Mathematics for Machine Technology [8 ed.] 1337798312, 9781337798310

Table of contents :
Title Page
Copyright Page
Contents
Preface
Section 1: Common Fractionsand Decimal Fractions
Unit 1: Introduction to Common Fractions and Mixed Numbers
Unit 2: Addition of Common Fractions and Mixed Numbers
Unit 3: Subtraction of Common Fractions and Mixed Numbers
Unit 4: Multiplication of Common Fractions and Mixed Numbers
Unit 5: Division of Common Fractions and Mixed Numbers
Unit 6: Combined Operations of Common Fractions and Mixed Numbers
Unit 7: Computing with a Calculator: Fractions and Mixed Numbers
Unit 8: Computing with a Spreadsheet: Fractions and Mixed Numbers
Unit 9: Introduction to Decimal Fractions
Unit 10: Rounding Decimal Fractions and Equivalent Decimal and Common Fractions
Unit 11: Addition and Subtraction of Decimal Fractions
Unit 12: Multiplication of Decimal Fractions
Unit 13: Division of Decimal Fractions
Unit 14: Powers
Unit 15: Roots
Unit 16: Table of Decimal Equivalents and Combined Operations of Decimal Fractions
Unit 17: Computing with a Calculator: Decimals
Unit 18: Computing with a Spreadsheet: Decimals
Unit 19: Achievement Review—Section One
Section 2: Ratio, Proportion, and Percentage
Unit 20: Ratio and Propor tion
Unit 21: Direct and Inverse Proportions
Unit 22: Introduction to Percents
Unit 23: Basic Calculations of Percentages, Percents, and Rates
Unit 24: Percent Practical Applications
Unit 25: Achievement Review—Section Two
Section 3: Linear Measurement: Customary (English) and Metric
Unit 26: Customary (English) Units of Measure
Unit 27: Metric Units of Linear Measure
Unit 28: Degree of Precision, Greatest Possible Error, Absolute Error, and Relative Error
Unit 29: Tolerance, Clearance, and Interference
Unit 30: Customary and Metric Steel Rules
Unit 31: Customary Vernier Calipers and Height Gages
Unit 32: Metric Vernier Calipers and Height Gages
Unit 33: Digital Calipers and Height Gages
Unit 34: Customary Micrometers
Unit 35: Metric Vernier Micrometers
Unit 36: Digital Micrometers
Unit 37: Customary and Metric Gage Blocks
Unit 38: Achievement Review—Section Three
Section 4: Fundamentals of Algebra
Unit 39: Symbolism and Algebraic Expressions
Unit 40: Signed Numbers
Unit 41: Algebraic Operations of Addition, Subtraction, and Multiplication
Unit 42: Algebraic Operations of Division, Powers, and Roots
Unit 43: Introduction to Equations
Unit 44: Solution of Equations by the Subtraction, Addition, and Division Principles of Equality
Unit 45: Solution of Equations by the Multiplication, Root, and Power Principles of Equality
Unit 46: Solution of Equations Consisting of Combined Operations and Rearrangement of Formulas
Unit 47: Applications of Formulas to Cutting Speed, Revolutions per Minute, and Cutting Time
Unit 48: Applications of Formulas to Spur Gears
Unit 49: Achievement Review—Section Four
Section 5: Fundamentals of Plane Geometry
Unit 50: Lines and Angular Measure
Unit 51: Protractors—Simple Semicircular and Vernier
Unit 52: Types of Angles and Angular Geometric Principles
Unit 53: Introduction to Triangles
Unit 54: Geometric Principles for Triangles and Other Common Polygons
Unit 55: Introduction to Circles
Unit 56: Arcs and Angles of Circles, Tangent Circles
Unit 57: Fundamental Geometric Constructions
Unit 58: Achievement Review—Section Five
Section 6: Geometric Figures: Areas and Volumes
Unit 59: Areas of Rectangles, Parallelograms, and Trapezoids
Unit 60: Areas of Triangles
Unit 61: Areas of Circles, Sectors, and Segments
Unit 62: Volumes of Prisms and Cylinders
Unit 63: Volumes of Pyramids and Cones
Unit 64: Volumes of Spheres and Composite Solid Figures
Section 7: Trigonometry
Unit 66: Introduction to Trigonometric Functions
Unit 67: Analysis of Trigonometric Functions
Unit 68: Basic Calculations of Angles and Sides of Right Triangles
Unit 69: Simple Practical Machine Applications
Unit 70: Complex Practical Machine Applications
Unit 71: The Cartesian Coordinate System
Unit 72: Oblique Triangles: Law of Sines and Law of Cosines
Unit 73: Achievement Review—Section Seven
Section 8: Compound Angles
Unit 74: Introduction to Compound Angles
Unit 75: Drilling and Boring Compound-Angular Holes: Computing Angles of Rotation and Tilt Using Given Lengths
Unit 76: Drilling and Boring Compound-Angular Holes: Computing Angles of Rotation and Tilt Using Given Angles
Unit 77: Machining Compound-Angular Surfaces: Computing Angles of Rotation and Tilt
Unit 78: Computing Angles Made by the Intersection of Two Angular Surfaces
Unit 79: Computing Compound Angles on Cutting and Forming Tools
Unit 80: Achievement Review—Section Eight
Section 9: Computer Numerical Control (CNC)
Unit 81: Introduction to Computer Numerical Control (CNC)
Unit 82: Control Systems, Absolute Positioning, Incremental Positioning
Unit 83: Location of Points: Polar Coordinate System
Unit 84: Binary Numeration System
Unit 85: Hexadecimal Numeration System
Unit 86: BCD (Binary Coded Decimal) Numeration Systems
Unit 87: An Introduction to G- and M-Codes for CNC Programming
Unit 88: Achievement Review—Section Nine
Appendix
Appendix A: United States Customary and Metric Units of Measure
Appendix B: Principles of Plane Geometry
Appendix C: Formulas for Areas (A ) of Plane Figures
Appendix D: Formulas for Volumes (V ) of Solid Figures
Appendix E: Trigonometry
Appendix F: Common G- and M-Codes
Answers to odd-numbered ApplicAtions
Index

Citation preview

MatheMatics

for Machine technology eighth edition

John C. Peterson Robert D. Smith

Australia

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Mathematics for Machine Technology, 8e

© 2020, 2016 Cengage Learning, Inc.

John C. Peterson Robert D. Smith

Unless otherwise noted, all content is © Cengage.

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Cengage is a leading provider of customized learning solutions with employees residing in nearly 40 different countries and sales in more than 125 countries around the world. Find your local representative at www .cengage.com. Cengage products are represented in Canada by Nelson Education, Ltd. To learn more about Cengage platforms and services, register or access your online learning solution, or purchase materials for your course, visit www.cengage.com. Notice to the Reader Publisher does not warrant or guarantee any of the products described herein or perform any independent analysis in connection with any of the product information contained herein. Publisher does not assume, and expressly disclaims, any obligation to obtain and include information other than that provided to it by the manufacturer. The reader is expressly warned to consider and adopt all safety precautions that might be indicated by the activities described herein and to avoid all potential hazards. By following the instructions contained herein, the reader willingly assumes all risks in connection with such instructions. The publisher makes no representations or warranties of any kind, including but not limited to, the warranties of fitness for particular purpose or merchantability, nor are any such representations implied with respect to the material set forth herein, and the publisher takes no responsibility with respect to such material. The publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or part, from the readers’ use of, or reliance upon, this material.

Printed in the United States of America Print Number: 01 Print Year: 2019

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Contents vi

PREFACE

1

sECtion 1

UNIT 1 UNIT 2 UNIT 3 UNIT 4 UNIT 5 UNIT 6 UNIT 7

Introduction to Common Fractions and Mixed Numbers / 1 Addition of Common Fractions and Mixed Numbers / 8 Subtraction of Common Fractions and Mixed Numbers / 14 Multiplication of Common Fractions and Mixed Numbers / 20 Division of Common Fractions and Mixed Numbers / 27 Combined Operations of Common Fractions and Mixed Numbers / 33 Computing with a Calculator: Fractions and Mixed Numbers / 40

UNIT 8 UNIT 9 UNIT 10

Computing with a Spreadsheet: Fractions and Mixed Numbers / 49 Introduction to Decimal Fractions / 57 Rounding Decimal Fractions and Equivalent Decimal and Common Fractions / 63 Addition and Subtraction of Decimal Fractions / 69 Multiplication of Decimal Fractions / 74 Division of Decimal Fractions / 78 Powers / 84 Roots / 91 Table of Decimal Equivalents and Combined Operations of Decimal Fractions / 97 Computing with a Calculator: Decimals / 103 Computing with a Spreadsheet: Decimals / 112 Achievement Review—Section One / 121

UNIT 11 UNIT 12 UNIT 13 UNIT 14 UNIT 15 UNIT 16 UNIT 17 UNIT 18 UNIT 19

126

sECtion 2

UNIT 20 UNIT 21 UNIT 22 UNIT 23 UNIT 24 UNIT 25

160

Common Fractions and Decimal Fractions

Ratio and Propor tion / 126 Direct and Inverse Proportions / 135 Introduction to Percents / 141 Basic Calculations of Percentages, Percents, and Rates / 145 Percent Practical Applications / 150 Achievement Review—Section Two / 157

sECtion 3

UNIT 26 UNIT 27 UNIT 28 UNIT 29

Ratio, Propor tion, and Percentage

Linear Measurement: Customary (English) and Metric

Customary (English) Units of Measure / 160 Metric Units of Linear Measure / 169 Degree of Precision, Greatest Possible Error, Absolute Error, and Relative Error / 178 Tolerance, Clearance, and Interference / 183 iii

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Contents

UNIT 30 UNIT 31 UNIT 32 UNIT 33 UNIT 34 UNIT 35 UNIT 36 UNIT 37 UNIT 38

244

sECtion 4

UNIT 39 UNIT 40 UNIT 41 UNIT 42 UNIT 43 UNIT 44 UNIT 45 UNIT 46 UNIT 47 UNIT 48 UNIT 49

347

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Fundamentals of Plane Geometry

Lines and Angular Measure / 347 Protractors—Simple Semicircular and Vernier / 363 Types of Angles and Angular Geometric Principles / 370 Introduction to Triangles / 380 Geometric Principles for Triangles and Other Common Polygons / 387 Introduction to Circles / 399 Arcs and Angles of Circles, Tangent Circles / 409 Fundamental Geometric Constructions / 420 Achievement Review—Section Five / 430

sECtion 6

UNIT 59 UNIT 60 UNIT 61 UNIT 62 UNIT 63 UNIT 64 UNIT 65

Fundamentals of Algebra

Symbolism and Algebraic Expressions / 244 Signed Numbers / 253 Algebraic Operations of Addition, Subtraction, and Multiplication / 267 Algebraic Operations of Division, Powers, and Roots / 274 Introduction to Equations / 287 Solution of Equations by the Subtraction, Addition, and Division Principles of Equality / 295 Solution of Equations by the Multiplication, Root, and Power Principles of Equality / 306 Solution of Equations Consisting of Combined Operations and Rearrangement of Formulas / 313 Applications of Formulas to Cutting Speed, Revolutions per Minute, and Cutting Time / 324 Applications of Formulas to Spur Gears / 334 Achievement Review—Section Four / 343

sECtion 5

UNIT 50 UNIT 51 UNIT 52 UNIT 53 UNIT 54 UNIT 55 UNIT 56 UNIT 57 UNIT 58

437

Customary and Metric Steel Rules / 192 Customary Vernier Calipers and Height Gages / 203 Metric Vernier Calipers and Height Gages / 212 Digital Calipers and Height Gages / 215 Customary Micrometers / 218 Metric Vernier Micrometers / 226 Digital Micrometers / 231 Customary and Metric Gage Blocks / 234 Achievement Review—Section Three / 238

Geometric Figures: Areas and Volumes

Areas of Rectangles, Parallelograms, and Trapezoids / 437 Areas of Triangles / 449 Areas of Circles, Sectors, and Segments / 454 Volumes of Prisms and Cylinders / 462 Volumes of Pyramids and Cones / 473 Volumes of Spheres and Composite Solid Figures / 481 Achievement Review—Section Six / 488

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Contents

494

sECtion 7

UNIT 66 UNIT 67 UNIT 68 UNIT 69 UNIT 70 UNIT 71 UNIT 72 UNIT 73

563

UNIT 76 UNIT 77 UNIT 78 UNIT 79 UNIT 80

610

Machining Compound-Angular Surfaces: Computing Angles of Rotation and Tilt / 581 Computing Angles Made by the Intersection of Two Angular Surfaces / 590 Computing Compound Angles on Cutting and Forming Tools / 598 Achievement Review—Section Eight / 608

Computer Numerical Control (CNC)

Introduction to Computer Numerical Control (CNC) / 610 Control Systems, Absolute Positioning, Incremental Positioning / 621 Location of Points: Polar Coordinate System / 631 Binary Numeration System / 638 Hexadecimal Numeration System / 645 BCD (Binary Coded Decimal) Numeration Systems / 650 An Introduction to G- and M-Codes for CNC Programming / 655 Achievement Review—Section Nine / 669

APPEndixEs

A B C D E F

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Compound Angles

Introduction to Compound Angles / 563 Drilling and Boring Compound-Angular Holes: Computing Angles of Rotation and Tilt Using Given Lengths / 567 Drilling and Boring Compound-Angular Holes: Computing Angles of Rotation and Tilt Using Given Angles / 573

sECtion 9

UNIT 81 UNIT 82 UNIT 83 UNIT 84 UNIT 85 UNIT 86 UNIT 87 UNIT 88

674

Introduction to Trigonometric Functions / 494 Analysis of Trigonometric Functions / 506 Basic Calculations of Angles and Sides of Right Triangles / 511 Simple Practical Machine Applications / 519 Complex Practical Machine Applications / 529 The Cartesian Coordinate System / 542 Oblique Triangles: Law of Sines and Law of Cosines / 546 Achievement Review—Section Seven / 558

sECtion 8

UNIT 74 UNIT 75

Trigonometry

United States Customary and Metric Units of Measure / 674 Principles of Plane Geometry / 676 Formulas for Areas (A ) of Plane Figures / 678 Formulas for Volumes (V ) of Solid Figures / 679 Trigonometry / 680 Common G- and M-Codes / 682

684

AnswERs to odd-numbEREd APPliCAtions

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Preface “I can’t think of any job in my sheet metal fabrication company where math is not important. If you work in a manufacturing facility, you use math every day; you need to compute angles and understand what happens to a piece of metal when it’s bent to a certain angle.”1— TRACI TAPANI, WyoMING MAChINE Mathematics for Machine Technology is written to overcome the often mechanical “plug in” approach found in many trade-related mathematics textbooks. An understanding of mathematical concepts is stressed in all topics ranging from general arithmetic processes to oblique trigonometry, compound angles, and numerical control. Both content and method are those that have been used by the authors in teaching applied machine technology mathematics classes for apprentices in machine, tool-anddie, and tool design occupations. Each unit is developed as a learning experience based on preceding units—making prerequisites unnecessary. Presentation of basic concepts is accompanied by realistic industry-related examples and actual industrial applications. The applications progress from the simple to those with solutions that are relatively complex. Many problems require the student to work with illustrations such as are found in machine technology handbooks and engineering drawings. Great care has been taken in presenting explanations clearly and in providing easy-tofollow procedural steps in solving exercise and problem examples. The book contains a sufficient number of exercises and problems to permit the instructor to selectively plan assignments. An analytical approach to problem solving is emphasized in the geometry, trigonometry, compound angle, and numerical control sections. This approach is necessary in actual practice in translating engineering drawing dimensions to machine working dimensions. Integration of algebraic and geometric principles with trigonometry by careful sequence and treatment of material also helps the student in solving industrial applications. The Instructor’s Guide provides answers and solutions for all problems. A majority of instructors state that their students are required to perform basic arithmetic operations on fractions and decimals prior to calculator usage. Thereafter, the students use the calculator almost exclusively in problem-solving computations. Calculator instructions and examples have been updated and expanded in this edition. The scientific calculator and the Machinist calc Pro 2 are introduced in the Preface. Extensive calculator instruction and examples are given directly following the units on fractions and mixed numbers and the units on decimals. Further calculator instruction and examples are given throughout the text wherever calculator applications are appropriate to the material presented. Source: Thomas Friedman, “If you’ve Got the Skills, She’s Got the Job,” New York Times, November 17, 2012, accessed November 18, 2012, http://www.nytimes.com/2012/11/18/opinion/sunday/Friedman-you-Got-the-Skills.html

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Many students and workers use tablet computers. one of the advantages of a tablet is that apps for scientific calculators, the Machinist calc Pro 2, and spreadsheets can be installed on a tablet. This text includes both spreadsheet and calculator instructions. As with the calculator, extensive spreadsheet instruction and examples are given directly following the unit on using calculators with fractions and mixed numbers and the unit on using calculators with decimals. Further spreadsheet instruction and examples are given throughout the text wherever spreadsheet applications are appropriate. Changes from the previous edition have been made to improve the presentation of topics and to update material. A survey of instructors using the seventh edition was conducted. Based on their comments and suggestions, changes were made. The result is an updated and improved eighth edition that includes the following revisions: Four major changes were made in this edition. ●●

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Spreadsheet instruction was added for tablets like the iPad, Microsoft Surface, and Samsung Galaxy. Many workers are using laptops on the job because of their convenience and portability. While tablets are bulkier than calculators, they have the advantage of having both spreadsheet and calculator apps, so a user needs only one machine. A unit on introductory G- ad M-codes for CNC programming was added. This unit is only designed to introduce students to the G-codes that prepare a machine to engage in a particular mode for machining and the M-codes that are used to turn on and off miscellaneous functions. Selected calculator instruction was included for the Machinist Calc™ Pro 2, Model 4088. This particular calculator can be used with material and tool settings combined with DoC (depth of cut) and WoC (width of cut) to solve speed and feed calculations for face, end, or slot milling plus turning, drilling and boring. It also gives step-saving drill and thread chart lookups, right triangle solutions, bolt-circle patterns, etc. It does have some limitations, particularly when working with fractions, and these limitations are pointed out in the text. Many existing calculator instructions were deleted. Students are familiar with calculators and have probably been using them since elementary school and should not need basic instruction in calculator usage. Since most new scientific calculators use a “natural display,” much of the duplicate instructions for calculator keystrokes has also been deleted. A supplement was prepared that includes Machinist Calc™ Pro 2, Model 4088 instructions for all places in the text where calculators are used. The supplement also has instructions that show how the calculator can be used for some topics that are not part of the text.

About the Authors John C. Peterson is a retired professor of mathematics at Chattanooga State Community College, Chattanooga, Tennessee. Before he began teaching, he worked in the manufacturing industry. he has taught at the middle school, high school, two-year college, and university levels. Dr. Peterson is the author or coauthor of four other Cengage Learning books: Introductory Technical Mathematics (with Robert D. Smith), Technical Mathematics, Technical Mathematics with Calculus, and Math for the Automotive Trade (with William J. deKryger). he was a member of the four-person team that revised and interpreted the

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viii

PrefaCe

two-year college portion of the 2015 College Board of Mathematical Sciences crosssectional survey of undergraduate mathematical science programs. In addition, he has had over 90 papers published in various journals, has given over 200 presentations, and has served as a vice president of the American Mathematical Association of Two-year Colleges. Robert D. Smith was Associate Professor Emeritus of Industrial Technology at Central Connecticut State University, New Britain, Connecticut. Mr. Smith had experience in the manufacturing industry as tool designer, quality control engineer, and chief manufacturing engineer. he also taught applied mathematics, physics, and industrial materials and processes on the secondary technical school level and machine technology applied mathematics for apprentices in machine, tool-and-die, and tool design occupations. he was the author of Technical Mathematics 4e, also published by Cengage.

Acknowledgments The publisher wishes to acknowledge the following instructors for their detailed reviews of this text: Jack Krikorian Tooling and Manufacturing Association Schaumburg, Illinois Larry Lichter Waukesha County Technical College Pewaukee, Wisconsin Michael A. organek Monroe Community College Rochester, New york In addition, the publisher and author acknowledge Linda Willey for her tireless commitment to the technical review of the text, examples, applications, answers, and solutions.

IntroductIon to the scIentIfIc cAlculAtor A scientific calculator is to be used in conjunction with the material presented in this textbook. Complex mathematical calculations can be made quickly, accurately, and easily with a scientific calculator. Although most functions are performed in the same way, there are some variations among different makes and models of scientific calculators. In this book, generally, where there are two basic ways of performing a function, or sequencing, both ways are shown. however, not all of the differences among the various makes and models of calculators can be shown. It is very important that you become familiar with the operation of your scientific calculator. An owner’s manual or user’s guide is included with the purchase of a scientific calculator; it explains the essential features and keys of the specific calculator, as well as providing information on the proper use. It is important that the owner’s manual or user’s guide be studied and referred to whenever there is a question regarding calculator usage. Also, information can be obtained from the manufacturer’s Internet website, which is often listed in the user’s guide. Most scientific calculator keys can perform more than one function. Depending on the calculator, generally the 2nd key or SHIFT key enable you to use alternate functions. The alternate functions are marked above the key. Alternate functions are shown and explained in the book where their applications are appropriate to specific content.

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IntroductIon to the Machinist calc Pro 2 The Machinist calc Pro 2 is more than just a calculator. It is a dedicated mathematics and reference tool that provides a complete assortment of machine shop solutions. It has more complete thread data than the Machinery’s Handbook, including step-saving thread and drill size chart lookups, speeds and feeds, right triangle solutions, and bolt patterns. It also provides hundreds of fast, precise machining-specific solutions for turning, drilling, boring, and face, end, and slot milling. Built-in tables for 20 materials, six processes and three tools means that you spend much less time looking up your most needed calculations on charts, in books, or on the Internet and more time machining. Although most functions are performed in the same way on the Machinist calc Pro 2 as on other calculators, there are some important differences. A User’s Guide is included with the purchase of the Machinist calc Pro 2 and explains the essential features and keys of the calculator, as well as providing detailed information on proper use. It is essential that the User’s Guide be studied and referred to whenever there is a question regarding calculator use. The Calculated Industries Website2 has some video tutorials that show how to use the calculator for some specific needs. In addition, apps for the iPhone, iPad, and Android mobile devices can be downloaded from the webpage. Because the Machinist calc Pro 2 is designed for machinists, it cannot perform all the mathematics functions needed for this text. however, in many cases, examples show how to use the Machinist calc Pro 2 in these situations.

decIsIons regArdIng cAlculAtor use The exercises and problems presented throughout the text are well suited for solutions using a calculator. however, it is felt that decisions regarding calculator usage should be left to the discretion of the course classroom or shop instructor. The instructor best knows the unique learning environment and objectives to be achieved by the students in a course. Judgments should be made by the instructor as to the degree of emphasis to be placed on calculator applications, when and where a calculator is to be used, and the selection of specific problems for solution by calculator. Therefore, exercises and problems in this text are not specifically identified as calculator applications. Calculator instruction and examples of the basic operations of addition, subtraction, multiplication, and division of fractions are presented in Unit 7. They are presented for decimals in Unit 17. Further calculator instruction and examples of mathematics operations and functions are given throughout the text wherever calculator applications are appropriate to the material presented.

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Courtesy of SharpCalculators.com

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sharp EL-W535TG

texas Instruments TI-30XS

Machinist calc™ Pro 2

IntrodutIon to spreAdsheets A spreadsheet is a computer application that simulates a paper accounting worksheet. It displays multiple cells in a grid formed by rows and columns. Each cell in the grid contains text, a numeric value, or a formula. A formula defines how the content of that cell is to be calculated from the contents of any other cell (or combination of cells) each time any cell is updated. The most popular spreadsheet is Microsoft’s Excel, which works on the Macintosh iPad, Windows-based tablets like the Surface and Asus Transformer Book Flip, and Androidbased machines like the Google Pixel, Samsung Galaxy Tab, and Sony Xperia. Some other programs similar to Excel are: ●● ●●

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Numbers, Apple Inc.’s spreadsheet software, is part of iWork. Google Sheets is free for all Google account owners. It is a cloud-based program that can be used on a phone, tablet, or computer. Apache OpenOffice Calc is a freely available, open-source program. LibreOffice, like Apache open office, was derived from the original openoffice.org program. Bime Analytics is a new way of expressing and analyzing data in a more presentable display. ThinkFree is a free Excel alternative that is quite similar to Google Drive Apps and lets you create several different types of online documents. Gnumeric is a simple open-source spreadsheet application that has all the basic Excel features. Like many other free Excel alternatives, it’s a fine tool for less complicated data analysis. BIRT Spreadsheet, formerly named Actuate e.Spreadsheet, is a downloadable application when you need to automate and centralize spreadsheet production, maintenance, archiving, and security. Zoho Sheet is another cloud-based free Excel alternative that empowers your productivity and enables working in an Excel-like environment, creating new spreadsheets and editing documents written in other spreadsheet applications.

The illustrations in this text were all made using Excel on an iPad.

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sectiON ONe

Common Fractions and Decimal Fractions

1

UNIT 1 Introduction to Common Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●● ●● ●●

Express fractions in lowest terms. Express fractions as equivalent fractions. Express mixed numbers as improper fractions. Express improper fractions as mixed numbers.

Most measurements and calculations made by a machinist are not limited to whole numbers. Dimensions are sometimes given as fractions and certain measuring tools are graduated in fractional units. The machinist must be able to make calculations using fractions and to measure fractional values.

Fractional Parts A fraction is a value that shows the number of equal parts taken of a whole quantity or unit. The symbols used to indicate a fraction are the bar (—) and the slash ( / ). Line segment AB as shown in Figure 1-1 is divided into 4 equal parts. 1 part 1 part 1 1 part 5 5 5 of the length of the line segment. total parts 4 parts 4 2 parts 2 parts 2 2 parts 5 5 5 of the length of the line segment. total parts 4 parts 4 3 parts 3 parts 3 3 parts 5 5 5 of the length of the line segment. total parts 4 parts 4 4 parts 4 parts 4 4 parts 5 5 5 5 1, or unity (four parts make up the whole). total parts 4 parts 4 1

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Section 1

4 4 3 4 2 4 1 4

Common FraCtions and deCimal FraCtions

= 1 OR UNITY (4 OF 4 PARTS)

(3 OF 4 PARTS)

(2 OF 4 PARTS)

(1 OF 4 PARTS)

A

B

Figure 1-1

Each of the 4 equal parts of the line segment AB in Figure 1-2 is divided into eight equal parts. There is a total of 4 3 8, or 32, parts. 1 part =

1 32

32 32

of the total length. 23 32

7 parts =

7 32

12 parts =

12 32

12 32

of the total length.

= 1 OR UNITY (32 OF 32 PARTS)

(23 OF 32 PARTS)

(12 OF 32 PARTS) 7 32

(7 OF 32 PARTS)

of the total length.

1 32

23 parts =

23 32

(1 OF 32 PARTS)

of the total length. A

1 2

32 parts =

32 32

of 1 part =

1 2

of the total length.

3

1 32

=

1 64

of the total length.

B 1 2

OF

8 32

1 32

OR

=

1 64

1 4

Figure 1-2

Note: 8 parts 5

8 1 8 1 of the total length and also of the total length. Therefore, 5 . 32 4 32 4

DeFinitions oF Fractions A fraction is a value that shows the number of equal parts taken of a whole quantity 3 5 99 17 or unit. Some examples of fractions are , , , and . These same fractions 4 8 100 12 written with a slash are 3@4, 5@8, 99@100, and 17@12. The denominator of a fraction is the total number of equal parts in the whole quantity. The denominator is written below the bar. The numerator of a fraction is the number that shows how many equal parts of the whole are taken. The numerator is written above the bar. The numerator and denominator are called the terms of the fraction. 3 d numerator 4 d denominator 5 3 13 A common fraction consists of two whole numbers. , , and are all examples of 5 7 4 common fractions.

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Unit 1

3

introduCtion to Common FraCtions and mixed numbers

A proper fraction has a numerator that is smaller than its denominator. Examples of 3 5 91 247 proper fractions are , , , and . 4 11 92 961 An improper fraction is a fraction in which the numerator is larger than or equal to the 3 5 11 6 17 denominator, as in , , , , and . 2 4 8 6 17 A mixed number is a number composed of a whole number and a fraction, as in 7 1 3 and 7 . 8 2 7 7 8 8 as seven and one-half.

1 2

1 2

Note: 3 means 3 1 . It is read as three and seven-eighths. 7 means 7 1 . It is read

Writing fractions with a slash can cause people to misread a number. For example, 11 1 some people might think that 11@4 means 11@4 5 rather than 1 . For this reason, the slash 4 4 notation for fractions will not be used in this book. A complex fraction is a fraction in which one or both of the terms are fractions or 3 3 7 4 14 4 32 8 4 16 mixed numbers, as in , 15 , , 2 , and 5 . 6 4 3 25 78

exPressing Fractions as equivalent Fractions The numerator and denominator of a fraction can be multiplied or divided by the same 1 134 4 number without changing the value. For example, 5 5 . Both the numerator 2 234 8 1 4 and denominator are multiplied by 4. Because and have the same value, they are equiv2 8 8 844 2 alent. Also, 5 5 . Both numerator and denominator are divided by 4. Since 12 12 4 4 3 8 2 and have the same value, they are equivalent. Equivalent fractions are necessary for 12 3 comparing two fractions or for addition and subtraction of fractions. A fraction is in its lowest terms when the numerator and denominator do not contain a 5 7 3 11 15 9 common factor, as in , , , , , and . Factors are the numbers used in multiplying. 9 8 4 12 32 11 For example, 2 and 5 are each factors of 10: 2 3 5 5 10. Expressing a fraction in lowest terms is often called reducing a fraction to lowest terms.

c Procedure

To reduce a fraction to lowest terms

Divide both numerator and denominator by the greatest common factor (GCF). 12 Example Reduce to lowest terms. 42 12 4 2 6 Both terms can be divided by 2. 5 42 4 2 21 ●●

Note: The fraction is reduced, but not to lowest terms. Further reduce

6 . 21

Both terms can be divided by 3.

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643 2 5 21 4 3 7

Ans

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4

Common FraCtions and deCimal FraCtions

Section 1

2 7 Six is the greatest common factor (GCF) of 12 and 42.

Note: The value may be obtained in one step if each term of 12 4 6 2 5 42 4 6 7

12 is divided by 2 3 3, or 6. 42

Ans

c Procedure To express a fraction as an equivalent fraction with an indicated denominator that is larger than the denominator of the fraction ●● ●●

Divide the indicated denominator by the denominator of the fraction. Multiply both the numerator and denominator of the fraction by the value obtained. 3 4

Example Express as an equivalent fraction with 12 as the denominator. Divide 12 by 4.

Multiply both 3 and 4 by 3.

12 4 4 5 3 333 9 5 4 3 3 12

Ans

exPressing MixeD nuMbers as iMProPer Fractions c Procedure ●● ●● ●●

To express a mixed number as an improper fraction

Multiply the whole number by the denominator. Add the numerator to obtain the numerator of the improper fraction. The denominator is the same as that of the original fraction.

1 2 Multiply the whole number by the denominator. Add the numerator to obtain the numerator for the improper fraction. The denominator is the same as that of the original fraction.

Example 1 Express 4 as an improper fraction.

43211 9 5 2 2

Example 2 Express 12

3 as an improper fraction. 16 12 3 16 1 3 195 5 16 16

43258 81159

Ans

Ans

exPressing iMProPer Fractions as MixeD nuMbers c Procedure ●● ●●

To express an improper fraction as a mixed number

Divide the numerator by the denominator. Express the remainder as a fraction.

Examples Express the following improper fractions as mixed numbers. 11 3 5 11 4 4 5 2 Ans 4 4 43 1 5 43 4 3 5 14 Ans 3 3 931 3 5 931 4 8 5 116 Ans 8 8

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5

introduCtion to Common FraCtions and mixed numbers

Unit 1

ApplicAtion Fractional Parts 1. Write the fractional part that each length, A through F, represents of the total shown on the scale in Figure 1-3. A5 B5

A

C5

B C

D5

D E

E5

F

F5

Figure 1-3

2. A welded support base is cut into four pieces as shown in Figure 1-4. What fractional part of the total length does each of the four pieces represent? All dimensions are in inches.

3

Piece 1:

4

Piece 2:

2

1

4 16

12

64

Piece 3: Piece 4:

Figure 1-4

The circle in Figure 1-5 is divided into equal parts. Write the fractional part represented by each of the following in Exercises 3 and 4: 3. a. 1 part b. 3 parts c. 7 parts d. 5 parts e. 16 parts

4. a. b. c. d. e.

1 of 1 part 2 1 of 1 part 3 3 of 1 part 4 1 of 1 part 10 1 of 1 part 16

Figure 1-5

expressing Fractions as equivalent Fractions 5. Reduce to halves. 4 a. 8 9 b. 18 100 c. 200 121 d. 242

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6. Reduce to halves. 25 a. 10 18 b. 12 126 c. 36 225 d. 50

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6

Section 1

Common FraCtions and deCimal FraCtions

Reduce numbers to lowest terms in Exercises 7 and 8. 7. a. b. c. d. e.

6 8 12 4 6 10 30 5 11 44

8. a. b. c. d. e.

14 6 24 8 65 15 25 150 14 105

Express the fractions in Exercises 9 and 10 as thirty-seconds. 1 4 3 b. 4 11 c. 8 7 d. 16

9. a.

21 16 19 b. 2 197 c. 16 21 d. 8

10. a.

In Exercises 11 and 12, express the given fractions as equivalent fractions with the indicated denominators. 11. a. b. c. d. e.

3 ? 5 4 8 7 ? 5 12 36 6 ? 5 15 60 17 ? 5 14 42 20 ? 5 9 45

14 ? 5 3 18 7 ? b. 5 16 128 13 ? c. 5 8 48 21 ? d. 5 16 160

12. a.

Mixed numbers and improper Fractions Express the mixed numbers in Exercises 13 and 14 as improper fractions. 2 3 7 b. 1 8 2 c. 5 5 3 d. 3 8 9 e. 5 32 3 f. 8 7

13. a. 2

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1 3 4 b. 9 5 1 c. 100 2 63 d. 4 64 3 e. 49 8 13 f. 408 16

14. a. 10

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Unit 1

7

introduCtion to Common FraCtions and mixed numbers

Express the improper fractions in Exercises 15 and 16 as mixed numbers. 5 127 15. a. 16. a. 3 32 21 57 b. b. 2 15 9 150 c. c. 8 9 87 235 d. d. 4 16 72 514 e. e. 9 4 127 401 f. f. 124 64 Express the mixed numbers in Exercises 17 and 18 as improper fractions. Then express the improper fractions as the equivalent fractions indicated. 1 ? 17. a. 2 5 2 8 3 ? b. 3 5 8 16 4 ? c. 7 5 5 15

2 ? 18. a. 12 5 3 18 7 ? b. 9 5 8 64 1 ? c. 15 5 2 128

19. Sketch and redimension the plate shown in Figure 1-6. Reduce all proper fractions to lowest terms. Reduce all improper fractions to lowest terms and express as mixed numbers. All dimensions are in inches. 9 4

37 32

40 32

DIA 56 64

44 64 156 128

65 32 11 8 18 32

4 8

70 64 22 16

24 64

DIA 3 HOLES

104 32

Figure 1-6

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8

Section 1

Common FraCtions and deCimal FraCtions

UNIT 2 Addition of Common Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●● ●●

Determine lowest common denominators. Express fractions as equivalent fractions having lowest common denominators. Add fractions and mixed numbers.

A machinist must be able to add fractions and mixed numbers in order to determine the length of stock required for a job, the distances between various parts of a machined piece, and the depth of holes and cutouts in a workpiece.

lowest coMMon DenoMinators Fractions cannot be added unless they have a common denominator. Common denominator 5 7 15 means that the denominators of each of the fractions are the same, as in , , and , 8 8 8 which all have a common denominator of 8. 3 1 7 In order to add fractions that do not have common denominators, such as 1 1 , 8 4 16 it is necessary to change to equivalent fractions with common denominators. Multiplying the denominators does give a common denominator, but it could be a very large number. We often find it easier to determine the lowest common denominator. The lowest common denominator is the smallest denominator that is evenly divisible by each of the denominators of the fractions being added. Stated in another way, the lowest common denominator is the smallest denominator into which each denominator can be divided without leaving a remainder.

c Procedure ●●

●●

To find the lowest common denominator

Determine the smallest number into which all denominators can be divided without leaving a remainder. Use this number as a common denominator.

3 1 7 . 8 4 16 The smallest number into which 8, 4, and 16 can be divided without leaving a remainder is 16. Write 16 as the lowest common denominator.

Example 1 Find the lowest common denominator of , , and

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Unit 2

9

addition oF Common FraCtions and mixed numbers

3 1 7 5 . 4 3 8 12 The smallest number into which 4, 3, 8, and 12 can be divided is 24. The lowest common denominator is 24.

Example 2 Find the lowest common denominator of , , , and

Note: In this example, denominators such as 48, 72, and 96 are common denominators because 4, 3, 8, and 12 divide evenly into these numbers, but they are not the lowest common denominators. Although any common denominator can be used when adding fractions, it is generally easier and faster to use the lowest common denominator.

exPressing Fractions as equivalent Fractions with the lowest coMMon DenoMinator c Procedure To change fractions into equivalent fractions having the lowest common denominator ●● ●●

Divide the lowest common denominator by each denominator. Multiply both the numerator and denominator of each fraction by the value obtained. 2 7 1 , and as equivalent fractions having a lowest common 3 15 2

Example 1 Express , denominator.

The lowest common denominator is 30. Divide 30 by each denominator. Multiply each term of the fraction by the value obtained.

2 3 10 20 5 3 3 10 30 732 14 30 4 15 5 2; 5 15 3 2 30 1 3 15 15 30 4 2 5 15; 5 2 3 15 30 30 4 3 5 10;

Ans Ans Ans

5 15 3 9 , , and to equivalent fractions having a lowest common 8 32 4 16

Example 2 Change , denominator.

The lowest common denominator is 32. 5 3 4 20 32 4 8 5 4; 5 Ans 8 3 4 32 15 3 1 15 32 4 32 5 1; 5 Ans 32 3 1 32

3 3 8 24 5 4 3 8 32 932 18 32 4 16 5 2; 5 16 3 2 32 32 4 4 5 8;

Ans Ans

aDDing Fractions c Procedure ●● ●● ●●

To add fractions

Express the fractions as equivalent fractions having the lowest common denominator. Add the numerators and write their sum over the lowest common denominator. Express an improper fraction as a mixed number when necessary and reduce the fractional part to lowest terms.

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10

Section 1

Common FraCtions and deCimal FraCtions

1 3 7 5 1 1 . 2 5 10 6 Express the fractions as equivalent fractions with 30 as the denominator.

Example 1 Add 1

Add the numerators and write their sum over the lowest common denominator, 30. Express the fraction as a mixed number.

1 15 5 2 30 3 18 5 5 30 7 21 5 10 30 5 25 1 5 6 30 15 1 18 1 21 1 25 5 30 5

79 19 52 30 30

Ans

Example 2 Determine the total length of the shaft shown in Figure 2-1. All dimensions are in inches.

3 32

15 16

29 32

7 8

1 4

Figure 2-1

Express the fractions as equivalent fractions with 32 as the denominator.

Add the numerators and write their sum over the lowest common denominator, 32. 98 as a mixed number and reduce to 32 lowest terms.

3 3 5 32 32 15 30 5 16 32 29 29 5 32 32 7 28 5 8 32 1 8 1 5 4 32 98 3 1 30 1 29 1 28 1 8 5 5 32 32

Express

Total length

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98 2 1 53 53 32 32 16 10 53 Ans 16 5

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Unit 2

11

addition oF Common FraCtions and mixed numbers

aDDing Fractions, MixeD nuMbers, anD whole nuMbers c Procedure ●● ●● ●●

To add fractions, mixed numbers, and whole numbers

Add the whole numbers. Add the fractions. Combine whole number and fraction.

1 1 5 19 12 . 3 2 12 24 Express the fractional parts as equivalent fractions with 24 as the denominator.

Example 1 Add 1 7 1 3 1

1 8 5 3 24 757 1 12 3 53 2 24 5 10 5 12 24 19 19 12 52 24 24 Add the whole numbers. 5 7 1 3 1 2 5 12 8 1 12 1 10 1 19 49 Add the fractions. 5 5 24 24 49 Combine the whole number and the fractions. 5 12 24 49 1 Express the answer in lowest terms. 5 12 5 14 Ans 24 24 1 2

Example 2 Find the distance between the two -inch diameter holes in the plate shown in Figure 2-2. All dimensions are in inches. 151 13 26 5 32 64 1 DIA 47 47 2 1 51 64 64 3 12 1 5 16 64 85 21 2 53 64 64

1 2

1

13 32

1 47 64

DIA

3 16

Figure 2-2

Distance 5 3

98310_sec_01_Unit01-04_ptg01.indd 11

210 64

Ans

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12

Section 1

Common FraCtions and deCimal FraCtions

ApplicAtion tooling Up 1. Reduce the fraction

15 to halves. 30

12 to lowest terms. 30 11 ? Express and as equivalent fractions. 8 32 3 Express the mixed number 7 as an improper fraction. 5 97 Express as a mixed number. 12 3 ? Express the mixed number 9 as an improper fraction and then express that improper fraction in the form . 5 15

2. Reduce 3. 4. 5. 6.

Lowest common Denominators Determine the lowest common denominators of the following sets of fractions. 2 1 5 5 7 3 19 7. , , 9. , , , 3 6 12 6 12 16 24 3 9 5 4 3 7 1 8. , , 10. , , , 5 10 6 5 4 10 2

equivalent Fractions with Lowest common Denominators Express these fractions as equivalent fractions having the lowest common denominator. 1 3 5 11. , , 2 4 12 7 3 1 12. , , 16 8 2

9 , 10 3 14. , 16 13.

1 3 1 , , 4 5 5 7 17 3 , , 32 64 4

Adding Fractions 15. Determine the dimensions A, B, C, D, E, and F of the profile gage in Figure 2-3. All dimensions are in inches. 11 64

1 2

A5 5 16

A

B5 F

C5 D5

9 16 21 64 35 64

3 8

31 32

1 8

B

C

15 32

7 16

1 4

E5 F5

D

E

Figure 2-3

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Unit 2

13

addition oF Common FraCtions and mixed numbers

16. Determine the overall length, width, and height of the casting in Figure 2-4. All dimensions are in inches. 29 64

length 5

9 16

width 5

5 32

height 5

1 2

3 8

63 64

17 32

21 32

7 16

1 4

Figure 2-4

Adding Fractions, Mixed numbers, and Whole numbers 17. Determine dimensions A, B, C, D, E, F, and G of the plate in Figure 2-5. Reduce to lowest terms where necessary. All dimensions are in inches. A5

G E 5 8

F 2 34

9 32

B5

3 1 64

C5

19 32

A

D5 1

7 1 16

E5 1 4

1 4

1 18

3 8

1 2 32

1 27 32

D

F5 G5

7 32

B C

Figure 2-5

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14

Section 1

Common FraCtions and deCimal FraCtions

18. Determine dimensions A, B, C, and D of the pin in Figure 2-6. All dimensions are in inches. A5

D C A

B5

B

C5 D5

3 32 3 32

5 1 64

1 1 16 1 8

1 4

9 32

Figure 2-6

19. The operation sheet for machining an aluminum housing specifies 1 hour for facing, 3 5 3 2 2 hours for milling, hour for drilling, hour for tapping, and hour for setting up. 4 6 10 5 What is the total time allotted for this job?

UNIT 3 Subtraction of Common Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●●

Subtract fractions. Subtract mixed numbers.

While making a part from an engineering drawing, a machinist often finds it necessary to express drawing dimensions as working dimensions. Subtraction of fractions and mixed numbers is sometimes required in order to properly position a part on a machine, to establish hole locations, and to determine depths of cut.

subtracting Fractions c Procedure ●● ●● ●● ●●

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To subtract fractions

Express the fractions as equivalent fractions having the lowest common denominator. Subtract the numerators. Write their difference over the lowest common denominator. Reduce the fraction to lowest terms.

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Unit 3

15

subtraCtion oF Common FraCtions and mixed numbers

3 8

Example 1 Subtract from

9 . 16

9 9 5 16 16 3 3 6 Express as 16ths. 2 5 8 8 16 926 3 Subtract the numerators. 5 16 16 Write their difference over the lowest common denominator.

The lowest common denominator is 16.

2 3 5 4 Express the fractions as equivalent fractions with 20 as the denominator.

Ans

Example 2 Subtract from .

3 15 5 4 20 2 8 2 5 5 20 7 15 2 8 5 5 20 20

Subtract the numerators and write the difference over the common denominator, 20.

Ans

Example 3 Find the distances x and y between the centers of the pairs of holes in the strap shown in Figure 3-1. All dimensions are in inches. To find distance x: To find distance y: 7 28 63 63 5 5 8 32 64 64 11 11 1 16 2 5 2 5 32 32 4 64 28 2 11 17 63 2 16 47 5 5 32 32 64 64 170 470 x5 Ans y5 32 64

11 32

x

1 4

y

7 8

63 64

Figure 3-1

Ans

subtracting MixeD nuMbers c Procedure ●● ●● ●●

To subtract mixed numbers

Subtract the whole numbers. Subtract the fractions. Combine whole number and fraction. 1 4

3 8

Example 1 Subtract 2 from 9 .

Subtract the whole numbers. Subtract the fractions. Combine the whole number and the fractions.

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3 3 9 59 8 8 1 2 22 52 4 8 592257 322 1 5 5 8 8 1 57 Ans 8

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16

Section 1

Common FraCtions and deCimal FraCtions

Example 2 Find the length of thread x of the bolt shown in Figure 3-2. All dimensions are in inches.

3 1 32

7 28 2 52 8 32 3 3 21 51 32 32 25 1 32 250 x51 32

x 2 78

Figure 3-2

Example 3 Subtract 7

Ans

15 5 from 12 . 16 8 5 10 26 12 5 12 5 11 8 16 16 15 15 15 27 5 7 5 7 16 16 16 4

11 16

Ans

15 10 cannot be subtracted from , one unit of the whole number 12 is 16 16 expressed as a fraction with the common denominator 16 and added to the fractional part of the mixed number.

Note: Since

Example 4 Subtract 52

31 from 75. 64 64 64 31 31 2 52 5 52 64 64 75 5 74

22

Ans

Example 5 Find dimension y of the counterbored block shown in Figure 3-3. All dimensions are in inches. 3 12 44 2 52 51 8 32 32 29 29 29 2 5 5 32 32 32

29 32

2 38 y

33 64

15 32 150 y51 32 1

Ans

Figure 3-3

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Unit 3

subtraCtion oF Common FraCtions and mixed numbers

17

ApplicAtion tooling Up 2 5 11 1. Determine the least common denominator of , , and . 3 8 12 1 5 7 2. Express , , and as equivalent fractions having the lowest common denominator. 4 6 12 29 3. Express as a mixed number. 8 1 4. Express the mixed number 5 as an improper fraction. 6 5 3 1 5. Add 1 1 . 12 4 6 3 3 1 6. Add 2 1 1 1 5 . 8 4 3

Subtracting Fractions Subtract each of the fractions in Exercises 7 through 9. Reduce to lowest terms where necessary. 5 9 7. a. 2 8 32 7 5 b. 2 8 8 9 19 c. 2 10 50 5 9 2 8 64 9 13 b. 2 16 64

8. a.

c.

19 3 2 24 16

7 5 2 4 8 3 3 b. 2 8 64 15 8 c. 2 32 64

9. a.

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18

Section 1

Common FraCtions and deCimal FraCtions

10. Determine dimensions A, B, C, and D of the casting in Figure 3-4. All dimensions are in inches.

A5 C

29 32

B5

15 16

1 2

C5

D

A 5 8

D5

3 4

B

7 32

63 64

Figure 3-4

11. Determine dimensions A, B, C, D, E, and F of the drill jig in Figure 3-5. All dimensions are in inches. 7 8

F

15 32

A5 9 32

17 32

3 4

B5 C5

23 32

E

D5 E5

D A

F5

1 2 3 16

B 25 32

3 8

C

5 16

49 64

Figure 3-5

Subtracting Mixed numbers 12. Determine dimensions A, B, C, D, E, F, and G of the tapered pin in Figure 3-6. All dimensions are in inches. 5 32 7 16

E

1 4

F

B5

G

11 1 32

A 1 78

D B

9 3 64

5 14

Figure 3-6

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C

1 38

23 64

A5

9 64

C5 D5 E5 F5 G5

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19

subtraCtion oF Common FraCtions and mixed numbers

Unit 3

13. Determine dimensions A, B, C, D, E, F, G, H, and I of the plate in Figure 3-7. All dimensions are in inches. 2 23 32 1

28

I

A5 A 1 1 64

B 2

3 8

B5 17 1 32

1 DIA 1 29 32

H

C 1 14

D5 E5

13 16

13 16

F G 3

1 32

E D

C5

1

1 2 32

F5 G5 H5

3 8

I5 1 2 16

17 2 32

Figure 3-7

14. Three holes are bored in a checking gage. The lower left edge of the gage is the reference point for the hole locations. Sketch the hole locations and determine the missing distances. From the reference point: 30 50 Hole #1 is 1 to the right, and 1 up. 32 8 10 30 Hole #2 is 2 to the right, and 2 up. 64 16 10 10 Hole #3 is 3 to the right, and 3 up. 4 2 Determine: a. The horizontal distance between hole #1 and hole #2. b. The horizontal distance between hole #2 and hole #3. c. The horizontal distance between hole #1 and hole #3. d. The vertical distance between hole #1 and hole #2. e. The vertical distance between hole #2 and hole #3. f. The vertical distance between hole #1 and hole #3.

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20

Section 1

Common FraCtions and deCimal FraCtions

UNIT 4 Multiplication of Common

Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●● ●●

Multiply fractions. Multiply mixed numbers. Divide by common factors (cancellation).

In machine technology, multiplying fractions and mixed numbers can be used to determine the area and volume of a piece of material or the amount of material that will be needed to produce a certain number of parts.

MultiPlying Fractions The answer to a multiplication problem is called the product.

c Procedure ●● ●● ●● ●●

To multiply two or more fractions

Multiply the numerators of the fractions to get the numerator of the product. Multiply the denominators of the fractions to get the denominator of the product. Write the product of the numerators over the product of the denominators. Reduce the resulting fraction to lowest terms.

Notice that a lowest common denominator is not needed because a lowest common denominator is only required for addition and subtraction. 3 8 Example 1 Multiply by . 4 9 Multiply the numerators. 3 3 8 5 24 Multiply the denominators. 4 3 9 5 36 24 Write the product of the numerators over the product of the denominators. 36 24 24 4 12 2 Reduce the resulting fraction to lowest terms. 5 5 Ans 36 36 4 12 3 2 5 3 Example 2 Multiply 3 3 . 3 6 10 23533 30 30 4 30 1 5 5 5 Ans 3 3 6 3 10 180 180 4 30 6 To multiply a whole number by a fraction, write the whole number as a fraction with a denominator of 1. 2 9

Example 3 Multiply 3 4. 2 2 4 8 345 3 5 9 9 1 9

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Unit 4

multipliCation oF Common FraCtions and mixed numbers

21

Example 4 Find the distance between the centers of the first and last holes shown in Figure 4-1. All dimensions are in inches. 7 6 7 637 42 Multiply 6 3 5 3 5 5 . 16 1 16 1 3 16 16 42 10 5 Reduce 52 52 . 16 16 8 7 50 16 TYPICAL 6 PLACES Distance 5 2 Ans 8 Figure 4-1 Note: The value of a number remains unchanged when the number is placed over a

6 denominator of 1. For Example, 6 5 . 1

DiviDing by coMMon Factors (cancellation) Problems involving multiplication of fractions are generally solved more quickly and easily if a numerator and denominator are divided by any common factors before the fractions are multiplied. This process of first dividing by common factors is commonly called cancellation. Cancellation allows you to avoid using large numbers in the numerator or denominator and reduces, or eliminates, the need to reduce the fraction after multiplying. 3 8 Example 1 Multiply by cancellation method 3 . 4 9 Divide by 3, which is the factor common to both the numerator 3 and the denominator 9. 1

34351 94353

3 8 3 8 3 5 3 4 9 4 93

Divide by 4, which is a factor common to both the denominator 4 and the numerator 8. 1

44451 84452

2

3 8 132 2 3 5 5 Ans 41 93 133 3

Multiply reduced fractions. 4 7

Example 2 Multiply 3

5 14 3 . 18 15

Divide 4 and 18 by 2. Divide 7 and 14 by 7. Divide 5 and 15 by 5. Multiply.

Example 3 Multiply

2

1

2

4 5 14 2 3 1 3 2 4 3 3 5 5 Ans 71 18 15 1 3 9 3 3 27 9 3 5 8 7 3 3 . 14 9 10 1

Divide 5 and 10 by 5. Divide 14 and 8 by 2.

2

/4

1

5 8 7 13231 2 3 3 5 5 Ans 14 9 10 13931 9 2 /7 1

/

1

The process is continued by dividing 7 and 7 by 7 and by dividing 2 and 4 by 2. Multiply.

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22

Section 1

Common FraCtions and deCimal FraCtions

MultiPlying MixeD nuMbers c Procedure ●● ●●

To multiply mixed numbers

Express the mixed numbers as improper fractions. Follow the procedure for multiplying proper fractions. 2 5

7 8

Example 1 Multiply 2 3 6 . 2 12 Write the mixed number 2 as the fraction . 5 5 7 55 Write the mixed number 6 as the fraction . 8 8 Divide 5 and 55 by 5. Divide 12 and 8 by 4. Multiply numerators. Multiply denominators. Express as a mixed number in lowest terms.

3

11

12 55 3 11 3 5 3 51 82 1 2 3 11 3 3 11 33 3 5 5 1 2 132 2 33 1 5 16 Ans 2 2

Example 2 The block of steel shown in Figure 4-2 is to be machined. The block mea3 9 7 sures 8 inches long, 4 inches wide, and inch thick. Find the volume of the block. All 4 16 8 dimensions are in inches. (Volume 5 length 3 width 3 thickness.) The volume will be in cubic inches. 3 9 7 35 73 7 35 3 73 3 7 8 34 3 5 3 3 5 4 16 8 4 16 8 4 3 16 3 8 17,885 477 5 5 34 512 512 477 Volume 5 34 cubic inches Ans 512

9 4 16

7 8

8 34

Figure 4-2

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Unit 4

23

multipliCation oF Common FraCtions and mixed numbers

ApplicAtion tooling Up 1 2 5 4 1. Express , , , and as equivalent fractions having the lowest common denominator. 3 5 6 9 1 2 5 4 2. Add 1 1 1 . 3 5 6 9 36 3. Reduce to lowest terms. 45 4. Determine the length A in Figure 4-3. All dimensions are in inches.

35 64

9 16

B 1 38

27

1 32

15 32

A

Figure 4-3

5. Determine the length of the template in Figure 4-3. All dimensions are in inches. 6. Determine the length B in Figure 4-3. All dimensions are in inches.

Multiplying Fractions Multiply the fractions in Exercises 7 through 9. Reduce to lowest terms where necessary. 2 1 5 2 7. a. 3 9. a. 3 3 6 4 3 1 1 7 5 b. 3 b. 4 3 3 2 4 8 21 5 13 5 4 c. 3 c. 3 3 3 8 64 9 15 3 3 2 3 3 4 5 3 9 b. 7 3 33 14 7 3 5 c. 3 3 15 8 7

8. a.

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24

Section 1

Common FraCtions and deCimal FraCtions

A5

10. Determine dimensions A, B, C, D, and E of the template shown in Figure 4-4. All dimensions are in inches.

B5 C5 D5 E5

7 64

3 8

TYPICAL 6 PLACES

TYPICAL 2 PLACES 3 16

A

C

9 16

TYPICAL 5 PLACES

E

7 32

TYPICAL 6 PLACES B

TYPICAL 5 PLACES

D

Figure 4-4

11. A special washer-faced nut is shown in Figure 4-5. All dimensions are in inches. DISTANCE ACROSS CORNERS

29 32

DISTANCE ACROSS FLATS

7 32

WASHER THICKNESS TOTAL THICKNESS

Figure 4-5

a. Determine the distance across flats. 55 Distance across flats 5 3 Distance across corners 64 b. Determine the washer thickness. 1 Washer thickness 5 3 Total thickness 8 12. The Unified Thread may have either a flat or rounded crest or root (Figure 4-6). If the sides of the Unified Thread are extended, a sharp V-thread is formed (Figure 4-7). In Figure 4-7, H is the height of a sharp V-thread. The pitch, P, is the distance between two adjacent threads.

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Unit 4

25

multipliCation oF Common FraCtions and mixed numbers

CREST

ROOT

DEPTH

INTERNAL THREAD ON NUT

PITCH MAJOR DIAMETER

EXTERNAL THREAD ON BOLT

MINOR OR ROOT DIAMETER

Figure 4-6 H (HEIGHT OF SHARP V-THREAD) A=

1 8

P (PITCH)

3H

CREST (FLAT OR ROUNDED)

60° B=

17 3 24

H

C=

1 8

3P

ROOT (FLAT OR ROUNDED)

Figure 4-7

Find dimensions A, B, and C as indicated. 70 a. H 5 , A5 16 30 b. H 5 , A5 8 150 c. H 5 , A5 16 210 d. H 5 , A5 32 30 e. H 5 , A5 4

,

B5

,

B5

,

B5

,

B5

,

B5

10 , 4 30 g. P 5 , 32 10 h. P 5 , 20 10 i. P 5 , 28 30 j. P 5 , 16 f.

P5

C5 C5 C5 C5 C5

Multiplying Mixed numbers 13. Multiply these mixed numbers. Reduce to lowest terms where necessary. 2 3 2 1 3 a. 1 3 6 d. 1 3 10 3 3 10 3 4 8 5 3 3 1 b. 3 37 e. 2 333 16 4 32 8 5 1 2 2 1 c. 4 3 2 f. 2 3 2 3 5 8 2 3 3 4 30 14. How many inches of drill rod are required in order to make 20 drills each 3 long? 16 30 Allow waste for each drill. 32

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26

Section 1

Common FraCtions and deCimal FraCtions

15. A hole is cut in a rectangular metal plate as shown in Figure 4-8. To find the area of a rectangle, multiply the length by the width. Determine the area of the plate after the hole has been removed. All dimensions are in inches. The area will be in square inches.

1 2

3 38

1 14

7 16

11 16

1 15 16

Figure 4-8

16. Six identical square holes are cut in a rectangular metal plate as shown in Figure 4-9. To find the area of a rectangle, multiply the length by the width. Determine the area of the plate after the holes have been removed. All dimensions are in inches. The area will be in square inches.

5

38 3 8

3 8

1

22

Figure 4-9

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Common Fractions and Decimal Fractions

1

UNIT 5

Division of Common fraCtions anD mixeD numbers

27

UNIT 5 Division of Common Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●●

Divide fractions. Divide mixed numbers.

In machine technology, division of fractions and mixed numbers can be used in determining production times and costs per machined unit, in calculating the pitch of screw threads, and in computing the number of parts that can be manufactured from a given amount of raw material.

DiviDing Fractions as the inverse oF Multiplying Fractions Division can be represented in several different ways. For example, 21 divided by 3 can 21 be shown as 21 4 3, 3q 21, , and 21/3. Here the number being divided, 21, is called the 3 dividend, the number used to divide, 3, is called the divisor, and the answer, 7, is called the 5 3 3 5 quotient. In the problem 4 , the divisor is and the dividend is . As you will see in 8 4 4 8 5 Example 1, the quotient is . 6 1 Division is the inverse of multiplication. Dividing by 2 is the same as multiplying by . 2 1 54252 2 1 1 53 52 2 2 1 542553 2 1 1 Two is the reciprocal, or multiplicative inverse, of , and is the reciprocal, or multiplica2 2 tive inverse, of 2. The reciprocal of a fraction is a fraction that has its numerator and denomi1 3 8 7 63 nator interchanged. The reciprocal of is , is the reciprocal of , is the reciprocal of 3 1 7 8 64 64 9 16 , and the reciprocal of is . 63 16 9

c Procedure ●● ●●

To divide fractions

Determine the reciprocal of the divisor. Multiply the dividend by the reciprocal of the divisor.

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28

SecTIoN 1

Example 1

Divide

Common fraCtions anD DeCimal fraCtions

5 3 by . 8 4

3 4 and its reciprocal is . 4 3 5 4 Multiply the dividend, , by . 8 3 The divisor is

1

Follow the procedure for multiplication.

5 3 5 4 5 4 5 3 5 8 4 82 3 6

Ans

10 . The pitch is the 16 70 distance between two adjacent threads. Find the number of threads in . All dimensions 8 1 are in inches. PITCH = 16 7 1 Divide by . 8 16

Example 2 The machine bolt shown in Figure 5-1 has a pitch of

2

7 1 7 16 4 5 3 5 14 Ans 8 16 81 1

7 8

figure 5-1

DiviDing MixeD nuMbers c Procedure

To divide mixed numbers

Express the mixed numbers as improper fractions. Follow the procedure for dividing fractions. 1 3 Example 1 Divide 7 by 2 . 2 8 1 3 1 3 15 19 Express 7 and 2 as improper fractions. 7 42 5 4 2 8 2 8 2 8 19 8 The reciprocal of is . 4 8 19 15 8 60 3 Multiply by the reciprocal. 3 5 53 21 19 19 19 ●● ●●

Ans

3 8 8 3 mixed number, first change the number to an improper fraction and then find the recip3 19 rocal of the improper fraction. Thus, to find the reciprocal of 2 , first rewrite it as . 8 8 3 8 The reciprocal of 2 is . 8 19 Example 2 A section of strip stock is shown in Figure 5-2 with five equally spaced holes. Determine the distance between two consecutive holes. All dimensions are in inches.

Note: The reciprocal of a mixed number such as 2 is NOT 2 . To find the reciprocal of a

Note: The number of spaces between the holes is one less than the number of holes.

x

x

x

x

4 38

figure 5-2

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UNIT 5

29

Division of Common fraCtions anD mixeD numbers

Express as improper fractions. Multiply by the reciprocal.

3 35 4 4 445 4 8 8 1 35 1 35 3 3 5 51 8 4 32 32 30 x51 Ans 32

ApplicAtion Tooling Up 1. Express the mixed number 7

3 as an improper fraction. 16

1 1 11 12 14 . 5 3 15 7 3 3. Subtract 2 . 16 32 5 2 4. Multiply 3 . 64 3

2. Add

5. Determine the length A in Figure 5-3. All dimensions are in inches.

5 16

B

TYPICAL 7 PLACES A

23 64

3 12

figure 5-3

6. Determine the length B in Figure 5-3. All dimensions are in inches.

Reciprocals Find the reciprocal of each of the fractions in Exercises 7 through 12. 7 7. 10. 6 8 1 3 8. 11. 6 4 4 25 2 9. 12. 3 8 7

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30

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

Dividing Fractions 13. This casting in Figure 5-4 shows seven tapped holes, A–G. The number of threads is determined by dividing the depth of the thread by the thread pitch. Find the number of threads in each of the tapped holes. All dimensions are in inches. HOLE A 1 PITCH 12

HOLE B 1 PITCH 9

HOLE C 1 PITCH 14 1 4

1 2

15 16

HOLE G 1 PITCH 5 5 8

13 32 9 16

HOLE D 1 PITCH 8

HOLE F 1 PITCH 16

HOLE E 1 PITCH 11

7 8

figure 5-4

A5

E5

B5

F5

C5

G5

D5 3 inch each time the stock turns once 64 3 (1 revolution). How many revolutions will the stock make when the tool advances inch? 4 15 15. A groove inch deep is to be milled in a steel plate. How many cuts are required if each 16 3 cut is inch deep? 16 14. Bar stock is being cut on a lathe. The tool feeds (advances)

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31

Division of Common fraCtions anD mixeD numbers

UNIT 5

Dividing Mixed Numbers 16. This sheet metal section shown in Figure 5-5 has five sets of drilled holes: A, B, C, D, and E. The holes within a set are equally spaced in the horizontal direction. Compute the horizontal distance between two consecutive holes for each set. All dimensions are in inches. 2

3 32

3

A5

15 32

B5

7

78 B

B

C5

E

B

E

A A

C A

D

E E

E C

A

5

5 8

15

5 16

D5 A

E5

A

C

C A

D

D

D

A

figure 5-5

17. The feed on a lathe is set for

1 inch. How many revolutions does the work make 64

3 when the tool advances 3 inches? 4

1 18. How many complete pieces can be blanked from a strip of steel 27 feet long if each 4 3 5 stamping requires 2 inches of material plus an allowance of inch at one end of 16 16 the strip? (12 inches 5 1 foot) 1 19. A slot is milled the full length of a steel plate that is 3 feet long. This operation takes 4 1 a total of 4 minutes. How many feet of steel are cut in 1 minute? 16 1 20. How many binding posts can be cut from a brass rod 42 inches long if each post is 2 7 3 1 inches long? Allow inch waste for each cut. 8 32 1 1 21. A bar of steel 23 feet long weighs 110 pounds. How much does a 1-foot length of 4 2 bar weigh?

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32

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

22. A single-threaded (or single-start) square-thread screw is shown in Figure 5-6. The lead of a screw is the distance that the screw advances in one turn (revolution). The lead is equal to the pitch in a single-threaded screw. Given the number of turns and the amount of screw advance, determine the leads. Screw Advance 10 4

Number of Turns

a.

2

b.

7

c.

70 2 16

1 6 2

d.

1

10 2

15

e.

6

30 10

37 0 64

Lead PITCH = LEAD

10

24

12

1 4 SINGLE-THREADED SQUARE-THREAD SCREW

figure 5-6

3 5

23. A double-threaded square-thread screw is shown in Figure 5-7. The pitch of a screw is the distance from the top of one thread to the same point on the top of the next thread. The lead is the distance the screw advances for each complete turn or revolution of the screw. In a double-threaded screw, the lead is twice the pitch. Given the number of turns and the amount of screw advance, determine the lead and pitch. Screw Advance

Number of Turns

a.

50 2 8

10

b.

6

c.

10

d.

3

61 0 64

22

1 4

50 16

16

1 2

90 16

98310_sec_01_Unit05-11_ptg01.indd 32

12

Lead

Pitch

Pitch

Lead

figure 5-7

10/31/18 4:33 PM

UNIT 6

CombineD operations of Common fraCtions anD mixeD numbers

33

UNIT 6 Combined Operations of Common Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●●

Solve problems that involve combined operations of fractions and mixed numbers. Solve complex fractions.

Before a part is machined, the sequence of machining operations, the machine setup, and the working dimensions needed to produce the part must be determined. In actual practice, calculations of machine setup and working dimensions require not only the individual operations of addition, subtraction, multiplication, and division but a combination of two or more of these operations.

orDer oF operations For coMbineD operations c Procedure ●●

Do all the work in the parentheses first. Parentheses are used to group numbers. In a problem expressed in fractional form, the numerator and the denominator are each considered as being enclosed in parentheses. Brackets, [ ], and braces, { }, are used for “nesting” one group within another. They are treated the same as parentheses. On your calculator, use the parentheses, ( ), symbols.

Example 1

Compute {5 1 [7 2 3s6 2 4d 1 2] 2 6} 1 1.

Begin with the innermost grouping symbols, the parentheses.

{5 1 [7 2 3s6 2 4d 1 2] 2 6} 1 1 5 {5 1 [7 2 3s2d 1 2] 2 6} 1 1 5 {5 1 [7 2 6 1 2] 2 6} 1 1

Next, work within the brackets.

5 {5 1 [3] 2 6} 1 1 5 {5 1 3 2 6} 1 1

Finally, do the operations inside the braces.

5 {2} 1 1 5 3 Ans

If an expression contains nested parentheses, do the work within the innermost parentheses first. ●● Do multiplication and division next. Perform multiplication and division in order from left to right. ●● Do addition and subtraction last. Perform addition and subtraction in order from left to right. Some people use the acronym PEMDAS to remember the order of operations. Here P represents parentheses and, since it is first, you should first do any calculations inside parentheses. The E stands for exponents or powers, which we will cover in Unit 13. Next is MD for multiplication and division. MD is considered as one step. Finally, do any AS, or addition and subtraction. As with MD, AS is a single step.

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34

SecTIoN 1

1

1

Common fraCtions anD DeCimal fraCtions

22

2 7 9 1 Find the value of 1 1 2 1 . 5 3 5 4 There are two sets of parentheses with one set nested inside the other. 7 9 8 Begin with the innermost parentheses: 2 5 3 5 15 2 8 1 The result is 1 1 1 . 5 15 4 2 8 14 Next, perform the operation in the remaining parentheses: 1 1 51 5 15 15 14 1 The result is 1 1 . 15 4 14 1 11 Add these two fractions: 1 1 52 Ans 15 4 60 Remember, in a problem expressed in fractional form, treat the numerator and denominator as if each was in parentheses.

Example 2

1

Example 3

2

3 1 4 2 4 2 Find the value of . 5 10 1 6 8 3 1 5 5 4 2 4 10 1 6 4 2 8

1

2 1

2

1 5 5 4 4 16 4 8 17 133 5 4 4 8 17 8 34 5 3 5 4 133 133

combining addition and subtraction 1 3 5 Find the value of 3 2 1 . 2 8 16 There are no parentheses and there is no multiplication or division. So, perform addition and subtraction in order from left to right. 3 1 1 3 1 Subtract from 3 . 3 2 53 HOLE #2 8 2 2 8 8 13 1 5 1 5 7 32 Add 3 to . 3 1 53 Ans 8 16 8 16 16

Example 1

Example 2

Find x, the distance from the base of the plate in Figure 6-1 to the center of hole #2. All dimensions are in inches. 90 1 0 130 x5 12 2 16 8 32 90 10 110 Add. 12 52 16 8 16 11 0 13 0 90 Subtract. 2 2 52 Ans 16 32 32

98310_sec_01_Unit05-11_ptg01.indd 34

2 18 x

9 16

figure 6-1

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UNIT 6

CombineD operations of Common fraCtions anD mixeD numbers

35

combining Multiplication and Division 2 1 3842 . 3 2 There are no parentheses. Perform multiplication and division in order from left to right. 2 2 3 8 16 Multiply. 385 5 3 331 3 16 1 16 5 16 2 32 2 Divide. 42 5 4 5 3 5 52 Ans 3 2 3 2 3 5 15 15

Example 1

Find the value of

Example 2 The stainless-steel plate shown in Figure 6-2 has slots that are of uniform 1 length and equally spaced within a distance of 33 inches. The time required to rough 2 7 and finish mill a 1-inch length of slot is minute. How many minutes are required for the 10 tool to cut all the slots? Disregard the time required to reposition the part. All dimensions are in inches.

3

4 16 TYPICAL 33 12

3

4 16 11 58

figure 6-2

10 1 3 5 33 4 4 . 2 2 16 7 5 The time required to cut 1 groove 5 3 11 . 10 8 Total time equals the number of grooves multiplied by the time for each groove. 1 3 7 5 33 4 4 3 3 11 2 16 10 8 1 3 67 16 Divide. 33 4 4 5 3 58 2 16 2 67 7 5 8 7 93 1 Multiply. 83 3 11 5 3 3 5 65 10 8 1 10 8 10 1 Total time 5 65 minutes Ans 10 The number of grooves in 33

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36

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

combining addition, subtraction, Multiplication, and Division 5 1 3 7 Find the value of 7 1 5 4 2 10 3 . 6 2 4 16 5 1 3 7 First, divide and multiply. 7 1 5 4 2 10 3 6 2 4 16 T T T 5 1 4 10 7 7 15 3 2 3 6 2 3 1 16

Example 1

T T 5 1 7 1 7 6 3

Next, add and subtract.

Example 2

1

2

T 3 19 4 5 10 8 24

Ans

2

5 1 3 7 Find the value of 7 1 5 4 2 10 3 . 6 2 4 16

1

2

5 1 5 3 1 First, do the work in parentheses. 7 1 5 5 7 1 5 5 13 6 2 6 6 3 1 3 40 4 160 7 3 5 5 17 Next, divide and multiply. 13 4 5 3 4 3 3 9 9 7 3 10 3 54 16 8 7 3 29 Then, subtract. 17 2 4 5 13 Ans 9 8 72

Note: This example is the same as the preceding example except for the parentheses.

complex Fractions A complex fraction is an expression in which either the numerator or denominator or both are fractions or mixed numbers. A fraction indicates a division operation. Therefore, complex fractions can be solved by dividing the numerator by the denominator. 5 9 5 1 5 4 1 9 3 3

Example

7 3 5 12 8 4 Find the value of . 15 1 3 21 16 8

Note: The complete numerator is divided by the complete denominator. Therefore, parentheses are used to indicate that addition in the numerator and subtraction in the denominator must be performed before division. 7 3 5 12 8 4 7 3 15 1 5 5 12 4 3 21 15 1 8 4 16 8 3 21 16 8 69 45 5 4 8 16

1

23

2 1

2

69 16 1 5 3 53 81 45 15 15

98310_sec_01_Unit05-11_ptg01.indd 36

2

Ans

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UNIT 6

37

CombineD operations of Common fraCtions anD mixeD numbers

ApplicAtion Tooling Up 3 7 1 5 , , , and as equivalent fractions having the lowest common denominator. 16 4 2 8 5 13 Subtract 7 2 1 . 16 32 1 1 Multiply 2 3 3 . 8 2 7 9 Divide 4 . 64 16 Determine the length A in Figure 6-3. All dimensions are in inches.

1. Express 2. 3. 4. 5.

A B

A

A

A

A

A

A

4 38

B

1

7 16

figure 6-3

6. Determine the length B in Figure 6-3. All dimensions are in inches.

order of operations for combined operations Solve the following examples of combined operations in Exercises 7 through 9. 1 3 1 1 2 2 16 4 7 3 3 b. 3 2 2 1 8 16 8 3 2 1 c. 18 23 10 5 25 2 1 d. 27 2 2 1 4 3 6

7. a.

1 3 3 e. 32 1 2 3 8 16 4 7 2 5 8. a. 3 13 9 3 6

1

2

1 1 3 b. 12 2 4 4 1 2 2 2 4 1 1 3 c. 16 2 4 4 1 5 2 2 8

1

98310_sec_01_Unit05-11_ptg01.indd 37

1

2 112 1 2 182

1 4 2 1 1 e. 15 3 1 1 2 4 3 3 2 1 9. a. 2 3 4 3 4 3 2 1 b. 2 3 4 3 4 d. 16 2 4

2 5 44 3 6

1

2 1 1 2 5 c. 4 3 13 1 2 2 2 2 3 8

1 1 1 d. 7 4 2 1 4 2 4 2 1 1 1 e. 7 4 2 1 4 2 4 2

1

2

2

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38

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

complex Fractions Find the value of the complex fractions in Exercises 10 through 12. 3 4 10. a. 1 2

1 5 1 3 6 11. a. 3 3 4

7 8 b. 5

1 2 2 2 5 12. a. 1 2 4 3 4 b. 1 1 2 21 4 2 1 1 2 15 3 2 c. 1 2 3 21 5 3

3 7 6 22 4 8 b. 1 1 3 11 2 16 1 1 10 3 2 2 c. 1 442 4

3

15 16 c. 1 2 8

Related Problems 13. Refer to the shaft shown in Figure 6-4. Determine the missing dimensions in the table using the dimensions given. All dimensions are in inches. G

A

B

C

E

F

D

figure 6-4 A

a.

1 2

b. c.

7 16

d.

5 8

e. f.

98310_sec_01_Unit05-11_ptg01.indd 38

B

3

13 32

4

3 32

3 11 16

4

3 4

3 16

C

D

1

3 8

6

1

5 8

5

E

3 4

37 64

1

7 16

5

31 32

1

11 16

6

1 32

F

G

15 16

7

3 8

4

3 8

3 4

5

1 8

27 32

7

1 32

7 8

7

15 16

7

3 64

4

61 64

25 32

5

3 16

7 8

10/31/18 4:33 PM

UNIT 6

39

CombineD operations of Common fraCtions anD mixeD numbers

14. The outside diameter of an aluminum tube is 3

1 5 inches. The wall thickness is inch. 16 32

What is the inside diameter? 15. Four studs of the following lengths in inches are to be machined from bar stock: 3 0 7 0 50 11 0 1 1 1 , 1 , 2 , and 1 . Allow inch waste for each cut and inch on each end of 4 8 16 32 8 32 each stud for facing. What is the shortest length of bar stock required so that only three cuts are needed? 16. Find dimensions A, B, C, and D of the idler bracket in Figure 6-5. All dimensions are in inches. 11 16

3 8

B

A5

DIA 4 HOLES

B5

3 58

C5 13

1 16

D5 7 16

29 32

DIA 2 HOLES

A

11 16

7 8

C 2 34

7 8

D

7 16 13 16

3 18

figure 6-5

1 17. How long does it take to cut a distance of 1 feet along a shaft that turns 150 revolutions 4 1 per minute with a tool feed of inch per revolution? 32 1 18. An angle iron 47 inches long has two drilled holes that are equally spaced from the 2 7 center of the piece. The center distance between the two holes is 19 inches. What is the 8 distance from each end of the piece to the center of the closest hole? 3 1 inch and a wall thickness of inch. The tube is to 4 16 be fitted in a drilled hole in a block. What diameter hole should be drilled in the block 1 to give inch total clearance? 64

19. A tube has an inside diameter of

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40

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

20. Two views of a mounting block are shown in Figure 6-6. Determine dimensions A–G. All dimensions are in inches. 3 16

DIA 5 HOLES EQUALLY SPACED

A 1 34

17 32

11 16

1 8 1 4

3 18

1 4

G

A5 2 12

B5

F

C5

3 16

D5

17 64

3 16

E5

B

F5

7 16

C

DIA 8 HOLES EQUALLY SPACED DIA 6 HOLES EQUALLY SPACED

G5 1 4

9 32

9 16

15 32

D E

figure 6-6

19 1 aluminum and copper. 20 50 The only other element in the alloy is magnesium. How many pounds of magnesium are required for casting 125 pounds of alloy?

21. The composition of an aluminum alloy by weight is

10 30 70 50 22. Pieces of the following lengths are cut from a 15-inch steel bar: 2 , 1 , 1 , and . 2 4 8 16 1 Allowing inch waste for each cut, what is the length of bar left after the pieces are cut? 8

UNIT 7 Computing with a Calculator:

Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●● ●●

Perform individual operations of addition, subtraction, multiplication, and division with fractions using a calculator. Perform combinations of operations with fractions using a calculator.

Fractions on a scientiFic calculator Depending on the calculator, the fraction key ( abc ) (or Ab/c ) is used when entering fractions and mixed numbers in a calculator. If your calculator has Ab/c , substitute it for all of the examples shown. The answers to expressions entered as fractions will be given as fractions or mixed numbers with the fraction in lowest terms.

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UNIT 7

41

Computing with a CalCulator: fraCtions anD mixeD numbers

Enter the numerator, press abc , and enter the denominator. The fraction is displayed with the symbol or between the numerator and denominator. 3 Example Enter . 4 3 b 3 ac 4, 3 4 or is displayed. We will use this newer, more realistic, notation for fractions 4 in calculator examples.

Fractions on a Machinist calc Pro 2 The fraction bar key, / , is used when entering fractions and mixed numbers in Machinist calc Pro 2. Both proper and improper fractions can be entered. For a fraction, enter the numerator, press / , and enter the denominator. The fraction is displayed. Results are always entered in typical dimensional fractional format. The default setting for the Machinist calc Pro 2 is to display fractional answers in the nearest 64th inch, perhaps reduced to lowest terms. You can change the Fractional Resolution to be displayed in other formats (e.g., 1/160, 1/320, etc.). 3 Example 1 Enter . 4

Solution Press

On/C On/C

to clear the calculator. Then press the calculator screen in Figure 7-1.

3

/

4

9

/

3

. You should see

figure 7-1

3 The screen indicates that the answer is inch. 4 9 Example 2 Enter . 31

Solution Press

On/C On/C

to clear the calculator. Then press should see the calculator screen in Figure 7-2.

1

and you

figure 7-2

Press 5 and the display will change to the one in Figure 7-3.

figure 7-3

Notice that the calculator rounds the fraction to the nearest 64th inch.

98310_sec_01_Unit05-11_ptg01.indd 41

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42

SecTIoN 1

Example 3 Enter Solution Press

Common fraCtions anD DeCimal fraCtions

7 . 15

On/C On/C

7

/

1

5

and you should see the result in Figure 7-4.

figure 7-4

Press 5 and the display will change to Figure 7-5.

figure 7-5

The calculator rounds the fraction to the nearest 64th inch and, in this case, reduces it to 30 15 32nds; that is, it reduces to . 64 32 11 Example 4 Enter the improper fraction . 4

Solution Press

On/C On/C

1

1

/

4

and you should see Figure 7-6.

figure 7-6

Press 5 and the display will change to the mixed number shown in Figure 7-7.

figure 7-7

inDiviDual arithMetic operations: Fractions The operations of addition, subtraction, multiplication, and division are performed with the four arithmetic keys and the equals key. The equals key completes all operations entered and readies the calculator for additional calculations. Certain makes and models of calculators have the execute key, EXE , or enter keys, ENTER and ENTER , instead of the equals key 5 . If your calculator has one of those keys, substitute it for 5 for all examples shown throughout this book. Examples of each of the four arithmetic operations of addition, subtraction, multiplication, and division are presented. Following the individual operation problems, combined operations expressions are given with calculator solutions. An answer to a problem should

98310_sec_01_Unit05-11_ptg01.indd 42

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43

Computing with a CalCulator: fraCtions anD mixeD numbers

UNIT 7

be checked by doing the problem a second time to ensure that improper data was not entered in its solution. Remember to clear or erase previously recorded data and calculations before doing a problem. Depending on the make and model of the calculator, generally, answers are displayed with either the symbol , /, or between the numerator and denominator. As previously mentioned, we will display the answers as fractions. 3 19 Example 1 Add 1 . 16 32 25 b b a a 3 c 16 1 19 c 32 5 Ans 32 7 8

Example 2 Subtract 2 7

abc

8 2 5

abc

64 5

Example 3 Multiply 3

abc

abc

32 3 11

3 11 3 . 32 16

16 5 5 8

13 . 15

Example 4 Divide 4 5

abc

8 4 13

abc

5 . 64 51 64

Ans

33 152

75 104

15 5

Ans

Ans

MixeD nuMbers on a scientiFic calculator Enter the whole number, press denominator. The symbol or 7 Example Enter 15 . 16 b b 15 ac 7 ac 16 7 15 is displayed. 16

abc

, enter the fraction numerator, press ac , and enter the is displayed between the whole number and fraction. b

MixeD nuMbers on a Machinist calc Pro 2 A mixed number can be entered in two ways. One method is to first enter the whole number part, press the Inch key, and then the fractional part with the numerator and denominator separated by the / key. The other method is to enter the fractional part, then the addition key, 1 , and then the whole number followed by the 5 or an operation key. 5 8

Example Enter the mixed number 7 using both methods just described. Solution MEthoD 1 Press

On/C On/C

7

Inch

5

/

8

and you get the image in Figure 7-8.

figure 7-8

MEthoD 2 Press On/C

On/C

5

/

8

1

7

5 and you again should see the image

in Figure 7-8.

98310_sec_01_Unit05-11_ptg01.indd 43

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SecTIoN 1

Common fraCtions anD DeCimal fraCtions

inDiviDual arithMetic operations: MixeD nuMbers The following examples are of mixed numbers with individual arithmetic operations. Depending on the particular calculator, generally, answers are displayed with either the symbol , , or between the whole number and fraction. The following examples are shown with . Substitute or for if necessary. 3 5 Example 1 Add 7 1 23 . 64 8 43 b b b b 7 ac 3 ac 64 1 23 ac 5 ac 8 5 30 Ans 64 7 8

Example 2 Subtract 43 2 36 abc

43

7

abc

8 2 36

abc

abc

29

5 6

Example 3 Multiply 38 3 14 abc

38

5

abc

6 3 14

abc

Example 4 Divide 159 159

abc

17

abc

13

29 . 32

abc

32 5

6

31 32

13 . 16 16 5

575

17 7 43 . 64 8

64 4 3

abc

7

abc

8 5

41

7 32

25 248

Ans

Ans

Ans

practice exercises, individual basic operations with Fractions and Mixed numbers Evaluate the following expressions. The expressions are basic arithmetic operations. Remember to check your answers by doing each problem twice. The solutions to the problems directly follow the practice exercises. Compare your answers to the given solutions. 5 11 7 63 1. 1 6. 125 2 67 8 16 8 64 31 7 13 1 2. 2 7. 62 3 47 32 8 16 6 9 5 27 3 3. 3 8. 785 42 16 8 32 4 23 4 59 27 4. 4 9. 1 46 25 5 64 32 7 3 3 45 5. 85 1 107 10. 37 2 64 4 8 64

solutions to practice exercises, individual basic operations with Fractions and Mixed numbers 1. 5 2. 31 3. 9 4. 23

98310_sec_01_Unit05-11_ptg01.indd 44

abc abc abc abc

8 1 11 32 2 7 16 3 5 25 4 4

abc abc abc abc

16 5 8 5 8 5 5 5

5 Ans 16 3 Ans 32 45 Ans 128 3 1 Ans 20 1

10/31/18 4:33 PM

5. 85

abc

abc

abc

7

abc

6. 125 7. 62

45

Computing with a CalCulator: fraCtions anD mixeD numbers

UNIT 7

abc

64 1 107

7

abc

8 2 67

abc

13

abc

16 3 47

3

abc

4 5

63

abc

64 5

1

abc

6 5

abc

55 Ans 64 57 57 Ans 64 21 2962 Ans 32 192

If your calculator showed the answer as 2692.65625, write the entire number on your paper. The whole number part of the answer is to the left of the decimal point, 2962. Now, for the fraction part of the answer, enter 0.65625 in your calculator and press the 2nd key and the f ➧➧●D key. (On a TI-30 calculator, the f ➧➧●D is over the PRB key.) Press ENTER . The cal21 culator should display 21 32, or . Combining the whole number and the fraction parts 32 21 b gives the mixed number 2962 . On a Sharp™ calculator, pressing the ac key toggles the 32 display between fraction and decimal expressions. abc

8. 785

27

abc

32 4 2

9. 59

abc

64 1 46

10. 37

abc

3

abc

abc

27

8 2 45

abc

3

abc

32 5

abc

abc

64 5

4 5 47 36

43 64

285 49 64

67 88

Ans

Ans Ans

coMbineD operations on a scientiFic calculator The expressions are solved by entering numbers and operations into the calculator in the same order as the expressions are written. Remember to check your answers by doing each problem twice. 17 7 3 Example 1 Evaluate 275 1 3 26 . 32 8 4 15 b b b b b a a a 1 3 c c c 275 17 32 7 8 26 ac 3 ac 4 5 298 Ans 16 7 3 Because the calculator has algebraic logic, the multiplication operation 3 26 was 8 4 17 performed before the addition operation adding 275 was performed. 32

1

1

2

35 5 2 1 18 4 10 64 8 b 8 1 18 4 10 ac

Example 2 Evaluate 35

abc

64 2 5

abc

1

2

2 . 3 2

abc

3 5

1

39 64

Ans

2

29 3 15 2 19 3 12. 32 16 64 As previously discussed in Unit 6, operations enclosed within parentheses are done first. A calculator with algebraic logic performs the operations within parentheses before performing other operations in a combined operations expression. If an expression contains parentheses, enter the expression in the calculator in the order in which it is written. The parentheses keys must be used. 27 b b b b b 380 ac 29 ac 32 2 ( 3 ac 16 1 9 ac 15 ac 64 ) 3 12 5 267 Ans 32

Example 3 Evaluate 380

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Common fraCtions anD DeCimal fraCtions

47 5 173 64 8 Example 4 Evaluate . 3 1 321 16 8 Recall that for a problem expressed in fractional form, the fraction bar is also used as a grouping symbol. The numerator and denominator are each considered as being enclosed in parentheses and these must be used when entering the problem in the calculator. You might find it helpful to write parentheses around the numerator and denominator before you use the calculator. For this example, you might want to rewrite the expression as 25

125 4764 1 7 3 582 1163 3 2 1 182 ( abc

abc

25 8

)

abc

64 1 7 3 5 7 60 Ans 32

47 5

abc

8

)

(

4

3

abc

16 3 2 1 1

Note: The expression may also be evaluated by using the 5 key to simplify the numerator without having to enclose the entire numerator in parentheses. However, parentheses must be used to enclose the denominator. 25

abc

47

abc

7 60 32

5

64 1 7 3 5

abc

(

8 5 4

3

abc

16 3 2 1 1

abc

8

)

Ans

coMbineD operations on a Machinist calc Pro 2 Fraction problems that combine addition or subtraction with multiplication or division cannot be done using the Machinist calc Pro 2 and results in a “Dimension error” message. The next example will illustrate this.

Example Evaluate 5

7 7 3 1 32 . 32 8 4

Solution On/C

On/C

5

Inch

7

/

3

2

1

7

/

8

3

2

Inch

3

/

4

5

The result is shown in Figure 7-9.

figure 7-9

This “Dimension error” message in Figure 7-9 often indicates that you have made an error, but in this case it indicates that the calculator cannot perform this operation as it has been entered. The Machinist calc Pro 2 is a dimensional calculator, which means that when you use the fraction bar ( / ), it interprets this as a measurement and that is why “INCH” is displayed at the bottom of the calculator’s screen. However, there is a way to “work around” dimension errors like this. A fraction is really 3 a division problem. That means that means the same numerically as 3 4 4. In a similar 4 3 way, a mixed number is the sum of a whole number and a fraction. We can think of 2 as 4

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UNIT 7

47

Computing with a CalCulator: fraCtions anD mixeD numbers

2 1 3 4 4. You must also remember the Order of Operations and apply them when using combined operations. Finally, you will need to use the calculator’s memory registers to store intermediate steps. Now we will rework the last example and show how it can be done on the Machinist calc Pro 2. This is not the only way to solve this problem. We are using this method to show how to use the calculator’s memory registers.

Example Evaluate 5

7 7 3 1 32 . 32 8 4

Solution We begin by applying the Order of Operations. The Order of Operations dictate that mul7 7 3 tiplication is performed before addition, so think of this problem as 5 1 3 2 . We 32 8 4 will perform the multiplication part first, save the result in the calculator’s memory, and then add the remaining part. However, one of the factors in this multiplication is a mixed number, so we write and store it first.

1

On/C

On/C

Now press

2

1

3

Conv Rcl 1

4

4

2

5

to save the result in Memory 1, as shown in Figure 7-10.

figure 7-10

7 We multiply this result by by pressing 8 in Figure 7-11.

7

4

8

1

5

1

Rcl 1

5 , with the result shown

figure 7-11

7 Finally, add 5 to this result by pressing 1 32 result shown in Figure 7-12.

7

4

3

2

5 , with the final

figure 7-12

You might be able to work the previous example without storing the intermediate results in the memory; however, we used this as an opportunity to introduce you to the means to store and recall number in the calculator memory. Fortunately, as a machinist, you will probably never need to work a problem this complicated, and for this reason we will not give any more examples. Notice that we didn’t use the “Inch” key for the whole numbers and used the 4 key rather than the / key for fraction but added the whole number to the fractional part of the mixed number.

98310_sec_01_Unit05-11_ptg01.indd 47

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48

Common fraCtions anD DeCimal fraCtions

SecTIoN 1

practice exercises, combined operations with Fractions and Mixed numbers Evaluate the following combined operations expressions. Remember to check your answers by doing each problem twice. The solutions to the problems directly follow the practice exercises. Compare your answers to the given solutions. 1.

11116 1 12 31322 4 18

12532 2 342 4 12 3 34 4 9 3 1 272 3 7. 50 3 128 2 17 5 10 5 6.

108 5 23 3 64 8 9 3 7 43 2 17 1 10 5 20 3. 5

2.

1

13 5 3 1 98 3 6 2 16 8 4 3 7 56 1 20 3 4 8 5. 2 43 3 4. 120

1 2 5 29 2 15 18 8. 1 64 32 1 8 1 7 270 2 175 3 2 8 9. 1 3 128 64 40

2

1

2

solutions to practice exercises, combined operations with Fractions and Mixed numbers Note: In each solution the first answer is with a scientific calculator and the second is with a Machinist calc Pro 2. 1 b b b b 1. ( 11 ac 16 1 12 ac 31 ac 32 ) 4 1 ac 8 5 109 Ans 4 On/C On/C 11 / 16 1 12 Inch 31 / 32 5 4 1 / 8 5 109.25 Ans 2. 108 4 3 On/C

3.

abc

On/C 5

/

(

abc

9

43

On/C

On/C

4

3

Inch

abc

8 2 3

64

abc

5

Conv

8

5

64 5 Rcl

10 2 17

abc

3

43

Inch

/

10

9

59 64

108

1

284.92188

abc

284

Ans 4

Inch

Rcl

5 2

1

3

Ans

abc

5 1 7

2

abc

Inch

17

20

)

4 5 5

3

/

5

1

5 7

33 100

/

20

Ans 5

21 Ans 64 Remember, the Machinist calc Pro 2 only expresses fraction answers with a denominator that is a power of 2: 2, 4, 8, 16, 32, or 64. 4

4. 120

5

5

abc

13

On/C 13

98310_sec_01_Unit05-11_ptg01.indd 48

On/C

/

16

5

abc

16 1 98 6

5

2

3

638

19 32

abc

4

5 4

abc

8 3

5 3

( 98

6 2 3 Inch

5

abc

4 /

) 8

5 1

638 120

19 32

Ans

Inch

Ans

10/31/18 4:33 PM

5.

56 ac 3 27 27 Ans 32 b

abc

4 1 20 3 7

abc

On/C

On/C

20

3

5 1

/

3

(

3

2

25

ac

4

6.

(

On/C

On/C Rcl

3

9.

On/C

3

50 abc

1

abc

5

12

1 64

On/C

On/C

ac b

ac

4

4

1

4

Rcl

2

5

ac

5 2 17

3

/

5

5 abc

2 4 1

1

/

ac b

1137

3

4

5

Rcl

Rcl

1

27.84375 Ans

3 64 4

1

1

5

Conv

5

2

4 5

b

)

Ans 3

2

4 4

3

0.0833333 Ans ac

)

10 1 27

b

9

2

5

abc

Conv

Rcl

ac

2 3 3

3

4

9

/

abc

64 1 8

1

/

29

/

32

4

(

1

Inch

17

abc

3 3 10

1

5 5 1137 5

Inch

27

Ans (

8 2

15

abc

5

abc

29

abc

)

32

Ans 40

15

Inch

2

Rcl

2

5

Inch

1

/

2

4

5

/

64

1

8

Inch

1 64

Ans

12.015625 5 12

270 2 175 7 58 Ans 32

abc

On/C

On/C

Inch

Inch

2

(

175

Rcl

Rcl

ac

2

Inch

1

4

4 1

b

/

3

4

1

abc

4 3 2

Inch

Rcl

2

(

4

56

32

28

Rcl

1

)

8

Rcl

)

4

b

4

8

Conv b

Rcl

2

7. 50 3 ( 28 1137 Ans On/C

5

4

25

/

7

32 2 3

b

Conv

8. 40

49

Computing with a spreaDsheet: fraCtions anD mixeD numbers

UNIT 8

2

1

5

1

abc

1

2 3 7

/

2

abc

3

Inch

1

)

8

7

/

8

/ 64 3 7 58.21875 5 58 Ans 32

Conv

Rcl

1

1

8

Conv

abc

Conv 128

5

Inch Rcl

Conv 2

Rcl

1

64 3 128

)

Rcl

Inch

Inch

1 Conv

270 Rcl

5

2

Rcl

UNIT 8 Computing with a Spreadsheet: Fractions and Mixed Numbers

Objectives After studying this unit you should be able to ●●

Enter fractions and mixed numbers in a spreadsheet. Perform individual operations of addition, subtraction, multiplication, and division with fractions using a spreadsheet.

●●

Perform combinations of operations with fractions using a spreadsheet.

●●

98310_sec_01_Unit05-11_ptg01.indd 49

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50

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

spreaDsheet basics When you open Excel on an iPad, you are presented with a screen similar to the one in Figure 8-1. Click on the circled Blank Workbook icon in the upper left-hand corner. This will open a blank spreadsheet. We begin with some basic terms for spreadsheets. Consider Figure 8-2. The spreadsheet is made up of columns (A, B, C, . . . ) and rows (1, 2, 3, . . . ). The intersection of a column and row is a cell. Cell A1 is identified in the figure. A row above the column headings includes the “Formula Bar” that displays the contents of the active cell. In Figure 8-2, the Formula Bar is blank.

figure 8-2

entering anD coMputing with Fractions in a spreaDsheet When entering fractions in a spreadsheet, think of the fraction as a division problem and replace the fraction bar, or vinculum, with a forward slash, /. 7 5 Example 1 Use a spreadsheet to compute 1 . 8 6 Solution The spreadsheet recognizes the Order of Operations, so this statement can be entered in one cell without using parentheses. Enter = 7/8 + 5/6 in Cell A1, as shown in Figure 8-3. When you press RETURN or ENTER , you get the result in Figure 8-4—a decimal approximation to the actual result.

figure 8-1

figure 8-3

98310_sec_01_Unit05-11_ptg01.indd 50

figure 8-4

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Computing with a spreaDsheet: fraCtions anD mixeD numbers

UNIT 8

51

If you want the result written in the form of a fraction, then you must reformat the cell that displays the answer. Highlight Cell A1, open the “Number Formatting” panel, click on “Fraction” in the menu, and then choose the required number of digits—watch the “Sample” until it appears to show the correct display (see Figure 8-5).

figure 8-5

You should get the result shown in Figure 8-5, which shows that

7 5 17 1 5 1 17/24 5 1 . 8 6 24

3 8 3 . 32 21 Solution Enter = 3/32 * 8/21 in Cell A2, as shown in Figure 8-6, and press RETURN . 1 The result, , is shown in Figure 8-7. Note that the * key is used as the multiplication 28 sign. The answer is in lowest terms.

Example 2 Use a spreadsheet to compute

figure 8-6

figure 8-7

coMputing with MixeD nuMbers Mixed numbers have to be treated differently than fractions. There is no fraction key in a spreadsheet as there is on many calculators. You have to treat a mixed 5 5 number as the sum of a whole number and a fraction. Thus, think of 1 as 1 1 . If you 16 16 are subtracting a mixed number, then you will need to put the number in parentheses after the subtraction sign.

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52

SecTIoN 1

3 5

Common fraCtions anD DeCimal fraCtions

7 8

Example Use a spreadsheet to compute 1 12 . 7 7 7 8 8 8 While we do not need parentheses when adding, we will always use them so that we don’t forget them when they are needed. The Formula Bar of Figure 8-8 shows that = 3/5 + (12 + 7/8) was entered in Cell A3. 19 When you have finished typing the expression, hit RETURN and the result 13 will be 40 displayed, as shown in Cell A3 of Figure 8-9.

Solution Here 12 is a mixed number. You need to remember that 12 means 12 1 .

figure 8-8

figure 8-9

The following examples are of mixed numbers with individual operations. 7 8

Example 1 Use a spreadsheet to subtract 23 2 7

3 . 16

Solution The parentheses around the mixed number are very important in this subtraction problem. Placing parentheses around the 7

3 means that you are subtracting both 16

3 . As shown in the Formula Bar of Figure 8-10, = (23 + 7/8) − (7 + 3/16) was 16 11 entered. The answer, 16 , is shown in Cell A4 of Figure 8-11. 16

the 7 and the

figure 8-10

98310_sec_01_Unit05-11_ptg01.indd 52

figure 8-11

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UNIT 8

53

Computing with a spreaDsheet: fraCtions anD mixeD numbers

7 8

1 2

Example 2 Use a spreadsheet to multiply 4 3 6 . Solution Again, don’t forget the parentheses around the mixed numbers. When you have finished typing the expression shown in the Formula Bar of Figure 8-12, hit 11 and the result, 31 , will be displayed as shown in Cell A5 of Figure 8-13. 16

figure 8-12

RETURN

figure 8-13

7 8

3 4

Example 3 Use a spreadsheet to work this division problem: 21 4 8 . Solution The typed expression is shown in the Formula Bar of Figure 8-14, and the 1 result, 2 , is shown in Cell A6 of Figure 8-15. 2

figure 8-14

98310_sec_01_Unit05-11_ptg01.indd 53

figure 8-15

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54

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

practice exercises: individual operations with Fractions and Mixed numbers Use a spreadsheet to evaluate the following expressions. The expressions are basic single arithmetic operations with fractions and mixed numbers. Remember to first estimate the answer and then check the way it was entered in the Formula Bar. The keystrokes and answers are given directly following the exercises. Compare your answers to the given solutions. 1. 2. 3. 4. 5.

5 7 1 8 24 25 5 2 32 8 8 5 3 15 9 4 20 4 15 21 8 1 2 1 1 15 5 3

7 3 1 187 64 4 5 15 275 2 74 8 16 3 1 35 3 21 16 4 1 3 64 4 24 8 4 13 35 2 18 78

6. 213 7. 8. 9. 10.

solutions to practice exercises: individual operations with Fractions and Mixed numbers 11 Ans 12 5 Ans = 25/32 − 5/8 RETURN 32 8 = 8/15 * 5/9 RETURN Ans 27 7 Ans = (4/15)/(20/21) RETURN 25 2 = 8/15 + 1/5 + 2/3 RETURN 1 Ans 5

1. = 5/8 + 7/24 2. 3. 4. 5.

RETURN

55 Ans 64 11 200 Ans = (275 + 5/8) − (74 + 15/16) RETURN 16 47 747 Ans = (35 + 3/16) * (21 + 1/4) RETURN 64 13 = (64 + 1/8)/(24 + 3/4) RETURN 2 Ans 16 5 16 Ans = 35 − (18 + 13/78) RETURN 6

6. = (213 + 7/64) + (187 + 3/4) 7. 8. 9. 10.

RETURN

400

coMbineD operations with Fractions anD MixeD nuMbers When there are combined operations, the numbers are entered into the spreadsheet in the same order as the expressions are written. Of course, care must be taken with mixed numbers to write them within parentheses.

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UNIT 8

Computing with a spreaDsheet: fraCtions anD mixeD numbers

Example 1 Use a spreadsheet to evaluate 32

55

5 5 1 1 35 . 16 8 2

Solution Enter the following in the Formula Bar: = (32 + 5/16) + 5/8 * (5 + 1/2). See Figure 8-16. Press

RETURN

3 . The result, 35 , is shown in Cell A7 of Figure 8-17. 4

figure 8-16

figure 8-17

Example 2 Use a spreadsheet to evaluate 130 Solution Think of this as 130

1

1

2

29 5 3 1 2 17 38 . 32 16 8 2

1 22 3 8 12. Notice that there are two right

29 5 3 2 1 7 32 16 8

3 parentheses after 7 . Some people use brackets [ ] when they write the problem to help 8 tell which grouping symbols belong together, but you still have to type parentheses and not brackets. Enter the following in the Formula Bar: = (130 + 29/32) − (5/16 + (7 + 3/8)) * (8 + 1/2). 9 See Figure 8-18. Press RETURN . The result, 16 , is shown in Cell A8 of Figure 8-19. 16

figure 8-18

98310_sec_01_Unit05-11_ptg01.indd 55

figure 8-19

10/31/18 4:33 PM

56

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

13 3 193 16 4 . 5 7 341 8 16

20

Example 3 Use a spreadsheet to evaluate

Solution Remember that, when you enter this in the Formula Bar, you will need to put the entire numerator in parentheses and also the entire denominator. What you enter in the Formula Bar should look like this: = ((20 + 13/16) + 9 * 3/4)/ 18 (5/8 * 4 + 7/16). See Figure 8-20. Press RETURN . The result, 9 , is shown in Cell A9 of 47 Figure 8-21.

figure 8-20

figure 8-21

practice exercises: combined operations with Fractions and Mixed numbers Evaluate the following expressions involving combined operations with fractions and mixed numbers. Remember to first estimate the answer and then check the way the expression was entered in the Formula Bar. The keystrokes and answers are given directly following the exercises. Compare your answers to the given solutions. 1 1 11 15 3 36 1 20 4 3 1. 1 15 4 4 3 32 16 8 5. 2 1 125 11 62 32 2. 14 3 2 5 16 4 3 1 3 2 16 6. 3 17 4 1 3 15 5 2 4 15 12 3 95 18 1 25 5 1 50 3. 36 2 5 13 4 4 7. 1 21 1 10 3 3 1 3 64 32 63 25 1 5 10 4 4 8 4. 12 2 7 9 4 8. 75 3 14 2 6 1 11 3 5 10 25 5

1

2

1

2

1

1

98310_sec_01_Unit05-11_ptg01.indd 56

2

2

10/31/18 4:33 PM

UNIT 9

57

introDuCtion to DeCimal fraCtions

solutions to practice exercises: combined operations with Fractions and Mixed numbers 1. = (11/32 + (15 + 15/16))/(3/8) 2. = 125/(5/16) + (4 + 11/16)

43

RETURN

RETURN

404

3. = ((95 + 12/25) + (8 + 3/5))/4 * (6 + 1/4) 4. = ((63 + 3/10) − (5 + 3/4) + 1/4)/12

5 12

11 16

Ans

Ans 162

RETURN

RETURN

4

5. = ((36 + 1/4) + 20/(3 + 1/3))/(6 − 2/3 * (2 + 1/2)) 6. = ((3 + 4/15) + (7 + 3/5))/(1/2) + 3/4 * 2/15

49 60

Ans

Ans 9

RETURN

RETURN

7. = ((50 + 1/2)/(5 + 3/8)) + (21 + 5/64) + (10 + 13/32) 8. = 75 * ((14 + 2/5) − (6 + 7/10) + (11 + 9/25)) * 4/5

5 8

21

5 6

RETURN RETURN

3 4

Ans

Ans 256 Ans 291 3 1143 Ans 5 40

UNIT 9 Introduction to Decimal Fractions Objectives After studying this unit you should be able to ●● ●●

●● ●●

Locate decimal fractions on a number line. Express common fractions having denominators of powers of 10 as equivalent decimal fractions. Write decimal numbers in word form. Write numbers expressed in word form as decimal fractions.

Most engineering drawings are dimensioned with decimal fractions rather than common fractions. The dials that are used in establishing machine settings and movement are graduated in decimal units. Tool speeds and travel are determined in decimal units and machined parts are usually measured in decimal units.

explanation oF DeciMal Fractions A decimal fraction is not written as a common fraction with a numerator and denominator. The denominator is omitted and replaced by a decimal point placed to the left of the numerator. Decimal fractions are equivalent to common fractions having denominators that are powers of 10, such as 10; 100; 1000; 10,000; 100,000; and 1,000,000. Powers of 10 are numbers that are obtained by multiplying 10 by itself a certain number of times.

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58

SecTIoN 1

Common fraCtions anD DeCimal fraCtions

Meaning oF Fractional parts The line segment shown in Figure 9-1 is one unit long. It is divided into ten equal smaller parts. The locations of common fractions and their decimal fraction equivalents are shown on the line. 1 UNIT COMMON FRACTIONS 1 10

2 10

3 10

4 10

5 10

6 10

7 10

8 10

9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

10 10

=1

0 1.0

DECIMAL FRACTIONS 1 UNIT LINE

figure 9-1

1 One of the ten equal small parts, or 0.1 of the 1 unit line, is shown enlarged in 10 1 Figure 9-2. The or 0.1 unit is divided into ten equal smaller units. The locations of com10 mon fractions and their decimal fraction equivalents are shown on the line in Figure 9-2. 1 10

OR 0.1 UNIT

COMMON FRACTIONS 1 100

2 100

3 100

4 100

5 100

6 100

7 100

8 100

9 100

10 100

=

1 10

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.1

DECIMAL FRACTIONS ENLARGED 0.1 UNIT LINE

figure 9-2

1 or 0.01 division is divided into ten equal smaller parts, the resulting parts are 100 1 2 3 9 10 1 or 0.001; or 0.002; or 0.003; . . . or 0.009; 5 or 0.01. 1000 1000 1000 1000 1000 100 1 ●● Each time the decimal point is moved one place to the left, a value or 0.1 times the 10 previous value is obtained. ●● Each time a decimal point is moved one place to the right, a value 10 times greater than the previous value is obtained. If the

Each time a decimal fraction is multiplied by 10 the decimal point is moved one place to the right. Each step in the following table shows both the decimal fraction and its equivalent common fraction. Decimal Fraction

Common Fraction

0.000003 3 10 5 0.00003

3/1,000,000 3 10 5 3/100,000

0.00003 3 10 5 0.0003

3/100,000 3 10 5 3/10,000

0.0003 3 10 5 0.003

3/10,000 3 10 5 3/1000

0.003 3 10 5 0.03

3/1000 3 10 5 3/100

0.03 3 10 5 0.3

3/100 3 10 5 3/10

0.3 3 10 5 3

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Note: The metric (or SI) system uses a small space rather than a comma to separate groups of three digits. In the metric system 10,000 is written 10 000. This applies to decimal fractions as well with 0.00003 written as 0.000 03.

reaDing anD writing DeciMal Fractions The following chart (Figure 9-3) gives the names of the parts of a number with respect to the positions from the decimal point. VALUES GREATER THAN OR EQUAL TO 1 (WHOLE NUMBERS) 7

6

5

4

VALUES LESS THAN 1 (DECIMALS) 3

2

1

.

1

2

3

4

5

6

Millionths Hundred-thousandths Ten-thousandths

Millions Hundred thousands Ten thousands Thousands Hundreds Tens Units

Thousandths Hundredths Tenths

figure 9-3

To read a decimal, read the number as a whole number. Then say the name of the decimal place of the last digit to the right.

Examples 1. 0.5 is read “five tenths.” 2. 0.07 is read “seven hundredths.” 3. 0.011 is read “eleven thousandths.” To write a decimal fraction from a word statement, write the number using a decimal point and zeros before the number as necessary for the given place value.

Examples 1. Two hundred nineteen ten-thousandths is written as 0.0219. 2. Forty-three hundred-thousandths is written as 0.00043. 3. Eight hundred seventeen millionths is written as 0.000817. A number that consists of a whole number and a decimal fraction is called a mixed decimal. To read a mixed decimal, read the whole number, read the word and at the decimal point, and read the decimal.

Examples 1. 3.4 is read “three and four tenths.” 2. 1.002 is read “one and two thousandths.” 3. 16.0793 is read “sixteen and seven hundred ninety-three ten-thousandths.” 4. 8.00032 is read “eight and thirty-two hundred-thousandths.”

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alternative way to reaD DeciMal Fractions Some workers always work in thousandths, and whenever a decimal number is read “thousandths” is used.

Examples 1. 7.025 is read “seven and twenty-five thousandths.” 2. 0.503 would be read as “five hundred three thousandths.” 3. 542.906 is read as “five hundred forty-two and nine hundred six thousandths.” 4. 14.37 as thousandths is 14.370 and is read as “fourteen and three hundred seventy thousandths.” 5. 96.2 can be thought of as 96.200 and read “ninety-six and two hundred thousandths.” In manufacturing, the ten-thousandths place is used but the word “tenths” is understood to mean “ten-thousandths.”

Examples 1. In manufacturing, 0.5678 is read “five hundred sixty-seven and eight tenths.” 2. Someone in manufacturing would read 3.7129 as “three and seven hundred twelve and nine tenths.” The fourth place is said as tenths (with the understanding that it is ten-thousandths—they just say tenths rather than ten-thousandths). 3. In a manufacturing setting, if someone says “two tenths” to you, then you should write “0.0002.”

siMpliFieD MethoD oF reaDing DeciMal Fractions Usually, a simplified method of reading decimal fractions is used in the machine trades. This method is generally quicker, easier, and less likely to be misinterpreted. A tool-anddie maker reads 0.0265 inches as point zero, two, six, five inches. A machinist reads 4.172 millimeters as four, point one, seven, two millimeters.

writing DeciMal Fractions FroM coMMon Fractions having DenoMinators that are powers oF 10 A common fraction with a denominator that is a power of 10 can be written as a decimal fraction. For a common fraction with a numerator smaller than the denominator, replace the denominator with a decimal point. The decimal point is placed to the left of the first digit of the numerator. There are as many decimal places as there are zeros in the denominator. When writing a decimal fraction, place a zero to the left of the decimal point.

Examples 9 5 0.9 Ans 10 381 2. 5 0.381 Ans 1000 7 3. 5 0.0007 Ans 10,000 1.

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There is 1 zero in 10 and 1 decimal place in 0.9. There are 3 zeros in 1000 and 3 decimal places in 0.381. There are 4 zeros in 10,000 and 4 decimal places in 0.0007. In order to maintain proper place value, 3 zeros are written between the decimal point and the 7.

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UNIT 9

ApplicAtion Tooling Up Use Figure 9-4 to answer Exercises 1–6. All dimensions are in inches. B 21

31

6 32 C

1 64

7

3 16 D

TYPICAL 6 PLACES

TYPICAL 5 PLACES

5 8

A E F 5 16

TYPICAL 5 PLACES

TYPICAL 6 PLACES

figure 9-4

1. Distance A 5

4. Distance D 5

2. Distance B 5

5. Distance E 5

3. Distance C 5

6. Distance F 5

Meaning of Fractional Parts 7. Find the decimal value of each of the distances A, B, C, D, and E in Figure 9-5. Note the total unit value of the line. 1 UNIT

A5 B5 C5

E A

D5

B

E5

C D

figure 9-5

8. Find the decimal value of each of the distances A, B, C, D, and E in Figure 9-6. Note the total unit value of the lines. 0.1 UNIT

A5 B5 C5

E A

D5

B

E5

C D

figure 9-6

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9. Find the decimal value of each of the distances A, B, C, D, and E in Figure 9-7. Note the total unit value of the line. 0.01 UNIT

A5 B5

E

C5

A

D5

B C

E5

D

figure 9-7

In each of the following exercises, the value on the left must be multiplied by one of the following numbers: 0.0001; 0.001; 0.01; 0.1; 10; 100; 1000; or 10,000 in order to obtain the value on the right of the equal sign. Determine the proper number. 10. 0.9 3

5 0.0009

15. 4 3

11. 0.7 3

5 0.007

16. 0.0643 3

5 0.000643

12. 0.03 3

5 0.3

17. 0.0643 3

5 6.43

13. 0.0003 3

5 0.003

14. 0.135 3

5 0.00135

5 0.4

18. 0.00643 3 19. 643 3

5 64.3 5 0.643

Reading and Writing Decimal Fractions Write these numbers as words. 20. 0.064

25. 1.5

21. 0.007

26. 10.37

22. 0.132

27. 16.0007

23. 0.0035

28. 4.0012

24. 0.108

29. 13.103

Write these words as numbers. 30. eighty-four ten-thousandths

34. thirty-five ten-thousandths

31. three tenths

35. ten and two tenths

32. forty-three and eight hundredths

36. five and one ten-thousandth

33. four and five hundred-thousandths

37. twenty and seventy-one hundredths

Alternative Method for Reading Decimal Fractions Write these numbers as words using the alternative method for reading decimal fractions. 38. 15.086

42. 0.208

39. 12.104

43. 0.715

40. 903.802

44. 380.1

41. 3047.59

45. 97.003

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63

rounDing DeCimal fraCtions anD equivalent DeCimal anD Common fraCtions

Write these words as numbers. 46. forty-three and eight thousandths 47. fourteen and five hundred thousandths 48. thirty-seven and twenty-five thousandths 49. one hundred six and fifty-three thousandths 50. seventy-six thousandths 51. four and one hundred five thousandths Each of the following common fractions has a denominator that is a power of 10. Write the equivalent decimal fraction for each. 52.

9 10

7 10,000 17 54. 100 53.

43 100 61 56. 1000 999 57. 10,000 55.

73 1000 1973 59. 100,000 47,375 60. 100,000 58.

UNIT 10 Rounding Decimal Fractions and Equivalent Decimal and Common Fractions

Objectives After studying this unit you should be able to ●● ●● ●●

Round decimal fractions to any required number of places. Express common fractions as decimal fractions. Express decimal fractions as common fractions.

When engineering drawing dimensions of a part are given in fractional units, a machinist is usually required to express these fractional values as decimal working dimensions. In computing material requirements and in determining stock waste and scrap allowances, it is sometimes more convenient to express decimal values as approximate fractional equivalents.

rounDing DeciMal Fractions When working with decimals, the computations and answers may contain more decimal places than are required. The number of decimal places needed depends on the degree of precision desired. The degree of precision depends on how the decimal value is going to be used. The tools, machines, equipment, and materials determine the degree of precision obtainable. For example, a length of 0.875376 inch cannot be cut on a

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milling machine. In cutting to the nearer thousandths of an inch, the machinist would consider 0.875376 inch as 0.875 inch. Rounding a decimal means expressing the decimal with a fewer number of decimal places.

c Procedure ●● ●●

●●

To round a decimal fraction

Determine the number of decimal places required in an answer. If the digit directly following the last decimal place required is less than 5, drop all digits that follow the required number of decimal places. If the digit directly following the last decimal place required is 5 or larger, add 1 to the last required digit and drop all digits that follow the required number of decimal places.

Example 1 Round 0.873429 to 3 decimal places. The digit following the third decimal place is 4. Because 4 is less than 5, drop all digits after the third decimal place.

0.873 4 29 0.873

Ans

Example 2 Round 0.36845 to 2 decimal places. The digit following the second decimal place is 8. Because 8 is greater than 5, add 1 to the 6.

0.36 8 45 0.37

Ans

Example 3 Round 18.738257 to 4 decimal places. The digit following the fourth decimal place is 5. Add 1 to the 2.

18.7382 5 7 18.7383 Ans

expressing coMMon Fractions as DeciMal Fractions A common fraction is an indicated division. For example,

3 5 is the same as 3 4 4; is the 4 16

99 is the same as 99 4 171. 171 Because both the numerator and the denominator of a common fraction are whole numbers, expressing a common fraction as a decimal fraction requires division with whole numbers. same as 5 4 16, and

c Procedure ●●

To express a common fraction as a decimal fraction

Divide the numerator by the denominator. A common fraction that divides without a remainder is called a terminating decimal. A common fraction that does not terminate is expressed as a repeating or nonterminating decimal. The division should be carried out to one more place than the number of places required in the answer, then rounded one place.

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65

2 3 Divide the numerator by the denominator. After the 2, add one more zero than the required number of decimal places. (Add 5 zeros.)

Example 1 Express as a 4-place decimal.

0.66666 3q 2.00000 0.6667 Ans

Round 0.66666 to 4 places. 5 7

Example 2 Express as a 2-place decimal. 0.714 7q 5.000 0.71 Ans

Add 3 zeros after the 5. Round to 2 places.

expressing DeciMal Fractions as coMMon Fractions c Procedure ●● ●●

●●

To express a decimal fraction as a common fraction

Write the number after the decimal point as the numerator of a common fraction. Write the denominator as 1 followed by as many zeros as there are digits to the right of the decimal point. Express the common fraction in lowest terms.

Example 1 Express 0.375 as a common fraction. Write 375 as the numerator. There are three digits to the right of the decimal point. Write the denominator as 1 followed by 3 zeros. The denominator is 1000. Reduce

375 to lowest terms. 1000

0.375 5

Example 2 Express 0.27 as a common fraction. The numerator is 27. The denominator is 100.

0.27 5

375 3 5 1000 8

27 100

Ans

Ans

Example 3 Express 0.03125 as a common fraction. The numerator is 3125. The denominator is 100,000. Reduce

3125 to lowest terms. 100,000

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0.03125 5

3125 1 5 100,000 32

Ans

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ApplicAtion Tooling Up 1. Find the decimal value of the distance A in Figure 10-1. Note the total unit value of the line. 1 UNIT

A

figure 10-1

2. Find the decimal value of the distance B in Figure 10-2. Note the total unit value of the line. 0.1 UNIT

B

figure 10-2

3. Find the decimal value of the distance C in Figure 10-3. Note the total unit value of the line. 0.01 UNIT

C

figure 10-3

Use Figure 10-4 to answer Exercises 4 and 5. All dimensions are in inches. E 11

1 4

13 32

2 32 D

7 16

figure 10-4

4. Find the length of D in Figure 10-4. 5. Determine the length of E in Figure 10-4. 1 3 6. Multiply 4 3 2 . 3 4

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rounDing DeCimal fraCtions anD equivalent DeCimal anD Common fraCtions

67

Rounding Decimal Fractions Round the following decimals to the indicated number of decimal places. 7. 0.63165 (3 places)

12. 0.90039 (2 places)

8. 0.1247 (2 places)

13. 0.72008 (4 places)

9. 0.23975 (3 places)

14. 0.0006 (3 places)

10. 0.01723 (3 places)

15. 0.0003 (3 places)

11. 0.03894 (2 places)

16. 0.099 (3 places)

express common Fractions as Decimal Fractions Express the common fractions as decimal fractions. Express the answer to 4 decimal places. 17. 18. 19. 20. 21. 22.

11 16 7 8 5 8 3 4 2 3 10 11

23. 24. 25. 26. 27. 28.

2 25 47 64 7 32 1 2 4 7 3 8

Solve the following. 29. In Figure 10-5, what decimal fraction of distance B is distance A? Express the answer to 4 decimal places. All dimensions are in inches.

A=1 B=3

figure 10-5

30. Five pieces are cut from the length of round stock shown in Figure 10-6. After the pieces are cut, the remaining length is thrown away. What decimal fraction of the original length of round stock (17 inches) is the length that is thrown away? All dimensions are in inches. #1

#2

#3

1

2 12

CUT

1 8

#5

15

2 34

3 16 1 8

#4

2 18

3 16 1 8

CUT

CUT

1 8

CUT

17

figure 10-6

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31. Dimensions in Figure 10-7 are in feet and inches. F = 69- 00 E = 90 H = 39- 30

G

figure 10-7

a. What decimal fraction of distance F is distance E? Note: Both the numerator and denominator of a common fraction must be in the same units before the value is expressed as a decimal fraction. Use 1 foot 5 12 inches. b. Determine distance G. c. What decimal fraction of distance H is distance G? Express the answer to 4 decimal places.

expressing Decimal Fractions as common Fractions Express the following decimal fractions as common fractions. Reduce to lowest terms. 32. 0.875

43. 0.4375

33. 0.125

44. 0.2113

34. 0.4

45. 0.8717

35. 0.75

46. 0.0005

36. 0.6

47. 0.03

37. 0.6875

48. 0.09375

38. 0.67

49. 0.237

39. 0.003

50. 0.45

40. 0.008

51. 0.045

41. 0.502

52. 0.0045

42. 0.99 Solve the following. 53. In Figure 10-8, what common fractional part of distance B is distance A? All dimensions are in inches.

A = 0.875 B = 1.000

figure 10-8

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69

aDDition anD subtraCtion of DeCimal fraCtions

54. In Figure 10-9, what common fractional part of diameter C is diameter D? All dimensions are in feet. DIA C = 1

DIA D = 0.38

figure 10-9

55. What common fractional part of distance A is each distance listed in Figure 10-10? All dimensions are in inches. E = 0.375

D = 1.00

a. Distance B b. Distance C c. Distance D d. Distance E e. Distance F

C = 5.625 A = 10.00

B = 2.50

F = 3.75

figure 10-10

UNIT 11 Addition and Subtraction of Decimal Fractions Objectives After studying this unit you should be able to ●● ●● ●● ●●

Add decimal fractions. Add combinations of decimals, mixed decimals, and whole numbers. Subtract decimal fractions. Subtract combinations of decimals, mixed decimals, and whole numbers.

Adding and subtracting decimal fractions are required at various stages in the production of most products and parts. It is necessary to add and subtract decimals in order to estimate machining costs and production times, to compute stock allowances and tolerances, to determine locations and lengths of cuts, and to inspect finished parts.

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aDDing DeciMal Fractions c Procedure ●● ●● ●●

To add decimal fractions

Arrange the numbers so that the decimal points are directly under each other. Add each column as with whole numbers. Place the decimal point in the sum directly under the other decimal points.

Example Add 7.35 1 114.075 1 0.3422 1 0.003 1 218.7. Note: To reduce the possibility of error, add zeros to decimals so that all the values have the same number of places to the right of the decimal point. Zeros added in this manner do not affect the value of the number. 7.3500 114.0750 0.3422 0.0030 1 218.7000 340.4702 Ans

Arrange the numbers so that the decimal points are directly under each other. Add each column as with whole numbers. Place the decimal point in the sum directly under the other decimal points.

As shown in the next example, the decimal point location of a whole number is directly to the right of the last digit.

Example 1 Add 15.4 1 27 1 9.21. Arrange the numbers so that the decimal points are directly under each other. Write 27 as 27.00.

15.40 27.00 9.21 51.61

Ans

Example 2 Find the length x of the swivel bracket shown in Figure 11-1. All dimensions are in millimeters. Add.

x 5 90.13 mm

8.78 25.40 12.80 30.00 3.90 1 9.25 90.13 Ans

30 3.9

8.78 25.40

9.25

12.8 R

x

figure 11-1

subtracting DeciMal Fractions c Procedure ●● ●● ●●

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To subtract decimal fractions

Arrange the numbers so that the decimal points are directly under each other. Subtract each column as with whole numbers. Place the decimal point in the difference directly under the other decimal points.

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UNIT 11

71

aDDition anD subtraCtion of DeCimal fraCtions

Example 1 Subtract 13.261 from 25.6. Arrange the numbers so that the decimal Subtract. 25.600 points are directly under each other. 213.261 Add 2 zeros to 25.6 so that it has the 12.339 Ans same number of decimal places as 13.261. Subtract each column as with whole numbers. Place the decimal point in the difference directly under the other decimal points.

Example 2 Determine dimensions A, B, C, and D of the support bracket shown in Figure 11-2. All dimensions are given in inches. Solve for A: A 5 0.505 2 0.18 A 5 0.3250 Ans

0.505 2 0.180 0.325

Solve for B: B 5 1.4 2 0.301 B 5 1.0990 Ans

1.400 2 0.301 1.099

Solve for C: C 5 1.74 2 0.365 C 5 1.3750 Ans

1.740 2 0.365 1.375

Solve for D: D 5 0.746 2 0.46 D 5 0.2860 Ans

0.746 2 0.460 0.286

D 0.746 0.46

C 1.74

0.18

0.505

A

0.365 B

0.301 1.4

figure 11-2

ApplicAtion Tooling Up 1. Round 0.53745 to 3 decimal places. 9 2. Express the common fraction as a decimal fraction. 16 3. Express the decimal fraction 0.2472 as a common fraction in lowest terms. 3 5 4. Determine the quotient of 2 4 1 . 4 8 Use Figure 11-3 to answer Exercises 5 and 6. All dimensions are in inches.

7

1 16

7

1 8

1 16 CUT

7

1 8

1 16 CUT 8 12

7

1 8

1 16 CUT

7

1 8

1 16 CUT

figure 11-3

5. Five pieces are cut from the length of round stock shown in Figure 11-3. After the pieces are cut, the remaining length is thrown away. Determine the length of stock used for the five pieces including the cuts. 6. After the five pieces have been cut from the round stock in Figure 11-3, what is the length that is thrown away?

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Adding Decimal Fractions Add the numbers in Exercises 7 through 9. 9. a. 4.2 + 3.42 + 52.4 + 5.906 b. 7 + 0.007 + 0.077 + 0.777 c. 0.0942 + 12.59 + 62.1 + 0.704 d. 548.02 + 406.709 + 39.807 + 72.604 e. 0.074 + 12.026 + 83 + 42.9

7. a. 0.375 1 10.4 1 5 b. 0.003 1 0.13795 c. 0.375 1 0.8 1 0.12 d. 4.187 1 0.932 1 0.01 e. 363.13 1 18.2 1 0.027 8. a. 4 1 0.4 1 0.04 1 0.004 b. 87 1 0.0239 1 7.23 c. 0.0001 1 0.1 1 0.01 d. 4.705 1 0.0937 1 0.98 e. 0.063 1 4.9 1 324

10. Determine dimensions A, B, C, D, E, and F of the profile gauge shown in Figure 11-4. All dimensions are in inches.

0.247 0.35

A5

E

B5

0.452 D F

0.257

C5 D5

0.375 C

E5

0.24

F5

0.362 0.47 0.2 A

1.51 B

figure 11-4

11. A sine plate is to be set to a desired angle by using size (gauge) blocks of the following thicknesses: 3.000 inches, 0.500 inch, 0.250 inch, 0.125 inch, 0.100 inch, 0.1007 inch, and 0.1001 inch. Determine the total height that the sine plate is raised. 12. Three cuts are required to turn a steel shaft. The depths of the cuts, in millimeters, are 6.25, 3.18, and 0.137. How much stock has been removed per side? Round answer to 2 decimal places.

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UNIT 11

73

aDDition anD subtraCtion of DeCimal fraCtions

13. A thickness or feeler gauge is shown in Figure 11-5. Thickness gauges are widely used in manufacturing and machine service and repair occupations. Find the smallest combination of gauge leaves that total each of the following thicknesses (more than one combination may total certain thicknesses): a. 0.014 0 b. 0.033 0

GAUGE LEAF

c. 0.021 0

f. 0.042 0

3

0.015

00

0.0

06

0.0

0.

e. 0.011 0

10 0.0

d. 0.038 0

02

0.0

04

g. 0.029 0

0.008

0.012

h. 0.049 0

CUSTOMARY FEELER GAUGE

figure 11-5

Subtracting Decimal Fractions

Subtract the numbers in Exercises 14 through 16. Where necessary, round answers to 3 decimal places. 14. a. 0.527 2 0.4136 b. 0.319 2 0.0127 c. 2.308 2 0.7859 d. 0.3 2 0.299 e. 0.4327 2 0.412

16. a. 153.63 2 86.47 b. 7.173 2 6.98 c. 47.4 2 0.95 d. 0.106 2 0.058 e. 49.004 2 9.504

15. a. 23.062 2 0.973 b. 0.313 2 0.2323 c. 4.697 2 0.0002 d. 5.923 2 3.923 e. 103.48 2 91.474 17. The front and right side views of a sliding shoe are shown in Figure 11-6. Determine dimensions A, B, C, D, E, and F. All dimensions are in millimeters. 9.98 R

A5

D

B5 42.80

C5 16.22

A

D5

22.80

E5

C 12.68 28.25

E

B

F

F5

23.26

55.27

30.65

figure 11-6

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18. Refer to the plate shown in Figure 11-7 and determine the following distances. All dimensions are in inches. 0.150 DIA

0.725 DIA 0.385 DIA 0.562 DIA

0.265 DIA

EDGE A

EDGE B 0.625

0.402 0.952

0.827 2.232 3.695

figure 11-7

a. The horizontal center distance between the 0.265 0 diameter hole and the 0.150 0 diameter hole. b. The horizontal center distance between the 0.385 0 diameter hole and the 0.150 0 diameter hole. c. The distance between edge A and the center of the 0.725 0 diameter hole. d. The distance between edge B and the center of the 0.385 0 diameter hole. e. The distance between edge B and the center of the 0.562 0 diameter hole.

UNIT 12 Multiplication of Decimal Fractions Objectives After studying this unit you should be able to ●● ●●

Multiply decimal fractions. Multiply combinations of decimals, mixed decimals, and whole numbers.

A machinist must readily be able to multiply decimal fractions for computing machine feeds and speeds, for determining tapers, and for determining lengths and stock sizes. Multiplication of decimal fractions is also required in order to solve problems that involve geometry and trigonometry.

Multiplying DeciMal Fractions c Procedure ●● ●●

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To multiply decimal fractions

Multiply using the same procedure as with whole numbers. Beginning at the right of the product, point off the same number of decimal places as there are in the multiplicand and the multiplier combined.

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Common Fractions and Decimal Fractions

Multiplication of Decimal Fractions

1 UNIT 12

UNIT 12

75

Multiplication of DeciMal fractions

Example 1 Multiply 50.123 by 0.87. Multiply the same as with whole numbers. Beginning at the right of the product, point off as many decimal places as there are in both the multiplicand and the multiplier.

Multiplicand S Multiplier S

50.123 (3 places) 3 0.87 (2 places) 3 50861 40 0984 S 43.60701 (5 places) Ans

Product

Example 2 Compute the lengths of thread on each end of the shaft shown in Figure 12-1. All dimensions are in inches. PITCH = 0.09

27.3 THREADS

PITCH = 0.125

19.8 THREADS

LENGTH B

LENGTH A

figure 12-1

Compute Length A: A 5 2.4570 Ans Compute Length B:

B 5 2.47500 Ans

27.3 (1 place) 3 0.09 (2 places) 2.457 (3 places) 19.8 (1 place) 3 0.125 (3 places) 990 396 1 98 2.4750 (4 places)

When multiplying certain decimal fractions, the product has a smaller number of digits than the number of decimal places required. For these products, add as many zeros to the left of the product as are necessary to give the required number of decimal places.

Example Multiply 0.0237 by 0.04. Round the answer to 5 decimal places. The multiplicand, 0.0237, has 4 decimal places, and the multiplier, 0.04, has 2 decimal places. Therefore, the product must have 6 decimal places. Add 3 zeros to the left of the product. Round 0.000948 to 5 places.

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Multiply.

0.0237 (4 places) 3 0.04 (2 places) 0.000948 (6 places) 0.00095

Ans

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ApplicAtion Tooling Up 1. Round 0.42538 to 2 decimal places. 2. Express the decimal fraction 0.056 as a common fraction in lowest terms. 3. Add 0.032 1 0.23 1 0.0032.

C

12.524

28.310

B

11

1 64

Use Figure 12-2 to answer Exercises 4 through 6. All dimensions are in inches.

7.849

26.325

32.678

12.435

A

figure 12-2

4. Determine the length of A. 5. Determine the length of B. 6. Determine the length of C.

Multiplying Decimal Fractions Multiply the numbers in Exercises 7 through 9. Where necessary, round the answers to 4 decimal places. 7. a. 4.693 3 0.012 b. 2.2 3 1.5 c. 40 3 0.15 d. 6.43 3 0.26

8. a. 12.5 3 1.4 b. 24.4 3 6.5 c. 32 3 4.5 d. 0.95 3 6.4

9. a. 0.84 3 0.25 b. 12.36 3 0.08 c. 0.082 3 9.05 d. 0.0074 3 12.05

10. A section of a spur gear is shown in Figure 12-3. Given the circular pitches for various gear sizes, determine the working depths, clearances, and tooth thicknesses. Round the answers to 4 decimal places. CIRCULAR PITCH

TOOTH THICKNESS

WORKING DEPTH

Working depth 5 0.6366 3 Circular pitch Clearance 5 0.05 3 Circular pitch Tooth thickness 5 0.5 3 Circular pitch

CLEARANCE

figure 12-3

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77

Multiplication of DeciMal fractions

Circular Pitch (inches)

a.

0.3925

b.

0.1582

c.

0.8069

d.

1.2378

e.

1.5931

Working Depth (inches)

Clearance (inches)

Tooth Thickness (inches)

11. Determine diameters A, B, C, D, and E of the shaft in Figure 12-4. All dimensions are in millimeters. DIA B

DIA A

DIA D

A5

DIA E

DIA C

B5 C5

37.937 DIA 7.915

9.458 6.282

D5 E5

7.683

3.087

figure 12-4

12. Determine dimension x for each of these figures. c. Round the answer to 3 decimal places. All dimensions are in inches.

a. All dimensions are in inches.

TAPPED HOLE 0.125 PITCH 25 THREADS 0.382 TYPICAL 32 PLACES x

x

b. All dimensions are in millimeters. 8.06 TYPICAL 55 PLACES

d. Round the answer to 3 decimal places. All dimensions are in inches. 3.20 TYPICAL 54 PLACES 3.75

3.75 x

x

DISTANCE ACROSS CORNERS = 1.938 DISTANCE ACROSS FLATS = 0.866 × DISTANCE ACROSS CORNERS

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13. The length, L, of the point on any standard 1188 included angle drill, as shown in Figure 12-5, can be calculated using the formula L 5 0.3O, where O represents the diameter of the drill. Determine the lengths of the following drill points with the given diameters. Round to 3 decimal places for inches and 1 decimal place for millimeters. 10 a. d. 10 mm 2 10 b. e. 25 mm 4 30 c. f. 45 mm 8

118°

14. The length, L, of the point on any standard 828 included angle drill can be calculated using the formula L 5 0.575O, where O represents the diameter of the drill. Determine the lengths of the following drill points with the given diameters. Round to 3 decimal places for inches and 1 decimal place for millimeters. 10 2 10 b. 4 30 c. 8 a.

L

figure 12-5

d. 10 mm e. 25 mm f. 45 mm

UNIT 13 Division of Decimal Fractions Objectives After studying this unit you should be able to ●● ●● ●●

Divide decimal fractions. Divide decimal fractions with whole numbers. Divide decimal fractions with mixed decimals.

Division with decimal fractions is used for computing the manufacturing cost and time per piece after total production costs and times have been determined. Division with decimal fractions is also required in order to compute thread pitches, gear tooth thicknesses and depths, cutting speeds, and depths of cut.

DiviDing Decimal Fractions Moving a decimal point to the right is equivalent to multiplying the decimal by a power of 10. 0.237 3 10 5 2.37 0.237 3 1000 5 237. 0.237 3 100 5 23.7

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0.237 3 10,000 5 2370.

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79

Division of DeciMal fractions

When dividing decimal fractions, the value of the answer (quotient) is not changed if the decimal points of both the divisor and the dividend are moved the same number of places to the right. It is the same as multiplying both divisor and dividend by the same number. 0.9375 4 0.612 5 s0.9375 3 1000d 4 s0.612 3 1000d 5 937.5 4 612. 14.203 4 6.87 5 s14.203 3 100d 4 s6.87 3 100d 5 1420.3 4 687.

c Procedure ●●

●●

●● ●● ●●

To divide decimal fractions

Move the decimal point of the divisor as many places to the right as are necessary to make the divisor a whole number. Move the decimal point of the dividend the same number of places to the right as were moved in the divisor. Place the decimal point in the quotient directly above the decimal point in the dividend. Add zeros to the dividend if necessary. Divide as with whole numbers.

Example 1 Divide 0.643 by 0.28. Round the answer to 3 decimal places. To make the divisor a whole number move the decimal point 2 places to the right, 28. The decimal point in the dividend is also moved 2 places to the right, 64.3. Add 3 zeros to the dividend. One extra place is necessary in order to round the answer to 3 decimal places. Place the decimal point of the quotient directly above the decimal point of the dividend. Divide as with whole numbers.

2.2964 < 2.296 Ans 28q 64.3000 56 83 56 270 252 180 168 120 112 8

Example 2 3.19 4 0.072. Round the answer to 2 decimal places. Move the decimal point 3 places to the right in the divisor, and 3 places to the right in the dividend. Add 3 zeros to the dividend. Place the decimal point of the quotient directly above the decimal point of the dividend. Divide.

44.305 < 44.31 Ans 72q 3190.000 288 310 288 220 216 400 360 40

When dividing a decimal fraction or a mixed decimal by a whole number, it is not necessary to move the decimal point of either the divisor or the dividend. Add zeros to the right of the dividend, if necessary, to obtain the desired number of decimal places in the answer.

Example 1 Divide 0.63 by 12 and round to 4 decimal places.

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0.0525 12q 0.6300 Ans

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6.794

Example 2 Divide 33.97 by 5 and round the quotient to 3 decimal places. 5q 33.970 Ans Example 3 Compute the pitch and the number of threads given the lengths of thread on each end of the shaft shown in Figure 13-1. All dimensions are in inches. 31.4 THREADS

PITCH A

B THREADS

2.669

PITCH = 0.105

3.759

figure 13-1

Compute Pitch A

Move the decimal point 1 place to the right in both the divisor and in the dividend. The pitch of the threads on the left end of the shaft is 0.085 inch. Ans

0.085 314q 26.690 2512 1570 1570 0

Compute B Threads

Move the decimal point 3 places to the right in both the divisor and in the dividend. The right end of the shaft has 35.8 threads. Ans

35.8 105q 3759.0 315 609 5250 840 840 0

ApplicAtion Tooling Up 1. Add 5

3 5 14 . 16 6

0.582 TYPICAL 5 PLACES

44 2. Express the common fraction 125 as a decimal fraction. C

3. Subtract 1.702 − 0.854. Use Figure 13-2 to answer Exercises 4 through 6. All dimensions are in millimeters (mm). 4. Determine the length of A. 5. Determine the length of B. 6. Determine the length of C.

1.587 TYPICAL 5 PLACES B

A

0.926

10.248

figure 13-2

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81

Division of DeciMal fractions

Dividing Decimal Fractions Divide the numbers in Exercises 7 through 9. Express the answers to the indicated number of decimal places. 7. a. 0.69 4 0.432 (3 places)

9. a. 1.023 4 0.09 (3 places) 16.3 b. (2 places) 3.8 37 c. (2 places) 0.273 0.005 d. (4 places) 0.81

b. 0.92 4 0.36 (2 places) c. 0.001 4 0.1 (4 places) d. 10 4 0.001 (3 places) 8. a. 12.45 ÷ 0.05 (2 places) b. 24.0016 ÷ 0.32 (3 places) 42 c. (4 places) 0.065 2.006 d. (4 places) 0.075

10. As indicated in Figure 13-3, rack sizes are given according to diametral pitch. Given four different diametral pitches, find the linear pitch and the whole depth of each rack to 4 decimal places. All dimensions are in inches. 3.1416 Diametral Pitch

Linear Pitch 5 Diametral Pitch

a.

6.75

b.

2.75

c.

7.25

d.

16.125

Linear Pitch

Whole Depth 5 Whole Depth

2.157 Diametral Pitch

LINEAR PITCH

WHOLE DEPTH

RACK

PITCH LINE

figure 13-3

11. Four sets of equally spaced holes are shown in the machined plate in Figure 13-4. Determine dimensions A, B, C, and D to 2 decimal places. All dimensions are in millimeters. DIM C TYPICAL 2 PLACES 24.92

A5 B5 C5

41.40

D5

DIM B TYPICAL 7 PLACES

43.78

DIM A TYPICAL 6 PLACES

DIM D TYPICAL 4 PLACES

70.52

figure 13-4

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12. A cross-sectional view of a bevel gear is shown in Figure 13-5. Given the diametral pitch and the number of gear teeth, determine the pitch diameter, the addendum, and the dedendum. Round the answers to 4 decimal places. Pitch Diameter 5

Number of Teeth Diametral Pitch

Addendum 5

1 Diametral Pitch

Dedendum 5

1.1570 Diametral Pitch

AXIS OF GEAR

DEDENDUM ADDENDUM

PITCH LINE

PITCH DIAMETER

figure 13-5

Diametral Pitch

Number of Teeth

a.

4

45

b.

6

75

c.

8

44

d.

3

54

Pitch Diameter (inches)

Addendum (inches)

Dedendum (inches)

13. How many complete bushings each 14.60 millimeters long can be cut from a bar of bronze that is 473.75 millimeters long? Allow 3.12 millimeters waste for each piece. 14. A shaft is being cut in a lathe. The tool feeds (advances) 0.015 inch each time the shaft turns once (1 revolution). How many revolutions will the shaft turn when the tool advances 3.120 inches? Round the answer to 2 decimal places. 15. How much stock per stroke is removed by the wheel of a surface grinder if a depth of 4.725 millimeters is reached after 75 strokes? Round the answer to 3 decimal places. 16. An automatic screw machine is capable of producing one piece in 0.02 minute. How many pieces can be produced in 1.25 hours? 17. The bolt in Figure 13-6 has 7.7 threads. Determine the pitch to 3 decimal places. All dimensions are in inches. PITCH

0.962

figure 13-6

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83

Division of DeciMal fractions

18. The block in Figure 13-7 has a threaded hole with a 0.0625-inch pitch. Determine the number of threads for the given depth to 1 decimal place. All dimensions are in inches.

0.718

figure 13-7

19. The length of a side of a square equals the distance from point A to point B divided by 1.4142. Determine the length of a side of the square plate in Figure 13-8 to 2 decimal places. All dimensions are in millimeters.

B 54.44

A x

figure 13-8

20. All sections of the block in Figure 13-9 are equal in length. Determine the length A to the nearest thousandth millimeter.

A

A

A

154.15 mm figure 13-9

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UNIT 14 Powers Objectives After studying this unit you should be able to ●● ●●

Raise numbers to indicated powers. Solve problems that involve combinations of powers with other basic operations.

Powers of numbers are used to compute areas of square plates and circular sections and to compute volumes of cubes, cylinders, and cones. Use of powers is particularly helpful in determining distances in problems that require applications of geometry and trigonometry.

Description oF powers Two or more numbers multiplied to produce a given number are factors of the given number. Two factors of 8 are 2 and 4. The factors of 15 are 3 and 5. A power is the product of two or more equal factors. The third power of 5 is 5 3 5 3 5, or 125. An exponent shows how many times a number is taken as a factor. It is written smaller than the number and above and to the right of the number. The expression 32 means 3 3 3. The exponent 2 shows that 3 is taken as a factor twice. It is read as 3 to the second power or 3 squared.

Examples Find the indicated powers. 1. 25 Two to the fifth power means 2 3 2 3 2 3 2 3 2, or 32. 2. 3

Ans

Three to the third power, or 3 cubed, means 3 3 3 3 3, or 27.

3

3. 0.72

Ans

0.72 to the second power, or 0.72 squared, means 0.72 3 0.72, or 0.5184. Ans

2

A 5 s2 is called a formula. A formula is a short method of expressing an arithmetic relationship by the use of symbols. Known values may be substituted for the symbols and other values can be found.

Example Determine the area of the square shown in Figure 14-1. The area of a square equals the length of a side squared. The answer is given in square units. All dimensions are in inches. A 5 s2 A5

1 2 7 in. 8

2

7 7 in. 3 in. 8 8 49 A5 sq in. Ans 64

A5

SIDE =

SIDE =

7 8

7 8

figure 14-1

The symbols in2, in.2, sq in, and sq in. are all used for square inches. Some people write the symbol for inch as “in.” so that the period will keep it from being confused with the word “in.” In this text we use in. as the symbol for inch.

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UNIT 14

85

powers

Note: In linear measurement we often use the symbol ’ for feet and ” for inches. For example, 39 5 3 ft 5 3 feet. These symbols are not to be used for area or volume. To indicate an area of 3 square feet, you may write 3 sq ft or 3 ft2, but never write 392. Example Find the volume of the cube shown in Figure 14-2. The volume of a cube equals the length of a side cubed. The answer is given in cubic units. All dimensions are in millimeters. Round answer to 1 decimal place. V 5 s3 V 5 s1.6 mmd3 V 5 1.6 mm 3 1.6 mm 3 1.6 mm V 5 4.096 mm3 or 4.1 mm3 Ans

SIDE = 1.6

The symbol mm3 is used for cubic millimeters.

SIDE = 1.6

SIDE = 1.6

figure 14-2

Use oF parentheses In this example, only the numerator is squared. 22 2 3 2 4 1 5 5 51 3 3 3 3

Ans

In this example, only the denominator is squared. 2 2 2 5 5 Ans 2 3 333 9 Parentheses are used as grouping symbols. Parentheses indicate that both the numerator and the denominator of a fraction are raised to the given power.

12 2 3

c Procedure

2

5

22 2 3 2 4 5 5 32 3 3 3 9

Ans

To solve problems that involve operations within parentheses

Perform the operations within the parentheses. Raise to the indicated power.

●● ●●

Examples 1. s1.2 3 0.6d2 5 0.722 5 0.72 3 0.72 5 0.5184 Ans 2. s0.5 1 2.4d2 5 2.92 5 2.9 3 2.9 5 8.41 Ans 3. s0.75 2 0.32d2 5 0.432 5 0.43 3 0.43 5 0.1849 Ans 5 4.5 5 4.5 3 4.5 5 20.25 114.4 3.2 2 2

4.

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2

Ans

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SecTIoN 1

When solving power problems that also require addition, subtraction, multiplication, or division, perform the power operation first.

Examples 1. 5 3 32 2 12 5 5 3 9 2 12 5 45 2 12 5 33 Ans 2. 33.5 2 5.52 1 8.7 5 33.5 2 30.25 1 8.7 5 11.95 Ans 3.

2.23 2 5.608 10.648 2 5.608 5.040 5 5 5 3.6 Ans 1.4 1.4 1.4

The symbol  (pi) represents a constant value used in mathematical relationships involving circles. Depending upon the specific problem to be solved, generally, the value of 1 pi used is 3 , 3.14, or 3.1416. Most calculators have a  key. 7

Example Compute the volume of the cylinder shown in Figure 14-3 to 2 decimal places. The answer is given in cubic units. All dimensions are in inches. V 5  3 r2 3 h V 5 3.14 3 s0.85 in.d2 3 1.25 in. V 5 3.14 3 0.7225 sq in. 3 1.25 in. V 5 2.8358 cu in.,

2.84 cu in.

Ans

HEIGHT (h) = 1.25

RADIUS (r ) = 0.85

figure 14-3

The symbols in3, in.3, cu in, and cu in. are all used for cubic inches. Many problems require the application of the same formula more than once or the application of two different formulas in the solutions.

Example Find the metal area of this square plate (Figure 14-4). Round the answer to 2 decimal places. All dimensions are in inches. A 5 s2. The metal area equals the area of the large square, A1, minus the area of the removed square, A2. A1 5 s5.250 in.d2 A1 5 27.5625 sq in. A2 5 s2.500 in.d2

5.250 2.500

A2 5 6.2500 sq in. A3 5 27.5625 sq in. 2 6.2500 sq in. < 21.31 sq in. Ans (rounded)

2.500 5.250

figure 14-4

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UNIT 14

87

powers

ApplicAtion Tooling Up 5 7 22 . 16 8 3 2. Multiply 7 3 2 . Express the result as a mixed number and as a decimal fraction. 8 1. Subtract 7

3. Multiply 1.702 3 2.35. Use Figure 14-5 to answer Exercises 4 through 6. All dimensions are in millimeters (mm).

DIM B TYPICAL 3 PLACES

30.084

DIM A TYPICAL 6 PLACES C

81.138

8.908

102.570

figure 14-5

4. Determine the length of A. 5. Determine the length of B. 6. Determine the length of C.

Raising a Number to a Power Raise the following numbers to the indicated power. 7. 3.43 8. 18 9. 1004

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12

2 3 23 11. 3 3 12. 3 4 10.

3

13. s0.3 3 7d2 14. s20.7 1 7.2d2 15.

1 2 28.8 7.2

3

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Related Problems In the following table, the lengths of the sides of squares are given. Determine the areas of the squares. Round the answers to 2 decimal places where necessary. Side

Side

Area

16.

1.25 in.

17.

23.070 mm

18.

0.17 in.

19.

10.70 mm2

23.

20.

0.02 in.

24.

21. 22.

25.

where

A 5 s2

Area

3 in. 4 7 in. 8 3 3 in. 4 13 in. 16 3 13 in. 4

A 5 area s 5 side

s

s

In the following table, the lengths of the sides of cubes are given. Determine the volumes of the cubes. Round answers to 2 decimal places where necessary. Side

26.

0.29 in.

27.

20.60 mm

28.

3.930 in.

29.

14.00 mm

30.

0.075 in.

Volume

Side

31. 32. 33. 34. 35.

Volume

V 5 s3

1 in. 3 7 in. 8 1 1 in. 2

where V 5 volume s 5 side

s

1 9 in. 8

s

s

3 in. 4

In the following table, the radii of circles are given. Determine the areas of the circles. Round the answers to the nearest whole number. Radius

36.

16.20 mm

37.

15.60 mm

38.

0.07 in.

39.

19.28 in.

40.

12.35 in.

A = π 3 R2

Area

where

A 5 area  5 3.14 R 5 radius

R

In the following table, the diameters of spheres are given. Determine the volumes of the spheres. Round the answers to 1 decimal place where necessary. Diameter

41.

0.65 in.

42.

6.500 mm

43.

0.75 in.

44.

10.80 mm

45.

7.060 mm

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Area

V5

 3 D3 6

where V 5 volume  5 3.14 D 5 diameter

D

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UNIT 14

89

powers

In the following table, the radii and heights of cylinders are given. Determine the volumes of the cylinders. Round the answers to the nearest whole number. Radius

Height

46.

5.00 mm

3.20 mm

47.

1.50 in.

2.30 in.

48.

2.25 in.

3.00 in.

49.

0.7 in.

6.70 in.

50.

7.81 mm

6.72 mm

Volume

V 5  3 r2 3 h where

V 5 volume  5 3.14 r 5 radius h 5 height

h

r

In the following table, the diameters and heights of cones are given. Find the volumes of the cones. Round the answers to the nearest whole number. Diameter

Height

Volume

51.

3.20 in.

4.00 in.

52.

3.00 in.

5.00 in.

53.

10.60 mm

13.10 mm

54.

9.90 mm

6.20 mm

55.

0.37 in.

0.96 in.

V 5 0.2618 3 d 2 3 h where V 5 volume d 5 diameter h 5 height

h

d

Solve the following problems. Use  5 3.14. Round answers to the nearest whole number. 56. Find the metal area of this washer. All dimensions are in millimeters. A 5  3 R2

9.38 R

21.87 R

57. Find the metal area of this spacer. All dimensions are in millimeters. Area of square 5 s2 4.20 R

Area of circle 5  3 R2 s = 19.60

s = 19.60

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58. Find the area of this plate. All dimensions are in millimeters.

Hint: The broken lines indicate one method of solution. A 5 s2

21.20

8.70 12.60

3.90 21.20

12.60

59. Find the metal volume of this bushing. All dimensions are in inches. V 5  3 R2 3 H

0.60 R

0.85 R H = 1.22

60. Find the volume of this pin. All dimensions are in inches. 0.65 R

Volume of cylinder 5  3 R2 3 H Volume of cone 5 0.2618 3 D2 3 H

1.30 D

0.50

6.87

61. A materials estimator finds the weight of aluminum needed for the casting shown. Aluminum weighs 0.0975 pound per cubic inch. Find, to the nearer pound, the weight of aluminum required for 15 castings. All measurements are in inches.

Hint: The broken line indicates a method of solution. V 5 s3

8.2

8.2

22.7 14.5

14.5

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14.5

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UNIT 15

91

roots

UNIT 15 Roots Objectives After studying this unit you should be able to ●● ●● ●●

Extract whole number roots. Determine the root of any positive number using a calculator. Solve problems that involve combinations of roots with other basic arithmetic operations.

The operation of extracting roots of numbers is used to determine lengths of sides and heights of squares and cubes and radii of circular sections when areas and volumes are known. The machinist uses roots in computing distances between various parts of machined pieces from given dimensions.

Description oF roots A root of a number is a quantity that is taken two or more times as an equal factor of the 3 number. The expression Ï 64 is called a radical. A radical is an indicated root of a number. The symbol Ï is called a radical sign and indicates a root of a number. The digit 3 is called the index. An index indicates the number of times that a root is to be taken as an equal factor to produce the given number. The index is written smaller than the number, above, and in the “V” of the radical sign. The given number 64 is called a radicand. A radicand is the number under the radical sign whose root is to be found. 3 index Ï 64 5 4 root radical sign radicand

The index 2 is omitted for an indicated square root. For example, the square root of 9 is written Ï9. The expression Ï9 means to find the number that can be multiplied by itself and equal 9. Since 3 3 3 5 9, 3 is a square root of 9.

Examples Find the indicated roots. 1. Ï36

Since 6 3 6 5 36, a square root of 36 is 6. Ans

2. Ï144

Since 12 3 12 5 144, a square root of 144 is 12. Ans

3 3. Ï 8

Since 2 3 2 3 2 5 8, a cube root of 8 is 2. Ans

3

4. Ï 125

Since 5 3 5 3 5 5 125, a cube root of 125 is 5. Ans

4 5. Ï 81

Since 3 3 3 3 3 3 3 5 81, a fourth root of 81 is 3. Ans

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Roots must be extracted in determining unknown dimensions represented in certain formulas.

Example 1 Compute the length of the side of the square shown in Figure 15-1. This square has an area of 25 square inches. Since A 5 s2, the length of a side of the square equals the square root of the area. s 5 ÏA s 5 Ï25 sq in.

s

s 5 Ï5 in. 3 5 in. s 5 5 inches

Ans s

figure 15-1

Example 2 Compute the length of the side of the cube shown in Figure 15-2. The volume of this cube equals 64 cubic inches. 3 s5Ï V 3 s5Ï 64 cu in.

s

3 s5Ï 4 in. 3 4 in. 3 4 in.

s 5 4 inches

Ans

s

s

figure 15-2

roots oF Fractions

In this example, only the root of the numerator is taken. 4 Ï16 Ï4 3 4 5 5 Ans 25 25 25 In this example, only the root of the denominator is taken. 16 16 16 1 Ans 5 5 53 5 5 Ï25 Ï5 3 5 A radical sign that encloses a fraction indicates that the roots of both the numerator and denominator are to be taken. The same answer is obtained by extracting both roots first and dividing second as by dividing first and extracting the root second.

Example Find

Î

36 . 9

Method 1:

Extract both roots, then divide.

Method 2:

36 Ï36 6 5 5 5 2 Ans 9 3 Ï9 Divide, then extract the root.

Î Î

36 5 Ï4 5 2 Ans 9

expressions encloseD within the raDical symbol The radical symbol is a grouping symbol. An expression consisting of operations within the radical symbol is done using the order of operations.

c Procedure To solve problems that involve operations within the radical symbol ●● ●●

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Perform the operations within the radical symbol first using the order of operations. Then find the root.

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93

roots

Examples Find the indicated roots. 1. Ï3 3 12 5 Ï36 5 Ï6 3 6 5 6 Ans 2. Ï5 1 59 5 Ï64 5 Ï8 3 8 5 8 Ans 3. Ï128 2 7 5 Ï121 5 Ï11 3 11 5 11 Ans Problems involving formulas may involve operations within a radical symbol.

Example Compute the length of the chord, C, of the circular segment shown in Figure 15-3. All dimensions are in inches. C 5 2 3 ÏH 3 s2 3 R 2 Hd where C 5 length of chord H 5 height of segment R 5 radius of circle

H = 1.5

C 5 2 3 ÏH 3 s2 3 R 2 Hd C 5 2 3 Ï1.5 3 s2 3 3.75 2 1.5d C 5 2 3 Ï1.5 3 6 C 5 2 3 Ï9 C5233 C56 Length of chord 5 6 inches Ans

C R = 3.75

figure 15-3

roots that are not whole nUmbers The root examples and exercises have all consisted of numbers that have whole number roots. These roots are relatively easy to determine by observation. Most numbers do not have whole number roots. For example, Ï259 5 16.0935 3 (rounded to 4 decimal places) and Ï 17.86 5 2.6139 (rounded to 4 decimal places). The root of any positive number can easily be computed with a calculator or spreadsheet. Calculator solutions to root expressions are given at the end of Unit 17. Spreadsheet solutions to root expressions are given at the end of Unit 18.

Î

V can be used to determine length of the radius,  3H R, of a cylinder if the volume, V, and the height, H, of the cylinder are known. What is the radius of a cylinder with a volume of 471.050 mm3 and a height of 8.5 mm?

Example The formula R 5

Î Î Î

R5

V  3H

5

471.050  3 8.5

≈

471.050 26.704

≈ Ï17.640 ≈ 4.20 The radius is about 4.20 mm.

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Fractional exponents Fractional exponents can be used to indicate roots. n ●● a1yn 5 Ï a n n m myn ●● a 5 sÏ adm 5 Ï a

Examples Find the indicated roots. 1. 491y2 5 Ï49 5 7 Ans 3 2. 1251y3 5 Ï 125 5 5 Ans 3 2 3 3. 82y3 5 Ï 8 5Ï 64 5 4 Ans

ApplicAtion Tooling Up 1. Raise (4 1 2.3)3 to the indicated power. 3 1 2. Compute 7 4 4 . Express the result as a common fraction or mixed number, whichever is appropriate. 5 2 3. Compute 4.891 4 1.34. 4. Write 0.275 as a common fraction.

1

2

3 9 1 5. Evaluate 7 2 12 . 4 16 2 1 6. The feed on a lathe is set for inch. How many revolutions does the work make when 64 1 the tool advances 5 inches? 8

Radicals That Are Whole Numbers The following problems have either whole number roots or numerators and denominators that have whole number roots. Determine these roots. 25 3 7. Ï 216 10. 13. Ï56.7 1 87.3 Ï36 8. 9.

Î

4 9

11.

Ï4 9

Î

3 3 3 4 4

14. Ï16.4 2 7.4

12. Ï0.5 3 18

15.

Î 3

428.8 6.7

The following problems have whole number square roots. Solve for the missing values in the tables. 16. The areas of squares are given in the following table. Determine the lengths of the sides. s 5 ÏA Area (A)

a.

225 mm2

b.

121 mm2

c.

64 mm

d.

81 sq in.

e.

49 sq in.

Side (s) s

2

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s

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roots

UNIT 15

17. The volumes of cubes are given in the following table. Determine the lengths of the sides. 3 s5Ï V

Volume (V )

a.

216 mm3

b.

64 cu in.

c.

512 cu in.

d.

1000 mm3

e.

1 cu in.

Side (s)

s s

s

18. The areas of circles are given in this table. Determine the lengths of the radii. Use  5 3.14. A R5 

Î

Area (A)

a.

50.24 sq in.

b.

12.56 sq in.

c.

314 mm

d.

28.26 sq in.

e.

153.86 mm3

Radius (R) R

2

19. The volumes of spheres are given in this table. Determine the lengths of the diameters. D5 Volume (V )

a.

14.1372 cu in.

b.

113.0976 mm3

c.

4.1888 cu in.

d.

0.5236 cu in.

e.

523.6 mm3

Î 3

V 0.5236

Diameter (D)

D

Radicals That Are Not Whole Numbers The following problems have square roots that are not whole numbers. They require calculator computations. Refer to Unit 17 for calculator root solutions. Compute these roots to the indicated number of decimal places. 20. Ï15.63

(3 places)

24. Ï0.07 3 28

(2 places)

21. Ï391

(2 places)

25. Ï15.82 1 3.71

(2 places)

(3 places)

26. Ï178.5 2 163.7 (3 places)

(3 places)

27.

22. 23.

Î Î

3 5

3

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1 2

Î

0.441 60

(4 places)

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Determine the following numbers with fractional exponents. 28. 251y2

32. 253y2

29. 361y2

33. 324y5

30. 1211y2

34. 283y4

(3 decimal places)

31. 196

35. 75

(3 decimal places)

1y2

5y3

The following problems have roots that are not whole numbers. Solve for the missing values in the tables. 36. The volumes of cylinders and their heights are given in the following table. Find the lengths of the radii to 2 decimal places. Use  5 3.14. V R5 3H Volume Height Radius (V )

(H )

a.

249.896 mm3

7.00 mm

b.

132.634 mm3

12.00 mm

c.

14.00 cu in.

29.00 in.

d.

10.00 cu in.

28.00 in.

Î

(R )

H

R

37. The volumes of cones and their heights are given in the following table. Compute the lengths of the diameters to 2 decimal places. V D5 0.262 3 H Volume Height Diameter (V )

(H )

a.

116.328 mm3

8.00 mm

b.

19.388 cu in.

2.00 in.

c.

1257.6 mm3

10.00 mm

d.

15 cu in.

50.00 in.

Î

(D)

H

D

Solve the following problems. 38. The pitch of broach teeth depends upon the length of cut, the depth of cut, and the material being broached. Minimum Pitch 5 3 3 ÏL 3 d 3 F

where

L 5 length of cut d 5 depth of cut F 5 a factor related to the type of material being broached

Find the minimum pitch, to 3 decimal places, for broaching cast iron where L 5 0.825 0, d 5 0.007 0, and F 5 5. 39. The dimensions of keys and keyways are determined in relation to the diameter of the shafts with which they are used. D5

Î

L3T 0.3

where

D 5 shaft diameter L 5 key length T 5 key thickness

What is the shaft diameter that would be used with a key where L 5 2.7099 and T 5 0.2599?

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table of DeciMal equivalents anD coMbineD operations of DeciMal fractions

UNIT 16 Table of Decimal Equivalents and Combined Operations of Decimal Fractions

Objectives After studying this unit you should be able to ●● ●● ●●

Write decimal or fraction equivalents using a decimal equivalent table. Determine nearer fraction equivalents of decimals by using the decimal equivalent table. Solve problems consisting of combinations of operations by applying the order of operations.

Generally, fractional engineering drawing dimensions are given in multiples of 64ths of an inch. A machinist is often required to express these fractional dimensions as decimal equivalents for machine settings. When laying out parts such as castings that have ample stock allowances, it is sometimes convenient to use a fractional steel scale and to express decimal dimensions to the nearer equivalent fractions. The amount of computation and the chances of error can be reduced by using the decimal equivalent table.

table oF Decimal eqUivalents Using a decimal equivalent table saves time and reduces the chance of error. Decimal equivalent tables are widely used in the manufacturing industry. They are posted as large wall charts in work areas and are carried as pocket-size cards. Skilled workers memorize many of the equivalents after using decimal equivalent tables. The decimals listed in the table are given to 6 places. For actual on-the-job uses, a decimal is rounded to the degree of precision required for a particular application. DeCimAL equiVALeNT TAbLe 1/64 ---- 0.015625

17/64 --- 0.256625

33/64 --- 0.515625

49/64 --- 0.765625

1/32 ----------- 0.03125

9/32 ---------- 0.28125

17/32 ---------- 0.53125

25/32 ---------- 0.78125

3/64 ---- 0.046875

19/64 --- 0.296875

35/64 --- 0.546875

51/64 --- 0.796875

1/16 ------------- 0.0625

5/16 ------------ 0.3125

9/16 ------------ 0.5625

13/16 ------------ 0.8125

5/64 ---- 0.078125

21/64 --- 0.328125

37/64 --- 0.578125

53/64 --- 0.828125

3/32 ----------- 0.09375

11/32 ---------- 0.34375

19/32 ---------- 0.59375

27/32 -------- 0.84375

7/64 ---- 0.109375

23/64 --- 0.359375

39/64 --- 0.609375

55/64 --- 0.859375

1/8 ---------------------- 0.125

3/8 ------------------------ 0.375

5/8 ------------------------ 0.625

7/8 ------------------------ 0.875

9/64 ---- 0.140625

25/64 --- 0.390625

41/64 --- 0.640625

57/64 --- 0.890625

5/32 ----------- 0.15625

13/32 ---------- 0.40625

21/32 ---------- 0.65625

29/32 ---------- 0.90625

11/64 ---- 0.171875

27/64 --- 0.421875

43/64 --- 0.671875

59/64 --- 0.921875

3/16 ------------- 0.1875

7/16 ------------ 0.4375

11/16 ------------ 0.6875

15/16 ------------ 0.9375

13/64 ---- 0.203125

29/64 --- 0.453125

45/64 --- 0.703125

61/64 --- 0.953125

7/32 ----------- 0.21875

15/32 ---------- 0.46875

23/32 ---------- 0.71875

31/32 ---------- 0.96875

15/64 ---- 0.234375

31/64 --- 0.484375

47/64 --- 0.734375

63/64 --- 0.984375

1/4 ------------------------ 0.25

1/2 ---------------------------- 0.5

3/4 -------------------------- 0.75

1 ------------------------------- 1.

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The following examples illustrate the use of the decimal equivalent table.

Example 1 Find the decimal equivalent of The decimal equivalent is shown directly

23 0 . 32 23 0 5 0.718750 32

to the right of the common fraction.

Ans

Example 2 Find the fractional equivalent of 0.31250. The fractional equivalent is shown directly to the left of the decimal fraction.

0.31250 5

50 16

Ans

Example 3 Find the nearer fractional equivalents of the decimal dimensions given on the casting shown in Figure 16-1. All dimensions are in inches. Compute dimension A. The decimal 0.757 lies between 0.750 and 0.765625. The difference between 0.757 and 0.750 is 0.007. The difference between 0.757 and 0.765625 is 0.008625. Since 0.007 is less than 0.008625, the 0.750 value is closer to 0.757. 30 The nearer fractional equivalent of 0.7500 is . Ans 4 Compute dimension B. The decimal 0.978 lies between 0.96875 and 0.984375. The difference between 0.978 and 0.96875 is 0.00925. The difference between 0.978 and 0.984375 is 0.006375. Since 0.006375 is less than 0.00925, the 0.984375 value is closer to 0.978. The nearer fractional equivalent of 0.9843750 is

63 0 . 64

A = 0.757

B = 0.978

figure 16-1

Ans

combineD operations oF Decimal Fractions In the process of completing a job, a machinist must determine stock sizes, cutter sizes, feeds and speeds, and roughing allowances as well as cutting dimensions. Usually most and sometimes all of the fundamental operations of mathematics must be used for computations in the manufacture of a part. Determination of powers and roots must also be considered in the order of operations. The following procedure incorporates all six fundamental operations. Study the following order of operations.

order of operations for combined operations of addition, subtraction, multiplication, Division, powers, and roots c Procedure ●●

●

●

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Do all the work in parentheses first. Parentheses are used to group numbers. In a problem expressed in fractional form, two or more numbers in the dividend (numerator) and/or divisor (denominator) should be considered as being enclosed in parentheses. 4.87 1 0.34 should be considered as s4.87 1 0.34d 4 s9.75 2 8.12d. 9.75 2 8.12 If an expression contains parentheses within parentheses or brackets, such as For example,

[5.6 3 s7 2 0.09d 1 8.8], do the work within the innermost parentheses first.

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UNIT 16

●●

●●

●●

99

table of DeciMal equivalents anD coMbineD operations of DeciMal fractions

Do powers and roots next. The operations are performed in the order in which they occur from left to right. If a root consists of two or more operations within the radical sign, perform all operations within the radical sign, then extract the root. Do multiplication and division next. The operations are performed in the order in which they occur from left to right. Do addition and subtraction last. The operations are performed in the order in which they occur from left to right.

Some people use the memory aid “Please Excuse My Dear Aunt Sally” to help them remember the order of operations. The P in “Please” stands for parentheses, the E for exponents (or raising to a power) and roots, M and D for multiplication and division, and the A and S for addition and subtraction.

Example 1 Find the value of 7.875 1 3.2 3 4.3 2 2.73. Add.

7.875 1 3.2 3 4.3 2 2.73

55

Multiply.

7.875 1 13.76

Subtract.

2 2.73

21.635

2 2.73 5 18.905 Ans

Example 2 Find the value of (27.34 2 4.82) 4 (2.41 3 1.78 + 7.89). Round the answer to 2 decimal places. Perform operations within parentheses. s27.34 2 4.82d 4 s2.41 3 1.78 1 7.89d

5

Subtract.

22.52

4 s2.41 3 1.78 1 7.89d

Add.

22.52

4 s4.2898 1 7.89d

Divide.

22.52

4

Example 3 Find the value of

55

Multiply.

12.1798 5 1.84896 5 1.85 Ans

13.79 1 s27.6 3 0.3d2 Ï23.04 1 0.875 2 3.76

.

Round the answer to 3 decimal places. Grouping symbol operations is done first. Consider the numerator and the denominator as if each were within parentheses. All of the operations are performed in the numerator and in the denominator before the division is performed. 13.79 1 s27.6 3 0.3d2

4 _Ï23.04 1 0.875 2 3.76+

Extract the square root.

[13.79 1 68.5584]

4 _Ï23.04 1 0.875 2 3.76+

Add.

[13.79 1 68.5584]

4

s4.8 1 0.875 2 3.76d

Add.

82.3484

4

s4.8 1 0.875

Subtract.

82.3484

4

Divide.

82.3484

4

Rewrite.

4 _Ï23.04 1 0.875 2 3.76+ 4 _Ï23.04 1 0.875 2 3.76+

5

5

Multiply.

5

Square.

f13.79 1 s27.6 3 0.3d2g f13.79 1 s27.6 3 0.3d2g f13.79 1 s8.28d2g

5

Ï23.04 1 0.875 2 3.76

2 3.76d

5

s5.675 2 3.76d 1.915

5 43.00178 5 43.002

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Example 4 Blanks in the shape of regular pentagons (5-sided figures) are punched from strip stock as shown in Figure 16-2. Determine the width of strip stock required, using the given dimensions and the formula for dimension R. Round the answer to 3 decimal places. All dimensions are in inches. Width 5 R 1 0.980 1 2 3 0.125 where R 5 Ïr2 1 s2 4 4 0.125

R WIDTH r = 0.980

s = 1.424

0.125

figure 16-2

Ï0.9802 1 1.4242 4 4 1 0.980 1 2 3 0.125 Ï0.9802 1 1.4242 4 4 1 0.980 1 2 3 0.125 Ï0.9604 1 1.4242 4 4 1 0.980 1 2 3 0.125 Ï0.9604 1 2.027776 4 4 1 0.980 1 2 3 0.125

55 5 h

h

Substitute the given values. Square. Square. Divide.

Ï0.9604 1 0.506944 1 0.980 1 2 3 0.125

Extract the square root. Multiply. Add.

Ï1.467344 1 0.980 1 2 3 0.125 1.211 1 0.980 1 2 3 0.125 1.211 1 0.980 1 0.250 5 2.441 Width 5 2.441 inches

5

Add.

Ans

Note: In solving expressions that consist of numerous multiplication and power operations, it is often necessary to carry out the work to 2 or 3 more decimal places than the number of decimal places required in the answer.

ApplicAtion Tooling Up

Î

V 1. A sphere has a volume, V, of 12.249 cu in. Use the formula D 5 3 to determine the diameter, D, of 0.5236 the sphere. 3 1 2. Compute 7 4 4 . Express the result as a common fraction or mixed number, whichever is appropriate. 5 2 3. Compute 4.891 4 1.34. 4. Write 0.275 as a common fraction.

1

2

3 9 1 5. Evaluate 7 2 12 . 4 16 2 6. The feed on a lathe is set for

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1 3 inch. How many revolutions does the work make when the tool advances 7 inches? 32 8

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table of DeciMal equivalents anD coMbineD operations of DeciMal fractions

Using the Decimal equivalent Table Find the fraction or decimal equivalents of these numbers using the decimal equivalent table. 25 32 7 8. 32 11 9. 32

13 16 5 11. 64

7.

10.

13. 0.3125 14. 0.28125

12. 0.671875

15. 0.203125

Find the nearer fraction equivalents of these decimals using the decimal equivalent table. 16. 0.541

18. 0.465

20. 0.209

17. 0.762

19. 0.498

21. 0.805

combined operations of Decimal Fractions Solve these examples of combined operations. Round the answers to 2 decimal places where necessary. 22. 0.5231 1 10.375 4 4.32 3 0.521

28. sÏ3.98 1 0.87 3 3.9d2

23. 81.07 4 12.1 1 2 3 3.7 56.050 24. 3 0.875 2 3.92 3.8

29. s3.29 3 1.7d2 4 s3.82 2 0.86d 30. 0.25 3

25. s24.78 2 19.32d 3 4.6 26. s14.6 4 4 2 1.76d2 3 4.5

3 3.87 1 18.3 1Ï64 8.32 3 5.13 2

2

31. 18.32 2

27. 27.16 4 Ï1.76 1 12.32

Î

7.86 3 13.5 3 0.7 3.52 2 0.52

Solve the following problems by using combined operations. 32. Figure 16-3 shows the three-wire method of checking screw threads. With proper diameter wires and a micrometer, very accurate pitch diameter measurements can be made. Using the formula given, determine the micrometer dimension over wires of the American (National) Standard threads in the following table. Round the answer to 4 decimal places. M 5 D 2 s1.5155 3 Pd 1 s3 3 W d major Wire Dimension Diameter Pitch Diameter Over Wires D (inches) P (inches) W (inches) M (inches)

a.

0.8750

0.1250

0.0900

b.

0.2500

0.0500

0.0350

c.

0.6250

0.1000

0.0700

d.

1.3750

0.16667

0.1500

e.

2.5000

0.2500

0.1500

W

M

D

PITCH DIAMETER

MICROMETER

P

figure 16-3

33. A bronze bushing with a diameter of 22.225 millimeters is to be pressed into a mounting plate. The assembly print calls for a bored hole in the plate to be 0.038 millimeter less in diameter than the bushing diameter. The hole diameter in the plate checks 22.103 millimeters. By how much must the diameter of the plate hole be increased in order to meet the print specification?

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34. A stamped sheet steel plate is shown in Figure 16-4. Compute dimensions A–F to 3 decimal places. All dimensions are in inches. 0.390 0.492

2.875 0.282 DIA 4 HOLES EQUALLY SPACED

0.470 0.625 R

0.897 0.450

0.400 0.312 D

3 SLOTS EQUALLY SPACED

E

C

A

0.392

B 1.008

0.224 DIA

F

0.390 DIA 3 HOLES

0.470 TYPICAL 3 PLACES

1.280

3.090

0.368 0.180 R TYPICAL 6 PLACES

figure 16-4

A5

C5

E5

B5

D5

F5

35. A flat is to be milled in three pieces of round stock each of a different diameter. The length of the flat is determined by the diameter of the stock and the depth of cut. The table gives the required length of flat and the stock diameter for each piece. Determine the depth of cut for each piece to 2 decimal places using this formula. C5

Diameter D

Length of Flat F

a.

34.80 mm

30.50 mm

b.

55.90 mm

40.60 mm

c.

91.40 mm

43.40 mm

D 2 0.5 3 2

Depth of Cut C

Î

43

C

1 2 2F D 2

2

2

F

D

36. A slot is machined in a circular plate with a 41.36-millimeter diameter. Two milling cuts, one 6.30 millimeters deep and the other 3.15 millimeters, are made. A grinding operation then removes 0.40 millimeter. What is the distance from the center of the plate to the bottom of the slot? All dimensions are in millimeters.

x

41.36 DIA

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UNIT 17

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coMputing with a calculator: DeciMals

37. A 608 slot has been machined in a fixture. The slot is checked by placing a pin in the slot and indicating the distance between the top of the fixture and the top of the pin as shown. Compute distance H to 3 decimal places by using this formula. All dimensions are in inches. H 5 1.5 3 D 2 0.866 3 W W = 1.210

D = 0.750

H

60°

UNIT 17 Computing with a Calculator: Decimals

Objectives After studying this unit you should be able to ●●

●●

Perform individual operations of addition, subtraction, multiplication, division, powers, and roots with decimals using a calculator. Perform combinations of operations with decimals using a calculator.

Decimals The decimal point key ( . ) is used when entering decimal values in a calculator. When entering a decimal fraction in a calculator, the decimal point key is pressed at the position of the decimal point in the number. For example, to enter the number 0.732, first press . and then enter 732. To enter the number 567.409, enter 567 . 409. In calculator examples and illustrations of operations with decimals in this text, the decimal key . will not be shown to indicate the entering of a decimal point. Wherever the decimal point occurs in a number, it is understood that the decimal point key . is pressed.

Decimals with basic operations of addition, subtraction, multiplication, and Division Example 1 Add 19.37 1 123.9 1 7.04. 19.37 1 123.9 1 7.04 5 150.31

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Example 2 Subtract 2876.78 2 405.052. 2876.78 2 405.052 5 2471.728

Ans

Example 3 Multiply 427.935 3 0.875 3 93.400 (round answer to 1 decimal place). 427.935 3 .875 3 93.4 5 34972.988 34,973.0 Ans Notice that the 2 zeros following the 4 are not entered. The final zero or zeros to the right of the decimal point may be omitted. Notice that the zero to the left of the decimal point is not entered. The leading zero is omitted.

Example 4 Divide 813.7621 4 6.466 (round answer to 3 decimal places). 813.7621 4 6.466 5 125.85247 125.852 Ans

powers Expressions involving powers and roots are readily computed with a scientific calculator. The square key is used to raise a number to the second power (to square a number).

powers on a scientific calculator To square a number, enter the number, press the square key (

Example To calculate 28.752, enter 28.75, press 28.75

x2

5 826.5625

x2

x2

), and press 5 or

ENTER

.

, and press 5 .

Ans

The universal power key, x y or ^ , depending on the calculator used, raises a number to a power. To raise a number to a power using the universal power key, do the following: Enter the number to be raised to a power. Press the universal power key,

xy

or

^.

Enter the power. Press the 5 or ENTER key.

Examples 1. Calculate 15.723. Enter 15.72, press x y or 15.72 x y (or ^ ) 3 5 3884.7012 Ans 2. Calculate 0.957. .95 x y (or ^ ) 7 5 0.6983373

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^ , enter 3, and press 5 .

Ans

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UNIT 17

powers on a Machinist calc Pro 2 Expressions involving powers and roots are readily computed with a Machinist calc Pro 2 calculator. The square key is used to raise a number to the second power (to square a number) indicated by the x2 over the % in the lower left-hand corner of the calculator. The Machinist calc Pro 2 does not have a universal power key, so it cannot be used for powers of 3, 5, 6, and so on. Since 4 5 22, the Machinist calc Pro 2 can be used to raise numbers to the fourth power. Notice that the x2 is printed in an orange color. Any Machinist calc Pro 2 function that is printed in orange is accessed by pressing Conv and then the key under the orange printing. To square a number, enter the number, press the square key (use the Conv % combination), and press 5 .

Example To calculate 28.752, enter 28.75 Conv Solution 28.75

Conv

%

%

.

826.5625 Ans

While it is not be necessary to press the 5 when you square a number, it is often a good idea to indicate that you have “finished the problem.”

roots To obtain the square root of any positive number, the square root key (

) is used.

roots on a scientific calculator c Procedure ●●

To obtain the square root of a positive number

Press the square root key (

), enter the number, and press 5 or

ENTER .

Example Calculate Ï27.038. Press , enter 27.038, and press 5 . 27.038 5 5.199807689 Ans ●●

If the square root key is an alternate function, press

2nd

first.

Example Calculate Ï27.038. 2nd

●●

27.038 5 5.199807689

Ans

The root of any positive number can be computed with a calculator. Generally, roots are alternate functions. On calculators with an x y key, press SHIFT x y to take the root. 5 Example Calculate Ï 475.19.

5

98310_sec_01_Unit12-19_ptg01.indd 105

SHIFT

xy

475.19 5 3.430626662

Ans

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coMMon fractions anD DeciMal fractions

SecTIoN 1

c Procedure ●●

Press

2nd

To use the alternate function

(or

SHIFT

) directly before pressing the root key (

x

).

5 Example Calculate Ï 475.19.

5

2nd

(or

SHIFT

)

475.19 5 3.430626662

x

Ans

Some calculators automatically insert a left parenthesis when the Ï key is pressed. If your calculator does this, then you need to place a right parenthesis to show you are finished taking the root.

Example Evaluate Ï67.24 2 5. Incorrect:

67.24 2 5 5 7.889233169

What you see on the calculator: Ï (67.24 2 5 7.889233169 Correct:

67.24

)

2 5 5 3.2, 3.2

Ans

What you see on the calculator: Ï (67.24) 2 5 3.2

roots on a Machinist calc Pro 2 The square root key is used to obtain the square root of any positive number. On the Machinist calc Pro 2 the square root is a second function and is over the key. So, for the square root, you have to press the Conv key combination. Because the Machinist calc Pro 2 does not have a universal power key, you cannot take roots other than square roots on the Machinist calc Pro 2.

c Procedure ●●

To obtain the square root of a positive number

Enter the number and then press

Conv

.

Example 1 Calculate Ï27.038. Solution On/C On/C

27.038

Conv

5.1998077 Ans

Example 2 Evaluate Ï67.24 2 5. Solution On/C On/C

67.24

Conv

2 5 5

3.2 Ans

Notice that it was necessary to press the 5 at the end of the problem to tell the calculator that you have finished with the subtraction.

Example 3 Evaluate 3 1 Ï6.25 2 15. Solution On/C On/C

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3 1 6.25

Conv

2 15 5

29.5 Ans

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UNIT 17

107

coMputing with a calculator: DeciMals

practice exercises, individual basic operations Evaluate the following expressions. The expressions are basic arithmetic operations including powers and roots. Remember to check your answers by doing each problem twice. The solutions to the problems directly follow the practice exercises. Compare your answers to the given solutions. Round each answer to the indicated number of decimal places. 1. 276.84 1 312.094 (2 places)

7. 54.419 4 6.7 (1 place)

2. 16.09 1 0.311 1 5.516 (1 place)

8. 0.9316 4 0.0877 (4 places)

3. 6704.568 2 4989.07 (2 places)

9. 36.222 (2 places)

4. 0.9244 2 0.0822 (3 places)

10. 7.0635 (1 place)

5. 43.4967 3 6.0913 (4 places)

11. Ï28.73721 (4 places)

6. 8.503 3 0.779 3 13.248 (3 places)

5 12. Ï 1068.470 (3 places)

solutions to individual basic operations 1. 276.84 1 312.094 5 588.934, 588.93

Ans

2. 16.09 1 .311 1 5.516 5 21.917, 21.9

Ans

3. 6704.568 2 4989.07 5 1715.498, 1,715.50 4. .9244 2 .0822 5 0.8422, 0.842

Ans

Ans

5. 43.4967 3 6.0913 5 264.9514487, 264.9514

Ans

6. 8.503 3 .779 3 13.248 5 87.752593, 87.753 7. 54.419 4 6.7 5 8.1222388, 8.1

Ans

8. .9316 4 .0877 5 10.622577, 10.6226 9. 36.22

x2

10. 7.063

xy

5 1311.8884, 1311.89 (or

12. 5

2nd

(or

Ans

Ans

^ ) 5 5 17577.052, 17,577.1 Ans

28.73721 5 5.3607098, 5.3607

11.

Ans

SHIFT

)

x

Ans

1068.47 5 4.03415394, 4.034

Ans

combineD operations on a scientiFic calcUlator The expressions are solved by entering numbers and operations into the calculator in the same order as the expressions are written.

Example 1 Evaluate 30.75 1 15 4 4.02 (round answer to 2 decimal places). 30.75 1 15 4 4.02 5 34.481343,

34.48

Ans

4 1 33.151 3 2.707 (round answer to 2 decimal places). 0.091 51.073 2 4 4 .091 1 33.151 3 2.707 5 96.856713, 96.86 Ans

Example 2 Evaluate 51.073 2

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coMMon fractions anD DeciMal fractions

Example 3 Evaluate 46.23 1 (5 1 6.92) 3 (56.07 2 38.5). As previously discussed in the order of operations, operations enclosed within parentheses are performed first. Calculators having algebraic logic perform the operations within parentheses before performing other operations in combined operations expressions. If an expression contains parentheses, enter the expression into the calculator in the order in which it is written. The parentheses keys ( and ) must be used. 46.23 1 (5 1 6.92 ) 3 ( 56.07 2 38.51 ) 5 255.6644 Ans 13.463 1 9.864 3 6.921 (round answer to 3 decimal places). 4.373 1 2.446 Recall that for problems expressed in fractional form, the fraction bar is also used as a grouping symbol. The numerator and denominator are each considered as being enclosed in parentheses. s13.463 1 9.864 3 6.921d 4 s4.373 1 2.446d ( 13.463 1 9.864 3 6.921 ) 4 ( 4.373 1 2.446 ) 5 11.985884, 11.986 Ans

Example 4 Evaluate

The expression may also be evaluated by using the 5 key to simplify the numerator without having to enclose the entire numerator in parentheses. However, parentheses must be used to enclose the denominator. 13.463 1 9.864 3 6.921 5 4 ( 4.373 1 2.446 ) 5 11.985884, 11.986 Ans 100.32 2 s16.87 1 13d (round answer to 2 decimal places). 111.36 2 78.47 100.32 2 s16.87 1 13d 5 s100.32 2 s16.87 1 13dd 4 s111.36 2 78.47d 111.36 2 78.47

Example 5 Evaluate

Observe these parentheses. To be sure that the complete numerator is evaluated before dividing by the denominator, enclose the complete numerator within parentheses. This is an example of an expression containing parentheses within parentheses. ( 100.32 2 ( 16.87 1 13 ) ) 4 ( 111.36 2 78.47 ) 5 2.1419884, 2.14 Ans Using the 5 key to simplify the numerator: 100.32 2 ( 16.87 1 13 ) 5 4 ( 111.36 2 78.47 ) 5 2.1419884, 2.14 Ans

Example 6 Evaluate [4.73 1 s0.24 2 5.16d2] 1 Ï12.45 (round answer to 3 decimal places). Use the parenthesis keys when there are brackets or braces. ( 4.73 1 ( .24 2 5.16 ) x 2 ) 1 12.45 5 32.46485575, 32.465

Example 7 Evaluate

Ans

873.03 1 12.123 3 41 (round answer to 2 decimal places). Ï16.43 2 266.76 4 107.88

873.03 1 12.12 x y 3 3 41 ) 4 ( (or 2nd ) 16.43 2 266.76 4 107.88 ) 5 46732.658, 46,732.66 Ans Using the 5 key to simplify the numerator: 873.03 1 12.12 x y 3 3 41 5 4 ( (or 2nd ) 16.43 2 266.76 4 107.88 46732.658, 46,732.66 Ans (

98310_sec_01_Unit12-19_ptg01.indd 108

)

5

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coMputing with a calculator: DeciMals

UNIT 17

combineD operations on a Machinist calc Pro 2 Because the Machinist calc Pro 2 obeys the order of operations, the expressions are solved by entering numbers and operations into the calculator in the same order as the expressions are written.

Example 1 Evaluate 30.75 + 15 ÷ 4.02 (round answer to 2 decimal places). Solution 30.75 1 15 4 4.02 5 Example 2 Evaluate 51.073 2

34.48

Ans

4 1 33.151 3 2.707 (round answer to 2 decimal places). 0.091

Solution You cannot use the / key when either the numerator or denominator is not a whole number. 4 So, enter as 4 4 .091. 0.091 51.073 2 4 4 .091 1 33.151 3 2.707 5 96.86 (rounded)

Ans

Example 3 Evaluate 46.23 1 s5 1 6.92d 3 s56.07 2 38.5d. Solution As previously discussed in the order of operations, any operations enclosed within parentheses are performed first. Calculators having algebraic logic perform the operations within parentheses before performing other operations in combined operations expressions. But the Machinist calc Pro 2 does not have parentheses, so perform the operations within parentheses and save them in some of the memory records of the calculator. KeYSTROKe On/C

On/C

DiSPLAY 0.

1. Compute and store (5 1 6.92) in Memory 1: 5

1

6.92 Conv

11.92

1

Rcl

2. Compute and store (56.07 2 38.5) in Memory 2: 56.07

2

38.5 Conv

Rcl

2

17.57

3. Evaluate 46.23 1 (5 1 6.92) 3 (56.07 2 38.5): 46.23

1

98310_sec_01_Unit12-19_ptg01.indd 109

Rcl

1

3

Rcl

2

5

255.6644 Ans

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110

SecTIoN 1

Example 4 Evaluate

coMMon fractions anD DeciMal fractions

13.463 1 9.864 3 6.921 (round answer to 3 decimal places). 4.373 1 2.446

Solution Recall that for problems expressed in fractional form, the fraction bar is also used as a grouping symbol. The numerator and denominator are each considered as being enclosed in parentheses. You should think of this as (13.463 1 9.864 3 6.921) 4 (4.373 1 2.446). KeYSTROKe On/C

DiSPLAY

On/C

0.

1. Compute and store the numerator, 13.463 1 9.864 3 6.921, in Memory 1: 13.463

1

9.864

3

6.921 Conv

Rcl

1

81.731744

2. Compute and store the denominator, 4.373 1 2.446 in Memory 2:

1

4.373

3. Evaluate Rcl

2.446 Conv

Rcl

6.819

2

13.463 1 9.864 3 6.921 : 4.373 1 2.446

4

1

Rcl

5

2

11.985884 11.986 Ans

Example 5 Evaluate

100.32 2 s16.87 1 13d (round answer to 2 decimal places). 111.36 2 78.47

Solution Again we add parentheses around both the numerator and denominator. 100.32 2 s16.87 1 13d 5 s100.32 2 s16.87 1 13dd 4 s111.36 2 78.47d 111.36 2 78.47 Observe these parentheses. To be sure that the complete numerator is evaluated before dividing by the denominator, enclose the complete numerator within parentheses. This is an example of an expression containing parentheses within parentheses. You will see in this example that we store 2 different values in Memory 1. The second value replaces the first value with the new one. KeYSTROKe On/C

On/C

DiSPLAY 0.

1. Compute and store the operation in the innermost parentheses, 16.87 1 13, in Memory 1: 16.87

1

13 Conv

Rcl

29.87

1

2. Evaluate the numerator, 100.32 2 M1 and store it in Memory 1: 100.32

2

Rcl

1

Conv

Rcl

1

70.45

3. Compute and store the denominator, 111.36 2 78.47 in Memory 2: 111.36

2

4. Evaluate Rcl

1

78.47 Conv

Rcl

2

32.89

100.32 2 s16.87 1 13d : 111.36 2 78.47

4

Rcl

2

5

2.1419884 2.14 Ans

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UNIT 17

111

coMputing with a calculator: DeciMals

Example 6 Evaluate [4.73 1 s0.24 2 5.16d2] 1 Ï12.45 (round answer to 3 decimal places).

Solution Brackets are just another grouping symbol like parentheses and are often used to make it easier to see when a particular grouping starts and stops. KeYSTROKe On/C

DiSPLAY

On/C

0.

1. Compute and store the operation in the parentheses, 0.24 − 5.16, and store in Memory 1: 0.24

2

5.16 Conv

Rcl

1

24.92

2. Square the stored value (0.24 2 5.16)2 and store it in Memory 1: Rcl

1 Conv

% Conv

Rcl

1

24.2064

3. Compute and store the value inside the brackets, [4.73 + (0.24 2 5.16)2] and store it in Memory 1: 4.73

1

Rcl

1

Conv

Rcl

28.9364

1

4. Compute the square root, Ï12.45, and add it to the value in Memory 1: 12.45 Conv

1

Rcl

1

32.464856

5

32.465 Ans

practice exercises, combined operations Evaluate the following combined operations expressions. Remember to check your answers by doing each problem twice. The solutions to the problems directly follow the practice exercises. Compare your answers to the given solutions. Round each answer to the indicated number of decimal places. 1. 503.97 2 487.09 3 0.777 1 65.14 (2 places) 5 2. 27.028 1 2 5.875 3 1.088 (3 places) 6.331 3. 23.073 3 (0.046 1 5.934 2 3.049) 2 17.071 (3 places) 4. 30.180 3 (0.531 1 12.939 2 2.056) 2 60.709 (3 places) 5.

643.72 2 18.192 3 0.783 470.07 2 88.33

6.

793.32 2 2.67 3 0.55 107.9 1 88.93

(2 places)

(1 place)

7. 2,446 1 8.9173 3 5.095 (3 places) 8. 679.07 1 (36 1 19.973 2 0.887)2 3 2.05 (1 place) 9. 43.71 2 Ï256.33 2 107 1 17.59 (2 places) 10.

5 3 282.608 Ï14.773 1 93.977 3 Ï 3.033

11.

3 1,202.03Ï 706.8 2 44.317 2 2.63 (1 place) s14.03 3 0.54 2 2.08d2

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(3 places)

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SecTIoN 1

coMMon fractions anD DeciMal fractions

solutions to practice exercises, combined operations 1. 503.97 2 487.09 3 .777 1 65.14 5 190.64107, 190.64

Ans

2. 27.028 1 5 4 6.331 2 5.875 3 1.088 5 21.425765, 21.426 3. 23.073 3

(

.046 1 5.934 2 3.049

4. 30.180 3

(

.531 1 12.939 2 2.056

5.

(

)

643.72 2 18.192 3 .783

4 (

(

793.32 2 2.67 3 .55

)

4

7. 2446 1 8.917 (

8. 679.07 1 10.

(

5

(or

2nd

(or

(

1202.03 3 3

2.08

)

x

2

)

2nd

) )

5 1.6489644, 1.65

Ans

5 1.6489644, 1.65

Ans

) )

5 4.0230224, 4.0

Ans

5 4.0230224, 4.0

Ans

)

)

3 2.05 5 6899.7282, 6,899.7

x2

1 17.59 5 49.079935, 49.08

14.773 1 93.977 3 3

x

3.033 5 203.89927, 203.899 11.

470.07 2 88.33

Ans

^ ) 3 3 5.095 5 6058.4387, 6,058.438 Ans

256.33 2 107 SHIFT

470.07 2 88.33

107.9 1 88.93

36 1 19.973 2 .887 (

9. 43.71 2

x

y

Ans

2 60.709 5 283.76552, 283.766

107.9 1 88.93

( (

or 793.32 2 2.67 3 .55 5 4

2 17.071 5 50.555963, 50.556 )

(

or 643.72 2 18.192 3 .783 5 4 6.

)

Ans

(or

2nd

(or

Ans

Ans

SHIFT

)

)

4

282.608

x

)

4

Ans

SHIFT

)

x

(

706.8 2 44.317

2 2.63 5 344.25205, 344.3

)

(

14.03 3 .54 2

Ans

UNIT 18 Computing with a Spreadsheet: Decimals

Objectives After studying this unit you should be able to ●●

●●

Perform individual operations of addition, subtraction, multiplication, and division with decimals using a spreadsheet. Perform combinations of operations with decimals using a spreadsheet.

Decimals Just as with a calculator, the decimal point, . , is used to enter a decimal fraction in a spreadsheet. The key for the decimal point is the same key that is used as a period to end a sentence. When entering a decimal fraction, the decimal point is pressed at the position of the decimal point in the number. For example, to enter the number 0.4723, because the leading 0 is to the left of the decimal point, it does not have to be typed, so first press . , then then type the digits “4723.” To enter the number 12.095, enter 12 . 095. The four basic operations of addition, subtraction, multiplication, and division are performed the same with decimals as with whole numbers.

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UNIT 18

113

coMputing with a spreaDsheet: DeciMals

In the spreadsheet examples and illustrations of operations with decimals in this book, the decimal key . will not be shown to indicate that a decimal point was typed. It should be understood that the decimal key is pressed wherever the decimal point occurs in a number.

examples of Using Decimals with the basic operations of addition, subtraction, multiplication, and Division Example 1 Use a spreadsheet to compute 49.37 1 562.8 1 5.06. Solution Enter 5 49.37 1 562.8 1 5.06 in Cell A1, as shown in Figure 18-1, and press RETURN

. You should get the result in Figure 18-2, which shows the decimal value of the result, 617.23.

figure 18-1

figure 18-2

figure 18-3

What if you got the fractional result shown in Figure 18-3 rather than the decimal value? Then you need to click on the cell or column you want in decimal format, press “Home,” and then press the number that is 4th from the right in the row above the Formula Bar. In Figure 18-4, the number is a 12, but your tablet may have a different number. This opens the “Number Formatting” menu shown in Figure 18-4. Press the i to the right of “Number,” and you should get the “Number Options” menu shown in Figure 18-5. Here you can select the number of decimal places you want for the answer, in this example we want 2 decimal places. You can also activate the “Separator,” which places a comma between every third digit of large numbers.

98310_sec_01_Unit12-19_ptg01.indd 113

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114

SecTIoN 1

figure 18-4

coMMon fractions anD DeciMal fractions

figure 18-5

Example 2 Subtract 2,954.21 2 675.034. Solution Enter 5 2,954.21 2 675.034 in Cell A2, as in Figure 18-6, and press

RETURN

. The result is shown in Figure 18-7. Notice that you can use a comma when keying in the numbers.

figure 18-7

figure 18-6

But we want the answer to have 3 decimal places. We need to follow the procedure we just used in Figures 18-3 through 18-5. Click on the 1 sign until you see a 3. You should now have the answer shown in Figure 18-8.

figure 18-8

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UNIT 18

115

coMputing with a spreaDsheet: DeciMals

Example 3 Multiply 528.32 3 42.5 3 1.2. Solution Enter = 528.32 * 42.5 * 1.2 in Cell A3 and press

RETURN

. The result of

26,944.32 is shown in Figure 18-9.

figure 18-9

Example 4 Divide 813.7621 ÷ 6.466 and round the answer to 3 decimal places. Solution Enter 5 813.7621/6.466 in Cell A4 and press

RETURN

.

The result is 125.852 as shown in Figure 18-10.

powers Expressions involving powers and roots are easily computed using a spreadsheet. There are two methods that can be used. One method uses the caret key ^ , found over the 6 on most keyboards, to raise a number to a power. The second method uses a function named “POWER” that is built into the spreadsheet. We will look at each in turn.

Example Use a spreadsheet and the ^ key to compute 53. Solution Enter 5 5^3 in Cell A1 (Figure 18-11) and press RETURN .

figure 18-10

The result, shown in Figure 18-12, is 125.

figure 18-11

98310_sec_01_Unit12-19_ptg01.indd 115

figure 18-12

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116

SecTIoN 1

coMMon fractions anD DeciMal fractions

With the built-in power function, you type POWER(a,n) to get the value of an.

Example 1 Use a spreadsheet and the power function to compute 53. Solution Begin by typing “5po” in Cell A3. A menu of Functions appears with three functions all beginning with the letters PO. Click on “POWER,” and you should get the display in Figure 18-13. Enter the number, 5, and the power, 3, to get Figure 18-14. Now press RETURN . The result, shown in Figure 18-15, is 125.

figure 18-13

figure 18-14

figure 18-15

Example 2 Use a spreadsheet and the ^ key to compute 12.524. Solution Enter 5 12.52 ^ 4 in Cell A4 and press

. The result, shown in Figure 18-16, is 24,570.68790016. Notice that this is printed with eight decimal places. On Excel you can get an answer with up to 30 decimal places. RETURN

figure 18-16

Example 3 Use a spreadsheet and the power function to compute 0.957. Solution Enter 5 POWER(.95, 7) in Cell A5 and press

RETURN . The result, shown in Figure 18-17, is 0.6983373. We could have expanded this to more decimal places, but that that is not necessary for this example.

figure 18-17

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UNIT 18

117

coMputing with a spreaDsheet: DeciMals

sqUare roots To use a spreadsheet to obtain the square root of a number, you use the built-in command SQRT. For roots other than square roots we use fractional exponents.

Example Use a spreadsheet and the SQRT command to compute Ï27.038 to nine decimal places. Solution Enter 5 SQRT(27.038) in Cell A7 as shown in Figure 18-18, and press

RETURN

.

The result, 5.19980769, is shown in Figure 18-19.

figure 18-18

figure 18-19

roots other than square roots To use a spreadsheet to obtain the cube root, fourth root, or any other root of a number, you use fractional exponents and the ^ or the POWER function. 4 Example Use a spreadsheet to compute Ï 27.038 to 4 decimal places. Solution As shown in Figure 18-20, enter 5 27.038 ^ (1/4) in Cell A8 and press

RETURN . The result, 2.28 , is shown in Figure 18-21. If you used the POWER function, the entry would look as shown in the Formula Bar of Figure 18-21.

figure 18-20

98310_sec_01_Unit12-19_ptg01.indd 117

figure 18-21

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118

coMMon fractions anD DeciMal fractions

SecTIoN 1

practice exercises: individual operations with Decimal Fractions Use a spreadsheet to evaluate the following expressions. The expressions are basic single arithmetic operations with decimal fractions. Remember to first estimate the answer and then check the way it was entered in the Formula Bar. The keystrokes and answers are given directly following the exercises. Compare your answers to the given solutions. Round each answer to the indicated number of decimal places. 1. 295.63 + 793.642 (2 decimal places) 2. 27.09 + 83.734 + 0.4165 (3 places) 3. 6,980.724 − 4,899.04 (2 places) 4. 0.7612 − 0.033 (3 places) 5. 43.1274 × 9.5445 (4 places) 6. 6.305 × 0.096 (3 places) 7. 48.962 ÷ 8.5 (1 place) 8. 0.8623 ÷ 0.0452 (4 places) 9. 52.272 (2 places) 10. 4.325 (1 place) 11. Ï17,424 (1 place) 12. Ï625.32 (3 places) 13. 424.361/4 (3 places) 14. 8,673.2951/5 (3 places)

solutions to practice exercises: individual operations with Decimal Fractions 1. = 295.63 + 793.642

2. = 27.09 + 83.734 + 0.4165 3. = 6980.724 − 4899.04 4. = 0.7612 − 0.033

7. = 48.962/8.5 9. = 52.27 ^ 2 10. = 4.32 ^ 5

5.8

RETURN

Ans Ans

Ans

Ans 19.0774

Ans

or = POWER(52.27, 2)

RETURN

or = POWER(4.32, 5)

11. = SQRT(17424)

RETURN

132.0

12. = SQRT(625.32)

RETURN

25.006

13. = 424.36 ^ (1/4)

RETURN

4.539

RETURN

Ans

Ans

411.6295 0.605

RETURN

14. = 8673.295 ^ (1/5)

98310_sec_01_Unit12-19_ptg01.indd 118

0.728

RETURN

RETURN

8. = 0.8623/0.0452

2081.68

RETURN

RETURN

Ans

111.241

RETURN

RETURN

5. = 43.1274 * 9.5445 6. = 6.305 * 0.096

889.27

RETURN

RETURN

RETURN

2732.15 1504.6

Ans

Ans

Ans Ans Ans

6.132

Ans

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UNIT 18

119

coMputing with a spreaDsheet: DeciMals

combineD operations with Decimal Fractions When there are combined operations, the numbers are entered into the spreadsheet in the same order as the expressions are written.

Example 1 Evaluate 45.795 2 15.28 4 5.4. Round the answer to three decimal places. Solution Enter 5 45.795 2 15.28/5.4 in Cell A1 and press

. After you set the cell to three decimal places, you get the result 42.965 as shown in Figure 18-22. RETURN

figure 18-22

Example 2 Use a spreadsheet to evaluate 53.87 2 answer to four decimal places.

8.91 1156.2 39.4. Round the 2.5

Solution Enter 5 53.87 − 8.91/2.5 1 156.2 * 9.4 in Cell A2 and press

RETURN

. The

result, shown in Figure 18-23, is 1,518.5860.

figure 18-23

25.46 1 7.182 by using a spreadsheet. Round the 8.22 2 5.31 answer to three decimal places. Notice the use of parentheses in working the problem.

Example 3 Evaluate Ï0.52 1 1.32 1

Solution In Cell A3 we enter 5 SQRT(0.5 ^ 2 1 1.3 ^ 2) 1 (25.46 1 7.182)/(8.22 − 5.31) and press RETURN . The result, 12.610, is shown in Figure 18-24. Don’t forget to enclose the numerator and denominator of the fraction in parentheses.

figure 18-24

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120

SecTIoN 1

coMMon fractions anD DeciMal fractions

251.43 1 21.213 3 25 by using a spreadsheet and round the 532.85 2 Ï255.34 4 4.8 answer to three decimal places.

Example 4 Evaluate

Solution In Cell A4 we enter 5 (251.43 1 21.21 ^ 3 * 25)/(532.85 2 SQRT(255.34)/4.8) and press

RETURN

. The result, 450.958, is shown in Figure 18-25.

figure 18-25

practice exercises: combined operations with Decimal Fractions Evaluate the following expressions involving combined operations with decimal fractions. Remember to first estimate the answer and then check the way the expression was entered in the Formula Bar. The keystrokes and answers are given directly following the exercises. Compare your answers to the given solutions. Round each answer to the indicated number of decimal places. 1.

1 45.822 4 24.5 s2 decimal placesd 175.32 1.6

2. 83.045 1

32 2 4.375 3 2.004 s4 placesd 6.2

3. 125.397 3 s0.0048 1 9.72 2 6.962d 2 80.052 s3 placesd 4. 42.950 3 0.54 1 3.025 2 263.488 s2 placesd 5.

973 2 4.57 3 0.95 s4 placesd 542.237 1 Ï9.42 3 6.743

6. 9.546 1 7.154 3 0.462 s3 placesd 7. 845.937 1 s54 2 23.159 1 478.652d2 3 4.005 s4 placesd 8.

Ï24.75 1 82.546 3 Ï465.852 1 98.5 s2 placesd 8.22

solutions to practice exercises: combined operations with Decimal Fractions 1. 5 (75.32/1.6 1 45.82)/24.5

RETURN

2. 5 83.045 1 (32/6.2) 2 4.375 * 2.004

3.79 RETURN

3. 5 125.397 * (0.0048 1 9.72 2 6.962) − 80.052 4. 5 42.95 * 0.54 1 3.02 ^ 5 2 263.488

98310_sec_01_Unit12-19_ptg01.indd 120

RETURN

Ans 79.4388 RETURN

10.91

Ans 263.887

Ans

Ans

10/31/18 6:16 PM

UNIT 19

121

achieveMent review—section one

5. 5 (973 2 4.57 * 0.95)/(542.237 1 SQRT(9.42 * 6.74 ^ 3)) 6. 5 9.546 1 7.15 ^ 4 * 0.462

RETURN

1,216.988

7. 5 845.937 + (54 2 23.159 1 478.652) ^ 2 * 4.005

RETURN

1.6254

Ans

Ans RETURN

8. 5 (SQRT(24.75) 1 82.546 * SQRT(465.852))/8.22 1 98.5

1,040,476.3208

RETURN

315.85

Ans Ans

UNIT 19 Achievement Review—Section One Objective You should be able to solve the exercises and problems in this Achievement Review by applying the principles and methods covered in Units 1–18.

1. Express each of the following fractions as equivalent fractions as indicated. 3 ? 1 ? a. 5 c. 5 8 32 4 64 7 ? 9 ? b. 5 d. 5 10 100 16 128 2. Express each of the following mixed numbers as improper fractions. 1 3 a. 3 d. 13 5 8 9 9 b. 2 e. 6 10 32 3 c. 5 4 3. Express each of the following improper fractions as mixed numbers. 5 115 a. d. 2 32 21 329 b. e. 5 64 75 c. 4 4. Express each of the following fractions as a fraction in lowest terms. 8 18 a. d. 16 64 12 28 b. e. 100 128 30 c. 32

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122

SecTIoN 1

coMMon fractions anD DeciMal fractions

5. Express the fractions in each of the following sets as equivalent fractions having the least common denominator. 1 3 9 7 3 9 13 a. , , c. , , , 4 16 32 10 4 25 20 7 5 9 b. , , 16 32 64 6. Add or subtract each of the following values. Express the answers in lowest terms. 11 7 1 5 f. 2 a. 1 16 16 8 8 17 3 7 15 g. 2 b. 1 20 5 16 16 49 3 5 13 h. 2 c. 1 64 8 8 32 11 7 49 i. 6 2 d. 3 1 16 10 100 1 7 9 1 21 j. 13 2 9 e. 1 1 8 32 32 4 64 7. Multiply or divide each of the following values. Express the answers in lowest terms. 1 5 3 2 8 3 4 2 b. 3 3 4 5 3 7 3 c. 5 3 32 8 1 1 d. 3 38 10 4 3 1 e. 3 20 3 5 16 2 a.

8. Perform each of the indicated combined operations. 3 5 3 a. 1 2 4 16 8 1 3 1 b. 20 1 3 3 2 8 8 3 7 3 c. 3 1 3 5 8 4

1

d. 18 2 5

2

3 1 7 4 13 4 2 8

3 4 10 14 g. 4 15

2 5 7 25 1 h. 16 4 3 17 9 i. 2 4 32 24 29 3 j. 2 48 32 4 f.

1 2 e. 1 1 12 2 8 1 1 5 3 4 2 f. 3 643 4 16 2 4

9. How many complete pieces can be blanked from a strip of aluminum 72 inches 3 3 long if each stamping requires 1 inches of material plus an allowance of inch 8 4 at one end of the strip? 3 10. How many inches of bar stock are needed to make 30 spacers each 1 -inches 16 1 long? Allow inch waste for each spacer. 8

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UNIT 19

123

achieveMent review—section one

1 inch per revolution. 32 How many minutes does it take to cut a distance of 50 inches along the shaft?

11. A shaft is turned at 200 revolutions per minute with a tool feed of

5 3 12. A shop order calls for 1800 steel pins each 1 inches long. If inch is allowed for 8 16 cutting off and facing each pin, how many complete 10-foot lengths of stock are needed for the order? 13. Compute dimensions A, B, C, D, and E of the support bracket shown in Figure 19-1. All dimensions are given in inches. 5 TYPICAL 1 32 3 PLACES

E

3 1 16

15 16

A5 B5 C5 D5

5 29 32

E5

D

1 1 32

2 18 A

B

TYPICAL 4 PLACES 11 34

C

19 18 64

figure 19-1

14. Write each of the following numbers as words. a. 0.6 b. 0.74 c. 0.147

d. 0.0086 e. 4.208 f. 16.0419

15. Write each of the following words as decimal fractions or mixed decimals. a. three tenths d. five and eighty-one ten-thousandths b. twenty-six thousandths c. nine and twenty-six thousandths 16. Round each of the following numbers to the indicated number of decimal places. a. 0.596 (2 places) c. 0.80729 (4 places) b. 5.0463 (3 places) d. 7.0005 (3 places)

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124

SecTIoN 1

coMMon fractions anD DeciMal fractions

17. Express each of the following common fractions as decimal fractions. Where necessary, round the answers to 3 decimal places. 3 2 a. d. 4 25 7 13 b. e. 8 20 2 c. 3 18. Express each of the following decimal fractions as common fractions in lowest terms. a. 0.7 d. 0.915 b. 0.525 e. 0.0075 c. 0.007 19. Add or subtract each of the following values. a. 0.875 1 0.712 b. 5.004 1 0.92 1 0.5034 c. 0.006 1 12.3 1 0.0009 d. 2.99 1 6.015 1 0.1003 e. 23 1 0.0007 1 0.007 1 0.4

f. 0.879 2 0.523 g. 0.1863 2 0.0419 h. 5.400 2 5.399 i. 0.009 2 0.0068 j. 14.001 2 13.999

20. Multiply or divide each of the following values. Round the answer to 4 decimal places where necessary. a. 0.923 3 0.6 f. 0.85 4 0.39 b. 3.63 3 2.30 g. 0.100 4 0.01 c. 4.81 3 0.07 h. 4.016 4 0.03 d. 0.005 3 0.180 i. 123 4 0.665 e. 12.123 3 0.001 j. 0.0098 4 5.036 21. Raise each of the following values to the indicated powers. a. 2.62

d.

b. 0.50

3

c. 0.006

2

2

12 3 5

3

1 2

20.8 e. 6.5

22. Determine the roots of each of the following values as indicated. a. Ï49 d. Ï46.83 1 17.17 3 b. Ï 64

c.

Î

e. Ï39.2 3 1.25

36 81

23. Determine the square roots of each of the following values to the indicated number of decimal places.

Î

2 (3 places) 5

a. Ï379 (2 places)

c.

b. Ï0.8736 (3 places)

d. Ï93.876 2 47.904 (3 places)

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UNIT 19

125

achieveMent review—section one

24. Find the decimal or fraction equivalents of each of the following numbers using the decimal equivalent table. 5 a. d. 0.65625 8 17 b. e. 0.671875 32 21 c. 64 25. Determine the nearer fractional equivalents of each of the following decimals using the decimal equivalent table. a. 0.465 c. 0.038 b. 0.769 d. 0.961 26. Solve each of the following combined operations expressions. Round answers to 2 decimal places. a. 0.4321 1 10.870 4 3.43 3 0.93 c. 35.98 4 Ï6.35 2 4.81 Ï81 3 4.03 b. s12.60 4 3 2 0.98d2 3 3.60 d. 6 3 2 1.72 3.30 3 2.75

1

2

27. The basic form of an ISO Metric Thread is shown in Figure 19-2. Given a thread pitch of 1.5 millimeters, compute thread dimensions A, B, C, D, E, and F to 3 decimal places. A5

PITCH (P) E = 0.125 × P

D = 0.108 25 × P

B5 C5

60° B = 0.541 27 × P

30°

A = 0.866 03 × P

MAJOR DIAMETER

D5 E5 F5

MINOR DIAMETER

F = 0.250 × P 90°

C = 0.216 51 × P

AXIS OF SCREW THREAD

figure 19-2

28. A combination of gage blocks is selected to provide a total thickness of 0.4573 inch. One block 0.250 inch thick and one block 0.118 inch thick are selected. What is the required thickness of the remaining blocks? 29. A piece of round stock is being turned to a 17.86-millimeter diameter. A machinist measures the diameter of the piece as 18.10 millimeters. What depth of cut should be made to turn the piece to the required diameter? 30. A plate 57.20 millimeters thick is to be machined to a thickness of 44.10 millimeters. The plate is to be rough cut with the last cut a finish cut 0.30 millimeter deep. If each rough cut is 3.20 millimeters deep, how many rough cuts are required? 31. A shaft is turned in a lathe at 120 revolutions per minute. The cutting tool advances 0.030 inch per revolution. How long is the length of cut along the shaft at the end of 3.50 minutes?

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sectiON tWO

2

Ratio, Propor tion, and Percentage

UNIT 20 Ratio and Propor tion Objectives After studying this unit you should be able to ●● ●● ●● ●●

Write comparisons as ratios. Express ratios in lowest terms. Solve for the unknown term of a proportion. Substitute given numerical values for symbols in a proportion and solve for the unknown term.

The ability to solve practical machine shop problems using ratio and proportion is a requirement for the skilled machinist. Ratio and proportion are used for calculating gear and pulley speeds and sizes, for computing thread cutting values on a lathe, for computing taper dimensions, and for determining machine cutting times.

Description of ratio Ratio is the comparison of two like quantities. Like quantities have to be in the same units.

Example 1 Two pulleys are shown in Figure 20-1. What is the ratio of the

3 DIA

5 DIA

diameter of the small pulley to the diameter of the larger pulley? All dimensions are in inches. The ratio is 3 to 5.

Ans Figure 20-1

126

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127

ratio and ProPor tion

Example 2 A triangle with given lengths of 3 meters, 4 meters, and 5 meters for sides a, b, and c is shown in Figure 20-2. a. What is the ratio of side a to side b? The ratio is 3 to 4.

c=5

Ans

a=3

b. What is the ratio of side b to side a? The ratio is 4 to 3.

Ans

b=4

c. What is the ratio of side a to side c? The ratio is 3 to 5.

Figure 20-2

Ans

d. What is the ratio of side b to side c? The ratio is 4 to 5.

Ans

The terms of a ratio are the two numbers that are compared. Both terms of a ratio must be expressed in the same units.

Example Two pieces of bar stock are shown in Figure 20-3. What is the ratio of the short piece to the long piece? The terms cannot be compared to a ratio until the 2-foot length is expressed as 24 inches. The ratio is 11 to 24.

110 29

Figure 20-3

Ans

It is impossible to express two quantities as ratios if the terms have unlike units that cannot be expressed as like units. Inches and pounds as shown in Figure 20-4 cannot be compared as ratios. 60

10 lb

Figure 20-4

Expressing Ratios. Ratios are expressed in the following ways. ●● With words, such as 4 to 7. ●● With a colon between the two terms, such as 4 : 7. The ratio 4 : 7 is read as 4 to 7. 4 ●● With a division sign separating the two numbers, such as 4 4 7, or as the fraction, 7 or 4y7.

orDer of terms The terms of a ratio must be compared in the order in which they are given. The first term is the numerator of a fraction, and the second is the denominator.

Examples 1 Ans 3 3 2. The ratio 3 to 1 5 3 4 1 5 Ans 1 x 3. The ratio x : y 5 x 4 y 5 Ans y y 4. The ratio y : x 5 y 4 x 5 Ans x 1. The ratio 1 to 3 5 1 4 3 5

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SecTIoN 2

ratio, ProPor tion, and Percentage

expressing ratios in Lowest terms Generally, a ratio should be expressed in lowest fractional terms.

Examples Each ratio is expressed in lowest terms. 3 1 5 Ans 9 3 40 8 2. 40 : 15 5 5 Ans 15 3 3 9 3 9 3 16 2 3. : 5 4 5 3 5 Ans 8 16 8 16 8 9 3 5 5 10 6 12 4. 10 : 5 10 4 5 3 5 Ans 6 6 1 5 1

1. 3 : 9 5

Description of proportions A proportion is an expression that states the equality of two ratios. Expressing Proportions. Proportions are expressed in the following ways: ●● 3 : 4 :: 6 : 8, which is read as “3 is to 4 as 6 is to 8.” ●● 3 : 4 5 6 : 8, which is read as “the ratio of 3 to 4 equals the ratio of 6 to 8” or “3 is to 4 as 6 is to 8.” 3 6 ●● 5 . This equation form is generally the way that proportions are written in practical 4 8 applications. A proportion consists of four terms. The first and the fourth term are called extremes and the second and third terms are called means.

Example 1 In the proportion 2 : 3 :: 4 : 6

extremes

2 and 6 are the extremes; 3 and 4 are the means. Ans

a:b = c :d

extremes or

a c = b d

means

means 5 10 6 12 5 and 12 are the extremes; 6 and 10 are the means. Ans

Example 2 In the proportion 5

In a proportion, the product of the means equals the product of the extremes. If the terms are cross multiplied, their products are equal.

Examples 1.

3 6 5 4 8 Cross multiply,

2. 3 → →6 4 ← ← 8 3385436 24 5 24

a c 5 b d Cross multiply,

a → →c b ← ← d a3d5b3c ad 5 bc

The method of cross multiplying is used in solving proportions that have an unknown term. You can check your answer by inserting it back in the original proportion.

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UNIT 20

129

ratio and ProPor tion

Example 1 Solve for the value of x. 3 x 5 4 16 Cross multiply.

3 x 5 4 16 4x 5 3(16) 4x 5 48

Divide both sides of the equation by 4.

4x 48 5 4 4

x 5 12 Ans 3 x Check. 5 4 16 3 12 5 4 16 3 3 5 Ck 4 4 Example 2 Solve for the value of x. 7 8 7 8 5 Check. 5 x x 15 15 8x 5 7(15) 7 8 8x 5 105 5 1 15 13 8 8x 105 5 8 8 1 8 8 x 5 13 Ans 5 Ck 8 15 15

Example 3 Solve for the value of x. x 23.4 x 23.4 5 Check. 5 7.5 20 7.5 20 20x 5 7.5(23.4) 20x 5 175.5 20x 175.5 8.775 23.4 5 5 Ck 20 20 7.5 20 x 5 8.775 Ans 1.17 5 1.17 Ck Solving by calculator: 7.5 3 23.4 4 20 5 8.775

Ans

Solving by spreadsheet: 5 7.5 * 23.4 / 20 5 8.775 Ans

RETURN

ApplicAtion Tooling Up 21 . 64 2 2. Compute (27.5 4 2.5 1 4.36) 31.75. Express the result as a decimal rounded to 3 decimal places. 1. Use a decimal equivalents table to find the decimal equivalent of

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130

SecTIoN 2

ratio, ProPor tion, and Percentage

Î

43V to determine the diameter, D, of a cylinder p3H with height H 5 4.3 mm. Round your answer to 2 decimal places.

3. A cylinder has a volume, V, of 32.85 mm3. Use the formula D 5

4. The volume V of a cube is V 5 s3, where s is the length of one side. Determine the volume of a cube with a side that measures 3.45 feet. Round your answer to 1 decimal place. 5. Multiply 15.32 3 8.75. 6. Express the decimal fraction 0.4275 as a common fraction. Reduce to lowest terms.

Ratios Express the following ratios in lowest fractional form. 7. 6 : 21

12. 3 lb : 21 lb

8. 21 : 6

13. 13 mi : 9 mi

9. 2 : 11

14. 156 mm : 200 mm 2 1 15. : 3 2 1 2 16. : 2 3

10. 7 : 21 11. 120 : 460

Related Ratio Problems 17. Length A in Figure 20-5 is 3 inches and length B is 2.5 feet. Determine the ratio of length A to length B in lowest fractional form.

A = 3 in.

B = 2.5 ft

Figure 20-5

18. The diameters of pulleys E, F, G, and H, shown in Figure 20-6, are given in the table. Determine the ratios in lowest fractional form. F

G

H

E

Figure 20-6 Diameter (inches)

98310_sec_02_ptg01.indd 130

E

F

G

H

a.

8

6

4

3

b.

10

8

5

4

c.

12

9

6

3

d.

15

12

10

6

ratios E F

E G

E H

F G

F H

G G H E

H F

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UNIT 20

131

ratio and ProPor tion

19. Refer to the hole locations given for the plate in Figure 20-7. Determine the ratios in lowest fractional form. All dimensions are in millimeters. 160 a. Dimension A to dimension B. 120 b. Dimension A to dimension C. 80 c. Dimension C to dimension D. 60 d. Dimension C to dimension E. 20 e. Dimension D to dimension F. f. Dimension F to dimension B. g. Dimension F to dimension C. B A h. Dimension E to dimension A. C D i. Dimension D to dimension B. E j. Dimension C to dimension F. F

Figure 20-7

20. In Figure 20-8, gear A is turning at 120 revolutions per minute and gear B is turning at 3.6 revolutions per second. Determine the ratio of the speed of gear A to the speed of gear B. GEAR B 3.6 r/s

GEAR A 120 r/min

Figure 20-8

Proportions Solve for the unknown value in each of the following proportions. Check each answer. Round the answers to 3 decimal places where necessary. x 6 5 2 24 3 15 22. 5 A 30 21.

30.

2.4 M 5 3 0.8

31.

4 8 5 4.1 L

23.

7 E 5 9 45

32.

3.4 1 5 y 9

24.

3 24 5 y 13

33.

24 3.2 5 5 A

25.

15 5 5 c 4

26.

P 1 5 27 3

6 15 27. 5 7 F 28.

12 4 5 H 25

29.

T 7.5 5 6.6 22.0

98310_sec_02_ptg01.indd 131

3 1 8 2 34. 5 N 4 3 5 5 1 F 4 7 G 8 36. 5 1 3 4 8 35.

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132

ratio, ProPor tion, and Percentage

SecTIoN 2

7 x 5 1 9 8 16 4 2.5 38. 5 R 12.5 11 E 39. 5 8 12

M 15 5 12 9 6.08 5.87 41. 5 H 12.53 E 0.36 42. 5 7.53 1.86

37.

40.

Related Proportion Problems A C 5 compares the sides of two triangles like those in Figure 20-9. B D Determine the missing values in the table.

43. The proportion

A

B

a.

180

4.50

b.

6

c.

10 2

1

25.8 mm

D 30

50 8

87.5 mm

d.

C

4

A

10 2

B

75.0 mm

62.5 mm

20.6 mm

16.4 mm

C

D

Figure 20-9

44.

Where machine parts are doweled in position, it is good practice to extend the pin 1 to 1 1 times its diameter into the mating part as shown in Figure 20-10. Use the following 2 proportion to determine the value of each unknown in the table. MATING Round the answers to 3 decimal places where necessary. PART N L 5 1 D

L

where N 5 the number of times the pin extension is greater than the pin diameter L 5 the length of the pin extension D 5 the pin diameter

D

Figure 20-10

a.

N

D

L

1.250

7.940 mm

f.

1.375

10 2

g.

1.250

0.8750

h.

1.500

3.680 mm

i.

1.125

j.

1.000

b.

1

1 4

c.

1

1 4

30 4

d.

1.375

8.730 mm

e.

1.250

98310_sec_02_ptg01.indd 132

N

D

L 1.0320

0.2810 7.500 mm

16.120 mm

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UNIT 20

133

ratio and ProPor tion

45. It is sometimes impractical to make engineering drawings full size. If the part to be drawn is very large or small, a scale drawing is generally made. The scale, which is shown on the drawing, compares the lengths of the lines on the 10 drawing to the dimensions on the part. A scale on a drawing that states 5 10 means the drawing is one-quarter 4 1 the size of the part. It is expressed as a ratio of 1 : 4 or . A scale drawing that states 20 5 10 means that the draw4 2 ing is double the size of the part. It is expressed as a ratio of 2 : 1 or . The actual dimensions of a steel support 1 are given in Figure 20-11. All dimensions are in inches. Using the proportion given, compute the lengths for each unknown in the table. Round the answers to 3 decimal places where necessary. Drawing Length Numerator of Scale Ratio 5 Denominator of Scale Ratio Part Dimension

H = 0.375 G = 0.281 F = 12.600 D = 8.875

E = 10.600

A = 1.872 C = 1.360 B = 16.062

Figure 20-11

scale

98310_sec_02_ptg01.indd 133

Drawing Length

a.

10 5 10 2

B5

b.

40 5 10

c.

scale

Drawing Length

i.

10 5 10 2

E5

G5

j.

60 5 10

G5

10 5 10 4

B5

k.

30 5 10 4

F5

d.

20 5 10

C5

e.

1

10 5 10 2

l.

A5

f.

30 5 10 4

E5

g.

30 5 10

H5

h.

10 51 8

F5

1

10 5 10 2

C5

m.

10 5 10 2

F5

n.

30 5 10

G5

o.

10 5 10 4

B5

p.

20 5 1

A5

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ratio, ProPor tion, and Percentage

46. Figure 20-12 shows the relationship of gears in a lathe using a simple gear train. The proportion given is used for lathe thread cutting computations using simple gearing. The fixed stud gear and the spindle gear have the same number of teeth. Determine the missing values for each of the following problems. NL NC

5

TS

where

NL 5 number of threads per inch on the lead screw NC 5 number of threads per inch to be cut TS 5 number of teeth on stud gear TL 5 number of teeth on lead screw gear

TL

Note: Intermediate gears only change direction. SPINDLE GEAR STUD GEAR

THREADS CUT

INTERMEDIATE OR IDLER GEAR

SPINDLE FIXED STUD GEAR

LEAD SCREW GEAR

LEAD SCREW

Figure 20-12

a. If NL 5 4, NC 5 8, and TS 5 32, find TL. b. If NL 5 7, TS 5 35, and NC 5 15, find TL. c. If NC 5 10, NL 5 6, and TL 5 40, find TS. d. If NL 5 8, TL 5 42, and TS 5 28, find NC. 47. A template is shown on the left of Figure 20-13. A drafter makes an enlarged drawing of the template as shown on the right. The original length of 1.80 inches on the enlarged drawing is 3.06 inches, as shown. Determine the lengths of A, B, C, and D. 0.600

B

0.800

C

D

0.900 0.500

A 1.800

3.060

Figure 20-13

A

B 10 5 39. 4 Determine the value of each missing value in the table.

C

48. A drawing has a scale of

scale Length

a. b. c. d.

98310_sec_02_ptg01.indd 134

D actual Length

10 8 70 16

1

69 90 169 1

10 16

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UNIT 21

direct and inverse ProPortions

135

UNIT 21 Direct and Inverse Proportions Objectives After studying this unit you should be able to ●● ●●

Analyze problems to determine whether quantities are directly or inversely proportional. Set up and solve direct and inverse proportions.

Many shop problems are solved by the use of proportions. A machinist may be required to express word statements or other given data as proportions. Generally, three of the four terms of a proportion must be known in order to solve the proportion. When setting up a proportion, it is important that the terms be placed in their proper positions.

Direct proportions In actual practice, word statements or other data must be expressed as proportions. When a proportion is set up, the terms of the proportion must be placed in their proper positions. A problem that is set up and solved as a proportion must first be analyzed in order to determine where the terms are placed. Depending on the position of the terms, proportions are either direct or inverse. Two quantities are directly proportional if a change in one produces a change in the other in the same direction. If an increase in one produces an increase in the other, or if a decrease in one produces a decrease in the other, the two quantities are directly proportional. The proportions discussed will be those that change at the same rate. An increase or decrease in one quantity produces the same rate of increase or decrease in the other quantity.

Note: When setting up a direct proportion in fractional form, the numerator of the first ratio must correspond to the numerator of the second ratio. The denominator of the first ratio must correspond to the denominator of the second ratio. Example 1 If a machine produces 120 parts in 2 hours, how many parts are produced in 3 hours? Analyze the problem. An increase in time (from 2 hours to 3 hours) will produce an increase in the number of pieces produced. Production increases as time increases. The proportion is direct. Set up the direct proportion. Let x represent the number of parts that are produced in 3 hours. The numerator of the first ratio must correspond to the numerator of the second ratio; 2 hours corresponds to 120 parts. The denominator of the first ratio must correspond to the denominator of the second ratio; 3 hours corresponds to x.

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136

SecTIoN 2

Solve for x. 2 hours 120 parts 5 x 3 hours 2x 5 3(120 parts) 2x 5 360 parts x 5 180 parts Ans

ratio, ProPor tion, and Percentage

Check. 2 hours 120 parts 5 x 3 hours 2 hours 120 parts 5 3 hours 180 parts 2 2 5 Ck 3 3

Example 2 A tapered shaft is one that varies uniformly in diameter along its length. The shaft shown in Figure 21-1 is 15.000 inches long with a 1.200-inch diameter on the large end. A 9.000-inch piece is cut from the shaft. Determine the diameter at the large end of the 9.000-inch piece. x DIA

1.200 DIA

9.000 15.000

Figure 21-1

Analyze the problem. As the length decreases from 15.000 inches to 9.000 inches, the diameter also decreases at the same rate. The proportion is direct. Set up the direct proportion. Let x represent the diameter at the large end of the 9.000-inch piece. The numerator of the first ratio must correspond to the numerator of the second ratio; the 15.000-inch piece has a 1.200-inch diameter at the large end. The denominator of the f irst ratio must correspond to the denominator of the second ratio; the 9.000-inch piece has a diameter of x at the large end. Solve for x. Length of Shaft Diameter of Shaft 5 Length of Piece Diameter of Piece 5 15.000 0 1.2000 DIA 5 9.000 0 x DIA 3 5x 5 3(1.2000) 5x 5 3.6000 x 5 0.7200 Ans

Check.

15.0000 1.2000 DIA 5 9.0000 x DIA 15.0000 1.2000 DIA 5 9.0000 0.7200 DIA 1.67 5 1.67 Ck

inverse proportions Two quantities are inversely or indirectly proportional if a change in one produces a change in the other in the opposite direction. If an increase in one produces a decrease in the other, or if a decrease in one produces an increase in the other, the two quantities are inversely proportional. For example, if one quantity increases by 4 times its original value, the other 1 quantity decreases by 4 times its value, or is of its original value. 4 4 1 Notice 4, or , inverted is . 1 4

Note: When setting up an inverse proportion in fractional form, the numerator of the first ratio must correspond to the denominator of the second ratio. The denominator of the first ratio must correspond to the numerator of the second ratio.

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UNIT 21

137

direct and inverse ProPortions

Example 1 Two gears in mesh are shown in Figure 21-2. The driver gear has 40 teeth and revolves at 360 revolutions per minute. Determine the number of revolutions per minute of a driven gear with 16 teeth. 360 r/min x r/min

A

DRIVEN GEAR (16 TEETH)

DRIVER GEAR (40 TEETH)

Figure 21-2

Analyze the problem. When the driver gear turns one revolution, 40 teeth pass point A. The same number of teeth on the driven gear must pass point A. Therefore, the driven gear turns more than one revolution for each revolution of the driver gear. The gear with 16 teeth (driven gear) revolves at greater revolutions per minute than the gear with 40 teeth (driver gear). A decrease in the number of teeth in the driven gear produces an increase in revolutions per minute. The proportion is inverse. Set up the inverse proportion. Let x represent the revolutions per minute of the gear with 16 teeth. The numerator of the first ratio must correspond to the denominator of the second ratio; the gear with 40 teeth revolves at 360 r/min. The denominator of the first ratio must correspond to the numerator of the second ratio; the gear with 16 teeth corresponds to x. Note that the product of the first gear’s speed and its number of teeth equals the product of the second gear’s speed and its number of teeth. Thus, the product 360 3 40 equals the number of teeth on the driver gear that pass point A in 1 minute. The corresponding product of the driven gear must also be 360. These are put on diagonals so that the crossproducts are equal. Solve for x. Teeth in Driver Gear Revolutions in Driven Gear 5 Teeth in Driven Gear Revolutions in Driver Gear 5 40 teeth x 5 16 teeth 360 r/min 2 2x 5 1800 r/min x 5 900 r/min Ans

Check.

40 teeth x 5 16 teeth 360 r/min 40 teeth 900 r/min 5 16 teeth 360 r/min 2.5 5 2.5 Ck

The driven gear revolves at the rate of 900 revolutions per minute.

Example 2 Five identical machines produce the same parts at the same rate. The five machines complete the required number of parts in 2.1 hours. How many hours does it take three machines to produce the same number of parts? Analyze the problem. A decrease in the number of machines (from five to three) requires an increase in time. Time increases as the number of machines decreases; therefore, the proportion is inverse.

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138

SecTIoN 2

ratio, ProPor tion, and Percentage

Set up the inverse proportion. Let x represent the time required by three machines to produce the parts. 5 Machines Time for 3 Machines 5 3 Machines Time for 5 Machines x 5 2.1 hours Notice that the numerator of the first ratio corresponds to the denominator of the second ratio; five machines corresponds to 2.1 hours. The denominator of the first ratio corresponds to the numerator of the second ratio for a product of 10.5 machine-hours; three machines corresponds to x. Solve for x. 5 x 5 3 2.1 hours

Check. 5 x 5 3 2.1 hours 5 3.5 hours 5 3 2.1 hours – – 1.6 5 1.6 Ck

3x 5 5(2.1 hours) 3x 10.5 hours 5 3 3 x 5 3.5 hours Ans

1 It will take 3 hours for three machines to produce as many parts as five machines did in 2 2.1 hours (2 hours 6 minutes).

ApplicAtion Tooling Up 3 11 5 for R. If necessary, round the answer to 3 decimal places. R 9 2. Find the nearer fraction equivalent of 0.647 using the decimal equivalent table. 1. Solve the proportion

3. Use a calculator to compute Ï0.42 3 5.49. Round your answer to 2 decimal places. 4. How many complete bushings each 12.80 mm long can be cut from a bar of bronze that is 527.45 mm long? Allow 2.91 mm waste for each piece. 5. Subtract : 15.32 2 8.755. 6. Write the words ninety-seven thousand eight hundred six and seventeen thousandths as a number.

Tapers Taper is the difference between the diameters at each end of a part. Tapers are expressed as the difference in diameters for a particular length along the centerline of a part (Figure 21-3 and Figure 21-4). Note: All dimensions are in millimeters.

20

Note: All dimensions are in inches.

7

1.187

0.885 4.250

300

20 mm 2 7 mm 5 13 mm taper per 300 mm Figure 21-3

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1.1870 2 0.8850 5 0.3020 taper per 4.2500 Figure 21-4

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UNIT 21

139

direct and inverse ProPortions

7. A plug gage tapers 3.10 mm along a 38.00 mm length, as shown in Figure 21-5. Set up a proportion and determine the amount of taper in the workpiece for each of the following problems. Express the answers to 2 decimal places.

WorKPieCe tHiCKNess

a.

18.40 mm

b.

31.75 mm

c.

14.28 mm

d.

28.58 mm

e.

23.60 mm

taPer iN WorKPieCe

ProPortioN

WORKPIECE THICKNESS

38.00 mm

Figure 21-5

8. A reamer tapers 0.1300 along a 4.2500 length (Figure 21-6). Set up a proportion and determine length A for each of the following problems. Express the answers to 3 decimal places.

taPer iN LeNGtH A

a.

0.030”

b.

0.108”

c.

0.068”

d.

0.008”

e.

0.093”

ProPortioN

LeNGtH A A 4.2500

Figure 21-6

9. A micrometer reading is made at dimension D on a tapered shaft (Figure 21-7). For each of the problems use the dimensions given in the table, compute the taper, set up a proportion, and determine diameter C to 3 decimal places. MICROMETER

DIA C

DIA B DIA A

D LENGTH OF SHAFT

Figure 21-7 LeNGtH oF sHaFt

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Diameter a Diameter B

DimeNsioN D Diameter C

a.

10.200”

1.500”

0.700”

6.500”

b.

8.750”

1.250”

0.375”

4.875”

c.

550.000 mm

106.250 mm

62.500 mm

337.500 mm

d.

147.500 mm

22.500 mm

10.000 mm

112.500 mm

e.

8.800”

1.325”

0.410”

8.620”

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140

SecTIoN 2

ratio, ProPor tion, and Percentage

Direct and Inverse Proportions Analyze each of the following problems to determine whether the problem is a direct or inverse proportion. Set up the proportion and solve. 1 1 10. A sheet of steel 8 feet long weighs 325 pounds. A piece 2 feet long is sheared from the 4 2 1 sheet. Determine the weight of the 2 -foot piece to the nearest whole pound. 2 11. If 1350 parts are produced in 6.75 hours, find the number of parts produced in 8.25 hours. 12. The production rate for each of three machines is the same. Using these three machines, 720 parts are produced in 1.6 hours. How many hours will it take two of these machines to produce 720 parts? 13. Two forgings are made of the same stainless steel alloy. A forging, which weighs 76.00 kilograms, contains 0.38 kilogram of chromium. How many kilograms of chromium does the second forging contain if it weighs 96.00 kilograms? Round the answer to 2 decimal places.

Gears and Pulleys 14. A 10.00-inch diameter pulley rotates at 160.0 rpm. A belt connects this 10.00-inch diameter pulley with a 6.5-inch diameter pulley. An 8.00-inch diameter pulley is fixed to the same shaft as the 6.50-inch pulley. A belt connects the 8.00-inch pulley with a 3.50-inch diameter pulley. Determine the revolutions per minute of the 3.50-inch diameter pulley. Round the answer to 1 decimal place. 15. Of two gears that mesh, the one that has the greater number of teeth is called the gear, and the one that has the fewer teeth is called the pinion (Figure 21-8). For each of the problems, set up a proportion, and determine the unknown value, x. Round the answers to 1 decimal place where necessary.

PINION

GEAR

Figure 21-8 NUmBer oF teetH oN Gear

NUmBer oF teetH oN PiNioN

sPeeD oF Gear (rpm)

a.

48

20

100.0

b.

32

24

c.

35

d.

x5

e.

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54

x5

sPeeD oF PiNioN (rpm) x5

x5

210.0 160.0

200.0

15

150.0

250.0

26

80.0

x 5Ò

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UNIT 22

141

introduction to Percents

16. Figure 21-9 shows a compound gear train. Gears B and C are keyed to the same shaft; therefore, they turn at the same speed. Gear A and gear C are driving gears. Gear B and gear D are driven gears. Set up a proportion for each problem and determine the unknown values, x, y, and z in the table. Round the answers to 1 decimal place where necessary.

GEAR A

GEAR C

GEAR D GEAR B

Figure 21-9 NUmBer oF teetH

sPeeD (rpm)

Gear a

Gear B

Gear C

Gear D

Gear a

a.

80

30

50

20

120.0

b.

60

c. d.

x5

x5

45 24

55

25

60 x5

y5

Gear B x5

100.0 36 15

144.0 y5

300.0 y5 z5

Gear C y5

Gear D z5

z5

450.0

z5

280.0

175.0

350.0

UNIT 22 Introduction to Percents Objectives After studying this unit you should be able to ●● ●●

Express decimal fractions and common fractions as percents. Express percents as decimal fractions and common fractions.

Percents are widely used in both business and nonbusiness fields. Merchandise selling prices and discounts, wage deductions, and equipment depreciation are determined by percentages. In manufacturing technology, percentage concepts have many applications, such as expressing production increases or decreases, power inputs and outputs, quality control product rejections, and material allowances for waste and nonconforming parts.

Definition of percent The percent symbol (%) indicates the number of hundredths of a whole. The large square shown in Figure 22-1 is divided into 100 equal parts. The whole (large square) contains 100 small parts, or 100% of the small squares. Each small square is one part of 100 parts or 1 1 of the large square in the figure. Therefore, each small square is of 100% or 1% of 100 100 the large square. 1 part of 100 parts 1 5 0.01 5 1% 100

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Figure 22-1

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SecTIoN 2

ratio, ProPor tion, and Percentage

Example What percent of the square in Figure 22-2 is shaded? The large square is divided into four equal smaller squares. Three of the smaller squares are shaded. 3 parts of 4 parts 3 5 0.75 5 75% Ans 4 Figure 22-2

expressing DecimaL fractions as percents A decimal fraction can be expressed as a percent by moving the decimal point two places to the right and inserting the percent symbol. Moving the decimal point two places to the right is actually multiplying by 100.

Example 1 Express 0.0152 as a percent. Move the decimal point 2 places to the right. Insert the percent symbol.

0.01 52 5 1.52%

Ans

3.87 6 5 387.6%

Ans

Example 2 Express 3.876 as a percent. Move the decimal point 2 places to the right. Insert the percent symbol.

expressing common fractions anD mixeD numbers as percents To express a common fraction as a percent, first express the common fraction as a decimal fraction. Then express the decimal fraction as a percent. If it is necessary to round, the decimal fraction must be 2 more decimal places than the desired number of places for the percent. 7 8

Example 1 Express as a percent. 7 as a decimal fraction. 8 Then express 0.875 as a percent.

First, express

7 5 0.875 8 0.875 5 87.5%

Ans

2 3

Example 2 Express 5 as a percent to 1 decimal place. 2 First, express 5 as a decimal fraction. 3 Next, express 5.667 as a percent.

2 5 5 5.667 3 5.667 5 566.7%

Ans

expressing percents as DecimaL fractions Expressing a percent as a decimal fraction can be done by dropping the percent symbol and moving the decimal point two places to the left. Moving the decimal point two places to the left is actually dividing by 100.

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UNIT 22

143

introduction to Percents

Examples 16 % as a decimal fraction. Round the answer to 4 decimal places. 21 16 Express 38 % as the approximate decimal fraction 38.76%. 21 Drop the percent symbol and move the decimal point 2 places to the left. 16 38 % 5 38.76% 5 0.3876 Ans 21

1. Express 38

Express each percent as a decimal fraction. Round the answers to 3 decimal places. 2. 0.48% 0.005 Ans 4. 7% 0.070 Ans 3 5. 300% 3.000 Ans 3. 15 % 0.158 Ans 4

expressing percents as common fractions A percent is expressed as a fraction by first finding the equivalent decimal fraction. The decimal fraction is then expressed as a common fraction.

Examples 1. Express 37.5% as a common fraction. Express 37.5% as a decimal fraction. Then express 0.375 as a common fraction.

37.5 375 3 5 5 100 1000 8 37.5% 5 0.375 375 3 0.375 5 5 Ans 1000 8

Express each percent as a common fraction. 10 1 2. 10% 10% 5 0.10 5 5 Ans 100 10 5 1 3. 0.5% 0.5% 5 0.005 5 5 Ans 1000 200 1 1 225 9 4. 222 % 222 % 5 222.5% 5 2.225 5 2 52 2 2 1000 40 1 5. 6 % 4

1 625 1 6 % 5 6.25% 5 0.0625 5 5 4 10,000 16

Ans

Ans

ApplicAtion Tooling Up 1. A sheet of steel 280 centimeters long weighs 165 kilograms. A sheet 92.4 centimeters long is sheared from the sheet. Determine the weight of the 92.4 centimeter piece of steel to the nearest half kilogram. 13 T 2. Solve the proportion 5 for T. If necessary, round the answer to 3 decimal places. 5 21 4 3. A sphere of radius R has a volume, V, given by the formula V = pR3. Determine the volume of a sphere with a 3 radius of 6.500 mm. Use p 5 3.14. If necessary, round the answer to 3 decimal places. 4. Multiply: 12.345 × 7.86. 5. Express 0.6125 as a common fraction.

1

6. Compute 15 2 3

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2

1 1 3 4 1 7 . Express the answer as a mixed number. 4 8 4

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144

SecTIoN 2

ratio, ProPor tion, and Percentage

Determining Percents Determine the percent of each figure that is shaded. 7.

8.

9.

10. 7. 8. 9. 10.

expressing Decimals and Fractions as Percents Express each value as a percent. 11. 0.35

18. 0.0639

12. 0.96

19. 0.0002

13. 0.04

20. 3.005

14. 0.062 15. 0.008 16. 1.33 17. 2.076

21.

1 4

22.

21 80

23.

3 20

37 50 17 25. 32 1 26. 250 24.

27. 1

59 100

7 25 5 29. 14 8 1 30. 3 200 28. 2

expressing Percents as Decimals Express each percent as a decimal fraction or mixed decimal. 3 % 4

31. 82%

36. 103%

32. 19%

37. 224.9%

33. 3%

38. 0.6%

42. 0.1%

34. 2.6%

39. 4.73% 1 40. 12 % 2

3 43. 2 % 8

46. 205

35. 27.76%

41.

44. 0.05% 1 45. 37 % 4 1 % 10

expressing Percents as Fractions Express each percent as a common fraction or mixed number. 47. 50%

50. 4%

53. 190%

56. 100.1%

48. 25%

51. 16%

54. 0.2%

57. 0.9%

49. 62.5%

52. 275%

55. 1.8%

58. 0.05%

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UNIT 23

Basic calculations oF Percentages, Percents, and rates

145

UNIT 23 Basic Calculations of Percentages, Percents, and Rates

Objectives After studying this unit you should be able to Determine the percentage, given the base and rate. Determine the percent (rate), given the percentage and base. Determine the base, given the rate and percentage.

●● ●● ●●

types of simpLe percentage probLems A simple percentage problem has three parts: the rate, the base, and the percentage. In the problem 10% of $80 5 $8, the rate is 10%, the base is $80, and the percentage is $8. The rate is the percent. The base is the number of which the rate or percent is taken. It is the whole or a quantity equal to 100%. The percentage is the quantity of the percent of the base. In solving problems, the rate, percentage, and base must be identified. Some people like to use the term “amount” for percentage. This is perfectly correct; however, you will often see percentage used in this book. In solving percentage problems, the words is and of are often helpful in identifying the three parts. The word is generally relates to the rate or percentage and the word of generally relates to the base. The following descriptions may help you recognize the rate, base, and percentage more quickly: Base: The total, original, or entire amount. The base usually follows the word “of.” Rate: The number with a % sign. Sometimes it is written as a decimal or fraction. Percentage: The value that remains after the base and rate have been determined. It is a portion of the base. The percentage is often close to the word “is.”

Examples 1. What is 25% of 120? Is relates to 25% (the rate) and of relates to 120 (the base). 2. What percent of 48 is 12? Is relates to 12 (the percentage) and of relates to 48 (the base). 3. 60 is 30% of what number? Is relates to 60 (the percentage) and 30% (the rate). Of relates to what number (the base). There are three types of simple percentage problems. The type used depends on which two quantities are given and which quantity must be found. The three types are as follows: ●●

●●

98310_sec_02_ptg01.indd 145

Finding the percentage, given the rate (percent) and the base. A problem of this type is, “What is 25% of 120?” If the rate is less than 100%, the percentage is less than the base. If the rate is greater than 100%, the percentage is greater than the base. Finding the rate (percent), given the base and the percentage. A problem of this type is, “What percent of 48 is 12?”

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146

SecTIoN 2

●●

ratio, ProPor tion, and Percentage

If the percentage is less than the base, the rate is less than 100%. If the percentage is greater than the base, the rate is greater than 100%. Finding the base, given the rate (percent) and the percentage. A problem of this type is, “60 is 30% of what number?” All three types of percentage problems can be solved using the following proportion: P R 5 B 100 where B is the base, P is the percentage or part of the base, and R is the rate or percent.

Practical applications involve numbers that have units or names of quantities called denominate numbers. The base and the percentage have the same unit or denomination. For example, if the base unit is expressed in inches, the percentage is expressed in inches. The rate is not a denominate number; it does not have a unit or denomination. Rate is the part to be taken of the whole quantity, the base.

finDing the percentage, given the base anD rate In some problems, the base and rate are given and the percentage must be found. First, express the rate (percent) as an equivalent decimal fraction. Then solve with the P R proportion 5 . B 100

Example 1 What is 15% of 60? 15 , so R 5 15. 100 The base, B, is 60. It is the number of which the rate is taken—the whole or a quantity equal to 100%.

Solution The rate is 15% 5

The percentage is to be found. It is the quantity of the percent of the base. The proportion is

P 15 5 . 60 100

Now, using cross-products and division: 100P 5 15 3 60 100P 5 900 900 P5 5 9 Ans 100

Example 2 Find 56

9 % of $183.76. 25

Solution The rate is 56

9 9 %, so R 5 56 . 25 25

The base, B, is $183.76. The percentage is to be found. 9 9 Express R, 56 , as a decimal: 56 5 56.36. 25 25

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UNIT 23

147

Basic calculations oF Percentages, Percents, and rates

P 56.36 5 . $183.76 100 Again, using cross-products and division: 100P 5 56.36 3 $183.76 100P < $10,357 $10,357 P5 5 $103.57 Ans 100

The proportion is

finDing the percent (rate), given the base anD percentage In some problems, the base and percentage are given, and the percent (rate) must be found.

Example 1 What percent of 12.87 is 9.620? Round the answer to 1 decimal place. Since a percent of 12.87 is to be taken, the base or whole quantity equal to 100% is 12.87. The percentage or quantity of the percent of the base is 9.620. The rate is to be found. Since the percentage, 9.620, is less than the base, 12.87, the rate must be less than 100%. 9.620 R The proportion is 5 . 12.87 100 Cross multiply Divide

9.620 3 100 5 12.87R 962 5 12.87R 962 5R 12.87 962 R5 5 74.7474 < 74.7% 12.87

Ans

Example 2 What percent of 9.620 is 12.87? Round the answer to 1 decimal place. Notice that although the numbers are the same as in Example 1, the base and percentage are reversed. Since a percent of 9.620 is to be taken, the base or whole quantity equal to 100% is 9.620. The percentage or quantity of the percent of the base is 12.87. Since the percentage, 12.87, is greater than the base, 9.620, the rate must be greater than 100%. 12.87 R B 5 9.620 and P 5 12.87. Thus, the proportion is 5 . 9.620 100 12.87 3 100 5 9.620R 1287 5 9.620R 1287 5R 9.620 1287 R5 5 133.78 < 133.8% Ans (rounded) 9.620

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148

SecTIoN 2

ratio, ProPor tion, and Percentage

finDing the base, given the percent (rate) anD the percentage In some problems, the percent (rate) and the percentage are given, and the base must be found.

Example 1 816 is 68% of what number? 816 68 5 . B 100 816 3 100 5 68B 81,600 5 68B 81,600 5B 68 81,600 B5 5 1200 Ans 68

R 5 68 and P 5 816. The proportion is

2 3

Example 2 $149.50 is 115 % of what value? $149.50 115.67 5 . B 100 $149.50 3 100 5 115.67B $14,950 5 115.67B $14,950 5B 115.67 $14,950 B5 5 $129.25 Ans 115.67

R 5 115.67 and P 5 $149.50 and the proportion is

ApplicAtion Tooling Up 1. Express 5.037 as a percent. 2. If 2100 parts can be produced in 5.25 hours, how long will it take to produce 26,000 parts? 3. Solve 57.12 3 Ï2.52 1 7.48. If necessary, round the answer to 2 decimal places. to the indicated power. 121.5 4.3 2 3

4. Raise

5. The length, L, of the standard 828 included angle drill can be calculated using the formula L 5 0.575[, where [ represents the diameter of the drill. Determine the length of a drill point with a diameter of 12.5 mm. Round your answer to 1 decimal place. 6. Write the number 107.2004 as words.

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UNIT 23

Basic calculations oF Percentages, Percents, and rates

149

Finding Percentage Find each percentage. Round the answers to 2 decimal places when necessary. 7. 20% of 80

18. 1.8% of 1240

8. 2.15% of 80

19. 39% of 18.3

9. 60% of 200

20. 0.42% of 50

10. 15.23% of 150 11. 25% of 312.6 12. 7% of 140.34 13. 156% of 65 14. 0.8% of 214 15. 12.7% of 295 16. 122% of 1.68 17. 140% of 280

21. 0.03% of 424.6 1 22. 8 % of 375 2 3 23. % of 132 4 24. 296.5% of 81 1 1 25. 15 % of 35 4 4 17 3 26. % of 139 50 10

Finding Percent (Rate) Find each percent (rate). Round the answers to 2 decimal places when necessary. 27. What percent of 8 is 4? 28. What percent of 20.7 is 5.6? 29. What percent of 100 is 37? 30. What percent of 84.37 is 70.93? 31. What percent of 70.93 is 84.37? 32. What percent of 258 is 97? 33. What percent of 132.7 is 206.3? 34. What percent of 19.5 is 5.5? 35. What percent of 1.25 is 0.5? 36. What percent of 0.5 is 1.25? 1 37. What percent of 6 is 2? 2

3 38. What percent of 134 is 156 ? 4 7 3 39. What percent of is ? 8 8 3 7 40. What percent of is ? 8 8 41. What percent of 3.08 is 4.76? 42. What percent of 0.65 is 0.09? 1 43. What percent of 12 is 3? 4 44. What percent of 312 is 400.9? 3 3 45. What percent of is ? 4 8 4 3 46. What percent of 13 is 6 ? 5 10

Finding Base Find each base. Round the answers to 2 decimal places when necessary.

50. 3.8 is 95.3% of what number?

1 55. 7 is 180% of what number? 2 1 56. 10 is 6 % of what number? 4 57. 190.75 is 70.5% of what number?

51. 13.6 is 8% of what number?

58. 6.6 is 3.3% of what number?

52. 123.86 is 88.7% of what number?

59. 88 is 205% of what number?

53. 203 is 110% of what number? 1 54. 44 is 60% of what number? 3

60. 1.3 is 0.9% of what number? 7 61. is 175% of what number? 8

47. 15 is 10% of what number? 48. 25 is 80% of what number? 49. 80 is 25% of what number?

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150

SecTIoN 2

1 1 is 1 % of what number? 10 5 63. 9.3 is 238.6% of what number? 62.

64. 0.84 is 2.04% of what number?

ratio, ProPor tion, and Percentage

1 65. 20 is 71% of what number? 2 3 66. is 123% of what number? 4

Finding Percentage, Percent, or Base Find each percentage, percent (rate), or base. Round the answers to 2 decimal places when necessary. 67. What percent of 24 is 18?

73. 72.4% of 212.7 is

68. What is 30% of 50?

74. What percent of 228 is 256?

69. What is 123.8% of 12.6?

75. 51.03 is 88% of what number? ? 76. 36.5 is % of 27.6. 1 ? 77. 2 % of 150 is . 4 ? 78. is 18% of 120.66.

70. 73 is 82% of what number? 1 71. What percent of 10 is 2? 2 ? 72. is 48% of 94.82.

?

.

UNIT 24 Percent Practical Applications Objectives After studying this unit you should be able to ●● ●●

Solve simple percentage practical applications in which two of the three parts are given. Solve more complex percentage practical applications in which two of the three parts are not directly given.

iDentifying rate, base, anD percentage in various types of practicaL appLications In solving simple problems, generally there is no difficulty in identifying the rate or percent. A common mistake is to incorrectly identify the percentage and the base. There is sometimes confusion as to whether a value is a percentage or a base; the base and percentage are incorrectly interchanged. The following statements summarize the information that was given when each of the three types of problems was discussed and solved. A review of the statements should be helpful in identifying the rate, percentage, and base. ●● The rate (percent) is the part taken of the whole quantity (base). ●● The base is the whole quantity or a quantity that is equal to 100%. It is the quantity of which the rate is taken. ●● The percentage is the quantity of the percent that is taken of the base. It is the quantity equal to the percent that is taken of the whole. ●● If the rate is 100%, the percentage and the base are the same quantity. If the rate is less than 100%, the percentage is less than the base. If the rate is greater than 100%, the percentage is greater than the base.

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UNIT 24

●●

●●

151

Percent Practical aPPlications

In practical applications, the percentage and the base have the same unit or denomination. The rate does not have a unit or denomination. The word is generally relates to the rate or percentage, and the word of generally relates to the base.

finDing percentage in practicaL appLications Example The cost of a production run of steel pins is estimated at $3275. Material cost is estimated as 35% of the total cost. What is the estimated material cost to the nearest dollar? Think the problem through to determine what is given and what is to be found. The rate is 35%. The base, B, is $3275. It is the total cost or the whole quantity. The percentage, P, which is the material cost, is to be found. The proportion is Cross multiply

P 35 5 . $3275 100

Divide The estimated cost of materials is $1146.

100P 5 35 3 $3275 100P 5 $114,625 $114,625 P5 5 $1146.25 Ans 100

finDing percent (rate) in practicaL appLications Example An inspector rejects 23 out of a total production of 630 electrical switches. What percent of the total production is rejected? Round the answer to 1 decimal place. Think the problem through to determine what is given and what is to be found. Since a percent of the total production of 630 switches is to be found, the base or whole quantity equal to 100% is 630 switches. 23 switches R 5 . 630 switches 100 23 switches 3 100 5 630 switches 3 R 2300 switches 5 630 switches 3 R 2300 switches 5R 630 switches 2300 switches R5 5 3.651, 3.7% 630 switches

The proportion is

Ans (rounded)

3.7% of the total production is rejected.

finDing the base in practicaL appLications Example A motor is said to be 80% efficient if the output (power delivered) is 80% of the input (power received). How many horsepower does a motor receive if it is 80% efficient with a 6.20-horsepower (hp) output? Think the problem through to determine what is given and what is to be found. The rate is 80%, so R 5 80. Since the output of 6.20 hp is the quantity of the percent of the base, the percentage is 6.20 hp (6.20 hp is 80% of the base).

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ratio, ProPor tion, and Percentage

The base to be found is the input; the whole quantity equal to 100%. 6.20 hp 80 5 . B 100 6.20 hp 3 100 5 80B 620 hp 5 80B 620 hp 5B 80 620 hp B5 5 7.75 hp 80

R 5 80 and P 5 6.20 hp, so the proportion is

Ans

The motor receives 7.75 hp.

more compLex percentage practicaL appLications In certain percentage problems, two of the three parts are not directly given. One or more additional operations may be required in setting up and solving a problem. Examples of these types of problems follow.

Example 1 By replacing high-speed steel cutters with carbide cutters, a machinist increases production by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced with high-speed steel cutters? Think the problem through. The base (100%) is the daily production using high-speed steel cutters. Since the base is increased by 35%, the carbide cutter production of 270 pieces is 100% 1 35%, or 135%, of the base. Therefore, the rate is 135% and the percentage is 270. The base is to be found. 270 pieces per day 135 5 . B 100 270 pieces per day 3 100 5 135B 27,000 pieces per day 5 135B 27,000 pieces per day 5B 135 27,000 pieces per day B5 5 200 pieces per day Ans 135

The proportion is

The machinist produced 200 pieces per day using high-speed cutters.

Example 2 A mechanic purchases a set of socket wrenches for $54.94. The purchase price is 33% less than the list price. What is the list price? Think the problem through. The base (100%) is the list price. Since the base is decreased by 33%, the purchase price, $54.94, is 100% 2 33%, or 67%, of the base. Therefore, the rate is 67% and the percentage is $54.94. The base is to be found. The proportion is

$54.94 67 5 . B 100 $54.94 3 100 5 67B $5494 5 67B $5494 5B 67 $5494 B5 5 $82 Ans 67

The list price for this set of socket wrenches is $82.

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153

Percent Practical aPPlications

Example 3 An aluminum bar measures 137.168 millimeters before it is heated. When heated, the bar measures 137.195 millimeters. What is the percent increase in length? Express the answer to 2 decimal places. Think the problem through. The base (100%) is the bar length before heating, 137.168 millimeters. The increase in length is 137.195 millimeters 2 137.168 millimeters, or 0.027 millimeter. Therefore, the percentage is 0.027 millimeter, and the base is 137.168 millimeters. The rate (percent) is to be found. 0.027 mm R The proportion is 5 . 137.168 mm 100 0.027 mm 3 100 5 s137.168 mmdR 2.7 mm 5 s137.168 mmdR 2.7 mm 5R 137.168 mm 2.7 mm R5 5 0.01968%, or 0.02% Ans sroundedd 137.168 mm Heating the aluminum bar increased its length by 0.02%.

ApplicAtion Tooling Up 1. What is 7.25% of 43.80? Round your answer to 2 decimal places. 2. Express 3.5% as a common fraction. 3. Solve the proportion 4. Compute

Î

A 27 5 . If necessary, round the answer to 2 decimal places. 12 45

12 . Round the answer to 3 decimal places. 453

5. Divide: 456.2 4 9.42. Round the answer to 3 decimal places. 6. Write the common fraction

123 as a decimal fraction rounded to 4 decimal places. 654

Finding Percentage, Percent, and Base in Practical Applications Solve the following problems. 7. The total amount of time required to machine a part is 12.5 hours. Milling machine operations take 7.0 hours. What percent of the total time is spent on the milling machine? 8. A casting, when first poured, is 17.875 centimeters long. The casting shrinks 0.188 centimeter as it cools. What is the percent shrinkage? Round the answer to 2 decimal places. 9. A machine operator completes a job in 80% of the estimated time. The estimated time 1 is 8 hours. How long does the job actually take? 2 10. A machine is sold for 42% of the original cost. If the original cost is $9255.00, find the selling price of the used machine.

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SecTIoN 2

ratio, ProPor tion, and Percentage

11. On a production run, 8% of the units manufactured are rejected. If 120 units are rejected, how many total units are produced? 12. An engine loses 4.2 horsepower through friction. The power loss is 6% of the total rated horsepower. What is the total horsepower rating? 13. A small manufacturing plant employs 130 persons. On certain days, 16 employees are absent. What percent of the total number of employees are absent? Round the answer to the nearest whole percent. 14. This year’s earnings of a company are 140% of last year’s earnings. The company earned $910,000 this year. How much did the company earn last year? 15. In 3 hours, 73.50 feet of railing are fabricated. This is 28% of a total order. How many feet of railing were ordered? 16. How many pounds of manganese bronze can be made with 955.0 pounds of copper if the manganese bronze is to contain 58% copper by weight? Round the answer to the nearest whole pound. 17. An alloy of manganese bronze is made up by weight of 58% copper, 40% tin, 1.5% manganese, and 0.5% other materials. How many pounds of each metal are there in 1250 pounds of alloy? Round the answers to the nearest whole pound. 18. A manufacturer estimates the following percent costs to produce a product: labor, 38%; materials, 45%; overhead, 17%. The total cost of production is $120,000. Determine each of the dollar costs. 1 inch per foot. What is the percentage shrinkage? 8 Round the answer to the nearest whole percent.

Copper: Tin: Manganese: Other: Labor: Materials: Overhead:

19. An iron casting shrinks

20. A hot brass casting when first poured in a mold is 9.25 inches long. The shrinkage is 1.38%. What is the length of the casting when cooled? Round the answer to 2 decimal places. 21. The following table shows the number of pieces manufactured in three consecutive days. The numbers of defective pieces are shown as rework and scrap for each day. Determine the percents of rework and scrap for each day. Round the answers to 1 decimal place. Number of Defective Pieces

Date

Number of Pieces manufactured

rework

scrap

9/16

1650

44

59

9/17

1596

29

48

9/18

1685

52

34

% Defective Pieces rework

scrap

22. The power output of a machine is equal to the product of the power input and the percent efficiency. Power Output 5 Power Input 3 Percent Efficiency. What is the output of a machine with an 8.0-horsepower motor running at full capacity and at 82% efficiency? Round the answer to 1 decimal place. 23. Material cost for a job is $1260. The cost is 38.6% of the total cost. What is the total cost? Round the answer to the nearest dollar.

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UNIT 24

155

Percent Practical aPPlications

24. A machinist’s weekly gross income is $745. The following percent deductions are made from the gross income: Federal Income Tax, 14.20% State Income Tax, 4.50% Social Security, 7.60% Determine the net income (take-home pay) after the deductions are made. 25. In the heat treatment of steel, a rough approximation of temperatures can be made by observing the color of the heated steel. At approximately 13008F (degrees Fahrenheit) the steel is dark red. What percent increase in temperature from the 13008F must be made for the heated steel to turn to each of the following colors? Round the answers to the nearest whole percent. a. Dull cherry-red at approximately 14708F b. Orange-yellow at approximately 22008F c. Brilliant white at approximately 27308F 26. The following table lists the percent of carbon by weight for various types of carbon steel tools. Determine the amount of carbon needed to produce 2.60 tons of carbon steel required in the production of each type of tool. Round the answers to the nearest pound. type of machinist’s tool

Percent Carbon

type of machinist’s tool

Percent Carbon

1. 2.

1. Twist Drill

1.15

4. Ordinary file

1.25

3.

2. Wrench

0.75

5. Machinist’s hammer

0.95

4.

3. Threading Die

1.05

6. Chuck jaw

0.85

5. 6.

27. A machine shop has 2600 castings in stock at the beginning of the month. At the end of the first week, 28.0% of the stock is used. At the end of the second week, 50.0% of the stock remaining is used. How many castings remain in stock at the end of the second week? 28. It is estimated that 125 meters of channel iron are required for a job. Channel iron is ordered, including an additional 20% allowance for scrap and waste. Actually, 175 meters of channel iron are used for the job. The amount actually used is what percent more than the amount ordered? Round the answer to the nearest whole percent. 29. An alloy of red brass is composed of 85% copper, 5% tin, 6% lead, and zinc. Find the number of pounds of zinc required to make 450 pounds of alloy. 30. The day shift of a manufacturing firm produces 6% defective pieces out of a total 1 production of 1638 pieces. The night shift produces 4 % defective pieces out of a 2 total of 1454 pieces. How many more acceptable pieces are produced by the day shift than by the night shift? 31. The following table shows the number of pieces of a product produced each day during one week. Also shown are the number of pieces rejected each day by the quality control department. Find the percent rejection for the week’s production. Round the answer to 1 decimal place.

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moN.

tUes.

WeD.

tHUr.

Fri.

Number of Pieces Produced

735

763

786

733

748

Number of Pieces Rejected

36

43

52

47

31

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ratio, ProPor tion, and Percentage

32. A manufacturer estimates that 15,500 pieces per day could be produced with the installation of new machinery. The machines now used produce 11,000 pieces per day. What percent increase in production would be gained by replacing the present machinery with new machinery? Round the answer to the nearest whole percent. 33. The average percent defective product of a manufacturing plant is 1.20%. On a particular day 50 pieces were rejected out of a total daily production of 2730 pieces. What is the percent increase of defective pieces for the day above the average percent defective? Round the answer to 2 decimal places. 34. Before machining, a steel forging weighs 7.8 pounds. A milling operation removes 1.5 pounds, drilling removes 0.7 pound, and grinding removes 0.5 pound. What percent of the original weight of the forging is the final machined forging? Round the answer to 1 decimal place. 35. A machine is 85% efficient and loses 1.3 horsepower through its drivetrain. Determine the horsepower input of the machine. Round the answer to 1 decimal place. 36. The cost of one dozen cutters is listed as $525. A multiple discount of 12% and 8% is applied to the purchase. Determine the net (selling) price of the cutters.

Note: With multiple discounts, the first discount is subtracted from the list price. The second discount is subtracted from the price computed after the first discount was subtracted. 37. A manufacturer’s production this week is 3620 pieces. This is 13.5% greater than last week’s production. Find last week’s production. Round the answer to the nearest whole piece. 38. Two machines are used to produce the same product. One machine has a capability of producing 750 pieces per 8-hour shift. It is operating at 80% of its capability. The second machine has a capability of producing 900 pieces per 8-hour shift. It is operating at 75% of its capability. Find the total number of pieces produced per hour with both machines operating. Round the answer to the nearest whole piece. 39. Allowing for scrap, a firm produced 1890 pieces. The number produced is 8% more than the number of pieces required for the order. How many pieces does the order call for? Round the answer to the nearest whole piece. 40. A manufacturing company receives $122,000 upon the completion of a job. Total expenses for the job are $110,400. What percent of the job is profit? Round the answer to 1 decimal place. 41. Manufacturing costs consist of labor costs, material costs, and overhead. Refer to the following table. What percent of the total manufacturing cost for each of Jobs 1, 2, and 3 is each manufacturing cost? Round the answer to the nearest whole percent. maNUFaCtUriNG Costs Job

Labor Costs

material Costs

overhead Costs

1

$1890

$ 875

$1240

2

$ 930

$1060

$ 880

3

$2490

$1870

$1600

1. Labor: Materials: Overhead: 2. Labor: Materials: Overhead: 3. Labor: Materials: Overhead:

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157

achievement review—section two

UNIT 25 Achievement Review— Section Two

Objective You should be able to solve the exercises and problems in this Achievement Review by applying the principles and methods covered in Units 20–24.

Express these ratios in lowest fractional form. 1 1 to 4 2 2 8. 16 to 3 1 9. h to 25 min 4

1. 15 : 32

7.

2. 46 : 12 3. 12 : 46 4. 27 mm : 45 mm 5. 21 ft : 33 ft

10. 2 ft to 8 in.

6. 45 in. : 27 in.

11. The cost and selling price of merchandise are listed in the following table. Determine the cost-to-selling price ratio and the cost-to-profit ratio.

Note: Profit 5 Selling Price 2 Cost.

Cost

selling Price

a.

$ 60

$ 96

b.

$105

$180

c.

$ 18

$ 33

d.

$204

$440

ratio of Cost to selling Price

ratio of Cost to Profit

12. Bronze is an alloy of copper, zinc, and tin with small amounts of other elements. Two types of bronze castings are listed in the table below with the percent composition of copper, tin, and zinc in each casting. Determine the ratios called for in the table. Percent Composition

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type of Casting

Copper

tin

Zinc

a.

Manganese Bronze

58

1

40

b.

Hard Bronze

86

10

2

ratios Copper to tin

tin to Zinc

Copper to zinc

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ratio, ProPor tion, and Percentage

13. Solve for the unknown value in each of the following proportions and check. Round the answers to 3 decimal places where necessary. P 3 5 12.8 2 3.6 E b. 5 0.9 2.7 1 H 2 c. 5 1 3 4 8 7 C d. 5 6 12 a.

e.

6.5 8.2 5 M 41

f.

22.517 1.297 5 x 13.503

g.

20.021 1.892 5 5.773T 4.518

h.

10.360 2.015 5 7.890 N

14. Analyze each of the following problems to determine whether the problem is a direct or inverse proportion and solve. Round the answers to 3 decimal places where necessary. a. A reamer tapers 0.0975 inch along a 3.2625-inch length. What is the amount of taper along a 2.1250-inch length? b. A machine produces 2550 parts in 8.5 hours. How many parts are produced by the machine in 10 hours? c. Of two gears that mesh, one gear with 12 teeth revolves at 420 rpm. What is the revolutions per minute of the other gear, which has 16 teeth? 15. Express each value as a percent. a. 1

b. 1

1 2

c. 2

16. Express each value as a percent. a. 0.72

b. 2.037

c.

3 4

d. 0.5

1 25

d. 0.0003

3 % 4

d. 310

17. Express each percent as a decimal fraction or mixed decimal. a. 19%

b. 0.7%

c.

18. Express each percent as a common fraction or mixed number. a. 30% b. 140% c. 12.5%

3 % 10

d. 0.65%

19. Find each percentage. Round the answers to 2 decimal places when necessary. a. 15% of 60 b. 3% of 42.3 c. 72.8% of 120 d. 0.7% of 812 e. 42.6% of 53.76

f. 130% of 212 1 g. 12 % of 32 2 1 h. % of 627.3 4

20. Find each percent (rate). Round the answers to 2 decimal places when necessary. a. What percent of 10 is 2? b. What percent of 2 is 10? c. What percent of 88.7 is 21.9? d. What percent of 275 is 108?

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e. What percent of 2.84 is 0.8? 1 f. What percent of 12 is 3? 4 g. What percent of 312 is 400.9?

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159

achievement review—section two

21. Find each base. Round the answers to 2 decimal places when necessary. a. 20 is 60% of what number? b. 4.1 is 24.9% of what number? c. 340 is 152% of what number? d. 44.08 is 73.5% of what number?

e. 9.3 is 238.6% of what number? f. 0.84 is 2.04% of what number? 3 g. is 123% of what number? 4

22. Find each percentage, percent (rate), or base. Round the answers to 2 decimal places when necessary. a. What percent of 24 is 18? ? g. 72.4% of 212.7 is b. What is 30% of 50?

. h. What percent of 228 is 256?

c. What is 123.8% of 12.6?

i. 51.08 is 88% of what number?

d. 73 is 82% of what number? 1 e. What percent of 10 is 2? 2 ? f. is 48% of 94.82.

? j. 36.5 is % of 27.6. 1 ? k. 2 % of 150 is . 4 ? l. is 18% of 120.66.

23. The carbon content of machine steel for gages usually ranges from 0.15% to 0.25%. Round the answers for a and b to 2 decimal places. a. What is the minimum weight of carbon in 250 kilograms of machine steel? b. What is the maximum weight of carbon in 250 kilograms of machine steel? 24. A piece of machinery is purchased for $8792. In 1 year, the machine depreciates 14.5%. By how many dollars does the machine depreciate in 1 year? Round the answer to the nearest dollar. 25. Engine pistons and cylinder heads are made of an aluminum casting alloy that contains 4% silicon, 1.5% magnesium, and 2% nickel. Round the answers to the nearest tenth kilogram. a. How many kilograms of silicon are needed to produce 575 kilograms of alloy? b. How many kilograms of magnesium are needed to produce 575 kilograms of alloy? c. How many kilograms of nickel are needed to produce 575 kilograms of alloy? 26. Before starting two jobs, a shop has an inventory of eighteen 15.0-foot lengths of flat stock. The first job requires 30% of the inventory. The second job requires 25% of the inventory remaining after the first job. How many feet of flat stock remain in inventory at the end of the second job? Round the answer to the nearest whole foot. 27. An alloy of stainless steel contains 73.6% iron, 18% chromium, 8% nickel, 0.1% carbon, and sulfur. How many pounds of sulfur are required to make 5800 pounds of stainless steel? Round the answer to the nearest whole pound. 28. Two machines together produce a total of 2015 pieces. Machine A operates for 1 6 hours and produces an average of 170 pieces per hour. Machine B operates for 2 7 hours. What percent of the average hourly production of Machine A is the average hourly production of Machine B? Round the answer to the nearest whole percent.

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3

Linear Measurement: Customary (English) and Metric

SECTION THREE

UNIT 26 Customary (English) Units of Measure ObjECTIvES After studying this unit you should be able to ●● ●●

Express customary lengths as larger or smaller customary linear units. Perform arithmetic operations with customary linear units and compound numbers.

The United States uses two systems of weights and measures: the American or U.S. customary system and the International System of Units, called the SI metric system. The American customary system is based on the English system of weights and measures and is sometimes called the “English” system. Throughout this book, American customary units are called “customary” units and SI metric units are called “metric” units. Both customary and metric systems include all types of units of measure, such as length, area, volume, and capacity. It is important that you have the ability to measure and compute with both customary and metric units. In the machine trades, linear or length measure is used most often. Throughout this book, linear measure is the primary type of measure presented. However, in Section 6, some fundamentals of area and volume and their applications are presented.

MeasureMent Definitions Measurement is the comparison of a quantity with a standard unit. A linear measurement is a means of expressing the distance between two points; it is the measurement of lengths. A linear measurement has two parts: a unit of length and a multiplier.

160

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UNIT 26

161

Customary (English) units of mEasurE

3.872 inches multiplier

27.18 millimeters

unit of length

multiplier

unit of length

The measurements 3.872 inches and 27.18 millimeters are examples of denominate numbers. A denominate number is a number that specifies a unit of measure.

CustoMary units of Linear Measure The yard is the standard unit of linear measure in the customary system. From the yard, other units such as the inch and foot are established. The smallest unit is the inch. Common customary units of length with their symbols are shown in the following table. CUSTOMARY UNITS OF LINEAR MEASURE 1 yard (yd) 5 3 feet (ft) 1 yard (yd) 5 36 inches (in.) 1 foot (ft) 5 12 inches (in.) 1 mile (mi) 5 1760 yards (yd) 1 mile (mi) 5 5280 feet (ft)

In the machine trades, customary linear units other than the inch are seldom used. Customary measure dimensions on engineering drawings are given in inches. Although customary linear units other than the inch are rarely required for on-the-job applications, you should be able to use any units in the system. Notice that most of the symbols, ft for foot, mi for mile, yd for yard, do not have periods at the end. That is because they are symbols and not abbreviations. The one exception is in. for inch. Many people prefer in. because the period helps you know that they do not mean the word “in.” Both in and in. are correct. We will use the symbol in. for inch in this text.

expressing equivaLent units of Measure When expressing equivalent units of measure, either of two methods can be used. Throughout the unit, examples are given using either of the two methods. Many examples show how both methods are used in expressing equivalent units of measure.

METHOD 1

This is a practical method used for many on-the-job applications. It is useful when simple unit conversions are made. In this method you multiply or divide the given unit of measure by a conversion factor.

METHOD 2 This method is called the unity fraction method. The unity fraction method eliminates the problem of incorrectly expressing equivalent units of measure. Using this method removes any doubt as to whether to multiply or divide when making a conversion. The unity fraction method is particularly useful in solving problems that involve a number of unit conversions. This method multiplies the given unit of measure by a fraction equal to 1. The unity fraction contains the given unit of measure and its equivalent expressed in the unit of measure to which the given unit is to be converted. The unity fraction is set up in such a way that the original unit cancels out and the unit you are converting to remains. Recall that canceling is the common term used when a numerator and a denominator are divided by a common factor.

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SecTIoN 3 linEar mEasurEmEnt: Customary (English) and mEtriC

expressing Larger CustoMary units of Linear Measure as sMaLLer units c Procedure ●●

●●

To express a larger unit as a smaller unit of length, either

Multiply the given length by the number of smaller units contained in one of the larger units (Method 1), or Multiply the given length by an appropriate unity fraction (Method 2).

Example 1 Express 2.28 yards as inches. METHOD 1 Since 36 inches equal 1 yard, the conversion factor is 36. Multiply 2.28 by 36.

2.28 3 36 5 82.08 2.28 yd ø 82.1 inches

METHOD 2

Ans

36 in. . 1 yd 36 in. 2.28 yd 3 5 2.28 3 36 in. 1 yd

Since 36 inches equal 1 yard, the unity fraction is Multiply 2.28 yd by the unity fraction.

< 82.1 in. Ans 1 Example 2 Express 2 ft as inches. 2

METHOD 1 Since 12 inches equal 1 foot, the conversion factor is 12. 1 1 Multiply 2 by 12. 2 3 12 5 30 2 2 1 2 feet 5 30 inches Ans 2

METHOD 2

12 in. . 1 ft 1 12 in. 1 2 ft 3 5 2 3 12 in. 5 30 inches Ans 2 1 ft 2

Since 12 inches equal 1 foot, the unity fraction is Multiply 2

1 ft by the unity fraction. 2

Example 3 How many inches are in 0.25 yard? METHOD 1 Since 36 inches equal 1 yard, multiply 0.25 by 36.

0.25 3 36 5 9 0.25 yard 5 9 inches

METHOD 2

Ans

36 in. . 1 yd 36 in. 0.25 yd 3 5 0.25 3 36 in. 1 yd

Since 36 inches equal 1 yard, the unity fraction is Multiply 0.25 yd by the unity fraction.

5 9 inches

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Ans

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UNIT 26

163

Customary (English) units of mEasurE

expressing sMaLLer CustoMary units of Linear Measure as Larger units c Procedure ●●

●●

To express a smaller unit as a larger unit of length, either

Divide the given length by the number of smaller units contained in one of the larger units (Method 1), or Multiply the given length by an appropriate unity fraction (Method 2).

Example 1 Express 67.2 inches as feet. METHOD 1 Since 12 inches equal 1 foot, divide 67.2 by 12.

67.2 4 12 5 5.6 67.2 inches 5 5.6 feet

Ans

METHOD 2 1 ft . 12 in. 1 ft 67.2 ft 67.2 in. 3 5 5 5.6 feet Ans 12 in. 12

Since 12 inches equal 1 foot, the unity fraction is Multiply 67.2 inches by the unity fraction.

Example 2 How many yards are in 122.4 inches? METHOD 1 Since 36 inches equal 1 yard, divide 122.4 by 36.

122.4 4 36 5 3.4 122.4 inches 5 3.4 yards

MeTHod 2 Since 36 inches equal 1 yard, the unity fraction is Multiply 122.4 inches by the unity fraction.

Ans

1 yd . 36 in.

122.4 in. 3

1 yd 122.4 yd 5 36 in. 36 53.4 yards Ans

CustoMary units of Linear Measure WitH tHe Machinist calc Pro 2 In this section we will introduce one calculator key and one key combination that are used with the Machinist Calc Pro 2 to indicate customary units. One of these, the Inch key, identifies that an entry is in inches and, with repeated presses, toggles among linear, area, and volume units. The Conv 7 key combination identifies that an entry is in feet. It is also used with Inch and / for entering feet-inch values. Notice that there is no key for yards.

expressing Customary units Example Enter 23980 in Machinist Calc Pro 2.

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164

SecTIoN 3 linEar mEasurEmEnt: Customary (English) and mEtriC

Solution Press On/C On/C to clear the calculator. Then press the calculator screen in Figure 26-1.

23 Conv

7

8

Inch

. You should see

figurE 26-1

To change one unit of customary linear measure to another, first enter the customary measure, then the Conv key, followed by the key of the desired measure. It is not necessary to press the 5 key.

Example 1 Convert 9 feet 8 inches to inches. Solution 9

Conv

7

8

Inch

Conv

Inch

5 116 inches

Ans

Example 2 Convert 237 inches to feet. Solution 237

Inch

Conv

7

19.75 feet

Ans

aritHMetiC operations WitH Linear units In order to add two measurements, they must be in the same units. The addition of unlike units cannot be performed unless one of the measurements is converted to the other unit. For example, 9 inches and 4 inches can be added. Both units are in inches. But 7 inches and 5 feet cannot be added unless 7 inches is expressed in feet or 5 feet is expressed in inches.

c Procedure following: ●● ●●

To add or subtract measures in the same units, do the

Add or subtract the numerical values. Leave the units unchanged.

Example 1 Add 5 in. and 13 in. 5 in. 13 in. __________ 18 in. Ans 1 2

1 4

Example 2 Subtract 7 in. from 21 in. 1 5 21 in. 5 20 in. 4 4 1 2 7 in. 5 7 in. 2 4 _______________________ 3 13 in. Ans 4

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addition and subtraction of Compound numbers A compound number is a quantity that is expressed in terms of two or more different units, such as 3 feet 4 inches. To add or subtract compound numbers, arrange like units in the same column, then add each column. When necessary, simplify the answer. 3 4 3 ft

1 2

Example Add 3 feet and 9 inches, 2 feet 8 inches, and 2 feet 10 inches. Arrange like units in the same column.

Add each column.

9 in. 3 2 ft 8 in. 4 1 2 ft 10 in. 2 ___________ 1 7 ft 28 in. 4

1 1 1 Simplify the sum. Divide 28 by 12 to express 28 inches as 2 feet 4 inches. 4 4 4 1 1 28 inches 5 2 feet 4 inches 4 4 1 1 Add. 7 feet 1 2 feet 4 inches 5 9 feet 4 inches Ans 4 4

Multiplication of Compound numbers To multiply compound numbers, multiply each unit of the compound number by the multiplier. When necessary, simplify the product.

Example Six pieces are to be cut from a piece of bar stock. Each piece is 1 foot

3 1 5 inches long. Allow inch for cutting each piece. Determine the total length of 4 8 the pieces and cuts that are removed. 3 1 7 Add the length of each piece and the cut. 1 foot 5 inches 1 inch 5 1 foot 5 inches 4 8 8 7 1 foot 5 inches 8 Multiply this sum by 6. 3 6 _____________________ 42 inches 8 1 6 feet 35 inches 4 1 8 feet 11 inches Ans 4 6 feet 30

Simplify the product.

Division of Compound numbers To divide compound numbers, divide each unit of the divisor starting at the left. If a unit is not exactly divisible, express the remainder as the next smaller unit and add it to the given number of smaller units. When necessary, simplify the quotient.

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Example The five holes in the angle iron shown in Figure 26-2 are equally spaced. Determine the distance between two consecutive holes. 149-60

figurE 26-2

Since there are four spaces between holes, divide 14 feet 6 inches by 4. Divide 14 feet by 4.

14 ft 4 4 5 3 ft (quotient) and a 2-ft remainder.

Express the 2-foot remainder as 24 inches.

2 ft 5 2 3 12 in. 5 24 in.

Add 24 inches to the 6 inches given in the problem.

24 in. 1 6 in. 5 30 in. 1 30 in. 4 4 5 7 in. (quotient) 2 1 3 ft 7 in. Ans 2

Divide 30 inches by 4. Collect quotients.

aritHMetiC operations WitH CustoMary Linear units on tHe Machinist calc Pro 2 In order to add two measurements, they must be in the same units. Normally 7 inches and 5 feet cannot be added unless either 7 inches is expressed in feet or 5 feet is expressed in inches. This is not necessary with the Machinist Calc Pro 2. 1 1 Example Add 7 inches and 21 inches. 2 4

Solution Use the following keystrokes on the Machinist Calc Pro 2. On/C

On/C

21

1 /

Inch

4 1 7

Inch

1 /

3 2 5 28 inches 4

Ans

addition and subtraction of Compound numbers A compound number is a quantity that is expressed in terms of two or more different units, such as 3 feet 4 inches. To add or subtract compound numbers on paper, arrange like units in the same column, then add each column. When necessary, simplify the answer. With the Machinist Calc Pro 2, this is not necessary.

Example 3 1 Add 3 feet 9 inches, 2 feet 8 inches, and 2 feet 10 inches. 4 2

Solution Conv

7

On/C

10

1 2 Conv 1 1 / 2 5 9 feet 4 inches Ans 4

On/C

Inch

3

Conv

7

9

Inch

7

8

Inch

3 / 4 1 2

The Machinist Calc Pro 2 also does a nice job of subtracting compound numbers, as is shown in the next example.

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1 8

Example Subtract 12 feet 3 inches 27 feet 5 Solution

On/C

On/C

/ 32 5 4 feet 9

12

Conv

23 inches 32

7

3

Inch

13 inches. 32 1 /

8 2 7

Conv

7

5

Inch

13

Ans

Multiplication of Compound numbers To multiply compound numbers, multiply each unit of the compound number by the multiplier. When necessary, simplify the product. Again, the Machinist calc Pro 2 does this nicely and simplifies the product automatically. 3 Example Six pieces are to be cut from a piece of bar stock. Each piece is 1 foot 5 4 1 inches long. Allow inch for cutting each piece. What is the shortest length of the bar 8 that is needed?

Solution Add the length of each piece and the cut. 1 Conv 7 5 Multiply this sum by 6.

On/C

On/C

1 3 6 5 8 feet 11 inches 4

Inch

3 /

4 1 1 / 8 5

Ans

Division of Compound numbers To divide compound numbers by hand, divide each unit of the divisor starting at the left. If a unit is not exactly divisible, express the remainder as the next smaller unit and add it to the given number of smaller units. When necessary, simplify the quotient. As with the other operations, the Machinist Calc Pro 2 does this easily and simplifies the quotient for you.

Example The five holes in the angle iron shown in Figure 26-2 of the text are equally spaced. Determine the distance between two consecutive holes.

Solution Since there are four spaces between holes, divide 14 feet 6 inches by 4. On/C On/C

14

Conv

7

6

Inch

1 4 4 5 3 ft 7 inches 2

Ans

ApplicAtion Tooling Up 1. A mechanic purchases a drill and impact driver kit that is listed for $139.95. He has a coupon for 15% off. a. How much is the discount? b. How much is the purchase price of the kit? 2. 53.6 is 82% of what number? If necessary, round the answer to 2 decimal places. 3. Of two gears that mesh, the one with the greater number of teeth is called the gear and the one with the fewer number of teeth is the pinion. In a certain setup, a gear has 36 teeth and the pinion has 22 teeth. If the speed of the gear is 140 rpm, what is the speed of the pinion? If necessary, round the answer to 1 decimal places.

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82.3 3 12.4 . Round the answer to 3 decimal places. 19.35 3 4.62 5. The volume V of a cone with height h and diameter d is given by V 5 0.2618 3 d2 3 h. What is the volume of a cone with diameter 72.4 cm and height 122.5 cm? Round the answer to 2 decimal places. 4. Compute Ï16.0025 1

6. Add: 432.78 1 43.278 1 4.3278.

customary Units of Linear Measure In Exercises 7 through 12, express each of the following lengths as indicated. 1 7. a. 96 inches as feet 10. a. yard as inches 3 b. 123 inches as feet b. 258 inches as feet 1 c. 3 feet as inches c. 42 feet as yards 2 1 d. 0.4 yard as inches d. 2 yards as inches 4 1 2 8. a. 1 yards as inches 11. a. 7 feet as inches 4 3 b. 144 inches as yards b. 0.20 yard as inches c. 75 inches as feet c. 140.25 feet as yards 3 d. 2 feet as inches d. 333 inches as yards 4 9. a. 8 yards as feet b. 4.2 yards as feet c. 27 feet as yards d. 51 feet as yards 13.

14. 15.

16.

12. a. 186 inches as feet 2 b. 20 yards as feet 3 c. 9.25 feet as inches d. 243 inches as yards 7 1 10 Pieces that are each 3 inches long are to be cut from an aluminum bar that is 4 feet long. Allow for cutoff for 8 2 16 each piece. How many complete pieces can be cut from this aluminum bar? 1 A 3 -inch diameter milling cutter revolving at 130.0 revolutions per minute has a cutting 2 speed of 120.0 feet per minute. What is the cutting speed in inches per minute? How many complete 6-foot lengths of round stock should be ordered to make 230 pieces 1 each 1.300 inches long? Allow 1 lengths of stock for cutoff and scrap. 2 Pieces each 3.25 inches long are to be cut from lengths of bar stock. Allowing 0.10 inch for cutoff per piece, how many complete pieces can be cut from twelve 8-foot lengths of stock?

Arithmetic operations with Linear Units In Exercises 17 through 20, add or subtract as indicated. Express each answer in the same unit as given in the exercise. 17. a. 3 in. 1 7 in. b. 12 in. 1 5 in. c. 15 in. 2 8 in. d. 12 in. 2 9 in. 1 3 18. a. 11 in. 1 4 in. 4 8 3 5 b. 7 in. 1 6 in. 4 8

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3 5 c. 7 in. 2 4 in. 4 8 1 11 d. 8 in. 2 4 in. 4 16 19. a. 5 ft 4 in. 1 2 ft 6 in. b. 3 ft 8 in. 1 4 ft 3 in. c. 3 ft 7 in. 2 2 ft 6 in. d. 5 ft 9 in. 2 2 ft 7 in.

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1 3 20. a. 7 ft 3 in. 1 5 ft 2 in. 2 8 3 7 b. 2 ft 5 in. 1 3 ft 4 in. 4 16

1 3 c. 7 ft 6 in. 2 6 ft 5 in. 2 4 7 5 d. 3 ft 9 in. 2 1 ft 7 in. 8 16

Arithmetic operations with compound Numbers In Exercises 21 and 22, perform the indicated operation. Express each answer in the same units as given in the exercise. Regroup the answer when necessary. 21. a. 3 ft 9 in. 1 4 ft 7 in. b. 6 ft 5 in. 2 4 ft 9 in. c. 7 ft 9 in. 3 3 d. 17 ft 11 in. 4 5 3 1 22. a. 6 ft 3 in. 1 4 ft 1 in. 8 2 3 1 8 ft 10 in. 4

3 7 b. 10 ft 1 in. 2 7 ft 9 in. 8 16 3 c. 12 ft 3 in. 3 5.25 4 1 d. 10 ft 6 in. 4 4 4

UNIT 27 Metric Units of Linear Measure ObjECTIvES After studying this unit you should be able to ●● ●● ●●

Express metric lengths as larger or smaller metric linear units. Express metric length units as customary length units. Express customary length units as metric length units.

An advantage of the metric system is that it allows easy and fast computations. Since metric units are based on powers of 10, conversions from one unit to another are simplified. To express a metric unit as a smaller or larger unit, all that is required is to move the decimal point a certain number of places to the left or right. The metric system does not require difficult conversions as with the customary system. For example, it is easier to remember that 1000 meters equal 1 kilometer than to remember that 1760 yards equal 1 mile. The meter is the standard unit of linear measure in the metric system. Other linear metric units are based on the meter. Metric measure dimensions on engineering drawings are given in millimeters. In the machine trades, metric linear units other than the millimeter are seldom used. However, you should be able to use any units in the system. Some metric units of length with their symbols are shown in this table. Observe that each unit is ten times greater than the unit directly above it.

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Metric Units of Linear Measure 1 millimeter (mm) 5 0.001 meter (m)

1000 millimeters (mm) 5 1 meter (m)

1 centimeter (cm) 5 0.01 meter (m)

100 centimeters (cm) 5 1 meter (m)

1 decimeter (dm) 5

0.1 meter (m)

1 meter (m) 5

1 meter (m)

10 decimeters (dm) 5 1 meter (m) 1 meter (m)

5 1 meter (m)

1 dekameter (dam) 5

10 meters (m)

0.1 dekameter (dam) 5 1 meter (m)

1 hectometer (hm) 5

100 meters (m)

0.01 hectometer (hm) 5 1 meter (m)

1 kilometer (km) 5 1000 meters (m)

0.001 kilometer (km)

5 1 meter (m)

The following metric power of 10 prefixes are used throughout the metric system: milli means one thousandth (0.001) centi means one hundredth (0.01) deci means one tenth (0.1)

deka means ten (10) hecto means hundred (100) kilo means thousand (1000)

The most frequently used metric units of length are the kilometer (km), meter (m), centimeter (cm), and millimeter (mm). In actual applications, the dekameter (dam) and hectometer (hm) are not used. The decimeter (dm) is seldom used. Periods are not used after the unit symbols. For example, write 1 mm, not 1 m.m. or 1 mm., when expressing the millimeter as a symbol. Some people spell the unit of measure metre so that it is not confused with a meter used as a measuring instrument. Be careful when using capital letters for metric units. For example, Mm and mm do not mean the same thing. A capital, or uppercase, M is the symbol for one million, so 5 Mm is the symbol for 5 000 000 meters. On the other hand, 5 mm means 5 millimeters, or 0.005 meter.

expressing equivaLent units WitHin tHe MetriC systeM To express a given unit of length as a larger unit, move the decimal point a certain number of places to the left. To express a given unit of length as a smaller unit, move the decimal point a certain number of places to the right. The procedure of moving decimal points is shown in the following examples. Refer to the table of metric units of linear measure.

Example 1 Express 72 millimeters (mm) as centimeters (cm). Since a centimeter is the next larger unit to a millimeter, move the decimal point 1 place to the left. (In moving the decimal point 1 place to the left, you are actually dividing by 10.)

72. 72 mm 5 7.2 cm

Ans

Example 2 Express 0.96 centimeter (cm) as millimeters (mm). Since a millimeter is the next smaller unit to a centimeter, move the decimal point 1 place to the right. (In moving the decimal point 1 place to the right, you are actually multiplying by 10.)

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0.96 0.96 cm 5 9.6 mm

Ans

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Example 3 Express 0.245 meter (m) as millimeters (mm). Since a millimeter is three units smaller than a meter, move the decimal point 3 places to the right. (In moving the decimal point 3 places to the right, you are actually multiplying by 10 3 or 1000.)

0.245 0.245 m 5 245 mm

Ans

Example 4 Add 0.3 meter (m) 1 12.6 centimeters (cm) 1 76 millimeters (mm). Express the answer in millimeters. Express each value in millimeters.

0.3 m 5 300 mm 12.6 cm 5 126 mm 76 mm 5 76 mm 502 mm Ans

MetriC units of Linear Measure WitH tHe Machinist calc Pro 2 In this section we will introduce one calculator key and two key combinations that are used with the Machinist Calc Pro 2 to indicate metric units. One of these, the mm key, identifies an entry as millimeters, and, with repeated presses, toggles among linear, area, and volume units. This key also converts dimensional value in other units to its equivalent value in millimeters and, with repeated presses, toggles between millimeters and meters. The Conv 5 key combination identifies that an entry is in centimeters and, with repeated presses, toggles among linear, area, and volume units. The Conv 9 key combination identifies that an entry is in meters and, with repeated presses, toggles among linear, area, and volume units.

expressing Metric units A metric linear measure is entered using either the millimeter key ( mm ) or the key combinations for centimeters ( Conv 5 ) or meters ( Conv 9 ).

Example Enter each of the following metric linear units in a Machinist calc Pro 2. a. 95 millimeters b. 8.7 centimeters c. 42 meters

Solution Don’t forget to begin work on each problem by pressing On/C On/C . Again, observe that the unit is displayed at the bottom of the calculator screen. Because this calculator cannot display lowercase letters, it shows mm, cm, and m as MM, CM, and M. Keystrokes

a.

Display

95 mm

(Continued )

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Keystrokes

b.

8.7 Conv

5

c.

42 Conv

9

Display

Metric measurement dimensions on engineering drawings are given in millimeters. In the machine trades, metric linear units other than the millimeter are seldom used. However, you should be able to use any units in the system.

expressing equivalent units Within the Metric system The Machinist Calc Pro 2 will convert among millimeters, centimeters, and meters by first entering the number in the given metric unit and then pressing the desired metric unit.

Example Express 572 centimeters (cm) as meters (m).

Solution The Machinist Calc Pro 2 handles this using the following keystrokes. 572 Conv 5 Conv 9 5.72 m Ans

arithmetic operations with Metric Linear units Normally, in order to add two measurements, they must be in the same units. Usually quantities such as 9 meters and 5 centimeters cannot be added unless 9 meters is expressed in centimeters or 5 centimeters is expressed in meters. This is not necessary with the Machinist Calc Pro 2.

Example 1 Add 5 meters and 13 meters and express the answer in both meters and millimeters.

Solution Use the following keystrokes to find the answer in meters. 5

Conv

1 13

9

And pressing

mm

Conv

9

5 18 m

Ans

will express the answer in millimeters.

18 000 mm

Ans

Example 2 Subtract 13 centimeters from 5 meters and express the answer in millimeters. Solution Use the following keystrokes for subtraction. 5

Conv

9

2 13

Conv

5

5 4.87 m

This answer is in meters, so press

mm

Ans

to get the answer in millimeters.

4870 mm

Ans

Example 3 Multiply 8.46 cm 3 5. Solution 8.46

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Conv

5

3 5 5 42.3 cm

Ans

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Example 4 Divide 47.275 m 4 6.2. Express the answer in centimeters. Solution 47.275 Conv 9 4 6.2 5 This gives a result of 7.625 m. Now, convert this result to centimeters. Conv 5 762.5 cm Ans

MetriC–CustoMary Linear equivaLents (Conversion faCtors) Since both the customary and metric systems are used in this country, it is sometimes necessary to express equivalents between systems. Dimensioning an engineering drawing with both customary and metric dimensions is called dual dimensioning. Since dual dimensioning tends to clutter a drawing and introduces additional opportunities for error, many companies do not use the system. Instead, some companies use metric dimensions only, with an inch–millimeter conversion table on or attached to the print. However, certain companies use dual dimensioning; it can be a practical method for industries that have plants in foreign countries. Examples of two types of dual dimensioning are shown in Figure 27-1. METRIC SYSTEM

METRIC SYSTEM 38.10 mm

38.10

1.500 in.

1.500 CUSTOMARY SYSTEM

CUSTOMARY SYSTEM

figurE 27 -1

The commonly used equivalent factors of linear measure are shown in this table. Equivalent factors are commonly called conversion factors. METRIC–CUSTOMARY LINEAR EQUIVALENTS (CONVERSION FACTORS) Metric to Customary Units

Customary to Metric Units

1 millimeter (mm) 5 0.03937 inch (in.)

1 inch (in.) 5 25.4 millimeters (mm)

1 centimeter (cm) 5 0.3937 inch (in.)

1 inch (in.) 5 2.54 centimeters (cm)

1 meter (m) 5 39.37 inches (in.) 1 meter (m) 5 3.2808 feet (ft) 1 kilometer (km) 5 0.6214 mile (mi)

1 foot (ft) 5 0.3048 meter (m) 1 yard (yd) 5 0.9144 meter (m) 1 mile (mi) 5 1.609 kilometers (km)

Metric–customary linear equivalents other than millimeter–inch equivalents are seldom used in the machine trades. However, you should be able to express any unit in one measuring system as a unit in the other system. The relationship between customary decimal inch units and metric millimeter units is shown in Figure 27-2 by comparing these scales.

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METRIC SCALE (MILLIMETERS) 1 mm = 0.03937 in. 10 mm

20 mm

30 mm

40 mm

1 in.

50 mm

60 mm

70 mm

2 in.

80 mm

3 in.

90 mm

100 m

4i

DECIMAL INCH SCALE

figurE 27-2

c Procedure To express a unit in one system as an equivalent unit in the other system, either use unity fractions or multiply the given measurement by the appropriate conversion factor in the Metric–Customary Linear Equivalents table. Example 1 Express 12.700 inches as millimeters. METHOD 1 Since 1 in. 5 25.4 mm, 12.700 3 25.4 mm 5 322.58 mm Ans

METHOD 2 Multiply by the unity fraction 12.700 in. 3

25.4 mm . 1 in.

25.4 mm 5 322.58 mm Ans 1 in.

Example 2 Express 6.780 centimeters as inches. Round the answer to 3 decimal places. METHOD 1 Since 1 cm 5 0.3937 in., 6.78 3 0.3937 in. 5 2.669 in. Ans

MeTHod 2 Multiply by the unity fraction 6.78 cm 3

1 in. . 2.54 cm

1 in. 6.78 in. 5 5 2.669 in. Ans 2.54 cm 2.54

Example 3 The template in Figure 27-3 is dimensioned in millimeters. Determine, in inches, the total length of the template. Round the answer to 3 decimal places. 96.73

120.30

138.15

figurE 27-3

Add the dimensions in millimeters as they are given and express the sum in inches.

MeTHod 1 Since 1 mm 5 0.03937 in., multiply the sum in millimeters by 0.03937. ( 96.73 1 120.3 1 138.15 ) 1 0.03937 5 13.983437, 13.983 in. Ans

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METHOD 2

1 in. . 25.4 mm (96.73 mm 1 120.3 mm 1 138.15 mm)

Multiply by the unity fraction

(

96.73 1 120.3 1 138.15

)

1 in. 25.4 mm

4 25.4 5 13.983465, 13.983 in.

Ans

Example 4 Express 453.7 millimeters as yards. There is no conversion factor in the table for millimeters to yards, so we will have to use two conversion factors. First, convert millimeters to inches and then convert inches to 1 yd 0.03937 in. yards. The conversion factors are and . 1 mm 36 in. 1 yd 453.7 3 0.03937 yd 453.7 mm 0.03937 in. 3 3 5 < 0.5 yd Ans 1 1 mm 36 in. 36 Another way to solve this problem is to first convert millimeters to meters and then convert 1 yd 1m meters to yards. The conversion factors in this case are and . 1000 mm 0.9144 m 1 yd 453.7 yd 453.7 mm 1m 3 3 5 < 0.5 yd Ans 1 1000 mm 0.9144 m 1000 3 0.9144

ApplicAtion Tooling Up 1. Express 198 inches as feet. 2. Of the 2478 pieces manufactured one day, the inspector rejected 37 of them as defective. What percent of the pieces manufactured that day were defective? Round your answer to 1 decimal point. 3. 179.3 is 132% of what number? If necessary, round the answer to 2 decimal places. 4 4. Use a calculator to determine Ï 5793.25. Round your answer to 4 decimal places.

5. A square with an area A has a side of length s given by s 5 ÏA. What is the length of a square with an area of 156 sq ft? Round your answer to 2 decimal places. 6. Multiply 97.125 3 0.26.

Metric Units of Linear Measure In Exercises 7 through 12, express each of the given lengths as indicated. 7. a. 2.9 centimeters as millimeters b. 15.78 centimeters as millimeters c. 219.75 millimeters as centimeters d. 97.83 millimeters as centimeters

9. a. 673 centimeters as meters b. 0.93 millimeter as centimeters c. 0.08 centimeter as millimeters d. 0.0086 meter as millimeters

8. a. 0.97 meter as centimeters b. 0.17 meter as millimeters c. 153 millimeters as meters d. 23.85 centimeters as millimeters

10. a. 1.046 meters as centimeters b. 30.03 centimeters as millimeters c. 2003.6 millimeters as meters d. 135.06 centimeters as meters

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176

11. a. 876.84 millimeters as centimeters b. 2039 millimeters as meters c. 3.47 centimeters as meters d. 0.049 meter as millimeters

SecTIoN 3 linEar mEasurEmEnt: Customary (English) and mEtriC

12. a. 7.321 meters as centimeters b. 6.377 centimeters as millimeters c. 0.934 meter as millimeters d. 31.75 meters as centimeters

In Exercises 13 through 22, perform the indicated operations. Express the answer in the indicated unit. 13. 25.73 mm 1 7.6 cm 5 ? mm

18. 0.793 m 2 523.8 mm 5 ? mm

14. 3.7 m 1 98 cm 5 ? m

19. 214 mm 1 87.6 cm 1 0.9 m 5 ? m

15. 59.6 cm 2 63.7 mm 5 ? cm

20. 0.056 m 1 4.93 cm 1 57.3 mm 5 ? mm

16. 184.8 mm 2 12.3 cm 5 ? mm

21. 54.4 mm 1 5.05 cm 1 204.3 mm 5 ? mm

17. 1.06 m 2 43.7 cm 5 ? cm

22. 3.927 m 2 812 mm 5 ? m

23. An aluminum slab 0.082 meter thick is machined with three equal cuts; each cut is 10 millimeters deep. Determine the finished thickness of the slab in millimeters. 24. A piece of sheet metal is 1.12 meters wide. Strips each 3.4 centimeters wide are cut. Allow 3 millimeters for cutting each strip. a. Determine the number of complete strips cut. b. Determine the width of the waste strip in millimeters.

Metric–customary Linear equivalents In Exercises 25 through 28, express each of the given customary units of length as the indicated metric unit of length. Where necessary round the answer to 3 decimal places. 25. a. 37.000 millimeters as inches b. 126.800 millimeters as inches c. 17.300 centimeters as inches d. 0.840 centimeter as inches

27. a. 736.00 millimeters as inches b. 34.050 millimeters as inches c. 56.300 centimeters as inches d. 2.000 meters as yards

26. a. 2.400 meters as inches b. 0.090 meter as inches c. 8.000 meters as feet d. 10.200 meters as feet

28. a. 45.000 centimeters as feet b. 780.000 millimeters as feet c. 203.6 millimeters as inches d. 135.06 centimeters as yards

In Exercises 29 through 32, express each of the given metric units of length as the indicated customary unit of length. Where necessary, round the answer to 2 decimal places. 29. a. 4.000 inches as millimeters b. 0.360 inch as millimeters c. 34.00 inches as centimeters d. 20.85 inches as centimeters 30. a. 6.00 feet as meters b. 0.75 foot as meters c. 3.50 yards as meters d. 1.30 yards as meters 31. a. 2.368 inches as millimeters b. 0.73 inch as centimeters

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c. 216.00 inches as meters 1 d. inch as millimeters 2 1 32. a. 3 inches as centimeters 4 3 b. 75 inches as meters 8 5 c. 4 feet as centimeters 8 d. 2.35 feet as millimeters

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33. Determine the total length of stock in inches required to make 35 bushings each 3 34.2 mm long. Allow inch waste for each bushing. Round the answer to 1 decimal place. 16 34. The part shown in Figure 27-4 is to be made in a machine shop using decimal-inch machinery and tools. All dimensions are in millimeters. Express each of the dimensions A–L in inches to 3 decimal places. 20.90 (E) 3.50 DIA (F)

6.00 (D)

14.25 DIA (G) 6.00 R (H)

28.60 (I) 16.00 (J) 5.88 (K)

9.22 (L) 8.25 (C) 49.78 (B) 70.00 (A)

figurE 27-4

A5 B5 C5 D5

E5 F5 G5 H5

I5 J5 K5 L5

35. The shaft shown in Figure 27-5 is dimensioned in inches. Express each dimension, A–J, in millimeters and round each dimension to 2 decimal places. 3 32 3 8 11 16

R (H) 1 2

DIA (J)

DIA (F)

63 64

DIA (I)

DIA (G)

0.130 (E)

0.620 (A) 1.125 (B)

12.875 (C)

0.988 (D)

figurE 27-5

A5 B5 C5 D5

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E5 F5 G5

H5 I5 J5

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UNIT 28 Degree of Precision, Greatest Possible Error, Absolute Error, and Relative Error

ObjECTIvES After studying this unit you should be able to ●● ●● ●●

Determine the degree of precision of any given number. Compute the greatest possible error of customary and metric length units. Compute absolute error and relative error.

Degree of preCision The cost of producing a part increases with the degree of precision called for; therefore, no greater degree of precision than is actually required should be specified on a drawing. The degree of precision specified for a particular machining operation dictates the type of machine, the machine setup, and the measuring instrument used for that operation. The exact length of an object cannot be measured. All measurements are approximations. By increasing the number of graduations on a measuring instrument, the degree of precision is increased. Increasing the number of graduations enables the user to get closer to the exact length. The precision of a measurement depends on the measuring instrument used. The degree of precision of a measuring instrument depends on the smallest graduated unit of the instrument. No measurement is truly correct, but some are more precise than others. Depending on the need for precision in a measurement, you might measure one object with a tape measure and something else with a micrometer, a caliper, or a dial bore gage. As a general rule, a measuring instrument should measure to a degree 10 times finer than the smallest unit it will be used to measure. Machinists often work to 0.001-inch or 0.02-millimeter precision. In the manufacture of certain products, very precise measurements to 0.00001 inch or 0.0003 millimeter and 0.000001 inch or 0.00003 millimeter are sometimes required. Various measuring instruments have different limitations on the degree of precision possible. There is always some “rounding” involved in measuring. If the edge of a part falls between two graduations on the measuring scale, you should round to the nearer one. The accuracy achieved in measurement does not depend only on the limitations of the measuring instrument. Accuracy can also be affected by errors of measurement. Errors can be caused by defects in the measuring instruments and by environmental changes such as differences in temperature. Perhaps the greatest cause of error is the inaccuracy of the person using the measuring instrument.

LiMitations of Measuring instruMents Following are the limitations on the degree of precision or the accuracy possible of some commonly used manufacturing measuring instruments.

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dEgrEE of PrECision, grEatEst PossiblE Error, absolutE Error, and rElativE Error

Steel rules:

10 (fractional-inch); 0.010 (decimal-inch); 0.5 mm (metric). 64

Micrometers: 0.0010 (decimal-inch) and 0.00010 (with vernier scale); 0.01 mm (metric) and 0.002 mm (with vernier scale). Digital micrometers: Accurate to 0.000050 (decimal-inch); 0.0001 mm (metric). Vernier and dial calipers:

0.0010 (decimal-inch); 0.02 mm (metric).

Dial indicators (comparison measurement): (decimal-inch); 0.002 mm (metric).

Graduations as small as 0.000050

Precision gage blocks (comparison measurement): Accurate to 0.0000020 (decimal-inch); 0.00006 mm (metric). The degree of precision of measurement is only as precise as the measuring instrument that is used with the blocks. High amplification comparators (mechanical, optical, pneumatic, electronic): Accurate to 0.0000060 (decimal-inch); 0.00001 mm (metric).

Degree of preCision of nuMbers The degree of precision of a number depends upon the unit of measurement. The degree of precision of a number increases as the number of decimal places increases. Suppose a ruler is marked in 0.1 inch. Then, any measurement we read covers a whole range of values. For example, as indicated by the “rounding funnels” in Figure 28-1, a reading of 1.2 inches means that the actual measurement is between 1.15 inches and 1.2499 . . . inches; a reading of 1.3 inches means the actual measurement is between 1.25 inches and 1.3499 . . . inches; and a reading of 1.4 inches means the actual measurement is between 1.35 inches and 1.4499 . . . inches. 1.15

1.25

1.2

1.35

1.3

1.45

1.4

Range of values

Measurement scale

figurE 28-1

Example 1 The degree of precision of 20 is to the nearer inch, as shown in Figure 28-2. The range of values includes all numbers equal to or greater than 1.50 or less than 2.50. RANGE OF VALUES FOR 2-INCH MEASUREMENT

1.5

2.49 ...

figurE 28-2

Example 2 The degree of precision of 2.00 is to the nearer 10th of an inch as shown in Figure 28-3. The range of values includes all numbers equal to or greater than 1.950 and less than 2.050.

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RANGE OF VALUES FOR 2.0-INCH MEASUREMENT

1.95 2.049 ...

figurE 28-3

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RANGE OF VALUES FOR 2.00-INCH MEASUREMENT

1.995 2.0049 ...

Example 3 The degree of precision of 2.000 is to the nearer 100th of an inch as shown in Figure 28-4. The range of values includes all numbers equal to or greater than 1.9950 and less than 2.0050.

Example 4 The degree of precision of 2.0000 is to the nearer 1000th of an inch. The range of values includes all numbers equal to or greater than 1.99950 and less than 2.00050.

figurE 28-4

greatest possibLe error The greatest possible error of a measurement is one-half the smallest graduated unit of the instrument used to make the measurement. Therefore, the greatest possible error is equal 1 to , or 0.5, of the degree of precision. 2

Example 1 A machinist reads a measurement of 36 millimeters on a steel rule. The smallest graduation on the rule used is 1 millimeter; therefore, the degree of precision is 1 millimeter. Since the greatest possible error is one-half of the smallest graduated unit, the greatest possible error is 0.5 3 1 mm, or 0.5 mm. The actual length measured is between 36 mm 2 0.5 mm and 36 mm 1 0.5 mm, or between 35.5 mm and 36.5 mm.

Example 2 A tool and die maker reads a measurement of 0.4754 inch on a vernier scale micrometer. The smallest graduation on the micrometer is 0.0001 inch; therefore, the degree of precision is 0.0001 inch. The greatest possible error is 0.5 3 0.00010 or 0.000050. The actual length measured is between 0.47540 2 0.000050 and 0.47540 1 0.000050, or between 0.475350 and 0.475450.

Example 3 A machinist reads a measurement of 114 mm on a steel rule. The smallest graduation on the rule is 0.5 mm.

Smallest possible length:

1 3 0.5 mm 5 0.5 3 0.5 mm 5 0.25 mm 2 114 mm 2 0.25 mm 5 113.75 mm

Largest possible length:

114 mm 1 0.25 mm 5 114.25 mm

Greatest possible error:

absoLute error anD reLative error Absolute error and relative error are commonly used to express the amount of error between an actual or true value and a measured value. Absolute error is the difference between a true value and a measured value. Since the measured value can be either a smaller or larger value than the true value, subtract the smaller value from the larger value. Absolute Error 5 True Value 2 Measured Value or Absolute Error 5 Measured Value 2 True Value Relative error is the ratio of the absolute error to the true value. It is expressed as a percent. Absolute Error Relative Error 5 3 100 True Value

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dEgrEE of PrECision, grEatEst PossiblE Error, absolutE Error, and rElativE Error

181

Example 1 The actual or true value of the diameter of a shaft is 1.7056 inches. The shaft is measured as 1.7040 inches. Compute the absolute and relative error. The true value is larger than the measured value, therefore: Absolute Error 5 True Value − Measured Value Absolute Error 5 1.7056 in. 2 1.7040 in. 5 0.0016 in. Ans Absolute Error Relative Error 5 3 100 True Value 0.0016 in. Relative Error 5 3 100 < 0.094% Ans (rounded) 1.7056 in.

Example 2 An inspector measured a taper angle as 3.01 degrees. The true value of the angle is 2.98 degrees. Compute the absolute and relative error. The measured value is larger than the true value; therefore: Absolute Error 5 Measured Value − True Value Absolute Error 5 3.018 2 2.988 5 0.038 Ans Relative Error 5

Absolute Error 3 100 True Value

Relative Error 5

0.038 3 100 ø 1.0% 2.988

Ans (rounded)

ApplicAtion Tooling Up 1. Express 42 792.6 millimeters as meters. 3 1 2. Add: 7 ft 9 in. 1 4 ft 5 in. 4 2 3. A machinist’s gross income is $875 a week. If the cost of living goes up 3.4%, how much must the gross income increase in order for the machinist to keep even? 4. Solve the proportion

1215 314 5 1. x 43

13.45 1 9.32 4 0.98. Round your answer to 2 decimal places. 6.41 3 2.92 6. Divide: 97.126 4 0.53. Round the answer to 4 decimal places.

5. Compute 52.75 2

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Degree of Precision For each measurement find: a. the degree of precision. b. the value that is equal to or less than the range of values. c. the value that is greater than the range of values. 7. 4.30

a.

b.

c.

8. 1.620

a.

b.

c.

9. 4.0780

a.

b.

c.

10. 6.070

a.

b.

c.

11. 15.8850

a.

b.

c.

12. 9.18370

a.

b.

c.

13. 11.0030

a.

b.

c.

14. 36.00

a.

b.

c.

15. 7.010

a.

b.

c.

16. 23.000

a.

b.

c.

17. 6.10

a.

b.

c.

18. 14.010700

a.

b.

c.

19. 26.87 mm

a.

b.

c.

20. 15.4 mm

a.

b.

c.

21. 117.06 mm

a.

b.

c.

22. 0.976 mm

a.

b.

c.

23. 48.01 mm

a.

b.

c.

24. 104.799 mm

a.

b.

c.

25. 7.00 mm

a.

b.

c.

26. 34.0825 mm

a.

b.

c.

27. 8.001 mm

a.

b.

c.

28. 14.0000 mm

a.

b.

c.

Greatest Possible error For each of the exercises in the following tables, the measurement made and the smallest graduation of the measuring instrument are given. Determine the greatest possible error and the smallest and largest possible actual lengths for each. CUSTOMARY SYSTEM

Measurement Made (inches)

Smallest Graduation of Measuring Instrument Used (inches)

29.

5.30

0.05 (steel rule)

30.

15.68

0.02 (steel rule)

31.

0.753

0.001 (vernier caliper)

32.

0.226

0.001 (micrometer)

33.

0.9369

0.0001 (vernier micrometer)

34.

5 3 8

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Actual Length Greatest Smallest Largest Possible Error (inches) Possible (inches) Possible (inches)

1 (steel rule) 64

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METRIC SYSTEM Measurement Made (millimeters)

35.

64

36.

105

Smallest Graduation of Measuring Instrument Used (millimeters)

Greatest Possible Error (millimeters)

Actual Length Smallest Possible (millimeters)

Largest Possible (millimeters)

1 (steel rule) 0.5 (steel rule)

37.

98.5

0.5 (steel rule)

38.

53.38

0.02 (vernier caliper)

39.

13.37

0.01 (micrometer)

40.

12.778

0.002 (vernier micrometer)

Absolute and Relative error For each of the values in the following table, the true value and measured value are given. Determine the absolute and relative error of each. Where necessary, round the answers to 3 decimal places. True Value

41.

38.720 in.

42.

0.530 mm

43.

12.700º

Measured Value

12.900º

0.485 in.

0.482 in.

45.

23.860 mm

24.000 mm

46.

6.056º

6.100º

47.

1.050 mm

1.020 mm

48.

0.9347 in.

0.9341 in.

49.

1.005º

1.015º

50.

27.200 in.

26.900 in.

51.

18.276 in.

18.302 in.

0.983 mm

Relative Error

0.520 mm

44.

52.

Absolute Error

38.700 in.

1.000 mm

UNIT 29 Tolerance, Clearance, and Interference ObjECTIvES After studying this unit you should be able to ●● ●● ●● ●●

Compute total tolerances and maximum and minimum limits of dimensions. Compute maximum and minimum clearances of mating parts. Compute maximum and minimum interferences of mating parts. Express unilateral tolerances as bilateral tolerances.

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toLeranCe Tolerance is the amount of variation permitted on the dimensions or surfaces of manufactured parts. Limits are the extreme permissible dimensions of a part. Tolerance is equal to the difference between the maximum and minimum limits of any specified dimension of a part.

Example The maximum limit of a hole diameter is 0.878 inch and the minimum limit is 0.872 inch. Find the tolerance. The tolerance is 0.8780 2 0.8720 5 0.0060

Ans

A basic dimension is the standard size from which the maximum and minimum limits are determined. Usually tolerances are given in such a way as to show the amount of variation and in which direction from the basic dimension these variations can occur. Unilateral tolerance means that the total tolerance is taken in one direction from the basic dimension. Bilateral tolerance means that the tolerance is divided partly plus (1) or above and partly minus (2) or below the basic dimension. A mean dimension is a value that is midway between the maximum and minimum limits. Where bilateral tolerances are used with equal plus and minus tolerances, the mean dimension is equal to the basic dimension.

Example 1 The part shown in Figure 29-1 is dimensioned with a unilat-

0.0000 3.7500 + – 0.0016

eral tolerance. The dimensions are given in inches. The basic dimension is 3.75000. The total tolerance is a minus (2) tolerance. Find the maximum permissible dimension (maximum limit) and minimum permissible dimension (minimum limit).

figurE 29-1

Maximum Limit:

3.75000 1 0.00000 5 3.75000 Ans

Minimum Limit:

3.75000 2 0.00160 5 3.74840 Ans

Example 2 The part shown in Figure 29-2 is dimensioned with a bilater-

62.79 ± 0.04

al tolerance. The dimensions are given in millimeters. The basic dimension is 62.79 mm. The tolerance is given in two directions, plus (1) and minus (2). Find the maximum limit and the minimum limit. figurE 29-2

Maximum Limit:

62.79 mm 1 0.04 mm 5 62.83 mm

Ans

Minimum Limit:

62.79 mm 2 0.04 mm 5 62.75 mm

Ans

Example 3 What is the mean dimension of a part if the maximum dimension (maximum limit) is 46.35 millimeters and the minimum dimension (minimum limit) is 46.27 millimeters? Subtract.

46.35 mm 2 46.27 mm 5 0.08 mm

Divide.

0.08 mm 4 2 5 0.04 mm

Subtract.

46.35 mm 2 0.04 mm 5 46.31 mm

Ans

Note: The mean dimension is midway between 46.35 mm and 46.27 mm. 46.31 mm

10.04 mm 5 46.35 mm sMax. limitd 20.04 mm 5 46.27 mm sMin. limitd

expressing uniLateraL toLeranCe as biLateraL toLeranCe In the actual processing of parts, given unilateral tolerances are sometimes changed to bilateral tolerances. A machinist may prefer to work to a mean dimension and take equal plus and minus tolerances while machining a part. The following example shows the procedure for expressing a unilateral tolerance as a bilateral tolerance.

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Example The part shown in Figure 29-3 is dimensioned with unilat-

0.0028 1.2590 + – 0.0000

eral tolerances. Express the unilateral tolerance as a bilateral tolerance. Dimensions are given in inches. Divide the total tolerance by 2.

0.00280 4 2 5 0.00140

Determine the mean dimension. 1.25900 1 0.00140 5 1.26040 Show as a bilateral tolerance.

1.26040 6 0.00140

figurE 29-3

Ans

fits of Mating parts Fits between mating parts, such as between shafts and holes, have wide application in the manufacturing industry. The tolerances applied to each of the mating parts determine the relative looseness or tightness of fit between parts. When one part is to move within another there is a clearance between the parts. A shaft made to turn in a bushing is an example of a clearance fit. The shaft diameter is less than the bushing hole diameter. When one part is made to be forced into the other there is interference between parts. A pin pressed into a hole is an example of an interference fit. The pin diameter is greater than the hole diameter. Allowance is the intentional difference in the dimensions of mating parts that provides for different classes of fits. Allowance is the minimum clearance or the maximum interference intended between mating parts. Allowance represents the condition of the tightest permissible fit.

Example 1 A mating shaft and a hole with a clearance fit dimensioned with bilateral tolerances is shown in Figure 29-4. All dimensions are in inches. Determine the following: a. Maximum shaft diameter

0.75020 1 0.00080 5 0.75100

Ans

b. Minimum shaft diameter

0.75020 2 0.00080 5 0.74940

Ans

c. Maximum hole diameter

0.75360 1 0.00080 5 0.75440

Ans

0.7502 ± 0.0008

d. Minimum hole diameter 0.75360 2 0.00080 5 0.75280 Ans e. Maximum clearance equals maximum hole diameter minus minimum shaft diameter 0.75440 2 0.74940 5 0.00500 Ans f. Minimum clearance equals minimum hole diameter minus maximum shaft diameter 0.75280 2 0.75100 5 0.00180 Ans

0.7536 ± 0.0008 BASIC SHAFT DIA = 0.7502 BASIC HOLE DIA = 0.7536

figurE 29-4

Since allowance is defined as the minimum clearance, the allowance 5 0.00180 Ans This example can be summarized by the following table:

Basic Dimension

Maximum Diameter (Max. Limit)

Minimum Diameter (Min. Limit)

Shaft

0.75020

0.75100

0.74940

Hole

0.75360

0.75440

0.75280

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Maximum Clearance

Minimum Clearance (Allowance)

0.00500

0.00180

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Example 2 A pin intended to be pressed into a hole is shown in Figure 29-5. This is an example of an interference fit dimensioned with unilateral tolerances. Dimensions are in millimeters. Determine the following: a. Maximum pin diameter:

13.860 mm 1 0.010 mm 5 13.870 mm

Ans

b. Minimum pin diameter:

13.860 mm 2 0.000 mm 5 13.860 mm

Ans

c. Maximum hole diameter:

13.855 mm 1 0.000 mm 5 13.855 mm

Ans

0.010 13.860 + – 0.000

d. Minimum hole diameter:

13.855 mm 2 0.010 mm 5 13.845 mm

Ans

0.000 13.855 + – 0.010

e. Minimum interference equals minimum pin diameter minus maximum hole diameter: 13.860 mm 2 13.855 mm 5 0.005 mm

Ans

f. Maximum interference equals maximum pin diameter minus minimum hole diameter: 13.870 mm 2 13.845 mm 5 0.025 mm

Ans

BASIC PIN DIA = 13.860 BASIC HOLE DIA = 13.855

figurE 29-5

Since allowance is defined as the maximum interference, the allowance 5 0.025 mm Ans This example can be summarized by the following table:

Basic Dimension

Maximum Diameter (Max. Limit)

Minimum Diameter (Min. Limit)

Pin

13.860 mm

13.870 mm

13.860 mm

Hole

13.855 mm

13.855 mm

13.845 mm

Maximum Interference (Allowance)

Minimum Interference

0.025 mm

0.005 mm

ApplicAtion Tooling Up 1. For the measurement 14.304 mm find (a) the degree of precision, (b) the value that is equal to or less than the range of values, and (c) the value that is greater than the range of values.

50.1 mm R

2. Express 7.43 meters as inches. Round the answer to 3 decimal places. 1 5 3. Subtract: 16 ft 5 in. 2 7 ft 9 in. 2 8 4. What percent of 258 is 196? Round the answer to 2 decimal places. 4 5. Use a calculator to determine Ï 23.74 3 3.52. Round the answer to 2 decimal places.

6. Find the metal area of the washer in Figure 29-6. Round the answer to the nearest tenth.

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87.5 mm R

figurE 29-6

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Tolerance, Maximum and Minimum Limits Refer to the following tables and determine the tolerance, maximum limit, or minimum limit as required for each problem. 7. Customary System

Tolerance

a.

Maximum Limit

Minimum Limit

70 16 10 799 16

130 32 150 798 16

5

b. c.

0.030

d.

0.0070

e. f.

8. Metric System

5

0.0040

Minimum Limit

50.7 mm

49.9 mm

b.

26.8 cm

26.6 cm

0.04 mm

0.9040

d.

0.12 mm

1.69980

e.

0.006 cm

10.9990

Maximum Limit

a.

c.

16.760

1.70010

Tolerance

f.

258.03 mm 79.65 mm 12.731 cm 4.01 mm

3.98 mm

Unilateral and Bilateral Tolerance

A

9. Refer to Figure 29-7. Dimension A with its tolerance is given in each of the following problems. Determine the maximum dimension (maximum limit) and the minimum dimension (minimum limit) for each.

figurE 29-7

a. Dimension A 5 4.6400 10.0030 20.0000 maximum minimum

mm f. Dimension A 5 28.16 mm 10.00 20.06 mm maximum minimum

b. Dimension A 5 5.9270 10.0050 20.0000 maximum minimum

mm g. Dimension A 5 43.94 mm 10.04 20.00 mm maximum minimum

c. Dimension A 5 2.0040 10.0000 20.0040 maximum minimum

mm h. Dimension A 5 118.66 mm 10.07 20.00 mm maximum minimum

d. Dimension A 5 4.67290 10.00000 20.00120 maximum minimum

mm i. Dimension A 5 73.398 mm 10.000 20.012 mm maximum minimum

e. Dimension A 5 1.08750 10.00090 20.00000 maximum minimum

mm j. Dimension A 5 45.106 mm 10.009 20.000 mm maximum minimum

10. The following dimensions are given with bilateral tolerances. For each value determine the maximum dimension (maximum limit) and the minimum dimension (minimum limit). a. 2.8120 6 0.0060 g. 43.46 mm 6 0.05 mm maximum minimum maximum minimum b. 3.0030 6 0.0040 h. 107.07 mm 6 0.08 mm maximum minimum maximum minimum c. 3.9710 6 0.0100 i. 62.04 mm 6 0.10 mm maximum minimum maximum minimum d. 4.05620 6 0.00120 j. 10.203 mm 6 0.024 mm maximum minimum maximum minimum e. 1.37990 6 0.00090 k. 289.005 mm 6 0.007 mm maximum minimum maximum minimum f. 2.00000 6 0.00070 l. 66.761 mm 6 0.015 mm maximum minimum maximum minimum

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11. Express each of the following unilateral tolerances as bilateral tolerances having equal plus and minus values.

c. 3.0000

10.0100 20.0000 10.0020 20.0000 10.0000 20.0040

d. 0.0730

10.0000 20.0080

a. 0.9380 b. 1.6860

i. 44.30 mm j. 10.06 mm k. 64.89 mm

10.02 mm 20.00 mm 10.00 mm 20.08 mm 10.06 mm 20.00 mm 10.056 mm 20.000 mm 10.000 mm 20.017 mm 10.000 mm 20.026 mm 10.009 mm 20.000 mm

l. 37.988 mm

e. 4.18730

10.00140 20.00000

m. 125.00 mm

f. 1.00210

10.00000 20.00740

n. 43.091 mm

g. 1.00100

10.00000 20.00080

o. 98.879 mm

h. 8.46490

10.00220 20.00000

Fits of Mating Parts The following problems require computations with both clearance fits and interference fits between mating parts. Find the missing values in the following tables. 12. Refer to Figure 29-8 to determine the values in the table. The answer to the first problem is given. Allowance is equal to the minimum clearance. All dimensions are in inches.

0.0000 DIA A + – 0.0030 0.0030 DIA B + – 0.0000

Maximum Minimum Minimum Basic Diameter Diameter Maximum Clearance Dimension (Max. Limit) (Min. Limit) Clearance (Allowance)

a. b. c.

DIA A

1.4580

1.4580

1.4550

DIA B

1.4610

1.4640

1.4610

DIA A

0.6345

DIA B

0.6365

DIA A

2.1053

DIA B

2.1078

0.0090

13. Refer to Figure 29-9 to determine the values in the table. Allowance is equal to the maximum interference. All dimensions are in millimeters.

figurE 29-8

0.0030

DIA A ± 0.02 DIA B ± 0.02

figurE 29-9 Maximum Minimum Maximum Basic Diameter Diameter Interference Minimum Dimension (Max. Limit) (Min. Limit) (Allowance) Interference

a. b. c.

DIA A

20.73

DIA B

20.68

DIA A

32.07

DIA B

32.01

DIA A

12.72

DIA B

12.65

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UNIT 29

14. Refer to Figure 29-10 to determine the values in the table. Allowance is equal to the maximum interference. All dimensions are in millimeters. DIA A ± 0.03 DIA B ± 0.03

Maximum Minimum Maximum Basic Diameter Diameter Interference Minimum Dimension (Max. Limit) (Min. Limit) (Allowance) Interference

a. b. c.

DIA A

87.58

DIA B

87.50

DIA A

9.94

DIA B

9.85

DIA A

130.03

DIA B

129.96

figurE 27-10

15. Refer to Figure 29-11 to determine the values in the table. Allowance is equal to minimum clearance. All dimensions are in inches.

A ± 0.0008 B ± 0.0008

figurE 29-11 Maximum Minimum Minimum Basic Dimension Dimension Maximum Clearance Dimension (Max. Limit) (Min. Limit) Clearance (Allowance)

a. b. c.

DIM A

0.9995

DIM B

1.0020

DIM A

2.0334

DIM B

2.0360

DIM A

1.4392

DIM B

1.4412

Related Problems 16. Spacers are manufactured to the mean dimension and tolerance shown in Figure 29-12. An inspector measures 10 spacers and records the following thicknesses: 0.3720 0.3760

0.3790 0.3750

0.3700 0.3730

0.3770 0.3780

0.3710 0.3800

0.375 ± 0.003

figurE 29-12

Which spacers are defective (above the maximum limit or below the minimum limit)? All dimensions are in inches.

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17. A tool and die maker grinds a pin to an 18.25-millimeter diameter as shown in Figure 29-13. The pin is to be pressed (an interference fit) into a hole. The minimum interference permitted is 0.03 millimeter. The maximum interference permitted is 0.07 millimeter. Determine the mean diameter of the hole. All dimensions are in millimeters.

18.25

MEAN DIAMETER

figurE 29-13

18. A piece is to be cut to the dimensions and tolerances shown in Figure 29-14. Determine the maximum permissible value of length A. All dimensions are in inches.

3 5 16 ±

1 32

6 34 ±

A 3 20 8 ±

1 32

1 16

figurE 29-14

19. Determine the maximum and minimum permissible wall thickness of the steel sleeve shown in Figure 29-15. All dimensions are in millimeters. WALL THICKNESS

20.50 ± 0.01 DIA 26.20 ± 0.05 DIA

figurE 29-15

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maximum minimum

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191

tolEranCE, ClEaranCE, and intErfErEnCE

20. Mating parts are shown in Figure 29-16. The pins in the top piece fit into the holes in the bottom piece. All dimensions are in inches. Determine the following: a. The mean pin diameters b. The mean hole diameters c. The maximum dimension A d. The minimum dimension A e. The maximum dimension B f. The minimum dimension B g. The maximum total clearance between dimension C and dimension D h. The minimum total clearance between dimension C and dimension D

C + 0.0000 0.3750 – 0.0008 DIA 2 PINS

A 2.0000 ± 0.0004 2.0000 ± 0.0004

B 0.0008 0.3762 + – 0.0000 DIA 2 HOLES D

figurE 29-16

21. Figure 29-17 gives the locations with tolerances of six holes that are to be drilled in a length of angle iron. A machinist drills the holes then checks them for proper locations from edge A. The actual locations of the drilled holes are shown in Figure 29-18. Which holes are drilled out of tolerance (located incorrectly)? 70 15 16 ±

10 8

18

EDGE A

±

10 8

#3

#2

#1

70 8

TYPICAL 5 PLACES

#4

#5

#6

figurE 29-17

Figure 29-17 shows specifications for the locations of holes. 150

108 16 90

30 8

30

72 16 50

53 16 34 15 EDGE A

30 8

10 2

#1

#2

#3

#4

#5

#6

figurE 29-18

Figure 29-18 shows actual locations of drilled holes.

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UNIT 30 Customary and Metric Steel Rules Objectives After studying this unit you should be able to ●● ●● ●● ●●

Read measurements on fractional-inch and decimal-inch steel rules. Measure lengths using fractional-inch and decimal-inch scales. Read measurements on metric steel rules. Measure lengths using metric scales.

Steel rules are widely used for machine shop applications that do not require a high degree of precision. The steel rule is often the most practical measuring instrument to use for checking dimensions where stock allowances for finishing are provided. Steel rules are also used for locating roughing cuts on machined pieces and for determining the approximate locations of parts for machine setups. Steel rules used in the machine shop are generally six inches long, although rules anywhere from a fraction of an inch to several inches in length are also used.

CorreCt ProCedure in the use of steel rules The end of a rule receives more wear than the rest of the rule. Therefore, the end should not be used as a reference point unless it is used with a knee (a straight block), as shown in Figure 30-1. If a knee is not used, errors can occur when attempting to measure from the end (Figure 30-2). CORRECT

1

2

3

4

5

INCORRECT

6 1

2

3

4

5

6

PART KNEE

PART

Figure 30-1

Figure 30-2

If a knee is not used, the 1-inch graduation of customary measure rules should be used as a reference point, as shown in Figure 30-3. The 1 inch must be subtracted from the measurement obtained. For metric measure rules, use the 10-millimeter graduation as the reference point, as in Figure 30-4. The 10 millimeters must be subtracted from the measurement obtained.

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Unit 30

CORRECT

CORRECT 1

2

3

4

5

6

0.5 mm 10 1 mm

20

30

40

50

60

70

PART 1 inch

PART

10 millimeters

Figure 30-3

Figure 30-4

The scale edge of the rule should be put on the part to be measured. Following the correct procedure shown in Figure 30-5 eliminates parallax error (error caused by the scale and the part being in different planes) like that in Figure 30-6. INCORRECT

CORRECT 1

2

3

4

5

6

1

SCALE EDGE IS ON THE PART

2

3

4

5

SCALE EDGE IS NOT ON THE PART

Figure 30-6

Figure 30-5

reading fraCtional-inCh rules 1 inch. An enlarged fractional-inch rule is shown. 64 The top scale in Figure 30-7 is graduated in 64ths of an inch and the bottom scale in 32nds of an inch. The staggered graduations are halves, quarters, eighths, sixteenths, and thirtyseconds of an inch. The smallest division of fractional rules is

10 8 10 16 10 32

10 4

10 64

10 2

10

64

D

1

32 A

B

C

E

F

ENLARGED FRACTIONAL-INCH RULE

Figure 30-7

Measurements can be read by noting the last complete inch unit and counting the number of fractional units past the inch unit. Generally, a short-cut method of reading measurements is used as illustrated by the following examples.

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Examples Read the following measurements on the enlarged fractional-inch rule shown in Figure 30-7. 10 10 1. Length A: Subtract one graduation from . 32 4 10 10 80 10 70 2 5 2 5 Ans 4 32 32 32 32 10 10 2. Length B: Add one graduation to . 16 2 10 10 80 10 90 1 5 1 5 Ans 2 16 16 16 16 10 3. Length C: Subtract one graduation from 10. 8 10 8 0 10 7 0 10 2 5 2 5 Ans 8 8 8 8 10 30 4. Length D: Add one graduation to 1 . 64 8 30 10 24 0 10 25 0 1 1 51 1 51 Ans 8 64 64 64 64 Often the edge of an object being measured does not fall exactly on a rule graduation. In these cases, read the measurement to the nearer rule graduation.

Examples Read the following measurements, to the nearer graduation, on the enlarged fractional-inch rule shown in Figure 30-7. 30 10 1. Length E: Since the measurement is nearer to 1 than 1 , 32 8 30 length E is read as 1 . Ans 32 30 11 0 2. Length F: Since the measurement is nearer to 1 than 1 , 8 32 30 length F is read as 1 . Ans 8

reading deCimal-inCh rules An enlarged decimal-inch rule is shown in Figure 30-8. The top scale is graduated in hundredths of an inch (0.010). The bottom scale is graduated in fiftieths of an inch (0.020). The staggered graduations are halves, tenths, and fiftieths of an inch. 10 = 2 10 10 10 50 10 100

10

0.50

= 0.10

= 0.020 = 0.010

D

100

1

50 A

B

C

E

F

ENLARGED DECIMAL-INCH RULE

Figure 30-8

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Examples Read the following measurements on the enlarged decimal-inch rule shown in Figure 30-8. 1. Length A:

Count two 0.10 graduations. 2 3 0.10 5 0.20

2. Length B:

Ans

Add two 0.10 graduations to the 0.50. 0.50 1 0.20 5 0.70

3. Length C:

Ans

Add three 0.020 graduations to 0.80. 0.80 1 0.060 5 0.860 Ans

4. Length D:

Add 10, plus three 0.10 graduations, plus five 0.010 graduations. 10 1 0.30 1 0.050 5 1.350 Ans

5. Length E:

Since the measurement is nearer to 1.180 than 1.160, length E is read as 1.180.

6. Length F:

Ans

Since the measurement is nearer 1.400 than 1.420, length F is read as 1.400.

Ans

reading a metriC rule An enlarged metric rule is shown in Figure 30-9. The top scale is graduated in half millimeters (0.5 mm). The bottom scale is graduated in millimeters (1 mm). The following examples show the method of reading measurements with a metric rule with 0.5 mm and 1 mm scales.

Examples Read the following measurements on the enlarged metric rule shown. C

0.5 mm 1 mm

10

A

20

30

40

50

60

B

70

D

ENLARGED METRIC RULE (1 mm and 0.5 mm)

Figure 30-9

1. Length A:

Add four 1 mm graduations to 10 mm. 10 mm 1 4 mm 5 14 mm

2. Length B:

Subtract one 1 mm graduation from 40 mm. 40 mm 2 1 mm 5 39 mm

3. Length C:

Ans Ans

Add 20 mm, plus two 1 mm graduations, plus one 0.5 mm graduation. 20 mm 1 2 mm 1 0.5 mm 5 22.5 mm

4. Length D:

Since the measurement is nearer 71 mm than 70 mm, length D is read as 71 mm.

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Ans

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ApplicAtion tooling Up 1. A shaft has a diameter of 0.45030 6 0.00060. What are the maximum and minimum diameters of the shaft? 2. For the measurement 26.3284 in. find (a) the degree of precision, (b) the value that is equal to or less than the range of values, and (c) the value that is greater than the range of values. 3 3. Express 42 inches as centimeters. Round the answer to 1 decimal place. 4 4. A machine shop charges a customer $415.65 for some materials. This price includes a mark-up of 72%. How much did these materials cost the shop? 738 412 5. Solve the proportion 5 2 . w 103 6. Express the number 0.253/2 as a decimal.

Fractional-inch Steel Rules 7. Read measurements a–d on the enlarged fractional rule shown in Figure 30-10. c

d

64

1

32 a

b

Figure 30-10

a.

b.

c.

d.

8. Read measurements e–h on the enlarged fractional rule shown in Figure 30-11. g

h

64

1

32 e

f

Figure 30-11

e.

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f.

g.

h.

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9. Read measurements m–p on the enlarged fractional rule shown in Figure 30-12. o

p 64

1

32 m

n

Figure 30-12

m.

n.

o.

p.

10. Read measurements w–z on the enlarged fractional rule shown in Figure 30-13. y

z

64

1

32 w

x

Figure 30-13

w.

x.

y.

z.

In Exercises 11 and 12, measure the length of each of the following line segments to the nearer a

b

d

c

e

f

h

i

11. a.

12. f.

b.

g.

c.

h.

d.

i.

e.

j.

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10 . 16

g j

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13. Measure the lengths of dimensions a–g in Figure 30-14 to the nearer

10 . 32 g

a

c

f

b

d

e

Figure 30-14

a.

c.

e.

b.

d.

f.

14. Measure the lengths of dimensions h–n in Figure 30-15 to the nearer

g.

10 . 32

k

i

m j

l

h n

Figure 30-15

h.

j.

l.

i.

k.

m.

n.

In Exercises 15 and 16, measure the length of each of the following line segments to the nearer a

10 . 64

b

c

d h

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e i

f

g j

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CustoMary and MetriC steeL ruLes

15. a. b. c. d. e.

16. f. g. h. i. j.

Decimal-inch Steel Rules 17. Read measurements a–d on the enlarged fractional rule shown in Figure 30-16. c

d

100

1

50

a

b

Figure 30-16

a.

b.

c.

d.

18. Read measurements e–h on the enlarged fractional rule shown in Figure 30-17. g

h

100

1

50

e

f

Figure 30-17

e.

f.

g.

h.

19. Read measurements m–p on the enlarged fractional rule shown in Figure 30-18. o

p

100

1

50

m

n

Figure 30-18

m.

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n.

o.

p.

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20. Read measurements w–z on the enlarged fractional rule shown in Figure 30-19. y

z

100

1

50

w

x

Figure 30-19

w.

x.

y.

z.

In Exercises 21 and 22, measure the length of each of the following line segments to the nearer fiftieth of an inch (0.02”). a

b

c

e

f

h

d g

i

j

k

21. a. b. c. d. e. f.

l

m

22. g. h. i. j. k. l. m.

23. Measure the diameters of the holes in the plate shown in Figure 30-20 to the nearer fiftieth of an inch (0.020). HOLE A

HOLE F

HOLE B

HOLE C

HOLE D

HOLE H

HOLE G

HOLE E

HOLE I

Figure 30-20

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Note: Measure to the inside of the hole line thickness. A5

D5

G5

B5

E5

H5

C5

F5

I5

Metric Steel Rules 24. Read measurements a–d on the enlarged fractional rule shown in Figure 30-21. c

0.5 mm

d

10

1 mm

20

30

40

a

50

60

70

b

Figure 30-21

a.

b.

c.

d.

25. Read measurements e–h on the enlarged fractional rule shown in Figure 30-22. g

0.5 mm

10

1 mm

h

20

30

40

50

e

60

70

f

Figure 30-22

e.

f.

g.

h.

26. Read measurements m–p on the enlarged fractional rule shown in Figure 30-23. o

0.5 mm 1 mm

p

10

20

m

30

40

50

60

70

n

Figure 30-23

m.

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n.

o.

p.

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27. Read measurements w–z on the enlarged fractional rule shown in Figure 30-24. y

0.5 mm 1 mm

10

20

z

30

40

50

60

70

w

x

Figure 30-24

w.

x.

y.

z.

In Exercises 28 and 29, measure the length of each of the following line segments to the nearer whole millimeter. a

b d

c

e

f

g

h j

i

k

l

28. a. b. c. d. e. f. g.

m

n

29. h. i. j. k. l. m. n.

30. Measure the lengths of dimensions a–f in Figure 30-25 to the nearer whole millimeter. e b f

c

d a

Figure 30-25

a.

c.

e.

b.

d.

f.

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CustoMary Vernier CaLipers and height gages

31. Measure the lengths of dimensions g–k in Figure 30-26 to the nearer whole millimeter.

i

j k

h

g

Figure 30-26

g.

i.

h.

j.

k.

UNIT 31 Customary Vernier Calipers and Height Gages

Objectives After studying this unit you should be able to ●● ●● ●● ●●

Read measurements set on a decimal-inch vernier caliper. Set given measurements on a decimal-inch vernier caliper. Read measurements set on a decimal-inch vernier height gage. Set given measurements on a decimal-inch vernier height gage.

Digital calipers and height gages are widely used. Measurements are read directly on a five-digit LCD readout display with instant inch/millimeter conversion. We will look at reading digital calipers and height gages in Unit 33. Vernier calipers and height gages have been largely replaced by digital instruments. However, many conventional (non-digital) customary vernier calipers and height gages that have been used for years are still in use. Therefore, these customary measuring instruments are retained in this edition of the book. Decimal-inch vernier calipers are used in machine shop applications when the degree of precision to thousandths of an inch is adequate. They are used for measuring lengths of parts, distances between holes, and both inside and outside diameters of cylinders. Vernier height gages are widely used on surface plates and on machine tables. The height gage with an indicator attachment is used for checking locations of surfaces and holes. The height gage with a scriber attachment is used to mark reference lines, locations, and stock allowances on castings and forgings.

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deCimal-inCh Vernier CaliPer The basic parts of a vernier caliper are a main scale, which is similar to a steel rule with a fixed jaw, and a sliding jaw with a vernier scale. The vernier scale slides parallel to the main scale and provides a degree of precision to 0.0010. Calipers are available in a wide range of lengths with different types of jaws and scale graduations. A vernier caliper, which is commonly used in machine shops, is shown in Figure 31-1. LOCKING SCREWS

MAIN SCALE

BEAM OUTSIDE 0

12345678

1

12

1234567

9

3

12 3 4 5 6 7 8 9

4

123456789

5

123456789

6

0 5 10152025

FINE ADJUSTMENT NUT

FIXED JAW

VERNIER SCALE SLIDING JAW OUTSIDE MEASUREMENT INSIDE MEASUREMENT

Figure 31-1

The main scale is divided into inches and the inches are divided into 10 divisions each equal to 0.10. The 0.10 divisions are divided into 4 parts each equal to 0.0250. The vernier scale consists of 25 divisions. A vernier scale is shown in Figure 31-2. The vernier scale has 25 divisions in a length equal to a length on the main scale that has 24 divisions. The difference between a main 1 scale division and a vernier division is of 0.0250, or 0.0010. 25 24 DIVISIONS ON MAIN SCALE

0

5

10

15

20

25

25 DIVISIONS ON VERNIER SCALE

Figure 31-2

reading and setting a measurement on a deCimal-inCh Vernier CaliPer A measurement is read by adding the thousandths reading on the vernier scale to the reading from the main scale.

c Procedure ●●

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To read a measurement on a decimal-inch vernier caliper

Read the number of 10 graduations, 0.10 graduations, and 0.0250 graduations on the main scale that are left of the zero graduation on the vernier scale.

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●●

205

CustoMary Vernier CaLipers and height gages

On the vernier scale, find the graduation that most closely coincides with a graduation on the main scale. Add this vernier reading, which indicates the number of 0.0010 graduations to the main scale reading.

Setting a given measurement is the reverse procedure of reading a measurement on the vernier caliper.

Example 1 Read the measurement set on the vernier caliper in Figure 31-3.

7

8

0

MAIN SCALE

3

9

5

1

10

2

15

3

20

4

25

VERNIER SCALE GRADUATION COINCIDES WITH MAIN SCALE GRADUATION

VERNIER SCALE

Figure 31-3

In reference to the zero division on the vernier scale, read two 10 divisions, seven 0.10 divisions, and three 0.0250 divisions on the main scale. (20 1 0.70 1 0.0750 5 2.7750) Observe which vernier scale graduation most closely coincides with a main scale graduation. The 8 vernier scale graduation coincides; therefore, 0.0080 is added to 2.7750. Measurement: 2.7750 1 0.0080 5 2.7830

Ans

Example 2 Set 1.2370 on a vernier caliper. Move the vernier zero graduation to 10 1 0.20 1 0.0250 on the main scale. An additional 0.0120 (1.2370 2 1.2250) is set by adjusting the sliding jaw until the 12 graduation on the vernier scale coincides with a graduation on the main scale. The 1.237-inch setting is shown in Figure 31-4. MAIN SCALE

1

1

2

3

0

4

5

5

6

7

10

15

20

VERNIER SCALE

8

25

SET 12 VERNIER GRADUATION

Figure 31-4

The accuracy of measurement obtainable with a vernier caliper depends on the user’s ability to align the caliper with the part that is being measured and the user’s “feel” when measuring. The line of measurement must be parallel to the beam of the caliper and lie in the same plane as the caliper. Care must be used to prevent a caliper setting that is too loose or too tight.

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The front side of the customary vernier caliper (25 divisions) is used for outside measurements, as shown in Figure 31-5. The reverse or back side is used for inside measurements, as shown in Figure 31-6.

Figure 31-5 measure an outside diameter the measurement is read on the front side of the caliper. (the L.s. starrett Company)

Figure 31-6 measure an inside diameter the measurement is read on the back side of the caliper. (the L.s. starrett Company)

deCimal-inCh Vernier height gage The vernier height gage and vernier caliper are similar in operation. The height gage also has a sliding jaw; the fixed jaw is the surface plate with which the height gage is usually used. The gage can be used with a scriber, a depth gage attachment, or an indicator. The indicator is the most widely used and, generally, the most accurate attachment. The parts of a vernier height gage are shown in Figure 31-7. Measurements on the vernier height gage are read and set using the same procedure as with the vernier caliper. COLUMN 10

3

MAIN SCALE

9

FINE ADJUSTMENT NUT SLIDE

2 1

LOCKING SCREWS

1

BASE

9 8 7 6 5

0

7 6 5 4 3 2 1

25 20 15 10 5 0

VERNIER SCALE

8 7 6 5 4 3 2 1

SLIDE ARM (SCRIBER, INDICATOR, OR DEPTH GAGE CAN BE ATTACHED)

DECIMAL-INCH HEIGHT GAGE

Figure 31-7

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CustoMary Vernier CaLipers and height gages

Example 1 Read the measurement set on the vernier height gage in Figure 31-8. In reference to the zero division on the vernier scale read 50, four 0.10 divisions, and two 0.0250 divisions on the main scale. (50 1 0.40 1 0.0500 5 5.4500) Observe which vernier scale graduation most closely coincides with the main scale graduation. The 21 vernier scale graduation coincides; therefore, 0.0210 is added to 5.4500. Measurement 5 5.4500 1 0.0210 5 5.4710

VERNIER SCALE GRADUATION COINCIDES WITH MAIN SCALE

1

25

6

20

9

15

VERNIER SCALE

8

10

7

5

6 5

0

Ans

4

MAIN SCALE

Figure 31-8

Example 2 Set 8.3980 on a vernier height gage. Move the vernier zero graduation to 80 1 0.30 1 0.0750 5 8.3750. Lock the upper side in place. An additional 0.0230 (8.3980 2 8.3750) is set by turning the fine adjustment screw until the 23 graduation on the vernier scale coincides with a graduation on the main scale.

SET 23 VERNIER GRADUATION

9

25 20 15

9 8 7

10

The 8.398-inch setting is shown in Figure 31-9. VERNIER SCALE

6

5

5

0

4

MAIN SCALE

3

Figure 31-9

ApplicAtion tooling Up 1. Measure this line segment to the nearest

10 . 64

mm 2. A hole has a diameter of 13.741mm 10.005 . What are the maximum and minimum diameters of the hole? 20.002 mm

3. For the measurement 74.38 mm, find (a) the degree of precision, (b) the value that is equal to or less than the range of values, and (c) the value that is greater than the range of values. 4. Express 87.6 inches as feet. 5. If 1450 parts can be produced in 15.35 hours, how long will it take to produce 16,500 parts? Round the answer to the nearest whole hour. 6. Use the order of operations to evaluate 3.252 1 19.21 3 Ï256.4 4 0.35 2 82.3. Round the answer to the nearest 2 decimal places.

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

Decimal-inch Vernier caliper Read the decimal-inch vernier caliper measurements the settings in Exercises 7 through 14. 11.

7. MAIN SCALE 6

7

MAIN SCALE

8

0

3

9

5

10

1

15

2

20

5

3

7

25

8

0

VERNIER SCALE

1

9

5

10

2

15

3

20

4

25

VERNIER SCALE

8.

12. MAIN SCALE

MAIN SCALE

4 3

4

0

5

6

5

10

7

8

15

9

20

25

3

1

0

VERNIER SCALE

2

5

3

10

4

5

15

6

20

7

25

VERNIER SCALE

9.

13. MAIN SCALE

2

9

MAIN SCALE

1

0

2

5

3

10

4

5

15

20

6

5

25

VERNIER SCALE

6

0

2

9

7

8

5

10

15

1

20

2

25

VERNIER SCALE

14.

10. MAIN SCALE 5

0

6

MAIN SCALE

7

5

9

8

10

VERNIER SCALE

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15

1

20

1

25

2

1

1

0

2

5

3

4

10

5

15

6

20

7

25

VERNIER SCALE

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Unit 31

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CustoMary Vernier CaLipers and height gages

The tables for Exercises 15 and 16 give the position of the zero graduation on the vernier scale in reference to the main scale and the vernier scale graduation that coincides with a main scale graduation. Determine the vernier caliper settings. The answer to the first part of Exercise 15 is given. 15.

16.

Zero Vernier Graduation Lies Between These Main Scale Graduations (inches)

Vernier Graduation That Coincides with a Main Scale Graduation

Zero Vernier Graduation Lies Between These Main Scale Graduations (inches)

Vernier Graduation That Coincides with a Main Scale Graduation

Vernier Caliper Setting (inches)

a.

1.875–1.900

19

1.894

a.

0.000–0.025

5

b.

3.025–3.050

21

b.

0.825–0.850

17

c.

0.050–0.075

6

c.

3.550–3.575

23

d.

5.775–5.800

11

d.

5.075–5.100

20

e.

1.225–1.250

7

e.

3.325–3.350

15

f.

0.075–0.100

16

f.

2.075–2.100

6

g.

3.000–3.025

4

g.

4.400–4.425

10

h.

2.650–2.675

9

h.

1.025–1.050

13

i.

1.000–1.025

13

i.

0.675–0.700

18

j.

5.975–6.000

18

j.

0.050–0.075

2

k.

2.825–2.850

8

k.

3.000–3.025

21

l.

4.950–4.975

1

l.

2.925–2.950

22

Vernier Caliper Setting (inches)

In Exercises 17 and 18, refer to the following sentence and to the following given vernier caliper settings to find values A, B, and C. “The zero vernier scale graduation lies between A and B on the main scale, and the vernier graduation C coincides with the main scale graduation.” The answer to the first part of Exercise 17 is given. 17.

18.

Vernier Caliper Setting (inches)

A (inches)

B (inches)

C

a.

3.242

3.225

3.250

17

b. c.

Vernier Caliper Setting (inches)

a.

1.646

2.877

b.

4.034

4.839

c.

0.022

d.

0.611

d.

3.333

e.

4.369

e.

5.999

f.

0.084

f.

0.278

g.

7.857

g.

0.965

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A (inches)

B (inches)

C

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The distance between the centers of two holes can be checked with a vernier caliper. The position of the caliper in measuring the inside distance between two holes is shown in Figure 29-10. To determine the setting on the caliper, subtract the radius of each hole (one-half the diameter) from the center distance. In Exercises 19 through 23, give the hole diameters and the distances between centers. For each, determine (1) the main scale setting and (2) the vernier scale setting. All dimensions are in inches.

INSIDE HOLE MEASUREMENT DISTANCE BETWEEN CENTERS

Figure 29-10

19.

22. Note: Hole tolerances are shown. Maximum and minimum vernier scale settings are required.

2 HOLES 0.232 DIA

+

+

0.750 ± 0.004 DIA

4.674

(1) 20.

0.478 ± 0.002 DIA

+

+

(2)

5.345

2 HOLES 0.186 DIA

(1) +

+

4.358

(1) 21.

23. Note: Hole tolerances and center distance tolerances are shown. Maximum and minimum vernier scale settings are required.

(2) 0.123 DIA

0.137 DIA

(1)

3.664

(1)

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0.375 ± 0.003 DIA

0.327 ± 0.005 DIA

+

+ 3.262 ± 0.003

+

+

(2)

(2)

(2)

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CustoMary Vernier CaLipers and height gages

Unit 31

Decimal-inch Height Gage Read height gage measurement settings in Exercises 24 and 25. 24.

MAIN SCALE

VERNIER SCALE 25

5

25

2

20

4

20

1

15 10

3

5 0

10

1

5

2 9

0

8

20

5

15

2

25

15

9

25 20 15

5 0

0

8

6 5

0

4

9

1

d.

25

2

20

1

15

8

10

7

10

5

6

5

a.

7

6

c.

0

8

5

2

8

15

1

20

9

9

5

3

3

3

1

20

7

25

1

4

10

4

10

5

2

b.

25.

25

10

8

5

a.

1

6

7 9

5

0

4

9

20

4

15

9

25 8

15

9

7

10

8

6

7

5

5

6

0

4 3

5

b.

d.

c.

26. The hole locations of the block in Figure 31-11 are checked by placing the block on a surface plate and indicating the bottom of each hole using a height gage with an indicator attachment. Determine the height gage settings from the bottom of the part to the bottom of the holes. Assume that the actual hole diameters and locations are the same as the given dimensions. The setting for Hole Number 1 is given.

HOLE #2

HOLE #3

HOLE #5

E

C

Hole Number

Hole Diameter (inches)

Given Locations to Centers of Holes (inches)

Main Scale Setting (inches)

Vernier Scale Setting (inches)

1

0.376

A 5 0.640

0.450–0.475

2

2

0.258

B 5 1.008

3

0.188

C 5 0.514

4

0.496

D 5 0.312

5

0.127

E 5 0.810

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HeiGHT GAGe SeTTiNGS

B

D

A

HOLE #4 HOLE #1 HEIGHT GAGE SETTING TAKEN FROM BOTTOM OF BLOCK

HEIGHT GAGE INDICATES BOTTOMS OF HOLES

Figure 31-11

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UNIT 32 Metric Vernier Calipers and Height Gages Objectives After studying this unit you should be able to ●● ●● ●● ●●

Read measurements set on a metric vernier caliper. Set given measurements on a metric vernier caliper. Read measurements set on a metric vernier height gage. Set given measurements on a metric vernier height gage.

Reading MetRic MicRoMeteRs The same principles are used in reading and setting metric vernier calipers as are used for customary vernier calipers. The main scale is divided in 1-millimeter divisions. Each millimeter division is divided in half, or into 0.5-millimeter divisions. A graduation is numbered every 10 millimeters in the sequence: 10 mm, 20 mm, 30 mm, and so on. The vernier scale has 1 25 divisions. Each division is of 0.5 millimeter, or 0.02 millimeter. 25 A measurement is read by adding the 0.02-millimeter reading on the vernier scale to the reading from the main scale. On the main scale, read the number of millimeter divisions and 0.5-millimeter divisions that are to the left of the zero graduation on the vernier scale. On the vernier scale, find the graduation that most closely coincides with a graduation on the main scale. Multiply the graduation by 0.02 millimeter and add the value obtained to the main scale reading.

Example Read the measurement set on the metric vernier caliper in Figure 32-1. MAIN SCALE 20

0 5 10 VERNIER SCALE

30

15

20

25 0.02 mm

VERNIER SCALE GRADUATION COINCIDES WITH MAIN SCALE GRADUATION

Figure 32-1

To the left of the zero graduation on the vernier scale, read the main scale reading: twenty-one 1-millimeter divisions and one 0.5-millimeter division. (21 3 1 mm 1 1 3 0.5 mm 5 21.5 mm) Observe which vernier scale graduation most closely coincides with a main scale graduation. The sixth vernier scale graduation coincides. Each vernier scale graduation represents 0.02 mm. Multiply to find the number of millimeters represented by six divisions. (6 3 0.02 mm 5 0.12 mm) Add the 0.12 millimeter to the main scale reading. (21.5 mm 1 0.12 mm 5 21.62 mm) Vernier caliper reading: 21.62 mm Ans

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Unit 32

213

MetriC Vernier CaLipers and height gages

MetRic VeRnieR HeigHt gage As with the customary versions, the metric vernier height gage and metric vernier caliper are similar in operation. The height gage also has a sliding jaw; the fixed jaw is the surface plate with which the height gage is usually used. The gage can be used with a scriber, a depth gage attachment, or an indicator. The indicator is the most widely used and, generally, the most accurate attachment. The parts of a vernier height gage are similar to those shown in Figure 31-7.

Example Read the measurement set on the metric height gage in Figure 32-2. 30 25 VERNIER SCALE GRADUATION COINCIDES WITH MAIN SCALE GRADUATION

20 15 10 0.02 mm

VERNIER SCALE

MAIN SCALE

20

5 0

Figure 32-2

In reference to the zero division on the main scale read sixteen 1-millimeter divisions and zero 0.5-millimeter division. (16 3 1 mm 1 0 3 0.5 mm 5 16.0 mm) Observe which vernier scale graduation most closely coincides with a main scale graduation. As indicated by the arrow, the nineteenth vernier scale graduation coincides. Each vernier scale graduation represents 0.02 mm. Multiply to find the number of millimeters represented by nineteen divisions. (19 3 0.02 mm 5 0.38 mm) Add the 0.38 mm to the main scale reading. (16.0 mm 1 0.38 mm 5 16.38 mm) Vernier height gage reading: 16.38 mm Ans

ApplicAtion tooling Up 1. Read the decimal-inch vernier caliper measurement for this setting. MAIN SCALE 6

0

7

8

5

9

10

3

15

1

20

2

3

25

VERNIER SCALE

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

2. Measure this line segment to the nearest

10 . 64

mm 3. A hole has a diameter of 24.649 mm10.002 . What are 20.004 mm

the maximum and minimum diameters of the hole? 4. Express 37.295 centimeters as inches. Round the answer to 3 decimal places. 5 5. Express 4 % as a decimal fraction. 8 4 3 6. Use a calculator to determine Ï 166.375 4 Ï 1.4641. Round the answer to 2 decimal places.

Metric Vernier calipers Read the metric vernier caliper measurements for the following settings. 7.

0 5 10 VERNIER SCALE

8.

10.

MAIN SCALE 30

40

15

20

15

20

25

70

15

20

25 0.02 mm

MAIN SCALE 50

0 5 10 VERNIER SCALE

0.02 mm

12.

MAIN SCALE 60

98310_sec_03_Unit32-38_ptg01.indd 214

0 5 10 15 VERNIER SCALE

11.

MAIN SCALE 10

0 5 10 VERNIER SCALE

25 0.02 mm

0 5 10 VERNIER SCALE

9.

20

MAIN SCALE 80

90

20

0.02 mm

60

15

20

25 0.02 mm

MAIN SCALE 20

0 5 10 15 VERNIER SCALE

25

30

20

25 0.02 mm

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Unit 33

215

digitaL CaLipers and height gages

Metric Height Gages Read the metric height gage measurements for the following settings. 13.

15.

25 20

15

15

40

10

10

5

5

5

16.

70

18.

0 30

25

20

20

15

15

60

15 10

90

10

10

5

5

0

0

0.02 mm

0.02 mm

0.02 mm

60

10

25

25

5

0

0.02 mm

0.02 mm

0.02 mm

40

14.

0

20

20

10

20

25

20

50

15

0

17.

25

50

UNIT 33 Digital Calipers and Height Gages Objectives After studying this unit you should be able to ●● ●●

Read measurements on an electronic or digital caliper. Read measurements on an electronic or digital height gage.

Reading digital calipeRs Many analog vernier calipers are being replaced with electronic digital display calipers on which the reading is displayed as a single value. Some digital calipers can be switched between centimeters/millimeters and inches. All provide for zeroing the display at any point along the slide, allowing the same sort of differential measurements as with dial calipers. Digital calipers may contain some sort of “reading hold” feature to allow the reading of dimensions in awkward locations where the display cannot be seen.

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

Digital calipers have two upper jaws and two lower jaws. One of the upper jaws is movable, as is one of the lower jaws. The smaller upper jaws are used to measure inside diameters, and the larger lower jaws are used to measure outside diameters. As the jaws are opened, a depth bar slides out from the end of the caliper body. This feature is used to make depth measurements. The caliper has two or three buttons: ON/OFF, ZERO, and IN/mm. On the caliper in Figure 33-1, the ON/OFF and ZERO buttons are the same. There is also a locking screw. The reading is electronically generated and digitally displayed on a large LCD display. Digital calipers require a battery. Depending on the size of the calipers, it can measure up to 12 or more inches.

Figure 33-1 a digital caliper. (The L. S. Starrett Company)

The are some advantages to digital calipers: (a) measurements are read is a single step on an LCD display, (b) the same instrument can be used for both metric or inch measurements, (c) a button allows for switching between metric and inch measurements, (d) some versions allow the data to be saved to a USB drive or printed, (e) some versions are designed so that data can be sent wirelessly or with a traditional cable, and (f) they can provide for SPC (statistical process control) analysis and documentation.

Use of tHe calipeRs Before you begin, loosen the locking screw on the caliper and use a clean cloth to dry and clean the object you are going to measure and the jaws of the caliper. Once the object being measured and the calipers are clean, do not touch any of the measuring surfaces with your hands since the oil or sweat on your hands might create an inaccurate measurement. (a) Press the ON/OFF button to turn on the power, and then press the IN/mm button to select whether the object will be measured in inches or millimeters. Close the jaws of the calipers and press the ZERO button to set the caliper to 0. (b) For an outside measurement, place the object between the lower jaws and tighten the jaws against the outside surfaces of the object. Read the numbers directly from the LCD display. For example, the object in Figure 33-2 measures 1.0020 in.

Figure 33-2

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Unit 33

217

digitaL CaLipers and height gages

(c) For an inside measurement, insert the upper jaws inside the object and then open the jaws until they fit snugly against the hole of the object. Again, read the numbers directly from the LCD display. For example, the object in Figure 33-3 measures 0.8930 in.

Figure 33-3

(d) To take a depth measurement, use the depth bar on the end of the digital caliper. Move the slider of the caliper and adjust the depth bar so that it is perpendicular to both the hard, flat plane and the internal side of the object. Next, move the slider until the bottom of the fixed scale touches the top surface of the object and read the LCD display directly. The object in Figure 33-4 has a depth of 21.55 mm.

Figure 33-4

ApplicAtion tooling Up 1. Read the metric vernier caliper measurement for this setting. MAIN SCALE 80

90

0

5

VERNIER SCALE

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10

15

20

25 0.02 mm

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

2. Read the decimal-inch vernier caliper measurement for this setting. MAIN SCALE 3

1

0

2

5

3

4

10

5

15

6

20

7

25

VERNIER SCALE

3. Measure this line segment to the nearest

10 . 64

4. Express 74.079 centimeters as inches. Round the answer to 3 decimal places. 7 5. Express 5 % as a decimal fraction. 8 4 3 6. Use a calculator to determine Ï 166.375 4 Ï 1.4641. Round the answer to 2 decimal places.

Digital calipers Determine the digital caliper measurements for the following objects. 7. The thickness, in inches, of a washer. 8. The thickness, in mm, of the barrel of your pen or pencil. 9. The width, in mm, of both ends of a combination wrench. 10. The length, in inches, of a wrench. 11. The depth, in inches, of the tread on an automobile or truck tire. 12. The depth and outside diameter of a PVC coupler.

UNIT 34 Customary Micrometers Objectives After studying this unit you should be able to ●● ●● ●●

Read settings from the barrel and thimble scales of a 0.001-inch micrometer. Set given dimensions on the scales of 0.001-inch and 0.0001-inch micrometers. Read settings from the barrel, thimble, and vernier scales of 0.0001-inch micrometers.

As with digital calipers, digital micrometers are widely used. Measurements are read directly on a five-digit LCD readout display with inch/millimeter selection. Also, as with calipers, many conventional (non-digital) customary micrometers are still in use. Therefore, these customary micrometers are retained in this edition.

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CustoMary MiCroMeters

Unit 34

Micrometers are basic measuring instruments used by machinists in the processing and checking of parts. Micrometers are available in a wide range of sizes and types. Outside micrometers are used to measure dimensions between parallel surfaces of parts and outside diameters of cylinders. Other types, such as depth micrometers, screw thread micrometers, disc and blade micrometers, bench micrometers, and inside micrometers, also have wide application in the machine shop. A few of the many types of non-digital micrometers are shown in Figure 34-1. (a)

(d)

Anvil Micrometer

(The L.S. Starrett Company)

(b) Micrometer Depth Gage (The L.S. Starrett Company)

(e) Bow Micrometer

(The L.S. Starrett Company)

(c) Screw Thread Micrometer

Inside Micrometer

(The L.S. Starrett Company)

(The L.S. Starrett Company)

Figure 34-1

tHe 0.001-incH MicRoMeteR A 0.001-inch outside micrometer is shown in Figures 34-2 and 34-3 with its principal parts labeled. ANVIL

SPINDLE

SLEEVE (BARREL)

THIMBLE

BARREL SCALE

THIMBLE SCALE

0

1

2

0.0010 DIVISION

3 0

READING LINE

0.0250 DIVISION (DISTANCE MOVED IN ONE REVOLUTION OF THIMBLE) 0.1000 DIVISION

Figure 34-2

Figure 34-3

(Courtesy of L.S. Starrett Company )

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

The part to be measured is placed between the anvil and the spindle. The barrel of a micrometer consists of a scale, which is one inch long. The one-inch length is divided into 10 divisions each equal to 0.100 inch. The 0.100-inch divisions are further divided into four divisions each equal to 0.025 inch. The thimble has a scale that is divided into 25 parts. One revolution of the thimble moves 0.025 inch on the barrel scale. Therefore, a movement of one graduation on the 1 thimble equals of 0.025 inch, or 0.001 inch, along the barrel. 25

Reading and setting a 0.001-incH MicRoMeteR A micrometer is read by observing the position of the bevel edge of the thimble in reference to the scale on the barrel. Observe the greatest 0.100-inch division and the number of 0.025-inch divisions on the barrel scale. To this barrel reading, add the number of the 0.001-inch divisions on the thimble that coincide with the horizontal line (reading line) on the barrel scale.

c Procedure ●● ●●

HORIZONTAL (READING) LINE

0

1

2

10

3

To read a 0.001-inch micrometer

Observe the greatest 0.100-inch division on the barrel scale. Observe the number of 0.025-inch divisions on the barrel scale. Add the thimble scale reading (0.001-inch division) that coincides with the horizontal line on the barrel scale.

●●

5

Example 1 Read the micrometer setting shown in Figure 34-4. Observe the greatest 0.100-inch division on the barrel scale. (three 0.1000 5 0.3000) Observe the number of 0.025-inch divisions between the 0.300-inch mark and the thimble. (two 0.0250 5 0.0500) Add the thimble scale reading that coincides with the horizontal line on the barrel scale. (eight 0.0010 5 0.0080) Micrometer reading: 0.3000 1 0.0500 1 0.0080 5 0.3580

Figure 34-4

Ans

Example 2 Read the micrometer setting shown in Figure 34-5. 0

1

2

On the barrel scale, two 0.1000 5 0.2000.

0

On the barrel scale, zero 0.0250 5 00. 20

On the thimble scale, twenty-three 0.0010 5 0.0230. Micrometer reading: 0.2000 1 0.0230 5 0.2230

Figure 34-5

c Procedure ●●

●●

7

8

9

Ans

To set a 0.001-inch micrometer to a given dimension

Turn the thimble until the barrel scale indicates the required number of 0.100-inch divisions plus the necessary number of 0.025-inch divisions. Turn the thimble until the thimble scale indicates the required additional 0.001-inch divisions.

Example 1 Set 0.949 inch on a micrometer. 0

Turn the thimble to nine 0.100-inch divisions plus one 0.025-inch division on the barrel scale. (9 3 0.1000 1 0.0250 5 0.9250)

20

Turn the thimble an additional twenty-four 0.001-inch thimble scale divisions. (0.9490 2 0.9250 5 0.0240)

Figure 34-6

The 0.949-inch setting is shown in Figure 34-6.

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CustoMary MiCroMeters

Unit 34

Example 2 Set 0.520 inch on a micrometer.

3

Turn the thimble to five 0.100-inch divisions on the barrel scale. (5 3 0.1000 5 0.5000)

4

5 20

Turn the thimble an additional twenty 0.001-inch divisions. (0.5200 2 0.5000 5 0.0200)

Figure 34-7

The 0.520-inch setting is shown in Figure 34-7.

tHe VeRnieR (0.0001-incH) MicRoMeteR The addition of a vernier scale on the barrel of a 0.001-inch micrometer increases the degree of precision of the instrument to 0.0001 inch. The barrel scale and thimble scale of a vernier micrometer are identical to that of a 0.001-inch micrometer. Figure 34-8 shows the relative positions of the barrel scale, thimble scale, and vernier scale of a 0.0001-inch vernier micrometer. The vernier scale consists of 10 divisions. Ten vernier divisions on the circumference of the barrel are equal in length to nine divisions of the thimble scale. The difference between one vernier division and one thimble division is 0.0001-inch. A flattened view of a vernier and a thimble scale is shown in Figure 34-9. THIMBLE SCALE

VERNIER SCALE

09876543210

210

5 0

1

2

3 0

10 VERNIER DIVISIONS (0.00010)

20 BARREL SCALE

10

5

9 THIMBLE DIVISIONS (0.0010)

0

VERNIER SCALE

THIMBLE SCALE

Figure 34-9

Figure 34-8

Reading and setting tHe VeRnieR (0.0001-incH) MicRoMeteR Reading a vernier micrometer is the same as reading a 0.001-inch micrometer except for the addition of reading the vernier scale. A particular vernier graduation coincides with a thimble scale graduation. This vernier graduation gives the number of 0.0001-inch divisions that are added to the barrel and thimble scale readings.

VERNIER SCALE 09876543210

Example 1 Read the vernier micrometer setting shown in the flattened view in Figure 34-10. Read the barrel scale reading. Three 0.1000 divisions plus three 0.0250 divisions 5 0.3750. Read the thimble scale. The reading is between the 0.0090 and 0.0100 divisions; therefore, the thimble reading is 0.0090.

15 0 1 2 3

BARREL SCALE

Read the vernier scale. The 0.00040 division of the vernier scale coincides with a thimble division.

20

VERNIER COINCIDES

10

THIMBLE SCALE

Figure 34-10

Vernier micrometer reading: 0.3750 1 0.0090 1 0.00040 5 0.38440

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Ans

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

09876543210

5

VERNIER COINCIDES

Example 2 Read the vernier micrometer setting shown in this flattened view in Figure 34-11. On the barrel scale read 0.2000.

0

On the thimble scale read 0.0200.

0 1 2 20

On the vernier scale read 0.00080. Figure 34-11

Vernier micrometer reading: 0.2000 1 0.0200 1 0.00080 5 0.22080 Ans Setting a vernier (0.0001-inch) micrometer is the same as setting a 0.001-inch micrometer except for the addition of setting the vernier scale.

09876543210

Example Set 0.2336 inch on a vernier micrometer.

20 15 0 1 2

VERNIER COINCIDES

10

Figure 34-12

Turn the thimble to two 0.100-inch divisions plus one 0.025-inch division on the barrel scale. (2 3 0.1000 1 0.0250 5 0.2250) Turn the thimble an additional eight 0.001-inch divisions. (0.23360 2 0.2250 5 0.00860) Turn the thimble carefully until a graduation on the thimble scale coincides with the 0.0006-inch division on the vernier scale. (0.23360 2 0.2330 5 0.00060) The 0.2336-inch setting is shown in Figure 34-12.

ApplicAtion tooling Up 1. Use an electronic vernier caliper to measure the length of a ballpoint pen. 2. Read the metric vernier depth gage measurement for this setting. 25

30

20 15 10 5 0.02 mm

20 0

3. Read the decimal-inch vernier caliper measurement for this setting. MAIN SCALE 50

0 5 10 VERNIER SCALE

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60

15

20

25

0.02 mm

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223

CustoMary MiCroMeters

Unit 34

4. A hole has a diameter of 1.578010.0020 . What are the maximum and minimum diameters of the hole? 20.0060 5. A manufacturer has kept a record of various expenses and knows that the costs to produce a certain product are 42% for labor, 39% for materials, and 19% for overhead. During the last year, the labor costs were $975,500. Determine the dollar cost for the materials and overhead. Round each answer to the nearest $10. 6. If 5275 parts can be produced in 12.5 hours, how long will it take to produce 21,500 parts? Round the answer to the nearest whole hour.

0.001-inch Micrometer Read the settings on the following 0.001-inch micrometer scales. 7.

8.

3

4

5

9. 15

0

0

15

10.

5

6

7

0

10

10

20

5

10

11.

12.

5

6

7

8

10

13.

1

2

3

15

15.

16.

1

4

5

6

7

10

5

0

14.

10

15

5

6

5

5

4

0

5

20

10

17.

4

3

18.

7

8

0

9

0

2

3

4

5 20

20

Given the following barrel scale and thimble scale settings of a 0.001-inch micrometer, determine the readings in the tables. The answer to the first problem is given. Barrel Scale Setting Is Between: (inches)

Thimble Scale Setting (inches)

Micrometer Reading (inches)

19.

0.425–0.450

0.016

0.441

20.

0.075–0.100

21.

0.150–0.175

22. 23.

Barrel Scale Setting Is Between: (inches)

Thimble Scale Setting (inches)

24.

0.000–0.025

0.023

0.009

25.

0.025–0.050

0.013

0.003

26.

0.750–0.775

0.017

0.875–0.900

0.012

27.

0.975–1.000

0.008

0.400–0.425

0.024

28.

0.625–0.650

0.016

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Given the following 0.001-inch micrometer readings, determine the barrel scale and thimble scale settings. The answer to the first problem is given.

Micrometer Reading (inches)

Barrel Scale Setting Is Between: (inches)

Thimble Scale Setting (inches)

29.

0.387

0.375–0.400

0.012

30.

Barrel Scale Setting Is Between: (inches)

Micrometer Reading (inches)

34.

0.998

0.841

35.

0.038

31.

0.973

36.

0.281

32.

0.002

37.

0.427

33.

0.079

38.

0.666

Thimble Scale Setting (inches)

the Vernier (0.0001-inch) Micrometer Read the settings on the following 0.0001-inch micrometer scales. The vernier, thimble, and barrel scales are shown in flattened views. 40.

39.

41.

VERNIER SCALE

20 0 1 2 3

15

BARREL SCALE

20

09876543210

09876543210

09876543210

0

15 0 1 2

10

10 5 0

0

THIMBLE SCALE

42.

43.

44.

5 0 1 2 3 4 5

0

20 15 0 1 2 3

46.

20

48.

20 15 0 1 2 3 4

50.

10 5 0 1 2

09876543210

09876543210

09876543210

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20

20 15

5

0

0

0 1 2 3 4 5

10

49.

0

09876543210

0 0 1

5

47. 09876543210

09876543210

5

10 0 1 2 3

10

45.

15

09876543210

09876543210

09876543210

10

20 15 0 1 2 3

0

10

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Unit 34

225

CustoMary MiCroMeters

Given the following barrel scale, thimble scale, and vernier scale settings of a 0.0001-inch micrometer, determine the micrometer readings in these tables. The answer to the first problem is given. Barrel Scale Setting Is Between: (inches)

Thimble Scale Setting Is Between: (inches)

Vernier Scale Setting (inches)

Micrometer Reading (inches)

51.

0.375–0.400

0.017–0.018

0.0008

0.3928

52.

0.125–0.150

0.008–0.009

0.0003

53.

0.950–0.975

0.021–0.022

0.0007

54.

0.075–0.100

0.011–0.012

0.0005

55.

0.300–0.325

0.000–0.001

0.0004

56.

0.625–0.650

0.021–0.022

0.0002

57.

0.000–0.025

0.000–0.001

0.0009

58.

0.275–0.300

0.020–0.021

0.0007

59.

0.850–0.875

0.009–0.010

0.0004

60.

0.225–0.250

0.014–0.015

0.0008

Given the following 0.0001-inch micrometer readings, determine the barrel scale, thimble scale, and vernier scale settings. The answer to the first problem is given. Micrometer Reading (inches)

Barrel Scale Setting Is Between: (inches)

Thimble Scale Setting Is Between: (inches)

Vernier Scale Setting (inches)

61.

0.7846

0.775–0.800

0.009–0.010

0.0006

62.

0.1035

63.

0.0083

64.

0.9898

65.

0.3001

66.

0.0012

67.

0.8008

68.

0.3135

69.

0.9894

70.

0.0479

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UNIT 35 Metric Vernier Micrometers Objectives After studying this unit you should be able to ●● ●●

Read settings on 0.01-millimeter metric micrometer scales. Read settings on 0.002-millimeter metric vernier micrometer scales.

Figure 35-1 shows a 0.01-millimeter outside micrometer.

Figure 35-1 (Courtesy of L.S. Starrett Company )

The barrel of a 0.01-millimeter micrometer consists of a scale that is 25 millimeters long. Refer to the barrel and thimble scales in Figure 35-2. The 25-millimeter barrel scale length is divided into 25 divisions each equal to 1 millimeter. Every fifth millimeter is numbered from 0 to 25 (0, 5, 10, 15, 20, 25). On the lower part of the barrel scale, each millimeter is divided in half (0.5 mm). TOP BARREL SCALE (1-mm DIVISIONS)

0

5

0.01-mm DIVISION

5 0 THIMBLE SCALE

0.5-mm DIVISION (DISTANCE MOVED IN ONE REVOLUTION OF THIMBLE)

Figure 35-2

The thimble has a scale that is divided into 50 parts. One revolution of the thimble moves 0.5 millimeter on the barrel scale. A movement of one graduation on the thimble 1 equals of 0.5 millimeter or 0.01 millimeter along the barrel. 50

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Unit 35

227

MetriC Vernier MiCroMeters

Reading a MetRic MicRoMeteR Procedures for reading a 0.01-millimeter micrometer: ●● Observe the number of 1-millimeter division on the top barrel scale. ●● Observe the number of 0.5-millimeter divisions (either 0 or 1) on the lower part of the barrel scale. ●● Add the thimble reading (0.01 division) that coincides the horizontal line on the barrel scale.

Example 1 Read the metric micrometer setting shown in Figure 35-3. Observe the number of 1-millimeter division on the top barrel scale. (4 3 1 mm 5 4 mm)

0

Observe the number of 0.5-millimeter divisions (either 0 or 1) on the lower part of the barrel scale. (0 3 0.5 mm 5 0 mm)

35 30

Add the thimble reading (0.01 division) that coincides with the horizontal line on the barrel scale. (33 3 0.01 mm 5 0.33 mm)

Figure 35-3

Micrometer reading: 4 mm 1 0.33 mm 5 4.33 mm

Ans

Example 2 Read the metric micrometer setting shown in Figure 35-4. On the top barrel scale read 17 millimeters.

5

On the lower barrel scale read 0.5 millimeter.

10

15

25

On the thimble scale read 0.26 millimeter. Micrometer reading: 17 mm 1 0.5 mm 1 0.26 mm 5 17.76 mm

Ans

30

20 Figure 35-4

tHe MetRic VeRnieR MicRoMeteR The addition of a vernier scale on the barrel of a 0.01-millimeter micrometer increases the degree of precision of the instrument to 0.002 millimeter. The barrel scale and the thimble scale of a vernier micrometer are identical to that of a 0.01-millimeter micrometer. Figure 35-5 shows the relative positions of the barrel scale, thimble scale, and vernier scale of 0.002-millimeter micrometer. VERNIER SCALE

THIMBLE SCALE

20

0

5

5 0

BARREL SCALE

Figure 35-5

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The vernier scale consists of five divisions. Each division equals one-fifth of a thimble 1 division, or of 0.01 millimeter, or 0.002 millimeter. Figure 35-6 shows a flattened view of 5 a vernier scale and a thimble scale.

0

10

8 6

5 VERNIER DIVISIONS (0.002 mm)

5

4

9 THIMBLE DIVISIONS (0.01 mm)

2 0

0

VERNIER SCALE

THIMBLE SCALE

Figure 35-6

Reading a MetRic VeRnieR MicRoMeteR Reading a metric vernier micrometer is the same as reading a 0.01-millimeter micrometer except for the addition of reading the vernier scale. Observe which division on the vernier scale coincides with a division on the thimble scale. If the vernier division that coincides is marked 2, add 0.002 millimeter to the barrel and thimble scale reading. Add 0.004 millimeter for a coinciding vernier division marked 4, and 0.006 for a division marked 6, and add 0.008 millimeter for a division marked 8.

Example 1 A flattened view of a metric vernier micrometer is shown in Figure 35-7. Read the setting. VERNIER SCALE 0 8 6

40

4 2 0

35

0

5

VERNIER COINCIDES

30 25

BARREL SCALE

THIMBLE SCALE

Figure 35-7

Read the barrel scale. (6 3 1 mm 1 0 3 0.5 mm 5 6 mm) Read the thimble scale. (26 3 0.01 mm 5 0.26 mm) Read the vernier scale. (0.004 mm) Vernier micrometer reading: 6 mm 1 0.26 mm 1 0.004 mm 5 6.264 mm Ans

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Unit 35

229

MetriC Vernier MiCroMeters

10

0 8

Example 2 A flattened view of a metric vernier micrometer is shown in Figure 35-8. Read the setting.

6

5

4 2 0

On the barrel scale read 9.5 mm. On the thimble scale read 0.43 mm.

0

On the vernier scale read 0.008 mm. Vernier micrometer reading:

VERNIER COINCIDES

0

5

45

9.5 mm 1 0.43 mm 1 0.008 mm 5 9.938 mm Ans

Figure 35-8

ApplicAtion tooling Up 09876543210

1. Read the setting on the following 0.001-inch micrometer scale. The vernier, thimble, and barrel scales are shown in flattened view.

20 15 0 1 2

10

2. Use a digital caliper to measure the thickness of the side of a drinking cup or coffee mug. 3. Read the metric vernier depth gage measurement for this setting.

90

25 20 15 10

0.02 mm

5

80

0

4. Measure this line segment to the nearest millimeter. 3 5 13 5. Add 9 ft 7 in. 1 11 ft 4 in. 1 3 ft 5 in. 8 16 32 6. Express 235% as a decimal fraction or mixed decimal.

Reading a Metric Micrometer Read the setting of these metric micrometer scales graduated in 0.01 mm. 7.

8.

9.

10. 25

15 0

5

10 5

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15

20

40 35

0

5

20 15

10

15

10 5

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11.

12.

13.

14. 40

0

5

0

30

5

25

0

20

25

15.

10

35

15

30 25

30

16.

17.

18.

10 0

15

0

10

15

5

20

5

45

10

10

5

40

0

Reading a Metric Vernier Micrometer Read the settings of these metric vernier micrometer scales graduated in 0.002 mm. In each case the arrow shows where the vernier division matches a thimble scale graduation.

0 8

35

6

0

6

40

21.

0 8

20.

VERNIER SCALE

0 8

19.

6 4

4

30

4 2

45

2

2

0

0

35

0

30

0

0

5

10

40

15

0

5

20

25 BARREL SCALE

THIMBLE SCALE

30

2

0 45

15 0

15

27.

5

8

10

6 4

4

4

2 2

2

40

6

6

30

5

0

5

35

0

0

0

25 0

10

0 8

35

0 8

0

26. 45

0

0

0

25.

5

20

4

2

4 2

4

20

0 0

25

6

25

6

6

5

24.

0 8

23.

0 8

10

0 8

22.

25

0 20

5

0

30

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Unit 36

4 0

35

0

0

5

0

2

20

2

2

0

5

6

4

4

40

0 8

25

6

6

45

30.

0 8

29.

0 8

28.

231

digitaL MiCroMeters

0

5

10

15 10

0

5

45 40

UNIT 36 Digital Micrometers Objective After studying this unit you should be able to ●●

Read digital micrometers in metric and customary units.

A digital micrometer, like a customary micrometer, is used to measure the length, diameter, or thickness of an object. Just as with a customary micrometer, a digital micrometer has an anvil, spindle, lock nut, sleeve, thimble, and ratchet. In addition, a digital micrometer has three buttons: ON/OFF, ZERO, and IN/mm. Some, like the micrometer in Figure 36-1, have a fourth button: SHIFT/SET. Because the reading is electronically generated and digitally displayed on an LCD display, a digital micrometer requires a battery. Digital micrometers come in various sizes. The most common sizes are 0–25 mm (0–1 in.), 25–50 mm (1–2 in.), 50–75 mm (2–3 in.), and 75–100 mm (3–4 in.).

Figure 36-1 a digital outside micrometer. (The L. S. Starrett Company)

The are some advantages to digital micrometers: (a) measurements are read in a single step on an LCD display, (b) a button allows for switching between metric and inch measurements, (c) some versions allow the data to be saved to a USB drive or printed, (d) some versions are designed so that data can be sent wirelessly or with a traditional cable, and (e) they can provide for SPC (statistical process control) analysis and documentation.

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Use of a MicRoMeteR Before you begin, use a clean cloth to dry and clean the object you are going to measure. Also clean and dry the micrometer including the measuring surface of both the anvil and the spindle of the micrometer. Once the object being measured and the micrometer are clean, do not touch any of the measuring surfaces with your hands since the oil or sweat on your hands might create an inaccurate measurement. (a) Turn the digital micrometer ON by pressing the ON/OFF button and push the IN/mm button to select the desired system of measurement. (b) Close the micrometer jaws and observe the LCD display. If it reads 0.000, you can begin to measure the object. If the display does not read 0.000, adjust the thimble and the ratchet stop until the display reads 0.000 or depress the ZERO button to set the LCD display reading to 0.000. (c) Fully close the digital micrometer and turn the thimble to make sure the thimble 0 line aligns with the center line on the sleeve. (d) Open the jaws of the micrometer by turning the thimble. Place the object to be measured against the anvil and rotate the ratchet until the spindle contacts the object. Make sure the micrometer is perpendicular to the surfaces being measured. (e) Rotate the ratchet stop until the spindle contacts the item. Use only enough pressure on the ratchet stop to allow the object to just fit between the anvil and the spindle. Read the measurement of the LCD display. In Figure 36-2, the object measured 3.077 mm. (The lock nut need not be “locked” unless the micrometer is in an unusual position where the reading cannot be immediately or directly observed. The lock is most often used when checking many similar components. Then it is used in the “locked” position as a “go, no-go” gauge.)

Figure 36-2

ApplicAtion tooling Up 1. Read the setting of this metric micrometer scale graduated in 0.01 mm.

0

20 15

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Unit 36

233

digitaL MiCroMeters

2. Read the setting on this customary vernier micrometer scale graduated to 0.00010. The vernier, thimble, and barrel scales are shown in flattened view. 09876543210

20 15 10

0 1 2 3

3. Use a digital vernier caliper to measure the diameter of the opening at the top of a drinking cup or coffee mug. 4. Read the decimal-inch vernier caliper measurement of this setting. MAIN SCALE 7 1 2

0

5

10

3

4

15

5

20

6

7

25

VERNIER SCALE

0.02 mm

5. Express 2.0276 meters as centimeters. 6. What percent of 92.4 is 12.35? Round the answer to 1 decimal place.

Reading a Digital Micrometer Use a digital micrometer to measure the indicated dimension of each listed object. 7. The thickness, in mm, of your calculator, not including the removable cover. 8. The thickness, in mm, of the removable cover for your calculator. 9. The width, in inches, of the wire of a paperclip. 10. The thickness, in inches, of the edge a 10¢ coin. 11. The width, in inches, of a 10¢ coin. 12. The length, in inches, of a roofing nail. 13. The length, in mm, of a push pin or thumb tack. 14. The thickness, in mm, of wire for a staple. 15. The thickness in both inches and millimeters of a metal washer. 16. The thickness in both inches and millimeters of a paper clip.

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

UNIT 37 Customary and Metric Gage Blocks Objective After studying this unit you should be able to ●●

Determine proper gage block combinations for specified customary or metric system dimensions.

Gage blocks are used in machine shops as standards for checking and setting (calibration) of micrometers, calipers, dial indicators, and other measuring instruments. Other applications of gage blocks are for layout, machine setups, and surface plate inspection.

descRiption of gage Blocks Gage blocks like those in Figure 37-1 are square- or rectangular-shaped hardened steel blocks that are manufactured to a high degree of accuracy, flatness, and parallelism. Gage blocks, when properly used, provide millionths of an inch accuracy with millionths of an inch precision. By wringing blocks (slipping blocks one over the other using light pressure), a combination of the proper blocks can be achieved to provide a desired length. Wringing the blocks produces a very thin air gap that is similar to liquid film in holding the blocks together. There are a variety of both customary unit and metric gage block sets available. The following tables list the thicknesses of blocks of a frequently used customary gage block set and the thicknesses of blocks of a commonly used metric gage block set. To reduce the possibility of error, it is customary to use the fewest number of blocks possible to achieve the stack.

Figure 37-1 a complete set of gage blocks (Courtesy of Brown & Sharpe Mfg. Co.)

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Unit 37

235

CustoMary and MetriC gage BLoCks

BLOCK THICKNESSES OF A CUSTOMARY GAGE BLOCK SET* 9 Blocks 0.0001” Series 0.1001

0.1002

0.1003

0.1004

0.1005

0.1006

0.1007

0.1008

0.1009

49 Blocks 0.001” Series 0.101

0.102

0.103

0.104

0.105

0.106

0.107

0.108

0.109

0.110

0.111

0.112

0.113

0.114

0.115

0.116

0.117

0.118

0.119

0.120

0.121

0.122

0.123

0.124

0.125

0.126

0.127

0.128

0.129

0.130

0.131

0.132

0.133

0.134

0.135

0.136

0.137

0.138

0.139

0.140

0.141

0.142

0.143

0.144

0.145

0.146

0.147

0.148

0.149

19 Blocks 0.050” Series 0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

0.450

0.500

0.550

0.600

0.650

0.700

0.750

0.800

0.850

0.900

0.950

4 Blocks 1.000” Series 1.000

2.000

3.000

4.000

*All thicknesses are in inches.

BLOCK THICKNESSES OF A METRIC GAGE BLOCK SET* 9 Blocks 0.001 mm Series 1.001

1.002

1.003

1.004

1.005

1.006

1.007

1.008

1.009

1.04

1.05

1.06

1.07

1.08

1.09

1.4

1.5

1.6

1.7

1.8

1.9

4

5

6

7

8

9

40

50

60

70

80

90

9 Blocks 0.01 mm Series 1.01

1.02

1.03

9 Blocks 0.1 mm Series 1.1

1.2

1.3

9 Blocks 1 mm Series 1

2

3

9 Blocks 10 mm Series 10

20

30

*All thicknesses are in millimeters.

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

deteRMining gage Block coMBinations Usually there is more than one combination of blocks that will give a desired length. The most efficient procedure for determining block combinations is to eliminate digits of the desired measurement from right to left. This procedure saves time, minimizes the number of blocks, and reduces the chances of error. The following examples show how to apply the procedure in determining block combinations. 0.1008 0.146 0.700 2.9468 2.000

Figure 37-2

0.1003 0.134 0.050 1.000

Choose the block that eliminates the last digit to the right, the 8. Choose the 0.10080 block. Subtract. (2.94680 2 0.10080 5 2.8460) Eliminate the last digit, 6, of 2.8460. Choose the 0.1460 block that eliminates the 4 as well as the 6. Subtract. (2.8460 2 0.1460 5 2.7000) Eliminate the last non-zero digit, 7, of 2.7000. Choose the 0.7000 block. Subtract. (2.7000 2 0.7000 5 2.0000) The 2.0000 block completes the required dimension as shown in Figure 37-2. Check. Add the blocks chosen. (0.10080 1 0.1460 1 0.7000 1 2.0000 5 2.94680)

Example 2 Determine a combination of gage blocks for 10.2843 inches. Refer to the gage block sizes given in the Table of Block Thicknesses for a Customary Gage Block Set. All dimensions are in inches.

2.000

Eliminate the 3. Choose the 0.10030 block. Subtract. (10.28430 2 0.10030 5 10.1840)

3.000

Eliminate the 4. Choose the 0.1340 block. Subtract. (10.1840 2 0.1340 5 10.0500)

10.2843

4.000

Figure 37-3

1.002 1.07 1.3 4

157.372

Example 1 Determine a combination of gage blocks for 2.9468 inches. Refer to the gage block sizes given in the Table of Block Thicknesses of a Customary Gage Block Set. All dimensions are in inches.

60

90

Eliminate the 5. Choose the 0.0500 block. Subtract. (10.0500 2 0.0500 5 10.0000) The 1.0000, 2.0000, 3.0000, and 4.0000 blocks complete the required dimensions as shown in Figure 37-3. Check.

(0.10030 1 0.1340 1 0.0500 1 1.0000 1 2.0000 1 3.0000 1 4.0000 5 10.28430)

Example 3 Determine a combination of gage blocks for 157.372 millimeters. Refer to the gage block sizes given in the Table of Block Thicknesses for a Metric Gage Block Set. All dimensions are in millimeters. Eliminate the 2. Choose the 1.002 mm block. Subtract. (157.372 mm 2 1.002 mm 5 156.37 mm) Eliminate the 7. Choose the 1.07 mm block. Subtract. (156.37 mm 2 1.07 mm 5 155.3 mm) Eliminate the 3. Choose the 1.3 mm block. Subtract. (155.3 mm 2 1.3 mm 5 154 mm) Eliminate the 4. Choose the 4 block. Subtract. (154 mm 2 4 mm 5 150 mm) The 60 and 90 block complete the required dimension as shown in Figure 37-4. Check. (1.002 mm 1 1.07 mm 1 1.3 mm 1 4 mm 1 60 mm 1 90 mm 5 157.372 mm)

Figure 37-4

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Unit 37

237

CustoMary and MetriC gage BLoCks

ApplicAtion tooling Up 1. Use a digital micrometer to measure the thickness of the cover of a book. 2. Read the settings of this metric vernier micrometer scale graduated in 0.002 mm. The arrow shows where the vernier division matches a thimble scale graduation.

0 8 6 4 2 0

25 20

0 5 10

15 10

3. Read the setting on this customary vernier micrometer scale graduated to 0.00010. The vernier, thimble, and barrel scales are shown in flattened view. 09876543210

0 20 0 1 23

15 10

4. Read the metric vernier caliper measurement for the following setting. MAIN SCALE 50

0 5 VERNIER SCALE

60

10

15

20

25

0.02 mm

5. For a measurement 19.700 using a steel tape with the smallest graduation of 0.050, determine, in inches, (a) the greatest possible error, (b) the smallest possible actual length, and (c) the greatest possible actual length. 6. The total amount of time needed to machine a part is 14.75 hours. Milling machine operations take 8.5 hours. What percent of the total time is spent on the milling machine? (Round the answer to 1 decimal place.)

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

customary Gage Blocks Using the Table of Block Thicknesses for a Customary Gage Block Set, determine a combination of gage blocks for each of the following dimensions.

Note: Usually more than one combination of blocks will give the desired dimension. 7. 3.86380

16. 9.0500

24. 0.67540

8. 1.87020

17. 4.87570

25. 7.77770

9. 3.12220

18. 1.00010

26. 10.01010

10. 0.63330

19. 0.26210

27. 9.43460

11. 0.27590

20. 2.73110

28. 4.82080

12. 5.80020

21. 5.0900

29. 6.0030

13. 7.9730

22. 6.08070

30. 10.00210

14. 0.99990

23. 2.97890

31. 0.69980

15. 10.2500

Metric Gage Blocks Using the Table of Block Thicknesses for a Metric Gage Block Set, determine a combination of gage blocks for each of the following dimensions.

Note: Usually more than one combination of blocks will give the desired dimension. 32. 43.285 mm

40. 157.08 mm

48. 41.87 mm

33. 14.073 mm

41. 13.86 mm

49. 2.007 mm

34. 34.356 mm

42. 28.727 mm

50. 107.23 mm

35. 156.09 mm

43. 6.071 mm

51. 193.03 mm

36. 213.9 mm

44. 85.111 mm

52. 73.061 mm

37. 43.707 mm

45. 39.099 mm

53. 10.804 mm

38. 9.999 mm

46. 134.44 mm

54. 149.007 mm

39. 76.46 mm

47. 67.005 mm

55. 55.555 mm

UNIT 38 Achievement Review— Section Three

Objectives You should be able to solve the exercises and problems in this Achievement Review by applying the principles and methods covered in Units 26–37.

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Unit 38

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aChieVeMent reView—seCtion three

1. Express each of the following lengths as indicated. d. 2.7 centimeters as millimeters

a. 81 inches as feet 1 b. 6 feet as inches 4 c. 9.6 yards as feet

e. 0.8 meter as millimeters f. 218 millimeters as centimeters

2. Holes are to be drilled in the length of angle iron as shown in Figure 38-1. What is the distance between two consecutive holes? 14 HOLES EQUALLY SPACED

8 50 8

8 50 8

5 12 FEET

Figure 38-1

3. How many complete 3-meter lengths of tubing are required to make 250 pieces each 54 millimeters long? Allow a total one-half length of tubing for cutoff and scrap. 4. Express each of the following lengths as indicated. When necessary, round the answer to 3 decimal places. a. 47 millimeters as inches b. 5.5 meters as feet c. 16.8 centimeters as inches

d. 4.75 inches as millimeters e. 31 inches as centimeters f. 4.5 feet as meters

5. For each of the exercises in the following table, the measurement made and the smallest graduation of the measuring instrument are given. Determine the greatest possible error and the smallest and largest possible actual length measure for each.

Measurement Made

Smallest Graduation of Measuring Instrument Used

a.

4.280

0.020 (steel rule)

b.

0.83670

0.00010 (vernier micrometer)

c.

46.16 mm

0.02 mm (vernier caliper)

d.

16.45 mm

0.01 mm (micrometer)

Greatest Possible Error

ACTUAL LENGTH Smallest Possible

Largest Possible

6. Compute the Absolute Error and Relative Error of each of the values in the following table. Where necessary, round the answers to 3 decimal places. True Value

Measured Value

True Value

Measured Value

a.

5.963 in.

5.960 in.

d.

0.1070 in.

0.0990 in.

b.

0.392 mm

0.388 in.

e.

0.8639 in.

0.8634 in.

c.

7.123º

7.200º

f.

0.713º

0.706º

a. Absolute Error Relative Error b. Absolute Error Relative Error c. Absolute Error Relative Error

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d. Absolute Error Relative Error e. Absolute Error Relative Error f. Absolute Error Relative Error

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

7. The following dimensions with tolerances are given. Determine the maximum dimension (maximum limit) and the minimum dimension (minimum limit) for each. a. 1.7140 6 0.0050 maximum minimum b. 4.06880 10.00000 20.00120 maximum minimum c. 5.90470 10.00080 20.00000 maximum minimum

d. 64.91 mm 60.08 mm maximum minimum mm e. 173.003 mm 10.000 20.013 mm maximum minimum

8. Express each of the following unilateral tolerances as bilateral tolerances having equal plus and minus values. a. 0.8760 10.0060 20.0000

mm c. 37.53 mm 10.00 20.03 mm

b. 5.26190 10.00000 20.00120

mm d. 78.909 mm 10.009 20.000 mm

9. The following problems require computations with both clearance fits and interference fits between mating parts. Determine the clearance or interference values as indicated. All dimensions are given in inches.

0.9980 ± 0.0007 1.0006 ± 0.0007

a. Find the maximum clearance. b. Find the minimum clearance.

+ 0.0005 – 0.0000 + 0.0000 1.6232 – 0.0005 1.6250

e. Find the maximum interference (allowance). f. Find the minimum interference.

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+ 0.0000 – 0.0007 + 0.0007 1.3010 – 0.0000

1.3004

c. Find the maximum clearance. d. Find the minimum clearance.

0.5967 ± 0.0003 0.5952 ± 0.0003

g. Find the maximum interference. (allowance). h. Find the minimum interference.

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aChieVeMent reView—seCtion three

10. Determine the minimum permissible length of distance A of the part shown in Figure 38-2. All dimensions are in millimeters.

10.38 ± 0.02

11.52 ± 0.02 A

38.54 ± 0.04

Figure 38-2

11. Read measurements a–h on the enlarged 32nds and 64th graduated fractional rule shown in Figure 38-3. e

g

f

h

64

1

32 a

b

c

d

Figure 38-3

a. b.

c. d.

e. f.

g. h.

12. Read measurements i–p on the enlarged 32nds and 64th graduated fractional rule shown in Figure 38-4. n

m

p

o

64

1

32 i

j

k

l

Figure 38-4

i. j.

k. l.

m. n.

o. p.

13. Read measurements a–h on the enlarged 50ths and 100th graduated decimal-inch rule shown in Figure 38-5. f

e

g

100

h

1

50 a

b

c

d

Figure 36-5

a. b.

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c. d.

e. f.

g. h.

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Section 3 Linear MeasureMent: CustoMary (engLish) and MetriC

14. Read measurements i–p on the enlarged 50ths and 100th graduated decimal-inch rule shown in Figure 38-6. m

n

o

p

100

1

50 i

j

k

l

Figure 36-6

i. j.

k. l.

m. n.

o. p.

15. Read measurements a–h on the enlarged 1 millimeter and 0.5 millimeter graduated metric rule shown in Figure 38-7. e

f

0.5 mm 10 1 mm

20

a

g

30

40

b

h

50

60

c

70

d

Figure 36-7

a. b.

c. d.

e. f.

g. h.

16. Read measurements i–p on the enlarged 1 millimeter and 0.5 millimeter graduated metric rule shown in Figure 38-8. m

n

0.5 mm 10 1 mm

o

20

i

30

p

40

50

j

60

k

70

l

Figure 36-8

i. j.

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k. l.

m. n.

o. p.

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aChieVeMent reView—seCtion three

17. Read the vernier caliper and height gage measurements for the following decimal-inch settings.

8

9

4

1

2

3

4

3

25

5

2

20

4

1

15

3

8 0

5

10

15

VERNIER SCALE

20

8

0.0010

a.

5 0

7

b.

20 15

2

10

9

25

25

10

1

3

5 9

0

0.0010

7

0.0010

MAIN SCALE

c.

18. Read the settings on the following 0.001 decimal-inch micrometer scales. 3

4

5

10

0

5

a.

5

15

6

5

6

c.

15 10

0

10

b.

4

5

d.

19. Read the settings on the following 0.0001 decimal-inch vernier micrometer scales.

a.

5

15 0 1 2 3

5 0

10

b.

10

0

0 1 2 3 4

0

c.

5

09876543210

0 1 2 3

20

09876543210

10

09876543210

09876543210

15

20

d.

20. Using the Table of Block Thicknesses for a Customary Gage Block Set found in Unit 37, determine a combination of gage blocks for each of the following dimensions.

Note: Usually more than one combination of blocks will give the desired dimension. a. 0.37840 b. 2.54860 c. 1.70620 d. 5.64670

e. 3.09010 f. 0.20090 g. 7.88950 h. 8.00140

21. Using the Table of Block Thicknesses for a Metric Gage Block Set found in Unit 37, determine a combination of gage blocks for each of the following dimensions.

Note: Usually more than one combination of blocks will give the desired dimension. a. 67.53 mm b. 125.22 mm c. 85.092 mm d. 13.274 mm

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e. 66.066 mm f. 43.304 mm g. 99.998 mm h. 107.071 mm

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4

Fundamentals of Algebra

UNIT 39 Symbolism and Algebraic Expressions Objectives After studying this unit you should be able to ●● ●● ●●

Express word statements as algebraic expressions. Express diagram dimensions as algebraic expressions. Evaluate algebraic expressions by substituting numbers for symbols.

sectiON FOUR

Algebra is a branch of mathematics in which letters are used to represent numbers. By the use of letters, general rules called formulas can be stated mathematically. Algebra is an extension of arithmetic; therefore, the rules and procedures that apply to arithmetic also apply to algebra. Many problems that are difficult or impossible to solve by arithmetic can be solved by algebra. The basic principles of algebra discussed in this text are intended to provide a practical background for machine shop applications. A knowledge of algebraic fundamentals is essential in the use of trade handbooks and for the solutions of many geometric and trigonometric problems.

SymboliSm Symbols are the language of algebra. Both arithmetic numbers and literal numbers are used in algebra. Arithmetic numbers are numbers that have definite numerical values, such 7 as 4, 5.17, and . Literal numbers are letters that represent arithmetic numbers, such as a, x, 8 V, and P. Depending on how it is used, a literal number can represent one particular arithmetic number, a wide range of numerical values, or all numerical values. Customarily, the multiplication sign s3d is not used in algebra, because it can be misinterpreted as the letter x. When a literal number is multiplied by a numerical value, or when two or more literal numbers are multiplied, no sign of operation is required. 244

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SymboliSm and algebraic expreSSionS

Examples 1. 5 times a is written 5a 2. 17 times c is written 17c 3. V times P is written VP 4. 6 times a times b times c is written 6abc Parentheses ( ) are often used in place of the multiplication sign (3) when numerical values are multiplied; 3 3 4 is written 3s4d; 18 3 3.4 3 52 is written 18s3.4ds52). A raised dot ? is also used by some people as a multiplication sign. Here, 3 3 4 is written 3 ? 4 and 6 3 7.25 as 6 ? 7.25. It is better to use parentheses if there is any chance that the raised dot might be confused for a decimal point. So, either 6(7.25) or (6)(7.25) is preferred to 6 ? 7.25. An algebraic expression is a word statement put into mathematical form by using literal numbers, arithmetic numbers, and signs of operation. The following are examples of algebraic expressions.

Example 1 As shown in Figure 39-1, a dimension is increased by 0.5 inch. How long is the increased dimension? All dimensions are in inches.

x

0.5 x + 0.5

Figure 39-1

If x is the original dimension, the increased dimension is x 1 0.50.

Ans

Example 2 The production rate of a new machine is four times as great as an old machine. Write an algebraic expression for the production rate of the new machine. If the old machine produced y parts per hour, the new machine produces 4y parts per hour. Ans

Example 3 As shown in Figure 39-2, a drill rod is cut into 3 equal pieces. How long is each piece? (Disregard waste.)

L 3

L 3

L 3

L

Figure 39-2

L If L is the length of the drill rod, the length of each piece is . Ans 3

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3 4 A and dimension C is twice dimension A. Find the total height of the block.

Example 4 In the step block shown in Figure 39-3, dimension B equals of dimension

C = 2d 3 B=

3 4

3 4

d

d

A=d

Figure 39-3

3 If d is the length of dimension A, dimension B is d and dimension C is 2d. The total height 4 3 3 is d 1 d 1 2d or 3 d. Ans 4 4

Note: If no arithmetic number appears before a literal number, it is assumed that the value is the same as if a one (1) appeared before the letter, d 5 1d. Example 5 A plate with eight drilled holes is shown in Figure 39-4. The distance from the left edge of the plate to hole 1 and the distance from the right edge of the plate to hole 8 are each represented by a. The distances between holes 1 and 2, holes 2 and 3, and holes 3 and 4 are each represented by b. The distances between holes 4 and 5, holes 5 and 6, holes 6 and 7, and holes 7 and 8 are each represented by c. Find the total length of the plate. All dimensions are in millimeters. 1

a

2

b

3

b

4

b

5

c

6

c

7

c

8

c

a

2a + 3b + 4c

Figure 39-4

The total length of the plate is a 1 b 1 b 1 b 1 c 1 c 1 c 1 c 1 a, or 2a 1 3b 1 4c. Ans

Note: Only like literal numbers may be arithmetically added.

Evaluation of algEbraic ExprESSionS Certain problems in this text involve the use of formulas. Some problems require substituting numerical values for letter values. The problems are solved by applying the order of operations of arithmetic as presented in Section 1, Unit 16.

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247

SymboliSm and algebraic expreSSionS

order of operations for combined operations of addition, Subtraction, multiplication, Division, powers, and roots ●●

●●

●●

●●

Do all the work in parentheses first. Parentheses are used to group numbers. In a problem expressed in fractional form, two or more numbers in the dividend (numerator) and/or divisor (denominator) may be considered as being enclosed in parentheses. 4.87 1 0.34 For example, may be considered as s4.87 1 0.34d 4 s9.75 2 8.12d. If an 9.75 2 8.12 algebraic expression contains parentheses within parentheses or brackets, such as [5.6 3 s7 2 0.09d 1 8.8], do the work within the innermost parentheses first. Do powers and roots next. The operations are performed in the order in which they occur from left to right. If a root consists of two or more operations within the radical sign, perform all the operations within the radical sign, then extract the root. Do multiplication and division next. The operations are performed in the order in which they occur from left to right. Do addition and subtraction last. The operations are performed in the order in which they occur from left to right.

Again, you can use the memory aid “Please Excuse My Dear Aunt Sally” to help remember the order of operations. The P in “Please” stands for parentheses, the E for exponents (or raising to a power) and roots, the M and D for multiplication and division, and the A and S for addition and subtraction.

Example 1 The formula for finding the perimeter of a rectangle is given. Find the perimeter of the rectangle shown in Figure 39-5. All dimensions are in millimeters. P 5 2L 1 2W P 5 2s50 mmd 1 2s30 mmd P 5 100 mm 1 60 mm P 5 160 mm Ans

P 5 2L 1 2W where P 5 perimeter L 5 length W 5 width L = 50

W = 30

Figure 39-5

Example 2 The formula for finding the area of a ring is given. Find the area of the ring shown in Figure 39-6. All dimensions are in inches. Round the answer to 2 decimal places. A 5 pR2 2 pr2

where

A 5 area R 5 outside radius r 5 inside radius

R = 5.126

r = 2.017

Figure 39-6

The symbol p (pi) represents a constant value used in mathematical relationships involving circles. It is described in Unit 53.

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FundamentalS oF algebra

Scientific calculators have a pi key, p . Depressing the pi key, p , enters the value of pi to 10 digits (3.141592654) on most calculators. On some calculators, p is an alternate function. You may have to press SHIFT or 2nd first. A 5 pR2 2 pr 2 A 5 p (5.126 in.)2 2 p (2.017 in.)2 p 3 5.126 x 2 2 p 3 2.017 x 2 5 69.76719217 A < 69.77 sq in. Ans (rounded) Spreadsheets use the command PI( ) for the value of pi. For example, the area would be entered as 5 PI()*5.126 ^ 22PI()*2.017 ^ 2. When you press RETURN , the result is 69.76719217. So A 5 69.77 Ans.

Example 3 The formula for the approximate perimeter of an ellipse is given. Find the perimeter of the ellipse shown in Figure 39-7. All dimensions are in inches. Round the answer to 2 decimal places. P 5 pÏ2sa2 1 b2d

where P 5 perimeter a 5 0.5 smajor axisd b 5 0.5 sminor axisd

P 5 pÏ2s8.502 1 6.422d p 3 ( 2 3 ( 8.5

x2

1 6.42

x2

)

)

5 47.32585933 P < 47.33 in. Ans (rounded) a = 8.50 b = 6.42

Figure 39-7

Example 4 Find the value of

3s2b 1 3dyd when b 5 6, d 5 4, and y 5 2. 4s7d 2 bd d

3[2s6d 1 3s4ds2d] 3s12 1 24d 3s36d 108 5 5 5 5 6.75 Ans 4[7s4d 2 6s4d] 4s28 2 24d 4s4d 16

Example 5 Find the value of 3m[4p 1 5sx 2 md 1 p]2 when m 5 2, p 5 3, and x 5 8. 3s2d[4s3d 1 5s8 2 2d 1 3]2 5 6[12 1 5s6d 1 3]2 5 6s12 1 30 1 3d2 5 6s45d2 5 6s2025d 5 12,150 Ans 6a abc 3 1 sa 2 12bd when a 5 5, b 5 10, and c 5 8. b 20 6s5d 5s10ds8d 3 30 400 1 [5 2 12s10d] 5 1 s125 2 120d 10 20 10 20 5 3 1 20s5d 5 3 1 100 5 103 Ans

Example 6 Find the value of

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UNIT 39

249

SymboliSm and algebraic expreSSionS

ApplicAtion Tooling Up 1. Use the Table of Block Thicknesses of a Customary Gage Block Set under the heading “Description of Gage Blocks“ in Unit 37 to determine a combination of gage blocks for 3.76420. 2. Use a digital micrometer to measure the length and the thickness of a key such as a house key. 3. Read the setting of the metric micrometer scale in Figure 39-8 graduated in 0.01 mm. 4. Use an electronic vernier caliper to measure the diameter of a 25¢ coin.

0

5

45

5. A hole has a diameter of 1.6214 in. 6 0.0007 in. Determine the maximum and minimum diameter.

40

6. 43 is 62% of what number? Round the answer to 2 decimal places.

Figure 39-8

Algebraic expressions Express each of the following problems as an algebraic expression. 7. The product of 6 and x increased by y. 8. The sum of a and 12. 9. Subtract b from 21.

11. Divide r by s. 12. Twice L minus one-half P. 13. The product of x and y divided by the square of m.

10. Subtract 21 from b. 14. In the part shown in Figure 39-9, all dimensions are in inches. a. What is the total length of this part? b. What is the length from point A to point B?

A

B

3

x

4

x

y

5

Figure 39-9

15. Find the distance between the indicated points of Figure 39-10. a. Point A to point B b. Point F to point C c. Point B to point C d. Point D to point E

F D

5 X 8

1 12 X

A X

B E C 3 X 4

Figure 39-10

16. What are the lengths of the following dimensions in Figure 39-11? All dimensions are in millimeters. a. Dimension A b. Dimension B c. Dimension C

L B

h

l

H A

C

12.5

Figure 39-11

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SecTIoN 4

FundamentalS oF algebra

17. Stock is removed from a block in two operations. The original thickness of the block is represented by n. The thickness removed by the milling operation is represented by p and the thickness removed by the grinding operation is represented by t. What is the final thickness of the block? 18. Given: In Figure 39-12, s is the length of a side of a hexagon, r is the radius of the inside circle, and R is the radius of the outside circle. a. What is the length of r if r equals the product of 0.866 and the length of a side of the hexagon? b. What is the length of R if R equals the product of 1.155 and the radius of the inside circle? c. What is the area of the hexagon if the area equals the product of 2.598 and the square of the radius of the outside circle?

s

r R

Figure 39-12

evaluation of Algebraic expressions Substitute the given numbers for letters and find the values of the following expressions. 19. If a 5 5 and c 5 3, find a. 5a 1 3c2 b. 5c 1 a

10c a a1c d. a2c

a 1 5c ac 1 a 5c 1 a f. 5c 2 a

c.

e.

20. If b 5 8, d 5 4, and e 5 2, find a.

b 1e23 d

b. bd s3 1 4d 2 bd c. 5b 2 sbd 1 3d 21. If x 5 12 and y 5 6, find a. 2xy 1 7 b. 3x 2 2y 1 xy c.

5xy 2 2y 8x 2 xy

d. 3esb 2 ed 2 d

1b22

12d 2 [3b 2 sd 1 ed 1 4] e b1d1e f. 1 bd2 2 de3 d2e

e.

4x 2 4y 3 e. 6x 2 3y 1 xy

d.

f. x2 2 y2

22. If m 5 5, p 5 4, and r 5 3, find a. m 1 mp2 2 r2

d.

p3 1 3p 2 12 m2 1 15

b. s p 1 2d2 sm 2 rd2

e.

r3 1 m2smp 2 6rd2 3p 2 9

f.

m3 2 p2 1 5 p 1 2m 1 r2 m1p1r

c.

s prd2 2 pr 1 m3 2

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UNIT 39

251

SymboliSm and algebraic expreSSionS

For Exercises 23 through 34, round the answers to 1 decimal place. 23. All dimensions in Figure 39-13 are in inches. a. Find the area (A) of this square. 1 A 5 d2 2

S

d = 15.00

b. Find the side (S) of this square. S 5 0.7071d S

Figure 39-13

24. All dimensions in Figure 39-14 are in millimeters.

l

a. Find the length of this arc (l). pRa l5 1808

 = 122.25°

b. Find the area of this sector (A). 1 A 5 Rl 2

R = 60.120

Figure 39-14

25. All dimensions in Figure 39-15 are in inches. Refer to the triangle shown. 1 a. Find S when S 5 sa 1 b 1 cd. 2 b. Find the area (A) when A 5 ÏSsS 2 adsS 2 bdsS 2 cd.

c = 12.360 b = 9.032

a = 7.930

Figure 39-15

26. All dimensions in Figure 39-16 are in millimeters. a. Find the radius of this circle. c2 1 4h2 r5 8h b. Find the length of the arc (l). l 5 0.0175ra

l h = 30.85 c = 120.06 r  = 106.80°

Figure 39-16

27. All dimensions are in inches. Find the shaded area of Figure 39-17. Area 5

a = 1.500

c = 1.008 b = 5.122

sH 1 hdb 1 ch 1 aH 2 H = 6.000

h = 4.076

Figure 39-17

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SecTIoN 4

28. All dimensions in Figure 39-18 are in inches. Find the length of belt on the pulleys. Length of belt 5 2C 1

FundamentalS oF algebra

D = 12.25

11D 1 11d sD 2 dd2 1 7 4C d = 8.00

C = 14.50

Figure 39-18

29. All dimensions are in millimeters. Find the shaded area of Figure 39-19. Area 5 dt 1 2a ss 1 nd

n = 1.06

t = 3.26

a = 2.25 s = 0.50 d = 6.18

Figure 39-19

30. All dimensions are in inches. Find the shaded area of Figure 39-20. Area 5 p sab 2 cd d d = 2.00

b = 3.00

c = 4.00 a = 5.00

Figure 39-20

31. All dimensions are in inches. Find the shaded area of Figure 39-21. Area 5

psR2 2 r2d 2 R = 8.103

r = 5.076

Figure 39-21

32. All dimensions are in centimeters. Find the shaded area of Figure 39-22.

a = 5.12

Area 5 t[b 1 2sa 2 td]

a = 5.12 t = 2.00 b = 15.15

Figure 39-22

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UNIT 40

253

Signed numberS

33. All dimensions in Figure 39-23 are in inches.

R = 5.500 S

a. Find the slant height (S). S 5 ÏsR 2 rd2 1 h2 b. Find the volume. Volume 5 1.05h sR2 1 Rr 1 r2d

h = 4.074

r = 2.512

Figure 39-23

34. All dimensions in Figure 39-24 are in inches. Find the volume. s2a 1 cdbh Volume 5 6

c = 12.043

h = 8.000

b = 7.126 a = 10.235

Figure 39-24

UNIT 40 Signed Numbers Objectives After studying this unit you should be able to ●● ●● ●●

●●

Compare signed numbers according to size and direction using the number scale. Determine absolute values of signed numbers. Perform basic operations of addition, subtraction, multiplication, division, powers, and roots using signed numbers. Solve expressions that involve combined operations of signed numbers.

Signed numbers are required for solving problems in mechanics and trigonometry. Positive and negative numbers express direction, such as machine table movement from a reference point. Signed numbers are particularly useful in programming machining operations for numerical control.

mEaning of SignED numbErS Plus and minus signs, which you have worked with so far in this book, have been signs of operation. These are signs used in arithmetic, with the plus sign (1) indicating the operation of addition and the minus sign (2) indicating the operation of subtraction. In algebra, plus and minus signs are used to indicate both operation and direction from a reference point or zero. A positive number is indicated either with no sign or with a plus sign (1) preceding the number. For example, 17 or 7 is a positive number that is 7 units

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FundamentalS oF algebra

greater than zero. A negative number is indicated with a minus sign (2) preceding the number. For example, 27 is a negative number that is 7 units less than zero. Positive and negative numbers are called signed numbers or directed numbers.

thE numbEr ScalE A number scale like the one in Figure 40-1 shows the relationship of positive and negative numbers. It shows both distance and direction between numbers. Considering a number as a starting point and counting to a number to the right represents positive (1) direction with numbers increasing in value. Counting to the left represents negative (2) direction with numbers decreasing in value. The number 0 does not have a sign, because 0 is neither positive nor negative. NEGATIVE NUMBERS –10 –9

–8

–7

–6

–5

–4

–3

POSITIVE NUMBERS –2

–1

0

+1

+2

+3

+4

+5

+6

+7

+8

+9 +10

POSITIVE DIRECTION NEGATIVE DIRECTION

Figure 40-1

Examples 1. Starting at 0 and counting to the right to 15 represents 5 units in a positive (1) direction; 15 is 5 units greater than 0. 2. Starting at 0 and counting to the left to 25 represents 5 units in a negative (2) direction; 25 is 5 units less than 0. 3. Starting at 22 and counting to the right to 16 represents 8 units in a positive (1) direction; 16 is 8 units greater than 22. 4. Starting at 16 and counting to the left to 22 represents 8 units in a negative (2) direction; 22 is 8 units less than a 16. 5. Starting at 23 and counting to the left to 210 represents 7 units in a (2) direction; 210 is 7 units less than 23. 6. Starting at 29 and counting to the right to 0 represents 9 units in a (1) direction; 0 is 9 units greater than 29.

opErationS uSing SignED numbErS In order to solve problems in algebra, you must be able to perform the basic operations using signed numbers. The following procedures and examples show how to perform operations of addition, subtraction, multiplication, division, powers, and roots with signed numbers.

absolute value The procedures for performing certain operations of signed numbers are based on an understanding of absolute value. The absolute value of a number is the number without regard to its sign. For example, the absolute value of 14 is 4, the absolute value of 24 is also 4. Therefore, the absolute value of 14 and 24 is the same value, 4. The absolute value of a number is indicated by placing the number between a pair of vertical bars, |1 4u 5 4, u24u 5 4. The absolute value of 0 is 0, that is, u0u 5 0.

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The absolute value of 220 is 15 greater than the absolute value of 15; 20 is 15 greater than 5.

aDDition of SignED numbErS c Procedure ●●

To add two or more positive numbers

Add the numbers as in arithmetic. Positive numbers do not require a positive sign as a prefix.

Examples Add the following numbers. 1. 13 15 18 Ans

c Procedure ●● ●●

2. 15 7 22

3. 2 1 9 1 13 5 24 Ans

Ans

4. 112 1 (115) 5 127

Ans

To add two or more negative numbers

Add the absolute values of the numbers. Prefix a minus sign to the sum.

Examples Add the following numbers. 1. 25 22 27

Ans

c Procedure ●● ●●

2. 213 24 215 232

3. 26 1 (25) 5 211

Ans

Ans

4. 28 1 (210) 1 (24) 1 (23) 5 225 Ans

To add a positive and a negative number

Subtract the smaller absolute value from the larger absolute value. Prefix the sign of the number having the larger absolute value to the difference.

Examples Add the following numbers. 1. 15 23 12 2. 25 13 22

Ans

Ans

c Procedure ●● ●● ●●

3. 217 117 0

Ans

4. 112 1 (28) 5 14

Ans

5. 212 1 (18) 5 24

Ans

To add more than two positive and negative numbers

Add all the positive numbers. Add all the negative numbers. Add their sums following the procedure for adding signed numbers.

Examples Add the following numbers. 1. 22 1 4 1 (210) 1 5 5 9 1 (212) 5 23

Ans

2. 8 1 7 1 (26) 1 4 1 (23) 1 (25) 1 10 5 29 1 (214) 5 15 Ans 3. 4 1 (26) 1 12 1 3 1 (27) 1 1 1 (25) 1 (22) 5 20 1 (220) 5 0 Ans

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Enter a negative number in a calculator by first pressing the negative sign key (2) , then enter the absolute value of the number.

Example 1 Add 225.873 1 (2138.029). (2) 25.873 1 (2) 138.029 5 2163.902

Ans

Example 2 Add 26.053 1 (20.072) 1 (215.763) 1 (20.009). (2) 6.053 1 (2) .072 1 (2) 15.763 1 (2) .009

5 221.897

Ans

With a spreadsheet, you do not have to use parentheses around a negative number that is being added to another number.

Example 1 Use a spreadsheet to add 225.873 1 (2138.029). Solution Here we will use parentheses: Enter 5 225.873 1 (2138.029) in Cell A1 and press

RETURN

.

2163.902

Ans

Example 2 Use a spreadsheet to add 26.053 1 (20.072) 1 (215.763) 1 (210.009). Solution Here we will omit the parentheses: Enter 5 26.053 1 20.072 1 215.763 1 20.009 in Cell A2 and press

RETURN

232.897

.

Ans

Subtraction of SignED numbErS c Procedure ●● ●●

To subtract signed numbers

Change the sign of the number subtracted (subtrahend) to the opposite sign. Follow the procedure for addition of signed numbers.

Note: When the sign of the subtrahend is changed, the problem becomes one of addition. Therefore, subtracting a negative number is the same as adding a positive number. Subtracting a positive number is the same as adding a negative number. Examples 1. Subtract 5 from 8.

8 2 (15) 5 8 1 (25) 5 3

2. Subtract 8 from 5.

5 2 (18) 5 5 1 (28) 5 23

3. Subtract 25 from 8. 4. Subtract 25 from 28.

Ans Ans

8 2 (25) 5 8 1 (15) 5 13

Ans

28 2 (25) 5 28 1 (15) 5 23

5. 23 2 (17) 5 23 1 (27) 5 210 6. 0 2 (214) 5 0 1 (114) 5 14 7. 0 2 (114) 5 0 1 (214) 5 214

Ans

Ans

Ans Ans

8. 214 2 (214) 5 214 1 (114) 5 0

Ans

Use a calculator to perform the following subtractions.

Example 1 Subtract 2163.94 2 (2150.65). (2) 163.94 2 (2) 150.65 5 213.29

Ans

Example 2 Subtract 227.55 2 (28.64 1 0.74) 2 (253.41). (2) 27.55 2 ( (2) 8.64 1 .74 ) 2 (2) 53.41 5 33.76

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Signed numberS

Use a spreadsheet for the next two subtraction examples.

Example 1 Subtract 2163.94 2 (2150.65). Solution Enter 5 2163.94 2 (2150.65) in Cell A3 and press

RETURN

1 4

. 213.29 Ans

1 382. Round

Example 2 Use a spreadsheet to evaluate 4.212s28.4d27 2 25 your answer to 2 decimal places.

Solution Notice how parentheses are used with the compound fractions. Enter 5 4.21 2 (28.4) 2 (7 1 1/ 4) 2 (2(5 1 3/ 8)) in Cell A4 and press 10.735. 10.74 Ans

RETURN

multiplication of SignED numbErS c Procedure

To multiply two or more signed numbers

Multiply the absolute values of the numbers. Count the number of negative signs. If there is an odd number of negative signs, the product is negative. If there is an even number of negative signs, the product is positive. If all numbers are positive, the product is positive. It is not necessary to count the number of positive values in an expression consisting of both positive and negative numbers. Count only the number of negative values to determine the sign of the product. ●● ●●

Examples 1. 4(23) 5 212

2. 24(23) 5 112

Ans

(There is one negative sign. Since one is an odd number, the product is negative.)

Ans

(There are two negative signs. Since two is an even number, the product is positive.)

3. (22)(24)(23)(21)(22)(21) 5 148 4. (22)(24)(23)(21)(22) 5 248 5. (2)(4)(3)(1)(2) 5 148

Ans

Ans

Ans

6. (2)(24)(23)(1)(22) 5 248 7. (22)(4)(23)(21)(22) 5 148

(Six negatives, even number, positive product) (Five negatives, odd number, negative product) (All positives, positive product)

Ans Ans

(Three negatives, odd number, negative product) (Four negatives, even number, positive product)

Note: The product of any number or numbers and 0 5 0; for example, 0(9) 5 0; 0(29) 5 0; 8(26)(0)(6) 5 0.

Example Use a calculator to multiply (28.61)(3.04)(21.85)(24.03)(0.162). Round the answer to 1 decimal place. (2) 8.61 3 3.04 3 (2) 1.85 3 (2) 4.03 3 .162 5 231.61320475, 231.6 Ans (rounded)

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Examples 1. Use a spreadsheet to evaluate 2.65 3 (24.1).

Solution Enter 5 2.65 * (24.1) in Cell A5 and press

RETURN

.

210.865 Ans

3 3 2. Use a spreadsheet to multiply 1 3 27 . 5 8

Solution Here we will omit the parentheses: Enter 5 (1 1 3/5) * (2(7 1 3/8)) in Cell A6 and press

RETURN

211.8

.

Ans

DiviSion of SignED numbErS c Procedure

To divide two signed numbers

Divide the absolute values of the numbers. Determine the sign of the quotient. If both numbers have the same sign (both negative or both positive) the quotient is positive. If the two numbers have unlike signs (one positive and one negative) the quotient is negative.

●● ●●

Examples 28 5 14 Ans 22 8 2. 5 14 Ans 2 3. 15 4 3 5 15 Ans

230 5 26 Ans 5 30 6. 5 26 Ans 25 7. 221 4 3 5 27 Ans

4. 23q 215 5 15 Ans

8. 23q 21 5 27 Ans

1.

5.

Example Divide: 31.875 4 (256.625). Round the answer to 3 decimal places. 31.875 4 (2) 56.625 5 20.562913907, 20.563 Ans (rounded)

Examples 1. Use a spreadsheet to evaluate 27.364 4 29.43. Round your answer to 3 decimal places.

Solution Enter 5 27.364/2 9.43 in Cell A7 and press 22.902

RETURN

22.9018028

Ans

5 2. Divide: 23 4 2.96. Express the answer as a compound fraction and as a decimal 6 rounded to 2 decimal places.

Solution Enter 5 2(3 1 5/6)/2.96 in Cell A8 and press 18 21 61

Ans

21.30

RETURN

21.295945

Ans

Note: Zero divided by any number equals zero. For example, 0 4 (13) 5 0, 0 4 (23) 5 0. Dividing by zero is undefined. For example, 13 4 0 and 23 4 0 are not defined.

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Signed numberS

powErS of SignED numbErS c Procedure ●●

To raise numbers with positive exponents to a power

Apply the procedure for multiplying signed numbers to raising signed numbers to powers.

Examples 1. 32 5 19

5. (23)2 5 (23)(23) 5 19

Ans

Ans

2. 33 5 127

Ans

6. (23) 5 (23)(23)(23) 5 227

3. 24 5 116

Ans

7. (22)4 5 (22)(22)(22)(22) 5 116

4. 25 5 132

Ans

8. (22)5 5 (22)(22)(22)(22)(22) 5 232

3

Ans Ans Ans

Note: ●● ●● ●● ●●

A positive number raised to any power is positive. A negative number raised to an even power is positive. A negative number raised to an odd power is negative. Use parentheses to enclose a negative number if it is to be raised to a power. As presented in Unit 17, the universal power key, positive number to a power.

xy

, or caret key

^ raises any

Example Solve 2.0735. Round the answer to 2 decimal places. 2.073 x y (or ^ ) 5 5 38.28216674, 38.28 Ans (rounded) The universal power key can also be used to raise a negative number to a power. The negative number to be raised must be enclosed within parentheses. Notice that this was done in Examples 5 through 8 above. The reason is that a negative number, such as 27, is considered to be 21 ? 7, so 272 5 21 ? 72 5 21 ? 49 5 249. But, s27d2 5 s27ds27d 5 149 5 49.

Example 1 Solve (23.874)4. Round the answer to 1 decimal place. ( (2)

3.874

)

xy

(or

^ ) 4 5 225.236342, 225.2 Ans (rounded)

Note: 23.874 must be enclosed within parentheses. Example 2 Solve (23.874)5. Round the answer to 1 decimal place. ( (2)

3.874

)

xy

(or

^ ) 5 5 2872.565589, 2872.6 Ans (rounded)

A number with a negative exponent is equal to the reciprocal of the number (the number inverted) with a positive exponent. If x represents a number and n represents an x2n 1 exponent, then 5 n. x 1

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c Procedure ●● ●●

To raise numbers with negative exponents to a power

Invert the number. Change the negative exponent to a positive exponent.

Examples 322 1 1 5 2 5 , or 0.111 Ans (rounded) 1 3 9 23 2 1 1 2. 223 5 5 3 5 , or 0.125 Ans 1 2 8 23 24 1 1 3. 2423 5 5 5 , or 20.016 Ans (rounded) 1 243 264

1. 322 5

A negative exponent is entered with the negative key (2) . The rest of the procedure is the same as used with positive exponents.

Example 1 Calculate 3.16223. Round the answer to 3 decimal places. 3.162

xy

(or

^ ) (2) 3 5 0.0316311078, 0.032 Ans (rounded)

Example 2 Calculate (23.162)23. Round the answer to 3 decimal places. ( (2)

3.162

)

xy

(or

^ ) (2) 3 5 20.031631108, 20.032 Ans (rounded)

Use a spreadsheet to complete the following Examples.

Example 1 Use a spreadsheet and the power function to evaluate s24.75d3. Round the answer to 3 decimal places.

Solution Enter 5 POWER(24.75, 3) in Cell A1 and press 2107.172

RETURN

2107.17188.

Ans

1 2 1

6

Example 2 Use a spreadsheet and the ^ key to compute 25 . Round the answer 3 to 2 decimal places.

Solution Enter 5 (2(5 1 1/3)) ^ 6 in Cell A8 and press 23,014.01

RETURN

23014.0137.

Ans

Example 3 Use a spreadsheet to evaluate 22.61224. Round the answer to 3 decimal places. Solution Enter 5 22.612 ^ (24) in Cell A3 and press

0.21

Ans

RETURN

0.021483612.

Notice that 22.61224 was considered to be s22.612d24 in the spreadsheet. This is different than when using a calculator.

rootS of SignED numbErS When either a positive number or a negative number is squared, a positive number results. For example, 32 5 9 and s23d2 5 9. Therefore, every positive number has two square roots, one positive root and one negative root. The square roots of 9 are 13 and 23. The expression Ï9 is used to indicate the positive or principal square root, 13 or 3. The expression 2Ï9 is used to indicate the negative square root, 23. The expression 6Ï9 indicates both the positive and negative square roots, 63. The principal cube root of 8 is 2, 3 3 8 5 2. The principal cube root of 28 is 22, Ï 28 5 22. In this book, only principal roots Ï are to be determined or used in problem solving.

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Examples 1. Ï36 5 Ïs6ds6d 5 6 Ans 4 4 2. Ï 16 5 Ï s2ds2ds2ds2d 5 2 Ans 3 3. Ï 227 5 Ïs23ds23ds23d 5 23 Ans 5 5 4. Ï 32 5 Ï s2ds2ds2ds2ds2d 5 2 Ans

5.

Î Î 3

28 5 27

3

s22ds22ds22d 22 5 s3ds3ds3d 3

Ans

As presented in Unit 18, generally roots are alternate functions. 4 Solve Ï 562.824.

4

2nd

(or

)

SHIFT

X

562.824 5 4.870719863

Ans

The following example shows the procedure for calculating roots of negative numbers. 5 Solve Ï 285.376.

5

2nd

(or

SHIFT

)

X

(2) 85.376 5 22.433700665

Ans

The square root of a negative number has no solution in the real number system. For example, Ï24 has no solution; Ï24 is not equal to Ïs22ds22d and is not equal to Ïs12ds12d. Any even root (even index) of a negative number has no solution in the real 4 6 number system. For example, Ï 216 and Ï 264 have no solution. When you use a calculator to take an even root of a negative number, it will give an error message to show that there is no solution in the real number system. For example, some calculators will give a message like Error or DoMAIN Error. A Machine calc Pro 2 gives a display of MATH Error.

ExprESSing numbErS with fractional ExponEntS aS raDicalS c Procedure ●● ●● ●●

To simplify numbers with fractional exponents

Write the numerator of the fractional exponent as the power of the radicand. Write the denominator of the fractional exponent as the root index of the radicand. Simplify.

Examples 2 1. 251@2 5 Ï 251 5 Ï25 5 Ïs5ds5d 5 5 Ans 3 1 3 2. 81@3 5 Ï 8 5Ï s2ds2ds2d 5 2 Ans 3 2 3 3 3. 82@3 5 Ï 8 5Ï 64 5 Ï s4ds4ds4d 5 4 Ans 1 1 1 1 4. 3621@2 5 1@2 5 5 5 Ans 36 Ï36 Ïs6ds6d 6

Use the universal power key, x y , or caret key, Enclose the fractional exponent in parentheses.

^ , with fractional exponents.

Example 1 Solve 8.7322@3. 8.732

xy

(or

^)

(

2

abc

3

5 4.240422706

)

Ans

Example 2 Solve 8.73222@3. 8.732

xy

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(or

^)

( (2)

2

abc

3

)

5 0.235825546

Ans

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Use a spreadsheet to complete the following Examples. 3 Example 1 Use a spreadsheet and the power function to evaluate Ï 245.32.

Round the answer to 3 decimal places.

Solution Enter 5 POWER(245.32, 1/3) in Cell A1 and press 23.565

Ans

Example 2 Use a spreadsheet and the ^ key to compute answer to 2 decimal places.

RETURN

Î 5

Solution Enter 5 (2(1425 1 2/3)) ^ (1/5) in Cell A2 and press 24.27

23.565304553.

2 21425 . Round the 3 RETURN

24.273695732.

Ans

4 Example 3 Use a spreadsheet to evaluate Ï 22.6. Round the answer to 3 decimal places.

Solution Enter 5 22.6 ^ (1/4) in Cell A3 and press

RETURN

#NUM!

Notice the last answer. It is not possible to take the square root or any even root of a negative number in the real number system. #NUM! is a message that a spreadsheet uses to tell you that an invalid value has been entered in the spreadsheet.

combinED opErationS of SignED numbErS Expressions consisting of two or more operations of signed numbers are solved using the same order of operations as in arithmetic.

Example Compute the value of 50 1 (22)[6 1 (22)3(4)]. 50 1 s22d[6 1 s22d3s4d] 5 50 1 s22d[6 1 s28ds4d] 5 50 1 s22d[6 1 s232d] 5 50 1 s22ds226d 5 50 1 52 5 102 Ans Use a calculator to complete the following Examples.

Example 1 Solve Ï38.44 2 s23d[8.2 2 s5.6d3s27d]. 38.44 2 (2) 3 3 3718.736 Ans

8.2 2 5.6

(

Example 2 Solve 18.32 2 s24.52d 1 places. 18.32 2 (2) 4.52 1 4

(or

2nd

(or

xy

^ ) 3 3 (2) 7

)

5

4 Ï93.724 2 6.023 . Round the answer to 2 decimal 21.2363

SHIFT

)

X

(

93.724

2 6.023 ) 4 (2) 1.236 x (or ^ ) 3 5 21.21932578, 21.22 Ans (rounded) y

ApplicAtion Tooling Up 1. If a 5 2.4, b 5 5.2, and c 5 10.25, what is the value of a2 1 bc 2 b2? 2. Use the Table of Block Thicknesses of a Metric Gage Block Set under the heading “Description of Gage Blocks” in Unit 37 to determine a combination of gage blocks for 42.196 mm.

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3. Use a digital micrometer to measure the thickness of a piece of sheet metal. If possible, compare your measure with the recorded thickness of this piece of sheet metal. 4. Use a micrometer to measure the thickness of a 25¢ coin. 5. Measure the length of the line segment in Figure 40-2 to the nearest fiftieth of an inch (0.020).

Figure 40-2

6. Express the length of the line segment in Figure 40-2 in millimeters.

The Number Scale In Exercises 7 and 8, refer to the number scale in Figure 40-3 and give the direction (1 or 2) and the number of units counted going from the first to the second number.

–11 –10 –9

–8

–7

–6

–5

–4

–3

–2

–1

0

+1

+2

+3

+4

+5

+6

+7

+8

+9 +10 +11

Figure 40-3

7. a. 211 to 22

8. a. 19 to 11

b. 28 to 23

b. 111 to 0

c. 26 to 0

c. 0 to 26

d. 22 to 28 e. 12 to 28 f. 13 to 110 g. 110 to 210 h. 110 to 0 i. 14 to 17

d. 27.5 to 110 e. 110 to 27.5 f. 210.8 to 24.3 g. 22.3 to 20.8 1 1 h. 17 to 2 2 4 3 i. 15 to 0 4

comparing Signed Numbers In Exercises 9 and 10, select the greater of the two signed values and indicate the number of units by which it is greater. 9. a. 15, 214 b. 17, 23 c. 28, 21 d. 114.3, 123 e. 21.8, 11.8

10. a. 18, 113 b. 120, 222 c. 216, 24 d. 117.6, 221.9 e. 9.75, 29.75

11. List the following signed numbers in order of increasing value starting with the smallest number. a. 117, 21, 12, 0, 218, 14, 225 b. 25, 15, 0, 113, 127, 221, 22, 219 c. 110, 210, 27, 17, 0, 125, 225, 114 d. 0, 15, 23.6, 22.5, 214.9, 117, 10.3 1 7 5 e. 216, 114 , 213 , 16, 23 8 8 8

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Absolute Value 12. Express each of the following pairs of signed numbers as absolute values and subtract the smaller absolute value from the larger absolute value. a. 123, 214 b. 217, 19

c. 26, 16 d. 125, 113

e. 216, 116 f. 233.7, 229.7

Note: For Exercises 13 through 62 that follow, round the answers to 3 decimal places wherever necessary.

Addition of Signed Numbers In Exercises 13 through 16, add the following signed numbers as indicated. 1 3 13. a. 115 1 (18) 15. a. 29 1 23 4 4 b. 7 1 (118) 1 5 5 3 b. 18 1 221 c. 0 1 (125) 8 4 d. 28 1 (215) e. 218 1 (24) 1 (211) 14. a. 112 1 (25) b. 118 1 (226) c. 220 1 (119) d. 223 1 17 e. 225 1 3

1 2 1 2 5 c. 213 1 12 2 16

d. 24.25 1 (27) 1 (23.22) e. 18.07 1 (217.64) 16. a. 16 1 (24) 1 (211) b. 253.07 1 (26.37) 1 19.82 c. 30.88 1 (20.95) 1 1.32 d. 212.77 1 (29) 1 (27.61) 1 0.48 e. 2.53 1 16.09 1 (254.05) 1 21.37

Subtraction of Signed Numbers In Exercises 17 through 20, subtract the following signed numbers as indicated. 17. a. 210 2 (24) b. 15 2 (213) c. 222 2 (214) d. 117 2 (16) e. 140 2 (140) 18. a. 240 2 (240) b. 240 2 (140) c. 0 2 (212) d. 252 2 (28) e. 16.5 2 (114.3)

19. a. 218.4 2 (214.3) b. 250.2 2 (151) c. 150.2 2 (251) d. 0.03 2 (10.06) 1 1 e. 210 2 27 2 4

1 2 7 1 20. a. 5 2 124 2 8 8 b. (6 1 10) 2 (27 1 9) c. (214 1 5) 2 (2 2 10) d. (7.23 2 6.81) 2 (210.73) e. [28.76 1 (25.83)] 2 [12.06 2 (20.97)]

Multiplication of Signed Numbers In Exercises 21 through 24, multiply the following signed numbers as indicated. 21. a. (24)(6) b. (24)(26) c. (110)(23) d. (210)(23) e. (25)(7)

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22. a. (22)(214) b. 0(216) c. (6.5)(25) d. (23.2)(20.1) e. (20.06)(20.60)

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265

Signed numberS

1 12212 342

23. a. 1

24. a. (23.86)(22.1)(27.85)(232.56) b. (8)(22.65)(0.5)(21)

1 b. s0d 4 c. (22)(22)(22) d. (22)(12)(12)

c. (26.3)(20.35)(2)(21)(0.05) d. (24.03)(20.25)(23)(20.127) e. (20.03)(2100)(20.10)

e. (8)(24)(3)(0)(21)

Division of Signed Numbers In Exercises 25 through 28, divide the following signed numbers as indicated. 1 1 25. a.210 4 (25) 27. a.2 4 2 2 2 b. 210 4 (12.5) 260 3 b. 264 c.118 4 (19) 20.5 4 d. 221 4 3 210 1 2 c. 1 4 2 22.5 3 3 e.230 4 (26) 217.92 26. a.148 4 (26) d. 3.28 b. 235 4 7 e.0.562 4 (20.821) 216 c. 28. a.229.96 4 5.35 24 b. 24.125 4 (20.75) 0 d. c.241.87 4 7.9 210 d. 220.47 4 0.537 248 e. e.244.876 4 (27.836) 28

1 2

1 2

Powers of Signed Numbers In Exercises 29 through 32, raise the following signed numbers to the indicated powers. 29. a. (22)2 b. 23 c. (22)3 d. (24)3 e. (22)4 30. a. (22)5 b. (26)2 c. (25)3

d. (22)6 e. (21.6)2 31. a. (20.4)3 b. 0.936 c. (21.58)2 d. (20.85)3 e. 0.733

3

1 2

2 32. a. 2 3 b. (21.038)25 c. 17.6622 d. (20.83)23 e. (26.087)24

Roots of Signed Numbers In Exercises 33 through 36, determine the indicated root of the following signed numbers. 3 33. a. Ï 64 3 b. Ï 264 3 c. Ï 227 3 d. Ï 21000 3 e. Ï 1000 5 34. a. Ï 232 3 b. Ï 125 5 c. Ï 132 3 d. Ï 11 3 e. Ï 21

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7 35. a. Ï 21 3 b. Ï 216

c. d. e.

Î Î Î 3

3

4

28 264

36. a.

Î 3

127 2125

3 b. Ï 2236.539 5 c. Ï 286.009

18 227

d.

11 116

e.

Î Î 3

3

297.326 123.592 289.096 217.323

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expressing Numbers with Fractional exponents as Radicals In Exercises 37 and 38, determine the value of the following: 37. a. 91@2 b. 811@2 c. 81@3 d. 641@3

e. 281@3 f. 161@4 38. a. 21251@3 b. 1251@3

c. 273.192@3 d. 41.67321@2 e. 8.0072@3 f. 67.72522@3

combined operations of Signed Numbers Solve each of the following problems using the proper order of operations. 39. 40. 41. 42.

19 2 (3)(22) 1 (25)2 4 2 5(8 2 10) 22(4 1 2) 1 3(5 2 7) 5 2 3(8 2 6) 2 [1 1 (26)] 2 s21ds23d 2 s6ds5d 43. 3 s7d 2 9 44. s23d3 1 33 2 s26ds3d 2

12622

3 45. 52 1 Ï 28 1 s24ds0ds23d

46. [42 1 (2)(5)(23)]2 1 2(23)3 47. s22d3 1 Ï16 2 s5ds3ds8d 48.

2 s25d2 s24d3 2 2 s5d 18 1 s22d

49. s22.87d3 1 Ï15.93 2 s5.63ds4ds25.26d3 50.

2s25.16d2 s24.66d3 2 3.07s4.98d 18.37 1 s22.02d

3 51. s22.46d3 1 Ï s23.86ds210.42d 2 s26.16d

52. 10.7822 1 [43.28 1 (9)(20.563)]23 Substitute the given numbers for letters in the following expressions and solve. 53. Find 6xy 1 5 2 xy when x 5 22 and y 5 7. 23ab 2 2bc 54. Find when a 5 23, b 5 10, and c 5 24. abc 2 35 55. Find (x 2 y)(3x 2 2y) when x 5 25 and y 5 27. d3 1 4 f 2 f h 56. Find 2 when d 5 22, f 5 24, and h 5 4. h 2 s2 1 d d x2 21 1 y3 57. Find 2 when n 5 5, x 5 25, and y 5 21. xy n 58. Find Ï6 sab 2 6d 2 sbd3 when a 5 26 and b 5 22. x2 21 1 y3 59. 2 ; n 5 5.31, x 5 25.67, y 5 21.87 xy n 60. Ï6sab 2 6d 2 scd3; a 5 26.07, b 5 22.91, c 5 1.56 3 61. 5Ï e 1 sef 2 d d 2 sd d3; d 5 210.55, e 5 8.26, f 5 27.09

62.

4 Ïsmpt 1 pt 1 19d ; m 5 2, p 5 22.93, t 5 25.86 t2 1 2p 2 7

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algebraic operationS oF addition, Subtraction, and multiplication

267

UNIT 41 Algebraic Operations of Addition, Subtraction, and Multiplication

Objectives After studying this unit you should be able to ●●

Perform the basic algebraic operations of addition, subtraction, and multiplication.

A knowledge of basic algebraic operations is essential in order to solve equations. For certain applications, formulas given in machine trade handbooks cannot be used directly as given, but must be rearranged. Formulas are rearranged by using the principles of algebraic operations.

DEfinitionS It is important to understand the following definitions in order to apply the procedures that are required for solving problems involving basic operations. A term of an algebraic expression is that part of the expression that is separated from ab the rest by a plus or a minus sign. For example, 4x 1 2 12 1 3ab2x 2 8aÏb is 2x ab an expression that consists of five terms: 4x, , 12, 3ab2x, and 8aÏb. 2x A factor is one of two or more literal and/or numerical values of a term that are multiplied. For example, 4 and x are each factors of 4x; 3, a, b2, and, x are each factors of 3ab2x; 8, a, and Ïb are each factors of 8aÏb.

Note: It is absolutely necessary that you distinguish between factors and terms. A numerical coefficient is the number factor of a term. The letter factors of a term are the literal factors. For example, in the term 5x, 5 is the numerical coefficient; x is 1 1 the literal factor. In the term ab2c3, is the numerical coefficient; a, b2, and c3 3 3 are the literal factors. Like terms are terms that have identical literal factors including exponents. The numerical coefficients do not have to be the same. For example, 6x and 13x are like 1 terms; 15 ab2c3, 3.2ab2c3, and ab2c3 are like terms. 8 Unlike terms are terms that have different literal factors or exponents. For example, 12x and 12y are unlike terms because they have different literal factors. The terms 15xy, 3x2y, and 4x2y2 are unlike terms. Although the literal factors are x and y in each of the terms, these literal factors are raised to different powers.

aDDition Only like terms can be added. The addition of unlike terms can only be indicated. As in arithmetic, like things can be added, but unlike things cannot be added. For example, 4 inches 1 5 inches 5 9 inches. Both values are inches; therefore, they can be added. But 4 inches 1 5 pounds cannot be added because they are unlike things.

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c Procedure

FundamentalS oF algebra

To add like terms

Add the numerical coefficients applying the procedure for addition of signed numbers. If a term does not have a numerical coefficient, the coefficient 1 is understood: x 5 1x, abc 5 1abc, n2rs3 5 1n2rs3. Leave the literal factors unchanged.

●●

●●

Examples Add the following like terms. 1. 3x 12x 15x 4.

2. Ans

6x2y3 213x2y3 27x2y3

5. Ans

c Procedure

x 214x 213x

3. 25xy2 15xy2 0

Ans

Ans

2sa 1 bd 23sa 1 bd 7sa 1 bd 6sa 1 bd Ans

To add unlike terms

The addition of unlike terms can only be indicated.

●●

Examples Add the following unlike terms. 1. 15

2. 7x 8x 7x 1 8y

x

15 1 x Ans 3.

3x 27x2 3x 1 (27x2)

c Procedure more terms ●● ●●

4. Ans

Ans

8a 26b 2c 8a 1 (26b) 1 2c

Ans

To add two or more expressions that consist of two or

Group the like terms in the same column. Add like terms and indicate the addition of unlike terms.

Examples Add the following expressions. 1. 12x 2 2xy 1 6x2y3 and 24x 2 7xy 1 5x2y3 Group like terms in the same column. Add like terms.

12x 2 2xy 1 6x2y3 24x 2 7xy 1 5x2y3 8x 2 9xy 1 11x2y3 Ans

2. 6a 2 7b and 18b 2 3ab 1 a and 214a 1 ab2 2 5ab Group like terms. 6a 2 7b a 1 18b 2 3ab 214a 2 5ab 1 ab2 Add like terms and indicate the 27a 1 11b 2 8ab 1 ab2 Ans addition of unlike terms.

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269

Subtraction As in addition, only like terms can be subtracted. The subtraction of unlike terms can only be indicated. The same principles apply in arithmetic. For example, 8 feet 2 3 feet 5 5 feet, but 8 feet 2 3 ounces cannot be subtracted because they are unlike things.

c Procedure ●●

●●

To subtract like terms

Subtract the numerical coefficients applying the procedure for subtraction of signed numbers. Leave the literal factors unchanged.

Examples Subtract the following like terms as indicated. 1. 18ab 2 7ab 5 11ab Ans 2. bx2y3 2 13bx2y3 5 212bx2y3 3. 25x y 2 8x y 5 213x y 2

2

2

Ans

Ans

4. 224dmr 2 s224dmrd 5 0 Ans

c Procedure ●●

To subtract unlike terms

The subtraction of unlike terms can only be indicated.

Examples Subtract the following unlike terms as indicated. 1. 3x2 2 s12xd 5 3x2 2 2x

Ans

2. 213abc 2 s18abc d 5 213abc 2 8abc2 2

3. 22xy 2 s27yd 5 22xy 1 7y

c Procedure ●● ●●

Ans

Ans

To subtract expressions that consist of two or more terms

Group like terms in the same column. Subtract like terms and indicate the subtractions of the unlike terms.

Note: Each term of the subtrahend is subtracted following the procedure for subtraction of signed numbers. Examples Subtract the following expressions as indicated. 1. Subtract 7a 1 3b 2 3d from 8a 2 7b 1 5d Group like terms in the same column. 8a 2 7b 1 5d 5 8a 2 7b 1 5d Change the sign of each term in the 2(7a 1 3b 2 3d ) 5 1(27a 2 3b 1 3d) subtrahend and follow the procedure a 2 10b 1 8d Ans for addition of signed numbers. 2. Subtract as indicated: s3x2 1 5x 2 12xyd 2 s7x2 2 x 2 3x3 1 6yd 3x2 1 5x 2 12xy 5 3x2 1 5x 2 12xy 2(7x2 2 x 2 3x3 1 6y) 5 1(27x2 1 x 1 3x3 2 6y) 24x2 1 6x 2 12xy 1 3x3 2 6y Ans

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FundamentalS oF algebra

multiplication It was shown that unlike terms cannot be added or subtracted. In multiplication, unlike terms can be multiplied. For example, x2 can be multiplied by x4. The term x2 means (x)(x). The term x 4 means (x)(x)(x)(x). sx2dsx4d 5 sxdsxdsxdsxdsxdsxd 5 x214 5 x6 Also, 2x can be multiplied by 4y. The product is 8xy.

c Procedure ●●

●● ●●

To multiply two or more terms

Multiply the numerical coefficients following the procedure for multiplication of signed numbers. Add the exponents of the same literal factors. Show the product as a combination of all numerical and literal factors.

Examples Multiply as indicated. 1. Multiply s23x2ds6x4d. Multiply numerical coefficients: s23ds6d 5 218 Add exponents of like literal factors: sx2dsx4d 5 x214 5 x6 Show product as combination of all numerical and literal factors. s23x2ds6x4d 5 218x6 Ans 2. s3a2b3ds7ab3d 5 s3ds7dsa211dsb313d 5 21a3b6 3. s24ads27b c ds22ac d d 5 s24ds27ds22dsa 2 2

3 3

Ans 111

dsb2dsc213dd3 5 256a2b2c5d3

Ans

c Procedure To multiply expressions that consist of more than one term within an expression ●● ●●

Multiply each term of one expression by each term of the other expression. Combine like terms.

Before applying the procedure to algebraic expressions, an example is given to show that the procedure is consistent with arithmetic.

Example in Arithmetic Multiply 3s4 1 2d. From arithmetic: From algebra: Multiply each term of one expression by each term of the other expression. Combine like terms.

3s4 1 2d 5 3s6d 5 18 Ans 3s4 1 2d 5 3s4d 1 3s2d 5 12 1 6 5 18 Ans

Examples in Algebra 1. 3a s6 1 2a2d 5 s3ads6d 1 3a s2a2d 5 18a 1 6a3

Ans

2. 25x ys3xy 2 4x y 1 5yd 5 25x y s3xyd 2 5x y s24x3y2d 2 5x2y s5yd 5 215x3y2 1 20x5y3 2 25x2y2 Ans 2

3 2

2

2

3. Multiply s3c 1 5d2ds4d2 2 2cd.

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271

algebraic operationS oF addition, Subtraction, and multiplication

This is an example in which both expressions have two terms. The solution illustrates a shortcut of the distributive property called the FOIL method.

Find the sum of the products of 1. the First terms: F 2. the Outer terms: O 3. the Inner terms: I 4. the Last terms: L Then combine like terms. O F s3c 1 5d 2ds4d 2 2 2cd 5 3cs4d 2d 5 12cd 2 I

1 1

3cs22cd s26c2d

1 1

5d 2s4d 2d 20d 4

1 1

5d 2s22cd s210cd 2d

L Product of First terms

Product of Outer terms

Product of Inner terms

Product of Last terms

F

O

I

L

Combine like terms. COMBINE 5 12cd 2 1 s26cd2 1 20d 4 1 s210cd 2d 5 2cd 2 1 s26c2d 1 20d 4 5 2cd 2 2 6c2 1 20d 4 Ans

ApplicAtion Tooling Up 1. Subtract the signed numbers 216 2 (8.4 2 5.2). x1y 2. If x 5 5.1, y 5 9.4, and z 5 12.6, what is the value of ? Give the answer as an improper fraction, mixed numz2x ber, and a decimal rounded to 2 decimal places. 3. Use the Table of Block Thicknesses for a Customary Gage Block Set, like those in Unit 37, to determine a combination of gage blocks for 3.46720. 4. Read the setting of the metric vernier micrometer scale graduated in Figure 41-1 in 0.001mm. 0 8 6 4 2 0 0

5

5 0 10

45

Figure 41-1

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FundamentalS oF algebra

5. Read the decimal-inch measurement on the vernier height gage in Figure 41-2. 25

5

20

4

15

3

10

2

5

1

8

0

9

Figure 41-2

6. Express 127.25 inches as centimeters. If necessary, round the answer to 2 decimal places.

Addition of Single Terms Add the terms in the following expressions. 7. 18y 1 y

21. 4P 1 s26Pd 1 P 1 12P

8. 15xy 1 7xy

22. 20.3dt 2 1 s21.7dt 2d 1 s2dt 2d

9. 215xy 1 s27xyd

23. 5P 1 2P2

10. 22m2 1 s2m2d

24. 2a3 1 2a2

11. 25x2y 1 5x2y

25. 7ab2 1 s22a2bd 1 s2a2b2d

12. 4c3 1 0

26. s2xyzd 1 x 2yz 1 s2xy2zd 1 5xyz 2

13. 29pt 1 s2ptd

27.

14. 0.4x 1 s20.8xd

1 7 xy 1 xy 1 xy 1 s24xyd 4 8

15. 8.3a2b 1 6.9a2b

28. 20.06D 1 s219.97Dd 1 s20.7Dd

16. 20.04y 1 0.07y

29. 6M 1 0.6M 1 0.06M 1 0.006M

17.

30. 23xy2 1 8xy2 1 7.8xy2

1 3 xy 1 xy 2 4

31. 5T 1 2T 2 1 s28T d

1

32. 2x2 1 5ax2 1 s27x2d

2

3 1 18. 2 c2d 1 23 c2d 4 8

33. 15ax2 1 3a2x 1 s215ax2d 1 s210a2xd

19. 22.06gh3 1 s20.85gh3d

34. 28abc 1 8ab2c 1 s28abc2d 1 s28ab2cd

20. 250.6abc 1 50.5abc

35. The machined plate distances shown in Figure 41-3 are dimensioned, in millimeters, in terms of x. Determine dimensions A–G. A 0.5x 0.8x x

A

C

B

B 2.1x

C

1.7x 0.9x

E

x

F 0.7x

0.2x

0.3x

D

0.6x 0.3x

3.7x D

1.9x

G

F G

1.3x

E

Figure 41-3

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algebraic operationS oF addition, Subtraction, and multiplication

UNIT 41

Addition of expressions with Two or More Terms Add the following expressions. 36. 25x 1 7xy 2 8y 29x 2 12xy 1 13y

39. s3xy2 1 x2y 2 x2y2d, s2x2y 1 x2y2d

37. 3a 2 11d 2 8m 2a 1 11d 2 3m

41. sx3 1 5d, s3x 2 7x2 1 7d, sx 2 3x3d

38. 26ab 2 5a2b2 2 3a2b 25ab 1 14a2b2 2 12a3b 29ab 2 7a2b2 1 a3b ab 2 2a3b

43. sx2 2 4xyd, s4xy 2 y2d, s2x2 1 y2d

40. s10a 2 5bd, s212a 2 7bd, s11a 1 bd

42. sb4 1 4b3c 2 2b2cd, s4b3c 2 7bcd

44. s1.3M 2 3N d, s28M 1 0.5N d, s20M 1 0.4N d 45. sc 1 3.6cd 2 4.9dd, s21.4c 1 8.6dd

Subtraction of Single Terms Subtract the following terms as indicated. 46. 7xy2 2 s218xy2d 47. 3xy 2 xy

56. 13a 2 9a2 57. 213a 2 s27a2d 58. 0.2xy 2 0.9xy2

48. 23xy 2 xy

59. 2ax2 2 ax2

49. 23xy 2 s2xyd 50. 9ab 2 s29abd 51. 25a2 2 s5a2d 52. 0.7a2b2 2 2.3a2b2

1

1 3 dt 2 2 dt 2 8

61.

1 2 2 1 d t 2 2 d 2t 2 2 2

53. 0 2 s212mn3d

62. 21 2 3x

54. 28mn3 2 0

63. 3x 2 21

55.

1

7 2 3 x 2 2 x2 8 8

2

60.

1

2

64. 23.2d 2 6.4d

2

65. 21.4xy 2 s21.4xyd

Subtraction of expressions with Two or More Terms Subtract the following expressions as indicated. 66. s2a2 2 3ad 2 s7a2 2 10ad

71. s2a3 2 0.3a2d 2 s2a3 1 a2 2 ad

67. s4x2 1 8xyd 2 s3x2 1 5xyd

72. s5x 1 3xy 2 7yd 2 s3y2 2 x2yd

68. s9b2 1 1d 2 s9b2 2 1d

73. s2d2 2 dt 1 dt2d 2 s24 1 dtd

69. s9b2 2 1d 2 s9b2 2 1d

74. s15L 2 12H d 2 s212L 1 6H 2 4d

70. sxy 2 x y 1 x y d 2 0

75. s9.08e 1 14.76f d 2 se 2 f 2 10.03d

2

2 2

2 2

Multiplication of Single Terms Multiply the following terms as indicated. 76. s25b2cds3b3d

79. s8ab2cds7a3bc2d

77. sxdsx2d

80. s2x3y3ds5a3bd

78. s23a2ds25a4d

81. s23xyds0d

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82. s7ab4ds3a4bd

90. sabc3dsc3d d

83. s23d5r4ds2d3d

91. s2x6y6ds2x2d

84. s23d5r4ds2d3ds21d

92.

85. s0.3x y ds0.7x d 2 4

86.

5

114 a 2138 a 2 3

FundamentalS oF algebra

12 23 mt21t 2 4

93. s7ab3ds27a3bd

2

94. s20.3a3b2ds24b3d

87. s25xds0ds25xd

95. s2x2yds2xyds2xd

88. sm2tdsst2d

96. sd 4m2ds21ds2m3d

89. s21.6bcds2.1d

Multiplication of expressions with Two or More Terms Multiply the following expressions as indicated and combine like terms where possible. 97. 25xy s2xy2 2 3x4d

102. sm2t3s4ds2m4s2 1 m 2 s5d

98. 3a2 s2a2 1 a3bd

103. s3x 1 7dsx2 1 9d

99. 22a3b2 s4ab3 2 b2 2 2d

104. s7x2 2 y3ds22x3 1 y2d

100. xy2 sx2 1 y3 1 xyd

105. s5ax3 1 bxds2a2x3 1 b2xd

101. 24 sdt 1 t2 2 1d

106. s23a2b3 1 5xy2ds4a2b3 2 5xyd

UNIT 42 Algebraic Operations of

Division, Powers, and Roots

Objectives After studying this unit you should be able to ●● ●● ●● ●● ●●

Perform the basic algebraic operations of division, powers, and roots. Remove parentheses that are preceded by a plus or minus sign. Simplify algebraic expressions that involve combined operations. Write decimal numbers in scientific notation. Compute expressions using scientific notation.

DiviSion As with multiplication, unlike terms can be divided. For example, x4 can be divided by x. x4 sxdsxdsxdsxd 5 5 x421 5 x3 x x

c Procedure ●● ●●

●●

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Ans

To divide two terms

Divide the numerical coefficients following the procedure for division of signed numbers. Subtract the exponents of the literal factors of the divisor from the exponents of the same literal factors of the dividend. Combine numerical and literal factors.

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algebraic operationS oF diviSion, powerS, and rootS

275

This division procedure is consistent with arithmetic.

Example in Arithmetic Divide

25 . 22

25 s2ds2ds2ds2ds2d 5 5 s2ds2ds2d 5 8 Ans 22 s2ds2d 25 5 2522 5 23 5 8 Ans 22

From arithmetic: From algebra:

Examples in Algebra 1. Divide 216x3 by 8x. Divide the numerical coefficients following the procedure for signed numbers.

216 4 8 5 22

Subtract the exponents of the literal factors in the divisor from the exponents of the same literal factors in the dividend.

x3 4 x 5 x321 5 x2 216x3 5 22x2 Ans 8x

Combine the numerical and literal factors. 2.

1 21a 21b 21c 2 5 6ab c

230a3b5c2 230 5 25a2b3 25

322

523

2

2 2

Ans

In arithmetic, any number except 0 divided by itself equals 1. For example, 4 4 4 5 1. Applying the division procedure, 4 4 4 5 4121 5 40. Therefore, 40 51. Any number except 0 raised to the zero power equals 1. 53 5 5323 5 50 5 1 Ans 53 a3b2c Example 2 3 2 5 sa323dsb222dsc121d 5 a0b0c0 5 s1ds1ds1d 5 1 Ans abc

Example 1

c Procedure To divide when the divisor consists of one term and the dividend consists of more than one term ●● ●●

Divide each term of the dividend by the divisor following the procedure just described. Combine terms.

Example in Arithmetic Divide

618 . 2

From arithmetic: From algebra:

618 5 2 618 5 2

14 5 7 Ans 2 6 8 1 5 3 1 4 5 7 Ans 2 2

Example in Algebra Divide

220xy2 1 15x2y3 1 35x3y . 25xy 220xy2 1 15x2y3 1 35x3y 220xy2 15x2y3 35x3y 5 1 1 25xy 25xy 25xy 25xy 2 2 5 4y 2 3xy 2 7x Ans

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FundamentalS oF algebra

powErS c Procedure ●●

●●

●●

To raise a single term to a power

Raise the numerical coefficients to the indicated power following the procedure for powers of signed numbers. Multiply each of the literal factor exponents by the exponent of the power to which it is raised. Combine numerical and literal factors.

This power procedure is consistent with arithmetic.

Example in Arithmetic s22d3 s22d3 5 s4d3 5 s4ds4ds4d 5 64 Ans s22d3 5 22s3d 5 26 5 s2ds2ds2ds2ds2ds2d 5 64 Ans

Raise to the indicated power. From arithmetic: From algebra:

Examples in Algebra s5x3d2

1. Raise to the indicated power. Raise the numerical coefficient to the indicated power following the procedure for powers of signed numbers.

52 5 25

Multiply each literal factor exponent by the exponent of the power to which it is to be raised.

sx3d2 5 x3s2d 5 x6

Combine numerical and literal factors.

s5x3d2 5 25x6

2. s23a b cd 5 s23d a b c 2 4

3

3

1 3. 2 x3syd 2d3r4 2

3 2s3d 4s3d 1s3d

5 227a b c

6 12 3

4 5 32 12x y d r 4 5 14 x y d r 2

Ans

Ans

2

3 3 6 4

6 6 12 8

Ans

Note: (x3)2 is not the same as x3 x2. (x3)2 5 (x3)(x3) 5 (x)(x)(x)(x)(x)(x) 5 x6 x3x2 5 (x)(x)(x)(x)(x) 5 x5 c Procedure ●●

To raise two or more terms to a power

Apply the procedure for multiplying expressions that consist of more than one term.

Example Raise to the indicated power. Solve.

s2x 1 5y3d2 s2x 1 5y3d2 5 s2x 1 5y3ds2x 1 5y3d F O I Step 1 Step 2 Step 3

2

1 (2x 1 5y 3) L Step 4

(2x 1 5y 3) 3

4

5 2xs2xd 1 2xs5y3d 1 5y3s2xd 1 5y3s5y3d 5 4x2 1 10xy3 1 10xy3 1 25y6 5 4x2 1 20xy3 1 25y6

Ans

Combine

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277

algebraic operationS oF diviSion, powerS, and rootS

rootS c Procedure

To extract the root of a term

Determine the root of the numerical coefficient following the procedure for roots of signed numbers. The roots of the literal factors are determined by dividing the exponent of each literal factor by the index of the root. Combine the numerical and literal factors.

●●

●●

●●

This procedure for extracting roots is consistent with arithmetic.

Example in Arithmetic Ï26 Ï26 5 Ïs2ds2ds2ds2ds2ds2d 5 Ï64 5 8 Ans Ï26 5 2642 5 23 5 s2ds2ds2d 5 8 Ans

Find the indicated root. From arithmetic: From algebra:

Examples in Algebra 1. Ï25a6b4c8 5 Ï25sa642dsb442dsc842d 5 5a3b2c4 3

3

2. Ï 227d x y 5 Ï 227sd 3.

Î 4

3 9 2

16 8 12 2 d t y 5 81

Î 4

343

dsx

943

3

Ans

3 2 dÏ y 5 23dx 3Ï y 2

Ans

16 844 1244 244 2 2 sd dst dsy d 5 d 2t 3y1/2 5 d 2t 3Ïy 81 3 3

Ans

Note: Roots of expressions that consist of two or more terms cannot be extracted by this procedure. For example, Ïx2 1 y2 consists of two terms and does not equal Ïx 2 1 Ïy 2. The mistake of considering the expressions equal is commonly made by students and must be avoided. This fact is consistent with arithmetic as shown. Ï32 1 42 5 Ï9 1 16 5 Ï25 5 5, but Ï32 1 Ï42 5 3 1 4 5 7 5 does not equal 7; therefore Ï32 1 42 Þ Ï32 1 Ï42.

rEmoval of parEnthESES In certain expressions, terms are enclosed within parentheses, which are preceded by a plus or minus sign. In order to combine like terms, it is necessary to first remove parentheses.

c Procedure ●●

●●

To remove parentheses preceded by a plus sign

Remove the parentheses without changing the signs of any terms within the parentheses. Combine like terms.

Example 5a 1 s4b 1 7a 2 3dd 5 5a 1 4b 1 7a 2 3d 5 12a 1 4b 2 3d

c Procedure ●● ●●

Ans

To remove parentheses preceded by a minus sign

Remove the parentheses and change the sign of each term within the parentheses. Combine like terms.

Example 2s7a2 1 b 2 3d 1 12 2 s2b 1 5d 5 27a2 2 b 1 3 1 12 1 b 2 5 5 27a2 1 10 Ans

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FundamentalS oF algebra

combinED opErationS c Procedure operations ●●

To solve expressions, consisting of two or more different

Apply the proper order of operations. The order of operations as presented in Units 17 and 39 is repeated as follows:

order of operations ●●

●● ●● ●●

First, do all operations within grouping symbols. Grouping symbols are parentheses ( ), brackets [ ], braces { }, and absolute value signs u u. Second, do powers and roots. Next, do multiplication and division operations in order from left to right. Last, do addition and subtraction operations in order from left to right.

Once again, you can use the memory aid “Please Excuse My Dear Aunt Sally” to help remember the order of operations. The P in “Please” stands for parentheses, the E for exponents (or raising to a power) and roots, the M and D for multiplication and division, and the A and S for addition and subtraction.

Example 1 10x 2 3x s2 1 x 2 4x2d 5 10x 2 6x 2 3x2 1 12x3 5 4x 2 3x2 1 12x3

Example 2 15a6b3 1 s2a2bd3 2

Ans

a7sb3d2 a7b6 6 3 6 3 5 15a b 1 8a b 2 ab3 ab3 6 3 6 3 5 15a b 1 8a b 2 a6b3 5 22a6b3

Ans

Example 3 24a [15 2 3s2a 1 abd 1 a] 2 2a2b 5 24a s15 2 6a 2 3ab 1 ad 2 2a2b 5 260a 1 24a2 1 12a2b 2 4a2 2 2a2b 5 260a 1 20a2 1 10a2b Ans

SciEntific notation In scientific applications and certain technical fields, computations with very large and very small numbers are required. The numbers in their regular or standard form are inconvenient to read, to write, and to use in computations. For example, copper expands 0.00000900 per unit of length per degree Fahrenheit. Scientific notation simplifies reading, writing, and computing with large and small numbers. In scientific notation, a number is written as a whole number or a decimal with an absolute value between 1 and 10 multiplied by 10 with a suitable exponent. For example, a value of 325,000 is written in scientific notation as 3.25 3 105. The effect of multiplying a number by 10 is to shift the position of the decimal point. Changing a number from the standard decimal form to scientific notation involves counting the number of decimal places the decimal point must be shifted.

Expressing Decimal (Standard form) numbers in Scientific notation A positive or negative number whose absolute value is 10 or greater has a positive exponent when expressed in scientific notation.

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Example 1 Rewrite 146,000 using scientific notation. a. Write the number as a value between 1 and 10: 1.46. b. To determine the exponent of 10, count the number of places the decimal point is shifted: 1 46000. The decimal point is shifted 5 places. The exponent of 10 is 5: 105. c. Multiply 1.46 3 105. 146,000 5 1.46 3 105

Ans

Example 2 Express 63,150,000 using scientific notation. 6 3,150,000. 5 6.315 3 107

Ans

Shift 7 places.

Example 3 Express 297.856 using scientific notation. 29 7.856 5 29.7856 3 101

Ans

Shift 1 place. A positive or negative number whose absolute value is less than 1 has a negative exponent when expressed in scientific notation.

Example 1 Rewrite 0.0289 using scientific notation. 0.02 89 5 2.89 3 1022 Shift 2 places.

Ans Observe that the decimal point is shifted to the right, resulting in a negative exponent.

Example 2 Rewrite 0.0000318 using scientific notation. 0.00003 18 5 3.18 3 1025

Ans

Shift 5 places.

Example 3 Rewrite 0.859 using scientific notation. 0.8 59 5 8.59 3 1021

Ans

Shift 1 place.

Expressing Scientific notation as Decimal (Standard form) numbers To express a number given in scientific notation as a decimal number, shift the decimal point in the reverse direction and attach required zeros. Move the decimal point according to the exponent of 10. With positive exponents the decimal point is moved to the right; with negative exponents it is moved to the left.

Examples Express the following values in decimal form. 1. 4.3 3 103 5 4,300. Ans Shift 3 places. Attach required zeros. 2. 8.907 3 10 5 8 90,700. Ans 5

3. 3.8 3 10

24

Shift 5 places. Attach required zeros. 5 0.0003 8 Ans Shift 4 places. Attach required zeros.

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A number in scientific notation can be typed directly into a spreadsheet, or the spreadsheet can be used to convert a number to scientific notation. To enter a number written in scientific notation into a spreadsheet, use the letter “E” to denote the power of 10.

Example In a spreadsheet enter 54.37 3 1013 in scientific notation.

Solution Enter: 5 54.37E 1 13 in Cell A1 as shown in Figure 42-1.

Figure 42-1

As you can see in Figure 42-2, after RETURN is pressed, the number in the cell is still displayed in scientific notation while the value is shown in the Formula Bar. What you see in the cell may depend on the number setting for that cell. In Figure 42-2, the cell was set to general number.

Figure 42-2

If you want to enter the result as a general number and have the spreadsheet change it to scientific notation, then you must reformat the cell that displays the answer. Much as we did when changing a decimal to a fraction, first highlight the cell, click the “Home” tab, then click on “Number.” In the popup menu, click on “Scientific”, and the number will be converted to scientific notation.

Example Have a spreadsheet convert 32,451,976,458.94 to scientific notation.

Solution Enter: 5 32451976458.94 in a cell as shown in Figure 42-3. Highlight the cell, click on the “Home” tab, under “Number” select “Scientific” from the popup menu, and the number will be converted to scientific notation. In Figure 42-3, the result is 3.25E 1 10.

Figure 42-3

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multiplication and Division using Scientific notation Scientific notation is used primarily for multiplication and division operations. The procedures presented in Unit 40 for the algebraic operations of multiplication and division are applied to operations involving scientific notation.

Examples Compute the following expressions. 1. s2.8 3 103d 3 s3.5 3 105d a. Multiply the decimals: 2.8 3 3.5 5 9.8 b. The product of the 10’s equals 10 raised to a power that is the sum of the exponents: 103 3 105 5 10315 5 108 c. Combine both parts (9.8 and 108d as a product: s2.8 3 103d 3 s3.5 3 105d 5 9.8 3 108 Ans 2. 340,000 3 7,040,000 Rewrite the numbers in scientific notation and solve: 340,000 3 7,040,000 5 s3.4 3 105d 3 s7.04 3 106d 5 23.936 3 1011. Notice that the decimal part is greater than 10. Rewrite the decimal part and solve: 23.936 5 2.3936 3 101. s2.3936 3 101d 3 1011 5 2.3936 3 1012 Ans 3. 2840,000 4 0.0006 2840,000 4 0.0006 5 s28.4 3 105d 4 s6 3 1024d 5 s28.4 4 6d 3 s105 4 1024d 5 21.4 3 105 2 s24d 5 21.4 3 109 Ans 4.

s3.4 3 1028d 3 s7.9 3 105d s22 3 106d 3.4 3 7.9 4 22 5 213.43 1028 3 105 4 106 5 10281526 5 1029 213.43 3 1029 5 s21.343 3 101d 3 1029 s21.343 3 101d 3 1029 5 21.343 3 10129 5 21.343 3 1028 Ans With 10-digit calculators, the number shown in the calculator display is limited to 10 digits. Calculations with answers that are greater than 9,999,999,999 or less than 0.000000001 are automatically expressed in scientific notation.

Example 1 80000000 3 400000 5 3.213 (Answer displayed as 3.213 or, on a SHARP, as 3.2 3 1013) The display shows the number (mantissa) and the exponent of 10; it does not necessarily show the 10. The displayed answer of 3.213 does not mean that 3.2 is raised to the thirteenth power. The display 3.213 means 3.2 3 1013; 80,000,000 3 400,000 5 3.2 3 1013.

Example 2 .0000007 3 .000002 5 1.4212 (Answer is displayed as 1.4212) The display 1.4212 means 1.4 3 10212; 0.0000007 3 0.000002 5 1.4 3 10212. Numbers in scientific notation can be directly entered in a calculator. For calculations whose answer does not exceed the number of digits in the calculator display, the answer is displayed in standard decimal form.

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The answer is displayed in decimal (standard) form on certain calculators with the exponent entry key, EE , or exponent key, EXP . Exponent entry is often an alternate 2nd function as shown.

Example Solve s3.86 3 103d 3 s4.53 3 104d. 3.86 2nd EE 3 3 4.53 2nd EE 4 5 174858000 Ans or 3.86 EXP 3 3 4.53 EXP 4 5 174858000 Ans The answer is displayed in standard form. For calculations with answers that exceed the number of digits in the calculator display, the answer is displayed in scientific notation. Both calculators with the EE key or EXP key display the answer in scientific notation. s21.96 3 107d 3 s2.73 3 105d . 8.09 3 1024 1. Using the EE key: (2) 1.96 2nd EE 7 3 2.73 2nd EE 5 4 8.09 2nd EE (2) 4 5

Examples Solve

26.614091471 3 1015, 26.614091471 3 1015 2. Using the EXP key: (2) 1.96 EXP 7 3 2.73

EXP

5 4 8.09

Ans

EXP (2)

26.614091471 3 10 , 26.614091471 3 10 15

15

4 5

Ans

Some calculators can be set so that all results are displayed in scientific notation. On some calculators, this can be set by pressing 2nd and then SCI/ENG. When these keys are pressed, the calculator displays something like that shown in Figure 42-4. The default setting for the calculator is FLO (Float). Press the ▶ until SCI is underlined and then press ENTER . To return to the default display, press 2nd SCI/ENG and press the ▶ until FLO is underlined and press ENTER .

Figure 42-4

On some calculators, you first press described above.

MODE

and then change to scientific notation as

s24.85 3 1014ds9.24 3 1028d . 6.37 3 1015 Solution Enter: 5 (24.85E 1 14) * (9.24E 2 8)/(6.37E 1 15) in a spreadsheet cell and press RETURN . 27.03516E209 Ans

Example 1 Use a spreadsheet to evaluate

Example 2 Use a spreadsheet to evaluate 72,900,000,000,000 4 0.0000000042 and express the answer in scientific notation.

Solution We could first convert each number to scientific notation. However, we will let the spreadsheet do the work for us. Enter: 5 790000000000/0.0000000042 in a spreadsheet cell and press RETURN . 1.88E120 Ans

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ApplicAtion Tooling Up 1. Add (9x2y 1 xy 2 5xy2), (23x2y 2 4xy 1 5xy2), and (7x2y 1 3xy). 2. Multiply the signed numbers 216.2, 12.3, and 24.5. 3. Use the proper order of operations to simplify 27.1 2 (62 3 2.5) 4 (215). 4. Use a digital micrometer to measure the thickness of your thumbnail in both inches and millimeters. 5. Read the metric vernier caliper measurements for the setting in Figure 42-5. MAIN SCALE 20

30

0

5

10

15

VERNIER SCALE

20

25

0.02 mm

Figure 42-5

6. For a measurement 37.260 mm, determine (a) the degree of precision, (b) the value that is equal to or less than the range of values, and (c) the value that is greater than the range of values.

Division of Single Terms Divide the following terms as indicated. 7. 8. 9. 10. 11.

4x2 2x 216a4b5 4ab3 FS2 2FS2 2FS2 2FS2 0 4 14mn

12. s242a5d2d 4 s26a2d2d 13. s23.6H2Pd 4 s0.6HPd 14. DM2 4 s21d 15. 3.7ab 4 ab 16. 0.8PV2 4 s20.2V d

1 2 3 1 c d 4 cd 2 4 4 1 1 18. 2 x3y3 4 x3 3 9 17. 1

1

2

1

2

3 19. 26g3h2 4 2 gh 4

20. 224x2y5 4 s20.5x2y4d 21. x2y3z4 4 xy3z4 22. 18a2bc2y 4 s2a2d 23. 0.25P2V 4 0.0625 24. 20.08xy 4 0.02y 3 25. 2 FS3 4 s23Sd 4 26. 29.6x2yz 4 s21.2xd

Division of expressions with Two or More Terms in the Dividend Divide the following expressions as indicated. 27. s8x3 1 12x2d 4 x

29. s9x6y3 2 6x2y5d 4 s23xy2d

28. s12x3y3 2 8x2y2d 4 4xy

30. s2x 2 4yd 4 4

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31. s15a2 1 25a5d 4 s2ad 32. s218a2b7 2 12a5c5d 4 s26a2b5d 33. s14cd 2 35c d 2 7d 4 s27d 2

34. s0.8x5y6 1 0.2x4y7d 4 s2x2y4d

FundamentalS oF algebra

36. s5y2 2 25xy2 2 10y4d 4 5y2 37.

112 a c 2 34 a c 2 ac 2 4 18 ac 2

3 2

3

38. s22.5e2f 2 0.5ef 2 1 e2f 2d 4 0.5f

35. s20.9a2x 2 0.3ax2 1 0.6d 4 s20.3d

Powers of Single Terms Raise the following terms to indicated powers. 39. s3abd2

50. s8C 3FH2d2

40. s24xyd3

51. s0.4x3yd3

41. s2x2yd3

52. s20.5c2d3ed3

42. s4a4b3d2

53. s4.3M2N2Pd2

43. s23c3d 2e4d3

3 54. abc3 4

1

44. s2MS2d2 45. s27x4y5d2

3

2

55. [28sa2b3d2c]3

46. s23N2P2T 3d4

56. [23x2sy2d2z3]3

47. sa3bc2d3

57. [0.6d3sef 2d3]2

48. s22a2bc3d3

58. [s22x2yd2sxy2d2]3

49. s2x4y5zd3

Powers of expressions of Two or More Terms Raise the following terms to the indicated powers and combine like terms where possible. 59. s3x2 2 5y3d2

64. s20.2x2y 2 y4d2

60. sa4 1 b3d2

2 2 3 65. c d 1 cd2 3 4

1

61. s5t2 2 6xd2 62. sa2b3 1 ab3d2 63. s0.4d t 2 0.2td 2 3

2

2

2

66. [sx2d3 2 sy3d2]2 67. [s2a4bd2 1 sx2yd3]2

Roots Determine the roots of the following terms. 68. Ï16c2d 6 69. Ïm6n4s2 3 70. Ï 64x3y9

76. 77.

71. Ï81x8y6 3 9 6 3 72. Ï ptw 3 73. Ï 227x6y12

74. Ï0.25h4y2 75. Ï0.16a8c2 f 6

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78.

Î Î Î 3

4 2 4 6 abc 9 1 2 2 xy 16 8 6 3 mn 27

3 79. Ï 264d 6t 9 4 80. Ï 16x4y8

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5 81. Ï 32h10

85.

82. Ï25ab2 3 83. Ï 64a3c

84.

Î 3

2

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UNIT 42

Î

9 2 2 a bc 16

3 86. Ï 27d 3e6f 2

1 3 6 2 xyz 64

5 87. Ï 232a5b3

Removal of Parentheses Simplify. 88. 6a 1 s3a 2 2a2 1 a3d

95. 15 2 sr2 1 rd 1 sr2 2 14d

89. 9b 2 s15b2 2 c 1 dd

96. 2sa2 1 b2d 1 sa2 1 b2d

90. 15 1 sx2 2 10d

97. 2s3x 1 xy 2 6d 1 18 1 sx 1 xyd

91. 2sab 1 a b 2 ad

98. 20 1 scd 2 c2d 1 dd 1 14 2 scd 1 d d

2

92. 210c3 2 s28c3 2 d 1 12d 93. 2s16 1 xy 2 xd 1 s2xd

99. 20 2 scd 2 c2d 1 d d 2 14 1 scd 1 d d

94. 225a2b 2 s22a2b 2 a 1 b2d

combined operations Simplify the following expressions. 100. 15 2 2s3xyd2 1 x2y2 2 8

106.

101. 5sa2 2 bd 1 a2 2 b 102. s2 2 c2d s2 1 c2d 1 2c 103.

1

ab 2a2b a3b 2 2 3 a a2 a

2

4 2 8x 1 16x2 3x4 1 2 2 x 8 16xy 105. 2 sy2d3 1 15 2xy2 104.

107. 108.

Ï25x2 s3xy3d 2 s210d 25

Î

64d6 4 d2 9

12x6 1 6x4y 2 s16x4y2d1/2 s2xd2

109. 25as28 1 sab2d3 2 12d 110. 5a[26 1 sab2d3 2 10] 111. s10 f 6 1 12 f 4hd 4 Ï4 f 4

Rewriting Numbers in Scientific Notation Rewrite the following standard form numbers in scientific notation. 112. 625

118. 0.00004

113. 80,000

119. 0.2

114. 1,320,000

120. 39

115. 976,000

121. 0.00039

116. 0.0073

122. 175,000

117. 0.015

123. 0.00175

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Rewriting Scientific Notation Values Rewrite the following scientific notation values in standard decimal form. 124. 3 3 103

130. 1.05 3 1023

125. 1.6 3 105

131. 3.123 3 1026

126. 8.5 3 102

132. 7.312 3 104

127. 5.09 3 106

133. 7.321 3 1024

128. 4.7 3 1021

134. 2.09 3 106

129. 6.32 3 1025

135. 2.09 3 1022

multiplying and Dividing in Scientific notation The following problems are given in scientific notation. Solve and leave answers in scientific notation. Round the answers (mantissas) to 2 decimal places. s8.76 3 1025d 3 s1.05 3 109d 136. s2.50 3 103d 3 s5.10 3 105d 142. s6.37 3 103d 137. s3.10 3 1023d 3 s5.20 3 1024d s5.50 3 104d 3 s26.00 3 106d 143. 138. s27.60 3 104d 3 s1.90 3 105d s6.92 3 1023d 26 3 139. s2.43 3 10 d 4 s7.60 3 10 d s8.46 3 1025d 144. 7 25 140. s8.51 3 10 d 4 s6.30 3 10 d s3.90 3 107d 3 s6.77 3 1023d 141.

s1.25 3 104d 3 s6.30 3 105d s7.83 3 103d

The following problems are given in decimal (standard) form. Calculate and give answers in scientific notation. Round the answers (mantissas) to 2 decimal places. 65,300 3 517,000 145. 1510 3 30,500 151. 0.00786 146. 0.000300 3 0.00210 20.000829 152. 147. 256,100 3 781,000 405,000 3 0.00312 148. 61,770 3 53,100 518,000 3 0.00612 153. 37,400 3 0.0000830 149. 0.0000821 4 2315 20.00623 3 742,000 150. 651,000 154. The amount of expansion of metal when heated is computed as follows: Expansion 5 Original Length 3 Linear Expansion per Unit of Length per Degree Fahrenheit 3 Temperature Change Calculate the amount of expansion for the metals shown in the table. Give the answers in decimal (standard) form to 3 decimal places.

Metal

a.

Aluminum

b.

Copper

c.

Carbon Steel

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Original Length of Metal 6.7520 in. 35.750 ft 3.0950 in.

Linear Expansion per Unit Length per Degree Fahrenheit

Original Temperature

Temperature to Which Heated

1.244 3 1025

68.0ºF

225.0ºF

9.000 3 10

26

35.0ºF

97.0ºF

6.330 3 1026

84.0ºF

743.0ºF

a. b. c.

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UNIT 43 Introduction to Equations Objectives After studying this unit you should be able to ●● ●● ●●

Express word problems as equations. Express problems given in graphic form as equations. Solve simple equations using logical reasoning.

It is essential that the skilled machine technician understand equations and their applications. The solution of equations is required to compute problems using trade handbook formulas. Often machine shop problems are solved using a combination of equations with elements of geometry and trigonometry.

ExprESSion of Equality An equation is a mathematical statement of equality between two or more quantities and always contains the equal sign (5). The value of all quantities on the left side of the equal sign equals the value of all quantities on the right side of the equal sign. A formula is a particular type of an equation that states a mathematical rule. The following are examples of simple equations: 12 7125514 1 2 3 5 5 18 2 4 3 10 10 3 3 5 5 16 3608 5 5 3 808 2 408 2 2 xy a1b5c1d 5x1y 2 Because it expresses the equality of the quantities on the left and on the right of the equal sign, an equation is a balanced mathematical statement. An equation may be considered similar to a balanced scale as illustrated in Figure 43-1(a). The total weight on the left side of the scale equals the total weight on the right side; therefore, the scale balances. 3 pounds 1 5 pounds 1 2 pounds 5 4 pounds 1 6 pounds 10 pounds 5 10 pounds 2 lb

3 lb

5 lb

4 lb

2 lb

6 lb

3 lb

5 lb 4 lb

(a)

6 lb

(b)

Figure 43-1

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When the 2-pound weight is removed from the scale, the scale is no longer in balance as illustrated in Figure 43-1(b). 3 pounds 1 5 pounds Þ 4 pounds 1 6 pounds 8 pounds Þ 10 pounds

thE unknown quantity In general, an equation is used to determine the numerical value of an unknown quantity. Although any letter or symbol can be used to represent the unknown quantity, the letter x is commonly used. The first letter of the unknown quantity is often used to represent a quantity. Some common letter designations are L A t D

to represent length to represent area to represent time to represent diameter

P F W h

to represent pressure to represent feed of cutter to represent weight to represent height

writing EquationS from worD StatEmEntS An equation with an unknown quantity asks a question. It asks for the value of the unknown, which makes the left side of the equation equal to the right side. The question asked may not be in equation form; instead it may be expressed in words. It is important to develop the ability to express word statements as mathematical symbols, or equations. A problem must be fully understood before it can be written as an equation. Whether the word problem is simple or complex, a definite logical procedure should be followed to analyze the problem. A few or all of the following steps may be required, depending on the complexity of the particular problem. ●● Carefully read the entire problem, several times if necessary. ●● Break the problem down into simpler parts. ●● It is sometimes helpful to draw a simple picture as an aid in visualizing the various parts of the problem. ●● Identify and list the unknowns. Give each unknown a letter name, such as x. ●● Decide where the equal sign should be, and group the parts of the problem on the proper side of the equal sign. ●● Check. Are the statements on the left equal to the statements on the right of the equal sign? ●● After writing the equation, check it against the original problem, step by step. Does the equation state mathematically what the problem states in words? The following examples illustrate the method of writing equations from given word statements. After each equation is written the value of the unknown quantity is obtained. No specific procedures are given at this time in solving for the unknowns. The unknown quantity values are determined by logical reasoning.

Example 1 What weight must be added to a 12-pound weight so that it will be in balance with a 20-pound weight? Ask the question: 12 pounds 1 what weight 5 20 pounds? To help visualize the problem, a picture is shown in Figure 43-2(a). Identify the unknowns.

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Let x represent the unknown weight. Write the equation.

12 lb 1 x 5 20 lb

Ask the question: What number added to 12 pounds equals 20 pounds? Since 8 pounds added to 12 pounds equals 20 pounds, x 5 8 lb Ans

12 lb

20 lb

12 lb

x

20 lb 8 lb

(a)

(b)

Figure 43-2

Check the answer by substituting 8 pounds for x in the original equation. 12 lb 1 8 lb 5 20 lb 20 lb 5 20 lb Ck The equation is balanced as shown in Figure 43-2(b) since the left side of the equation equals the right side. 1 2 the unused piece. Make no allowance for thickness of the cut.

Example 2 A 9 -inch piece is cut from a 12-inch length of bar stock. Find the length of

1 Ask the question: What number subtracted from 12 inches 5 9 inches? 2 A picture of the problem is shown in Figure 43-3. All dimensions are in inches. Let x represent the number of inches cut off.

= 12

x

9 12

Figure 43-3

Express the problem as an equation. 120 2 x 5 9

10 2

1 1 Since 2 inches subtracted from 12 inches is equal to 9 inches, as shown in Figure 43-4, 2 2 10 x52 Ans 2

– 12

= 2 12

9 12

Figure 43-4

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1 Check the answer by substituting 2 inches for x in the original equation. 2 10 10 59 2 2 10 10 9 59 Ck 2 2

120 2 2

The equation is balanced.

Example 3 The sum of two angles equals 908. One angle is twice as large as the other. What is the size of the smaller angle? An angle 1 an angle twice as large 5 908. A picture of the problem is shown in Figure 43-5.

x

+

x

2x x

90°

=

Figure 43-5

Let x represent the smaller angle. Let 2x represent the larger angle. Express the problem as an equation. x 1 2x 5 908 or 3x 5 908 Ask the question: What number multiplied by 3 5 908? Since 3 multiplied by 308 5 908, x 5 308. The smaller angle is x 5 308. Ans The larger angle is 2x 5 608. Ans Figure 43-6 shows the sizes of both angles and also shows the solution.

30°

+

30° 60° 30°

60°

=

90° 30°

Figure 43-6

Check the answer by substituting 308 for x in the original equation. 308 1 2 s308d 5 908 908 5 908 Ck The equation is balanced.

Example 4 Three gage blocks are used to tilt a sine plate. The total height of the three blocks is 2.75 inches. The bottom block is four times as thick as the middle block. The middle block is twice as thick as the top block. How thick is each block? Convert the problem from word form to equation form. Let x represent the thickness of the thinnest block, the top block. The middle block is twice as thick as the top block, or 2x.

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The bottom block is four times as thick as the middle block, or (4)(2x) 5 8x. The sum of the three blocks 5 2.750. Therefore, x 1 2x 1 8x 5 2.750, or 11x 5 2.750. A picture of the problem is shown in Figure 43-7. All dimensions are in inches. x 2x 8x

=

11x

2.75

Figure 43-7

Ask the question: What number multiplied by 11 5 2.750? Since 11 3 0.250 5 2.750, x 5 0.250. The top block is x or 0.250.

Ans

The middle block is 2x or 2(0.250) 5 0.500.

Ans

The bottom block is 8x or 8(0.250) 5 2.000.

Ans

The thickness of each block is shown in Figure 43-8. All dimensions are in inches. 0.25 0.50 2.00

=

2.75

Figure 43-8

Check the answer by substituting 0.250 for x in the original equation: x 1 2x 1 4(2x) 5 2.750 .25 1 2 3 .25 1 4 3 2 3 .25 5 2.75 2.75 5 2.75

Ck

The equation is balanced. In many cases the problems to be solved in actual machine shop applications will be more difficult than the preceding examples. It is essential, therefore, to be able to use the procedure shown to analyze the problem, determine the unknowns, and set up the equation. If the solution to a problem is a rounded value, the check may result in a very small difference between both sides.

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FundamentalS oF algebra

chEcking thE Equation In the final step in each of the preceding examples, the value found for the unknown was substituted in the original equation to prove that it was the correct value. If an equation is properly written and if both sides of the equation are equal, the equation is balanced and the solution is correct. All work in a machine shop should be checked and rechecked to prevent errors. It is important that you check your computations. When working with equations on the job, checking your work is essential. Errors in computation can often be costly in terms of time, labor, and materials.

ApplicAtion Tooling Up 1. Raise the term (2x2 2 5y3)2 to the indicated power and combine any like terms. 2. Subtract as indicated: (14a2 1 ab 2 3b2) 2 (9a2 1 4ab 2 5b2). 3. Use the proper order of operations to simplify 132.6511/3 2

s25d3 1 s242d. 7 1 8s23d 2

4. Use the Table of Block Thicknesses of a Metric Gage Block Set under the heading “Description of Gage Blocks” in Unit 37 to determine a combination of gage blocks for 123.578 mm. 5. Use a digital caliper to measure the indicated measurement of the thickness of this page in both inches and millimeters. 0.06 mm 6. Express the unilateral tolerance 42.59 mm 21 0.00 as a bilateral tolerance with equal plus and minus values. mm

Writing equations from Word Statements Express each of the word problems in Exercises 7 through 19 as equations. Let the unknown number equal x and by logical reasoning solve for the value of the unknown. Check the equation by comparing it to the word problem. Does the equation state mathematically what the problem states in words? Check whether the equation is balanced by substituting the value of the unknown in the equation. 7. A number plus 20 equals 32. Find the number. 8. A number less 7 equals 15. Find the number. 9. Five times a number equals 55. Find the number. 10. A number divided by 4 equals 9. Find the number. 11. Thirty-two divided by a number equals 8. Find the number. 12. A number plus twice the number equals 36. Find the number. 13. Five times a number minus the number equals 48. Find the number. 14. Seven times a number plus eight times the number equals 60. Find the number. 15. Sixty divided by the product of 3 and a number equals 4. Find the number. 16. A piece of bar stock 32 inches long is cut into two unequal lengths. One piece is 3 times as long as the other. How long is each piece? 17. Three blocks are used to tilt a sine plate. The total height of the three blocks is 4.5 inches. The first block is 3 times as thick as the second block. The second block is twice as thick as the third block. How thick is each block?

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UNIT 43

293

introduction to equationS

18. Five holes are drilled in a steel plate on a bolt circle as shown in Figure 43-9. There are 3008 between hole 1 and hole 5. The number of degrees between any two consecutive holes doubles in going from hole 1 to hole 5. Find the number of degrees between the indicated holes. 1

a. 1 and 2 b. 2 and 3

2 3

5

c. 3 and 4

300° 4

d. 4 and 5 Figure 43-9

19. The total amount of stock milled off an aluminum casting in two cuts is 8.58 millimeters. The roughing cut is 6.35 millimeters greater than the finish cut. What is the depth of the finish cut? In each of the following problems, refer to the corresponding figure. Write an equation, solve for x, and check. 20. All dimensions are in inches. x5

23.

x5

x

4x

2

x

80°

20°

6 14

21. All dimensions are in millimeters. x5

24. All dimensions are in inches. x5

2x

x

4x

150

3x

2 23

22. All dimensions are in inches. x5

25.

x5

x 2x x

2x

0.5x

3x 18

1.5x 180°

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FundamentalS oF algebra

SecTIoN 4

26. All dimensions are in millimeters. x5

27. All dimensions are in inches. x5 x

30

x 8x

2x

x 250

x 4x 6x

4x

11

40

For each of the following problems, refer to the given figure, solve for the unknowns, and check. 28. Find the distances between the indicated holes. All dimensions are in millimeters. a. Hole 1 to hole 2 b. Hole 2 to hole 3 c. Hole 3 to hole 4 d. Hole 4 to hole 5 e. Hole 5 to hole 6 f. Hole 2 to hole 4 g. Hole 3 to hole 6

1

2

3

4

5

L

6

L

2L

3L

2L

180

29. Find the distances between the indicated points. All dimensions are in inches. a. A and B

2h + 34

E

b. C and D h + 14

c. E and F

C

3 h

A

F D

B

30. Find the value of each of the four angles. a. b. c. d.

∠1 ∠2 ∠3 ∠4

180° 3y – 20° 2y

3 2

y

4y – 40° 4

1

Solve for the unknown values in the following equations. 31. x 1 9x 5 30

34. 32 5 17 1 y

32. x 1 3 5 12

35. 18 2 a 5 12

33. 2y 1 5y 1 3y 5 70

36. b 2 13 5 80

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UNIT 44

Solution oF equationS by the Subtraction, addition, and diviSion principleS oF equality

37. 3b 1 5b 2 2b 5 96 6s30d 38. 3s5ad 5 2 1 39. x 5 42 2 x 40. 5 15 4 27 41. 59 x

42.

295

d 1459 6

18 1 30 4 44. 6(2.5x ) 1 5x 5 80 43. 0.75x 2 0.5x 5

2y 1 4y 1 6y 5 80 3 46. 27 2 (3)(6) 5 b 1 3 45.

UNIT 44 Solution of Equations by the Subtraction,

Addition, and Division Principles of Equality

Objectives After studying this unit you should be able to ●● ●● ●● ●●

Solve equations using the subtraction principle of equality. Solve equations using the addition principle of equality. Solve equations using the division principle of equality. Solve equations using transposition.

principlES of Equality In actual practice, equations cannot usually be solved by inspection or common sense. There are specific procedures for solving equations using the fundamental principles of equality. The principles of equality that will be presented in this unit are those of subtraction, addition, and division. Multiplication, root, and power principles are presented in Unit 45.

Solution of EquationS by thE Subtraction principlE of Equality The subtraction principle of equality states that if the same number is subtracted from both sides of an equation, the sides remain equal, and the equation remains balanced. The subtraction principle is used to solve an equation in which a number is added to the unknown, such as x 1 15 5 20. The values on each side of an equation are equal and an equation is balanced. If the same value is subtracted from both sides, the equation remains balanced. The equation 8 pounds 1 4 pounds 5 12 pounds is pictured in Figure 44-1(a). If 4 pounds are removed from the left side only, the scale is not in balance as shown in Figure 44-1(b). If 4 pounds are removed from both the left and right sides, the scale remains in balance as in Figure 44-1(c).

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SecTIoN 4

8 lb

=

4 lb

8 lb

12 lb

12 lb

8 lb + 4 lb = 12 lb 12 lb = 12 lb

8 lb + 4 lb – 4 lb 8 lb

12 lb 12 lb

FundamentalS oF algebra

8 lb

8 lb

8 lb + 4 lb – 4 lb = 12 lb – 4 lb 8 lb = 8 lb

(b)

(a)

=

(c)

Figure 44-1

(a)

x

+ 4

=

9

c Procedure unknown ●● ●●

x

(b)

=

To solve an equation in which a number is added to the

Subtract the number that is added to the unknown from both sides of the equation. Check.

5

Example 1 x 1 4 5 9. Solve for x.

9

(c)

In the equation, the number 4 is added to x as in Figure 44-2(a). To solve, subtract 4 from both sides of the equation.

=

9

Check. Figure 44-2

x1459 24 5 24 x 5 5 Ans x1459 51459 9 5 9 Ck

Example 2 In the part shown in Figure 44-3, determine dimension y.

y

5.5

All dimensions are in inches. Write an equation. Subtract 5.50 from both sides.

17

Figure 44-3

Check.

5.50 1 y 5 170 25.50 5 25.50 y 5 11.50 Ans 5.50 1 y 5 170 5.50 1 11.50 5 170 170 5 170 Ck

Example 3 239 5 P 1 18. Solve for P. 239 5 P 1 18 218 5 218 257 5 P Ans 3 4 3 3 24 5 24 4 4 1 W57 Ans 4

Check. 239 5 P 1 18 239 5 257 1 18 239 5 239 Ck

Example 4 W 1 4 5 12. Solve for W. 3 Check. W 1 4 5 12 4 1 3 7 1 4 5 12 4 4 12 5 12 Ck

tranSpoSition With your instructor’s permission, an alternate method of solving certain equations may be used. The alternate method is called transposition. Transposition or transposing a term means that a term is moved from one side of an equation to the opposite side with the sign of the term changed.

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Solution oF equationS by the Subtraction, addition, and diviSion principleS oF equality

297

Transposition is not a mathematical process, although it is based on the addition and subtraction principles of equality. Transposition should only be used after the principles of equality are fully understood and applied. Transposition is a quick and convenient means of solving equations in which a term is added to or subtracted from the unknown. The purpose of using transposition is the same as that of using the addition and subtraction principles of equality. Both methods involve getting the unknown term to stand alone on one side of the equation in order to determine the value of the unknown. The following example is solved by applying the subtraction principle of equality and transposition. Notice that when applying the subtraction principle of equality, a term is eliminated on one side of the equation and appears on the other side with the sign changed.

Example Solve for x. x 1 15 5 25

Method 1

The Subtraction Principle of Equality

x 1 15 5 25 2 15 5 215 x 5 25 2 15 x 5 10 Ans

← Observe that 115 is eliminated from the left side of the equation and appears as 215 on the right side.

Method 2

Transposition x 1 15 5 25 x 1 15 5 25 2 15 x 5 25 2 15 x 5 10 Ans

← Observe that this expression is identical to the expression obtained when applying the subtraction principle of equality.

The following examples are solved by transposition.

Example 1 y 1 10.7 5 18. Solve for y. Move 110.7 from the left side of the equation to the right side and change to 210.7.

y 1 10.7 5 18 y 1 10.7 5 18 2 10.7 y 5 18 2 10.7 y 5 7.3 Ans

1 8

Example 2 T 1 6 5 219. Solve for T. 1 Move 16 from the left side 8 of the equation to the right side 1 and change to 26 . 8

1 T 1 6 5 219 8 1 T 5 219 2 6 8 1 T 5 225 Ans 8

Example 3 Solve 13.5 5 w 1 6.4 for w. Move 1 6.4 from the right side of the equation to the left side and change to 26.4.

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13.5 5 w 1 6.4 13.5 2 6.4 5 w 7.1 5 w w 5 7.1 Ans

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SecTIoN 4

FundamentalS oF algebra

Solution of EquationS by thE aDDition principlE of Equality The addition principle of equality states that if the same number is added to both sides of an equation, the sides remain equal and the equation remains balanced. The addition principle is used to solve an equation in which a number is subtracted from the unknown, such as x 2 17 5 30.

c Procedure the unknown ●● ●●

To solve an equation in which a number is subtracted from

Add the number that is subtracted from the unknown to both sides of the equation. Check.

Example 1 x 2 6 5 15. Solve for x. x – 6

=

15

x

=

21

15

=

15

In the equation, the number 6 is subtracted from the x, as shown in Figure 44-4. To solve, add 6 to both sides of the equation. x 2 6 5 15 1 6 5 16 x 5 21 Ans Check. x 2 6 5 15 21 2 6 5 15 15 5 15 Ck

Example 2 A 7-inch piece is cut from the height of a block as shown in Figure 44-5. The

Figure 44-4

remaining block is 10 inches high. What is the height of the original block? All dimensions are in inches. Make no allowance for thickness of cut. Let y 5 the height of the original block. Write an equation. y 2 70 5 100 Add 70 to both sides 1 70 5 170 of the equation. y 5 170 Ans

7 y

=

10

Check. Figure 44-5

y 2 70 5 100 170 2 70 5 100 100 5 100 Ck

Example 3 235 5 P 2 20.4. Solve for P.

235 5 P 2 20.4 120.4 5 120.4 214.6 5 P Ans

Check.

235 5 P 2 20.4 235 5 214.6 2 20.4 235 5 235 Ck

The following examples are solved by transposition.

Example 1 x 2 4 5 19. Solve for x. Move 24 from the left side of the equation to the right and change to 14.

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x 2 4 5 19 x 5 19 1 4 x 5 23 Ans

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UNIT 44

299

Solution oF equationS by the Subtraction, addition, and diviSion principleS oF equality

Example 2 y 2 16.9 5 30. Solve for y. Move 216.9 from the left side of the equation to the right and change to 116.9.

y 2 16.9 5 30 y 5 30 1 16.9 y 5 46.9

Ans

Example 3 Solve 212.4 5 p 2 18.5 for p. Move 218.5 from the right side of the equation to the left side and change to 118.5.

212.4 5 p 2 18.5 212.4 1 18.5 5 p 6.1 5 p p 5 6.1 Ans

Solution of EquationS by thE DiviSion principlE of Equality The division principle of equality states that if both sides of an equation are divided by the same number, the sides remain equal and the equation remains balanced. The division principle is used to solve an equation in which a number is multiplied by the unknown, such as 3x 5 18.

c Procedure a number ●● ●●

To solve an equation in which the unknown is multiplied by

Divide both sides of the equation by the number that multiplies the unknown. Check.

Example 1 6x 5 24. Solve for x. In the equation shown at the top of Figure 44-6, x is multiplied by 6. To solve, divide both sides of the equation by 6. 6x 524 6x 24 5 6 6 x 54 Ans Check.

6x 524 6s4d 5 24 24 524 Ck

4y 5 280 mm

Divide both sides of the equation by 4.

4y 280 5 mm 4 4

Check.

y 5 70 mm Ans

=

24

x

=

4

24

=

24

Figure 44-6

Example 2 A part is shown in Figure 44-7. Solve for y. All dimensions are in millimeters. Write an equation.

6x

TYPICAL 4 PLACES

y

280

Figure 44-7

4y 5280 4s70 mmd 5 280 mm 280 mm 5 280 mm Ck

Example 3 214.4 5 3.2F Solve for F. 214.4 5 3.2F 214.4 3.2F 5 3.2 3.2 24.5 5 F Ans

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Check. 214.4 53.2 214.4 5 3.2s24.5d 214.4 5 214.4 Ck

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300

SecTIoN 4

1 3 4 4 1 3 7 A 521 4 4 1 3 7 A 21 4 4 5 1 1 7 7 4 4 A 53 Ans

FundamentalS oF algebra

Example 4 7 A 521 . Solve for A. Check.

1 3 7 A 521 4 4 1 3 7 s3d 5 21 4 4 3 3 21 5 21 Ck 4 4

ApplicAtion Tooling Up 1. Write the expression “A number divided by 7 plus twice the number equals 75” as an equation with the unknown as x. s6.325 3 108d 3 s9.4 3 109d 2. Use scientific notation to solve . Leave the answer in scientific notation with the 2 5 3 1025 answer (mantissa) rounded to 2 decimal places. 3. Use the proper order of operations to solve 72 1 Ï6.25 2 2 decimal places.

s25.1d2 9.605 1 Ï11.56

. If necessary, round the answer to

a 1 4b ? If necessary, round the answer to 2 decimal places. b 2 2a 5. Read the setting in Figure 44-8 of the customary vernier micrometer scale graduated in 0.00010.

4. If a 5 7.5 and b 5 9.45, what is the value of

09876543210

0 20 0 1 2 3

15

Figure 44-8

6. Measure this line segment to the nearest 0.5 mm.

Solution by the Subtraction Principle of equality Solve each of the following equations using the subtraction principle of equality. Check each answer. 7. P 1 15 5 22

11. 13 5 T 1 9

8. x 1 18 5 27

12. 37 5 D 1 2

9. M 1 24 5 43

13. 62 5 a 1 19

10. y 1 48 5 82

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UNIT 44

301

Solution oF equationS by the Subtraction, addition, and diviSion principleS oF equality

7 3 25. 2 5 x 1 8 4 3 1 26. 20 5 A 1 17 16 8 5 7 27. 39 5 y 1 40 8 8 7 9 28. 1 5W1 16 16 1 29. x 1 13 5 210 8

15. y 1 30 5 223 16. x 1 63 5 17 17. 10 1 R 5 53 18. 51 5 48 1 E 19. 236 5 14 1 x 20. H 1 7.6 5 14.7 21. 22.5 5 L 1 3.7 22. 236.2 5 y 1 6.2 23. T 1 9.07 5 9.07 1 1 24. H 1 3 5 6 4 2

30. 0.023 5 1.009 1 H

Write an equation for each of the following problems, solve for the unknown, and check. 31. All dimensions are in inches. Find x. ________ x 5 ________

34. All dimensions are in inches. Find T. ________ T 5 ________ T

2.125

9

x

2.563

27

32. All dimensions are in millimeters. Find y. ________ y 5 ________

35. All dimensions are in millimeters. Find x. ________ x 5 ________ 90.65 DIA

y

x DIA

97.23 38.50

33. All dimensions are in inches. Find r. ________ r 5 ________ r 5 8

128.26

36. All dimensions are in inches. Find H. ________ H 5 ________ 0.387

7

1 16 H

1.015

37. The height of two gage blocks is 0.8508 inch. One block is 0.750 inch thick. What is the thickness of the other block?

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SecTIoN 4

FundamentalS oF algebra

38. Three holes are drilled on a horizontal line in a housing. The center distance between the first hole and the second hole is 193.75 millimeters, and the center distance between the first hole and the third hole is 278.12 millimeters. What is the distance between the second hole and the third hole? 5 9 39. A metal bar is 7 inches long. If inch is cut off one end, how long is the bar 8 32 after the cut? 40. A shaft rotates in a bearing that is 0.3968 inch in diameter. The total clearance between the shaft and bearing is 0.0008 inch. What is the diameter of the shaft? For each of the following problems, substitute the given values in the formula and solve for the unknown. Check. 41. One of the formulas used in computing spur gear dimensions is DO 5 D 1 2a. Determine D when a 5 0.1429 inch and DO 5 4.7144 inches. 42. A formula used to compute the dimensions of a ring is D 5 d 1 2T. Determine d when D 5 52.0 millimeters and T 5 9.40 millimeters. 43. A formula used in relation to the depth of a gear tooth is WD 5 a 1 d. Determine d when WD 5 0.3082 inch and a 5 0.1429 inch. 44. A sheet metal formula used in computing the size of a stretch-out is L.S. 5 4s 1 W. 1 Determine W when s 5 3 inches and L.S. 5 12 inches. 8

Solution by the Addition Principle of equality Solve each of the following equations using the addition principle of equality. Check each answer. 45. 25 5 d 2 9

60. 5.07 5 r 2 3.07

46. T 2 12 5 34

61. 230.003 5 x 2 29.998

47. x 2 9 5 219

62. 91.96 5 L 2 13.74

48. B 2 4 5 9

63. x 2 8.12 5 213.01 1 1 64. D 2 5 2 2 7 3 65. y 2 5 2 8 8 5 7 66. 15 5 H 2 2 8 8 3 15 67. 246 5 x 2 29 32 16

49. P 2 48 5 87 50. y 2 23 5 220 51. 16 5 M 2 12 52. 240 5 E 2 21 53. 47 5 R 2 36 54. h 2 8 5 12 55. 39 5 F 2 39 56. W 2 18 5 33 57. N 2 2.4 5 6.9 58. A 2 0.8 5 0.3 59. x 2 10.09 5 213.78

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68. C 2 5

7 7 5 25 16 16

69. W 2 10.0039 5 8.0481 15 7 70. 214 5 y 2 14 32 16

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UNIT 44

Solution oF equationS by the Subtraction, addition, and diviSion principleS oF equality

303

Write an equation for each of the following problems, solve for the unknown, and check. 11 71. The point of a conical workpiece has been faced off to a 1 inch length of the tapered 16 3 portion. If 6 inches of the original length was removed, what was the original length x 4 of the tapered portion? 150 16 30 4

110

1 16

x

72. The bushing shown has a body diameter of 44.45 millimeters, which is 14.29 millimeters less than the head diameter. What is the size of the head diameter? All dimensions are in millimeters. x

44.45

1 3 73. The flute length of the reamer shown is 1 inches, which is 3 inches less than the 8 8 shank length. How long is the shank? All dimensions are in inches.

1

x

18

74. A hole is countersunk as shown to a depth of 0.250 inch. The depth of the countersink is 1.650 inches less than the depth of the 0.625-inch hole. Find depth x. All dimensions are in inches. 0.250 x

0.625

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SecTIoN 4

FundamentalS oF algebra

For each of the following problems, substitute the given values in the formula and solve for the unknown. Check each answer. 75. The total taper of a shaft equals the diameter of the large end minus the diameter of the small end, T 5 D 2 d. Determine D when T 5 22.5 millimeters and d 5 30.8 millimeters. 76. Using the spur gear formula, DR 5 D 2 2d, compute the pitch diameter (D) when the root diameter (DR) 5 3.0118 inches and the dedendum (d) 5 0.1608 inch. 77. Using a sheet metal formula, W 5 L.S. 2 4S, determine the length size (L.S.) when W 5 382 millimeters and S 5 112 millimeters. 1 is used to determine the size of a hole into which threads will be N tapped. If N 5 10 threads per inch and d 5 0.65 inch, what is D, the outside diameter of the threading tool?

78. The formula d 5 D 2

Solution by the Division Principle of equality Solve each of the following equations using the division principle of equality. Check each answer. 79. 5A 5 115

94. 13.2W 5 0

80. 4D 5 32

95. 2x 5 219.75

81. 7x 5 221

91. 0.6L 5 12

96. 0.125P 5 1.500 1 97. D 5 8 4 3 98. 24 5 B 8 1 99. 2 y 5 36 2 5 3 100. 1 L 5 9 8 4 3 3 101. 248 5 10 x 8 4 7 7 102. 2 52 y 16 16

92. 22.7x 5 23.76

103. 50.98W 5 10.196

93. 0.1y 5 20.18

104. 0.0621 5 0.027t

82. 15M 5 75 83. 54 5 9P 84. 227 5 3y 85. 54 5 6x 86. 10y 5 0.80 87. 18T 5 41.4 88. 12x 5 254 89. 25C 5 0 90. 7.1E 5 21.3

Write an equation for each of the following problems, and solve for the unknown. 105. The depth, D, of an American Standard thread is given by the formula 0.6495D 5 P, where P is the pitch. Compute the depth of the thread shown in the figure. PITCH = 0.5000

DEPTH (D)

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UNIT 44

305

Solution oF equationS by the Subtraction, addition, and diviSion principleS oF equality

106. All dimensions are in millimeters. Find x. ________ x 5 ________

107. Find x.

________ x 5 ________

x x 63.09

x

TYPICAL 3 PLACES

87° x

x

108. The feed of a drill is the depth of material that the drill penetrates in one revolution. The total depth of penetration equals the product of the number of revolutions and the feed. Compute the feed of a drill that cuts to a depth of 3.300 inches while turning 500.0 revolutions. x = FEED (DEPTH IN 1 REVOLUTION) 3.300 (DEPTH IN 500.0 REVOLUTIONS)

For each of the following problems, substitute the given values in the formula and solve for the unknown. Check each answer. Round the answers to 2 decimal places. 109. The circumference of a circle (C) equals p (approximately 3.1416) times the diameter (d) of the circle, C 5 p d. Determine d when C 5 392.50 millimeters. 110. The depth (d) of a sharp V-thread is equal to 0.866 times the pitch (p), or d 5 0.866p. Determine p when d 5 0.125 inch. 111. The length of cut (L) in inches of a workpiece in a lathe is equal to the product of the cutting time (T) in minutes, the tool feed (F) in inches per revolution, and the number of revolutions per minute (N) of the workpiece: L 5 TFN. Determine N when L 5 9.50 inches, T 5 3.00 minutes, and F 5 0.050 inch per revolution. 112. The length of cut, L, in inches, of a workpiece in a lathe is equal to the product of the cutting time, in minutes, T; the tool feed, F, in inches per revolution; and N, the number of revolutions per minute of the workpiece: L 5 TFN. Determine N when L 5 18 inches, T 5 2.5 minutes, and F 5 0.050 inch per revolution. Solve each of the following equations using either the addition, subtraction, or division principle of equality. Check each answer. 3 1 113. T 2 19 5 25 123. 2 5 21 y 16 16 114. 20.006x 5 4.938 115. 9.37R 5 103.07

124. 20.66x 5 4.752

116. 222 5 x 2 31

125. P 2 0.20 5 0.07

117. C 1 34 5 12 118. x 1 6 5 213

7 13 126. G 2 59 5 48 8 16

119. E 2 29.8936 5 18.3059

127. 2x 5 19

120. A 2 16.37 5 9.03

128. 0 5 7H

121. 78.09 5 x 1 61.95

129. 214.067 5 3.034 1 x

122. F 1 0.007 5 1.006

130. 20.863 5 D 1 25.942

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306

Section 4

Fundamentals oF algebra

UNIT 45 Solution of Equations by the Multiplication, Root, and Power Principles of Equality

Objectives After studying this unit you should be able to ●● ●● ●●

Solve equations using the multiplication principle of equality. Solve equations using the root principle of equality. Solve equations using the power principle of equality.

Solution of EquationS by thE Multiplication principlE of Equality The multiplication principle of equality states that if both sides of an equation are multiplied by the same number, the sides remain equal and the equation remains balanced. The multiplication principle is used to solve an equation in which the unknown is x divided by a number, such as 5 10. 4

c Procedure number ●● ●●

Multiply both sides of the equation by the number that divides the unknown. Check.

Example 1 x 3

=

7

x

=

21

=

7

7

4.5 TYPICAL 5 PLACES y

x 5 7. Solve for x in Figure 45-1. 3

To solve, multiply both sides of the equation by 3. Check.

Figure 45-1

To solve an equation in which the unknown is divided by a

x 57 3

13x 2 5 3s7d

3

x 5 21 Ans x 57 3 21 57 3 7 5 7 Ck

Example 2 The length of bar stock shown in Figure 45-2 is cut into five equal pieces. Each piece is 4.5 inches long. Find y, the length of the bar before it was cut. All dimensions are in inches. Make no allowance for thickness of cuts. y Write the equation. 5 4.50 5 y Multiply both sides of the 5 5 5s4.5d0 5 equation by 5. y 5 22.50 Ans

12

Figure 45-2

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Unit 45

307

solution oF equations by the multiplication, root, and power principles oF equality

y 5 4.50 5 22.50 5 4.50 5 4.50 5 4.50 Ck

Check.

1 8

Example 3 6 5

F . Solve for F. 25 1 F 6 5 8 25 1 F 25 6 5 25 8 25 5 230 5 F Ans 8 1 F 6 5 8 25 5 230 1 8 6 5 8 25 1 1 6 56 Ck 8 8

1 2

Check.

1 2

Solution of EquationS by thE root principlE of Equality The root principle of equality states that if the same root of both sides of an equation is taken, the sides remain equal and the equation remains balanced. The root principle is used to solve an equation that contains an unknown that is raised to a power, such as x2 5 36.

c Procedure ●●

●●

To solve an equation in which an unknown is raised to a power

Extract the root of both sides of the equation that leaves the unknown with an exponent of 1. Check.

Example 1 x2 5 9 Solve for x in Figure 45-3.

Check.

=

9

x

=

3

9

=

9

x2 5 9 Ïx2 5 Ï9 x 5 3 Ans

To solve, extract the square root of both sides of the equation.

x2

x2 5 9 32 5 9 9 5 9 Ck

Example 2 The area of a square piece of sheet steel shown in Figure 45-4 equals 16 Figure 45-3

square feet. What is the length of each side (s)? Write an equation. Extract the square root of both sides of the equation. Check.

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s 5 16 sq ft Ïs2 5 Ï16 sq ft s 5 4 ft Ans 2

s2 5 16 sq ft s4 ftd2 5 16 sq ft 16 sq ft 5 16 sq ft Ck

s

s

Figure 45-4

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Fundamentals oF algebra

Section 4

Example 3 Solve for T.

Check.

T 3 5 264 3 3 3 T 5Ï 264 Ï T 5 24 Ans

T 3 5 264 s24d3 5 264 264 5 264 Ck

Example 4 Solve for V. V2 5 ÏV 2 5 V5

9 64

Î 3 8

Check.

V2 5

9 64

2

1 2 5 649

9 64

3 8

9 9 5 64 64

Ans

Ck

Solution of EquationS by thE powEr principlE of Equality The power principle of equality states that if both sides of an equation are raised to the same power, the sides remain equal and the equation remains balanced. The power principle is used to solve an equation that contains a root of the unknown, such as Ïx 5 8. x

=

8

c Procedure ●●

●●

To solve an equation that contains a root of the unknown

Raise both sides of the equation to the power that leaves the unknown with an exponent of 1. Check.

Example 1 Ïx 5 8. Solve for x in Figure 45-5. x

64

=

In the equation, x is expressed as a root. Ïx 5 8 2

_Ïx + 5 82

To solve, square both sides of the equation. 8

=

8

Figure 45-5

x 5 64 Ans Ïx 5 8 Ï64 5 8 8 5 8 Ck

Check.

Example 2 The length of a side of the cube shown in Figure 45-6 equals 2.8620 inches. The cube root of the volume equals the length of a side. Find the volume of the cube. Round the answer to 3 decimal places. Let V 5 the volume of the cube. Write the equation.

2.8620

3 V 5 2.8620 in. Ï 3

3 V + 5 s2.8620 in.d3 _Ï

Cube both sides of the equation.

2.8620

2.8620

Figure 45-6

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Check.

V 5 23.443 cu in. Ans 3

Ï V 5 2.8620 in. 23.443 cu in 5 2.8620 in. Ï 2.8620 in 5 2.8620 in. Ck 3

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309

solution oF equations by the multiplication, root, and power principles oF equality

ApplicAtion tooling Up 1. Solve A 2

3 1 5 22 using the addition principle of equality. 4 8

2. Solve the equation 4x 2 5x 1 7x 5 54 for the unknown value x. 3. Write 0.0000275 in scientific notation. 3 4. Use the proper order of operations to solve u32 1 Ï 2125 2 5s22d5u.

5. Read the setting in Figure 45-7 of the metric micrometer scale graduated in 0.01 mm. 0

5

40

25

2

20

1

4

15 10

35 Figure 45-7

6. Read the customary height gage measurement for this setting in Figure 45-8.

9 8

5

7

0

6

Figure 45-8

Solution by the Multiplication Principle of equality Solve each of the following equations using the multiplication principle of equality. Check each answer. P 56 5 M 8. 55 12 7.

9. D 4 9 5 7 10. 3 5 L 4 8 11. 3 5 W 4 9 N 12. 5 22 12 C 13. 50 14 x 14. 59 210 F 15. 55 4.3 A 16. 5 24 20.5

21. 0 5 H 4 (23.8) 22. M 4 9.5 5 212 H 23. 1.04 5 0.06 24.

1 25. V 4 1 5 3 4 26.

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x 1 52 3 4 8

1 161 2 5 232

27. D 4 2

1 782

28. 4 5 y 4 2

17. S 4 (7.8) 5 3 18. x 4 (20.3) 5 16 y 19. 220 5 0.3 T 20. 5 2.4 21.8

B 57 1 2

29.

1 T 5 2 1 1 2

30. H 4 s22d 5 7

9 16

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Section 4

Fundamentals oF algebra

Write an equation for each of the following problems, solve for the unknown, and check. 31. All dimensions are in millimeters. Find x. ________ x 5 ________

32. Find x.

________ x 5 ________ 30° 30°

x

x

108.78 TYPICAL 4 PLACES

30° 30° 30°

33. A 10-inch sine plate is tilted at an angle of 458 as shown. The gage block height divided by 10 equals 0.70711 inch. Compute the height of the gage blocks. All dimensions are in inches.

x 45°

34. The width of a rectangular sheet of metal shown is equal to the area of the sheet divided 1 1 by its length. Compute the area of a sheet that is 3 feet wide and 5 feet long. 4 2 1

34 1

52

35. The depth of an American Standard thread shown divided by 0.6495 is equal to the pitch. Compute the depth of a thread with a 0.0500-inch pitch. All dimensions are in inches. Round the answer to 3 decimal places. PITCH = 0.0500

x

For each of the following problems, substitute the given values in the formula and solve for the unknown. Check each answer. 36. In mechanical energy applications, force (F) in pounds equals work (W) in foot-pounds divided W by distance (D) in feet, F 5 . Determine W when F 5 150.0 pounds and D 5 7.500 feet. D

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solution oF equations by the multiplication, root, and power principles oF equality

C 37. The diameter (D) of a circle equals the circle circumference (C) divided by 3.1416, D 5 . 3.1416 Determine C when D 5 52.14 millimeters. Round the answer to 1 decimal place. 38. The pitch (P) of a spur gear equals the number of gear teeth (N) divided by the pitch N diameter (D), P 5 . Determine N when P 5 5 teeth per inch and D 5 5.6000 inches. D

Solution by the Root Principle of equality Solve each of the following equations using the root principle of equality. Round the answers to 3 decimal places where necessary. 1 52. M 3 5 39. S 2 5 16 64 40. P 2 5 81 1 53. 2 5 y3 41. 81 5 M 2 8 2 42. 49 5 B 64 54. D3 5 3 27 43. D 5 64 2 55. E 5 0.04 44. x3 5 264 45. 144 5 F 2

56. 0.64 5 H 2

46. 264 5 y3

57. W 2 5 2.753

47. 10,000 5 L2

58. 0.0017 5 R2

48. 2125 5 x3 59. N 3 5 0.123 9 60. 20.123 5 x3 49. 5 W2 25 61. 7.843 5 F 4 1 50. C 2 5 62. T 2 5 7.056 16 16 51. P2 5 49 Write an equation for each of the following problems, solve for the unknown, and check. 63. The area of a square equals the length of a side squared, A 5 s2. For each area of a square given, compute the length of a side. Round the answers to 3 decimal places where necessary. a. 36 square inches 25 b. square foot 64 c. 1.44 square meters d. 64.700 square meters e. 0.049 square foot

s5 s5

s

s5 s5 s5

s

64. The volume of a cube equals the length of a side cubed, V 5 s3. For each volume of a cube given, compute the length of a side. Round the answers to 3 decimal places where necessary. a. 125 cubic inches 27 b. cubic foot 216 c. 0.642 cubic meter d. 92.76 cubic millimeters e. 0.026 cubic foot

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s5 s5 s5 s5 s5

s

s

s

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Section 4

Fundamentals oF algebra

Solution by the Power Principle of equality Solve each of the following equations using the power principle of equality. Check all answers. Round the answers to 3 decimal places where necessary. 65. ÏC 5 6

4 78. Ï P 5 0.1

66. ÏT 5 12

73. ÏA 5 0

3 79. 0.1 5 Ï B 1 80. 5 ÏA 4 1 4 81. Ï F5 4 3 3 82. 2 5 Ï y 5 5 83. 5 ÏH 8 84. ÏP 5 1.256

5 74. Ï N51

3 85. Ï B 5 2.868

5 75. 22 5 Ï y

5 86. Ï x 5 21.090

4 76. 0.3 5 Ï D

3 87. 0.7832 5 Ï y

3 77. Ï x 5 20.6

3 88. 0.364 5 Ï y

67. ÏP 5 1.2 68. 0.8 5 ÏM 69. 0.82 5 ÏF 3 70. Ï V53 3 71. Ï H 5 1.7 3 72. Ï x 5 24

Write an equation for each of the following problems, solve for the unknown, and check. Round the answers to 3 decimal places where necessary. 89. The length of a side of a square equals the square root of the area, s 5 ÏA. For each side of a square given, compute the area. a. 3.40 b. 0.759 c. 0.652 m d. 2.162 mm e. 1.2900

A5 A5 A5 A5 A5

s

s

3 90. The length of a side of a cube equals the cube root of the volume, s 5 Ï V. For each side of a cube given, compute the volume. Round the answers to 2 decimal places where necessary.

a. 3.3000 b. 0.9009 c. 0.62 m d. 4.073 mm e. 1.2810

V5 V5 V5 V5 V5

s

s

s

Solve each of the following equations using either the multiplication, root, or power principle of equality. Check each answer. 91. L3 5 2125

94. 13 5 y 4 s25d

92. T 5 0 E 93. 5 218 22

3 95. 20.1 5 Ï y

5

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4 96. Ï M53

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Unit 46

solution oF equations consisting oF combined operations and rearrangement oF Formulas

y 5 20.01 20.1 R 98. 5 0.002 12.6 3 99. ÏR 5 8 97.

102. x3 5

264 125

3 103. Ï x 5 22.9631 5 104. Ï x 5 0.797 3 105. y 5 0.0393

106. 22.127 5 y5 M 107. 5 100 0.009 108. x 4 6.004 5 20.2125

2 100. Ï V 5 3 3

101. G 3 5

313

64 125

UNIT 46 Solution of Equations Consisting of Combined Operations and Rearrangement of Formulas

Objectives After studying this unit you should be able to ●● ●● ●● ●●

Solve equations involving several operations. Rearrange formulas in terms of any letter value. Substitute values in formulas and solve for unknowns. Solve for the unknown term of a proportion.

Often in actual occupational applications, the formulas used result in complex equations. These equations require the use of two or more principles of equality for their solutions. For example, 0.13x 2 4.73(x 1 6.35) 5 5.06x 2 2.87 requires a definite procedure in determining the value of x. Use of proper procedure results in the unknown standing alone on one side of the equation with its value on the other.

procEdurE for Solving EquationS conSiSting of coMbinEd opErationS It is essential that the steps used in solving an equation be taken in the following order. Some or all of these steps may be used depending upon the particular equation. Not every equation will require every step. ●● Remove parentheses. ●● Combine like terms on each side of the equation.

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314

Section 4

●●

●● ●● ●●

Fundamentals oF algebra

Apply the addition and subtraction principles of equality to get all unknown terms on one side of the equation and all known terms on the other side. Combine like terms. Apply the multiplication and division principles of equality. Apply the power and root principles of equality.

Note: Always solve for a positive unknown. A positive unknown may equal a negative value, but a negative unknown is not a solution. For example, x 5 210 is correct, but 2x 5 10 is incorrect. When solving equations where the unknown remains a negative value, multiply both sides of the equation by 21. Multiplying a negative unknown by 21 results in a positive unknown. For example, multiplying both sides of 2x 5 10 by 21 gives (21) (2x) 5 (21)(10), with the result x 5 210. The following are examples of equations consisting of combined operations.

Example 1 5x 1 7 5 22. Solve for x. The operations involved are multiplication and addition. Follow the procedure for solving equations consisting of combined operations. Apply the subtraction principle. 5x 1 7 5 22 Subtract 7 from both sides of the 5x 1 7 2 7 5 22 2 7 equation. 5x 5 15 5x 15 Apply the division principle. 5 5 5 Divide both sides of the equation by 5. x 5 3 Ans Check.

5x 1 7 5 22 5s3d 1 7 5 22 15 1 7 5 22 22 5 22 Ck

Example 2 6x 1 4x 5 3x 2 5x 1 19 1 5. Solve for x. Combine like terms on each side of the equation. Apply the addition principle. Add 2x to both sides of the equation. Apply the division principle. Divide both sides of the equation by 12. Check.

6x 1 4x 5 3x 2 5x 1 19 1 5 10x 5 22x 1 24 10x 1 2x 5 22x 1 24 1 2x 12x 5 24 12x 24 5 12 12 x 5 24 4 12 x 5 2 Ans 6x 1 4x 5 3x 2 5x 1 19 1 5 6s2d 1 4s2d 5 3s2d 2 5s2d 1 19 1 5 12 1 8 5 6 2 10 1 19 1 5 20 5 20 Ck

Example 3 9x 1 7(x 1 3) 5 25. Solve for x. Remove parentheses. Combine like terms. Apply the subtraction principle. Subtract 21 from both sides of the equation.

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9x 1 7sx 1 3d 5 25 9x 1 7x 1 21 5 25 16x 1 21 5 25 16x 1 21 2 21 5 25 2 21 16x 5 4

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Unit 46

solution oF equations consisting oF combined operations and rearrangement oF Formulas

16x 4 5 16 16

Apply the division principle. Divide both sides of the equation by 16. Check.

1 Ans 4 9x 1 7sx 1 3d 5 25 1 1 9 1 7 1 3 5 25 4 4 1 3 2 1 22 5 25 4 4 25 5 25 Ck x5

12 1

Example 4 2x 5 14. Solve for x.

x2 2 32 5 223. Solve for x. 4

Apply the addition principle. Add 32 to both sides of the equation. Apply the multiplication principle. Multiply both sides of the equation by 4.

s21ds2xd 5 s21ds14d x 5 214 Ans 2x 5 14 2s214d 5 14 14 5 14 Ck 2 x 2 32 5 223 4

x2 2 32 1 32 5 223 1 32 4 x2 59 4 x2 4 5 4 s9d 4

12

x2 5 36 Ïx2 5 Ï36

Apply the root principle. Extract the square root of both sides of the equation.

x 5 6 Ans x2 2 32 5 223 4 2 6 2 32 5 223 4 9 2 32 5 223 223 5 223 Ck

Check.

3 3 Example 6 6 Ï x 5 4sÏ x 1 1.5d. Solve for x.

Remove parentheses. Apply the subtraction principle. 3 Subtract 4 Ï x from both sides of the equation. Apply the division principle. Divide both sides of the equation by 2. Apply the power principle. Raise both sides of the equation to the third power. Check.

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2

2x 5 14

Apply the multiplication principle. Multiply both sides of the equation by 21. Check.

Example 5

315

3 3 6Ï x 5 4 sÏ x 1 1.5d 3 3 6Ï x 5 4Ï x16 3 3 3 6 Ïx 2 4 Ïx 5 4 Ïx 1 6 2 4 Ï x 3 2 Ïx 5 6 3

3 2Ï x 6 5 2 2 3 x 5 3 Ï 3 sÏ xd3 5 33 x 5 27 Ans

3 3 6Ï 27 5 4 sÏ 27 1 1.5d 6 s3d 5 4 s3 1 1.5d 18 5 4 s4.5d 18 5 18 Ck

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Section 4

Fundamentals oF algebra

SubStituting valuES and Solving forMulaS Manufacturing applications often require solving formulas in which all but one numerical value for letter values is known. The unknown letter value can appear anywhere within the formula. To determine the numerical value of the unknown, write the original formula, substitute the known number values for their respective letter values, and simplify. Then follow the procedure given for solving equations consisting of combined operations.

Example 1 An open belt pulley system is shown in Figure 46-1. The number of inches between the pulley centers is represented by x. The larger pulley diameter (D) is 6.25 inches and the smaller pulley diameter (d) is 4.25 inches. The belt length is 56.0 inches. Find the distance between pulley centers using this formula found in a trade handbook. Round the answer to 1 decimal place. L 5 3.14(0.5D 1 0.5d) 1 2x

BELT

d = 4.250 D = 6.250

x

where L 5 belt length D 5 the diameter of the larger pulley d 5 the diameter of the smaller pulley x 5 the distance between pulley centers

Figure 46-1

Write the formula. L 5 3.14s0.5D 1 0.5dd 1 2x Substitute the known 56.0 in. 5 3.14[0.5(6.25 in.) 1 0.5(4.25 in.)] 1 2x numerical values for 3.14 3 ( .5 3 6.25 1 their respective letter .5 3 4.25 ) 5 16.485 values and simplify. 56.0 in. 5 16.485 in. 1 2x Apply the subtraction 56.0 in. 5 16.485 in. 1 2x principle. Subtract 16.485 56.0 in. 2 16.485 in. 5 16.485 in. 1 2x 2 16.485 in. inches from both sides. 39.515 in. 5 2x 39.515 in. 2x Apply the division 5 2 2 principle. Divide both sides by 2. 19.7575 in. 5 x, x 5 19.8 in. Ans (rounded) Check.

L 5 3.14(0.5D 1 0.5d) 1 2x 56 5 3.14 3 ( .5 3 6.25 1 .5 3 4.25 1 2 3 19.7575 5 56 56 in. 5 56 in. Ck

)

Example 2 Use a spreadsheet to solve the exercise in Example 1. With a spreadsheet you can have cells for each of the variables and a cell for the formula. This allows you to check several problems that use this same formula. We begin by entering the known values in different cells and then evaluating L. This will save having to remember these values and will make it easier to correct mistakes. In Cell A1 write “D5.” Notice that we did not begin this entry with an 5 sign, so, after we press RETURN , we will see D5 in Cell A1. In Cell B1, enter the value of D, in this case 6.25. In Cell A2, enter “d5” and in Cell B2, the value of d, 4.25. Finally, in Cell A3, key in “x5” and in Cell B3, the value of x, 19.7575. Your spreadsheet should look something like Figure 46-2.

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solution oF equations consisting oF combined operations and rearrangement oF Formulas

317

Figure 46-2

Now, in Cell A5, enter “L5” and in Cell B5, enter 53.14*(0.5*B110.5*B2)12*B3. Notice that we began the entry in Cell B5 with an 5 sign because we want the value of L. When we wanted the value of D, we clicked on Cell B1, for d we clicked on Cell B2, and for x we clicked on Cell B3. When RETURN is pressed, we see in Cell B5 that L 5 56 (Figure 46-3).

Figure 46-3

rEarranging forMulaS A formula that is used to find a particular value must sometimes be rearranged to solve for another value. Consider the letter to be solved for as the unknown term and the other letters in the formula as the known values. The formula must be rearranged so that the unknown term is on one side of the equation and all other values are on the other side. A formula is rearranged by using the same procedure that is used for solving equations consisting of combined operations. Problems are often solved more efficiently by first rearranging formulas than by directly substituting values in the original formula and solving for the unknown. This is particularly true in solving more complex formulas that involve many operations. Also, it is sometimes necessary to solve for the same unknown after a formula has been rearranged using different known values. Since the formula has been rearranged in terms of the specific unknown, solutions are more readily computed. First rearranging formulas and then substituting known values enables you to solve for the unknown using a calculator for continuous operations. This is illustrated in Example 4. Given the formulas in the following Examples, rearrange and solve for the designated letter.

Example 1 A 5 bh. Solve for h. Apply the division principle. Divide both sides of the equation by b.

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A 5 bh A bh 5 b b A 5 h Ans b

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Section 4

Fundamentals oF algebra

Example 2 In the figure shown in Figure 46-4, L 5 a 1 b. Solve for a.

a

L5a1b L2b5a1b2b L 2 b 5 a Ans

Apply subtraction principle. Subtract b from both sides of the equation.

b L

Example 3 A screw thread is checked using a micrometer and three wires as shown in

Figure 46-4

Figure 46-5. The measurement is checked using the following formula. Solve the formula for W.

THREAD CHECKING

M 5 D 2 1.5155P 5 3W where

D M W P

Figure 46-5

W A = 20.20 mm

M 5 measurement over the wires D 5 major diameter P 5 pitch W 5 wire size Apply subtraction principle. M 5 D 2 1.5155P 1 3W Subtract D from both sides M 2 D 5 D 2 1.5155P 1 3W 2 D of the equation. M 2 D 5 21.5155P 1 3W Apply addition principle. M 2 D 1 1.5155P 5 21.5155P 1 3W 1 1.5155P Add 1.5155P to both sides M 2 D 1 1.5155P 5 3W of the equation. Apply division principle. M 2 D 1 1.5155P 3W Divide both sides of the 5 3 3 equation by 3. M 2 D 1 1.5155P 5 W Ans 3

Example 4 A slot is cut in the circular piece shown in Figure 46-6. The piece has a radius (R) of 97.60 millimeters. The number of millimeters in the width is represented by W. Dimension A is 20.20 millimeters. This formula is found in a machine trade handbook. A 5 R 2 ÏR2 2 0.2500W 2

R = 97.60 mm

Figure 46-6

Solve for W. Apply the subtraction principle. Subtract R from both sides of the equation.

sA 2 Rd2 5 _2ÏR2 2 0.2500W 2 +2 sA 2 Rd2 5 R2 2 0.2500W 2

Apply the power principle. Square both sides of the equation. Apply the subtraction principle. Subtract R2 from both sides of the equation.

sA 2 Rd2 5 R2 2 0.2500W 2 2R2 5 2R2 2 sA 2 Rd 2 R2 5 20.2500W 2

Apply the division principle. Divide both sides of the equation by 20.2500.

sA 2 Rd2 2 R2 20.2500W 2 5 20.2500 20.2500

Simplify. Apply the root principle. Take the square root of both sides.

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A 5 R 2 ÏR2 2 0.2500W 2 2R 5 2R A 2 R 5 2ÏR2 2 0.2500W 2

Î Î

sA 2 Rd2 2 R2 5 W2 20.2500 sA 2 Rd2 2 R2 5 ÏW 2 20.2500 sA 2 Rd2 2 R2 5W 20.2500

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solution oF equations consisting oF combined operations and rearrangement oF Formulas

319

Substitute the given numerical values for their respective letter values and find dimension W. W5 W5

(

(

(

Î

s20.20 2 97.60d2 2 97.602 20.2500

20.2 2 97.6

118.911732 W 5 118.9 mm

)

x2

2 97.6

x2

)

4 (2) .25

)

5

Ans (rounded)

Check. Substitute numerical values in the original formula. ( 97.6 x 2 2 .25 3 118.911732 20.20 5 97.6 2 20.20000001 20.20 5 20.20 Ck

x2

)

5

Example 5 Use a spreadsheet to solve the exercise in Example 4. We will use the same method to check our answer as we used with the belt pulley system. Enter the labels for R and W in Cells A1 and A2 and their values in Cells B1 and B2. In Cell B4, type “5B1-SQRT(B1^2-0.25*B2^2)” and you should see that A 5 20.20 when rounded to 2 decimal places. (See Figure 46-7.)

Figure 46–7

proportions As we saw in Unit 20, in a proportion the product of the means equals the product of the extremes. If the terms are cross multiplied, their products are equal. Cross multiplying is used to solve proportions that have an unknown term. Since a proportion is an equation, the principles used for solving equations are applied in determining the value of the unknown after the terms have been cross multiplied. a c Example 1 Solve the proportion 5 for the value of x. b x a c 5 b x Cross multiply. ax 5 bc ax bc Apply the division principle 5 a a of equality. Divide both sides of the equation by a. bc x5 Ans a

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320

Section 4

a 5 b a 5 b

c x c c a or 3 bc 1 bc a a a 5 Ck b b

Check.

4 R

Example 2 Solve the proportion 5

Cross multiply. Apply the division principle of equality. Divide both sides of the equation by 2. Apply the root principle. Take the square root of both sides. Check.

Fundamentals oF algebra

2R for R. 12.5 4 2R 5 R 12.5 2R2 5 s4ds12.5d 2R2 5 50 2R2 50 5 2 2 R2 5 25 ÏR2 5 Ï25 R 5 5 Ans 4 2R 5 R 12.5 4 2 s5d 5 5 12.5 4 10 5 5 12.5 0.8 5 0.8 Ck

ApplicAtion tooling Up 25

1 3 1. Solve Ï x 5 . Express the answer as both a common fraction and a decimal fraction. 5 2. Solve the equation 275.4 5 5.2C for the unknown value C. b 3. Find the unknown value of b in the equation 2 6 5 12.1. 5 4. Multiply (4x2 1 3)(4x2 2 3).

6. Read the metric height gage measurement for this setting in Figure 46-8.

equations consisting of combined operations

15 10

10

5 0.02 mm

5. Use a digital micrometer to measure the thickness of one of your hairs. Be careful not to crush the hair.

20

0

Figure 46-8

Solve for the unknown and check each of the following combined operations equations. Round answers to 2 decimal places where necessary. 7. 5x 2 33 5 12

10. 4B 2 7 5 B 1 21

8. 10M 1 5 1 4M 5 89

11. 7T 2 14 5 0

9. 8E 2 14 5 2E 1 28

12. 6N 1 4 5 84 1 N

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Unit 46

solution oF equations consisting oF combined operations and rearrangement oF Formulas

13. 2.5A 1 8 5 15 2 4.5 14. 12 2 (2x 1 8) 5 18 15. 3H 1 (2 2 H) 5 20

26. x s4 1 xd 1 20 5 x2 2 sx 2 5d 27.

3

1 2 1 34 5 42 b 2

16. 12 5 2(2 1 C) 2 (4 1 2C)

28. 3F 3 1 F sF 1 8d 5 8F 1 F 2 1 81

17. 25(R 1 6) 5 10(R 2 2)

29. 9 1 y2 5 s y 2 4ds y 2 1d

18. 0.29E 5 9.39 2 0.01E 19. 7.2F 1 5(F 2 8.1) 5 0.6F 1 15.18 P 20. 1 8 5 6.3 7 1 3 21. W 1 sW 2 8d 5 4 4 1 1 22. D 2 3 sD 2 7d 5 5 D 2 3 8 8

30.

1 1 s2B 2 12d 1 B2 5 B 1 22 4 2

31. 24 sy 2 1.5d 5 2Ïy 2 4y 32. 14 Ïx 5 6 sÏx 1 8d 1 16 33. 8.12P2 1 6.83P 1 5.05 5 16.7P2 1 6.83P 34. 7.3 Ïx 5 3 sÏx 1 8.06d 2 4.59

23. 0.58y 5 18.3 2 0.02y

35. ÏB2 2 2.53B 5 22.53 sB 2 3.95d

24. 2H 2 20 5 sH 1 4dsH 2 4d

36. s2yd3 2 2.80 s5.89 1 3yd 5 23.87 2 8.40y

2

25. 4A2 1 3A 1 36 5 8A2 1 3A

321

Substituting Values and Solving Formulas The following formulas are used in the machine trades. Substitute the given values in each formula and solve for the unknown. Round the answers to 3 decimal places where necessary. 37. F 5 2.380P 1 0.250 Given: F 5 2.375. Solve for P.

45. M 5 E 2 0.866P 1 3W Given: M 5 3.3700, E 5 3.2000, P 5 0.125. Solve for W.

38. a 5 3H 4 8 Given: a 5 0.1760. Solve for H.

46. S 5

39. H.P. 5 0.000016MN Given: H.P. 5 22, N 5 50.8. Solve for M. 40. N 5 0.707DPn Given: N 5 24, Pn 5 8. Solve for D. 1.732 41. S 5 T 2 N Given: S 5 0.4134, N 5 20. Solve for T. D2 2 D1 42. a 5 2 Given: a 5 0.250, D1 5 0.875. Solve for D2. 0.290W t2 Given: S 5 1000, t 5 0.750. Solve for W.

43. S 5

44. W 5 St s0.55d2 2 0.25dd Given: W 5 1150, d 5 0.750. Solve for St.

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3

4

L1 1 sD 2 D2d L2 2 1 1 1 Given: S 5 , L1 5 16, L 2 5 4, D2 5 2 . 4 2 Solve for D1. pDN 12 Given: C 5 210, D 5 6, p 5 3.1416. Solve for N.

47. C 5

48. Do 5

Pc sN 1 2d

p Given: Do 5 4.3750, p 5 3.1416, Pc 5 0.3927.

Solve for N. 49. S 5

Î

d2 1 h2 4

Given: S 5 12.700, d 5 6. Solve for h. 50. C 5 2 Ïh s2r 2 hd Given: C 5 7.600, h 5 3.750. Solve for r.

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Fundamentals oF algebra

Section 4

Rearranging Formulas The following formulas are used in machine trade calculations. Rearrange the formulas in terms of the designated letters. 51. The dimensions shown can be found using these two formulas. (1) A 5 ab (2) d 5 Ïa2 1 b2

54. A 5 p sR2 2 r2d R

a

d

r

a. Solve for R. b. Solve for r.

b

a. Solve formula (1) for a. b. Solve formula (1) for b. c. Solve formula (2) for a. d. Solve formula (2) for b.

55. M 5 D 2 1.5155P 1 3W W

52. The radii shown in this figure can be found using these two formulas. (1) R 5 1.155r (2) A 5 2.598R2

D

M

R P

a. Solve for D. b. Solve for P. c. Solve for W.

r

56. /A 1 /B 1 /C 5 1808

a. Solve formula (1) for r. b. Solve formula (2) for R. 53. The dimensions shown can be found using these two formulas. (1) FW 5 ÏD2o 2 D2 (2) Do 5 2C 2 d 1 2a a

C B

A

a. Solve for /A. b. Solve for /B. c. Solve for /C.

d WORM

57. L 5 3.14 s0.5D 1 0.5d d 1 2x C PULLEYS–OPEN BELT D Do WORM GEAR

a. Solve formula (1) for Do. b. Solve formula (1) for D. c. Solve formula (2) for d. d. Solve formula (2) for a.

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D

x

d

a. Solve for D. b. Solve for d. c. Solve for x.

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Unit 46

323

solution oF equations consisting oF combined operations and rearrangement oF Formulas

58. By sF 2 1d 5 Cx

59. Ca 5 SsC 2 Fd BEVEL GEAR

PLANETARY GEARING x y

DRIVER

C

Ca F

C B S FIXED

FOLLOWER

a. Solve for S. b. Solve for C.

a. Solve for x. b. Solve for B. c. Solve for C.

For applications 60 and 61, rearrange each formula for the designated letter and solve. 60. The horsepower of an electric motor is found with this formula. 6.2832Tsrpmd hp 5 where hp 5 horsepower 33,000 T 5 torque in pound-feet (lb-ft) rpm 5 revolutions per minute Solve for T when hp 5 1.50 and rpm 5 2250. Round the answer to 2 decimal places. 61. A tapered pin is shown. S

h

R

r

a. Solve for h when R 5 2.38 cm, r 5 1.46 cm, and V 5 69.5 cm3. Round the answer to 2 decimal places. V 5 1.05 sR 1 Rr 1 r 2d h b. Solve for h when S 5 0.875 in, R 5 0.420 in, and r 5 0.200 in. Round the answer to 3 decimal places. S 5 ÏsR 2 rd2 1 h2

Proportions Solve for the unknown value in each of the following proportions. Check each answer. Round the answers to 3 decimal places where necessary. 10 5x 5 x 8 A 4.761 63. 5 12.5 5A 11 E 1 3 64. 5 8 12

62.

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M 2 5 15 5 12 8 5 2 B B 1 9.2 66. 5 4.38 11.71 E 2 15 0.36 67. 5 E 1 7.53 1.86 65.

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Section 4

6.08 5.87 5 3H 2 2 12 12.53 12.8 7.43 69. 5 2 5 2 3C 231.29 8.62M 1 23.30 7.62M 1 0.05 70. 5 12.36 0.86 P2 2 186.73 223.30P 71. 5 5.65P 3.04 68.

Fundamentals oF algebra

B11 B13 5 B22 B25 C14 C21 73. 5 C23 C16 N3 2 4 N3 74. 5 3 6 3 2A 1 7A 2 5 A3 1 5A 1 8 75. 5 7 5

72.

UNIT 47 Applications of Formulas to

Cutting Speed, Revolutions per Minute, and Cutting Time

Objectives After studying this unit you should be able to ●●

●●

Solve cutting speed, revolutions per minute, and cutting time problems by substitution in given formulas. Solve production time and cutting feed problems by rearranging and combining formulas.

In order to perform cutting operations efficiently, a machine must be run at the proper cutting speed. Proper cutting speed is largely determined by the type of material that is being cut, the feed and depth of cut, the cutting tool, and the machine characteristics. The machinist must be able to determine proper cutting speeds by using trade handbook data and formulas.

cutting SpEEd uSing cuStoMary unitS of MEaSurE Cutting speeds or surface speeds for lathes, drills, milling cutters, and grinding wheels are computed using the same formula. On the lathe, the workpiece revolves. On drill presses, milling machines, and grinders, the tool revolves. Speeds are computed in reference to the tool rather than the workpiece. The speed of a revolving object equals the product of the circumference times the number of revolutions per minute made by the object. Generally, when using customary units, diameters are expressed in inches and cutting speeds are in feet per minute (fpm). In order to express inches per minute as feet per minute, it is necessary to divide by 12.

Note: p 5 3.1416 rounded to 4 decimal places. When solving problems with a calculator use the p key. On certain calculators, you must use the 2nd key to get p.

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applications oF Formulas to cutting speed, revolutions per minute, and cutting time

Unit 47

C5

3.1416DN 12

325

where C 5 cutting speed in feet per minute (fpm) D 5 diameter in inches

pDN or C 5 12

N 5 revolutions per minute (rpm)

lathe The cutting speed of a lathe is the number of feet that the revolving workpiece travels past the cutting edge of the tool in 1 minute.

Example A steel shaft 2.500 inches in diameter is turned in a lathe at 184.0 rpm. Determine the cutting speed to 1 decimal place. C5

p DN p s2.500ds184.0d 5 12 12

p 3 2.5 3 184 4 12 5 120.4277184, 120.4 fpm

Ans (rounded)

Milling Machine, drill press, and grinder The cutting speed or surface speed of a drill press, milling machine, and grinder is the number of feet that a point on the circumference of the tool travels in 1 minute.

Example 1 A 10-inch diameter grinding wheel runs at 1910 rpm. Determine the surface speed to the nearer whole number. C5

3.1416DN 3.1416s10ds1910d 5 < 5000.38, 5000 fpm 12 12

Ans (rounded) 1 2

Example 2 Determine the cutting speed to the nearer whole number of a 3 -inch diameter milling cutter revolving at 120 rpm. p DN p s3.5ds120d C5 5 12 12

p 3 3.5 3 120 4 12 5 109.9557429, 110 fpm

Ans (rounded)

rEvolutionS pEr MinutE uSing cuStoMary unitS of MEaSurE The cutting speed formula is rearranged in terms of N in order to determine the revolutions per minute of a workpiece or tool. C5

3.1416DN 12

12C 5 3.1416DN 12C 12C 12C 5 N or N 5 or N 5 3.1416D 3.1416D pD

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Section 4

Fundamentals oF algebra

lathe Example An aluminum cylinder with a 6.000-inch outside diameter is turned in a lathe at a cutting speed of 225 feet per minute. Determine the revolutions per minute to the nearest whole revolution. N5

12s225d 12C 5 < 143.239, 143 rpm Ans (rounded) 3.1416D 3.1416s6d

Milling Machine, drill press, and grinder 1 2 Determine the revolutions per minute to the nearest whole revolution. 12s60d 12C N5 5 < 458.365, 458 rpm Ans (rounded) 3.1416D 3.1416s0.5d

Example 1 A -inch diameter twist drill has a cutting speed of 60.0 feet per minute.

Example 2 A 6.00-inch diameter grinding wheel operates at a cutting speed of 6000 feet per minute. Determine the revolutions per minute to the nearest whole revolution. 12C 12s6000d N5 5 pD p s6.00d 12 3 6000 4 ( p 3 6 ) 5 3819.7186, 3820 rpm

Ans (rounded)

cutting tiME uSing cuStoMary unitS of MEaSurE The same formula is used to compute cutting times for machines that have a revolving workpiece, such as the lathe, as is used for machines that have a revolving tool, such as the milling machine, drill press, and grinder. Cutting time is determined by the length or depth to be cut in inches, the revolutions per minute of the revolving workpiece or revolving tool, and the tool feed in inches for each revolution of the workpiece or tool. T5

L FN

where

T 5 cutting time per cut in minutes L 5 length of cut in inches F 5 tool feed in inches per revolution N 5 speed of revolving workpiece or tool in revolutions per minute

lathe Example 1 How many minutes are required to take one cut 22.00 inches in length on a steel shaft when the lathe feed is 0.050 inch per revolution and the shaft turns 152 rpm? Round the answer to 1 decimal place. T5

L 22.00 5 < 2.895, 2.9 min FN 0.050s152d

Ans (rounded)

Example 2 A 3.250-inch-diameter cast iron sleeve that is 20.00 inches long is turned in a lathe to a 2.450-inch diameter. Roughing cuts are each made to a 0.125-inch depth of cut. One finish cut using a 0.025-inch depth of cut is made. The feed is 0.100 inch per revolution for roughing and 0.030 inch for finishing. Roughing cuts are made at 150 rpm and the finish cut at 200 rpm. What is the total cutting time required?

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Unit 47

applications oF Formulas to cutting speed, revolutions per minute, and cutting time

Compute the total depth of cut.

3.2500 2 2.4500 0.8000 5 5 0.4000 2 2

Compute the number of roughing cuts required.

0.4000 2 0.0250 53 0.1250

Compute the time required for one roughing cut.

T5

Compute the total time for roughing. Compute the time required for finishing. Compute the total cutting time.

327

20.00 5 1.33 min s0.100ds150d 3 3 1.33 min 5 4.0 min

T5

20.00 5 3.3 min s0.030ds200d

4.0 min 1 3.3 min 5 7.3 min

Ans

Milling Machine, drill press, and grinder Example 1 Determine the cutting time required to drill through a workpiece that is 3.600 inches thick with a drill revolving 300 rpm and a feed of 0.025 inch per revolution. L 3.600 T5 5 5 0.48 min Ans FN 0.025s300d

Example 2 A milling machine cutter makes 460 rpm with a table feed of 0.020 inch per revolution. Four cuts are required to mill a slot in an aluminum plate 28.68 inches long. Compute the total cutting time. Round the answer to 1 decimal place. L 28.68 Total Cutting Time 5 345 34 FN 0.020s460d 28.68 4 ( .02 3 460 ) 3 4 5 12.46956, 12.5 min Ans (rounded)

cutting SpEEd uSing MEtric unitS of MEaSurE Diameters are expressed in millimeters. Cutting speeds are expressed in meters per minute. The symbol for meters per minute is m/min. In order to express speed in millimeters per minute as meters per minute, it is necessary to divide by 1000 or to move the decimal point 3 places to the left. C5

3.1416DN 1000

pDN or C 5 1000

where C 5 cutting speed in meters per minute D 5 diameter in millimeters N 5 revolutions per minute

Example A medium-steel shaft is cut in a lathe using a high-speed tool. The shaft has a diameter of 55 millimeters and is turning at 260 revolutions per minute. Determine the cutting speed to the nearest whole number. C5

3.1416DN 3.1416s55ds260d 5 < 45 m/min 1000 1000

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Ans (rounded)

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Section 4

Fundamentals oF algebra

rEvolutionS pEr MinutE uSing MEtric unitS of MEaSurE In the metric system, the symbol for revolutions per minute is r/min. The cutting speed formula is rearranged in terms of N in order to determine the revolutions per minute of a workpiece or tool. 3.1416DN 1000 1000C 5 3.1416DN 1000C 1000C 1000C 5 N or N 5 or N 5 3.1416D 3.1416D pD C5

Example A high-speed steel milling cutter with a 45 millimeter diameter and a cutting speed of 12 meters per minute is used for a roughing operation on an annealed chromium-nickel steel workpiece. Determine the revolutions per minute to the nearest whole number. 1000C 1000s12d N5 5 pD p s45d 1000 3 12 4 ( p 3 45 ) 5 84.88264, 85 r/min Ans (rounded)

cutting tiME uSing MEtric unitS of MEaSurE Cutting time is determined by the length or depth to be cut in millimeters, the revolutions per minute of the revolving workpiece or revolving tool, and the tool feed in millimeters for each revolution of the workpiece or tool. L T5 where T 5 cutting time per cut in minutes FN L 5 length of cut in millimeters F 5 tool feed in millimeters per revolution N 5 r/min of revolving workpiece or tool

Example An 88-millimeter diameter cast iron cylinder is turned in a lathe at 260 revolutions per minute. Each length of cut is 700 millimeters and 5 cuts are required. A carbide tool is fed into the workpiece at 0.40 millimeter per revolution. What is the total cutting time? Round the answer to the nearest minute. Calculate the time required for one cut. L 700 T5 5 < 6.73 min FN 0.40s260d Calculate the total cutting time. 5(6.73 min) 5 33.65 min, 34 min Ans (rounded)

uSing data froM a cutting SpEEd tablE Tables of cutting speeds have been developed and are used in determining machine spindle speed (revolutions per minute) settings. The tables take into consideration the material to be cut and the tool material.

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329

applications oF Formulas to cutting speed, revolutions per minute, and cutting time

In addition to the material being cut and the type of tool used, other factors must be taken into consideration. Variables are considered, such as the depth and width of cut, the design of the cutting tool, the rate of feed, the coolant used, and the finish required. Because cutting speed depends upon many factors, data given in cutting speed tables should be considered as recommended values. Generally it is not possible to set machines to an exact calculated spindle speed. Therefore, a simplified spindle speed formula is used in computing revolutions per minute. In the simplified formula, 3.1416 is rounded to 3. 12C 4C N5 5 3D D Comprehensive detailed cutting speed tables are available that list cutting speeds for specific materials based on material alloy composition, hardness, and condition. Some tables list cutting speeds separately for rough and finish cuts. Selected materials are listed in the following table with their respective cutting speeds using high-speed steel and carbide tools. CUTTING SPEEDS: FEET PER MINUTE (fpm) Turning

Milling

Carbide Tool

100

350

70–130

200–400

70

40

175

Alloy Steel (4320), BHN 220–275

70

300

50–100

225–450

60

40

150

Malleable Cast Iron (32510), BHN 110–160

200

600

130–225

400–800

130

90

240

60

200

50–80

175–275

40

25

100

Aluminum (5052)

600

1200

500–800

1000–1800

250

250

700

Brass, annealed

300

650

250–450

500–900

160

160

320

Manganese Bronze, cold drawn

250

550

200–350

450–650

140

120

275

Beryllium Copper, annealed

100

200

80–140

180–275

60

50

180

Carbon Steel (1020), BHN 175–225

Stainless Steel (305), BHN 225–275

Carbide Tool

High-Speed Steel Tool

Reaming

High-Speed Steel Tool

Material

High-Speed Steel Tool

Drilling

High-Speed Steel Tool

Carbide Tool

Revolutions per minute are generally computed using table cutting speeds with the simplified revolutions per minute formula. Where a range of cutting speed table values is listed, use the average of the low and high speeds given. For example, the cutting speed for milling the alloy steel shown in the table with a high-speed steel cutter is listed as 50–100 feet per minute. Use the average cutting speed of: 75 feet per minute

150 12 100 5 752

After revolutions per minute are calculated, generally, the machine spindle speed is set to the closest spindle speed below the calculated revolutions per minute. The spindle speed may then be increased or decreased depending on the performance of the operation.

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Section 4

Fundamentals oF algebra

Example 1 Calculate the revolutions per minute required to turn a 3.500-inch diameter piece of stainless steel using a carbide toolbit. Express the answer to the nearest revolution per minute. Refer to the table of cutting speeds. The recommended cutting speed is 200 feet per minute. N5

4C 4s200d 5 < 228.571, 229 rpm Ans (rounded) D 3.500

Example 2 A carbon steel plate is milled using a 2.75-inch diameter high-speed steel cutter. Compute, to the nearest whole number, the revolutions per minute. Refer to the table of cutting speeds. The recommended cutting speed is 100 feet per minute.

170 12 130 5 1002

N5

4C 4s100d 5 < 145.455, 145 rpm Ans (rounded) D 2.75

ApplicAtion tooling Up 3 1 1. Solve E 1 s18 2 Ed 5 2 E 1 12. 4 4 1 2. Solve Y 3 5 2 . Express the answer as both a common fraction and a decimal fraction. 125 3. Solve the equation 4.035 2 T 5 9.635 for the unknown value T. 4. Determine the root of the term

Î

9 4 8 12 x bc . 19

5. Use the Table of Block Thicknesses of a Customary Gage Block Set under the heading “Description of Gage Blocks” in Unit 37 to determine a combination of gage blocks for 4.32970. 6. Use a digital height gage to measure the height of your textbook in both customary and metric units.

cutting Speeds Given the workpiece or tool diameters and the revolutions per minute, determine the cutting speeds in the following 3.1416DN pDN 3.1416DN pDN tables to the nearest whole number. Use C 5 or for customary units and C 5 or for 12 12 1000 1000 metric units. Workpiece or Tool Diameter

Workpiece or Tool Diameter

Revolutions Cutting per Minute Speed (fpm)

Revolutions per Minute

7.

0.475”

460

12.

190.00 mm

59

8.

2.750”

50

13.

53.98 mm

764

9.

4.000”

86

14.

3.25 mm

1525

10.

0.850”

175

15.

133.35 mm

254

11.

1.750”

218

16.

6.35 mm

4584

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Cutting Speed (m/min)

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applications oF Formulas to cutting speed, revolutions per minute, and cutting time

Revolutions per Minute Given the cutting speed and the tool or workpiece diameter, determine the revolutions per minute in the following 12C 12C 1000C 1000C tables to the nearest whole number. Use N 5 or for customary units and N 5 or for metric 3.1416D pD 3.1416D pD units. Cutting Workpiece or Revolutions Speed (fpm) Tool Diameter per Minute

Cutting Workpiece or Revolutions Speed (m/min) Tool Diameter per Minute

17.

70

2.400”

22.

130

25.50 mm

18.

120

0.750”

23.

100

66.70 mm

19.

90

8.000”

24.

30

6.35 mm

20.

180

8.000”

25.

180

15.80 mm

21.

200

0.375”

26.

150

114.30 mm

cutting time Given the number of cuts, the length of cut, the revolutions per minute of the workpiece or tool, and the tool feed, determine the total cutting time in the table to 1 decimal place. L Use T 5 . FN Number of Cuts

Feed (per Revolution)

Length of Cut

Revolutions per Minute

27.

1

0.002”

20”

28.

1

0.12 mm

925 mm

610

29.

4

0.008”

8”

350

Tool Cutting Time (Minutes)

2100

cutting Speed and Surface Speed Problems Compute the following problems. Express the answers to the nearer whole number. Use C 5 3.1416DN pDN customary units and C 5 or for metric units. 1000 1000 1 30. A 3 -inch diameter high-speed steel cutter, running at 55 rpm, is used to rough 2 mill a steel casting. What is the cutting speed?

3.1416DN pDN or for 12 12

31. A 50-millimeter diameter carbon steel drill running at 286 r/min is used to drill an aluminum plate. Find the cutting speed. 32. What is the surface speed of a 16-inch diameter surface grinder wheel running at 1194 rpm? 33. A medium-steel shaft is cut in a lathe using a high-speed steel tool. Determine the cutting speed if the shaft is 2.125 inches in diameter, and is turning at 275 rpm. 34. A finishing cut is taken on a brass workpiece using a 100-millimeter diameter carbon steel milling cutter. What is the cutting speed when the cutter is run at 86 r/min?

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Section 4

Fundamentals oF algebra

Revolutions per Minute Problems 12C 12C Compute the following problems. Express the answers to the nearest whole number. Use N 5 or for 3.1416D pD 1000C 1000C customary units and N 5 or for metric units. 3.1416D pD 35. Grooves are cut in a stainless steel plate using a 3.750-inch-diameter carbide milling cutter with a cutting speed of 180 feet per minute. Determine the revolutions per minute. 36. An annealed cast iron housing is drilled with a cutting speed of 20 meters per minute using a 22-millimeter diameter carbon steel drill. Find the revolutions per minute. 37. A grinding operation is performed using a 150-millimeter diameter wheel with a cutting speed of 1800 meters per minute. Determine the revolutions per minute. 38. Determine the revolutions per minute of an aluminum alloy rod 1.250 inches in diameter with a cutting speed of 550 feet per minute. 39. A high-speed steel milling cutter with a 1.750-inch diameter and a cutting speed of 40 feet per minute is used for a roughing operation on an annealed chromium-nickel steel workpiece. Find the revolutions per minute.

cutting time Problems Compute the following problems. Express the answers to 1 decimal place. Use: L T5 FN 1 40. Cast iron, 3 inches in diameter, is turned in a lathe at 270 rpm. Each length 4 of cut is 27.00 inches and five cuts are required. A carbide tool is fed into the work at 0.015 inch per revolution. What is the total cutting time? 41. A slot 812.00 millimeters long is cut into a carbon steel baseplate with a feed of 0.80 millimeter per revolution. Find the cutting time using a 75-millimeter diameter carbide milling cutter running at 640 r/min. 42. Fifteen 3.20-millimeter diameter holes each 57.15 millimeters deep are drilled in an aluminum workpiece. The high-speed steel drill runs at 9200 r/min with a feed of 0.05 millimeter per revolution. Determine the total cutting time. 43. Thirty 2-inch diameter stainless steel shafts are turned in a lathe at 250 rpm. Two cuts each 14.5 inches long are required using a feed of 0.020 inch per revolution. Setup and handling time averages 3 minutes per piece. Calculate the total production time. 44. Seven brass plates 9.00 inches wide and 21.00 inches long are machined with a milling cutter along the length of the plates. The entire top face of each plate is 1 milled. The width of each cut allowing for overlap is 2 inches. Using a feed of 4 0.020 inch per revolution and 525 rpm, determine the total cutting time.

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Unit 47

applications oF Formulas to cutting speed, revolutions per minute, and cutting time

333

complex Problems The solution of the following problems requires more than one formula and the rearrangement of formulas. 3.1416DN pDN or for customary units and 12 12 3.1416DN pDN C5 or for metric units. 1000 1000 12C 12C N5 or for customary units and 3.1416D pD 1000C 1000C N5 or for metric units. 3.1416D pD L T5 FN

Use C 5

45. A 3.000-inch diameter cylinder is turned for an 11.300-inch length of cut. The cutting speed is 300 feet per minute and the cutting time is 1.02 minutes. Calculate the tool feed in inches per revolution. Round the answer to 3 decimal places. 46. A combination drilling and countersinking operation on bronze round stock is 3 performed on an automatic screw machine. The length of cut per piece is 1 inches. 4 1 The total cutting time for 2300 pieces is 6 hours running at 1600 rpm. What is the 2 tool feed in inches per revolution? Round the answer to 3 decimal places. 1 47. Steel shafts, 1 inches in diameter, are turned on an automatic machine. One 4 finishing operation is required for a 16.5-inch length of cut. The tool feed is 0.015 inch per revolution using a cutting speed of 200 feet per minute. Determine the number of hours of cutting time required for 1500 shafts. Round the answer to the nearest hour. 48. A carbide milling cutter is used for machining a 560.00-millimeter length of stainless steel. The cutting time is 11.95 minutes, the cutting speed is 60.000 meters per minute, and the feed is 0.250 millimeter per revolution. What is the diameter of the carbide milling cutter? Round the answer to 1 decimal place. 5 1 49. Aluminum baseplates are produced that are 1 inches thick. Six -inch-diameter 8 4 holes are drilled in each plate using a feed of 0.004 inch per revolution and a cutting speed of 300 feet per minute. Setup and handling time is estimated at 0.5 minute per piece. What is the total number of hours required to produce 850 aluminum baseplates? Round the answer to 1 decimal place.

cutting Speed table Refer to the cutting speed table under the heading “Using Data from a Cutting Speed Table” earlier in this unit. Use the 4C table values and the simplified revolutions per minute formula, N 5 . Compute the revolutions per minute to the D nearer revolution for each problem in the following table.

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Section 4

Material Machined

Cutting Operation

Fundamentals oF algebra

Tool Material

Tool or Workpiece Diameter (inches)

50.

Aluminum (5052)

Milling

High-Speed Steel

3.500

51.

Stainless Steel (305), BHN 225–275

Turning

Carbide

5.200

52.

Alloy Steel (4320), BHN 220–275

Reaming

High-Speed Steel

0.480

53.

Manganese Bronze, cold drawn

Drilling

High-Speed Steel

0.375

54.

Brass, annealed

Milling

High-Speed Steel

4.000

55.

Carbon Steel (1020), BHN 175–225

Turning

Carbide

6.100

56.

Beryllium Copper, annealed

Drilling

High-Speed Steel

1.100

57.

Malleable Cast Iron (32510), BHN 110–160

Milling

Carbide

3.000

58.

Alloy Steel (4320), BHN 220–275

Milling

Carbide

2.500

59.

Aluminum (5052)

Turning

High-Speed Steel

5.800

60.

Carbon Steel (1020), BHN 175–225

Milling

Carbide

4.500

61.

Brass, annealed

Turning

High-Speed Steel

2.750

62.

Stainless Steel (305), BHN 225–275

Reaming

Carbide

0.620

63.

Malleable Cast Iron (32510), BHN 110–160

Turning

Carbide

7.000

64.

Carbon Steel (1020), BHN 175–225

Drilling

High-Speed Steel

0.375

Speed (rpm)

UNIT 48 Applications of Formulas to Spur Gears Objectives After studying this unit you should be able to ●● ●● ●● ●●

Identify the proper gear formula to use depending on the unknown and the given data. Compute gear part dimensions by substituting known values directly into formulas. Compute gear part dimensions by rearranging given formulas in terms of the unknowns. Compute gear part dimensions by the application of two or more formulas in order to determine an unknown.

Gears have wide application in machine technology. They are basic to the design and operation of machinery. Most machine shops are equipped to cut gears, and some shops specialize in gear design and manufacture. It is essential that the machinist and drafter have an understanding of gear parts and the ability to determine gear dimensions by the use of trade handbook formulas.

dEScription of gEarS Gears are used for transmitting power by rotary motion between shafts. Gears are designed to prevent slippage and to ensure positive motion while maintaining a high degree of accuracy of the speed ratios between driving and driven gears. The shape of the gear tooth is of primary importance in providing a smooth transmission of motion. The shape of most

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applications oF Formulas to spur gears

gear teeth is an involute curve. This curve is formed by the path of a point on a straight line as it rolls along a circle. Spur gears are gears that are in mesh between parallel shafts. Of two gears in mesh, the smaller gear is called the pinion and the larger gear is called the gear.

Spur gEar dEfinitionS Spur gears and the terms that apply to these gears are shown in Figures 48-1 and 48-2. It is essential to study the figures and gear terms before computing gear problems by the use of formulas.

OUTSIDE DIAMETER PITCH DIAMETER ROOT DIAMETER

Pitch Circles are the imaginary circles of two meshing gears that make contact with each other. The circles are the basis of gear design and gear calculations.

ROOT CIRCLE

Pitch Diameter is the diameter of the pitch circle.

PITCH CIRCLE

Root Circle is a circle that coincides with the bottoms of the tooth spaces.

Figure 48-1

Root Diameter is the diameter of the root circle. Outside Diameter is the diameter measured to the tops of the gear teeth. Addendum is the height of the tooth above the pitch circle. Dedendum is the depth of the tooth space below the pitch circle. Whole Depth is the total depth of the tooth space. It is equal to the addendum plus the dedendum. Working Depth is the total depth of mating teeth when two gears are in mesh. It is equal to twice the addendum. Clearance is the distance between the top of a tooth and the bottom of the mating tooth space of two gears in mesh. It is equal to the whole depth minus the working depth. Tooth Thickness (Circular) is the length of the arc, on the pitch circle, between the two sides of a tooth. Circular Pitch is the length of the arc measured on the pitch circle between the centers of two adjacent teeth. It is equal to the circumference of the pitch circle divided by the number of teeth on the gear. Diametral Pitch (Pitch) is the ratio of the number of gear teeth to the number of inches of pitch diameter. It is equal to the number of gear teeth for each inch of pitch diameter. PITCH CIRCLE TOOTH THICKNESS CIRCULAR PITCH CLEARANCE

CLEARANCE

WHOLE DEPTH WORKING DEPTH

PITCH CIRCLE

DEDENDUM ADDENDUM

Figure 48-2

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When the pitch of a gear is mentioned, the reference is to diametral pitch, rather than circular pitch. For example, if a gear has 28 teeth and a pitch diameter of 4 inches, it 28 has a pitch (diametral pitch) of , or seven. It has seven teeth per inch of pitch diameter, 4 and it is called a 7-pitch gear. It will only mesh with other 7-pitch gears. Gears must have the same pitch in order to mesh.

gEaring–diaMEtral pitch SyStEM The diametral pitch system is the system of gear design that is generally applied to decimalinch dimensional gears. The following table lists the symbols and formulas used in the diametral pitch system. Remember to use the p key when using these formulas with a calculator or PI() when using these formulas in a spreadsheet. In the table, N represents the number of teeth, D the pitch diameter, and P the pitch. DECIMAL-INCH SPUR GEARS (American National Standard) Term

Symbol

Pitch (diametral pitch)

P

Circular Pitch

PC

Pitch Diameter

D

Formulas N D 3.1416 P5 PC P5

3.1416D N 3.1416 PC 5 P PC 5

D5 D5

Outside Diameter

DO

N P

NPC

3.1416

DO 5 DO 5

N12 P PC sN 1 2d

3.1416 DO 5 D 1 2a Root Diameter

DR

DR 5 D 2 2d

Addendum

a

1 a5 * P a 5 0.3183PC*

Dedendum

d

d5

1.157 * P

d 5 0.3683PC* Whole Depth

WD

2.157 * P WD 5 0.6866PC* WD 5

WD 5 a 1 d* Working Depth

Clearance

WD

c

2.000 * P WD 5 0.6366PC* WD 5

0.157 * P c 5 0.050PC* c5

(Continued)

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DECIMAL-INCH SPUR GEARS (American National Standard) (continued) Term

Symbol

Tooth Thickness

T

Number of Teeth

N

Formulas T5

1.5708 P

N 5 PD N5

3.1416D PC

*Note: Formulas for 14 12 -degree Involute and Composite Full-Depth Teeth and 20-degree Involute Full-Depth Teeth.

gEar calculationS Most gear calculations are made by identifying the proper formula, which is given in terms of the unknown, and substituting the known dimensions. It is sometimes necessary to rearrange a formula in terms of a particular unknown. The solution of a problem may require the substitution of values in two or more formulas. Refer to the Decimal-Inch Spur Gears Table under the previous heading “Gearing-Diametral Pitch System” to complete the following Examples.

Example 1 Determine the pitch diameter of a 5-pitch gear that has 28 teeth. Identify the formula whose parts consist of pitch diameter, pitch, and number of teeth.

D5

N P

This is a 5-pitch gear, so P 5 5, and since the gear has 28 teeth, N 5 28. 28 Solve. D5 5 D 5 5.6000 inches Ans

Example 2 Determine the outside diameter of a gear that has 16 teeth and a circular pitch of 0.7854 inch. PC sN 1 2d Identify the formula whose DO 5 3.1416 parts consist of outside diameter, number of teeth, and circular pitch. Here the number of teeth is 16, so N 5 16, and with a circular pitch of 0.7854 in., PC 5 0.7854. 0.7854s16 1 2d Solve. DO 5 3.1416 .7854 3 ( 16 1 2 ) 4 p 5 4.50001523 DO 5 4.5000 inches Ans (rounded)

Example 3 Determine the circular pitch of a gear with a whole depth dimension of 0.3081 inch. Identify the formula whose parts WD 5 0.6866PC consist of circular pitch and whole depth. WD The formula must be rearranged in PC 5 0.6866 terms of circular pitch. 0.3081 Solve. PC 5 0.6866 PC < 0.4487 inch

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Ans (rounded)

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Fundamentals oF algebra

Example 4 Determine the addendum of a gear that has an outside diameter of 3.0000 inches and a pitch diameter of 2.7500 inches. Identify the formula whose parts consist of addendum, outside diameter, and pitch diameter. The formula must be rearranged in terms of the addendum.

DO 5 D 1 2a

DO 2 D 5 2a a5

DO 2 D

2 3.0000 2 2.7500 a5 2 a 5 0.1250 inch Ans

Solve.

Example 5 Determine the working depth of a gear that has 46 teeth and a pitch diameter of 11.5000 inches. There is no single formula in the table that consists of working depth, number of teeth, and pitch diameter. Therefore, it is necessary to substitute in two formulas in order to solve the problem. 2.0000 N Observe WD 5 . P5 P D 46 The pitch must be found first. P5 11.5000 P54 2.000 Solve for WD. WD 5 P 2.000 WD 5 4 WD 5 0.5000 inch Ans

gEaring–MEtric ModulE SyStEM The module system of gear design is generally the system that is used with metric system units of measure. The module of a gear equals the pitch diameter divided by the number of teeth. In the metric system, the module of a gear means the pitch diameter in millimeters is divided by the number of teeth. Module is an actual dimension in millimeters, not a ratio as with diametral pitch. For example, if a gear has 20 teeth and a 50-millimeter pitch diameter, the module is 2.5 millimeters (50 mm 4 20). A module of 2.5 millimeters means that there are 2.5 millimeters of pitch diameter per tooth. A partial list of a standard series of modules (in millimeters) is listed as follows:

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1

2

3

4

6

1.25

2.25

3.25

4.5

6.5

9

1.5

2.5

3.5

5

7

11

1.75

2.75

3.75

5.5

8

12

10

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The relation between module and various gear parts using a metric module system is shown in the following table. METRIC SPUR GEARS Term

Symbol

Module

m

Circular Pitch

Formulas m5

D N

PC

PC 5

m 0.3183

Pitch Diameter

D

D 5 mN

Outside Diameter

DO

Addendum

a

a5m

Dedendum

d

d 5 1.157m* d 5 1.167m**

Whole Depth

WD

Working Depth

WD

Clearance

c

c 5 0.157m* c 5 0.1667m**

Tooth Thickness

T

T 5 1.5708m

DO 5 msN 1 2d

WD 5 2.157m* WD 5 2.167m** WD 5 2m

*When clearance 5 0.157 3 module **When clearance 5 0.1667 3 module

Examples Refer to the Metric Spur Gears Table. 1. Determine the circular pitch of a 6-millimeter module gear. Circular Pitch 5 Module 4 0.3183 Circular Pitch 5 6 mm 4 0.3183 ≈ 18.850 mm Ans (rounded) 2. Determine the outside diameter of a 3.5-millimeter module gear with 20 teeth. Outside Diameter 5 Module 3 (Number of teeth 1 2) Outside Diameter 5 3.5 mm 3 (20 1 2) 5 3.5 mm 3 22 5 77.000 mm Ans 3. Compute the dedendum of a 4.5-millimeter module gear designed with a clearance of 0.157 3 module. When Clearance 5 0.157 3 Module, the Dedendum 5 1.157 3 Module Dedendum 5 1.157 3 4.5 mm ≈ 5.207 mm Ans (rounded) 4. Compute the whole depth of 7-millimeter module gear designed with a clearance of 0.1667 3 module. When Clearance 5 0.1667 3 Module, the Whole Depth 5 2.167 3 Module Whole Depth 5 2.167 3 7 mm 5 15.169 mm Ans

ApplicAtion tooling Up 3.1416DN pDN or to 12 12 determine the cutting speed in feet per minute (fpm). Round your answer to the nearest whole number.

1. A workpiece or tool with a diameter of 0.7500 turns at 350 revolutions per minute. Use C 5

1 2. Use the formula N 5 Ï2DPn with D 5 5 and N 5 12 to determine Pn. Round your answer to 3 decimal places. 2

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N 3. The pitch (P) of a spur gear equals the number (N) of gear teeth divided by the pitch diameter (D): P 5 . D Determine N when P 5 5 and D 5 5.6000 inches. 27 2 15 4. Solve 1.95x 2 0.75x 5 . 5 p2 2 qr 5. If p 5 5, q 5 3.4, and r 5 9, what is the value of ? 4sp 2 qd 6. Read the setting on the following 0.001-inch micrometer scale. The vernier, thimble, and barrel scales are shown in flattened view. 09876543210

0 20

0 1 2 3 4 5 6 15

Gearing–Diametral Pitch System Refer to the Decimal-Inch Spur Gears Table under the heading “Gearing-Diametral Pitch System” earlier in this unit for each of the following gearing problems. GIVEN VALUES

FIND

Circular Pitch 5 1.57080

Pitch

Pitch 5 10

Circular Pitch

Pitch Diameter 5 5.20000 Number of Teeth 5 26

Circular Pitch

10.

Pitch Diameter 5 12.57140 Number of Teeth 5 44

Pitch

11.

Pitch 5 7 Number of Teeth 5 26

Pitch Diameter

12.

Circular Pitch 5 0.31420 Number of Teeth 5 12

Pitch Diameter

13. 14.

Pitch 5 18

Circular Pitch

Pitch Diameter 5 0.72730 Number of Teeth 5 16

Pitch

15.

Pitch 5 12 Pitch Diameter 5 1.16670

Number of Teeth

16.

Circular Pitch 5 0.62830 Pitch Diameter 5 8.40000

Number of Teeth

17.

Number of Teeth 5 56 Pitch 5 8

Outside Diameter

18. 19.

Pitch 5 14

Addendum

Pitch Diameter 5 1.33330 Dedendum 5 0.06430

Root Diameter

20. 21. 22. 23. 24.

Pitch 5 3.4

Whole Depth

Circular Pitch 5 0.28560

Working Depth

Circular Pitch 5 1.46500

Clearance

Pitch 5 20

Tooth Thickness

Pitch Diameter 5 3.50000 Addendum 5 0.18180

Outside Diameter

7. 8. 9.

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ANSWER

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applications oF Formulas to spur gears

GIVEN VALUES

FIND

25. 26. 27.

Circular Pitch 5 0.09540

Dedendum

Circular Pitch 5 0.89760

Addendum

Addendum 5 0.14290 Dedendum 5 0.16530

Whole Depth

28. 29. 30. 31.

Pitch 5 4

Dedendum

Circular Pitch 5 0.30760

Whole Depth

Pitch 5 17

Clearance

Pitch 5 9

Working Depth

ANSWER

Refer to the Decimal-Inch Spur Gears Table under the heading “Gearing-Diametral Pitch System” earlier in this unit. The formula in terms of the unknown is not given. Choose the formula that consists of the given parts, rearrange in terms of the unknown, and solve. GIVEN VALUES

FIND

32. 33. 34.

Addendum 5 0.08570

Circular Pitch

Addendum 5 0.06660

Pitch

Addendum 5 0.20000 Outside Diameter 5 4.80000

Pitch Diameter

35.

Outside Diameter 5 2.71440 Number of Teeth 5 17

Pitch

36.

Outside Diameter 5 4.37500 Circular Pitch 5 0.39270

Number of Teeth

37. 38. 39.

Working Depth 5 0.07690

Pitch

Working Depth 5 0.45000

Circular Pitch

Outside Diameter 5 4.71440 Pitch Diameter 5 4.42860

Addendum

ANSWER

Refer to the Decimal-Inch Spur Gears Table under the heading “Gearing-Diametral Pitch System” earlier in this unit. No single formula is given that consists of the given parts and the unknown. Two or more formulas, some in rearranged form, must be used in solving these problems. GIVEN VALUES

FIND

40.

Number of Teeth 5 72 Pitch Diameter 5 6.00000

Addendum

41.

Number of Teeth 5 44 Pitch Diameter 5 3.66670

Dedendum

42.

Number of Teeth 5 10 Pitch Diameter 5 2.50000

Whole Depth

43.

Number of Teeth 5 90 Pitch Diameter 5 12.85710

Working Depth

44.

Pitch Diameter 5 1.06250 Pitch 5 16

Outside Diameter

45.

Pitch Diameter 5 2.91670 Pitch 5 12

Root Diameter

46.

Number of Teeth 5 29 Pitch Diameter 5 2.07140

Root Diameter

47.

Number of Teeth 5 75 Pitch Diameter 5 6.81820

Clearance

48.

Addendum 5 0.14290

Tooth Thickness

49.

Pitch Diameter 5 1.04550 Addendum 5 0.04550

Number of Teeth

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ANSWER

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Backlash is the amount that a tooth space is greater than the engaging tooth on the pitch circles of two gears. Determine the average backlash of each of the following using this formula. Average backlash 5

0.030 P

50. A 7-pitch gear 51. A 20-pitch gear 52. A 3.5-pitch gear

PITCH CIRCLE

PITCH CIRCLE

53. A gear with a whole depth of 0.26960

BACKLASH

54. A gear with a working depth of 0.11760 55. A gear with a pitch diameter of 4.8000 and 24 teeth The center distance of a pinion and a gear is the distance between the centers of the pitch circles. Determine the center distance of each of the following using this formula. Center Distance 5

Pitch Diameter of Gear 1 Pitch Diameter of Pinion 2

56. A pinion with a pitch diameter of 2.8300 inches and a gear with a pitch diameter of 4.1667 inches 57. A pinion with a pitch diameter of 4.8889 inches and a gear with a pitch diameter of 8.6752 inches 58. A 9-pitch pinion and gear; the pinion has 23 teeth and the gear has 38 teeth

PITCH CIRCLE

59. A 16-pitch pinion and gear; the pinion has 18 teeth and the gear has 44 teeth

PITCH DIAMETER

60. A gear and pinion with a circular pitch of 0.1745 inch; the gear has 55 teeth and the pinion has 37 teeth

CENTER DISTANCE

gearing–Metric Module System Refer to the Metric Spur Gears Table under the heading "GearingMetric Module System" earlier in this unit and determine the values in the following table. Module

Number of Teeth

61.

6.5 mm

18

62.

9 mm

24

63.

2.5 mm

10

64.

3.75 mm

16

65.

10 mm

26

a. Pitch Diameter

b. Circular Pitch

c. Outside Diameter

PITCH DIAMETER PITCH CIRCLE

d. Addendum

e. Working Depth

f. Tooth Thickness

Solve the following metric module system gearing problems. Certain problems require rearranging the data given in the Metric Spur Gears Table under the heading “Gearing-Metric Module System” earlier in this unit. 66. Compute the whole depth of a 5-millimeter module gear designed with a clearance of 0.157 3 module.

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67. What is the number of teeth on a 4-millimeter module gear with a pitch diameter of 120 millimeters? 68. What is the module of a gear that has a working depth of 13 millimeters? 69. Compute the dedendum of a 7-millimeter module gear designed with a clearance of 0.1667 3 module. 70. What is the module of a gear with 38 teeth and an outside diameter of 220 millimeters?

UNIT 49 Achievement Review—Section Four Objective You should be able to solve the exercises and problems in this Achievement Review by applying the principles and methods covered in Units 39–48.

Express each of the problems in Exercises 1 and 2 as an algebraic expression. 1. a. The sum of x and y reduced by c. b. The product of a and b divided by d. 2. a. Twice M minus the square of P. b. The sum of A and B divided by the square root of D. In Exercises 3 and 4, substitute the given numbers for letters and find the value for each of the expressions. Round the answers to 3 decimal places where necessary. 3. a. Find (5a 1 6b) 4 4b when a 5 4 and b 5 5. b. Find 3xy 2 (2x 1 y) when x 5 6 and y 5 3. 2

b. Find Ïhr 1 5p

2

1 2 when e 5 5.125, f 5 3.062, and m 5 6.127.

m 4. a. Find e 1 s2f d 2 f 2

15hp 1 h2 when h 5 10.26, p 5 8.00, and r 5 6.59.

Perform the operation or operations as indicated for each of Exercises 5 and 6. Round the answers to 3 decimal places where necessary. 5. a.225 1 (212) b. 24 1 (28) c.21.8 2 (12.6) d. 18(24) e.(20.3)(22.6) f.(25)3 3

g. Ï 227

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6. a. 218 4 3 b. 212.8 4 (20.4) c. (26)2 2

1 2

d. 2 e.

1 4

Î 3

27.063 8.920

3 f. s24.02d2 1 Ï 8.96 2 s3.86ds25.66d

g. [4(210.66)(0.37)] 4 (12 1 18.95)

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The expressions in Exercises 7 and 8 consist of literal terms. Perform the indicated operations. 7. a. 28P 1 21P b. 20.05H2 2 1.13H2 c. 12d 1 8d2 2 7d 1 14d2 d. (5a 2 2a2) 2 (6a 2 4a2) e. (210x)(9x2y) f. (25.9e2f2)(2f2) g. (16x2 2 4x3) 4 0.5x h. (0.6f 2g3 2 fg 2 2f 2) 4 0.2f i.

Î

25 4 2 egd 64

8. a. (24a3bc2)3 b. [(xy2)3 2 (x2y2)]2 3 c. Ï 227a6b3c9 d. (6x2 2 y3)(23x3 1 y2) e. 26(x2 1 y 2 2x) f. 36 2 (m3 1 m) 1 (m3 2 12) g. 23as2ad2 1 Ï36a4 h. 9y 2 x[28 1 (xy)2 2 y] 1 12x i. b(b 1 7m)2 2 b(b 2 7m)2

In Exercises 9 and 10, solve for the unknown in each of the equations using one of the six principles of equality. Check each answer. Round the answers to 3 decimal places where necessary. x 9. a. x 1 12 5 33 10. a. 5 8.48 20.8 b. y 2 15 5 23 b. 6.75 5 220.25x 3 1 c. 232 5 B 2 46 c. x 5 2 4 4 d. H 1 11.7 5 43.9 d. s2 5 81 28 e. 14.3 5 x 1 53.6 e. x3 5 27 f. R 2 7.8 5 29.2 f. ÏM 5 12.892 3 g. 7y 5 284 g. Ï V 5 5.873 h. 20.0284 5 y3 3.866 3 i. Ï B5 4.023

h. 1.3E 5 7.54 L i. 11.22 5 6.6

In Exercises 11 and 12, solve for the unknown and check each of the combined operations equations. Round the answers to 3 decimal places where necessary. 11. a. 32 2 (2P 1 18) 5 45 b. 10(M 2 4) 5 25(M 2 4) c. 7.1E 1 3(E 2 6) 5 0.5E 1 22.8 H d. 1 7.8 5 13.6 4 1 1 3 e. F 1 6 F 2 5 15 4 2 4

1

2

12. a. 12x2 2 53 5 (x 2 3)(x 1 3) b. 28.53(G 2 3.67) 5 5.7(218.36) c. (T 2 7.8)(T 2 8) 5 T2 1 0.3 d. 59.66Ïx 5 8.71s1.07Ïx 1 55.32d e. H 1 5.023ÏH 5 29.777ÏH 1 H

In each of the formulas in Exercises 13 and 14, substitute the given numerical values for letter values, solve for the unknown, and check. Round the answers to 3 decimal places where necessary. 13. a. N 5 0.707DPn nE b. I 5 R 1 nr 0.290W c. S 5 t2

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Solve for D when N 5 36 and 5 12. Solve for E when I 5 0.3, n 5 5, R 5 6, and r 5 4. Solve for W when S 5 600 and t 5 0.375.

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achievement review—section Four

14. a. c 5 Ïa2 1 b2 b. V 5 1.570h(R2 1 r2) c. P 5 2.4TÏ9 1 W2

Solve for b when a 5 8.053 and c 5 10.096. Solve for R when V 5 105.823, h 5 5.897, and r 5 2.023. Solve for W when P 5 4.344 and T 5 0.25.

In Exercises 15 and 16, rearrange each of the formulas in terms of the designated letter. 15. a. E 5 I(R 1 r) b. DO 5 2C 2 d 1 2a c. M 5 D 2 1.5155P 1 3W

Solve for r. Solve for a. Solve for W.

16. a. r 5 Ïx2 1 y2 b. L 5 3.14(0.5D 1 0.5d) 1 2x D2N c. HP 5 2.5

Solve for y. Solve for x. Solve for D.

In Exercises 17 and 18, solve for the unknown value in each of the proportions and check. Round the answers to 3 decimal places where necessary. P 3 5 12.8 2 3.6 E b. 5 0.9 2.7

17. a.

1 H21 2 c. 5 1 3 4 8 7 C17 d. 5 6 12 10.360 2.015 e. 5 7.890 N 2 2.515

10 1 D 5 5 D 9 G15 G b. 5 G16 G12 R24 R26 c. 5 R12 R18

18. a.

d.

n3 1 6 n3 2 8 5 10 8

e.

T 3 1 3T 2 0.7 T 3 1 5T 1 9.1 5 3 5

19. The following problems are given in scientific notation. Solve and leave answers in scientific notation. Round the answers (mantissas) to 2 decimal places. a. (3.76 3 104) 3 (2.87 3 103)

c.

s9.76 3 1025d 3 s1.77 3 109d s5.87 3 103d

b. (8.63 3 107) 4 (5.77 3 1025)

d.

s9.09 3 1025d s4.72 3 106d 3 s6.15 3 1023d

20. The following problems are given in decimal (standard) form. Calculate and give answers in scientific notation. Round the answers (mantissas) to 2 decimal places. a. 0.00021 3 0.00039 b. 1,476,000 3 20.0000373

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0.0000287 3 216,000,000 0.00981 0.00503 3 0.000406 d. 416,000 3 0.00392 c.

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21. Refer to the Decimal-Inch Spur Gears Table under the heading “Gearing-Diametral Pitch System” earlier in this unit for each of the following. The formula in terms of the unknown may not be given. Given Values

Find

Answer

a.

Circular Pitch 5 0.57120 Number of Teeth 5 18

Pitch Diameter

b.

Pitch 5 1.5740

Addendum

c.

Dedendum 5 0.0520

Pitch

d.

Clearance 5 0.02810

Circular Pitch

e.

Pitch Diameter 5 4.320 Number of Teeth 5 36

Tooth Thickness

f.

Pitch Diameter 5 1.03460 Dedendum 5 0.08550

Number of Teeth

22. Solve the following cutting speed and gear problems. a. A steel shaft 2.300 inches in diameter is turned in a lathe at 250 rpm. Determine the cutting speed to the nearer whole number. C5

3.1416DN 12

b. Determine the revolutions per minute to the nearer whole number of an aluminum cylinder 40.00 millimeters in diameter with a cutting speed of 160 meters per minute. N5

1000C 3.1416D

c. Twenty 5.50-millimeter diameter holes each 58.00 millimeters deep are drilled in a workpiece. The drill turns at 6500 r/min with a feed of 0.05 millimeter per revolution. Determine the total cutting time to the nearest hundredth minute. T5

L FN

d. What is the pitch of a gear with 44 teeth and a pitch diameter of 12.5714 inches? P5

N D

e. Determine the whole depth to 4 decimal places of a gear with 20 teeth and a pitch diameter of 5.0000 inches. P5

N 2.157 and WD 5 D P

f. Determine the center distance of a pinion with a pitch diameter of 6.4445 inches and a gear with a pitch diameter of 10.7533 inches.

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Fundamentals of Plane Geometry

5

UNIT 50 Lines and Angular Measure ObjECTIVES After studying this unit you should be able to ●● ●● ●●

Express decimal degrees as degrees, minutes, and seconds. Express degrees, minutes, and seconds as decimal degrees. Add, subtract, multiply, and divide angles in terms of degrees, minutes, and seconds.

The fundamental principles of geometry generally applied to machine shop problems are those used to make the calculations required for machining parts from engineering drawings. An engineering drawing is an example of applied geometry.

Plane Geometry Plane geometry is the branch of mathematics that deals with points, lines, and various figures that are made of combinations of points and lines. The figures lie on a flat surface, or plane. Examples of plane geometry are the views of a part as shown on an engineering drawing. Since geometry is fundamental to machine technology, it is essential to understand the definitions and terms of geometry. It is equally important to be able to apply the geometric principles in problem solving. The methods and procedures used in problem solving are the same as those required for the planning, making, and checking of machined parts.

●● ●● ●● ●●

To solve a geometry problem

Study the figure. Relate it to the principle or principles that are needed for the solution. Base all conclusions on fact: given information and geometric principles. Do not assume that something is true because of its appearance or because of the way it is drawn.

Note: The same requirements are applied in reading engineering drawings. 347

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axioms and Postulates In the study of geometry, certain basic statements called axioms or postulates are assumed to be true without requiring proof. Axioms or postulates may be compared to the rules of a game. Some axioms or postulates are listed. Others will be given as they are required for problem solving. ●● Quantities equal to the same quantity, or to equal quantities, are equal to each other. A quantity may be substituted for an equal quantity. ●● If equals are added to or subtracted from equals, the sums or remainders are equal. ●● If equals are multiplied or divided by equals, the products or quotients are equal. ●● The whole is equal to the sum of its parts. ●● Only one straight line can be drawn between two given points. ●● Through a given point, only one line can be drawn parallel to a given line. ●● Two different or distinct straight lines can intersect at only one point.

Points and lines A point has no size or form; it has only location. A point is shown as a dot. Each point is usually named by a capital letter as shown.

A

B C D

A line extends without end in two directions. A line has no width. It has an infinite number of points. A ray starts with a point on one end, called an endpoint, and continues indefinitely. Sometimes a ray is referred to as a half-line. A line, as it is used in this book, always means a straight line. A line other than a straight line, such as a curved line, is identified. Formally, arrowheads are used in drawing a line to show that there are no endpoints. In this book, arrowheads are not used to identify lines. A line segment is that part of a line that lies between two definite points. Line segments are often named by placing a bar over the endpoint letters. For example, segment AB may be shown as AB. In this book, segments are shown without a bar. Segment AB is shown as AB. In this book, no distinction is made between naming a line and a line segment. Parallel lines do not meet regardless of how far they are extended. They are the same distance apart (equidistant) at all points. The symbol i means parallel. In Figure 50-1, line AB is parallel to line CD; therefore, AB and CD are equidistant (distance x) at all points. B

A x

x

C

D

FiGure 50-1

Perpendicular lines meet or intersect at a right, or 908, angle. The symbol ' means perpendicular. Figure 50-2 shows three examples of perpendicular lines. 90° 90°

90°

FiGure 50-2

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Unit 50

Oblique lines meet or intersect at an angle other than 908. They are neither parallel nor perpendicular. Three examples of oblique lines are shown in Figure 50-3.

FiGure 50-3

anGles An angle is a figure that consists of two lines that meet at a point called the vertex. The symbol / means angle. The size of an angle is determined by the number of degrees one side of the angle is rotated from the other. The length of the side does not determine the size of the angle. For example, in Figure 50-4, /1 is equal to /2. The rotation of side AC from side AB is equal to the rotation of side DF from side DE although the lengths of the sides are not equal. F C 1

A

=

2

D

B

E

FiGure 50-4

units of anGular measure The degree is the basic unit of angular measure. The symbol for degree is 8. A radius that is rotated one revolution makes a complete circle or 3608. A circle may be thought of as a ray with a fixed endpoint. The ray is rotated. One rotation makes a complete circle of 3608, as shown in Figure 50-5.

120°

110°

100° 90°

80° 70° 60°

130°

50°

140°

40°

150°

30°

160°

20°

170°

10°

RAY

360°

180° 190°

350°

200°

340°

210°

330°

220°

320°

230°

310° 240°

250°

260° 270° 280°

290°

300°

FiGure 50-5

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The degree of precision required in computing and measuring angles depends on how the angle is used. Some manufactured parts are designed and processed to a very high degree of precision. In metric calculations, the decimal degree is generally the preferred unit of measurement. In the customary system, angular measure is expressed in these ways. ●● As decimal degrees, such as 6.5 degrees and 108.274 degrees. 1 1 ●● As fractional degrees, such as 12 degrees and 53 degrees. Fractional degrees are 4 10 seldom used. ●● As degrees, minutes, and seconds, such as 37 degrees, 18 minutes and 123 degrees, 46 minutes, 53 seconds. In the customary system, angular measure is generally expressed this way. Decimal and fractional degrees are added, subtracted, multiplied, and divided in the same way as any other numbers.

units of anGular measure in deGrees, minutes, and seconds A degree is divided into 60 equal parts called minutes. The symbol for minute is 9. A minute is divided into 60 equal parts called seconds. The symbol for second is 0. The relationship between degrees, minutes, and seconds is shown in the following chart. 1 Circle 5 360 Degrees (8) 1 Degree (8) 5 60 Minutes (9) 1 Minute (9) 5 60 Seconds (0)

1 of a Circle 360 1 1 Minute (9) 5 of a Degree (8) 60 1 Degree (8) 5

1 Second (0) 5

1 of a Minute (9) 60

exPressinG deGrees, minutes, and seconds as decimal deGrees Often an angle given in degrees and minutes is to be expressed as decimal degrees. This is often the case when computations involve metric system units of measure.

c Procedure ●● ●●

To express degrees and minutes as decimal degrees

Divide the minutes by 60 to obtain the decimal degree. Combine whole degrees and the decimal degree. Round the answer to 2 decimal places.

Example Express 768299 as decimal degrees. Divide 299 by 60 to obtain decimal degree.

299 4 60 5 0.488

Combine whole degrees with the decimal degree.

768 1 0.488 5 76.488 Ans

When working with customary and metric units of measure, it may be necessary to express angles given in degrees, minutes, and seconds as angles in decimal degrees.

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lines and anGular measure

To express degrees, minutes, and seconds as decimal degrees

Divide the seconds by 60 in order to obtain the decimal minute. Add the decimal minute to the given number of minutes. Divide the sum of the minutes by 60 in order to obtain the decimal degree. Add the decimal degree to the given number of degrees. Round the answer to 4 decimal places.

Example Express 238189440 as decimal degrees. Divide 440 by 60 to obtain the decimal minute. Add the decimal minute s0.73339d to the given minutes s189d. Divide the sum of the minutes (18.73339) by 60 to obtain the decimal degree. Add the decimal degree (0.31228) to the given degrees (238).

440 4 60 5 0.73339 189 1 0.73339 5 18.73339 18.73339 4 60 5 0.31228 238 1 0.31228 5 23.31228 Ans

using a calculator to express degrees, minutes, and seconds as decimal degrees There are two basic procedures used in entering degrees, minutes, seconds and then converting them to decimal degrees. Depending on the make and model of your calculator, one of the two procedures should apply. The degrees, minutes, seconds key, 8 9 0 , is used for both procedures.

c Procedure 1 Enter degrees, press 8 9 0 , enter minutes, press 8 9 0 , enter . seconds, press 8 9 0 , press 5 , press SHIFT , press Note:

is the alternate function of the primary function key 8 9 0 .

Example Convert 538479250 to decimal degrees. 53 8 9 0 47 8 9 0 25 8 9 0 5 Degrees, minutes, seconds are entered

→ 53.79027778

SHIFT

Ans

Conversion to decimal degrees

c Procedure 2 The scroll or cursor key is used with 8 9 0 in order to enter minutes and seconds. It is pressed twice with seconds. enter degrees, press 8 9 0 , enter minutes, press 8 9 0 , press , press ENTER . press 8 9 0 , press , press , press ENTER 5 5

, enter seconds,

Example Convert 538479250 to decimal degrees. 53 8 9 0 47 8 9 0 ENTER

5

25 8 9 0

53.790277788

ENTER

5

Ans

Conversion to decimal degree

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using a spreadsheet to express degrees, minutes, and seconds as decimal degrees We are going to show two ways to use a spreadsheet to convert degrees, minutes, and seconds to decimal degrees. The first is rather straightforward and will use the previous two examples. The other method will create a spreadsheet file that you can save and use. It will also be demonstrated with the previous two examples. A spreadsheet does not have a built-in function for converting degrees, minutes, and seconds to decimal degrees. However, with one addition, you can use the preceding procedures with your spreadsheet to quickly get the conversions. Since there are 60 minutes in 1 degree and 60 seconds in 1 minute, there are 3600 seconds in one degree. Thus, we have the following procedures: ●● ●● ●●

Divide the minutes by 60 to obtain the decimal degree. Divide the seconds by 3600 to obtain the decimal degree. Add the decimal degrees to the given number of degrees.

The next two examples are identical to the two we just worked, so the answers should be the same.

Example 1 Use a spreadsheet to express 788439 as decimal degrees. Solution Enter: 5 78 1 43/60 in Cell A1 and press

RETURN

with the result

78.71666667.

Example 2 Use a spreadsheet to express 135879390 as decimal degrees. Solution Enter: 5 135 1 7/60 1 39/3600 in Cell A2 and press

RETURN

. 135.1275

Ans

We next create a spreadsheet to convert degrees, minutes, and seconds to decimal degrees. We can save this spreadsheet and use it as needed. In Cell A1, we type in the title of what this file does: “Convert degrees, minutes, seconds to decimal degrees.” In Cells A3, A4, and A5, type “Degrees,” “Minutes,” and “Seconds,” respectively. We will enter the values of the degrees, minutes, and seconds in Cells B3, B4, and B5. In Cell A7, type “Decimal degrees,” and in Cell B7 enter “5 B3 1 B4/60 1 B5/3600”. Your spreadsheet should look like Figure 50-6. There is a 0 in Cell B7 because we have not entered any values in Cells B3, B4, or B5.

FiGure 50-6

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Save this spreadsheet file as DMSToDecimalDegrees, and you are ready to use it to convert degrees, minutes, seconds to decimal degrees.

Example 1 Use the DMSToDecimalDegrees spreadsheet file to express 788439 as decimal degrees.

Solution This is the first of the previous two examples. In Cell B3, enter the number of degrees, 78, and in Cell B4 the number of minutes, 43. Press RETURN , and the result 78.716667 is shown in Figure 50-7.

FiGure 50-7

Example 2 Use the DMSToDecimalDegrees spreadsheet file to express 135879390 in decimal degrees.

Solution In Cell B3, enter the number of degrees, 135, in Cell B4 the number of minutes, 7, and in Cell B5 the number of seconds, 39. Press RETURN , and the result 135.1275 is shown in Figure 50-8.

fiGure 50-8

exPressinG decimal deGrees as deGrees, minutes, and seconds The measure of an angle given in the form of decimal degrees, such as 47.19388, must often be expressed as degrees, minutes, and seconds.

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c Procedure ●● ●●

●●

Fundamentals oF Plane Geometry

To express decimal degrees as degrees, minutes, and seconds

Multiply the decimal part of the degrees by 609 in order to obtain minutes. If the number of minutes obtained is not a whole number, multiply the decimal part of the minutes by 600 in order to obtain seconds. Round to the nearer whole second if necessary. Combine degrees, minutes, and seconds.

Example Express 47.19388 as degrees, minutes, and seconds. Multiply 0.1938 by 609 to obtain minutes.

609 3 0.1938 5 11.62809

Multiply 0.6280 by 600 to obtain seconds.

600 3 0.6280 5 37.680, 380

Round to the nearer whole second. Combine degrees, minutes, and seconds.

478 1 119 1 380 5 478119380 Ans

There are two basic procedures used in converting decimal degrees to degrees, minutes, and seconds. Depending on the make and model of your calculator, one of the two procedures should apply.

c Procedure 1

Enter decimal degrees, press 5 , press

SHIFT

, press

.

Example Convert 23.30758 to degrees, minutes, and seconds. 23.3075 5

23818827, 238189270

SHIFT

Ans

Note: Minutes are not displayed on the calculator with the symbol 9 and seconds are not displayed with the symbol 0. c Procedure 2 The scroll or cursor key, , is used with the degrees, minutes, seconds key, 8 9 0 , in converting to degrees, minutes, and seconds. Enter decimal degrees, press 8 9 0 , press

, press

ENTER

5

, press

ENTER

5

.

Example Convert 23.30758 to degrees, minutes, and seconds. 23.3075 8 9 0

ENTER

5

ENTER

5

We are going to modify the DMSToDecimalDegrees spreadsheet file so that it will directly convert decimal degrees to degrees, minutes, and seconds. In Cell A10 enter “Convert decimal degrees to degrees, minutes, seconds. In Cell A12, enter “Decimal degrees,” and in Cells B14, C14, and D14 enter “Degrees,”“Minutes,” and “Seconds,” respectively. Finally, in Cell A15, we enter “Degrees and minutes,” and in Cell A16, we type “Degrees, minutes, seconds.” Before we begin, we will need a special Excel function. The INT function rounds a positive number down to the nearest whole number. For example, INT(7.83) 5 7. For the degrees part, enter “5 INT(B12)” in both Cell B15 and Cell B16. For the minutes, enter “5 INT((B12 − B15) * 60)” in Cell C15 and “5 INT((B12 − B16) * 60)” in Cell C16. For the seconds part, enter “5 (B12 − B16 − C16/60) * 3600” in Cell D16. You may want to display the number of seconds to 2 decimal points. Save the file. Now let’s try this new DMSToDecimalDegrees spreadsheet file on the last two examples.

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Example 1 Use the DMSToDecimalDegrees spreadsheet file to convert 72.658 to degrees and minutes. Solution Enter 72.65 in Cell B12 and press

. Since we just want degrees and minutes, we look in Cells B15 and C15 of Figure 50-9 and see that 72.658 5 728 399. The 2.038E?11 in Cell C16 is just one of the ways that Excel indicates 0 seconds. RETURN

FiGure 50-9

Example 2 Use the DMSToDecimalDegrees spreadsheet file to express 63.74328 as degrees, minutes, and seconds.

Solution In Cell B12, enter 63.7432, and press

. Since we want degrees, minutes, and seconds, we look in Cells B16, C16, and D16 of Figure 50-10 and see that 63.74328 5 63844935.52,0 which, to the nearest whole second, rounds off to 638449360. RETURN

FiGure 50-10

arithmetic oPerations on anGular measure in deGrees, minutes, and seconds The division of degrees into minutes and seconds permits very precise computations and measurements. In machining operations, dimensions at times are computed to seconds in order to ensure the proper functioning of parts. When computing with degrees, minutes, and seconds, it is sometimes necessary to exchange units. When exchanging units, keep in mind that 1 degree equals 60 minutes and 1 minute equals 60 seconds. These examples illustrate adding, subtracting, multiplying, and dividing angles in degrees, minutes, and seconds. Arithmetic operations with degrees, minutes, and seconds computed with a calculator may require entering an angle as degrees, minutes, and seconds, converting to decimal degrees, and converting back to degrees, minutes, and seconds. Calculator and spreadsheet applications for each of the arithmetic operations are shown.

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addinG anGles exPressed in deGrees, minutes, and seconds Example 1 Determine the size of /1 shown in Figure 50-11. /1 5 158189 1 638379 158189

158189 1 638379 788559 Ans

638379

FiGure 50-11

Example 2 Determine the size of /2 shown in Figure 50-12. /2 5 438379 1 828549 438379 1 828549 1258919 5 1268319 Ans

828549

438379

FiGure 50-12

Note: Express 919 as degrees and minutes. 919 5 609 1 319 5 18319. Add 1258 1 18319 5 1268319 Ans Example 3 Determine /3 shown in Figure 50-13. /3 5 788439270 1 298389520 788439270

298389520

1 298389520 1078819790 5 1078829190 5 1088229190 Ans

788439270

FiGure 50-13

Note: 790 5 600 1 190 5 19190 therefore, 1078819790 5 1078829190. 829 5 609 1 229 5 18229; therefore, 1078829190 5 1088229190.

Example 4 Determine /3 shown in Figure 50-13. /3 5 788439270 1 298389520 78 8 9 0 43 8 9 0 27 8 9 0 1 29 8 9 0 38 8 9 0 52 8 9 0 5 108822819, 1088229190 Ans

subtractinG anGles exPressed in deGrees, minutes, and seconds Example 1 Determine /1 shown in Figure 50-14.

1238479320

/1 5 1238479320 2 868139070 1238479320 2 868139070 378349250 Ans

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868139070

FiGure 50-14

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Example 2 Determine /2 shown in Figure 50-15.

978129

/2 5 978129 2 458269

458269

978129 5 968729 2 458269 5 458269 518469 Ans

FiGure 50-15

Since 269 cannot be subtracted from 129, 18 is exchanged for 609. 978129 5 968 1 18 1 129 5 968 1 609 1 129 5 968729

Example 3 Determine /3 shown in Figure 50-16. 578139280

/3 5 578139280 2 448199420 578139280 5 568739280 5

568729880

448199420

2 448199420 5 448199420 5 2 448199420 128539460 Ans

FiGure 50-16

Note: Since 199 cannot be subtracted from 139, and 420 cannot be subtracted from 280, 18 is exchanged for 609 and 19 is exchanged for 600. 578139280 5 568 1 18 1 139 1 280 5 568739280 5 568729 1 19 1 280 5 568729880 Example 4 Determine /3 shown in Figure 50-16. /3 5 578139280 2 448199420 57 8 9 0 13 8 9 0 28 8 9 0 2 44 8 9 0 19 8 9 0 42 8 9 0 5 12853846, 128539460 Ans

multiPlyinG anGles exPressed in deGrees, minutes, and seconds Example 1 Five holes are drilled on a circle as shown in Figure 50-17. The angular measure between two consecutive holes is 328189. Determine the angular measure, /1, between hole 1 and hole 5. /1 5 4s328189d 328189

+

+

328189

328189 328189

+

34 1288729 5 1298129 Ans

+

HOLE 1

Note: 729 5 18 129

328189 + HOLE 5

FiGure 50-17

Example 2 Determine the size of /2 shown in Figure 50-18 when x 5 418279420. /2 5 5x 5 5s418279420d 418 279 420 35 205813592100 5 20581389300 5 2078189300 Ans

Note: 2100 5 39300 and 1389 5 28189.

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x x

x x

x 2

FiGure 50-18

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Example 3 Determine /2 in Figure 50-18 when x 5 418279420. /2 5 5x 5 5s418279420d 5 3 41 8 9 0 27 8 9 0 42 8 9 0 5 207818830, 2078189300 Ans

dividinG anGles exPressed in deGrees, minutes, and seconds Example 1 /1 and /2 are the same size. Determine the size of /1 and /2 shown in Figure 50-19. 1048589

/1 5 /2 5 1048589 4 2 528299 Ans 2q 1048589

FiGure 50-19

Example 2 Determine the size of /1, /2, and /3 shown in Figure 50-20 if each is the same size. /1 5 /2 5 /3 5 1288379210 4 3

1288 4 3 5 428 plus a remainder of 28.

428 3q 1288 126 28

Add the 28 s1209d to the 379. 1209 1 379 5 1579 Divide 1579 by 3. 1579 4 3 5 529 plus a remainder of 19.

529 3q 1579 156 19

Add 19 s600d to the 210. 600 1 210 5 810 Divide 810 by 3. 810 4 3 5 270 Combine.

270 3q 810

Divide 1288 by 3.

1288379210

FiGure 50-20

428 1 529 1 270 5 428529270 Ans

Example 3 Determine /1, /2, and /3 in Figure 50-20. /1 5 /2 5 /3 5 1288379210 4 3 128 8 9 0 37 8 9 0 21 8 9 0 4 3 5 42852827, 428529270 Ans

ApplicAtion tooling Up 1. Refer to the Decimal-Inch Spur Gears Table under the heading “Gearing-Diametral Pitch System” in Unit 48 to determine the pitch diameter of an 8-pitch gear that has 45 teeth. 1000C 1000C 2. If the cutting speed is 160 m/min and the workpiece or tool diameter is 45.27 mm, use N 5 or N 5 3.1416D D to determine the revolutions per minute to the nearest whole number.

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3. Solve the equation 28 2 A2 5 (5 1 A)(2 2 A) for A.

5

4. Use the addition of equality to solve p 2 2.0053 5 4.7907 22a2b 1 4c 5. If a 5 25.2, b 5 4.8, and c 5 7.25, what is the value of ? If necessary, round abc 2 9a your answer to 3 decimal places.

1

10 5

FiGure 50-21

6. Read the setting in Figure 50-21 of the metric micrometer scale graduated in 0.01 mm.

Definitions and terms 7. Refer to Figure 50-22 and identify each of the following as parallel, perpendicular, or oblique lines. a. Line AB and line CD b. Line AB and EF c. Line CD and GH

H 32° C

8. a. How many degrees are in a circle? b. How many minutes are in 1 degree? c. How many seconds are in 1 minute? d. How many seconds are in 1 degree? e. How many minutes are in a circle?

y A

y F

B 90°

E

FiGure 50-22

9. Write the symbols for the following words. a. parallel b. perpendicular

D

G

d. minute e. second

c. degree

expressing Decimal Degrees as Degrees and Minutes Express the following decimal degrees as degrees and minutes. When necessary, round the answer to the nearest whole minute. 10. 13.508

15. 93.158

11. 67.858

16. 81.088

12. 48.108

17. 6.478

13. 117.708

18. 125.918

14. 18.608

19. 77.678

expressing Decimal Degrees as Degrees, Minutes, and Seconds Express the following decimal degrees as degrees, minutes, and seconds. When necessary, round the answer to the nearest whole second. 20. 52.13808

25. 103.00908

21. 212.07108

26. 37.93658

22. 7.92508

27. 89.90568

23. 44.44408

28. 182.06928

24. 73.93308

29. 19.89738

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expressing Degrees and Minutes as Decimal Degrees Express the following degrees and minutes as decimal degrees. Round the answer to 2 decimal places. 30. 228409

35. 568489

31. 1078459

36. 878379

32. 68109

37. 28199

33. 878169

38. 328089

34. 1228079

39. 798599

expressing Degrees, Minutes, and Seconds as Decimal Degrees Express the following degrees, minutes, and seconds as decimal degrees. Round the answer to 4 decimal places. 40. 288189300

44. 1768279180

41. 578089450

45. 28079130

42. 1308509100

46. 198499590

43. 988209250

47. 618129060

Adding Angles expressed in Degrees, Minutes, and Seconds 48. Determine /1.

51. Determine /1 1 /2 1 /3. 1

1 = 818519

68°

2 = 738479

31° 3 = 888309

29°

49. Determine /2.

52. Determine /5.

158189270 178349 598189 438329060

53. Determine /6.

50. Determine /3. 198339

258029

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718129

98299530

378129190

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54. Determine /7 1 /8 1 /9.

55. Determine /1 1 /2 1 /3 1 /4 1 /5.

7 = 238329110 2 = 888139480 3 = 518379

4 = 2228199540 9 = 308159410

1 = 908

5 = 878499180

8 = 558509470

Subtracting Angles expressed in Degrees, Minutes, and Seconds Subtract the angles in each of the following exercises. 56. 1148 2 898

59. 1228369170 2 138159080

57. 928359 2 768269

60. 498349120 2 198139420

58. 638239 2 328589 61. Determine /1.

63. Determine /2. 121° 488139420

1 77°

268149070

62. Determine /3 2 /2.

64. Determine /1 2 /2. 3 = 848129

1 = 378569110

2 = 148559470

2 = 518179

65. In the figure shown, /6 5 7208 2 s/1 1 /2 1 /3 1 /4 1 /5d. Determine /6.

Note: 7208 5 7198599600 2 = 908

3 = 1188179350 4 = 2418429260

1 = 908

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5 = 828569

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Section 5

Multiplying Angles expressed in Degrees, Minutes, and Seconds Multiply the angles in each of the following exercises. 66. 7(158)

69. 5(228109130)

67. 3(298199)

70. 8(438239280)

68. 2(438439) 71. In the figure shown, /1 5 /2 5 428. Determine /3.

73. In the figure shown, /1 5 /2 5 /3 5 /4 5 /5 5 548039. Determine /6.

2

6 3

1

4

2

3

5

1

72. If x 5 398149, find /4.

x x

x 4

Dividing Angles expressed in Degrees, Minutes, and Seconds Divide the angles in each of the following exercises. 74. 948 4 2

79. Determine y.

75. 878 4 2 76. 1058209 4 4

y

77. If /1 5 /2, find /1.

y

y

1

y y

y

2

y

137°

80. If /1 5 /2 5 /3 5 /4, find /1.

78. Determine x. 109° x x

49°

x x

4

1

2

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81. The sum of the angles in the figure shown equals 14408. If /1 5 /2 5 /3 5 /4 5 /5 5 /6 and /7 5 /8 5 /9 5 /10 5 1188149230, find /1. 8

7 1

4 5

2 6

3 9

10

UNIT 51 Protractors—Simple Semicircular and Vernier

ObjECTIVES After studying this unit you should be able to ●● ●● ●● ●●

Lay out (construct) angles with a simple protractor. Measure angles with a simple protractor. Read settings on a vernier bevel protractor. Compute complements and supplements of angles.

Protractors are used for measuring, drawing, and laying out angles. Various types of protractors are available, such as the simple semicircular protractor, swinging blade protractor, and bevel protractor. The type of protractor used depends on its application and the degree of precision required. Protractors have wide occupational use, particularly in the metal and woodworking trades.

simPle semicircular Protractor A simple semicircular protractor like the one in Figure 51-1 has two scales, each graduated from 08 to 1808 so that it can be read from either the left or right side. The vertex of the angle to be measured or drawn is located at the center of the base of the protractor.

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80 100

100 80

90

110 70

12 0 60

13 0 50

4 14 0 0

0 14 40

50 0 13

70 110

60 0 12

30 15 0

0 15 30

20 160

160 20

10 170

170 10

SCALE USED FOR READINGS FROM THE LEFT

SCALE USED FOR READINGS FROM THE RIGHT

LOCATION OF VERTEX ANGLE

PROTRACTOR BASE

FiGure 51-1

c Procedure ●● ●● ●●

●●

●●

To lay out (construct) a given angle

Draw a baseline. On the baseline mark a point as the vertex. Place the protractor base on the baseline with the center of the base of the protractor on the vertex. If the angle rotates from the right, choose the scale that has a zero degree reading on the right side of the protractor. If the angle rotates from the left, choose the scale that has a zero degree reading on the left side of the protractor. At the scale reading for the angle being drawn, mark a point. Remove the protractor and connect the two points.

Example Lay out an angle of 1058. Draw baseline AB. On AB mark point O as the vertex. Place the protractor base on AB with the center of the base of the protractor on point O as shown in Figure 51-2. The angle is rotated from the right. The inside scale has a zero degree reading on the right side of the protractor. Use the inside scale and mark a point (point P) at the scale reading of 1058. Remove the protractor and connect points P and O. POINT P

50

80 100

90

100 80

110 70

12 0 60

4 14 0 0 30 15 0 20 160

160 20

105°

0 15 30

10 170

170 10

A

13 0 50 0 14 40

0 13

60 0 12

70 110

B

O CENTER OF PROTRACTOR VERTEX OF ANGLE

BASELINE

FiGure 51-2

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Unit 51

c Procedure ●●

●●

365

Protractors—simPle semicircular and Vernier

To measure a given angle

Place the protractor base on one side of the angle with the center of the base of the protractor on the angle vertex. If the angle rotates from the right, choose the scale that has the zero degree reading on the right side of the protractor. If the angle rotates from the left, choose the scale that has the zero degree reading on the left side of the protractor. Read the measurement where the side crosses the protractor scale.

Example 1 Measure /AOB 5 /1 in Figure 51-3a.

20 0 16

30 150

50 130

40 140

60 120

70 110

80 100

90 0 10 0 8

1 17 0 0

Extend the sides OA and OB of /1 as shown by the dashed lines in Figure 51-3b.

0 11 70 120 60

SCALE READIN

G

130 50

O

150 30

O

140 40

A

VERTE X OF A NG CENTE R OF P LE ROTRA CTO

A

R

160 20

1

1

170 10

B

B

FiGure 51-3a

FiGure 51-3b

Place the protractor base on side OB with the protractor center on the angle vertex, point O. Angle 1 is rotated from the right. The angle measurement is read from the inside scale since the inside scale has a zero degree (08) reading on the right side of the protractor base. Read the measurement where the extension of side OA crosses the protractor scale. Angle 1 5 408 Ans

Example 2 Measure /DOF = /2 as shown in Figure 51-4a. D 10 170

VERTEX OF ANGLE CENTER OF PROTRACTOR

O

20 160 30 150

D 170 10

40 140

2 0 16 20

6 12 0 0

80 100

90

100 80

110 70

120 60

30 130 50

14 0 40

15 0

F

50 13 0

2

F 70 0 11

O

SCALE READING

FiGure 51-4a

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Fundamentals oF Plane Geometry

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Extend the sides OD and OF of /2 as shown in Figure 51-4b. The protractor is positioned upside down. Place the protractor base on the side OD with the center of the protractor base on the angle vertex point O. Angle 2 is rotated from the right. The angle measurement is read from the outside scale since the outside scale has a zero degree (08) reading on the right side of the protractor base. Read the measurement where the extension of side OF crosses the protractor scale. Angle 2 5 1258 Ans

bevel Protractor with vernier scale The bevel protractor is the most widely used vernier protractor in the machine shop. A vernier bevel protractor is shown in Figure 51-5. DIAL OR MAIN SCALE 50

60

70

90

80

60

30

VERNIER SCALE

40

20

60

70

50

30

40

80

20

10

60

0

30

30

0

20

40

10

30

0

20

10

10

BLADE CLAMP

50

30 40

50

60

70

80

90

80

70

60

ADJUSTABLE BLADE

BASE

FiGure 51-5

A vernier bevel protractor consists of a fixed dial or main scale. The main scale is divided into four sections, each from 08 to 908. The vernier scale rotates within the main scale. A blade that can be adjusted to required positions is rotated to a desired angle. 1 The vernier scale permits accurate readings to degree or 5 minutes. The vernier scale 12 is divided into 24 units, with 12 units on each side of zero. The divisions on the vernier scale are in minutes. Each division is equal to 5 minutes. The left vernier scale is used when the vernier zero is to the left of the dial zero. The right vernier scale is used when the vernier zero is to the right of the dial zero.

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Unit 51

Example Read the setting on the vernier protractor shown in Figure 51-6. DIAL (MAIN SCALE) SMALLEST DIVISION = 18

10

20

30

0

40

50

10

60

20

DIAL READING (308)

30

0

30

30

60

VERNIER READING (359)

40

VERNIER SCALE SMALLEST DIVISION = 59

50

NOTE: THE VERNIER SCALE ZERO IS TO THE RIGHT OF THE DIAL ZERO

FiGure 51-6

The zero mark on the vernier scale is just to the right of the 308 division of the dial scale. The vernier zero is to the right of the dial zero; therefore, the right vernier scale is read. The 359 vernier graduation coincides with a dial graduation. The protractor reading is 308359. Ans

comPlements and suPPlements of scale readinGs When using the bevel protractor, the machinist must determine whether the desired angle of the part being measured is the actual reading on the protractor or the complement or the supplement of the protractor reading. Particular caution must be taken when measuring angles close to 458 and 908. Two angles are complementary when their sum is 908. For example, in Figure 51-7 438 1 478 5 908. Therefore, 438 is the complement of 478 and 478 is the complement of 438. Two angles are supplementary when their sum is 1808. For example, in Figure 51-8 928 1 888 5 1808. Therefore, 928 is the supplement of 888 and 888 is the supplement of 928.

180°

90° 43° 47°

88°

92°

COMPLEMENTARY ANGLES

SUPPLEMENTARY ANGLES

FiGure 51-7

FiGure 51-8

ApplicAtion tooling Up 1. Express 159.35968 as degrees, minutes, and seconds. If necessary, round the answer to the nearest whole second. 2. Refer to the Decimal-Inch Spur Gears Table under the heading “Gearing-Diametral Pitch System” in Unit 48 to determine the pitch of a spur gear with an addendum of 0.25000 and an outside diameter of 5.6000.

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3. A combination drilling and countersinking operation on bronze round stock is performed on an automatic screw 3 1 machine. The length of cut per piece is 2 inches. The total cutting time for 3200 pieces is 9 hours running at 8 4 1800 rpm. What is the tool feed in inches per revolution? Round the answer to 3 decimal places. p2 4. Solve 5 1210. 0.1 5. Subtract 3x2 2 5xy 1 9y3 from 9x2 1 4x2y 2 6y3. 6. Use a digital micrometer to measure the thickness in millimeters of the textbook for this course.

Simple Protractor 7. Write the values of angles A–E on the protractor scale shown. E D

A

100 80 70 100 90 80 7110 1 0 0 2 1 0 1 6 0 60 0 1 12 5030 5030 1

170 160 10 0 20 15 0 30 1440

B

10 170 120 60 30 15 0 40 14 0

C

A5 B5 C5 D5 E5

8. Write the values of angles F–J on the preceding protractor. J

100 80 70 100 90 80 7110 1 0 0 2 1 0 6 0 1 60 0 1 12 0 5030 5 30 1

170 160 10 0 20 15 0 30 1440

10 170 120 60 30 15 0 40 14 0

I

H

F5 G5 G F

H5 I5 J5

9. Using a protractor, lay out the angles in Exercises 9 and 10. a. 198 b. 658

c. 808 d. 128

e. 48

10. a. 988

c. 1238 d. 1508

e. 1668

b. 978

11. Lay out a three-sided closed figure (triangle) of any size containing angles of 478 and 1058. Measure the third angle. How many degrees are contained in the third angle? 12. Lay out a four-sided figure (quadrilateral) of any size containing angles of 898, 698, and 1248. Measure the fourth angle. How many degrees are contained in the fourth angle?

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Protractors—simPle semicircular and Vernier

Unit 51

In Exercises 13 through 16, measure each of the angles, /12/19, to the nearer degree. Extend the sides of the angles if necessary. 13.

/1 = 

17.

1

10

/2 = 

11

14.

/3 =  /4 = 

12

2

/5 =  13

9

/6 = 

14

/7 =  15.

/8 = 

18. 3

/9 = 

4 18

/11 = 

15

5

/10 =  /12 = 

19

/13 = 

17

16

16.

/14 =  /15 = 

8

/16 =  /17 = 

7

6

/18 =  /19 = 

Vernier Protractor Write the values of the settings on the following vernier protractor scales. 19.

22. 10

60

0

10

30

0

20

30

30

60

70

60

60

20.

30

50

0

40

30

30

60

23. 30

40

60

30

20

0

10

30

0

60

21.

70

80

60

60

30

0

10

20

30

0

50

30

40

60

24. 10

60

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20

30

30

0

40

30

50

60

0

60

30

30

40

60

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Section 5

25.

Fundamentals oF Plane Geometry

27. 30

60

20

10

30

0

10

20

0

30

10

40

30

60

60

30

50

0

60

30

70

60

26. 0

60

30

0

30

30

40

60

complementary and Supplementary Angles Write the complements of the angles in Exercises 28 and 29. In Exercises 30 and 31, write the supplements of the following angles. 28. a. 438 b. 768 c. 178 d. 58 e. 678499 f. 458199

30. a. 138 b. 658 c. 918 d. 1798599 e. 08499 f. 898599

29. a. 798 b. 278 c. 148379 d. 218439 e. 788199270 f. 59809590

31. a. 1268 b. 428 c. 958279 d. 28439200 e. 688219290 f. 1338329080

UNIT 52 Types of Angles and Angular Geometric Principles

ObjECTIVES After studying this unit you should be able to ●● ●●

Identify different types of angles. Determine unknown angles in geometric figures using the principles of parallel lines and opposite, alternate interior, corresponding, and perpendicular angles.

Solving a practical application may require working with a number of different angles. To avoid confusion, angles must be properly named and their types identified. Determination of required unknown angular and linear dimensions is often based on the knowledge and understanding of angular geometric principles and their practical applications.

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Unit 52

naminG anGles Angles are named by a number, a letter, or three letters. When an angle is named with three letters, the vertex must be the middle letter. For example, the angle shown in Figure 52-1 can be called /1, /C, /ACB, or /BCA. In cases where a point is the vertex of more than one angle, a single letter cannot be used to name an angle. For example, in Figure 52-2, point E is the vertex of three different angles; the single letter E cannot be used in naming the angle. H

/1 is called /GEH or /HEG. /2 is called /FEG or /GEF. /3 is called /FEH or /HEF

B 1 C

A

FiGure 52-1

3 G 1 2 F

E

FiGure 52-2

tyPes of anGles An acute angle is an angle that is less than 908. Angle 1 in Figure 52-3 is acute. A right angle is an angle of 908. Angle A in Figure 52-4 is a right angle. Right angles are often marked with a small square. An obtuse angle is an angle greater than 908 but less than 1808. Angle ABC in Figure 52-5 is an obtuse angle. 180°

A 90° 90°

90° 1 A

FiGure 52-3

B

FiGure 52-4

C

FiGure 52-5

A straight angle is an angle of 1808. A straight line is a straight angle. Line EFG in Figure 52-6 is a straight angle. A reflex angle is an angle greater than 1808 and less than 3608. Angle 3 in Figure 52-7 is a reflex angle. 180°

180° 3

E

F

FiGure 52-6

G

FiGure 52-7

adjacent angles Two angles are adjacent if they have a common side and a common vertex. Angle 1 and angle 2 shown in Figure 52-8 are adjacent since they both contain the common side BC and

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Fundamentals oF Plane Geometry

Section 5

the common vertex B. Angle 4 and angle 5 shown in Figure 52-9 are not adjacent. The angles do not have a common vertex. A

1

C

5 2

4

D

B

FiGure 52-8

FiGure 52-9

angles formed by a transversal A transversal is a line that intersects (cuts) two or more lines. Line EF in Figure 52-10 is a transversal since it cuts lines AB and CD. Alternate interior angles are pairs of interior angles on opposite sides of the transversal. The angles have different vertices. For example, in Figure 52-10, angles 3 and 5 are alternate interior angles. Angles 4 and 6 are also alternate interior angles. Corresponding angles are pairs of angles, one interior and one exterior with both angles on the same side of the transversal. The angles have different vertices. For example, in Figure 52-10, angles 1 and 5, 2 and 6, 3 and 7, and 4 and 8 are pairs of corresponding angles. E

A

1

2

4

3

B

5

C

6 8

7

D

F

FiGure 52-10

Geometric PrinciPles In this book, geometric postulates, theorems, and corollaries are grouped together and are called geometric principles. Geometric principles are statements of truth that are used as geometric rules. The principles will not be proved, but they will be used as the basis for problem solving.

Note: Angles that are referred to or shown as equal are angles of equal measure (m). For example, /A 5 /B means m/A 5 m/B. Also, line segments that are referred to or shown as equal are segments of equal length. For example, AB 5 CD means length AB 5 length CD.

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Unit 52

373

tyPes oF anGles and anGular Geometric PrinciPles

■ Principle 1 If two lines intersect, the opposite or vertical angles are equal. Conclusion: /1 5 /3 and /2 5 /4.

B

4

C

Given: AB intersects CD in Figure 52-11.

3

1

D

2

A

FiGure 52-11

■ Principle 2 If two parallel lines are intersected by a transversal, the alternate interior angles are equal. E B

Given: AB i CD as shown in Figure 52-12. Conclusion: /3 5 /5 and /4 5 /6.

A

3

4

6

5

D

C F

FiGure 52-12

● If two lines are intersected by a transversal and a pair of alternate interior angles are equal, the lines are parallel.

B

Given: /1 5 /2 as in Figure 52-13. Conclusion: AB i CD.

D 1

E

F

2 A C

FiGure 52-13

■ Principle 3 If two parallel lines are intersected by a transversal, the corresponding angles are equal. Conclusion: /1 5 /5, /2 5 /6,

B

2

Given: In Figure 52-14, AB i CD.

3

E

D

6

7

/3 5 /7, and /4 5 /8.

1 A

F

5

4 C

8

FiGure 52-14

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Section 5

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● If two lines are intersected by a transversal and a pair of corresponding angles are equal, the lines are parallel. Given: In Figure 52-15, /1 5 /2. Conclusion: AB i CD.

C

A 2

1

F

E D

B

FiGure 52-15

■ Principle 4 Two angles are either equal or supplementary if their corresponding sides are parallel. Given: AB i FG and BC i DE as in Figure 52-16. Conclusion: /1 5 /3 and /1 and /2 are A

supplementary (/1 1 /2 5 1808).

F

2

1 B

C

D

3 E

G

FiGure 52-16

■ Principle 5 Two angles are either equal or supplementary if their corresponding sides are perpendicular. E

Given: In Figure 52-17, AB ' DH and BC ' EF.

A

(/1 1 /3 5 1808).

2

D

Conclusion: /1 5 /2; /1 and /3 are supplementary 1 B

C

H 3 F

FiGure 52-17

The following example illustrates the method of solving angular measure problems. Values of angles are determined by applying angular geometric principles and the fact that a straight angle (straight line) contains 1808.

Example Given: In Figure 52-18, AB uu CD, EF uu GH, /1 5 1158, and /2 5 828. Determine the values of /3 through /9.

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Unit 52

375

tyPes oF anGles and anGular Geometric PrinciPles

E 1 = 115°

2 = 82°

A

B 5

4 C

3

7

G 9

8

F D

6 H

FiGure 52-18

Solve for /3. Apply Principle 1. If two lines intersect, the opposite or vertical angles are equal. /3 5 /1 5 1158 Ans Solve for /4. Apply either Principle 2 or Principle 3. Applying Principle 2: If two parallel lines are intersected by a transversal, the alternate interior angles are equal. /4 5 /3 5 1158 Ans or Applying Principle 3: If two parallel lines are intersected by a transversal, the corresponding angles are equal. /4 5 /1 5 1158 Ans Solve for /5. Since a straight angle (straight line) contains 1808, /5 and /1 are supplementary. /5 5 1808 2 /1 5 1808 2 1158 5 658 Ans Solve for /6. Apply Principle 3. If two parallel lines are intersected by a transversal, the corresponding angles are equal. /6 5 /5 5 658 Ans Solve for /7. Apply Principle 1. If two lines intersect, the opposite or vertical angles are equal. /7 5 /6 5 658 Ans Solve for /8. Apply Principle 4. Two angles are either equal or supplementary if their corresponding sides are parallel. /8 5 /2 5 828 Ans Solve for /9. Since a straight angle (straight line) contains 1808, /8 and /9 are supplementary. /9 5 1808 2 /8 5 1808 2 828 5 988 Ans

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Section 5

Fundamentals oF Plane Geometry

ApplicAtion tooling Up 1. What is the complement of a 278419190 angle? 2. Express 728129520 as decimal degrees. Round the answer to 4 decimal places. 3. Refer to the Metric Spur Gears Table under the “Gearing-Metric Module System” heading in Unit 48 to determine the circular pitch of a 14-millimeter module gear. 4.25F 2 21.75 15.4 2 5.45F 4. Solve for F in the proportion 5 . 42.36 21.16 4x6y2 5. Simplify 4 (xy). 25

Î

6. Use the Table of Block Thicknesses of a Metric Gage Block Set under the “Description of Gage Blocks” heading in Unit 37 to determine a combination of gage blocks for 47.537 mm.

naming Angles

E

7. Name each of the following angles in three additional ways. a. /1 b. /2 c. /C d. /D e. /E f. /F

5

F 6

4 C

D 3

1

2 B

A

8. Name each of the following angles in two additional ways. a. /1 b. /CBF c. /3 d. /ECB e. /5 f. /BCD

3 D

C 1 6

F

2

E

5 A

B

4

types of Angles 9. Identify each of the following angles as acute, obtuse, right, straight, or reflex. a. /BAF g. /ABC C b. /ABF h. /BCD c. /CBF i. /AED d. /DCA j. /1 e. /BFA k. /DFA f. /BFD

B

A

90°

90° 180°

90° F D

90° E

1

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Unit 52

377

tyPes oF anGles and anGular Geometric PrinciPles

10. Name all pairs of adjacent angles shown in the figure. 4 3

2

5 1 11

8

10

6

9

7

11. Alternate interior angles and corresponding angles are shown in the figure. a. Name all pairs of alternate interior angles. b. Name all pairs of corresponding angles.

1 4

6 7

2 3

5 8

Applications of Geometric Principles

388 638149

Solve the following exercises: 12. Determine the values of /1 through /5. /1 5 /2 5

788469

/3 5 /4 5 /5 5

978 1618469 5

13. Determine the values of /2, /3, and /4 for these given values of /1. a. /1 5 328 /2 5

2

1

/3 5

4

b. /1 5 358199 /2 5

3

/4 5

/4 5

/3 5

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Fundamentals oF Plane Geometry

Section 5

14. Given: AB i CD. Determine the values of /2 through /8 for these given values of /1. a. /1 5 688 /2 5 /3 5 /4 5 /5 5 b. /1 5 528559

2

/6 5 /7 5 /8 5

3

A

B 1

4

6

/2 5 /3 5 /4 5 /5 5

5

C

/6 5 /7 5 /8 5

D 7

8

15. Given: Hole centerlines EF i GH and MP i KL. Determine the values of /1 through /15 for these values of /16. a. /16 5 718 /1 5 /2 5 /3 5 /4 5 /5 5 b. /16 5 868529 /1 5 /2 5 /3 5 /4 5 /5 5

/6 /7 /8 /9 /10

5 5 5 5 5

/11 /12 /13 /14 /15

5 5 5 5 5

/6 /7 /8 /9 /10

5 5 5 5 5

/11 /12 /13 /14 /15

5 5 5 5 5 F 2

G

K

3

5 6

4 1

8

15 M

16

9 10

14 13 E

L 7

12

P 11

H

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Unit 52

379

tyPes oF anGles and anGular Geometric PrinciPles

16. Given: Hole centerlines AB i CD and EF i GH. Determine the values of /1 through /22 for these given values of /23, /24, and /25. a. /23 5 978, /24 5 348, and /25 5 1028 b. /23 5 1128239, /24 5 278539, and /25 5 958189 G F 15

6

5

13

24

B

A 7

23

12

14 17

1

4

16

11

22 25

18

8

19 D

C 3

2

9

10

E

20

21

H

a. /1 5

/7 5

/13 5

/19 5

/2 5

/8 5

/14 5

/20 5

/3 5

/9 5

/15 5

/21 5

/4 5

/10 5

/16 5

/22 5

/5 5

/11 5

/17 5

/6 5

/12 5

/18 5

b. /1 5

/7 5

/13 5

/19 5

/2 5

/8 5

/14 5

/20 5

/3 5

/9 5

/15 5

/21 5

/4 5

/10 5

/16 5

/22 5

/5 5

/11 5

/17 5

/6 5

/12 5

/18 5

17. Given: AB i CD, AC i ED. Determine the value of /2 and /3 for these values of /1. a. /1 5 678 b. /1 5 748129

/2 5 /2 5

/3 5 /3 5

E

D

B

2

3

1 A

18. Given: FH i GS i KM and FG i HK. Determine the values of /1, /2, and /3 for these values of /4. a. /4 5 1168 H /1 5 /3 5 /2 5 3 b. /4 5 1078439 F 1 /1 5 /3 5 2 /2 5

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C

4 S

M

K

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380

Section 5

Fundamentals oF Plane Geometry

UNIT 53 Introduction to Triangles Objectives After studying this unit you should be able to ●● ●● ●●

Identify different types of triangles. Determine unknown angles based on the principle that all triangles contain 1808. Identify corresponding parts of triangles.

A polygon is a closed plane figure formed by three or more straight line segments. A triangle is a three-sided polygon; it is the simplest kind of polygon. The symbol n means triangle. Triangles are widely applied in engineering and manufacturing. The triangle is a rigid figure that is the basic figure in many designs. Machine technicians and drafters require a knowledge of triangles in laying out work.

Types of Triangles A scalene triangle has three unequal sides. It also has three unequal angles. In Figure 53-1, triangle ABC is scalene. Sides AB, AC, and BC are unequal and angles A, B, and C are unequal.

B

A

C

SCALENE TRIANGLE

FiGure 53-1

An isosceles triangle has two equal sides. The equal sides are called legs. The third side is called the base. An isosceles triangle also has two equal base angles. Base angles are the angles that are opposite the legs. Figure 53-2 shows isosceles triangle RST, with side RT 5 side ST and /R 5 /S.

T LEG

LEG

R

S BASE ANGLES

BASE

ISOSCELES TRIANGLE

FiGure 53-2

An equilateral triangle has three equal sides. It also has three equal angles. In Figure 53-3 is equilateral triangle DEF, with sides DE 5 DF 5 EF and /D 5 /E 5 /F. Because an equilateral triangle has three equal angles, it may also be called an equiangular triangle.

E

D

F

EQUILATERAL TRIANGLE

FiGure 53-3

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introduction to trianGles

A right triangle has a right or 908 angle. The symbol for a right angle is a small square placed at the vertex of the angle. The side opposite the right angle is called the hypotenuse. The other two sides are called legs. Right triangle HJK is shown in Figure 53-4 with /H 5 908 and JK the hypotenuse. J

HYPOTENUSE (SIDE OPPOSITE THE RIGHT ANGLE)

SYMBOL FOR A RIGHT ANGLE

K

H RIGHT TRIANGLE

FiGure 53-4

angles of a Triangle ■ Principle 6 The sum of the angles of any triangle is equal to 1808. This principle is applied in many practical applications.

C 87°559270

Example 1 Angles A, B, and C in Figure 53-5 are hole centerline angles. Angle A 5 488359520 and Angle C 5 878559270. 48°359520

Determine /B. /B 5 1808 2 s488359520 1 878559270d 180 2 ( 48 8 9 0 35 8 9 0 52 8 9 0 1 87 8 9 0 55 8 9 0 27 8 9 0

)

5

SHIFT

43828841, 438289410

B

A

Ans

FiGure 53-5

Example 2 In isosceles triangle EFG shown in Figure 53-6, EF 5 EG and

F

/E 5 338189. Determine /F and /G. /E 1 /F 1 /G 5 1808 1808 2 /E 5 /F 1 /G 1808 2 338189 5 /F 1 /G 1468429 5 /F 1 /G

E

33°189 G

1468429 Since /F 5 /G, then /F and /G each 5 5 738219 2

FiGure 53-6

Ans

Example 3 In Figure 53-7, triangle HJK is equilateral. Determine /H, /J, and /K. /H 1 /J 1 /K 5 1808 1808 Since /H 5 /J 5 /K, each angle 5 5 608 3

H

J

Ans

K

FiGure 53-7

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Fundamentals oF Plane Geometry

Section 5

Corresponding parTs of Triangles It is essential to develop the ability to identify corresponding angles and sides of two or more triangles. Corresponding sides and angles between triangles are not determined by the positions of the triangles. The smallest angle of a triangle lies opposite the shortest side and the largest angle of a triangle lies opposite the longest side. Corresponding angles between two triangles are determined by comparing the lengths of the sides that lie opposite the angles. Corresponding sides between two triangles are determined by comparing the sizes of the angles that lie opposite the sides.

Example 1 In triangle ABC shown in Figure 53-8, determine the longest, next longest, and shortest sides.

C 43°

The longest side is CB since it lies opposite the largest angle, 1078. Ans The next longest side is AB since it lies opposite the next largest angle, 438. Ans

107°

The shortest side is AC since it lies opposite the smallest angle, 308. Ans

30°

A

B

FiGure 53-8

Example 2 In triangle DEF shown in Figure 53-9, determine the largest, next largest, and smallest angle. All dimensions are in inches. 4

7

E

D

F 10

FiGure 53-9

The largest angle is /E since it lies opposite the longest side, 10 inches.

Ans

The next largest angle is /D since it lies opposite the next longest side, 7 inches. The smallest angle is /F since it lies opposite the shortest side, 4 inches.

Ans

Ans

Example 3 In triangles ABC and FED shown in Figure 53-10, determine the pairs of corresponding angles between the two triangles. All dimensions are in millimeters. F C

37.5 98 B

50

70 A

70

E D

52.5

FiGure 53-10

Angle C corresponds to /D since each angle lies opposite the longest side of each triangle. Angle B corresponds to /E since each angle lies opposite the next longest side of each triangle. Angle A corresponds to /F since each angle lies opposite the shortest side of each triangle.

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introduction to trianGles

Unit 53

ApplicAtion tooling Up 1. Determine the values of /2, /3, and /4 if /1 is 1138.

1 2 4 3

2. Use a protractor to measure the angle to the nearest degree. Extend the sides of the angle if necessary.

1

3. Express 191.53268 as degrees, minutes, and seconds. If necessary, round the answer to the nearest whole second. 4. Cast iron 10 cm in diameter is turned in a lathe at 350 rpm. Each length of cut is 65 cm and eight cuts are required. L A carbide tool is fed into the work at 0.0425 cm 5 0.425 mm per revolution. Use the formula T 5 to find the FN total cutting time. Round the answer to 1 decimal place. 5. Solve

4t 2 7t 2 22 1 5 t. 16 2

m2 2 4p 1 Ï3r pr 2 m 1 . Give the answer as both an improper fraction and a mp mr decimal fraction rounded to 2 decimal places.

6. If m 5 5, p 5 22, and r 5 12, find

types of triangles Identify each of the triangles 7 through 14 as scalene, isosceles, equilateral, or right. 7.

8. All dimensions are in inches. 6.8

3.9

9.5 70°

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Section 5

9.

Fundamentals oF Plane Geometry

12. All dimensions are in millimeters.

80° 80 46°

54°

80

10. All dimensions are in millimeters. 13.

120

60° 120

60°

105 60°

11. 14. All dimensions are in inches. 2.7 2.7 2.7

Angles of a triangle Solve the following exercises: 15. Find the value of /A 1 /B 1 /C.

B

C

A

16. Find the value of the unknown angles for these given angle values.

3

1

a. If /1 5 568 and /2 5 868, find /3. b. If /2 5 818 and /3 5 468, find /1.

2

17. Find the value of the unknown angles for these given angle values. a. If /4 5 328439 and /5 5 1198179, find /6. b. If /5 5 1238179130 and /6 5 278, find /4. 18. Find the value of the unknown angles for these given angle values.

4

A

5

6

C

a. If /A 5 198439, find /B. b. If /B 5 678589, find /A. B

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introduction to trianGles

C

19. In triangle ABC, BC 5 17.3 inches. a. Find AB. b. Find AC.

60° 60°

60°

A

20. In triangle EFG, find the value of the unknown angles for these given angle values. All dimensions are in inches.

B

G

13.4

a. If /E 5 818, find /G. b. If /G 5 838279, find /F.

F 13.4

E

21. Find the value of the unknown angles for these given angle values. All dimensions are in millimeters.

150 3

a. If /3 5 178, find /1. b. If /3 5 258199, find /2.

2 1

150

22. All dimensions are in inches. a. Find /3. b. Find /4.

3.4 3.4 4

3

23. Find the value of the unknown angles for these given angle values.

A

1

a. If /1 5 268 and /3 5 488, find /2. b. If /1 5 288 and /2 5 158, find /3.

3

2

33°

24. Hole centerlines AB i CD. a. If /1 5 868329, find /2. b. If /2 5 678479, find /1.

A 23°309 C

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B

1 2

D

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Fundamentals oF Plane Geometry

Section 5

25. Find the value of the unknown angles listed. F

a. /3 b. /4

G

97° D

B 59°

35°

4

C

3

A

E

26. AB i DE, BC is an extension of AB.

E

a. If /E 5 668439, find /A. b. If /A 5 198079, find /E.

D C

B A

F

corresponding Parts of triangles Determine the answers to the following exercises which are based on corresponding parts. 27. All dimensions are in inches.

3.2

C

a. Find the largest angle. b. Find the next largest angle. c. Find the smallest angle.

1.8

B

2.9 A

28. Refer to triangle EFG.

F

a. Find the shortest side. b. Find the next shortest side. c. Find the longest side.

E

46°

22°

112° G

29. Identify the angle that corresponds with each angle listed. All dimensions are in millimeters. 28

a. /A b. /B c. /1

42

D

B 26

2

1

13 A

21

E

56

30. Identify the angle that corresponds with each angle listed. All dimensions are in inches. a. /F b. /G c. /H

0.83

L

G

1.56 0.75

F

0.83 K

0.88 H

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Unit 54

387

Geometric PrinciPles For trianGles and other common PolyGons

UNIT 54 Geometric Principles for Triangles and Other Common Polygons

Objectives After studying this unit you should be able to ●● ●● ●●

Identify similar triangles and compute unknown angles and sides. Compute angles and sides of isosceles, equilateral, and right triangles. Determine the sizes of interior angles of any polygon.

CongruenT Triangles Two triangles are congruent if they are identical in size and shape. If one congruent triangle is placed on top of the other, they fit together exactly. The symbol > means congruent.

Corresponding parts of Congruent Triangles are equal. Example 1 In Figure 54-1, nABC > nDEF. C

F

A

B

D

E

FiGure 54-1

Corresponding parts of congruent triangles are equal. /A 5 /D, /B 5 /E, and /C 5 /F AB 5 DE, AC 5 DF, and BC 5 EF

Example 2 In Figure 54-2, nGHI > nJKL. I

G

L

H

K

J

FiGure 54-2

Notice that in order to get the triangles to fit together exactly, one of them will need to be “flipped over.” Corresponding parts of these two congruent triangles are: /G 5 /J, /H 5 /K, and /I 5 /L GH 5 JK, GI 5 JL, and HI 5 KL

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Section 5

Fundamentals oF Plane Geometry

similar figures Stated in a general way, similar figures are figures that are alike in shape but different in size. For example, a photograph is similar to the object that is photographed. In machine technology, engineering drawings made to scale are similar to the objects they represent. Often, scale drawings are in the form of similar polygons or combinations of similar polygons. Recall that a polygon is a closed plane figure formed by three or more straight line segments. Similar polygons have the same number of sides, equal corresponding angles, and proportional corresponding sides. The symbol ~ means similar. A triangle is a three-sided polygon; it is the simplest kind of polygon.

similar Triangles Two triangles are similar if their corresponding angles are equal; their corresponding sides will also be proportional.

Example 1 Triangles ABC and DEF in Figure 54-3 have equal corresponding angles. D

C

70°

80°

80°

30°

E

70°

30°

F

A

B

FiGure 54-3

Two triangles are similar if their corresponding angles are equal. In Figure 54-3, /A 5 /D 5 808, /B 5 /E 5 308, and /C 5 /F 5 708. nABC , nDEF

Example 2 The lengths of the sides of triangles HJK and LMN in Figure 54-4 are given in inches. HJ JK HK 5 5 LM MN LN 2 4 5 5 5 4 8 10 1 1 1 5 5 2 2 2

2

H 5

J 4 K

10

L 4

N 8

M

FiGure 54-4

The corresponding sides are proportional. nHJK ~ nLMN

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Geometric PrinciPles For trianGles and other common PolyGons

Example 3 In Figure 54-5, nPRS ~ nTWY. All linear dimensions are in millimeters. R W

53.48°

53.48°

18.75

15.00

11.25

36.52° P

20.00

36.52° S

T

Y

FiGure 54-5

a. Determine the length of side PR. b. Determine the length of side TY. Set up proportions and solve for the unknown sides, PR and TY. WY WT WY TY a. 5 b. 5 RS PR RS PS 11.25 mm 18.75 mm 11.25 mm TY 5 5 15.00 mm PR 15.00 mm 20.00 mm s11.25 mmd PR 5 15.00 mm s18.75 mmd s15.00 mmd TY 5 20.00 mms11.25 mmd 15.00 mm s18.75 mmd 20.00 mm s11.25 mmd PR 5 TY 5 11.25 mm 15.00 mm TY 5 15.00 mm Ans PR 5 25.00 mm Ans

■ Principle 7

C

Two triangles are similar if their sides are respectively parallel.

F

Given: AB i DE, AC i DF, and BC i EF in Figure 54-6. Conclusion: nABC ~ nDEF.

E

D

B

A

FiGure 54-6 L

● Two triangles are similar if their sides are respectively perpendicular.

K

Given: HJ ⊥ LM, HK ⊥ LP, and JK ⊥ MP in Figure 54-7. Conclusion: nHJK ~ nLMP.

P H

J M

FiGure 54-7

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Fundamentals oF Plane Geometry

Section 5

● Within a triangle, if a line is drawn parallel to one side,

C

the triangle formed is similar to the original triangle.

E

Given: DE i BC in Figure 54-8. Conclusion: nADE ~ nABC. B D A

FiGure 54-8

● In a right triangle, if a line is drawn from the vertex of

H

the right angle perpendicular to the opposite side, the two triangles formed and the original triangle are similar.

L

Given: In Figure 54-9, we have right △HFG, FL ⊥ HG. Conclusion: nFLH ~ nGLF ~ nGFH.

F

G

FiGure 54-9

isosCeles, equilaTeral, and righT Triangles ■ Principle 8 In an isosceles triangle, an altitude to the base bisects the base and the vertex angle. An altitude is a line drawn from a vertex perpendicular to the opposite side.

C ALTITUDE 2

1 A

B

D

To bisect means to divide into two equal parts.

FiGure 54-10

Given: Isosceles nABC in Figure 54-10 with AC 5 CB and line CD the altitude to base AB. Conclusion: AD 5 BD and /1 5 /2.

● In an equilateral triangle, an altitude to any side bisects the side and the vertex angle.

F

Given: Equilateral nEFG in Figure 54-11 with EH the altitude to FG.

H

3

Conclusion: FH 5 GH and /3 5 /4.

4 E

G

FiGure 54-11

■ Principle 9 (Pythagorean Theorem) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides or legs. If two sides of a right triangle are known, the third side can be calculated. This principle, called the Pythagorean Theorem, is often used for solving machine technology problems.

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391

Geometric PrinciPles For trianGles and other common PolyGons

Example 1 In the right △ABC shown in Figure 54-12, dimensions a and b are given; the centerlines meet at right angles at C. To determine the distance between holes A and B, distance c must be computed. Dimensions are given in inches. Side c is the hypotenuse. Substitute the given values for sides a and b and solve for distance c. c2 5 a2 1 b2 B

c 5 Ïa2 1 b2

c

c 5 Ïs6.027 in.d2 1 s8.139 in.d2 c 5 Ï36.3247 in.2 1 66.2433 in.2

a = 6.027

c 5 Ï102.5680 in.2 c 5 10.128 in.

C

Ans sroundedd

A b = 8.139

FiGure 54-12

c 5 Ïs6.027 in.d2 1 s8.139 in.d2 Ïa ( 6.027 x 2 1 8.139 x 2 10.128 in. Ans (rounded)

)

5 10.12758856,

Example 2 In the right nEFG shown in Figure 54-13, f 5 5.800 inches and hypotenuse g 5 7.200 inches. Determine side e. Side g is the hypotenuse.

F

g2 5 e 2 1 f 2 s7.200 in.d2 5 e2 1 s5.800 in.d2 51.840 sq in. 5 e2 1 33.640 sq in. 18.200 sq in. 5 e2

Substitute the given values, rearrange the equation, and solve for e.

Ï18.200 sq in. 5 e e 5 4.266 in.

Ans sroundedd

g = 7.200 e

E

G f = 5.800

7.2002 5 e2 1 5.8002

FiGure 54-13

Rearrange the equation in terms of e: Ï7.2002 2 5.8002 5 e Ïa

(

7.2

x2

e 5 4.266 in.

2 5.8

x2

)

5 4.266145802,

Ans sroundedd

polygons oTher Than Triangles The types of polygons most common to machine trade applications in addition to triangles are the quadrilaterals of squares, rectangles, and parallelograms. Quadrilaterals are foursided polygons. Regular hexagons also have wide application. A regular polygon is one that has equal sides and equal angles. A square is a regular four-sided polygon (quadrilateral). Each angle equals 908. In the square ABCD shown in Figure 54-14, AB 5 BC 5 CD 5 AD and /A 5 /B 5 /C 5 /D 5 908

D

C

A

B

FiGure 54-14

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Fundamentals oF Plane Geometry

Section 5

A rectangle is a four-sided polygon (quadrilateral) with opposite sides parallel and equal. Each angle equals 908. In the rectangle EFGH shown in Figure 54-15, EF i GH, EH i FG; EF 5 GH, EH 5 FG; /E 5 /F 5 /G 5 /H 5 908.

F

G

E

H

FiGure 54-15

A parallelogram is a four-sided polygon (quadrilateral) with opposite sides parallel and equal. Opposite angles are equal. In the parallelogram ABCD shown in Figure 54-16, AB i CD, AD i BC; AB 5 CD, AD 5 BC; /A 5 /C, /B 5 /D.

B

C

A

D

FiGure 54-16

A rhombus is a four-sided polygon (quadrilateral) with opposite sides parallel and all four sides equal. Opposite angles are equal. In the rhombus EFGH in Figure 51-17, EF i GH, EH i FG; EF 5 FG = GH = HE; /E 5 /G /F = /H.

F

E

G

H

FiGure 54-17 B

A regular hexagon is a six-sided figure with all sides equal and all angles equal. In the regular hexagon ABCDEF shown in Figure 54-18, AB 5 BC 5 CD 5 DE 5 EF 5 AF, and /A 5 /B 5 /C 5 /D 5 /E 5 /F.

C

A

D

F

E

FiGure 54-18

■ Principle 10 The sum of the interior angles of a polygon of N sides is equal to (N 2 2) times 1808. Example 1 In Figure 54-19 is quadrilateral EFGH, with /E 5 728, /F 5 958, /G 5 1088. Determine /H. Since EFGH has four sides, N 5 4. The sum of the four angles 5 s4 2 2d1808 5 2s1808d 5 3608. Add the three given angles and subtract from 3608 to find /H. F

/H 5 3608 2 s/E 1 /F 1 /Gd /H 5 3608 2 s728 1 958 1 1088d /H 5 3608 2 2758 /H 5 858 Ans

G

95° 108° 72° E

H

FiGure 54-19

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393

Geometric PrinciPles For trianGles and other common PolyGons

Example 2 Refer to polygon ABCDEF in Figure 54-20 and determine /1. Since ABCDEF has six sides, N 5 6. The sum of the 6 angles 5 s6 2 2d1808 5 4s1808d 5 7208. Find /2. /2 5 3608 2 114.028 5 245.988 Add the 5 known interior angles and subtract from 7208 to find /1. /1 5 7208 2 s57.658 1 245.988 1 40.188 1 77.268 1 908d /1 5 7208 2 511.078 /1 5 208.938 Ans 720 2 ( 57.65 1 245.98 1 40.18 1 77.26 1 90 /1 5 208.938 Ans

)

5 208.93

C 40.18°

77.26° D

114.02°

2

B

1 57.65°

E F

A

FiGure 54-20

ApplicAtion tooling Up 1. Determine the size of /1.

1

109°

47°

2. If AB i CD, BC i DE, and /1 5 278259, what are the sizes of /2 and /3? B

A 2

1 C

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E

3 D

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Section 5

Fundamentals oF Plane Geometry

3. What is the supplement of a 1058139440 angle? 4. Determine the center diameter of a pinion gear with a pitch diameter of 71.1 mm and a gear with a pitch diameter of 105.8 mm. 7 3 5. Solve 25 5 2 x. 8 4 6. Determine the value of 406.442−1/3. Round the answer to 3 decimal places.

Similar triangles 7. Determine which of the following pairs of triangles (A through F) are similar. All linear dimensions are in inches. The similar pairs of triangles are 9.50 30°

110° 40°

30°

6.30

8.50

4.25

12.60

40° 110°

4.75 PAIR A

PAIR D

10

64° 64°

5 1

32

7 PAIR E

PAIR B

2.4

61°

61°

3.2

4.8

4.2

2.8

3.6 PAIR C

PAIR F

Solve the following exercises: F

8. In nABC and nDEF, /A 5 /D, /B 5 /E, /C 5 /F. All dimensions are in inches.

C

15

A 12

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10

6

a. Find AC. b. Find DE.

D

E

B

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Geometric PrinciPles For trianGles and other common PolyGons

9. In the figure, /H 5 /P, /J 5 /M, /K 5 /L. All dimensions are in millimeters. Round the answers to 2 decimal places.

J 70.82

a. Find HK. b. Find LM.

54.68 95.87

H

K

P

L

121.55 M

10. In △ABC and △DEF, AB i DE, AC i DF, BC i EF.

23°

A

a. Find /A. b. Find /F. c. Find /B. d. Find /E.

C

56°

D

F B E

11. Use the figure to find the value of the following angles. 37°189

a. Find /1. b. Find /2.

2 1

12. In nHJK, PM i JK.

H

a. Find /HPM. b. Find /PMK.

M

76°109

25°279

P

K

J

13. Refer to the figure to find these angles. a. /1 b. /2 c. /3

1 2

3 79°

14. Refer to the figure to find these dimensions. All dimensions are in millimeters. Round the answers to 1 decimal place. a. Find dimension A. b. Find dimension B.

425.0

90° 260.00 A

300.00 90°

B 498.2

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396

Section 5

15. In this figure, AB i DE and CB i EF. All dimensions are in inches. Round the answers to 3 decimal places.

2.200

a. Find x. b. Find y.

Fundamentals oF Plane Geometry

4.000 A

C

D

F

y

2.700 B

x

4.800

E

isosceles, equilateral, and Right triangles Solve the following exercises: 16. All dimensions are in inches. a. Find x. b. Find /1.

12.40

12.40

38°409

1 x 18.4

17. All dimensions are in millimeters. a. Find x. b. Find y.

72.5

x

38.2°

38.2° 56.8 y

18. All dimensions are in inches. 2

a. Find /1. b. Find /2.

1 17.93

17.93

64°589

19. All dimensions are in millimeters. C

a. Find x. b. Find y.

118.30 60°

y

B 60° x A

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Unit 54

397

Geometric PrinciPles For trianGles and other common PolyGons

20. All dimensions are in inches.

7.86

E

a. Find /1. b. Find x.

G 30° 1 7.86 x F

21. Refer to this figure. Using the given values, find the values of x.

x

a. If d 5 90 and e 5 120, find x. b. If d 5 30 and e 5 40, find x. d

e

22. Using the figure and these given values, find the values of y. Round the answers to the nearest whole millimeter.

g

a. If g 5 108 mm and m 5 123 mm, find y. b. If g 5 153.70 mm and m 5 170 mm, find y.

y m

23. Using the figure and these given values, find the values of y.

T

a. If radius A 5 360.00 mm and x 5 480.00 mm, find y. b. If radius A 5 216.00 mm and x 5 288.00 mm, find y.

x

y

P S

RADIUS A

24. Three holes are drilled in the plate shown. All dimensions are in inches. Determine dimensions A and B to 3 decimal places. a. A 5 b. B 5

7.000

17.823

M

30.263

18.090

18.090

B A

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Fundamentals oF Plane Geometry

Section 5

25. All dimensions are in inches. Round the answers to 3 decimal places. a. If y 5 2.8000, find x. b. If y 5 3.0000, find x.

1.850

y

1.700 0.800 1.375 x

26. All dimensions are in inches. Round the answers to 3 decimal places. a. If y 5 2.1450, find x. b. If y 5 2.2650, find x.

x

1.100 RADIUS

1.100

y

0.685 2.100 5.820

other Polygons Solve the following exercises: 27. A template is shown. All dimensions are in millimeters. Determine length x and length y to 2 decimal places. x y

124.00

60.00

50.00

50.00 90.00 y x

28. Refer to polygon ABCD. a. If /2 5 87.08, find /1. b. If /1 5 114.08, find /2.

C 62.05°

D 1

109.62° A

2 B

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Unit 55

399

introduction to circles

29. Use the angle values given.

30. Use the angle values given to find /2.

a. If /1 5 1148, find /2. b. If /2 5 838, find /1.

a. If /1 5 378, find /2. b. If /1 5 298, find /2. 1 90°

64°

2

52°

78°

83°

151°

96° 1

84° 2

UNIT 55 Introduction to Circles Objectives After studying this unit you should be able to ●● ●●

Identify parts of a circle. Solve problems by using geometric principles that involve chords, arcs, central angles, perpendiculars, and tangents.

Circles are the simplest of all closed curves and their basic properties are readily understood. Holes are often laid out on bolt circles, and rotary tables move in a circular motion. Machines operate by the circular motion of gears. Parts are machined with cutting tools and/or work pieces revolving in a circular path.

definiTions A circle is a closed curve on which every point is equally distant from a fixed point called the center. Refer to Figure 55-1 for the following definitions:

CHORD C O

The circumference is the length of the curved line that forms the circle. A chord is a straight line segment that joins two points on the circle. AB is a chord. A diameter is a chord that passes through the center of a circle. CD is a diameter.

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CIRCUMFERENCE

A

B DIAMETER D

E

RADIUS

FiGure 55-1

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400

Fundamentals oF Plane Geometry

Section 5

A radius (plural radii) is a straight line segment that connects the center of the circle with a point on the circle. The radius is one-half the diameter. OE is a radius. Refer to Figure 55-2 for the following definitions: An arc is that part of a circle between any two points on the circle. The symbol C C written above the letters means arc. AB is an arc. Any two points on a circle divide the circle into two arcs. A semicircle is an arc that is one-half of a circle. If the arcs are not equal, the smaller is a minor arc and the larger is a major arc. In Figure 55-2, ACB is the minor arc and APB is the major arc. Unless otherwise stated, ACB means the minor arc ACB. A tangent to a circle is a straight line that touches the circle at one point only. The point on the circle touched by the tangent is called the point of tangency or tangent point. CD is a tangent and point P is a tangent point. A secant is a straight line that passes through a circle and intersects the circle at two points. EF is a secant.

A

E

ARC B

C F

TANGENT POINT

P

D

TANGENT

B

ARC

Refer to Figure 55-3 for the following definitions:

SECANT

FiGure 55-2 C CHORD

SEGMENT

A segment is a figure formed by an arc and the chord joining the endpoints of the arc. The shaded figure ABC is a segment. A sector is a figure formed by two radii and the arc intercepted by the radii. The shaded figure EOF is a sector.

A

RADIUS

O

F SECTOR

RADIUS

Figure 55-3 M

Refer to Figure 55-4 for the following definitions: A central angle is an angle whose vertex is at the center of a circle and whose sides are radii. Angle MON is a central angle. An inscribed angle is an angle in a circle whose vertex is on the circle and whose sides are chords. Angle SRT is an inscribed angle.

ARC

E

N CENTRAL ANGLE S

O R T

INSCRIBED ANGLE

FiGure 55-4

CirCumferenCe formula A polygon is inscribed in a circle when each vertex of the polygon is a point of the circle. In Figure 55-5, regular polygons are inscribed in circles. As the number of sides increases, the perimeter increases and approaches the circumference.

A

B

C

D

FiGure 55-5

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introduction to circles

An important relationship exists between the circumference and the diameter of a circle. As the number of sides of an inscribed polygon increases, the perimeter approaches a certain number times the diameter. This number is called pi. The symbol for pi is p. No matter how many sides an inscribed polygon has, the value of p cannot be expressed exactly with digits. Pi is called an irrational number. The circumference of a circle is equal to pi (p) times the diameter or two pi times the radius. Generally, for the degree of precision required in machining applications, a value of 3.1416 is used for p if a calculator is not available. C 5 pd where C 5 circumference or p 5 pi C 5 2pr d 5 diameter r 5 radius

Example 1 Compute the circumference of a circle with a 50.70-mm diameter. C 5 pd 5 3.1416s50.70 mmd 5 159.28 mm

Ans sroundedd

As presented under EVALUATION OF ALGEBRAIC EXPRESSIONS in Unit 39, depressing the pi key ( p ) displays the value of pi to 10 digits (3.141592654) on most calculators. Recall that p is an alternate function on many calculators. C 5 p d, C 5 p s50.70 mmd p 3 50.7 5 159.2787475 C 5 159.28 mm

Ans sroundedd

Example 2 Determine the radius of a circle that has a circumference of 14.860 inches. C 5 2pr 14.860 in. 5 2s3.1416dsrd r 5 2.365 in. Ans sroundedd 14.860 in. r5 2p 14.86 4 ( 2 3 p ) 5 2.365042454, r 5 2.365 in. Ans sroundedd

geomeTriC prinCiples ■ Principle 11

M

E

In the same circle or in equal circles, equal chords cut off equal arcs. Given: In Figure 55-6, Circle A 5 Circle B and chords CD 5 EF 5 GH 5 MS. Conclusion: CD 5 EF 5 GH 5 MS.

F

S H

D

C

G CIRCLE A

CIRCLE B

FiGure 55-6

■ Principle 12 In the same circle or in equal circles, equal central angles cut off equal arcs.

A

B H

1

Given: In Figure 55-7, Circle D 5 Circle E and /1 5 /2 5 /3 5 /4.

K

Conclusion: AB5 FG 5 HK 5 MP and AG5 BF 5 HM 5 KP.

M

4

2 G

P F

CIRCLE D

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3

CIRCLE E

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Fundamentals oF Plane Geometry

Section 5

■ Principle 13 In the same circle or in equal circles, two central angles have the same ratio as the arcs that are cut off by the angles. Example In Figure 55-8, Circle A 5 Circle B. If /COD 5 908, /EOF 5 508, CD 5 1.4000, and GH 5 2.1000, determine (a) the length of EF and (b) the size of /GOH.

2.100

C

1.400 90° O

E

G

H

D

O

50° F CIRCLE B

CIRCLE A

FiGure 55-8

a. Set up a proportion between CD and EF with their respective central angles. Solve for EF . /COD CD 5 /EOF EF 908 1.4000 5 508 EF 90EF 5 50s1.4000d 50s1.4000d EF 5 90 EF 5 0.7780 Ans b. Set up a proportion between CD and GH with their central angles. Solve for /GOH. /COD CD 5 /GOH GH 908 1.4000 5 /GOH 2.1000 1.400s/GOHd 5 908s2.100d 908s2.100d /GOH 5 1.400 /GOH 5 1358 Ans The perpendicular bisector of a line segment AB is the line through the midpoint and perpendicular to AB. For example, in Figure 55-9, CD is the perpendicular bisector of AB.

■ Principle 14 A line drawn from the center of a circle perpendicular to a chord bisects the chord and the arc cut off by the chord. B Given: In Figure 55-9, diameter DE ' chord AB.

D C

Conclusion: AC 5 BC and AD 5 BD and AE 5 BE.

● The perpendicular bisector of a chord passes through the

O

A

center of a circle.

Given: In Figure 55-9, DE is the perpendicular bisector of chord AB.

E

FiGure 55-9

Conclusion: DE passes through the center, O, of the circle.

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introduction to circles

The use of Principle 14 with the Pythagorean Theorem (Principle 9) has wide practical application in the machine trades.

250.00 A

Example Holes A, B, and C are to be drilled in the plate shown in Figure 55-10. The centers of holes A and C lie on a 280.00-mm-diameter circle. The center of hole B lies on the intersection of chord AC and segment OB, which is perpendicular to AC. Compute working dimensions F, G, and H. All dimensions are in millimeters.

B

O H

C 280.00 DIA

180.00

Compute dimension F: Applying Principle 14, AC is bisected by OB.

F

AB 5 BC 5 250.00 mm 4 2 5 125.00 mm: F 5 200.00 mm 2 125.00 mm 5 75.00 mm Ans

200.00 G

Compute dimension G.

FiGure 55-10

G 5 200.00 mm 1 125.00 mm 5 325.00 mm Ans Compute dimension H. In right nABO, AB 5 125.00 mm, AO 5 280.00 mm 4 2 5 140.00 mm Compute OB by applying the Pythagorean Theorem (Principle 9). AO2 5 OB2 1 AB2 s140.00 mmd2 5 OB2 1 s125.00 mmd2 OB 5 Ïs140.00 mmd2 2 s125.000 mmd2 OB 5 63.05 mm H 5 180.00 mm 1 63.05 mm 5 243.05 mm

Ans

■ Principle 15 A line perpendicular to a radius at its extremity is tangent to the circle. A tangent is perpendicular to a radius at its tangent point. Example 1 Given: Line AB ' to

D

radius CO at point C in Figure 55-11. Conclusion: Line AB is a tangent.

Example 2 Given: In Figure 55-11, tangent DE passes through point F of radius FO.

A

F

O

E

C TANGENT POINT

Conclusion: Tangent DE 'radius FO.

TANGENT POINT

B

FiGure 55-11

■ Principle 16 Two tangents drawn to a circle from a point outside the circle are equal. The angle at the outside point is bisected by a line drawn from the point to the center of the circle. A P

Example 1 Given: In Figure 55-12, tangents AP and BP are drawn to the circle from point P. Conclusion: AP 5 BP.

O B

FiGure 55-12

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Section 5

Fundamentals oF Plane Geometry

Example 2 Given: In Figure 55-12, line OP extends from outside point P to center O. Conclusion: /APO 5 /BPO.

■ Principle 17 If two chords intersect inside a circle, the product of the two segments of one chord is equal to the product of the two segments of the other chord. Example 1 Given: Chords AC and DE intersect at point B in Figure 55-13.

A

Conclusion: AB(BC) 5 BD(BE).

E

Example 2 In Figure 55-13, if AB 5 7.5 inches, BC 5 2.8 inches, and BD 5 2.1 inches, determine the length of BE. ABsBCd 5 BDsBEd 7.5s2.8d 5 2.1sBEd 21.0 5 2.1BE BE 5 10.0 inches

B D

C

FiGure 55-13

Ans

ApplicAtion tooling Up 1. Determine the length of a. Round the answer to 1 decimal place.

mm

a

56.8

63.5 mm

2. Determine the size of /1.

43°289

1

59°379

3. Identify /1 in the preceding figure as an acute, right, obtuse, straight, or reflex angle. 4. Express 1328279430 in decimal degrees. Round the answer to 4 decimal places.

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Unit 55

405

introduction to circles

3 5. Solve Ï x 5 4.3.

6. Add (4x2 1 xy2 2 7y3), (−7x2 1 3x2y2 1 9y3), and (5x2 1 12xy2 2 8x2y2 2 4y3).

Definitions Name each of the parts of circles for the following exercises. 7. a. AB b. CD c. EO d. Point O

9. a. M b. P c. SO d. TO e. RW f. RW

B D O A C

E

H G

8. a. GF b. HK c. LM d. GF e. Point P

K

L

F P

M

10. a. /1 b. /2 c. AO d. CD e. CE f. AB

S

T

M O

R

W P

B 1

D

O

A C

2 E

circumference Formula Use C 5 p d or C 5 2p r

where C 5 circumference p 5 3.1416 if not using a calculator d 5 diameter r 5 radius

11. Determine the unknown value for each of the following exercises. Round the answers to 3 decimal places. a. If d 5 6.5000, find C. b. If d 5 30.000 mm, find C. c. If r 5 18.600 mm, find C. d. If r 5 2.9300, find C.

e. If C 5 35.0000, find d. f. If C 5 218.000 mm, find d. g. If C 5 327.000 mm, find r. h. If C 5 7.6800, find r.

12. Determine the length of wire, in feet, in a coil of 60.0 turns. The average diameter of the coil is 30.0 inches. Round the answer to the nearest whole foot. 13. A pipe with a wall thickness of 6.00 millimeters has an outside diameter of 79.20 millimeters. Compute the inside circumference of the pipe. Round the answer to 2 decimal places. 14. The flywheel of a machine has a 0.80-meter-diameter and revolves 240.0 times per minute. How many meters does a point on the outside of the flywheel rim travel in 5.0 minutes? Round the answer to the nearest whole meter.

Geometric Principles Solve the following exercises based on Principles 11–14, although an exercise may require the application of two or more of any of the principles. Round the answers to 3 decimal places where necessary unless otherwise stated.

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Section 5

Fundamentals oF Plane Geometry

15. Determine the length of belt required to connect the two pulleys shown. All dimensions are in inches. Round the answer to 2 decimal places.

62.00 14.00 DIA B

16. nABC is equilateral. All dimensions are in inches. a. Find AB. b. Find BC. A

C 2.67

4.090

17. All dimensions are in inches. a. Find AB. b. Find BC.

C 3.980 A

B E

18. a. If EF 5 160 mm, find HP. b. If HP 5 284 mm, find EF. Round the answer to the nearest whole millimeter.

80° F H

P

140°

19. a. If SW 5 4.8000 and TM 5 5.7600, find /1. b. If TM 5 4.1280 and SW 5 2.0640, find /1.

W

S

1

T

120°09 M

20. a. If AB 5 5.3780 and AC 5 3.7820, find (1) DB and (2) ACB. (1) (2) b. If DB 5 3.0170 and ACB 5 7.3080, find (1) AB and (2) CB. (1) (2)

B D C A

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407

introduction to circles

21. Find HK when EF 5 21.23 mm. E

H

86°

F

214°

22. All dimensions are in inches.

K

B

1.300 RADIUS

A

a. If /1 5 240809, find ABC. b. If ABC 5 2.3000, find /1.

27°

O

101°

C O 1 D

23. All dimensions are in inches. a. If x 5 5.1000, find /1. b. If x 5 4.7500, find /1.

63°09

x 1

4.500

4.500 R

24. a. If radius x 5 7.5000 and y 5 4.5000, find PM. b. If radius x 5 8.0000 and y 5 4.8000, find PM.

P y

RADIUS x

M

25. The circumference of this circle is 14.4000. x

1

a. If x 5 3.2000, find /1. b. If /1 5 36809, find x.

26. Determine the centerline distance between hole A and hole B for these values. a. Radius x 5 8.0000 and DO 5 2.1000. b. Radius x 5 1.2000 and DO 5 0.7000.

HOLE B

D O HOLE A

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RADIUS x

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Section 5

Fundamentals oF Plane Geometry

Solve the following exercises based on Principles 15–17, although an exercise may require the application of two or more of any of the principles. Round the answers to 3 decimal places where necessary unless otherwise stated. 27. Point P is a tangent point and /1 5 1078189.

P

E

a. If /2 5 418219, find (1) /E and (2) /F. (1) (2) b. If /2 5 488209, find (1) /E and (2) /F. (1) (2)

F 1

2 O

28. AB and CB are tangents.

y A

a. If y 5 137.20 mm and /ABC 5 67.08, find (1) /1 and (2) x. (1) (2)

B 1 x

b. If x 5 207.70 mm and /1 5 33.88, find (1) /ABC and (2) y. (1) (2)

C

29. Point A is a tangent point of the V-groove cut and pin shown. All dimensions are in inches.

x

0.800 RADIUS

a. If y 5 1.4000, find x. b. If y 5 1.8000, find x.

A 2.842 y 0.544

30. Points E, G, and F are tangent points.

D

a. If /1 5 1098, find /2. b. If /1 5 1188459, find /2.

C F

E

O

58°

2 1 A

B

G

31. All dimensions are in millimeters. Round the answers to 2 decimal places.

E

75.00 F

a. If EK 5 150.00 mm, find GK. b. If GK 5 120.00 mm, find EK. K 337.50

G

H

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409

arcs and anGles oF circles, tanGent circles

32. All dimensions are in inches.

T S

a. If PT 5 1.8000, find x. b. If PT 5 2.0000, find x.

E

1.200

P x 0.614 2.584 DIA CIRCLE M

UNIT 56 Arcs and Angles of Circles, Tangent Circles Objectives After studying this unit you should be able to ●● ●●

Solve problems by using geometric principles that involve angles formed inside, on, and outside a circle. Solve problems by using geometric principles that involve internally and externally tangent circles.

The geometric principles of arcs and angles of circles and tangent circles have wide application in machine technology. For example, circle arc and angle principles are used in computing working dimension locations for machining holes and in determining angle rotations for positioning workpieces.

angles formed inside a CirCle

A

■ Principle 18 A central angle is equal to its intercepted arc.

O

(An intercepted arc is an arc that is cut off by a central angle.)

78°

78°

B

Given: AB 5 788 in Figure 56-1.

FiGure 56-1

Conclusion: /AOB 5 788.

● An angle formed by two chords that intersect inside a circle is equal to one-half the sum of its two intercepted arcs. Example 1 Given: Chords CD and EF intersect at point P in Figure 56-2. Conclusion: /EPD 5

1 sCF 1 DEd. 2

C E P D

Example 2 In Figure 56-2, if CF 5 1068 and ED 5 428, determine /EPD. 1 /EPD 5 s1068 1 428d 5 748 2

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Ans

F

FiGure 56-2

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410

Section 5

Fundamentals oF Plane Geometry

Example 3 In Figure 56-2, if /EPD 5 648129 and CF 5 958589, determine DE. 1 sCF 1 DEd 2 1 648129 5 s958589 1 DEd 2 1 648129 5 478599 1 DE 2 1 168139 5 DE 2 /EPD 5

DE 5 328269 Ans DE 5 2 3 648129 2 958589 2 3 64 8 9 0 12 8 9 0 2 95 8 9 0 58 8 9 0 5 32826800, DE 5 328269

Ans

■ An inscribed angle is equal to one-half of its intercepted arc. A

Given: AC 5 1058 in Figure 56-3. Conclusion: /ABC 5

1 1 AC 5 s1058d 5 528309. 2 2

B 105°

C

FiGure 56-3

arC lengTh formula

Consider a complete circle as an arc of 3608. The ratio of the number of degrees of an arc to 3608 is the fractional part of the circumference that is used to find the length of an arc. The ratio of the length of an arc to the circumference of a circle is the same as the ratio of the number of degrees of the arc to 360º. Arc Length Arc Degrees 5 Circumference 3608

or

Arc Length Central Angle 5 Circumference 3608

If you know the radius of the circle, these two formulas can be rewritten as: Arc Length 5

Arc Degrees s2prd 3608

or

Arc Length 5

Central Angle s2prd 3608

Example 1 AC 5 130.008 and the radius is 120.00 mm in Figure 56-4. Determine the arc length AC to 2 decimal places. Arc Degrees s2prd 3608 130.008 Arc Length 5 [2s3.1416ds120.00 mmd] 3608

130.00°

Arc Length 5

Arc Length 5 272.27 mm Ans (rounded) 130 4 360 3 2 3 p 3 120 5 272.2713633, 272.27 mm Ans (rounded)

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A

C

120.00 mm RADIUS

FiGure 56-4

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Unit 56

411

arcs and anGles oF circles, tanGent circles

Example 2 The arc length of DF is 8.4260 and the radius is 5.0210 in Figure 56-5. Determine /1. All dimensions are in inches. Give the answer in degrees and minutes.

8.426 E D 1

Arc Length Central Angle 5 Circumference 3608 Arc Length /1 5 2pr 3608 8.4260 /1 5 2s3.1416ds5.0210d 3608 s8.4260ds3608d 5 /1 2s3.1416ds5.0210d /1 5 968099

F

O

5.021 RADIUS

FiGure 56-5

Ans sroundedd

8.4260s3608d 5 Central Angle 2ps5.0210d 8.426 3 360 4 ( 2 3 p 3 5.021 968983.65, 96899 Ans (rounded)

)

5

SHIFT

angles formed on a CirCle

TANGENT POINT D

■ Principle 19 An angle formed by a tangent and a chord at the tangent point is equal to one-half of its intercepted arc.

A

C

Example 1 In Figure 56-6, tangent CD meets chord AB at tangent point A and AEB 5 1108. Determine /CAB.

E 110°

1 1 /CAB 5 AEB 5 s1108d 5 558 2 2

B

Ans FiGure 56-6

Example 2 In Figure 56-7, the centers of three holes lie on line ABC. Line ABC is tangent to circle O at hole-center B. The hole-center D, of a fourth hole, lies on the circle. Determine /ABD. A central angle is equal to its intercepted arc (Principle 18).

D

O

DB 5 /DOB 5 1328

132°

?

Apply Principle 19. /ABD 5

1 1 DB 5 s1328d 5 668 2 2

Ans

A

B

C

FiGure 56-7

angles formed ouTside a CirCle ■ Principle 20 An angle formed at a point outside a circle by two secants, two tangents, or a secant and a tangent is equal to one-half the difference of the intercepted arcs.

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Section 5

Fundamentals oF Plane Geometry

Two secants Example 1 Given: Secants AP and DP meet at point P and intercept BC and AD in Figure 56-8. Conclusion: /P 5

1 sAD 2 BCd. 2

Example 2 In Figure 56-8, if AD 5 85840900 and BC 5 39817900, find /P. 1 1 1 /P 5 sAD 2 BCd 5 s85840900 2 39817900d 5 s46823900d 2 2 2 5 238119300 Ans

P B C

A

Example 3 If /P 5 288 and BC 5 408 in Figure 56-8, determine AD. 1 sAD 2 BCd 2 1 288 5 sAD 2 408d 2 AD 5 968 Ans

D

/P 5

FiGure 56-8

Two Tangents Example 1 Given: Tangents DP and EP meet at point P in Figure 56-9 and intercept DE and DCE. Conclusion: /P 5

1 sDCE 2 DEd. 2

C

D

Example 2 If DCE 5 2538379 and DE 5 1068239 in Figure 56-9, determine /P. 1 1 1 /P 5 sDCE 2 DEd 5 s2538379 2 1068239d 5 s1478149d 2 2 2 5 738379 Ans

P

E

FiGure 56-9

a Tangent and a secant Example 1 Given: Tangent AP and secant CP meet at point P and intercept AC and AB in Figure 56-10. Conclusion: /P 5

1 sAC 2 ABd. 2

Example 2 In Figure 56-10, if AC 5 1268389 and /P 5 288509, determine AB. 1 sAC 2 ABd 2 1 288509 5 s1268389 2 ABd 2 1 288509 5 638199 2 AB 2 1 AB 5 638199 2 288509 2 AB 5 2s638199 2 288509d AB 5 688589 Ans /P 5

98310_sec_05_Unit_53-56_ptg01.indd 412

A

P

B

C

FiGure 56-10

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Unit 56

413

arcs and anGles oF circles, tanGent circles

2 3 ( 63 8 9 0 19 8 9 0 2 28 8 9 0 50 8 9 0 6885880, AB 5 688589 Ans

)

5

inTernally and exTernally TangenT CirCles Two circles that are tangent to the same line at the same point are tangent to each other. Circles can be either internally or externally tangent.

TANGENT POINT

Internally tangent—Two circles are internally tangent if both circles are on the same side of the common tangent line. (See Figure 56-11.) Externally tangent—Two circles are externally tangent if the circles are on opposite sides of the common tangent line. (See Figure 56-12.)

INTERNALLY TANGENT CIRCLES

FiGure 56-11

■ Principle 21 If two circles are either internally or externally tangent, a line connecting the centers of the circles passes through the point of TANGENT tangency and is perpendicular to the tangent line. POINT

inTernally TangenT CirCles Example Given: Circle D and Circle E in Figure 56-13 are internally tangent at point C. D is the center of Circle D and E is the center of Circle E. Line AB is tangent to both circles at point C.

EXTERNALLY TANGENT CIRCLES

FiGure 56-12

Conclusion: An extension of line DE passes through tangent point C and line CDE ⊥ tangent line AB. Principle 21 is often used as the basis for computing dimensions of parts on which two or more radii blend to give a smooth curved surface. This type of application is illustrated by the following example.

C

the two radii will result in a smooth curve from point A to point B.

Refer to Figure 56-15. The 12.0000-radius arc and the 25.0000-radius arc are internally tangent. Apply Principle 21. A line connecting arc centers F and H passes through tangent point C.

D E

A

Example A part is to be machined as shown in Figure 56-14. The proper locations of Note: The curve from A to B is not an arc of one circle; it is made up of arcs from two different sized circles. In order to make the part, the location to the center of the 12.000-inch radius (dimension x) must be determined. Compute x. All dimensions are in inches.

CIRCLE D CIRCLE E

B

FiGure 56-13 B

12.000 R

21.000

25.000 R A

x

FiGure 56-14

Tangent point C is the endpoint of the 25.0000 radius, CH 5 25.0000. Tangent point C is the endpoint of the 12.0000 radius, CF 5 12.0000.

B

FH 5 25.0000 2 12.0000 5 13.0000

C

Since BE is vertical and AH is horizontal, /FEH 5 908. In right nFEH, FH 5 13.0000, FE 5 21.0000 2 BF 5 9.0000. Apply the Pythagorean Theorem (Principle 9) to compute EH. FH2 5 EH2 1 FE2 s13.000 in.d2 5 EH2 1 s9.000 in.d2 EH 5 9.381 in. x 5 EH 5 9.381 in. Ans

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F

A

E

21.000

H

FiGure 56-15

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414

Section 5

Fundamentals oF Plane Geometry

externally Tangent Circles A

Example 1 Given: Circle D and Circle E are externally tangent at point C in Figure 56-16. D is the center of Circle D and E is the center of Circle E. Line AB is tangent to both circles at point C.

E C

D

Conclusion: Line DE passes through tangent point C and line DE ⊥ tangent line AB at point C.

CIRCLE E

B CIRCLE D

Figure 56-16

Example 2 Three holes are to be bored in a steel plate as shown in Figure 56-17. The 42.00-mm diameter and 61.40-mm diameter holes are tangent at point D. CD is the common tangent line. Determine the distances between hole centers (AB, AC, and BC). All dimensions are in millimeters. Round the answers to 2 decimal places. Compute AB. Apply Principle 21. Since AB connects the centers of two tangent circles, AB passes through tangent point D. AB 5 AD 1 DB 5 21.00 mm 1 30.70 mm 5 51.70 mm Ans Compute AC and BC. Since AB connects the centers of two tangent circles, AB ⊥ tangent line DC. Triangle ADC and triangle BDC are right triangles. Apply the Pythagorean Theorem (Principle 9).

61.40 DIA HOLE

42.00 DIA HOLE

A

D

B

76.80

In right nADC, AD 5 21.00 mm and DC 5 76.80 mm.

C 28.40 DIA HOLE

FiGure 56-17

AC2 5 AD2 1 DC2 AC2 5 s21.00 mmd2 1 s76.80 mmd2 AC 5 79.62 mm Ans AC 5 Ïs21.00 mmd2 1 s76.80 mmd2 21 x 2 1 76.8 x 2 ) 5 79.61934438 AC 5 79.62 mm Ans sroundedd Ïa

(

In right nBDC, DB 5 30.70 mm and DC 5 76.80 mm. BC2 5 DB2 1 DC2 BC2 5 s30.70 mmd2 1 s76.80 mmd2 BC 5 82.71 mm Ans BC 5 Ïs30.70 mmd2 1 s76.80 mmd2 Ïa

(

30.7

x2

BC 5 82.71 mm

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1 76.8

x2

)

5 82.70870571

Ans sroundedd

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arcs and anGles oF circles, tanGent circles

ApplicAtion tooling Up 1. A pipe has an inside circumference of 82.50 mm and an outside diameter of 28.04 mm. What is the wall thickness of the pipe in millimeters? If necessary, round the answer to 2 decimal places. 2. Determine the length of AB, AC, and ED. Round the answer to 2 decimal places. C

58 37.

mm 60°

B

E

D

60°

m

6m

59.2

A

3. Identify the angle that corresponds with /A, /D, and /ACB. A 46 36

D

32

C

18 64

B

23

E

4. What is the complement of a 728219470 angle? 5. Solve R 5

KL for K d2

6. Use scientific notation to compute

s9.1 3 1027d 3 s4.2 3 1012d . Round your answer to 2 decimal places. s4.3 3 108d 3 s5.5 3 10215d

Arc Length Formula

B

A

Determine the unknown value for each of the following exercises. Round the answers to 3 decimal places.

1

7. ABC 5 90809 and r 5 3.500 in. Find arc length ABC.

O

C

r

8. ABC 5 85.008 and r 5 60.000 mm. Find arc length ADC. 9. Arc length ABC 5 510.000 mm and r 5 120.000 mm. Find /1.

D

10. Arc length ADC 5 22.700 in and r 5 5.200 in. Find /1. 11. Arc length ABC 5 18.750 in and /1 5 72809. Find r. 12. Arc length ABC 5 620.700 mm and /1 5 69.308. Find r.

Geometric Principles Solve the following exercises based on Principles 18 through 21, although an exercise may require the application of two or more of any of the principles. Where necessary, round linear answers in inches to 3 decimal places and millimeters to 2 decimal places. Round angular answers in decimal degrees to 2 decimal places and degrees and minutes to the nearest minute.

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Section 5

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13. a. If /1 5 76.008, find: (1) DC (2) /EOD (3) AC b. If /1 5 63.768, find: (1) DC (2) /EOD (3) BD

E

D 107.00° 1

O 36.

A

00°

98.00°

C

B

14. a. If /1 5 638, find: (1) HK (2) HM

M 1 O

b. If /1 5 598479, find: (1) HK (2) HM

K

H

15. a. If PS 5 468, find: (1) /1 (2) /2

P 1

S

2

b. If PS 5 398, find: (1) /1 (2) /2

36°

16. a. If DC 5 358, find AB.

B

b. If AB 5 1278, find DC. 82°

C

A

17. a. If /3 5 478 and GH 5 328, find: (1) EF (2) /4

D

E

G

3 4

b. If /4 5 178539 and EF 5 1038, find: (1) /3 (2) GH

H

F

18. a. If /1 5 258 and MPT 5 958, find: (1) KTP (2) PT (3) MP b. If /1 5 178309 and MPT 5 1038, find: (1) KPT (2) PT (3) MP

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M 1

P

K T

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arcs and anGles oF circles, tanGent circles

19. a. If AB 5 1168, find: (1) /1 (2) /2

A

b. If AB 5 1128569, find: (1) /1 (2) /2

C

2

20. a. If EF 5 848, find: (1) /EFD (2) HF (3) /1

1

H

E

b. If EF 5 798, find: (1) /EFD (2) HF (3) /1

87°

D

F

21. a. If ST 5 208189 and SM 5 388079, find: (1) /1 (2) /2 b. If ST 5 258179 and SM 5 358249, find: (1) /1 (2) /2

B

1

P T 1

P S M

113°089 2 95°239

22. a. If AB 5 728209 and CD 5 508189, find: (1) /1 (2) /2 (3) /3

2 B A

b. If CD 5 438159 and AD 5 1068059, find: (1) /1 (2) /2 (3) /3

C 3

1 D

23. a. If /1 5 24.008 and /2 5 60.008, find: (1) DH (2) EH b. If /1 5 29.008 and /2 5 64.008, find: (1) DH (2) EH

1088199

144.00° 85.00°

E

2 1

D H

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Section 5

Fundamentals oF Plane Geometry

24. a. If Dia A 5 3.7560 and Dia B 5 1.6220, find x. b. If x 5 0.9750 and Dia B 5 1.0260, find Dia A. DIA A x

DIA B CIRCLE B

25. a. If x 5 24.93 mm and y 5 28.95 mm, find Dia A. b. If x 5 78.36 mm and y 5 114.48 mm, find Dia A.

CIRCLE A

x

y

DIA A

26. a. If /1 5 678009 and /2 5 938009, find: (1) AB (2) DE

F

P E D 70°009

b. If /1 5 758009 and /2 5 858009, find: (1) AB (2) DE

A

M

G H

2

1

C

B 62°009

27. All dimensions are in inches.

1.200 DIA

a. If Dia A 5 1.0000, find x. b. If Dia A 5 0.8000, find x.

x

DIA A 1.600 DIA

28. Determine the length of x for Gage A and Gage B. All dimensions are in inches.

CL

a. Gage A: y 5 0.3500, find x. b. Gage B: y 5 0.4100, find x.

1.500 R 0.520 R

1.300

0.520 R y

x

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arcs and anGles oF circles, tanGent circles

29. AC is a diameter.

A

a. If /2 5 228009, find /1. b. If /2 5 308549, find /1.

D 2

C

E

1

B

30. Three posts are mounted on the fixture shown. Each post is tangent to the arc made by the 0.650-inch radius. Determine (a) dimension A and (b) dimension B.

0.650 RADIUS

Note: The fixture is symmetrical (identical) on each side of the horizontal centerline (CL). All dimensions are in inches. a. b.

CL

B

0.270 A

31. Points A, B, C, D, and E are tangent points. a. If AB 5 46.008 and DE 5 66.008, find /1.

C

b. If AB 5 53.008 and DE 5 70.008, find /1.

3 POSTS 0.260 DIA

D 38°

A CL

E

1 B

32. Three holes are to be located on the layout shown. The 72.40-mm diameter and 30.80-mm diameter holes are tangent at point T, and TA is the common tangent line between the two holes. Determine (a) dimension C and (b) dimension D. a. b.

95.00 mm 72.40 mm DIA C B

T

A

C D

30.80 mm DIA

41.60 mm DIA

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Section 5

Fundamentals oF Plane Geometry

UNIT 57 Fundamental Geometric Constructions Objectives After studying this unit you should be able to ●● ●●

Make constructions that are basic to the machine trades. Lay out typical machine shop problems using the methods of construction.

A knowledge of basic geometric constructions done with a compass or dividers and a steel rule is required of a machinist in laying out work. The constructions are used in determining stock allowances and reference locations on castings, forgings, and sheet stock. For certain jobs where wide dimensional tolerances are permissible, the most practical and efficient way of producing a part may be by scribing and centerpunching locations. Layout dimensions are sometimes used as a reference for machining complex parts that require a high degree of precision. Locations lightly scribed on a part are used as a precaution to ensure that the part or table movement is in the proper direction. It is particularly useful in operations that require part rotation or repositioning. Some common marking tools are shown in Figure 57-1. There are many geometric constructions, some of which are relatively complex. The constructions presented in this book are those that are most basic and common to a wide range of practical applications.

Pocket Scriber (Courtesy of L. S. Starrett Company)

Center Punch (Courtesy of L. S. Starrett Company)

Dividers (Courtesy of L. S. Starrett Company)

Trammels (Courtesy of L. S. Starrett Company)

FiGure 57-1

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Fundamental Geometric constructions

Unit 57

ConstruCtion 1 to ConstruCt a PerPendiCular BiseCtor of a line segment Required: Construct a perpendicular bisector to line segment AB.

c Procedure ●●

●●

●●

With endpoint A as a center and using a radius equal to more than half AB, draw arcs above and below AB as in Figure 57-2(a). With endpoint B as a center and with the same radius used at A, draw arcs above and below AB that intersect the first pair of arcs as shown in Figure 57-2(b). Draw a connecting line between the intersection of the arcs above and below AB. Line CD in Figure 57-2(c) is perpendicular to AB and point O is the midpoint of AB. C

A

B

A

A

B

O

B

D (a)

(c)

(b)

FiGure 57-2

Practical application Locate the center of a circle.

Solution: The perpendicular bisector of a chord passes through the center of the circle. The center of a circle is located by drawing two chords and constructing a perpendicular bisector to each chord. The intersection of the two perpendicular bisectors locates the center of the circle. The construction lines are shown in Figure 57-3.

CENTER OF CIRCLE

FiGure 57-3

ConstruCtion 2 to ConstruCt a PerPendiCular to a line segment at a given Point on the line segment Required: Construct a perpendicular at point O on line segment AB.

c Procedure ●●

●●

●●

With given point O as a center, and with a radius of any convenient length, draw arcs intersecting AB at points C and D as in Figure 57-4(a). With C as a center, and with a radius greater than OC, draw an arc. With D as a center, and with the same radius used at C, draw an arc that intersects the first arc at E as shown in Figure 57-4(b). Draw a line connecting point E and point O. Line EO in Figure 57-4(c) is perpendicular to line AB at point O. E

E

A

C

O (a)

D

B

A

C

O (b)

D

B

A

C

O

D

B

(c)

FiGure 57-4

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Section 5

Fundamentals oF Plane Geometry

Practical application A triangular piece is to be scribed and cut. As shown in Figure 57-5, the piece is laid out as follows: The 22-inch base is measured and marked off. 7 The 10 -inch distance is measured and 64 marked off at point A on the baseline.

B 1

7 –2

From point A, a perpendicular to the baseline is constructed.

10

A

7 64

The construction lines are shown. 22 1 FiGure 57-5 The 7 -inch distance is measured and 2 marked off at point B on the constructed perpendicular. Lines are scribed connecting vertex B with the endpoints of the baseline.

ConstruCtion 3 to ConstruCt a line Parallel to a given line at a given distanCe Required: Construct a line parallel to line AB at a given distance of 1 inch.

c Procedure ●● ●●

●●

Set the compass to the required distance (1 inch) as in Figure 57-6(b). With any points C and D as centers on AB, draw arcs with the given distance (1 inch) as the radius as shown in Figure 57-6(c). Draw a line, EF, that touches each arc at one point (the tangent point). Line EF in Figure 57-6(d) is parallel to line AB and EF is 1 inch from AB.

E

A

B (a)

0

1

A

C

D

(b)

B

F

A

C

D

B

(d)

(c)

FiGure 57-6

Practical application The cutout shown in the drawing in Figure 57-7 is laid out on a sheet as follows:

40

40

140 CL

CL

40

310

40

FiGure 57-7

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Unit 57

423

Fundamental Geometric constructions

All dimensions are in millimeters. The centerline (CL) is scribed and the 310-mm distance is marked off. Points A and B are the endpoints of the 310-mm segment.

E CL

F

J

From points A and B perpendiculars are constructed. The perpendiculars are extended more than 70 mm (140 mm 4 2) above and below AB (see Figure 57-8).

A

C

D

K

B

G

CL

H

FiGure 57-8

With points C and D as centers on AB, 70-mm radius arcs are drawn above and below AB. A line is scribed above and a line is scribed below AB touching the pairs of arcs. The lines are extended to intersect the perpendiculars constructed. The points of intersection are E, F, G, and H. From point A and from point B on AB, 40-mm distances are marked off. Point J and point K are the endpoints. Lines are scribed connecting J to E and G and connecting K to F and H. Scribed figure JEFKHG is the required cutout.

ConstruCtion 4

to BiseCt a given angle

Required: Bisect /ABC.

c Procedure ●●

●●

●●

With point B as the center, draw an arc intersecting sides BA and BC at points D and E as in Figure 57-9(a). With D as the center and with a radius equal to more than half the distance DE, draw an arc. With E as the center, and with the same radius, draw an arc. The intersection of the two arcs is point F as shown in Figure 57-9(b). Draw a line from point B to point F. Line BF in Figure 57-9(c) is the bisector of /ABC. A

A

A

D

D

D

F

F

B E

C

(a)

B E

C

B E

(b)

C

(c)

FiGure 57-9

Practical application 1 The centers of the three -inch-diameter holes in the mounting plate 2 1 shown in Figure 57-10 are located and center punched. Two -inch 4 diameter holes are located by constructing the bisector of /ABC as 3 1 shown and marking and center punching the 1 -inch and 4 -inch 8 4 hole center locations on the bisector.

3 HOLES 1 – 2 DIA

2 HOLES 1 – 4 DIA

A 110° 55°

B

1

4 –4 C

3 1 –8

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Section 5

Fundamentals oF Plane Geometry

ConstruCtion 5 to ConstruCt tangents to a CirCle from an outside Point Required: Construct tangents to given circle O from given outside point P.

c Procedure ●●

●●

●●

Draw a line segment connecting center O and point P. Bisect OP. Point A is the midpoint of OP as in Figure 57-11(a). With point A as the center and AP as a radius, draw arcs intersecting circle O at points B and C as shown in Figure 57-11(b). Points B and C are tangent points. Connect points B and P, and C and P. Line segments BP and CP in Figure 57-11(c) are tangents. B

A O

B A

P

A

P

O

P

O

C

C (b)

(a)

(c)

FiGure 57-11

Practical application A piece is to be made as shown in the drawing in Figure 57-12. All dimensions are in millimeters. The piece is laid out as follows and shown in Figure 57-13. A baseline is scribed and AB (170 mm) is marked off. Distance OA (152 mm) is set on dividers and, with OA as the radius, an arc is scribed. Distance OB (104 mm) is set on dividers and, with OB as the radius, an arc is scribed to intersect with the OA-radius arc. The intersection of the arcs locates center O of the 42-mm radius circle. Dividers are set to the 42-mm radius dimension, and the circle is scribed from center O. Tangents to the circle from points A and B are constructed resulting in tangent points C and D and tangent line segments AC and BD. The piece is now laid out and ready to be cut to the scribed lines. 42 RADIUS 152

C O

O

D

104

A 170

FiGure 57-12

98310_sec_05_Unit_57-58_ptg01.indd 424

B

A

B

FiGure 57-13

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Unit 57

425

Fundamental Geometric constructions

ConstruCtion 6 to divide a line segment into a given numBer of equal Parts Required: Divide line segment AB into three equal parts.

c Procedure ●●

●●

●●

●●

From point A, draw line AC forming any convenient angle with AB as in Figure 57-14(a). On AC, with a compass, lay off any three equal segments, AD, DE, and EF (Figure 57-14(a)). Connect point F with point B as in Figure 57-14(b). With centers at points F, E, and D, draw arcs of equal radii. The arc with a center at point F intersects AC at point G and BF at point H. Set distance GH on the compass and mark off this distance on the other two arcs. The points of intersection are K and M. Connect points E and K, and D and M, extending the lines past AB. Line AB in Figure 57-14(c) is divided into three equal segments; AP 5 PS 5 SB.

Note: Line segment AB can be divided into any required number of equal segments by laying off the required number of equal segments on AC and following the procedure given. A

B

M

A

B

M

A

K D

D

H E

E F

F

D

H

C

(b)

(a)

B

K

E

G

C

S

P

G

F

C

(c)

FiGure 57-14

Practical application 11 inches. Since six holes are 16 11 required, there will be five equal spaces between holes. Dividing 2 inches by 5 results in 16 8.6 fractional distances that are difficult to accurately measure or transfer, such as inch, 16 17.2 34.4 inch, or inch. By careful construction, the hole centers are accurately located as 32 64 shown in Figure 57-15. Six holes are to be equally spaced within a distance of 2

11

2 16

FiGure 57-15

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Section 5

Fundamentals oF Plane Geometry

ApplicAtion tooling Up 1. If /A 5 308 and BD 5 588269, find the measure of CD. D A B C

2. Determine the circumference of a circle with a radius of 4.250. Round the answer to 2 decimal places. 3. Determine the size of / ACB.

A

42°529

42°529

B

C

4. Determine the size of /2, /3, and /4.

3 2 23°159 4

5. A carbide milling cutter is used for machining a 27.25-inch length of stainless steel. The cutting time is 12.35 minutes, the cutting speed is 240 feet per minute, and the feed is 0.01 inches per revolution. What is the diameter of the carbide milling cutter? Round the answer to 2 decimal places. 6. Solve 4y 2 6 5 9y 1 28 for y.

construction 1 and 2 Applications Show construction lines and arcs for each of these exercises. 7. Trace each line segment in Exercises a through d and construct perpendicular bisectors to each segment. (a)

(b)

(c) P

M

(d) S

N

T

O

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Q

R

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Unit 57

427

Fundamental Geometric constructions

8. Trace each line in Exercises a through c and construct perpendiculars to each line at the given points on the lines. (a)

(b)

(c) G C

A

D E

B

F

9. With a compass, draw a circle 2 inches in diameter. By construction, locate the center of the circle. 10. Lay out a figure as follows: 1 a. Draw a horizontal line and mark off a distance of 2 inches. Label the left endpoint of the 2 1 2 -inch line segment point A and label the right endpoint point D. 2 b. From point A and above point A, construct a perpendicular to AD. Mark off a distance of 7 1 inches on the perpendicular from point A. Label the top endpoint point B. 8 c. From point B and to the right of point B, construct a perpendicular to AB. Mark off a distance 1 of 2 inches on the perpendicular from point B. Label the right endpoint point C. 2 d. From point C and below point C, construct a perpendicular to BC. Mark off a distance of 7 7 1 inches on the perpendicular from point C. If your constructions are accurate, the 1 -inch 8 8 distance marked off coincides with point D. What kind of a figure is formed by this construction?

construction 3 and 4 Applications Show construction lines and arcs for each of these exercises. 11. Trace each of the lines in Exercises a through d and construct a line parallel to each line at a 1 distance of 1 inches. 2 (a)

(b)

a

(c)

(d)

c

b

d

12. Trace each of the angles a through c and construct a bisector to each. (a)

(c)

(b)

b a

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c

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Section 5

Fundamentals oF Plane Geometry

13. Lay out the following angles by construction. Check the angle with a protractor but do not lay out angles with a protractor. 18 38 a. 458 b. 228309 c. 678309 d. 157 e. 168 2 4 14. Lay out the plate shown. Make the layout full size using construction methods. Use a protractor only for checking. All dimensions are in inches. 3 8

DIA, 6 HOLES

7

18

6

1

11°159

24

67°309

45°

3

18 1

12

5 10

15. Lay out the gage shown. Make the layout full size using construction methods. Use a protractor only for checking. All dimensions are in inches. 1.00 R 0.75

45°

1.50 2.00 2.00

6.25

2.50 0.75

1.50 R

22°309 5.00 8.75

construction 5 and 6 Applications Show construction lines and arcs for each of these exercises. 16. Trace each circle and point in Exercises a through c and construct tangents to the circles from the given points. (a)

(b)

(c)

B1 B2

A

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C1

C2

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Unit 57

429

Fundamental Geometric constructions

17. Trace each line segment of Exercises a, b, and c. Divide the given lines into the designated number of segments by means of construction. (a) M

P

(b)

N

(c) Q

3 SEGMENTS O

R

5 SEGMENTS

4 SEGMENTS

18. Lay out the template shown. Make the layout full size using construction methods. All dimensions are in inches. 5

1 8 RADIUS 3

54

1

42

5

19. Lay out the cutout shown. Make the layout full size using construction methods. All dimensions are in inches. 15

116 RADIUS 1

78 1

4 16

3

68

15

116 RADIUS

20. Trace the plate shown on next page. Lay out three sets of holes by construction. Follow the given directions. Directions: 3 7 ●● Bisect /A and construct four equally spaced -inch diameter holes. Make the first hole inch from point A 16 8 7 and the last hole 2 inches from point A. 16 1 3 ●● Bisect /B and construct eight equally spaced -inch diameter holes. Make the first hole inch from point B 4 4 11 and the last hole 3 inches from the first hole. 16

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430

●●

Section 5

Fundamentals oF Plane Geometry

3 9 Bisect /C and construct four equally spaced -inch diameter holes. Make the first hole inch from point C and 16 16 11 the last hole 2 inches from point C. 16

C

A

B

UNIT 58 Achievement Review—Section Five Objective You should be able to solve the exercises in this Achievement Review by applying the principles and methods covered in Units 50–57.

1. Add, subtract, multiply, or divide each of the following exercises as indicated. a. 378189 1 868239 b. 388469 1 238439 c. 1368369280 2 948179150 d. 588149 2 448589

e. 4s278239d f. 3s78239430d g. 878 4 2 h. 1038209 4 4

2. Determine / A.

3. Given: The sum of all angles 5 7208009000 /3 5 /4 5 /5 5 /6. /1 5 /2 5 688429180. Determine /3.

81°559

9 °17

24

46°209

A A A

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A

3

4 2

1 5

6

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431

achievement review—section Five

Unit 58

4. Express 68.858 as degrees and minutes. 5. Express 64.14208 as degrees, minutes, and seconds. 6. Express 378239 as decimal degrees to 2 decimal places. 7. Express 1038389430 as decimal degrees to 4 decimal places. 8. Using a simple protractor, measure each of the angles, /1 through /7, to the nearer degree. It may be necessary to extend sides of angles. /1 5 2

/2 5

3

4

/3 5

5

7

/4 5

6

/5 5

1

/6 5 /7 5

9. Write the values of the settings shown in the following vernier protractor scales. a. b. c. 0

10

60

30

10

0

20

30

30

60

20

30

40

60

30

10

0

30

0

60

30

20

10

60

0

30

40

30

50

60

10. Write the complement of each of the following angles. a. 678

b. 178419

c. 548479530

11. Write the supplement of each of the following angles. a. 418

b. 998329

c. 1038039270

12. Given: AB i CD and EF i GH. Determine the value of each angle, /1 through /10, to the nearer minute. E /1 5

G

6

/2 5

A

/3 5

5

10

/4 5 /6 5

2

/10 5

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8

C

9 1

/8 5 /9 5

3

81°009

/5 5 /7 5

B

4

D

142°009 94°409

F

7 H

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432

Section 5

13. a. Determine: (1) /1 (2) Side a

Fundamentals oF Plane Geometry

1.30 ft

1.30 ft 1

39°439 a 2.00 ft

b. Determine: (1) /1 (2) Side b (3) Side c

b 1 30°009 30°009 9.600 c

c. Determine: (1) /1 (2) /2

14.00 cm

35.00°

1

2 14.00 cm

14. a. Given: a 5 8.4000 and b 5 9.2000. Find c. b. Given: b 5 90.00 mm and c 5 150.00 mm. Find a.

c a 90°

b

15. Compute /1. 31° 77° 1 38° 80°

16. Determine the circumference of a circle that has a 5.360-inch radius. Round the answer to 3 decimal places.

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17. Determine the diameter of a circle that has a 360.00-millimeter circumference. Round the answer to 2 decimal places. 18. a. Given: CD 5 184 mm and CCE 5 118 mm.

E

C

CD. Determine CF and C

F

CF 5

D

CC D5

C b. Given: FD 5 26 mm and C D 5 78 mm. Determine CD and C ED. CD 5

C ED 5

19. a. Given: EB 5 5.1500.

3.0000

Determine AE. b. Given: AE 5 4.2000.

8.6000

C

Determine AB.

C 20. Given: Points A and E are tangent points. EB is a diameter. A E 5 1568, C C CE 5 1408, and ED 5 608. Determine angles /1 through /10.

E

A

B

O

D

/1 5 /2 5 /3 5

A

P

B

1

/4 5 /6 5

3

/7 5

10

4

/9 5

E

/10 5

9

O

6

F

/8 5

C

7

2

/5 5

D

8 5

C 21. a. Given: A C 5 1108 and r 5 4.7000.

C Compute arc length A C to 3 decimal places.

C B

b. Given: Arc length ABC = 478.60 mm and r 5 105.00 mm.

1

Compute /1 to 2 decimal places.

r O

A D

DIA M

22. a. Given: Dia H 5 14.5200 and d 5 8.3000. Compute Dia M. b. Given: Dia M 5 36.9000, e 5 15.8400, and d 5 12.6200.

f DIA H

DIA T

Compute f. A

B

d

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e

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Fundamentals oF Plane Geometry

Section 5

23. Given: /CAD 5 388, /BEC 5 408, AC 5 1308, and CE 5 1348. Determine angles /1 through /10. /1 5 E /2 5 /3 5 3 2 /4 5 D 9 1 /5 5 6 4 /6 5 7 8 A /7 5 /8 5 5 /9 5 10 C /10 5 B

24. a. Given: x 5 3.60 inches and y 5 5.10 inches. Compute Dia A to 2 decimal places. b. Given: Dia A 5 8.76 inches and x 5 10.52 inches. Compute t to 2 decimal places. a. b. DIA B

t

x y

DIA A

25. Points A and C are tangent points, DC is a diameter, AC 5 1168, EC 5 1408, EF 5 648, and CH 5 428. Determine angles /1 through /10. /1 5 /2 5 /3 5 /4 5 /5 5 /6 5 /7 5 /8 5 /9 5 /10 5

8

F

9

10

5

E 7

O 4

H

2

3

D 1 A

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C

6

B

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26. A flat is cut on a circular piece as shown. Determine the distance from the center of the circle to the flat, dimension x. 26.8 mm DIA

x

21.5 mm

27. A spur gear is shown. Pitch circles of spur gears are the imaginary circles of meshing gears that make contact with each other. A pitch diameter is the diameter of a pitch circle. Circular pitch is the length of the arc measured on the pitch circle between the centers of two adjacent teeth. Determine the circular pitch of a spur gear that has 26 teeth and a pitch diameter of 4.1250 inches. Express the answer to 4 decimal places. Pitch Circle

Circular Pitch Pitch Diameter

28. Determine the arc length from point C to point D on the template shown. 50.00 mm RADIUS D 70.00°

C

67.00 mm 96.00 mm

29. In the layout shown, points E, F, and G are tangent points. Determine lengths OA, OB, and OC. OA 5 C

20.00 cm Radius

E 87.00 cm

OB 5 OC 5

F O

A

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68.00 cm 98.00 cm

G

B

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Fundamentals oF Plane Geometry

30. Determine dimension x to 3 decimal places. 1.900 in. T

S

1.200 in.

x P

E 0.614 in.

O

M

2.620 in. DIA

31. Refer to the drill jig shown. Determine /1. 83°259

51°109 1

90° 36°329

150°189

32. Circle O has a diameter AB of 14.50 inches and a chord CD of 8.00 inches. Chord CD is perpendicular to diameter AB. How far is chord CD from the center O of the circle? Express the answer to 2 decimal places.

Note: It is helpful to sketch and label this exercise. 33. Refer to the figure shown. Determine dimension x. x 6.7 in. T

O

M

S

1.4 in. P 8.6 in. DIA

3.8 in.

34. Lay out the template shown. Make the layout full size using construction methods. Do not use a protractor. All dimensions are in inches. 1 2

1 – RADIUS 6

1 2

2–

1 8

4–

67°309 3 7– 8

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6

sectiON siX

Geometric Figures: Areas and Volumes UNIT 59 Areas of Rectangles, Parallelograms, and Trapezoids

Objectives After studying this unit you should be able to ●● ●● ●● ●● ●● ●● ●●

Express given customary area measures in larger and smaller units. Express given metric area measures in larger and smaller units. Convert between customary area measures and metric area measures. Compute areas, lengths, and widths of rectangles. Compute areas, bases, and heights of parallelograms. Compute areas, both bases, and heights of trapezoids. Compute areas of more complex figures (composite figures) that consist of two or more common polygons.

As previously stated, in machine technology linear or length measure is used more often than area and volume measure. However, the ability to compute areas and volumes is required in determining job-material quantities and costs. Often, before a product is manufactured, part weights are computed. Volumes of simple geometric figures and combinations of figures (composite figures) must be calculated before weights can be determined. Section 6 presents area and volume measure of two-dimensional and three-dimensional geometric figures and practical area and volume applications.

Customary and metriC units of surfaCe measure (area) A surface is measured by determining the number of surface units contained in it. A surface is two dimensional. It has a length and width, but no thickness. Both length and width must be expressed in the same unit of measure. Area is computed as the 437

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Section 6

1 ft = 12 in.

AREA = 1 sq ft 144 sq in.

1 in.

1 in. 1 sq in.

1 ft = 12 in.

product of two linear measures and is expressed in square units. For example, 2 inches 3 4 inches 5 8 square inches (8 sq in. or 8 in.2). The surface enclosed by a square that is 1 inch on a side is 1 square inch (1 sq in. or 1 in.2). The surface enclosed by a square that is 1 foot on a side is 1 square foot (1 sq ft or 1 ft2). The reduced drawing in Figure 59-1 shows a square inch and a square foot. Observe that 1 linear foot 5 12 linear inches, but 1 square foot 5 12 inches 3 12 inches or 144 square inches. The following table has the common customary units of surface measure that might be used in a machine shop. CUSTOMARY UNITS OF AREA MEASURE

FiGure 59-1

1 square foot (sq ft or ft2) 5 144 square inches (sq in. or in.2) 1 square yard (sq yd or yd2) 5 9 square feet (sq ft or ft2) 1 square yard (sq yd or yd2) 5 1296 square inches (sq in. or in.2)

Expressing Customary Area Measure Equivalents To express a given customary unit of area as a larger customary unit of area, either divide the given area by the number of square units contained in one of the smaller units or multiply by a unit ratio.

Example Express 728 square inches as square feet. METHOD 1 Since 144 sq in. 5 1 sq ft, divide 728 by 144. 728 4 144 < 5.06; 728 sq in. < 5.06 sq ft Ans

METHOD 2

1 sq ft . 144 sq in. 1 sq ft 728 sq ft 728 sq in. 3 5 < 5.06 sq ft 144 sq in. 144 The appropriate unit ratio is

Ans

To express a given customary unit of area as a smaller customary unit of area, either multiply the given area by the number of square units contained in one of the larger units or multiply by a unit ratio.

Example Express 1.612 square yards as square inches. Two methods could be used.

Method 1 Multiply 1.612 square yards by the two unit ratios 1.612 sq yd 3

9 sq ft 144 sq in. 3 < 2089 sq in. 1 sq yd 1 sq ft

Method 2 Use the unit ratio yards by this unit ratio. 1.612 sq yd 3

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9 sq ft 144 sq in. and . 1 sq yd 1 sq ft

Ans

1296 sq in. . Here you would multiply 1.612 square 1 sq yd

1296 sq in. ≈ 2089 sq in. 1 sq yd

Ans

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metric units of surface measure (area) The method of computing surface measure is the same in the metric system as it is in the customary system. The product of two linear measures produces square measure. The only difference is that metric units are used rather than customary units. For example, 2 millimeters 3 4 millimeters 5 8 square millimeters. Surface measure symbols are expressed as linear measure symbols with an exponent of 2. For example, 4 square meters is written as 4 m2, and 25 square millimeters is written 25 mm2. The basic metric unit of area is the square meter. The surface enclosed by a square that is 1 meter on a side is 1 square meter (1 m2). The surface enclosed by a square that is 1 millimeter on a side is 1 square millimeter (1 mm2). One linear meter 5 1000 linear millimeters, but 1 square meter (1 m2) 5 1000 millimeters 3 1000 millimeters, or 1 000 000 square millimeters (1 000 000 mm2). Also, since 1 linear meter 5 100 linear centimeters, 1 square meter 5 100 centimeters 3 100 centimeters or 10 000 square centimeters (10 000 cm2). To make numbers easier to read they may be put into groups of three, separated by spaces (or thin spaces), as in 12 345, but not commas or points. This applies to digits on both sides of the decimal marker (0.901 234 56). Numbers with four digits may be written either with the space (5 678) or without it (5678). This practice not only makes large numbers easier to read but also allows all countries to keep their custom of using either a point or a comma as decimal marker. For example, engine size in the United States is written as 3.2 L and in Germany as 3,2 L. The space prevents possible confusion and sources of error. The following table has the common metric units of surface measure that might be used in a machine shop. METRIC UNITS OF AREA MEASURE 1 square millimeter (mm2) 5 0.000 001 square meter (m2) 1 square millimeter (mm2) 5 0.01 square centimeter (cm2) 1 square centimeter (cm2) 5 0.000 1 square meter (m2) 1000 000 square millimeters (mm2) 5 1 square meter (m2) 100 square millimeters (mm2) 5 1 square centimeter (cm2) 10 000 square centimeters (cm2) 5 1 square meter (m2)

Expressing Metric Area Measure Equivalents To express a given metric unit of area as a larger metric unit of area, divide the given area by the number of square units contained in the smaller unit or multiply by the appropriate unit ratio.

Example Express 840.5 square centimeters as square meters. METHOD 1 Since 10 000 cm2 5 1 m2, divide 840.5 by 10 000. 840.5 4 10 000 5 0.08405; 840.5 cm2 < 0.08 m2 Ans

METHOD 2

1 m2 . 10 000 cm2 1 m2 840.5 m2 840.5 cm2 3 5 50.084 05 m2 < 0.08 m2 2 10 000 cm 10 000 The appropriate unit ratio is

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Ans

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To express a given metric unit of area as a smaller metric unit of area, multiply the given area by the number of square units contained in one of the larger units.

Example Express 47.6 square centimeters (cm2) as square millimeters (mm2). From the table, we see that 1 cm2 5 100 mm2. 100 mm2 47.6 cm2 3 5 47.6 3 100 mm2 5 4760 mm2; 47.6 cm2 5 4760 mm2 1 cm2

Ans

Conversion Between metric and Customary systems In technical work it is sometimes necessary to change from one measurement system to the other. Use the following metric-customary conversions for the area of an object. Since 1 inch is defined to be 2.54 cm, the conversion between square inches and square centimeters is exact. The other conversions are approximations. METRIC-CUSTOMARY AREA CONVERSIONS 1 square inch (sq in. or in.2) 5 6.4516 cm2 1 square foot (sq ft or ft2) < 0.0929 m2 1 square yard (sq yd or yd2) < 0.8361 m2

Example Convert 12.75 ft2 to square centimeters. Since 12.75 ft2 is to be expressed in square centimeters, multiply by the unit ratios 0.0929 m2 10 000 cm2 and . 2 1 ft 1 m2 0.0929 m2 10 000 cm2 12.75 ft2 512.75 ft2 3 3 511 844.75 cm2, 11 840 cm2 Ans 1 ft2 1 m2

areas of reCtangles A rectangle is a four-sided polygon with opposite sides equal and parallel and with each angle equal to a right angle. The area of a rectangle is equal to the product of the length and width. A 5 lw where A 5 area l 5 length w 5 width

Example 1 A rectangle cross-section of a steel bar is 24 millimeters long and 14 millimeters wide. Find the cross-sectional area of the bar. A 5 lw A 5 24 mm 3 14 mm A 5 336 mm2 Ans

Example 2 A metal stamping is shown in Figure 59-2. All dimensions are in inches. Determine the area of the stamping. Divide the figure into rectangles. One way of dividing the figure is shown in Figure 59-3. To find the cross-sectional area, compute the area of each rectangle and add the three areas. ●● Area of rectangle 1 A 5 10.500 in. 3 4.000 in. A 5 42 sq in. ●● Area of rectangle 2 Length 5 10.500 in. 1 7.500 in. 5 18.000 in. Width 5 9.625 in. 2 4.000 in. 5 5.625 in. A 5 18.000 in. 3 5.625 in. 5 101.25 sq in.

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Area of rectangle 3 A 5 11.190 in. 3 4.500 in. 5 50.355 sq in.

●●

Total cross-sectional area 42 sq in. 1 101.25 sq in. 1 50.355 sq in. < 193.6 sq in. Ans (rounded) 4.500

9.625

5.625

10.500

10.500

1

2

18.000

11.190

11.190

3

4.000

7.500

4.500

4.000

FiGure 59-2

FiGure 59-3

areas of Parallelograms A parallelogram is a four-sided polygon with opposite sides parallel and equal. The area of a parallelogram is equal to the product of the base and height. An altitude is a segment perpendicular to the line containing the base drawn from the side opposite the base. The height is the length of the altitude. Since a rhombus is a parallelogram with all four sides the same length, these formulas also apply to the rhombus. A 5 bh

where A 5 area b 5 base h 5 height

D

A

C

E

B

FiGure 59-4 D

In Figure 59-4, AB is a base, and DE is a height of parallelogram ABCD.

C

Area of parallelogram ABCD 5 AB(DE) In Figure 59-5, BC is a base, and DF is a height of parallelogram ABCD. Area of parallelogram ABCD 5 BC(DF)

F A

B

FiGure 59-5

Example 1 What is the area of a parallelogram with a 152.3-millimeter base and a 40.5-millimeter height? A 5 152.3 mm 3 40.5 mm 5 6170 mm2 Ans (rounded)

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Example 2 A drawing of a baseplate is shown in Figure 59-6. The plate is made of number 2 gage (thickness) aluminum, which weighs 3.4 pounds per square foot. Find the weight of the plate to the nearest tenth pound. All dimensions are in inches. 14.00

10.00

6.00

47°

47°

6.00

30.00

FiGure 59-6

The area of the plate must be found. By studying the drawing, one method for finding the area is to divide the figure into a rectangle and a parallelogram as shown in Figure 59-7. 14.00

4.00 6.00

30.00

FiGure 59-7 ●●

●●

●●

Find the area of the rectangle. A 5 14.00 in. 3 4.00 in. A 5 56 sq in. Find the area of the parallelogram. A 5 30.00 in. 3 6.00 in. A 5 180 sq in. Find the total area of the plate. Total area 5 56 sq in. 1 180 sq in. 5 236 sq in.

Compute the weight. ●● Find the area in square feet. Since 1 foot equals 12 inches, 1 square foot equals 12 inches squared. (12 in.)2 5 144 in.2 or 144 sq in. 1 sq ft 5 144 sq in. 236 sq in. 4 144 sq in. < 1.64, 1.64 sq ft Weight of plate < 1.64 sq ft 3 3.4 lb/sq ft < 5.6 lb Ans

areas of traPezoids A trapezoid is a four-sided polygon that has only two sides parallel. The parallel sides are called bases. The area of a trapezoid is equal to one-half the product of the height and the sum of the bases. 1 A 5 h(b1 1 b2) where A 5 area 2 h 5 height b1 and b2 5 bases

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In Figure 59-8, DE is the height, and AB and DC are the bases of trapezoid ABCD.

D

C

1 Area of trapezoid ABCD 5 DE (AB 1 DC) 2

Example 1 What is the area of a trapezoid that has bases of 7.000 inches and 3.800 inches and a height of 4.200 inches? 1 A 5 (4.200 in.)(7.000 in. 1 3.800 in.) 2 1 A 5 (4.200 in.)(10.800 in.) 2 A 5 22.68 sq in. Ans

A

E

B

FiGure 59-8

Example 2 The area of a trapezoid is 376.58 square centimeters. The height is 16.25 centimeters, and one base is 35.56 centimeters. Find the other base. Substitute values in the formula for the area of a trapezoid and solve. 1 376.58 cm2 5 (16.25 cm)(35.56 cm 1 b2) 2 2 376.58 cm 5 8.125 cm (35.56 cm 1 b2) 376.58 cm2 5 288.925 cm2 1 8.125 cm(b2) 87.655 cm2 5 8.125 cm(b2) b2 < 10.79 cm Ans

Example 3 Find the area of the template shown in Figure 59-9. All dimensions are in inches. Express the answer to 1 decimal place. Divide the template into simpler figures. One way is to divide the template into two rectangles and a trapezoid as shown in Figure 59-10. ●●

●●

●●

●●

Find area 1 (rectangle). A 5 4.20 in. 3 2.00 in. A 5 8.40 sq in. Find area 2 (trapezoid). Height 5 8.30 in. 2 6.50 in. 5 1.80 in. First base 5 6.80 in. Second base 5 2.00 in. 1 A 5 (1.80 in.)(6.80 in. 1 2.00 in.) 2 1 A 5 (1.80 in.)(8.80 in.) 2 A 5 7.92 sq in. Find area 3 (rectangle). A 5 6.50 in. 3 6.80 in. A 5 44.2 sq in. Find the total area of template. Total Area 5 8.40 sq in. 1 7.92 sq in. 1 44.2 sq in. Total Area < 60.5 sq in. Ans (rounded)

8.30 6.50 4.20 6.80

2.00 4.20 6.50 8.30

FiGure 59-9

2.00

1

2

3

1.80

6.50

6.80

4.20

FiGure 59-10

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ApplicAtion tooling Up 1. Trace this line segment two times. On one copy construct a perpendicular bisector of the segment. On the other copy divide the segment into three equal segments.

2. Find the length of x. Round the answer to 2 decimal places. 15 mm DIA

x

5 mm DIA 10 mm DIA 20 mm DIA

3. What is the diameter of a circle with a circumference of 425.5 cm? Round the answer to 2 decimal places. 4. Determine the size of /BDC and /ABD. A

358

1018

C

218

D 178

B

5. What is the module of a gear with a working depth of 17 millimeters? 5 13 6. Solve H 2 32 5 41 . 8 16

equivalent customary Units of Area Measure Express each area as indicated. Round each answer to the same number of significant digits as in the original quantity. 7. 196 square inches as square feet 8. 1085 square inches as square feet 9. 45.8 square feet as square yards

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10. 2.02 square feet as square yards 11. 2300 square inches as square yards 12. 0.624 square foot as square inches 13. 4.30 square yards as square feet 14. 0.612 square yard as square inches

equivalent Metric Units of Area Measure Express each area as indicated. Round each answer to the same number of significant digits as in the original quantity. 15. 500 square millimeters as square centimeters 16. 2470 square millimeters as square centimeters 17. 38 250 square centimeters as square meters 18. 7520 square centimeters as square meters 19. 2.3 square meters as square millimeters 20. 5.74 square centimeters as square millimeters 21. 0.902 square centimeters as square millimeters 22. 0.0075 square meters as square millimeters

conversion Between Metric and customary Units of Area Measure Express each area as indicated. Round each answer to the same number of significant digits as in the original quantity. 23. 18.5 square feet as square centimeters 24. 47.75 square inches as square millimeters 25. 3.9 square yards as square meters 26. 1.20 square feet as square millimeters 27. 680 square millimeters as square inches 28. 370.8 square inches as square centimeters 29. 18.75 m2 as square feet 30. 18.75 m2 as square yards

Areas of Rectangles Find the unknown area, length, or width for each of the rectangles 31 through 42. Where necessary, round the answers to 1 decimal place. Length

Width

31.

8.5 in.

6.0 in.

37.

32.

5.0 m

9.0 m

38.

33.

2.6 in. 23 mm

34. 35. 36.

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0.4 m 0.086 in.

Area

Length

Width

26.2 mm 39.8 in.

11.7 sq in.

39.

64.2 mm

200.1 mm2

40.

2.95 in.

0.2 m2

41.

7.4 ft

0.136 sq in.

42.

Area 366.8 mm2 31.7 sq in. 3762 mm2

0.76 in. 6.7 sq ft 125.0 ft

26,160 sq ft

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Solve these exercises. Round the answers to 2 decimal places unless otherwise specified. 43. A rectangular strip is 9.00 inches wide and 6.50 feet long. Find the area of the strip in square feet. 44. A sheet metal square contains 729 square inches. Find the length of each side of the square sheet. 45. The cost of a rectangular plate of aluminum 3900 wide and 4900 long is $45.00. Find the cost of a rectangular plate 6900 wide and 8900 long, using the same stock. 46. Find the area of the sheet metal piece shown.

29.00 in.

36.00 in. 24.00 in. 10.00 in.

25.00 in. 14.00 in.

47. The support base shown has an area of 6350 square millimeters. Determine dimension x to the nearest tenth millimeter.

74.0 mm

81.0 mm x 94.0 mm

48. The rectangular cross-sectional area of a metal brace is to have an area of 2250 square millimeters. The length is to be one and one-half times the width. Compute the length and width dimensions to the nearest millimeter.

Areas of Parallelograms and composite Figures Find the unknown area, base, or height for each of the parallelograms 49 through 60. Where necessary, round the answers to 1 decimal place. Base

Height

49.

20.00 mm

5.20 mm

55.

50.

6.00 in.

9.80 in.

56.

51.

26.0 in. 37.4 mm

52. 53.

0.07 m

54.

24.0 in.

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Area

Height

18.5 in. 0.60 ft

57.

56.00 mm

6.80 mm

2057 mm

58.

17.00 in.

18.30 in.

2

2

38.0 mm

59. 60.

0.38 m

Area 312.5 sq in.

486.2 sq in.

0.014 m 4.50 in.

Base

5.1 sq ft

1887.2 mm2

0.266 m

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Solve these exercises. Round the answers to 2 decimal places unless otherwise specified. 61. The cross section of the piece of tool steel shown is in the shape of a parallelogram. Find the cross-sectional area. All dimensions are in inches.

1.250

0.500 1.500

62. Two cutouts in the shape of parallelograms are stamped in a strip of metal as shown. Segment AB is parallel to segment CD, and dimension E equals dimension F. Compare the areas of the two cutouts. A

B

C

D E

F

63. An oblique groove is cut in a block as shown. Before the groove was cut, the top of the block was in the shape of a rectangle. Determine the area of the top after the groove is cut. 35.0 mm

26.0 mm

14.0 mm

64. Find the area of the template shown. 85 mm 28°

28°

28° 290 mm

215 mm 100 mm

100 mm

310 mm

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Areas of trapezoids and composite Figures Find the unknown area, height, or base for each of the trapezoids 65 through 76. Where necessary, round the answers to 1 decimal place. Bases

Bases

Height (h)

b1

b2

65.

8.00 in.

16.00 in.

10.00 in.

66.

28.0 mm

47.0 mm

38.0 mm

8.00 ft

4.00 ft 5.5 ft

67. 68.

1.2 ft

69.

0.6 m

0.8 m 56.0 m

70.

Height (h)

b1

b2

71.

18.70 in.

36.00 in.

28.40 in.

72.

38.0 mm

64.0 sq ft

73.

0.1 mm

7.7 sq ft

74.

0.4 m

75.

0.3 m

0.8 m

738.4 m2

76.

14.00 in.

20.00 in.

Area (A)

2

48.00 m

8.7 mm 1.2 m

0.6 m

66.37 in.

43.86 in.

Area (A) 210.9 mm2

2125 sq in. 0.2 m2

3.200 in.

77. A cross section of an aluminum bar in the shape of a trapezoid is shown. All the dimensions are in inches. a. Find the cross-sectional area of the bar. b. Find the length of side AB. Round the answer to the nearest hundredth inch. 2.25 A

6.00 B 5.25

78. Determine the area of the metal plate shown. All dimensions are in inches. Round the answer to the nearest tenth square foot. 28.00 12.50

6.500

6.50 9.00

14.00

46.00 14.00

9.00

16.00 6.50 11.00 42.00

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79. An industrial designer decided that the front plate of an appliance should be in the shape of an isosceles trapezoid with an area of 0.420 square meters. To give the desired appearance, the lower base dimension is to be equal to the height dimension, and the upper base dimension is to be equal to three-quarters of the lower base dimension. Compute the dimensions of the height and each base. Round answers to the nearest tenth millimeter. 80. One of the examples showed how to find the area of the metal stamping in Figure 59-2. Another method is to find the area of the large rectangle and subtract the areas of the rectangles that are removed. In the following figure, the rectangles that will be removed are marked 1 and 2 . Use this subtraction method to determine the area of the stamping. 14.125 4.500

1

18.000

11.190 7.500

2

4.000

UNIT 60 Areas of Triangles Objectives After studying this unit you should be able to ●● ●● ●● ●●

Compute areas of triangles given the base and height. Compute areas of triangles given three sides. Compute bases and heights given triangle areas. Compute areas of more complex figures (composite figures) that consist of two or more common polygons.

Machined products are often in the shape of a triangle or can be divided into triangles, rectangles, parallelograms, and trapezoids.

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D

C

Geometric FiGures: AreAs And Volumes

areas of triangles given the Base and height In parallelogram ABCD shown in Figure 60-1, segment DE is the altitude to the base AB. Diagonal DB divides the parallelogram into two congruent triangles.

A

E

n ABD > nCDB

B

FiGure 60-1

Parallelogram ABCD and triangles ABD and CDB have equal bases and equal heights. The area of either triangle is equal to one-half the area of the parallelogram. The area of 1 parallelogram ABCD 5 AB(DE ). Therefore, the area of nABD or nCDB 5 AB sDEd. The area 2 of a triangle is equal to one-half the product of the base and height. 1 A 5 bh 2

where

A 5 area b 5 base h 5 height

Example Find the area of the triangle shown in Figure 60-2. 1 A 5 s22.0 mmd s19.0 mmd 2 A 5209 mm2 Ans

19.0 mm

22.0 mm

FiGure 60-2

areas of triangles given three sides Often, three sides of a triangle are known, but a height is not known. A height can be determined by applying the Pythagorean theorem and a system of equations. However, with a calculator, it is quicker and easier to compute areas of triangles, given three sides using a formula called Hero’s (or Heron’s) formula.

hero’s (heron’s) formula A 5 Ïs ss 2 ad ss 2 bd ss 2 cd

where A 5 area a, b, and c 5 sides 1 s 5 sa 1 b 1 cd 2

Example Refer to the flat triangular brace shown in Figure 60-3. a. Find the area of the brace. b. Find the height JK.

Solutions

10.20 in.

J 5.12 in.

a. Compute the area by using Hero’s formula. 1 7.84 in. K s 5 s7.84 in. 1 5.12 in. 1 10.20 in.d 2 FiGure 60-3 s 5 11.58 in. A 5 Ïs11.58 in.d s11.58 in. 2 7.84 in.d s11.58 in. 2 5.12 in.d s11.58 in. 2 10.20 in.d A 5 Ïs11.58 in.d s3.74 in.d s6.46 in.d s1.38 in.d A < Ï386.093 in.4 < 19.65 sq in. Ans

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1 b. Compute height JK from the formula A 5 bh. 2 1 19.65 sq in. < s10.20 in.d sJKd 2 19.65 sq in. < s5.10 in.d sJKd JK < 3.85 in. Ans

ApplicAtion tooling Up 1. Find the area of this figure. Round your answer to 2 decimal places.

3.5240

0.5240

0.4180

0.8340 150

2. Three circles are equally spaced on a bolt circle 8.4 cm in diameter. Use construction techniques to lay out the centers for the holes. 3. The four holes in the figure are 0.8 cm in diameter and equally spaced around the bolt circle. Find the distances x and y rounded to 2 decimal places.

20 cm

y

x

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Section 6

4. Plans call for triangle △ BCD to be sheared off a corner of a rectangle as shown in the figure. A decision is made to move one vertex of the triangle 1.5 in. from B to B9. In order for the new smaller triangle (△B9CD9) to be similar to the planned triangle, what is the length of CD9? 18 in. D9

D

E

17 in.

C

B9

6 in.

B

A

F

5. Add 458199420 1 218359260 1 548279310. 6. Solve 125x3 5 64.

Areas of triangles and composite Figures Find the unknown area, base, or height for each of the triangles 7 through 18. Where necessary, round the answers to 1 decimal place.

7.

Base

Height

21.0 mm

17.0 mm 6.0 ft

8. 9.

0.2 m 1.40 in.

10.

Area

Base

13. 78.2 sq ft

14.

0.02 m2

15.

3.22 sq in.

16.

11.

0.8 m

0.4 m

17.

12.

18.5 in.

7.25 in.

18.

Height

Area

30.5 mm

427 mm2 38.0 mm

17.0 in.

1919 mm2

9.8 in. 0.8 ft

16 sq ft

45.41 in.

249.7 sq in. 3.43 in.

1.76 sq in.

Given three sides, find the area of each of the triangles 19 through 26. Where necessary, round the answers to 1 decimal place. Side a

Side b

Side c

19.

4.00 in.

6.00 in.

8.00 in.

20.

2.0 ft

5.0 ft

21.

3.5 in.

22.

20.0 mm

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Side a

Side b

Side c

23.

3.2 ft

3.6 ft

0.8 ft

6.0 ft

24.

9.10 in.

30.86 in.

28.57 in.

4.0 in.

2.5 in.

25.

7.20 mm

10.00 mm

9.00 mm

15.0 mm

25.0 mm

26.

0.5 m

1.0 m

0.8 m

19. 20. 21. 22. 23. 24. 25. 26.

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Solve these exercises. 27. Find the cross-sectional area of metal in the triangular tubing shown. Round the answer to the nearest whole square millimeter.

33.3 mm 22.4 mm

25.8 mm 38.4 mm

28. Two triangular pieces are sheared from the aluminum sheet shown. After the triangular pieces are cut, the sheet is discarded. Find the number of square feet of aluminum wasted. Round the answer to 1 decimal place. 59-60 39-00 49-00

39-60

79-00 89-00

29. Determine the area of the figure shown. Round the answer to the nearest square millimeter. 55.3 mm

90° 68.0 mm 50.0 mm 90° 72.0 mm

30. Find the area of the template shown. All dimensions are in inches.

18.00 10.00 5.00

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31. The area of the irregularly shaped sheet metal piece in the following figure on the left is to be determined. The longest diagonal is drawn on the figure as shown in the figure on the right. Perpendiculars are drawn to the diagonal from each of the other vertices. The perpendicular segments are measured as shown. From the measurements, the areas of each of the common polygons are computed. This is one method often used to compute areas of irregular figures. Compute the area of the sheet metal piece. 25.00 cm

20.00 cm

62.00 cm

12.00 cm 23.00 cm A

B

18.00 cm 21.00 cm 13.00 cm

70.00 cm

UNIT 61 Areas of Circles, Sectors, and Segments Objectives After studying this unit you should be able to ●● ●● ●● ●●

Compute areas, radii, and diameters of circles. Compute areas, radii, and central angles of sectors. Compute areas of segments. Compute areas of more complex figures that consist of two or more simple figures.

Computations of areas of circular objects are often made when planning for the production of a product. Also, many industrial material-strength calculations are based on circular cross-sectional areas of machined parts.

areas of CirCles The area of a circle is equal to the product of p and the square of the radius. where

A 5 pr2

A 5 area r 5 radius p < 3.1416

The formula for the area of a circle can be expressed in terms of the diameter. Since the d radius is one-half the diameter, can be substituted in the formula for r. 2 A5p Since

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2

1 2 or A 5 p d4 d 2

2

p < 0.7854, the formula A < 0.7854 d 2 is often used. 4

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Example 1 Find the area of a circle that has a radius of 6.500 inches. Substitute the values in the formula and solve. A 5 pr2 < 3.1416 s6.500 in.d2 < 3.1416 s42.25 sq in.d < 132.7 sq in. Ans

Example 2 A circular hole is cut in a square metal plate as shown in Figure 61-1. The plate weighs 8.3 pounds per square foot. What is the weight of the plate after the hole is cut? All dimensions are in inches. Compute the area of the square. A 5 s10.30 in.d2 5 106.09 sq in.

7.00 DIA

Compute the area of the hole. A < 0.7854 s7.00 in.d2 < 38.48 sq in. Compute the area of the plate. A < 106.09 sq in. 2 38.48 sq in. < 67.61 sq in. Compute the weight of the plate. 67.61 sq in. 4 144 sq in./sq ft < 0.470 sq ft

10.30

Ans

10.30

FiGure 61-1

areas of seCtors A sector of a circle is a figure formed by two radii and the arc intercepted by the radii. To find the area of sector of a circle, first find the fractional part of a circle represented by the central angle. Then multiply the fraction by the area of the circle. A5

u spr2d 3608

where

A 5 area u 5 central angle p < 3.1416 r 5 radius

Example 1 A base plate in the shape of a sector is shown in Figure 61-2. Find the area of the plate to the nearest hundredth square foot. Find the area of the sector in square inches. A5

135.08 s3.1416ds16.25 in.d2 < 311.09 sq in. 3608

135.08 16.25 in. Radius

FiGure 61-2

Find the area in square feet. Since 1 sq ft 5 144 sq in., 311.09 4 144 < 2.16 5 2.16 sq ft Ans (rounded)

Example 2 Pieces in the shape of sectors are to be stamped from sheet stock. Each piece is to have an area of 1080 square millimeters and a central angle of 222 degrees. Compute the length of the straight sides (radii) of a piece. 222 s3.1416d r2 360 1080 mm2 < 1.93732r2 r2 < 557.471 mm2 r < 23.6 mm Length of straight sides < 23.6 mm Ans 1080 mm2 5

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areas of segments

B

A segment of a circle is a figure formed by an arc and the chord joining the end points of the arc. In the circle shown in Figure 61-3, the area of segment ACB is found by subtracting the area of triangle AOB from the area of sector OACB.

O C

Example Segment ACB is cut from the circular plate shown in Figure 61-4. Find the area of the segment. Find the area of the sector.

A

FiGure 61-3

A