 𝑚 moves from 1 to 2 following the path ℒ1-2 . We wish to determine the work done by the weight force 𝑚𝑔 during this motion, so it is the only force shown, though there may be other forces acting on the particle. To determine the work done by 𝑚𝑔, we apply Eq. (14.7) to obtain ) ( 𝑈1-2 𝑔 =

∫ℒ

𝐹⃗𝑔 ⋅ 𝑑⃗𝑟 =

1-2

∫ℒ

−𝑚𝑔 𝚥̂ ⋅ (𝑑𝑥 𝚤̂ + 𝑑𝑦 𝚥̂) = −𝑚𝑔

1-2

∫𝑦

𝑦2

𝑦

2

ℒ1-2 𝑦

1

𝑑𝑦,

(14.12)

𝑦1

𝑥

1

where we have used 𝐹⃗𝑔 = −𝑚𝑔 𝚥̂, 𝑑⃗𝑟 = 𝑑𝑥 𝚤̂ + 𝑑𝑦 𝚥̂, and the fact that 𝑚𝑔 is constant. Carrying out the final integral, we obtain the work done by 𝑚𝑔 as ( ) ( ) 𝑈1-2 𝑔 = −𝑚𝑔 𝑦2 − 𝑦1 . (14.13) Equation (14.13) says that the work done by gravity on a particle moving along an arbitrary path is equal to the particle’s weight times the particle’s change in height 𝑦2 − 𝑦1 . The path followed between 1 and 2 is arbitrary because the result in Eq. (14.13) depends only on the endpoints 1 and 2 rather than the shape of ℒ1-2 . We now consider another important example of this type of force, namely, a central force, which includes as special cases both the force of a spring and gravity as described by the universal law of gravitation.

Figure 14.6 A particle of mass 𝑚 travels between 1 and 2 along the path ℒ1-2 . The force 𝑚𝑔 is the only force considered in the work calculation in Eq. (14.12), but it need not be the only force acting on the particle. 1

ℒ1-2 𝐹⃗𝑐

A central force 𝐹⃗𝑐 (𝑟) acting on a particle 𝑃 is a force whose line of action always passes through the same point 𝑂 (see Fig. 14.7) and whose magnitude is a function of the distance 𝑟 from 𝑂 to 𝑃 . Referring again to Fig. 14.7, we wish to determine the work done by 𝐹⃗𝑐 (𝑟) on the particle 𝑃 as it moves along the path ℒ1-2 from 1, at time 𝑡1 , to 2, at time 𝑡2 . Figure 14.8 shows the polar coordinate system we will use to describe the motion of 𝑃 and the central force 𝐹⃗𝑐 . Using the component system shown and applying Eq. (14.10) on p. 863, the work done by 𝐹⃗𝑐 is given by ∫𝑡

𝑡2 1

𝐹⃗ ⋅ 𝑣⃗ 𝑑𝑡 =

∫𝑡

𝑡2

( ) 𝐹𝑐 (𝑟) 𝑢̂ 𝑟 ⋅ 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 𝑑𝑡,

∫𝑟

1

𝐹𝑐 (𝑟) 𝑑𝑟.

at time 𝑡2

𝐹⃗𝑐

Figure 14.7 Particle 𝑃 moving under the action of a central force 𝐹⃗𝑐 . 𝑢̂ 𝑟 𝑣⃗

𝑢̂ 𝜃 ℒ1-2

𝑃 𝐹⃗𝑐 𝑟

(14.14) 𝜃

where we have used 𝐹⃗𝑐 = 𝐹𝑐 (𝑟) 𝑢̂ 𝑟 and Eq. (12.68) on p. 720, which is the expression of the velocity in polar coordinates. Expanding the last dot product in Eq. (14.14) and noting that 𝑟̇ 𝑑𝑡 = (𝑑𝑟∕𝑑𝑡) 𝑑𝑡 = 𝑑𝑟, we obtain 𝑟2

2

𝑃

1

( ) 𝑈1-2 𝑐 =

at time 𝑡1

𝑃

𝑂

Work of a central force

( ) 𝑈1-2 𝑐 =

𝑦2

𝑚𝑔

(14.15)

𝑂 Figure 14.8 A particle 𝑃 under the action of a central force 𝐹⃗𝑐 . The figure also shows the velocity vector of 𝑃 , as well as the polar coordinate system used to describe 𝐹⃗𝑐 and the motion of 𝑃 .

878

Chapter 14

Energy Methods for Particles

Work done by a spring force. Figure 14.9(a) shows a linear spring connecting a particle 𝑃 to the fixed point 𝑂 while 𝑃 moves in a plane from point 1 to 2 along the path ℒ1-2 . Figure 14.9(b) shows that the spring force always acts along the line 𝑂𝑃 . For a spring whose current length is 𝑟 and whose unstretched length is 𝐿0 , the 2

1

ℒ1-2

𝑃

𝑃

2

1

𝑃

𝑃 𝐹𝑒 (𝑟2 )

𝐹𝑒 (𝑟1 )

𝑘

spring

𝑂

𝑂

(a)

(b)

Figure 14.9. (a) A particle 𝑃 moving along the path ℒ1-2 from 1 to 2. One end of the linear spring is attached to 𝑃 and the other to the fixed point 𝑂. (b) Direction of the spring force as 𝑃 moves along ℒ1-2 . Note that the figure is showing a situation in which 𝑟 > 𝐿0 .

( ) spring’s stretch is 𝛿 = 𝑟 − 𝐿0 and the corresponding force is 𝐹⃗𝑒 = −𝑘 𝑟 − 𝐿0 𝑢̂ 𝑟 . Thus, using Eq. (14.15), we have [( 𝑟2 ( ( ) ) )2 ( )2 ] 𝑈1-2 𝑒 = −𝑘 𝑟 − 𝐿0 𝑑𝑟 = − 12 𝑘 𝑟2 − 𝐿0 − 𝑟1 − 𝐿0 , (14.16) ∫𝑟 1

or ( ) ( ) 𝑈1-2 𝑒 = − 12 𝑘 𝛿22 − 𝛿12 ,

(14.17)

where 𝛿1 = 𝑟1 − 𝐿0 and 𝛿2 = 𝑟2 − 𝐿0 . As with the work done by the constant force of gravity, the work done by a spring force acting on a particle only depends on the endpoints of the path followed by the particle. 𝐵, 𝑚𝐵 𝑟1 𝐹⃗𝐵𝐴 𝑢̂ 𝑟 𝐴, 𝑚𝐴

ℒ1-2

Work done by the force of gravity. Figure 14.10 shows the gravitational force on a particle 𝐵 of mass 𝑚𝐵 exerted by a particle 𝐴 of mass 𝑚𝐴 whose position is fixed. This force is given by Newton’s universal law of gravitation [see Eq. (11.5) on p. 621], that is, 𝐺𝑚𝐴 𝑚𝐵 𝑢̂ 𝑟 . (14.18) 𝐹⃗𝐵𝐴 = − 𝑟2 Equation (14.15) then gives the work on 𝐵 as

𝑟2 Figure 14.10 The force of gravity 𝐹⃗𝐵𝐴 on 𝐵 due to its gravitational attraction to 𝐴.

ISTUDY

( ) 𝑈1-2 𝐺 = which yields

∫𝑟

𝑟2 1

−

𝐺𝑚𝐴 𝑚𝐵 𝑟2

𝑑𝑟 = −𝐺𝑚𝐴 𝑚𝐵

∫𝑟

( ) ( ) 1 1 𝑈1-2 𝐺 = −𝐺𝑚𝐴 𝑚𝐵 − + . 𝑟2 𝑟1

𝑟2 1

𝑑𝑟 , 𝑟2

(14.19)

(14.20)

Again, we see that the work done depends only on the initial and final positions of the particle.

Conservative forces and potential energy In computing the work done on a particle by spring and gravitational forces, we found that this work depended only on its initial and final positions, as opposed to depending on the path in some intrinsic way. As we saw in the marginal note “Line

ISTUDY

Section 14.2

Conservative Forces and Potential Energy

integral of an exact differential” on p. 862, if a path or line integral depends on only the limits of integration, then the integrand must be the exact differential of some function. Therefore, for these special forces, we can write the work as 𝑈1-2 =

∫ℒ

1-2

𝐹⃗ ⋅ 𝑑⃗𝑟 = −

∫ℒ

( )] ( ) [ ( ) 𝑑𝑉 = − 𝑉 𝑟⃗2 − 𝑉 𝑟⃗1 = − 𝑉2 − 𝑉1 , (14.21)

1-2

where 𝑟⃗1 and 𝑟⃗2 are the position vectors corresponding to 1 and 2, respectively, and where 𝑉 is a scalar function of position called the potential energy of the force 𝐹⃗ . Forces for which Eq. (14.21) is true are called conservative forces. If all of the forces doing work are conservative, then the statement of the workenergy principle can be given a form reflecting this. Substituting Eq. (14.21) into Eq. (14.9) on p. 862, we obtain the form of the work-energy principle called conservation of mechanical energy: 𝑇 1 + 𝑉 1 = 𝑇2 + 𝑉 2 .

879

Interesting Fact The minus sign in Eq. (14.21)? The minus sign is there so that the potential energy 𝑉 can be viewed as a measure of the potential or capacity to do positive work in returning to the zero potential energy state. This can be seen by noticing that Eq. (14.21) states that 𝑈1-2 = 𝑉1 − 𝑉2 so that the work is positive if 𝑉 decreases and negative if 𝑉 increases.

(14.22)

Systems for which Eq. (14.22) holds are called conservative systems. Equation (14.22) states that the total mechanical energy, that is, the potential plus kinetic energy, is conserved between any two points on the path ℒ1-2 , as long as all forces doing work are conservative. We can now use Eq. (14.21) to determine 𝑉 for the conservative forces we have encountered. Potential energy of a constant gravitational force. In Eq. (14.13), we saw that the work done by a constant gravitational force on a particle of mass 𝑚 is 𝑈1-2 = ( ) −𝑚𝑔 𝑦2 − 𝑦1 . Comparing this expression with Eq. (14.21) shows that 𝑉𝑔 = 𝑚𝑔𝑦,

(14.23)

where 𝑉𝑔 is the potential energy corresponding to the constant gravitational force 𝑚𝑔 and 𝑦 is the vertical position measured from the 𝑦 = 0 or datum line,∗ with 𝑦 increasing in the direction opposite to gravity. Since it is only the change in height that determines the work done by a constant gravitational force, the vertical position of the datum line is entirely arbitrary. Potential energy of a spring force. The work done by a spring force was found ) ( to be 𝑈1-2 = − 12 𝑘 𝛿22 − 𝛿12 . Comparing this with Eq. (14.21), we see that 𝑉𝑒 = 12 𝑘𝛿 2 ,

(14.24)

where 𝑉𝑒 is the elastic potential energy of the linear spring force and 𝛿 is the distance the spring is stretched or compressed from its unstretched length. Potential energy of the force of gravity. In Eq. (14.20), we determined the work done by the force of gravity on a particle 𝐵 due to its interaction with a particle 𝐴 to ( ) be 𝑈1-2 = −𝐺𝑚𝐴 𝑚𝐵 −1∕𝑟2 + 1∕𝑟1 . Comparing this expression with Eq. (14.21), we see that the potential energy associated with a gravitational force between 𝐴 and 𝐵 is given by 𝐺𝑚𝐴 𝑚𝐵 , 𝑉𝐺 = − (14.25) 𝑟 where 𝑟 is the distance between 𝐴 and 𝐵. ∗ The

word datum comes from the Latin verb dare, which means to give, and it is used in surveying, mapping, and geology to designate a given point, line, or surface used as a reference.

Common Pitfall Elastic potential energy is never negative. It may be tempting to write 𝑉𝑒 = − 12 𝑘𝛿 2 when 𝛿 is negative, i.e., when the spring is compressed. However, it is important to remember that the potential energy of a linear elastic spring is never negative because a spring always has the potential to do positive work when it returns to its unstretched length from a stretched or compressed condition. This is reflected in the fact that 𝛿 appears in the expression for 𝑉 as 𝛿 2 , which must always be positive or zero.

880

ISTUDY

Chapter 14

Energy Methods for Particles

Work-energy principle for any type of force The work-energy principle, as given by Eq. (14.9) on p. 862, applies to any system, conservative or nonconservative. When all the forces doing work on a system are conservative, then we can use changes in potential energy to find the work done by those forces [see Eq. (14.21)] instead of computing the work from its integral definition [see Eq. (14.7)]. When a problem involves both conservative and nonconservative forces, it would still be nice to take advantage of Eq. (14.21) to compute the work of ( ) ) ( the conservative forces. To do this, we let 𝑈1-2 c and 𝑈1-2 nc be the work done by conservative and nonconservative forces, respectively. Then, noting that their sum is the total work done, we can write the work-energy principle as ( ) ( ) 𝑇1 + 𝑈1-2 c + 𝑈1-2 nc = 𝑇2 . (14.26) ( ( ) ) Using Eq. (14.21), we can write 𝑈1-2 c = − 𝑉2 − 𝑉1 so that Eq. (14.26) becomes ( ) 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 .

(14.27)

Equation (14.27) is just as general as Eq. (14.9), but it is easier to apply since the work done by conservative forces, such as spring and gravitational forces, can be included in the 𝑉1 and 𝑉2 terms.

When is a force conservative? Is there a way to tell whether or not a force is conservative? That is, can we determine whether or not an associated potential energy exists and can we find it? Also, given a potential energy, can we find the associated force? Equation (14.21) implies that the infinitesimal work of a conservative force 𝐹⃗ can be written as 𝐹⃗ ⋅ 𝑑⃗𝑟 = −𝑑𝑉 . (14.28) Since 𝑉 = 𝑉 (𝑥, 𝑦, 𝑧) in Cartesian components, we can write the differential of 𝑉 as 𝑑𝑉 =

𝜕𝑉 𝜕𝑉 𝜕𝑉 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧

(14.29)

and 𝐹⃗ ⋅ 𝑑⃗𝑟 as ) ( ) ( 𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝐹𝑥 𝚤̂ + 𝐹𝑦 𝚥̂ + 𝐹𝑧 𝑘̂ ⋅ 𝑑𝑥 𝚤̂ + 𝑑𝑦 𝚥̂ + 𝑑𝑧 𝑘̂ = 𝐹𝑥 𝑑𝑥 + 𝐹𝑦 𝑑𝑦 + 𝐹𝑧 𝑑𝑧.

(14.30)

Inserting Eqs. (14.29) and (14.30) into Eq. (14.28), we see that 𝐹𝑥 𝑑𝑥 + 𝐹𝑦 𝑑𝑦 + 𝐹𝑧 𝑑𝑧 = −

Interesting Fact The gradient operator. The vector differen⃗ should be read as nabla or tial operator ∇ del. The gradient of the scalar function 𝑉 is often written grad 𝑉 . Note that the gradient operator in cylindrical coordinates can be written as ⃗ = 𝜕 𝑢̂ + 1 𝜕 𝑢̂ + 𝜕 𝑢̂ . ∇ 𝜕𝑟 𝑟 𝑟 𝜕𝜃 𝜃 𝜕𝑧 𝑧

𝜕𝑉 𝜕𝑉 𝜕𝑉 𝑑𝑥 − 𝑑𝑦 − 𝑑𝑧. 𝜕𝑥 𝜕𝑦 𝜕𝑧

(14.31)

Since 𝑑𝑥, 𝑑𝑦, and 𝑑𝑧 are independent of each other, for Eq. (14.31) to be satisfied, it must be true that 𝜕𝑉 𝜕𝑉 𝜕𝑉 𝐹𝑥 = − , 𝐹𝑦 = − , and 𝐹𝑧 = − . (14.32) 𝜕𝑥 𝜕𝑦 𝜕𝑧 You may recognize that, except for the sign, the right-hand sides of Eqs. (14.32) are the components of the gradient of 𝑉 , so we can say ( ) 𝜕𝑉 𝜕𝑉 𝜕𝑉 ̂ ⃗ , 𝐹⃗ = − 𝚤̂ + 𝚥̂ + 𝑘 = −∇𝑉 (14.33) 𝜕𝑥 𝜕𝑦 𝜕𝑧

ISTUDY

Section 14.2

Conservative Forces and Potential Energy

⃗ is the gradient of 𝑉 (also written as grad 𝑉 ). Equation (14.33) answers where ∇𝑉 one of our questions; that is, given a potential 𝑉 , we can find 𝐹⃗ by computing the negative of the gradient of 𝑉 . ⃗ 𝐹⃗ (also written as curl 𝐹⃗ ), Now, if we take the curl of 𝐹⃗ , which is defined to be ∇× in Cartesian components, we obtain ) ( ( ) 𝜕 𝜕 ̂ 𝜕 ⃗ ⃗ 𝚤̂ + 𝚥̂ + 𝑘 × 𝐹𝑥 𝚤̂ + 𝐹𝑦 𝚥̂ + 𝐹𝑧 𝑘̂ ∇×𝐹 = 𝜕𝑥 𝜕𝑦 𝜕𝑧 ( ) ( ) ( 𝜕𝐹 ) 𝜕𝐹𝑧 𝜕𝐹𝑦 𝜕𝐹𝑥 𝜕𝐹𝑧 𝜕𝐹 𝑦 ̂ = − − − 𝑥 𝑘. 𝚤̂ + 𝚥̂ + (14.34) 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 Substituting Eqs. (14.32) into Eq. (14.34) gives ⃗ × 𝐹⃗ = ∇

(

𝜕2𝑉 𝜕2𝑉 − + 𝜕𝑧𝜕𝑦 𝜕𝑦𝜕𝑧

)

( 2 ) 𝜕 𝑉 𝜕2𝑉 𝚤̂ + − + 𝚥̂ 𝜕𝑥𝜕𝑧 𝜕𝑧𝜕𝑥 ) ( 2 𝜕2𝑉 𝜕 𝑉 ⃗ + 𝑘̂ = 0, + − 𝜕𝑦𝜕𝑥 𝜕𝑥𝜕𝑦

(14.35)

where the equality to zero holds as long as the partial derivatives in Eq. (14.35) are continuous so that we can interchange the order of differentiation of the second derivatives of 𝑉 , that is, 𝜕2𝑉 𝜕2𝑉 = , 𝜕𝑧 𝜕𝑦 𝜕𝑦 𝜕𝑧

𝜕2𝑉 𝜕2𝑉 = , 𝜕𝑥 𝜕𝑧 𝜕𝑧 𝜕𝑥

and

𝜕2𝑉 𝜕2𝑉 = . 𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦

(14.36)

Equation (14.35) helps us answer the other question we asked because it shows that the curl of a conservative force is necessarily equal to zero. It is possible to prove that the converse of this statement is also true, that is, if the curl of a force is equal to zero, then the force is conservative.

End of Section Summary In this section we discovered that there are forces whose work, in taking a particle from an initial to a final position, does not depend on the path connecting these positions. Work of a constant gravitational force. For a particle of mass 𝑚 moving in a constant gravitational field from 1 to 2, the work done on the particle is Eq. (14.13), p. 877 ( ( ) ) 𝑈1-2 𝑔 = −𝑚𝑔 𝑦2 − 𝑦1 , for which gravity must act in the −𝑦 direction. Work done by a spring force. For a particle subject to the force of a linear elastic spring with constant 𝑘, the work done on the particle is Eq. (14.17), p. 878 ( ( ) ) 𝑈1-2 𝑒 = − 12 𝑘 𝛿22 − 𝛿12 , where 𝛿1 and 𝛿2 are the stretch of the spring at 1 and 2, respectively.

881

882

ISTUDY

Chapter 14

Energy Methods for Particles

Conservative systems. A force is said to be conservative if the work done by it depends on only the initial and final position of its point of application, but is otherwise independent of the path connecting these positions. The work of a conservative force can be characterized by a scalar potential energy function 𝑉. A conservative system is one for which all the forces doing work are conservative. For conservative systems, the work-energy principle becomes the conservation of mechanical energy, which is given by Eq. (14.22), p. 879 𝑇1 + 𝑉 1 = 𝑇2 + 𝑉 2 . Important conservative forces include elastic spring forces and gravitational forces. The potential energy of a linear elastic spring force is given by Eq. (14.24), p. 879 𝑉𝑒 = 12 𝑘𝛿 2 , the potential energy of a constant gravitational force is given by Eq. (14.23), p. 879 𝑉𝑔 = 𝑚𝑔𝑦, and the potential energy of the force of gravity is given by Eq. (14.25), p. 879 𝑉𝐺 = −

𝐺𝑚𝐴 𝑚𝐵 𝑟

.

Finally, if we have a system in which there are both conservative and nonconservative forces doing work, we can use Eq. (14.9) on p. 862, that is, 𝑇1 + 𝑈1-2 = 𝑇2 , or we can take advantage of the potential energies of the conservative forces by writing the work-energy principle as Eq. (14.27), p. 880 ( ) 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 , ( ) where 𝑈1-2 nc is the work done by nonconservative forces. The value of Eq. (14.22) is that it enables a far more efficient solution to problems. Unlike many solution procedures used in Chapter 13, there is no need to integrate along the path of a particle to determine its final state. When only conservative forces are present, the end state can be inferred exclusively from the initial state and the form of the potential functions. The solution is path-independent. When a nonconservative force is present, as in Eq. (14.27), a path-dependent calculation is required only for that nonconservative force.

ISTUDY

Section 14.2

883

Conservative Forces and Potential Energy

E X A M P L E 14.5

Demolishing a Building: Speed of a Wrecking Ball

A 2500 lb wrecking ball 𝐴 is released from rest at 27◦ with respect to the vertical as shown in Fig. 1. When the ball hits the structure to be demolished, the cable holding the wrecking ball forms an angle of 11◦ . If the length 𝐿 of the cable is 30 f t, determine the speed with which the wrecking ball hits the structure. 27◦ 𝐿

SOLUTION Road Map & Modeling This problem requires that we relate a change in the position of the wrecking ball to its change in speed, and therefore, the work-energy principle should lead to a solution. Referring to Fig. 2, the wrecking ball is modeled as a particle subject to its own weight and the cable tension. We assume that the cable is inextensible and that the end at 𝑂 is fixed. Referring to Fig. 3, we denote the release position as 1 and the position right before impact with the structure as 2. The only force doing work is gravity, which is a conservative force.

𝐴

Figure 1 𝑢̂ 𝜃 𝑢̂ 𝑟

Governing Equations Balance Principles We now apply the principle of conservation of mechanical energy between 1 and 2, which reads

𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 , and 𝑇2 = 12 𝑚𝑣22 ,

(2)

Force Laws

Recalling that only gravity does work, and choosing the datum line as shown in Fig. 3, we have

Kinematic Equations

and 𝑉2 = −𝑚𝑔𝐿 cos 𝜃2 .

(3)

𝐴 𝑚𝑔 Figure 2 FBD of the wrecking ball. 𝑂

datum

Since the system is released from rest, we have 𝑣1 = 0.

Computation

𝜃

𝑃

and where 𝑚 is the mass of the wrecking ball and 𝑣1 and 𝑣2 are its speed at 1 and 2, respectively.

𝑉1 = −𝑚𝑔𝐿 cos 𝜃1

𝐿

𝑂

(1)

where 𝑇1 = 21 𝑚𝑣21

11◦

(4)

𝜃1

Substituting Eqs. (2)–(4) into Eq. (1), we have −𝑚𝑔𝐿 cos 𝜃1 =

1 𝑚𝑣22 2

− 𝑚𝑔𝐿 cos 𝜃2 ,

𝐿

(5)

1

which, after solving for 𝑣2 , yields √ 𝑣2 =

( ) 2𝑔𝐿 cos 𝜃2 − cos 𝜃1 = 13.23 f t∕s,

(6)

where we have used 𝜃1 = 27◦ and 𝜃2 = −11◦ to obtain the numerical result. Discussion & Verification

𝐴

The dimensions of the term under the square root sign are length squared over time squared. Therefore, our final result is dimensionally correct, and it has been expressed with correct units. ( ) A Closer Look Notice that the quantity 𝐿 cos 𝜃2 − cos 𝜃1 corresponds to the vertical drop of the ball. Hence, the speed achieved by the(ball is precisely) what the ball would achieve if it were dropped from a height equal to 𝐿 cos 𝜃2 − cos 𝜃1 .

𝜃2 2

Figure 3 Definition of 1 and 2, as well as of the datum line for gravitational potential energy.

884

Chapter 14

Energy Methods for Particles

E X A M P L E 14.6

Bungee Jumping: Conservation of Mechanical Energy The mechanics of bungee jumping are rather straightforward, but a miscalculation can have dire consequences. Let’s consider the highest bungee jumping location in the world — the Verzasca dam in southern Switzerland. The jump height off of the Verzasca dam is 722 f t. Given a jumper weighing 170 lb, determine the relationship between the stiffness 𝑘 of the bungee cord and its unstretched length 𝐿0 ; i.e., find 𝑘 as a function of 𝐿0 , so that the jumper has zero speed at the bottom of the dam. In addition, determine the unstretched length so that the acceleration of the jumper does not exceed 4𝑔 during the jump.

SOLUTION Road Map & Modeling Referring to the FBD in Fig. 2, we model the jumper as a particle subject to gravity and the force 𝐹𝑏 of the bungee cord, which we model as a linear elastic spring. The jumper begins the jump at 1 with zero speed and ends the jump at 2 with zero speed (see Fig. 2). Since we know the jumper’s speed at 1 and 2 and we know all the forces doing work on the jumper, we can apply the work-energy principle to the jumper to determine the relationship between the bungee stiffness and its unstretched length. We will then apply Newton’s second law to the jumper to determine the maximum acceleration so that we can find 𝑘 and 𝐿0 for the bungee cord. Note that all the forces acting on the jumper are conservative and that 𝐹𝑏 is zero until the jumper falls a distance equal to the unstretched length of the cord. GFC Collection/Alamy Stock Photo

Figure 1 The Verzasca dam in southern Switzerland, which was the location for the bungee jumping scene in the 1995 James Bond film GoldenEye.

𝑦

1

FBD of jumper

bung ee cord

𝐹𝑏

Governing Equations Balance Principles

𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 ,

where 𝑚 is the jumper’s mass and 𝑣1 and 𝑣2 are the jumper’s speed at 1 and 2, respectively. Since we also need to determine the jumper’s acceleration, we will write Newton’s second law for the jumper in the 𝑦 direction as ∑

bungee jumper

𝐹𝑦∶ 𝑚𝑔 − 𝐹𝑏 = 𝑚𝑎𝑦 ,

(3)

where we note that 𝐹𝑏 = 0 until the bungee cord engages. Force Laws datum

If we place the datum line for gravitational potential energy at 2, then

2

Figure 2 Profile of the Verzasca dam showing the jump platform as well as the bungee jumper. 1 and 2 are also defined. The blue inset shows the FBD after the jumper has fallen a distance greater than the unstretched length of the cord.

ISTUDY

(1)

where we have used the fact that all forces doing work are conservative. The kinetic energies can be written as 𝑇1 = 21 𝑚𝑣21 and 𝑇2 = 12 𝑚𝑣22 , (2)

722 f t 𝑚𝑔

Applying the work-energy principle between 1 and 2, we have

𝑉1 = 𝑚𝑔ℎ

)2 ( and 𝑉2 = 21 𝑘 ℎ − 𝐿0 ,

(4)

where ℎ = 722 f t, 𝑚𝑔 = 170 lb, and we have accounted for the potential energy of the bungee cord in 𝑉2 . We will also need the force law for the bungee cord, which is given by ( ) 𝐹𝑏 = 𝑘𝛿 = 𝑘 𝑦 − 𝐿0 ,

(5)

where we note that 𝐹𝑏 = 0 when 𝑦 ≤ 𝐿0 . Kinematic Equations

Since the jumper starts and ends the jump with zero speed, we

have 𝑣1 = 𝑣2 = 0.

(6)

ISTUDY

Section 14.2

Substituting Eqs. (2), (4), and (6) into Eq. (1) and solving for 𝑘, we obtain 2𝑚𝑔ℎ 𝑘= ( )2 . ℎ − 𝐿0

(7)

Equation (7) gives the desired 𝑘 as a function of 𝐿0 , a plot of which is shown in Fig. 3. We now want to design the bungee system so that the maximum acceleration of a jumper weighing 170 lb does not exceed 𝑎max = 4𝑔. As with any design problem, there are infinitely many solutions that will satisfy the criteria that the jumper have zero speed at 2 and that the jumper’s acceleration not exceed 4𝑔. Referring to the FBD inset in Fig. 2, we know that until the spring engages, the jumper will be in free fall and his acceleration will be 𝑔 downward. Once the spring engages, the acceleration is determined by solving Eqs. (3) and (5) for 𝑎𝑦 , which gives 𝑎𝑦 = 𝑔 −

) 𝑘( 𝑦 − 𝐿0 . 𝑚

(8)

Recall that Eq. (7) provides a relation between 𝑘 and 𝐿0 that ensures that the jumper has zero speed at 2. Therefore, using Eq. (7), Eq. (8) becomes ( ) ( ) 𝑎𝑦 2ℎ 𝑦 − 𝐿0 2𝑔ℎ 𝑦 − 𝐿0 =1− ( or (9) 𝑎𝑦 = 𝑔 − ( )2 ) 2 , 𝑦 > 𝐿0 . 𝑔 ℎ − 𝐿0 ℎ − 𝐿0 We will choose the following value of the unstretched length of the bungee cord:

250 200 𝑘 (lb∕f t)

Computation

885

Conservative Forces and Potential Energy

150 100 50 200

400 𝐿0 (f t)

600

Figure 3 Bungee stiffness 𝑘 as a function of its unstretched length 𝐿0 required to achieve zero speed at the bottom of the jump (dark green curve). The red vertical line is at 𝐿0 = 722 ft.

1 0 𝑎𝑦 𝑔 −1 −2

𝐿0 = ℎ∕2 = 361.0 f t.

(10)

We now need to verify that the design criterion requiring 𝑎𝑦 < 4𝑔 is always met. To do so, using the chosen 𝐿0 and referring to Fig. 4, we plot 𝑎𝑦 ∕𝑔 versus 𝑦. The plot shows that the chosen value of 𝐿0 is such that our design goal is met. Discussion & Verification

The right-hand side of Eq. (7) has dimensions of force over length and therefore, has the proper dimensions for 𝑘. Equation (7) also implies that the stiffness of the cord must be proportional to the jumper’s weight, which is to be expected. In addition, Eq. (7) and the corresponding plot in Fig. 3 demonstrate an interesting aspect of the required stiffness 𝑘. If the bungee cord has a very short unstretched length (i.e., the denominator in Eq. (7) will be large), then the stiffness required to stop the jumper after falling 722 f t is very small, and it increases very slowly until 𝐿0 reaches about 500 f t. As 𝐿0 gets closer to the jump height of 722 f t [i.e., the denominator in Eq. (7) will be small], the spring has to get very stiff to be able to stop the jumper in time and, at 𝐿0 = 722 f t, the stiffness becomes infinite. Hence, overall our solution seems to be correct. As far as the value of 𝐿0 is concerned, we have already verified that it satisfies the required criteria. A Closer Look Figure 4 shows the acceleration of the jumper as a function of distance from the top of the dam for 𝐿0 = 361.0 f t. This figure demonstrates that the largest acceleration experienced by the jumper is 3𝑔, and this occurs at the very end of the jump (i.e., at 𝑦 = 722 f t, the acceleration is 3𝑔 upward). As 𝐿0 increases, the maximum acceleration experienced by the jumper increases so that, for example, when 𝐿0 = 650 f t, we find that the maximum acceleration of the jumper is over 19𝑔! An acceleration of 19𝑔 is about twice the maximum permissible acceleration of a fighter pilot or astronaut under their most extreme environments and therefore completely unacceptable. Although we can generate a stiffness for any unstretched length, as shown in Figure 3, a significant portion of the parameter space, perhaps 𝐿0 > 500 f t, should be declared “out-of-bounds” because of the associated unacceptable maximum acceleration.

−3

0

200

400 𝑦 (f t)

600

Figure 4 The acceleration of the bungee jumper as a function of 𝑦 for 𝐿0 = ℎ∕2 = 361.0 ft.

Interesting Fact Bungee cord stiffness and unstretched length. We have not taken into account the fact that a real bungee cord has limits to the amount that it can stretch. For example, for 𝐿0 = 1 𝖿 𝗍, Eq. (7) tells us that 𝑘 = 0.4722 𝗅𝖻∕𝖿 𝗍 for ℎ = 722 𝖿 𝗍 and 𝑚𝑔 = 170 𝗅𝖻. In fact, increasing 𝐿0 to 400 𝖿 𝗍 only increases 𝑘 to 2.368 𝗅𝖻∕𝖿 𝗍. Of course, the 1 𝖿 𝗍 bungee cord would have to stretch 721 𝖿 𝗍 or 72,100%, and the 400 𝖿 𝗍 bungee cord would only have to stretch 300 𝖿 𝗍 or 80.5%. While a rubber band can be easily stretched to a little less than twice its original length, we are not aware of a rubber band that can stretch to over 700 times its original length!

886

Chapter 14

Energy Methods for Particles

E X A M P L E 14.7

Pole Vaulting: Turning Speed into Height

𝐵

𝐴

Figure 1 Sequence of images showing the positions of a vaulter during a vault. The pole has been shortened in some of the frames for clarity.

A pole vaulter relies on several skills to achieve maximum height during a vault, but one of the most important is running speed. Speed is important because it is vital that the vaulter turn kinetic energy (running speed) into potential energy (vault height). Figure 1 shows the final moments of a vault during which the forward speed of the vaulter at 𝐴 is turned into height at 𝐵. Modeling the vaulter as a particle, determine the maximum possible height that can be achieved by the “world’s fastest pole vaulter,” using the following data and assumptions: 1. At the time this book was written, the men’s world record in the 100 m dash was 9.58 s, which was established by Usain Bolt of Jamaica on August 16, 2009; the record in the 200 m was 19.19 s, which was set on August 20, 2009, also by Usain Bolt. 2. The pole vaulter is 6 f t tall with mass center at 55% of body height as measured from the ground. 3. All of the vaulter’s kinetic energy is converted to potential energy during the vault, and no energy is lost from the system consisting of the pole and the vaulter.

𝐺 𝑚𝑣 𝑔

4. The vaulter does no work during the vault, and his velocity is zero when he reaches the peak height of the vault.

SOLUTION Road Map & Modeling

𝑅𝑥 𝑅𝑦 Figure 2 FBD of the vaulter and pole during the vault. The weight force acting on the vaulter acts at his mass center 𝐺.

ISTUDY

Interesting Fact Top human speed. It is believed that the fastest recorded speed achieved by a human was by Donovan Bailey of Canada, who set the world record in the 100 𝗆 dash on July 27, 1996, at the Atlanta Olympic Games with a time of 9.84 𝗌. Near the 60 𝗆 mark, a radar gun recorded his speed to be 12.1 𝗆∕𝗌 = 27.1 𝗆𝗉𝗁. At the 1997 World Championships in Athens, via more careful measurements, both Maurice Greene and Bailey were clocked at 11.87 𝗆∕𝗌 (Greene ran 9.86 𝗌 to Bailey’s 9.91 𝗌). At the same championship, Bailey was clocked at 11.91 𝗆∕𝗌 in the 𝟦 × 100 𝗆 relay. See J. R. Mureika, “A Simple Model for Predicting Sprint-Race Times Accounting for Energy Loss on the Curve,” Canadian Journal of Physics, 75(11), 1997, pp. 837–851.

We will assume that the vault begins when the vaulter has reached maximum speed. To determine that top speed, we will use the 100 and 200 m dash data as follows: We assume that the vaulter starts from rest and accelerates to top speed with an acceleration that is approximately the same for both races. Since the runner has reached top speed by the time he reaches 100 m, we will assume that he runs at a uniform speed between 100 and 200 m. Therefore, we can approximate his maximum speed as 𝑣max =

200 − 100 m∕s = 10.41 m∕s. 19.19 − 9.58

(1)

To complete our model, we need to determine the forces acting on the vaulter. Referring to Fig. 2, we have treated the vaulter as a particle, and we have neglected air resistance as well as any moment that might be applied to the bottom of the pole by the box during the vault.∗ Note that in the FBD we have not included the weight of the pole. Observe that the point of application of the forces 𝑅𝑥 and 𝑅𝑦 is fixed so that these forces do no work. Hence, the only force doing work is gravity, which is a conservative force for which we know the potential energy function. Consequently, given that we want to relate changes in speed to changes in position, we can solve this problem by applying the work-energy principle. Governing Equations Balance Principles

All the forces doing work are conservative, and the work-energy

principle is 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 .

(2)

Referring to Fig. 3, 1 corresponds to the instant the vaulter plants the pole and 2 corresponds to the peak height of the vault. The vaulter’s kinetic energies at 1 and 2 are given by (3) 𝑇1 = 12 𝑚𝑣 𝑣21 and 𝑇2 = 12 𝑚𝑣 𝑣22 , where 𝑚𝑣 is the vaulter’s mass and 𝑣1 and 𝑣2 are the vaulter’s speed at 1 and 2, respectively. ∗ The

box is the 20 cm depression in which the vaulter plants the pole to begin the vault.

ISTUDY

Section 14.2

Conservative Forces and Potential Energy

887

2

ℎ

1

𝑑

datum

Figure 3. Sequence showing 1 and 2 within the vault. Force Laws

We will include all potential energy and work terms here. Recalling that the vaulter’s weight is the only force doing work, we have (see Fig. 3) 𝑉1 = 𝑚𝑣 𝑔𝑑

and 𝑉2 = 𝑚𝑣 𝑔(𝑑 + ℎ),

(4)

Interesting Fact

where we have used the ground as our datum line for gravitational potential energy. Kinematic Equations

Based on the Road Map & Modeling discussion, and ignoring any horizontal component of velocity the vaulter might have at 2, we have 𝑣1 = 𝑣max Computation

and 𝑣2 = 0.

(5)

Substituting Eqs. (3)–(5) into Eq. (2) and simplifying, we have 1 2 𝑣 2 max

= 𝑔ℎ

⇒

ℎ=

𝑣2max 2𝑔

= 5.519 m,

(6)

where we have used the value of 𝑣max in Eq. (1). Since at 1 the vaulter’s mass center is a distance 𝑑 = 0.55(6) ft = 3.3 f t = 1.006 m above the ground, we can add the distance 𝑑 to ℎ to find that the maximum height achievable by a pole vaulter is ℎmax = 6.526 m.

(7)

Discussion & Verification

Given that the right-hand side of Eq. (6) has dimensions of length, we can say that our result has the correct dimensions. As far as the correctness of the value we have obtained is concerned, it is larger than the current world record but not far from it (see discussion below). Hence, overall, our solution appears to be correct. A Closer Look At the time this book was written, the world record in the pole vault was 6.18 m (see marginal note entitled “World records”). The height we obtained is not far from this (it is 5.6% higher). There are many factors that we have not included in our analysis. Here are a few of these factors, which may contribute to our result being too high.

• Vaulters cannot take advantage of all of their speed at the pole plant since they must leap at a takeoff angle of 15–20◦ . • While elite male vaulters run faster than 9.0 m∕s during the last 5 m of their approach sprint, our estimate of 10.41 m∕s was too high. • Vaulters cannot have zero velocity at the maximum height because they need some horizontal speed to clear the bar. On the other hand, we have not taken into account the fact that the vaulter can do work during the vault between 1 and 2, for example, by pushing with the arms against the pole. This allows the vaulter to go higher than our prediction would suggest, and this additional work can generally add about 0.8 m to the vault.

Fiberglass poles. Fiberglass poles were introduced in the early 1960s. Before that, aluminum, steel, or bamboo poles were used. With fiberglass poles, vault heights and records increased dramatically. From 1940 to 1960, the world record increased from 4.7 to 4.8 𝗆, whereas from 1960 to 1963 it increased from 4.8 to 5.1 𝗆. Fiberglass poles were thought to “catapult” the vaulter over the bar, but studies showed otherwise. There are two reasons why fiberglass poles facilitate performances. First, a flexible pole allows for a higher grip height on the pole. A vaulter’s grip height is determined by the kinetic energy at takeoff — the higher the kinetic energy, the longer the pole vaulter can rotate to vertical. Second, flexible poles allow vaulters to have a lower takeoff angle since flexible poles can accept and return much more energy than a stiff pole.

Interesting Fact World records. At the time of writing, the world record in the pole vault is 6.18 m = 20 ft 31⁄4 in., set by Armand Duplantis of Sweden in 2020. For comparison, the high jump world record was 2.45 m = 8 ft 1⁄2 in., set by Javier Sotomayor of Cuba in 1993. If all that mattered were kinetic energy, these two records would be the same. However, without a pole this energy cannot be used to gain elevation, and that’s why horizontal speed does not play a strong role in the high jump.

888

E X A M P L E 14.8

Sliding With and Without Friction A crate is unloaded from the top of a recreational vehicle (RV) using the ramp shown in Fig. 1. Determine the speed of the crate as it hits the ground for each of the following two cases if the crate is released from rest in the position shown and if

crate 11 f t

Chapter 14

Energy Methods for Particles

19 f t ramp

(a) there is negligible friction between the crate and the ramp; and

Figure 1 The RV with the ramp and its dimensions.

(b) the kinetic friction coefficient between the crate and the ramp is 𝜇𝑘 = 0.32. Assume that static friction is insufficient to prevent slipping.

SOLUTION 𝐹 𝚥̂ 𝚤̂ 𝑁 𝑚𝑔

𝜃

Figure 2. FBD of the crate as it slides down the ramp. Road Map & Modeling

1

crate

ramp 2

datum

Figure 3 Definition of 1 and 2 for the work-energy principle.

ISTUDY

We will model the crate as a particle subject to its own weight, the normal reaction between the crate and the slide, and the friction force 𝐹 (Fig. 2). The force 𝐹 will be set to zero when we consider the case of frictionless contact between crate and slide. Since we need to relate a change in the crate’s position to a corresponding change in speed, we will apply the work-energy principle. However, since one of the forces doing work is the friction force 𝐹 , we will also apply Newton’s second law in the direction perpendicular to the slide to determine the relation between the normal reaction 𝑁 and the weight. We define 1 to be the point of release of the crate and 2 to be just before the crate hits the ground (see Fig. 3). The length of the ramp 𝑙 is 19 f t, and the vertical drop ℎ is 11 f t. We will solve Parts (a) and (b) at the same time since the solution for Part (a) is a special case of the solution to case (b). Finally, since it will be used later in the calculations, we note that the angle 𝜃 appearing in the FBD of Fig. 2 is such that sin 𝜃 = 11∕19

⇒

𝜃 = 35.38◦ .

(1)

Governing Equations Balance Principles

The work-energy principle, applied between 1 and 2, reads 𝑇1 + 𝑉1 + (𝑈1-2 )nc = 𝑇2 + 𝑉2 .

(2)

The kinetic energies can be written as 𝑇1 = 21 𝑚𝑣21

and 𝑇2 = 12 𝑚𝑣22 ,

(3)

where 𝑚 is the crate’s mass and 𝑣1 and 𝑣2 are the crate’s speed at 1 and 2, respectively. Referring to the FBD in Fig. 2, Newton’s second law in the 𝑦 direction gives ∑

𝐹𝑦∶ 𝑁 − 𝑚𝑔 cos 𝜃 = 𝑚𝑎𝑦 .

(4)

ISTUDY

Section 14.2

Force Laws

Conservative Forces and Potential Energy

By setting the datum line at 2, the potential energies in Eq. (2) are 𝑉1 = 𝑚𝑔ℎ

and 𝑉2 = 0.

(5)

The work done by the friction force 𝐹 is (𝑈1-2 )nc =

∫ℒ

−𝐹 𝚤̂ ⋅ 𝑑⃗𝑟 =

1-2

∫0

𝑙

−𝐹 𝚤̂ ⋅ 𝑑𝑥 𝚤̂ =

∫0

𝑙

−𝐹 𝑑𝑥,

(6)

where 𝐹 is related to the normal force 𝑁 by the Coulomb friction law for sliding, i.e., 𝐹 = 𝜇𝑘 𝑁. Kinematic Equations

(7)

Since the crate slides along the ramp, we have 𝑎𝑦 = 0.

(8)

To compute the kinetic energies at 1 and 2, we note that 𝑣1 = 0

(9)

and that 𝑣2 is the main unknown of the problem. Computation

Using Eqs. (4), (7), and (8), we have 𝐹 = 𝜇𝑘 𝑚𝑔 cos 𝜃.

(10)

Substituting Eq. (10) into Eq. (6), we then have (𝑈1-2 )nc = −𝜇𝑘 𝑚𝑔𝑙 cos 𝜃.

(11)

Finally, substituting Eqs. (3), (5), (9), and (11) into Eq. (2), we obtain 𝑚𝑔ℎ − 𝜇𝑘 𝑚𝑔𝑙 cos 𝜃 = 12 𝑚𝑣22 .

(12)

For case (a), we have 𝜇𝑘 = 0, so Eq. (12) yields 𝑣2 =

√

2𝑔ℎ = 26.62 f t∕s.

(13)

For case (b), 𝜇𝑘 is different from zero and is equal to the given value of 0.32, so that, from Eq. (12), we have √ 𝑣2 =

2𝑔(ℎ − 𝜇𝑘 𝑙 cos 𝜃) = 19.73 f t∕s.

(14)

Discussion & Verification

Given that the terms under the square root sign in Eqs. (13) and (14) have dimensions of acceleration times length, our results are dimensionally correct. In addition, the result in Eq. (13) is larger than that in Eq. (14), as expected given that the effect of friction is to oppose the motion of the crate. Therefore, our solution is reasonable. A Closer Look The result in Eq. (13) corresponds to the speed that the crate would have if the crate had been dropped from the top of the RV. While perhaps surprising, this result is indeed correct. In fact, in Part (a), the only two forces acting on the crate are the weight force 𝑚𝑔 and the normal force 𝑁. The normal force does no work, and the work done by the weight force can be easily computed as the force times the component of displacement in the direction of the force (since the weight, in this problem, is a constant force); this component is the vertical drop ℎ. Therefore, the work done by 𝑚𝑔 is 𝑚𝑔ℎ, and that is exactly what it would be if we had dropped the crate from the top of the RV.

889

890

Chapter 14

Energy Methods for Particles

E X A M P L E 14.9

Cyclic Work to Promote Swinging Motion

𝑂

𝐷 𝐴

𝐶

1. At 1, that is, at the top of a backswing, the speed is zero and you are squatting.

𝐵 Figure 1 One possible stand-squat sequence during a half swing. You are squatting at 𝐴, standing at 𝐵, squatting at 𝐶, and then standing again at the top of the swing at 𝐷. The orange dots indicate the position of your mass center, and the blue line is the trajectory of your mass center. The two dashed lines indicate arcs of nearest and farthest distance of your mass center from point 𝑂.

3. You next swing from 3 to 4 and then back to 5 while standing upright. 4. At 5, you then instantaneously squat and another cycle begins. The goal is to find 𝜃5 as a function of 𝜃1 due to the pumping motion described above. When you are standing, your mass center is a distance 𝑙 from the fixed point 𝑂; and when you are squatting, your mass center is a distance 𝑙 + 𝛿𝑙 from 𝑂.

Road Map & Modeling

𝜃

5

𝑙 1 3

2. At 2, which occurs at the lowest point in the swing, you stand upright to move to 3. This motion is assumed to occur instantaneously.

SOLUTION

𝑂 4

Figure 1 shows you swinging in the standing and squatting positions. As you probably know, if you raise (stand) and lower (squat) yourself as you swing, you can increase the amplitude of swinging with each cycle. While the analysis of the motion shown in Fig. 1 can be very involved,∗ we can create a simple model to investigate how pumping can increase the amplitude of each swing cycle. We will analyze the pumping motion depicted in Fig. 2. This motion consists of the following sequence of events:

𝛿𝑙

2

Figure 2 The five key positions of the pumping motion, along with the dimensions of the movement of the mass center.

As you alternately stand and squat, you will be doing work, and it is this work that will increase the swing angle with each cycle. We will write the work-energy principle to relate the potential energy at the beginning and end of the swing to the work done during the swing. You will be modeled as a particle located at your mass center, which moves as described in Fig. 2. In applying the work-energy principle between 1 and 5, we will need to calculate the work done as you go from squatting to standing from 2 to 3. Your speed at both 1 and 5 is zero. We will ignore any drag and dissipative forces. Your FBD in an arbitrary position is shown in Fig. 3. This figure shows us that the weight force 𝑚𝑔 must do work (which we will compute in potential energy terms). Although perhaps not obvious, the force 𝑅 also does work because it is responsible for lifting your mass center from 𝑙 + 𝛿𝑙 up to 𝑙 in going from 2 to 3. Governing Equations Balance Principles

𝑅

𝑢̂ 𝑡

𝑇1 + 𝑉1 + (𝑈1-5 )nc = 𝑇5 + 𝑉5 .

𝑢̂ 𝑛

(1)

The kinetic energies are given by 𝑚𝑔 𝜃

Figure 3 FBD of you on the swing at an arbitrary angle.

ISTUDY

The work-energy principle, applied between 1 and 5, reads

𝑇1 = 21 𝑚𝑣21

and 𝑇5 = 12 𝑚𝑣25 ,

(2)

where 𝑚 is the mass of the person and 𝑣1 and 𝑣5 are the speed of the person at 1 and 5, respectively. In addition, referring to Fig. 3, Newton’s second law applied at 2 (i.e., at 𝜃 = 0) gives ∑ ∗ See,

𝐹𝑛∶ 𝑅 − 𝑚𝑔 = 𝑚𝑎𝑛 .

(3)

for example, W. B. Case and M. A. Swanson, “The Pumping of a Swing from the Seated Position,” American Journal of Physics, 58(5), 1990, pp. 463–467; or W. B. Case, “The Pumping of a Swing from the Standing Position,” American Journal of Physics, 64(3), 1996, pp. 215–220.

ISTUDY

Section 14.2

Force Laws

Conservative Forces and Potential Energy

Choosing the datum for the gravitational potential energy at 2, we have ( ) ( ) 𝑉1 = 𝑚𝑔(𝑙 + 𝛿𝑙) 1 − cos 𝜃1 and 𝑉5 = 𝑚𝑔𝑙 1 − cos 𝜃5 . (4)

In addition, recalling that (𝑈1-5 )nc is the work done by 𝑅 in lifting the mass center from 2 to 3, we have (𝑈1-5 )nc = 𝑅 𝛿𝑙, (5) where we assume that 𝑅 is constant throughout the lift. Kinematic Equations

At 1 and 5, we have 𝑣1 = 0 and 𝑣5 = 0.

(6)

In addition, the acceleration 𝑎𝑛 at 2 is given by 𝑎𝑛 = Computation

𝑣22 𝑙 + 𝛿𝑙

.

(7)

Combining Eqs. (3)–(7) with Eq. (1), we obtain

( ) ( 𝑚𝑔(𝑙 + 𝛿𝑙) 1 − cos 𝜃1 + 𝑚 𝑔 +

𝑣22 𝑙 + 𝛿𝑙

) ) ( 𝛿𝑙 = 𝑚𝑔𝑙 1 − cos 𝜃5 .

891

Helpful Information Is it correct to assume that 𝑹 is constant? Equations (3) and (7) imply that 𝑅 cannot be constant between 2 and 3 because 𝑅 depends on 𝑎𝑛 . In turn, 𝑎𝑛 depends on the distance to point 𝑂 and on 𝑣2 , both of which vary as the particle is lifted from 2 to 3. The distance from the particle to 𝑂 changes from 𝑙 + 𝛿𝑙 to 𝑙. In addition, applying the work-energy principle from 2 to 3, we know that there must be a change in speed between these two positions because 𝑅 does work between them. However, it can be argued that the assumption that 𝑅 is constant between 2 and 3 is acceptable as long as 𝛿𝑙 is small compared to 𝑙.

(8)

Applying the work-energy principle between 1 and 2, during which only the weight force does work, we can relate 𝑣2 to 𝜃1 as follows: ( ) 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 ⇒ 𝑚𝑔(𝑙 + 𝛿𝑙) 1 − cos 𝜃1 = 21 𝑚𝑣22 , (9) 120◦

which, by solving for 𝑣22 ∕(𝑙 + 𝛿𝑙), yields 𝑣22 𝑙 + 𝛿𝑙

( ) = 2𝑔 1 − cos 𝜃1 .

90◦

(10)

𝜃5 60◦

Substituting Eq. (10) into Eq. (8) and solving for 𝜃5 , we obtain the final result [ ] ) 𝛿𝑙 ( −1 𝜃5 = cos cos 𝜃1 + 3 cos 𝜃1 − 4 . 𝑙

30◦ 0◦

(11)

Discussion & Verification

The argument of the inverse cosine function in Eq. (11) is nondimensional, as it should be. In addition, recalling that the cosine function varies only between −1 and 1, we observe that the term 3 cos 𝜃1 − 4 is always negative. In turn this causes the value of 𝜃5 to be greater than 𝜃1 whenever 𝛿𝑙 is positive. This result confirms that the pumping action will result in an increased swing angle. Now that we have the angle at the end of a swing cycle 𝜃5 in terms of the angle at the beginning of a swing cycle 𝜃1 , we can plug in some numbers to see if our pumping scheme works. For example, say that 𝑙 = 1.8 m, 𝛿𝑙 = 0.2 m, and 𝜃1 = 40◦ . With these numbers, Eq. (11) tells us that 𝜃5 = 54.8◦ , thus confirming that we get an increase in swing angle for every cycle (see Figs. 4 and 5). However, notice that Figs. 4 and 5 tell us that if 𝜃1 = 0◦ , then by simply standing up (while swinging at zero speed), you will somehow swing to 𝜃5 ≈ 27◦ ! This result is clearly at odds with what happens in reality, and it indicates that our model has some strong limitations. In particular, as discussed in the Helpful Information marginal note, our model cannot be expected to be accurate unless the ratio 𝛿𝑙∕𝑙 is sufficiently small. In addition, our model implicitly assumes that 𝜃1 ≠ 0; i.e., that our motion does not start from static equilibrium. The lesson to learn here is that when we construct a physical model, we must be aware of the limitations of the model when interpreting the model’s predictions.

0◦

30◦

60◦

90◦

𝜃1 Figure 4 The angle 𝜃5 as a function of 𝜃1 . This figure, along with Fig. 5, shows that our model gives physically unrealistic results when 𝜃1 is small. 30◦

𝜃5 − 𝜃1

A Closer Look

20◦ 10◦ 0◦

0◦

30◦

60◦

90◦

𝜃1 Figure 5 This plot shows the increase in swing amplitude 𝜃5 − 𝜃1 as a function of 𝜃1 . Our model gives physically unrealistic results when 𝜃1 is small.

892

Chapter 14

Energy Methods for Particles

E X A M P L E 14.10

A Conservative Force Field The force field shown in Fig. 1 is mathematically represented by

6

𝑦

𝐹⃗ = 𝑦2 𝚤̂ + 2𝑥𝑦 𝚥̂.

4

Determine the potential energy of 𝐹⃗ .

2

SOLUTION

0

Road Map & Modeling

(1)

By computing the curl of the given force field, we can easily ⃗ Therefore, we know that there is a potential 𝑉 whose gradient verify that curl 𝐹⃗ = 0. is −𝐹⃗ . With this in mind, our strategy will be to use Eqs. (14.32) on p. 880 since they tell us how the components of 𝐹⃗ relate to 𝑉 . We can view each of these equations as a simple first-order differential equation that we integrate.

−2 −4

Governing Equations −6 −4

−2

0 𝑥

2

4

From Eq. (1), the components of 𝐹⃗ are 𝐹𝑥 = 𝑦 2

6

Figure 1 Graphical representation of the force field given in Eq. (1). If each red dot were a particle, then the arrow on each of those dots would represent the force vector on it. In addition, the length of each arrow is proportional to the magnitude of the force.

and 𝐹𝑦 = 2𝑥𝑦.

(2)

Using Eqs. (14.32), the relation between the components of 𝐹⃗ and 𝑉 is 𝐹𝑥 = − Computation

𝜕𝑉 𝜕𝑥

and 𝐹𝑦 = −

𝜕𝑉 . 𝜕𝑦

(3)

Equating the first of Eqs. (2) with the first of Eqs. (3), we have 𝑦2 = −

𝜕𝑉 , 𝜕𝑥

(4)

which we integrate to obtain 𝑉 = −𝑥𝑦2 + 𝑓 (𝑦),

where 𝑓 (𝑦) is an arbitrary function of 𝑦. Next, we equate the second of Eqs. (2) with the second of Eqs. (3) to obtain 𝜕𝑉 2𝑥𝑦 = − , (6) 𝜕𝑦 which we integrate to obtain 𝑉 = −𝑥𝑦2 + 𝑔(𝑥), (7)

6 4 2

where 𝑔(𝑥) is an arbitrary function of 𝑥. Comparing Eqs. (5) and (7), we see that

𝑦 0

𝑉 = −𝑥𝑦2 + 𝑓 (𝑦) and 𝑉 = −𝑥𝑦2 + 𝑔(𝑥),

(8)

and the only way that both of these can be true is if 𝑉 = −𝑥𝑦2 + constant. Since only differences in potential energies matter, we can set this constant equal to zero to obtain

−2 −4

𝑉 = −𝑥𝑦2 .

−6 −2

0 𝑥

2

4

6

Figure 2 Figure 1 with constant values of 𝑉 superimposed (i.e., the blue lines).

(9)

Equation (9) for different values of 𝑉 has been plotted on top of Fig. 1, and the result is shown in Fig. 2. Notice that the blue lines, that is, lines of constant 𝑉 , are perpendicular to the force vectors. This is what we should expect since the force field 𝐹⃗ is the negative of the gradient of 𝑉 . Recall from calculus that the gradient of a scalar function gives, at each point, the direction of greatest change of that function. This agrees with our result since the direction of greatest change of 𝑉 must be orthogonal to lines of constant 𝑉 . Discussion & Verification

−4

ISTUDY

(5)

ISTUDY

Section 14.2

893

Conservative Forces and Potential Energy

Problems Problem 14.34 The pendulum shown is put in motion with a speed 𝑣0 when 𝜃 = 0◦ . Letting 𝐿 = 2 f t, determine 𝑣0 if the pendulum first comes to a stop at 𝜃 = 47◦ . 𝑣0 𝐴 𝜌 𝜃

𝐿

𝐵 𝐶

Figure P14.34

Figure P14.35

Problem 14.35 Point 𝐴 is the highest point along the roller coaster ride section shown. The inscribed circle at 𝐴 has radius 𝜌 = 25 m and center at 𝐶. If point 𝐵 is on a horizontal line going through 𝐶 and if the roller coaster has a speed 𝑣0 = 45 km∕h at 𝐴, neglecting friction and air drag and treating the roller coaster as a particle, determine the speed of the roller coaster at 𝐵.

Problem 14.36 Assuming that the plunger of a pinball machine has negligible mass and that friction is negligible, determine the spring constant 𝑘 such that a 2.85 oz ball is released with a speed 𝑣 = 15 f t∕s, after pulling back the plunger 2 in. from its rest position, i.e., from the position in which the spring is uncompressed.

𝑘

Figure P14.36

Problems 14.37 and 14.38 Consider a 3300 lb car whose speed is increased by 35 mph over a distance of 200 f t while traveling up a rectilinear incline with a 15% grade. Model the car as a particle, assume that the tires do not slip, and neglect all sources of frictional losses and drag. 15 100 Figure P14.37 and P14.38 Problem 14.37

Determine the work done on the car by the engine if the car starts

from rest. Problem 14.38

Determine the work done on the car by the engine if the car has an initial speed of 30 mph.

Problem 14.39 A classic car is driving down an incline at 60 km∕h when its brakes are applied. Treating the car as a particle, neglecting all forces except gravity and friction, and assuming that the tires slip, determine the coefficient of kinetic friction if the car comes to a stop in 55 m and 𝜃 = 20◦ .

𝜃 Figure P14.39

𝑚

894

Chapter 14

Energy Methods for Particles

Problem 14.40 A 75 kg skydiver is falling at a speed of 250 km∕h when, at a height of 245 m, the parachute is deployed, allowing the skydiver to land at a speed of 4 m∕s. Modeling the skydiver as a particle and assuming that the skydiver follows a perfectly vertical trajectory, determine the average force exerted by the parachute from the moment of deployment until landing.

Problems 14.41 and 14.42 Each of the two rubber tubes of the wrist rocket has an unstretched length 𝐿0 = 1 f t. They are symmetrically pulled back so that 𝓁 = 3 f t and then are released from rest. The pellet 𝑃 that is launched by the wrist rocket weighs 0.145 oz. Neglect the mass of the rubber tubes and any change in height of the pellet while it is in contact with the rubber tubes. Finally, let 𝑑 = 1.5 in.

Figure P14.40

𝑑 𝑑

Problem 14.41 If the desired launch speed of the pellet 𝑃 is 100 mph, determine the required stiffness of each rubber tube.

𝓁 𝑃

Problem 14.42 If the stiffness of each rubber tube is 𝑘 = 5 lb∕f t, determine the range of the pellet if it is launched at 45◦ with respect to the horizontal and if it lands at the same height from which it was launched. Neglect air resistance.

Problems 14.43 through 14.45

Figure P14.41 and P14.42

ISTUDY

The collar 𝐶 of mass 𝑚 = 2 kg slides freely on the rod 𝐸𝐹 . At the instant shown, the collar 𝐶 is moving downward with speed 𝑣1 = 2.5 m∕s and it is 1∕3 of the way between 𝐸 and 𝐹 . The dimensions are 𝐿 = 2.1 m, 𝑤 = 1.9 m, ℎ = 2.3 m, and 𝑑 = 1.2 m. 𝑧 𝐸 𝐶 𝑣1

𝑔

ℎ

𝑘 𝐹

𝑥

𝐴 𝑤

𝐵

𝑦

𝐷 𝑑

𝐿

Figure P14.43–P14.45

If the unstretched length of the spring is 𝐿0 = 1 m, and the spring stiffness is 𝑘 = 50 N∕m, determine the speed of the collar when it reaches 𝐹 . Assume that the attachment point at 𝐴 does not impede the spring.

Problem 14.43

Problem 14.44 If the unstretched length of the spring is 𝐿0 = 4 m and the spring stiffness is 𝑘 = 50 N∕m, determine the speed of the collar when it reaches 𝐹 . Assume that the attachment point at 𝐴 does not impede the spring. Problem 14.45 If the unstretched length of the spring is 𝐿0 = 1 m and the spring stiffness is 𝑘 = 170 N∕m, show that the collar does not reach 𝐹 . Assume that the attachment point at 𝐴 does not impede the spring.

ISTUDY

Section 14.2

895

Conservative Forces and Potential Energy

Problems 14.46 and 14.47 The crate of mass 𝑚 = 35 kg is attached to the springs with constants 𝑘1 = 1 kN∕m and 𝑘2 = 2 kN∕m. Both springs are unstretched when 𝛿 = 0 and the box is centered between the two vertical walls.

𝛿

𝑘1

𝑚

𝑘2

Problem 14.46

If the crate is released from rest after being displaced a distance 𝛿 = 2 m and friction between the crate and the surface on which it slides is negligible, determine the speed of the crate when it returns to 𝛿 = 0.

Figure P14.46 and P14.47

If the crate is released from rest after being displaced a distance 𝛿 = 2 m and the coefficient of kinetic friction between the crate and the surface on which it slides is 𝜇𝑘 = 0.2, determine the speed of the crate when 𝛿 = 0.

Problem 14.47

Problem 14.48 Packages for transporting delicate items (e.g., a laptop or glass) are designed to “absorb” some of the energy of the impact in order to protect their contents. These energy absorbers can get very complicated (e.g., the mechanics of Styrofoam peanuts can be complex), but we can begin to understand how they work by modeling them as a linear elastic spring of constant 𝑘 that is placed between the contents (an expensive vase) of mass 𝑚 and the package 𝑃 . Assume that the vase’s mass is 3 kg and that the box is dropped from rest from a height of 1.5 m. Treating the vase as a particle and neglecting all forces except for gravity and the spring force, determine the value of the spring constant 𝑘 so that the maximum displacement of the vase relative to the box is 0.15 m. Assume that the spring relaxes after the box is dropped and that it does not oscillate. DELIVERY

𝑚

𝜃

𝑘 𝑃 Figure P14.48

Figure P14.49

Problem 14.49 The pendulum is released from rest when 𝜃 = 0◦ . If the string holding the pendulum bob breaks when the tension is twice the weight of the bob, at what angle does the string break? Treat the pendulum as a particle, ignore air resistance, and let the string be inextensible and massless.

Problem 14.50 The force acting on a stationary electric charge 𝑞𝐴 interacting with a charge 𝑞𝐵 is described by Coulomb’s law and takes the form 𝑞 𝑞 𝐹⃗ = 𝑘 𝐴 2 𝐵 𝑢̂ 𝑟 , 𝑟 9 2 2 where 𝑘 = 8.9875 × 10 N ⋅ m ∕C (C is the symbol for coulomb, the unit used to measure electric charge) is a constant and 𝑟 is the distance between 𝐴 and 𝐵. This force law is mathematically very similar to Newton’s universal gravitation law. With this in mind, determine an expression of the electrostatic potential energy, choosing the datum at infinity, i.e., such that the potential energy is equal to zero when the two charges are separated by an infinite distance.

𝐴

𝐵

𝑢̂ 𝑟 𝑟

Figure P14.50

896

Chapter 14

Energy Methods for Particles

Problem 14.51 The force-compression profile of a rubber bumper 𝐵 is given by 𝐹𝐵 = 𝛽𝑥3 , where 𝛽 = 3.5×106 lb∕f t 3 and 𝑥 is the bumper’s compression measured in the horizontal direction. Determine the expression for the potential energy of the bumper 𝐵. In addition, if the cruiser 𝐶 weighs 70,000 lb and impacts 𝐵 with a speed of 5 f t∕s, determine the compression required to bring 𝐶 to a stop. Model 𝐶 as a particle, and neglect 𝐶’s vertical motion as well as the drag force between the water and the cruiser 𝐶. 5

𝐶

𝐹𝐵 ( × 105 lb)

𝐵

4 3 2 1 0 0

0.1

satellite

0.2 0.3 𝑥 (f t)

0.4

0.5

Figure P14.51 𝑃 , perigee

𝑅𝑃

𝐴, apogee

𝑅𝐴

Problem 14.52 A satellite orbits the Earth along the orbit shown. The minimum and maximum distances from the center of the Earth are 𝑅𝑃 = 4.5×107 m and 𝑅𝐴 = 6.163×107 m, respectively, where the subscripts 𝑃 and 𝐴 stand for perigee (the point on the orbit closest to Earth) and apogee (the point on the orbit farthest from Earth), respectively. Modeling the satellite as a particle and assuming that the center of the Earth can be chosen as the origin of an inertial frame of reference, if the satellite’s speed at 𝑃 is |𝑣| ⃗ 𝑃 = 3.2×103 m∕s, determine the satellite’s speed at 𝐴. Problems 14.53 through 14.55

Figure P14.52

ISTUDY

The arm 𝐴𝐵 can rotate freely about the pin at 𝐴. The spring with stiffness 𝑘 = 500 N∕m is designed so that the system is in static equilibrium when 𝜃 = 0◦ . Let 𝐿 = 18.2 cm, ℎ = 24.6 cm, and the mass of the ball 𝐵 be 5 kg. Neglect the mass of arm 𝐴𝐵. Hint: Sketch an FBD of the ball and the arm together. For Probs. 14.53 and 14.54, let 𝓁 be the distance between 𝐶 and 𝐷 and choose 𝓁 as the primary unknown. 𝐷 ℎ

𝑘 𝐴 𝐶 𝐵

𝜃 𝐿

𝐿 Figure P14.53–P14.55

If the system is released from rest when 𝜃 = −30◦ , determine the maximum angle 𝜃 reached by the arm 𝐴𝐵. Problem 14.53

If the system is released from rest when 𝜃 = −30◦ , determine the maximum speed achieved by the ball 𝐵, and determine the angle at which it occurs. Problem 14.54

If the arm 𝐴𝐵 is released from rest when 𝜃 = 90◦ such that it swings to the left, determine the speed of the ball 𝐵 when the arm reaches 𝜃 = 0◦ . Problem 14.55

ISTUDY

Section 14.2

Conservative Forces and Potential Energy

Problem 14.56

𝐶

The package handling system is designed to launch the small package of mass 𝑚 from 𝐴 by using a compressed linear spring of constant 𝑘. After launch, the package slides along the track until it lands on the conveyor belt at 𝐵. The track has small, well-oiled rollers so that you can neglect any energy loss due to the movement of the package along the track. Modeling the package as a particle, determine the minimum initial compression of the spring so that the package gets to 𝐵 without separating from the track at 𝐶. Finally, determine the speed with which the package reaches the conveyor at 𝐵.

3 𝑟 2

track 𝑘

Problem 14.57 Compressed gas is used in many circumstances to propel objects within tubes. For example, you can still find pneumatic tubes in use in many banks to receive and send items to drive-through tellers,∗ and it is compressed gas that propels a bullet out of a gun barrel. Let the cross-sectional area of the tube be given by 𝐴 and the position of the cylinder by 𝑠, and assume that the compressed gas is an ideal gas at constant temperature so that the pressure 𝑃 times the volume Ω is a constant, i.e., 𝑃 Ω = constant. Show that the potential energy of this compressed gas is given by 𝑉 = −𝑃0 𝑠0 𝐴 ln(𝑠∕𝑠0 ), where 𝑃0 is the initial pressure and 𝑠0 is the initial value of 𝑠. Model the cylinder as a particle, and assume that the forces resisting the motion of the cylinder are negligible.

𝑟

𝐵

𝑚 𝐴

Figure P14.56

tube

𝑠

Problem 14.58 When a gun fires a bullet, the gun barrel acts as the tube, and the bullet acts as the cylinder in Prob. 14.57. Using the assumptions and result of that problem, determine the velocity of a bullet at the end of a 24 in. gun barrel, given that the bore diameter is 0.458 in., the bullet weight is 300 gr (7000 gr = 1 lb), the initial firing pressure is 27,000 psi, and the initial distance between the back of the bullet and the back wall of the firing chamber (i.e., 𝑠0 in Prob. 14.57) is

compressed gas

cylinder

Figure P14.57 and P14.58

(a) 1.855 in. (this distance is realistic and accurate), (b) 1.5 in., and (c) explain why the velocity of the bullet at the end of the gun barrel is lower in Part (b) when compared to Part (a).

Problem 14.59 The resistance of a material to fracture is assessed with a fracture test. One such test is the Charpy impact test, in which the fracture toughness is assessed by measuring the energy required to break a specimen of a specified geometry. This is done by releasing a heavy pendulum from rest at an angle 𝜃𝑖 and by measuring the maximum swing angle 𝜃𝑓 reached by the pendulum after the specimen is broken. Suppose that in an experiment 𝜃𝑖 = 45◦ , 𝜃𝑓 = 23◦ , the weight of the pendulum’s bob is 3 lb, and the length of the pendulum is 3 f t. Neglecting the mass of any other component of the testing apparatus, assuming that the pendulum’s pivot is frictionless, and treating the pendulum’s bob as a particle, determine the fracture energy of the specimen tested. Assume that the fracture energy is the energy required to break the specimen. ∗ Pneumatic tubes were used extensively by postal services in the late 19th and early 20th centuries. By the

early 20th century, the cities of Philadelphia, New York, Boston, and Chicago had pneumatic networks, as did Paris, Berlin, and London. Their use for mail did not last past the end of World War I (1918), but almost every department store in the United States had tubes carrying cash and paperwork in the 1920s and 1930s. They still find use in hospitals for the delivery of medication and other items. See Robin Pogrebin, “Underground Mail Road,” New York Times, May 7, 2001.

𝜃𝑓 𝜃𝑖

Figure P14.59

897

898

Chapter 14

Energy Methods for Particles Problems 14.60 and 14.61

𝐿∕2

𝜃

𝐿

A pendulum with mass 𝑚 = 1.4 kg and length 𝐿 = 1.75 m is released from rest at an angle 𝜃𝑖 . Once the pendulum has swung to the vertical position (i.e., 𝜃 = 0), its cord runs into a small fixed obstacle. In solving this problem, neglect the size of the obstacle, model the pendulum’s bob as a particle, model the pendulum’s cord as massless and inextensible, and let gravity and the tension in the cord be the only relevant forces. Problem 14.60

What is the maximum height, measured from its lowest point, reached by the pendulum if 𝜃𝑖 = 20◦ ?

𝜙

Problem 14.61

If the bob is released from rest at 𝜃𝑖 = 90◦ , at what angle 𝜙 does the

cord go slack? Figure P14.60 and P14.61

Problem 14.62

bungee cord

While the stiffness of an elastic cord can be nearly constant over a large range of deformation, as a bungee cord is stretched, it tends to get less stiff as it gets longer. Assume a softening force-displacement relation of the form 𝑘𝛿 − 𝛽𝛿 3 , where 𝑘 = 2.58 lb∕f t, 𝛽 = 0.000013 lb∕f t 3 , and 𝛿 (measured in ft) is the displacement of the cord from its unstretched length. For a bungee cord whose unstretched length is 150 f t, determine (a) the expression of the cord’s potential energy as a function of 𝛿;

400 f t bungee jumper

(b) the velocity at the bottom of a 400 f t tower of a bungee jumper weighing 170 lb and starting from rest; (c) the maximum acceleration, expressed in 𝑔’s, felt by the bungee jumper in question.

Problems 14.63 through 14.65 Figure P14.62

ISTUDY

A crate, initially traveling horizontally with a speed of 18 f t∕s, is made to slide down a 14 f t chute inclined at 35◦ . The surface of the chute has a coefficient of kinetic friction 𝜇𝑘 , and at its lower end, it smoothly lets the crate onto a horizontal trajectory. The horizontal surface at the end of the chute has a coefficient of kinetic friction 𝜇𝑘2 . Model the crate as a particle, and assume that gravity and the contact forces between the crate and the sliding surface are the only relevant forces.

14 f t

𝜇𝑘

𝜇𝑘2

35◦

𝑘

5 ft Figure P14.63–P14.65

If 𝜇𝑘 = 0.35, what is the speed with which the crate reaches the bottom of the chute (immediately before the crate’s trajectory becomes horizontal)?

Problem 14.63

Problem 14.64 Find 𝜇𝑘 such that the crate’s speed at the bottom of the chute (immediately before the crate’s trajectory becomes horizontal) is 15 f t∕s.

Let 𝜇𝑘 = 0.5 and suppose that, once the crate reaches the bottom of the chute and after sliding horizontally for 5 f t, the crate runs into a bumper. If the weight of the crate is 𝑊 = 110 lb, 𝜇𝑘2 = 0.33, modeling the bumper as a linear spring with constant 𝑘, and neglecting the mass of the bumper, determine the value of 𝑘 so that the crate comes to a stop 2 f t after impacting with the bumper. Problem 14.65

ISTUDY

Section 14.2

899

Conservative Forces and Potential Energy

The Lennard-Jones force law between two atoms can be represented as ( 6 ) 2𝜎 12 𝜎 𝑓𝑖𝑗 = 24𝜖 7 − 13 , 𝑟𝑖𝑗 𝑟𝑖𝑗 where 𝑟𝑖𝑗 is the distance between atoms 𝑖 and 𝑗, and 𝜖 and 𝜎 are material-specific parameters with dimensions of energy and length, respectively. Using this equation, determine the potential between two Ni atoms. Assume the potential between the two atoms is zero when the distance between those two atoms is infinite.

Force 𝑓𝑖𝑗 ( × 10−9 N)

Problem 14.66

attractive

1 0

𝑟𝑖𝑗

−1 repulsive 0

3 4 5 1 2 Distance 𝑟𝑖𝑗 ( × 10−10 m)

Figure P14.66

Problems 14.67 and 14.68 Spring scales work by measuring the displacement of a spring that supports both the platform and the object, of mass 𝑚, whose weight is being measured. Neglect the mass of the platform on which the mass sits, and assume that the spring is uncompressed before the mass is placed on the platform. In addition, assume that the spring is linear elastic with spring constant 𝑘. You may have solved these same problems using Newton’s second law when doing Probs. 13.27 and 13.28 — here use the work-energy principle to solve them.

ACME

𝑚

platform

Problem 14.67 If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the maximum reading on the scale after the mass is released. Figure P14.67 and P14.68

If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the expression for maximum velocity attained by the mass 𝑚 as the spring compresses. Problem 14.68

Problem 14.69 A 6 lb collar is constrained to travel along a rectilinear and frictionless bar of length 𝐿 = 5 f t. The springs attached to the collar are identical, and they are unstretched when the collar is at 𝐵. Treating the collar as a particle, neglecting air resistance, and knowing that at 𝐴 the collar is moving to the right with a speed of 11 f t∕s, determine the linear spring constant 𝑘 so that the collar reaches 𝐷 with zero speed. Points 𝐸 and 𝐹 are fixed. 𝐿∕4

𝐿∕2 𝐴

𝐵 𝑘

𝑘

𝐸

𝐷

𝐿

𝐿∕2

𝐹

𝐴

Figure P14.69 𝑅

Problem 14.70 A 3 kg collar is constrained to travel in the horizontal plane along a frictionless ring of radius 𝑅 = 0.75 m. The spring attached to the collar has a spring constant 𝑘 = 21 N∕m. Treating the collar as a particle, neglecting air resistance, and knowing that at 𝐴 the collar is at rest, determine the spring’s unstretched length if the collar is to reach point 𝐵 with a speed of 2 m∕s.

5𝑅∕3 𝐵

Figure P14.70

6

900

Chapter 14

Energy Methods for Particles

Design Problems

ISTUDY

Design Problem 14.1 In practice, during a jump, bungee cords stretch to two to four times their unstretched length, and a jumper feels no more than 2.5–3.5𝑔 of acceleration. Assume the force in the bungee cord has the mathematical form 𝑘𝛿 − 𝛽𝛿 3 , where 𝑘 and 𝛽 are constants and 𝛿 is the amount of stretch in the cord past its unstretched length. Design a cord (that is, design the constants 𝑘 and 𝛽) so that the bungee cord stretches 2.5 times its unstretched length, the acceleration of the jumper does not exceed 3𝑔, and the bungee cord has zero stiffness at the bottom of a 400 f t drop.

bungee cord 400 f t bungee jumper

tbkmedia.de/Alamy Stock Photo

Figure DP14.1

ISTUDY

Section 14.3

14.3

901

Work-Energy Principle for Systems of Particles

Work-Energy Principle for Systems of Particles

This section presents the application of the work-energy principle to systems of particles. We will discover a fundamental new concept pertaining to systems of particles that is absent in the single-particle case — internal work.

Internal work and work-energy principle for a system Referring to Fig. 14.11, consider a system of 𝑛 particles. The 𝑖th particle, whose mass is 𝑚𝑖 and position is 𝑟⃗𝑖 , is acted upon by a total external force 𝐹⃗𝑖 and interacts with the other 𝑛 − 1 particles through internal forces labeled 𝑓⃗𝑖𝑗 (𝑗 = 1, … , 𝑛, 𝑖 ≠ 𝑗). Applying Newton’s second law to the 𝑖th particle, we have 𝐹⃗𝑖 +

𝑛 ∑

𝑖 ≠ 𝑗.

𝑓⃗𝑖𝑗 = 𝑚𝑖 𝑎⃗𝑖 ,

𝑦 𝐹⃗𝑗 𝐹⃗𝑖

𝑓⃗𝑖𝑗

As we did for a single particle, we take the dot product of this equation with 𝑑⃗𝑟𝑖 and integrate along (ℒ1-2 )𝑖 , the path traveled by particle 𝑖 in going from 1 to 2. There will be 𝑛 of these equations (one for each particle). Summing all 𝑛 of them gives ) ( 𝑛 𝑛 𝑛 ∑ ∑ ∑ 𝑑 𝑟⃗̇ 𝑓⃗𝑖𝑗 ⋅ 𝑑⃗𝑟𝑖 = (14.38) 𝑚𝑖 𝑖 ⋅ 𝑑⃗𝑟𝑖 , 𝐹⃗𝑖 + ∫ ∫ 𝑑𝑡 𝑖=1 (ℒ1-2 )𝑖 𝑗=1 𝑖=1 (ℒ1-2 )𝑖 where we have written 𝑎⃗𝑖 = 𝑑 𝑟⃗̇ 𝑖 ∕𝑑𝑡. Proceeding as we did for a single particle, we can write the right-hand side of Eq. (14.38) as

𝑖=1

∫

𝑚𝑖

𝑑 𝑟⃗̇ 𝑖 𝑑𝑡

⋅ 𝑑⃗𝑟𝑖 =

𝑛 ∑ 𝑖=1

∫

𝑛 ∑ 1

|2 𝑚𝑖 𝑣⃗𝑖 ⋅ 𝑣⃗𝑖 | |1 𝑖=1 ( 𝑛 ) ( 𝑛 ) ∑1 ∑1 2 2 = 𝑚𝑣 − 𝑚𝑣 , 2 𝑖 𝑖 2 𝑖 𝑖

𝑚𝑖 𝑣⃗𝑖 ⋅ 𝑑 𝑣⃗𝑖 =

2

𝑖=1

2

𝑖=1

(14.39)

1

where each integral is performed over the path (ℒ1-2 )𝑖 . We define the kinetic energy of the particle system to be the sum of the kinetic energies of each part of the system. Denoting the kinetic energy of the system by 𝑇 , we have 𝑇 =

𝑛 ∑ 1 𝑖=1

2

𝑚𝑖 𝑣2𝑖 .

(14.40)

Using 𝑇 , Eq. (14.39) can be rewritten as 𝑛 ∑ 𝑖=1

∫(ℒ

1-2 )𝑖

𝑚𝑖

𝑑 𝑟⃗̇ 𝑖 𝑑𝑡

⋅ 𝑑⃗𝑟𝑖 = 𝑇2 − 𝑇1 .

(14.41)

Going back to Eq. (14.38), the left-hand side of this equation can be written as

𝑛 ∑ 𝑖=1

∫

( ) 𝑛 ∑ 𝐹⃗𝑖 + 𝑓⃗𝑖𝑗 ⋅ 𝑑⃗𝑟𝑖 𝑗=1

( ( ) ) 𝑈1-2 ext 𝑈1-2 int ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ ⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 𝑛 𝑛 𝑛 ∑ ∑ ∑ 𝐹⃗𝑖 ⋅ 𝑑⃗𝑟𝑖 + 𝑓⃗ ⋅ 𝑑⃗𝑟𝑖 , = ∫ ∫ 𝑖𝑗 𝑖=1

𝑓⃗𝑗𝑖

𝑚𝑗

𝑚𝑖

(14.37)

𝑟⃗𝑗

𝑗=1

𝑛 ∑

(ℒ1-2 )𝑖

𝑖=1 𝑗=1 𝑗≠𝑖

(14.42)

(ℒ1-2 )𝑗

𝑟⃗𝑖

𝑥 Figure 14.11 Particles 𝑖 and 𝑗 of a system of 𝑛 particles.

902

Chapter 14

Energy Methods for Particles

( ) ( ) where 𝑈1-2 ext and 𝑈1-2 int represent the work done on the system by external and internal forces, respectively, and each integral is performed over the path (ℒ1-2 )𝑖 . In discussing Newton’s second law for systems, we saw that terms similar to the double sum in Eq. (14.42) vanish because of Newton’s third law. However, in this case the double sum does not vanish, and it remains an important contribution to the workenergy principle. In fact, for each pair of particles 𝑖 and 𝑗, the double sum that defines ( ) the term 𝑈1-2 int contains the terms ( ) 𝑓⃗𝑖𝑗 ⋅ 𝑑⃗𝑟𝑖 + 𝑓⃗𝑗𝑖 ⋅ 𝑑⃗𝑟𝑗 = 𝑓⃗𝑖𝑗 ⋅ 𝑑⃗𝑟𝑖 − 𝑑⃗𝑟𝑗 = 𝑓⃗𝑖𝑗 ⋅ 𝑑⃗𝑟𝑖∕𝑗 ,

(14.43)

where we have used the fact that 𝑓⃗𝑖𝑗 = −𝑓⃗𝑗𝑖 by Newton’s third law and where, using relative motion kinematics, 𝑑⃗𝑟𝑖∕𝑗 = 𝑑⃗𝑟𝑖 −𝑑⃗𝑟𝑗 is the infinitesimal relative displacement of particle 𝑖 with respect to particle 𝑗. Equation (14.43) tells us that for internal forces to do no work, either there must be no relative motion or the component of 𝑑⃗𝑟𝑖∕𝑗 along 𝑓⃗ must be zero for all particle pairs. 𝑖𝑗

Summarizing, we use Eqs. (14.41) and (14.42) to write the work-energy principle for a system of particles as ( ) ( ) 𝑇1 + 𝑈1-2 ext + 𝑈1-2 int = 𝑇2 ,

(14.44)

( )ext ( )int 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 ,

(14.45)

or as

when there are also conservative forces doing work on the system. We see that in applying the work-energy principle to particle systems, we must account for the work of internal forces. Referring to the discussion following Eq. (14.43), there are situations in which the work of the internal forces vanishes even when there is relative motion, such as in pulley systems with inextensible cables.

Kinetic energy for a system of particles 𝑦

We now write the kinetic energy of a particle system in a way that will help us to understand the expression for the kinetic energy of a rigid body in Chapter 18. We have defined the kinetic energy for a system of particles as

𝐹⃗𝑗 𝐹⃗𝑖 𝑚𝑖

𝑓⃗𝑗𝑖

𝑓⃗𝑖𝑗

𝑚𝑗

𝑇 =

𝑖=1

𝑟⃗𝑖∕𝐺 𝑟⃗𝑖

𝐺 𝑟⃗𝐺 𝑥

Figure 14.12 Particles 𝑖 and 𝑗 of a system of 𝑛 particles with center of mass at 𝐺.

ISTUDY

𝑛 ∑ 1 2

𝑚𝑖 𝑣⃗𝑖 ⋅ 𝑣⃗𝑖 .

(14.46)

Now, referring to Fig. 14.12, notice that 𝑟⃗𝑖 = 𝑟⃗𝐺 + 𝑟⃗𝑖∕𝐺 , where 𝑟⃗𝐺 and 𝑟⃗𝑖∕𝐺 are the position of the center of mass 𝐺 and the position of particle 𝑖 relative to 𝐺, respectively. Differentiating this expression for 𝑟⃗𝑖 with respect to time gives 𝑣⃗𝑖 = 𝑣⃗𝐺 + 𝑣⃗𝑖∕𝐺 , which allows us to write Eq. (14.46) as

𝑇 =

𝑛 ∑ 1 𝑖=1

2

( ) ( ) 𝑚𝑖 𝑣⃗𝐺 + 𝑣⃗𝑖∕𝐺 ⋅ 𝑣⃗𝐺 + 𝑣⃗𝑖∕𝐺 =

𝑛 ∑ 1 𝑖=1

2

( ) 𝑚𝑖 𝑣⃗𝐺 ⋅ 𝑣⃗𝐺 + 2𝑣⃗𝐺 ⋅ 𝑣⃗𝑖∕𝐺 + 𝑣⃗𝑖∕𝐺 ⋅ 𝑣⃗𝑖∕𝐺 ,

(14.47)

ISTUDY

Section 14.3

Work-Energy Principle for Systems of Particles

or

( 𝑇 =

1 𝑚 𝑣⃗𝐺 2

)

𝑛 ∑

⋅ 𝑣⃗𝐺 + 𝑚𝑖 𝑣⃗𝑖∕𝐺 ⋅ 𝑣⃗𝐺 + ⏟⏟⏟ 𝑖=1 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ 𝑣2𝐺 𝑛 𝑑 ∑ 𝑚 𝑟⃗ 𝑑𝑡 𝑖=1 𝑖 𝑖∕𝐺

𝑛 ∑ 1 𝑖=1

2

𝑚𝑖 𝑣⃗𝑖∕𝐺 ⋅ 𝑣⃗𝑖∕𝐺 , ⏟⏞⏞⏟⏞⏞⏟ 𝑣2𝑖∕𝐺

903

(14.48)

where 𝑚 is the system’s total mass. Referring to the second term on the last line of Eq. (14.48), using the definition of center of mass in Eq. (13.35) on p. 838, and recalling that 𝑟⃗𝑖∕𝐺 describes the position of each particle relative to the mass center, we have 𝑛 ) 𝑑 ∑ 𝑑( 𝑚 𝑟⃗𝐺∕𝐺 . 𝑚𝑖 𝑟⃗𝑖∕𝐺 = (14.49) 𝑑𝑡 𝑖=1 𝑑𝑡 Since the term 𝑟⃗𝐺∕𝐺 is the position of the mass center relative to the mass center itself, the left-hand side of Eq. (14.49) must always be equal to zero. In conclusion, Eq. (14.48) takes on the form 𝑇 = 12 𝑚𝑣2𝐺 +

1 2

𝑛 ∑ 𝑖=1

𝑚𝑖 𝑣2𝑖∕𝐺 ,

(14.50)

which states that the kinetic energy of a system of particles depends on the motion of the mass center of the system of particles, as well as the motion of all the particles relative to the mass center.

End of Section Summary In this section, we developed the work-energy principle for a system of particles, and we discovered that internal forces play an important role in writing this principle. We defined the kinetic energy of a particle system as Eq. (14.40), p. 901 𝑇 =

𝑛 ∑ 1 𝑖=1

2

𝑚𝑖 𝑣2𝑖 ,

where 𝑚𝑖 and 𝑣𝑖 are the mass and the speed of the 𝑖th particle in the system, respectively. Then the work-energy principle for a system of particles was found to be Eq. (14.45), p. 902 ( )ext ( )int 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 , where 𝑉 is the potential energy of the system, which may have contributions from both external and internal forces. The kinetic energy of a system can also be expressed in terms of the speed of the mass center and the relative velocity of the particles in the system with respect to the mass center, that is, Eq. (14.50), p. 903 𝑇 = 12 𝑚𝑣2𝐺 +

1 2

𝑛 ∑ 𝑖=1

𝑚𝑖 𝑣2𝑖∕𝐺 .

Concept Alert The kinetic energy of a spinning system. Equation (14.50) tells us that a system can have a nonzero kinetic energy even when the center of mass is not moving. A typical example of this idea is the flywheel, which, even with its center of mass remaining stationary, can store a remarkable amount of kinetic energy via its spinning motion.

904

ISTUDY

Energy Methods for Particles

Chapter 14

Note that for a rigid body experiencing pure translation, 𝑣𝑖∕𝐺 → 0 and 𝑇 = 12 𝑚𝑣2𝐺 . In this case, the rigid body’s kinetic energy looks like that of a particle. If the rigid body experiences rotation in addition to translation, 𝑣𝑖∕𝐺 ≠ 0, and the second term results in rotational kinetic energy. The details of this are developed in Chapter 18.

ISTUDY

Section 14.3

E X A M P L E 14.11

A System of Particles with Friction

The crates 𝐴 and 𝐵 of mass 𝑚 and 2𝑚, respectively, are connected by a pulley system (Fig. 1). If the system is released from rest and friction between 𝐴 and the horizontal surface is insufficient to prevent slipping, determine the speed of crate 𝐴 after 𝐵 drops a distance ℎ.

SOLUTION

Figure 1 The cables in the pulley system are inextensible, and the coefficient of kinetic friction between the crate 𝐴 and the horizontal surface is 𝜇𝑘 .

2𝑃 𝚤̂

𝐴

𝐹

𝐵

𝑃

𝚥̂

2𝑚𝑔

𝑁

Figure 2. The FBDs of crates 𝐴 and 𝐵.

forces that do work are the friction force 𝐹 , the cable tension 𝑃 , and the weight 2𝑚𝑔. Since the cables are inextensible, the work done by the cable tension 𝑃 on 𝐴 is equal and opposite to the work done by the cable tension 2𝑃 on 𝐵. This will also be seen shortly in the relative kinematics of 𝐴 and 𝐵; a displacement Δ𝑦𝐵 will be shown to result in Δ𝑥𝐴 = −2Δ𝑦𝐵 . Cable tension 𝑃 does positive work on 𝐴 through distance 2ℎ, 2𝑃 does negative work on 𝐵 through distance ℎ, and the net result for the system is zero work. Therefore, we can apply the work-energy principle to 𝐴 and 𝐵 together to find the speed of 𝐴 with only 𝐹 and 2𝑚𝑔 doing work. Governing Equations

Letting 1 be when the system is at rest and 2 be when 𝐵 has dropped a distance ℎ, the work-energy principle is )int )ext ( ( 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 , (1)

Balance Principles

where the kinetic energies are 𝑇1 = 21 𝑚𝑣2𝐴1 + 𝑚𝑣2𝐵1

and 𝑇2 = 12 𝑚𝑣2𝐴2 + 𝑚𝑣2𝐵2 .

(2)

The weight force on 𝐵 is conservative, so the potential energies are 𝑉1 = 0

and 𝑉2 = −2𝑚𝑔ℎ.

(3)

Since we are treating the cable tensions as external forces, we have ( )int 𝑈1-2 nc = 0.

(4)

The work of friction and the tension 𝑃 are included in the nonconservative external work term, which gives ∫𝑥

𝑥𝐴2 𝐴1

𝐴

The FBDs of each of the two crates are shown in Fig. 2. The 𝑚𝑔

( )ext 𝑈1-2 nc =

𝜇𝑘

𝐵

Road Map & Modeling

Force Laws

905

Work-Energy Principle for Systems of Particles

(𝐹 − 𝑃 ) 𝑑𝑥𝐴 −

∫𝑦

𝑦𝐵2

2𝑃 𝑑𝑦𝐵

𝐵1

= 𝜇𝑘 𝑚𝑔(𝑥𝐴2 − 𝑥𝐴1 ) −

∫𝑥

𝑥𝐴2 𝐴1

𝑃 𝑑𝑥𝐴 − 2

∫𝑦

𝑦𝐵2

𝑃 𝑑𝑦𝐵 ,

(5)

𝐵1

where 𝑥𝐴2 − 𝑥𝐴1 is the displacement of crate 𝐴 and we have used the fact that 𝐹 = 𝜇𝑘 𝑁 = 𝜇𝑘 𝑚𝑔 is constant (𝑁 = 𝑚𝑔 because 𝐴 does not move vertically). Kinematic Equations Since the system starts from rest, 𝑣𝐴1 = 𝑣𝐵1 = 0.

(6)

906

𝑥𝐴1 1

Chapter 14

Energy Methods for Particles

𝐴

2

Referring to Fig. 3, since the cables are inextensible, we must have 𝑥𝐴2

𝑥𝐴 + 2𝑦𝐵 = constant.

(7)

Taking the differential of Eq. (7) so that we can relate the integrals in Eq. (5), we obtain 𝑦𝐵1 𝐵

𝑑𝑥𝐴 + 2𝑑𝑦𝐵 = 0

𝑦𝐵2 1

, datum

ℎ 2

Figure 3 The positions of crates 𝐴 and 𝐵 in 1 and 2.

ISTUDY

𝑑𝑥𝐴 = −2𝑑𝑦𝐵 .

⇒

(8)

In addition, this tells us that if 𝐵 falls a distance ℎ, then 𝐴 must move to the right a distance 2ℎ. Therefore, 𝑦𝐵2 − 𝑦𝐵1 = ℎ and 𝑥𝐴2 − 𝑥𝐴1 = −2ℎ. (9) Differentiating Eq. (7) with respect to time so that we can relate the velocities of 𝐴 and 𝐵 in the expression for kinetic energy, we obtain 𝑣𝐴𝑥 + 2𝑣𝐵𝑦 = 0

⇒

| | |𝑣𝐵𝑦 | = |− 21 𝑣𝐴𝑥 | | |

⇒

𝑣𝐵2 = 12 𝑣𝐴2 ,

(10)

where we have evaluated the speed relation in 2 to obtain the final result. Computation

Using Eq. (8) to make a change of variables in the first integral in Eq. (5), this equation becomes (𝑈1-2 )ext nc = 𝜇𝑘 𝑚𝑔(𝑥𝐴2 − 𝑥𝐴1 ) −

∫𝑦

𝑦𝐵2

𝑃 (−2𝑑𝑦𝐵 ) − 2

𝐵1

= 𝜇𝑘 𝑚𝑔(𝑥𝐴2 − 𝑥𝐴1 ) + 2

∫𝑦

𝑦𝐵2

𝑃 𝑑𝑦𝐵

𝐵1

∫𝑦

𝑦𝐵2

𝑃 𝑑𝑦𝐵 − 2

𝐵1

∫𝑦

𝑦𝐵2

𝑃 𝑑𝑦𝐵

𝐵1

= 𝜇𝑘 𝑚𝑔(𝑥𝐴2 − 𝑥𝐴1 ).

(11)

Substituting Eqs. (2), (3), (4), (6), and (9)–(11) into Eq. (1), we obtain 𝜇𝑘 𝑚𝑔(−2ℎ) = 12 𝑚𝑣2𝐴2 + 𝑚

(1

𝑣 2 𝐴2

)2

− 2𝑚𝑔ℎ,

(12)

where we have used the fact that (𝑈1-2 )int nc = 0. Solving for 𝑣𝐴2 , we obtain √ 𝑣𝐴2 = Discussion & Verification

8 𝑔ℎ(1 3

− 𝜇𝑘 ).

(13)

The right side of Eq. (13) has the dimension of speed, which is correct. Equation (13) also says that 𝑣𝐴 increases as ℎ increases, which is reasonable. In addition, notice that the work done by the cables cancels out of Eq. (12), which it should since the cables are inextensible.

ISTUDY

Section 14.3

Work-Energy Principle for Systems of Particles

E X A M P L E 14.12

907

Conservation of Energy in a Particle System

The simple catapult in Fig. 1 consists of the two blocks 𝐴 and 𝐵 connected by the pulley at 𝑃 . The falling weight 𝐵 and stretched spring to which it is attached will launch 𝐴. Mass 𝐵, with 𝑚𝐵 = 8 kg, is released from rest at a distance ℎ = 2 m above the ground. At the time of release, 𝐴, with 𝑚𝐴 = 1 kg, is in the position shown in Fig. 1. The spring attached to 𝐵 is linear with constant 𝑘 = 40 N∕m, and it is unstretched when 𝐵 reaches the ground. Determine the speeds of 𝐴 and 𝐵 immediately before 𝐵 hits the ground.

𝑑 = 2.5 m 𝐷

𝑃

𝑙 = 4m

SOLUTION

𝐵

Road Map & Modeling

By applying the work-energy principle, we can determine the speed of 𝐴 at the position of interest. The FBD of the system in Fig. 2 reflects the fact

𝑘

ℎ = 2m

𝐴 𝑃𝑦 𝚥̂

0.25 m

𝑃𝑥

Figure 1 Proposed catapult system with masses 𝐴 and 𝐵 connected by the pulley at 𝑃 . The dimensions shown are those at release.

𝚤̂ 𝑚𝐵 𝑔

𝐹𝑠

𝑁𝐴

𝑁𝐵

𝑚𝐴 𝑔 Figure 2. An FBD of the system shown in Fig. 1.

that we are modeling 𝐴 and 𝐵 as particles, that we are ignoring any friction, and that we are neglecting the weight of the chain connecting 𝐴 and 𝐵 by the pulley 𝑃 . The only forces doing work as 𝐵 drops the distance ℎ are the spring force 𝐹𝑠 and the two weight forces. The reactions at 𝑃 do no work since their point of application is fixed. While the chain provides an internal force, for it to do work the two ends of the chain would need to move relative to one another in the direction of the tension [see discussion of Eq. (14.43) on p. 902]. That is, the chain would have to stretch. Since the chain is inextensible, its tension does no work. We define 1 to be at release (from rest) and 2 to be the position at which 𝐵 hits the ground and the spring becomes unstretched. 𝑑

Governing Equations

)int )ext ( ( Balance Principles From the discussion above, we see that 𝑈1-2 nc and 𝑈1-2 nc in Eq. (14.45) are both zero. Therefore, the work-energy principle between 1 and 2 becomes 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 .

(1)

The system’s kinetic energies can be written as 𝑇1 =

1 𝑚 𝑣2 2 𝐴 𝐴1

+

1 𝑚 𝑣2 2 𝐵 𝐵1

and 𝑇2 =

𝑙 1 𝑚 𝑣2 2 𝐴 𝐴2

+

1 𝑚 𝑣2 . 2 𝐵 𝐵2

Force Laws

The potential energies in Eq. (1) are made up of the gravitational potential energies of 𝑚𝐴 and 𝑚𝐵 , as well as the elastic potential energy of the spring. Therefore, referring to Figs. 1 and 3, we have 𝑉1 = 12 𝑘ℎ2 + 𝑚𝐴 𝑔𝑦𝐴1 + 𝑚𝐵 𝑔𝑦𝐵1

and 𝑉2 = 𝑚𝐴 𝑔𝑦𝐴2 + 𝑚𝐵 𝑔𝑦𝐵2 ,

where we have used the fact that the spring is unstretched at 2.

𝐴

(2)

(3)

𝐵

𝑦𝐴 datum

𝑦𝐵

Figure 3 Definition of 𝑦𝐴 , 𝑦𝐵 , and the datum line for the gravitational potential energy.

908

ISTUDY

Chapter 14

Energy Methods for Particles

Kinematic Equations

At 1 we have 𝑣𝐴1 = 0 and 𝑣𝐵1 = 0.

(4)

As far as 2 is concerned, we note that 𝑣𝐴2 and 𝑣𝐵2 are the unknowns of the problem. We also note that both the position and speed of 𝐴 are related to the position and speed of 𝐵. To determine these relations, referring to Fig. 3, we observe that the chain’s length is √ ) )2 ( ( 𝐿= (5) 𝑙 − 𝑦𝐴 + 𝑑 2 + 𝑙 − 𝑦𝐵 . Since the chain is inextensible, differentiating Eq. (5) with respect to time, we have ( ) − 𝑙 − 𝑦𝐴 𝑣𝐴𝑦 𝑙 − 𝑦𝐴 | | |𝑣 = 𝑣 , − 𝑣𝐵𝑦 ⇒ ||− √ 0= √ | 𝐴 𝐵 ( ( )2 )2 | | 𝑙 − 𝑦𝐴 + 𝑑 2 𝑙 − 𝑦𝐴 + 𝑑 2

Common Pitfall Chains can become slack. Having assumed that the chain is inextensible means that the chain’s length 𝐿 must be a constant. By enforcing this constraint via Eq. (5) we are also saying that not only does 𝐿 remain constant, but also remains taut. The solution should be verified by checking whether or not the chain becomes slack. This verification is one element of Prob. 14.96.

(6)

where the introduction of the absolute value in the second equation allows us to replace 𝑣𝐴𝑦 and 𝑣𝐵𝑦 with the speeds 𝑣𝐴 and 𝑣𝐵 , respectively. To determine the relationship between the displacement of 𝐴 and the displacement of 𝐵, we can evaluate Eq. (5) at 1 and 2 as follows: √ √ ) ) )2 )2 ( ( ( ( (7) 𝑙 − 𝑦𝐴1 + 𝑑 2 + 𝑙 − 𝑦𝐵1 = 𝑙 − 𝑦𝐴2 + 𝑑 2 + 𝑙 − 𝑦𝐵2 . 𝐿= The second equality in Eq. (7) implies that √ √ ( ( )2 )2 𝑦𝐵1 − 𝑦𝐵2 = 𝑙 − 𝑦𝐴1 + 𝑑 2 − 𝑙 − 𝑦𝐴2 + 𝑑 2 .

(8)

Recalling that 𝑦𝐵1 − 𝑦𝐵2 = ℎ and 𝑦𝐴1 is known, and solving Eq. (8) for 𝑦𝐴2 , we have √ √ )2 )2 ( ( (9) 𝑦𝐴2 = 𝑙 − ℎ2 + 𝑙 − 𝑦𝐴1 − 2ℎ 𝑙 − 𝑦𝐴1 + 𝑑 2 . Computation 1 𝑘ℎ2 2

Substituting Eqs. (2)–(4) into Eq. (1), we obtain + 𝑚𝐴 𝑔𝑦𝐴1 + 𝑚𝐵 𝑔𝑦𝐵1 = 12 𝑚𝐴 𝑣2𝐴2 + 21 𝑚𝐵 𝑣2𝐵2 + 𝑚𝐴 𝑔𝑦𝐴2 + 𝑚𝐵 𝑔𝑦𝐵2 .

(10)

Recalling again that 𝑦𝐵1 − 𝑦𝐵2 = ℎ, substituting Eqs. (6) into Eq. (10), and solving for 𝑣𝐴2 yield √ ( ) √ √ 𝑘ℎ2 + 2𝑚𝐴 𝑔 𝑦𝐴1 − 𝑦𝐴2 + 2𝑚𝐵 𝑔ℎ √ (11) 𝑣𝐴2 = √ ]. )2 )2 /[( ( 𝑙 − 𝑦𝐴2 + 𝑑 2 𝑚𝐴 + 𝑚𝐵 𝑙 − 𝑦𝐴2 Recalling that 𝑙 = 4 m, 𝑦𝐴1 = 0.25 m, ℎ = 2 m, and 𝑑 = 2.5 m, from Eq. (9) we see that 𝑦𝐴2 = 3.814 m. Using this result, along with the given data, from Eqs. (11) and (6) we see that 𝑣𝐴2 = 19.67 m∕s and 𝑣𝐵2 = 1.462 m∕s. (12) The term multiplying 𝑣𝐴 in Eq. (6) is nondimensional, which implies that Eq. (6) is dimensionally correct. The argument of the square root in Eq. (11) has dimensions of energy divided by mass, i.e., speed squared. Therefore, Eq. (11) is also dimensionally correct. To assess whether or not the value for 𝑣𝐴2 (and therefore, the corresponding value of 𝑣𝐵2 ) in Eq. (12) is reasonable, we can say that 𝑣𝐴2 should not exceed the speed that we would obtain if all of the potential energy of 𝐵 were turned into the kinetic energy of 𝐴. We can find such a value as follows: Discussion & Verification

1 𝑘ℎ2 2

+ 𝑚𝐵 𝑔ℎ = 21 𝑚𝐴 𝑣2𝐴2

⇒

𝑣𝐴2 = 21.77 m∕s.

(13)

Because the result in Eq. (12) is less than that of Eq. (13), we can say that our answer behaves as expected. Interestingly, even without the spring (i.e., for 𝑘 = 0), we achieve 𝑣𝐴2 = 15.29 m∕s with 𝑣𝐵2 = 1.137 m∕s.

ISTUDY

Section 14.3

Work-Energy Principle for Systems of Particles

E X A M P L E 14.13

909

Why Does a Snapped Towel Hurt?

We now analyze a simple model to explain the behavior of a snapped towel, that is, the sting it produces when hitting someone and the loud crack that can be heard at the end of the snap. The model of the towel is shown in Fig. 2(a). Using this string-based model, we attach one end to the ceiling, fold the string back on itself, and then release the unattached end and let it fall freely. For an inextensible string of length 𝐿 and mass 𝑚, determine the speed of the free end of the string as a function of its distance below the ceiling.

SOLUTION Road Map & Modeling

Since we are interested in the speed of the free end of the string after it has fallen a given distance, we apply the work-energy principle. The string is not a system of particles (its mass is distributed in space), but we can apply the concepts for systems of particles by applying the work-energy principle to the equivalent mass center of the system. We assume that the string is inextensible, and we will view it as a system of two particles (whose mass sums to that of the string), one particle being the left branch and the other the right [see Fig. 2(b)]. These particles will be placed at the midpoint or 𝑅 𝑦 𝐿∕2 𝐿∕2

𝓁𝐿

𝑚𝐿 𝑔 𝑚𝑅 𝑔

𝑦∕2 (a)

𝓁𝑅 = 𝐿∕2 − 𝑦∕2

(b)

Gary L. Gray

Figure 2. (a) The initial configuration of the string. (b) FBD and kinematics of the string after the free end has fallen a distance 𝑦, where 𝑚𝐿 𝑔 is the weight of the left segment and 𝑚𝑅 𝑔 is the weight of the right.

mass center of their corresponding branches. Figure 2(b) shows the string in an arbitrary position, which we will call 2, and we will let 1 be at release. The weight forces 𝑚𝐿 𝑔 and 𝑚𝑅 𝑔 are the only forces doing work on the string, so the system is conservative. Governing Equations Balance Principles

Since the system is conservative, the work-energy principle is 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 .

(1)

The kinetic energies are 𝑇1 = 12 𝑚𝐿1 𝑣2𝐿1 + 21 𝑚𝑅1 𝑣2𝑅1

and 𝑇2 = 12 𝑚𝐿2 𝑣2𝐿2 + 21 𝑚𝑅2 𝑣2𝑅2 ,

(2)

where 𝑚𝐿 and 𝑚𝑅 are the masses of the left and right branches of the string, respectively, and where 𝑣𝐿 and 𝑣𝑅 are the speeds of the mass centers of the left and right branches of the string, respectively. In all cases, the second subscript indicates the position. Force Laws Assuming that the changing bend in the string does not dissipate any energy, letting 𝓁𝐿 and 𝓁𝑅 be the lengths of the left and right branches of the string, respectively, and defining the datum for the gravitational potential energy to be at the top of the fixed portion of string, we have ( ) 𝑉1 = −𝑚𝐿1 𝑔𝐿∕4 − 𝑚𝑅1 𝑔𝐿∕4 = − 𝑚𝐿1 + 𝑚𝑅1 𝑔𝐿∕4, (3)

Figure 1 A sequence of photos showing a snapped towel. The exposure time of each photo is 0.0008 s. Top frame: Preparing for the snap. Middle frame: The towel is halfway through the forward part of the snap. The towel is bent like the initial state of our string. Bottom frame: The towel has reached its farthest extent. Notice that the last few inches of the towel are blurry due to the high speed at which this part of the towel is moving.

910

Chapter 14

Energy Methods for Particles ( ) 𝑉2 = −𝑚𝐿2 𝑔𝓁𝐿 ∕2 − 𝑚𝑅2 𝑔 𝑦 + 𝓁𝑅 ∕2 ,

(4)

where, letting 𝜌 = 𝑚∕𝐿 be the mass per unit length of string, the masses of the left and right branches of the string at 1 and 2 are given by 𝑚𝐿1 = 𝜌𝐿∕2,

𝑚𝑅1 = 𝜌𝐿∕2,

𝑚𝐿2 = 𝜌𝓁𝐿 ,

𝑚𝑅2 = 𝜌𝓁𝑅 ,

(5)

so that 𝑉1 and 𝑉2 become 𝑉1 = − 41 𝜌𝑔𝐿2

) ( and 𝑉2 = −𝜌𝑔𝓁𝐿2 ∕2 − 𝜌𝑔𝓁𝑅 𝑦 + 𝓁𝑅 ∕2 .

(6)

Referring to Fig. 2(b), the total length of the string 𝐿 and the fall distance 𝑦 are related to the length of the left and right segments as follows:

Kinematic Equations

𝐿 = 𝓁𝐿 + 𝓁𝑅

and 𝓁𝐿 = 𝑦 + 𝓁𝑅 .

(7)

Solving Eqs. (7) for 𝓁𝐿 and 𝓁𝑅 , we obtain 𝓁𝐿 = 𝐿∕2 + 𝑦∕2 and

𝓁𝑅 = 𝐿∕2 − 𝑦∕2,

(8)

which are shown in Fig. 2(b). Since the string is released from rest, at 1 we have 𝑣𝐿1 = 0 and 10 𝑣𝑅2 6 √ 𝑔𝐿 4

𝑣𝐿2 = 0.

2

(10)

Computation 0

0.2

0.4 0.6 𝑦∕𝐿

0.8

1

Figure 3 Nondimensional speed of the falling string (red line) and a particle in free fall (blue line) as a function of nondimensional distance. The orange vertical line represents 𝑦 = 𝐿.

ISTUDY

(9)

Due to the inextensibility of the string, all points on the left segment of string must have a common velocity (as must all points of the right segment). Therefore, since the left branch is attached to the fixed ceiling, we must have

8

0

𝑣𝑅1 = 0.

Interesting Fact Cracking a whip. When a whip is cracked, the tip reaches a supersonic velocity for a duration of 1.2 𝗆𝗌. In a distance of 45 𝖼𝗆, the tip is accelerated from 𝑀 = 1 to 𝑀 = 2.2 (𝑀 is the Mach number, which is the ratio of the object’s speed to the speed of sound in the surrounding medium). Assuming a uniform acceleration during this period and the speed of sound in air of 345 𝗆∕𝗌, the acceleration of the tip exceeds 51,000 𝗀! See P. Krehl, S. Engemann, and D. Schwenkel, “The Puzzle of Whip Cracking— Uncovered by a Correlation of Whip-Tip Kinematics with Shock Wave Emission,” Shock Waves, 8(1), 1998, pp. 1–9.

By substituting Eqs. (5), (8), (9), and (10) into Eq. (2) and then simplifying, the kinetic energies become 𝑇1 = 0 and 𝑇2 = 12 𝜌𝓁𝑅 𝑣2𝑅2 = 41 𝜌(𝐿 − 𝑦)𝑣2𝑅2 .

(11)

Substituting Eqs. (8) into the second of Eqs. (6) and simplifying, we obtain the final form for 𝑉2 to be ) 𝜌𝑔 ( 2 𝑉2 = − 𝐿 + 2𝐿𝑦 − 𝑦2 . (12) 4 Substituting the first of Eqs. (6), Eqs. (11), and Eq. (12) into Eq. (1) and solving for 𝑣𝑅2 , we obtain √ ( ) 2𝐿 − 𝑦 𝑣𝑅2 = 𝑔𝑦 (13) . 𝐿−𝑦 Discussion & Verification

The argument of the square root in Eq. (13) has dimensions of acceleration times length, which is the same as speed squared. Therefore, the answer in Eq. (13) is dimensionally correct. In addition, notice that the value of the speed increases as 𝑦 increases, as should be expected. A Closer Look When we let 𝑦 → 𝐿 in Eq. (13), that is, as the string approaches the end of its fall, we see that 𝑣𝑅2 → ∞ (see Fig. 3). This is an amazing result! Simply take a string, bend it, and drop it as we have done here, and the speed of the tip of the string becomes infinite at the end of the fall. While we know that an infinite speed is not achieved in reality, what the model does imply is that the tip of the string is going very fast. In fact, when a real snapped towel is aided by the person snapping it, the loud crack produced is a result of the shock wave generated by the tip of the towel when the speed of the tip becomes greater than the speed of sound.

ISTUDY

Section 14.3

911

Work-Energy Principle for Systems of Particles

E X A M P L E 14.14

Internal Work Due to Friction

The two blocks 𝐴 and 𝐵 of mass 𝑚𝐴 = 4 kg and 𝑚𝐵 = 1 kg, respectively, in Fig. 1 are connected by an inextensible cord and the pulley system shown. There is negligible friction between 𝐴 and the 𝜃 = 30◦ incline, and the coefficient of kinetic friction between 𝐴 and 𝐵 is 𝜇𝑘 = 0.1. Assuming that 𝜇𝑠 is insufficient to prevent slipping and that the system is released from rest, determine the velocity of 𝐴 and 𝐵 after 𝐵 has moved up the incline a distance 𝑑 = 0.35 m relative to 𝐴.

𝜇𝑠 , 𝜇𝑘

𝐵

SOLUTION

𝐴 frictionless

Road Map & Modeling

Since we need to relate changes in position to changes in velocity, we will apply the work-energy principle. In Fig. 2, we have sketched the system’s FBD, which includes 𝐴 and 𝐵 modeled as particles, the cord, and the pulleys. We ignore friction in the pulleys and assume the cord is inextensible and has negligible mass. The cord tension does no work because the cord is inextensible. However, since 𝐴 and 𝐵 slide relative to one another, we need to include the work done by the internal friction force between 𝐴 and 𝐵, so we have also drawn the FBD of 𝐵 in Fig. 2 so that we can find this friction force. Finally, we will let 1 be when the system is released and 2 be when 𝐵 has moved the distance 𝑑 up the incline relative to 𝐴.

Figure 1 The two-block system connected by pulleys.

𝚥̂

𝑅𝑥

𝑚𝐵 𝑔 𝜃

The work-energy principle for the system in Fig. 2 is )int ( 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 ,

𝑇1 = 21 𝑚𝐴 𝑣2𝐴1 + 12 𝑚𝐵 𝑣2𝐵1

and 𝑇2 = 12 𝑚𝐴 𝑣2𝐴2 + 21 𝑚𝐵 𝑣2𝐵2 ,

𝑚𝐵 𝑔 𝜃

(2)

where 𝑣𝐴 and 𝑣𝐵 are the speed of 𝐴 and 𝐵, respectively, and the second subscript on the speeds indicates position. Again referring to Fig. 2, we sum forces in the 𝑦 direction on 𝐵 and obtain ∑ 𝐹𝑦∶ 𝑁𝐵 − 𝑚𝐵 𝑔 cos 𝜃 = 0, (3) where we have set 𝑎𝐵𝑦 = 0 because there is no motion in the 𝑦 direction. For the system’s potential energy we can write 𝑉1 = 𝑉𝐴1 + 𝑉𝐵1 = 0,

(4)

𝑉2 = 𝑉𝐴2 + 𝑉𝐵2 = −𝑚𝐴 𝑔 Δ𝑥𝐴 sin 𝜃 − 𝑚𝐵 𝑔 Δ𝑥𝐵 sin 𝜃,

(5)

where we have placed the datum for the potential energy at 1 and where Δ𝑥𝐴 and Δ𝑥𝐵 are the displacements parallel to the incline of 𝐴 and 𝐵, respectively, between 1 and 2. Both Δ𝑥𝐴 and Δ𝑥𝐵 are unknown, and we will use kinematics to relate them to the given distance 𝑑. The work of the internal friction force 𝐹𝐵 is given by ( )int 𝑈1-2 nc = −

∫0

𝑑

𝐹𝐵 𝑑𝑥.

(6)

Finally, since we know that 𝜇𝑠 is insufficient to prevent slipping, we know that 𝐹 𝐵 = 𝜇 𝑘 𝑁𝐵 .

𝑃𝑦

(1)

where (𝑈1-2 )int nc is the work done by friction. The kinetic energies are given by

Force Laws

𝑅𝑦

𝚤̂

𝑃𝑥

Governing Equations Balance Principles

𝜃

(7)

𝑚𝐴 𝑔

𝜃

𝑁

𝐹𝑐 𝐹𝐵

Figure 2 FBD of the system and of just 𝐵.

𝑁𝐵

912

Chapter 14

Energy Methods for Particles

Kinematic Equations 𝑥𝐵

Since the system starts from rest, we have 𝑣𝐴1 = 0 and 𝑣𝐵1 = 0.

𝑥𝐴

(8)

In addition, we can relate the motions of 𝐴 and 𝐵 using pulley kinematics (see Section 12.6 on p. 738). Referring to Fig. 3, we see that for arbitrary 𝑥𝐴 and 𝑥𝐵 3𝑥𝐴 + 𝑥𝐵 = 𝐿

𝐵 𝐴

ISTUDY

3𝑣𝐴 = −𝑣𝐵 ,

(9)

where 𝐿 is constant and where 𝑣𝐴 and 𝑣𝐵 represent the velocity of 𝐴 and 𝐵, respectively, since the motion is one-dimensional. From Eq. (9), at 2 we have 3𝑣𝐴2 = −𝑣𝐵2

Figure 3 Definition of 𝑥𝐴 and 𝑥𝐵 for the two blocks.

⇒

and 3Δ𝑥𝐴 = −Δ𝑥𝐵 .

(10)

Finally, we know that 𝐴 displaces a distance 𝑑 relative to 𝐵. Therefore, we can write Δ𝑥𝐴∕𝐵 = 𝑑 = Δ𝑥𝐴 − Δ𝑥𝐵 .

(11)

Solving Eqs. (10) and (11) for Δ𝑥𝐴 and Δ𝑥𝐵 , we obtain Δ𝑥𝐴 = 14 𝑑 Computation

and Δ𝑥𝐵 = − 34 𝑑.

(12)

Substituting Eqs. (3) and (7) into Eq. (6), we have ( )int 𝑈1-2 nc = −𝜇𝑘 𝑚𝐵 𝑔𝑑 cos 𝜃.

(13)

Next, we substitute the first of Eqs. (10) into Eq. (2) to obtain ( ( ) )2 𝑇2 = 12 𝑚𝐴 𝑣2𝐴2 + 12 𝑚𝐵 −3𝑣𝐴2 = 21 𝑚𝐴 + 9𝑚𝐵 𝑣2𝐴2 , and then substitute Eqs. (12) into Eq. (5) to obtain ( ) 𝑉2 = 14 3𝑚𝐵 − 𝑚𝐴 𝑔𝑑 sin 𝜃.

(14)

(15)

We then take Eqs. (4), (8), and (13)–(15) and substitute them into Eq. (1) to obtain one equation for 𝑣𝐴2 ( ( ) ) −𝜇𝑘 𝑚𝐵 𝑔𝑑 cos 𝜃 = 12 𝑚𝐴 + 9𝑚𝐵 𝑣2𝐴2 + 41 3𝑚𝐵 − 𝑚𝐴 𝑔𝑑 sin 𝜃, (16)

Helpful Information What does the ± sign in Eq. (17) mean? In this example we need to determine a velocity. Since the motion is one-dimensional, the velocity we are seeking can be equal to the speed or opposite to the speed. The ± sign in Eq. (17) is there to remind us that we need to determine the sign of the velocity.

which gives

√ 𝑣𝐴2 = ±

] ) 2𝑔𝑑 [ 1 ( sin 𝜃 − 𝜇 𝑚 cos 𝜃 . 𝑚 − 3𝑚 𝑘 𝐵 𝐵 𝑚𝐴 + 9𝑚𝐵 4 𝐴

(17)

Observing that 𝐴 moves down the incline and 𝐵 moves up the incline, using Eqs. (17) and (9), as well as the given data, we have 𝑣𝐴2 = 0.1424 m∕s and 𝑣𝐵2 = −0.4273 m∕s. Discussion & Verification

(18)

The dimensions of the argument of the square root in Eq. (17) are energy divided by mass, i.e., speed squared. Hence, the expression for 𝑣𝐴2 in Eq. (17) is dimensionally correct. The expression of 𝑣𝐵 in Eqs. (9) is also dimensionally correct. As far as the values in Eqs. (18) are concerned, since 𝐵 acts as a counterweight for 𝐴, a way to check the reasonableness of our result is to compute the speed that 𝐴 would have if it were disconnected from 𝐵 and moved √a distance 𝑑∕4 down the incline without friction. This speed, given by (𝑣𝐴2 )no friction = 2(𝑑∕4)𝑔 sin 𝜃 = 0.9265 m∕s, is an upper bound for 𝑣𝐴2 . That is, 𝑣𝐴2 must be less than 0.9265 m∕s, as indeed it is.

ISTUDY

Section 14.3

Work-Energy Principle for Systems of Particles

913

Problems Problem 14.71 The truck comes to a stop under the action of a constant braking force. During braking, either the crate slides or it does not. Considering the work-energy principle applied to the truck during braking, will the truck stop in a shorter distance (or time) if the crate slides, or will the distance (or time) be shorter if it does not slide? Assume that the truck bed is long enough that you don’t have to worry about whether or not the crate hits the truck. Justify your answer.

𝐵

𝐴

Figure P14.71

Problem 14.72 Consider a pulley system in which bodies 𝐴 and 𝐵 have masses 𝑚𝐴 = 2 kg and 𝑚𝐵 = 10 kg. If the system is released from rest, neglecting all sources of friction, as well as the inertia of the pulleys, determine the speeds of 𝐴 and 𝐵 after 𝐵 has displaced a distance of 0.6 m downward. Treat all cable segments as purely vertical.

Problem 14.73 A 700 lb floating platform is at rest when a 200 lb crate is thrown onto it with a horizontal speed 𝑣0 = 12 f t∕s. Once the crate stops sliding relative to the platform, the platform and the crate move with a speed 𝑣 = 2.667 f t∕s. Neglecting the vertical motion of the system, as well as any resistance due to the relative motion of the platform with respect to the water, determine the distance that the crate slides relative to the platform if the coefficient of kinetic friction between the platform and the crate is 𝜇𝑘 = 0.25.

𝐵 𝐴 Figure P14.72

𝑣0 𝑘

𝐴 𝐵

Figure P14.73

Figure P14.74

Problem 14.74 Blocks 𝐴 and 𝐵, weighing 7 and 15 lb, respectively, are released from rest when the spring is unstretched. If all sources of friction are negligible and 𝑘 = 12 lb∕f t, determine the maximum vertical displacement of 𝐵 from the release position, assuming that 𝐴 never leaves the horizontal surface shown and the cord connecting 𝐴 and 𝐵 is inextensible.

Problems 14.75 through 14.77 Crates 𝐴 and 𝐵 of mass 50 kg and 75 kg, respectively, are released from rest. The linear elastic spring has stiffness 𝑘 = 500 N∕m. Neglect the mass of the pulleys and cables and neglect friction in the pulley bearings. If 𝜇𝑘 = 0 and the spring is initially unstretched, determine the speed of 𝐵 after 𝐴 slides 4 m.

Problem 14.75

𝑘

𝐴 𝜇𝑘

If 𝜇𝑘 = 0.25 and the spring is initially unstretched, determine the speed of 𝐵 after 𝐴 slides 4 m.

Problem 14.76

If 𝜇𝑘 = 0.25 and the spring is initially stretched 1 m, how far does 𝐴 slide before the system momentarily comes to rest?

𝐵

Problem 14.77

Figure P14.75–P14.77

914

Chapter 14

Energy Methods for Particles Problems 14.78 through 14.80 𝐴

𝜇𝑘

𝐵

The crates 𝐴 and 𝐵 of weight 𝑊𝐴 = 50 lb and 𝑊𝐵 = 75 lb, respectively, are connected by a pulley system. The system is released from rest and friction between 𝐴 and the horizontal surface is insufficient to prevent slipping. The cables in the pulley system are inextensible, and the coefficient of kinetic friction between the crate 𝐴 and the horizontal surface is 𝜇𝑘 . Problem 14.78

If 𝜇𝑘 = 0, determine the speed of 𝐵 after 𝐴 slides 10 f t.

Problem 14.79

Determine the required value of 𝜇𝑘 so that the speed of 𝐵 is 27 f t∕s

after it drops 15 f t.

Figure P14.78–P14.80

Problem 14.80

If 𝜇𝑘 = 0.25, determine the speed of 𝐴 after 𝐵 drops 15 f t.

Problem 14.81 𝑘

𝐴 𝐵

Blocks 𝐴 and 𝐵 are released from rest when the spring is unstretched. Block 𝐴 has a mass 𝑚𝐴 = 2 kg, and the linear spring has stiffness 𝑘 = 7 N∕m. If all sources of friction are negligible, determine the mass of block 𝐵 such that 𝐵 has a speed 𝑣𝐵 = 1.5 m∕s after moving 1.2 m downward, assuming that 𝐴 never leaves the horizontal surface shown and the cord connecting 𝐴 and 𝐵 is inextensible.

Figure P14.81

Problem 14.82 Spring scales work by measuring the displacement of a spring that supports both the platform of mass 𝑚𝑝 and the object of mass 𝑚, whose weight is being measured. In your solution, note that most scales read zero when no mass 𝑚 has been placed on them; that is, they are calibrated so that the weight reading neglects the mass of the platform 𝑚𝑝 . Assume that the spring is linear elastic with spring constant 𝑘. If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the maximum reading on the scale after the mass is released. ACME

𝑚

𝐴

𝑚𝑝 𝑑

𝑅 𝐵 𝐷 Figure P14.82

Figure P14.83

Problem 14.83 Two small spheres 𝐴 and 𝐵, each of mass 𝑚, are attached at either end of a stiff light rod of length 𝑑. The system is released from rest in the position shown. Determine the speed of the spheres when 𝐴 has reached point 𝐷, and determine the normal force between sphere 𝐴 and the surface on which it is sliding immediately before it reaches point 𝐷. Neglect friction, treat the spheres as particles (assume that their diameter is negligible), and neglect the mass of the rod.

𝜇𝑠 , 𝜇𝑘

𝐵 𝐴 frictionless 𝜃 Figure P14.84

ISTUDY

Problem 14.84 Solve Example 14.14 by applying the work-energy principle to each block individually, and show that the net work done by the cord on the two blocks is zero.

ISTUDY

Section 14.3

Work-Energy Principle for Systems of Particles

915

Problem 14.85 Consider a simple elevator design in which a 15,000 kg car 𝐴 is connected to a 12,000 kg counterweight 𝐵. Suppose that a failure of the drive system occurs (the failure does not affect the rope connecting 𝐴 and 𝐵) when the car is at rest and 50 m above its buffer, causing the elevator car to fall. Model the car and the counterweight as particles and the cord as massless and inextensible, and model the action of the emergency brakes using a Coulomb friction model with kinetic friction coefficient 𝜇𝑘 = 0.5 and a normal force equal to 35% of the car’s weight. Determine the speed with which the car impacts the buffer.

car 𝐴

counterweight 𝐵

car buffer

Problem 14.86 Two identical balls, each of mass 𝑚, are connected by a string of negligible mass and length 2𝑙. A short string is attached to the first at its middle and is pulled vertically with a constant force 𝑃 (exerted by the hand). If the system starts at rest when 𝜃 = 𝜃0 , determine the speed of the two balls as 𝜃 approaches 90◦ . Neglect the size of the balls as well as friction between the balls and the surface on which they slide.

counterweight buffer

Figure P14.85

5 ft 𝐴 𝑙

𝑙

𝜃

𝜃

Figure P14.86

10 f t 35◦ 𝐵

Figure P14.87

Problem 14.87 Consider the simple catapult shown in the figure with an 800 lb counterweight 𝐴 and a 150 lb projectile 𝐵. If the system is released from rest as shown, determine the speed of the projectile after the arm rotates (counterclockwise) through an angle of 110◦ . Model 𝐴 and 𝐵 as particles, neglect the mass of the catapult’s arm, and assume that friction is negligible. The catapult’s frame is fixed with respect to the ground, and the projectile does not separate from the arm during the motion considered.

Problems 14.88 and 14.89 Two blocks 𝐴 and 𝐵 weighing 123 and 234 lb, respectively, are released from rest as shown. At the moment of release the spring is unstretched. In solving these problems, model 𝐴 and 𝐵 as particles, neglect air resistance, and assume that the cord is inextensible.

𝑘

𝐴 𝛼

Problem 14.88 Determine the maximum speed attained by block 𝐵 and the distance from the floor where the maximum speed is achieved if 𝛼 = 20◦ , the contact between 𝐴 and the incline is frictionless, and the spring constant is 𝑘 = 30 lb∕f t.

𝐵 2 ft

Problem 14.89 Determine the maximum displacement of block 𝐵 if 𝛼 = 20◦ , the contact between 𝐴 and the incline is frictionless, and the spring constant is 𝑘 = 300 lb∕f t.

Figure P14.88 and P14.89

916

Chapter 14

Energy Methods for Particles

Problem 14.90 Revisit Example 14.7 on p. 886 and determine the maximum height reached by the pole vaulter, but this time include the mass of the pole in your analysis. To solve the problem, use the following data: the maximum speed achieved by the pole vaulter and pole at the time the vault begins is 𝑣max = 34.19 f t∕s; the pole vaulter is 6 f t tall with mass center at 55% of body height (as measured from the ground) and weighs 190 lb; the pole is uniform, has a weight of 5.8 lb, and is 17.1 f t long. In addition, assume that as the pole vaulter sprints before the vault, the pole is carried horizontally at the same height as the vaulter’s center of mass relative to the ground. Explain why the vault height you find when the pole is included is higher than the vault height determined in Example 14.7, when the pole was not included.

Figure P14.90

Problems 14.91 through 14.93 𝐶

𝐴

𝜇𝑘

The two crates 𝐴 and 𝐵 of mass 𝑚𝐴 = 100 kg and 𝑚𝐵 = 70 kg, respectively, are connected by a system of pulleys. The system is initially at rest, when the man 𝐶 starts pushing on 𝐴 with a constant force 𝑃 . Neglect the mass of the cables and the pulleys and neglect friction in the pulley bearings. Assume that cable segments are long enough so that crate 𝐴 does not hit any of the pulleys. Problem 14.91

𝐵

Problem 14.92 Figure P14.91–P14.93

ISTUDY

If 𝜇𝑘 = 0 and 𝑃 = 375 N, determine the speed of each crate after 𝐴

slides 6 m. If 𝜇𝑘 = 0.2 and 𝑃 = 375 N, determine the speed of each crate after 𝐴

slides 6 m. If 𝜇𝑘 = 0.3 and 𝑃 = 350 N, determine the distance the man needs to push the crate 𝐴 to achieve a speed of 3 m∕s.

Problem 14.93

Problem 14.94 The rope of mass 𝑚 and length 𝑙 is released from rest with a very small amount of it hanging over the edge (i.e., 𝑠 > 0, but it is very close to zero). Determine the speed of the rope as a function of 𝑠. Assume the surface is smooth.

𝑠

Figure P14.94

ISTUDY

Section 14.3

Work-Energy Principle for Systems of Particles

Problem 14.95 Consider a system of four identical particles 𝐴, 𝐵, 𝐷, and 𝐸, each with a mass of 2.3 kg. These particles are mounted on a wheel of negligible mass whose hub is at 𝑂 and whose radius is 𝑅 = 0.75 m. The wheel is mounted on a cart 𝐻 of negligible mass. Suppose that the wheel is spinning counterclockwise with an angular speed 𝜔 = 3 rad∕s and that the cart is moving to the right with a speed of 𝑣𝐻 = 11 m∕s. Compute the kinetic energy of the system, using (a) Eq. (14.46) on p. 902, (b) Eq. (14.50) on p. 903. Show that the result is the same in both cases.

𝐴

𝐸

𝜔

𝑂 𝐵

𝐷 𝐻

Figure P14.95

Problem 14.96 Reexamine Example 14.12 by choosing 1 to be the same as that in the example and 2 to be an arbitrary position after release. Determine the speed of 𝐴 as a function of 𝑦𝐴 , and plot the kinetic energy of 𝐴 as a function of 𝑦𝐴 . Use this plot to argue that the cord connecting 𝐴 and 𝐵 never goes slack between the instant of release and the impact of 𝐵 with the ground. Hint: To argue that the cord remains taut between the positions of interest, you may want to look at concept Prob. 14.29 on p. 876. 𝑑 = 2.5 m 𝑃

𝐷

𝑙 = 4m

𝐵

𝑘 𝐴 0.25 m Figure P14.96

ℎ = 2m

917

918

Chapter 14

Energy Methods for Particles

Design Problems

ISTUDY

Design Problem 14.2 The plunger (the rod and rectangular block attached to the end of it) of a pinball machine has weight 𝑊𝑝 = 5 oz. The weight of the ball is 𝑊𝑏 = 2.85 oz, and the machine is inclined at 𝜃 = 8◦ with the horizontal. Design the springs 𝑘1 and 𝑘2 (i.e., their stiffnesses and unstretched lengths) so that the ball separates from the plunger with a speed 𝑣 = 15 f t∕s after pulling back the plunger 2 in. from its rest position and so that the plunger comes no closer than 0.5 in. from the stop. Note that as part of your design, you will also need to specify reasonable values for the dimensions 𝓁1 and 𝓁2 . stop 𝓁1 𝑘1

𝓁2

𝑊𝑏

𝑘2 𝑊𝑝

𝜃 Figure DP14.2

ISTUDY

Section 14.4

14.4

Power and Efficiency

919

Power and Efficiency

Power developed by a force The concept of mechanical power arises from a need to quantify the ability of a motor or other machine to do mechanical work or provide energy in a given amount of time. Referring to Fig. 14.14, we consider a force 𝐹⃗ whose point of application undergoes an infinitesimal displacement 𝑑⃗𝑟 in a corresponding infinitesimal time interval 𝑑𝑡. The infinitesimal work done by 𝐹⃗ during the time interval 𝑑𝑡 is 𝑑𝑈 = 𝐹⃗ ⋅ 𝑑⃗𝑟.

(14.51)

The power 𝑃 developed by the force 𝐹⃗ is defined as the rate at which 𝐹⃗ does work, that is, 𝑑𝑈 power = 𝑃 = . (14.52) 𝑑𝑡 Substituting Eq. (14.51) into Eq. (14.52), we have 𝑃 =

𝑑⃗𝑟 𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝐹⃗ ⋅ . 𝑑𝑡 𝑑𝑡

(14.53)

Recalling that the velocity of the point of application of 𝐹⃗ is 𝑣⃗ = 𝑑⃗𝑟∕𝑑𝑡, we can then rewrite Eq. (14.53) as 𝑃 = 𝐹⃗ ⋅ 𝑣. ⃗ (14.54) Dimensions and units of power Because of its definition, the power developed by a force is a scalar quantity with dimensions of (force) × (length)∕(time) or, equivalently, (mass) × (length)2 ∕(time)3 . In the SI system, the unit of measure for power is called the watt, which is abbreviated by the symbol W and is defined as follows: 1 W = 1 N⋅(1 m∕s) = 1 J∕s.

(14.55)

In the U.S. Customary system, the unit of measure for power is called horsepower, which is abbreviated by the symbol hp and is defined as follows: 1 hp = 550 f t ⋅lb∕s.

(14.56)

The conversion from horsepower to watt is 1 hp = 745.7 W.

Efficiency Efficiency is a measure of the performance of a motor or a machine. It is the ratio between its power output and the power needed for its operation. Efficiency, for which we use the Greek letter 𝜖 (epsilon), is defined as: 𝜖=

power output . power input

(14.57)

Efficiency is a nondimensional scalar quantity. Since some of the power supplied to a motor or a machine is used to move internal components and to overcome internal frictional resistance, the power output of a motor or machine is always smaller than the power input. Therefore, the efficiency of a motor or a machine is a number that is always smaller than 1.

David Acosta Allely/Shutterstock.com

Figure 14.13 The 2009 Ferrari Formula One race car. Machines such as this car produce a tremendous amount of power for their size.

𝐹⃗ 𝑑⃗𝑟 Figure 14.14 A force acting on a point moving along a path.

920

Chapter 14

Energy Methods for Particles

E X A M P L E 14.15

Computing the Power of an Aircraft Carrier Catapult In Example 14.2 on p. 866, we considered the takeoff of a 50,000 lb F/A-18 Hornet (see Fig. 1) from a stationary aircraft carrier. The F/A-18 goes from 0 to 165 mph in a distance of 300 f t, assisted by a catapult and with each of its two engines generating a constant thrust of 22,000 lb. Neglecting aerodynamic forces,( in Example 14.2 we determined ) that the work done on the aircraft by the catapult was 𝑈1-2 catapult = 3.227 × 107 f t ⋅lb. Continuing that example, determine the power developed by the thrust generated by the engines at the beginning and at the end of the takeoff. Also determine the average power supplied to the aircraft by the catapult, knowing that the takeoff occurs in roughly 2 s.

U.S. Navy photo by Mass Communication Specialist 3rd Class Torrey W. Lee

Figure 1 An F/A-18 taking off from an aircraft carrier.

𝑣2

3

𝐹𝑇 NAVY

SOLUTION Road Map & Modeling

Denoting the beginning and the end of the takeoff by 1 and 2, respectively, we know the thrust of the engines and the velocity of the aircraft at 1 and 2. Therefore, we can compute the power developed by the engines’ thrust with the relation expressing the power developed by a force in terms of the force itself and the velocity vector. To compute the average power of the catapult, we will compute the time average of the power developed by the force exerted by the catapult on the aircraft and relate that to the work done by the catapult.

Computation Recall that the quantity 𝐹⃗ ⋅ 𝑣⃗ is the power developed by a force 𝐹⃗ whose ⃗ point of application moves with velocity 𝑣. ⃗ Since the plane starts from rest, i.e., 𝑣⃗1 = 0, letting 𝑃𝑇 1 denote the power developed by the engines’ thrust at 1, we must have

𝑃𝑇 1 = 0.

𝚥̂ 𝚤̂

(1)

Referring to Fig. 2, at 2 we will assume that the plane is moving horizontally with a velocity 𝑣⃗2 = 𝑣2 𝚤̂, where 𝑣2 = 165 mph = 242.0 f t∕s. In addition, assuming that the engines’ thrust is also horizontal, the total thrust of the engines is 𝐹⃗𝑇 = 𝐹𝑇 𝚤̂ with 𝐹𝑇 = 44,000 lb. Therefore, letting 𝑃𝑇 2 be the power developed by 𝐹⃗𝑇 at 2, we have 𝑃𝑇 2 = 𝐹𝑇 𝚤̂ ⋅ 𝑣2 𝚤̂ = 𝐹𝑇 𝑣2 = 19,360 hp,

(2)

( ) where we have used 1 hp = 550 f t ⋅lb∕s. Finally, letting 𝑃cat and 𝑃avg cat be the instantaneous and the average power supplied to the aircraft by the catapult between 1 and 2, respectively, we have Figure 2 Diagram showing the aircraft’s velocity vector and the engines’ thrust vector at 2.

ISTUDY

( ) 𝑃avg cat =

𝑡2 𝑡2 𝑑𝑈 (𝑈 ) 1 1 cat 𝑑𝑡 = 1-2 cat . 𝑃cat 𝑑𝑡 = 𝑡2 − 𝑡1 ∫𝑡1 𝑡2 − 𝑡1 ∫𝑡1 𝑑𝑡 𝑡2 − 𝑡1

(3)

) ( Recalling that 𝑡2 − 𝑡1 = 2 s and that 𝑈1-2 cat = 3.227×107 f t ⋅lb, we have (

𝑃avg

) cat

= 29,340 hp,

(4)

where we have again used 1 hp = 550 f t ⋅lb∕s. Discussion & Verification

The results in Eqs. (1) and (2) are obtained as the direct application of a known formula. Hence, we need only verify that we used the correct units, which we have since we expressed our result in horsepower and the data was provided in U.S. Customary units. The result in Eq. (3) was obtained by applying the definition of time average. Keeping this in mind, intuition suggests that the notion of “average power” should correspond to the ratio between the work done by the catapult and the time taken to do such work, which is exactly what our calculation yielded.

ISTUDY

Section 14.4

921

Power and Efficiency

E X A M P L E 14.16

Assessing Lift Speed Given the Efficiency

An electric motor with an efficiency of 85% drives the pulley system shown to lift a 400 kg crate. After the motor is switched on and the system has reached a constant lifting speed, the motor is drawing 15 kW. Assume that the motor is operating at its rated efficiency, that the frictional losses in the pulley system are negligible, and that the weights of the pulleys 𝐴 and 𝐵 are negligible. Determine the speed with which the crate is being lifted by considering

𝐴

𝐵 𝑣𝑐

(a) an FBD of the system that includes the crate and both pulleys 𝐴 and 𝐵,

𝐻

(b) an FBD of the crate that has been separated from the system at the hook 𝐻.

crate

motor

Neglect the size of the pulleys in your solution and treat the two cable segments between 𝐴 and 𝐵 as purely vertical.

Figure 1 Pulley system lifting a crate with speed 𝑣𝑐 .

SOLUTION Road Map & Modeling We can calculate the output power of the motor since we know its efficiency and input power. This output power is the power used by the pulley system to lift the crate. Since there are no power losses within the pulley system, the output power can be viewed as either the power of the cable between pulley 𝐴 and the motor or the power developed by the tension in the hook 𝐻 attached to the crate. Parts (a) and (b) of the problem reflect these two viewpoints. Hence, we need to use Newton’s second law to determine the tension in the cable attached to the motor, and then we will be able to compute the lifting speed by dividing the power developed by this force by the magnitude of the force. We will perform a similar calculation for the hook attached to the crate.

𝐴𝑦 𝐴𝑥

𝐴

𝚥̂ 𝚤̂

𝐵 𝑇

Solution for Part (a)

𝑚𝑔

Governing Equations Balance Principles

The FBD of the pulleys and crate is shown in Fig. 2, in which the crate has been modeled as a particle and where 𝑇 is the tension in the cable being pulled in by the motor. While it is the power developed by the tension 𝑇 that we want to find, this FBD does not allow us to find 𝑇 itself. Instead, applying Newton’s second law to the FBD in Fig. 3(a), we find ∑ 𝐹𝑦∶ 2𝑇 − 𝑚𝑔 = 𝑚𝑎𝑦 . (1) Force Laws

𝚥̂

𝐴

𝚤̂ 𝑇

All forces are accounted for on the FBD.

Kinematic Equations

Figure 2 FBD of the pulleys and crate.

𝓁𝑚

𝓁𝑐

𝑇

Since the crate is moving with constant velocity, we have

𝐵

𝐵

𝑎𝑦 = 0.

(2)

To find the speed of the point of application of the tension in the cable between the motor and the pulley at 𝐴, we need to refer to the pulley kinematics in Fig. 3(b) and write 2𝓁𝑐 + 𝓁𝑚 = constant

⇒

2𝓁̇𝑐 = −𝓁̇𝑚

⇒

2𝑣𝑐 = 𝓁̇𝑚 ,

(3)

where we have used the fact that 𝑣𝑐 = −𝓁̇𝑐 and we note that when 𝓁̇𝑚 > 0, the motor is winding in cable. Computation

Substituting Eq. (2) into Eq. (1), we find that 𝑇 = 𝑚𝑔∕2.

(4)

We need to compute the power provided by the motor to the pulley system. Letting 𝑃𝑖 and 𝑃𝑜 be the motor power input and output, respectively, we have 𝑃𝑖 = 15 kW and 𝑃𝑜 = 𝜖𝑃𝑖 = 12.75 kW.

(5)

𝑢̂ 𝑚

𝐻 𝑚𝑔

(a)

crate

𝑣𝑐 (b)

Figure 3 (a) FBD of the crate and the pulley at 𝐵. (b) Kinematics of the pulley system connecting the motor to the crate.

922

Chapter 14

Energy Methods for Particles

Recalling that 𝑃𝑜 is the power developed by the tension 𝑇 , and that 𝑇 is in the same direction as the velocity of the cable winding up on the motor, we then must have 𝑃𝑜 = 𝑇 𝑢̂ 𝑚 ⋅ 𝓁̇𝑚 𝑢̂ 𝑚 = 𝑇 𝓁̇𝑚 =

𝑚𝑔 ( ) 2𝑣𝑐 = 𝑚𝑔𝑣𝑐 , 2

(6)

where we have used Eqs. (3) and (4). Solving Eq. (6) for 𝑣𝑐 , we have 𝑣𝑐 =

𝑃𝑜 𝑚𝑔

= 3.249 m∕s,

(7)

where we have used 𝑃𝑜 from Eqs. (5). Solution for Part (b) Governing Equations Balance Principles

𝐹𝐻 𝚥̂

The FBD of the crate is shown in Fig. 4, in which the crate has been modeled as a particle subject to its own weight and the tension 𝐹𝐻 in the hook. Using the FBD in Fig. 4, Newton’s second law yields ∑ 𝐹𝑦∶ 𝐹𝐻 − 𝑚𝑔 = 𝑚𝑎𝑦 . (8) Force Laws

𝑚𝑔

All forces are accounted for on the FBD.

Kinematic Equations

Since the crate is moving with constant velocity, we have 𝑎𝑦 = 0.

Figure 4 FBD of the crate.

ISTUDY

Computation

(9)

Substituting Eq. (9) into Eq. (8) and solving for 𝐹𝐻 , we have 𝐹𝐻 = 𝑚𝑔.

(10)

Again, we need to compute the power provided by the motor to the pulley system. Recalling that 𝑃𝑜 is the power developed by the tension 𝐹𝐻 , and that 𝐹𝐻 is in the same direction as the velocity of the crate, we then must have 𝑃𝑜 = 𝐹𝐻 𝚥̂ ⋅ 𝑣𝑐 𝚥̂ = 𝐹𝐻 𝑣𝑐 ,

(11)

where 𝑣𝑐 is the speed of the crate. Solving Eq. (11) for 𝑣𝑐 , we have 𝑣𝑐 =

𝑃𝑜 𝐹𝐻

=

𝑃𝑜 𝑚𝑔

= 3.249 m∕s,

(12)

where we have used 𝑃𝑜 from Eqs. (5). In Parts (a) and (b), 𝑃𝑜 has dimensions of force times length divided by time, so the results in Eqs. (7) and (12) have the expected dimensions of length over time. Also, given that the data was provided in SI units, the results are expressed using appropriate units. As far as the value of 𝑣𝑐 is concerned, as expected, it is smaller than the maximum possible value of lifting speed given by 𝑃𝑖 ∕𝑚𝑔 = 3.823 m∕s. Hence, our solution appears to be correct.

Discussion & Verification

A Closer Look Notice that in Part (a) of this example, we had to analyze the pulley kinematics, but we did not have to do so in Part (b). That is because in pulley systems, the power supplied determines the speed of the payload, regardless of the pulley arrangement. This stems from the fact that power is the product of force and velocity, and the nature of pulley systems is to keep that product constant.

ISTUDY

Section 14.4

923

Power and Efficiency

Problems Problem 14.97 Consider a battery that can power a machine whose operation requires 200 W of power. If the battery can keep the machine operating for 8 h and supply energy at a constant rate, determine the amount of energy initially stored in the battery pack.

Problem 14.98

Figure P14.97

A person takes 16 s to lift a 200 lb crate to a height of 30 f t. Determine the average power supplied by the person.

𝐹

𝐹 𝐵

𝐵 (b)

(a) Figure P14.98

Figure P14.99

Problem 14.99 The weights 𝐵 in the pulley systems shown are identical. Assume that the two systems are released from rest, there are no energy losses in the pulleys, and the same constant force 𝐹 is applied. After 1 s, will the power developed by the force 𝐹 for system (a) be smaller than, equal to, or greater than that of system (b)?

Problem 14.100 The weights 𝐵 in the pulley systems shown are identical and are being lifted at the same constant speed 𝑣. Assume there are no energy losses in the pulleys. Is the power developed by force 𝐹1 in (a) smaller than, equal to, or greater than that developed by force 𝐹2 in (b)?

𝐹1 𝐵

Problem 14.101 For the two pulley systems shown, the block 𝐵 has a mass 𝑚𝐵 = 254 kg and is being lifted with a constant speed 𝑣 = 3 m∕s. Determine the power developed by the forces 𝐹1 and 𝐹2 .

𝐵 (b)

(a) Figure P14.100 and P14.101

𝑣𝑡

Problem 14.102 A truck crane with a mass of 40,000 kg has an engine with a rated power of 300 kW. Assuming that all power is transmitted to the wheels and that they do not slip, determine the maximum climb angle 𝜃 if the truck is to maintain a constant speed of 𝑣𝑡 = 25 km∕h.

𝐹2

𝑣

𝑣

𝜃 Figure P14.102

924

Chapter 14

Energy Methods for Particles

Problem 14.103 The 185 lb man climbs a distance ℎ = 13 f t to the top of the ladder in 9 s. Determine his average power output.

ℎ

Problem 14.104 A cyclist is riding with a speed 𝑣 = 18 mph over an inclined road with 𝜃 = 15◦ . Neglecting aerodynamic drag, if the cyclist were to keep his power output constant, what speed would he attain if 𝜃 were equal to 20◦ ?

Figure P14.103

𝑣

𝜃 Figure P14.104

15 100 Figure P14.105

Problem 14.105 A 1500 kg car is traveling up the slope shown at a constant speed. Knowing that the maximum power output of the car is 160 hp, at what speed can the car travel if air resistance is negligible? Also, knowing that 1 L of regular gasoline provides 34.8 MJ of energy, how many liters of gasoline will be required in 1 h if the engine has an efficiency 𝜖 = 0.20?

Problem 14.106 A 2600 lb car on a straight horizontal road goes from 0 to 60 mph in 7 s. Neglecting aerodynamic drag, if the force propelling the car is constant during the acceleration phase, determine the power developed by this force 7 s after the beginning of the motion. Gary L. Gray

Figure P14.106

ISTUDY

Problem 14.107 The height ℎ of the aerobic stepper shown is 25 cm. A 65 kg woman goes up and down the step once every 2 s. Accounting only for the work done in lifting her body and assuming a muscle efficiency of 25%, determine the calories (1 C = 4.184 kJ) she burns in 1 h.

ℎ Figure P14.107

ISTUDY

Section 14.4

Power and Efficiency

Problem 14.108 The motor powering the pulley system shown is drawing 6 kW of power. The crate 𝐵 has a mass of 250 kg. Determine the speed of the crate if the crate is lifted at a constant speed and the motor’s efficiency is 𝜖 = 0.87.

Problems 14.109 and 14.110 The motor 𝐵 is used to raise and lower the crate 𝐶 by a pulley system. At the instant shown, the cable is being retracted by the motor with the constant speed 𝑣𝑐 = 5 f t∕s. The weight of the crate is 𝑊𝐶 = 450 lb.

𝐴

𝐵

𝐵 motor

Figure P14.108

𝑣𝑐

𝐶 Figure P14.109 and P14.110

If the power meter 𝐴 shows a power input to the motor of 1.36 hp, determine the overall efficiency of the system.

Problem 14.109

Problem 14.110

If the overall efficiency of the system is 0.82, determine the power input to the motor that the meter at 𝐴 would read.

Problem 14.111 Assuming that the motor shown has an efficiency 𝜖 = 0.80, determine the power to be supplied to the motor if it is to pull a 250 lb crate up the incline with a constant speed 𝑣 = 7 f t∕s. Assume that the kinetic friction coefficient between the slide and the crate is 𝜇𝑘 = 0.25 and that 𝜃 = 28◦ . 𝜃

Problem 14.112 Assume that the motor shown has an efficiency 𝜖 = 0.82 and that it draws 6 hp. Determine the constant speed with which the motor can pull a 300 lb crate up the incline if the kinetic friction coefficient between the slide and the crate is 𝜇𝑘 = 0.45 and 𝜃 = 32◦ .

Figure P14.111 and P14.112

925

926

Chapter 14

Energy Methods for Particles

14.5 C h a p t e r R e v i e w In this chapter we introduced the concepts of work of a force, kinetic energy of a particle, and kinetic energy of a particle system. We showed that the work done on a system is equal to the change in kinetic energy of the system. This relationship is referred to as the work-energy principle, and it is a direct consequence of Newton’s second law. Finally, we introduced the concepts of power developed by a force and efficiency of a motor or a machine.

Work-energy principle for a particle 𝑃2

The work-energy principle is expressed by the following equation:

ℒ1-2

Eq. (14.9), p. 862

𝐹⃗ 𝑃1 𝑚

𝑇1 + 𝑈1-2 = 𝑇2 ,

𝑣⃗

𝑢̂ 𝑡

Figure 14.15 Particle of mass 𝑚 moving along the path ℒ1-2 while being subjected to the force 𝐹⃗ .

ISTUDY

where 𝑈1-2 is the work done by the force 𝐹⃗ (see Fig. 14.15) and is defined as Eq. (14.7), p. 862 𝑈1-2 =

∫ℒ

𝐹⃗ ⋅ 𝑑⃗𝑟,

1-2

Helpful Information Other forms of the work of a force. By recalling that position and velocity are related by the expression 𝑑⃗𝑟 = 𝑣⃗ 𝑑𝑡 and that, in normal-tangential components, the velocity can be expressed as 𝑣⃗ = 𝑣 𝑢̂ 𝑡 , we can obtain the following two useful forms of the work of a force: 𝑈1-2 =

∫𝑡

𝑡2 1

𝐹⃗ ⋅ 𝑣⃗ 𝑑𝑡 =

∫𝑠

𝑠2

𝐹⃗ ⋅ 𝑢̂ 𝑡 𝑑𝑠,

1

where 𝑡 is time, 𝑠 is the arc length along the path, and 𝑢̂ 𝑡 is the unit vector tangent to the path and pointing in the direction of motion.

and 𝑇 is the particle’s kinetic energy, which is defined as Eq. (14.8), p. 862 𝑇 = 21 𝑚𝑣2 , where 𝑚 is the mass of the particle and 𝑣 is its speed. Work and kinetic energy have the following basic properties: • Work and kinetic energy are scalar quantities. • The work depends on only the component of 𝐹⃗ in the direction of motion, that is, on 𝐹⃗ ⋅ 𝑢̂ , and its sign is determined by the sign of 𝐹⃗ ⋅ 𝑢̂ . 𝑡

𝑡

• The kinetic energy is never negative.

Conservative forces and potential energy In Section 14.2, we discovered that there are forces whose work, in taking a particle from an initial to a final position, does not depend on the path connecting these positions. Specifically, we derived the following expressions:

Work of a constant gravitational force. For a particle of mass 𝑚 moving in a constant gravitational field from 1 to 2, the work done on the particle is Eq. (14.13), p. 877 ) ) ( ( 𝑈1-2 𝑔 = −𝑚𝑔 𝑦2 − 𝑦1 , for which gravity must act in the −𝑦 direction.

ISTUDY

Section 14.5

Chapter Review

927

Work done by a spring force. For a particle subject to the force of a linear elastic spring with constant 𝑘, the work done on the particle is Eq. (14.17), p. 878 ) ( ( ) 𝑈1-2 𝑒 = − 12 𝑘 𝛿22 − 𝛿12 , where 𝛿1 and 𝛿2 are the amount the spring is stretched (or compressed) at 1 and 2, respectively.

Conservative systems. A force is said to be conservative if its work depends on only the initial and final positions of its point of application, but is otherwise independent of the path connecting these positions. The work of a conservative force can be characterized by a scalar potential energy function. A conservative system is one for which all the forces doing work are conservative. For conservative systems, the work-energy principle becomes the conservation of mechanical energy, which is given by Eq. (14.22), p. 879 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 . Important conservative forces include elastic spring forces and gravitational forces (constant or not). The potential energy of a linear elastic spring force is given by Eq. (14.24), p. 879 𝑉𝑒 =

1 𝑘𝛿 2 . 2

The potential energy of a constant gravitational force is given by Eq. (14.23), p. 879 𝑉𝑔 = 𝑚𝑔𝑦, where the positive 𝑦 direction is opposite to gravity. The potential energy of the force of gravity is given by Eq. (14.25), p. 879 𝑉𝐺 = −

𝐺𝑚𝐴 𝑚𝐵 𝑟

.

In a system for which there are both conservative and nonconservative forces doing work, we can use Eq. (14.9) on p. 862 or we can take advantage of the potential energies of the conservative forces by writing the work-energy principle as Eq. (14.27), p. 880 ( ) 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 , ( ) where 𝑈1-2 nc is the work done by nonconservative forces.

Work-energy principle for systems of particles In Section 14.3, we developed the work-energy principle for a system of particles, and we discovered that internal forces play an important role in writing this principle. We defined

Concept Alert Elastic potential energy is never negative. Whether a spring is elongated or compressed, i.e., whether 𝛿 is positive or negative, the potential energy of a spring is never negative!

928

ISTUDY

Chapter 14

Energy Methods for Particles

the kinetic energy of a particle system as the sum of the individual kinetic energies of each particle, that is, Eq. (14.40), p. 901 𝑇 =

𝑛 ∑ 1 𝑖=1

2

𝑚𝑖 𝑣2𝑖 ,

where 𝑚𝑖 and 𝑣𝑖 are the mass and the speed of the 𝑖th particle in the system. The workenergy principle for a system of particles was found to be Eq. (14.45), p. 902

Common Pitfall Internal work due to nonconservative forces. Don’t forget that internal work due to nonconservative (forces )intneeds to be accounted for in the 𝑈1-2 nc term when we write the work-energy principle. The most important example of this type of work is due to internal friction.

)int )ext ( ( 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 , where 𝑉 is the potential energy of the system, which may have contributions from both external and internal forces. The kinetic energy of a system can also be expressed in terms of the speed of the mass center and the speed of each of the particles relative to the mass center. Specifically, we have Eq. (14.50), p. 903

Common Pitfall Internal work due to conservative forces. Don’t forget that internal work due to conservative forces (e.g., springs) is not included ( )int in the 𝑈1-2 nc term — it is accounted for by the potential energy 𝑉 , where, for a system, the potential energy 𝑉 includes the potential energies of both the external and the internal conservative forces.

𝑇 = 12 𝑚𝑣2𝐺 +

1 2

𝑛 ∑

𝑚𝑖 𝑣2𝑖∕𝐺 .

𝑖=1

Power and efficiency In Section 14.4, we defined the power 𝑃 developed by a force 𝐹⃗ whose point of application moves with a velocity 𝑣⃗ as Eq. (14.54), p. 919 𝑃 = 𝐹⃗ ⋅ 𝑣, ⃗ and we observed that power has units of work per unit time. In addition to power, Section 14.4 introduced the notion of efficiency, which is an important concept used to assess the performance of motors and/or machines. Given a machine or a motor whose operation requires a certain power input, efficiency, usually denoted by the Greek letter 𝜖 (epsilon), is defined as Eq. (14.57), p. 919 𝜖=

power output , power input

where power output is the work done by the machine or motor per unit time.

ISTUDY

Section 14.5

Chapter Review

929

Review Problems Problem 14.113 Rubber bumpers are commonly used in marine applications to keep boats and ships from getting damaged by docks. Consider a boat with a mass of 35,000 kg and a bumper with a force compression profile given by 𝐹𝐵 = 𝛽𝑥3 , where 𝛽 is a constant and 𝑥 is the compression of the bumper. Treating the boat 𝐶 as a particle, neglecting its vertical motion, and neglecting the drag force between the water and the boat 𝐶, determine 𝛽 so that if the boat were to impact the bumper with a speed of 4 m∕s, the maximum compression of the bumper would be 18 cm. 5

𝐶

𝐵

𝐹𝐵 ( × 105 lb)

4

3 2 1

0 0

0.1

0.2 0.3 𝑥 (f t)

0.4

0.5

Figure P14.113

Problems 14.114 through 14.116 The crates 𝐴 and 𝐵 of mass 𝑚𝐴 = 50 kg and 𝑚𝐵 = 75 kg, respectively, are connected by a pulley system. The system is released from rest on the inclined surfaces for which 𝜃 = 30◦ . Friction between 𝐴 and the inclined surface on which it slides is insufficient to prevent slipping, and friction between 𝐵 and the incline on which it slides is negligible. The cables in the pulley system are inextensible, and the coefficient of kinetic friction between crate 𝐴 and the horizontal surface is 𝜇𝑘 . Problem 14.114

𝜃

Letting 𝜇𝑘 = 0, determine the speed of 𝐵 after 𝐴 slides 5 m up the

incline. Figure P14.114–P14.116

Determine the required value of 𝜇𝑘 so that the speed of 𝐵 is 7 m∕s after it slides 5 m down the incline. Problem 14.115

Problem 14.116

Letting 𝜇𝑘 = 0.25, determine the speed of 𝐴 after 𝐵 slides 5 m down

the incline.

Problem 14.117 Strength of materials tells us that if a load 𝑃 is applied(at the )free end of a cantilevered beam, then the tip displacement 𝛿 is given by 𝛿 = 𝑃 𝐿3 ∕ 3𝐸𝐼cs , where 𝐿 is the length of the beam and 𝐸 and 𝐼cs are constants that depend on the material makeup and the geometry of the cross section, respectively. Determine an expression for the potential energy of a cantilevered beam loaded as shown. 𝑂 𝛿

𝑃 𝐿 Figure P14.117

𝐴

𝐵

𝜇𝑘

930

Chapter 14

Energy Methods for Particles

Problem 14.118 Starting from the position shown, each horse moves to the right in such a way that the tension in the cord is the same in cases (a) and (b) and remains constant. Knowing that 𝛽 < 𝛾, and that in both cases (a) and (b), the horse advances by an equal amount 𝐿, determine which of the following statements is true: (1) the tension in the cord does more work in (a) than in (b); (2) the tension in the cord does exactly the same amount of work in (a) as in (b); (3) the tension in the cord does less work in (a) than in (b). 𝐴

𝐴

𝛾

𝛽

(b)

(a) Figure P14.118

Problem 14.119 A metal ball with mass 𝑚 = 0.15 kg is released from rest in a fluid. The magnitude of the resistance due to the fluid is given by 𝐶𝑑 𝑣, where 𝐶𝑑 is a drag coefficient and 𝑣 is the ball’s speed. If 𝐶𝑑 = 2.1 kg∕s, determine the total work done on the ball from the moment of release until the ball achieves 99% of terminal velocity.

Problem 14.120

Figure P14.119 and P14.120

ISTUDY

A metal ball weighing 0.2 lb is released from rest in a fluid. If the magnitude of the resistance due to the fluid is given by 𝐶𝑑 𝑣, where 𝐶𝑑 = 0.5 lb⋅s∕f t is a drag coefficient and 𝑣 is the ball’s speed, determine the work done by the drag force during the first 2 s of the ball’s motion.

Problem 14.121 A 7 lb collar is constrained to travel along a frictionless vertical ring of radius 𝑅 = 1 f t. The spring attached to the collar has a spring constant 𝑘 = 20 lb∕f t. Treating the collar as a particle, neglecting air resistance, and knowing that, while at rest at 𝐴, the collar is displaced gently to the left, determine the spring’s unstretched length if the collar is to reach point 𝐵 with a speed of 15 f t∕s. 𝐴 𝑅

5𝑅∕3 𝐵

Figure P14.121

ISTUDY

Section 14.5

Chapter Review

𝐿∕2

Problem 14.122 An 11 kg collar is constrained to travel along a rectilinear and frictionless bar of length 𝐿 = 2 m that lies in the vertical plane. The springs attached to the collar are identical, and they are unstretched when the collar is at 𝐵. Treating the collar as a particle, neglecting air resistance, and knowing that at 𝐴 the collar is moving upward with a speed of 23 m∕s, determine the spring constant 𝑘 so that the collar reaches 𝐷 with zero speed. Points 𝐸 and 𝐹 are fixed.

𝐷

2m

𝐴

𝐵

35◦

𝐴

𝐿∕4

𝐿

𝑘

𝐴

𝐸 Figure P14.122

𝑘2

𝛿

4m

0.7 m

𝑘

𝐿∕2

Problem 14.123 Consider the catapult shown in the figure with a 1200 kg counterweight 𝐴 and a 330 kg projectile 𝐵. If the system is released from rest as shown, determine the speed of the projectile after the arm rotates (counterclockwise) through an angle of 110◦ . Model 𝐴 and 𝐵 as particles; assume that the catapult’s arm has negligible mass and that friction is negligible. In addition, assume that the cord has negligible mass, is inextensible, and is always vertical. The catapult’s frame is fixed with respect to the ground, and the projectile does not separate from the arm during the motion considered.

𝐹

𝑘1 𝐵

Figure P14.123

𝜃 Figure P14.124 and P14.125

Problems 14.124 and 14.125 The crate 𝐴 weighs 100 lb and is attached to the springs with constants 𝑘1 = 150 lb∕f t and 𝑘2 = 300 lb∕f t. Both springs are unstretched when 𝛿 = 0 and the box is centered between the two walls. The angle of the incline is 𝜃 = 30◦ . Problem 14.124 If the crate is released from rest after being displaced a distance 𝛿 = 6 f t up the incline, and friction between the crate and the inclined surface on which it slides is negligible, determine the speed of the crate when it returns to 𝛿 = 0. Problem 14.125 If the crate is released from rest after being displaced a distance 𝛿 = 6 f t and the coefficient of kinetic friction between the crate and the surface on which it slides is 𝜇𝑘 = 0.2, determine the speed of the crate when 𝛿 = 0.

Problem 14.126 Two blocks 𝐴 and 𝐵 weighing 123 and 234 lb, respectively, are released from rest as shown. At the moment of release the spring is unstretched. Model 𝐴 and 𝐵 as particles, neglect air resistance, and assume that the cord is inextensible. Determine the maximum speed attained by block 𝐵 and the distance from the floor where the maximum speed is achieved if 𝛼 = 35◦ , the static and kinetic friction coefficients are 𝜇𝑠 = 0.25 and 𝜇𝑘 = 0.2, respectively, and the spring constant is 𝑘 = 25 lb∕f t.

𝑘

𝐴 𝛼 𝐵 2 ft

Figure P14.126

931

932

Chapter 14

Energy Methods for Particles

Problem 14.127 Spring scales work by measuring the displacement of a spring that supports both the platform of mass 𝑚𝑝 and the object, of mass 𝑚, whose weight is being measured. In your solution, note that most scales read zero when no mass 𝑚 has been placed on them; that is, they are calibrated so that the weight reading neglects the mass of the platform 𝑚𝑝 . Assume that the spring is linear elastic with spring constant 𝑘. If the mass 𝑚 is gently placed on the spring scale (i.e., it is released from zero height above the scale), determine the maximum velocity attained by the mass 𝑚 as the spring compresses. ACME

𝑚

𝑚𝑝

Figure P14.127

Problem 14.128 A cyclist is riding with a constant speed 𝑣 = 25 km∕h over an inclined road with 𝜃 = 15◦ . The combined mass of the cyclist and bicycle is 85 kg. Neglecting aerodynamic drag, determine the number of calories (1 C = 4.184 kJ) burned by the cyclist in the course of 15 min if his muscle efficiency is 25%. 𝑣

𝑅 𝑑

𝐵

𝐴

ℎ

𝑘 𝜃 Figure P14.128

𝜃

𝐶 Figure P14.129

Problem 14.129

𝐹

𝐹

The mass 𝐴 is pressed against the linear elastic spring with stiffness 𝑘 = 500 N∕m such that the spring compresses a distance 𝛿 = 15 cm. The mass is then released in the position shown and slides a distance 𝑑 = 1.2 m up the incline. The mass leaves the incline at 𝐵 and becomes a projectile. The angle of the incline is 𝜃 = 30◦ , and the height ℎ = 0.9 m. Neglecting friction between 𝐴 and the incline while 𝐴 is on the incline, determine the range 𝑅 where the mass first lands on the horizontal surface 𝐶. 𝐴 has mass 𝑚 = 0.25 kg.

Problem 14.130 𝐵 (a) Figure P14.130

ISTUDY

𝐵 (b)

For the two pulley systems shown, the same block 𝐵, weighing 150 lb, is being lifted by applying the same constant force 𝐹 = 80 lb. If the two systems are released from rest, determine the power developed by the force 𝐹 in the two cases 1 s after release.

ISTUDY

15

Momentum Methods for Particles

The chapter begins by discussing the concepts of momentum and impulse. We will derive a new balance law, the impulse-momentum principle, that integrates Newton’s second law for a particle with respect to time. This balance law will be essential for the study of impacts. Next, we will study a new concept that turns out to be fundamental to mechanics— the concept of angular momentum. We will see not only that the concept of angular momentum is useful for solving problems like those found in orbital mechanics, but also that it is fundamental in extending the applicability of Newton’s laws to rigid bodies. Finally, we close with the study of mass flows, which is again a topic for which the idea of momentum will be essential.

NASA

The Space Shuttle Atlantis taking off from launch pad 39A at the Kennedy Space Center. The acceleration of the Shuttle due to the thrust of its rocket engines can be computed starting with the impulse-momentum principle.

15.1

Momentum and Impulse

Impulse-momentum principle We begin by looking at how a change in velocity is related to the time interval over which that change takes place. To do this, we integrate Newton’s second law with respect to time over a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 for a particle of mass 𝑚: ∫𝑡

𝑡2 1

𝐹⃗ 𝑑𝑡 =

∫𝑡

𝑡2

𝑚𝑎⃗ 𝑑𝑡.

(15.1)

1

Since 𝑎⃗ 𝑑𝑡 = 𝑑 𝑣⃗ (and 𝑚, for a particle, is constant), we obtain ∫𝑡

𝑡2

⃗ 1 ). 𝐹⃗ 𝑑𝑡 = 𝑚𝑣(𝑡 ⃗ 2 ) − 𝑚𝑣(𝑡

(15.2)

1

933

934

Chapter 15

Momentum Methods for Particles

𝑚

𝑚𝑣(𝑡 ⃗ 1)

𝐹⃗ (𝑡1 )

𝑚𝑣(𝑡 ⃗ 2)

𝐹⃗ (𝑡2 )

Equation (15.2) states that a change in the quantity 𝑚𝑣⃗ over a certain time interval is related to the time integral of the force over that time interval. To better understand the meaning of Eq. (15.2) and to see how it can help us deal with phenomena such as collisions, we will introduce some basic terminology. Linear momentum of a particle

path Figure 15.1 A force 𝐹⃗ (𝑡) on the mass 𝑚 acting over a time interval changes the quantity 𝑚𝑣. ⃗

Since the quantity 𝑚𝑣⃗ plays a prominent role in Eq. (15.2), we give this quantity its own name and symbol. Let the vector quantity 𝑝(𝑡) ⃗ be defined as 𝑝(𝑡) ⃗ = 𝑚𝑣(𝑡). ⃗

(15.3)

The quantity 𝑝(𝑡) ⃗ is called linear momentum or simply momentum of the particle of mass 𝑚. By definition, the momentum has dimensions of mass times length divided by time: (15.4) [𝑝⃗ ] = [𝑀] [𝐿] [𝑇 ]−1 . Consequently, the units in the SI system are kg⋅m/s and in the U.S. Customary system are lb⋅s or slug⋅f t/s. Linear impulse of a force The quantity on the left-hand side of Eq. (15.2), that is, ∫𝑡

𝐹 (𝑡)

𝑡2

𝐹 (𝑡) 𝑑𝑡

𝑡1

𝑡1

𝑡2

𝑡

Figure 15.2 Geometric interpretation of the concept of impulse of a force.

ISTUDY

𝑡2

𝐹⃗ (𝑡) 𝑑𝑡,

(15.5)

1

is called the linear impulse (or simply impulse) of the force 𝐹⃗ (𝑡) between times 𝑡1 and 𝑡2 . Because it is an integral over time, any component of the impulse can be given a graphical interpretation as the area under the curve of a force versus time plot, as shown in Fig. 15.2. The impulse has dimensions of force times time or mass times length divided by time. Therefore, the units of impulse are the same as for momentum. However, it is common to express the impulse using units of N⋅s in the SI system and lb⋅s in the U.S. Customary system. Impulse-momentum principle for a particle Using the definition of momentum, Eq. (15.2) can be written as ∫𝑡

𝑡2

⃗ 1 ) or 𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡

𝑝(𝑡 ⃗ 1) +

1

∫𝑡

𝑡2

𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ).

(15.6)

1

Equation (15.6) states that a particle’s change in momentum over an interval of time is equal to the impulse imparted to that particle during that same time interval. The expression in Eq. (15.6) is called the impulse-momentum principle. Equation (15.6) can be written in differential form as 𝐹⃗ = 𝑝, ⃗̇ which can be viewed as a restatement of Newton’s second law.

(15.7)

ISTUDY

Section 15.1

Average force

Momentum and Impulse

935

0

6

10

𝐹⃗avg =

∫𝑡

𝑡2

6 4 2 0

1

2

𝐹⃗ (𝑡) 𝑑𝑡

1

𝑡2 − 𝑡1

.

(15.8)

.

(15.9)

Therefore, using Eqs. (15.6), the average force is 𝐹⃗avg =

8 Force

Knowing the change of momentum of a particle of mass 𝑚 between two instants 𝑡1 and 𝑡2 is, in general, not enough to allow the reconstruction of the force 𝐹⃗ (𝑡) for every instant in time between 𝑡1 and 𝑡2 . However, we can determine the average force that caused the change of momentum in question. Figure 15.3 shows a time-varying force and its average over the same time interval. By definition, the average force over the time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 is

⃗ 1) 𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡 𝑡2 − 𝑡1

3 4 Time

5

Figure 15.3 A force (the black line) and its average (the red line) over a certain time interval. The area under the force curve (light brown) equals the area under the average force curve (dashed area).

The ability to calculate the average force needed to accomplish a change in momentum is useful. Since the magnitude of the actual force must be at least as large as the magnitude of the average force during some part of the time interval, the magnitude of the average force provides a lower bound for the magnitude of the maximum force. Impulse-momentum principle for systems of particles To extend the impulse-momentum principle to a system of particles, we need to distinguish between closed and open systems. Closed systems are systems that do not exchange mass with their surroundings, whereas open systems do exchange mass with their surroundings. The mass of a closed system is necessarily constant, whereas an open system can have constant or variable mass. The distinction between these two types of systems becomes important when we talk about systems of particles and, in particular, about mass flows and variable mass systems, which are open by definition (see Section 15.5). We consider a closed system of 𝑁 particles, and we view the total force on each particle as the sum of two parts (see Fig. 15.4): 1. an external force 𝐹⃗𝑖 due to the interaction of particle 𝑖 with the physical bodies that do not belong to the system, ∑ ⃗ 2. an internal force 𝑁 𝑗=1 𝑓𝑖𝑗 due to the interaction between particle 𝑖 and all of the other particles in the system (𝑓⃗ = 0⃗ since a particle does not exert a force on 𝑖𝑖

itself). Applying Eq. (15.7), we have 𝐹⃗𝑖 +

𝑁 ∑

𝑓⃗𝑖𝑗 = 𝑝⃗̇ 𝑖 ,

(15.10)

𝑗=1

where 𝑝⃗𝑖 = 𝑚𝑖 𝑣⃗𝑖 is the momentum of particle 𝑖. Equation (15.10) represents 𝑁 equations, one for each particle. Let 𝐹⃗ =

𝑁 ∑ 𝑖=1

𝐹⃗𝑖

and

𝑝⃗ =

𝑁 ∑ 𝑖=1

𝑝⃗𝑖 ,

(15.11)

𝑧

𝑣⃗1 𝑓⃗1 𝑖

𝑚1 𝐹⃗1 𝑟⃗1

𝑣⃗𝑖

𝑚𝑖 𝑓⃗1 2 𝑟⃗𝑖

𝑟⃗2 𝑂

𝐹⃗𝑖

𝑓⃗𝑖 1

𝑟⃗𝐺

𝐺

𝑓⃗𝑖 2 𝑓⃗2 𝑖

𝑓⃗2 1 𝑚2

𝑣⃗2 𝐹⃗2

𝑦

𝑥 Figure 15.4 A system of particles under the action of internal and external forces. The mass center is at 𝐺.

6

Chapter 15

Momentum Methods for Particles

be the total external force acting on and the total momentum of the(system of)particles, ∑ ∑𝑁 ⃗ ⃗ respectively. Recalling that Newton’s third law requires that 𝑁 𝑖=1 𝑗=1 𝑓𝑖𝑗 = 0, we see that the sum of all 𝑁 of Eq. (15.10), along with Eqs. (15.11), gives

Concept Alert

ternal forces do not change momentum. ternal forces cannot change the momenm of a system. This statement is a fundaental axiom of mechanics.

𝑤

(15.12)

We encountered this relationship in Eq. (13.37) on p. 838. Equation (15.12) states that internal forces play no role in changing the total momentum of a particle system. In addition, using the definition of center of mass of a closed system of particles, we have 𝑁 𝑁 ∑ ∑ 𝑝⃗𝑖 = 𝑚𝑖 𝑣⃗𝑖 = 𝑚𝑣⃗𝐺 , (15.13) 𝑝⃗ = 𝑖=1

𝑖=1

where 𝑚 is the total mass of the system and 𝑣⃗𝐺 is the velocity of the mass center of the system. Using Eq. (15.13), we can write Eq. (15.12) as ) 𝑑( 𝐹⃗ = 𝑚𝑣⃗𝐺 = 𝑚𝑎⃗𝐺 , 𝑑𝑡

(15.14)

where we have used the fact that the mass of a closed system of particles is constant. Equation (15.14) states that we can view a system of particles as a single particle, whose mass is equal to the total mass of the system of particles, moving with the mass center of the system of particles. Newton’s second law only applies to a single particle. Leonhard Euler postulated that Eq. (15.12) holds for any body whose mass is constant and for any internal force system. For this reason, Eq. (15.12) is generally considered to be a more fundamental axiom governing the motion of bodies than Newton’s second law, and it is referred to as Euler’s first law. As argued earlier in this section, to directly investigate the relationships between changes in momentum and the applied forces, we can integrate both sides of Eq. (15.12) with respect to time, which yields ∫𝑡

𝐹⃗5 𝑢̂ 𝑞 𝑞

FBD of the 5 ball FBD of both balls

ure 15.5 o billiard balls hitting one another, as well as FBD of the 5 ball and the FBD of both balls ing the impact. ISTUDY

𝐹⃗ = 𝑝. ⃗̇

𝑡2

𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡 ⃗ 1 ),

(15.15)

1

where the left-hand side of Eq. (15.15) is the total external impulse exerted on the system between 𝑡1 and 𝑡2 . Even though Eqs. (15.6) and (15.15) are visually identical to one another, it is important to keep in mind that the definitions of 𝐹⃗ and 𝑝⃗ are different in the two equations.

Conservation of linear momentum Figure 15.5 shows two billiard balls hitting one another, as well as the FBD of the 5 ball during impact and the FBD of both balls during impact. Let 𝑞 be the line tangent to the impact surface and parallel to the billiard table. Let 𝑤 be the line perpendicular to the impact surface. If we ignore friction between the two surfaces during impact, the only force in the 𝑞𝑤 plane acting on the 5 ball is directed along the 𝑤 axis, and there is no net external force on the two balls taken together. If the impact begins at time 𝑡1 and ends at time 𝑡2 , and if we define our system to consist of both balls, then during the interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 , the FBD of both balls indicates that ⃗ 𝐹⃗ext (𝑡) = 0,

(15.16)

ISTUDY

Section 15.1

Momentum and Impulse

937

and therefore, by Eq. (15.15), the total momentum of the two-ball system is conserved between 𝑡1 and 𝑡2 , ⃗ 2) 𝑝(𝑡 ⃗ 1 ) = 𝑝(𝑡

⇒

𝑚5 𝑣⃗5 (𝑡1 ) + 𝑚12 𝑣⃗12 (𝑡1 ) = 𝑚5 𝑣⃗5 (𝑡2 ) + 𝑚12 𝑣⃗12 (𝑡2 ),

(15.17)

where we have used Eq. (15.13) and the numeric subscript indicates the ball. In Eq. (15.17), the impulse-momentum principle has become conservation of momentum (we saw conservation of energy in Section 14.1). The two billiard balls are an example of an isolated system, which is a system that does not exchange mass with its surroundings and which is not acted upon by external forces. Momentum is always conserved for isolated systems. In other applications, including impact problems like the billiard balls, it is common to have a fixed direction in which the force is zero. For example, if we now define the system to consist of only the 5 ball, for 𝑡1 ≤ 𝑡 ≤ 𝑡2 , the upper FBD in Fig. 15.5 indicates that 𝐹⃗5 (𝑡) ⋅ 𝑢̂ 𝑞 = 0 ⇒ 𝐹5𝑞 = 0. (15.18) Since the 𝑞 component of the force on the 5 ball is zero, the 𝑞 component of Eq. (15.6) indicates that the 𝑞 component of momentum is conserved for the 5 ball, that is, 𝑝5𝑞 (𝑡1 ) = 𝑝5𝑞 (𝑡2 ) ⇒ 𝑚5 𝑣5𝑞 (𝑡1 ) = 𝑚5 𝑣5𝑞 (𝑡2 ) ⇒ 𝑣5𝑞 (𝑡1 ) = 𝑣5𝑞 (𝑡2 ),

(15.19)

during the impact. Similarly, the momentum and velocity of the 12 ball are conserved in the 𝑞 direction. The first of Eqs. (15.19) states that, as long as the total external force has a component that is zero throughout the time interval, the corresponding component of the momentum remains constant. This holds for systems, as well as for individual particles.

Concept Alert Motion of a system with zero external forces. Since a system’s total momentum can be expressed as the product of the system’s total mass and the velocity of the system’s center of mass, we can conclude that the center of mass of a system with zero external forces is either at rest or moving with a constant velocity.

End of Section Summary We learned in Chapter 13 that forces lead to changes in velocities since forces cause accelerations. We began this section by learning that forces acting over time change momentum (not just velocity). By integrating Newton’s second law, we obtained the impulse-momentum principle, which is given by Eq. (15.6), p. 934 𝑝(𝑡 ⃗ 1) +

∫𝑡

𝑡2

𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ),

1

where the linear momentum (or momentum) was defined to be Eq. (15.3), p. 934 𝑝(𝑡) ⃗ = 𝑚𝑣(𝑡), ⃗ and a force acting over some time interval was called the impulse (or linear impulse) and is given by Eq. (15.5), p. 934 ∫𝑡

𝑡2 1

𝐹⃗ (𝑡) 𝑑𝑡.

Helpful Information When should we use the impulse-momentum principle? The impulse-momentum principle provides a natural approach to problems in which we need to relate velocity, force, and time since it relates forces acting over time to changes in momentum.

938

ISTUDY

Chapter 15

Momentum Methods for Particles

In addition, we found that without detailed knowledge of the force acting on a particle at every instant in time, we could not determine the change in momentum. On the other hand, knowing just the change in momentum allows us to determine the average force acting on a particle during the corresponding time interval, that is, Eq. (15.9), p. 935 𝐹⃗avg =

𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡 ⃗ 1) 𝑡2 − 𝑡1

.

Impulse-momentum principle for systems of particles. When dealing with closed systems of particles, we discovered that we can write the impulse-momentum principle as Eqs. (15.12) and (15.15), p. 936 𝐹⃗ = 𝑝⃗̇ and

∫𝑡

𝑡2

⃗ 1 ), 𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡

1

∑ where 𝐹⃗ is the total external force on the particle system and 𝑝⃗ = 𝑁 ⃗𝑖 is the 𝑖=1 𝑚𝑖 𝑣 total momentum of the system of particles. Using the definition of the mass center of a system of particles, the impulse-momentum principle can also be written as Eq. (15.14), p. 936 ) 𝑑( 𝐹⃗ = 𝑚𝑣⃗𝐺 = 𝑚𝑎⃗𝐺 , 𝑑𝑡 where 𝑚 is the total mass of the system of particles, 𝑣⃗𝐺 is the velocity of its mass center, and 𝑎⃗𝐺 is the acceleration of its mass center. Conservation of linear momentum. When there is a direction in which the external force on a system of particles is zero, then the momentum in that direction is constant and is said to be conserved. If the total external force on a system of particles is zero, that is, 𝐹⃗ = 0, then the momentum in every direction is constant, and the mass center of the system of particles will move with constant velocity.

ISTUDY

Section 15.1

Momentum and Impulse

E X A M P L E 15.1

939

Average Force on an Aircraft Due to the Arresting Cable

To successfully land on an aircraft carrier, a pilot must use the airplane’s tailhook to catch one of four steel arresting cables stretched across the landing deck (see Fig. 1). The hydraulic system activated by the motion of the arresting cable can stop a 54,000 lb aircraft traveling at a speed of 150 mph in a little less than 2 s. Using this information, estimate the average braking force exerted on the plane by the arresting cable.

SOLUTION Road Map & Modeling

We model the plane as a particle, and we will consider the time between when the tailhook first engages one of the arresting cables 𝑡𝑒 and the time at which the plane comes to a stop 𝑡𝑠 . Referring to Fig. 2, we only include the force applied by the arresting cable, gravity, and the contact force between the plane and the landing deck. Since we are given a change of velocity over a known time interval and we need to estimate the value of a force, we can apply the impulse-momentum principle. Since we want to determine the average braking force, which is the force that would act on the plane if this force were constant, we will apply the impulse-momentum principle under the assumption that the force exerted on the plane by the arresting cable is constant.

PH3 Christopher Mobley/U.S. Navy

Figure 1 Military airplane landing on an aircraft carrier with a detailed view of the arresting cable.

Governing Equations Balance Principles

Applying Eq. (15.6) in component form to the FBD in Fig. 2, we

obtain

𝑚𝑔 𝚥̂

∫𝑡

𝑥 momentum ∶

𝚤̂

𝑡𝑠

𝐹 (𝑡) 𝑑𝑡 = 𝑝𝑥 (𝑡𝑠 ) − 𝑝𝑥 (𝑡𝑒 ),

(1)

(𝑁 − 𝑚𝑔) 𝑑𝑡 = 𝑝𝑦 (𝑡𝑠 ) − 𝑝𝑦 (𝑡𝑒 ),

(2)

𝐹 𝑁

𝑒

𝑦 momentum ∶

∫𝑡

𝑡𝑠

𝑒

Figure 2 FBD of an airplane landing on a carrier. The force 𝐹 is the force exerted by the tailhook.

where 𝑝𝑥 = 𝑚𝑣𝑥 and 𝑝𝑦 = 𝑚𝑣𝑦 . Force Laws

All forces that we are modeling are accounted for on the FBD.

Kinematic Equations

Since we are assuming that the plane only moves in a straight line parallel to the landing deck, Fig. 2 tells us that the motion is only along the 𝑥 axis with 𝑣𝑥 (𝑡𝑒 ) = 150 mph = 220.0 f t∕s. Since the plane comes to a stop at time 𝑡𝑠 , we have 𝑣𝑥 (𝑡𝑒 ) = 220.0 f t∕s,

𝑣𝑥 (𝑡𝑠 ) = 0 f t∕s,

𝑣𝑦 (𝑡𝑒 ) = 0 f t∕s,

𝑣𝑦 (𝑡𝑠 ) = 0 f t∕s.

(3)

The last two of Eqs. (3) tell us there is no motion in the 𝑦 direction, so Eq. (2) tells us that the force 𝑁 and 𝑚𝑔 cancel each other throughout the time interval of interest. Hence, treating 𝐹 as a constant, Eqs. (1) and (3) give Computation

𝐹 (𝑡𝑠 − 𝑡𝑒 ) = 𝑚𝑣𝑥 (𝑡𝑒 )

⇒

𝐹 =

𝑚𝑣𝑥 (𝑡𝑒 ) 𝑡𝑠 − 𝑡𝑒

= 184,500 lb.

(4)

Discussion & Verification The acceleration of the airplane is 𝑎𝑥 = 𝐹 ∕𝑚 = 110.0 f t∕s2 = 3.416𝑔. Carrier landings are usually considered to be maneuvers with roughly 3𝑔. Since we have computed an average force value, the plane will have accelerations in excess of 3.416𝑔. However, these higher accelerations would not be sustained for the entire duration of the maneuver (i.e., 2 s).

Interesting Fact Carrier landing systems. The arresting cables are about 1.5 in. in diameter and are stretched 2–5 in. above the landing deck at 35–40 ft intervals. When one of the cables is caught by an airplane’s tailhook, the cable’s motion results in the activation of a hydraulic system housed below the deck, whose task is to dissipate the plane’s kinetic energy. On carriers, planes land at speeds between 130 and 150 mph, and they are brought to a halt in roughly 300 ft. The limit of human tolerance for sustained acceleration (for more than 2 s) is approximately 8𝑔.

940

Chapter 15

Momentum Methods for Particles

E X A M P L E 15.2

Impulse-Momentum Applied to a Racquetball Hitting a Wall Figure 1 shows a 1.4 oz racquetball hitting a wall at 85 mph with an angle 𝜃1 = 65◦ . The duration of the impact is 2.5 ms, and the rebound speed is 72.5 mph with the rebound angle 𝜃2 = 61.9◦ . Determine the change in momentum, the impulse, and the average force applied to the ball during the impact.

𝑣1

𝜃1

SOLUTION Road Map & Modeling

𝜃2 𝑣2 Figure 1 Racquetball approaching a wall at speed 𝑣1 and then rebounding at speed 𝑣2 .

Governing Equations

𝐹 (𝑡)

𝚤̂

We are given the mass of the ball and its pre- and postimpact velocities, so we can readily compute its pre- and postimpact momenta, 𝑝⃗1 and 𝑝⃗2 , respectively. This will allow us to compute the change in momentum and impulse using Eq. (15.6) and the average force using Eq. (15.9). We model the ball as a particle, and we will ignore the effects of gravity during the 2.5 ms impact. The FBD of the ball during the impact in Fig. 2 reflects this model. We have broken the contact force into a normal force 𝑁 and a friction force 𝐹 , both of which are a function of time.

Balance Principles

Applying Eq. (15.6), the impulse-momentum principle, to the ball during impact, we obtain

𝚥̂

Δ𝑝⃗ =

𝑁(𝑡)

∫𝑡

𝑡2

𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝⃗2 − 𝑝⃗1 ,

(1)

1

where Δ𝑝⃗ is the change in momentum, ∫𝑡 2 𝐹⃗ (𝑡) 𝑑𝑡 is the impulse applied to the racquet1 ball, 𝑝⃗1 = 𝑚𝑣⃗1 is the momentum just before impact, and 𝑝⃗2 = 𝑚𝑣⃗2 is the momentum just after impact. 𝑡

Figure 2 FBD of the racquetball during its impact with the wall.

Force Laws

All forces are accounted for on the FBD.

Kinematic Equations

From Fig. 1, we can see the pre- and postimpact velocities of the

racquetball are ( ) 𝑣⃗1 = 𝑣1 − sin 𝜃1 𝚤̂ + cos 𝜃1 𝚥̂ , ( ) 𝑣⃗2 = 𝑣2 sin 𝜃2 𝚤̂ + cos 𝜃2 𝚥̂ .

(2) (3)

Computation

Substituting Eqs. (2) and (3) into Eq. (1) and evaluating the result using the given quantities, which, when converted, are 𝑚 = 0.002717 slug, 𝑣1 = 124.7 f t∕s, and 𝑣2 = 106.3 f t∕s, we obtain the impulse and change in momentum as ) ] ) ( [( 𝑝⃗2 − 𝑝⃗1 = 𝑚 𝑣2 sin 𝜃2 + 𝑣1 sin 𝜃1 𝚤̂ + 𝑣2 cos 𝜃2 − 𝑣1 cos 𝜃1 𝚥̂ ( ) = 0.5619 𝚤̂ − 0.007071 𝚥̂ lb⋅s.

(4)

The average force is found by applying Eq. (15.9) to obtain 0.7210◦ −𝑥 𝑦

direction of the impulse and 𝐹⃗avg

Figure 3 The direction of the change in momentum, impulse, and average force acting on the racquetball during its impact with the wall. This vector deviates from the 𝑥 axis by less than 1◦ .

ISTUDY

𝐹⃗avg = Discussion & Verification

𝑝⃗2 − 𝑝⃗1 𝑡2 − 𝑡1

( ) = 224.8 𝚤̂ − 2.828 𝚥̂ lb.

(5)

Figure 3 shows the direction of the change in momentum of the ball. This is also the direction of the impulse acting on the ball and of the average force on the ball during its impact with the wall. Equation (4) tells us that most of the change in momentum occurs in the 𝑥 direction. Since the momentum changes very little in the direction parallel to the wall, the frictionless impact assumption we often use for impacts is a good one. The magnitude of the average force is 224.8 lb, which seems about right if you have ever been hit by a racquetball.

ISTUDY

Section 15.1

Momentum and Impulse

E X A M P L E 15.3

941

A Person Pulling a Commercial Airliner

An event at the World’s Strongest Man (WSM) competition involves the timed pull of a Boeing 737 airliner over a distance of 82 f t. Given that a Boeing 737 weighs about 75 ton, that the competitor starts pulling with a force of 850 lb, that his strength decreases linearly by 30% during the course of the pull, and that he completes the pull in 40 s, determine how fast he is moving at the end of the pull.

SOLUTION

𝐵

𝐴

Figure 1 A man at 𝐴 pulling a Boeing 737 airliner using both his arms (with the rope 𝐴𝐵) and his legs.

Road Map & Modeling

We are given enough information to determine the force as a function of time with which the competitor pulls. We know that the system starts from rest and that we want to find its final speed after a given amount of time, so this problem lends itself to the application of the impulse-momentum principle. We neglect the rolling resistance of the airplane and the mass of the competitor, treat the plane as a particle, and assume that the force with which the person pulls the plane is parallel to the motion. Using these assumptions, the FBD of the plane is shown in Fig. 2.

𝚥̂ 𝚤̂

𝑚𝑔

𝐹𝑃 𝑁

Governing Equations Balance Principles Applying the impulse-momentum principle in the 𝑥 direction to the FBD in Fig. 2 from the time that the person begins pulling until he pulls for 40 s gives

𝑥 momentum ∶ 𝑝1 +

∫𝑡

𝑡2

𝐹𝑃 𝑑𝑡 = 𝑝2 ,

(1)

Figure 2 FBD of the 737 as it is being pulled by the WSM competitor.

1

Force Laws To derive the force law 𝐹𝑃 (𝑡) for the WSM competitor, we note that he starts pulling with a force of 850 lb and it decreases linearly by 30% over the course of the pull. Therefore, the force versus time curve must be as shown in Fig. 3, and the equation for the curve is 0.3(850) 𝐹𝑃 = − 𝑡 + 850 = (−6.375𝑡 + 850) lb. (2) 40 Kinematic Equations The kinematic equation for this problem is that the competitor starts from rest, so 𝑣1 = 0. (3) Computation Substituting Eqs. (2) and (3) into Eq. (1), integrating, and using 𝑚 = 4658 slug, we obtain the final speed 𝑣2 :

∫0

40

(−6.375𝑡 + 850) 𝑑𝑡 = 𝑚𝑣2

Discussion & Verification

⇒

𝑣2 = 6.204 f t∕s = 4.230 mph.

(4)

The competitor and the plane are both moving at a little over 4 mph at the end of the 82 f t pull. This is in the range of typical walking speeds, so it is not an unreasonable result. Thus, our choice of 𝐹𝑃 in Eq. (2) seems appropriate. Note that a 75 ton Boeing 737 would have substantial rolling resistance, which we have ignored. It is also noted that a person really did pull a 75 ton Boeing 737 a distance of 82 f t in 40 s during a WSM competition. Therefore, the initial force generated by the WSM competitor must have actually been larger than the 850 lb that we have estimated.

Helpful Information Why have we ignored the mass of the competitor? To include the mass of the competitor in our solution, we would have to realize that the force 𝐹𝑃 with which the competitor is pulling the airplane is also pulling him. Of course, adding in a 300 lb strongman barely changes the result since he is only 0.2% of the weight of the airplane.

850 𝐹𝑃 (lb)

where 𝑡1 = 0 s is the time at which he starts pulling, 𝑡2 = 40 s is the time at which he stops pulling, 𝑝1 = 𝑚𝑣1 is the momentum of the plane at time 𝑡1 , and 𝑝2 = 𝑚𝑣2 is the momentum of the plane at time 𝑡2 .

750 650 550 0

10

20 Time (s)

30

40

Figure 3 The force 𝐹𝑃 versus time curve for the WSM competitor.

942

Chapter 15

Momentum Methods for Particles

E X A M P L E 15.4

Walking on a Floating Platform: Conservation of Momentum

platform

pier 𝐵

𝐴

A man of mass 𝑚𝑝 is at end 𝐴 of a floating platform of mass 𝑚fp and length 𝐿fp . The man and the platform are initially at rest. The platform is touching the pier (as shown in Fig. 1) when the man starts moving toward 𝐵 with a constant speed 𝑣0 , relative to the platform. Determine the distance between the platform and the pier when the man reaches the other end of the platform at 𝐵.

SOLUTION Road Map & Modeling

We model both the man and the platform as particles and ignore any vertical motion and the drag force between the platform and the water. These assumptions imply the FBD shown in Fig. 2, in which 𝐹𝑏 is the buoyancy force exerted by the water on the platform. We will let 𝑡𝐴 and 𝑡𝐵 be the times at which the man is at 𝐴 and 𝐵, respectively, whereas 𝑡 will denote a generic time instant.

𝐿fp Figure 1 Man on a floating platform. 𝑚𝑝 𝑔

𝚥̂ 𝚤̂ 𝑚fp 𝑔

Governing Equations Balance Principles The FBD in Fig. 2 shows no external forces in the 𝑥 direction. This implies that the linear momentum in the 𝑥 direction is conserved (see discussion starting on p. 936). Using Eq. (15.13) on p. 936, we express this fact by writing

𝑚𝑣𝐺𝑥 (𝑡𝐴 ) = 𝑚𝑣𝐺𝑥 (𝑡)

𝑣𝐺𝑥 (𝑡𝐴 ) = 𝑣𝐺𝑥 (𝑡),

⇒

where 𝑚 = 𝑚𝑝 + 𝑚fp is the total mass of the system and 𝑡 ≥ 𝑡𝐴 .

𝐹𝑏

Force Laws

Figure 2 FBD of the system, i.e., the man + platform. The force 𝐹𝑏 is the buoyancy force.

All forces are accounted for on the FBD.

Kinematic Equations

Both the man and platform start from rest, so 𝑣𝐺𝑥 (𝑡𝐴 ) = 0.

Computation

(1)

(2)

Substituting Eq. (2) into the last of Eqs. (1), we obtain

𝑥𝑝

𝑣𝐺𝑥 (𝑡) = 0.

(3)

Let 𝑥𝐺 denote the 𝑥 coordinate of the system’s center of mass. Since 𝑣𝐺𝑥 = 𝑥̇ 𝐺 , Eq. (3) implies that 𝑥𝐺 is constant, so 𝑥𝐺 (𝑡𝐴 ) = 𝑥𝐺 (𝑡𝐵 ). Referring to Fig. 3, we then have

𝑥 𝑥fp

) 1( ) 1( 𝑚 𝑥 + 𝑚𝑝 𝑥𝑝𝐴 = 𝑚 𝑥 + 𝑚𝑝 𝑥𝑝𝐵 , 𝑚 fp fp𝐴 𝑚 fp fp𝐵

(4)

where (𝑚fp 𝑥fp + 𝑚𝑝 𝑥𝑝 )∕𝑚 = 𝑥𝐺 , and where 𝐴 and 𝐵 in the subscripts stand for “at 𝑡𝐴 ” and “at 𝑡𝐵 ,” respectively. We now observe that 𝑥fp and 𝑥𝑝 are such that Figure 3 Kinematics diagram showing the origin of the 𝑥 axis as well as the definition of the positions of the man and the platform.

ISTUDY

𝑥fp𝐴 = 0,

𝑥𝑝𝐴 = 𝐿fp ,

and 𝑥𝑝𝐵 = 𝑥fp𝐵 .

(5)

Substituting Eqs. (5) in Eq. (4), we obtain an equation in 𝑥fp whose solution is ( 𝑥fp𝐵 =

𝑚𝑝 𝑚𝑝 + 𝑚fp

) 𝐿fp .

(6)

Discussion & Verification

The final result has dimensions of length, as it should. Equation (6) tells us that the greater 𝑚𝑝 , the farther from the pier the platform will end up. In addition, we observe that if something with very little mass (e.g., a flea) walks across the platform, the platform will barely move. Probably both of these observations agree with your experience. A Closer Look The answer is independent of 𝑣0 , that is, regardless of how fast the man walks, the platform will always end up at the same final location.

ISTUDY

Section 15.1

943

Momentum and Impulse

Problems Problem 15.1 An airplane performs a turn at constant speed and elevation so as to change its course by 180◦ . Let 𝐴 and 𝐵 designate the beginning and endpoints of the turn. Assuming that the change in mass of the plane due to fuel consumption is negligible, is the airplane’s momentum at 𝐴 different from the airplane’s momentum at 𝐵? In addition, again neglecting the change in mass between 𝐴 and 𝐵, is the total work done on the plane between 𝐴 and 𝐵 positive, negative, or equal to zero?

𝐴

𝐵

Figure P15.1

Problem 15.2 In a simple force-controlled experiment, two curling stones 𝐴 and 𝐵 are made to slide over a sheet of ice. Initially, 𝐴 and 𝐵 are at rest on the start line. Then they are acted upon by identical and constant forces 𝐹⃗ , which continually push 𝐴 and 𝐵 all the way to the finish line. Let 𝑝⃗𝐴𝑓 and 𝑝⃗𝐵𝑓 denote the momentum of 𝐴 and 𝐵 at the finish line, respectively. Assume that the forces 𝐹⃗ are the only nonnegligible forces acting in the plane of motion. If 𝑚𝐴 < 𝑚𝐵 , which of the following statements is true?

finish

𝑑

(a) |𝑝⃗𝐴𝑓 | < |𝑝⃗𝐵𝑓 |.

𝐴

(b) |𝑝⃗𝐴𝑓 | = |𝑝⃗𝐵𝑓 |.

𝐵

start

(c) |𝑝⃗𝐴𝑓 | > |𝑝⃗𝐵𝑓 |. 𝐹

(d) There is not enough information given to make a comparison between |𝑝⃗𝐴𝑓 | and |𝑝⃗𝐵𝑓 |. Figure P15.2

Problem 15.3 A train is moving at a constant speed 𝑣𝑡 relative to the ground, when a person who is initially at rest (relative to the train) starts running and gains a speed 𝑣0 (relative to the train) after a time interval Δ𝑡. Had the person started from rest on the ground (as opposed to on the moving train), would the magnitude of the total impulse exerted on the person during Δ𝑡 be smaller than, equal to, or larger than the impulse needed to cause the same change in relative velocity in the same amount of time on the moving train? Assume that the person always moves in the direction of motion of the train.

𝑣0

𝑣𝑡

Figure P15.3 and P15.4

Problem 15.4 A train is decelerating at a constant rate, when a person who is initially at rest (relative to the train) starts running and gains a speed 𝑣0 (again relative to the train) after a time interval Δ𝑡. Had the person started from rest on the ground (as opposed to on the moving train), would the magnitude of the total impulse exerted on the person during Δ𝑡 be smaller than, equal to, or larger than the impulse needed to cause the same change in velocity in the same amount of time on the moving train? Assume that the person always moves in the direction of motion of the train and that the train does not reverse its motion during the time interval Δ𝑡.

𝐹

944

Chapter 15

Momentum Methods for Particles

Problem 15.5 A car of mass 𝑚 collides head-on with a truck of mass 50𝑚. What is the ratio between the magnitude of the impulse provided by the car to the truck and the magnitude of the impulse provided by the truck to the car during the collision?

Problem 15.6 The spacecraft shown is out in space and is far enough from any other mass (e.g., planets, etc.) so as not to be affected by any gravitational influence (i.e., the net external force on the rocket is approximately zero). The system (i.e., the spacecraft and all its fuel) is at rest when it starts at 𝐴, and it thrusts all the way to 𝐵 along the straight line shown using internal chemical rockets (which work by ejecting the fuel mass at very high speeds out the tail of the rocket). We are given that the mass of the system at 𝐴 is 𝑚 and that it has ejected half of its mass in thrusting from 𝐴 to 𝐵. What will be the location of the system’s mass center when the spacecraft reaches 𝐵? 𝐴

𝐵 𝐿 𝐹 = −0.3𝑡(𝑡 − 2) Figure P15.6

𝐹 (lb)

3 2

𝐹 = 0.3(1 − 𝑒−2𝑡 )

1

Problem 15.7

𝐹 =1−𝑡 0.5

−1

1.0 𝑡 (s)

Figure P15.7

1.5

2.0

Use the definition of impulse given in Eq. (15.5) to compute the impulse of the forces shown during the interval 0 ≤ 𝑡 ≤ 2 s.

Problem 15.8 The mass of the Earth is 𝑚𝑒 = 5.9736 × 1024 kg. Modeling the Earth (with everything in and on it) as an isolated system and assuming that the center of the Earth is also the center of mass of the Earth, determine the displacement of the center of the Earth due to (a) a 2 m jump off the surface by an 85 kg person; (b) the Space Shuttle, with a mass of 124,000 kg, reaching an orbit of 200 km; (c) 170,000 km3 of water being elevated 50 m (these numbers are estimates based on publicly available information about the Aswan Dam at the border between Egypt and Sudan). Use 1 g∕cm3 for the density of water.

Problems 15.9 and 15.10 A 200 lb skydiver deploys the parachute after 10 s of free fall and concludes the jump by touching down with a speed of 15 f t∕s. Model the motion of the skydiver as being a vertical drop from rest.

𝚥̂

Problem 15.9 Neglecting the change in acceleration due to gravity with elevation, determine the impulse provided by gravity to the skydiver during free fall. 𝑔 Problem 15.10 Figure P15.9 and P15.10

ISTUDY

Determine the impulse provided by all the forces acting on the skydiver from the beginning of the jump to touchdown.

ISTUDY

Section 15.1

Momentum and Impulse

Problem 15.11 Consider an elevator that moves with an operating speed of 2.5 m∕s. Suppose that a person who boards the elevator on the ground floor gets off on the fifth floor. Assuming that the elevator has achieved operating speed by the time it reaches the second floor and that it is still moving at its operating speed as it passes the fourth floor, determine the momentum change of a person with a mass of 80 kg between the second and fourth floors if each floor is 4 m high. In addition, determine the impulse of the person’s weight during the same time interval.

Problem 15.12

iurii/Shutterstock

A 180 gr (7000 gr = 1 lb) bullet goes from rest to 3300 f t∕s in 0.0011 s. Determine the magnitude of the impulse imparted to the bullet during the given time interval. In addition, determine the magnitude of the average force acting on the bullet.

Figure P15.11

𝜃 Figure P15.12

Figure P15.13

Problem 15.13 A 3400 lb car is parked as shown. Determine the impulse of the normal reaction force acting on the car during the span of an hour if 𝜃 = 15◦ .

Problems 15.14 and 15.15 A 3850 lb sports car (driver’s weight included), driving along a horizontal rectilinear stretch of road, goes from 0 to 62 mph in 4.2 s. 𝑣

Figure P15.14 and P15.15 Problem 15.14

Determine the magnitude of the average force that needs to be applied to the car for such an acceleration to occur. ( ) Problem 15.15 If the magnitude of the force propelling the car has the form 𝐹0 1−𝑒−𝑡∕𝜏 , with 𝜏 = 0.5 s, determine 𝐹0 . 𝑃

Problem 15.16 A 75 lb crate is initially at rest when a force 𝑃 = 40 lb is applied to the pulley system as shown. Use the impulse-momentum principle to determine the speed of the crate after 2 s. Neglect the inertia of the rope and of the pulleys, and assume that all the cable segments are purely vertical.

𝐴 Figure P15.16

945

946

Chapter 15

Momentum Methods for Particles Problems 15.17 and 15.18

The pulley system shown is at rest when the motor 𝑀 starts pulling in rope so as to lift the 120 lb cargo 𝐶. For the first second of operation, the motor can produce a tension in the rope of the form 𝐹0 (1 + 𝑡∕𝜏), where 𝐹0 = 40 lb. Neglect the inertia of the pulleys and of the rope.

𝑀

If 𝜏 = 0.5 s, use the impulse-momentum principle to determine the speed of the crate after 1 s.

Problem 15.17

Problem 15.18 Use the impulse-momentum principle to find the value of 𝜏 needed for the cargo 𝐶 to travel upward with a speed of 10 f t∕s after 1 s.

𝐶

Problem 15.19 Figure P15.17 and P15.18

A box comes off of a conveyor belt with a speed 𝑣0 = 3 m∕s and then slides over a lowfriction surface. Determine the coefficient of kinetic friction between the box and floor if the box slides 1.5 s before coming to a stop. 𝑣0

4 mph 𝑥

Figure P15.19

Figure P15.20

Problem 15.20 A 60 ton railcar and its cargo, a 27 ton trailer, are moving to the right at 4 mph when they come into contact with a bumper that can bring the system to a stop in 0.78 s. Determine the magnitude of the average force exerted on the railcar by the bumper.

Problem 15.21 A 30,000 lb airplane is flying on a horizontal trajectory with a speed 𝑣0 = 650 mph when, at point 𝐴, it maneuvers so that at point 𝐵 it is on a steady climb with 𝜃 = 40◦ and a speed of 600 mph. If the change in mass of the plane between 𝐴 and 𝐵 is negligible, determine the impulse that had to be exerted on the plane in going from 𝐴 to 𝐵. 𝜃 𝑣0

𝐵

𝑦

𝜃

𝐴 𝑣⃗0

Figure P15.21

Figure P15.22

𝚥̂ 𝑂

𝚤̂

𝑥

Problem 15.22 A crate starts sliding from rest down an incline with 𝜃 = 35◦ . Determine the speed of the crate after 2.5 s if the coefficient of kinetic friction between the crate and the incline is 𝜇𝑘 = 0.25. Express the result in SI units.

Problem 15.23

Figure P15.23

ISTUDY

A trebuchet launches a projectile with an initial velocity 𝑣⃗0 such that the projectile takes 3 s to achieve its maximum height with a corresponding speed of 145 f t∕s. Use the impulsemomentum principle to determine 𝑣⃗0 .

ISTUDY

Section 15.1

947

Momentum and Impulse

Problem 15.24 A 15 oz football is kicked straight up in the air such that it takes 2.7 s (after separating from the kicker’s foot) to achieve its maximum height. Determine the impulse imparted by the kicker to the football, assuming that right before the kick the football is held stationary and that the weight of the football can be neglected while it is in contact with the foot. Also, if the contact between the football and the foot lasts 8 × 10−3 s, determine the average force exerted by the kicker on the football.

𝚥̂

Problems 15.25 and 15.26 The takeoff runway on carriers is much too short for a modern jetplane to take off on its own. For this reason, the takeoff of carrier planes is assisted by hydraulic catapults (Fig. A). The catapult system is housed below the deck except for a relatively small shuttle that slides along a rail in the middle of the runway (Fig. B). The front landing gear of carrier planes is equipped with a tow bar that, at takeoff, is attached to the catapult shuttle (Fig. C). When the catapult is activated, the shuttle pulls the airplane along the runway and helps the plane reach its takeoff speed. The takeoff runway is approximately 300 f t long, and most modern carriers have three or four catapults.

U.S. Navy photo by Photographer’s Mate 2nd Class H. Dwain Willis

A

PHAN James Farrally II, U.S. Navy

Figure P15.24

U.S. Navy photo by Photographer’s Mate 3rd Class (AW) J. Scott Campbell

B

C

Figure P15.25 and P15.26 Problem 15.25 In a catapult-assisted takeoff, assume that a 45,000 lb plane goes from 0 to 165 mph in 2 s while traveling along a rectilinear and horizontal trajectory. Also assume that throughout the takeoff the plane’s engines are providing 32,000 lb of thrust.

(a) Determine the average force exerted by the catapult on the plane. (b) Now suppose that the takeoff order is changed so that a small trainer aircraft must take off first. If the trainer’s weight and thrust are 13,000 and 5850 lb, respectively, and if the catapult is not reset to match the takeoff specifications for the smaller aircraft, estimate the average acceleration to which the trainer’s pilots would be subjected and express the answer in terms of 𝑔. What do you think would happen to the trainer’s pilot? Problem 15.26 If the carrier takeoff of a 45,000 lb plane subject to the 32,000 lb thrust of its engines were not assisted by a catapult, estimate how long it would take for a plane to safely take off, i.e., to reach a speed of 165 mph starting from rest. Also, how long a runway would be needed under these conditions?

Problem 15.27 A 10 lb box is released from rest at a distance 𝑑 = 5 f t from the bottom of a smooth chute with 𝜃 = 30◦ . The friction between the chute and the box is negligible, and so is the change in speed of the box in going from the inclined to the horizontal part of the chute. If the duration of the transition between inclined and horizontal motion takes 0.02 s, determine the average force acting on the box during this transition.

𝑑

𝚥̂ 𝚤̂ Figure P15.27

𝜃

948

Chapter 15

Momentum Methods for Particles

Problem 15.28 A ball with a mass 𝑚 = 0.25 kg is dropped from rest from a height ℎ1 = 1.5 m above an incline and rebounds off the incline at a height ℎ2 = 0.3 m above the ground. After the rebound, the ball lands at a horizontal distance 𝑑 = 5 m from the point of contact between the ball and the incline. If the ball takes 0.57 s to cover the distance 𝑑, and if the ball remains in contact with the incline for 0.01 s during the rebound, determine the average force exerted by the incline onto the ball during the rebound. Express the force using the component system shown. Also, neglect the weight of the ball while it is in contact with the incline. 𝑑 ℎ1 𝚥̂ ℎ2

𝚤̂

Figure P15.28

Figure P15.29

Problem 15.29 A 1600 kg car, when on a rectilinear and horizontal stretch of road and when the tires do not slip, can go from rest to 100 km∕h in 5.5 s. Assuming that the car travels on a straight stretch of road with a 40% slope and that no slip occurs, determine how long it would take to attain a speed of 100 km∕h if the car were propelled by the same maximum average force that can be generated on a horizontal road.

Problem 15.30 A 1600 kg car, when on a rectilinear and horizontal stretch of road, can go from rest to 100 km∕h in 5.5 s. (a) Assuming that the car travels on such a road, estimate the average value of the force acting on the car for the car to match the expected performance.

Figure P15.30

(b) Recalling that the force propelling a car is caused by the friction between the driving wheels and the road, and again assuming that the car travels on a rectilinear and horizontal stretch of road, estimate the average value of the friction force acting on the car for the car to match the expected performance. Also estimate the coefficient of friction required to generate such a force.

Problem 15.31

𝛼

A 5 81 oz baseball traveling at 80 mph rebounds off a bat with a speed of 160 mph. The ball is in contact with the bat for roughly 10−3 s. The incoming velocity of the ball is horizontal, and the outgoing trajectory forms an angle 𝛼 = 31◦ with respect to the incoming trajectory. (a) Determine the impulse provided to the baseball by the bat. (b) Determine the average force exerted by the bat on the ball.

Figure P15.31

ISTUDY

(c) Determine how much the angle 𝛼 would change (with respect to 31◦ ) if we were to neglect the effects of the force of gravity on the ball.

ISTUDY

Section 15.1

949

Momentum and Impulse

Problem 15.32 In an unfortunate incident, a 2.75 kg laptop computer is dropped onto the floor from a height of 1 m. Assuming that the laptop starts from rest, that it rebounds off the floor up to a height of 5 cm, and that the contact with the floor lasts 10−3 s, determine the impulse provided by the floor to the laptop and the average acceleration to which the laptop is subjected when in contact with the floor (express this result in terms of 𝑔, the acceleration of gravity).

1m

Figure P15.32

Problem 15.33 Three space-junk fragments with masses 𝑚1 = 7.45 kg, 𝑚2 = 3.22 kg, and 𝑚3 = 8.45 kg were the only masses generated from an explosion that split a single body apart. The fragments are traveling as shown with 𝑣1 = 7701 m∕s, 𝑣2 = 6996 m∕s, and 𝑣3 = 6450 m∕s. Assume the velocity vectors of the fragments are coplanar, 𝜃 = 25◦ , and 𝜙 = 55◦ . The angles 𝜃 and 𝜙 are measured with respect to lines that are perpendicular to the velocity 𝑣⃗2 . If the system is isolated, determine the mass and velocity of the single body before it exploded.

𝜙

𝑣1

𝑣3 𝜃

𝑣2

Problem 15.34 Figure P15.33

A 175 lb man, initially at rest at 𝐴 on a floating platform, starts walking to the right with a constant speed 𝑣0 = 3 f t∕s relative to the platform until he reaches the right end of the platform at 𝐵. The platform is attached to the pier by a rope that is initially slack but that becomes taut before the man reaches 𝐵. Neglecting any resistance to the motion of the platform due to the water, determine the impulse provided to the man/platform system by the rope when the rope becomes taut. 𝚥̂ rope 𝐴

𝚤̂

𝑣0

pier

platform 𝐵

𝐴 Figure P15.34

𝐵

Figure P15.35

Problem 15.35 Collar 𝐵 is at rest on a smooth horizontal guide when it is impacted by an identical collar 𝐴 traveling at a speed 𝑣0 = 12 f t∕s. The collars are designed to lock onto each other on impact, so 𝐴 and 𝐵 travel together after impact. Neglecting friction, determine the common velocity of 𝐴 and 𝐵 after impact. 𝑣𝐴

Problem 15.36 A 260 gr (1 lb = 7000 gr) bullet 𝐵 is fired into an 8600 lb SUV 𝐴, which is initially moving to the left with a speed 𝑣𝐴 = 15 mph. If the bullet becomes embedded into the SUV, how fast would the impact speed of the bullet 𝑣𝐵 need to be if the SUV is to be stopped cold?

𝐵

𝑣𝐵

Figure P15.36

𝐴

950

Chapter 15

Momentum Methods for Particles Problems 15.37 through 15.39 𝐴

𝐶

𝐵

𝐵

Figure P15.37–P15.39

These problems are an introduction to perfectly plastic impact (which we will cover in Section 15.2). In each problem, model the vehicles 𝐴 and 𝐶 as particles and treat the swarm of bugs 𝐵 hitting the vehicles as a single particle. Also assume that the swarm of bugs sticks perfectly to each vehicle (this is what is meant by a perfectly plastic impact). Problem 15.37 An 80,000 lb semitruck 𝐴 (the maximum weight allowed in many states) is traveling at 70 mph when it encounters a swarm of mosquitoes 𝐵. The swarm is traveling at 1 mph in the opposite direction of the truck. Assuming that the entire swarm sticks to the truck, the mass of each mosquito is 2 mg, and that all of these mosquitoes do not significantly damage the truck, how many mosquitoes must have hit the truck if it slows down by 2 mph on impact? If the same number of mosquitoes hit a small SUV 𝐶 weighing 3000 lb and traveling at 70 mph, by how much would the SUV slow down? Problem 15.38 An 80,000 lb semitruck 𝐴 (the maximum weight allowed in many states) is traveling at 70 mph when it encounters a swarm of worker bees 𝐵. The swarm is traveling at 12 mph in the opposite direction of the truck. Assuming that the entire swarm sticks to the truck, the mass of each bee is 0.1 g, and that all of these bees do not significantly damage the truck, how many bees must have hit the truck if it slows down by 2 mph on impact? If the same number of bees hit a small SUV 𝐶 weighing 3000 lb and traveling at 70 mph, by how much would the SUV slow down? Problem 15.39 An 80,000 lb semitruck 𝐴 (the maximum weight allowed in many states) is traveling at 70 mph when it encounters a swarm of dragonflies 𝐵. The swarm is traveling at 33 mph in the opposite direction of the truck. Assuming that the entire swarm sticks to the truck, the mass of each dragonfly is 0.25 g, and that all of these dragonflies do not significantly damage the truck, how many dragonflies must have hit the truck if it slows down by 2 mph on impact? If the same number of dragonflies hit a small SUV 𝐶 weighing 3000 lb and traveling at 70 mph, by how much would the SUV slow down?

Problem 15.40 Two canoeists 𝐴 and 𝐵 are drifting downstream with a common speed 𝑣0 = 8 m∕s. At some point, 𝐴 and 𝐵 use a rope to reduce the distance between them. If 𝐴 and 𝐵 can reduce their distance at a rate of 1 m∕s, determine the velocity of 𝐴 and 𝐵 when they finally come together. Let the masses of 𝐴 and 𝐵 (including their respective canoes) be 𝑚𝐴 = 90 kg and 𝑚𝐵 = 75 kg, respectively. In addition, neglect the drag acting on the canoes due to the water. 𝑣0 𝚥̂

𝐴 𝐵

𝚤̂

𝐴 Figure P15.40

𝐵 𝚥̂

𝚤̂

Figure P15.41

ISTUDY

Problem 15.41 Suppose that a 180 lb person 𝐴 (the weight includes the rifle and ammunition before firing) were to fire a 180 gr (7000 gr = 1 lb) bullet 𝐵 with a muzzle velocity of 3300 f t∕s. If the person fires while resting on ice, so that the friction between the person and the ground is negligible, determine the final velocity of both the person and the bullet.

ISTUDY

Section 15.1

Momentum and Impulse

Problems 15.42 and 15.43 Blocks 𝐴 and 𝐵, with masses 𝑚𝐴 = 400 kg and 𝑚𝐵 = 90 kg, respectively, are initially at rest when block 𝐴 starts sliding down the incline with 𝜃 = 30◦ .

𝚥̂ 𝚤̂

If friction between the two blocks and between block 𝐴 and the incline is negligible, use the impulse-momentum principle to determine the velocities of 𝐴 and 𝐵 1.5 s after release.

Problem 15.42

𝐵 𝐴

Let 𝜇𝑘 = 0.15 be the kinetic coefficient of friction between block 𝐴 and the incline. If friction between the two blocks is negligible, determine the velocities of 𝐴 and 𝐵 1.5 s after release.

Problem 15.43

Problems 15.44 through 15.46

𝜃 Figure P15.42 and P15.43

Two persons 𝐴 and 𝐵 weighing 140 and 180 lb, respectively, jump off a floating platform (in the same direction) with a velocity relative to the platform that is completely horizontal and with magnitude 𝑣0 = 6 f t∕s for both 𝐴 and 𝐵. The floating platform weighs 800 lb. Assume that 𝐴, 𝐵, and the platform are initially at rest. 𝑣0 𝐴

𝐵

Figure P15.44–P15.46

Neglecting the water resistance to the horizontal motion of the platProblem 15.44 form, determine the speed of the platform after 𝐴 and 𝐵 jump at the same time. Problem 15.45 Neglecting the water resistance to the horizontal motion of the platform, and knowing that 𝐵 jumps first, determine the speed of the platform after both 𝐴 and 𝐵 have jumped. Problem 15.46 Neglecting the water resistance to the horizontal motion of the platform, and knowing that 𝐴 jumps first, determine the speed of the platform after both 𝐴 and 𝐵 have jumped.

Problem 15.47 A 180 lb man 𝐴 and a 40 lb child 𝐶 are at the opposite ends of a 250 lb floating platform 𝑃 with a length 𝐿fp = 15 f t. The man, child, and platform are initially at rest at a distance 𝛿 = 1 f t from a mooring dock. The child and the man move toward each other with the same speed 𝑣0 relative to the platform. If the drag force due to the water is negligible, determine the distance 𝑑 from the mooring dock where the child and man will meet.

𝛿 𝐴 𝐶 𝑃

Problem 15.48 A man 𝐴, with a mass 𝑚𝐴 = 85 kg, and a child 𝐶, with a mass 𝑚𝐶 = 18 kg, are at the opposite ends of a floating platform 𝑃 , with a mass 𝑚𝑃 = 150 kg and a length 𝐿fp = 6 m. Assume that the man, child, and platform are initially at rest and that the resistance due to the water to the horizontal motion of the platform is negligible. Suppose that the man and child start moving toward each other in such a way that the platform does not move relative to the water. Determine the distance covered by the child to meet the man.

𝐿fp Figure P15.47 and P15.48

951

952

Chapter 15

Momentum Methods for Particles

Problem 15.49 The 28,000 lb A-10 Thunderbolt is flying at a constant speed of 375 mph when it fires a 4 s burst from its forward-facing seven-barrel Gatling gun. The gun fires 13.2 oz projectiles at a rate of 4200 rounds∕min. The muzzle velocity of each projectile is 3250 f t∕s. Assuming that each of the plane’s two jet engines maintains a constant thrust of 9000 lb, that the plane is subject to a constant air resistance while the gun is firing (equal to that before the burst), and that the plane flies straight and level, determine the plane’s change in velocity at the end of the 4 s burst. 𝐴

30 mm rounds

Gatling gun Figure P15.49

Problem 15.50 A man 𝑃 on a cart on rails is receiving packages from two men 𝑃𝐴 and 𝑃𝐵 standing on a stationary platform. Assume that 𝑃 and the cart have a combined weight of 350 lb and start from rest. In addition, suppose that 𝑃𝐴 throws a package 𝐴 weighing 60 lb, which is received by 𝑃 with a horizontal speed 𝑣𝐴 = 4.5 f t∕s. After 𝑃 has received the package from 𝑃𝐴 , 𝑃𝐵 throws a package 𝐵 weighing 80 lb, which is received by 𝑃 with a horizontal speed relative to 𝑃 and in the same direction as the velocity of 𝑃 of 5.25 f t∕s. Determine the final velocity of 𝑃 and the cart. Neglect any friction or air resistance acting on 𝑃 and the cart and assume that the three men are big enough that they can handle throwing around 60 and 80 lb packages.

𝑃 𝐵

𝐴

𝑃𝐵

𝑃𝐴 Figure P15.50

Problems 15.51 and 15.52 Box 𝐴, which weighs 47 lb, is released from rest in the position shown in the top figure, where ℎ = 3 f t. The box slides down the fixed incline with negligible friction and it lands on a cart 𝐵, which is initially at rest and weighs 12 lb. When 𝐴 reaches the bottom of the incline, the velocity of 𝐴 is completely horizontal, and 𝐴 starts sliding on the cart. The coefficient of kinetic friction between 𝐴 and 𝐵 is 𝜇𝑘 = 0.6.

𝐴 ℎ

𝐵

𝐵

𝐴

Determine the common speed of 𝐴 and 𝐵 after 𝐴 stops sliding relative Problem 15.51 to 𝐵 and the time it takes 𝐴 to stop sliding relative to 𝐵.

𝑑

Problem 15.52 Determine the distance 𝑑 from the right end of the cart at which the box 𝐴 stops relative to the cart.

Figure P15.51 and P15.52

Problems 15.53 and 15.54 Blocks 𝐴 and 𝐵, with masses 𝑚𝐴 = 5 kg and 𝑚𝐵 = 3 kg, respectively, are connected by a spring of stiffness 𝑘 = 20 N∕m and unstretched length 𝐿0 = 0.75 m. The blocks are at rest and separated by the distance 𝐿0 when block 𝐵 is given a velocity to the right with magnitude 𝑣0 = 15 m∕s. The friction between the blocks and the horizontal surface on which they rest is negligible.

𝚥̂ 𝚤̂

𝑑 𝑘, 𝐿0

𝐴 Figure P15.53 and P15.54

ISTUDY

𝑣0 𝐵

Problem 15.53 Determine the maximum value of the distance 𝑑 between 𝐴 and 𝐵 that will be achieved during the motion. Problem 15.54 Assuming that the spring can be squeezed so that 𝑑 = 0, verify that the blocks 𝐴 and 𝐵 will collide with one another in the ensuing motion by computing the velocities of the blocks when 𝑑 = 0.

ISTUDY

Section 15.1

953

Momentum and Impulse

Problem 15.55 Energy storage devices that use spinning flywheels to store energy are starting to become available. To store as much energy as possible, it is important that the flywheel spin as fast as possible. Unfortunately, if it spins too fast, internal stresses in the flywheel cause it to come apart catastrophically. Therefore, it is important to keep the speed at the edge of the flywheel below about 1000 m∕s. It is also critical that the flywheel be almost perfectly balanced to avoid the tremendous vibrations that would otherwise result. With this in mind, let the flywheel 𝐷, whose diameter is 0.3 m, rotate at 𝜔 = 60,000 rpm. Assume that the cart 𝐵 is constrained to move rectilinearly along the smooth guide tracks. Given that the flywheel is not perfectly balanced, that the unbalanced weight 𝐴 has mass 𝑚𝐴 , and that the total mass of the flywheel 𝐷, cart 𝐵, and electronics package 𝐸 is 𝑚𝐵 , determine the following as a function of 𝜃, the masses, the diameter, and the angular speed of the flywheel:

𝜔

𝐴 𝐷

𝐸 𝐵 guide tracks

top view

𝜃

(a) the amplitude of the motion of the cart, (b) the maximum speed achieved by the cart. Neglect the mass of the wheels, assume that initially everything is at rest, and assume that the unbalanced mass is at the edge of the flywheel. Finally, evaluate your answers to Parts (a) and (b) for 𝑚𝐴 = 1 g (about the mass of a paper clip) and 𝑚𝐵 = 70 kg (the mass of the flywheel might be about 40 kg).

Figure P15.55

𝐴 𝐿

Problem 15.56 The 135 lb woman 𝐴 sits atop the 90 lb cart 𝐵, both of which are initially at rest. If the woman slides down the frictionless incline of length 𝐿 = 11 f t, determine the velocity of both the woman and the cart when she reaches the bottom of the incline. Ignore the mass of the wheels on which the cart rolls and any friction in their bearings. The angle 𝜃 = 26◦ . Hint: A number of principles can be brought to bear to solve this problem, including relative motion of the woman along the incline, conservation of energy (a scalar principle), and conservation of linear momentum (a vector principle).

𝜃

Figure P15.56

Problem 15.57 An Apollo Lunar Module 𝐴 and Command and Service Module 𝐵 are moving through space far from any other bodies (so that their gravitational effects can be ignored). When 𝜃 = 30◦ , the two craft are separated using an internal linear elastic spring whose constant is 𝑘 = 200,000 N∕m and is precompressed 0.5 m. Noting that the mass of the Command and Service Module is about 29,000 kg and that the mass of the Lunar Module is about 15,100 kg, determine their postseparation velocities if their common preseparation velocity is 11,000 m∕s. 𝐵

𝜃

motion

𝐴

Separation

Preseparation

internal spring 𝑘

Postseparation Figure P15.57

𝐵

954

Chapter 15

Momentum Methods for Particles Problems 15.58 through 15.61

𝐵

𝜃

𝐿 𝐴

In the ride shown, child 𝐴 sits in a seat attached by a cable of length 𝐿 to a freely moving trolley 𝐵 of mass 𝑚𝐵 . The total mass of the child and seat is 𝑚𝐴 . The trolley is constrained by the beam to move only in the horizontal direction. The system is released from rest at the angle 𝜃 = 𝜃0 , and it is allowed to swing in the vertical plane. Neglect the mass of the cable and treat the child and seat as a single particle. Hint: Observe that there is no external force on the system in the horizontal direction. Energy conservation (a scalar principle) can be used in conjunction with momentum conservation (a vector principle) to address the questions that follow. Problem 15.58 Determine expressions for the velocities of the trolley and the rider the first time that 𝜃 = 0◦ . Evaluate your solution for 𝑊𝐴 = 100 lb, 𝑊𝐵 = 20 lb, 𝐿 = 15 f t, and 𝜃0 = 70◦ .

Figure P15.58–P15.61

Problem 15.59 Determine expressions for the velocities of the trolley and the rider the first time that 𝜃 = 0◦ . After doing so, for given 𝑔, 𝐿, 𝑚𝐴 , and 𝜃0 , determine the maximum speed achievable by the rider at 𝜃 = 0◦ and the corresponding value of 𝑚𝐵 . Evaluate your solution for 𝑊𝐴 = 100 lb, 𝐿 = 15 f t, and 𝜃0 = 70◦ . What would be the motion of 𝐵 for this value of 𝑚𝐵 ? Problem 15.60

Determine the velocity of the trolley and the speed of the rider for any arbitrary value of 𝜃. Problem 15.61 Determine the equations needed to find the velocity of the trolley and the rider for any arbitrary value of 𝜃. Clearly label all equations and list the corresponding unknowns, showing that you have as many equations as you have unknowns. Solve the equations for the unknowns, and then plot the velocity of the trolley and the speed of the rider as a function of the angle 𝜃 for both halves of a full swing of the rider. Use 𝑊𝐴 = 100 lb, 𝑊𝐵 = 20 lb, 𝐿 = 15 f t, and 𝜃0 = 70◦ .

16 f t 𝜃̇ 𝐴

Problem 15.62

42 f t

𝐵 16 f t

A tower crane is lifting a 10,000 lb object 𝐵 at a constant rate of 7 f t∕s while rotating at a constant rate of 𝜃̇ = 0.15 rad∕s. 𝐵 is also moving outward with a radial velocity of 1.5 f t∕s. Assume that the object 𝐵 does not swing relative to the crane (i.e., it always hangs vertically) and that the crane is fixed to the ground at 𝑂. (a) Determine the radial velocity required of the 20 ton counterweight 𝐴 to prevent the horizontal motion of the system’s center of mass. (b) Find the total force acting on 𝐴 and on 𝐵.

25 f t Figure P15.62

ISTUDY

(c) Determine the velocity and acceleration of the mass center of the system when 𝐴 moves as determined in Part (a).

ISTUDY

Section 15.2

15.2

Impact

955

Impact

Impacts are short, dramatic events Figure 15.6 shows a standard 1.4 oz. racquetball hitting a wall. The impact may only

𝑡 = 1.34 ms

𝑡 = 1.57 ms

𝑡 = 1.93 ms

Loren M. Winters, Durham, NC Loren M. Winters, Durham, NC Loren M. Winters, Durham, NC

Figure 15.6. High-speed photography sequence capturing a racquetball hitting a wall.

last for 2.5 ms, and yet the velocity completely reverses its direction during that time. That is a large change in momentum in a very short time, and there must be a correspondingly large impulse causing that change. The force acting on the ball during the impact, as estimated in Example 15.2, has a component normal to the wall that is almost 100 times larger than the tangent component and more than 1000 times larger than the weight of the ball. Hence, the force dominating the impact is not the weight of the ball or the friction with the wall, it is the normal component of the contact force between the ball and the wall. These observations indicate that impacts occur over time scales so small that they can be modeled as instantaneous. In addition, an impact is an event so dominated by constraint forces that all the other forces we normally account for in FBDs can be neglected. We now formalize these ideas into a theory of impacts for particles.

Definition of impact and notation We will model impacts as events that span infinitesimal time intervals. For this reason we adopt the notation that if 𝑡 is the time of impact, then 𝑡− and 𝑡+ will be the time instants right before and after the impact, respectively. The superscripts − and + will denote the pre- and postimpact values of all quantities in the model. We will also assume that for 𝑡− ≤ 𝑡 ≤ 𝑡+ the velocities of impacting objects can undergo a finite change, while we neglect any change in position.

Line of impact and contact force between impacting objects In Fig. 15.7, we assume that contact in an impact occurs at a single point at which we can identify the plane tangent to the contact. The 𝑞 and 𝑤 directions on this figure are the same as those shown in Fig. 15.5 on p. 936. The line through the point of contact and normal to the tangent plane is called line of impact (LOI). We assume that the LOI is also the line of action of the contact forces between the objects, which is equivalent to neglecting friction at the point of contact. This assumption implies that impacts affect the velocity of impacting objects only along the LOI.

plane tangent to the contact

line of impact: line perpendicular to the contact

Gary L. Gray

Figure 15.7 Photograph of two billiard balls in contact.

956

Chapter 15

Momentum Methods for Particles Classification of impacts

Solution strategies for impact problems usually depend on the orientation of the preimpact velocities relative to the LOI. Hence, it is convenient to have a corresponding classification of impacts. Referring to Fig. 15.8, an impact is called oblique if one of the preimpact velocities is not parallel to the LOI; otherwise it is called direct. An impact is called central if the LOI contains the mass centers of the impacting bodies; otherwise it is called eccentric. Particles coincide with their own mass centers and can only have central impacts. Eccentric impacts are discussed in Section 18.3.

𝑣− 𝐵

𝑦

𝚥̂

𝑣− 𝐵

𝚥̂ 𝐴

𝐵

𝐴 𝚤̂

𝑦

𝐵

𝚤̂ 𝑣− 𝐴

𝑣− 𝐴

𝑥

𝑥

(b)

(a)

Figure 15.8. (a) Oblique central impact. (b) Direct central impact.

Impulsive forces and impact-relevant FBDs A force is said to be impulsive if it causes a finite change of momentum, and hence velocity, over an infinitesimal time interval. The idea of average force from Section 15.1 then says that the magnitude of an impulsive force must therefore tend to infinity. Because of this, in FBDs of impacting objects, which we call impact-relevant FBDs, we neglect any force that is not impulsive. The concept of an impulsive force is only a modeling tool since we know that forces with infinite magnitude do not exist. In practice, we model a force as impulsive if its magnitude can grow as large as necessary to guarantee that given constraints are met. For example, the contact force between two billiard balls becomes as large as necessary to prevent them from passing through one another, and it is therefore modeled as impulsive. By contrast, forces whose magnitude must remain finite, such as weight forces, spring forces, and forces that depend on velocity, cannot be impulsive. An impact is called constrained if the system of impacting objects is subject to external impulsive forces; otherwise it is called unconstrained. We will present the solution of a constrained impact problem in Example 15.9 on p. 968.

Coefficient of restitution 𝑣𝐴 𝐴

𝑣𝐵 𝐵

Figure 15.9 Car 𝐴 has a speed greater than that of car 𝐵 and therefore collides with it.

ISTUDY

Figure 15.9 shows two cars 𝐴 and 𝐵, with mass 𝑚𝐴 and 𝑚𝐵 , respectively, with preim− − − pact speeds 𝑣𝐴 and 𝑣𝐵 , with 𝑣−𝐵 < 𝑣𝐴 . Car 𝐴 rear-ends 𝐵, and we assume that they separate after the collision. We wish to predict the postimpact velocities of 𝐴 and 𝐵. For simplicity, we assume the LOI to be parallel to the ground. This implies that neither car will push the other toward the ground and that the reaction forces between the cars and the ground have no reason to grow larger. Thus, we model these reaction

ISTUDY

Section 15.2

Impact

forces as nonimpulsive, and we model the overall impact with the impact-relevant FBD in Fig. 15.10, where we have neglected the weights of the cars because they are nonimpulsive. Since there are no external forces in the FBD, the momentum of the system is conserved. Hence, in the 𝑥 direction, which is the LOI, we have 𝑚𝐴 𝑣−𝐴𝑥 + 𝑚𝐵 𝑣−𝐵𝑥 = 𝑚𝐴 𝑣+𝐴𝑥 + 𝑚𝐵 𝑣+𝐵𝑥 ,

(15.20)

− − − where 𝑣𝐴𝑥 = 𝑣−𝐴 and 𝑣𝐵𝑥 = 𝑣𝐵 because 𝐴 and 𝐵 are moving in the 𝑥 direction + before impact. Equation (15.20), which is the balance law, has two unknowns, 𝑣𝐴𝑥 + and 𝑣𝐵𝑥 . We need a second equation in these unknowns to solve the problem. Since we have already written the kinematic equations by relating the speeds to the velocity components of the cars, the second equation will be a force law. Figure 15.11 shows the impact-relevant FBD of each of the cars individually. The contact force 𝑃 between 𝐴 and 𝐵 is the force that we need to describe in some way. While we cannot describe 𝑃 as a function of time (impacts span an infinitesimal time interval), we can describe the effect of 𝑃 on 𝐴 and 𝐵 by how 𝑃 measurably affects the pre- and postimpact velocities of 𝐴 and 𝐵 along the LOI, which is the line of action of 𝑃 . With this in mind, experience tells us that:

𝚥̂

𝐴

LOI

𝐵

𝚤̂

957

𝑥

Figure 15.10 Impact-relevant FBD of the cars 𝐴 and 𝐵 as a system.

𝚥̂ 𝚤̂

𝐴

LOI

𝑃

𝐵

𝑃

𝑥

Figure 15.11 Impact-relevant FBD of the cars 𝐴 and 𝐵 individually. Recall that we have assumed that the contact force exerted by 𝐴 and 𝐵 on one another is only directed along the LOI.

2. The “quality” of the rebound is described by the postimpact relative velocity, also called the separation velocity. The typical plot we obtain by correlating the separation velocity with the approach velocity measured in collision experiments is shown in Fig. 15.12. This plot shows that the separation velocity depends on the approach velocity and that, for higher approach velocities, there is a lack of proportionality between pre- and postimpact relative velocities (this is because more permanent damage is caused by the collision for higher approach velocities). Figure 15.12 also shows that there is a regime (near the origin) in which the approach and separation velocities are linearly related to one another. Confining our theory to impacts in their linear range, we therefore adopt the following impact force law: 𝑒=

+ + separation velocity 𝑣𝐵𝑥 − 𝑣𝐴𝑥 . = − − approach velocity 𝑣𝐴𝑥 − 𝑣𝐵𝑥

(15.21)

Equation (15.21) is called the coefficient of restitution equation, and the constant 𝑒, which is determined experimentally, is called the coefficient of restitution (COR). The COR equation is the force law we were looking for. Properties of the COR Referring to Eq. (15.21), we note the following: 1. The COR 𝑒 is dimensionless.

+ − − 2. Since 𝑣𝐴∕𝐵𝑥 = 𝑣−𝐴𝑥 − 𝑣𝐵𝑥 and 𝑣+𝐵∕𝐴𝑥 = 𝑣+𝐵𝑥 − 𝑣𝐴𝑥 have the same sign, 𝑒 ≥ 0 for any impact.

3. For materials without any energy-producing elements, the separation speed is never greater than the approach speed. This implies that 0 ≤ 𝑒 ≤ 1.

(15.22)

Separation velocity

1. The severity of an impact depends on the relative preimpact velocity, also called the approach velocity. linear approximation

𝑒 1

typical qualitative result

Approach velocity Figure 15.12 Qualitative trend usually found in collision experiments when plotting the post- versus preimpact relative velocity (black curve).

958

Chapter 15

Momentum Methods for Particles

Table 15.1 Impact type as a function of COR value. COR value

Impact type

𝑒 = 0a 0 𝑣0 , as is indeed the case. Hence, overall our solution appears to be correct. A Closer Look This example was deemed advanced because we had to recognize that part of the solution, Eq. (12), was applicable outside the assumptions underlying the rest of the solution. It was this realization that allowed us to determine how the initial conditions affected the motion at time 𝑡2 without having to explicitly compute the motion that takes the system from 𝑡1 to 𝑡2 .

ISTUDY

Section 15.3

999

Angular Momentum

Problems Problem 15.110 Consider the situation depicted in the figure. At the instant shown, how are the angular momenta of particle 𝑃 with respect to 𝑂 and 𝑄 related? 𝑂 ℎ

𝑣𝑃 𝑃

𝑃

𝑣𝑃 ℎ

ℎ 𝑄 Figure P15.110

𝑂

𝑄 Figure P15.111

Problem 15.111 Consider the situation depicted in the figure. At the instant shown, how are the angular momenta of particle 𝑃 with respect to 𝑂 and 𝑄 related?

Problems 15.112 through 15.114 At the instant shown, a truck 𝐴, of weight 𝑊𝐴 = 31,000 lb, and a car 𝐵, of weight 𝑊𝐵 = 3970 lb, are traveling with speeds 𝑣𝐴 = 35 mph and 𝑣𝐵 = 34 mph, respectively. Problem 15.112 Choosing point 𝑂 as the moment center, determine the angular momentum (with respect to 𝑂) of 𝐴 and 𝐵 individually at this instant.

𝐵 𝑣𝐵 13.0 f t 2.0 f t

𝑣𝐴

𝑂

𝐴

37.5 f t 27.5 f t

Problem 15.113 Choosing point 𝑂 as the moment center, determine the angular momentum (with respect to 𝑂) of the particle system formed by 𝐴 and 𝐵 at this instant.

Choosing point 𝑄 as the moment center, determine the angular momentum (with respect to 𝑄) of the particle system formed by 𝐴 and 𝐵 at this instant.

Problem 15.114

Figure P15.112–P15.114

Problems 15.115 and 15.116 At time 𝑡1 = 0, particle 𝐴 weighing 5 oz goes through point 𝑃 of (𝑥, 𝑦, 𝑧) coordinates ̂ f t∕s. At time 𝑡 , 𝐴 goes through a point (0, 2, 3) f t with a velocity 𝑣(𝑡 ⃗ 1 ) = (3 𝚤̂ + 5 𝚥̂ + 7 𝑘) 2 ̂ f t∕s. 𝑄 of coordinates (1, 0, 1) f t, and with a velocity 𝑣(𝑡 ⃗ 2 ) = (1 𝚤̂ + 2 𝚥̂ + 3 𝑘) Problem 15.115

Determine the angular momentum of 𝐴 with respect to the origin 𝑂

at times 𝑡1 and 𝑡2 . Problem 15.116 Determine the time rate of change of the angular momentum of 𝐴 relative to the origin 𝑂 at times 𝑡1 and 𝑡2 if the acceleration of 𝐴 at times 𝑡1 and 𝑡2 is ̂ f t∕s2 and 𝑎(𝑡 ̂ f t∕s2 , respectively. 𝑎(𝑡 ⃗ 1 ) = (1 𝚤̂ + 2 𝚥̂ + 3 𝑘) ⃗ 2 ) = (9 𝚤̂ + 8 𝚥̂ + 7 𝑘)

Problem 15.117 Referring to Probs. 15.115 and 15.116, would it be possible to calculate the velocity and acceleration of 𝐴 if the angular momentum and the time rate of change of angular momentum were given in place of the velocity and acceleration, respectively?

𝑄 3.0 f t

1000

Chapter 15

Momentum Methods for Particles

𝑧

𝑧

𝑃

𝐵 𝑣⃗𝐵

𝑄 𝚤̂

𝑘̂

𝑘̂ 𝑂

𝚥̂

𝚤̂ 𝐴 𝑥

𝑥

𝑂

𝚥̂

𝐴

𝑦

𝑦 𝑣⃗𝐴 Figure P15.115–P15.117

Figure P15.118 and P15.119

Problems 15.118 and 15.119 Particles 𝐴 and 𝐵 have masses 𝑚𝐴 = 3 kg and 𝑚𝐵 = 1.3 kg, respectively. At the instant shown, the (𝑥, 𝑦, 𝑧) coordinates of 𝐴 and 𝐵 are (3, 2, 0) m and (3, 0, 4) m, respectively. In ̂ m∕s, addition, the velocities of 𝐴 and 𝐵 are 𝑣⃗𝐴 = (7 𝚤̂ + 10 𝚥̂) m∕s and 𝑣⃗𝐵 = (−4 𝚤̂ − 3 𝑘) respectively. Determine the angular momentum of the system formed by 𝐴 and 𝐵 relative to the origin 𝑂 at the instant shown.

Problem 15.118

Determine the angular momentum of particle 𝐴 relative to particle 𝐵 at the instant shown.

Problem 15.119

Problem 15.120 A rotor consists of four horizontal blades each of length 𝐿 = 4 m and mass 𝑚 = 90 kg cantilevered from a vertical shaft. The rotor is initially at rest when it is subjected to a moment 𝑀 = 𝛽𝑡, with 𝛽 = 60 N⋅m∕s. Modeling each blade as having its mass concentrated at its midpoint, determine the angular speed of the rotor after 10 s. 𝑀

Figure P15.120

𝑣𝑐 𝐿2 2

𝑂 𝜃2 𝜃1

𝐿1

1

Figure P15.121

ISTUDY

Problem 15.121 The simple pendulum shown oscillates in the vertical plane (the plane of the figure) as the pendulum cord is being steadily retracted through the opening at 𝑂 with a constant speed 𝑣𝑐 . Consider the oscillations of the pendulum between positions 1 and 2, where 𝜃1 and 𝜃2 are the maximum swing angles of the pendulum at 1 and 2, respectively. Let 𝐿1 and 𝐿2 denote the lengths of the cord at 1 and 2, respectively, and let 𝑚 denote the mass of the pendulum bob. Neglecting all forces except gravity and the tension in the cord, determine the angular impulse relative to 𝑂 provided to the pendulum bob in going from 1 to 2.

ISTUDY

Section 15.3

1001

Angular Momentum

𝑧

Problem 15.122

𝑑

The object shown is called a speed governor, a mechanical device for the regulation and control of the speed of mechanisms. The system consists of two arms of negligible mass at the ends of which are attached two spheres, each of mass 𝑚. The upper end of each arm is attached to a fixed collar 𝐴. The system is then made to spin with a given angular speed 𝜔0 at a set opening angle 𝜃0 . Once it is in motion, the opening angle of the governor can be varied by adjusting the position of the collar 𝐶 (by the application of some force). Let 𝜃 represent the generic value of the governor opening angle. If the arms are free to rotate, that is, if no moment is applied to the system about the spin axis after the system is placed in motion, determine the expression of the angular velocity 𝜔 of the system as a function of 𝜔0 , 𝜃0 , 𝑚, 𝑑, and 𝐿, where 𝐿 is the length of each arm and 𝑑 is the distance of the top hinge point of each arm from the spin axis. Neglect any friction at 𝐴 and 𝐶.

𝐴 𝐿

𝜃 𝐶 𝑚

𝜔

Figure P15.122

Problems 15.123 and 15.124 ⃗ = 25 𝑘̂ kg⋅m2 ∕s be the angular momentum of a particle 𝐴 about the origin 𝑂 of Let ℎ 𝑂 an inertial (𝑥, 𝑦, 𝑧) reference frame at the instant shown. Let the mass of 𝐴 be 𝑚 = 0.75 kg, and let the coordinates of 𝐴 at the instant shown be (1, 2, 0) m. 𝑧

𝑘̂ 𝚤̂ 𝑥

𝑂

𝚥̂

𝐴

𝑦

Figure P15.123 and P15.124

If 𝐴 moves only in the 𝑥𝑦 plane, determine the vector component of the velocity of 𝐴 perpendicular to the position vector of 𝐴 at the instant shown.

Problem 15.123

Problem 15.124 If 𝐴 moves only in the 𝑥𝑦 plane, determine the vector component of the acceleration of 𝐴 perpendicular to the position vector of 𝐴 at the instant shown if, at ⃗ = 2 𝑘̂ N⋅m. this instant, the moment acting on 𝐴 relative to the origin 𝑂 is 𝑀 𝑂

Problems 15.125 through 15.127 The projectile 𝑃 of mass 𝑚𝑃 = 18.5 kg is shot with an initial speed 𝑣𝑃 = 1675 m∕s as shown in the figure. Ignore aerodynamic drag forces on the projectile. Problem 15.125 Compute the projectile’s angular momentum with respect to the point 𝑂 as a function of time from the time it exits the barrel until the time it hits the ground. Problem 15.126 Choose point 𝑂 as moment center. Then verify the validity of the angular impulse-momentum principle as given in Eq. (15.36) by showing that the time derivative of the angular momentum does, in fact, equal the moment.

Knowing that the helicopter 𝐸 happens to have the same horizontal coordinate as the projectile at the instant the projectile leaves the gun and that it moves at a constant speed 𝑣𝐸 = 15 m∕s as shown, and treating 𝐸 as a moving moment center, verify the angular impulse-momentum principle as given in Eq. (15.35). Problem 15.127

𝑣𝐸

𝑣𝑃 20◦ 𝑂 3.2 m

Figure P15.125–P15.127

10 m

1002

Chapter 15

Momentum Methods for Particles Problems 15.128 and 15.129 𝑂

The simple pendulum in the figure is released from rest when 𝜃 = 33◦ . Problem 15.128 Knowing that the bob’s weight is 𝑊 = 2 lb and that the length is 𝐿 = 4 f t, determine the bob’s angular momentum with respect to 𝑂 as a function of 𝜃.

𝐿

𝜃

Problem 15.129

Use the angular impulse-momentum principle in Eq. (15.36) to determine the equations of motion of the pendulum bob.

𝑊 Figure P15.128 and P15.129

Problem 15.130 At the lowest and highest points on its trajectory, the pendulum cord, with a length 𝐿 = 2 f t, forms angles 𝜙1 = 15◦ and 𝜙2 = 50◦ with the vertical direction, respectively. Determine the speed of the pendulum bob corresponding to 𝜙1 and 𝜙2 . Hint: Energy conservation and angular momentum conservation are both useful in solving this problem. Because the points of interest, 𝜙1 and 𝜙2 , represent the extrema of the orbit, the velocity vector at these locations consists of a single component. Begin by representing the velocity in a spherical component system centered at 𝑂 and find the special form the velocity vector takes at the extrema of the orbit.

𝑂 𝐿 𝜙0

Figure P15.130

ISTUDY

𝑣0

Problems 15.131 and 15.132 A collar with mass 𝑚 = 2 kg is mounted on a rotating arm of negligible mass that is initially rotating with an angular velocity 𝜔0 = 1 rad∕s. The collar’s initial distance from the 𝑧 axis is 𝑟0 = 0.5 m and 𝑑 = 1 m. At some point, the restraint keeping the collar in place is removed so that the collar is allowed to slide. Assume that the friction between the arm and the collar is negligible. Problem 15.131 If no external forces and moments are applied to the system, with what speed relative to the arm will the collar impact the end of the arm? Hint: Since friction is negligible, no radial force exists on the collar. When combined with angular momentum conservation, this should enable you to write 𝑟̈𝑑𝑟 = 𝑟𝑑 ̇ 𝑟.̇ Problem 15.132

Compute the moment that must be applied to the arm, as a function of position along the arm, to keep the arm rotating at a constant angular velocity while the collar travels toward the end of the arm. 𝑧

𝑟0

𝑑

𝑧

𝑟0

𝑑

cord 𝑚

𝜔0

Figure P15.131 and P15.132

𝑚

𝑀

Figure P15.133 and P15.134

Problems 15.133 and 15.134 A collar of mass 𝑚 is initially at rest on a horizontal arm when a constant moment 𝑀 is applied to the system to make it rotate. Assume that the mass of the horizontal arm is negligible and that the collar is free to slide without friction.

ISTUDY

Section 15.3

Angular Momentum

1003

Problem 15.133 Derive the equations of motion of the system, taking advantage of the angular impulse-momentum principle. Hint: Applying the angular impulse-momentum principle yields only one of the needed equations of motion. Problem 15.134 Continue Prob. 15.133 by integrating the collar’s equations of motion and determine the time the collar takes to reach the end of the arm. Assume that the collar weighs 1.2 lb and that 𝑀 = 20 f t ⋅lb. Also, at the initial time let 𝑟0 = 1 f t and 𝑑 = 3 f t.

Problem 15.135

𝑧

In a simple model of orbital motion under a central force, a disk 𝐷 slides with no friction over a horizontal surface while connected to a fixed point 𝑂 by a linear elastic cord of constant 𝑘 and unstretched length 𝐿0 . Let the mass of 𝐷 be 𝑚 = 0.45 kg and 𝐿0 = 1 m. Suppose that when 𝐷 is at its maximum distance from 𝑂, this distance is 𝑟0 = 1.75 m and the corresponding speed of 𝐷 is 𝑣0 = 4 m∕s. Determine the elastic cord constant 𝑘 such that the minimum distance between 𝐷 and 𝑂 is equal to the unstretched length 𝐿0 .

𝑟 𝑣 𝑘, 𝐿0

𝐷

𝜃

𝑂

Figure P15.135

Problem 15.136

spin axis

The body of the satellite shown has a weight that is negligible with respect to the two spheres 𝐴 and 𝐵 that are rigidly attached to it, which weigh 150 lb each. The distance between 𝐴 and 𝐵 from the spin axis of the satellite is 𝑅 = 3.5 f t. Inside the satellite there are two spheres 𝐶 and 𝐷 weighing 4 lb mounted on a motor that allows them to spin about the axis of the cylinder at a distance 𝑟 = 0.75 f t from the spin axis. Suppose that the satellite is released from rest and that the internal motor is made to spin up the internal masses at an absolute constant time rate of 5.0 rad∕s2 (measured relative to an inertial observer) for a total of 10 s. Treating the system as isolated, determine the angular speed of the satellite at the end of spin-up.

𝐵

𝐴

𝑟

𝑅

𝐶 𝐷

internal rotating element

Problem 15.137

Figure P15.136

A disk 𝐴 with mass 𝑚 moves on a frictionless horizontal surface. The disk is attached to point 𝑂 with an elastic cord. The disk follows the trajectory shown between 1 and 2. The coordinates of 𝐴 at 1 and 2 are (𝑥1 , 𝑦1 ) = (−1.5, 5.0) cm and (𝑥2 , 𝑦2 ) = (4.0, −3.0) cm, respectively. If the velocity of 𝐴 at 1 is parallel to the 𝑦 axis and has a magnitude 𝑣1 = 2.0 m∕s, determine 𝑣2 , the speed of 𝐴 at 2, knowing that 𝜃 = 35◦ . 1

𝑣1

𝑦

𝐴 𝑥

𝑂 2

𝜃 Figure P15.137

𝑣2

1004

Chapter 15

Momentum Methods for Particles

Problem 15.138 Masses 𝐴 and 𝐵, each of mass 𝑚, are attached to one another at the ends of a string that runs through the small hole in the center of a circular cone, for which 𝜃 = 30◦ . (a) If ball 𝐵 is hanging a constant distance below the cone and ball 𝐴 is moving in a horizontal circle with constant speed 𝑣0 inside the cone, determine, as a function of 𝑣0 and 𝑔, the distance 𝓁. (b) Now a force 𝑃 , which increases from zero to a constant value, is applied to 𝐵 to pull it down a distance 𝓁∕2 and it is then held at that position. When 𝐵 is in this position and 𝐴 is moving in a new circular path, determine the speed of 𝐴 as a function of 𝑣0 . (c) Determine, as a function of 𝑚 and 𝑣0 , the work done by 𝑃 in moving 𝐵 to its new position. Neglect friction between 𝐴 and the cone and between the string and the hole. Use sin 𝜃 = √ 1∕2 and cos 𝜃 = 3∕2.

Figure P15.138

Problem 15.139 Ball 𝐴 with mass 𝑚 and ball 𝐵 with mass 2𝑚, are attached to one another at the ends of a string that runs through the small hole in the center of a circular cone, for which 𝜃 = 30◦ . (a) If ball 𝐵 is hanging a constant distance below the cone and ball 𝐴 is moving in a horizontal circle with constant angular speed 𝜔1 on the outside of the cone, determine, as a function of 𝜔1 and 𝑔, the distance 𝓁. (b) Now a force 𝑃 , which increases from zero to a constant value, is applied to 𝐵 to pull it down a distance 𝓁∕2 and it is then held at that position. When 𝐵 is in this position and 𝐴 is moving in a new circular path, determine the speed of 𝐴 as a function of 𝜔1 . (c) Determine, as a function of 𝑚 and 𝜔1 , the work done by 𝑃 in moving 𝐵 to its new position. Neglect friction between 𝐴 and the cone and between the string and the hole. Use sin 𝜃 = √ 1∕2 and cos 𝜃 = 3∕2.

Figure P15.139

Problem 15.140 A particle of mass 𝑚 is attached to a light string that runs through a smooth hole in a circularly shaped bowl that lies in the vertical plane. A force 𝑃 is applied to the other end of the string so that the mass can rotate in a horizontal circle with constant speed 𝑣1 . Use √ sin 30◦ = 1∕2 and cos 30◦ = 3∕2 in your calculations. (a) If the tension in the string is 𝑃1 = 𝑚𝑔 when the mass is in 1, determine its speed 𝑣1 as a function of 𝑔 and 𝑅. (b) If the force is increased to 𝑃2 so that the mass moves to the new circular path shown in 2, determine the new speed of the mass 𝑣2 . (c) Determine, as a function of 𝑚 and 𝑔, the required tension in the string 𝑃2 in 2. Use

Figure P15.140

sin 15◦ =

√

3−1 √ 2 2

and cos 15◦ =

√ 3+1 √ . 2 2

𝑧 𝜙 𝑣0 ℎ ℎ 0

Figure P15.141

ISTUDY

Problem 15.141 A sphere of mass 𝑚 slides over the outer surface of a cone with angle 𝜙 and height ℎ. The sphere was released at a height ℎ0 with a velocity of magnitude 𝑣0 and a direction that was completely horizontal. Assume that the opening angle of the cone and the value of 𝑣0 are such that the sphere does not separate from the surface of the cone once put in motion. In addition, assume that the friction between the sphere and the cone is negligible. Determine the vertical component of the sphere’s velocity as a function of the vertical position 𝑧 (measured from the base of the cone), 𝑣0 , ℎ, ℎ0 , and 𝜙.

ISTUDY

Section 15.3

Angular Momentum

Problem 15.142 Consider a planet orbiting the Sun, and let 𝑃1 , 𝑃2 , 𝑃3 , and 𝑃4 be the planet’s position at four corresponding time instants 𝑡1 , 𝑡2 , 𝑡3 , and 𝑡4 such that 𝑡2 − 𝑡1 = 𝑡4 − 𝑡3 . Letting 𝑂 denote the position of the Sun, determine the ratio between the areas of the orbital sectors 𝑃1 𝑂𝑃2 and 𝑃3 𝑂𝑃4 . Hint: The area of triangle 𝑂𝐴𝐵 defined by the two planar vectors 𝑐⃗ ⃗ and 𝑑⃗ as shown is given by Area(𝑂𝐴𝐵) = |⃗ 𝑐 × 𝑑|∕2. 𝑑⃗ 𝐴

𝐵 𝑃1 𝑐⃗

𝑃4

𝑂 𝑃3 𝑃2 Figure P15.142

1005

1006

Chapter 15

Momentum Methods for Particles

Design Problems

ISTUDY

Design Problem 15.2 The device shown is designed to remotely rotate a video camera 𝐶 by counterrotating the two equal masses 𝑚 with an internal motor that couples the masses and the camera. Assume that the system is mounted on bearings that allow the counterrotating masses and the camera to rotate freely on the vertical mounting post. Model the video camera as two particles, each weighing 3.1 lb and each at a distance of 4.1 in. from the rotation axis. Design the radius of the counterrotating masses, their mass, and their maximum angular velocity so that the angular velocity of the camera never exceeds 10 rpm and so that the camera rotates 90◦ when the masses have rotated 360◦ . Treat the counterrotating masses as particles, and neglect the mass of the rod on which they are mounted. 𝑟𝑚

𝑟𝑚 𝐶

motor 𝑚

𝑚

bearings mounting post

Figure DP15.2

ISTUDY

Section 15.4

15.4

Orbital Mechanics

Orbital Mechanics

Central forces have a special role in dynamics. In Example 13.10 on p. 822, we saw that central force motion is related to an aspect of the motion that is constant, and in Section 15.3 we discovered that this constant is called angular momentum. In Section 14.2 on p. 877, we saw that the work done by a central force, such as a spring force or gravity, depends only on the initial and final positions of the point of application of the force, and so those forces are conservative. We will now study one special central force, namely, the central force that arises from Newton’s universal law of gravitation. The solutions to the problem of one body orbiting another are provided by combining Newton’s second law and this central force. These solutions form the foundation that allows us to predict the motion of planets, Earth satellites (Fig. 15.23), high-altitude rockets, and interplanetary probes.

Determination of the orbit

NASA

Figure 15.23 The International Space Station (ISS) as of June 2008. The equations for two-body orbital motion developed in this section form the basis for accurately predicting the orbit of the ISS about the Earth.

We are interested in the motion of a satellite 𝑆 of mass 𝑚 that is subject to Newton’s universal law of gravitation, 𝐺𝑚𝐵 𝑚 , (15.59) 𝐹 = 𝑟2 where 𝑚𝐵 is the mass of the primary, central, or attracting body, 𝐺 is the universal gravitational constant,∗ and 𝑟 is the distance between the centers of mass of the two bodies (see Fig. 15.24). We begin by making three important assumptions: 1. The attracting bodies 𝐵 and 𝑆 are treated as particles. This assumption is exactly correct if each body has a spherically symmetric mass distribution. It is a good approximation if the distance between the two bodies is large compared with their dimensions, which is the case for many planets and artificial satellites in high orbits. Because the Earth is oblate and has a nonuniform mass distribution, this assumption starts to break down for artificial satellites in low Earth orbit. 2. The only force acting on the satellite 𝑆 is the gravitational force 𝐹 . 3. The primary body 𝐵 is fixed in space. This assumption works well when 𝑚𝐵 is very large compared with 𝑚. Note that relaxing this assumption only slightly changes the two-body equations we will derive (see the “Interesting Fact” on the next page). Applying Newton’s second law to 𝑆 in the polar component system shown in Fig. 15.24, we obtain ∑ 𝐺𝑚𝐵 𝑚 ) ( (15.60) = 𝑚 𝑟̈ − 𝑟𝜃̇ 2 , 𝐹𝑟∶ − 2 𝑟 ∑ ( ) 𝐹𝜃∶ 0 = 𝑚 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ , (15.61) where we have substituted in the kinematic equations in polar components and have also used Eq. (15.59). In Section 15.3, we saw that Eq. (15.61) is equivalent to the conservation of angular momentum of 𝑆 about 𝐵, that is, ℎ𝐵 = 𝑚𝑟2 𝜃̇ = constant.

(15.62)

that the generally accepted value of this constant is 𝐺 = 6.674 × 10−11 m3 ∕(kg⋅s2 ) = 3.439 × 10−8 f t 3 ∕(slug⋅s2 ).

∗ Recall

1007

orbit of 𝑆

𝑣

𝑆 𝐹 focus of elliptical orbit

𝑟

𝑑𝜃 𝑢̂ 𝜃

𝑢̂ 𝑟

𝐵 Figure 15.24 Satellite 𝑆 of mass 𝑚 orbiting a primary body 𝐵 of mass 𝑚𝐵 under the action of Newton’s universal law of gravitation.

Helpful Information Evaluating 𝑮𝒎𝑩 for the Earth. If the primary body 𝐵 is the Earth, then from Eq. (11.10) on p. 622 we can write 𝐺𝑚𝐵 = 𝐺𝑚𝑒 = 𝑔𝑟2𝑒 , where 𝑚𝑒 is the mass of the Earth, 𝑔 is the acceleration due to gravity, and 𝑟𝑒 is the radius of the Earth, which is 6371 𝗄𝗆, or 3959 𝗆𝗂.

1008

ISTUDY

Chapter 15

Momentum Methods for Particles

Interesting Fact The primary mass really moves. In reality, the satellite 𝑆 and the primary body 𝐵 orbit about their common mass center. If this motion of 𝐵 is taken into account, Eqs. (15.60) and (15.61) no longer apply, and the governing equations become −

𝐺𝑚𝐵 𝑚 𝑟2

𝑚𝐵 𝑚 (̈𝑟 − 𝑟𝜃̇ 2 ), 𝑚𝐵 + 𝑚 𝑚𝐵 𝑚 ̇ 0= (𝑟𝜃̈ + 2𝑟̇ 𝜃), 𝑚𝐵 + 𝑚 =

where 𝑟 is the distance between mass centers of 𝐵 and 𝑆. Notice that if 𝑚𝐵 ≫ 𝑚, then these two equations are approximated by Eqs. (15.60) and (15.61). See Probs. 12.82 and 12.83 for a one-dimensional version of these two equations.

Equation (15.62) is also a reflection of Kepler’s second law of planetary motion. To see this, refer to Fig. 15.24 and note that the area 𝑑𝐴 swept by the radial line of length 𝑟 during the time 𝑑𝑡 (i.e., the yellow area) is equal to 12 (𝑟 𝑑𝜃)𝑟. Therefore, the areal velocity, which is the time rate at which area is swept by the radial line 𝑟, is constant according to Eq. (15.62) and is given by 𝑑𝐴 1 2 ̇ = 2 𝑟 𝜃 = constant. 𝑑𝑡

(15.63)

This is Kepler’s second law, which states that the line joining a planet to the Sun sweeps out equal areas in equal times as the planet travels around the Sun. Solution of the governing equations To find the trajectory of 𝑆, we need to solve Eqs. (15.60) and (15.61). This is traditionally done by eliminating 𝑡 and then solving for 𝑟(𝜃). To begin, we first rewrite Eqs. (15.60) and (15.61) as 𝑟̈ − 𝑟𝜃̇ 2 = −

𝐺𝑚𝐵 𝑟2

,

𝑟2 𝜃̇ = 𝜅,

(15.64) (15.65)

where 𝜅 (the Greek letter kappa) is the angular momentum per unit mass of the satellite 𝑆, which is equal to ℎ𝐵 ∕𝑚 from Eq. (15.62). The chain rule allows us to write 𝑟̇ and 𝑟̈ as ( ) 𝑑𝑟 𝑑𝜃 𝑑𝑟 𝜅 𝑑 1 𝑟̇ = = = −𝜅 , (15.66) 𝑑𝜃 𝑑𝑡 𝑑𝜃 𝑟2 𝑑𝜃 𝑟 [ ( )] ( ) 𝑑 𝑟̇ 𝜅 𝑑 1 𝜅2 𝑑2 1 𝜅 𝑑 𝑑 𝑟̇ 𝑑𝜃 = −𝜅 =− . (15.67) = 𝑟̈ = 𝑑𝜃 𝑑𝑡 𝑑𝜃 𝑟2 𝑑𝜃 𝑟 𝑟2 𝑑𝜃 𝑟2 𝑑𝜃 2 𝑟 Using Eqs. (15.65) and (15.67), Eq. (15.64) becomes 𝜅2 𝑑2 − 𝑟2 𝑑𝜃 2

( ) ( )2 𝐺𝑚𝐵 1 𝜅 −𝑟 =− . 2 𝑟 𝑟 𝑟2

(15.68)

Letting 𝑢 = 1∕𝑟, Eq. (15.68) becomes −𝜅 2 𝑢2

𝑑2𝑢 − 𝜅 2 𝑢3 = −𝐺𝑚𝐵 𝑢2 , 𝑑𝜃 2

(15.69)

which, upon canceling 𝑢2 and rearranging, becomes 𝐺𝑚𝐵 𝑑2𝑢 +𝑢= . 2 𝑑𝜃 𝜅2

(15.70)

As we will see again in Section 19.2, this is a second-order, constant-coefficient, nonhomogeneous, differential equation whose solution can be verified by direct substitution to be 𝐺𝑚𝐵 1 𝑢 = = 𝐶 cos(𝜃 − 𝛽) + , (15.71) 𝑟 𝜅2 where 𝐶 and 𝛽 are constants of integration. Equation (15.71) determines the unpowered or free-flight trajectory of the satellite. It is the polar coordinate representation of a conic section, which is defined as

ISTUDY

Section 15.4

Orbital Mechanics

1009

𝑝 conic section

periapsis directrix 𝐷

focus 𝐵 𝜃

apoapsis

𝑟

𝑑

Figure 15.25. Geometry used to define a conic section. In this case, the conic section is an ellipse with eccentricity 𝑒 = 0.661.

follows: Given a point 𝐵, called the focus, and a line 𝐷, called the directrix, a conic section is defined as the locus of points for which the ratio (see Fig. 15.25) 𝑒=

𝑟 distance to 𝐵 = distance to 𝐷 𝑑

(15.72)

is a constant. As we shall soon see, circles, ellipses, parabolas, and hyperbolas are all conic sections. The ratio in Eq. (15.72) is called the eccentricity of the conic section. We now need to obtain the constants of integration in Eq. (15.71). The constant 𝛽 in Eq. (15.71) can be eliminated by letting the 𝜃 = 0 axis be at periapsis, that is, at the point where the orbital radius 𝑟 is a minimum. At periapsis, 𝑟̇ = 0, which implies that 𝑑𝑢∕𝑑𝜃 = 0 from Eq. (15.66). Applying this condition, we find that 𝛽 = 0, which simplifies Eq. (15.71) to 𝐺𝑚𝐵 1 = 𝐶 cos 𝜃 + . 𝑟 𝜅2

(15.73)

Referring to Eq. (15.72) and Fig. 15.25, we can use the definition of eccentricity 𝑒 to write 𝑟 = 𝑒𝑑 = 𝑒(𝑝 − 𝑟 cos 𝜃), (15.74) where 𝑝 is the focal parameter. Rearranging this equation, we obtain 1 1 1 = cos 𝜃 + . 𝑟 𝑝 𝑒𝑝

(15.75)

Comparing Eqs. (15.73) and (15.75), we see that 1 , 𝐶

(15.76)

𝐶𝜅 2 . 𝐺𝑚𝐵

(15.77)

𝑝= and that 𝑒=

Using Eq. (15.77), we can write (15.73) as 1 𝐺𝑚𝐵 (1 + 𝑒 cos 𝜃). = 𝑟 𝜅2

(15.78)

Equations (15.73) and (15.78) are equivalent to one another. In the former, we need to determine the constants 𝐶 and 𝜅, and in the latter we need to determine the constants 𝑒 and 𝜅 [Eq. (15.77) provides the link between these three constants]. In either case, these constants are determined by knowing the position and velocity of the satellite at some point on its trajectory.

Helpful Information Periapsis and apoapsis. An apsis is the point of maximum or minimum distance from the focus containing the center of attraction in an elliptical orbit. The point at which the orbital radius is a minimum is called the periapsis, and the point at which the orbital radius is a maximum is called the apoapsis. For specific celestial bodies, names that apply specifically to orbits about those bodies are generally used rather than periapsis and apoapsis. The following table lists some of these. Body

Minimum 𝑟

Maximum 𝑟

Earth Moon Sun Mercury Venus Mars Jupiter Saturn

perigee perilune perihelion perihermion pericythe periareion perijove perichron

apogee apolune aphelion apohermion apocythe apoareion apojove apochron

1010

Chapter 15

Momentum Methods for Particles

𝑣𝐴 = 𝑣min

Launch conditions at periapsis. Consider a satellite in orbit about a primary body, as shown in Fig. 15.26. If instead of knowing the launch conditions at an arbitrary position 𝑆, we know the launch conditions at periapsis (point 𝑃 ), then 𝜙𝑃 = 90◦ and 𝜃𝑃 = 0◦ . Under these conditions, Eq. (15.65) tells us that the angular momentum per unit mass 𝜅 becomes 𝜅 = 𝑟2 𝜃̇ = 𝑟 𝑣 . (15.79)

𝐴

𝑟𝐴 = 𝑟max

𝑃 𝑃

𝐵

𝑣 𝜙

𝑟

𝑟𝑃 = 𝑟min

𝜃

𝑃 𝑃

To determine 𝐶, we evaluate Eq. (15.73) at 𝑃 to find that ( 1 𝐶= 𝑟𝑃

𝑆

1−

𝐺𝑚𝐵

)

𝑟𝑃 𝑣2𝑃

,

(15.80)

𝑣𝑃 = 𝑣max

𝑃

Figure 15.26 A satellite 𝑆 in orbit about a primary body 𝐵. The launch conditions are at periapsis 𝑃 and are 𝑟𝑃 and 𝑣𝑃 . The angle 𝜙 is defined to be the angle between the radial line 𝐵𝑆 and the velocity vector of the satellite. At 𝑃 and 𝐴, 𝜙 = 90◦ .

where we have used 𝜅 from Eq. (15.79). Substituting Eqs. (15.79) and (15.80) into Eq. (15.73), we obtain the following equation for the trajectory of the satellite: ( 1 1 = 𝑟 𝑟𝑃

1−

𝐺𝑚𝐵

)

𝑟𝑃 𝑣2𝑃

cos 𝜃 +

𝐺𝑚𝐵 𝑟2𝑃 𝑣2𝑃

.

(15.81)

Conic sections hyperbola, 𝑒 > 1 parabola, 𝑒 = 1 ellipse, 0 < 𝑒 < 1

𝑣𝑃

circle, 𝑒 = 0 periapsis

𝑟𝑃

𝐵

apoapsis

Figure 15.27 The four conic sections, showing 𝑟𝑃 and 𝑣𝑃 .

ISTUDY

We have already mentioned that the solution 𝑟(𝜃) to the governing orbital equations is a conic section. The type of conic section depends on the eccentricity of the trajectory 𝑒 (see Fig. 15.27), which, in turn, depends on 𝐶 and 𝜅 via Eq. (15.77). We will now look at each of the four possible cases: 𝑒 = 0, 0 < 𝑒 < 1, 𝑒 = 1, and 𝑒 > 1. Circular orbit (𝒆 = 𝟎). If 𝑟𝑃 and 𝑣𝑃 are chosen so that 𝑒 = 0, then Eq. (15.77) tells us that 𝐶𝜅 2 = 0, which implies that 𝐶 = 0 since 𝜅 cannot be zero. If 𝐶 = 0, then Eq. (15.80) tells us that the speed 𝑣𝑐 in a circular orbit of radius 𝑟𝑃 is given by ( 1 𝑟𝑃

1−

𝐺𝑚𝐵 𝑟𝑃 𝑣2𝑐

) =0

1=

⇒

𝐺𝑚𝐵 𝑟𝑃 𝑣2𝑐

√ 𝑣𝑐 =

⇒

𝐺𝑚𝐵 𝑟𝑃

.

(15.82)

Elliptical orbit (𝟎 < 𝒆 < 𝟏). For an elliptical orbit, 0 < 𝑒 < 1, and when evaluated at periapsis (𝜃 = 0◦ ), Eq. (15.81) gives 𝑟 = 𝑟𝑃 as expected since 𝑟𝑃 was chosen at 𝜃 = 0◦ . If we instead evaluate Eq. (15.81) at 𝜃 = 180◦ , we will find 𝑟 at apoapsis, that is, 𝑟𝐴 . Doing so gives ( ) 2𝐺𝑚𝐵 − 𝑟𝑃 𝑣2𝑃 𝐺𝑚𝐵 𝐺𝑚𝐵 1 −1 = 1− + = , (15.83) 𝑟𝐴 𝑟𝑃 𝑟𝑃 𝑣2𝑃 𝑟2𝑃 𝑣2𝑃 𝑟2𝑃 𝑣2𝑃 or, by simplifying and rearranging, 𝑟𝐴 =

𝑟𝑃 , /( ) 2𝐺𝑚𝐵 𝑟𝑃 𝑣2𝑃 − 1

(15.84)

where we recall that the launch conditions are at periapsis. An additional set of relations for elliptical orbits is obtained by evaluating Eq. (15.75) at periapsis (𝜃 = 0◦ ) and apoapsis (𝜃 = 180◦ ) to obtain 1 1 1 = + 𝑟𝑃 𝑝 𝑒𝑝

⇒

𝑟𝑃 =

𝑝𝑒 , 1+𝑒

(15.85)

ISTUDY

Section 15.4

Orbital Mechanics

and −1 1 1 = + 𝑟𝐴 𝑝 𝑒𝑝

𝑟𝐴 =

⇒

𝑝𝑒 , 1−𝑒

1011

(15.86)

respectively, where 𝑝 is the focal parameter shown in Fig. 15.25. Taking the ratio of Eqs. (15.85) and (15.86), it is easy to show that 𝑟𝐴 can be written in terms of 𝑒 as ) ( 1+𝑒 . (15.87) 𝑟𝐴 = 𝑟𝑃 1−𝑒 Referring to Fig. 15.28, and using Eqs. (15.85) and (15.86), we obtain

𝑟𝑃 = 𝑎(1 − 𝑒)

𝑟𝐴 = 𝑎(1 + 𝑒)

periapsis

𝑏

apoapsis

𝑏

𝑎

𝑎

Figure 15.28. The semimajor axis 𝑎 and semiminor axis 𝑏 of an ellipse.

(

1 1 2𝑎 = 𝑟𝑃 + 𝑟𝐴 = 𝑝𝑒 + 1+𝑒 1−𝑒

) ⇒

𝑎=

𝑝𝑒 , 1 − 𝑒2

(15.88)

where 𝑎 is referred to as the semimajor axis of the ellipse. Solving Eq. (15.88) for 𝑝 and substituting the result into Eq. (15.75) give 1 1 + 𝑒 cos 𝜃 = ( ), 𝑟 𝑎 1 − 𝑒2

(15.89)

which, when evaluated at periapsis and apoapsis, gives 𝑟𝑃 = 𝑎(1 − 𝑒)

and

𝑟𝐴 = 𝑎(1 + 𝑒).

(15.90)

Equation (15.89) is a statement of Kepler’s first law, which states that orbits of planets are ellipses with the Sun at one focus. To find the period of an elliptical orbit, we refer to the definition of areal velocity in Eq. (15.63) and integrate that equation over the entire ellipse to find that 𝜅 𝑑𝐴 1 2 ̇ = 2𝑟 𝜃 = 𝑑𝑡 2

⇒

∫0

𝐴

𝜏

𝑑𝐴 =

𝜅 𝑑𝑡 ∫0 2

⇒

𝐴=

𝜅 𝜏. 2

(15.91)

̇ which is Here 𝜏 is the orbital period, and we have used Eq. (15.65) to write 𝜅 = 𝑟2 𝜃, constant. For an ellipse, analytical geometry tells us that its area is equal to 𝜋𝑎𝑏, where 𝑏 is the semiminor axis of the ellipse (see Fig. 15.28). So we can write Eq. (15.91) as 𝜋𝑎𝑏 =

𝜅 𝜏 2

⇒

𝜏=

2𝜋𝑎𝑏 . 𝜅

(15.92)

Atlaspix/Alamy Stock Photo

Figure 15.29 A portrait of Johannes Kepler painted in 1610.

1012

ISTUDY

Chapter 15

Momentum Methods for Particles

Now, it can be shown that (see Prob. 15.143) √ 𝑏 = 𝑟𝑃 𝑟 𝐴 ,

(15.93)

and since Eq. (15.88) tells us that 𝑎 = (𝑟𝑃 + 𝑟𝐴 )∕2, we can write Eq. (15.92) as 𝜏=

)√ 𝜋( 𝑟 𝑃 + 𝑟𝐴 𝑟𝑃 𝑟𝐴 , 𝜅

(15.94)

where 𝜅 can be found by using Eq. (15.79). Kepler’s third law states that the square of the orbital period is proportional to the cube of the semimajor axis of that orbit. To show this, we start with Eq. (15.94) and substitute in Eqs. (15.90) to obtain ]√ 2𝜋 2 √ 𝜋[ 𝑎(1 − 𝑒) + 𝑎(1 + 𝑒) 𝑎(1 − 𝑒)𝑎(1 + 𝑒) = 𝑎 1 − 𝑒2 . (15.95) 𝜏= 𝜅 𝜅 Squaring both sides and noting from Eq. (15.88) that 𝑎(1 − 𝑒2 ) = 𝑝𝑒, we have 𝜏2 =

4𝜋 2 3 𝑎 𝑝𝑒. 𝜅2

(15.96)

Using Eqs. (15.76) and (15.77), we can write the product 𝑝𝑒 as 𝜅 2 ∕(𝐺𝑚𝐵 ), which means that Eq. (15.96) becomes 𝜏2 =

4𝜋 2 3 𝑎 , 𝐺𝑚𝐵

(15.97)

which is Kepler’s third law. Parabolic trajectory (𝒆 = 𝟏). If 𝑟𝑃 and 𝑣𝑃 are chosen so that the trajectory is parabolic, then 𝑒 = 1. Figure 15.27 indicates that the parabolic trajectory is one that divides those trajectories that are periodic from those that are not. That is, it divides trajectories that return to their initial starting point from those that do not. For a given 𝑟𝑃 , the launch velocity 𝑣par required to achieve a parabolic trajectory can be found by letting 𝑒 = 1 in Eq. (15.77), which gives 𝐺𝑚𝐵 = 𝐶𝜅 2 , and then substituting 𝐶 from Eq. (15.80) into that result to obtain ( ) 𝐺𝑚𝐵 ( )2 1 1− 𝑟𝑃 𝑣par , (15.98) 𝐺𝑚𝐵 = 2 𝑟𝑃 𝑟𝑃 𝑣par which, when solved for 𝑣par , becomes

Interesting Fact Crashing into the Earth to escape velocity. Comparing Eqs. (15.82) √ and (15.99), we see that only a factor of 2 separates an orbit that crashes into the Earth (𝑣 < 𝑣𝑐 ) with one that never returns (𝑣 > 𝑣esc ).

√ 𝑣par = 𝑣esc =

2𝐺𝑚𝐵 𝑟𝑃

.

(15.99)

The speed given in Eq. (15.99) is often referred to as escape velocity since it is the speed required to completely escape the influence of the primary body 𝐵. Hyperbolic trajectory (𝒆 > 𝟏). For a hyperbolic trajectory, the governing equations given by Eqs. (15.77)–(15.81) are used for values of 𝑒 > 1. Hyperbolic trajectories are unbound, and they are crucial to understanding interplanetary trajectories. By definition, these trajectories require a satellite to escape the orbit of a departure planet, fly under the influence of the Sun for a period of time, and then settle into an

ISTUDY

Section 15.4

1013

Orbital Mechanics

orbit around an arrival planet. Some of the quantities we’ve introduced for circular and elliptical orbits need to be redefined to account for 𝑒 > 1. For instance, the focal parameter 𝑝 and the perigee radius in a hyperbolic trajectory become: 𝑝𝑒 = 𝑎(𝑒2 − 1),

𝑟𝑃 = 𝑎(𝑒 − 1),

and

𝑒 > 1.

(15.100)

We can also define an apogee radius, although it doesn’t have the same meaning as one in a bound orbit. Because hyperbolic trajectories require additional development, almost all of our attention in this chapter will be devoted to bound (circular or elliptical) orbits.

Energy considerations

apoapsis

The motion of a satellite 𝑆 that is governed by Eqs. (15.64) and (15.65) not only conserves angular momentum about the primary body 𝐵, but also conserves total mechanical energy. We know this because the only force we are modeling is gravity, and Eq. (14.25) on p. 879 reminds us that this force is conservative. Therefore, the work-energy principle tells us that (see Fig. 15.30) 𝐺𝑚𝐵 𝑚

𝑇𝑃 + 𝑉𝑃 = 12 𝑚𝑣2𝑃 −

𝑟𝑃

= constant,

(15.101)

𝑟𝐴

orbit of 𝑆

2𝑎 𝐵

which, after dividing through by 𝑚, gives 1 2 𝑣 2 𝑃

−

𝐺𝑚𝐵 𝑟𝑃

=𝐸

or

𝐺𝑚𝐵 𝜅2 − = 𝐸, 2 𝑟𝑃 2𝑟𝑃

where we used the first of Eqs. (15.90) for 𝑟𝑃 . Substituting Eq. (15.103) and the first of Eqs. (15.90) into Eq. (15.102), we obtain ( ) 𝐺𝑚𝐵 𝑎 1 − 𝑒2 𝐺𝑚𝐵 𝐺𝑚𝐵 − 𝐸=− (15.104) ⇒ , 𝐸= 2 2 𝑎(1 − 𝑒) 2𝑎 2𝑎 (1 − 𝑒) where we see that the total energy of the satellite depends on only the semimajor axis 𝑎 of the orbit. Now that we have Eq. (15.104), we can apply the work-energy principle at an arbitrary position in the orbit to obtain 𝐺𝑚𝐵 2𝑎

= 12 𝑣2 −

𝐺𝑚𝐵 𝑟

.

(15.105)

Solving this equation for 𝑣, we have √ 𝑣=

𝐺𝑚𝐵

(

) 2 1 − , 𝑟 𝑎

𝑟𝑃

(15.102)

where 𝐸 is the mechanical energy per unit mass (sometimes called the specific energy) of the satellite. We used Eq. (15.79) to introduce 𝜅. Next, we evaluate Eq. (15.78) at periapsis (i.e., 𝜃 = 0) and solve for 𝜅 to obtain ( ) 𝜅 2 = 𝐺𝑚𝐵 (1 + 𝑒)𝑟𝑃 = 𝐺𝑚𝐵 (1 + 𝑒)𝑎(1 − 𝑒) = 𝐺𝑚𝐵 𝑎 1 − 𝑒2 , (15.103)

𝐸=−

𝑣

𝑟

(15.106)

which is very useful for solving orbital transfer problems and for finding the speed of a satellite at arbitrary positions in its orbit.

periapsis

𝑆

𝜃 𝑣𝑃

Figure 15.30 A satellite 𝑆 in an elliptical orbit about a primary body 𝐵. The orbital parameters 𝑎, 𝑟𝑃 , 𝑟𝐴 , 𝑟, and 𝜃 are shown.

1014

Chapter 15

Momentum Methods for Particles

End of Section Summary In this section, we studied the motion of a satellite 𝑆 of mass 𝑚 that is subject to Newton’s universal law of gravitation due to a body 𝐵 of mass 𝑚𝐵 , which is the primary or attracting body (see Fig. 15.31). We began with these important assumptions: 1. The primary body 𝐵 and the satellite 𝑆 are both treated as particles. apoapsis

2. The only force acting on the satellite 𝑆 is the force of mutual attraction between 𝐵 and 𝑆. orbit of 𝑆

3. The primary body 𝐵 is fixed in space. Determination of the orbit 𝐵

𝑣

𝑟 𝑟𝑃

periapsis

𝜃

𝑆

𝐺𝑚𝐵 1 = 𝐶 cos 𝜃 + or 𝑟 𝜅2 1 𝐺𝑚𝐵 = (1 + 𝑒 cos 𝜃), 𝑟 𝜅2

𝑣𝑃

Figure 15.31 A satellite 𝑆 of mass 𝑚 orbiting a primary body 𝐵 of mass 𝑚𝐵 on a conic section, which, in this case, is an ellipse. The angle 𝜃 and the launch conditions 𝑟𝑃 and 𝑣𝑃 are defined from periapsis.

ISTUDY

Solving the governing equations in polar coordinates, we found that the trajectory of the satellite 𝑆 under these assumptions is a conic section, whose equation can be written as Eqs. (15.73) and (15.78), p. 1009

where 𝑟 is the distance between the centers of mass of 𝑆 and 𝐵; 𝜃 is the orbital angle measured relative to periapsis; 𝐺 is the universal gravitational constant; 𝜅 is the angular momentum per unit mass of the satellite 𝑆 measured about 𝐵; 𝐶 is a constant to be determined; and 𝑒 is the eccentricity of the trajectory, which can be written as Eq. (15.77), p. 1009 𝑒=

𝐶𝜅 2 . 𝐺𝑚𝐵

If, as is generally the case here, the orbital conditions are known at periapsis, then 𝜅 is Eq. (15.79), p. 1010 𝜅 = 𝑟𝑃 𝑣𝑃 , the constant 𝐶 is Eq. (15.80), p. 1010 ( ) 𝐺𝑚𝐵 1 𝐶= 1− , 𝑟𝑃 𝑟𝑃 𝑣2𝑃 and the equation describing the trajectory becomes Eq. (15.81), p. 1010 ( ) 𝐺𝑚𝐵 𝐺𝑚𝐵 1 1 1− cos 𝜃 + = . 2 𝑟 𝑟𝑃 𝑟𝑃 𝑣𝑃 𝑟2𝑃 𝑣2𝑃

ISTUDY

Section 15.4

Orbital Mechanics

Conic sections. Equations (15.73), (15.78), and (15.81) represent equivalent conic sections in polar coordinates. The type of conic section depends on the eccentricity of the trajectory 𝑒 (see Fig. 15.32), which, in turn, depends on 𝐶 and 𝜅 via Eq. (15.77). There are four types of conic sections, which are determined by the value of the eccentricity 𝑒, that is, 𝑒 = 0, 0 < 𝑒 < 1, 𝑒 = 1, and 𝑒 > 1. For a circular orbit (𝑒 = 0), the radius is 𝑟𝑃 , and the speed in the orbit is equal to Eq. (15.82), p. 1010 √ 𝐺𝑚𝐵 𝑣𝑐 = . 𝑟𝑃

hyperbola, 𝑒 > 1 parabola, 𝑒 = 1 ellipse, 0 < 𝑒 < 1

𝑣𝑃

circle, 𝑒 = 0

For an elliptical orbit (0 < 𝑒 < 1), as expected, the radius at periapsis is 𝑟𝑃 . The radius at apoapsis is given by periapsis

Eq. (15.84), p. 1010 𝑟𝐴 =

1015

𝑟𝑃

𝐵

apoapsis

𝑟𝑃 . /( ) 2𝐺𝑚𝐵 𝑟𝑃 𝑣2𝑃 − 1

Referring to Fig. 15.33, additional relationships between the semimajor axis 𝑎 of an

𝑟𝑃 = 𝑎(1 − 𝑒)

𝑟𝐴 = 𝑎(1 + 𝑒)

periapsis

𝑏

apoapsis

𝑏

𝑎

𝑎

Figure 15.33. The semimajor axis 𝑎 and semiminor axis 𝑏 of an ellipse.

elliptical orbit, the eccentricity of the orbit, and the radii at periapsis and apoapsis are, respectively, Eqs. (15.90), p. 1011 𝑟𝑃 = 𝑎(1 − 𝑒) and

𝑟𝐴 = 𝑎(1 + 𝑒).

The period of an elliptical orbit 𝜏 can be written in the following two ways: Eq. (15.92), p. 1011, and Eq. (15.94), p. 1012 𝜏=

)√ 2𝜋𝑎𝑏 𝜋 ( 𝑟𝑃 + 𝑟𝐴 𝑟𝑃 𝑟𝐴 , = 𝜅 𝜅

Figure 15.32 The four conic sections, showing 𝑟𝑃 and 𝑣𝑃 .

1016

ISTUDY

Chapter 15

Momentum Methods for Particles

or, reflecting Kepler’s third law, the orbital period can be written as Eq. (15.97), p. 1012 𝜏2 =

4𝜋 2 3 𝑎 . 𝐺𝑚𝐵

A parabolic trajectory (𝑒 = 1) is that which divides periodic orbits (which return to their starting location) from trajectories that are not periodic. For a given 𝑟𝑃 , the speed 𝑣par required to achieve a parabolic trajectory is given by Eq. (15.99), p. 1012 √ 2𝐺𝑚𝐵 𝑣par = 𝑣esc = , 𝑟𝑃 which is also referred to as the escape velocity since it is the speed required to completely escape the influence of the primary body 𝐵. For a hyperbolic trajectory (𝑒 > 1), the governing equations given by Eqs. (15.77)– (15.81) are used for values of 𝑒 > 1. Energy considerations Using the work-energy principle, we discovered that the total mechanical energy in an orbit depends on only the semimajor axis 𝑎 of the orbit and is given by Eq. (15.104), p. 1013 𝐸=−

𝐺𝑚𝐵 2𝑎

,

where 𝐸 is the mechanical energy per unit mass of the satellite. Applying the workenergy principle at an arbitrary location within the orbit, we found that the speed can be written as Eq. (15.106), p. 1013 √ ) ( 2 1 . − 𝑣 = 𝐺𝑚𝐵 𝑟 𝑎

ISTUDY

Section 15.4

Orbital Mechanics

E X A M P L E 15.16

1017

Apogee and Perigee Speeds for a Given Orbit

An artificial satellite is launched from an altitude of 500 km with a velocity that is parallel to the surface of the Earth (Fig. 1). Requiring that the altitude at apogee be 20,000 km and using 6371 km for the radius of the Earth, determine (a) the required speed at perigee 𝑣𝑃 ,

Earth

𝑣𝑃

20,000 km

(b) the speed at apogee 𝑣𝐴 , and

500 km

(c) the period of the orbit.

SOLUTION This is an Earth orbit, and we are given 𝑟𝑃 and the required 𝑟𝐴 , so Eq. (15.84) will allow us to determine the speed at perigee 𝑣𝑃 . Knowing 𝑟𝑃 and having found 𝑣𝑃 , we see that Eq. (15.79) allows us to find 𝜅, which will then allow us to find 𝑣𝐴 given that 𝜅 is conserved and will allow us to find the orbital period 𝜏 using Eq. (15.94). Note that the orbital eccentricity mentioned in the caption of Fig. 1 can be found from either of Eqs. (15.90).

Road Map & Modeling

Computation

Referring to Fig. 1, we see that an altitude of 500 km at perigee corre-

sponds to 𝑟𝑃 = (500 + 6371) km = 6871×103 m.

(1)

Similarly, the altitude of 20,000 km at apogee means 𝑟𝐴 = (20,000 + 6371) km = 26,370×103 m. Using Eq. (11.10) on p. 622 in Eq. (15.84) and solving for 𝑣𝑃 , we find that √ 𝑟2𝑃 𝑣2𝑃 2𝑔𝑟𝐴 𝑟2𝑒 𝑟𝐴 = ⇒ 𝑣𝑃 = ( ) 2 2 2𝑔𝑟𝑒 − 𝑟𝑃 𝑣𝑃 𝑟𝐴 + 𝑟𝑃 𝑟𝑃 ⇒

𝑣𝑃 = 9589 m∕s = 34,520 km∕h,

(2)

(3) (4)

where 𝑟𝑒 is the radius of the Earth. To find the speed at apogee, we note from Eq. (15.79) that the angular momentum per unit mass 𝜅 is constant and so 𝑟 𝜅 = 𝑟𝑃 𝑣𝑃 = 𝑟𝐴 𝑣𝐴 ⇒ 𝑣𝐴 = 𝑃 𝑣𝑃 (5) 𝑟𝐴 ⇒

𝑣𝐴 = 2498 m∕s = 8994 km∕h.

(6)

To determine the period of the orbit, we can use Eq. (15.94) after substituting in 𝜅 from Eq. (5) as follows: 𝜏=

)√ 𝜋 ( 𝑟𝑃 + 𝑟𝐴 𝑟𝑃 𝑟𝐴 = 21,340 s = 5.927 h. 𝑟𝑃 𝑣𝑃

Discussion & Verification

(7)

It can be seen that both Eqs. (3) and (5) have dimensions of velocity, as they should. Equation (7) has the dimension of time, as it should. We also note that the speed at apogee is less than the speed at perigee, as it should be, since 𝜅 is constant.

Figure 1 An elliptical orbit about Earth with a launch altitude of 500 km and an apogee altitude of 20,000 km. It can be shown that the orbit has an eccentricity 𝑒 of 0.5866. The figure is drawn to scale.

1018

Chapter 15

Momentum Methods for Particles

Problems For the problems in this section, use 6371 km or 3959 mi for the radius of the Earth. Not all orbits or objects are drawn to scale. 𝑎 𝐷 𝑏 𝑂

Using the lengths shown, as well as the property of an ellipse that states that the sum of the distances from each of the foci (i.e., points 𝑂 and 𝐵) to any point on the ellipse is a constant, prove Eq. (15.93), that is, that the length of the semiminor axis can be related to √ the periapsis and apoapsis radii via 𝑏 = 𝑟𝑃 𝑟𝐴 .

𝓁

𝓁

𝑃

Problem 15.143

𝐶

𝐵

𝐴

Problem 15.144 𝑟𝑃

Using the last of Eqs. (15.103), along with Eq. (15.104), solve for the eccentricity 𝑒 as a function of 𝐸, 𝜅, and 𝐺𝑚𝐵 .

𝑟𝐴

(a) Using that result, along with fact that 𝑒 ≥ 0, show that 𝐸 < 0 corresponds to an elliptical orbit, 𝐸 = 0 corresponds to a parabolic trajectory, and 𝐸 > 0 corresponds to a hyperbolic trajectory.

Figure P15.143

(b) Show that for 𝑒 = 0, the expression you found for 𝑒 leads to Eq. (15.82). ellipse, 𝐸 0 Figure P15.144

Problem 15.145 Assuming that the Sun is the only significant body in the solar system (the mass of the Sun accounts for 99.8% of the mass of the solar system), determine the escape velocity from the Sun as a function of the distance 𝑟 from its center. What is the value of the escape velocity (expressed in km/h) when 𝑟 is equal to the radius of Earth’s orbit? Use 1.989×1030 kg for the mass of the Sun and 150×106 km for the radius of Earth’s orbit.

parabolic orbit

Problem 15.146

𝑃

The S-IVB third stage of the Saturn V rocket, which was used for the Apollo missions, would burn for about 2.5 min to place the spacecraft into a “parking orbit.”∗ Then, after several orbits, it would burn for about 6 min to accelerate the spacecraft to escape velocity to send it to the Moon. Assuming a circular parking orbit with an altitude of 170 km, determine the change in speed needed at 𝑃 to go from the parking orbit to escape velocity. Assume that the change in speed occurs instantaneously so that you need not worry about changes in orbital position during the engine thrust.

170 0 km parking orbit

Problem 15.147

Figure P15.146

In 1705, Edmund Halley (1656–1742), an English astronomer, claimed that the comet sightings of 1531, 1607, and 1682 were all the same comet. He predicted this comet would return again in 1758. Halley did not live to see the comet’s return, but it did return late in

ISTUDY

∗A

parking orbit is a temporary orbit of an artificial satellite or spacecraft in preparation for thrusting to another orbit or trajectory.

ISTUDY

Section 15.4

1758 and reached perihelion in March 1759. In honor of his prediction, this comet was named “Halley.” Each elliptical orbit of Halley is slightly different, but the average value of the semimajor axis 𝑎 is about 17.95 AU.∗ Using this value, along with the fact that its orbital eccentricity is 0.967 (the orbit is drawn to scale, but the Sun is shown to be 36 times bigger than it should be), determine

orbit of Halley’s comet Sun

𝐴

Figure P15.147

(a) the orbital period in years of Halley’s comet, and (b) its distance, in AU, from the Sun at perihelion 𝑃 and at aphelion 𝐴. (c) Look up the orbits of the planets of our solar system on the Web. What planetary orbits is Halley near to at perihelion and aphelion? Use 1.989×1030 kg for the mass of the Sun.

Problems 15.148 and 15.149 Explorer 7 was launched on October 13, 1959, with an apogee altitude above the Earth’s surface of 1073 km and a perigee altitude of 573 km above the Earth’s surface. Its orbital period was 101.4 min. Problem 15.148

with 𝑔𝑟2𝑒 .

Using this information, calculate 𝐺𝑚𝑒 for the Earth and compare it

Problem 15.149 Determine the eccentricity of the Explorer 7’s orbit, as well as its speeds at perigee and apogee.

Problem 15.150 A geosynchronous equatorial orbit is a circular orbit above the Earth’s equator that has a period of 1 day (these are sometimes called geostationary orbits). These geostationary orbits are of great importance for telecommunications satellites because a satellite orbiting with the same angular rate as the rotation rate of the Earth will appear to hover in the same point in the sky as seen by a person standing on the surface of the Earth. Using this information, determine the altitude ℎ𝑔 and radius 𝑟𝑔 of a geostationary orbit (in miles). In addition, determine the speed 𝑣𝑔 of a satellite in such an orbit (in miles per hour). 𝑣𝐵 𝐵 ℎ𝑔

Earth

𝑣𝑃

𝑟𝑔 20,000 km 500 km

Earth 𝑣𝑔

Figure P15.150

Figure P15.151

Problem 15.151 An artificial satellite is launched from an altitude of 500 km with a velocity 𝑣𝑃 that is parallel to the surface of the Earth. Requiring that the altitude at apogee be 20,000 km, determine the velocity at 𝐵, that is, the position in the orbit when the velocity is first orthogonal to the launch velocity. ∗ One

1019

Orbital Mechanics

astronomical unit (AU) is the distance between the center of mass of the Earth and that of the Sun and is approximately 1.496×108 km = 9.296×107 mi.

NASA

Figure P15.148 and P15.149

𝑃

1020

Chapter 15

Momentum Methods for Particles

Problem 15.152 The mass of the planet Jupiter is 318 times that of Earth, and its equatorial radius is 71,500 km. If a space probe is in a circular orbit about Jupiter at the altitude of the Galilean moon Callisto (orbital altitude 1.812×106 km), determine the change in speed Δ𝑣 needed in the outer orbit so that the probe reaches a minimum altitude at the orbital radius of the Galilean moon Io (orbital altitude 3.502 × 105 km). Assume that the probe is at the maximum altitude in the transfer orbit when the change in speed occurs and that the change in speed is impulsive; that is, it occurs instantaneously.

Jupiter

71,500 km

Callisto

Io

3.502 × 105 km 𝛥𝑣

1.812 × 106 km

transfer orbit Figure P15.152

Problem 15.153 The on-orbit assembly of the International Space Station (ISS) began in 1998 and continues today. The ISS has an apogee altitude above the Earth’s surface of 341.9 km and a perigee altitude of 331.0 km above the Earth’s surface. Determine its maximum and minimum speeds in orbit, its orbital eccentricity, and its orbital period. Research its actual orbital period and compare it with your calculated value.

NASA

Figure P15.153

ISTUDY

Problems 15.154 through 15.156 The optimal way (from an energy standpoint) to transfer from one circular orbit about a primary body 𝐵 to another circular orbit is via the so-called Hohmann transfer, which involves transferring from one circular orbit to another using an elliptical orbit that is tangent to both at the periapsis and apoapsis of the ellipse. The ellipse is uniquely defined because we know 𝑟𝑃 (the radius of the inner circular orbit) and 𝑟𝐴 (the radius of the outer circular orbit), and therefore we know the semimajor axis 𝑎 by Eq. (15.88) and the eccentricity 𝑒 by Eq. (15.87) or Eqs. (15.90). Performing a Hohmann transfer requires two maneuvers, the first to leave the inner (outer) circular orbit and enter the transfer ellipse and the second to leave the transfer ellipse and enter the outer (inner) circular orbit. Problem 15.154 A spacecraft 𝑆1 needs to transfer from circular low Earth parking orbit with altitude 120 mi above the surface of the Earth to a circular geosynchronous orbit with altitude 22,240 mi. Determine the change in speed Δ𝑣𝑃 required at perigee 𝑃 of the elliptical transfer orbit and the change in speed Δ𝑣𝐴 required at apogee 𝐴. In addition, compute the time required for the orbital transfer. Assume that the changes in speed are impulsive; that is, they occur instantaneously.

ISTUDY

Section 15.4

Problem 15.155 A spacecraft 𝑆2 must transfer from a circular Earth orbit whose period is 12 h (i.e., it is overhead twice per day) to a low Earth circular orbit with an altitude of 110 mi. Determine the change in speed Δ𝑣𝐴 required at apogee 𝐴 of the elliptical transfer orbit and the change in speed Δ𝑣𝑃 required at perigee 𝑃 . In addition, compute the time required for the orbital transfer. Assume that the changes in speed are impulsive; that is, they occur instantaneously.

𝐴

𝑟𝐴 𝑆2

𝑆1

𝐵

A spacecraft 𝑆1 is transferring from circular low Earth parking orbit with altitude 100 mi to a circular orbit with radius 𝑟𝐴 . Plot, as a function of 𝑟𝐴 for 𝑟𝑃 ≤ 𝑟𝐴 ≤ 100𝑟𝑃 , the change in speed Δ𝑣𝑃 required at perigee of the elliptical transfer orbit, as well as the change in speed Δ𝑣𝐴 required at apogee. In addition, plot the time as a function of 𝑟𝐴 , again for 𝑟𝑃 ≤ 𝑟𝐴 ≤ 100𝑟𝑃 , required for the orbital transfer. Assume that the changes in speed are impulsive; that is, they occur instantaneously. Problem 15.156

𝑟𝑃 𝑃

Problem 15.157 Referring to the description given for Probs. 15.154–15.156, for a Hohmann transfer from an inner circular orbit to an outer circular orbit, what would you expect to be the signs on the change in speed at periapsis and at apoapsis?

1021

Orbital Mechanics

inner circular orbit outer circular orbit elliptical transfer orbit Figure P15.154–P15.158

Problem 15.158 Referring to the description given for Probs. 15.154–15.156, for a Hohmann transfer from an outer circular orbit to an inner circular orbit, what would you expect to be the signs on the change in speed at periapsis and at apoapsis? 𝐴

Problem 15.159 During the Apollo missions, while the astronauts were on the Moon with the lunar module (LM), the command module (CM) would fly in a circular orbit around the Moon at an altitude of 60 mi. After the astronauts were done exploring the Moon, the LM would launch from the Moon’s surface (at 𝐿) and undergo powered flight until burnout at 𝑃 . This occurred when the LM was approximately 15 mi above the surface of the Moon with its velocity 𝑣bo parallel to the surface of the Moon (i.e., at periapsis). It would then fly under the influence of the Moon’s gravity until reaching apoapsis 𝐴, at which point it would rendezvous with the CM. The radius of the Moon is 1079 mi, and its mass is 0.0123 times that of the Earth. Assume that the changes in speed occur instantaneously. (a) (b) (c) (d)

Determine the required speed 𝑣bo at burnout 𝑃 . What is the change in speed Δ𝑣LM required of the LM at the rendezvous point 𝐴? Determine the time it takes the LM to travel from 𝑃 to 𝐴. In terms of the angle 𝜃, where should the CM be when the LM reaches 𝑃 so that they can rendezvous at 𝐴?

𝛥𝑣LM

ℎ𝐴

Moon 𝐵

LM

𝐿

ℎ𝑃

CM

𝑣bo

𝑃

𝜃 Figure P15.159

Problem 15.160 Use the work-energy principle applied between periapsis 𝑃 and 𝑟 = ∞, along with the potential energy of the force of gravity given in Eq. (14.25). (a) Show that a satellite on a hyperbolic trajectory arrives at 𝑟 = ∞ with speed √ 𝑟𝑃 𝑣2𝑃 − 2𝐺𝑚𝐵 . 𝑣∞ = 𝑟𝑃 (b) In addition, using Eqs. (15.77) and (15.80), show that for a hyperbolic trajectory, 𝑟𝑃 𝑣2𝑃 > 2𝐺𝑚𝐵 , which means that the square root in the above equation must always yield a real value.

𝐵 𝑟𝑃 𝑣𝑃 Figure P15.160

𝑃

hyperbolic trajectory

1022

ISTUDY

Chapter 15

Momentum Methods for Particles

Problem 15.161 One option when traveling to Mars from the Earth is to use a Hohmann transfer orbit like that described in Probs. 15.154–15.158. Assuming that the Sun is the primary gravitational influence and ignoring the gravitational influence of Earth and Mars (since the Sun accounts for 99.8% of the mass of the solar system), determine the change in speed required at the Earth Δ𝑣𝑒 (perihelion in the transfer orbit) and the required change in speed at Mars Δ𝑣𝑚 (aphelion in the transfer orbit) to accomplish the mission to Mars using a Hohmann transfer. In addition, determine the amount of time 𝜏 it would take for orbital transfer. Use 1.989 × 1030 kg for the mass of the Sun, assume that the orbits of Earth and Mars are circular, and assume that the changes in speed are impulsive; that is, they occur instantaneously. In addition, use 150 × 106 km for the radius of Earth’s orbit and 228×106 km for the radius of Mars’s orbit.

𝑅𝑚 𝑅𝑒 Earth Sun transfer orbit

Figure P15.161

Mars

ISTUDY

Section 15.5

15.5

Mass Flows

1023

Mass Flows

We now apply the impulse-momentum principles to physical systems that exchange mass with their surrounding environment. While we will focus on problems involving the motion of fluids, the ideas we develop are also applicable to other systems that, while not fluids, move in a fluidlike fashion. For example, in some cases the steady motion of bottles along the bottling line shown in Fig. 15.34 can be modeled as a continuous flow of mass. We will see that impulse-momentum principles offer a direct way to compute the forces present in simple fluid motions. Open and closed systems Before we begin, it is important to recall the distinction between open and closed systems (which were introduced on p. 935). A closed system does not exchange mass with its environment, whereas an open system does exchange mass with its environment. The mass of a closed system is constant. An open system can have constant or variable mass. If a physical system is referred to as a variable mass system, then such a system is necessarily open. The impulse-momentum principles presented earlier in the chapter are only applicable to closed systems.

Sean Gallup/Getty Images

Figure 15.34 Bottles moving along a bottling line. The bottles can be modeled as individual particles/bodies or as mass elements in a continuous mass flow.

Steady flows Figure 15.35 shows part of a pipe filled with a fluid in motion. To simplify our analysis, we assume that the fluid flow is steady. By this we mean that the velocity of a fluid particle going through a certain location within the pipe depends only on that location but is otherwise independent of time. On the other hand, the velocity of a particle can change as the particle moves from a point to another point within the pipe. In our analysis we also assume that the velocity of the particles at a given cross section is the same. If the velocity of the fluid particles changes, then these particles are accelerating, and by Newton’s second law, we conclude that the fluid must be subject to a force, which we now proceed to compute. Because a pipe can be a large structure, we determine only the force exerted on the fluid contained in a given portion of the pipe. We will refer to the chosen pipe portion as a control volume CV. Referring to Fig. 15.36, we consider a CV defined by the two cross sections 𝐴 and 𝐵. We assume that the

𝑣𝐴

𝑣𝐵

control volume 𝐵

𝐴

Figure 15.36. A CV corresponding to the portion of a pipe between cross sections 𝐴 and 𝐵.

fluid enters the CV at 𝐴 with velocity 𝑣⃗𝐴 and exits at 𝐵 with velocity 𝑣⃗𝐵 . We assume that the flow is such that 𝑣⃗𝐴 and 𝑣⃗𝐵 are perpendicular to the cross sections 𝐴 and 𝐵, respectively. Since mass flows in and out of the CV, a CV is an open system. When the flow is steady, then the mass of the fluid in the CV remains constant. This implies that if

𝑣𝐴 𝐴

𝐵 𝑣𝐵 Figure 15.35 A fluid flowing through a curved pipe with variable cross section.

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𝑡

𝑣𝐴

𝐵 𝐴 𝛥𝑚𝑓

𝛥𝑚𝑓

𝑣𝐵

𝑡 + 𝛥𝑡 𝐵 𝐴

Figure 15.37 A fluid body flowing through a CV defined by the cross sections 𝐴 and 𝐵 and behaving as a closed system.

Δ𝑚𝑓 (where the subscript 𝑓 stands for flux) is the mass that flows into the CV at 𝐴 during any time interval Δ𝑡, then Δ𝑚𝑓 is also the mass of the fluid that flows out of the CV at 𝐵 during the same time interval. The tools we have to relate forces and motion are the impulse-momentum principles, which only apply to closed systems. We will first apply the impulse-momentum principle to a closed system containing the open system of interest, and then we will “shrink” this closed system to make it coincide with the open system contained therein on an instant-by-instant basis. First, referring to Fig. 15.37, we select a body of fluid which at time 𝑡 fills the chosen CV and has a small element of mass Δ𝑚𝑓 about to enter the CV at 𝐴. This will be our closed system. The fluid element that is about to enter the CV is chosen to be small enough to allow us to assume that, at time 𝑡, 𝑣⃗𝐴 is the velocity of all of its particles. Second, we let Δ𝑡 be the time taken by the fluid element that is entering at 𝐴 to flow into the CV. Because the flow is steady, at time 𝑡 + Δ𝑡, the body will completely fill the CV between 𝐴 and 𝐵 and will also include an element of mass Δ𝑚𝑓 , which has exited the CV at 𝐵 (see Fig. 15.37). Since Δ𝑚𝑓 is small, we can assume that all of the particles to the right of 𝐵 have the same velocity 𝑣⃗𝐵 . Since the selected body is a closed system, we can apply to it Eq. (15.15) on p. 936, which yields ∫𝑡

𝑡+Δ𝑡

𝐹⃗ 𝑑𝑡 = 𝑝(𝑡 ⃗ + Δ𝑡) − 𝑝(𝑡), ⃗

(15.107)

where 𝐹⃗ is the total external force acting on the system and, using the stated assumptions, the linear momenta 𝑝(𝑡) ⃗ and 𝑝(𝑡 ⃗ + Δ𝑡) can be written as 𝑝(𝑡) ⃗ = Δ𝑚𝑓 𝑣⃗𝐴 + 𝑝⃗cv (𝑡),

(15.108)

𝑝(𝑡 ⃗ + Δ𝑡) = 𝑝⃗cv (𝑡 + Δ𝑡) + Δ𝑚𝑓 𝑣⃗𝐵 .

(15.109)

The quantity 𝑝⃗cv denotes the momentum of the fluid within the CV. Because the flow is steady, we must have (15.110) 𝑝⃗cv (𝑡) = 𝑝⃗cv (𝑡 + Δ𝑡). Substituting Eqs. (15.108)–(15.110) into Eq. (15.107), simplifying, and dividing all terms by Δ𝑡, we have 1 Δ𝑡 ∫𝑡

Common Pitfall The mass in the CV is constant. Often 𝑚̇ 𝑓 is misinterpreted as the time rate of change of the mass contained in the CV. However, the mass of the fluid within the CV is constant because the flow is steady. The quantity 𝑚̇ 𝑓 is simply a measure of the rate at which mass flows into and out of the CV.

𝑡+Δ𝑡

𝐹⃗ 𝑑𝑡 =

Δ𝑚𝑓 ( ) 𝑣⃗𝐵 − 𝑣⃗𝐴 . Δ𝑡

(15.111)

Equation (15.111) holds for a closed system that occupies a volume larger than the selected CV. Recalling that Δ𝑡 is the time it takes Δ𝑚𝑓 to flow into and out of the CV, we see that letting Δ𝑡 go to zero implies that Δ𝑚𝑓 must also go to zero and the fluid in the closed system at time 𝑡 will fill the CV exactly! Therefore, by letting Δ𝑡 go to zero (and using the Fundamental Theorem of calculus), Eq. (15.111) yields ( ) (15.112) 𝐹⃗ = 𝑚̇ 𝑓 𝑣⃗𝐵 − 𝑣⃗𝐴 , where the quantity 𝑚̇ 𝑓 = lim

Δ𝑡→0

Δ𝑚𝑓 Δ𝑡

(15.113)

is called the mass flow rate or mass flux; it measures the amount of mass flowing into and out of the chosen CV per unit time. The result in Eq. (15.112) applies to open systems and was possible because, in going from Eq. (15.111) to Eq. (15.112), we took a limit that forced the chosen closed system to coincide with the open system we wanted to characterize at time 𝑡.

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Section 15.5

1025

Mass Flows

Volumetric flow rate

𝛥𝓁𝐴

In addition to the mass flux, there is another commonly used measure of the amount of fluid moving through a CV called the volumetric flow rate. Referring to Fig. 15.38, we see that, at time 𝑡, the volume occupied by the fluid element of mass Δ𝑚𝑓 is approximately given by Δ𝓁𝐴 𝑆𝐴 , where Δ𝓁𝐴 is the pipe length occupied by Δ𝑚𝑓 and 𝑆𝐴 is the area of the cross section at 𝐴. Similarly, at time 𝑡 + Δ𝑡, the volume occupied by the fluid element of mass Δ𝑚𝑓 is Δ𝓁𝐵 𝑆𝐵 . Because the fluid motion is steady, the quantity Δ𝑚𝑓 is the same at 𝐴 and 𝐵. Therefore, letting 𝜌𝐴 and 𝜌𝐵 be the fluid density at 𝐴 and 𝐵, respectively, we have Δ𝑚𝑓 = 𝜌𝐴 Δ𝓁𝐴 𝑆𝐴 = 𝜌𝐵 Δ𝓁𝐵 𝑆𝐵 .

𝐵 𝑆𝐴

𝐴

Δ𝑡→0

Δ𝓁𝐴 Δ𝑡

= 𝑣𝐴

and

lim

Δ𝑡→0

Δ𝓁𝐵 Δ𝑡

= 𝑣𝐵 ,

𝑣𝐵

𝑡 + 𝛥𝑡

(15.114) 𝐵

(15.115)

where 𝑣𝐴 and 𝑣𝐵 are the values of the speed of the fluid at 𝐴 and 𝐵. If 𝑆 is the area of a generic cross section along the pipe and if 𝑣 is the fluid speed at that cross section, we define the volumetric flow rate as the quantity 𝑄 = 𝑣𝑆.

𝛥𝑚𝑓

𝛥𝑚𝑓

Since we have assumed that 𝑣⃗𝐴 and 𝑣⃗𝐵 are perpendicular to the cross sections 𝐴 and 𝐵, respectively, we have lim

𝑡

𝑣𝐴

𝛥𝓁𝐵

𝑆𝐵 𝐴 Figure 15.38 Volumes occupied by the fluid element with mass Δ𝑚𝑓 upon entering (top) and exiting (bottom) a chosen CV.

(15.116)

Dividing Eq. (15.114) by Δ𝑡, letting Δ𝑡 → 0, and using the definition in Eq. (15.116), we have 𝑚̇ 𝑓 = 𝜌𝐴 𝑄𝐴 = 𝜌𝐵 𝑄𝐵 , (15.117) where 𝑄𝐴 and 𝑄𝐵 are the volume flow rates at 𝐴 and 𝐵. Moment acting on the fluid Sometimes it is useful to relate the change in the fluid’s angular momentum, computed with respect to a chosen moment center, to the corresponding moment acting on the fluid. Referring to Fig. 15.39, we choose as moment center the fixed point 𝑃 . To compute the moment, we select a body of fluid as we did for the force calculation, and then we apply the angular impulse-momentum principle given in Eq. (15.51) on p. 985 between times 𝑡 and 𝑡 + Δ𝑡. Doing so gives ⃗ (𝑡) + ℎ 𝑃

∫𝑡

𝑡+Δ𝑡

⃗ (𝑡 + Δ𝑡), ⃗ 𝑑𝑡 = ℎ 𝑀 𝑃 𝑃

(15.118)

⃗ is the angular momentum of the selected fluid body and 𝑀 ⃗ is the moment where ℎ 𝑃 𝑃 we intend to compute. Because we have assumed that all of the particles in the volume elements of mass Δ𝑚𝑓 are moving with velocity 𝑣⃗𝐴 at time 𝑡 and 𝑣⃗𝐵 at time 𝑡 + Δ𝑡, we have ( ) ⃗ ⃗ (𝑡) = 𝑟⃗ ⃗𝐴 + ℎ (15.119) ℎ 𝑃 𝐶∕𝑃 × Δ𝑚𝑓 𝑣 𝑃 cv , ( ) ⃗ (𝑡 + Δ𝑡) = ℎ ⃗ ℎ ⃗𝐷∕𝑃 × Δ𝑚𝑓 𝑣⃗𝐵 , (15.120) 𝑃 𝑃 cv + 𝑟 where 𝐶 and 𝐷 are the centers of the cross sections 𝐴 and 𝐵, respectively, 𝑟⃗𝐶∕𝑃 ( ) ⃗ and 𝑟⃗ are the positions of 𝐶 and 𝐷 with respect to 𝑃 , respectively, and ℎ 𝐷∕𝑃

𝑃 cv

𝑣𝐴

𝑡 𝐵

𝐶 𝐴

𝑟⃗𝐶∕𝑃

𝛥𝑚𝑓

𝑃 𝛥𝑚𝑓 𝐷

𝑣𝐵

𝑡 + 𝛥𝑡 𝐵 𝑟⃗𝐷∕𝑃 𝐴 𝑃 Figure 15.39 Choice of moment center 𝑃 for the determination of the moment acting on the fluid contained in the CV (shaded area).

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is the angular momentum with respect to 𝑃 of the fluid contained in the CV. Since ( ) ⃗ the flow is steady, ℎ 𝑃 cv is a constant. Substituting Eqs. (15.119) and (15.120) into Eq. (15.118), simplifying, and rearranging terms, we have 1 Δ𝑡 ∫𝑡

𝑡+Δ𝑡

⃗ 𝑑𝑡 = 𝑀 𝑃

Δ𝑚𝑓 ( ) 𝑟⃗𝐷∕𝑃 × 𝑣⃗𝐵 − 𝑟⃗𝐶∕𝑃 × 𝑣⃗𝐴 , Δ𝑡

(15.121)

where we have divided all terms by Δ𝑡. By proceeding as in the case of the force calculation, i.e., letting Δ𝑡 → 0, Eq. (15.121) yields ( ) ⃗ = 𝑚̇ 𝑟⃗ 𝑀 ⃗𝐵 − 𝑟⃗𝐶∕𝑃 × 𝑣⃗𝐴 . 𝑃 𝑓 𝐷∕𝑃 × 𝑣 𝑣𝐴

𝐴

𝐵 𝑣𝐵 Figure 15.40 A rocket 𝐴 being propelled by the ejection of combustion gases, which we have labeled 𝐵.

(15.122)

Variable mass flows and propulsion Figure 15.40 shows a body 𝐴 (the rocket) propelled by the continuous ejection of some material 𝐵 (combustion gas). Since 𝐵 used to be part of 𝐴 before ejection, the mass of 𝐴 changes with time so that 𝐴 is a variable mass system.∗ We want to determine the force acting on 𝐴, and we will do so using CVs. That is, first we will apply the impulse-momentum principle to a closed system containing the open system of interest, and then we will “shrink” this closed system to make it coincide with the open system contained therein on an instant-by-instant basis. We now consider a body with mass 𝑚(𝑡) at time 𝑡 such that almost all of its par𝑣(𝑡) ⃗ 𝑡 𝑚(𝑡) 𝛥𝑚𝑜

𝑚(𝑡) − 𝛥𝑚𝑜 𝑡 + 𝛥𝑡 𝑣(𝑡 ⃗ + 𝛥𝑡)

𝑣(𝑡 ⃗ + 𝛥𝑡) + 𝑣⃗𝑜 (𝑡 + 𝛥𝑡) Figure 15.41. A variable mass system that ejects a material element of mass Δ𝑚𝑜 over the time interval Δ𝑡.

ticles travel with a velocity 𝑣(𝑡) ⃗ (Fig. 15.41). Some particles, which at time 𝑡 have a total mass that is negligible with respect to 𝑚(𝑡), are being ejected from the body. After an amount of time Δ𝑡, the body will have lost an amount of mass Δ𝑚𝑜 (the subscript 𝑜 stands for outflow), and we write 𝑚(𝑡 + Δ𝑡) = 𝑚(𝑡) − Δ𝑚𝑜 .

(15.123)

We assume that all the particles contributing to Δ𝑚𝑜 have the same inertial velocity 𝑣⃗ + 𝑣⃗𝑜 , where 𝑣⃗𝑜 is the relative velocity of the particles in question with respect to the main body. As long as the physical system we analyze consists of both the particles of mass Δ𝑚𝑜 and the main body of mass 𝑚, our system is a closed system. Applying to this ∗ As

mentioned at the beginning of this section on p. 1023, a variable mass system is necessarily open.

ISTUDY

Section 15.5

Mass Flows

1027

system the impulse-momentum principle given in Eq. (15.15) on p. 936, between times 𝑡 and 𝑡 + Δ𝑡, we have ∫𝑡

𝑡+Δ𝑡

𝐹⃗ 𝑑𝑡 = 𝑝(𝑡 ⃗ + Δ𝑡) − 𝑝(𝑡), ⃗

(15.124)

where 𝐹 (𝑡) is the total external force acting on the system and 𝑝(𝑡) ⃗ is the total momentum of the system. Using the stated assumptions, we have 𝑝(𝑡) ⃗ = 𝑚(𝑡)𝑣(𝑡), ⃗

[ ] ⃗ + Δ𝑡) + 𝑣⃗𝑜 (𝑡 + Δ𝑡) . 𝑝(𝑡 ⃗ + Δ𝑡) = 𝑚(𝑡 + Δ𝑡)𝑣(𝑡 ⃗ + Δ𝑡) + Δ𝑚𝑜 𝑣(𝑡

(15.125) (15.126)

Substituting Eq. (15.123) into Eq. (15.126), we have ] [ ] [ 𝑝(𝑡 ⃗ + Δ𝑡) = 𝑚(𝑡) − Δ𝑚𝑜 𝑣(𝑡 ⃗ + Δ𝑡) + Δ𝑚𝑜 𝑣(𝑡 ⃗ + Δ𝑡) + 𝑣⃗𝑜 (𝑡 + Δ𝑡) = 𝑚(𝑡)𝑣(𝑡 ⃗ + Δ𝑡) + Δ𝑚𝑜 𝑣⃗𝑜 (𝑡 + Δ𝑡).

(15.127)

Substituting Eqs. (15.125) and (15.127) into Eq. (15.124) and collecting the term 𝑚(𝑡), we have ∫𝑡

𝑡+Δ𝑡

[ ] 𝐹⃗ 𝑑𝑡 = 𝑚(𝑡) 𝑣(𝑡 ⃗ + Δ𝑡) − 𝑣(𝑡) ⃗ + Δ𝑚𝑜 𝑣⃗𝑜 (𝑡 + Δ𝑡).

(15.128)

Dividing Eq. (15.128) by Δ𝑡, we obtain 1 Δ𝑡 ∫𝑡

𝑡+Δ𝑡

Δ𝑚𝑜 𝑣(𝑡 ⃗ + Δ𝑡) − 𝑣(𝑡) ⃗ + 𝑣⃗ (𝑡 + Δ𝑡). 𝐹⃗ 𝑑𝑡 = 𝑚(𝑡) Δ𝑡 Δ𝑡 𝑜

(15.129)

By the definition of time derivative, we have lim

Δ𝑡→0

𝑣(𝑡 ⃗ + Δ𝑡) − 𝑣(𝑡) ⃗ = 𝑎(𝑡) ⃗ and Δ𝑡

lim

Δ𝑡→0

Δ𝑚𝑜 Δ𝑡

= 𝑚̇ 𝑜 (𝑡),

(15.130)

where 𝑎(𝑡) ⃗ is the acceleration of the main body at time 𝑡 and 𝑚̇ 𝑜 (𝑡) (with 𝑚̇ 𝑜 ≥ 0) is the rate at which mass flows out of the main body. In addition, by the fundamental theorem of calculus, we have 1 Δ𝑡→0 Δ𝑡 ∫𝑡 lim

𝑡+Δ𝑡

𝐹⃗ 𝑑𝑡 = 𝐹⃗ .

(15.131)

Therefore, taking the limit as Δ𝑡 → 0 of the terms in Eq. (15.129) and using Eqs. (15.130) and (15.131), we obtain 𝐹⃗ = 𝑚𝑎⃗ + 𝑚̇ 𝑜 𝑣⃗𝑜 ,

(15.132)

where all the terms in Eq. (15.132) are evaluated at time 𝑡. Equation (15.132) applies only to systems that lose mass. But if we follow steps analogous to those that gave us Eq. (15.132) and refer to Fig. 15.42, we can show that if the system also gains mass at the rate 𝑚̇ 𝑖 (the subscript 𝑖 stands for inflow), with 𝑚̇ 𝑖 ≥ 0 and with the inflowing mass having a velocity 𝑣⃗𝑖 relative to the main body, Eq. (15.132) can be generalized to 𝐹⃗ = 𝑚𝑎⃗ + 𝑚̇ 𝑜 𝑣⃗𝑜 − 𝑚̇ 𝑖 𝑣⃗𝑖 ,

(15.133)

where the contribution of the inflowing mass has a sign opposite to that of the outflowing mass.

𝑣⃗𝑖

𝑣⃗𝑜

Figure 15.42 A plane with a jet engine. The airplane is taking in air with a mass flow rate 𝑚̇ 𝑖 while combustion gases are ejected with a mass flow rate 𝑚̇ 𝑜 . The vector 𝑣⃗𝑖 is the velocity of the inflowing air relative to the plane. The vector 𝑣⃗𝑜 is the velocity of the outflowing combustion gases relative to the plane.

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Common Pitfall The impulse-momentum principles apply only to closed systems. The impulsemomentum principle can be written as 𝐹⃗ = 𝑝, ⃗̇ with 𝑝⃗ = 𝑚𝑣⃗𝐺 , where 𝐺 is the system’s center of mass. As we have repeatedly mentioned in this section, this principle applies only to closed systems. Because these systems must have constant mass, 𝐹⃗ = 𝑝⃗̇ implies that 𝐹⃗ = 𝑚𝑎⃗𝐺 + 𝑚̇ 𝑣⃗𝐺 = 𝑚𝑎⃗𝐺 , given that 𝑚̇ = 0. Unfortunately, sometimes the impulsemomentum principle is erroneously applied to variable mass systems by writing 𝐹⃗ = 𝑚𝑎⃗𝐺 + 𝑚̇ 𝑣⃗𝐺 and claiming that the term 𝑚̇ 𝑣⃗𝐺 (which is different from zero for variable mass systems) describes the effect of the mass change. To see that this conclusion is incorrect, consider the case of a rocket engine fired while held fixed in a test rig. ⃗ and if In this case, 𝑎⃗𝐺 = 0⃗ and 𝑣⃗𝐺 = 0, ⃗ the statement 𝐹 = 𝑚𝑎⃗𝐺 + 𝑚̇ 𝑣⃗𝐺 were applicable to variable mass systems, we would conclude that no force is needed to restrain the rocket motor, thus contradicting common experience. The correct force balance for a variable mass system is that shown in Eq. (15.133).

Equation (15.133) is an important result that can be viewed as the generalization of Newton’s second law to an open system with variable mass. We arrived at Eq. (15.133) starting from a balance principle applied to a closed system, whose mass can only be constant. This was possible because, in going from Eq. (15.129) to Eq. (15.132), we took a limit that forced the chosen closed system to coincide with the variable mass system we wanted to characterize at the time instant 𝑡. In the field of rocket propulsion, 𝑚̇ 𝑖 = 0, and it is often common to move the term 𝑚̇ 𝑜 𝑣⃗𝑜 in Eq. (15.133) to the left-hand side of the equation and then to refer to the term −𝑚̇ 𝑜 𝑣⃗𝑜 as the thrust force provided by the propulsion system. In jet propulsion, we have both mass outflow and mass inflow so that the thrust is given by the term 𝑚̇ 𝑖 𝑣⃗𝑖 − 𝑚̇ 𝑜 𝑣⃗𝑜 . In the case of a rocket, as in Fig. 15.40, 𝑚̇ 𝑜 is enormous and 𝑣⃗𝑜 is large and in the negative 𝑦-direction, providing upward thrust to counter the force due to gravity. Once the rocket lifts off, its acceleration increases even under constant thrust, as the mass of the rocket is diminishing with time. In jet propulsion, we have mass inflow and outflow, so the thrust is 𝑇⃗ = 𝑚̇ 𝑖 𝑣⃗𝑖 − 𝑚̇ 𝑜 𝑣⃗𝑜 . Now there is a small difference between the mass inflow (air) and outflow (air plus combustion gases) as fuel is mixed with air in the combustion chamber. The big source of thrust, however, is the difference between the modest air inlet velocity 𝑣⃗𝑖 and the much larger exhaust velocity, 𝑣⃗𝑜 . (In Fig. 15.42, a realistic 𝑣⃗𝑜 would be drawn with a much larger vector length than 𝑣⃗𝑖 .) Thrust is still dominated by the −𝑚̇ 𝑜 𝑣⃗𝑜 term, and in Fig. 15.42, large 𝑣⃗𝑜 to the right propels the jet to the left. Sometimes, additional fuel is added to the hot exhaust in the exit nozzle. The subsequent combustion and energy increase results in exhaust leaving the nozzle at even higher speed, increasing the thrust even further. This process is called “afterburning.”

End of Section Summary

𝑣𝐴

𝑣𝐵

control volume 𝐵

𝐴 Figure 15.43 A CV corresponding to the portion of a pipe between cross sections 𝐴 and 𝐵.

In this section, we have considered mass flows. Specifically, we have considered (1) steady mass flows, in which a fluid moves through a conduit with a velocity that depends only on the position within the conduit, and (2) variable mass flows, such as the flow of combustion gases out of a rocket. Steady flows. Given the control volume (CV) shown in Fig. 15.43, where by control volume we mean a portion of a conduit delimited by two cross sections, we showed that, in the case of a steady flow, the total external force 𝐹⃗ acting on the fluid in the CV is Eq. (15.112), p. 1024 ( ) 𝐹⃗ = 𝑚̇ 𝑓 𝑣⃗𝐵 − 𝑣⃗𝐴 . Here the cross sections are perpendicular to the flow velocity, 𝑚̇ 𝑓 is the mass flow rate, i.e., the amount of mass flowing through a cross section per unit time, and 𝑣⃗𝐴 and 𝑣⃗𝐵 are the flow velocities at the cross sections 𝐴 and 𝐵, respectively. In addition to the mass flow rate, we defined the volumetric flow rate as the quantity Eq. (15.116), p. 1025 𝑄 = 𝑣𝑆,

ISTUDY

Section 15.5

1029

Mass Flows

where 𝑣 is the speed of the fluid at a given cross section and 𝑆 is the area of the cross section in question. We showed that Eq. (15.117), p. 1025 𝑚̇ 𝑓 = 𝜌𝐴 𝑄𝐴 = 𝜌𝐵 𝑄𝐵 , where 𝜌𝐴 and 𝜌𝐵 are the values of the mass density of the fluid at 𝐴 and 𝐵, respectively. Referring to Fig. 15.44, we also showed that, given a fixed point 𝑃 , the total ⃗ acting on the fluid in the CV is moment 𝑀 𝑃

( ) ⃗ = 𝑚̇ 𝑟⃗ ⃗𝐵 − 𝑟⃗𝐶∕𝑃 × 𝑣⃗𝐴 , 𝑀 𝑃 𝑓 𝐷∕𝑃 × 𝑣 where 𝐶 and 𝐷 are the centers of the cross sections 𝐴 and 𝐵, respectively. With reference to Fig. 15.45, for a body with time varying

𝑣⃗𝑜

𝑣⃗𝑖

𝑣𝐴

control volume

𝐷 𝐵

𝐶

Eq. (15.122), p. 1026

Variable mass flows.

𝑣𝐵

Figure 15.45. A plane with a jet engine. The airplane is taking in air with a mass flow rate 𝑚̇ 𝑖 while combustion gases are ejected with a mass flow rate 𝑚̇ 𝑜 . The vector 𝑣⃗𝑖 is the velocity of the inflowing air relative to the plane. The vector 𝑣⃗𝑜 is the velocity of the outflowing combustion gases relative to the plane.

mass 𝑚(𝑡) due to an inflow of mass with rate 𝑚̇ 𝑖 and an outflow of mass with the rate 𝑚̇ 𝑜 , the total external force acting on the body is given by Eq. (15.133), p. 1027 𝐹⃗ = 𝑚𝑎⃗ + 𝑚̇ 𝑜 𝑣⃗𝑜 − 𝑚̇ 𝑖 𝑣⃗𝑖 , where 𝑎⃗ is the acceleration of the main body, 𝑣⃗𝑜 is the relative velocity of the outflowing mass with respect to the main body, and 𝑣⃗𝑖 is the relative velocity of the inflowing mass, again relative to the main body.

𝐴

𝑟⃗𝐷∕𝑃

𝑟⃗𝐶∕𝑃 𝑃

Figure 15.44 A fluid flowing through a CV along with a choice of moment center 𝑃 for the calculation of angular momenta and moments.

ISTUDY

1030

Chapter 15

Momentum Methods for Particles

E X A M P L E 15.17

Force of an Open Water Jet 𝑣0

𝑣𝑤

𝜃 𝜇𝑘

Figure 1

A water jet is let out of a nozzle attached to the ground. The jet has a constant mass flow rate and a speed 𝑣𝑤 = 65 f t∕s relative to the nozzle. The jet strikes a 25 lb incline and causes it to slide at a constant speed 𝑣0 = 5.5 f t∕s. The kinetic coefficient of friction between the incline and the ground is 𝜇𝑘 = 0.43. Neglecting the effect of gravity and air resistance on the water flow, as well as friction between the water jet and the incline, determine the mass flow rate of the water jet at the nozzle if 𝜃 = 50◦ .

SOLUTION 𝑅𝑦 𝑅𝑥

𝚥̂

Road Map & Modeling

𝑚𝑔 𝐹

𝚤̂ 𝑁 Figure 2 FBD of the incline. The force 𝐹 is the friction force due to the sliding relative to the ground.

Modeling the incline as a particle and referring to the FBD in Fig. 2, we represent the effect of the water jet on the incline by the reaction forces 𝑅𝑥 and 𝑅𝑦 . To solve the problem, we need to relate 𝑅𝑥 and 𝑅𝑦 to the mass flow rate out of the nozzle and then use these relations when applying Newton’s second law to the incline. In this way, we will be able to relate the mass flow rate to the friction force opposing the motion of the incline. The key to determining 𝑅𝑥 and 𝑅𝑦 is to realize that if the friction between the water jet and the incline is negligible, then there is no force that will slow down the flow of water over the incline. This fact, along with the fact that the incline moves at a constant velocity, allows us to model the flow of water over the incline as steady and with a constant speed relative to the incline.

Determination of 𝑹𝒙 and 𝑹𝒚 in terms of mass flow rate Governing Equations

𝑅𝑥 𝚥̂

𝑅𝑦 𝚤̂

Figure 3 FBD of the water jet. The forces 𝑅𝑥 and 𝑅𝑦 are equal and opposite to those indicated in Fig. 2 to comply with Newton’s third law.

𝑣𝐵 𝐵

𝜃

𝚥̂ 𝚤̂ 𝑣𝐴

Balance Principles Here we apply the force balance relation for CVs in Eq. (15.112) on p. 1024. The terms in this equation must be measured using an inertial frame of reference. Because the incline moves at a constant velocity, a reference frame attached to the incline is inertial.∗ Choosing such a frame and referring to the water jet FBD in Fig. 3, we choose our CV to be the volume occupied by the water flowing over the top surface of the incline. Although this CV is moving with respect to the ground, our choice is acceptable because such a CV is stationary with respect to the chosen inertial frame and, as discussed above, the water flow is steady over the incline. Then Eq. (15.112) in component form yields (see Fig. 4) ∑ ( ) (1) 𝐹𝑥∶ −𝑅𝑥 = 𝑚̇ 𝑓 𝑣𝐵𝑥 − 𝑣𝐴𝑥 , ∑ ) ( (2) 𝐹𝑦∶ 𝑅𝑦 = 𝑚̇ 𝑓 𝑣𝐵𝑦 − 𝑣𝐴𝑦 ,

where 𝑚̇ 𝑓 is the mass flow rate that goes past the cross section at 𝐴, while 𝑣⃗𝐴 and 𝑣⃗𝐵 are the velocities of the water flow at 𝐴 and 𝐵, respectively. We note, again, that 𝑚̇ 𝑓 , 𝑣⃗𝐴 , and 𝑣⃗𝐵 are measured by the inertial observer who moves with the incline. Force Laws

All forces are accounted for on the FBD.

Kinematic Equations 𝜃

𝐴

Figure 4 Velocities of the steady flow at 𝐴 and 𝐵 as perceived by an observer moving with the incline.

Using relative kinematics, an observer moving with the incline

measures 𝑣𝐴𝑥 = 𝑣𝑤 − 𝑣0

and 𝑣𝐴𝑦 = 0,

(3)

where 𝑣𝑤 and 𝑣0 are the speeds of the water jet and of the incline measured relative to the ground. Since we are neglecting the friction between the water and the incline, we must have |𝑣⃗𝐴 | = |𝑣⃗𝐵 |. Hence, the components of the velocity of the water at 𝐵 are ( ) ( ) 𝑣𝐵𝑥 = 𝑣𝑤 − 𝑣0 cos 𝜃 and 𝑣𝐵𝑦 = 𝑣𝑤 − 𝑣0 sin 𝜃. (4) Let (𝑚̇ 𝑓 )nz be the mass flow rate measured at the nozzle. This quantity is the unknown we want to determine. We assume that the water jet has a constant cross section even when ∗ This

statement is based on the assumption that the ground over which the incline slides can be chosen as an inertial reference frame (see discussion on p. 794).

ISTUDY

Section 15.5

Mass Flows

in contact with the incline. Then, letting 𝑆 denote the flow cross-sectional area, we see from Eq. (15.116) that the volumetric flow rate at the nozzle is 𝑄nz = 𝑣𝑤 𝑆, and therefore the mass flow rate at the nozzle is ( ) (5) 𝑚̇ 𝑓 nz = 𝜌𝑆𝑣𝑤 , where 𝜌 denotes the mass density of the water. Similarly, the mass flow rate 𝑚̇ 𝑓 measured by an observer moving with the incline is ) ) ( ) ( ( (6) 𝑚̇ 𝑓 = 𝜌𝑆𝑣𝐴 = 𝜌𝑆 𝑣𝑤 − 𝑣0 ⇒ 𝑚̇ 𝑓 = 𝑚̇ 𝑓 nz 𝑣𝑤 − 𝑣0 ∕𝑣𝑤 , where we have used Eq. (5) in the first of Eqs. (6). Computation

Substituting Eqs. (3), (4), and the last of Eqs. (6) into Eqs. (1) and (2), ( ) ( ) 𝑚̇ 𝑓 nz 𝑚̇ 𝑓 nz ( ( )2 )2 𝑅𝑥 = (1 − cos 𝜃) 𝑣𝑤 − 𝑣0 sin 𝜃 𝑣𝑤 − 𝑣0 . (7) and 𝑅𝑦 = 𝑣𝑤 𝑣𝑤

Discussion & Verification Recalling that the mass flow rate has dimensions of mass over time, we know that Eqs. (7) are dimensionally correct. In addition, given that the right-hand sides of Eqs. (7) have a positive sign under all circumstances, we see that the directions of 𝑅𝑥 and 𝑅𝑦 are as expected. ( ) 𝑭⃗ = 𝒎⃗ 𝒂 for the incline and determination of 𝒎̇ 𝒇 nz Governing Equations Balance Principles

Using the FBD in Fig. 2, the application of Newton’s second law to

the incline yields ∑ ∑ Force Laws

𝐹𝑥∶

𝑅𝑥 − 𝐹 = 𝑚𝑎𝑥 ,

(8)

𝐹𝑦∶

−𝑅𝑦 − 𝑚𝑔 + 𝑁 = 𝑚𝑎𝑦 .

(9)

Since the incline is sliding, we have 𝐹 = 𝜇𝑘 𝑁.

Kinematic Equations

(10)

Because the incline moves with constant velocity, we have 𝑎𝑥 = 0 and 𝑎𝑦 = 0.

(11)

Substituting Eqs. (11) into Eqs. (8) and (9), solving for 𝐹 and 𝑁, and then substituting the result into Eq. (10) yield

Computation

𝑅𝑥 = 𝜇𝑘 (𝑅𝑦 + 𝑚𝑔).

(12)

Substituting Eqs. (7) into Eq. (12) and solving for (𝑚̇ 𝑓 )nz , we have 𝜇𝑘 𝑚𝑔𝑣𝑤 ( ) 𝑚̇ 𝑓 nz = ( = 7.096 slug∕s. )2 𝑣𝑤 − 𝑣0 (1 − cos 𝜃 − 𝜇𝑘 sin 𝜃)

(13)

Since the terms 𝜇𝑘 and 1 − cos 𝜃 − 𝜇𝑘 sin 𝜃 in Eq. (13) are /( )2 nondimensional, and since the term 𝑔𝑣𝑤 𝑣𝑤 − 𝑣0 has dimensions of 1 over time, our result is dimensionally correct. Our result is directly proportional to the weight of the incline, as well as to the friction coefficient. This is reasonable since, keeping 𝑣𝑤 and 𝑣0 fixed, we expect that more water mass per unit time is needed to move a heavier incline over a rougher surface. We can also imagine that if the incline is too shallow, 𝜃 ≪ 1 rad, and the surface is too rough, 𝜇𝑘 ∼ 1, there is no amount of nozzle mass flow that moves the incline. Under these circumstances, the 1 − cos 𝜃 − 𝜇𝑘 sin 𝜃 becomes negative.

Discussion & Verification

1031

1032

Chapter 15

Momentum Methods for Particles

E X A M P L E 15.18

Geometry of Fluid Motion and Structural Loads

oatjo/Shutterstock

Figure 1 An example of intricate pipeline geometry in a refinement plant.

Pipelines can be quite intricate (see Fig. 1). The fluid going through a bend and/or a change in cross section can exert significant structural loads on the line. Referring to Fig. 2, consider two straight pipes connected by a flanged diverter/reducer of length 𝓁 = 88 in., height ℎ = 62 in., and internal volume 𝑉 = 34 f t 3 . Suppose that the flow is steady, the fluid specific weight is 𝛾 = 42.5 lb∕f t 3 (this is typical of gasoline), and the cross sections at 𝐴 and 𝐵 are circular with radii 𝑅𝐴 = 13 in. and 𝑅𝐵 = 9 in., respectively. Assume that the center of mass 𝐺 of the fluid between 𝐴 and 𝐵 is located as shown with 𝑑 = 𝓁∕2 and 𝑞 = 33 in. In addition, let 𝑝𝐴 = 1400 psi and 𝑝𝐵 = 1390 psi be known measurements of the static pressure at 𝐴 and 𝐵, respectively. If the fluid speed at 𝐴 is 𝑣𝐴 = 6 f t∕s, determine the loads that the fluid exerts on the diverter/reducer. Finally, neglect the weight of the diverter/reducer and sketch its FBD, showing the internal forces at 𝐴 and 𝐵. 𝑅𝐵

diverter/reducer 𝐷

𝑣𝐵

𝑞 ℎ

𝑑 𝑅𝐴

𝐵 𝐺

𝑣𝐴 𝐶 𝑚cv 𝑔

𝑝𝑑

𝑝𝐵

𝐵

𝐴

Figure 2. A diverter/reducer connecting two straight pipe segments in the vertical plane. Points 𝐶 and 𝐷 indicate the centers of the cross sections 𝐴 and 𝐵, respectively.

𝐺 𝑝𝐴

𝑝𝑑

𝐴

Figure 3 FBD of the fluid moving through the chosen CV, which was taken to be the interior volume of the diverter. 𝐵 𝑚cv 𝑔

𝚥̂ 𝚤̂ 𝐴

𝐺

𝐶

𝐷

𝐹𝐵

ℎ

𝐹𝐴

𝑅𝑥 (𝑀𝐶 )𝑝

𝑅𝑦 𝑑

𝑑 𝓁

Figure 4 FBD of the fluid within the CV obtained using the concept of equivalent force system.

ISTUDY

𝓁

SOLUTION Road Map & Modeling The CV is the region occupied by the fluid between the cross sections 𝐴 and 𝐵. Referring to Fig. 3, the fluid in question is subject to the pressure distributions 𝑝𝐴 and 𝑝𝐵 due to the fluid outside the CV. We assume that 𝑝𝐴 and 𝑝𝐵 are uniform over 𝐴 and 𝐵, respectively. The fluid in the CV is also subject to the pressure distribution 𝑝𝑑 due to the contact with the inner walls of the diverter. Finally, the fluid in the CV is subject to gravity, which is represented by the weight 𝑚cv 𝑔, applied at the fluid’s center of mass 𝐺. Although we do not have a detailed knowledge of 𝑝𝑑 , we can describe the overall effect of 𝑝𝑑 using the concept of equivalent force system.∗ Using this concept, the force system acting on the fluid in the CV can be represented as shown in Fig. 4. The forces 𝐹𝐴 and 𝐹𝐵 , applied as shown, are equivalent to the pressure distributions 𝑝𝐴 and 𝑝𝐵 , respectively. The force system equivalent to the pressure distribution 𝑝𝑑 consists of the forces 𝑅𝑥 and 𝑅𝑦 , as well as the moment (𝑀𝐶 )𝑝 , where 𝐶 has been chosen as the 𝑑 moment center because its location is known (we could have chosen some other convenient reference point such as 𝐺 or 𝐷). It is this force system that we need to compute, and we will do so by applying the force and moment balance for steady flows. Because 𝑅𝑥 , 𝑅𝑦 , and (𝑀𝐶 )𝑝 describe the action of the diverter on the fluid in the CV, in sketching the 𝑑 FBD of the diverter we need to include these forces and moment, but with opposite sign to abide by Newton’s third law. ∗ See

your statics textbook for the concept of equivalent force system.

ISTUDY

Section 15.5

1033

Mass Flows

Governing Equations Balance Principles Referring to the FBD in Fig. 4, choosing 𝐶 as the moment center, and recalling that 𝑚vc 𝑔 = 𝛾𝑉 , the force and impulse-momentum principles in Eqs. (15.112) and (15.122), in component form, give ∑ ( ) 𝐹𝑥∶ 𝑅𝑥 + 𝐹𝐴 − 𝐹𝐵 = 𝑚̇ 𝑓 𝑣𝐵𝑥 − 𝑣𝐴𝑥 , (1) ∑ ( ) 𝐹𝑦∶ 𝑅𝑦 − 𝛾𝑉 = 𝑚̇ 𝑓 𝑣𝐵𝑦 − 𝑣𝐴𝑦 , (2) ∑ ( ) 𝑀𝐶 ∶ (𝑀𝐶 )𝑝 − 𝛾𝑉 𝑑 + 𝐹𝐵 ℎ = 𝑚̇ 𝑓 𝑣𝐵𝑦 𝓁 − 𝑣𝐵𝑥 ℎ . (3) 𝑑

Since the cross sections at 𝐴 and 𝐵 are circular with radii 𝑅𝐴 and 𝑅𝐵 , respectively, and since we have assumed that the pressure distributions over 𝐴 and 𝐵 are uniform, we have 𝐹𝐴 = 𝜋𝑅2𝐴 𝑝𝐴 and 𝐹𝐵 = 𝜋𝑅2𝐵 𝑝𝐵 . (4)

Force Laws

Kinematic Equations

Based on the flow depicted in Fig. 2, we have

𝑣𝐴𝑥 = 𝑣𝐴 ,

𝑣𝐴𝑦 = 0,

𝑣𝐵𝑥 = 𝑣𝐵 ,

𝑣𝐵𝑦 = 0.

(5)

Furthermore, applying Eqs. (15.116) and (15.117), we must have 𝑚̇ 𝑓 = (𝛾∕𝑔)𝜋𝑅2𝐴 𝑣𝐴 = (𝛾∕𝑔)𝜋𝑅2𝐵 𝑣𝐵

⇒

𝑣𝐵 = 𝑣𝐴 𝑅2𝐴 ∕𝑅2𝐵 .

(6)

Computation

Substituting Eqs. (4)–(6) into Eqs. (1)–(3), the force system that is equivalent to the pressure distribution 𝑝𝑑 is given by ) ( ) 𝛾𝜋𝑣2𝐴 𝑅2𝐴 𝑅2𝐴 − 𝑅2𝐵 ( 2 2 = −389.4×103 lb, 𝑅𝑥 = 𝜋 𝑝𝐵 𝑅𝐵 − 𝑝𝐴 𝑅𝐴 + 𝑔𝑅2𝐵

(7)

𝑅𝑦 = 𝛾𝑉 = 1445 lb,

(8)

(𝑀𝐶 )𝑝 = 𝛾𝑉 𝑑 − 𝑑

𝜋𝑝𝐵 𝑅2𝐵 ℎ

−

𝛾𝜋𝑣2𝐴 𝑅4𝐴 ℎ 𝑔𝑅2𝐵

6

= −1.824×10 f t ⋅lb.

(9)

Now that we have 𝑅𝑥 , 𝑅𝑦 , and (𝑀𝐶 )𝑝 , by applying Newton’s third law, the FBD for the 𝑑 diverter is that given in Fig. 5, where the internal force system over the cross section 𝐴 consists of the forces 𝑁𝐴 (tension), 𝑉𝐴 (shear force), and 𝑀𝐶𝑖 (bending moment). Similarly, the internal force system at 𝐵 is given by 𝑁𝐵 , 𝑉𝐵 , and 𝑀𝐷𝑖 . Discussion & Verification

Since the dimensions of pressure and mass density are force per unit area and mass over length cubed, respectively, our results are dimensionally correct. The fact that 𝑅𝑥 is negative in Eq. (7) is consistent with the idea that the fluid motion in the 𝑥 direction is hindered by the presence of the bend in the line. The result in Eq. (8) also makes sense since it confirms that the diverter is supporting the weight of the fluid in the CV. To explain the result in Eq. (9), recall that if there is no flow (i.e., 𝑣𝐴 = 0), then the diverter must provide a positive moment to balance the weight. However, if 𝑣𝐴 ≠ 0 and we neglect the weight, then common experience tells us that the flow would cause a counterclockwise rotation of the system, and the diverter must exert a moment in the clockwise direction to prevent the rotation in question. This means that both positive and negative moments are to be expected, and the sign of the result in Eq. (9) tells us that, in our case, the moment due to the fluid motion has the greater effect. A Closer Look Referring to Fig. 5, if end 𝐵 of the diverter were free, then the internal forces at 𝐵 would be equal to zero and an elementary equilibrium calculation would show that we must have 𝑁𝐴 = −𝑅𝑥 , 𝑉𝐴 = 𝑅𝑦 , and (𝑀𝐶 )𝑖 = −(𝑀𝐶 )𝑝 . That is, if one end 𝐴 𝐴 𝑑 is free, we can compute the internal forces at the other end directly in terms of forces computed from the force balance for CVs.

𝑀𝐷𝑖

𝐵

𝚥̂

𝐷 𝚤̂

𝑁𝐵

𝑉𝐴

𝑀𝐶𝑖

𝑅𝑥

𝐶 𝑁𝐴

(𝑀𝐶 )𝑝

𝑉𝐵

𝐴

𝑅𝑦 𝑑

𝓁

Figure 5 FBD of the diverter/reducer.

ℎ

ISTUDY

1034

Chapter 15

Momentum Methods for Particles

E X A M P L E 15.19

Hovering Using a Jet Pack A jet pack is a rocket propulsion device worn on a person’s back that allows the person to become airborne and fly (see Fig. 1). Suppose that the jet pack can hold 75 lb of fuel. Suppose further that when there is no fuel in the pack, the combined weight of the pilot and the jet pack is 180 lb. It is assumed that, in operation, the outflow speed 𝑣𝑜 of the ejected material relative to the pack is constant. Neglecting the amount of time it takes for the pilot to start hovering a few feet off the ground, and assuming that the jets are oriented in the direction of gravity, determine 𝑣𝑜 so that the fuel will be completely spent after the pilot hovers for 45 s.

SOLUTION Road Map & Modeling

Simon Holdcroft/Alamy Stock Photo

Figure 1 A pilot with a jet pack.

The pilot and the pack form a simple variable mass system. We will, therefore, apply to this system the force balance given in Eq. (15.133) while enforcing the requirement that the mass outflow rate be equal to the time rate of decrease of the system’s mass. By doing so, we will be able to relate the speed 𝑣𝑜 to the time it takes to exhaust all of the fuel. When we use Eq. (15.133), the thrust due to the ejection of matter from the pack is not considered an external force. Hence, given that the pilot is simply hovering, the system’s FBD is that shown in Fig. 2, in which we have only included the system’s weight. After solving the problem, we will discuss another approach to the solution of propulsion problems according to which the thrust acting on the system is shown on the FBD and, at the same time, the force balance law is made to take on the form 𝐹⃗ = 𝑚𝑎. ⃗ Governing Equations Balance Principles Observing that there are no forces acting in the horizontal direction, we see that the only significant component of the force balance law is that in the 𝑦 direction. Hence, we have −𝑚𝑔 = 𝑚𝑎𝑦 + 𝑚̇ 𝑜 𝑣⃗𝑜 ⋅ 𝚥̂, (1)

𝚥̂ 𝑚𝑔

where 𝑚 is the time-varying combined mass of the pilot and of the pack, 𝑚̇ 𝑜 is the time rate at which mass is being ejected from the pack, 𝑣⃗𝑜 ⋅ 𝚥̂ is the 𝑦 component of the velocity of the ejected matter relative to the main system, and we have accounted for the fact that the system does not gain mass (i.e., 𝑚̇ 𝑖 = 0). Force Laws

All forces are accounted for on the FBD.

Kinematic Equations

Since the pilot (with the pack) is hovering, the system is stationary with respect to the ground, which is chosen as our inertial frame. Therefore, we must have 𝑎𝑦 = 0 and 𝑣⃗𝑜 = −𝑣𝑜 𝚥̂. (2)

Figure 2 FBD of the system consisting of the pilot and the jet pack.

In addition, as already observed, the mass of the system decreases at the rate at which mass is ejected from the pack, so we have 𝑚̇ = −𝑚̇ 𝑜 . Computation

(3)

Substituting Eqs. (2) and (3) into Eq. (1), we have −𝑚𝑔 = 𝑣𝑜 𝑚. ̇

(4)

Recalling that 𝑚̇ = 𝑑𝑚∕𝑑𝑡, Eq. (4) can be written as −

𝑔 𝑑𝑚 . 𝑑𝑡 = 𝑣𝑜 𝑚

(5)

ISTUDY

Section 15.5

Mass Flows

1035

Integrating this equation from 𝑡 = 0 to the final time 𝑡𝑓 = 45 s, we have −

∫0

𝑡𝑓

𝑚(𝑡𝑓 ) 𝑔 𝑑𝑚 𝑑𝑡 = ∫𝑚(0) 𝑚 𝑣𝑜

⇒

−

𝑚(𝑡𝑓 ) 𝑔 . 𝑡𝑓 = ln 𝑣𝑜 𝑚(0)

Thrust = 𝑚̇ 𝑜 𝑣⃗𝑜

(6)

Recalling that 𝑚(0) = (180 + 75) lb is the combined mass of the pilot, the pack, and 75 lb of fuel, and that 𝑚(𝑡𝑓 ) = 180 lb is the combined mass of the pilot and the empty pack, we can solve Eq. (6) for 𝑣𝑜 to obtain 𝑔𝑡𝑓 𝑔𝑡𝑓 𝑣𝑜 = − [ ] = [ ] = 4160 f t∕s. ln 𝑚(𝑡𝑓 )∕𝑚(0) ln 𝑚(0)∕𝑚(𝑡𝑓 )

(7)

Discussion & Verification

Since the argument of the natural logarithm in Eq. (7) is nondimensional, and since the product of acceleration and time has the dimensions of length over time, the result in Eq. (7) has the correct dimensions. As far as the numerical value obtained for 𝑣𝑜 is concerned, this result is not far from what is obtained from simple monopropellant rocket engines whose exhaust speeds are typically on the order of 5600 f t∕s (although they can get close to 10,000 f t∕s). A Closer Look The problem discussed in this example can be approached by writing the force balance law as 𝐹⃗ = 𝑚𝑎, ⃗ which is meant to resemble Newton’s second law.∗ If we approach the force balance for a variable mass system using the expression 𝐹⃗ = 𝑚𝑎, ⃗ then the force 𝐹⃗ includes both those forces that would be considered external according to a strict interpretation of the impulse-momentum principle and those forcelike terms that result from the inflow and outflow of mass. Therefore, our FBD would have been that in Fig. 3, where −𝑚𝑜 𝑣⃗𝑜 is the thrust force provided by the rocket engine.

𝐹⃗ = 𝑚𝑎⃗ for variable mass systems cannot be considered to be the same as applying Newton’s second law. This is so because Newton’s second law cannot be applied to variable mass systems. If 𝐹⃗ = 𝑚𝑎⃗ is applied to a variable mass system, the only correct interpretation that can be given is that what is being applied is actually Eq. (15.133), with 𝐹⃗ = 𝐹⃗ext − 𝑚̇ 𝑜 𝑣⃗𝑜 + 𝑚̇ 𝑖 𝑣⃗𝑖 , where 𝐹⃗ext is the total external force applied to the system according to a strict interpretation of the impulse-momentum principle.

∗ Writing

𝚥̂ 𝑚𝑔

Figure 3 Alternate FBD of the system. The force balance law that must be used with this system is 𝐹⃗ = 𝑚𝑎, ⃗ where 𝐹⃗ = −𝑚𝑔 𝚥̂ − 𝑚̇ 𝑜 𝑣⃗𝑜 .

ISTUDY

1036

E X A M P L E 15.20

Forces in a Falling String

release position 𝑦 𝓁𝐿 =

Chapter 15

Momentum Methods for Particles

𝐿+𝑦 2 𝓁𝑅 =

𝐿−𝑦 2

Figure 1 A string falling vertically down.

In Example 14.13 on p. 909, we discovered that the velocity of the free end of a falling inextensible string of length 𝐿, released from rest, is given by (see Fig. 1) √ 2𝐿 − 𝑦 𝑦̇ = 𝑔𝑦 , (1) 𝐿−𝑦 where 𝑔 is the acceleration due to gravity and 𝑦 is the position of the free end of the string. Letting 𝜌 be the string’s mass density per unit length, use Eq. (1) to determine the reaction force 𝑅 at the ceiling as a function of 𝑦 by modeling the whole string as a closed system and by modeling the two branches to the right and left of the bend as variable mass systems.

SOLUTION Modeling the whole string as a closed system Road Map & Modeling If we model the whole string as a closed system, then the string’s FBD is that shown in Fig. 2, where the only force other than 𝑅 is the string’s weight (this is 𝑅

consistent with the solution of Example 14.13). Since the system is closed, we can apply the impulse-momentum principle as given in Eq. (15.14) on p. 936. 𝑦

𝑦𝐺

Governing Equations Balance Principles

𝐺

𝚥̂

Using the FBD in Fig. 2 and Eq. (15.14), we obtain ∑ 𝐹𝑦∶ 𝜌𝐿𝑔 − 𝑅 = 𝜌𝐿𝑎𝐺 ,

(2)

where 𝑎𝐺 is the acceleration of the string’s mass center and 𝜌𝐿 is the string’s mass. Force Laws 𝑚𝑔 = 𝜌𝐿𝑔 Figure 2 FBD of the string as a whole. The weight of the string has been placed at the string’s center of mass, which has been denoted by 𝐺.

All forces are accounted for on the FBD.

Since 𝑎𝐺 = 𝑦̈𝐺 , we first find 𝑦𝐺 via Eq. (13.35) on p. 838. Recalling that the mass of the string is 𝜌𝐿 and referring to Fig. 1, we have

Kinematic Equations

𝜌𝐿𝑦𝐺 = 𝜌𝓁𝐿 (𝓁𝐿 ∕2) + 𝜌𝓁𝑅 (𝑦 + 𝓁𝑅 ∕2)

⇒

𝑦𝐺 =

) 1 ( 2 𝐿 + 2𝐿𝑦 − 𝑦2 , 4𝐿

(3)

where 𝜌𝓁𝐿 and 𝜌𝓁𝑅 are the masses of the left and right branches of the string, respectively. Differentiating the final result in Eq. (3) twice with respect to time, we obtain 𝑎𝐺 =

] 1 [ 𝑦(𝐿 ̈ − 𝑦) − 𝑦̇ 2 . 2𝐿

Using Eq. (1) along with the chain rule, we have ) √ ( 2 𝑔 2𝐿 − 2𝐿𝑦 + 𝑦2 𝑑 𝑦̇ 𝑦̇ ⇒ 𝑦̇ = 𝑦̈ = √ 𝑑𝑦 2(𝐿 − 𝑦)3∕2 𝑦(2𝐿 − 𝑦)

] [ 𝑔 𝐿2 𝑦̈ = . 1+ 2 (𝐿 − 𝑦)2

(4)

(5)

Substituting Eq. (1) and the final result in Eq. (5) into Eq. (4), after simplifying we obtain ( ) 𝑦 𝑔 𝐿 𝑎𝐺 = 3−3 − . (6) 4 𝐿 𝐿−𝑦 Computation

Substituting Eq. (6) into Eq. (2) and solving for 𝑅, we have ( ) 𝑦 𝜌𝐿𝑔 𝐿 1+3 + . 𝑅= 4 𝐿 𝐿−𝑦

(7)

ISTUDY

Section 15.5

1037

Mass Flows Variable mass systems modeling 𝑅

Road Map & Modeling

We can model the left and right branches of the string as variable mass systems that exchange mass with each other. Specifically, the left branch gains mass at the expense of the right branch. In this case, separating these systems with a cut at the bend, we have the FBDs in Fig. 3 (see the Helpful Information marginal note for further comments on these FBDs). Then we can apply to each branch the force balance for variable mass systems given in Eq. (15.133).

𝑦 𝚥̂

𝓁𝐿 𝜌𝓁𝐿 𝑔

Governing Equations Balance Principles

Using the FBDs in Fig. 3 along with the force balance for variable mass systems, for the left and right branches of the string we have, respectively, ∑ 𝐹𝑦𝐿∶ 𝜌𝓁𝐿 𝑔 − 𝑅 = 𝜌𝓁𝐿 𝑎𝑦𝐿 − 𝑚̇ 𝑖 𝑣⃗𝑖 ⋅ 𝚥̂, (8) ∑ 𝐹𝑦𝑅∶ 𝜌𝓁𝑅 𝑔 = 𝜌𝓁𝑅 𝑎𝑦𝑅 + 𝑚̇ 𝑜 𝑣⃗𝑜 ⋅ 𝚥̂, (9) where 𝑚̇ 𝑖 is the time rate of mass gain of the left branch, 𝑣⃗𝑖 is the velocity of the mass joining the left branch relative to the velocity of the left branch itself, 𝑚̇ 𝑜 is the time rate of mass loss of the right branch, and 𝑣⃗𝑜 is the velocity of the mass leaving the right branch relative to the right branch itself.

Force Laws

All forces are accounted for on the FBDs.

Kinematic Equations Because of inextensibility, all the mass elements on the left branch must move with the same velocity. The same is true for the mass elements on the right branch. Observing that one point on the left branch is fixed to the ceiling and that the acceleration of the top end of the right branch is 𝑦, ̈ we must have

𝑎𝑦𝐿 = 0

and 𝑎𝑦𝑅 = 𝑦. ̈

(10)

Referring to Fig. 1, the time derivatives of the lengths of the two branches are 𝓁̇ 𝐿 = 𝑦∕2 ̇ and 𝓁̇ 𝑅 = −𝑦∕2. ̇ Hence, since 𝑚𝐿 = 𝜌𝓁𝐿 and 𝑚𝑅 = 𝜌𝓁𝑅 , we have ̇ and 𝑚̇ 𝑜 = −𝑚̇ 𝑅 = −𝜌𝓁̇ 𝑅 = 𝜌𝑦∕2. ̇ 𝑚̇ 𝑖 = 𝑚̇ 𝐿 = 𝜌𝓁̇ 𝐿 = 𝜌𝑦∕2

(11)

The velocity of the mass elements joining the left branch and leaving the right branch must match the time rate of lengthening and shortening of these branches, i.e., 𝑣⃗𝑖 = 𝓁̇ 𝐿 𝚥̂ = 12 𝑦̇ 𝚥̂ and 𝑣⃗𝑜 = 𝓁̇ 𝑅 𝚥̂ = − 12 𝑦̇ 𝚥̂. Computation

𝜌𝓁𝑅 𝑔

𝓁𝑅

Figure 3 FBDs of the left and right branches of the falling string modeled as variable mass systems (see the Helpful Information marginal note for further comments on these FBDs).

Helpful Information Is something missing from the FBDs in Fig. 3? When we cut some structure and we sketch the FBD of the cut structure, we place on the FBD those forces that act internally to the structure at the location of the cut. However, here we are modeling the two branches as variable mass systems, and in this case, we do not include the forces at the cut because of how we derived Eq. (15.133). Specifically, the external forces that appear in Eq. (15.133) do not include any effects due to the exchange of mass.

(12)

Substituting Eqs. (10)–(12) into Eq. (8) and solving for 𝑅, we have 𝑅 = 𝜌𝓁𝐿 𝑔 + 41 𝜌𝑦̇ 2 .

(13)

Recalling that 𝑦̇ is given in Eq. (1) and 𝓁𝐿 = (𝐿 + 𝑦)∕2, after simplification, we have ( ) 𝑦 𝜌𝐿𝑔 𝐿 1+3 + . 𝑅= (14) 4 𝐿 𝐿−𝑦 Discussion & Verification

Since we have obtained the same result with two very different methods, we can be confident that our final result is correct. We present a plot of 𝑅 as a function of 𝑦 in Fig. 4. Notice that as the string becomes vertical, i.e., as 𝑦 → 𝐿, 𝑅 goes to infinity. This is so because the free end of the string moves with infinite speed when the string is almost completely vertical (i.e., 𝑦̇ → ∞ as 𝑦 → 𝐿), and therefore 𝑅 must become impulsive to bring the string to a complete stop as soon as the string becomes vertical. Finally, we note that we did not take advantage of Eq. (9). The reason for this is that substituting Eqs. (10)–(12) into Eq. (9) yields an equation whose solution coincides with Eq. (1) (see Prob. 15.198). If we had not been given Eq. (1), we would have had to use Eq. (9) to obtain the velocity of the free end as a function of 𝑦. A Closer Look

15 12

𝑅 9 𝜌𝐿𝑔 6 3 0

0

0.2

0.6 0.4 𝑦∕𝐿

0.8

1

Figure 4 The reaction 𝑅 (nondimensionalized with respect to the string weight 𝜌𝐿𝑔) at the top of the string as a function of the nondimensional fall distance (𝑦∕𝐿). The vertical red line corresponds to the end of the fall, and the horizontal green line corresponds to the weight of the string.

ISTUDY

1038

Chapter 15

Momentum Methods for Particles

Problems Problem 15.162 A fluid is in steady motion in the conduit shown. The lines depicted are tangent to the velocity of the fluid particles in the conduit (these lines are called streamlines). Explain whether or not the control volume defined by the cross sections 𝐴 and 𝐵 in the figure is consistent with the assumptions laid out in this section. 𝑣𝐵 control control volume

𝑣𝐴

𝐵

𝐴 Figure P15.162

Figure P15.163

Problem 15.163 A hydraulic system is being used to actuate the control surfaces of a plane. Suppose that there is a time interval during which (a) the speed of the hydraulic fluid within a particular line is constant relative to the line itself and (b) the plane is performing a turn. Explain whether or not the force balance for control volumes presented in this section is applicable to the analysis of the hydraulic fluid in question.

Problem 15.164 The cross sections labeled 𝐴 and 𝐵 in case (a) are identical to the corresponding cross sections in case (b). Assume that, in both (a) and (b), a fluid in steady motion flows through 𝐴 with speed 𝑣1 and exits the system at 𝐵 with a speed 𝑣2 . If the pipe sections are to remain stationary and if the mass flow rate is identical in the two cases, determine whether the magnitude of the horizontal force acting on the pipes due to the water flow in case (a) is smaller than, equal to, or larger than that in case (b). In addition, for both (a) and (b), establish the direction of the force. 𝑣2 𝐵 𝐵

(b)

(a) 𝐴

𝐴 𝑣1

𝑣1

𝑣2

Figure P15.164

𝑣𝑤

𝐴 𝐵 Figure P15.165

Problem 15.165 Experience tells us that when a steady water jet comes out of a nozzle 𝐵, the hose line 𝐴 attached to the nozzle is in tension, that is, the nozzle exerts a force on the hose that is in the direction of the flow. If the end of the nozzle at 𝐵 were capped to stop the water flow, would the force exerted by the nozzle on the hose decrease, stay the same, or increase?

ISTUDY

Section 15.5

1039

Mass Flows

Problem 15.166 Revisit Example 15.17 and use the numerical result in Eq. (13) of the example, along with the fact that the specific weight of water is 62.4 lb∕f t 3 , to determine the volumetric flow rate at the nozzle and the nozzle diameter. 𝑣0

𝑣𝑤

𝜃 𝜇𝑘 Figure P15.166

Problem 15.167 The rocket shown has 7 lb of propellant with a burnout time (time required to burn all the fuel) of 7 s. Assume that the mass flow rate is constant and that the speed of the exhaust relative to the rocket is also constant and equal to 6500 f t∕s. If the rocket is fired from rest, determine the initial weight of the rocket’s body if the rocket is to experience an initial acceleration of 6𝑔. Figure P15.167

Problem 15.168 The tip 𝐵 of a nozzle is 1.5 in. in diameter, whereas the diameter at 𝐴 where the hose is attached is 3 in. If water flows through the nozzle at 200 gpm (“gpm” stands for gallons per minute; 1 U.S. gallon is defined as 231 in.3 ) and the water static pressure in the line is 300 psi, determine the force necessary to hold the nozzle stationary. Recall that the specific weight of water is 𝛾 = 62.4 lb∕f t 3 , and neglect the atmospheric pressure at 𝐵. 𝑣𝑤

𝐴 𝐵

𝐴

Figure P15.168

Problem 15.169 An intubed fan is mounted on a cart connected to a fixed wall by a linear elastic spring with constant 𝑘 = 50 lb∕f t. Assume that in a test the fan draws air at 𝐴 with negligible speed and that the outgoing flow causes the cart to displace to the left so that the spring is stretched by 0.5 f t from its unstretched position. Assuming that the specific weight of the air 𝛾 = 7.5×10−2 lb∕f t 3 is constant, and letting the diameter of the tube at 𝐵 be 𝑑 = 4 f t (the cross section is assumed to be circular), determine the airspeed at 𝐵.

𝐵 𝑑

𝑘

Figure P15.169

Problem 15.170 A test is conducted in which an 80 kg person sitting in a 15 kg cart is propelled by the jets emitted by two household fire extinguishers with a combined initial mass of 18 kg. The cross section of the exhaust nozzles is 3 cm in diameter, and the density of the exhaust is 𝜌 = 1.98 kg∕m3 . The vehicle starts from rest, and it is determined that the initial acceleration of the “jet cart” is 1.8 m∕s2 . Recalling that the mass flow rate out of the nozzle is given by 𝑚̇ 𝑜 = 𝜌𝑆𝑣𝑜 , where 𝑆 is the area of the nozzle cross section and 𝑣𝑜 is the exhaust speed, determine 𝑣𝑜 at the initial time. Ignore any resistance to the horizontal motion of the cart.

𝑣𝐵

Figure P15.170

ISTUDY

1040

Chapter 15

Momentum Methods for Particles

Problem 15.171 Consider a rocket in space so that it can be assumed that no external forces act on the rocket. Let 𝑣𝑜 be the constant speed of the exhaust gases relative to the rocket. In addition, let 𝑚𝑏 + 𝑚𝑓 be the total mass of the rocket and its fuel at the initial time, and let 𝑚𝑏 be the mass of the body after all the fuel is burned. If the rocket is fired from rest, determine an expression for the maximum speed that the rocket can achieve.

Figure P15.171

Problem 15.172 A stationary 4 cm diameter nozzle emits a water jet with a speed of 30 m∕s. The water jet impinges on a vane with a mass of 15 kg. Recalling that water has a mass density of 1000 kg∕m3 , determine the minimum static friction coefficient with the ground such that the vane does not move if 𝜙 = 20◦ and 𝜃 = 30◦ . Neglect the weight of the water layer in contact with the vane, as well as friction between the water and the vane. 𝑣𝑤 𝜃

𝜙

𝜇𝑠 Figure P15.172

Problem 15.173 𝑣𝑤

𝑘 𝜃

Figure P15.173 𝑣1

A diffuser is attached to a structure whose rigidity in the horizontal direction can be modeled by a linear spring with constant 𝑘. The diffuser is hit by a water jet issued with a speed 𝑣𝑤 = 55 f t∕s from a 2 in. diameter nozzle. Assume that the friction between the jet and the diffuser is negligible and that the diffuser’s motion in the vertical direction can be neglected. Recalling that the specific weight of water is 𝛾 = 62.4 lb∕f t 3 , if the opening angle of the diffuser is 𝜃 = 40◦ , determine 𝑘 such that the horizontal displacement of the diffuser does not exceed 0.25 in. from the diffuser’s rest position. Assume that the water jet splits symmetrically over the diffuser.

𝑣𝑤

Problem 15.174 𝑣2

𝜃

Figure P15.174

A water jet with a mass flow rate 𝑚̇ 𝑓 at the nozzle impinges with a speed 𝑣𝑤 on a fixed flat vane inclined at an angle 𝜃 with respect to the horizontal. Assuming that there is no friction between the water jet and the vane, the jet will split into two flows with mass flow rates 𝑚̇ 𝑓 1 and 𝑚̇ 𝑓 2 . Neglecting the weight of the water, determine how 𝑚̇ 𝑓 1 and 𝑚̇ 𝑓 2 depend on 𝑚̇ 𝑓 , 𝑣𝑤 , and 𝜃. Hint: Due to the no-friction assumption, there is no force that slows down the water in the direction tangent to the vane, and this implies that the momentum in that direction is conserved.

Problem 15.175

Simon Holdcroft/Alamy Stock Photo

Figure P15.175

A person wearing a jet pack lifts off from rest and ascends along a straight vertical trajectory. Let 𝑀 denote the initial combined mass of the pilot and the equipment, including the fuel in the pack. Assume that the mass flow rate 𝑚̇ 𝑜 and exhaust gas speed 𝑣𝑜 are known constants and that the pilot can take off as soon as the rocket engine is started. If the engine exhaust is completely directed in the direction of gravity, determine the expression of the pilot’s speed as a function of time, 𝑀, 𝑚̇ 𝑜 , 𝑣𝑜 , and 𝑔 (the acceleration due to gravity) while the pack is providing a thrust. Neglect air resistance and assume that gravity is constant.

ISTUDY

Section 15.5

1041

Mass Flows

Problem 15.176 A 28,000 lb A-10 Thunderbolt is flying at a constant speed of 375 mph when it fires a 4 s burst from its forward-facing seven-barrel Gatling gun. The gun fires 13.2 oz projectiles at a constant rate of 4200 rounds∕min. The muzzle velocity of each projectile is 3250 f t∕s. Assume that each of the plane’s two jet engines maintains a constant thrust of 9000 lb, that the plane is subject to a constant air resistance while the gun is firing (equal to that before the burst), and that the plane flies straight and level during that time. Determine the plane’s change in velocity at the end of the 4 s burst, modeling the airplane’s change of mass due to firing as a continuous mass loss.

𝐴

30 mm rounds

Gatling gun Figure P15.176

Problem 15.177 A faucet is letting out water at a rate of 15 L∕min. Assume that the internal diameter 𝑑 of the faucet is uniform and equal to 1.5 cm, the distance 𝓁 = 20 cm, and the static water pressure at the wall is 0.30 MPa. Neglecting the weight of the water inside the faucet, as well as the weight of the faucet itself, determine the forces and the moment that the wall exerts on the faucet. Recall that the density of water is 𝜌 = 1000 kg∕m3 , and neglect the atmospheric pressure at the spout. Hint: Define your control volume using a section along the wall.

𝑣𝑤

𝑑∕2

𝓁 Figure P15.177 streamline

𝑝atm

Problem 15.178 Consider a wind turbine with a diameter 𝑑 = 110 m and the airflow streamlines shown, which are symmetric relative to the axis of the turbine. Since the airflow is tangent to the streamlines (by definition), these lines can be taken to define the top and bottom surfaces of a control volume. Suppose that pressure measurements indicate that the flow experiences atmospheric pressure at the cross sections 𝐴 and 𝐵 (as well as outside the control volume) where the wind speed is 𝑣𝐴 = 7 m∕s and 𝑣𝐵 = 2.5 m∕s, respectively. Furthermore, assume that the average pressure along the streamlines defining the control volume is also atmospheric. Finally, assume that 𝑂 is on the line of action of the overall weight of the turbine and that the diameter of the flow cross section at 𝐴 is 85% of the rotor diameter and that the rotor hub is at a distance ℎ = 75 m above the ground. If the density of air is constant and equal to 𝜌 = 1.25 kg∕m3 , determine the force exerted by the air on the wind turbine and the reaction moment at the base of the support.

𝑝atm 𝑑

𝑣𝐴

𝑣𝐵

ℎ 𝐴 𝑂

streamline

𝐵

Figure P15.178

Problem 15.179

𝑣𝐴

𝐵

Let 𝑝𝐴 and 𝑝𝐵 be given static pressure measurements at the cross sections 𝐴 and 𝐵 in the air duct shown. Assume that any cross section between 𝐴 and 𝐵 is circular with diameter 𝑑. Assume that the flow is steady and that the mass density 𝜌𝐴 at 𝐴 is known along with 𝑣𝐴 , the speed of the flow at 𝐴, and 𝑣𝐵 , the speed of the flow at 𝐵. Determine the expression of mass density at 𝐵 and the expression of the force 𝐹 acting on the fan.

𝑣𝐵

𝑑

Figure P15.179

𝑑

𝐴

ISTUDY

1042

Chapter 15

Momentum Methods for Particles

Problem 15.180 A rope with weight per unit length of 0.1 lb∕f t is lifted at a constant upward speed 𝑣0 = 8 f t∕s. Treating the rope as inextensible, determine the force applied to the top end of the rope after it is lifted 9 f t. Assume that the top end of the rope is initially at rest and on the floor. In addition, disregard the horizontal motion associated with the uncoiling of the rope.

Problem 15.181 A rope with mass per unit length of 0.05 kg∕m is lifted at a constant upward acceleration 𝑎0 = 6 m∕s2 . Treating the rope as inextensible, determine the force that must be applied at the top end of the rope after it is lifted 3 m. Assume that the top end of the rope is initially at rest and on the floor. In addition, disregard the horizontal motion associated with the uncoiling of the rope.

Figure P15.180–P15.182

Problem 15.182 A rope with mass per unit length of 0.05 kg∕m is lifted by applying a constant vertical force 𝐹 = 10 N. Treating the rope as inextensible, plot the velocity and position of the top end of the rope as a function of time for 0 ≤ 𝑡 ≤ 3 s. Assume that the top end of the rope is initially at rest and 1 mm off the floor. In addition, disregard the horizontal motion associated with the uncoiling of the rope.

Problem 15.183 An amateur rocket with a body weight of 6.5 lb is equipped with a rocket engine holding 2.54 lb of solid propellant with a burnout time (time required to burn all the fuel) of 5.25 s (this is the typical data made available by amateur rocket engine manufacturers). The initial thrust is 68 lb. Assuming that the mass flow rate and the speed of the exhaust relative to the rocket remain constant, determine the exhaust mass flow rate 𝑚̇ 𝑜 and the speed relative to the rocket 𝑣𝑜 . In addition, determine the maximum speed achieved by the rocket 𝑣max if the rocket is launched from rest. Neglect air resistance, and assume that gravity does not change with elevation.

Problem 15.184

Figure P15.183 and P15.184

Continue Prob. 15.183 and determine the maximum height reached by the rocket, again neglecting air resistance and changes of gravity with elevation. Hint: For 0 < 𝑡 < 𝑡0 , ( ) [ ( )] 𝑡 𝑡 ln 1 − 𝑑𝑡 = (𝑡0 − 𝑡) 1 − ln 1 − + 𝐶. ∫ 𝑡0 𝑡0

Problem 15.185 A Pelton impulse wheel, as shown in Fig. P15.185(a), is typically found in hydroelectric power plants and consists of a wheel with a series of buckets attached at the periphery. As shown in Fig. P15.185(b), water jets impinge on the buckets and cause the wheel to spin about its axis (labeled 𝑂). Let 𝑣𝑤 and (𝑚̇ 𝑓 )nz be the speed and the mass flow rate of the water jets at the nozzles (the nozzles are stationary), respectively. As the wheel spins, a given water jet will impinge on a given bucket only for a very small portion of the bucket’s trajectory. This fact allows us to model the motion of a bucket relative to a given jet (during the time the bucket interacts with that jet) as essentially rectilinear and with constant relative speed, as was done in Example 15.17. Although each bucket moves away from the jet, the fact that they are arranged in a wheel results in an effective

ISTUDY

Section 15.5

Mass Flows

mass flow rate experienced by the vanes of (𝑚̇ 𝑓 )nz instead of the reduced mass flow rate computed in Eq. (6) of Example 15.17. With this in mind, consider a bucket, as shown in Fig. P15.185(c), that is moving with a speed 𝑣0 horizontally away from a fixed nozzle, but subject to a mass flow rate (𝑚̇ 𝑓 )nz . The inside of the bucket is shaped so as to redirect the water jet laterally out (away from the plane of the wheel). The angle 𝜃 describes the orientation of the velocity of the fluid relative to the (moving) bucket at 𝐵, the point at which the water leaves the bucket. Determine 𝜃 and 𝑣0 such that the power transmitted by the water to the wheel is maximum. Express 𝑣0 in terms of 𝑣𝑤 .

𝑣𝐵 𝜃

𝐵

𝑂

𝐴 𝑣0

𝑣𝑤

kenary820/Shutterstock

(a)

(b) Figure P15.185

(c)

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Chapter 15

Momentum Methods for Particles

15.6 C h a p t e r R e v i e w Momentum and impulse Helpful Information When should you use the impulse-momentum principle? The impulse-momentum principle provides a natural approach to problems in which you need to relate velocity, force, and time, since it relates forces acting over time to changes in momentum.

We learned in Chapter 13 that forces lead to changes in velocities since forces cause accelerations. We began this section by learning that forces acting over time change momentum (not just velocity). By integrating Newton’s second law, we obtained the impulsemomentum principle, which is given by Eq. (15.6), p. 934 𝑝(𝑡 ⃗ 1) +

𝑡2

∫𝑡

𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ),

1

where the linear momentum (or momentum) was defined to be Eq. (15.3), p. 934 𝑝(𝑡) ⃗ = 𝑚𝑣(𝑡), ⃗ and a force acting over some time interval was called the impulse (or linear impulse) and is given by Eq. (15.5), p. 934 ∫𝑡

𝑡2

𝐹⃗ (𝑡) 𝑑𝑡.

1

We also found that without detailed knowledge of the force acting on a particle at every instant in time, we could not determine the change in momentum. On the other hand, knowing just the change in momentum allows us to determine the average force acting on a particle during the corresponding time interval, that is, Eq. (15.9), p. 935 𝐹⃗avg =

𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡 ⃗ 1) 𝑡2 − 𝑡1

.

Impulse-momentum principle for systems of particles. For closed systems of particles, we discovered that we can write the impulse-momentum principle as Eqs. (15.12) and (15.15), p. 936 𝐹⃗ = 𝑝⃗̇ and

∫𝑡

𝑡2

⃗ 1 ), 𝐹⃗ (𝑡) 𝑑𝑡 = 𝑝(𝑡 ⃗ 2 ) − 𝑝(𝑡

1

∑𝑁 where 𝐹⃗ is the total external force on the particle system and 𝑝⃗ = 𝑖=1 𝑚𝑖 𝑣⃗𝑖 is the total momentum of the system of particles. Using the definition of the mass center of a system of particles, the impulse-momentum principle can also be written as Eq. (15.14), p. 936 ) 𝑑( 𝑚𝑣⃗𝐺 = 𝑚𝑎⃗𝐺 , 𝐹⃗ = 𝑑𝑡 where 𝑚 is the total mass of the system of particles, 𝑣⃗𝐺 is the velocity of its mass center, and 𝑎⃗𝐺 is the acceleration of its mass center.

ISTUDY

Section 15.6

Chapter Review

1045

Conservation of linear momentum. When there is a direction in which the external force on a system of particles is zero, then the momentum in that direction is constant and is said to be conserved. If the total external force on a system of particles is zero, that is, 𝐹⃗ = 0, then the momentum in every direction is constant, and the mass center of the system of particles will move with constant velocity.

Impact In this section, we idealized particle impact as an event spanning an infinitesimal time interval in which objects can experience a finite change in velocity at fixed position. The model is based on the assumptions summarized in Table 15.3. We discovered that there are two key elements to every impact problem: (1) the application of the impulse-momentum principle and (2) a force law telling us how the colliding objects rebound. When applying the impulse-momentum principle during an impact, only impulsive forces play a role, so they are the only forces included in impact-relevant FBDs. Problems involving the impact between two particles generally involve four unknowns, so four equations are needed. The geometry of an unconstrained impact (for which there are no external impulsive forces) between two particles is shown in Fig. 15.46, and the four equations come from 1. Conservation of momentum of the two particles together along the LOI:

Table 15.3 Assumptions used in our impact model. Impact assumption

Physical characteristic duration of impact

infinitesimal

displacement of particle

zero

force on particle

infinite

change in momentum

instantaneous

Eq. (15.25), p. 958

𝑣− 𝐵

𝑦

𝑚𝐴 𝑣−𝐴𝑦 + 𝑚𝐵 𝑣−𝐵𝑦 = 𝑚𝐴 𝑣+𝐴𝑦 + 𝑚𝐵 𝑣+𝐵𝑦 .

𝚥̂

2. Conservation of momentum for particle 𝐴 in the 𝑥 direction:

𝑚𝐴 𝑣𝐴𝑥 = 𝑚𝐴 𝑣𝐴𝑥 −

+

𝚤̂

𝑣𝐴𝑥 = 𝑣𝐴𝑥 . −

⇒

𝐵

𝐴

Eq. (15.26), p. 958 +

LOI

𝑣− 𝐴

𝑥

3. Conservation of momentum for particle 𝐵 in the 𝑥 direction: Figure 15.46 Geometry of two impacting particles.

Eq. (15.27), p. 958 𝑚𝐵 𝑣−𝐵𝑥 = 𝑚𝐵 𝑣+𝐵𝑥

𝑣−𝐵𝑥 = 𝑣+𝐵𝑥 .

⇒

4. COR equation applied along the LOI: Eq. (15.21), p. 957 + + separation velocity 𝑣𝐵𝑦 − 𝑣𝐴𝑦 = − 𝑒= . approach velocity 𝑣𝐴𝑦 − 𝑣−𝐵𝑦

The coefficient of restitution 𝑒 determines the nature of the rebound between the two particles. When 𝑒 = 0, the impact is called plastic; when 0 < 𝑒 < 1, the impact is called elastic; and when 𝑒 = 1, it is called perfectly elastic. In an unconstrained, direct central impact, a plastic collision (𝑒 = 0) results in the objects sticking together postimpact. This is not necessarily the case in an oblique central impact as velocity components along the plane of contact may be different.

Impact and energy. If 𝑒 < 1, mechanical energy is lost during the impact. The energy loss is often indicated as a percentage of the preimpact total kinetic energy: Eq. (15.29), p. 959 Percentage of energy loss =

𝑇− − 𝑇+ 𝑇−

× 100%,

where 𝑇 − and 𝑇 + are the pre- and postimpact total kinetic energies, respectively.

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Chapter 15

Momentum Methods for Particles

Angular momentum

𝑧 𝑄’s trajectory

𝑣⃗𝑄

In this section, we developed the concept of angular momentum for a single particle and for systems of particles. We also derived the moment-angular momentum relations for a particle and for a system of particles.

𝑄 𝑦 𝑘̂

Definition of angular momentum of a particle. Referring to Fig. 15.47, the angular momentum of a particle 𝑄 with respect to the moment center 𝑃 is given by

𝑟⃗𝑄∕𝑃 𝚥̂

𝑃

𝑂

Eq. (15.31), p. 982

𝚤̂

⃗ = 𝑟⃗ ℎ ⃗𝑄 = 𝑟⃗𝑄∕𝑃 × 𝑚𝑣⃗𝑄 , 𝑃 𝑄∕𝑃 × 𝑝 𝑥 Figure 15.47 A particle 𝑄 in motion relative to a point 𝑃 . Point 𝑃 need not be a stationary point.

where 𝑃 can be fixed or moving; 𝑟⃗𝑄∕𝑃 is the position of 𝑄 relative to 𝑃 ; 𝑚 and 𝑣⃗𝑄 are the mass and the velocity of 𝑄, respectively; and 𝑝⃗𝑄 = 𝑚𝑣⃗𝑄 is the linear momentum of 𝑄.

Moment-angular momentum relation for a particle. The moment-angular momentum relation for a single particle is given by Eq. (15.35), p. 982 ⃗̇ + 𝑣⃗ × 𝑚𝑣⃗ , ⃗ =ℎ 𝑀 𝑃 𝑃 𝑃 𝑄 ⃗ is the moment with respect to 𝑃 of all the forces acting on 𝑄. If either one of where 𝑀 𝑃 the following conditions is satisfied: ⃗ 1. The reference point 𝑃 is fixed, i.e., if 𝑣⃗𝑃 = 0. ⃗ 2. 𝑣⃗𝑃 is parallel to 𝑣⃗𝑄 , i.e., 𝑣⃗𝑃 × 𝑚𝑣⃗𝑄 = 0. the moment-angular momentum relation for a single particle can be simplified to Eq. (15.36), p. 982 ⃗ =ℎ ⃗̇ . 𝑀 𝑃 𝑃 If either condition (1) or (2) is satisfied for 𝑡1 ≤ 𝑡 ≤ 𝑡2 , then the moment-angular momentum relation for a single particle can be integrated with respect to time to obtain the angular impulse-momentum principle as

𝑧

𝑣⃗1

𝑣⃗𝑃 𝑚1

𝑃 𝐹⃗1 𝑟⃗𝑃

𝐹⃗𝑖

𝑟⃗𝐺 𝑟⃗2

𝑣⃗𝑖

⃗ + ℎ 𝑃1

𝑚𝑖

𝑟⃗𝑖

𝑟⃗1

𝑂

𝑓⃗𝑖1

𝑓⃗1𝑖 𝑓⃗12

Eq. (15.37), p. 983

𝐺

⃗ =ℎ ⃗ (𝑡 ) and ℎ ⃗ =ℎ ⃗ (𝑡 ). where ℎ 𝑃1 𝑃 1 𝑃2 𝑃 2

𝑓⃗2𝑖 𝑣⃗2 𝑚2

⃗ 𝑑𝑡 = ℎ ⃗ , 𝑀 𝑃 𝑃2

1

𝑓⃗𝑖 2

𝑓⃗21

∫𝑡

𝑡2

𝐹⃗2

Moment-angular momentum relation for a system of particles. For a closed system of particles, the moment-angular momentum relation can be given the form 𝑦

Eq. (15.49), p. 984 𝑥 Figure 15.48 A system of particles under the action of internal and external forces. Point 𝑃 , in general, is a moving point.

⃗ =ℎ ⃗̇ + 𝑣⃗ × 𝑚𝑣⃗ . 𝑀 𝑃 𝑃 𝑃 𝐺 ⃗ is the moment with respect to 𝑃 of only the Here, with reference to Fig. 15.48, 𝑀 𝑃 ∑𝑁 external forces acting on the system, 𝑚 = 𝑖=1 𝑚𝑖 is the total mass of the system, 𝐺 is the

ISTUDY

Section 15.6

1047

Chapter Review

⃗ is the total angular momentum, system’s center of mass, 𝑣⃗𝐺 is the velocity of 𝐺, and ℎ 𝑃 which is defined as Eq. (15.48), p. 984 𝑁 ∑

⃗ = ℎ 𝑃

⃗ . ℎ 𝑃𝑖

𝑖=1

If any one of the following conditions is satisfied: ⃗ 1. 𝑃 is a fixed point, i.e., when 𝑣⃗𝑃 = 0, ⃗ 2. 𝐺 is a fixed point, i.e., when 𝑣⃗𝐺 = 0, 3. 𝑃 coincides with the center of mass and therefore 𝑣⃗𝑃 = 𝑣⃗𝐺 , or 4. vectors 𝑣⃗𝑃 and 𝑣⃗𝐺 are parallel, then the moment-angular momentum relation for a closed system of particles simplifies to Eq. (15.50), p. 985 ⃗ =ℎ ⃗̇ . 𝑀 𝑃 𝑃

Furthermore, if any of the above conditions are satisfied over a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 , then the moment-angular momentum relation for a closed system of particles can be integrated with respect to time to obtain the angular impulse-momentum principle as Eq. (15.51), p. 985 ⃗ + ℎ 𝑃1

∫𝑡

𝑡2

apoapsis

⃗ 𝑑𝑡 = ℎ ⃗ . 𝑀 𝑃 𝑃2

1

orbit of 𝑆

Orbital mechanics In orbital mechanics, we studied the motion of a satellite 𝑆 of mass 𝑚 that is subject to Newton’s universal law of gravitation due to a body 𝐵 of mass 𝑚𝐵 , which is the primary or attracting body (see Fig. 15.49). We began with these important assumptions:

𝐵 𝑟𝑃

2. The only force acting on the satellite 𝑆 is the force of mutual attraction between 𝐵 and 𝑆. 3. The primary body 𝐵 is fixed in space.

Determination of the orbit. Solving the governing equations in polar coordinates, we found that the trajectory of the satellite 𝑆 under these assumptions is a conic section, whose equation can be written as Eqs. (15.73) and (15.78), p. 1009 𝐺𝑚 1 = 𝐶 cos 𝜃 + 2𝐵 or 𝑟 𝜅 1 𝐺𝑚𝐵 (1 + 𝑒 cos 𝜃), = 𝑟 𝜅2 where 𝑟 is the distance between the centers of mass of 𝑆 and 𝐵; 𝜃 is the orbital angle measured relative to periapsis; 𝐺 is the universal gravitational constant; 𝜅 is the angular

𝑣

𝑟

1. The primary body 𝐵 and the satellite 𝑆 are both treated as particles.

periapsis

𝜃

𝑆 𝑣𝑃

Figure 15.49 A satellite 𝑆 of mass 𝑚 orbiting a primary body 𝐵 of mass 𝑚𝐵 on a conic section, which in this case is an ellipse. The angle 𝜃 and the launch conditions 𝑟𝑃 and 𝑣𝑃 are defined from periapsis.

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Chapter 15

Momentum Methods for Particles

momentum per unit mass of the satellite 𝑆 measured about 𝐵; 𝐶 is a constant to be determined; and 𝑒 is the eccentricity of the trajectory, which can be written as Eq. (15.77), p. 1009 𝑒=

𝐶𝜅 2 . 𝐺𝑚𝐵

If, as is generally the case here, the orbital conditions are known at periapsis, then 𝜅 is Eq. (15.79), p. 1010 𝜅 = 𝑟𝑃 𝑣𝑃 , the constant 𝐶 is Eq. (15.80), p. 1010 ) ( 𝐺𝑚𝐵 1 , 𝐶= 1− 𝑟𝑃 𝑟𝑃 𝑣2𝑃 and the equation describing the trajectory becomes hyperbola, 𝑒 > 1

parabola, 𝑒 = 1 ellipse, 0 < 𝑒 < 1

𝑣𝑃

circle, 𝑒 = 0 periapsis

𝑟𝑃

𝐵

apoapsis

Eq. (15.81), p. 1010 ) ( 𝐺𝑚 𝐺𝑚𝐵 1 1 cos 𝜃 + 2 𝐵2 . = 1− 2 𝑟 𝑟𝑃 𝑟𝑃 𝑣𝑃 𝑟𝑃 𝑣𝑃

Conic sections. Equations (15.73), (15.78), and (15.81) represent equivalent conic sections in polar coordinates. The type of conic sections depends on the eccentricity of the trajectory 𝑒 (see Fig. 15.50), which, in turn, depends on 𝐶 and 𝜅 via Eq. (15.77). There are four types of conic sections, which are determined by the value of the eccentricity 𝑒, that is, 𝑒 = 0, 0 < 𝑒 < 1, 𝑒 = 1, and 𝑒 > 1. For a circular orbit (𝑒 = 0), the radius is 𝑟𝑃 and the speed in the orbit is equal to Eq. (15.82), p. 1010 √ 𝐺𝑚𝐵 𝑣𝑐 = . 𝑟𝑃 For an elliptical orbit (0 < 𝑒 < 1), as expected, the radius at periapsis is 𝑟𝑃 . The radius at apoapsis is given by Eq. (15.84), p. 1010

Figure 15.50 The four conic sections, showing 𝑟𝑃 and 𝑣𝑃 .

𝑟𝐴 =

𝑟𝑃 . ( 2) 2𝐺𝑚𝐵 ∕ 𝑟𝑃 𝑣𝑃 − 1

Referring to Fig. 15.51, additional relationships between the semimajor axis 𝑎 of an elliptical orbit, the eccentricity of the orbit, and the radii at periapsis and apoapsis are, respectively, Eqs. (15.90), p. 1011 𝑟𝑃 = 𝑎(1 − 𝑒)

and 𝑟𝐴 = 𝑎(1 + 𝑒).

The period of an elliptical orbit 𝜏 can be written in the following two ways: Eq. (15.92), p. 1011, and Eq. (15.94), p. 1012 𝜏=

)√ 2𝜋𝑎𝑏 𝜋 ( = 𝑟 + 𝑟𝐴 𝑟𝑃 𝑟𝐴 , 𝜅 𝜅 𝑃

ISTUDY

Section 15.6

Chapter Review

𝑟𝑃 = 𝑎(1 − 𝑒)

𝑟𝐴 = 𝑎(1 + 𝑒)

periapsis

𝑏

apoapsis

𝑏

𝑎

𝑎

Figure 15.51. The semimajor axis 𝑎 and semiminor axis 𝑏 of an ellipse.

or, reflecting Kepler’s third law, the orbital period can be written as Eq. (15.97), p. 1012 𝜏2 =

4𝜋 2 3 𝑎 . 𝐺𝑚𝐵

A parabolic trajectory (𝑒 = 1) is one that divides periodic orbits (which return to their starting location) from trajectories that are not periodic. For a given 𝑟𝑃 , the speed 𝑣par required to achieve a parabolic trajectory is given by Eq. (15.99), p. 1012 √ 2𝐺𝑚𝐵 , 𝑣par = 𝑣esc = 𝑟𝑃 which is also referred to as the escape velocity since it is the speed required to completely escape the influence of the primary body 𝐵. For a hyperbolic trajectory (𝑒 > 1), the governing equations given by Eqs. (15.77)– (15.81) are used for values of 𝑒 > 1.

Energy considerations. Using the work-energy principle, we discovered that the total mechanical energy in an orbit depends on only the semimajor axis 𝑎 of the orbit and is given by Eq. (15.104), p. 1013 𝐸=−

𝐺𝑚𝐵 2𝑎

,

where 𝐸 is the mechanical energy per unit mass of the satellite. Applying the work-energy principle at an arbitrary location within the orbit, we found that the speed can be written as Eq. (15.106), p. 1013 √ ( ) 2 1 𝑣 = 𝐺𝑚𝐵 . − 𝑟 𝑎

1049

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Chapter 15

Momentum Methods for Particles

Mass flows In Section 15.5, we considered mass flows. We considered (1) steady mass flows, in which a fluid moves through a conduit with a velocity that depends only on the position within the conduit; and (2) variable mass flows, such as the flow of combustion gases out of a rocket.

Steady flows. Given the control volume (CV) shown in Fig. 15.52 (a control volume is a portion of a conduit delimited by two cross sections), we showed that, in the case of a steady flow, the total external force 𝐹⃗ acting on the fluid in the CV is 𝑣𝐴

𝑣𝐵

control volume

Eq. (15.112), p. 1024 𝐵

( ) 𝐹⃗ = 𝑚̇ 𝑓 𝑣⃗𝐵 − 𝑣⃗𝐴 ,

𝐴 Figure 15.52 A CV corresponding to the portion of a pipe between cross sections 𝐴 and 𝐵.

where, as long as the cross sections are perpendicular to the flow velocity, 𝑚̇ 𝑓 is the mass flow rate, i.e., the amount of mass flowing through a cross section per unit time, and 𝑣⃗𝐴 and 𝑣⃗𝐵 are the flow velocities at the cross sections 𝐴 and 𝐵, respectively. In addition to the mass flow rate, we defined the volumetric flow rate as the quantity Eq. (15.116), p. 1025 𝑄 = 𝑣𝑆, where 𝑣 is the speed of the fluid at a given cross section and 𝑆 is the area of the cross section in question. We showed that

𝑣𝐵 𝑣𝐴

control volume

𝐵

𝐶 𝐴

Eq. (15.117), p. 1025

𝐷

where 𝜌𝐴 and 𝜌𝐵 are the values of the mass density of the fluid at 𝐴 and 𝐵, respectively. ⃗ Referring to Fig. 15.53, we also showed that, given a fixed point 𝑃 , the total moment 𝑀 𝑃 acting on the fluid in the CV is

𝑟⃗𝐷∕𝑃

𝑟⃗𝐶∕𝑃

𝑚̇ 𝑓 = 𝜌𝐴 𝑄𝐴 = 𝜌𝐵 𝑄𝐵 ,

𝑃 Figure 15.53 A fluid flowing through a CV along with a choice of moment center 𝑃 for the calculation of angular momenta and moments.

Eq. (15.122), p. 1026 ) ( ⃗ = 𝑚̇ 𝑟⃗ ⃗𝐵 − 𝑟⃗𝐶∕𝑃 × 𝑣⃗𝐴 , 𝑀 𝑃 𝑓 𝐷∕𝑃 × 𝑣 where 𝐶 and 𝐷 are the centers of the cross sections 𝐴 and 𝐵, respectively.

𝑣⃗𝑖

𝑣⃗𝑜

Figure 15.54 A plane with a jet engine. The airplane is taking in air with a mass flow rate 𝑚̇ 𝑖 , while combustion gases are ejected with a mass flow rate 𝑚̇ 𝑜 . The vector 𝑣⃗𝑖 is the velocity of the inflowing air relative to the plane. The vector 𝑣⃗𝑜 is the velocity of the outflowing combustion gases relative to the plane.

Variable mass flows. With reference to Fig. 15.54, for a body with time varying mass 𝑚(𝑡) due to an inflow of mass with rate 𝑚̇ 𝑖 and an outflow of mass with the rate 𝑚̇ 𝑜 , the total external force acting on the body is given by Eq. (15.133), p. 1027 𝐹⃗ = 𝑚𝑎⃗ + 𝑚̇ 𝑜 𝑣⃗𝑜 − 𝑚̇ 𝑖 𝑣⃗𝑖 , where 𝑎⃗ is the acceleration of the main body, 𝑣⃗𝑜 is the relative velocity of the outflowing mass with respect to the main body, and 𝑣⃗𝑖 is the relative velocity of the inflowing mass, again relative to the main body.

ISTUDY

Section 15.6

Chapter Review

1051

Review Problems Problem 15.186 In Major League Baseball, a pitched ball has been known to hit the head of the batter (sometimes unintentionally and sometimes not). Let the pitcher be, for example, Nolan Ryan who can throw a 5 18 oz baseball that crosses the plate at 100 mph.∗ Studies have shown that the impact of a baseball with a person’s head has a duration of about 1 ms. So using Eq. (15.9) on p. 935 and assuming that the rebound speed of the ball after the collision is negligible, determine the magnitude of the average force exerted on the person’s head during the impact.

Problem 15.187 A 0.6 kg ball that is initially at rest is dropped on the floor from a height of 1.8 m and has a rebound height of 1.25 m. If the ball spends a total of 0.01 s in contact with the ground, determine the average force applied to the ball by the ground during the rebound. In addition, determine the ratio between the magnitude of the impulse provided to the ball by the ground and the magnitude of the impulse provided to the ball by gravity during the time interval that the ball is in contact with the ground. Neglect air resistance.

1800 mm

Figure P15.187

Problem 15.188 A person 𝑃 is initially standing on a cart on rails, which is moving to the right with a speed 𝑣0 = 2 m∕s. The cart is not being propelled by any motor. The combined mass of person 𝑃 , the cart, and all that is being carried on the cart is 270 kg. At some point a person 𝑃𝐴 standing on a stationary platform throws to person 𝑃 a package 𝐴 to the right with a mass 𝑚𝐴 = 50 kg. Package 𝐴 is received by 𝑃 with a horizontal speed 𝑣𝐴∕𝑃 = 1.5 m∕s. After receiving the package from 𝐴, person 𝑃 throws a package 𝐵 with a mass 𝑚𝐵 = 45 kg toward a second person 𝑃𝐵 . The package intended for 𝑃𝐵 is thrown to the right, i.e., in the direction of the motion of 𝑃 , and with a horizontal speed 𝑣𝐵∕𝑃 = 4 m∕s relative to 𝑃 . Determine the final velocity of the person 𝑃 . Neglect any friction or air resistance acting on 𝑃 and the cart. 𝑣0 𝐵 𝐴

𝑃

𝑃𝐴

𝑃𝐵 Figure P15.188

Problem 15.189 A Ford Excursion 𝐴, with a mass 𝑚𝐴 = 3900 kg, traveling with a speed 𝑣𝐴 = 85 km∕h, collides head-on with a Mini Cooper 𝐵, with a mass 𝑚𝐵 = 1200 kg, traveling in the opposite direction with a speed 𝑣𝐵 = 40 km∕h. Determine the postimpact velocities of the two cars if the impact’s coefficient of restitution is 𝑒 = 0.22. In addition, determine the percentage of kinetic energy loss. ∗ This

means that he must have thrown the ball at about 108 mph since the ball loses speed at the rate of 1 mph for every 7 f t it travels.

𝑣𝐵 𝐵

Figure P15.189

𝑣𝐴 𝐴

ISTUDY

1052

Chapter 15

Momentum Methods for Particles

𝐵 𝐴

Problem 15.190

𝑣− 𝐵 𝛽

𝛼 𝑣− 𝐴 Figure P15.190

The two spheres, 𝐴 and 𝐵, with masses 𝑚𝐴 = 1.35 kg and 𝑚𝐵 = 2.72 kg, respectively, collide with 𝑣−𝐴 = 26.2 m∕s and 𝑣−𝐵 = 22.5 m∕s. Let 𝛼 = 45◦ . Compute the value of 𝛽 if the component of the postimpact velocity of 𝐵 along the LOI is equal to zero and if the COR is 𝑒 = 0.63.

Problem 15.191 A 31,000 lb truck 𝐴 and a 3970 lb sports car 𝐵 collide at an intersection. At the moment of the collision, the truck and the sports car are traveling with speeds 𝑣−𝐴 = 60 mph and 𝑣−𝐵 = 50 mph, respectively. Assume that the entire intersection forms a horizontal surface. Letting the line of impact be parallel to the ground and rotated counterclockwise by 𝛼 = 20◦ with respect to the preimpact velocity of the truck, determine the postimpact velocities of 𝐴 and 𝐵 if the contact between 𝐴 and 𝐵 is frictionless and the COR 𝑒 = 0.1. Furthermore, assuming that the truck and the car slide after impact and that the coefficient of kinetic friction is 𝜇𝑘 = 0.7, determine the position at which 𝐴 and 𝐵 come to a stop relative to the position they occupied at the instant of impact.

𝑣− 𝐵 𝑧 𝐵

𝑣− 𝐴

𝑟0

𝑑 cord 𝑚

𝐴 𝜔0

Figure P15.191

Figure P15.192

Problem 15.192

17,500 mph ℎ𝐴 450 mi

Consider a collar with mass 𝑚 that is free to slide with no friction along a rotating arm of negligible mass. The system is initially rotating with a constant angular velocity 𝜔0 while the collar is kept at a distance 𝑟0 from the 𝑧 axis. At some point, the restraint keeping the collar in place is removed so that the collar is allowed to slide. Determine the expression for the moment that you need to apply to the arm, as a function of time, to keep the arm rotating at a constant angular velocity while the collar travels toward the end of the arm. Hint: ) ( √ 1 𝑑𝑥 = ln 𝑥 + 𝑥2 − 1 + 𝐶. √ ∫ 𝑥2 − 1

Earth

Problem 15.193

Figure P15.193

A satellite is launched parallel to the Earth’s surface at an altitude of 450 mi with a speed of 17,500 mph. Determine the apogee altitude ℎ𝐴 above the Earth’s surface, as well as the period of the satellite.

ISTUDY

Section 15.6

1053

Chapter Review

Problem 15.194 A spacecraft is traveling at 19,000 mph parallel to the surface of the Earth at an altitude of 250 mi, when it fires a retrorocket to transfer to a different orbit. Determine the change in speed Δ𝑣 necessary for the spacecraft to reach a minimum altitude of 110 mi during the ensuing orbit. Assume that the change in speed is impulsive; that is, it occurs instantaneously. Earth 19,000 mph

250 mi

Figure P15.194

Problems 15.195 and 15.196 The optimal way (from an energy standpoint) to transfer from one circular orbit about a primary body (in this case, the Sun) to another circular orbit is via the Hohmann transfer, which involves transferring from one circular orbit to another using an elliptical orbit that is tangent to both at the periapsis and apoapsis of the ellipse. This ellipse is uniquely defined because we know the perihelion radius 𝑟𝑒 (the radius of the inner circular orbit) and the aphelion radius 𝑟𝑗 (the radius of the outer circular orbit), and therefore we know the semimajor axis 𝑎 via Eq. (15.88) and the eccentricity 𝑒 via Eq. (15.87) or Eqs. (15.90). Performing a Hohmann transfer requires two maneuvers, the first to leave the inner (outer) circular orbit and enter the transfer ellipse and the second to leave the transfer ellipse and enter the outer (inner) circular orbit. Assume that the orbits of Earth and Jupiter are circular, use 150×106 km for the radius of Earth’s orbit, use 779×106 km for the radius of Jupiter’s orbit, and note that the mass of the Sun is 333,000 times that of the Earth. A space probe 𝑆1 is launched from Earth to Jupiter via a Hohmann transfer orbit. Determine the change in speed Δ𝑣𝑒 required at the radius of Earth’s orbit of the elliptical transfer orbit (perihelion) and the change in speed Δ𝑣𝑗 required at the radius of Jupiter’s orbit (aphelion). In addition, compute the time required for the orbital transfer. Assume that the changes in speed are impulsive; that is, they occur instantaneously.

aphelion

𝑟𝑗

𝑆2

Earth 𝑆1

Sun 𝑟𝑒 Jupiter

perihelion

Problem 15.195

Figure P15.195 and P15.196

A space probe 𝑆2 is at Jupiter and is required to return to the radius of Earth’s orbit about the Sun so that it can return samples taken from one of Jupiter’s moons. Assuming that the mass of the probe is 722 kg, determine the change in kinetic energy required at Jupiter Δ𝑇𝑗 for the maneuver at aphelion. In addition, determine the change in kinetic energy required at Earth Δ𝑇𝑒 for the perihelion maneuver. Finally, what is the change in potential energy Δ𝑉 of the spacecraft in going from Jupiter to the Earth?

Problem 15.196

𝑣0

Problem 15.197 A water jet is emitted from a nozzle attached to the ground. The jet has a constant mass flow rate (𝑚̇ 𝑓 )nz = 15 kg∕s and a speed 𝑣𝑤 relative to the nozzle. The jet strikes a 12 kg incline and causes it to slide at a constant speed 𝑣0 = 2 m∕s. The kinetic coefficient of friction between the incline and the ground is 𝜇𝑘 = 0.25. Neglecting the effect of gravity and air resistance on the water flow, as well as friction between the water jet and the incline, determine the speed of the water jet at the nozzle if 𝜃 = 47◦ .

𝑣𝑤

𝜃 𝜇𝑘

Figure P15.197

ISTUDY

1054

release position 𝑦

𝓁𝐿 =

Chapter 15

Momentum Methods for Particles

𝐿+𝑦 2 𝓁𝑅 =

Figure P15.198

𝐿−𝑦 2

Problem 15.198 Revisit Example 15.20 and derive the equation of motion of the free end of the string starting from the force balance for the right branch of the string modeled as a variable mass system.

Problem 15.199 An intubed fan (a fan rotating within a tube or other conduit) is mounted on a cart that is connected to a fixed wall by a linear elastic spring with constant 𝑘 = 70 N∕m. Assume that in a particular test the fan draws air that enters the tube at 𝐴 with a speed 𝑣𝐴 . The outgoing flow at 𝐵 has a speed 𝑣𝐵 . The flow of air through the tube causes the cart to displace to the left so that the spring is stretched by 0.25 m from its unstretched position. Assume that the density of air is constant throughout the tube and equal to 𝜌 = 1.25 kg∕m3 . In addition, let the tube’s cross section be circular, and let the cross-sectional diameters at 𝐴 and 𝐵 be 𝑑𝐴 = 3 m and 𝑑𝐵 = 1.5 m, respectively. Determine the velocities of the airflow at 𝐴 and 𝐵. 𝐴 𝐵 𝑑𝐴

𝑣𝐴

𝑑𝐵

𝑣𝐵 𝑘

Figure P15.199

ISTUDY

Planar Rigid Body Kinematics

16 This chapter begins the study of rigid body motion by developing the planar kinematics of a rigid body. As we did in Chapter 12, we will describe motion without addressing what causes the motion. We will assume that (1) the body is rigid and its mass is distributed over a region of space (see Section 11.2, p. 626) and (2) the velocity of each of the body’s points is parallel to a common plane. This may appear daunting because the description of a body’s motion requires that we know the motion of each point in the body. However, the rigidity assumption alleviates this difficulty, and we will discover that we can completely characterize the planar motion of a rigid body using only three functions of time. To understand how rigidity helps us develop a kinematics that is so efficient, we begin by giving a qualitative description of rigid body motions, and then we proceed with a quantitative analysis. General 3D rigid body motions are examined in Chapter 10.

John Peter Photography/Alamy Stock Photo

The Falkirk Wheel in central Scotland. This is a rotating lift system that connects the Union Canal with the Forth and Clyde Canal. Fixed-axis rotations are an important special case of rigid body motions.

16.1

Fundamental Equations, Translation, and Rotation About a Fixed Axis

Crank, connecting rod, and piston motion Figure 16.1 shows the interior of an internal combustion engine. While we have learned to describe the motion of individual points, we have not learned how to characterize the motion of entire bodies. For example, we can find the velocity of a point on one of the connecting rods, but we also want to be able to describe the motion of the connecting rod as a whole. Figure 16.2 shows three sequential views of what we would see in one of the cylinders if we looked down the crankshaft (i.e., the crankshaft axis is perpendicular to the page). Point 𝐴 represents the axis of the crankshaft, and points 𝐵 and 𝐶 represent the pin connections between the connecting rod and the crank and between the connecting rod and the piston, respectively. Because of these connections, the motion of the piston causes a rotation of the crankshaft. As the crankshaft goes through a complete rotation, the piston reverses its motion and slides back up the cylinder so that the overall motion is repeated. This is an

connecting rod piston

crankshaft Archivea Inc dba Ford Images

Figure 16.1 Interior view of the EcoBoost engine from Ford, revealing the pistons, connecting rods, and crankshaft.

1055

1056

Chapter 16

Planar Rigid Body Kinematics

Figure 16.2. Three different positions of the mechanical system formed by a piston, connecting rod, and crank in a typical internal combustion engine. The axis of the crankshaft is perpendicular to the page, and it is represented by the points labeled 𝐴.

example of a slider-crank mechanism.∗ The dimensions and relative position of the piston, connecting rod, and crankshaft influence the motion of these elements and contribute to the overall engine performance. We will use this system to illustrate most of the kinematics we will present in this chapter.

Qualitative description of rigid body motion Translation

Figure 16.3 Left: a lift with an articulated boom and a platform holding a worker. Right: a schematic of the lift’s range of motion (the gray area) showing that the platform remains parallel to the ground while moving along a curved path.

ISTUDY

Assume that the cylinder within which the piston moves in Fig. 16.2 is stationary. As the piston moves, we see that the piston rings (appearing as horizontal bands at the top of the piston) always remain horizontal. Therefore, the velocity of point 𝐶 on the piston is the same as that of any other point on the piston. This type of motion is called translation, and it is defined by saying that any line segment connecting two points in the body maintains its original orientation throughout the motion. Since a translating body does not necessarily move in a straight line (see Fig. 16.3), we classify translations into one of two categories, rectilinear translations or curvilinear translations, based on whether the trajectory of each point is a straight line or not. The motion of the piston in Fig. 16.2 is a rectilinear translation, whereas the motion of the platform in Fig. 16.3 is a curvilinear translation. Rotation about a fixed axis Figure 16.4 emphasizes the motion of the crank in Fig. 16.2. Point 𝐴 on the axis of the shaft does not move. Any point on the crank lying on the axis perpendicular to the plane of the figure and going through point 𝐴 does not move. Because the crank is rigid, off-axis points on the crank (e.g., point 𝐵) can only move in a circle centered at 𝐴. Any segment connecting points to the shaft axis rotates with the same angular ∗A

mechanism is “a system of elements arranged to transmit motion in a predetermined fashion.” From R. L. Norton, Design of Machinery, 4th ed., McGraw-Hill, New York, 2008.

ISTUDY

Section 16.1

Fundamental Equations, Translation, and Rotation About a Fixed Axis

1057

Figure 16.4. A modification of Fig. 16.2 emphasizing the motion of the crank.

velocity as any other segment. This type of motion, where there is a line of points with zero velocity functioning as an axis of rotation, is a rotation about a fixed axis. General planar motion Looking at the connecting rod, which is emphasized in Fig. 16.5, this motion does not correspond to either of the two cases discussed earlier since (1) we cannot find an axis that is fixed throughout the motion and (2) no two points define a segment that does not change its orientation. The only distinguishing feature of the rod’s motion is that none of its points has a component of velocity perpendicular to the plane of the page, which is called the plane of motion. This kind of motion is called general plane motion or general planar motion. This motion can be viewed as the composition of a translation with a rotation about an axis perpendicular to the plane of motion. Note that both the piston’s translation and the crankshaft’s rotation are also planar motions.

Figure 16.5 A modification of Fig. 16.2 emphasizing the motion of the connecting rod.

General motion of a rigid body

𝐵

For a rigid body, we can describe the motion of all points by describing (1) the motion of a single point and (2) the rate of change of the body’s orientation. Figure 16.6 shows the general plane motion of an aircraft carrier from above. We can express the velocity of point 𝐵 in terms of the velocity of point 𝐴 using Eq. (12.84) on p. 739 as (16.1) 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 . Noting that 𝑣⃗𝐵∕𝐴 = 𝑟⃗̇ 𝐵∕𝐴 and writing 𝑟⃗𝐵∕𝐴 as |⃗𝑟𝐵∕𝐴 | 𝑢̂ 𝐵∕𝐴 , we can use Eq. (12.60) on p. 718 for the time derivative of a vector, to obtain 𝑣⃗𝐵∕𝐴 =

) 𝑑|⃗𝑟𝐵∕𝐴 | 𝑑( |⃗𝑟𝐵∕𝐴 | 𝑢̂ 𝐵∕𝐴 = 𝑢̂ 𝐵∕𝐴 + |⃗𝑟𝐵∕𝐴 | 𝑢̂̇ 𝐵∕𝐴 𝑑𝑡 𝑑𝑡 = |⃗𝑟𝐵∕𝐴 |𝜔 ⃗ 𝐴𝐵 × 𝑢̂ 𝐵∕𝐴 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , (16.2)

where we have used the fact that 𝑑|⃗𝑟𝐵∕𝐴 |∕𝑑𝑡 = 0 since the distance between 𝐴 and 𝐵 is constant when they are both on the same rigid body. Substituting Eq. (16.2) into Eq. (16.1) gives 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 ,

(16.3)

where we note that when 𝐴 and 𝐵 are two points on the same rigid body, 𝑣⃗𝐵∕𝐴 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 . Equation (16.3), which relates the motion of points 𝐴 and 𝐵, depends on the angular velocity of the line segment 𝐴𝐵. Figure 16.7 shows that as the aircraft carrier moves from position 1 to position 2, the three line segments 𝐴𝐵, 𝐵𝐶, and

𝑣⃗𝐴

𝑣⃗𝐵

𝑟⃗𝐵∕𝐴 𝑢̂ 𝐵∕𝐴

𝐴

Figure 16.6 Aerial view of an aircraft carrier performing a maneuver. We model the carrier’s motion as a planar rigid body motion.

Concept Alert ⃗ should we use? The angular velocWhat 𝝎 ity is a property of the body, so referring to Fig. 16.7, all of the following equations can be written ⃗ body × 𝑟⃗𝐴∕𝐵 , 𝑣⃗𝐴 = 𝑣⃗𝐵 + 𝜔 ⃗ body × 𝑟⃗𝐵∕𝐶 , 𝑣⃗𝐵 = 𝑣⃗𝐶 + 𝜔 𝑣⃗𝐶 = 𝑣⃗𝐴 + 𝜔 ⃗ body × 𝑟⃗𝐶∕𝐴 , where the 𝜔 ⃗ body in these equations is the angular velocity of the body.

1058

ISTUDY

Chapter 16

Planar Rigid Body Kinematics

𝐴

𝐵

𝐶 position 2 𝐴

position 1

𝐵 𝐶

Figure 16.7. Aerial view of an aircraft carrier moving from position 1 to position 2 showing that all line segments rotate the same amount as a rigid body moves.

𝐶𝐴 all rotate the same amount. This implies that any line segment on the aircraft carrier will have the same rate of rotation as any other line segment, and so the angular ⃗ 𝐴𝐵 is a property of the body as a whole and not any particular velocity of a rigid body 𝜔 part of it. This means that Eq. (16.3) applies to any two points 𝐴 and 𝐵 on the same rigid body. The subscripts on angular velocities (and soon, angular accelerations) should be viewed as labels referring to a particular body. For example, in Eq. (16.3), the subscript 𝐴𝐵 on 𝜔 ⃗ tells us that it is the angular velocity of the body containing points 𝐴 and 𝐵. The equation relating the acceleration of two points on the same rigid body is found by differentiating Eq. (16.3) with respect to time to obtain ⃗̇ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗̇ 𝐵∕𝐴 . 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝜔

(16.4)

The quantity 𝜔 ⃗̇ 𝐴𝐵 is the angular acceleration of the body and is denoted by 𝛼⃗𝐴𝐵 . ̇ ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , Eq. (16.4) can be written as Since 𝑟⃗𝐵∕𝐴 = 𝑣⃗𝐵∕𝐴 = 𝜔 ( ) 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 ,

(16.5)

where we note that when 𝐴 and 𝐵 are two points on a rigid body, 𝑎⃗𝐵∕𝐴 = 𝛼⃗𝐴𝐵 × ( ) 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 . Note that no explicit functional dependence appears in the angular motions 𝜔 ⃗ 𝐴𝐵 and 𝛼⃗𝐴𝐵 . Angular motions are often functions of time, but they cannot be functions of space. Considering again our aircraft carrier in Fig. 16.7, the carrier may be turning into the wind at position 1 in preparation to launch fighter jets in position 2. Its angular motion has a beginning and end. The angular motions can’t be functions of space, however, as different angular motions of line segments 𝐴𝐵 and 𝐵𝐶 would imply the width of the carrier deck 𝐵𝐶 is widening or narrowing during the turn. In considering the slider-crank of Fig. 16.2, the crank is assumed to rotate counterclockwise, and the connecting rod experiences clockwise rotation for all states where point 𝐵 is above point 𝐴 (the left and center parts of the figure). It experiences counterclockwise rotation for all states where point 𝐵 is below point 𝐴 (the right part of the figure). At the instants where 𝐴 and 𝐵 are at the same level (the 3 o’clock and 9 o’clock positions), the angular velocity of the connecting rod is zero. The connecting rod’s angular velocity and angular acceleration are periodic functions of time. This will be examined in greater detail in Example 16.6.

ISTUDY

Section 16.1

Fundamental Equations, Translation, and Rotation About a Fixed Axis

Applying Eqs. (16.3) and (16.5) Equations (16.3) and (16.5) tell us that we can know the motion of all the points on a rigid body by knowing the motion of a single point (e.g., 𝑣⃗𝐴 and 𝑎⃗𝐴 ) and by knowing the rotation of the body (e.g., 𝜔 ⃗ 𝐴𝐵 and 𝛼⃗𝐴𝐵 ). Since we will apply Eqs. (16.3) and (16.5) to analyze mechanisms consisting of several rigid bodies, we must always relate points 𝐴 and 𝐵 that belong to the same rigid body. For example, in Fig. 16.8, we can use Eqs. (16.3) and (16.5) to relate the velocities and accelerations, respectively, of points 𝐴 and 𝐵 because these points are both on the crank. We can do the same for points 𝐵 and 𝐶 because they are both on the connecting rod. Since point 𝐵 is on both the crank and the connecting rod, it allows us to relate the motions of the connecting rod and the crank. We cannot use Eqs. (16.3) and (16.5) to relate the motion of points 𝐴 and 𝐶 because these points belong to two distinct bodies—the crank and the piston, respectively.

Graphical interpretation of Eqs. (16.3) and (16.5) As we said earlier, a general plane motion can be viewed as the composition of a translation with a rotation about an axis perpendicular to the plane of motion. Figure 16.9 shows this idea for the relative velocity equation given in Eq. (16.3). This

𝐵

𝐵

= 𝑣⃗𝐵

𝑣⃗𝐵∕𝐴

𝐴

(a)

(b)

𝑟⃗𝐵∕𝐴

+

𝑣⃗𝐴

𝐴

𝐵

𝜔 ⃗ 𝐴𝐵

𝑣⃗𝐴 𝑣⃗𝐵

𝐴 𝑣⃗𝐵∕𝐴

𝑣⃗𝐴

(c)

(d)

𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 Figure 16.9. A graphical representation of the equation 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 .

figure graphically demonstrates that we can add the translation of the body with point 𝐴 [i.e., 𝑣⃗𝐴 in Fig. 16.9(b)] to the pure rotation of point 𝐵 about point 𝐴 [i.e., 𝑣⃗𝐵∕𝐴 = 𝜔 ⃗ 𝐴𝐵 ×⃗𝑟𝐵∕𝐴 in Fig. 16.9(c)] to obtain the velocity of 𝐵 [i.e., 𝑣⃗𝐵 in Figs. 16.9(a) and 16.9(d)]. A similar graphical argument can be applied to the relative acceleration equation given in Eq. (16.5). Referring to Fig. 16.10, we again see that we can add the translation of the body with point 𝐴 [i.e., 𝑎⃗𝐴 in Fig. 16.10(b)] to the pure rotation of point 𝐵 about point 𝐴 [i.e., 𝑎⃗𝐵∕𝐴 = (𝑎⃗𝐵∕𝐴 )𝑡 + (𝑎⃗𝐵∕𝐴 )𝑛 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × (𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 ) in Fig. 16.10(c)] to obtain the acceleration of 𝐵 [i.e., 𝑎⃗𝐵 in Figs. 16.10(a) and 16.10(d)].

Elementary rigid body motions: translations Figure 16.11 shows the deployment of a basketball goal. The backboard 𝐴𝐵 and hoop are attached to an arm, which, in turn, is hinged to two parallel bars 𝐶𝐷 and 𝐸𝐹 . The design is meant to ensure that the hoop remains parallel to the floor. Therefore, the backboard, hoop, and supporting arm are in translation. This means that the angular

Figure 16.8 Slider-crank mechanism.

1059

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Chapter 16

Planar Rigid Body Kinematics

=

+

( Figure 16.10. A graphical representation of the equation 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 ×⃗𝑟𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × ) 𝑟⃗𝐵∕𝐴 . The subscripts 𝑡 and 𝑛 stand for the tangential and normal components, respectively, of the relative acceleration 𝑎⃗𝐵∕𝐴 .

velocity and angular acceleration of a translating rigid body are equal to zero, that is, 𝜔 ⃗ 𝐴𝐵 = 𝜔 ⃗ body = 0⃗ and

⃗ 𝛼⃗𝐴𝐵 = 𝛼⃗body = 0.

(16.6)

Substituting Eqs. (16.6) into Eqs. (16.3) and (16.5) gives 𝑣⃗𝐵 = 𝑣⃗𝐴

Figure 16.11 The deployment of a portable basketball goal (see inset photo).

ISTUDY

and

𝑎⃗𝐵 = 𝑎⃗𝐴 ,

(16.7)

where 𝐴 and 𝐵 are any two points on the body. Equations (16.7) say that the motion of a translating rigid body is characterized by one velocity and one acceleration.

Elementary rigid body motions: rotation about a fixed axis Figure 16.12 shows us that the crank is rotating about a fixed axis through point 𝐴

Figure 16.12. Crank motion in a slider-crank mechanism.

and perpendicular to the plane of motion. Referring to Fig. 16.13, this means that ⃗ 𝑎⃗ = 0, ⃗ and Eqs. (16.3) and (16.5) then give 𝑣⃗𝐴 = 0, 𝐴 𝑣⃗𝐵 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 ,

(16.8) (

⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔

) × 𝑟⃗𝐵∕𝐴 .

(16.9)

ISTUDY

Section 16.1

Fundamental Equations, Translation, and Rotation About a Fixed Axis

1061

Since the motion is planar, 𝜔 ⃗ 𝐴𝐵 and 𝛼⃗𝐴𝐵 can be written in terms of the body’s orientation by noting that points 𝐴 and 𝐵 are in the plane of motion and that the body’s orientation 𝜃 can be defined using the line connecting 𝐴 and 𝐵 (Fig. 16.13). Therefore, the vectors 𝜔 ⃗ 𝐴𝐵 and 𝛼⃗𝐴𝐵 can be written as: 𝜔 ⃗ 𝐴𝐵 = 𝜔𝐴𝐵 𝑘̂ = 𝜃̇ 𝑘̂

and

̂ 𝛼⃗𝐴𝐵 = 𝛼𝐴𝐵 𝑘̂ = 𝜃̈ 𝑘,

(16.10)

where 𝑘̂ = 𝑢̂ 𝑟 × 𝑢̂ 𝜃 = 𝚤̂ × 𝚥̂ is perpendicular to the plane of motion, and 𝜔𝐴𝐵 and 𝛼𝐴𝐵 denote the components of the angular velocity and acceleration, respectively, in the 𝑘̂ direction. Using the polar component system shown in Fig. 16.13, letting 𝑟⃗𝐵∕𝐴 = 𝑅 𝑢̂ 𝑟 , and substituting Eqs. (16.10) into Eqs. (16.8) and (16.9), 𝑣⃗𝐵 and 𝑎⃗𝐵 are 𝑣⃗𝐵 = 𝑅𝜃̇ 𝑢̂ 𝜃

and

𝑎⃗𝐵 = 𝑅𝜃̈ 𝑢̂ 𝜃 − 𝑅𝜃̇ 2 𝑢̂ 𝑟 ,

(16.11)

which is not surprising since 𝐵 is in circular motion about 𝐴. The acceleration coming from the term 𝜔 ⃗ 𝐴𝐵 × (𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 ) is equal to −𝑅𝜃̇ 2 𝑢̂ 𝑟 . Noting that 𝜃̇ 2 = 𝜔2𝐴𝐵 and that −𝑅 𝑢̂ 𝑅 = −⃗𝑟𝐵∕𝐴 , for planar motion, we can write ( ) ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = −𝑅𝜃̇ 2 𝑢̂ 𝑟 = −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 . 𝜔 ⃗ 𝐴𝐵 × 𝜔

Figure 16.13 Detailed view of the crank of a slider-crank mechanism.

(16.12)

We can see this geometrically if we note that 𝑟⃗𝐵∕𝐴 is always in the plane of motion and that 𝜔 ⃗ 𝐴𝐵 is always perpendicular to it (see Fig. 16.14). Taking their cross product

Figure 16.14. Geometric demonstration of the equivalence of 𝜔 ⃗ 𝐴𝐵 × (𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 ) and −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 for planar motion.

𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 results in a vector that is in the plane of motion and perpendicular to both 𝜔 ⃗ 𝐴𝐵 and 𝑟⃗𝐵∕𝐴 . Finally, taking 𝜔 ⃗ 𝐴𝐵 × (𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 ) (the cross product of the perpendicular purple vectors in Fig. 16.14) results in the vector −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 (the green vector in Fig. 16.14). Using Eq. (16.12), we can write Eq. (16.9) as 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 .

(16.13)

This form can save computation in finding accelerations for planar problems. Graphical interpretation of Eq. (16.8). Referring to the crank in Fig. 16.15, consider the velocity of points 𝐻, 𝐵, and 𝑄 lying on the radial line 𝓁 with origin at the center of rotation 𝐴. Equation (16.8), or the first of Eqs. (16.11), implies that 𝑣⃗𝐻 , 𝑣⃗𝐵 , and 𝑣⃗𝑄 are all perpendicular to 𝓁 (and parallel to one another) and have a magnitude proportional to their distance from 𝐴. The constant of proportionality is 𝜔𝐴𝐵 . Therefore, the distribution of the velocities of points on radial lines can be represented graphically by the triangle as shown.

Figure 16.15 Graphical representation of the velocities of points on radial lines originating at the center of rotation.

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Planar motion in practice For general planar motions, we just saw that it is always possible to express the term ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 ) as −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 . Equation (16.5) can then be written as 𝜔 ⃗ 𝐴𝐵 × (𝜔 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 . 𝑣⃗𝐵∕𝐴 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴

(16.14)

When relating the motion of two points 𝐴 and 𝐵 on the same rigid body, we will generally use this version of the acceleration equation.

𝐵 𝜔 ⃗ 𝐴𝐵

End of Section Summary −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 𝑟⃗𝐵∕𝐴 𝛼⃗𝐴𝐵 𝐴

Figure 16.16 A rigid body showing the quantities used in Eqs. (16.3), (16.5), and (16.14).

ISTUDY

This section began our study of the dynamics of rigid bodies. As with particles, we start with the study of kinematics, which is the focus of this chapter. The key kinematic idea is that a rigid body has only one angular velocity and one angular acceleration at any given time; i.e., each is a property of the body as a whole. This allowed us to use the relative velocity equation [Eq. (12.84) on p. 739] and the relation for the time derivative of a vector [Eq. (12.60) on p. 718] to relate the velocities of two points 𝐴 and 𝐵 on the same rigid body using (see Fig. 16.16) Eq. (16.3), p. 1057 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 . A similar approach allowed us to relate the accelerations of two points 𝐴 and 𝐵 on a rigid body as: Eq. (16.5), p. 1058 ( ) 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 , which for planar motion becomes Eq. (16.14), p. 1062 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 . Translation. For this motion, the angular velocity and angular acceleration of the body are equal to zero, i.e., Eq. (16.6), p. 1060 𝜔 ⃗ body = 0⃗ and

⃗ 𝛼⃗body = 0,

so the velocity and acceleration relations for the body reduce to Eq. (16.7), p. 1060 𝑣⃗𝐵 = 𝑣⃗𝐴

and

𝑎⃗𝐵 = 𝑎⃗𝐴 ,

where 𝐴 and 𝐵 are any two points on the body.

ISTUDY

Section 16.1

Fundamental Equations, Translation, and Rotation About a Fixed Axis

Rotation about a fixed axis. For this special motion, there is an axis of rotation perpendicular to the plane of motion that does not move. All points not on the axis of rotation can only move in a circle about that axis. If the axis of rotation is at point 𝐴, then the velocity of 𝐵 is given by Eq. (16.8), p. 1060 𝑣⃗𝐵 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , and its acceleration is Eq. (16.9), p. 1060, and Eq. (16.13), p. 1061 ( ) 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 , where the vectors 𝜔 ⃗ 𝐴𝐵 and 𝛼⃗𝐴𝐵 are the angular velocity and angular acceleration of the body, respectively, and 𝑟⃗𝐵∕𝐴 is the position of 𝐵 relative to 𝐴.

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E X A M P L E 16.1

Engine Pulleys: Fixed-Axis Rotation Most car engines have a number of belts connecting pulleys on the engine (Fig. 1). The belts are not supposed to slip relative to the pulleys they connect, and they are used to transmit, as well as synchronize, motion between engine parts. For the belt connecting pulley 𝐴, which rotates with the crankshaft, with pulley 𝐵, which drives the alternator, determine the angular speed of pulley 𝐵 if the crankshaft is spinning at 2550 rpm and the radii of pulleys 𝐴 and 𝐵 are 𝑅𝐴 = 4.25 in. and 𝑅𝐵 = 2.5 in., respectively.

SOLUTION Road Map Referring to Fig. 2, the no-slip condition between the belt and pulleys means that any two points on the belt and pulley that are in contact at a given instant, e.g., 𝐶 and ⃗ Combining this 𝐷 or 𝑃 and 𝑄, must have the same velocity (i.e., 𝑣⃗𝐶∕𝐷 = 𝑣⃗𝑃 ∕𝑄 = 0). observation with the fact that pulleys 𝐴 and 𝐵 rotate about fixed axes and the assumption that the belt is inextensible (all points on the belt must have the same speed) will allow us to solve the problem.

𝐴 belt Archivea Inc dba Ford Images

Figure 1 Typical car engine with several belts.

Computation

From Fig. 2, the inextensibility of the belt implies that |𝑣⃗𝐶 | = |𝑣⃗𝑃 |.

𝑅𝐵 𝑅𝐴 𝐴

𝜔𝐵

𝜔𝐴

Using this result and recalling the no-slip condition, we find

𝐵

𝑣⃗𝐶 = 𝑣⃗𝐷

and

𝑣⃗𝑃 = 𝑣⃗𝑄

⇒

|𝑣⃗𝐷 | = |𝑣⃗𝑄 |.

(2)

Referring to Fig. 2, pulleys 𝐴 and 𝐵 undergo fixed-axis rotation about their respective centers. Applying Eq. (16.8) to each of the pulleys to obtain 𝑣⃗𝐷 and 𝑣⃗𝑄 , we find

𝑄

𝐷

(1)

belt

𝑣⃗𝐷 = 𝜔 ⃗ 𝐴 × 𝑟⃗𝐷∕𝐴 Figure 2 Schematic of the system consisting of pulleys 𝐴 and 𝐵 and the belt connecting them. The pulleys are attached to the engine, which is assumed to be stationary.

and 𝑣⃗𝑄 = 𝜔 ⃗ 𝐵 × 𝑟⃗𝑄∕𝐵 ,

(3)

and since 𝜔 ⃗ and 𝑟⃗ are perpendicular to each other in each case, the corresponding speeds are |𝑣⃗𝐷 | = |𝜔 ⃗ 𝐴 |𝑅𝐴 and |𝑣⃗𝑄 | = |𝜔 ⃗ 𝐵 |𝑅𝐵 . (4) Substituting Eq. (4) into the last of Eqs. (2), we obtain |𝜔 ⃗ 𝐴 |𝑅𝐴 = |𝜔 ⃗ 𝐵 |𝑅𝐵

⇒

|𝜔 ⃗𝐵| =

𝑅𝐴 𝑅𝐵

|𝜔 ⃗ 𝐴 | = 4335 rpm,

(5)

where we have plugged in |𝜔 ⃗ 𝐴 | = 2550 rpm, 𝑅𝐴 , and 𝑅𝐵 . 𝜃𝐴 𝛥𝐿

𝑅𝐴

𝑅𝐵

Discussion & Verification

𝛥𝐿

Figure 3 Measure of the linear length of belt going around the pulleys under the no-slip condition.

ISTUDY

To verify the result in Eq. (5), Fig. 3 shows that the no-slip condition between the belt and pulley 𝐴 implies that if pulley 𝐴 rotates through an angle 𝜃𝐴 , then the belt must move an amount Δ𝐿 = 𝜃𝐴𝑅𝐴 around pulley 𝐴. Since the belt does not slip with respect to pulley 𝐵, we must also have Δ𝐿 = 𝜃𝐵𝑅𝐵 , where 𝜃𝐵 is the angle through which pulley 𝐵 rotates. Therefore, we must have 𝜃𝐵 = (𝑅𝐴 ∕𝑅𝐵 )𝜃𝐴 . This relation implies that the rates of change of the angle 𝜃𝐵 will be proportional to 𝜃𝐴 via the ratio of the pulley radii 𝑅𝐴 ∕𝑅𝐵 — this is what we obtained in Eq. (5).

ISTUDY

Section 16.1

Fundamental Equations, Translation, and Rotation About a Fixed Axis

E X A M P L E 16.2

1065

Gears: Fixed-Axis Rotation

An electric motor with a top angular speed of 3450 rpm is used to spin a centrifuge. The motor’s motion is transmitted to the centrifuge by two gears 𝐴 and 𝐵, with radii 𝑅𝐴 and 𝑅𝐵 , respectively (see Fig. 1). Determine the ratio 𝑅𝐴 ∕𝑅𝐵 if the centrifuge is to achieve 6000 rpm as its top angular speed. In addition, determine the ratio between the magnitude of the angular acceleration of 𝐴 to that of 𝐵 during spin-up.

centrifuge

𝜔𝐴

𝐵

SOLUTION

𝐴

Road Map Gear teeth are designed so that the gears behave as two wheels rolling without slip on each other. The radius of a gear is the radius of the wheel that the gear is designed to represent (see the dashed red line in Fig. 2), as opposed to the inner or outer radius of the teeth. Therefore, referring to Fig. 2, the no-slip condition between the gears tells us that the velocity of points 𝑃 and 𝑄 must be the same when they are in contact. Furthermore, gears 𝐴 and 𝐵 are rotating about the fixed axes of the motor shaft and centrifuge shaft, respectively. Computation The velocity of points 𝑃 and 𝑄, which are moving in circular motion about 𝐴 and 𝐵, respectively, can be found by applying Eq. (16.8), which gives ) ( 𝑣⃗𝑃 = 𝜔 ⃗ 𝐴 × 𝑟⃗𝑃 ∕𝐴 = 𝜔𝐴 𝑘̂ × −𝑅𝐴 𝚤̂ = −𝜔𝐴 𝑅𝐴 𝚥̂, (1)

𝑣⃗𝑄 = 𝜔 ⃗ 𝐵 × 𝑟⃗𝑄∕𝐵 = 𝜔𝐵 𝑘̂ × 𝑅𝐵 𝚤̂ = 𝜔𝐵 𝑅𝐵 𝚥̂,

(2)

where we have assumed that the angular velocities of 𝐴 and 𝐵 are both in the positive 𝑘̂ direction. Enforcing the no-slip condition between the gears, we have 𝑣⃗𝑃 = 𝑣⃗𝑄

⇒

−𝑅𝐴 𝜔𝐴 𝚥̂ = 𝑅𝐵 𝜔𝐵 𝚥̂

⇒

−𝑅𝐴 𝜔𝐴 = 𝑅𝐵 𝜔𝐵 ,

(3)

motor

Figure 1 A centrifuge driven by an electric motor via a gear system. 𝜔𝐴 𝚥̂ 𝚤̂

Figure 2 Two meshing gears. Note that the gears’ radii are the radii of the circumferences indicated by the dashed lines.

where the minus sign tells us that the two gears spin in opposite directions since 𝑅𝐴 and 𝑅𝐵 are both positive. Since |𝜔𝐴 | = 3450 rpm and |𝜔𝐵 | = 6000 rpm, we can solve Eq. (3) for 𝑅𝐴 ∕𝑅𝐵 to obtain | 𝜔 | 6000 rpm | | = | 𝐵| = = 1.739. 𝑅𝐵 || 𝜔𝐴 || 3450 rpm 𝑅𝐴

(4)

To find the ratio of the angular acceleration of gear 𝐴 to that of gear 𝐵, we can obtain the desired information by taking the time derivative of the result in Eq. (3) to obtain −𝑅𝐴 𝛼𝐴 = 𝑅𝐵 𝛼𝐵

⇒

| 𝛼 | 𝑅 1 | 𝐴| 𝐵 = 0.5750. = |− | = | 𝛼𝐵 | 𝑅𝐴 1.739 | |

(5)

Discussion & Verification These results are dimensionally correct and are consistent with the result in Example 16.1; that is, Eq. (5) indicates that the angular speed and the magnitude of the angular accelerations of two gears are proportional to each other by the ratio of the gears’ radii. A Closer Look The result in Eq. (4) could be found using ideas developed in Chapter 12, where we learned that the speed of a point in circular motion is equal to the radius of the path times the angular speed. Therefore, from Fig. 2, we have 𝑣𝑃 = 𝑅𝐴 |𝜔𝐴 | and 𝑣𝑄 = 𝑅𝐵 |𝜔𝐵 |. Determining the gear ratio 𝑅𝐴 ∕𝑅𝐵 is then a matter of once again setting 𝑣⃗𝑃 = 𝑣⃗𝑄 , which implies 𝑣𝑃 = 𝑣𝑄 . In turn, this last relation yields Eq. (4).

Interesting Fact Why use gears? Gears, like belts, are used to transmit rotary motion. Gears are essential when rotary motion must be transmitted without slip (e.g., in a watch). Gears are often used to provide gear reduction, that is, to connect a shaft with one angular speed with a second shaft with a different angular speed.

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E X A M P L E 16.3

A Carnival Ride: Translation Motion platforms, which are used in many of today’s amusement parks, are a type of carnival ride in which a platform with seats is made to move always parallel to the ground. Figure 1 shows an elementary type of motion platform, more often found in traveling carnivals than in big amusement parks. Given that the motion platform in the figure (see also Fig. 2) is designed so that the rotating arms 𝐴𝐵 and 𝐶𝐷 are of equal length 𝐿 = 10 f t and remain parallel to each other, determine the velocity and acceleration of a person 𝑃 onboard the ride when 𝜔𝐴𝐵 is constant and equal to 1.25 rad∕s.

SOLUTION

economic images/Alamy Stock Photo

Figure 1 A carnival ride consisting of a motion platform. 𝑃 𝜔𝐴𝐵

𝐵

𝐶

𝑢̂ 𝑟 𝑢̂ 𝜃

𝐴

𝜃 𝐷

Road Map

To simplify the problem, we will assume that the platform and the persons on it form a single rigid body. This means that the two rotating arms and the platform form a four-bar linkage.∗ Because the arms 𝐴𝐵 and 𝐶𝐷 are identical in size and are always parallel to each other, the four-bar linkage 𝐴𝐵𝐶𝐷 always forms a parallelogram, and so the platform 𝐵𝐶 does not change its orientation and has zero angular velocity. Knowing this and applying the kinematics of fixed-axis rotation will allow us to determine 𝑣⃗𝑃 and 𝑎⃗𝑃 . Again, from Fig. 2, we see that arms 𝐴𝐵 and 𝐶𝐷 are always parallel to each other, and so platform 𝐵𝐶 does not change its orientation. From the discussion on p. 1059, this means the angular velocity of the platform is zero, that is, Computation

𝐿

𝜔𝐵𝐶 = 0. Figure 2 Coordinate definition and geometry for the carnival ride.

ISTUDY

(1)

Also, observe that points 𝐵 and 𝐶 move along circles centered at 𝐴 and 𝐷, respectively, where 𝐴 and 𝐷 are fixed points. Given the fact that the trajectories of points on 𝐵𝐶 are not a straight line, 𝐵𝐶’s motion is a curvilinear translation. Consequently, all points on 𝐵𝐶, or any rigid extension of it, i.e., any of the passengers, share the same value of velocity, as well as acceleration. Therefore, we have 𝑣⃗𝑃 = 𝑣⃗𝐵 = 𝜔𝐴𝐵 𝐿 𝑢̂ 𝜃 = 12.50 f t∕s 𝑢̂ 𝜃 , 𝑎⃗𝑃 = 𝑎⃗𝐵 =

−𝜔2𝐴𝐵 𝐿 𝑢̂ 𝑟

2

and

= −15.62 f t∕s 𝑢̂ 𝑟 ,

(2) (3)

where the velocity and acceleration of 𝐵 were computed using circular motion formulas in Eqs. (16.11). Discussion & Verification

The dimensions and units in Eqs. (2) and (3) are correct, and the magnitudes of the velocity and acceleration of 𝑃 are reasonable. In particular, the acceleration is not far from those found in general public (as opposed to extreme) carnival rides, and it amounts to a little less than 50% of the acceleration of gravity. A Closer Look It is important to remember that the direction of 𝑣⃗𝑃 in Eq. (2) and the direction of 𝑎⃗𝑃 in Eq. (3) are those shown at point 𝐴 in Fig. 2. That is, since 𝑃 is not in fixed-axis rotation about 𝐴, the 𝑢̂ 𝑟 and 𝑢̂ 𝜃 in Eqs. (2) and (3) are not those of 𝑃 relative to 𝐴.

∗A

four-bar linkage is a mechanism with four members or links in which one of the links is fixed (link AD in this example) and the other three move in a predetermined fashion (links AB, BC, and CD in this example).

ISTUDY

Section 16.1

1067

Fundamental Equations, Translation, and Rotation About a Fixed Axis

Problems Problem 16.1 The velocities of points 𝐴 and 𝐵 on a disk, which is undergoing planar motion, are such that |𝑣⃗𝐴 | = |𝑣⃗𝐵 |. Is it possible for the disk to be rotating about a fixed axis going through the center of the disk at 𝑂? Explain. 𝐴

𝐴 𝑣⃗𝐴

𝑣⃗𝐴

𝐵

𝑂

𝑂

𝑣⃗𝐵

𝑎⃗𝐵 𝐵

𝜔𝑂

𝜔𝑂 Figure P16.1

Figure P16.2 and P16.3

Problem 16.2 The velocity of a point 𝐴 and the acceleration of a point 𝐵 on a disk undergoing planar motion are shown. Is it possible for the disk to be rotating about a fixed axis going through the center of the disk at 𝑂? Explain.

Problem 16.3 Assuming that the disk shown is rotating about a fixed axis going through its center at 𝑂, determine whether the disk’s angular velocity is constant, increasing, or decreasing. 𝜔bg

Problem 16.4 Do points 𝐴 and 𝐵 on the surface of the bevel gear (bg), which rotates with angular velocity 𝜔bg , move relative to each other? At what rate does the distance between 𝐴 and 𝐵 change?

𝐴

𝐵

Problem 16.5 Letting 𝑅𝐴 = 8 in., 𝑅𝐵 = 4.2 in., and 𝑅𝐶 = 6.5 in., determine the angular velocity of gears 𝐵 and 𝐶 when gear 𝐴 has the angular speed |𝜔𝐴 | = 945 rpm in the direction shown. Each gear should be assumed to be mounted on a fixed shaft through its center.

𝚥̂

𝐶 𝚤̂

𝑅𝐶

𝐵 𝑅𝐵

Lawrence Manning/ Getty Images

Figure P16.4

𝑅𝐴 𝜔𝐴 𝐴

Figure P16.5

Problem 16.6 Letting 𝑅𝐴 = 7.2 in. and 𝑅𝐵 = 4.6 in., and assuming that the belt does not slip relative to pulleys 𝐴 and 𝐵, determine the angular velocity and angular acceleration of pulley 𝐵 when pulley 𝐴 rotates at 340 rad∕s while accelerating at 120 rad∕s2 .

𝑅𝐴 𝐴

Figure P16.6

𝜔𝐴

𝑅𝐵 𝛼𝐵 𝐵

𝛼𝐴

𝜔𝐵

1068

Chapter 16

Planar Rigid Body Kinematics

Problem 16.7 Letting 𝑅𝐴 = 203 mm, 𝑅𝐵 = 107 mm, 𝑅𝐶 = 165 mm, and 𝑅𝐷 = 140 mm, determine the angular acceleration of gears 𝐵, 𝐶, and 𝐷 when gear 𝐴 has an angular acceleration with magnitude |𝛼𝐴 | = 47 rad∕s2 in the direction shown. Note that gears 𝐵 and 𝐶 are mounted on the same shaft and they rotate together as a unit.

𝑅𝐴

𝑅𝐷

𝚥̂

𝑅𝐵

𝛼𝐴

Problem 16.8 The bevel gears 𝐴 and 𝐵 have nominal radii 𝑅𝐴 = 20 mm and 𝑅𝐵 = 5 mm, respectively, and their axes of rotation are mutually perpendicular. If the angular speed of gear 𝐴 is 𝜔𝐴 = 150 rad∕s, determine the angular speed of gear 𝐵. Assume both gears are mounted on fixed shafts.

Figure P16.7

𝐵 𝑃 𝑂 𝐴 Sergey Ryzhov/Shutterstock

NASA

Figure P16.8

Figure P16.9

Problem 16.9 A rotor with a fixed spin axis through point 𝑂 is accelerated from rest with a constant angular acceleration 𝛼𝑟 = 0.5 rad∕s2 . If the rotor’s diameter is 𝑑 = 15 f t, determine the time it takes for the point 𝑃 on the outer edge of the rotor to reach a speed 𝑣𝑃 = 300 f t∕s and the magnitude of the acceleration of 𝑃 when the speed 𝑣𝑃 is achieved.

Problem 16.10 A Pelton turbine (a type of turbine used in hydroelectric power generation) is spinning at 1100 rpm when the water jets acting on it are shut off, thus causing the turbine to slow down. Assuming that the angular deceleration rate is constant and equal to 1.31 rad∕s2 , determine the time it takes for the turbine to stop. In addition, determine the number of revolutions of the turbine during the spin-down.

kenary820/Shutterstock

𝜔𝑠

𝑑 𝑂

Figure P16.11

ISTUDY

Figure P16.10 𝑑

𝐴

Courtesy of Voith Hydro, Germany

Problem 16.11 𝐵

The sprinkler shown consists of a pipe 𝐴𝐵 mounted on a hollow vertical shaft. The water comes in the horizontal pipe at 𝑂 and goes out the nozzles at 𝐴 and 𝐵, causing the pipe to rotate. Letting 𝑑 = 7 in., determine the angular velocity of the sprinkler 𝜔 ⃗ 𝑠 , and |𝑎⃗𝐵 |, the magnitude of the acceleration of 𝐵, if 𝐵 is moving with a constant speed 𝑣𝐵 = 20 f t∕s. Assume that the sprinkler does not roll on the ground.

ISTUDY

Section 16.1

1069

Fundamental Equations, Translation, and Rotation About a Fixed Axis

Problem 16.12 In a carnival ride, two gondolas spin in opposite directions about a fixed axis. If 𝓁 = 4 m, determine the maximum constant angular speed of the gondolas if the magnitude of the acceleration of point 𝐴 is not to exceed 2.5𝑔.

𝓁

𝐴

𝑂

Gary L. Gray

Figure P16.12 𝐴

Problem 16.13 The tractor shown is stuck with its right track off the ground, and therefore the track is able to move without causing the tractor to move. Letting the diameter of sprocket 𝐴 be 𝑑 = 2.5 f t and the diameter of sprocket 𝐵 be 𝓁 = 2 f t, determine the angular speed of sprocket 𝐵 if the sprocket 𝐴 is rotating at 1 rpm.

Figure P16.13

Problem 16.14 A battering ram is suspended in its frame with bars 𝐴𝐷 and 𝐵𝐶, which are identical and pinned at their endpoints. At the instant shown, point 𝐸 on the ram moves with a speed 𝑣0 = 15 m∕s. Letting 𝐻 = 1.75 m and 𝜃 = 20◦ , determine the magnitude of the angular velocity of the ram at this instant.

𝐻

𝜃

𝜃 𝐸

𝐻

𝐵

𝐵

𝐴 𝐺

ram

ram

𝐴

Figure P16.14

Figure P16.15

Problem 16.15

𝑧

𝑘̂

A battering ram is suspended in its frame with bars 𝑂𝐴 and 𝑂𝐵, which are pinned at 𝑂. At the instant shown, point 𝐺 on the ram is moving to the right with a speed 𝑣0 = 15 m∕s. Letting 𝐻 = 1.75 m, determine the angular velocity of the ram at this instant.

𝚤̂

𝐴

𝚥̂ 𝐶

ℎ

𝑑

Problems 16.16 through 16.18 The bent rod rotates about an axis connecting points 𝐴 and 𝐸. All bends in the rod are 90◦ angles, and the given dimensions are ℎ = 21 cm, 𝓁 = 14.5 cm, 𝑑 = 21 cm, and 𝑏 = 7.6 cm. Express all your answers using the component system shown. Hint: Because these are general 3D motions, the angular velocity vector is not necessarily normal to all relative position vectors.

𝓁

𝐷

𝐵 𝑏

𝑏

𝐸

𝑥 Figure P16.16–P16.18

𝑦

1070

Chapter 16

Planar Rigid Body Kinematics

Problem 16.16

If, at the instant shown, the rod is rotating with a constant angular speed of 25 rad∕s and the rotation is counterclockwise as viewed from 𝐴 looking toward 𝐸, determine the velocity and acceleration of point 𝐶.

Problem 16.17 If, at the instant shown, the rod is rotating with an angular speed of 20 rad∕s, it is decreasing at a rate of 14 rad∕s2 , and the rotation is clockwise as viewed from 𝐴 looking toward 𝐸, determine the velocity and acceleration of point 𝐵. Problem 16.18 If, at the instant shown, the rod is rotating with an angular speed of 15 rad∕s, it is increasing at a rate of 8 rad∕s2 , and the rotation is clockwise as viewed from 𝐴 looking toward 𝐸, determine the velocity and acceleration of point 𝐶.

Problems 16.19 through 16.21 𝐸

𝑑 2

𝑑

𝓁 2

𝐵

𝐶 ℎ 𝐷

𝚥̂

𝑘̂

The rectangular block is attached to a rod that runs through the block along a diagonal. The rod is mounted in bearings at 𝐴 and 𝐵 that allow it to rotate about its own axis. The given dimensions are ℎ = 8.5 cm, 𝓁 = 20 cm, and 𝑑 = 22 cm. Express all your answers using the component system shown. Hint: Because these are general 3D motions, the angular velocity vector is not necessarily normal to all relative position vectors.

𝑥 𝐴

𝓁

Figure P16.19–P16.21

ISTUDY

𝚤̂

Problem 16.19 If, at the instant shown, the rod is rotating with an angular speed of 18 rad∕s, it is increasing at a rate of 3 rad∕s2 , and the rotation is counterclockwise as viewed from 𝐴 looking toward 𝐵, determine the velocity and acceleration of point 𝐸. Problem 16.20 If, at the instant shown, the rod is rotating with an angular speed of 35 rad∕s, it is decreasing at a rate of 5 rad∕s2 , and the rotation is counterclockwise as viewed from 𝐴 looking toward 𝐵, determine the velocity and acceleration of point 𝐷. Problem 16.21 If, at the instant shown, the rod is rotating with an angular speed of 50 rad∕s, it is increasing at a rate of 10 rad∕s2 , and the rotation is counterclockwise as viewed from 𝐴 looking toward 𝐵, determine the velocity and acceleration of point 𝐶.

Problem 16.22 At the instant shown, the paper is being unrolled with a speed 𝑣𝑝 = 7.5 m∕s and an acceleration 𝑎𝑝 = 1 m∕s2 . If, at this instant, the outer radius of the roll is 𝑟 = 0.75 m, determine the angular velocity 𝜔𝑠 and acceleration 𝛼𝑠 of the roll.

𝑟

𝜔𝑠

𝛼𝑠 𝑣𝑝

Pressmaster/Shutterstock

Figure P16.22

𝑎𝑝

ISTUDY

Section 16.1

1071

Fundamental Equations, Translation, and Rotation About a Fixed Axis

Problem 16.23 At the instant shown, the propeller is rotating with an angular velocity 𝜔𝑝 = 400 rpm in the positive 𝑧 direction and it is slowing down at 2 rad∕s2 . The 𝑧 axis is aligned with the spin axis of the propeller. Compute the velocity and acceleration of 𝑄, which is 14 f t away from the spin axis. Express your answer in the cylindrical component system shown, which has its origin at 𝑂 on the 𝑧 axis and its unit vector 𝑢̂ 𝑅 pointing radially toward point 𝑄.

𝜔𝑝 𝑘̂

𝑢̂ 𝑅

𝑄

𝑂 𝑢̂ 𝜃 propeller disk

Problems 16.24 and 16.25 The hammer of a Charpy impact toughness test machine has the geometry shown, where 𝐺 is the mass center of the hammer head. Use Eqs. (16.8) and (16.13) and write your answers in terms of the component system shown.

Figure P16.23 𝑢̂ 𝜃

Determine the velocity and acceleration of 𝐺, assuming 𝓁 = 500 mm, ℎ = 65 mm, 𝑑 = 25 mm, 𝜃̇ = −5.98 rad∕s, and 𝜃̈ = −8.06 rad∕s2 .

Problem 16.24

𝑂 𝑢̂ 𝑟

Determine the velocity and acceleration of 𝐺 as a function of the ̈ geometric parameters shown, 𝜃̇ and 𝜃.

𝓁

Problem 16.25

𝜃

ℎ 𝐺

Problem 16.26 The bucket of a backhoe is being operated while holding the arm 𝑂𝐴 fixed. At the instant shown, point 𝐵 has a horizontal component of velocity 𝑣0 = 0.25 f t∕s and is vertically aligned with point 𝐴. Letting 𝓁 = 0.9 f t, 𝑤 = 2.65 f t, and ℎ = 1.95 f t, determine the velocity of point 𝐶. In addition, assuming that, at the instant shown, point 𝐵 is not accelerating in the horizontal direction, compute the acceleration of point 𝐶. Express your answers using the component system shown.

𝑑

Figure P16.24 and P16.25

𝐵 𝚥̂

𝓁 𝚤̂

ℎ

𝐴 𝑤

𝐶

Figure P16.26

𝜙

𝐴

Problem 16.27

𝜃

In a contraption built by a fraternity, a person is sitting at the center of a swinging platform with length 𝐿 = 12 f t that is suspended by two identical arms each of length 𝐻 = 10 f t. Determine the angle 𝜃 and the angular speed of the arms if the person is moving upward and to the left with a speed 𝑣𝑝 = 25 f t∕s at the angle 𝜙 = 33◦ .

𝐵 Figure P16.27

Problem 16.28 The wheel 𝐴, with diameter 𝑑 = 5 cm, is mounted on the shaft of the motor shown and is rotating with a constant angular speed 𝜔𝐴 = 250 rpm. The wheel 𝐵, with center at the fixed point 𝑂, is connected to 𝐴 with a belt, which does not slip relative to 𝐴 or 𝐵. The radius of 𝐵 is 𝑅 = 12.5 cm. At point 𝐶 the wheel 𝐵 is connected to a saw. If point 𝐶 is at distance 𝓁 = 10 cm from 𝑂, determine the velocity and acceleration of 𝐶 when 𝜃 = 20◦ . Express your answers using the component system shown.

𝐷 𝐻

𝐶

1072

Chapter 16

Planar Rigid Body Kinematics

𝐶 𝓁 𝑅

𝑂

𝜃 𝚥̂

𝐵

𝜔𝐴

𝚤̂

Figure P16.28

Problems 16.29 and 16.30 The mechanism shown is designed to move the tool at 𝐻 while keeping it oriented vertically. To do so, the rotor in the motor 𝑀 is attached to the gear 𝐴, which drives the gear 𝐵. In turn, gear 𝐵 drives the gear 𝐶, which is rigidly attached to the arm 𝐸𝐹 . Arms 𝐸𝐹 and 𝐷𝐺 both have length 𝐿 and are parallel to one another. The radii of gears 𝐴, 𝐵, and 𝐶 are 𝑟𝐴 , 𝑟𝐵 , and 𝑟𝐶 , respectively.

𝐴 𝜔𝐴

𝐻

𝐵 𝐿

𝐹

𝜃

𝐶

𝐿 𝐷

𝜃

Problem 16.29 If the motor 𝐴 rotates with a constant angular speed 𝜔𝐴 (i.e., 𝛼𝐴 = 0) in the direction shown, determine the velocity and acceleration of the tool 𝐻 as functions of the angle 𝜃.

𝐺 Problem 16.30 If the motor 𝐴 rotates with angular speed 𝜔𝐴 and angular acceleration 𝛼𝐴 in the directions shown, determine the velocity and acceleration of the tool 𝐻 as functions of the angle 𝜃.

Figure P16.29 and P16.30

ISTUDY

Problem 16.31 An acrobat lands at the end 𝐴 of a board and, at the instant shown, point 𝐴 has a downward vertical component of velocity 𝑣0 = 5.5 m∕s. Letting 𝜃 = 15◦ , 𝓁 = 1 m, and 𝑑 = 2.5 m, determine the vertical component of velocity of point 𝐵 at this instant if the board is modeled as a rigid body.

𝑣0 𝐵

𝐴

𝜃

Earth 𝑣𝑠

𝓁 𝑑 Figure P16.31

Figure P16.32

Problem 16.32 A geosynchronous equatorial orbit is a circular orbit above the Earth’s equator that has a period of 1 day (these are sometimes called geostationary orbits). These geostationary orbits are of great importance for telecommunications satellites because a satellite orbiting with the same angular rate as the rotation rate of the Earth will appear to hover in the same point in the sky as seen by a person standing on the surface of the Earth. Using this information, modeling a geosynchronous satellite as a rigid body, and noting that the

ISTUDY

Section 16.1

1073

Fundamental Equations, Translation, and Rotation About a Fixed Axis

satellite has been stabilized so that the same side always faces the Earth, determine the angular speed 𝜔𝑠 of the satellite.

Problem 16.33

𝜔𝐴

Wheels 𝐴 and 𝐶 are mounted on the same shaft and rotate together. Wheels 𝐴 and 𝐵 are connected by a belt, and so are wheels 𝐶 and 𝐷. The axes of rotation of all the wheels are fixed, and the belts do not slip relative to the wheels they connect. If, at the instant shown, wheel 𝐴 has an angular velocity 𝜔𝐴 = 2 rad∕s and an angular acceleration 𝛼𝐴 = 0.5 rad∕s2 , determine the angular velocity and acceleration of wheels 𝐵 and 𝐷. The radii of the wheels are 𝑅𝐴 = 1 f t, 𝑅𝐵 = 0.25 f t, 𝑅𝐶 = 0.6 f t, and 𝑅𝐷 = 0.75 f t.

𝐴

𝐷

𝐵 Figure P16.33

Problem 16.34 𝐶 𝓁 𝐷

At the instant shown, 𝐴 is moving upward with a speed 𝑣0 = 5 f t∕s and acceleration 𝑎0 = 0.65 f t∕s2 . Assuming that the rope that connects the pulleys does not slip relative to the pulleys, and letting 𝓁 = 6 in. and 𝑑 = 4 in., determine the angular velocity and angular acceleration of pulley 𝐶. Hint: Because the right rope segment, connected directly to the ceiling, is inextensible, the velocity and acceleration of all material points in this segment must be zero.

Problems 16.35 through 16.37

Figure P16.34

A bicycle has wheels 700 mm in diameter and a gear set with the dimensions given in the table below. S8 S9

S5

S3

C2

S2 S1

C1

Gary L. Gray

Figure P16.35–P16.37 Crank Sprocket No. of cogs Radius (mm)

C1 26 52.6

C2 36 72.8

C3 48 97.0

Cassette (9 speeds) Sprocket No. of cogs Radius (mm)

S1 11 22.2

S2 12 24.3

S3 14 28.3

S4 16 32.3

S5 18 36.4

S6 21 42.4

S7 24 48.5

S8 28 56.6

S9 34 68.7

Problem 16.35

If a cyclist has a cadence of 1 Hz, determine the angular speed of the rear wheel in rpm when using the combination of C3 and S2. In addition, knowing that the speed of the cyclist is equal to the radius of the wheel times its angular speed, determine the cyclist’s speed in m∕s.

Problem 16.36

If a cyclist has a cadence of 68 rpm, determine which combination of chain ring (a sprocket mounted on the crank) and (rear) sprocket would most closely

𝑑 𝐵

1074

Chapter 16

Planar Rigid Body Kinematics

make the rear wheel rotate with an angular speed of 127 rpm. Having found a chain ring/sprocket combination, determine the wheel’s exact angular speed corresponding to the chosen chain ring/sprocket combination and the given cadence. Problem 16.37

If a cyclist is pedaling so that the rear wheel rotates with an angular speed of 16 rad∕s, determine all possible (rear) sprocket/chain ring (crank-mounted sprocket) combinations that would allow him or her to pedal with a frequency within the range 1.00–1.25 Hz.

Problem 16.38 𝐴 𝜙 𝚥̂

Problem 16.39

𝐵 𝑂

Martin Child/Getty Images

Figure P16.38 and P16.39

ISTUDY

At the instant shown, the angle 𝜙 = 30◦ , |𝑣⃗𝐴 | = 292 f t∕s, and the turbine is rotating clockwise. Letting 𝑂𝐴 = 𝑅, 𝑂𝐵 = 𝑅∕2, 𝑅 = 182 f t, and treating the blades as being equally spaced, determine the velocity of point 𝐵 at the given instant and express it using the component system shown.

𝚤̂

At the instant shown, the angle 𝜙 = 30◦ , the turbine is rotating clockwise, and 𝑎⃗𝐵 = (70.8 𝚤̂ − 12.8 𝚥̂) m∕s2 . Letting 𝑂𝐴 = 𝑅, 𝑂𝐵 = 𝑅∕2, 𝑅 = 55.5 m, and assuming the blades are equally spaced, determine the angular velocity and angular acceleration of the turbine blades, as well as the acceleration of point 𝐴 at the given instant.

ISTUDY

Section 16.2

16.2

Planar Motion: Velocity Analysis

1075

Planar Motion: Velocity Analysis

In this section, we will develop three different approaches to the velocity analysis of any planar rigid body motion: the vector approach, differentiation of constraints, and instantaneous center of rotation. We will see how each of these approaches applies to the slider-crank mechanism we introduced in Section 16.1 (see Fig. 16.17).

𝐶 𝐶

𝐶

Vector approach

𝐵

𝐵

The connecting rod in the slider-crank mechanism in Fig. 16.17 is, in general, planar motion, which means that we can describe the velocity of any of its points by knowing the velocity of one point on the rod and the angular velocity of the rod [see the discussion under Applying Eqs. (16.3) and (16.5) on p. 1059]. On p. 1060, we found the velocity of point 𝐵, which is in fixed-axis rotation about the centerline 𝐴 of the crank (see Fig. 16.18). Points 𝐵 and 𝐶 are on the same rigid body, so we can relate the velocity of 𝐶 to that of 𝐵 using Eq. (16.3) on p. 1057 in Section 16.1, which gives 𝑣⃗𝐶 = 𝑣⃗𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 .

𝐵

Figure 16.17 Schematic of a slider-crank mechanism emphasizing the motion of the connecting rod.

𝚥̂

(16.15)

In planar motion, Eq. (16.15) is a vector equation that represents two scalar equations, ⃗ 𝐵𝐶 . This is so because 𝑣⃗𝐵 is known, the direcwhich allow us to determine 𝑣⃗𝐶 and 𝜔 tion of 𝑣⃗𝐶 is known (the motion of the piston 𝐶 is rectilinear along the 𝑦 axis), 𝑟⃗𝐶∕𝐵 can be found in terms of the crank angle 𝜃 (see Fig. 16.18), and the axis of rotation for 𝜔 ⃗ 𝐵𝐶 is known (it is perpendicular to the plane of motion). Therefore, the components 𝑣𝐶 and 𝜔𝐵𝐶 are the only unknowns in these two scalar equations. We will go through this solution in Example 16.6.

𝚤̂

𝐶 𝐿

𝜙

𝑦𝐶

𝑑

𝐷 𝐵

𝜔𝐴𝐵

Rolling without slip: velocity analysis 𝐴

Many applications in engineering involve the motion of disks or wheels rolling over a surface—for example, car wheels rolling on a road, train wheels rolling on tracks, or meshed gears rolling over each other. An important special case of rolling motion is called rolling without slip (also called rolling without slipping or rolling without sliding). Consider a wheel 𝑊 rolling over a surface 𝑆 (𝑆 can be moving), as shown in Fig. 16.19. If the wheel and the surface remain in contact during the motion and if 𝜔𝑊

𝑊 𝑢̂ 𝑛 𝑃

𝑢̂ 𝑡

𝓁

𝑄 𝑆

Figure 16.19. A wheel 𝑊 rolling over a surface 𝑆. At the instant shown, the line 𝓁 is tangent to the two bodies at their point of contact.

the contact points 𝑃 and 𝑄 (on 𝑊 and 𝑆, respectively) move relative to each other, then the only direction in which relative motion can occur at any instant is along the

𝜃

𝑅

Figure 16.18 Definitions of the geometric parameters used in the slider-crank analysis.

1076

Chapter 16

Planar Rigid Body Kinematics

line 𝓁 that is tangent to both 𝑊 and 𝑆. To say that 𝑊 is rolling without slip over 𝑆 means that points 𝑃 and 𝑄 do not move relative to each other. In terms of velocity, this definition implies that

𝜔𝑊

𝑣⃗𝑃 ∕𝑄 = 0⃗ 𝑂 𝑟⃗𝑃 ∕𝑂

𝚥̂ 𝚤̂

𝑃

Figure 16.20 A wheel rolling without slip on a horizontal fixed surface.

ISTUDY

⇒

𝑣⃗𝑃 = 𝑣⃗𝑄 .

(16.16)

𝑣𝑂

As an application of Eq. (16.16), consider a wheel of radius 𝑅 rolling without slipping on a flat stationary surface (Fig. 16.20). To find the wheel’s angular velocity when the center of the wheel is moving with speed 𝑣𝑂 , we can relate the velocities of points 𝑂 and 𝑃 using ⃗ 𝑊 × 𝑟⃗𝑃 ∕𝑂 , 𝑣⃗𝑃 = 𝑣⃗𝑂 + 𝜔

(16.17)

where, at the instant shown, 𝑃 is the point on the wheel in contact with the ground, ⃗ 𝑊 = 𝜔𝑊 𝑘̂ is the angular velocity of the wheel, and 𝑟⃗𝑃 ∕𝑂 = −𝑅 𝚥̂. 𝑣⃗𝑂 = 𝑣𝑂 𝚤̂, 𝜔 Enforcing the no-slip condition given in Eq. (16.16), we have ⃗ 𝑣⃗𝑃 = 𝑣⃗𝑄 = 0,

Common Pitfall 𝑷 is not a fixed point. Do not interpret Eq. (16.18) as saying that 𝑃 is fixed because 𝑣⃗𝑃 is equal to zero. Equation (16.18) only holds at the instant when 𝑃 is touching the ground. In Section 16.3, we will discover that although 𝑣⃗𝑃 = 0⃗ when 𝑃 is touching the ground, 𝑎⃗𝑃 ≠ 0⃗ at that instant. That is, 𝑃 only stops for an instant while it is accelerating away from its current position, so that some other point on the wheel can become the part of the wheel touching the ground.

(16.18)

where 𝑄 is the point on the ground in contact with 𝑃 at this instant and 𝑣⃗𝑄 = 0⃗ because the ground is stationary. Substituting Eq. (16.18) into Eq. (16.17) and simplifying, we have 0⃗ = 𝑣𝑂 𝚤̂ + 𝜔𝑊 𝑘̂ × (−𝑅 𝚥̂) = 𝑣𝑂 𝚤̂ + 𝜔𝑊 𝑅 𝚤̂

⇒

𝜔𝑊 = −

𝑣𝑂 𝑅

.

(16.19)

Equation (16.19) tells us that motion of the wheel center to the right requires clockwise rotation under no-slip conditions.

Differentiation of constraints Referring to Fig. 16.18, we can find 𝜔 ⃗ 𝐵𝐶 and 𝑣⃗𝐶 as functions of 𝜃 by writing the appropriate constraint equations and then differentiating them with respect to time. We introduced this idea in Section 12.6 on p. 739, and we now apply it to rigid bodies. Noting that the piston 𝐶 only moves in the 𝑦 direction, 𝑣⃗𝐶 = 𝑦̇ 𝐶 𝚥̂. Then, we can write a constraint equation for 𝑦𝐶 as 𝑦𝐶 = 𝑅 sin 𝜃 + 𝐿 cos 𝜙,

(16.20)

which can be differentiated with respect to time to find 𝑦̇ 𝐶 = 𝑅𝜃̇ cos 𝜃 − 𝐿𝜙̇ sin 𝜙.

(16.21)

̇ we note If 𝑅, 𝐿, and 𝜃(𝑡) are known, we see that 𝜃̇ = 𝜔𝐴𝐵 . To determine 𝜙 and 𝜙, that the orientation 𝜙 of the connecting rod and the crank’s orientation 𝜃 are related by 𝑅 (16.22) 𝑅 cos 𝜃 = 𝐿 sin 𝜙 ⇒ sin 𝜙 = cos 𝜃. 𝐿 ̇ where Equation (16.22) can then be differentiated with respect to time to find 𝜙, 𝜙̇ 𝑘̂ = 𝜔 ⃗ 𝐵𝐶 , as a function of 𝜃 and 𝜃̇ (see Prob. 16.87), thus concluding our analysis.

ISTUDY

Section 16.2

Planar Motion: Velocity Analysis

Instantaneous center of rotation

𝑣⃗𝐶

When a rigid body rotates about a fixed point 𝑄 (where 𝑄 is on the body or an extension of the body; see Fig. 16.21), the velocity of any point 𝐶 on the body is given by ⃗ body × 𝑟⃗𝐶∕𝑄 , (16.23) 𝑣⃗𝐶 = 𝜔 ⃗ If 𝑣⃗ = 0⃗ for all time, we call 𝑄 the center of rotation. If 𝑣⃗ = 0⃗ only because 𝑣⃗𝑄 = 0. 𝑄 𝑄 at a particular instant in time, then we call 𝑄 the instantaneous center of rotation or instantaneous center (IC).∗ If the motion is planar, then 𝜔 ⃗ body and 𝑟⃗𝐶∕𝑄 are mutually perpendicular, and we can write Eq. (16.23) as |𝑣⃗𝐶 | = |𝜔 ⃗ body ||⃗𝑟𝐶∕𝑄 | = |𝜔 ⃗ body ||⃗𝑟𝐶∕IC |,

(16.24)

that is, the speed of 𝐶 is proportional to its distance from the IC. Because the IC can always be found in planar motion, this formula provides a convenient tool. We now describe the three different possibilities. Given two nonparallel velocities on a body Applying the idea of IC to the connecting rod in Fig. 16.22, Eq. (16.23) implies that 𝑣⃗𝐶 instantaneous center of rotation 𝑄

𝐶 line of action of 𝑣⃗𝐶

𝐵

𝚥̂ 𝑅 𝐴

𝚤̂

𝜃 𝑥 line of action of 𝑣⃗𝐵

Figure 16.22. Graphical construction for the determination of the IC.

is perpendicular to both 𝜔 ⃗ 𝐵𝐶 and 𝑟⃗𝐶∕𝑄 . That means that the IC for the connecting rod must lie in the plane of motion and at the intersection of the line that passes through point 𝐶 and is perpendicular to 𝑣⃗𝐶 (line 𝓁𝐶 ) with the line that passes through 𝐵 and is perpendicular to 𝑣⃗𝐵 (line 𝓁𝐵 ). Thus, at the instant shown, the IC of the connecting rod must be at 𝑄, which is the intersection of 𝓁𝐶 and 𝓁𝐵 . Looking at Fig. 16.22, it is as though, at this instant, we have a rigid triangular plate bounded by 𝐵𝐶, 𝓁𝐶 and 𝓁𝐵 pivoting about IC 𝑄. An instant later, 𝐵𝐶 will have changed its orientation, and it will appear as though a differently shaped triangular plate is pivoting about a different IC. Nevertheless, the general plane motion of the connecting rod can be thought of as a continuous set of incremental pivots about a continuously varying IC. Thus, at the instant shown, the IC of the connecting rod must be at 𝑄, which is the intersection of 𝓁𝐶 and 𝓁𝐵 . ∗ The

instantaneous center is also sometimes called the instantaneous center of zero velocity.

1077

𝐶 𝑟⃗𝐶∕𝑄 𝑄 𝜔 ⃗ body

Figure 16.21 A rigid body 𝐵 rotating about a fixed point 𝑄.

1078

Chapter 16

Planar Rigid Body Kinematics Given two parallel velocities on a body

The graphical procedure just described does not work if lines 𝓁𝐵 and 𝓁𝐶 are parallel, of which there are two cases: (1) 𝓁𝐵 and 𝓁𝐶 are parallel and distinct and (2) 𝓁𝐵 and 𝓁𝐶 are parallel and coincide.

line of action of 𝑣⃗𝐶 𝓁𝐶

𝐶

IC → ∞ ⇒ 𝜔𝐵𝐶 = 0

𝚥̂ 𝑅

𝐵 𝐴

𝓁𝐵

𝚤̂

Figure 16.23 At this instant, 𝑣⃗𝐵 and 𝑣⃗𝐶 are parallel and 𝓁𝐵 and 𝓁𝐶 are parallel and distinct. The IC of 𝐵𝐶 is at infinity, and 𝜔𝐵𝐶 = 0.

Case 1: 𝓁𝐵 and 𝓁𝐶 are parallel and distinct. At the instant when the line 𝐴𝐵 is perpendicular to 𝐴𝐶, 𝑣⃗𝐵 and 𝑣⃗𝐶 are parallel, and 𝓁𝐵 and 𝓁𝐶 are parallel and distinct (see Fig. 16.23). In this case, 𝓁𝐵 and 𝓁𝐶 intersect at infinity, and the IC is infinitely far away. From Eq. (16.24), the only way that 𝐵 and 𝐶 can have finite velocities while being infinitely far from the IC is for the angular velocity of the body to be zero! Thus, if our geometric construction tells us that lines 𝓁𝐵 and 𝓁𝐶 are parallel and distinct, then 𝜔 ⃗ 𝐵𝐶 = 0⃗ at that instant and 𝑣⃗𝐵 = 𝑣⃗𝐶 . Case 2: 𝓁𝐵 and 𝓁𝐶 coincide. Figure 16.24 shows a wheel rolling without slip while in contact with two parallel surfaces 𝑆1 and 𝑆2 moving with the velocities shown. The no-slip condition at 𝐵 and 𝐶 causes 𝑣⃗𝐵 and 𝑣⃗𝐶 to be parallel because they must match the velocities of the surfaces 𝑆1 and 𝑆2 . The geometrical procedure for the determination of the IC tells us that 𝓁𝐵 and 𝓁𝐶 coincide. If all we know are the directions of 𝑣⃗𝐵 and 𝑣⃗𝐶 , then the IC cannot be determined because every point on 𝓁𝐵 and 𝓁𝐶 is an intersection point for these lines. However, if the values of 𝑣1 = 𝑣𝐵 and 𝑣2 = 𝑣𝐶 are known, then we can find the IC, as well as the body’s angular velocity. Referring to Fig. 16.25, recall from Fig. 16.15 on p. 1061 that the speed of points

𝑣1

𝑆1

𝑣1

𝑣𝐵 ℎ1

𝑂

𝐵

𝑅

2𝑅

𝑂

𝑊

IC

𝑆2

Figure 16.24 Wheel 𝑊 rolling without slipping over two parallel surfaces.

ISTUDY

𝑊 𝜔𝑊

ℎ2

𝑣𝐶 𝑣2

𝑣𝐵 = 𝑣1 𝑆1

𝑣𝐶 = 𝑣2

𝑆2

𝑣2

𝐶

Figure 16.25. Determination of the IC for a wheel 𝑊 rolling without slipping between two parallel surfaces. The top surface is moving to the right, and the bottom is moving to the left.

on line 𝓁𝐵𝐶 can be graphically represented by right triangles with a vertex at the center of rotation. Therefore, the IC must be at the intersection of 𝓁𝐵𝐶 , the (radial) line containing 𝐵 and 𝐶, and 𝓁𝑣 , the line representing the velocity profile of points on 𝓁𝐵𝐶 . Since the velocity of points on 𝓁𝐵𝐶 is proportional to the distance from the IC via the angular speed, we can calculate the angular speed using Eq. (16.11) on p. 1061 along with similar triangles as (see Fig. 16.25) 𝑣𝐵 ℎ1

=

𝑣𝐶 ℎ2

= 𝜔𝑊 .

(16.25)

Since ℎ1 + ℎ2 = 2𝑅, Eq. (16.25) becomes 𝑣𝐵 𝜔𝑊

+

𝑣𝐶 𝜔𝑊

= 2𝑅

⇒

𝜔𝑊 =

𝑣𝐵 + 𝑣𝐶 2𝑅

.

(16.26)

ISTUDY

Section 16.2

Since the motion is planar, we can assign a direction to 𝜔𝑊 so that our geometrical calculation based on similar triangles allows us to compute the angular velocity of the body. When both surfaces move in the same direction with known speeds, 𝑣⃗𝐵 and 𝑣⃗𝐶 are again parallel and the lines perpendicular to them that run through points 𝐵 and 𝐶 coincide in the single line 𝓁𝐵𝐶 (see Fig. 16.26). The same similar-triangles argument used for the case in Fig. 16.25 tells us that 𝑣𝐵 2𝑅 + ℎ

=

𝑣𝐶 ℎ

= 𝜔𝑊 .

(16.27)

Solving the first equality for ℎ and then substituting that into the second, we get ℎ=

2𝑅𝑣𝐶

𝜔𝑊 =

⇒

𝑣𝐵 − 𝑣𝐶

𝑣𝐵 − 𝑣𝐶 2𝑅

.

𝑣𝐵 = 𝑣 1 𝑆 1

𝐵

𝑅 𝑂

2𝑅

𝜔𝑊 𝑣2

𝐶

𝑣𝐶 = 𝑣 2

𝑆2

ℎ

IC

This case is depicted in Fig. 16.27 for the crank 𝐶 in an internal combustion engine, 𝑣𝐵 𝐵 𝑅

IC 𝜔𝐶

𝑣1

(16.28)

Given a velocity on a body and the body’s angular velocity

𝐶

Figure 16.27. Determination of the IC for the case in which the velocity of a point on a body and the body’s angular velocity are known.

for which we know the velocity of point 𝐵, as well as the angular velocity 𝜔𝐶 of the crank. In this case, the IC is located on line 𝓁𝐵 such that the distance from 𝐵 to the IC is 𝑅 = 𝑣𝐵 ∕𝜔𝐶 . We can determine on which side of 𝑣𝐵 the IC lies by considering the direction of rotation of the rigid body. In this case, it lies to the left of 𝑣𝐵 since the rotation is counterclockwise.

End of Section Summary This section presents three different ways to analyze the velocities of a rigid body in planar motion: the vector approach, differentiation of constraints, and instantaneous center of rotation. Vector approach.

1079

Planar Motion: Velocity Analysis

We saw in Section 16.1 that the equation Eq. (16.15), p. 1075 𝑣⃗𝐶 = 𝑣⃗𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 ,

relates the velocity of two points on a rigid body, 𝑣⃗𝐵 and 𝑣⃗𝐶 , by their relative position 𝑟⃗𝐶∕𝐵 and the angular velocity of the body 𝜔 ⃗ 𝐵𝐶 (see Fig. 16.28).

Figure 16.26 Determination of the IC for a wheel 𝑊 rolling without slipping over two parallel surfaces. Both surfaces are moving to the right, but the top is moving faster than the bottom.

1080

Chapter 16

Planar Rigid Body Kinematics

𝐵 𝑟⃗𝐶∕𝐵

𝜔𝑊

𝑊

𝐶 𝑣⃗𝐶

𝑢̂ 𝑛

rigid body

𝑃

𝑢̂ 𝑡

𝜔 ⃗ 𝐵𝐶

𝓁

𝑄

Figure 16.28. A rigid body on which we are relating the velocity of two points 𝐵 and 𝐶.

𝑆 Figure 16.29 A wheel 𝑊 rolling over a surface 𝑆. At the instant shown, the line 𝓁 is tangent to the two bodies at their point of contact.

Rolling without slip. When a body rolls without slip over another body, then the two points on the bodies that are in contact at any instant, points 𝑃 and 𝑄, must have the same velocity (see, for example, the wheel 𝑊 rolling over the surface 𝑆 in Fig. 16.29). Mathematically, this means that Eqs. (16.16), p. 1076 𝑣⃗𝑃 ∕𝑄 = 0⃗

𝑣⃗𝑃 = 𝑣⃗𝑄 .

⇒

If a wheel 𝑊 of radius 𝑅 is rolling without slipping over a flat, stationary surface, then the point 𝑃 on the wheel in contact with the surface must have zero velocity (see Fig. 16.30). The consequence is that the angular velocity of the wheel 𝜔𝑊 is related to the velocity of the center 𝑣𝑂 and the radius of the wheel 𝑅 according to Eq. (16.19), p. 1076 Figure 16.30 A wheel 𝑊 rolling without slip on a horizontal fixed surface.

ISTUDY

𝜔𝑊 = −

𝑣𝑂 𝑅

,

in which the positive direction for 𝜔𝑊 is taken to be the positive 𝑧 direction. Differentiation of constraints. As we discovered in Section 12.6, it is often convenient to write an equation describing the position of a point of interest, which can then be differentiated with respect to time to find the velocity of that point. For planar motion of rigid bodies, this idea can also apply for describing the position and velocity of a point on a rigid body, as well as for describing the body’s orientation. IC 𝐵 𝑣⃗𝐵

IC

𝐵 𝑣⃗𝐶 𝐶

𝐵

𝑣⃗𝐵 𝐶

𝑣⃗𝐶

𝑣⃗𝐵

𝑣⃗𝐶 𝐶

Figure 16.31. Three different possible motions for determining the IC. (a) Two nonparallel velocities are known. (b) The lines of action of two velocities are parallel and distinct; the IC is at infinity and the body is translating. (c) The lines of action of two parallel velocities coincide.

Instantaneous center of rotation. The point on a body (or imaginary extension of the body) whose velocity is zero at a particular instant is called the instantaneous center of rotation or instantaneous center (IC). The IC can be found geometrically if the velocity is known for two distinct points on a body or if a velocity on the body and the body’s angular velocity are known. The three possibilities are shown in Fig. 16.31.

ISTUDY

Section 16.2

1081

Planar Motion: Velocity Analysis

E X A M P L E 16.4

A Carnival Ride: Instantaneous Center Analysis

Motion platforms, which are used in many of today’s amusement parks, are a type of carnival ride in which a platform with seats is made to move always parallel to the ground. Figure 1 shows an elementary type of motion platform, found more often in traveling carnivals than in big amusement parks. Given that the motion platform in the figure (see also Fig. 2) is designed so that the rotating arms 𝐴𝐵 and 𝐶𝐷 are of equal length 𝐿 = 10 f t and remain parallel to each other, determine the velocity and acceleration of a person 𝑃 onboard the ride when 𝜔𝐴𝐵 is constant and equal to 1.25 rad∕s.

SOLUTION

economic images/Alamy Stock Photo

Road Map

To simplify the problem, we will assume that the platform and the persons on it form a single rigid body. This means that the two rotating arms and the platform form a four-bar linkage. Because the arms 𝐴𝐵 and 𝐶𝐷 are identical in size and are always parallel to each other, the four-bar linkage 𝐴𝐵𝐶𝐷 always forms a parallelogram. We will use this fact, along with the concept of instantaneous center of rotation, to determine the angular velocity of the platform. Knowing this and applying the kinematics of fixed-axis rotation will allow us to determine 𝑣⃗𝑃 and 𝑎⃗𝑃 . Referring to Fig. 2, we see that the arms 𝐴𝐵 and 𝐶𝐷 are always parallel to each other. Also observe that points 𝐵 and 𝐶 move along circles centered at 𝐴 and 𝐷, respectively, where 𝐴 and 𝐷 are fixed points. Hence, going through the geometrical procedure to identify the IC of the element 𝐵𝐶, we see that lines 𝓁𝐴𝐵 and 𝓁𝐶𝐷 , which are perpendicular to 𝑣⃗𝐵 and 𝑣⃗𝐶 , respectively, are parallel and distinct. This means that the IC of 𝐵𝐶 is at infinity, and therefore 𝜔𝐵𝐶 = 0. (1) Computation

Since the result in Eq. (1) is independent of the value of the angle 𝜃 of the arms 𝐴𝐵 and 𝐶𝐷, the motion of element 𝐵𝐶 is translation. Given the fact that the trajectories of points on 𝐵𝐶 are not straight lines, 𝐵𝐶’s motion is a curvilinear translation. Consequently, all points on 𝐵𝐶, or any rigid extension of it, i.e., any of the passengers, share the same value of velocity, as well as acceleration. In view of this fact, we have 𝑣⃗𝑃 = 𝑣⃗𝐵 = 𝑣⃗𝐶 = 𝜔𝐴𝐵 𝐿 𝑢̂ 𝜃 = (12.50 f t∕s) 𝑢̂ 𝜃 , and ( ) 𝑎⃗𝑃 = 𝑎⃗𝐵 = 𝑎⃗𝐶 = −𝜔2𝐴𝐵 𝐿 𝑢̂ 𝑟 = −15.62 f t∕s2 𝑢̂ 𝑟 ,

(2) (3)

where the velocity and acceleration of 𝐵 were computed using circular motion formulas [see Eqs. (16.11) on p. 1061]. Discussion & Verification The dimensions and units in Eqs. (2) and (3) are correct. The solution of this problem is very elementary; it is performed in a conceptual manner to illustrate the concept of curvilinear translation and its relation to the concept of the IC. As far as the acceleration values are concerned, these are not far from those found in actual general public (as opposed to extreme) carnival rides, and they amount to a little less than 50% of the acceleration of gravity.

Figure 1

𝑃 𝜔𝐴𝐵

𝐵 𝑢̂ 𝑟

𝑢̂ 𝜃

𝐶 𝜃 𝐿

Figure 2 Coordinate definition and geometry for the carnival ride.

1082

Chapter 16

Planar Rigid Body Kinematics

E X A M P L E 16.5

Planetary Gears Rolling Without Slip: Vector Approach sun gear ear

Planetary gear systems (Fig. 1) are used to transmit power between two shafts (a common application is in car transmissions). The center gear is called the sun, the outer gear is called the ring, and the inner gears are called planets. The planets are mounted on a component called the planet carrier (which is not shown in Fig. 1). Referring to Fig. 2, let 𝑅𝑆 = 2 in., 𝑅𝑃 = 0.67 in., the ring be fixed, and the angular velocity of the sun gear be 𝜔𝑆 = 1500 rpm. Determine 𝜔 ⃗ 𝑃 and 𝜔 ⃗ 𝑃 𝐶 , the angular velocities of the planet 𝑃 , and of the planet carrier (PC), respectively.

SOLUTION To compute 𝜔 ⃗ 𝑃 , we need to determine the velocity of two points on 𝑃 . Two promising candidates are points 𝐴′ and 𝑄′ because their velocities are completely determined by the rolling without slip condition at the 𝐴-𝐴′ and 𝑄-𝑄′ contacts and because 𝑣⃗𝐴 and 𝑣⃗𝑄 are easily computed. Once 𝜔 ⃗ 𝑃 is known, 𝜔 ⃗ 𝑃 𝐶 can be found by finding the velocity of two points on the planet carrier. We will choose 𝑂, because its velocity is zero, and 𝐶, because it is shared with the planet gear 𝑃 .

Road Map ring ge ring gear ar

planet gears planet gears VladimirV/Shutterstock

Figure 1 Photo of a planetary gear system.

Computation 𝐴

𝑣⃗𝑄′ = 𝑣⃗𝑄

𝐴

𝐶

𝑃

𝑄 𝑄

𝜔𝑆

𝚥̂

𝑅𝑃 𝐶

𝑂

Enforcing the rolling without slip condition at the 𝐴-𝐴′ and 𝑄-𝑄′ con-

tacts yields

𝚤̂

⃗ and 𝑣⃗𝐴′ = 𝑣⃗𝐴 = 0,

where 𝑣⃗𝐴 = 0⃗ because the ring is fixed. Next, because the sun gear rotates about the fixed point 𝑂, we can write 𝑣⃗𝑄 = 𝜔 ⃗ 𝑆 × 𝑟⃗𝑄∕𝑂 = −𝜔𝑆 𝑅𝑆 𝚤̂, (2) where we have used 𝜔 ⃗ 𝑆 = 𝜔𝑆 𝑘̂ and 𝑟⃗𝑄∕𝑂 = 𝑅𝑆 𝚥̂. In addition, writing 𝑣⃗𝑄′ using 𝐴′ as a reference point yields ⃗ 𝑃 × 𝑟⃗𝑄′ ∕𝐴′ = 2𝜔𝑃 𝑅𝑃 𝚤̂, 𝑣⃗𝑄′ = 𝑣⃗𝐴′ + 𝜔

𝑅𝑃

𝑅𝑆

ISTUDY

(3)

where we have used 𝜔 ⃗ 𝑃 = 𝜔𝑃 𝑘̂ and 𝑟⃗𝑄′ ∕𝐴′ = −2𝑅𝑃 𝚥̂, and we have let 𝑣𝐴′ = 0 from Eqs. (1). Substituting Eqs. (2) and (3) into the first of Eqs. (1) gives −𝜔𝑆 𝑅𝑆 = 2𝜔𝑃 𝑅𝑃

Figure 2 Schematic of a planetary gear system with three planets and a fixed ring. The planet carrier’s rotational axis is coincident with that of the sun gear (through 𝑂), but the planet carrier and sun gear are not directly connected. This permits the planet carrier and sun gear to have different angular velocities.

(1)

⇒

𝜔𝑃 = −

𝑅𝑆 2𝑅𝑃

𝜔𝑆 = −2239 rpm.

(4)

Now observe that 𝐶 is shared by both planet gear 𝑃 and the planet carrier. This means 𝑣⃗𝐶 = 𝑣⃗𝐴′ + 𝜔𝑃 𝑘̂ × 𝑟⃗𝐶∕𝐴′ = 𝜔𝑃 𝑅𝑃 𝚤̂,

(5)

𝑣⃗𝐶 = 𝜔 ⃗ 𝑃 𝐶 × 𝑟⃗𝐶∕𝑂 = −𝜔𝑃 𝐶 𝑅𝑃 𝐶 𝚤̂,

(6)

and

̂ and we have enforced the second of Eqs. (1). The where 𝑅𝑃 𝐶 = 𝑅𝑃 + 𝑅𝑆 , 𝜔 ⃗ 𝑃 𝐶 = 𝜔𝑃 𝐶 𝑘, two expressions for 𝑣⃗𝐶 must be equal to each other, that is, 𝜔𝑃 𝐶 = −

𝑅𝑃 𝑅𝑃 𝐶

𝜔𝑃 =

𝑅𝑆 2𝑅𝑃 𝐶

𝜔𝑆 = 561.8 rpm,

(7)

where we have taken advantage of the solution for 𝜔𝑃 in Eq. (4). Discussion & Verification To verify the correctness of our results, observe that since ⃗ point 𝐴′ is the IC for the planet gear 𝑃 . Hence, the speeds of point 𝐶 and 𝑄′ are 𝑣⃗𝐴′ = 0, |𝜔𝑃 |𝑅𝑃 and |𝜔𝑃 |2𝑅𝑃 , respectively, which is confirmed by Eqs. (5) and (3).

ISTUDY

Section 16.2

1083

Planar Motion: Velocity Analysis

E X A M P L E 16.6

Completing the Velocity Analysis of the Connecting Rod

On p. 1075 we outlined the vector approach for the velocity analysis of the connecting rod (CR) and piston in the slider-crank mechanism shown in Fig. 1. We will complete that analysis here. Referring to Fig. 2, we are given the radius of the crank 𝑅, the length of the CR 𝐿, the position of the mass center of the CR 𝐻, the angular velocity of the crank 𝜔𝐴𝐵 , and the crank angle 𝜃. Determine the angular velocity of the CR 𝜔 ⃗ 𝐵𝐶 , the velocity of the piston 𝑣⃗𝐶 , and the velocity of the mass center of the CR 𝑣⃗𝐷 .

SOLUTION Road Map

Figure 1

The road map for this problem was presented on p. 1075.

Computation We begin by recalling that, in Section 16.1, we found 𝑣⃗𝐵 to be [see Eq. (16.11) on p. 1061 and Fig. 2]

𝑣⃗𝐵 = 𝑅𝜃̇ 𝑢̂ 𝜃 = 𝑅𝜔𝐴𝐵 (−sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂),

(1) 𝐶

where we have used 𝜃̇ = 𝜔𝐴𝐵 and 𝑢̂ 𝜃 = −sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂. Since 𝐵 and 𝐶 are both points on the CR, we can relate their velocities using Eq. (16.3) on p. 1057, which gives 𝑣⃗𝐶 = 𝑣⃗𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 .

(2)

As discussed on p. 1075, Eq. (2) represents two scalar equations in the two unknowns 𝑣𝐶 and 𝜔𝐵𝐶 . Let’s now work out the details to see how. As for 𝜔 ⃗ 𝐵𝐶 , we can write it as ̂ 𝜔 ⃗ 𝐵𝐶 = 𝜔𝐵𝐶 𝑘.

(3)

𝑟⃗𝐶∕𝐵 = 𝐿(−sin 𝜙 𝚤̂ + cos 𝜙 𝚥̂),

(4)

where we note that the orientation 𝜙 of the CR and the crank’s orientation 𝜃 are related by 𝑅 cos 𝜃 = 𝐿 sin 𝜙, that is, √ 𝐿2 − 𝑅2 cos2 𝜃 𝑅 . (5) sin 𝜙 = cos 𝜃 and cos 𝜙 = 𝐿 𝐿 Finally, enforcing the condition 𝑣𝐶𝑥 = 0, we can write 𝑣⃗𝐶 as 𝑣⃗𝐶 = 𝑣𝐶𝑦 𝚥̂.

(6)

Substituting Eq. (1) and Eqs. (3)–(6) into Eq. (2) and carrying out the cross product give √ ) ( ) ( (7) 𝑣𝐶𝑦 𝚥̂ = − 𝑅𝜔𝐴𝐵 sin 𝜃 + 𝜔𝐵𝐶 𝐿2 − 𝑅2 cos2 𝜃 𝚤̂ + 𝑅 𝜔𝐴𝐵 − 𝜔𝐵𝐶 cos 𝜃 𝚥̂. Equation (7) represents the two scalar equations √ 𝑅𝜔𝐴𝐵 sin 𝜃 + 𝜔𝐵𝐶 𝐿2 − 𝑅2 cos2 𝜃 = 0, ( ) 𝑅 𝜔𝐴𝐵 − 𝜔𝐵𝐶 cos 𝜃 = 𝑣𝐶𝑦 ,

(8) (9)

in the unknowns 𝑣𝐶𝑦 and 𝜔𝐵𝐶 . Solving, we obtain 𝜔𝐵𝐶 = √

(𝐿∕𝑅)2 − cos2 𝜃

[ and

𝑣𝐶𝑦 = 1 + √

] sin 𝜃 (𝐿∕𝑅)2 − cos2 𝜃

𝜙 𝚥̂ 𝑢̂ 𝜃 𝑅

𝐴

𝐷

𝐻

𝑢̂ 𝑟 𝐵

𝜃 𝑥

𝚤̂ 𝜔𝐴𝐵 = 𝜃̇

To complete the right-hand side of Eq. (2), we can write 𝑟⃗𝐶∕𝐵 as

−𝜔𝐴𝐵 sin 𝜃

𝐿

𝑅𝜔𝐴𝐵 cos 𝜃, (10)

Figure 2 Detailed view of the connecting rod component of a slider-crank mechanism.

1084

Chapter 16

Planar Rigid Body Kinematics

where the solutions have been written to emphasize that, at least for 𝜔𝐵𝐶 , the geometry of the mechanism matters only through the ratio 𝐿∕𝑅. The vectors 𝜔 ⃗ 𝐵𝐶 and 𝑣⃗𝐶 are then given by Eqs. (3) and (6), respectively, as 𝜔 ⃗ 𝐵𝐶 = − √

𝜔𝐴𝐵 sin 𝜃 (𝐿∕𝑅)2 − cos2 𝜃 [

𝑣⃗𝐶 = 𝑅𝜔𝐴𝐵 cos 𝜃 1 + √

̂ 𝑘,

(11) ] sin 𝜃

(𝐿∕𝑅)2 − cos2 𝜃

𝚥̂.

(12)

⃗ 𝐵𝐶 are now both known, 𝑣⃗𝐷 is readily found To find 𝑣⃗𝐷 , we note that since 𝑣⃗𝐵 and 𝜔 by using Eq. (16.3) on p. 1057 to relate the motion of 𝐷 to that of 𝐵 as 𝑣⃗𝐷 = 𝑣⃗𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐷∕𝐵 .

(13)

Writing 𝑟⃗𝐷∕𝐵 = 𝐻(−sin 𝜙 𝚤̂+cos 𝜙 𝚥̂), substituting Eqs. (1), (3), (5), and (11) into Eq. (13), carrying out the cross product, and simplifying, we obtain { 𝑣⃗𝐷 = 𝑅𝜔𝐴𝐵

] } [ ) 𝐻 𝐻 sin 𝜃 𝚥̂ . sin 𝜃 − 1 𝚤̂ + cos 𝜃 1 + √ 𝐿 𝐿 (𝐿∕𝑅)2 − cos2 𝜃 (

(14)

Discussion & Verification

The answers in Eqs. (11), (12), and (14) are somewhat complicated, but we can see that they have some expected behavior. For example, we expect the piston’s velocity to be zero when 𝜃 = 90◦ and 𝜃 = 270◦ (when it reaches extreme positions along the 𝑦 axis), and Eq. (12) tells us that it is. In addition, we expect the angular velocity of the CR to be zero at 𝜃 = 0◦ and 𝜃 = 180◦ since its rotation changes direction at those points—Eq. (11) verifies that this is true. Finally, inspection of our three final results tells us that they are all dimensionally correct.

𝜔𝐵𝐶 (rad∕s)

100 50 0

𝐿∕𝑅 = 3.0 𝐿∕𝑅 = 3.3 𝐿∕𝑅 = 3.6

−50

−100 0

90

180

270

360

𝑣𝐶𝑦 (m∕s)

Figure 3 Plot of the angular velocity of the connecting rod for 𝜔𝐴𝐵 = 3500 rpm, 𝐿 = 150 mm, and three values of 𝐿∕𝑅 commonly found in practice. 𝐿∕𝑅 = 3.0 𝐿∕𝑅 = 3.3 𝐿∕𝑅 = 3.6

10 0 −10 −20

0

90

180

270

360

Figure 4 Plot of the piston’s velocity for 𝜔𝐴𝐵 = 3500 rpm, 𝐿 = 150 mm, and three values of 𝐿∕𝑅 commonly found in practice.

ISTUDY

A Closer Look Our results are general because they apply for any value of 𝜃 and 𝜔𝐴𝐵 , as well as for any possible values of 𝑅, 𝐿, and 𝐻, i.e., the geometry of the mechanism. General relations, such as these, are useful because they allow us to know 𝜔𝐵𝐶 , 𝑣𝐶𝑦 , and 𝑣⃗𝐷 for all values for the parameters 𝜃, 𝜔𝐴𝐵 , 𝑅, 𝐿, and 𝐻 when designing these machine components. This ability to see how one or more quantities change as parameters are changed is called parametric analysis. We now present plots of 𝜔𝐵𝐶 and 𝑣𝐶𝑦 for the operating conditions that are typical in car engines. ⇒ Observe that 𝜔𝐵𝐶 and 𝑣𝐶𝑦 are periodic functions of 𝜃, so we only need to plot them for one full crank rotation, that is, for 0 ≤ 𝜃 ≤ 360◦ . In addition, since 𝜔𝐵𝐶 and 𝑣𝐶𝑦 are directly proportional to 𝜔𝐴𝐵 , the plots obtained for one value of 𝜔𝐴𝐵 can be rescaled to obtain plots for other 𝜔𝐴𝐵 values. Finally, we see that the geometry of the mechanism appears in the equations primarily through the ratio 𝐿∕𝑅. It is this ratio that is usually found in the analysis of car engine performance. The plots of the functions in Eqs. (11) and (12) are presented in Figs. 3 and 4, respectively, for 𝜔𝐴𝐵 = 3500 rpm (e.g., highway cruising), 𝐿 = 150 mm (values for small block engines typically range between 140 and 155 mm), and three commonly found values of 𝐿∕𝑅. We see that the smaller 𝐿∕𝑅, the larger are 𝜔𝐵𝐶 and 𝑣𝐶𝑦 . In addition, as we discussed above, the piston’s velocity is zero when 𝜃 = 90◦ and 𝜃 = 270◦ . Finally, notice that 𝑣𝐶𝑦 looks different for 0◦ ≤ 𝜃 < 180◦ when compared with 180◦ ≤ 𝜃 ≤ 360◦ . This difference is even more pronounced in the behavior of the piston’s acceleration, discussed in Section 16.3. This lack of symmetry tends to disappear for larger values of 𝐿∕𝑅. ⇐

ISTUDY

Section 16.2

Planar Motion: Velocity Analysis

E X A M P L E 16.7

1085

A Mechanism with a Slider: Differentiation of Constraints

Figure 1 shows a variant of the slider-crank mechanism called a swinging block slider crank. First used in various steam locomotive engines in the 1800s, this mechanism is often found in door closing systems (see Fig. 2). Referring to Fig. 3, note that the slider 𝑆 is directly connected to the crank, and it slides within a swinging block that is free to swing about the pivot at 𝑂. For this mechanism we want to derive the relation between the angular velocity of the slider and that of the crank. Also, for 𝑅 = 8 in., 𝐻 = 25 in., 𝜃 = 20◦ , and 𝜃̇ = 265 rad∕s, we want to determine the velocity of the point 𝑃 on the slider that is underneath 𝑂 at this instant. 𝐵

crank

𝚤̂

bloc

𝚥̂

𝑅 𝜃

𝐴

𝑦

𝜙

𝑢̂ 𝑆

𝑂

slider 𝑆

Figure 3. Representation of a swinging block slider-crank mechanism with a sliding contact at 𝑂.

SOLUTION Because 𝜃 describes the crank’s orientation, the crank’s angular velocity is ̂ Similarly, the slider’s angular velocity is 𝜔 ̂ where the given by 𝜔 ⃗ 𝐴𝐵 = 𝜃̇ 𝑘. ⃗ 𝑆 = −𝜙̇ 𝑘, ̇ minus sign accounts for the fact that if 𝜙 > 0, the slider rotates clockwise. We can find 𝜙̇ by first relating 𝜙 to 𝜃 and then differentiating the resulting equation with respect to time, which is the differentiation of constraint method of solution. Once 𝜔 ⃗ 𝑆 is known, the ⃗ 𝑆 × 𝑟⃗𝑃 ∕𝐵 , velocity of any point 𝑃 on the slider can be found via the relation 𝑣⃗𝑃 = 𝑣⃗𝐵 + 𝜔 where point 𝐵 is on both the crank and the slider.

Jon Reis

Figure 1 Model illustrating the components of a swinging block slider-crank mechanism.

Road Map

Computation

pneumatic closer

Focusing on the triangle 𝐴𝑂𝐵, throughout the motion we must have McGraw Hill

𝑅 sin 𝜃 = (𝐻 − 𝑅 cos 𝜃) tan 𝜙.

(1)

Differentiating Eq. (1) with respect to time, we obtain 𝑅𝜃̇ cos 𝜃 = 𝑅𝜃̇ sin 𝜃 tan 𝜙 + (𝐻 − 𝑅 cos 𝜃)𝜙̇ sec2 𝜙.

(2)

Solving Eq. (2) for 𝜙̇ yields cos 𝜃 − sin 𝜃 tan 𝜙 ̇ 𝑅𝜃, 𝜙̇ = (𝐻 − 𝑅 cos 𝜃) sec2 𝜙

(3)

which can be simplified to read 𝜙̇ =

𝑅(𝐻 cos 𝜃 − 𝑅)𝜃̇ + 𝑅2 − 2𝐻𝑅 cos 𝜃

𝐻2

⇒

𝜔 ⃗𝑆 =

𝑅(𝑅 − 𝐻 cos 𝜃)𝜃̇ ̂ 𝑘, + 𝑅2 − 2𝐻𝑅 cos 𝜃

𝐻2

(4)

where we used Eq. (1) to write tan 𝜙 = 𝑅 sin 𝜃∕(𝐻 − 𝑅 cos 𝜃), as well as the identity sec2 𝜙 = 1 + tan2 𝜙. For the calculation of 𝑣⃗𝑃 , let 𝑡0 be the instant when 𝜃 = 20◦ . At this instant, we can write 𝑣⃗𝑃 (𝑡0 ) = 𝑣⃗𝐵 (𝑡0 ) + 𝜔 ⃗ 𝑆 (𝑡0 ) × 𝑟⃗𝑃 ∕𝐵 (𝑡0 ), (5)

Figure 2 The pneumatic door closers found on typical storm or screen doors are equivalent to the mechanism shown in Fig. 1.

1086

Chapter 16

Planar Rigid Body Kinematics

where 𝑃 is the point on the slider that, at 𝑡 = 𝑡0 , coincides with point 𝑂. Since 𝐵 rotates about the fixed point 𝐴, we must have

Helpful Information Why is 𝑷 not shown in Fig. 3? The point 𝑃 cannot be seen in Fig. 3 because it is inside the swinging block pivoted at 𝑂 and because it happens to coincide with 𝑂 itself.

̇ ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝜃̇ 𝑘̂ × 𝑅(sin 𝜃 𝚤̂ − cos 𝜃 𝚥̂) = 𝑅𝜃(cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂) 𝑣⃗𝐵 = 𝜔 ⇒

𝑣⃗𝐵 (𝑡0 ) = (166.0 𝚤̂ + 60.42 𝚥̂) f t∕s,

(6)

̇ and 𝑅. As for 𝑟⃗ (𝑡 ), since 𝑃 where we have substituted in the given data for 𝜃, 𝜃, 𝑃 ∕𝐵 0 ⃗ so 𝑟⃗ (𝑡 ) = 𝑟⃗ (𝑡 ) − 𝑟⃗ (𝑡 ) = coincides with the origin, we must have 𝑟⃗𝑃 (𝑡0 ) = 0, 𝑃 ∕𝐵 0 𝑃 0 𝐵 0 −⃗𝑟𝐵 (𝑡0 ). Hence, since 𝑟⃗𝐵 = 𝑅 sin 𝜃 𝚤̂ + (𝐻 − 𝑅 cos 𝜃) 𝚥̂, we have 𝑟⃗𝑃 ∕𝐵 (𝑡0 ) = −⃗𝑟𝐵 (𝑡0 ) = (−0.2280 𝚤̂ − 1.457 𝚥̂) f t.

(7)

Next, using Eq. (4), we have

𝜔𝑆 ∕𝜃̇

̂ rad∕s. 𝜔 ⃗ 𝑆 (𝑡0 ) = (−104.9 𝑘) 0.4 0.3 0.2 0.1 0.0 −0.1 −0.2

Finally, substituting the results in Eqs. (6), (7), and (8) into Eq. (5), we obtain 𝑣⃗𝑃 (𝑡0 ) = (13.20 𝚤̂ + 84.34 𝚥̂) f t∕s.

(9)

̇ for Intuitively, we would expect |𝜔𝑆 | to be smaller than |𝜃| ̇ ⇒ all 𝜃. This is the case for the result in Eq. (8). By plotting the function 𝜔𝑆 ∕𝜃 (see ̇ < 1 for any 𝜃, that is, |𝜔 | behaves as expected. ⇐ Fig. 4), we see that |𝜔𝑆 ∕𝜃| 𝑆

Discussion & Verification 0

90

180

270

360

Figure 4 Plot of the function 𝜔𝑆 ∕𝜃̇ as a function of 𝜃 over an entire cycle. The expected behavior of 𝜔𝑆 is to always be smaller than 𝜃̇ in absolute value.

ISTUDY

(8)

A Closer Look Referring to Fig. 3, notice that the axes of the slider and of the swinging block must always coincide; otherwise the mechanism would jam. We can express this condition by saying that, on an instant-by-instant basis, given a point 𝑄 on the slider having the same 𝑥 and 𝑦 coordinates of a corresponding point 𝑄′ on the swinging block, we must have 𝑣⃗𝑄∕𝑄′ = 𝑣𝑄∕𝑄′ 𝑢̂ 𝑆 with 𝑢̂ 𝑆 = sin 𝜙 𝚤̂ + cos 𝜙 𝚥̂, (10)

where 𝑢̂ 𝑆 is a unit vector identifying the orientation of the slider’s axis (see Fig. 3). Recall that at time 𝑡0 , 𝑃 has the same 𝑥 and 𝑦 coordinates as point 𝑂, which is a fixed point. Therefore, rewriting Eq. (10) for points 𝑃 and 𝑂, we have 𝑣⃗𝑃 ∕𝑂 (𝑡0 ) = 𝑣⃗𝑃 (𝑡0 ) − 0⃗ = 𝑣𝑃 (𝑡0 ) 𝑢̂ 𝑆 (𝑡0 ).

(11)

Equation (11) says that the vectors 𝑢̂ 𝑆 (𝑡0 ) and 𝑣⃗𝑃 (𝑡0 ) must be parallel. This gives the opportunity to check our calculations by comparing the direction of these two vectors. From the second of Eqs. (10), the direction of 𝑢̂ 𝑆 can be expressed as ( ) 𝑢̂ 𝑆 (𝑡0 ) 𝑥 sin 𝜙(𝑡0 ) = tan 𝜙0 = 0.1565, (12) ( ) = cos 𝜙(𝑡0 ) 𝑢̂ 𝑆 (𝑡0 ) 𝑦 where we used Eq. (1) to compute tan 𝜙 and evaluate it at 𝑡 = 𝑡0 . Repeating the calculation for 𝑣⃗𝑃 (𝑡0 ), using Eq. (9), we have ( ) 𝑣𝑃 (𝑡0 ) 𝑥 13.20 = 0.1565, (13) ( ) = 84.34 𝑣𝑃 (𝑡0 ) 𝑦

which implies that 𝑣⃗𝑃 has the direction we expected.

ISTUDY

Section 16.2

Planar Motion: Velocity Analysis

E X A M P L E 16.8

1087

Velocity Analysis of a Four-Bar Linkage: Vector Approach

Figure 1 shows three views of a prosthetic leg with an artificial knee joint. The primary kinematic component of this artificial knee joint is a four-bar linkage (see the right panel in Fig. 1 and the system 𝐴𝐵𝐶𝐷 in Fig. 2). The four-bar linkage consists of the four segments 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, and 𝐷𝐴, which are pin-connected and can therefore rotate relative to one another. The mechanism is built in such a way that, given the motion of two of its segments, the motion of the other two is uniquely determined. By varying the relative proportions of its elements, a four-bar linkage system can provide a large variety of controlled motions, and for this reason, four-bar linkages have myriad applications, including engines, sport machines, prosthesis components, drafting tools, and carnival rides. For the mechanism in Fig. 2, assume that the segment 𝐴𝐷 is fixed, and using the information given in Table 1 for the instant shown, determine the angular velocity of segment 𝐵𝐶 (the lower leg) if segment 𝐴𝐵 rotates counterclockwise with a rate |𝜔 ⃗ 𝐴𝐵 | = 1.5 rad∕s, i.e., as if walking forward (negative 𝑥 direction). The acceleration analysis is presented in Example 16.12 on p. 1107.

Courtesy of Otto Bock HealthCare, Germany

Figure 1

Table 1. Approximate values of the coordinates of the pin centers 𝐴, 𝐵, 𝐶, and 𝐷 for the system shown in Fig. 2 at the time instant considered. Points Coordinates (mm)

𝐴

𝐵

𝐶

𝐷

(0.0, 0.0)

(−27.0, 120)

(26.0, 124)

(30.0, 15.0)

𝐴

𝚤̂ 𝚥̂

SOLUTION

𝑥 𝐷

𝜔 ⃗ 𝐴𝐵

Road Map

This linkage system is a kinematic chain, i.e., a system in which motion information is passed from one component to the next along the chain. This means we will start from a point whose motion is known, say, 𝐴 because it is fixed, and then compute the velocity of point 𝐵, which is the next point along the chain 𝐴𝐵𝐶𝐷, using the equation 𝑣⃗𝐵 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 . We repeat this process for segments 𝐵𝐶 and 𝐶𝐷. In doing this, we will generate enough equations to determine the angular velocity of each element along the chain. It is important to keep in mind that the calculations performed in this example hold only at a given instant in time. Computation

−

𝐵

𝐹+

𝐶

For the velocity of 𝐵, we have 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔𝐴𝐵 𝑘̂ × 𝑟⃗𝐵∕𝐴 .

(1)

Letting 𝜔𝐴𝐵 = −1.5 rad∕s, observing that 𝑟⃗𝐵∕𝐴 = 𝑟⃗𝐵 = (−27 𝚤̂ + 120 𝚥̂) mm, and recalling ⃗ Eq. (1) yields that 𝑣⃗ = 0, 𝐴

𝑣⃗𝐵 = (180.0 𝚤̂ + 40.50 𝚥̂) mm∕s.

(2)

Since point 𝐶 is shared by both segments 𝐵𝐶 and 𝐶𝐷, we can express the velocity of 𝐶 in two independent ways as follows: 𝑣⃗𝐶 = 𝑣⃗𝐵 + 𝜔𝐵𝐶 𝑘̂ × 𝑟⃗𝐶∕𝐵 ,

(3)

𝑣⃗𝐶 = 𝑣⃗𝐷 + 𝜔𝐶𝐷 𝑘̂ × 𝑟⃗𝐶∕𝐷 .

(4)

and

Observing that 𝑟⃗𝐶∕𝐵 = 𝑟⃗𝐶 − 𝑟⃗𝐵 = (53.00 𝚤̂ + 4.000 𝚥̂) mm,

(5)

𝑟⃗𝐶∕𝐷 = 𝑟⃗𝐶 − 𝑟⃗𝐷 = (−4.000 𝚤̂ + 109.0 𝚥̂) mm,

(6)

Figure 2 Geometry of the four-bar linkage in the prosthetic leg. Note that for the coordinate system defined in this problem, 𝑘̂ is positive into the page, so a counterclockwise angular velocity is negative.

1088

ISTUDY

Chapter 16

Planar Rigid Body Kinematics

⃗ and noting that the 𝑣⃗ obtained from Eq. (3) must be the same as recalling that 𝑣⃗𝐷 = 0, 𝐶 that obtained from Eq. (4), we obtain [( ] [( ] ) ) 180.0 mm − (4.000 mm)𝜔𝐵𝐶 𝚤̂ + 40.50 mm + (53.00 mm)𝜔𝐵𝐶 𝚥̂ s s = −(109.0 mm)𝜔𝐶𝐷 𝚤̂ − (4.000 mm)𝜔𝐶𝐷 𝚥̂.

(7)

Equation (7) is a vector equation equivalent to a linear system of two scalar equations in the unknowns 𝜔𝐵𝐶 and 𝜔𝐶𝐷 . These equations are ) ( − (4.000 mm)𝜔𝐵𝐶 = −(109.0 mm)𝜔𝐶𝐷 , (8) 180.0 mm s ( mm ) (9) 40.50 s + (53.00 mm)𝜔𝐵𝐶 = −(4.000 mm)𝜔𝐶𝐷 , which can be solved to obtain 𝜔𝐵𝐶 = −0.6378 rad∕s and 𝜔𝐶𝐷 = −1.675 rad∕s. Discussion & Verification

(10)

The solution we have obtained seems reasonable in that both segment 𝐵𝐶 and segment 𝐶𝐷 rotate counterclockwise, as one would expect, when attempting to walk forward. What is interesting is that the proportions of the segments in the mechanism are such that, in the configuration shown, point 𝐵 moves down and to the right, that is, in a direction that would cause the foot to move into the ground, which, again, is consistent with what happens when we begin to walk forward from a standing position.

ISTUDY

Section 16.2

Planar Motion: Velocity Analysis

Problems Problem 16.40 A carrier is maneuvering so that, at the instant shown, |𝑣⃗𝐴 | = 25 knots (1 kn is exactly equal to 1.852 km∕h) and 𝜙 = 33◦ . Letting the distance between 𝐴 and 𝐵 be 220 m and 𝜃 = 22◦ , determine 𝑣⃗𝐵 at the given instant if the ship’s turning rate at this instant is 𝜃̇ = 2◦ ∕s clockwise.

𝜙 𝐵 𝚥̂

𝑣⃗𝐴 𝜃

𝐴

Problem 16.41 At the instant shown, the pinion is rotating between two racks with an angular velocity 𝜔𝑃 = 55 rad∕s. If the nominal radius of the pinion is 𝑅 = 4 cm and if the lower rack is moving to the right with a speed 𝑣𝐿 = 1.2 m∕s, determine the velocity of the upper rack.

𝑟⃗𝐵∕𝐴

𝚤̂

Figure P16.40

Problem 16.42 At the instant shown, the lower rack is moving to the right with a speed of 𝑣𝐿 = 4 f t∕s, while the upper rack is fixed. If the nominal radius of the pinion is 𝑅 = 2.5 in., determine 𝜔𝑃 , the angular velocity of the pinion, as well as the velocity of point 𝑂, i.e., the center of the pinion.

Problem 16.43

𝚥̂

𝑅

𝜔𝑃

𝑂

𝚤̂

Figure P16.41 and P16.42

A bar of length 𝐿 = 2.5 m is pin-connected to a roller at 𝐴. The roller is moving along a horizontal rail as shown with 𝑣𝐴 = 5 m∕s. If, at a certain instant, 𝜃 = 33◦ and 𝜃̇ = 0.4 rad∕s, compute the velocity of the bar’s midpoint 𝐶. 𝐴

𝜃

𝐶

𝐿

𝚥̂ 𝚤̂

𝐵

Figure P16.43 and P16.44

Problem 16.44 If the motion of the bar is planar, what would the speed of 𝐴 need to be for 𝑣⃗𝐶 to be perpendicular to the bar 𝐴𝐵? Why? 𝐷

Problems 16.45 and 16.46 At the instant shown, the disk of radius 𝑟, whose center is at 𝑂, is unwinding from the rope, which is attached to the fixed point at 𝐷, with the angular speed 𝜔𝑑 . Problem 16.45

𝐵, and 𝐶. Problem 16.46

and 𝜔𝑑 .

If 𝑟 = 6 in. and 𝜔𝑑 = 20 rad∕s, determine the velocities of points 𝐴, Determine the velocities of points 𝐴, 𝐵, and 𝐶 as functions of 𝑟

𝐴 𝐵

𝑟 𝑂

𝜔𝑑

Figure P16.45 and P16.46

1089

1090

Chapter 16

Planar Rigid Body Kinematics

Problem 16.47 𝐵

Points 𝐴 and 𝐵 are both on the trailer part of the truck. If the relative velocity of point 𝐵 with respect to 𝐴 is as shown, is the body undergoing a planar rigid body motion? 𝑣⃗𝐵∕𝐴

Problem 16.48 A truck is moving to the right with a speed 𝑣0 = 12 km∕h, while the pipe section with radius 𝑅 = 1.25 m and center at 𝐶 rolls without slipping over the truck’s bed. The center of the pipe section 𝐶 is moving to the right at 2 m∕s relative to the truck. Determine the angular velocity of the pipe section and the absolute velocity of 𝐶.

𝐴

𝑅

𝚥̂

𝐶

𝚤̂

Figure P16.47 Figure P16.48

Problem 16.49 A wheel 𝑊 of radius 𝑅𝑊 = 7 mm is connected to point 𝑂 via the rotating arm OC, and it rolls without slip over the stationary cylinder 𝑆 of radius 𝑅𝑆 = 15 mm. If, at the instant shown, 𝜃 = 47◦ and 𝜔𝑂𝐶 = 3.5 rad∕s, determine the angular velocity of the wheel and the velocity of point 𝑄, where point 𝑄 lies on the edge of 𝑊 and along the extension of the line OC. 𝐴 𝚥̂

𝑊

𝑢̂ 𝜃

𝜔𝑂𝐶 𝑢̂ 𝑟

𝑅𝑊

𝜔𝑊

𝐿

𝚤̂ 𝐿

𝜃

𝑅𝑆

𝐶

𝐵 𝑆

𝜃

𝐵 Figure P16.49

Figure P16.50

𝐿

Problem 16.50

𝜔𝑐

At the instant shown, bars 𝐴𝐵 and 𝐵𝐶 are perpendicular to each other, and bar 𝐵𝐶 is rotating counterclockwise at 20 rad∕s. Letting 𝐿 = 2.5 f t and 𝜃 = 45◦ , determine the angular velocity of bar 𝐴𝐵, as well as the velocity of the slider 𝐶. 𝑅

𝐴

𝚥̂

𝜃

𝑂

𝚤̂ 𝐻

Figure P16.51

ISTUDY

Problem 16.51 For the slider-crank mechanism shown, let 𝑅 = 20 mm, 𝐿 = 80 mm, and 𝐻 = 38 mm. Use the concept of instantaneous center of rotation to determine the values of 𝜃, with 0◦ ≤ 𝜃 ≤ 360◦ , for which 𝑣𝐵 = 0. Also, determine the angular velocity of the connecting rod at these values of 𝜃 for 𝜔𝑐 = 10 rad∕s in the direction shown.

ISTUDY

Section 16.2

1091

Planar Motion: Velocity Analysis

Problems 16.52 and 16.53 A ball of radius 𝑅𝐴 = 3 in. is rolling without slip in a stationary spherical bowl of radius 𝑅𝐵 = 8 in. Assume that the ball’s motion is planar. Express your answers using the component system shown.

𝜃

𝐴

𝑅𝐴

𝐵

𝑢̂ 𝑟 𝑢̂ 𝜃

𝑅𝐵

Problem 16.52 If the speed of the center of the ball is 𝑣𝐴 = 1.75 f t∕s and if the ball is moving down and to the right, determine the angular velocity of the ball. Problem 16.53

If the angular speed of the ball |𝜔𝐴 | = 4 rad∕s is counterclockwise, determine the velocity of the center of the ball.

Figure P16.52 and P16.53

Problems 16.54 through 16.56 In the mechanism shown, the block 𝐵 is constrained to move vertically and is attached to the bar 𝐵𝐷. The point 𝐴 on the bar 𝐴𝐷 is fixed. Express all your answers in the component system shown.

𝚥̂

At the instant shown, bar 𝐴𝐷 is rotating counterclockwise at the angular speed 𝜔𝐴𝐷 = 13 rad∕s, 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 24 in. and 𝑑 = 16 in., determine the velocity of the block 𝐵 at this instant.

𝐴

𝐶

At the instant shown, the block 𝐵 is moving downward at 2.5 f t∕s, 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 12 in. and 𝑑 = 8 in., determine the angular velocity of bar 𝐴𝐷 at this instant. Problem 16.54

Problem 16.55

𝚤̂

𝜙

𝜃 𝓁

𝐷

𝑑

Figure P16.54–P16.56

Problem 16.56 At the instant shown, the block 𝐵 is moving downward at 1.5 m∕s, 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 1.2 m and 𝑑 = 0.8 m, determine the angular velocity of bar 𝐴𝐷 and the velocity of point 𝐶 at this instant. yoke

Problem 16.57 One way to convert rotational motion into linear motion and vice versa is with the use of a mechanism called a Scotch yoke, which consists of a crank 𝐶 that is connected to a slider 𝐵 by a pin 𝐴. Crank 𝐶 is mounted on a fixed bearing at its center. The pin rotates with the crank while sliding within the yoke, which, in turn, rigidly translates with the slider. This mechanism has been used, for example, to control the opening and closing of valves in pipelines. Letting the radius of the crank be 𝑅 = 1.5 f t, determine the angular velocity 𝜔𝐶 of the crank so that the maximum speed of the slider is 𝑣𝐵 = 90 f t∕s.

𝐵 slider 𝚥̂

𝜔𝐶

𝑅 𝐶 𝚤̂

Figure P16.57

Problems 16.58 through 16.60 The system shown consists of a wheel of radius 𝑅 = 14 in. rolling without slip on a horizontal surface. A bar 𝐴𝐵 of length 𝐿 = 40 in. is pin-connected to the center of the wheel and to a slider 𝐴 that is constrained to move along a vertical guide. Point 𝐶 is the bar’s midpoint. If, when 𝜃 = 72◦ , the wheel is moving to the right so that 𝑣𝐵 = 7 f t∕s, determine the angular velocity of the bar, as well as the velocity of the slider 𝐴.

𝐴 𝐿 𝐶

Problem 16.58

If, when 𝜃 = 53◦ , the slider is moving downward with a speed 𝑣𝐴 = 8 f t∕s, determine the velocity of points 𝐵 and 𝐶. Problem 16.59

Problem 16.60 If the wheel rolls without slip with a constant counterclockwise angular velocity of 10 rad∕s, determine the velocity of the slider 𝐴 when 𝜃 = 45◦ .

𝜃 𝐵 𝑅

Figure P16.58–P16.60

𝚥̂

𝚤̂

1092

Chapter 16

Planar Rigid Body Kinematics Problems 16.61 and 16.62

As the circular cam whose center is at 𝐴 rotates, it causes the follower 𝐵 to move back and forth. The cam angle is 𝜃, the radius of the cam is 𝑅, and the angular speed of the cam is 𝜃̇ = 𝜔𝑂𝐴 . The cam is pin-connected to the fixed point 𝑂.

𝑅 𝜔𝑂𝐴

2 𝑅 3

𝑂

𝐵

𝐴 𝜃

follower

Problem 16.61 Using the given 𝑥 coordinate, determine the velocity of the follower at the instant 𝜃 = 30◦ if 𝑅 = 1.5 in., and 𝜃̇ = 𝜔𝑂𝐴 = 1000 rpm. Problem 16.62 Using the given 𝑥 coordinate, determine the velocity of the follower as a function of the cam angle 𝜃, the radius of the cam 𝑅, and the given angular speed of the cam 𝜔𝑂𝐴 .

Figure P16.61 and P16.62

Problem 16.63 At the instant shown, the lower rack is moving to the right with a speed of 2.7 m∕s while the upper rack is moving to the left with a speed of 1.7 m∕s. If the nominal radius of the pinion 𝑂 is 𝑅 = 0.25 m, determine the angular velocity of the pinion, as well as the position of the pinion’s instantaneous center of rotation relative to point 𝑂. 𝜙 𝐵 𝚥̂

𝚥̂

𝑟⃗𝐵∕𝐴 𝜃

𝑅 𝑂

𝚤̂ 𝐶

𝑣⃗𝐴

Figure P16.63

𝐴

𝚤̂

Figure P16.64

𝐿

Problem 16.64

𝜙 𝐻

𝐷 𝐵

A carrier is maneuvering so that, at the instant shown, |𝑣⃗𝐴 | = 22 kn, 𝜙 = 35◦ , and |𝑣⃗𝐵 | = 24 kn (1 kn is equal to 1 nautical mile (nml) per hour or 6076 f t∕h). Letting 𝜃 = 19◦ and the distance between 𝐴 and 𝐵 be 720 f t, determine the ship’s turning rate at the given instant if the ship is rotating clockwise.

𝜃

𝑅 𝐴

𝚥̂ 𝜔𝐴𝐵

Problems 16.65 and 16.66

𝚤̂

For the slider-crank mechanism shown, let 𝑅 = 1.9 in., 𝐿 = 6.1 in., and 𝐻 = 1.2 in. Also, at the instant shown, let 𝜃 = 27◦ and 𝜔𝐴𝐵 = 4850 rpm.

Figure P16.65 and P16.66

𝑊

𝑢̂ 𝜃

𝜔𝑂𝐶 𝑢̂ 𝑟

𝑅𝑆 𝑆 Figure P16.67

ISTUDY

𝑅𝑊 𝜃

𝜔𝑊

Problem 16.65

Determine the velocity of the piston at the instant shown.

Problem 16.66

Determine 𝜙̇ and the velocity of point 𝐷 at the instant shown.

Problem 16.67 A wheel 𝑊 of radius 𝑅𝑊 = 7 mm is connected to point 𝑂 via the rotating arm OC, and it rolls without slip over the stationary cylinder 𝑆 of radius 𝑅𝑆 = 15 mm. If, at the instant shown, 𝜃 = 63◦ and 𝜔𝑊 = 9 rad∕s, determine the angular velocity of the arm 𝑂𝐶 and the velocity of point 𝑃 , where point 𝑃 lies on the edge of 𝑊 and is vertically aligned with point 𝐶.

ISTUDY

Section 16.2

Planar Motion: Velocity Analysis

Problem 16.68 At the instant shown, the center 𝑂 of a spool with inner and outer radii 𝑟 = 1 m and 𝑅 = 2.2 m, respectively, is moving up the incline at speed 𝑣𝑂 = 3 m∕s. If the spool does not slip relative to the ground or relative to the cable 𝐶, determine the rate at which the cable is wound or unwound, that is, the length of cable being wound or unwound per unit time. 𝐶 𝚥̂

𝑅

𝑅 𝑣𝑂 𝑂

𝚥̂

𝑟

𝑂

𝚤̂

𝑣𝑂

𝚤̂ 𝑟 𝐶

Figure P16.68

Figure P16.69

Problem 16.69 At the instant shown, the center 𝑂 of a spool with inner and outer radii 𝑟 = 3 f t and 𝑅 = 7 f t, respectively, is moving down the incline at a speed 𝑣𝑂 = 12.2 f t∕s. If the spool does not slip relative to the rope and if the rope is fixed at one end, determine the velocity of point 𝐶 (the point on the spool that is in contact with the incline), as well as the rope’s unwinding rate, that is, the length of rope being unwound per unit time. Hint: If the rope is fixed at the top of the incline and is inextensible, then all material points along the rope have no velocity.

Problem 16.70 The bucket of a backhoe is the element 𝐴𝐵 of the four-bar linkage system 𝐴𝐵𝐶𝐷. Assume that the points 𝐴 and 𝐷 are fixed and that, at the instant shown, point 𝐵 is vertically aligned with point 𝐴, point 𝐶 is horizontally aligned with point 𝐵, and point 𝐵 is moving to the right with a speed 𝑣𝐵 = 1.2 f t∕s. Determine the velocity of point 𝐶 at the instant shown, along with the angular velocities of elements 𝐵𝐶 and 𝐶𝐷. Let ℎ = 0.66 f t, 𝑒 = 0.46 f t, 𝓁 = 0.9 f t, and 𝑤 = 1.0 f t. 𝐴 𝚥̂ 𝐶 𝚥̂

𝐴

𝚤̂ 𝐷

𝚤̂

𝐿

𝑣𝐵

𝐵

𝓁

𝐿

𝑒

𝐶 ℎ

𝐵 𝐷

Figure P16.70

Figure P16.71

𝜃

𝐿∕2

1093

1094

Chapter 16

Planar Rigid Body Kinematics

Problem 16.71 Bar 𝐴𝐵 is rotating counterclockwise with an angular velocity of 15 rad∕s. Letting 𝐿 = 1.25 m, determine the angular velocity of bar 𝐶𝐷 when 𝜃 = 45◦ . Hint: The method of IC provides an efficient solution procedure.

Problem 16.72 𝚥̂

At the instant shown, bars 𝐴𝐵 and 𝐶𝐷 are vertical and point 𝐶 is moving to the left with a speed of 35 f t∕s. Letting 𝐿 = 1.5 f t and 𝐻 = 0.6 f t, determine the velocity of point 𝐵. Hint: The method of IC provides an efficient solution procedure.

𝚤̂

𝐴

𝐵

𝐷

Problem 16.73 Collars 𝐴 and 𝐵 are constrained to slide along the guides shown and are connected by a bar with length 𝐿 = 0.75 m. Letting 𝜃 = 45◦ , determine the angular velocity of the bar 𝐴𝐵 at the instant shown if, at this instant, 𝑣𝐵 = 2.7 m∕s.

Figure P16.72

ISTUDY

𝑣𝐵 𝚥̂

𝐴

𝚤̂

𝜃

𝐵 𝜃

Figure P16.73

Problem 16.74 At the instant shown, an overhead garage door is being shut with point 𝐵 moving to the left within the horizontal part of the door guide at a speed of 5 f t∕s, while point 𝐴 is moving vertically downward. Determine the angular velocity of the door and the velocity of the counterweight 𝐶 at this instant if 𝐿 = 6 f t and 𝑑 = 1.5 f t. 𝐵 𝐵

𝑑 𝐴

𝐸

𝐿

𝐿𝐴𝐵

𝐶 𝐿𝐶𝐷 𝚥̂ 𝐷

𝚥̂

𝐶 floor Figure P16.74

𝜙

𝚤̂

𝜃 𝐴

𝚤̂ Figure P16.75

Problem 16.75 In the four-bar linkage system shown, the lengths of the bars 𝐴𝐵 and 𝐶𝐷 are 𝐿𝐴𝐵 = 46 mm and 𝐿𝐶𝐷 = 25 mm, respectively. In addition, the distance between points 𝐴 and 𝐷 is 𝑑𝐴𝐷 = 43 mm. The dimensions of the mechanism are such that when the angle 𝜃 = 132◦ , the angle 𝜙 = 69◦ . For 𝜃 = 132◦ and 𝜃̇ = 27 rad∕s, determine the angular velocity of bars 𝐵𝐶 and 𝐶𝐷, as well as the velocity of the point 𝐸, the midpoint of bar 𝐵𝐶. Note that the figure is drawn to scale and that bars 𝐵𝐶 and 𝐶𝐷 are not collinear.

ISTUDY

Section 16.2

1095

Planar Motion: Velocity Analysis

Problem 16.76

𝑣𝐴

A spool with inner radius 𝑅 = 1.5 m rolls without slip over a horizontal rail as shown. If the cable on the spool is unwound at a rate 𝑣𝐴 = 5 m∕s, in such a way that the unwound cable remains perpendicular to the rail, determine the angular velocity of the spool and the velocity of the spool’s center 𝑂.

𝐴 𝚥̂ 𝚤̂

𝑅

𝑂

Problem 16.77 A person is closing a heavy gate with rusty hinges by pushing the gate with car 𝐴. If 𝑤 = 24 m and 𝑣𝐴 = 1.2 m∕s, determine the angular velocity of the gate when 𝜃 = 15◦ .

Figure P16.76

𝐴

𝑣𝐴

𝚥̂ 𝚤̂ 𝜃 𝑂 Figure P16.77 𝐶

Problems 16.78 through 16.80

𝑅

In the four-bar linkage system shown, let the circular guide with center at 𝑂 be fixed and such that, when 𝜃 = 0◦ , the bars 𝐴𝐵 and 𝐵𝐶 are vertical and horizontal, respectively. In addition, let 𝑅 = 2 f t, 𝐿 = 3 f t, and 𝐻 = 3.5 f t.

𝐿 𝛽

𝜃 𝑂

𝐵 𝛾

When 𝜃 = 0◦ , the collar at 𝐶 is sliding downward with a speed of 23 f t∕s. Determine the angular velocities of the bars 𝐴𝐵 and BC at this instant. Hint: Use the method of IC to examine bar 𝐵𝐶.

Problem 16.78

𝐻

𝚥̂ 𝚤̂

𝐴

Figure P16.78–P16.80

When 𝜃 = 37◦ , 𝛽 = 25.07◦ , 𝛾 = 78.71◦ , and the collar is sliding clockwise with a speed 𝑣𝐶 = 23 f t∕s. Determine the angular velocities of the bars 𝐴𝐵 and 𝐵𝐶. Hint: Use the method of IC to examine bar 𝐵𝐶. Problem 16.79

Problem 16.80 Determine the general expression for the angular velocities of bars ̇ 𝐴𝐵 and 𝐵𝐶 as a function of 𝜃, 𝛽, 𝛾, 𝑅, 𝐿, 𝐻, and 𝜃. 𝐴 𝑅

Problem 16.81 The crank 𝐴𝐵, pinned at 𝐴, is rotating counterclockwise at a constant angular velocity of 12 rad∕s while the pin 𝐵 slides within the slot in the bar 𝐶𝐷, pinned at 𝐶. Letting 𝑅 = 0.5 m, ℎ = 1 m, and 𝑑 = 0.25 m, determine the angular velocity of 𝐶𝐷 at the instant shown (with points 𝐴, 𝐵, and 𝐶 vertically aligned), as well as the velocity of the horizontal bar to which bar 𝐶𝐷 is connected. Hint: For arbitrary orientations (other than 𝐴, 𝐵 and 𝐶 being vertically aligned) this becomes a more complicated problem due to the sliding of 𝐵 within slotted member 𝐶𝐷. At this particular instant, however, the relative velocity of 𝐵 within the slot is zero, and the interaction between the crank and the slotted member looks similar to that between two interlocking gears.

𝐵

ℎ

𝚥̂ 𝐶

𝚤̂

𝑑 bar

Figure P16.81

1096

Chapter 16

Planar Rigid Body Kinematics Problems 16.82 through 16.86 𝐵

For the slider-crank mechanism shown, let 𝑅 = 20 mm, 𝐿 = 80 mm, and 𝐻 = 38 mm. If 𝜃̇ = 1700 rpm, determine the angular velocity of the connecting rod 𝐴𝐵 and the speed of the slider 𝐵 for 𝜃 = 90◦ . Hint: Use the method of IC to examine connecting rod 𝐴𝐵. Problem 16.82

𝐿

Determine the angular velocity of the crank 𝑂𝐴 when 𝜃 = 20◦ and the slider is moving downward at 15 m∕s. Hint: Use the method of IC to examine connecting rod 𝐴𝐵. Problem 16.83

𝜔𝑐

𝐴

𝑅

𝚥̂

𝜃

𝑂

Problem 16.84 Determine the general expression for the velocity of the slider 𝐵 as a ̇ and the geometrical parameters 𝑅, 𝐻, and 𝐿, using the vector approach. function of 𝜃, 𝜃,

𝚤̂

Problem 16.85 Determine the general expression for the velocity of the slider 𝐵 as ̇ and the geometrical parameters 𝑅, 𝐻, and 𝐿, using differentiation of a function of 𝜃, 𝜃, constraints.

𝐻 Figure P16.82–P16.86

Problem 16.86 Plot the velocity of the slider 𝐵 as a function of 𝜃, for 0 ≤ 𝜃 ≤ 360◦ , and for 𝜃̇ = 1000 rpm, 𝜃̇ = 3000 rpm, and 𝜃̇ = 5000 rpm. 𝚥̂

𝐿

𝜙

𝑦𝐶

𝐵

𝜔𝐴𝐵 𝐴

Figure P16.87

ISTUDY

Problem 16.87 𝚤̂

𝐶

𝜃

𝑅

Complete the velocity analysis of the slider-crank mechanism, using differentiation of constraints that was outlined beginning on p. 1076. That is, determine the velocity of the piston 𝐶 and the angular velocity of the connecting rod as a function of the given quantities 𝜃, 𝜔𝐴𝐵 , 𝑅, and 𝐿. Use the component system shown for your answers.

ISTUDY

Section 16.3

16.3

Planar Motion: Acceleration Analysis

1097

Planar Motion: Acceleration Analysis

Figure 16.32 shows the slider-crank mechanism for which we now wish to perform an acceleration analysis. We will develop two different approaches to the problem that are applicable to the acceleration analysis of any planar rigid body motion: the vector approach and differentiation of constraints.

𝐶 𝐶 𝐵

𝐵

Vector approach The connecting rod in the slider-crank mechanism (Fig. 16.32) is, in general, planar motion, which means that we can describe the acceleration of any of its points by knowing the acceleration of one point on the rod and the rotational motion of the rod (i.e., its angular velocity and angular acceleration). The velocity analysis was done in Example 16.6 on p. 1083, and so we know the angular velocity of the rod 𝜔 ⃗ 𝐵𝐶 . The acceleration analysis proceeds in much the same way. On p. 1060, we found the acceleration of point 𝐵, which is in fixed-axis rotation about the centerline of the crank 𝐴 (see Fig. 16.33). Points 𝐵 and 𝐶 are on the same rigid body, so we can relate the acceleration of 𝐶 to that of 𝐵 using either Eq. (16.5) on p. 1058 or Eq. (16.14) on p. 1062 in Section 16.1, which give ( ) 𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 ,

𝐶

or

𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 − 𝜔2𝐵𝐶 𝑟⃗𝐶∕𝐵 ,

Figure 16.32 A slider-crank mechanism emphasizing the motion of the connecting rod.

𝚥̂

(16.30)

Differentiation of constraints Referring to Fig. 16.33, we can also determine 𝛼⃗𝐵𝐶 and 𝑎⃗𝐶 as functions of 𝜃 by writing the appropriate constraint equations and then differentiating them with respect to time. We found the slider-crank velocities in Section 16.2 on p. 1075; accelerations simply require an additional time derivative. As was done in Eq. (16.20) on p. 1076, we can write the constraint equation for the 𝑦 coordinate of 𝐶 as (16.31) 𝑦𝐶 = 𝑅 sin 𝜃 + 𝐿 cos 𝜙, which can be differentiated twice with respect to time to find the acceleration of 𝐶 as (16.32)

As with the velocity analysis, quantities 𝑅, 𝐿, and 𝜃(𝑡) are assumed to be known. This implies that we also know 𝜃̇ = 𝜔𝐴𝐵 and 𝜃̈ = 𝛼𝐴𝐵 , though we see that 𝑎𝐶 is also a ̇ and 𝜙. ̈ In the velocity analysis using differentiation of constraints, function of 𝜙, 𝜙, we said that we can determine 𝜙 and 𝜙̇ in terms of known quantities by relating the orientation 𝜙 of the connecting rod to the crank’s orientation 𝜃, using the second

𝚤̂

𝐶 𝐿

(16.29)

⃗ 𝐵𝐶 is known. In respectively, where 𝛼⃗𝐵𝐶 is the angular acceleration of bar 𝐵𝐶 and 𝜔 planar motion, either of the above equations is a vector equation that represents two scalar equations. These two scalar equations allow us to determine 𝑎⃗𝐶 and 𝛼⃗𝐵𝐶 since 𝑎⃗𝐵 is known, the direction of 𝑎⃗𝐶 is known (the motion of the piston 𝐶 is rectilinear along the 𝑦 axis), 𝑟⃗𝐶∕𝐵 can be found in terms of the crank angle 𝜃 (see Fig. 16.33), and the axis of rotation for 𝛼⃗𝐵𝐶 is known (it is perpendicular to the plane of motion). Therefore, the components 𝑎𝐶 and 𝛼𝐵𝐶 are the only unknowns in these two scalar equations. We will complete this solution in Example 16.10.

𝑦̈𝐶 = 𝑎𝐶 = 𝑅𝜃̈ cos 𝜃 − 𝑅𝜃̇ 2 sin 𝜃 − 𝐿𝜙̈ sin 𝜙 − 𝐿𝜙̇ 2 cos 𝜙.

𝐵

𝜙

𝑦𝐶

𝑑

𝐷 𝐵

𝜔𝐴𝐵 𝐴

𝜃

𝑅

Figure 16.33 The geometric parameters used in the slidercrank analysis. The crank’s angular velocity 𝜃̇ = 𝜔𝐴𝐵 is constant, though the solution methodology is valid whether or not 𝜔𝐴𝐵 is constant.

1098

Chapter 16

Planar Rigid Body Kinematics

𝜔𝑊

𝑊

𝛼𝑊

constraint equation sin 𝜙 = (𝑅∕𝐿) cos 𝜃. This equation was differentiated once with ̇ It can be differentiated twice to respect to time to find 𝜙̇ as a function of 𝜃 and 𝜃. ̇ ̈ ̈ ̂ obtain 𝜙, where 𝜙 𝑘 = 𝛼⃗𝐵𝐶 , as a function of 𝜃, 𝜃, and 𝜃̈ (see Prob. 16.137), which then completes the analysis. We will see this method again in Example 16.11. Before continuing, it is helpful to compare the methods available in acceleration analysis to those in velocity analysis. In particular, you will note that there is no “instantaneous center of zero acceleration” to complement the IC of zero velocity. This is due to the form of relative acceleration in Eqs. (16.29) or (16.30). The presence of the last term, the −𝜔2𝐵𝐶 𝑟⃗𝐶∕𝐵 in planar problems, means that there is a normal component in addition to a tangent component of relative acceleration. For this reason, the relative acceleration cannot simply be described as sweeping out a circular arc, and there is no corresponding instantaneous center. For acceleration analysis, our methods are confined to the vector approach and differentiation of constraints. The other important point about acceleration analysis is that it must be preceded by velocity analysis because angular velocity plays an important role in acceleration analysis. Acceleration analysis is dependent on velocity analysis.

Rolling without slip: acceleration analysis

𝑢̂ 𝑡

𝑃 𝓁

𝑄 𝑆

𝑎⃗𝑃 ∕𝑄 ⋅ 𝑢̂ 𝑡 = 0,

Figure 16.34 A wheel 𝑊 rolling over a surface 𝑆. At the instant shown, the line 𝓁 is tangent to the path of 𝑃 and 𝑄.

𝛼𝑊

𝑊 𝑎𝑂

𝜔𝑊 𝑂 𝑣𝑂 𝑟⃗𝑃 ∕𝑂

𝚥̂ 𝚤̂

𝑃

Figure 16.35 A wheel 𝑊 rolling without slip on a horizontal fixed surface.

ISTUDY

Referring to Fig. 16.34, the definition of rolling without slip given on p. 1075 stated that if the body 𝑊 is rolling without slip over the surface 𝑆 (𝑆 can be moving), then the contact points 𝑃 and 𝑄 (on 𝑊 and 𝑆, respectively) have the same velocity, ⃗ This condition implies that the component of 𝑎⃗ 𝑣⃗𝑃 = 𝑣⃗𝑄 or 𝑣⃗𝑃 ∕𝑄 = 0. 𝑃 ∕𝑄 tangent to the contact must be equal to zero, that is, (16.33)

where 𝑢̂ 𝑡 is a unit vector parallel to 𝓁, the line tangent to both 𝑊 and 𝑆 at their contact. In component form, Eq. (16.33) takes on the form ( ) 𝑎𝑃 ∕𝑄 𝑡 = 0

⇒

𝑎𝑃 𝑡 = 𝑎𝑄𝑡 .

(16.34)

Compare Eq. (16.34) to the corresponding velocity constraint in Eq. (16.16). There we required both tangent and normal velocity components to be zero. The tangent requirement corresponded to a no-slip condition, and the normal requirement was necessary to maintain contact between wheel and surface. Why is there no comparable requirement on normal acceleration? It may help to think about a specific material point on the wheel approaching the contact point 𝑄 on the surface. By necessity, this material point must move down to the surface, then lift off the surface as the wheel rolls. But such a change in velocity direction defines an acceleration. If the wheel is rolling, there must be a nonzero relative normal acceleration to enable touching down and then lifting off the contact surface. A quantitative expression for this normal acceleration is obtained below. As an application of Eq. (16.34), we can now determine the acceleration of the point in contact with the ground for a wheel 𝑊 rolling without slip on a flat stationary surface. If the wheel’s center 𝑂 moves as shown in Fig. 16.35, then 𝑣⃗𝑂 = 𝑣𝑂 𝚤̂ and

𝑎⃗𝑂 = 𝑎𝑂 𝚤̂.

(16.35)

̂ so the acceleraWe have already discovered that 𝑣⃗𝑃 = 0⃗ and that 𝜔 ⃗ 𝑊 = −(𝑣𝑂 ∕𝑅) 𝑘, tion of point 𝑃 is given by 𝑎⃗𝑃 = 𝑎⃗𝑂 + 𝛼𝑊 𝑘̂ × 𝑟⃗𝑃 ∕𝑂 − 𝜔2𝑊 𝑟⃗𝑃 ∕𝑂 ,

(16.36)

ISTUDY

Section 16.3

1099

Planar Motion: Acceleration Analysis

which, using 𝑟⃗𝑃 ∕𝑂 = −𝑅 𝚥̂, can be written as ( 𝑣 )2 𝑎⃗𝑃 = 𝑎𝑂 𝚤̂ + 𝛼𝑊 𝑘̂ × (−𝑅) 𝚥̂ − − 𝑂 (−𝑅 𝚥̂) 𝑅 ( 2) 𝑣𝑂 ) ( = 𝑎𝑂 + 𝛼𝑊 𝑅 𝚤̂ + 𝚥̂. 𝑅

(16.37)

Observe that the tangent to the wheel at the contact point 𝑃 is in the 𝑥 direction. Hence, the application of Eq. (16.34) reads 𝑎𝑃 𝑥 = 𝑎𝑄𝑥 = 0

𝛼𝑊 = −

⇒

𝑎𝑂 𝑅

,

(16.38)

since the point 𝑄 on the ground is stationary. This implies that the point 𝑃 on the wheel is accelerating along the 𝑦 axis according to 𝑎⃗𝑃 =

𝑣2𝑂 𝑅

𝚥̂.

(16.39)

This result tells us that although 𝑃 has zero velocity at the instant considered, it is accelerating, i.e., it is in the process of gaining velocity. This, in turn, will allow 𝑃 to move away from its current position and thus allow some other point to become the next contact point.

End of Section Summary This section presents two different ways to analyze the accelerations of a rigid body in planar motion: the vector approach and differentiation of constraints. Vector approach. equations

𝑎⃗𝐵

𝛼⃗𝐵𝐶

𝐵

We saw in Section 16.1 that, for planar motion, either of the

𝑟⃗𝐶∕𝐵 𝐶

Eqs. (16.29) and (16.30), p. 1097 ) ( 𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 , 𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 − 𝜔2𝐵𝐶 𝑟⃗𝐶∕𝐵 , relates the acceleration of two points on a rigid body, 𝑎⃗𝐵 and 𝑎⃗𝐶 , via their relative position 𝑟⃗𝐶∕𝐵 , the angular acceleration of the body 𝛼⃗𝐵𝐶 , and the angular velocity of the body 𝜔 ⃗ 𝐵𝐶 (see Fig. 16.36). Differentiation of constraints. As we discovered in Section 12.6, it is often convenient to write an equation describing the position of a point of interest, which can then be differentiated once with respect to time to find the velocity of that point and twice with respect to time to find the acceleration. For the planar motion of rigid bodies, this idea can also apply for describing the position, velocity, and acceleration of a point on a rigid body, as well as for describing the orientation of a rigid body, for which the first and second time derivatives provide its angular velocity and angular acceleration, respectively (we saw this again in Section 16.2 for velocities). Rolling without slip. Referring to Fig. 16.37, when a body rolls without slip over another body, the two points on the bodies that are in contact at any instant, points 𝑃 and 𝑄, must have the same velocity. For accelerations, this means that the component

𝑎⃗𝐶

𝜔 ⃗ 𝐵𝐶

rigid body Figure 16.36 A rigid body on which we are relating the acceleration of two points 𝐵 and 𝐶.

𝛼𝑊

𝜔𝑊

𝑊

𝑢̂ 𝑡

𝑃 𝓁

𝑄 𝑆

Figure 16.37 A wheel 𝑊 rolling over a surface 𝑆. At the instant shown, the line 𝓁 is tangent to the path of 𝑃 and 𝑄.

1100

Chapter 16

Planar Rigid Body Kinematics

of 𝑎⃗𝑃 ∕𝑄 tangent to the contact must be equal to zero, that is, 𝛼𝑊

𝑊 𝑎𝑂

𝜔𝑊 𝑂 𝑣𝑂 𝑟⃗𝑃 ∕𝑂

𝚥̂ 𝚤̂

𝑃

Figure 16.38 A wheel 𝑊 rolling without slip on a horizontal fixed surface.

ISTUDY

Eqs. (16.34), p. 1098 ( ) 𝑎𝑃 ∕𝑄 𝑡 = 0

𝑎𝑃 𝑡 = 𝑎𝑄𝑡 .

⇒

If a wheel of radius 𝑅 is rolling without slip over a flat, stationary surface, then the point 𝑃 on the wheel in contact with the surface must have zero velocity (see Fig. 16.38). The consequence is that the angular acceleration of the wheel 𝛼𝑊 is related to the acceleration of the center 𝑎𝑂 and the radius of the wheel 𝑅 according to Eq. (16.38), p. 1099 𝛼𝑊 = −

𝑎𝑂 𝑅

,

in which the positive direction for 𝛼𝑊 is taken to be the positive 𝑧 direction. The point 𝑃 on the wheel that is in contact with the ground at this instant is accelerating along the 𝑦 axis according to Eq. (16.39), p. 1099 𝑎⃗𝑃 =

𝑣2𝑂 𝑅

𝚥̂.

Even though 𝑃 has zero velocity at the instant considered, it is accelerating.

ISTUDY

Section 16.3

1101

Planar Motion: Acceleration Analysis

E X A M P L E 16.9

Planetary Gears Rolling Without Slip: Vector Approach sun gear

Recall from Example 16.5 on p. 1082 that planetary gear systems (see Fig. 1) are used to transmit power between two shafts. The center gear is called the sun, the outer gear is called the ring, and the three inner gears are called planets. The planets are mounted on a component called the planet carrier (which is not shown in Fig. 1). Referring to Fig. 2, let 𝑅𝑆 = 2 in., 𝑅𝑃 = 0.67 in., the ring be fixed, 𝜔𝑆 = 1500 rpm (the same as in Example 16.5), and 𝛼𝑆 = 2.7 rad∕s2 . Determine the angular acceleration of the planet gear 𝛼⃗𝑃 , the angular acceleration of the planet carrier 𝛼⃗𝑃 𝐶 , and the accelerations of points 𝐴′ and 𝑄′ .

SOLUTION Road Map

The solution of this problem has two essential elements: (1) the enforcement of the rolling without slip conditions for the 𝐴-𝐴′ and the 𝑄-𝑄′ contacts and (2) the enforcement of the constraint that point 𝐶 on the planet gear 𝑃 must move along a circle of radius 𝑅𝑃 𝐶 = 𝑅𝑆 +𝑅𝑃 . We carry out element 1 by first writing the accelerations of points 𝐴, 𝐴′ , 𝑄, and 𝑄′ . To enforce 2, we write 𝑎⃗𝐶 twice, the first time viewing 𝐶 as part of the planet carrier and the second time viewing 𝐶 as part of 𝑃 . We then force the two resulting expressions to be equal to each other. In general, before doing the acceleration analysis, we first need to find the angular velocities of all components in the system. This was done in Example 16.5 on p. 1082, in which we discovered that 𝜔𝑃 = −(𝑅𝑆 ∕2𝑅𝑃 )𝜔𝑆 = −2239 rpm and 𝜔𝑃 𝐶 = (𝑅𝑆 ∕2𝑅𝑃 𝐶 )𝜔𝑆 = 561.8 rpm. Computation

VladimirV/Shutterstock

Figure 1 Photo of a planetary gear system.

𝐴

𝑎⃗𝐴′ = 𝑎𝐴′ 𝑥 𝚤̂ + 𝑎𝐴′ 𝑦 𝚥̂.

⇒

𝑎𝐴′ 𝑥 = 0.

(1)

𝚥̂

𝑅𝑃 𝐶

𝛼𝑆 𝑂

𝑅𝑆

𝛼𝑆 = −4.030 rad∕s2

) ( ̂ 𝛼⃗𝑃 = −4.030 rad∕s2 𝑘.

(6) 2𝑅𝑃 As for the acceleration of point 𝐶, when viewed as part of the planet carrier, point 𝐶 is rotating about point 𝑂, so that 𝑎⃗𝐶 can be given the form ⇒

𝑎⃗𝐶 = 𝛼⃗𝑃 𝐶 × 𝑟⃗𝐶∕𝑂 − 𝜔2𝑃 𝐶 𝑟⃗𝐶∕𝑂 = −𝛼𝑃 𝐶 𝑅𝑃 𝐶 𝚤̂ − 𝜔2𝑃 𝐶 𝑅𝑃 𝐶 𝚥̂,

𝑅𝑃

𝑅𝑆

(3)

We now enforce the rolling without slip condition at the 𝑄-𝑄′ contact by observing that the tangent line to the contact is parallel to the 𝑥 axis. Therefore, from Eqs. (3) and (4), we have 𝑎𝑄𝑥 = 𝑎𝑄′ 𝑥 ⇒ −𝛼𝑆 𝑅𝑆 = 2𝑅𝑃 𝛼𝑃 , (5) 𝛼𝑃 = −

𝚤̂

(2)

where we have set 𝛼⃗𝑆 = 𝛼𝑆 𝑘̂ and 𝑟⃗𝑄∕𝑂 = 𝑅𝑆 𝚥̂. For 𝑄′ , using 𝐴′ as a reference point, we can write 𝑎⃗𝑄′ = 𝑎⃗𝐴′ + 𝛼⃗𝑃 × 𝑟⃗𝑄′ ∕𝐴′ − 𝜔2𝑃 𝑟⃗𝑄′ ∕𝐴′ ( ( ) ) = 𝑎𝐴′ 𝑦 𝚥̂ + 𝛼𝑃 𝑘̂ × −2𝑅𝑃 𝚥̂ − 𝜔2𝑃 −2𝑅𝑃 𝚥̂ ) ( (4) = 2𝑅𝑃 𝛼𝑃 𝚤̂ + 𝑎𝐴′ 𝑦 + 2𝑅𝑃 𝜔2𝑃 𝚥̂.

or

𝜔𝑆

𝑄

To find the accelerations of 𝑄 and 𝑄′ , since the sun gear is rotating about the (fixed) 𝑧 axis, we can express 𝑎⃗𝑄 as follows: 𝑎⃗𝑄 = 𝛼⃗𝑆 × 𝑟⃗𝑄∕𝑂 − 𝜔2𝑆 𝑟⃗𝑄∕𝑂 = −𝛼𝑆 𝑅𝑆 𝚤̂ − 𝜔2𝑆 𝑅𝑆 𝚥̂,

𝑃

𝑄

Observe that the line tangent to the 𝐴-𝐴′ contact is parallel to the 𝑥 direction, so that the rolling without slip condition between 𝑃 and the ring gear implies 𝑎𝐴𝑥 = 𝑎𝐴′ 𝑥

𝐴

𝐶

The accelerations of 𝐴 (a fixed point) and 𝐴′ can be expressed as 𝑎⃗𝐴 = 0⃗ and

ring gear

planet gears

(7)

Figure 2 Schematic of a planetary gear system with three planets and a fixed ring. The planet carrier’s rotational axis is coincident with that of the sun gear, but they are not directly connected.

1102

ISTUDY

Chapter 16

Planar Rigid Body Kinematics

̂ and we note that 𝑅 = 𝑅 + 𝑅 . where we have set 𝑟⃗𝐶∕𝑂 = 𝑅𝑃 𝐶 𝚥̂ and 𝛼⃗𝑃 𝐶 = 𝛼𝑃 𝐶 𝑘, 𝑃𝐶 𝑆 𝑃 Viewing point 𝐶 as part of the planet gear 𝑃 and relating its acceleration to 𝐴′ , we have 𝑎⃗𝐶 = 𝑎⃗𝐴′ + 𝛼⃗𝑃 × 𝑟⃗𝐶∕𝐴′ − 𝜔2𝑃 𝑟⃗𝐶∕𝐴′ = 𝑎𝐴′ 𝑦 𝚥̂ + 𝛼𝑃 𝑘̂ × 𝑟⃗𝐶∕𝐴′ − 𝜔2𝑃 𝑟⃗𝐶∕𝐴′ ) ( = 𝛼𝑃 𝑅𝑃 𝚤̂ + 𝑎𝐴′ 𝑦 + 𝜔2𝑃 𝑅𝑃 𝚥̂.

(8)

The expressions for 𝑎⃗𝐶 in Eqs. (7) and (8) must be equal to one another. Therefore, −𝛼𝑃 𝐶 𝑅𝑃 𝐶 = 𝛼𝑃 𝑅𝑃

− 𝜔2𝑃 𝐶 𝑅𝑃 𝐶 = 𝑎𝐴′ 𝑦 + 𝜔2𝑃 𝑅𝑃 .

and

(9)

Equations (9) can be solved for the unknowns 𝛼𝑃 𝐶 and 𝑎𝐴′ 𝑦 to obtain 𝛼𝑃 𝐶 = −

𝑅𝑃 𝑅𝑃 𝐶

and 𝑎𝐴′ 𝑦 = −𝜔2𝑃 𝑅𝑃 − 𝜔2𝑃 𝐶 𝑅𝑃 𝐶 .

𝛼𝑃

(10)

Using Eq. (6) for 𝛼𝑃 , the expressions for 𝜔𝑃 and 𝜔𝑃 𝐶 from the road map, and Eq. (2) for 𝑎𝐴′ 𝑥 , Eqs. (10) become 𝛼⃗𝑃 𝐶 =

𝑅𝑆 2𝑅𝑃 𝐶

𝑎⃗𝐴′ = −

( ) ̂ 𝛼𝑆 𝑘̂ = 1.011 rad∕s2 𝑘,

𝑅2𝑆 4𝑅𝑃

( 1+

𝑅𝑃 𝑅𝑃 𝐶

)

) ( 𝜔2𝑆 𝚥̂ = −3839 f t∕s2 𝚥̂.

(11) (12)

Finally, substituting the results in Eqs. (6) and (12) and 𝜔𝑃 from the road map into Eq. (4), we obtain 𝑎⃗𝑄′ = −𝛼𝑆 𝑅𝑆 𝚤̂ +

𝑅2𝑆 4𝑅𝑃

Discussion & Verification

( 1−

𝑅𝑃 𝑅𝑃 𝐶

) 𝜔2𝑆 𝚥̂ = (−0.4500 𝚤̂ + 2299 𝚥̂) ft∕s2 .

(13)

The dimensions of the symbolic answers are all as they should be, and so the units of the numerical answers are also correct. The results in Eqs. (6) and (11) are not hard to verify by differentiating the corresponding angular velocity equations (we will see this in Example 16.11 on p. 1105). While the procedure we have used in our solution is applicable in general, obtaining the component of an angular acceleration by simply differentiating the corresponding component of the angular velocity can be done only under special circumstances, such as when the components in question are with respect to a fixed axis (in our case, the 𝑧 axis). As far as the acceleration of point 𝐴′ is concerned, based on our discussion of the rolling without slip condition earlier in this section, because 𝐴′ was in contact with a stationary surface, we should have expected 𝑎⃗𝐴′ to be completely in the negative 𝑦 direction and proportional to 𝜔2𝑃 . This is exactly what we obtained, given that 𝜔𝑃 is proportional to 𝜔𝑆 . As far as 𝑎⃗𝑄′ is concerned, our expectation was that the 𝑥 component had to match the motion of the sun gear (and therefore be in the negative 𝑥 direction) while the 𝑦 component had to be in the positive 𝑦 direction and, again, be proportional to 𝜔2𝑃 , i.e., 𝜔2𝑆 . Again, these expectations match the obtained results.

ISTUDY

Section 16.3

1103

Planar Motion: Acceleration Analysis

E X A M P L E 16.10

Completing the Acceleration Analysis of the Connecting Rod

On p. 1097 we outlined the vector approach for the acceleration analysis of the connecting rod (CR) and piston in the slider-crank mechanism shown in Fig. 1. We will complete that analysis here. Referring to Fig. 2, we are given the radius of the crank 𝑅, the length of the CR 𝐿, the distance from 𝐵 to the mass center of the CR 𝐻, the constant angular velocity of the ̇ and the crank angle 𝜃. Determine the angular acceleration of the CR 𝛼⃗ crank 𝜔𝐴𝐵 = 𝜃, 𝐵𝐶 and the acceleration of the piston 𝑎⃗𝐶 .

𝐶

𝐶

𝐶 𝐵 𝐵

𝐴

𝐴

𝐴

𝐵

SOLUTION Road Map

Figure 1

The road map for this problem was laid out on p. 1097.

Computation

As with the velocity analysis, we begin by determining the motion of point 𝐶, which is constrained to move along the 𝑦 axis, and so 𝑎𝐶𝑥 = 0. Applying Eq. (16.14) on p. 1062 to the CR and using point 𝐵 as a reference point for the body, we have 𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 − 𝜔2𝐵𝐶 𝑟⃗𝐶∕𝐵 , (1)

𝐶

where 𝛼⃗𝐵𝐶 = 𝛼𝐵𝐶 𝑘̂ is the angular acceleration of the CR. Recall that in Example 16.6 on p. 1083, we found the angular velocity of the CR to be 𝜔𝐵𝐶 = − √

𝜔𝐴𝐵 sin 𝜃 (𝐿∕𝑅)2

−

cos2

𝜃

,

(3)

𝐵

𝚥̂ 𝐴

𝜃 𝑥

𝚤̂ 𝜔𝐴𝐵 = 𝜃̇

where sin 𝜙 and cos 𝜙 are found from the equations √ and

𝐻

𝐷 𝑅

𝑟⃗𝐶∕𝐵 = −𝐿 sin 𝜙 𝚤̂ + 𝐿 cos 𝜙 𝚥̂,

𝑅 cos 𝜃 𝐿

𝜙

(2)

and the position of 𝐶 relative to 𝐵 is given by

sin 𝜙 =

𝐿

cos 𝜙 =

𝐿2 − 𝑅2 cos2 𝜃 . 𝐿

(4)

Since 𝐵 is in fixed-axis rotation about 𝐴 with 𝜔𝐴𝐵 = 𝜃̇ = constant, its acceleration is given by (see p. 1060) 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 = −𝑅𝜔2𝐴𝐵 (cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂),

Figure 2 A slider-crank mechanism showing the relevant dimensions.

(5)

where we have used 𝛼𝐴𝐵 = 𝜃̈ = 0. Substituting Eqs. (3)–(5) into Eq. (1) and simplifying, we obtain ] [ √ ( ) 𝑎⃗𝐶 = 𝑅 cos 𝜃 𝜔2𝐵𝐶 − 𝜔2𝐴𝐵 − 𝛼𝐵𝐶 (𝐿∕𝑅)2 − cos2 𝜃 𝚤̂ ] [ √ (6) − 𝑅 𝛼𝐵𝐶 cos 𝜃 + 𝜔2𝐵𝐶 (𝐿∕𝑅)2 − cos2 𝜃 + 𝜔2𝐴𝐵 sin 𝜃 𝚥̂. Enforcing the condition 𝑎𝐶𝑥 = 0 in Eq. (6) and solving for 𝛼𝐵𝐶 yields ) ( cos 𝜃 𝜔2𝐵𝐶 − 𝜔2𝐴𝐵 𝛼𝐵𝐶 = √ . (𝐿∕𝑅)2 − cos2 𝜃

Helpful Information (7)

Substituting Eq. (2) into Eq. (7) and simplifying gives 𝛼𝐵𝐶 = [

[ ] 1 − (𝐿∕𝑅)2 cos 𝜃 2 ]3∕2 𝜔𝐴𝐵 (𝐿∕𝑅)2 − cos2 𝜃

[

⇒

𝛼⃗𝐵𝐶 = [

] 1 − (𝐿∕𝑅)2 cos 𝜃 2 ̂ ]3∕2 𝜔𝐴𝐵 𝑘, (𝐿∕𝑅)2 − cos2 𝜃

(8)

Another way to compute 𝜶𝑩𝑪 . The method used to obtain Eqs. (8) was laborious, but it only requires a series of algebraic steps. We could have obtained 𝛼𝐵𝐶 by differentiating 𝜔𝐵𝐶 in Eq. (2) with respect to time (this is true here since the motion is planar).

1104

Chapter 16

Planar Rigid Body Kinematics

̂ Substituting Eqs. (2) and (8) into Eq. (6), we obtain where we used 𝛼⃗𝐵𝐶 = 𝛼𝐵𝐶 𝑘.

Common Pitfall

} ] 1 − (𝐿∕𝑅)2 cos2 𝜃 sin2 𝜃 + sin 𝜃 𝚥̂, [ ]3∕2 + √ (𝐿∕𝑅)2 − cos2 𝜃 (𝐿∕𝑅)2 − cos2 𝜃

{[ 𝑎⃗𝐶 =

−𝑅𝜔2𝐴𝐵

𝑎𝐶𝑦 ∕𝑔

400 200 0 −200 −400 −600 −800

Discussion & Verification

The method we illustrated here,( i.e., based on ) the application of the (general) equation 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , while at times laborious, only involves a series of algebraic steps, as opposed to computing accelerations directly through differentiation with respect to time. There are many situations in which it is indeed simpler to differentiate with respect to time than it is to apply the acceleration formula for a rigid body. This strategy for the calculation of accelerations will be demonstrated in Example 16.11 on p. 1105. A Closer Look We can now finish the parametric analysis begun in Example 16.6 ⇒ We plot the angular acceleration of the CR 𝛼𝐵𝐶 , as well as the accelon p. 1083. eration of the piston 𝑎𝐶𝑦 , for the same conditions considered in Example 16.11. What is remarkable is the magnitude of the accelerations of the CR. Referring to Figs. 3 and 4, we

0

90

180

270

360

Figure 4 The piston’s acceleration for 𝜃̇ = 3500 rpm, 𝐿 = 150 mm, and three values of 𝐿∕𝑅.

ISTUDY

4 2 0 −2 −4 0

𝐿∕𝑅 = 3.0 𝐿∕𝑅 = 3.3 𝐿∕𝑅 = 3.6

(9)

where we have used 𝑎𝐶𝑥 = 0.

𝛼𝐵𝐶 ∕104 (rad∕s2 )

Finding velocities and accelerations by time differentiation. We mentioned that we could have computed Eqs. (8) by time differentiating Eq. (2). In doing this, a common mistake is to take the derivative at a particular instant rather than the general function of time. For example, suppose that we had computed 𝜔𝐵𝐶 at a particular instant 𝑡0 to be 𝜔𝐵𝐶 (𝑡0 ) = 1234 rad/s. We shouldn’t say that 𝛼𝐵𝐶 (𝑡0 ) = 0 because the time derivative of the number 1234 rad/s equals zero. While it is true that the time derivative of a constant is equal to zero, we must first take a time derivative of the function 𝜔𝐵𝐶 (𝑡) and then evaluate the result at the time of interest 𝑡0 . This concept is illustrated in Example 16.11.

90

𝐿∕𝑅 = 3.0 𝐿∕𝑅 = 3.3 𝐿∕𝑅 = 3.6 180 270

360

Figure 3. The angular acceleration of the CR for 𝜃̇ = 3500 rpm, 𝐿 = 150 mm, and three values of 𝐿∕𝑅.

see that, for a value of 𝜃̇ = 3500 rpm, the angular acceleration reaches values in excess of 40,000 rad∕s2 and the acceleration of the piston reaches values that are 900 times the acceleration of gravity 𝑔! Since the accelerations in question are proportional to 𝜃̇ 2 , if the engine’s angular velocity is increased by a factor of, for example, 2, the accelerations we plotted increase by a factor of 4! Thus, for an engine running at 7000 rpm, the piston’s accelerations can easily reach 3600𝑔. In Chapter 17, we will learn how to use this information to compute the forces and moments that a mechanism, such as the slider-crank, must be able to sustain. We will then understand why connecting rods are typically made of high-grade steel and have a cross-section that looks like an I-beam. As a final remark on Fig. 4, notice that the piston’s acceleration has two different behaviors: one for 0◦ ≤ 𝜃 < 180◦ and another for 180◦ ≤ 𝜃 ≤ 360◦ . We already observed this lack of symmetry during the velocity analysis, and we see now that it is even more pronounced in the acceleration behavior. It is the geometry of the mechanism that generates this lack of symmetry, and the behavior becomes more symmetric as the ratio 𝐿∕𝑅 is increased. ⇐

ISTUDY

Section 16.3

E X A M P L E 16.11

Motion of a Propped Ladder: Differentiation of Constraints

A person is propping up a ladder against a wall by pushing the end 𝐴 to the right along the ground (see Fig. 1). If, at the instant shown, the speed of 𝐴 is constant and equal to 𝑣𝐴 = 0.8 m∕s, the length 𝐿 = 6 m, the height 𝐻 = 4 m, and the distance 𝑑 = 1.57 m, determine the angular velocity and angular acceleration of the ladder. In addition, determine the acceleration of the midpoint of the ladder 𝐶 and the acceleration of the point 𝑃 on the ladder that, at the given instant, is in contact with the wall.

𝑃

𝐿 𝐶

𝐻 𝑣𝐴

SOLUTION

𝐴

Road Map Referring to Fig. 2, we can use 𝜃 to describe the ladder’s orientation so that we can write 𝜔 ⃗ 𝐿 = −𝜃̇ 𝑘̂ and 𝛼⃗𝐿 = −𝜃̈ 𝑘̂ (the minus signs are needed to reconcile the positive direction of 𝜃 with the component system used). Thus, we will need to relate 𝜃̇ and 𝜃̈ to the given data (𝑣𝐴 , 𝐻, and 𝑑). This can be done by differentiating the constraint relating 𝜃 to the position of 𝐴 with respect to time. We will compute 𝑎⃗𝐶 using differentiation of constraints and 𝑎⃗𝑃 using the vector approach to relate 𝑎⃗𝑃 to the acceleration of a known point on the ladder. Computation

1105

Planar Motion: Acceleration Analysis

Figure 1

𝑃 𝐿 𝐶

Looking at the triangle 𝐴𝑂𝑃 in Fig. 2, during the ladder’s motion we

have 𝑥𝐴 tan 𝜃 = 𝐻

𝑥̇ 𝐴 tan 𝜃 + 𝑥𝐴 𝜃̇ sec2 𝜃 = 0,

⇒

(1)

where the second equation was obtained by differentiating the first with respect to time. By using sec2 𝜃 = 1 + tan2 𝜃, tan 𝜃 = 𝐻∕𝑥𝐴 , and 𝑥̇ 𝐴 = −𝑣𝐴 , the second of Eqs. (1) can be solved for 𝜃̇ to obtain 𝜃̇ =

𝑣𝐴 𝐻 𝑥2𝐴

⇒

+ 𝐻2

̇ ) = 0.1733 rad∕s, 𝜃(𝑡 0

(2)

𝑥

𝜃

𝐻

𝑣𝐴

𝐴

𝑂

Figure 2 Coordinate system for the ladder kinematics. Note that for the coordinate system defined here, 𝑘̂ is positive into the page, so counterclockwise angular velocity and acceleration are negative.

̇ we can now obtain where 𝑡0 is the time at which 𝑥𝐴 = 𝑑 = 1.57 m. Having computed 𝜃, ̈𝜃 by time differentiating the first of Eqs. (2) with respect to time to obtain 𝜃̈ = (

2𝑥𝐴 𝑣2𝐴 𝐻 )2 𝑥2𝐴 + 𝐻 2

⇒

̈ ) = 0.02358 rad∕s2 , 𝜃(𝑡 0

(3)

̂ Eqs. (2) where we used the fact that 𝑣𝐴 is constant. Since 𝜔 ⃗ 𝐿 = −𝜃̇ 𝑘̂ and 𝛼⃗𝐿 = −𝜃̈ 𝑘, and (3) imply that 𝜔 ⃗ 𝐿 (𝑡0 ) = −0.1733 𝑘̂ rad∕s, 𝛼⃗ (𝑡 ) = −0.02358 𝑘̂ rad∕s2 . 𝐿 0

(4) 𝑟⃗𝑃 ∕𝐴

(5)

We now compute 𝑎⃗𝐶 (𝑡0 ) by differentiation of constraints. This means that we need to take two time derivatives of the general constraint equations for the position of 𝐶. Referring to Fig. 3, we can write ) ( 𝐿 𝐿 𝑟⃗𝐶 = 𝑥𝐴 − cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂. 2 2

(6)

( ) 𝐿 𝐿 𝑣⃗𝐶 = −𝑣𝐴 + 𝜃̇ sin 𝜃 𝚤̂ + 𝜃̇ cos 𝜃 𝚥̂, 2 2 and 𝑎⃗𝐶 =

) ( ) ] 𝐿 [( ̈ 𝜃 sin 𝜃 + 𝜃̇ 2 cos 𝜃 𝚤̂ + 𝜃̈ cos 𝜃 − 𝜃̇ 2 sin 𝜃 𝚥̂ , 2

𝐶 𝐿∕2

𝑥

𝑟⃗𝐶 𝜃(𝑡0 )

𝐴

Figure 3 Definition of 𝑟⃗𝐶 and 𝑟⃗𝑃 ∕𝐴 .

Taking one and then two time derivatives of Eq. (6), we obtain (7)

(8)

𝑃

𝑂

1106

ISTUDY

Chapter 16

Planar Rigid Body Kinematics

where we used the fact that 𝑣𝐴 is constant. We now find 𝜃(𝑡0 ) by substituting 𝑥𝐴 (𝑡0 ) = 𝑑 = 1.57 m and 𝐻 = 4 m into the first of Eqs. (1) to obtain 𝜃(𝑡0 ) = tan−1 (4∕1.57) = 68.57◦ .

(9)

Substituting the results in Eqs. (2), (3), and (9) into Eq. (8), we obtain ( ) 𝑎⃗𝐶 (𝑡0 ) = 0.09876 𝚤̂ − 0.05803 𝚥̂ m∕s2 .

(10)

Now we compute 𝑎⃗𝑃 by using the vector method to relate the acceleration of 𝑃 to that of 𝐴. Choosing 𝐴 as a reference point is convenient because 𝑎⃗𝐴 = 0⃗ (𝐴 is moving at a constant velocity). Since we need to compute 𝑎⃗𝑃 at the instant shown, we can write 𝑎⃗𝑃 (𝑡0 ) = 𝛼⃗𝐿 (𝑡0 ) × 𝑟⃗𝑃 ∕𝐴 (𝑡0 ) − 𝜔2𝐿 (𝑡0 )⃗𝑟𝑃 ∕𝐴 (𝑡0 ).

(11)

Referring to Fig. 3, at time 𝑡0 we have 𝑟⃗𝑃 ∕𝐴 (𝑡0 ) = −𝑑 𝚤̂ + 𝑑 tan 𝜃(𝑡0 ) 𝚥̂.

(12)

Substituting Eq. (12), along with the results in Eqs. (4), (5), and (9), into Eq. (11), we have ( ) 𝑎⃗𝑃 (𝑡0 ) = 0.1415 𝚤̂ − 0.08312 𝚥̂ m∕s2 .

(13)

Discussion & Verification

The dimensions and therefore the units of all of our results are as they should be. The signs for the angular velocity and acceleration of the ladder match our expectation, given that the ladder is rotating counterclockwise and the 𝑘̂ direction is into the page. The values of acceleration are harder to verify without using an alternative solution strategy.

A Closer Look Without actually performing the calculations, if we used the formula 𝑎⃗𝐶 = 𝛼⃗𝐿 × 𝑟⃗𝐶∕𝐴 − 𝜔2𝐿 𝑟⃗𝐶∕𝐴 , we would readily see that the vectors 𝛼⃗𝐿 × 𝑟⃗𝐶∕𝐴 and −𝜔2𝐿 𝑟⃗𝐶∕𝐴 have positive 𝑥 components, thus matching the fact that we obtained a positive value for 𝑎𝐶𝑥 (𝑡0 ). As far as their 𝑦 components are concerned, we would find that the terms 𝛼⃗𝐿 ×⃗𝑟𝐶∕𝐴 and −𝜔2𝐿 𝑟⃗𝐶∕𝐴 have positive and negative 𝑦 components, respectively. However, given that 𝜔𝐿 is larger than 𝛼𝐿 in absolute value, we expect that the term with 𝜔2𝐿 would dominate with respect to the 𝛼𝐿 term. Therefore, overall we expect 𝑎𝐶𝑦 (𝑡0 ) to be negative, which is exactly what we found. A similar logic can be applied to the discussion of the 𝑎⃗𝑃 result. Again, this example is meant to show that the techniques we learned in Chapter 12 are still relevant to the study of rigid bodies and can be used together with the vector method discussed in this chapter.

ISTUDY

Section 16.3

Planar Motion: Acceleration Analysis

E X A M P L E 16.12

1107

Acceleration Analysis of a Four-Bar Linkage: Vector Approach

We now continue the kinematic analysis of a prosthetic leg with an artificial knee joint presented in Example 16.8 on p. 1087 (see Fig. 1). The primary kinematic component of the artificial knee joint shown is the four-bar linkage system highlighted in Fig. 2. Since the determination of accelerations is crucial in the determination of the forces and moments acting on a mechanism, we will determine the angular accelerations of each link in the system in Fig. 2, at the instant shown, assuming that the 𝐴𝐷 segment is fixed. We will also determine the acceleration of point 𝑃 , which is the midpoint of the link 𝐵𝐶. As was done in Example 16.8, we will use the coordinates of points 𝐴, 𝐵, 𝐶, and 𝐷 given in Table 1. The segment 𝐴𝐵 is rotating counterclockwise at 1.5 rad∕s, and that rate is decreasing at 0.8 rad∕s2 . Table 1. Approximate values of the coordinates of the pin centers 𝐴, 𝐵, 𝐶, and 𝐷 for the system shown in Fig. 2 at the time instant considered. Points Coordinates (mm)

𝐴

𝐵

𝐶

𝐷

(0.0, 0.0)

(−27.0, 120)

(26.0, 124)

(30.0, 15.0)

Courtesy of Otto Bock HealthCare, Germany

Figure 1

SOLUTION

𝑥 𝐴

Road Map

We did the velocity analysis for this linkage in Example 16.8 on p. 1087, and we will use the angular velocities found there, which were 𝜔𝐵𝐶 = −0.6378 rad∕s and 𝜔𝐶𝐷 = −1.675 rad∕s. The calculation of accelerations is similar to the velocity analysis; that is, we will compute the acceleration of 𝐵 and then the acceleration of 𝐶. Because point 𝐶 is shared by both links 𝐵𝐶 and 𝐶𝐷, we will obtain two independent statements for 𝑎⃗𝐶 , which we will then require to be equal. This will allow us to obtain two equations for the angular accelerations of the links 𝐵𝐶 and 𝐶𝐷. Once the angular accelerations are known, then we will compute the acceleration of point 𝑃 . Computation

Recalling that 𝐴 is fixed, 𝑎⃗𝐵 is given by

𝑎⃗𝐵 = 𝛼𝐴𝐵 𝑘̂ × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 = −(35.25 𝚤̂ + 291.6 𝚥̂) mm∕s2 ,

(1)

𝚤̂

𝐷

𝚥̂

𝐵

𝑃

𝐶

where we set 𝛼𝐴𝐵 = 0.8 rad∕s2 and used 𝑟⃗𝐵∕𝐴 = (−27 𝚤̂ + 120 𝚥̂) mm from Table 1. Next, since 𝐶 is shared by both segments 𝐵𝐶 and 𝐶𝐷, we can express 𝑎⃗𝐶 in the following two independent ways: 𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼𝐵𝐶 𝑘̂ × 𝑟⃗𝐶∕𝐵 − 𝜔2𝐵𝐶 𝑟⃗𝐶∕𝐵 ,

(2)

𝑎⃗𝐶 = 𝑎⃗𝐷 + 𝛼𝐶𝐷 𝑘̂ × 𝑟⃗𝐶∕𝐷 − 𝜔2𝐶𝐷 𝑟⃗𝐶∕𝐷 ,

(3)

𝑟⃗𝐶∕𝐵 = 𝑟⃗𝐶 − 𝑟⃗𝐵 = (53.00 𝚤̂ + 4.000 𝚥̂) mm,

(4)

𝑟⃗𝐶∕𝐷 = 𝑟⃗𝐶 − 𝑟⃗𝐷 = (−4.000 𝚤̂ + 109.0 𝚥̂) mm,

(5)

and where

⃗ Substituting Eqs. (1), (4), (5), and the known angular velocities into Eqs. (2) and 𝑎⃗𝐷 = 0. and (3) and setting two expressions for 𝑎⃗𝐶 equal to one another, we obtain the following vector equation: [ ] [ ] mm + (4.000 mm)𝛼 + (53.00 mm)𝛼 − 56.81 mm 𝚤 ̂ + −293.2 𝚥̂ 2 2 𝐵𝐶 𝐵𝐶 s s [ ] [ ] mm 𝚤 ̂ − 305.7 − (109.0 mm)𝛼 + (4.000 mm)𝛼 = 11.22 mm 𝐶𝐷 𝐶𝐷 𝚥̂. s2 s2

(6)

Figure 2 Geometry of the four-bar linkage in the prosthetic leg. Note that for the coordinate system defined here, 𝑘̂ is positive into the page, so counterclockwise angular velocity and acceleration are negative.

1108

ISTUDY

Chapter 16

Planar Rigid Body Kinematics

Equating 𝚤̂ components and equating 𝚥̂ components yield the linear system of two equations −56.81 mm − (4.000 mm)𝛼𝐵𝐶 = 11.22 mm − (109.0 mm)𝛼𝐶𝐷 , s2 s2 −293.2

mm s2

+ (53.00 mm)𝛼𝐵𝐶 = −305.7

mm s2

− (4.000 mm)𝛼𝐶𝐷 ,

(7) (8)

in the two unknowns 𝛼𝐵𝐶 and 𝛼𝐶𝐷 , whose solution is 𝛼𝐵𝐶 = −0.2823 rad∕s2

and 𝛼𝐶𝐷 = 0.6137 rad∕s2 .

(9)

Now that the angular accelerations are known, we can find 𝑎⃗𝑃 by using 𝑎⃗𝑃 = 𝑎⃗𝐵 + 𝛼𝐵𝐶 𝑘̂ × 𝑟⃗𝑃 ∕𝐵 − 𝜔2𝐵𝐶 𝑟⃗𝑃 ∕𝐵 .

(10)

Since 𝑃 is the midpoint between 𝐵 and 𝐶, we have 𝑟⃗𝑃 =

𝑟⃗𝐵 + 𝑟⃗𝐶 2

⇒

𝑟⃗𝑃 ∕𝐵 = 𝑟⃗𝑃 − 𝑟⃗𝐵 =

𝑟⃗𝐶 − 𝑟⃗𝐵 2

= (26.50 𝚤̂ + 2.000 𝚥̂) mm.

(11)

Using Eqs. (1), (9), (11), and the previously computed value of 𝜔𝐵𝐶 , from Eq. (10) we obtain 𝑎⃗𝑃 = −(45.46 𝚤̂ + 299.9 𝚥̂) mm∕s2 . (12) Discussion & Verification

As a first check, we see that the dimensions, and thus the units of all results, are as they should be. Another way to argue that the results we obtained are reasonable is to observe that, in the position shown, this four-bar linkage is such that links 𝐴𝐵 and 𝐵𝐶 are nearly parallel to each other, while link 𝐵𝐶 is oriented such that point 𝐶 has a larger 𝑦 coordinate than point 𝐵. Thus, in the position shown, the behavior of this four-bar linkage should not be that different from a similarly sized parallelogram. Therefore, in the position shown, we would expect the angular acceleration of 𝐶𝐷 to have the same sign as 𝛼𝐴𝐵 . By the same token, we would expect the angular acceleration of 𝐵𝐶 to have a sign opposite to that of 𝐴𝐵. This is exactly what we obtained. However, obtaining an intuitive understanding of the signs and/or magnitudes of accelerations is not as easy as for velocities. Therefore, when it comes to accelerations, double-checking our calculations is more important than having an intuitive understanding of the mechanism’s motion.

ISTUDY

Section 16.3

1109

Planar Motion: Acceleration Analysis

Problems Problem 16.88 𝑦

A truck on an exit ramp is moving in such a way that, at the instant shown, |𝑎⃗𝐴 | = 17 f t∕s2 , 𝜃̇ = −0.3 rad∕s, and 𝜃̈ = −0.1 rad∕s2 . If the distance between points 𝐴 and 𝐵 is 𝑑𝐴𝐵 = 12 f t, 𝜃 = 57◦ , and 𝜙 = 13◦ , determine 𝑎⃗𝐵 .

𝑟⃗𝐵∕𝐴

Problems 16.89 through 16.91

𝐵 𝜃

𝚥̂

Let 𝐿 = 4 f t, let point 𝐴 travel parallel to the guide shown, and let 𝐶 be the midpoint of the bar. If point 𝐴 is accelerating to the right with 𝑎𝐴 = 27 f t∕s2 and 𝜃̇ = 7 rad∕s = constant, determine the acceleration of point 𝐶 when 𝜃 = 24◦ . Problem 16.89

𝐴

𝚤̂

𝑥 𝜙 𝑎⃗ 𝐴

Figure P16.88

Problem 16.90 If point 𝐴 is accelerating to the right with 𝑎𝐴 = 27 f t∕s2 , 𝜃̇ = 7 rad∕s, and 𝜃̈ = −0.45 rad∕s2 , determine the acceleration of point 𝐶 when 𝜃 = 26◦ . 𝐴

If, when 𝜃 = 0◦ , 𝐴 is accelerating to the right with 𝑎𝐴 = 27 f t∕s2 and ⃗ determine 𝜃̇ and 𝜃. ̈ 𝑎⃗𝐶 = 0, Problem 16.91

𝐶 𝜃

Problem 16.92 A wheel 𝑊 of radius 𝑅𝑊 = 5 cm rolls without slip over the stationary cylinder 𝑆 of radius 𝑅𝑆 = 12 cm, and the wheel is connected to point 𝑂 via the arm 𝑂𝐶. If 𝜔𝑂𝐶 = constant = 3.5 rad∕s, determine the acceleration of point 𝑄, which lies on the edge of 𝑊 and along the extension of the line 𝑂𝐶.

𝐿

𝚥̂

𝐵

𝚤̂ Figure P16.89–P16.91

𝑊 𝜔𝑂𝐶 𝑢̂ 𝑟

𝑢̂ 𝜃

𝑅𝑊

𝜔𝑊

𝜃

𝑅𝑆 𝑆 Figure P16.92

Problems 16.93 through 16.96 In the mechanism shown, the block 𝐵 is constrained to move vertically and is attached to the bar 𝐵𝐷. The point 𝐴 on the bar 𝐴𝐷 is fixed. Express all your answers in the component system shown. At the instant shown, the block 𝐵 is moving downward at a constant speed of 2.5 f t∕s, 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 12 in. and 𝑑 = 8 in., determine the angular acceleration of bar 𝐴𝐷 at this instant.

𝚥̂

𝐴

𝐶

Problem 16.93

Problem 16.94 At the instant shown, bar 𝐴𝐷 is rotating counterclockwise at a constant angular speed 𝜔𝐴𝐷 = 13 rad∕s, 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 24 in. and 𝑑 = 16 in., determine the acceleration of the block 𝐵 at this instant.

𝚤̂

𝜙

𝜃 𝓁

𝐷

Figure P16.93–P16.96

𝑑

1110

Chapter 16

Planar Rigid Body Kinematics

At the instant shown, block 𝐵 is moving downward at 1.5 m∕s and is increasing at 3 m∕s2 , 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 1.2 m and 𝑑 = 0.8 m, determine the angular acceleration of bar 𝐴𝐷 and the acceleration of point 𝐶 at this instant. Problem 16.95

At the instant shown, bar 𝐴𝐷 is rotating counterclockwise with angular speed 𝜔𝐴𝐷 = 30 rad∕s, which is decreasing at 4 rad∕s2 , 𝜙 = 45◦ , and 𝜃 = 30◦ . If 𝓁 = 24 cm and 𝑑 = 16 cm, determine the acceleration of the block 𝐵 at this instant. Problem 16.96

yoke 𝑣𝐵

𝐵

Problem 16.97 𝐴

𝜃

slider

𝑅 𝐶

𝚥̂

𝜔𝐶

𝚤̂

Figure P16.97

One way to convert rotational motion into linear motion and vice versa is by the use of a mechanism called the Scotch yoke, which consists of a crank 𝐶 that is connected to a slider 𝐵 by a pin 𝐴. The crank 𝐶 is mounted to a fixed bearing. The pin rotates with the crank while sliding within the yoke, which, in turn, rigidly translates with the slider. This mechanism has been used, for example, to control the opening and closing of valves in pipelines. Letting the radius of the crank be 𝑅 = 25 cm, determine the angular velocity 𝜔 ⃗ 𝐶 and the angular acceleration 𝛼⃗𝐶 of the crank at the instant shown if 𝜃 = 25◦ and the slider is moving to the right with a constant speed 𝑣𝐵 = 40 m∕s.

Problem 16.98 Collar 𝐶 moves along a circular guide with radius 𝑅 = 2 f t with a constant speed 𝑣𝐶 = 18 f t∕s. At the instant shown, the bars 𝐴𝐵 and 𝐵𝐶 are vertical and horizontal, respectively. Letting 𝐿 = 4 f t and 𝐻 = 5 f t, determine the angular accelerations of the bars 𝐴𝐵 and 𝐵𝐶 at this instant. 𝐿

𝑅 𝑂

𝐵

𝐶 𝑣𝐶

𝐻 𝚥̂ 𝚤̂

𝐴 Figure P16.98

Problems 16.99 and 16.100

𝑅𝑆 𝑃

𝐶

𝜃 𝑆

𝐵

𝑢̂ 𝑟 𝑢̂ 𝜃

𝑅𝐵

A sphere 𝑆 of radius 𝑅𝑆 = 5 in. is rolling without slip inside a stationary spherical bowl 𝐵 of radius 𝑅𝐵 = 17 in. Assume that the motion of the sphere is planar. The center of the sphere is at 𝐶, and the point of contact between the sphere and the bowl is at 𝑃 . If, at the instant shown, the center of the sphere 𝐶 is traveling counterclockwise with a speed 𝑣𝐶 = 32 f t∕s and such that 𝑣̇ 𝐶 = 0, determine the acceleration of 𝐶, as well as the acceleration of point 𝑃 , which is the point on the sphere that is in contact with the bowl at this instant. Problem 16.99

Figure P16.99 and P16.100

ISTUDY

If, at the instant shown, the center of the sphere 𝐶 is traveling counterclockwise with a speed 𝑣𝐶 = 32 f t∕s and such that 𝑣̇ 𝐶 = 24 f t∕s2 , determine the acceleration of 𝐶, as well as the acceleration of 𝑃 , which is the point on the sphere that is in contact with the bowl at this instant. Problem 16.100

ISTUDY

Section 16.3

1111

Planar Motion: Acceleration Analysis

Problem 16.101 𝑦

A truck on an exit ramp is moving in such a way that, at the instant shown, |𝑎⃗𝐴 | = 6 m∕s2 and 𝜙 = 13◦ . Let the distance between points 𝐴 and 𝐵 be 𝑑𝐴𝐵 = 4 m. If, at this instant, the truck is turning clockwise, 𝜃 = 59◦ , 𝑎𝐵𝑥 = 6.3 m∕s2 , and 𝑎𝐵𝑦 = −2.6 m∕s2 , determine the angular velocity and angular acceleration of the truck.

𝑟⃗𝐵∕𝐴

𝐵 𝜃

Problems 16.102 and 16.103

𝚥̂

As the circular cam whose center is at 𝐴 rotates, it causes the follower 𝐵 to move back and forth. The cam angle is 𝜃, the radius of the cam is 𝑅, the angular speed of the cam is 𝜃̇ = 𝜔𝑂𝐴 , and the angular acceleration of the cam is 𝜃̈ = 𝛼𝑂𝐴 . The cam is pin-connected to the fixed point 𝑂. 𝛼𝑂𝐴

𝜙 𝑎⃗ 𝐴

Figure P16.101

𝑅 2 𝑅 3

𝜔𝑂𝐴

𝑥

𝐴

𝚤̂

𝑂

𝐴 𝜃

𝐵 follower

Figure P16.102 and P16.103

Using the given 𝑥 coordinate, determine the acceleration of the follower at the instant 𝜃 = 30◦ if 𝑅 = 1.5 in., 𝜃̇ = 𝜔𝑂𝐴 = 1000 rpm, and 𝜃̈ = 𝛼𝑂𝐴 = 25 rad∕s2 . Problem 16.102

Problem 16.103 Using the given 𝑥 coordinate, determine the acceleration of the follower as a function of the cam angle 𝜃, the radius of the cam 𝑅, the given angular speed of the cam 𝜔𝑂𝐴 , and the given angular acceleration of the cam 𝛼𝑂𝐴 .

Problem 16.104 A bar of length 𝐿 = 2.5 m is falling so that, when 𝜃 = 34◦ , 𝑣𝐴 = 3 m∕s and 𝑎𝐴 = 8.7 m∕s2 . At this instant, determine the angular acceleration of the bar 𝐴𝐵 and the acceleration of point 𝐷, where 𝐷 is the midpoint of the bar.

𝐴 𝑎𝐴 𝜃

𝐷 𝐿

Problem 16.105 𝚥̂

A bar of length 𝐿 = 8 f t and midpoint 𝐷 is falling so that, when 𝜃 = 27◦ , |𝑣⃗𝐷 | = 18 f t∕s, and the vertical acceleration of point 𝐷 is 23 f t∕s2 downward. At this instant, compute the angular acceleration of the bar and the acceleration of point 𝐵.

𝐵 Figure P16.104–P16.106

Problem 16.106 Assuming that, for 0◦ ≤ 𝜃 ≤ 90◦ , 𝑣𝐴 is constant, compute the expression for the acceleration of point 𝐷, the midpoint of the bar, as a function of 𝜃 and 𝑣𝐴 .

𝚥̂ 3𝑅

Problem 16.107 The wheel 𝑊 of radius 𝑅 is rolling without slip with angular speed 𝜔𝑊 and angular acceleration 𝛼𝑊 . The bar 𝐴𝐵 has length 3𝑅 and is pinned to the outer edge of the wheel at 𝐴; the end 𝐵 remains in contact with the surface. At the instant shown, determine the acceleration of point 𝐵. Express your answer in the component system shown.

𝜔𝑊

𝑊

𝚤̂

𝑂 𝑅

𝐵 Figure P16.107

𝛼𝑊

1112

Chapter 16

Planar Rigid Body Kinematics

Problem 16.108 At the instant shown, point 𝐵 is moving with speed 𝑣𝐵 and acceleration 𝑎𝐵 . The bar 𝐴𝐵 has length 3𝑅; it is pinned to the outer edge of the wheel at 𝐴, and the end 𝐵 remains in contact with the surface. If the wheel rolls without slip, determine the angular velocity 𝜔 ⃗𝑊 and angular acceleration 𝛼⃗𝑊 of the wheel at the instant shown. Express all your answers in the component system shown. 𝚥̂

𝚥̂

𝑊

𝚤̂ 3𝑅

𝜔𝑊

𝑊

𝚤̂

𝑅

𝑂

𝑂

𝑣𝐵

𝐴 𝛼𝑊

𝑅 𝑎𝐵

3𝑅

𝐵

Figure P16.108

Figure P16.109

Problem 16.109 The wheel 𝑊 of radius 𝑅 is rolling without slip with angular speed 𝜔𝑊 and angular acceleration 𝛼𝑊 . The bar 𝐴𝐵 has length 3𝑅 and is pinned to the outer edge of the wheel at 𝐴, and the end 𝐵 remains in contact with the surface. At the instant shown, determine the acceleration of point 𝐵. Express your answer in the component system shown.

Problem 16.110 At the instant shown, point 𝐵 is moving with speed 𝑣𝐵 and acceleration 𝑎𝐵 . The bar 𝐴𝐵 has length 3𝑅 and is pinned to the outer edge of the wheel at 𝐴, and the end 𝐵 remains in contact with the surface. If the wheel rolls without slip, determine the angular velocity 𝜔 ⃗𝑊 and angular acceleration 𝛼⃗𝑊 of the wheel at the instant shown. Express all your answers in the component system shown. 𝚥̂

𝑣𝐵

𝚤̂

𝑅

𝑂

𝐴

3𝑅

𝐵 𝑎𝐵

Figure P16.110

Problems 16.111 through 16.113 𝐴

The wheel 𝑊 of radius 𝑅 = 1.4 m rolls without slip on a horizontal surface. A bar 𝐴𝐵 of length 𝐿 = 3.7 m is pin-connected to the center of the wheel and to a slider 𝐴 constrained to move along a vertical guide. Point 𝐶 is the bar’s midpoint.

𝐿 𝐶

Problem 16.111

If the wheel is rolling clockwise with a constant angular speed of 2 rad∕s, determine the angular acceleration of the bar when 𝜃 = 72◦ .

𝑊

𝜃 𝐵

𝑅

𝚥̂

Problem 16.112 If the slider 𝐴 is moving downward with a constant speed 3 m∕s, determine the angular acceleration of the wheel when 𝜃 = 53◦ . 𝚤̂ Problem 16.113

Figure P16.111–P16.113

ISTUDY

Determine the general relation expressing the acceleration of the slider 𝐴 as a function of 𝜃, 𝐿, 𝑅, the angular velocity of the wheel 𝜔𝑊 , and the angular acceleration of the wheel 𝛼𝑊 .

ISTUDY

Section 16.3

1113

Planar Motion: Acceleration Analysis

Problem 16.114 A spool with inner radius 𝑅 = 5 f t is made to roll without slip over a horizontal rail as shown. If the cable on the spool is unwound in such a way that the free or vertical portion of cable remains perpendicular to the rail, determine the angular acceleration of the spool and the acceleration of the spool’s center 𝑂. The vertical component of the velocity of point 𝐴 is 𝑣𝐴 = 12 f t∕s, and the vertical component of its acceleration is 𝑎𝐴 = 2 f t∕s2 .

𝑣𝐴

𝑎𝐴

𝐴 𝚥̂ 𝚤̂

𝑅

𝑂

Problems 16.115 through 16.118 For the slider-crank mechanism shown, let 𝑅 = 0.75 m and 𝐻 = 2 m, and let the length of bar 𝐵𝐶 be 𝐿𝐵𝐶 = 3.25 m. Hint: For the problems below, differentiation of constraints ̇ is recommended. Consult Example 16.7 for an expression for 𝜙.

Figure P16.114

𝐵 𝑅 𝑦

𝐴

𝜃

𝜙

𝑂

𝐶 Figure P16.115–P16.118 Problem 16.115 Assume that 𝜃̇ = 50 rad∕s = constant and compute the angular acceleration of the slider for 𝜃 = 27◦ .

Assume that, at the instant shown, 𝜃 = 27◦ , 𝜃̇ = 50 rad∕s, and 𝜃̈ = Compute the angular acceleration of the slider at this instant, as well as the acceleration of point 𝐶.

Problem 16.116

15 rad∕s2 .

Assuming that 𝜃̇ is constant, determine the expression of the angular acceleration of the slider as a function of 𝜃 and 𝜃̇ (and the accompanying geometrical parameters).

Problem 16.117

Letting 𝜃̇ = 300 rpm = constant, plot the angular acceleration of the slider as a function of 𝜃 for 0◦ ≤ 𝜃 ≤ 360◦ . In addition, plot the speed of point 𝐶 for the same range of 𝜃. Problem 16.118

Problems 16.119 through 16.121 A wheel 𝑊 of radius 𝑅𝑊 = 2 in. rolls without slip over the stationary cylinder 𝑆 of radius 𝑅𝑆 = 5 in., and the wheel is connected to point 𝑂 via the arm 𝑂𝐶.

𝛼𝑊 𝜔𝑂𝐶

Determine the acceleration of the point on the wheel 𝑊 that is in contact with 𝑆 for 𝜔𝑂𝐶 = 7.5 rad∕s and 𝛼𝑂𝐶 = 2 rad∕s2 .

If, at the instant shown, 𝜃 = 63◦ , 𝜔𝑊 = 9 rad∕s, and 𝛼𝑊 = −1.3 rad∕s2 , determine the angular acceleration of the arm 𝑂𝐶 and the acceleration of point 𝑃 , where 𝑃 lies on the edge of 𝑊 and is aligned vertically with point 𝐶.

𝑢̂ 𝜃 𝑅𝑆

𝑢̂ 𝑟

𝐶 𝑅𝑊

𝛼𝑂𝐶

Determine the acceleration of the point on the wheel 𝑊 that is in contact with 𝑆 for 𝜔𝑂𝐶 = 7.5 rad∕s = constant.

Problem 16.119

Problem 16.120

𝑊

𝜃

𝑂 𝑆

Problem 16.121

Figure P16.119–P16.121

𝜔𝑊

1114

Chapter 16

Planar Rigid Body Kinematics

Problem 16.122 𝐴

At the instant shown, bars 𝐴𝐵 and 𝐵𝐶 are perpendicular to each other while the slider 𝐶 has a velocity 𝑣𝐶 = 24 m∕s and an acceleration 𝑎𝐶 = 2.5 m∕s2 in the directions shown. If 𝐿 = 1.75 m and 𝜃 = 45◦ , determine the angular acceleration of bars 𝐴𝐵 and 𝐵𝐶. Hint: Begin by finding the angular velocities of bars 𝐴𝐵 and 𝐵𝐶. The geometry of this problem lends itself to the IC of zero velocity.

𝚥̂ 𝐿

𝚤̂ 𝐿

𝑣𝐶 𝐶 𝑎𝐶

𝐵

Problem 16.123 A flood gate is controlled by the hydraulic cylinder 𝐴𝐵. If the length of the cylinder is increased with a constant time rate of 2.5 f t∕s, determine the angular acceleration of the gate when 𝜙 = 0◦ . Let 𝓁 = 10 f t, ℎ = 2.5 f t, and 𝑑 = 5 f t. Hint: Differentiation of constraints is recommended. Begin by writing expressions for lengths 𝓁 and ℎ in terms of 𝜃, 𝜙, 𝑑 and 𝑅𝐴𝐵 , the length of the hydraulic cylinder.

𝜃 Figure P16.122

𝐴 𝚥̂ 𝚤̂ 𝐿

𝑑

𝐿

𝚥̂

𝐵 𝜙

𝚤̂ 𝜃

𝐴

ℎ 𝓁

Figure P16.123

𝐶

𝐵

𝛼𝐶𝐷 𝜔𝐶𝐷 𝐷

𝐶 𝜃 𝐿∕2

Figure P16.124

Problem 16.124 𝚥̂

At the instant shown, bar 𝐶𝐷 is rotating with an angular velocity 20 rad∕s and with angular acceleration 2 rad∕s2 in the directions shown. Furthermore, at this instant 𝜃 = 45◦ . If 𝐿 = 2.25 f t, determine the angular accelerations of bars 𝐴𝐵 and 𝐵𝐶. Hint: Begin by finding the angular velocities of bars 𝐴𝐵 and 𝐵𝐶. The geometry of this problem lends itself to the IC of zero velocity.

𝚤̂ 𝐿 𝐴

𝐷

𝐿

Problem 16.125 𝐵

At the instant shown, bars 𝐴𝐵 and 𝐶𝐷 are vertical. In addition, point 𝐶 is moving to the left with an increasing speed of 4 m∕s2 , and the magnitude of the acceleration of 𝐶 is 55 m∕s2 . If 𝐿 = 0.5 m and 𝐻 = 0.2 m, determine the angular accelerations of bars 𝐴𝐵 and 𝐵𝐶. Hint: Begin by finding the angular velocity of bar 𝐵𝐶. The geometry of this linkage lends itself to the IC of zero velocity.

𝐻

Figure P16.125 𝑤 𝐶 𝚥̂

𝐵 𝐴

𝚤̂ 𝐷

ℎ

Figure P16.126

ISTUDY

𝓁 𝑒

Problem 16.126 The bucket of a backhoe is the element 𝐴𝐵 of the four-bar linkage system 𝐴𝐵𝐶𝐷. Assume that the points 𝐴 and 𝐷 are fixed and that the bucket rotates with a constant angular velocity 𝜔𝐴𝐵 = 0.25 rad∕s. In addition, suppose that, at the instant shown, point 𝐵 is aligned vertically with point 𝐴, and 𝐶 is aligned horizontally with 𝐵. Determine the acceleration of point 𝐶 at the instant shown, along with the angular accelerations of the elements 𝐵𝐶 and 𝐶𝐷. Let ℎ = 0.66 f t, 𝑒 = 0.46 f t, 𝓁 = 0.9 f t, and 𝑤 = 1.0 f t.

ISTUDY

Section 16.3

Planar Motion: Acceleration Analysis

Problem 16.127 The disk 𝐷 of radius 𝑅 rolls without slipping inside the fixed ring whose inner radius is 2𝑅. Bar 𝐴𝐵 is pin-connected to the center of the disk at one end and is pin-connected to bar 𝐵𝐶 at the other end. The other end of 𝐵𝐶 is pin-connected to the fixed support at 𝐶. At the instant shown, the disk is at the lowest position in the ring, bar 𝐵𝐶 is horizontal, and the pin at 𝐵 is moving with constant speed 𝑣0 as shown. Using the given component system, at this instant, compute the angular velocities and angular accelerations of the disk 𝐷 and the bar 𝐴𝐵. Figure P16.127

Problem 16.128 The bar 𝐴𝐵 is pin-connected at one end to the slider at 𝐴 and is pin-connected at the other end at 𝐵, which is on the disk 𝐷 at the location shown. Disk 𝐷 rolls without slipping over the horizontal surface. At the instant shown, slider 𝐴 is moving in the direction shown with constant speed 𝑣0 . At this instant, determine, as functions of 𝑅 and 𝑣0 , the angular velocities and angular accelerations of bar 𝐴𝐵 and disk 𝐷.

Problem 16.129 Arm 𝐴𝐶𝐵 rotates counterclockwise about the fixed point 𝐶 with constant angular speed 𝜔0 in the direction shown. Two friction disks 𝐷 and 𝐸 are pinned at their centers to arm 𝐴𝐶𝐵 as shown. Knowing that the disks roll without slipping at all contact surfaces, determine the angular velocities of disks 𝐷 and 𝐸 as functions of 𝜔0 . Also, determine the accelerations of points 𝑃 and 𝑄, which are the points on disks 𝐷 and 𝐸, respectively, that are in contact with one another at this instant.

Figure P16.129

Figure P16.128

Figure P16.130

Problem 16.130 Arm 𝐴𝐶𝐵 rotates clockwise about the fixed point 𝐶, which is at the midpoint between points 𝐴 and 𝐵, with constant angular speed 𝜔0 . Two friction disks 𝐷 and 𝐸 are pinned at their centers to arm 𝐴𝐶𝐵 as shown. Knowing that the disks roll without slipping at all contact surfaces, determine the angular velocities of disks 𝐷 and 𝐸 as functions of 𝜔0 . Also, determine the accelerations of points 𝑃 and 𝑄, which are the points on disks 𝐷 and 𝐸, respectively, that are in contact with one another at this instant.

Problem 16.131 Disk 𝐷 rolls without slipping over the horizontal surface. At the instant shown, point 𝐴 on bar 𝑂𝐴, which is pinned to the fixed point at 𝑂 at its left end, is moving straight down

Figure P16.131

1115

1116

Chapter 16

Planar Rigid Body Kinematics

with constant speed 𝑣0 . End 𝐵 of bar 𝐴𝐵 is pinned to the disk at point 𝐵. At this instant, determine, as functions of 𝑅 and 𝑣0 , the angular velocities and angular accelerations of bar 𝐴𝐵 and disk 𝐷.

Problems 16.132 through 16.135 For the slider-crank mechanism shown, let 𝑅 = 1.9 in., 𝐿 = 6.1 in., and 𝐻 = 1.2 in. Assuming that 𝜔𝐴𝐵 = 4850 rpm and is constant, determine the angular acceleration of the connecting rod 𝐵𝐶 and the acceleration of point 𝐶 at the instant when 𝜃 = 27◦ .

Problem 16.132

𝐶 𝐿 𝜙

Problem 16.133 Assuming that 𝜔𝐴𝐵 = 4850 rpm and is constant, determine the acceleration of point 𝐷 at the instant when 𝜙 = 10◦ .

𝐻

𝐷 𝐵

Assuming that, at the instant shown, 𝜃 = 31◦ , 𝜔𝐴𝐵 = 4850 rpm, and 𝛼𝐴𝐵 = 𝜔̇ 𝐴𝐵 = −280 rad∕s2 , determine the angular acceleration of the connecting rod and the acceleration of point 𝐶.

𝜃

𝑅

Problem 16.134

𝐴 𝜔𝐴𝐵

𝚥̂ 𝚤̂

Problem 16.135 Determine the general expression of the acceleration of the piston ̇ and 𝛼 = 𝜃. ̈ 𝐶 as a function of 𝐿, 𝑅, 𝜃, 𝜔𝐴𝐵 = 𝜃, 𝐴𝐵

Figure P16.132–P16.135

Problem 16.136 In the four-bar linkage system shown, let the circular guide with center at 𝑂 be fixed and such that, for 𝜃 = 0◦ , the bars 𝐴𝐵 and 𝐵𝐶 are vertical and horizontal, respectively. In addition, let 𝑅 = 0.6 m, 𝐿 = 1 m, and 𝐻 = 1.25 m. When 𝜃 = 37◦ , 𝛽 = 25.07◦ , and 𝛾 = 78.71◦ , assume collar 𝐶 is sliding clockwise with a speed 7 m∕s. Assuming that, at the instant in question, the speed is increasing and that |𝑎⃗𝐶 | = 93 m∕s2 , determine the angular accelerations of the bars 𝐴𝐵 and 𝐵𝐶. Hint: Because 𝐶 is mounted on a circular guide, it is easier to formulate its acceleration tangent and normal to the circle. Once 𝑣̇ 𝐶 has been obtained from the information given, the components of 𝑎⃗𝐶 should then be transformed into the Cartesian frame shown. An alternative expression for 𝑎⃗𝐶 can then be established in terms of 𝜔𝐴𝐵 , 𝜔𝐵𝐶 , 𝛼𝐴𝐵 and 𝛼𝐵𝐶 using relative acceleration analyses. 𝐶 𝑅

𝜃

𝐿 𝛽

𝑂

𝐵

𝚥̂ 𝛾

𝚤̂

𝐶

𝐻

𝚥̂ 𝚤̂

𝐿

𝐴 Figure P16.136

𝜙

𝑦𝐶

Problem 16.137 𝐵

𝜔𝐴𝐵 𝐴

Figure P16.137

ISTUDY

𝜃

𝑅

Complete the acceleration analysis of the slider-crank mechanism using differentiation of constraints that was outlined on pp. 1097 and 1098. That is, determine the acceleration of the piston 𝐶 and the angular acceleration of the connecting rod as a function of the given quantities 𝜃, 𝜔𝐴𝐵 , 𝑅, and 𝐿. Assume that 𝜔𝐴𝐵 is constant, and use the component system shown for your answers.

ISTUDY

Section 16.3

Planar Motion: Acceleration Analysis

Design Problems Design Problem 16.1 In some high-performance mountain bikes, one element of the frame is attached to the rest via a four-bar linkage system. Referring to Fig. DP16.1, the four-bar linkage system is defined by the points 𝐴, 𝐵, 𝐶, and 𝐷. Notice that this system is also connected to the shock absorber pinned at points 𝐸 and 𝐹 . Research the available commercial literature (this information is readily available on the web), and select a bicycle frame containing a four-bar linkage system, such as that shown below. For the frame you select, obtain the necessary geometric information (again, this information is often made available on the web by manufacturers), and assume that the points attached to the front part of the frame, which in Fig. DP16.1 are points 𝐴 and 𝐷, are fixed. Then calling 𝓁 the length of the shock absorber, which in Fig. DP16.1 is the distance between 𝐸 and 𝐹 , analyze the kinematics of the linkage system and determine 𝓁̇ and 𝓁̈ as a function of both the angular velocity and the angular acceleration of the part of the frame to which the wheel is attached. In Fig. DP16.1, this is the part to the left of points 𝐵 and 𝐶.

𝐹

𝐵 𝐸

𝐴 𝐷 𝐶 3DMI/Shutterstock

Figure DP16.1

1117

1118

Chapter 16

Planar Rigid Body Kinematics

16.4

𝑦

𝑎⃗𝐴 𝑣⃗𝐴 𝑧

𝐵 𝑃

𝑟⃗𝑃 ∕𝐴 = 𝜌⃗ 𝐴

𝑌

𝑂

𝑟⃗𝑃

𝑋

Figure 16.39 Reference frame, position vector, velocity, and acceleration definitions used in describing the motion of a point 𝑃 relative to a rigid body 𝐵.

ISTUDY

In problems with multiple bodies or with bodies that have components moving relative to one another, the application of the kinematic relations developed so far is rarely straightforward. For these problems, it is convenient to study the motion of the various parts of the system using more than one frame of reference. For this reason, we now develop a vector-based approach to the study of kinematics that uses both a primary reference frame and a secondary reference frame that can rotate and translate relative to the primary frame.

The general kinematic equations for the motion of a point relative to a rotating reference frame

𝑥 𝑟⃗𝐴

Rotating Reference Frames

Referring to Fig. 16.39, we will determine the motion of point 𝑃 using two different reference frames: (1) a reference frame 𝑥𝑦𝑧 that translates and rotates, which we will call the rotating reference frame,∗ and (2) a reference frame 𝑋𝑌𝑍 with respect to which we are measuring the motion of 𝑃 and that we will call the primary reference frame. The primary frame will be inertial in kinetics problems. In addition, when the rotating reference frame is attached to a moving rigid body (as it is in Fig. 16.39), we will often call it the body-fixed frame. With this as background, we use the following to describe the motion of point 𝑃 : 1. The motion of the origin (point 𝐴) of the rotating reference frame. ⃗ and angular acceleration Ω ⃗̇ of the rotating reference 2. The angular velocity Ω frame. 3. How 𝑃 is seen to be moving by an observer attached to the rotating reference frame. To begin, we note from Fig. 16.39 that the position of 𝑃 can be written as ⃗ 𝑟⃗𝑃 = 𝑟⃗𝐴 + 𝑟⃗𝑃 ∕𝐴 = 𝑟⃗𝐴 + 𝜌,

(16.40)

where 𝑟⃗𝐴 is the position of the origin of the rotating frame and 𝑟⃗𝑃 ∕𝐴 = 𝜌⃗ is the position of 𝑃 relative to the rotating frame (using 𝜌⃗ makes the development more compact). ̂ Eq. (16.40) becomes† Writing 𝜌⃗ in terms of the 𝑥𝑦𝑧 frame as 𝜌⃗ = 𝜌𝑥 𝚤̂ + 𝜌𝑦 𝚥̂ + 𝜌𝑧 𝑘, ̂ 𝑟⃗𝑃 = 𝑟⃗𝐴 + 𝜌𝑥 𝚤̂ + 𝜌𝑦 𝚥̂ + 𝜌𝑧 𝑘.

(16.41)

To find velocities and accelerations, we need to differentiate Eq. (16.41) with respect to time. As we do this, we need to keep in mind that the 𝑥𝑦𝑧 frame is attached to the body 𝐵, and so 𝚤̂ and 𝚥̂ rotate with 𝐵 and, therefore, are not constant. Velocity using a rotating frame Differentiating Eq. (16.41) with respect to time, we get ̂̇ 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝜌̇ 𝑥 𝚤̂ + 𝜌𝑥 𝚤̂̇ + 𝜌̇ 𝑦 𝚥̂ + 𝜌𝑦 𝚥̂̇ + 𝜌̇ 𝑧 𝑘̂ + 𝜌𝑧 𝑘, ∗ Other † For

common names are the secondary reference frame or moving reference frame. planar motion, we can always consider the plane of motion to be the 𝑧 = 0 plane.

(16.42)

ISTUDY

Section 16.4

1119

Rotating Reference Frames

where 𝑣⃗𝑃 is the velocity of 𝑃 and 𝑣⃗𝐴 is the velocity of the origin 𝐴 of the rotating frame. We can now rewrite this expression, using our knowledge of the time derivative of a unit vector from Eq. (12.58) on p. 718, that is, 𝚤̂̇ = 𝜔 ⃗ 𝑥𝑦𝑧 × 𝚤̂,

𝚥̂̇ = 𝜔 ⃗ 𝑥𝑦𝑧 × 𝚥̂,

and

̂ 𝑘̂̇ = 𝜔 ⃗ 𝑥𝑦𝑧 × 𝑘,

(16.43)

̂ Since the 𝑥𝑦𝑧 frame is body-fixed, where 𝜔 ⃗ 𝑥𝑦𝑧 is the angular velocity of 𝚤̂, 𝚥̂, and 𝑘. ⃗ and so we know that all the unit vectors rotate with the body at angular velocity Ω, ( ) ( ) ( ) ⃗ × 𝚤̂ + 𝜌 Ω ⃗ × 𝚥̂ + 𝜌 Ω ⃗ × 𝑘̂ 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝜌̇ 𝑥 𝚤̂ + 𝜌̇ 𝑦 𝚥̂ + 𝜌̇ 𝑧 𝑘̂ + 𝜌𝑥 Ω 𝑦 𝑧 ( ) ⃗ × 𝜌 𝚤̂ + 𝜌 𝚥̂ + 𝜌 𝑘̂ = 𝑣⃗𝐴 + 𝜌̇ 𝑥 𝚤̂ + 𝜌̇ 𝑦 𝚥̂ + 𝜌̇ 𝑧 𝑘̂ + Ω 𝑥 𝑦 𝑧 ⃗ × 𝜌⃗, = 𝑣⃗𝐴 + 𝜌̇ 𝑥 𝚤̂ + 𝜌̇ 𝑦 𝚥̂ + 𝜌̇ 𝑧 𝑘̂ + Ω ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ 𝑣⃗𝑃 rel ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 𝜌⃗̇

(16.44)

𝐵

𝑦

or ⃗ × 𝑟⃗ , 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝐴

𝑣⃗𝐴

(16.45)

where 𝑣⃗𝑃 rel = 𝜌̇ 𝑥 𝚤̂ + 𝜌̇ 𝑦 𝚥̂ + 𝜌̇ 𝑧 𝑘̂

(16.46)

𝑥

𝑧

is the velocity of 𝑃 relative to the rotating or body-fixed frame (i.e., as seen by an observer moving with the body 𝐵) and we have replaced 𝜌⃗ with 𝑟⃗𝑃 ∕𝐴 . Equation (16.45) is an important development since it allows us to relate the velocities of two points that are not on the same rigid body—in this case 𝑃 and 𝐴. In words, and referring to Fig. 16.40, Eq. (16.45) tells us that the velocity of 𝑃 can be found by using

𝑟⃗𝐴

𝑟⃗𝑃

𝑌

𝑣⃗𝐴 = velocity of the origin of the rotating or body-fixed reference frame (point 𝐴 in Fig. 16.40), 𝑣⃗𝑃 rel = velocity of 𝑃 as seen by an observer moving with the rotating reference frame 𝑥𝑦𝑧, ⃗ = angular velocity of the rotating reference frame 𝑥𝑦𝑧, Ω

𝑃

𝑟⃗𝑃 ∕𝐴 = 𝜌⃗

𝐴

𝑂

𝑋

Figure 16.40 The essential ingredients for finding velocities in rotating reference frames.

𝑟⃗𝑃 ∕𝐴 = vector from the origin of the rotating reference frame to point 𝑃 . Before looking at accelerations, let’s apply Eq. (16.45) in a short example. Mini-Example Referring to Fig. 16.41, find the velocity of the point 𝑃 that is moving in the slot in the disk. The angular velocity of the disk 𝜔0 is constant, and 𝑠(𝑡) is known. Solution This is a sliding contact (𝑃 is not a point on the rotating disk), so we can’t apply the kinematic equations developed in Section 16.1. We will apply Eq. (16.45) as ⃗ × 𝑟⃗ 𝑣⃗𝑃 = 𝑣⃗𝑂 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝑂 ,

(16.47)

where we have attached the rotating 𝑥𝑦𝑧 frame to the disk with its origin at the center 𝑂, as shown in Fig. 16.41. Interpreting each of these terms, 𝑣⃗𝑂 = 0⃗ since it is the velocity of the origin of the rotating 𝑥𝑦𝑧 frame, which is not moving. The term 𝑣⃗𝑃 rel is the velocity of 𝑃 as seen by an observer rotating with the disk. If we

𝑌 𝑦

𝑥 𝑠

𝜔0

𝑂

𝑃

𝜃 𝑋

Figure 16.41 A particle moving in the radial slot of a rotating rigid disk.

1120

ISTUDY

Chapter 16

Planar Rigid Body Kinematics

Common Pitfall The velocity and acceleration can be expressed in either the primary or rotating reference frame. We have expressed our final answers in terms of components in the rotating or body-fixed reference frame. Any vector can be expressed in terms of the components of either the rotating frame or the primary frame. For example, in terms of the 𝑋𝑌𝑍 frame, we can write the velocity of 𝑃 in the mini-example as ( ) 𝑣⃗𝑃 = 𝑠̇ cos 𝜃 𝐼̂ + sin 𝜃 𝐽̂ ( ) + 𝑠𝜔0 − sin 𝜃 𝐼̂ + cos 𝜃 𝐽̂ = (𝑠̇ cos 𝜃 − 𝑠𝜔0 sin 𝜃) 𝐼̂ + (𝑠̇ sin 𝜃 + 𝑠𝜔0 cos 𝜃) 𝐽̂ where 𝐼̂ and 𝐽̂ are the unit vectors associated with the 𝑋𝑌𝑍 frame.

are sitting on the disk, we see 𝑃 moving in just the 𝑥 direction, and so 𝑣⃗𝑃 rel = 𝑠̇ 𝚤̂. ⃗ is the angular velocity of the rotating frame, which is 𝜔 𝑘, ̂ and 𝑟⃗ Finally, Ω 0 𝑃 ∕𝑂 is the vector from the origin of the rotating frame to 𝑃 , which is 𝜌⃗ = 𝑠 𝚤̂. Putting this all in Eq. (16.47), we get 𝑣⃗𝑃 = 𝑠̇ 𝚤̂ + 𝜔0 𝑘̂ × 𝑠 𝚤̂ = 𝑠̇ 𝚤̂ + 𝑠𝜔0 𝚥̂.

(16.48)

Referring to Fig. 16.42, this problem could have been handled using polar coordi𝑦

𝑌 𝑢̂ 𝜃 𝑠

𝑢̂ 𝑟 𝑃

𝑂

𝜔0

𝑥 𝜃 𝑋

Figure 16.42. The spinning disk of Fig. 16.41 with polar coordinates defined.

nates by writing 𝑣⃗𝑃 as

𝑣⃗𝑃 = 𝑟̇ 𝑢̂ 𝑟 + 𝑟𝜃̇ 𝑢̂ 𝜃 .

(16.49)

Notice that 𝑟 = 𝑠, 𝑟̇ = 𝑠,̇ 𝜃̇ = 𝜔0 , and the 𝚤̂ and 𝚥̂ are in the same direction as 𝑢̂ 𝑟 and 𝑢̂ 𝜃 , respectively. That means that Eqs. (16.48) and (16.49) are giving identical results (as they should!).

Acceleration using a rotating frame To find the acceleration of 𝑃 , we start by differentiating Eq. (16.44) with respect to time [we use Eq. (16.44) so that we have the component form of 𝑣⃗𝑃 rel ] to get ⃗̇ × 𝜌⃗ + Ω ⃗ × 𝜌⃗̇ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝜌̈𝑥 𝚤̂ + 𝜌̇ 𝑥 𝚤̂̇ + 𝜌̈𝑦 𝚥̂ + 𝜌̇ 𝑦 𝚥̂̇ + 𝜌̈𝑧 𝑘̂ + 𝜌̇ 𝑧 𝑘̂̇ + Ω ( ) ( ) ( ) ⃗ × 𝚤̂ + 𝜌̇ Ω ⃗ × 𝚥̂ + 𝜌̇ Ω ⃗ × 𝑘̂ = 𝑎⃗𝐴 + 𝜌̈𝑥 𝚤̂ + 𝜌̈𝑦 𝚥̂ + 𝜌̈𝑧 𝑘̂ + 𝜌̇ 𝑥 Ω 𝑦 𝑧 ) ( ⃗̇ × 𝜌⃗ + Ω ⃗ × 𝑣⃗ ⃗ ⃗ +Ω 𝑃 rel + Ω × 𝜌

(16.51)

⃗ × (𝜌̇ 𝚤̂ + 𝜌̇ 𝚥̂ + 𝜌̇ 𝑘̂ ) = 𝑎⃗𝐴 + 𝜌̈𝑥 𝚤̂ + 𝜌̈𝑦 𝚥̂ + 𝜌̈𝑧 𝑘̂ + Ω 𝑥 𝑦 𝑧 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ 𝑎⃗𝑃 rel 𝑣⃗𝑃 rel ( ) ⃗̇ × 𝜌⃗ + Ω ⃗ × 𝑣⃗ ⃗ ⃗ +Ω 𝑃 rel + Ω × 𝜌 ) ( ⃗ × 𝜌⃗ , ⃗ × 𝑣⃗ ⃗̇ ⃗× Ω = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω ⃗+Ω 𝑃 rel + Ω × 𝜌

(16.52)

(16.50)

where we have used the time derivative of the unit vectors from Eq. (16.43) and ⃗ × 𝜌⃗ from Eq. (16.44) to go from Eq. (16.50) to Eq. (16.51). In summary, 𝜌⃗̇ = 𝑣⃗𝑃 rel + Ω ( ) ⃗ ⃗ ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 + Ω × Ω × 𝑟 𝑃 ∕𝐴 ,

(16.53)

where we have replaced 𝜌⃗ with 𝑟⃗𝑃 ∕𝐴 , and 𝑎⃗𝑃 rel = 𝜌̈𝑥 𝚤̂ + 𝜌̈𝑦 𝚥̂ + 𝜌̈𝑧 𝑘̂

(16.54)

is the acceleration of 𝑃 relative to the rotating frame (i.e., as seen by an observer moving with the body 𝐵). Equation (16.53) looks complicated, but it is readily applied

ISTUDY

Section 16.4

1121

Rotating Reference Frames

as long as each term is considered individually. In words, the terms in Eq. (16.53) are (see Fig. 16.43):

𝑦 𝑎⃗𝐴

𝑎⃗𝐴 = acceleration of the origin of the rotating reference frame, 𝑎⃗𝑃 rel = acceleration of 𝑃 as seen by an observer moving with the rotating reference frame, ⃗ = angular velocity of the rotating reference frame 𝑋𝑌𝑍, Ω

𝑧

𝑃

𝑟⃗𝑃 ∕𝐴

𝐴

𝑥

𝑣⃗𝑃 rel = velocity of 𝑃 as seen by an observer moving with the rotating reference frame 𝑥𝑦𝑧, ̇⃗ Ω = angular acceleration of the rotating reference frame 𝑥𝑦𝑧,

𝐵

𝑟⃗𝐴

𝑟⃗𝑃

𝑌

𝑟⃗𝑃 ∕𝐴 = vector from the origin of the rotating reference frame to point 𝑃 . 𝑂

Now, let’s revisit the mini-example on p. 1119 and find the acceleration of 𝑃 . Mini-Example Referring to Fig. 16.44, find the acceleration of the point 𝑃 that is moving in the slot in the disk. The disk is rotating with angular velocity 𝜔0 and angular acceleration 𝛼0 , and 𝑠(𝑡) is known.

Figure 16.43 The essential quantities needed for finding accelerations in rotating reference frames. 𝑌 𝑦

Solution We will apply Eq. (16.53) as

𝑥 𝑠

) ( ⃗ ⃗ ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝑂 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝑂 + Ω × Ω × 𝑟 𝑃 ∕𝑂 ,

0

𝑃

𝜃 𝑋

𝑂

(16.55)

where we have attached the rotating 𝑥𝑦𝑧 frame to the disk with its origin at the center 𝑂, as shown in Fig. 16.44. First, 𝑎⃗𝑂 = 0⃗ since the origin of the rotating 𝑥𝑦𝑧 frame is not moving. Second, 𝑎⃗𝑃 rel is the acceleration of 𝑃 as seen by an observer rotating with the disk, which means that 𝑎⃗𝑃 rel = 𝑠̈ 𝚤̂. Using similar rea⃗ is the angular velocity of the rotating frame, which soning, 𝑣⃗𝑃 rel = 𝑠̇ 𝚤̂. Finally, Ω ̇ ⃗ is the angular acceleration of the rotating frame, which is 𝛼 𝑘; ̂ Ω ̂ and 𝑟⃗ is 𝜔 𝑘; 0

𝑋

𝛼0

𝜔0

Figure 16.44 A particle moving in the radial slot of a rotating rigid disk.

𝑃 ∕𝑂

is the vector from the origin of the rotating frame to 𝑃 , which is 𝑠 𝚤̂. Putting this all in Eq. (16.55), we get ( ) 𝑎⃗𝑃 = 𝑠̈ 𝚤̂ + 2𝜔0 𝑘̂ × 𝑠̇ 𝚤̂ + 𝛼0 𝑘̂ × 𝑠 𝚤̂ + 𝜔0 𝑘̂ × 𝜔0 𝑘̂ × 𝑠 𝚤̂ ( ) ( ) = 𝑠̈ − 𝑠𝜔20 𝚤̂ + 𝑠𝛼0 + 2𝑠𝜔 ̇ 0 𝚥̂. (16.56) Using polar coordinates gives us the same result (see Prob. 16.138).

Useful tidbits when using rotating reference frames • Equations (16.45) and (16.53) apply in a 2D or 3D context. For planar motion, the acceleration given in Eq. (16.53) can be written as 2 ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗𝑃 ∕𝐴 , 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 − Ω 𝑟

𝑣𝐵

𝜔𝑚

(16.57)

⃗ where Ω = |Ω|. • The terms 𝑣⃗𝑃 rel and 𝑎⃗𝑃 rel in Eqs. (16.45) and (16.53) tell us the apparent motion of something as seen by a moving observer. Referring to Fig. 16.45, the

𝐴 𝑟

𝐵 ℎ

𝜃 𝑑

merry-go-round Figure 16.45 A person 𝐴 standing on a spinning merry-goround as the person 𝐵 walks on a sidewalk in the direction shown.

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person at 𝐴 is not moving relative to the merry-go-round, which is rotating at the constant rate 𝜔𝑚 , and the person at 𝐵 is walking at a constant speed along the sidewalk in the direction shown. The velocity and acceleration of 𝐵 as seen by 𝐴 are not 𝑣⃗𝐵∕𝐴 and 𝑎⃗𝐵∕𝐴 , respectively; they are instead

𝐵

𝑦

𝑟⃗𝑃 ∕𝐴

𝐴

𝑃

𝑥

𝑧

𝑟⃗𝑃

𝑟⃗𝐴

𝑋

Figure 16.46 Relating the velocity of 𝑃 to that of 𝐵 when 𝑃 is a point on the rigid body.

𝜔𝑒 = 𝜔𝑥𝑦𝑧 = 1 rev∕day

Figure 16.47 An object 𝑃 moving northeast in London, England, with speed 𝑣𝑃 relative to the surface of the Earth. The 𝑥𝑦𝑧 frame is attached to the surface of the Earth with 𝑥 pointing east, 𝑦 north, and 𝑧 is the local vertical. The latitude of London is 51◦ north.

ISTUDY

𝑎⃗𝐵rel = 𝑎⃗𝐵 − 𝑎⃗𝐴 − 2𝜔 ⃗ 𝑚 × 𝑣⃗𝑃 rel − 𝛼⃗𝑚 × 𝑟⃗𝐵∕𝐴 + 𝜔2𝑚 𝑟⃗𝐵∕𝐴 ,

(16.59)

• When deciding where to attach the rotating frame, choose a frame that makes it easy to determine each of the terms in Eqs. (16.45) and (16.53). • If 𝑃 is a point on the body 𝐵, as shown in Fig. 16.46, then 𝑣⃗𝑃 rel = 0⃗ in Eq. (16.45), and we have ⃗ × 𝑟⃗ 𝑣⃗𝑃 = 𝑣⃗𝐴 + Ω ⃗𝐴 + 𝜔 ⃗ 𝐴𝑃 × 𝑟⃗𝑃 ∕𝐴 , 𝑃 ∕𝐴 = 𝑣

𝑌

𝑂

(16.58)

respectively. In the given situation, 𝛼⃗𝑚 and 𝑎⃗𝐵 would both be zero. We will explore this idea in Example 16.15.

𝑣⃗𝐴 = 𝜔𝐴𝑃

𝑣⃗𝐵rel = 𝑣⃗𝐵 − 𝑣⃗𝐴 − 𝜔 ⃗ 𝑚 × 𝑟⃗𝐵∕𝐴 ,

which is just Eq. (16.3) on p. 1057. A corresponding simplification applies to the acceleration in Eq. (16.53) leading to Eq. (16.5).

Coriolis component of acceleration ⃗ × 𝑣⃗ The term 2Ω 𝑃 rel in Eq. (16.53) is known as the Coriolis acceleration of 𝑃 ; this term results from two equal, but different effects: (1) the change in direction of 𝑣⃗𝑃 rel ⃗ and (2) the effect of Ω ⃗ on the change in magnitude of 𝑟⃗ due to Ω 𝑃 ∕𝐴 relative to the rotating reference frame. It is named after Gaspard-Gustave Coriolis (1792–1843), who studied mechanics and mathematics in France. He gave the terms work and kinetic energy their present scientific meaning, and in an 1835 publication, Coriolis showed that Newton’s laws may be applied in a rotating reference frame, as long as an “extra” acceleration is added to the equations of motion. The phenomenon in which a person feels as though she or he is being thrown sideways when moving on a rotating platform is a consequence of this acceleration. In fact, this acceleration gives rise to the ccw circulation around low-pressure systems and cw circulation around high-pressure systems in the Northern Hemisphere. This circulation around high- and low-pressure systems can be understood if we consider the motion of an object on the surface of the Earth as seen by an observer moving with the Earth. Figure 16.47 depicts an√ object 𝑃 moving both east and north in London, England, at a constant speed 𝑣𝑃 = 𝑣2𝑃 𝑥 + 𝑣2𝑃 𝑦 relative to the surface of the Earth. The 𝑥𝑦𝑧 frame is attached to the surface of the Earth with its origin in London (which we will call point 𝐴) and with 𝑥 pointing east, 𝑦 pointing north, and 𝑧 pointing radially outward from the center of the Earth. The angular velocity of the Earth is 𝜔𝑒 = 1 rev∕day (assumed to be constant), but this is also the angular velocity of the rotating frame 𝜔𝑥𝑦𝑧 since it is attached to the Earth. Referring to Eq. (16.53) and noting that the origin of the rotating frame, which we are calling point 𝐴, is also the location of London, we have that 𝑎⃗𝐴 = 𝑎⃗London , 𝑣⃗𝑃 rel = 𝑣𝑃 𝑥 𝚤̂ + 𝑣𝑃 𝑦 𝚥̂,

⃗ 𝑎⃗𝑃 rel = 0, ⃗̇ = 0, ⃗ Ω

⃗ =𝜔 Ω ⃗ 𝑒, ⃗ 𝑟⃗𝑃 ∕𝐴 = 0.

(16.60) (16.61)

ISTUDY

Section 16.4

Rotating Reference Frames

1123

𝑦 (north) 𝑃 𝜆 𝑂

Figure 16.48. A different perspective of Fig. 16.47 showing the latitude 𝜆 of the object 𝑃 in London, England.

Referring to Fig. 16.48, we can write the nonzero quantities in the 𝑥𝑦𝑧 frame as ( ) ( ) ⃗𝑒 × 𝜔 ⃗ 𝑒 × 𝑟⃗𝐴∕𝑂 = 𝜔 ⃗𝑒 × 𝜔 ⃗ 𝑒 × 𝑟⃗𝐴∕𝑂 𝑎⃗𝐴 = 𝑎⃗𝑂 + 𝛼⃗𝑒 × 𝑟⃗𝐴∕𝑂 + 𝜔 ] ( ) [ ( ) = 𝜔𝑒 cos 𝜆 𝚥̂ + sin 𝜆 𝑘̂ × 𝜔𝑒 cos 𝜆 𝚥̂ + sin 𝜆 𝑘̂ × 𝑅𝑒 𝑘̂ ( ) (16.62) = 𝑅𝑒 𝜔2𝑒 cos 𝜆 sin 𝜆 𝚥̂ − cos 𝜆 𝑘̂ , ( ) ⃗ = 𝜔 cos 𝜆 𝚥̂ + sin 𝜆 𝑘̂ , Ω (16.63) 𝑒

The first term in this equation is the normal acceleration due to the rotation of the Earth, and it is zero at the poles and largest on the equator. The second term is the Coriolis acceleration and, in this case, it is given by ( ) ̂ (16.65) 𝑎⃗Coriolis = 2𝜔𝑒 sin 𝜆 −𝑣𝑃 𝑦 𝚤̂ + 𝑣𝑃 𝑥 𝚥̂ − 2𝜔𝑒 𝑣𝑃 𝑥 cos 𝜆 𝑘. Notice that the (vector) component of the Coriolis acceleration in the 𝑥𝑦 plane has, in turn, a component that is perpendicular to 𝑣⃗𝑃 , and it points to the left of 𝑣⃗𝑃 if one is facing in the direction of the relative motion (see Fig. 16.49). Referring back to Example 13.11 on p. 824, this means that 𝑃 will be deflected to the right if no external forces prevent it from doing so.∗ Now, low- and high-pressure systems are so named because the air pressure in them is lower and higher, respectively, than the surrounding air pressure. Since air flows from areas of higher pressure to areas of lower pressure, we see that air will try to flow into the low from its surroundings and away from the high to its surroundings (the black arrows in Fig. 16.50). As air flows into the low, the Coriolis component of acceleration causes it to deflect to the right (the red arrows in Fig. 16.50). Since every “particle” of air is deflected to the right, the overall motion becomes that of a circulation around the low—a circulation in the ccw direction. A similar argument tells us that the circulation around a high must be in the cw direction. In the Southern Hemisphere, for an Earth-fixed 𝑥𝑦𝑧 frame oriented as it was in the Northern Hemisphere, ∗ Equation

(16.65) also tells us that the object will be deflected upward, but that is not a relevant effect for this discussion.

𝚥̂ 2𝜔𝑒 sin 𝜆𝑣𝑃 𝑥

where 𝑅𝑒 is the radius of the Earth, and where we have used the fact that points 𝐴 and 𝑃 coincide at this instant. Substituting Eqs. (16.60)–(16.63) into Eq. (16.53), the acceleration of 𝑃 is ( ) ( ) ( ) 𝑎⃗𝑃 = 𝑅𝑒 𝜔2𝑒 cos 𝜆 sin 𝜆 𝚥̂−cos 𝜆 𝑘̂ +2𝜔𝑒 cos 𝜆 𝚥̂+sin 𝜆 𝑘̂ × 𝑣𝑃 𝑥 𝚤̂+𝑣𝑃 𝑦 𝚥̂ . (16.64)

𝚤̂ (𝑎Coriolis )𝑥𝑦 𝑣𝑃

𝑃 −2𝜔𝑒 sin 𝜆𝑣𝑃 𝑦

𝑣𝑃 𝑦

𝑣𝑃 𝑥

Figure 16.49 The velocity of 𝑃 relative to the surface of the Earth and the component of the resulting Coriolis acceleration that lies in the 𝑥𝑦 plane.

1 rev∕day

Figure 16.50 Air flow around low- and high-pressure systems in the Northern Hemisphere.

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Interesting Fact Toilets flushing. You may have heard that it is the Coriolis force or Coriolis acceleration that causes the draining water in a toilet or in a kitchen sink to rotate in one direction in the Northern Hemisphere and in the opposite direction in the Southern Hemisphere. As we shall see in Chapter 20 when we do a careful three-dimensional analysis, the effect of the Coriolis component of acceleration is tiny when compared with other factors, such as the shape of the toilet bowl, the angle of the water jets in the toilet, etc. Even in something as large as a swimming pool, it is negligible.

⃗ in Eq. (16.63) will reverse, which means that the the sign on the 𝑘̂ component of Ω sign on the 𝑥𝑦 component of the Coriolis acceleration will reverse. Therefore, in the Southern Hemisphere, by arguments similar to those for the Northern Hemisphere, circulation will reverse direction—that is, around a low-pressure system it is clockwise, and around a high it is counterclockwise. So, why don’t you have to worry about the Coriolis component of acceleration when you play baseball or basketball? It is there, it is just that the angular speed of the Earth is really small. It is easy to show that 1 rev∕day = 0.00007272 rad∕s. Although it is present when you shoot a basketball, this acceleration would need to act for a long time to have a visible effect—the sort of times (days) experienced by a particle of air moving through the atmosphere. That is not to say that engineers don’t need to worry about the Coriolis component of acceleration—they do. However, if the kinematic analysis of a system is carried out using the ideas presented in this section, the Coriolis component of acceleration will always be taken into account “automatically.”

End of Section Summary In this section, we developed the kinematic equations that allow us to relate the motion of two points that are not on the same rigid body. Referring to Fig. 16.51, this means we can relate the velocity of point 𝑃 to that of 𝐴 using the relation 𝑦 𝑎⃗𝐴

Eq. (16.45), p. 1119 𝑣⃗𝐴 𝑟⃗𝑃 ∕𝐴

𝐴 𝑧

where

𝑥

𝑣⃗𝐴 = velocity of the origin of the rotating or body-fixed reference frame (point 𝐴 in Fig. 16.51),

𝐵

𝑟⃗𝐴 𝑌

𝑂

𝑟⃗𝑃

𝑋

Figure 16.51 The essential ingredients for finding velocities and accelerations using a rotating reference frame.

ISTUDY

⃗ × 𝑟⃗ , 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝐴

𝑃

𝑣⃗𝑃 rel = velocity of 𝑃 as seen by an observer moving with the rotating reference frame, ⃗ Ω = angular velocity of the rotating reference frame, 𝑟⃗𝑃 ∕𝐴 = position of point 𝑃 relative to 𝐴. We also found that we can relate the acceleration of point 𝑃 to that of 𝐴 by using Eq. (16.53), p. 1120 ( ) ⃗ ⃗ ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 + Ω × Ω × 𝑟 𝑃 ∕𝐴 , where, in addition to the terms defined for 𝑣⃗𝑃 , we have 𝑎⃗𝐴 = acceleration of the origin of the rotating or body-fixed reference frame (point 𝐴 in Fig. 16.51), 𝑎⃗𝑃 rel = acceleration of 𝑃 as seen by an observer moving with the rotating reference frame, ⃗ × 𝑣⃗ 2Ω 𝑃 rel = Coriolis acceleration of 𝑃 , ⃗̇ = angular acceleration of the rotating reference frame. Ω

ISTUDY

Section 16.4

1125

Rotating Reference Frames

E X A M P L E 16.13

A Particle in a Rotating Slot: Equation of Motion and Forces

Figure 1 shows a small object 𝑃 sliding in a slot relative to a rotating disk 𝐷. The disk is rotating in the horizontal plane with angular velocity 𝜔0 and angular acceleration 𝛼0 . The particle 𝑃 is constrained to move in the slot, which is a distance 𝑑 from the center of the disk. In addition, a linear elastic spring of constant 𝑘 is attached to the particle, such that the spring is undeformed when the particle is at 𝑠 = 0. Determine the equation(s) of motion of the particle and the normal force between the particle and the slot.

Road Map & Modeling

Newton’s second law will provide the equation of motion for the particle and the normal force between the particle and the slot. Since 𝑃 is sliding relative to the disk, we will attach a rotating reference frame to the disk to find 𝑎⃗𝑃 . The FBD of the particle at an arbitrary position 𝑠 is shown in Fig. 2, where 𝐹𝑠 is the spring force acting on 𝑃 and 𝑁 is the normal force acting on 𝑃 due to the slot. We neglect friction between the particle and the slot. In addition, we attach a rotating 𝑥𝑦𝑧 frame to the disk with its origin at point 𝑂, which is the center of the disk. Since this system has one degree of freedom, we expect to obtain one equation of motion.

𝑃

𝑘 𝑂 𝛼0

SOLUTION

𝑑

𝑠

𝜔0

𝐷

Figure 1 𝑦 𝑠

𝑑

𝑃

𝑥

𝑁 𝐹𝑠 𝑂 𝚥̂

𝚤̂

Governing Equations Balance Principles

gives

∑

Referring to Fig. 2, Newton’s second law applied to the particle 𝐹𝑥∶ −𝐹𝑠 = 𝑚𝑎𝑃 𝑥 ,

and

∑

𝐹𝑦∶ 𝑁 = 𝑚𝑎𝑃 𝑦 ,

(1)

where 𝑎𝑃 𝑥 and 𝑎𝑃 𝑦 are the 𝑥 and 𝑦 components, respectively, of the acceleration of 𝑃 .

Figure 2 The FBD of the particle, as well as the definition of the rotating reference frame. The unit vectors 𝚤̂ and 𝚥̂ rotate with the disk.

The spring is undeformed at 𝑠 = 0, so the spring force law is

Force Laws

𝐹𝑠 = 𝑘𝑠. Kinematic Equations

(2)

To determine 𝑎𝑃 𝑥 and 𝑎𝑃 𝑦 , we apply Eq. (16.57), which gives

2 ⃗̇ × 𝑟⃗ ⃗ × 𝑣⃗ + Ω 𝑎⃗𝑃 = 𝑎𝑃 𝑥 𝚤̂ + 𝑎𝑃 𝑦 𝚥̂ = 𝑎⃗𝑂 + 𝑎⃗𝑃 rel + 2Ω ⃗𝑃 ∕𝑂 , 𝑃 rel 𝑃 ∕𝑂 − Ω 𝑟

(3)

⃗ = 𝜔 𝑘, ⃗̇ = 𝛼 𝑘, ̂ Ω ̂ where 𝑎⃗𝑂 = 0⃗ since point 𝑂 is not moving, 𝑎⃗𝑃 rel = 𝑠̈ 𝚤̂, 𝑣⃗𝑃 rel = 𝑠̇ 𝚤̂, Ω 0 0 and 𝑟⃗𝑃 ∕𝑂 = 𝑠 𝚤̂ + 𝑑 𝚥̂. Substituting these terms into Eq. (3), we obtain 𝑎⃗𝑃 = 𝑠̈ 𝚤̂ + 2𝜔0 𝑘̂ × 𝑠̇ 𝚤̂ + 𝛼0 𝑘̂ × (𝑠 𝚤̂ + 𝑑 𝚥̂) − 𝜔20 (𝑠 𝚤̂ + 𝑑 𝚥̂) ) ) ( ( ̇ 0 + 𝑠𝛼0 − 𝑑𝜔20 𝚥̂, = 𝑠̈ − 𝑑𝛼0 − 𝑠𝜔20 𝚤̂ + 2𝑠𝜔

(4) 𝑑𝜔20 .

(5)

Substituting Eqs. (2) and (5) into Eqs. (1), we obtain ) ) ( ( and 𝑁 = 𝑚 2𝑠𝜔 ̇ 0 + 𝑠𝛼0 − 𝑑𝜔20 . −𝑘𝑠 = 𝑚 𝑠̈ − 𝑑𝛼0 − 𝑠𝜔20

(6)

⇒ 𝑎𝑃 𝑥 = 𝑠̈ − 𝑑𝛼0 −

𝑠𝜔20

and 𝑎𝑃 𝑦 = 2𝑠𝜔 ̇ 0 + 𝑠𝛼0 −

Computation

Interesting Fact

Rearranging the first of Eqs. (6), we obtain the equation of motion of the particle as ) ( 𝑘 − 𝜔20 𝑠 = 𝑑𝛼0 . 𝑠̈ + (7) 𝑚 The second of Eqs. (6) tells us that once we know 𝑠(𝑡) from the solution of Eq. (7), then we can find the normal force between the particle and the slot from ) ( 𝑁 = 𝑚 2𝑠𝜔 ̇ 0 + 𝑠𝛼0 − 𝑑𝜔20 . (8) Discussion & Verification

Note that the dimension of each term in Eq. (7) is that of acceleration, and the dimension of each term in Eq. (8) is that of force, so both results are dimensionally consistent.

Analysis of Eq. (7) if 𝜶𝟎 = 𝟎. If you have taken a differential equations course, you may recognize that if 𝛼0 = 0 in Eq. (7), it becomes a second-order, constant coefficient, homogeneous linear ordinary differential equation. The solutions are harmonic functions (i.e., sines and cosines) if 𝑘∕𝑚 > 𝜔20 , and they are growing exponentials if 𝑘∕𝑚 < 𝜔20 . This property would allow us to use a system like this to monitor when the disk starts rotating faster than some critical value, since the particle would then hit the end of the slot.

1126

Chapter 16

Planar Rigid Body Kinematics

E X A M P L E 16.14

Analysis of a Reciprocating Rectilinear Motion Mechanism

𝜔𝐴𝐵

The reciprocating rectilinear motion mechanism shown in Fig. 1 consists of a disk pinned at its center at 𝐴 that rotates with a constant angular velocity 𝜔𝐴𝐵 , a slotted arm 𝐶𝐷 that is pinned at 𝐶, and a bar that can oscillate within the guides at 𝐸 and 𝐹 . As the disk rotates, the peg at 𝐵 moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between 𝐶 and 𝐵. For the position shown (𝜃 = 30◦ ), determine the

𝐷 𝑅

disk

𝐴 𝜃 𝐵

ℎ

(a) angular velocity and angular acceleration of the slotted arm 𝐶𝐷, (b) velocity and acceleration of the bar. Evaluate your results for 𝜔𝐴𝐵 = 60 rpm, 𝑅 = 0.1 m, ℎ = 0.2 m, and 𝑑 = 0.12 m.

𝐶 𝑑 bar 𝐸

𝐹

Road Map

This is a mechanism with a sliding contact, so we cannot use Eqs. (16.3) and (16.5) to relate the motion of the disk to that of the slotted arm 𝐶𝐷. Therefore, we will use a rotating reference frame to relate the motion of the pin at 𝐵 to the pivot point at 𝐶, and we will attach the frame to arm 𝐶𝐷, as shown in the schematic of the mechanism in Fig. 2.

Figure 1 𝐷

𝑌 𝑦

𝐴

𝑅 60◦

ℎ

Computation

𝚥̂ 𝐵

⃗ 𝑣⃗𝐵 = 𝑣⃗𝐶 + 𝑣⃗𝐵rel + Ω ⃗𝐵∕𝐶 , 𝐶𝐷 × 𝑟

𝐽̂ 𝑋

𝑑

𝐼̂

𝑥 𝑄 Figure 2 The primary and rotating reference frame for the reciprocating rectilinear motion mechanism. Note that the right angle between 𝐴𝐵 and 𝐵𝐶 is a consequence of the given lengths and the angle between 𝐴𝐶 and 𝐵𝐶 at this instant. The angle between 𝐴𝐵 and 𝐵𝐶 is not a right angle in general.

ISTUDY

Since we are relating the motion of 𝐵 to 𝐶, for velocities we can write

𝚤̂

30◦ 𝐶

SOLUTION

(1)

⃗ where 𝑣⃗𝐵rel is the velocity of 𝐵 as seen by an observer in the rotating frame and Ω 𝐶𝐷 = 𝜔 ⃗ 𝐶𝐷 is both the angular velocity of the rotating frame and the angular velocity of the arm 𝐶𝐷. Now, since 𝐵 is rotating in a circle about 𝐴, 𝑣⃗𝐵 is found using Eq. (16.8) to be 𝑣⃗𝐵 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝜔𝐴𝐵 𝑘̂ × 𝑅 𝚤̂ = 𝑅𝜔𝐴𝐵 𝚥̂.

(2)

In addition, we can see that ⃗ 𝑣⃗ ⃗ ̂ 𝑣⃗𝐶 = 0, Ω 𝑟⃗𝐵∕𝐶 = 𝑟𝐵∕𝐶 𝚥̂, (3) 𝐵rel = 𝑟̇ 𝐵∕𝐶 𝚥̂, 𝐶𝐷 = Ω𝐶𝐷 𝑘, √ where 𝑟𝐵∕𝐶 = ℎ2 − 𝑅2 = 0.1732 m at this instant. Substituting Eqs. (2) and (3) into Eq. (1), we obtain 𝑅𝜔𝐴𝐵 𝚥̂ = 𝑟̇ 𝐵∕𝐶 𝚥̂ + Ω𝐶𝐷 𝑘̂ × 𝑟𝐵∕𝐶 𝚥̂, (4) which is equivalent to two scalar equations for Ω𝐶𝐷 and 𝑟̇ 𝐵∕𝐶 . Solving, we find that 𝑟̇ 𝐵∕𝐶 = 𝑅𝜔𝐴𝐵 = 0.6283 m∕s and Ω𝐶𝐷 = 0 rad∕s

⇒

⃗ Ω ⃗ 𝐶𝐷 = 0⃗ rad/s. 𝐶𝐷 = 𝜔

(5)

For the acceleration analysis, we apply Eq. (16.53) on p. 1120, ⃗̇ ⃗ 𝑎⃗𝐵 = 𝑎⃗𝐶 + 𝑎⃗𝐵rel + 2Ω ⃗𝐵rel + Ω ⃗𝐵∕𝐶 − Ω2𝐶𝐷 𝑟⃗𝐵∕𝐶 𝐶𝐷 × 𝑣 𝐶𝐷 × 𝑟 ⃗̇ ⃗𝐵∕𝐶 , = 𝑎⃗𝐶 + 𝑎⃗𝐵rel + Ω 𝐶𝐷 × 𝑟

(6) (7)

where to go from Eq. (6) to Eq. (7) we have used Eq. (5). Since 𝐵 is moving in a circle centered at 𝐴, we can use Eq. (16.13) to find 𝑎⃗𝐵 as 𝑎⃗𝐵 = −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 = −𝜔2𝐴𝐵 𝑅 𝚤̂.

(8)

⃗̇ In addition, 𝑎⃗𝐵rel and Ω 𝐶𝐷 in Eq. (7) are given by ⃗̇ ̇ ̂ 𝑎⃗𝐵rel = 𝑟̈𝐵∕𝐶 𝚥̂ and Ω 𝐶𝐷 = Ω𝐶𝐷 𝑘.

(9)

ISTUDY

Section 16.4

Rotating Reference Frames

Finally, noting that 𝑎⃗𝐶 = 0⃗ and substituting the last of Eqs. (3) and Eqs. (8) and (9) into Eq. (7), we obtain −𝜔2𝐴𝐵 𝑅 𝚤̂ = 𝑟̈𝐵∕𝐶 𝚥̂ + Ω̇ 𝐶𝐷 𝑘̂ × 𝑟𝐵∕𝐶 𝚥̂, (10) which is equivalent to two scalar equations for 𝑟̈𝐵∕𝐶 and Ω̇ 𝐶𝐷 . Solving, we find that 𝑟̈𝐵∕𝐶 = 0 and 𝑅 2 Ω̇ 𝐶𝐷 = 𝜔 = 22.79 rad∕s2 ⇒ 𝑟𝐵∕𝐶 𝐴𝐵

( ) ⃗̇ ̂ Ω ⃗𝐶𝐷 = 22.79 rad∕s2 𝑘. 𝐶𝐷 = 𝛼

(11)

Referring to Fig. 2, now that we have the angular velocity and angular acceleration of the slotted arm 𝐶𝐷, we can find the velocity of the bar using ⃗ 𝑣⃗bar = 𝑣⃗𝑄 = 𝜔 ⃗ 𝐶𝐷 × 𝑟⃗𝑄∕𝐶 = 0,

(12)

since 𝜔 ⃗ 𝐶𝐷 = 0⃗ at this instant. In addition, the acceleration of the bar is given by ( ) ( ) ̂ = 𝛼 𝐾̂ × −𝑑 𝐽̂ = 𝑑𝛼 𝐼, 𝑎⃗ = 𝑎⃗ ⋅ 𝐼̂ 𝐼̂ = 𝛼⃗ × 𝑟⃗ bar

𝑄

𝐶𝐷

𝑄∕𝐶

𝐶𝐷

𝐶𝐷

(13)

that is, 𝑎⃗bar is just the tangential component of acceleration of 𝑄. Therefore, the acceleration of the bar at this instant is ( ) ̂ 𝑎⃗bar = 𝑑𝛼𝐶𝐷 𝐼̂ = 2.735 m∕s2 𝐼. (14) Discussion & Verification The dimensions and thus the units of the final results are as they should be. Looking more deeply, we note that the angular velocity of the slotted arm 𝐶𝐷 is zero at this instant. This makes sense since the line 𝐴𝐵 on the rotating disk is perpendicular to the slot in the arm at this instant. Therefore, at this instant, the velocity of 𝐵 is parallel to the slot, and so it is not inducing any rotation of the arm containing the slot. Consequently, the velocity of the bar must be zero at this instant since it is the angular velocity of the slotted arm that imparts a velocity to the bar. On the other hand, we note that the slotted arm does have an angular acceleration. This also makes sense since an instant before the arm is in this position, and an instant after, the slotted arm must have an angular velocity, and therefore there must be an angular acceleration causing this change in angular velocity. In Probs. 16.158 and 16.159, we will see why this mechanism is sometimes called a quick-return mechanism.

1127

1128

E X A M P L E 16.15

Actual Versus Perceived Motion: Finding 𝑣⃗𝐵rel and 𝑎⃗𝐵rel A person 𝐴 is standing on a merry-go-round (i.e., not moving relative to it), which is rotating at the constant rate 𝜔𝑚 , and a person 𝐵 is walking in a straight line at constant speed 𝑣𝐵 along a sidewalk in a stationary frame (see Fig. 1). Determine the velocity and acceleration of 𝐵 as seen by 𝐴.∗ Evaluate the results for 𝑣𝐵 = 4 mph, 𝜔𝑚 = 3 rad∕s, 𝜃 = 45◦ , 𝑑 = 30 f t, and ℎ = 10 f t.

𝑣𝐵

𝜔𝑚

𝐴 𝑟

𝐵 ℎ

𝜃 𝑑

SOLUTION

merry-go-round

Since 𝐴 is not moving relative to the merry-go-round, the motion of 𝐵 as seen by 𝐴 is equivalent to the motion of 𝐵 as seen by an observer rotating with the merrygo-round. As we mention on pp. 1121 and 1122, the velocity of 𝐵 as seen by 𝐴 is not equal to 𝑣⃗𝐵∕𝐴 since that quantity gives the velocity of 𝐵 as seen by 𝐴 only if 𝐴 is not rotating. Therefore, recall that the term 𝑣⃗𝐵rel in Eq. (16.45) is the velocity of 𝐵 as seen by an observer moving with the rotating frame—this is exactly what we want (along with 𝑎⃗𝐵rel for the acceleration of 𝐵 as seen by 𝐴). Road Map

Figure 1 𝐽̂ 𝚤̂

𝚥̂

𝐼̂

𝑌 𝜔𝑚

Chapter 16

Planar Rigid Body Kinematics

𝑟

𝑣𝐵

𝑥

𝑦

Computation Referring to Fig. 2, the primary 𝑋𝑌𝑍 frame is as shown, and the rotating 𝑥𝑦𝑧 frame is attached to the merry-go-round with its origin at 𝐴. We want to compute 𝑣⃗𝐵rel , which is given by Eq. (16.45) to be

𝐵

𝐴 𝜃

ℎ 𝑋 𝑑

merry-go-round

𝑣⃗𝐵rel = 𝑣⃗𝐵 − 𝑣⃗𝐴 − 𝜔 ⃗ 𝑚 × 𝑟⃗𝐵∕𝐴 ,

where 𝜔 ⃗ 𝑚 is also the angular velocity of the rotating 𝑥𝑦𝑧 frame since the frame is attached to the merry-go-round. Evaluating the three terms above, we find 𝑣⃗𝐵 = 𝑣𝐵 𝐽̂ = 𝑣𝐵 (sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂),

Figure 2 Definition of the primary and rotating frames of reference.

𝑣⃗𝐴 = −𝑟𝜔𝑚 𝚥̂, ( ) 𝜔 ⃗ 𝑚 × 𝑟⃗𝐵∕𝐴 = −𝜔𝑚 𝑘̂ × −𝑟 𝚤̂ + 𝑑 𝐼̂ + ℎ 𝐽̂ [ ] = −𝜔𝑚 𝑘̂ × −𝑟 𝚤̂ + 𝑑(cos 𝜃 𝚤̂ − sin 𝜃 𝚥̂) + ℎ(sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂) = 𝜔𝑚 (ℎ cos 𝜃 − 𝑑 sin 𝜃) 𝚤̂ − 𝜔𝑚 (𝑑 cos 𝜃 + ℎ sin 𝜃 − 𝑟) 𝚥̂.

𝑣⃗𝐵rel , 𝑣⃗𝐵rel = 100.5 f t∕s

(1)

(2) (3)

(4)

Substituting Eqs. (2)–(4) into Eq. (1), we get 𝑣⃗𝐵rel = (𝑣𝐵 sin 𝜃 − ℎ𝜔𝑚 cos 𝜃 + 𝑑𝜔𝑚 sin 𝜃) 𝚤̂ + (𝑣𝐵 cos 𝜃 + ℎ𝜔𝑚 sin 𝜃 + 𝑑𝜔𝑚 cos 𝜃) 𝚥̂ = (46.57 𝚤̂ + 89.00 𝚥̂) f t∕s,

(6)

where the numerical result was obtained using 𝑣𝐵 = 4 mph = 5.867 f t∕s, ℎ = 10 f t, 𝑑 = 30 f t, 𝜔𝑚 = 3 rad∕s, and 𝜃 = 45◦ . Figure 3 shows the actual velocity of 𝐵, i.e., 𝑣⃗𝐵 , as well as the velocity 𝐵 as seen by 𝐴, i.e., 𝑣⃗𝐵rel . Now computing 𝑎⃗𝐵rel using Eq. (16.57), we have

17.4◦

𝑎⃗𝐵rel = 𝑎⃗𝐵 − 𝑎⃗𝐴 − 2𝜔 ⃗ 𝑚 × 𝑣⃗𝐵rel − 𝛼⃗𝑚 × 𝑟⃗𝐵∕𝐴 + 𝜔2𝑚 𝑟⃗𝐵∕𝐴 , 𝑣⃗𝐵 , 𝑣⃗𝐵 = 5.867 f t∕s

(7)

where, in the given situation, 𝛼⃗𝑚 and 𝑎⃗𝐵 are both zero. Computing the other terms in Eq. (7), we have 𝑎⃗𝐴 = −𝑟𝜔2𝑚 𝚤̂, [ ] 2𝜔 ⃗ 𝑚 × 𝑣⃗𝐵rel = 2𝜔𝑚 ℎ𝜔𝑚 sin 𝜃 + (𝑣𝐵 + 𝑑𝜔𝑚 ) cos 𝜃 𝚤̂ [ ] + 2𝜔𝑚 ℎ𝜔𝑚 cos 𝜃 − (𝑣𝐵 + 𝑑𝜔𝑚 ) sin 𝜃 𝚥̂,

Figure 3 A comparison of the velocity of 𝐵 in the stationary frame, i.e., 𝑣⃗𝐵 , with the velocity 𝐵 as seen by 𝐴, i.e., 𝑣⃗𝐵rel .

ISTUDY

(5)

∗ Person

(8) (9)

𝐴 cannot turn his or her head to follow 𝐵 as the merry-go-round rotates, or else we will have introduced yet another rotation.

ISTUDY

Section 16.4

Rotating Reference Frames

[ ] 𝜔2𝑚 𝑟⃗𝐵∕𝐴 = 𝜔2𝑚 (ℎ sin 𝜃 + 𝑑 cos 𝜃 − 𝑟) 𝚤̂ + (ℎ cos 𝜃 − 𝑑 sin 𝜃) 𝚥̂ .

(10)

Substituting Eqs. (8)–(10), as well as 𝛼⃗𝑚 = 0⃗ and 𝑎⃗𝐵 = 0⃗ into Eq. (7), we obtain [ ] 𝑎⃗𝐵rel = −𝜔𝑚 (2𝑣𝐵 + 𝑑𝜔𝑚 ) cos 𝜃 + ℎ𝜔𝑚 sin 𝜃 𝚤̂ ] [ + 𝜔𝑚 (2𝑣𝐵 + 𝑑𝜔𝑚 ) sin 𝜃 − ℎ𝜔𝑚 cos 𝜃 𝚥̂ 2

= (−279.4 𝚤̂ + 152.2 𝚥̂) f t∕s ,

1129

𝑎⃗𝐵 = 0⃗ 16.4◦

(11) (12)

where the numerical result was obtained using 𝑣𝐵 = 4 mph = 5.867 f t∕s, ℎ = 10 f t, 𝑑 = 30 f t, 𝜔𝑚 = 3 rad∕s, and 𝜃 = 45◦ . Figure 4 shows the acceleration of 𝐵 as perceived by 𝐴, i.e., 𝑎⃗𝐵rel , as well as the actual acceleration of 𝐵, which is zero, so only a dot is shown. Discussion & Verification The dimensions of both 𝑣⃗𝐵rel and 𝑎⃗𝐵rel are as they should be, that is, velocity and acceleration, respectively. More importantly, this example illustrates an important idea—the motion of an object (i.e., its velocity and acceleration) that one perceives when on a rotating frame is very different from the actual motion of the object. Figure 3 shows the vast difference between the actual velocity vector of 𝐵 and the velocity of 𝐵 as seen by 𝐴 who is rotating with the merry-go-round. Person 𝐵 is walking at 4 mph, but 𝐴 sees 𝐵 moving at 68.49 mph. As far as acceleration is concerned, 𝐵 is not accelerating at all, yet 𝐴 sees 𝐵 moving at 318.2 f t∕s2 = 9.882𝑔! A Closer Look Notice that the relative velocity 𝑣⃗𝐵rel and acceleration 𝑎⃗𝐵rel of 𝐵 as seen by 𝐴 do not depend on 𝑟, which is the radius of the circular path of 𝐴. This happens in this example because the person 𝐴, who is observing the relative quantities 𝑣⃗𝐵rel and 𝑎⃗𝐵rel , is not moving relative to the moving reference frame. That is, 𝐴 and the moving reference frame are a single rigid body. Because of this, the motion of 𝐴 always cancels with part of the motion of 𝐵 as seen by 𝐴. For example, referring to the velocities, we see that 𝑣⃗𝐴 = −𝑟𝜔𝑚 𝚥̂ and that part of 𝜔 ⃗ 𝑚 × 𝑟⃗𝐵∕𝐴 is given by 𝑟𝜔𝑚 𝚥̂. Therefore, when these terms are combined in Eq. (1), they cancel. A similar cancellation occurs for the relative acceleration.

𝑎⃗𝐵rel , 𝑎⃗𝐵rel = 318.2 f t∕s2 Figure 4 The acceleration of 𝐵 as perceived by 𝐴, i.e., 𝑎⃗𝐵rel , as well as the actual acceleration of 𝐵, which is zero, so only a dot is shown.

1130

Chapter 16

Planar Rigid Body Kinematics

Problems Problem 16.138

𝑌 𝑦 𝑠

𝜃 𝑃

𝑋

𝑂 𝛼0

Using polar coordinates, obtain the acceleration of point 𝑃 of the mini-example on p. 1121; that is, obtain Eq. (16.56).

𝑥

𝜔0

Problems 16.139 through 16.141 The disk 𝐷 is rotating with angular speed 𝜔𝐷 about its central point 𝑂 in the direction shown. The particle 𝑃 moves in the slot with known motion 𝑠(𝑡), which is measured relative to the disk. Express your answers using the 𝑥𝑦𝑧 frame that is attached to the disk 𝐷 with its origin at 𝑂 as shown.

Figure P16.138

Determine the velocity of 𝑃 as a function of 𝑑, 𝑠 and its time deriva-

Problem 16.139

tives, and 𝜔𝐷 . Problem 16.140 If 𝜔𝐷 is constant, determine the Coriolis acceleration of 𝑃 and its total acceleration as functions of 𝑑, 𝑠 and its time derivatives, and 𝜔𝐷 . Problem 16.141 If 𝜔𝐷 is increasing at the rate 𝜔̇ 𝐷 , determine the Coriolis acceleration of 𝑃 and its total acceleration as functions of 𝑑, 𝑠 and its time derivatives, 𝜔𝐷 , and 𝜔̇ 𝐷 . 𝑦

𝑦 𝑠 𝑃

𝑠̈

𝑠

𝐵

𝑥

𝐷

𝑑

𝜔2

𝐶

𝑥

𝑠̇

𝑑

𝑂 𝜔𝐷 𝐷 Figure P16.139–P16.141

𝜔1 𝐴 Figure P16.142 and P16.143

Problems 16.142 and 16.143 Bar 𝐴𝐵 rotates about point 𝐴 with constant angular speed 𝜔1 . Bar 𝐵𝐶 is pinned to bar 𝐴𝐵 at 𝐵 and rotates with constant angular speed 𝜔2 relative to bar 𝐴𝐵 about the pin at 𝐵. The position of collar 𝐷 relative to bar 𝐵𝐶 is given by the coordinate 𝑠, which is a known function of time. Express your answers in the 𝑥𝑦𝑧 frame that is attached to bar 𝐵𝐶 with its origin at 𝐵. Hint: As we will see in Eq. (20.14), 𝜔 ⃗ 𝐵𝐶 = 𝜔 ⃗ 𝐴𝐵 + 𝜔 ⃗ 𝐵𝐶∕𝐴𝐵 = 𝜔 ⃗1 + 𝜔 ⃗ 2.

𝐷

𝓁 𝛼𝐴𝐵 𝐴𝜔

𝐵

Problem 16.142 Determine the velocity of the collar 𝐷 in terms of 𝑑, 𝜔1 , 𝜔2 , and the position coordinate 𝑠 and its time derivatives. Problem 16.143 Determine the acceleration of the collar 𝐷 in terms of 𝑑, 𝜔1 , 𝜔2 , and the position coordinate 𝑠 and its time derivatives.

𝐴𝐵

Problems 16.144 and 16.145 𝑠 𝐶

𝜃

Figure P16.144 and P16.145

ISTUDY

Bar 𝐴𝐵 is pinned to the fixed support at 𝐴, and the collar 𝐵 is pinned to the bar at its opposite end. The bar 𝐶𝐷 can slide freely through the collar at 𝐵. At the instant shown, bar 𝐴𝐵 is horizontal, 𝓁 = 1.2 m, 𝑠 = 1.07 m, 𝜃 = 60◦ , and 𝜔𝐴𝐵 = 40 rad∕s. Problem 16.144 If 𝛼𝐴𝐵 is zero at the instant shown, determine the angular velocity and angular acceleration of the bar 𝐶𝐷.

ISTUDY

Section 16.4

Rotating Reference Frames

1131

Problem 16.145 If 𝛼𝐴𝐵 = 15 rad∕s2 at the instant shown, determine the angular velocity and angular acceleration of the bar 𝐶𝐷.

Problems 16.146 and 16.147 Bar 𝐴𝐵 is pinned to the fixed support at 𝐴, and the pin 𝑃 is fixed to the disk at radius 𝑅𝑖 . The disk with outer radius 𝑅𝑜 rolls without slipping over the horizontal flat surface with angular speed 𝜔𝑑 and angular acceleration 𝛼𝑑 in the directions shown. At the instant shown, 𝑠 = 35 in., 𝑅𝑖 = 8.3 in., 𝑅𝑜 = 11 in., and 𝜔𝑑 = 15 rad∕s. 𝐵 𝛼𝑑

𝑃

𝑠

𝑅𝑜 𝑅𝑖 disk

𝜔𝑑

𝐴 Figure P16.146 and P16.147

If 𝛼𝑑 is zero at the instant shown, determine the angular velocity and angular acceleration of the bar 𝐴𝐵.

Problem 16.146

If 𝛼𝑑 = 30 rad∕s2 at the instant shown, determine the angular velocity and angular acceleration of the bar 𝐴𝐵. Problem 16.147

𝑧, 𝑍

𝑑

Problem 16.148 A vertical shaft has a base 𝐵 that is stationary relative to an inertial reference frame with vertical axis 𝑍. Arm 𝑂𝐴 is attached to the vertical shaft and rotates about the 𝑍 axis with an angular velocity 𝜔𝑂𝐴 = 5 rad∕s and an angular acceleration 𝛼𝑂𝐴 = 1.5 rad∕s2 . The 𝑧 axis is coincident with the 𝑍 axis, but is part of a reference frame that rotates with the arm 𝑂𝐴. The 𝑥 axis of the rotating reference frame coincides with the axis of the arm 𝑂𝐴. At the instant shown, the 𝑦 axis is perpendicular to the page and directed into the page. At this instant, the collar 𝐶 is sliding along 𝑂𝐴 with a constant speed 𝑣𝐶 = 5 f t∕s and is at a distance 𝑑 = 1.2 f t from the 𝑧 axis. Compute the inertial acceleration of the collar, and express it relative to the rotating coordinate system.

𝑣𝐶

𝐴

𝑂 𝜔𝑂𝐴

𝑥

𝐶

𝛼𝑂𝐴 𝐵 Figure P16.148

Problem 16.149 Let frames 𝐴 and 𝐵 be the frames with origins at points 𝐴 and 𝐵, respectively. Point 𝐵 does not move relative to point 𝐴. The velocity and acceleration of point 𝑃 relative to frame 𝐵 are 𝑣⃗𝑃 rel = (−6.14 𝚤̂𝐵 + 23.7 𝚥̂𝐵 ) f t∕s

𝑣⃗𝑃 rel 𝑦𝐴 𝑦𝐵

and 𝑎⃗𝑃 rel = (3.97 𝚤̂𝐵 + 4.79 𝚥̂𝐵 ) f t∕s2 .

Knowing that, at the instant shown, frame 𝐵 rotates relative to frame 𝐴 at a constant angular velocity 𝜔𝐵 = 1.2 rad∕s, that the position of point 𝑃 relative to frame 𝐵 is 𝑟⃗𝑃 ∕𝐵 = (8 𝚤̂𝐵 + 4.5 𝚥̂𝐵 ) f t, and that frame 𝐴 is fixed, determine the velocity and acceleration of 𝑃 at the instant shown and express the results using the frame 𝐴 component system.

𝑎⃗𝑃 rel 𝑟⃗𝑃 ∕𝐵 𝚥̂𝐵 𝚥̂𝐴

𝐵 𝐴

Problem 16.150 Repeat Prob. 16.149, but express the results using the component system of frame 𝐵.

𝜔𝐵

𝑥𝐵

𝑃 23◦

𝚤̂𝐵

𝚤̂𝐴

Figure P16.149 and P16.150

𝑥𝐴

1132

Chapter 16

Planar Rigid Body Kinematics

Problem 16.151

𝑧, 𝑍

𝑑 𝑣𝐶

𝐷 𝓁

𝑂

𝐶

𝜔𝑂𝐴

𝐴

A vertical shaft has a base 𝐵 that is stationary relative to an inertial reference frame with vertical axis 𝑍. Arm 𝑂𝐴 is attached to the vertical shaft and rotates about the 𝑍 axis with an angular velocity 𝜔𝑂𝐴 = 5 rad∕s and an angular acceleration 𝛼𝑂𝐴 = 1.5 rad∕s2 . The 𝑧 axis is coincident with the 𝑍 axis but is part of a reference frame that rotates with the arm 𝑂𝐴. The 𝑥 axis of the rotating reference frame coincides with the axis of the arm 𝑂𝐴. At the instant shown, the 𝑦 axis is perpendicular to the page and directed into the page. At this instant, the collar 𝐶 is sliding along 𝑂𝐴 with a constant speed 𝑣𝐶 = 3.32 m∕s and is rotating with a constant angular velocity 𝜔𝐶 = 2.3 rad∕s relative to the arm 𝑂𝐴. At the instant shown, point 𝐷 happens to be in the 𝑥𝑧 plane and is at a distance 𝓁 = 0.05 m from the 𝑥 axis and at a distance 𝑑 = 0.75 m from the 𝑧 axis. Compute the inertial acceleration of point 𝐷, and express it relative to the rotating reference frame.

𝑥

𝜔𝐶

𝛼𝑂𝐴 𝐵 Figure P16.151

𝐷

𝑦 𝜔2

𝑣0 𝑅

𝛼2

𝐵

Problems 16.152 and 16.153

𝑎0

𝐶 𝑑 𝛼1

𝑥

Bar 𝐴𝐵 rotates about point 𝐴 with angular speed 𝜔1 and angular acceleration 𝛼1 , both in the directions shown. The curved bar 𝐵𝐶 is pinned to bar 𝐴𝐵 at 𝐵 and rotates with angular speed 𝜔2 and angular acceleration 𝛼2 about the pin at 𝐵 in the directions shown, relative to bar 𝐴𝐵. At this instant, collar 𝐷 is at the top of the curved bar 𝐵𝐶. The speed and acceleration of 𝐷 relative to bar 𝐵𝐶 are given by 𝑣0 and 𝑎0 , respectively. Express your answers in the 𝑥𝑦𝑧 frame that is attached to bar 𝐵𝐶 with its origin at 𝐵. Hint: As we will see in Eq. (20.14), 𝜔 ⃗ 𝐵𝐶 = 𝜔 ⃗ 𝐴𝐵 + 𝜔 ⃗ 𝐵𝐶∕𝐴𝐵 = 𝜔 ⃗1 + 𝜔 ⃗ 2.

𝜔1

Problem 16.152

𝐴

𝑣0 . Problem 16.153

Figure P16.152 and P16.153

Determine the velocity of the collar 𝐷 in terms of 𝑑, 𝜔1 , 𝜔2 , 𝑅, and

Determine the acceleration of the collar 𝐷 in terms of 𝑑, 𝑅, 𝜔1 , 𝜔2 ,

𝛼1 , 𝛼2 , 𝑣0 , and 𝑎0 .

Problems 16.154 through 16.158

𝜔𝐴𝐵

𝐷 𝑅

disk

𝐴 𝐵

The reciprocating rectilinear motion mechanism consists of a disk pinned at its center at 𝐴 that rotates with a constant angular velocity 𝜔𝐴𝐵 , a slotted arm 𝐶𝐷 that is pinned at 𝐶, and a bar that can oscillate within the guides at 𝐸 and 𝐹 . As the disk rotates, the peg at 𝐵 moves within the slotted arm, causing it to rock back and forth. As the arm rocks, it provides a slow advance and a quick return to the reciprocating bar due to the change in distance between 𝐶 and 𝐵. Problem 16.154

For the position shown, determine

ℎ

(a) the angular velocity of the slotted arm 𝐶𝐷 and the velocity of the bar. 𝑑

(b) the angular acceleration of the slotted arm 𝐶𝐷 and the acceleration of the bar.

𝐶 bar

𝐸 Figure P16.154

ISTUDY

𝐹

Evaluate your results for 𝜔𝐴𝐵 = 120 rpm, 𝑅∕ℎ = 0.5, and 𝑑 = 0.12 m. Problem 16.155

For the position shown, determine

(a) the angular velocity of the slotted arm 𝐶𝐷 and the velocity of the bar. (b) the angular acceleration of the slotted arm 𝐶𝐷 and the acceleration of the bar. Evaluate your results for 𝜔𝐴𝐵 = 90 rpm, 𝑅∕ℎ = 0.5, and 𝑑 = 0.12 m.

ISTUDY

Section 16.4

1133

Rotating Reference Frames

𝐷

𝐷 𝐵 disk

𝐴

𝑅

𝜔𝐴𝐵

disk

𝜔𝐴𝐵 𝐴 𝑅

ℎ

𝐵

ℎ 𝐶

𝐶

𝑑

𝑑 bar

bar 𝐹

𝐸

𝐸

Figure P16.155 Problem 16.156

𝐹 Figure P16.156

For the position shown, determine

(a) the angular velocity of the slotted arm 𝐶𝐷 and the velocity of the bar. (b) the angular acceleration of the slotted arm 𝐶𝐷 and the acceleration of the bar. Evaluate your results for 𝜔𝐴𝐵 = 60 rpm, 𝑅∕ℎ = 0.5, and 𝑑 = 0.12 m. Problem 16.157

For the arbitrary position shown, determine 𝐷

𝜔𝐴𝐵

(a) the angular velocity of the slotted arm 𝐶𝐷 and the velocity of the bar. (b) the angular acceleration of the slotted arm 𝐶𝐷 and the acceleration of the bar

𝑅

as functions of 𝜃, 𝛿 = 𝑅∕ℎ, 𝑑, and 𝜔𝐴𝐵 . Problem 16.158 Determine the angular velocity and angular acceleration of the slotted arm 𝐶𝐷 as functions of 𝜃, 𝛿 = 𝑅∕ℎ, 𝑑, and 𝜔𝐴𝐵 . After doing so, plot the velocity and acceleration of the bar as a function of the disk angle 𝜃 for one full cycle of the disk’s motion and for 𝜔𝐴𝐵 = 90 rpm, 𝑑 = 0.12 m, and

𝜃

𝐵

disk 𝐴

ℎ

𝐶

(a) 𝛿 = 𝑅∕ℎ = 0.1.

𝑑

(b) 𝛿 = 𝑅∕ℎ = 0.3. (c) 𝛿 = 𝑅∕ℎ = 0.6. (d) 𝛿 = 𝑅∕ℎ = 0.9. Explain why this mechanism is often referred to as a quick-return mechanism.

Problem 16.159 The reciprocating rectilinear motion mechanism of Probs. 16.154–16.158 is often referred to as a quick-return mechanism since it can move much more quickly in one direction than the other. To see this, we will determine the velocity of the bar in each of the two positions shown (position 1 when 𝐵 is farthest from 𝐶 and position 2 when 𝐵 is closest to 𝐶) under the assumption that the disk, whose center is at 𝐴, rotates with a constant angular velocity 𝜔𝐴𝐵 = 120 rpm. Find the velocity of the bar in positions 1 and 2 for 𝑑 = 0.12 m and for (a) 𝑅∕ℎ = 0.3. (b) 𝑅∕ℎ = 0.8. Comment on which of the two 𝑅∕ℎ values would provide a better quick return and why.

bar 𝐸 Figure P16.157 and P16.158

𝐹

1134

Chapter 16

Planar Rigid Body Kinematics

𝐷

𝐷 𝐵

disk

𝜔𝐴𝐵

𝑅

𝐴

disk

𝜔𝐴𝐵 𝐴 𝑅

ℎ

𝐵

ℎ 𝐶

𝐶

𝑑

𝑑 bar

𝐸

𝐹

position 1

𝐸

𝐹

position 2

Figure P16.159

Problem 16.160 The wheel 𝐷 rotates with a constant angular velocity 𝜔𝐷 = 14 rad∕s about the fixed point 𝑂, which is assumed to be stationary relative to an inertial frame of reference. The 𝑥𝑦𝑧 frame rotates with the wheel. Collar 𝐶 slides along the bar 𝐴𝐵 with a constant velocity 𝑣𝐶 = 4 f t∕s relative to the 𝑥𝑦𝑧 frame. Letting 𝓁 = 0.25 f t, determine the inertial velocity and acceleration of 𝐶 when 𝜃 = 25◦ . Express the result with respect to the 𝑥𝑦𝑧 frame. 𝜔𝐷 𝑣𝐶 𝐴

𝐶

𝑌 𝑣𝐶

𝑦 𝐵

𝜃

𝑂

𝐴

𝑦 𝐵

𝐶

𝜃

𝓁

𝑂

𝓁 𝑥

𝑥 𝜔𝐷 𝐷

Figure P16.160

𝑅 𝐷 𝑋 Figure P16.161

Problem 16.161 The wheel 𝐷 rotates without slipping over a flat surface. The 𝑋𝑌𝑍 frame shown is inertial, whereas the 𝑥𝑦𝑧 frame is attached to 𝐷 at 𝑂 and rotates with it at a constant angular velocity 𝜔𝐷 = 14 rad∕s. Collar 𝐶 slides along the bar 𝐴𝐵 with a constant velocity 𝑣𝐶 = 4 f t∕s relative to the 𝑥𝑦𝑧 frame. Letting 𝓁 = 0.25 f t and 𝑅 = 1 f t, determine the inertial velocity and acceleration of 𝐶 when 𝜃 = 25◦ and the 𝑥𝑦𝑧 frame is parallel to the 𝑋𝑌𝑍 frame as shown. Express your result in both the 𝑥𝑦𝑧 and 𝑋𝑌𝑍 frames.

𝜔𝐵𝐶 𝑑 𝐵 𝐴

ℎ

𝜃 𝓁

Figure P16.162

ISTUDY

Problem 16.162

𝜙 𝐶

A floodgate is controlled by the motion of the hydraulic cylinder 𝐴𝐵. If the gate 𝐵𝐶 is to be lifted with a constant angular velocity 𝜔𝐵𝐶 = 0.5 rad∕s, determine 𝑑̇𝐴𝐵 and 𝑑̈𝐴𝐵 , where 𝑑𝐴𝐵 is the distance between points 𝐴 and 𝐵 when 𝜙 = 0. Let 𝓁 = 10 f t, ℎ = 2.5 f t, and 𝑑 = 5 f t.

ISTUDY

Section 16.4

1135

Rotating Reference Frames

Problem 16.163 The wheel 𝐷 rotates without slipping over a flat surface. The 𝑋𝑌𝑍 frame shown is inertial, whereas the 𝑥𝑦𝑧 frame is attached to 𝐷 and rotates with it at a constant angular velocity 𝜔𝐷 . Collar 𝐶 slides along the bar 𝐴𝐵 with a velocity 𝑣𝐶 relative to the 𝑥𝑦𝑧 frame. Suppose that 𝓁, 𝜃, and 𝑅 are given and that we want to determine the inertial acceleration of 𝐶 when the 𝑥𝑦𝑧 frame is parallel to the 𝑋𝑌𝑍 frame as shown. Would the expression of the inertial acceleration of the collar in the two frames be different or the same? 𝑌 𝑣𝐶 𝐴

𝜔𝐷

𝑦 𝐵

𝐶

𝜃

𝑂

𝓁 𝑥

𝑅 𝐷 𝑋 Figure P16.163

Problem 16.164 The Pioneer 3 spacecraft was a spin-stabilized spacecraft launched on December 6, 1958, by the U.S. Army Ballistic Missile agency in conjunction with NASA. It was designed with a despin mechanism consisting of two equal masses 𝐴 and 𝐵 that could be spooled out to the end of two wires of variable length 𝓁(𝑡) when triggered by a hydraulic timer.∗ As a prelude to Probs. 17.95, 17.96, and 18.117, we will find the velocity and acceleration of each of the two masses. To do this, assume that masses 𝐴 and 𝐵 are initially at positions 𝐴0 and 𝐵0 , respectively. After the masses are released, they begin to unwind symmetrically, and the length of the cord attaching each mass to the spacecraft of radius 𝑅 is 𝓁(𝑡). Given that the angular velocity of the spacecraft at each instant is 𝜔𝑠 (𝑡), determine the velocity and acceleration of mass 𝐴 in components expressed in the rotating reference frame whose origin is at 𝑄, as well as 𝑅, 𝓁(𝑡), and 𝜔𝑠 (𝑡). Note that the rotating frame is always aligned with the unwinding cord, and 𝑄 is the point on the cord that is about to unwind at time 𝑡. Hint: The point 𝑄 moves around the periphery of the spacecraft with angular speed 𝜃̇ ̈ It is not fixed to the spacecraft. and angular acceleration 𝜃.

Problem 16.165 At the instant shown, the wheel 𝐷 rolls without slipping over a flat surface with an angular velocity 𝜔𝐷 = 14 rad∕s and an angular acceleration 𝛼𝐷 = 1.1 rad∕s2 . The 𝑋𝑌𝑍 frame shown is inertial, whereas the 𝑥𝑦𝑧 frame is attached to 𝐷. At the instant shown, the collar 𝐶 is sliding along the bar 𝐴𝐵 with a velocity 𝑣𝐶 = 4 f t∕s and acceleration 𝑎𝐶 = 7 f t∕s2 , both relative to the 𝑥𝑦𝑧 frame. Letting 𝓁 = 0.25 f t and 𝑅 = 1 f t, determine the inertial velocity and acceleration of 𝐶 when 𝜃 = 25◦ and the 𝑥𝑦𝑧 frame is parallel to the 𝑋𝑌𝑍 frame as shown. Express your result in both the 𝑥𝑦𝑧 and the 𝑋𝑌𝑍 frames.

𝑧

𝑥

𝜔𝑠

𝐴

𝓁 𝑦

𝜔𝑠

𝑄 𝐵0

𝜃 𝑂

𝐴0

𝑅

top view 𝐵 NASA

Figure P16.164 𝜔𝐷

𝛼𝐷

𝑌

𝐴

𝐶

𝑣𝐶

𝑦 𝐵

𝑎𝐶

𝜃

𝑂

𝓁 𝑥

𝑅 𝐷 𝑋

∗ The

spacecraft was intended as a lunar probe, but it failed to go past the Moon and into a heliocentric (sun-centered) orbit as planned. It did reach an altitude of 107,400 km before falling back to Earth. It was a cone-shaped probe 58 cm high and 25 cm diameter at its base and was designed with a despin mechanism consisting of two 7 g masses that could be spooled out to the end of two 150 cm wires when triggered by a hydraulic timer 10 h after launch. The masses would slow the spacecraft spin from 400 to 6 rpm, and then the masses and wires would be released.

Figure P16.165

1136

Chapter 16

Planar Rigid Body Kinematics

16.5 C h a p t e r R e v i e w Fundamental equations, translation, and rotation about a fixed axis 𝑣⃗𝐵∕𝐴 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 𝐵 𝜔 ⃗ 𝐴𝐵 −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 𝑟⃗𝐵∕𝐴 𝛼⃗𝐴𝐵 𝐴

This section began our study of the dynamics of rigid bodies. As with particles, we started with the study of kinematics, which is the focus of this chapter. The key kinematic idea is that a rigid body has only one angular velocity and one angular acceleration; i.e., each is a property of the body as a whole. This allowed us to use the relative velocity equation [Eq. (12.84) on p. 739] and the relation for the time derivative of a vector [Eq. (12.60) on p. 718] to relate the velocities of two points 𝐴 and 𝐵 on the same rigid body using (see Fig. 16.52) Eq. (16.3), p. 1057 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝑣⃗𝐴 + 𝑣⃗𝐵∕𝐴 , and the accelerations using Eq. (16.5), p. 1058 ( ) 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 ,

Figure 16.52 A rigid body showing the quantities used in Eqs. (16.3), (16.5), and (16.14).

ISTUDY

which for planar motion becomes Eq. (16.14), p. 1062 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 .

Translation. For this motion, the angular velocity and angular acceleration of the body are equal to zero, i.e., Eqs. (16.6), p. 1060 ⃗ 𝜔 ⃗ body = 0⃗ and 𝛼⃗body = 0, so that the velocity and acceleration relations for the body reduce to Eqs. (16.7), p. 1060 𝑣⃗𝐵 = 𝑣⃗𝐴

and 𝑎⃗𝐵 = 𝑎⃗𝐴 ,

where 𝐴 and 𝐵 are any two points on the body.

Rotation about a fixed axis. For this special motion, there is an axis of rotation perpendicular to the plane of motion that does not move. All points not on the axis of rotation can only move in a circle about that axis. If the axis of rotation is at point 𝐴, then the velocity of 𝐵 is given by Eq. (16.8), p. 1060 𝑣⃗𝐵 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , and its acceleration is Eq. (16.9), p. 1060, and Eq. (16.13), p. 1061 ( ) 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 , where the vectors 𝜔 ⃗ 𝐴𝐵 and 𝛼⃗𝐴𝐵 are the angular velocity and angular acceleration of the body, respectively, and 𝑟⃗𝐵∕𝐴 is the vector that describes the position of 𝐵 relative to 𝐴.

ISTUDY

Section 16.5

1137

Chapter Review 𝑣⃗𝐵

Planar motion: velocity analysis 𝐵

This section presented three different ways to analyze the velocities of a rigid body in planar motion: the vector approach, differentiation of constraints, and instantaneous center of rotation.

𝐶 𝑣⃗𝐶

Vector approach. We saw in Section 16.1 that the equation Eq. (16.15), p. 1075 𝑣⃗𝐶 = 𝑣⃗𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 ,

𝑟⃗𝐶∕𝐵 𝜔 ⃗ 𝐵𝐶

rigid body Figure 16.53 A rigid body on which we are relating velocity of two points 𝐵 and 𝐶.

relates the velocity of two points on a rigid body 𝑣⃗𝐵 and 𝑣⃗𝐶 via their relative position 𝑟⃗𝐶∕𝐵 and the angular velocity of the body 𝜔 ⃗ 𝐵𝐶 (see Fig. 16.53).

𝜔𝑊

𝑊

Rolling without slip. When a body rolls without slip over another body (see, for example, the wheel 𝑊 rolling over the surface 𝑆 in Fig. 16.54), then the two points on the bodies that are in contact at any instant, points 𝑃 and 𝑄, must have the same velocity. Mathematically, this means that

𝑢̂ 𝑛 𝑃

𝑢̂ 𝑡

𝓁

𝑄

Eqs. (16.16), p. 1076 𝑣⃗𝑃 ∕𝑄 = 0⃗

⇒

𝑣⃗𝑃 = 𝑣⃗𝑄 .

If a wheel of radius 𝑅 is rolling without slip over a flat, stationary surface, then the point 𝑃 on the wheel in contact with the surface must have zero velocity (see Fig. 16.55). The consequence is that the angular velocity of the wheel 𝜔𝑊 is related to the velocity of the center 𝑣𝑂 and the radius of the wheel 𝑅 according to

𝑆 Figure 16.54 A wheel 𝑊 rolling over a surface 𝑆. At the instant shown, the line 𝓁 is tangent to the path of points 𝑃 and 𝑄.

𝜔𝑊

Eq. (16.19), p. 1076 𝜔𝑊 = −

𝑣𝑂 𝑅

𝑅

in which the positive direction for 𝜔𝑊 is taken to be the positive 𝑧 direction using the right-hand rule.

𝑣𝑂

𝑂

,

𝑊

𝚥̂ 𝚤̂

𝑃 𝑄

Differentiation of constraints. As we discovered in Section 12.6, it is often convenient to write an equation describing the position of a point of interest, which can then be differentiated with respect to time to find the velocity of that point. For planar motion of rigid bodies, this idea can also apply for describing the position and velocity of a point on a rigid body, as well as for describing the orientation of a rigid body, for which the time derivative provides its angular velocity. Instantaneous center of rotation. The point on a body (or imaginary extension of the body) whose velocity is zero at a particular instant is called the instantaneous center of rotation or instantaneous center (IC). The IC can be found geometrically if the velocity is known for two distinct points on a body or if a velocity on the body and the body’s angular velocity are known. The three possible geometric constructions are shown in Fig. 16.56.

Planar motion: acceleration analysis This section presented two different ways to analyze the accelerations of a rigid body in planar motion: the vector approach and differentiation of constraints.

Figure 16.55 A wheel rolling without slip on a horizontal fixed surface.

1138

Chapter 16

Planar Rigid Body Kinematics

(b)

(a)

(c) IC

𝑣⃗𝐵 𝓁𝐶

𝐵

𝐵

IC 𝓁𝐵

𝐵

𝑣⃗𝐵

𝑣⃗𝐶

𝐶

𝐶

𝑣⃗𝐶 𝐶

𝑣⃗𝐶

𝓁𝐶

𝓁𝐵

𝑣⃗𝐵

Figure 16.56. The three different possible motions for determining the IC. (a) Two nonparallel velocities are known. (b) The lines of action of two velocities are parallel and distinct; in this case, the IC is at infinity and the body is translating. (c) The lines of action of two parallel velocities coincide.

Vector approach. For planar motion, either of the equations 𝑎⃗𝐵

𝛼⃗𝐵𝐶

Eqs. (16.29) and (16.30), p. 1097 ( ) 𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 + 𝜔 ⃗ 𝐵𝐶 × 𝜔 ⃗ 𝐵𝐶 × 𝑟⃗𝐶∕𝐵 ,

𝐵 𝑟⃗𝐶∕𝐵 𝐶 𝑎⃗𝐶

𝜔 ⃗ 𝐵𝐶

rigid body Figure 16.57 A rigid body on which we are relating the acceleration of two points 𝐵 and 𝐶.

𝛼𝑊

𝜔𝑊

𝑊

𝑎⃗𝐶 = 𝑎⃗𝐵 + 𝛼⃗𝐵𝐶 × 𝑟⃗𝐶∕𝐵 − 𝜔2𝐵𝐶 𝑟⃗𝐶∕𝐵 , relates the acceleration of two points on a rigid body 𝑎⃗𝐵 and 𝑎⃗𝐶 via their relative position 𝑟⃗𝐶∕𝐵 , the angular acceleration of the body 𝛼⃗𝐵𝐶 , and the angular velocity of the body 𝜔 ⃗ 𝐵𝐶 (see Fig. 16.57).

Differentiation of constraints. It is often convenient to write an equation describing the position of a point of interest, which can then be differentiated with respect to time to find the velocity and acceleration of that point. For planar motion of rigid bodies, this idea can also apply for describing the position, velocity, and acceleration of a point on a rigid body, as well as for describing the orientation of a rigid body, for which time derivatives provide its angular velocity and angular acceleration. Rolling without slip. When a body rolls without slip over another body (see Fig. 16.58), then the two points on the bodies that are in contact at any instant, points 𝑃 and 𝑄, must have the same velocity. For accelerations, this means that the component of 𝑎⃗𝑃 ∕𝑄 tangent to the contact must be equal to zero, that is,

𝑢̂ 𝑡

𝑃

Eqs. (16.34), p. 1098

𝓁

𝑄

(𝑎𝑃 ∕𝑄 )𝑡 = 0

𝑆 Figure 16.58 A wheel 𝑊 rolling over a surface 𝑆. At the instant shown, the line 𝓁 is tangent to the path of 𝑃 and 𝑄.

ISTUDY

𝑎𝑃 𝑡 = 𝑎𝑄𝑡 .

⇒

If a wheel of radius 𝑅 is rolling without slip over a flat, stationary surface, then the point 𝑃 on the wheel in contact with the surface must have zero velocity (see Fig. 16.59). The consequence is that the angular acceleration of the wheel 𝛼𝑊 is related to the

𝛼𝑊

𝑊

𝜔𝑊

𝑎𝑂 𝑂

𝑣𝑂

𝑟⃗𝑃 ∕𝑂

𝚥̂ 𝚤̂

𝑃 𝑄

Figure 16.59. A wheel rolling without slip on a horizontal fixed surface.

ISTUDY

Section 16.5

Chapter Review

1139

acceleration of the center 𝑎𝑂 and the radius of the wheel 𝑅 according to Eq. (16.38), p. 1099 𝛼𝑊 = −

𝑎𝑂 𝑅

,

in which the positive direction for 𝛼𝑊 is taken to be the positive 𝑧 direction. The point 𝑃 on the wheel that is in contact with the ground at this instant is accelerating along the 𝑦 axis according to Eq. (16.39), p. 1099 𝑎⃗𝑃 =

𝑣2𝑂 𝑅

𝚥̂.

Even though 𝑃 has zero velocity at the instant considered, it is accelerating.

Rotating reference frames In this section, we developed the kinematic equations that allow us to relate the motion of two points that are not on the same rigid body. Referring to Fig. 16.60, this means we can relate the velocity of point 𝑃 to that of 𝐴 using the relation

𝑦 𝑎⃗𝐴

𝛺

Eq. (16.45), p. 1119

𝑣⃗𝐴

𝑧

⃗ × 𝑟⃗ , 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝐴

𝑥

𝛺̇

where

𝑃

𝑟⃗𝑃 ∕𝐴

𝐴

𝐵

𝑟⃗𝐴

𝑣⃗𝐴 = velocity of the origin of the rotating or body-fixed reference

𝑌

frame (point 𝐴 in Fig. 16.60),

𝑟⃗𝑃

𝑣⃗𝑃 rel = velocity of 𝑃 as seen by an observer moving with the rotat-

ing reference frame, ⃗ = angular velocity of the rotating reference frame, Ω 𝑟⃗𝑃 ∕𝐴 = position of point 𝑃 relative to 𝐴. We also found that we can relate the acceleration of point 𝑃 to that of 𝐴 using Eq. (16.53), p. 1120 ( ) ⃗̇ ⃗ ⃗ × 𝑣⃗ ⃗ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 + Ω × Ω × 𝑟 𝑃 ∕𝐴 , where, in addition to the terms defined for 𝑣⃗𝑃 , we have 𝑎⃗𝐴 = acceleration of the origin of the rotating or body-fixed ref-

erence frame (point 𝐴 in Fig. 16.60), 𝑎⃗𝑃 rel = acceleration of 𝑃 as seen by an observer moving with the

rotating reference frame, ⃗ × 𝑣⃗ 2Ω 𝑃 rel = Coriolis acceleration of 𝑃 , ⃗̇ = angular acceleration of the rotating reference frame. Ω

𝑂

𝑋

𝑍 Figure 16.60 The essential ingredients for finding velocities and accelerations using a rotating reference frame.

1140

Chapter 16

Planar Rigid Body Kinematics

Review Problems Problem 16.166 The manually operated road barrier shown has a total length 𝑙 = 15.7 f t and has a counterweight 𝐶 whose position is identified by the distances 𝑑 = 2.58 f t and 𝛿 = 1.4 f t. The barrier is pinned at 𝑂 and can move in the vertical plane. If the barrier is raised and then released so that the end 𝐵 of the barrier hits the support with a speed 𝑣𝐵 = 1.5 f t∕s, determine the speed of the counterweight 𝐶 when 𝐵 hits 𝐴. 𝐵

𝑙

𝐴 𝐶

𝐵

𝑂

𝑂

𝑑 𝛿 𝐶

𝐴

Figure P16.166

Problems 16.167 and 16.168 Bar 𝐴𝐵 rotates about point 𝐴 with angular speed 𝜔1 and angular acceleration 𝛼1 , both in the directions shown. Bar 𝐵𝐶 rotates about the pin at 𝐵 relative to bar 𝐴𝐵 with angular speed 𝜔2 and angular acceleration 𝛼2 , both in the directions shown. The position of collar 𝐷 relative to bar 𝐵𝐶 is given by the coordinate 𝑠, which is a known function of time. Express your answers in the 𝑥𝑦𝑧 frame that is attached to bar 𝐵𝐶 with its origin at 𝐵. Hint: As we will see in Eq. (20.14), 𝜔 ⃗ 𝐵𝐶 = 𝜔 ⃗ 𝐴𝐵 + 𝜔 ⃗ 𝐵𝐶∕𝐴𝐵 = 𝜔 ⃗1 + 𝜔 ⃗ 2. 𝑦 𝑠 𝐵 𝛼2 𝑑

𝜔2

𝑠̈ 𝐷

𝐶

𝑥

𝑠̇

𝜔1 𝛼1

𝐴 Figure P16.167 and P16.168

Problem 16.167

Determine an expression for the Coriolis acceleration of the collar

𝐷. Determine an expression for the acceleration of the collar 𝐷 in terms of 𝑑, 𝜔1 , 𝜔2 , 𝛼1 , 𝛼2 , and the position coordinate 𝑠 and its time derivatives.

Problem 16.168

Problem 16.169

𝚥̂ 𝚤̂ 𝜔𝐴 𝐴

𝑅

𝑑 Figure P16.169

ISTUDY

𝐶 𝓁 𝜃 𝑂

𝐿 𝐵

𝜙

𝐻 𝐷

At the instant shown, the hammer head 𝐻 is moving to the right with a speed 𝑣𝐻 = 45 f t∕s and the angle 𝜃 = 20◦ . Assuming that the belt does not slip relative to wheels 𝐴 and 𝐵 and assuming that wheel 𝐴 is mounted on the shaft of the motor shown, determine the angular velocity of the motor at the instant shown. The diameter of wheel 𝐴 is 𝑑 = 0.25 f t, the radius of wheel 𝐵 is 𝑅 = 0.75 f t, and point 𝐶 is at a distance 𝓁 = 0.72 f t from 𝑂, which is the center of the wheel 𝐵. Finally, let 𝐶𝐷 have a length 𝐿 = 2 f t, and assume that, at the instant shown, 𝜙 = 25◦ .

ISTUDY

Section 16.5

1141

Chapter Review

Problems 16.170 and 16.171

𝑥

The bar 𝐴𝐵 is pinned to the ground at 𝐴, and its angle 𝜃 with respect to the vertical is a known function of time. The bar 𝐵𝐶 is pinned to the bar 𝐴𝐵 at 𝐵, and its orientation 𝜙 is measured with respect to the bar 𝐴𝐵. The position of the collar 𝐷 is measured relative to the bar 𝐵𝐶 by the coordinate 𝑠, which is also a known function of time. The length of the bar 𝐴𝐵 is 𝑑 = 2 f t. At the instant shown, 𝜃 = 30◦ , 𝜙 = 30◦ , 𝜃̇ = 5 rad∕s, 𝜃̈ = 3 rad∕s2 , 𝜙̇ = −5 rad∕s, 𝜙̈ = 4 rad∕s2 , 𝑠 = 2.5 f t, 𝑠̇ = 10 f t∕s, and 𝑠̈ = −15 f t∕s2 . Express all your answers in the 𝑥𝑦𝑧 frame that is attached to the bar 𝐵𝐶 and whose origin is at 𝐵. Hint: As we will see in Eq. (20.14), 𝜔 ⃗ 𝐵𝐶 = 𝜔 ⃗ 𝐴𝐵 + 𝜔 ⃗ 𝐵𝐶∕𝐴𝐵 . Problem 16.170

Determine the velocity of the collar 𝐷 at this instant.

Problem 16.171

Determine the acceleration of the collar 𝐷 at this instant.

𝐶 𝑠̇ 𝑠̈

𝐷 𝑠 𝜙

𝑦 𝐵 𝜃 𝑑

𝐴

Problem 16.172 Assuming that the rope does not slip relative to any of the pulleys in the system, determine the velocity and acceleration of 𝐴 and 𝐷, knowing that the angular velocity and acceleration of pulley 𝐵 are 𝜔𝐵 = 7 rad∕s and 𝛼𝐵 = 3 rad∕s2 , respectively. The diameters of pulleys 𝐵 and 𝐶 are 𝑑 = 25 cm and 𝓁 = 34 cm, respectively.

Figure P16.170 and P16.171

𝐶

At the instant shown, bar 𝐴𝐵 rotates with a constant angular velocity 𝜔𝐴𝐵 = 24 rad∕s. Letting 𝐿 = 0.75 m and 𝐻 = 0.85 m, determine the angular acceleration of bar 𝐵𝐶 when bars 𝐴𝐵 and 𝐶𝐷 are as shown, i.e., parallel and horizontal. 𝐿 𝜔𝐴𝐵

𝑑 𝑂 𝐿

𝐵 𝐴 𝐻

𝚥̂

𝐷

𝑟

𝐿

Figure P16.173

𝐶

𝓁

𝑑

𝐷 𝛼𝐷 𝑣𝐷

𝐵

𝛼𝐵

𝐸

𝜔𝐵

𝐺 𝑣

𝚤̂ 𝐷

𝛼𝐶

𝜔𝐶

Problem 16.173

𝑎𝐴

𝐵

𝐴

𝐴

𝑣𝐴

Figure P16.172

Figure P16.174

Problem 16.174 A slender bar 𝐴𝐵 of length 𝐿 = 1.45 f t is mounted on two identical disks 𝐷 and 𝐸 pinned at 𝐴 and 𝐵, respectively, and of radius 𝑟 = 1.5 in. The bar is allowed to move within a cylindrical bowl with center at 𝑂 and diameter 𝑑 = 2 f t. At the instant shown, the center 𝐺 of the bar is moving with a speed 𝑣 = 7 f t∕s. Determine the angular velocity of the bar at the instant shown. 𝐶

Problem 16.175 The bucket of a backhoe is the element 𝐴𝐵 of the four-bar linkage system 𝐴𝐵𝐶𝐷. The bucket’s motion is controlled by extending or retracting the hydraulic arm 𝐸𝐶. Assume that the points 𝐴, 𝐷, and 𝐸 are fixed and that the bucket is made to rotate with a constant angular velocity 𝜔𝐴𝐵 = 0.25 rad∕s. In addition, suppose that, at the instant shown, point 𝐵 is vertically aligned with point 𝐴, and point 𝐶 is horizontally aligned with 𝐵. Letting 𝑑𝐸𝐶 denote the distance between points 𝐸 and 𝐶, determine 𝑑̇𝐸𝐶 and 𝑑̈𝐸𝐶 at the instant shown. Let ℎ = 0.66 f t, 𝑒 = 0.46 f t, 𝓁 = 0.9 f t, 𝑤 = 1.0 f t, 𝑑 = 4.6 f t, and 𝑞 = 3.2 f t.

𝜔𝐴𝐵

𝑤 𝐵 𝐴 𝑞

𝓁 𝑒

𝐷 ℎ 𝚥̂

𝐸 𝑑 Figure P16.175

𝚤̂

1142

Chapter 16

Planar Rigid Body Kinematics Problems 16.176 and 16.177

horizontal guide

fully open

An overhead fold-up door with height 𝐻 = 30 f t consists of two identical sections hinged at 𝐶. The roller at 𝐴 moves along a horizontal guide, whereas the rollers at 𝐵 and 𝐷, which are the midpoints of sections 𝐴𝐶 and 𝐶𝐸, move along a vertical guide. The door’s operation is assisted by a counterweight 𝑃 . Express your answers using the component system shown.

𝐴 𝜃

𝐵

𝐶

If at the instant shown, the angle 𝜃 = 55◦ and 𝑃 is moving upward with a speed 𝑣𝑃 = 15 f t∕s, determine the velocity of point 𝐸, as well as the angular velocities of sections 𝐴𝐶 and 𝐶𝐸.

Problem 16.176

vertical guide

𝐻 𝐷

𝑃

If at the instant shown, 𝜃 = 45◦ , and 𝐴 is moving to the right with a speed 𝑣𝐴 = 2 f t∕s while decelerating at a rate of 1.5 f t∕s2 , determine the acceleration of point 𝐸.

Problem 16.177

𝐸

𝚥̂ 𝚤̂

fully closed floor

Problem 16.178 A vertical shaft has a base 𝐵 that is stationary relative to an inertial reference frame with vertical axis 𝑍. Arm 𝑂𝐴 is attached to the vertical shaft and rotates about the 𝑍 axis with an angular velocity 𝜔𝑂𝐴 and an angular acceleration 𝛼𝑂𝐴 . The 𝑥𝑦𝑧 frame is attached at 𝐶, such that 𝑍 and 𝑧 are always parallel and the 𝑥 axis of the rotating reference frame coincides with the axis of the arm 𝑂𝐴. At this instant, the collar 𝐶 is at a distance 𝑑 from the 𝑍 axis, is sliding along 𝑂𝐴 with a constant speed 𝑣𝐶 (relative to the arm 𝑂𝐴), and is rotating with a constant angular velocity 𝜔𝐶 (relative to the 𝑥𝑦𝑧 frame). Point 𝐷 is attached to the collar and is at a distance 𝓁 from the 𝑥 axis. At the instant shown, point 𝐷 happens to be in the plane that is rotated by an angle 𝜙 from the 𝑥𝑧 plane. Compute expressions for the inertial velocity and acceleration of point 𝐷 at the instant shown in terms of the parameters given, and express them relative to the rotating coordinate system.

Figure P16.176 and P16.177

𝑍

𝑑

𝑧

𝑧

𝐷

𝑣𝐶

𝑂 𝜔𝑂𝐴

𝐶

𝐷 𝜙 𝐴 𝑥

𝜔𝐶

𝓁 𝑂

𝜔𝐶

𝑦

𝛼𝑂𝐴 𝐵 𝑣⃗𝑃

Figure P16.178

𝑦𝐴 𝑦𝐵

Problem 16.179 𝑎⃗𝑃 𝑟⃗𝑃 ∕𝐵 𝚥̂𝐵

𝚥̂𝐴

𝜔𝐵 𝐵

𝐴

𝚤̂𝐴

Figure P16.179

ISTUDY

𝑥𝐵

𝑃 23◦

𝚤̂𝐵

Let frames 𝐴 and 𝐵 have their origins at points 𝐴 and 𝐵, respectively. Point 𝐵 does not move relative to point 𝐴. The velocity and acceleration of point 𝑃 relative to frame 𝐴, which is fixed, are 𝑣⃗𝑃 = (−14.9 𝚤̂𝐴 + 19.4 𝚥̂𝐴 ) f t∕s

𝑥𝐴

and 𝑎⃗𝑃 = (1.78 𝚤̂𝐴 + 5.96 𝚥̂𝐴 ) f t∕s2 .

Knowing that frame 𝐵 rotates relative to frame 𝐴 at a constant angular velocity 𝜔𝐵 = 1.2 rad∕s and that the position of point 𝑃 relative to frame 𝐵 is 𝑟⃗𝑃 ∕𝐵 = (8 𝚤̂𝐵 + 4.5 𝚥̂𝐵 ) f t, determine the velocity and acceleration of 𝑃 relative to frame 𝐵 at the instant shown. Express the results using the frame 𝐴 component system.

ISTUDY

Section 16.5

Chapter Review

Problem 16.180 The wheel 𝐷 rotates without slip over a curved cylindrical surface with constant radius of curvature 𝐿 = 1.6 f t and center at the fixed point 𝐸. The 𝑋𝑌 frame is attached to the rolling surface, and it is inertial. The 𝑥𝑦 frame is attached to 𝐷 at 𝑂 and rotates with it at a constant angular velocity 𝜔𝐷 = 14 rad∕s. Collar 𝐶 slides along the bar 𝐴𝐵 with a constant velocity 𝑣𝐶 = 4 f t∕s relative to the 𝑥𝑦 frame. At the instant shown, points 𝑂 and 𝐸 are vertically aligned. Letting 𝓁 = 0.25 f t and 𝑅 = 1 f t, determine the inertial velocity and acceleration of 𝐶 at the instant shown when 𝜃 = 25◦ and the 𝑥𝑦 frame is parallel to the 𝑋𝑌 frame. Express your result in the 𝑥𝑦 and 𝑋𝑌 frames. 𝜔𝐷

𝑌

𝐸 𝑦

𝑣𝐶 𝐴

𝐵

𝐶

𝜃

𝑂

𝓁

𝐿

𝑥

𝑅 𝐷 𝑋 Figure P16.180

Problem 16.181 The wheel 𝐷 rotates without slip over a curved cylindrical surface with constant radius of curvature 𝐿 = 1.6 f t and center at the fixed point 𝐸. The 𝑋𝑌 frame is attached to the rolling surface, and it is inertial. The 𝑥𝑦 frame is attached to 𝐷 at 𝑂 and rotates with it at an angular velocity 𝜔𝐷 = 14 rad∕s and an angular acceleration 𝛼𝐷 = 1.3 rad∕s2 . Collar 𝐶 slides along the bar 𝐴𝐵 with a constant velocity 𝑣𝐶 = 4 f t∕s relative to the 𝑥𝑦 frame. Letting 𝓁 = 0.25 f t and 𝑅 = 1 f t, determine the inertial velocity and acceleration of 𝐶, at the instant shown, when 𝜃 = 25◦ and the 𝑥𝑦 frame is parallel to the 𝑋𝑌 frame. Express your result in the 𝑥𝑦 and 𝑋𝑌 frames. 𝛼𝐷

𝑌

𝜔𝐷 𝐸 𝑦

𝑣𝐶 𝐴

𝐵

𝐶

𝜃

𝑂

𝓁 𝑥

𝐿

𝑅 𝐷 𝑋 Figure P16.181

1143

ISTUDY

ISTUDY

17

Newton-Euler Equations for Planar Rigid Body Motion

We now begin the study of rigid body kinetics. In this chapter, we develop the balance principles for forces and moments, this time for a rigid body. In Chapter 18, we will study energy and momentum methods for rigid bodies. By the time we complete this chapter, we will know how to combine the NewtonEuler equations for a rigid body with the force laws and the kinematic equations to obtain the equations governing the body’s motion. In rigid body kinetics we will be writing force and moment equations. Therefore, in the spirit of the discussion in Section 15.3, when referring to the force and moment equations for a rigid body, we refer to these equations as Newton-Euler equations instead of Newton’s second law. Quinn Rooney/Getty Images

Casey Stoner of the Ducati MotoGP Team, performing a turn with his racing motorcycle. By modeling the motorcycle as a rigid body we can determine the maximum acceleration of the motorcycle for which it does not slip or tip.

17.1

Newton-Euler Equations: Bodies Symmetric with Respect to the Plane of Motion

We now develop and apply the force and moment balance laws for rigid bodies. The moment balance law accounts for the fact that, in general, we do not have concurrent force systems acting on rigid bodies and we must now account for the rotation or tendency for rotation of objects.

Linear momentum: translational equations To obtain the translational equations of motion, we use Euler’s first law, which is given by Eq. (15.57) on p. 986 and applies to any system with constant mass: 𝐹⃗ = 𝑚𝑎⃗𝐺 ,

(17.1)

where 𝐹⃗ is the resultant of all external forces, 𝑚 is the total mass of the system, and 𝑎⃗𝐺 is the inertial acceleration of its mass center, which is given by 𝑎⃗𝐺 =

𝑑 2 𝑟⃗𝐺 𝑑𝑡2

.

(17.2)

1145

1146

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Referring to Fig. 17.1, 𝑟⃗𝐺 is the position of the mass center of body 𝐵 relative to point 𝑄 and is given by 1 𝑟⃗ 𝑑𝑚, (17.3) 𝑟⃗𝐺 = 𝑚 ∫𝐵 𝑑𝑚

𝑦

𝑑𝑚

where the integral is performed over the body 𝐵 and 𝑟⃗𝑑𝑚 is the position of the infinitesimal element of mass 𝑑𝑚.

body 𝐵 𝑟⃗𝑑𝑚 𝐺

Angular momentum: rotational equations

𝑟⃗𝐺 𝑥

𝑄

Figure 17.1 Vectors needed for the determination of the position of the mass center 𝑟⃗𝐺 of a rigid body.

Since Euler’s second law relates moments to angular momentum, it provides the rotational equations of motion. We begin with the moment–angular momentum relation given by Eq. (15.58) on p. 986, which is ⃗̇ + 𝑣⃗ × 𝑚𝑣⃗ , ⃗ =ℎ 𝑀 𝑃 𝑃 𝑃 𝐺

(17.4)

where ⃗ is the moment of all external forces about point 𝑃 , which is an arbitrary • 𝑀 𝑃 point in space.

Helpful Information Are there any restrictions on 𝑷 ? No, point 𝑃 can be any point in space. On p. 1149 we will show what the equations look like when we restrict 𝑃 to be on the body (or an arbitrary extension of the body).

⃗ is the angular momentum of the entire system with respect to point 𝑃 . • ℎ 𝑃 • 𝑣⃗𝑃 is the velocity of the arbitrary point 𝑃 . • 𝑣⃗𝐺 is the velocity of the mass center of the system. • 𝑚 is the total mass of the system. When applying Eq. (17.4) to a rigid body, all terms are easy to interpret except for ⃗̇ . To find ℎ ⃗̇ , recall that, for a system of particles, ℎ ⃗ was computed using the ℎ 𝑃 𝑃 𝑃 definitions in Eqs. (15.31) and (15.48) (on pp. 982 and 984, respectively), which yield ⃗ = ℎ 𝑃

𝑁 ∑

𝑟⃗𝑖∕𝑃 × 𝑚𝑖 𝑣⃗𝑖 ,

(17.5)

𝑖=1

𝑦

𝑣⃗𝑑𝑚

𝑃

𝑟⃗𝑑𝑚∕𝑃 𝑑𝑚 𝑎⃗𝑑𝑚

where 𝑁 is the number of particles in the system, 𝑟⃗𝑖∕𝑃 is the position of particle 𝑖 relative to the point 𝑃 , and 𝑚𝑖 𝑣⃗𝑖 is the linear momentum of particle 𝑖. For a rigid body whose mass is distributed in space, we can generalize Eq. (17.5) by replacing the summation over the number of particles with an integral over the body. When we do so, particle 𝑖 of mass 𝑚𝑖 becomes the infinitesimal element of mass 𝑑𝑚, and Eq. (17.5) becomes ⃗ = ℎ 𝑟⃗ × 𝑣⃗𝑑𝑚 𝑑𝑚, (17.6) 𝑃 ∫𝐵 𝑑𝑚∕𝑃

𝑟⃗𝑃 body 𝐵

𝑟⃗𝑑𝑚 𝐺

where 𝑣⃗𝑑𝑚 is the velocity of the element of mass 𝑑𝑚 (see Fig. 17.2). Since we need ⃗̇ , we differentiate Eq. (17.6) with respect to time to obtain ℎ

𝑟⃗𝐺 𝑄

𝑃

Figure 17.2 Points and corresponding position vectors needed to develop the moment–angular momentum relationship for a rigid body.

ISTUDY

⃗̇ = ℎ 𝑃

𝑥

∫𝐵

𝑣⃗𝑑𝑚∕𝑃 × 𝑣⃗𝑑𝑚 𝑑𝑚 +

∫𝐵

𝑟⃗𝑑𝑚∕𝑃 × 𝑎⃗𝑑𝑚 𝑑𝑚.

Using 𝑣⃗𝑑𝑚∕𝑃 = 𝑣⃗𝑑𝑚 − 𝑣⃗𝑃 , Eq. (17.7) becomes ⃗̇ = ℎ 𝑃

∫𝐵

( ) 𝑣⃗𝑑𝑚 − 𝑣⃗𝑃 × 𝑣⃗𝑑𝑚 𝑑𝑚 +

∫𝐵

𝑟⃗𝑑𝑚∕𝑃 × 𝑎⃗𝑑𝑚 𝑑𝑚

(17.7)

ISTUDY

Section 17.1

1147

Newton-Euler Equations for Bodies Symmetric with Respect to the Plane of Motion

𝑣⃗ × 𝑣⃗𝑑𝑚 𝑑𝑚 + 𝑟⃗ × 𝑎⃗𝑑𝑚 𝑑𝑚. (17.8) ∫𝐵 𝑃 ∫𝐵 𝑑𝑚∕𝑃 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

=−

integral 𝐴

integral 𝐵

Since 𝑣⃗𝑃 does not depend on 𝑑𝑚, integral 𝐴 can be written as −

∫𝐵

𝑣⃗𝑃 × 𝑣⃗𝑑𝑚 𝑑𝑚 = −𝑣⃗𝑃 ×

∫𝐵

Helpful Information

𝑑 𝑟⃗ 𝑑𝑚 𝑑𝑡 ∫𝐵 𝑑𝑚 ) 𝑑( = −𝑣⃗𝑃 × 𝑚⃗𝑟𝐺 = −𝑣⃗𝑃 × 𝑚𝑣⃗𝐺 , (17.9) 𝑑𝑡

𝑣⃗𝑑𝑚 𝑑𝑚 = −𝑣⃗𝑃 ×

where we have used the definition of the mass center of a rigid body given in Eq. (17.3). Substituting Eq. (17.9) into Eq. (17.8) gives ⃗̇ = −𝑣⃗ × 𝑚𝑣⃗ + ℎ 𝑃 𝑃 𝐺

𝑟⃗ × 𝑎⃗𝑑𝑚 𝑑𝑚. ∫𝐵 𝑑𝑚∕𝑃 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

(17.10)

Which terms can be pulled outside the integral signs? In going from Eq. (17.8) to Eq. (17.9), 𝑣⃗𝑃 is pulled out of the integral. We can pull 𝑣⃗𝑃 out since it is a property of a single point and does not vary with 𝑑𝑚. We will see this again in Eq. (17.21) where, for example, the same is true for 𝑎𝐺𝑦 , which is also a property of a particular point, and for 𝛼𝐵 , which is a property of the body as a whole.

integral 𝐵

Substituting Eq. (17.10) into Eq. (17.4), we are left with just integral 𝐵, that is, ⃗ = 𝑀 𝑃

∫𝐵

𝑟⃗𝑑𝑚∕𝑃 × 𝑎⃗𝑑𝑚 𝑑𝑚,

(17.11)

since −𝑣⃗𝑃 × 𝑚𝑣⃗𝐺 in Eq. (17.10) cancels with 𝑣⃗𝑃 × 𝑚𝑣⃗𝐺 in Eq. (17.4). We will now interpret the integral in Eq. (17.11) for the case in which the rigid body is symmetric with respect to the plane of motion (Chapter 20 covers the case when it is not). Bodies symmetric with respect to the plane of motion

𝑦

Referring to Fig. 17.3, since the body 𝐵 is rigid, we can relate 𝑎⃗𝑑𝑚 to 𝑎⃗𝐺 via Eq. (16.13) on p. 1061 as 𝑎⃗𝑑𝑚 =

𝑎⃗𝐺 + 𝛼⃗𝐵 × 𝑞⃗ − 𝜔2𝐵 𝑞, ⃗

𝑑𝑚

(17.12) 𝑟⃗𝑑𝑚

(17.14)

𝑞⃗ = 𝑞𝑥 𝚤̂ + 𝑞𝑦 𝚥̂,

(17.15)

𝜔 ⃗ 𝐵 = 𝜔𝐵 𝑘̂

and

̂ 𝛼⃗𝐵 = 𝛼𝐵 𝑘.

(17.16) (17.17)

Note that if loads and position vectors are in the 𝑥𝑦 plane, there will be no external moments in the 𝑥 or 𝑦 directions, but we include 𝑀𝑃 𝑥 and 𝑀𝑃 𝑦 for completeness and then show below that they are both equal to zero. Equations (17.14) and (17.17) reflect the constraint that the body’s motion is planar, and Eqs. (17.15) and (17.16) reflect our assumption that the body is symmetric with respect to the plane of motion. Substituting Eqs. (17.14), (17.15), and (17.17) into Eq. (17.12), we obtain 𝑎⃗𝑑𝑚 in component form as ( ) ( ) 𝑎⃗𝑑𝑚 = 𝑎𝐺𝑥 − 𝛼𝐵 𝑞𝑦 − 𝜔2𝐵 𝑞𝑥 𝚤̂ + 𝑎𝐺𝑦 + 𝛼𝐵 𝑞𝑥 − 𝜔2𝐵 𝑞𝑦 𝚥̂. (17.18)

𝑟⃗𝐺∕𝑃

𝜔𝐵 , 𝛼𝐵

𝑎⃗𝑑𝑚 𝐺

(17.13)

𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂, ( ) ( ) 𝑟⃗𝑑𝑚∕𝑃 = 𝑟⃗𝐺∕𝑃 + 𝑞⃗ = 𝑥𝐺∕𝑃 + 𝑞𝑥 𝚤̂ + 𝑦𝐺∕𝑃 + 𝑞𝑦 𝚥̂,

𝑟⃗𝑃 𝑞⃗

where 𝑞⃗ is the position of 𝑑𝑚 relative to 𝐺. It is now convenient to write all vectors in Cartesian components as: ⃗ = 𝑀 𝚤̂ + 𝑀 𝚥̂ + 𝑀 𝑘, ̂ 𝑀 𝑃 𝑃𝑥 𝑃𝑦 𝑃𝑧

𝑃

𝑟⃗𝑑𝑚∕𝑃

𝑟⃗𝐺 𝑄

𝐵 𝑎⃗𝐺 𝑥

Figure 17.3 The definitions of all needed kinematic and kinetic quantities for a rigid body moving in planar motion.

1148

ISTUDY

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Substituting Eqs. (17.13), (17.16), and (17.18) into Eq. (17.11), expanding the cross product, and equating components, we obtain the following three expressions for the rotational equations of motion: 𝑀𝑃 𝑥 = 0,

(17.19)

𝑀𝑃 𝑦 = 0, 𝑀𝑃 𝑧 = 𝛼𝐵

(17.20) ∫𝐵

( 2 ) ( ) 𝑞𝑥 + 𝑞𝑦2 𝑑𝑚 + 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥

∫𝐵

𝑑𝑚

( ) + 𝑎𝐺𝑦 + 𝛼𝐵 𝑥𝐺∕𝑃 + 𝜔2𝐵 𝑦𝐺∕𝑃

( + −𝑎𝐺𝑥 + 𝛼𝐵 𝑦𝐺∕𝑃

𝑞 𝑑𝑚 ∫𝐵 𝑥 ) − 𝜔2𝐵 𝑥𝐺∕𝑃 𝑞 𝑑𝑚, ∫𝐵 𝑦

(17.21)

where we have taken all terms that do not depend on the element of mass 𝑑𝑚 outside of the integrals. Observe that there are no moments about the 𝑥 or 𝑦 axes when the body is symmetric with respect to the plane of motion. This is a direct consequence of our assumption that the body is symmetric with respect to the plane of motion — it is not true if that assumption is violated. The second integral in Eq. (17.21) is the total mass of the body, that is, ∫𝐵

Concept Alert Developing intuition for mass moments of inertia. Just as mass is a measure of an object’s resistance to linear acceleration, the mass moment of inertia in Eq. (17.27) is a measure of an object’s resistance to angular acceleration by accounting for the distribution of mass of a body with respect to the rotation axis. For example, it is much more difficult to spin a long stick perpendicular to its long axis (𝑥-spin) than parallel to its long axis (𝑦-spin). 𝑥-spin 𝑥

𝐺 𝑦-spin 𝑦

This is reflected in the mass moments of inertia of a thin stick, for which 𝐼𝐺𝑥 = 121 𝑚𝑙2 and 𝐼𝐺𝑦 ≈ 0, where 𝑚 is the mass of the stick and 𝑙 is its length.

𝑑𝑚 = 𝑚.

(17.22)

Recalling that Eq. (17.3) defines the position of the mass center relative to point 𝑄 (see Fig. 17.3), in component form, Eq. (17.3) becomes 𝑟⃗𝐺 = 𝑥𝐺∕𝑄 𝚤̂ + 𝑦𝐺∕𝑄 𝚥̂ = or 𝑚𝑥𝐺∕𝑄 =

∫𝐵

𝑥𝑑𝑚∕𝑄 𝑑𝑚

( ) 1 𝑥𝑑𝑚∕𝑄 𝚤̂ + 𝑦𝑑𝑚∕𝑄 𝚥̂ 𝑑𝑚, 𝑚 ∫𝐵

(17.23)

∫𝐵

(17.24)

and

𝑚𝑦𝐺∕𝑄 =

𝑦𝑑𝑚∕𝑄 𝑑𝑚.

Since 𝑞⃗ = 𝑞𝑥 𝚤̂ + 𝑞𝑦 𝚥̂ defines the position of each element of mass 𝑑𝑚 relative to the mass center 𝐺, the component form of the definition of the mass center of a rigid body in Eq. (17.24) tells us that (see Fig. 17.3) ∫𝐵

𝑞𝑥 𝑑𝑚 =

∫𝐵

𝑥𝑑𝑚∕𝐺 𝑑𝑚 = 𝑚𝑥𝐺∕𝐺 = 0,

(17.25)

∫𝐵

𝑞𝑦 𝑑𝑚 =

∫𝐵

𝑦𝑑𝑚∕𝐺 𝑑𝑚 = 𝑚𝑦𝐺∕𝐺 = 0,

(17.26)

where we see that these integrals just define the position of the mass center relative to the mass center. There is only one integral left to interpret, which is ∫𝐵

( 2 ) 𝑞𝑥 + 𝑞𝑦2 𝑑𝑚 = 𝐼𝐺 .

(17.27)

This integral defines the mass moment of inertia of the rigid body about an axis perpendicular to the plane of motion passing through the mass center 𝐺, which we will label as 𝐼𝐺 . Because the motion is planar, we will usually refer to the term in question simply as the mass moment of inertia, and the subscript on 𝐼 will uniquely identify the axis with respect to which 𝐼 is calculated (see Appendix C). Using Eqs. (17.22)–(17.27), we can see that Eq. (17.21) becomes ( ) 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 ,

(17.28)

ISTUDY

Section 17.1

1149

Newton-Euler Equations for Bodies Symmetric with Respect to the Plane of Motion

where 𝑀𝑃 𝑧 = 𝑀𝑃 to simplify the notation for motion. Equation (17.28) is the most general rotational equation for the planar motion of a rigid body that is symmetric with respect to the plane of motion. Equation (17.28) and the two equations represented by Eq. (17.1) give the three equations needed to describe the planar motion of a rigid body. Now that we have the rotational equation we need, note that • If the body is not symmetric with respect to the 𝑥𝑦 plane of motion, then moments in the 𝑥 and/or 𝑦 directions will be required to maintain planar motion. • Equation (17.28) can be written using vector notation as ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ ⃗𝐺 , 𝑀 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

(17.29)

Concept Alert Why do very different bodies have the same mass moments of inertia? Referring to the table inside the back of the book after the Index and the figures below, notice that for a thin plate (𝐼𝑥 )plate = 121 𝑚𝑏2 , which is the

same as that of a thin rod, (𝐼𝑥 )rod = 121 𝑚𝑙2 (if they have the same length, i.e., 𝑙 = 𝑏). 𝑎

̂ 𝛼⃗ = 𝛼 𝑘, ̂ 𝑟⃗ ⃗ = 𝑀 𝑘, where, since the motion is planar, 𝑀 𝑃 𝑃 𝐵 𝐵 𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂, and 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂. Again, Eqs. (17.28) and (17.29) apply when 𝑃 is an arbitrary point in space, that is, it does not have to be a point on the rigid body 𝐵.

𝑏

𝑙∕2

𝐺

𝐺 𝑥

𝑥

𝑙∕2

• If any one of the following conditions is true: ⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0.

Why is this? Looking down the 𝑥 axis, the thin plate resembles a thin rod. As long as the thin plate extends uniformly behind its projection (e.g., it doesn’t taper in or out) and it has the same mass as a thin rod, it must have the same mass moment of inertia 𝐼𝐺 .

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 is fixed or moves with constant velocity). 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 . then Eq. (17.28) reduces to 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 .

(17.30) 𝑦

What if the moment center is on the rigid body? Equation (17.29) is valid for any possible choice of moment center 𝑃 . We now derive a version of Eq. (17.29) that is applicable when the moment center is a point 𝑂 on the rigid body or on an arbitrary extension of the rigid body. Referring to Fig. 17.4, if points 𝑂 and 𝐺 are both on the rigid body, we can write the acceleration of 𝐺 as 𝑎⃗𝐺 =

𝑎⃗𝑂 + 𝛼⃗𝐵 × 𝑟⃗𝐺∕𝑂 − 𝜔2𝐵 𝑟⃗𝐺∕𝑂 ,

𝑂 𝜔 𝐵 , 𝛼𝐵 𝑟⃗𝐺∕𝑂 𝑟⃗𝑂

𝑟⃗𝐺

and use the parallel axis theorem (see Appendix C) to write 𝐼𝐺 as 𝐼𝐺 = 𝐼𝑂 − 𝑚𝑟2𝐺∕𝑂 ,

𝐺

(17.31)

(17.32)

where 𝐼𝑂 is the mass moment of inertia of the body about an axis perpendicular to the plane of motion passing through point 𝑂. Substituting Eqs. (17.31) and (17.32) into Eq. (17.29), we obtain ( ) ( ) ⃗ = 𝐼 − 𝑚𝑟2 𝛼⃗ + 𝑟⃗𝐺∕𝑂 × 𝑚 𝑎⃗𝑂 + 𝛼⃗𝐵 × 𝑟⃗𝐺∕𝑂 − 𝜔2𝐵 𝑟⃗𝐺∕𝑂 (17.33) 𝑀 𝑂 𝑂 𝐺∕𝑂 𝐵 ( ) ) ( = 𝐼𝑂 − 𝑚𝑟2𝐺∕𝑂 𝛼⃗𝐵 + 𝑟⃗𝐺∕𝑂 × 𝑚𝑎⃗𝑂 + 𝑚 𝑟⃗𝐺∕𝑂 ⋅ 𝑟⃗𝐺∕𝑂 𝛼⃗𝐵 (17.34) ( ) − 𝑚 𝑟⃗𝐺∕𝑂 ⋅ 𝛼⃗𝐵 𝑟⃗𝐺∕𝑂 ( ) = 𝐼𝑂 − 𝑚𝑟2𝐺∕𝑂 𝛼⃗𝐵 + 𝑟⃗𝐺∕𝑂 × 𝑚𝑎⃗𝑂 + 𝑚𝑟2𝐺∕𝑂 𝛼⃗𝐵 , (17.35)

𝑄

𝐵 𝑥

Figure 17.4 An arbitrary rigid body with the point 𝑂 being a point on the rigid body.

1150

ISTUDY

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

( ) ( ) ( ) where we have used the vector identity 𝑎⃗ × 𝑒⃗ × 𝑐⃗ = 𝑎⃗ ⋅ 𝑐⃗ 𝑒⃗ − 𝑎⃗ ⋅ 𝑒⃗ 𝑐⃗, 𝑟⃗𝐺∕𝑂 × ⃗ 𝑟⃗ 𝑟⃗𝐺∕𝑂 = 0, ⃗𝐺∕𝑂 = 𝑟2𝐺∕𝑂 , and the fact that 𝑟⃗𝐺∕𝑂 is orthogonal to 𝛼⃗𝐵 (so that 𝐺∕𝑂 ⋅ 𝑟 𝑟⃗𝐺∕𝑂 ⋅ 𝛼⃗𝐵 = 0). Making the final simplification by canceling the two terms involving 𝑚𝑟2𝐺∕𝑂 𝛼⃗𝐵 , Eq. (17.35) becomes ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ ⃗𝑂 . 𝑀 𝑂 𝑂 𝐵 𝐺∕𝑂 × 𝑚𝑎

(17.36)

Because of the assumptions in going from Eq. (17.29) to Eq. (17.36), Eq. (17.36) is subject to the restriction that point 𝑂 must be a point on the rigid body 𝐵 or an extension of the rigid body. Finally, we note that if any of the following is true: ⃗ 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗𝐺∕𝑂 = 0. 2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑂 is fixed or moves with constant velocity). 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 . then Eq. (17.36) becomes 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 ,

(17.37)

where we have used the scalar form to reflect the fact that the motion is planar. Equation (17.37) is particularly useful when bodies are undergoing fixed-axis rotation about an axis through 𝑂.

Graphical interpretation of the equations of motion Equation (17.29) has a graphical interpretation that can help us remember it and understand it physically. Unfortunately, this interpretation is also very easy to misapply, and therefore we need to be careful. With this warning in mind, it does provide a convenient way of applying and remembering Eq. (17.29). We begin by referring to Fig. 17.5. The left side of the figure shows a rigid body

FBD 𝐹1

= 𝐹2

𝑄1

𝐼 𝐺 𝛼𝐵

=

𝐺 𝑄2

𝑟⃗𝑄

1 ∕𝑃

𝑃

KD

𝐵 𝑟⃗𝑄

2 ∕𝑃

𝑀1

𝑚𝑎⃗𝐺

𝐺 𝑟⃗𝐺∕𝑃

𝐵

𝑃

Figure 17.5. The free body diagram and kinetic diagram of a general rigid body. Equating them and writing the associated equations always give the correct equations of motion, i.e., Eqs. (17.1) and (17.29).

that is being acted upon by a number of external forces and moments — it is the FBD of the rigid body. The right side of the figure introduces a new diagram called a kinetic diagram (KD). The kinetic diagram always contains the 𝐼𝐺 𝛼⃗𝐵 vector and the 𝑚𝑎⃗𝐺 vector, which, although it has units of force, we will color green since these vectors originate from accelerations. Now, if we always draw the KD in this way,

ISTUDY

Section 17.1

1151

Newton-Euler Equations for Bodies Symmetric with Respect to the Plane of Motion

then by setting the FBD equal to the KD, we always recover Eqs. (17.1) and (17.29), which are the Newton-Euler equations for a rigid body. We can see this by noting that the left-hand side of Eq. (17.1) (i.e., 𝐹⃗𝑅 ) and the left-hand side of Eq. (17.29) ⃗ ) are readily obtained from the FBD in Fig. 17.5. The right-hand side of (i.e., 𝑀 𝑃 Eq. (17.1) is simply 𝑚𝑎⃗𝐺 , which is what one obtains from the KD. The right-hand side of Eq. (17.29) comes from taking moments about 𝑃 on the KD in Fig. 17.5. Thus, if we always draw the KD by including the 𝑚𝑎⃗𝐺 vector (written in a convenient component system) and the 𝐼𝐺 𝛼𝐵 vector, and we equate forces and moments on the FBD with forces and moments on the KD, we will always end up with the correct Newton-Euler equations for that rigid body.

Common Pitfall The KD must be consistent with the kinematics. The positive directions of 𝐼𝐺 𝛼𝐵 and 𝑚𝑎⃗𝐺 on the KD must be consistent with the positive directions for 𝛼𝐵 and 𝑎⃗𝐺 in the kinematic equations. If they are not, sign errors will end up polluting the problem solution.

End of Section Summary In this section, we developed the Newton-Euler equations (equations of motion) for a rigid body. We began by showing that the translational equations are given by Euler’s first law, which is Eq. (17.1), p. 1145

𝑃

𝑦

𝑟⃗𝐺∕𝑃

𝐹⃗ = 𝑚𝑎⃗𝐺 , where 𝐹⃗ is the resultant of all external forces, 𝑚 is the mass of the rigid body, and 𝑎⃗𝐺 is the inertial acceleration of its mass center (see Fig. 17.6). Bodies symmetric with respect to the plane of motion. For rigid bodies, we also need rotational equations of motion. Applying Euler’s second law, i.e., the moment-angular momentum relationship for a system of particles, we were able to show that, for a rigid body that is symmetric with respect to the plane of motion, the most general form of the rotational equations of motion is given by (see Fig. 17.6) Eq. (17.28), p. 1148, and Eq. (17.29), p. 1149 ( ) 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 , ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝐺 , 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

where the second equation is simply the vector form of the first, and • 𝑀𝑃 is the total moment about 𝑃 in the 𝑧 direction, • 𝐼𝐺 is the mass moment of inertia of the body about an axis perpendicular to the plane of motion and through its mass center 𝐺, • 𝛼𝐵 is the angular acceleration of the body, • 𝑚 is the total mass of the body, • 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂ is the acceleration of the mass center, • 𝑟⃗𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂ is the position of the mass center 𝐺 relative to the moment center 𝑃 . Now, in addition, if any one of the following conditions is true: ⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0,

𝜔𝐵 , 𝛼𝐵 𝐺

𝑎⃗𝐺

𝑄

𝐵 𝑥

Figure 17.6 The relevant kinematic quantities for the NewtonEuler equations for a rigid body.

1152

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

𝑦 𝑂

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 moves with constant velocity),

𝑎⃗𝑂

3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 ,

𝑟⃗𝐺∕𝑂

𝛼𝐵

then Eqs. (17.28) and (17.29) reduce to

𝐺

Eq. (17.30), p. 1149 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 .

𝐵

𝑄

𝑥

If the moment center is a point 𝑂 on the rigid body or an arbitrary extension of the rigid body, then an alternate form of Eq. (17.29) is (see Fig. 17.7)

Figure 17.7 The relevant kinematic quantities for the rotational equations of motion of a rigid body when the moment center 𝑂 is a point on the rigid body.

ISTUDY

Eq. (17.36), p. 1150 ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝑂 , 𝑂 𝑂 𝐵 𝐺∕𝑂 × 𝑚𝑎 where 𝐼𝑂 is the mass moment of inertia of the body about an axis perpendicular to the plane of motion passing through point 𝑂 and 𝑎⃗𝑂 is the acceleration of point 𝑂. If any one of the following is true: ⃗ 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗𝐺∕𝑂 = 0, 2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑃 moves with constant velocity), 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 , then Eq. (17.36) becomes Eq. (17.37), p. 1150 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 . Graphical interpretation of the equations of motion. Referring to Fig. 17.8, there is a graphical/visual way of obtaining Eqs. (17.1) and (17.29). We draw the FBD of the rigid body, including all forces and moments, then draw the KD (kinetic diagram) of the rigid body, which includes the vectors 𝐼𝐺 𝛼𝐵 and 𝑚𝑎⃗𝐺 . As shown in

FBD 𝐹1

= 𝐹2

𝑄1

𝐼 𝐺 𝛼𝐵

=

𝐺 𝑄2

𝑟⃗𝑄

1 ∕𝑃

𝑃

KD

𝐵 𝑟⃗𝑄

2 ∕𝑃

𝑀1

𝑚𝑎⃗𝐺

𝐺 𝑟⃗𝐺∕𝑃

𝐵

𝑃

Figure 17.8. Free body and kinetic diagrams of a general rigid body. Equating them and writing the associated equations always give the correct equations of motion, i.e., Eqs. (17.1) and (17.29).

Fig. 17.8, we graphically equate these two diagrams, and we write the equations generated by that equation. In doing so, we automatically obtain Eqs. (17.1) and (17.29).

ISTUDY

Section 17.2

17.2

1153

Newton-Euler Equations: Translation

direction of motion

Newton-Euler Equations: Translation

Our first application of the equations developed in Section 17.1 is to rigid bodies that translate in the plane of motion. Figure 17.9 shows a motorcycle accelerating to the right. If we treat the motorcycle and rider as a single rigid body, and we ignore the inertia of the tires, then we can model the system using the FBD shown in Fig. 17.10, where the mass center of the system (i.e., rider plus motorcycle) is at 𝐺, the mass of the system is 𝑚, 𝐹𝐴 is the friction force between the rear tire and the road, and the relevant dimensions are as shown. Since all points on the rigid body move in parallel straight horizontal lines, the system is only translating, which means that all angular velocities and angular accelerations are zero. Therefore, the first two Newton-Euler equations are the two scalar components of Eq. (17.1), which is (see Fig. 17.11)

Figure 17.9 A motorcycle and rider undergoing only translation. Only the tires are not translating, though we are neglecting their rotational inertia. 𝑑

𝚥̂

𝐹⃗ = 𝑚𝑎⃗𝐺 .

(17.38)

Referring to Fig. 17.11, the third Newton-Euler equation is given by a rotational equation of motion. If the moment center is an arbitrary point 𝑃 , then the rotational equation is given by ⃗ = 𝑟⃗ × 𝑚𝑎⃗ , (17.39) 𝑀 𝑃

𝐺∕𝑃

𝐺

𝚤̂ 𝑚𝑔 𝐺 𝐷 𝐹𝐴

where we have used the fact that 𝛼⃗𝐵 = 0⃗ for pure translation. If the moment center is an arbitrary point 𝑃 and any one of the following conditions is true: ⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0,

ℎ 𝐵

𝐴 𝑤

𝑁𝐴

𝑁𝐵

Figure 17.10 The FBD of the rider and motorcycle in Fig. 17.9. There is no friction force at 𝐵 on the front tire since we are ignoring its rotational inertia.

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 , then, when written in scalar form, Eq. (17.39) reduces to 𝑀𝑃 = 0.

𝑦

(17.40)

Referring to Fig. 17.11, if the moment center is a point 𝑂 on the translating rigid body, then we can apply Eq. (17.36), which is ⃗ = 𝑟⃗ ⃗𝑂 , 𝑀 𝑂 𝐺∕𝑂 × 𝑚𝑎

𝑟⃗𝐺∕𝑂 𝐺

𝑎⃗𝐺

(17.41) 𝑟⃗𝐺∕𝑃

⃗ If, in addition to the moment center being on the where we have again used 𝛼⃗𝐵 = 0. translating rigid body, any of the following is true: ⃗ 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗𝐺∕𝑂 = 0,

𝐵 𝑃 𝑄

𝛼⃗𝐵 = 0⃗

𝑥

Figure 17.11 Illustration of the important points and kinematic quantities for a rigid body in translation for which 𝑎⃗𝐺 = 𝑎⃗𝑂 .

2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑂 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 , then, when written in scalar form, Eq. (17.41) becomes 𝑀𝑂 = 0.

𝑎⃗𝑂

𝑂

(17.42)

Note that the graphical interpretation of the equations of motion that equates the KD with the FBD is always applicable. See p. 1150.

1154

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.1

Maximum Acceleration of a Motorcycle direction of motion

For the motorcycle and a rider in Fig. 1 with a combined weight of 650 lb, determine the largest possible acceleration, such that the motorcycle does not pop a wheelie, and find the minimum value of 𝜇𝑠 compatible with this motion. Treat the rider and motorcycle as a single rigid body, assume that the motorcycle is driven by its rear wheel, ignore the rotational inertia of the front wheel, and use the dimensions shown in Fig. 2.

SOLUTION Road Map & Modeling The FBD of the system is shown in Fig. 3. Since we are looking for the largest possible acceleration, we will solve for the condition in which the front tire is just about to leave the ground, and so 𝑁𝐵 = 0 in our solution.

Figure 1

Governing Equations 𝑑

Balance Principles

∑

∑

𝐺

∑

ℎ 𝐴

𝐵

Applying Eqs. (17.38) and (17.39), we obtain

𝐹𝑥 ∶

𝐹𝐴 = 𝑚𝑎𝐺𝑥 ,

(1)

𝐹𝑦 ∶

𝑁𝐴 + 𝑁𝐵 − 𝑚𝑔 = 𝑚𝑎𝐺𝑦 ,

(2)

𝑀𝐴 ∶ 𝑁𝐵 𝑤 − 𝑚𝑔(𝑤 − 𝑑) = 𝑚𝑎𝐺𝑦 (𝑤 − 𝑑) − 𝑚𝑎𝐺𝑥 ℎ.

Force Laws Since we want the maximum acceleration and the minimum value of 𝜇𝑠 (i.e., slip must be impending), we have

𝑤 Figure 2 Relevant dimensions for the motorcycle shown in Fig. 1. The mass center of the system is at 𝐺, 𝑤 = 57.5 in., ℎ = 22.5 in., and 𝑑 = 27.8 in.

𝑁𝐵 = 0 and Kinematic Equations

Computation

𝐹 𝐴 = 𝜇 𝑠 𝑁𝐴 .

and 𝑎𝐺𝑦 = 0.

𝜇𝑠 𝑁𝐴 = 𝑚𝑎max ,

𝑚𝑔

𝑁𝐴 − 𝑚𝑔 = 0, 𝐺 𝐹𝐴

𝐵

𝐴 𝑁𝐴

𝑤

Figure 3 The FBD of the motorcycle in Fig. 2.

ISTUDY

−𝑚𝑔(𝑤 − 𝑑) = −𝑚𝑎max ℎ,

ℎ

𝑁𝐵

(5)

Substituting Eqs. (4) and (5) into Eqs. (1)–(3), we obtain

𝚤̂

𝐷

(4)

The kinematic equations are 𝑎𝐺𝑥 = 𝑎max

𝑑

𝚥̂

(3)

(6) (7) (8)

which are three equations for the unknowns 𝑁𝐴 , 𝜇𝑠 , and 𝑎max . Solving, we obtain 𝑁𝐴 = 𝑚𝑔 = 650 lb and 𝑤−𝑑 𝜇𝑠 = (9) = 1.320, ) (ℎ 𝑤−𝑑 (10) 𝑔 = 42.50 f t∕s2 . 𝑎max = ℎ Discussion & Verification

The dimensions of all three final answers are correct in that 𝑁𝐴 has the dimensions of force, 𝜇𝑠 is dimensionless, and 𝑎max has the dimensions of acceleration. A coefficient of static friction of 1.320 may seem high. However, it is not unusual to achieve coefficients this high for good “sticky” tires on asphalt.

ISTUDY

Section 17.2

E X A M P L E 17.2

Hang Angle of an Accelerating Disk

The pin 𝐴, which is rigidly attached to the uniform disk of radius 𝑅, is accelerating to the right in the overhead track at 𝑎0 . If the mass of the disk is 𝑚, determine the constant angle 𝜃 at which the disk will hang with the prescribed motion. In addition, determine the angle that the reaction force at 𝐴 on the disk forms with the vertical and plot it and 𝜃 as a function of 𝑎0 .

𝐴 𝑎0 𝐺

Since 𝐴 is moving horizontally and the angle 𝜃 is constant, we conclude that the disk is only translating. Referring to Fig. 2, we can use the prescribed motion of the system and the Newton-Euler equations to determine the angles 𝜃 and 𝛽, the latter of which describes the orientation of the reaction force 𝑅𝐴 at 𝐴. Road Map & Modeling

∑

∑

𝓁

Figure 1

𝛽

𝚥̂ 𝚤̂

Governing Equations Balance Principles

(17.39), we obtain ∑

𝜃

𝑅

SOLUTION

Force Laws

1155

Newton-Euler Equations: Translation

𝐴

𝑅𝐴

Applying the Newton-Euler equations given by Eqs. (17.38) and 𝑅

𝐹𝑥 ∶

𝑅𝐴 sin 𝛽 = 𝑚𝑎𝐺𝑥 ,

(1)

𝐹𝑦 ∶

𝑅𝐴 cos 𝛽 − 𝑚𝑔 = 𝑚𝑎𝐺𝑦 ,

(2)

𝑀𝐴 ∶

𝑚𝑔𝓁 sin 𝜃 = 𝑚𝑎𝐺𝑥 𝓁 cos 𝜃 − 𝑚𝑎𝐺𝑦 𝓁 sin 𝜃.

(3)

All forces are accounted for on the FBD.

Kinematic Equations

Since the motion is only horizontal, we have that 𝑎𝐺𝑥 = 𝑎0

and 𝑎𝐺𝑦 = 0,

𝐺

𝜃 𝓁 𝑚𝑔

Figure 2 The FBD of the disk in Fig. 1 showing the angle 𝛽 of the resultant force on the pin at 𝐴.

(4)

where 𝑎0 is the given acceleration of point 𝐴. Computation

Substituting Eqs. (4) into Eqs. (1)–(3), we obtain 𝑅𝐴 sin 𝛽 = 𝑚𝑎0 ,

(5)

𝑅𝐴 cos 𝛽 − 𝑚𝑔 = 0,

(6)

𝑔 sin 𝜃 = 𝑎0 cos 𝜃.

(7)

From Eq. (7), we see that the hang angle of the disk is given by 𝜃 = tan−1

(

𝑎0 𝑔

)

.

60◦

To determine the angle 𝛽, we eliminate 𝑅𝐴 from Eqs. (5) and (6) to obtain (

𝑚𝑔 cos 𝛽

)

sin 𝛽 = 𝑚𝑎0

⇒

𝛽 = tan−1

90◦

(8)

(

𝑎0 𝑔

)

.

𝜃, 𝛽

(9)

30◦ 0◦

A plot of 𝜃 and 𝛽 as functions of 𝑎0 ∕𝑔 can be seen in Fig. 3. Discussion & Verification

We first note that the dimensions of the results in Eqs. (8) and (9) are as they should be, that is, the arguments of the inverse tangent functions are dimensionless, as they should be. The results in Eqs. (8) and (9), which are plotted in Fig. 3, tell us that the reaction force on the disk at the pin 𝐴 is always aligned with the line 𝐴𝐺. This seems reasonable since there should not be a moment about the mass center 𝐺 if the disk does not have an angular acceleration.

0

1

2 𝑎0 ∕𝑔

3

4

Figure 3 Plot of 𝜃 and 𝛽 as functions of 𝑎0 ∕𝑔. The two curves are identical.

1156

E X A M P L E 17.3

Translation: Slipping Versus Tipping with Friction

3 ft 𝜇𝑠 , 𝜇𝑘

2.95 f t

5 ft Figure 1 A person pushing a crate across a level surface.

ISTUDY

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

A uniform flat crate is pushed with a constant horizontal force of 95 lb across a rough surface (Fig. 1). The force is applied 2.95 f t above the floor, and the crate is 5 f t long, 3 f t high, and weighs 120 lb. The coefficients of static and kinetic friction between the crate and the surface are 𝜇𝑠 = 0.4 and 𝜇𝑘 = 0.35, respectively. Verify that the crate slips and does not tip, and determine its acceleration.

SOLUTION The FBD of the crate is shown in Fig. 2, where 𝑃 is the pushing

Road Map & Modeling

𝑤 𝑃

𝑚𝑔 𝐺

𝑑

ℎ∕2 𝑂

𝚥̂ 𝚤̂

𝓁

𝑁

𝐹

Figure 2. FBD of the crate shown in Fig. 1. The mass center is at 𝐺.

Helpful Information How do we know the crate slips? We can answer this question by solving the statics problem associated with the FBD in Fig. 2. Writing the equilibrium equations, we see that Eqs. (1)–(3) are still valid, except that all accelerations are now zero. Solving these modified versions of Eqs. (1)–(3), we obtain 𝑁 = 𝑚𝑔 = 120.0 𝗅𝖻, 𝐹 = 𝑃 = 95.00 𝗅𝖻, 𝑑𝑃 = 2.335 𝖿 𝗍. 𝓁= 𝑚𝑔

force applied at a height 𝑑 above the floor, 𝐹 is the friction force, and 𝑁 is the equivalent normal force on the crate due to the ground. The normal force 𝑁 is that point force that is equivalent to the actual distributed normal force on the bottom of the crate. Since we don’t know the exact form of that distributed force, we place 𝑁 at an unknown position 𝓁, which will be determined as part of the solution.∗ In addition, 𝑤, ℎ, and 𝑚𝑔 are the crate’s width, height, and weight, respectively. We are to verify that the crate does not tip, so we will first solve the problem by assuming just that. During the verification, we will discuss what to look for if the crate were to tip. Governing Equations Balance Principles Applying Eqs. (17.1) and (17.30), the Newton-Euler equations for the FBD in Fig. 2 are

∑

For the crate to slip, 𝐹 ≥ 𝜇𝑠 𝑁, or

∑

∑

𝐹 = 95.00 𝗅𝖻 ≥ (0.4)(120.0 𝗅𝖻) ?

= 48.00 𝗅𝖻 = 𝜇𝑠 𝑁. We can see that 𝐹 is, in fact, greater than 𝜇𝑠 𝑁, and so the crate does slip.

𝐹𝑥 ∶

𝑃 − 𝐹 = 𝑚𝑎𝐺𝑥 ,

(1)

𝐹𝑦 ∶

𝑁 − 𝑚𝑔 = 𝑚𝑎𝐺𝑦 ,

(2)

𝑀𝐺 ∶ 𝑁𝓁 − 𝑃 (𝑑 − ℎ∕2) − 𝐹 ℎ∕2 = 𝐼𝐺 𝛼𝑐 ,

(3)

where 𝑎𝐺𝑥 and 𝑎𝐺𝑦 are the 𝑥 and 𝑦 components of the acceleration of the mass center 𝐺, respectively, 𝛼𝑐 is the crate’s angular acceleration, and 𝐼𝐺 is the crate’s mass moment of inertia, which is given by ) 1 ( 2 𝑚 𝑤 + ℎ2 . (4) 𝐼𝐺 = 12 The friction force can be related to 𝑁 using the Coulomb law for sliding friction, which is 𝐹 = 𝜇𝑘 𝑁. (5)

Force Laws

∗ To

review this idea, see Sections 5.2 and 9.1 of M. E. Plesha, G. L. Gray, R. J. Witt, and F. Costanzo, Engineering Mechanics: Statics, 3rd ed., McGraw Hill Publishing, New York, 2023.

ISTUDY

Section 17.2

Newton-Euler Equations: Translation

1157

Kinematic Equations Letting 𝜔𝑐 denote the crate’s angular velocity, we can relate the acceleration of 𝐺 to 𝑂 using 𝑎⃗𝐺 = 𝑎⃗𝑂 + 𝛼⃗𝑐 × 𝑟⃗𝐺∕𝑂 − 𝜔2𝑐 𝑟⃗𝐺∕𝑂 , and since we are assuming that the crate slips and does not tip, we have that 𝑎⃗𝑂 is only in the 𝑥 direction. Therefore, we can write 𝜔𝑐 = 0, 𝛼𝑐 = 0, and 𝑎𝐺𝑦 = 0. (6) Computation

Plugging Eqs. (4)–(6) into Eqs. (1)–(3), we obtain the following three equations for the unknowns 𝑁, 𝑎𝐺𝑥 , and 𝓁: 𝑃 − 𝜇𝑘 𝑁 = 𝑚𝑎𝐺𝑥 ,

(7)

𝑁 − 𝑚𝑔 = 0,

(8)

𝑁𝓁 − 𝑃 (𝑑 − ℎ∕2) − ℎ𝜇𝑘 𝑁∕2 = 0.

(9)

Solving Eqs. (7)–(9), we obtain 𝑁 = 𝑚𝑔 = 120.0 lb,

(10) 2

𝑎𝐺𝑥 = 𝑃 ∕𝑚 − 𝜇𝑘 𝑔 = 14.22 f t∕s , 𝑃 (𝑑 − ℎ∕2) = 1.673 f t, 𝓁 = 12 ℎ𝜇𝑘 + 𝑚𝑔

(11) (12)

where we have used the given data to obtain the numerical results. We have the acceleration of the crate, but we need to verify that it doesn’t tip. The key is that the equivalent normal force 𝑁 needs to be located within the crate; that is, it can’t be located outside the right or left edges of the crate. The idea behind this criterion is that we have assumed that the crate does not tip, and so our solution must be compatible with that assumption. A normal force outside the boundaries of the crate would mean that a wider base (all other parameters being the same) would be required to prevent tipping. With this said, since 𝓁 ≤ 𝑤∕2 (i.e., 1.673 f t ≤ 2.5 f t), the crate does not tip. Discussion & Verification

• The dimensions of the solutions in Eqs. (10)–(12) are all as they should be. • It is reasonable that a crate with the given dimensions would not tip under the given circumstances. • Although it is hard to know whether 0.44𝑔 is reasonable for 𝑎𝐺𝑥 , it is certainly true that its direction is as expected. 3

postslip

A Closer Look

• Figure 3 shows how 𝓁 (i.e., the distance of 𝑁 to the right of the centerline of the crate) varies as 𝑃 (the applied load) is increased from 0 to 200 lb. As the statics solution tells us, before 𝑃 reaches 𝜇𝑠 𝑁, the crate does not even move, and so we get the dark red preslip curve. Once the crate starts to slip, there is a small sudden drop in 𝓁 since there is a small sudden drop in the friction force as it goes from 𝜇𝑠 𝑁 to 𝜇𝑘 𝑁, and we get the dark green postslip curve. The red lines indicate those values of 𝓁 and 𝑃 at which the normal force reaches the edge of the crate; at this point, the crate would also start to tip. • We assumed that the crate slipped, but did not tip. We were then able to verify that assumption. It is important to realize that we can assume any motion we like and that the correctness of our assumption can always be verified using our solution. For example, under the given conditions, if we assumed that the crate tips and slips, we would discover that an impossible motion is obtained, and we would then be able to rule out that possibility. We will explore this possibility and others in the exercises.

2 𝓁 (f t) 1 preslip 0

0

50

100 𝑃 (lb)

150

200

Figure 3 The normal force offset 𝓁 as a function of the applied load 𝑃 .

1158

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problems Problem 17.1 𝐺 17.0 in. 91.7 in. Gary L. Gray

Figure P17.1

The roadster weighs 2750 lb, its mass is evenly distributed between its front and rear wheels, and it accelerates from 0 to 60 mph in 7.0 s. If the acceleration is uniform and if the rear wheels do not slip, determine the forces on each of the front and rear wheels due to the pavement. Also determine the minimum coefficient of static friction compatible with this motion. Assume that the mass is evenly distributed between the right and left sides of the car, neglect the rotational inertia of the front wheels, and assume that the front wheels roll freely.

Problem 17.2 The conveyor is moving the cans at a constant speed 𝑣0 = 18 f t∕s when, to proceed to the next step in packaging, the cans are transferred onto a stationary surface at 𝐴. If each can weighs 0.95 lb, 𝑤 = 2.71 in., ℎ = 5 in., and 𝜇𝑘 = 0.3 between the cans and the stationary surface, determine the time and distance it takes for each can to stop. In addition, show that the cans don’t tip. Treat each can as a uniform circular cylinder. 𝑎0 𝑤

𝑤 𝐴

ℎ

ℎ 𝑣0

cleat 𝜃

Figure P17.2

Problem 17.3

𝐺

ℎ

Determine the maximum acceleration 𝑎0 of the conveyor so that the cans do not tip over the cleats. The cleats completely prevent slipping, but are not tall enough to dynamically influence tipping. Treat each can as a uniform circular cylinder of mass 𝑚.

𝑑

𝑤

Problem 17.4

Figure P17.4 𝑎0 𝐴

𝐿

𝜃 𝑑 𝐵 ℎ 𝐶

Figure P17.3

The file cabinet, which weighs 230 lb, is being pushed to the right with a horizontal force of 70 lb, which is applied a distance ℎ from the floor. If the mass center 𝐺 of the file cabinet is 𝑑 = 24 in. from the floor and the width of the file cabinet is 𝑤 = 15 in., determine the maximum height ℎ at which the file cabinet can be pushed so that it does not tip, and determine the corresponding acceleration of the file cabinet. Assume that friction between the file cabinet and the floor is negligible.

Problems 17.5 and 17.6 The uniform slender bar 𝐴𝐵 has a weight 𝑊𝐴𝐵 = 150 lb while the crate’s weight is 𝑊𝐶 = 500 lb. The bar 𝐴𝐵 is rigidly attached to the cage containing the crate. Neglect the mass of the cage, and assume that the mass of the crate is uniformly distributed. Furthermore, let 𝐿 = 8.5 f t, 𝑑 = 2.5 f t, ℎ = 4 f t, and 𝑤 = 6 f t. If the trolley is accelerating with 𝑎0 = 11 f t∕s2 , determine 𝜃 so that the bar-crate system translates with the trolley. Problem 17.5

𝑤

If the bar-crate system is translating with the trolley so that 𝜃 = 26◦ , determine the acceleration 𝑎0 of the trolley.

Problem 17.6 Figure P17.5 and P17.6

ISTUDY

ISTUDY

Section 17.2

1159

Newton-Euler Equations: Translation

Problem 17.7 A person is pushing a lawnmower of mass 𝑚 = 38 kg and with ℎ = 0.75 m, 𝑑 = 0.25 m, 𝓁𝐴 = 0.28 m, and 𝓁𝐵 = 0.36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at 𝐺, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled 𝐴) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

ℎ 𝐺 𝐵

𝐴

Problem 17.8 The mower is self-propelled through its rear wheels. If a person were to apply a purely horizontal force, would this force help or hinder the rear wheels’ contribution to the forward motion of the mower? That is, would the rear wheels slip less easily or more easily?

𝑑

𝓁𝐴 𝓁𝐵 Figure P17.7 and P17.8

Problem 17.9 The stationary wheel loader is lifting the load at 𝐵 vertically with acceleration 𝑎0 = 10 f t∕s2 . If the weight of the load 𝐵 is 10,000 lb, the weight of the wheel loader is 33,000 lb, 𝓁 = 118 in., ℎ = 90 in., and 𝑑 = 27 in., determine the reactions on each of its four wheels if the wheel loader is laterally symmetric. In addition, compute the static load on each of the four wheels (i.e., with 𝑎0 = 0) and compare it with the corresponding dynamic load. The center of mass of the wheel loader is at 𝐴.

𝑎0

𝐴

𝐵 𝑑

ℎ 𝓁

Figure P17.9 and P17.10

Problem 17.10 The stationary wheel loader is lifting the load at 𝐵 vertically with acceleration 𝑎0 . If the weight of the load 𝐵 is 10,000 lb, the weight of the wheel loader is 33,000 lb, 𝓁 = 118 in., ℎ = 90 in., and 𝑑 = 27 in., determine the largest value of 𝑎0 for which the reaction on each of the rear wheels does not drop below 1000 lb. The center of mass of the wheel loader is at 𝐴.

Problem 17.11 The door is at rest when the man 𝑀 starts pushing it to the right with a constant horizontal force of 40 lb. If the weight of the door is 300 lb, its width and height are both 𝓁 = 10 f t, its center of mass 𝐺 is at its geometric center, and the man pushes at a height ℎ = 4 f t, determine the speed of the door after it has moved 10 f t and determine the reactions at the rollers 𝐴 and 𝐵.

𝐴

𝓁 2

𝓁 2

𝓁∕2 𝑀 𝐺

Problem 17.12 The 300 lb door is at rest when the man 𝑀 starts pushing it to the right. If the width and height of the door are both 𝓁 = 10 f t, its center of mass 𝐺 is at its geometric center, and the man pushes at a height ℎ = 4 f t, determine the horizontal force with which he must push if he is to move the door 10 f t to the right in 3 s, and determine the reactions at the rollers 𝐴 and 𝐵.

𝐵

ℎ

Figure P17.11 and P17.12

𝓁∕2

1160

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problem 17.13 The portable hydraulic crane with mass 𝑚𝐺 = 60 kg is stationary as it lifts the crate with mass 𝑚𝐶 = 110 kg vertically with acceleration 𝑎𝐶 = 2.5 m∕s2 . The mass center of the crate is at 𝐶 and the mass center of the crane is at 𝐺. If 𝑑 = 0.145 m, 𝓁 = 1.3 m, and ℎ = 0.27 m, determine the reaction on each of the two front 𝐹 and the two rear 𝑅 wheels.

𝐺 𝐶

Problem 17.14 𝐹

𝑅 𝑑

ℎ 𝓁

Figure P17.13 and P17.14

The portable hydraulic crane with mass 𝑚𝐺 = 60 kg is stationary as it lifts the crate with mass 𝑚𝐶 = 110 kg vertically with acceleration 𝑎𝐶 . The mass center of the crate is at 𝐶 and the mass center of the crane is at 𝐺. If 𝑑 = 0.145 m, 𝓁 = 1.3 m, and ℎ = 0.27 m, determine the maximum value of the acceleration 𝑎𝐶 , such that no wheel reaction exceeds 400 N. Assume that the crane is laterally symmetric so that the loads on the front wheels are equal and the loads on the rear wheels are equal.

Problems 17.15 and 17.16 During takeoff, the thrust from each of the two engines of a Boeing 737 is 𝑇 = 95 kN. The mass center of the airplane is located at 𝐺, the horizontal distance between the main landing gear and 𝐺 is 𝑑 = 5.15 m, the horizontal distance between the front landing gear and the main landing gear is 𝓁 = 11.1 m, and the height of the mass center above the ground is ℎ = 2.36 m. The mass of the plane is 𝑚 = 74,200 kg, and the thrust from each engine is a distance 𝛿 = 1.15 m above the ground.

𝐿 𝐺

ℎ

𝑇

𝛿

𝑑 𝓁 Figure P17.15 and P17.16

As the airplane begins its takeoff, the lift 𝐿 from its wings can be neglected. Determine the normal reaction on the front landing gear and each of the two wheels of the main landing gear at the start of takeoff.

Problem 17.15

Problem 17.16

Assuming that the wings are the only surfaces generating lift during takeoff, determine the lift 𝐿 generated by each wing when the normal force on the main landing gear is 25% of its initial takeoff value. The line of action of 𝐿 goes through 𝐺.

Problems 17.17 and 17.18 A file cabinet weighing 230 lb is being pushed to the right with a horizontal force 𝑃 applied a distance ℎ from the floor. The width of the file cabinet is 𝑤 = 15 in., its mass center 𝐺 is a distance 𝑑 = 2 f t above the floor, and static friction is insufficient to prevent slipping between the cabinet and the floor. 𝐺

ℎ 𝑑

Problem 17.17 If 𝑃 = 70 lb and the coefficient of kinetic friction between the cabinet and the floor is 𝜇𝑘 = 0.28, determine the maximum height ℎ at which the cabinet can be pushed so that it does not tip over, and find the corresponding acceleration of the cabinet. Problem 17.18

𝑤 Figure P17.17 and P17.18

ISTUDY

If the coefficient of kinetic friction between the file cabinet and the floor is 𝜇𝑘 = 0.15 and the horizontal force 𝑃 is applied at a height ℎ = 42 in., determine the maximum value of 𝑃 that can be applied so that the cabinet does not tip over, and find the corresponding acceleration of the cabinet.

ISTUDY

Section 17.2

1161

Newton-Euler Equations: Translation

Problem 17.19 A conveyor belt must accelerate the cans from rest to 𝑣 = 18 f t∕s as quickly as possible. Treating each can as a uniform circular cylinder weighing 1.1 lb, find the minimum possible time to reach 𝑣 so that the cans do not tip or slip on the conveyer. Assume that the acceleration is uniform, and use 𝑤 = 2.71 in., ℎ = 5 in., and 𝜇𝑠 = 0.5. 𝑤 ℎ

𝑣 𝑎0 Figure P17.19 𝐴

𝐿

Problem 17.20 The uniform slender bar 𝐴𝐵, with mass 𝑚𝐴𝐵 = 75 kg and length 𝐿 = 4.5 m, is pinconnected at 𝐴 to a trolley accelerating with 𝑎0 = 3 m∕s2 along a horizontal rail. A crate with uniformly distributed mass 𝑚𝐶 = 250 kg, height ℎ = 1.5 m, and width 𝑤 = 2 m is contained in a cage with negligible mass that is pin-connected to 𝐴𝐵 at 𝐵. The distance between 𝐵 and the top of the crate is 𝑑 = 0.75 m. Determine the angles 𝜙 and 𝜃 so that the bar and the crate translate with the trolley.

𝜃 ℎ 𝐵

𝐶

𝑤

𝑑

Problem 17.21

𝜙

The system shown lies in the vertical plane. The trolley 𝐴 is moving to the right with a constant acceleration 𝑎𝐴 . Attached to the trolley is a rope 𝐴𝐵 of negligible mass. Attached to the end of the rope is a thin uniform bar 𝐵𝐶 of length 𝐿 and mass 𝑚. When the trolley 𝐴 is accelerating at constant 𝑎𝐴 , the angles 𝜃 and 𝜙 are both constant. Determine these two constant angles in this steady (state as)functions of one or more of the given quantities (i.e., 𝐿, 𝑚, and 𝑎𝐴 ). Note that cos 𝜋2 − 𝑥 = sin 𝑥.

Figure P17.20

𝑎𝐴

𝐴

𝜋 2

−𝜙

𝜃

𝜃

𝐵 𝐺 𝐶

𝜙

𝑚, 𝐿

Problem 17.22 The stationary excavator is vertically lifting the load at 𝐴 with acceleration 𝑎0 = 2.5 m∕s2 . If 𝓁 = 4 m, 𝑤 = 3.65 m, 𝑑 = 0.82 m, the mass of the load at 𝐴 is 𝑚𝐴 = 8300 kg, and the total mass of the excavator is 𝑚𝐺 = 20,000 kg, determine the equivalent normal force acting on each of the two tracks and its location relative to the rear of the track 𝛿. The center of mass of the excavator is at 𝐺.

𝑑

𝑎0

Figure P17.21

𝐺

𝐴 𝑤 𝓁 Figure P17.22

𝛿

1162

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

𝑎𝐴

𝐴

𝜋 2

−𝜙

𝜃

𝐷

The system shown lies in the vertical plane. The trolley 𝐴 is moving to the right with a constant acceleration 𝑎𝐴 . Attached to the trolley is a rope 𝐴𝐵 of negligible mass. Attached to the end of the rope is a T-bar consisting of two thin uniform bars 𝐵𝐶 and 𝐷𝐸, each of which has length 𝐿 and mass 𝑚 (bar 𝐵𝐶 is attached at the midpoint of bar 𝐷𝐸). When the trolley 𝐴 is accelerating at constant 𝑎𝐴 , the angles 𝜃 and 𝜙 (are both ) constant. Determine these two constant angles in this steady state. Note that cos 𝜋2 − 𝑥 = sin 𝑥 and the mass center of the T-bar is at 𝐺.

𝜃

𝐵 𝐿 4

𝑚, 𝐿

Problem 17.23

𝜙

Problem 17.24

𝐺 𝐶

The 3300 lb front-wheel-drive car whose mass center is at 𝐴 is pulling a 4300 lb trailer whose mass center is at 𝐵. The car and trailer start from rest and accelerate uniformly to 60 mph in 18 s. Determine the forces on all tires, as well as the total force acting on the car due to the trailer. In addition, determine the friction required so that the wheels of the car do not slip. Assume that the car and trailer are laterally symmetric and that the rotational inertia of the wheels is negligible. Note that the mass center of the trailer is directly above the axle of the rear wheel. Hint: With front-wheel-drive in towing mode, assume all of the traction force is developed at the front tires at 𝐶.

𝑚, 𝐿 𝐸 Figure P17.23

13 𝐴

𝐵

𝐷

25 𝐶

41

𝐸 48

52

𝐹 37

110

Figure P17.24 and P17.25

Problem 17.25

𝑎𝐴 𝐴

The 3300 lb front-wheel-drive car, which is pulling a 4300 lb trailer, is traveling 60 mph and applies its brakes to come to a stop. Assuming that all four wheels of the car assist in the braking and that 𝜇𝑠 = 0.85, determine the minimum possible stopping distance, and find the forces on all tires, as well as the total force acting on the car due to the trailer. Assume that the car and trailer are laterally symmetric. Note that the mass center of the trailer is directly above the axle of the rear wheel.

Problem 17.26 𝜋 2

𝜃

−𝜙

𝜃

𝐵 𝜙 𝑚, 𝐿 𝐶 Figure P17.26

ISTUDY

𝑚, 𝐿

The system shown lies in the vertical plane. The trolley 𝐴 is moving to the right with a constant acceleration 𝑎𝐴 . Attached to the trolley by a pin is a thin uniform bar 𝐴𝐵 of mass 𝑚 and length 𝐿. Attached to the end of the bar 𝐴𝐵 by a pin is a thin uniform bar 𝐵𝐶 of mass 𝑚 and length 𝐿. When the trolley 𝐴 is accelerating at constant 𝑎𝐴 , the angles 𝜃 and 𝜙 are both constant. Determine these two constant angles in this steady state ( as )functions of one or more of the given quantities (i.e., 𝐿, 𝑚, and 𝑎𝐴 ). Note that cos 𝜋2 − 𝑥 = sin 𝑥.

ISTUDY

Section 17.3

17.3

1163

Newton-Euler Equations: Rotation About a Fixed Axis

𝑂

Newton-Euler Equations: Rotation About a Fixed Axis

We now apply the equations developed in Section 17.1 to the motion of rigid bodies that are rotating about a fixed axis. Consider, for example, the paper cutter shown in Fig. 17.12. The axis about which the arm rotates is fixed, and the equations we derived in Section 17.1 can be easily specialized to deal with systems like this one. Modeling the arm as a uniform slender bar of mass 𝑚 and length 𝐿, and neglecting all friction and other moments (e.g., torsional springs) at the axis of rotation, we obtain the model and FBD shown in Fig. 17.13. As usual, the first two Newton-Euler equations are the two scalar components of Eq. (17.1), which is 𝐹⃗ = 𝑚𝑎⃗𝐺 .

axis of rotation

𝜙

Figure 17.12 A paper cutter. The arm containing the cutting blade rotates about a fixed axis at 𝑂. The opening angle of the arm is defined by the angle 𝜙.

(17.43)

Referring to Fig. 17.14, for rotation about a fixed axis through 𝑂, we can see that, 𝑚, 𝐿 𝐺 𝑂

𝜙

𝐺

𝑂𝑦

𝚥̂ 𝚤̂

𝑂

𝜙

𝑚𝑔

𝑂𝑥

Figure 17.13. A model (left) of the cutting arm of the cutting board in Fig. 17.12, along with the FBD of that model (right). 𝑦 𝐵

since 𝐺 is moving in a circle centered at 𝑂, the acceleration of 𝐺 can be written as 𝑎⃗𝐺 = 𝛼⃗𝐵 × 𝑟⃗𝐺∕𝑂 − 𝜔2𝐵 𝑟⃗𝐺∕𝑂 .

𝛼⃗𝐵 × 𝑟⃗𝐺∕𝑂

(17.44) 𝑎⃗𝐺

Equation (17.44) implies that, for a rotation about a fixed axis, 𝑎⃗𝐺 is completely determined by the angular velocity and angular acceleration. Referring again to Fig. 17.14, the third Newton-Euler equation is given by a rotational equation of motion. If the moment center is an arbitrary point 𝑃 , then the rotational equation is given by

𝐺 𝑟⃗𝐺∕𝑂

𝑟⃗𝐺∕𝑃

−𝜔2𝐵 𝑟⃗𝐺∕𝑂

𝑃

𝑂

⃗ = 𝐼 𝛼⃗ + 𝑟⃗ ⃗𝐺 , 𝑀 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

(17.45)

where again, the acceleration of 𝐺 is determined by Eq. (17.44). If the moment center is an arbitrary point 𝑃 and any one of the following conditions is true: ⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0,

𝑄

𝜔𝐵

𝑥

Figure 17.14 Illustration of the important points and kinematic quantities for a rigid body rotating about a fixed axis.

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 , then, when written in scalar form, Eq. (17.45) reduces to 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 .

𝛼𝐵

(17.46)

1164

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

When a rigid body is rotating about a fixed axis through point 𝑂, then we typically want to use that axis as the moment center. In this case, we can apply Eq. (17.37), which is 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 , (17.47) where we have used the fact that 𝑎⃗𝑂 = 0⃗ in this case. 𝑎

radius = 𝑘𝑎

Radius of gyration

𝐺

𝑐

𝑏

Figure 17.15 A thin ring whose mass is all a distance 𝑘𝑎 from the 𝑎 axis. The radius of gyration of this object with respect to the 𝑎 axis would be 𝑘𝑎 since its mass moment of inertia would be 𝐼𝑎 = 𝑚𝑘2𝑎 .

In some handbooks and other references, the property of mass moment of inertia is described in terms of the radius of gyration. The radius of gyration 𝑘𝑎 , with respect to an axis 𝑎, is defined in terms of the mass moment of inertia about that axis as √ 𝐼𝑎 𝑘𝑎 = , (17.48) 𝑚 where 𝑚 is the mass of the body in question. Therefore, if a body of mass 𝑚 had all its mass concentrated at a distance 𝑘𝑎 from the axis 𝑎, its mass moment of inertia would be 𝐼𝑎 (see Fig. 17.15).∗ The radius of gyration has the dimension of length.

End of Section Summary Referring to Fig. 17.16, for a body in fixed-axis rotation about a point, the three Newton-Euler equations of motion consist of the two scalar components of

𝑦 𝐵

Eq. (17.43), p. 1163

𝛼⃗𝐵 × 𝑟⃗𝐺∕𝑂

𝐹⃗ = 𝑚𝑎⃗𝐺 ,

𝑎⃗𝐺 𝐺 𝑟⃗𝐺∕𝑂

along with a moment equation. If the moment center 𝑃 is arbitrary, then the moment equation is given by

𝑟⃗𝐺∕𝑃

−𝜔2𝐵 𝑟⃗𝐺∕𝑂

Eqs. (17.45) and (17.46), p. 1163

𝑃

𝑂

⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝐺 , 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

𝛼𝐵 𝑄

𝜔𝐵

𝑀𝑃 = 𝐼𝐺 𝛼𝐵 , 𝑥

Figure 17.16 Illustration of the important points and kinematic quantities for a rigid body rotating about a fixed axis.

ISTUDY

⃗ If, as is typical with fixedwhere the second equation applies when 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0. axis rotation, the moment center 𝑂 is on the axis of rotation, then the moment equation is Eq. (17.47), p. 1164 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 .

∗ The radius of gyration can also be interpreted in terms of the statistical measure known as standard devi-

ation. In fact, the radius of gyration can be thought of as the “standard deviation of the mass distribution.”

ISTUDY

Section 17.3

1165

Newton-Euler Equations: Rotation About a Fixed Axis

E X A M P L E 17.4

Angular Speed of the Paper Cutter Arm When Released from Rest

For the paper cutter model shown in Fig. 1, determine the angular speed with which the cutting arm will reach the horizontal position if it is released from rest at 𝜙0 = 30◦ . As shown in Fig. 1, model the cutting arm as a uniform slender bar of length 𝐿 and mass 𝑚. Neglect friction in the pin at 𝑂.

𝑂

axis of rotation

𝜙

SOLUTION Road Map & Modeling The FBD of the cutting arm at an arbitrary angle 𝜙 is shown in Fig. 2. Applying the Newton-Euler equations, we can find 𝜙̈ as a function of 𝜙 and then integrate that to find the angular speed 𝜙̇ when the cutting arm reaches the horizontal position. Governing Equations Balance Principles

Applying Eqs. (17.43) and (17.47) to the FBD in Fig. 2, we obtain ∑ 𝐹𝑥 ∶ 𝑂𝑥 = 𝑚𝑎𝐺𝑥 , (1) ∑ 𝐹𝑦 ∶ 𝑂𝑦 − 𝑚𝑔 = 𝑚𝑎𝐺𝑦 , (2) ∑ (3) 𝑀𝑂 ∶ −𝑚𝑔 𝐿2 cos 𝜙 = 𝐼𝑂 𝛼arm ,

𝑚, 𝐿 𝐺 𝜙

𝑂

Figure 1 A paper cutter (top) and the slender bar model of the cutting arm of the cutter (bottom).

where, via the parallel axis theorem, 𝐼𝑂 = 31 𝑚𝐿2 . Force Laws

All forces are accounted for on the FBD. ̈ the kinematic From Eq. (17.44) and from the fact that 𝛼arm = 𝜙,

Kinematic Equations

equations must be

𝑂

𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂ = 𝜙̈ 𝑘̂ × or 𝑎𝐺𝑥 = − Computation

𝐿 𝐿 (cos 𝜙 𝚤̂ + sin 𝜙 𝚥̂) − 𝜙̇ 2 (cos 𝜙 𝚤̂ + sin 𝜙 𝚥̂), 2 2

) 𝐿( ̈ 𝜙 sin 𝜙 + 𝜙̇ 2 cos 𝜙 2

and 𝑎𝐺𝑦 =

(4)

) 𝐿( ̈ 𝜙 cos 𝜙 − 𝜙̇ 2 sin 𝜙 . 2

(5)

̈ Eqs. (1)–(3) become Using Eqs. (5), along with 𝛼arm = 𝜙, ) 𝐿( ̈ 𝜙 sin 𝜙 + 𝜙̇ 2 cos 𝜙 , 2 ) 𝐿( 𝑂𝑦 − 𝑚𝑔 = 𝑚 𝜙̈ cos 𝜙 − 𝜙̇ 2 sin 𝜙 , 2 ̈ −𝑚𝑔 𝐿 cos 𝜙 = 1 𝑚𝐿2 𝜙. 𝑂𝑥 = −𝑚

2

(6) (7) (8)

3

Rearranging Eq. (8) and cancelling an 𝑚𝐿, we obtain 3𝑔 cos 𝜙 𝜙̈ = − 2𝐿

𝑑 𝜙̇ ̇ 3𝑔 cos 𝜙 𝜙=− 𝑑𝜙 2𝐿

⇒ ⇒

∫0

𝜙̇ 𝑓

0◦

3𝑔 cos 𝜙 𝑑𝜙 𝜙̇ 𝑑 𝜙̇ = − 2𝐿 ∫30◦ ⇒

3𝑔 1 ̇2 𝜙 = 2 𝑓 4𝐿

⇒

√

𝜙̇ 𝑓 = −

3𝑔 , 2𝐿

(9)

where we have used the chain rule and integrated from the point of release to the horizontal position of the arm and we have chosen the minus sign when taking the square root since the arm is rotating clockwise. Discussion & Verification

should be.

𝑂𝑦

The dimension of the final result in Eq. (9) is 1∕time, as it

𝐺 𝜙

𝚥̂ 𝑚𝑔

𝚤̂

𝑂𝑥

Figure 2 The FBD of the paper cutting arm model in Fig. 1 at an arbitrary angle 𝜙.

1166

E X A M P L E 17.5 𝐺

𝑂

𝐴

The Sweet Spot of a Baseball Bat 𝐵

𝑃

𝛿 𝓁 𝑑 Figure 1 A baseball bat showing the location of its mass center 𝐺, pivot point 𝑂, and the point of impact with a baseball.

𝚥̂

𝑂𝑦 𝐴 𝑂 𝛿

𝑂𝑥

𝐺

𝑃

In baseball or softball, you may have experienced that feeling when the ball is hit “just right,” that is, you can barely feel the bat hit the ball, but the ball goes a long way. This may happen when you hit the ball near the “sweet spot” of the bat. The sweet spot is thought to involve the vibrational modes of the bat, as well as a point called the center of percussion.∗ Figure 1 shows a ball hitting a bat at a distance 𝑑 from the knob 𝐴 when the batter has “choked up” a distance 𝛿. Assuming that the batter swings at his or her wrists (i.e., the pivot point is 𝑂), determine the distance 𝑑 at which the ball should be hit so that, no matter how large the force applied at 𝑃 to the bat by the ball, the lateral force (i.e., perpendicular to the bat) at 𝑂 is zero. That point 𝑃 defines the center of percussion, and its location depends on the pivot point 𝑂. Assume the bat has mass 𝑚, its mass center is at 𝐺, and 𝐼𝐺 is its mass moment of inertia. Use 𝛿 = 2 in., and for a typical bat used in Major League Baseball whose stated weight is 32 oz and whose length is 34 in., 𝓁 = 22.5 in., 𝑚 = 0.0630 slug, and 𝐼𝐺 = 0.0413 slug⋅f t 2 .

SOLUTION

𝚤̂ 𝐵

𝑅

𝓁 𝑑

Figure 2 FBD of a bat as it is striking a ball. Point 𝑂 is considered to be fixed since the batter is swinging at the wrists.

ISTUDY

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Road Map & Modeling The FBD of the bat as it is striking the ball is shown in Fig. 2. We have neglected the moment that the batter applies to the bat because its effect is negligible next to that of the forces due to the collision of the ball against the bat. We have neglected the weight of the bat for the same reason. Since the goal is to find the distance 𝑑, such that the lateral reaction force 𝑂𝑦 is zero, we will want to determine 𝑂𝑦 as a function of that distance 𝑑. Governing Equations Balance Principles Applying Eqs. (17.43) and (17.46) to the FBD in Fig. 2 and summing moments about 𝐺, we obtain the following Newton-Euler equations: ∑ 𝐹𝑥∶ 𝑂𝑥 = 𝑚𝑎𝐺𝑥 , (1) ∑ 𝐹𝑦∶ 𝑂𝑦 + 𝑅 = 𝑚𝑎𝐺𝑦 , (2) ∑ 𝑀𝐺∶ 𝑅(𝑑 − 𝓁) − 𝑂𝑦 (𝓁 − 𝛿) = 𝐼𝐺 𝛼bat , (3)

where 𝐼𝐺 is given. Force Laws

All forces are accounted for on the FBD.

Kinematic Equations Since the axis of the bat is parallel to the 𝑥 axis, the 𝑥 component of acceleration is simply the normal acceleration of 𝐺 toward 𝑂, and the 𝑦 component of acceleration is the tangential component, that is

𝑎𝐺𝑥 = −(𝓁 − 𝛿)𝜔2bat

and

𝑎𝐺𝑦 = (𝓁 − 𝛿)𝛼bat ,

(4)

where we are assuming positive directions for 𝜔bat and 𝛼bat . Computation

Substituting Eqs. (4) into Eqs. (1)–(3), we obtain 𝑂𝑥 = −𝑚(𝓁 − 𝛿)𝜔2bat , 𝑂𝑦 + 𝑅 = 𝑚(𝓁 − 𝛿)𝛼bat , 𝑅(𝑑 − 𝓁) − 𝑂𝑦 (𝓁 − 𝛿) = 𝐼𝐺 𝛼bat .

(5) (6) (7)

Eliminating 𝛼bat from Eqs. (6) and (7) and solving for 𝑂𝑦 , we find that 𝑂𝑦 = ∗ See R. Cross,

𝑚𝑅(𝑑 − 𝓁)(𝓁 − 𝛿) − 𝑅𝐼𝐺 𝐼𝐺 + 𝑚(𝓁 − 𝛿)2

.

(8)

“The Sweet Spot of a Baseball Bat,” American Journal of Physics, 66(9), 1998, pp. 772–779.

ISTUDY

Section 17.3

Newton-Euler Equations: Rotation About a Fixed Axis

This is the 𝑂𝑦 as a function of 𝑑 that we desired. Since we want the distance 𝑑 for which 𝑂𝑦 = 0, we set this result equal to zero and then solve for the distance 𝑑 to obtain 𝑑=

𝐼𝐺 + 𝑚𝓁(𝓁 − 𝛿) 𝑚(𝓁 − 𝛿)

= 2.259 f t = 27.10 in.,

(9)

where we have used 𝛿 = 2 in. = 0.1667 f t, 𝓁 = 22.5 in. = 1.875 f t, 𝑚 = 0.0630 slug, and 𝐼𝐺 = 0.0413 slug⋅f t 2 . Discussion & Verification The dimension of the final result in Eq. (9) is length, as it should be. In addition, we see that the location of the sweet spot depends on the distance 𝓁 − 𝛿 from the mass center to the pivot point, that is, it depends on the location of the pivot point and is not an inherent property of the bat. A Closer Look How is the center of percussion useful? Knowing the location of the center of percussion is important in pendulum-type impact test machines, which are designed to measure the failure resistance of a material to an impulsive force. This is done by measuring the impact energy, which is the energy absorbed by the test specimen prior to failure. If the pendulum arm that strikes the specimen does so at the arm’s center of percussion, the force transmitted to the frame of the machine during the test will be minimized.

1167

1168

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.6

Rotation About a Fixed Axis: Moments About a Fixed Point rotor cavity

𝜔𝑟 axis of rotation

test tube

Roman Zaiets/Shutterstock

(a)

𝑑

(b)

Figure 1. (a) A tabletop ultracentrifuge. (b) Swinging-bucket centrifuge rotor that holds 12 sample tubes.

Centrifuges like the one shown in Fig. 1(a) can generate accelerations exceeding 1 million 𝑔. With the swinging-bucket rotor shown in Fig. 1(b), the centrifuge can spin at 60,000 rpm and achieve an acceleration of 485,000𝑔 at the ends of the buckets. As the rotor spins up, the buckets that hang from the bottom of the rotor swing up and eventually assume the nearly horizontal position shown in Fig. 2.∗

𝑟𝑖 𝑟𝑜 Figure 2 Cross-section of one-half of the centrifuge rotor shown in Fig. 1(b). 𝑟𝑖 = 63.1 mm, 𝑟𝑜 = 120.5 mm, 𝑑 = 11.0 mm, and 𝜔𝑟 = 60,000 rpm.

(a) Determine the radial force and the moment parallel to the axis of rotation required to hold the test tube in place when the rotor is spinning at its maximum rated speed of 60,000 rpm. (b) What are the implications of these loads on the rotor bearings? Assume that the test tube and its contents (e.g., blood) can be modeled as a uniform circular cylinder with mass 10 g, and ignore gravity.

SOLUTION top view of rotor

𝚥̂ 𝚤̂

axis of rotation 𝑂

𝜔𝑟

𝑀𝑡

𝑅𝑥

𝑑

𝐺 𝑅𝑦

𝑟𝑖 𝑟𝑜

FBD Figure 3 Top view of the FBD of the test tube in Fig. 2. The mass center of the test tube is at 𝐺.

ISTUDY

Road Map & Modeling The forces shown on the FBD of the test tube in Fig. 3 form the equivalent force-couple system to the system of forces that are actually acting on the tube. The FBD could also show a force in the 𝑧 direction, 𝑅𝑧 , as well as moments in both the 𝑥 and 𝑦 directions, 𝑀𝑂𝑥 and 𝑀𝑂𝑦 , respectively. Summing forces in the 𝑧 direction would simply tell us that 𝑅𝑧 = 𝑚𝑔. Since the body is symmetric with respect to the plane of motion, it follows that 𝑀𝑂𝑥 = 𝑀𝑂𝑦 = 0, and so those moment equations become part of a statics problem. In this case, we are only interested in 𝑅𝑥 , 𝑅𝑦 , and 𝑀𝑡 , so we have chosen the FBD given in Fig. 3. We will approximate the test tube as a uniform circular cylinder, thus ignoring any motion of the fluid within the tube and the nonuniform shape and mass distribution of the tube. Since we know the motion of the test tube, we can find all the velocities and accelerations needed to use Eq. (17.43), as well as all the moment equations we have developed. Therefore, the forces and moments will fall right out of the Newton-Euler equations once the kinematics is included. Governing Equations Balance Principles Equating the FBD and KD of the test tube shown in Figs. 3 and 4, respectively [this is equivalent to applying Eqs. (17.43) and (17.45) to the FBD in Fig. 3], ∗ The

final orientation cannot be exactly horizontal, but at these high accelerations, it is so close to horizontal that we are treating it as such.

ISTUDY

Section 17.3

the Newton-Euler equations for the test tube are ∑ 𝐹𝑥∶ 𝑅𝑥 = 𝑚𝑎𝐺𝑥 , ∑ 𝐹𝑦∶ 𝑅𝑦 = 𝑚𝑎𝐺𝑦 , ( ) ∑ 𝑟𝑜 + 𝑟𝑖 ( ) 𝑀𝑂∶ 𝑀𝑡 + 𝑅𝑦 = 𝐼𝐺 𝛼𝑟 + 𝑚 𝑥𝐺∕𝑂 𝑎𝐺𝑦 − 𝑦𝐺∕𝑂 𝑎𝐺𝑥 , 2

(1) (2)

= 2.821×10

−6

2

kg⋅m ,

top view of rotor

𝚥̂

(3)

where 𝐼𝐺 is the mass moment of inertia of the test tube and 𝑎𝐺𝑥 and 𝑎𝐺𝑦 are the 𝑥 and 𝑦 components of the acceleration of the mass center 𝐺, respectively. Modeling the test tube as a uniform circular cylinder, we calculate its mass moment of inertia as [ ] ) 1 ( 2 1 𝐼𝐺 = 12 𝑚 3𝑟 + ℎ2 = 12 (0.01 kg) 3(0.0055 m)2 + (0.0574 m)2

(4)

𝚤̂ axis of rotation 𝑂

𝜔𝑟

𝑑

𝑚𝑎𝐺𝑥

𝑟𝑖 𝑟𝑜

where we have used 𝑚 = 0.01 kg, 𝑟 = 𝑑∕2 = 0.0055 m, and ℎ = 𝑟𝑜 − 𝑟𝑖 = 0.0574 m. Force Laws

1169

Newton-Euler Equations: Rotation About a Fixed Axis

𝐺 𝑚𝑎𝐺𝑦

KD

All forces are accounted for on the FBD.

Kinematic Equations

As for the accelerations on the right-hand side of Eqs. (1)–(3), since we know the motion of the rotor, these are readily found by relating 𝑎⃗𝐺 to 𝑎⃗𝑂 , where point 𝑂 is on an arbitrary rigid body extension of the test tube. Doing this, we obtain 𝑎⃗𝐺 = 𝑎⃗𝑂 + 𝛼⃗𝑟 × 𝑟⃗𝐺∕𝑂 − 𝜔2𝑟 𝑟⃗𝐺∕𝑂 ,

𝐼𝐺 𝛼𝑟

Figure 4 Top view of the KD of the test tube in Fig. 2. Equating this KD to the FBD in Fig. 3, we obtain the test tube’s Newton-Euler equations.

(5)

in which we note that 𝑎⃗𝑂 = 0⃗ since it is on the axis of rotation and 𝛼⃗𝑟 = 0⃗ since the rotor has reached its constant final speed. Substituting in 𝜔𝑟 = 60,000 rpm = 6283 rad∕s and 𝑟⃗𝐺∕𝑂 = (𝑟𝑜 + 𝑟𝑖 )∕2 𝚤̂ = 0.0918 m 𝚤̂, we have ( ) 𝑎⃗𝐺 = −(6283 rad∕s)2 (0.0918 m 𝚤̂) = −3.624×106 m∕s2 𝚤̂. (6) Computation

Substituting Eqs. (4), (6), 𝑟⃗𝐺∕𝑂 , and 𝛼𝑟 = 0 into Eqs. (1)–(3), we obtain

( ) 𝑅𝑥 = −(0.01 kg) 3.624×106 m∕s2 = −36,240 N,

𝑅𝑦 = 0 N, ( ) 0.1205 m + 0.0631 m 𝑀𝑡 + 𝑅𝑦 =0 2

⇒

𝑀𝑡 = 0 N⋅m,

(7)

Interesting Fact

(8) (9)

where we have substituted 𝑅𝑦 = 0 from Eq. (8) to obtain the final result in Eq. (9). Discussion & Verification

The force needed to keep the test tube in place is 36,240 N (8147 lb), even though the test tube only has a mass of 10 g (an average paper clip has a mass of about 1 g). Note that an equal and opposite force is acting on the rotor bearing and that force is rotating around the bearing 60,000 times per minute or 1000 times per second! Therefore, not only is the rotor subject to failure due to a huge load imbalance, but also it is subject to failure due to fatigue loading (see marginal note). This means that balancing the rotor is essential to safely operate the centrifuge. A Closer Look When using kinetic diagrams to obtain the equations of motion for a rigid body, we are always applying Eq. (17.29) [which becomes Eq. (17.45) for fixed-axis rotation]. So, for example, we could have written the moment equation as 𝑀𝑂 = 𝐼𝑂 𝛼𝑟 by using Eq. (17.47), which is equivalent to Eq. (17.29), but it is not applying the graphical idea of equating the FBD to the KD. Also note that even though 𝑅𝑥 is very large after the test tube has been spun up, that state may not be the critical one from a design perspective. For example, during spin-up, the angular velocity will not be at its maximum value, but the angular acceleration may be very large, which would make 𝑅𝑦 and 𝑀𝑡 nonzero. These will induce shear and bending loads that must also be accounted for.

Cyclic loading and fatigue. The fact that, under the given conditions, the rotor bearing experiences a cyclic load 1000 times per second means that it will quickly experience a large number of load cycles. It turns out that even a rather low stress can cause an object to break after millions of load cycles. The higher the stress, the smaller the number of cycles required. This mechanism of failure is called fatigue. Since the number of load cycles on the rotor bearing of a centrifuge grows quickly, even a small imbalance can cause failure due to fatigue. To learn more about fatigue, see W. D. Callister, Jr., Materials Science and Engineering: An Introduction, 7th ed., John Wiley & Sons, 2006.

1170

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.7 𝐴

𝐵

Analysis of a Composite Rigid Body +

𝑅

𝐿

𝐷

A composite body consisting of a uniform thin bar 𝐴𝐵 of length 𝐿 and a uniform disk 𝐷 of radius 𝑅 = 𝐿∕5 is pinned at 𝐴 and lies in the vertical plane. The mass of the bar is 𝑚𝐴𝐵 = 𝑚, and the mass of the disk is 𝑚𝐷 = 3𝑚. If the system is released from rest in the position shown, determine its initial angular acceleration.

Figure 1

SOLUTION 𝚥̂ 𝐴𝑦

𝐷

𝚤̂

𝐴𝑥

𝐵

𝐶

𝐴

𝐸 𝑅

𝐿 2

𝐿 2

𝑚𝐷 𝑔

𝑚𝐴𝐵 𝑔

Figure 2 FBD of the composite rigid body in Fig. 1. Point 𝐶 is the mass center of the thin bar and point 𝐸 is the mass center of the disk.

Road Map & Modeling We will demonstrate two ways to solve this problem. For both methods, the FBD of the system is as shown in Fig. 2. Since we are interested in the initial angular acceleration, we will apply the Newton-Euler equations to determine it. In the first solution method, we will sum moments about point 𝐴 and apply Eq. (17.47). In the second solution method, we will sum moments about point 𝐴, but we will make use of the kinetic diagram (KD). We will, of course, obtain the same answer, but it will be informative to compare the methods. First solution method Governing Equations Balance Principles

∑

∑

∑

Applying Eqs. (17.43) and (17.47) to the FBD in Fig. 2, we obtain

𝐹𝑥 ∶

𝐴𝑥 = 𝑚𝐴𝐵 𝑎𝐶𝑥 + 𝑚𝐷 𝑎𝐸𝑥 ,

(1)

𝐹𝑦 ∶

𝐴𝑦 − 𝑚𝐴𝐵 𝑔 − 𝑚𝐷 𝑔 = 𝑚𝐴𝐵 𝑎𝐶𝑦 + 𝑚𝐷 𝑎𝐸𝑦 ,

(2)

𝑀𝐴 ∶

−𝑚𝐴𝐵 𝑔 𝐿2

− 𝑚𝐷 𝑔(𝐿 + 𝑅) = 𝐼𝐴 𝛼,

(3)

where 𝛼 is the angular acceleration of the composite body and where the parallel axis theorem gives 𝐼𝐴 for the composite body as 𝐼𝐴 = (𝐼𝐴 )bar + (𝐼𝐴 )disk = = =

( )2 1 𝑚 𝐿2 + 𝑚𝐴𝐵 𝐿2 + 21 𝑚𝐷 𝑅2 + 12 𝐴𝐵 ] [ 1 𝑚 𝐿2 + 𝑚𝐷 12 𝑅2 + (𝐿 + 𝑅)2 3 𝐴𝐵

𝑚𝐷 (𝐿 + 𝑅)2 (4)

707 𝑚𝐿2 , 150

(5)

where we have used 𝑚𝐴𝐵 = 𝑚, 𝑚𝐷 = 3𝑚, and 𝑅 = 𝐿∕5 to obtain the final result. Force Laws

All forces are accounted for on the FBD.

Kinematic Equations

Since Eq. (3) is sufficient to determine the initial angular acceleration of the composite body, there is no need to find 𝑎⃗𝐶 and 𝑎⃗𝐸 . Therefore, no additional equations are needed for the kinematics.

𝚥̂ 𝚤̂

𝐴

𝑚𝐴𝐵 𝑎𝐶𝑦

𝐼𝐶 𝛼 𝐿 2

𝐵 𝐸 𝐶 𝑚𝐴𝐵 𝑎𝐶𝑥 𝑅 𝐿 2

𝑚𝐷 𝑎𝐸𝑦 𝐼𝐸 𝛼 𝑚𝐷 𝑎𝐸𝑥 𝐷

Figure 3 The KD for the composite body in Fig. 1. Note that, as usual, all unknown acceleration components have been drawn in their positive directions, even though we may intuitively see that the initial accelerations of the bar will not be in these directions. Our solution will account for this and will tell us when a component is negative.

ISTUDY

Computation

Substituting Eq. (5) into Eq. (3), we obtain 41 𝑚𝑔𝐿 = − 10

707 𝑚𝐿2 𝛼 150

⇒

𝛼⃗ = −

615 𝑔 ̂ 𝑘, 707 𝐿

(6)

where we have used 𝑚𝐴𝐵 = 𝑚, 𝑚𝐷 = 3𝑚, and 𝑅 = 𝐿∕5 in Eq. (3). Second solution method Governing Equations Balance Principles Equating the FBD in Fig. 2 to the KD in Fig. 3, we obtain the following Newton-Euler equations for the composite body: ∑ 𝐹𝑥 ∶ 𝐴𝑥 = 𝑚𝐴𝐵 𝑎𝐶𝑥 + 𝑚𝐷 𝑎𝐸𝑥 , (7)

ISTUDY

Section 17.3 ∑

∑

Newton-Euler Equations: Rotation About a Fixed Axis

𝐹𝑦 ∶

𝐴𝑦 − 𝑚𝐴𝐵 𝑔 − 𝑚𝐷 𝑔 = 𝑚𝐴𝐵 𝑎𝐶𝑦 + 𝑚𝐷 𝑎𝐸𝑦 ,

𝑀𝐴 ∶

−𝑚𝐴𝐵 𝑔 𝐿2

− 𝑚𝐷 𝑔(𝐿 + 𝑅) = 𝐼𝐶 𝛼 + 𝐼𝐸 𝛼 +

(8)

𝑚𝐴𝐵 𝑎𝐶𝑦 𝐿2

+ 𝑚𝐷 𝑎𝐸𝑦 (𝐿 + 𝑅),

(9)

where the mass moments of inertia are given by 𝐼𝐶 =

1 𝑚 𝐿2 12 𝐴𝐵

=

1 𝑚𝐿2 12

and

𝐼𝐸 = 12 𝑚𝐷 𝑅2 =

3 𝑚𝐿2 , 50

(10)

where we have used 𝑚𝐴𝐵 = 𝑚, 𝑚𝐷 = 3𝑚, and 𝑅 = 𝐿∕5. Force Laws

All forces are accounted for on the FBD.

Kinematic Equations With this second solution method, we need to find 𝑎𝐶𝑦 and 𝑎𝐸𝑦 since they appear in the moment equation. Since the composite body is released from rest, all velocities are zero. With this in mind, we can relate the accelerations of points 𝐶 and 𝐸 to the angular acceleration of the composite body as follows:

𝑎⃗𝐶 = 𝛼⃗ × 𝑟⃗𝐶∕𝐴 = 𝛼 𝑘̂ ×

𝐿 2

𝚤̂ =

𝐿 𝛼 𝚥̂, 2

(11)

𝑎⃗𝐸 = 𝛼⃗ × 𝑟⃗𝐸∕𝐴 = 𝛼 𝑘̂ × (𝐿 + 𝑅) 𝚤̂ = (𝐿 + 𝑅)𝛼 𝚥̂ =

6 𝐿𝛼 𝚥̂, 5

(12)

where we have used 𝑅 = 𝐿∕5. Computation Substituting Eqs. (10)–(12) into Eq. (9) and using 𝑚𝐴𝐵 = 𝑚, 𝑚𝐷 = 3𝑚, and 𝑅 = 𝐿∕5, we obtain

− 41 𝑚𝑔𝐿 = 10

707 𝑚𝐿2 𝛼 150

⇒

𝛼⃗ = −

615 𝑔 ̂ 𝑘, 707 𝐿

(13)

which is identical to the result obtained in Eq. (6). Discussion & Verification

As expected, both solution methods give the same result. This gives us confidence that our final result is correct. With the first method, we were able to avoid computing the accelerations of points 𝐶 and 𝐸 in terms of the angular acceleration of the composite body. With the second method, we were able to avoid the computation of the overall mass moment of inertia using the parallel axis theorem. Which method you choose is really a matter of convenience. A Closer Look The second solution method used in this example demonstrates the utility of the kinetic diagram when applied to composite rigid bodies. We will continue to see this in the rest of this chapter.

1171

1172

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problems Problem 17.27 𝑂 𝐴 𝐶 𝛿

ℎ

𝐵

𝐺

𝑃

𝓁 𝑑

Figure P17.27

Following up on Example 17.5 on p. 1166, now assume that the pivot point 𝑂 is close to the center of mass of the batter so that it is a distance 𝛿 = 3 in. from the knob of the bat at 𝐴. Determine the location of the center of percussion 𝑃 relative to the knob at 𝐴, and show that the position of 𝑃 is independent of the location of point 𝐶 at which the batter grips the bat. Recall that 𝑃 is the point at which the ball should be hit so that, no matter how large the force applied at 𝑃 to the bat by the ball, the lateral force (i.e., perpendicular to the bat) felt by the batter at the grip 𝐶 is zero. Assume the bat has mass 𝑚, its mass center is at 𝐺, and 𝐼𝐺 is its mass moment of inertia. Evaluate your answer for a typical bat used in Major League Baseball whose weight is 32 oz and whose length is 34 in., 𝓁 = 22.5 in., 𝑚 = 0.0630 slug, and 𝐼𝐺 = 0.0413 slug⋅f t 2 . Ignore the weight force on the bat.

Problem 17.28 For the centrifuge rotor and test tube given in Example 17.6 on p. 1168, assume that all the test tubes are locked into their horizontal position and that the rotor is uniformly accelerated from rest to 60,000 rpm in 9.5 min. Determine, as a function of time, the forces and moments on one of the test tubes during this spin-up phase of motion. Assume that each test tube and its contents can be modeled as a uniform circular cylinder with a mass of 10 g, and ignore gravity. 𝜔𝑟 = 60,000 rpm axis of rotation

test tube

11.0 mm

63.1 mm 120.5 mm Figure P17.28

Problems 17.29 and 17.30 The spool is pinned at its center at 𝑂, about which it can spin freely. The radius of the spool is 𝑅 = 0.15 m, its radius of gyration is 𝑘𝑂 = 0.11 m, and the mass of the spool is 𝑚𝑠 = 5 kg. The mass 𝐵 is suspended from the periphery of the spool by a chain of negligible mass that moves over the spool without slip. The mass of 𝐵 is 𝑚𝐵 = 7 kg.

𝑅 𝑂

Problem 17.29

If the system is released from rest, determine the angular acceleration of the spool and the tension in the chain. 𝐵 Figure P17.29 and P17.30

ISTUDY

Problem 17.30 If the system is released from rest, determine the number of revolutions of the spool and the time it takes for mass 𝐵 to achieve a speed of 10 m∕s.

ISTUDY

Section 17.3

1173

Newton-Euler Equations: Rotation About a Fixed Axis

Problems 17.31 through 17.33 The uniform disk of radius 𝑅 = 0.8 f t and weight 𝑊 = 20 lb is pin-connected to the link 𝐴𝐵 and is pulled on its periphery by a force 𝑃 via a rope that is wrapped around the disk. The coefficient of kinetic friction between the disk and the surface on which it sits is 𝜇𝑘 = 0.5. Neglect the mass of link 𝐴𝐵 and assume that 𝜇𝑠 is insufficient to prevent slipping. For 𝑃 = 15 lb and 𝜃 = 0◦ , determine the angular acceleration of the disk 𝛼𝑑 and the time it takes to achieve an angular velocity of 𝜔𝑑 = 35 rad∕s, assuming that the disk starts from rest.

disk, 𝑑 𝜃

𝑅

𝐴

𝑃

𝐵

𝜇𝑘

Problem 17.31

Figure P17.31–P17.33

For 𝑃 = 25 lb and 𝜃 = 90◦ , determine the angular acceleration of the disk 𝛼𝑑 and the number of revolutions for it to achieve an angular velocity of 𝜔𝑑 = 45 rad∕s, assuming that the disk starts from rest. Problem 17.32

Problem 17.33 For 𝑃 = 25 lb, determine the angular acceleration of the disk 𝛼𝑑 as a function of the pull angle 𝜃.

Problem 17.34 A classic balsa wood toy airplane is powered by a rubber band that winds up when its four-bladed propeller is rotated by hand. When released, the rubber band unwinds and the propeller starts spinning, thus propelling the plane around the room. Model each propeller blade as a slender rod of length 𝐿 = 6 cm and mass 𝑚 = 2 g, and neglect air resistance. If we model the rubber band as a linear elastic torsional spring with constant 𝑘𝑡 = 6 × 10−7 N⋅m∕rad, determine the angular speed of the propeller if it is initially wound up 20 revolutions.

𝐿 𝑘𝑡

Problem 17.35 The driveway gate is hinged at its right end and is pushed open with a force 𝑃 . Where should the force 𝑃 be applied (i.e., where should 𝐴 be located) so that the force acting on the hinge due to the gate always acts along a line parallel to the gate and in the plane of the gate during the entire time the gate is opening? Neglect the weight force acting on the gate, and model the gate as a uniform thin bar as shown below the photo. 𝑑 𝑃

𝐴

𝑃

hing hingee line

𝐺

𝑤∕2 𝑤 ∕2

𝐴

Figure P17.34

ℎ

𝑤∕2 𝑤 ∕2 𝑤 𝑂 𝑑

𝑂

Courtesy of Amazing Gates of America

Figure P17.35

𝑊 ,𝐿

Problem 17.36 Assuming the helicopter is on the ground, and modeling each of its four blades as a slender rod with weight 𝑊 = 400 lb and length 𝐿 = 24 f t, determine the constant moment that must be applied by the engine to the mast at 𝑂 to spin the blades from rest to an angular speed of 289 rpm in 90 s.

Figure P17.36

1174

Newton-Euler Equations for Planar Rigid Body Motion

𝐴

Chapter 17

Problem 17.37

𝐵

𝑅

+

𝜔𝑏

The composite body lies in the vertical plane and is rotating with angular speed 𝜔𝑏 = 11 rad∕s at the instant shown. The mass of the disk 𝐷 is 𝑚𝑏 = 4 kg, the mass of the bar 𝐴𝐵 is 𝑚𝐴𝐵 = 1.5 kg, the length of the bar 𝐴𝐵 is 𝐿 = 50 cm, and the radius of the disk is 𝑅 = 10 cm. At the instant shown, determine the angular acceleration of the composite body and the horizontal and vertical reactions on the body at 𝐴.

𝐷

𝐿 Figure P17.37

Problems 17.38 through 17.40 The bar 𝐴𝐵 is pinned to a fixed support at 𝐴 at one end and to the center of the bike wheel 𝐵 at its other end. The bike wheel is spinning with angular speed 𝜔0 = 50 rad∕s when it is gently placed on the horizontal surface. The mass of the bike wheel is 𝑚𝑤 = 2.5 kg, its radius is 𝑅 = 33 cm, and its mass moment of inertia is 𝐼𝐵 = 0.2486 kg⋅m2 .

𝓁 𝐴 𝓁

Problem 17.38 If the mass of bar 𝐴𝐵 is negligible and the coefficient of kinetic friction between the wheel and the surface is 𝜇𝑘 = 0.6, determine the time it takes for the wheel to come to a complete stop, and find the reactions at 𝐴 and 𝐵 on bar 𝐴𝐵. Hint: If the mass of bar 𝐴𝐵 is negligible, bar 𝐴𝐵 is a two-force member.

𝑅 𝜔0

𝐵

wheel, 𝑤 Figure P17.38–P17.40

Problem 17.39 If the mass of bar 𝐴𝐵 is 𝑚𝐴𝐵 = 1.5 kg and the coefficient of kinetic friction between the wheel and the surface is 𝜇𝑘 = 0.6, determine the time it takes for the wheel to come to a complete stop, and find the reactions at 𝐴 and 𝐵 on bar 𝐴𝐵. Hint: If the mass of bar 𝐴𝐵 is significant, bar 𝐴𝐵 is not a two-force member.

If the mass of bar 𝐴𝐵 is 𝑚𝐴𝐵 = 1.5 kg, determine the coefficient of kinetic friction between the wheel and the surface so that it takes 2 s to come to a complete stop. In addition, find the reactions at 𝐴 and 𝐵 on bar 𝐴𝐵 for these conditions.

Problem 17.40

𝑤 𝑂

Problem 17.41 𝑚, 𝐿

𝑀𝑂

Assume that the plane is on the ground, and model each of its two propeller blades as a rectangular plate of mass 𝑚 = 29 kg, length 𝐿 = 137 cm, and width 𝑤 = 20 cm. If the engine torque 𝑀𝑂 applied to the propeller shaft is 410 N⋅m, determine the time it takes for the blades to achieve an angular speed of 2000 rpm if the blades start from rest. What would be the required time if, instead, the blades were modeled as thin rods of length 𝐿 and mass 𝑚?

Figure P17.41

Problems 17.42 and 17.43

𝑑 𝑃

𝐴

𝐺

𝑤∕2 𝑤∕2

𝐴 𝑃

ℎ

𝑤∕2 𝑤∕2

Given that a force of 𝑃 = 20 lb is applied at the center of mass of the gate (i.e., 𝑑 = 𝑤∕2), determine the reactions at the hinge 𝑂 after the force 𝑃 has been continuously applied for 2 s.

Problem 17.42

𝑤 𝑂 𝑑

Courtesy of Amazing Gates of America

Figure P17.42 and P17.43

ISTUDY

hing hingee line

The driveway gate is hinged at its right end and can swing freely in the horizontal plane. The gate is pushed open by the force 𝑃 that always acts perpendicular to the plane of the gate at point 𝐴, which is a horizontal distance 𝑑 from the gate hinge. The weight of the gate is 𝑊 = 215 lb, and its mass center is at 𝐺, which is a distance 𝑤∕2 from each end of the gate, where 𝑤 = 16 f t. Assume that the gate is initially at rest, and model the gate as a uniform thin bar as shown below the photo. Hint: Given that force 𝑃 always acts perpendicular to the plane of the gate, a polar coordinate system with origin at 𝑂 is recommended.

Problem 17.43 Given that a force of 𝑃 = 20 lb is applied at the center of percussion of the gate, determine the reactions at the hinge 𝑂 after the force 𝑃 has been continuously applied for 2 s.

ISTUDY

Section 17.3

Newton-Euler Equations: Rotation About a Fixed Axis

Problems 17.44 and 17.45

1175

𝓁

The uniform thin bar 𝐴𝐵 of length 𝐿 and mass 𝑚 is released from rest in the horizontal position shown. Determine the distance 𝓁 at which the pin 𝑂 should be located from the end of the bar so that it has the maximum possible angular acceleration 𝛼max , and find that angular acceleration. Problem 17.44

𝐴

𝐵

𝑂

𝐿

Figure P17.44 and P17.45

Problem 17.45 Determine the distance 𝓁 at which the pin should be located from the end of the bar so that it has the maximum possible angular acceleration 𝛼max , and find that angular acceleration. In addition, determine the angular acceleration 𝛼0 of the bar when 𝓁 = 0, and then find the ratio 𝛼max ∕𝛼0 .

𝑂

axis of rotation

Problems 17.46 and 17.47 The cutting arm of the paper cutter is pinned about a fixed axis at 𝑂, and its angle relative to the horizontal is measured by 𝜙. A linear elastic torsional spring at 𝑂 with constant 𝑘𝑡 keeps the arm from falling when not in use. Model the cutting arm as a uniform slender bar of length 𝐿 = 20 in. and weight 𝑊 = 2.5 lb. Neglect friction in the pin at 𝑂.

𝜙

𝑚, 𝐿

Problem 17.46

Determine the angular speed with which the cutting arm will reach the horizontal position if it is released from rest at 𝜙𝑖 = 70◦ with 𝑘𝑡 = 1.6 f t ⋅lb∕rad. Assume that the torsional spring is undeformed when 𝜙 = 90◦ .

Determine the value of the torsional spring constant 𝑘𝑡 so that when the cutting arm is released from rest at 𝜙𝑖 = 70◦ , it reaches 𝜙𝑓 = 15◦ with zero angular speed. Assume that the torsional spring is undeformed when 𝜙 = 90◦ . Problem 17.47

𝐺 𝑘𝑡

𝜙

𝑂 Figure P17.46 and P17.47

Problem 17.48 The uniform thin platform 𝐴𝐵 of length 𝐿 and mass 𝑚𝑝 is pinned both at 𝐴 and at 𝐷. A uniform crate of height ℎ, width 𝑤, and mass 𝑚𝑐 is placed at the end of the platform a distance 𝓁 from the pin at 𝐴. The system is at rest when the pin at 𝐴 breaks. Determine the angular acceleration of the platform and crate, as well as the force on the platform due to the pin at 𝐷, immediately after the pin at 𝐴 breaks. Assume that the crate and the platform do not separate immediately after the pin fails and that friction is sufficient to prevent slipping between the platform and crate. 𝑚𝑐

𝓁 𝐺

ℎ 𝐵

𝑚𝑝

𝐷

𝑑 𝐴

𝑤 𝐵

𝐿 Figure P17.48 𝑚, 𝐿

Problems 17.49 and 17.50 The ladder of mass 𝑚 and length 𝐿 is released from rest at the angle 𝜃0 . Model the ladder as a uniform slender bar. If the friction at 𝑂 between the ladder and the ground is sufficient to prevent slipping, and the ladder is given a slight nudge from rest at 𝜃 = 90◦ , determine the angular speed of the ladder when it reaches 𝜃 = 0. Problem 17.49

𝑂

𝜃

Figure P17.49 and P17.50

1176

Newton-Euler Equations for Planar Rigid Body Motion

Chapter 17

Problem 17.50 If the friction at 𝑂 between the ladder and the ground is sufficient to prevent slipping, and the ladder is given a slight nudge from rest at 𝜃 = 90◦ , determine the normal and frictional forces at 𝑂 as a function of the angle 𝜃, and find the minimum coefficient of static friction that is compatible with this motion.

Problems 17.51 and 17.52 𝐵

The T-bar consists of two thin rods, 𝑂𝐴 and 𝐵𝐷, each of length 𝐿 = 1.5 m and mass 𝑚 = 12 kg, that are connected to the frictionless pin at 𝑂. The rods are welded together at 𝐴 and lie in the vertical plane.

𝐴

Problem 17.51

If the rods are released from rest in the position shown, determine the force on the pin at 𝑂, as well as the angular acceleration of the rods immediately after release.

𝐿

𝑂 𝐿

Problem 17.52 If, at the instant shown, the system is rotating clockwise with angular velocity 𝜔0 = 7 rad∕s, determine the force on the pin at 𝑂, as well as the angular acceleration of the rods.

𝐷 Figure P17.51 and P17.52

Problem 17.53

𝐴

𝑑 𝐵

𝐺

𝐶

Figure P17.53

𝐴

Problems 17.54 through 17.56

𝐵 𝜔𝑏

The composite rigid body lies in the vertical plane and consists of the uniform block 𝐷 and the L-shaped bar 𝐴𝐵𝐶. The block is rigidly attached to the L-shaped bar, which is uniform. Each segment of the L-shaped bar has length 𝐿 = 0.75 m and mass 𝑚𝐴𝐵 = 𝑚𝐵𝐶 = 2 kg, the mass of the block is 𝑚𝐷 = 5 kg, and the width and height of the block are ℎ = 0.3 m. Hint: Begin by finding the center of mass of the whole system. Components 𝑎𝐺𝑥 and 𝑎𝐺𝑦 can be expressed in terms of the system’s angular velocity and acceleration. Be careful using the parallel axis theorem to find 𝐼𝐴 as distances between parallel axes perpendicular to the page include both 𝑥 and 𝑦 components.

𝐿 𝐿

𝐶 +𝐷

ℎ Figure P17.54–P17.56

ISTUDY

One way to measure the mass moment of inertia of any body relative to its mass center 𝐺 is to horizontally suspend it by a string at one end 𝐴, and to support it by a scale 𝐶 at the other end 𝐵. The location of points 𝐴 and 𝐵 is not important as long as we know the distance 𝑑 between them and the mass 𝑚 of the body. When the body is statically supported as described, we take note of the reading on the scale. We then cut the string at 𝐴 and note the new reading on the scale immediately after the string is cut. Knowing the mass 𝑚 of the body, the distance 𝑑, and the reading on the scale immediately before (scale reading 𝑁𝑏 ) and after (scale reading 𝑁𝑎 ) the string is cut, determine 𝐼𝐺 for the body.

ℎ

Problem 17.54 If the system is released from rest in the position shown, determine the initial angular acceleration of the composite body and the reaction at the pin 𝐴 on the composite body.

If the system is rotating clockwise with the angular speed 𝜔𝑏 = 10 rad∕s in the position shown, determine the reaction at the pin 𝐴 on the composite body. Problem 17.55

If the system is rotating clockwise with the angular speed 𝜔𝑏 = 10 rad∕s in the position shown, determine the internal reaction at the rigid joint 𝐵 on the bar 𝐵𝐶 of the composite body. Problem 17.56

ISTUDY

Section 17.4

17.4

1177

Newton-Euler Equations: General Plane Motion

Newton-Euler Equations: General Plane Motion

We now apply the equations developed in Section 17.1 to the motion of rigid bodies that are in general planar motion, that is, they are both translating and rotating.

Newton-Euler equations for general plane motion Referring to Fig. 17.17, the first two Newton-Euler equations are always the two scalar components of Euler’s first law, Eq. (17.1), which is 𝐹⃗ = 𝑚𝑎⃗𝐺 .

(17.49)

Newton-Euler equations for an arbitrary moment center Again referring to Fig. 17.17, when the moment center is an arbitrary point 𝑃 , the third Newton-Euler equation is given by Eq. (17.29), which we recall here as ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝐺 , 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

𝑦 𝑃 𝑂

(17.50)

𝑟⃗𝐺∕𝑂

⃗ = 𝑀 𝑘, ̂ 𝛼⃗ = 𝛼 𝑘, ̂ 𝑟⃗ where, since the motion is planar, 𝑀 𝑃 𝑃 𝐵 𝐵 𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂, and 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂. If any one of the following conditions is true: 𝑎⃗𝐺 𝑎⃗𝑂

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 ,

𝑄

⃗ and Eq. (17.50) reduces to then 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0, (17.51)

where we have written the result in scalar form. Newton-Euler equations when the moment center is on the rigid body Referring to Fig. 17.17, when the moment center is a point 𝑂 on the rigid body, Eq. (17.36) on p. 1150 can be used to obtain the third Newton-Euler equation, which is ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝑂 . (17.52) 𝑂 𝑂 𝐵 𝐺∕𝑂 × 𝑚𝑎 ⃗ This occurs if The more useful form of this equation is obtained if 𝑟⃗𝐺∕𝑂 × 𝑚𝑎⃗𝑂 = 0. any of the following is true: ⃗ 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗𝐺∕𝑂 = 0, 2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑂 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 . Then Eq. (17.52) becomes 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 ,

𝜔 𝐵 , 𝛼𝐵

𝐺

⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0,

𝑀𝑃 = 𝐼𝐺 𝛼𝐵 ,

𝑟⃗𝐺∕𝑃

(17.53)

where we have used the scalar form to reflect the fact that the motion is planar.

𝐵 𝑥

Figure 17.17 A general rigid body defining all relevant quantities needed to write its Newton-Euler equations.

1178

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.8

𝑂

Analysis of a Falling Rigid Body As part of a movie stunt, a long, thin, 388 lb platform whose length is 𝐿 = 39 f t has been rigged across a ravine using two ropes 𝑂𝐴 and 𝐵𝐷. Rope 𝑂𝐴 of length 𝑑 = 13.4 f t is securely tied to the tree, but rope 𝐵𝐷 has been tied to a carabiner at 𝐷 that has not been adequately fastened to the rock face. After everything is set up as shown in Fig. 1, the carabiner at 𝐷 breaks free, and the platform starts to fall. Determine the angular acceleration of the platform and the tension in the rope 𝑂𝐴 immediately after the rope 𝐵𝐷 breaks free. The initial value of 𝜃 is 39◦ .

𝐷

𝜃 𝑑

𝐵 𝐴

Chapter 17

𝐿

SOLUTION Road Map & Modeling

Figure 1

FBD 𝜃 𝐺 𝐴

𝐿∕2

𝐵

𝑚𝑔

The FBD of the platform immediately after the carabiner breaks is shown in Fig. 2. We model the platform as a slender rod, and our key assumption is that, immediately after the rope 𝐵𝐷 breaks, all velocities are zero. In addition, we will neglect the mass of each rope and assume that they are inextensible. If the ropes are massless, then neither gravity nor acceleration will affect the behavior of the ropes. Therefore, assuming that rope 𝑂𝐴 is in tension, it will behave as a straight-line segment with constant length. That is, at the time instant considered, 𝑂𝐴 can be treated as if it were a massless rigid body. Of course, we will need to verify that it doesn’t go slack. Using this model, applying the Newton-Euler equations to the platform will allow us to determine the forces and accelerations once we have determined the kinematics of the platform immediately after rope 𝐵𝐷 breaks. Governing Equations Balance Principles

𝚥̂ 𝚤̂ 𝐿∕2

𝑚𝑎𝐺𝑦

KD

𝐺 𝐴

𝑚𝑎𝐺𝑥

𝐵

Figure 2 FBD (top) and KD (bottom) of the platform in Fig. 1.

ISTUDY

By equating the FBD and KD of the platform shown in Fig. 2 [this is equivalent to applying Eqs. (17.1) and (17.29) to the FBD in Fig. 2], the Newton-Euler equations for the platform are ∑ 𝐹𝑥∶ −𝑇 sin 𝜃 = 𝑚𝑎𝐺𝑥 , (1) ∑ 𝐹𝑦∶ 𝑇 cos 𝜃 − 𝑚𝑔 = 𝑚𝑎𝐺𝑦 , (2) ∑ 𝐿 𝑀𝐺 ∶ − 𝑇 cos 𝜃 = 𝐼𝐺 𝛼𝐴𝐵 , (3) 2 where 𝑇 is the tension in the rope and 𝐼𝐺 is the mass moment of inertia of the platform, which is given by 1 𝑚𝐿2 . (4) 𝐼𝐺 = 12 Force Laws All forces are accounted for on the FBD, although we must verify that 𝑇 > 0 to make sure that the rope doesn’t go slack. If the rope does go slack, we will solve the problem with the knowledge that 𝑇 = 0.

We can see from Eqs. (1)–(3) that we need to relate 𝑎⃗𝐺 to the angular acceleration of the platform, subject to the constraint that point 𝐴 moves in a circle about 𝑂 and all velocities are zero immediately after release. Relating 𝐴 to 𝑂, we get 𝑎⃗𝐴 = 𝑎⃗𝑂 + 𝛼⃗𝑂𝐴 × 𝑟⃗𝐴∕𝑂 − 𝜔2𝑂𝐴 𝑟⃗𝐴∕𝑂

Kinematic Equations

= 𝛼𝑂𝐴 𝑘̂ × 𝑑(sin 𝜃 𝚤̂ − cos 𝜃 𝚥̂) = 𝑑𝛼𝑂𝐴 cos 𝜃 𝚤̂ + 𝑑𝛼𝑂𝐴 sin 𝜃 𝚥̂,

(5)

since 𝑎⃗𝑂 = 0⃗ and all velocities are zero. Relating 𝐺 to 𝐴, we get 𝑎⃗𝐺 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐺∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐺∕𝐴 = 𝑑𝛼𝑂𝐴 (cos 𝜃 𝚤̂ + sin 𝜃 𝚥̂) + 𝛼𝐴𝐵 𝑘̂ × (𝐿∕2) 𝚤̂ ( ) = 𝑑𝛼𝑂𝐴 cos 𝜃 𝚤̂ + 𝑑𝛼𝑂𝐴 sin 𝜃 + 𝐿𝛼𝐴𝐵 ∕2 𝚥̂,

where we have used the expression for 𝑎⃗𝐴 from Eq. (5) and set all velocities to zero.

(6)

ISTUDY

Section 17.4

Newton-Euler Equations: General Plane Motion

Computation

We can substitute Eqs. (4) and (6) into Eqs. (1)–(3) to obtain the following three equations in the three unknowns 𝑇 , 𝛼𝐴𝐵 , and 𝛼𝑂𝐴 : −𝑇 sin 𝜃 = 𝑚𝑑𝛼𝑂𝐴 cos 𝜃, ( ) 𝐿 𝑇 cos 𝜃 − 𝑚𝑔 = 𝑚 𝑑𝛼𝑂𝐴 sin 𝜃 + 𝛼𝐴𝐵 , 2 𝐿 1 𝑚𝐿2 𝛼𝐴𝐵 . − 𝑇 cos 𝜃 = 12 2

(7) (8) (9)

Solving these three equations for the three unknowns, we obtain 2𝑚𝑔 cos 𝜃 = 107.2 lb, 5 + 3 cos(2𝜃) 12𝑔 cos2 𝜃 =− = −1.064 rad∕s2 , 𝐿[5 + 3 cos(2𝜃)] 2𝑔 sin 𝜃 = −0.5378 rad∕s2 , =− 𝑑[5 + 3 cos(2𝜃)]

𝑇 = 𝛼𝐴𝐵 𝛼𝑂𝐴

(10) (11) (12)

where we have used 𝐿 = 39 f t, 𝑑 = 13.4 f t, 𝜃 = 39◦ , 𝑚 = (388 lb)∕(32.2 f t∕s2 ) = 12.05 slug, and 𝑔 = 32.2 f t∕s2 to obtain the final numerical results. Discussion & Verification

• The dimensions and the units of the final results in Eqs. (10)–(12) are all as they should be. • The initial angular accelerations of both the rope 𝑂𝐴 and the bar 𝐴𝐵 are negative, as expected, since they should both initially rotate clockwise. • It is difficult to have a sense of the magnitude of the tension in the rope 𝑂𝐴, but it should certainly be positive, which it is. The fact that 𝑇 > 0 confirms that our use of Eq. (5) was correct. In Eq. (5) we treated the rope as a massless rigid body, and this is acceptable only as long as the rope does not go slack. • Referring to Fig. 2, right after the carabiner fails, we expect point 𝐺 to accelerate downward, i.e., 𝑎𝐺𝑦 < 0. Since from Eq. (2) we have 𝑇 = 𝑚(𝑔 + 𝑎𝐺𝑦 )∕ cos 𝜃, the expectation that 𝑎𝐺𝑦 < 0 implies the expectation that 𝑇 < 𝑚𝑔∕ cos 𝜃 = 499.3 lb. The result in Eq. (10) is consistent with this expectation.

1179

1180

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.9

Rolling Without Slip When accelerating from 0 to 60 mph, the rear-wheel drive roadster shown in Fig. 1 has the loads shown in Fig. 2 applied to each of the two rear wheels from the rear axle of the 2570 lb car. Given that each wheel weighs 47 lb, has a mass moment of inertia 𝐼𝐺 of 0.989 slug⋅f t 2 , and has a diameter of 24.3 in., determine (a) The normal and frictional forces between the wheel and the ground.

Gary L. Gray

(b) The minimum coefficient of static friction required for the wheel to roll without slip.

Figure 1

(c) The time it takes for the car to reach 60 mph. Assume that the car accelerates uniformly while moving over a flat and level surface. 689 lb

SOLUTION Road Map & Modeling

485 lb

𝐺 522 f t ⋅lb lb

Figure 2 The loads applied by the axle to one of the rear wheels of the roadster shown in Fig. 1.

The FBD of the wheel is shown in Fig. 3, where, from Fig. 2, the vertical force is 𝑉 , the horizontal force is 𝐻, and the moment is 𝑀𝑎 . Since we know all the loads causing the wheel to move and the inertia properties of the wheel, we should be able to determine how the wheel moves by solving the Newton-Euler equations. We will use the normal and friction forces that we find to determine the minimum friction coefficient required for rolling without slip. Governing Equations Balance Principles

∑

Interesting Fact Where do the forces shown in Fig. 2 come from? In Prob. 17.85, one can find the forces on the rear wheels due to the axle by first performing an analysis of the entire car to get the friction and normal forces on the rear wheels, and then isolating one of the rear wheels and analyzing it.

∑

∑

𝑚=

𝚥̂ 𝑀𝑎

𝐻

𝐺 𝑚𝑔

Force Laws 𝑟

𝐹 𝑁 Figure 3 The FBD of one of the rear wheels of the car shown in Fig. 1.

ISTUDY

𝐹𝑥∶

𝐹 − 𝐻 = 𝑚𝑎𝐺𝑥 ,

(1)

𝐹𝑦∶

𝑁 − 𝑉 − 𝑚𝑔 = 𝑚𝑎𝐺𝑦 ,

(2)

𝐹 𝑟 − 𝑀𝑎 = 𝐼𝐺 𝛼𝑤 ,

(3)

𝑀𝐺 ∶

where 𝑎𝐺𝑥 and 𝑎𝐺𝑦 are the 𝑥 and 𝑦 components of the acceleration of the mass center 𝐺, the friction force acting at the bottom of the wheel is 𝐹 , the normal force between the ground and the wheel is 𝑁, the radius of the wheel is 𝑟 = (24.3∕2) in. = 1.012 f t, and 𝛼𝑤 is the angular acceleration of the wheel. Also, the inertia properties of the wheel are its mass 𝑚 and its mass moment of inertia 𝐼𝐺 , which in this case are

𝑉 𝚤̂

Based on the FBD in Fig. 3, the Newton-Euler equations are

47 lb = 1.460 slug and 𝐼𝐺 = 0.9890 slug⋅f t 2 . 32.2 f t∕s2

(4)

The inequality that must be satisfied for the wheel to roll without slip is |𝐹 | ≤ 𝜇𝑠 |𝑁|,

(5)

where 𝜇𝑠 is the coefficient of static friction between the wheel and the ground. Kinematic Equations Since the car is on a flat surface, the center of the wheel cannot undergo any vertical motion, and since we are assuming that the wheel is rolling without slip, we have the following two kinematic constraints:

𝑎𝐺𝑦 = 0 and 𝑎𝐺𝑥 = −𝑟𝛼𝑤 ,

(6)

where the minus sign comes from the fact that 𝛼𝑤 has been assumed to be positive in the positive 𝑧 direction.

ISTUDY

Section 17.4

Computation

Newton-Euler Equations: General Plane Motion

Substituting Eqs. (6) into Eqs. (1)–(3), we obtain the following three

equations: 𝐹 − 𝐻 = −𝑚𝑟𝛼𝑤 ,

(7)

𝑁 − 𝑉 − 𝑚𝑔 = 0,

(8)

𝐹 𝑟 − 𝑀𝑎 = 𝐼𝐺 𝛼𝑤 ,

(9)

for the three unknowns 𝑁, 𝐹 , and 𝛼𝑤 . Solving, we obtain 𝑁 = 𝑉 + 𝑚𝑔 = 736.0 lb, 𝐼 𝐻 + 𝑚𝑟𝑀𝑎 = 503.4 lb, 𝐹 = 𝐺 𝐼𝐺 + 𝑚𝑟2 𝑟𝐻 − 𝑀𝑎 = −12.45 rad∕s2 , 𝛼𝑤 = 𝐼𝐺 + 𝑚𝑟2

(10) (11) (12)

where we have used 𝑉 = 689 lb, 𝐻 = 485 lb, 𝑚 = (47 lb)∕(32.2 f t∕s2 ) = 1.460 slug, 𝑔 = 32.2 f t∕s2 , 𝐼𝐺 = 0.9890 slug⋅f t 2 , 𝑟 = (12.15 in.)∕(12 in.∕f t) = 1.012 f t, and 𝑀𝑎 = 522 f t ⋅lb, to obtain the final numerical results. Now that we know 𝐹 and 𝑁, we can find the minimum value of 𝜇𝑠 that is compatible with the no-slip assumption by simply using the equality in Eq. (5), that is, |𝐹 | 𝜇𝑠 ≥ || || |𝑁 |

⇒

( ) 𝜇𝑠 min = 0.6840.

(13)

Finally, to determine the time it takes for the car to reach 60 mph, we first find 𝑎𝐺𝑥 from the second of Eqs. (6) as ( ) 𝑎𝐺𝑥 = −(1.012 f t) −12.45 rad∕s2 = 12.60 f t∕s2 , (14)

and then we apply Eq. (12.32) on p. 670 since we are assuming the acceleration is uniform, that is, ( ) 𝑣 = 𝑣0 + 𝑎𝐺𝑥 𝑡 ⇒ 88 f t∕s = 12.60 f t∕s2 𝑡 ⇒ 𝑡 = 6.982 s. (15) Discussion & Verification

• The dimensions of each of the results in Eqs. (10)–(12) are correct. • The value of static friction found in Eq. (13) is reasonable for a tire on asphalt. • The “0 to 60” time we found in Eq. (15) is consistent with times found in the product literature for a roadster like the one analyzed here. A Closer Look We are given all the nonconstraint forces acting on a wheel, which allows us to then determine the motion of the wheel and whether it slips. In this case, since one of the things we were looking for was the minimum 𝜇𝑠 for rolling without slip, we could assume no slip and then find the 𝜇𝑠 compatible with that assumption. Had we not been told whether the wheel slips, then we could have assumed no slip and verified that assumption in the usual way by comparing the needed friction with the available friction (we would need to be given the static friction coefficient). If we discovered that the wheel slips, then Eq. (5) would become 𝐹 = 𝜇𝑘 𝑁 and the second of Eqs. (6) would no longer be valid.

1181

1182

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.10 𝜃 = 0 at release

Example 13.9 Revisited

roll without slip

𝑅

In Example 13.9, we released a small sphere from the top of a semicylinder and, by modeling it as a particle, we determined that it separated from the semicylinder at 𝜃 = 48.2◦ (see Fig. 1). Here we wish to determine the value of 𝜃 at which the small sphere separates if we treat it as a uniform sphere of radius 𝜌 and mass 𝑚. We release the sphere from the top of the semicylinder by giving it a slight nudge to the right, and we assume that there is sufficient friction between the semicylinder and the sphere for the sphere to roll without slip.

𝑅

𝜃

SOLUTION Road Map & Modeling

As with Example 13.9, the key is to find the normal force between the sphere and the semicylinder as a function of 𝜃 and then say that the sphere separates at the location where this force becomes zero. The FBD of the sphere as it slides down the semicylinder is shown in Fig. 2. The FBD has been drawn at an arbitrary angle 𝜃 since we need to find that angle at which 𝑁 becomes zero, and so we need to find 𝑁 for any 𝜃. Since the motion of the mass center of the sphere is along a circular path until it separates from the surface, we will use polar coordinates for the solution. Note that the friction force 𝐹 has been drawn in the indicated direction since the sphere is passively rolling down the semicylinder, i.e., it is not driven.

Figure 1

𝜃 𝑚𝑔 FBD 𝑠

𝐺

𝐹

Governing Equations Balance Principles

𝑃 𝑁 𝑢̂ 𝑟 𝑢̂ 𝜃 𝜃 𝑚𝑎𝐺𝑟

where 𝑚 is the mass of the sphere, 𝐹 is the friction force between the sphere and semicylinder, and the mass moment of inertia 𝐼𝐺 of the sphere is

KD 𝐺 𝐼𝐺 𝛼𝑠

𝑃

Equating the FBD and KD of the sphere shown in Fig. 2 [this is equivalent to applying Eqs. (17.1) and (17.29) to the FBD in Fig. 2], the Newton-Euler equations for the sphere are ∑ 𝐹𝑟 ∶ 𝑁 − 𝑚𝑔 cos 𝜃 = 𝑚𝑎𝐺𝑟 , (1) ∑ 𝐹𝜃 ∶ −𝐹 + 𝑚𝑔 sin 𝜃 = 𝑚𝑎𝐺𝜃 , (2) ∑ 𝑀𝑃 ∶ 𝑚𝑔𝜌 sin 𝜃 = 𝐼𝐺 𝛼𝑠 + 𝜌𝑚𝑎𝐺𝜃 , (3)

𝑠 𝑚𝑎𝐺𝜃

𝐼𝐺 = 52 𝑚𝜌2 . Force Laws

Figure 2 FBD of the sphere 𝑠 and the polar component system drawn at an arbitrary angle 𝜃. Note that, in this component system, positive 𝑧 points into the page.

ISTUDY

(4)

To ensure that the sphere rolls without slip, we must have |𝐹 | ≤ 𝜇𝑠 |𝑁|.

(5)

̈ + 𝜌) 𝑢̂ , 𝑎⃗𝐺 = −𝜃̇ 2 (𝑅 + 𝜌) 𝑢̂ 𝑟 + 𝜃(𝑅 𝜃

(6)

If we want 𝑁 as a function of 𝜃, then the accelerations need to be expressed as functions of 𝜃. We begin by writing 𝑎⃗𝐺 using polar coordinates as

Kinematic Equations

since 𝐺 is moving in a circle centered at 𝑂. Equation (6) implies that 𝑎𝐺𝑟 = −𝜃̇ 2 (𝑅 + 𝜌), ̈ + 𝜌). 𝑎 = 𝜃(𝑅 𝐺𝜃

(7) (8)

Now that we have 𝑎𝐺𝑟 and 𝑎𝐺𝜃 as a function of 𝜃, we need 𝛼𝑠 (𝜃). We can get this by finding 𝜔𝑠 (𝜃), the sphere’s angular velocity, and then differentiating with respect to time. Relating 𝑣⃗𝐺 to 𝑣⃗𝑃 , we obtain 𝑣⃗𝐺 = 𝑣⃗𝑃 + 𝜔 ⃗ 𝑠 × 𝑟⃗𝐺∕𝑃

⇒

̇ + 𝜌) 𝑢̂ = 𝑣⃗ + 𝜔 𝜃(𝑅 ⃗ 𝑠 × 𝑟⃗𝐺∕𝑃 , 𝜃 𝑃

(9)

ISTUDY

Section 17.4

Newton-Euler Equations: General Plane Motion

where 𝑣𝐺 has been written using polar coordinates. Noting that 𝑣⃗𝑃 = 0⃗ and using components, Eq. (9) becomes ( ) 𝑅+𝜌 ̇ ̇ + 𝜌) 𝑢̂ = 𝜔 𝑢̂ × 𝜌 𝑢̂ 𝜃(𝑅 ⇒ 𝜔 = 𝜃, (10) 𝜃 𝑠 𝑧 𝑟 𝑠 𝜌 where we have used 𝑢̂ 𝑧 × 𝑢̂ 𝑟 = 𝑢̂ 𝜃 . Differentiating Eq. (10), we obtain ) ( 𝑅+𝜌 ̈ 𝛼𝑠 = 𝜃. 𝜌

(11)

Computation

We get the equations of motion for the sphere by substituting Eqs. (4), (7), (8), and (11) into Eqs. (1)–(3), which gives 𝑁 − 𝑚𝑔 cos 𝜃 = −𝑚𝜃̇ 2 (𝑅 + 𝜌), ̈ + 𝜌), −𝐹 + 𝑚𝑔 sin 𝜃 = 𝑚𝜃(𝑅 𝑔 sin 𝜃 =

7 (𝑅 5

̈ + 𝜌)𝜃,

(12) (13) (14)

which are three equations to solve for 𝑁, 𝐹 , and 𝜃 (a differential equation must be solved to get 𝜃). However, all we really want is 𝑁(𝜃). To get 𝑁(𝜃), we can see from Eq. (12) that we will need to get 𝜃̇ as a function of 𝜃 — we can do this by using the chain rule, i.e., ̇ 𝜃̈ = 𝜃̇ 𝑑 𝜃∕𝑑𝜃, and then integrating Eq. (14) as follows: ∫0

𝜃̇

𝜃̇ 𝑑 𝜃̇ =

𝜃 5𝑔 sin 𝜃 𝑑𝜃 7(𝑅 + 𝜌) ∫0

⇒

𝜃̇ 2 =

10𝑔 (1 − cos 𝜃). 7(𝑅 + 𝜌)

1183

Common Pitfall Can we really satisfy Eq. (5)? We stated at the beginning that there is enough friction between the surface and the sphere to prevent the sphere from slipping on the surface. Is this possible? It isn’t, and let’s quickly see why. To obtain 𝐹 (𝜃), we can solve Eq. (14) for ̈ substitute the result into Eq. (13), and then 𝜃, solve for 𝐹 . Now 𝑁(𝜃) is found in Eq. (16). Taking the ratio of the two as given by Eq. (5), we obtain | |𝐹 | | 2 sin 𝜃 |. 𝜇𝑠 = || || = || | 𝑁 17 cos 𝜃 − 10 | | | |

Notice that as the sphere rolls down the semicylinder and 𝜃 approaches the separation position, the denominator 17 cos 𝜃 − 10 goes to zero [see Eq. (17)], and so 𝜇𝑠 goes to ∞. This actually tells us that it isn’t possible for the sphere to roll without slipping until it separates from the semicylinder since infinite friction would be required.

(15)

Substituting Eqs. (15) into Eq. (12), 𝑁 as a function of 𝜃 is 𝑁 = 71 𝑚𝑔(17 cos 𝜃 − 10).

(16)

Therefore, calling 𝜃sep the separation angle, the sphere separates from the surface, i.e., 𝑁 becomes zero, when 17 cos 𝜃sep − 10 = 0

⇒

𝜃sep = ±53.97◦ + 𝑛360◦ , 𝑛 = 0, ±1, … , ±∞.

Since we are only interested in 0◦ ≤ 𝜃 ≤ 90◦ , the only acceptable answer is 𝜃sep = 53.97◦ .

(17)

(18)

Discussion & Verification

When we compare this example with Example 13.9, we see that when rotary inertia plays a role in the dynamics, as it does in this example, the object separates from the surface almost 6◦ farther down the cylinder, independent of 𝑅, 𝜌, 𝑔, and 𝑚. Given that this result is “in the same ballpark” as the 48.19◦ separation angle for a particle, it helps build some confidence that the result for a sphere is correct. A Closer Look In Example 13.9, we treated the object sliding down the semicylinder as a particle. A finite-sized sphere would act as a particle if the contact interface were frictionless. Therefore, we should be able to recover the result of Example 13.9 if, in this example, (1) we let the friction go to zero and (2) we account for the fact that the mass center is 𝑅 + 𝜌 from the center of the semicylinder at 𝑂. We will see this in Prob. 17.68.

Helpful Information Why does the sphere go 6◦ farther than the particle? The answer lies in the speed of the objects as they fall down the cylinder. We haven’t covered the work-energy principle for rigid bodies, but we know that in conservative systems, a decrease in potential energy leads to a corresponding increase in kinetic energy. As the particle and sphere move down the cylinder, for a given height change, they each experience the same increase in kinetic energy. For the particle, all the kinetic energy goes into its speed. For the sphere, some goes into its translational speed, but some also goes into the energy associated with its rotation. Either the sphere or the particle separates from the cylinder when they are moving fast enough that their 𝑣2 ∕𝜌 acceleration overcomes the normal component of 𝑚𝑔. It takes the sphere a little longer to get up to that speed since some of its energy goes into rotation, so it separates at a larger angle.

1184

E X A M P L E 17.11

A System with Multiple Rigid Bodies A man starts pushing the lawn roller shown in Fig. 1, such that the angle 𝜃 remains at a constant 40◦ and the center of the lawn roller at 𝐵 accelerates to the right at a constant 0.45 m∕s2 . Given that the mass of the roller is 100 kg, the mass of the handle is 4 kg, 𝜌 = 25 cm, and 𝐿 = 1.1 m, determine the force at 𝐴 that the man must apply to the handle to achieve this motion and the minimum necessary coefficient of static friction between the roller and ground if the roller is to roll without slip. Treat the roller as a uniform circular cylinder and the handle as a thin rod.

𝐿

𝐴

𝜃

𝜌

SOLUTION

𝐵 Figure 1 𝐴𝑦

𝚥̂

𝜃 𝐴𝑥

𝚤̂

𝐿∕2

𝐿∕2

𝑚ℎ 𝑔

𝐵𝑦 𝜃

Figure 2 FBD of the handle of the lawn roller shown in Fig. 1.

𝐵𝑦 𝜌

𝐵𝑥 𝐵

𝚥̂ 𝚤̂

𝑚𝑟 𝑔 𝐹

Road Map & Modeling The FBD of the handle is shown in Fig. 2, and the FBD of the roller is shown in Fig. 3. We have let 𝑚𝑟 be the mass of the roller, 𝑚ℎ be the mass of the handle, and 𝐹 and 𝑁 be the friction and normal forces, respectively, between the roller and the ground. Since we are treating the bar and roller as uniform, we have placed their mass centers at their geometric centers. These FBDs tell us that this is a problem in which we must analyze two rigid bodies. To do so, we will write a set of Newton-Euler equations for each, which will result in a set of six equations. Since the kinematics are entirely known, the unknowns will then turn out to be six forces in the system. Governing Equations

𝐺

Balance Principles The Newton-Euler equations corresponding to the FBD of the handle in Fig. 2 are ∑ 𝐹𝑥 ∶ 𝐴𝑥 + 𝐵𝑥 = 𝑚ℎ 𝑎𝐺𝑥 , (1) ∑ 𝐹𝑦 ∶ 𝐴𝑦 + 𝐵𝑦 − 𝑚ℎ 𝑔 = 𝑚ℎ 𝑎𝐺𝑦 , (2) ∑ 𝐿 𝐿 𝐿 𝐿 𝑀𝐺 ∶ 𝐵𝑥 sin 𝜃 + 𝐵𝑦 cos 𝜃 − 𝐴𝑥 sin 𝜃 − 𝐴𝑦 cos 𝜃 = 𝐼𝐺 𝛼𝐴𝐵 , (3) 2 2 2 2 1 where 𝐼𝐺 = 12 𝑚ℎ 𝐿2 . The Newton-Euler equations corresponding to the FBD of the roller in Fig. 3 are ∑ 𝐹𝑥 ∶ −𝐵𝑥 − 𝐹 = 𝑚𝑟 𝑎𝐵𝑥 , (4) ∑ 𝐹𝑦 ∶ 𝑁 − 𝐵𝑦 − 𝑚𝑟 𝑔 = 𝑚𝑟 𝑎𝐵𝑦 , (5) ∑ 𝑀𝐵 ∶ −𝐹 𝜌 = 𝐼𝐵 𝛼𝑟 , (6)

where 𝐼𝐵 = 21 𝑚𝑟 𝜌2 and 𝛼𝑟 is the angular acceleration of the roller. 𝑁

Figure 3 FBD of the roller of the lawn roller shown in Fig. 1.

ISTUDY

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Force Laws The force law for this system is the friction inequality that must be satisfied for the roller to roll without slip, that is,

|𝐹 | ≤ 𝜇𝑠 |𝑁|.

(7)

All other forces are accounted for on the FBD.

Kinematically, we know 𝜃 is a constant, and so the bar 𝐴𝐵 is in pure translation. This implies that

Kinematic Equations

𝜔𝐴𝐵 = 𝛼𝐴𝐵 = 0

⇒

𝑎⃗𝐺 = 𝑎⃗𝐵 .

(8)

In addition, since the roller is rolling without slip over a flat surface, we can say that 𝑎𝐵𝑥 is known and is equal to the given acceleration of 0.45 m∕s2 and that 𝑎𝐵𝑦 = 0

and 𝛼𝑟 = −

𝑎𝐵𝑥 𝜌

.

(9)

ISTUDY

Section 17.4

Computation

Newton-Euler Equations: General Plane Motion

Substituting Eqs. (7)–(9) into Eqs. (1)–(6), we obtain the six equations 𝐴𝑥 + 𝐵𝑥 = 𝑚ℎ 𝑎𝐵𝑥 ,

(10)

𝐴𝑦 + 𝐵𝑦 − 𝑚ℎ 𝑔 = 0, [( ] ) ( ) 𝐿 𝐵𝑥 − 𝐴𝑥 sin 𝜃 + 𝐵𝑦 − 𝐴𝑦 cos 𝜃 = 0, 2 −𝐵𝑥 − 𝐹 = 𝑚𝑟 𝑎𝐵𝑥 ,

(11)

𝑁 − 𝐵𝑦 − 𝑚𝑟 𝑔 = 0, 𝐹𝜌 =

1 𝑚 𝜌𝑎𝐵𝑥 , 2 𝑟

(12) (13) (14)

1185

Helpful Information Solving six simultaneous equations. While these equations are not difficult to solve by hand, mathematical software packages, such as Mathematica, WolframAlpha (webbased), Maple, MATHCAD, or MATLAB are invaluable for quickly solving systems like this.

(15)

which we can solve for the six unknowns 𝐴𝑥 , 𝐴𝑦 , 𝐵𝑥 , 𝐵𝑦 , 𝐹 , and 𝑁. This system is “weakly coupled,” which means that each equation involves only a small number of the six unknowns. This makes it easier to solve the equations by hand. For example, since we know 𝑎𝐵𝑥 , we can immediately find 𝐹 using Eq. (15). We can then substitute 𝐹 and 𝑎𝐵𝑥 into Eq. (13) to find 𝐵𝑥 . We can then substitute 𝐵𝑥 and 𝑎𝐵𝑥 into Eq. (10) to find 𝐴𝑥 . Continuing similarly, we find all six solutions as ( ) (16) 𝐴𝑥 = 𝑚ℎ + 23 𝑚𝑟 𝑎𝐵𝑥 = 69.30 N, ( ) ] 1 [ (17) 𝐴𝑦 = 2 𝑚ℎ 𝑔 − 𝑚ℎ + 3𝑚𝑟 𝑎𝐵𝑥 tan 𝜃 = −37.77 N, 𝐵𝑥 = − 23 𝑚𝑟 𝑎𝐵𝑥 = −67.50 N, [ ( ) ] 𝐵𝑦 = 21 𝑚ℎ 𝑔 + 𝑚ℎ + 3𝑚𝑟 𝑎𝐵𝑥 tan 𝜃 = 77.01 N, 𝐹 =

𝑁=

1 𝑚 𝑎 = 22.50 N, 2 𝑟 𝐵𝑥 ) ( 1 [( 𝑚ℎ + 2𝑚𝑟 𝑔 + 𝑚ℎ 2

)

]

+ 3𝑚𝑟 𝑎𝐵𝑥 tan 𝜃 = 1058 N.

The force that must be applied at 𝐴 is given by

𝐴⃗ = (69.30 𝚤̂ − 37.77 𝚥̂) N,

(18)

(19) (20) (21)

(22)

√ | | or |𝐴⃗| = 𝐴2𝑥 + 𝐴2𝑦 = 78.92 N at the angle shown in Fig. 4. | | Now that we have the friction and normal forces, we can use the friction inequality in Eq. (7) to determine how much friction is needed to ensure that the roller rolls without slipping, that is, |𝐹 | 𝜇𝑠 ≥ || || = 0.02127. (23) |𝑁 |

𝐴

28.59◦ 𝐴⃗

Discussion & Verification

The dimension of each final result in Eqs. (16)–(21) is correct, and the magnitude of the required force at 𝐴 is reasonable. Notice that Eq. (23) tells us that not much friction is needed for the roller to roll without slip. This is so because of the (considerable) weight of the roller in relation to the small value of acceleration that is being imparted to the roller by the person pushing it. A Closer Look

The force the man must apply to the handle of the lawn roller is given either by Eqs. (16) and (17) or by Eq. (22). Notice that the force that must be applied at 𝐴 is not parallel to the handle. The reason is that the handle has mass — if we were to let 𝑚ℎ be zero in our model, we would find that the angle in Eq. (22) would be −40◦ and the force at 𝐴 would be directed along the handle. If we now compute the magnitude of the force at 𝐵, we obtain √ | ⃗| (24) |𝐵 | = 𝐵𝑥2 + 𝐵𝑦2 = 102.4 N, | | where we have also shown this force on the handle at 𝐵 in Fig. 4. Notice that even though the bar is pin-connected at each end, it is not a two-force member. This is so for two distinct reasons. The first is that the weight of the bar has not been neglected, and the second is that the center of mass of the bar is accelerating.

𝐺 𝐵⃗ 131.2◦ 40◦ Figure 4 The forces at the ends of the lawn roller handle.

1186

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

E X A M P L E 17.12 𝑂

Derivation of Equations of Motion

10 N∕m 𝐴

0.4 m

𝐵

Placing a thin rod at the end of a spring suspended from the ceiling, and then letting the rod swing freely, results in very complicated motions. Write the equations of motion for this system, and then study the motion of the rod for two different sets of initial conditions using computer simulations. Use a rod 0.4 m long with a mass of 0.1 kg. Assume the spring is linear elastic with constant 10 N∕m and with an unstretched length of 0.2 m. In the first ̇ ̇ simulation, use 𝜃(0) = 30◦ , 𝜙(0) = 60◦ , 𝑟(0) = 0.2 m, and 𝜃(0) = 𝜙(0) = 𝑟(0) ̇ = 0; and in the second, use the same conditions except let 𝑟(0) = 0.4 m.

SOLUTION

Figure 1

Road Map & Modeling

Figure 2 shows the coordinates 𝑟, 𝜃, and 𝜙 used to define the

𝑟 𝜃 𝐴 𝜙

𝐿

Figure 2. Definition of the coordinates 𝑟, 𝜃, and 𝜙 used to define the position of the rod.

𝜃 𝑢̂ 𝜃 𝑢̂ 𝑟

𝚤̂

Governing Equations

𝚥̂ 𝐴

Balance Principles 𝜙

𝐺

Figure 3 The FBD of the thin rod in Fig. 1.

ISTUDY

position of the rod, where 𝑟 is the length of the spring, 𝐿 is the length of the rod, and 𝜃 and 𝜙 define the angle of the spring and of the rod with respect to the vertical, respectively. Ignoring the mass of the spring, the hanging rod has three degrees of freedom. Therefore, we will need to derive three equations of motion. We will obtain these by writing the Newton-Euler equations for the rod using the FBD shown in Fig. 3, where 𝐹𝑠 is the force on the rod due to the spring. Notice that we are using two coordinate systems in the FBD — a global Cartesian system and a polar coordinate system aligned with the spring (and thus the spring force) that we will use to describe the motion of point 𝐴.

The Newton-Euler equations corresponding to the FBD in Fig. 3

are given by ∑

∑ 1 𝑚𝐿2 , 12

where 𝐼𝐺 =

𝑟⃗𝐴∕𝐺 =

∑

𝐹𝑥∶

−𝐹𝑠 sin 𝜃 = 𝑚𝑎𝐺𝑥 ,

(1)

𝐹𝑦∶

𝑚𝑔 − 𝐹𝑠 cos 𝜃 = 𝑚𝑎𝐺𝑦 ,

(2)

𝑀𝐺 ∶

𝐿 (− sin 𝜙 𝚤̂ − 2

cos 𝜙 𝚥̂)

̂ 𝑟⃗𝐴∕𝐺 × 𝐹⃗𝑠 = 𝐼𝐺 𝛼𝐴𝐵 𝑘,

and 𝐹⃗𝑠 = 𝐹𝑠 (− sin 𝜃 𝚤̂ − cos 𝜃 𝚥̂),

(3)

(4)

and 𝛼𝐴𝐵 is the angular acceleration of the rod. Substituting Eqs. (4) into Eq. (3), Eq. (3) becomes 1 𝐿𝐹𝑠 sin(𝜙 − 𝜃) = 𝐼𝐺 𝛼𝐴𝐵 , (5) 2 where we have used the trigonometric identity sin 𝜙 cos 𝜃 − sin 𝜃 cos 𝜙 = sin(𝜙 − 𝜃). Force Laws

The one force that has not been accounted for in Fig. 3 is that of the linear elastic spring, which is 𝐹𝑠 = 𝑘(𝑟 − 𝑟0 ), (6)

where 𝑟0 is the unstretched length of the spring.

ISTUDY

Section 17.4

1187

Newton-Euler Equations: General Plane Motion

Kinematic Equations Since we are using 𝑟, 𝜃, and 𝜙 as the three coordinates to define the position of the rod, we need to write 𝑎⃗𝐺 and 𝛼𝐴𝐵 in terms of those coordinates and their derivatives. We can do this by relating 𝑎⃗𝐺 to 𝑎⃗𝐴 as follows:

𝑎⃗𝐺 = 𝑎⃗𝐴 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐺∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐺∕𝐴 ,

(7)

−0.4 −0.2

0.2

0.4

𝑥 (m)

path h of of 𝐴 pat

where 𝜔𝐴𝐵 is the angular velocity of the rod and we can write ̂ 𝛼⃗𝐴𝐵 = −𝜙̈ 𝑘,

̇ 𝜔𝐴𝐵 = −𝜙,

and

𝑟⃗𝐺∕𝐴 = −⃗𝑟𝐴∕𝐺 =

𝐿 (sin 𝜙 𝚤̂ + 2

0.2

cos 𝜙 𝚥̂).

Using the polar coordinate system defined in Fig. 3, we can write 𝑎⃗𝐴 as ( ) ( ) 𝑎⃗𝐴 = 𝑟̈ − 𝑟𝜃̇ 2 𝑢̂ 𝑟 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝑢̂ 𝜃 ,

(8)

0.4 0.6

(9) path of path of 𝐵

in which

𝑢̂ 𝑟 = sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂ and 𝑢̂ 𝜃 = cos 𝜃 𝚤̂ − sin 𝜃 𝚥̂. Substituting Eqs. (8)–(10) into Eq. (7), the components of 𝑎⃗𝐺 become ( ) ( ) 𝑎𝐺𝑥 = 𝑟̈ − 𝑟𝜃̇ 2 sin 𝜃 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ cos 𝜃 + 𝐿2 𝜙̈ cos 𝜙 − 𝐿2 𝜙̇ 2 sin 𝜙, ( ) ( ) 𝑎𝐺𝑦 = 𝑟̈ − 𝑟𝜃̇ 2 cos 𝜃 − 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ sin 𝜃 − 𝐿2 𝜙̈ sin 𝜙 − 𝐿2 𝜙̇ 2 cos 𝜙.

(10)

𝑦 (m) −0.4 −0.2

(11)

0.8

0

0.2

0.4

𝑥 (m)

path of of 𝐴 path

(12)

0.2

Computation

Now that we have assembled all the pieces, the equations of motion are obtained by substituting Eqs. (6), (8), (11), and (12) into Eqs. (1), (2), and (5) to obtain the three equations of motion: ( ) ( ) 𝑘 𝑟̈ − 𝑟𝜃̇ 2 sin 𝜃 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ cos 𝜃 + 𝐿2 𝜙̈ cos 𝜙 − 𝐿2 𝜙̇ 2 sin 𝜙 + (𝑟 − 𝑟0 ) sin 𝜃 = 0, (13) 𝑚 ( ) ( ) 𝑘 𝑟̈ − 𝑟𝜃̇ 2 cos 𝜃 − 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ sin 𝜃 − 𝐿2 𝜙̈ sin 𝜙 − 𝐿2 𝜙̇ 2 cos 𝜙 + (𝑟 − 𝑟0 ) cos 𝜃 = 𝑔, 𝑚 (14) 𝐿 ̈ 𝑘 (15) 𝜙 + (𝑟 − 𝑟0 ) sin(𝜙 − 𝜃) = 0. 6 𝑚

0.4 0.6 path of of 𝐵 0.8 path

Figure 4 Paths of ends 𝐴 and 𝐵 for the first (top) and second (bottom) sets of initial conditions.

⇒ Computer simulations of the rod motion are shown in Figs. 4 and 5. ⇐ Discussion & Verification

• Each term in Eqs. (13) and (14) has been divided by 𝑚, so each term should have the units of acceleration, which it does.

initial

• Each term in Eq. (15) has been divided by 𝑚𝐿, so each term should have the units of acceleration, which it does. A Closer Look Figure 4 shows the trajectories of the ends of the bar for the first 10 s after release for the first (top) and second (bottom) sets of initial conditions. The first set of initial conditions releases the bar when the spring is unstretched, and the second set has the bar stretched to twice its unstretched length at release (everything else is equal). In the second case, adding that additional initial energy to the system dramatically changes the ensuing motion; that is, it goes from a fairly regular pattern to one in which the bar is moving very irregularly. Figure 5 shows stroboscopic images of the movement of the thin rod for the first 10 s of motion. In each case, the earliest image is the one in light gray (labeled “initial”), and each successive image becomes a darker purple (with the last one labeled “final”). The time between successive images is 0.5 s. These figures demonstrate the regular motion associated with the first set of initial conditions and the irregular motion associated with the second set. Systems, such as this one, whose motion is described by a set of nonlinear differential equations can be very sensitive to how the system is put in motion. This is one of the subjects of chaos theory.∗ ∗ See, for example, S. H. Strogatz, Nonlinear Dynamics and Chaos: With Applications to Physics, Biology,

Chemistry and Engineering, Perseus Books, Reading, Mass., 1994.

final 𝑂

initial

final

Figure 5 Stroboscopic image sequence of the rod for the first (top) and second (bottom) sets of initial conditions.

1188

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problems Problem 17.57 The sphere, cylinder, and thin ring each have mass 𝑚 and radius 𝑟. Each is released from rest on identical inclines. Assuming they all roll without slipping, which will have the largest initial angular acceleration? Which one will reach the bottom of the incline first?

+

𝑟

+

𝑟

𝑚

+

𝑟

𝑚

𝑚

cylinder

sphere

thin ring

Figure P17.57

Problems 17.58 and 17.59 A bowling ball of radius 𝑟, mass 𝑚, and radius of gyration 𝑘𝐺 is released from rest on a rough surface that is inclined at the angle 𝜃 with respect to the horizontal. The coefficients of static and kinetic friction between the ball and the incline are 𝜇𝑠 and 𝜇𝑘 , respectively. Assume that the mass center 𝐺 is at the geometric center.

𝐺 𝑟 𝑚 𝜇𝑠 , 𝜇𝑘

𝜃

Problem 17.58 Assuming the ball rolls without slip, determine expressions for the angular acceleration of the ball and the friction and normal force between the ball and the incline. In addition, find the minimum value of 𝜇𝑠 that is compatible with this motion. Problem 17.59

Figure P17.58 and P17.59

ISTUDY

Let the weight of the ball be 14 lb, the radius be 4.25 in., and the radius of gyration be 𝑘𝐺 = 2.6 in. If the incline is 10 f t long, determine the time it takes the ball to reach the bottom of the incline and the speed of 𝐺 when it reaches the bottom. Use 𝜃 = 40◦ , 𝜇𝑠 = 0.2, and 𝜇𝑘 = 0.15.

Problems 17.60 and 17.61 A bowling ball is thrown onto a lane with a backspin 𝜔0 and forward velocity 𝑣0 . The mass of the ball is 𝑚, its radius is 𝑟, its radius of gyration is 𝑘𝐺 , and the coefficient of kinetic friction between the ball and the lane is 𝜇𝑘 . Assume the mass center 𝐺 is at the geometric center. 𝜔0 𝜇𝑘

𝐺 𝑟

𝑣0 𝑚

Figure P17.60 and P17.61 Problem 17.60

Find the acceleration of 𝐺 and the ball’s angular acceleration while

the ball is slipping. For a 14 lb ball with 𝑟 = 4.25 in., 𝑘𝐺 = 2.6 in, 𝜔0 = 10 rad∕s, and 𝑣0 = 17 mph, determine the time it takes for the ball to start rolling without slip and its speed when it does so. In addition, determine the distance it travels before it starts rolling without slip. Use 𝜇𝑘 = 0.10. Hint: It should be possible to solve for a constant linear deceleration, 𝑣(𝑡) = 𝑣𝑜 − 𝑎𝑜 𝑡, as well as a constant angular deceleration, 𝜔(𝑡) = 𝜔𝑜 − 𝛼𝑜 𝑡. The time required for rolling without slip occurs when 𝑣(𝑡) + 𝑟𝜔(𝑡) = 0. Problem 17.61

ISTUDY

Section 17.4

1189

Newton-Euler Equations: General Plane Motion

Problems 17.62 and 17.63 A shop sign, with a uniformly distributed mass 𝑚 = 30 kg, ℎ = 1.5 m, 𝑤 = 2 m, and 𝑑 = 0.6 m, is at rest when cord 𝐴𝐵 suddenly breaks.

𝐶 𝐷

𝐵 𝐴

Modeling 𝐴𝐵 and 𝐶𝐷 as inextensible and with negligible mass, determine the tension in cord 𝐶𝐷 and the acceleration of the sign’s center of mass immediately after 𝐴𝐵 breaks.

𝑑

Problem 17.62

ℎ

Problem 17.63 Modeling 𝐴𝐵 and 𝐶𝐷 as elastic cords with negligible mass and stiffness 𝑘 = 8000 N∕m, determine the tension in cord 𝐶𝐷 and the acceleration of the sign’s center of mass immediately after 𝐴𝐵 breaks.

𝑤 Figure P17.62 and P17.63

Problems 17.64 and 17.65 Solve Example 17.3 on p. 1156 using the given assumptions about the motion and show that this motion is not possible for the given conditions since part of your solution will not be physically reasonable. Problem 17.64

Assume that the crate slips and tips.

Problem 17.65

Assume that the crate just tips. 𝐴 𝚥̂ 𝚤̂ 𝑅

𝜇𝑘 3 f t

2.95 f t

𝐵 Figure P17.64 and P17.65

𝐷 Figure P17.66

Problem 17.66 The cord, which is wrapped around the inner radius of the spool of mass 𝑚, is pulled vertically at 𝐴 by a constant force 𝑃 , causing the spool to roll over the horizontal bar 𝐵𝐷. Assuming that the cord is inextensible and of negligible mass, that the spool rolls without slip, and that its radius of gyration is 𝑘𝐺 , determine the angular acceleration of the spool and the total force between the spool and the bar. Hint: The force 𝑃 tends to impart clockwise rotation, so that a material point on the spool adjacent to 𝐵𝐷 tends to move to the left. This should inform the direction of the friction force acting on the spool.

sphere diam. = 2𝜌

sphere rolls without slip

𝜃 = 0 at release

Problem 17.67 Refer to the systems in Example 13.9 on p. 820 (particle separating from semicylinder) and Example 17.10 on p. 1182 (sphere separating from semicylinder).

𝑅 𝜃

𝑅

(a) Determine the speed of the particle and that of the sphere when each separates from the semicylinder. (b) Compare their speeds of separation and explain the sources of any difference. (c) Determine the value of 𝜌 such that the sphere and the particle separate at the same speed.

𝑂 Figure P17.67 and P17.68

1190

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problem 17.68 Referring to the systems in Example 13.9 on p. 820 (particle separating from semicylinder) and Example 17.10 on p. 1182 (sphere separating from semicylinder), show that the sphere dynamically behaves just as a particle if the interface between the sphere and the semicylinder is frictionless. In this case, that will mean that the sphere separates from the semicylinder at the same location as the particle.

Problem 17.69 Referring to the system in Example 17.10 on p. 1182 (and conveniently ignoring the Pitfall on p. 1183 so that we can assume that the object rolls without slip), how would the results change if one were to release a uniform cylinder, of mass 𝑚 and radius 𝜌, instead of a sphere? 2𝜌

2𝜌 𝜃 = 0 at release

roll without slip 𝑅 𝜃

𝜃 = 0 at release

roll without slip 𝑅

𝑅

𝑂

𝜃

𝑅

𝑂 Figure P17.69

Figure P17.70

Problem 17.70 Referring to the system in Example 17.10 on p. 1182 (and conveniently ignoring the Pitfall on p. 1183 so that we can assume that the object rolls without slip), how would the results change if one were to release a uniform thin ring, of mass 𝑚 and radius 𝜌, instead of a sphere? 𝜌

Problem 17.71

𝐺 𝑅

A spool of mass 𝑚 = 300 kg, inner and outer radii 𝜌 = 1.5 m and 𝑅 = 2 m, respectively, and radius of gyration 𝑘𝐺 = 1.8 m, is placed on an incline with 𝜃 = 43◦ . The cable that is wrapped around the spool and attached to the wall is initially taut. If the static and kinetic friction coefficients between the incline and the spool are 𝜇𝑠 = 0.35 and 𝜇𝑘 = 0.3, respectively, determine the acceleration of 𝐺, the angular acceleration of the spool, and the tension in the cable once the spool is released from rest.

𝜃 Figure P17.71

Problem 17.72 24.3 in.

Gary L. Gray

Figure P17.72

ISTUDY

𝐺

In Prob. 17.1 you were told to neglect the rotational inertia of the front wheels — would including it really make a difference? Let’s see. A certain roadster can go from 0 to 60 mph in 7.0 s. The weight of the car (including the two front wheels) is 2750 lb, the weight of each of its front wheels is 47 lb, and they each have a mass moment of inertia 𝐼𝐺 of 0.989 slug⋅f t 2 . To determine the effect of the rotational inertia of the front wheels, perform the following analysis: (a) Isolate one of the front wheels and determine the friction force that must be acting on the wheel for it to accelerate as given. Hint: The weight of the car on the front wheel is not known, but it is not needed to find the friction force since we are assuming that friction is sufficient to prevent slipping of the front wheels.

ISTUDY

Section 17.4

1191

Newton-Euler Equations: General Plane Motion

(b) Note that it is the friction force that makes the rotational motion of each front wheel possible, and if the mass moment of inertia 𝐼𝐺 of the front wheels were zero, then the friction force would be zero. Therefore, by neglecting the rotational inertia of the front wheels, the car would not be slowed by the friction forces found in (a). In other words, when we do account for the rotational inertia of the front wheels, we can then conclude that there is a force equal to twice the friction force that is “retarding” the motion of the car. Use this fact, and your result from (a), to determine the 0 to 60 mph time of this same roadster with front wheels that have no rotational inertia.

Problems 17.73 through 17.76 The uniform thin rod 𝐴𝐵 couples the slider 𝐴, which moves along a frictionless guide, to the wheel 𝐵, which rolls without slip over a horizontal surface. While not required, the use of computer algebra software is recommended for Probs. 17.74–17.76. Assuming that 𝐴 and 𝐵 have negligible mass, that the mass of 𝐴𝐵 is 𝑚𝐴𝐵 , and that the system is released from rest at the angle 𝜃, determine, immediately after release, the angular acceleration of the rod 𝐴𝐵, the acceleration of the center of the wheel at 𝐵, and the angular acceleration of the wheel. Problem 17.73

𝐴 𝐿 𝐺

Assuming that 𝐴 has negligible mass, 𝐵 is a uniform disk of mass 𝑚𝐵 , the mass of 𝐴𝐵 is 𝑚𝐴𝐵 , and the system is released from rest at the angle 𝜃, determine, immediately after release, the angular acceleration of the rod 𝐴𝐵, the acceleration of the center of the wheel at 𝐵, and the angular acceleration of the wheel.

Problem 17.74

Assuming that 𝐴𝐵 has negligible mass, the mass of 𝐴 is 𝑚𝐴 , 𝐵 is a uniform disk of mass 𝑚𝐵 , and the system is released from rest at the angle 𝜃, determine, immediately after release, the angular acceleration of the rod 𝐴𝐵, the acceleration of the center of the wheel at 𝐵, and the angular acceleration of the wheel. Hint: If 𝐴𝐵 is massless, 𝐴𝐵 can be treated as a two-force member.

𝜃 𝐵 𝑅

Problem 17.75

Figure P17.73–P17.76

Assuming that the mass of 𝐴 is 𝑚𝐴 , 𝐵 is a uniform disk of mass 𝑚𝐵 , the mass of 𝐴𝐵 is 𝑚𝐴𝐵 , and that the system is released from rest at the angle 𝜃, determine, immediately after release, the angular acceleration of the rod 𝐴𝐵, the acceleration of the center of the wheel at 𝐵, and the angular acceleration of the wheel. Problem 17.76

Problem 17.77 A spool of mass 𝑚 = 220 kg, inner and outer radii 𝜌 = 1.75 m and 𝑅 = 2.25 m, respectively, and radius of gyration 𝑘𝐺 = 1.9 m, is being lowered down an incline with 𝜃 = 29◦ . There is no slip between the spool and the cable as the spool moves down the incline. If the static and kinetic friction coefficients between the incline and the spool are 𝜇𝑠 = 0.4 and 𝜇𝑘 = 0.35, respectively, determine the acceleration of 𝐺, the angular acceleration of the spool, and the tension in the cable if the system is released from rest. Hint: The velocity of all material points on the cable is zero. If there is no slip between spool and cable, a material point on the spool tangent to the cable represents an IC of zero velocity.

𝜌 𝐺

𝑅

𝜃 Figure P17.77

Problem 17.78 An inextensible cord of negligible mass is wound around a homogeneous circular object. Assume that the cord is pulled to the right while remaining horizontal, and determine the value of the object’s mass moment of inertia 𝐼𝐺 , such that the object rolls without slip no matter how large the tension in the cord. What is the shape of such an object? Hint: Assume the friction force on the object acts to the left. After finding an expression for the friction force, note that the criterion can only be satisfied if the friction force remains finite as the applied force becomes infinite.

𝑅 𝐺

Figure P17.78

𝚥̂

𝚤̂

1192

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problem 17.79

𝑟𝑖

𝑟𝑜 𝐺

𝑎𝑐

The spool of mass 𝑚, radius of gyration 𝑘𝐺 , inner radius 𝑟𝑖 , and outer radius 𝑟𝑜 is placed on a horizontal conveyer belt. The cable that is wrapped around the spool and attached to the wall is initially taut. Both the spool and the conveyer belt are initially at rest when the conveyer belt starts moving with acceleration 𝑎𝑐 . If the coefficient of static friction between the conveyer belt and spool is 𝜇𝑠 , determine (a) The maximum acceleration of the conveyer belt so that the spool rolls without slipping on the belt.

Figure P17.79

(b) The initial tension in the cable that attaches the spool to the wall. (c) The angular acceleration of the spool. Evaluate your answers for 𝑚 = 500 kg, 𝑘𝐺 = 1.3 m, 𝜇𝑠 = 0.5, 𝑟𝑖 = 0.8 m, and 𝑟𝑜 = 1.6 m.

Problems 17.80 through 17.82 The thin uniform bar 𝐴𝐵 of mass 𝑚 and length 𝐿 hangs from a wheel at 𝐴, which rolls freely on the horizontal bar 𝐷𝐸. In the following problems, neglect the mass of the wheel and assume that the wheel never separates from the horizontal bar. 𝑥 𝐴

𝐷

𝐸

𝜃 𝐿 𝐵

Figure P17.80–P17.82 Problem 17.80 If the bar is released from rest at the angle 𝜃, determine, immediately after release, the angular acceleration of the bar, the force on the bar at 𝐴, and the acceleration of end 𝐴. Problem 17.81 Find the equation(s) of motion of the bar, using the coordinates 𝑥 and 𝜃 shown on the figure as the dependent variables. Problem 17.82 Find the equation(s) of motion of the bar using the coordinates 𝑥 and 𝜃 shown on the figure as the dependent variables and then simulate the system’s behavior by numerically solving the equations of motion for 5 s, using 𝑚 = 2 kg, 𝐿 = ̇ 0.6 m, 𝑥(0) = 0 m, 𝑥(0) ̇ = 0 m∕s, 𝜃(0) = 60◦ , and 𝜃(0) = 0 rad∕s. Plot 𝑥 and 𝜃 for 0 ≤ 𝑡 ≤ 5 s. 𝐵 𝑚, 𝐿

A uniform thin rod is slightly nudged at 𝐵 from the 𝜃 = 0 position so that it falls to the right. The coefficient of static friction between the rod and the floor is 𝜇𝑠 .

𝜃

Problem 17.83

𝐴

(a) Determine as a function of 𝜃 the normal force (𝑁) and the frictional force (𝐹 ) exerted by the ground on the rod as the rod falls over.

Figure P17.83 and P17.84

ISTUDY

Problems 17.83 and 17.84

(b) Knowing that the rod will slip when |𝐹 ∕𝑁| exceeds 𝜇𝑠 , determine whether the rod will slip as it falls.

ISTUDY

Section 17.4

Newton-Euler Equations: General Plane Motion

1193

Problem 17.84

(a) Determine as a function of 𝜃 the normal force (𝑁) and the frictional force (𝐹 ) exerted by the ground on the rod as the rod falls over. (b) Knowing that the rod will slip when |𝐹 ∕𝑁| exceeds 𝜇𝑠 , determine whether the rod will slip as it falls. (c) Plot 𝐹 ∕(𝑚𝑔), 𝑁∕(𝑚𝑔), and |𝐹 ∕𝑁| as a function of 𝜃 for 0 ≤ 𝜃 ≤ 𝜋∕2 rad. Use those plots to show that for smaller values of 𝜇𝑠 , end 𝐴 of the rod slips to the left, and for larger values of 𝜇𝑠 , it slips to the right.

Problem 17.85 The roadster weighs 2570 lb, and its mass is evenly distributed between its front and rear wheels. It can accelerate from 0 to 60 mph in 6.98 s. The rear wheel, shown in the blowup above the roadster, weighs 47 lb, its mass center is at its geometric center, and its mass moment of inertia 𝐼𝐵 is 0.989 slug⋅f t 2 . With this in mind, we want to determine the forces on the rear wheel shown in Fig. 2 of Example 17.9.

𝐵

24.3 in.

(a) Assuming that its acceleration is uniform, determine the forces on the front and rear wheels due to the pavement. 𝐺

(b) Now that you have the normal and friction forces between the rear wheels and the pavement, isolate one of the rear wheels and determine the forces and moments exerted by the axle on that rear wheel. Assume that the mass is evenly distributed between the right and left sides of the car and that friction is sufficient to prevent slipping of the wheels.

17.0 in. 91.7 in. Gary L. Gray

Figure P17.85

Problem 17.86 A spool of mass 𝑚, radius 𝑟, and radius of gyration 𝑘𝐺 rolls without slipping on the incline, whose angle with respect to the horizontal is 𝜃. A linear elastic spring with constant 𝑘 and unstretched length 𝐿0 connects the center of the spool to a fixed wall. Determine the equation(s) of motion of the spool, using the 𝑥 coordinate shown.

𝑥 𝑘

Problem 17.87 A spool of mass 𝑚 = 200 kg, radius 𝑟 = 0.8 m, and radius of gyration 𝑘𝐺 = 0.65 m rolls without slipping on the incline, whose angle with respect to the horizontal is 𝜃 = 38◦ . A linear elastic spring with constant 𝑘 = 500 N∕m and unstretched length 𝐿0 = 1.5 m connects the center of the spool to a fixed wall. Determine the equation(s) of motion of the spool, using the 𝑥 coordinate shown; solve them for 15 s, using the initial conditions 𝑥(0) = 2.5 m and 𝑥(0) ̇ = 0 m∕s; and then plot 𝑥 versus 𝑡. What is the approximate period of oscillation of the spool?

Problem 17.88 The uniform bar 𝐴𝐵 of mass 𝑚 and length 𝐿 is leaning against the corner with 𝜃 ≈ 0 when end 𝐵 is given a slight nudge so that end 𝐴 starts sliding down the wall as 𝐵 slides along the floor. Assuming that friction is negligible between the bar and the two surfaces against which it is sliding, determine the angle 𝜃 at which end 𝐴 will lose contact with the vertical wall. Hint: Establish the relationship between components of 𝑎⃗𝐴 , 𝜔𝐴𝐵 , 𝛼𝐴𝐵 , 𝐿 and 𝜃 by requiring that 𝐵 must move only in the horizontal direction.

𝐺 𝜃 Figure P17.86 and P17.87

𝑟

1194

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

𝐴

𝐴

𝜃 𝑚, 𝐿

𝑚, 𝐿

𝐵 Figure P17.88

𝜙

𝜃

𝐵

Figure P17.89

Problem 17.89 The uniform thin bar, which is leaning on the incline, is released from rest in the position shown and slides in the vertical plane. The contacts between the bar and the surface at ends 𝐴 and 𝐵 have negligible friction. Determine the angular acceleration of the bar immediately after it is released. Evaluate your answer for 𝑚 = 3 kg, 𝐿 = 0.75 m, 𝜙 = 45◦ , and 𝜃 = 30◦ . Hint: Establish the relationship between components of 𝑎⃗𝐴 , 𝛼𝐴𝐵 , 𝐿, 𝜃, and 𝜙 at 𝑡 = 0 by requiring that 𝐵 must move only in the horizontal direction.

Problems 17.90 through 17.93 The uniform ball of radius 𝜌 and mass 𝑚 is gently placed in the bowl 𝐵 with inner radius 𝑅 and is released. The angle 𝜙 measures the position of the center of the ball at 𝐺 with respect to a vertical line through 𝑂. Assume that the system lies in the vertical plane. Hint: In working the following problems, use the 𝑟𝜙 coordinate system shown.

𝑢̂ 𝜙 𝑂 𝑅

𝑢̂ 𝑟 𝜌

𝜙 𝐺

Problem 17.90 Assuming that the ball rolls without slip, determine the acceleration of the center of the ball at 𝐺, the angular acceleration of the ball, and the force on the ball due to the bowl immediately after the ball is released. Problem 17.91 Assuming that the ball rolls without slip, that it weighs 3 lb, is at the position 𝜙 = 40◦ , and is moving clockwise at 10 f t∕s, determine the acceleration of the center of the ball at 𝐺 and the normal and friction force between the ball and the bowl. Use 𝑅 = 4 f t and 𝜌 = 1.2 f t.

𝐵 Figure P17.90–P17.93

Problem 17.92 Assuming that friction is sufficient to prevent slipping, derive the equation(s) of motion of the ball in terms of the angle 𝜙. Problem 17.93

Assume that friction is sufficient to prevent slipping.

(a) Derive the equation(s) of motion of the ball in terms of the angle 𝜙. (b) Determine the friction force as a function of 𝜙. ̇ (c) Letting 𝜙(0) = 𝜙0 and 𝜙(0) = 0, with 0◦ < 𝜙0 < 90◦ , integrate the equation(s) of motion to determine the normal force as a function of 𝜙. (d) Using the results of Parts (b) and (c), and given a value for 𝜇𝑠 , determine the maximum value of 𝜙(0) = 𝜙0 so that the ball does not slip.

ℎ

Figure P17.94

ISTUDY

+

𝑟

Problem 17.94 An important problem in billiards or pool is the determination of the height at which you should hit the cue ball to give it backspin, topspin, or no spin. With this in mind, at what height ℎ should the cue hit the ball so that the ball always rolls without slip, regardless of how hard the ball is hit and how much friction is available? Assume a uniform ball of

ISTUDY

Section 17.4

1195

Newton-Euler Equations: General Plane Motion

mass 𝑚 and radius 𝑟. You can determine this position without having to worry about the impact between the cue and the ball by studying an arbitrary horizontal force applied to a ball at height ℎ. Hint: Assume the friction force on the cue ball acts to the left. After finding an expression for the friction force, note that the criterion can only be satisfied if the friction force remains finite as the applied force becomes infinite.

Problems 17.95 and 17.96 The Pioneer 3 spacecraft was a spin-stabilized spacecraft launched on December 6, 1958, by the U.S. Army Ballistic Missile agency in conjunction with NASA. It was designed with a despin mechanism consisting of two equal masses 𝐴 and 𝐵, each of mass 𝑚, that could be spooled out to the ends of two wires of variable length 𝓁(𝑡) when triggered by a hydraulic timer. As the masses unwound, they would slow the spacecraft’s spin from an initial angular velocity 𝜔𝑠 (0) to the final angular velocity (𝜔𝑠 )final , and then the weights and wires would be released. Assume that masses 𝐴 and 𝐵 are initially at positions 𝐴0 and 𝐵0 , respectively, before the wire begins to unwind, that the mass moment of inertia of the spacecraft is 𝐼𝑂 (this does not include the two masses 𝐴 and 𝐵), and that gravity and the mass of each wire are negligible. Hint: Refer to Prob. 16.164 if you need help with the kinematics.

NASA

𝑧

𝑥

𝜔𝑠 𝓁

Problem 17.95

Derive the equation(s) of motion of the system in terms of the dependent variables 𝓁(𝑡) and 𝜔𝑠 (𝑡).

𝑦 𝜔𝑠

𝑄 Problem 17.96 Derive the equation(s) of motion of the system in terms of the dependent variables 𝓁(𝑡) and 𝜔𝑠 (𝑡). After doing so:

𝐵0

(a) Use a computer to solve the equations of motion for 0 ≤ 𝑡 ≤ 4 s, using 𝑅 = 12.5 cm, 𝑚 = 7 g, 𝐼𝑂 = 0.0277 kg⋅m2 , and the initial conditions 𝜔𝑠 (0) = 400 rpm, 𝓁(0) = ̇ 0.01 m, and 𝓁(0) = 0 m∕s.

(b) Determine the time at which the angular velocity of the spacecraft becomes zero (this can be done by plotting the solution for 𝜔𝑠 and then estimating the time or by using numerical root finding to determine when 𝜔𝑠 = 0).

𝐴

𝜃 𝑂

𝑅

top view 𝐵 NASA

Figure P17.95 and P17.96

(c) Determine the length 𝓁 of each of the wires at the instant that the angular velocity of the spacecraft becomes zero.

Problem 17.97 An SUV is pushing a large drum to the right with force 𝑃 , using its front bumper. The drum has mass 𝑚 and radius of gyration 𝑘𝐺 . The static and kinetic friction coefficients between the drum and the ground and between the drum and the SUV are 𝜇𝑠 and 𝜇𝑘 , respectively. (a) Assuming that there is no slipping between the drum and the ground, determine the acceleration of the drum and the minimum value of 𝜇𝑠 that is consistent with this motion. (b) Determine the acceleration of point 𝐺 and the angular acceleration of the drum if 𝑃 is increased so that the drum slips relative to the ground.

Problems 17.98 and 17.99 The disk 𝐴 rolls without slipping on a horizontal surface. End 𝐵 of bar 𝐵𝐶 is pinned to the edge of the disk 𝐴, and end 𝐶 of the bar 𝐵𝐶 can slide freely along the horizontal surface. In addition, bar 𝐵𝐶 is pushed by the force 𝑃 = 𝑚𝑔 at its left end. The mass of bar 𝐵𝐶 is 𝑚𝐵𝐶 , and the mass of the disk 𝐴 is 𝑚𝐴 . The system is initially at rest. Hint: Kinematic

𝐴0

𝑚 𝐺 𝑅 𝜇𝑠 , 𝜇𝑘 Figure P17.97

1196

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

entities can be related by equating expressions such as 𝑎⃗𝐵 = 𝑎⃗𝐴 + 𝑎⃗𝐵∕𝐴 = 𝑎⃗𝐶 + 𝑎⃗𝐵∕𝐶 .

5 3

𝑅

𝐵

𝐴+

𝑅

𝑃 𝐶 Figure P17.98 and P17.99 Problem 17.98

Determine 𝑎⃗𝐴 and 𝛼𝐵𝐶 immediately after the force 𝑃 is applied if

𝑚𝐴 = 𝑚𝐵𝐶 = 𝑚. Problem 17.99 Determine 𝑎⃗𝐴 and 𝛼𝐵𝐶 immediately after force 𝑃 is applied if 𝑚𝐵𝐶 = 𝑚 and the mass of the disk 𝑚𝐴 is negligible.

Problem 17.100 The composite body consists of a uniform thin bar 𝐴𝐵 with mass 2𝑚 that is rigidly attached to a uniform disk 𝐷 with mass 3𝑚. The body is released from rest on the corner of a table in the position shown in the top figure. If the coefficient of static friction between bar 𝐴𝐵 and the table is 𝜇𝑠 , determine (a) The normal force between the body and the table as a function of 𝑚, 𝑔, and 𝜃. (b) The friction force between the body and the table as a function of 𝑚, 𝑔, and 𝜃. (c) The angle 𝜃 at which the body will start to slip relative to the table as a function of 𝜇𝑠 .

Figure P17.100

ISTUDY

Problem 17.101 A uniform sphere of mass 𝑚 and radius 𝑅 is at rest on a sharp corner 𝑂 with 𝜃 = 0. The sphere is given a tiny nudge to the left, causing the sphere to rotate about the corner. If the coefficient of static friction between the sphere and the corner is 𝜇𝑠 , determine 𝜇𝑠 as a function of 𝜃 at which the sphere slips. If 𝜇𝑠 = 12 , determine the angle at which the sphere slips.

Figure P17.101

Figure P17.102

Problem 17.102 A uniform cylinder of mass 𝑚 and radius 𝑅 is at rest on a sharp corner 𝑂 with 𝜃 = 0. The cylinder is given a tiny nudge to the right, causing the cylinder to rotate about the corner. If the coefficient of static friction between the cylinder and the corner is 𝜇𝑠 , determine 𝜇𝑠 as a function of 𝜃 at which the cylinder slips. If 𝜇𝑠 = 41 , determine the angle at which the cylinder slips.

ISTUDY

Section 17.4

Newton-Euler Equations: General Plane Motion

Problem 17.103 A thin plate with lateral dimension 2𝑅 is welded to a thin pipe section of radius 𝑅. The system is released from rest in the position shown, with the plate horizontal, and the pipe section rolls without slipping on the flat horizontal surface. If both the pipe section and the plate each have mass 𝑚, determine, as a function of 𝑔 and 𝑅, the initial angular acceleration of the system, and determine the minimum value of 𝜇𝑠 compatible with this motion.

Problem 17.104 A cylinder of radius 𝑅∕2 is welded to a thin pipe section of radius 𝑅. The system is released from rest with the center of the cylinder at 𝑂 at the same height as the center of the pipe section 𝐶. The pipe section rolls without slipping on the flat horizontal surface. If both the pipe section and the cylinder each have mass 𝑚, determine, as a function of 𝑔 and 𝑅, the initial angular acceleration of the system, and determine the minimum value of 𝜇𝑠 compatible with this motion.

Figure P17.103

Problem 17.105 The crank 𝐴𝐵 in the slider-crank mechanism is rotating counterclockwise with constant angular velocity 𝜔𝐴𝐵 . The crankshaft radius is 𝑅, the length of the connecting rod 𝐵𝐶 is 𝐿, and the distance from the mass center of the connecting rod 𝐷 to the end of the crank at 𝐵 is 𝑑. The mass of the connecting rod is 𝑚𝐷 , the mass moment of inertia of the connecting rod is 𝐼𝐷 , and the mass of the piston is 𝑚𝐶 . (a) Using the component system shown, determine the 𝑥 and 𝑦 components of the forces on the connecting rod at 𝐵 and 𝐶 as functions of the crank angle 𝜃. (b) Using 𝜔𝐴𝐵 = 5700 rpm, 𝑅 = 48.5 mm, 𝐿 = 141 mm, 𝑑 = 36.4 mm, 𝑚𝐷 = 0.439 kg, 𝐼𝐷 = 0.00144 kg⋅m2 , and 𝑚𝐶 = 0.434 kg, plot each of the four force components, the magnitude of the forces at 𝐵 and 𝐶, and the moment acting on the connecting rod about point 𝐷, all as a function of 𝜃, for one full rotation of the crank. Hint: The kinematics of this problem have been considered in Example 16.10 on p. 1103.

𝐶 blowup of connecting rod and piston

𝐶

𝐴 𝜔𝐴𝐵

𝐿

𝐵

𝑑

𝐷 𝚥̂

𝑅

𝚤̂

Figure P17.105

𝐵 𝜃

Figure P17.104

1197

1198

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Design Problems Design Problem 17.1 Revisit the calculations done at the beginning of the chapter concerning the determination of the maximum acceleration that can be achieved by a motorcycle without causing the front wheel to lift off the ground. Specifically, construct a new model of the motorcycle by selecting a real-life motorcycle and researching its geometry and inertia properties, including the inertia properties of the wheels. Then analyze your model to determine how the maximum acceleration in question depends on the horizontal and vertical positions of the center of mass with respect to the points of contact between the ground and the wheels. Include in your analysis a comparison of results that account for the inertia of the front wheel with results that neglect the inertia of the front wheel.

Guitar Studio/Shutterstock

Figure DP17.1 direction of motion of belt

Design Problem 17.2 seat belt

𝐴 ratchet wheel pawl direction of motion of car

𝐵 weight

𝑘, 𝐿0

Figure DP17.2

ISTUDY

One end of a seat belt on passenger cars is wound around a ratchet wheel that can be locked when the deceleration of the car exceeds a set value. In the sketch shown, consider a ratchet wheel that can rotate about the fixed point 𝐴. The ratchet wheel has cogs and is positioned above a weight. The weight can pivot about point 𝐵 and is rigidly connected to a pawl that, for a sufficiently large rotation, will lock the motion of the ratchet wheel. Research realistic dimensions for the ratchet wheel, and design a locking mechanism, such that the belt’s motion will be stopped for horizontal decelerations greater than 0.5𝑔, where 𝑔 is the acceleration due to gravity. In your design you may use a spring to limit the motion of the weight.

ISTUDY

Section 17.5

1199

Chapter Review

17.5 C h a p t e r R e v i e w Newton-Euler Equations: Bodies Symmetric with Respect to the Plane of Motion

𝑃

𝑦

For a rigid body, the translational Newton-Euler equations are given by Euler’s first law, which is Eq. (17.1), p. 1145

𝑟⃗𝐺∕𝑃

𝜔𝐵 , 𝛼𝐵

𝐹⃗ = 𝑚𝑎⃗𝐺 , 𝐺

where 𝐹⃗ is the resultant of all external forces, 𝑚 is the mass of the rigid body, and 𝑎⃗𝐺 is the inertial acceleration of its mass center (see Fig. 17.18).

The moment center is an arbitrary point 𝑃 . For a rigid body that is symmetric with respect to the plane of motion, the third Newton-Euler equation is the rotational equation of motion given by (see Fig. 17.18) Eq. (17.28), p. 1148, and Eq. (17.29), p. 1149 ) ( 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 ,

𝑎⃗𝐺

𝐵 𝑥

𝑄

Figure 17.18 The relevant kinematic quantities for the NewtonEuler equations for a rigid body.

⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝐺 , 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

where the second equation is simply the vector form of the first, and • 𝑀𝑃 is the total moment about 𝑃 in the 𝑧 direction. • 𝐼𝐺 is the mass moment of inertia of the body about its mass center 𝐺. • 𝛼𝐵 is the angular acceleration of the body. • 𝑚 is the total mass of the body. • 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂ is the acceleration of the mass center. • 𝑟⃗𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂ is the position of the mass center 𝐺 relative to the moment center 𝑃 . ⃗ If any one of the following conditions is true, that is, if 𝑟⃗ × 𝑚𝑎⃗ = 0: 𝐺∕𝑃

1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃

𝐺

⃗ = 0,

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 moves with constant velocity), 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 ,

𝑦

then Eqs. (17.28) and (17.29) reduce to

𝑂

𝑎⃗𝑂

Eq. (17.30), p. 1149 𝑟⃗𝐺∕𝑂

𝑀𝑃 = 𝐼𝐺 𝛼𝐵 .

𝐺

The moment center is a point 𝑂 on the body. If the moment center is a point 𝑂 on the rigid body or an arbitrary extension of the rigid body, then an alternate form of Eq. (17.29) is (see Fig. 17.19) Eq. (17.36), p. 1150 ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ 𝑀 ⃗𝑂 , 𝑂 𝑂 𝐵 𝐺∕𝑂 × 𝑚𝑎 where 𝐼𝑂 is the mass moment of inertia of the body about an axis perpendicular to the plane of motion passing through point 𝑂 and 𝑎⃗𝑂 is the acceleration of point 𝑂. If any one ⃗ of the following is true, that is, if 𝑟⃗𝐺∕𝑂 × 𝑚𝑎⃗𝑂 = 0:

𝛼𝐵

𝐵

𝑄

𝑥

Figure 17.19 The relevant kinematic quantities for the rotational equations of motion of a rigid body when the moment center 𝑂 is a point on the rigid body.

1200

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

⃗ 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗𝐺∕𝑂 = 0, 2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑃 moves with constant velocity), 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 , then Eq. (17.36) becomes Eq. (17.37), p. 1150 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 .

Graphical interpretation of the equations of motion. Referring to Fig. 17.20, there is a graphical/visual way of obtaining Eqs. (17.1) and (17.29). We begin by drawing the FBD of the rigid body, including all forces and moments, and then we draw the KD (kinetic diagram) of the rigid body, which includes the vectors 𝐼𝐺 𝛼𝐵 and 𝑚𝑎⃗𝐺 . As shown in Fig. 17.20, we graphically equate these two diagrams, and we write the equations generated by that equation. In doing so, we automatically obtain Eqs. (17.1) and (17.29).

=

FBD 𝐹1

𝐹2

𝑄1

𝐼𝐺 𝛼𝐵

=

𝐺 1 ∕𝑃

𝐵

𝑟⃗𝐺∕𝑃

𝑀1

𝑟⃗𝑄

𝑚𝑎⃗𝐺

𝐺

𝑄2

𝑟⃗𝑄

𝑃

KD

𝐵

𝑃

2 ∕𝑃

Figure 17.20. The free body diagram and kinetic diagram of a general rigid body. Equating them and writing the associated equations always give the correct equations of motion, i.e., Eqs. (17.1) and (17.29).

Newton-Euler Equations: Translation For a body that is only translating, the first two Newton-Euler equations are the two scalar components of Euler’s first law, which is (see Fig. 17.21) 𝑦

Eq. (17.38), p. 1153

𝑎⃗𝑂

𝑂

𝐹⃗ = 𝑚𝑎⃗𝐺 .

𝑟⃗𝐺∕𝑂 𝐺

𝑎⃗𝐺

If the moment center is an arbitrary point 𝑃 , since 𝛼⃗𝐵 = 0⃗ for a body in pure translation, then the rotational equation is given by Eq. (17.39), p. 1153

𝑟⃗𝐺∕𝑃 𝐵 𝑃 𝑄

𝛼⃗𝐵 = 0⃗

⃗ = 𝑟⃗ 𝑀 ⃗𝐺 . 𝑃 𝐺∕𝑃 × 𝑚𝑎 𝑥

Figure 17.21 Illustration of the important points and kinematic quantities for a rigid body in translation.

ISTUDY

If the moment center is an arbitrary point 𝑃 and any one of the following conditions is ⃗ true, thus making 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0: ⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0, 2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 ,

ISTUDY

Section 17.5

1201

Chapter Review

then, when written in scalar form, Eq. (17.39) reduces to Eq. (17.40), p. 1153 𝑀𝑃 = 0. Referring to Fig. 17.21, if the moment center is a point 𝑂 on the translating rigid body, ⃗ which is then we can apply Eq. (17.36) with 𝛼⃗𝐵 = 0, Eq. (17.41), p. 1153 ⃗ = 𝑟⃗ 𝑀 ⃗𝑂 , 𝑂 𝐺∕𝑂 × 𝑚𝑎 ⃗ If, in addition to the moment center being on the where we have again used 𝛼⃗𝐵 = 0. translating rigid body, any of the following is true: ⃗ 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗ = 0, 𝐺∕𝑂

2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑂 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 , then, when written in scalar form, Eq. (17.41) becomes Eq. (17.42), p. 1153 𝑀𝑂 = 0.

Newton-Euler Equations: Rotation About a Fixed Axis Referring to Fig. 17.22, for a body in fixed-axis rotation about a point, the three NewtonEuler equations of motion consist of the two scalar components of

𝑦 𝐵 𝛼⃗𝐵 × 𝑟⃗𝐺∕𝑂

Eq. (17.43), p. 1163 𝐹⃗ = 𝑚𝑎⃗𝐺 ,

𝑎⃗𝐺 𝐺

along with a moment equation. If the moment center 𝑃 is arbitrary, then the moment equation is given by Eqs. (17.45) and (17.46), p. 1163

𝑟⃗𝐺∕𝑂

⃗ = 𝐼 𝛼⃗ + 𝑟⃗ ⃗𝐺 , 𝑀 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

−𝜔2𝐵 𝑟⃗𝐺∕𝑂

Eq. (17.47), p. 1164 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 .

Newton-Euler Equations: General Plane Motion Referring to Fig. 17.23, when a body is translating and rotating, the first two Newton-Euler equations are always the two scalar components of Euler’s first law, which is Eq. (17.49), p. 1177 𝐹⃗ = 𝑚𝑎⃗𝐺 .

𝑃

𝑂

𝑀𝑃 = 𝐼𝐺 𝛼𝐵 , ⃗ If, as is typical with fixed-axis where the second equation applies when 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0. rotation, the moment center 𝑂 is on the axis of rotation, then the moment equation is

𝑟⃗𝐺∕𝑃

𝛼𝐵 𝑄

𝜔𝐵

𝑥

Figure 17.22 Illustration of the important points and kinematic quantities for a rigid body in fixed-axis rotation.

1202

Newton-Euler equations for an arbitrary moment center. Again referring to Fig. 17.23, when the moment center is an arbitrary point 𝑃 , the third Newton-Euler equation is given by Eq. (17.50), p. 1177

𝑦 𝑃 𝑂 𝑟⃗𝐺∕𝑂

𝑟⃗𝐺∕𝑃

𝜔 𝐵 , 𝛼𝐵

⃗ = 𝐼 𝛼⃗ + 𝑟⃗ ⃗𝐺 , 𝑀 𝑃 𝐺 𝐵 𝐺∕𝑃 × 𝑚𝑎

𝐺

⃗ = 𝑀 𝑘, ̂ 𝛼⃗ = 𝛼 𝑘, ̂ 𝑟⃗ where, since the motion is planar, 𝑀 𝑃 𝑃 𝐵 𝐵 𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂, and 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂. If any one of the following conditions is true:

𝑎⃗𝐺 𝑎⃗𝑂 𝑄

𝐵

⃗ 1. Point 𝑃 is the mass center 𝐺, so that 𝑟⃗𝐺∕𝑃 = 0, 𝑥

Figure 17.23 A general rigid body defining all relevant quantities needed to write its Newton-Euler equations.

ISTUDY

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

2. 𝑎⃗𝐺 = 0⃗ (i.e., 𝐺 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 , ⃗ and Eq. (17.50) reduces to then 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0, Eq. (17.51), p. 1177 𝑀𝑃 = 𝐼𝐺 𝛼𝐵 , where we have written the result in scalar form.

Newton-Euler equations when the moment center is on the rigid body. Referring to Fig. 17.23, when the moment center is a point 𝑂 on the rigid body, Eq. (17.36) on p. 1150 can be used to obtain the third Newton-Euler equation, which is Eq. (17.52), p. 1177 ⃗ = 𝐼 𝛼⃗ + 𝑟⃗ ⃗𝑂 . 𝑀 𝑂 𝑂 𝐵 𝐺∕𝑂 × 𝑚𝑎 ⃗ This occurs if any The more useful form of this equation is obtained if 𝑟⃗𝐺∕𝑂 × 𝑚𝑎⃗𝑂 = 0. of the following is true: ⃗ = 0, 1. Point 𝑂 is the mass center 𝐺 so that 𝑟⃗ 𝐺∕𝑂

2. 𝑎⃗𝑂 = 0⃗ (i.e., 𝑂 is fixed or moves with constant velocity), 3. 𝑟⃗𝐺∕𝑂 is parallel to 𝑎⃗𝑂 , and then Eq. (17.52) becomes Eq. (17.53), p. 1177 𝑀𝑂 = 𝐼𝑂 𝛼𝐵 , where we have used the scalar form to reflect the fact that the motion is planar.

ISTUDY

Section 17.5

Chapter Review

1203

Review Problems Problems 17.106 through 17.111 The figure shows a weapon called a battering ram (modern large battering rams are typically mounted on armored vehicles). The ram has a weight 𝑊𝑟 = 2500 lb, center of mass at 𝐸, and radius of gyration 𝑘𝐸 = 6.5 f t. Also let the distance between points 𝐴 and 𝐵 and between points 𝐶 and 𝐷 be 6 f t. In addition, let ℎ = 4.5 f t and 𝑑 = 3 f t. Finally, let the connections at points 𝐴, 𝐵, 𝐶, and 𝐷 be pin connections, and assume that the cart does not move while the ram swings.

𝐴

𝐵 𝐸

Assuming that the ram is suspended by inextensible cords of negligible mass, determine the tension in the cords and the acceleration of 𝐸 immediately after the ram is released from rest at 𝜃 = 75◦ . Let the ram be at rest with 𝜃 = 0◦ . Assume that the cords 𝐴𝐵 and 𝐶𝐷 are inextensible and of negligible mass. Also assume that cord 𝐴𝐵 breaks suddenly. Determine the tension in cord 𝐶𝐷 and the acceleration of 𝐸 immediately after 𝐴𝐵 breaks.

𝐶 𝜃

Problem 17.106

Problem 17.107

𝑑

ℎ

𝐷

Figure P17.106–P17.111

Problem 17.108 Assume that 𝐴𝐵 and 𝐶𝐷 are inextensible cords with negligible mass. In addition, assume that, at the instant shown, 𝜃 = 10◦ and the ram is swinging forward with |𝑣⃗𝐸 | = 7 f t∕s. At this instant, determine the acceleration of 𝐸, as well as the reaction forces at points 𝐴 and 𝐶. Problem 17.109 Assume that 𝐴𝐵 and 𝐶𝐷 are uniform thin rods weighing 100 lb each. If the ram is released from rest when 𝜃 = 63◦ , determine the acceleration of 𝐸, as well as the reaction forces at points 𝐴 and 𝐶 immediately after release. Problem 17.110 Assume that 𝐴𝐵 and 𝐶𝐷 are uniform thin rods weighing 100 lb each and that the ram is at rest with 𝜃 = 0◦ . Assume that bar 𝐶𝐷 breaks suddenly, and determine the acceleration of 𝐸 and the forces at 𝐴 immediately after 𝐶𝐷 breaks. Problem 17.111 Assume that 𝐴𝐵 and 𝐶𝐷 are uniform thin rods weighing 100 lb each. Assume that, at the instant shown, 𝜃 = 10◦ and the ram is swinging forward with |𝑣⃗𝐸 | = 7 f t∕s. At this instant, determine the acceleration of the ram, as well as the reaction forces at points 𝐴 and 𝐶.

Problems 17.112 through 17.114 A spool has a weight 𝑊 = 450 lb, outer and inner radii 𝑅 = 6 f t and 𝜌 = 4.5 f t, respectively, radius of gyration 𝑘𝐺 = 4.0 f t, and mass center at 𝐺. The spool is being pulled to the right as shown, and the cable wrapped around the spool is inextensible and of negligible mass. Problem 17.112 Assume that the spool rolls without slipping with respect to both the cable and the ground. If the pickup truck pulls the cable with a force 𝑃 = 125 lb, determine the acceleration of the center of the spool and the minimum value of the static friction coefficient between the spool and the ground that is compatible with this motion. Problem 17.113

Assume that the static friction coefficient between the spool and the ground is 𝜇𝑠 = 0.75, and determine the maximum value of the force that the truck could exert on the cable without causing the spool to slip relative to the ground.

Problem 17.114

Assume that the static and kinetic friction coefficients between the spool and the ground are 𝜇𝑠 = 0.25 and 𝜇𝑘 = 0.2. Furthermore, assume that the spool is initially at rest and that the pickup truck pulls the spool with a force of 550 lb. Determine the initial acceleration of the center of the spool.

𝜌 𝑅

𝐺

Figure P17.112–P17.114

ram

1204

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion Problems 17.115 and 17.116

𝑣0

A bowling ball is thrown onto a lane with a forward spin 𝜔0 and forward velocity 𝑣0 . The mass of the ball is 𝑚, its radius is 𝑟, its radius of gyration is given by 𝑘𝐺 , and the coefficient of kinetic friction between the ball and the lane is 𝜇𝑘 . Assume the mass center 𝐺 is at the geometric center.

Figure P17.115 and P17.116

Problem 17.115 Assuming that 𝑣0 > 𝑟𝜔0 , determine the acceleration of the center of the ball and the angular acceleration of the ball until it starts rolling without slip.

𝜔0 𝐺 𝑟

𝜇𝑘

𝑚

Assuming that 𝑣0 < 𝑟𝜔0 , determine the acceleration of the center of the ball and the angular acceleration of the ball until it starts rolling without slip.

Problem 17.116

Problems 17.117 and 17.118 The car, as seen from the front, is traveling at a constant speed 𝑣𝑐 on a turn of constant radius 𝑅 that is banked at an angle 𝜃 with respect to the horizontal. The coefficient of static friction between the tires and the road is 𝜇𝑠 . The car’s center of mass is at 𝐺. 𝑅

𝐺

𝜃

𝑤∕2

ℎ

𝑤∕2

Figure P17.117 and P17.118

Determine the bank angle 𝜃 so that there is no tendency to slip or Problem 17.117 tip, i.e., so that no friction is required to keep the car on the road. Problem 17.118 For a given bank angle 𝜃, and assuming that the car does not tip, find the maximum speed 𝑣𝑚 that the car can achieve without slipping.

horizontal guide 𝐴

fully open

𝐵

𝑘

𝜃

𝐶

vertical guide

ℎ 𝐷 𝐸

An overhead fold-up door, with height ℎ and mass 𝑚, consists of two identical sections hinged at 𝐶. The roller at 𝐴 moves along a horizontal guide, whereas the rollers at 𝐵 and 𝐷, which are the midpoints of sections 𝐴𝐶 and 𝐶𝐸, move along a vertical guide. The door’s operation is assisted by two identical springs attached to the horizontally moving rollers (only one of the two springs is shown). The springs are stretched an amount 𝛿0 when the door is fully open. Problem 17.119 Let ℎ = 10 m and 𝑚 = 380 kg. In addition, let 𝑘 = 2400 N∕m and 𝛿0 = 0.15 m. Assuming that the door is released from rest when 𝜃 = 10◦ and that all sources of friction are negligible, determine the angular acceleration of each section of the door right after release. Problem 17.120

Assuming that friction between the rollers and the guide can be neglected, determine the equation(s) of motion of the system.

fully closed floor Figure P17.119–P17.121

ISTUDY

Problems 17.119 through 17.121

Problem 17.121 Let ℎ = 10 m and 𝑚 = 320 kg. In addition, let 𝑘 = 2400 N∕m and 𝛿0 = 0.15 m. Assuming that the door is released from rest when 𝜃 = 5◦ , determine the time the door will take to close and the speed of 𝐸 at closing.

ISTUDY

Section 17.5

1205

Chapter Review

Problems 17.122 through 17.124 A wheel with center 𝑂, radius 𝑅, weight 𝑊 , radius of gyration 𝑘𝐺 , and center of mass 𝐺 at a distance 𝜌 from 𝑂 is released from rest on a rough incline. The angle 𝜙 is the angle between the segment 𝑂𝐺 (which rotates with the wheel) and the horizontal.

𝜌

𝐺 𝜙

𝑂

Let 𝑅 = 1.5 f t, 𝜌 = 0.8 f t, 𝑘𝐺 = 0.6 f t, 𝑊 = 4 lb, and 𝜃 = 25◦ . In addition, let 𝜙 = at the instant of release. Determine the minimum coefficient of static friction so that the wheel starts moving while rolling without slip. In addition, determine the corresponding angular acceleration right after release. Problem 17.122

35◦

Problem 17.123 Assuming that there is enough friction for the wheel to roll without slip, determine the equation(s) of motion of the wheel, as well as the constraint force equations, that is, those equations that would allow you to compute the reaction forces at the contact point with the incline if the motion were known.

𝑅 𝜃 Figure P17.122–P17.124

Let 𝑅 = 1.5 f t, 𝜌 = 0.8 f t, 𝑘𝐺 = 0.6 f t, 𝑊 = 4 lb, and 𝜃 = 25◦ , and let 𝜙 = 60 at the instant of release. Assuming that there is sufficient friction for the wheel to roll without slip and that the incline is sufficiently long that we need not worry about the wheel reaching the end of the incline, determine the equation(s) of motion of the wheel and expressions for the friction and normal forces at the point of contact between the wheel and the incline. Then, integrate the equation(s) of motion as a function of time for 0 ≤ 𝑡 ≤ 2 s. Plot the normal force as a function of time over the given time interval and determine if and when the wheel loses contact with the incline. Problem 17.124 ◦

𝑥𝐴

Problems 17.125 and 17.126 The uniform slender bar 𝐴𝐵 has mass 𝑚𝐴𝐵 and length 𝐿. The crate has a uniformly distributed mass 𝑚𝐶 and dimensions ℎ and 𝑤. Bar 𝐴𝐵 is pin-connected to the trolley at 𝐴 and to the crate at 𝐵. The trolley is constrained to move along the horizontal guide shown. Point 𝑂 on the trolley’s guide is a fixed reference point. Neglect the mass of the trolley and friction.

ℎ 𝐵

Derive the equations of motion of the system and express them in terms of the variables 𝑥𝐴 , 𝜃, and 𝜙 along with their time derivatives. Let 𝑚𝐴𝐵 = 75 kg, 𝐿 = 4.5 m, 𝑚𝐶 = 250 kg, 𝑑 = 0.5 m, ℎ = 1.5 m, and 𝑤 = 2 m. Finally, assume that the system is released from rest when 𝑥𝐴 = 0, 𝜃 = 30◦ , and 𝜙 = 45◦ . Plot 𝑥𝐴 , 𝜃, and 𝜙 as functions of time for 0 ≤ 𝑡 ≤ 15 s.

𝑤

𝐶 𝑑 𝜙

Figure P17.125 and P17.126

Problem 17.127 A drum of mass 𝑚𝑑 , radius 𝑅, radius of gyration 𝑘𝐺 , and with center of mass at 𝐺 is placed on a cart of mass 𝑚𝑐 for transport. The system is initially at rest, and the cart is pushed to the right with the force 𝑃 . The coefficient of static friction between the cart and the drum is 𝜇𝑠 . Neglecting the mass of the wheels, determine the maximum force 𝑃 that can be applied to the cart so that the drum does not slip on the cart, and find the corresponding acceleration of the cart and of point 𝐺. drum 𝑚𝑑 𝐺 𝑃 cart 𝑚𝑐 Figure P17.127

𝑅

𝐴

𝜃

Problem 17.125

Problem 17.126

𝑂

𝐿

1206

small box

Chapter 17

Newton-Euler Equations for Planar Rigid Body Motion

Problem 17.128

𝐴

The uniform bar 𝐴𝐵 of mass 𝑚 and length 𝐿 is leaning against the corner with 𝜃 ≈ 0. A small box is placed on top of the bar at 𝐴. End 𝐵 of the bar is given a slight nudge so that end 𝐴 starts sliding down the wall as 𝐵 slides along the floor. Assuming that friction is negligible between the bar and the two surfaces against which it is sliding, and neglecting the weight of the box, determine the angle 𝜃 at which the box will lose contact with the bar.

𝜃 𝑚, 𝐿

𝐵

Problem 17.129 Bars 𝐴𝐵 and 𝐵𝐶 are uniform with masses 𝑚𝐴𝐵 = 2 kg and 𝑚𝐵𝐶 = 1 kg, respectively. Their lengths are 𝐿 = 1.25 m and 𝐻 = 0.75 m. Bar 𝐵𝐶 is pin-connected to a fixed support at 𝐶 that is a distance 𝛿 = 0.2 m from the ground. Bar 𝐴𝐵 is pin-connected at 𝐴 to a uniform wheel with radius 𝑅 = 0.60 m and mass 𝑚𝑂𝐴 = 5 kg. Note that 𝐴 is at a distance 𝜌 from the center of the wheel. At the instant shown, 𝐴 is vertically aligned with 𝑂; in addition, 𝐴𝐵 and 𝐵𝐶 are parallel and perpendicular to the ground, respectively. At this instant, bar 𝐵𝐶 is rotating clockwise with an angular velocity of 2 rad∕s and an angular acceleration of 1.2 rad∕s2 . Assuming that the wheel rolls without slip, determine the force 𝑃 that is applied to the wheel at the instant shown. In addition, determine the minimum static coefficient of friction necessary for the wheel not to slip.

Figure P17.128

𝐿

𝐵

𝐴 𝜌 𝑃

𝚥̂

𝐻

𝚤̂

𝑂 𝑅

𝐶 𝛿

Figure P17.129

Problem 17.130 𝐵

𝐴

𝜃

𝑃

Two identical uniform bars are pinned together at one end, and each bar has a roller at its other end. The rollers can roll freely along the horizontal surface as shown. Each bar has mass 𝑚 and length 𝐿, and a horizontal force 𝑃 is applied to bar 𝐵𝐶 at 𝐵. Although the bars can rotate relative to each other, for a given value of 𝑃 there exists a corresponding value of 𝜃 such that the system moves with 𝜃 constant. (a) Find the forces on the bars at 𝐴 and 𝐵 and show that they are independent of 𝑃 .

𝐶 Figure P17.130

ISTUDY

(b) Determine 𝜃 as a function of 𝑃 and find 𝜃0 as 𝑃 → 0 and 𝜃∞ as 𝑃 → ∞. Neglect the mass of the rollers and any friction in their bearings.

ISTUDY

18

Energy and Momentum Methods for Rigid Bodies

In this chapter, we apply three fundamental balance laws to rigid bodies: the work-energy principle, the linear impulse-momentum principle, and the angular impulse-momentum principle. Each of these balance principles stems from the Newton-Euler equations we studied in Chapter 17, and they make it easier for us to solve many problems that would be far more challenging if solved using the Newton-Euler approach as presented in Chapter 17.

NASA

Astronaut Dave Williams working at the installation of a new control moment gyroscope on the International Space Station. The concept of angular momentum is essential to understand the working principles of gyroscopes.

18.1

Work-Energy Principle for Rigid Bodies

𝑦

𝜔𝐵

Kinetic energy of rigid bodies in planar motion Referring to Fig. 18.1, consider an infinitesimal mass element 𝑑𝑚 with velocity 𝑣⃗𝑑𝑚 and kinetic energy 𝑑𝑇𝑑𝑚 = 12 (𝑑𝑚)𝑣2𝑑𝑚 in a body 𝐵. The kinetic energy of the body 𝐵 is therefore 1 2 𝑣 𝑑𝑚. (18.1) 𝑇 = ∫𝐵 2 𝑑𝑚 Because 𝐺 and 𝑑𝑚 are two points on the same rigid body, we can write ⃗ 𝐵 × 𝑞, ⃗ 𝑣⃗𝑑𝑚 = 𝑣⃗𝐺 + 𝜔

𝑣⃗𝑑𝑚 𝑞⃗ 𝑣⃗𝐺

𝑟⃗𝑑𝑚 𝐺

(18.2)

where 𝐺 and 𝜔 ⃗ 𝐵 are the mass center and angular velocity of 𝐵, respectively, and 𝑞⃗ is the position of 𝑑𝑚 relative to 𝐺. In planar motion, 𝜔 ⃗ 𝐵 = 𝜔𝐵 𝑘̂ and 𝑞⃗ = 𝑞𝑥 𝚤̂ + 𝑞𝑦 𝚥̂, so 𝜔 ⃗ 𝐵 × 𝑞⃗ = 𝜔𝐵 (−𝑞𝑦 𝚤̂ + 𝑞𝑥 𝚥̂). Using Eq. (18.2), we can then write 𝑣2𝑑𝑚 = 𝑣⃗𝑑𝑚 ⋅ 𝑣⃗𝑑𝑚

( ( ) ) = 𝑣⃗𝐺 ⋅ 𝑣⃗𝐺 + 2𝑣⃗𝐺 ⋅ 𝜔𝐵 −𝑞𝑦 𝚤̂ + 𝑞𝑥 𝚥̂ + 𝜔2𝐵 𝑞𝑥2 + 𝑞𝑦2 ( ) ( ) = 𝑣2𝐺 + 2𝜔𝐵 −𝑣𝐺𝑥 𝑞𝑦 + 𝑣𝐺𝑦 𝑞𝑥 + 𝜔2𝐵 𝑞𝑥2 + 𝑞𝑦2 ,

𝑑𝑚

𝚥̂ 𝑂

𝑟⃗𝐺 𝚤̂

𝐵 𝑥

Figure 18.1 A rigid body in planar motion.

(18.3)

1207

1208

Chapter 18

Energy and Momentum Methods for Rigid Bodies

where we have used 𝑣⃗𝐺 = 𝑣𝐺𝑥 𝚤̂ + 𝑣𝐺𝑦 𝚥̂. Substituting Eq. (18.3) into Eq. (18.1) gives [ ( ) 1 2( 2 )] 1 2 2 𝑇 = 𝑑𝑚. (18.4) 𝑣 + 𝜔 𝑞 + 𝑣 𝑞 𝜔 + 𝑞 −𝑣 + 𝑞 𝐵 𝐺𝑥 𝑦 𝐺𝑦 𝑥 𝑦 2 𝐵 𝑥 ∫𝐵 2 𝐺 We can simplify Eq. (18.4), starting with the term 12 𝑣2𝐺 , as ∫𝐵

1 2 𝑣 2 𝐺

𝑑𝑚 = 12 𝑣2𝐺

∫𝐵

𝑑𝑚 = 12 𝑚𝑣2𝐺 ,

(18.5)

where 𝑚 = ∫𝐵 𝑑𝑚 is the total mass of 𝐵. The last term in the integral of Eq. (18.4) can be written as ) ( 2 ) 1 2( 2 𝜔𝐵 𝑞𝑥 + 𝑞𝑦2 𝑑𝑚 = 12 𝜔2𝐵 (18.6) 𝑞𝑥 + 𝑞𝑦2 𝑑𝑚 = 12 𝐼𝐺𝑧 𝜔2𝐵 , 2 ∫𝐵 ∫𝐵 where we have used Eq. (17.27) on p. 1148 to introduce the moment of inertia. The second term in the integral of Eq. (18.4) can be simplified as ∫𝐵

( ) 𝜔𝐵 −𝑣𝐺𝑥 𝑞𝑦 + 𝑣𝐺𝑦 𝑞𝑥 𝑑𝑚 = −𝜔𝐵 𝑣𝐺𝑥

∫𝐵

𝑞𝑦 𝑑𝑚 + 𝜔𝐵 𝑣𝐺𝑦

∫𝐵

𝑞𝑥 𝑑𝑚 = 0, (18.7)

where we have used the fact that both ∫𝐵 𝑞𝑥 𝑑𝑚 and ∫𝐵 𝑞𝑦 𝑑𝑚 are zero because they measure the position of 𝐺 with respect to 𝐺 [see Eqs. (17.25) and (17.26) on p. 1148]. Substituting Eqs. (18.5)–(18.7) into Eq. (18.4) yields

Concept Alert

𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺 𝜔2𝐵 ,

Kinetic energy of a rigid body. The kinetic energy of a rigid body consists of two parts. The first part, 12 𝑚𝑣2𝐺 , is due to motion of the center of mass (translational kinetic energy), and the second, 12 𝐼𝐺 𝜔2𝐵 , is due to the rotational motion of points relative to the center of mass (rotational kinetic energy). A rigid body can have kinetic energy even when the center of mass does not move.

(18.8)

where we have written the shorthand 𝐼𝐺 for 𝐼𝐺𝑧 (as in Chapter 17). From Eq. (18.8), the kinetic energy of a rigid body in planar motion consists of 1. The term 12 𝑚𝑣2𝐺 , often called the translational kinetic energy, which is the kinetic energy of the body in translation 2. The term 12 𝐼𝐺 𝜔2𝐵 , often called the rotational kinetic energy, which is the kinetic energy of the body in fixed-axis rotation about the mass center 𝐺 and with the axis of rotation perpendicular to the plane of motion Kinetic energy: rotation about a fixed axis

𝑂 𝜔𝐵 ℎ 𝐵

𝐺

Figure 18.2 shows a platform 𝐵 supported by two bars that are both pinned at 𝑂 and rigidly connected to 𝐵. The platform 𝐵 rotates about the fixed axis perpendicular to the plane of the figure and going through 𝑂. If 𝜔𝐵 is the angular speed of 𝐵, rigid body kinematics tells us that the center of mass 𝐺 of 𝐵 has speed 𝑣𝐺 = 𝜔𝐵 ℎ, where ℎ is the distance between 𝐺 and 𝑂. Equation (18.8) then tells us that the kinetic energy of 𝐵 is (18.9) 𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺 𝜔2𝐵 = 12 𝑚𝜔2𝐵 ℎ2 + 12 𝐼𝐺 𝜔2𝐵 , which simplifies to

Figure 18.2 A platform 𝐵 swinging about the pin at 𝑂.

ISTUDY

𝑇 =

1( 𝑚ℎ2 2

) + 𝐼𝐺 𝜔2𝐵 = 12 𝐼𝑂 𝜔2𝐵 ,

(18.10)

where, using the parallel axis theorem, 𝐼𝑂 = 𝑚ℎ2 + 𝐼𝐺 is the mass moment of inertia of 𝐵 relative to the axis of rotation (see Appendix C). This calculation shows that the kinetic energy of a rigid body 𝐵 in a fixed-axis rotation can always be written as 𝑇 = 12 𝐼𝑂 𝜔2𝐵 ,

(18.11)

where 𝐼𝑂 is the mass moment of inertia relative to the axis of rotation. This result is useful because fixed-axis rotations are common in applications.

ISTUDY

Section 18.1

1209

Work-Energy Principle for Rigid Bodies

Using the IC to find kinetic energy. Since the kinetic energy only depends on the speeds, we can also use Eq. (18.11) to determine the kinetic energy of a rigid body using the IC as point 𝑂. Referring to Fig. 18.3, which shows a uniform thin bar 𝐴𝐵 of mass 𝑚 and length 𝐿 sliding down a corner, we see that the IC is easy to locate and that it is a distance 𝓁 = 𝐿∕2 from the mass center 𝐺 of the bar. If we know the bar’s orientation 𝜃 and either 𝑣𝐴 or 𝑣𝐵 , then we can find 𝜔𝐴𝐵 using either of the following expressions: 𝑣𝐴 𝑣𝐵 or 𝜔𝐴𝐵 = . (18.12) 𝜔𝐴𝐵 = 𝐿 sin 𝜃 𝐿 cos 𝜃 Knowing 𝜔𝐴𝐵 , we can then compute the kinetic energy of bar 𝐴𝐵 as 𝑇 = 12 𝐼IC 𝜔2𝐴𝐵 =

1( 𝐼 2 𝐺

) + 𝑚𝓁 2 𝜔2𝐴𝐵 ,

(18.13)

𝐴

IC 𝐿 2

𝑣𝐴 𝜃

𝓁 𝐿 2

𝐺 𝜔𝐴𝐵

𝐵

𝑣𝐵

Figure 18.3 A bar sliding down a corner, identifying the kinematic parameters needed to find its kinetic energy using its IC.

where 𝐼𝐺 is the mass moment of inertia of the bar. Example 18.1 uses the IC to compute the kinetic energy of a bicycle.

Work-energy principle for a rigid body Since a rigid body can be viewed as a rigid system of mass elements, the form of the work-energy principle for a rigid body is the same as that obtained for a system of particles, which is given in Eq. (14.44) on p. 902 as ( ) ( ) 𝑇1 + 𝑈1-2 ext + 𝑈1-2 int = 𝑇2 ,

(18.14)

( ) ( ) where 𝑇 is the system’s kinetic energy and 𝑈1-2 ext and 𝑈1-2 int are the work done in going from 1 to 2 by the forces that are external and internal to the system, respectively. Since we have already discussed the kinetic energy 𝑇 , here we focus on ) ) ( ( the term 𝑈1-2 int , leaving the discussion about 𝑈1-2 ext for later. As discussed in Section 14.3, the internal forces do work only when the system deforms. In the case of a rigid body, regardless of the internal forces, the rigidity of the body prevents deformation, and therefore the internal forces do no work, i.e., ) ( 𝑈1-2 int = 0.

(18.15)

Substituting Eq. (18.15) into Eq. (18.14), we get the work-energy principle for a rigid body 𝑇1 + 𝑈1-2 = 𝑇2 ,

(18.16) 𝑦

where 𝑈1-2 is only the work of the external forces and couples. Equation (18.16) shows that the work-energy principle for a rigid body has the same form as that for a single particle. Next, we see how to compute term 𝑈1-2 in Eq. (18.16).

𝐹⃗1 𝐹⃗3 𝑟⃗1

Work done on rigid bodies In statics we learned that a general force system consists of both forces and couples. We will briefly review how to compute the work of forces, and then we will learn how to compute the work of a couple. Figure 18.4 shows a rigid body acted upon by 𝑛 forces, labeled 𝐹⃗𝑖 (𝑖 = 1, … , 𝑛). The work of this force system is given by 𝑈1-2 =

𝑛 ∑ 𝑖=1

𝑟⃗2

1-2 )𝑖

𝐹⃗𝑖 ⋅ 𝑑⃗𝑟𝑖 ,

(18.17)

𝑟⃗3

𝐵 𝑟⃗𝑖

𝐹⃗𝑖

𝚥̂ 𝑂

∫(ℒ

(ℒ1-2 )𝑖

𝐹⃗2

𝑟⃗𝑛 𝚤̂

(ℒ1-2 )𝑛

𝑥

Figure 18.4 A rigid body under the action of a force system.

1210

𝑦

𝜔𝐴𝐵 𝑟⃗𝐴∕𝐵 𝐹⃗𝐵

𝐴

𝜃

where (ℒ1-2 )𝑖 is the path of the point of application of force 𝐹⃗𝑖 and 𝑟⃗𝑖 is the position of the point of application force 𝐹⃗𝑖 . Equation (18.17) states that the work done by a force system is the sum of work contributions due to each force, where each contribution is computed by applying what we learned in Chapter 14. Figure 18.5 shows a rigid body subject to a couple consisting of two equal and opposite forces 𝐹⃗𝐴 and 𝐹⃗𝐵 with parallel lines of action separated by the distance ℎ. The moment of this couple is

𝐹⃗𝐴

⃗ = 𝑟⃗ ⃗ 𝑀 ⃗𝐵∕𝐴 × 𝐹⃗𝐵 , 𝐴∕𝐵 × 𝐹𝐴 = 𝑟

𝐵

𝑂

(18.18)

where 𝐴 and 𝐵 are any two arbitrarily chosen points on the lines of action of 𝐹⃗𝐴 and 𝐹⃗𝐵 , respectively. Applying Eq. (18.17) to the case of 𝐹⃗𝐴 and 𝐹⃗𝐵 , we have

ℎ 𝑥

𝑈1-2 =

Figure 18.5 Rigid body subject to a couple.

ISTUDY

Chapter 18

Energy and Momentum Methods for Rigid Bodies

∫(ℒ

1-2 )𝐴

𝐹⃗𝐴 ⋅ 𝑑⃗𝑟𝐴 +

∫(ℒ

1-2 )𝐵

𝐹⃗𝐵 ⋅ 𝑑⃗𝑟𝐵 .

(18.19)

Since 𝑑⃗𝑟𝐴 = 𝑣⃗𝐴 𝑑𝑡 and 𝑑⃗𝑟𝐵 = 𝑣⃗𝐵 𝑑𝑡, Eq. (18.19) can be rewritten as 𝑈1-2 =

∫𝑡

𝑡2 (

) 𝐹⃗𝐴 ⋅ 𝑣⃗𝐴 + 𝐹⃗𝐵 ⋅ 𝑣⃗𝐵 𝑑𝑡 =

1

∫𝑡

𝑡2

𝐹⃗𝐴 ⋅ (𝑣⃗𝐴 − 𝑣⃗𝐵 ) 𝑑𝑡,

(18.20)

1

where we have used the fact that 𝐹⃗𝐵 = −𝐹⃗𝐴 . Because the body is rigid, 𝑣⃗𝐴 − 𝑣⃗𝐵 = 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐴∕𝐵 , so Eq. (18.20) becomes 𝑈1-2 =

∫𝑡

𝑡2

) ( ⃗ 𝐴𝐵 × 𝑟⃗𝐴∕𝐵 𝑑𝑡 = 𝐹⃗𝐴 ⋅ 𝜔

1

∫𝑡

𝑡2

( ) 𝜔 ⃗ 𝐴𝐵 ⋅ 𝑟⃗𝐴∕𝐵 × 𝐹⃗𝐴 𝑑𝑡,

(18.21)

1

where the identity 𝑎⃗ ⋅ (𝑏⃗ × 𝑐⃗) = 𝑏⃗ ⋅ (⃗ 𝑐 × 𝑎) ⃗ has been used. The term 𝑟⃗𝐴∕𝐵 × 𝐹⃗𝐴 is ⃗ from Eq. (18.18). Also, 𝜔 ⃗ = 𝑀 𝑘. ̂ Hence, Eq. (18.21) equal to 𝑀 ⃗ = 𝜔 𝑘̂ and 𝑀 𝐴𝐵

𝐴𝐵

becomes 𝑈1-2 =

∫𝑡

𝑡2 1

𝑀𝜔𝐴𝐵 𝑑𝑡 =

∫𝜃

𝜃2

𝑀 𝑑𝜃,

(18.22)

1

since in planar rigid body motions 𝑑𝜃 = 𝜔𝐴𝐵 𝑑𝑡, where 𝑑𝜃 is the infinitesimal angular displacement of the body. If 𝑀 is constant, Eq. (18.22) simplifies to 𝑈1-2 = 𝑀(𝜃2 − 𝜃1 ), where 𝜃2 − 𝜃1 is the angular displacement of the body between 1 and 2. The angles in Eq. (18.22) must be measured in radians. We see that, while forces do work through displacements, moments do work through rotations. As is the case of forces acting through displacements, determination of positive or negative work depends on the alignment of the moment with the angular displacement. [It may be best to keep in mind that the integrand of the first ̂ If moment and angular displaceintegral in Eq. (18.22) arises from 𝑀 𝑘̂ ⋅ 𝜔𝐴𝐵 𝑘.] ment are both counterclockwise or both clockwise, the work done by the moment is positive. If the two entities are in opposing directions, one clockwise and the other counterclockwise, the work done is negative.

Potential energy and conservation of energy If some of the forces acting on a rigid body are conservative, we can account for their work using their associated potential energy.∗ Therefore, the work-energy principle ∗ See

Section 14.2 on p. 877 to review the concepts of conservative forces and potential energy.

ISTUDY

Section 18.1

Work-Energy Principle for Rigid Bodies

for a rigid body moving between 1 and 2 under a general force system can be given the form ( ) 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 , (18.23) ( ) where 𝑉 is the total potential energy of the body and where 𝑈1-2 nc is the work of nonconservative forces, i.e., forces for which we do not have a potential energy function. Equation (18.23) has the same form as Eq. (14.27) on p. 880, which is the general form of the work-energy principle for a particle. If the entire force system consists of conservative forces, then Eq. (18.23) reduces to the familiar statement of conservation of mechanical energy, which is 𝑇 1 + 𝑉 1 = 𝑇2 + 𝑉 2 .

(18.24)

Potential energy of a torsional spring Torsional springs are designed to provide a restoring moment in response to an angular displacement (Fig. 18.6). A typical application of a torsional spring is shown in Fig. 18.7(a), where we see a bar pinned at one end with a spring. If the spring is unstretched at 𝜃 = 0 radians, and the bar is subject to an angular displacement 𝜃, then the deformation in the torsional spring causes a restoring moment to be applied to the bar in the direction opposing the angular displacement, as shown in Fig. 18.7(b). The 𝑦 𝑀 = −𝑘𝑡 𝜃 𝑘𝑡

𝜃

Courtesy of Peninsula Spring

Figure 18.6 A collection of torsional springs.

𝑥

(a)

(b)

Figure 18.7. Bar with linear torsional spring (a) and spring’s reaction moment (b).

magnitude of the moment generated is a function of the angular displacement. For example, for a linear torsional spring, the moment–angular displacement relation is 𝑀 = −𝑘𝑡 𝜃,

(18.25)

where 𝑘𝑡 is the torsional spring constant. The dimensions of 𝑘𝑡 are those of force × length∕angle. Therefore, the SI units of 𝑘𝑡 are N⋅m∕rad, and in U.S. Customary units they are f t⋅lb∕rad. Because radians are a dimensionless measure of angular displacement, the units of 𝑘𝑡 are sometimes just listed as N⋅m or ft⋅lb. Applying Eq. (18.22) to compute the work done by a torsional spring, we obtain ( ) 𝑈1-2 torsional spring = −

∫𝜃

𝜃2

( ) 𝑘𝑡 𝜃 𝑑𝜃 = − 12 𝑘𝑡 𝜃22 − 12 𝑘𝑡 𝜃12 .

(18.26)

1

Equation (18.26) tells us that the work of a torsional spring depends only on the initial and final positions of the spring, which means that we can account for its work using a potential energy function. Since the relation between work and potential energy is 𝑈1-2 = −(𝑉2 − 𝑉1 ), the potential energy of a linear torsional spring is 𝑉torsional spring = 12 𝑘𝑡 𝜃 2 .

(18.27)

1211

1212

Chapter 18

Energy and Momentum Methods for Rigid Bodies 𝑔

𝑦

Potential energy of a constant gravitational force

2

𝐺

2

𝑚𝑔

1

𝐴

𝐺

𝑦2 1

𝑚𝑔 𝐺

𝐴

𝐴 𝑚𝑔

𝐴

𝑦1 = 𝑦2

𝑦1

A particle occupies a single point, so finding its change in height to compute the work done on it by gravity is straightforward. For rigid bodies, it may not be immediately obvious what point should be used to compute the change in height. Referring to the left side of Fig. 18.8, we see that, for example, point 𝐴 moves to 𝐴′ when the rigid body moves from 1 to 2, but the net force of gravity does not act at 𝐴, it acts at 𝐺. Therefore, when computing the change in height of a rigid body for purposes of computing its gravitational potential energy, we still use Eq. (14.23) on p. 879,

𝑥

𝑂

Figure 18.8 Left: A rigid body that rotates and whose mass center 𝐺 has changed its height. Right: A rigid body that rotates and whose mass center 𝐺 has not changed its height.

𝑉𝑔 = 𝑚𝑔𝑦,

(18.28)

where 𝑦 now measures the height of the mass center with respect to the arbitrarily chosen datum line. Referring to the right side of Fig. 18.8, we see that a rigid body that simply rotates about its mass center 𝐺 has no change in gravitational potential, even though infinitely many points on the body change their height (e.g., 𝐴 to 𝐴′ ).

Work-energy principle for systems Whether we view a physical system as consisting of particles, rigid bodies, or a mixture of the two, the statement of the work-energy principle takes on a form that is always the same. Therefore, avoiding unnecessary rederivations, we simply state the work-energy principle for any physical system as )ext ( )int ( 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 ,

(18.29)

where • 𝑇 is the total kinetic energy, given by the sum of the kinetic energy of each individual part;

Common Pitfall Work of internal forces for rigid bodies. The ( )int internal work term 𝑈1-2 nc in Eq. (18.29) does not refer to work internal to a single rigid body — it refers to the work done between two or more rigid bodies in a system of rigid bodies.

• 𝑉 is the total potential energy, consisting of contributions from all conservative forces whether external or internal to the system; ( )ext • 𝑈1-2 nc is the work of all external forces without a potential energy; and )int ( • 𝑈1-2 nc is the work of all internal forces without a potential energy. Equation (18.29) has the same form as Eq. (14.45) on p. 902.

Power 𝑦

𝜔𝐴𝐵 𝑟⃗𝐴∕𝐵 𝐹⃗𝐵

𝐴

𝜃

𝐹⃗𝐴

𝑑𝑈 = 𝐹⃗ ⋅ 𝑣. ⃗ (18.30) Power developed by a force 𝐹⃗ = 𝑑𝑡 Referring to Fig. 18.9, the power developed by a couple is found by a direct application of the fundamental theorem of calculus to Eq. (18.21), i.e.,

𝐵 ℎ

𝑂 Figure 18.9 Rigid body subject to a couple.

ISTUDY

We first discussed the power developed by a force in Section 14.4 on p. 919. Whether we model an object as a particle or as a rigid body, the power developed by the force 𝐹⃗ as its point of application moves with a velocity 𝑣⃗ is the work done by the force per unit time and is

𝑥

( ) Power developed 𝑑𝑈 ⃗ ⋅𝜔 = =𝜔 ⃗ 𝐴𝐵 ⋅ 𝑟⃗𝐴∕𝐵 × 𝐹⃗𝐴 = 𝑀 ⃗ 𝐴𝐵 , by a couple 𝑑𝑡 ⃗ is the moment of the couple. where 𝑀

(18.31)

ISTUDY

Section 18.1

Work-Energy Principle for Rigid Bodies

1213

End of Section Summary The kinetic energy 𝑇 of a rigid body 𝐵 in planar motion is given by Eq. (18.8), p. 1208 𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺 𝜔2𝐵 , or by Eq. (18.11), p. 1208 𝑇 = 12 𝐼𝑂 𝜔2𝐵 , where 𝑚, 𝐺, and 𝐼𝐺 are the body’s mass, center of mass, and mass moment of inertia, respectively. The second equation applies to fixed-axis rotation, where 𝐼𝑂 is the mass moment of inertia relative to the axis of rotation. The work-energy principle for a rigid body is Eq. (18.16), p. 1209 𝑇1 + 𝑈1-2 = 𝑇2 , where 𝑈1-2 is the work done on the body in going from 1 to 2 by only the external forces. If we make use of the potential energy 𝑉 of conservative forces, the workenergy principle can also be written as Eqs. (18.23) and (18.24), p. 1211 ) ( 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 (general systems), 𝑇1 + 𝑉 1 = 𝑇2 + 𝑉 2

(conservative systems),

where the second expression applies only to a conservative system and states that the total mechanical energy of the system is conserved. For systems of rigid bodies or mixed systems of rigid bodies and particles, the work-energy principle is Eq. (18.29), p. 1212 ( )ext ( )int 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 . Here, 𝑉 is the total potential energy (of both external and internal conservative forces). ( ( )ext )int 𝑈1-2 nc and 𝑈1-2 nc are the work contributions due to external and internal forces for which we do not have a potential energy, respectively. Referring to Fig. 18.10, in planar rigid body motion, the work of a couple is

𝑦 𝑟⃗𝐴∕𝐵 𝐹⃗𝐵

Eq. (18.22), p. 1210 𝑈1-2 =

∫𝑡

𝑡2 1

𝑀𝜔𝐴𝐵 𝑑𝑡 =

∫𝜃

𝜃2

𝜔𝐴𝐵 𝐴

𝜃

𝐹⃗𝐴

𝐵 ℎ

𝑀 𝑑𝜃.

1

Here, 𝑑𝜃 is the body’s infinitesimal angular displacement, and 𝑀 is the component of the moment of the couple in the direction perpendicular to the plane of motion, taken to be positive in the direction of positive 𝜃, and 𝜃 must be measured in radians. The work done by the moment is positive if the moment and angular displacement are

𝑂 Figure 18.10 Rigid body subject to a couple.

𝑥

1214

ISTUDY

Chapter 18

Energy and Momentum Methods for Rigid Bodies

in the same direction (both clockwise or both counterclockwise) and negative if the moment and angular displacement are in opposing directions (one clockwise and the other counterclockwise). In addition, the power developed by a couple with moment ⃗ is computed as 𝑀 Eq. (18.31), p. 1212 Power developed by a couple = where 𝜔 ⃗ 𝐴𝐵 is the body’s angular velocity.

𝑑𝑈 ⃗ ⋅𝜔 =𝑀 ⃗ 𝐴𝐵 , 𝑑𝑡

ISTUDY

Section 18.1

Work-Energy Principle for Rigid Bodies

E X A M P L E 18.1

Kinetic Energy Computation with Rolling Without Slip

A cyclist rides on a horizontal road at a constant speed 𝑣0 = 35 km∕h. The mass of the frame (the bicycle without the wheels) is 𝑚𝑓 = 2.25 kg. The front and the rear wheels have the same diameter 𝑑 = 0.7 m. The masses of the front and rear wheels are 𝑚fw = 0.737 kg and 𝑚rw = 0.933 kg, respectively, and the mass moments of inertia of the front and rear wheels are 𝐼𝐶 = 0.0510 kg⋅m2 and 𝐼𝐷 = 0.0501 kg⋅m2 , respectively (see Fig. 2). Assuming the wheels roll without slip, compute the kinetic energy of just the bicycle. Neglect the kinetic energy of the pedals, chain, and other components not specifically mentioned.

SOLUTION Road Map

We are given a system’s motion, and we are asked to compute the system’s kinetic energy, where the system consists of a bicycle frame and two wheels. The kinetic energy of the bicycle is found by computing the kinetic energy of each part and then adding the individual contributions. The solution does not require any knowledge of the forces acting on the system. Therefore, as in kinematics problems, our solution will consist of only the computation step. Computation

Letting 𝑇 denote the kinetic energy of the system, we have 𝑇 = 𝑇𝑓 + 𝑇fw + 𝑇rw ,

(1)

where 𝑇𝑓 , 𝑇fw , and 𝑇rw are the kinetic energies of the frame, front wheel, and rear wheel, respectively. Because the system is traveling along a horizontal road, the frame is translating in the 𝑥 direction with speed 𝑣0 (see Fig. 2). Therefore, since the angular velocity 𝑣0

𝚥̂ 𝚤̂

𝐺 𝜔fw

𝐷

𝐶

𝑅

𝑅

𝐴

𝜔rw

𝐵

Figure 2. Bicycle moving at constant speed 𝑣0 . The points 𝐺, 𝐶, and 𝐷 are the centers of mass of the frame, front wheel, and rear wheel, respectively.

of the frame is equal to zero, applying Eq. (18.8) on p. 1208, we have 𝑇𝑓 = 21 𝑚𝑓 𝑣20 = 106.3 J.

(2)

Applying Eq. (18.8) to the front wheel, we have 𝑇fw = 12 𝑚fw 𝑣2𝐶 + 12 𝐼𝐶 𝜔2fw ,

(3)

where 𝑣𝐶 and 𝜔fw are the speed of the center of mass and the angular speed of the front wheel, respectively. Because the wheel rolls without slip, we have 𝑣𝐶 = 𝑣0 = 𝑅𝜔fw

⇒

𝜔fw = 𝜔0 =

𝑣0 𝑅

= 27.78 rad∕s,

(4)

where 𝜔0 is the angular speed of both wheels since they have the same radius 𝑅 = 𝑑∕2 = 0.35 m. Substituting Eq. (4) into Eq. (3) gives ( ) 𝑇fw = 21 𝑚fw 𝑅2 𝜔20 + 21 𝐼𝐶 𝜔20 = 21 𝐼𝐶 + 𝑚fw 𝑅2 𝜔20 = 54.51 J. (5)

Polka Dot/Getty Images

Figure 1 A cyclist riding on a horizontal road.

1215

1216

ISTUDY

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Because the rear wheel is undergoing the same motion as the front wheel, the expression for 𝑇rw has the same form as that in Eq. (5). Therefore, we have ( ) 𝑇rw = 21 𝐼𝐷 + 𝑚rw 𝑅2 𝜔20 = 63.42 J. (6) Substituting the results in Eqs. (2), (5), and (6) into Eq. (1), the total kinetic energy of the bicycle is 𝑇 = 224.3 J.

(7)

Discussion & Verification

Our answer was obtained by combining contributions from Eqs. (2), (5), and (6), each of which is dimensionally correct. The units used in the final answer are appropriate since the problem’s data were given in SI units. One way to check whether or not our result is reasonable is to verify that the computed kinetic energy is larger than what we would compute if the system were just translating. If the wheels did not rotate, their rotational kinetic energy would not contribute to the system’s total kinetic energy, and therefore the corresponding system’s kinetic energy would have to be smaller than that in Eq. (7). (Computing the kinetic energy as if the ) system were just translating gives 𝑇translation = 21 𝑚𝑓 + 𝑚fw + 𝑚rw 𝑣20 = 185.3 J, which is less than our computed value, as expected. A Closer Look An important observation about our calculation is that we solved the problem by applying Eq. (18.8) on p. 1208, i.e., the general formula for the kinetic energy of a rigid body in planar motion. In so doing, for the wheels, we ended up with the following two expressions: ( ) ( ) 𝑇fw = 12 𝐼𝐶 + 𝑚fw 𝑅2 𝜔20 and 𝑇rw = 21 𝐼𝐷 + 𝑚rw 𝑅2 𝜔20 . (8)

The first term in parentheses is equivalent to applying the parallel axis theorem to find the mass moment of inertia of the front wheel about 𝐴, that is, 𝐼𝐴 = 𝐼𝐶 + 𝑚fw 𝑅2 ,

(9)

where, referring to Fig. 2, point 𝐴 is the contact point between the wheel and the ground. What is special about point 𝐴 is that it is the instantaneous center of rotation of the front wheel. Similarly, for the rear wheel we have 𝐼𝐷 + 𝑚rw 𝑅2 = 𝐼𝐵 , which is the mass moment of inertia of the rear wheel about 𝐵, where point 𝐵 is the IC of the rear wheel. These observations point to the fact that we could have computed the kinetic energy of the wheels using Eq. (18.11) on p. 1208 as 𝑇fw = 21 𝐼𝐴 𝜔20

and 𝑇rw = 12 𝐼𝐵 𝜔20 .

(10)

ISTUDY

Section 18.1

1217

Work-Energy Principle for Rigid Bodies

E X A M P L E 18.2

The Work-Energy Principle for Rotation About a Fixed Axis

The manually operated road barrier shown in Fig. 1 is easily opened and closed by hand due to the counterweight. Referring to Fig. 2, model the arm of the barrier as a uniform thin counterweight

arm 𝑙 𝐵

𝑙 2

𝑚𝑎

𝑂

𝑑 𝑚𝑐

Figure 1. A manually operated road barrier. Note the counterweight on the right end.

𝐴

bar of mass 𝑚𝑎 that is pinned at 𝑂 and with mass center at 𝐴, and model the counterweight as a particle of mass 𝑚𝑐 . Determine the angular velocity of the arm and the speed of end 𝐵 as the arm reaches the horizontal position if it is nudged from rest when it is vertical and falls freely. Evaluate your answers for 𝑙 = 15.7 f t, 𝑑 = 2.58 f t, 𝛿 = 1.4 f t, a 45 lb arm, and a 160 lb counterweight.

Figure 2 The relevant dimensions of the road barrier. The arm has mass 𝑚𝑎 and the counterweight has mass 𝑚𝑐 .

SOLUTION Road Map & Modeling Since we want to relate the change in speed of the arm to its displacement, we will apply the work-energy principle. As can be seen in the FBD of the arm in Fig. 3, only weight forces do work, so this is a conservative system. We will assume that there are no losses due to friction or drag. 1

𝐵

𝑙 2

𝐴 2

𝑚𝑎 𝑔 datum

𝑂𝑦 𝑂𝑥

𝜃

𝑑

𝛿 𝑚𝑐 𝑔 Figure 3. The FBD of the road barrier as it is falling from the vertical position.

Governing Equations Balance Principles

Since the system is conservative, the work-energy principle gives 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 ,

(1)

where 1 is when the arm is vertical (𝜃 = 𝜋∕2 rad) and 2 is when it is horizontal (𝜃 = 0). At 1, the kinetic energy is zero, and at 2, the kinetic energy must account for the arm

𝛿

1218

ISTUDY

Chapter 18

Energy and Momentum Methods for Rigid Bodies

and the counterweight, and so + 12 𝐼𝑂 𝜔2𝑎2 , ⏟⏟⏟ ⏟⏟⏟ 1 𝑚 𝑣2 2 𝑐 𝑐2

𝑇1 = 0 and 𝑇2 =

counterweight

(2)

arm

where 𝑣𝑐2 is the speed of the counterweight at 2, 𝐼𝑂 is the mass moment of inertia of the arm with respect to point 𝑂, 𝜔𝑎2 is the angular speed of the arm at 2, and we have used Eq. (18.11) to compute the kinetic energy of the arm. Using the parallel axis theorem, the mass moment of inertia of the arm with respect to point 𝑂 is given by 𝐼𝑂 = Force Laws

1 𝑚 𝑙2 12 𝑎

+ 𝑚𝑎

(1 2

)2 𝑙−𝑑 .

The potential energies of the system at 1 and 2 are ) ( 𝑉1 = 𝑚𝑎 𝑔 12 𝑙 − 𝑑 − 𝑚𝑐 𝑔𝛿 and 𝑉2 = 0.

(3)

(4)

Kinematic Equations Since we want to solve for 𝜔𝑎2 , we will need to write 𝑣𝑐2 in terms of 𝜔𝑎2 , which is readily done as 𝑣𝑐2 = 𝛿𝜔𝑎2 . (5) Computation

Substituting Eqs. (2)–(5) into Eq. (1), we obtain [ ( ) )2 ] 1 𝑚𝑎 𝑔 2 𝑙 − 𝑑 − 𝑚𝑐 𝑔𝛿 = 12 𝑚𝑐 (𝛿𝜔𝑎2 )2 + 21 12 𝑚𝑎 𝑙2 + 𝑚𝑎 21 𝑙 − 𝑑 𝜔2𝑎2 , (1

(6)

which, solving for 𝜔𝑎2 , gives

𝜔𝑎2

√ √ √ =√

[ ] 2𝑔 𝑚𝑎 (𝑙∕2 − 𝑑) − 𝑚𝑐 𝛿 ] = 0.5835 rad∕s, [ 𝑚𝑐 𝛿 2 + 𝑚𝑎 𝑙2 ∕12 + (𝑙∕2 − 𝑑)2

(7)

and so the speed of the end of the arm at 𝐵 is 𝑣𝐵2 = (𝑙 − 𝑑)𝜔𝑎2 = 7.655 f t∕s.

(8)

The dimensions in Eqs. (7) and (8) are correct. If 𝑚𝑐 𝛿 is increased, the argument of the square root in Eq. (7) becomes negative, and the solution is no longer meaningful. This makes sense because we expect that if 𝑚𝑐 𝛿 exceeds a critical value, the barrier will never reach the horizontal position. Different design configurations for the arm can be explored in the exercises and design problems.

Discussion & Verification

A Closer Look Problems 18.55 and 18.56 take a closer look at the problem examined in this example by taking into account the rotational inertia (and thus rotational kinetic energy) of the counterweight. Even without solving those problems, we can predict what the inclusion of rotational inertia will do to our results in Eqs. (7) and (8). It won’t change the potential energy of the counterweight, but it will add a kinetic energy term to the right side of Eq. (6). Once this kinetic energy term is written in terms of 𝜔𝑎2 , we see that the denominator in Eq. (7) will get larger and so, all other things being equal, 𝜔𝑎2 will decrease.

ISTUDY

Section 18.1

1219

Work-Energy Principle for Rigid Bodies

E X A M P L E 18.3

Work-Energy Principle and Rolling Without Slip

A rear-wheel-drive car accelerates from rest to a final speed 𝑣𝑓 over a distance 𝐿 along a horizontal stretch. The front wheels do not slip, and we neglect rolling resistance, bearing friction, and air drag. If the geometric center of each wheel is also the wheel’s mass center 𝐺, determine those forces on each front wheel that do work, and express that work in terms of the given speed 𝑣𝑓 , the wheel’s radius 𝑟, mass 𝑚𝑤 , and mass moment of inertia 𝐼𝐺 . Can this expression for work be used to find the average friction force acting on the wheel? Gary L. Gray

Figure 1 A roadster on a horizontal stretch of road.

SOLUTION Road Map & Modeling

The work-energy principle tells us that the work done on a wheel is the difference between the wheel’s final and initial kinetic energy. Since we have enough information to determine these kinetic energies, the problem is solved by computing their difference. As for calculating the average friction force at the ground, we will need to see if and how the friction force appears in the expression for the work done on the wheel. Since the work-energy principle accounts for the work of all forces acting on the wheel, it is important to sketch the wheel’s FBD, as shown in Fig. 2. The only friction we consider is that at the ground, and no couple is assumed to be acting on the wheel because the front wheels are not the driving wheels. The force 𝐻 comes from the fact that the front axle of the car is pushing the wheel forward, and the force 𝑅 is due to the weight of the car pushing down on the tire.

𝑦 𝑅 𝚥̂

𝐻

𝐺

𝑟

𝑄 𝐹

Governing Equations Balance Principles

𝚤̂

𝑚𝑤 𝑔 𝑥

𝑁

The work-energy principle for each front wheel is 𝑇1 + 𝑈1-2 = 𝑇2 ,

(1)

where the roadster’s speed is zero at 1 and 𝑣0 at 2. The kinetic energy 𝑇 can then be written as 1 1 𝑇1 = 0 and 𝑇2 = 𝑚𝑤 𝑣2𝐺2 + 𝐼𝐺 𝜔2𝑤2 , (2) 2 2 where 𝑣𝐺2 is the speed of 𝐺 and 𝜔𝑤2 is the angular speed of the wheel, both at 2. Force Laws To express the work of each force appearing on the FBD, recall that if the point of application of a force 𝑃⃗ displaces by 𝑑⃗𝑟, the corresponding work dU is

𝑑𝑈 = 𝑃⃗ ⋅ 𝑑⃗𝑟 with

𝑑⃗𝑟 = 𝑣⃗ 𝑑𝑡,

(3)

where 𝑣⃗ is the velocity of the point of application of 𝑃⃗ . Referring to Fig. 2, 1. Forces 𝑅 and 𝑚𝑤 𝑔 are oriented vertically and do no work because their points of application move horizontally. ⃗ 2. 𝐹 and 𝑁 do no work because the rolling-without-slip condition demands that 𝑣⃗𝑄 = 0, where 𝑄 is the point of application of 𝐹 and 𝑁. Therefore, the only force doing work is the force 𝐻, and we can write 𝑈1-2 =

∫(ℒ

1-2 )𝐺

𝐻 𝚤̂ ⋅ 𝑑⃗𝑟𝐺 =

∫𝑥

𝑥𝐺2

𝐻 𝑑𝑥𝐺 , 𝐺1

where 𝐺 is the point of application of 𝐻 and 𝑥𝐺 is the horizontal position of 𝐺.

(4)

Figure 2 FBD of one of the front wheels. The point 𝐺 identifies both the geometric center of the wheel and the wheel’s center of mass.

1220

ISTUDY

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Kinematic Equations

At 2 we have

Interesting Fact

𝑣𝐺2 = 𝑣𝑓 .

If friction does no work, then what is “rolling resistance”? Rolling resistance occurs when one object rolls without slip over another object and one or both of the objects deform. Not only does the deformation itself dissipate energy, but also the equivalent normal force between the object and the ground impedes the motion. cylinder

𝜔cyl

To find 𝜔𝑤2 , we apply the kinematics of rolling without slip using 𝑣⃗𝐺 = 𝑣⃗𝑄 + 𝜔𝑤 𝑘̂ × 𝑟 𝚥̂ with

𝑣⃗𝐺2 = 𝑣𝑓 𝚤̂ = 𝜔𝑤2 𝑘̂ × 𝑟 𝚥̂ Computation

(6)

𝑣𝑓

.

(7)

( 𝑣 )2 𝑓 1 1 . 𝑚𝑤 𝑣2𝑓 + 𝐼𝐺 − 2 2 𝑟

(8)

⇒

𝜔𝑤2 = −

𝑟

Combining Eqs. (5) and (7) with Eq. (2), we have 𝑇1 = 0 and

𝐹

⃗ 𝑣⃗𝑄 = 0,

so that at 2 we have

𝑚𝑔

𝐺

(5)

𝑇2 =

Substituting Eqs. (8) into Eq. (1) and solving for 𝑈1-2 , we have 𝑈1-2 =

∫𝑥

𝑥𝐺2 𝐺1

( 𝑣 )2 𝑓 1 1 2 𝐻 𝑑𝑥𝐺 = 𝑚𝑤 𝑣𝑓 + 𝐼𝐺 − , 2 2 𝑟

(9)

where we have also made use of Eq. (4). Equation (9) can be simplified to 𝑃stat deformable surface

𝑃dyn 𝑁stat

𝑁dyn

The figure above shows a rigid cylinder on a deformable surface. When 𝐹 = 0 and the object is at rest, the pressure distribution on the object due to the deformation of the surface is given by 𝑃stat , and the equivalent normal force is given by the vertical force 𝑁stat . On the other hand, when a force 𝐹 is applied to the cylinder so that it rolls with a constant angular speed 𝜔cyl , then the pressure distribution is that given by 𝑃dyn , and the equivalent normal force is 𝑁dyn . Notice that there is a horizontal component of 𝑁dyn that is impeding the forward motion of the cylinder. Since the cylinder is moving with a constant velocity, the horizontal component of 𝑁dyn must be equal and opposite to 𝐹 .

𝑈1-2 =

∫𝑥

𝑥𝐺2

𝐻 𝑑𝑥𝐺 = 𝐺1

( ) ) 𝑣𝑓 2 1( . 𝐼𝐺 + 𝑚𝑤 𝑟2 2 𝑟

(10)

Now that we have found the expression for 𝑈1-2 , we see that the friction force does not appear in it. Therefore, we cannot calculate the average value of the friction force acting on the wheel directly from Eq. (10). The term 𝑣𝑓 ∕𝑟 in Eq. (10) has dimensions of 1∕time. Since 2 𝐼𝐺 is a mass moment of inertia, it has dimensions of mass × length . Therefore, overall, ( ) 2 the result in Eq. (10) has dimensions of mass × length∕time (length), i.e., dimensions of work, as it should. In general, rolling without slip requires a friction force to act on the rolling body at the point of contact between the body in question and the rolling surface. However, this friction force does not dissipate any energy because the absence of slip prevents the friction force from doing work (i.e., the friction is being applied at a point that is not moving at that instant). Therefore, if a rigid body is rolling without slip on a rigid surface, the mechanical energy of the body will be conserved during the motion. This is an important result because in many problems involving rolling without slip, we can apply conservation of energy even if a friction force does appear on the body’s FBD. Discussion & Verification

ISTUDY

Section 18.1

1221

Work-Energy Principle for Rigid Bodies

E X A M P L E 18.4

Computing Engine Power Output and Wheel Torque

The 2600 lb∗ rear-wheel-drive roadster shown in Fig. 1 accelerates from rest to 60 mph in 7 s. Each of the four wheels has a diameter 𝑑 = 24.3 in., weighs 47 lb, and has a mass moment of inertia 𝐼𝐺 = 0.989 slug⋅f t 2 with respect to its mass center. Assuming that the car accelerates uniformly, that we neglect rolling resistance, bearing friction, and air drag, and that the wheels do not slip relative to the ground, estimate the average engine power output, as well as the torque provided to the rear wheels. Gary L. Gray

SOLUTION Road Map & Modeling We can use the work-energy principle to compute the total work done on the car as the difference between the final and initial kinetic energies. Referring to the FBD in Fig. 2, we see that none of the external forces does work since the wheels roll without slip and the car’s trajectory is horizontal. All the work done on the car is due to internal forces, namely, the torque provided by the engine. The corresponding average power output is then computed by dividing this work by the given time interval. To measure the average torque provided by the engine, we need to consider the forces internal to the car, depicted in Fig. 3. The work done by the forces 𝑅𝑥 and 𝑅𝑦 applied at 𝐺 on a rear 𝚥̂

(𝑚 − 2𝑚𝑤 )𝑔

𝚤̂

𝐴

𝑅𝑥

𝐷

𝑀

𝑅𝑥

𝐸 𝐵

𝑅𝑦

𝑁𝐴

𝐹𝐵

𝑁𝐵

Gary L. Gray

Figure 3. FBDs of the rear wheels and of the rest of car. Point 𝐸 is the mass center of the car without the rear wheels. Points 𝐺 and 𝐷 are coincident: 𝐺 is part of the wheel, whereas 𝐷 is part of the axle on which the wheel is mounted.

wheel is equal and opposite to the work done by 𝑅𝑥 and 𝑅𝑦 applied at 𝐷 because 𝐺 and 𝐷 do not move relative to one another. By contrast, the torque 𝑀 does work because the rear wheels rotate relative to the rest of the car. Therefore, we can estimate 𝑀 by dividing the work of the torque by the angle swept by the wheels during the car’s motion. Governing Equations Balance Principles

The general form of the work-energy principle for a system is ( )ext ( )int 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 , (1)

where the car’s speed is equal to zero at 1 and is equal to 60 mph = 88.00 f t∕s at 2. The system’s kinetic energy at 1 and 2 can be written as ( ) (2) 𝑇1 = 0 and 𝑇2 = 12 𝑚𝑣22 + 4 21 𝐼𝐺 𝜔2𝑤2 = 12 𝑚𝑣22 + 2𝐼𝐺 𝜔2𝑤2 , ⏟⏟⏟ ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ tr KE

rot KE

where 𝑚 is the mass of the roadster (including its wheels), tr KE is the translational kinetic energy of the car, rot KE is the rotational kinetic energy of all four wheels, 𝑣2 is the translational speed of the car and the wheels, and 𝜔𝑤2 is the angular speed of the wheels. ∗ This

𝑄 𝐴 𝑁𝐴

𝐹𝐵 𝐵

𝚥̂ 𝚤̂

𝑁𝐵

Gary L. Gray

𝐺 2𝑚𝑤 𝑔

𝐹𝐴

𝑚𝑔

𝐹𝐴

𝑅𝑦 𝑀

Figure 1 A roadster on a horizontal stretch of road.

weight includes the wheels, the fuel, and the passenger.

Figure 2 FBD of the car as a whole. 𝑄 is the center of mass of the whole car, and 𝐴 and 𝐵 are the contact points between the ground and the rear and front wheels, respectively.

1222

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Force Laws

The external forces do no work, so we can write ( )ext 𝑈1-2 nc = 0, 𝑉1 = 0, and 𝑉2 = 0.

Helpful Information Work of internal forces. The work done by 𝑅𝑥 and 𝑅𝑦 on the rear wheels is equal and opposite to the work done by these forces on the rest of the car because the relative displacement of points 𝐺 and 𝐷 (see Fig. 3) is equal to zero. Therefore, the overall work of the internal forces 𝑅𝑥 and 𝑅𝑦 is equal to zero. If the displacement of 𝐺 relative to 𝐷 were not equal to zero, the car would be broken into two parts. By contrast, the rear wheels rotate relative to the car, so the relative angular displacement between these wheels and the rest of the car is not equal to zero. This allows the torque 𝑀 to do (internal) work.

( )int As for 𝑈1-2 nc , referring to Fig. 3, because the points 𝐺 and 𝐷 do not move relative to one another, 𝑅𝑥 and 𝑅𝑦 do no work. Consequently, all the internal work is done by the torque 𝑀, which can be written as )int ( 𝑈1-2 nc =

∫𝜃

𝜃2

( ) 𝑀 𝑑𝜃 = 𝑀 𝜃2 − 𝜃1 ,

where 𝜃 measures the rotation of the rear wheels (see Fig. 4), and we have assumed that 𝑀 is constant between 1 and 2. The work done is positive as the torque and angular displacement are both clockwise. Kinematic Equations At 2, the translational speed 𝑣2 is given, and the corresponding 𝜔𝑤2 can be found by enforcing the rolling-without-slip condition, which gives

ISTUDY

𝑣2 𝑑∕2

= 86.91 rad∕s.

(5)

Note that to compute 𝑀, we will need the value of 𝜃2 − 𝜃1 , which can be calculated if the distance 𝐿 traveled by the car between 1 and 2 is known. The distance 𝐿 can be computed since we have assumed that the car is uniformly accelerating. Letting Δ𝑡 = 7 s, the constant acceleration of the car is 𝑎 = 𝑣2 ∕Δ𝑡 = 12.57 f t∕s2 . Therefore, using constant acceleration equations, we have 𝐿=

Figure 4 Definition of the angle 𝜃 measuring the rotation of the rear wheels. The red line is an arbitrarily chosen reference line that rotates with the wheels.

(4)

1

𝑣2 = 88.00 f t∕s and 𝜔𝑤2 =

𝜃

(3)

Computation

1 𝑎Δ𝑡2 = 308.0 f t 2

𝜃2 − 𝜃1 =

⇒

𝐿 = 304.2 rad. 𝑑∕2

(6)

Combining Eqs. (5) with Eqs. (2) and substituting in known values, we

have 𝑇1 = 0 and 𝑇2 = 327,600 f t ⋅lb.

(7)

Substituting Eqs. (7) into Eq. (1) and using Eqs. (3), for the average power 𝑃avg we have (

𝑈1-2

)int nc

= 327,600 f t ⋅lb ⇒ 𝑃avg =

)int ( 𝑈1-2 nc Δ𝑡

= 46,800

f t ⋅lb = 85.09 hp. s

(8)

Using Eq. (4), along with the first of Eqs. (8) and the result in Eq. (6), we have ( ) ( )int 𝑀 𝜃2 − 𝜃1 = 𝑈1-2 nc Discussion & Verification

⇒

𝑀=

)int ( 𝑈1-2 nc 𝜃2 − 𝜃1

= 1077 f t ⋅lb.

(9)

The answers are all dimensionally correct. As far as the acceptability of each value is concerned, the calculated power may seem low with respect to the power of typical sports car engines (e.g., 170 hp at an engine speed of 6000 rpm). However, the power indicated in ads and product literature is a peak value corresponding to a specific angular velocity of the crankshaft. In our case, we have calculated an average power value, and, as such, it is not surprising that it is smaller than the peak values seen in advertisements. As far as the torque is concerned, the value we have obtained is not that unusual if we keep in mind that it measures the torque provided to the driving wheels. That is, the computed torque value should not be confused with the typically much smaller value of torque reported in product literature, which is the torque output at the crankshaft, i.e., before the transmission gets it to the wheels.

ISTUDY

Section 18.1

1223

Work-Energy Principle for Rigid Bodies

E X A M P L E 18.5

An Overhead Fold-Up Door: Conservation of Energy

An overhead door, with height ℎ = 30 f t and weight 𝑊 = 800 lb, consists of two identical sections hinged at 𝐶. The roller at 𝐴 moves along a horizontal guide, whereas the rollers at 𝐵 and 𝐷, which are the midpoints of sections 𝐴𝐶 and 𝐶𝐸, move along a vertical guide. The door’s operation is assisted by two identical springs attached to the horizontally moving rollers (only one of the two springs is shown). The springs are stretched 0.25 f t when the door is fully open. Determine the minimum value of the spring constant 𝑘 if the roller at 𝐴 is to strike the left end of the horizontal guide with a maximum speed of 1.5 f t∕s after the door is released from rest in the fully open position.

SOLUTION

horizontal guide 𝐴

fully open

𝑘

1

𝐵

𝐶

vertical guide

ℎ

Since the desired value of 𝑘 is found by relating changes in speed to changes in position, we will use the work-energy principle as a solution method. We will treat the two identical sections 𝐴𝐶 and 𝐶𝐸 as uniform thin rigid bodies, neglect the inertia of the rollers, neglect the mass of the springs, and neglect all friction. Letting 1 and 2 be the fully open and fully closed positions, respectively, the system’s FBD for a generic position between 1 and 2 is that shown in Fig. 2. The force in each of the two springs is 𝐹𝑠 , and 𝑁𝐴 , 𝑁𝐵 , and 𝑁𝐷 are the reactions at each of the rollers 𝐴, 𝐵, and 𝐷, respectively. Note that the reactions at the rollers do no work because each roller moves in a direction perpendicular to the reaction force acting on it. The remaining forces 𝑊 and 𝐹𝑠 are conservative, so we have conservation of mechanical energy. Our solution strategy will be to find the value of 𝑘 for which the speed of 𝐴 is 𝑣max = 1.5 f t∕s and then show that larger values of 𝑘 result in values of the speed of 𝐴 at 2 that are less than 𝑣max . Road Map & Modeling

𝐷 𝐸 fully closed 2

floor Figure 1 Side view of an overhead door at various positions during its operation. 2𝐹𝑠

𝐴

Governing Equations Balance Principles

2𝑁𝐵

Since mechanical energy is conserved, we can write 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2 ,

(1) 𝐶

where, since the system is at rest at 1, 𝑇1 and 𝑇2 can be expressed as 𝑇1 = 0 and 𝑇2 = 21 𝑚𝐴𝐶 𝑣2𝐵2 + 12 𝐼𝐵 𝜔2𝐴𝐶2 + 21 𝑚𝐶𝐸 𝑣2𝐷2 + 12 𝐼𝐷 𝜔2𝐶𝐸2 ,

Force Laws Choosing the datum for gravitational potential energy as shown in Fig. 3 and recalling that there are two springs, we have ( ( ) ) 𝑊 𝑊 𝑦𝐵2 + 𝑦𝐷2 + 2 21 𝑘𝛿22 , 𝑉1 = 2 12 𝑘𝛿12 and 𝑉2 = (4) 2 2

where 𝛿1 = 0.25 f t, and where 𝑦𝐷2 = −3ℎ∕4,

and 𝛿2 = 𝛿1 + ℎ∕4.

(5)

Kinematic Equations When at 2, 𝐵 and 𝐷 are at the lower limit of their respective motion ranges. In addition, given that 𝐵 and 𝐷 cannot move in the 𝑥 direction, it must therefore be true that 𝑣⃗𝐵2 = 0⃗ = 𝑣⃗𝐷2 , and so

𝑣𝐵2 = 0

and 𝑣𝐷2 = 0.

𝑊 ∕2

𝚥̂ 𝚤̂

(2)

where 𝑚𝐴𝐶 and 𝐼𝐵 are the mass and mass moment of inertia of 𝐴𝐶, respectively, and 𝑚𝐶𝐸 and 𝐼𝐷 are the mass and mass moment of inertia of 𝐶𝐸, respectively. Since 𝐴𝐶 and 𝐶𝐸 are identical, we have ( )( )2 𝑊 ∕2 𝑊 ℎ2 1 𝑊 ℎ and 𝐼𝐵 = 𝐼𝐷 = . (3) 𝑚𝐴𝐶 = 𝑚𝐶𝐸 = = 𝑔 12 2𝑔 2 96𝑔

𝑦𝐵2 = −ℎ∕4,

2𝑁𝐴 𝐵

(6)

2𝑁𝐷 𝑊 ∕2

𝐷

𝐸

Figure 2 Side view of the system’s FBD for a generic position between 1 and 2. The factor of 2 in front of 𝐹𝑠 , 𝑁𝐴 , 𝑁𝐵 , and 𝑁𝐷 is necessary because there are two springs and two sets of rollers on the door (in the view shown only one set is visible). Note that the centers of mass of sections 𝐴𝐶 and 𝐶𝐸 coincide with points 𝐵 and 𝐷, respectively.

1224

Chapter 18

Energy and Momentum Methods for Rigid Bodies

𝑦 𝑂

𝐴 datum

𝑥

ℎ∕4 𝐵 𝐶

⃗ at 2 we can write In addition, since 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔𝐴𝐶 𝑘̂ × 𝑟⃗𝐵∕𝐴 and since 𝑣⃗𝐵2 = 0,

𝜃 ℎ∕4

𝜃 𝐷

ℎ∕4 𝐸

2

Referring to Fig. 3, observe that 𝐴𝐶 and 𝐶𝐸 rotate in opposite directions while remaining mirror images of one another relative to the line bisecting the angle 𝐴𝐶𝐸. Therefore, we must have 𝜔𝐴𝐶 = −𝜔𝐶𝐸 ⇒ 𝜔𝐴𝐶2 = −𝜔𝐶𝐸2 . (7) ( ) [( ) ] 𝑣⃗𝐵2 = 0⃗ = 𝑣𝐴𝑥 2 𝚤̂ + 𝜔𝐴𝐶2 𝑘̂ × (−ℎ∕4) 𝚥̂ = 𝑣𝐴𝑥 2 + ℎ𝜔𝐴𝐶2 ∕4 𝚤̂,

where we have used the fact that 𝐴 can only move in the horizontal direction. Solving Eq. (8) for 𝜔𝐴𝐶2 and using Eq. (7), we obtain ( ) ( ) 𝜔𝐴𝐶2 = −4 𝑣𝐴𝑥 2 ∕ℎ and 𝜔𝐶𝐸2 = 4 𝑣𝐴𝑥 2 ∕ℎ. (9) Computation

Using Eq. (2), along with Eqs. (3), (6), and (9), we have

ℎ∕4

𝑇1 = 0 Figure 3 Coordinate system used with the indication of the datum choice. Note that, for convenience, only 2 is shown. When the system is at 1, sections 𝐴𝐶 and 𝐶𝐸 lie on the 𝑥 axis since their thickness has been neglected.

ISTUDY

(8)

( )2 ( )2 𝑊 𝑣𝐴𝑥 2 𝑊 ℎ2 16 𝑣𝐴𝑥 2 = . and 𝑇2 = 96𝑔 6𝑔 ℎ2

Substituting Eqs. (5) into Eqs. (4) and simplifying, we have ) ( 𝛿1 ℎ ℎ 2 𝑊ℎ 2 2 − + . 𝑉1 = 𝑘𝛿1 and 𝑉2 = 𝑘 𝛿1 + 2 16 2

(10)

(11)

Substituting Eqs. (10) and (11) into Eq. (1) and simplifying, we obtain 0=

( )2 𝑊 𝑣𝐴𝑥 2 6𝑔

( +𝑘

𝛿1 ℎ

ℎ2 + 2 16

) −

𝑊ℎ , 2

(12)

which can be solved for 𝑘 to obtain ( )2 𝑊 3ℎ𝑔 − 𝑣𝐴𝑥 2 𝑘=8 = 199.8 lb∕f t, 𝑔 24𝛿1 ℎ + 3ℎ2

(13)

where we have let (𝑣𝐴𝑥 )2 = 𝑣max = 1.5 f t∕s. The result in Eq. (13) is consistent with the fact that 𝑘 has dimensions of force over length, and the units used to express the numerical result are therefore appropriate. In addition, the result matches our expectation that the slower (𝑣𝐴𝑥 )2 is, the stiffer the spring must be. Therefore, the value of 𝑘 = 199.8 lb∕f t is the value of 𝑘 we were looking for.

Discussion & Verification

A Closer Look If we were to use the value of 𝑘 in Eq. (13), each of the springs would be subject to 𝐹𝑠2 = 𝑘𝛿2 = 1549 lb when at 2. If this value of force is judged to be too high, we can design a door with a smaller value of the spring constant (and therefore of force) using a system of counterweights that are lowered when the door opens and that are lifted when the door closes. A properly designed counterweight system can make the use of springs unnecessary.

ISTUDY

Section 18.1

1225

Work-Energy Principle for Rigid Bodies

E X A M P L E 18.6

Application of the Work-Energy Principle and 𝐹⃗ = 𝑚𝑎⃗

Let’s revisit Example 17.10 (on p. 1182), in which a solid uniform sphere was released from the top of a semicylinder and we determined where it separated from the semicylinder as it rolled without slipping. As was done in Example 17.10, we assume that the sphere is uniform; has mass 𝑚 and radius 𝜌; and is released from rest by giving it a slight nudge to the right so that it begins to roll along the surface (Fig. 1). Assuming that there is sufficient friction between the surface and the sphere, such that the sphere will not slip on the surface, we want to determine the angle 𝜃 at which the sphere separates from the surface. The purpose of this example is to show how the work-energy principle can be combined with the methods studied in Chapter 17 to obtain a solution without direct integration of the system’s equations of motion.

2𝜌 𝜃 = 0 at release

roll without slip

𝑅 𝜃

SOLUTION

𝑅

𝑂

Road Map & Modeling

The FBD of the sphere is shown in Fig. 2. The sphere will begin to separate from the cylinder when the reaction 𝑁 becomes zero. Therefore, we need to write 𝐹⃗ = 𝑚𝑎⃗𝐺 in the direction of 𝑁 (which is the radial direction) and solve for the value of 𝜃 at which 𝑁 = 0. This equation will involve 𝑁, the sphere’s weight, and 𝑎𝐺𝑟 . As long as the sphere does not separate, 𝐺 moves in a circle so that 𝑎𝐺𝑟 depends on 𝜃̇ but ̈ This observation is important because it tells us that to find 𝜃 such that 𝑁 = 0, not 𝜃. we must relate 𝜃̇ to 𝜃, i.e., position to velocity, and this problem is ideal for the workenergy principle. Therefore, our solution strategy will be to combine 𝐹𝑟 = 𝑚𝑎𝐺𝑟 with the work-energy principle.

Figure 1

𝜃 𝑚𝑔 𝑢̂ 𝑟

𝐺

𝐹

𝑢̂ 𝜃

𝑃

Governing Equations Balance Principles Letting 1 be when 𝜃 = 0 and 2 be at an arbitrary angle 𝜃 = 𝜃2 , the Newton-Euler equations in the radial direction at 2 and the work-energy principle applied to the sphere between 1 and 2 are, respectively, ∑ ( ) 𝐹𝑟∶ 𝑁2 − 𝑚𝑔 cos 𝜃2 = 𝑚 𝑎𝐺𝑟 2 , (1) ( ) 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 , (2)

where 𝑇1 and 𝑇2 can be written as 𝑇1 = 0

and 𝑇2 = 21 𝑚𝑣2𝐺2 + 12 𝐼𝐺 𝜔2𝑠2 .

𝑁 Figure 2 FBD of the sphere drawn at an arbitrary angle 𝜃 = 𝜃2 , which corresponds to 2. Note that 𝑘̂ points into the page, so clockwise rotations are positive in this problem.

(3)

Here, 𝑣𝐺2 and 𝜔𝑠2 are the speed of the mass center of the sphere and the angular speed of the sphere, respectively, at 2. The sphere’s mass moment of inertia is given by 𝐼𝐺 = 52 𝑚𝜌2 .

(4)

Referring to Fig. 2, since the sphere rolls without slip, both 𝐹 and 𝑁 do no work, so the only force doing work is the sphere’s weight. Referring to the datum line in Fig. 3, we have ) ( (5) 𝑈1-2 nc = 0, 𝑉1 = 0, and 𝑉2 = −𝑚𝑔(𝑅 + 𝜌)(1 − cos 𝜃2 ).

Force Laws

(𝑅 + 𝜌)(1 − cos 𝜃2 ) 1

datum 𝐺

𝐺

𝑅

𝜃2

Recalling that we are using polar coordinates and that 𝐺 moves along a circle of radius 𝑅 + 𝜌, we have ( ) 𝑎𝐺𝑟 2 = −(𝑅 + 𝜌)𝜃̇ 22 . (6) Kinematic Equations

Next, we focus on the information needed to compute the kinetic energy. We know the kinetic energy at 1, and at 2, since 𝐺 moves in the 𝜃 direction along a circle of radius 𝑅 + 𝜌, we have 𝑣𝐺2 = (𝑅 + 𝜌)𝜃̇ 2 . (7)

2

𝑅 𝑅+𝜌

𝑂 Figure 3 Placement of the datum line coincides with the location of the center of the sphere at 1.

1226

Chapter 18

Energy and Momentum Methods for Rigid Bodies

2

𝐺 𝑃 𝑅

𝜃2

To compute 𝜔𝑠2 we need to apply the kinematics of rolling without slip. Rolling without slip implies that 𝑣⃗𝑃 2 = 0⃗ (see Fig. 4), and so we can write

𝑢̂ 𝑟

1

𝐺

𝑢̂ 𝜃

𝑣⃗𝑃 2 = 0⃗ = 𝑣⃗𝐺2 + 𝜔𝑠2 𝑘̂ × (−𝜌 𝑢̂ 𝑟 ).

(8)

From Eq. (7), we know that 𝑣⃗𝐺2 = (𝑅 + 𝜌)𝜃̇ 2 𝑢̂ 𝜃 , so Eq. (8) yields

𝑅

(𝑅 + 𝜌)𝜃̇ 2 𝑢̂ 𝜃 − 𝜌𝜔𝑠2 𝑢̂ 𝜃 = 0⃗

𝑅+𝜌

𝜔𝑠2 =

⇒

𝑅+𝜌 ̇ 𝜃2 , 𝜌

(9)

̂ where we recall that 𝑘̂ is defined such that 𝑢̂ 𝑟 × 𝑢̂ 𝜃 = 𝑘. 𝑂 Figure 4 Component system at 2.

ISTUDY

Computation

Combining Eqs. (7) and (9) with Eq. (3) yields 𝑇1 = 0

and 𝑇2 =

( )2 𝑅 + 𝜌 ̇2 1 1 𝑚(𝑅 + 𝜌)2 𝜃̇ 22 + 𝐼𝐺 𝜃2 , 2 2 𝜌

(10)

which, using Eq. (4), can be simplified to 𝑇1 = 0

and 𝑇2 =

7 𝑚(𝑅 10

+ 𝜌)2 𝜃̇ 22 .

(11)

Substituting Eqs. (5) and (11) into Eq. (2) gives 0 = −𝑚𝑔(𝑅 + 𝜌)(1 − cos 𝜃2 ) +

7 𝑚(𝑅 10

+ 𝜌)2 𝜃̇ 22 ,

(12)

which can be solved for 𝜃̇ 22 to obtain 𝜃̇ 22 =

10𝑔(1 − cos 𝜃2 ) 7(𝑅 + 𝜌)

.

(13)

Substituting Eq. (6) into Eq. (1) and solving for 𝑁2 , we obtain 𝑁2 = 𝑚𝑔 cos 𝜃2 − 𝑚(𝑅 + 𝜌)𝜃̇ 22 ,

(14)

𝑁2 = 71 𝑚𝑔(17 cos 𝜃2 − 10).

(15)

which, using Eq. (13), yields

We want to solve for the value of 𝜃 that corresponds to 𝑁 = 0. Therefore, letting 𝑁2 = 0 gives ( ) 𝜃2 = cos−1 10 (16) = 53.97◦ , 17 where, recalling that cos−1 (10∕17) = ±53.97◦ + 𝑛360◦ , 𝑛 = 0, ±1, … , ±∞, we have selected the only physically meaningful solution for 𝜃2 . Discussion & Verification

As expected, our solution matches that found in Example 17.10 on p. 1182. What is important here is to notice that we were able to solve the same problem given in Example 17.10 by using a mixed solution strategy that combined both the use of 𝐹⃗ = 𝑚𝑎⃗𝐺 and the work-energy principle. The use of this principle allowed us to completely bypass the derivation and subsequent integration of the equation of motion in the 𝜃 direction.

ISTUDY

Section 18.1

1227

Work-Energy Principle for Rigid Bodies

Problems Problem 18.1

𝑤

A conveyor is moving cans at a constant speed 𝑣0 when, to proceed to the next step in packaging, the cans are transferred onto a stationary surface at 𝐴. The cans each have mass 𝑚, width 𝑤, and height ℎ. Assuming that there is friction between each can and the stationary surface, under what conditions would we be able to compute the stopping distance of the cans, using the work-energy principle for a particle?

ℎ

𝐴

𝑣0

Figure P18.1

Problem 18.2 At the instant shown, the centers of the two identical uniform disks 𝐴 and 𝐵 are moving to the right with the same speed 𝑣0 . In addition, disk 𝐴 is rolling clockwise with an angular speed 𝜔0 , while disk 𝐵 has a backspin with angular speed equal to 𝜔0 . Letting 𝑇𝐴 and 𝑇𝐵 be the kinetic energies of 𝐴 and 𝐵, respectively, state which of the following statements is true and why: (a) 𝑇𝐴 < 𝑇𝐵 ; (b) 𝑇𝐴 = 𝑇𝐵 ; (c) 𝑇𝐴 > 𝑇𝐵 .

At the instant shown, the centers of the two identical uniform disks 𝐴 and 𝐵, each with mass 𝑚 and radius 𝑅, are moving to the right with the same speed 𝑣0 = 4 m∕s. In addition, disk 𝐴 is rolling clockwise with an angular speed 𝜔0 = 5 rad∕s, while disk 𝐵 has a backspin with angular speed 𝜔0 = 5 rad∕s. Letting 𝑚 = 45 kg and 𝑅 = 0.75 m, determine the kinetic energy of each disk.

Problem 18.4 Two identical battering rams are mounted in two different ways on their respective frames as shown. Bars BC and AD are identical and pinned at 𝐵 and 𝐶 and at 𝐴 and 𝐷, respectively. Bars FO and HO are rigidly attached to the ram and are pinned at 𝑂. At the instant shown, the mass centers of rams 1 and 2, at 𝐸 and 𝐺, respectively, are moving horizontally with speed 𝑣0 . Letting 𝑇1 and 𝑇2 be the kinetic energies of rams 1 and 2, respectively, state which of the following statements is true and why: (a) 𝑇1 < 𝑇2 ; (b) 𝑇1 = 𝑇2 ; (c) 𝑇1 > 𝑇2 . 𝐿

𝐿 𝐶

𝐷

𝑂 𝐻

𝑣0 𝐴

𝜔0

𝐸

𝐵

𝐹

ram 1

𝑣0 𝐺

𝐻 ram 2

Figure P18.4 and P18.5

Problem 18.5 Two identical battering rams are mounted in two different ways on their respective frames as shown. Bars BC and AD are identical and pinned at 𝐵 and 𝐶 and at 𝐴 and 𝐷, respectively. Bars FO and HO are rigidly attached to the ram and are pinned at 𝑂. At the instant shown, the centers of mass of rams 1 and 2, at 𝐸 and 𝐺, respectively, are moving horizontally with a speed 𝑣0 = 20 f t∕s. Treating the rams as slender bars with length 𝐿 = 10 f t and weight 𝑊 = 1250 lb, and letting 𝐻 = 3 f t, compute the kinetic energy of the two rams.

𝑣0

𝑅 𝐴

Problem 18.3

𝐻

𝑣0

𝜔0 𝑚

Figure P18.2 and P18.3

𝑅 𝐵

𝑚

1228

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Problem 18.6

𝑂

A pendulum consists of a uniform disk 𝐴 of diameter 𝑑 = 0.15 m and mass 𝑚𝐴 = 0.35 kg attached at the end of a uniform bar 𝐵 of length 𝐿 = 0.75 m and mass 𝑚𝐵 = 0.8 kg. At the instant shown, the pendulum is swinging with an angular speed 𝜔 = 0.24 rad∕s clockwise. Determine the kinetic energy of the pendulum at this instant using Eq. (18.8) on p. 1208.

𝐵 𝐿 𝜔 𝑑 𝐴

Problem 18.7

Figure P18.6

A 2570 lb car (this includes the weight of the wheels) is traveling on a horizontal flat road at 60 mph. If each wheel has a diameter 𝑑 = 24.3 in. and a mass moment of inertia with respect to its mass center equal to 0.989 slug⋅f t 2 , determine the kinetic energy of the car. Neglect the rotational energy of all parts except for the wheels, which roll without slip.

𝜔 axis of rotation

Gary L. Gray

Figure P18.7 test tube 𝐺

Problem 18.8 In Example 17.6 on p. 1168, we analyzed the forces acting on a test tube in an ultracentrifuge. Recalling that the center of mass 𝐺 of the test tube was assumed to be at a distance 𝑟 = 0.0918 m from the centrifuge’s spin axis, and that the test tube had a mass 𝑚 = 0.01 kg and a mass moment of inertia 𝐼𝐺 = 2.821 × 10−6 kg⋅m2 , determine the kinetic energy of the test tube when it is spun at 𝜔 = 60,000 rpm. If you were to convert the computed kinetic energy to gravitational potential energy, at what height, in meters, relative to the ground could you lift a 10 kg mass?

𝑟

Problem 18.9

Figure P18.8

The uniform thin bars 𝐴𝐵, 𝐵𝐶, and 𝐶𝐷 have masses 𝑚𝐴𝐵 = 2.3 kg, 𝑚𝐵𝐶 = 3.2 kg, and 𝑚𝐶𝐷 = 5.0 kg, respectively. The connections at 𝐴, 𝐵, 𝐶, and 𝐷 are pinned joints. Letting 𝑅 = 0.75 m, 𝐿 = 1.2 m, and 𝐻 = 1.55 m, and 𝜔𝐴𝐵 = 4 rad∕s, compute the kinetic energy 𝑇 of the system at the instant shown. 𝐿

𝐵 𝑅

𝐶

𝜔𝐴𝐵 𝐴

𝐻

𝐵 𝐷 𝐴 𝑂

𝐿

Figure P18.9

𝐿

Problem 18.10 𝐷 Figure P18.10

ISTUDY

A T-bar consisting of two uniform bars, each of length 𝐿 = 5 f t, is released from rest in the position shown. Neglecting friction, determine the angular speed of the T-bar when point 𝐴 is directly below point 𝑂.

ISTUDY

Section 18.1

1229

Work-Energy Principle for Rigid Bodies

Problem 18.11 One of the basement doors is left open in the vertical position when it is given a nudge and allowed to freely fall to the closed position. Given that the door has mass 𝑚 and that it is modeled as a uniform thin plate of width 𝑤 and length 𝑑, determine its angular velocity when it reaches the closed position. Hint: Assume that the door is symmetric with respect to a plane of motion in which the acceleration due to gravity is 𝑔 cos 𝜃 rather than 𝑔.

𝑔

𝑑

ℎ 𝜃

𝑤

𝑙 McGraw Hill

𝑂 𝜃

Figure P18.11

Problem 18.12 𝐿

The L-bar consisting of two uniform bars each of length 𝐿 is released from rest when 𝜃 = 90◦ . Neglecting friction, determine the smallest value ( achieved ) by 𝜃. Hint: The equation sin 𝜃 + 𝐴 cos 𝜃 = 𝐵 admits the solution 𝜃 = sin−1 𝐵 cos 𝜙 − 𝜙, with 𝜙 = tan−1 𝐴, if |𝐵 cos 𝜙| ≤ 1.

𝐴

Problem 18.13 𝐿

A turbine rotor with weight 𝑊 = 3000 lb, center of mass at the fixed point 𝐺, and radius of gyration 𝑘𝐺 = 15 f t is brought from rest to an angular speed 𝜔 = 1500 rpm in 20 revolutions by applying a constant torque 𝑀. Neglecting friction, determine the value of 𝑀 needed to spin up the rotor as described. 𝑀

NASA

Figure P18.12

𝜔

𝑀

𝐺

𝐵

𝐺

NASA

Figure P18.13

Figure P18.14

Problem 18.14 A turbine rotor with weight 𝑊 = 3000 lb, center of mass at the fixed point 𝐺, and radius of gyration 𝑘𝐺 = 15 f t is spinning with an angular speed 𝜔 = 1200 rpm when a braking system is engaged that applies a constant torque 𝑀 = 3000 f t ⋅lb. Determine the number of revolutions needed to bring the rotor to a stop.

𝑔 𝑂

𝐺

Problem 18.15 The uniform rectangular plate of length 𝓁, height ℎ, and mass 𝑚 lies in the vertical plane and is pinned at one corner. If the plate is released from rest in the position shown, determine its angular velocity when the center of mass 𝐺 is directly below the pivot 𝑂. Neglect any friction at the pin at 𝑂.

𝓁

Figure P18.15

ℎ

1230

Chapter 18

Energy and Momentum Methods for Rigid Bodies Problems 18.16 and 18.17

A door 𝐴𝐵 weighing 80 lb is pinned at 𝐴 and swings in the horizontal plane. The spring 𝐶𝐷 has stiffness 𝑘 and is unstretched when 𝜃 = 0◦ . Let 𝐿 = 1.5 f t and ℎ = 0.5 f t. 𝐷 𝑘

ℎ

𝐴

𝐶 𝜃

𝐿

𝐵 𝐿

Figure P18.16 and P18.17

If the door is rotating counterclockwise and the speed of 𝐵 is 10 f t∕s when 𝜃 = 0, determine 𝑘, such that the door temporarily stops when 𝜃 = 90◦ . Assume that the spring does not impinge on the mount at 𝐴.

Problem 18.16

If the door is released from rest when 𝜃 = 45◦ and 𝑘 = 50 lb∕f t, determine the speed of 𝐵 when 𝜃 = 0.

𝐴

𝜃

Problem 18.17

Problem 18.18

𝐿

A uniform thin bar 𝐴𝐵 of length 𝐿 = 4 f t is released from rest at an angle 𝜃 = 𝜃1 . As the bar slides, the ends 𝐴 and 𝐵 maintain contact with the surfaces on which they slide. Neglecting friction and knowing that the end 𝐴 has a speed of 18 f t∕s right before hitting the floor, determine 𝜃1 .

𝐵 Figure P18.18

Problem 18.19 A uniform thin bar 𝐴𝐵 of length 𝐿 = 3 f t is released from rest at an angle 𝜃 = 30◦ . As the bar slides, the ends 𝐴 and 𝐵 maintain contact with the surfaces on which they slide. The inclination of the wall is 𝜙 = 50◦ . Neglecting friction, determine the angular speed of the bar right before the end 𝐴 hits the floor. 𝐴 𝐿 𝜙

𝜃

𝐵

Figure P18.19

Problem 18.20 𝐷

𝑅 𝐺 𝑀

Figure P18.20

ISTUDY

𝑘

The disk 𝐷, which has mass 𝑚, center of mass 𝐺, and radius of gyration 𝑘𝐺 , is at rest on a flat horizontal surface when the constant moment 𝑀 is applied to it. The disk is attached at its center to a vertical wall by a linear elastic spring of constant 𝑘. The spring is unstretched when the system is at rest. Assuming that the disk rolls without slipping and that it has not yet come to a stop, determine an expression for the angular velocity of the disk after its center 𝐺 has moved a distance 𝑑. After doing so, determine the distance 𝑑𝑠 that the disk moves before it comes to a stop.

ISTUDY

Section 18.1

Work-Energy Principle for Rigid Bodies

1231

Problems 18.21 and 18.22 An automobile wheel test rig consists of a uniform disk 𝐴, of mass 𝑚𝐴 = 5000 kg and radius 𝑟𝐴 = 1.5 m, that can rotate freely about its fixed center 𝐶 and over which the wheel 𝐵 of an automobile is made to roll. The wheel 𝐵, with center and center of mass at 𝐷, is mounted on a shaft (not shown) that holds 𝐷 fixed while allowing the wheel to rotate about 𝐷. The wheel has diameter 𝑑 = 0.62 m, mass 𝑚𝐵 = 21.5 kg, and mass moment of inertia about its mass center 𝐼𝐷 = 44 kg⋅m2 . Both 𝐴 and 𝐵 are initially at rest when 𝐵 is subject to a constant torque 𝑀 that causes 𝐵 to roll without slip over 𝐴.

𝑀

𝐷

𝑑 𝐵 𝐴 𝑟𝐴

If 𝑀 = 1500 N⋅m, determine the number of revolutions of 𝐵 needed to reach conditions simulating a car speed of 100 km∕h.

Problem 18.21

𝐶

Determine 𝑀 if it takes 100 revolutions of the wheel 𝐵 to achieve conditions simulating a car speed of 60 km∕h.

Problem 18.22

Problems 18.23 and 18.24 An electric motor drawing 15 kW and with an efficiency of 85% lifts a 400 kg crate 𝐵 with a constant speed 𝑣𝑐 . Pulley 𝐴 has radius 𝑟𝑝 = 15 cm, and the center of mass of 𝐴 is also the center of 𝐴. The cord is inextensible and does not slip relative to the pulley. Assume that the friction at the pulley bearings results in a moment about the pulley’s center with magnitude 𝛽|𝜔𝑝 | opposing the rotation of the pulley, where 𝛽 is a constant and |𝜔𝑝 | is the angular speed of the pulley. Hint: Review Example 14.16 on p. 921. Also note that friction in the pulley bearings causes the tension in the cord to be different on the two sides of the pulley. Problem 18.23

Determine 𝛽 if 𝑣𝑐 = 3 m∕s.

Problem 18.24

Determine 𝑣𝑐 if 𝛽 = 2 kg⋅m2 ∕s.

Figure P18.21 and P18.22

𝑟𝑝 motor

𝐴

𝐺

ℎ

𝑂 𝑅

𝑣𝑐 𝐵 Figure P18.23 and P18.24

Figure P18.25

Problem 18.25 An eccentric wheel with weight 𝑊 = 250 lb, mass center 𝐺, and radius of gyration 𝑘𝐺 = 1.32 f t is initially at rest in the position shown. Letting 𝑅 = 1.75 f t and ℎ = 0.8 f t, and assuming that the wheel is gently nudged to the right and rolls without slip, determine the speed of 𝑂 when 𝐺 is closest to the ground. 𝑅

Problem 18.26 A cord is wound around a uniform disk of mass 𝑚 = 2.5 kg and radius 𝑅 = 10 cm. A person pulls on the cord to the right with a constant horizontal force 𝑃 . The disk is initially at rest and rolls without slip. Determine 𝑃 if the center of the disk has a speed of 0.5 m∕s after the hand pulling the cord displaces horizontally to the right by 20 cm.

𝐺

Figure P18.26

1232

Energy and Momentum Methods for Rigid Bodies

Chapter 18

Problems 18.27 and 18.28 𝑂

The pendulum consists of a thin bar of length 𝐿 = 1.5 m and mass 𝑚𝑏 = 2 kg, at the end of which is rigidly attached a uniform solid sphere of radius 𝑟 = 0.25 m and mass 𝑚𝑠 = 3 kg. The pendulum is pinned at the top end 𝑂 of the bar. The pendulum is initially at 𝜃 = 0 when the center of the sphere 𝐶 is given a speed 𝑣1 = 4 m∕s to the right. The quantity 𝑑 denotes the distance between 𝑂 and the center of mass of the bar 𝐺. Neglect friction.

𝑑 𝐿

𝐺 𝜃 𝐴

Problem 18.27 If the bar is uniform (i.e., 𝑑 = 𝐿∕2), determine the maximum swing angle of the pendulum.

2𝑟 𝐶

If the radius of gyration of the bar is 𝑘𝐺 = 0.35 m, determine 𝑑 so that the maximum swing angle is 55◦ .

Figure P18.27 and P18.28

Problem 18.28

Problems 18.29 and 18.30 A 14 lb bowling ball is thrown onto a lane with a backspin angular speed 𝜔0 = 10 rad∕s and forward speed 𝑣0 = 17 mph. After a few seconds, the ball starts rolling without slip and moving forward with a speed 𝑣𝑓 = 17.2 f t∕s. Let 𝑟 = 4.25 in. be the radius of the ball, and let 𝑘𝐺 = 2.6 in. be its radius of gyration. Problem 18.29 Determine the work done by friction on the ball from the initial time until the time that the ball starts rolling without slip.

𝜔0 𝐺

𝜇𝑘

𝑣0

𝑟

Problem 18.30

Knowing that the coefficient of kinetic friction between the lane and the ball is 𝜇𝑘 = 0.1, determine the length 𝐿𝑓 over which the friction force acts in order to slow down the ball from 𝑣0 to 𝑣𝑓 . Does 𝐿𝑓 also represent the distance traveled by the center of the ball? Explain.

𝑚

Figure P18.29–P18.31

Problem 18.31 A bowling ball is thrown onto a lane with a forward speed 𝑣0 and no angular speed (𝜔0 = 0). Because of friction between the lane and the ball, after a short time, the ball starts rolling without slip and moving forward with speed 𝑣𝑓 . Let 𝐿𝐺 be the distance traveled by the center of the ball while slowing down from 𝑣0 to 𝑣𝑓 . In addition, let 𝐿𝑓 be the length over which the friction force had to act in order to slow down the ball from 𝑣0 to 𝑣𝑓 . State which of the following relations is true and why: (a) 𝐿𝐺 < 𝐿𝑓 ; (b) 𝐿𝐺 = 𝐿𝑓 ; (c) 𝐿𝐺 > 𝐿𝑓 .

Problems 18.32 and 18.33

𝐷

ℎ

𝑘

The uniform sphere 𝐵, of radius 𝑟 = 6 cm and mass 𝑚𝐵 = 5 kg, is rigidly attached to the uniform thin bar 𝐴𝐵, which is pinned at 𝐴 and has a mass 𝑚𝐴𝐵 = 8 kg. The system rotates in the vertical plane. The spring 𝐶𝐷 has stiffness 𝑘 = 2000 N∕m and is designed so that the system is in static equilibrium when 𝜃 = 0◦ . Let 𝐿 = 18.2 cm and ℎ = 24.6 cm.

𝐴 𝐶

𝜃

𝐵

𝐿 𝐿 2𝑟

Figure P18.32 and P18.33

ISTUDY

If the system is released from rest when 𝜃 = −30◦ , determine the Problem 18.32 angular speed of the arm for 𝜃 = 0◦ . Problem 18.33 If the system is released from rest when 𝜃 = −30◦ , determine the maximum angle 𝜃 reached by the arm 𝐴𝐵. Hint: The stretched length of the spring can be found using the law of cosines. Note cos(𝜋∕2 + 𝜃) = − sin 𝜃. The solution is easier if sin 𝜃 is written in terms of the stretched and equilibrium lengths (𝜃 = 0◦ ) of the spring.

ISTUDY

Section 18.1

1233

Work-Energy Principle for Rigid Bodies

Problems 18.34 and 18.35 A 500 lb spool with inner and outer radii 𝜌 = 4 f t and 𝑅 = 6 f t, respectively, is released from rest on an incline with 𝜃 = 30◦ . The center of the spool 𝐺 coincides with its center of mass, and the radius of gyration of the spool is 𝑘𝐺 = 5 f t. Assume that the only forces acting on the spool after its release are the spool’s weight, the tension in the cord, and the contact force between the spool and the incline.

𝜌 𝑅

If the angular speed of the spool is 𝜔𝑠2 = 1.2 rad∕s after 𝐺 has displaced a distance 𝑑 = 10 f t from the release position, determine the work done by friction from the instant of release to when 𝜔𝑠2 is achieved.

𝐺

Problem 18.34

Problem 18.35

If the spool starts moving immediately after release and the coefficient of kinetic friction between the spool and the incline is 𝜇𝑘 = 0.25, determine the speed of 𝐺 after 𝐺 has displaced a distance 𝑑 = 10 f t down the incline. Hint: The work done by friction depends on the distance slipped along the incline, which in turn is the difference between the distance traveled by a material point at the outer radius 𝑅 and the distance traveled by 𝐺.

𝜃 Figure P18.34 and P18.35

Problems 18.36 and 18.37 A 400 lb uniform disk with center 𝐺 and radius 𝑟 = 3 f t is connected by a pulley system to a counterweight 𝐴 weighing 75 lb. The system is initially at rest when 𝐴 is allowed to drop, thus causing the disk to roll without slipping to the right.

𝑟 𝐶 𝐺

Neglecting the inertia of the pulley system, determine the speed of 𝐺 after 𝐴 has dropped 2 f t.

Problem 18.36

Problem 18.37 Neglect the inertia of pulley 𝐵 and the cord, but model pulley 𝐶 as a uniform disk with radius 𝑟𝐶 = 0.8 f t and weight 𝑊𝐶 = 50 lb. Assuming the cord does not slip relative to pulley 𝐶, determine the speed of 𝐺 after 𝐴 has dropped 2 f t.

Problems 18.38 and 18.39 In a contraption built by a fraternity, a person sits at the center of a swinging platform with mass 𝑚 = 400 kg and length 𝐿 = 4 m suspended by two identical arms of length 𝐻 = 3 m. 𝐴

𝐷 𝜃

𝐻

𝐶

𝐵 𝐿 Figure P18.38 and P18.39

Problem 18.38 Neglecting the mass of the arms and of the person, neglecting friction, and assuming that the platform is released from rest when 𝜃 = 180◦ , compute the speed of the person as a function of 𝜃 for 0◦ ≤ 𝜃 ≤ 180◦ . In addition, find the speed of the person for 𝜃 = 0◦ . Problem 18.39 Neglecting the mass of the person, neglecting friction, letting the mass of each arm be 𝑚𝐴 = 150 kg, and assuming that the platform is released from rest when 𝜃 = 180◦ , compute the speed of the person as a function of 𝜃 for 0◦ ≤ 𝜃 ≤ 180◦ . In addition, find the speed of the person for 𝜃 = 0◦ .

𝐵 𝐴 Figure P18.36 and P18.37

1234

Chapter 18

Energy and Momentum Methods for Rigid Bodies

𝐿

𝐵

Problem 18.40 𝐶

𝑅

𝜔𝐴𝐵

𝐻

𝐴

𝜙 𝐷

The weights of the uniform thin pin-connected bars 𝐴𝐵, BC, and CD are 𝑊𝐴𝐵 = 4 lb, 𝑊𝐵𝐶 = 6.5 lb, and 𝑊𝐶𝐷 = 10 lb, respectively. Letting 𝜙 = 47◦ , 𝑅 = 2 f t, 𝐿 = 3.5 f t, and 𝐻 = 4.5 f t, and knowing that bar 𝐴𝐵 rotates at an angular speed 𝜔𝐴𝐵 = 4 rad∕s, compute the kinetic energy 𝑇 of the system at the instant shown. Hint: Identify the IC of bar 𝐵𝐶. The system kinetic energy is then the sum of three rotational kinetic energy terms (𝐴𝐵 about 𝐴, 𝐶𝐷 about 𝐷, and 𝐵𝐶 about its IC).

Figure P18.40

Problem 18.41 A payload 𝐵 of mass 𝑚𝐵 = 50 kg is lifted via the pulley system shown by the application of a constant force 𝐹 = 300 N. The pulleys are identical and can be modeled as uniform disks of radius 𝑟𝑝 = 10 cm and mass 𝑚𝑝 = 8 kg. The cord does not slip relative to the pulleys. Modeling the cord as inextensible and neglecting friction at the pulley bearings, determine the speed of 𝐵 after 𝐵 has been lifted a height ℎ = 1 m from its initial rest position. Treat all cable segments as being purely vertical.

𝐴

𝐷

𝐵

𝐹

𝐶

𝐵 Figure P18.41

Figure P18.42

Problem 18.42 A winch drawing 9 hp powers a pulley system lifting a 600 lb crate 𝐶 with a constant speed 𝑣𝑐 . The pulleys are all identical and have a radius 𝑟𝑝 = 1.25 f t. The cord is inextensible and does not slip relative to the pulleys. For each pulley, the friction at the pulley bearings produces a moment about the pulley’s center with magnitude 𝜅|𝜔𝑝 | opposing the rotation of the pulley, where 𝜅 = 1.5 lb⋅f t ⋅s and |𝜔𝑝 | is the angular speed of the pulley. Neglecting the inertia of the pulleys and of the cord, and treating segments of cord that do not touch the pulleys as being vertical, determine 𝑣𝑐 if the motor’s efficiency is 𝜖 = 0.87. Hint: Adapt to this problem the solution in Part (a) of Example 14.16 on p. 921, observing that friction at the pulley bearings causes the tension in the cord on the two sides of a pulley to be different.

Problem 18.43 𝑅

𝐿 𝐵

𝑃 𝐶 Figure P18.43

ISTUDY

𝐴

A 10 lb uniform thin bar 𝐵𝐶 of length 𝐿 = 10 f t is pinned at 𝐵 to the edge of a 20 lb uniform disk of radius 𝑅 = 3.5 f t. The system is initially at rest in the position shown when a constant horizontal force 𝑃 = 60 lb is applied to the end 𝐶. Assume that the disk rolls without slip. In addition, neglect the friction between the end 𝐶 of the bar and the ground, as well as the friction at the pin. Determine the angular speed of the disk when point 𝐵 is directly above the center of the disk. Hint: To determine the displacement of point 𝐶, keep in mind that the overall displacement of point 𝐴 is 𝜋𝑅∕2.

ISTUDY

Section 18.1

1235

Work-Energy Principle for Rigid Bodies

Problem 18.44 The two blocks 𝐴 and 𝐵 weighing 30 lb and 25 lb, respectively, are released from rest. The pulleys are identical and can be modeled as uniform disks of radius 𝑟𝑝 = 0.75 f t and weight 𝑊𝑝 = 8 lb. Modeling the cord as inextensible and neglecting friction at the pulley bearings, determine the speed of 𝐴 after 𝐵 has dropped a height ℎ = 3 f t. 𝐵

Problem 18.45 An eccentric wheel with mass 𝑚 = 150 kg, mass center 𝐺, and radius of gyration 𝑘𝐺 = 0.4 m is placed on the incline shown, such that the wheel’s center of mass 𝐺 is vertically aligned with 𝑃 , which is the point of contact with the incline. If the wheel rolls without slip once it is gently nudged away from its initial placement, letting 𝑅 = 0.55 m, ℎ = 0.25 m, 𝜃 = 25◦ , and 𝑑 = 0.5 m, determine whether the wheel arrives at 𝐵 and, if yes, determine the corresponding speed of the center 𝑂. Note that the angle 𝑃 𝑂𝐺 is not equal to 90◦ at release. Hint: Conservation of energy can be used to find an expression for the speed of 𝑂 at 𝐵. If 𝑣2𝑂 is positive, that confirms arrival at 𝐵 is possible.

𝐴 Figure P18.44

ℎ 𝐺

𝑂

𝐵 𝑅 𝐴

𝑃

𝑑

𝜃 Figure P18.45

Problems 18.46 and 18.47 A spool of mass 𝑚𝑠 = 150 kg and inner and outer radii 𝜌 = 0.8 m and 𝑅 = 1.2 m, respectively, is connected to a counterweight 𝐴 of mass 𝑚𝐴 = 50 kg by a pulley system whose cord, at one end, is wound around the inner hub of the spool. The center 𝐺 of the spool is also the center of mass of the spool, and the radius of gyration of the spool is 𝑘𝐺 = 1 m. The system is at rest when the counterweight is released, causing the spool to move to the right. Assume that the spool rolls without slip.

𝐷 𝐶

𝜌 𝑅

𝐺 𝐵

Problem 18.46

Neglecting the inertia of the pulley system, determine the angular speed of the spool after the counterweight has dropped 0.5 m.

Assume that the inertia of the cord and of pulleys 𝐵 and 𝐷 can be neglected, but model pulley 𝐶 as a uniform disk of mass 𝑚𝐶 = 15 kg and radius 𝑟𝐶 = 0.3 m. If the cord does not slip relative to pulley 𝐶, determine the angular speed of the spool after 𝐴 drops 0.5 m. Problem 18.47

𝐴 Figure P18.46 and P18.47

𝑂

Problem 18.48 The uniform slender bar 𝐴𝐵 has length 𝐿 = 1.45 f t and weight 𝑊𝐴𝐵 = 20 lb. Rollers 𝐷 and 𝐸, which are pinned at 𝐴 and 𝐵, respectively, can be modeled as two identical uniform disks, each with radius 𝑟 = 1.5 in. and weight 𝑊𝑟 = 0.35 lb. Rollers 𝐷 and 𝐸 roll without slip on the surface of a cylindrical bowl with center at 𝑂 and radius 𝑅 = 1 f t. Determine the system’s kinetic energy when 𝐺 (the center of mass of bar 𝐴𝐵) moves with a speed 𝑣 = 7 f t∕s.

𝑅

𝐸

𝐿 𝑟

𝐷

𝐴 Figure P18.48

𝐺 𝑣

𝐵

1236

Energy and Momentum Methods for Rigid Bodies

Chapter 18

Problems 18.49 and 18.50 The Charpy impact test is one test that measures the resistance of a material to fracture. In this test, the fracture toughness is assessed by measuring the energy required to break a specimen of a given geometry. This is done by releasing a heavy pendulum from rest at an angle 𝜃𝑖 and then measuring the maximum swing angle 𝜃𝑓 reached by the pendulum after the specimen is broken.

𝑂 𝐿𝐴

𝐴 𝜃𝑓 𝜃𝑖

Consider a test rig in which the striker 𝑆 (the pendulum’s bob) can be Problem 18.49 modeled as a uniform disk of mass 𝑚𝑆 = 19.5 kg and radius 𝑟𝑆 = 150 mm, and the arm can be modeled as a thin rod of mass 𝑚𝐴 = 2.5 kg and length 𝐿𝐴 = 0.8 m. Neglecting friction and noting that the striker and the arm are rigidly connected, determine the fracture energy (i.e., the kinetic energy lost in breaking the specimen) in an experiment where 𝜃𝑖 = 158◦ and 𝜃𝑓 = 43◦ .

𝑆 𝑄

𝑟𝑆

Figure P18.49 and P18.50 Problem 18.50 Consider a test rig in which the striker 𝑆 (the pendulum’s bob) can be modeled as a uniform disk of weight 𝑊𝑆 = 40 lb and radius 𝑟𝑆 = 6 in., and the arm can be modeled as a thin rod of weight 𝑊𝐴 = 5.5 lb and length 𝐿𝐴 = 2.75 f t. If the release angle of the striker is 𝜃𝑖 = 158◦ and if the striker impacts the specimen when the pendulum’s arm is vertical, determine the speed of the point 𝑄 on the striker immediately before the striker impacts with the specimen. Neglect friction, and observe that the striker and the arm are rigidly connected. ℎ

𝑏

Problem 18.51

𝑂 𝐺

A crate, with weight 𝑊 = 155 lb and mass center 𝐺, is placed on a slide and released from rest as shown. The lower part of the slide is circular, with radius 𝑅 = 6 f t. Model the crate as a uniform body with 𝑏 = 3.6 f t and ℎ = 2 f t. Take into account the gap between the crate and the slide when the crate is in its lowest position, and assume that when the crate is in its lowest position on the slide, the crate’s center of mass is moving to the left with a speed 𝑣𝐺 = 12 f t∕s. Determine the work done by friction on the crate as the crate moves from the release point to the lowest point on the slide.

𝑅 𝐺

Figure P18.51

Problems 18.52 and 18.53 𝐷

𝐴

𝑀 𝜃

𝐵

𝐿

Figure P18.52 and P18.53

ISTUDY

𝐻

In a contraption built by a fraternity, a person sits at the center of a swinging platform with weight 𝑊𝑝 = 800 lb and length 𝐿 = 12 f t suspended by two identical arms each of length 𝐻 = 10 f t and weight 𝑊𝑎 = 200 lb. The platform, which is at rest when 𝜃 = 0◦ , is put in motion by a motor that pumps the ride by exerting a constant moment 𝑀 in the direction shown, during each upswing, whenever 0 ≤ 𝜃 ≤ 𝜃𝑝 , while exerting zero moment otherwise.

𝐶 Problem 18.52 Neglecting the mass of the person, neglecting friction, letting 𝑀 = 900 f t ⋅lb, and letting 𝜃𝑝 = 25◦ , find the minimum number of swings necessary to achieve 𝜃 > 90◦ and the ensuing speed achieved by the person at the lowest point in the swing. Model the arms 𝐴𝐵 and 𝐶𝐷 as uniform thin bars. Problem 18.53 Neglecting the mass of the person, neglecting friction, and letting 𝜃𝑝 = 20◦ , determine the value of 𝑀 required to achieve a maximum value of 𝜃 equal to 90◦ in 6 full swings. Model the arms 𝐴𝐵 and 𝐶𝐷 as uniform thin bars.

ISTUDY

Section 18.1

1237

Work-Energy Principle for Rigid Bodies

Problem 18.54 The disk 𝐷, which has weight 𝑊 , mass center 𝐺 coinciding with the disk’s geometric center, and radius of gyration 𝑘𝐺 , is at rest on an incline when the constant moment 𝑀 is applied to it. The disk is attached at its center to a wall by a linear elastic spring of constant 𝑘. The spring is unstretched when the system is at rest. Assuming that the disk rolls without slipping and that it has not yet come to a stop, determine an expression for the angular velocity of the disk after its center 𝐺 has moved a distance 𝑑 down the incline. After doing so, using 𝑘 = 5 lb∕f t, 𝑅 = 1.5 f t, 𝑊 = 10 lb, and 𝜃 = 30◦ , determine the value of the moment 𝑀 for the disk to stop after rolling 𝑑𝑠 = 5 f t down the incline.

𝑘 𝐷

𝑅 𝐺 𝑀 𝜃

Problem 18.55

Figure P18.54

In Example 18.2 on p. 1217, we ignored the rotational inertia of the counterweight. Let’s revisit that example and remove that simplifying assumption. Assume that the arm 𝐴𝐷 is still a uniform thin bar of length 𝐿 = 15.7 f t and weight 45 lb. The hinge 𝑂 is still 𝑑 = 2.58 f t from the right end of the arm, and the 160 lb counterweight 𝐶 is still 𝛿 = 1.4 f t from the hinge. Now model the counterweight as a uniform block of height ℎ = 14 in. and width 𝑤 = 9 in. With this new assumption, solve for the angular velocity of the arm as it reaches the horizontal position after being nudged from the vertical position. Determine the percent change in angular velocity compared with that found in Example 18.2. counterweight

𝐿 𝐴

𝑑 𝑂 𝛿

𝑙

arm

𝐵

𝑤

ℎ 𝐷 𝐶

Figure P18.55 and P18.56

Problem 18.56 For the barrier gate shown, assume that the arm consists of a section of aluminum tubing from 𝐴 to 𝐵 of length 𝑙 = 11.6 f t and weight 20 lb and a steel support section from 𝐵 to 𝐷 of weight 40 lb. The overall length of the arm is 𝐿 = 15.7 f t. In addition, the 120 lb counterweight 𝐶 is placed a distance 𝛿 from the hinge at 𝑂, and the hinge is 𝑑 = 2.58 f t from the right end of section BD. Model the two sections 𝐴𝐵 and BD as uniform thin bars, and model the counterweight as a uniform block of height ℎ = 14 in. and width 𝑤 = 9 in. Using these new assumptions, determine the distance 𝛿 so that the angular speed of the arm is 0.25 rad∕s as it reaches the horizontal position after being nudged from the vertical position.

𝑘 𝐵

𝑑 𝐴

Problem 18.57 The figure shows the cross section of a garage door with length 𝐿 = 9 f t and weight 𝑊 = 175 lb. At 𝐴 and 𝐵, there are rollers of negligible mass constrained to move in the guide whose horizontal portion is at a distance 𝐻 = 11 f t from the floor. The door’s motion is assisted by two springs, each with constant 𝑘 (only one spring is shown). The door is released from rest when 𝑑 = 26 in. and the spring is stretched 4 in. Neglecting friction, knowing that, when 𝐴 touches the floor, 𝐵 is in the vertical portion of the guide, and modeling the door as a uniform thin plate, determine the minimum value of 𝑘 so that 𝐴 will strike the ground with a speed no greater than 1 f t∕s.

𝐿 𝐻

𝐿

floor Figure P18.57

1238

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Problem 18.58 𝐵 𝑑 𝐴

𝐿 𝐻

𝐿

𝐶

The figure shows the cross section of a garage door with length 𝐿 = 2.5 m and mass 𝑚 = 90 kg. At the ends 𝐴 and 𝐵, there are rollers of negligible mass constrained to move in the guide whose horizontal portion is at a distance 𝐻 = 3 m from the floor. The door’s motion is assisted by two counterweights 𝐶, each of mass 𝑚𝐶 (only one counterweight is shown). If the door is released from rest when 𝑑 = 53 cm, neglecting friction and modeling the door as a uniform thin plate, determine the minimum value of 𝑚𝐶 so that 𝐴 will strike the ground with a speed no greater than 0.25 m∕s.

Problems 18.59 through 18.61

floor

The uniform thin rod 𝐴𝐵 is pin-connected to the slider 𝑆, which moves along the frictionless guide, and to the disk 𝐷, which rolls without slip over the horizontal surface. The pins at 𝐴 and 𝐵 are frictionless, and the system is released from rest. Neglect the vertical dimension of 𝑆.

Figure P18.58

𝑆 𝐴 𝐿 𝐺 𝐷 𝜃 𝐵

𝑅

Figure P18.59–P18.61

Letting 𝐿 = 1.75 m and 𝑅 = 0.6 m, assuming that 𝑆 and 𝐷 are of negligible mass, that the mass of rod 𝐴𝐵 is 𝑚𝐴𝐵 = 7 kg, and that the system is released from the angle 𝜃0 = 65◦ , determine the speed of the slider 𝑆 when it strikes the ground. Problem 18.59

horizontal guide 𝐴

fully open

𝐵

Problem 18.60 Letting 𝐿 = 4.5 f t and 𝑅 = 1.2 f t, assuming that AB is of negligible mass, the weight of 𝑆 is 𝑊𝑆 = 3 lb, 𝐷 is a uniform disk of weight 𝑊𝐷 = 9 lb, and the system is released from the angle 𝜃0 = 67◦ , determine the speed of the slider 𝑆 when it strikes the ground.

Letting 𝐿 = 1.75 m and 𝑅 = 0.6 m, assuming that the mass of 𝑆 is 𝑚𝑆 = 4.2 kg, 𝐷 is a uniform disk of mass 𝑚𝐷 = 12 kg, the mass of AB is 𝑚𝐴𝐵 = 7 kg, and that the system is released from the angle 𝜃 = 69◦ , determine the speed and the direction of motion of point 𝐵 when the slider 𝑆 strikes the ground. Problem 18.61

𝐶

vertical guide

𝐻 𝐷

𝑃 𝐸 fully closed floor

Figure P18.62

ISTUDY

Problem 18.62 Revisit Example 18.5 on p. 1223 and replace the two springs with a system of two counterweights 𝑃 (only one counterweight is shown) each of weight 𝑊𝑃 . Recalling that the door’s weight is 𝑊 = 800 lb and that the total height of the door is 𝐻 = 30 f t, if the door is released from rest in the fully open position and friction is negligible, determine the minimum value of 𝑊𝑃 so that 𝐴 will strike the left end of the horizontal guide with a speed no greater than 0.5 f t∕s.

ISTUDY

Section 18.1

Work-Energy Principle for Rigid Bodies

Problems 18.63 and 18.64

Neglecting the mass of the pulley 𝑃 , determine the speed of the crate 𝐶 and the angular velocity of the pulley 𝐷 after the crate has dropped a distance ℎ = 2 m.

Problem 18.63

Assuming that the pulley 𝑃 has a mass of 1.5 kg and a radius of gyration 𝑘𝐴 = 3.5 cm, determine the speed of the crate 𝐶 and the angular velocity of the pulley 𝐷 after the crate has dropped a distance ℎ = 2 m. Hint: Remember to include the change in potential energy associated with pulley 𝑃 . Pulley 𝑃 also has both translational and rotational kinetic energy.

𝐷

𝑟𝑜

The double pulley 𝐷 has mass of 15 kg, center of mass 𝐺 coinciding with its geometric center, radius of gyration 𝑘𝐺 = 10 cm, outer radius 𝑟𝑜 = 15 cm, and inner radius 𝑟𝑖 = 7.5 cm. It is connected to the pulley 𝑃 with radius 𝑅 by a cord of negligible mass that unwinds without slip from the inner and outer spools of the double pulley 𝐷. The crate 𝐶, which has a mass of 20 kg, is released from rest, and the inner and outer parts of the double pulley rotate together as a single unit.

𝐺

𝑃

𝐴 𝑅

Problem 18.64

𝐶

Figure P18.63 and P18.64

Problems 18.65 through 18.67 Torsional springs provide a simple propulsion mechanism for toy cars. When the rear wheels are rotated as if the car were moving backward, they cause a torsional spring (with one end attached to the axle and the other to the body of the car) to wind up and store energy. Therefore, a simple way to charge the spring is to place the car onto a surface and to pull it backward, making sure that the wheels roll without slipping. Note that the torsional spring can only be wound by pulling the car backward; that is, the forward motion of the car unwinds the spring. Let the weight of the car (body and wheels) be 𝑊 = 5 oz, the weight of each of the wheels be 𝑊𝑤 = 0.15 oz, and the radius of the wheels be 𝑟 = 0.25 in., where the wheels roll without slip and can be treated as uniform disks. Neglecting friction internal to the car and letting the car’s torsional spring be linear with constant 𝑘𝑡 = 0.0002 f t ⋅lb∕rad, determine the maximum speed achieved by the car if it is released from rest after pulling it back a distance 𝐿 = 0.75 f t from a position in which the spring is unwound.

Problem 18.65

Let the weight of the car (body and wheels) be 𝑊 = 5 oz, the weight of each of the wheels be 𝑊𝑤 = 0.15 oz, and the radius of the wheels be 𝑟 = 0.25 in., where the wheels roll without slip and can be treated as uniform disks. In addition, let the torque 𝑀 provided by the nonlinear torsional spring be given by 𝑀 = −𝛽𝜃 3 , where 𝛽 = 0.5×10−6 f t ⋅lb∕rad3 , 𝜃 is the angular displacement of the rear axle, and the minus sign in front of 𝛽 indicates that 𝑀 acts opposite to the direction of 𝜃. Neglecting any friction internal to the car, determine the maximum speed achieved by the car if it is released from rest after pulling it back a distance 𝐿 = 0.75 f t from a position in which the spring is unwound.

Problem 18.66

Let the mass of the car (body and wheels) be 𝑚 = 120 g, the mass of each of the wheels be 𝑚𝑤 = 5 g, and the radius of the wheels be 𝑟 = 6 mm, where the wheels roll without slip and can be treated as uniform disks. In addition, let the car’s torsional spring be linear with constant 𝑘𝑡 = 0.00025 N⋅m∕rad. Neglecting any friction internal to the car, if the angle of the incline is 𝜙 = 25◦ and the car is released from rest after pulling it back a distance 𝐿 = 25 cm from a position in which the spring is unwound, determine the maximum distance 𝑑max that the car will travel up the incline (from its release point), the maximum speed 𝑣max achieved by the car, and the distance 𝑑𝑣max (from the release point) at which 𝑣max is achieved.

𝑟𝑖

Figure P18.65 and P18.66

Problem 18.67

𝜙 Figure P18.67

1239

1240

Energy and Momentum Methods for Rigid Bodies

Chapter 18

Problem 18.68 𝐵 𝐴

𝐿

𝐿

𝐶

floor

The figure shows the cross section of a garage door with length 𝐿 = 2.5 m and mass 𝑚 = 90 kg. At the ends 𝐴 and 𝐵, there are rollers of negligible mass constrained to move in a vertical and a horizontal guide, respectively. The door’s motion is assisted by two counterweights (only one counterweight is shown), each of mass 𝑚𝐶 = 22 kg. If the door is released from rest when horizontal, neglecting friction and modeling the door as a uniform thin plate, determine the speed with which 𝐵 strikes the left end of the horizontal guide. Hint: At the instant 𝐵 strikes the end of the horizontal guide, point 𝐴 is at the bottom of the vertical guide and is acting as an IC.

Problem 18.69

Figure P18.68 𝐵 𝜃 𝐺

𝐿

𝐴

A stick of length 𝐿 and mass 𝑚 is in equilibrium while standing on its end 𝐴 when end 𝐵 is gently nudged to the right, causing the stick to fall. Model the stick as a uniform slender bar, and assume that there is friction between the stick and the ground. Under these assumptions, there is a value of 𝜃, let’s call it 𝜃max , such that the stick must start slipping before reaching 𝜃max for any value of the coefficient of static friction 𝜇𝑠 . To find the value of 𝜃max , follow the steps below. (a) Letting 𝐹 and 𝑁 be the friction and normal forces, respectively, between the stick and the ground, draw the FBD of the stick as it falls. Then set the sum of forces in the horizontal and vertical directions equal to the corresponding components of 𝑚𝑎⃗𝐺 . ̇ and 𝜃. ̈ Finally, express 𝐹 and 𝑁 as Express the components of 𝑎⃗𝐺 in terms of 𝜃, 𝜃, ̇ and 𝜃. ̈ Hint: Assume 𝐴 tends to slip to the left, so that friction acts functions of 𝜃, 𝜃, to the right. (b) Use the work-energy principle to find an expression for 𝜃̇ 2 (𝜃). Differentiate the eẍ pression for 𝜃̇ 2 (𝜃) with respect to time, and find an expression for 𝜃(𝜃).

Figure P18.69

̈ (c) Substitute the expressions for 𝜃̇ 2 (𝜃) and 𝜃(𝜃) into the expressions for 𝐹 and 𝑁 to obtain 𝐹 and 𝑁 as functions of 𝜃. For impending slip, |𝐹 ∕𝑁| must be equal to the coefficient of static friction. Use this fact to determine 𝜃max .

Problem 18.70 𝐵 𝜃 𝐺 𝐴 Figure P18.70

ISTUDY

𝐿

A stick of length 𝐿 and mass 𝑚 is in equilibrium while standing on its end 𝐴 when the end 𝐵 is gently nudged to the right, causing the stick to fall. Letting 𝜇𝑠 be the coefficient of static friction between the stick and the ground and modeling the stick as a uniform slender bar, find the largest value of 𝜇𝑠 for which the stick slides to the left, as well as the corresponding value of 𝜃 at which sliding begins. To solve this problem, follow the steps below. (a) Let 𝐹 and 𝑁 be the friction and normal forces, respectively, between the stick and the ground, and let 𝐹 be positive to the right and 𝑁 positive upward. Draw the FBD of the stick as it falls. Then set the sum of forces in the horizontal and vertical directions equal to the corresponding components of 𝑚𝑎⃗𝐺 . Express the components of 𝑎⃗𝐺 in ̇ and 𝜃. ̈ Finally, express 𝐹 and 𝑁 as functions of 𝜃, 𝜃, ̇ and 𝜃. ̈ terms of 𝜃, 𝜃, (b) Use the work-energy principle to find an expression for 𝜃̇ 2 (𝜃). Differentiate the eẍ pression for 𝜃̇ 2 (𝜃) with respect to time, and find an expression for 𝜃(𝜃). ̈ into the expressions for 𝐹 and 𝑁 to ob(c) Substitute the expressions for 𝜃̇ 2 (𝜃) and 𝜃(𝜃) tain 𝐹 and 𝑁 as functions of 𝜃. When slip is impending (i.e., when |𝐹 | = 𝜇𝑠 |𝑁|), |𝐹 ∕𝑁| must be equal to the static coefficient of friction. Therefore, compute the maximum value of |𝐹 ∕𝑁| by differentiating it with respect to 𝜃 and setting the resulting derivative equal to zero.

ISTUDY

Section 18.1

Problem 18.71 The uniform thin pin-connected bars 𝐴𝐵, 𝐵𝐶, and 𝐶𝐷 have masses 𝑚𝐴𝐵 = 2.3 kg, 𝑚𝐵𝐶 = 3.2 kg, and 𝑚𝐶𝐷 = 5.0 kg, respectively. In addition, 𝑅 = 0.75 m, 𝐿 = 1.2 m, and 𝐻 = 1.55 m. When bars 𝐴𝐵 and 𝐶𝐷 are vertical, 𝐴𝐵 is rotating with angular speed 𝜔𝐴𝐵 = 4 rad∕s in the direction shown. At this instant, the motor connected to 𝐴𝐵 starts to exert a constant torque 𝑀 in the direction opposite to 𝜔𝐴𝐵 . If the motor stops 𝐴𝐵 after 𝐴𝐵 has rotated 90◦ counterclockwise, determine 𝑀 and the maximum power output of the motor during the stopping phase. In the final position, 𝜙 = 64.36◦ and 𝜓 = 29.85◦ .

Problem 18.72

𝜔𝐴𝐵 𝑅

𝐿

𝐵

𝜓 𝐴

(a) Letting 𝐹 and 𝑁 be the friction and normal forces, respectively, between the stick and the ground, draw the FBD of the stick as it falls. Then set the sum of forces in the horizontal and vertical directions equal to the corresponding components of 𝑚𝑎⃗𝐺 . ̇ and 𝜃. ̈ Finally, express 𝐹 and 𝑁 as Express the components of 𝑎⃗𝐺 in terms of 𝜃, 𝜃, ̇ and 𝜃. ̈ functions of 𝜃, 𝜃,

𝐶

𝑀 𝐻

𝜙 𝐷 Figure P18.71

A stick of length 𝐿 and mass 𝑚 is in equilibrium while standing on its end 𝐴 when end 𝐵 is gently nudged to the right, causing the stick to fall. Letting the coefficient of static friction between the stick and the ground be 𝜇𝑠 = 0.7 and modeling the stick as a uniform slender bar, find the value of 𝜃 at which end 𝐴 of the stick starts slipping, and determine the corresponding direction of slip. As part of the solution, plot the absolute value of the ratio between the friction and normal force as a function of 𝜃. To solve this problem, follow the steps below.

𝐵 𝜃 𝐺

𝐿

𝐴 Figure P18.72

(b) Use the work-energy principle to find an expression for 𝜃̇ 2 (𝜃). Differentiate the eẍ pression for 𝜃̇ 2 (𝜃) with respect to time, and find an expression for 𝜃(𝜃). 2 ̈ (c) After substituting the expressions for 𝜃̇ (𝜃) and 𝜃(𝜃) into the expressions for 𝐹 and 𝑁, plot |𝐹 ∕𝑁| as a function of 𝜃. For impending slip, |𝐹 ∕𝑁| must be equal to 𝜇𝑠 . Therefore, the desired value of 𝜃 corresponds to the intersection of the plot of |𝐹 ∕𝑁| with the horizontal line intercepting the vertical axis at the value 0.7. After determining the desired value of 𝜃, the direction of slip can be found by determining the sign of 𝐹 evaluated at the 𝜃 computed. 𝐶

Problems 18.73 and 18.74

Letting 𝜔𝐴𝐵 = 2500 rpm, compute the kinetic energy of the connecting rod for 𝜃 = 90◦ and for 𝜃 = 180◦ .

Problem 18.73

Plot the kinetic energy of the connecting rod as a function of the crank angle 𝜃 over one full cycle of the crank for 𝜔𝐴𝐵 = 2500 rpm, 5000 rpm, and 7500 rpm.

𝐿

𝜙

For the slider-crank mechanism shown, let 𝐿 = 141 mm, 𝑅 = 48.5 mm, and 𝐻 = 36.4 mm. In addition, observing that 𝐷 is the center of mass of the connecting rod, let the mass moment of inertia of the connecting rod be 𝐼𝐷 = 0.00144 kg⋅m2 and the mass of the connecting rod be 𝑚 = 0.439 kg.

Problem 18.74

1241

Work-Energy Principle for Rigid Bodies

𝐷

𝐻 𝐵

𝜃

𝑅 𝐴 𝜔𝐴𝐵

Figure P18.73 and P18.74

1242

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Design Problems

ISTUDY

Design Problem 18.1 The opening and closing of the manually operated road barrier is assisted by the counterweight 𝐶 and linear elastic torsional spring with constant 𝑘𝑡 that is mounted at 𝑂. Assume that the length of the arm is 𝐿 = 15.7 f t and that it consists of a section of aluminum tubing from 𝐴 to 𝐵 of length 𝑙 = 11.6 f t and weight 20 lb and a steel support section from 𝐵 to 𝐷 of weight 40 lb. Model both sections of the arm as uniform thin rods. Model the counterweight as a uniform rectangular rigid body of weight 𝑊𝑐 , height ℎ, and width 𝑤, and let the hinge 𝑂 be a distance 𝑑 = 2.58 f t from the right end of section 𝐵𝐷. Using these assumptions, design the unspecified parameters 𝛿, ℎ, 𝑤, 𝑊𝐶 , and the torsional spring (its stiffness 𝑘𝑡 and the position at which it is undeformed), so that a small nudge will close the barrier from the vertical position and so that the arm will reach the closed position with an angular speed that is less than 0.25 rad∕s. In addition, make sure that the barrier is still easy to open. counterweight

arm

𝐿 𝑑 𝑙

𝑂

𝐴

𝐵

𝑘𝑡

𝛿

ℎ 𝐷

𝑤 𝐶

Figure DP18.1

ISTUDY

Section 18.2

18.2

1243

Momentum Methods for Rigid Bodies

Momentum Methods for Rigid Bodies

In this section, we develop both the linear and the angular impulse-momentum principles for rigid bodies. This is a departure from what was done in Chapter 15, where we devoted individual sections to each of these principles. The reason for this different approach is that, for rigid bodies, the linear and angular impulse-momentum principles must often be applied together to get a complete picture of a body’s motion, as shown in the following example. 𝑦

Impulse-momentum principle for a rigid body

𝐹⃗1

A rigid body’s mass center moves according to Eq. (17.1) on p. 1145, i.e., 𝐹⃗ = 𝑚𝑎⃗𝐺 ,

body 𝐵

(18.32)

where, referring to Fig. 18.11, 𝐹⃗ = 𝐹⃗1 + 𝐹⃗2 + ⋯ + 𝐹⃗𝑁 is the total force acting on the body and 𝑚 and 𝑎⃗𝐺 are the body’s mass and acceleration of the mass center, respectively. Integrating Eq. (18.32) over time for 𝑡1 ≤ 𝑡 ≤ 𝑡2 , we obtain ∫𝑡

𝑡2

𝐹⃗ 𝑑𝑡 =

1

∫𝑡

𝑡2

𝑚𝑎⃗𝐺 𝑑𝑡 = 𝑚𝑣⃗𝐺 (𝑡2 ) − 𝑚𝑣⃗𝐺 (𝑡1 ).

∫𝑡

𝑡2

𝐺

(18.33)

𝐹⃗ 𝑑𝑡 = 𝑚𝑣⃗𝐺2

or

𝑝⃗1 +

1

∫𝑡

𝑡2

𝐹⃗ 𝑑𝑡 = 𝑝⃗2 ,

𝑣⃗𝐺

𝐹⃗𝑁

1

Here, using concepts introduced in Section 15.1, the first term in Eq. (18.33) is the total linear impulse acting on the body, and 𝑚𝑣⃗𝐺 (𝑡) is the body’s linear momentum, which we denote by 𝑝. ⃗ As we have done for a particle [see Eq. (15.6) on p. 934], we can rewrite Eq. (18.33) as 𝑚𝑣⃗𝐺1 +

𝑎⃗𝐺

𝐹⃗2

𝑂

𝑥

Figure 18.11 A rigid body under the action of a system of forces.

(18.34)

1

where the subscripts 1 and 2 indicate the values of a quantity at 𝑡1 and 𝑡2 , respectively. Equations (18.34) express the linear impulse-momentum principle for a rigid body. Extension of Eq. (18.34) for a system

𝑦

Recalling that 𝐹⃗ is the sum of only the external forces, Eqs. (18.34) can be applied to a system of rigid bodies if we properly compute 𝑝. ⃗ For a system of 𝑁 rigid bodies (see Fig. 18.12), the system’s total momentum is 𝑝⃗ =

𝑁 ∑

𝑚𝑖 𝑣⃗𝐺𝑖 (𝑡),

𝑣⃗𝐺2 𝐺2 𝐺𝑖

(18.35) 𝐺1

𝑖=1

where 𝑚𝑖 and 𝑣⃗𝐺𝑖 (𝑖 = 1, … , 𝑁) are the mass and the velocity of the mass center, respectively, of body 𝑖. Conservation of linear momentum

or

𝑣⃗𝐺1

𝑂

If 𝐹⃗ = 0⃗ for 𝑡1 ≤ 𝑡 ≤ 𝑡2 , Eqs. (18.34) reduce to 𝑚𝑣⃗𝐺1 = 𝑚𝑣⃗𝐺2

𝑣⃗𝐺𝑖

Figure 18.12 A system of rigid bodies.

𝑝⃗1 = 𝑝⃗2 ,

(18.36)

which states that the system’s momentum is conserved for 𝑡1 ≤ 𝑡 ≤ 𝑡2 . In many applications the total external force 𝐹⃗ is not equal to zero over the time interval

𝑥

1244

Chapter 18

Energy and Momentum Methods for Rigid Bodies

considered, but there is a direction, say 𝑞, along which the component 𝐹𝑞 = 0 for 𝑡1 ≤ 𝑡 ≤ 𝑡2 . In this case, we can write ( ) ( ) (18.37) 𝑚 𝑣𝐺𝑞 1 = 𝑚 𝑣𝐺𝑞 2 or 𝑝𝑞1 = 𝑝𝑞2 ; that is, the momentum is conserved in the 𝑞 direction.

Angular impulse-momentum principle for a rigid body 𝑦

The moment-angular momentum equation governing the motion of a rigid body is Eq. (17.4) on p. 1146, i.e.,

𝑣⃗𝑃 𝑃

𝑟⃗𝑑𝑚∕𝑃 𝑑𝑚

𝑟⃗𝑃

𝑟⃗𝑑𝑚 𝐺 𝑟⃗𝐺

𝜔𝐵

𝑟⃗𝐺∕𝑃

𝑣⃗𝑑𝑚

𝑂

⃗ =ℎ ⃗̇ + 𝑣⃗ × 𝑚𝑣⃗ , 𝑀 𝑃 𝑃 𝑃 𝐺

𝑣⃗𝐺

⃗ = ℎ 𝑃

Figure 18.13 The quantities needed to obtain the angular momentum relationships for a rigid body.

ISTUDY

where 𝑃 is an arbitrarily chosen moment center (see Fig. 18.13), 𝑣⃗𝑃 is the velocity of ⃗ is the total moment relative to 𝑃 due to the external force system acting 𝑃 , and 𝑀 𝑃 ⃗ is the body’s angular momentum relative to 𝑃 ; it was on the body. The quantity ℎ 𝑃 defined in Eq. (17.6) on p. 1146 as

𝐵 𝑥

(18.38)

∫𝐵

𝑟⃗𝑑𝑚∕𝑃 × 𝑣⃗𝑑𝑚 𝑑𝑚.

(18.39)

While Eqs. (18.38) and (18.39) are valid for any type of body and for any motion, the applications we focus on in this chapter concern the planar motion of rigid bodies that are symmetric with respect to the plane of motion. Therefore, as shown in Eq. (D.25) ⃗ can be given the following compact form: of App. D, ℎ 𝑃 ⃗ =𝐼 𝜔 ℎ ⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , 𝑃 𝐺 ⃗𝐵 + 𝑟

(18.40)

where 𝐼𝐺 and 𝜔 ⃗ 𝐵 are the body’s mass moment of inertia and angular velocity, respectively. If the point 𝑃 is in the same plane as 𝐺 and any one of the following is true: ⃗ or 1. 𝑃 is a fixed point (i.e., 𝑣⃗𝑃 = 0); ⃗ or 2. 𝑃 coincides with 𝐺 (i.e., 𝑣⃗𝑃 = 𝑣⃗𝐺 ⇒ 𝑣⃗𝑃 × 𝑣⃗𝐺 = 0); ⃗ 3. 𝑃 and 𝐺 move parallel to one another (i.e., 𝑣⃗𝑃 × 𝑣⃗𝐺 = 0); Eq. (18.38) simplifies to ⃗̇ . ⃗ =ℎ 𝑀 𝑃 𝑃

(18.41)

If the assumptions underlying Eq. (18.41) hold throughout a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 , then integrating Eq. (18.41) over this time interval, we obtain the traditional form of the angular impulse-momentum principle as ⃗ + ℎ 𝑃1

∫𝑡

𝑡2

⃗ , ⃗ 𝑑𝑡 = ℎ 𝑀 𝑃 𝑃2

(18.42)

1

⃗ are the values of ℎ ⃗ at times 𝑡 and 𝑡 , respectively. If the moment ⃗ and ℎ where ℎ 𝑃1 𝑃2 𝑃 1 2 ⃗ and then using Eq. (18.40) in center is taken to be the mass center 𝐺, then 𝑟⃗𝐺∕𝑃 = 0, Eq. (18.42), the angular impulse-momentum principle becomes 𝐼𝐺1 𝜔𝐵1 +

∫𝑡

𝑡2 1

𝑀𝐺𝑧 𝑑𝑡 = 𝐼𝐺2 𝜔𝐵2 .

(18.43)

ISTUDY

Section 18.2

Momentum Methods for Rigid Bodies

Here, 𝑀𝐺𝑧 is the 𝑧 component of the moment about 𝐺, subscripts 1 and 2 indicate the value of a quantity at 𝑡1 and 𝑡2 , respectively, and Eq. (18.43) has been written in scalar form because, under the current assumptions, the only nonzero component of Eq. (18.42) is perpendicular to the plane of motion. Equation (18.43) is valid even if 𝐼𝐺 changes with time (see Example 18.9). If a body 𝐵 is in fixed-axis rotation about a point 𝑂 (Fig. 18.14), then 𝑣⃗𝐺 = 𝜔 ⃗ 𝐵 ×⃗𝑟𝐺∕𝑂 , and by choosing the center of rotation 𝑂 as our moment center, Eq. (18.40) becomes ( ) ⃗ =𝐼 𝜔 ⃗𝐺∕𝑂 × 𝑚 𝜔 ⃗ 𝐵 × 𝑟⃗𝐺∕𝑂 ℎ 𝑂 𝐺 ⃗𝐵 + 𝑟 ( ) ⃗ 𝐵 = 𝐼𝑂 𝜔 ⃗𝐵, (18.44) = 𝐼𝐺 + 𝑚|⃗𝑟𝐺∕𝑂 |2 𝜔

𝑦

𝐵

𝑣⃗𝐺 𝑟⃗𝐺∕𝑂

𝐺

𝜔𝐵

𝑥

𝑂

where, by the parallel axis theorem, 𝐼𝑂 = 𝐼𝐺 +𝑚|⃗𝑟𝐺∕𝑂 |2 is the body’s mass moment of inertia about the axis of rotation. Using Eq. (18.44), the angular impulse-momentum principle for a body that is symmetric with respect to the plane of motion and under fixed-axis rotation takes on the form 𝐼𝑂1 𝜔𝐵1 +

∫𝑡

𝑡2

𝑀𝑂𝑧 𝑑𝑡 = 𝐼𝑂2 𝜔𝐵2 ,

(18.45)

1245

𝐵 Figure 18.14 A rigid body in a fixed-axis rotation.

1

⃗ . We have written where 𝑂 is the center of rotation and 𝑀𝑂𝑧 is the 𝑧 component of 𝑀 𝑂 Eq. (18.45) in scalar form because, under the current assumptions, the only nonzero component of Eq. (18.42) is normal to the plane of motion. 𝑦

Angular impulse-momentum principle for a system

𝑣⃗𝐺2

Equations (18.42) and (18.43) apply to systems of rigid bodies if the assumptions underlying these equations are satisfied by each element of the system. Referring to Fig. 18.15, for a system of 𝑁 rigid bodies, in which body 𝑖 has angular velocity 𝜔 ⃗ 𝑖, ⃗ mass center 𝐺𝑖 , mass 𝑚𝑖 , and mass moment of inertia 𝐼𝐺𝑖 , ℎ𝑃 is ⃗ = ℎ 𝑃

𝑁 ∑

𝜔2

𝑟⃗𝐺

𝐺𝑖

2 ∕𝑃

𝑟⃗𝐺

( ) ⃗ 𝑖 + 𝑟⃗𝐺𝑖∕𝑃 × 𝑚𝑖 𝑣⃗𝐺𝑖 , 𝐼𝐺𝑖 𝜔

𝜔𝑖

𝐺2

1 ∕𝑃

𝑃

𝑟⃗𝐺 ∕𝑃 𝑖

𝑣⃗𝐺𝑖

𝐺1

(18.46)

𝑖=1

𝜔1

where 𝑣⃗𝐺𝑖 is the velocity of 𝐺𝑖 and 𝑟⃗𝐺𝑖∕𝑃 is the position of 𝐺𝑖 relative to 𝑃.

𝑂

Conservation of angular momentum

Figure 18.15 A system of rigid bodies.

⃗ = 0⃗ for 𝑡 ≤ 𝑡 ≤ 𝑡 , Eq. (18.42) implies that If 𝑀 𝑃 1 2 ⃗ = constant, ⃗ =ℎ ℎ 𝑃1 𝑃2

(18.47)

which states that the body’s angular momentum relative to 𝑃 is conserved. Another ⃗ but there is a fixed direction, say 𝑞, along ⃗ ≠ 0, useful result is obtained when 𝑀 𝑃 which 𝑀𝑃 𝑞 = 0. In this case, we can write ( ) ( ) ℎ𝑃 𝑞 1 = ℎ𝑃 𝑞 2 = constant, and apply conservation of angular momentum in the 𝑞 direction.

𝑣⃗𝐺1

(18.48)

𝑥

1246

Chapter 18

Energy and Momentum Methods for Rigid Bodies

End of Section Summary

𝑦 𝐹⃗1

The linear impulse-momentum principle for a rigid body is (see Fig. 18.16)

body 𝐵

Eqs. (18.34), p. 1243

𝑎⃗𝐺

𝐹⃗2

𝐺

𝑚𝑣⃗𝐺1 + 𝑣⃗𝐺

𝐹⃗ 𝑑𝑡 = 𝑚𝑣⃗𝐺2

𝑝⃗1 +

1

∫𝑡

𝑡2

𝐹⃗ 𝑑𝑡 = 𝑝⃗2 ,

1

𝑥

Eqs. (18.36), p. 1243

Figure 18.16 A rigid body under the action of a system of force.

𝑦

𝑑𝑚

𝑚𝑣⃗𝐺1 = 𝑚𝑣⃗𝐺2

𝑃

𝑝⃗1 = 𝑝⃗2 ,

Eq. (18.40), p. 1244

𝑟⃗𝑃 𝜔𝐵

𝑟⃗𝐺∕𝑃

𝑣⃗𝑑𝑚 𝑟⃗𝑑𝑚 𝐺

𝑣⃗𝐺 𝐵

𝑟⃗𝐺

𝑥

𝑂

Figure 18.17 The quantities needed to obtain the angular momentum relationships for a rigid body.

⃗ =𝐼 𝜔 ℎ ⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , 𝑃 𝐺 ⃗𝐵 + 𝑟 where 𝐼𝐺 is the mass moment of inertia of 𝐵, 𝜔 ⃗ 𝐵 is the angular velocity of 𝐵, and 𝑟⃗𝐺∕𝑃 is the position of 𝐺 relative to 𝑃. If 𝑃 is chosen so that (1) 𝑃 is fixed or (2) 𝑃 ⃗ =ℎ ⃗̇ , where coincides with 𝐺 or (3) 𝑃 and 𝐺 move parallel to one another, then 𝑀 𝑃 𝑃 ⃗ is the moment relative to 𝑃 of the external force system acting on 𝐵. When this 𝑀 𝑃 equation holds, by integrating with respect to time over a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 , we have Eq. (18.42), p. 1244 ⃗ + ℎ 𝑃1

𝑦

∫𝑡

𝑡2

⃗ 𝑑𝑡 = ℎ ⃗ . 𝑀 𝑃 𝑃2

1

When 𝑃 coincides with 𝐺 or if the body undergoes a fixed-axis rotation about a point 𝑂 as shown in Fig. 18.18, then the above equation becomes 𝑣⃗𝐺

𝐵

Eq. (18.43), p. 1244, and Eq. (18.45), p. 1245

𝐺 𝜔𝐵

or

and we say that the body’s momentum is conserved. If 𝑃 in Fig. 18.17 is a moment center coplanar with 𝐺 and if 𝐵 is symmetric relative to the plane of motion, the angular momentum of 𝐵 relative to 𝑃 is

𝑣⃗𝑃 𝑟⃗𝑑𝑚∕𝑃

𝐼𝐺1 𝜔𝐵1 +

𝑟⃗𝐺∕𝑂

𝑂

Figure 18.18 A rigid body in a fixed-axis rotation.

ISTUDY

or

where 𝐹⃗ is the total external force on 𝐵, 𝑝⃗ = 𝑚𝑣⃗𝐺 is 𝐵’s linear momentum, and 𝑣⃗𝐺 ⃗ we have is the velocity of 𝐵’s center of mass. If 𝐹⃗ = 0,

𝐹⃗𝑁

𝑂

∫𝑡

𝑡2

𝑥

𝐼𝑂1 𝜔𝐵1 +

∫𝑡

𝑡2

𝑀𝐺𝑧 𝑑𝑡 = 𝐼𝐺2 𝜔𝐵2 ,

1

∫𝑡

𝑡2

𝑀𝑂𝑧 𝑑𝑡 = 𝐼𝑂2 𝜔𝐵2 ,

1

respectively, where 𝐼𝑂 is the mass moment of inertia about the fixed axis of rotation.

ISTUDY

Section 18.2

Momentum Methods for Rigid Bodies

E X A M P L E 18.7

A Rolling Wheel: Computing Angular Momentum

The wheel 𝑤 shown in Fig. 1 has radius 𝑟 and mass 𝑚. Point 𝐺 is both the wheel’s center of mass and the wheel’s geometric center. The radius of gyration is 𝑘𝐺 . The wheel is rolling without slip with 𝐺 moving to the right with a speed 𝑣0 . Compute the wheel’s angular momentum relative to both 𝐺 and 𝑄, which is the point on the wheel in contact with the ground.

𝑤

𝑣0

𝐺

SOLUTION Road Map

This problem is solved by a direct application of the expression for the angular momentum of a rigid body given in Eq. (18.40) on p. 1244.

Figure 1

Computation Referring to Fig. 2 and applying Eq. (18.40), the angular momentum of the wheel relative to its center of mass is

⃗ =𝐼 𝜔 ⃗𝐺∕𝐺 × 𝑚𝑣⃗𝐺 , ℎ 𝐺 𝐺 ⃗𝑤 + 𝑟

(1)

𝑤 𝐺

where 𝑟⃗𝐺∕𝐺 = 0⃗ and 𝐼𝐺 = 𝑚𝑘2𝐺 . Recalling that because of rolling without slip we must ̂ Eq. (1) can be rewritten as have 𝜔 ⃗ 𝑤 = −(𝑣0 ∕𝑟) 𝑘, ⃗ = −𝑚𝑘2 ℎ 𝐺 𝐺

(𝑣 ) 0

𝑟

̂ 𝑘.

(2)

Again referring to Fig. 2 and applying Eq. (18.40), the angular momentum of the wheel relative to 𝑄 is ⃗ =𝐼 𝜔 ℎ ⃗𝐺∕𝑄 × 𝑚𝑣⃗𝐺 . (3) 𝑄 𝐺 ⃗𝑤 + 𝑟 In this case, we have 𝑟⃗𝐺∕𝑄 = 𝑟 𝚥̂ and 𝑣⃗𝐺 = 𝑣0 𝚤̂

⇒

̂ 𝑟⃗𝐺∕𝑄 × 𝑚𝑣⃗𝐺 = −𝑚𝑟𝑣0 𝑘.

(4)

̂ we ⃗ 𝑤 = −𝑚𝑘2𝐺 (𝑣0 ∕𝑟) 𝑘, Substituting the last of Eqs. (4) into Eq. (3) and recalling that 𝐼𝐺 𝜔 have ( ) 𝑘2 𝑣 ⃗ = −𝑚𝑟𝑣 𝑘̂ − 𝑚𝑘2 0 𝑘̂ = −𝑚𝑣 𝑟 + 𝐺 𝑘. ̂ ℎ (5) 0 𝑄 0 𝐺 𝑟 𝑟 Discussion & Verification

To verify that the results in Eqs. (2) and (5) are dimensionally correct, recall that angular momentum has the dimensions of moment of the momentum, i.e., of mass × velocity × length, which is what we see in Eqs. (2) and (5). A Closer Look Using the parallel axis theorem, we can give the expression in Eq. (5) a much more compact form. Let 𝐼𝑄 denote the wheel’s mass moment of inertia relative to 𝑄. Using the parallel axis theorem, we have ) ( (6) 𝐼𝑄 = 𝐼𝐺 + 𝑚𝑟2 = 𝑚 𝑘2𝐺 + 𝑟2 .

Going back to Eq. (5) and factoring 1∕𝑟 out of the term in parentheses, we can rewrite this equation as 𝑣 ( ) ⃗ = −𝑚 0 𝑟2 + 𝑘2 𝑘̂ = 𝐼 𝜔 (7) ℎ 𝑄 ⃗ 𝑤, 𝑄 𝐺 𝑟 ̂ Comparwhere we have taken advantage of Eq. (6) and the fact that 𝜔 ⃗ 𝑤 = −(𝑣0 ∕𝑟) 𝑘. ing Eq. (7) with Eq. (18.44) on p. 1245, we could interpret the result by saying that the wheel appears as though it were in a fixed-axis rotation about 𝑄. This interpretation is appropriate because 𝑄 is the wheel’s instantaneous center of rotation.

𝑣0 𝑟

Figure 2 Wheel rolling without slip.

1247

1248

Chapter 18

Energy and Momentum Methods for Rigid Bodies

E X A M P L E 18.8

A Rolling Pipe: Application of Impulse and Momentum

𝐴 2𝑟

𝐺 𝑣0

Figure 1 A pipe section of radius 𝑟 lowered very gently over a conveyor belt.

A pipe section 𝐴 of radius 𝑟, center 𝐺, and mass 𝑚 is gently placed (i.e., with zero velocity) on a conveyor belt moving with a constant speed 𝑣0 to the right, as shown in Fig. 1. The friction between the belt and pipe will cause the pipe to move to the right, as well as to rotate and, eventually, to roll without slip. Determine the velocity of 𝐺 and the angular velocity of the pipe when it rolls without slip.

SOLUTION Because the pipe 𝐴 is stationary when it is placed on the conveyor belt, 𝐴 must slip relative to the belt until rolling without slip begins. Modeling 𝐴 as a uniform rigid body, until rolling without slip begins, the FBD of 𝐴 is that shown in Fig. 2. Assuming that 𝐺 does not move in the vertical direction, the motion of the body is determined by the impulse provided by the friction force, which is the only force acting in the horizontal direction. We can then solve the problem by applying the linear and angular impulse-momentum principles, along with the kinematic relations that describe rolling without slip over a moving surface. Road Map & Modeling

𝑚𝑔

𝐴

𝐺 𝚥̂ 𝑄 𝚤̂

𝐹

𝑁

Figure 2 FBD of the pipe section at an instant between the time at which the pipe is placed over the conveyor and the time at which the pipe starts rolling without slip.

ISTUDY

Governing Equations

Let 𝑡1 denote the time at which 𝐴 is placed on the conveyor and 𝑡2 denote the time at which the pipe starts rolling without slip. The impulse-momentum principle in the 𝑥 direction is

Balance Principles

( ) 𝑚 𝑣𝐺𝑥 1 +

∫𝑡

𝑡2

( ) 𝐹 𝑑𝑡 = 𝑚 𝑣𝐺𝑥 2 .

(1)

1

Choosing the mass center 𝐺 as moment center, the angular impulse-momentum principle applied between 𝑡1 and 𝑡2 is ⃗ + ℎ 𝐺1

𝑡2 (

) ⃗ , 𝑟⃗𝑄∕𝐺 × 𝐹 𝚤̂ 𝑑𝑡 = ℎ 𝐺2

∫𝑡

(2)

1

where, using Eq. (18.40) on p. 1244 and modeling the pipe section as a thin ring, 2 ⃗ =𝐼 𝜔 ̂ ⃗ ̂ ⃗ 𝐴2 = 𝑚𝑟2 𝜔𝐴2 𝑘. ℎ 𝐺1 𝐺 ⃗ 𝐴1 = 𝑚𝑟 𝜔𝐴1 𝑘 and ℎ𝐺2 = 𝐼𝐺 𝜔

Force Laws

All forces are accounted for on the FBD.

Kinematic Equations

Helpful Information Is the friction force 𝑭 constant? In the integral in Eq. (7), we left the friction force 𝐹 inside the integral because we did not know 𝐹 as a function of time. The important point to understand here is that the final solution to this problem does not require that we know 𝐹 as a function of time. With this said, starting from the system’s FBD and what we learned in Chapter 17, we can show that in this problem 𝐹 is constant for 𝑡1 < 𝑡 < 𝑡2 , and it then becomes equal to zero as soon as the pipe section starts rolling without slip.

(3)

At time 𝑡1 , 𝐴 is stationary, so we have ( ) 𝑣𝐺𝑥 1 = 0 and 𝜔𝐴1 = 0.

(4)

At time 𝑡2 , 𝐴 rolls without slip over the moving belt, which means that 𝑣⃗𝑄2 = 𝑣0 𝚤̂. For 𝐺 we have ( ) 𝑣⃗𝐺2 = 𝑣⃗𝑄2 + 𝜔𝐴2 𝑘̂ × 𝑟⃗𝐺∕𝑄 ⇒ 𝑣⃗𝐺2 = 𝑣0 − 𝑟𝜔𝐴2 𝚤̂. (5) Computation

After expanding the cross-product, the integrand in the second term of Eq. (2) can be written as follows: ̂ 𝑟⃗𝑄∕𝐺 × 𝐹 𝚤̂ = −𝑟 𝚥̂ × 𝐹 𝚤̂ = 𝐹 𝑟 𝑘.

(6)

Substituting Eqs. (3), the second of Eqs. (4), and Eq. (6) into Eq. (2), we have 𝑟

∫𝑡

𝑡2

𝐹 𝑑𝑡 = 𝑚𝑟2 𝜔𝐴2 ,

1

where we have pulled 𝑟 outside the integral because it is constant.

(7)

ISTUDY

Section 18.2

Momentum Methods for Rigid Bodies

Substituting the first of Eqs. (4) and the last of Eqs. (5) into Eq. (1), we have ∫𝑡

𝑡2

) ( 𝐹 𝑑𝑡 = 𝑚 𝑣0 − 𝑟𝜔𝐴2 .

(8)

1

Substituting Eq. (8) into Eq. (7), we obtain ( ) 𝑚𝑟 𝑣0 − 𝑟𝜔𝐴2 = 𝑚𝑟2 𝜔𝐴2

⇒

𝜔𝐴2 =

𝑣0 2𝑟

.

(9)

Substituting 𝜔𝐴2 from Eq. (9) into the last of Eqs. (5), we have 𝑣⃗𝐺2 = 21 𝑣0 𝚤̂.

(10)

Discussion & Verification The result we obtained is dimensionally correct and consistent with the FBD in that friction will cause the pipe section to move to the right and rotate

counterclockwise. A Closer Look The problem’s solution does not depend on the mass of the object, only on its shape. That is, we would have obtained a different result had we modeled the pipe section as, say, a cylinder. Note that we could have obtained the solution by enforcing the conservation of angular momentum about point 𝑄 without invoking the linear impulse-momentum principle. To see this, referring to the system FBD in Fig. 2, observe that the moment of the external forces about 𝑄 is equal to zero. Normally this observation does not help much since 𝑄 is neither a fixed point nor the system’s center of mass. However, referring to the list preceding Eq. (18.41) on p. 1244, point 𝑄 does move parallel to the center of mass 𝐺. ⃗ Eq. (18.41) implies that ℎ ⃗ = 0, ⃗ =ℎ ⃗ . Furthermore, since the pipe Therefore, since 𝑀 𝑄 𝑄1 𝑄2 ⃗ ⃗ = 0. section was stationary when it was placed on the conveyor belt, we must have ℎ 𝑄1

This fact, along with Eq. (18.40) on p. 1244, yields 𝐼𝐺 𝜔𝐴2 𝑘̂ + 𝑟⃗𝐺∕𝑄 × 𝑚𝑣⃗𝐺2 = 0⃗

⇒

( ) 𝑚𝑟2 𝜔𝐴2 − 𝑚𝑟 𝑣𝐺𝑥 2 = 0.

(11)

The result in Eq. (11), along with the rolling-without-slip condition in Eq. (5), yields the same solution we derived in Eqs. (9) and (10).

1249

1250

Chapter 18

Energy and Momentum Methods for Rigid Bodies

E X A M P L E 18.9

A Spinning Skater: Conservation of Angular Momentum

Jill Braaten

Figure 1 Three snapshots of a forward spin. By extending and retracting her arms and leg, the skater controls her spin rate.

The skater in Fig. 1 begins to spin with her arms completely stretched out and then brings her arms close to her body to increase her spin rate. In Example 15.11 on p. 989, we modeled the skater using a single particle. Here, we revisit the problem by modeling the skater as a system of rigid bodies, as shown in Fig. 2. Except for her arms,∗ her body is modeled as a cylinder of radius 𝑟𝑏 = 0.55 f t, weight 𝑊𝑏 = 102 lb, and radius of gyration 𝑘𝐺 = 0.3 f t, where 𝐺 is her body’s center of mass. Each arm has weight 𝑊𝑎 = 7.4 lb and length 𝓁 = 2.2 f t and is divided into an upper arm and a forearm. The upper arm and forearm are each modeled as a uniform thin rod weighing 𝑊𝑎 ∕2 and with length 𝓁∕2. Assuming that the skater starts spinning with a rate 𝜔0 = 60 rpm, as shown, and that her arms are stretched out, determine her spin rate (a) if her upper arms are kept stretched out and her forearms are folded so as to overlap with her upper arms; and (b) if her arms are placed vertically downward next to her body.

SOLUTION

𝜔𝑠

Neglecting friction between the skater and ice, the skater’s FBD for 𝑡1 ≤ 𝑡 ≤ 𝑡2 is shown in Fig. 3, where 𝑡1 is the time at which the spin begins and 𝑡2 is the time at which one of the positions corresponding to (a) or (b) is achieved. None of the external forces in the FBD contributes to a moment about the 𝑧 axis. Assuming that the spin axis coincides with the 𝑧 axis for 𝑡1 ≤ 𝑡 ≤ 𝑡2 , the condition 𝑀𝑧 = 0 causes the angular momentum about this axis to be conserved. Since we need to find only one scalar unknown, namely, the angular velocity at 𝑡2 , satisfying this conservation statement will lead us to the solution. Road Map & Modeling

𝓁∕2

𝐺

𝓁∕2

𝑟𝑏

Figure 2 Model of the skater as a system of rigid bodies.

Governing Equations Balance Principles

Referring to Fig. 3, since the skater’s body is in a fixed-axis rotation about the 𝑧 axis with 𝑀𝑂𝑧 = 0, Eq. (18.45) on p. 1245 implies 𝐼𝑂1 𝜔𝑠1 = 𝐼𝑂2 𝜔𝑠2 ,

𝐺 𝑘̂

𝓁∕2 𝑚𝑏 𝑔

𝚤̂

𝑟𝑏 𝑂

𝑥 𝑁

Figure 3 FBD of a spinning skater.

ISTUDY

𝓁∕2

(1)

where 𝜔𝑠 is the angular velocity of the skater and 𝐼𝑂 is the mass moment of inertia of the skater about 𝑂 (or any other point along the 𝑧 axis). In this problem, the mass moment of inertia changes as the skater moves her arms! Referring to Fig. 2, when completely outstretched, the upper arm and forearm can be viewed as forming a single uniform thin rod of mass 𝑚𝑎 and length 𝓁 with mass center 𝑟𝑏 + 𝓁∕2 away from the 𝑧 axis. Therefore, applying the parallel axis theorem, we have [ )2 ] ( 1 𝐼𝑂1 = 𝑚𝑏 𝑘2𝐺 + 2 12 (2) 𝑚𝑎 𝓁 2 + 𝑚𝑎 𝑟𝑏 + 12 𝓁 , ⏟⏟⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ body

each arm

where 𝑚𝑏 is the mass of her body. Equation (2) can be simplified as ) ( 𝐼𝑂1 = 𝑚𝑏 𝑘2𝐺 + 2𝑚𝑎 𝑟2𝑏 + 𝑟𝑏 𝓁 + 31 𝓁 2 .

(3)

Referring to Fig. 4(a), for case (a), when the skater folds her forearms horizontally, using the parallel axis theorem again, at time 𝑡2 we have [ ( )2 ( )2 ] 𝑚𝑎 ( ) 𝓁 1 𝑚𝑎 𝓁 2 , (4) 𝑟𝑏 + + 𝐼𝑂2 out = 𝑚𝑏 𝑘𝐺 + 4 12 2 2 2 4 ⏟⏟⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ body each upper arm and each forearm ∗ We use arm according to its common meaning, i.e., everything from the shoulder to the tip of the fingers.

However, in medical anatomy, an arm is only what lies between shoulder and elbow.

ISTUDY

Section 18.2

where the subscript out indicates that the arms are partially stretched out. Equation (4) can be simplified to ( ) ( ) 𝓁2 𝐼𝑂2 out = 𝑚𝑏 𝑘2𝐺 + 𝑚𝑎 2𝑟2𝑏 + 𝑟𝑏 𝓁 + . (5) 6 When her arms are folded completely downward, as in Fig. 4(b), we have ( ) 𝐼𝑂2 in = 𝑚𝑏 𝑘2𝐺 + 2𝑚𝑎 𝑟2𝑏 , ⏟⏟⏟ ⏟⏟⏟ body

(6)

(a)

𝐺

arms partially out 𝚤̂

𝑟𝑏

each arm

𝑂

𝑥 𝑧

All forces are accounted for on the FBD.

Kinematic Equations

𝓁∕2

𝑘̂

where the subscript in indicates that the arms are completely downward. Force Laws

1251

Momentum Methods for Rigid Bodies

We know that the skater is initially spinning at 𝜔0 and so 𝜔𝑠1 = 𝜔0 .

(b)

(7)

Substituting Eq. (7) into Eq. (1) and solving for 𝜔𝑠2 , we have 𝜔𝑠2 = (𝐼𝑂1 ∕𝐼𝑂2 )𝜔0 , and therefore for the two cases considered we have

𝐺

Computation

( 𝑚𝑏 𝑘2𝐺 + 2𝑚𝑎 𝑟2𝑏 + 𝑟𝑏 𝓁 + ( ) 𝜔𝑠2 out = ( 𝑚𝑏 𝑘2𝐺 + 𝑚𝑎 2𝑟2𝑏 + 𝑟𝑏 𝓁 + ( 𝑚𝑏 𝑘2𝐺 + 2𝑚𝑎 𝑟2𝑏 + 𝑟𝑏 𝓁 + ( ) 𝜔𝑠2 in = 𝑚𝑏 𝑘2𝐺 + 2𝑚𝑎 𝑟2𝑏

𝓁2 3 𝓁2 6 𝓁2 3

) ) 𝜔0 = 116.4 rpm

and

(8)

) 𝜔0 = 243.6 rpm,

(9)

arms in 𝑘̂ 𝚤̂

𝑟𝑏

Figure 4 Configuration of the skater’s arms for the two cases we are considering.

where we have used Eqs. (3), (5), and (6) and where we have plugged in the given data to obtain the numerical results. Discussion & Verification

In each of Eqs. (8) and (9) the final spin rate is greater than the initial spin rate 𝜔0 , as expected. In addition, the result in Eq. (9) is larger than that in Eq. (8); i.e., the increase in spin rate for the case where the skater’s arms are completely against her body is larger than that when only the forearms are folded, again as expected. Finally, the spin rates that we have obtained are certainly within reach of professional skaters (see the Interesting Fact in the margin). A Closer Look The particle solution to this problem was given in Eq. (5) on p. 990 in Example 15.11, which, in terms of the current variables, is ) ( 𝜔𝑠2 = 𝑟21 ∕𝑟22 𝜔0 , (10)

where 𝑟1 and 𝑟2 are the distances between the arms and the spin axis at times 𝑡1 and 𝑡2 , respectively. The simplicity of Eq. (10) is due to the fact that we ignored the dimensions of the body and its parts and only considered the mass of the arms. However, it is important to understand that the particle and rigid body solutions are not that different in spirit. We can rewrite Eq. (10) as ( ) 𝐼𝑂1 𝑝 2𝑚𝑎 𝑟21 2𝑚𝑎 𝑟21 ( ) ( ) 𝜔 = 𝜔 = (11) 𝜔𝑠2 = ( ) 𝜔0 ⇒ 𝐼𝑂1 𝑝 𝜔0 = 𝐼𝑂2 𝑝 𝜔𝑠2 , 0 0 2𝑚𝑎 𝑟2 2𝑚𝑎 𝑟2 𝐼𝑂2 2

2

𝑝

where (𝐼𝑂1 )𝑝 and (𝐼𝑂2 )𝑝 represent the mass moments of inertia relative to the spin axis for the particle model. Recalling that 𝜔0 = 𝜔𝑠1 and comparing the last of Eqs. (11) with Eq. (1), we see that the only difference between the two models is how the mass moments of inertia are calculated.

Interesting Fact How fast can a skater spin? Russian-born Natalia Kanounnikova established a world record on March 27, 2006, by spinning at 308 rpm while skating at Rockefeller Center in New York City.

1252

Chapter 18

Energy and Momentum Methods for Rigid Bodies

E X A M P L E 18.10

Space Shuttle Docking with ISS: Conservation of Momentum Figure 1 shows the Space Shuttle docked with the International Space Station (ISS). To explore what docking entails, we consider the simplified 2D scenario in Fig. 2, in which the Shuttle 𝐴 docks to the ISS 𝐵 with a speed 𝑣0 = 0.03 m∕s. We wish to determine the velocities of 𝐴 and 𝐵 immediately after they dock, assuming that no spacecraft attitude controls are exerted on 𝐴 or 𝐵 and assuming that, after docking, 𝐴 and 𝐵 form a single rigid body. Referring to Fig. 2, we will use the following data: the mass and mass moment of inertia of 𝐴 are 𝑚𝐴 = 120 ×103 kg and 𝐼𝐶 = 14×106 kg⋅m2 , respectively; the mass and mass moment of inertia of 𝐵 are 𝑚𝐵 = 180 × 103 kg and 𝐼𝐷 = 34 × 106 kg⋅m2 , respectively; the dimensions are 𝓁 = 24 m and ℎ = 8 m. Note that we are not assuming that 𝐴 and 𝐵 are rectangles in Fig. 2. Since a body’s mass and mass moment of inertia completely describe it, these rectangles are used only to describe the relative position of points 𝐶 and 𝐷.

European Space Agency

Figure 1 Artist’s rendition of the Space Shuttle Discovery docked to the International Space Station.

𝐷 ℎ

docking location 𝐶

𝑣0

Figure 2 Relative positions of the mass centers 𝐶 and 𝐷 of 𝐴 and 𝐵, respectively, at docking. The rectangles shown are not physical models, they are used only to describe the relative position of 𝐶 and 𝐷.

SOLUTION Road Map & Modeling Since we are assuming that 𝐴 and 𝐵 join to form a single rigid body, we can use rigid body kinematics to describe the motion of the 𝐴-𝐵-body via the motion of only two points, namely, 𝐶 and 𝐷, provided we know their relative position, which is given in Fig. 2. We will neglect all gravitational effects and assume that 𝐵 is initially at rest relative to an inertial frame. Since we want the motion immediately after docking, we can assume that the positions of 𝐴 and 𝐵 are still the same as those at the time of docking. This allows us to make no distinction between the positions of the system immediately before and after docking. Finally, recalling that no attitude controls are used, the FBD of the system right before and right after docking is that in Fig. 3, so the system’s linear and angular momenta are conserved. These conservation statements give three scalar equations which, when combined with the assumption that 𝐴 and 𝐵 form a single rigid body, are sufficient to solve the problem. Governing Equations Balance Principles

𝓁

𝐵

𝑂 𝐷

𝑥 ℎ

𝐶

𝚥̂ 𝚤̂

𝐴

Figure 3 FBD of 𝐴 and 𝐵 right before and after docking. The coordinate system shown is fixed in space, and 𝐴 and 𝐵 can move relative to it.

ISTUDY

In components, the conservation of total linear momentum is ( ) ( ) ( ) ( ) 𝑚𝐴 𝑣𝐶𝑥 1 + 𝑚𝐵 𝑣𝐷𝑥 1 = 𝑚𝐴 𝑣𝐶𝑥 2 + 𝑚𝐵 𝑣𝐷𝑥 2 , (1) ( ) ( ) ( ) ( ) (2) 𝑚𝐴 𝑣𝐶𝑦 1 + 𝑚𝐵 𝑣𝐷𝑦 1 = 𝑚𝐴 𝑣𝐶𝑦 2 + 𝑚𝐵 𝑣𝐷𝑦 2 ,

where the subscripts 1 and 2 indicate right before and right after docking, respectively. Choosing the fixed point 𝑂 as the moment center, the conservation of total angular momentum is ( ) ( ) ( ) ( ) ⃗ ⃗ ⃗ ⃗ ℎ (3) 𝑂 𝐴1 + ℎ𝑂 𝐵1 = ℎ𝑂 𝐴2 + ℎ𝑂 𝐵2 , where, because 𝐴 and 𝐵 do not move significantly between times 𝑡1 and 𝑡2 , we have ( ) ( ) ⃗ ⃗ ⃗ 𝐵1 + 𝑟⃗𝐷∕𝑂 × 𝑚𝐵 𝑣⃗𝐷1 , (4) ⃗ 𝐴1 + 𝑟⃗𝐶∕𝑂 × 𝑚𝐴 𝑣⃗𝐶1 , ℎ ℎ 𝑂 𝐵1 = 𝐼𝐷 𝜔 𝑂 𝐴1 = 𝐼𝐶 𝜔 ( ) ( ) ⃗ ⃗ ℎ ⃗ 𝐴2 + 𝑟⃗𝐶∕𝑂 × 𝑚𝐴 𝑣⃗𝐶2 , ℎ ⃗ 𝐵2 + 𝑟⃗𝐷∕𝑂 × 𝑚𝐵 𝑣⃗𝐷2 . (5) 𝑂 𝐴2 = 𝐼𝐶 𝜔 𝑂 𝐵2 = 𝐼𝐷 𝜔 Force Laws

All forces are accounted for on the FBD.

Kinematic Equations

Before docking, ( ) ( ) 𝑣𝐶𝑥 1 = −𝑣0 , 𝑣𝐶𝑦 1 = 0, 𝜔𝐴1 = 0, ( ( ) ) 𝑣𝐷𝑥 1 = 0, 𝑣𝐷𝑦 1 = 0, 𝜔𝐵1 = 0.

(6) (7)

After docking, 𝐴 and 𝐵 form a single rigid body, so we have 𝜔𝐴2 = 𝜔𝐵2 = 𝜔𝐴𝐵

and

𝑣⃗𝐶2 = 𝑣⃗𝐷2 + 𝜔𝐴𝐵 𝑘̂ × 𝑟⃗𝐶∕𝐷 ,

(8)

ISTUDY

Section 18.2

1253

Momentum Methods for Rigid Bodies

where 𝜔𝐴𝐵 is the common angular velocity of 𝐴 and 𝐵 immediately after docking. The relative position vectors in Eqs. (4), (5), and (8) are given by 𝑟⃗𝐶∕𝑂 = −ℎ 𝚥̂, Computation

𝑟⃗𝐷∕𝑂 = −𝓁 𝚤̂,

and 𝑟⃗𝐶∕𝐷 = 𝓁 𝚤̂ − ℎ 𝚥̂.

(9)

Substituting the first two of Eqs. (6) and (7) into Eqs. (1) and (2), we have ( ) ( ) −𝑚𝐴 𝑣0 = 𝑚𝐴 𝑣𝐶𝑥 2 + 𝑚𝐵 𝑣𝐷𝑥 2 , (10) ( ) ( ) (11) 0 = 𝑚𝐴 𝑣𝐶𝑦 2 + 𝑚𝐵 𝑣𝐷𝑦 2 .

Referring to Fig. 4 and substituting Eqs. (6), (7), the first of Eqs. (8), and the first two of Eqs. (9) into Eqs. (4) and (5), we have ( ) [ ( ) ] ( ) ̂ ̂ ⃗ ⃗ (12) ℎ ℎ 𝑂 𝐴2 = 𝐼𝐶 𝜔𝐴𝐵 + 𝑚𝐴 ℎ 𝑣𝐶𝑥 2 𝑘, 𝑂 𝐴1 = −𝑚𝐴 ℎ𝑣0 𝑘, ( ) [ ( ) ] ( ) ⃗ ⃗ ̂ ⃗ ℎ (13) ℎ 𝑂 𝐵1 = 0, 𝑂 𝐵2 = 𝐼𝐷 𝜔𝐴𝐵 − 𝑚𝐴 𝓁 𝑣𝐷𝑦 2 𝑘. Substituting Eqs. (12) and (13) into Eq. (3), we obtain ( ) ( ) −𝑚𝐴 ℎ𝑣0 = (𝐼𝐶 + 𝐼𝐷 )𝜔𝐴𝐵 + 𝑚𝐴 ℎ 𝑣𝐶𝑥 2 − 𝑚𝐵 𝓁 𝑣𝐷𝑦 2 .

𝑦 𝓁

𝐵

𝑂 𝐷

𝑥 ℎ

𝐶

𝚥̂ 𝚤̂

𝜔𝐴𝐵

(14)

Equation (14) is in scalar form because the only nonzero component of Eq. (3) is in the 𝑧 direction. Finally, substituting the last of Eqs. (9) into the second of Eqs. (8), expanding the cross-product, and expressing the result in components, we have ( ) ( ) ) ( ( ) (15) 𝑣𝐶𝑥 2 = 𝑣𝐷𝑥 2 + 𝜔𝐴𝐵 ℎ and 𝑣𝐶𝑦 2 = 𝑣𝐷𝑦 2 + 𝜔𝐴𝐵 𝓁.

Figure 4 Sketch of the velocity components of the system right after docking.

Equations ( ) ( (10), ) (11), ) ( and)(15) form a system of five equations in the five unknowns ( (14), 𝑣𝐶𝑥 2 , 𝑣𝐶𝑦 2 , 𝑣𝐷𝑥 2 , 𝑣𝐷𝑦 2 , and 𝜔𝐴𝐵 . The solution to these five equations is found to be ( ) 𝑚 𝑚 −𝑚𝐴 𝐼 + 𝑚𝐵 ℎ2 + 𝐴𝑚 𝐵 𝓁 2 𝑣0 ( ) 𝑣𝐶𝑥 2 = = −0.01288 m∕s, (16) 𝑚𝐴 𝑚𝐵 𝑑 2 + 𝑚𝐼 𝑚 𝑚

−𝑚𝐵 𝐴𝑚 𝐵 ℎ𝓁𝑣0 ( ) 𝑣𝐶𝑦 2 = = −0.002645 m∕s, 𝑚𝐴 𝑚𝐵 𝑑 2 + 𝑚𝐼 ) ( 𝑚 𝑚 −𝑚𝐴 𝐼 + 𝐴𝑚 𝐵 𝓁 2 𝑣0 ) ( 𝑣𝐷𝑥 2 = = −0.01141 m∕s, 𝑚𝐴 𝑚𝐵 𝑑 2 + 𝑚𝐼

(17) (18)

𝑚 𝑚

𝑚𝐴 𝐴𝑚 𝐵 ℎ𝓁𝑣0 ( ) 𝑣𝐷𝑦 2 = = 0.001763 m∕s, 𝑚𝐴 𝑚𝐵 𝑑 2 + 𝑚𝐼 −𝑚𝐴 𝑚𝐵 ℎ𝑣0 = −0.0001837 rad∕s, 𝜔𝐴𝐵 = 𝑚𝐴 𝑚𝐵 𝑑 2 + 𝑚𝐼 √ where 𝑚 = 𝑚𝐴 + 𝑚𝐵 , 𝑑 = ℎ2 + 𝓁 2 , and 𝐼 = 𝐼𝐶 + 𝐼𝐷 .

(19) (20)

Discussion & Verification The results appear reasonable since the computed velocities are comparable to 𝑣0 . In addition, the signs appear correct in that, after docking, we expect both 𝐴 and 𝐵 to move to the left and the 𝐴𝐵-body to rotate clockwise. This rotation then causes 𝐶 and 𝐷 to move slightly downward and upward, respectively. A Closer Look We assumed that 𝐴 and 𝐵 form a rigid body after docking because we did not know the exact position of the docking location. A better assumption is that 𝐴 and 𝐵 become pinned to each other after docking. In this way, we better capture the effect of the local flexibility of the docking location. This possibility is considered in Prob. 18.134 on p. 1281.

Interesting Fact Mass of the ISS. The assembly of the ISS began in 1998. Eventually the ISS mass will be about 472,000 kg. The mass and moment of inertia given in the problem are rough estimates based on stage 9A.1 of the ISS assembly process (see J. A. Wojtowicz, “Dynamic Properties of the International Space Station throughout the Assembly Process,” Report N. A282843, Air Force Institute of Technology, Wright-Patterson AFB, Ohio, 1998).

1254

Energy and Momentum Methods for Rigid Bodies

Chapter 18

Problems 𝜔0

Problem 18.75

𝜔0 𝑂

𝐶

𝐷 𝐵

𝐴 Figure P18.75

Problem 18.76

𝑣0

𝐺 𝐵

Disks 𝐴 and 𝐵 have identical masses and mass moments of inertia about their respective mass centers. Point 𝐶 is both the geometric center and center of mass of disk 𝐴. Points 𝑂 and 𝐷 are the geometric center and center of mass of disk 𝐵, respectively. If, at the instant shown, the two disks are rotating about their centers with the same angular velocity 𝜔0 , |(⃗ ) | |(⃗ ) | determine which of the following statements is true and why: (a) | ℎ 𝐶 𝐴 || < || ℎ𝑂 𝐵 ||, | |(⃗ ) | |(⃗ ) | |( ⃗ ) | |( ⃗ ) | (b) | ℎ |=| ℎ |, (c) | ℎ |>| ℎ |. | 𝐶 𝐴| | 𝑂 𝐵 | | 𝐶 𝐴| | 𝑂 𝐵 |

𝐸

𝑃

Body 𝐵 has mass 𝑚 and mass moment of inertia 𝐼𝐺 , where 𝐺 is the mass center of 𝐵. If 𝐵 is translating as shown, determine which of the following statements is true and why: |(⃗ ) | |(⃗ ) | |(⃗ ) | |(⃗ ) | |( ⃗ ) | |( ⃗ ) | (a) | ℎ || ℎ |. | 𝐸 𝐵| | 𝑃 𝐵| | 𝐸 𝐵| | 𝑃 𝐵| | 𝐸 𝐵| | 𝑃 𝐵|

Figure P18.76

Problem 18.77 The rear-wheel-drive car can go from rest to 60 mph in Δ𝑡 = 8 s. Assume that the wheels are all identical and that their geometric centers coincide with their mass centers. Let 𝑀rear be the average moment applied to one of the rear wheels during Δ𝑡 and computed relative to the wheel’s center. Finally, let 𝑀front be the average moment applied to one of the front wheels during Δ𝑡 and computed relative to the wheel’s center. Modeling the wheels as rigid bodies, determine which of the following statements is true and why: | | | | | | | | | | | | (a) |𝑀rear | < |𝑀front |, (b) |𝑀rear | = |𝑀front |, (c) |𝑀rear | > |𝑀front |. | | | | | | | | | | | |

Problem 18.78

Gary L. Gray

The rear-wheel-drive car can go from rest to 60 mph in Δ𝑡 = 8 s. Assume that its wheels are identical, with their geometric centers coinciding with their mass centers. Let 𝐹avg be the average friction force acting on the system during Δ𝑡 due to contact with the ground. Modeling the car and the wheels as rigid bodies, does the value of 𝐹avg change whether or not we account for the rotational inertia of the wheels? Why?

Figure P18.77 and P18.78

𝚥̂

𝑅 𝚤̂

𝑣𝐺 𝐺

Problem 18.79

𝑂

A uniform disk of mass 𝑚 and radius 𝑅 rolls to the right without slip, such that the speed of the center of mass is 𝑣𝐺 . Provide an expression for the linear momentum of the disk in terms of the given quantities. In addition, provide an expression for the angular momentum of the disk relative to 𝑂, the point of contact with the ground. Express your answers using the component system shown.

𝐺

Problem 18.80

𝑂

At the instant shown, the eccentric wheel with center at 𝑂 and center of mass at 𝐺 is rotating counterclockwise without slip with an angular speed 𝜔𝑤 = 10 rad∕s. The weight of the wheel is 𝑊 = 90 lb. In addition, let 𝑅 = 2 f t, ℎ = 1 f t, and the radius of gyration 𝑘𝐺 = 1.45 f t. At the instant shown, determine the linear momentum of the disk and the angular momentum about 𝐶, the point of contact with the ground. Express your answers in the component system shown.

Figure P18.79

𝜔𝑤 𝚥̂ 𝚤̂

𝐶 Figure P18.80

ISTUDY

ISTUDY

Section 18.2

1255

Momentum Methods for Rigid Bodies

Problem 18.81 The top of the Space Needle in Seattle, Washington, hosts a revolving restaurant that goes through one full revolution every 47 min under the action of a motor with a power output of 1.5 hp. The portion of the restaurant that rotates is a ring-shaped turntable with internal and external radii 𝑟𝑖 = 33.3 f t and 𝑟𝑜 = 47.3 f t, respectively, and approximate weight 𝑊 = 125 tons (1 ton = 2000 lb). Use the given values of power output and angular speed to estimate the torque 𝑀 that the engine provides. Then, assuming that the motor can provide a constant torque equal to 𝑀, neglecting all friction, and modeling the turntable as a uniform body, determine the time 𝑡𝑠 that it takes to spin up the revolving restaurant from rest to its working angular speed.

𝑀

𝑂 𝑟 𝑖 𝑟𝑜

Problem 18.82 A rotor, spinning freely about the fixed point 𝑂, consists of a thin uniform bar 𝐴𝐵 that functions as a hub and two identical blades pinned at 𝐴 and 𝐵, respectively. The dimensions of the system are: 𝑑 = 0.5 m, 𝓁 = 5 m, and 𝑤 = 0.3 m. The bar 𝐴𝐵 has a mass 𝑚𝐴𝐵 = 30 kg, and each of the blades has a mass 𝑚𝑏 = 20 kg. Each blade can be modeled as a uniform thin plate. The angular speed is 𝜔𝑟 = 100 rpm when the angle 𝜃 is equal to 90◦ . At some point, an internal mechanism causes the blades to change orientation relative to 𝐴𝐵 in such a way that 𝜃 becomes constant and equal to 180◦ . Neglecting aerodynamic forces and friction at the bearings at 𝑂, determine the angular speed of the rotor after 𝜃 = 180◦ .

Golden Gate Images/Alamy Stock Photo

Figure P18.81

𝐴 𝑑 𝑑

𝜃 𝜃 𝐵

𝑂 𝑤∕2

Figure P18.82

Problems 18.83 and 18.84 A uniform disk of mass 𝑚 = 20 kg and radius 𝑅 = 0.75 m is being pulled to the left with a constant horizontal force 𝑃 by the cord wrapped around it. Assume that the disk starts from rest and that it rolls without slip. 𝑅

If 𝑃 = 30 N, apply the impulse-momentum principles to determine the angular speed of the disk after 4 s.

Problem 18.83

Problem 18.84 Apply the impulse-momentum principles to determine 𝑃 if the center of the disk achieves a speed 𝑣𝐺 = 5 m∕s after 3 s.

𝐺

Figure P18.83 and P18.84

Problem 18.85 Moving on a straight and horizontal stretch of road, the rear-wheel-drive car shown can go from rest to 60 mph in Δ𝑡 = 8 s. The car weighs 2570 lb (the weight includes the wheels). Each wheel has diameter 𝑑 = 24.3 in., mass moment of inertia relative to its own center of mass 𝐼𝐺 = 0.989 slug⋅f t 2 , and the center of mass of each wheel coincides with its geometric center. Determine the average friction force 𝐹avg acting on the car during Δ𝑡. In addition, if the wheels roll without slip, for each wheel, determine the average moment 𝑀avg , computed relative to the wheel’s center, that is applied to the wheel during Δ𝑡.

Gary L. Gray

Figure P18.85

1256

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Problem 18.86 A spool with radius 𝑅 = 3 f t is released from rest on an incline with 𝜃 = 30◦ , and its center, which coincides with its center of mass, is observed to reach a speed of 8 f t∕s two seconds after release. If the spool rolls without slip, use the impulse-momentum principles to determine the radius of gyration of the spool 𝑘𝐺 .

𝐺

𝑅

𝜃 Figure P18.86

Problem 18.87 An eccentric wheel 𝐵 weighing 150 lb has its mass center 𝐺 at a distance 𝑑 = 4 in. from the wheel’s center 𝑂. The wheel is in the horizontal plane and is spun from rest by applying a constant torque 𝑀 = 32 f t ⋅lb. Determine the wheel’s radius of gyration 𝑘𝐺 if it takes 2 s to spin up the wheel to 140 rpm. Neglect all possible sources of friction.

𝑀 𝑑 𝑂

𝐺

𝐵

Problem 18.88 Figure P18.87

The uniform thin pin-connected bars AB, BC, and CD have masses 𝑚𝐴𝐵 = 2.3 kg, 𝑚𝐵𝐶 = 3.2 kg, and 𝑚𝐶𝐷 = 5.0 kg, respectively. Letting 𝑅 = 0.75 m, 𝐿 = 1.2 m, and 𝐻 = 1.55 m, and knowing that bar 𝐴𝐵 rotates at a constant angular velocity 𝜔𝐴𝐵 = 4 rad∕s, compute the angular momentum of bar 𝐴𝐵 about 𝐴, of bar 𝐵𝐶 about 𝐴, and bar CD about 𝐷 at the instant shown.

𝐵

𝜔𝐴𝐵

𝐶

𝐴

𝜔𝐴𝐵 𝐴 𝜙 𝐷

𝐷 Figure P18.88

Figure P18.89

𝑊 𝜔𝑊

𝑅𝑊

𝜔𝑂𝐶

Problem 18.89 The weights of the uniform thin pin-connected bars 𝐴𝐵, BC, and CD are 𝑊𝐴𝐵 = 4 lb, 𝑊𝐵𝐶 = 6.5 lb, and 𝑊𝐶𝐷 = 10 lb, respectively. Letting 𝜙 = 47◦ , 𝑅 = 2 f t, 𝐿 = 3.5 f t, and 𝐻 = 4.5 f t, and knowing that bar 𝐴𝐵 rotates at a constant angular velocity 𝜔𝐴𝐵 = 4 rad∕s, compute the magnitude of the linear momentum of the system at the instant shown.

𝑅𝑆 𝑆

Problem 18.90

Figure P18.90

A uniform disk 𝑊 of radius 𝑅𝑊 = 7 mm and mass 𝑚𝑊 = 0.15 kg is connected to point 𝑂 via the rotating arm 𝑂𝐶. Disk 𝑊 also rolls without slip over the stationary cylinder 𝑆 of radius 𝑅𝑆 = 15 mm. Assuming that 𝜔𝑊 = 25 rad∕s, determine the angular momentum of 𝑊 about its own center of mass 𝐶, as well as about point 𝑂.

𝑀

Problem 18.91

𝐺

𝐵 NASA

Figure P18.91

ISTUDY

A rotor 𝐵 with center of mass 𝐺, weight 𝑊 = 3000 lb, and radius of gyration 𝑘𝐺 = 15 f t is spinning with an angular speed 𝜔𝐵 = 1200 rpm when a braking system is applied to it, providing a time-dependent torque 𝑀 = 𝑀0 (1 + 𝑐𝑡), with 𝑀0 = 3000 f t ⋅lb and 𝑐 = 0.01 s−1 . If 𝐺 is also the geometric center of the rotor and is a fixed point, determine the time 𝑡𝑠 that it takes to stop the rotor.

ISTUDY

Section 18.2

1257

Momentum Methods for Rigid Bodies

Problem 18.92

𝐴

The uniform bar 𝐴𝐵 has length 𝐿 = 4.5 f t and weight 𝑊𝐴𝐵 = 14 lb. At the instant shown, 𝜃 = 67◦ and 𝑣𝐴 = 5.8 f t∕s. Determine the magnitude of the linear momentum of 𝐴𝐵, as well as the angular momentum of 𝐴𝐵 about its mass center 𝐺 at the instant shown.

𝑣𝐴

𝐿 𝐺

Problem 18.93 A uniform pipe section 𝐴 of radius 𝑟, mass center 𝐺, and mass 𝑚 is gently placed (i.e., with zero velocity) on a conveyor belt moving with a constant speed 𝑣0 to the right. Friction between the belt and pipe causes the pipe to move to the right and eventually to roll without slip. If 𝜇𝑘 is the coefficient of kinetic friction between the pipe and the conveyor belt, find an expression for 𝑡𝑟 , the time it takes for 𝐴 to start rolling without slip. Hint: Using the methods of Chapter 17, we can show that the force between the pipe section and the belt is constant. 2𝑟

𝐺

𝜃 𝐵 Figure P18.92

𝐴 𝑣0

Figure P18.93 𝑀

Problems 18.94 and 18.95

𝐷 𝐵

An automobile wheel test rig consists of a uniform disk 𝐴, of mass 𝑚𝐴 = 5000 kg and radius 𝑟𝐴 = 1.5 m, that can rotate freely about its fixed center 𝐶 and over which the wheel of an automobile is made to roll. A wheel 𝐵, whose center and center of mass coincide at 𝐷, is mounted on a shaft (not shown) that holds 𝐷 fixed while it allows the wheel to rotate about 𝐷. The wheel has diameter 𝑑 = 0.62 m, mass 𝑚𝐵 = 21.5 kg, and mass moment of inertia about its mass center 𝐼𝐷 = 44 kg⋅m2 . Both 𝐴 and 𝐵 are initially at rest when 𝐵 is subject to a constant torque 𝑀 that causes 𝐵 to roll without slip on 𝐴.

𝐴 𝑟𝐴

𝐶

If 𝑀 = 1500 N⋅m, use the angular impulse-momentum principle to determine how long it takes to reach conditions simulating a car speed of 100 km∕h.

Problem 18.94

Use the angular impulse-momentum principle and determine 𝑀 if it takes 15 seconds to achieve conditions simulating a car speed of 60 km∕h.

Problem 18.95

Problems 18.96 and 18.97 A spool of mass 𝑚𝑠 = 150 kg and inner and outer radii 𝜌 = 0.8 m and 𝑅 = 1.2 m, respectively, is connected to a counterweight 𝐴 of mass 𝑚𝐴 = 50 kg by a pulley system whose cord, at one end, is wound around the inner hub of the spool. The center 𝐺 and the center of mass of the spool coincide, and the radius of gyration of the spool is 𝑘𝐺 = 1 m. The system is at rest when the counterweight is released, causing the spool to move to the right. The spool rolls without slip, and the cord unwinds from the spool without slip.

Figure P18.94 and P18.95

𝐷 𝜌

𝐶

Problem 18.96

Neglecting the inertia of the pulley system, use the impulse-momentum 𝐺 𝑅 principles to determine the angular speed of the spool 3 s after release. Hint: Use the inextensible nature of the cord to determine the relationship between the speed of the counterweight and the speed of the cord unraveling from the spool. Note that this cord speed is not the same as 𝑣𝐺 . A linear momentum principle for the counterweight as well as both linear and angular momentum principles for the spool should prove useful. Figure P18.96 and P18.97 Assume that the inertia of the cord and of pulleys 𝐵 and 𝐷 can be neglected, but model pulley 𝐶 as a uniform disk mass 𝑚𝐶 = 15 kg and radius 𝑟𝐶 = 0.3 m.

Problem 18.97

𝐵 𝐴

1258

Chapter 18

Energy and Momentum Methods for Rigid Bodies

If the cord does not slip relative to pulley 𝐶, use the impulse-momentum principles to determine the angular speed of the spool 3 s after release.

Problem 18.98 𝑒 𝜔0 2𝑟0 𝐺

Figure P18.98

ISTUDY

2𝑟𝑖

𝐴

An 0.8 lb collar with center of mass at 𝐺 and a uniform cylindrical horizontal arm 𝐴 of length 𝐿 = 1 f t, radius 𝑟𝑖 = 0.022 f t, and weight 𝑊𝐴 = 1.5 lb are rotating as shown with 𝜔0 = 1.5 rad∕s while the collar’s mass center is at a distance 𝑑 = 0.44 f t from the 𝑧 axis. The vertical shaft has radius 𝑒 = 0.03 f t and negligible mass. After the cord restraining the collar is cut, the collar slides with no friction relative to the arm. Assuming that no external forces and moments are applied to the system, determine the collar’s impact speed with the end of 𝐴 if (a) the collar is modeled as a particle coinciding with its own mass center (in this case, neglect the collar’s dimensions), and (b) the collar is modeled as a uniform hollow cylinder with length 𝓁 = 0.15 f t, inner radius 𝑟𝑖 , and outer radius 𝑟𝑜 = 0.048 f t. Hint: Conservation of angular momentum in the 𝑧-direction allows us to construct an ̇ expression for 𝜃(𝑟). The absence of radial force implies 𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 = 0, which allows us to construct 𝑟̈(𝑟).

Problem 18.99 The uniform disk 𝐴, of mass 𝑚𝐴 = 1.2 kg and radius 𝑟𝐴 = 0.25 m, is mounted on a vertical shaft that can translate along the horizontal guide 𝐶. The uniform disk 𝐵, of mass 𝑚𝐵 = 0.85 kg and radius 𝑟𝐵 = 0.38 m, is mounted on a fixed vertical shaft. Both disks 𝐴 and 𝐵 can rotate about their own axes, namely, 𝓁𝐴 and 𝓁𝐵 , respectively. Disk 𝐴 is initially spun with 𝜔𝐴 = 1000 rpm and then brought into contact with 𝐵, which is initially stationary. The contact is maintained by a spring, and due to friction between 𝐴 and 𝐵, disk 𝐵 starts spinning and eventually 𝐴 and 𝐵 will stop slipping relative to one another. Neglecting any friction except at the contact between the two disks, determine the angular velocities of 𝐴 and 𝐵 when slipping stops.

𝜔𝐵

𝑟𝐵

𝑟𝐴 𝜔𝐴

𝐵 𝐵 𝐴

𝐶 𝜔𝐶

𝐴 𝐸

𝑑

Figure P18.99

Figure P18.100

Problem 18.100 The uniform disk 𝐴, of mass 𝑚𝐴 = 1.2 kg and radius 𝑟𝐴 = 0.25 m, is mounted on a vertical shaft that can translate along the horizontal arm 𝐸. The uniform disk 𝐵, of mass 𝑚𝐵 = 0.85 kg and radius 𝑟𝐵 = 0.18 m, is mounted on a vertical shaft that is rigidly attached to arm 𝐸. Disk 𝐴 can rotate about axis 𝓁𝐴 , disk 𝐵 can rotate about axis 𝓁𝐵 , and the arm 𝐸, along with disk 𝐶, can rotate about the fixed axis 𝓁𝐶 . Disk 𝐶 has negligible mass and is rigidly attached to 𝐸 so that they rotate together. While keeping both 𝐵 and 𝐶 stationary, disk 𝐴 is spun to 𝜔𝐴 = 1200 rpm. Disk 𝐴 is then brought in contact with disk 𝐶 (contact is maintained by a spring), and 𝐵 and 𝐶 (and the arm 𝐸) are then allowed to freely rotate. Due to friction between 𝐴 and 𝐶, disks 𝐶 (and arm 𝐸) and 𝐵 start spinning. Eventually, 𝐴 and 𝐶 stop slipping relative to one another. Disk 𝐵 always rotates without

ISTUDY

Section 18.2

Momentum Methods for Rigid Bodies

slip over 𝐶. Let 𝑑 = 0.27 m and 𝑤 = 0.95 m. If the only elements of the system that have mass are 𝐴 and 𝐵, and if all friction in the system can be neglected except for that between 𝐴 and 𝐶 and between 𝐶 and 𝐵, determine the angular speeds of 𝐴 and 𝐶 when they stop slipping relative to one another. Hint: It is helpful to sketch this system from above; it looks a bit like the planet/sun gear configuration of some transmission problems. No external moments act on the system, so system angular momentum is conserved normal to the plane of the disks.

Problems 18.101 and 18.102 The double pulley 𝐷 has mass of 𝑚𝐷 = 15 kg, center of mass 𝐺 coinciding with its geometric center, radius of gyration 𝑘𝐺 = 10 cm, outer radius 𝑟𝑜 = 15 cm, and inner radius 𝑟𝑖 = 7.5 cm. It is connected to the pulley 𝑃 with radius 𝑅 = (𝑟𝑜 − 𝑟𝑖 )∕2 by a cord of negligible mass that unwinds from the inner and outer spools of the double pulley 𝐷. The crate 𝐶, which has a mass 𝑚𝐶 = 20 kg, is released from rest. The cord does not slip relative to the pulleys, and the inner and outer pulleys rotate as a single unit. Neglecting the mass of the pulley 𝑃 , use the impulse-momentum principles to determine the speed of the crate 4 s after release.

Problem 18.101

Assuming that the pulley 𝑃 has a mass of 1.5 kg and a radius of gyration 𝑘𝐴 = 3.5 cm, use the impulse-momentum principles to determine the speed of the crate 4 s after release.

Problem 18.102

𝐷

𝑟𝑜 𝐺

𝑃

𝑟𝑖

𝐴 𝑅

𝐶

Figure P18.101 and P18.102

Problem 18.103 Some pipe sections of radius 𝑟 and mass 𝑚 are being unloaded and placed in a row against a wall. The first of these pipe sections, 𝐴, is made to roll without slipping into a corner with an angular velocity 𝜔0 as shown. Upon touching the wall, 𝐴 does not rebound, but slips against the ground and against the wall. Modeling 𝐴 as a uniform thin ring with center at 𝐺 and letting 𝜇𝑔 and 𝜇𝑤 be the coefficients of kinetic friction of the contacts between 𝐴 and the ground and between 𝐴 and the wall, respectively, determine an expression for the angular velocity of 𝐴 as a function of time from the moment 𝐴 touches the wall until it stops. Hint: Using the methods learned in Chapter 17, we can show that the friction forces at the ground and at the wall are constant.

𝐴 2𝑟

Figure P18.103

𝜔0 𝐺

𝜇𝑘

𝑣0

𝐺 𝑟 𝑚

Figure P18.104

Problem 18.104 A 14 lb bowling ball is thrown onto a lane with a backspin 𝜔0 = 9 rad∕s and forward velocity 𝑣0 = 18 mph. Point 𝐺 is both the geometric center and the mass center of the ball. After a few seconds, the ball starts rolling without slip. Let 𝑟 = 4.25 in., and let the radius of gyration of the ball be 𝑘𝐺 = 2.6 in. If the coefficient of kinetic friction between the ball and the floor is 𝜇𝑘 = 0.1, determine the speed 𝑣𝑓 that the ball will achieve when it starts rolling without slip. In addition, determine the time 𝑡𝑟 the ball takes to achieve 𝑣𝑓. Hint: Using the methods of Chapter 17, we can show that the force between the ball and the floor is constant.

1259

1260

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Problem 18.105

𝐷 𝐶

A crate 𝐴 with weight 𝑊𝐴 = 250 lb is hanging from a rope wound around a uniform drum 𝐷 of radius 𝑟 = 1.2 f t, weight 𝑊𝐷 = 125 lb, and center 𝐶. The system is initially at rest when the restraining system holding the drum stationary fails, thus causing the drum to rotate, the rope to unwind, and, consequently, the crate to fall. Assuming that the rope does not stretch or slip relative to the drum and neglecting the inertia of the rope, determine the speed of the crate 1.5 s after the system starts to move. Hint: A linear impulse equation for the crate combined with an angular impulse equation for the drum is most useful.

𝑟

𝐴

Figure P18.105

Problem 18.106 A toy helicopter consists of a rotor 𝐴, a body 𝐵, and a small ballast 𝐶. The axis of rotation of the rotor goes through 𝐺, which is the center of mass of the body 𝐵 and ballast 𝐶. While holding the body (and ballast) fixed, the rotor is spun as shown with a given angular velocity 𝜔0 . If there is no friction between the helicopter’s body and the rotor’s shaft, will the body of the helicopter start spinning once the toy is released? 𝐴

𝜔0 𝐶

𝐺

𝐵

Figure P18.106 and P18.107

Problem 18.107 A toy helicopter consists of a rotor 𝐴 with diameter 𝑑 = 10 in. and weight 𝑊𝑟 = 0.09× 10−3 oz, a thin body 𝐵 of length 𝓁 = 12 in. and weight 𝑊𝐵 = 0.144 × 10−3 oz, and a small ballast 𝐶 placed at the front end of the body and with weight 𝑊𝐶 = 0.0723 × 10−3 oz. The ballast’s weight is such that the axis of the rotation of the rotor goes through 𝐺, which is the center of mass of the body 𝐵 and ballast 𝐶. While holding the body (and ballast) fixed, the rotor is spun as shown with 𝜔0 = 150 rpm. Neglecting aerodynamic effects, the weights of the rotor’s shaft and the body’s tail, and assuming there is friction between the helicopter’s body and the rotor’s shaft, determine the angular velocity of the body once the toy is released and the angular velocity of the rotor decreases to 120 rpm. Model the body as a uniform thin rod and the ballast as a particle. Assume that the rotor and the body remain horizontal after release.

Problem 18.108

𝐴

𝚥̂ 𝚤̂

𝐵 Figure P18.108

ISTUDY

𝐺

𝑅

𝐷

A cord, which is wrapped around the inner radius of the spool of mass 𝑚 = 35 kg, is pulled vertically at 𝐴 by a constant force 𝑃 = 120 N (the cord is pulled in such a way that it remains vertical), causing the spool to roll over the horizontal bar BD. The inner radius of the spool is 𝑅 = 0.3 m, and the center of mass of the spool is at 𝐺, which also coincides with the geometric center of the spool. The spool’s radius of gyration is 𝑘𝐺 = 0.18 m. Assuming that the spool starts from rest, that the cord’s inertia and extensibility can be neglected, and that the spool rolls without slip, determine the speed of the spool’s center 3 s after the application of the force. In addition, determine the minimum static friction coefficient for rolling without slip to be maintained during the time interval in question.

Problem 18.109 A spool has weight 𝑊 = 450 lb, outer and inner radii 𝑅 = 6 f t and 𝜌 = 4.5 f t, respectively, center of mass 𝐺 coinciding with its geometric center, and radius of gyration 𝑘𝐺 = 4.0 f t. The spool is at rest when it is pulled to the right as shown. The cable wrapped

ISTUDY

Section 18.2

Momentum Methods for Rigid Bodies

around the spool can be modeled as being inextensible and of negligible mass. Assume that the spool rolls without slip relative to both the cable and the ground. If the cable is pulled with a force 𝑃 = 125 lb, determine the speed of the center of the spool after 2 s and the minimum value of the static friction coefficient between the spool and the ground necessary to guarantee rolling without slip.

Problem 18.110

1261

𝜌 𝑅

𝐺

Figure P18.109

The wind turbine in the figure consists of three equally spaced blades that are rotating as shown about the fixed point 𝑂 with an angular velocity 𝜔0 = 30 rpm. Suppose that each 38,000 lb blade can be modeled as a narrow uniform rectangle of length 𝑏 = 182 f t, width 𝑎 = 12 f t, and negligible thickness, with one of its corners coinciding with the center of rotation 𝑂. The orientation of each blade can be controlled by rotating the blade about an axis going through the center 𝑂 that coincides with the blade’s leading edge. Neglecting aerodynamic forces and any source of friction, and assuming that the turbine is freely rotating, determine the turbine’s angular velocity 𝜔𝑓 after each blade has been rotated 90◦ about its own leading edge.

𝑏 𝑎

𝑂

𝜔0

leading edge of blade

Problem 18.111 Cars 𝐴 and 𝐵 collide as shown. Neglecting the effect of friction, what would be the angular velocity of 𝐴 and 𝐵 immediately after impact if 𝐴 and 𝐵 were to form a single rigid body as a result of the collision? In solving the problem, let 𝐶 and 𝐷 be the mass centers of 𝐴 and 𝐵, respectively, the weight of 𝐴 is 𝑊𝐴 = 3130 lb, the radius of gyration of 𝐴 is 𝑘𝐶 = 34.5 in., the speed of 𝐴 right before impact is 𝑣𝐴 = 12 mph, the weight of 𝐵 is 𝑊𝐵 = 3520 lb, the radius of gyration of 𝐵 is 𝑘𝐷 = 39.3 in., the speed of 𝐵 right before impact is 𝑣𝐵 = 15 mph, 𝑑 = 19 in., and 𝓁 = 144 in. Finally, assume that while 𝐴 and 𝐵 form a single rigid body right after impact, the mass center of the rigid body formed by 𝐴 and 𝐵 coincides with the mass center of the 𝐴-𝐵 system right before impact.

Martin Child/Getty Images

𝐵

Figure P18.110

𝐷 𝑣𝐵 𝐶

𝑑 𝑣𝐴

𝐴

Figure P18.111

Problems 18.112 through 18.115 For Problems 18.112–18.115, when the body hits the step, its motion changes almost instantaneously from rolling without slip (Problems 18.112 and 18.113), or sliding (Problems 18.114 and 18.115), on the ground to a fixed-axis rotation about the corner of the step. Model this transition, using the ideas presented in Section 15.2 on p. 955. That is, assume that there is an infinitesimal time interval right after the impact between a body and the step in which the body does not change its position significantly, the body loses contact with the ground, and its weight is negligible relative to the contact forces between the body and the step. Problem 18.112 Some pipe sections are gently nudged from rest down an incline and roll without slipping all the way to a step of height 𝑏. Assume that each pipe section does not slide or rebound against the step, so that the pipes move as if hinged at the corner of the step. Modeling a pipe as a uniform thin ring of mass 𝑚 and radius 𝑟, and letting 𝑑 be the height from which the pipes are released, determine the minimum value of 𝑑 so that the pipes can roll over the step.

2𝑟

Figure P18.112

𝑏

1262

Chapter 18

Energy and Momentum Methods for Rigid Bodies

The disk of mass 𝑚 and radius 𝑅 is rolling without slipping with √ angular speed 𝜔1 = 53 𝑔∕𝑅 on the horizontal surface when it encounters a step of height 𝑅∕4. Assuming that the disk does not rebound when it hits the step and that it rotates about it, determine

Problem 18.113

(a) The normal force between the disk and the corner immediately after the impact. (b) The percentage of energy lost during the impact. (c) If the disk will make it to the top of the step. Figure P18.113

ISTUDY

Problem 18.114 A uniform equilateral triangle of mass 𝑚 and lateral dimensions 𝐿 is sliding with negligible friction on the horizontal surface with speed 𝑣0 when it impacts a small step at 𝐴. Assuming that the triangle does not rebound from the step after impact and assuming that the dimension of the step is very small compared with the lateral dimension of the triangle, determine

(a) The angular velocity of the triangle immediately after the impact as a function of 𝑣0 and 𝐿. √ (b) The angle 𝜃 to which the triangle will rotate about 𝐴 after the impact if 𝑣0 = 𝑔𝐿. The mass moment of inertia of an equilateral triangle is 𝐼𝐺 =

Figure P18.114

1 𝑚𝐿2 . 12

Figure P18.115

A uniform square block of mass 𝑚 and lateral dimension 𝐿 is sliding with negligible friction on the horizontal surface with speed 𝑣0 when it impacts a small step at 𝐴. Assuming that the block does not rebound from the step after impact and assuming that the dimension of the step is very small compared with the lateral dimension of the block, determine

Problem 18.115

(a) The angular velocity of the block immediately after the impact as a function of 𝑣0 and 𝐿. √ (b) The angle 𝜃 to which the block will rotate about 𝐴 after the impact if 𝑣0 = 𝑔𝐿.

Problem 18.116 A crane has a boom 𝐴 of mass 𝑚𝐴 and length 𝓁 that can rotate in the horizontal plane about a fixed point 𝑂. A trolley 𝐵 of mass 𝑚𝐵 is mounted on one side of 𝐴, such that the mass center of 𝐵 is always at a distance 𝑒 from the longitudinal axis of 𝐴. The position of 𝐵 is controlled by a cable and a system of pulleys. Both 𝐴 and 𝐵 are initially at rest in the position shown, where 𝑑 is the initial distance of 𝐵 from 𝑂 measured along the longitudinal axis of 𝐴. The boom 𝐴 is free to rotate about 𝑂 and, for a short time interval 0 ≤ 𝑡 ≤ 𝑡𝑓 , 𝐵 moves with constant acceleration 𝑎0 without reaching the end of 𝐴. Letting 𝐼𝑂 be the mass moment of inertia of 𝐴, modeling 𝐵 as a particle, and accounting only for the inertia of 𝐴 and 𝐵, determine the direction of rotation of 𝐴 and the angle 𝜃 swept by

ISTUDY

Section 18.2

Momentum Methods for Rigid Bodies

𝐴 from 𝑡 = 0 to 𝑡 = 𝑡𝑓 . Neglect the mass of the cable and of the pulleys. Hint: System angular momentum is conserved in the out-of-plane direction. The trolley’s angular momentum about 𝑂 includes contributions from both its radial and circumferential velocity components acting through circumferential and radial displacements, respectively. 𝐴

𝑒

𝑂 𝐵

𝑎0

Figure P18.116

Problem 18.117 Following up on parts (b) and (c) of the Pioneer 3 despin in Prob. 17.96, it turns out that we can analytically determine the length of the unwound wire needed to achieve any value of 𝜔𝑠 by using conservation of energy and conservation of angular momentum. In doing so, let the masses of 𝐴 and 𝐵 each be 𝑚, and the mass moment of inertia of the spacecraft ̇ body be 𝐼𝑂 . Let the initial conditions of the system be 𝜔𝑠 (0) = 𝜔0 , 𝓁(0) = 0, and 𝓁(0) = 0, and neglect gravity and the mass of each wire. 𝐴 𝓁 𝑦 𝜔𝑠

𝑄 𝐵0

𝜃 𝑂

𝑅

𝐴0

top view 𝐵 NASA

NASA

Figure P18.117

(a) Find the velocity of each of the masses 𝐴 and 𝐵 as a function of the wire length 𝓁(𝑡), the angular velocity of the spacecraft body 𝜔𝑠 (𝑡), and the radius of the spacecraft 𝑅. Hint: This part of the problem involves just kinematics — refer to Prob. 16.164 if you need help with the kinematics. (b) Apply the work-energy principle to the spacecraft system between the time just before the masses start to unwind and any arbitrary later time. You should obtain an ̇ 𝜔 , and constants. Hint: No external work is done on the expression relating 𝓁, 𝓁, 𝑠 system. (c) Since no external forces act on the system, its total angular momentum must be conserved about point 𝑂. Relate the angular momentum for this system between the time just before the masses start unwinding and any arbitrary later time. As with Part (b), ̇ 𝜔 , and constants. you should obtain an expression relating 𝓁, 𝓁, 𝑠 (d) Solve the energy and angular momentum equations obtained in Parts (b) and (c), respectively, for 𝓁̇ and 𝜔𝑠 . Now, letting 𝜔𝑠 = 0, show that the length of the unwound wire when the angular velocity of the spacecraft body is zero is given by 𝓁𝜔 =0 = 𝑠 √ (𝐼𝑂 + 2𝑚𝑅2 )∕(2𝑚). (e) From your solutions for 𝓁̇ and 𝜔𝑠 in Part (d), find the equations for 𝓁(𝑡) and 𝜔𝑠 (𝑡). These are the general solutions to the nonlinear equations of motion found in Prob. 17.95.

1263

1264

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Design Problems

ISTUDY

Design Problem 18.2 The Pioneer 3 spacecraft was a spin-stabilized spacecraft launched on December 6, 1958, by the U.S. Army Ballistic Missile agency in conjunction with NASA. It was designed with a despin mechanism consisting of two equal masses 𝐴 and 𝐵, each of mass 𝑚, that could be spooled out to the ends of two wires of variable length 𝓁(𝑡) when triggered by a hydraulic timer. As the masses unwound, they would slow the spacecraft’s spin from an initial angular velocity 𝜔𝑠 (0) to the final angular velocity (𝜔𝑠 )final , and then the weights and wires would be released. Assume that masses 𝐴 and 𝐵 are initially at positions 𝐴0 and 𝐵0 , respectively, before the wire begins to unwind, that the mass moment of inertia of the spacecraft is 𝐼𝑂 (this does not include the two masses 𝐴 and 𝐵), and neglect gravity and the mass of each wire. With this in mind, you are given the task of despinning the spacecraft body from an angular velocity of 400 rpm to any angular velocity in the range −400 rpm < 𝜔𝑠 < 400 rpm. To do this, a transducer is used that senses the tension in one of the wires at every instant. Design the total length of the wires to achieve the desired range of angular velocities, and determine the tension in the wires as a function of the angular velocity of the spacecraft so that the sensor can know when the spacecraft has achieved the desired velocity and can release the wires and masses. Use 𝑅 = 12.5 cm, 𝑚 = 7 g, 𝐼𝑂 = 0.0277 kg⋅m2 , and the initial conditions 𝓁(0) = ̇ 0.01 m and 𝓁(0) = 0 m∕s, with the despin mechanism shown. Hint: Refer to Prob. 16.164 on p. 1135 if you need help with the kinematics. 𝐴 𝓁 𝑦 𝜔𝑠

𝑄 𝐵0

𝜃 𝑂

𝑅

𝐴0

top view 𝐵 NASA

NASA

Figure DP18.2

ISTUDY

Section 18.3

18.3

Impact of Rigid Bodies

1265

Impact of Rigid Bodies

In this section we continue the study of impacts, begun in Section 15.2 on p. 955, by considering impacts between rigid bodies. We will use the notation and concepts introduced in Section 15.2. As a reminder, • We denote the value of a quantity right before and right after a collision with + denotes the speed of a the superscripts − and +, respectively (for example, 𝑣𝐴 point 𝐴 right after impact). • Our modeling of impacts is based on the concept of impulsive force (see p. 956). As in Section 15.2, this modeling assumption is reflected in our impact-relevant FBDs, which include only impulsive forces. 𝐴

The difference between rigid body and particle impact Figure 18.19 shows a collision between two bumper cars 𝐴 and 𝐵. Assume that we know the COR (coefficient of restitution) 𝑒 for the collision, the masses of 𝐴 and 𝐵, the location of their mass centers, their preimpact velocities, and their mass moments of inertia about their respective mass centers. Since the motion is planar, the postimpact velocities of 𝐴 and 𝐵 are described by six pieces of information: the velocity components of the mass centers of 𝐴 and 𝐵 (four unknowns) and the angular velocities of 𝐴 and 𝐵 (two unknowns). Consequently, we will need six scalar equations to solve for the six unknowns. Solving this rigid body impact problem will not require any new theory. Figure 18.20 shows the FBDs of 𝐴 and 𝐵 during the impact. Since there are no 𝑦, LOI

𝐴 𝐶 𝐴

(a)

𝐶

FBD of 𝐴

𝐸

and 𝐵 as a system

(b)

𝐸 𝑂

𝑁 𝑂

𝑥

𝑑𝑥

𝑄 𝑑𝑦 𝐵

𝐷 𝑑𝑥

𝑁 𝑄

individual 𝑥

FBDs of

𝐴 and 𝐵 𝐵

𝑑𝑦 𝐷

Figure 18.20. (a) Impact-relevant FBD of 𝐴 and 𝐵 as a system, and (b) the individual FBDs of 𝐴 and 𝐵. The contact points 𝐸 and 𝑄 belong to 𝐴 and 𝐵, respectively, and coincide with 𝑂.

external impulsive forces, the system’s total linear and angular momenta are conserved. As with particle impact, conservation of linear momentum will provide three equations from these FBDs, and the application of the COR equation along the LOI will provide a fourth equation. The final two equations are obtained by conserving angular momentum about point 𝑂 for each of the two bumper cars. Therefore, we see that the analysis of rigid body impact adds two conservation of angular momentum equations to the four equations we used in particle impact.

𝑣0 𝐵 Figure 18.19 A collision between two bumper cars. Before impact, car 𝐴 is at rest and 𝐵 is moving with speed 𝑣0 in the direction shown.

1266

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Rigid body impact: basic nomenclature and assumptions As in the case of particle impacts (see Section 15.2 on p. 955), a rigid body impact is called plastic if the COR 𝑒 = 0. We call a rigid body impact perfectly plastic if the colliding bodies form a single rigid body after impact. An impact is elastic if 0 < 𝑒 < 1, and the ideal case with 𝑒 = 1 is called perfectly elastic. In addition, an impact is unconstrained if the impacting objects are not acted upon by any external impulsive forces; otherwise the impact is called constrained. Rigid body impacts can be very complex, and we will consider only those cases, such as the bumper car collision problem, in which the following assumptions hold: 1. The impact involves only two bodies in planar motion where no impulsive force has a component perpendicular to the plane of motion. 2. Contact between any two rigid bodies occurs only at one point, and at this point we can clearly define the LOI. 3. The contact between the bodies is frictionless. 𝑣− 𝐶

𝐴

Classification of impacts

𝐶 𝐸

𝑂 𝑄

𝑑𝑦

𝑣− 𝐷

𝑥 𝐵

𝐷

Figure 18.21 Preimpact system configuration of an oblique eccentric impact.

ISTUDY

In Section 15.2, we classified impacts based on (1) the position of the mass centers of the colliding objects in relation to the LOI and (2) the orientation of the preimpact velocities in relation to the LOI (see Table 18.1). Based on Table 18.1, the bumper car collision in Fig. 18.19 is a direct eccentric impact because the cars’ mass centers are not on the LOI and because the preimpact velocity of the mass center of 𝐵 is parallel to the LOI while the velocity of 𝐴 is zero. If, right before impact, car 𝐴 were moving as shown in Fig. 18.21, then one of the mass centers would have a preimpact velocity not parallel to the LOI, and the impact would be oblique and eccentric. Table 18.1. Classification of impacts.

Impact geometry criteria Preimpact velocities

Centers of mass

parallel to LOI parallel to LOI not parallel to LOI not parallel to LOI

on LOI not on LOI on LOI not on LOI

Impact type

direct central direct eccentric oblique central oblique eccentric

Central impact In a central impact, the centers of mass of the two colliding objects lie on the LOI. There are two types of central impacts: direct and oblique (see Table 18.1). No matter the type, under the assumptions introduced above, rigid body central impacts have two important characteristics: 1. The angular velocities are conserved through the impact. 2. The COR equation can be written directly in terms of the velocity components along the LOI of the mass centers.

ISTUDY

Section 18.3

1267

Impact of Rigid Bodies

To illustrate these properties, consider the collision of two identical billiard balls 𝐴 and 𝐵, as shown in Fig. 18.22. Ball 𝐴 is initially stationary, while 𝐵 is rolling without slip to the left at a speed 𝑣0 . For a perfectly elastic collision, we want to determine the postimpact velocities of the two balls. We begin by sketching the system’s impactrelevant FBD [Fig. 18.23(a)], and the impact-relevant FBDs of the individual bodies [Fig. 18.23(b)]. The FBDs should show the LOI and a coordinate system aligned with the LOI. Points 𝐸 and 𝑄 belong to 𝐴 and 𝐵, respectively. The origin 𝑂 of the chosen coordinate system coincides with 𝐸 and 𝑄 at the time of impact, but is otherwise understood to be a point fixed in space. The FBDs in Fig. 18.23 do not show any reaction force between the balls and the table, which indicates that we have regarded these forces as nonimpulsive. This is because we have assumed that the contact between the two balls is frictionless, which implies that no impulsive friction force can be generated along the 𝑦 direction in response to the sliding of point 𝑄 relative to point 𝐸. The absence of a vertical impulsive force at the point of contact implies that no corresponding impulsive reaction force is generated in the vertical direction at the supporting surface. Because the impact is central, the impulsive force 𝑁 has no moment with respect to 𝐶 and 𝐷. Summing moments about the mass center of each ball using Eq. (18.43) + + and 𝐼𝐷 𝜔−𝐵 = 𝐼𝐷 𝜔𝐵 , where 𝐼𝐶 and on p. 1244, we can then say that 𝐼𝐶 𝜔−𝐴 = 𝐼𝐶 𝜔𝐴 𝐼𝐷 are the mass moments of inertia of 𝐴 and 𝐵, respectively. Therefore, we conclude that − − 𝜔𝐴 = 𝜔+𝐴 and 𝜔𝐵 = 𝜔+𝐵 , (18.49)

𝑣0 𝐷

𝐶

2𝑅

Figure 18.22 Example of a direct central impact.

𝐴 LOI

𝑂 𝐵 𝐷

𝐶 𝐸

𝑥

𝑄 (a) 𝑦 𝐵

𝐴 𝐸

LOI

𝑁

𝑁

𝑄

𝑂

𝐶

𝑥

𝐷

Figure 18.23 Impact-relevant FBDs (a) of the system as a whole and (b) of the two balls individually.

that is, the angular velocities of 𝐴 and 𝐵 are conserved through the impact. Because there are no impulsive forces in the direction perpendicular to the LOI, the components of velocity in that direction do not change through the impact. Thus, + + − = 0 and 𝑣𝐷𝑦 = 𝑣𝐷𝑦 = 0, and so the only remaining unknowns are we have 𝑣−𝐶𝑦 = 𝑣𝐶𝑦 + + 𝑣𝐶𝑥 and 𝑣𝐷𝑥 . To find these unknowns, we enforce the conservation of the system’s linear momentum along the LOI, i.e., 𝑚𝐴 𝑣−𝐶𝑥 + 𝑚𝐵 𝑣−𝐷𝑥 = 𝑚𝐴 𝑣+𝐶𝑥 + 𝑚𝐵 𝑣+𝐷𝑥 , and the COR equation for the contacting points 𝐸 and 𝑄, that is, ) ( + − 𝑣+𝑄𝑥 = 𝑒 𝑣−𝑄𝑥 − 𝑣−𝐸𝑥 . 𝑣𝐸𝑥

(18.50)

(18.51) 𝑘̂ × 𝑟⃗𝐸∕𝐶 𝜔± 𝐴

Using rigid body kinematics to relate the motion of points 𝐸 and 𝑄 to the mass centers on their respective bodies, we obtain ± ± ̂ 𝑘 × 𝑟⃗𝐸∕𝐶 = 𝑣⃗±𝐶 + 𝜔𝐴 𝑣⃗𝐸

and

± ̂ 𝑘 × 𝑟⃗𝑄∕𝐷 , 𝑣⃗±𝑄 = 𝑣⃗±𝐷 + 𝜔𝐵

(18.52)

where we have used the fact that 𝑟⃗𝐸∕𝐶 and 𝑟⃗𝑄∕𝐷 do not change during the impact. Figure 18.24 shows that 𝑟⃗𝐸∕𝐶 = 𝑅 𝚤̂ and 𝑟⃗𝑄∕𝐷 = −𝑅 𝚤̂, and so 𝑘̂ × 𝑟⃗𝐸∕𝐶 = 𝜔± 𝜔± 𝑅 𝚥̂ and 𝐴 𝐴

𝑘̂ × 𝑟⃗𝑄∕𝐷 = −𝜔± 𝜔± 𝑅 𝚥̂. 𝐵 𝐵

(18.53)

Comparing Eqs. (18.52) and (18.53), we see that the 𝑥 components, i.e., those along ± and 𝑣⃗±𝑄 must be identical to the 𝑥 components of 𝑣⃗±𝐶 and 𝑣⃗±𝐷 , respecthe LOI, of 𝑣⃗𝐸 tively, i.e., ± ± 𝑣𝐸𝑥 = 𝑣±𝐶𝑥 and 𝑣±𝑄𝑥 = 𝑣𝐷𝑥 . (18.54) Equations (18.54) allow us to write the COR equation directly in terms of the velocity of the mass centers, i.e., ( ) 𝑣+𝐶𝑥 − 𝑣+𝐷𝑥 = 𝑒 𝑣−𝐷𝑥 − 𝑣−𝐶𝑥 .

(18.55)

𝐶

LOI

𝑦

𝑟⃗𝐸∕𝐶

𝜔𝐵

𝑂 𝑄

𝐷

𝐸

𝑟⃗𝑄∕𝐷

𝐴 𝜔𝐴

𝑥 𝐵

𝜔± 𝑘̂ × 𝑟⃗𝑄∕𝐷 𝐵

Figure 18.24 Kinematics of a central impact.

Concept Alert Central impacts and COR equation. In a central impact the COR equation can be written directly in terms of the velocity components of the mass centers.

1268

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Going back to the billiard ball impact problem, recalling the given preimpact conditions and that 𝑒 = 1, we obtain the following solution: 𝜔+𝐴 = 0,

+ = −𝑣0 , 𝑣𝐶𝑥

+ 𝑣𝐷𝑥 = 0,

and

𝜔+𝐵 = 𝑣0 ∕𝑅.

(18.56)

+ Even though 𝑣𝐷𝑥 = 0, because 𝜔+𝐵 = 𝑣0 ∕𝑅, ball 𝐵 does not stop after impact. The + + = 𝑣0 ∕𝑅 taken together imply that, right after impact, the equations 𝑣𝐷𝑥 = 0 and 𝜔𝐵 mass center of ball 𝐵 has zero velocity for an instant, while ball 𝐵 slips against the pool table. With friction between the table and balls, the friction force due to sliding will cause the center of mass of 𝐵 to start moving again to the left, as if to chase ball 𝐴. Had we modeled the balls as particles, we would have concluded that 𝐵 simply stopped after impact.

Eccentric impact LOI

𝜔𝐴

𝜔𝐵 𝐵

𝐴

𝑣⃗𝐷 𝑣⃗𝐶

Figure 18.25 Collision between two boats.

ISTUDY

Figure 18.25 shows an eccentric impact between two boats. It is eccentric because at least one of the mass centers of the colliding bodies is not on the LOI. An eccentric impact not only affects the velocities of the mass centers, but it also affects the bodies’ angular velocities. For the two boats colliding in Fig. 18.25, as usual, we start the solution with the FBD of the system and the FBDs of the individual bodies at the time of impact, which are shown in Fig. 18.26(a) and (b), respectively. On these diagrams we clearly indicate LOI

𝑥

𝐸

𝐴

LOI

𝑄

𝐸

𝑁 𝚤̂

𝑂

𝑂

𝐴

𝐵

𝑁 𝐵

𝑦

𝑄

𝚥̂

𝑦

Figure 18.26. FBDs of the colliding bodies (a) as a system and (b) individually.

the LOI and the chosen coordinate system. In addition, we choose the origin 𝑂 so as to coincide with the points of contact 𝐸 and 𝑄 between the two rigid bodies at the time of impact, and we keep in mind that 𝑂 is a fixed point. The reason for choosing 𝑂 as stated is that it is a convenient point to use as the moment center when enforcing the angular impulse-momentum principle. The solution of an unconstrained rigid body eccentric impact problem is governed by six scalar equations. The first of these equations enforces the conservation of linear momentum for the system along the LOI, that is, − − + 𝑚𝐴 𝑣𝐶𝑥 + 𝑚𝐵 𝑣𝐷𝑥 = 𝑚𝐴 𝑣𝐶𝑥 + 𝑚𝐵 𝑣+𝐷𝑥 .

(18.57)

The second and third of these equations are − + 𝑣𝐶𝑦 = 𝑣𝐶𝑦

and

+ 𝑣−𝐷𝑦 = 𝑣𝐷𝑦 ,

(18.58)

which follow from the frictionless contact assumption (assumption 3 on p. 1266). This assumption implies that the linear momentum of each colliding body is conserved in the direction perpendicular to the LOI.

ISTUDY

Section 18.3

Impact of Rigid Bodies

1269

The fourth equation is the COR equation, which is first written in terms of the components of the velocities of the contact points along the LOI, i.e., ( ) 𝑣+𝐸𝑥 − 𝑣+𝑄𝑥 = 𝑒 𝑣−𝑄𝑥 − 𝑣−𝐸𝑥 .

(18.59)

It is then rewritten in terms of the velocity components of the mass centers, making sure to satisfy rigid body kinematics, which requires that 𝑣⃗𝐸 = 𝑣⃗𝐶 + 𝜔𝐴 𝑘̂ × 𝑟⃗𝐸∕𝐶 ±

±

±

and

𝑣⃗𝑄 = 𝑣⃗𝐷 + 𝜔𝐵 𝑘̂ × 𝑟⃗𝑄∕𝐷 . ±

±

±

𝑥

𝐴

LOI

𝐸

𝑄

𝐵

𝑟⃗𝑄∕𝐷

(18.60)

Referring to Fig. 18.27, because 𝐸, 𝑄, and 𝑂 coincide at the time of impact, 𝑟⃗𝐸∕𝐶 = −⃗𝑟𝐶 = −𝑥𝐶 𝚤̂ − 𝑦𝐶 𝚥̂ and 𝑟⃗𝑄∕𝐷 = −⃗𝑟𝐷 = −𝑥𝐷 𝚤̂ − 𝑦𝐷 𝚥̂.

𝑟⃗𝐸∕𝐶

𝑂

(18.61)

Then, substituting Eqs. (18.61) into Eqs. (18.60), carrying out the cross products, and rearranging terms, we obtain ) ( ± ) ( ± ± ± (18.62) = 𝑣±𝐶𝑥 + 𝜔𝐴 𝑦𝐶 𝚤̂ + 𝑣𝐶𝑦 − 𝜔𝐴 𝑥𝐶 𝚥̂, 𝑣⃗𝐸 ( ± ) ( ± ) ± ± ± 𝑣⃗𝑄 = 𝑣𝐷𝑥 + 𝜔𝐵 𝑦𝐷 𝚤̂ + 𝑣𝐷𝑦 − 𝜔𝐵 𝑥𝐷 𝚥̂. (18.63)

𝑦

𝚤̂ 𝚥̂

Figure 18.27 Relative position vectors of 𝐸 and 𝑄 with respect to 𝐶 and 𝐷, respectively.

Using Eqs. (18.62) and (18.63), Eq. (18.59) can be written as: ( − ) + + − 𝑣+𝐶𝑥 + 𝜔+𝐴 𝑦𝐶 − 𝑣𝐷𝑥 − 𝜔𝐵 𝑦𝐷 = 𝑒 𝑣𝐷𝑥 + 𝜔𝐵 𝑦𝐷 − 𝑣−𝐶𝑥 − 𝜔−𝐴 𝑦𝐶 .

(18.64)

Since the line of action of the contact forces between 𝐴 and 𝐵 goes through fixed point 𝑂, the remaining two equations say that each of the angular momenta of 𝐴 and 𝐵 relative to 𝑂 is conserved, that is, ( −) ( +) ⃗ ⃗ ℎ 𝑂 𝐴 = ℎ𝑂 𝐴

and

( −) ( +) ⃗ ⃗ ℎ 𝑂 𝐵 = ℎ𝑂 𝐵 ,

(18.65)

( ) ( ) ⃗ ⃗ where ℎ 𝑂 𝐴 and ℎ𝑂 𝐵 are the angular momenta of 𝐴 and 𝐵 relative to 𝑂, respec( ±) ( ±) ⃗ ⃗ tively. Applying Eq. (18.40) on p. 1244, ℎ and ℎ can be written as 𝑂 𝐴

𝑂 𝐵

( ±) ⃗ ℎ = 𝐼𝐶 𝜔 ⃗ ±𝐴 + 𝑟⃗𝐶∕𝑂 × 𝑚𝐴 𝑣⃗±𝐶 , 𝑂 𝐴 ( ±) ± ⃗ = 𝐼𝐷 𝜔 ⃗ ±𝐵 + 𝑟⃗𝐷∕𝑂 × 𝑚𝐵 𝑣⃗𝐷 , ℎ 𝑂 𝐵

(18.66) (18.67)

where 𝐼𝐶 and 𝐼𝐷 are the mass moments of inertia of 𝐴 and 𝐵, respectively. In Eqs. (18.66) and (18.67), the relative position vectors 𝑟⃗𝐶∕𝑂 and 𝑟⃗𝐷∕𝑂 are assumed to be constant through the impact, and therefore do not need the superscript ±. Referring to Fig. 18.26(b), the physical justification for Eqs. (18.65) is that the impulsive forces acting on 𝐴 and 𝐵 provide no moment about the fixed point 𝑂. Observe that Eqs. (18.65) yield only two scalar equations because the only nonzero component of these equations is perpendicular to the plane of motion. Equations (18.57), (18.58), (18.64), and (18.65) are all that is needed to solve the most general case of unconstrained oblique eccentric impact of two rigid bodies (under the assumptions stated on p. 1266).

Constrained eccentric impact In a constrained impact one or both of the colliding bodies are subject to external impulsive forces (see Example 15.9, p. 968). Modeling these impacts can be challenging, and we consider only a simple case in which one of the impacting bodies is constrained to move in fixed-axis rotation.

Concept Alert Conservation of angular momentum does not mean conservation of angular velocity. Equations (18.66) and (18.67) make clear that conservation of angular momentum does not imply that the angular velocity is conserved. Equations (18.65) demand that the left-hand sides of Eqs. (18.66) and (18.67) remain constant (through the impact), and this is possible if the corresponding angular velocities and mass center velocities on the right-hand sides of Eqs. (18.66) and (18.67) change in concert.

1270

Energy and Momentum Methods for Rigid Bodies

𝑂 𝜔

𝐻 𝐿 𝐵

ℎ

𝑣0 Figure 18.28 A ballistic pendulum.

𝑅𝑦 𝑂

𝚥̂

𝑅𝑥

𝑥

𝚤̂ 𝐵 Figure 18.29 FBD of the ballistic pendulum and bullet system at the time of impact.

ISTUDY

Chapter 18

Referring to Fig. 18.28, consider a ballistic pendulum consisting of a uniform thin rod of mass 𝑚𝑟 and length 𝐿 pinned at 𝑂 and a target block of mass 𝑚𝑡 , width 𝑤, and height ℎ. Suppose that a bullet of mass 𝑚𝑏 is fired horizontally with a speed 𝑣0 as shown. Assuming that the bullet becomes embedded in the block, we wish to determine the postimpact velocity of the pendulum-bullet system. Since the only possible motion of the pendulum is a fixed-axis rotation about 𝑂, the only piece of information we need to describe the system’s postimpact behavior is 𝜔𝑝+ , the postimpact angular velocity of the pendulum. As usual, we sketch the system’s FBD at the time of impact, given in Fig. 18.29, making sure to include only impulsive forces. The pin reactions 𝑅𝑥 and 𝑅𝑦 appear on the FBD because they will take on whatever value is required of them to keep 𝑂 from moving, and, as such, they are impulsive. The presence of 𝑅𝑥 and 𝑅𝑦 makes the impact a constrained impact. Observe that 𝑅𝑥 and 𝑅𝑦 , while impulsive, provide no moment about the fixed point 𝑂, so the system’s total angular momentum relative to 𝑂 must be conserved through ( ) ( ) ⃗ ⃗ the impact. Calling ℎ 𝑂 𝑏 and ℎ𝑂 𝑝 the angular momenta relative to 𝑂 of the bullet and the pendulum, respectively, we have ( −) ( +) ( +) ( −) ⃗ ⃗ ⃗ ⃗ (18.68) ℎ 𝑂 𝑏 + ℎ 𝑂 𝑝 = ℎ𝑂 𝑏 + ℎ𝑂 𝑝 , where, modeling the bullet as a particle and recalling that the pendulum can move only in a fixed-axis rotation about 𝑂 [see Eq. (18.44) on p. 1245], ( ±) ( ) ± ( ±) ± ⃗ ⃗ = 𝑟⃗𝐵∕𝑂 × 𝑚𝑏 𝑣⃗𝐵 and ℎ = 𝐼𝑂 𝑝 𝜔 ⃗𝑝 , (18.69) ℎ 𝑂 𝑏 𝑂 𝑝 ( ) where 𝐼𝑂 𝑝 is the mass moment of inertia of the pendulum relative to 𝑂 and where the relative position vector 𝑟⃗𝐵∕𝑂 does not have the superscript ± because it is treated as a constant through the impact. Since the pendulum is initially stationary and the bullet becomes embedded in the block, substituting Eqs. (18.69) into Eq. (18.68), carrying out the cross products, and simplifying gives ] [( ) (18.70) 𝑚𝑏 𝑣0 𝐻 = 𝐼𝑂 𝑝 + 𝑚𝑏 𝐻 2 𝜔+𝑝 . Solving Eq. (18.70) for the postimpact angular velocity of the pendulum-bullet system, we have 𝑚 𝑣 𝐻 . (18.71) 𝜔𝑝+ = ( ) 𝑏 0 𝐼𝑂 𝑝 + 𝑚𝑏 𝐻 2 The problem just discussed illustrates a key element of the solution of most constrained impact problems, namely, the identification of a fixed point about which the system’s angular momentum is conserved. Examples 18.12 and 18.13 demonstrate the use of this strategy in the case of more involved physical situations.

End of Section Summary In this section we studied planar rigid body impacts. We learned that, contrary to particle impacts, rigid bodies can experience eccentric impacts. These are collisions in which at least one of the mass centers of the impacting bodies does not lie on the LOI. We also learned that the basic concepts used in particle impacts are applicable to rigid body impacts, and we reviewed solution strategies for a variety of situations. As with particle impacts, we say that a rigid body impact is plastic if the COR 𝑒 = 0. A rigid body impact will be called perfectly plastic if the colliding bodies form a single rigid body after impact. An impact is elastic if the COR 𝑒 is such that

ISTUDY

Section 18.3

1271

Impact of Rigid Bodies

0 < 𝑒 < 1. Finally, the ideal case with 𝑒 = 1 is referred to as a perfectly elastic impact. In addition, an impact is unconstrained if the system consisting of the two impacting objects is not subject to external impulsive forces; otherwise, the impact is called constrained. We have considered only impacts satisfying the following assumptions: 1. The impact involves only two bodies in planar motion where no impulsive force has a component perpendicular to the plane of motion. 2. Contact between any two rigid bodies occurs at only one point, and at this point we can clearly define the LOI. 3. The contact between the bodies is frictionless.

LOI

We recommend organizing the solution of any impact problem as follows: • Begin with an FBD of the impacting bodies as a system and FBDs for each of the colliding bodies. Neglect nonimpulsive forces. • Choose a coordinate system with the origin coincident with the points that come into contact at the time of impact. Recall that the origin of such a coordinate system is a fixed point.

𝑄

𝐸 𝑂

𝐵

𝑦 𝑥

𝚤̂

LOI

𝐸

𝚥̂ (b)

• Enforce the linear and/or the angular impulse-momentum principles for the system and/or for the individual bodies. In applying the angular impulsemomentum principle for the whole system, the moment center should be a fixed point, whereas for an individual body, the moment center should be a fixed point or the body’s mass center. • For plastic, elastic, and perfectly elastic impacts, the COR equation is first written using the velocity components along the LOI of the points that actually come into contact. For example, for the impact shown in Fig. 18.30, the COR equation is first written as Eq. (18.59), p. 1269 ( − ) + − 𝑣+𝐸𝑥 − 𝑣𝑄𝑥 = 𝑒 𝑣𝑄𝑥 − 𝑣𝐸𝑥 , where the COR 𝑒 is such that 0 ≤ 𝑒 ≤ 1. The COR equation must then be rewritten in terms of the colliding bodies’ angular velocities and velocities of the mass centers using rigid body kinematics. For the situation in Fig. 18.30, ± ± ̂ 𝑘× + 𝜔𝐴 this means rewriting the COR equation using the relations 𝑣⃗±𝐸 = 𝑣⃗𝐶 ± ± ± ̂ 𝑟⃗𝐸∕𝐶 and 𝑣⃗𝑄 = 𝑣⃗𝐷 +𝜔𝐵 𝑘× 𝑟⃗𝑄∕𝐷 . Notice that the relative position vectors 𝑟⃗𝐸∕𝐶 and 𝑟⃗𝑄∕𝐷 do not have the ± superscript because they are treated as constants during the impact. • In perfectly plastic impacts, kinematic constraint equations must be enforced that express the fact that two bodies form a single rigid body after impact.

𝑁 𝐴 𝑦

𝑂

𝑁 𝑄

𝐵

Figure 18.30 FBDs of the colliding bodies (a) as a system and (b) individually.

1272

E X A M P L E 18.11 𝐴

𝑣0

𝐶

Rigid Body Central Impact

𝐵 𝜔0

𝐷 𝛽

Figure 1 Two colliding hockey pucks.

𝑥 𝐴 𝑦

𝐵 LOI

𝐶

𝐷 𝛽

Two identical hockey pucks sliding over ice collide as shown in Fig. 1. If 𝛽 = 43◦ , the COR is 𝑒 = 0.95, 𝐴 is initially at rest, and puck 𝐵 is moving with speed 𝑣0 = 30 m∕s and angular speed 𝜔0 = 4 rad∕s as shown, determine the postimpact velocities of 𝐴 and 𝐵.

SOLUTION Road Map & Modeling We start with the FBD of the system and the FBDs of the colliding bodies separately, shown in Figs. 2 and 3, respectively, where we also indicate the LOI and a convenient coordinate system. Because 𝐶 and 𝐷 lie on the LOI, the impact is central. As discussed earlier in the section, we assume that the contact between the pucks is frictionless. This assumption and the fact that the impact is central allow us to immediately state that the angular velocities of the bodies are unaffected by the impact and that the COR equation can be expressed directly in terms of velocity components of the mass centers. These simplifications make the problem solution very similar to that of a particle impact problem. 𝑥

Figure 2 Impact-relevant FBD of the two pucks as a system.

ISTUDY

Chapter 18

Energy and Momentum Methods for Rigid Bodies

𝐴 𝑦

𝐵 𝑁

𝑁

LOI

𝐶

𝐷 𝛽

Figure 3. Impact-relevant FBDs of each of two pucks.

Governing Equations Balance Principles

There are no external impulsive forces acting on the system (Fig. 2). Therefore, the system’s linear momentum is conserved along the LOI, which gives 𝑚𝐴 𝑣−𝐶𝑦 + 𝑚𝐵 𝑣−𝐷𝑦 = 𝑚𝐴 𝑣+𝐶𝑦 + 𝑚𝐵 𝑣+𝐷𝑦 .

(1)

No impulsive force acts on 𝐴 or 𝐵 in the direction perpendicular to the LOI (Fig. 3). Therefore, we can write 𝑣−𝐶𝑥 = 𝑣+𝐶𝑥

and

𝑣−𝐷𝑥 = 𝑣+𝐷𝑥 .

(2)

Since the impact is central (and the contact between the pucks is frictionless), the angular impulse-momentum principle for 𝐴 and 𝐵 individually yields the following two equations: 𝜔+𝐴 = 𝜔−𝐴

and 𝜔+𝐵 = 𝜔−𝐵 .

(3)

Force Laws

The COR equation can be expressed directly in terms of the velocity components along the LOI of the mass centers, so we have ( ) 𝑣+𝐶𝑦 − 𝑣+𝐷𝑦 = 𝑒 𝑣−𝐷𝑦 − 𝑣−𝐶𝑦 .

Kinematic Equations

(4)

The preimpact velocities of 𝐴 and 𝐵 are

𝜔−𝐴 = 0,

𝑣−𝐶𝑥 = 0,

𝑣−𝐶𝑦 = 0,

(5)

𝜔−𝐵 = 𝜔0 ,

𝑣−𝐷𝑥 = 𝑣0 sin 𝛽,

𝑣−𝐷𝑦 = 𝑣0 cos 𝛽.

(6)

ISTUDY

Section 18.3

Impact of Rigid Bodies

Computation

Equations (1)–(4) form a system of six equations in the six unknowns 𝜔+𝐴 , 𝑣+𝐶𝑥 , 𝑣+𝐶𝑦 , 𝜔+𝐵 , 𝑣+𝐷𝑥 , and 𝑣+𝐷𝑦 . Recalling that 𝑚𝐴 = 𝑚𝐵 , this system of equations can be solved to obtain 1+𝑒 𝑣 cos 𝛽, 2 0 1−𝑒 = 𝑣 cos 𝛽, 2 0

𝜔+𝐴 = 0,

𝑣+𝐶𝑥 = 0,

𝑣+𝐶𝑦 =

(7)

𝜔+𝐵 = 𝜔0 ,

𝑣+𝐷𝑥 = 𝑣0 sin 𝛽,

𝑣+𝐷𝑦

(8)

which, upon substitution of the given data, yields the following numerical answer: 𝜔+𝐴 = 0 rad∕s,

𝑣+𝐶𝑥 = 0 m∕s,

𝑣+𝐶𝑦 = 21.39 m∕s,

(9)

𝜔𝐵 = 4 rad∕s,

𝑣𝐷𝑥 = 20.46 m∕s,

𝑣𝐷𝑦 = 0.5485 m∕s.

(10)

+

+

+

Discussion & Verification

The solution appears to be reasonable in that, as expected, the center of mass of puck 𝐴 moves only along the LOI: this is the expected behavior of any impact in which there is no friction between the colliding bodies. Since the impact is central (and the contact between the pucks is frictionless), the angular velocities of the colliding bodies are conserved. In addition to these considerations, we should check that the postimpact kinetic energy of the system is smaller than the preimpact kinetic energy since the COR was less than 1. Going through this verification is a bit involved in this problem because neither the masses of 𝐴 and 𝐵 nor their respective mass moments of inertia, 𝐼𝐶 and 𝐼𝐷 , were given. Using Eqs. (5) and (6), the preimpact kinetic energy 𝑇 − is ( )2 ( )2 ( )2 ( )2 𝑇 − = 21 𝑚𝐴 𝑣−𝐶 + 12 𝐼𝐶 𝜔−𝐴 + 21 𝑚𝐵 𝑣−𝐷 + 12 𝐼𝐷 𝜔−𝐵 = 21 𝑚𝐵 𝑣20 + 12 𝐼𝐷 𝜔20 .

(11)

Using Eqs. (7) and (8), the postimpact kinetic energy 𝑇 + is ( )2 ( )2 ( )2 ( )2 𝑇 + = 21 𝑚𝐴 𝑣+𝐶 + 12 𝐼𝐶 𝜔+𝐴 + 21 𝑚𝐵 𝑣+𝐷 + 12 𝐼𝐷 𝜔+𝐵 = 81 𝑚𝐴 (1 + 𝑒)2 𝑣20 cos2 𝛽 [ ] + 18 𝑚𝐵 4 sin2 𝛽 + (1 − 𝑒)2 cos2 𝛽 𝑣20 + 21 𝐼𝐷 𝜔20 .

(12)

Since the two pucks are identical, 𝐼𝐶 = 𝐼𝐷 . Letting 𝑚 = 𝑚𝐴 = 𝑚𝐵 , subtracting Eq. (12) from Eq. (11), and simplifying, we have ( ) (13) 𝑇 − − 𝑇 + = 41 1 − 𝑒2 𝑚𝑣20 cos2 𝛽, where we have used the fact that 12 𝑚𝐵 𝑣20 − 12 𝑚𝐵 𝑣20 sin2 𝛽 = 21 𝑚𝑣20 cos2 𝛽. Finally, since 𝑒 < 1, we have 1−𝑒2 > 0, and consequently the right-hand side of Eq. (13) is positive, i.e., 𝑇− − 𝑇+ > 0 as expected.

⇒

𝑇 + < 𝑇 −,

(14)

1273

1274

Chapter 18

Energy and Momentum Methods for Rigid Bodies

E X A M P L E 18.12 𝐵 2𝑟

𝐶

Constrained Impact of Two Rigid Bodies 𝐴

𝑣0

𝐷

𝑄

ℎ

Figure 1 Pipe sections being placed horizontally. Points 𝐶 and 𝐷 are the centers of 𝐵 and 𝐴, respectively. Point 𝑄 is the point on 𝐴 in contact with the block of height ℎ.

𝐵 𝐶

LOI

Identical pipe sections of radius 𝑟 = 1.5 f t and weight 𝑊 = 200 lb are being unloaded and aligned horizontally. A block of height ℎ = 6 in. is fixed to the ground and used to hold in place 𝐴, the first pipe in the row (see Fig. 1). If the next pipe section 𝐵 rolls without slip to the right with a speed 𝑣0 and bumps into 𝐴, the COR for the 𝐴-𝐵 collision is 𝑒 = 0.85, and 𝐴 does not rebound off the block or slide relative to it, then determine the smallest value of 𝑣0 that would make 𝐴 roll over the block.

SOLUTION Road Map

The solution can be organized in two parts. In the first part, we will study the collision of 𝐴 and 𝐵 and determine the postimpact motion of 𝐴 and 𝐵. Once the motion of 𝐴 is described as a function of 𝑣0 , in the second part we will find the smallest value of 𝑣0 that makes 𝐴 roll over the block by relating the postimpact kinetic energy of 𝐴 to the amount of work needed to move 𝐴 by a vertical distance ℎ against the direction of gravity.

𝐴

The impact of 𝐴 and 𝐵 Modeling

𝐻𝐷

𝐸 𝑂

𝑥 𝑅𝑥

𝑄

𝚥̂

ℎ 𝑅𝑦

Figure 2 Impact-relevant FBD of 𝐴 and 𝐵 combined at the time of impact.

The FBD of the system is shown in Fig. 2, and the FBDs of each colliding body are shown in Fig. 3. At the time of impact, the origin of the coordinate system 𝑂, which is a fixed point, coincides with the contact points 𝐸 and 𝐻, belonging to 𝐵 and 𝐴, respectively. Notice that we are dealing with a constrained impact due to the presence of external impulsive forces at 𝑄. These forces exist because 𝐴 is initially in contact with a fixed block and does not slide relative to it. This also means that 𝐴 moves as if in a fixed-axis rotation about 𝑄. Governing Equations Balance Principles

The external impulsive forces in Fig. 2 provide no moment about the fixed point 𝑄, and therefore the system’s angular momentum about 𝑄 is conserved, ( −) ( +) ( −) ( +) ⃗ ⃗ ⃗ ⃗ ℎ (1) 𝑄 𝐴 + ℎ 𝑄 𝐵 = ℎ𝑄 𝐴 + ℎ𝑄 𝐵 ,

𝑦 𝐵 𝐶

𝐸

𝑟

𝑁

LOI

𝑂

𝚥̂

𝑥

𝑦 𝐴

𝚤̂ LOI

𝐻𝐷

𝑁 𝑂

𝑟

𝑥 𝑄 𝑅𝑦

Figure 3 Impact-relevant FBD of 𝐴 and 𝐵 individually at the time of impact.

(2) (3)

𝐼𝑄 and 𝐼𝐶 are the mass moments of inertia of 𝐴 relative to 𝑄 and of 𝐵 relative to 𝐶, respectively. Note that, in calculating 𝐼𝑄 , we used the parallel axis theorem. The first of Eqs. (2) holds because 𝐴 moves as if pinned at 𝑄. The impulsive force 𝑁, which acts on 𝐵, points toward 𝐶. Therefore, since 𝑚 (the mass of 𝐵) and 𝐼𝐶 are constant, the 𝑦 component of the linear momentum of 𝐵, as well as the angular momentum of 𝐵 relative to 𝐶 are conserved, that is,

𝑅𝑥 ℎ

ISTUDY

where, viewing both pipe sections as uniform thin rings of radius 𝑟, ( ±) ⃗ = 𝐼𝑄 𝜔 ⃗ ±𝐴 , with 𝐼𝑄 = 𝑚𝑟2 + 𝑚𝑟2 = 2𝑚𝑟2 , ℎ 𝑄 𝐴 ( ±) ⃗ ℎ = 𝐼𝐶 𝜔 ⃗ ±𝐵 + 𝑟⃗𝐶∕𝑄 × 𝑚𝑣⃗±𝐶 , with 𝐼𝐶 = 𝑚𝑟2 . 𝑄 𝐵

𝑣−𝐶𝑦 = 𝑣+𝐶𝑦

and 𝜔−𝐵 = 𝜔+𝐵 .

(4)

Note that the impulsive force system acting on 𝐴 does not allow us to write relations for 𝐴 analogous to those in Eqs. (4) for 𝐵. Force Laws Since the contact points between 𝐴 and 𝐵 are 𝐸 and 𝐻, the COR equation reads ) ( (5) 𝑣+𝐻𝑥 − 𝑣+𝐸𝑥 = 𝑒 𝑣−𝐸𝑥 − 𝑣−𝐻𝑥 . Kinematic Equations Before impact, 𝐴 is stationary and 𝐵 rolls without slip, so 𝑣−𝐻𝑥 = 0,

𝜔−𝐴 = 0,

𝑣−𝐶𝑦 = 0,

and 𝜔−𝐵 = −𝑣0 ∕𝑟.

(6)

To express 𝑣𝐸𝑥 and 𝑣𝐻𝑥 in Eq. (5) in terms of 𝑣⃗𝐶 , 𝜔𝐵 , and 𝜔𝐴 , recall that 𝑣⃗𝐸 = 𝑣⃗𝐶 + 𝜔 ⃗𝐵 × 𝑟⃗𝐸∕𝐶 and 𝑣⃗𝐻 = 𝜔 ⃗ 𝐴 × 𝑟⃗𝐻∕𝑄 (𝐴 rotates about 𝑄). Consequently, we have ±

+

𝑣−𝐸𝑥 = 𝑣0 ,

+

𝑣+𝐸𝑥 = 𝑣+𝐶𝑥 ,

and

+

+

𝑣+𝐻𝑥 = −𝜔+𝐴 (𝑟 − ℎ).

(7)

ISTUDY

Section 18.3

1275

Impact of Rigid Bodies

Computation

Substituting Eqs. (2) and (3) into Eq. (1), taking advantage of Eqs. (4) and the last two of Eqs. (6), expanding the cross products, and simplifying, we have −𝑣0 (𝑟 − ℎ) = −𝑣+𝐶𝑥 (𝑟 − ℎ) + 2𝑟2 𝜔+𝐴 ,

(8)

where we have written this equation in scalar form because the only nonzero component of Eq. (1) is in the 𝑧 direction. Substituting the first of Eqs. (6) and all of Eqs. (7) into Eq. (5), we have −𝜔+𝐴 (𝑟 − ℎ) − 𝑣+𝐶𝑥 = 𝑒𝑣0 .

(9)

Equations (8) and (9) form a system of two equations for 𝑣𝐶𝑥 and 𝜔𝐴 . Solving these equations, we get the postimpact motion of 𝐴 and 𝐵 in terms of 𝑣0 : +

𝜔+𝐴 = −

(1 + 𝑒)(𝑟 − ℎ)𝑣0 (𝑟 − ℎ)2 + 2𝑟2

and 𝑣+𝐶𝑥 =

+

ℎ2 − 2ℎ𝑟 + (1 − 2𝑒)𝑟2 𝑣0 . (𝑟 − ℎ)2 + 2𝑟2

(10)

The work-energy principle applied to 𝐴

After the impact, 𝐴 rolls over the block as if pinned at 𝑄, and 𝐴’s FBD prior to reaching the top of the block is that in Fig. 4. Since 𝑄 is fixed, the reactions at 𝑄 do no work, so energy is conserved during the postimpact motion of 𝐴. Modeling

Governing Equations

𝚥̂

datum 𝑟 𝚤̂

Choosing 1 to be immediately after impact and 2 when 𝐴 gets to the top of the block with zero velocity (because we want to compute the minimum 𝑣0 ), we have

𝐴

𝑊 𝐷 𝑄

𝑅𝑥 𝑅𝑦

ℎ 𝑥

Balance Principles

𝑇𝐴1 + 𝑉𝐴1 = 𝑇𝐴2 + 𝑉𝐴2 ,

where

𝑇𝐴1 = 21 𝐼𝑄 𝜔2𝐴1

and 𝑇𝐴2 = 12 𝐼𝑄 𝜔2𝐴2 ,

(11)

and where 𝐼𝑄 = 2(𝑊 ∕𝑔)𝑟2 . Force Laws

Choosing the datum line as shown in Fig. 4, we have 𝑉𝐴1 = 0

Kinematic Equations

and 𝑉𝐴2 = 𝑊 ℎ.

(12)

Based on our modeling assumptions, we have 𝜔𝐴1 = 𝜔+𝐴

and 𝜔𝐴2 = 0.

(13)

Combining Eqs. (11)–(13), recalling that 𝜔+𝐴 is given by the first of Eqs. (10), and solving for 𝑣0 , we have

Computation

𝑣0 =

√ [ ] 𝑔ℎ (𝑟 − ℎ)2 + 2𝑟2 𝑟(1 + 𝑒)(𝑟 − ℎ)

= 7.953 f t∕s.

(14)

The solution seems reasonable since the computed 𝑣0 , along with Eqs. (10), gives 𝜔+𝐴 = −2.675 rad∕s and 𝑣+𝐶𝑥 = −4.085 f t∕s, i.e., after impact 𝐴 rotates clockwise and 𝐵 moves to the left, as one would expect. In fact, the first of Eqs. (10) indicates that the rotation will always be clockwise, as long as 𝑟 > ℎ. Furthermore, the √ value of 𝑣0 was expected to be larger than 𝑔ℎ = 4.012 f t∕s, which is the 𝑣0 needed by 𝐵 to get to a height equal to ℎ off the ground by simply rolling without slip along any path and without any collisions. Finally, using the computed value of 𝑣0 to compute the preand postimpact values of the system’s kinetic energy, we have 𝑇 − = 392.8 f t ⋅lb > 𝑇 + = 348.2 f t ⋅lb, as it should. Discussion & Verification

Figure 4 FBD of 𝐴 following the impact with 𝐵.

1276

Chapter 18

Energy and Momentum Methods for Rigid Bodies

E X A M P L E 18.13

Modeling a Catch as a Rigid Body Impact Using the model shown in Fig. 2, what is the average force the catcher needs to apply to catch the flyer if the catcher is in an ideal position (i.e., has zero velocity) and the flyer has a free-fall speed 𝑣0 = 2 m∕s (after dropping about 20 cm)? In this model, the catcher/trapeze system is viewed as the nonuniform slender bar of mass 𝑚𝑐 = 90 kg, length 𝐿 = 4 m, mass center at 𝐸, mass moment of inertia 𝐼𝐸 = 30 kg⋅m2 , and 𝑤 = 1 m. The flyer is modeled as a uniform slender bar of length 𝐻 = 2 m and mass 𝑚𝑓 = 70 kg. We model the grasp between catcher and flyer as a pin connection and assume that, at the instant shown, catcher and flyer both have zero angular velocity. Finally, we assume that catcher and flyer can establish a firm grasp in 0.15 s (the best athletes’ reaction times are 120–160 ms), and since the catch happens quickly, we model it as an impact.

David Madison/Getty Images

Figure 1 A midair catch. The acrobat hanging at the knee from the trapeze is called the catcher, and the other acrobat is called the flyer.

𝐴

catcher 𝑐 𝐸

𝜃 = 20◦ 𝐿

𝑤

𝐶 𝐵

𝑄 𝑣0

Figure 2. Model of the catch between two trapeze artists. The situation shown is at the time of the catch.

SOLUTION 𝑅𝑥

𝐴 𝑅𝑦

𝐿

𝚤̂

𝐻

𝐸

𝜃

𝚥̂

Road Map & Modeling

𝑦

𝑤

𝐶 𝐵𝑂

𝑄

𝑥

Figure 3 Impact-relevant FBD of the catcher-flyer system. Points 𝐵, 𝐶, and 𝑂 are coincident. 𝐴

Governing Equations

catcher 𝑐 𝑦

𝑅𝑦

𝐿

𝑤

Referring to Fig. 4 and to Eq. (18.34) on p. 1243, applying the linear impulse-momentum principle to 𝑓 gives

𝑁𝑥

𝑥

𝐵 𝑂

𝚥̂ 𝚤̂

Balance Principles

𝑁𝑦

𝐸

𝜃

𝑚𝑓 𝑣⃗−𝑄 +

𝑦 flyer 𝑓 𝑁𝑥

𝐶 𝑂

𝑄

𝑥

𝑁𝑦 Figure 4 Impact-relevant FBD of 𝑐 and 𝑓 individually.

ISTUDY

By modeling the catch as an impact, we can (1) relate the preand postcatch velocities using the linear impulse-momentum principle and compute the impulse exerted by 𝑐 on 𝑓 , and (2) divide this impulse by the reaction time to compute the desired average force. Note that the impact we are modeling cannot be classified into any of the elastic or plastic categories found on p. 1266, since the two bodies attach but can rotate relative to one another. As usual, we consider the system’s FBD and the FBDs of 𝑐 and 𝑓 individually, shown in Figs. 3 and 4, respectively. At the time of catch, the fixed origin 𝑂 of the coordinate system coincides with points 𝐵 and 𝐶 at which 𝑐 and 𝑓 grasp one another. Since 𝑐 can only move in a fixed-axis rotation about 𝐴, the postcatch velocities consist of four quantities: the angular velocities of 𝑐 and 𝑓 and the two components of 𝑣⃗𝑄 .

𝑡+ (

∫𝑡−

) 𝑁𝑥 𝚤̂ + 𝑁𝑦 𝚥̂ 𝑑𝑡 = 𝑚𝑓 𝑣⃗+𝑄 .

(1)

The forces 𝑅𝑥 and 𝑅𝑦 have no moment about the fixed point 𝐴 (see Fig. 3), and so the system’s angular momentum about 𝐴 is conserved, that is, ( −) ( −) ( +) ( +) ⃗ ⃗ ⃗ ⃗ (2) ℎ 𝐴 𝑐 + ℎ𝐴 𝑓 = ℎ𝐴 𝑐 + ℎ𝐴 𝑓 , where [ ] ± ( ±) ⃗ = 𝐼𝐸 + 𝑚𝑐 (𝐿 − 𝑤)2 𝜔 ⃗𝑐 ℎ 𝐴 𝑐

and

( ±) ⃗ = 𝐼𝑄 𝜔 ⃗ ±𝑓 + 𝑟⃗𝑄∕𝐴 × 𝑚𝑓 𝑣⃗±𝑄 . ℎ 𝐴 𝑓

(3)

Note that the first of Eqs. (3) accounts for the fixed-axis rotation of 𝑐 about 𝐴. Next, since 𝐶 and 𝑂 coincide at the time of catch (see Fig. 4), 𝑁𝑥 and 𝑁𝑦 provide no moment on 𝑓 ( ) ⃗ about 𝑂. Thus, the angular momentum of 𝑓 relative to 𝑂, ℎ 𝑂 𝑓 , must be conserved, that is, ( +) ( ±) ( −) ⃗ ⃗ ⃗ (4) where ℎ = 𝐼𝑄 𝜔 ⃗ ±𝑓 + 𝑟⃗𝑄∕𝑂 × 𝑚𝑓 𝑣⃗±𝑄 . ℎ 𝑂 𝑓 = ℎ𝑂 𝑓 𝑂 𝑓

ISTUDY

Section 18.3

Impact of Rigid Bodies

Force Laws All forces are accounted for on the FBD. Also, in this impact we do not have a COR equation. Kinematic Equations

After the catch 𝑣⃗+𝐵 = 𝑣⃗+𝐶 , so we must have ⃗ +𝑓 × 𝑟⃗𝐶∕𝑄 , 𝜔 ⃗ +𝑐 × 𝑟⃗𝐵∕𝐴 = 𝑣⃗+𝑄 + 𝜔

(5)

where Eq. (5) enforces both rigid body kinematics and the fact that 𝐴 is a fixed point. Computation

Substituting Eqs. (3) into Eq. (2), we have

[ ] [ ] − 𝑚𝑓 𝑣0 𝐿 cos 𝜃 + (𝐻∕2) = 𝐼𝐸 + 𝑚𝑐 (𝐿 − 𝑤)2 𝜔+𝑐 + 𝐼𝑄 𝜔+𝑓

[ ] + 𝑚𝑓 𝑣+𝑄𝑥 𝐿 sin 𝜃 + 𝑚𝑓 𝑣+𝑄𝑦 𝐿 cos 𝜃 + (𝐻∕2) ,

(6)

where Eq. (6) is in scalar form since the only nonzero component of Eq. (2) is in the 𝑧 direction. Proceeding similarly with Eqs. (4), we have −𝑚𝑓 𝑣0 (𝐻∕2) = 𝐼𝑄 𝜔+𝑓 + 𝑚𝑓 𝑣+𝑄𝑦 (𝐻∕2).

(7)

Expanding the products in Eq. (5) and expressing the result in components, we have 𝜔+𝑐 𝐿 sin 𝜃 = 𝑣+𝑄𝑥

and 𝜔+𝑐 𝐿 cos 𝜃 = 𝑣+𝑄𝑦 − 𝜔+𝑓 (𝐻∕2).

(8)

Equations (6)–(8) form a system of four equations in the four unknowns 𝜔+𝑐 , 𝜔+𝑓 , 𝑣+𝑄𝑥 , and 𝑣+𝑄𝑦 . Recalling that 𝐼𝑄 = 𝑚𝑓 𝐻 2 ∕12, the solution of this system of equations is 𝐿𝑚𝑓 𝑣0 cos 𝜃 𝜔+𝑐 = − [ ( ] ) = −0.1080 rad∕s, 4 𝐼𝐸 + 𝑚𝑐 (𝐿 − 𝑤)2 + 𝐿2 𝑚𝑓 1 + 3 sin2 𝜃 𝜔+𝑓 = −

2𝐼𝐸 + 2𝑚𝑓 𝐿2 sin2 𝜃 + 2𝑚𝑐 (𝐿 − 𝑤)2 ] ( ) = −1.196 rad∕s, 𝐻 4 𝐼 + 𝑚 (𝐿 − 𝑤)2 + 𝐿2 𝑚 1 + 3 sin2 𝜃 𝐸 𝑐 𝑓

3𝑣0

[

(9)

(10)

𝐿2 𝑚𝑓 𝑣0 cos 𝜃 sin 𝜃 𝑣+𝑄𝑥 = − [ ] ( ) = −0.1477 m∕s, 4 𝐼𝐸 + 𝑚𝑐 (𝐿 − 𝑤)2 + 𝐿2 𝑚𝑓 1 + 3 sin2 𝜃

(11)

3𝐼𝐸 + 𝑚𝑓 𝐿2 (1 + 2 sin2 𝜃) + 3𝑚𝑐 (𝐿 − 𝑤)2 𝑣+𝑄𝑦 = −𝑣0 [ ] ( ) = −1.601 m∕s. 4 𝐼𝐸 + 𝑚𝑐 (𝐿 − 𝑤)2 + 𝐿2 𝑚𝑓 1 + 3 sin2 𝜃

(12)

Recalling that 𝑡+ − 𝑡− = 0.15 s, using the definition of average force over the time interval +( ) ( ⃗ = 1 ∫ −𝑡 𝑁 𝚤̂ + 𝑁 𝚥̂ 𝑑𝑡 = 𝑚𝑓 𝑣⃗+ − 𝑡+ − 𝑡− , and employing Eq. (1), we have 𝑁 + − 𝑡− + − 𝑡− avg 𝑥 𝑦 𝑡 𝑄 𝑡 𝑡 ) 𝑣⃗−𝑄 , which gives ( ) ( ) 𝑁𝑥 avg = −68.94 N and 𝑁𝑦 avg = 186.0 N Discussion & Verification

⇒

|⃗ | |𝑁avg | = 198.3 N. | |

(13)

Each of the results in Eqs. (9)–(12) has a negative value. This is as expected since it makes physical sense that, after the catch, the angular velocities of both the catcher and the flyer are clockwise, and that the flyer moves down and to the left. In addition, as expected, 𝑣+𝑄𝑦 < 𝑣0 , i.e., the catcher slows down the flyer’s drop. As far as the forces are concerned, the signs are also as expected, and their value is consistent with the velocity result and the given reaction time.

1277

1278

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Problems Problem 18.118 𝑣0

A stop shot is a pool shot in which the cue ball (white) stops upon striking the object ball (aqua). Modeling the collision between the two balls as a perfectly elastic collision of two rigid bodies with frictionless contact, determine which condition must be true for the preimpact angular velocity of the cue ball in order to properly execute a stop shot: (a) 𝜔0 < 0; (b) 𝜔0 = 0; (c) 𝜔0 > 0.

Figure P18.118

Problem 18.119 The cue ball (white) is rolling without slip to the left, and its center is moving with a speed 𝑣0 = 6 f t∕s while the object ball (aqua) is stationary. The diameter 𝑑 of the two balls is the same and is equal to 2.25 in. The coefficient of restitution of the impact is 𝑒 = 0.98. Let 𝑊𝑐 = 6 oz and 𝑊𝑜 = 5.5 oz be the weights of the cue ball and object ball, respectively. Let 𝑃 and 𝑄 be the points on the cue ball and on the object ball, respectively, that are in contact with the table at the time of impact. Assuming that the contact between the two balls is frictionless and modeling the balls as uniform spheres, determine the postimpact velocities of 𝑃 and 𝑄.

Figure P18.119

Problem 18.120 Consider the impact-relevant FBD of a car involved in a collision. Assume that, at the time of impact, the car was stationary. In addition, assume that the impulsive force 𝐹 , with line of action 𝓁, is the only impulsive force acting on the car at the time of impact. The point 𝑃 at the intersection of 𝓁 and the line perpendicular to 𝓁 and passing through 𝐺, the center of mass of the car, is sometimes referred to as the center of percussion (for an alternative definition of center of percussion, see Example 17.5 on p. 1166). Is it true that, at the time of impact, the instantaneous center of rotation of the car lies on the same line as 𝑃 and 𝐺?

𝐹 𝐺 𝑃 Figure P18.120

Problem 18.121 A basketball with mass 𝑚 = 0.6 kg is rolling without slipping as shown when it hits a small step with 𝓁 = 7 cm. Letting the ball’s radius be 𝑟 = 12.0 cm, modeling the ball as a thin spherical shell (the mass moment of inertia of a spherical shell about its mass center is 32 𝑚𝑟2 ), and assuming that the ball does not rebound off the step or slip relative to it, determine 𝑣0 such that the ball barely makes it over the step.

𝑣0

𝑑

𝓁

Problem 18.122

𝐵 𝐴 Figure P18.122

ISTUDY

Figure P18.121

𝜃

A bullet 𝐵 of mass 𝑚𝑏 is fired with a speed 𝑣0 as shown against a uniform thin rod 𝐴 of length 𝓁 and mass 𝑚𝑟 that is pinned at 𝑂. Determine the distance 𝑑, such that no horizontal reaction is felt at the pin when the bullet strikes the rod.

ISTUDY

Section 18.3

1279

Impact of Rigid Bodies

Problem 18.123 A bullet 𝐵 weighing 147 gr (1 lb = 7000 gr) is fired with a speed 𝑣0 as shown and becomes embedded in the center of a rubber block of dimensions ℎ = 4.5 in. and 𝑤 = 6 in. weighing 𝑊rb = 2 lb. The rubber block is attached to the end of a uniform thin rod 𝐴 of length 𝐿 = 1.5 f t and weight 𝑊𝑟 = 5 lb that is pinned at 𝑂. After the impact, the rod (with the block and the bullet embedded in it) swings upward to an angle of 60◦ . Determine the speed of the bullet right before impact.

𝜔

𝐿 𝐵

ℎ

𝑣0 Figure P18.123

Problem 18.124 Solve the problem in Example 17.5 on p. 1166 using momentum methods and the concept of impulsive force. Specifically, consider a ball hitting a bat at a distance 𝑑 from the handle when the batter has “choked up” a distance 𝛿. Find the “sweet spot” 𝑃 (more properly called the center of percussion∗ ) of the bat 𝐵 by determining the distance 𝑑 at which the ball should be hit so that the lateral force (i.e., perpendicular to the bat) at 𝑂 is zero. Assume that the bat is pinned at 𝑂, it has mass 𝑚, the mass center is at 𝐺, and the mass moment of inertia is 𝐼𝐺 . 𝑄 𝜔0

𝛿 𝓁 Figure P18.124

𝑂 𝛿 𝓁

𝑣𝑏

Figure P18.125

Problem 18.125 A batter is swinging a 34 in. long bat with weight 𝑊𝐵 = 32 oz, mass center 𝐺, and mass moment of inertia 𝐼𝐺 = 0.0413 slug⋅f t 2 . The center of rotation of the bat is point 𝑄. Compute the distance 𝑑 identifying the position of point 𝑃 , the bat’s “sweet spot” or center of percussion, such that the batter will not feel any impulsive forces at 𝑂 where he is grasping the bat. In addition, knowing that the ball, weighing 5 oz, is traveling at a speed 𝑣𝑏 = 90 mph and that the batter is swinging the bat with an angular velocity 𝜔0 = 45 rad∕s, determine the speed of the ball and the angular velocity of the bat immediately after impact. To solve the problem, use the following data: 𝛿 = 6 in., 𝜌 = 14 in., 𝓁 = 22.5 in., and COR 𝑒 = 0.5.

Problem 18.126 A thin homogeneous bar 𝐴 of length 𝓁 = 1.75 m and mass 𝑚 = 23 kg is translating as shown with a speed 𝑣0 = 12 m∕s when it collides with the fixed obstacle 𝐵. Modeling the contact between the bar and obstacle as frictionless, letting 𝛽 = 32◦ , and letting the distance 𝑑 = 0.46 m, determine the angular velocity of the bar immediately after the collision, knowing that the COR for the impact is 𝑒 = 0.74. Hint: The LOI of the collision is normal to the axis of the thin bar.

Problem 18.127

𝐴 𝛽 𝑣0

𝐵

Figure P18.126

𝐴

A uniform bar 𝐴 with a hook 𝐻 at the end is dropped from rest as shown from a height 𝑑 = 3 f t over a fixed pin 𝐵. Letting the weight and length of 𝐴 be 𝑊 = 100 lb and 𝓁 = 7 f t, respectively, determine the angle 𝜃 that the bar will sweep through if the bar becomes hooked with 𝐵 and does not rebound. Although bar 𝐴 becomes hooked with 𝐵, ∗ See R. Cross,

“The Sweet Spot of a Baseball Bat,” American Journal of Physics, 66(9), 1998, pp. 772–779.

𝐺

𝑑 𝜃

Figure P18.127

𝐵

1280

Chapter 18

Energy and Momentum Methods for Rigid Bodies

assume that there is no friction between the hook and the pin. Hint: Conservation of energy before and after the collision provide useful additions to impulse-momentum principles.

Problem 18.128 A drawbridge of length 𝓁 = 30 f t and weight 𝑊 = 600 lb is released in the position shown and freely pivots clockwise until it strikes the right end of the moat. If the COR for the collision between the bridge and the ground is 𝑒 = 0.45 and if the contact point between the bridge and the ground is effectively 𝓁 away from the bridge’s pivot point, determine the angle to which the bridge rebounds after the collision. Neglect any possible source of friction. 𝑣0 𝓁

𝐵

𝜃 𝐺

Figure P18.128

Figure P18.129

Problem 18.129 A stick 𝐴 with length 𝓁 = 1.55 m and mass 𝑚𝐴 = 6 kg is in static equilibrium as shown when a ball 𝐵 with mass 𝑚𝐵 = 0.15 kg traveling at a speed 𝑣0 = 30 m∕s strikes the stick at distance 𝑑 = 1.3 m from the lower end of the stick. If the COR for the impact is 𝑒 = 0.85, determine the velocity of the mass center 𝐺 of the stick, as well as the stick’s angular velocity right after the impact. Hint: Since the stick is simply standing (not pinned) on the floor and the collision is parallel to the floor, any reactions that might develop between the stick and the floor are non-impulsive.

Problem 18.130 A gymnast on the uneven parallel bars has a vertical speed 𝑣0 and no angular speed when she grasps the upper bar. Model the gymnast as a single uniform rigid bar 𝐴 of weight 𝑊 = 92 lb and length 𝓁 = 6 f t. Neglecting all friction, letting 𝛽 = 12◦ , and assuming that the upper bar 𝐵 does not move after the gymnast grasps it, determine the minimum speed 𝑣0 for the gymnast to swing (counterclockwise) into the horizontal position on the other side of the bar. Assume that, during the motion, the friction between the gymnast’s hands and the upper bar is negligible.

𝛽 𝐴

𝑣0

𝐺

𝐵

Corbis/VCG/Getty Images

2𝑟

Figure P18.130

𝐴 𝑑

Problem 18.131 𝛽 Figure P18.131

ISTUDY

A uniform thin ring 𝐴 of mass 𝑚 = 7 kg and radius 𝑟 = 0.5 m is released from rest as shown and rolls without slip until it meets a step of height 𝓁 = 0.45 m. Letting 𝛽 = 12◦ and assuming that the ring does not rebound off the step or slip relative to it, determine

ISTUDY

Section 18.3

Impact of Rigid Bodies

1281

the distance 𝑑, such that the ring barely makes it over the step. Hint: A careful sketch of the ring as it impacts the step is an important place to begin. Note that the LOI is parallel to the step surface and does not go through the center of mass of the ring.

Problem 18.132 Two identical uniform bars 𝐴𝐵 and BD are pin-connected at 𝐵, and bar BD has a hook at the free end. The two bars are dropped as shown from a height 𝑑 = 3 f t over a fixed pin 𝐸 (shown in cross section). Letting the weight and length of each bar be 𝑊 = 100 lb and 𝓁 = 7 f t, respectively, determine the angular velocities of 𝐴𝐵 and BD immediately after bar BD becomes hooked on 𝐸 and does not rebound. Hint: The angular momentum of bar AB is conserved about 𝐵 during impact.

𝐴

𝐵

𝐷 𝐸

Figure P18.132

Problem 18.133 Cars 𝐴 and 𝐵 collide as shown. Determine the angular velocities of 𝐴 and 𝐵 immediately after impact if the COR is 𝑒 = 0.35. In solving the problem, let 𝐶 and 𝐷 be the mass centers of 𝐴 and 𝐵, respectively. In addition, enforce assumption 3 on p. 1266 and use the following data: 𝑊𝐴 = 3130 lb (weight of 𝐴), 𝑘𝐶 = 34.5 in. (radius of gyration of 𝐴), 𝑣𝐶 = 12 mph (speed of the mass center of 𝐴), 𝑊𝐵 = 3520 lb (weight of 𝐵), 𝑘𝐷 = 39.3 in. (radius of gyration of 𝐵), 𝑣𝐷 = 15 mph (speed of the mass center of 𝐵), 𝑑 = 19 in., 𝓁 = 79 in., 𝛿 = 7.1 in., 𝜌 = 65 in., and 𝛽 = 12◦ .

𝐵

𝐷 𝑑

LOI

𝑣𝐷 𝛿

𝑂 𝑣𝐶

𝐶

𝐴

Figure P18.133

Problem 18.134 Consider the collision of two rigid bodies 𝐴 and 𝐵, which, referring to Example 18.10 on p. 1252, models the docking of the Space Shuttle (body 𝐴) to the International Space Station (body 𝐵). As in Example 18.10, we assume that 𝐵 is stationary relative to an inertial frame of reference and that 𝐴 translates as shown. In contrast to Example 18.10, here we assume that 𝐴 and 𝐵 join at point 𝑄 but, due to the flexibility of the docking system, can rotate relative to one another. Determine the angular velocities of 𝐴 and 𝐵 right after docking if 𝑣0 = 0.03 m∕s. In solving the problem, let 𝐶 and 𝐷 be the centers of mass of 𝐴 and 𝐵, respectively. In addition, let the mass and mass moment of inertia of 𝐴 be 𝑚𝐴 = 120×103 kg and 𝐼𝐶 = 14×106 kg⋅m2 , respectively, and the mass and mass moment of inertia of 𝐵 be 𝑚𝐵 = 180 × 103 kg and 𝐼𝐷 = 34 × 106 kg⋅m2 , respectively. Finally, use the following dimensions: 𝓁 = 24 m, 𝑑 = 8 m, 𝜌 = 2.6 m, and 𝛿 = 2.4 m.

𝐵 (ISS)

𝜌 𝐷

𝛿

𝑄 𝐶 𝑣0

Figure P18.134

𝐴 (shuttle)

1282

Chapter 18

Energy and Momentum Methods for Rigid Bodies

18.4 C h a p t e r R e v i e w Work-energy principle for rigid bodies In this section, we discovered that the kinetic energy 𝑇 of a rigid body 𝐵 in planar motion is given by Eq. (18.8), p. 1208 𝑇 = 12 𝑚𝑣2𝐺 + 21 𝐼𝐺 𝜔2𝐵 , where 𝑚, 𝐺, and 𝐼𝐺 are the body’s mass, center of mass, and mass moment of inertia, respectively, or by Eq. (18.11), p. 1208 𝑇 = 12 𝐼𝑂 𝜔2𝐵 , which applies to fixed-axis rotation, and where 𝐼𝑂 is the mass moment of inertia relative to the axis of rotation. The work-energy principle for a rigid body is Eq. (18.16), p. 1209 𝑇1 + 𝑈1-2 = 𝑇2 , where 𝑈1-2 is the work done on the body in going from 1 to 2 by only the external forces. If we make use of the potential energy 𝑉 of conservative forces, the work-energy principle can also be written as Eqs. (18.23) and (18.24), p. 1211 ) ( 𝑇1 + 𝑉1 + 𝑈1-2 nc = 𝑇2 + 𝑉2 (general systems), 𝑇1 + 𝑉1 = 𝑇2 + 𝑉2

(conservative systems),

where the second expression applies only to a conservative system and states that the total mechanical energy of the system is conserved. For systems of rigid bodies or mixed systems of rigid bodies and particles, the work-energy principle is 𝜔𝐴𝐵 𝑟⃗𝐴∕𝐵 𝐹⃗𝐵

𝐴

𝜃

𝐹⃗𝐴

Eq. (18.29), p. 1212 ( )ext ( )int 𝑇1 + 𝑉1 + 𝑈1-2 nc + 𝑈1-2 nc = 𝑇2 + 𝑉2 , where 𝑉 is the total potential energy (of both external and internal conservative forces) ( )ext )int ( and 𝑈1-2 nc and 𝑈1-2 nc are the work contributions due to external and internal forces for which we do not have a potential energy, respectively. Referring to Fig. 18.31, in planar rigid body motion, the work of a couple is

𝐵 ℎ

Eq. (18.22), p. 1210

𝑥 Figure 18.31 Rigid body subject to a couple.

ISTUDY

𝑈1-2 =

∫𝑡

1

𝑡2

𝑀𝜔𝐴𝐵 𝑑𝑡 =

∫𝜃

𝜃2

𝑀 𝑑𝜃,

1

where 𝑑𝜃 is the body’s infinitesimal angular displacement, 𝑀 is the component of the moment of the couple in the direction perpendicular to the plane of motion, taken to be positive in the direction of positive 𝜃, and 𝜃 must be measured in radians. In addition, ⃗ is computed as the power developed by a couple with moment 𝑀 Eq. (18.31), p. 1212 Power developed by a couple =

𝑑𝑈 ⃗ ⋅𝜔 =𝑀 ⃗ 𝐴𝐵 . 𝑑𝑡

ISTUDY

Section 18.4

1283

Chapter Review

Momentum methods for rigid bodies

𝐹⃗1

In this section, we derived the linear and the angular impulse-momentum principles for rigid bodies. Referring to Fig. 18.32, the linear impulse-momentum principle for a rigid body reads Eqs. (18.34), p. 1243 𝑚𝑣⃗𝐺1 +

∫𝑡

𝑡2

𝐹⃗ 𝑑𝑡 = 𝑚𝑣⃗𝐺2

or

𝑝⃗1 +

1

∫𝑡

𝑡2

body 𝐵 𝑎⃗𝐺

𝐹⃗2

𝐹⃗ 𝑑𝑡 = 𝑝⃗2 ,

𝑣⃗𝐺

𝐺

1

where 𝐹⃗ is the total external force on 𝐵, 𝑝⃗ = 𝑚𝑣⃗𝐺 is 𝐵’s linear momentum, and 𝑣⃗𝐺 is the ⃗ we have velocity of 𝐵’s center of mass. If 𝐹⃗ = 0, Eqs. (18.36), p. 1243 𝑚𝑣⃗𝐺1 = 𝑚𝑣⃗𝐺2

or

𝑝⃗1 = 𝑝⃗2 ,

𝐹⃗𝑁

Figure 18.32 A rigid body under the action of a system of forces.

and we say that the body’s momentum is conserved. If 𝑃 in Fig. 18.33 is a moment center coplanar with 𝐺 and if 𝐵 is symmetric relative to the plane of motion, the angular momentum of 𝐵 relative to 𝑃 is 𝑑𝑚

⃗ =𝐼 𝜔 ⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , ℎ 𝑃 𝐺 ⃗𝐵 + 𝑟

Eq. (18.42), p. 1244 ⃗ + ℎ 𝑃1

∫𝑡

𝑡2

⃗ . ⃗ 𝑑𝑡 = ℎ 𝑀 𝑃 𝑃2

𝑃

𝑟⃗𝑑𝑚∕𝑃

Eq. (18.40), p. 1244

where 𝐼𝐺 is the mass moment of inertia of 𝐵, 𝜔 ⃗ 𝐵 is the angular velocity of 𝐵, and 𝑟⃗𝐺∕𝑃 is the position of 𝐺 relative to 𝑃 . If 𝑃 is chosen so that (1) 𝑃 is fixed or (2) 𝑃 coincides ⃗ =ℎ ⃗̇ , where 𝑀 ⃗ is the with 𝐺 or (3) 𝑃 and 𝐺 move parallel to one another, then 𝑀 𝑃 𝑃 𝑃 moment relative to 𝑃 of the external force system acting on 𝐵. When this equation holds, by integrating with respect to time over a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2 , we have

𝑥

𝑟⃗𝑃 𝜔𝐵

𝑟⃗𝐺∕𝑃

𝑣⃗𝑑𝑚 𝑟⃗𝑑𝑚 𝐺

𝑣⃗𝐺 𝐵

𝑟⃗𝐺

𝑥 Figure 18.33 The quantities needed to obtain the angular momentum relationships for a rigid body.

1

When 𝑃 coincides with 𝐺, or if the body undergoes a fixed-axis rotation about a point 𝑂 as shown in Fig. 18.34, then the above equation becomes Eq. (18.43), p. 1244, and Eq. (18.45), p. 1245 𝐼𝐺1 𝜔𝐵1 +

∫𝑡

𝑡2

∫𝑡

𝑡2

𝑣⃗𝐺

𝑀𝐺𝑧 𝑑𝑡 = 𝐼𝐺2 𝜔𝐵2 ,

1

𝐼𝑂1 𝜔𝐵1 +

𝐵 𝐺

𝑀𝑂𝑧 𝑑𝑡 = 𝐼𝑂2 𝜔𝐵2 ,

1

respectively, where 𝐼𝑂 is the mass moment of inertia about the fixed axis of rotation.

𝜔𝐵

𝑟⃗𝐺∕𝑂

𝑂

Impact of rigid bodies In this section, we studied planar rigid body impacts. We learned that, contrary to particle impacts, rigid bodies can experience eccentric impacts. These are collisions in which at least one of the mass centers of the impacting bodies does not lie on the LOI. We also learned that the basic concepts used in particle impacts are applicable to rigid body impacts, and we reviewed solution strategies for a variety of situations.

Figure 18.34 A rigid body in a fixed-axis rotation.

𝑥

1284

Chapter 18

Energy and Momentum Methods for Rigid Bodies

As with particle impacts, we say that a rigid body impact is plastic if the COR 𝑒 = 0. A rigid body impact will be called perfectly plastic if the colliding bodies form a single rigid body after impact. An impact is elastic if the COR 𝑒 is such that 0 < 𝑒 < 1. Finally, the ideal case with 𝑒 = 1 is referred to as a perfectly elastic impact. In addition, an impact is unconstrained if the system consisting of the two impacting objects is not subject to external impulsive forces; otherwise, the impact is called constrained. We have considered only impacts satisfying the following assumptions: 1. The impact involves only two bodies in planar motion where no impulsive force has a component perpendicular to the plane of motion. 2. Contact between any two rigid bodies occurs at only one point, and at this point we can clearly define the LOI. 3. The contact between the bodies is frictionless. We recommend organizing the solution of any impact problem as follows: • Begin with an FBD of the impacting bodies as a system and FBDs for each of the colliding bodies. Neglect nonimpulsive forces. • Choose a coordinate system with the origin coincident with the points that come into contact at the time of impact. Recall that the origin of such a coordinate system is a fixed point. • Enforce the linear and/or the angular impulse-momentum principles for the system and/or for the individual bodies. In applying the angular impulse-momentum principle for the whole system, the moment center should be a fixed point, whereas for an individual body, the moment center should be a fixed point or the body’s mass center.

LOI

𝑄

𝐶

𝐸

• For plastic, elastic, and perfectly elastic impacts, the COR equation is first written using the velocity components along the LOI of the points that actually come into contact. For example, for the impact shown in Fig. 18.35, the COR equation is first written as Eq. (18.59), p. 1269

𝐷 𝑂

𝐵

𝑦 𝑥

𝚤̂

LOI

𝐶

𝐸

) ( 𝑣+𝐸𝑥 − 𝑣+𝑄𝑥 = 𝑒 𝑣−𝑄𝑥 − 𝑣−𝐸𝑥 ,

𝚥̂ (b)

𝑁 𝐴 𝑦

𝑂

𝑁 𝐷 𝑄

𝐵

Figure 18.35 FBDs of the colliding bodies (a) as a system and (b) individually.

ISTUDY

where the COR 𝑒 is such that 0 ≤ 𝑒 ≤ 1. The COR equation must then be rewritten in terms of the angular velocities and velocities of the mass centers using rigid body kinematics. For the situation in Fig. 18.35, this means rewriting the COR equation using the relations 𝑣⃗±𝐸 = 𝑣⃗±𝐶 + 𝜔±𝐴 𝑘̂ × 𝑟⃗𝐸∕𝐶 and 𝑣⃗±𝑄 = 𝑣⃗±𝐷 + 𝜔±𝐵 𝑘̂ × 𝑟⃗𝑄∕𝐷 . Notice that the relative position vectors 𝑟⃗𝐸∕𝐶 and 𝑟⃗𝑄∕𝐷 do not have the ± superscript because they are treated as constants during the impact. • In perfectly plastic impacts, kinematic constraint equations must be enforced that express the fact that two bodies form a single rigid body after impact.

ISTUDY

Section 18.4

1285

Chapter Review

Review Problems Problem 18.135 A uniform thin ring 𝐴 and a uniform disk 𝐵 roll without slip as shown. Letting 𝑇𝐴 and 𝑇𝐵 be the kinetic energies of 𝐴 and 𝐵, respectively, if the two objects have the same mass and radius and if their centers are moving with the same speed 𝑣0 , state which of the following statements is true and why: (a) 𝑇𝐴 < 𝑇𝐵 ; (b) 𝑇𝐴 = 𝑇𝐵 ; (c) 𝑇𝐴 > 𝑇𝐵 .

+

𝑅

𝑅

𝐴

+

𝐵

𝑚

𝑚

Figure P18.135

Problem 18.136 At the instant shown, the disk 𝐷, which has mass 𝑚 and radius of gyration 𝑘𝐺 , is rolling without slip down the flat incline with angular velocity 𝜔0 . The disk is attached at its center to a wall by a linear elastic spring of constant 𝑘. If, at the instant shown, the spring is unstretched, determine the distance 𝑑 down the incline that the disk rolls before coming to a stop. Use 𝑘 = 65 N∕m, 𝑅 = 0.3 m, 𝑚 = 10 kg, 𝑘𝐺 = 0.25 m, 𝜔0 = 60 rpm, and 𝜃 = 30◦ .

𝑘

𝜔0

𝑅 𝐺 𝐷

Problem 18.137

𝜃

A pendulum consists of a uniform disk 𝐴 of diameter 𝑑 = 5 in. and weight 𝑊𝐴 = 0.25 lb attached at the end of a uniform bar 𝐵 of length 𝐿 = 2.75 f t and weight 𝑊𝐵 = 1.3 lb. At the instant shown, the pendulum is swinging with an angular velocity 𝜔 = 0.55 rad∕s clockwise. Determine the kinetic energy of the pendulum at this instant, using Eq. (18.11) on p. 1208. 𝐷 𝑂 𝐵 𝜔

𝜔𝑂𝐶

Figure P18.136

𝐶 𝑅𝐷

𝜔𝐷

𝑅𝑆 𝑂

𝑆 Figure P18.137

Figure P18.138

Problem 18.138 A uniform disk 𝐷 of radius 𝑅𝐷 = 7 mm and mass 𝑚𝐷 = 0.15 kg is connected to point 𝑂 via the rotating arm OC and rolls without slip over the stationary cylinder 𝑆 of radius 𝑅𝑆 = 15 mm. Assuming that 𝜔𝐷 = 25 rad∕s, and treating the arm OC as a uniform slender bar of length 𝐿 = 𝑅𝐷 + 𝑅𝑆 and mass 𝑚𝑂𝐶 = 0.08 kg, determine the kinetic energy of the system.

𝐵 𝐴

Problem 18.139 The figure shows the cross section of a garage door with length 𝐿 = 9 f t and weight 𝑊 = 175 lb. At the ends 𝐴 and 𝐵 there are rollers of negligible mass constrained to move in a vertical and a horizontal guide, respectively. The door’s motion is assisted by two springs (only one spring is shown), each with constant 𝑘 = 9.05 lb∕f t. If the door is released from rest when horizontal and the spring is stretched 4 in., neglecting friction, and modeling the door as a uniform thin plate, determine the speed with which 𝐵 strikes the left end of the horizontal guide.

floor Figure P18.139

1286

Chapter 18

Energy and Momentum Methods for Rigid Bodies

Problem 18.140 Body 𝐵 has mass 𝑚 and mass moment of inertia 𝐼𝐺 , where 𝐺 is the mass center of 𝐵. If 𝐵 is in fixed-axis rotation about its center of mass 𝐺, determine which of( the) fol|(⃗ ) | |( ⃗ ) | | ⃗ |( ⃗ ) | | < ℎ , (b) ℎ lowing statements is true and why: (a) | ℎ | | | | | = | ℎ |, | 𝑃 𝐵| | 𝐸 𝐵| | 𝑃 𝐵| | 𝐸 𝐵| ( ( ) ) | ⃗ | | ⃗ | (c) | ℎ𝐸 𝐵 | > | ℎ𝑃 𝐵 |. | | | | 𝐵 𝐵

𝐶

𝜔𝐵

𝜔𝐴𝐵

𝐺

𝐴 𝜙

𝐸

𝐷

𝑃

Figure P18.140

Figure P18.141

Problem 18.141 The weights of the uniform thin pin-connected bars 𝐴𝐵, BC, and CD are 𝑊𝐴𝐵 = 4 lb, 𝑊𝐵𝐶 = 6.5 lb, and 𝑊𝐶𝐷 = 10 lb, respectively. Letting 𝜙 = 47◦ , 𝑅 = 2 f t, 𝐿 = 3.5 f t, and 𝐻 = 4.5 f t, and knowing that bar 𝐴𝐵 rotates at a constant angular velocity 𝜔𝐴𝐵 = 4 rad∕s, compute the angular momentum of the system about 𝐷 at the instant shown.

𝑂

Problem 18.142

𝑑

Consider Prob. 18.87 on p. 1256 in which an eccentric wheel 𝐵 is spun from rest under the action of a known torque 𝑀. In that problem, it was said that the wheel was in the horizontal plane. Is it possible to solve Prob. 18.87 by just applying Eq. (18.42) on p. 1244 if the wheel is in the vertical plane? Why?

𝐺

Figure P18.142

Problem 18.143 𝜔𝐵 𝐵

𝜔𝐴 𝐴

𝐶 𝜔𝐶

Figure P18.143

𝐸

The uniform disk 𝐴 of mass 𝑚𝐴 = 1.2 kg and radius 𝑟𝐴 = 0.25 m is mounted on a vertical shaft that can translate along the horizontal rod 𝐸. The uniform disk 𝐵, of mass 𝑚𝐵 = 0.85 kg and radius 𝑟𝐵 = 0.18 m, is mounted on a vertical shaft that is rigidly attached to 𝐸. Disk 𝐶 has a negligible mass and is rigidly attached to 𝐸; i.e., 𝐶 and 𝐸 form a single rigid body. Disk 𝐴 can rotate about the axis 𝓁𝐴 , disk 𝐵 can rotate about the axis 𝓁𝐵 , and the arm 𝐸 along with 𝐶 can rotate about the fixed axis 𝓁𝐶 . While keeping both 𝐵 and 𝐶 stationary, disk 𝐴 is initially spun with 𝜔𝐴 = 1200 rpm. Disk 𝐴 is then brought in contact with 𝐶 (contact is maintained by a spring), and at the same time, both 𝐵 and 𝐶 (and the arm 𝐸) are free to rotate. Due to friction between 𝐴 and 𝐶, 𝐶 along with 𝐸 and disk 𝐵 start spinning. Eventually 𝐴 and 𝐶 will stop slipping relative to one another. Disk 𝐵 always rotates without slip over 𝐶. Let 𝑑 = 0.27 m and 𝑤 = 0.95 m. Assuming that the only elements of the system that have mass are 𝐴, 𝐵, and 𝐸 and that 𝑚𝐸 = 0.3 kg, and assuming that all friction in the system can be neglected except for that between 𝐴 and 𝐶 and between 𝐶 and 𝐵, determine the angular speeds of 𝐴, 𝐵, and 𝐶 (the angular velocity of 𝐶 is the same as that of 𝐸 since they form a single rigid body), when 𝐴 and 𝐶 stop slipping relative to one another.

Problem 18.144

Figure P18.144

ISTUDY

A billiard ball is rolling without slipping with a speed 𝑣0 = 6 f t∕s as shown when it hits the rail. According to regulations, the nose of the rail is at a height from the table bed of 63.5% of the ball’s diameter (i.e., 𝓁∕(2𝑟) = 0.635). Model the impact with the rail as

ISTUDY

Section 18.4

Chapter Review

perfectly elastic, neglect friction between the ball and the rail, as well as between the ball and the table, and neglect any vertical motion of the ball. Based on the stated assumptions, determine the velocity of the point of contact between the ball and the table right after impact. The diameter of the ball is 2𝑟 = 2.25 in., and the weight of the ball is 𝑊 = 5.5 oz. Hint: The LOI goes through the nose and the ball’s mass center.

Problem 18.145 A basketball with mass 𝑚 = 0.6 kg is rolling without slipping as shown when it hits a small step with 𝓁 = 7 cm. Letting the ball’s radius be 𝑟 = 12.0 cm, modeling the ball as a thin spherical shell (the mass moment of inertia of a spherical shell about its mass center is 32 𝑚𝑟2 ), and assuming that the ball does not rebound off the step or slip relative to it, determine the maximum value of 𝑣0 for which the ball will roll over the step without losing contact with it.

𝑣0

Figure P18.145

Problem 18.146 A bullet 𝐵 weighing 147 gr (1 lb = 7000 gr) is fired with a speed 𝑣0 = 2750 f t∕s as shown against a thin uniform rod 𝐴 of length 𝓁 = 3 f t, weight 𝑊𝑟 = 35 lb, and pinned at 𝑂. If 𝑑 = 1.5 f t and the COR for the impact is 𝑒 = 0.25, determine the bar’s angular velocity immediately after the impact. In addition, determine the maximum value of the angle 𝜃 to which the bar swings after impact.

𝑑

𝜃 𝓁

𝐵 𝐴 Figure P18.146

Problem 18.147 An airplane is about to crash-land on only one wheel with a vertical component of speed 𝑣0 = 2 f t∕s and zero roll, pitch, and yaw. Determine the vertical component of velocity of the center of mass of the airplane 𝐺, as well as the airplane’s angular velocity immediately after touching down, assuming that (1) the only available landing gear is rigid and rigidly attached to the airplane, (2) the coefficient of restitution between the landing gear and the ground is 𝑒 = 0.1, (3) the airplane can be modeled as a rigid body, (4) the mass center 𝐺 and the point of first contact between the landing gear and the ground are in the same plane perpendicular to the longitudinal axis of the airplane, and (5) friction between the landing gear and the ground is negligible. In solving the problem use the following data: 𝑊 = 2500 lb (weight of the airplane), 𝐺 is the mass center of the airplane, 𝑘𝐺 = 3 f t is the radius of gyration of the airplane, and 𝑑 = 5.08 f t.

𝑣0 Figure P18.147

1287

ISTUDY

ISTUDY

Mechanical Vibrations

19 A vibration is a type of dynamic behavior in which a system or part of a system oscillates about an equilibrium position. Vibrations occur in mechanical, electrical, thermal, and fluid systems, but we will consider only those occurring in mechanical systems. Vibrations in mechanical systems can be undesirable (e.g., vibration in structures can lead to failure and can create unwanted noise), or they can be desirable (e.g., vibration in fluid systems can dissipate unwanted energy, and music would be impossible without vibrations).

Bob Parent/Hulton Archive/Getty Images

Thelonious Monk and friends in 1953 at the Open Door in New York City (from left to right, Charles Mingus, Roy Haynes, Thelonious Monk, and Charlie Parker). The music we hear is the result of vibrations induced in air by musical instruments.

19.1

Undamped Free Vibration

Oscillation of a railcar after coupling Recall Example 13.5 on p. 802 in which a railcar ran into a large spring that was designed to stop it (see Fig. 19.1). In that example, we were interested in the maximum compression of the spring and the time it took to stop the railcar. After maximum compression was reached, the spring would push the railcar back. If the railcar were to couple to the spring, the railcar would overshoot the equilibrium position of the spring and the railcar would start oscillating back and forth on the tracks. Let’s look at this motion. In that example, we found the equation of motion of the railcar to be 𝑥̈ +

𝑘 𝑥 = 0, 𝑚

(19.1)

𝑥

Figure 19.1 A railcar hitting a spring. The coordinate 𝑥 measures the displacement of the spring from its equilibrium position. Recall from Example 13.5 that the railcar and its load weigh 87 tons and are moving at 4 mph at impact. Also, recall that we found 𝑘 = 22,270 lb∕f t. We assume that the trailer does not move relative to the railcar.

where 𝑘 is the spring constant, 𝑚 is the mass of the railcar and its load, and 𝑥 is measured from the equilibrium position of the spring (see Fig. 19.1). Using 𝑥(0) = 𝑥𝑖 and 𝑥(0) ̇ = 𝑣𝑖 for initial conditions, we were able to integrate this equation of motion to obtain time as a function of position [see Eq. (11) on p. 803], which can be inverted

1289

1290

Chapter 19

Mechanical Vibrations

to obtain

√ 𝑥(𝑡) =

( √ )] [√ 𝑣2𝑖 + 𝑚𝑘 𝑥2𝑖 𝑥𝑖 𝑘∕𝑚 𝑘 −1 sin 𝑡 + tan . √ 𝑚 𝑣𝑖 𝑘∕𝑚

(19.2)

Here, we have treated both 𝑥𝑖 and 𝑣𝑖 as positive quantities and then used the trigono) ( √ metric identity sin−1 1∕ 𝑧2 + 1 = tan−1 (1∕𝑧). Equation (19.2) looks complicated, but it is really of the form ( ) 𝑥(𝑡) = 𝐶 sin 𝜔𝑛 𝑡 + 𝜙 , (19.3) where

√ 𝜔𝑛 =

𝑘 , 𝑚

√ 𝑣2𝑖 ∕𝜔2𝑛 + 𝑥2𝑖 , 𝑥𝜔 tan 𝜙 = 𝑖 𝑛 . 𝑣𝑖 𝐶=

𝐶 sin 𝜙 𝜋 2

− 𝜋2

𝜋

3𝜋 2

2𝜋

𝑡

𝜙 𝜔𝑛

Figure 19.2 Plot of 𝑥(𝑡) in Eq. (19.3) showing the amplitude 𝐶, phase angle 𝜙, and period 𝜏 of a harmonic oscillator.

ISTUDY

(19.4) (19.5) (19.6)

The quantity 𝜔𝑛 is a constant called the natural frequency∗ of vibration, and it is expressed in rad∕s in both the SI and U.S. Customary unit systems. The quantities 𝐶 and 𝜙, called the amplitude and phase angle of vibration, respectively, are constants that depend on the initial conditions and 𝜔𝑛 . Consistent with Eq. (19.5), the amplitude of vibration 𝐶 is understood to be a positive quantity. The angle 𝜙 can be determined via Eq. (19.6) for 𝑣𝑖 ≠ 0. When 𝑣𝑖 = 0, 𝜙 can be chosen to be equal to −𝜋∕2 or 𝜋∕2 rad, depending on whether 𝑥𝑖 < 0 or 𝑥𝑖 > 0, respectively. Equation (19.3), which is a solution of Eq. (19.1), describes a harmonic motion with natural frequency 𝜔𝑛 . For this reason, the physical system modeled by Eq. (19.1) is called a harmonic oscillator. Equation (19.1) represents an undamped vibration because there are no terms that depend on 𝑥, ̇ which would occur with viscous damping or in some models of aerodynamic drag. Equation (19.1) also represents a free vibration since it is homogeneous; that is, there are only terms containing the dependent variable 𝑥 and no terms that are functions of time or are constant.† The function in Eq. (19.3) is plotted in Fig. 19.2. The oscillator represented by Eq. (19.3) completes one cycle in the time 𝜏=

2𝜋 . 𝜔𝑛

(19.7)

The quantity 𝜏 (the Greek letter tau) is called the period of the vibration (see Fig. 19.2). Finally, the number of cycles of vibration per unit of time is called the frequency, and it is defined as 1 𝜔 𝑓 = = 𝑛. (19.8) 𝜏 2𝜋 Generally the frequency 𝑓 is expressed in cycles per second or hertz (Hz). Applying these ideas to the railcar, we see that if it couples to the spring, its natural frequency is 𝜔𝑛 = 2.030 rad∕s, its period is 𝜏 = (2𝜋 rad)∕(2.030 rad∕s) = 3.095 s, and its frequency is 𝑓 = 0.3231 Hz, where we have used 𝑘 = 22,270 lb∕f t and 𝑚 = 5404 slug. The amplitude of vibration is 𝐶 = 2.890 f t and the phase angle is 𝜙 = 0, where we have used 𝑥𝑖 = 0 f t and 𝑣𝑖 = 5.867 f t∕s. ∗ The

natural frequency is also sometimes called the circular frequency. see in Section 19.2 that this is equivalent to saying that there is no forcing function in Eq. (19.1).

† We will

ISTUDY

Section 19.1

Undamped Free Vibration

Mini-Example A jumper whose mass is 𝑚 = 65 kg is hanging in equilibrium from a linear elastic bungee cord with constant 𝑘 = 200 N∕m. The jumper is then pulled down 5 m and released from rest (see Fig. 19.3). Determine the equation governing the ensuing vibration, the period, the amplitude, and the phase angle of the vibration. Treat the jumper as a particle. Solution The FBD of the jumper after being pulled a distance 𝑦 below the 𝑦 = 0 static equilibrium position is shown in Fig. 19.4. Summing forces in the 𝑦 direction gives ∑

𝐹𝑦∶ 𝑚𝑔 − 𝐹𝑠 = 𝑚𝑎𝑦 ,

1291

𝑘 𝑔

𝑚 Figure 19.3 A bungee jumper of mass 𝑚 hanging from a bungee cord of stiffness 𝑘.

(19.9)

where 𝐹𝑠 is the force in the bungee cord and 𝑎𝑦 = 𝑦. ̈ Since 𝑦 is measured from the static equilibrium position, the force in the bungee cord must be (see Fig. 19.5) 𝐹𝑠 = 𝑚𝑔 + 𝑘𝑦.

𝑦

𝐹𝑠

(19.10)

Substituting Eq. (19.10) into Eq. (19.9), we obtain 𝑚𝑔 − (𝑚𝑔 + 𝑘𝑦) = 𝑚𝑦̈

⇒

𝑦̈ +

𝑘 𝑦 = 0. 𝑚

Equation (19.11) is of the same form √ as Eq. (19.1), and so the jumper’s natural frequency of vibration is 𝜔𝑛 = 𝑘∕𝑚 = 1.754 rad∕s, and the corresponding period of vibration is 𝜏 = 2𝜋∕𝜔𝑛 = 3.582 s. Since 𝑣𝑖 = 0 and 𝑥𝑖 = 5 m, Eq. (19.5) tells us that the jumper’s amplitude of vibration is 5 m, i.e., as expected, the jumper oscillates about the static equilibrium position. Finally, since 𝑣𝑖 = 0 and 𝑥𝑖 > 0, we can choose the phase angle to be 𝜙 = 𝜋∕2 rad. One of the lessons of this mini-example is that it is convenient to choose the origin of the displacement variable to be at the equilibrium position of the system rather than at the position of zero spring deflection. Doing this allows us to ignore the equal and opposite forces associated with equilibrium.

Standard form of the harmonic oscillator Based on the preceding development, we define a harmonic oscillator to be any one degree of freedom (DOF, see definition on p. 794) system whose equation of motion can be given the form 𝑥̈ + 𝜔2𝑛 𝑥 = 0,

(19.12)

where 𝜔𝑛 is the natural frequency∗ of the oscillator and 𝑥 is the coordinate for the one DOF system. Equation (19.12) is called the standard form of the harmonic oscillator equation. As we have seen in Eq. (19.3), we know the complete vibrational solution for any system whose equation of motion can be put in the form of Eq. (19.12). We also note ∗ In

the case of a spring-mass system, 𝜔𝑛 takes the form in Eq. (19.4).

𝑚𝑔

(19.11)

Figure 19.4 FBD of the bungee jumper, along with the coordinate system used in the mini-example.

equilibrium value of 𝐹𝑠

slope = 𝑘 𝑚𝑔

Figure 19.5 The force in the bungee cord as a function of 𝑦. The position 𝑦 = 0 corresponds to static equilibrium.

1292

ISTUDY

Chapter 19

Mechanical Vibrations

that an alternative form of the solution to Eq. (19.12) is 𝑥(𝑡) = 𝐴 cos 𝜔𝑛 𝑡 + 𝐵 sin 𝜔𝑛 𝑡,

(19.13)

where the solution in Eq. (19.3) is recovered from Eq. (19.13) if we let 𝐶=

√ 𝐴2 + 𝐵 2

tan 𝜙 =

and

𝐴 . 𝐵

(19.14)

̇ = 𝑣𝑖 , we find 𝐴 = 𝑥𝑖 and 𝐵 = 𝑣𝑖 ∕𝜔𝑛 , and so Noting again that 𝑥(0) = 𝑥𝑖 and 𝑥(0) Eq. (19.13) becomes 𝑥(𝑡) = 𝑥𝑖 cos 𝜔𝑛 𝑡 +

𝑣𝑖 𝜔𝑛

sin 𝜔𝑛 𝑡.

(19.15)

Linearizing nonlinear systems Not all vibrating one-DOF systems are harmonic oscillators described by Eq. (19.12). However, many systems can be approximated as harmonic oscillators. As an example, consider the uniform thin bar in Fig. 19.6(a) that is pinned and suspended at one end. Using the FBD in Fig. 19.6(b) and the methods of Chapter 17, the equation of motion

𝑂𝑦 𝑂𝑥

𝑂

𝑂 𝜃

𝑚, 𝐿

𝐿 2

𝜃 𝐺 𝑚𝑔

Helpful Information Mathematical justification for small-angle approximations. Figure 19.7 graphically demonstrates that sin 𝑥 behaves as 𝑥 for small 𝑥. We can also see this with the Taylor series. The Taylor series expansion of sin 𝑥 about 𝑥 = 0 is sin 𝑥 = 𝑥 −

𝑥3 𝑥5 + − ⋯. 3! 5!

If 𝑥 is small, then 𝑥3 , 𝑥5 , and higher-order terms will be so small that sin 𝑥 ≈ 𝑥. Similarly, the Taylor series expansion of cos 𝑥 about 𝑥 = 0 is 2

cos 𝑥 = 1 −

4

𝑥 𝑥 + − ⋯. 2! 4!

If 𝑥 is small, then 𝑥2 , 𝑥4 , and higher-order terms will be so small that cos 𝑥 ≈ 1.

Figure 19.6. (a) A uniform swinging bar. (b) The FBD of the swinging bar.

for this bar is 3𝑔 𝜃̈ + sin 𝜃 = 0. 2𝐿

(19.16)

This equation is nonlinear in 𝜃 because of the presence of sin 𝜃, which is a nonlinear function of 𝜃 (in general, nonlinear equations are more challenging to solve than linear equations). Equation (19.16) can be written in the standard form 𝜃̈ + 𝜔2𝑛 𝜃 = 0 if we consider only vibrations for small values of 𝜃, although this creates an approximate version of the original equation. As seen in Fig. 19.7, when 𝜃 is small, sin 𝜃 behaves as 𝜃, and Eq. (19.16) becomes 3𝑔 𝜃̈ + 𝜃 = 0, (19.17) 2𝐿 √ which is in standard form with 𝜔𝑛 = 3𝑔∕(2𝐿). This process, in which a nonlinear ordinary differential equation is approximated to be linear, is called linearization. We will explore linearization further in the example problems.

ISTUDY

Section 19.1

1293

Undamped Free Vibration

𝑥 𝑓 (𝑥) 1.0

sin 𝑥

𝑓 (𝑥) 0.2

sin 𝑥

0.1

0.5

0.1

0.2

0.3

Figure 19.7. Plots of 𝑓 (𝑥) = sin 𝑥 and 𝑓 (𝑥) = 𝑥 for two different ranges of 𝑥. The plot on the left shows that for large 𝑥, the two curves diverge. The plot on the right shows that for 𝑥 ≲ 0.3 rad the two curves are almost indistinguishable.

Energy method The equations of motion for all three of the systems considered in this section were derived by applying the Newton-Euler equations to the FBD of the particle or body of interest. In addition, all three of these systems are conservative; that is, all forces doing work are conservative (the spring on the railcar, the bungee cord and gravity on the bungee jumper, and gravity on the swinging bar). For systems like these, the fact that energy is conserved can provide a way to derive the equation of motion. Finding the equation of motion To see how to find the equation of motion using conservation of mechanical energy, consider the bungee jumper mini-example on p. 1291. Since the bungee jumper is a conservative system, we know that the work-energy principle gives 𝑇1 + 𝑉 1 = 𝑇 2 + 𝑉 2 ,

(19.18)

where 1 is at release and 2 is any subsequent position. This implies that 𝑇 + 𝑉 = constant

⇒

𝑑 (𝑇 + 𝑉 ) = 0, 𝑑𝑡

(19.19)

where we have dropped the use of the subscript 2 to reinforce the idea that 2 is any position following 1. If we now compute 𝑇 and 𝑉 at an arbitrary position for the bungee jumper, we find that the potential energy is given by (see Fig. 19.8) ( )2 𝑉 = 𝑉𝑒 + 𝑉𝑔 = 12 𝑘 𝑦 + 𝛿st − 𝑚𝑔𝑦, (19.20) where 𝑦 is measured from the static equilibrium position of the jumper and 𝛿st is the amount of stretch in the bungee cord at the static equilibrium position. The kinetic energy is given by (19.21) 𝑇 = 12 𝑚𝑦̇ 2 . Substituting Eqs. (19.20) and (19.21) into Eq. (19.19) and taking the time derivative, we find that ( ) 𝑑 (19.22) (𝑇 + 𝑉 ) = 𝑘 𝑦 + 𝛿st 𝑦̇ − 𝑚𝑔 𝑦̇ + 𝑚𝑦̇ 𝑦̈ = 0. 𝑑𝑡 [ ( ) ] Rewriting Eq. (19.22) as 𝑘 𝑦 + 𝛿st − 𝑚𝑔 + 𝑚𝑦̈ 𝑦̇ = 0, we see that this equation is satisfied for any value of 𝑦̇ if and only if ) ( (19.23) 𝑘 𝑦 + 𝛿st − 𝑚𝑔 + 𝑚𝑦̈ = 0. Since 𝑘𝛿st = 𝑚𝑔, we recover the harmonic oscillator equation 𝑚𝑦̈ + 𝑘𝑦 = 0, which is equivalent to Eq. (19.11) for the same bungee jumper.

(19.24)

𝐿0 𝛿st 𝑘

𝑦=0 (static equilibrium) 𝑦 𝑔

𝑚 Figure 19.8 The unstretched length of the bungee cord 𝐿0 , the static equilibrium position of the jumper (𝑦 = 0), and the jumper at an arbitrary 𝑦 position.

1294

ISTUDY

Chapter 19

Mechanical Vibrations The energy method and linearization

When using the energy method to find the linearized equations of motion of a mechanical system, we can begin by approximating the kinetic and potential energies as quadratic functions of position and velocity before taking their time derivative. Then the time derivative of the quadratic approximation of the energy yields equations of motion that are automatically linear. The reason for starting with the quadratic approximation of the energy is that, in some cases, linearizing the equations of motion obtained from the nonapproximated form of the energy is more involved than developing the quadratic approximation of the energy. To see what we mean by “approximating the kinetic and potential energies as quadratic functions of position and velocity,” consider again the uniform thin bar in Fig. 19.6. Writing the kinetic and potential energies at an arbitrary angle 𝜃, we find that 𝑇 = 12 𝐼𝑂 𝜃̇ 2 = 16 𝑚𝐿2 𝜃̇ 2 ,

) ( 𝑉 = − 12 𝑚𝑔𝐿 cos 𝜃 ≈ − 12 𝑚𝑔𝐿 1 − 12 𝜃 2 .

Helpful Information Natural frequency, stiffness, and mass. For a simple spring-mass system whose 𝑘

𝑚

equation of motion is 𝑚𝑥̈ + 𝑘𝑥 √ = 0, the natural frequency of vibration is 𝑘∕𝑚 (𝑥 is measured from the position of 𝑚 when the spring is undeformed). Notice in Eq. (19.27) that the “𝑚” is 13 𝑚𝐿2 and the “𝑘” is 12 𝑚𝑔𝐿. This is common in harmonic oscillators, so it is useful to introduce the notions of effective mass and effective stiffness. The effective mass is an inertial quantity that is the coefficient of the second time derivative of the position, but it is not always just 𝑚. The effective stiffness is a restoring quantity that is the coefficient of the position, but it is not always just 𝑘.

(19.25) (19.26)

Notice that the kinetic energy 𝑇 in Eq. (19.25) is a quadratic function of the angular ̇ and therefore it does not need to be approximated in any way. By contrast, velocity 𝜃, notice that the potential energy 𝑉 = − 12 𝑚𝑔𝐿 cos 𝜃 is not a quadratic function of 𝜃, but we could approximate it as such by using the first two terms in the Taylor series expansion of cos 𝜃 (see the Helpful Information marginal note on p. 1292). Using 𝑑 (𝑇 + 𝑉 ) = 0, we obtain Eqs. (19.25) and (19.26) in 𝑑𝑡 1 𝑚𝐿2 𝜃̇ 𝜃̈ 3

+ 12 𝑚𝑔𝐿𝜃 𝜃̇ = 0

⇒

3𝑔 𝜃 = 0, 𝜃̈ + 2𝐿

(19.27)

which is exactly what we obtained in Eq. (19.17). The simple message here is that when we use the energy method to derive the equation of motion, it is important to approximate the kinetic and potential energies as quadratic functions of position and velocities before taking their time derivatives. For future reference, we note that in linearizing the sine function, and in approximating the cosine function as a quadratic function of its argument, we use the following relations: sin 𝜃 ≈ 𝜃

and

cos 𝜃 ≈ 1 − 𝜃 2 ∕2,

(19.28)

respectively. In addition, we note that since the quadratic term in the power series expansion of the sine function is equal to zero (see the Helpful Information marginal note on p. 1292), the linearized form of the sine function can also be viewed as the quadratic approximation of the sine function.

ISTUDY

Section 19.1

Undamped Free Vibration

1295

End of Section Summary Any one DOF system whose equation of motion is of the form Eq. (19.12), p. 1291 𝐶 sin 𝜙

𝑥̈ + 𝜔2𝑛 𝑥 = 0, is called a harmonic oscillator, and the above expression is referred to as the standard form of the harmonic oscillator equation. The solution of this equation can be written as (see Fig. 19.9) Eq. (19.3), p. 1290

where 𝜔𝑛 is the natural frequency, 𝐶 is the amplitude, and 𝜙 is the phase angle of vibration. A simple example of a harmonic oscillator is a system consisting of a block of mass 𝑚 attached at the free end of a spring with constant 𝑘 and with the other end fixed (see Fig. 19.10). The natural frequency of such a system is given by Eq. (19.4), p. 1290 √ 𝑘 𝜔𝑛 = . 𝑚 In addition, the amplitude 𝐶 and the phase angle 𝜙 are given by, respectively, Eqs. (19.5) and (19.6), p. 1290 √ 𝑣2𝑖 ∕𝜔2𝑛 + 𝑥2𝑖

tan 𝜙 =

and

𝑥𝑖 𝜔 𝑛 𝑣𝑖

,

where, by letting 𝑡 = 0 be the initial time, 𝑥𝑖 = 𝑥(0) (i.e., 𝑥𝑖 is the initial position) ̇ (i.e., 𝑣𝑖 is the initial velocity). If 𝑣𝑖 = 0, then 𝜙 can be chosen equal and 𝑣𝑖 = 𝑥(0) to −𝜋∕2 or 𝜋∕2 rad for 𝑥𝑖 < 0 and 𝑥𝑖 > 0, respectively. An alternative form of the solution to Eq. (19.12) is given by Eq. (19.15), p. 1292 𝑥(𝑡) = 𝑥𝑖 cos 𝜔𝑛 𝑡 +

𝑣𝑖 𝜔𝑛

sin 𝜔𝑛 𝑡.

The period of the oscillation is given by Eq. (19.7), p. 1290 Period = 𝜏 =

2𝜋 , 𝜔𝑛

and the frequency of vibration is Eq. (19.8), p. 1290 Frequency = 𝑓 =

1 𝜔𝑛 = . 𝜏 2𝜋

𝜋

3𝜋 2

2𝜋

𝑡

𝜙 𝜔𝑛

Figure 19.9 Plot of Eq. (19.3) showing the amplitude 𝐶, phase angle 𝜙, and period 𝜏 of a harmonic oscillator.

( ) 𝑥(𝑡) = 𝐶 sin 𝜔𝑛 𝑡 + 𝜙 ,

𝐶=

𝜋 2

− 𝜋2

𝑘

𝑚

Figure 19.10 A simple spring-mass harmonic oscillator whose equation of motion √ is given by 𝑚𝑥̈ + 𝑘𝑥 = 0 and for which 𝜔𝑛 = 𝑘∕𝑚. The position 𝑥 is measured from the location of 𝑚 when the spring is undeformed.

1296

ISTUDY

Chapter 19

Mechanical Vibrations

Energy method. For conservative systems, the work-energy principle tells us that the quantity 𝑇 + 𝑉 is constant, and so its time derivative must be zero. This provides a convenient way to obtain the equations of motion via Eq. (19.19), p. 1293 𝑑 (𝑇 + 𝑉 ) = 0 𝑑𝑡

⇒

equations of motion.

When we apply the energy method to determine the linearized equations of motion, it is often convenient to first approximate the kinetic and potential energies as quadratic functions of position and velocity and then take derivatives with respect to time. This process yields equations of motion that are linear. In approximating the sine and cosine functions as quadratic functions of their arguments, we use the relations Eq. (19.28), p. 1294 sin 𝜃 ≈ 𝜃

and

cos 𝜃 ≈ 1 − 𝜃 2 ∕2.

Since the quadratic term in the power series expansion of the sine function is equal to zero, the linearized form of the sine function can also be viewed as the quadratic approximation of the sine function.

ISTUDY

Section 19.1

1297

Undamped Free Vibration

E X A M P L E 19.1

Finding the Moment of Inertia of a Rigid Body

When the connecting rod shown in Fig. 1 is suspended from the knife-edge at point 𝑂 and displaced slightly so that it oscillates like a pendulum, its period of oscillation is 0.77 s. The mass center 𝐺 is located a distance 𝐿 = 110 mm from 𝑂, and the mass of the connecting rod is 661 g. Using this information, determine the mass moment of inertia of the connecting rod about 𝐺.

𝐿

SOLUTION 𝐺 Road Map & Modeling

If we write the equation of motion of the connecting rod for small angles, then, as with a pendulum, we should be able to write it in standard form and extract the natural frequency of vibration. The natural frequency will depend on the mass moment of inertia of the connecting rod, which should then allow us to solve for the moment of inertia. The FBD of the connecting rod is shown in Fig. 2. Governing Equations Balance Principles

Summing moments about the fixed point 𝑂 (see Fig. 2), we obtain ∑ 𝑀𝑂∶ −𝑚𝑔𝐿 sin 𝜃 = 𝐼𝑂 𝛼cr , (1)

Figure 1 A connecting rod hinged on a knife-edge at 𝑂 and allowed to oscillate freely in the plane of the page. Point 𝐺 is the rod’s mass center. 𝑂𝑦

where 𝐼𝑂 is the moment of inertia of the connecting rod with respect to point 𝑂, and 𝛼cr is the angular acceleration of the connecting rod. Force Laws All forces are accounted for on the FBD. ̈ Kinematic Equations The only kinematic equation is 𝛼cr = 𝜃. Computation

𝑂𝑥

𝑂

𝚥̂ 𝚤̂

Substituting the kinematic equation into Eq. (1) and rearranging, we ob-

tain

𝐿

𝑚𝑔𝐿 𝜃̈ + sin 𝜃 = 0. 𝐼𝑂 For small 𝜃, sin 𝜃 ≈ 𝜃 and so Eq. (2) becomes 𝑚𝑔𝐿 𝜃 = 0. 𝜃̈ + 𝐼𝑂

(2)

𝑚𝑔𝐿 𝐼𝑂

⇒

𝐼𝑂 =

𝑚𝑔𝐿 𝜔2𝑛

𝐺

(3) 𝑚𝑔

Equation (3) is in standard form, so we know that 𝜔2𝑛 =

𝜃

⇒

𝐼𝑂 =

𝑚𝑔𝐿𝜏 2 , 4𝜋 2

(4)

where we have used 𝜔2𝑛 = 4𝜋 2 ∕𝜏 2 from Eq. (19.7). Noting that the parallel axis theorem states that 𝐼𝐺 = 𝐼𝑂 − 𝑚𝐿2 , we find ) ( 𝑚𝑔𝐿𝜏 2 𝑔𝜏 2 2 2 𝐼𝐺 = 𝐼𝑂 − 𝑚𝐿2 = − 𝑚𝐿 ⇒ 𝐼 = 𝑚𝐿 − 1 , (5) 𝐺 4𝜋 2 4𝜋 2 𝐿 which, when evaluated for 𝜏 = 0.77 s, 𝐿 = 0.11 m, and 𝑚 = 0.661 kg, yields 𝐼𝐺 = 0.002714 kg⋅m2 .

(6)

Discussion & Verification The dimensions of Eq. (5) are mass times length squared, as they should be. In addition, the first of Eqs. (4) tells us that the natural frequency of vibration is inversely proportional to the square root of the mass moment of inertia (as it is inversely proportional to the mass), which agrees with our intuition. A Closer Look This simple measurement is tremendously useful. We typically talk about evaluating the mass moment of inertia from either its integral definition or by using the method of composite bodies (see App. C). A complicated shape like this connecting rod does not lend itself to either method, but here we have a simple experiment that allows us to determine the mass moment of inertia.

Figure 2 FBD of the connecting rod during oscillation.

1298

Chapter 19

Mechanical Vibrations

E X A M P L E 19.2

Vibration of a Silicon Nanowire Modeled as a Rigid Bar The natural frequency of a vibrating spring-mass √system is a function of the equivalent mass and stiffness according to the relation 𝜔𝑛 = 𝑘eq ∕𝑚eq . Therefore, if the cantilevered

2 𝜇m Courtesy of Theresa Mayer

Figure 1 A field-emission scanning electronic microscope image of a silicon (Si) nanowire. From Mingwei Li et al., “Bottom-up Assembly of Large-Area Nanowire Resonator Arrays,” Nature Nanotechnology, 3(2), 2008, pp. 88–92. 2𝑟

𝑚

𝑂

Figure 2 A rigid bar and torsional spring model of a flexible nanowire.

𝛿

𝑃

silicon nanowire (SiNW) shown in Fig. 1 were to vibrate, it would have a different natural frequency as shown than if additional mass were added to the end of the wire. Vibrating nanoelectromechanical systems (NEMS), such as this SiNW, have been proposed for use in chip-based sensor arrays as ultrasensitive mass detectors to detect masses in the zeptogram (zg) range. Such a wire would be capable of detecting small numbers of viruses! Given a uniform Si nanowire with a circular cross section that is 9.8 𝜇m long and 330 nm in diameter, compute its natural frequency, using a rigid bar model with all the flexibility lumped in a linear torsional spring at the base of the wire (see Fig. 2). Use 𝜌 = 2330 kg∕m3 for the density of silicon and 𝐸 = 152 GPa for its modulus of elasticity.∗

SOLUTION Road Map & Modeling

To obtain the linear torsional spring constant 𝑘𝑡 , we will employ a result from mechanics of materials, which states that the deflection of a cantilevered bar subjected to a load 𝑃 at its end is (see Fig. 3) 𝑃 =

3𝐸𝐼cs 𝐿3

(1)

𝛿,

where 𝐸 is its modulus of elasticity and 𝐼cs = 41 𝜋𝑟4 is the centroidal area moment of inertia of the beam’s cross section. Using Eq. (1), we will find the value of the torsional spring constant in Fig. 2 that gives this same deflection for a given load 𝑃 . Once we have 𝑘𝑡 , we can apply the Newton-Euler equations to then obtain the equation of motion in the form of Eq. (19.12). Finding the torsional spring constant

Figure 3 A flexible cantilever beam subject to a load 𝑃 at its end. The deflection is given by Eq. (1).

Governing Equations Balance Principles Referring to the FBD in Fig. 4, taking moments about point 𝑂, and noting that this is a statics problem for the purpose of finding 𝑘𝑡 , we obtain

∑

𝚥̂ 𝚤̂ 𝑀𝑡 𝛿

𝑃

𝜃

𝑂𝑦 𝑂𝑥

ISTUDY

𝑃 𝐿 − 𝑀𝑡 = 0,

(2)

where 𝑀𝑡 is the moment due to the torsional spring, and we note that the moment arm for the load 𝑃 is the distance 𝐿 for small 𝜃. Force Laws

Figure 4 FBD of the rigid bar for finding the equivalent torsional spring constant 𝑘𝑡 .

𝑀𝑂 ∶

For a torsional spring, we have 𝑀𝑡 = 𝑘𝑡 𝜃.

Kinematic Equations

For small 𝜃, we relate 𝛿 and 𝜃 using 𝛿 = 𝐿𝜃.

Computation

Substituting the force law and the kinematic relation into Eq. (2) and solving the resulting equation for 𝑘𝑡 , we obtain 𝑘𝑡 =

𝑃 𝐿2 𝛿

⇒

𝑘𝑡 =

3𝐸𝐼cs 𝐿

,

(3)

where we have used Eq. (1) in going from the first to the second expression for 𝑘𝑡 . ∗ The

modulus of elasticity is a material property representing material flexibility. Recall that 1 Pa = 1 N∕m2 .

ISTUDY

Section 19.1

Undamped Free Vibration

1299

The equation of motion for the rigid bar 𝚥̂

Governing Equations

𝚤̂

Balance Principles

Now that we have 𝑘𝑡 , for the bar in Fig. 2, we draw the FBD shown in Fig. 5, and we sum moments about point 𝑂 to obtain ∑ 𝑀𝑂∶ −𝑀𝑡 = 𝐼𝑂 𝛼bar , (4)

where the mass moment of inertia of the bar with respect to 𝑂 is 𝐼𝑂 = 13 𝑚𝐿2 . Force Laws

𝜃

𝑂𝑦 𝑂𝑥

Figure 5 FBD of the bar while vibrating.

The expression for 𝑀𝑡 is unchanged and is given by 𝑀𝑡 = 𝑘𝑡 𝜃.

Kinematic Equations

𝛿

𝑀𝑡

The kinematic equation relating 𝛼bar to 𝜃 is ̈ 𝛼bar = 𝜃.

(5)

Computation

Substituting Eq. (3), the force law, and Eq. (5) into Eq. (4), we obtain the equation of motion as −

3𝐸𝐼cs 𝐿

𝜃=

1 𝑚𝐿2 𝜃̈ 3

⇒

𝜃̈ +

9𝐸𝐼cs 𝑚𝐿3

𝜃 = 0,

(6)

where we recall that 𝐼cs = 41 𝜋𝑟4 . Comparing Eq. (6) to Eq. (19.12), we see that the natural frequency is √ 𝐸𝐼cs . 𝜔𝑛 = 3 (7) 𝑚𝐿3 To obtain a numerical value for 𝜔𝑛 , we find that the volume of the nanowire is 𝜋𝑟2 𝐿 and that the mass is then 𝑚 = 𝜌𝜋𝑟2 𝐿 = 1.953×10−15 kg. The area moment of inertia is 𝐼cs = 14 𝜋𝑟4 = 5.821 × 10−28 m4 . Using these results, along with 𝐿 = 9.8 × 10−6 m and 𝐸 = 152×109 N∕m2 , we find that 𝜔𝑛 = 2.081×107 rad∕s

and

𝑓 = 3.313 MHz.

(8)

Discussion & Verification

The quantity under the square root in Eq. (7) has dimensions of 1 over time squared. Hence, the dimensions of 𝜔𝑛 are 1 over time, as they should be. While it is hard to know what the frequency of vibration of a cantilevered bar of this size should be, in the Closer Look below, we will see that our model actually is quite good. A Closer Look From the theory of vibration of continuous systems, one can show that a continuous system, like this nanowire, vibrates with infinitely many natural frequencies. The first (i.e., the smallest) of these frequencies provides the relevant comparison and is given by √

( ) 𝜔𝑛 exact = 3.516

𝐸𝐼cs 𝑚𝐿3

,

(9)

which, when compared with Eq. (7), tells us that our model is only off by about 15%. Evaluating Eq. (9) numerically, we find that ( ) 𝜔𝑛 exact = 2.439×107 rad∕s and 𝑓exact = 3.883 MHz. (10) In Problem 19.17, we have the opportunity to examine another model for a cantilevered wire such as this and see how the natural frequency changes with the addition of a few zeptograms of virus to the end of the wire.

Interesting Fact Natural frequency of a wooden yardstick. As a comparison, it is interesting to compute the natural frequency of a wooden yardstick using Eq. (9). Using properties typical of a wooden yardstick, that is, 𝐸 = 12 GPa, 𝑚 = 0.0614 kg, a 28 × 4 mm cross section (which gives 𝐼cs = 1.493 × 10−𝟣𝟢 m4 ), and 𝐿 = 0.9144 m, we find that 𝜔𝑛 = 21.72 rad/s, which corresponds to 𝑓 = 3.457 Hz. The nanowire’s natural frequency is 1.1 million times higher!

1300

Chapter 19

Mechanical Vibrations

E X A M P L E 19.3

Energy Method: Equation of Motion of a Diving Board A diver is causing the end of the diving board shown in Fig. 1 to oscillate. Referring to Fig. 2, we will model the board as a thin, uniform, rigid plate of mass 𝑚𝑏 and length 𝐿 and assume that the board is pinned at 𝑂. To model the elastic response of the board, we assume that the board oscillates due to a spring of stiffness 𝑘 attached to the board at what was the fulcrum at 𝐴. In addition, we will model the diver as a point mass of mass 𝑚𝑑 standing at the end of the board. Use the energy method to find the equation of motion for small rotations of the board.

𝑂 𝐴

SOLUTION

Figure 1

Road Map & Modeling

𝑚𝑑 𝑚𝑏

𝐴

𝑂

The system is conservative since only the spring and the two weight forces do work as the system oscillates. This allows us to write the sum of the kinetic and potential energies of the system at an arbitrary position and then differentiate that sum with respect to time to obtain the equation of motion. Governing Equations Balance Principles Since energy is conserved, we can write that the sum of the kinetic and potential energies is constant, so

𝑘 ℎ

𝑇 + 𝑉 = constant Figure 2 The model used to analyze the oscillation of the diving board and diver.

𝑂𝑥

𝑑 (𝑇 + 𝑉 ) = 0. 𝑑𝑡

(1)

The kinetic energy is given by 𝑇 = 21 𝐼𝑂 𝜃̇ 2 + 12 𝑚𝑑 𝑣2𝑑 ,

(2)

where 𝐼𝑂 = 31 𝑚𝑏 𝐿2 is the mass moment of inertia of the diving board with respect to point 𝑂 and 𝑣𝑑 is the speed of the diver.

𝐹𝑠

𝑂

⇒

Force Laws

𝐴 ℎ

𝜃 𝑚𝑏 𝑔 𝑚𝑑 𝑔

𝐿∕2

Figure 3 FBD of the board and diver as they oscillate. Note that the dimensions shown apply only when 𝜃 is small, so cos 𝜃 ≈ 1.

The potential energy of the system at an arbitrary angle 𝜃 is (see Fig. 4) ( )2 (3) 𝑉 = 21 𝑘 𝛿st + ℎ sin 𝜃 − 𝑚𝑏 𝑔 𝐿2 sin 𝜃 − 𝑚𝑑 𝑔𝐿 sin 𝜃 ( ) 2 ≈ 21 𝑘 𝛿st + ℎ𝜃 − 𝑚𝑏 𝑔 𝐿2 𝜃 − 𝑚𝑑 𝑔𝐿𝜃, (4)

where 𝛿st is the compression of the spring when the diving board is in static equilibrium with the diver on it, i.e., when 𝜃 = 0, and where we have approximated 𝑉 as a quadratic function of 𝜃 (i.e., position) in approximating Eq. (3) as Eq. (4). Kinematic Equations

Since the diver is rotating about the fixed point at 𝑂, the diver’s

̇ speed is 𝑣𝑑 = 𝐿𝜃. Computation

𝜃 𝐴

𝐿 2

datum sin 𝜃

Figure 4 The vertical displacements of the relevant points on the diving board as it rotates.

ISTUDY

Substituting the potential energy in Eq. (4), the kinetic energy in Eq. (2), and the kinematic equation into Eq. (1), and then taking the time derivative, we obtain ) ( ) ( (5) 𝐼𝑂 + 𝑚𝑑 𝐿2 𝜃̇ 𝜃̈ + 𝑘 𝛿st + ℎ𝜃 ℎ𝜃̇ − 12 𝑚𝑏 𝑔𝐿𝜃̇ − 𝑚𝑑 𝑔𝐿𝜃̇ = 0.

̇ we note that the term 𝑘ℎ𝛿 is the moment about 𝑂 due to the spring force Canceling 𝜃, st needed to hold the system in equilibrium (i.e., at 𝜃 = 0). This moment is equal and opposite to the moment about 𝑂 created by the two weight forces, that is, − 12 𝑚𝑏 𝑔𝐿 − 𝑚𝑑 𝑔𝐿. Therefore, Eq. (5) reduces to the final equation of motion (we have used 𝐼𝑂 = 31 𝑚𝑏 𝐿2 ) 𝜃̈ + ( 1

𝑘ℎ2

) 𝜃 = 0. 𝑚 + 𝑚 𝐿2 𝑏 𝑑 3

Discussion & Verification

squared, as it should.

(6)

The coefficient of 𝜃 in Eq. (6) has dimensions of 1 over time

ISTUDY

Section 19.1

1301

Undamped Free Vibration

Problems Problem 19.1 Show that Eq. (19.15) is equivalent to Eq. (19.3) if 𝐶 =

√

𝐴2 + 𝐵 2 and tan 𝜙 = 𝐴∕𝐵.

Problem 19.2

𝑂 𝐿

Derive the formula for the mass moment of inertia of an arbitrarily shaped rigid body about its mass center based on the body’s period of oscillation 𝜏 when suspended as a pendulum. Assume that the mass of the body 𝑚 is known and that the location of the mass center 𝐺 is known relative to the pivot point 𝑂.

Problem 19.3

𝐺 𝑚

Figure P19.2

The thin ring of radius 𝑅 and mass 𝑚 is suspended by the pin at 𝑂. Determine its period of vibration if it is displaced a small amount and released. 𝑂

𝑂 𝑚

𝑚

𝑅

Figure P19.3

Figure P19.4

Problem 19.4 The thin square hoop has mass 𝑚 and is suspended by the pin at 𝑂. Determine its period of vibration if it is displaced a small amount and released. vibrating mass

Problem 19.5 The swinging bar and the vibrating block of mass 𝑚 are made to vibrate on Earth, and their respective natural frequencies are measured. The two systems are then taken to the Moon and are again allowed to vibrate at their respective natural frequencies. How will the natural frequency of each system change when compared with that on the Earth, and which of the two systems will experience the larger change in natural frequency?

𝑔 𝜃

𝑘

𝑚, 𝐿

swinging bar

𝑚

Figure P19.5

Problem 19.6

𝐴

The uniform disk of radius 𝑅 and thickness 𝑡 is attached to the thin shaft of radius 𝑟, length 𝐿, and negligible mass. The end 𝐴 of the shaft is fixed. From mechanics of materials, it can be shown that if a torque 𝑀𝑧 is applied to the free end of the shaft, then it can be related to the twist angle 𝜃 via 𝑀 𝐿 𝜃= 𝑧 , 𝐺𝐽 where 𝐺 is the shear modulus of elasticity of the shaft and 𝐽 = 𝜋𝑟4 ∕2 is the polar moment of inertia of the cross-sectional area of the shaft. Let 𝜌 be the mass density of the disk. Using the given relationship between 𝑀𝑧 and 𝜃, determine the natural frequency of vibration of the disk in terms of the given dimensions and material properties when it is given a small angular displacement 𝜃 in the plane of the disk.

2𝑟

𝑡

𝑅 𝜃 Figure P19.6

1302

Chapter 19

Mechanical Vibrations Problems 19.7 and 19.8 𝑂

A rigid body of mass 𝑚, mass center at 𝐺, and mass moment of inertia 𝐼𝐺 is pinned at an arbitrary point 𝑂 and allowed to oscillate as a pendulum.

𝐺

Problem 19.7 By writing the Newton-Euler equations, determine the distance 𝓁 from 𝐺 to the pivot point 𝑂 so that the pendulum has the highest possible natural frequency of oscillation.

𝓁

𝑚

Problem 19.8

Using the energy method, determine the distance 𝓁 from 𝐺 to the pivot point 𝑂 so that the pendulum has the highest possible natural frequency of oscillation.

Figure P19.7 and P19.8

Problem 19.9 A block of mass 𝑚 = 3 kg is in equilibrium when a hammer hits it, imparting a velocity 𝑣0 of 2 m∕s to it. If 𝑘 is 120 N∕m, determine the amplitude of the ensuing vibration and find the maximum acceleration experienced by the block. 𝑘

Figure P19.9

𝑚

Problem 19.10 A construction worker 𝐶 is standing at the midpoint of a 14 f t long pine board that is simply supported. The board is a standard 2 × 12, so its cross-sectional dimensions are as shown. Assuming the worker weighs 180 lb and he flexes his knees once to get the board oscillating, determine his vibration frequency. Neglect the weight of the beam, and use the fact that a load 𝑃 applied to a simply supported beam will deflect the center of the beam 𝑃 𝐿3 ∕(48𝐸𝐼cs ), where 𝐿 is the length of the beam, 𝐸 is its modulus of elasticity, and 𝐼cs is the area moment of inertia of the cross section of the beam. The elastic modulus of pine is 1.8×106 psi. 𝑚𝑡

1.5 in. 11.25 in.

𝑚𝑠

𝐵

𝐴

Figure P19.10

Figure P19.11

Problem 19.11 A truck drives onto a deck scale to be weighed, thus causing the truck and scale to vibrate vertically at the natural frequency of the system. The empty truck weighs 74,000 lb, the scale platform weighs 51,000 lb, and the platform is supported by eight identical springs (four of which are shown), each with constant 𝑘 = 3.6×105 lb∕f t. Modeling the truck, its contents, and the concrete deck as a single particle, if a vibration frequency of 3.3 Hz is measured, what is the weight of the payload being carried by the truck?

Problem 19.12

Mark Allen Robinson/Shutterstock

Figure P19.12

ISTUDY

The buoy in the photograph can be modeled as a circular cylinder of diameter 𝑑 and mass 𝑚. If the buoy is pushed down in the water, which has density 𝜌, it will oscillate vertically. Determine the frequency of oscillation. Evaluate your result for 𝑑 = 1.2 m, 𝑚 = 900 kg, and surface seawater, which has a density of 𝜌 = 1027 kg∕m3 . Hint: Use Archimedes’ principle, which states that a body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid.

ISTUDY

Section 19.1

Undamped Free Vibration

1303

Problems 19.13 through 19.16 The L-shaped bar lies in the vertical plane and is pinned at 𝑂. One end of the bar has a linear elastic spring with constant 𝑘 attached to it, and attached at the other end is a sphere of mass 𝑚 and negligible size. The angle 𝜃 is measured from the equilibrium position of the system, and it is assumed to be small.

𝐴 𝜃

Problem 19.13

Assuming that the L-shaped bar has negligible mass, determine the natural period of vibration of the system by writing the Newton-Euler equations. Problem 19.14 Assuming that the L-shaped bar has negligible mass, determine the natural period of vibration of the system using the energy method.

𝑚

𝑂 𝜃

Problem 19.15

Assuming that the L-shaped bar has mass per unit length 𝜌, determine the natural period of vibration of the system by writing the Newton-Euler equations.

Figure P19.13–P19.16

Problem 19.16

Assuming that the L-shaped bar has mass per unit length 𝜌, determine the natural period of vibration of the system using the energy method.

Problem 19.17

𝑚

For the silicon nanowire in Example 19.2, use the lumped mass model shown, in which a point mass 𝑚 is connected to a rod of negligible mass and length 𝐿 that is pinned at 𝑂, to determine the natural frequency 𝜔𝑛 and frequency 𝑓 of the nanowire. Use the values given in Example 19.2 for the mass of the lumped mass, the length of the massless rod, and the parameters used to determine the spring constant 𝑘 = 3𝐸𝐼cs ∕𝐿3 . You may use either 𝛿 or 𝜃 as the position variable in your solution. Assume that the displacement of 𝑚 is small so that it moves vertically.

𝑂

𝑘 = 3𝐸𝐼cs ∕𝐿3 𝛿

𝜃

Figure P19.17

Problem 19.18 The small sphere 𝐴 has mass 𝑚 and is fixed at the end of the arm 𝑂𝐴 of negligible mass, which is pinned at 𝑂. If the linear elastic spring has stiffness 𝑘, determine the equation of motion for small oscillations, using (a) The vertical position of the mass 𝐴 as the position coordinate. (b) The angle formed by the arm 𝑂𝐴 with the horizontal as the position coordinate. 𝐴 𝑂

𝑘 Figure P19.18

Problems 19.19 and 19.20 𝐴

The uniform cylinder rolls without slipping on a flat surface. Let 𝑘1 = 𝑘2 = 𝑘 and 𝑟 = 𝑅∕2. Assume that the horizontal motion of 𝐺 is small. Problem 19.19

Determine the equation of motion for the cylinder by writing its NewtonEuler equations. Use the horizontal position of the mass center 𝐺 as the degree of freedom. Problem 19.20 Determine the equation of motion for the cylinder using the energy method. Use the horizontal position of the mass center 𝐺 as the degree of freedom.

𝑅 𝐺 𝑟 𝐵 Figure P19.19 and P19.20

𝑘2

1304

Chapter 19

Mechanical Vibrations Problems 19.21 and 19.22

Grandfather clocks keep time by advancing the hands a set amount per oscillation of the pendulum. Therefore, the pendulum needs to have a very accurate period for the clock to keep time accurately. As a fine adjustment of the pendulum’s period, many grandfather clocks have an adjustment nut on a bolt at the bottom of the pendulum disk. Screwing this nut inward or outward changes the mass distribution of the pendulum by moving the pendulum disk closer to or farther from the axis of rotation at 𝑂. Model the pendulum as a uniform disk of radius 𝑟 and mass 𝑚𝑝 at the end of a rod of negligible mass and length 𝐿 − 𝑟, and assume that the oscillations of 𝜃 are small. Let 𝑚𝑝 = 0.7 kg and 𝑟 = 0.1 m. 𝑂

pendulum disk 𝑟

𝐿 𝑚𝑝

𝜃 pendulum disk

adjustment nut

Figure P19.21 and P19.22 Problem 19.21

If the pendulum disk is initially at a distance 𝐿 = 0.85 m from the pin at 𝑂, how much would the period of the pendulum change if the adjustment nut with a lead∗ of 0.5 mm was rotated four complete rotations closer to the disk? In addition, how much time would the clock gain or lose in a 24 h period if this were done? Problem 19.22 The clock is running slow so that it is losing 2 minutes every 24 hours (i.e., the clock takes 1442 minutes to complete a 1440-minute day). If the pendulum disk is at 𝐿 = 0.85 m, how many turns of the adjustment nut would be needed, and in what direction, to correct the pendulum’s period if the screw lead∗ is 0.5 mm?

Problems 19.23 and 19.24 𝑘

The uniform cylinder of mass 𝑚 and radius 𝑅 rolls without slipping on the inclined surface. The spring with constant 𝑘 wraps around the cylinder as it rolls. 𝑅

𝜃

Problem 19.23 Determine the equation of motion for the cylinder by writing its NewtonEuler equations. Determine the numerical value of the period of oscillation of the cylinder using 𝑘 = 30 N∕m, 𝑚 = 10 kg, 𝑅 = 30 cm, and 𝜃 = 20◦ . Problem 19.24

Determine the equation of motion for the cylinder using the energy method. Determine the numerical value of the period of oscillation of the cylinder using 𝑘 = 30 N∕m, 𝑚 = 10 kg, 𝑅 = 30 cm, and 𝜃 = 20◦ .

Figure P19.23 and P19.24

Problem 19.25 𝐴 𝜔0 Figure P19.25

ISTUDY

𝐵 𝜔0

A uniform bar of mass 𝑚 is placed off-center on two counter-rotating drums 𝐴 and 𝐵. Each drum is driven with constant angular speed 𝜔0 , and the coefficient of kinetic friction between the drums and the bar is 𝜇𝑘 . Determine the natural frequency of oscillation of the bar on the rollers. Hint: Measure the horizontal position of 𝐺 relative to the midpoint between the two drums, and assume that the drums rotate sufficiently fast so that the drums are always slipping relative to the bar. ∗ The lead of a screw is the axial distance a nut would advance along the screw during one complete rotation

of the nut.

ISTUDY

Section 19.1

1305

Undamped Free Vibration

Problem 19.26

𝐴

The uniform cylinder 𝐴 of radius 𝑟 and mass 𝑚 is released from a small angle 𝜃 inside the large cylinder of radius 𝑅. Assuming that it rolls without slipping, determine the natural frequency and period of oscillation of 𝐴.

Problem 19.27

𝜃

𝑟 𝐺

𝐵

The uniform sphere 𝐴 of radius 𝑟 and mass 𝑚 is released from a small angle 𝜃 inside the large cylinder of radius 𝑅. Assuming that it rolls without slipping, determine the natural frequency and period of oscillation of the sphere.

Figure P19.26

𝐴 𝜃

𝑟 𝐺

𝐵 Figure P19.27

Problem 19.28 The U-tube manometer lies in the vertical plane and contains a fluid of density 𝜌 that has been displaced a distance 𝑦 and oscillates in the tube. If the cross-sectional area of the tube is 𝐴 and the total length of the fluid in the tube is 𝐿, determine the natural period of oscillation of the fluid, using the energy method. Hint: As long as the curved portion of the tube is always filled with liquid (i.e., the oscillations do not get large enough to empty part of it), the contribution of the liquid in the curved portion to the potential energy is constant.

Problem 19.29 The uniform semicylinder of radius 𝑅 and mass 𝑚 rolls without slip on the horizontal surface. Using the energy method, determine the period of oscillation for small 𝜃.

𝐺

𝑅 𝜃

Figure P19.29

𝑅 𝜃

Figure P19.30

Problem 19.30 The thin-shell semicylinder of radius 𝑅 and mass 𝑚 rolls without slip on the horizontal surface. Using the energy method, determine the period of oscillation for small 𝜃.

𝑦

Figure P19.28

equilibrium line

1306

Chapter 19

Mechanical Vibrations

Design Problems

ISTUDY

Design Problem 19.1 As part of a manufacturing process, a uniform bar is placed on a centering device consisting of two counter-rotating drums 𝐴 and 𝐵 and a frictional slider 𝐶 that provides light damping. When the bar is placed off-center on the two counter-rotating drums, it oscillates back and forth due to the sliding friction between each drum and the bar, and it eventually settles into the centered position due to friction between the block at 𝐶 and the bar. Once the bar has been centered, the next step in the manufacturing process can begin. Assume that the steel bar has length 𝐿 = 1.2 m, radius 𝑟 = 22.5 mm, and density 𝜌 = 7.85 g∕cm3 . Assuming that the initial misalignment (i.e., the initial distance between 𝐺 and 𝐶) of each bar can be up to 0.25 m and that sliding between the drums and the bar never ceases, design the drums (i.e., their radius and what they are made of) and determine their placement so that sliding is maintained for the entire range of motion. Since the damping is light, it can be neglected. In addition, assume that you want to drive each drum with a constant angular speed 𝜔0 .

𝐺 𝑅 𝜔0

2𝑟

𝑅

𝐶

𝐴

𝐵 ℎ∕2

Figure DP19.1

𝜔0

ISTUDY

Section 19.2

19.2

Undamped Forced Vibration

1307

Undamped Forced Vibration

Many systems are forced to vibrate by an external excitation. This section is devoted to the forced vibration of mechanical systems.

Standard form of the forced harmonic oscillator

𝑘

A standard forced harmonic oscillator is shown in Fig. 19.11, in which the block of mass 𝑚 is attached to a fixed support by a linear spring of constant 𝑘 and is also being driven by the time-dependent force 𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡. Modeling the block as a particle, its FBD is as shown in Fig. 19.12, where 𝐹𝑠 is the spring force acting on the block. Summing forces in the 𝑥 direction, we obtain ∑ 𝐹𝑥 ∶ 𝑃 (𝑡) − 𝐹𝑠 = 𝑚𝑎𝑥 , (19.29)

Figure 19.11 A forced harmonic oscillator whose equation √ of motion is given by Eq. (19.31) with 𝜔𝑛 = 𝑘∕𝑚. The position 𝑥 is measured from the equilibrium position of the system when 𝐹0 = 0.

where the force law is given by 𝐹𝑠 = 𝑘𝑥 and the kinematic equation is 𝑎𝑥 = 𝑥. ̈ Substituting these relations, as well as 𝑃 (𝑡) into Eq. (19.29), we obtain 𝐹0 sin 𝜔0 𝑡 − 𝑘𝑥 = 𝑚𝑥̈

𝐹 𝑘 𝑥̈ + 𝑥 = 0 sin 𝜔0 𝑡. 𝑚 𝑚

⇒

𝑥̈ + 𝜔2𝑛 𝑥 =

𝑚

sin 𝜔0 𝑡,

(19.31)

which is the standard form of the forced harmonic oscillator equation. It is a nonhomogeneous version of Eq. (19.12) on p. 1291 as a result of the term (𝐹0 ∕𝑚) sin 𝜔0 𝑡. The term on the right-hand side of Eq. (19.31) is a function of only the independent variable 𝑡. It is often called a forcing function because it forces the system to vibrate. This particular type of forcing is harmonic because it is a harmonic function of time. The theory of differential equations tells us that the general solution of Eq. (19.31) is the sum of the complementary solution 𝑥𝑐 (𝑡) and a particular solution 𝑥𝑝 (𝑡). The complementary solution∗ is the solution of the associated homogeneous equation [i.e., Eq. (19.12)] given in Eq. (19.3) [or in Eq. (19.13)]. The particular solution is any solution of Eq. (19.31). One way to obtain a particular solution is to guess its form and then verify whether or not the guess is correct. Since it seems reasonable that the response of a forced harmonic oscillator should resemble the forcing, we conjecture that the particular solution 𝑥𝑝 is of the form 𝑥𝑝 = 𝐷 sin 𝜔0 𝑡,

(19.32)

where 𝐷 is a constant to be determined. We can verify whether our guess is correct by substituting Eq. (19.32) into Eq. (19.31). Doing so yields −𝐷𝜔20 sin 𝜔0 𝑡 + 𝜔2𝑛 𝐷 sin 𝜔0 𝑡 =

𝐹0 𝑚

sin 𝜔0 𝑡.

(19.33)

Canceling sin 𝜔0 𝑡 and solving for 𝐷, we obtain 𝐷= ∗ The

𝐹0 ∕𝑚 𝜔2𝑛

− 𝜔20

=

𝐹0 ∕𝑘 ( )2 , 1 − 𝜔0 ∕𝜔𝑛

complementary solution is sometimes called the homogeneous solution.

𝚤̂

𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡

𝐹𝑠 𝑁

Noting that 𝜔2𝑛 = 𝑘∕𝑚, this last equation becomes 𝐹0

𝑚𝑔 𝚥̂

(19.30)

𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡

𝑚

(19.34)

Figure 19.12 FBD of the forced harmonic oscillator in Fig. 19.11.

Interesting Fact How practical is harmonic forcing? The answer to this question lies in an amazing result due to Jean Baptiste Joseph Fourier (1768–1830) and later contributors. It says that any periodic piecewise smooth function can be represented by an infinite series of sines and cosines (called Fourier series in honor of Fourier). This means that any periodic forcing can be regarded as the sum of harmonic functions! In addition, given the nature of the left side of Eq. (19.31) (i.e., it is linear), it turns out that the overall particular solution for a sum of harmonic forcing terms is simply the sum of the particular solutions for each individual harmonic forcing term. These results taken together allow engineers to easily obtain the solutions to problems with any periodic forcing as a sum of simple forced harmonic oscillator solutions. Because periodic forcing is ubiquitous in engineering systems, this is one of the most important results in applied mathematics.

1308

Chapter 19

Mechanical Vibrations

where we have assumed that 𝜔0 ≠ 𝜔𝑛 and have used the fact that 𝜔2𝑛 = 𝑘∕𝑚. Choosing 𝐷 as in Eq. (19.34), we see that the guess in Eq. (19.32) is correct, and the corresponding particular solution is 𝑥𝑝 =

𝐹0 ∕𝑘 ( )2 sin 𝜔0 𝑡. 1 − 𝜔0 ∕𝜔𝑛

(19.35)

Combining the complementary solution in Eq. (19.13) with the particular solution in Eq. (19.35), the general solution to Eq. (19.31) is 𝑥 = 𝑥𝑐 + 𝑥𝑝 = 𝐴 sin 𝜔𝑛 𝑡 + 𝐵 cos 𝜔𝑛 𝑡 +

𝐹0 ∕𝑘 )2 sin 𝜔0 𝑡, ( 1 − 𝜔0 ∕𝜔𝑛

(19.36)

where, as usual, 𝐴 and 𝐵 are constants determined by enforcing the initial conditions. Equation (19.36) tells us that the vibration of a forced harmonic oscillator is composed of two parts: the complementary solution 𝑥𝑐 that describes the free vibration of the system and the particular solution 𝑥𝑝 that describes the forced vibration due to 𝐹0 sin 𝜔0 𝑡. As we will see in Section 19.3, the free vibration corresponding to 𝑥𝑐 will die out with any amount of damping or energy dissipation, which is always present in real physical systems. For this reason, free vibration is often referred to as transient vibration. On the other hand, the forced vibration corresponding to 𝑥𝑝 will be there as long as the forcing is there, and so it is often called steady-state vibration.∗ When a vibration is forced, it is important to know the amplitude of the motion since that will determine the deformation and the deformation-related forces that the system has to endure. Equation (19.35) tells us that the amplitude of the steady-state vibration, which is given by 4 𝜔 𝜔 𝑛

−3 0

1

2

MF =

𝐹0 ∕𝑘

=

1 ( )2 , 1 − 𝜔0 ∕𝜔𝑛

(19.38)

3

Figure 19.13 MF as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 .

ISTUDY

𝑥amp

a plot of which is shown in Fig. 19.13. Notice that when 𝜔0 ∕𝜔𝑛 ≪ 1, MF → 1. That is, for low-frequency forcing, the motion of the block is in the direction of the forcing (they are said to be in phase with one another). We expect this to be the case since, in the limit as 𝜔0 ∕𝜔𝑛 → 0, we obtain the static deflection. As the forcing frequency 𝜔0 approaches the natural frequency of the system 𝜔𝑛 , MF increases dramatically and goes to infinity as 𝜔0 ∕𝜔𝑛 → 1. The situation in which 𝜔0 ≈ 𝜔𝑛 is called resonance, and it results in very large vibration amplitudes. Resonances in engineering systems are generally undesirable because they result in large displacements and deformations, often leading to premature failure. However, there do exist beneficial resonances in engineering systems, such as in amplifiers or in devices designed to ∗ We

do not discuss the solution of Eq. (19.31) for 𝜔0 = 𝜔𝑛 because such a solution does not describe a steady-state vibration.

ISTUDY

Section 19.2

Undamped Forced Vibration

1309

aid sputum clearance of the airways of respiratory patients.∗ As we will see in Section 19.3, in a system with even a small amount of energy dissipation or damping, resonance does not result in infinite amplitudes, but the amplitudes can still grow to be very large. Looking back at the system in Fig. 19.11, when 𝜔0 > 𝜔𝑛 , the MF is negative and the forcing is out of phase with the motion of the block. Finally, when 𝜔0 ≫ 𝜔𝑛 , the force changes direction so rapidly compared to the natural frequency of the system that the system remains almost stationary, and the MF → 0. Harmonic excitation of the support If the support that couples the spring to the block is excited harmonically rather than the block itself, we obtain a modified form of the forced harmonic oscillator equation. To see this, consider the system shown in Fig. 19.14, which shows a block of mass 𝑚 that is coupled to a harmonically excited support by a linear spring of constant 𝑘. The position of the oscillating support is given by 𝑥𝑠 = 𝑋𝑠 sin 𝜔𝑠 𝑡, and the position of the block is given by 𝑥. The FBD of this system is shown in Fig. 19.15. Summing forces in the 𝑥 direction gives ∑

𝐹𝑥∶

−𝐹𝑠 = 𝑚𝑎𝑥 ,

(19.39)

𝑥 𝑥𝑠 = 𝑋𝑠 sin 𝜔𝑠 𝑡 𝑘

Figure 19.14 A simple harmonic oscillator whose support is being harmonically excited.

where 𝐹𝑠 is the force in the spring. Since both ends of the spring are moving, its force law is given by 𝐹𝑠 = 𝑘(𝑥 − 𝑥𝑠 ) = 𝑘(𝑥 − 𝑋𝑠 sin 𝜔𝑠 𝑡), so Eq. (19.39) becomes ( ) −𝑘 𝑥 − 𝑋𝑠 sin 𝜔𝑠 𝑡 = 𝑚𝑎𝑥 = 𝑚𝑥, ̈

𝑚

𝑚𝑔 𝐹𝑠

𝚥̂ 𝚤̂

(19.40)

𝑁

̈ Upon rearranging, this where we have substituted in the kinematic equation 𝑎𝑥 = 𝑥. becomes 𝑘𝑋𝑠 sin 𝜔𝑠 𝑡, (19.41) 𝑥̈ + 𝜔2𝑛 𝑥 = 𝑚

Figure 19.15 FBD of the block for the system shown in Fig. 19.14.

where 𝜔2𝑛 = 𝑘∕𝑚. Equation (19.41) is of the same form as Eq. (19.31) with the term 𝐹0 replaced by the term 𝑘𝑋𝑠 . The results given in Eqs. (19.35)–(19.38) are valid with that same replacement.

End of Section Summary When a harmonic oscillator is subject to harmonic forcing, the standard form of the equation of motion is Eq. (19.31), p. 1307 𝑥̈ + 𝜔2𝑛 𝑥 =

𝐹0 𝑚

𝑘

sin 𝜔0 𝑡,

where 𝐹0 is the amplitude of the forcing and 𝜔0 is its frequency (see Fig. 19.16). The general solution to this equation consists of the sum of the complementary solution and a particular solution. The complementary solution 𝑥𝑐 is the solution of the associated homogeneous equation, which is given by, for example, Eq. (19.13). For ∗ See

L. C. de Lima et al., “Mechanical Evaluation of a Respiratory Device,” Medical Engineering & Physics, 27, 2005, pp. 181–187.

𝑚

𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡

Figure 19.16 A forced harmonic oscillator whose equation √ of motion is given by Eq. (19.31) with 𝜔𝑛 = 𝑘∕𝑚. The position 𝑥 is measured from the equilibrium position of the block.

1310

Chapter 19

Mechanical Vibrations

𝜔0 ≠ 𝜔𝑛 , a particular solution was found to be Eq. (19.35), p. 1308 𝑥𝑝 =

𝐹0 ∕𝑘 ( )2 sin 𝜔0 𝑡, 1 − 𝜔0 ∕𝜔𝑛

and so the general solution is given by Eq. (19.36), p. 1308 𝑥 = 𝑥𝑐 + 𝑥𝑝 = 𝐴 sin 𝜔𝑛 𝑡 + 𝐵 cos 𝜔𝑛 𝑡 + 4 𝜔 𝜔 𝑛

−3 0

1

2

3

𝐹0 ∕𝑘 ( )2 , 1 − 𝜔0 ∕𝜔𝑛

which means that the corresponding magnification factor MF is Eq. (19.38), p. 1308

Figure 19.17 MF as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 .

𝑥amp 𝐹0 ∕𝑘

=

1 )2 . ( 1 − 𝜔0 ∕𝜔𝑛

The plot of the MF in Fig. 19.17 illustrates the phenomenon of resonance, which occurs when 𝜔0 ≈ 𝜔𝑛 and which results in very large vibration amplitudes.

𝑥 𝑥𝑠 = 𝑋𝑠 sin 𝜔𝑠 𝑡 𝑘

𝑚

Figure 19.18 A harmonic oscillator whose support is being excited harmonically.

ISTUDY

MF =

Harmonic excitation of the support. If the support of a structure is excited harmonically rather than the structure itself (see Fig. 19.18), then Eq. (19.31) is still the governing equation, except that 𝐹0 is replaced by the spring constant 𝑘 times the amplitude of the support vibration 𝑋𝑠 . All solutions described previously are then valid with that same replacement.

ISTUDY

Section 19.2

E X A M P L E 19.4

Motion of a Weight on a Vibrating Spring Scale

The block 𝐴 is resting on the platform 𝑃 of a spring scale when the lab bench to which it is rigidly attached begins vibrating vertically. The block and the platform are coupled to the base 𝐵 of the scale by a linear elastic spring that is internal to the scale (Fig. 1). If the combined mass of the block and platform is 𝑚𝐴 = 1.5 kg, the spring constant is 𝑘 = 50 N∕m, and the vibration is sinusoidal with a frequency of 15 Hz and amplitude of 5 mm, determine the vertical motion of the platform and block as a function of time. Assume that the block 𝐴 does not separate from the platform 𝑃 during the vibration.

𝐴

ACME platform 𝑃

internal spring 𝑘

vibrating base 𝐵

SOLUTION lab bench

Road Map & Modeling

The FBD of the block and platform is shown in Fig. 2, where 𝐹𝑠 is the force exerted by the spring on the platform. Since this an example of harmonic excitation of a support, we can apply Newton’s second law in the 𝑦 direction to obtain the equation of motion and then use the results of this section to find the motion. Governing Equations Balance Principles

Force Laws

1311

Undamped Forced Vibration

Summing forces in the 𝑦 direction in Fig. 2, we obtain ∑ 𝐹𝑦 ∶ −𝐹𝑠 − 𝑚𝐴 𝑔 = 𝑚𝐴 𝑎𝐴𝑦 .

Figure 1 A scale sitting on a lab bench that is vibrating vertically. The base of the scale 𝐵 is rigidly attached to the lab bench. The platform 𝑃 and the mass 𝐴 are coupled to the base of the scale by an internal spring.

(1)

and 𝑦𝐵 = 𝑌𝐵 sin 𝜔𝐵 𝑡,

(3)

where 𝜔𝐵 = 15 Hz = 94.25 rad∕s is the frequency of vibration of the base 𝐵 and 𝑌𝐵 is the amplitude of vibration of the base.

𝚤̂ 𝑦𝐴

𝐹𝑠

where 𝑦𝐴 is the vertical position of the block measured from the static equilibrium position of the block, 𝑦𝐵 is the 𝑦 position of the base of the scale, and 𝛿𝑠 is the deflection of the spring when the system is in static equilibrium. Kinematic Equations For the kinematic equations, we need to express 𝑎𝐴𝑦 and 𝑦𝐵 in terms of motion of the base 𝐵 and the block 𝐴, which gives 𝑎𝐴𝑦 = 𝑦̈𝐴

𝚥̂

𝑚𝐴 𝑔 ACME

Since both ends of the spring are moving, the force law for the spring is ( ) (2) 𝐹𝑠 = 𝑘 𝑦 𝐴 − 𝑦 𝐵 − 𝛿𝑠 ,

𝑦𝐴 = 0 static equilibrium

𝑦𝐵

𝑘

Figure 2 FBD of the block 𝐴 and platform 𝑃 in Fig. 1, as well as the definitions of 𝑦𝐴 and 𝑦𝐵 .

Computation

Substituting Eqs. (2) and (3) into Eq. (1), we get the equation of motion 𝑘𝑌 ) ( 𝑘 𝑦 = 𝐵 sin 𝜔𝐵 𝑡, (4) −𝑘 𝑦𝐴 − 𝑌𝐵 sin 𝜔𝐵 𝑡 − 𝛿𝑠 − 𝑚𝐴 𝑔 = 𝑚𝐴 𝑦̈𝐴 ⇒ 𝑦̈𝐴 + 𝑚𝐴 𝐴 𝑚𝐴

where we have used the fact that 𝑘𝛿𝑠 = 𝑚𝐴 𝑔. Equation (4) is of the same form as Eq. (19.31), so we know that the solution must be given by Eq. (19.36) with 𝐹0 = 𝑘𝑌𝐵 , 𝑌𝐵 ( )2 sin 𝜔𝐵 𝑡, 1 − 𝜔𝐵 ∕𝜔𝑛

(5)

√ where 𝜔𝑛 = 𝑘∕𝑚𝐴 = 5.774 rad∕s. Since the system starts from rest, the initial conditions are 𝑦𝐴 (0) = 0 and 𝑦̇ 𝐴 (0) = 0. Applying the first initial condition to Eq. (5), we obtain 𝐷 = 0. Applying the second initial condition gives 𝑦(0) ̇ = 0 = 𝐶𝜔𝑛 +

𝑌𝐵 𝜔𝐵 ( )2 1 − 𝜔𝐵 ∕𝜔𝑛

⇒

𝐶=

𝑌𝐵 𝜔𝐵 ∕𝜔𝑛 ( )2 . 1 − 𝜔𝐵 ∕𝜔𝑛

(6)

Substituting Eq. (6), 𝑌𝐵 = 5 mm, and the values of 𝜔𝑛 and 𝜔𝐵 into Eq. (5), we obtain ( ) 𝑦𝐴 = −0.3075 sin 5.774𝑡 − 0.01884 sin 94.25𝑡 mm. (7) Discussion & Verification

The plot of Eq. (7) in Fig. 3 shows the high-frequency forcing 𝜔𝐵 superimposed on the much lower-frequency natural vibration of the scale platform.

𝑦𝐴 (mm)

𝑦𝐴 = 𝐶 sin 𝜔𝑛 𝑡 + 𝐷 cos 𝜔𝑛 𝑡 +

0.3 0.2 0.1 0.0 −0.1 −0.2 −0.3 0

1

Figure 3 A plot of 𝑦𝐴 in Eq. (7).

2

3

4

5

1312

Chapter 19

Mechanical Vibrations

E X A M P L E 19.5

Equations of Motion for an Unbalanced Motor

platform 𝑘𝑠

𝑘𝑠

𝑘𝑠

Figure 1 An unbalanced motor mounted on a platform that is elastically suspended on six springs, three of which are shown.

A motor is not perfectly balanced when the mass center of the rotor is not on its spin axis. When this happens, as the rotor spins, it transmits time-varying forces to the housing. In turn, these forces cause the motor and the table or platform on which it is mounted to vibrate. Figure 1 shows a motor of mass 𝑚𝑚 mounted on a platform whose mass is 𝑚𝑝 . The platform is supported on six linear elastic springs, each with constant 𝑘𝑠 , whose equivalent spring constant is 𝑘eq = 6𝑘𝑠 . The rotor spins inside the motor at a constant angular velocity 𝜔𝑟 , and the effect of the unbalance is equivalent to an unbalanced mass 𝑚𝑢 located a distance 𝜀 from the axis of rotation (the mass of the motor 𝑚𝑚 includes the unbalanced mass 𝑚𝑢 ). Using this information, show that the equation of motion of the system consisting of platform and rotor is of the form of the standard harmonically forced harmonic oscillator, that is, of the form 𝑥̈ + 𝜔2𝑛 𝑥 =

𝑚𝑢 𝑔 𝑅𝑥 𝚥̂ 𝚤̂ 𝑅𝑦

𝑚

sin 𝜔0 𝑡.

(1)

SOLUTION Road Map & Modeling

The FBDs of the unbalanced mass and of the motor and platform combination are shown in Figs. 2 and 3, respectively, where we have isolated the unbalanced mass 𝑚𝑢 from the motor. By separating the unbalanced mass from the rest of the system, we will be able to determine the forces required to rotate the unbalanced mass, and the equal and opposite forces will be applied to the rest of the system.

Figure 2 FBD of the unbalanced mass 𝑚𝑢 . The forces 𝑅𝑥 and 𝑅𝑦 are the forces exerted by the motor on the unbalanced mass.

ISTUDY

𝐹0

𝑅𝑦

(𝑚𝑚 − 𝑚𝑢 )𝑔

𝚥̂ 𝚤̂

𝑅𝑥 𝑚𝑝 𝑔

𝑁

𝐹𝑠 Figure 3. FBD of the motor and platform with the unbalanced mass 𝑚𝑢 removed. The forces 𝑅𝑥 and 𝑅𝑦 are the forces exerted by the particle on the motor and, which, by Newton’s third law, are equal and opposite to the corresponding forces in Fig. 2. 𝑁 is the net normal force on the platform due to the vertical walls.

Governing Equations Balance Principles Applying Newton’s second law in the 𝑦 direction to the unbalanced mass 𝑚𝑢 in Fig. 2, we obtain (∑ ) 𝐹𝑦 ∶ −𝑅𝑦 − 𝑚𝑢 𝑔 = 𝑚𝑢 𝑎𝑢𝑦 , (2) Fig 2

where 𝑎𝑢𝑦 is the 𝑦 component of the acceleration of the unbalanced mass. Applying Newton’s second law to the FBD of the rotor and platform with the unbalanced mass removed, and observing that the motor and platform can move only in the 𝑦 direction (see Fig. 3), we obtain (∑ ) ( ) ( ) 𝐹𝑦 ∶ 𝑅𝑦 − 𝑚𝑚 − 𝑚𝑢 𝑔 − 𝑚𝑝 𝑔 − 𝐹𝑠 = 𝑚𝑚 − 𝑚𝑢 + 𝑚𝑝 𝑎𝑚𝑦 , (3) Fig 3

where 𝑎𝑚𝑦 is the 𝑦 component of acceleration of the motor/platform combination and 𝐹𝑠 is the force on the motor/platform due to the springs.

ISTUDY

Section 19.2

Force Laws

Undamped Forced Vibration

1313

The spring force is given by ( ) 𝐹𝑠 = 𝑘eq 𝑦𝑚 − 𝛿𝑠 ,

(4)

where 𝛿𝑠 is the deflection of the spring when the system is in static equilibrium. Kinematic Equations

Referring to Fig. 4, we denote the 𝑦 coordinate of the motor by 𝜀

𝚥̂

𝑦𝑢

𝜃

𝚤̂

𝑦𝑚

𝑦𝑚 = 0

Figure 4. Kinematics of the motor/platform and the unbalanced mass.

𝑦𝑚 , which is measured from the static equilibrium position of the system. Again referring to Fig. 4, the kinematic equation for the unbalanced mass is 𝑦𝑢 = 𝑦𝑚 + 𝜀 sin 𝜃

⇒

𝑦̈𝑢 = 𝑦̈𝑚 − 𝜀𝜃̇ 2 sin 𝜃

⇒

𝑎𝑢𝑦 = 𝑦̈𝑚 − 𝜀𝜔2𝑟 sin 𝜃,

(5)

where 𝜃̇ = 𝜔𝑟 , 𝑦̈𝑢 = 𝑎𝑢𝑦 , and 𝜃̈ = 0 since 𝜔𝑟 is constant. For the motor/platform, we have 𝑎𝑚𝑦 = 𝑦̈𝑚 .

(6)

Computation Substituting Eqs. (4)–(6) into Eqs. (2) and (3) and then eliminating 𝑅𝑦 from the resulting two equations, we obtain

) ( ( ) − 𝑚𝑢 𝑔 − 𝑚𝑢 𝑦̈𝑚 − 𝜀𝜔2𝑟 sin 𝜃 − 𝑚𝑚 − 𝑚𝑢 𝑔 − 𝑚𝑝 𝑔 ( ) ( ) − 𝑘eq 𝑦𝑚 − 𝛿𝑠 = 𝑚𝑚 − 𝑚𝑢 + 𝑚𝑝 𝑦̈𝑚 . Canceling terms and rearranging, we obtain ) ) ( ( 𝑚𝑚 + 𝑚𝑝 𝑦̈𝑚 + 𝑘eq 𝑦𝑚 + 𝑚𝑚 + 𝑚𝑝 𝑔 − 𝑘eq 𝛿𝑠 = 𝑚𝑢 𝜀𝜔2𝑟 sin 𝜃.

(7)

(8)

Noting that 𝜃 = 𝜔𝑟 𝑡,∗ and that the definition of the static deflection 𝛿𝑠 implies that (𝑚𝑚 + 𝑚𝑝 )𝑔 = 𝑘eq 𝛿𝑠 , this equation becomes 𝑦̈𝑚 +

𝑘eq 𝑚𝑚 + 𝑚𝑝 ⏟⏞⏟⏞⏟

𝑦𝑚 =

𝑚𝑢 𝜀𝜔2𝑟 𝑚𝑚 + 𝑚𝑝 ⏟⏞⏟⏞⏟

sin 𝜔𝑟 𝑡,

(9)

𝐹0 ∕𝑚

𝜔2𝑛

which is the equation of motion for the unbalanced motor and platform. Discussion & Verification As indicated, Eq. (9) is of the same form as Eq. (1), with the correspondence shown in Table 1. We will complete the analysis of this system in Example 19.6. Table 1. Parameters in Eq. (1) and the corresponding parameters in Eq. (9). In Eq. (1), we are assuming 𝜔2𝑛 = 𝑘∕𝑚. Eq. (1) Eq. (9) 𝑘 𝑚 𝐹0 ∗ We

have assumed that 𝜃(0) = 0.

𝑘eq 𝑚𝑚 + 𝑚𝑝 𝑚𝑢 𝜀𝜔2𝑟

Helpful Information Obtaining the equation of motion using the center of mass of the system. Note that Eq. (9) can be obtained more directly by applying Newton’s second law to the center of mass of the system shown in Fig. 1. This is how Prob. 19.35 asks you to obtain the equation of motion.

1314

Chapter 19

Mechanical Vibrations

E X A M P L E 19.6 𝜔𝑟

motor, 𝑚𝑚

eccentric mass, 𝑚𝑢

platform, 𝑚𝑝 𝑘𝑠

Response and MF for the Unbalanced Motor

𝑘𝑠

𝑘𝑠

Find the general solution for the displacement, 𝑦𝑚 (𝑡), of the unbalanced motor (Fig. 1) subject to the initial conditions 𝑦𝑚 (0) = 0 and 𝑦̇ 𝑚 (0) = 𝑣𝑚0 . After doing so, plot the solution for 0 < 𝑡 < 1 s, using 𝑚𝑚 = 40 kg, 𝑚𝑝 = 15 kg, 𝑘eq = 6𝑘𝑠 = 420,000 N∕m, 𝜀 = 15 cm, 𝜔𝑟 = 1200 rpm, 𝑦𝑚 (0) = 0 m, 𝑦̇ 𝑚 (0) = 0.4 m∕s, and three different values of 𝑚𝑢 : 10 g, 100 g, and 1000 g. From the plot, find the approximate maximum amplitude of the vibration. Finally, determine and plot the MF for the unbalanced rotor.

SOLUTION

Figure 1

Road Map & Modeling

The equation of motion for our model of the unbalanced motor and platform was derived in Example 19.5 as Eq. (9), so we need only apply the general solution in Eq. (19.36) to determine and plot the response. For the MF, as found in Eq. (19.38), we will need to find a function of 𝜔𝑟 ∕𝜔𝑛 where 𝜔𝑟 is the angular velocity of the unbalanced rotor inside the motor. 𝜀 𝜃=𝜔 𝑡 𝑟

𝚥̂ 𝚤̂

𝑦𝑚

𝑦𝑢

Governing Equations 𝑦𝑚 = 0

For convenience, we repeat the equation of motion for the unbalanced motor and platform, which was found in Eq. (9) of Example 19.5 to be 𝑦̈𝑚 +

Figure 2 Kinematic definitions for the unbalanced motor and platform.

ISTUDY

𝑘eq 𝑚 𝑚 + 𝑚𝑝

𝑦𝑚 =

𝑚𝑢 𝜀𝜔2𝑟

sin 𝜔𝑟 𝑡,

𝑚 𝑚 + 𝑚𝑝

(1)

where 𝑦𝑚 is measured from the static equilibrium position of the system, as shown in Fig. 2. Computation

Equation (1) is of the same form as Eq. (19.31), which is repeated below

for convenience 𝑥̈ + 𝜔2𝑛 𝑥 =

𝐹0 𝑚

sin 𝜔0 𝑡,

(2)

where, comparing Eqs. (1) and (2), we have 𝜔2𝑛 =

𝑘eq 𝑚𝑚 + 𝑚𝑝

,

𝐹0 𝑚

=

𝑚𝑢 𝜀𝜔2𝑟 𝑚𝑚 + 𝑚𝑝

,

𝜔0 = 𝜔𝑟 .

and

(3)

Therefore, the general solution to Eq. (1) can be found using Eq. (19.36), which gives 𝑦𝑚 = 𝐴 sin 𝜔𝑛 𝑡 + 𝐵 cos 𝜔𝑛 𝑡 +

𝑚𝑢 𝜀𝜔2𝑟 ∕𝑘eq ( )2 sin 𝜔𝑟 𝑡. 1 − 𝜔𝑟 ∕𝜔𝑛

(4)

For 𝑡 = 0, Eq. (4) gives 𝑦𝑚 (0) = 𝐵. Therefore, recalling that we must have 𝑦𝑚 (0) = 0, we have 𝐵 = 0. (5) Differentiating 𝑦𝑚 in Eq. (4) with respect to time, we obtain 𝑦̇ 𝑚 = 𝐴𝜔𝑛 cos 𝜔𝑛 𝑡 − 𝐵𝜔𝑛 sin 𝜔𝑛 𝑡 +

𝑚𝑢 𝜀𝜔3𝑟 ∕𝑘eq ( )2 cos 𝜔𝑟 𝑡, 1 − 𝜔𝑟 ∕𝜔𝑛

(6)

and then applying the initial condition 𝑦̇ 𝑚 (0) = 𝑣𝑚0 , we get 𝐴𝜔𝑛 +

𝑚𝑢 𝜀𝜔3𝑟 ∕𝑘eq ( )2 = 𝑣𝑚0 1 − 𝜔𝑟 ∕𝜔𝑛

⇒

𝐴=

𝑣𝑚0 𝜔𝑛

−

2 𝜔𝑟 𝑚𝑢 𝜀𝜔𝑟 ∕𝑘eq . 𝜔𝑛 1 − (𝜔 ∕𝜔 )2 𝑟 𝑛

(7)

Combining Eqs. (4), (5), and (7), the general solution becomes 2 ⎡𝑣 𝑚𝑢 𝜀𝜔2𝑟 ∕𝑘eq 𝜔 𝑚𝑢 𝜀𝜔𝑟 ∕𝑘eq ⎤ ⎥ sin 𝜔 𝑡 + 𝑦𝑚 = ⎢ 𝑚0 − 𝑟 ( )2 sin 𝜔𝑟 𝑡, 𝑛 ⎢ 𝜔𝑛 𝜔𝑛 1 − (𝜔 ∕𝜔 )2 ⎥ 1 − 𝜔𝑟 ∕𝜔𝑛 ⎣ 𝑟 𝑛 ⎦

(8)

ISTUDY

Section 19.2

Undamped Forced Vibration

1315

a plot of which is shown in Fig. 3. From this figure, we can see that the maximum ampli𝑚𝑢 = 10 g 𝑚𝑢 = 100 g 𝑚𝑢 = 1000 g 0.015 0.010 0.005 𝑦𝑚 (m)

0.000 −0.005 −0.010 −0.015 0.0

0.2

0.4

0.6

0.8

1.0

Figure 3. The response 𝑦𝑚 of the motor and platform for three values of the unbalanced mass 𝑚𝑢 .

tude of vibration for 𝑚𝑢 = 10 g is about 5 mm, for 𝑚𝑢 = 100 g it is about 6 mm, and for 𝑚𝑢 = 1 kg it is about 17 mm. To compute MF, we take the amplitude of the particular solution and rearrange it so that 𝜔𝑟 and 𝜔𝑛 always appear as their ratio. Doing this gives 𝑚𝑢 𝜀𝜔2𝑟 ∕𝑘eq |𝑦𝑚𝑝 | = )2 ( 1 − 𝜔𝑟 ∕𝜔𝑛

⇒

)2 /( ) ( 𝑚𝑢 𝜀 𝜔𝑟 ∕𝜔𝑛 𝑚𝑚 + 𝑚 𝑝 |𝑦𝑚𝑝 | = . )2 ( 1 − 𝜔𝑟 ∕𝜔𝑛

(9)

𝜔𝑟 < 𝜔 𝑛 2

MF

0

Therefore, MF is given by

MF =

( ) |𝑦𝑚𝑝 | 𝑚𝑚 + 𝑚𝑝 𝑚𝑢 𝜀

)2 ( 𝜔𝑟 ∕𝜔𝑛 = )2 , ( 1 − 𝜔𝑟 ∕𝜔𝑛

−2

(10)

a plot of which is shown in Fig. 4. In our case, since 𝜔𝑟 = 1200 rpm = 125.7 rad∕s and 𝜔𝑛 = 87.39 rad∕s, we have MF = −1.937.

(11)

Discussion & Verification

The dimensions of all the terms in the general solution in Eq. (8) are length, as they should be. The amplitude of the vibration found by examining Fig. 3 is a few millimeters in all cases, which seems reasonable given the masses and stiffnesses involved. Finally, MF is dimensionless, again as it should be. A Closer Look It is interesting to compare the MF of the unbalanced motor plotted in Fig. 4, with the MF in Fig. 19.13, which applies to a particle that is forced directly (see Fig. 19.11). Notice that for small 𝜔𝑟 , the MF in Fig. 4 approaches 0 rather than 1, as it does in Fig. 19.13. This means that when the unbalanced rotor within the motor is spinning very slowly, it does not shake the motor and platform by any appreciable amount, which agrees with our intuition. On the other hand, when a particle is harmonically forced directly as in Fig. 19.11 and the forcing frequency is very small, the particle moves with the forcing, and so the MF is 1.

𝜔𝑟 > 𝜔 𝑛 −4 0

1

2

3

Figure 4 The MF for response of the unbalanced motor as given by Eq. (10).

1316

Chapter 19

Mechanical Vibrations

Problems Problem 19.31 The magnification factor for a forced (undamped) harmonic oscillator is measured to be equal to 5. Determine the driving frequency of the forcing if the natural frequency of the system is 100 rad∕s. 𝑥 𝑘

Problem 19.32 𝐹0 cos 𝜔0 𝑡

𝑚

Figure P19.32

Suppose that the equation of motion of a forced harmonic oscillator is given by 𝑥̈ + 𝜔2𝑛 𝑥 = (𝐹0 ∕𝑚) cos 𝜔0 𝑡. Obtain the expression for the response of the oscillator, and compare it to the response presented in Eq. (19.36) [which is for a forced harmonic oscillator with the equation of motion given in Eq. (19.31)].

𝑦(𝑡)

Problem 19.33

𝑂

A uniform bar of mass 𝑚 and length 𝐿 is pinned to a slider at 𝑂. The slider is forced to oscillate horizontally according to 𝑦(𝑡) = 𝑌 sin 𝜔𝑠 𝑡. The system lies in the vertical plane.

𝜃

𝑚, 𝐿

(a) Derive the equation of motion of the bar for small angles 𝜃. (b) Determine the amplitude of steady-state vibration of the bar.

Problem 19.34

Figure P19.33

Determine the amplitude of vibration of the unbalanced motor we studied in Example 19.6 if the forcing frequency of the motor is 0.95𝜔𝑛 . 𝜔𝑟

motor, 𝑚𝑚

unbalanced mass, 𝑚𝑢

platform, 𝑚𝑝 𝑘𝑠

𝑘𝑠

𝑘𝑠

Figure P19.34 and P19.35

Problem 19.35 Derive the equations of motion for the unbalanced motor introduced in this section by applying Newton’s second law to the center of mass of the system shown in Fig. 1 of Example 19.5.

Welcome to Minnesota

Problem 19.36 𝜃̇ 𝐿

Figure P19.36

ISTUDY

𝑑𝑖 𝑑𝑜 cross section of the pole

Consider a sign mounted on a circular hollow steel pole of length 𝐿 = 5 m, outer diameter 𝑑𝑜 = 5 cm, and inner diameter 𝑑𝑖 = 4 cm. Aerodynamic forces due to wind provide a harmonic torsional excitation with frequency 𝑓0 = 3 Hz and amplitude 𝑀0 = 10 N⋅m about the 𝑧 axis. The mass center of the sign lies on the central axis 𝑧 of the pole. The mass moment of inertia of the(sign is 𝐼)𝑧 = 0.1 kg⋅m2 . The torsional stiffness of the pole can be estimated as 𝑘𝑡 = 𝜋𝐺st 𝑑𝑜4 − 𝑑𝑖4 ∕(32𝐿), where 𝐺st is the shear modulus of steel, which is 79 GPa. Neglecting the inertia of the pole, calculate the amplitude of steady-state vibration of the sign.

ISTUDY

Section 19.2

1317

Undamped Forced Vibration

Problem 19.37 An unbalanced motor is mounted at the tip of a rigid beam of mass 𝑚𝑏 and length 𝐿. The beam is restrained by a torsional spring of stiffness 𝑘𝑡 and an additional support of stiffness 𝑘 located at the half length of the beam. In the static equilibrium position, the beam is horizontal, and the torsional spring does not exert any moment on the beam. The mass of the motor is 𝑚𝑚 , and the unbalance results in a harmonic excitation 𝐹 (𝑡) = 𝐹0 sin 𝜔0 𝑡 in the vertical direction. Derive the equation of motion for the system, assuming that 𝜃 is small.

motor, 𝑚𝑚 𝜃 𝑘𝑡

beam, 𝑚𝑏 𝑘

Figure P19.37

Problem 19.38 Revisit Example 19.6 and discuss whether it is possible to obtain the equation of motion of the system via the energy method. 𝜔𝑟

motor, 𝑚𝑚

unbalanced mass, 𝑚𝑢

platform, 𝑚𝑝 𝑘𝑠

𝑘𝑠

𝑘𝑠

Figure P19.38

Problem 19.39 A fatigue-testing machine for electronic components consists of a platform with an unbalanced motor. Assume that the rotor in the motor spins at 𝜔𝑟 = 3000 rpm, the mass of the platform is 𝑚𝑝 = 20 kg, the mass of the motor is 𝑚𝑚 = 15 kg, the unbalanced mass is 𝑚𝑢 = 0.5 kg, and the equivalent stiffness of the platform suspension is 𝑘eq = 𝑛𝑘𝑠 = 5 × 106 N∕m, where 𝑛 is the number of springs. For the testing machine, the distance 𝜀 between the spin axis of the rotor and the location at which 𝑚𝑢 is placed can be varied to obtain the desired vibration level. Calculate the range of values of 𝜀 that would provide amplitudes of the particular solution ranging from 0.1 mm to 2 mm.

Problem 19.40 At time 𝑡 = 0, a forced harmonic oscillator occupies position 𝑥(0) = 0.1 m and has a velocity 𝑥(0) ̇ = 0. The mass of the oscillator is 𝑚 = 10 kg, and the stiffness of the spring is 𝑘 = 1000 N∕m. Calculate the motion of the system if the forcing function is 𝐹 (𝑡) = 𝐹0 sin 𝜔0 𝑡, with 𝐹0 = 10 N and 𝜔0 = 200 rad∕s. 𝑥

𝑥 𝑘

𝑚

Figure P19.40

𝐹0 sin 𝜔0 𝑡

𝑘

𝑘 𝑘

𝑘

𝑚

𝐹0 sin 𝜔0 𝑡

Figure P19.41

Problem 19.41 The forced harmonic oscillator shown has a mass 𝑚 = 10 kg. In addition, the harmonic excitation is such that 𝐹0 = 150 N and 𝜔0 = 200 rad∕s. If all sources of friction can be neglected, determine the spring constant 𝑘 such that the magnification factor MF = 5.

𝜀 motor, 𝑚𝑚

𝜔𝑟 mass, 𝑚𝑢

platform, 𝑚𝑝 𝑘𝑠 Figure P19.39

𝑘𝑠

𝑘𝑠

1318

Chapter 19

Mechanical Vibrations

Problem 19.42

𝐹 (𝑡) 𝑚 𝑦(𝑡) 𝑘

𝑘

𝑠(𝑡) Figure P19.42 and P19.43

ISTUDY

A ring of mass 𝑚 is attached by two linear elastic cords with elastic constant 𝑘 and unstretched length 𝐿0 < 𝐿 to a support, as shown. Assuming that the pretension in the cords is large, so that the cords’ deflection due to the ring’s weight can be neglected, find the linearized equation of motion for the case where 𝐹 (𝑡) = 𝐹0 sin 𝜔0 𝑡 and 𝑠(𝑡) = 0 (i.e., the support is stationary). In addition, find the response of the system for 𝑦(0) = 0 and 𝑦(0) ̇ = 0.

Problem 19.43 A ring of mass 𝑚 is attached by two linear elastic cords with elastic constant 𝑘 and unstretched length 𝐿0 < 𝐿 to a support, as shown. Assuming that the pretension in the cords is large, so that the cords’ deflection due to the ring’s weight can be neglected, find the linearized equation of motion for the case where 𝐹 (𝑡) = 0 and 𝑠(𝑡) = 𝑠0 sin 𝜔𝑡. In addition, find the response of the system for 𝑦(0) = 0 and 𝑦(0) ̇ = 0.

Problem 19.44 Modeling the beam as a rigid uniform thin bar, ignoring the inertia of the pulleys, assuming that the system is in static equilibrium when the bar is horizontal, and assuming that the cord is inextensible and does not go slack, determine the linearized equation of motion of the system in terms of 𝑥, which is the position of 𝐴. Finally, determine an expression for the amplitude of the steady-state vibration of block 𝐴.

𝑚𝐵

𝐵

𝐴

𝑥

𝑚𝐴 𝐹 (𝑡) = 𝐹0 sin 𝜔0 𝑡

𝑘

Figure P19.44 and P19.45

Problem 19.45 For the system in Prob. 19.44, determine the maximum forcing frequency 𝜔0 for steadystate motion, such that the cord does not go slack.

Problems 19.46 and 19.47 One of the propellers on the Beech King Air 200 is unbalanced, such that the unbalanced weight 𝑊𝑢 is a distance 𝑅 from the spin axis of the propeller. The propellers spin at a constant rate 𝜔𝑝 , and the weight of each engine is 𝑊𝑒 (this includes the mass of the propeller). Assume that the wing is a uniform beam that is cantilevered at 𝐴, has weight 𝑊𝑤 and bending stiffness 𝐸𝐼, and whose mass center is at 𝐺. Treat the engine as a point mass and evaluate your answers for 𝑊𝑢 = 3 oz, 𝑊𝑒 = 450 lb, 𝑅 = 5.1 f t, 𝜔𝑝 = 2000 rpm, 𝐸𝐼 = 1.13 × 1011 lb⋅in.2 , 𝑑 = 8.7 f t, and ℎ = 10.9 f t. In addition, ignore the angular motion of the wing in computing the angular velocity and angular acceleration of the propeller, and ignore the time-dependent inertia term in the final equation of motion.

ISTUDY

Section 19.2

Undamped Forced Vibration

𝜔𝑝

𝑅

𝑚𝑢

𝑚𝑒

𝐴

𝐺

𝑚𝑤 , 𝐸𝐼

𝑑 Figure P19.46 and P19.47 Problem 19.46

Neglect the mass of the wing, and model the wing as was done in Example 19.2. Determine the resonance frequency of the system, and find the MF for the given parameters. Problem 19.47

Let the mass of the wing be 𝑚𝑤 = 350 lb∕𝑔, and model the wing as was done in Example 19.2. Determine the resonance frequency of the system, and find the MF for the given parameters.

1319

1320

Chapter 19

Mechanical Vibrations

Design Problems Design Problem 19.2 The device shown can detect when the angular velocity and angular acceleration of a rigid body 𝐵 achieve a combination of specified values. The device works using the principle that the vibration amplitude of the sphere 𝑃 depends on both the angular velocity and angular acceleration of the rigid body. When the angular velocity 𝜔𝐵 and angular acceleration 𝛼𝐵 reach the appropriate combination, 𝑃 will contact the touch sensor, thus signaling that the specified values have been reached. With this as background, assume that the rigid body rotates in the horizontal plane with angular velocity 𝜔𝐵 and angular acceleration 𝛼𝐵 , and that 𝑃 is constrained to move in the slot, which is at a distance 𝑑 from the center of the disk. In addition, a linear elastic spring of constant 𝑘 is attached to the mass, such that the spring is undeformed when the mass is at 𝑠 = 0.

𝑑 𝑠 𝑃

𝑘 𝑂

(a) Derive the equation of motion for 𝑃 with 𝑠 as the dependent variable. rigid body 𝐵

𝛼𝐵 Figure DP19.2

ISTUDY

𝜔𝐵

(b) Assuming that 𝑃 is released from rest at 𝑠 = 0, find the solution to the equation of motion found in (a), knowing that the solution to ordinary differential equations of the type 𝑠̈ + 𝜔2𝑛 𝑠 = 𝐷, is given by 𝑠(𝑡) =

𝐷 + 𝐶1 cos 𝜔𝑛 𝑡 + 𝐶2 sin 𝜔𝑛 𝑡, 𝜔2𝑛

where 𝐶1 and 𝐶2 are constants determined from the initial conditions and 𝐷 is a known constant. (c) Using the solution for 𝑠(𝑡) found above, for given values of 𝑑, 𝑘, 𝑚, 𝜔𝐵 , and 𝛼𝐵 , determine the maximum distance from 𝑠 = 0 that 𝑃 achieves in one cycle. (d) For a disk-shaped rigid body 𝐵 whose diameter is 1.5 m, specify the mass of 𝑃 (treat it as a particle), the spring constant 𝑘, the length ℎ, and the distance 𝑑 so that the touch sensor can detect when the rigid body reaches an angular velocity 𝜔crit = 100 rpm for a constant angular acceleration 𝛼𝐵 = 1 rad∕s2 .

ISTUDY

Section 19.3

19.3

Viscously Damped Vibration

1321

Viscously Damped Vibration upper mount

All mechanical systems exhibit some energy dissipation or damping due to air drag, viscous fluids, friction, and other effects. If the damping is small enough, the undamped solutions obtained in Sections 19.1 and 19.2 will be in close agreement with the damped solution for a short period of time. On the other hand, if we need a solution for a longer period of time or if there is more damping, we need to resort to the solution of the equations that model damped mechanical systems. In this section, we will consider linear viscous damping. This is damping in which the damping force is directly proportional and opposite in sign to the velocity of a body. This type of damping tends to occur when the energy dissipation is due to a fluid (e.g., oil, water, or air), as seen in the shock absorber in Fig. 19.19. In addition, even when the damping is due to other physical mechanisms, linear viscous damping can still be an effective model.

piston rod oil reserve cylinder pressure tube base valve lower mount

Figure 19.19 A cutaway view of a typical shock absorber, which is used in many suspension systems.

Viscously damped free vibration The effect of viscous damping is usually modeled by an element called a dashpot, a schematic of which is shown in Fig. 19.20. Damping in a dashpot occurs when the cylinder 𝑥̇

𝚤̂

𝐹𝑑 piston

viscous fluid

Figure 19.20. Schematic diagram of a dashpot illustrating its basic operation.

piston moves within the fluid-filled cylinder and forces the fluid to flow either around the piston or through one or more holes in it. Because the fluid within the piston is viscous, this motion results in energy dissipation. Referring to Fig. 19.20, if the relative velocity of the two ends of the piston is 𝑥, ̇ then the damping force 𝐹𝑑 on the piston is equal to 𝐹𝑑 = 𝑐 𝑥, ̇ (19.42) where 𝑐 is a constant called the coefficient of viscous damping. This damping coefficient depends on the physical properties of the fluid and the geometry of the dashpot. The coefficient of viscous damping is expressed in lb⋅ s∕f t in U.S. Customary units and N⋅s∕m in SI units. Revisiting Example 13.5 on p. 802 in which a railcar runs into a large spring that is designed to stop it, a dashpot has now been added in parallel with the spring (see Fig. 19.21). Assuming that the railcar couples with the spring and dashpot after it hits them, we obtain the FBD of the railcar shown in Fig. 19.22. Summing forces in the 𝑥 direction, we find ∑ 𝐹𝑥∶ −𝐹𝑑 − 𝐹𝑠 = 𝑚𝑥, ̈ (19.43) where 𝐹𝑑 is the force due to the dashpot, 𝐹𝑠 is the force due to the spring, and 𝑥 measures the displacement of the spring from its equilibrium position. The forces 𝐹𝑑

𝑥 𝑐

Figure 19.21 Railcar hitting a spring and dashpot. Recall from Example 13.5 that the railcar weighs 87 tons and is moving at 4 mph at impact, and 𝑘 = 22,270 lb∕f t. 𝑚𝑔 𝚥̂ 𝚤̂

𝐹𝑑

𝐹𝑠 𝑁

Figure 19.22 FBD of the railcar after it has hit and coupled with the spring and dashpot.

1322

Chapter 19

Mechanical Vibrations

and 𝐹𝑠 can be written as 𝐹𝑑 = 𝑐 𝑥̇

and

𝐹𝑠 = 𝑘𝑥,

(19.44)

which allows us to write Eq. (19.43) as 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 0.

(19.45)

This is the standard form of the viscously damped harmonic oscillator. The theory of differential equations tells us that Eq. (19.45) is a linear, second-order, homogeneous, constant coefficient differential equation, and as such, it has solutions of the form 𝑥 = 𝑒𝜆𝑡 ,

(19.46)

where 𝜆 (the Greek letter lambda) is a constant to be determined. Replacing 𝑥 in Eq. (19.45) with Eq. (19.46), we find 𝑚𝜆2 𝑒𝜆𝑡 + 𝑐𝜆𝑒𝜆𝑡 + 𝑘𝑒𝜆𝑡 = 0. Factoring out 𝑒𝜆𝑡 in Eq. (19.47), we have ( ) 𝑒𝜆𝑡 𝑚𝜆2 + 𝑐𝜆 + 𝑘 = 0.

(19.47)

(19.48)

Since 𝑒𝜆𝑡 never vanishes, to have a solution to Eq. (19.45) we must have 𝑚𝜆2 + 𝑐𝜆 + 𝑘 = 0.

(19.49)

If 𝜆 is a root of this quadratic equation, called the characteristic equation, then 𝑒𝜆𝑡 is a solution to Eq. (19.45). The two roots of Eq. (19.49) are given by √ √ ( )2 ( )2 𝑐 𝑘 𝑐 𝑘 𝑐 𝑐 + − and 𝜆2 = − − − . (19.50) 𝜆1 = − 2𝑚 2𝑚 𝑚 2𝑚 2𝑚 𝑚

1.0

𝑐 = 𝑐𝑐

0.8 𝑥 (f t)

Referring to Eqs. (19.50), the theory of differential equations tells us that the general solution of Eq. (19.45) takes on one of three possible forms determined by the values of 𝜆1 and 𝜆2 . Observe that the character of 𝜆1 and 𝜆2 depends on whether the term (𝑐∕2𝑚)2 − 𝑘∕𝑚 is positive, zero, or negative. Therefore, we introduce a special value of the damping coefficient called the critical damping coefficient, which we denote by 𝑐𝑐 and define as the value of 𝑐 that makes the term (𝑐∕2𝑚)2 − 𝑘∕𝑚 equal to zero, that is, √ ( 𝑐 )2 𝑘 𝑘 𝑐 − =0 ⇒ 𝑐𝑐 = 2𝑚 = 2𝑚𝜔𝑛 , (19.51) 2𝑚 𝑚 𝑚 √ where 𝜔𝑛 = 𝑘∕𝑚. We now distinguish the three cases mentioned above based on whether 𝑐 > 𝑐𝑐 , 𝑐 = 𝑐𝑐 , or 𝑐 < 𝑐𝑐 .

0.6

Overdamped system (𝒄 > 𝒄𝒄 )

0.4 𝑐 > 𝑐 𝑐 0.2 0.0

0

1

2

3

4

Figure 19.23 The position of the railcar as a function of time after it impacts the spring and dashpot. The blue curve is overdamped with 𝑐 = 35,000 lb⋅s∕f t, and the red curve is critically damped with 𝑐 = 𝑐𝑐 ≈ 21,950 lb⋅s∕f t.

ISTUDY

If 𝑐 > 𝑐𝑐 , the term (𝑐∕2𝑚)2 − 𝑘∕𝑚 is positive. Thus, 𝜆1 and 𝜆2 are real, distinct, and negative. In this case, the general solution of Eq. (19.45) is [ √ ] √ 2 2 𝑥 = 𝑒−(𝑐∕2𝑚)𝑡 𝐴𝑒𝑡 (𝑐∕2𝑚) −𝑘∕𝑚 + 𝐵𝑒−𝑡 (𝑐∕2𝑚) −𝑘∕𝑚 , (19.52) where 𝐴 and 𝐵 are constants that are determined from the initial conditions of the system. The motion represented by Eq. (19.52) is characterized by decaying exponentials, the system does not vibrate, and there is no period associated with the motion (see Fig. 19.23). This type of system is said to be overdamped.

ISTUDY

Section 19.3

Viscously Damped Vibration

1323

Critically damped system (𝒄 = 𝒄𝒄 ) If 𝑐 = 𝑐𝑐 , the term (𝑐∕2𝑚)2 − 𝑘∕𝑚 is zero. Thus 𝜆1 = 𝜆2 = −𝑐𝑐 ∕2𝑚. In this case, the general solution of Eq. (19.45) is 𝑥 = (𝐴 + 𝐵𝑡)𝑒−𝜔𝑛 𝑡 ,

(19.53)

where, again, 𝐴 and 𝐵 are constants that are determined from the initial conditions. If 𝑐 = 𝑐𝑐 , then 𝑐 has the smallest value for which no vibration occurs, and the system is said to be critically damped. Referring to Fig. 19.23, notice that a critically damped system approaches equilibrium faster than an overdamped system. Critically damped systems are of great interest in engineering applications since they approach equilibrium in the minimum possible time. Underdamped system (𝒄 < 𝒄𝒄 ) If 𝑐 < 𝑐𝑐 , the term (𝑐∕2𝑚)2 − 𝑘∕𝑚 is negative. Thus, 𝜆1 and 𝜆2 are complex since they involve the square root of a negative quantity. In this case, the general solution of Eq. (19.45) is ( ) 𝑥 = 𝑒−(𝑐∕2𝑚)𝑡 𝐴 sin 𝜔𝑑 𝑡 + 𝐵 cos 𝜔𝑑 𝑡 ,

(19.54)

Interesting Fact Landing on an aircraft carrier. A carrierbased aircraft has a tailhook, which is attached to an 8 𝖿 𝗍 bar extending from the rear of the aircraft. When the aircraft lands on the deck of the carrier, the tailhook is supposed to catch one of four steel arresting cables that are stretched across the deck.

where 𝐴 and 𝐵 are determined from the initial conditions, and 𝜔𝑑 is the damped natural frequency, which is given by √

√ ( )2 ( )2 𝑘 𝑐 − 𝜔𝑑 = = 𝜔𝑛 1 − 𝑐∕𝑐𝑐 , (19.55) 𝑚 2𝑚 √ and where we recall that 𝜔𝑛 = 𝑘∕𝑚 and 𝑐𝑐 = 2𝑚𝜔𝑛 . A system for which 𝑐 < 𝑐𝑐 is said to be underdamped, and for any such system, Eq. (19.55) implies that 𝜔𝑑 is always less than 𝜔𝑛 . Using the definition of 𝜔𝑑 given in Eq. (19.55), the period of damped vibration is given by 𝜏𝑑 = 2𝜋∕𝜔𝑑 .

(19.56) PH3 Christopher Mobley/U.S. Navy

Note that the solution for 𝑥 given in Eq. (19.54) can also be written as 𝑥 = 𝐷𝑒

−(𝑐∕2𝑚)𝑡

( ) sin 𝜔𝑑 𝑡 + 𝜙 ,

(19.57)

where 𝐷 and 𝜙 are constants determined by the initial conditions. The solution in Eq. (19.57) has been plotted in Fig. 19.24 using the data specified in Fig. 19.21 and a value of 𝑐 that makes the system underdamped. Damping ratio In practice, the three cases just discussed are often classified in terms of a nondimensional parameter called the damping ratio or damping factor, which is usually denoted by the symbol 𝜁 (the Greek letter zeta) and is defined as 𝜁 = 𝑐∕𝑐𝑐 .

(19.58)

Using 𝜁 , the standard form of the viscously damped harmonic oscillator in Eq. (19.45) is rewritten as 𝑥̈ + 2𝜁 𝜔𝑛 𝑥 + 𝜔2𝑛 𝑥 = 0. (19.59)

Each 1.375 𝗂𝗇. thick arresting cable connects to a hydraulic cylinder below deck that acts as a huge dashpot. When the tailhook snags one of the wires, the wire pulls a piston within the fluid-filled hydraulic cylinder. As the piston moves through the cylinder, hydraulic fluid is forced through the small holes at the end of the cylinder, which dissipates the kinetic energy of the aircraft. This system can stop a 54,000 𝗅𝖻 aircraft traveling at 150 𝗆𝗉𝗁 in less than 350 𝖿 𝗍. This system can handle a landing every 45 𝗌.

1324

Chapter 19

Mechanical Vibrations

3 𝐷𝑒−(𝑐∕2𝑚)𝑡

2 𝐷 1

𝜏𝑑 2

−1 −2

4

6

8

10

12

𝑡 (s)

−𝐷𝑒−(𝑐∕2𝑚)𝑡 𝐷𝑒−(𝑐∕2𝑚)𝑡 sin 𝜔𝑑 𝑡 + 𝜙

Figure 19.24. The position of the railcar as a function of time after it impacts the spring and dashpot. The red curve is the underdamped solution for 𝑐 = 3000 lb⋅s∕f t.

Overdamped system (𝜁 > 1). In terms of 𝜁 , an overdamped system is characterized by 𝜁 > 1, and the general solution in Eq. (19.52) is rewritten as ( √ ) √ 2 2 𝑥 = 𝑒−𝜁𝜔𝑛 𝑡 𝐴𝑒 𝜁 −1 𝜔𝑛 𝑡 + 𝐵𝑒− 𝜁 −1 𝜔𝑛 𝑡 . (19.60) Critically damped system (𝜁 = 1). In terms of 𝜁 , a critically damped system is characterized by 𝜁 = 1, and the general solution in Eq. (19.53) is unchanged. Underdamped system (𝜁 < 1). In terms of 𝜁 , an underdamped system is characterized by 𝜁 < 1, and the solutions in Eqs. (19.54) and (19.57) are rewritten as ( ) ( ) 𝑥 = 𝑒−𝜁𝜔𝑛 𝑡 𝐴 sin 𝜔𝑑 𝑡 + 𝐵 cos 𝜔𝑑 𝑡 = 𝐷𝑒−𝜁𝜔𝑛 𝑡 sin 𝜔𝑑 𝑡 + 𝜙 , (19.61) respectively, where, referring to Eq. (19.55), 𝜔𝑑 is expressed in terms of 𝜁 as √ 𝜔𝑑 = 𝜔 𝑛 1 − 𝜁 2 .

Viscously damped forced vibration

𝑐

𝐹0 sin 𝜔0 𝑡 𝑘

𝑚

Figure 19.25 A simple damped harmonic oscillator that is harmonically forced.

ISTUDY

(19.62)

Here, we consider the case of the vibration of a one degree of freedom system that is both damped and forced, for example, the simple system shown in Fig. 19.25, in which we have simply added a dashpot to the system in Fig. 19.11 on p. 1307. The equation of motion for this system is given by 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 𝐹0 sin 𝜔0 𝑡,

(19.63)

where 𝑥 is measured from the equilibrium position of the block. Using the damping ratio 𝜁 , Eq. (19.63) can be written as 𝑥̈ + 2𝜁 𝜔𝑛 𝑥̇ + 𝜔2𝑛 𝑥 =

𝐹0 𝑚

sin 𝜔0 𝑡.

(19.64)

If, rather than the harmonic forcing being applied to 𝑚, the support in Fig. 19.25 is displaced harmonically according to 𝑌 sin 𝜔0 𝑡, we cannot simply replace 𝐹0 with 𝑘𝑌 , as we did in Eq. (19.41) in Section 19.2 (see Example 19.7 and Prob. 19.63 for

ISTUDY

Section 19.3

1325

Viscously Damped Vibration

ways to handle this situation). On the other hand, if the dashpot remains attached to a fixed support and we harmonically displace the support to which the spring is attached (see Fig. 19.26), then the equation of motion is of the form 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 𝑘𝑌 sin 𝜔0 𝑡,

𝑐

𝑘 𝑚

(19.65)

so we can replace 𝐹0 with 𝑘𝑌 in all the corresponding solutions. As was the case with undamped forced vibration in Section 19.2, the solution to Eq. (19.63) or Eq. (19.64) is the sum of the complementary solution and a particular solution. As a reminder, the complementary solution 𝑥𝑐 is the solution to the associated homogeneous equation, which is Eq. (19.45) and for which we found the solution to depend on the level of damping (i.e., whether 𝑐 is greater than, equal to, or less than the critical damping 𝑐𝑐 ). Regardless of the level of damping, this complementary solution will die out in time, and thus its contribution is transient (see p. 1308). As discussed in Section 19.2 on p. 1308, the forced vibration associated with a particular solution 𝑥𝑝 will exist as long as the forcing does, and it is thus a steady-state vibration. Since we have already found the complementary solution, it will be the particular solution that we focus on here. As with undamped forced vibration, the response of a damped forced system should also resemble the forcing. Therefore, we will assume a particular solution of either of the following forms: ( ) 𝑥𝑝 = 𝐴 sin 𝜔0 𝑡 + 𝐵 cos 𝜔0 𝑡 = 𝐷 sin 𝜔0 𝑡 − 𝜙 ,

𝑥 𝑦

Figure 19.26 Harmonic support displacement of a viscously damped harmonic oscillator.

(19.66)

where, in the first expression, 𝐴 and 𝐵 are constants to be determined and, in the second expression, 𝐷 and 𝜙 are constants to be determined and 𝐷 is assumed to be a positive quantity.∗ While either of these expressions can be used, the latter gives a more easily interpreted result since the amplitude 𝐷 and phase 𝜙 are immediately apparent, and so we will use it and proceed to determine 𝐷 and 𝜙. Substituting the second expression for 𝑥𝑝 from Eq. (19.66) into Eq. (19.63), we obtain ( ) ( ) ( ) −𝐷𝑚𝜔20 sin 𝜔0 𝑡−𝜙 +𝐷𝑐𝜔0 cos 𝜔0 𝑡−𝜙 +𝐷𝑘 sin 𝜔0 𝑡−𝜙 = 𝐹0 sin 𝜔0 𝑡. (19.67) Using the trigonometric identities sin(𝛼 − 𝛽) = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽 and cos(𝛼 − ) ( 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 cos 𝛽 and then collecting coefficients of sin 𝜔0 𝑡 − 𝜙 and ( ) cos 𝜔0 𝑡 − 𝜙 , we obtain ( ) 𝐷 −𝑚𝜔20 cos 𝜙 + 𝑐𝜔0 sin 𝜙 + 𝑘 cos 𝜙 sin 𝜔0 𝑡 ) ( + 𝐷 𝑚𝜔20 sin 𝜙 + 𝑐𝜔0 cos 𝜙 − 𝑘 sin 𝜙 cos 𝜔0 𝑡 = 𝐹0 sin 𝜔0 𝑡.

(19.68)

Since this equation must be true for all time, we can equate the coefficients of sin 𝜔0 𝑡 and cos 𝜔0 𝑡 to obtain two equations for the unknowns 𝐷 and 𝜙. Doing this for cos 𝜔0 𝑡 allows us to solve for tan 𝜙 as ( )( ) 2 𝑐∕𝑐𝑐 𝜔0 ∕𝜔𝑛 2𝜁 𝜔0 ∕𝜔𝑛 tan 𝜙 = = ( )2 = ( )2 , 2 𝑘 − 𝑚𝜔0 1 − 𝜔0 ∕𝜔𝑛 1 − 𝜔0 ∕𝜔𝑛 𝑐𝜔0

𝜙 (rad)

Eq. (19.66) we have used a sine function of the form sin(𝜔0 𝑡 − 𝜙) instead of sin(𝜔0 𝑡 + 𝜙) because it results in a more convenient expression for tan 𝜙.

𝜁 = 0.05 𝜁 = 0.2 𝜁 = 0.4

𝜁 =1

𝜋 2

𝜋 4

0

(19.69)

where the definitions of 𝜔𝑛 , 𝑐𝑐 , and 𝜁 , in Eqs. (19.4), Eq. (19.51), and Eq. (19.58), respectively, have been used. The phase angle 𝜙, which is plotted in Fig. 19.27 as a ∗ In

3𝜋 4

0

1

2

3

4

Figure 19.27 The phase angle 𝜙 as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 .

1326

Chapter 19

Mechanical Vibrations

function of the frequency ratio 𝜔0 ∕𝜔𝑛 for different values of 𝜁 , represents the amount of a cycle by which the response of the system lags the forcing applied to it. Equating the coefficients of sin 𝜔0 𝑡, we obtain 𝐷=

𝐹0 ∕ cos 𝜙 𝑘 − 𝑚𝜔20

+ 𝑐𝜔0 tan 𝜙

=

𝐹0 ∕ cos 𝜙 𝑐𝜔0 ∕ tan 𝜙 + 𝑐𝜔0 tan 𝜙

,

(19.70)

where, to get the second expression for 𝐷, we have used Eq. (19.69). Now multiplying the numerator and denominator by tan 𝜙 and noting that 1 + tan2 𝜙 = 1∕ cos2 𝜙, we obtain

𝐷=

𝐹0 sin 𝜙 𝑐𝜔0

⎧ ⎫ 2𝜁 𝜔0 ∕𝜔𝑛 𝐹0 ⎪ ⎪ = √[ ⎨ ⎬, 𝑐𝜔0 ⎪ ( )2 ]2 ( )2 ⎪ 1 − 𝜔0 ∕𝜔𝑛 + 2𝜁 𝜔0 ∕𝜔𝑛 ⎭ ⎩

(19.71)

−1 where √ we have used the trigonometric identity that if 𝛼 = tan 𝑥, then sin 𝛼 = 𝑥∕ 1 + 𝑥2 . Again using the definitions of 𝜔𝑛 , 𝑐𝑐 , and 𝜁 as above, 𝐷 simplifies to

𝐹0 ∕𝑘 𝐷= √ . [ ( )2 ]2 ( )2 1 − 𝜔0 ∕𝜔𝑛 + 2𝜁 𝜔0 ∕𝜔𝑛

Now that we have the particular solution, we will, as was done in Section 19.2, define a magnification factor for a damped forced harmonic oscillator as the ratio of the amplitude of steady-state vibration 𝐷 to the static deflection 𝐹0 ∕𝑘. Thus, we obtain

𝜁 =0 4 𝜁 = 0.125 3

MF =

𝜁 = 0.2

MF

(19.72)

2

𝐷 =√ [ 𝐹0 ∕𝑘

( 1 − 𝜔0 ∕𝜔𝑛

1 ) ]2 2

)2 ( + 2𝜁 𝜔0 ∕𝜔𝑛

.

(19.73)

𝜁 = 0.4 1 𝜁 =1 0

0

1

2

3

Figure 19.28 The MF as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 for various values of the damping ratio 𝜁.

ISTUDY

A plot of the MF is shown in Fig. 19.28 for various values of the damping ratio 𝜁 . Figure 19.28 illustrates some important features of the behavior of a damped forced harmonic oscillator: • The magnitude of the oscillation can be small in two ways: by keeping the forcing frequency 𝜔0 away from the natural frequency and/or by increasing the amount of damping, i.e., increasing 𝜁 . • As the amount of damping is increased, the peak in the MF moves farther to the left, away from 𝜔0 ∕𝜔𝑛 = 1. The peak for any given value of 𝜁 can be found by using the usual technique from calculus to find the maximum value of a function (see Prob. 19.52). • Comparing Fig. 19.28 with Fig. 19.13, observe that the effect of damping, while quite noticeable near the resonance frequency, becomes very small for values of 𝜔0 ∕𝜔𝑛 away from 1.

ISTUDY

Section 19.3

Viscously Damped Vibration

End of Section Summary Viscously damped free vibration. The standard form of the equation of motion for a one degree of freedom viscously damped harmonic oscillator is Eq. (19.45), p. 1322 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 0, where 𝑚 is the mass, 𝑐 is the coefficient of viscous damping, and 𝑘 is the linear spring constant. The character of the solution to this equation depends on the amount of damping relative to a specific amount of damping called the critical damping coefficient, which is defined as Eq. (19.51), p. 1322 √ 𝑘 𝑐𝑐 = 2𝑚 = 2𝑚𝜔𝑛 , 𝑚 √ where 𝜔𝑛 = 𝑘∕𝑚. In particular, if 𝑐 ≥ 𝑐𝑐 , then the motion is nonoscillatory, whereas if 𝑐 < 𝑐𝑐 , then the motion is oscillatory. If 𝑐 > 𝑐𝑐 , the system is said to be overdamped, and the solution is given by Eq. (19.52), p. 1322 [

𝑥 = 𝑒−(𝑐∕2𝑚)𝑡 𝐴𝑒𝑡

√ (𝑐∕2𝑚)2 −𝑘∕𝑚

+ 𝐵𝑒−𝑡

√

(𝑐∕2𝑚)2 −𝑘∕𝑚

] ,

where 𝐴 and 𝐵 are constants to be determined from the initial conditions. If 𝑐 = 𝑐𝑐 , the system is said to be critically damped, and the solution is given by Eq. (19.53), p. 1323 𝑥 = (𝐴 + 𝐵𝑡)𝑒−𝜔𝑛 𝑡 , where, again, 𝐴 and 𝐵 are constants to be determined from the initial conditions. Finally, if 𝑐 < 𝑐𝑐 , the system is said to be underdamped, and the solution is given by Eq. (19.54), p. 1323 ( ) 𝑥 = 𝑒−(𝑐∕2𝑚)𝑡 𝐴 sin 𝜔𝑑 𝑡 + 𝐵 cos 𝜔𝑑 𝑡 , or, equivalently, by Eq. (19.57), p. 1323 ( ) 𝑥 = 𝐷𝑒−(𝑐∕2𝑚)𝑡 sin 𝜔𝑑 𝑡 + 𝜙 , where 𝐴 and 𝐵 are constants to be determined from the initial conditions in the first solution, 𝐷 and 𝜙 are analogous constants in the second solution, and 𝜔𝑑 is the damped natural frequency, which is given by Eq. (19.55), p. 1323, and Eq. (19.62), p. 1324 √ √ ( )2 √ ( )2 𝑘 𝑐 𝜔𝑑 = − = 𝜔𝑛 1 − 𝑐∕𝑐𝑐 = 𝜔𝑛 1 − 𝜁 2 , 𝑚 2𝑚 where 𝜁 = 𝑐∕𝑐𝑐 is the damping ratio.

1327

1328

Chapter 19

Mechanical Vibrations Viscously damped forced vibration. forced harmonic oscillator is

The standard form of a viscously damped

Eq. (19.63), p. 1324 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 𝐹0 sin 𝜔0 𝑡, where 𝐹0 is the amplitude of the forcing function and 𝜔0 is the frequency of the forcing function. When expressed using the damping ratio 𝜁 , the equation above takes on the form Eq. (19.64), p. 1324 𝑥̈ + 2𝜁 𝜔𝑛 𝑥̇ + 𝜔2𝑛 𝑥 =

𝐹0 𝑚

sin 𝜔0 𝑡.

The general solution of either of these equations is the sum of the complementary solution and a particular solution. The complementary solution is transient; i.e., it vanishes as time increases. The particular or steady-state solution is of the form Eq. (19.66), p. 1325 ( ) 𝑥𝑝 = 𝐷 sin 𝜔0 𝑡 − 𝜙 , where 𝜙 and 𝐷 are given by 𝜁 =0

Eq. (19.69), p. 1325, and Eq. (19.72), p. 1326 ( )( ) 2 𝑐∕𝑐𝑐 𝜔0 ∕𝜔𝑛 2𝜁 𝜔0 ∕𝜔𝑛 𝑐𝜔0 tan 𝜙 = = )2 = )2 , ( ( 2 𝑘 − 𝑚𝜔0 1 − 𝜔0 ∕𝜔𝑛 1 − 𝜔0 ∕𝜔𝑛 𝐹0 ∕𝑘 . 𝐷= √ [ ( )2 ]2 ( )2 1 − 𝜔0 ∕𝜔𝑛 + 2𝜁 𝜔0 ∕𝜔𝑛

4 𝜁 = 0.125 3 𝜁 = 0.2

MF 2

𝜁 = 0.4 1

The magnification factor for a damped, forced harmonic oscillator is

𝜁 =1 0

0

1

2

Figure 19.29 The MF as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 for various values of the damping ratio 𝜁.

ISTUDY

Eq. (19.73), p. 1326

3

MF =

𝐷 1 , =√ [ 𝐹0 ∕𝑘 ( )2 ]2 ( )2 1 − 𝜔0 ∕𝜔𝑛 + 2𝜁 𝜔0 ∕𝜔𝑛

a plot of which is shown in Fig. 19.29 for various values of the damping ratio 𝜁 .

ISTUDY

Section 19.3

Viscously Damped Vibration

E X A M P L E 19.7

1329

Motion of a Weight on a Viscously Damped Vibrating Spring Scale

The scale shown in Fig. 1 uses an internal spring that couples the platform 𝑃 to the base 𝐵 of the scale to determine the weight of an object placed on the platform. So that oscillations of the platform die out quickly after an object is placed on the scale, it also includes an internal dashpot in parallel with the spring to dissipate energy. If the combined mass of the block and platform is 𝑚𝑃 = 1.5 kg and the spring constant is 𝑘 = 50 N∕m, determine the critical viscous damping coefficient 𝑐𝑐 for the dashpot so that oscillations dissipate quickly.

𝐴

ACME platform 𝑃

internal spring

𝑐

internal dashpot

𝑘 base 𝐵

SOLUTION Road Map & Modeling

The FBD of the block and platform during oscillation of the platform is shown in Fig. 2. We will apply Newton’s second law in the 𝑦 direction to obtain the equation of motion, from which we should be able to extract 𝑐𝑐 . Governing Equations Balance Principles

Summing forces in the 𝑦 direction in Fig. 2, we obtain ∑ 𝐹𝑦 ∶ −𝐹𝑠 − 𝐹𝑑 − 𝑚𝑃 𝑔 = 𝑚𝑃 𝑎𝑃 𝑦 .

(1)

Figure 1 A scale used to measure the mass of an object 𝐴 that has been placed on it. The platform 𝑃 and the mass 𝐴 are coupled to the base of the scale via an internal spring and dashpot. The block 𝐴 does not separate from the platform 𝑃 during the vibration.

Force Laws The force laws for the spring and dashpot are determined by the motion of the platform 𝑃 , so we obtain ) ( 𝐹𝑠 = 𝑘 𝑦𝑃 − 𝛿𝑠 and 𝐹𝑑 = 𝑐 𝑦̇ 𝑃 , (2)

where 𝑦𝑃 is the vertical position of the platform measured from the static equilibrium position of the platform, and 𝛿𝑠 is the deflection of the spring when the system is in static equilibrium (see Fig. 2).

𝑚𝑃 𝑔 AC CME 𝚥̂ 𝚤̂

𝐹𝑠

𝐹𝑑

Kinematic Equations

For the kinematic equations, we need to express 𝑎𝑃 𝑦 in terms of motion of the platform, which gives 𝑎𝑃 𝑦 = 𝑦̈𝑃 . Computation

as

(3)

Substituting Eqs. (2) and (3) into Eq. (1), we get the equation of motion

( ) −𝑘 𝑦𝑃 − 𝛿𝑠 − 𝑐 𝑦̇ 𝑃 − 𝑚𝑃 𝑔 = 𝑚𝑃 𝑦̈𝑃

⇒

𝑚𝑃 𝑦̈𝑃 + 𝑐 𝑦̇ 𝑃 + 𝑘𝑦𝑃 = 0,

(4)

where we have used the fact that 𝑘𝛿𝑠 = 𝑚𝑃 𝑔. Since Eq. (4) is in the standard form of the viscously damped harmonic oscillator in Eq. (19.45), we can find the critical damping coefficient by using Eq. (19.51), which gives √ 𝑘 = 2𝑚𝑃 𝜔𝑛 ⇒ 𝑐𝑐 = 17.32 kg∕s, (5) 𝑐𝑐 = 2𝑚𝑃 𝑚𝑃 where 𝑚𝑃 = 1.5 kg and 𝜔𝑛 = Discussion & Verification

√

𝑘∕𝑚𝑃 = 5.774 rad∕s.

The units of 𝑐𝑐 are kg∕s, which is equivalent to the SI units for the coefficient of viscous damping stated earlier in this section, that is, N⋅s∕m.

Figure 2 FBD of the block 𝐴 and platform 𝑃 in Fig. 1. 𝐹𝑠 and 𝐹𝑑 are the force exerted by the spring and the force exerted by the dashpot, respectively, on the platform.

1330

Chapter 19

Mechanical Vibrations

E X A M P L E 19.8

Critically Damped Free Vibration of a Gate The carnival ride in Fig. 1 has a gate, which is shown in Fig. 2, that is used to count people entering the ride. The gate has a linear elastic torsional spring of stiffness 𝑘𝑡 and a torsional damper with constant 𝑐𝑡 at the pin 𝑂 that control how it returns to the closed position after being opened. Determine 𝑘𝑡 and 𝑐𝑡 so that the gate is critically damped and returns to within 𝜃 = 4◦ of the closed position less than 2.5 s after starting from rest at 𝜃 = 80◦ . Model the gate as a thin bar of mass 𝑚𝑏 and length 𝐿 with a point mass 𝑚 at its end. Neglect friction at the pin 𝑂, as well as air resistance.

𝑚

Gary L. Gray

𝑚𝑏

Figure 1 A carnival ride with a gate that counts riders.

𝑘𝑡 𝑂

𝜃 𝑐𝑡

Figure 2. A carnival ride gate for which 𝐿 = 42 in., the weight of the thin bar is 4 lb, and the weight of the point mass is 3 lb. The gate lies in the horizontal plane.

SOLUTION Road Map & Modeling

We will first need to derive the equation of motion for the gate in the standard form of Eq. (19.45), using the FBD of the arm shown in Fig. 3. Once we have the equation of motion, we can find its response using Eq. (19.53). From the response, we can determine the value of 𝑘𝑡 needed to get the arm closed in the required time and then use that to find the 𝑐𝑡 needed to critically damp the arm.

𝐿 𝑚 𝑚𝑏

𝑂𝑦

𝑂

𝑂𝑥

𝜃

Figure 3 FBD of the carnival ride gate in Fig. 2.

ISTUDY

𝚥̂ 𝚤̂

Governing Equations Balance Principles

Summing moments about point 𝑂, we obtain ∑ 𝑀𝑂∶ −𝑀𝑠 − 𝑀𝑑 = 𝐼𝑂 𝛼gate ,

(1)

where 𝑀𝑠 is the restoring moment due to the spring, 𝑀𝑑 is the damping moment due to the dashpot, and 𝛼gate is the angular acceleration of the gate. The mass moment of inertia of the gate with respect to 𝑂 is computed as ( ) 1 𝑚𝑏 𝐿2 + 𝑚𝑏 (𝐿∕2)2 = 31 𝑚𝑏 + 𝑚 𝐿2 . (2) 𝐼𝑂 = 𝑚𝐿2 + 12 Force Laws

The moment laws for the spring and dashpot are 𝑀𝑠 = 𝑘 𝑡 𝜃

Kinematic Equations

̇ and 𝑀𝑑 = 𝑐𝑡 𝜃.

(3)

The angular acceleration of the gate can be written in terms of 𝜃

̈ as 𝛼gate = 𝜃. Computation

Substituting the kinematic relation, as well as Eqs. (2) and (3) into Eq. (1) and rearranging, we obtain the equation of motion of the gate as (1 ) (4) 𝑚 + 𝑚 𝐿2 𝜃̈ + 𝑐𝑡 𝜃̇ + 𝑘𝑡 𝜃 = 0. 3 𝑏 This implies that the natural frequency 𝜔𝑛 is given by √ √ √ 𝑘𝑡 √ = 0.7788 𝑘𝑡 rad∕s. 𝜔𝑛 = √ ( 1 ) 𝑚 + 𝑚 𝐿2 3 𝑏

(5)

ISTUDY

Section 19.3

Viscously Damped Vibration

1331

To determine 𝑘, we will enforce the condition requiring the gate to be within 4◦ of the closed position in 2.5 s or less to the solution of Eq. (4), which is given by 𝜃 = (𝐴+𝐵𝑡)𝑒−𝜔𝑛 𝑡 for critical damping. But first we need to find 𝐴 and 𝐵. Enforcing the condition that 𝜃(0) = 80◦ = 1.396 rad, we obtain 𝜃(0) = 𝐴 = 1.396 rad. (6) ̇ We now enforce the condition that 𝜃(0) = 0 rad∕s, which gives ( ) 𝜃̇ = 𝐵𝑒−𝜔𝑛 𝑡 + (𝐴 + 𝐵𝑡) −𝜔𝑛 𝑒−𝜔𝑛 𝑡

⇒

̇ 𝜃(0) = 𝐵 − 𝐴𝜔𝑛 = 0 ⇒

𝐵 = 1.396𝜔𝑛 .

Substituting Eqs. (5)–(7) into the critically damped solution for 𝜃, we obtain √ ) ( √ 𝜃 = 1.396 + 1.087𝑡 𝑘𝑡 𝑒−0.7788𝑡 𝑘𝑡 . To obtain 𝑘𝑡 , we will say that we want 𝜃(2.5) = 4◦ = 0.06981 rad, which gives √ ) ( √ 0.06981 = 1.396 + 2.719 𝑘𝑡 𝑒−1.947 𝑘𝑡 ,

(7)

(8)

60◦

(9)

which is a transcendental equation for 𝑘𝑡 . This equation can be solved iteratively or by using a mathematical software package (e.g., Mathematica or Matlab). Solving Eq. (9) gives 𝑘𝑡 = 5.936 f t ⋅lb∕rad.

80◦

(10)

Now that we have 𝑘𝑡 , the condition for critical damping in Eq. (19.51) tells us that the damping coefficient must be given by ( ) 𝑐𝑡𝑐 = 2 13 𝑚𝑏 + 𝑚 𝐿2 𝜔𝑛 ⇒ 𝑐𝑡𝑐 = 6.256 f t ⋅lb⋅s. (11) A plot of the solution in Eq. (8) [using 𝑘𝑡 from Eq. (10)] can be found in Fig. 4. Discussion & Verification

The dimensions of each of the two torsional constants found in Eqs. (10) and (11) are as they should be. A Closer Look Note that the gate could be returned to the closed position even more quickly by increasing the value of 𝑘𝑡 (compare the curves for two different values of 𝑘𝑡 in Fig. 4). Unfortunately, this has potentially undesirable consequences. Referring to Fig. 5, we see that larger values of 𝑘𝑡 result in the angular velocity of the gate being larger for every value of 𝜃. This means that someone going through the gate before it closes will get hit harder by the gate (compare 𝜃̇ at points 𝐴 and 𝐵 in Fig. 5). In addition, referring to Eq. (11), we see that larger values of 𝑘𝑡 mean that the damping coefficient must be larger to achieve critical damping. This would likely imply that the damping mechanism must be more substantial and more expensive.

𝜃 40◦ 20◦

4◦

0◦

1.0

0.0

2.0

3.0

Figure 4 Plot of response of the gate for two different values of the torsional spring constant.

0.0

𝐵 𝐴

−0.5 −1.0 4◦ −1.5 0◦

20◦

40◦

60◦

80◦

Figure 5 A plot of the angular velocity of the gate 𝜃̇ as a function of its position 𝜃 for two different values of the torsional spring constant.

1332

E X A M P L E 19.9 motor

𝜀

𝑚 𝑘

Response and MF for a Rotating Unbalanced Mass

𝑚𝑢 platform

𝑐

Figure 1 An unbalanced motor on a platform that is elastically supported, and whose vertical motion is damped by dashpots. The values of the system parameters are 𝑚 = 55 kg, 𝑘 = 420,000 N∕m, 𝑐 = 4000 N⋅s∕m, 𝜀 = 15 cm, 𝜔𝑟 = 1200 rpm, and 𝑚𝑢 is 10 g, 100 g, or 1000 g.

ISTUDY

Chapter 19

Mechanical Vibrations

If we add dashpots in parallel with the springs that are supporting the platform to the unbalanced motor we studied in Examples 19.5 and 19.6, we obtain the system shown in Fig. 1. We will assume that all the springs supply a total spring constant 𝑘, the dashpots provide a total damping coefficient 𝑐, and the combined mass of the motor and of the platform is 𝑚. From Example 19.5, we can show that the equation of motion for the unbalanced motor is (see Prob. 19.54) 𝑦̈ + 2𝜁𝜔𝑛 𝑦̇ + 𝜔2𝑛 𝑦 =

𝑚𝑢 𝜀𝜔2𝑟

sin 𝜔𝑟 𝑡, (1) 𝑚 where 𝑦 is the vertical position of the motor measured from its static equilibrium position, 𝑚𝑢 is the eccentric mass (𝑚 includes 𝑚𝑢 ), 𝜀 is the distance from the unbalanced mass to the rotor axis, and 𝜔𝑟 is the angular velocity of the rotor. Using 𝑐 = 4000 N⋅s∕m, determine and plot the steady-state solution, using the parameters given in Example 19.6. In addition, determine and plot the MF for the unbalanced motor and compare it with the MF shown in Fig. 19.28, which applies to Eq. (19.64).

SOLUTION Road Map & Modeling

The steady-state solution to an equation of the form in Eq. (1) is given by Eqs. (19.66), (19.69), and (19.72). We need to interpret the forcing amplitude on the right-hand side of Eq. (1) to obtain the steady-state solution. The MF is found from the amplitude of the steady-state solution, so we will look at the expression for 𝐷 that we obtain from Eq. (19.72) after interpreting the right-hand side of Eq. (1). Once we write the amplitude as a function of 𝜔𝑟 ∕𝜔𝑛 and 𝜁, we will have the desired MF. Governing Equations

As discussed above, the steady-state solution is given by

Eq. (19.66), i.e., 𝑦ss = 𝐷 sin(𝜔𝑟 𝑡 − 𝜙),

(2)

where 𝐷 is given by Eq. (19.72), i.e., 𝐷= √ [

𝐹0 ∕𝑘 ) 2 ]2

( 1 − 𝜔0 ∕𝜔𝑛

( )2 + 2𝜁𝜔0 ∕𝜔𝑛

,

(3)

and 𝜙 is given by Eq. (19.69), i.e., tan 𝜙 =

𝑐𝜔0 𝑘 − 𝑚𝜔20

=

2𝜁𝜔0 ∕𝜔𝑛 ( )2 , 1 − 𝜔0 ∕𝜔𝑛

(4)

√ in which 𝜔𝑛 = 𝑘∕𝑚 = 87.39 rad∕s and 𝜁 = 𝑐∕(2𝑚𝜔𝑛 ) = 0.4161. Comparing Eq. (1) with Eq. (19.64) on p. 1324, we see that 𝜔0 = 𝜔𝑟

and

𝐹0 = 𝑚𝑢 𝜀𝜔2𝑟 .

(5)

Computation

The steady-state solution is now found by substituting Eqs. (3)–(5) into Eq. (2). Doing this and then substituting in all given parameters, we find 𝑦ss = 0.003519𝑚𝑢 sin(125.7𝑡 + 0.8423) m,

(6)

which is plotted in Fig. 2 for the three given values of 𝑚𝑢 . To find the MF, we substitute Eqs. (5) into Eq. (3) to obtain 𝐷= √ [

𝑚𝑢 𝜀𝜔2𝑟 ∕𝑘 . ( ) 2 ]2 ( )2 1 − 𝜔𝑟 ∕𝜔𝑛 + 2𝜁𝜔𝑟 ∕𝜔𝑛

(7)

ISTUDY

Section 19.3

Viscously Damped Vibration

1333

𝑚𝑢 = 100 g 𝑚𝑢 = 1000 g 3 2 1 𝑦𝑠𝑠 (mm)

0 −1 −2 −3 0.00

0.05

0.10

0.15

0.20

0.25

Figure 2. The steady-state solution of the unbalanced motor for three different values of the unbalanced mass 𝑚𝑢 .

Focusing on the numerator, we see that we can write it as 𝑚𝑢 𝜀𝜔2𝑟 𝑘

=

𝑚𝑢 𝜀𝜔2𝑟 𝑚 𝑚𝑢 𝜀 𝑚 2 𝑚𝑢 𝜀 𝜔2𝑟 = 𝜔 = , 𝑘 𝑚 𝑚 𝑘 𝑟 𝑚 𝜔2𝑛

(8) 5

where, to obtain the last equality, we have used the fact that 𝑚∕𝑘 = 1∕𝜔2𝑛 . Substituting Eq. (8) into Eq. (7) and moving 𝑚𝑢 𝜀∕𝑚 to the left-hand side, we obtain the MF as ( )2 𝜔𝑟 ∕𝜔𝑛 𝑚𝐷 , = √ MF = [ 𝑚𝑢 𝜀 )2 )2 ]2 ( ( + 2𝜁𝜔𝑟 ∕𝜔𝑛 1 − 𝜔𝑟 ∕𝜔𝑛

4

𝜁 = 0.125

MF 3

𝜁 = 0.2

(9)

a plot of which is shown in Fig. 3 for various values of 𝜁. Discussion & Verification

Careful examination of the steady-state response given in Eq. (2), with Eqs. (3)–(5) substituted in, reveals that it has the dimension of length, as should be expected. In addition, we see in Fig. 2 that as we increase the amount of the unbalanced mass 𝑚𝑢 , the oscillation amplitude increases as expected. The MF in Eq. (9) is dimensionless, as it should be. A Closer Look

𝜁 =0

Comparing Fig. 3 with Fig. 19.28 on p. 1326, we see that for low

forcing frequencies: • When a harmonic oscillator is forced by applying the forcing directly to the oscillator, the oscillator follows the forcing (MF → 1 in Fig. 19.28). • When a harmonic oscillator is forced by an internal rotating unbalanced mass, the oscillator barely moves (MF → 0 in Fig. 3). For high forcing frequencies: • When a harmonic oscillator is forced by applying the forcing directly to the oscillator, the oscillator barely moves (MF → 0 in Fig. 19.28). • When a harmonic oscillator is forced by an internal rotating unbalanced mass, the oscillator moves with the forcing (MF → 1 in Fig. 3).

2

𝜁 = 0.4

1 𝜁 =1 0

0

1

2

3

Figure 3 The unbalanced motor MF as a function of the frequency ratio 𝜔𝑟 ∕𝜔𝑛 for various values of the damping ratio 𝜁 .

1334

Chapter 19

Mechanical Vibrations

Problems shock absorber and spring steering link

Problem 19.48 In the design of a MacPherson strut suspension, what would you choose for the damping ratio 𝜁? Explain your answer in terms of automotive ride and comfort.

Problem 19.49

lower control arm

Figure P19.48

ISTUDY

car frame

For identical systems, one with damping and the other without, would you expect the period of damped vibration to be greater, less than, or equal to the period of undamped vibration? Explain your answer.

Problem 19.50 A vibration test is performed on a structure, in which both the magnification factor MF and the phase angle 𝜙 are recorded as a function of excitation frequency 𝜔0 . After the test, it is discovered that, for some unfortunate reason, the recording of the magnification factor data is corrupted so that only the phase angle data is available for analysis. Is it possible to determine the resonant frequency from the available data? What can be inferred about the amount of damping in the system from the phase data?

Problem 19.51 Suppose that the equation of motion of a damped forced harmonic oscillator is given by 𝑥̈ + 2𝜁𝜔𝑛 𝑥̇ + 𝜔2𝑛 𝑥 = (𝐹0 ∕𝑚) cos 𝜔0 𝑡, where 𝑥 is measured from the equilibrium position of the system. Obtain the expression for the amplitude of the steady-state response of the oscillator, and compare it with the expression presented in Eq. (19.72) [which is for a system with equation of motion 𝑥̈ + 2𝜁𝜔𝑛 𝑥̇ + 𝜔2𝑛 𝑥 = (𝐹0 ∕𝑚) sin 𝜔0 𝑡].

𝑐

𝐹0 cos 𝜔0 𝑡 𝑘

𝑚

Figure P19.51

Problem 19.52 Differentiate Eq. (19.73) with respect to 𝜔0 ∕𝜔𝑛 , and set the result equal to zero to determine the frequency 𝜔0 at which peaks in the MF curve occur as a function of 𝜁 and 𝜔𝑛 . Use this result to show that the peak always occurs at 𝜔0 ∕𝜔𝑛 ≤ 1. Finally, determine the value of 𝜁 for which the MF has no peak.

Problem 19.53 Calculate the response described by the equations listed below, in which 𝑥 is measured in feet and time is measured in seconds. (a) 5𝑥̈ + 10𝑥̇ + 100𝑥 = 0, with 𝑥(0) = 0.1 and 𝑥(0) ̇ = −0.1 (b) 3𝑥̈ + 15𝑥̇ + 12𝑥 = 0, with 𝑥(0) = 0 and 𝑥(0) ̇ = 0.5 (c) 𝑥̈ + 10𝑥̇ + 25𝑥 = 0, with 𝑥(0) = 0.15 and 𝑥(0) ̇ =0 (d) 25𝑥̈ + 200𝑥̇ + 1500𝑥 = 0, with 𝑥(0) = 0.01 and 𝑥(0) ̇ =0

ISTUDY

Section 19.3

1335

Viscously Damped Vibration

Problem 19.54 Derive the equation of motion given in Eq. (1) of Example 19.9 for the system in that example. The independent variable 𝑦 is measured from the equilibrium position of the system, 𝑚 is the mass of the motor and platform, 𝑐 is the total damping coefficient of the dashpots, 𝑘 is the total constant of the linear elastic springs, 𝜔𝑟 is the angular velocity of the unbalanced rotor, 𝜀 is the distance of the eccentric mass from the rotor axis, and 𝑚𝑢 is the eccentric mass. Note that 𝑚 includes the eccentric mass so that the nonrotating mass is equal to 𝑚 − 𝑚𝑢 .

𝜀

𝑚𝑢

𝑚 𝑐

𝑘 Figure P19.54

Problem 19.55 A module with sensitive electronics is mounted on a panel that vibrates due to excitation from a nearby diesel generator. To prevent fatigue failure, the module is placed on vibration-absorbing mounts. The displacement of the panel is measured to be 𝑦𝑝 (𝑡) = 𝑦0 sin 𝜔0 𝑡, where 𝑦0 = 0.001 m, 𝜔0 = 300 rad∕s, and the time 𝑡 is measured in seconds. Letting the mass of the electronic module be 𝑚 = 0.5 kg, calculate the amplitude of the vibration of the module if the equivalent stiffness and damping coefficients for all the mounts combined are 𝑘 = 10,000 N∕m and 𝑐 = 40 N⋅s∕m, respectively. 𝑦𝑚

𝑦𝑝

𝑐

𝑘

Figure P19.55

Problem 19.56 A hard drive arm undergoes flow-induced vibration caused by the vortices of air produced by a platter that rotates at 𝜔0 = 10,000 rpm. The arm has length 𝐿 = 0.037 m and mass 𝑚 = 0.00075 kg, and it is made from aluminum with a modulus of elasticity 𝐸 = 70 GPa. In addition, assume that the cross section of the arm has an area moment of inertia 𝐼cs = 8.5 × 10−14 m4 . Following the steps in Example 19.2 on p. 1298, the arm can be modeled as a rigid rod that is pinned at one end and is restrained by a torsional spring with equivalent spring constant 𝑘𝑡 = 3𝐸𝐼cs ∕𝐿. In addition to the torsional spring, assume that the arm’s motion is affected by a torsional damper with torsional damping coefficient 𝑐𝑡 . Assuming that the damping ratio is 𝜁 = 0.02 and that the vortices produce an aerodynamic force with the same frequency as the rotation of the platter, determine the amplitude of the aerodynamic force needed to cause a steady-state vibration amplitude of 0.0001 m at the tip of the arm. Assume that the aerodynamic force is applied at the midpoint 𝐵 of the hard drive. What vibration amplitude will result if the same excitation is applied to a hard drive arm assembly with the damping ratio of 0.05?

𝜔0

hard drive disk hard drive arm

spin axis 𝑐𝑡

𝑦 𝐵 aerodynamic force

𝑘𝑡

Figure P19.56

Problem 19.57 The mechanism consists of a disk 𝐷 pinned at 𝐺, which is both the geometric center of the disk and its mass center. The outer circumference of the disk has radius 𝑟𝑜 = 0.1 m and is connected to an element consisting of a linear spring with stiffness 𝑘1 = 100 N∕m in parallel with a dashpot with damping coefficient 𝑐 = 50 N⋅s∕m. The disk has a hub of radius 𝑟𝑖 = 0.05 m that is connected to a linear spring with constant 𝑘2 = 350 N∕m. Knowing that for 𝜃 = 0 the disk is in static equilibrium and that the mass moment of inertia of the disk is 𝐼𝐺 = 0.001 kg⋅m2 , derive the linearized equation of motion of the disk in terms of 𝜃. In addition, calculate the resulting vibrational motion if the system is released from rest with an initial angular displacement 𝜃𝑖 = 0.05 rad.

𝐷 𝑘1

𝑟𝑜 𝐺

𝜃 Figure P19.57

𝑟𝑖

𝑘2

1336

Chapter 19

Mechanical Vibrations

Problem 19.58 A box of mass 0.75 kg is thrown on a scale, causing both the scale and the box to move vertically downward with an initial speed of 0.5 m∕s. Before the box lands on the scale, the scale is in equilibrium. The total mass of the scale’s moving platform and the box is 𝑚 = 1.25 kg. Modeling the platform’s support as a spring and dashpot with stiffness 𝑘 = 1000 N∕m and damping coefficient 𝑐 = 70.7 N⋅s∕m, find the response of the scale. Hint: Place the origin of the 𝑦 axis at the position of the platform corresponding to the equilibrium configuration of the platform and box together. box ACME

scale platform box 𝑦 𝑐

𝑘

Figure P19.58

Problem 19.59 Consider a simple viscously damped harmonic oscillator governed by Eq. (19.45), and analyze the case in which the damping coefficient 𝑐 is negative. Calculate the general expression for the response (without taking into account specific initial conditions), using 𝑚 = 1 kg, 𝑐 = −1 N⋅s∕m, and 𝑘 = 10 N∕m. Comment on the system’s response. 𝑐

𝑚 𝑘

Figure P19.59

𝑐

𝑚

𝐹0 sin 𝜔0 𝑡

𝑘 Figure P19.60

Problem 19.60 The MF for a harmonically excited spring-mass-damper system at 𝜔0 ∕𝜔𝑛 ≈ 1 is equal to 5. Calculate the damping ratio of the system. What would the damping ratio be if the MF were equal to 10? Sketch the magnification factor at 𝜔0 ∕𝜔𝑛 ≈ 1 as a function of the damping ratio.

𝑥 𝜃

𝐹

𝑐 𝜃

Figure P19.61

ISTUDY

Problem 19.61

𝑘

𝑘

A slider moves in the horizontal plane under the action of the harmonic forcing 𝐹 (𝑡) = 𝐹0 sin 𝜔0 𝑡. The slider is connected to two identical linear springs, each of which has constant 𝑘. When 𝑡 = 0, 𝑥(0) = 0, the springs are unstretched, 𝜃 = 45◦ , and 𝐿 = 𝐿0 . The slider is also connected to a damper with damping coefficient 𝑐. Treating 𝐹0 , 𝑘, 𝑐, and 𝐿0 as known quantities, neglecting friction, and letting 𝑥(0) ̇ = 𝑣𝑖 , (a) derive the equations of motion of the system, (b) derive the linearized equations of motion about the initial position, and (c) determine the amplitude of the steady-state vibrations for the linearized equations of motion.

ISTUDY

Section 19.3

1337

Viscously Damped Vibration

Problem 19.62 The mechanism shown is a pendulum consisting of a pendulum bob 𝐵 with mass 𝑚 and a T-bar, which is pinned at 𝑂 and has negligible mass. The horizontal portion of the T-bar is connected to two supports, each of which has an identical spring and dashpot system, each with spring constant 𝑘 and damping coefficient 𝑐. The springs are unstretched when 𝐵 is vertically aligned with the pin at 𝑂. Modeling 𝐵 as a particle, derive the linearized equations of motion of the system. In addition, assuming that the system is underdamped, derive the expression for the damped natural frequency of vibration of the system. 𝑘

𝑐

𝑐

𝑘

𝑂 0.3 𝐿 0.3 𝐿

𝜃

𝐵 Figure P19.62

Problem 19.63 The block of mass 𝑚 is coupled to the support 𝐴, which is displacing harmonically according to 𝑦 = 𝑌 sin 𝜔0 𝑡, by the linear elastic spring with constant 𝑘 and the dashpot with constant 𝑐. (a) Derive its equation of motion, using 𝑥 as the independent variable, and explain in what way the resulting equation of motion is not in the form of Eq. (19.65). (b) Next, let 𝑧 = 𝑥 − 𝑦 and substitute it into the equation of motion found in Part (a). After doing so, show that you obtain an equation of motion in 𝑧 that is of the same form as Eq. (19.65).

𝑥 𝑌 sin 𝜔0 𝑡

𝑘

𝑚

𝐴 𝑐 𝑦 Figure P19.63

(c) Find the steady-state solution to the equation of motion found in Part (b) and then using that, determine the steady-state solution for 𝑥.

Problems 19.64 and 19.65 The block 𝐴 and the platform 𝑃 of a spring scale are at rest when the lab bench to which the scale is rigidly attached begins vibrating sinusoidally with a frequency of 15 Hz and amplitude of 5 mm. The block and the platform are coupled to the base 𝐵 of the scale by a linear elastic spring and a viscous damper that are internal to the scale. The combined mass of the block and platform is 𝑚𝐴 = 1.5 kg, the spring constant is 𝑘 = 50 N∕m, and the viscous damping coefficient is 𝑐 = 7.5 N⋅s∕m.

𝐴

ACME platform 𝑃

internal spring

𝑐 𝑘

Problem 19.64

Determine the vertical motion of the platform and block as a function of time. The base of the scale 𝐵 is rigidly attached to the lab bench, and the block 𝐴 does not separate from the platform 𝑃 during the vibration. Hint: Parts (a)–(c) of Prob. 19.63 will be helpful. Problem 19.65

Determine and plot for 10 s the vertical motion of the platform and block as a function of time. The base of the scale 𝐵 is rigidly attached to the lab bench, and the block 𝐴 does not separate from the platform 𝑃 during the vibration. Hint: Parts (a)–(c) of Prob. 19.63 will be helpful.

Figure P19.64 and P19.65

internal dashpot vibrating base 𝐵

1338

Chapter 19

Mechanical Vibrations

piping to the fuel and oxygen tanks

𝑘

Problem 19.66

𝑐

𝑐

𝑘

𝑚

structural elements of the engine mount

𝑥

The engine in the rocket shown is supposed to provide a constant thrust of 5000 kN. The turbopump in the engine nominally operates at 7000 rpm, and as a result of a design issue, the actual thrust provided by the engine oscillates harmonically with an amplitude of 10 kN at the same rotational frequency of the turbopump. The mass of the engine is 𝑚 = 5000 kg. The rest of the rocket is much heavier than the engine and can be treated as being fixed. The engine is mounted to the rocket via two structural members, each of which can be modeled as consisting of a linear spring of stiffness 𝑘 in parallel with a dashpot with linear viscous damping coefficient 𝑐. If 𝑘 = 2×108 N∕m and 𝑐 = 100 N⋅s∕m, determine the amplitude of vibration in the nominal operating regime. Ignore the stiffness and damping due to the piping. Hint: If 𝑥 is measured from the equilibrium position of the engine that results from the combined effect of the thrust and gravity, then the engine is subject to an externally applied forcing equal to (10 kN) sin 𝜔0 𝑡, where 𝜔0 is the rotational frequency of the turbopump.

Problem 19.67

Figure P19.66

A simple model for a ship rolling on waves∗ treats the waves as sinusoids. Using this model, it can be shown that a linear model for the roll angle 𝜃 is given by 𝐼𝐺 𝜃̈ + 𝑐 𝜃̇ + 𝑚𝑔𝜃 = −

𝐼𝐺 𝐴𝜔20

sin 𝜔0 𝑡, 𝜆 where 𝐺 denotes the mass center of the ship, 𝐼𝐺 is the ship’s mass moment of inertia, 𝑐 is a rotational viscous damping constant coefficient, 𝑚 is the mass of the ship, 𝐴 is the wave amplitude, and 𝜆 is the wavelength of the waves.

𝐴 cos(𝑘𝑦 − 𝜔0 𝑡) 𝐺

wave

Figure P19.67

(a) What is the natural frequency of the system? (b) Find the magnification factor for the system. (c) Assuming that the damping is negligible (i.e., 𝑐 ≈ 0), if the maximum amplitude of oscillation that the ship can undergo without capsizing is 𝜃max = 1 rad, find the maximum 𝐴 so that the crew remains safe.

Problem 19.68 𝑚 𝑘

𝑦(𝑡) 𝑐

vibration isolator 𝑢(𝑡) = 𝐴 sin 𝜔0 𝑡

Figure P19.68

ISTUDY

A delicate instrument of mass 𝑚 must be isolated from excessive vibration of the ground, which is described by the function 𝑢(𝑡) = 𝐴 sin 𝜔0 𝑡. To do so, we need to design a vibration isolating mount, modeled by the spring and dashpot system shown. (a) Find the equation of motion of the instrument and reduce it to standard form. (b) Find the steady-state response 𝑦(𝑡). (c) Find the displacement transmissibility, i.e., the response amplitude 𝐷 divided by 𝐴, where 𝐴 is the amplitude of the ground’s vibration. ∗ See

J. M. T. Thompson, R. C. T. Rainey, and M. S. Soliman, “Mechanics of Ship Capsize under Direct and Parametric Wave Excitation,” Philosophical Transactions of the Royal Society of London A, 338(1651), 1992, pp. 471–490.

ISTUDY

Section 19.3

Viscously Damped Vibration

Design Problems Design Problem 19.3 As a result of firing a projectile, a 300 kg naval gun assembly gains momentum in the 𝑥 direction. We can consider the motion of the assembly to start from the equilibrium position 𝑥(0) = 0 with an initial velocity 𝑥(0) ̇ = 50 m∕s. Choose values of spring stiffness 𝑘 and damping coefficient 𝑐 to provide the fastest return of the assembly to its equilibrium position without oscillation. In addition, make sure that the maximum displacement of the assembly does not exceed 0.1 m. Finally, estimate the time it takes for the gun assembly to return to within 1% of its maximum displacement.

𝑥

𝑐 𝑚 𝑐

Figure DP19.3

1339

1340

Chapter 19

Mechanical Vibrations

19.4 C h a p t e r R e v i e w In this chapter, we studied the vibration or oscillation of mechanical systems about their equilibrium position. We considered only harmonic oscillators, although we did study the effects of viscous damping and harmonic forcing on the response of a harmonic oscillator.

Undamped free vibration Any one DOF system whose equation of motion is of the form

𝐶 sin(𝜔𝑛 𝑡 + 𝜙)

Eq. (19.12), p. 1291 𝐶 sin 𝜙 𝜋 2

− 𝜋2

𝜋

3𝜋 2

2𝜋

𝜙 𝜔𝑛

Figure 19.30 Plot of Eq. (19.3) showing the amplitude 𝐶, phase angle 𝜙, and period 𝜏 of a harmonic oscillator. 𝑥 𝑘

𝑚

Figure 19.31 A simple spring-mass harmonic oscillator whose equation of motion √ is given by 𝑚𝑥̈ + 𝑘𝑥 = 0 and for which 𝜔𝑛 = 𝑘∕𝑚.

ISTUDY

𝑥̈ + 𝜔2𝑛 𝑥 = 0,

𝑡

is called a harmonic oscillator, and the above expression is referred to as the standard form of the harmonic oscillator equation. The solution of this equation can be written as (see Fig. 19.30) Eq. (19.3), p. 1290 ( ) 𝑥(𝑡) = 𝐶 sin 𝜔𝑛 𝑡 + 𝜙 , where 𝜔𝑛 is the natural frequency, 𝐶 is the amplitude, and 𝜙 is the phase angle of vibration. A simple example of a harmonic oscillator is a system consisting of a mass 𝑚 attached at the free end of a spring with constant 𝑘 and with the other end fixed (see Fig. 19.31). The natural frequency of such a system is given by Eq. (19.4), p. 1290 √ 𝑘 . 𝜔𝑛 = 𝑚 In addition, the amplitude 𝐶 and the phase angle 𝜙 are given by, respectively, Eqs. (19.5) and (19.6), p. 1290 𝐶=

√ 𝑣2𝑖 ∕𝜔2𝑛 + 𝑥2𝑖

tan 𝜙 =

and

𝑥𝑖 𝜔𝑛 𝑣𝑖

,

where we let 𝑡 = 0 be the initial time, 𝑥𝑖 = 𝑥(0) (i.e., 𝑥𝑖 is the initial position), and 𝑣𝑖 = 𝑥(0) ̇ (i.e., 𝑣𝑖 is the initial velocity). If 𝑣𝑖 = 0, then 𝜙 can be chosen equal to −𝜋∕2 or 𝜋∕2 rad for 𝑥𝑖 < 0 and 𝑥𝑖 > 0, respectively. An alternative form of the solution to Eq. (19.12) is given by Eq. (19.15), p. 1292 𝑥(𝑡) = 𝑥𝑖 cos 𝜔𝑛 𝑡 +

𝑣𝑖 𝜔𝑛

sin 𝜔𝑛 𝑡.

The period of the oscillation is given by Eq. (19.7), p. 1290 Period = 𝜏 =

2𝜋 , 𝜔𝑛

and the frequency of vibration is Eq. (19.8), p. 1290 Frequency = 𝑓 =

𝜔 1 = 𝑛. 𝜏 2𝜋

ISTUDY

Section 19.4

Chapter Review

1341

Energy method. For conservative systems, the work-energy principle tells us that the quantity 𝑇 + 𝑉 is constant, and so its time derivative must be zero. This provides a convenient way to obtain the equations of motion via Eq. (19.19), p. 1293 𝑑 (𝑇 + 𝑉 ) = 0 𝑑𝑡

⇒

equations of motion.

When we apply the energy method to determine the linearized equations of motion, it is often convenient to first approximate the kinetic and potential energies as quadratic functions of position and velocity, and then take derivatives with respect to time. This process yields equations of motion that are linear. In approximating the sine and cosine functions as quadratic functions of their arguments, we use the relations: Eqs. (19.28), p. 1294 sin 𝜃 ≈ 𝜃

and

cos 𝜃 ≈ 1 − 𝜃 2 ∕2.

Since the quadratic term in the power series expansion of the sine function is identically equal to zero, the linearized form of the sine function can also be viewed as the quadratic approximation of the sine function.

Undamped forced vibration When a harmonic oscillator is subject to harmonic forcing, the standard form of the equation of motion is Eq. (19.31), p. 1307 𝑥̈ + 𝜔2𝑛 𝑥 =

𝐹0 𝑚

Eq. (19.35), p. 1308 𝐹0 ∕𝑘 ( )2 sin 𝜔0 𝑡, 1 − 𝜔0 ∕𝜔𝑛

and so the general solution is given by Eq. (19.36), p. 1308 𝑥 = 𝑥𝑐 + 𝑥𝑝 = 𝐴 sin 𝜔𝑛 𝑡 + 𝐵 cos 𝜔𝑛 𝑡 +

𝐹0 ∕𝑘 ( )2 sin 𝜔0 𝑡, 1 − 𝜔0 ∕𝜔𝑛

where 𝐴 and 𝐵 are constants determined by the initial conditions. The amplitude of the steady-state vibration is Eq. (19.37), p. 1308 𝑥amp =

𝑘

𝑚

𝑃 (𝑡) = 𝐹0 sin 𝜔0 𝑡

sin 𝜔0 𝑡,

where 𝐹0 is the amplitude of the forcing and 𝜔0 is its frequency (see Fig. 19.32). The general solution to this equation consists of the sum of the complementary solution and a particular solution. The complementary solution 𝑥𝑐 is the solution of the associated homogeneous equation, which is given by, for example, Eq. (19.13). For 𝜔0 ≠ 𝜔𝑛 , a particular solution is

𝑥𝑝 =

𝑥

𝐹0 ∕𝑘 ( )2 , 1 − 𝜔0 ∕𝜔𝑛

Figure 19.32 A forced harmonic oscillator whose equation √ of motion is given by Eq. (19.31) with 𝜔𝑛 = 𝑘∕𝑚. The position 𝑥 is measured from the equilibrium position of the mass.

1342

Chapter 19

Mechanical Vibrations

which means that the corresponding magnification factor MF is

4 𝜔 𝜔𝑛

−3 0

1

2

3

Figure 19.33 MF as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 .

ISTUDY

𝑥amp

Harmonic excitation of the support. If the support of a structure is excited harmonically rather than the structure itself (see Fig. 19.34), then Eq. (19.31) is still the governing equation, except that 𝐹0 is replaced by the spring constant 𝑘 times the amplitude of the support vibration 𝑋𝑠 . All solutions previously described are then valid with that same replacement. 𝑥 𝑥𝑠 = 𝑋𝑠 sin 𝜔𝑠 𝑡 𝑘

𝑚

Figure 19.34. A harmonic oscillator whose support is being excited harmonically.

Viscously damped vibration Viscously damped free vibration. The standard form of the equation of motion for a one degree of freedom viscously damped harmonic oscillator is Eq. (19.45), p. 1322 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 0, where 𝑚 is the mass, 𝑐 is the coefficient of viscous damping, and 𝑘 is the linear spring constant. The character of the solution to this equation depends on the amount of damping relative to a specific amount of damping called the critical damping coefficient, which is defined as Eq. (19.51), p. 1322 √ 𝑘 = 2𝑚𝜔𝑛 , 𝑐𝑐 = 2𝑚 𝑚 √ where 𝜔𝑛 = 𝑘∕𝑚. In particular, if 𝑐 ≥ 𝑐𝑐 , then the motion is nonoscillatory, whereas if 𝑐 < 𝑐𝑐 , then the motion is oscillatory. If 𝑐 > 𝑐𝑐 , the system is said to be overdamped, and the solution is given by Eq. (19.52), p. 1322 [ √ ] √ 2 2 𝑥 = 𝑒−(𝑐∕2𝑚)𝑡 𝐴𝑒𝑡 (𝑐∕2𝑚) −𝑘∕𝑚 + 𝐵𝑒−𝑡 (𝑐∕2𝑚) −𝑘∕𝑚 , where 𝐴 and 𝐵 are constants to be determined from the initial conditions. If 𝑐 = 𝑐𝑐 , the system is said to be critically damped, and the solution is given by Eq. (19.53), p. 1323 𝑥 = (𝐴 + 𝐵𝑡)𝑒−𝜔𝑛 𝑡 ,

ISTUDY

Section 19.4

Chapter Review

where, again, 𝐴 and 𝐵 are constants to be determined from the initial conditions. Finally, if 𝑐 < 𝑐𝑐 , the system is said to be underdamped, and the solution is given by Eq. (19.54), p. 1323 ( ) 𝑥 = 𝑒−(𝑐∕2𝑚)𝑡 𝐴 sin 𝜔𝑑 𝑡 + 𝐵 cos 𝜔𝑑 𝑡 , or, equivalently, by Eq. (19.57), p. 1323 ( ) 𝑥 = 𝐷𝑒−(𝑐∕2𝑚)𝑡 sin 𝜔𝑑 𝑡 + 𝜙 , where 𝐴 and 𝐵 are constants to be determined from the initial conditions in the first solution, 𝐷 and 𝜙 are analogous constants in the second solution, and 𝜔𝑑 is the damped natural frequency, which is given by Eq. (19.55), p. 1323, and Eq. (19.62), p. 1324 √ √ ( )2 √ ( )2 𝑐 𝑘 = 𝜔𝑛 1 − 𝑐∕𝑐𝑐 = 𝜔𝑛 1 − 𝜁 2 , − 𝜔𝑑 = 𝑚 2𝑚 where 𝜁 = 𝑐∕𝑐𝑐 is the damping ratio.

Viscously damped forced vibration. The standard form of a viscously damped forced harmonic oscillator is Eq. (19.63), p. 1324 𝑚𝑥̈ + 𝑐 𝑥̇ + 𝑘𝑥 = 𝐹0 sin 𝜔0 𝑡, where 𝐹0 is the amplitude of the forcing function and 𝜔0 is the frequency of the forcing function. When expressed using the damping ratio 𝜁, the equation above takes on the form Eq. (19.64), p. 1324 𝑥̈ + 2𝜁𝜔𝑛 𝑥̇ + 𝜔2𝑛 𝑥 =

𝐹0 𝑚

sin 𝜔0 𝑡.

The general solution to either of these equations is the sum of the complementary solution and a particular solution. The complementary solution is transient; i.e., it vanishes as time increases. The particular or steady-state solution is of the form Eq. (19.66), p. 1325 ( ) 𝑥𝑝 = 𝐷 sin 𝜔0 𝑡 − 𝜙 , where 𝜙 and 𝐷 are given by Eq. (19.69), p. 1325, and Eq. (19.72), p. 1326 ) )( ( 2 𝑐∕𝑐𝑐 𝜔0 ∕𝜔𝑛 2𝜁𝜔0 ∕𝜔𝑛 𝑐𝜔0 tan 𝜙 = = )2 = )2 , ( ( 2 𝑘 − 𝑚𝜔0 1 − 𝜔0 ∕𝜔𝑛 1 − 𝜔0 ∕𝜔𝑛 𝐹0 ∕𝑘 𝐷= √ . [ ( ) 2 ]2 ( )2 + 2𝜁𝜔0 ∕𝜔𝑛 1 − 𝜔0 ∕𝜔𝑛

1343

1344

Chapter 19

Mechanical Vibrations

The magnification factor for a damped, forced harmonic oscillator is

5 𝜁 =0

Eq. (19.73), p. 1326

4 𝜁 = 0.125

MF =

3

( 1 − 𝜔0 ∕𝜔𝑛

1 )2 ]2

)2 ( + 2𝜁𝜔0 ∕𝜔𝑛

,

𝜁 = 0.2

MF 2

a plot of which is shown in Fig. 19.35 for various values of the damping ratio 𝜁. 𝜁 = 0.4

1 𝜁 =1 0

0

1

2

3

Figure 19.35 The MF as a function of the frequency ratio 𝜔0 ∕𝜔𝑛 for various values of the damping ratio 𝜁.

ISTUDY

𝐷 = √ [ 𝐹0 ∕𝑘

ISTUDY

Section 19.4

Chapter Review

1345

Review Problems Problem 19.69 When the connecting rod shown is suspended from the knife-edge at point 𝑂 and displaced slightly so that it oscillates as a pendulum, its period of oscillation is 0.77 s. In addition, it is known that the mass center 𝐺 is located a distance 𝐿 = 110 mm from 𝑂 and that the mass of the connecting rod is 661 g. Using the energy method, determine the mass moment of inertia of the connecting rod 𝐼𝐺 .

𝐿

𝐺

Problem 19.70 Derive the equation of motion for the system, in which the springs with constants 𝑘1 and 𝑘2 connecting 𝑚 to the wall are joined in series. Neglect the mass of the small wheels, and assume that the attachment point 𝐴 between the two springs has negligible mass. Hint: The force in the two springs must be the same; use this fact, along with the fact that the total deflection of the mass must equal the sum of the deflections of the springs, to find an equivalent spring constant 𝑘eq . 𝑘2

𝑘1

Figure P19.69

𝑚

𝐴 Figure P19.70

Problem 19.71 Revisit Example 19.2 and compute the natural frequency of the silicon nanowire, using the energy method. Use a uniform Si nanowire with a circular cross section that is 9.8 𝜇m long and 330 nm in diameter and with all its flexibility lumped in a torsional spring at the base of the wire. In addition, use 𝜌 = 2330 kg∕m3 for the density of silicon and 𝐸 = 152 GPa for its modulus of elasticity.

Problem 19.72 Structural health monitoring technology detects damage in civil, aerospace, and other structures. Structural damage is usually comprised of cracking, delaminations, or loose fasteners, which result in the reduction of stiffness. Many structural health monitoring methods are based on tracking changes in natural frequencies. Modeling a structure as a one DOF harmonic oscillator, calculate the change in stiffness needed to cause a 3% reduction in the natural frequency of the structure being monitored.

Problems 19.73 through 19.75 When the electric motor is resting on the beam, the static deflection of the beam is 𝛿𝑠 = 15 mm. The motor is not perfectly balanced, so when it is operating the unbalanced mass is equivalent to a mass 𝑚𝑢 = 200 g at a distance of 𝜀 = 150 mm from the axis of the rotor. The combined mass of the motor and sprung mass of the beam is 𝑚𝑐 = 40 kg. Problem 19.73

Determine the angular speed of the rotor for resonance to occur. Express your answer in revolutions per minute.

Problem 19.74

Determine the amplitude of steady-state vibration of the motor if the rotor is spinning at 𝜔𝑟 = 150 rpm.

2 𝜇m Courtesy of Theresa Mayer

2𝑟

Figure P19.71

𝑚

𝑂

1346

Chapter 19

Mechanical Vibrations

Problem 19.75

It is determined that the largest allowable vibration amplitude of the motor is 10 mm. Determine the angular speed of the rotor at which this will occur. 𝜔𝑟

motor, 𝑚𝑚

mass, 𝑚𝑢

Figure P19.73–P19.75

Problem 19.76

𝑥 𝑐

𝑚 𝑘

Figure P19.76

ISTUDY

𝐹0 sin 𝜔0 𝑡

The harmonic oscillator shown has a mass 𝑚 = 5 kg, a spring with constant 𝑘 = 4000 N∕m, and a dashpot with a damping coefficient 𝑐 = 20 N⋅s∕m. Calculate the amplitude 𝐹0 of the sinusoidal excitation force that is necessary to produce a steady-state vibration with a velocity amplitude of 10 m∕s at resonance. What is the corresponding amplitude of the acceleration?

Problem 19.77 Modeling the beam as a uniform thin bar, ignoring the inertia of the pulleys, assuming that the system is in static equilibrium when the bar is horizontal, and assuming that the cord is inextensible and does not go slack, determine the linearized equation of motion of the system. In addition, determine the system’s natural frequency of vibration. Treat the parameters shown in the figure as known.

𝐴

𝑚𝐵

𝐵 𝑥 𝑘

𝑚𝐴

Figure P19.77

Problem 19.78 Revisit Example 19.9 and obtain the expression for the force transmitted to the floor, using the expression for the steady-state response of the unbalanced motor. 𝑚𝑢

motor 𝑚 𝑘

platform 𝑐

Figure P19.78

ISTUDY

Section 19.4

Chapter Review

1347

Problem 19.79 The system shown is released from rest when both springs are unstretched and 𝑥 = 0. Neglecting the inertia of the pulley 𝑃 and assuming that the disk rolls without slip, derive the equation of motion of the system in terms of 𝑥. Assume that point 𝐺 is both the mass center of the disk and its geometric center. Treat the quantities 𝑘1 , 𝑘2 , 𝑐, 𝑚1 , 𝑚2 , and 𝐼𝐺 as known, where 𝐼𝐺 is the mass moment of inertia of the disk. Finally, assuming that the system is underdamped, derive an expression for the damped natural frequency of the system. 𝑘1

𝑐

𝑚2 , 𝐼𝐺 𝑟𝑜

𝑃 𝐺

𝑘2

𝑟𝑖

𝑚1

𝑥

Figure P19.79

Problem 19.80 A ring of mass 𝑚 is attached by two linear elastic cords to the vertical supports as shown. The cords have elastic constant 𝑘 and unstretched length 𝐿0 < 𝐿. Assuming that the pretension in the cords is large enough that the deflection of the cords due to the ring’s weight can be neglected, find the nonlinear equation of motion for the mass 𝑚.

𝑚 𝑦(𝑡) 𝑘

Problem 19.81 A ring of mass 𝑚 is attached by two linear elastic cords to the vertical supports as shown. The cords have elastic constant 𝑘 and unstretched length 𝐿0 < 𝐿. Assuming that the pretension in the cords is large enough that the deflection of the cords due to the ring’s weight can be neglected, use Newton’s second law to find the linearized equation of motion about 𝑦 = 0 for the mass 𝑚. In addition, determine the natural frequency of the ring’s vibration.

Problem 19.82 Solve Prob. 19.81 by finding the linearized equations of motion using the energy method.

Figure P19.80–P19.82

𝑘

ISTUDY

ISTUDY

20

Three-Dimensional Dynamics of Rigid Bodies

Until now, we have only considered rigid bodies in planar motion that are symmetric with respect to the plane of motion. In this chapter, we will relax these assumptions and allow for not only nonsymmetric bodies, but also for bodies moving in all three dimensions. In Chapter 17, we had three Newton-Euler equations (two force and one moment equation) for a rigid body. For threedimensional motion, we will find that we need six Newton-Euler equations—three moment equations and three force equations. Before we derive these equations, as with planar motion, we need to begin by understanding the kinematics of rigid bodies moving in three dimensions. We do that in Section 20.1. In Section 20.2, we derive the Newton-Euler equations for three-dimensional motion. aragami12345s/Shutterstock

The HH/MH-65 Multi-Mission Cutter Helicopter used by the United States Coast Guard for search and rescue and law enforcement missions. The understanding of three-dimensional dynamics is essential for modeling its rotor blades.

20.1

𝑦

Three-Dimensional Kinematics of Rigid Bodies

𝐶

In Chapter 16, we considered the planar kinematics of rigid bodies, but most of the equations we derived apply equally well to three-dimensional motion. Equations (16.3) and (16.5) on pp. 1057 and 1058, which are given by 𝑣⃗𝐶 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐵 × 𝑟⃗𝐶∕𝐴 ,

) 𝑎⃗𝐶 = 𝑎⃗𝐴 + 𝛼⃗𝐵 × 𝑟⃗𝐶∕𝐴 + 𝜔 ⃗𝐵 × 𝜔 ⃗ 𝐵 × 𝑟⃗𝐶∕𝐴 , (

(20.1)

) ( ⃗̇ ⃗ ⃗ × 𝑣⃗ ⃗ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 + Ω × Ω × 𝑟 𝑃 ∕𝐴 ,

𝐴

𝑌

𝑧 𝑟⃗𝑃 ∕𝐴

𝑄

𝐵

𝑋

𝑥

𝑃

(20.2)

where 𝜔 ⃗ 𝐵 and 𝛼⃗𝐵 are the angular velocity and angular acceleration of the body 𝐵, respectively, both apply for the three-dimensional motion of rigid bodies, as long as 𝐴 and 𝐶 are two points on the same rigid body (see Fig. 20.1). The equations we derived for rotating reference frames, which were useful for sliding contacts in planar motion, that is, Eqs. (16.45) on p. 1119, and (16.53) on p. 1120, also apply in three-dimensional motion, and we repeat them here for convenience: ⃗ × 𝑟⃗ , 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝐴

𝑟⃗𝐶∕𝐴

Figure 20.1 Rigid body 𝐵 displaying the points and kinematic quantities needed to apply Eqs. (20.1)–(20.4). Frame 𝑋𝑌𝑍 is the primary reference frame and 𝑥𝑦𝑧 is the secondary, body-fixed, or rotating reference frame.

(20.3) (20.4)

1349

1350

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

where 𝑣⃗𝑃 rel and 𝑎⃗𝑃 rel are the velocity and acceleration, respectively, of the point 𝑃 ⃗ is the angular velocity as seen by an observer in the rotating 𝑥𝑦𝑧 reference frame, Ω ̇⃗ of the rotating frame, and Ω is the angular acceleration of the rotating frame. We will see that, even though the kinematic equations are familiar, the added difficulty in three-dimensional motion concerns angular velocities and angular accelerations. One angular velocity can cause another angular velocity to change direction, and this adds a dimension to the angular acceleration that we haven’t seen before.

Computation of angular accelerations

𝐵

When analyzing the kinematics of rigid bodies in three dimensions, we will often need to use a second reference frame that rotates relative to the primary frame. The reference frame is chosen in such a way as to make the analysis of the motion as easy as possible. Sometimes the rotating frame will be attached to a rigid body and will rotate with it. Alternatively, the rotating reference frame may rotate relative to both the primary frame and the rigid body. Referring to Fig. 20.2, the primary reference frame is 𝑋𝑌𝑍 and the rotating frame is 𝑥𝑦𝑧. The angular velocity and angular acceleration of the rotating reference frame 𝑥𝑦𝑧 relative to the primary reference frame are given ⃗ and Ω, ⃗̇ respectively. The angular velocity and angular acceleration of the rigid by Ω ⃗̇ , respectively. body 𝐵 relative to the primary reference frame are given by 𝜔 ⃗ and 𝜔

𝜔 ⃗𝐵 𝛼⃗𝐵 = 𝜔 ⃗̇ 𝐵 𝑦

𝑌 𝑧 𝑄

𝐴

𝑋

𝐵

Figure 20.2 The primary reference frame, rotating reference frame, and rigid body 𝐵.

𝐵

We now express the angular velocity of the rigid body 𝐵 in terms of its components in the rotating reference frame as ̂ 𝜔 ⃗ 𝐵 = 𝜔𝐵𝑥 𝚤̂ + 𝜔𝐵𝑦 𝚥̂ + 𝜔𝐵𝑧 𝑘.

(20.5)

⃗̇ 𝐵 = 𝛼⃗𝐵 , we obtain Differentiating 𝜔 ⃗ 𝐵 to find the angular acceleration 𝜔 ⃗ ×𝜔 𝜔 ⃗̇ 𝐵 = 𝛼⃗𝐵 = 𝜔̇ 𝐵𝑥 𝚤̂ + 𝜔̇ 𝐵𝑦 𝚥̂ + 𝜔̇ 𝐵𝑧 𝑘̂ + Ω ⃗𝐵, 𝜔 ⃗𝐵

𝐷

𝑦

𝑟⃗𝐷∕𝐴

𝑧

𝐴

𝐶

(20.6)

where we have applied Eq. (12.60) on p. 718 to take the time derivative of 𝜔 ⃗ 𝐵 . Equation (20.6) is invaluable for computing the angular acceleration of rigid bodies in three-dimensional motion. Notice that if the rotating frame is body-fixed, that is, if ⃗ =𝜔 Ω ⃗ 𝐵 , then ̂ 𝜔 ⃗̇ 𝐵 = 𝜔̇ 𝐵𝑥 𝚤̂ + 𝜔̇ 𝐵𝑦 𝚥̂ + 𝜔̇ 𝐵𝑧 𝑘. (20.7)

𝑌

Summing angular velocities

𝜔 ⃗ 𝐵rel 𝑄

𝑋

Figure 20.3 A body with angular velocity 𝜔 ⃗ 𝐵 and a moving ⃗ The anreference frame with angular velocity Ω. gular velocity of the body relative to the moving reference frame is 𝜔 ⃗ 𝐵rel .

ISTUDY

Referring to Fig. 20.3, to understand how to sum angular velocities with one another, we will relate the velocity of point 𝐷 to point 𝐴 and point 𝐶 using ⃗ × 𝑟⃗ 𝑣⃗𝐷 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐵 × 𝑟⃗𝐷∕𝐴 = 𝑣⃗𝐶 + 𝑣⃗𝐷rel + Ω 𝐷∕𝐶 ,

(20.8)

where 𝑣⃗𝐷rel is the velocity of 𝐷 relative to the 𝑥𝑦𝑧 frame. Since points 𝐴 and 𝐷 are two points on the same rigid body, an observer in the moving reference frame would see the velocity of 𝐷 as ⃗ 𝐵rel × 𝑟⃗𝐷∕𝐴 , 𝑣⃗𝐷rel = 𝑣⃗𝐴rel + 𝜔

(20.9)

where 𝜔 ⃗ 𝐵rel is the angular velocity of the body as seen by an observer in the moving ⃗ × 𝑟⃗ , we can rearrange it as reference frame. Since 𝑣⃗𝐴 = 𝑣⃗𝐶 + 𝑣⃗𝐴rel + Ω 𝐴∕𝐶 ⃗ × 𝑟⃗ . 𝑣⃗𝐴rel = 𝑣⃗𝐴 − 𝑣⃗𝐶 − Ω 𝐴∕𝐶

(20.10)

ISTUDY

Section 20.1

1351

Three-Dimensional Kinematics of Rigid Bodies

Substituting Eq. (20.10) into Eq. (20.9), we obtain ⃗ × 𝑟⃗ ⃗ 𝐵rel × 𝑟⃗𝐷∕𝐴 , 𝑣⃗𝐷rel = 𝑣⃗𝐴 − 𝑣⃗𝐶 − Ω 𝐴∕𝐶 + 𝜔

(20.11)

which can be substituted into Eq. (20.8) to obtain ⃗ × 𝑟⃗ ⃗ × 𝑟⃗ 𝜔 ⃗ 𝐵 × 𝑟⃗𝐷∕𝐴 = −Ω ⃗ 𝐵rel × 𝑟⃗𝐷∕𝐴 + Ω 𝐴∕𝐶 + 𝜔 𝐷∕𝐶 ,

(20.12)

where we have canceled out 𝑣⃗𝐴 and 𝑣⃗𝐶 . This equation can be rewritten as ( ) ⃗ × 𝑟⃗ ⃗𝐴∕𝐶 + 𝜔 ⃗ 𝐵rel × 𝑟⃗𝐷∕𝐴 𝜔 ⃗ 𝐵 × 𝑟⃗𝐷∕𝐴 = Ω 𝐷∕𝐶 − 𝑟 ⃗ × 𝑟⃗ =Ω ⃗ 𝐵rel × 𝑟⃗𝐷∕𝐴 𝐷∕𝐴 + 𝜔 ( ) ⃗ +𝜔 ⃗ 𝐵rel × 𝑟⃗𝐷∕𝐴 . = Ω

(20.13)

Since Eq. (20.13) must hold for every pair of points 𝐴 and 𝐷 in the body 𝐵, we obtain the following important result, ⃗ +𝜔 𝜔 ⃗𝐵 = Ω ⃗ 𝐵rel ,

(20.14)

⃗ of a rotating (usually body-fixed) frame is which states that if the angular velocity Ω known and the angular velocity of a body 𝐵 relative to that frame 𝜔 ⃗ 𝐵rel is known, then the angular velocity of the body 𝜔 ⃗ 𝐵 is the sum of those two angular velocities. Mini-Example The platform shown in Fig. 20.4 is rotating with constant angular speed 𝜔𝑝 in the direction shown. On the platform is mounted a motor that is spinning with constant angular speed 𝜔𝑠 relative to the platform in the direction shown. Given the dimensions shown, determine the acceleration of point 𝑄, which is on the periphery of the motor. Solution We have attached a moving reference frame 𝑥𝑦𝑧 to the platform with origin at point 𝑂. To find the acceleration of point 𝑄 on the periphery of the motor, we first make use of Eq. (20.2) to relate the acceleration of 𝑄 to the acceleration of 𝐴 as ) ( (20.15) ⃗𝑚 × 𝜔 ⃗ 𝑚 × 𝑟⃗𝑄∕𝐴 , 𝑎⃗𝑄 = 𝑎⃗𝐴 + 𝛼⃗𝑚 × 𝑟⃗𝑄∕𝐴 + 𝜔 ⃗ 𝑚 are the angular acceleration and angular velocity of the motor, where 𝛼⃗𝑚 and 𝜔 respectively. Since point 𝐴 is moving with constant angular velocity in a circle of radius 𝑅, we can write its acceleration as 𝑎⃗𝐴 = −𝑅𝜔2𝑝 𝚤̂.

(20.16)

To determine 𝜔 ⃗ 𝑚 , we can apply Eq. (20.14) by saying that the angular velocity of the motor is the angular velocity of the rotating 𝑥𝑦𝑧 frame (which rotates with the platform) plus the angular velocity of the motor as seen by an observer moving with the platform, that is, ⃗𝑝 + 𝜔 ⃗ 𝑚rel 𝜔 ⃗𝑚 = 𝜔

⇒

𝜔 ⃗ 𝑚 = 𝜔𝑝 𝑘̂ + 𝜔𝑠 𝚤̂,

(20.17)

where 𝜔 ⃗ 𝑚rel is the angular velocity of the motor as seen by an observer on the platform. To find the angular acceleration of the motor, we apply Eq. (20.6) to obtain ) ( ⃗𝑝 × 𝜔 ⃗ 𝑚 = 𝜔𝑝 𝑘̂ × 𝜔𝑝 𝑘̂ + 𝜔𝑠 𝚤̂ = 𝜔𝑝 𝜔𝑠 𝚥̂, (20.18) 𝛼⃗𝑚 = 𝜔̇ 𝑝 𝑘̂ + 𝜔̇ 𝑚 𝚤̂ + 𝜔

𝜔𝑝 𝑘̂

𝚥̂ 𝚤̂

platform

𝜔𝑠 𝑘̂

𝑥

𝜃 𝑟 𝑄

𝐴

𝚥̂

Figure 20.4 A motor that is spinning with angular speed 𝜔𝑠 , mounted on a platform that is spinning with angular speed 𝜔𝑝 .

1352

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

⃗ where we have used the fact that 𝜔𝑝 and 𝜔𝑚 are constant and that 𝑘̂ × 𝑘̂ = 0. ( ) Substituting Eqs. (20.16)–(20.18) and 𝑟⃗𝑄∕𝐴 = 𝑟 sin 𝜃 𝚥̂ + cos 𝜃 𝑘̂ into Eq. (20.15), we obtain ( ) 𝑎⃗𝑄 = −𝑅𝜔2𝑝 𝚤̂ + 𝜔𝑝 𝜔𝑠 𝚥̂ × 𝑟 sin 𝜃 𝚥̂ + cos 𝜃 𝑘̂ ( ) [( ) ( )] + 𝜔𝑠 𝚤̂ + 𝜔𝑝 𝑘̂ × 𝜔𝑠 𝚤̂ + 𝜔𝑝 𝑘̂ × 𝑟 sin 𝜃 𝚥̂ + cos 𝜃 𝑘̂ ( ) ( ) ̂ = 2𝑟𝜔𝑠 𝜔𝑝 cos 𝜃 − 𝑅𝜔2𝑝 𝚤̂ − 𝑟 sin 𝜃 𝜔2𝑠 + 𝜔2𝑝 𝚥̂ − 𝑟𝜔2𝑠 cos 𝜃 𝑘.

The expression for the acceleration in Eq. (20.19) is written in terms of the components in the rotating frame, but it represents the acceleration of 𝑄 in the primary frame. Therefore, it is inertial if the primary frame is inertial. In addition, because the angular velocity of the platform causes the angular velocity of the motor to change direction, we again see a gyroscopic term in this acceleration, i.e., the term 2𝑟𝜔𝑠 𝜔𝑝 cos 𝜃. Equation (20.18) can also be interpreted using Eq. (12.60) on p. 718. That is, the time derivative of 𝜔 ⃗ 𝑚 is the sum of its change in magnitude, that is, 𝜔̇ 𝑚 𝚤̂, and its change in direction, that is, 𝜔 ⃗𝑝 × 𝜔 ⃗ 𝑚 . Notice that 𝜔 ⃗ 𝑝 is the angular velocity of 𝜔 ⃗ 𝑚.

𝑦 𝐶

𝑟⃗𝐶∕𝐴 𝐴

𝑌

𝐵

𝑧 𝑟⃗𝑃 ∕𝐴

𝑄

𝑋

𝑥

𝑃

Figure 20.5 Rigid body 𝐵 displaying the points and kinematic quantities needed to apply Eqs. (20.1)–(20.4). Frame 𝑋𝑌𝑍 is the primary reference frame and 𝑥𝑦𝑧 is the secondary, body-fixed, or rotating reference frame.

ISTUDY

(20.19)

End of Section Summary Referring to Fig. 20.5, for two points 𝐴 and 𝐶 on the same rigid body 𝐵, we can relate their motion using Eqs. (20.1) and (20.2), p. 1349 𝑣⃗𝐶 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐵 × 𝑟⃗𝐶∕𝐴 ,

( ) ⃗𝐵 × 𝜔 ⃗ 𝐵 × 𝑟⃗𝐶∕𝐴 , 𝑎⃗𝐶 = 𝑎⃗𝐴 + 𝛼⃗𝐵 × 𝑟⃗𝐶∕𝐴 + 𝜔

where 𝜔 ⃗ 𝐵 and 𝛼⃗𝐵 are the angular velocity and angular acceleration, respectively, of ( ⃗ 𝐴𝐵 × the body. Note that for three-dimensional motion, we cannot write 𝜔 ⃗ 𝐴𝐵 × 𝜔 ) 𝑟⃗𝐵∕𝐴 as −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 . Referring to Fig. 20.5, if we use a rotating or secondary reference frame with its origin at 𝐴, then we can relate the motion of points 𝑃 and 𝐴, which are not necessarily on the same body, using Eqs. (20.3) and (20.4), p. 1349 ⃗ × 𝑟⃗ , 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝐴

( ) ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 + Ω × Ω × 𝑟 𝑃 ∕𝐴 ,

where 𝑣⃗𝑃 rel and 𝑎⃗𝑃 rel are the velocity and acceleration, respectively, of 𝑃 as seen by ⃗ and Ω ⃗̇ are the angular velocity and anguan observer in the rotating frame; and Ω lar acceleration, respectively, of the rotating reference frame. The rotating frame is usually attached to a rigid body, though sometimes the rotating reference frame will rotate relative to both the primary frame and the rigid body.

ISTUDY

Section 20.1

Referring to Fig. 20.6, if the rotating frame is not attached to the rigid body 𝐵, then the angular acceleration of 𝐵 is given by

𝐵

Eq. (20.6), p. 1350

𝜔 ⃗𝐵

⃗ ×𝜔 ⃗̇ 𝐵 = 𝛼⃗𝐵 = 𝜔̇ 𝐵𝑥 𝚤̂ + 𝜔̇ 𝐵𝑦 𝚥̂ + 𝜔̇ 𝐵𝑧 𝑘̂ + Ω ⃗𝐵, 𝜔

𝛼⃗𝐵 = 𝜔 ⃗̇ 𝐵

⃗ is the angular velocity of the rotating frame relative to the primary frame where Ω and 𝜔 ⃗ 𝐵 is the angular velocity of 𝐵 relative to the primary frame. Referring to Fig. 20.7, if the angular velocity of the rigid body 𝐵 is expressed relative to the rotating frame as 𝜔 ⃗ 𝐵rel and the angular velocity of the rotating frame ⃗ then the angular velocity of 𝐵 relative to the prirelative to the primary frame is Ω, mary frame, 𝜔 ⃗ 𝐵 , is Eq. (20.14), p. 1351

𝑌

𝐵

𝑧

𝑋

𝐴

𝑦

𝑌 𝑧 𝑄

𝑋

𝐴

Figure 20.6 The primary reference frame 𝑋𝑌𝑍, rotating reference frame 𝑥𝑦𝑧, and rigid body 𝐵.

⃗ +𝜔 𝜔 ⃗𝐵 = Ω ⃗ 𝐵rel .

𝑄

1353

Three-Dimensional Kinematics of Rigid Bodies

𝜔 ⃗ 𝐵rel

⃗ and a rigid body rotating with 𝜔 Figure 20.7. A secondary frame rotating with Ω ⃗ 𝐵rel relative to the rotating frame.

1354

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

E X A M P L E 20.1

Motion of a Disk Rolling Without Slip The bent arm in Fig. 1 rotates with constant angular speed 𝜔arm in the direction shown. The wheel of radius 𝑅 with center at 𝐴 rotates relative to the bent arm as it rolls without slipping over the horizontal surface. Given the dimensions shown, determine expressions for the angular velocity and angular acceleration of the wheel relative to a primary frame attached to the horizontal surface. Express the results in a frame attached to the bent arm.

𝜔arm

𝑂

𝐵

𝛾

SOLUTION Road Map

Referring to Fig. 2, we begin by defining a rotating reference frame that is attached to the arm 𝑂𝐵𝐴, with origin at 𝐵, and that is aligned as shown. In addition, we let 𝜔 ⃗ 𝑤rel be the angular velocity of the wheel relative to the bent arm. The key kinematic constraints in this problem are that: (i) Point 𝐴 moves in a circle centered on the vertical part of the bent arm with constant angular velocity 𝜔 ⃗ arm .

Figure 1

(ii) The wheel rolls without slipping at point 𝐶. 𝜔arm

Computation

𝑦 𝛾

𝑂

𝜔 ⃗𝑤 = 𝜔 ⃗ arm + 𝜔 ⃗ 𝑤rel ,

𝑑 𝐵

The angular velocity of the wheel can be written as (1)

where 𝜔 ⃗ 𝑤rel = 𝜔𝑤rel 𝚤̂. We can write 𝜔 ⃗ arm in the rotating frame as

𝛾

𝜔 ⃗ arm = 𝜔arm (− sin 𝛾 𝚤̂ + cos 𝛾 𝚥̂). 𝑅 𝑅 𝐶

𝐴

To determine 𝜔 ⃗ 𝑤rel , we note that the wheel rolls without slipping, and we relate the velocity of 𝐴 to that of 𝐶, 𝜔𝑤rel 𝑥

Figure 2 The definition of the rotating reference frame 𝑥𝑦𝑧, as well as 𝜔 ⃗ 𝑤rel , which is the angular velocity of the wheel relative to the bent arm.

ISTUDY

(2)

𝑣⃗𝐴 = 𝑣⃗𝐶 + 𝜔 ⃗ 𝑤 × 𝑟⃗𝐴∕𝐶 = 𝜔 ⃗ 𝑤 × 𝑟⃗𝐴∕𝐶 [ ] = 𝜔arm (− sin 𝛾 𝚤̂ + cos 𝛾 𝚥̂) + 𝜔𝑤rel 𝚤̂ × 𝑅 𝚥̂ ( ) ̂ =𝑅 𝜔 − 𝜔 sin 𝛾 𝑘, 𝑤rel

(3)

arm

where, in the first line, we have taken advantage of the fact that 𝑣⃗𝐶 = 0⃗ since the wheel rolls without slipping over a stationary surface. Since the velocity of point 𝐴 is easily seen to be given by ̂ 𝑣⃗𝐴 = −(𝑑 + 𝓁 cos 𝛾)𝜔arm 𝑘, (4) we can equate Eqs. (3) and (4) to obtain ( ) − (𝑑 + 𝓁 cos 𝛾)𝜔arm = 𝑅 𝜔𝑤rel − 𝜔arm sin 𝛾 ⇒

𝜔𝑤rel =

(

𝑑 + 𝓁 cos 𝛾 sin 𝛾 − 𝑅

)

𝜔arm .

(5)

Substituting Eqs. (2) and (5) into Eq. (1) and noting that 𝜔𝑤rel is only in the 𝑥 direction, we obtain ( ) 𝑑 + 𝓁 cos 𝛾 𝜔 ⃗𝑤 = − (6) 𝚤̂ + cos 𝛾 𝚥̂ 𝜔arm . 𝑅 To find the angular acceleration of the wheel, we apply Eq. (20.6) in the form ⃗ ×𝜔 𝛼⃗𝑤 = 𝜔̇ 𝑤𝑥 𝚤̂ + 𝜔̇ 𝑤𝑦 𝚥̂ + 𝜔̇ 𝑤𝑧 𝑘̂ + Ω ⃗ 𝑤,

(7)

ISTUDY

Section 20.1

Three-Dimensional Kinematics of Rigid Bodies

⃗ = 𝜔 where Ω ⃗ arm is the angular velocity of the rotating frame. Since all dimensions, as well as 𝜔arm are constant, we can say that 𝜔̇ 𝑤𝑥 = 𝜔̇ 𝑤𝑦 = 𝜔̇ 𝑤𝑧 = 0, and so ( ) 𝛼⃗𝑤 = 𝜔 ⃗ arm × 𝜔 ⃗ arm + 𝜔 ⃗ 𝑤rel = 𝜔 ⃗ arm × 𝜔 ⃗ 𝑤rel ( ) 𝑑 + 𝓁 cos 𝛾 = 𝜔arm (− sin 𝛾 𝚤̂ + cos 𝛾 𝚥̂) × sin 𝛾 − 𝜔arm 𝚤̂, 𝑅

(8) (9)

or, carrying out the cross products, 𝛼⃗𝑤 =

(

) 𝑑 + 𝓁 cos 𝛾 ̂ − sin 𝛾 𝜔2arm cos 𝛾 𝑘. 𝑅

(10)

Discussion & Verification

The dimensions of the results in Eqs. (6) and (10) are one over time and one over time squared, respectively, as they should be. A Closer Look Referring to Eq. (10), notice that the angular acceleration of the wheel is always parallel to the 𝑧 direction, which means that it is always perpendicular to the plane formed by the bent arm. We can see why this must be so if, referring to Fig. 2, we note that 𝜔 ⃗ arm causes the tip of 𝜔 ⃗ 𝑤rel to move away from us, which is in the −𝑧 direction. We will see that this angular acceleration plays an important role when we write the Newton-Euler equations for a rigid body moving in three dimensions.

1355

1356

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

E X A M P L E 20.2 yaw 𝜔𝑧

Motion of the Tip of an Airplane Propeller Blade

pitch 𝜔𝑦

Referring to Fig. 1, assume that the velocity and acceleration of point 𝐶 are known as ̂ respectively, and that the airplane 𝑣⃗𝐶 = 𝑣𝐶𝑥 𝚤̂+𝑣𝐶𝑦 𝚥̂+𝑣𝐶𝑧 𝑘̂ and 𝑎⃗𝐶 = 𝑎𝐶𝑥 𝚤̂+𝑎𝐶𝑦 𝚥̂+𝑎𝐶𝑧 𝑘, is pitching at a known rate 𝜔𝑦 (𝑡), but that the roll and yaw rates are both zero. Given the constant angular speed of the propeller in the direction shown, determine expressions for the velocity and acceleration of a point 𝑄, which is a radial distance 𝑅 from the hub of the propeller at 𝐵, as a function of the angle 𝜃 (see Fig. 2).

𝑦

𝑃 𝜔prop roll 𝜔𝑥

𝑥 𝜃

𝑘̂

Figure 1 A secondary or moving 𝑥𝑦𝑧 reference frame attached to an airplane. The primary 𝑋𝑌𝑍 reference frame (not shown) is attached to the Earth.

𝚥̂

𝜔̇ 𝑦

𝚤̂

𝑄 𝜔prop

𝜔𝑦 𝐶

𝑘̂

𝐵

𝑥

𝑦

Figure 2. The geometry needed to determine the velocity and acceleration of point 𝑄.

SOLUTION The secondary frame 𝑥𝑦𝑧 is attached to the airplane and has its origin at 𝐶. We observe that the motion of point 𝐶 is known and that the axis of the propeller lies on the 𝑥 axis. This makes the motion of point 𝑄, as seen by an observer in the 𝑥𝑦𝑧 frame, straightforward to describe because the motion of 𝑄 relative to the 𝑥𝑦𝑧 frame is circular.

Road Map & Modeling

Since the 𝑥𝑦𝑧 frame with origin at 𝐶 is a body-fixed frame, we can apply Eq. (16.45) to obtain ⃗ × 𝑟⃗ , 𝑣⃗𝑄 = 𝑣⃗𝐶 + 𝑣⃗𝑄rel + Ω (1) 𝑄∕𝐶

Computation

𝜃 𝑢̂ 𝑟 𝑢̂ 𝜃

𝑘̂

𝑢̂ 𝑟

𝚥̂ 𝑢̂ 𝜃

⃗ is the angular velocity of the body-fixed reference frame and 𝑣⃗ where Ω 𝑄rel is the velocity of point 𝑄 as seen by an observer in the body-fixed reference frame. Since we know that 𝜔𝑥 = 𝜔𝑧 = 0, then the angular velocity of the rotating frame, expressed in the rotating frame, is ⃗ = 𝜔 𝚥̂. Ω (2) 𝑦 As seen by the 𝑥𝑦𝑧 frame, the point 𝑄 is just moving in a circle centered on the 𝑥 axis and so, using the polar frame shown in Fig. 3, we obtain

𝜔prop

𝑣⃗𝑄rel = 𝑅𝜔prop 𝑢̂ 𝜃 .

𝑦

Figure 3 The definition of an additional polar component system to allow us to easily find the velocity and acceleration of 𝑄 relative to the rotating frame.

ISTUDY

(3)

Substituting Eqs. (2) and (3), along with 𝑣⃗𝐶 = 𝑣𝐶𝑥 𝚤̂ + 𝑣𝐶𝑦 𝚥̂ + 𝑣𝐶𝑧 𝑘̂ and 𝑟⃗𝑄∕𝐶 = 𝑑 𝚤̂ + 𝑅 𝑢̂ 𝑟 , into Eq. (1), we obtain ( ) 𝑣⃗𝑄 = 𝑣𝐶𝑥 𝚤̂ + 𝑣𝐶𝑦 𝚥̂ + 𝑣𝐶𝑧 𝑘̂ + 𝑅𝜔prop 𝑢̂ 𝜃 + 𝜔𝑦 𝚥̂ × 𝑑 𝚤̂ + 𝑅 𝑢̂ 𝑟 ( ) = 𝑣𝐶𝑥 𝚤̂ + 𝑣𝐶𝑦 𝚥̂ + 𝑣𝐶𝑧 𝑘̂ + 𝑅𝜔prop − cos 𝜃 𝚥̂ − sin 𝜃 𝑘̂ [ ( )] + 𝜔𝑦 𝚥̂ × 𝑑 𝚤̂ + 𝑅 cos 𝜃 𝑘̂ − sin 𝜃 𝚥̂ , (4) or

( ) ( ) 𝑣⃗𝑄 = 𝑣𝐶𝑥 + 𝑅𝜔𝑦 cos 𝜃 𝚤̂ + 𝑣𝐶𝑦 − 𝑅𝜔prop cos 𝜃 𝚥̂ ) ( ̂ + 𝑣𝐶𝑧 − 𝑅𝜔prop sin 𝜃 − 𝑑𝜔𝑦 𝑘,

(5)

ISTUDY

Section 20.1

Three-Dimensional Kinematics of Rigid Bodies

1357

̂ carried out the cross products, and where we have written 𝑢̂ 𝑟 and 𝑢̂ 𝜃 in terms of 𝚥̂ and 𝑘, collected terms. To determine the acceleration of 𝑄, we now apply ) ( ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗ ⃗ ⃗ 𝑎⃗𝑄 = 𝑎⃗𝐶 + 𝑎⃗𝑄rel + 2Ω 𝑄rel + Ω × 𝑟 𝑄∕𝐶 + Ω × Ω × 𝑟 𝑄∕𝐶 ,

(6)

⃗̇ is the angular acceleration of the body-fixed reference frame and 𝑎⃗ where Ω 𝑄rel is the acceleration of point 𝑄 as seen by an observer in the body-fixed reference frame. We ⃗̇ and 𝑎⃗ . For the former, we know or have determined everything in Eq. (6) except for Ω simply express the angular acceleration of the 𝑥𝑦𝑧 frame as

𝑄rel

⃗̇ = 𝜔̇ 𝚥̂. Ω 𝑦

(7)

For the relative acceleration, since the angular speed of the propeller is constant, point 𝑄 only has a normal component of relative acceleration, which gives 𝑎⃗𝑄rel = −𝑅𝜔2prop 𝑢̂ 𝑟 .

(8)

̂ into Eq. (6), Substituting Eqs. (2), (3), (7), and (8), along with 𝑎⃗𝐶 = 𝑎𝐶𝑥 𝚤̂ + 𝑎𝐶𝑦 𝚥̂ + 𝑎𝐶𝑧 𝑘, we obtain 𝑎⃗𝑄 = 𝑎𝐶𝑥 𝚤̂ + 𝑎𝐶𝑦 𝚥̂ + 𝑎𝐶𝑧 𝑘̂ − 𝑅𝜔2prop 𝑢̂ 𝑟 + 2𝜔𝑦 𝚥̂ × 𝑅𝜔prop 𝑢̂ 𝜃 [ ( ) ( )] + 𝜔̇ 𝑦 𝚥̂ × 𝑑 𝚤̂ + 𝑅 𝑢̂ 𝑟 + 𝜔𝑦 𝚥̂ × 𝜔𝑦 𝚥̂ × 𝑑 𝚤̂ + 𝑅 𝑢̂ 𝑟 ( ) = 𝑎𝐶𝑥 𝚤̂ + 𝑎𝐶𝑦 𝚥̂ + 𝑎𝐶𝑧 𝑘̂ − 𝑅𝜔2prop cos 𝜃 𝑘̂ − sin 𝜃 𝚥̂ ( ) [ ( )] + 2𝜔𝑦 𝚥̂ × 𝑅𝜔prop − cos 𝜃 𝚥̂ − sin 𝜃 𝑘̂ + 𝜔̇ 𝑦 𝚥̂ × 𝑑 𝚤̂ + 𝑅 cos 𝜃 𝑘̂ − sin 𝜃 𝚥̂ { [ ( )]} + 𝜔𝑦 𝚥̂ × 𝜔𝑦 𝚥̂ × 𝑑 𝚤̂ + 𝑅 cos 𝜃 𝑘̂ − sin 𝜃 𝚥̂ , (9) ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 𝑅𝜔𝑦 cos 𝜃 𝚤̂ − 𝑑𝜔𝑦 𝑘̂

or ( ) 𝑎⃗𝑄 = 𝑎𝐶𝑥 − 2𝑅𝜔𝑦 𝜔prop sin 𝜃 + 𝑅𝜔̇ 𝑦 cos 𝜃 − 𝑑𝜔2𝑦 𝚤̂ ( ) + 𝑎𝐶𝑦 + 𝑅𝜔2prop sin 𝜃 𝚥̂ ] [ ( ) ̂ + 𝑎𝐶𝑧 − 𝑅 𝜔2prop + 𝜔2𝑦 cos 𝜃 − 𝑑 𝜔̇ 𝑦 𝑘,

Interesting Fact (10)

̂ carried out the cross products, where we have again written 𝑢̂ 𝑟 and 𝑢̂ 𝜃 in terms of 𝚥̂ and 𝑘, and collected terms. Discussion & Verification

The final results for velocity and acceleration are dimensionally correct. We also see that, even though the motion of point 𝑄 on the propeller is very complicated as seen by an inertial observer, it is straightforward to determine using a rotating reference frame. We also note that the velocity and acceleration given in Eqs. (5) and (10), respectively, are measured with respect to some primary 𝑋𝑌𝑍 frame (not shown in Figures 1 or 2), and are therefore inertial if 𝑋𝑌𝑍 is inertial, but they are expressed in the body-fixed 𝑥𝑦𝑧 frame.

Gyroscopic terms in the acceleration. Terms in the acceleration that contain the product of two different angular velocities [e.g., the term containing 𝜔𝑦 𝜔prop in Eq. (10)] are often called gyroscopic terms. They result from the changing direction of angular velocities. This is really the only new result encountered in this section. In this example, the pitch angular velocity of the plane 𝜔𝑦 causes the angular velocity of the propeller to change direction. Thus, the time derivative of 𝜔 ⃗ prop is not equal to zero because it changes direction due to 𝜔𝑦 .

1358

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

E X A M P L E 20.3

Angular Velocity and Angular Acceleration of a Bar in 3D The mechanism in Fig. 1 consists of a disk of radius 𝑅 that rotates with constant angular speed 𝜔𝑑 about the 𝑥 axis in the direction shown. Attached by a ball joint to the disk at 𝐵 is the bar 𝐴𝐵. End 𝐴 of bar 𝐴𝐵 is attached by a ball joint to a collar that slides along the bar 𝐶𝐷. Bar 𝐶𝐷 lies in the 𝑦𝑧 plane and is inclined at the angle 𝜃 with respect to the 𝑦 axis. At the instant shown, the point 𝐵 lies in the 𝑥𝑦 plane. At this instant, determine expressions for the 𝑥, 𝑦, and 𝑧 components of the angular velocity of the bar 𝐴𝐵, as well as the velocity of the slider 𝐴. Treat 𝑟 and 𝐿 as given.

𝐷

𝑧

𝑦 𝐴 𝜃

𝑘̂

𝚥̂ 𝚤̂

𝜔𝑑

𝐶 𝑂 𝐵

𝑅

𝑥

Figure 1

SOLUTION The velocity and acceleration of point 𝐵 are easily determined since it is moving in a circle centered at 𝑂. Given that point 𝐴 is constrained to move in a known direction, we can use Eq. (20.1) to relate the velocity of point 𝐴 to that of point 𝐵, which should lead us to the angular velocity of the bar 𝐴𝐵.

Road Map & Modeling

𝑘̂

𝑧

𝚥̂

Computation

At this instant, in terms of the given component system, the velocity of point 𝐵 can be written as ̂ 𝑣⃗𝐵 = −𝑅𝜔𝑑 𝑘. (1)

𝚤̂

𝐷 𝑢̂ 𝐷∕𝐶

𝑦

Applying Eq. (20.1) to points 𝐴 and 𝐵, we obtain

𝐴

⃗ 𝐴𝐵 × 𝑟⃗𝐴∕𝐵 , 𝑣⃗𝐴 = 𝑣⃗𝐵 + 𝜔

𝜃

𝜔𝑑

where 𝜔 ⃗ 𝐴𝐵 is the angular velocity of the bar 𝐴𝐵. Since point 𝐴 is constrained by the collar to move along bar 𝐶𝐷, we can write 𝑣⃗𝐴 as (see Fig. 2) ( ) (3) 𝑣⃗𝐴 = 𝑣𝐴 𝑢̂ 𝐷∕𝐶 = 𝑣𝐴 cos 𝜃 𝚥̂ + sin 𝜃 𝑘̂ ,

𝐶 𝑂 𝐵

𝑅

Figure 2 The mechanism showing the unit vector 𝑢̂ 𝐷∕𝐶 .

ISTUDY

(2)

𝑥

where 𝑣𝐴 is the component of 𝑣⃗𝐴 in the direction of 𝑢̂ 𝐷∕𝐶 . The vector 𝑟⃗𝐴∕𝐵 can be formed using ( ) 𝑟⃗𝐴∕𝐵 = 𝑟⃗𝐴 − 𝑟⃗𝐵 = 𝑟 cos 𝜃 𝚥̂ + sin 𝜃 𝑘̂ − (𝐿 𝚤̂ − 𝑅 𝚥̂) ̂ = −𝐿 𝚤̂ + (𝑅 + 𝑟 cos 𝜃) 𝚥̂ + 𝑟 sin 𝜃 𝑘.

(4)

Noting that we can write the angular velocity of bar 𝐴𝐵 as ̂ 𝜔 ⃗ 𝐴𝐵 = 𝜔𝐴𝐵𝑥 𝚤̂ + 𝜔𝐴𝐵𝑦 𝚥̂ + 𝜔𝐴𝐵𝑧 𝑘,

(5)

ISTUDY

Section 20.1

Three-Dimensional Kinematics of Rigid Bodies

we can substitute Eqs. (1), (3), (4), and (5) into Eq. (2) to obtain ( ) ( ) 𝑣𝐴 cos 𝜃 𝚥̂ + sin 𝜃 𝑘̂ = −𝑅𝜔𝑑 𝑘̂ + 𝜔𝐴𝐵𝑥 𝚤̂ + 𝜔𝐴𝐵𝑦 𝚥̂ + 𝜔𝐴𝐵𝑧 𝑘̂ [ ] × −𝐿 𝚤̂ + (𝑅 + 𝑟 cos 𝜃) 𝚥̂ + 𝑟 sin 𝜃 𝑘̂ .

(6)

Carrying out the cross product and equating components, we obtain the following three equations: 0 = 𝑟𝜔𝐴𝐵𝑦 sin 𝜃 − (𝑅 + 𝑟 cos 𝜃)𝜔𝐴𝐵𝑧 ,

(7)

𝑣𝐴 cos 𝜃 = −𝐿𝜔𝐴𝐵𝑧 − 𝑟𝜔𝐴𝐵𝑥 sin 𝜃,

(8)

𝑣𝐴 sin 𝜃 = (𝑅 + 𝑟 cos 𝜃)𝜔𝐴𝐵𝑥 + 𝐿𝜔𝐴𝐵𝑦 − 𝑅𝜔𝑑 .

(9)

These three equations have four unknowns, that is, 𝜔𝐴𝐵𝑥 , 𝜔𝐴𝐵𝑦 , 𝜔𝐴𝐵𝑧 , and 𝑣𝐴 . With this said, there is one thing that we have not taken into account—any rotation of the bar 𝐴𝐵 about its own axis does not affect the motion of either point 𝐴 or point 𝐵, so that component of angular velocity is arbitrary as far as Eq. (2) is concerned. Therefore, we can set the component of the angular velocity along the bar 𝐴𝐵 to zero to obtain the fourth equation (this is equivalent to saying that 𝜔 ⃗ 𝐴𝐵 is orthogonal to the bar 𝐴𝐵), that is, 𝜔 ⃗ 𝐴𝐵 ⋅ 𝑢̂ 𝐴∕𝐵 = 0

⇒

( ) 𝑟⃗ − 𝑟⃗𝐵 𝜔𝐴𝐵𝑥 𝚤̂ + 𝜔𝐴𝐵𝑦 𝚥̂ + 𝜔𝐴𝐵𝑧 𝑘̂ ⋅ 𝐴 = 0, |⃗𝑟𝐴 − 𝑟⃗𝐵 |

(10)

or, substituting in for 𝑟⃗𝐴 − 𝑟⃗𝐵 from Eq. (4), we obtain ( ) −𝐿 𝚤̂ + (𝑅 + 𝑟 cos 𝜃) 𝚥̂ + 𝑟 sin 𝜃 𝑘̂ 𝜔𝐴𝐵𝑥 𝚤̂ + 𝜔𝐴𝐵𝑦 𝚥̂ + 𝜔𝐴𝐵𝑧 𝑘̂ ⋅ √ = 0. 𝐿2 + (𝑅 + 𝑟 cos 𝜃)2 + 𝑟2 sin2 𝜃

(11)

Expanding the dot product and simplifying, this equation becomes −𝐿𝜔𝐴𝐵𝑥 + (𝑅 + 𝑟 cos 𝜃)𝜔𝐴𝐵𝑦 + 𝑟 sin 𝜃𝜔𝐴𝐵𝑧 = 0.

(12)

Solving Eqs. (7)–(9) and (12) for 𝜔𝐴𝐵𝑥 , 𝜔𝐴𝐵𝑦 , 𝜔𝐴𝐵𝑧 , and 𝑣𝐴 , we obtain 𝜔𝐴𝐵𝑥 𝜔𝐴𝐵𝑦 𝜔𝐴𝐵𝑧

( ) 𝑅𝜔𝑑 cos 𝜃 𝑟2 + 𝑅2 + 2𝑟𝑅 cos 𝜃 = ( ), (𝑟 + 𝑅 cos 𝜃) 𝐿2 + 𝑟2 + 𝑅2 + 2𝑟𝑅 cos 𝜃 𝐿𝑅𝜔𝑑 cos 𝜃(𝑅 + 𝑟 cos 𝜃) = ( ), (𝑟 + 𝑅 cos 𝜃) 𝐿2 + 𝑟2 + 𝑅2 + 2𝑟𝑅 cos 𝜃 𝐿𝑟𝑅𝜔𝑑 cos 𝜃 sin 𝜃 = ( ), (𝑟 + 𝑅 cos 𝜃) 𝐿2 + 𝑟2 + 𝑅2 + 2𝑟𝑅 cos 𝜃

and 𝑣𝐴 = −

𝑟𝑅𝜔𝑑 sin 𝜃 𝑟 + 𝑅 cos 𝜃

⇒

𝑣⃗𝐴 = −

𝑟𝑅𝜔𝑑 sin 𝜃 ( ) cos 𝜃 𝚥̂ + sin 𝜃 𝑘̂ . 𝑟 + 𝑅 cos 𝜃

(13) (14) (15)

(16)

Discussion & Verification

The dimension of each of the four results is correct. In addition, given the direction of the angular velocity of the disk, we would expect the collar at 𝐴 to be moving down the bar 𝐶𝐷, and it is. A Closer Look

apply

If we wanted to find the angular acceleration of the bar 𝐴𝐵, we could ) ( 𝑎⃗𝐴 = 𝑎⃗𝐵 + 𝛼⃗𝐴𝐵 × 𝑟⃗𝐴∕𝐵 + 𝜔 ⃗ 𝐴𝐵 × 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐴∕𝐵 ,

(17)

along with a condition similar to Eq. (10) on the angular acceleration 𝛼⃗𝐴𝐵 . This is the objective of Problems 20.17 and 20.18.

1359

1360

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

Problems Problems 20.1 and 20.2 𝜔1

𝐵

The radar dish can rotate about the vertical 𝑧 axis at rate 𝜔1 and about the horizontal 𝑦 ̇ The distance between the center of rotation at 𝑂 axis (not shown in the figure) at rate 𝜃. and the subreflector at 𝐵 is 𝓁. Problem 20.1 If 𝜔1 and 𝜃̇ are both constant, determine the velocity and acceleration of the subreflector 𝐵 in terms of the elevation angle 𝜃.

𝜃 𝑂

̇ are known functions of time, determine the velocity If 𝜔1 (𝑡) and 𝜃(𝑡) and acceleration of the subreflector 𝐵 in terms of the elevation angle 𝜃. Problem 20.2

𝑥

Problems 20.3 and 20.4 The truncated cone rolls without slipping on the 𝑥𝑦 plane. At the instant shown, the angular speed about the 𝑧 axis is 𝜔1 , and it is changing at 𝜔̇ 1 .

Figure P20.1 and P20.2

𝜔1 𝜃 𝑂

𝑥 Figure P20.3 and P20.4

Problem 20.3 Determine expressions for the angular velocity and angular acceleration ̇ Express your answers in the rotating component of the cone in terms of 𝓁, 𝑑, 𝜃, 𝜔1 , and 𝜔. system shown.

Determine expressions for the velocity and acceleration of the point 𝐴, which is at the highest point on the cone at this instant, in terms of 𝓁, 𝑑, 𝜃, 𝜔1 , and 𝜔. ̇ Express your answers in the rotating component system shown. Problem 20.4

Problems 20.5 through 20.8 The bent arm rotates with angular speed 𝜔arm and angular acceleration 𝛼arm in the directions shown. The wheel of radius 𝑅 with center at 𝐴 rotates relative to the bent arm as it rolls without slipping over the stationary horizontal surface. At the instant shown, the line 𝑃 𝑄 is perpendicular to the line 𝐸𝐹 , which is parallel to the horizontal surface (i.e., the line 𝑃 𝑄 lies in the 𝑥𝑦 plane). Express your answers using the 𝑥𝑦𝑧 reference frame that is attached to the arm 𝑂𝐵𝐴. Treat 𝑑, 𝓁, 𝑅, and 𝛾 as known.

𝛼arm 𝑦

𝜔arm

𝑂 𝑧

𝛾

Problem 20.5 Assuming that 𝛼arm = 0 at the instant shown, determine expressions for the velocity and acceleration of point 𝑃 . 𝐹 𝑥 𝑄

Problem 20.6 Assuming that 𝛼arm = 0 at the instant shown, determine expressions for the velocity and acceleration of point 𝐸.

Assuming that 𝛼arm ≠ 0 at the instant shown, determine expressions for the velocity and acceleration of point 𝑃 .

Problem 20.7

Assuming that 𝛼arm ≠ 0 at the instant shown, determine expressions for the velocity and acceleration of point 𝐸.

Problem 20.8 Figure P20.5–P20.8

ISTUDY

ISTUDY

Section 20.1

1361

Three-Dimensional Kinematics of Rigid Bodies

Problems 20.9 through 20.11

𝐵

The fire truck ladder can rotate about the vertical 𝑧 axis at known rate 𝜔1 (𝑡), elevate about ̇ the horizontal 𝑥 axis at known rate 𝜃(𝑡), and the ladder can change its length (moving the ̇ bucket at 𝐵 outward or inward) with known rate 𝓁(𝑡).

𝓁(𝑡)

𝑧 𝜔1

Problem 20.9 If 𝜔1 and 𝜃̇ are constant and 𝓁̇ = 0, determine expressions for the velocity and acceleration of the bucket at 𝐵 as functions of the elevation angle 𝜃. Express your answer in the given rotating 𝑥𝑦𝑧 frame.

𝜃(𝑡)

̇ and 𝓁̇ are each constant, determine expressions for the veIf 𝜔1 , 𝜃, locity and acceleration of the bucket at 𝐵 as functions of the elevation angle 𝜃. Express your answer in the given rotating 𝑥𝑦𝑧 frame. Problem 20.10

̇ as functions of time, determine expressions ̇ Problem 20.11 Given 𝜔1 (𝑡), 𝜃(𝑡), and 𝓁(𝑡) for the velocity and acceleration of the bucket at 𝐵 as functions of the elevation angle 𝜃. Express your answer in the given rotating 𝑥𝑦𝑧 frame.

𝑥

𝑂

Figure P20.9–P20.11

Problems 20.12 and 20.13 The bar 𝐴𝐵 rotates at the rate 𝜔𝑏 about the fixed 𝑥 axis as shown. The bar 𝐶𝐷 is attached perpendicularly to 𝐴𝐵 to form a T-bar. The disk of radius 𝑅 centered at 𝐶 rotates at the rate 𝜔𝑑 relative to the arm 𝐶𝐷 in the direction shown. Assume that the 𝑥𝑦𝑧 reference frame is attached to the T-bar and that its origin is at 𝐷. In addition, assume that the angular rates 𝜔𝑏 and 𝜔𝑑 are not constant. Treat ℎ and 𝑅 as known.

𝐴 𝑃 𝛽 𝑅

Problem 20.12

Determine expressions for the angular velocity and angular acceleration of the disk at 𝐶. Express your answer in the rotating 𝑥𝑦𝑧 reference frame.

𝜔𝑏

𝑧

𝐵

𝐷

𝑥

𝐶

Problem 20.13

Determine expressions for the velocity and acceleration of the point 𝑃 , which lies at the edge of the disk at an arbitrary angle 𝛽 with respect to the 𝑦 axis. Express your answer in the rotating 𝑥𝑦𝑧 reference frame.

Figure P20.12 and P20.13

Problems 20.14 through 20.16 The cone rolls without slipping over the 𝑥𝑧 plane and around the 𝑦 axis with angular speed 𝜔0 and angular acceleration 𝛼0 in the directions shown. At the instant shown, the line 𝐵𝐶 is parallel to the surface on which the cone is rolling and the line 𝐴𝐷 lies on the base of the cone and is perpendicular to the line 𝐵𝐶. Treat 𝐿 and 𝛽 as known. 𝛼0

Problem 20.14

Using the component system shown, determine expressions for the angular velocity 𝜔 ⃗ 𝑐 and angular acceleration 𝛼⃗𝑐 of the cone. Problem 20.15 Using the component system shown, determine expressions for the velocity and acceleration of point 𝐴 at this instant.

𝚥̂ 𝑘̂

𝑧

The mechanism consists of a disk of radius 𝑅 that rotates with angular speed 𝜔𝑑 and angular acceleration 𝛼𝑑 about the 𝑥 axis in the directions shown. Attached by a ball joint to the disk at 𝐵 is the bar 𝐴𝐵. End 𝐴 of bar 𝐴𝐵 is attached by a ball joint to a collar that slides along the bar 𝐶𝐷. Bar 𝐶𝐷 lies in the 𝑦𝑧 plane and is inclined at the angle 𝜃 with respect to the 𝑦 axis. At the instant shown, the point 𝐵 lies in the 𝑥𝑦 plane. Use 𝑅 = 1 f t,

𝐴 𝐶

𝑂 𝐵 𝐷

Problem 20.16 Using the component system shown, determine expressions for the velocity and acceleration of point 𝐵 at this instant.

Problems 20.17 and 20.18

𝚤̂

𝜔0

Figure P20.14–P20.16

𝑥

1362

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

𝐿 = 3 f t, 𝑟 = 1.4 f t, 𝜃 = 25◦ , and 𝜔𝑑 = 20 rad∕s. 𝐷

𝑧

𝑦 𝐴

𝑘̂

𝚥̂ 𝚤̂

𝜃 𝜔𝑑 𝐶 𝑂 𝐵

𝑅

𝑥

Figure P20.17 and P20.18

Assuming that 𝛼𝑑 = 0 at this instant, determine the angular acceleration of the bar 𝐴𝐵, as well as the acceleration of the slider 𝐴.

Problem 20.17

Assuming that 𝛼𝑑 = 35 rad∕s2 at this instant, determine the angular acceleration of the bar 𝐴𝐵, as well as the acceleration of the slider 𝐴. Problem 20.18

Problems 20.19 and 20.20 𝐶

The T-bar support structure is mounted in bearings at 𝐶 and 𝐷 and spins with angular velocity 𝜔 ⃗ 𝑠 and angular acceleration 𝛼⃗𝑠 in the directions shown. Bar 𝐴𝐵 of length 𝐿 is pinned at 𝐴 to the T-bar support, and its position in the vertical plane is defined by the variable angle 𝛽. As we will see in the next section, the Newton-Euler equations for the bar will require that we know the acceleration of 𝐺, the angular velocity of the bar 𝐴𝐵, and the angular acceleration of the bar 𝐴𝐵 relative to the primary frame 𝑋𝑌𝑍, which is assumed to be inertial. Express your answers in the 𝑥𝑦𝑧 frame shown, which is attached to the T-bar support at 𝐴. Treat 𝑑 and 𝐿 as known.

𝑦 𝐴

𝑂

𝑥 𝛽

𝜔𝑠

𝐺

𝛼𝑠 𝐵

Figure P20.19 and P20.20

ISTUDY

Problem 20.19

Find expressions for 𝑎⃗𝐺 , 𝜔 ⃗ 𝐴𝐵 , and 𝛼⃗𝐴𝐵 assuming that 𝛼𝑠 = 0.

Problem 20.20

Find expressions for 𝑎⃗𝐺 , 𝜔 ⃗ 𝐴𝐵 , and 𝛼⃗𝐴𝐵 assuming that 𝛼𝑠 ≠ 0.

Problems 20.21 through 20.24 The mechanism consists of a disk of radius 𝑅 that rotates with angular speed 𝜔𝑑 and angular acceleration 𝛼𝑑 about the 𝑥 axis in the directions shown. Attached by a ball joint to the disk at 𝐵 is the bar 𝐴𝐵. End 𝐴 of bar 𝐴𝐵 is attached by a clevis joint to a collar that slides along the bar 𝐶𝐷. Bar 𝐶𝐷 lies in the 𝑦𝑧 plane and is inclined at the angle 𝜃 with respect to the 𝑦 axis. At the instant shown, the point 𝐵 lies in the 𝑥𝑦 plane. Use 𝑅 = 0.2 m, 𝐿 = 0.5 m, 𝑟 = 0.3 m, 𝜃 = 25◦ , and 𝜔𝑑 = 30 rad∕s. Hint: The clevis joint constrains the rotation of arm 𝐴𝐵 relative to the collar at 𝐴 to be perpendicular to the plane formed by bar 𝐶𝐷 and arm 𝐴𝐵. Therefore, the angular velocity of arm 𝐴𝐵 is the sum of the angular velocity of the collar at 𝐴 and the angular velocity associated with the change in the angle 𝛽.

ISTUDY

Section 20.1

1363

Three-Dimensional Kinematics of Rigid Bodies

𝐷

𝑧

𝑦 𝐴

𝑘̂

𝛽

𝚥̂ 𝚤̂

𝜃 𝜔𝑑 𝐶 𝑂 𝐵

𝑥

𝑅 𝛼𝑑

Figure P20.21–P20.24 Problem 20.21

Using the component system shown, determine the velocity of the

collar at 𝐴. Assuming that 𝛼𝑑 = 15 rad∕s2 at this instant, determine the angular velocity and angular acceleration of the arm 𝐴𝐵. Use the component system shown. Problem 20.22

Assuming that 𝛼𝑑 = 0 at this instant, determine the acceleration of the collar at 𝐴. Use the component system shown.

Problem 20.23

Assuming that 𝛼𝑑 = 15 rad∕s2 at this instant, determine the acceleration of the collar at 𝐴. Use the component system shown. Problem 20.24

Problems 20.25 and 20.26 Rod 𝐴𝐵 is attached to the collar at 𝐵 by a ball joint and to the slider at 𝐴 by a clevis joint. Collar 𝐴 slides along the fixed bar 𝑂𝐷, which lies in the 𝑦𝑧 plane and is inclined at the angle 𝜃 with respect to the 𝑦 axis. At the instant shown, collar 𝐵 lies in the 𝑥𝑦 plane, and it is moving with the velocity and acceleration shown. Bar 𝐶𝐸 is fixed and parallel to the 𝑧 axis. Use 𝑟 = 0.3 m, 𝜃 = 25◦ , 𝐿 = 0.8 m, 𝑑 = 0.1 m, 𝑣𝐵 = 2.5 m∕s, and 𝑎𝐵 = 1.5 m∕s2 . Hint: The clevis joint constrains the rotation of arm 𝐴𝐵 relative to the collar at 𝐴 to be perpendicular to the plane formed by bar 𝑂𝐷 and arm 𝐴𝐵. Therefore, the angular velocity of arm 𝐴𝐵 is the sum of the angular velocity of the collar at 𝐴 and the angular velocity associated with the change in the angle 𝛽. Problem 20.25

Using the component system shown, determine the angular velocity and angular acceleration of the bar 𝐴𝐵.

Problem 20.26

Using the component system shown, determine the velocity and acceleration of the collar at 𝐴.

𝐷

𝑧 𝐴

𝑦

𝛽 𝜃

𝐶

𝑂 𝑘̂

𝚥̂ 𝚤̂

𝐵 𝑣𝐵 𝐸

Figure P20.25 and P20.26

Problems 20.27 through 20.32 Bar 𝐴𝐵 of length 𝐿𝐴𝐵 = 2.5 m is attached by ball joints to a collar at 𝐴 and to a disk at 𝐵. The disk lies in the 𝑥𝑦 plane and its center at 𝐸 lies on the 𝑦 axis in the 𝑦𝑧 plane. The disk rotates about a vertical axis at the constant angular rate 𝜔𝑑 = 100 rpm. The dimensions 𝑑 = 1.2 m, ℎ = 0.9 m, and 𝑅 = 0.75 m are given.

𝑑 𝑎𝐵

𝑥

1364

ISTUDY

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

𝓁 𝐴 ℎ 𝐿𝐴𝐵

𝑥

𝑂

𝑑 𝜃 𝐵

𝐸 𝑦

𝑅 𝜔𝑑

Figure P20.27–P20.32

For the disk position shown, that is, 𝜃 = 90◦ , determine the angular velocity of the bar. Express your answer in the given component system, and assume that the angular velocity of the bar is orthogonal to it.

Problem 20.27

For the disk position shown, that is, 𝜃 = 90◦ , determine the angular acceleration of the bar. Express your answer in the given component system, and assume that the angular velocity and angular acceleration of the bar are orthogonal to it.

Problem 20.28

For the disk position 𝜃 = 0◦ , determine the velocity of the collar at 𝐴. Express your answer in the given component system, and assume that the angular velocity of the bar is orthogonal to it. Problem 20.29

For the disk position 𝜃 = 0◦ , determine the acceleration of the collar at 𝐴. Express your answer in the given component system, and assume that the angular velocity and angular acceleration of the bar are orthogonal to it. Problem 20.30

Determine the angular velocity of the bar 𝐴𝐵 and the velocity of the collar at 𝐴 for any position 𝜃 of the disk. Express your answers in the given component system, and assume that the angular velocity of the bar is orthogonal to it. Problem 20.31

Problem 20.32 Determine the angular acceleration of the bar 𝐴𝐵 and the acceleration of the collar at 𝐴 for any position 𝜃 of the disk. Express your answers in the given component system, and assume that the angular velocity and angular acceleration of the bar are orthogonal to it.

Problems 20.33 through 20.35 The robotic arm shown is used to drill holes during an assembly line manufacturing process. The base of the arm rotates about the vertical axis relative to the platform on which it is mounted at the rate 𝜔𝑏 . The bent arm rotates about the 𝑦 axis at the rate 𝜔𝑎 relative to the base. To extend and retract the drill bit, the telescoping shaft moves in and out of the bent arm with the speed 𝑣𝑡 and acceleration 𝑎𝑡 relative to the arm. The angular speed of the drill bit relative to the telescoping arm is 𝜔𝑑 . At this instant, the 𝑧 axis is perpendicular to the platform. Express all answers using the 𝑥𝑦𝑧 reference frame that is attached to the bent arm at point 𝑂, and do not assume that the angular rates 𝜔𝑏 , 𝜔𝑎 , and 𝜔𝑑 are constant. Treat 𝓁, 𝑟𝑡 , and ℎ as known.

ISTUDY

Section 20.1

Three-Dimensional Kinematics of Rigid Bodies

Problem 20.33

Determine expressions for the angular velocity and angular accelera-

tion of the drill bit. Determine an expression for the velocity of point 𝐵 at the end of the

Problem 20.34

drill bit. Determine an expression for the acceleration of point 𝐵 at the end of

Problem 20.35

the drill bit.

𝑎𝑡 𝐵 𝜔𝑑

𝑣𝑡 arm telescoping shaft

𝑂

drill bit

𝜔𝑎

platform 𝑥

Figure P20.33–P20.35

base

𝑦

1365

1366

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

20.2 𝑃

⃗ 𝑀 𝑃

Newton-Euler equations for three-dimensional motion

𝑣⃗𝑃 𝑟⃗𝐺∕𝑃 𝑘̂ 𝚤̂

𝑎⃗𝐺

Referring to Fig. 20.8, to obtain the Newton-Euler equations for a rigid body, we begin with Euler’s first law as given by Eq. (15.57) on p. 986, which is

𝑣⃗𝐺 𝐺

Three-Dimensional Kinetics of Rigid Bodies

𝑦

𝚥̂

𝐹⃗ = 𝑚𝑎⃗𝐺 ,

(20.20)

𝐵 𝑥

𝜔 ⃗𝐵

Figure 20.8 Rigid body 𝐵 showing quantities used in obtaining the Newton-Euler rotational equations of motion for three-dimensional motion.

ISTUDY

where 𝐹⃗ is the total external force acting on the rigid body, 𝑚 is the mass of the rigid body, and 𝑎⃗𝐺 is the acceleration of the mass center of the rigid body. Referring to Fig. 20.8, we next apply Euler’s second law as given by Eq. (15.58) on p. 986, that is, ⃗̇ + 𝑣⃗ × 𝑚𝑣⃗ , ⃗ =ℎ 𝑀 𝑃 𝑃 𝑃 𝐺

(20.21)

⃗ is the total external moment acting on the rigid body about an arbitrary where 𝑀 𝑃 ⃗ is the angular momentum of the rigid body about 𝑃 , 𝑣⃗ is the moment center 𝑃 , ℎ 𝑃 𝑃 velocity of the point 𝑃 , and 𝑣⃗𝐺 is the velocity of the mass center of the rigid body. To make this equation useful, we need an expression for the angular momentum of a rigid body. Equations (D.15) on p. A-77 and (D.18) on p. A-78 in Appendix D provide two such expressions, both of which we repeat below for convenience. ( ) ⃗ = 𝐼 𝜔 −𝐼 𝜔 −𝐼 𝜔 ℎ 𝑃 𝐺𝑥 𝐵𝑥 𝐺𝑥𝑦 𝐵𝑦 𝐺𝑥𝑧 𝐵𝑧 𝚤̂ ( ) + −𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 + 𝐼𝐺𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑧 𝚥̂ ( ) + −𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 + 𝐼𝐺𝑧 𝜔𝐵𝑧 𝑘̂ + 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , (20.22) {ℎ𝑃 } = [𝐼𝐺 ]{𝜔𝐵 } + 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 ,

(20.23)

where:

Helpful Information The 𝒙𝒚𝒛 reference frame is not, in general, attached to 𝑩. The angular velocity of the body 𝜔 ⃗ 𝐵 and the moments and products of inertia are all written in the components of the 𝑥𝑦𝑧 frame, which is rotating with angular ⃗ Note that, at this point, the body velocity Ω. can be rotating relative to the 𝑥𝑦𝑧 frame so that the angular velocity of the body and the angular velocity of the frame are, in general, not the same.

̂ • The column vector {ℎ𝑃 } = ℎ𝑃 𝑥 𝚤̂ + ℎ𝑃 𝑦 𝚥̂ + ℎ𝑃 𝑧 𝑘. • 𝐼𝐺𝑥 , 𝐼𝐺𝑦 , and 𝐼𝐺𝑧 are the mass moments of inertia of the rigid body with respect to its mass center 𝐺 (see Appendix C). • 𝐼𝐺𝑥𝑦 , 𝐼𝐺𝑥𝑧 , and 𝐼𝐺𝑦𝑧 are the mass products of inertia of the rigid body with respect to its mass center 𝐺 (see Appendix C). • [𝐼𝐺 ] is the inertia matrix of 𝐵 with respect to the 𝑥𝑦𝑧 axes, and its entries consist of the moments and products of inertia as follows: ⎡ 𝐼 ⎢ 𝐺𝑥 [𝐼𝐺 ] = ⎢−𝐼𝐺𝑥𝑦 ⎢−𝐼 ⎣ 𝐺𝑥𝑧

−𝐼𝐺𝑥𝑦 𝐼𝐺𝑦 −𝐼𝐺𝑦𝑧

−𝐼𝐺𝑥𝑧 ⎤ ⎥ −𝐼𝐺𝑦𝑧 ⎥ . 𝐼𝐺𝑧 ⎥⎦

(20.24)

• 𝜔𝐵𝑥 , 𝜔𝐵𝑦 , and 𝜔𝐵𝑧 are the angular velocity components of the rigid body writ̂ ten in the 𝑥𝑦𝑧 frame and {𝜔𝐵 } = 𝜔 ⃗ 𝐵 = 𝜔𝐵𝑥 𝚤̂ + 𝜔𝐵𝑦 𝚥̂ + 𝜔𝐵𝑧 𝑘. ̂ • 𝑟⃗𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂ + 𝑧𝐺∕𝑃 𝑘. ̂ • 𝑣⃗𝐺 = 𝑣𝐺𝑥 𝚤̂ + 𝑣𝐺𝑦 𝚥̂ + 𝑣𝐺𝑧 𝑘.

ISTUDY

Section 20.2

Three-Dimensional Kinetics of Rigid Bodies

To obtain the rotational Newton-Euler equations represented by Eq. (20.21), we ⃗ with respect to time. Starting with the time derivative of the last now differentiate ℎ 𝑃 term on the right-hand side of Eq. (20.22), we obtain ) 𝑑( 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 = 𝑟⃗̇ 𝐺∕𝑃 × 𝑚𝑣⃗𝐺 + 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 𝑑𝑡 ( ) = 𝑣⃗𝐺 − 𝑣⃗𝑃 × 𝑚𝑣⃗𝐺 + 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = −𝑣⃗𝑃 × 𝑚𝑣⃗𝐺 + 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 .

(20.25)

When taking the time derivative of the first three terms in Eq. (20.22), we need to differentiate all terms that are time varying. This includes not only the angular velocity components and Cartesian unit vectors, but also moments and products of inertia if the 𝑥𝑦𝑧 frame is allowed to rotate relative to the rigid body 𝐵.∗ To greatly simplify the equations, while still allowing us to write the rotational equations for almost any rigid body, we will assume that the 𝑥𝑦𝑧 frame is attached to the rigid body 𝐵 so that the moment and products of inertia are constant with respect to that reference frame. Therefore, the time derivative of the first three terms of Eq. (20.22) is ) ) ( 𝑑( [𝐼 ]{𝜔𝐵 } = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 − 𝐼𝐺𝑥𝑦 𝜔̇ 𝐵𝑦 − 𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑧 𝚤̂ 𝑑𝑡 𝐺 ( ) + −𝐼𝐺𝑥𝑦 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔̇ 𝐵𝑧 𝚥̂ ) ( + −𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 𝑘̂ ) )( ( ⃗ × 𝚤̂ + 𝐼𝐺𝑥 𝜔𝐵𝑥 − 𝐼𝐺𝑥𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑥𝑧 𝜔𝐵𝑧 Ω ( )( ) ⃗ × 𝚥̂ + −𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 + 𝐼𝐺𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑧 Ω ( ) )( ⃗ × 𝑘̂ , + −𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 + 𝐼𝐺𝑧 𝜔𝐵𝑧 Ω

(20.26)

⃗ is the angular velocity of the 𝑥𝑦𝑧 reference frame. Since the 𝑥𝑦𝑧 frame is where Ω ⃗ =𝜔 ̂ so now attached to the rigid body, we have that Ω ⃗ 𝐵 = 𝜔𝐵𝑥 𝚤̂ + 𝜔𝐵𝑦 𝚥̂ + 𝜔𝐵𝑧 𝑘, that, after expanding the cross products and collecting terms, Eq. (20.26) becomes ) ( ) ) [ ( 𝑑( [𝐼𝐺 ]{𝜔𝐵 } = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑦 𝜔𝐵𝑧 + 𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 𝜔𝐵𝑧 − 𝜔̇ 𝐵𝑦 𝑑𝑡 ( ) ( )] −𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 𝜔𝐵𝑦 + 𝜔̇ 𝐵𝑧 − 𝐼𝐺𝑦𝑧 𝜔2𝐵𝑦 − 𝜔2𝐵𝑧 𝚤̂ [ ( ) ( ) + 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝐼𝐺𝑦𝑧 𝜔𝐵𝑥 𝜔𝐵𝑦 − 𝜔̇ 𝐵𝑧 ( ) ( )] −𝐼𝐺𝑥𝑧 𝜔2𝐵𝑧 − 𝜔2𝐵𝑥 − 𝐼𝐺𝑥𝑦 𝜔̇ 𝐵𝑥 + 𝜔𝐵𝑦 𝜔𝐵𝑧 𝚥̂ [ ( ) ( ) + 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑥 𝜔𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝜔̇ 𝐵𝑦 ( ) ( )] ̂ (20.27) −𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑥 − 𝜔𝐵𝑦 𝜔𝐵𝑧 − 𝐼𝐺𝑥𝑦 𝜔2𝐵𝑥 − 𝜔2𝐵𝑦 𝑘. Substituting Eqs. (20.25) and (20.27) into Eq. (20.21), and writing ⃗ = 𝑀 𝚤̂ + 𝑀 𝚥̂ + 𝑀 𝑘, ̂ 𝑀 𝑃 𝑃𝑥 𝑃𝑦 𝑃𝑧

(20.28)

̂ 𝑟⃗𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂ + 𝑧𝐺∕𝑃 𝑘,

(20.29)

̂ 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂ + 𝑎𝐺𝑧 𝑘, ∗ See

the discussion under Practical use of Eqs. (D.15) and (D.16) in Appendix D on p. A-77.

(20.30)

1367

1368

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

then the three components of Eq. (20.21) become ( ) ( ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 + 𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 𝜔𝐵𝑧 − 𝜔̇ 𝐵𝑦 ( ) ( ) − 𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 𝜔𝐵𝑦 + 𝜔̇ 𝐵𝑧 − 𝐼𝐺𝑦𝑧 𝜔2𝐵𝑦 − 𝜔2𝐵𝑧 ) ( + 𝑚 𝑦𝐺∕𝑃 𝑎𝐺𝑧 − 𝑧𝐺∕𝑃 𝑎𝐺𝑦 , ( ) ( ) 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝐼𝐺𝑦𝑧 𝜔𝐵𝑥 𝜔𝐵𝑦 − 𝜔̇ 𝐵𝑧 ( ) ( ) − 𝐼𝐺𝑥𝑧 𝜔2𝐵𝑧 − 𝜔2𝐵𝑥 − 𝐼𝐺𝑥𝑦 𝜔̇ 𝐵𝑥 + 𝜔𝐵𝑦 𝜔𝐵𝑧 ) ( + 𝑚 𝑧𝐺∕𝑃 𝑎𝐺𝑥 − 𝑥𝐺∕𝑃 𝑎𝐺𝑧 , ( ) ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝜔̇ 𝐵𝑦 ( ) ( ) − 𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑥 − 𝜔𝐵𝑦 𝜔𝐵𝑧 − 𝐼𝐺𝑥𝑦 𝜔2𝐵𝑥 − 𝜔2𝐵𝑦 ) ( + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 ,

Concept Alert Recognizing principal body axes. While mass moments and products of inertia are discussed in detail in Appendix C, for all the rigid bodies we will consider here, the principal axes of inertia can be easily recognized. In particular, all products of inertia containing a coordinate whose corresponding axis is perpendicular to a plane of symmetry for a body must be zero, as long as the origin of the coordinate system lies in that plane of symmetry.

(20.31)

(20.32)

(20.33)

where we have canceled −𝑣⃗𝑃 ×𝑚𝑣⃗𝐺 in Eq. (20.25) with 𝑣⃗𝑃 ×𝑚𝑣⃗𝐺 in Eq. (20.21). These moment equations are rather complicated due to the nonzero products of inertia, but they can be greatly simplified if the 𝑥𝑦𝑧 axes that are attached to the body 𝐵 are principal body axes (see Appendix C). In that case, all the products of inertia become zero, and Eqs. (20.31)–(20.33) become ( ) ( ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 + 𝑚 𝑦𝐺∕𝑃 𝑎𝐺𝑧 − 𝑧𝐺∕𝑃 𝑎𝐺𝑦 , (20.34) ) ) ( ( 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝑚 𝑧𝐺∕𝑃 𝑎𝐺𝑥 − 𝑥𝐺∕𝑃 𝑎𝐺𝑧 , (20.35) ) ) ( ( 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 . (20.36) If any one of the following special cases can be applied: ⃗ (1) 𝑃 is the mass center 𝐺 of the body 𝐵 so that 𝑟⃗𝑃 ∕𝐺 = 0, ⃗ that is, the mass center of the body moves with constant velocity, or (2) 𝑎⃗𝐺 = 0, ⃗ (3) 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 so that the cross product 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0, then Eqs. (20.34)–(20.36) become

( ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 , ) ( 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 , ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 .

𝑃

⃗ 𝑀 𝑃

𝑟⃗𝐺∕𝑃 𝑘̂ 𝚤̂

𝑎⃗𝐺

𝐺

𝑦

𝚥̂ 𝐵

𝑥

𝜔 ⃗𝐵

Figure 20.9 Illustration of the kinetic and kinematic quantities needed to understand and apply Eqs. (20.31)– (20.39).

ISTUDY

(20.37) (20.38) (20.39)

Equations (20.37)–(20.39) are the famous Euler’s equations of rotational motion for a rigid body. Equations (20.31)–(20.33), or Eqs. (20.34)–(20.36), or Eqs. (20.37)–(20.39), along with Eq. (20.20), give us the six Newton-Euler equations of motion that we need for a rigid body in three-dimensional motion. Before moving on, let’s summarize the assumptions and conditions for each of the three sets of rotational equations (see Fig. 20.9). Conditions for applying Eqs. (20.31)–(20.33) • Point 𝑃 is arbitrary, and the 𝑥𝑦𝑧 reference frame is attached to the body 𝐵. • The moments of inertia are expressed relative to a set of axes that are parallel to 𝑥𝑦𝑧 and whose origin is at 𝐺, the center of mass of the body. • The angular velocities and their time derivatives are inertial, but are expressed in the rotating 𝑥𝑦𝑧 frame.

ISTUDY

Section 20.2

Three-Dimensional Kinetics of Rigid Bodies

Conditions for applying Eqs. (20.34)–(20.36) All the conditions for applying Eqs. (20.31)–(20.33) plus • The 𝑥𝑦𝑧 axes are principal axes of inertia so that all products of inertia are zero. Conditions for applying Eqs. (20.37)–(20.39) All the conditions for applying Eqs. (20.34)–(20.36) plus ⃗ which can happen in any of the following three ways: • 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0, ⃗ ⋄ 𝑃 is the mass center 𝐺 of the body 𝐵 so that 𝑟⃗𝐺∕𝑃 = 0, ⃗ that is, the mass center of the body moves with constant velocity, ⋄ 𝑎⃗𝐺 = 0, ⃗ ⋄ 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 so that the cross product 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0. Rotational equations of motion for a body with an axis of radial symmetry Many mechanical systems have axisymmetric parts that spin about their axis of symmetry. For these systems, any reference frame that has an axis coinciding with the axis of symmetry of that body will be a principal one (i.e., its products of inertia will be zero; see Appendix C). For example, referring to Fig. 20.10, we introduce 𝑦𝐵

𝜔 ⃗𝑠 𝜔 ⃗𝐵 𝚥̂ 𝑃

𝑘̂

𝐵 𝜔 ⃗𝑠

𝐺 𝚤̂

𝑥, 𝑥𝐵

𝑧 Figure 20.10. A rigid body 𝐵 spinning about its axis of radial symmetry 𝑥𝐵 relative to the ⃗ relative to an 𝑥𝑦𝑧 frame with angular velocity 𝜔 ⃗ 𝑠 . The angular velocity of the 𝑥𝑦𝑧 frame is Ω, underlying inertial frame.

the principal body axes 𝑥𝐵 𝑦𝐵 𝑧𝐵 with the 𝑥𝐵 axis coinciding with the axis of radial ⃗ the angular symmetry of the body 𝐵. The 𝑥𝑦𝑧 frame rotates with angular velocity Ω, velocity of the body is 𝜔 ⃗ 𝐵 , and the body spins relative to the 𝑥𝑦𝑧 frame about its axis of symmetry with angular velocity 𝜔 ⃗ 𝑠 . With these definitions, we can write that angular velocity of the body as ⃗ +𝜔 𝜔 ⃗𝐵 = Ω ⃗ 𝑠,

(20.40)

or, in component form relative to the 𝑥𝑦𝑧 frame as 𝜔𝐵𝑥 = Ω𝑥 + 𝜔𝑠 ,

𝜔𝐵𝑦 = Ω𝑦 ,

and

𝜔𝐵𝑧 = Ω𝑧 .

(20.41)

We will again apply Euler’s second law as given by Eq. (20.21). Referring back to Eq. (20.22), [𝐼𝐺 ]{𝜔𝐵 } for the rigid body in this case is [𝐼𝐺 ]{𝜔𝐵 } = 𝐼𝐺𝑥 𝜔𝐵𝑥 𝚤̂ + 𝐼𝐺𝑦 𝜔𝐵𝑦 𝚥̂ + 𝐼𝐺𝑦 𝜔𝐵𝑧 𝑘̂ ( ) ̂ = 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 𝚤̂ + 𝐼𝐺𝑦 Ω𝑦 𝚥̂ + 𝐼𝐺𝑦 Ω𝑧 𝑘,

(20.42)

1369

1370

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

where we have used the fact that all products of inertia are zero, have substituted in Eq. (20.41), and have used the fact that 𝐼𝐺𝑧 = 𝐼𝐺𝑦 . Differentiating Eq. (20.42) with respect to time, we obtain ) ( ) 𝑑( [𝐼 ]{𝜔𝐵 } = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 𝚤̂ + 𝐼𝐺𝑦 Ω̇ 𝑦 𝚥̂ 𝑑𝑡 𝐺 ) ( )( ⃗ × 𝚤̂ + 𝐼𝐺𝑦 Ω̇ 𝑧 𝑘̂ + 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω ) ) ( ( ⃗ × 𝚥̂ + 𝐼 Ω Ω ⃗ × 𝑘̂ . + 𝐼𝐺𝑦 Ω𝑦 Ω 𝐺𝑦 𝑧

(20.43)

Expanding the cross products and collecting terms, this equation becomes ) ( ) 𝑑( [𝐼𝐺 ]{𝜔𝐵 } = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 𝚤̂ 𝑑𝑡 [ ] ( ) + 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 𝚥̂ [ ] ( ) ̂ + 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 𝑘.

(20.44)

⃗̇ is the sum of Eqs. (20.25) and (20.44), and writing Noting that ℎ 𝑃 ⃗ = 𝑀 𝚤̂ + 𝑀 𝚥̂ + 𝑀 𝑘, ̂ 𝑀 𝑃 𝑃𝑥 𝑃𝑦 𝑃𝑧 𝑟⃗𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂, ̂ 𝑎⃗𝐺 = 𝑎𝐺𝑥 𝚤̂ + 𝑎𝐺𝑦 𝚥̂ + 𝑎𝐺𝑧 𝑘,

(20.45) (20.46) (20.47)

then the three components of Eq. (20.21) become ( ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 , 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 (Ω𝑥 + 𝜔𝑠 )Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 − 𝑚𝑥𝐺∕𝑃 𝑎𝐺𝑧 ,

(20.48)

𝑀𝑃 𝑧 = 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 (Ω𝑥 + 𝜔𝑠 )Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 + 𝑚𝑥𝐺∕𝑃 𝑎𝐺𝑦 .

(20.50)

(20.49)

⃗ then these equations simplify to If 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0, ( ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 , 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 (Ω𝑥 + 𝜔𝑠 )Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 ,

(20.51)

= 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 (Ω𝑥 + 𝜔𝑠 )Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 .

(20.53)

𝑀𝑃 𝑧

Planar motion of bodies not symmetric with respect to the plane of motion

𝑧

𝐴

𝑂

𝛽 𝜔𝑠 𝛼𝑠

𝐺 𝑘̂

𝑥 𝐵

𝚤̂ Figure 20.11 An example of a rigid body undergoing planar motion in which the rigid body is not symmetric with respect to the plane of motion.

ISTUDY

(20.52)

In Chapter 17, we emphasized that the Newton-Euler rotational equations we derived there were for the planar motion of bodies symmetric with respect to the plane of motion. What if we have a rigid body undergoing planar motion, but the body is not symmetric with respect to the plane of motion? The translational equation, i.e., Eq. (20.20), doesn’t change in this case. The rotational or moment equations are given by Eqs. (20.31)–(20.33), but a number of the terms can be eliminated. To see which terms we can eliminate, refer to the system in Fig. 20.11 in which the bar 𝐴𝐵 is rigidly attached to the T-bar support (i.e., the angle 𝛽 is constant), which is, in turn, spinning about the vertical axis with angular velocity 𝜔𝑠 and angular acceleration 𝛼𝑠 . The 𝑥𝑦𝑧 frame has its origin at 𝐺, and it rotates with the arm 𝐴𝐵 and the support.

ISTUDY

Section 20.2

1371

Three-Dimensional Kinetics of Rigid Bodies

• 𝜔𝐵𝑥 , 𝜔𝐵𝑦 , 𝜔̇ 𝐵𝑥 , and 𝜔̇ 𝐵𝑦 are zero since the motion is planar in the 𝑥𝑦 plane. • 𝑎𝐺𝑧 is zero since 𝐺 does not move in the 𝑧 direction. • 𝑧𝐺∕𝑃 is zero since the mass center lies at the origin of the 𝑥𝑦𝑧 frame. With these simplifications, Eqs. (20.31)–(20.33) become

Helpful Information 𝑀𝑃 𝑥 = −𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦𝑧 𝜔2𝐵𝑧 ,

(20.54)

𝑀𝑃 𝑦 = −𝐼𝐺𝑦𝑧 𝜔̇ 𝐵𝑧 − 𝐼𝐺𝑥𝑧 𝜔2𝐵𝑧 , ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 .

(20.55) (20.56)

Equations (20.54)–(20.56) are the most general rotational equations for the planar motion of a rigid body. As is discussed in Appendix C, 𝐼𝐺𝑧 , 𝐼𝐺𝑥𝑧 , and 𝐼𝐺𝑦𝑧 are measures of the distribution of mass of a body with respect to the indicated axes, with 𝐼𝐺𝑥𝑧 and 𝐼𝐺𝑦𝑧 being measures of the symmetry of the body with respect to those axes. • We now see that if a body is not symmetric about the 𝑥𝑦 plane of motion, then moments in the 𝑥 and/or 𝑦 directions will be required to maintain planar motion.

The utility of Eqs. (20.54)–(20.56). While Eqs. (20.54)–(20.56) apply to systems like the one shown in Fig. 20.11, they are usually not the most convenient set of rotational equations to use. The reason is that one needs to compute products of inertia to apply Eqs. (20.54)–(20.56), which is usually a tedious task. We will see in Example 20.6 that using Eqs. (20.34)–(20.36) or Eqs. (20.37)–(20.39) is not conceptually difficult and eliminates the need to calculate products of inertia.

• If the body is symmetric with respect to the 𝑥𝑦 plane of motion, then 𝐼𝐺𝑥𝑧 = 𝐼𝐺𝑦𝑧 = 0, and we are left with ) ( 𝑀𝑃 𝑧 = 𝐼𝐺 𝛼𝐵 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 ,

(20.57)

where we have again let 𝐼𝐺𝑧 = 𝐼𝐺 for the symmetric case and have let 𝜔̇ 𝐵𝑧 = 𝛼𝐺 . This equation is identical to Eq. (17.28) on p. 1148 in Chapter 17. In Example 20.6, we will analyze the bar 𝐴𝐵 in Fig. 20.11 to determine the required forces and moments at 𝐴 for the arm to move as described.

Kinetic energy of a rigid body in three-dimensional motion The work-energy principle for a rigid body is the same whether its motion is twodimensional or three-dimensional. Therefore, the work-energy principle as given by any of Eqs. (18.16) on p. 1209, (18.23) on p. 1211, (18.24) on p. 1211, or (18.29) on p. 1212 is valid for three-dimensional motion. However, the expression for the kinetic energy of a rigid body changes dramatically when we consider three-dimensional motion. Referring to Fig. 20.12, to derive the kinetic energy for a rigid body 𝐵 moving in three-dimensional motion, we begin as we did in Section 18.1 and write the kinetic energy as 𝑇 =

∫𝐵

1 2 𝑣 𝑑𝑚, 2 𝑑𝑚

𝑣⃗𝑑𝑚

𝑑𝑚

𝑞⃗

𝑣⃗𝐺

𝑟⃗𝑑𝑚

𝐺 𝑟⃗𝐺

𝑘̂

𝐵

(20.58) 𝚥̂

where 𝑑𝑚 is an element of mass in the body 𝐵, 𝑣𝑑𝑚 is the speed of the mass element 𝑑𝑚, and the integral is over the entire body 𝐵. Because the mass center 𝐺 and 𝑑𝑚 are two points on the same rigid body, we can write ⃗ 𝐵 × 𝑞, ⃗ 𝑣⃗𝑑𝑚 = 𝑣⃗𝐺 + 𝜔

𝜔 ⃗𝐵

(20.59)

𝑂

𝚤̂

Figure 20.12 The quantities needed for deriving the kinetic energy of a rigid body in three-dimensional motion.

1372

ISTUDY

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

where 𝜔 ⃗ 𝐵 is the angular velocity of 𝐵 and 𝑞⃗ is the position of 𝑑𝑚 relative to 𝐺. To evaluate the integral in Eq. (20.58), we write the vectors in component form as ̂ 𝑣⃗𝐺 = 𝑣𝐺𝑥 𝚤̂ + 𝑣𝐺𝑦 𝚥̂ + 𝑣𝐺𝑧 𝑘,

(20.60)

̂ 𝜔 ⃗ 𝐵 = 𝜔𝐵𝑥 𝚤̂ + 𝜔𝐵𝑦 𝚥̂ + 𝜔𝐵𝑧 𝑘,

(20.61)

̂ 𝑞⃗ = 𝑞𝑥 𝚤̂ + 𝑞𝑦 𝚥̂ + 𝑞𝑧 𝑘.

(20.62)

Substituting Eqs. (20.60)–(20.62) into Eq. (20.59) and noting that 𝑣2𝑑𝑚 = 𝑣⃗𝑑𝑚 ⋅ 𝑣⃗𝑑𝑚 , we find that 𝑣2𝑑𝑚 = 𝑣⃗𝑑𝑚 ⋅ 𝑣⃗𝑑𝑚 =

(

𝑣2𝐺𝑥

)

(

)

(20.63)

+ 𝑣2𝐺𝑦 + 𝑣2𝐺𝑧 + 𝜔2𝐵𝑥 𝑞𝑦2 + 𝑞𝑧2 + 𝜔2𝐵𝑦 𝑞𝑥2 + 𝑞𝑧2 ( ) + 𝜔2𝐵𝑧 𝑞𝑥2 + 𝑞𝑦2 − 2𝑞𝑥 𝑞𝑦 𝜔𝐵𝑥 𝜔𝐵𝑦 − 2𝑞𝑥 𝑞𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧

( ) − 2𝑞𝑦 𝑞𝑧 𝜔𝐵𝑦 𝜔𝐵𝑧 + 2 𝑣𝐺𝑦 𝜔𝐵𝑧 − 𝑣𝐺𝑧 𝜔𝐵𝑦 𝑞𝑥 ( ) ( ) + 2 𝑣𝐺𝑧 𝜔𝐵𝑥 − 𝑣𝐺𝑥 𝜔𝐵𝑧 𝑞𝑦 + 2 𝑣𝐺𝑥 𝜔𝐵𝑦 − 𝑣𝐺𝑦 𝜔𝐵𝑥 𝑞𝑧 .

(20.64)

Noting that 𝑣2𝐺𝑥 + 𝑣2𝐺𝑦 + 𝑣2𝐺𝑧 = 𝑣2𝐺 , Eq. (20.58) becomes [ ( 2 ) ( 2 ) 1 2 𝑞 + 𝑞𝑧2 𝑑𝑚 + 𝜔2𝐵𝑦 𝑞 + 𝑞𝑧2 𝑑𝑚 𝑑𝑚 + 𝜔2𝐵𝑥 𝑇 = 𝑣 ∫𝐵 𝑦 ∫𝐵 𝑥 2 𝐺 ∫𝐵 ( 2 ) + 𝜔2𝐵𝑧 𝑞𝑥 + 𝑞𝑦2 𝑑𝑚 − 2𝜔𝐵𝑥 𝜔𝐵𝑦 𝑞 𝑞 𝑑𝑚 ∫𝐵 ∫𝐵 𝑥 𝑦 − 2𝜔𝐵𝑥 𝜔𝐵𝑧

∫𝐵

𝑞𝑥 𝑞𝑧 𝑑𝑚 − 2𝜔𝐵𝑦 𝜔𝐵𝑧

+ 2(𝑣𝐺𝑦 𝜔𝐵𝑧 − 𝑣𝐺𝑧 𝜔𝐵𝑦 )

∫𝐵

+ 2(𝑣𝐺𝑧 𝜔𝐵𝑥 − 𝑣𝐺𝑥 𝜔𝐵𝑧 )

∫𝐵

+2(𝑣𝐺𝑥 𝜔𝐵𝑦 − 𝑣𝐺𝑦 𝜔𝐵𝑥 )

∫𝐵

∫𝐵

𝑞𝑦 𝑞𝑧 𝑑𝑚

𝑞𝑥 𝑑𝑚 𝑞𝑦 𝑑𝑚 ] 𝑞𝑧 𝑑𝑚 .

(20.65)

The first integral defines the mass of the rigid body. The last three integrals are zero since they measure the position of the mass center relative to the mass center. From Eqs. (C.1)–(C.3) in Appendix C on pp. A-65–A-66, we see that the second, third, and fourth integrals define the mass moments of inertia of the rigid body with respect to the 𝑥𝑦𝑧 axes whose origin is at the mass center of the body; that is, they define 𝐼𝐺𝑥 , 𝐼𝐺𝑦 , and 𝐼𝐺𝑧 , respectively. From Eqs. (C.4)–(C.6) on p. A-66 of Appendix C, the fifth, sixth, and seventh integrals define the mass products of inertia of the rigid body with respect to the indicated axes whose origin is at the mass center of the body; that is, they define 𝐼𝐺𝑥𝑦 , 𝐼𝐺𝑥𝑧 , and 𝐼𝐺𝑦𝑧 , respectively. Substituting these definitions for the mass properties into Eq. (20.65), we obtain the final form of the kinetic energy of a rigid body for three-dimensional motion, 𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺𝑥 𝜔2𝐵𝑥 + 12 𝐼𝐺𝑦 𝜔2𝐵𝑦 + 12 𝐼𝐺𝑧 𝜔2𝐵𝑧 − 𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 𝜔𝐵𝑦 − 𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 𝜔𝐵𝑧 .

(20.66)

End of Section Summary Newton-Euler equations in 3D. There are six Newton-Euler equations for a rigid body in three-dimensional motion. Three are given by the component form of Euler’s

ISTUDY

Section 20.2

1373

Three-Dimensional Kinetics of Rigid Bodies

first law, which is Eq. (20.20), p. 1366 𝐹⃗ = 𝑚𝑎⃗𝐺 , where 𝐹⃗ is the total external force acting on the rigid body, 𝑚 is the mass of the rigid body, and 𝑎⃗𝐺 is the acceleration of the mass center of the rigid body (see Fig. 20.13). The other three equations are given by Euler’s second law. We derived three different versions of these equations, each successive version with more restrictions on their application. The last two versions are the most useful in practice, the first of which is given by Eqs. (20.34)–(20.36), p. 1368 ( ( ) ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 + 𝑚 𝑦𝐺∕𝑃 𝑎𝐺𝑧 − 𝑧𝐺∕𝑃 𝑎𝐺𝑦 , ( ) ( ) 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝑚 𝑧𝐺∕𝑃 𝑎𝐺𝑥 − 𝑥𝐺∕𝑃 𝑎𝐺𝑧 , ( ( ) ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 , where (see Fig. 20.13): • Point 𝑃 is arbitrary, the 𝑥𝑦𝑧 reference frame is attached to the body 𝐵, and 𝑥𝑦𝑧 are principal axes of inertia so that all products of inertia are zero. • The moments of inertia are expressed relative to a set of axes that are parallel to 𝑥𝑦𝑧 and whose origin is at 𝐺, the center of mass of the body. • The angular velocities and their time derivatives are inertial, but are expressed in the rotating 𝑥𝑦𝑧 frame. The final set of equations, also known as Euler’s equations of rotational motion for a rigid body, are given by Eqs. (20.37)–(20.39), p. 1368 ) ( 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 , ( ) 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 , ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 , where, in addition to the conditions for Eqs. (20.34)–(20.36), we also have that: ⃗ which can happen in any of the following three ways: • 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0, ⃗ ⇒ 𝑃 is the mass center 𝐺 of the body 𝐵 so that 𝑟⃗𝐺∕𝑃 = 0, ⃗ that is, the mass center of the body moves with constant velocity, ⇒ 𝑎⃗𝐺 = 0, ⃗ ⇒ 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 so that the cross product 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0. Rotational equations of motion for a body with an axis of radial symmetry. For a rigid body with an axis of radial symmetry, any coordinate system that has an axis coinciding with the axis of symmetry of that body will be a principal coordinate system. Referring to Fig. 20.14, 𝑥𝐵 𝑦𝐵 𝑧𝐵 are principal body axes with the 𝑥𝐵 axis coinciding with the axis of radial symmetry of the body 𝐵. The 𝑥𝑦𝑧 frame rotates

⃗ 𝑀 𝑃

𝑃 𝑟⃗𝐺∕𝑃

𝐹⃗ 𝑘̂ 𝚤̂

𝑎⃗𝐺

𝐺

𝑦

𝚥̂ 𝐵

𝑥

𝜔 ⃗𝐵

Figure 20.13 Illustration of the kinetic and kinematic quantities needed to understand and apply Eqs. (20.20) and (20.31)–(20.39).

1374

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Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

𝑦𝐵

𝜔 ⃗𝑠 𝜔 ⃗𝐵 𝚥̂ 𝑃

𝑘̂

𝐵 𝐺

𝜔 ⃗𝑠

𝚤̂

𝑥, 𝑥𝐵

𝑧 Figure 20.14. A rigid body 𝐵 spinning about its axis of radial symmetry 𝑥𝐵 relative to the ⃗ relative to an 𝑥𝑦𝑧 frame with angular velocity 𝜔 ⃗ 𝑠 . The angular velocity of the 𝑥𝑦𝑧 frame is Ω, underlying inertial frame.

⃗ the angular velocity of the body is 𝜔 with angular velocity Ω, ⃗ 𝐵 , and the body spins relative to the 𝑥𝑦𝑧 frame about its axis of symmetry with angular velocity 𝜔 ⃗ 𝑠 . With these definitions, the rotational equations of motion of the body are

𝑀𝑃 𝑥 𝑀𝑃 𝑦 𝑀𝑃 𝑧

Eqs. (20.48)–(20.50), p. 1370 ( ) = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 − 𝑚𝑥𝐺∕𝑃 𝑎𝐺𝑧 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 + 𝑚𝑥𝐺∕𝑃 𝑎𝐺𝑦 .

⃗ then these equations become If 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0,

𝑀𝑃 𝑥 𝑀𝑃 𝑦 𝑀𝑃 𝑧

Eqs. (20.51)–(20.53), p. 1370 ( ) = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 .

Planar motion of bodies not symmetric with respect to the plane of motion. In Chapter 17, we emphasized that the Newton-Euler rotational equations derived there were for the planar motion of bodies symmetric with respect to the plane of motion. If the rigid body is undergoing planar motion, but the body is not symmetric with respect to the plane of motion, the governing rotational equations are: Eqs. (20.54)–(20.56), p. 1371 𝑀𝑃 𝑥 = −𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦𝑧 𝜔2𝐵𝑧 , 𝑀𝑃 𝑦 = −𝐼𝐺𝑦𝑧 𝜔̇ 𝐵𝑧 − 𝐼𝐺𝑥𝑧 𝜔2𝐵𝑧 , ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 , where the plane of motion is the 𝑥𝑦 plane and the moments and products of inertia are measured with respect to a set of axes whose origin is at the mass center 𝐺. Kinetic energy of a rigid body in 3D. The work-energy principle for a rigid body is the same whether its motion is two-dimensional or three-dimensional. Therefore,

ISTUDY

Section 20.2

Three-Dimensional Kinetics of Rigid Bodies

the work-energy principle, as given by any of Eqs. (18.16) on p. 1209, (18.23) on p. 1211, (18.24) on p. 1211, or (18.29) on p. 1212, is valid for the three-dimensional motion of this chapter. The expression for the kinetic energy of a rigid body in threedimensional motion is: Eq. (20.66), p. 1372 𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺𝑥 𝜔2𝐵𝑥 + 12 𝐼𝐺𝑦 𝜔2𝐵𝑦 + 12 𝐼𝐺𝑧 𝜔2𝐵𝑧 − 𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 𝜔𝐵𝑦 − 𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 𝜔𝐵𝑧 , where the moments and products of inertia are measured with respect to axes whose origin is at the mass center of the rigid body.

1375

1376

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

E X A M P L E 20.4

Gyroscopic Moment on a Spinning Disk Referring to Fig. 1, the T-bar support spins about the line 𝐶𝐷 with constant angular speed 𝜔𝑠 . The disk at 𝐵 spins with constant angular speed 𝜔𝑑 relative to the T-bar. If the mass of the disk is 𝑚𝑑 and the mass of the T-bar is negligible, determine the reactions at the support 𝐴 required for this motion.

𝜔𝑠

SOLUTION 𝐴

𝑅

The FBD of the disk and the arm 𝐴𝐵 is shown in Fig. 2. If we attach a rotating 𝑥𝑦𝑧 frame to the T-bar as shown in Fig. 2, then the disk will spin about an axis of symmetry relative to that frame. Therefore, to find reactions at 𝐴, we can apply Eqs. (20.20) and (20.51)–(20.53) for the force and moment equations. By taking 𝐴 as the moment center, we will obtain the reactions at 𝐴. Road Map & Modeling

𝐵 𝜔𝑑 𝐶

Governing Equations Balance Principles

obtain

Figure 1

∑

𝐴𝑧 𝑀𝐴𝑥

𝑀𝐴𝑧

𝑧

𝐴

𝐴𝑥

𝑀𝐴𝑦

𝐵

∑

𝐴𝑦

𝑦 𝑚𝑑 𝑔 Figure 2 The FBD of the disk and bar 𝐴𝐵 for the system in Fig. 1. The mass of the bar 𝐴𝐵 is negligible.

𝐹𝑥 ∶

𝐴𝑥 = 𝑚𝑑 𝑎𝐵𝑥 ,

(1)

𝐹𝑦 ∶

𝐴𝑦 = 𝑚𝑑 𝑎𝐵𝑦 ,

(2)

𝐹𝑧 ∶

𝐴𝑧 − 𝑚𝑑 𝑔 = 𝑚𝑑 𝑎𝐵𝑧 , ( ) 𝑀𝐴𝑥 ∶ 𝑀𝐴𝑥 = 𝐼𝐵𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑑 , ∑ ( ) 𝑀𝐴𝑦 ∶ 𝑀𝐴𝑦 + 𝑚𝑑 𝑔𝓁 = 𝐼𝐵𝑦 Ω̇ 𝑦 + 𝐼𝐵𝑥 Ω𝑥 + 𝜔𝑑 Ω𝑧 − 𝐼𝐵𝑦 Ω𝑥 Ω𝑧 , ∑ ) ( 𝑀𝐴𝑧 ∶ 𝑀𝐴𝑧 = 𝐼𝐵𝑦 Ω̇ 𝑧 − 𝐼𝐵𝑥 Ω𝑥 + 𝜔𝑑 Ω𝑦 + 𝐼𝐵𝑦 Ω𝑥 Ω𝑦 . ∑

𝑥

ISTUDY

∑

Applying Eqs. (20.20) and (20.51)–(20.53) to the FBD in Fig. 2, we

Force Laws

(3) (4) (5) (6)

All forces are accounted for on the FBD.

Kinematic Equations

Determining the acceleration of 𝐵, we obtain 𝑎⃗𝐵 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵∕𝐴 − 𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 = −𝜔2𝑠 𝓁 𝚤̂,

(7)

⃗ For the angular velocity and angular accelwhere we have used the fact that 𝛼⃗𝐴𝐵 = 0. ⃗ eration components, noting that Ω is the angular velocity of the rotating 𝑥𝑦𝑧 frame, we have Ω𝑥 = Ω𝑦 = 0, Ω𝑧 = 𝜔𝑠 , Ω̇ 𝑥 = Ω̇ 𝑦 = Ω̇ 𝑧 = 0, 𝜔̇ 𝑑 = 0. (8) Computation

Substituting Eqs. (7) and (8) into the right sides of Eqs. (1)–(6), we obtain

𝐴𝑥 = −𝑚𝑑 𝓁𝜔2𝑠 , 𝑀𝐴𝑥 = 0,

𝐴𝑦 = 0, 𝑀𝐴𝑦 = 𝑚𝑑

(1 2

2

)

𝑅 𝜔𝑑 𝜔𝑠 − 𝑔𝓁 ,

𝐴𝑧 = 𝑚𝑑 𝑔, 𝑀𝐴𝑧 = 0,

(9) (10)

where we have used 𝐼𝐵𝑥 = 21 𝑚𝑑 𝑅2 . Discussion & Verification

The dimensions of each result in Eqs. (9) and (10) are as they should be. The force reactions given in Eq. (9) are all as expected, that is, we should expect a reaction opposing the weight of the disk, and we should expect a tension in the bar due to the normal acceleration resulting from the circular path of the mass center of the disk 𝐵 centered at 𝐴. We also see that no moments are required in the 𝑥 and 𝑧 directions to maintain the given motion. Part of the moment in the 𝑦 direction is due to the weight of the disk, but the other part is not intuitive since it is a gyroscopic moment due to the fact that the rotation of the support 𝜔𝑠 is causing the rotation of the disk 𝜔𝑑 to change direction.

ISTUDY

Section 20.2

1377

Three-Dimensional Kinetics of Rigid Bodies

E X A M P L E 20.5

Forces and Moments at the Base of a Rotating Propeller

The airplane shown in Fig. 1 is pitching about the body-fixed 𝑦′ axis. The acceleration of point 𝐶 on the plane is 𝑎⃗𝐶 = 𝑎𝐶𝑥′ 𝚤̂′ + 𝑎𝐶𝑧′ 𝑘̂ ′ , and the constant angular speed of the propeller 𝜃̇ is measured relative to the airplane. Referring to Fig. 2, consider the propeller blade with center of mass at 𝐺. If 𝑚 is the mass of the propeller blade, determine the forces and moments applied to its base 𝐵 required for the blade to undergo the prescribed motion. Assume that the blade is a uniform thin rod of length 𝐿. The 𝑥′ 𝑦′ 𝑧′ frame is attached to the airplane at 𝐶 with the 𝑥′ axis aligned with the axis of rotation of the propeller. Ignore gravity.

pitch 𝜔𝑦

𝑃

𝑦

𝑥 Figure 1 The secondary or moving 𝑥′ 𝑦′ 𝑧′ reference frame is attached to an airplane at 𝐶.

SOLUTION The 𝑥𝑦𝑧 frame is attached to the propeller blade and rotates relative to the frame as shown in Fig. 2. The FBD of the propeller blade is as shown in Fig. 3, where 𝐵𝑥 , 𝐵𝑦 , 𝐵𝑧 are the resultant forces acting on the blade and 𝑀𝐵𝑥 , 𝑀𝐵𝑦 , 𝑀𝐵𝑧 are the resultant moments measured relative to point 𝐵. Since the motion of the propeller is specified, we can write the six Newton-Euler equations to solve for the three forces and three moments applied at 𝐵 necessary to sustain the given motion.

𝜔̇ 𝑦

Road Map & Modeling

𝑥′ 𝑦′ 𝑧′

𝜔𝑦 𝐶

𝑧

𝐵

𝑥

𝑧

𝜃

Governing Equations Balance Principles

Summing forces on the propeller blade, we obtain 𝐵𝑥 = 𝑚𝑎𝐺𝑥 ,

𝐵𝑦 = 𝑚𝑎𝐺𝑦 ,

and 𝐵𝑧 = 𝑚𝑎𝐺𝑧 .

(1)

Since the 𝑥𝑦𝑧 frame is attached to the propeller blade and aligned with its principal axes, we can apply either Eqs. (20.34)–(20.36) or Eqs. (20.37)–(20.39). If we sum moments about 𝐺, we can apply Eqs. (20.37)–(20.39), but then we will have to include the moments due to 𝐵𝑥 and 𝐵𝑦 . If we sum moments about 𝐵, then we need to apply Eqs. (20.34)–(20.36) because the moment center is accelerating. Since we will have to compute 𝑎⃗𝐺 as part of the solution anyway, let’s use Eqs. (20.34)–(20.36), which give ∑ ( ) ( ) 𝑀𝐵𝑥 ∶ 𝑀𝐵𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝑝𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝑝𝑧 𝜔𝑝𝑦 + 𝑚 𝑦𝐺 𝑎𝐺𝑧 − 𝑧𝐺 𝑎𝐺𝑦 , (2) ∑ ( ) ( ) 𝑀𝐵𝑦 ∶ 𝑀𝐵𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝑝𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝑝𝑥 𝜔𝑝𝑧 + 𝑚 𝑧𝐺 𝑎𝐺𝑥 − 𝑥𝐺 𝑎𝐺𝑧 , (3) ∑ ( ) ( ) 𝑀𝐵𝑧 ∶ 𝑀𝐵𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝑝𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝑝𝑦 𝜔𝑝𝑥 + 𝑚 𝑥𝐺 𝑎𝐺𝑦 − 𝑦𝐺 𝑎𝐺𝑥 , (4)

𝑦 𝐵 𝜃̇ Figure 2 A side view of the airplane and a front view of the propeller showing the two moving reference frames 𝑥′ 𝑦′ 𝑧′ and 𝑥𝑦𝑧, as well as the relevant dimensions. The 𝑥𝑦𝑧 frame is attached to the propeller with the 𝑧 axis aligned with the axis of the blade in question and with the 𝑥 and 𝑥′ axes always parallel to one another.

1 𝑚𝐿2 and 𝐼𝐺𝑧 = 0 are computed where the mass moments of inertia 𝐼𝐺𝑥 = 𝐼𝐺𝑦 = 12 ̂ and where 𝜔 relative to the mass center, 𝑟⃗𝐺∕𝐵 = 𝑥𝐺 𝚤̂ + 𝑦𝐺 𝚥̂ + 𝑧𝐺 𝑘, ⃗ 𝑝 and 𝜔 ⃗̇ 𝑝 are the angular velocity and angular acceleration of the propeller, respectively.

Force Laws

𝑦

𝐺

𝑧 𝜃

All forces are accounted for on the FBD.

Kinematic Equations We need to determine the acceleration of 𝐺, the angular velocity and angular acceleration of the propeller, 𝜔 ⃗ 𝑝 and 𝜔 ⃗̇ 𝑝 , respectively, and the vector 𝑟⃗𝐺∕𝐵 . Beginning with the acceleration of 𝐺, we first write ( ) ⃗ plane × 𝜔 ⃗ plane × 𝑟⃗𝐵∕𝐶 𝑎⃗𝐵 = 𝑎⃗𝐶 + 𝛼⃗plane × 𝑟⃗𝐵∕𝐶 + 𝜔 ( ) = 𝑎𝐶𝑥′ 𝚤̂′ + 𝑎𝐶𝑧′ 𝑘̂ ′ + 𝜔̇ 𝑦′ 𝚥̂′ × 𝑑 𝚤̂′ + 𝜔𝑦′ 𝚥̂′ × 𝜔𝑦′ 𝚥̂′ × 𝑑 𝚤̂′ ) ( ) ( (5) = 𝑎𝐶𝑥′ − 𝑑𝜔2𝑦′ 𝚤̂ + 𝑎𝐶𝑧′ − 𝑑 𝜔̇ 𝑦′ 𝑘̂ ′ ,

where 𝛼⃗plane = 𝜔̇ 𝑦′ 𝚥̂′ , the acceleration of 𝐶 was given as 𝑎𝐶𝑥′ 𝚤̂′ +𝑎𝐶𝑧′ 𝑘̂ ′ , and we have used the fact that 𝚤̂ = 𝚤̂′ since the 𝑥 and 𝑥′ axes coincide. We can now relate the acceleration of 𝐺 to that of 𝐵 using ( ) 𝑎⃗𝐺 = 𝑎⃗𝐵 + 𝜔 ⃗̇ 𝑝 × 𝑟⃗𝐺∕𝐵 + 𝜔 ⃗𝑝 × 𝜔 ⃗ 𝑝 × 𝑟⃗𝐺∕𝐵 , (6)

𝐺 𝑀𝐵𝑦 𝐵 𝑀𝐵𝑥 𝐵𝑥

𝐵𝑦

𝑦

𝑀𝐵𝑧 𝐵𝑧

Figure 3 FBD of one of the propeller blades modeled as a uniform thin rod when it is at an arbitrary angle 𝜃 with respect to the 𝑧′ axis.

1378

ISTUDY

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

where 𝑎⃗𝐵 is given by Eq. (5) and the position of 𝐺 relative to 𝐵 is 𝑟⃗𝐺∕𝐵 = 𝑥𝐺 𝚤̂ + 𝑦𝐺 𝚥̂ + 𝑧𝐺 𝑘̂ =

𝐿 2

̂ 𝑘.

(7)

To find 𝜔 ⃗ 𝑝 and 𝜔 ⃗̇ 𝑝 , we note that the angular velocity of the propeller is equal to the sum of the angular velocity of the plane and the angular velocity of the propeller relative to the plane. Therefore, ( ) 𝜔 ⃗𝑝 = 𝜔 ⃗ plane + 𝜔 ⃗ 𝑝∕plane = 𝜔𝑦′ 𝚥̂′ + 𝜃̇ 𝚤̂ = 𝜔𝑦′ cos 𝜃 𝚥̂ − sin 𝜃 𝑘̂ + 𝜃̇ 𝚤̂ ̂ = 𝜃̇ 𝚤̂ + 𝜔𝑦′ cos 𝜃 𝚥̂ − 𝜔𝑦′ sin 𝜃 𝑘,

(8)

̂ To find 𝜔 where we have used 𝚥̂′ = cos 𝜃 𝚥̂ − sin 𝜃 𝑘. ⃗̇ 𝑝 , we differentiate 𝜔 ⃗ 𝑝 to obtain ) ( ) ( 𝜔 ⃗̇ 𝑝 = 𝜔̇ 𝑦′ cos 𝜃 − 𝜔𝑦′ 𝜃̇ sin 𝜃 𝚥̂ − 𝜔̇ 𝑦′ sin 𝜃 + 𝜔𝑦′ 𝜃̇ cos 𝜃 𝑘̂ ( ) ( ) ( ) + 𝜃̇ 𝜔 ⃗ 𝑝 × 𝚤̂ + 𝜔𝑦′ cos 𝜃 𝜔 ⃗ 𝑝 × 𝚥̂ − 𝜔𝑦′ sin 𝜃 𝜔 ⃗ 𝑝 × 𝑘̂ (9) ( ) ( ) = 𝜔̇ 𝑦′ cos 𝜃 − 𝜔𝑦′ 𝜃̇ sin 𝜃 𝚥̂ − 𝜔̇ 𝑦′ sin 𝜃 + 𝜔𝑦′ 𝜃̇ cos 𝜃 𝑘̂ ) ( ) ) ( ( ⃗ 𝑝 × 𝜔𝑦′ sin 𝜃 𝑘̂ (10) ⃗ 𝑝 × 𝜔𝑦′ cos 𝜃 𝚥̂ − 𝜔 + 𝜔 ⃗ 𝑝 × 𝜃̇ 𝚤̂ + 𝜔 ( ) ( ) = 𝜔̇ 𝑦′ cos 𝜃 − 𝜔𝑦′ 𝜃̇ sin 𝜃 𝚥̂ − 𝜔̇ 𝑦′ sin 𝜃 + 𝜔𝑦′ 𝜃̇ cos 𝜃 𝑘̂ ) ( (11) +𝜔 ⃗ 𝑝 × 𝜃̇ 𝚤̂ + 𝜔𝑦′ cos 𝜃 𝚥̂ − 𝜔𝑦′ sin 𝜃 𝑘̂ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 𝜔 ⃗𝑝

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⃗ 𝑝 =0⃗ 𝜔 ⃗𝑝 × 𝜔

) ( ) ( ̂ = 𝜔̇ 𝑦′ cos 𝜃 − 𝜔𝑦′ 𝜃̇ sin 𝜃 𝚥̂ − 𝜔̇ 𝑦′ sin 𝜃 + 𝜔𝑦′ 𝜃̇ cos 𝜃 𝑘,

(12)

where we have used the fact that 𝜃̇ is constant and that the angular velocity of the 𝑥𝑦𝑧 frame is 𝜔 ⃗ 𝑝 . Substituting Eqs. (5), (7), (8), and (12) into Eq. (6), we obtain ) ( ̇ ′ sin 𝜃 𝚤̂ 𝑎⃗𝐺 = 𝑎𝐶𝑥′ − 𝑑𝜔2𝑦′ + 𝐿2 𝜔̇ 𝑦′ cos 𝜃 − 𝐿𝜃𝜔 𝑦 ( ) + 𝑎𝐶𝑧′ − 𝑑 𝜔̇ 𝑦′ − 𝐿2 𝜔2𝑦′ cos 𝜃 sin 𝜃 𝚥̂ ] [( ) ̂ + 𝑎𝐶𝑧′ − 𝑑 𝜔̇ 𝑦′ cos 𝜃 − 𝐿2 𝜃̇ 2 − 𝐿2 𝜔2𝑦′ cos2 𝜃 𝑘. Computation

Substituting Eq. (13) into Eq. (1) gives the forces at 𝐵, which are ( ) ̇ ′ sin 𝜃 , 𝐵𝑥 = 𝑚 𝑎𝐶𝑥′ − 𝑑𝜔2𝑦′ + 𝐿2 𝜔̇ 𝑦′ cos 𝜃 − 𝐿𝜃𝜔 𝑦 ) ( 𝐿 2 𝐵𝑦 = 𝑚 𝑎𝐶𝑧′ − 𝑑 𝜔̇ 𝑦′ − 2 𝜔𝑦′ cos 𝜃 sin 𝜃, ] [( ) 𝐵𝑧 = 𝑚 𝑎𝐶𝑧′ − 𝑑 𝜔̇ 𝑦′ cos 𝜃 − 𝐿2 𝜃̇ 2 − 𝐿2 𝜔2𝑦′ cos2 𝜃 .

(13)

(14) (15) (16)

Substituting Eqs. (8), (12), and (13) into Eqs. (2)–(4), we obtain the following three expressions for the moments at the base of the propeller ) ( 𝑚𝐿 −3𝑎𝐶𝑧′ + 3𝑑 𝜔̇ 𝑦′ + 2𝐿𝜔2𝑦′ cos 𝜃 sin 𝜃, 6 ( ) 𝑚𝐿 ̇ ′ sin 𝜃 , = 3𝑎𝐶𝑥′ − 3𝑑𝜔2𝑦′ + 2𝐿𝜔̇ 𝑦′ cos 𝜃 − 4𝐿𝜃𝜔 𝑦 6 = 0.

𝑀𝐵𝑥 =

(17)

𝑀𝐵𝑦

(18)

𝑀𝐵𝑧

Discussion & Verification

(19)

The dimensions of each term in Eqs. (14)–(19) are as they should be. In addition, we would not expect that a moment about the 𝑧 axis on the propeller would be required, and indeed that moment is zero.

ISTUDY

Section 20.2

E X A M P L E 20.6

Planar Motion of a Rigid Body That Is Not Symmetric with Respect to the Plane of Motion

The bar 𝐴𝐵 is rigidly attached to the T-bar support (i.e., the angle 𝛽 is constant), which is, in turn, spinning about the vertical axis with angular velocity 𝜔 ⃗ 𝑠 and angular acceleration 𝛼⃗𝑠 as shown. The 𝑥𝑦𝑧 frame has its origin at 𝐺, and it is attached to the arm 𝐴𝐵. Assuming that the arm 𝐴𝐵 lies in the vertical plane, determine the forces and moments exerted by the T-bar onto the bar 𝐴𝐵 at 𝐴 required for this motion, and express them in the 𝑥′ 𝑦′ 𝑧′ frame that is attached to the T-bar with its origin at 𝐴.

𝐶 𝑧 𝐴

𝑂

𝑥

𝛽

𝑧

𝜔𝑠

SOLUTION

Governing Equations

Summing forces in the three coordinate directions in Fig. 2, we

obtain ∑ ∑ ∑

𝐹𝑥 ∶

𝐴𝑥 + 𝑚𝑔 sin 𝛽 = 𝑚𝑎𝐺𝑥 ,

(1)

𝐹𝑦 ∶

𝐴𝑦 = 𝑚𝑎𝐺𝑦 ,

(2)

𝐹𝑧 ∶

𝐴𝑧 − 𝑚𝑔 cos 𝛽 = 𝑚𝑎𝐺𝑧 ,

(3)

where 𝑎𝐺𝑥 , 𝑎𝐺𝑦 , and 𝑎𝐺𝑧 are the components of 𝑎⃗𝐺 in the indicated coordinate directions. Applying Eqs. (20.34)–(20.36), we can sum moments about 𝐴 to obtain ∑ ) ( 𝑀𝐴𝑥 ∶ 𝑀𝐴𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐴𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐴𝐵𝑧 𝜔𝐴𝐵𝑦 ) ( (4) + 𝑚 𝑦𝐺∕𝐴 𝑎𝐺𝑧 − 𝑧𝐺∕𝐴 𝑎𝐺𝑦 , ∑ ( ) 𝐿 𝑀𝐴𝑦 ∶ 𝑀𝐴𝑦 + 𝑚𝑔 2 cos 𝛽 = 𝐼𝐺𝑦 𝜔̇ 𝐴𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐴𝐵𝑥 𝜔𝐴𝐵𝑧 ( ) (5) + 𝑚 𝑧𝐺∕𝐴 𝑎𝐺𝑥 − 𝑥𝐺∕𝐴 𝑎𝐺𝑧 , ∑ ( ) 𝑀𝐴𝑧 ∶ 𝑀𝐴𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐴𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐴𝐵𝑦 𝜔𝐴𝐵𝑥 ( ) + 𝑚 𝑥𝐺∕𝐴 𝑎𝐺𝑦 − 𝑦𝐺∕𝐴 𝑎𝐺𝑥 , (6) where all moments of inertia are computed with respect to the mass center 𝐺, 𝜔 ⃗ 𝐴𝐵 is the ̇ angular velocity of the bar 𝐴𝐵, and 𝜔 ⃗ 𝐴𝐵 is the angular acceleration of bar 𝐴𝐵. The mass moments of inertia are given by 𝐼𝐺𝑥 = 0 and 𝐼𝐺𝑦 = 𝐼𝐺𝑧 = Force Laws

1 𝑚𝐿2 . 12

(7)

All forces are accounted for on the FBD.

Kinematic Equations

The components of the position of 𝐺 relative to 𝐴 are 𝑥𝐺∕𝐴 = 𝐿∕2 and 𝑦𝐺∕𝐴 = 𝑧𝐺∕𝐴 = 0.

(8)

Since 𝐺 is moving in a circle centered on the vertical line 𝐶𝐷, its acceleration is readily found to be ( ) ) ( (9) 𝑎⃗𝐺 = − 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 𝚤̂′ + 𝑑 + 𝐿2 cos 𝛽 𝛼𝑠 𝚥̂′ ) ( ( ) 2( ) 𝐿 𝐿 = − 𝑑 + 2 cos 𝛽 𝜔𝑠 cos 𝛽 𝚤̂ + sin 𝛽 𝑘̂ + 𝑑 + 2 cos 𝛽 𝛼𝑠 𝚥̂ (10) ) ( )( 2 𝐿 2 (11) = − 𝑑 + 2 cos 𝛽 𝜔𝑠 cos 𝛽 𝚤̂ − 𝛼𝑠 𝚥̂ + 𝜔𝑠 sin 𝛽 𝑘̂ .

𝐺

𝛼𝑠

An FBD of the arm 𝐴𝐵 is shown in Fig. 2. The motion of the bar 𝐴𝐵 is specified, so we can apply the six Newton-Euler equations to the bar to determine the six unknown forces and moments at 𝐴 required to sustain this motion. Road Map & Modeling

Balance Principles

1379

Three-Dimensional Kinetics of Rigid Bodies

𝐵

𝑔

𝑥 Figure 1 A rigid body undergoing planar motion in which the rigid body is not symmetric with respect to the plane of motion.

𝑀𝐴𝑧

𝑀𝐴𝑥

𝑀𝐴𝑦

𝐴 𝐿 2

𝐴𝑦

𝑧

𝛽

𝐺 𝑚𝑔

𝐿 2

𝐵 Figure 2 FBD of the bar 𝐴𝐵 in Fig. 1.

𝑦

1380

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

The angular velocity of the bar is given to be ( ) 𝜔 ⃗ 𝐴𝐵 = 𝜔𝑠 𝑘̂ ′ = 𝜔𝑠 −sin 𝛽 𝚤̂ + cos 𝛽 𝑘̂ ,

(12)

and the angular acceleration of the bar is

( ) 𝜔 ⃗̇ 𝐴𝐵 = 𝛼𝑠 𝑘̂ ′ = 𝛼𝑠 −sin 𝛽 𝚤̂ + cos 𝛽 𝑘̂ .

Computation

Substituting Eq. (11) into Eqs. (1)–(3) and solving for the forces at 𝐴, we

obtain 𝑀𝐴𝑧

) ( 𝐴𝑥 = −𝑚𝑔 sin 𝛽 − 𝑚 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 cos 𝛽, ( ) 𝐴𝑦 = 𝑚 𝑑 + 𝐿2 cos 𝛽 𝛼𝑠 , ) ( 𝐴𝑧 = 𝑚𝑔 cos 𝛽 − 𝑚 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 sin 𝛽.

𝐴𝑧 𝑀𝐴𝑧

𝛽

Referring to Fig. 3, these forces can be written in terms of the 𝐴𝑥

𝐴 𝛽

𝑀𝐴𝑥 Figure 3

ISTUDY

𝑀𝐴𝑥

(13)

(14) (15) (16)

𝑥′ 𝑦 ′ 𝑧′

directions as

𝐴𝑥′ = 𝐴𝑥 cos 𝛽 + 𝐴𝑧 sin 𝛽 [ ] ( ) = −𝑚𝑔 sin 𝛽 − 𝑚 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 cos 𝛽 cos 𝛽 ] [ ) ( + 𝑚𝑔 cos 𝛽 − 𝑚 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 sin 𝛽 sin 𝛽 ( = −𝑚 𝑑 +

𝐿 2

)

cos 𝛽 𝜔2𝑠

( 𝐴𝑦′ = 𝐴𝑦 = 𝑚 𝑑 +

𝐿 2

( 𝐴𝑥′ = −𝑚 𝑑 +

⇒

) cos 𝛽 𝛼𝑠

⇒

(17)

(18)

) 2 𝐿 cos 𝛽 𝜔𝑠 , 2

( 𝐴𝑦′ = 𝑚 𝑑 +

𝐿 2

) cos 𝛽 𝛼𝑠 ,

𝐴𝑧′ = −𝐴𝑥 sin 𝛽 + 𝐴𝑧 cos 𝛽 ] [ ) ( = − −𝑚𝑔 sin 𝛽 − 𝑚 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 cos 𝛽 sin 𝛽 [ ] ( ) + 𝑚𝑔 cos 𝛽 − 𝑚 𝑑 + 𝐿2 cos 𝛽 𝜔2𝑠 sin 𝛽 cos 𝛽 = 𝑚𝑔

(20) (21)

(22)

𝐴𝑧′ = 𝑚𝑔.

⇒

(19)

(23)

Substituting Eqs. (7), (8), and (11)–(13) into Eqs. (4)–(6), and then solving for 𝑀𝐴𝑥 , 𝑀𝐴𝑦 , and 𝑀𝐴𝑧 , we obtain 𝑀𝐴𝑥 = 0, 𝑀𝐴𝑦 = 𝑀𝐴𝑧 =

[ 1 𝑚𝐿 −3𝑔 cos 𝛽 + 𝜔2𝑠 (3𝑑 6 1 𝑚𝐿(3𝑑 + 2𝐿 cos 𝛽) 𝛼𝑠 . 6

]

+ 2𝐿 cos 𝛽) sin 𝛽 ,

(24) (25) (26)

Substituting Eqs. (24)–(26) into the relationships relating the moments in the two different reference frames, 𝑀𝐴𝑥′ = 𝑀𝐴𝑥 cos 𝛽 + 𝑀𝐴𝑧 sin 𝛽, 𝑀𝐴𝑦′ = 𝑀𝐴𝑦 , and 𝑀𝐴𝑧′ = −𝑀𝐴𝑧 sin 𝛽 + 𝑀𝐴𝑧 cos 𝛽, gives 𝑀𝐴𝑥′ = 16 𝑚𝐿(3𝑑 + 2𝐿 cos 𝛽) 𝛼𝑠 sin 𝛽, [ ] 𝑀𝐴𝑦′ = 16 𝑚𝐿 −3𝑔 cos 𝛽 + 𝜔2𝑠 (3𝑑 + 2𝐿 cos 𝛽) sin 𝛽 ,

(27)

𝑀𝐴𝑧′ = 16 𝑚𝐿(3𝑑 + 2𝐿 cos 𝛽) 𝛼𝑠 cos 𝛽.

(29)

Discussion & Verification

(28)

The final results for the forces and moments are all dimensionally as we expect them to be. In addition, given the motion of the bar 𝐴𝐵, the directions of the forces are as expected. As for the moments, we would expect a negative moment in the 𝑦′ direction due to the weight of the bar, and that is what we found. In addition, notice that there is a moment in the positive 𝑦′ direction that is proportional to the square of the angular velocity. This moment results from the tendency of the bar 𝐴𝐵 to rotate upward when it is spinning. To keep the angle 𝛽 constant, a moment needs to be applied in the positive 𝑦′ direction.

ISTUDY

Section 20.2

Three-Dimensional Kinetics of Rigid Bodies

E X A M P L E 20.7

1381

Computing the Kinetic Energy of a Bar Moving in 3D

At the instant shown, collar 𝐴 is moving downward with speed 𝑣𝐴 . The uniform thin bar 𝐴𝐵 has mass 𝑚 and is attached to collars 𝐴 and 𝐵 with ball-and-socket joints. Neglecting the dimensions of the collars, determine the kinetic energy of bar 𝐴𝐵 at this instant. Assume that the angular velocity of the rod 𝐴𝐵 is orthogonal to the line 𝐴𝐵.

𝐶 𝑣𝐴

𝐴

SOLUTION Road Map & Modeling

The kinetic energy of the bar can be computed using Eq. (20.66). Since we would like to avoid computing products of inertia, we will attach a set of principal body axes to the bar 𝐴𝐵 as shown in Fig. 2, with the 𝑥 axis aligned with the bar and the 𝑦 axis perpendicular to the plane defined by the bar 𝐴𝐵 and the bar 𝐶𝐷.

𝐷

𝐸

𝑋

Governing Equations Balance Principles

Using the principal axes shown in Fig. 2, the kinetic energy of the

bar is given by

𝐹

𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺𝑥 𝜔2𝐴𝐵𝑥 + 21 𝐼𝐺𝑦 𝜔2𝐴𝐵𝑦 + 21 𝐼𝐺𝑧 𝜔2𝐴𝐵𝑧 , where 𝐼𝐺𝑥 = 0, 𝐼𝐺𝑦 = 𝐼𝐺𝑧 =

(1) Figure 1

1 𝑚𝐿2 , 12

and 𝐿 is the length of the bar 𝐴𝐵, which is given by √ (2) 𝐿 = 𝑑 2 + ℎ2 + 𝓁 2 .

𝐶

𝑥 Force Laws

𝑌

𝐵

We are not applying the work-energy principle here, so there are no force

laws.

𝐴

Kinematic Equations

Referring to Eq. (1), we need to determine the speed of the mass center of the bar 𝐴𝐵, as well as the components of its angular velocity expressed in the 𝑥𝑦𝑧 frame. Starting with the angular velocity, we can relate the velocity of 𝐵 to that of 𝐴 using 𝑣⃗𝐵 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐵∕𝐴 , ( ) ( ) 𝑣𝐵 𝐼̂ = −𝑣𝐴 𝐾̂ + 𝜔𝐴𝐵𝑋 𝐼̂ + 𝜔𝐴𝐵𝑌 𝐽̂ + 𝜔𝐴𝐵𝑍 𝐾̂ × 𝓁 𝐼̂ + 𝑑 𝐽̂ − ℎ 𝐾̂ .

𝑧 𝑦

(3) (4)

𝐺 𝐷

𝐸

𝑋 𝐵

𝑌

Expanding and equating coefficients, Eq. (4) becomes 𝑣𝐵 = −ℎ𝜔𝐴𝐵𝑌 − 𝑑𝜔𝐴𝐵𝑍 ,

(5)

0 = ℎ𝜔𝐴𝐵𝑋 + 𝓁𝜔𝐴𝐵𝑍 ,

(6)

0 = −𝑣𝐴 + 𝑑𝜔𝐴𝐵𝑋 − 𝓁𝜔𝐴𝐵𝑌 .

(7)

Implementing the constraint that the angular velocity of the rod 𝐴𝐵 be perpendicular to 𝐴𝐵, we obtain 𝜔 ⃗ 𝐴𝐵 ⋅ 𝑢̂ 𝐵∕𝐴 = 0

⇒

⇒

( ) 𝑟⃗𝐵∕𝐴 𝜔𝐴𝐵𝑋 𝐼̂ + 𝜔𝐴𝐵𝑌 𝐽̂ + 𝜔𝐴𝐵𝑍 𝐾̂ ⋅ =0 |⃗𝑟𝐵∕𝐴 | ( ) ( ) 𝓁 𝐼̂ + 𝑑 𝐽̂ − ℎ 𝐾̂ ̂ ̂ ̂ 𝜔𝐴𝐵𝑋 𝐼 + 𝜔𝐴𝐵𝑌 𝐽 + 𝜔𝐴𝐵𝑍 𝐾 ⋅ = 0, 𝐿

(8)

which, upon expanding and multiplying through by 𝐿, becomes 𝓁𝜔𝐴𝐵𝑋 + 𝑑𝜔𝐴𝐵𝑌 − ℎ𝜔𝐴𝐵𝑍 = 0.

(9)

Solving Eqs. (5)–(7) and Eq. (9) for 𝜔𝐴𝐵𝑋 , 𝜔𝐴𝐵𝑌 , 𝜔𝐴𝐵𝑍 , and 𝑣𝐵 , we obtain 𝜔𝐴𝐵𝑋 =

𝑑 𝑣 , 𝐿2 𝐴

𝜔𝐴𝐵𝑌 = −

ℎ2 + 𝓁 2 𝑣𝐴 , 𝐿2 𝓁

𝜔𝐴𝐵𝑍 = −

𝑑ℎ 𝑣 , 𝐿2 𝓁 𝐴

(10)

𝐹 Figure 2 The system in Fig. 1 showing the body-fixed 𝑥𝑦𝑧 frame and the plane (in light blue) to which the 𝑦 axis is perpendicular.

1382

ISTUDY

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

and 𝑣𝐵 = ℎ𝑣𝐴 ∕𝓁. We now need to transform the components of the angular velocity vector from the 𝑋𝑌𝑍 frame to the 𝑥𝑦𝑧 frame. Referring to Fig. 2, we see that ) −1 ( ̂ 𝓁 𝐼 + 𝑑 𝐽̂ − ℎ 𝐾̂ , 𝐿 𝐾̂ × 𝑢̂ 𝐴∕𝐵 𝑑 𝐼̂ − 𝓁 𝐽̂ 𝚥̂ = = , |̂ | √ 2 |𝐾 × 𝑢̂ 𝐴∕𝐵 | 𝑑 + 𝓁2 | | [ ( ) ] ̂𝑘 = 𝚤̂ × 𝚥̂ = √ 1 ℎ𝓁 𝐼̂ + 𝑑ℎ 𝐽̂ + 𝑑 2 + 𝓁 2 𝐾̂ . 𝐿 𝑑 2 + 𝓁2 𝚤̂ = −𝑢̂ 𝐵∕𝐴 = 𝑢̂ 𝐴∕𝐵 =

Common Pitfall An easier way to compute 𝑻 ? The special condition we applied to the angular velocity of the bar is what led to the result that 𝜔𝐴𝐵𝑥 = 0, as it should since 𝑥 is aligned with the axis of the bar. This means that the angular velocity of the bar must be perpendicular to it at every instant, and so it is like the bar is in planar motion at every instant. This means that we could have computed the kinetic energy of the bar using the equation

(12) (13)

We can now find each of the three components of 𝜔 ⃗ 𝐴𝐵 expressed in the 𝑥𝑦𝑧 frame by dotting it with each of the three unit vectors in the 𝑥𝑦𝑧 frame, that is, 𝜔𝐴𝐵𝑥 = 𝜔 ⃗ 𝐴𝐵 ⋅ 𝚤̂ ) ( ) −1 ( 𝓁 𝐼̂ + 𝑑 𝐽̂ − ℎ 𝐾̂ = 0, = 𝜔𝐴𝐵𝑋 𝐼̂ + 𝜔𝐴𝐵𝑌 𝐽̂ + 𝜔𝐴𝐵𝑍 𝐾̂ ⋅ 𝐿 𝜔𝐴𝐵𝑦 = 𝜔 ⃗ 𝐴𝐵 ⋅ 𝚥̂

𝜔𝐴𝐵𝑧

𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺 𝜔2𝐴𝐵 , where 𝐼𝐺 is the moment of inertia of the bar about any axis perpendicular to the bar through 𝐺 and 𝜔𝐴𝐵 is the angular speed of the bar. While that would have been much easier, it is a very special case, and it is better to learn how to compute 𝑇 the way we did here so that we can do it in more general situations.

(11)

𝑣 ( ) 𝑑 𝐼̂ − 𝓁 𝐽̂ , = √ 𝐴 = 𝜔𝐴𝐵𝑋 𝐼̂ + 𝜔𝐴𝐵𝑌 𝐽̂ + 𝜔𝐴𝐵𝑍 𝐾̂ ⋅ √ 𝑑 2 + 𝓁2 𝐿 2 − ℎ2 =𝜔 ⃗ 𝐴𝐵 ⋅ 𝑘̂ [ ( ) 1 ℎ𝓁 𝐼̂ = 𝜔𝐴𝐵𝑋 𝐼̂ + 𝜔𝐴𝐵𝑌 𝐽̂ + 𝜔𝐴𝐵𝑍 𝐾̂ ⋅ √ 𝐿 𝑑 2 + 𝓁2 ( ) ] + 𝑑ℎ 𝐽̂ + 𝑑 2 + 𝓁 2 𝐾̂ =−

𝑑ℎ𝑣𝐴 , √ 𝐿𝓁 𝑑 2 + 𝓁 2

(14) (15) (16) (17) (18)

(19) (20)

where we have substituted in Eq. (10) to get the final results. Now that we have the angular velocity components in the 𝑥𝑦𝑧 frame, all that is left is to find 𝑣𝐺 . This can now easily be done using 𝑣⃗𝐺 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐴𝐵 × 𝑟⃗𝐺∕𝐴 [ ] ( 2 ) ℎ + 𝓁2 𝑣𝐴 ( ) 𝑑ℎ ̂ ̂ ̂ ̂ = −𝑣𝐴 𝐾 + 2 𝑑 𝐼 − 𝐽− 𝐾 × 21 𝓁 𝐼̂ + 𝑑 𝐽̂ − ℎ 𝐾̂ 𝓁 𝓁 𝐿 =

ℎ 1 ̂ 𝑣 𝐼̂ − 𝑣𝐴 𝐾, 2𝓁 𝐴 2

(21) (22) (23)

and so 𝑣2𝐺 = 𝑣⃗𝐺 ⋅ 𝑣⃗𝐺 =

ℎ2 + 𝓁 2 2 𝑣𝐴 . 4𝓁 2

(24)

Computation

Substituting the moments of inertia, Eqs. (15), (17), (20), and (24) into Eq. (1) and simplifying, we obtain 𝑇 = where we have used the 𝐿 =

Discussion & Verification

√

) ( 2 ℎ + 𝓁2 6𝓁 2

𝑚𝑣2𝐴 ,

𝑑 2 + ℎ2 + 𝓁 2 .

The dimensions of the kinetic energy are as expected. We see that the kinetic energy depends on the square of the only speed in the problem 𝑣𝐴 , so that seems reasonable.

ISTUDY

Section 20.2

1383

Three-Dimensional Kinetics of Rigid Bodies

Problems Problems 20.36 and 20.37 The angled bar 𝐶𝐷𝐸 is rigidly attached to the horizontal shaft 𝐴𝐵, which can rotate freely in the bearings at 𝐴 and 𝐵. The system is released from rest when the segment 𝐷𝐸 is vertical. Segment 𝐶𝐷 has mass 𝑚 and length 𝐿. Segments 𝐶𝐷 and 𝐷𝐸 have the same linear density. Problem 20.36

𝑦

𝑚, 𝐿

𝑥 𝐵

𝜃

𝐸

Determine expressions for the angular velocity of the system when it

has rotated 180◦ .

𝐶

𝐴

Problem 20.37 Determine expressions for the reactions at the bearings 𝐴 and 𝐵 when the system has rotated 90◦ .

Figure P20.36 and P20.37

Problem 20.38 The system is at rest when a time-dependent moment 𝑀(𝑡) = 3𝑡3∕2 N⋅m is applied to the shaft 𝐴𝐵 starting at 𝑡 = 0. If the mass of the plate is 𝑚 = 10 kg, its width 𝑤 = 0.5 m, and its height ℎ = 0.25 m, determine the angular speed of the system after 10 s. Neglect the mass of the horizontal shaft 𝐴𝐵.

𝑌

𝑤

ℎ 𝐴

𝐺

𝐵 𝑋

Figure P20.38

Problems 20.39 and 20.40 The uniform bar 𝐴𝐵 of length 𝐿 and mass 𝑚 is attached to the T-bar support by a frictionless pin at 𝐴. The 𝑥′ 𝑦′ 𝑧′ frame is attached to the T-bar at 𝐴 and is aligned as shown. The support rotates with angular speed 𝜔𝑠 , and the angle between the bar 𝐴𝐵 and the 𝑥′ axis is 𝛽. Find the equation of motion of the bar 𝐴𝐵 in terms of the angle 𝛽 for the case where 𝜔𝑠 is constant.

Problem 20.39

Find the equation of motion of the bar 𝐴𝐵 in terms of the angle 𝛽 for the case where 𝜔𝑠 is not constant, that is, 𝜔̇ 𝑠 = 𝛼𝑠 .

Problem 20.40

𝐶 𝑧 𝐴

𝑂 𝜔𝑠

The uniform bar 𝐴𝐵 of length 𝐿 and mass 𝑚 is attached to the T-bar support by a frictionless pin at 𝐴. The mass of the T-bar is negligible, and it rotates freely in the bearings at 𝐶 and 𝐷. The 𝑥′ 𝑦′ 𝑧′ frame is attached to the T-bar at 𝐴 and is aligned as shown. The angle between the bar 𝐴𝐵 and the 𝑥′ axis is 𝛽, and the angle measuring the orientation of the T-bar is 𝜙. Determine the equations of motion of the system in terms of the angles 𝛽 and 𝜙.

𝑧 𝐺

𝛼𝑠 𝑔 Figure P20.39 and P20.40

Problem 20.41

𝑥

𝛽

𝐵

1384

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

𝑧 𝐴

𝑂

𝛽 𝜙̇

𝜔𝑠

𝑥 𝑧 𝐺

𝑂

drum

𝜙̈

𝑟 𝑔

Figure P20.41

𝐵

𝜔𝑑 𝑥

Figure P20.42

Problem 20.42 The uniform drum of length 𝐿, radius 𝑟, and mass 𝑚 is spinning about its axis of symmetry with constant angular speed 𝜔𝑑 relative to the T-bar support. The T-bar support is rotating with constant angular speed 𝜔𝑠 about the vertical 𝑧 axis. Neglecting the mass of the horizontal support of length ℎ, determine the forces and moments at 𝑂 acting on the horizontal bar required to sustain this motion.

𝑧 𝐴

𝑂

Problem 20.43

𝑥

𝛽

𝑧

𝜔𝑠

The uniform bar 𝐴𝐵 of length 𝐿 and mass 𝑚 is attached to the T-bar support by a frictionless pin at 𝐴. The 𝑥′ 𝑦′ 𝑧′ frame is attached to the T-bar at 𝐴 and is aligned as shown. The support rotates with angular speed 𝜔𝑠 , and the angle between the bar 𝐴𝐵 and the 𝑥′ axis is a known function of time 𝛽(𝑡). Determine the kinetic energy of the bar 𝐴𝐵.

𝐺 𝐵

Problems 20.44 through 20.47 Figure P20.43

When ℎ1 = 0.6 m, 𝓁1 = 0.5 m, and 𝑑 = 0.9 m, collar 𝐴 is moving downward at speed 𝑣𝐴1 = 3 m∕s. The uniform thin bar 𝐴𝐵 has mass 𝑚𝐴𝐵 = 4 kg and length 𝐿 and is attached to collars 𝐴 and 𝐵 with ball-and-socket joints. Collars 𝐴 and 𝐵 slide smoothly on rods 𝐶𝐷 and 𝐸𝐹 , respectively, and collar 𝐵 is attached to the stop at 𝐸 by a linear elastic spring with constant 𝑘 = 200 N∕m and unstretched length 2𝓁1 . Neglect the dimensions of the collars and assume that the angular velocity of the rod 𝐴𝐵 is such that 𝐴𝐵 does not spin about its axis.

𝐶 𝑣𝐴

𝐴

Problem 20.44 Assuming that the spring is absent and that the masses of the collars at 𝐴 and 𝐵 are negligible, determine the speed of the collar 𝐴 when it reaches 𝐷.

𝐿 𝐷 𝑋

𝑘 𝐵 𝐹

Figure P20.44–P20.47

ISTUDY

Assuming that the masses of the collars at 𝐴 and 𝐵 are negligible, determine the speed of the collar 𝐴 when it reaches 𝐷.

Problem 20.45

𝐸 𝑌

Assuming that the spring is absent and that the mass of collar 𝐴 is 𝑚𝐴 = 0.5 kg and the mass of collar 𝐵 is 𝑚𝐵 = 0.5 kg, determine the speed of the collar 𝐴 when it reaches 𝐷.

Problem 20.46

Problem 20.47 Assuming that the mass of collar 𝐴 is 𝑚𝐴 = 0.5 kg and the mass of collar 𝐵 is 𝑚𝐵 = 0.5 kg, determine the speed of the collar 𝐴 when it reaches 𝐷.

Problem 20.48 The thin disk of mass 𝑚 and radius 𝑅 rotates with constant angular speed 𝜓̇ about the pin at 𝐴. The pin itself twists with the light shaft 𝑂𝐴 of length 𝐿 at a constant rate 𝜃̇

ISTUDY

Section 20.2

1385

Three-Dimensional Kinetics of Rigid Bodies

about the horizontal 𝑦 axis. In turn, the mechanism at 𝑂 allows the horizontal bar to be ̇ All motion occurs with driven about the vertical axis 𝐵𝐶 at a constant precession rate 𝜙. negligible friction. Determine the total moment exerted on the disk at 𝐴, and express it in the given rotating reference frame whose origin is at 𝐴 and that is attached to the bar 𝑂𝐴. The position shown (when the disk is vertically aligned) corresponds to 𝜃 = 0.

𝐵 𝜙̇

𝑚

𝜃 𝑅

𝑂

𝐴

𝑦 𝜃̇

𝐴

𝐺

𝑅

𝑅

𝜔𝑠 𝑥

𝑋

𝜓̇

𝐶

Figure P20.48

Figure P20.49

Problem 20.49 The thin uniform disk of radius 𝑅 and mass 𝑚 is mounted on the horizontal shaft, such that the mass center of the disk is on the axis of rotation. Due to an error in manufacturing, the disk has a misalignment angle 𝜃 relative to the shaft, such that the position shown occurs only once for each revolution of the shaft. If the shaft is rotating with a constant angular speed 𝜔𝑠 , determine the reactions at the bearing 𝐴 on the shaft. 𝜃

Problem 20.50 𝐴

The thin uniform disk of radius 𝑅 and mass 𝑚 is mounted on the horizontal shaft, such that the mass center of the disk is on the axis of rotation. Due to an error in manufacturing, the disk has a misalignment angle 𝜃 relative to the shaft, such that the position shown occurs only once for each revolution of the shaft. If the shaft is rotating with a constant angular speed 𝜔𝑠 , determine the reactions at the bearings 𝐴 and 𝐵 on the shaft.

𝐺

𝑅

𝐵

𝜔𝑠

𝑋

𝑥

𝑅

Figure P20.50

Problems 20.51 and 20.52 Unlike the rotor shown in Example 17.6, centrifuge rotors that spin at very high speeds (> 100,000 rpm) don’t have swinging buckets to hold the test tubes. For these high-speed rotors, the buckets are fixed at an oblique angle. With this in mind, even if we model the test tube as a uniform circular cylinder, the tube itself is no longer symmetric with respect to the plane of motion, and so the products of inertia in Eqs. (20.54) and (20.55) are not all zero. For this fixed-angle rotor, assume a maximum angular velocity of 130,000 rpm, and assume that the angle of each test tube is fixed at 𝜃 = 35◦ relative to the spin axis of the rotor. Model the test tube as a uniform circular cylinder of diameter 𝑑 = 11 mm, so that the mass center is at the midpoint between 𝑟𝑜 = 120.5 mm and 𝑟𝑖 = 63.1 mm. In addition, use the following inertia properties: 𝑚 = 10 g, (𝐼𝑧𝑧 )𝐶 = 2.87×10−6 kg⋅m2 , (𝐼𝐺𝑥𝑧 )𝐶 = −3.89×10−6 kg⋅m2 , and (𝐼𝐺𝑦𝑧 )𝐶 = 0.00 kg⋅m2 . Problem 20.51

Assuming that the rotor has reached its terminal speed of 130,000 rpm, use Eqs. (20.54)–(20.56) to determine the net forces and moments acting at the mass center of the test tube to sustain this motion. Problem 20.52

Assume that the rotor accelerates uniformly from rest and takes 9.5 min to reach its terminal speed of 130,000 rpm. Determine, as a function of time, the net

fixed-angle centrifuge rotor

cross section of half the rotor

𝑟𝑖

𝜃 𝐺 𝑘̂ 𝚤̂ 𝑟𝑜

𝑑

axis of rotation Rattiya Thongdumhyu/Shutterstock

Figure P20.51 and P20.52

1386

ISTUDY

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

forces and moments acting at the mass center of the test tube to sustain this motion. Use Eqs. (20.54)–(20.56) for your solution.

Problem 20.53 The horizontal shaft 𝐴𝐵 is spinning with a constant angular speed 𝜔𝑠 in the direction shown. If the mass of the uniform rectangular plate is 𝑚 and it has the dimensions shown, determine the reactions on the shaft 𝐴𝐵 due to the bearings at 𝐴 and 𝐵. Neglect the mass of the shaft 𝐴𝐵. 𝑌

𝐿

𝐿 2

𝐴

𝐵

𝐺

𝜔𝑠

𝑋

Figure P20.53

Problem 20.54 The L-shaped bar 𝑂𝐶𝐷 is pin-connected at 𝑂 to the vertical bar 𝐴𝐵. The segments 𝑂𝐶 and 𝐶𝐷 are uniform, and each has mass 𝑚 and length 𝐿. The bar 𝐴𝐵 rotates about its own axis at the constant speed 𝜔𝑠 . Determine the angular speed 𝜔𝑠 required to keep the L-shaped bar in the position shown. 𝜔𝑠

𝜔𝑠 𝐵

𝐵 𝐿

𝑂

𝐶

𝐿

𝑂

𝐶 𝑃

𝐸

ℎ

𝐹

𝐿

𝐴

𝐷

Figure P20.54

𝐿

𝐴

𝐷

Figure P20.55

Problem 20.55 The L-shaped bar 𝑂𝐶𝐷 is pin-connected at 𝑂 to the vertical bar 𝐴𝐵. The segments 𝑂𝐶 and 𝐶𝐷 are uniform, and each has mass 𝑚 and length 𝐿. The bar 𝐴𝐵 rotates about its own axis at the constant speed 𝜔𝑠 . The horizontal bar 𝐸𝐹 is rigidly attached to the bar 𝐴𝐵 at 𝐸, and there is a string attaching the bar 𝐸𝐹 to the bar 𝑂𝐶 at a distance ℎ from the spin axis. (a) Determine the angular speed 𝜔𝑠 required to keep the L-shaped bar in the position shown, such that there is zero tension in the string 𝐹𝑃 . (b) For angular speeds 𝜔𝑡 greater than 𝜔𝑠 found in Part (a), determine the tension in the string 𝐹𝑃 as a function of 𝜔𝑡 .

ISTUDY

Section 20.2

1387

Three-Dimensional Kinetics of Rigid Bodies

Problem 20.56 A thin uniform disk of radius 𝑅 and mass 𝑚 rolls without slipping over the horizontal surface. The disk, whose center is at 𝐶, can rotate freely relative to the bent shaft 𝐴𝐵𝐶, which precesses with constant angular speed 𝜔0 about the vertical axis. Friction in the pin connecting the bent shaft to the vertical bar at 𝐴 is negligible. Neglecting the mass of the bent shaft, determine an expression for the magnitude of the normal force exerted on the disk by the horizontal surface.

𝐴

𝐵

𝛽

𝜔0 𝐶

Figure P20.56

Problem 20.57 The uniform cone of mass 𝑚, length 𝐿, and apex angle 2𝛽 rolls without slipping over the horizontal surface. The cone rotates at constant 𝜔0 about a fixed vertical axis intersecting the apex 𝐴. Determine an expression for the maximum value of 𝜔0 for which the cone will not tip over the rim at 𝐵. Hint: Model the distributed normal force acting on the cone as a concentrated normal force acting at an unknown distance 𝑑 from 𝐴 and find an expression for that distance as a function of 𝜔0 . 𝜔0

𝐵

𝐴 Figure P20.57

Problem 20.58 The uniform sphere of mass 𝑚 and radius 𝑅 can spin freely relative to the shaft 𝐴𝐵, whose mass is negligible. The shaft 𝐴𝐵 precesses about the vertical axis with constant angular speed 𝜔0 . Assuming that the friction between the sphere and the horizontal surface above is sufficient to prevent slipping and neglecting friction in the pin at 𝐴, determine an expression for the minimum value of 𝜔0 for which this motion is possible.

𝐴 𝜔0

𝑅 𝐵

Figure P20.58

1388

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

20.3 C h a p t e r R e v i e w Three-dimensional kinematics of rigid bodies Referring to Fig. 20.15, for two points 𝐴 and 𝐶 on the same rigid body 𝐵, we can relate their motion using Eqs. (20.1) and (20.2), p. 1349

𝑦 𝐶

𝑟⃗𝐶∕𝐴 𝐵

𝐴

𝑌

𝑧 𝑟⃗𝑃 ∕𝐴

𝑄

𝑋

𝑥

𝑃

Figure 20.15 Rigid body 𝐵 displaying the points and kinematic quantities needed to apply Eqs. (20.1)–(20.4). Frame 𝑋𝑌𝑍 is the primary reference frame and 𝑥𝑦𝑧 is the secondary, body-fixed, or rotating reference frame.

𝑣⃗𝐶 = 𝑣⃗𝐴 + 𝜔 ⃗ 𝐵 × 𝑟⃗𝐶∕𝐴 ,

( ) 𝑎⃗𝐶 = 𝑎⃗𝐴 + 𝛼⃗𝐵 × 𝑟⃗𝐶∕𝐴 + 𝜔 ⃗𝐵 × 𝜔 ⃗ 𝐵 × 𝑟⃗𝐶∕𝐴 ,

where 𝜔 ⃗ 𝐵 and 𝛼⃗𝐵 are the angular velocity and angular acceleration, respectively, of )the ( ⃗ 𝐴𝐵 ×⃗𝑟𝐵∕𝐴 as rigid body. Note that for three-dimensional motion, we cannot write 𝜔 ⃗ 𝐴𝐵 × 𝜔 −𝜔2𝐴𝐵 𝑟⃗𝐵∕𝐴 . Referring to Fig. 20.15, if we use a rotating or secondary reference frame with its origin at 𝐴, then we can relate the motion of points 𝑃 and 𝐴, which are not necessarily on the same body, using Eqs. (20.3) and (20.4), p. 1349 ⃗ × 𝑟⃗ , 𝑣⃗𝑃 = 𝑣⃗𝐴 + 𝑣⃗𝑃 rel + Ω 𝑃 ∕𝐴

( ) ⃗ × 𝑣⃗ ⃗̇ ⃗ ⃗ ⃗ ⃗ 𝑎⃗𝑃 = 𝑎⃗𝐴 + 𝑎⃗𝑃 rel + 2Ω 𝑃 rel + Ω × 𝑟 𝑃 ∕𝐴 + Ω × Ω × 𝑟 𝑃 ∕𝐴 ,

where 𝑣⃗𝑃 rel and 𝑎⃗𝑃 rel are the velocity and acceleration, respectively, of 𝑃 as seen by an ⃗̇ are the angular velocity and angular acceler⃗ and Ω observer in the rotating frame; and Ω ation, respectively, of the rotating reference frame. The rotating frame is usually attached to a rigid body, though sometimes the rotating reference frame will rotate relative to both the primary frame and the rigid body. Referring to Fig. 20.16, if the rotating frame is not attached to the rigid body 𝐵, then the angular acceleration of 𝐵 is given by

𝐵 𝜔 ⃗𝐵

Eq. (20.6), p. 1350 𝛼⃗𝐵 = 𝜔 ⃗̇ 𝐵

⃗ ×𝜔 𝜔 ⃗̇ 𝐵 = 𝛼⃗𝐵 = 𝜔̇ 𝐵𝑥 𝚤̂ + 𝜔̇ 𝐵𝑦 𝚥̂ + 𝜔̇ 𝐵𝑧 𝑘̂ + Ω ⃗𝐵, 𝑦

𝑌 𝑧 𝑄

𝑋

𝐴

Figure 20.16 The primary reference frame 𝑋𝑌𝑍, rotating reference frame 𝑥𝑦𝑧, and rigid body 𝐵.

ISTUDY

⃗ is the angular velocity of the rotating frame relative to the primary frame and 𝜔 where Ω ⃗𝐵 is the angular velocity of 𝐵 relative to the primary frame. Referring to Fig. 20.17, if the angular velocity of the rigid body 𝐵 is expressed relative to the rotating frame as 𝜔 ⃗ 𝐵rel and the angular velocity of the rotating frame relative to the ⃗ then the angular velocity of 𝐵 relative to the primary frame, 𝜔 primary frame is Ω, ⃗ 𝐵 , is Eq. (20.14), p. 1351 ⃗ +𝜔 𝜔 ⃗𝐵 = Ω ⃗ 𝐵rel .

Kinetics: equations of motion and kinetic energy Newton-Euler equations in 3D. There are six Newton-Euler equations for a rigid body in three-dimensional motion. Three are given by the component form of Euler’s first law, which is Eq. (20.20), p. 1366 𝐹⃗ = 𝑚𝑎⃗𝐺 ,

ISTUDY

Section 20.3

1389

Chapter Review

𝑌

𝐵

𝑧 𝑄

𝑋

𝐴

𝜔 ⃗ 𝐵rel

⃗ and a rigid body rotating with 𝜔 Figure 20.17. A secondary frame rotating with Ω ⃗ 𝐵rel relative to the rotating frame.

where 𝐹⃗ is the total external force acting on the rigid body, 𝑚 is the mass of the rigid body, and 𝑎⃗𝐺 is the acceleration of the mass center of the rigid body (see Fig. 20.18). The other three equations are given by Euler’s second law. We derived three different versions of these equations under three corresponding sets of assumptions, each set more restrictive than the next. The last two versions are the most useful in practice, the first of which is given by Eqs. (20.34)–(20.36), p. 1368 ( ) ( ) 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 + 𝑚 𝑦𝐺∕𝑃 𝑎𝐺𝑧 − 𝑧𝐺∕𝑃 𝑎𝐺𝑦 , ) ) ( ( 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 + 𝑚 𝑧𝐺∕𝑃 𝑎𝐺𝑥 − 𝑥𝐺∕𝑃 𝑎𝐺𝑧 , ) ) ( ( 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 , where (see Fig. 20.13): • Point 𝑃 is arbitrary, the 𝑥𝑦𝑧 reference frame is attached to the body 𝐵, and 𝑥𝑦𝑧 are principal axes of inertia, so all products of inertia are zero. • The moments of inertia are expressed relative to a set of axes that are parallel to 𝑥𝑦𝑧 and whose origin is at 𝐺, the center of mass of the body. • The angular velocities and their time derivatives are inertial, but are expressed in the rotating 𝑥𝑦𝑧 frame. The final set of equations, also known as Euler’s equations of rotational motion for a rigid body, are given by Eqs. (20.37)–(20.39), p. 1368 ) ( 𝑀𝑃 𝑥 = 𝐼𝐺𝑥 𝜔̇ 𝐵𝑥 + 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝜔𝐵𝑧 𝜔𝐵𝑦 , ) ( 𝑀𝑃 𝑦 = 𝐼𝐺𝑦 𝜔̇ 𝐵𝑦 + 𝐼𝐺𝑥 − 𝐼𝐺𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 , ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦 − 𝐼𝐺𝑥 𝜔𝐵𝑦 𝜔𝐵𝑥 , where, in addition to the conditions for Eqs. (20.34)–(20.36), we also have that: ⃗ which can happen in any of the following three ways: • 𝑟⃗ × 𝑚𝑎⃗ = 0, 𝐺∕𝑃

𝐺

⃗ ⇒ 𝑃 is the mass center 𝐺 of the body 𝐵 so that 𝑟⃗𝐺∕𝑃 = 0, ⃗ that is, the mass center of the body moves with constant velocity, ⇒ 𝑎⃗𝐺 = 0, ⃗ ⇒ 𝑟⃗𝐺∕𝑃 is parallel to 𝑎⃗𝐺 so that the cross product 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0.

Rotational equations of motion for a body with an axis of radial symmetry. For a rigid body with an axis of radial symmetry, any coordinate system that has an axis coinciding with the axis of symmetry of that body will be a principal coordinate system. Referring to Fig. 20.19, 𝑥𝐵 𝑦𝐵 𝑧𝐵 are principal body axes with the 𝑥𝐵 axis coinciding with the axis of radial symmetry of the body 𝐵. The 𝑥𝑦𝑧 frame rotates with angular velocity ⃗ the angular velocity of the body is 𝜔 Ω, ⃗ 𝐵 , and the body spins relative to the 𝑥𝑦𝑧 frame

⃗ 𝑀 𝑃

𝑃 𝑟⃗𝐺∕𝑃

𝐹⃗ 𝑘̂ 𝚤̂

𝑎⃗𝐺

𝐺

𝑦

𝚥̂ 𝐵

𝑥

𝜔 ⃗𝐵

Figure 20.18 Illustration of the kinetic and kinematic quantities needed to understand and apply Eqs. (20.20) and (20.31)–(20.39).

1390

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Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

𝑦𝐵

𝜔 ⃗𝑠 𝜔 ⃗𝐵 𝚥̂

𝐵 𝐺

𝑃

𝜔 ⃗𝑠

𝚤̂

𝑘̂

𝑥, 𝑥𝐵

𝑧 Figure 20.19. A rigid body 𝐵 spinning about its axis of radial symmetry 𝑥𝐵 relative to the ⃗ relative to an 𝑥𝑦𝑧 frame with angular velocity 𝜔 ⃗ 𝑠 . The angular velocity of the 𝑥𝑦𝑧 frame is Ω, underlying inertial frame.

about its axis of symmetry with angular velocity 𝜔 ⃗ 𝑠 . With these definitions, the rotational equations of motion of the body are

𝑀𝑃 𝑥 𝑀𝑃 𝑦 𝑀𝑃 𝑧

Eqs. (20.48)–(20.50), p. 1370 ) ( = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 − 𝑚𝑥𝐺∕𝑃 𝑎𝐺𝑧 , ) ( = 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 + 𝑚𝑥𝐺∕𝑃 𝑎𝐺𝑦 .

⃗ then these equations become If 𝑟⃗𝐺∕𝑃 × 𝑚𝑎⃗𝐺 = 0,

𝑀𝑃 𝑥 𝑀𝑃 𝑦 𝑀𝑃 𝑧

Eqs. (20.51)–(20.53), p. 1370 ( ) = 𝐼𝐺𝑥 Ω̇ 𝑥 + 𝜔̇ 𝑠 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑦 + 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑧 − 𝐼𝐺𝑦 Ω𝑥 Ω𝑧 , ( ) = 𝐼𝐺𝑦 Ω̇ 𝑧 − 𝐼𝐺𝑥 Ω𝑥 + 𝜔𝑠 Ω𝑦 + 𝐼𝐺𝑦 Ω𝑥 Ω𝑦 .

Planar motion of bodies not symmetric with respect to the plane of motion. In Chapter 17, we emphasized that the Newton-Euler rotational equations derived there were for the planar motion of bodies symmetric with respect to the plane of motion. If the rigid body is undergoing planar motion, but the body is not symmetric with respect to the plane of motion, the governing rotational equations are: Eqs. (20.54)–(20.56), p. 1371 𝑀𝑃 𝑥 = −𝐼𝐺𝑥𝑧 𝜔̇ 𝐵𝑧 + 𝐼𝐺𝑦𝑧 𝜔2𝐵𝑧 , 𝑀𝑃 𝑦 = −𝐼𝐺𝑦𝑧 𝜔̇ 𝐵𝑧 − 𝐼𝐺𝑥𝑧 𝜔2𝐵𝑧 , ( ) 𝑀𝑃 𝑧 = 𝐼𝐺𝑧 𝜔̇ 𝐵𝑧 + 𝑚 𝑥𝐺∕𝑃 𝑎𝐺𝑦 − 𝑦𝐺∕𝑃 𝑎𝐺𝑥 , where the plane of motion is the 𝑥𝑦 plane and the moments and products of inertia are measured with respect to a set of axes whose origin is at the mass center 𝐺.

Kinetic energy of a rigid body in 3D. The work-energy principle for a rigid body is the same whether its motion is two-dimensional or three-dimensional. Therefore, the work-energy principle as given by any of Eqs. (18.16) on p. 1209, (18.23) on p. 1211, (18.24) on p. 1211, or (18.29) on p. 1212 is valid for the three-dimensional motion of this chapter. The expression for the kinetic energy of a rigid body in three-dimensional motion

ISTUDY

Section 20.3

Chapter Review

is: Eq. (20.66), p. 1372 𝑇 = 12 𝑚𝑣2𝐺 + 12 𝐼𝐺𝑥 𝜔2𝐵𝑥 + 21 𝐼𝐺𝑦 𝜔2𝐵𝑦 + 21 𝐼𝐺𝑧 𝜔2𝐵𝑧 − 𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 𝜔𝐵𝑦 − 𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 𝜔𝐵𝑧 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 𝜔𝐵𝑧 , where the moments and products of inertia are measured with respect to axes whose origin is at the mass center of the rigid body.

1391

1392

Chapter 20

Three-Dimensional Dynamics of Rigid Bodies

Review Problems Problems 20.59 through 20.62

𝑦 ℎ 𝑃

𝐷 𝛽 𝜔𝑏

𝑅 𝑧

𝐶 disk

Figure P20.59–P20.62

ISTUDY

The bar 𝐴𝐵 rotates about the vertical 𝑦 axis with angular speed 𝜔𝑏 (𝑡). The disk, whose center is at 𝐶, can rotate freely relative to the arm 𝐶𝐷 as it rolls without slipping on the horizontal surface.

𝐵

𝐴

𝑥

Problem 20.59 If 𝜔𝑏 is constant, determine the angular velocity 𝜔 ⃗ 𝑑 of the disk. Express your answer in the rotating 𝑥𝑦𝑧 component system shown. Problem 20.60 If 𝜔𝑏 (𝑡) is a known function of time, determine the angular acceleration 𝛼⃗𝑑 of the disk. Express your answer in the rotating 𝑥𝑦𝑧 component system shown. Problem 20.61 If 𝜔𝑏 is constant, determine the velocity 𝑣⃗𝑃 of the point 𝑃 on the periphery of the disk as a function of the angle 𝛽. Express your answer in the rotating 𝑥𝑦𝑧 component system shown.

If 𝜔𝑏 (𝑡) is a known function of time, determine the acceleration 𝑎⃗𝑃 of Problem 20.62 the point 𝑃 on the periphery of the disk as a function of the angle 𝛽. Express your answer in the rotating 𝑥𝑦𝑧 component system shown.

Problems 20.63 and 20.64 The shaft 𝐴𝐵 rotates with angular speed 𝜔0 (𝑡) about the vertical axis. The uniform thin rod 𝐶𝐷 of length 𝐿 is rigidly attached to the end of the horizontal arm 𝑂𝐺, which is rigidly attached to 𝐴𝐵. The rod 𝐶𝐷 is tilted from the vertical position through the angle 𝜃 in the 𝑋𝑌 plane about the −𝑍 axis. Express your answers in the 𝑋𝑌𝑍 frame. Problem 20.63 If the angular speed 𝜔0 of the vertical shaft is constant, determine the reaction at 𝑂 on the horizontal shaft required for this motion. Problem 20.64 If the angular speed 𝜔0 (𝑡) of the vertical shaft is not constant, determine the reaction at 𝑂 on the horizontal shaft required for this motion. 𝜔0 (𝑡) 𝐵

𝐿 𝐺

𝑂

𝐿 2

𝐿 2

𝐴

𝜔0 (𝑡)

𝑌 𝑦 𝜃 𝐷

𝐵 𝑋 𝑥

𝐶

Figure P20.63 and P20.64

𝑌 𝑦 𝜃 𝐷 𝐿 𝐺

𝑂

𝑅 𝐴

𝑅 𝑋 𝑥

𝐶

Figure P20.65 and P20.66

Problems 20.65 and 20.66 The shaft 𝐴𝐵 rotates with angular speed 𝜔0 (𝑡) about the vertical axis. The uniform thin disk 𝐶𝐷 of radius 𝑅 is rigidly attached to the end of the horizontal arm 𝑂𝐺, which is rigidly attached to 𝐴𝐵. The disk 𝐶𝐷 is tilted from the vertical position through the angle 𝜃 in the 𝑋𝑌 plane about the −𝑍 axis. Express your answers in the 𝑋𝑌𝑍 frame. Problem 20.65 If the angular speed 𝜔0 of the vertical shaft is constant, determine the reaction at 𝑂 on the horizontal shaft required for this motion. Problem 20.66 If the angular speed 𝜔0 (𝑡) of the vertical shaft is not constant, determine the reaction at 𝑂 on the horizontal shaft required for this motion.

ISTUDY

Section 20.3

Chapter Review

1393

Problems 20.67 and 20.68

𝑦 𝑚, 𝐿

If 𝑀 = 𝑀(𝜙) = 3𝜙1∕2 , determine an expression for the angular velocity of the system after it has undergone three revolutions. Problem 20.67

If 𝑀 = 𝑀(𝑡) = 5𝑡1∕3 , determine an expression for the angular velocity of the system when 𝑡 = 15 s.

Problem 20.68

𝜃

𝑀 𝐴

𝐶 𝑧

Figure P20.67 and P20.68

Problems 20.69 through 20.74 Bar 𝐴𝐵 of length 𝐿𝐴𝐵 = 2.5 m is attached by a fork and clevis joint to the collar at 𝐴 and by a ball joint to the disk at 𝐵. The disk lies in the 𝑥𝑦 plane, and its center at 𝐸 lies on the 𝑦 axis in the 𝑦𝑧 plane. The disk rotates about a vertical axis at the constant angular rate 𝜔𝑑 = 100 rpm. The dimensions 𝑑 = 1.2 m, ℎ = 0.9 m, and 𝑅 = 0.75 m are given. Hint: The clevis joint constrains the rotation of arm 𝐴𝐵 relative to the collar at 𝐴 to be perpendicular to the plane formed by bar 𝐶𝐷 and arm 𝐴𝐵. Therefore, the angular velocity of arm 𝐴𝐵 is the sum of the angular velocity of the collar at 𝐴 and the angular velocity associated with the change in the angle 𝛽, which lies in the plane formed by bars 𝐶𝐷 and 𝐴𝐵. 𝑧 𝓁 𝐷

𝐶 𝐴 𝛽

ℎ 𝐿𝐴𝐵 𝑥 𝑂 𝑑 𝜃 𝐵

𝐸 𝑦

𝑅 𝜔𝑑

Figure P20.69–P20.74

For the disk position shown, that is, 𝜃 = 90◦ , determine the angular velocity of the bar. Express your answer in the given component system, and assume that the angular velocity of the bar is orthogonal to it.

Problem 20.69

For the disk position shown, that is, 𝜃 = 90◦ , determine the angular acceleration of the bar. Express your answer in the given component system, and assume that the angular velocity and angular acceleration of the bar are orthogonal to it.

Problem 20.70

Problem 20.71 For the disk position 𝜃 = 0◦ , determine the velocity of the collar at 𝐴. Express your answer in the given component system, and assume that the angular velocity of the bar is orthogonal to it.

𝜙

𝐷

The moment 𝑀 is applied to the shaft 𝐴𝐵 at 𝑡 = 0 and when 𝜙 = 0. The angled bar 𝐶𝐷𝐸 is rigidly attached to the shaft 𝐴𝐵. Segment 𝐶𝐷 has mass 𝑚 and length 𝐿. Segments 𝐶𝐷 and 𝐷𝐸 have the same linear density. The rotational inertia of the shaft is negligible, as is friction in the bearings at 𝐴 and 𝐵.

𝑥 𝐵 𝐸

1394

ISTUDY

Three-Dimensional Dynamics of Rigid Bodies

Chapter 20

Problem 20.72 For the disk position 𝜃 = 0◦ , determine the acceleration of the collar at 𝐴. Express your answer in the given component system, and assume that the angular velocity and angular acceleration of the bar are orthogonal to it.

Determine an expression for the angular velocity of the bar 𝐴𝐵 and the velocity of the collar at 𝐴 for any position 𝜃 of the disk. Express your answers in the given component system, and assume that the angular velocity of the bar is orthogonal to it.

Problem 20.73

Determine the angular acceleration of the bar 𝐴𝐵 and the acceleration of the collar at 𝐴 for any position 𝜃 of the disk. Express your answers in the given component system, and assume that the angular velocity and angular acceleration of the bar are orthogonal to it.

Problem 20.74

ISTUDY

Technical Writing In this appendix we use a short report as an example to discuss technical writing. The report is written by Bucky Badger, mascot of the University of Wisconsin–Madison.

A

A Short Guide to Technical Writing January 1, 2022 by Bucky Badger Dept. of Engineering Physics University of Wisconsin–Madison Abstract This document is a brief overview of technical writing. It is written in a form you can use as a model for short reports that are up to a few pages in length. The abstract should briefly describe the problem you are considering and should summarize important results. For short reports, the abstract is normally only a few sentences in length. 1. Introduction Effective written reports are necessary for communication and documentation of your analysis or design and are an essential part of your work as an engineer. Your best ideas are only as good as your ability to communicate those ideas to others. Furthermore, reports archive your ideas and may be needed in the future by you or others to show how your design was developed, to aid in modifications, to help support patent rights, to use in litigation, and so on. 2. Technical Writing Tips A technical report should be written using clear, simple prose with correct grammar, punctuation, and spelling. It should be typewritten using a font size and line spacing that are easy to read. All pages should be numbered. The report should not be one long narrative, but rather should use sections and possibly subsections to break the material into smaller logical units. You should evaluate the intended audience and gauge the technical level of your report accordingly. The main body of your report should clearly summarize the assumptions you made, your methods of analysis, and important observations. It should also discuss value decisions and economic and/or manufacturing considerations, if appropriate, and it should clearly spell out your design recommendations and specifications. You should provide convincing justification to the reader that your findings and/or recommendations are accurate and reasonable. If equations are included in the main body of your report, they should be numbered. However, often most of the equations, detailed calculations, and/or raw data should be placed in an appendix and referenced in the main body of the report. Any articles, books, reports, published data, or similar sources that you use in your work should be referenced in the report using a standard format. References to

A-1

A-2

ISTUDY

Appendix A

Technical Writing

Internet websites are permissible, but they may be short-lived as the addresses and content of websites change often. Additional details on technical writing can be found in Beer and McMurrey (2019). Use the spell checker in your word processor to check spelling, but be sure to thoroughly proofread your report, including the final draft, before you submit it. 3. Figures, Graphs, and Tables Figures, graphs, and tables can convey a considerable amount of information. These should always be numbered so they can be easily referred to in your report, and they should have comprehensive captions that explain what the figure, graph, or table shows. 3.1 Figures. The time-honored Chinese proverb that “a picture is worth ten thousand words” applies here. It is considerably more effective to use a figure than to use a paragraph of text to describe what is probably self-evident in the figure. For example, imagine you are designing a wooden clothespin and must specify dimensions and the stiffness for the torsional spring. By including a sketch of the clothespin where dimensions and important parts are identified, as shown in Fig. A.1, your discussion will be considerably more straightforward and easier to comprehend. 𝐶

𝐴 𝐵

𝐷

Figure A.1. Clothespin design with important parts and dimensions identified.

3.2 Graphs. Graphs should always have labeled axes, including appropriate units. If the graph contains multiple curves, each of these should be labeled, perhaps by use of a legend. 3.3 Tables. Rows and columns of a table should always be labeled, including appropriate units. 4. Summary and Conclusions In this final section of your report you should very clearly summarize your findings and how you reached your conclusions. As a guide, if a person were to read only this section, he or she should have a reasonable idea of your work and recommendations. References Beer, D. F., and McMurrey, D. A., A Guide to Writing as an Engineer, 5th ed., Wiley, 2019. Appendix The appendix contains detailed calculations, raw data, and other ancillary information. This is information the reader may choose not to consult, but it should be included to help support your remarks and claims. It can be neatly handwritten.

ISTUDY

Appendix A

End of Section Summary This appendix used an example report to present guidelines and tips for writing short technical reports. Following is a summary of some of the most common mistakes that students make when writing technical reports. Technical Writing Pitfalls • Your report should not be one long narrative. Use sections and possibly subsections to break your report into logical units. • Do not embellish your report with fictional company names or talk about yourself as a statics/engineering student. Write your report as a professional document—short and to the point. • Provide formal references for all sources of data (follow the format given in the example report). • The vast majority of the time (almost always!), a figure that helps describe the problem you are working on is very effective. It should be in the main report. • Very prominently and succinctly state your findings and/or design recommendations in the main body of the report. The reader should not have to search for this information. • Do not overreference the appendix. In a short report that is only a few pages long, you should probably reference the appendix no more than a few times. Anything that is central to your discussion in the main report (e.g., a figure, table, graph, or equation) should be present in the main report. • Figures, graphs, and tables should always be labeled. • Graphs should have labeled axes, including units, and a legend if the graph contains multiple curves. • Your report should be easy to read and convincing. Ideally, a person should be able to comprehend your report with one reading. • Proofread your entire report one last time before submitting it. Do not rely completely on your word processor’s spell checker.

Technical Writing

A-3

ISTUDY

ISTUDY

Answers to Even-Numbered Problems

B

Chapter 1 1.2 1.4 1.6 1.8 1.10 1.12 1.14 1.16

Answers given in problem statement (a) 𝑙 = 13.90 m; (b) 𝑚 = 9.279×105 kg; (c) 𝐹 = 98.30 kN; (d) 𝑀 = 10.39 kN⋅m (a) 𝑙 = 25.4 𝜇m; (b) 𝑚 = 53.4 kg; (c) 𝐹 = 11.4 kN; (d) 𝐼mass = 2.60 N⋅m⋅s2 (a) 𝑙 = 60.37 f t; (b) 𝑚 = 0.2961 slug; (c) 𝐹 = 476.6 lb; (d) 𝑀 = 7656 in.⋅lb (a) 𝑙 = 60.2 in.; (b) 𝑚 = 4.46 slug; (c) 𝐹 = 20.1 lb; (d) 𝑀 = 291 in.⋅lb (a) 𝑝 = 3.63 lb∕in.2 ; (b) 𝐸 = 29.0×106 lb∕in.2 ; (c) 𝐼area = 5.65 in.4 ; (d) 𝐼mass =109 in.⋅lb⋅s2 𝑚 = 0.115 kg 𝑔theory = 9.822 m∕s2

1.18 𝐹 = 2.029×1020 N 1.20 𝐹 = 0.5649 lb 1.22 (a) 𝛾 = 65.7 kN∕m3 , 𝜌 = 6.70×103 kg∕m3 ; (b) 𝛾 = 20.9 kN∕m3 , 𝜌 = 2.13×103 kg∕m3 ; (c) 𝛾 = 0.314 kN∕m3 , 𝜌 = 32.0 kg∕m3 ; (d) 𝛾 = 21.4 kN∕m3 , 𝜌 = 2.19×103 kg∕m3 1.24 𝑏 = 7.087 in.; ℎ1 = 23.62 in.; ℎ2 = 19.69 in.; 𝐿 = 6.562 f t; 𝑊 = 3239 lb 1.26 0.217 lb 1.28 (a) 𝜃 = 0.621 rad; (b) 𝜃 = 0.0188 mrad; (c) 𝜃 = 266◦ ; (d) 𝜃 = 0.0146◦

Chapter 2 2.2 (a) 𝑅⃗ = 255 mm @ 36.0◦ ; (b) 𝑅⃗ = 1.10 kip @ −142◦ 2.4 (a) 𝑅⃗ = 88.39 N @ −118.1◦ , or if desired, this vector may be stated using a positive angle as 𝑅⃗ = 88.39 N @ 241.9◦ ; (b) 𝑅⃗ = 138.5 mm @ 14.67◦ 2.6 (a) 𝑅⃗ = 243.6 lb @ −34.80◦ , or if desired, this vector may be stated using a positive angle as 𝑅⃗ = 243.6 lb @ 325.2◦ ; (b) 𝑅⃗ = 9.129 in. @ 100.3◦ 2.8 (a) 𝑅⃗ = 6.32 m @ 71.6◦ ; (b) 𝑅⃗ = 7.21 m @ −56.3◦ ; (c) 𝑅⃗ = 17.0 m2 @ 45.0◦ ; (d) 𝑅⃗ = 1.41 @ 45.0◦ 2.10 𝐹2 = 1035 N, 𝑅 = 267.9 N 2.12 𝛼 = 16.0◦ or 104◦ 𝐵 𝑟𝐴𝐵 = 300 mm 𝐵

𝑟𝐴𝐵 = 300 mm

𝛽 𝑟𝑂𝐶 = 650 mm

𝛼

𝐴

𝛼

60◦

𝛽 𝐴

𝑟𝑂𝐴 = 400 mm

60◦

𝑂

𝑟𝑂𝐶 = 650 mm

60◦

𝑟𝑂𝐴 = 400 mm 60◦

𝑂

2.14 (a) 𝑅⃗ = 12.14 kN @ −19.23◦ , or if desired, this vector may be stated using a positive angle as 𝑅⃗ = 12.14 kN @ 340.8◦ ; (b) 𝑅⃗ = 7.134 in. @ 76.20◦

A-5

A-6

2.16 𝐹 = 1.60 kN, 𝜃 = 90◦ Possible choices for 𝐹 2 kN

60◦

30◦

𝑎

𝑎

2 kN sin 30◦ 3 kN

3 kN sin 60◦

2.18 (a) 364 N @ 147◦

; (b) 239 N @ 257◦

; (c) 𝑃 = 607 N; (d) 𝛼 = 0◦ , 𝑃 = 304 N

2.20 𝐹1 = 19.28 kN, 𝛼 = 0◦ 2.22 𝐹1 = 246 N, 𝛼 = 60◦ Possible choices for 𝐹1 𝑎 𝛼

200 N 60◦ 400 N

2.24 (a) |𝐹⃗𝑎 | = 25.9 lb, |𝐹⃗𝑏 | = 96.6 lb; (b) |𝐹⃗𝑎 | = 29.9 lb, |𝐹⃗𝑏 | = 112 lb , 𝐹⃗𝑏 = 352.5 N @ 90◦

2.26 𝐹⃗𝑎 = 164.4 N @ −50◦ 2.28 𝛽 = 90◦ , 𝐹𝑂𝐶 ′ = 200 lb 2.30 𝑅⃗ = (−18 𝚤̂ + 35 𝚥̂) kN 2.32 𝐹⃗ = 1.60 𝚥̂ kN

2.34 (a) 𝑅⃗ = (−304 𝚤̂ + 200 𝚥̂) N; (b) 𝑅⃗ = (−53.6 𝚤̂ − 233 𝚥̂) N; (c) 𝑃 = 607 N; (d) 𝛼 = 0◦ , 𝑃 = 304 N 2.36 𝛼 = 60◦ , 𝐹1 = 246 N, 𝐹⃗1 = (−213 𝚤̂ + 123 𝚥̂) N 2.38 (a) 𝑅⃗ = (−15 𝚤̂ − 4 𝚥̂) kN, 𝑅 = 15.52 kN; (b) 𝑅⃗ = (3 𝚤̂ − 28 𝚥̂) kN, 𝑅 = 28.16 kN; (c) 𝑠 = 0.6667; (d) 𝑅̂ = 0.1483 𝚤̂ + 0.9889 𝚥̂ 2.40 (a) 𝑟⃗𝐴𝐵 = (4 𝚤̂ − 3 𝚥̂) m; (b) 𝑟⃗𝐵𝐴 = (−4 𝚤̂ + 3 𝚥̂) m; (c) 𝑢̂ 𝐴𝐵 = (e) 𝐹⃗𝐴𝐵 = (9.6 𝚤̂ − 7.2 𝚥̂) kN; (f) 𝐹⃗𝐵𝐴 = (−9.6 𝚤̂ + 7.2 𝚥̂) kN

4 5

𝚤̂ −

3 5

𝚥̂; (d) 𝑢̂ 𝐵𝐴 = − 45 𝚤̂ +

⃗ = 549 N, 𝑅⃗ = 549 N @ 99.8◦ 2.42 𝑅⃗ = (−93.6 𝚤̂ + 541 𝚥̂) N, |𝑅| 2.44 𝑅⃗ = (−168.9 𝚤̂ − 207.0 𝚥̂) N = 267.2 N @ 230.8◦

, 𝑅 = 267.2 N

2.46 𝑟⃗𝐴𝐵 = (3.235 𝚤̂ + 12.07 𝚥̂) m, 𝑟⃗𝐵𝐶 = (2.585 𝚤̂ + 1.077 𝚥̂) m, 𝑟⃗𝐶𝐷 = (5.734 𝚤̂ − 4.015 𝚥̂) m, 𝑟⃗𝐷𝐸 = (2.236 𝚤̂ − 1.118 𝚥̂) m, 𝑟⃗𝐴𝐸 = (13.79 𝚤̂ + 8.018 𝚥̂) m 2.48 𝐹2 ∕𝐹1 = 1.50 2.50 𝛼 = 30.65◦ , 𝑅⃗ = (−735.9 𝚤̂ + 436.1 𝚥̂) lb, 𝑅 = 855.4 lb 400 lb 𝐹 = 400 lb 𝐵 4 5

800 lb

𝛼 𝑦 0 80 lb

ISTUDY

Appendix B

Answers to Even-Numbered Problems

3 5

800 lb

𝑅⃗ 𝛼 𝐴 600 lb 𝑂 600 lb

𝑥

3 5

𝚥̂;

ISTUDY

Appendix B

Answers to Even-Numbered Problems

2.52 𝛼 = 33.7◦ , 𝑅 = 6.42 kN 𝑦 Possible choices for 𝐹 𝑅

𝑂

8 kN

𝛼 12 kN

𝐴

𝑥

2.54 (a) Not safe; (b) Safe (the working load multiplier is 70%, determined using linear interpolation); (c) Safe (the working load multiplier is 96%, determined using linear interpolation) 2.56 𝑅𝑥 = 100 N, 𝑅𝑦 = −200 N, 𝑅𝑡 = −13.40 N, 𝑅𝑛 = −223.2 N 𝑦 𝑛

𝑡 30◦

𝑥 100 N = 26.57◦ 200 N 𝛼 = 30◦ − 26.57◦ = 3.43◦

tan−1 200 N

100 N

2.58 (a) 𝑅𝑥 = 21.61 N, 𝑅𝑦 = 48.30 N; (b) 𝑅𝑡 = 8.375 N, 𝑅𝑛 = 52.25 N 2.60 (a) 𝑅𝑥 = −213 N, 𝑅𝑦 = 51.7 N; (b) 𝑅𝑡 = −187 N, 𝑅𝑛 = −114 N 2.62 𝑟⃗𝐴𝐵 = (−20 𝚤̂ + 48 𝚥̂) mm, 𝑥𝐵 = 165 mm, 𝑦𝐵 = 48 mm 2.64 𝑟⃗𝐴𝐶 = (−50 𝚤̂ + 120 𝚥̂) mm, 𝑥𝐶 = 135 mm, 𝑦𝐶 = 120 mm 2.66 𝑟⃗𝑂𝐴 = (95 𝚤̂ + 56 𝚥̂) mm, 𝑟⃗𝑂𝐵 = (68 𝚤̂ + 92 𝚥̂) mm, 𝑟⃗𝑂𝐶 = (17 𝚤̂ + 160 𝚥̂) mm, 𝑥𝐵 = 68 mm, 𝑦𝐵 = 92 mm 2.68 𝜃𝑥 = 36◦ , 144◦ 𝑧

−0.8090

0.5

0.8090

−0.3090

𝑥

𝑦 0.5 𝑢̂ 2 −0.3090 𝑢̂ 1

√ 2.70 With 𝜃 = 𝜃𝑥 = 𝜃𝑦 = 𝜃𝑧 , 𝜃 = 54.74◦ , 125.3◦ , cos 𝜃 = ± 13 = ± 0.5774

( ) ( ) ⃗ = 114.3 𝚤̂ + 228.6 𝚥̂ − 157.1 𝑘̂ N, 2.72 𝐹⃗ = −59.24 𝚤̂ − 162.8 𝚥̂ + 100 𝑘̂ N, 𝑄 ( ) 𝑅⃗ = 55.05 𝚤̂ + 65.81 𝚥̂ − 57.14 𝑘̂ N

( ) ( ) ( ) ⃗ = 39.8 𝚤̂ + 30.0 𝚥̂ − 33.4 𝑘̂ lb, 𝑃⃗ = 50.0 𝚤̂ − 70.7 𝚥̂ + 50.0 𝑘̂ lb, 2.74 𝐹⃗ = −4.59 𝚤̂ + 26.8 𝚥̂ + 12.6 𝑘̂ lb, 𝑄 ( ) 𝑅⃗ = 85.2 𝚤̂ − 13.9 𝚥̂ + 29.2 𝑘̂ lb ( ) ( ) ( ) ⃗ = −2.67 𝚤̂ + 2.67 𝚥̂ + 1.33 𝑘̂ kN, 𝐹⃗ = 2.12 𝚤̂ + 1.50 𝚥̂ + 1.50 𝑘̂ kN, 2.76 𝑃⃗ = −1.33 𝚤̂ + 0.766 𝚥̂ + 1.29 𝑘̂ kN, 𝑄 ( ) 𝑅⃗ = −1.87 𝚤̂ + 4.93 𝚥̂ + 4.12 𝑘̂ kN ( ) ( ) 2.78 𝐹⃗ = 12.5 𝚤̂ − 14.5 𝚥̂ + 16.1 𝑘̂ N where 𝜃𝑥 = 60◦ , 𝜃𝑦 = 125◦ , 𝜃𝑧 = 50◦ ; 𝑃⃗ = 49.4 𝚤̂ + 53.0 𝚥̂ + 19.4 𝑘̂ N ( ) where 𝜃 = 48.8◦ , 𝜃 = 45◦ , 𝜃 = 75◦ ; 𝑅⃗ = 61.9 𝚤̂ + 38.5 𝚥̂ + 35.5 𝑘̂ N where 𝑥

𝑦

𝑧

𝜃𝑥 = 40.3◦ , 𝜃𝑦 = 61.6◦ , 𝜃𝑧 = 64.0◦ 2.80 4.16 km

A-7

A-8

ISTUDY

Answers to Even-Numbered Problems

Appendix B

( ) ( ) ( ) ( ) 2.82 (a) 𝜃𝑥 = ± cos−1 sin 𝜃ℎ cos 𝜃𝑣 , 𝜃𝑦 = ± cos−1 cos 𝜃ℎ cos 𝜃𝑣 , 𝜃𝑧 = ± cos−1 sin 𝜃𝑣 = ± 90◦ − 𝜃𝑣 ; (b) 𝜃𝑥 = 75.5◦ , 𝜃𝑦 = 64.3◦ , 𝜃𝑧 = 30◦ 𝑧 𝑟⃗

𝑟𝑥

𝑟𝑧

𝜃𝑣 𝑟𝑦 𝜃ℎ

𝑦 𝑟𝑎

𝑎

𝑥

2.84 𝜃ℎ = 33.7◦ , 𝜃𝑣 = 59.0◦ ( ) 2.86 𝑟⃗𝐴𝐷 = −80.09 𝚤̂ − 150.2 𝚥̂ − 90.00 𝑘̂ cm ( ) ) 2.88 𝐹⃗𝐵𝐸 = −44.4 𝚤̂ + 77.8 𝚥̂ − 44.4 𝑘̂ lb, 𝐹⃗𝐸𝐵 = (44.4 𝚤̂ − 77.8 𝚥̂ + 44.4 𝑘̂ lb ( ) 2.90 𝐹⃗𝐵𝐴 = 23.62 𝚤̂ − 16.24 𝚥̂ − 8.857 𝑘̂ lb, 𝐹⃗𝐵𝐶 = −1500 lb 𝚥̂ ( ) 2.92 𝐹⃗𝐵𝐴 = 16.60 𝚤̂ − 24.88̂𝚥 − 2.262 𝑘̂ lb, 𝐹⃗𝐵𝐶 = −1500 lb 𝚥̂ ( ) 2.94 (𝑥𝐸 , 𝑦𝐸 , 𝑧𝐸 ) = (80, 60, 120) mm, 𝐹⃗𝐸𝐺 = −47.1 𝚤̂ + 52.9 𝚥̂ + 70.6 𝑘̂ N ( ) 2.96 𝐹⃗𝐶𝐷 = −2.4 𝚤̂ + 1.8 𝑘̂ kN, 𝐹⃗𝐵𝐸 = −5 kN 𝑘̂ ( ) 2.98 𝐹⃗𝐶𝐷 = −0.8208 𝚤̂ − 2.255 𝚥̂ + 1.800 𝑘̂ kN, 𝐹⃗𝐵𝐸 = −5 kN 𝑘̂ 2.100 𝜃𝑥 = cos−1 (sin 𝜙 cos 𝜃), 𝜃𝑦 = cos−1 (sin 𝜙 sin 𝜃), 𝜃𝑧 = 𝜙 2.102 (a) 𝐹𝐵 ∕𝐹𝐴 = 1.51; (b) Spokes on side 𝐵 are more severely loaded; (c) Essay-type answer ( ) 2.104 (a) 128◦ ; (b) 𝐴‖ = −4.33 lb, 𝐴⊥ = 5.50 lb; (c) 𝐴⃗‖ = 4.04 𝚤̂ + 0.578 𝚥̂ − 1.44 𝑘̂ lb, ( ) 𝐴⃗ = 1.96 𝚤̂ − 2.58 𝚥̂ + 4.44 𝑘̂ lb ⊥

̂ N, 2.106 (a) 145.7◦ ; (b) 𝐴∥ = −10.73 N, 𝐴⊥ = 7.335 N; (c) 𝐴⃗∥ = (1.431 𝚤̂ − 3.578 𝚥̂ + 10.02 𝑘) ( ) 𝐴⃗ = 2.569 𝚤̂ + 6.578 𝚥̂ + 1.982 𝑘̂ N ⊥

2.108 𝑑 = 5.77 cm , 𝜃 = 41.4◦ 2.110 𝛼 = 4.70◦ , 𝛽 = 9.43◦ , 𝑎 = 12.2 in., 𝑏 = 12.0 in. 2.112 𝐹‖ = −67.4 N, 𝐹⊥ = 73.9 N. The bead slides toward 𝐶. 2.114 Answer given in problem statement ( ) 2.116 𝐹 = −1.79 kip, 𝐹⃗ = 0.808 𝚤̂ + 0.0659 𝚥̂ − 1.60 𝑘̂ kip 2.118 𝐹 = 5.83 N 2.120 𝐹‖ = (0.937)𝐹 , 𝐹⊥ = (0.349)𝐹 2.122 𝑊 = 45.0 N 2.124 (a) 𝐹∥ = 44.25 N, 𝐹⊥ = 23.28 N; (b) 𝐹∥ = 41.31 N, 𝐹⊥ = 28.17 N 2.126 460.0 lb 2.128 593.5 lb

ISTUDY

Appendix B

Answers to Even-Numbered Problems

( ) ( ) 2.130 𝑇⃗∥ = −3.537 𝚤̂ + 1.179 𝚥̂ − 1.769 𝑘̂ lb, 𝑇⃗⊥ = 4.648 𝚤̂ + 3.265 𝚥̂ − 7.120 𝑘̂ lb. The bead slides toward 𝐵. 2.132 𝐹‖ = −1.71 N, 𝐹⊥ = 60.0 N. The bead slides toward 𝐴. 2.134 86.93 km∕h 2.136 The shortest distance is 209.5 m. The point where the road is closest lies outside of segment 𝐶𝐷. 2.138 𝑣 = 1.149𝑠 2.140 51.7 mm 2.142 12.0 in. 2.144 (a) and (b) 350 𝑘̂ N⋅mm ( ) ( ) 2.146 (a) −120 𝚤̂ − 2 𝚥̂ − 26 𝑘̂ mm2 ; (b) 120 𝚤̂ + 2 𝚥̂ + 26 𝑘̂ mm2 ; (c) Essay-type answer; (d) Answer given in problem statement ( ) ( ) 2.148 (a) −118 𝚤̂ − 426 𝚥̂ − 360 𝑘̂ in.⋅lb; (b) 118 𝚤̂ + 426 𝚥̂ + 360 𝑘̂ in.⋅lb; (c) The results of Parts (a) and (b) are vectors with equal magnitude but opposite direction; (d) Answer given in problem statement Concept Problem ⃗ = (−19300 𝚤̂ + 13600 𝚥̂) in.⋅lb 2.152 (a) Yes; (b) 𝑀 𝑂

2.150

̂ (c) 364 cm2 2.154 (a) Essay-type answer; (b) 𝑢̂ = 0.192 𝚤̂ + 0.192 𝚥̂ + 0.962 𝑘; 2.156 0.5147◦ . The surface is not sufficiently level. ( ) ( ) 2.158 𝑣⃗𝑛 = −2.92 𝚤̂ − 7.87 𝚥̂ − 6.30 𝑘̂ km/s, 𝑣⃗𝑡 = −3.62 𝚤̂ − 1.95 𝚥̂ + 4.11 𝑘̂ km/s, |𝑣⃗𝑛 | = 10.5 km∕s, |𝑣⃗𝑡 | = 5.82 km∕s ̂ 2.160 (a) Since 𝑟⃗1 ⋅ 𝑟⃗𝐴𝐵 = 0, the vectors 𝑟⃗1 and 𝑟⃗𝐴𝐵 are perpendicular; (b) 𝑟̂2 = 0.8920 𝚤̂ − 0.03678 𝚥̂ − 0.4506 𝑘; (c) 𝑃𝐴𝐵 = 50.55 N, 𝑃1 = 281.3 N, 𝑃2 = −958.3 N 2.162 (a) Since 𝑟⃗𝐴𝐵 ⋅ 𝑟⃗𝐶𝐷 = 0, the vectors 𝑟⃗𝐴𝐵 and 𝑟⃗𝐶𝐷 are perpendicular; ̂ (c) 𝑃 = 412.2 lb, 𝑃 = 162.4 lb, 𝑃 = −404.6 lb (b) 𝑟̂ = 0.8909 𝚤̂ − 0.4483 𝚥̂ − 0.07284 𝑘; 𝐴𝐵 𝐶𝐷 𝑟 2.164 6.57 mm 2.166 (a) and (b) 23.4 N⋅m

2.168 −8.8 kN ≤ 𝑃 ≤ 4.6 kN 𝑅𝑛

1.6 kN

5 kN − 1.6 kN = 3.4 kN

𝑃 (if𝑃 > 0) 1.6 kN

𝑅𝑡

𝐹

1.2 kN 10 kN − 1.2 kN = 8.8 kN

𝑃 (if𝑃 < 0)

( ) 2.170 (a) 𝑅⃗ = −2.29 𝚤̂ + 2.72 𝚥̂ − 1.45 𝑘̂ kN; (b) 𝑇⃗ = (2.29 𝚤̂) kN 2.172 The distance from point 𝐴 to point 𝐵 is 3058 f t and the length of the power line is 2796 f t. 2.174 𝑃⊥ = (−0.970)𝑃 , 𝑃‖ = (0.243) 𝑃 2.176 1.672◦ . The surface meets the drainage specification. ( ) ( ) 2.178 𝑃‖ = 7.66 kN, 𝑃⊥ = 6.43 kN, 𝑃⃗‖ = −3.48 𝚤̂ − 3.48 𝚥̂ − 5.87 𝑘̂ kN, 𝑃⃗⊥ = 3.48 𝚤̂ + 3.48 𝚥̂ − 4.13 𝑘̂ kN ̂ (c) 𝑃 = 12.08 N 2.180 (a) 𝜃𝑥 = 125.5◦ ; (b) 𝑟̂ = −0.2632 𝚤̂ − 0.7291 𝚥̂ + 0.6318 𝑘; 𝑟 ◦ ◦ ̂ ̂ 2.182 (a) 𝜃 = 107 , 𝜃 = 114 ; (b) 0.520 𝚤̂ + 0.693 𝚥̂ + 0.500 𝑘 (or −0.520 𝚤̂ − 0.693 𝚥̂ − 0.500 𝑘) 𝑥

𝑦

2.184 900 f t 2.186 Answers given in problem statement

A-9

A-10

ISTUDY

Appendix B

Answers to Even-Numbered Problems

Chapter 3 3.2 𝑇𝐴𝐵 = 217.0 lb (tension), 𝑇𝐵𝐶 = 176.9 lb (tension), 𝑇𝐵𝐷 = 200 lb (tension) 3.4 𝑇𝐴𝐵 = 92.65 N (tension), 𝑇𝐵𝐶 = −54.50 N (compression), 𝑇𝐵𝐷 = 49.05 N (tension) 3.6 𝛼 = 50◦ , 𝑇𝐴𝐵 = 128.6 lb, 𝑇𝐵𝐶 = 153.2 lb, 𝑇𝐵𝐷 = 200 lb 3.8 𝑅𝐶 = 𝑅𝐷 = 31.8 lb 3.10 (a) 𝑇1 = 251 lb, 𝑅 = 141 lb; (b) 𝑇2 = 149 lb, 𝑅 = 141 lb 3.12 (a) 𝐹 = 17.9 N, 𝑅𝐴 = 17.9 N, 𝑅𝐵 = 98.1 N, 𝑅 = 52.2 N. (b) The system is in equilibrium, and blocks 𝐴 and 𝐵 may move with constant velocity, but they will not accelerate. (c) The system is not in equilibrium, block 𝐵 will accelerate to the right, and block 𝐴 will accelerate downward. 𝑊 = 49.05 N 𝑦

𝑅𝐴

𝐴

𝑥

20◦

𝑅

𝑊 = 49.05 N 𝑅 𝐵

𝐹

𝑅𝐵

3.14 𝑇𝐴𝐵 = 5.13 kN, 𝑇𝐴𝐶 = 0.340 kN 3.16 𝐹𝐴𝐵 = 1580 N, 𝐹𝐴𝐶 = −1500 N 10◦

𝐴

𝑃 = 800 N

𝑦

60◦ 𝐹𝐴𝐵

𝐹𝐴𝐶

𝑥

3.18 𝑇𝐴𝐵 = 259.1 N 3.20 𝑇𝐴𝐶 = 2.294 lb 3.22 (a) 𝑇𝐴𝐵 = 175 N; (b) 𝑇𝐶𝐷 = 𝑇𝐶𝐸 = 109 N; (c) Essay-type answer 3.24 𝑇𝐴𝐵𝐶 = 169.8 N, 𝑅 = 268.8 N 3.7

3.5

1.2 𝑅

𝑇𝐴𝐵𝐶

𝑦 𝑥

𝐷 𝐶

𝐸

3.7

1.2

3.5 400 N

3.26 (a) Answer provided in problem description; (b) 𝐹 = 2𝑇 cos 𝜃; (c) Essay-type answer 3.28 𝑇𝐴𝐶𝐵 = 2620 N, 𝑇𝐶𝐷𝐸 = 4530 N 𝐹 = 7848 N

30◦ 𝑇𝐶𝐷𝐸

𝑦

𝑦

60◦

𝑥

𝐶 𝑥

𝑇𝐴𝐶𝐵 𝑇𝐴𝐶𝐵

𝐷

30◦ 𝑇𝐶𝐷𝐸

30◦

𝑇𝐶𝐷𝐸

3.30 𝑚 = 547 kg 𝑦

𝑊

𝑦

60◦ 𝑇𝐶𝐷𝐸

𝑥 𝐶 30◦ 𝑇𝐴𝐶𝐵

𝑄 45◦ 𝑇𝐴𝐶𝐵

3.32 𝑄max = 2041 lb

𝑥 60◦ 𝑇𝐶𝐷𝐸

𝐷 60◦ 𝑇𝐶𝐷𝐸

ISTUDY

Appendix B

Answers to Even-Numbered Problems

3.34 𝑇𝐶𝐷 = 5710 N, 𝑇𝐵𝐴 = 3710 N, 𝑇𝐶𝐵 = 4290 N, 𝑊 = 1140 N 3.36 𝑊max = 2.4 kN 3.38 (a) System (a): 𝑇 = 𝑊 , system (b): 𝑇 = 12 𝑊 , system (c): 12 𝑊 , system (d): 13 𝑊 , system (e): 13 𝑊 , system (f): 14 𝑊 ; (b) Essay-type answer

3.40 𝑊 = 218 lb 3.42 𝑇𝐵𝐻 = 540 lb, 𝑇𝐻𝐸 = 540 lb, 𝑇𝐵𝐶 = 717 lb, 𝑇𝐸𝐹 = 586 lb, 𝑊𝐴 = 322 lb, 𝑊𝐺 = 369 lb, 𝑊𝐷 = 108 lb 3.44 𝜃 = 46.8◦ , 𝑇 = 86.2 N 2𝑇

𝑇

70◦

𝜃

𝑦

𝐴 𝑥 𝑊 = 225 N

3.46 𝑊1 = 11.3 N, 𝑊2 = 32.6 N Computer Problem. 𝑑 = 45 in., 𝑇𝐴𝐵𝐶 = 54.17 lb (5 lb)(cos 30◦ + sin 30◦ ) 3.50 𝛼 = −30◦ ; intermediate answer: 𝐹𝐵𝐷 = cos 30◦ cos 𝛼 − sin 30◦ sin 𝛼 3.52 𝑇𝐴𝐶 = 3460 lb, 𝑇𝐴𝐸 = 6000 lb, 𝑇𝐴𝐺 = 4240 lb, 𝛿 = 6.00 in.

3.48

𝐶 𝛿

𝑇𝐴𝐶 30◦ 𝑇𝐴𝐸 45◦

𝐴

12 kip 𝑦

𝑅𝐴

𝐹𝐶 = 500 lb∕in. 𝛿 𝑅𝐶 𝐸

𝐹𝐸 = 1000 lb∕in. 𝛿 𝑅𝐸

𝑇𝐴𝐺 𝐺

𝑥

𝐹𝐺 = 500 lb∕in. 𝛿

𝑅𝐺

3.54 𝛿𝐴 = 14.72 mm, 𝛿𝐵 = 27.80 mm 3.56 𝛿𝐴 = 5.2 in., 𝛿𝐵 = 4.9 in., 𝛿𝐶 = 4 in., 𝛿𝐷 = 2.4 in. 3.58 (a) 𝐿𝐴𝐵 = 94.3 cm, 𝐿𝐴𝐷 = 120 cm; (b) 𝑥 = 104 cm, 𝑦 = 46.7 cm 3.60 𝑃 = 10.39 lb, 𝑅 = −5.196 lb 𝑦

𝐹𝑠

𝑃

𝑥 60◦ 𝑅

𝑃 5 lb

3.62 (a) 10.0 N, 8.00 N; (b) 7.00 N, 5.00 N 4N 𝐹𝑠

𝐵 𝑦 𝐹𝑠

𝐶

𝛿(𝑡)

𝑥 𝐴 𝛥

𝐴

𝐹𝑠 𝑅

3.64 𝑘 = 167 lb∕in., 𝐿0 = 21.0 in. 3.66 𝛿 = 0.587 in., 𝐹1 = 128 lb, 𝐹2 = 345 lb, 𝐹3 = 128 lb 3.68 (a) 𝑃 = 6000 N, 𝑇𝐴𝐵 = 8000 N; (b) 𝑥𝐴 = 229.2 mm, 𝑦𝐴 = −71.11 mm 3.70 𝛿 = 3.42 mm 3.72 (a) 𝑑 = 95.49 in., 𝐹1 = −1082 lb (compression), 𝐹2 = 1082 lb (tension); (b) 𝑑 = 86.40 in., 𝐹1 = −3264 lb (compression), 𝐹2 = −1736 lb (compression)

A-11

A-12

Appendix B

3.74 𝛿𝐴 = 4.355 mm, 𝛿𝐵 = 5.547 mm, 𝐹1 = 119.2 N, 𝐹2 = −665.6 N, 𝐹3 = −609.7 N 3.76 (a) 𝛿 = 1.17 in.; (b) 𝛿 = 1.99 in. 3.78 𝑊 = 13.33 lb Computer Problem. 𝜃 = 83.62◦ √ √ (5 mm)2 +(2 mm−𝛿)2 − 29mm √ 3.82 (a) 𝐹 = −2(2 N∕mm) (2 mm − 𝛿) + (0.3 N∕mm)𝛿; (b) Answer not provided; 3.80

(5 mm)2 +(2 mm−𝛿)2

(c) 𝐹max = 0.600 N; (d) Essay-type answer 𝑦

𝐹

𝑥 𝛼

𝛼

𝐹𝑠

𝐹𝑠 𝑃 = (2 N∕mm) 𝛿

3.84 𝐹max = 3.333 kN 3.86 𝑚𝐷 = 91.7 kg 𝑧 2m 0.9 m

𝐶 𝐹⃗𝐴𝐶 𝐴

1.2 m

𝐵 𝐹⃗𝐴𝑂

⃗ 𝑊

𝑥

𝐹⃗𝐴𝐵 𝑂 1.2 m 𝑦

3.88 𝑊 = 500 N 3.90 𝑇𝐴𝑂 = −698.5 lb, 𝑇𝐴𝐵 = 833.3 lb, 𝑇𝐴𝐶 = 541.7 lb, 𝑇𝐴𝐷 = 1000 lb ) ( 𝑑; (b) 𝑑 = 16.9 in. 3.92 (a) 𝑊 = 3𝑘 1 − √ 𝑟 𝑟2 +𝑑 2

𝑧 𝑊

√ 𝑟2

𝑑 𝛼 𝑟

+ 2

𝐹𝑠

𝐹𝑠

𝑑

ISTUDY

Answers to Even-Numbered Problems

𝐹𝑠 𝑦 𝑥

3.94 𝑃 = 533 lb 3.96 𝑃 = 532 lb 𝐷 𝑧 12 ft 8 ft 𝐵 6 ft

𝐹⃗𝐴𝐷 𝐹⃗𝐴𝐵 𝐴

𝐹⃗𝐴𝐶

1 ft

𝐹⃗𝐴𝐸 = 𝑃⃗

6 ft 𝐶

𝑦

𝑥

ISTUDY

Appendix B

Answers to Even-Numbered Problems

3.98 𝑃 = 1450 lb 𝑧 𝐹⃗𝐴𝐶

𝐶

𝐴

𝐹⃗𝐴𝐵 𝐹⃗𝐴𝑂

𝐵

𝑃⃗ 𝐷

𝑂

𝑦

𝑥

3.100 𝑇𝐴𝐵 = 365.8 lb, 𝑇𝐵𝐷 = 292.3 lb, 𝑇𝐵𝐸 = 339.8 lb, 𝑇𝐵𝐻 = 1200 lb 3.102 (a) 𝐹𝐴𝐸 = −420 lb, 𝐹𝐴𝐶 = −120 lb, 𝐹𝐴𝐷 = −120 lb; (b) 𝐹𝐸𝐵 = −120 lb 𝐴

𝐴 𝐹𝐴𝐷

𝐹𝐴𝐸

𝑧

𝑧

𝐹𝐴𝐶

𝑇1 𝑇2

𝐸

𝐸 𝐹𝐸𝐵

𝐷

𝑦

𝐹𝐴𝐸 𝑇1

𝐹 𝐵

𝑥

𝐷

𝑦 𝐹

𝐶

𝐵

𝑥

𝐶

3.104 𝐹𝐴𝐵 = 𝐹𝐴𝐶 = 780 lb, 𝑅 = 800 lb ̂ N, 𝑅⃗ = (−68.3 𝚤̂ + 22.3 𝚥̂ − 11.8 𝑘) ̂ N 3.106 𝑊 = 400 N, 𝑅1 = 291 N, 𝑅2 = 72.8 N, 𝑅⃗ 1 = (28.2 𝚤̂ + 198 𝚥̂ + 212 𝑘) 2 ̂ N 3.108 𝑃 = 107 N, 𝑅⃗ = (32.0 𝚤̂ + 42.6 𝚥̂ − 107 𝑘) 𝑦 𝑅 𝐵 𝐶 150 mm

𝑃

4 𝐹𝐶𝐷 3

𝑥

𝑂

𝐴 120 mm 𝐷

150 mm

𝑧

3.110 𝑊 = 68.52 lb 3.112 151.5 N 3.114 𝐹𝐸𝐷 = 280 lb, 𝐹𝐸𝐹 = 360 lb, 𝐹𝐹 𝐵 = 210 lb, 𝐹𝐸𝐴 = 380 lb, 𝐹𝐹 𝐶 = 210 lb. 60 lb must be added to traffic light 𝐸 3.116 (a) 𝑇𝐴𝐵 = −96.75 lb, 𝑇𝐴𝐶 = 114.3 lb; (b) 𝑘𝐴𝐵 = 297.4 lb∕in., 𝑘𝐴𝐶 = 461.4 lb∕in. 3.118 Essay-type answer 3.120 For scenario (a): 𝑇𝐴𝐵 = 4000 lb, for scenario (b): 𝑇𝐴𝐷 = 2611 lb 3.122 𝑇𝐴𝐸 = 44.6 lb, 𝑇𝐴𝐷 = 69.6 lb, 𝑇𝐷𝐹 = 192 lb, 𝑇𝐷𝐺 = 274 lb, 𝑇𝐴𝐵𝐶𝐷 = 100 lb 𝑇𝐷𝐺 3

𝑦

4

𝐷

𝑇𝐷𝐹

5 12

𝑊 ∕2

𝑇𝐴𝐷

𝑇𝐴𝐸

12

𝑥

5

3

4

𝐴 𝑊 ∕2

3.124 𝑇𝐴𝐵 = −2625 lb, 𝑇𝐵𝐶 = −1250 lb, 𝑇𝐵𝐷 = 2125 lb, 𝑇𝐶𝐷 = 750 lb, 𝑇𝐶𝐸 = 1000 lb 3.126 𝜃 = 19.1◦ , 𝑇𝐴𝐵 = 153 N 3.128 𝐹𝑆 = 1.5

A-13

ISTUDY

Appendix B

Answers to Even-Numbered Problems

3.130 𝐹 = 90.1 N 3.132 𝛿1 = 10.0 mm, 𝛿2 = 10.0 mm 100 N 𝑛

𝑇2

𝑡

30◦

100 N 30◦

𝑇1

30◦

30◦ 𝑇3

𝑅2

𝑅1

3.134 𝑇𝐴𝐵 = 9.33 kN, 𝑇𝐴𝐶 = 9.33 kN, 𝑇𝐴𝐷 = 16.0 kN, 𝑇𝐴𝐸 = 8.00 kN 3.136 𝑇𝐴𝐵 = 8.40 kN, 𝑇𝐴𝐶 = 4.40 kN, 𝑇𝐴𝐷 = 8.00 kN, 𝑇𝐴𝐸 = 8.00 kN 3.138 𝑇𝐴𝐵 = 275.4 lb, 𝑅𝑦 = 130.9 lb, 𝑅𝑛 = 479.8 lb 3.140 𝑃 = 30 lb, 𝑇𝐵𝐸 = 70 lb, reactions for collar 𝐵: 𝑅𝐵𝑥 = 60 lb, 𝑅𝐵𝑦 = −30 lb, reactions for collar 𝐸: 𝑅𝐸𝑥 = −60 lb, 𝑅𝐸𝑧 = 40 lb 3.142 The bead slides toward 𝐵 3.144 𝑇𝐴𝐶 = 1650 lb, 𝑇𝐵𝐶 = 106 lb, 𝑇𝐶𝐷 = 2330 lb, 𝐹𝐶𝐸 = −1471 lb, 𝐹𝐶𝐹 = −1471 lb, 𝑇𝐵𝐺 = 600 lb, 𝑃 = 106 lb 3.146 (a) 𝑇𝐴𝐵 = −5000 lb, 𝑇𝐴𝐶 = 7071 lb, 𝑑 = 5.063 in.; (b) 𝑃 = 3884 lb; (c) Essay-type answer

Chapter 4 Note: Throughout Chapter 4, unless otherwise stated, answers for two-dimensional problems are reported using positive values for counterclockwise moments and negative values for clockwise moments. Answers for three-dimensional problems are reported using positive values for moments that act in the positive coordinate directions and negative values for moments that act in the negative coordinate directions. 4.2 (a) 𝑑 = 6.00 in., 𝑀𝐵 = −150 in.⋅lb; (b) 𝑀𝐵 = −150 in.⋅lb; (c) 𝑀𝐵 = −150 in.⋅lb; (d) 𝑀𝐵 = −150 in.⋅lb; ⃗ = −150 𝑘̂ in.⋅lb (e) 𝑀 𝐵 4.4 (a) 𝑑 = 14.1 mm, 𝑀𝐵 = −42.4 N⋅mm; (b) 𝑀𝐵 = −42.4 N⋅mm; (c) 𝑀𝐵 = −42.4 N⋅mm; ⃗ = −42.4 𝑘̂ N⋅mm (d) 𝑀𝐵 = −42.4 N⋅mm; (e) 𝑀 𝐵 2 sin 𝛼 + 3 cos 𝛼 5 kN⋅m 4.6 𝑀𝐴 = √ 14 − 6 sin 𝛼 + 4 cos 𝛼 6 𝐷

4

𝑀𝐴 (kN⋅m)

A-14

2 𝐸

0 −2 −4 −6

0◦

60◦

𝛼

120◦

180◦

4.8 (a) 𝐹 = 20.1 lb; (b) 𝐹 = 56.1 lb; (c) 𝑘 = 13.0 lb∕in. 4.10 𝐹 = 640 lb, 𝛼 = 140◦ 4.12 (a) 𝑀𝐴 = 30.0 kN⋅m; (b) 𝑀𝐴 = 35.0 kN⋅m; (c) 𝑀𝐴 = 38.0 kN⋅m 4.14 (a) 𝑄 = 10.0 kN; (b) 𝑊 = 5.00 kN, 𝛼 = 53.1◦ Hint: In Part (b), the value of 𝑊 you determine must not produce a moment about point 𝐴 that exceeds 20 kN⋅m for any possible value of 𝛼. ( ) ⃗ = 194.2 𝚥̂ + 120 𝑘̂ in. ⋅ lb, 4.16 Open-end wrench: 𝑀 𝑂( ) ⃗ = −121.4 𝚤̂ + 194.2 𝚥̂ + 73.65 𝑘̂ in.⋅lb; The open-end wrench is probably more ratchet wrench: 𝑀 𝑂

effective.

ISTUDY

Appendix B

Answers to Even-Numbered Problems

4.18 𝐹 = −100 N

( ) ⃗ = 184.6 𝚤̂ + 3.077 𝚥̂ − 129.2 𝑘̂ kN⋅m 4.20 (a) and (b) 𝑀 𝐴 ( ) ( ) ⃗ = 3570 𝚤̂ + 286 𝚥̂ − 3000 𝑘̂ in.⋅lb; (b) 𝑀 ⃗ = 8030 𝚤̂ + 514 𝚥̂ − 3000 𝑘̂ in.⋅lb 4.22 (a) 𝑀 𝐵 𝑂 ⃗ = (−7200 𝚤̂ + 7200 𝚥̂) N⋅mm; (b) 𝑀 ⃗ = (29,600 𝚥̂) N⋅mm 4.24 (a) 𝑀 𝐴 𝐵 𝑧 𝐵 𝐹⃗𝐵𝐸 120 mm

𝐹⃗𝐵𝐷

𝐴 𝐹⃗𝐴𝐶 80 mm

𝐹⃗𝐴𝐷

𝑦

𝐸

150 mm

𝑂 𝐷 𝐶

60 mm

150 mm

𝑥

( ) ( ) ⃗ = −4600 𝚤̂ − 2880 𝑘̂ ft⋅lb; (b) 𝑀 ⃗ = −2300 𝚤̂ + 1920 𝚥̂ − 2880 𝑘̂ ft⋅lb; 4.26 (a) 𝑀 𝐵 𝐴 ( ) ⃗ = −3840 𝚤̂ + 7680 𝚥̂ − 2880 𝑘̂ ft⋅lb (c) 𝑀 𝑂 4.28 𝐹 = 1.75 kN ⃗ = 288 𝚤̂ N⋅m; (c) Essay-type answer 4.30 (a) 𝐹 = 240 N; (b) 𝑀 𝑂 4.32 Points 𝐴 and 𝐵 lie on a line that is parallel to the line of action of the force. ( ) ( ) 4.34 (a) 𝑀 = 1900 N⋅mm, 𝐹⃗ = −3.85 𝚤̂ + 9.23 𝑘̂ N, 𝐹⃗ = 3.85 𝚤̂ − 9.23 𝑘̂ N is also acceptable; 𝐴𝐵

(b) 𝑀𝐴𝐵 = 472 N⋅mm; (c) 𝑀𝐴𝐵 = 0 4.36 𝐹 = 92.6 lb 4.38 𝐹 = 115 N. Twisting occurs at both fittings on pipe 𝑂𝐴. dimensions in mm 350 𝐴 𝑦 30◦ 𝑂 350 𝐵 𝑑1 𝑑2 = 300 30◦ 𝐶 𝐹 𝑥

Strength

4.40 𝑀𝐴𝐵

Strength

= 8 in.⋅lb, 𝑀𝐵𝐶 = 4.992 in.⋅lb ( ) ( ) ⃗ = 12.0 𝚤̂ − 6.00 𝑘̂ N⋅mm, ⃗ = 2.29 𝚤̂ + 10.3 𝚥̂ + 2.29 𝑘̂ N⋅mm, 𝑀 = −8.00 N⋅mm; (b) 𝑀 4.42 (a) 𝑀 𝐴 𝐴𝐵 𝐵 𝑀𝐴𝐵 = −8.00 N⋅mm; (c) Essay-type answer

4.44 𝑀𝑎 = −19,850 N⋅cm 4.46 𝑀𝐴𝐵 = −817.5 N⋅cm 4.48 𝑀𝐵𝐶𝐷 = −263.8 in.⋅lb 4.50 𝑀𝑂𝐵 = 12.5 N⋅m ( ) ⃗ = 9830 𝚤̂ + 1940 𝚥̂ + 7770 𝑘̂ N⋅mm, 𝑀 = 11300 N⋅mm; (b) 𝑀 ⃗ = 14000 𝚤̂ N⋅mm, 4.52 (a) 𝑀 𝐴 𝑎 𝑂 𝑀𝑎 = 11300 N⋅mm ( ) 4.54 𝐹⃗ = 13.4 𝚥̂ + 6.71 𝑘̂ lb, 𝑀𝐺𝐻 = 537 in.⋅lb ( ) ⃗ = 6 𝚤̂ + 6 𝚥̂ + 3 𝑘̂ lb, 𝑀 = −290 in.⋅lb 4.56 𝑄 𝐵𝐶 4.58 𝑀𝑎 = −17,400 ft⋅lb, 𝛼 = 90◦

A-15

ISTUDY

Appendix B

Answers to Even-Numbered Problems

4.60 𝑀𝑂𝐵𝐴 = cos 𝛼 sin 𝛼 (1.8 kN⋅m) 1.0 0.8 𝑀𝑂𝐵𝐴 (kN⋅m)

A-16

0.6 0.4 0.2 0.0

0◦

20◦ 𝛼

10◦

30◦

40◦

4.62 𝑃max = 1546 N 4.64 𝑀 = 144 in.⋅lb 4.66 (a) 𝑄 = 120 N; (b) The answer does not change. 4.68 𝐹𝐵 = 80 lb, 𝛽 = 70◦ , 𝑀 = 3363 in.⋅lb 4.70 (a) 𝑀 = 96.6 N⋅mm; (b) 𝑀 = 96.6 N⋅mm 4.72 (a) 𝐹𝐴 = 200 N, 𝛼 = 22.62◦ , 𝑀 = 2400 N⋅cm; (b) 𝐹1 = 𝐹2 = −123.8 N; (c) 𝑄1 = 𝑄2 = 133.3 N; (d) 𝑀𝐴 = 2400 N⋅cm 4.74 𝐹𝐵 = 18.93 lb ( ) ⃗ = 200 𝚤̂ − 600 𝑘̂ N⋅m 4.76 𝑀 4.78 𝑑 = 1.92 m ⃗ = (−800 𝚤̂ − 200 𝚥̂) N⋅m; (b) Essay-type answer 4.80 (a) 𝑀 4.82 The forces 𝐹 are resolved into horizontal and vertical components as shown. The two vertical forces 𝐹 cos 30◦ constitute one couple. The two horizontal forces 𝐹 sin 30◦ and the horizontal force 𝐹 constitute another couple. 𝑦

𝐹 𝐹 cos 30◦

𝑥

𝐹 sin 30◦

𝐹 sin 30◦ 𝐹 cos 30◦

4.84 Force systems (a) and (c) are equivalent. 4.86 Force systems (a) and (d) are equivalent, and force systems (b) and (c) are equivalent. 4.88 For force system (b), 𝐹 = 2 kip and 𝑃 = 4 kip; for force system (c), there are no values for 𝐹 and 𝑃 such that force systems (a) and (c) will be equivalent; for force system (d), 𝐹 = 5 kip and 𝑃 = −1 kip. 4.90 𝐹𝑅𝑥 = −1490 lb, 𝐹𝑅𝑦 = −102 lb, 𝑀𝑅 −5110 in.⋅lb. The original force system and the force system at point 𝑂 are shown below. 𝑦 200 lb cos 20◦ 200 lb sin 20◦ 𝐴 𝐷 400 lb cos 20◦ 300 lb 400 lb sin 20◦ 𝑥 𝑂 ◦ ◦ 300 lb 400 lb sin 20 400 lb cos 20 𝐶 200 lb sin 20◦ 𝐵

𝑦 102 lb 1490 lb 𝑂 5110 in.⋅lb

𝑥

200 lb cos 20◦

( ) ( ) ⃗ ̂ 4.92 𝐹⃗𝑅 = 144.5 𝚤̂ + 2091 𝚥̂ + 17,000 𝑘̂ lb, 𝑀 𝑅𝑂 = −31,650 𝚤̂ + 2402 𝚥̂ − 795.2 𝑘 ft⋅lb 4.94 𝐹𝑅𝑥 = −0.500 kN, 𝐹𝑅𝑦 = −3.87 kN, 2.58 m to the right of the bearing at 𝐶

ISTUDY

Appendix B

Answers to Even-Numbered Problems

4.96 𝐹𝑅𝑥 = −49.24 lb, 𝐹𝑅𝑦 = −41.32 lb, 𝑥 = 9.342 in. ( ) ( ) ⃗ ⃗ ̂ 4.98 (a) 𝐹⃗𝑅 = −1800 𝑘̂ N, 𝑀 𝑅𝑂 = (6.94 𝚤̂ − 12.0 𝚥̂) N⋅m; (b) 𝐹𝑅 = −1800 𝑘 N, (𝑥, 𝑦) = (−3.86 mm, −6.67 mm) ( ) ( ) ⃗ 4.100 (a) 𝐹⃗𝑅 = −5.00 𝚤̂ + 50.0 𝚥̂ − 10.0 𝑘̂ nN, 𝑀 = 120 𝚤̂ − 60.0 𝑘̂ nN⋅𝜇m; 𝑅𝐵 ( ) ( ) ⃗ ̂ (b) 𝐹⃗𝑅 = −5.00 𝚤̂ + 50.0 𝚥̂ − 10.0 𝑘̂ nN, 𝑀 𝑅𝑂 = 220 𝚤̂ + 800 𝚥̂ + 3890 𝑘 nN⋅𝜇m ( ) ⃗ ̂ 4.102 (a) 𝐹⃗𝑅 = (10.0 𝚥̂) kN, 𝑀 𝑅𝐴 = 25.0 𝚤̂ + 4.00 𝚥̂ + 20.0 𝑘 kN⋅m; (b) 𝐹𝑅𝑥 = 0, 𝐹𝑅𝑦 = 10.0 kN, 𝐹𝑅𝑧 = 0, 𝑀𝑅𝐴𝑥 = 25.0 kN⋅m, 𝑀𝑅𝐴𝑦 = 4.00 kN⋅m, 𝑀𝑅𝐴𝑧 = 20.0 kN⋅m ̂ 𝑀 ⃗ = 3𝑃 𝑟 𝑘, ̂ (𝑥, 𝑦) = (0, 𝑟) 4.104 𝐹⃗𝑅 = −2𝐹 𝑘, 𝑅 ̂ N, 𝑀 ⃗ ̂ 4.106 𝐹⃗𝑅 = (2.00 𝚤̂ + 3.00 𝚥̂ − 4.00 𝑘) 𝑅𝐴 = (9.00 𝚤̂ + 8.00 𝚥̂ − 4.00 𝑘) N⋅m ( ) ( ) ( ) ⃗ = 𝐹 𝑎 1 −̂𝚥 + 𝑘̂ , (𝑥, 𝑦) = 1 𝑎, 𝑏 4.108 𝐹⃗𝑅 = 𝐹 −̂𝚥 + 𝑘̂ , 𝑀 𝑅 2 2 ( ) ⃗ = 150 𝚤̂ + 104 𝚥̂ − 200 𝑘̂ N⋅m 4.110 𝑀 𝑅 4.112 𝐹 = 444.4 N 4.114 (a) 𝑀𝑎 = 256 N⋅m; (b) 𝑢̂ = −0.530 𝚤̂ + 0.662 𝚥̂ + 0.530 𝑘̂ 4.116 𝐹 = 300 lb 4.118 𝑀𝑎 = −311.0 in.⋅lb 4.120 𝑀 = −9638 N⋅mm 4.122 𝑃 = −10 lb, 𝑄 = −5 lb, 𝑅 = 0 , 𝑆 = −20 lb, 𝑀𝐷 = −10 in.⋅lb 4.124 (a) 𝐹𝑅𝑥 = −0.880 lb, 𝐹𝑅𝑦 = −0.660 lb, 𝑀𝑂 = 0.552 in.⋅lb; (b) 𝐹𝑅𝑥 = −0.880 lb, 𝐹𝑅𝑦 = −0.660 lb, 𝑦 = 0.627 in. 𝑦

𝑦

0.660 lb 0.660 lb 𝑂 0.880 lb

𝑦 𝑥

0.880 lb 𝑂

𝑥

0.522 in.⋅lb

force system 2

force system 3

4.126 (a) 𝐹𝑅𝑥 = 0, 𝐹𝑅𝑦 = −1800 lb, 𝐹𝑅𝑧 = 0, 𝑀𝐴𝑥 = 8700 ft⋅lb, 𝑀𝐴𝑦 = 0, 𝑀𝐴𝑧 = 26,700 ft⋅lb; (b) 𝐹𝑅𝑥 = 0, 𝐹𝑅𝑦 = −1800 lb, 𝐹𝑅𝑧 = 0, 𝑥 = 12.2 f t, 𝑧 = 7.83 f t ( ) ( ) ⃗ ̂ 4.128 𝐹⃗𝑅 = −30.0 𝚤̂−120 𝚥̂−22.5 𝑘̂ lb, 𝑀 𝑅𝑂 = −990 𝚤̂+60.0 𝚥̂+1180 𝑘 in.⋅lb

Chapter 5 Note: Throughout Chapter 5, unless otherwise stated, the following conventions apply. Answers for two-dimensional problems are referred to an 𝑥𝑦 coordinate system, where 𝑥 is horizontal with positive to the right and 𝑦 is vertical with positive upward. Answers are reported so that positive forces act in positive coordinate directions, and positive moments are counterclockwise. For three-dimensional problems, moments are reported using positive values for moments that act in the positive coordinate directions. 5.2 𝐴 = 23.5 N, 𝐵 = 107 N 55 N 45 N 30 N

𝑦 𝑥 𝐴

𝐵

A-17

A-18

ISTUDY

Appendix B

Answers to Even-Numbered Problems

5.4 𝐹 = 120 lb, 𝐵 = 213 lb, 𝐴 = −53.5 lb, essay-type answer 200 lb

𝐹 𝐴

𝑦 𝐵 𝑥

12 9 15

5.6 𝑇𝐷𝐸 = −258.1 lb, 𝑁 = 377.6 lb, 𝑇𝐵𝐶 = 77.30 lb 5.8 𝐵 = 42.9 N (force supported by one link), 𝐴𝑥 = 68.6 N, 𝐴𝑦 = −31.4 N

5.10 If 0 ≤ 𝛽 < 51.32◦ , the spool will roll to the left, and if 51.32◦ < 𝛽 ≤ 180◦ , the spool will roll to the right. 5.12 𝑑 = 3.10 f t 5.14 (a) 𝐴𝑦 = 23 𝐹 + 13 𝑃 , 𝐴𝑥 = 0, 𝐷𝑦 = 13 𝐹 + 23 𝑃 ; (b) 𝐴𝑥 = 0.192𝐹 + 0.385𝑃 , 𝐴𝑦 = 𝑀𝐷 = 𝐴𝑥 𝐴𝑦

2 𝐹 + 13 𝑃 , 𝐷 3 1 𝐿(𝐹 + 2𝑃 ); 3 𝑃 𝐹 𝑦 𝑥

= 0.385𝐹 + 0.770𝑃 ; (c) 𝐴𝑦 = 𝐹 + 𝑃 , 𝐷𝑥 = 0, (d) 𝐴𝑥 = 0, 𝐴𝑦 = 𝐹 + 𝑃 , 𝑀𝐴 = 13 𝐿(𝐹 + 2𝑃 ) 𝐹

𝐴𝑥

𝐷𝑦

𝐴𝑦

𝑃 𝑦

𝐹 30◦

𝑥

𝐷

𝐴𝑦

𝑃

𝑀𝐷

𝐴𝑥

𝐷𝑥

𝑦 𝑥

𝑀𝐴

𝐴𝑦

𝐹

𝑃 𝑦 𝑥

5.16 𝐴𝑦 = 1.88 kN, 𝐸𝑦 = 2.12 kN, 𝐸𝑥 = −2.00 kN 5.18 𝐴𝑥 = 0, 𝐴𝑦 = 3.00 kip, 𝑃𝑦 = 1.00 kip 5.20 𝐶𝑦 = 11,300 lb, 𝐷𝑦 = 6170 lb 5.22 𝐴𝑥 = 0, 𝐵𝑦 = 𝑃 , 𝑀𝐴 = 𝑃 𝑎 5.24 𝐶𝑦 = 1516 lb, 𝐴𝑥 = 845.2 lb, 𝐴𝑦 = 582.5 lb 5.26 (a) 𝑇 = 75 lb, 𝐵𝑥 = 112.4 lb, 𝐵𝑦 = 49.41 lb; (b) 𝐹𝐶𝐷 = 134.2 lb, 𝐵𝑥 = 12.44 lb, 𝐵𝑦 = 139.4 lb 5.28 𝐶𝑛 = 11,940 N, 𝐵𝑥 = 10,880 N, 𝐵𝑦 = 6856 N 5.30 (a) 𝑃 = 50 N; (b) Rollers 𝐵 and 𝐶 make contact, 𝐴𝑦 = 0, 𝐵𝑦 = −10 N, 𝐶𝑦 = 100 N, 𝐷𝑦 = 0 5.32 Rollers 𝐵 and 𝐶 make contact, 𝐴𝑦 = 0, 𝐵𝑦 = −40.8 lb, 𝐶𝑦 = 231 lb, 𝐷𝑦 = 0 5.34 Rollers 𝐴 and 𝐶 make contact, 𝐴𝑦 = 193 lb, 𝐵𝑦 = 0, 𝐶𝑦 = 198 lb, 𝐷𝑦 = 0 5.36 (a) 𝐴 = 498 lb; (b) Essay-type answer 5.38 (a) 𝛼 > 75.5◦ ; (b) 𝛼 > 90.0◦ 5.40

Concept Problem

5.42 𝐹𝑥 = 9.08 lb, 𝐹𝑦 = −6.91 lb, 𝐸𝑦 = 19.1 lb 5.44 𝑇𝑡 = 759.7 N, 𝐵𝑥 = −777.3 N, 𝐵𝑦 = 389.1 N 5.46 𝐺 = 4.66 N, 𝑁 = 17.4 N, 𝜃 = 17.0◦ 5.48 𝐴𝑥 = 0, 𝐴𝑦 = 200 lb, 𝑇 = 600 lb 5.50 𝐴𝑥 = 21.92 lb, 𝐴𝑦 = −54.22 lb, 𝑇 = 28.61 lb 5.52 𝐹𝐵𝐸 = −4.472 kN, 𝐶𝑥 = 2 kN, 𝐶𝑦 = −1 kN 5.54 (a) 𝐴𝑥 = 2.59 kN, 𝐴𝑦 = −35.2 kN, 𝐵𝑦 = 54.8 kN; (b) 𝐴𝑥 = 7.59 kN, 𝐴𝑦 = −41.2 kN, 𝐵𝑦 = 59.5 kN 5.56 𝐹𝐵𝐸 = −143.5 lb, 𝐴𝑥 = −64.62 lb, 𝐴𝑦 = −89.42 lb

ISTUDY

Appendix B

Answers to Even-Numbered Problems

5.58 𝑇 = 1.479 kip (tension in cable 𝐵𝐶𝐷), 𝐴𝑥 = −1.474 kip, 𝐴𝑦 = 0.8711 kip, 𝑇𝐷𝐸 = 2.456 kip, 𝐹𝐷𝐹 = 2.094 kip 5◦ 1 kip

𝑇 𝑇

5◦

𝑇 𝑦

𝐴𝑥

10◦

𝑇

𝑦

𝐴𝑦

3 4

𝑥

𝐹𝐷𝐹 𝑇𝐷𝐸

𝑥

5.60 (a) Complete fixity, statically determinate; (b) 𝑇 = 10.68𝐹 ; Additional results: 𝐶𝑥 = −2.000𝐹 , 𝐶𝑦 = −11.32𝐹 √ 5

𝑦 1

2

𝑇

𝑥

𝑇

√ 2 1

𝐶𝑥

𝐶𝑦

𝐹

1

5.62 (a) Partial fixity, statically indeterminate; (b) The cable tension cannot be determined. Additional result: The equations of static equilibrium require 𝐹 = 0. If 𝐹 ≠ 0, then the equations of static equilibrium cannot be satisfied, and dynamic motion occurs. √ 2

𝐹

1

1

√ 2 1

1

𝑇

𝑇

𝑦

𝐶𝑥

𝐶𝑦 𝑥

5.64 𝐹 = 13.6 lb 5.66 𝑄 = 191.5 N, 𝐶𝑥 = −63.40 N, 𝐶𝑦 = 170.6 N 5.68 𝐹𝑠 = 200 lb, 𝑃 = 81.84 lb, 𝐴𝑥 = 127.4 lb, 𝐴𝑦 = −77.75 lb 5.70 The pretwist of the torsional spring is 0.7003 turns, 𝑃 = 11 lb. 5.72 (a) 𝜃 = 4.58◦ ; (b) 𝜃 = 36.7◦ ; (c) Essay-type answer 𝐹 𝑦 𝑥 𝜃 𝑀 = 𝑘𝜃

𝐴𝑥 𝐴𝑦

5.74 Essay-type answer [ 5.76 𝐴𝑥 = 0, 𝐴𝑦 = 𝐹 1 −

[ ] ] 𝑘𝐿2 𝐹 𝑘𝐿2 𝐹𝐿 𝑘𝐿2 1− , 𝑀𝐴 = , 𝐶𝑦 = 2 2(𝑘𝑡 + 𝑘𝐿2 ) 2(𝑘𝑡 + 𝑘𝐿2 ) 𝑘𝑡 + 𝑘𝐿2 5.78 For structure (a), full fixity and statically determinate; for structure (b), full fixity and statically indeterminate; for structure (c), full fixity and statically determinate; for structure (d), partial fixity and statically determinate; for structure (e), full fixity and statically determinate; for structure (f), partial fixity and statically indeterminate 5.80 𝐷 = 0.314 N, 𝐿 = 1.34 N 5.82 𝛿 = 2.40 mm, 𝜃 = 1.49◦

5.84 (a) 𝑇𝐸𝐴 = 1.33 kip, 𝐵𝑥 = 0, 𝐵𝑦 = 3.33 kip; (b) 𝑇𝐸𝐴 = 12.3 kip, 𝐵𝑥 = 0, 𝐵𝑦 = 20.3 kip; (c) 𝑇𝐸𝐴 = 15.9 kip, 𝐵𝑥 = 0, 𝐵𝑦 = 25.7 kip 5.86 𝐴𝐵𝐶 is a three-force member and 𝐵𝐷 is a two-force member. 5.88 𝐴𝐵𝐷𝐸 is a multiforce member and 𝐵𝐶 is a two-force member. 5.90 Three-force member 5.92 Three-force member 5.94 𝐴𝐵𝐶𝐷 is a multiforce member and 𝐷𝐸 is a 2-force member.

A-19

A-20

ISTUDY

Answers to Even-Numbered Problems

Appendix B

5.96 Multiforce member 5.98 Three-force member 5.100 𝐹𝐴𝐵 = 318 kN to begin opening the dump when 𝜃 = 0◦ ; 𝐹𝐴𝐵 = 0 when the center of gravity 𝐺 of the dump and its contents is immediately above point 𝑂. 5.102 𝐶𝑥 = −0.333 kN, 𝐶𝑦 = 0.667 kN, 𝑇𝐴𝐷 = 1.25 kN, 𝑀𝐶𝑥 = 0, 𝑀𝐶𝑦 = 2.00 kN⋅m, 𝑀𝐶𝑧 = 3.33 kN⋅m 𝑧 𝑀𝐶 𝑦

𝑀𝐶 𝑧

𝐶𝑦

𝐷 𝐶𝑥

𝑀𝐶𝑥 𝑇𝐴𝐷

𝑥

𝑦

ring inee Engibrary L

1 kN

5.104 𝐴𝑥 = 0, 𝐴𝑦 = 0, 𝐴𝑧 = 10 lb, 𝑀𝐴𝑥 = 120 in.⋅lb, 𝑀𝐴𝑧 = 0, 𝐹𝑧 = 20 lb 5.106 (a) Partial fixity and statically determinate; (b) 𝑇 = 27 lb, 𝐴𝑥 = 18.0 lb, 𝐴𝑧 = 9.00 lb, 𝑀𝐴𝑥 = 54.0 in.⋅lb, 𝑀𝐴𝑧 = 0 𝑦 𝐷

18 lb

𝑀𝐴𝑥

𝐴𝑥

𝐵

𝐴𝑧

𝐶

𝑀𝐴𝑧

𝑧

𝑥

𝑇

5.108 𝑇𝐴𝐷 = 50.75 kN Concept Problem. (b) 2, 0, 4, 1, 3

5.110

𝑧 𝐸

𝑀2

𝐴

𝐹2 𝐹 1

𝑀1

𝐵

𝐹𝐴𝐶

𝐶

𝑦

𝑇𝐷𝐸 𝑥

200 lb

5.112 𝐴𝑥 = 0, 𝐴𝑦 = 0, 𝐴𝑧 = 2𝐹 , 𝑀𝐴𝑥 = −2𝐹 𝑟, 𝑀𝐴𝑦 = 0, 𝑀𝐴𝑧 = −2𝑃 𝑟 𝑧

𝐹 𝑃

𝐹

𝑟

𝑃

𝑦 𝐴

𝑥

𝐴𝑦

𝐴𝑥

𝑀𝐴𝑦

𝐴𝑧 𝑀𝐴𝑥

𝑀𝐴𝑧

5.114 𝐵𝑥 = −0.01 lb, 𝐵𝑦 = −0.001429 lb, 𝐵𝑧 = 0.1150 lb, 𝑀𝐵𝑥 = −0.08571 in.⋅lb, 𝐸𝑦 = 0.001429 lb, 𝐸𝑧 = 0.0850 lb 5.116 𝐹 = 3.21 kN, 𝐶𝑥 = −3.00 kN, 𝐶𝑦 = −0.429 kN, 𝐶𝑧 = 2.07 kN, 𝑀𝐶𝑥 = 0.343 kN⋅m, 𝑀𝐶𝑧 = −0.600 kN⋅m

ISTUDY

Appendix B

Answers to Even-Numbered Problems

5.118 𝐴 = 5.77 kN, 𝐶𝑥 = 2.89 kN, 𝐷𝑦 = 5.00 kN, 𝐷𝑧 = 0, 𝑀𝐷𝑦 = 0, 𝑀𝐷𝑧 = −9.23 kN⋅m 𝑦 𝑀𝐷𝑦 𝑀𝐷𝑧

𝐷𝑦

𝐷𝑧

𝑃 = 10 kN

𝑥

𝐶𝑥 𝑧

𝐴𝑥 = 𝐴 sin 30◦

𝐴𝑦 = 𝐴 cos 30◦ 30◦ 𝐴

5.120 Statically determinate, 𝑄 = 833.3 lb, 𝐴𝑥 = 1667 lb, 𝐴𝑦 = 0, 𝐴𝑧 = 4313 lb, 𝐶𝑥 = −2500 lb, 𝐶𝑧 = −162.5 lb 5.122 (a) Statically indeterminate; (b) 𝑇𝐶𝐹 = 2289 N 5.124 (a) 𝐴𝑥 = 38 kN, 𝐴𝑦 = 50 kN, 𝑀𝐴 = −1281 kN⋅m; (b) 𝑇𝐵𝐹 = 153.8 kN; (c) 𝑇𝐵𝐹 = 93.76 kN Concept Problem. Partial answer: The propellers rotate counterclockwise.

5.126

5.128 𝑇 = 163 N, 𝐴 = −83.2 N, 𝐵 = 163 N 80 N 80 N cos 30◦

30◦

𝑦

80 N sin 30◦

𝑥

30◦

20 mm

𝑂

30◦

60◦

30◦

𝐵 30◦

𝑇

𝐴

5.130 𝐴𝑥 = 7.00 N, 𝐴𝑦 = 26.1 N, 𝑀𝐴 = −276 N⋅cm 5.132 𝑇 = 16.74 lb, 𝑇𝐽 𝐾 = 14 lb 5.134 𝑘𝑡 = 1.034 in.⋅lb∕rad, 𝜃0 = 3.142 rad 5.136 (a) 𝐹 = 3.29 N; (b) 𝐹 = 11.5 N; (c) 𝐹 = 20.1 N 5.138 (a) Statically indeterminate; (b) 𝐹𝐴𝐸 = 833.3 lb (compression), 𝑇𝐶𝐺 = 1167 lb (tension), 𝐵𝑥 = −1732 lb, 𝐵𝑦 = 666.7 lb 5.140 For structure (a), full fixity and statically determinate; for structure (b), partial fixity and statically indeterminate; for structure (c), full fixity and statically indeterminate; for structure (d), partial fixity and statically determinate; for structure (e), partial fixity and statically determinate; for structure (f), full fixity and statically indeterminate

(a)

(b)

(c)

(d)

(e)

(f )

A-21

A-22

ISTUDY

Appendix B

Answers to Even-Numbered Problems

5.142 (a) Essay-type answer; (b) 𝑇1 = 15.7 lb, 𝑇2 = 21.5 lb, 𝑂𝑥 = 13.3 lb, 𝑂𝑦 = 11.5 lb, 𝑂𝑧 = −19.4 lb 5.144 𝑃 = 444.4 lb, 𝐴𝑥 = 307.2 lb, 𝐴𝑦 = 204.5 lb, 𝐴𝑧 = −355.6 lb, 𝐵𝑥 = 209.5 lb, 𝐵𝑦 = −54.50 lb 5.146 (a) Full fixity and statically determinate; (b) 𝐴𝑦 = 0, 𝐴𝑧 = 60.0 lb, 𝐵𝑥 = −40.0 lb, 𝐵𝑧 = 40.0 lb, 𝑀𝐴𝑦 = 200 in.⋅lb, 𝑀𝐴𝑧 = −80.0 in.⋅lb 𝑧

𝐵𝑥

100 lb

𝑦 𝑀𝐴𝑦

𝐴𝑦 𝐵𝑧

𝐴𝑧

𝑥

40 lb

𝑀𝐴𝑧

5.148

Concept Problem. d – Object 1 could never be in equilibrium; Object 2 could be in equilibrium.

5.150

Concept Problem. c – Object 1 could be in equilibrium; Object 2 could never be in equilibrium.

Chapter 6 6.2 𝑇𝐴𝐵 = −500 lb, 𝑇𝐴𝐶 = 0, 𝑇𝐵𝐶 = 625 lb, 𝑇𝐵𝐷 = −375 lb, 𝑇𝐶𝐷 = 625 lb, 𝑇𝐶𝐸 = 0, 𝑇𝐷𝐸 = −500 lb 6.4 𝑃 = 3200 lb 6.6 𝑃 = 8000 lb 6.8 𝑇𝐶𝐸 = 3.75 kN, 𝑇𝐷𝐹 = −3.66 kN 6.10 𝑇𝐴𝐵 = 4.00 kN, 𝑇𝐴𝐶 = 0, 𝑇𝐵𝐶 = −7.21 kN, 𝑇𝐵𝐷 = 6.00 kN, 𝑇𝐶𝐷 = −5.00 kN, 𝑇𝐶𝐸 = −6.00 kN, 𝑇𝐷𝐸 = 7.21 kN 𝑄 𝐷

𝑇𝐵𝐷

𝐵

2

√ 13 3

𝑇𝐵𝐶

𝑇𝐴𝐵

𝑦 𝑥

𝐶

𝑇𝐴𝐶

𝐴

𝑇𝐷𝐸

𝑇𝐶𝐷

𝑃

𝐸

𝑇𝐶𝐸

𝑃

𝑄 + 2𝑃

6.12 𝑃 = 10.0 kN 6.14 𝑃 = 6.66 kN 6.16 1.92 6.18 1.77 6.20 𝑇𝐴𝐵 = −𝐹 , 𝑇𝐴𝐸 =

√ √ 2𝐹 , 𝑇𝐶𝐷 = 0, 𝑇𝐶𝐸 = 0, 𝑇𝐷𝐵 = − 2𝐹 , 𝑇𝐷𝐸 = 𝐹 , 𝑇𝐸𝐵 = −𝐹

6.22 𝑇𝐴𝐵 = 3333 lb, 𝑇𝐴𝐶 = −2667 lb, 𝑇𝐵𝐶 = −4667 lb, 𝑇𝐵𝐸 = 3771 lb, 𝑇𝐶𝐸 = −2667 lb 6.24 𝑇𝐴𝐵 = −5.66 kN, 𝑇𝐴𝐷 = 4.00 kN, 𝑇𝐵𝐶 = −7.07 kN, 𝑇𝐵𝐷 = 2.24 kN, 𝑇𝐵𝐸 = 3.00 kN, 𝑇𝐵𝐹 = 4.47 kN, 𝑇𝐶𝐹 = 5.00 kN, 𝑇𝐷𝐸 = 3.00 kN, 𝑇𝐸𝐹 = 3.00 kN 𝐵 𝑦 𝑇𝐴𝐵 𝑇𝐵𝐷

𝑥

𝐴𝑥 = 0

𝐴

𝑇𝐴𝐵

𝑇𝐵𝐷

45◦ 𝐷 𝑇𝐴𝐷

63.43◦

𝐴𝑦 = 4 kN

6.26 𝐵𝐷, 𝐶𝐷, 𝐺𝐻 6.28 𝐴𝐵, 𝐶𝐷, 𝐸𝐹

2 kN

𝑇𝐵𝐸 𝑇𝐵𝐸 𝐸

𝑇𝐵𝐶 𝑇𝐵𝐹 𝑇𝐵𝐹

𝑇𝐵𝐶

63.43◦

𝐹

𝑇𝐸𝐹

𝑇𝐷𝐸 3 kN

45◦

𝐶

𝑇𝐶𝐹 4 kN

𝐶𝑦 = 5 kN

ISTUDY

Appendix B

Answers to Even-Numbered Problems

6.30 𝐵𝐷, 𝐶𝐷, 𝐸𝐹 6.32 (a) 𝐹 𝐺, 𝐷𝐺; (b) 𝑇𝐹 𝐺 = 0, 𝑇𝐷𝐺 = 0, 𝑇𝐴𝐶 = −12.0 kN, 𝑇𝐴𝐵 = 10.4 kN, 𝑇𝐶𝐵 = 4.00 kN, 𝑇𝐵𝐷 = 10.4 kN, 𝑇𝐺𝐻 = −8.00 kN, 𝑇𝐹 𝐻 = 6.93 kN, 𝑇𝐸𝐺 = −8.00 kN, 𝑇𝐶𝐸 = −8.00 kN, 𝑇𝐷𝐸 = 8.00 kN, 𝑇𝐶𝐷 = −4.00 kN, 𝑇𝐷𝐹 = 6.93 kN 6.34 𝑇𝐴𝐵 = −3 kip, 𝑇𝐴𝐶 = −6.25 kip, 𝑇𝐴𝐷 = 3.75 kip, 𝑇𝐵𝐷 = −4 kip, 𝑇𝐶𝐷 = −5 kip, 𝑇𝐶𝐸 = −15 kip, 𝑇𝐶𝐹 = 6.25 kip, 𝑇𝐷𝐹 = −6.75 kip 6.36 (a) 𝐵𝐺, 𝐷𝐼, 𝐸𝐽 , 𝐷𝐸, 𝐶𝐷, 𝐶𝐻; (b) 𝑇𝐵𝐺 = 0, 𝑇𝐶𝐻 = 0, 𝑇𝐴𝐵 = −3.00 kip, 𝑇𝐴𝐹 = −5.20 kip, 𝑇𝐵𝐶 = −3.00 kip, 𝑇𝐶𝐹 = −7.21 kip, 𝑇𝐹 𝐺 = 6.00 kip, 𝑇𝐶𝐺 = 5.00 kip, 𝑇𝐺𝐻 = 3.00 kip; (c) 𝑇𝐷𝐼 = 0, 𝑇𝐸𝐽 = 0, 𝑇𝐷𝐸 = 0, 𝑇𝐶𝐷 = 0, 𝑇𝐶𝐽 = −14.4 kip, 𝑇𝐼𝐽 = 9.00 kip, 𝑇𝐶𝐼 = 10.0 kip, 𝑇𝐻𝐼 = 3.00 kip 6.38 𝑇𝐴𝐵 = 0, 𝑇𝐴𝐷 = −20 lb, 𝑇𝐵𝐶 = 0, 𝑇𝐵𝐷 = −29.15 lb, 𝑇𝐵𝐹 = −29.15 lb, 𝑇𝐶𝐹 = −30 lb, 𝑇𝐷𝐸 = 15 lb, 𝑇𝐷𝐺 = −65 lb, 𝑇𝐸𝐹 = 15 lb, 𝑇𝐸𝐺 = −29.15 lb, 𝑇𝐸𝐻 = −29.15 lb, 𝑇𝐹 𝐻 = −85 lb, 𝑇𝐺𝐻 = 15 lb 6.40 𝑇𝐵𝐷 = −0.577𝑃 , 𝑇𝐶𝐷 = 0.577𝑃 , 𝑇𝐶𝐸 = 0.289𝑃 6.42 𝑇𝐶𝐸 = 5.667 kN, 𝑇𝐷𝐸 = −4.807 kN, 𝑇𝐷𝐹 = −9.000 kN 6.44 𝐹𝐶𝐷 = −485.3 lb, 𝐹𝐶𝐼 = −159.1 lb, 𝐹𝐼𝐽 = 970.6 lb 6.46 (a) 𝐷𝐸, 𝐻𝐼, 𝐽 𝐾, 𝐿𝑀, 𝑁𝑂; (b) Eliminate 𝐷𝐸, 𝐻𝐼, 𝐿𝑀, essay-type answer; (c) 𝑇𝐺𝐻 = 58 𝑄 6.48 (a) 𝐻𝐼, 𝐽 𝐾, 𝐿𝑀; (b) Eliminate 𝐽 𝐾, essay-type answer; (c) 𝑇𝐹 𝐺 = 2.00 kN 6.50 (a) Statically determinate; (b) 𝑇𝐶𝐷 = −8.00 kip, 𝑇𝐷𝐸 = 0, 𝑇𝐴𝐷 = 10.4 kip, 𝑇𝐵𝐷 = −10.4 kip 6.52 𝑇𝐶𝐸 = −18,667 lb, 𝑇𝐷𝐸 = 5833 lb, 𝑇𝐸𝐹 = −3000 lb, 𝑇𝐸𝐺 = −14,000 lb 6.54 𝑇𝐺𝐼 = −10,000 lb, 𝑇𝐻𝐼 = 4167 lb, 𝑇𝐼𝐽 = 0, 𝑇𝐼𝐾 = −4000 lb, 𝑇𝐼𝐿 = −3333 lb 6.56 (a) 𝑇𝐽 𝑇 = 40.2 kN; (b) 𝑇𝐻𝐽 = 30.0 kN, 𝑇𝐼𝐽 = 14.1 kN, 𝑇𝐽 𝐾 = 0, 𝑇𝐽 𝑀 = 0.563 kN, 𝑇𝐽 𝐿 = 0.796 kN 6.58 (a) Statically indeterminate; (b) Mechanism; (c) Statically indeterminate; (d) Statically determinate 6.60 𝑇𝐻𝐽 = −12 kip, 𝑇𝐻𝐾 = 0, 𝑇𝐼𝐾 = 0, 𝑇𝐼𝐿 = 12 kip 6.62 (a) For truss (a): 𝑇𝐴𝐵 = −1.41𝑊 , 𝑇𝐵𝐷 = −𝑊 , 𝑇𝐷𝐹 = −1.41𝑊 , 𝑇𝐴𝐶 = 𝑊 , 𝑇𝐶𝐸 = 𝑊 , 𝑇𝐸𝐹 = 𝑊 , 𝑇𝐵𝐶 = 𝑊 , 𝑇𝐶𝐷 = 0, 𝑇𝐷𝐸 = 𝑊 ; for truss (b): 𝑇𝐴𝐵 = −1.12𝑊 , 𝑇𝐵𝐷 = −0.500𝑊 , 𝑇𝐷𝐹 = −1.12𝑊 , 𝑇𝐴𝐶 = 0.500𝑊 , 𝑇𝐶𝐸 = 0.500𝑊 , 𝑇𝐸𝐹 = 0.500𝑊 , 𝑇𝐵𝐶 = 𝑊 , 𝑇𝐶𝐷 = 0, 𝑇𝐷𝐸 = 𝑊 ; (b) Essay-type answer; (c) Essay-type answer; (d) Essay-type answer 6.64 𝑃crit = 𝑘𝑡 ∕𝐿 6.66 (a) Partial fixity; (b) Statically determinate; (c) 𝑇𝐴𝐵 = 0, 𝑇𝐴𝐶 = 0, 𝑇𝐴𝐷 = 4.00 kN, 𝑇𝐵𝐶 = 0.417 kN, 𝑇𝐵𝐷 = −2.13 kN, 𝑇𝐵𝐸 = −0.838 kN, 𝑇𝐶𝐷 = −2.13 kN, 𝑇𝐶𝐸 = −0.838 kN, 𝑇𝐷𝐸 = 1.33 kN 6.68 𝑇𝐴𝐶 = −76.04 N, 𝑇𝐵𝐶 = 0, 𝑇𝐶𝐷 = 0, 𝑇𝐶𝐹 = 77.71 N, 𝑇𝐶𝐺 = 0, 𝑇𝐶𝐻 = −72.5 N 6.70 Statically determinate 6.72 𝑇𝑀𝐽 = −10.0 kN 6.74 𝑇𝐽 𝐺 = −15.0 kN 6.76

𝑄 𝐺𝑦 𝑦

𝐺𝑥 𝑥

𝐹𝐸𝐺

𝐹𝐸𝐺 𝐹𝐸𝐺 𝐹𝐸𝐺 𝐷𝑥 𝐷𝑦

Neglect the weight of individual members.

𝐹𝐶 𝐹𝐶

5N

𝐵𝑥

𝐵𝑦

A-23

A-24

ISTUDY

Appendix B

Answers to Even-Numbered Problems

6.78 𝐴𝑥 = 430.8 N, 𝐴𝑦 = 1984 N, 𝑇𝐷𝐻 = 1259 N, 𝑇𝐵𝐸 = −2333 N, 𝐶𝑥 = 1400 lb, 𝐶𝑦 = 1067 N 0.3 m 0.4 m 0.2 m 𝑦 𝐷 20◦

𝐶𝑦

𝐷 𝑥

𝐹

𝐶 𝐸

𝐺

𝐶𝑥

𝑇𝐷𝐻 = 1259 N

1.3 m 𝑇𝐷𝐻

𝑊 = 800 N

𝐵

𝑇𝐵𝐸

𝐶𝑦 𝐵

𝐸 𝑇𝐵𝐸

𝐹

𝑊 = 800 N

20◦

𝐴𝑥

𝐴 𝐴𝑦

0.5 0.4 0.3

𝐴

𝐴𝑥 = 430.8 N

𝐺

𝑦 𝑥

𝑇𝐵𝐸

𝐴𝑦 = 1984 N

Concept Problem

6.80

6.82 Horizontal force applied by the operator is 𝑄 = 20.8 lb to the right, 𝑇𝐴𝐸 = −6.67 lb; For the following results, absolute values are reported: 𝐵𝑥 = 5.77 lb, 𝐵𝑦 = 6.67 lb, 𝐶𝑥 = 26.6 lb, 𝐶𝑦 = 6.67 lb. 6.84 Horizontal force applied by the operator is 𝑄 = 47.7 N to the right, 𝐹𝐵𝐸 = 64.0 N; For the following results, absolute values are reported: 𝐴𝑥 = 55.7 N, 𝐴𝑦 = 10.7 N, 𝐶𝑥 = 103 N, 𝐶𝑦 = 10.7 N. 6.86 350 N when ℎ = 0.9 m 6.88 𝐷𝑦 = 440.7 lb, 𝑇𝐵𝐸 = −1740 lb 6.90 𝐷𝑦 = 640.5 lb, 𝑇𝐵𝐸 = 1131 lb 6.92 𝑇𝐵𝐺 = 200 lb, 𝑇𝐶𝐺 = −240 lb, 𝑇𝐷𝐺 = 200 lb 6.94 (b) 𝐹𝐴𝐷 = −3390 lb 𝐴

𝑇

𝐶

𝐹 𝐴

𝑇

𝐴

𝑥

45◦ 𝐹

𝐹𝐴𝐷 𝐹

𝐹

𝑁

𝐶

𝑇 𝐷

𝐵 𝑉

𝑇

𝑦

𝐷𝑦

𝐵

𝐷𝑥

𝐷

𝐸

𝐷

𝐷𝑥

𝐷𝑦

2000 lb

𝑀

6.96 (b) 𝑇1 = 283 N, 𝑇2 = 283 N; (c) 𝐴1𝑥 = 260 N, 𝐴1𝑦 = −225 N, 𝐴2𝑥 = −260 N, 𝐴2𝑦 = 225 N 𝑇2

𝑦

𝐺

𝑥

𝑇1 50 N

𝐴2𝑦

𝐴1𝑦 𝐸 𝐴

𝐷

200 N

𝐴

𝐴1𝑥

50 N

𝐻𝑦

200 N

𝐴2𝑥 𝐴

𝐻𝑥

𝐴1𝑥 𝐴1𝑦

𝐴2𝑦

𝐴2𝑥

𝐹

ISTUDY

Appendix B

Answers to Even-Numbered Problems

6.98 The chair is safe from collapsing. Intermediate answer: 𝐻𝑦 = 21.33 N (compression) 6.100 (b) 𝐹𝐶𝐹 = 27,200 lb 𝐴𝑦 𝐸𝑦

𝐴𝑥

𝑦 𝐸

𝐷

𝑥

𝐹𝐶𝐹

𝐴

𝐸𝑥

𝐹𝐺 45◦

𝐹𝐶𝐷 𝐷

45◦

𝐹

𝐶

45◦

2000 1b

𝐺

𝐵𝑦 𝐴𝑦

𝐹𝐶𝐹

𝐹𝐶𝐷

45◦

𝐶 𝐵

𝐶

𝐹𝐶𝐹

𝐺

45◦

𝐹𝐶𝐷

𝐴𝑥

𝐵𝑥 𝐴

6.102 𝑃 = 27.30 lb 6.104 𝐹𝐹 𝐻 = 22.5 N, 𝐹𝐵𝐹 = −70.8 N 6.106 (a) The structure is not a truss, essay-type answer; (b) 𝐶𝑦 = 5 kN, 𝐴𝑥 = 0, 𝐴𝑦 = 4 kN, 𝑇𝐷𝐺 = 2.00 kN, 𝑇𝐵𝐸 = 3.00 kN, 𝑇𝐹 𝐻 = 4.00 kN, 𝑇𝐴𝐷 = 3.00 kN, 𝑇𝐷𝐸 = 3.00 kN, 𝑇𝐸𝐹 = 3.00 kN, 𝑇𝐶𝐹 = 3.00 kN, 𝐵𝑥 = −3.00 kN, 𝐵𝑦 = 1.00 kN 𝑦

𝐵

𝐺

𝐴

𝐴𝑥

𝐵

𝐵𝑥

𝑇𝐷𝐺 𝑇𝐵𝐸

𝑥

𝐵𝑦

𝐵𝑦

𝐻

𝑇𝐵𝐸

𝑇𝐹 𝐻

𝐶𝑦

𝐴𝑦 𝑇𝐵𝐸

𝑇𝐷𝐺

𝑇𝐹 𝐻

𝐹

𝐸

𝐷 𝑇𝐷𝐸

𝑇𝐴𝐷

𝑇𝐴𝐷

𝐶

𝑇𝐶𝐹

𝑇𝐴𝐷

𝑇𝐸𝐹

2 kN

3 kN

𝑇𝐴𝐷 𝑇𝐷𝐸

𝑇𝐷𝐸 𝑇𝐸𝐹

𝑇𝐶𝐹 4 kN 𝑇𝐶𝐹

𝑇𝐸𝐹 𝑇𝐶𝐹

6.108 (a) Statically indeterminate; (b) 𝐻𝑄 and 𝐼𝑄; (c) 𝐹𝐷𝐺 =

√

2 kN

6.110 𝑇𝐵𝑃 = 813 lb, 𝑇𝐵𝑂 = 813 lb, 𝑇𝐶𝐷 = 100 lb 6.112 𝑇𝐵𝐺 = −36.46 lb, 𝑇𝐶𝐹 = −1620 lb 6.114 (a) The structure is not a truss; (b) Statically indeterminate; (c) 𝐸𝑥 = 0, 𝐸𝑦 = 1.25 kN, 𝐻𝑦 = 1.75 kN, 𝑇𝐵𝐹 = 0, 𝑇𝐶𝐻 = −1.094 kN, 𝑇𝐷𝐻 = −1.094 kN 6.116 𝑇𝐴𝐷 = −163 lb, 𝑇𝐴𝐵 = 115 lb, 𝑇𝐵𝐸 = 163 lb, 𝑇𝐵𝐷 = −115 lb, 𝑇𝐺𝐼 = 0, 𝑇𝐻𝐼 = 85.0 lb, 𝑇𝐸𝐻 = 184 lb, 𝑇𝐷𝐺 = −184 lb, 𝑇𝐸𝐺 = −35.0 lb 𝐼𝑦 = 85 lb 𝐼 𝐼 =0 45◦ 𝑇𝐺𝐼 𝑇𝐻𝐼 𝑥

𝑦 𝐼𝑦

120 lb

120 lb

𝑥

𝐼

𝐼𝑥

80 lb

𝐺

𝐹 80 lb

𝑇𝐴𝐷 𝑇 𝐵𝐷 𝑇𝐵𝐸

𝑥 𝐴𝑦

𝐷

𝐶

𝑦

𝐴

𝑇𝐷𝐺 𝑇 𝐸𝐺 𝑇𝐸𝐻

45◦ 𝐴

𝑇𝐴𝐵

𝐴𝑦 = 115 lb

𝐵

𝐸

𝐻

A-25

A-26

ISTUDY

Appendix B

Answers to Even-Numbered Problems

6.118 (b) 𝐹𝐴𝐸 = −555.2 lb 600 lb 𝐺𝑦

𝑦 𝐷

𝑥

𝐶

𝐶𝑥

𝐹𝐷 𝐸

𝐷 37

12

𝐺𝑥

𝐺 13 12 𝐹𝐷 5

𝐹𝐴𝐸 𝐵

35

𝐹𝐸𝐹

𝐹𝐸𝐹

𝐸

𝐶𝑦 𝐺𝑦

𝐵𝑥

𝐹𝐴𝐸

𝐸

𝐵𝑦

37

𝐴

𝐹𝐸𝐹

𝐹

𝐺

𝐺𝑥

𝐹

12

50 lb

35

𝐹𝐴𝐸

𝐹𝐸𝐹

Chapter 7 7.2 (𝑥, ̄ 𝑦) ̄ = (0, 65.0) mm 7.4 (𝑥, ̄ 𝑦) ̄ = (1.18, 1.18) in. 7.6 (𝑥, ̄ 𝑦) ̄ = (60.8, 30.0) mm 7.8 (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ = (2.31, 2.12, 1.42) in. (9 9) 7.10 (𝑥, ̄ 𝑦) ̄ = 20 , 20 in. (6 3) 7.12 (𝑥, ̄ 𝑦) ̄ = 5, 2 m ( 5 83 ) , 110 in. 7.14 (𝑥, ̄ 𝑦) ̄ = 11 7.16 (𝑥, ̄ 𝑦) ̄ = (5.602, 5.167) in. 7.18 (𝑥, ̄ 𝑦) ̄ = (48.0, 16.0) in. 7.20 𝑦1 = 2𝑥, 𝑦2 = −92 + 4𝑥, (𝑥, ̄ 𝑦) ̄ = (16.44, 13.16) cm 7.22 (𝑥, ̄ 𝑦) ̄ = (3.429, 1.429) m 7.24 Answer given in problem statement. √ ( 3 ) ̄ 𝑦) ̄ = 12 𝑎, 𝑏 7.26 𝑐1 = 𝑏∕ 𝑎, 𝑐2 = 𝑏∕𝑎3 , (𝑥, 25 7

√ 12 cm ∫3 cm 𝑥 13 10 𝑑𝑥 1√ 10 𝑑𝑥 = 9.487 cm, 𝑥̄ = 7.28 𝐿 = = 7.5 cm, ∫3 cm (3 9.487 cm ) √ 12 cm ∫3 cm 2 + 𝑥3 13 10 𝑑𝑥 = 4.5 cm 𝑦̄ = 9.487 cm √ √ 1 in. 1 in. ∫0 𝑥 1 + 9𝑥4 𝑑𝑥 ∫0 𝑥3 1 + 9𝑥4 𝑑𝑥 7.30 (a) 𝑥̄ = , 𝑦̄ = ; √ √ 1 in. 1 in. ∫0 ∫0 1 + 9𝑥4 𝑑𝑥 1 + 9𝑥4 𝑑𝑥 √ √ )2 )2 ( ( 1 in. 1∕3 1 in. 1 −2∕3 ∫0 𝑦 ∫0 𝑦 1 + 13 𝑦−2∕3 𝑑𝑦 1 + 3𝑦 𝑑𝑦 (b) 𝑥̄ = , 𝑦̄ = ; (c) (𝑥, ̄ 𝑦) ̄ = (0.609, 0.366) in. √ √ ) ) ( ( 12 cm

∫0

1 in.

√

1+

1 −2∕3 𝑦 3

2

𝑑𝑦

∫0

1 in.

1+ √

1 −2∕3 𝑦 3

2

𝑑𝑦

( )2 ( )2 √ 𝑟 𝑟 ∫−𝑟 𝑥 1 + √ −𝑥 ∫−𝑟 𝑟2 − 𝑥2 1 + √ −𝑥 𝑑𝑥 𝑑𝑥 ( ) 𝑟2 −𝑥2 𝑟2 −𝑥2 , 𝑦̄ = ; (b) (𝑥, ̄ 𝑦) ̄ = 0, 𝜋2 𝑟 7.32 (a) 𝑥̄ = √ √ )2 )2 ( ( 𝑟 𝑟 ∫−𝑟 1 + √ −𝑥 ∫−𝑟 1 + √ −𝑥 𝑑𝑥 𝑑𝑥 𝑟2 −𝑥2

𝑟2 −𝑥2

ISTUDY

Appendix B

Answers to Even-Numbered Problems

7.34 𝑉 = 60.32 in.3 , (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ = (1.389, 0, 0) in. 7.36 𝑥̄ =

11 𝑅 28

7.38 𝑑𝐴 = 4938 m, 𝑑𝐵 = 3.951 × 105 km 7.40 The center of gravity is located on the horizontal member of the antenna, at a distance of 133 in. from the left-hand end. 7.42 (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ = (2.98, 0, 0) in. 7.44 𝑧̄ = 2.675 cm 7.46 𝑚𝑐 = 680 g 𝑦

𝑚1 𝑥 =

𝑂

counterweight (𝑚4 = 𝑚𝐶 ) 𝑚2 𝑚 𝑚4 3 𝑚5 + + bolt hole for bolt

7.48 (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ = (1.07, 0, 0.876) in. 7.50 The center of mass is located 1.46 m to the right of point 𝐴 and 1 m above point 𝐴, 𝑇 = 639 N, 𝐵𝑥 = 553 N, 𝐵𝑦 = 118 N 𝑦 30◦

𝐵𝑦

𝑇𝐶𝐹

𝐵𝑥 𝐺 𝑊

𝑥

1.461 m

106 𝜋𝜌0 5

in.3 ) ) ( ( 𝑟∕2 𝜌 𝑟 ∫0 𝑥 20 𝜋 𝑟2 − 𝑥2 𝑑𝑥 + ∫𝑟∕2 𝑥𝜌0 𝜋 𝑟2 − 𝑥2 𝑑𝑥 , 𝑦̄ = 0, 𝑧̄ = 0; (b) 𝑥̄ = 7.54 (a) 𝑥̄ = ( ) ( ) 𝑟∕2 𝜌 𝑟 ∫0 20 𝜋 𝑟2 − 𝑥2 𝑑𝑥 + ∫𝑟∕2 𝜌0 𝜋 𝑟2 − 𝑥2 𝑑𝑥 ( ) 2 2 𝑎 ∫0 𝑥𝜌0 𝜋ℎ2 1 − 𝑥𝑎2 𝑑𝑥 5 , 𝑦̄ = 0, 𝑧̄ = 0; (b) 16 𝑎 7.56 (a) 𝑥̄ = ( ) 2 2 𝑎 𝑥 ∫0 𝜌0 𝜋ℎ2 1 − 𝑎2 𝑑𝑥 [ ( ) )2 ] ( 2∕3 𝐿 𝑥 𝑅𝑥 2 ∫0 𝑥𝛾0 𝜋 𝑅 𝐿 − 2𝐿 𝑑𝑥 75 , 𝑦̄ = 0, 𝑧̄ = 0; (b) 𝑥̄ = 124 7.58 (a) 𝑥̄ = 𝐿 [ ( ) )2 ] ( 2∕3 𝐿 ∫0 𝛾0 𝜋 𝑅2 𝐿𝑥 − 𝑅2 𝐿𝑥 𝑑𝑥 7.52 𝑥̄ = 1.447 in., 𝑚 =

7.60 (a) 𝑦̄ =

∫0

1 cm 1 2

( ) ( ) 1 + 𝑥 + 𝑥2 𝛾0 2𝜋𝑥 1 + 𝑥2 − 𝑥 𝑑𝑥 ; (b) 𝑥̄ = ( ) 1 cm ∫0 𝛾0 2𝜋𝑥 1 + 𝑥2 − 𝑥 𝑑𝑥

11 10

25 𝑟 56

cm

7.62 𝑉 = 283 mm3 , 𝐴 = 226 mm2 7.64 𝑉 = 1.78×103 mm3 , 𝐴 = 1.98×103 mm2 √ ) ( 7.66 𝑉 = 8𝜋ℎ3 , 𝐴 = 𝜋ℎ2 11 + 8 2 7.68 𝑉 = 19.8 in.3 , 𝑊 = 0.118 lb 7.70 (a) 𝑉 = 4.40×105 mm3 ; (b) 𝐴outside = 34.6×103 mm2 ; (c) 𝐴inside = 12.2×103 mm2 √ ) ( 3 , 𝐴 = 𝜋 11 + 7.72 𝑉 = 7𝜋 𝑅 5 𝑅2 12 4

A-27

A-28

ISTUDY

Answers to Even-Numbered Problems

Appendix B

7.74 (a) 𝑤 = 2.78 lb∕in. for 0 ≤ 𝑥 ≤ 36 in.; (b) 𝑤 = 1.85 lb∕in. for 0 ≤ 𝑥 ≤ 18 in., 𝑤 = 3.70 lb∕in. for 18 in. ≤ 𝑥 ≤ 36 in.; (c) 𝑤 = 1.85 lb∕in. + (0.0514 lb∕in.2 )𝑥 for 0 ≤ 𝑥 ≤ 36 in.; (d) 𝑤 = 1.85 lb∕in. + (0.103 lb∕in.2 )𝑥 for 0 ≤ 𝑥 ≤ 18 in., 𝑤 = 5.56 lb∕in. − (0.103 lb∕in.2 )𝑥 for 18 in. ≤ 𝑥 ≤ 36 in. 7.76 𝐴𝑥 = 0, 𝐴𝑦 = 33.3 lb, 𝐵𝑦 = 66.7 lb 7.78 𝐴𝑥 = 0, 𝐴𝑦 = 49.1 kN, 𝑀𝐴 = 98.1 kN⋅m 7.80 (a) 𝑥̄ = 21.0 in.; (b) & (c) 𝐴𝑥 = 0, 𝐴𝑦 = 22.2 lb, 𝐵𝑦 = 77.8 lb 7.82 (a) 𝑥̄ = 1.67 m; (b) & (c) 𝐴𝑥 = 0, 𝐴𝑦 = 49.1 kN, 𝑀𝐴 = 81.8 kN⋅m lb lb , 𝑏 = − 15 ; (b) 𝑤(𝑥 = 0) = 4.375 lb∕in., 𝑤(𝑥 = 4 in.) = 0.625 lb∕in.; (c) Essay-type answer; 7.84 (a) 𝑎 = 35 8 in. 16 in.2 (d) 𝐴 = 17.96 lb, 𝐵 = 17.04 lb 7.86 (a) 𝑎 = 800 N∕m, 𝑏 = 0, 𝑐 = −8.00 N∕m3 ; (b) 𝑑 = 800 N∕m, 𝑓 = 20𝜋m

7.88 𝐴𝑥 = 0, 𝐴𝑦 = −4.00 kN, 𝐵𝑦 = 16.0 kN 7.90 𝐴𝑥 = 0, 𝐴𝑦 = −40.0 kip, 𝐵𝑦 = 40.0 kip 7.92 𝐴𝑥 = 0, 𝐴𝑦 = 6.08 kip, 𝐵𝑦 = 4.92 kip 7.94 𝐶𝑥 = 0, 𝐶𝑦 = 12.0 kN, 𝑀𝐶 = −24.0 kN⋅m 7.96 𝐴𝑥 = 0, 𝐴𝑦 = −80 kip, 𝐶𝑦 = 80 kip

( ) 7.98 𝐴𝑥 = 0, 𝐴𝑦 = 𝑤1 𝑎 + 𝑤2 (𝐿 − 𝑎), 𝑀𝐴 = 12 𝑤1 𝑎2 + 12 𝑤2 𝐿2 − 𝑎2 7.100

Computer Problem. 𝑎 = 0.524 kN∕m, 𝑏 = 1.21 kN∕m2 , 𝑐 = −0.0556 kN∕m3 , 𝐹 = 23.0 kN, 𝑥̄ = 4.05 m

7.102 𝑘𝑡 = 87.71 kN⋅m∕degree 7.104 𝐴𝑥 = 0, 𝐴𝑦 = 0, 𝐴𝑧 = 𝑊 , 𝑀𝐴𝑥 = −𝑊 𝑅(1 − 𝜋2 ), 𝑀𝐴𝑦 = − 𝜋2 𝑊 𝑅, 𝑀𝐴𝑧 = 0 7.106 𝑇 = 𝛾𝑑 3 7.108 𝑇 = 39.0 lb 7.110 𝜃0 = 0.235 rad = 13.5◦ . 𝑦 𝐵𝑦

𝑥 15 cm

𝑝𝐵

𝐵𝑥

𝑃1

20 cm

𝑀𝐵

𝑃2

𝐴𝑥 = 0 𝑝𝐴

7.112 𝑇𝐵𝐶 = 13,333 lb, 𝑇𝐷𝐹 = 12,000 lb 7.114 (a) 𝑇 = 40.7 lb; (b) 𝑇 = 323 lb 7.116 𝑇𝐵 = 414 N, 𝑇𝐶 = 414 N, 𝑇𝐷 = 684 N 7.118 𝐹𝐸𝐺 = 381 lb 7.120 (𝑥, ̄ 𝑦) ̄ = (4.487, 8.153) cm 7.122 (𝑥, ̄ 𝑦) ̄ = (0, 20.2) mm ) ( ,1 m 7.124 (𝑥, ̄ 𝑦) ̄ = 56 25 7.126 (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ = (34.7, 0, 0) mm 7.128 (a) Let 𝜌𝑜 = 0.002 g∕mm3 , 𝜌𝑖 = 0.003 g∕mm3 , { ( [( )2 )2 ( )2 ]} 6 mm ∫0 𝑥 𝜌𝑖 𝜋 2 − 16 𝑥 + 𝜌𝑜 𝜋 3 + 16 𝑥 − 2 − 16 𝑥 𝑑𝑥 , 𝑦̄ = 0, 𝑧̄ = 0; 𝑥̄ = { ( [ ]} )2 )2 ( )2 ( 6 mm ∫0 𝜌𝑖 𝜋 2 − 16 𝑥 + 𝜌𝑜 𝜋 3 + 16 𝑥 − 2 − 16 𝑥 𝑑𝑥 (b) (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ = (3.20, 0, 0) mm 7.130 𝑥̄ =

4 3

m

ISTUDY

Appendix B

Answers to Even-Numbered Problems ∫0

( )2 𝑥𝜋 1 + 𝑥2 𝑑𝑥 7.132 (a) 𝑥̄ = , 𝑦̄ = 𝑧̄ = 0; (b) 𝑥̄ = 58 cm )2 1 cm ( 2 ∫0 𝜋 1 + 𝑥 𝑑𝑥 √ √ 2 km 2 km ∫−1 km 𝑥2 1 + 4𝑥2 𝑑𝑥 ∫−1 km 𝑥 1 + 4𝑥2 𝑑𝑥 , 𝑦̄ = ; 7.134 (a) 𝑥̄ = √ √ 2 km 2 km ∫−1 km 1 + 4𝑥2 𝑑𝑥 ∫−1 km 1 + 4𝑥2 𝑑𝑥 (b) (𝑥, ̄ 𝑦) ̄ = (0.801, 1.48) km 1 cm

7.136 (𝑥, ̄ 𝑦) ̄ = (42.00, −11.03) in., 𝐴𝑥 = −508.6 lb, 𝐴𝑦 = 0, 𝐵𝑥 = 508.6 lb, 𝐵𝑦 = 218.0 lb 7.138 𝑉 =

4𝜋 3 𝑟 , 3

𝐴 = 4𝜋𝑟2

7.140 𝑉 = 3180 cm3 , 𝐴 = 867 cm2 7.142 𝐴𝑥 = 0, 𝐴𝑦 = 1700 lb, 𝐵𝑦 = 1300 lb 7.144 𝐵𝑦 = 640 lb, 𝐶𝑥 = 0, 𝐶𝑦 = 1460 lb where pin 𝐶 is located 6 f t from the left-hand end of the beam. 7.146 𝐴𝑥 = 0, 𝐴𝑦 = 2.17 kN, 𝐶𝑦 = 3.83 kN where roller 𝐶 is located 2 m from the right-hand end of the beam. 7.148 𝐴𝑥 = −46.8 lb, 𝐴𝑦 = 281 lb, 𝐶𝑥 = 187 lb 𝑊

𝐶𝑥

𝑃

𝑝𝐴 𝑦 𝑥

𝐴𝑦

1 ft 𝐴𝑥

7.150 𝑑 = −12.5 mm

Chapter 8 8.2 𝑁𝐻 = −34.7 lb, 𝑉𝐻 = −197 lb, 𝑀𝐻 = 1180 in.⋅lb, 𝑁𝐽 = −480 lb, 𝑉𝐽 = 0, 𝑀𝐽 = 0 8.4 𝑁𝐸 = 200 lb, 𝑉𝐸 = 346 lb, 𝑀𝐸 = 0, 𝑁𝐹 = 200 lb, 𝑉𝐹 = 346 lb, 𝑀𝐹 = −8310 in.⋅lb 8.6 𝑁𝐹 = −3.20 kN, 𝑉𝐹 = 4.27 kN, 𝑀𝐹 = 0, 𝑁𝐺 = −3.20 kN, 𝑉𝐺 = 4.27 kN, 𝑀𝐺 = 10.7 kN⋅m, 𝑁𝐻 = −0.800 kN, 𝑉𝐻 = 1.07 kN, 𝑀𝐻 = 10.7 kN⋅m, 𝑁𝐼 = −0.800 kN, 𝑉𝐼 = 1.07 kN, 𝑀𝐼 = 13.3 kN⋅m 8.8 𝑁𝐹 = −11.2 kN, 𝑉𝐹 = 1.60 kN, 𝑀𝐹 = 0, 𝑁𝐺 = −11.2 kN, 𝑉𝐺 = 1.60 kN, 𝑀𝐺 = 4.00 kN⋅m, 𝑁𝐻 = −8.80 kN, 𝑉𝐻 = −1.60 kN, 𝑀𝐻 = 4.00 kN⋅m, 𝑁𝐼 = −8.80 kN, 𝑉𝐼 = −1.60 kN, 𝑀𝐼 = 0 8.10 𝑁𝐷 = 0, 𝑉𝐷 = 100 lb, 𝑀𝐷 = 0, 𝑁𝐸 = 0, 𝑉𝐸 = 100 lb, 𝑀𝐸 = −150 in.⋅lb, 𝑁𝐹 = −70.7 lb, 𝑉𝐹 = 70.7 lb, 𝑀𝐹 = −238 in.⋅lb 8.12 𝑁𝐻 = −2.00 kN, 𝑉𝐻 = 2.00 kN, 𝑀𝐻 = 0, 𝑁𝐼 = −2.00 kN, 𝑉𝐼 = 2.00 kN, 𝑀𝐼 = −5.60 kN⋅m 8.14 𝑁𝐿 = −2.00 kN, 𝑉𝐿 = −2.00 kN, 𝑀𝐿 = −3.20 kN⋅m, 𝑁𝑂 = −2.00 kN, 𝑉𝑂 = −2.00 kN, 𝑀𝑂 = 0 8.16 𝑁𝑅 = −5.83 kN, 𝑉𝑅 = 0, 𝑀𝑅 = 0, 𝑁𝐴 = −4.00 kN, 𝑉𝐴 = 0, 𝑀𝐴 = 4.20 kN⋅m 8.18 𝑁𝐻 = −400 lb, 𝑉𝐻 = −693 lb, 𝑀𝐻 = −1390 ft⋅lb, 𝑁𝐼 = −400 lb, 𝑉𝐼 = −693 lb, 𝑀𝐼 = 0 8.20 𝑁𝐿 = 400 lb, 𝑉𝐿 = −693 lb, 𝑀𝐿 = 1390 ft⋅lb, 𝑁𝑀 = 0, 𝑉𝑀 = −800 lb, 𝑀𝑀 = 0 8.22

Concept Problem

8.24 𝑉𝐴𝑥 = 0, 𝑉𝐴𝑦 = 150 lb, 𝑁𝐴𝑧 = 0, 𝑀𝐴𝑥 = −2400 in.⋅lb, 𝑀𝐴𝑦 = 0, 𝑀𝐴𝑧 = 3600 in.⋅lb 8.26 𝑁𝐵𝑥 = 0, 𝑉𝐵𝑦 = 0, 𝑉𝐵𝑧 = 120 N, 𝑀𝐵𝑥 = 8.40 N⋅m, 𝑀𝐵𝑦 = −12.6 N⋅m, 𝑀𝐵𝑧 = 0 8.28 𝑉𝐷𝑥 = 0, 𝑁𝐷𝑦 = 0, 𝑉𝐷𝑧 = 120 N, 𝑀𝐷𝑥 = 13.8 N⋅m, 𝑀𝐷𝑦 = −14.4 N⋅m, 𝑀𝐷𝑧 = 0

A-29

A-30

ISTUDY

Appendix B

Answers to Even-Numbered Problems

8.30 𝑉 = 2.50 kN, 𝑀 = (2.50 kN)𝑥 for 0 ≤ 𝑥 ≤ 3 m; 𝑉 = −7.50 kN, 𝑀 = (7.50 kN)(4 m − 𝑥) for 3m ≤ 𝑥 ≤ 4m 𝑉 (kN)

𝑀 (kN⋅m)

5.000

10.00

2.500

7.500

0

5.000

−2.500

2.500

−5.000

0

−7.500

−2.500

−10.00

0

2 𝑥 (m)

1

3

−5.000

4

2 𝑥 (m)

1

0

4

3

8.32 (a) 𝑉 = 𝑃 , 𝑀 = 𝑃 𝑥 for 0 ≤ 𝑥 ≤ 13 𝐿; 𝑉 = 0, 𝑀 = 13 𝑃 𝐿 for 13 𝐿 ≤ 𝑥 ≤ 23 𝐿; 𝑉 = −𝑃 , 𝑀 = 𝑃 (𝐿 − 𝑥) for 2 𝐿 3

≤ 𝑥 ≤ 𝐿; (b) Essay-type answer

8.34 𝑉 = 125 lb, 𝑀 = (125 lb) 𝑥 for 0 ≤ 𝑥 ≤ 8 f t; 𝑉 = 125 lb, 𝑀 = (−125 lb)(16 f t − 𝑥) for 8 f t ≤ 𝑥 ≤ 16 f t 𝑉 (lb)

𝑀 (ft⋅lb)

125

1000 𝑥 (ft)

0

𝑥 (ft)

0

16

8

16

8 −1000

8.36 𝑉 = 1000 lb, 𝑀 = (−1000 lb)(20 f t − 𝑥) 8.38 𝑉 = (−8 kN∕m) 𝑥, 𝑀 = (−4 kN∕m) 𝑥2 8.40 𝑉 = 1200 N − (600 N∕m) 𝑥 + (50 N∕m2 ) 𝑥2 , 𝑀 = (1200 N) 𝑥 − (300 N∕m) 𝑥2 + (16.7 N∕m2 ) 𝑥3 𝑀 (N⋅m)

𝑉 (N) 1500 1200 1000 600

500 𝑥 (m)

0 2

4

𝑥 (m)

0

6

2

4

6

−600 zero slope

Concept Problem

8.42

8.44 𝑤0 = 550.5 lb∕f t

8.46 𝑉 = 3.00 N, 𝑀 = −8.00 N⋅mm + (3.00 N) 𝑥 for 0 ≤ 𝑥 ≤ 2 mm; 𝑉 = −1.00 N, 𝑀 = −(1.00 N) 𝑥 for 2 mm ≤ 𝑥 ≤ 4 mm; 𝑉 = 2.00 N, 𝑀 = −12 N⋅mm + (2.00 N) 𝑥 for 4 mm ≤ 𝑥 ≤ 6 mm 𝑉 (N)

𝑀 (N⋅ mm)

4

8

3

6

2

4

1

2

0

0

−1

−2

−2

−4

−3

−6

−4

0

2

4 𝑥 (mm)

6

−8

0

2

4 𝑥 (mm)

6

8.48 𝑉 = 3.00 N, 𝑀 = −8.00 N⋅mm + (3.00 N) 𝑥 for 0 ≤ 𝑥 ≤ 2 mm; 𝑉 = 0, 𝑀 = 0 for 2 mm ≤ 𝑥 ≤ 6 mm 8.50

Concept Problem

ISTUDY

Appendix B

Answers to Even-Numbered Problems

[ ] 8.52 𝑉 = −100 lb∕ft(6 f t − 𝑥) + 1000 lb, 𝑀 = −50 lb∕ft (12 f t)𝑥 − 𝑥2 + (1000 lb) 𝑥 for 0 ≤ 𝑥 ≤ 6 f t; ] [ 𝑉 = −100 lb∕ft(6 f t − 𝑥) − 1000 lb, 𝑀 = −50 lb∕ft (12 f t)𝑥 − 𝑥2 + (1000 lb)(12 f t − 𝑥) for 6 f t ≤ 𝑥 ≤ 12 f t 𝑉 (lb)

𝑀 (ft⋅lb)

1000

4200

400 𝑥 (ft)

0 6

𝑥 (ft)

0 6

12

12

−400

−1000

8.54 𝑉 = 1000 N, 𝑀 = (1000 N) 𝑥 for 0 ≤ 𝑥 ≤ 1 m; 𝑉 = 100 N, 𝑀 = (100 N)(9 m + 𝑥) for 1 m ≤ 𝑥 ≤ 2 m; 𝑉 = −1100 N, 𝑀 = (1100 N)(3 m − 𝑥) for 2 m ≤ 𝑥 ≤ 3 m 8.56 𝑉 = (0.202 kip∕in.) 𝑥, 𝑀 = (0.101 kip∕in.)𝑥2 for 0 ≤ 𝑥 ≤ 20 in.; 𝑉 = −10.0 kip + (0.202 kip∕in.) 𝑥, 𝑀 = 200 kip⋅in. − (10.0 kip) 𝑥 + (0.101 kip∕in.)𝑥2 for 20 in. ≤ 𝑥 ≤ 79 in.; 𝑉 = (−0.202 kip∕in.)(99.0 in. − 𝑥), 𝑀 = (0.101 kip∕in.)(99.0 in. − 𝑥)2 for 79 in. ≤ 𝑥 ≤ 99 in. 𝑉 (kip)

𝑀 (kip⋅in.) 5.96

4.04

𝑥 (in.)

0 20

40.4

20

79

𝑥 (in.)

0

99

79

40.4

99

−4.04 −47.5 −5.96 zero slope

8.58 𝑉 = (−8 kN∕m) 𝑥, 𝑀 =

(−4 kN∕m) 𝑥2

8.60 𝑉 = 36 kN − (8 kN∕m) 𝑥 +

(1

) ( ) 2 𝑥2 , 𝑀 = −96 kN⋅m + (36 kN) 𝑥 − (4 kN∕m) 𝑥2 + 1 kN∕m2 𝑥3 kN∕m 3 9

𝑉 (kN)

𝑀 (kN⋅m) slope = −8 kN∕m

6 𝑥 (m)

0

36 slope = −4 kN∕m

slope = 0 𝑥 (m)

0 6

slope = 36 kN −96

[ ] [ ] 8.62 𝑉 = 4000 lb − (15 lb∕ft) (40 f t)𝑥 − 𝑥2 , 𝑀 = (4000 lb) 𝑥 − (5 lb∕ft2 ) (60 f t)𝑥2 − 𝑥3 , 𝑀max = 15,400 ft⋅lb at 𝑥 = 8.45 f t 𝑉 (lb)

𝑀 (ft⋅lb) 𝑀∗

4000 slope = −600 lb∕ft

20 0

𝑥 (ft) 𝑥∗

−2000

0

𝑥 (ft) 𝑥∗

20 slope = −2000 lb

slope = 0

slope = 4000 lb

A-31

A-32

ISTUDY

Appendix B

Answers to Even-Numbered Problems ( ) ( 8.64 𝑉 = 16 𝑤0 𝐿 − 𝑤0 𝑥 − 𝐿1 𝑥2 , 𝑀 = 𝑤0 𝐿6 𝑥 − 12 𝑥2 + 𝑀max = (−0.0160)𝑤0 𝐿2 at 𝑥 = (0.789)𝐿 8.66 𝑉 = 2𝑃 , 𝑀 = − 32 𝑃 𝐿 + 2𝑃 𝑥 for 0 ≤ 𝑥 ≤

𝐿 ; 2

1 3) 𝑥 , 3𝐿

𝑀max = (0.0160)𝑤0 𝐿2 at 𝑥 = (0.211)𝐿 and

( ) 𝑉 = 𝑃 , 𝑀 = − 12 𝑃 𝐿 + 𝑃 𝑥 − 12 𝐿 for

𝐿 2

≤𝑥≤𝐿

( ) lb∕ft2 𝑥3 for 0 ≤ 𝑥 ≤ 4 f t; 𝑉 = 1800 lb − (25 lb∕ft2 ) 𝑥2 , 8.68 𝑉 = (−25 lb∕ft2 ) 𝑥2 , 𝑀 = − 25 3 ( ) 𝑀 = −7200 ft⋅lb + (1800 lb) 𝑥 − 25 lb∕ft2 𝑥3 for 4 f t ≤ 𝑥 ≤ 12 f t 3 𝑉 (lb)

𝑀 (ft⋅lb)

slope = −200 lb∕ft

𝑀∗

1400

slope = −1800 lb 𝑥 (ft)

0 𝑥∗

4

−400

4

12

𝑥 (ft)

0 −533.3

𝑥∗

12

slope = 0 −1800

slope = 1400 lb slope = −400 lb

slope = −600 lb∕ft

8.70 𝑉 = (−1.00 kN∕m) 𝑥, 𝑀 = (−0.500 kN∕m) 𝑥2 for 0 ≤ 𝑥 ≤ 1.3 m; 𝑉 = −6.86 kN − (1.00 kN∕m)(𝑥 − 1.30 m), 𝑀 = 7.22 kN⋅m − (5.56 kN) 𝑥 − (0.500 kN∕m) 𝑥2 for 1.3 m ≤ 𝑥 ≤ 4 m; 𝑉 = −14.0 kN − (1.00 kN∕m)(𝑥 − 4.00 m), 𝑀 = 25.0 kN⋅m − (10.0 kN) 𝑥 − (0.500 kN∕m) 𝑥2 for 4 m ≤ 𝑥 ≤ 5.5 m 𝑉 (kN ( )

𝑀 (kN⋅m) 1.3

4

1.3

5.5 𝑥 (m)

0

4

5.5 𝑥 (m)

0

−1.3 −0.845

−6.856 −9.556

−23

−14 −15.5 −45.13

8.72 𝑉max = 4𝑃 at 𝑥 = 0, 𝑀max = −10𝑃 𝑑 at 𝑥 = 0 𝑀

𝑉

𝑑

2𝑑

3𝑑

4𝑑 𝑥

0

4𝑃 3𝑃 2𝑃

−(5)𝑃 𝑑 𝑃 𝑥

0 𝑑

2𝑑

3𝑑

4𝑑

𝐵

𝐶

𝐷

𝐸

−(10)𝑃 𝑑 𝐵

𝐶

𝐷

𝐸

8.74 𝑉max = −45.0 N just to the left of 𝑥 = 30 mm, 𝑀max = −1130 N⋅mm at 𝑥 = 30 mm 𝑉 ( N)

𝑀 (N⋅mm) 35

40

slope = −30 N

20

80 𝑥 (mm)

0 𝑥 (mm)

0 30 −20

30

10

−30

−40 −45

80 Note: The slopees of both lines are the same and are equal to −0.5 N∕mm.

slope = 10 N

−500

slope = 35 N

−1000 −1500

−1125 N⋅mm slope = −45 N

ISTUDY

Appendix B

Answers to Even-Numbered Problems

8.76 𝑉max = 10.0 lb just to the right of 𝑥 = 17 in., 𝑀max = −16.0 in.⋅lb at 𝑥 = 4 in. 𝑉 (lb) 10

9.9

10

slope = 1.9 lb

𝑀 (in.⋅lb) slope = −4 lb∕in.

5

slope = −1 lb∕in.

1.9 0 4

𝑥 (in.)

𝑥∗

8

17

𝑀∗

10

7.6

slope = 0

slope = 0 0 4

21

8

𝑥 (in.)

𝑥∗

17

21

−5 −7.1

−8 −10

𝐷

𝐴

−10

𝐸

𝐵

slope = 10 lb

slope = 9.9 lb −15.8

−16 −20

𝐴

𝐵

𝐷

slope = −8 lb

𝐸

slope = −7.1 lb

8.78 Beam 1: Shear diagram (b), Moment diagram (g); Beam 2: Shear diagram (a), Moment diagram (g); Beam 3: Shear diagram (d), Moment diagram (f) 8.80 𝑁𝐹 = −6.19 kip, 𝑉𝐹 = 0.940 kip, 𝑀𝐹 = 0, 𝑁𝐺 = −6.19 kip, 𝑉𝐺 = 0.940 kip, 𝑀𝐺 = 4.00 kip⋅ft 8.82 𝑁𝐸 = −833 N, 𝑉𝐸 = 833 N, 𝑀𝐸 = 0, 𝑁𝐹 = −1180 N, 𝑉𝐹 = 0, 𝑀𝐹 = −69.0 N⋅m, 𝑁𝐺 = 833 N, 𝑉𝐺 = −833 N, 𝑀𝐺 = 0 8.84 𝑉 = 12 kN − (0.750 kN∕m2 ) 𝑥2 , 𝑀 = (12 kN) 𝑥 − (0.250 kN∕m2 ) 𝑥3 for 0 ≤ 𝑥 ≤ 4 m; 𝑉 = 36 kN − (12 kN∕m) 𝑥 + (0.750 kN∕m2 ) 𝑥2 , 𝑀 = −32 kN⋅m + (36 kN) 𝑥 − (6 kN∕m) 𝑥2 + (0.250 kN∕m2 ) 𝑥3 for 4 m ≤ 𝑥 ≤ 8 m 𝑀 (kN⋅m)

𝑉 (kN) slope = 0

12

slope = −6 kN∕m 𝑥 (m)

0 4 −12

32

𝑥 (m)

0 4

8

8

slope = 0

8.86 𝑉 = 12 kN − (0.750 kN∕m2 ) 𝑥2 , 𝑀 = (12 kN) 𝑥 − (0.250 kN∕m2 ) 𝑥3 for 0 ≤ 𝑥 ≤ 4 m; 𝑉 = 36 kN − (12 kN∕m) 𝑥 + (0.750 kN∕m2 ) 𝑥2 , 𝑀 = −32 kN⋅m + (36 kN) 𝑥 − (6 kN∕m) 𝑥2 + (0.250 kN∕m2 ) 𝑥3 for 4 m ≤ 𝑥 ≤ 8 m; see 𝑉 and 𝑀 plots for Prob. 8.84 8.88 𝑉 = 60 N, 𝑀 = −12,000 N⋅mm + (60 N) 𝑥 8.90 𝑉 = 60 N, 𝑀 = −21,000 N⋅mm + (60 N) 𝑥 for 0 ≤ 𝑥 ≤ 150 mm; 𝑉 = 240 N, 𝑀 = −48,000 N⋅mm + (240 N) 𝑥 for 150 mm ≤ 𝑥 ≤ 200 mm 𝑀 (N⋅mm)

𝑉 (N) 300

240

200 60

100 200 𝑥 (mm)

0 100

0

𝑥 (mm)

100

200 −10000 −20000

−12000 −21000

A-33

A-34

ISTUDY

Appendix B

Answers to Even-Numbered Problems 8.92 𝑉max = −1750 lb for 15 f t ≤ 𝑥 ≤ 20 f t, 𝑀max = 8750 ft⋅lb at 𝑥 = 15 f t 𝑀 (ft⋅lb) 𝑉 (lb)

10000

8750

2000

7500 1250

1000

5000

250

slope = 250 lb

𝑥 (ft)

0 5

15

10

20

−1000

𝑥 (ft)

0

−2000

5

−1750 𝐶 𝐷

𝐵

10

15

20

slope = 1250 lb

(1

) ( ) 8.94 (a) 𝑉 = −4 kN − 3 kN∕m2 𝑥2 ; (b) 𝑀 = (−4 kN) 𝑥 − 19 kN∕m2 𝑥3 ; (c) 𝑉max = 9.00 kN at 𝑥 = 3 m, 𝑀max = −15.0 kN⋅m at 𝑥 = 3 m 𝑥 (m)

0

𝑥 (m)

0

3

slope = 0

𝑀 (kN⋅m)

slope = −2 kN∕m

𝑉 (kN) 9

3

6

−15

6

slope = 9 kN

slope = −4 kN∕m

8.96 (a) 𝑀 = (−2 kip∕ft) 𝑥2 ; (b) 𝑉 = 56 kip − (8 kip∕ft) 𝑥 + 𝑀max = −72.0 ft⋅kip at 𝑥 = 6 f t

(1 3

) kip∕ft2 𝑥2 ; (c) 𝑉max = 20.0 kip at 𝑥 = 6 f t,

slope = −4 kip∕ft 𝑀 (ft⋅ kip) 6 12

𝑉 (kip) slope = 0

20

0

𝑥 (ft)

slope = 8 kip

8 0

𝑥 (ft)

12

6

−72 slope = 20 kip

8.98 (a) 𝑀 = 132 kN⋅m − (17 kN) 𝑥 + (0.5 kN∕m) 𝑥2 ; (b) 𝑉 = 16 kN − (4 kN∕m) 𝑥 + slope = −4 kN∕m slope = 4 kN 𝑉 (kN)

𝑀 (kN⋅m)

slope = 0

16

12

4 0 −5 −11

slope = −11 kN

48 slope = −5 kN

𝑥 (m) 0

slope = 1 kN∕m

𝑥 (m) 6

12

slope = 16 kN

( ) ( ) 8.100 (a) 𝑉 = − 25 lb∕ft2 𝑥2 ; (b) 𝑀 = −800 ft⋅lb − 25 lb∕ft2 𝑥3 ∕3; (c) 𝑀 (ft⋅lb)

𝑉 (lb) Slope = −200 lb∕ft

1500

0

−1700

4 Slope = −600 lb∕ft

12

𝑥 (ft)

0

−1330

4

𝑥 (ft) 12 Slope = −1700 lb Slope = 1500 lb

Chapter 9 9.2 𝑁 = 52.1 lb, 𝐹 = 10.3 lb. The box remains at rest on the ramp. 9.4 𝑃 = 471.4 N 9.6 𝜃 = tan−1 𝜇 9.8 ℎ = 60.0 cm

) 2 𝑥2 ; (c) kN∕m 3

(1

ISTUDY

Appendix B

Answers to Even-Numbered Problems

9.10 𝑁 = 1.43×106 lb, 𝐹 = 8.58×104 lb 9.12 𝜇 ≥ 0.0939 2(1 ft) 𝜋

4(1 ft) 3𝜋

𝐴 𝑊𝐴𝐵 = 60 lb

𝑇 1 ft + 2 (1 ft) 3 𝑊𝑤

𝑃

𝑦

1 (1 3

𝐵𝑥

𝑥

ft)

𝐵𝑦

9.14 The dam will not slide or tip when the reservoir is completely full. 9.16 The dam will not tip and will fail by sliding. 𝑦 𝑥

1 ft 1 ft 𝑊1 𝑊 2

3 ft

𝑃

2 ft 𝐴 𝑝

𝐹 𝑁 𝓁

9.18 𝑃 = 47.4 N 9.20 𝑇 = 1.940 N 9.22 (a) 𝜇 = 1.103, safe from tipping; (b) 𝜇 = 0.3540, safe from tipping; (c) 𝜇 = 0.2680, safe from tipping 9.24 (a) 𝜇min = 0.289; (b) The answer to Part (a) will change, essay-type answer; (c) The answer to Part (a) does not change, essay-type answer 9.26 𝑃1 = 0.6344 lb 9.28 𝑄 = 0.0915 lb 9.30 𝜇𝑠 = 0.750 9.32 𝑃 = 0.900 lb will cause all books to slide together on the table. 9.34 𝜇𝐴 = 𝜇𝐵 = 0.6867 9.36 Nine books may be lifted. 9.38 𝜇𝑠 = 0.417 4 in. 𝐶𝑥

7 lb 𝐷

𝐶 𝐹2

𝐶𝑦

𝑁2 𝐹2 𝐵

7 lb

𝑦 𝑥

𝐴 𝐹1 𝑁1

9.40 𝑄 = 7.70 lb 9.42 𝑃 = 10.6 N 9.44 The truck is not capable of pulling the dumpster. 9.46 𝑃 = 7.00 kN 9.48 𝜇𝑠 =

5 3

9.50 𝑇𝐵𝐷 = 789.9 lb

A-35

A-36

ISTUDY

Answers to Even-Numbered Problems

Appendix B

9.52 (a) 𝑃 = 4.56 N; (b) 𝑇𝐶𝐷 = 6.58 N, essay-type answer 9.54 (a), (b), (c): 73.0 lb

9.56 8.494 lb ≤ 𝑃 ≤ 16.95 lb 9.58 𝑀𝐴 = 23.1 N⋅m 𝑦

𝑇1 20 cm

𝑥

𝐸𝑦

𝐴 𝐸𝑥

𝐶

𝑇2

𝑃 = 90 N 15 cm

𝐵

𝑇1

20 cm

𝐴𝑥 𝐴 𝐴𝑦

30 cm

9.60 𝑀𝐴 = 575 in.⋅lb 9.62 (a) 𝑃 = 14.1 N; (b) 𝑃 = 4.27 N 9.64 𝑊 = 35.3 lb 9.66 𝑊𝐵 = 53.8 lb, surface 𝐶 will stick while motion is impending at 𝐴 and 𝐵. 9.68 (a) 𝑇𝐶𝐷 = 89.86 lb; (b) The assumption of no tipping is correct. 9.70 𝜇 = 0.0741 𝑦 𝑁1

𝐴

𝐹1 𝑁2

𝑥 𝐵

𝐶

𝐹2 𝑃

9.72 𝑁 = 156 kN, 𝐹 = 14.7 kN, 𝜇 = 0.0943; 𝑁 and 𝐹 are located 0.730 m to the left of point 𝐴. 9.74 ℎ = 9 mm 9.76 𝑃 = 1.60 lb, the roll of paper will tip. 9.78 (a) It is not possible to descend the hill with uniform speed; (b) It is possible to descend the hill with uniform speed, 𝑄 = 16.03 lb. 9.80 𝑃 = 56.2 lb, slip is impending at 𝐸 and there is no slip at 𝐵 and 𝐶. 9.82 𝑊 = 4402 N, 𝑀𝐴 = 553.2 N⋅m 9.84 𝑇0 = 57.4 lb 𝑦 𝑥 𝑀𝐴

10◦ 200 in.⋅lb

𝑇0 − 𝛥 𝑇

𝐵

𝐴 𝑇0 + 𝛥 𝑇

Chapter 10 10.2

Concept Problem

10.4 𝐼𝑥 = 568 𝑎4 , 𝐼𝑦 = 328 𝑎4 10.6 𝐼𝑥 = 0.0404 in.4 , 𝐼𝑦 = 0.0570 in.4 10.8 𝐼𝑥 =

1 𝑏ℎ3 , 𝐼𝑦 12

=

1 ℎ𝑏3 12

10.10 Answer provided in problem statement 𝜋 4 𝑟 , 16 0 𝜋 4 𝐽𝐴 = 4 𝑟 𝑜

10.12 𝐼𝑦 = 10.14 10.16

𝐽𝑂 = 𝜋8 𝑟4𝑜 , essay-type answer

Concept Problem

ISTUDY

Appendix B

Answers to Even-Numbered Problems

10.18 (a) 𝐼𝑥 = 2.74 mm4 ; (b) 𝐼𝑦 = 11.0 mm4 10.20 (a) 𝐼𝑥 = 0.305 mm4 ; (b) 𝐼𝑦 = 1.60 mm4 10.22 (a) 𝐼𝑥 = 0.500 in.4 ; (b) 𝐼𝑦 = 0.375 in.4 √ 1 10.24 𝑐1 = 𝑏∕ 𝑎, 𝑐2 = 𝑏∕𝑎3 , 𝐼𝑥 = 10 𝑎𝑏3 10.26 𝐼𝑥 = 0.305 mm4 10.28 𝐼𝑥 = 4.64×107 mm4 , 𝐼𝑦 = 4.58×106 mm4 , 𝑘𝑥 = 105 mm, 𝑘𝑦 = 33.1 mm 10.30 𝐼𝑥 = 117 in.4 , essay-type answer 10.32 𝑘𝑥 = 23.1 in. 10.34 For the rectangular cross section: 𝐼𝑥 = 108 in.4 , 𝐼𝑦 = 108 in.4 ; For the I cross section: 𝐼𝑥 = 428 in.4 , 𝐼𝑦 = 76.0 in.4 10.36 𝐼𝑥 = 4.06×105 mm4 , 𝐼𝑦 = 1.83×105 mm4 10.38 𝐼𝑥 = 37.6 in.4 , 𝐼𝑦 = 10.9 in.4 10.40 (a) 𝐼𝑥 = 1820 mm4 ; (b) 𝐼𝑥 = 2550 mm4 1

2

10.42 (a) 𝑑 = 4.00 in.; (b) 𝐼𝑥 = 128 in.4 ; (c) 𝐼𝑦 = 704 in.4 10.44 (a) 𝑑 = 2.00 mm; (b) 𝐼𝑥 = 88.0 mm4 ; (c) 𝐼𝑦 = 64.0 mm4 10.46 (a) 𝑑 = 65.0 mm; (b) 𝐼𝑥 = 1.45×106 mm4 ; (c) 𝐼𝑦 = 1.13×105 mm4 10.48 Answer provided in problem statement 10.50 𝐼𝑦 = 14 𝑚𝑟2 where 𝑚 = 𝜌𝜋𝑟2 𝑡 ) ( 10.52 𝐼𝑦 = 𝑚 14 𝑅2 + 13 𝐿2 where 𝑚 = 𝜌𝜋𝑅2 𝐿 ( ) 10.54 𝐼𝑧 = 35 𝑚 14 𝑅2 + 𝐿2 where 𝑚 = 13 𝜌𝜋𝑅2 𝐿 ( ) 10.56 𝐼𝑦 = 35 𝑚 13 𝑏2 + ℎ2 where 𝑚 = 43 𝜌𝑎𝑏ℎ Concept Problem 𝑟∕2 𝜌 𝑟 )2 ( )2 1 1 0 ( 2 10.60 (a) 𝐼𝑥 = 𝜋 𝑟 − 𝑥2 𝑑𝑥 + 𝜌0 𝜋 𝑟2 − 𝑥2 𝑑𝑥; (b) 𝐼𝑥 = 2 ∫0 2 2 ∫𝑟∕2 10.58

103 𝜋𝑟5 𝜌0 640

10.62 𝐼𝑥 = 4.28×10−5 slug⋅in.2 10.64 𝐼𝑦 = 13 𝑚𝑎2 where 𝑚 = 12 𝜌𝜋ℎ𝑎2 [ ] [( ) ( )2 ] 𝐿 𝐿 𝜌𝜋 4 ( 𝑥 )4∕3 ( 𝑥 )4 𝑥 2∕3 𝑥 10.66 (a) 𝐼𝑦 = − 𝜌𝜋𝑅2 𝑥2 − 𝑅 𝑑𝑥 + 𝑑𝑥; ∫0 4 ∫0 𝐿 2𝐿 𝐿 2𝐿 16464 𝐿2 + 7689 𝑅2 (b) 𝐼𝑦 = 𝑚 where 𝑚 = 31 𝜌𝜋𝐿𝑅2 60 38192 √ √ ) ( )3 ( 𝐿 𝑅𝑥 𝑅2 𝑥 9 𝜋 10.68 (a) 𝐼𝑥 = 2 𝜌𝑡0 2 − 1+ 𝑑𝑥; (b) 𝐼𝑥 = 20 𝑚𝑅2 where 𝑚 = 43 𝜌𝑡0 𝜋𝑅𝐿 1 + ∫0 𝐿 𝐿 𝐿2 −3 2 10.70 𝐼𝑦 = 1.50×10 kg⋅cm 10.72 𝐼𝑥 = 10.74 𝐼𝑧 =

25 𝑚𝑅2 56 7 𝑚𝑅2 16

where 𝑚 = where 𝑚 =

7 𝜌𝜋𝑅3 12 7 𝜌𝜋𝑅3 12

10.76 𝐼𝑂𝑧 = 0.0217 slug⋅ft2 , essay-type answer ) ( 10.78 𝐼𝑦 = 13 𝑚 𝑎2 + 𝑐 2 10.80 𝐼𝐵 = 20,700 kg⋅mm2 10.82 𝐼𝐴 = 9.24×10−4 slug⋅in.2 10.84 𝐼𝑥 = 0.497 slug⋅in.2

𝑅2 𝐿2

A-37

A-38

ISTUDY

Appendix B

Answers to Even-Numbered Problems

10.86 𝐼𝐵 = 0.0778 kg⋅m2 10.88 𝐼𝑥 = 1.61 slug⋅in.2 10.90 𝐼𝑧 = 0.627 slug⋅in.2 10.92 𝐼𝑦 = 3.53 slug⋅in.2 10.94 𝐼𝑦 = 5.10×10−3 slug⋅ft2 ( ) 10.96 𝐽𝑂 = 𝜋4 𝑟4𝑜 − 𝑟4𝑖 (√ ) 4m 10.98 (a) 𝐼𝑦 = ∫0 𝑥2 𝑥 − 14 𝑥 𝑑𝑥; (b) 𝐼𝑦 = 20.6 m4 10.100 𝑐1 = 6, 𝑐2 = − 13 , 𝑐3 = 1; (a) 𝐼𝑦 = 10.102 (a) 𝐼𝑦 = 2

∫0

∫0

9 in.

√ ) ( 1 𝑥2 6 − 𝑥 − 𝑥 𝑑𝑥; (b) 𝐼𝑦 = 286 in.4 3

1 in.

𝑥2 (𝑥 − 1)2 𝑑𝑥; (b) 𝐼𝑦 = 0.0667 in.4

10.104 𝐼𝑥 = 2.98×106 mm4 , 𝐼𝑦 = 3.25×105 mm4 10.106 (a) 𝑑 = 1.183 in.; (b) 𝐼𝑥 = 5.56 in.4 10.108 𝐼𝑧 =

3 𝑚𝑟2 10

where 𝑚 = 13 𝜌𝜋𝑟2 ℎ

10.110 𝐼𝑥 = 1700 kg⋅mm2 10.112 𝐼𝑧 = 3040 kg⋅mm2 10.114 𝐼𝑥 = 0.0113 kg⋅mm2 ℎ

10.116 𝐼𝑥 =

1 1 𝜌 𝜋𝑥2 𝑑𝑥 + 2 ∫0 1 2 ∫ℎ

2ℎ

𝜌2 𝜋𝑥2 𝑑𝑥

10.118 𝐼𝑦 = 1.79×105 kg⋅m2 10.120 (a) 𝐼𝑥 = 1.43×108 mm4 ; (b) 𝐼𝑥 = 5.45×105 kg⋅mm2 10.122 𝐼𝑥 = 13.3 slug⋅in.2 10.124 𝐼𝑧 = 16.7 slug⋅in.2

Chapter 11 11.2 (𝑟𝐵∕𝐴 )𝓁 = 3.883 f t | | | | | | 11.4 𝑟⃗𝐵∕𝐴 | = (4.000 𝚤̂ − 1.000 𝚥̂) ft, 𝑟⃗𝐵∕𝐴 | = (3.313 𝑢̂ 𝑝 − 2.455 𝑢̂ 𝑞 ) ft, |𝑟⃗𝐵∕𝐴 | = |𝑟⃗𝐵∕𝐴 | = 4.123 f t | |𝑥𝑦 |𝑝𝑞 |𝑥𝑦 | |𝑝𝑞 11.6 𝑣⃗ = (79.68 𝚤̂ + 7.190 𝚥̂) ft∕s 11.8 𝜃 = 3.229 rad, 𝜙 = 0.08731 rad 11.10 𝑣𝑟 = 504.6 f t∕s, 𝑣𝜃 = −353.3 f t∕s 11.12 𝜙 = 101.1◦ 11.14 𝑥𝑃 2 = −0.7679 f t, 𝑦𝑃 2 = 5.330 f t 11.16 𝑣⃗𝐴 = −(20.02 𝚤̂ + 12.81 𝚥̂) ft∕s, 𝑎⃗𝐴 = (−1.414 𝚤̂ + 4.243 𝚥̂) ft∕s2 11.18 𝑟 = 26.36 mm 11.20 [𝐼𝑥𝑥 ] = [𝐼𝑦𝑦 ] = [𝐼𝑧𝑧 ] = 𝑀𝐿2 . Units of 𝐼𝑥𝑥 , 𝐼𝑦𝑦 , and 𝐼𝑧𝑧 in the SI system are kg⋅m2 . Units of 𝐼𝑥𝑥 , 𝐼𝑦𝑦 , and 𝐼𝑧𝑧 in the U.S. Customary system are slug⋅ft2 = lb⋅s2 ⋅ft. 11.22

Concept Problem

( ) 11.24 The units of 𝐸 are kg∕ m⋅s2 . 11.26 The units of 𝑞 are km⋅s. The units of ℎ are km2 /s.

ISTUDY

Appendix B

Answers to Even-Numbered Problems

Chapter 12 12.2

Concept Problem

12.4

Concept Problem

( ( ) ) 12.6 Δ⃗𝑟1 = 8.747 𝑢̂ 𝑟 m, Δ⃗𝑟2 = 13.73 𝑢̂ 𝑟 m, 𝑣⃗avg 1 = 5.467 𝑢̂ 𝑟 m∕s, 𝑣⃗avg 2 = 8.579 𝑢̂ 𝑟 m∕s 12.8 𝑣⃗ = (−154.0 𝚤̂ + 266.7 𝚥̂) ft∕s 12.10 𝑣⃗ = (145.4 𝚤̂ + 69.81 𝚥̂) ft∕s 12.12 Δ𝑣⃗1 = −(0.4623 𝚤̂ + 0.08155 𝚥̂) m∕s, Δ𝑣⃗2 = −(0.002550 𝚤̂ + 0.0004499 𝚥̂) m∕s 12.14 𝜃1 = 12.21◦ , 𝜃2 = −17.30◦ , 𝜙1 = 102.2◦ , 𝜙2 = 72.70◦ ( ) 3 + 4𝑥 −1 12.16 𝜙(𝑥) = cos √ 10 + 24𝑥 + 16𝑥2 ⃗ s) = (−0.001154 𝚤̂ − 0.003175 𝚥̂) ft∕s2 12.18 𝑎⃗avg = (−0.04588 𝚤̂ + 3.886 𝚥̂) ft∕s2 , 𝑎⃗avg − 𝑎(5 ⃗ 𝑣⃗ = 0, ⃗ 𝑑 = 1.394 f t, 𝑣 = 0.3485 f t∕s 12.20 Δ⃗𝑟 = 0, avg avg Computer Problem

12.22

12.24 𝑣max = 2𝑣0 = 58.67 f t∕s, 𝑣min = 0 f t∕s, 𝑦𝑣

min

𝑣20

= 0 ft, 𝑦𝑣

max

𝑣20

= 2𝑅 = 2.300 f t,

𝚥̂ = (748.2 f t∕s2 ) 𝚥̂, 𝑎⃗𝑣 = − 𝚥̂ = (−748.2 f t∕s2 ) 𝚥̂ max 𝑅 𝑅 2 2 3 2 8𝑣0 𝑎 −4𝑣0 𝑎 𝑦 12.26 𝑥̈ = ( )2 , 𝑦̈ = ( )2 𝑦2 + 4𝑎2 𝑦2 + 4𝑎2 𝑎⃗𝑣

min

=

12.28 𝑣⃗ = (203.2 𝚤̂ + 117.3 𝚥̂) ft∕s, 𝑎⃗ = (−27.53 𝚤̂ + 47.69 𝚥̂) ft∕s2 12.30 𝑣max = 80.82 f t∕s 12.32 𝑣⃗ = (51.32 𝚤̂ − 1.046 𝚥̂) ft∕s, 𝑎⃗ = −(0.3873 𝚤̂ + 19.00 𝚥̂) ft∕s2 √ ( )2 √ 𝑥3 𝑥 | | | 2 2 2 12.34 𝑣 = 𝜔 𝑑 − 𝑥 1+ℎ − , 𝑣| = 0.5236 f t∕s, 𝑣| = 0.4554 f t∕s, 𝑣| =0 2 4 | | |𝑥=1 𝑥=0 𝑥=0.5 𝑑 𝑑 12.36 𝑣⃗𝐴 = −(72.00 𝚤̂ + 96.00 𝚥̂) ft∕s, 𝑎⃗𝐴 = (460.8 𝚤̂ − 345.6 𝚥̂) ft∕s2 12.38

Computer Problem

12.40 𝑣̇ = 9.108 m∕s2 12.42 𝑡braking = 7.854 s

( 12.44 |𝑎|max = 3.600 m∕s2 , 𝑠|𝑎

) max | 1

( = 12.84 m, 𝑠|𝑎

) max | 2

= 128.5 m

√ 12.46 Largest distance traveled in 1 s is 𝑑 = 0.2667 m, corresponding to 𝑎 = 𝛽1 𝑡. 12.48 𝑣(0) = −0.08000 f t∕s 12.50 𝑣0 = 10.10 m∕s 12.52 𝑎𝑐 = 22.36 𝑔 12.54 𝑡stop = 2.880 s

12.56 𝜂 = 98.10 s−1 { } 12.58 𝑥̇ = 7.000 sin[(1.000 rad∕s)𝑡] + 10.50 cos[(0.5000 rad∕s)𝑡] − 10.50 m∕s, { } 𝑥 = 7.000 − 7.000 cos[(1.000 rad∕s)𝑡] + 21.00 sin[(0.5000 rad∕s)𝑡] − (10.50 s−1 )𝑡 m∕s2 12.60 𝑡stop = 0.2233 s 12.62 𝑣term = 4.998 m∕s 12.64 |𝑣|max = 1.128 m∕s, 𝑠|𝑣| = 0.1250 m, 𝑠|𝑣| = −0.1250 m max max ) 𝑚𝑔 𝑚𝑔 ( −𝐶𝑑 𝑡∕𝑚 1−𝑒 , 𝑣term = 12.66 𝑣(𝑡) = 𝐶𝑑 𝐶𝑑

A-39

A-40

ISTUDY

Appendix B

Answers to Even-Numbered Problems

12.68 𝑣𝑓 = 3.563 m∕s 12.70 12.72 12.74 12.76 12.78 12.80 12.82

Computer Problem (𝑠wet − 𝑠dry ) 𝑠wet = 26.31 m, (100%) = 65.21% 𝑠dry √ ) ( ̇𝜃(𝜃) = ± 𝜃̇ 2 + 2 𝑔 cos 𝜃 − cos 𝜃 0 0 𝐿 𝜃̇ min = 4.930 rad∕s √ ( ) 𝑘𝐿0 ) 𝑘( 2 𝑥̇ = ± 𝑣20 + 2 𝑔 + (𝑥 − 𝑥0 ) − 𝑥 − 𝑥20 𝑚 𝑚 𝑡𝑥 = 0.2433 s max √ √ ( ) 𝑟0 − 𝑟 (a) 𝑟̇ = − 2𝐺 𝑚𝐴 + 𝑚𝐵 ; (b) (i) 𝑟̇ = −5.980 × 10−5 ft∕s, (ii) 𝑟̇ → −∞ 𝑟𝑟0

12.84 𝜇𝑘 = 0.3012 | | ,𝛼 | = 0.001956 rad∕s2 , 𝛼𝑠 | = 0.03056 rad∕s2 |𝑟=𝑟2 2𝜋𝑟3 𝑠 |𝑟=𝑟1 Computer Problem

12.86 𝛼𝑠 = 12.88

ℎ𝑣2𝑝

12.90 𝜃opt = 48.13◦ , 𝑣0 = 105.6 f t∕s 12.92 𝑑 = 358.1 f t 12.94 𝑣⃗gun = (14.65 f t∕s) 𝚤̂

𝚥̂

12.96 𝑦 = ℎ + tan 𝛽(𝑥 − 𝑤) −

𝚤̂

𝑔

2𝑣20 cos2 𝛽

(𝑥 − 𝑤)2

12.98 The golfer’s chip shot is successful. 12.100 𝑣𝐴 = 52.82 f t∕s

12.102 30.43◦ ≤ 𝜃 ≤ 56.84◦ and 69.41◦ ≤ 𝜃 ≤ 74.50◦ ; range seen by the observer 𝛽 = 21.32◦ 12.104 𝑑𝐴𝐵 = 81.96 m 12.106 𝜃 = 20.62◦ , 𝑣0 = 62.52 f t∕s Computer Problem √ 𝑔𝑅 cos 𝜃 12.110 𝑣0 = √ 2 sin 𝛽 cos(𝛽 − 𝜃) 12.108

12.112 𝑡𝐶 = 3.052 s, 𝑎𝑥 = −7.002 f t∕s2 , 𝑡𝐵 = 1.275 s ( )2 = 5.211 s, 𝑅max = 437.1 f t, 𝜃𝑅 12.114 𝑅2 = 𝑣20 𝑡2𝑓 − ℎ + 12 𝑔𝑡2𝑓 , 𝑡𝑓 𝑅max

max

= 45.66◦

12.116 𝑡max = 3.000 s, 𝐻max = 49.21 m, percent increase in height with no air resistance = 2.363% 12.118

Computer Problem

12.120

Concept Problem

12.122

Concept Problem

12.124 𝜌 = 49.69 f t 12.126 𝑣̇ = 3.213 m∕s2 12.128 𝜌S¨udkurve = 50.56 m, 𝜌Nordkurve = 92.54 m 12.130 𝜌 = 327.3 f t 12.132 𝑑 = 831.0 f t 12.134 𝑎⃗ = 0⃗ 12.136 𝑎⃗ = (−39.40 𝚤̂ + 42.88 𝚥̂) m∕s2

ISTUDY

Appendix B

Answers to Even-Numbered Problems

12.138 Concept Problem 12.140 ||𝑎⃗|| = (33.69×10−3 cos 𝜙) m∕s2 12.142 𝑎⃗ = (121.2 𝚤̂ + 230.0 𝚥̂) ft∕s2 ( ) [( ) ( ) ] 12.144 𝑎⃗ = − 18.96 f t∕s2 𝑢̂ 𝑡 + 69.91 f t∕s2 − 0.2917 s−2 𝑠 𝑢̂ 𝑛 12.146 𝑣̇ = −5.343 m∕s2 , 𝜌 = 17.50 m 12.148 𝜌 = 282.2 f t 12.150 𝜌min = 564.5 f t, 𝑡𝑓 − 𝑡0 = 4.742 s √ ( ( ) )2 4𝑣20 2𝑣 2𝑣 1 12.152 |𝑎| ⃗ = 𝑣0 − 0 𝑠 + 𝑣0 − 0 𝑠 𝜌 𝜋𝜌 𝜋𝜌 𝜋2 12.154 𝑑 = 304.4 f t, 𝑡𝑓 = 3.106 s 12.156 𝜌 = 0.2 m, 𝑎⃗ = −5 m∕s2 𝚥̂ ( ) ( ) 12.158 (a) 𝑎⃗ ⋅ 𝑏⃗ = 1 ⋅ −6 + 2 ⋅ 3 + 3 ⋅ 0 = 0; (b) 𝑎⃗ × 𝑎⃗ × 𝑏⃗ = 84 𝚤̂ − 42 𝚥̂ + 0 𝑘̂ ; (c) they are the same 12.160

Concept Problem

Concept Problem 12.164 𝑟̇ = 684.1 f t∕s, 𝜃̇ = −0.01157 rad∕s, 𝑟̈ = 4.944 f t∕s2 , 𝜃̈ = 0.0004281 rad∕s2 12.162

12.166 𝑣 = 1.816 m∕s, |𝑎| ⃗ = 4.573 m∕s2 12.168 𝑟̈ = 1.599 m∕s2 12.170 𝜃0 = 0, 𝑟0 = 0.01486 m ) ) ) ( ( ( 12.172 𝑣⃗ = (−1.3 m∕s) 𝑢̂ 𝑟 + 0.22 s−1 𝑟 𝑢̂ 𝜃 , 𝑎⃗ = − 0.04840 s−2 𝑟 𝑢̂ 𝑟 − 0.5720 m∕s2 𝑢̂ 𝜃 12.174 𝑣⃗ = (0.1500 𝑢̂ 𝑟 + 1.732 𝑢̂ 𝜃 ) ft∕s, 𝑎⃗ = (−0.3638 𝑢̂ 𝑟 + 0.1125 𝑢̂ 𝜃 ) ft∕s2 12.176 𝑟⃗ = 30.20 𝑢̂ 𝑟 ft, 𝑣⃗ = (3 𝑢̂ 𝑟 + 7.550 𝑢̂ 𝜃 ) ft∕s, 𝑎⃗ = (−1.888 𝑢̂ 𝑟 + 1.500 𝑢̂ 𝜃 ) ft∕s2 12.178 𝜂 = 5.602 𝜇m, 𝜔 = 2630 rad∕s = 25,120 rpm 12.180 𝑣 = 5490 f t∕s, |𝑎| ⃗ = 1.005×106 f t∕s2 𝜅𝑣0 𝑣 (𝑟 + 𝜅𝜃) 12.182 𝑣⃗ = √ 𝑢̂ 𝑟 + √ 0 0 𝑢̂ 𝜃 , 𝜅 2 + (𝑟0 + 𝜅𝜃)2 𝜅 2 + (𝑟0 + 𝜅𝜃)2 [ { ]} 𝑣20 (𝑟0 + 𝜅𝜃)3 𝜅𝑣20 (𝑟0 + 𝜅𝜃)2 + 2 𝜅 2 + (𝑟0 + 𝜅𝜃)2 𝑎⃗ = − [ 𝑢̂ 𝜃 ]2 𝑢̂ 𝑟 + ]2 𝑟20 + 2𝑟0 𝜅𝜃 + (1 + 𝜃 2 )𝜅 2 [𝜅 2 + (𝑟0 + 𝜅𝜃)2 12.184 𝜃̇ 𝐴 = −0.001157 rad∕s, 𝜃̇ 𝐻 = −4.287 rad∕s, 𝜃̈𝐴 = −1.684×10−4 rad∕s2 , 𝜃̈𝐻 = −0.07547 rad∕s2 ( ) [ ( ) ] 12.186 𝑎⃗̇ = ⃛𝑟 − 3𝑟𝜃̇ 𝜃̈ − 3𝑟̇ 𝜃̇ 2 𝑢̂ + 𝑟 𝜃⃛ − 𝜃̇ 3 + 3̈𝑟𝜃̇ + 3𝑟̇ 𝜃̈ 𝑢̂ 𝑟

𝜃

12.188 𝜙̇ = 0, 𝜙̈ = −15.83 rad∕s2 12.190 12.192

12.194 12.196

𝑣20 cos3 𝜃 𝑣0 cos2 𝜃 𝑎0 cos2 𝜃 2𝑣20 cos3 𝜃 sin 𝜃 ̇ ̈ 𝑟 = ℎ∕ cos 𝜃, 𝑟̇ = 𝑣0 sin 𝜃, 𝜃 = , 𝑟̈ = 𝑎0 sin 𝜃 + ,𝜃 = − ℎ ℎ ℎ ℎ2 √ 2 2 𝑟 = (ℎ + 𝜌 sin 𝜙) + (𝑑 + 𝜌 cos 𝜙) , 𝑣0 (𝜌 + 𝑑 cos 𝜙 + ℎ sin 𝜙) 𝑣0 (ℎ cos 𝜙 − 𝑑 sin 𝜙) 𝑟̇ = √ , 𝜃̇ = , 2 2 2 𝑑 + ℎ2 + 𝜌2 + 2𝑑𝜌 cos 𝜙 + 2ℎ𝜌 sin 𝜙 (𝑑 + 𝜌 cos 𝜙) + (ℎ + 𝜌 sin 𝜙) ) ( 𝑣2 (𝜌 + 𝑑 cos 𝜙 + ℎ sin 𝜙) 𝑑 2 + ℎ2 + 𝑑𝜌 cos 𝜙 + ℎ𝜌 sin 𝜙 𝑟̈ = − 0 , ( )3∕2 𝜌 𝑑 2 + ℎ2 + 𝜌2 + 2𝑑𝜌 cos 𝜙 + 2ℎ𝜌 sin 𝜙 ( ) 𝑣2 (ℎ cos 𝜙 − 𝑑 sin 𝜙) 𝑑 2 + ℎ2 − 𝜌2 ̈𝜃 = 0 𝜌 (𝑑 2 + ℎ2 + 𝜌2 + 2𝑑𝜌 cos 𝜙 + 2ℎ𝜌 sin 𝜙)2 𝑟 𝜔 √ 𝑟̇ = 0 0 𝑟2 − 𝑟20 𝑟 𝑟̇ = 8.825 f t∕s, 𝜙̇ = −0.7746 rad∕s, 𝑟̈ = −18.23 f t∕s2 , 𝜙̈ = −9.779 rad∕s2

A-41

A-42

ISTUDY

Appendix B

Answers to Even-Numbered Problems

12.198 𝑟̇ = −438.5 f t∕s, 𝜃̇ = 347.4×10−6 rad∕s, 𝑟̈ = 14.95 f t∕s2 , 𝜃̈ = −10.59×10−6 rad∕s2 12.200

Concept Problem

12.202

Concept Problem

( ) 𝑟⃗𝐵∕𝐴 12.204 ROS = 𝑣⃗𝐵 − 𝑣⃗𝐴 ⋅ |⃗𝑟𝐵∕𝐴 | 12.206 ROS𝐴𝐵 = −163.8 f t∕s, ROS𝐶𝐵 = −16.50 f t∕s 12.208 (𝑣𝑃 )avg = 12.50 f t∕s 12.210 𝑡 = 3.500 s, 𝑥 = 24.50 f t 12.212 Distance = 47.12 m 12.214 𝑣𝑃 = 442.7 mph 2

12.216 𝑣𝐴∕𝐵 = 5.517 m∕s 12.218 𝑑 = 1.229 m 12.220 Δ𝑡 = 0.3752 s, distance traveled by 𝑃 = 110.9 m 12.222

Computer Problem 𝚤̂

12.224 𝑣⃗𝐵 = −0.7500 𝚥̂ m∕s

𝚥̂

12.226 𝑣⃗𝐵 = −0.5000 𝚥̂ m∕s

𝚥̂

𝚤̂

12.228 𝑣bicycle = 0.1042 f t∕s 12.230 𝑣𝐵 = 2.000 m∕s 𝚤̂

12.232 𝑣⃗𝐺 = −0.6923 𝚥̂ ft∕s, 𝑎⃗𝐺 = −0.2308 𝚥̂ ft∕s2

𝚥̂

12.234 𝑣𝐴 = 2.136 f t∕s 12.236 𝑣⃗𝐶 = −55.52 𝚥̂ ft∕s ( ) 17.82 cos 𝜃 12.238 𝑣⃗𝐶 = − sin 𝜃 61.09 + √ 𝚥̂ ft∕s 2 0.1951 − 0.08507 sin 𝜃 { ( ) 0.2917 cos 𝜃 4 cos 𝜃 1 + √ 𝑎⃗𝐶 = −1.279×10 0.1951 − 0.08507 sin2 𝜃 [ ]} 2 0.02481 cos 𝜃 sin 𝜃 0.2917 sin 𝜃 + sin 𝜃 ( 𝚥̂ ft∕s2 )3∕2 − √ 2 2 0.1951 − 0.08507 sin 𝜃 0.1951 − 0.08507 sin 𝜃 𝑣20 (𝜌 + 𝑑 cos 𝜙 + ℎ sin 𝜙)(𝑑 2 + ℎ2 + 𝜌𝑑 cos 𝜙 + ℎ𝜌 sin 𝜙) 12.240 𝑟̈ = − , [ ]3∕2 𝜌 (𝑑 + 𝜌 cos 𝜙)2 + (ℎ + 𝜌 sin 𝜙)2 𝑣2 (𝑑 2 + ℎ2 − 𝜌2 )(𝑑 sin 𝜙 − ℎ cos 𝜙) 𝜃̈ = − 0[ ]2 𝜌 (𝑑 + 𝜌 cos 𝜙)2 + (ℎ + 𝜌 sin 𝜙)2 1 1 12.242 𝑣𝐴 = 𝑣0 , 𝑣𝐵∕𝐶 = − 𝑣0 4 2 12.244 𝑣⃗𝐴 = −3.078 𝚤̂ ft∕s, 𝑎⃗𝐴 = −0.8931 𝚤̂ ft∕s2 12.246

Concept Problem

12.248

Concept Problem

𝚤̂ 𝚥̂

Concept Problem | | 12.252 |𝑎⃗| = 17,280 f t∕s2 ( ) ( ) ( ) 12.254 𝑎⃗ = −𝐿 𝜙̇ 2 + 𝜃̇ 2 sin2 𝜙 𝑢̂ 𝑟 + 𝐿 𝜙̈ − 𝜃̇ 2 sin 𝜙 cos 𝜙 𝑢̂ 𝜙 + 𝐿 𝜃̈ sin 𝜙 + 2𝜙̇ 𝜃̇ cos 𝜙 𝑢̂ 𝜃 12.250

12.256 𝑥land = 3.345 m, 𝑦land = 1.420 m 12.258 𝑟̇ = 0, 𝜙̇ = 0, 𝜃̇ = −0.01768 rad∕s, 𝑟̈ = 11.27 f t∕s2 , 𝜙̈ = 78.10×10−6 rad∕s2 , 𝜃̈ = 0

ISTUDY

Appendix B

Answers to Even-Numbered Problems

( ) 𝐾 𝐾2 12.260 𝑣⃗ = 𝑧̇ tan 𝛽 𝑢̂ 𝑅 + 𝑢̂ + 𝑧̇ 𝑢̂ 𝑧 , 𝑎⃗ = 𝑧̈ tan 𝛽 − 𝑢̂ 𝑅 + 𝑧̈ 𝑢̂ 𝑧 𝑧 tan 𝛽 𝜃 𝑧3 tan3 𝛽 12.262 𝑥fw = 0, 𝑦fw = 1.833 f t, 𝑧fw = 5.067 f t 12.264

Concept Problem Computer Problem

12.266

12.268 𝑣⃗avg = (0.4495 𝚤̂ + 20.67 𝚥̂) ft∕s, 𝑡̄ = 4.949 s. In general, it is not possible to find a time instant in an interval [𝑡1 , 𝑡2 ] for which the velocity and the average velocity are equal. 𝑣𝑠 𝜆 12.270 𝑥̇ = √ ( ) 𝜆2 + 4ℎ2 𝜋 2 cos2 2𝜋𝑥 𝜆 Computer Problem

12.272

12.274 𝑡𝑓 = 5.839 s, stopping time percent increase = 129.1% 12.276 𝑣0 = 64.19 f t∕s, 𝑑 = 202.1 f t, 𝑣𝑖 = 98.08 f t∕s, impact angle = 26.51◦ measured clockwise from the slope ( ) | | = 17.28 𝚥̂ − 0.4400 𝑘̂ ft∕s, 𝑎⃗𝐴 | = −108.7 𝚥̂ ft∕s2 12.278 𝑣⃗𝐴 | |𝛽=0 |𝛽=0 12.280 𝑎𝑃 𝑛 = 𝜔2𝐸 𝑅𝐸 cos 𝜆 = 0.03369 cos 𝜆 m∕s2 , 𝑎𝐸𝑛 = 𝜔2𝑂 𝑅𝑂 = 0.005942 m∕s2 , 𝜆 = 79.84◦ 12.282 𝜌𝑓 = 629.2 m, 𝑡𝑓 = 5.238 s 12.284 𝜔max = 230.9 rad∕s, 𝜃min = 54.74◦ 12.286 𝑎𝑡 = 3.330 f t∕s2 12.288 |𝑎| ⃗ = 1.240 f t∕s2 12.290 𝜔 ⃗ 𝐵𝐶 = (−1169 rpm) 𝑘̂ 12.292 𝑦̇ 𝐴 (𝑡𝑓 ) = −64.22 m∕s, 𝑦̈𝐴 (𝑡𝑓 ) = −22,080 m∕s2 12.294 𝜌 =

(𝑣20 + 𝑔 2 𝑡2 − 2𝑣0 𝑔𝑡 sin 𝛽)3∕2 𝑔𝑣0 cos 𝛽

Chapter 13 13.2

Concept Problem

13.4 𝑎⃗𝐴 = 2.147 𝚥̂ ft∕s2

𝚥̂ 𝚤̂

13.6 𝑎max = 17.75 f t∕s2 13.8 𝑎𝐵

max

13.10 𝑥̈ + 13.12 𝑎𝐵

= 0.1650 m∕s2 = 0.5417 f t∕s2

𝐸𝐴 𝑥=0 𝑚𝐿 = 5.048 m∕s2 = 16.57 f t∕s2

max

13.14 𝑑 = 6.630 f t 13.16 𝑑 = 417.5 f t. If an entire train weighing 30×106 lb was skidding to a stop, instead of a locomotive, we would have the same stopping distance because the weight does not appear in our solution. ) ( 𝐶𝑑 2 𝚥̂ 𝑣 − 𝑔 𝚥̂ 13.18 𝑎⃗ = 𝚤̂ 𝑚 13.20 (𝜇𝑠 )min = 0.3436 13.22 𝑣𝑖 = 5.297 f t∕s 13.24

Computer Problem

13.26 𝑣𝑖 = 32.68×10−6 m∕s √ 𝑚 13.28 𝑣max = 𝑔 𝑘

A-43

A-44

ISTUDY

Answers to Even-Numbered Problems

13.30 Spring compression = 0.7367 f t 13.32 𝑘 = 5.084×104 lb∕f t 13.34 𝛿max = 0.1675 m, (𝐹𝑠 )max = 586.1 N 13.36

Computer Problem

13.38 𝑥stop = 0.1597 m 13.40 𝑘 = 5.930 lb∕f t 13.42 𝑘 = 9034 kg∕s2 √ [ ( )] √ 𝑘 2 𝑥0 + 2𝐿0 𝐿 − 𝑥20 + 𝐿2 13.44 𝑣(0) = 2 𝑚 13.46

Concept Problem

13.48

Concept Problem

13.50

Concept Problem

13.52 𝜌 = 1.401×104 f t 13.54 No value of 𝜔𝑐 exists that would cause the person to slide up the wall. 𝑚𝑔 13.56 (𝐹𝑂𝐴 )before release = , (𝐹𝑂𝐴 )after release = 𝑚𝑔 cos 𝜃, (% change in 𝐹𝑂𝐴 )𝜃=30◦ = −25.00% cos 𝜃 13.58 |𝑎| ⃗ = 3.620 𝑔, |𝐹𝐿 | = 43,060 N ) ) ( ( 𝑟𝑢 𝑟𝑢 𝑘 𝑘 = 0, 𝑦̈ + 𝑦 1 − √ =0 13.60 𝑥̈ + 𝑥 1 − √ 𝑚 𝑚 𝑥2 + 𝑦2 𝑥2 + 𝑦2 ) 𝑘( 13.62 𝑟̈ − 𝑟𝜃̇ 2 + 𝑟 − 𝑟𝑢 − 𝑔 cos 𝜃 = 0, 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ + 𝑔 sin 𝜃 = 0 𝑚 13.64 𝑡𝑠 = 0.7466 s ( ( ) )2 𝑚2 𝑔 𝑚 𝑒𝐶𝑑 𝑥∕𝑚 − 1 tan 𝜃0 𝑒𝐶𝑑 𝑥∕𝑚 − 1 − 13.66 𝑦 = 𝐶𝑑 2𝐶𝑑2 𝑣20 cos2 𝜃0 𝑔(𝜇𝑠 + tan 𝜃)

= −120.1 f t∕s2 , 1 − 𝜇𝑠 tan 𝜃 𝜇𝑠 𝑚𝑔 𝑚𝑔 𝐹𝜙 = 0, 𝐹 = = (83.16 f t∕s2 ) 𝑚, 𝑁 = = (92.40 f t∕s2 ) 𝑚, cos 𝜃 − 𝜇𝑠 sin 𝜃 cos 𝜃 − 𝜇𝑠 sin 𝜃 √ 𝜇𝑠 + tan 𝜃 √ 𝑣max = 𝜌𝑔 = 363.4 f t∕s = 247.8 mph 1 − 𝜇𝑠 tan 𝜃

13.68 𝑎𝜙 = 0 = 𝑎𝑧 , 𝑎𝑅 = −

13.70 𝐹 = 5971 lb 13.72 (𝜔𝑐 )max = 2.733 rad∕s 13.74 (a) 𝑎⃗ = (318.8×103 𝑢̂ 𝑟 + 603.0 𝑢̂ 𝜃 ) m∕s2 ; (b) 𝑃 = 5.898×106 N; (c) 𝑅 = 11.33×103 N 13.76 𝜔 = 9.939 rad∕s 13.78 𝜃𝐵 = 41.81◦ Concept Problem 𝑚 13.82 𝑟̈ − 𝑟𝜃̇ 2 + 𝐺 𝑒 = 0, 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 0 𝑟2 √ √ 5𝑚𝑔𝑟 13.84 𝛿min = , 𝑣𝐵 = 2𝑔𝑟 𝑘 13.86 Computer Problem 13.80

13.88 ℎmax = 0.3618 f t

Appendix B

ISTUDY

Appendix B

Answers to Even-Numbered Problems

( ) ( ) √ 𝑥̇ 2 𝐿2 − 𝑦2 + 𝑦̇ 2 𝐿2 − 𝑥2 + 2𝑥𝑦𝑥̇ 𝑦̇ ( 2 ) 2 13.90 𝑥̈ 𝐿 − 𝑦 + 𝑦𝑥𝑦 + 𝑔𝑥 𝐿2 − 𝑥2 − 𝑦2 = 0, ̈ +𝑥 2 2 2 ( )𝐿 − 𝑥 ( −𝑦 ) 2 𝐿2 − 𝑦2 + 𝑦̇ 2 𝐿2 − 𝑥2 + 2𝑥𝑦𝑥̇ 𝑦̇ √ 𝑥 ̇ ) ( ̈ +𝑦 + 𝑔𝑦 𝐿2 − 𝑥2 − 𝑦2 = 0 𝑦̈ 𝐿2 − 𝑥2 + 𝑥𝑥𝑦 𝐿2 − 𝑥2 − 𝑦2 2 13.92 𝜙̈ − 𝜃̇ sin 𝜙 cos 𝜙 + (𝑔∕𝐿) sin 𝜙 = 0, 𝜃̈ sin 𝜙 + 2𝜙̇ 𝜃̇ cos 𝜙 = 0 13.94 𝑟̈ − 𝑟𝜃̇ 2 = 0, 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ = 0 √ √ √ 13.96 𝑀𝑧 = 2𝑚𝜔20 𝑟 𝑟2 − 𝑟20 , |𝑣𝑟 | = 𝜔0 𝑑 2 + 2𝑑𝑟0 , 𝑣 = 𝜔0 2𝑑 2 + 4𝑑𝑟0 + 𝑟20 Concept Problem 5 13.100 𝜃 = cos−1 = 33.56◦ 6 ( 13.98

13.102 𝜃 = tan 13.104

−1

−1

𝜇𝑠 − sin

𝜇𝑠 𝑣20 √ 𝜌𝑔 1 + 𝜇𝑠2

) = 𝜃 = 15.59◦

Concept Problem

Concept Problem ( ) 13.108 𝐹 = 2𝑚𝐴 𝑔 − 𝑎𝐴𝑦 = 1650 N 13.106

13.110 𝐹 = { 1901 N 49.29 lb 13.112 𝑇 = 47.33 lb

for 0 ≤ 𝑡 ≤ 1 s, for 𝑡 > 1 s.

13.114 𝑇 = 47.91 lb ( ) 13.116 𝜇𝑠 min = 0.4245 𝚥̂

13.118 𝑎⃗𝐴 = 8.050 𝚤̂ ft∕s2 , 𝑎⃗𝐵 = 145.0 𝚤̂ ft∕s2 𝚤̂ √ ) ( 𝑃 𝑙 | = 13.120 𝑣| 1 − sin 𝜃0 |𝜃=90◦ 𝑚 13.122 𝑎⃗𝐴 = 37.32 𝚤̂ ft∕s2 , 𝑎⃗𝐵 = −53.60 𝚥̂ ft∕s2 , |𝐹𝑏 | = 6.979 lb in tension 𝑚𝑔 13.124 𝑣max = √ 𝑘(𝑚 + 𝑚𝑝 ) √ 𝑘𝑑 2 13.126 Distance traveled for separation = 𝑑, 𝑣𝑠 = 𝑚𝐴 + 𝑚𝐵 13.128 𝑡contact = 78.50×103 s = 21.80 h 13.130 𝑇1 = 179.7 N, 𝑇2 = 108.9 N √ ( ) √ 𝑔 𝑑2 𝑑 1 2 2 13.132 𝑣𝐷 = 𝑅 − 𝑅 − 4𝑑 2− 𝑅 2𝑅2 𝑅2 13.134 𝑣⃗𝐴 = −(3.159 𝚤̂ + 3.133 𝚥̂) m∕s, 𝑣⃗𝐵 = 3.264 𝚤̂ m∕s

𝚥̂ 𝚤̂

13.136 𝑑𝐵 = 2(𝑚𝐴 sin 𝛼 + 𝑚𝐵 )𝑔∕𝑘 = 18.40 f t; Since the above value is larger than 2 f t, we conclude that the max actual maximum value of 𝑑𝐵 is 𝑑𝐵 = 2 f t; 𝑣𝐵 = 9.422 f t∕s max max ] [ 13.138 𝑑𝐵 = 2 𝑚𝐴 (sin 𝛼 − 𝜇𝑘 cos 𝛼) + 𝑚𝐵 𝑔∕𝑘 = 22.75 f t; Since the above value is larger than 2 f t, we max conclude that the actual maximum value of 𝑑𝐵 is 𝑑𝐵 = 2 f t; 𝑣𝐵 = 9.674 f t∕s max

13.140

Computer Problem

13.142 𝑎⃗𝐺 = −9.976 𝚥̂ ft∕s2 , 𝑇𝐵𝐶 = 1213 lb 13.144

Computer Problem

𝚤̂ 𝚥̂

max

A-45

A-46

ISTUDY

Appendix B

Answers to Even-Numbered Problems

13.146 𝑡𝑠 = 1.805 s 13.148 𝑣contact = 273.1×10−6 f t∕s 13.150 𝑎𝑥 = 11.01 f t∕s2 , 𝑎𝑦 = 0, 𝑎𝑧 = −1.942 f t∕s2

𝑘̂ 𝚤̂

, 𝜌 = 99,720 f t = 18.89 mi

13.152 Proof Computer Problem

13.154

13.156 𝑊𝐴 = 161.0 lb 𝑣2𝑏 − 𝑣2𝑐 ( ) (𝑣 − 𝑣𝑐 )2 , 𝑑ss rel = 𝑏 𝜇𝑘 𝑔 2𝜇𝑘 𝑔 2𝜇𝑘 𝑔 2 ̈ ̈ ̇ 13.160 𝐿1 𝜃 cos 𝜃 + 𝐿2 𝜙 + 𝐿1 𝜃 sin(𝜙 − 𝜃) + 𝑔 sin 𝜃 = 0, ] [ 𝑚1 𝐿1 𝜃̈ − 𝐿1 𝜃̇ 2 cos(𝜙 − 𝜃) − 𝐿2 𝜙̇ 2 − 𝑔 cos 𝜙 = 0 sin(𝜙 − 𝜃) + 𝑚2 sin(𝜙 − 𝜃)

13.158 𝑡ss =

𝑣𝑏 − 𝑣𝑐

, 𝑑ss =

13.162 |𝑎⃗𝐴 | = 14.86 f t∕s2 up, |𝑎⃗𝐵 | = 7.431 f t∕s2 down, 𝑇 = 5.769 lb

Chapter 14 Concept Problem ( ) ( ) ) | |( ) | |( 𝑈1-2 𝑁 = 1810 J, 𝑈1-2 𝐹 = −1570 J, | 𝑈1-2 𝑁 | ≠ | 𝑈1-2 𝐹 | because 𝑁 ≠ 𝑚𝑔 due to the fact that the | | | 𝑔 𝑔| man is accelerating. 14.6 𝑈1-2 = 360.7×103 f t ⋅lb √ 14.8 𝑣2 = 𝑑 𝑘∕𝑚 14.2 14.4

14.10 𝑈1-2 = 117.2 kJ 14.12 𝑈1-2 = 288.5 f t ⋅lb ( ) 14.14 𝑈1-2 friction = 1870 J 14.16 𝑑 = 25.22 m 14.18

Concept Problem

14.20 𝑣2 = 35.62 f t∕s Computer Problem

14.22

14.24 Energy lost to permanent deformation = 4.984×10−11 J 14.26 𝑣1 = 7.093 f t∕s Computer Problem

14.28

14.30 𝛽 = 1.293×10−5 lb∕f t 3 14.32 𝑘 = 273.9×103 N∕m 14.34 𝑣0 = 6.400 f t∕s 14.36 𝑘 = 44.81 lb∕f t ( ) 14.38 𝑈1-2 engine = 464.4×103 f t ⋅lb 14.40 (𝐹𝑝 )avg = 1471 N 14.42 𝑅 = 4425 f t 14.44 𝑣2 = 6.348 m∕s 14.46 𝑣2 = 18.52 m∕s 14.48 𝑘 = 4316 N∕m 𝑞 𝑞 14.50 𝑉 = 𝑘 𝐴 𝐵 𝑟 14.52 𝑣𝐴 = 2336 m∕s 14.54 𝜃(𝑣

𝐵 )max

= 0, (𝑣𝐵 )max = 0.3822 m∕s

ISTUDY

Appendix B

Answers to Even-Numbered Problems √

√ 5𝑚𝑔𝑟 , 𝑣𝐵 = 2𝑔𝑟 𝑘 14.58 (a) 𝑣2 = 1626 f t∕s. (b) 𝑣2 = 1522 f t∕s. (c) The form of the potential energy allows us to interpret the work done by the expanding gas as the work of the force 𝑃0 𝐴 along an effective distance 𝑠0 log 𝑠2 ∕𝑠0 . The force 𝑃0 𝐴 is the same in both cases. However, the decrease in 𝑠0 in Part (b) is such that the effective distance over which the force acts is smaller, thus causing 𝑣2 to be smaller in Part (b). 14.60 ℎmax = 0.1055 m

14.56 𝛿min =

14.62 (a) 𝑉𝑐 = 12 𝑘𝛿 2 − 14 𝛽𝛿 4 ; (b) 𝑣⃗bottom = −5.161 𝚥̂ ft∕s

𝚥̂

𝚤̂ ; (c)

|𝑎| ⃗ max = 1.599𝑔

14.64 𝜇𝑘 = 0.8343 [( ) ( )6 ] 12 𝜎 𝜎 14.66 𝑉 = 4𝜖 − 𝑟𝑖𝑗 𝑟𝑖𝑗 √ 𝑚 14.68 𝑣max = 𝑔 𝑘 14.70 𝐿0 = 1.060 m 14.72 𝑣𝐴2 = 1.093 m∕s, 𝑣𝐵2 = 3.278 m∕s 14.74 𝑑𝐵

max

= 2.500 f t

14.76 𝑣𝐵2 = 5.648 m∕s 14.78 𝑣𝐵2 = 33.23 f t∕s 14.80 𝑣𝐴2 = 13.77 f t∕s 14.82 𝑊 = 2𝑚𝑔 14.84 The work done by the tension in the cord on 𝐴 is equal and opposite to the corresponding work done by the tension in the cord on 𝐵. Thus the net work done by the tension in the cord on the system is equal to zero. 𝚥̂

𝑣⃗𝐴2 = 0.1424 𝚤̂ m∕s, 𝑣⃗𝐵2 = −0.4273 𝚤̂ m∕s √ 14.86 𝑣2 = 𝑃 𝑙(1 − sin 𝜃0 )∕𝑚

𝚤̂

@𝜃

14.88 Distance of 𝐵 from the floor = 0, 𝑣max = 9.422 f t∕s 14.90 ℎmax = 21.85 f t 14.92 𝑣𝐴2 = 7.001 m∕s, 𝑣𝐵2 = 14.00 m∕s √ 𝑔 14.94 𝑣 = 𝑠 𝑙 14.96 Computer Problem 14.98 𝑃avg = 375.0 f t ⋅lb∕s = 0.6818 hp 14.100

Concept Problem

14.102 𝜃 = 6.321◦ 14.104 𝑣2 = 13.62 mph 14.106 𝑃 = 162.4 hp 14.108 𝑣𝐵 = 2.128 m∕s 14.110 𝑃𝑖 = 1.247 hp 14.112 𝑣 = 9.895 f t∕s 14.114 𝑣𝐵2 = 10.84 m∕s 14.116 𝑣𝐴2 = 3.631 m∕s 14.118 14.120

Concept Problem ( ) 𝑈1-2 𝑑 = −0.1585 f t ⋅lb

A-47

A-48

ISTUDY

Appendix B

Answers to Even-Numbered Problems

14.122 𝑘 = 9034 N∕m 14.124 𝑣2 = 73.55 f t∕s 14.126 𝑣max = 9.674 f t∕s and is achieved when the distance of 𝐵 from the ground is zero. 14.128 𝐸𝑏 = 1290 C { 4637 f t ⋅lb∕s, 14.130 𝑃 = 343.5 f t ⋅lb∕s,

for case (a), for case (b)

Chapter 15 15.2

Concept Problem

15.4

Concept Problem

15.6

Concept Problem

15.8 𝑑 = 28.46×10−24 m, 𝑑 = 4.152×10−15 m, 𝑑 = 1.423×10−6 m 15.10

∫𝑡

𝑡2

𝐹⃗ 𝑑𝑡 = −93.17 𝚥̂ lb⋅s

1

| 𝑡2 | | | | | 15.12 | 𝐹⃗ 𝑑𝑡| = 2.635 lb⋅s, |𝐹⃗avg | = 2396 lb | |∫𝑡 | | | 1 | 15.14 𝐹avg = 2589 lb | = 4.293 f t∕s 15.16 𝑣| |𝑡=2 s 15.18 𝜏 = 1.610 s | | 15.20 |𝐹⃗𝑏 | = 40,640 lb | |avg | = 9.045 m∕s 15.22 𝑣| |𝑡=2.5 s ( ) 15.24 Impulse imparted to the ball by kicker = 2.531 𝚥̂ lb⋅s, 𝐹⃗𝑘 avg = 316.4 𝚥̂ lb 15.26 𝑡2 = 10.57 s, 𝑑 = 1279 f t ( ) 15.28 𝐹⃗𝑐 avg = (219.3 𝚤̂ + 192.4 𝚥̂) N

| | 15.30 (a) |𝐹⃗avg | = 8081 N; (b) 𝜇𝑠 = 0.5148 | |

( ) 15.32 Impulse provided by floor = (14.93 N⋅s) 𝚥̂, 𝑎⃗avg = 5420 m∕s2 𝚥̂ = 552.5𝑔 𝚥̂

15.34 Impulse of the rope = 16.30 𝚤̂ lb⋅s 15.36 𝑣⃗𝐵 = 5.094×106 𝚤̂ ft∕s

𝚥̂ 𝚤̂

15.38 Number of worker bees = 9.062×106 , slowdown = 48.11 f t∕s = 32.80 mph 15.40 𝑣⃗𝐴 = 8.455 𝚤̂ m∕s, 𝑣⃗𝐵 = 7.455 𝚤̂ m∕s | | 15.42 𝑣⃗𝐴 | = 0.7905 𝚤̂ m∕s, 𝑣⃗𝐵 | = −2.371 𝚤̂ m∕s |𝑡=1.5 s |𝑡=1.5 s 15.44 𝑣𝑃 2 = 1.714 f t∕s 15.46 𝑣𝑃 3 = 1.852 f t∕s 15.48 𝑑 = 4.951 m 15.50 (𝑣⃗𝑃 )final = (1.516 f t∕s) 𝚤̂

𝚥̂ 𝚤̂

15.52 𝑑 = 1.017 f t 15.54 𝑣⃗𝐴 = 11.17 𝚤̂ m∕s, 𝑣⃗𝐵 = −3.624 𝚤̂ m∕s 15.56 𝑣⃗𝐴2 = (−8.826 𝚤̂ − 10.76 𝚥̂) ft∕s, 𝑣⃗𝐵2 = 13.24 𝚤̂ ft∕s 15.58 𝑣⃗𝐴2 = (−10.29 f t∕s) 𝚤̂, 𝑣⃗𝐵2 = (51.46 f t∕s) 𝚤̂

𝚥̂ 𝚤̂

𝚥̂ 𝚤̂

𝚥̂ 𝚤̂

ISTUDY

Appendix B

Answers to Even-Numbered Problems √ √ 𝑚𝐴 cos 𝜃 2𝑔𝐿 cos 𝜃 − cos 𝜃0

𝚥̂

𝚤̂ 15.60 𝑣⃗𝐵 = ± √ , 𝚤̂ (𝑚𝐴 − 𝑚𝐵 )2 − 𝑚𝐴 (𝑚𝐴 − 3𝑚𝐵 ) cos2 𝜃 √ √ √ (𝑚𝐴 − 𝑚𝐵 )2 − 𝑚𝐴 (𝑚𝐴 − 2𝑚𝐵 ) cos2 𝜃 𝑣𝐴 = 2𝑔𝐿 cos 𝜃 − cos 𝜃0 (𝑚𝐴 − 𝑚𝐵 )2 − 𝑚𝐴 (𝑚𝐴 − 3𝑚𝐵 ) cos2 𝜃 15.62 (a) 𝑣𝐴𝑅 = 0.3750 f t∕s, (b) 𝐹⃗𝐴 = −[(447.2 lb) + (10.48 lb∕s)𝑡] 𝑢̂ 𝑅 + (139.8 lb) 𝑢̂ 𝜃 , 𝐴

𝐴

𝐹⃗𝐵 = −[(174.7 lb) + (10.48 lb∕s)𝑡] 𝑢̂ 𝑅 + (139.8 lb) 𝑢̂ 𝜃 𝐵 𝐵 ̂ 𝑎⃗ = (−0.1755 f t∕s) 𝑢̂ (c) 𝑣⃗ = (1.170 f t∕s) 𝑢̂ + (1.400 f t∕s) 𝑘, 𝐺

15.64

𝜃𝐺

𝐺

𝑅𝐺

Concept Problem 𝚥̂

15.66 𝑣⃗+ = 𝑣⃗+ = (−38.09 mph) 𝚤̂ 𝐴 𝐵

𝚤̂

15.68 𝛿 = 0.04175 f t 15.70 𝑣−𝐵 = 209.7 m∕s 15.72 𝑣+𝐴 = 2.346 f t∕s, 𝑣+𝐵 = 4.154 f t∕s 15.74 𝑒 = 𝑚𝐴 ∕𝑚𝐵

15.76 0.7280 ≤ 𝑒 ≤ 0.7616 15.78 Δ𝑡𝐴 = 0.5184 s, Δ𝑡𝐵 = 0.6600 s 15.80 𝑑 = 2.887 m 15.82 𝑊max = 818.6 lb + + + = 0, 𝑣𝐵𝑥 = 0, 𝑣𝐶𝑥 = 6 f t∕s 15.84 𝑣𝐴𝑥

15.86 Treat each impact as only involving two balls. Because the COR 𝑒 = 1 and the masses are identical, we see from the solution to Prob. 15.74 that ball 1 will come to a complete stop after striking ball 2. We also see that ball 2 will have a postimpact velocity identical to the preimpact velocity of ball 1. Each ball in the train is tangent to the next, so it will not appear to move at all during its impact with the next ball. Ball 4 impacts ball 5, which is free to move. Ball 5 will have a postimpact velocity equal to the preimpact velocity of ball 1. The work-energy principle tells us that ball 5 will stop moving when it has reached the initial height ball 1 was released from. Finally, since the lengths of the pendulums are identical, the maximum swing angle of ball 5 is equal to the initial release angle of ball 1. 15.88 Balls 3, 4, and 5 will swing up as a single unit until reaching the height that balls 1, 2, and 3 were released from while balls 1 and 2 will hang motionless. 15.90 𝑣⃗+𝐴 = 53.19 𝚥̂ mph = 78.01 𝚥̂ ft∕s, 𝑣⃗+𝐵 = (50 𝚤̂ + 53.19 𝚥̂) mph = (73.33 𝚤̂ + 78.01 𝚥̂) ft∕s, 𝑟⃗𝐴 = 135.0 𝚥̂ ft, 𝑟⃗𝐵 = (174.2 𝚤̂ + 185.3 𝚥̂) ft

𝚥̂ 𝚤̂

15.92 Given the assumption that the masses are not identical, it is not possible to have a moving ball 𝐴 hit a stationary ball 𝐵 so that 𝐴 stops right after the impact. 15.94 𝑣⃗+𝐵 = 0.7071 𝚥̂ m∕s 𝚥̂ 𝚤̂ @ − 45◦ 15.96 𝑣−𝐵 = 0.6886 f t∕s 15.98 𝛽 = tan−1 (cot 𝛼) 15.100 𝑣⃗+𝐴 = (−8.309 𝚤̂ + 18.53 𝚥̂) m∕s, 𝑣⃗+𝐵 = (−8.309 𝚤̂ − 6.202 𝚥̂) m∕s 15.102 𝑑 = 5.266 f t 𝑚 [1 + 𝑒(1 + cos 𝜃)] − 𝑚𝐴 cos 𝜃 ( +) =− 𝐵 𝑣0 15.104 𝑣+𝐴𝑥 = 𝑣𝐴 LOI 𝑚𝐴 + 𝑚𝐵 𝑚 [cos 𝜃 + 𝑒(1 + cos 𝜃)] − 𝑚𝐵 𝑚 ( ) 𝑣+𝐵𝑥 = 𝑣+𝐵 LOI = 𝐴 𝑣0 , 𝐴 > 4 𝑚𝐴 + 𝑚 𝐵 𝑚𝐵

𝚥̂ 𝚤̂

A-49

A-50

ISTUDY

Appendix B

Answers to Even-Numbered Problems √ √ 𝑣0 + 4− 3 5− 3 15.106 𝑣𝐴 = √ , 𝑣⃗𝐵 = √ 𝑣0 𝚤̂, 𝑒 = √ = 0.6906 3 2 3 3+ 3 ( )√ 15.108 𝑣+𝑁 = 2𝑁 − 1 2𝑔ℎ +

Concept Problem ( ) ( ) 3̂ 2 ⃗ ⃗ 15.112 ℎ𝑂 𝐴 = 593.0×103 𝑘̂ slug⋅ft2 ∕s, ℎ 𝑂 𝐵 = −79.93×10 𝑘 slug⋅ft ∕s 15.110

15.114 15.116 15.118 15.120

15.126

15.128

15.130 15.132

𝚤̂

⃗ = −1.439×106 𝑘̂ slug⋅ft2 ∕s 𝚥̂ ℎ 𝑄 𝚤̂ ( ) ⃗ℎ̇ (𝑡 ) = −0.07764 𝚤̂ + 0.01941 𝚥̂ + 0.07764 𝑘̂ slug⋅ft2 ∕s2 𝑂 2 ( ) ⃗ = −9.100 𝚥̂ + 48.00 𝑘̂ kg⋅m∕s ℎ 𝑂 ̇ ) = 2.083 rad∕s 𝜃(𝑡 2

15.122 𝜔 = 15.124

𝚥̂

(𝐿 sin 𝜃0 + 𝑑)2

𝜔 (𝐿 sin 𝜃 + 𝑑)2 0 𝑎⃗𝐴𝑛 = (−1.067 𝚤̂ + 0.5333 𝚥̂) m∕s2 ⃗̇ + 𝑣⃗ × 𝑚 𝑣⃗ = 𝑚 𝑔𝑡[𝑣 − 𝑣 (0) cos 𝜃] 𝑘̂ and 𝑀 ⃗ = 𝑚 𝑔𝑡[𝑣 − 𝑣 (0) cos 𝜃] 𝑘, ̂ Since ℎ 𝐸 𝐸 𝑃 𝑃 𝑃 𝐸 𝑃 𝐸 𝑃 𝐸 𝑃 ̇ 𝚥 ̂ ⃗ + 𝑣⃗ × 𝑚 𝑣⃗ . ⃗ =ℎ it is true that 𝑀 𝐸 𝐸 𝐸 𝑃 𝑃 𝚤̂ √ 3 ⃗ = ±𝑚𝑔 2𝐿 (cos 𝜃 − cos 33◦ ) 𝑘̂ ℎ 𝑂 𝑔 √ √ 2𝑔𝐿(cos 𝜙2 − cos 𝜙1 ) 2𝑔𝐿(cos 𝜙2 − cos 𝜙1 ) 𝑣1 = sin 𝜙2 = 6.854 f t∕s, 𝑣2 = sin 𝜙1 = 2.316 f t∕s 2 2 sin 𝜙1 − sin 𝜙2 sin2 𝜙1 − sin2 𝜙2 √ 𝑀𝐴 = 2𝑚𝜔20 𝑟 𝑟2 − 𝑟20

Computer Problem (𝑚 + 𝑚𝐷 )𝑟2 15.136 𝜔𝑠2 = − 𝐶 𝛼𝜏 (𝑚𝐴 + 𝑚𝐵 )𝑅2 𝑖 15.134

⇒

| | ⃗ | = 0.06122 rad∕s |𝜔 | 𝑠2 |

2𝑣20

) ( , 𝑣2 = 2𝑣0 , 𝑈1-2 𝑃 = 𝑚𝑣20 √ ( √ ) 2 54 − 3 √ √ 𝑚𝑔 15.140 𝑣1 = 3𝑔𝑅, 𝑣2 = 3 𝑔𝑅, 𝑃2 = √ 3+ 3 15.142 Area(𝑃1 𝑂𝑃2 )∕Area(𝑃3 𝑂𝑃4 ) = 1 √ 2𝐸𝜅 2 15.144 𝑒 = 1 + (𝐺𝑚𝐵 )2 2𝐸𝜅 2 (a) 𝐸 < 0 ⇒ 1 + 1 ⇒ 𝑒>1 ⇒ (𝐺𝑚𝐵 )2 √ 𝐺𝑚𝐵 (b) 𝑣𝑐 = , which agrees with Eq. (15.82) 𝑟𝑐 15.138 𝓁 =

3𝑔

15.146 Δ𝑣 = 3232 m∕s 15.148 𝐺𝑚𝑒 = 3.971×1014 m3 ∕s2

elliptical orbit parabolic trajectory hyperbolic orbit

ISTUDY

Appendix B

Answers to Even-Numbered Problems

15.150 𝑟𝑔 = 1.386×108 f t, ℎ𝑔 = 1.177×108 f t, 𝑣𝑐 = 1.008×104 f t∕s 15.152 Δ𝑣 = −3240 m∕s = −11,660 km∕h 15.154 Δ𝑣𝑃 = 8064 f t∕s = 5498 mph, Δ𝑣𝐴 = 4850 f t∕s = 3307 mph, 𝑡 = 18,930 s = 5.258 h Computer Problem

15.156

Concept Problem √ 𝑟𝑃 𝑣2𝑃 − 2𝐺𝑚𝐵 15.160 (a) 𝑣∞ = ; (b) 𝑟𝑃 𝑣2𝑃 > 2𝐺𝑚𝐵 𝑟𝑃 15.158

15.162

Concept Problem

15.164

Concept Problem

√

3

15.166 𝑄nz = 3.662 f t ∕s, 𝑑 = 2 15.168 𝑅 =

1 𝑝 𝜋𝑑𝐴2 4 𝐴

4𝛾𝑄2 + 𝜋𝑔

(

𝑄nz

= 0.2678 f t

𝑣𝑤 𝜋

1 1 − 𝑑𝐴2 𝑑𝐵2

) = 2097 lb

15.170 𝑣𝑜 = 269.6 m∕s 15.172 𝜇𝑠 = 0.07578 15.174 𝑚̇ 𝑓 1 = 12 𝑚̇ 𝑓 (1 + cos 𝜃), 𝑚̇ 𝑓 2 = 12 𝑚̇ 𝑓 (1 − cos 𝜃) 15.176 Δ𝑣𝑥 = −26.92 f t∕s 15.178 𝐹𝑅 = 270.4 kN, 𝑀𝑂 = 𝐹𝑅 ℎ = 20.28×106 N⋅m 15.180 𝐹 = 1.099 lb Computer Problem

15.182

15.184 𝑦max = 30,470 f t | | 15.186 |𝐹⃗avg | = 1459 lb | | ( ) 15.188 𝑣⃗𝑃 final = (1.672 m∕s) 𝚤̂

𝚥̂ 𝚤̂

15.190 𝛽 = 14.26◦

( ) 15.192 𝑀 = 12 𝑚𝜔20 𝑟20 𝑒2𝜔0 𝑡 − 𝑒−2𝜔0 𝑡 15.194 Δ𝑣 = −2.919×103 f t∕s = −1990 mph 15.196 Δ𝑇𝑗 = −4.160×1010 J, Δ𝑇𝑒 = −2.161×1011 J, Δ𝑉 = −5.153×1011 J 15.198 (𝐿 − 𝑦)(𝑦̈ − 𝑔) − 12 𝑦̇ 2 = 0

Chapter 16 16.2

Concept Problem

Concept Problem 16.6 𝜔 ⃗ 𝐵 = 532.2 𝑘̂ rad∕s, 𝛼⃗𝐵 = 187.8 𝑘̂ rad∕s2 𝑅 16.8 𝜔𝐵 = 𝐴 𝜔𝐴 = 600.0 rad∕s 𝑅𝐵 16.10 𝑡stop = 87.94 s, Δ𝜃pt = 806.1 rev 16.4

𝚥̂ 𝚤̂

16.12 (𝜔𝑂𝐴 )max = 2.476 rad∕s 16.14 |𝜔 ⃗ ram | = 0 ( ) ( ) 16.16 𝑣⃗𝐶 = 71.88 𝚤̂ − 287.6 𝚥̂ + 198.6 𝑘̂ cm∕s, 𝑎⃗𝐶 = 7244 𝚤̂ + 4061 𝚥̂ + 3260 𝑘̂ cm∕s2 ( ) ( ) 16.18 𝑣⃗𝐶 = −43.13 𝚤̂ + 172.6 𝚥̂ − 119.2 𝑘̂ cm∕s, 𝑎⃗𝐶 = 2585 𝚤̂ + 1554 𝚥̂ + 1110 𝑘̂ cm∕s2 ) ( ) ( 16.20 𝑣⃗ = 498.0 𝚥̂ + 211.7 𝑘̂ cm∕s, 𝑎⃗ = −1.331×104 𝚤̂ − 5341 𝚥̂ + 1.237×104 𝑘̂ cm∕s2 𝐷

𝐷

A-51

A-52

ISTUDY

Appendix B

Answers to Even-Numbered Problems

𝚥̂

16.22 𝜔 ⃗ 𝑠 = −10.00 𝑘̂ rad∕s, 𝛼⃗𝑠 = −1.333 𝑘̂ rad∕s2

𝚤̂

16.24 𝑣⃗𝐺 = (−0.1495 𝑢̂ 𝑟 − 3.379 𝑢̂ 𝜃 ) m∕s, 𝑎⃗𝐺 = (−20.41 𝑢̂ 𝑟 − 3.660 𝑢̂ 𝜃 ) m∕s2 16.26 𝑣⃗𝐶 = −(0.2917 𝚤̂ + 0.7361 𝚥̂) ft∕s, 𝑎⃗𝐶 = (−0.2045 𝚤̂ + 0.08102 𝚥̂) ft∕s2 16.28 𝑣⃗𝐶 = (−0.1791 𝚤̂ + 0.4920 𝚥̂) m∕s, 𝑎⃗𝐶 = −(2.576 𝚤̂ + 0.9377 𝚥̂) m∕s2 𝑟 𝐿 𝚥̂ 16.30 𝑣⃗𝐻 = 𝐴 𝜔𝐴 (− sin 𝜃 𝚤̂ + cos 𝜃 𝚥̂) , 𝚤̂ 𝑟𝐶 [( ) ( ) ] 𝑟𝐴 𝐿 𝑟𝐴 2 𝑟𝐴 2 −𝛼𝐴 sin 𝜃 − 𝜔𝐴 cos 𝜃 𝚤̂ + 𝛼𝐴 cos 𝜃 − 𝜔𝐴 sin 𝜃 𝚥̂ 𝑎⃗𝐻 = 𝑟𝐶 𝑟𝐶 𝑟𝐶 −6

16.32 𝜔𝑠 = 72.72×10

𝚥̂ 𝚤̂

rad∕s

16.34 𝜔 ⃗ 𝐶 = 40.00 𝑘̂ rad∕s, 𝛼⃗𝐶 = 5.200 𝑘̂ rad∕s2

𝚥̂ 𝚤̂

16.36 Chain ring/sprocket combination: C1/S3, 𝜔𝑤 = 𝜔𝑠 =

𝑅𝑐 𝑅𝑠

𝜔𝑐 = 126.4 rpm

16.38 𝑣⃗𝐵 = 146.0 𝚥̂ ft∕s 16.40 𝑣⃗𝐵 = (9.881 𝚤̂ + 3.666 𝚥̂) m∕s 16.42 𝜔 ⃗ = 9.600 𝑘̂ rad∕s, 𝑣⃗ = 2.000 𝚤̂ ft∕s 𝑃

16.44

𝑂

Concept Problem

16.46 𝑣⃗𝐴 = −𝑟𝜔𝑑 (̂𝚤 + 𝚥̂), 𝑣⃗𝐵 = −2𝑟𝜔𝑑 𝚥̂, 𝑣⃗𝐶 = 𝑟𝜔𝑑 (̂𝚤 − 𝚥̂) 16.48 𝜔 ⃗ = −1.600 𝑘̂ rad∕s, 𝑣⃗ = 5.333 𝚤̂ m∕s 𝑃

𝚥̂ 𝚤̂

𝐶

⃗ 𝐴𝐵 = 20.00 𝑘̂ rad∕s 16.50 𝑣⃗𝐶 = (50.00 𝚤̂ + 50.00 𝚥̂) ft∕s, 𝜔 16.52 𝜔 ⃗ = −7.000 𝑘̂ rad∕s 𝐴

16.54 𝜔 ⃗ 𝐴𝐷 = 2.745 𝑘̂ rad∕s 16.56 𝜔 ⃗ 𝐴𝐷 = 1.373 𝑘̂ rad∕s, 𝑣⃗𝐶 = (0.2745 𝚤̂ − 1.225 𝚥̂) m∕s 16.58 𝜔 ⃗ = 2.208 𝑘̂ rad∕s, 𝑣⃗ = −2.274 𝚥̂ ft∕s 𝐴𝐵

𝐴

16.60 𝑣⃗𝐴 = 11.67 𝚥̂ ft∕s 𝚥̂

16.62 𝑣⃗𝐵 = − 23 𝑅𝜔𝑂𝐴 sin 𝜃 𝚤̂ 16.64 𝜔 ⃗ = −0.06805 𝑘̂ rad∕s

𝚤̂

𝐴𝐵

16.66 𝜙̇ = −74.76 rad∕s, 𝑣⃗𝐷 = (−29.33 𝚤̂ + 73.73 𝚥̂) ft∕s 16.68 The cable is unwinding at 1.364 m∕s. ⃗ 𝐵𝐶 = −0.3000 𝑘̂ rad∕s, 𝑣⃗𝐶 = (1.200 𝚤̂ + 0.3000 𝚥̂) ft∕s 16.70 𝜔 ⃗ 𝐶𝐷 = −0.8824 𝑘̂ rad∕s, 𝜔 16.72 𝑣⃗𝐵 = −35.00 𝚤̂ ft∕s 16.74 𝜔 ⃗ 𝐴𝐵 = 3.333 𝑘̂ rad∕s, 𝑣⃗𝐶 = 19.36 𝚥̂ ft∕s 16.76 𝜔 ⃗ = −3.333 𝑘̂ rad∕s, 𝑣⃗ = 5.000 𝚤̂ rad∕s 𝑠

𝑂

⃗ 𝜔 ⃗ 𝐵𝐶 = 7.667 𝑘̂ rad∕s 16.78 𝜔 ⃗ 𝐴𝐵 = 0, 𝑅𝜃̇ sin(𝛽 + 𝜃) ̂ 𝑅𝜃̇ sin(𝛾 + 𝜃) ̂ 16.80 𝜔 ⃗ 𝐴𝐵 = − 𝑘, 𝜔 ⃗ 𝐵𝐶 = 𝑘 𝐻 sin(𝛽 − 𝛾) 𝐿 sin(𝛽 − 𝛾) 16.82 𝜔 ⃗ 𝐴𝐵 = −50.57 𝑘̂ rad∕s, 𝑣𝐵 = 1.922 m∕s ] [ ̇ 𝑅𝜃(𝐻 − 𝑅 cos 𝜃) sin 𝜃 ̇ 16.84 𝑣⃗𝐵 = 𝑅𝜃 cos 𝜃 − √ 𝚥̂ 𝐿2 − (𝐻 − 𝑅 cos 𝜃)2 16.86 Computer Problem 16.88 𝑎⃗𝐵 = (16.98 𝚤̂ − 5.383 𝚥̂) ft∕s2 16.90 𝑎⃗𝐶 = (−16.77 𝚤̂ + 87.69 𝚥̂) ft∕s2 16.92 𝑎⃗𝑄 = −9.163 𝑢̂ 𝑟 m∕s2

ISTUDY

Appendix B

Answers to Even-Numbered Problems

16.94 𝑎⃗𝐵 = −188.7 𝚥̂ ft∕s2 16.96 𝑎⃗𝐵 = −11,970 𝚥̂ cm∕s2 16.98 𝛼⃗𝐴𝐵 = 48.60 𝑘̂ rad∕s2 , 𝛼⃗𝐵𝐶 = 0⃗ 16.100 𝑎⃗𝑃 = −3482 𝑢̂ 𝑟 ft∕s2 , 𝑎⃗𝐶 = −1024 𝑢̂ 𝑟 ft∕s2 16.102 𝑎⃗𝐵 = −792.5 𝚤̂ ft∕s2

𝚥̂ 𝚤̂

16.104 𝑎⃗𝐷 = (−3.845 𝚤̂ − 4.350 𝚥̂) m∕s2 𝑣2𝐴

16.106 𝑎⃗𝐷 = −

𝚤̂ 2𝐿 sin3 𝜃 ) ( 𝑣2𝐵 𝑣𝐵 𝑎𝐵 ̂ 𝛼⃗ = − 𝑘̂ 𝑘, =− + √ 𝑊 2𝑅 4 5𝑅2 2𝑅 √ ) √ ) √ ( ( 98 4 + 2 𝑅𝑎𝐵 + 479 + 202 2 𝑣2𝐵 2 2 𝑣𝐵 ̂ 𝛼⃗ = − = ( 𝑘̂ √ ) 𝑘, 𝑊 343𝑅2 𝑅 1−2 2 = 7.971 𝑘̂ rad∕s2

16.108 𝜔 ⃗𝑊 16.110 𝜔 ⃗𝑊 16.112 𝛼⃗𝑊

16.114 𝛼⃗𝑠 = −0.4000 𝑘̂ rad∕s2 , 𝑎⃗𝑂 = 2.000 𝚤̂ ft∕s2 16.116 𝛼⃗ = 1633 𝑘̂ rad∕s2 , 𝑎⃗ = (4636 𝚤̂ + 1683 𝚥̂) m∕s2 𝐵𝐶

16.118

𝐶

Computer Problem

16.120 𝑎⃗𝑄 = 82.03 𝑢̂ 𝑟 ft∕s2 16.122 𝛼⃗𝐴𝐵 = 93.03 𝑘̂ rad∕s2 , 𝛼⃗𝐵𝐶 = −95.05 𝑘̂ rad∕s2 16.124 𝛼⃗𝐴𝐵 = −92.13 𝑘̂ rad∕s2 , 𝛼⃗𝐵𝐶 = −190.7 𝑘̂ rad∕s2 16.126 𝛼⃗𝐵𝐶 = −0.01749 𝑘̂ rad∕s2 , 𝛼⃗𝐶𝐷 = 0.004516 𝑘̂ rad∕s2 , 𝑎⃗𝐶 = (0.003164 𝚤̂ − 0.03876 𝚥̂) ft∕s2 𝑣0

𝑣0

11𝑣20

̂ 𝜔 𝑘, ⃗𝐷 =

̂ 𝛼⃗ = − 𝑘, 𝐴𝐵

̂ 𝛼⃗ = 𝑘, 𝐷

119𝑣20

𝑘̂ 2𝑅 3𝑅 72𝑅2 108𝑅2 5 ̂ 15 ̂ 185 15 16.130 𝜔 ⃗ 𝐸 = 𝜔0 𝑘, 𝑅𝜔20 𝚤̂, 𝑎⃗𝑄 = − 𝑅𝜔20 𝚤̂ 𝜔 ⃗ 𝐷 = − 𝜔0 𝑘, 𝑎⃗𝑃 = − 8 4 16 16 16.132 𝛼⃗𝐵𝐶 = −72,900 𝑘̂ rad∕s2 , 𝑎⃗𝐶 = −10,990 𝚥̂ ft∕s2 16.134 𝛼⃗𝐵𝐶 = −72,900 𝑘̂ rad∕s2 , 𝑎⃗𝐶 = −10,990 𝚥̂ ft∕s2 16.136 𝛼⃗𝐴𝐵 = 87.67 𝑘̂ rad∕s2 , 𝛼⃗𝐵𝐶 = −9.974 𝑘̂ rad∕s2 ( ) ( ) 16.138 𝑎⃗𝑃 = 𝑠̈ − 𝑠𝜔20 𝚤̂ + 𝑠𝛼0 + 2𝑠𝜔 ̇ 0 𝚥̂ ( ) ( ) ( ) 16.140 𝑎⃗𝑃 = 𝑠̈ − 𝑠𝜔2𝐷 𝚤̂ − 2𝑠𝜔 ̇ 𝐷 + 𝑑𝜔2𝐷 𝚥̂, 𝑎⃗𝑃 Coriolis = −2𝑠𝜔 ̇ 𝐷 𝚥̂ ( ) ( ) 16.142 𝑣⃗𝐷 = 𝑠̇ − 𝑑𝜔1 𝚤̂ + 𝑠 𝜔1 − 𝜔2 𝚥̂ 16.128 𝜔 ⃗ 𝐴𝐵 = −

𝚥̂

16.144 𝜔 ⃗ 𝐶𝐷 = 22.43 𝑘̂ rad∕s, 𝛼⃗𝐶𝐷 = −202.0 𝑘̂ rad∕s2 16.146 𝜔 ⃗ 𝐴𝐵 = 4.561 𝑘̂ rad∕s, 𝛼⃗𝐴𝐵 = 18.43 𝑘̂ rad∕s2

𝚤̂

𝚥̂

𝚤̂

16.148 𝑎⃗𝐶 = (−30.00 𝚤̂ + 51.80 𝚥̂) ft∕s2 16.150 𝑣⃗𝑃 = (−11.54 𝚤̂𝐵 + 33.30 𝚥̂𝐵 ) ft∕s, 𝑎⃗𝑃 = (−64.43 𝚤̂𝐵 − 16.43 𝚥̂𝐵 ) ft∕s2 [ ] 16.152 𝑣⃗𝐷 = 𝑣0 − 𝑑𝜔1 − 𝑅(𝜔1 − 𝜔2 ) 𝚤̂ + 𝑅(𝜔1 − 𝜔2 ) 𝚥̂ 16.154 (a) 𝜔 ⃗ 𝐶𝐷 = 2.513 𝑘̂ rad∕s, 𝑣⃗bar = 0.3016 𝐼̂ m∕s (b) 𝛼⃗𝐶𝐷 = 37.90 𝑘̂ rad∕s2 , 𝑎⃗bar = 4.548 𝐼̂ m∕s2

𝐽̂ 𝐽̂

16.156 (a) 𝜔 ⃗ 𝐶𝐷 = −6.283 𝑘̂ rad∕s, 𝑣⃗bar = −0.7540 𝐼̂ m∕s (b) 𝛼⃗𝐶𝐷

⃗ 𝑎⃗ = 0⃗ = 0, bar

𝐽̂, 𝚥̂ ̂ 𝚤̂ 𝐼,

𝚥̂ 𝚤̂

𝐼̂ 𝚥̂

𝐼̂ 𝚤̂ 𝐽̂, 𝚥̂ ̂ 𝚤̂ 𝐼,

A-53

A-54

ISTUDY

Appendix B

Answers to Even-Numbered Problems

Computer Problem ( ) ( ) 16.160 𝑣⃗𝐶 = 7.500 𝚤̂ + 7.506 𝚥̂ ft∕s, 𝑎⃗𝐶 = 105.1 𝚤̂ − 161.0 𝚥̂ ft∕s2 16.162 𝑑̇𝐴𝐵 = 1.118 f t∕s, 𝑑̈𝐴𝐵 = 2.012 f t∕s2 ( [ ] ) ] [ ( )2 ) 𝓁̇ 𝓁 (̇ 1 16.164 𝑣⃗𝐴 = −𝑅𝜔𝑠 𝚤̂ + 𝓁 𝜔𝑠 + 𝚥̂, 𝑎⃗𝐴 = − 𝓁 + 𝑅𝜔𝑠 + 𝑅𝜔̇ 𝑠 𝚤̂ + 𝓁 𝓁̈ + 𝓁̇ 2 − 𝑅𝜔2𝑠 + 𝓁 𝜔̇ 𝑠 𝚥̂ 𝑅 𝑅 𝑅2 16.166 𝑣𝐶 = 0.1601 f t∕s [ [ ( ( )2 ] ) ( ) ] 𝚤̂ + 2 𝜔1 − 𝜔2 𝑠̇ + 𝑠 𝛼1 − 𝛼2 − 𝑑𝜔21 𝚥̂ 16.168 𝑎⃗𝐷 = 𝑠̈ − 𝑑𝛼1 − 𝑠 𝜔1 − 𝜔2 16.158

16.170 𝑣⃗𝐷 = (1.340 𝚤̂ + 5.000 𝚥̂) ft∕s 16.172 𝑣⃗𝐴 = 0.8750 𝚥̂ m∕s, 𝑎⃗𝐴 = 0.3750 𝚥̂ m∕s2 , 𝑣⃗𝐷 = −1.750 𝚥̂ m∕s, 𝑎⃗𝐷 = −0.7500 𝚥̂ m∕s2 16.174 𝜔 ⃗ bar

𝚤̂

𝚥̂

𝚥̂

= −14.29 𝑘̂ rad∕s

𝚤̂

16.176 𝜔 ⃗ 𝐴𝐶 = 2.442 𝑘̂ rad∕s, 𝜔 ⃗ 𝐶𝐸 = −2.442 𝑘̂ rad∕s, 𝑣⃗𝐸 = (−15.00 𝚤̂ − 42.01 𝚥̂) ft∕s ( ) ( ) ̂ 16.178 𝑣⃗𝐷 = 𝑣𝐶 + 𝓁𝜔𝑂𝐴 sin 𝜙 𝚤̂ + 𝑑𝜔𝑂𝐴 − 𝓁𝜔𝐶 cos 𝜙 𝚥̂ − 𝓁𝜔𝐶 sin 𝜙 𝑘, ( ) [ ( ) ] 𝑎⃗𝐷 = 2𝓁𝜔𝐶 𝜔𝑂𝐴 cos 𝜙 + 𝓁𝛼𝑂𝐴 sin 𝜙 − 𝑑𝜔2𝑂𝐴 𝚤̂ + 𝑑𝛼𝑂𝐴 + 2𝑣𝐶 𝜔𝑂𝐴 + 𝓁 𝜔2𝐶 + 𝜔2𝑂𝐴 sin 𝜙 𝚥̂ − 𝓁𝜔2𝐶 cos 𝜙 𝑘̂ ( ) 16.180 𝑣⃗𝐶 = (21.50 𝚤̂ + 7.506 𝚥̂) ft∕s = 21.50 𝐼̂ + 7.506 𝐽̂ ft∕s, ( ) 𝑎⃗ = (105.1 𝚤̂ + 165.7 𝚥̂) ft∕s2 = 105.1 𝐼̂ + 165.7 𝐽̂ ft∕s2 𝐶

Chapter 17 17.2 𝑡stop = 1.863 s, 𝑑stop = 16.77 f t 17.4 ℎmax = 4.054 f t, 𝑎𝐺𝑥 = 9.800 f t∕s2

𝚥̂ 𝚤̂

17.6 𝑎0 = 15.70 f t∕s2 17.8

Concept Problem

17.10 𝑎0 = 66.80 f t∕s2 17.12 𝑃 = 20.70 lb, 𝑁𝐴 = 152.1 lb, 𝑁𝐵 = 147.9 lb 17.14 𝑎⃗𝐶 = −2.875 𝚥̂ m∕s2

𝚥̂ 𝚤̂

17.16 𝐿 = 2.875×105 N 17.18 𝑎𝐺𝑥 = 2.147 f t∕s2 , 𝑃 = 49.83 lb

𝚥̂ 𝚤̂

17.20 𝜃 = 17.00◦ , 𝜙 = 17.00◦ 17.22 𝑁 = 1.492×105 N, 𝛿 = 1.909 m 17.24 𝑁𝑓 = 783.7 lb, 𝑁𝑟 = 783.2 lb, 𝑁𝐵 = 2233 lb, 𝐻 = 673.7 lb, 𝐹 = 576.9 lb, (𝜇𝑠 )min = 0.7362 ( ) ( ) 𝑎𝐴 𝑎𝐴 −1 −1 17.26 𝜙 = tan , 𝜃 = tan 𝑔 𝑔 17.28 𝑅𝑟 = −0.1115𝑡2 N∕s2 , 𝑅𝜃 = 0.01012 N, 𝑀𝑧 = 3.110×10−5 N⋅m, where 𝑧 is the axis of rotation and 𝑟 is radially outward 17.30 𝑡𝑓 = 1.411 s, 𝑛 = 7.485 rev (cw) 17.32 𝛼⃗𝑑 = −10.06 𝑘̂ rad∕s2

𝚥̂ 𝚤̂

, 𝑛 = 16.01 rev (cw)

17.34 𝜔𝑓 = 31.42 rad∕s = 300.0 rpm 17.36 𝑀 = 3208 f t ⋅lb 17.38 𝐹⃗𝐴 = 9.197 (−̂𝚤 + 𝚥̂) N, 𝐹⃗𝐵 = 9.197 (̂𝚤 − 𝚥̂) N, 𝑡𝑠 = 4.096 s

𝚥̂ 𝚤̂

17.40 𝜇𝑘 = 1.443, 𝐹⃗𝐴 = (−18.83 𝚤̂ + 26.19 𝚥̂) N, 𝐹⃗𝐵 = (18.83 𝚤̂ − 11.48 𝚥̂) N 17.42 𝑂𝑟 = −16.85 lb, 𝑂𝜃 = −5.000 lb

𝑢̂ 𝜃

𝑢̂ 𝑟

𝚥̂ 𝚤̂

ISTUDY

Appendix B

Answers to Even-Numbered Problems

√ ( √ ) 3𝑔 for 𝓁 = 16 3 ± 3 𝐿 17.44 |𝛼max | = 𝐿 17.46 𝜙̇ = 1.503 rad∕s ] [ 3𝑚𝑐 𝑔ℎ 2 (𝑑 − 𝓁) 𝑚𝑐 + (2𝑑 − 𝐿) 𝑚𝑝 𝚥̂ 17.48 𝐷𝑥 = [ ] ( ) 𝚤̂ 𝑤2 + 4ℎ2 + 12 (𝑑 − 𝓁)2 𝑚𝑐 + 4 3𝑑 2 − 3𝑑𝐿 + 𝐿2 𝑚𝑝 ( [ ( ) )] 𝐿2 𝑚2𝑝 + 4ℎ2 + 𝑤2 𝑚2𝑐 + 4ℎ2 + 𝑤2 + 4 𝐿2 − 3𝓁𝐿 + 3𝓁 2 𝑚𝑐 𝑚𝑝 𝐷𝑦 = 𝑔 ] [ ( ) 4 3𝑑 2 − 3𝑑𝐿 + 𝐿2 𝑚𝑝 + 4ℎ2 + 𝑤2 + 12 (𝑑 − 𝓁)2 𝑚𝑐 ] [ 6𝑔 2 (𝓁 − 𝑑) 𝑚𝑐 + (𝐿 − 2𝑑) 𝑚𝑝 𝛼𝑝 = 𝛼𝑐 = ( [ ] ) 4 3𝑑 2 − 3𝑑𝐿 + 𝐿2 𝑚𝑝 + 4ℎ2 + 𝑤2 + 12 (𝑑 − 𝓁)2 𝑚𝑐 ( ) ( ) 3 1 3 sin 𝜃 − 1 , 𝑁 = 𝑚𝑔 (1 − 3 sin 𝜃)2 , 𝜇𝑠 min = ∞ 17.50 𝐹𝑓 = 𝑚𝑔 cos 𝜃 2 2 4 𝚥̂ 2 ⃗ ̂ 17.52 𝛼 = −6.925 𝑘 rad∕s , 𝐹 = (−1323 𝚤̂ + 48.47 𝚥̂) N 𝑇

𝑂

𝚥̂

17.56 𝐹⃗𝐵 = (−560.1 𝚤̂ + 558.6 𝚥̂) N, 𝑀𝐵 = −423.8 𝑘̂ N⋅m

𝚥̂

17.58 𝐹 =

𝑘2𝐺 + 𝑟2

𝑚𝑔 sin 𝜃, 𝑁 = 𝑚𝑔 cos 𝜃, 𝛼𝑏 = −

17.60 𝑎⃗𝐺 = −𝜇𝑘 𝑔 𝚤̂, 𝛼⃗𝑏 = −

𝑔𝑟𝜇𝑘 𝑘2𝐺

𝚤̂

𝚤̂

17.54 𝐹⃗𝐴 = (−35.07 𝚤̂ + 48.22 𝚥̂) N, 𝛼⃗𝑏 = −6.679 𝑘̂ rad∕s2 𝑘2𝐺

𝚥̂

𝑘̂

𝑟𝑔 sin 𝜃 𝑘2𝐺 + 𝑟2

𝚤̂ 𝚤̂

, (𝜇𝑠 )min =

𝑘2𝐺

tan 𝜃

𝑘2𝐺 + 𝑟2

𝚥̂ 𝚤̂

17.62 𝑇𝐶𝐷 = 100.8 N, 𝑎⃗𝐺 = −6.450 𝚥̂ m∕s2

𝚥̂ 𝚤̂

17.64 The sign of the angular acceleration 𝛼𝑐 contradicts our assumption that the crate tips and slips, since the only physically meaningful tipping would be in the clockwise direction. Therefore, the solution shows that the crate cannot tip and slip. ) ( 𝑅2 𝑃𝑅 𝚥̂ ⃗ 17.66 𝐹𝑠 = 𝑃 𝚤̂ + (𝑚𝑔 − 𝑃 ) 𝚥̂, 𝛼⃗𝑠 = − ( ) 𝑘̂ 𝚤̂ 2 2 2 2 𝑅 + 𝑘𝐺 𝑚 𝑅 + 𝑘𝐺 17.68 𝜃sep = 48.19◦ 𝜋 = 60.00◦ 3 ( ) 17.72 𝐹 = 12.13 lb, 𝑡𝑓 new = 6.845 s 17.70 𝜃sep =

6𝑚𝐴𝐵 𝑔 cos 𝜃 6𝑚𝐴𝐵 𝑔 sin 𝜃 cos 𝜃 6𝑚 𝑔 sin 𝜃 cos 𝜃 ̂ 𝑎⃗ = 𝚤̂, 𝛼⃗𝑤 = − ( 𝐴𝐵 ( ) 𝑘, ) 𝑘̂ 𝐵 2 2 𝐿 4𝑚𝐴𝐵 + 18𝑚𝐵 sin 𝜃 4𝑚𝐴𝐵 + 18𝑚𝐵 sin 𝜃 𝑅 4𝑚𝐴𝐵 + 18𝑚𝐵 sin2 𝜃 6𝑔(2𝑚𝐴 + 𝑚𝐴𝐵 ) cos 𝜃 6𝑔(2𝑚𝐴 + 𝑚𝐴𝐵 ) sin 𝜃 cos 𝜃 ̂ 𝑎⃗ = ( 17.76 𝛼⃗𝐴𝐵 = ( ) 𝑘, ) 𝚤̂, 𝐵 2 2 2𝐿 6𝑚𝐴 cos 𝜃 + 2𝑚𝐴𝐵 + 9𝑚𝐵 sin 𝜃 2 6𝑚𝐴 cos2 𝜃 + 2𝑚𝐴𝐵 + 9𝑚𝐵 sin2 𝜃 6𝑔(2𝑚𝐴 + 𝑚𝐴𝐵 ) sin 𝜃 cos 𝜃 𝛼⃗𝑤 = − ( ) 𝑘̂ 2𝑅 6𝑚𝐴 cos2 𝜃 + 2𝑚𝐴𝐵 + 9𝑚𝐵 sin2 𝜃

17.74 𝛼⃗𝐴𝐵 =

17.78 𝐼𝐺 = 𝑚𝑅2 6𝑔 sin 𝜃 𝑚𝑔 3𝑔 sin 𝜃 cos 𝜃 ̂ 𝐹⃗ = 17.80 𝛼⃗𝑏 = − ( 𝚥̂, 𝑎⃗𝐴 = 𝚤̂ ) 𝑘, 𝐴 2 2 𝐿 1 + 3 sin 𝜃 1 + 3 sin 𝜃 1 + 3 sin2 𝜃 17.82

Computer Problem

17.84

Computer Problem

𝚥̂ 𝚤̂

A-55

A-56

ISTUDY

Appendix B

Answers to Even-Numbered Problems (

𝑘2𝐺

)

𝑥̈ + 𝑘𝑥 = 𝑚𝑔 sin 𝜃 + 𝑘𝐿0 𝑟2 ( ) 2 = 48.19◦ 17.88 𝜃 = cos−1 3 5𝑔 sin 𝜙 ̂ 17.90 𝐹⃗due to bowl = −𝑚𝑔 cos 𝜙 𝑢̂ 𝑟 + 27 𝑚𝑔 sin 𝜙 𝑢̂ 𝜙 , 𝛼⃗𝑏 = 𝑘, 𝑎⃗𝐺 = − 57 𝑔 sin 𝜙 𝑢̂ 𝜙 7𝜌 7 (𝑅 − 𝜌)𝜙̈ + 𝑔 sin 𝜙 = 0 17.92 5 17.94 ℎ = 75 𝑟 17.86 𝑚 1 +

17.96 17.98 17.100 17.102 17.104 17.106

Computer Problem 60𝑔 45𝑔 ̂ 𝚥̂ 𝚤̂, 𝛼⃗𝑏 = 𝑘 𝑎⃗𝐴 = 𝚤̂ 181 181𝑅 ( ) 105 255 7 𝑁= 𝑚𝑔 cos 𝜃, 𝐹 = 𝑚𝑔 sin 𝜃, 𝜃 = tan−1 𝜇𝑠 31 31 17 sin 𝜃 ◦ 𝜇𝑠 = , 𝜃 = 30.51 7 cos 𝜃 − 4 8𝑔 ̂ ( ) 8 𝛼⃗ = 𝑘, 𝜇𝑠 min = 33𝑅 29 𝑊 𝑑 cos 𝜃 𝑊 ℎ cos 𝜃 = 258.8 lb, 𝑇𝐶𝐷 = 𝑟 = 388.2 lb, 𝑎 = 𝑔 sin 𝜃 = 31.10 f t∕s2 𝑇𝐴𝐵 = 𝑟 𝑑+ℎ 𝑑+ℎ 2

𝚥̂

2

17.108 𝑎⃗𝐸 = (6.925 𝚤̂ + 7.072 𝚥̂) ft∕s , 𝑇𝐴𝐵 = 1238 lb, 𝑇𝐶𝐷 = 1858 f t∕s

𝚤̂

𝚥̂

2

17.110 𝑎⃗𝐸 = −10.43 𝚥̂ ft∕s , 𝐴𝑥 = 0, 𝐴𝑦 = 1790 lb |𝐹 | ( ) 17.112 𝑎⃗𝐺 = 10.84 𝚤̂ ft∕s2 , 𝜇𝑠 min = || || = 0.05876 |𝑁 |

𝚤̂ 𝚥̂ 𝚤̂

𝚥̂

17.114 𝑎⃗𝐺 = 45.80 𝚤̂ ft∕s2 𝚤̂ 𝜇𝑘 𝑟𝑔 𝚥̂ 𝑘̂ 17.116 𝑎⃗𝐺 = 𝜇𝑘 𝑔 𝚤̂, 𝛼⃗𝑏 = 𝚤̂ 𝑘2𝐺 √ 𝑔𝑅(𝜇𝑠 cos 𝜃 + sin 𝜃) 17.118 𝑣𝑚 = cos 𝜃 − 𝜇𝑠 sin 𝜃 ( ) ( ) 24𝑚𝑔 cos 𝜃 − 6𝑘 ℎ + 4𝛿0 sin 𝜃 + 3ℎ 2𝑘 + 5𝑚𝜃̇ 2 sin 𝜃 cos 𝜃 ̈ 17.120 𝜃 = ( ) 𝑚ℎ 1 + 15 cos2 𝜃 ( ) |𝐹 | 17.122 𝜇𝑠 min = || || = 0.4793, 𝛼⃗𝑤 = −0.1879 𝑘̂ rad∕s2 𝚥̂ 𝚤̂ |𝑁 | 17.124 Computer Problem 17.126

Computer Problem

17.128 𝜃 = 36.42◦ 17.130 𝐴𝑦 = 34 𝑚𝑔, 𝐵𝑦 = 54 𝑚𝑔, 𝜃 = 2 tan−1

Chapter 18 18.2

Concept Problem

18.4

Concept Problem

18.6 𝑇 = 0.01121 J 18.8 𝑇 = 1719 J, ℎ = 17.52 m 18.10 𝜔𝑏2 = 3.693 rad∕s 18.12 𝜃min = 53.13◦

(

𝑃 𝑚𝑔

) , 𝜃0 = 0, 𝜃∞ = 𝜋

𝚥̂ 𝚤̂

𝚥̂ 𝚤̂

ISTUDY

Appendix B

Answers to Even-Numbered Problems

18.14 Δ𝜃 = 8781 rev 18.16 𝑘 = 472.0 lb∕f t 18.18 𝜃1 = 33.02◦ √ (2𝑀𝑑∕𝑅) − 𝑘𝑑 2 2𝑀 18.20 𝜔𝑑2 = ( 2 ) , 𝑑𝑠 = 2 𝑘𝑅 𝑚 𝑘𝐺 + 𝑅 18.22 𝑀 = 653.8 N⋅m 18.24 𝑣𝑐 = 3.040 m∕s 18.26 𝑃 = 2.344 N 18.28 𝑑 = 0.4990 m

and

𝑑 = 1.102 m

18.30 𝐿𝑓 = 34.13 f t 18.32 𝜔𝑠2 = 2.640 rad∕s 18.34 (𝑈1-2 )nc = −2042 f t ⋅lb 18.36 𝑣𝐺2 = 3.951 f t∕s √ | 18.38 𝑣person = 2𝑔𝐻(1 + cos 𝜃), 𝑣person | ◦ = 10.85 m∕s |𝜃=0 18.40 𝑇 = 15.85 f t ⋅lb 18.42 𝑣𝑐 = 6.291 f t∕s 18.44 𝑣𝐴2 = 2.709 f t∕s, 𝑣𝐵2 = 5.417 f t∕s 18.46 𝜔𝑠2 = 1.124 rad∕s 18.48 𝑇 = 28.88 f t ⋅lb 18.50 𝑣𝑄2 = 20.45 f t∕s 18.52 𝑛min = 26, 𝑣person = 26.54 f t∕s √ 𝑑[(2𝑀∕𝑅) + 2𝑚𝑔 sin 𝜃 − 𝑑𝑘] ̂ 𝑘 18.54 𝜔 ⃗𝑑 = ( ) 𝑚 𝑅2 + 𝑘2𝐺

𝚥̂

𝚤̂

@ 𝜃, 𝑀 = 11.25 f t ⋅lb

18.56 𝛿 = 1.031 f t 18.58 𝑚𝐶 = 26.96 kg 18.60 𝑣𝑆2 = 16.00 f t∕s 18.62 𝑊𝑃 = 799.5 lb ⃗ 𝐷 = (44.61 rad∕s) 𝑘̂ 18.64 𝑣𝐶 = 5.019 m∕s, 𝑎𝜔

𝚥̂ 𝚤̂

18.66 𝑣max = 6.389 f t∕s 18.68 (𝑣𝐵 )final = 1.279 m∕s ( ) ( ) 𝚥̂ 18.70 (a) 𝑎𝐺𝑥 = 12 𝐿 𝜃̈ cos 𝜃 − 𝜃̇ 2 sin 𝜃 , 𝑎𝐺𝑦 = − 12 𝐿 𝜃̈ sin 𝜃 + 𝜃̇ 2 cos 𝜃 , [ ]𝚤̂ ( ) ( ) 𝐿 ̈ 𝐹 = 12 𝑚𝐿 𝜃̈ cos 𝜃 − 𝜃̇ 2 sin 𝜃 , 𝑁 = 𝑚𝑔 1 − 𝜃 sin 𝜃 + 𝜃̇ 2 cos 𝜃 2𝑔 3𝑔 3𝑔 (b) 𝜃̇ 2 = (1 − cos 𝜃), 𝜃̈ = sin 𝜃 𝐿 2𝐿 3 (c) 𝐹 = 4 𝑚𝑔 sin 𝜃(3 cos 𝜃 − 2), 𝑁 = 14 𝑚𝑔(1 − 3 cos 𝜃)2 , (𝜇𝑠 )max = 0.3706, 𝜃slide = 35.10◦ 18.72

Computer Problem

18.74

Computer Problem

18.76

Concept Problem

18.78

Concept Problem

⃗ = (310.3 f t ⋅lb⋅s) 𝑘̂ 18.80 𝑝⃗𝑤 = (−83.85 lb⋅s) 𝚤̂, ℎ 𝐶 18.82 |𝜔 ⃗ 𝑟2 | = 77.59 rpm

A-57

A-58

ISTUDY

Appendix B

Answers to Even-Numbered Problems

18.84 𝑃 = 25.00 N 18.86 𝑘𝐺 = 5.218 f t ( ) 𝚥̂ 1 2 2 ̂ ̂ ⃗ 18.88 ℎ 𝚤̂ , 𝐴 𝐴𝐵 = 3 𝑚𝐴𝐵 𝑅 𝜔𝐴𝐵 𝑘 = (1.725 kg⋅m ∕s) 𝑘 ( ) 𝚥 ̂ 2̂ 2 ⃗ ̂ ℎ 𝚤̂ , 𝐴 𝐵𝐶 = 𝑚𝐵𝐶 𝜔𝐴𝐵 𝑅 𝑘 = (7.200 kg⋅m ∕s) 𝑘 ( ) 𝚥̂ 1 2 ̂ ̂ ⃗ ℎ 𝚤̂ 𝐷 𝐶𝐷 = 3 𝑚𝐶𝐷 𝐻𝑅𝜔𝐴𝐵 𝑘 = (7.750 kg⋅m ∕s) 𝑘 ( ) ( ) −6 2 −6 2 ⃗ ̂ ⃗ ̂ 18.90 ℎ 𝐶 𝑊 = (91.88×10 kg⋅m ∕s) 𝑘, ℎ𝑂 𝑊 = (669.4×10 kg⋅m ∕s) 𝑘 𝑊 𝐿𝑣 𝚥̂ | | 𝑊 𝑣 ⃗ = 𝐴𝐵 𝐴 𝑘̂ = (2.420 f t ⋅lb⋅s) 𝑘̂ 18.92 |𝑝⃗𝐴𝐵 | = 𝐴𝐵 𝐴 = 3.227 lb⋅s, ℎ 𝐺 | | 2𝑔 cos 𝜃 12𝑔 cos 𝜃 18.94 Δ𝑡 = 16.98 s 18.96 |𝜔 ⃗ 𝑠2 | = 2.527 rad∕s 18.98 (a) Collar modeled as a particle: 𝑣impact = 0.9905 f t∕s (b) Collar modeled as a rigid body: 𝑣impact = 0.9432 f t∕s 18.100 |𝜔𝐴 |after slip stops = |𝜔𝐵 |after slip stops = 18.70 rad∕s 18.102 𝑣𝐶2 = 25.19 m∕s 18.104 𝑣𝑓 = 18.34 f t∕s, 𝑡𝑟 = 2.502 s

𝑢̂ 𝜃

𝑢̂ 𝑟 @𝜃

𝚤̂

Concept Problem 18.106 | 18.108 𝑣𝐺 | = 7.563 m∕s, (𝜇𝑠 )min = 0.3951 |𝑡=3 s 18.110 𝜔𝑓 = 30.13 rpm 4𝑟2 𝑏 (2𝑟 − 𝑏)2 6𝑣 ̂ (b) 𝜃 = 12.31◦ 18.114 (a) 𝜔 ⃗ + = √ 0 𝑘, 5 3𝐿 18.112 𝑑 =

[ ( 𝑑 + 1 𝑎 𝑡2 ) ( )] 𝑑 2 0 𝑓 −1 tan − tan−1 √ 18.116 The system rotates clockwise. 𝜃 = √ √ 𝐼𝑂 ∕𝑚𝐵 + 𝑒2 𝐼𝑂 ∕𝑚𝐵 + 𝑒2 𝐼𝑂 ∕𝑚𝐵 + 𝑒2 𝑒

Concept Problem Concept Problem 2 18.122 𝑑 = 𝓁 3 𝐼𝐺 + 𝑚𝓁(𝓁 − 𝛿) 18.124 𝑑 = 𝑚(𝓁 − 𝛿) 18.118 18.120

= (10.74 rad∕s) 𝑘̂ 18.126 𝜔 ⃗+ 𝑏 18.128 𝜃rebound = 78.32◦ 18.130 𝑣0 = 7.482 f t∕s

𝚥̂ 𝚤̂

+ ̂ 𝜔 18.132 𝜔 ⃗ 𝐴𝐵 = (−0.8510 rad∕s) 𝑘, ⃗ +𝐵𝐷 = (2.553 rad∕s) 𝑘̂

𝚥̂ 𝚤̂

+ + ̂ 𝜔 18.134 𝜔 ⃗𝐴 = (−0.0007152 rad∕s) 𝑘, ⃗𝐵 = (−0.00002414 rad∕s) 𝑘̂ 18.136 𝑑 = 1.978 m 18.138 𝑇 = 0.003854 J 18.140 Concept Problem 18.142 Concept Problem

18.144 𝑣⃗+𝑃 = (12 f t∕s) 𝚤̂

𝚥̂ 𝚤̂

18.146 𝜔 ⃗ 𝐴 = (1.031 rad∕s) 𝑘̂ +

𝚥̂

𝚤̂ , 𝜃

= 14.76◦

𝚥̂ 𝚤̂

ISTUDY

Appendix B

Answers to Even-Numbered Problems

Chapter 19 𝑚𝑔𝐿𝜏 2 𝑚𝑔𝐿𝜏 2 or 𝐼𝐺 = − 𝑚𝐿2 2 2 4𝜋 4𝜋 √ 7𝐿 𝜏 = 2𝜋 6𝑔 √ 𝐺𝑟4 𝜔𝑛 = 𝜌𝐿𝑅4 𝑡 √ 𝐼𝐺 𝓁= 𝑚 𝑓 = 1.771 Hz √ 𝜔 𝜌𝑔 𝑑 𝑓= 𝑛 = = 0.5663 Hz 2𝜋 4 𝜋𝑚 √ 2𝜋𝑑 𝑚 𝜏= ℎ 𝑘 √ √1 ( √ 𝜌 ℎ3 + 𝑑 3 ) + 𝑚𝑑 2 √3 𝜏 = 2𝜋 √ ) √ ( 𝑘 − 12 𝜌𝑔 ℎ2

19.2 𝐼𝑂 = 19.4 19.6 19.8 19.10 19.12 19.14

19.16

19.18 𝑚𝑦̈ + 𝑘𝑦 = 0, 𝑚𝜃̈ + 𝑘𝜃 = 0 17𝑘 19.20 𝑥̈ 𝐺 + 𝑥 =0 6𝑚 𝐺 19.22 Moving the disk up 𝑛 = 4.778 turns 8𝑘 19.24 𝑥̈ 𝐺 + 𝑥𝐺 = 0, 𝜏 = 2.221 s 3𝑚 √ √ 2𝑔 , 𝜏 = 2𝜋 3(𝑅 − 𝑟) √ 𝐿 19.28 𝜏 = 2𝜋 2𝑔 √ (3𝜋 − 2)𝑅 19.30 𝜏 = 2𝜋 3𝑔

19.26 𝜔𝑛 =

3(𝑅 − 𝑟) 2𝑔

𝐹0 ∕𝑘 ( )2 cos 𝜔0 𝑡 1 − 𝜔0 ∕𝜔𝑛 √ 19.34 𝑚𝑢 = 0.01 kg ⇒ |𝑦𝑚 | = 2.190×10−6 cos(4.369𝑡) + 1.888×10−5 m, √ 𝑚𝑢 = 0.1 kg ⇒ |𝑦𝑚 | = 1.100×10−5 cos(4.369𝑡) + 1.112×10−5 m, √ 𝑚𝑢 = 1 kg ⇒ |𝑦𝑚 | = 9.798×10−4 cos(4.369𝑡) + 1.014×10−3 m 19.36 𝜃amp = 0.001758 rad

19.32 𝑥(𝑡) = 𝐴 sin 𝜔𝑛 𝑡 + 𝐵 cos 𝜔𝑛 𝑡 +

19.38 Concept Problem ( ) 19.40 𝑥(𝑡) = 0.0005013 sin 10𝑡 + 0.1000 cos 10𝑡 − 2.506×10−5 sin 200𝑡 m ] ( ) [ 𝐹0 ∕𝑘eq 𝜔0 𝐿0 19.42 𝑚𝑦̈ + 2𝑘 1 − 𝑦 = 𝐹0 sin 𝜔0 𝑡, 𝑦(𝑡) = ( )2 sin 𝜔0 𝑡 − 𝜔 sin 𝜔𝑛 𝑡 , 𝐿 𝑛 1 − 𝜔0 ∕𝜔𝑛 √ ( ( ) ) 𝐿 𝐿 2𝑘 where 𝑘eq = 2𝑘 1 − 0 and 𝜔𝑛 = 1− 0 𝐿 𝑚 𝐿 ) ( 𝐹0 ( ) 2 19.44 𝑚 + 12 𝑚𝐴 𝑥̈ 𝐴 + 2𝑘𝑥𝐴 = 12 𝐹0 sin 𝜔0 𝑡, 𝑥𝐴 amp = ) ( 3 𝐵 4𝑘 − 43 𝑚𝐵 + 𝑚𝐴 𝜔20

A-59

A-60

ISTUDY

Appendix B

Answers to Even-Numbered Problems √

𝜔 3𝑔𝐸𝐼 = 505.7 rad∕s, 𝑓 = 𝑛 = 80.48 Hz, ( ) 3 2𝜋 𝑊 𝑢 + 𝑊𝑒 𝑑 ) ( )2 | |( |𝜃𝑝 | 𝑊𝑢 + 𝑊𝑒 𝑑 𝜔𝑝 ∕𝜔𝑛 MF = | | = )2 = 0.2071 ( 𝑊𝑢 𝑅 1 − 𝜔 ∕𝜔

19.46 𝜔𝑛 =

𝑝

19.48

Concept Problem

19.50

Concept Problem

19.52 No peak in MF for 𝜁 ≥ 19.54 𝑦̈ + 2𝜁 𝜔𝑛 𝑦̇ + 𝜔2𝑛 𝑦 =

𝑛

√ 1∕2

𝑚𝑢 𝜀𝜔2𝑟

sin 𝜔𝑟 𝑡, where 𝑐∕𝑚 = 2𝜁 𝜔𝑛 and 𝑘∕𝑚 = 𝜔2𝑛 𝑚 19.56 𝐹0 = 0.01584 N, |𝑦| = 0.00009410 m ( ) 19.58 𝑦(𝑡) = 0.5𝑡𝑒−28.28𝑡 m 19.60 𝜁 =

1 1 = 0.1 for MF = 5, 𝜁 = = 0.05 10 20

for MF = 10

19.62 𝑚𝐿𝜃̈ + 0.18𝑐𝐿𝜃̇ + (𝑚𝑔 + 0.72𝑘𝐿) 𝜃 = 0, 𝜔𝑑 =

√

𝑔 𝑘 𝑐2 + 0.72 − 0.0081 𝐿 𝑚 𝑚2

) ] [ ( 𝜔0 𝐸 2𝑘 𝑘 − 𝑚𝐴 𝜔20 + 𝑐 2 𝜔20 𝑚𝐴 𝑐𝜔30 𝐸 −(𝑐∕2𝑚𝐴 )𝑡 19.64 𝑦𝐴 = − sin 𝜔 𝑡 + 𝑒−(𝑐∕2𝑚𝐴 )𝑡 cos 𝜔𝑑 𝑡 𝑒 [( ] ( ) 𝑑 )2 2 2 + 𝑐 2 𝜔2 2 2 2 𝑘 − 𝑚 𝜔 2𝜔𝑑 𝑘 − 𝑚𝐴 𝜔0 + 𝑐 𝜔0 𝐴 0 0 𝑚𝐴 𝜔20 𝐸 [( ) ] +( 𝑘 − 𝑚𝐴 𝜔20 sin 𝜔0 𝑡 − 𝑐𝜔0 cos 𝜔0 𝑡 + 𝐸 sin 𝜔0 𝑡 ) 2 𝑘 − 𝑚𝐴 𝜔20 + 𝑐 2 𝜔20 [ ] = −0.0002665 cos(94.25𝑡) + 𝑒−2.500𝑡 0.0002665 cos(5.204𝑡) + 0.0002121 sin(5.204𝑡) − 4.642×10−6 sin(94.25𝑡) 19.66 𝐷 = 4.373×10−6 m 19.68 𝑚𝑠̈ + 𝑐 𝑠̇ + 𝑘𝑠 = 𝑚𝐴𝜔20 sin 𝜔0 𝑡, where 𝑠(𝑡) = 𝑦(𝑡) − 𝑢(𝑡), √ ( ) √ 3 ⎤ ⎡ ⎡ 𝑚𝐴𝜔2 𝑘 − 𝑚𝜔2 ⎤ √ 𝑘2 + 𝑐 2 𝜔20 𝑚𝑐𝐴𝜔0 √ 0 0 ⎢ ⎥ ⎥ ⎢ √ 𝑦(𝑡) = ( + 𝐴 𝑡 − 𝑡, DT = cos 𝜔 sin 𝜔 ( )2 0 0 ⎢ 𝑘 − 𝑚𝜔2 )2 + 𝑐 2 𝜔2 ⎥ ⎢ (𝑘 − 𝑚𝜔2 )2 + 𝑐 2 𝜔2 ⎥ 𝑘 − 𝑚𝜔20 + 𝑐 2 𝜔20 ⎣ ⎦ ⎣ 0 0 0 0⎦ 19.70 𝑚𝑥̈ +

𝑘1 𝑘2 𝑘1 + 𝑘2

𝑥=0

19.72 Δ𝑘 = −5.910% ( ) 19.74 𝑦𝑐 amp = 0.0004544 m = 0.4544 mm ̈ = 282.0 m∕s2 19.76 𝐹0 = 200.1 N, |𝑥| 19.78 𝑚𝑢 = 0.01 kg ∶ 𝐹𝑡 = 14.77 sin(125.7𝑡 + 0.8422) + 17.67 cos(125.7𝑡 + 0.8422) N, 𝑚𝑢 = 0.1 kg ∶ 𝐹𝑡 = 147.7 sin(125.7𝑡 + 0.8422) + 176.7 cos(125.7𝑡 + 0.8422) N, 𝑚𝑢 = 1 kg ∶ 𝐹𝑡 = 1477 sin(125.7𝑡 + 0.8422) + 1767 cos(125.7𝑡 + 0.8422) N ) ( 𝐿0 𝑦=0 19.80 𝑚𝑦̈ + 2𝑘 1 − √ 𝑦2 + 𝐿2 ( ) 𝐿0 19.82 𝑚𝑦̈ + 2𝑘 1 − 𝑦=0 𝐿

ISTUDY

Appendix B

Answers to Even-Numbered Problems

Chapter 20

( ) 20.2 𝑣⃗𝐵 = 𝓁 −𝜔1 sin 𝜃 𝚤̂ + 𝜔1 cos 𝜃 𝚥̂ + 𝜃̇ cos 𝜃 𝑘̂ , ( 2 ) ( ) 𝑎⃗𝐵 = −𝓁 𝜃̇ cos 𝜃 − 𝜔21 cos 𝜃 − 𝜔̇ 1 sin 𝜃 𝚤̂ + 𝓁 𝜔̇ 1 cos 𝜃 − 𝜔21 sin 𝜃 𝚥̂ + 𝓁 𝜃̈ cos 𝜃 𝑘̂ 20.4 𝑣⃗𝐴 = 2(𝓁 + 𝑑)𝜔1 cos2 𝜃 𝚥̂, 𝑎⃗𝐴 = −2(𝓁 + 𝑑)𝜔21 cos2 𝜃 𝚤̂ + 2(𝓁 + 𝑑)𝜔̇ 1 cos2 𝜃 𝚥̂ − (𝓁 + 𝑑)𝜔21 cos 𝜃 𝑘̂ ̂ 20.6 𝑣⃗𝐸 = 𝑅𝜔arm cos 𝛾 𝚤̂ + (𝑑 + 𝓁 cos 𝛾)𝜔arm 𝚥̂ − (𝑑 + 𝓁 cos 𝛾)𝜔arm 𝑘, 2 ] 𝜔 [ ( ) 𝑎⃗𝐸 = −𝜔2arm (𝑑 + 𝓁 cos 𝛾) 𝚤̂ − 𝜔2arm sin 𝛾(𝑑 + 𝓁 cos 𝛾) 𝚥̂ − arm 𝑑 2 + 2𝑑𝓁 cos 𝛾 + 𝓁 2 + 𝑅2 cos2 𝛾 𝑘̂ 𝑅

̂ 20.8 𝑣⃗𝐸 = 𝑅𝜔arm cos 𝛾 𝚤̂ + (𝑑 + 𝓁 cos 𝛾)𝜔arm 𝚥̂ − (𝑑 + 𝓁 cos 𝛾)𝜔arm 𝑘, [ ] ( ) 2 𝑎⃗𝐸 = 𝑅𝛼arm − 𝜔arm (𝑑 + 𝓁 cos 𝛾) cos 𝛾 𝚤̂ + (𝑑 + 𝓁 cos 𝛾) 𝛼arm − 𝜔2arm sin 𝛾 𝚥̂ [ ( ] ) ( ) ( ) 1 𝑑 𝑅𝛼arm + 𝑑𝜔2arm + 𝓁 𝑅𝛼arm + 2𝑑𝜔2arm cos 𝛾 + 𝓁 2 + 𝑅2 𝜔2arm cos2 𝛾 𝑘̂ − 𝑅 ) ( ( ) ̂ 20.10 𝑣⃗𝐵 = 𝓁̇ cos 𝜃 − 𝓁 𝜃̇ sin 𝜃 𝚤̂ + 𝓁𝜔1 cos 𝜃 𝚥̂ + 𝓁 𝜃̇ cos 𝜃 + 𝓁̇ sin 𝜃 𝑘, [ ( 2 ) ] ( ) ( ) 2 ̇ ̇ ̇ ̇ ̇ 𝑎⃗𝐵 = −𝓁 𝜃 + 𝜔1 cos 𝜃 − 2𝓁 𝜃 sin 𝜃 𝚤̂ + 2𝜔1 𝓁 cos 𝜃 − 𝓁 𝜃 sin 𝜃 𝚥̂ + 𝜃̇ 2𝓁̇ cos 𝜃 − 𝓁 𝜃̇ sin 𝜃 𝑘̂ ̂ 20.12 𝜔 ⃗ disk = 𝜔𝑏 𝚤̂ + 𝜔𝑑 𝑘, 𝛼⃗disk = 𝜔̇ 𝑏 𝚤̂ − 𝜔𝑑 𝜔𝑏 𝚥̂ + 𝜔̇ 𝑑 𝑘̂ 20.14 𝜔 ⃗ 𝑐 = −𝜔0

) cos 𝛽 cos 𝛽 ( 𝚤̂, 𝛼⃗𝑐 = −𝛼0 𝚤̂ + 𝜔20 𝑘̂ sin 𝛽 sin 𝛽

̂ 20.16 𝑣⃗𝐵 = −𝐿𝜔0 cos 𝛽 𝚥̂ − 𝐿𝜔0 cos2 𝛽 𝑘, ( ) 𝑎⃗𝐵 = −𝐿𝜔20 cos2 𝛽 𝚤̂ + 𝐿𝛼0 cos 𝛽 𝚥̂ − 𝐿 cos 𝛽 cot 𝛽 𝜔20 + 𝛼0 sin 𝛽 𝑘̂ ( ) ( ) 20.18 𝑎⃗𝐴 = 231.3 𝚥̂ + 107.9 𝑘̂ ft∕s2 , 𝛼⃗𝐴𝐵 = 18.29 𝚤̂ + 15.08 𝚥̂ + 34.90 𝑘̂ rad∕s2 ̂ 𝛼⃗ = 𝛽𝜔 ̇ 𝚤̂ − 𝛼 𝚥̂ − 𝛽̈ 𝑘, ̂ 20.20 𝜔 ⃗ 𝐴𝐵 = −𝜔𝑠 𝚥̂ − 𝛽̇ 𝑘, 𝑠 [ (𝐴𝐵 )𝑠 ] ) 𝐿 ̇2 𝐿 ( ̇2 𝐿 𝐿̈ 2 𝑎⃗𝐺 = − 𝛽 sin 𝛽 + 𝑑 + cos 𝛽 𝜔𝑠 + 𝛽 cos 𝛽 𝚤̂ + 𝛽 sin 𝛽 − 𝛽̈ cos 𝛽 𝚥̂ 2 [( 2) 2 ] 2 𝐿 ̇ sin 𝛽 𝑘̂ + 𝑑 + cos 𝛽 𝛼𝑠 − 𝐿𝛽𝜔 𝑠 2 ( ) ( ) 20.22 𝜔 ⃗ 𝐴𝐵 = 0.3139 𝚤̂ + 10.37 𝚥̂ + 2.786 𝑘̂ rad∕s, 𝛼⃗𝐴𝐵 = −35.35 𝚤̂ + 135.2 𝚥̂ + 161.9 𝑘̂ rad∕s2 ( ) 20.24 𝑎⃗𝐴 = 101.8 𝚥̂ + 47.49 𝑘̂ m∕s2 ( ) ( ) 20.26 𝑣⃗𝐴 = −0.7354 𝚥̂ − 0.3429 𝑘̂ m∕s and 𝑎⃗𝐴 = −12.49 𝚥̂ − 5.825 𝑘̂ m∕s2 ( ) 20.28 𝛼⃗𝐴𝐵 = −6.040 𝚥̂ − 8.054 𝑘̂ rad∕s2 20.30 𝑎⃗𝐴 = −124.2 𝚤̂ m∕s2

[ ( 4.187 20.32 𝛼⃗𝐴𝐵 = −11.84 cos 𝜃 𝚤̂ + 26.25 cos 𝜃 + 0.7500 19.53 cos 2𝜃 − 5.400 cos 3𝜃 + 0.5625 cos 4𝜃 − 3∕2 𝑍[ )] ( )] 101.2 24.55 𝚥̂ − −7.438 cos 𝜃 + 0.7500 −1.200 − 3.600 cos 2𝜃 + 0.7500 cos 3𝜃 𝑘̂ rad∕s2 , 𝑍 3∕2 where 𝑍 = (7.438 + 3.600 cos 𝜃 − 0.5625 cos 2𝜃) m2 , [ √ ( ) 29.08 −31.37 + 25.92 cos 𝜃 2 + 1.688 cos 2𝜃 − 4.800 cos 𝜃 1.688 cos 2𝜃 − 6.312 − 22.63 𝑍 sin 𝜃 + 𝑎⃗𝐴 = 𝑍 3∕2 ( √ √ )] 0.7500 0.5625 cos 4𝜃 + 2 2(0.7500 cos 𝜃 − 2.400) 𝑍 sin 2𝜃 𝚤̂ m∕s2 , where 𝑍 = (7.438 + 3.600 cos 𝜃 − 0.5625 cos 2𝜃) m2 ( ) ( ) ( ) 20.34 𝑣⃗𝐵 = 𝑣𝑡 + ℎ𝜔𝑎 𝚤̂ + 𝓁 + 𝑟𝑡 𝜔𝑏 𝚥̂ − 𝓁 + 𝑟𝑡 𝜔𝑎 𝑘̂

A-61

A-62

ISTUDY

Answers to Even-Numbered Problems

Appendix B

√ 24𝑔 (1 + sin 𝜃) 𝚤̂ 𝐿 sin 𝜃 (5 + 3 sin 𝜃) 20.38 𝜔10 = 4554 rad∕s ( ) 20.40 13 𝐿𝛽̈ + 13 𝐿 cos 𝛽 + 12 𝑑 𝜔2𝑠 sin 𝛽 − 12 𝑔 cos 𝛽 = 0 ( ( ) ) 20.42 𝑂𝑥 = −𝑚 ℎ + 𝐿2 𝜔2𝑠 , 𝑂𝑦 = 0, 𝑂𝑧 = 𝑚𝑔, 𝑀𝑂𝑥 = 0, 𝑀𝑂𝑦 = −𝑚𝑔 ℎ + 𝐿2 + 12 𝑚𝑟2 𝜔𝑑 𝜔𝑠 , 𝑀𝑂𝑧 = 0

20.36 𝜔 ⃗ bar = ±

20.44 𝑣𝐴2 = 6.294 m∕s 20.46 𝑣𝐴2 = 6.165 m∕s 1 20.48 𝑀𝐴𝑥 = − 𝑚𝑅2 𝜃̇ 𝜙̇ cos 𝜃, 2 ( ) 1 1 𝑀𝐴𝑦 = 𝑚𝑅2 𝜙̇ 2 cos 𝜃 sin 𝜃 + 𝑚𝑅2 𝜙̇ cos 𝜃 𝜓̇ − 𝜙̇ sin 𝜃 , 4 2 ( ) 1 1 𝑀𝐴𝑧 = − 𝑚𝑅2 𝜃̇ 𝜙̇ sin 𝜃 − 𝑚𝑅2 𝜃̇ 𝜓̇ − 𝜙̇ sin 𝜃 2 2 𝑚 2 2 𝑚 2 2 𝑅 𝜔𝑠 cos 𝜃 sin 𝜃, 𝐴𝑍 = 0, 𝐵𝑌 = 𝑅 𝜔𝑠 cos 𝜃 sin 𝜃, 20.50 𝑀𝑋 = 0, 𝐴𝑌 = 8𝐿 8𝐿 𝐵𝑍 = 0, where 𝑋𝑌 𝑍 is attached to the shaft 20.52 𝑅𝑥 = −0.5236𝑡2 N, 𝑅𝑦 = 0.02193 N, 𝑅𝑧 = 0.09810 N, 𝑀𝑥 = 0.00009291 N⋅m, 𝑀𝑦 = 0.002219𝑡2 N⋅m, 𝑀𝑧 = 0.00006855 N⋅m √ 3𝑔 20.54 𝜔𝑠 = 𝐿 ( ) 8𝑚𝑔 (1 + cos 𝛽) + 𝑚𝑅𝜔20 4 cos 𝛽 + 4 cos2 𝛽 − 16 sin 𝛽 − 17 cos 𝛽 sin 𝛽 20.56 𝑁 = 4 (2 + 2 cos 𝛽 − sin 𝛽) √ ( ) 5𝑔 20.58 𝜔0 min = 7𝑅 ℎ 2 ℎ 20.60 𝛼⃗𝑑 = − 𝜔𝑏 𝚤̂ + 𝜔̇ 𝑏 𝚥̂ − 𝜔̇ 𝑏 𝑘̂ [𝑅 ( 2𝑅 2 ) ] [ ( 2) ] ℎ +𝑅 ℎ 20.62 𝑎⃗𝑃 = ℎ𝜔̇ 𝑏 cos 𝛽 + 𝜔2𝑏 sin 𝛽 𝚤̂ + ℎ𝜔̇ 𝑏 sin 𝛽 − 𝜔2𝑏 cos 𝛽 𝚥̂ 𝑅 𝑅 [ ] + 𝑅𝜔̇ sin 𝛽 − ℎ𝜔2 (1 + 2 cos 𝛽) 𝑘̂ 𝑏

20.64 𝑂𝑋 =

𝑏

1 = 𝑚𝑔, 𝑂𝑍 = −𝑚𝐿𝜔̇ 0 , 𝑀𝑋 = 𝑚𝐿2 𝜔̇ 0 cos 𝜃 sin 𝜃, 12 ) 1 sin2 𝜃 1− , 𝑀𝑍 = 𝑚𝑔𝐿 − 𝑚𝐿2 𝜔20 cos 𝜃 sin 𝜃 12 12

−𝑚𝐿𝜔20 , 𝑂𝑌

𝑀𝑌 = 𝑚𝐿2 𝜔̇ 0

(

20.66 𝑂𝑋 = −𝑚𝐿𝜔20 , 𝑂𝑌 = 𝑚𝑔, 𝑂𝑍 = −𝑚𝐿𝜔̇ 0 , ) ( 3 1 1 𝑚𝐿2 𝜔̇ 0 cos 𝜃 sin 𝜃, 𝑀𝑌 = 𝑚𝐿2 𝜔̇ 0 17 − 3 sin2 𝜃 , 𝑀𝑍 = 𝑚𝑔𝐿 − 𝑚𝐿2 𝜔20 cos 𝜃 sin 𝜃 𝑀𝑋 = 16 16 16 675 × 151∕3 csc2 𝜃 20.68 𝜙̇ = 4𝑚𝐿2 (1 + sin 𝜃) [ ] 2 𝑑 4 + ℎ2 (𝓁 − 𝑅)2 + 𝑑 2 (ℎ + 𝑅 − 𝓁)(ℎ − 𝑅 + 𝓁) 𝜔2 2 2 ℎ𝑅 2𝑑ℎ𝑅 𝜔𝑑 𝑑 20.70 𝛼⃗𝐴𝐵 = − ( 𝚥̂ )2 𝚤̂ − ( )2 3 𝑑 2 + ℎ2 𝑑 2 + ℎ2 (𝓁 − 𝑅) [ ] 𝑑𝑅2 𝑑 4 + 𝑑 2 ℎ2 + 2ℎ2 (𝓁 − 𝑅)2 𝜔2𝑑 − 𝑘̂ ( )2 𝑑 2 + ℎ2 (𝓁 − 𝑅)3 √ ( ) = −26.32 𝚤̂ − 0.9870 𝚥̂ − 17.77 𝑘̂ rad∕s2 , where 𝓁 = 𝐿2𝐴𝐵 − 𝑑 2 − ℎ2 + 𝑅 [ ] √ 𝑅 𝑑 2 𝑅 + 𝓁 (𝓁 − 𝑅)2 𝜔2𝑑 20.72 𝑎⃗𝐴 = − 𝚤̂ = −124.2 𝚤̂ m∕s2 , where 𝓁 = 𝐿2𝐴𝐵 − 𝑑 2 − ℎ2 + 𝑅 (𝓁 − 𝑅)3

ISTUDY

Appendix B

Answers to Even-Numbered Problems

[ ( ) ( ) 2𝑑 2 𝑅 − 6𝑅𝓁 2 − 𝑑 𝑅2 − 4𝓁 2 cos 𝜃 + 2𝑅 −𝑑 2 + 𝓁 2 cos 2𝜃 + 𝑑𝑅2 cos 3𝜃 3 4 (𝓁 − 𝑅 sin 𝜃) ] + 5𝑅2 𝓁 sin 𝜃 + 4𝓁 3 sin 𝜃 − 4𝑑𝑅𝓁 sin 2𝜃 + 𝑅2 𝓁 sin 3𝜃 𝚤̂ [ 20.56 −31.87 + 35.62 cos 𝜃 + 9.184 cos 2𝜃 + 0.6750 cos 3𝜃 + 90.92 sin 𝜃 =− (2.750 − 0.7500 sin 𝜃)3 √ ] − 9.900 sin 2𝜃 + 1.547 sin 3𝜃 rad∕s2 , where 𝓁 = 𝐿2𝐴𝐵 − 𝑑 2 − ℎ2 + 𝑅 [( ( ) ) ] ℎ𝑅𝜔2𝑑 4 𝑑 2 + ℎ2 + 5𝑅2 cos 𝜃 − 𝑅 (8𝑑 + 𝑅 cos 3𝜃) 𝛼⃗𝐴𝐵 = 𝚤̂ ]2 [ 4 𝑑 2 + ℎ2 + 𝑅 cos 𝜃 (𝑅 cos 𝜃 − 2𝑑) { ℎ𝑅𝜔2𝑑 +[ 𝑅3 𝓁 cos4 𝜃 (2𝑅 sin 𝜃 − 𝓁) ]2 𝑑 2 + ℎ2 + 𝑅 cos 𝜃 (𝑅 cos 𝜃 − 2𝑑) (𝑅 sin 𝜃 − 𝓁)3 ] [ ( ) − 𝑑𝑅2 cos3 𝜃 −3𝓁 2 + 𝑅 sin 𝜃 (6𝓁 + 𝑅 sin 𝜃) − 𝑅 cos2 𝜃 3𝑑 2 + ℎ2 𝓁 2 )]] [ [ ( ) ( − 𝑅 cos2 𝜃 𝑅 sin 𝜃 −2 3𝑑 2 + ℎ2 𝓁 + 𝑅 sin 𝜃 𝓁 2 − 3𝑑 2 + 𝑅 sin 𝜃 (𝑅 sin 𝜃 − 2𝓁) [( 2 ) [ ( ) ( ( ))]] + 𝑑 cos 𝜃 𝑑 + ℎ2 𝓁 2 + 𝑅 sin 𝜃 −2 𝑑 2 + ℎ2 𝓁 + 𝑅 sin 𝜃 −3𝑑 2 − ℎ2 + 2𝑅 sin 𝜃 𝑅 sin 𝜃 − 2𝓁 } [ ( ( ) ( ) ( )) ] + 𝑅 sin 𝜃 sin 𝜃 𝑑 4 + ℎ2 𝓁 2 + 𝑑 2 ℎ2 − 𝓁 2 + 𝑑 2 − ℎ2 𝑅 sin 𝜃 2𝓁 − 𝑅 sin 𝜃 + 𝑑𝑅𝓁 2 sin 2𝜃 𝚥̂ { 𝑅𝜔2𝑑 (𝑅 cos 𝜃 − 𝑑) 2ℎ2 𝑅2 sin3 𝜃 (2𝓁 − 𝑅 sin 𝜃) +[ ]2 3 2 2 𝑑 + ℎ + 𝑅 cos 𝜃 (𝑅 cos 𝜃 − 2𝑑) (𝑅 sin 𝜃 − 𝓁) [ ] + 𝓁 2 cos 𝜃 (𝑅 cos 𝜃 − 𝑑) 𝑑 2 + ℎ2 + 𝑅 cos 𝜃 (𝑅 cos 𝜃 − 2𝑑) ( ) + 2𝑅𝓁 cos 𝜃 (𝑑 − 𝑅 cos 𝜃) 𝑑 2 + ℎ2 + 𝑅 cos 𝜃 (𝑅 cos 𝜃 − 2𝑑) sin 𝜃 } ( ) ( )] [ + 𝑅 −𝑑 2 𝑑 2 + ℎ2 − 2ℎ2 𝓁 2 + 𝑑𝑅 cos 𝜃 3𝑑 2 + ℎ2 + 𝑅 cos 𝜃 (𝑅 cos 𝜃 − 3𝑑) sin2 𝜃 𝑘̂

20.74 𝑎⃗𝐴 = −

𝑅𝜔2𝑑

−133.2 + 218.6 cos 𝜃 − 10.41 cos 3𝜃 =[ ]2 𝚤̂ 2.250 + (−1.800 + 0.5625 cos 𝜃) cos 𝜃 [ (−2.750 + 0.7500 sin 𝜃)−3 +[ ]2 −1229 − 1142 cos 2𝜃 + 149.8 cos 3𝜃 − 18.22 cos 4𝜃 2.250 + (−1.800 + 0.5625 cos 𝜃) cos 𝜃 + 5.270 cos 5𝜃 − 0.5489 cos 6𝜃 ] + cos 𝜃 (2490 − 1340 sin 𝜃) + 434.1 sin 𝜃 + 289.8 sin 3𝜃 − 25.76 sin 4𝜃 𝚥̂ [ (−2.750 + 0.7500 sin 𝜃)−3 +[ ]2 −1889 − 3302 cos 2𝜃 + 1170 cos 3𝜃 2.250 + (−1.800 + 0.5625 cos 𝜃) cos 𝜃 − 199.7 cos 4𝜃 + 19.03 cos 5𝜃 − 0.7319 cos 6𝜃 + cos 𝜃 (4329 − 1880 sin 𝜃) ] + 278.2 sin 𝜃 + 815.8 sin 3𝜃 − 277.5 sin 4𝜃 + 42.94 sin 5𝜃 − 3.355 sin 6𝜃 𝑘̂ rad∕s2

A-63

ISTUDY

ISTUDY

Mass Moments of Inertia

C

Mass moments and products of inertia are measures of how the mass is distributed within a body. Mass moments and products of inertia of a body depend on its geometry (size and shape), the density of the material at each point in the body, and the axes selected for measuring them.

NASA

The International Space Station as it existed in June 2008. Accurately determining its mass moments and products of inertia requires sophisticated computer codes, but the principles used in those codes are no different than what we study in this book.

Mass moments of inertia and mass products of inertia are measures of how the mass is distributed within a body. Mass moments and products of inertia appear in the rotational equations of motion for a rigid body (Chapter 17), the kinetic energy of a rigid body (Chapter 18), the angular momentum of a rigid body (Chapter 18 and Appendix D), and the three-dimensional dynamics of rigid bodies.

Definition of mass moments and products of inertia The mass moment of inertia of the element of mass 𝑑𝑚 within the body 𝐵 about any of the three coordinate axes 𝑥, 𝑦, or 𝑧, is defined as the product of the mass of the element and the square of the shortest distance from the element to the axis in question. For example, referring to Fig. C.1, the mass moment of inertia of 𝑑𝑚 about the 𝑦 axis is 𝑑𝐼𝑃 𝑦 = 𝑟2𝑦 𝑑𝑚, where we include the origin 𝑃 of the coordinate system in the subscript so that different sets of axes can be identified. The mass moments of inertia of the entire body 𝐵 with respect to the three axes are found by integrating the differential moments of inertia over the entire body. Therefore, the mass moments of inertia for the body 𝐵 shown in Fig. C.1 are defined as 𝐼𝑃 𝑥 =

∫𝐵

𝑟2𝑥 𝑑𝑚 =

∫𝐵

(

) 𝑦2 + 𝑧2 𝑑𝑚,

𝑑𝑚 = 𝜌 𝑑𝑉

𝑟𝑦 𝑦 𝑟 𝑧 𝑃

𝑥

𝑟𝑥 𝑧

𝑥

Figure C.1 A body 𝐵 with mass 𝑚, density 𝜌, and volume 𝑉 . The scalar quantities 𝑟𝑥 , 𝑟𝑦 , and 𝑟𝑧 are radial distances from the 𝑥, 𝑦, and 𝑧 axes, respectively, to the mass element 𝑑𝑚.

(C.1)

A-65

A-66

Appendix C

Mass Moments of Inertia

𝐼𝑃 𝑦 = 𝐼𝑃 𝑧 =

∫𝐵

𝑟2𝑦 𝑑𝑚 =

∫𝐵

∫𝐵

𝑟2𝑧 𝑑𝑚 =

∫𝐵

( 2 ) 𝑥 + 𝑧2 𝑑𝑚,

(C.2)

( 2 ) 𝑥 + 𝑦2 𝑑𝑚,

(C.3)

where: 𝑟𝑥 , 𝑟𝑦 , and 𝑟𝑧 are shown in Fig. C.1 and are the radial distances (i.e., moment arms) from the 𝑥, 𝑦, and 𝑧 axes, respectively, to the mass element 𝑑𝑚. 𝑥, 𝑦, and 𝑧 are shown in Fig. C.1 and are the coordinates of the mass element 𝑑𝑚. 𝐼𝑃 𝑥 , 𝐼𝑃 𝑦 , and 𝐼𝑃 𝑧 are the mass moments of inertia of the body 𝐵 about the 𝑥, 𝑦, and 𝑧 axes, respectively. 𝑑𝑚 = 𝜌 𝑑𝑉 𝑥

𝑃

𝑦

𝑥

Figure C.2 A body 𝐵 with mass 𝑚 demonstrating the mass product of inertia 𝑑𝐼𝑃 𝑥𝑦 = 𝑥𝑦 𝑑𝑚 of the element of mass 𝑑𝑚.

ISTUDY

The three mass products of inertia for the element of mass 𝑑𝑚 are defined with respect to the three possible pairs of orthogonal planes as the perpendicular distance from each pair of planes to the mass element. For example, referring to Fig. C.2, the mass product of inertia of the element 𝑑𝑚 with respect to the two planes 𝑥𝑧 and 𝑦𝑧 is 𝑑𝐼𝑃 𝑥𝑦 = 𝑥𝑦 𝑑𝑚. The mass products of inertia of the entire body 𝐵 with respect to the three pairs of planes are found by integrating the differential products of inertia over the entire body. The mass products of inertia for the body 𝐵 shown in Fig. C.2 are defined as 𝐼𝑃 𝑥𝑦 = 𝐼𝑃 𝑦𝑥 =

∫𝐵

𝑥𝑦 𝑑𝑚,

(C.4)

𝐼𝑃 𝑦𝑧 = 𝐼𝑃 𝑧𝑦 =

∫𝐵

𝑦𝑧 𝑑𝑚,

(C.5)

𝐼𝑃 𝑥𝑧 = 𝐼𝑃 𝑧𝑥 =

∫𝐵

𝑥𝑧 𝑑𝑚,

(C.6)

where: 𝐼𝑃 𝑥𝑦 , 𝐼𝑃 𝑦𝑧 , and 𝐼𝑃 𝑥𝑧 are the products of inertia of the mass with respect to the 𝑥𝑧-𝑦𝑧, 𝑥𝑦-𝑥𝑧, and 𝑥𝑦-𝑦𝑧 plane pairs, respectively. Remarks • When referring to mass moments of inertia, we often omit the word “mass” when it is obvious from the context that we are dealing with mass moments of inertia as opposed to area moments of inertia. • In each of Eqs. (C.1)–(C.3), two equivalent integral expressions are provided, and each is useful depending on the geometry of the object under consideration. • The moments of inertia in Eqs. (C.1)–(C.6) measure the second moment of the mass distribution. That is, to determine 𝐼𝑃 𝑥 , 𝐼𝑃 𝑦 , and 𝐼𝑃 𝑧 in Eqs. (C.1)–(C.3), the moment arms 𝑟𝑥 , 𝑟𝑦 , and 𝑟𝑧 are squared. The second integral in each of these expressions is obtained by noting that 𝑟2𝑥 = 𝑦2 + 𝑧2 , and similarly for 𝑟2𝑦 and 𝑟2𝑧 . For the products of inertia 𝐼𝑃 𝑥𝑦 , 𝐼𝑃 𝑦𝑧 , and 𝐼𝑃 𝑥𝑧 in Eqs. (C.4)–(C.6), the product of two different moment arms is used. • In Eqs. (C.1)–(C.6), 𝑥, 𝑦, and 𝑧 have dimensions of length, and 𝑑𝑚 has the dimension of mass. Hence, all mass moments of inertia have dimensions of (mass)(length)2 and are expressed in slug⋅f t 2 and kg⋅m2 in the U.S. Customary and SI unit systems, respectively.

ISTUDY

Appendix C

Mass Moments of Inertia

A-67

• When the 𝑥, 𝑦, and 𝑧 axes pass through the center of mass of an object, we denote these axes as 𝑥′ , 𝑦′ , and 𝑧′ , and we refer to the moments of inertia associated with these axes as mass center moments of inertia with the designations 𝐼𝐺𝑥 , 𝐼𝐺𝑦 , etc. • The quantities 𝐼𝑃 𝑥 , 𝐼𝑃 𝑦 , and 𝐼𝑃 𝑧 are never negative. The products of inertia 𝐼𝑃 𝑥𝑦 , 𝐼𝑃 𝑦𝑧 , and 𝐼𝑃 𝑥𝑧 may be positive, zero, or negative, as discussed below. • Evaluation of moments of inertia using composite shapes is possible using the parallel axis theorem, as discussed later in this appendix.

How are mass moments of inertia used? It is useful to discuss why there are six mass moments of inertia, how they differ from each other, and how they are used.

𝑦 𝑧

Moments of inertia 𝐼𝑃 𝑥 , 𝐼𝑃 𝑦 , and 𝐼𝑃 𝑧 . In Fig. C.3, the International Space Station with a docked Space Shuttle is shown. Modeling the system as rigid, if a moment 𝑀𝑥 about the 𝑥 axis is applied to it, the system will begin to undergo an angular acceleration about the 𝑥 axis. The value of the angular acceleration is proportional to the mass moment of inertia about the 𝑥 axis 𝐼𝑃 𝑥 . Furthermore, the larger 𝐼𝑃 𝑥 is, the lower the angular acceleration will be for a given value of 𝑀𝑥 . Similar remarks apply to moments applied about the 𝑦 and 𝑧 axes and the influence that moments of inertia 𝐼𝑃 𝑦 and 𝐼𝑃 𝑧 have on angular accelerations about these axes. Products of inertia 𝐼𝑃 𝑥𝑦 , 𝐼𝑃 𝑦𝑧 , and 𝐼𝑃 𝑥𝑧 . Products of inertia measure the asymmetry of a body’s mass distribution with respect to the 𝑥𝑦, 𝑦𝑧, and 𝑥𝑧 planes. Products of inertia can have a positive, zero, or negative value, depending on the shape and mass distribution of an object, the selection of the 𝑥, 𝑦, and 𝑧 axes, and the location of the origin of the 𝑥𝑦𝑧 coordinate system. Figure C.4 shows the cross section of a uniform body at some arbitrary 𝑧 coordinate. The tan shaded region shows that part of the cross section that is symmetric about the 𝑦 axis. The two mass elements 𝐴 and 𝐵 (shown in orange) are an equal distance 𝑑 from the 𝑦𝑧 plane, and both have the same 𝑦 coordinate, i.e., 𝑦 = −ℎ. Referring to the integrals for 𝐼𝑃 𝑥𝑦 and 𝐼𝑃 𝑥𝑧 in Eqs. (C.4) and (C.6), 𝑥𝑦 𝑑𝑚 and 𝑥𝑧 𝑑𝑚 for the left element 𝐴 have the opposite sign of the analogous quantities for the right element 𝐵. Therefore, the mass in the tan region does not contribute to the products of inertia 𝐼𝑃 𝑥𝑦 and 𝐼𝑃 𝑥𝑧 . By contrast, the mass in the gray shaded region has no corresponding region that is symmetric with respect to the 𝑦𝑧 plane, and since the product 𝑥𝑦 for all points in that region is positive, it contributes positively to 𝐼𝑃 𝑥𝑦 . Using a similar argument, the pink shaded region contributes negatively to 𝐼𝑃 𝑥𝑦 . If the body in Fig. C.4 had a uniform mass distribution, it would have 𝐼𝑃 𝑥𝑦 > 0 since the gray region is larger than the pink. Note that nothing can be said about the sign of 𝐼𝑃 𝑥𝑧 since we don’t know if the cross section in Fig. C.4 is at a positive or negative 𝑧 coordinate. The preceding arguments lead us to the conclusion that if the body in Fig. C.4 consisted of only the area shaded in tan, then both 𝐼𝑃 𝑥𝑦 and 𝐼𝑃 𝑥𝑧 would be zero. This argument implies that: All products of inertia containing a coordinate that is perpendicular to a plane of symmetry for a body must be zero, as long as the origin of the coordinate system lies in that plane of symmetry.

𝑥

European Space Agency

Figure C.3 The International Space Station with a Space Shuttle docked to it. 𝑦

+ 𝐺 ℎ

𝐴 𝑑

𝐵

ℎ

−

𝑥

𝑑

Figure C.4 A uniform body and a coordinate system used to understand products of inertia.

A-68

Appendix C

Mass Moments of Inertia

Figure C.5 If the mallet’s geometry and mass distribution are both symmetric about the 𝑥𝑦 plane, then the mallet is said to be a symmetric object. If the mallet is also symmetric about the 𝑦𝑧 plane, it may be called a doubly symmetric object.

For example, if an object is symmetric about the 𝑥𝑦 plane, such as the mallet shown in Fig. C.5, then 𝐼𝑃 𝑥𝑧 = 𝐼𝑃 𝑦𝑧 = 0. If an object is symmetric about at least two of the 𝑥𝑦, 𝑦𝑧, and 𝑥𝑧 planes, then all of the products of inertia are zero. For example, the products of inertia are zero for a uniform solid of revolution if one of the coordinate directions coincides with the axis of revolution. Objects that have one or more nonzero products of inertia may display complicated behavior in three-dimensional motions. Forcing such a body to undergo planar motion generally requires the application of moments along directions in that plane of motion. For example, the Space Station in Fig. C.3 is nonsymmetric about the 𝑥𝑦𝑧 axes shown, and thus it has nonzero products of inertia. If a moment 𝑀𝑥 about the 𝑥 axis is applied, the Space Station, in addition to rotating about the 𝑥 axis, will also rotate about the 𝑦 and/or 𝑧 axes (or moments about those axes will be required to prevent it from doing so).

Radius of gyration Rather than using mass moments of inertia to quantify the distribution of mass within a body, the radii of gyration are often used. For a body of mass 𝑚, the radii of gyration are directly related to the mass moments of inertia, and are defined as √ 𝑘𝑃 𝑥 =

√ 𝐼𝑃 𝑥 𝑚

,

𝑘𝑃 𝑦 =

𝐼𝑃 𝑦 𝑚

√ ,

𝑘𝑃 𝑧 =

𝐼𝑃 𝑧 𝑚

,

(C.7)

where 𝑘𝑃 𝑥 , 𝑘𝑃 𝑦 , and 𝑘𝑃 𝑧 are called the radii of gyration of the body about the 𝑥, 𝑦, and 𝑧 axes, respectively. The radii of gyration have units of length.

Parallel axis theorem 𝑚 𝐺 𝑟⃗𝑃 ∕𝐺

𝑥

𝑞⃗ 𝑑𝑚

𝑘̂ 𝚤̂ 𝑥

𝑦

𝑟⃗𝑑𝑚∕𝑃

𝑃

𝐵

𝚥̂

Figure C.6 An object with mass 𝑚 and center of mass at point 𝐺. The 𝑥 and 𝑥′ axes, the 𝑦 and 𝑦′ axes, and the 𝑧 and 𝑧′ axes are parallel to one another, respectively.

ISTUDY

The parallel axis theorem relates mass moments and products of inertia 𝐼𝑃 𝑥 , 𝐼𝑃 𝑦 , 𝐼𝑃 𝑧 , 𝐼𝑃 𝑥𝑦 , 𝐼𝑃 𝑦𝑧 , and 𝐼𝑃 𝑥𝑧 to the mass center moments and products of inertia 𝐼𝐺𝑥 , 𝐼𝐺𝑦 , 𝐼𝐺𝑧 , 𝐼𝐺𝑥𝑦 , 𝐼𝐺𝑦𝑧 , and 𝐼𝐺𝑥𝑧 . Chapters 17–20 demonstrate how important the parallel axis theorem is for the dynamics of rigid bodies. Referring to the body 𝐵 and coordinate systems in Fig. C.6, the 𝑥𝑦𝑧 axes are parallel to the 𝑥′ 𝑦′ 𝑧′ axes, respectively, the origin of the 𝑥𝑦𝑧 axes is at an arbitrary point 𝑃 , and the origin of the 𝑥′ 𝑦′ 𝑧′ system is at the mass center 𝐺 of 𝐵. Parallel axis theorem for moments of inertia Referring to Fig. C.6, we begin by noting that the mass moment of inertia of the body 𝐵 about the 𝑥 axis is given by Eq. (C.1), which is repeated here for convenience as 𝐼𝑃 𝑥 =

∫𝐵

( 2 ) 𝑦 + 𝑧2 𝑑𝑚.

(C.8)

In addition, we can write the position of a mass element 𝑑𝑚 relative to 𝐺 as 𝑞⃗ = 𝑟⃗𝑃 ∕𝐺 + 𝑟⃗𝑑𝑚∕𝑃 .

(C.9)

Writing Eq. (C.9) in component form, we obtain 𝑥′ 𝚤̂ + 𝑦′ 𝚥̂ + 𝑧′ 𝑘̂ =

[( ) ( ) ( ) ] 𝑟𝑃 ∕𝐺 𝑥 𝚤̂ + 𝑟𝑃 ∕𝐺 𝑦 𝚥̂ + 𝑟𝑃 ∕𝐺 𝑧 𝑘̂

( ) + 𝑥 𝚤̂ + 𝑦 𝚥̂ + 𝑧 𝑘̂ .

(C.10)

ISTUDY

Appendix C

A-69

Mass Moments of Inertia

Substituting Eq. (C.10) into Eq. (C.8) results in 𝐼𝑃 𝑥 = =

∫𝐵 ∫𝐵

[ ] (𝑦′ − (𝑟𝑃 ∕𝐺 )𝑦 )2 + (𝑧′ − (𝑟𝑃 ∕𝐺 )𝑧 )2 𝑑𝑚,

(C.11)

[( ( ′2 ) )2 ( )2 ] 𝑦 + 𝑧′2 𝑑𝑚 + 𝑟𝑃 ∕𝐺 𝑦 + 𝑟𝑃 ∕𝐺 𝑧 ( ) − 2 𝑟𝑃 ∕𝐺 𝑦

∫𝐵

𝑑𝑚 ∫𝐵 ( ) 𝑦′ 𝑑𝑚 − 2 𝑟𝑃 ∕𝐺 𝑧

∫𝐵

𝑧′ 𝑑𝑚.

(C.12)

The first term in Eq. (C.12) is the mass center moment of inertia about the 𝑥′ axis 𝐼𝐺𝑥′ = 𝐼𝐺𝑥 [see Eq. (C.1)]. The second integral in Eq. (C.12) is the mass 𝑚 of 𝐵. Finally, since 𝑥′ , 𝑦′ , and 𝑧′ measure the position of 𝑑𝑚 relative to 𝐺, the last two terms in Eq. (C.12) measure the position of the mass center of 𝐵 relative to 𝐺, and so they must both be zero. Therefore, Eq. (C.12) becomes [( )2 ( )2 ] (C.13) 𝐼𝑃 𝑥 = 𝐼𝐺𝑥 + 𝑚 𝑟𝑃 ∕𝐺 𝑦 + 𝑟𝑃 ∕𝐺 𝑧 .

𝑧 𝑧 𝑟⃗𝑃 ∕𝐺

)2 ( )2 ( Referring to Fig. C.7, we see that 𝑟𝑃 ∕𝐺 𝑦 + 𝑟𝑃 ∕𝐺 𝑧 is the square of the perpendicular distance 𝑑𝑥 between the 𝑥 and 𝑥′ axes, i.e., )2 ( )2 ( 𝑑𝑥2 = 𝑟𝑃 ∕𝐺 𝑦 + 𝑟𝑃 ∕𝐺 𝑧 ,

(C.14)

𝐼𝑃 𝑥 = 𝐼𝐺𝑥 + 𝑚𝑑𝑥2 .

(C.15)

𝑥

𝐺 𝑦

𝑑𝑦 (𝑟𝑃 ∕𝐺 )𝑧

𝑑𝑥 𝑃

𝑥

(𝑟𝑃 ∕𝐺 )𝑦

𝑑𝑧 (𝑟𝑃 ∕𝐺 )𝑥

𝑦

so that Eq. (C.13) becomes Similarly, substituting Eq. (C.10) into Eqs. (C.2) and (C.3), we obtain the following: [( )2 ( )2 ] 𝐼𝑃 𝑦 = 𝐼𝐺𝑦 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 + 𝑟𝑃 ∕𝐺 𝑧 = 𝐼𝐺𝑦 + 𝑚𝑑𝑦2 , (C.16) ] [( )2 ( )2 (C.17) 𝐼𝑃 𝑧 = 𝐼𝐺𝑧 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 + 𝑟𝑃 ∕𝐺 𝑦 = 𝐼𝐺𝑧 + 𝑚𝑑𝑧2 ,

Figure C.7 The 𝑥𝑦𝑧 and 𝑥′ 𝑦′ 𝑧′ axes are parallel with separation distances 𝑑𝑥 , 𝑑𝑦 , and 𝑑𝑧 , respectively.

where, referring to Fig. C.7, we have used the fact that )2 ( )2 ( 𝑑𝑦2 = 𝑟𝑃 ∕𝐺 𝑥 + 𝑟𝑃 ∕𝐺 𝑧

and

)2 ( )2 ( 𝑑𝑧2 = 𝑟𝑃 ∕𝐺 𝑥 + 𝑟𝑃 ∕𝐺 𝑦 .

(C.18)

Summarizing these results, we have the parallel axis theorem for moments of inertia, which relates the mass moments of inertia with respect to the 𝑥, 𝑦, and 𝑧 axes to the mass center moments of inertia as follows: 𝐼𝑃 𝑥 𝐼𝑃 𝑦 𝐼𝑃 𝑧

[( )2 ( )2 ] = 𝐼𝐺𝑥 + 𝑚𝑑𝑥2 = 𝐼𝐺𝑥 + 𝑚 𝑟𝑃 ∕𝐺 𝑦 + 𝑟𝑃 ∕𝐺 𝑧 , [( )2 ( )2 ] = 𝐼𝐺𝑦 + 𝑚𝑑𝑦2 = 𝐼𝐺𝑦 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 + 𝑟𝑃 ∕𝐺 𝑧 , [( )2 ( )2 ] = 𝐼𝐺𝑧 + 𝑚𝑑𝑧2 = 𝐼𝐺𝑧 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 + 𝑟𝑃 ∕𝐺 𝑦 .

𝑚

(C.19)

𝐺

(C.20)

𝚤̂

𝑥𝑦 𝑑𝑚.

(C.22)

𝑥

𝑦 𝑑𝑚

𝑘̂

Referring to Fig. C.8, an 𝑥′ 𝑦′ 𝑧′ coordinate system is defined, with origin at the center of mass 𝐺 of the object 𝐵, and the 𝑥, 𝑦, and 𝑧 axes are parallel to the 𝑥′ , 𝑦′ , and 𝑧′ axes, respectively. The product of inertia of the body 𝐵 about the 𝑥 and 𝑦 axes is ∫𝐵

𝑞⃗

(C.21)

Parallel axis theorem for products of inertia

𝐼𝑃 𝑥𝑦 =

𝑟⃗𝑃 ∕𝐺

𝑥

𝑟⃗𝑑𝑚∕𝑃

𝑃

𝐵

𝚥̂

Figure C.8 An object with mass 𝑚 and center of mass at point 𝐺. The 𝑥 and 𝑥′ axes, the 𝑦 and 𝑦′ axes, and the 𝑧 and 𝑧′ axes are parallel to one another, respectively.

A-70

ISTUDY

Appendix C

Mass Moments of Inertia Substituting in Eq. (C.10) for 𝑥 and 𝑦, Eq. (C.22) becomes 𝐼𝑃 𝑥𝑦 = =

Parallel axis theorem. Consider the body whose center of mass is at point 𝐺, as well as the parallel axes 𝑥1 , 𝑥2 , and mass center axis 𝑥′ . 𝑥

𝑥2

𝑄 𝑃

∫𝐵

) ( ) ( 𝑥′ 𝑦′ 𝑑𝑚 + 𝑟𝑃 ∕𝐺 𝑥 𝑟𝑃 ∕𝐺 𝑦 ) ( − 𝑟𝑃 ∕𝐺 𝑦

∫𝐵 ∫𝐵

(C.23)

𝑑𝑚 ( ) 𝑥′ 𝑑𝑚 − 𝑟𝑃 ∕𝐺 𝑥

∫𝐵

𝑦′ 𝑑𝑚.

(C.24)

The last two integrals are zero by definition of center of mass. The first integral is the mass center product of inertia about the 𝑥′ 𝑦′ axes, and the second integral defines the mass of the body 𝐵. Therefore, Eq. (C.24) becomes

Common Pitfall

𝑥1

∫𝐵

[ ) ][ ) ] ( ( 𝑥′ − 𝑟𝑃 ∕𝐺 𝑥 𝑦′ − 𝑟𝑃 ∕𝐺 𝑦 𝑑𝑚

) ( ) ( 𝐼𝑃 𝑥𝑦 = 𝐼𝐺𝑥𝑦 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 𝑟𝑃 ∕𝐺 𝑦 .

(C.25)

Using a similar approach, we can find 𝐼𝑃 𝑦𝑧 and 𝐼𝑃 𝑥𝑧 in terms of the mass center products of inertia. In summary, the parallel axis theorem for products of inertia is ) ( ) ( 𝐼𝑃 𝑥𝑦 = 𝐼𝐺𝑥𝑦 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 𝑟𝑃 ∕𝐺 𝑦 , ( ) ( ) 𝐼𝑃 𝑥𝑧 = 𝐼𝐺𝑥𝑧 + 𝑚 𝑟𝑃 ∕𝐺 𝑥 𝑟𝑃 ∕𝐺 𝑧 , ) ( ) ( 𝐼𝑃 𝑦𝑧 = 𝐼𝐺𝑦𝑧 + 𝑚 𝑟𝑃 ∕𝐺 𝑦 𝑟𝑃 ∕𝐺 𝑧 ,

(C.26) (C.27) (C.28)

𝐺

If 𝐼𝑃 𝑥 is known, a common error is to use the parallel axis theorem to directly determine 𝐼𝑄𝑥 . In the parallel axis theorem, one of the axes is always a mass center axis. Thus, with 𝐼𝑃 𝑥 known, the parallel axis theorem must be employed twice, the first time to determine 𝐼𝐺𝑥 from 𝐼𝑃 𝑥 , and the second time to use 𝐼𝐺𝑥 to determine 𝐼𝑄𝑥 .

Helpful Information The column matrix representation of a vector. In representing the vector 𝜔 ⃗ 𝐵 as ⎧𝜔 ⎫ ⎪ 𝐵𝑥 ⎪ {𝜔𝐵 } = ⎨𝜔𝐵𝑦 ⎬ ⎪𝜔𝐵𝑧 ⎪ ⎭ ⎩ the first row contains the 𝑥 component, the second row the 𝑦 component, and the third row the 𝑧 component of the vector. The same interpretation applies to {ℎ𝐺 }. When we multiply [𝐼𝐺 ] by {𝜔𝐵 }, we obtain the vector {ℎ𝐺 } whose first, second, and third rows are the 𝑥, 𝑦, and 𝑧 components of the vector, respectively.

where we recall that 𝐼𝐺𝑥𝑦 , 𝐼𝐺𝑥𝑧 , and 𝐼𝐺𝑦𝑧 are the mass center products of inertia and ) ( ) ) ( ( 𝑟𝑃 ∕𝐺 𝑥 , 𝑟𝑃 ∕𝐺 𝑦 , and 𝑟𝑃 ∕𝐺 𝑧 are defined in Eq. (C.10) and can be seen in Fig. C.7.

Principal moments of inertia The moments and products of inertia for a rigid body 𝐵 depend on both the origin and orientation of the coordinate axes used to compute them. For a given origin, we can find a unique orientation of the coordinate axes, such that all the products of inertia are zero when computed with respect to those axes. These special axes are called the principal axes of inertia, and the corresponding moments of inertia are called the principal moments of inertia 𝐼̄𝑥 , 𝐼̄𝑦 , and 𝐼̄𝑧 . Let’s see how we find these axes and the corresponding moments of inertia. Equation (D.16) in Appendix D states that the angular momentum of a rigid body ⃗ , can be written as with respect to its mass center 𝐺, that is ℎ 𝐺 ( ) ⃗ = 𝐼 𝜔 −𝐼 𝜔 −𝐼 𝜔 ℎ 𝐺 𝐺𝑥 𝐵𝑥 𝐺𝑥𝑦 𝐵𝑦 𝐺𝑥𝑧 𝐵𝑧 𝚤̂ ( ) + −𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 + 𝐼𝐺𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑧 𝚥̂ ) ( ̂ + −𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 + 𝐼𝐺𝑧 𝜔𝐵𝑧 𝑘, = {ℎ𝐺 } = [𝐼𝐺 ]{𝜔𝐵 },

(C.29) (C.30)

⃗ , and [𝐼 ] where {𝜔𝐵 } = 𝜔 ⃗ 𝐵 is the angular velocity of the rigid body 𝐵, {ℎ𝐺 } = ℎ 𝐺 𝐺 is the inertia tensor or inertia matrix, which is written as ⎡ 𝐼 ⎢ 𝐺𝑥 [𝐼𝐺 ] = ⎢−𝐼𝐺𝑥𝑦 ⎢−𝐼 ⎣ 𝐺𝑥𝑧

−𝐼𝐺𝑥𝑦 𝐼𝐺𝑦 −𝐼𝐺𝑦𝑧

−𝐼𝐺𝑥𝑧 ⎤ ⎥ −𝐼𝐺𝑦𝑧 ⎥ . 𝐼𝐺𝑧 ⎥⎦

(C.31)

ISTUDY

Appendix C

Mass Moments of Inertia

A-71

We now assume that the rigid body is spinning about just one of its principal axes ̄ If this is the case, then the angular momentum of inertia whose moment of inertia is 𝐼. can be written as ⃗ = 𝐼̄𝜔 ̂ ̄ ̄ ̄ ℎ ⃗ 𝐵 = 𝐼𝜔 𝐺 𝐵𝑥 𝚤̂ + 𝐼𝜔𝐵𝑦 𝚥̂ + 𝐼𝜔𝐵𝑧 𝑘.

(C.32)

⃗ given in Eq. (C.29) and Eq. (C.32) must be equal to one anThe expressions for ℎ 𝐺 other. Since the Cartesian basis vectors are linearly independent, we can equate the coefficients of 𝚤̂, 𝚥̂, and 𝑘̂ in these two equations to obtain a linear system of equations in the unknowns 𝜔𝐵𝑥 , 𝜔𝐵𝑦 , and 𝜔𝐵𝑧 , which can be expressed in matrix notation as ⎡𝐼 − 𝐼̄ ⎢ 𝐺𝑥 ⎢ −𝐼𝐺𝑥𝑦 ⎢ −𝐼 ⎣ 𝐺𝑥𝑧

−𝐼𝐺𝑥𝑦 𝐼𝐺𝑦 − 𝐼̄ −𝐼𝐺𝑦𝑧

−𝐼𝐺𝑥𝑧 ⎤ ⎧𝜔𝐵𝑥 ⎫ ⎧0⎫ ⎥⎪ ⎪ ⎪ ⎪ −𝐼𝐺𝑦𝑧 ⎥ ⎨𝜔𝐵𝑦 ⎬ = ⎨0⎬ . ⎪ ⎪0⎪ 𝐼𝐺𝑧 − 𝐼̄⎥⎦ ⎪ ⎩𝜔𝐵𝑧 ⎭ ⎩ ⎭

(C.33)

This system of equations has a nonzero solution if and only if the determinant of the coefficient matrix is zero, which implies that ( ) ( 2 2 2 + 𝐼𝐺𝑥𝑧 + 𝐼𝐺𝑦𝑧 − 𝐼𝐺𝑥 𝐼𝐺𝑦 𝐼 − 𝐼𝐺𝑥 + 𝐼𝐺𝑦 + 𝐼𝐺𝑧 𝐼̄2 − 𝐼𝐺𝑥𝑦 ) ( 2 2 2 − 𝐼𝐺𝑥 𝐼𝐺𝑧 − 𝐼𝐺𝑦 𝐼𝐺𝑧 𝐼̄ + 𝐼𝐺𝑥 𝐼𝐺𝑦𝑧 + 𝐼𝐺𝑦 𝐼𝐺𝑥𝑧 + 𝐼𝐺𝑧 𝐼𝐺𝑥𝑦 ) − 𝐼𝐺𝑥 𝐼𝐺𝑦 𝐼𝐺𝑧 + 2𝐼𝐺𝑥𝑦 𝐼𝐺𝑦𝑧 𝐼𝐺𝑥𝑧 = 0. ̄3

(C.34)

Interesting Fact Finding principal moments of inertia and their directions is an eigenvalue problem. If you have studied linear or matrix algebra, you may recognize that the principal moments of inertia are the eigenvalues of the [𝐼𝐺 ] matrix and the corresponding directions are its eigenvectors.

Equation (C.34) is called the characteristic equation. The three roots of this equation represent the three principal moments of inertia 𝐼̄𝐺1 , 𝐼̄𝐺2 , and 𝐼̄𝐺3 . The corresponding directions of those principal moments of inertia are found by substituting each in turn into ⎡𝐼 − 𝐼̄ 𝐺𝑖 ⎢ 𝐺𝑥 ⎢ −𝐼𝐺𝑥𝑦 ⎢ −𝐼 𝐺𝑥𝑧 ⎣

−𝐼𝐺𝑥𝑦 𝐼𝐺𝑦 − 𝐼̄𝐺𝑖 −𝐼𝐺𝑦𝑧

⎤ ⎧𝑥 ⎫ ⎧0⎫ ⎥ ⎪ 𝑖⎪ ⎪ ⎪ ⎥ ⎨𝑦𝑖 ⎬ = ⎨0⎬ , ⎥ ⎪𝑧 ⎪ ⎪0⎪ 𝐺𝑖 ⎦ ⎩ 𝑖 ⎭ ⎩ ⎭

−𝐼𝐺𝑥𝑧 −𝐼𝐺𝑦𝑧 𝐼 − 𝐼̄ 𝐺𝑧

𝑖 = 1, 2, 3

(C.35)

which then represents three equations in terms of the unknowns 𝑥𝑖 , 𝑦𝑖 , and 𝑧𝑖 . Each of the three solutions are then the components of a vector in the principal direction corresponding to the principal moment of inertia used to obtain it. Mini-Example Given the system of particles lying in the 𝑧 = 0 plane in Fig. C.9, determine the principal moments of inertia and their corresponding directions. Since the system is discrete, the moments and products of inertia are computed using sums instead of integrals as 𝐼𝑂𝑥 =

4 ∑ 𝑖=1

𝐼𝑂𝑦 =

4 ∑ 𝑖=1

2 4

2 −2

(C.36)

( ) 𝑚𝑖 𝑥2𝑖 + 𝑧2𝑖

= 𝑚(𝓁)2 + 2𝑚(−2𝓁)2 + 3𝑚(−2𝓁)2 + 4𝑚(2𝓁)2 = 37𝑚𝓁 2 ,

3

0

( ) 𝑚𝑖 𝑦2𝑖 + 𝑧2𝑖

= 𝑚(−2𝓁)2 + 2𝑚(−𝓁)2 + 3𝑚(4𝓁)2 + 4𝑚(𝓁)2 = 58𝑚𝓁 2 ,

4

(C.37)

−2

0

2

Figure C.9 Four point masses in the 𝑧 = 0 plane. The coordinates are dimensionless, and the masses are in terms of an arbitrary 𝑚.

A-72

ISTUDY

Appendix C

Mass Moments of Inertia

𝐼𝑂𝑧 =

4 ∑

( ) 𝑚𝑖 𝑥2𝑖 + 𝑦2𝑖

𝑖=1 [ ] [ ] = 𝑚 (𝓁)2 + (−2𝓁)2 + 2𝑚 (−2𝓁)2 + (−𝓁)2 [ ] [ ] + 3𝑚 (−2𝓁)2 + (4𝓁)2 + 4𝑚 (2𝓁)2 + (𝓁)2 = 95𝑚𝓁 2 ,

𝐼𝑂𝑥𝑦 =

4 ∑

(C.38)

𝑚𝑖 𝑥𝑖 𝑦𝑖

𝑖=1

= 𝑚(𝓁)(−2𝓁) + 2𝑚(−2𝓁)(−𝓁) + 3𝑚(−2𝓁)(4𝓁) + 4𝑚(2𝓁)(𝓁) = −14𝑚𝓁 2 ,

(C.39)

𝐼𝑂𝑥𝑧 =

4 ∑

𝑚𝑖 𝑥𝑖 𝑧𝑖 = 0,

(C.40)

𝐼𝑂𝑦𝑧 =

𝑖=1 4 ∑

𝑚𝑖 𝑦𝑖 𝑧𝑖 = 0.

(C.41)

𝑖=1

Substituting the results from Eqs. (C.36)–(C.41) into Eq. (C.31) as

Eq. (C.34) becomes

⎡58𝑚𝓁 2 [𝐼𝑂 ] = ⎢14𝑚𝓁 2 ⎢ ⎣ 0

14𝑚𝓁 2 37𝑚𝓁 2 0

0 ⎤ 0 ⎥, ⎥ 95𝑚𝓁 2 ⎦

𝐼̄3 − 190𝑚𝓁 2 𝐼̄2 + 10,975𝑚2 𝓁 4 𝐼̄ − 185,250𝑚3 𝓁 6 = 0.

(C.42)

(C.43)

The three roots of this equation are 𝐼̄𝑂1 = 95𝑚𝓁 2 ,

𝐼̄𝑂2 = 65𝑚𝓁 2 ,

and

𝐼̄𝑂3 = 30𝑚𝓁 2 ,

(C.44)

which are the three principal moments of inertia. Substituting 𝐼̄𝑂1 and Eq. (C.42) into Eq. (C.35), we obtain −37𝑚𝓁 2 𝑥1 + 14𝑚𝓁 2 𝑦1 = 0, 2

2

14𝑚𝓁 𝑥1 − 58𝑚𝓁 𝑦1 = 0, 0𝑧1 = 0,

(C.45) (C.46) (C.47)

the only solution of which is 𝑥1 = 𝑦1 = 0 and 𝑧1 is arbitrary. Therefore, the principal direction corresponding to the principal moment of inertia 𝐼̄𝑂1 is the 𝑧 direction (the blue axis in Fig. C.10). Similarly substituting 𝐼̄𝑂2 , we obtain −7𝑚𝓁 2 𝑥2 + 14𝑚𝓁 2 𝑦2 = 0, 2

(C.48)

2

(C.49)

2

(C.50)

14𝑚𝓁 𝑥2 − 28𝑚𝓁 𝑦2 = 0, 30𝑚𝓁 𝑧2 = 0, 1 𝑥 2 2

the solution of which is 𝑦2 = and 𝑧2 = 0. This direction is shown in Fig. C.10 as the line labeled 𝐼̄𝑂2 . Finally, similarly substituting 𝐼̄𝑂3 , we obtain 28𝑚𝓁 2 𝑥3 + 14𝑚𝓁 2 𝑦3 = 0, 2

2

(C.51)

14𝑚𝓁 𝑥3 + 7𝑚𝓁 𝑦3 = 0,

(C.52)

65𝑚𝓁 2 𝑧3 = 0,

(C.53)

the solution of which is 𝑦3 = −2𝑥3 and 𝑧3 = 0. This direction is shown in Fig. C.10 as the line labeled 𝐼̄𝑂3 . Notice that the principal directions are mutually orthogonal and, in this case, they happen to go through the point masses themselves.

ISTUDY

Appendix C

Mass Moments of Inertia

4𝓁

3𝑚 (−2𝓁, 4𝓁, 0) (−2

2𝓁

A-73

𝑦

𝑦 0

4𝑚 (2𝓁, 1𝓁, 0) (2

𝐼̄𝑂1

̄ 𝐼̄𝑂2

−2𝓁 𝑂 𝑧

𝑚 m (1 −2𝓁, (1𝓁, −2 0)

2𝑚 (−2 −1 (−2𝓁, −1𝓁, 0)

1𝓁 0 −1𝓁

𝑥

𝐼̄𝑂3

−2𝓁 0

Figure C.10. The locations of the four masses showing the three principal directions.

Moment of inertia about an arbitrary axis

𝑚

𝓁

Referring to Fig. C.11, suppose we know all the components of the inertia tensor relative to the 𝑥𝑦𝑧 axes shown and that we wish to find the moment of inertia of the body about the arbitrarily oriented axis 𝓁. The direction of 𝓁 is defined by the unit vector 𝑢̂ 𝓁 . We know that the definition of mass moment of inertia states the moment of inertia about the 𝓁 axis is given by 𝐼𝑃 𝓁 =

∫𝐵

( )2 |⃗𝑟𝑑𝑚∕𝑃 | sin 𝜃 𝑑𝑚,

|⃗𝑟𝑑𝑚∕𝑃 × 𝑢̂ 𝓁 | = |⃗𝑟𝑑𝑚∕𝑃 ||𝑢̂ 𝓁 | sin 𝜃 = |⃗𝑟𝑑𝑚∕𝑃 | sin 𝜃,

(C.55)

)2 ( ) ( ) ( |⃗𝑟𝑑𝑚∕𝑃 | sin 𝜃 = 𝑟⃗𝑑𝑚∕𝑃 × 𝑢̂ 𝓁 ⋅ 𝑟⃗𝑑𝑚∕𝑃 × 𝑢̂ 𝓁 ,

(C.56)

and therefore that

then we can write the integral in Eq. (C.54) as ∫𝐵

(

) ( ) 𝑟⃗𝑑𝑚∕𝑃 × 𝑢̂ 𝓁 ⋅ 𝑟⃗𝑑𝑚∕𝑃 × 𝑢̂ 𝓁 𝑑𝑚.

(C.57)

If we now write 𝑟⃗𝑑𝑚∕𝑃 and 𝑢̂ 𝓁 in component form as ̂ 𝑟⃗𝑑𝑚∕𝑃 = 𝑥 𝚤̂ + 𝑦 𝚥̂ + 𝑧 𝑘, ̂ 𝑢̂ 𝓁 = 𝑢𝓁𝑥 𝚤̂ + 𝑢𝓁𝑦 𝚥̂ + 𝑢𝓁𝑧 𝑘,

(C.58) (C.59)

so that ( ) ( ) ( ) ̂ 𝑟⃗𝑑𝑚∕𝑃 × 𝑢̂ 𝓁 = 𝑢𝓁𝑧 𝑦 − 𝑢𝓁𝑦 𝑧 𝚤̂ + 𝑢𝓁𝑥 𝑧 − 𝑢𝓁𝑧 𝑥 𝚥̂ + 𝑢𝓁𝑦 𝑥 − 𝑢𝓁𝑥 𝑦 𝑘, then Eq. (C.57) becomes [( )2 ( )2 ( )2 ] 𝑑𝑚 𝑢𝓁𝑧 𝑦 − 𝑢𝓁𝑦 𝑧 + 𝑢𝓁𝑥 𝑧 − 𝑢𝓁𝑧 𝑥 + 𝑢𝓁𝑦 𝑥 − 𝑢𝓁𝑥 𝑦 𝐼𝑃 𝓁 = ∫𝐵

(C.60)

(C.61)

𝑑𝑚

𝑢̂ 𝓁 𝜃

(C.54)

where 𝜃 is the angle between the positive directions of the 𝑟⃗𝑑𝑚∕𝑃 and 𝑢̂ 𝓁 vectors and |⃗𝑟𝑑𝑚∕𝑃 | sin 𝜃 is the perpendicular distance from 𝓁 to 𝑑𝑚. If we now notice that

𝐼𝑃 𝓁 =

𝑟⃗𝑑𝑚∕𝑃 sin 𝜃

𝑟⃗𝑑𝑚∕𝑃

𝑃 𝑥

𝑦

Figure C.11 The quantities needed to find the moment of inertia of a body 𝐵 about an arbitrarily oriented axis 𝓁.

A-74

ISTUDY

Appendix C

Mass Moments of Inertia

= 𝑢2𝓁𝑥

∫𝐵

( 2 ) 𝑦 + 𝑧2 𝑑𝑚 + 𝑢2𝓁𝑦

− 2𝑢𝓁𝑥 𝑢𝓁𝑦

∫𝐵

∫𝐵

( 2 ) 𝑥 + 𝑧2 𝑑𝑚 + 𝑢2𝓁𝑧

𝑥𝑦 𝑑𝑚 − 2𝑢𝓁𝑥 𝑢𝓁𝑧

∫𝐵

∫𝐵

( 2 ) 𝑥 + 𝑦2 𝑑𝑚

𝑥𝑧 𝑑𝑚 − 2𝑢𝓁𝑦 𝑢𝓁𝑧

∫𝐵

𝑦𝑧 𝑑𝑚.

(C.62)

The integrals in Eq. (C.62) are the known components of the inertia tensor relative to the 𝑥𝑦𝑧 axes, so we can write Eq. (C.62) as 𝐼𝑃 𝓁 = 𝑢2𝓁𝑥 𝐼𝑃 𝑥 + 𝑢2𝓁𝑦 𝐼𝑃 𝑦 + 𝑢2𝓁𝑧 𝐼𝑃 𝑧 − 2𝑢𝓁𝑥 𝑢𝓁𝑦 𝐼𝑃 𝑥𝑦 − 2𝑢𝓁𝑥 𝑢𝓁𝑧 𝐼𝑃 𝑥𝑧 − 2𝑢𝓁𝑦 𝑢𝓁𝑧 𝐼𝑃 𝑦𝑧 .

(C.63)

Thus, if we know the moments and products of inertia for a body 𝐵 about a given set of axes 𝑥𝑦𝑧 (i.e., the inertia tensor relative to these axes), we can find the moment of inertia of that body about an arbitrarily oriented axis 𝓁 using Eq. (C.63), where we note that the components of the unit vector 𝑢̂ 𝓁 , namely, 𝑢𝓁𝑥 , 𝑢𝓁𝑦 , and 𝑢𝓁𝑧 , are also called the direction cosines of 𝑢̂ 𝓁 relative to the 𝑥, 𝑦, and 𝑧 axes, respectively.

Evaluation of moments of inertia using composite shapes The parallel axis theorem written for composite shapes is 𝐼𝑥 =

𝑛 ∑ ( ) 𝐼𝐺𝑥 + 𝑑𝑥2 𝑚 𝑖 ,

(C.64)

𝑖=1

where 𝑛 is the number of shapes, 𝐼𝐺𝑥 is the mass moment of inertia for shape 𝑖 about its mass center 𝑥′ axis, 𝑑𝑥 is the shift distance for shape 𝑖 (i.e., the distance between the 𝑥 axis and the 𝑥′ axis for shape 𝑖), and 𝑚 is the mass for shape 𝑖. Similar expressions may be written for 𝐼𝑦 and 𝐼𝑧 . To use Eq. (C.64), it is necessary to know the mass moment of inertia for each of the composite shapes about its mass center axis, and this generally must be obtained by integration or, when possible, by consulting a table of moments of inertia for common shapes, such as the Table of Properties of Solids after the Index at end of this book. A common error is to use the parallel axis theorem to relate moments of inertia between two parallel axes where neither of them is a mass center axis.

ISTUDY

D

Angular Momentum of a Rigid Body

In Section 18.2, we used the angular momentum of a rigid body to derive the angular impulse-momentum principle for a rigid body. We will now show how we obtained Eq. (18.40) on p. 1244 for the angular momentum of a rigid body. Along the way, we will derive the angular momentum for a rigid body in three-dimensional motion since that result will be useful in Chapter 20.

NASA

Hurricane Isabelle on September 15, 2003, as seen from the International Space Station. Even though the air particles near the eye of a hurricane have a small moment arm relative to the center of the eye, they contribute significantly to the total angular momentum because they have the highest speed.

⃗ , as given by Eq. (18.40) We now obtain the angular momentum of a rigid body ℎ 𝑃 on p. 1244, which we repeat here as ⃗ =𝐼 𝜔 ⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , ℎ 𝑃 𝐺 ⃗𝐵 + 𝑟

𝑣⃗𝑃 𝑃

(D.1) 𝑟⃗𝑃

where 𝐼𝐺 and 𝜔 ⃗ 𝐵 are the body’s mass moment of inertia and angular velocity, respectively. We could derive Eq. (D.1) directly, but it will be useful to derive the general three-dimensional form of the angular momentum of a rigid body and then simplify the result to Eq. (D.1) since the general result will be useful in Chapter 20.

𝑞⃗

𝐺

Angular momentum of a rigid body undergoing three-dimensional motion

𝑟⃗𝐺

Referring to Fig. D.1, recall that the angular momentum of a rigid body 𝐵 about the arbitrary point 𝑃 was defined in Eq. (17.6) on p. 1146 to be ⃗ = ℎ 𝑃

∫𝐵

𝑟⃗𝑑𝑚∕𝑃 × 𝑣⃗𝑑𝑚 𝑑𝑚.

(D.2)

The vectors 𝑟⃗𝑑𝑚∕𝑃 and 𝑣⃗𝑑𝑚 indicate the position of the infinitesimal mass element 𝑑𝑚 relative to 𝑃 and the velocity of 𝑑𝑚, respectively. Letting 𝑞⃗ indicate the position of

𝑟⃗𝐺∕𝑃

𝑂

𝑣⃗𝐺 𝐵

𝜔 ⃗𝐵

Figure D.1 A rigid body in motion with the definitions of the vectors needed to characterize the body’s angular momentum.

A-75

A-76

Appendix D

Angular Momentum of a Rigid Body

𝑑𝑚 relative to the mass center 𝐺, and noting that 𝐺 and 𝑑𝑚 are two points on the same rigid body, we can write 𝑟⃗𝑑𝑚∕𝑃 = 𝑞⃗ + 𝑟⃗𝐺∕𝑃

and

𝑣⃗𝑑𝑚 = 𝑣⃗𝐺 + 𝜔 ⃗ 𝐵 × 𝑞. ⃗

(D.3)

Substituting Eqs. (D.3) into Eq. (D.2) and expanding the cross products, we have ⃗ = ℎ 𝑃

∫𝐵

∫𝐵

( ) 𝑞⃗ × 𝜔 ⃗ 𝐵 × 𝑞⃗ 𝑑𝑚 + +

𝑞⃗ × 𝑣⃗𝐺 𝑑𝑚

∫𝐵

( ) 𝑟⃗𝐺∕𝑃 × 𝜔 ⃗ 𝐵 × 𝑞⃗ 𝑑𝑚 +

∫𝐵

𝑟⃗𝐺∕𝑃 × 𝑣⃗𝐺 𝑑𝑚. (D.4)

To simplify the right-hand side of Eq. (D.4), recall that 𝑞⃗ measures the position of each mass element 𝑑𝑚 relative to the mass center 𝐺 of the body. Therefore, the definition of center of mass requires that ∫𝐵

⃗ 𝑞⃗ 𝑑𝑚 = 𝑚⃗𝑟𝐺∕𝐺 = 0.

(D.5)

Since the quantities 𝑣⃗𝐺 , 𝑟⃗𝐺∕𝑃 , and 𝜔 ⃗ 𝐵 are not a function of position within 𝐵, the last three terms on the right-hand side of Eq. (D.4) become (

∫𝐵

𝑟⃗𝐺∕𝑃

and ∫𝐵 𝑃

⃗ = ℎ 𝑃

𝑘̂ 𝚤̂

𝑣⃗𝐺 𝐺

𝑦

𝚥̂

𝜔 ⃗𝐵

Figure D.2 ⃗ Vector definitions needed in the derivation of ℎ 𝑃 for a rigid body.

ISTUDY

∫𝐵

𝑑𝑚 = 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 .

(D.7)

(D.8)

∫𝐵

( ) 𝑞⃗ × 𝜔 ⃗ 𝐵 × 𝑞⃗ 𝑑𝑚 + 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 .

(D.9)

To interpret the integral on the right side of Eq. (D.9), write 𝑞⃗ and 𝜔 ⃗ 𝐵 in Cartesian components as (see Fig. D.2)

𝐵 𝑥

( ) 𝑟⃗𝐺∕𝑃 × 𝑣⃗𝐺 𝑑𝑚 = 𝑟⃗𝐺∕𝑃 × 𝑣⃗𝐺

(D.6)

Substituting Eqs. (D.6)–(D.8) into Eq. (D.4) then gives

𝑟⃗𝐺∕𝑃

𝑞⃗

) ⃗ 𝑞⃗ × 𝑣⃗𝐺 𝑑𝑚 = 𝑞⃗ 𝑑𝑚 × 𝑣⃗𝐺 = 0, ∫𝐵 ∫𝐵 [ ( )] ( ) ⃗ × 𝜔 ⃗ 𝐵 × 𝑞⃗ 𝑑𝑚 = 𝑟⃗𝐺∕𝑃 × 𝜔 ⃗𝐵 × 𝑞⃗ 𝑑𝑚 = 0, ∫𝐵

𝑞⃗ = 𝑞𝑥 𝚤̂ + 𝑞𝑦 𝚥̂ + 𝑞𝑧 𝑘̂

and

̂ 𝜔 ⃗ 𝐵 = 𝜔𝐵𝑥 𝚤̂ + 𝜔𝐵𝑦 𝚥̂ + 𝜔𝐵𝑧 𝑘,

(D.10)

and then substitute them into the integral on the right side of Eq. (D.9) to obtain ∫𝐵

( ) 𝑞⃗ × 𝜔 ⃗ 𝐵 × 𝑞⃗ 𝑑𝑚 =

∫𝐵

[( ] ) 𝑞𝑦2 + 𝑞𝑧2 𝜔𝐵𝑥 − 𝑞𝑥 𝑞𝑦 𝜔𝐵𝑦 − 𝑞𝑥 𝑞𝑧 𝜔𝐵𝑧 𝑑𝑚 𝚤̂

+

∫𝐵

[

+

∫𝐵

[

] ( ) −𝑞𝑥 𝑞𝑦 𝜔𝐵𝑥 + 𝑞𝑥2 + 𝑞𝑧2 𝜔𝐵𝑦 − 𝑞𝑦 𝑞𝑧 𝜔𝐵𝑧 𝑑𝑚 𝚥̂

] ( ) ̂ −𝑞𝑥 𝑞𝑧 𝜔𝐵𝑥 − 𝑞𝑦 𝑞𝑧 𝜔𝐵𝑦 + 𝑞𝑥2 + 𝑞𝑦2 𝜔𝐵𝑧 𝑑𝑚 𝑘.

(D.11)

Bringing the components of 𝜔 ⃗ 𝐵 outside of the integrals since they are not a function of position within 𝐵, and then using the definitions of moment and product of

ISTUDY

Appendix D

Angular Momentum of a Rigid Body

A-77

inertia given in Eqs. (C.1)–(C.6) on pp. A-65–A-66, we obtain ( ) ( ) 𝑞⃗ × 𝜔 ⃗ 𝐵 × 𝑞⃗ 𝑑𝑚 = 𝐼𝐺𝑥 𝜔𝐵𝑥 − 𝐼𝐺𝑥𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑥𝑧 𝜔𝐵𝑧 𝚤̂ ∫𝐵 ( ) + −𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 + 𝐼𝐺𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑧 𝚥̂ ( ) ̂ + −𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 + 𝐼𝐺𝑧 𝜔𝐵𝑧 𝑘,

(D.12)

where the moments and products of inertia are with respect to the mass center 𝐺 since 𝑞⃗ measures the position of 𝑑𝑚 with respect to 𝐺. Writing 𝑟⃗𝐺∕𝑃 and 𝑣⃗𝐺 in Cartesian components as ̂ 𝑟⃗𝐺∕𝑃 = 𝑥𝐺∕𝑃 𝚤̂ + 𝑦𝐺∕𝑃 𝚥̂ + 𝑧𝐺∕𝑃 𝑘,

(D.13)

̂ 𝑣⃗𝐺 = 𝑣𝐺𝑥 𝚤̂ + 𝑣𝐺𝑦 𝚥̂ + 𝑣𝐺𝑧 𝑘,

(D.14)

and then substituting Eqs. (D.12)–(D.14) into Eq. (D.9), the angular momentum of a rigid body is given by ( ) ⃗ = 𝐼 𝜔 −𝐼 𝜔 −𝐼 𝜔 ℎ 𝑃 𝐺𝑥 𝐵𝑥 𝐺𝑥𝑦 𝐵𝑦 𝐺𝑥𝑧 𝐵𝑧 𝚤̂ ) ( + −𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 + 𝐼𝐺𝑦 𝜔𝐵𝑦 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑧 𝚥̂ ) ( + −𝐼 𝜔 − 𝐼 𝜔 + 𝐼 𝜔 𝑘̂ + 𝑟⃗ 𝐺𝑥𝑧 𝐵𝑥

𝐺𝑦𝑧 𝐵𝑦

𝐺𝑧 𝐵𝑧

𝐺∕𝑃

× 𝑚𝑣⃗𝐺 .

(D.15)

If any of the following conditions is satisfied: ⃗ 1. The reference point 𝑃 is the mass center 𝐺 of the rigid body so that 𝑟⃗𝐺∕𝑃 = 0. ⃗ 2. The mass center 𝐺 of the rigid body is a fixed point so that 𝑣⃗𝐺 = 0. 3. The velocity of the mass center 𝐺, 𝑣⃗𝐺 , is parallel to the position of 𝐺 relative to 𝑃 , 𝑟⃗𝐺∕𝑃 . then Eq. (D.15) simplifies to ( ) ( ⃗ = 𝐼 𝜔 −𝐼 𝜔 −𝐼 𝜔 ℎ 𝑃 𝐺𝑥 𝐵𝑥 𝐺𝑥𝑦 𝐵𝑦 𝐺𝑥𝑧 𝐵𝑧 𝚤̂ + −𝐼𝐺𝑥𝑦 𝜔𝐵𝑥 + 𝐼𝐺𝑦 𝜔𝐵𝑦 ) ( ) ̂ − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑧 𝚥̂ + −𝐼𝐺𝑥𝑧 𝜔𝐵𝑥 − 𝐼𝐺𝑦𝑧 𝜔𝐵𝑦 + 𝐼𝐺𝑧 𝜔𝐵𝑧 𝑘, ̂ = ℎ𝑃 𝑥 𝚤̂ + ℎ𝑃 𝑦 𝚥̂ + ℎ𝑃 𝑧 𝑘.

(D.16) (D.17)

Conditions 1 and 2 are very common in practice, so Eq. (D.16) is frequently used.

𝑦

𝑦

Practical use of Eqs. (D.15) and (D.16) The 𝑥𝑦𝑧 reference frame referred to in Eqs. (D.15) and (D.16) can be any frame (inertial or not) as long as velocity and angular velocity components are computed with respect to an inertial frame. With that in mind, if a body rotates relative to the 𝑥𝑦𝑧 frame, the moments and products of inertia in Eqs. (D.15) and (D.16) will be time dependent. Referring to Fig. D.3, notice that as the rectangular body moves in the 𝑥𝑦 plane, its distribution of mass changes relative to the nonmoving 𝑥𝑦𝑧 reference frame, and therefore its moments and products of inertia will be time dependent relative to that frame. On the other hand, the moments and products of inertia relative to the 𝑥′𝑦′𝑧′ frame that is attached to the body will be constant. We will see in applications that attaching a reference frame to the body will be the preferred strategy for solving problems. We will refer to these frames as body-fixed.

𝑥

𝑦

𝐴 𝑥 𝑦

𝐵 𝑂

𝑥

𝑥

Figure D.3 A rectangular rigid body rotating in planar motion relative to the fixed 𝑥𝑦 frame. The 𝑥′𝑦′ is attached to the body.

A-78

ISTUDY

Appendix D

Angular Momentum of a Rigid Body A compact way to write Eqs. (D.15) and (D.16)

Using a combination of matrix and vector notation, Eq. (D.15) can be written as {

Helpful Information The column matrix representation of a ⃗ as vector. In representing the vector ℎ 𝑃 {

ℎ𝑃

}

⎧ℎ ⎫ }𝑇 ⎪ 𝑃 𝑥⎪ { = ⎨ ℎ𝑃 𝑦 ⎬ = ℎ 𝑃 𝑥 ℎ𝑃 𝑦 ℎ𝑃 𝑧 , ⎪ ℎ𝑃 𝑧 ⎪ ⎩ ⎭

the first row in the middle expression contains the 𝑥 component, the second row the 𝑦 component, and the third row the 𝑧 component of the vector. The representation on the right is the same column matrix, where the “𝑇 ” represents the transpose to turn the row matrix into the column{matrix. } The same interpretation applies to 𝜔𝐵 . Similarly, when [ ] { } we multiply 𝐼𝐺 by 𝜔𝐵 , we obtain a vector whose first, second, and third rows are the 𝑥, 𝑦, and 𝑧 components of the vector, respectively.

ℎ𝑃

}

[ ]{ } = 𝐼𝐺 𝜔𝐵 + 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 ,

(D.18)

where {ℎ𝑃 } is the angular momentum vector of the body 𝐵 with respect to 𝑃 , {𝜔𝐵 } is the angular velocity vector of 𝐵, and [𝐼𝐺 ] is the inertia matrix or inertia tensor of 𝐵 with respect to the 𝑥𝑦𝑧 axes. In matrix form and vector form, these quantities are given by { } { }𝑇 ⃗ = ℎ 𝚤̂ + ℎ 𝚥̂ + ℎ 𝑘, ̂ ℎ𝑃 = ℎ 𝑃 𝑥 ℎ 𝑃 𝑦 ℎ 𝑃 𝑧 = ℎ (D.19) 𝑃 𝑃𝑥 𝑃𝑦 𝑃𝑧 { } { }𝑇 ̂ (D.20) 𝜔𝐵 = 𝜔𝐵𝑥 𝜔𝐵𝑦 𝜔𝐵𝑧 = 𝜔 ⃗ 𝐵 = 𝜔𝐵𝑥 𝚤̂ + 𝜔𝐵𝑦 𝚥̂ + 𝜔𝐵𝑧 𝑘, and

⎡ 𝐼 −𝐼𝐺𝑥𝑦 −𝐼𝐺𝑥𝑧 ⎤ [ ] ⎢ 𝐺𝑥 ⎥ 𝐼𝐺𝑦 −𝐼𝐺𝑦𝑧 ⎥ . 𝐼𝐺 = ⎢−𝐼𝐺𝑥𝑦 ⎢−𝐼 𝐼𝐺𝑧 ⎥⎦ ⎣ 𝐺𝑥𝑧 −𝐼𝐺𝑦𝑧 { } [ ] Standard matrix multiplication of 𝐼𝐺 and 𝜔𝐵 , with the addition of the vector cross product 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , then gives {ℎ𝑃 } as given in either Eq. (D.15) or (D.18). With this as background, we see that Eq. (D.16) can be written much more compactly as { } [ ]{ } ℎ𝑃 = 𝐼𝐺 𝜔𝐵 . (D.21)

Angular momentum of a rigid body in planar motion We are now in a position to develop Eq. (D.1), which we recall is the angular momentum of a rigid body in planar motion that is symmetric with respect to the plane of motion. If the rigid body 𝐵 is in planar motion and the motion is occurring in the 𝑥𝑦 plane, then (D.22) 𝜔𝐵𝑥 = 𝜔𝐵𝑦 = 0, 𝑣𝐺𝑧 = 0, and 𝑧𝐺∕𝑃 = 0, which means that Eq. (D.15) becomes ⃗ = −𝐼 𝜔 𝚤̂ − 𝐼 𝜔 𝚥̂ ℎ 𝑃 𝐺𝑥𝑧 𝐵𝑧 𝐺𝑦𝑧 𝐵𝑧 [ ( )] ̂ + 𝐼𝐺𝑧 𝜔𝐵𝑧 + 𝑚 𝑥𝐺∕𝑃 𝑣𝐺𝑦 − 𝑦𝐺∕𝑃 𝑣𝐺𝑥 𝑘,

(D.23)

where, again, the mass moments and products of inertia are computed with respect to the mass center 𝐺 of the rigid body. Finally, if the 𝑥𝑦 plane is also a plane of symmetry for the body, then 𝐼𝐺𝑥𝑧 = 0 and 𝐼𝐺𝑦𝑧 = 0, and so Eq. (D.23) can be written as [ ( )] ̂ ⃗ = 𝐼 𝜔 +𝑚 𝑥 𝑘. (D.24) 𝑣 − 𝑦 𝑣 ℎ 𝑃 𝐺𝑧 𝐵𝑧 𝐺∕𝑃 𝐺𝑦 𝐺∕𝑃 𝐺𝑥 Noting that for planar motion, 𝐼𝐺𝑧 = 𝐼𝐺 , 𝜔𝐵𝑧 𝑘̂ = 𝜔 ⃗ 𝐵 , and that 𝑟⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 = ( ) 𝑚 𝑥𝐺∕𝑃 𝑣𝐺𝑦 − 𝑦𝐺∕𝑃 𝑣𝐺𝑥 , we can write Eq. (D.24) as ⃗ =𝐼 𝜔 ℎ ⃗𝐺∕𝑃 × 𝑚𝑣⃗𝐺 , 𝑃 𝐺 ⃗𝐵 + 𝑟 ⃗ used in Eq. (18.40). which is the form of ℎ 𝑃

(D.25)

ISTUDY

INDEX

A Absolute location, 29 Absolute pressure, 470 Absolute values, 32, 38, 39, 50 Acceleration defined, 7, 23 due to gravity, 16–17, 24, 450 (See also Gravitational force) to calculate well depth, 676–677 defined, 621–622 effects on projectile motions, 68, 689 as function of position, 668–670, 675, 775–776 as function of time, 673 as function of velocity, 674, 678–679 Acceleration analysis Coriolis component, 1122–1124 example problems, 1101–1108, 1154, 1168–1171, 1178–1181 rigid bodies in three dimensions, 1350 rigid bodies in two dimensions, 1097–1098, 1137–1139 rotating reference frames, 1120–1121, 1139 Acceleration vectors form in Cartesian component systems, 653 normal-tangential components, 703, 704, 706–709 notation, 626 polar coordinates, 719–720 properties, 652, 775 Accessory belts, 562–563 Accuracy of numbers, 13, 632 Acrobatic catch example, 1276–1277 Active guidance systems, 693 Addition of vectors, 628. See also Vector addition Aerodynamic drag, 677, 679, 816–817 Air hockey, 966–967 Airbus A300 failure, 21, 22 Aircraft, person pulling, 941 Aircraft carrier landings, 939, 1323 Aircraft carrier launch forces, 866–867, 920, 947 Aircraft landing gear, 325, 326 Aircraft service vehicles, 410, 411 Aircraft tracking examples, 722, 769 Airplane window, 95 Algebraic equations for particles in three-dimensional equilibrium, 166–167 Allowable loads, 183–185, 196 Alloys, coefficients of friction, 536 All-terrain vehicle (ATV), 231 Altitude, acceleration due to gravity and, 622 Aluminum alloy, 20 Aluminum channels, 438 American Airlines Flight 587, 21, 22 American Concrete Institute (ACI), 346 American Gear Manufacturers Association (AGMA), 277 American Institute of Aeronautics and Astronautics (AIAA), 346

American Institute of Steel Construction (AISC), 346 American National Standards Institute (ANSI), 346–347 American Society of Mechanical Engineers (ASME), 346 Ames Research Center, 666, 708 Amontons, Guillaume, 533–534 Amplitude, 1290, 1340 Anchoring blocks for cranes, 552–553 Andretti, Marco, 812 Angle of friction, 537, 567 Angle of twist, 576 Angle of wrap, 560, 563 Angles between vectors, 85, 89–90 Angular acceleration. See also Acceleration analysis particles in planar motion, 58–59, 670, 680–681 rigid bodies in planar motion, 1058, 1170–1171 rigid bodies in three dimensions, 1350, 1352, 1354–1359, 1388 Angular impulse with respect to P, 982 Angular impulse-momentum principle for impacts between rigid bodies, 1265–1270 for particles, 982, 1046 in place of Newton’s third law, 620 for rigid bodies, 1244–1245, 1248–1249 for systems of particles, 983–986 Angular measure, units of, 11–12. See also Small angle approximations Angular momentum. See also Conservation of angular momentum computing for rigid bodies, 1247, A-75–A-78 conservation for particles, 723, 985–986 conservation for rigid bodies, 1245, 1250–1253, 1269, 1270 defined for particles, 982–983, 1046 example problems, 989–998 per unit mass, 1008 rigid bodies in planar motion, A-78 rigid bodies in three dimensions, 1366, A-70–A-71, A-75–A-78 rotational equations of motion, 1146–1150 Angular speed, 680, 1079, 1165 Angular velocity defined, 670 rigid bodies in planar motion, 1057, 1058 rigid bodies in three dimensions, 1350–1352, 1354–1359, 1388 in rigid body collisions, 1267 of rotations, 708–709 Anisotropic friction, 538 Answers, to even-numbered problems, A-5–A-63 Anticommutative property of vector cross product, 628 Apoapses, 1009, 1010–1011 Apogee, 832, 896, 991–994, 1017 Apollo 16 Lunar Module, 177 Apollo missions, 1021 Apparent motions, 1121–1122, 1128–1129

Approach velocity, 957 Apses, 1009 Arbitrary axes, moments of inertia about, A-73–A-74 Arbitrary moment center, 1177, 1202 Arc length, 703 Archimedes, 4 Arcs, centroids of, 442–443 Area moments of inertia applications, 575–576 basic concepts, 573–575, 612 composite shape examples, 588–589 evaluating, 577 integration examples, 579–581, 587 parallel axis theorem, 585–586, 612–613 radii of gyration and, 576–577 thin flat plate example, 599 Areal velocity, 1008 Areas centroids of, 433–434, 436, 438, 440–441 surfaces of revolution, 461, 487 Argument, 630 Aristotle, 4 Arresting cables, 939, 1323 Articulated dump trucks, 336–337 Artificial knee joint examples, 1087–1088, 1107–1108 Artillery, 698 Asteroids, diverting, 27 Atlantis shuttle, 933 Atmospheric pressure, 478 Atomic force microscope (AFM), 215, 261 Atoms, forces between, 899 Authalic mean radius of Earth, 621 Automobile engines, 1055–1056, 1064 Average braking force, 939 Average force defined, 935, 1044 relating to kinetic energy, 866–867 Average speed, 651 Average velocity vectors, 650 Axes of revolution moments of inertia about, 600 producing solids of revolution, 439, 462, 487 producing surfaces of revolution, 461, 487 Axes of rotation, 718 Axes of symmetry, 435, 576 Axial forces in slender members, 494, 495 Axial springs, 299–300, 308, 353 Axioms, 789 Azimuthal angles, 82

B Backpack hoist example, 139 Bailey, Donovan, 886 Balance principles, 790 Balancing centrifuges, 1169 Ball bearings, 702, 981

I-1

Ball-and-socket joints, 399, 400 Ballast, 513 Ballistic pendulums, 973, 1270 Baltimore trusses, 385 Bars allowable loads, 183–185 basic modeling of, 131–133, 195 common assumptions about, 152 internal and external forces on, 68–69 three-dimensional example problems, 170–173 two-dimensional example problems, 136–137, 140–141 as two-force members, 304, 306 Base dimensions, 630 Base units, 9, 630 Baseball bats, 1166–1167 Baseball trajectory example, 694–695 Basketball hoops, 416–417 Basketballs, 974 Bassinet, 96 Bay Bridge Skyway, 573 Bead-and-cord examples, 92–94 Beads, 98–99 Beams area moments of inertia, 574, 579 design criteria, 518 distributed forces on, 468–469, 475, 487 equilibrium approach to internal force calculations, 503–509, 527 equivalent forces on, 252 Galileo’s studies, 4–5 incorrect free body diagram example, 327, 328 integration approach to internal force calculations, 515–517, 519–523, 527 leveling with wedges, 554–555 moments of couples on, 237 in multispan bridges, 414 nonsymmetric, 576 shear, moment, and distributed force relations, 514, 517, 527 strengthening, 588 two-dimensional idealization example, 284 Bearings, 322–324, 334–335 Belt tensioner examples, 214, 562–563 Belts, 560–563, 568 Belts on pulleys, 1064 Bending, 369, 510 Bending moments, 494, 495 Bernoulli, Daniel, 5 Bernoulli, Johann, 5 Bicycle wheels, 83 Bifurcations, 386 Billiard ball standards, 977 Bilz precision leveling wedges, 538 Binormal unit vectors, 704 Bodies defined, 6 Body forces, 8, 468, 487 Body-fixed frames, 1118, A-77 Boeing 737 airliner, 941 Boeing CH-47 Chinook helicopter, 260 Bollman trusses, 378 Bolt, Usain, 668, 886 Bonanza 35 airplane, 268 Bookshelf example, 475

I-2 ISTUDY

Booms failure criteria examples, 136–137, 170–171 pin and cable support example, 332–333 socket and cable support example, 330–331 space trusses as, 402, 404 Bottle openers, 497 Boundary lubrication, 537 Bowling balls, 19 Box-end wrenches, 241 Brakes (metal fabrication), 226 Braking moment, 565 Braking test example, 675 Bridges example problems, 414–415 examples of failure, 21, 387 truss types, 364, 365 Briefcase latch example, 156–157 Buckling. See also Failure determining potential, 385–386 relation to member length, 185, 386, 427 zero-force members and, 387 Built-in supports, common reaction forces, 323 Bulletin board diagrams, 278 Bulletin board free body diagrams, 277 Bumper cars, 1265, 1266 Bungee jumping examples, 884–885, 1291, 1293

C Cable car example, 138 Cable tension, 815 Cables allowable loads, 183–185, 186–187 basic approaches to analyzing, 298–299 basic modeling of, 131–133, 195 common assumptions about, 151 contacting cylindrical surfaces, 560–561, 568 internal and external forces on, 68–69 modeling as springs, 314 structure of, 68, 185 tensile force vector example, 71 three-dimensional example problems, 170–171, 330–335 in trusses, 374–375 two-dimensional example problems, 136–141 as two-force members, 304, 306 Cable-supported cantilevers, 325, 326 Calculations, accuracy in, 13 Calculus, 16 Canada Place, 493 Canning, 478 Canoe examples, 840–841 Cantilevered bar deflection, 1298 Cantilevers cable-supported, 325, 326 distributed forces on, 468–469 highest internal forces, 495 shear and moment, 508–509 subject to couples, 237 superposition example, 311 two-dimensional idealization example, 284 Capacity ratings, 348–349 Car engines, 1055–1056, 1064 Carets, 626

Carnival ride examples, 1066, 1081, 1330–1331 Cart handle example, 312–313 Cartesian components, 9–10, 626–627, 629–630. See also Components Cartesian coordinate system basic concepts, 6, 625, 775 basic features, 6, 48 for cross products, 102–103 Newton’s second law applied, 812, 813 radii of curvature, 704 relation to components of vectors, 32–33, 652–653 representing vectors in, 626–627 three dimensions in, 759, 779 for vectors in three dimensions, 66–69, 117 for vectors in two dimensions, 48–52, 54–59, 116–117 Cartesian representation, 49, 85, 102–103 Castings, equivalent forces on, 255 Catapult systems aircraft carrier, 866–867, 920, 947 conservation of energy example, 907–908 Catch example, 1276–1277 Caterpillar Ultra High Demolition machine, 62 Cauchy, Augustin, 5 Center for Gravitational Biology Research, 666, 708 Center of gravity basic concepts, 431–433 defined, 17, 432, 449, 486 determining, 450–451, 486 example problems, 454–456 Center of mass defined, 433, 449, 486 determining, 449–450, 486 example problems, 453 moments of inertia about, 594 Center of percussion, 1166–1167, 1278 Center of pressure, 432 Centers of mass (particle systems) calculating motions, 855 in closed versus open systems, 984 defining position, 838 use in determining kinetic energy, 902–903 Centers of mass (rigid bodies), 1266, 1267 Central forces example problems, 822–823 gravitational, 1007 work on particles, 877–878 Central impacts between particles, 956 between rigid bodies, 1266–1268, 1272–1273 Centrifuges Center for Gravitational Biology Research, 666, 708–709 electric motors for, 1065 fixed angle, 1385 with swinging-bucket rotors, 1168–1169 for uranium enrichment, 713 Centroidal moments of inertia, 575, 585, 589 Centroids center of gravity versus, 433, 449 of cones, 456 defined, 433, 449, 486

ISTUDY

example problems, 438–444 methods of determining, 433–436, 486 Cessna 172 airplane, 293, 341 Chain rule, 669 Chameleons, 796 Channels (freshwater), 476–477 Chaos theory, 1187 Characteristic equation, 1322, A-71 Charpy impact test, 897, 1236 Circular frequency, 1290 Circular motion example problems, 708–709, 820–821 one-dimensional relations, 670–671, 776 Circular orbits, 993, 1010, 1048 Clamped supports, 323 Clamping devices, 282–283 Clockwise, counterclockwise versus, 213 Closed force polygons, 140–141 Closed systems defined, 935, 1023 impulse-momentum principle application, 1023, 1026, 1028 modeling control volumes as, 1023, 1026 Coast Guard icebreakers, 545 Coast Guard rescue craft example, 58–59 Codes, 3, 346–347 Coefficient of kinetic friction, 535, 536, 567, 791, 855 Coefficient of restitution (COR) in particle collisions, 702, 956–958, 970–971 in rigid body collisions, 1265, 1267, 1269, 1284 Coefficient of static friction, 535, 536, 567, 791, 854 Coefficient of viscous damping, 1321, 1342 Coefficients of friction, 535–536, 567 Coil springs, 152. See also Springs Collars about, 97–98 position and force vector example, 76–77 reaction forces, 140–141, 273 summing forces on, 172–173 Collector plates, 767–768 Collisions. See Impacts Comet Halley, 1018–1019 Commercially manufactured beams, 518 Complementary solutions, 1307, 1308, 1341, 1343 Complete fixity in plane trusses, 384 in rigid bodies, 300, 302–303, 354 Complex trusses, 384–385, 427 Component form of Newton’s second law, 790 Component systems, 812–814 tangent-normal-binormal, 759–761, 779 Components. See Scalar components; Vector components Components of vectors, 9–10, 32–33, 629–630, 635, 652–653 Composite rigid bodies, 1170–1171 Composite shapes method about, A-74 for area moments of inertia, 586, 588–589 center of mass and gravity examples, 453–455 determining centroids via, 433–434, 435, 438–439

for mass moments of inertia, 597, 605 use with distributed load problems, 469 Compound trusses, 384, 385, 427 Compressive forces from fluid and gas pressure, 469–473 in free body diagrams, 132, 133 Compressive loads allowable in bars and pipes, 183, 184, 185 failure criteria examples, 136–137 Computer aided design, 181 Computer numerical control (CNC), 65 Concrete block walls, 522–523 Concrete smoother example, 240 Concrete traffic barriers, 541 Concurrent design, 344 Concurrent force systems defined, 125, 249 three-force members subjected to, 305, 306 Cones, center of gravity and centroid, 456 Conic sections, 1008–1009, 1010–1012, 1048–1049 Connecting rods. See also Slider-crank mechanisms acceleration analysis, 1097–1098, 1103–1104 basic functions, 1055–1056 example, 604 general planar motions, 1057 oscillation example, 1297 velocity analysis examples, 1083–1084 Conservation of angular momentum for particles, 723, 985–986, 989–998 for rigid bodies, 1245, 1250–1253 in rigid body collisions, 1265, 1269, 1270 Conservation of linear momentum example problems, 840–841, 942, 1252–1253 for rigid bodies, 1243–1244 in rigid body collisions, 1265, 1268 for systems of particles, 936–937, 1045 Conservation of mechanical energy defined, 879, 927 example problems, 884–885, 907–908, 1223–1224 harmonic oscillation, 1293 rigid bodies in planar motion, 1210–1211 Conservative forces defined, 879, 926 example problems, 892 identifying, 880–881 in potential energy, 928 Conservative systems, 879, 927 Constant acceleration in one-dimensional problems, 670–671 of projectile motion, 68, 689 propeller example, 58–59, 680–681 Constant forces, work of, 863–864 Constant of universal gravitation, 621 Constant velocity, 4, 722, 769 Constrained eccentric impacts, 1269–1270 Constrained impacts between particles, 956, 968–969, 1045 between rigid bodies, 1266, 1269–1270, 1274–1275, 1284 Constrained systems example problems, 746–749 relative motion in, 739–741, 778 Constraint force equations, 1076 Consumer Product Safety Commission (CPSC), 347

Contact forces, 8–9, 625 Contact surfaces multiple, 537, 542–543, 551–555 roughness, 533, 534–535 Continuous bodies, 986 Control volumes, 1023–1024 Conversion factors, 631 Coordinate lines, 653 Coordinate systems. See Component systems, tangent-normal-binormal absolute location and, 29 basic concepts, 6, 625 Cartesian system, 6, 625, 626–627, 759, 767–768, 775 common uses, 6 cylindrical, 761–762, 766, 779–780 in free body diagrams, 129, 130 polar, 719–720, 776–778 spherical, 82–83, 762–763, 769, 780 for structures in equilibrium, 275 tangent-normal-binormal, 759–761, 779 three-dimensional, 66–69, 117 two-dimensional, 48–52, 116–117 Coordinates, defined, 6, 48, 49 Coplanar force systems, 208, 249–250 Coriolis, Gaspard-Gustave, 1122 Coriolis acceleration, 1122–1124 Cosine law, 34, 36–37, 116 Coulomb, Charles Augustin de, 5, 534, 791 Coulomb friction model, 791–793, 799, 801, 854–855 Coulomb’s law elements of, 535, 567 negated by lubrication, 537 for tensile forces on cylindrical surfaces, 560–561 in three dimensions, 538 Counterclockwise, clockwise versus, 213 Couples about, 1210, 1212 moments of, 232–235, 264–265 Cover plates, 588 Cranes about, 766 anchoring blocks for, 552–553 direction angle example problem, 70 space trusses in, 402, 404 support reactions of pulleys in, 309–310 tower, 125 Crankshafts basic functions, 1055–1056 in Ford EcoBoost engine, 1055 planar motion, 1057 Critical damping coefficient, 1322, 1342 Critically damped systems, 1323, 1324, 1330–1331, 1342 Crop planting positioners, 327, 328 Cross products basic concepts, 101–106, 118–119 evaluation examples, 107–110 example problems, 111–115 finding for two vectors, 628, 630 Cross sections area moments of inertia, 574, 576, 579 centroids, 438, 440–441 Cubes, moments of couples on, 238–239

I-3

Curvature of a path, 703 Curve geometry, 704 Curved lines, centroids of, 442–443 Curved members centroids, 440–441 fluid forces on, 472, 473 in trusses, 365, 369, 372–373 as two-force members, 304, 306 Curvilinear motion, 812–814 Curvilinear translations, 1056, 1066, 1081 Cuts, 127, 380 Cyclic loading, 1169 Cylinder volume and surface area, 465 Cylindrical coordinates basic concepts, 761–762, 779–780 example problems, 766 Newton’s second law applied, 813–814 Cylindrical gates, 477 Cylindrical surfaces, belt friction, 560–563, 568

D D ring vector example, 36–37 Da Vinci, Leonardo, 4, 533 d’Alembert, Jean le Rond, 5 Damped natural frequency, 1323, 1343 Damped vibrations, 1321–1326, 1329–1333, 1342–1344 Damping factor, 1323 Damping ratio, 1323, 1343 Dams, 431–432 Darboux vector, 760 Dashpots, 1321, 1324–1325, 1330 Datum lines, 879 Dead load, 314 Deer Isle bridge, 21 Deformation about, 970, 1220 defined, 151 internal forces, 494–495 measuring, 8 negating force transmissibility, 247 representing as springs, 152–155 Degree measure, 11–12 Degrees of freedom defined, 794, 854 in harmonic oscillators, 1291 rotating reference frame examples, 1125 Del, 880 Density defined, 18, 24 of solids, surfaces, and wires, 450 specific weight and, 17–18, 19, 20, 24 Derivatives, time. See Time derivatives of vectors Derived dimensions, 630 Derived units, 9, 630 Design. See Engineering design Design, role of dynamics in, 646–647 Design analysis, 346 Design codes, 3, 346, 347 Despin mechanisms, 1135, 1195, 1263 Determinants, 103–104, 119 Determinants, finding cross products with, 630 Die springs, 158 Differential forces causing tides, 31

I-4 ISTUDY

Differentiation of constraints acceleration analysis by, 1097–1098, 1105–1106, 1138 in relating speed to position, 869 relative motion analysis by, 739–741, 778 velocity analysis by, 1076, 1085–1086, 1137 Dimensional analysis, 638–640 Dimensional homogeneity, 631, 638 Direct central impacts, 958, 961–965 Direct eccentric impacts, 1266 Direct impacts, 956, 1266 Direction angles examples determining, 70, 72–73, 74–75 methods of determining, 66–67, 117 Direction cosines, 67, A-74 Direction of rotation, 718 Directions of sliding motions, 551 Directrices, 1009 Discontinuities in shear and moment diagrams, 505 Displacement vectors, 650, 775 Distributed forces contact forces as, 8 erroneous replacement by equivalent force, 507 example problems, 475–478 idealizing as point forces, 468–473 major types, 468–469, 487 relation to shear and moment in beams, 514, 517, 527 Distributive property, 102 Diverters, 1032–1033 Diving boards, 1300 Docking example, 1252–1253 Documenting designs, 346 Door closers, 1085 Dot products about, 628 basic concepts, 84–88, 117–118 example problems, 89–94, 173 for summing forces in three dimensions, 168, 173, 196 Double-headed arrows, 205, 627 Doubly symmetric objects, 595, A-68 Drag projectile motion with, 816–817 role in terminal velocity, 679 when to disregard, 677 Drag coefficients, 816 Drop hammers, 158, 159 Drum mixers, 287–288 Dual arch bridge, 1 Duffing’s equation, 811 Dump trucks, 336–337 Duplantis, Armand, 887 Dynamics basic concepts, 2, 619–620, 625–632 basic role in design, 646–647

E Earth acceleration due to gravity, 621–622 Coriolis acceleration, 1122–1124 motion due to impacts, 963 as particle, 991 radius, 621

Earth’s gravity, 17 Eccentric impacts, 956, 1268–1270, 1283 Eccentricity of conic sections, 1009, 1048 EcoBoost engine, 1055 Edinburgh Castle grandstand, 166 Effective mass and stiffness, 1294 Efficiency, 919, 921–922, 928 Eiffel Tower, 399 Eigenvectors, A-71 Elastic cord force examples, 92–94 Elastic impacts, 958, 1045, 1266, 1267, 1284 Elastic modulus, 153 Elastic potential energy, 879 Elasticity, 152 Electric motor examples, 921–922 Electric transmission tower example, 390–391 Electrostatic precipitators, 767–768 Elementary motions in particle kinematics, 668–670, 775–776 Elevation angles in projectile motion, 70–73, 692–695 Elliptical orbits, 1010–1012, 1048–1049 Empirical equations, 536 Energy method, 1293–1294, 1300, 1341 Engine accessory belts, 562–563 Engine connecting rods, 604 Engineering design basic concepts, 181–185, 344–346 cable selection example, 186–187 codes and standards, 346–347 criteria for beams, 518 defined, 181 forklift example, 348–349 friction considerations, 538 objectives, 181–182 process steps, 344–346 trusses, 384–387 water cooler support example, 188–189 Engines, 1055–1056, 1064, 1221–1222. See also Slider-crank mechanisms Equals sign, 631 Equation counting for plane trusses, 383–384, 426–427 for rigid bodies, 301, 302–303, 354 for space trusses, 400–401 Equations of motion, defined, 793–794, 854 Equilibrium equations for rigid bodies, 271–272, 322, 353 geometry of structures in, 151 of particles in three dimensions, 166–168, 195 of particles in two dimensions, 125–135 rigid bodies in three dimensions, 322–328 rigid bodies in two dimensions, 272–280, 298–306 Equilibrium approach, internal force determination, 503, 516, 527 Equivalent couples, 234, 265 Equivalent force systems, 1032 erroneous replacement of distributed force by, 507 example problems, 252–257 overview, 232, 234–235, 246–248, 265–266 rationale, 250–251 special types, 248–250, 266 in trusses, 382

ISTUDY

Equivalent vectors, 32 Error checking, 167 Escambia Bay bridge, 21 Escape velocity, 1012, 1049 Euler, Leonhard, 5, 620, 936 Euler’s first law, 936, 986, 1145, 1199, 1366, 1388–1389 Euler’s rotational equations for rigid bodies, 1368, 1389 Euler’s second law, 986, 1146, 1366, 1389 Evaluation in design, 344–345 Exact differentials, 862 Exam scores, 573–574 Exercise machines, 374–375 Experimental determination of mass moments of inertia, 604 Explorer 7, 1019 Expo Center, 399 External effects of forces, 246 External forces in particle systems defined, 837 impulse-momentum principle, 935–936 work-energy principle, 901–902 on structural members, 68–69 Extruded aluminum channels, 438 Eyebolts, 54–57

F F/A-18 Hornet aircraft, 866, 867, 920 Factors of safety, 184–185, 196 Failure. See also Buckling bars and cables in three dimensions, 170–171 bars and cables in two dimensions, 136–137 factors of safety and, 184–185 famous examples, 21–22, 387 house-on-slope example, 108–109 multiple criteria in free body diagrams, 136–137 in statically determinate structures, 387 Failure loads, 183–184 Falkirk Wheel, 1055 Falling rigid body example, 1178–1179 Falling string examples, 909–910, 1036–1037 Fatigue, 1169 Feasibility studies, 345 Ferrari 250 GTO sports car, 203 Ferrari Formula One, 919 Fiberglass poles, 887 Field forces, 8–9 Figure skating, 593 Final designs, 346 Finite element method, 110, 400 Fink trusses, 385 First law of motion (Newton), 7, 23, 620 First moments, 433–434 Fishing rod, 97 Fixed axis rotation. See Rotation about a fixed axis Fixed-angle centrifuge rotors, 1385 Fixity determining in three dimensions, 330 determining in two dimensions, 302–303 elements of, 300–301, 354 in plane trusses, 384 in space trusses, 400–401

Flat surfaces, fluid forces on, 471–472 Flexibility of cables, 151 Flight time, 691 Floating platform examples, 942 Fluid flows, 1023–1028, 1030–1033 Fluid pressure example problems, 476–477 forces of, 469–473, 487–488 Flywheels, 851, 903, 953 Foam packaging, 153, 154 Focal parameters, 1009 Foci of ellipses, 1009 Folding desk free body diagrams, 278, 279 Force. See also Equivalent force systems; Moments defined, 6 internal and external, 68–69 major types, 8–9 transmissibility, 246–247 Force concepts, 6, 625 Force due to gravity, 16 Force fields, 892 Force laws, 790, 791–793 Force polygons basic concepts, 36, 40, 41–42 example problems, 140–141 Force systems, work on rigid bodies, 1209–1210 Force vectors denoting in figures, 31–32 examples determining, 71, 74–77 lines of action, 30, 87 position vectors with, 68 Force vectors in free body diagrams, 790 Forced vibrations damped, 1324–1326, 1343 undamped, 1307–1310, 1341–1342 Forces, power and efficiency, 919 Forcing functions, 1307 Ford Excursion, 974, 1051 Forklifts, 348–349 Formula 1 racing, 706 Forward spin, 1250 Four-bar linkages, 1066, 1081, 1087–1088, 1107–1108 Fourier, Jean Baptiste Joseph, 1307 Four-point bending, 510 Fracture toughness tests, 897 Frames defined, 408, 427 example problems, 414–419 free body diagrams, 408–413 when to model as, 365, 400 Frames of reference. See Reference frames Free body diagrams (FBDs) in dynamics, 789–790 fluid and gas pressures, 471, 472, 473 for frames and machines, 408–413 impact-relevant, 956 for particles in equilibrium, 126–130, 195 procedures for drawing, 127, 274, 409 for rigid bodies in three dimensions, 325–328 for rigid bodies in two dimensions, 274–276, 277–280 Free vectors, 234, 235, 265 Free vibrations damped, 1321–1324, 1342–1344

example problems, 1297–1300, 1330–1331 undamped, 1289–1294, 1340–1341 Frequency, 1290, 1340 Frequency distributions, 573–574 Frequency ratio, 1308 Freshwater channel examples, 476–477 Friction, system of particles with, 905–906 Friction forces belts and cables on cylindrical surfaces, 560–561, 568 Coulomb model, 791–793, 799, 801, 854–855 defined, 8 design considerations, 538 example problems, 540–543, 552–555, 562–563 factors determining, 534–538, 567 as function of time, 1248 impending slip examples, 798–799, 818–819 internal work examples, 911–912 with multiple contact surfaces, 537, 551 rolling without slip examples, 1180–1181, 1219–1220 sliding examples, 844–845, 888–889 on wedges, 538 Full fixity in plane trusses, 384 in rigid bodies, 300, 302–303, 354 in space trusses, 400 Full Moon, 31 Functionally graded material, 456 Furnace door example, 334–335

G Gage pressure, 470, 471, 478, 487 Galilei, Galileo, 4–5, 8 Gambrel trusses, 396 Gas pressure example problems, 478 forces of, 469–470, 473, 487–488 Gas springs, 152 Gates in freshwater channels, 476–477 Gear reduction, 1065 Gears, 277, 287–288, 1065, 1082 Gearshift lever, 97 General multiforce members, 304, 355 General planar motion defined, 1057, 1059 Newton-Euler equations, 1177, 1178–1187, 1201–1202 rigid body kinetics, 1201–1202 General solution, equations of motion for harmonic forcing, 1307, 1308, 1341 Generating areas, 462, 487 Generating curves, 461, 487 Geometrical constraints, 739–741, 778 Geometry of curves, 704 Geostationary orbits, 1019, 1072 Geosynchronous equatorial orbits, 1019, 1072 Gildemeister NEF 330 metal cutting lathe, 538 Golf balls, 816–817 Governing equations, types, 790, 854 Gradient operator, 880 Grain silo, 96 Grand Canyon Skywalk, 495 Grandfather clocks, 1304

I-5

Gravitation, Newton’s law, 620–622, 723, 1007 Gravitational attraction example problems, 19 overview of Newton’s law, 16–18, 24 Gravitational force orbital mechanics, 1007–1013, 1047–1049 potential energy, 879, 1212 work on particles, 877, 878, 926 Gravity gradient satellites, 450 Gravity tractors, 27 Greene, Maurice, 886 Grinding, 538 Guldin, Paul (Guldinus), 461, 462, 487 Gun tackles, 754–755 Gyration, radii of, 1164 Gyratory compactors, 261 Gyroscopic moment on spinning disk, 1376 Gyroscopic terms, 1357

H Halley, Edmund, 1018–1019 Hammer mass moment of inertia example, 605 Hand saws, 498 Hand truck free body diagrams, 279, 280 Hang angle example, 1155 Hanging on basketball rims, 416 Harmonic motion, 1290 Harmonic oscillators defined, 1290, 1340 forced, 1307–1310, 1333, 1341, 1343 natural frequency, 1294 standard form, 1291–1292, 1340 viscously damped, 1322, 1323, 1343 Haynes, Roy, 1289 Hazards, minimizing in design, 181–182 Heads of vectors, 30 Head-to-tail addition of vectors, 33 Helicopters, 260 Hertz, 1290 High-pressure systems, 1123–1124 High-tension side of cables and belts, 560, 561, 568 Hinges, 323, 325, 326 History of statics, 3–5 Hockenheim racing track, 711 Hockey pucks, 1272–1273 Hohmann transfers, 1020, 1053 Homogeneity, dimensional, 631, 638 Hooke, Robert, 5 Hoover Dam, 431–432 Horizontal area elements, centroids of, 440–441 Horsepower, 919 House-on-slope example, 108–109 Howe trusses, 364 Howitzers, 698 Hubble Space Telescope, 996 Hurricane Ivan, 21 Hydraulic cylinders, 336–337 Hydrostatic pressure, 470 Hyperbolic trajectories, 1012–1013, 1049

I I beams. See also Beams minimizing deflection with, 574 strengthening, 588

I-6 ISTUDY

support reaction example, 314 trolley mounted on, 286 Icebreakers, 545 Idler pulleys, 562 Impact force law, 957 Impact tests, 897, 1167 Impact-relevant FBDs, 956 Impacts classifying, 956, 1266–1270 coefficient of restitution in, 956–958 constrained, 956 energy of, 959, 1045 example problems, 961–971, 1272–1277 modeling between particles, 955, 1045 of rigid bodies, 1265–1270, 1283–1284 Impending motion, 534 Impending slip, 792, 798–799, 818–819 Impulse, 934 Impulse-momentum principle angular, for impacts between rigid bodies, 1265–1270 angular, for particles, 982 angular, for rigid bodies, 1244–1245, 1248–1249, 1283 collisions between particles, 1045 as conservation law, 936–937 example problems, 939–942, 962–963, 969, 970–971 linear, for particles, 933–937, 1044 linear, for rigid bodies, 1243–1244, 1248–1249, 1283 for mass flows, 1023–1028, 1050 Impulsive forces, 956, 1265 Indentors, 874 Inertia concepts, 625 Inertia matrix, A-70, A-78 Inertia properties, 625 Inertia tensor, A-70, A-78 Inertial forces, designing for, 188 Inertial reference frames basic properties, 794 example problems, 798–799 rigid bodies in planar motion, A-77 in water jet example, 1030 Inextensibility, 151, 152 Infinity Bridge, 1 Instantaneous center of rotation, 1077–1079, 1137, 1209 Integration in elementary motion calculations, 668–669 Integrations for center of gravity example, 456 determining centroids with, 434, 435, 440–444 determining internal forces in beams with, 515–517, 519–523, 527–528 evaluating area moments of inertia with, 577, 579–581, 586, 587 evaluating mass moments of inertia with, 596–597 use with distributed load problems, 469 Interaction diagrams, 120 Internal combustion engines, 1055–1056, 1221–1222. See also Slider-crank mechanisms Internal effects of forces, 246

Internal forces basic concepts, 68–69, 493–495, 527 equilibrium approach to determining, 503–509, 516 integration approach to determining, 515–517, 519–523, 527–528 three-dimensional example problems, 499 two-dimensional example problems, 497–498 Internal forces in particle systems in closed systems, 935–936 defined, 837, 838 work-energy principle, 901–902 Internal work example problems, 911–912 in systems of particles, 901–902, 928 International Organization for Standardization (ISO), 347 International Space Station, 594, 595 altitude above Earth, 1020 control moment gyroscope installation, 1207 June 2008 photo, 1007, A-65 mass, 1253 Space Shuttle docking, 1252, A-67 Interstate 35W bridge, 387 Intrinsic triad, 704 Inverse cosine function (calculators), 70 Isolated systems, 937 Isotropic friction, 538 Iterative nature of design, 344, 345 Ivan, Hurricane, 21

J Jerk, 733 Jet packs, 1034–1035 Jet propulsion, 681, 1028 John Deere 2720 tractor, 342 Joints. See also Method of joints defined, 364, 426 in space trusses, 399 treating forces not at joints, 382 Joules, 863 Jupiter, 621

K K trusses, 364 Kanounnikova, Natalia, 1251 Kansas City Hyatt Regency Hotel, 22 Kepler, Johannes, 620–621, 1011 Kepler’s first law of motion, 1011 Kepler’s second law of motion, 1008 Kepler’s third law of motion, 1012 Keweenaw Peninsula, 58 Kinematic chains, 1087 Kinematic equations, 790 Kinematics, 537, 551, 649–653. See also Particle kinematics; Rigid body kinematics Kinetic diagrams (KDs), 1150 Kinetic energy approximating as quadratic function of position and velocity, 1294 bicycle example, 1215–1216 converting to potential energy, 886–887 of impacts, 959, 1045

ISTUDY

relating to average force, 866–867 rigid bodies in planar motion, 1207–1208 rigid bodies in three-dimensional motion, 1371–1372, 1381–1382, 1390–1391 for systems of particles, 901, 902–903, 928 work and, 862–863, 926 Kinetic friction coefficient, 535, 536, 791, 855 Kinetic friction force, 534 Kinetics, 789. See also Rigid body kinetics King Kong, 683 Kites, 453

L Ladders, 418–419, 748–749, 1105–1106 Lagrange, Joseph-Louis, 5 Landing gear, 325, 326 Latches, 156–157 Lateral G-force, 706 Law of cosines, 34, 36–37, 116 Law of sines, 34, 36–37, 116 Law of universal gravitational attraction, 16–18 Lawn roller example, 1184–1185 Lead of screw, 1304 Leaf springs, 154–155 Leibniz, Gottfried Wilhelm, 16n Lennard-Jones force law, 899 Leonard P. Zakim Bunker Hill Bridge, 29 Leonardo da Vinci, 4, 533 Leveling wedges, 538 Lift coefficient, 644 Lift speed example, 921–922 Lifting machine free body diagrams, 278, 279 Line forces, 468, 487 Line integrals, 434, 862 Linear elastic springs, 793, 855 Linear functions, expressions for, 456 Linear impulse, 934, 1044, 1243 Linear impulse-momentum principle for particles, 933–937, 1044 for rigid bodies, 1243–1244, 1248–1249 Linear momentum. See also Conservation of linear momentum conservation law, 936–937 defined, 1044 of particles, 934 of rigid bodies, 1243–1244 translational equations of motion, 1145–1146 Linear torsional spring constant, 1298 Linear viscous damping, 1321 Linearity of springs, 152, 195–196 Linearization, 1292, 1294 Linearly independent equations, 73 Lines, centroids of, 434, 436, 442–443 Lines of action centers of gravity and pressure, 432 defined, 30 determining angle between, 89–90 determining moments at different points along, 210–211 effects of deformation on, 247 equilibrium of objects and, 87 in free body diagrams, 790

for moments defined, 205 in two- and three-force members, 304 Lines of impact (LOI), 955–956 Live load, 314 Load center, 348 Load paths, 158 Location of objects, 29 Lock washers, 153 Loose-fitting gears example, 287 Low-pressure systems, 1123–1124 Low-tension side of cables and belts, 560, 561, 568 Lubrication, 537 Lucky Peak Reservoir, 364 Lumped-mass models, 995

M M777 lightweight 155mm howitzer, 698 Machine control free body diagrams, 279, 280 Machine handle moment example, 210–211 Machines defined, 363, 408, 427 free body diagrams, 408–413 Magnification factors damped, forced harmonic oscillators, 1326, 1343 determining for unbalanced motor, 1311–1315 forced, undamped vibrations, 1308–1309, 1341–1342 Magnitude of vectors about, 627, 717 defined, 30 in three dimensions, 66, 117 in two dimensions, 49–51, 116–117 Mars orbiter, 631 Mass basic concepts, 625, 631 center of, 433, 449–450, 453, 486 defined, 6 effective, 1294 weight versus, 10, 108, 138 Mass center moments of inertia, 594, 595–596, A-67, A-69 Mass center motions. See Centers of mass (particle systems) Mass centers of rigid bodies, 1146 Mass detectors, 1298 Mass flow rate, 1024 Mass flows example problems, 1030–1037 impulse-momentum principle, 1023–1028, 1050 Mass flux, 1024 Mass moments of inertia about arbitrary axis, A-73–A-74 in angular impulse-momentum principle for rigid bodies, 1245 applications, 594–595, A-67–A-68 basic concepts, 593–594, 613–614 in centrifuge example, 1169 defining, 573, 593, 1148, A-65–A-67 determining from oscillations, 1297 evaluating, 596–597 example problems, 598–605 parallel axis theorem, 595–596, 614, A-68–A-70 principal moments, A-70–A-73

as radii of gyration, 1164, A-68 for spinning skater, 1250 Mass points, 986 Mass products of inertia, A-66, A-67–A-68, A-69–A-70 Material frame indifference principle, 620 Matrices, 103–104 McIlroy, Rory, 816 Mechanical advantage, 542 Mechanical energy per unit mass, 1013, 1049 Mechanical power, 919 Mechanical vibrations. See Vibrations Mechanicians, 646 Mechanics of materials, 2 Mechanisms in trusses, 384, 385, 400 Meridional Earth radius, 621 Merry-go-round examples, 824–825 Method of joints plane truss analysis by, 365–367, 370–375, 426 space truss analysis by, 402–403 Method of sections plane truss analysis by, 380–382, 389–391, 426 space truss analysis by, 404–405 Micro spiral pumps, 731 Microelectromechanical system machines, 408 Midair catch example, 1276–1277 Mingus, Charles, 1289 Mini Cooper, 974, 1051 Minivan rear door example, 285 Minnesota cities example, 636–637 Minors, expanding determinants with, 104 Misuse, designing for, 182 Mixer gears, 287–288 Modeling, 130, 131, 647 Molecular dynamics (MD), 837 Moment arm belt tensioner example, 214 defined, 204, 263 Moment center, 982, 1164, 1199–1200 Moment diagrams. See Shear and moment diagrams Moment equilibrium equations, 272, 276–277 Moment of inertia about arbitrary axis, A-73–A-74 Moment-angular momentum relation Euler’s second law, 986, 1146 for particles, 982–983, 1046 for systems of particles, 984, 1046–1047 Moment-resisting collars, 273 Moments. See also Area moments of inertia; Mass moments of inertia about a line or direction, 220–221, 223–225, 264 couple example problems, 237–240 of couples, 232–235, 264–265 direction conventions, 331 eliminating in bearings, 322–324 errors in summing, 332–333 maximizing example, 214 overview, 203–204 relation to shear and distributed forces in beams, 514, 517, 527 scalar and vector approaches, 204–207, 208, 263–264 selecting summation points, 276 three-dimensional example problems, 212–213, 238–239 two-dimensional example problems, 209–211

I-7

Moments—cont. use of cross product to determine, 101 Varignon’s theorem, 207–208, 263 of weight distributions, 433–434 Moments acting on fluids, 1025–1026 Moments of couples, 1210 Moments of inertia. See Mass moments of inertia Momentum of particles, 934, 936–937. See also Impulse-momentum principle Monaco Grand Prix, 706 Monk, Thelonius, 1289 Moon describing location, 29–30 mass and mean distance from Earth, 27 Mores Creek bridge, 364 Motion laws, 5, 620 Motion laws of Newton, 7–8 Motion platforms, 1066, 1081 Motorcycles, 1154 Motors, unbalanced, 1311–1315, 1332–1333 Moving frames, 738 Moving targets, 744–745 Mud slides, 108–109 Multiple contact surfaces, 537, 542–543, 551–555 Multistory buildings as springs, 153, 154 Mutual attraction, forces of, 19

N Nabla, 880 Nanoelectromechanical systems, 1298 Nanoindentation tests, 874 Nanotechnology, 837, 1298–1299 National Transportation Safety Board (NTSB), 21 NATO 7.62 mm ammunition, 973 Natural frequency, 1290, 1294, 1340 Navier, Claude Louis Marie Henri, 5 Negative square roots, 70 Negative wrenches, 250 Neptune, 621 Newton, Isaac, 1–2, 5, 7–8, 619, 723 Newton-Euler equations application to rigid bodies in planar motion, 1145–1151, 1199–1200 example problems using, 1178–1187, 11541157 fundamental importance, 986 general planar motion, 1177, 1201–1202 from kinetic diagrams, 1150 rotation about a fixed axis, 1163–1164, 1201 three-dimensional motions, 1366–1371, 1388–1389 translation in plane of motion, 1153, 1200–1201 Newtonian physics, 1–2 Newtons (SI units), 9, 631 Newton’s cradle, 976 Newton’s laws. See also Second law of motion (Newton); Third law of motion (Newton) about, 5, 620 free body diagrams and, 127 overview, 7–8, 23 static equilibrium of particles under, 126 universal gravitational attraction, 16–18, 24 in vector and scalar form, 168 Nodes of finite elements, 110 Nonconservative forces, 880, 928

I-8 ISTUDY

Nonflat surfaces, fluid forces on, 472, 473 Nonlinear responses of springs, 154–155 Nonlinear systems, 1292 Nonsymmetric shapes, 576 Normal direction, 105, 108, 110 Normal forces in slender members, 494, 495 Normal-tangential components, particle paths in planar motion, 703–704, 706–709, 719 No-slip conditions, 791, 1064, 1065. See also Rolling without slip Nuclear reactors, 713 Number of degrees of freedom, 794

O Objects of revolution, 461–462, 464–465, 596 Oblique central impacts, 958–959, 966–969 Oblique impacts, 956, 958–959, 966–969, 1266 Obtuse angles, 34 Occupational Safety and Health Administration (OSHA), 22, 347 Ocean tides, 29–30 Octahedral planes, 78 Octopus ride, 772 One-dimensional motions, 668–671 Open systems, 935, 1023–1024 Open-end wrenches, 241 Optimal angle in projectile motion, 695 Optimization, 41–42 Orbital mechanics example problems, 1017 major concepts, 1007–1013, 1047–1048 Orbital periods, 1011–1012 Orbits of satellites acceleration analysis, 723 determining, 1007–1013, 1047–1048 energy considerations, 1013 equations of motion, 832 geostationary, 1019, 1072 speed, 896, 991–994, 1017 Orientation of vectors, 49–51, 116–117 Origins, 625 Origins, coordinate system, 48, 49 Orthogonal vectors, resolution into components, 38 Oscillators. See also Vibrations example problems, 1297–1300 forced harmonic, 1307–1310 railcar illustration, 1289–1291 simple, 644 Osculating circles, 704 Output power, electric motor, 921–922 Overdamped systems, 1322, 1324, 1342 Overdetermined problems, 73 Overhead door example, 1223–1224

P Packaging, energy absorbing, 809, 875, 895 Paper cutters, 181–182, 1163, 1165 Pappus of Alexandria, 461, 462, 487 Parabolic trajectories, 658, 1012, 1049 Parallel axis theorem about, A-68–A-70, A-74 with area moments of inertia, 575, 585–586, 612–613

example problems, 587–589 with mass moments of inertia, 595–596, 597, 614 Parallel force systems defined, 249, 250 three-force members subjected to, 304, 305 Parallel vector components, 40, 86, 118 Parallelogram addition of vectors, 33 Parallelogram areas, 105 Parametric analysis, 1084, 1104 Parker, Charlie, 1289 Parking orbits, 1018 Parthenon, 3 Partial fixity in plane trusses, 384 in rigid bodies, 300, 302–303, 354 in space trusses, 400, 401 Particle kinematics basic concepts, 649–653 differentiation of geometrical constraints, 739–741 elementary motions, 668–670, 775–776 major concepts, 775 normal-tangential components of planar motion, 703–704, 776–778 polar coordinates in planar motion, 719–720 projectile motion, 689–695, 776 rectilinear motions, 789–794 relative motion concepts, 738–739, 778 three-dimensional coordinate systems, 759–763 time derivative of a vector, 649–650 Particles angular momentum, 982–983 basic concepts, 626 conservative forces and potential energy, 877–881 defined, 6 linear impulse-momentum principle, 933–937 systems of, 837–839, 901–903 three-dimensional equilibrium, 166–168 treating Earth as, 991 two-dimensional equilibrium, 125–135 work-energy principle, 861–864, 926 Particular solutions, 1307–1308, 1343 Pascals, 468 Path component system, 812, 813 Paths. See also Trajectories defined, 650 normal-tangential components, 703–704, 706–709 relation to work, 877, 878–879 Pelton impulse wheels, 1042–1043 Pelton turbines, 1068 Pendulums ballistic, 973, 1270 Charpy impact test, 897 clock, 1304 conservation of angular momentum example, 997–998 Newton’s cradle, 976 spherical, 771 Perceived motions, 1128–1129 Perfect flexibility, 151 Perfectly aligned bearings, 322, 324

ISTUDY

Perfectly elastic impacts between particles, 958, 966–967, 1045 between rigid bodies, 1266, 1267, 1284 Perfectly plastic impacts between particles, 950, 958 between rigid bodies, 1266, 1283–1284 Periapses, 1009, 1010–1011 Perigean spring tides, 31 Perigee, 832, 896, 991–994, 1017 Periodic forcing, 1307 Periods of elliptical orbits, 1011–1012 Periods of vibration, 1290, 1323, 1340 Perpendicular vector components, 40, 87, 118 Phase angle, 1290, 1340 Philosophers, 3 Pickup tools, 410, 411 Pickup truck free body diagrams, 279, 280 Pin and cable support example, 332–333 Pin supports example problems, 332–333 free body diagrams in frames and machines, 418 reaction forces in three dimensions, 323 reaction forces in two dimensions, 273 Pioneer 3 spacecraft, 1135, 1195, 1263, 1264 Pipe, 184, 185, 454 Pipeline example, 1032–1033 Pistons basic functions, 1055–1056 characterizing motions, 1056, 1057, 1075 maximum accelerations, 1104 Pitch radii, 277 Planar motion. See also Rigid body kinematics; Rigid body kinetics bodies not symmetric with respect to plane of motion, 1370–1371, 1379–1380, 1390 general, 1057, 1059 normal-tangential components, 703–704, 776–778 polar coordinates, 719–720 relation to time derivative of a vector, 717–719 Plane angles, 631 Plane trusses, 364, 366, 426. See also Trusses Planes of motion, 1057 Planes of symmetry, 336 Planet carriers, 1101 Planetary gear system examples, 1082, 1101–1102 Plastic impacts, 958, 1045, 1266, 1283–1284 Platform bench scales, 975 Play structure, 95 Play structure examples, 38–39 Pliers, 246 Pneumatic door closers, 1085 Pneumatic tires as springs, 153, 154 Pneumatic tubes, 897 Pogo sticks, 191 Point forces, 468–473 Point mass, 6 Polar coordinates basic application to planar motion, 719–720 example problems in planar motion, 722–727 Newton’s second law applied, 812–814 Polar moment of inertia, 574, 575, 576, 612 Polar radius of gyration, 576, 612 Polar vector representation, 33, 35

Polar vector representation in two dimensions, 116 Pole vaulting examples, 886–887 Pool ladders, 418–419 Position acceleration as function, 668–670, 675, 775–776 coordinate systems and, 29–30, 52 defined, 775 relating to speed, 868–870 relation to velocity and acceleration, 7, 23 Position vectors applications, 58–59, 71–73 basic concepts, 52, 53 denoting in figures, 31–32 example problems, 636–637 form in Cartesian component systems, 652 notation, 7, 626 polar coordinates, 719–7205 properties, 650 in three dimensions, 67–68, 117 Positive wrenches, 250 Potential energy approximating as quadratic function of position and velocity, 1294 converting kinetic energy to, 886–887 defined, 879 rigid bodies in planar motion, 1210–1212 of various forces, 879, 926 Power basic concepts, 919, 928, 1212 computing engine output, 1221–1222 of couples, 1212 example problems, 920–922 Pratt trusses, 364 Precision of calculations, 13 Prefixes, 10–11, 11–12, 631–632 Preimpact velocities, 1266 Pressure angles, 277 Pressure cookers, 478 Primary reference frames, 1118 Primitive concepts, 625 Prince Felipe Science Museum, 363 Principal axes of inertia, A-70 Principal body axes, 1368, 1369 Principal moments of inertia, 644, A-70–A-73 Principal unit normals, 703–704 Principia, 5 Principle of material frame indifference, 620 Principle of moments, 207–208, 263 Principle of transmissibility, 246–247, 265 Problem evaluation in design, 344–345 Problem identification in design, 344 Problem solving, 131 Products of inertia about, A-66, A-67–A-68, A-69–A-70 in area moments of inertia, 574, 612 in mass moments of inertia, 593, 613 Projectile motion example problems, 70–73, 691–695, 707, 726–727, 816–817 modeling, 68, 689, 776 Propellers, 58–59, 680–681, 1356–1357, 1377–1378 Propulsion, mass flows in, 1026–1028 Prosthetic leg examples, 1087–1088, 1107–1108

Pruners, 411–413 Pulleys basic approaches to analyzing, 298–299 basic modeling of, 133–134, 195 belt friction against, 562–563 effects on cable strength, 185 examples, 132 fixed axis rotation, 1064 forces on trusses, 402–405 free body diagrams in frames, 411, 412 geometrical constraints illustrated by, 740–741, 746–747 support reactions of, 309–310 system examples, 842–843 in trusses, 374–375 in winch and block systems, 868–870 Pullout tests, 874 Pump turbine blades, 440–441, 581 Pumping while swinging, 890–891 “Punkin Chunkin” competition, 707 Pythagorean theorem, 434

Q Quadratic mean radius of Earth, 621 Queen Mary II, 260 Quick-return mechanisms, 1127, 1133

R Racquetballs, 940, 955 Radar gun examples, 742 Radial components of acceleration, 720 Radial components of velocity, 720 Radial symmetry, 1369–1370, 1389–1390 Radians, 11–12, 631 Radii of curvature, 704, 706, 707 Radii of gyration about, 1164, A-68 for areas, 576–577, 579, 612 for masses, 595, 605, 614 Radius of Earth, 621 Radius of torsion, 760 Railcar oscillation, 1289–1291, 1321–1322 Ramps, 540 Range-of-motion analysis, 336–337 Rascasse, 706 Ratchet pruners, 411–413 Rate of separation, 750, 751 RCS jets, 996 Reaction forces basic modeling of, 134 cantilevered beam example, 284 defined, 126 directions in free body diagrams, 275, 332 irrelevance to equivalent external force determinations, 252 for particles in three-dimensional equilibrium, 166 for particles in two-dimensional equilibrium, 128, 140–141, 156 on rigid bodies in three dimensions, 322, 323 on rigid bodies in two dimensions, 140–141, 272–273, 306 Reaction wheels, 996

I-9

Reciprocating rectilinear motions, 126–1127, 1132 Rectangular cross sections, area moments of inertia, 574, 579 Rectilinear motion force laws, 791–793 Newton’s second law applied, 789–791 polar coordinates example, 724–725 reciprocating, 126–1127, 1132 relations in particle kinematics, 668–670 Rectilinear translations, 1056 Red Arrows aerobatic team, 789 Reference frames inertial, 794, 798–799, 1030, A-77 rotating, 1118–1124, 1125–1129, 1139, 1349–1352, 1388 Reference points for position vectors, 650 Reference points of vectors, 29–30 Refrigerator supports, 542–543 Relative motion basic concepts, 738–739, 778 in constrained systems, 739–741, 778 example problems, 742–749 Relative slip, 560, 561 Reports, technical, 187 Request for proposals, 344 Rescue craft position vector example, 58–59 Resolution into vector components, 34–35, 38–39 Resonances, 1308–1309, 1342 Restaurant sign example, 454–455 Restitution coefficient. See Coefficient of restitution Restitution in particle collisions, 970 Resultant couple moments, 235, 240, 265 RFPs (request for proposals), 344 Right triangles, resolution into vector components, 38 Right-hand Cartesian coordinate system, 66, 625 Right-hand rule about, 625 for cross products, 101 for moment vectors, 205, 275, 331 Rigid bodies about, 626 defined, 6 equations in static equilibrium, 271–272, 353 equilibrium in three dimensions, 322–328 equilibrium in two dimensions, 272–280, 298–306 idealizing structures as, 247, 271, 298 transmissibility of forces in, 246–247, 265 Rigid body kinematics acceleration analysis in planar motion, 1097–1098 basic planar motion concepts, 1055–1062, 1136 rotating reference frames, 1118–1124 in three dimensions, 1349–1352, 1388 velocity analysis in planar motion, 1075–1079 Rigid body kinetics angular impulse-momentum principle, 1244–1245, 1283 angular momentum equations, 1146–1150, 1199–1200, A-75–A-78 general planar motion, 1177 impacts between bodies, 1265–1270, 1283–1284

I-10 ISTUDY

linear impulse-momentum principle, 1243–1244, 1283 linear momentum equations, 1145–1146 in planar motion, 1207–1208 rotation about a fixed axis, 1163–1164, 1201 in three dimensions, 1366–1372, 1388–1391, A-75–A-78 translation in plane of motion, 1153 work-energy principle, 1207–1212, 1282 Rigid body motion, 301 Ring gears, 1082, 1101 RMS Queen Mary II, 260 Road barrier example, 1217–1218 Robotic arm examples, 724–725 Rockers, 273 Rocket propulsion, 1026–1028, 1034–1035 Rocket sled, 673 Rollers, 273, 286 Rolling resistance, 1220 Rolling without slip acceleration analysis for, 1098–1099, 1138–1139 disk motion in three dimensions example, 1354–1355 friction examples, 1180–1181 impulse-momentum principle examples, 1248–1249 planetary gear examples, 1082, 1101–1102 velocity analysis for, 1075–1076, 1080, 1137 work-energy principle examples, 1215–1216, 1219–1220, 1225–1226 Roof, 100 Roof trusses, 364 Rotating reference frames, 1118–1124, 1125–1129, 1139, 1349–1352, 1388 Rotation about a fixed axis angular impulse-momentum principle, 1245 example problems, 1064–1065, 1168–1169, 1217–1218 kinetic energy of rigid bodies, 1208–1209 Newton-Euler equations, 1163–1164, 1201 qualitative description, 1056–1057 velocity and acceleration with, 1060–1061, 1136 Rotational kinetic energy, 1208 Rotational motions angular momentum equations for rigid bodies, 1146–1150 Euler’s equations for rigid bodies, 1368, 1389 Rough surfaces, common reaction forces, 323 Ryan, Nolan, 1051

S Safety as design priority, 181–182 Safety factors, 184–185, 196, 390–391 Sandia National Laboratory, 570 Satellites about, 450 acceleration during orbital motion, 723 angular momentum examples, 991–995 apogee and perigee, 832, 896, 991–994 orbital mechanics concepts, 1007–1013, 1047–1049 speed and orbital period example, 1017

Saturn V rocket, 1018 Saws, 498 Scalar approaches equilibrium of rigid bodies, 324–325 moment about a line, 221, 264 moment about a point, 204–205, 206, 207, 212–213, 263 moments of couples, 233, 264–265 Scalar components, 49, 53, 116 Scalar form, 126, 168 Scalar product, 84, 628. See also Dot products Scalar triple products, 105, 119 Scalars about, 6–7, 626–628 defined, 6, 23 multiplying vectors by, 33 School of Athens, 619 Scotch yokes, 1091, 1110 Scrapers, 411, 412 Screws, 478, 538 Second law of motion (Newton) about, 7–8, 23 applied to curvilinear motion, 723, 812–814 applied to rectilinear motion of particles, 789–794 as balance principle, 790 Euler’s first law compared, 986 framework for kinetics applications, 854 friction and spring forces, 791–793 impulse-momentum principle and, 933–934 inertial reference frame requirement, 794 scalar form, 630 statement of, 620, 625 in systems of particles, 837–839 work-energy principle and, 861–863 Second moments of an area, 575 Sections. See Method of sections Selective reinforcement of beams, 588 Self-aligning bearings, 322, 324 Self-locking, 558, 569 Semimajor axes, 1011 Separation velocity, 957 Sequences of collisions, 964–965 Serret-Frenet formulas, 760 Serret-Frenet triad, 704 Shackles, 133 Shafts, 576 Shear and moment diagrams basic features, 503 discontinuities in, 505 equations for, 514 graphical method for drawing, 516, 522–523 tips for drawing, 517–518 Shear forces relation to moment and distributed forces in beams, 514, 517, 527 in slender members, 494, 495 Sheet metal fabrication examples, 90, 158–159 Shift distance, 586, 597, 613, A-74 Shifting forks, 327, 328, 499 Shock absorbers, 743, 1321 Shotgun, 97 Shuttles on aircraft carrier catapults, 947 SI system, 11–12, 630–631, 632, 862–863 SI unit system, 9

ISTUDY

Sign conventions in cable and bar free body diagrams, 132, 133 for internal force problems, 494–495, 503 for moment problems, 205, 207, 322, 331 spring law, 153, 156, 195, 300 Significant digits, 13, 632 Silicon fracture, 837 Silicon nanowire example, 1298–1299 Simple space trusses, 400 Simple trusses, 384, 385, 427 Sine function, linearizing, 1294 Sine law, 34, 36–37, 116 Sinusoids, 1338 Skater’s spin, 989–990, 1250–1251 Ski jumpers, 698 Skydiver descent example, 678–679 Skysurfing, 759 Slackness in chains, 908 Slender members, 369, 493–495. See also Beams Slider-crank mechanisms acceleration analysis, 666, 1097–1098 types of rigid-body motions, 1055–1057, 1060–1061 velocity analysis, 666, 1075–1079, 1083–1086 Sliding motions, 540–543, 551, 552–555, 844–845, 888–889 Slip about, 560–563 example problems, 1156–1157 friction forces in, 791–792, 844–845, 888–889 Slope failure example, 108–109 Slugs, 9, 630 Small angle approximations, 12, 24 Smallest-distance problems, 94 Snapped towels, 909, 910 Society of Automotive Engineers (SAE), 346 Sockets, 323, 330–331 Solid-fuel rocket motors, 464–465 Solids, centers of gravity and mass, 449, 450, 451 Solids of revolution defined, 439 determining volume, 462, 487 example problems, 464–465, 600–601 Solution checking, 167 Sotomayor, Javier, 887 Sound speed, 676–677, 680, 681 Space, 5–6, 625 Space, defined, 5 Space curves, 650 Space Needle, 1255 Space Shuttle, 933, 996, 1252 Space Shuttle Columbia, 271 Space Station, 594, 595 Space trusses basic features, 399–400, 427 design considerations, 400–401 example problems, 402–405 Space X Dragon spacecraft position vector example, 72–73 Spacecraft Apollo 16 Lunar Module, 177 gravity tractors, 27 Space Shuttle Columbia, 271 Specific energy of satellites, 1013 Specific weight, 17–18, 20, 24, 456

Speed about, 99 converting to height, 886–887 defined, 651, 775 relating to position, 868–870 relating to velocity and acceleration, 38–39, 658–659 Speed governors, 1001 Spheres, particles versus, 1183 Spherical coordinate systems, 82–83 Spherical coordinates basic concepts, 762–763, 780 example problems, 769 Newton’s second law applied, 814 Spherical pendulums, 771 Spin rate of satellite, 995–996 Spinning motions, 903, 989–990 Spiral pumps, 731 Spring and rod example, 1186–1187 Spring constant, 793, 855 Spring law, 153, 195–196, 299, 300 Spring scales, 808, 850, 1311, 1329 Spring stiffness, 153, 300, 354 Spring tides, 31 Springs basic approaches to analyzing, 299–300 basic properties, 152–155, 195–196, 353–354 crate-on-incline example, 797 example problems, 156–159, 308 forces on, 793, 855 linear spring and disk example, 822–823 modeling cables as, 314 modeling packaging as, 809, 875, 895 in nanowire example, 1298–1299 potential energy, 879, 1211 railcar oscillation, 1289–1291, 1321–1322 railcar stopping example, 802–803 work on particles, 878, 927 Spur gears, 277 Square roots, negative, 70 Stability in space trusses, 400–401 Standard deviation, 573, 574 Standard form of forced harmonic oscillator equation, 1307–1310 of harmonic oscillator equation, 1291–1292, 1340 Standards engineering design, 346–347 for gears, 277 use in determining safety factors, 184 Static equilibrium. See also Equilibrium conditions of particles in, 126, 195 defined, 8, 23, 126 Static friction coefficient, 535, 536, 791, 854 Static friction force, 534 Statically determinate structures defined, 301, 354 identifying bodies as, 302–303, 354 plane trusses as, 383–384, 387, 389, 426–427 space trusses as, 400, 401 Statically indeterminate structures defined, 158, 301, 354 example problems, 158–159, 314

identifying bodies as, 302–303, 354 plane trusses as, 383–384, 387, 426–427 space trusses as, 400, 401 Static-kinetic friction transitions, 800–801 Statics defined, 1, 2 historical development, 3–5 Stationary frames, 738 Steady flows, 1023–1026, 1050 Steady-state vibration, 1308 Stealth fighter, 268 Steel bars, 183. See also Bars Steel cables, 68, 183. See also Cables Steel pipe, 184 Steering linkage example, 225 Steering wheels, 203–204, 220, 232 Stick, nonsliding condition, 534 Stiffness of springs, 153, 300, 354, 1294 Stomp rockets, 700 Stoner, Casey, 1145 Stop shots, 1278 Storage chests, 325, 326 Strength, internal forces and, 495 Stretch, 793 Strike plates, 702, 981 Strings as models, 640, 909–910, 1036–1037 Strong form of Newton’s third law, 620 Structural design codes, 346 Structural members, internal and external forces on, 68–69 Structured problem solving approach, 131 Structures, defined, 363 Struts, 285 Subtraction of vectors, 34 Summation of forces in directions other than x, y, or z, 167–168, 172–173 Summation points of moments, 276 Sun, 621, 1018 Sun gears, 1082, 1101 Sun’s gravitational force on Earth, 31 Super Balls, 28 Superposition, 300, 311, 508–509 Supersonic effects on propellers, 680, 681 Support structures, harmonic excitation, 1309 Supports basic reaction force concepts, 134 common three-dimensional reactions, 322, 323 common two-dimensional reactions, 272–273, 306 example problems, 330–337, 454–455 fixity of, 300–301, 354 Supports, harmonic excitation, 1342 Surface area, objects of revolution, 464–465 Surface forces, 468, 487 Surfaces center of gravity for, 451 center of mass for, 449, 450 of revolution, 461, 487 Sweet spots, 1166–1167 Swimming pool ladders, 418–419 Swinging block slider cranks, 735, 1085–1086 Swinging motion examples, 890–891 Swinging-bucket rotors, 1168 Symbols for vectors, 7, 620, 626

I-11

Symmetric geometry about, 1369–1370, 1389–1390 centroids, 435 moments of inertia with, 576, 595, 597, 600–603 simplified analysis with, 336 Symmetric objects, A-68 System modeling, 647 Systems (physical), work-energy principle, 1212 Systems of equations, solving, 166–167, 171 Systems of particles angular impulse-momentum principle, 983–986, 1047 closed versus open, 935, 1023 linear impulse-momentum principle, 935–936, 1044 Newton’s second law applied, 837–839, 855 work-energy principle, 901–903, 927–928, 1209 Systems of rigid bodies, 1245

T Tables, equivalent forces on, 254 Tacoma Narrows bridge, 21, 646 Tailhooks, 939, 1323 Tails of vectors, 30 Talladega Superspeedway, 818 Tangential components of acceleration, 703, 704, 707, 709 Tangent-normal-binomial components, 759–761, 779 Tapered prism example, 602–603 Taylor series expansions, 1292 Technical reports, 187 Technical writing, A-1–A-3 Telstar satellite, 995 Tennis balls, 974 Tensile forces in belts and cables on cylindrical surfaces, 560–563, 568 deformation under, 151 position and force vector examples, 71, 76–77 sign conventions, 132, 139, 153 Tensile loads in cables and bars, 151–152, 183, 184, 185 Tension, notation for, 139 Terminal velocity, 679 Test scores, 573–574 Testing for coefficients of friction, 536 Thelonious Monk Quartet, 1289 Theodolites, 79, 80 Thin, flat plate-type objects, 598–599 Thin disk volume elements, 444, 596, 600, 601 Thin shell volume elements, 444, 596, 600 Thin spaces, 632 Third law of motion (Newton) about, 7–8, 23, 71 effects on systems of particles, 838, 984 statement of, 620 Three-dimensional component systems, 813–814 Three-dimensional dynamics coordinate systems, 759–763 example problems, 1354–1359, 1376–1382 rigid body kinematics, 1349–1352, 1388 rigid body kinetics, 1366–1372, 1388–1391 Three-dimensional trusses, 399–401

I-12 ISTUDY

Three-force members example problems, 312–313 features of, 304, 305, 306, 355 Thrust bearings, 324 Thrust force, 1028 Tides, 29–30 Time about, 5–6 acceleration as function, 673 basic concepts, 5–6, 625 Time derivatives of vectors major concepts, 649–650, 717–719 notation, 29–30, 649–650 velocity and acceleration with, 653 Time differentiation, 1104 Time of impact in projectile motion, 691 Time-varying processes, 619–620 Tipping motions, 541, 1156–1157 Tires as springs, 153, 154 Toilet flushing, 1124 Top-slewing cranes, 766 Torque, 495, 1221–1222 Torsion, 760 Torsional spring constant, 1211, 1298 Torsional spring potential energy, 1211 Torsional springs, 300, 308, 354 Tow bars, 947 Towel snapping, 909, 910 Tower cranes, 125 Tracking, 722, 769 Traction (medical), 64 Tractions, 468, 487 Tractor removal examples, 40–42, 91 Traffic barriers, 541 Traffic signal cables, 186–187 Traffic signal poles, 409, 410 Trajectories. See also Paths; Projectile motion defined, 650, 775 normal-tangential components, 703–704, 706–709 parabolic, 658, 1012, 1049 relation to acceleration vectors, 652, 657, 658–659 relation to velocity vectors, 651 Transformation of vectors, 57, 63–64 Transient vibration, 1308 Translation defined, 1056 example problems, 1066, 1156–1157 rigid bodies in planar motion, 1153, 1200–1201 velocity and acceleration with, 1059–1060, 1136 Translational kinetic energy, 1208 Transmissibility of a force, 246–247, 265 Transmission tower example, 390–391 Trapeze catch example, 1276–1277 Trebuchets, 707 Tribology, 533–534. See also Friction forces Trigonometric identities, 869 Trivial solutions, 386 Trolley examples, 286 Tropicana Casino parking garage, 22 Trusses approximating structures as, 372–373, 389 basic features, 364–365, 426 designing, 384–387

example problems, 370–375, 389–391, 402–405 forces not at joints, 382 idealizing structures as, 365 method of joints, 365–367, 370–375, 426 method of sections, 380–382, 389–391, 426 static determinacy and indeterminacy, 383–384 three-dimensional, 399–401 typical members in, 369 zero-force members, 367–369 Turbine blades, 440–441, 581 Turbines, 1068 Two-dimensional Cartesian representation, 48–52, 54–59, 116–117 Two-dimensional component systems, 812–813 Two-dimensional problems, moments in, 206, 208, 209–211 Two-force members common types, 69 features of, 304–306, 355 identifying in FBDs, 409, 414 in trusses, 364–365

U Unbalanced motors, 1311–1315, 1332–1333 Unconstrained eccentric impacts, 1268–1269 Unconstrained impacts between particles, 956, 958–959, 968–969, 1045 between rigid bodies, 1266, 1268–1269, 1284 Undamped vibrations defined, 1290 forced, 1307–1310, 1341–1342 free, 1289–1294, 1340–1341 Underdamped systems, 1323, 1324, 1342–1344 Uniform gravity fields, 450 Unit conversions, 20, 631, 640 Unit tangent vectors, 704 Unit vectors acceleration and, 635 defined, 48, 116, 626 with dot products, 86–87 as functions of time, 726 in three dimensions, 66, 67 time derivatives, 717–718 in two dimensions, 48, 49 Units basic concepts, 9–13 conversion, 10, 14–15 omitting, 14 of power, 919 systems of, 10–12, 630–632 work and energy, 862–863 Universal gravitation, 620–622, 723, 1007. See also Gravitational attraction Universal gravitational constant, 15, 621, 1007 Unstretched length of a spring, 793 Uranium enrichment, 713 U.S. Customary system, 9, 11–12, 630–631, 862–863, 919

V V tails, 268 V-22 Osprey aircraft, 680 Variable mass flows, 1026–1028, 1034–1037, 1050

ISTUDY

Variable mass systems, 935, 1023 Variance, 573, 574 Varignon’s theorem, 207–208, 263 Vector addition about, 628 basic approaches, 32–33 with laws of sines and cosines, 34, 36–37, 116 using Cartesian components, 51–52, 54–59 Vector approaches acceleration analysis, 1097, 1101–1104, 1107–1108, 1138 equilibrium of rigid bodies, 324–325 moment about a line, 220–221, 264 moment about a point, 205–207, 208, 263 moments of couples, 233, 264 velocity analysis, 1075–1076, 1137 Vector components about, 627 Cartesian representation, 49, 116–117 defined, 34 optimization, 41–42 parallel and perpendicular, 40 resolution into, 34–35, 38–39 using dot product, 86–88, 91–93, 118 Vector differential operator, 880 Vector form, 126, 129, 168 Vector polygons, 33, 36, 40, 41–42 Vector product. See Cross products Vector transformation, 57, 63–64 Vectors basic concepts, 29–35 basic concepts in kinematics, 649–653 Cartesian representation, 6–7, 626–627 Cartesian three-dimensional representation, 66–69, 117 Cartesian two-dimensional representation, 48–52, 116–117 cross product concepts, 101–106, 118–119 defined, 6, 23 dot product concepts, 84–88, 117–118 finding components, 9–10, 629–630, 635 free, 234, 235, 265 notation, 7, 620, 626 notation for, 32, 232 operations defined, 628 Velocity about, 7, 23 acceleration as function, 674 determining from acceleration, 678–679

Velocity analysis example problems, 1081–1088 in rigid body planar motion, 1075–1079, 1137 rotating reference frames, 1118–1120, 1139 Velocity vectors example problems, 36–42, 655–659 form in Cartesian component systems, 653 normal-tangential components, 703, 704, 707 notation, 626 polar coordinates, 719–720 properties, 650–651, 775 Vertical area elements, centroids of, 440, 441 Verzasca dam, 884 Vibrations defined, 1289 example problems, 1297–1300, 1311–1315, 1329–1333 undamped and forced, 1307–1310, 1341–1342 undamped and free, 1289–1294, 1340–1341 viscously damped, 1321–1326, 1342–1344 Vibrations, isolating with springs, 158 Vlasi’c, Blanka, 861 Volume forces, 468, 487 Volumes centroids, 435, 439, 444 solids of revolution, 462, 464–465, 487 Volumetric flow rate, 1025, 1050 Volumetric radius of Earth, 621

W Wall-mounted jib cranes, 82 Warren trusses, 364 Water cooler support example, 188–189 Water jet example, 1030–1031 Watts (power), 919 Watts, Naomi, 683 Wedges, 538, 542–543, 554–555 Weight about, 631 calculating, 16, 19, 24 designing for inertial effects on, 188 mass versus, 10, 108, 138 Weldments, 270 Well depth measurement, 676–677 Wheel torque, 1221–1222 Whip cracking, 910 Whip-upon-whip purchases, 755 Whistle, 95

Williams, Dave, 1207 Winch, 98 Winch-and-block systems, 868–870 Wings lift coefficients, 644 modeling, 647 propellers as, 681 Wires center of gravity for, 451 center of mass for, 449, 450 internal and external forces on, 68–69 as two-force members, 304 Woods, Tiger, 817 Work calculating for a force, 863–864 of central forces on particles, 877–878 of force systems on rigid bodies, 1209–1210 relation to kinetic energy, 862–863, 926 Work-energy principle conservative forces and potential energy, 877–881 example problems, 865–870, 883–892, 905–912, 962, 1215–1226 for impacts, 962, 1275 particle applications, 861–864, 926 in rigid body kinetics, 1207–1212, 1282, 1371–1372, 1390–1391 for systems of particles, 901–903, 927–928 Working loads, 54–55, 183–185 World Championship “Punkin Chunkin” competition, 707 World records, 886, 887 World’s Strongest Man competition, 941 Wrecking balls, 815, 883 Wrench force systems, 250, 256–257, 266

Y Yardstick natural frequency, 1299 Yielding, 184. See also Failure Young, Thomas, 5

Z Zectron, 28 Zero vectors, 49, 103 Zero-force members benefits in trusses, 386–387, 426 defined, 304, 355 identifying in trusses, 367–369, 370, 426

I-13

ISTUDY

U.S. Customary and SI unit systems

System of Units Base Dimension

U.S. Customary

SI

force

pound (lb)

newtona (N) ≡ kg⋅m∕s2

mass

sluga ≡ lb⋅s2 ∕ft

kilogram (kg)

length

foot (ft)

meter (m)

time

second (s)

second (s)

a Derived

unit.

Conversion factors between U.S. Customary and SI unit systems U.S. Customary

length

force

mass

SI

1 in.

=

0.0254 m (2.54 cm, 25.4 mm)a

1 f t (12 in.)

=

0.3048 ma

1 mi (5280 f t)

=

1.609 km

1 lb

=

4.448 N

1 kip (1000 lb)

=

4.448 kN

1 slug (1 lb⋅s2 ∕f t)

=

14.59 kg

a Exact.

Common prefixes used in the SI unit systems Multiplication Factor 1 000 000 000 000 000 000 000 000 1 000 000 000 000 000 000 000 1 000 000 000 000 000 000 1 000 000 000 000 000 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100 10 0.1 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001 0.000 000 000 000 001 0.000 000 000 000 000 001 0.000 000 000 000 000 000 001 0.000 000 000 000 000 000 000 001

24

10 1021 1018 1015 1012 109 106 103 102 101 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24

Prefix

Symbol

yotta zetta exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto zepto yocto

Y Z E P T G M k h da d c m 𝜇 n p f a z y

ISTUDY

Properties of lines and areas. Length 𝐿, area 𝐴, centroid 𝐶, and area moments of inertia. Quarter & Semicircular Arcs 𝐿=

𝜋 𝑟 2

Segment of Circular Arc 𝐿 = 2𝛼𝑟,

𝐿 = 𝜋𝑟

𝑟 𝐶

2𝑟 𝜋

𝐶

𝛼 𝛼

𝑟

2𝑟 𝜋

𝑦

ℎ 2

𝑦

𝑏 3

= =

𝑥

1 ℎ𝑏3 , 12 1 ℎ𝑏3 3

ℎ 3

𝑏 𝐼𝑥 = 𝐼𝑥 =

𝐶 𝐼𝑥 = 𝐼𝑦 = 𝜋4 𝑟4 𝐽𝐶 = 𝜋2 𝑟4

1 𝑏ℎ3 , 𝐼𝑦 36 1 𝑏ℎ3 , 𝐼𝑦 12

= =

1 ℎ𝑏3 , 36 1 ℎ𝑏3 4

𝑥

𝑟 𝐼𝑥 =

𝜋 8

4𝑟 3𝜋

𝑥

−

𝑟4 , 𝐼𝑦 = 81 𝜋𝑟4 , 𝐼𝑦 = 18 𝜋𝑟4

𝐼𝑥 =

1 𝑏ℎ3 , 36 1 𝑏ℎ3 12

Quarter Circular Area 𝑦 𝑦

4𝑟 3𝜋

𝑥 8 9𝜋

𝐼𝑥 =

𝐴 = 41 𝜋𝑟2

𝑦, 𝑦

𝐶

𝐼𝑥 =

𝑥

ℎ 3

𝑏

Semicircular Area 𝐴 = 12 𝜋𝑟2

𝑦

𝑟

𝑥

𝐶

𝑥

Circular Area 𝐴 = 𝜋𝑟2

𝑦

ℎ

𝑥

𝐶

𝑏 1 𝑏ℎ3 , 𝐼𝑦 12 𝐼𝑥 = 13 𝑏ℎ3 , 𝐼𝑦

𝐼𝑥 =

Triangular Area 𝐴 = 12 𝑏ℎ

𝑦

ℎ

𝑥

𝐶

ℎ

Right Triangular Area 𝐴 = 12 𝑏ℎ

𝑦

𝑏 2

𝐶

𝑟 sin 𝛼 𝛼

Rectangular Area 𝐴 = 𝑏ℎ

𝛼 in radians

𝑟

𝐶 𝐼𝑥 = 𝐼𝑦 =

4𝑟 3𝜋

𝑟

𝑥 4 − 9𝜋 1 𝜋𝑟4 16

𝜋 16

𝐼 𝑥 = 𝐼𝑦 =

𝑥 𝑟4 ,

ISTUDY

Properties of solids. Volume 𝑉, center of mass and centroid 𝐺, cross-sectional area 𝐴, and mass moments of inertia for homogeneous bodies. For thin solids, thickness t or rod diameter d is negligible compared to other dimensions. Thin Quarter-Circular Rod

Thin Rod 𝑉 = 𝐴𝑙

𝑧

𝑉 = 21 𝜋𝑟𝐴

𝑧 𝑙∕2

2𝑟 𝜋

𝑟

2𝑟 𝜋

𝐴 𝐺

𝑦

𝐺 𝑟

1 𝑚𝑙2 12

𝐼 𝑥 = 𝐼𝑧 =

𝑚𝑟2

𝑧

𝑧

𝐼𝑥 = 𝐼𝑦 = 𝐼𝑧 = 25 𝑚𝑟2

𝐼 𝑥 = 𝐼𝑦 =

83 𝑚𝑟2 , 𝐼𝑧 320

𝑐 𝑥

𝑥

𝐼𝑦 = 𝐼𝑧 =

+ 𝑐2

𝑎2

+ 𝑏2

1 𝑚 3𝑟2 12 𝐼𝑧 = 12 𝑚𝑟2

𝐼 𝑥 = 𝐼𝑦 =

,

𝑉 = 31 𝜋𝑟2 ℎ

𝑧

𝑎

𝑉 =

𝐺 𝑦

1 𝑚ℎ2 + 𝑚𝑟2 14 − 162 12 9𝜋 1 𝐼𝑦 = 12 𝑚ℎ2 + 14 𝑚𝑟2 , 𝐼𝑧 = 𝑚𝑟2 12 − 162 9𝜋

𝑥

𝑡

𝑏 4

ℎ

1 1 𝑚𝑎2 , 𝑚𝑏2 , 𝐼𝑦 = 12 𝐼𝑥 = 12 1 𝐼𝑧 = 12 𝑚 𝑎2 + 𝑏 2

𝐼𝑥 =

𝑦

ℎ 2

Right Tetrahedron 𝑧 𝑉 = 16 𝑎𝑏𝑐

𝐺 𝑥

ℎ 2

𝑥

𝐼𝑥 =

+ ℎ2 ,

𝑏

𝑦

4𝑟 3𝜋

𝑧

𝑐

𝐺 𝑥

1 2 𝜋𝑟 ℎ 2

ℎ 2

Cone 𝑧

Thin Rectangular Plate 𝑉 = 𝑎𝑏𝑡

𝐼𝑥 = 𝐼𝑦 = 41 𝑚𝑟2 , 𝐼𝑧 = 21 𝑚𝑟2 Half Cylinder

ℎ 2

𝑦 𝑏2 + 𝑐 2 ,

𝑦

𝑟 𝐺

𝐺

𝐺

𝑡

𝑟

𝑏

𝑎2

𝑥

= 25 𝑚𝑟2

Cylinder 𝑧

𝑉 = 𝜋𝑟2 ℎ

𝑧

1 𝑚 12 1 𝑚 12 1 𝑚 12

𝑟 3 𝑟 8

𝑦

Rectangular Prism

𝐼𝑥 =

𝑧

𝑉 = 𝜋𝑟2 𝑡

𝑟

𝑥

𝑦

=

𝑚𝑟2

Thin Circular Disk

𝐺

𝑥

𝑎

𝐼 𝑥 = 𝐼𝑦 =

,

1 𝑚𝑟2 , 𝐼𝑧 2

8 𝜋2

Hemisphere

𝐺

𝑉 = 𝑎𝑏𝑐

4 𝜋2

−

Sphere

𝑟

𝑦

𝑥 1 2

𝐼𝑦 = 𝑚𝑟2 1 −

𝑉 = 32 𝜋𝑟3

𝐴

𝐺

𝑦

𝑥

𝑥

𝑉 = 43 𝜋𝑟3

𝑧

𝑉 = 2𝜋𝑟𝐴

𝐴

𝑙∕2

𝐼𝑥 = 0, 𝐼𝑦 = 𝐼𝑧 =

Thin Ring

𝑟 3 𝑚 4𝑟2 𝐼𝑦 = 80 3 𝑚𝑟2 𝐼𝑧 = 10

ℎ 4

+ ℎ2

𝑦 𝑦

,

𝐺

𝑐 4

𝑎 4

𝑏

𝑎 𝐼𝑥 = 𝐼𝑦 = 𝐼𝑧 =

3 𝑚 80 3 𝑚 80 3 𝑚 80

𝑏2 + 𝑐 2 , 𝑎2 + 𝑐 2 , 𝑎2 + 𝑏2

,