Engineering Mechanics: Statics & Dynamics, Instructor’s Solutions Manual [14 ed.]

Table of contents :
M12_HIBB0000_14_ISM_C12_1-45
M12_HIBB0000_14_ISM_C12_46-114
M12_HIBB0000_14_ISM_C12_115-138)
M12_HIBB0000_14_ISM_C12_139-203
M12_HIBB0000_14_ISM_C12_204-235
M13_HIBB0000_14_ISM_C13_ok
M13_HIBB0000_14_ISM_C13_96-132
M16_HIBB0000_14_ISM_C16(1-78)
M16_HIBB0000_14_ISM_C16(79-152)

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Instructor’s Solutions Manual

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1–1. What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg?

Solution Ans.

(a)  W = 9.81(8) = 78.5 N (b)  W = 9.81(0.04) ( 10 - 3 ) = 3.92 ( 10 - 4 ) N = 0.392 mN

Ans.

(c)  W = 9.81(760) ( 10

Ans.

3

) = 7.46 ( 10 ) N = 7.46 MN 6

Ans: W = 78.5 N W = 0.392 mN W = 7.46 MN 1

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1–2. Represent each of the following combinations of units in the correct SI form: (a) >KN>ms, (b) Mg>mN, and (c) MN>(kg # ms).

Solution (a)  kN>ms = 103N> ( 10 - 6 ) s = GN>s

Ans.

(b)  Mg>mN = 106g>10 - 3 N = Gg>N

Ans.

(c)  MN>(kg # ms) = 10 N>kg(10 6

-3

s) = GN>(kg # s)

Ans.

Ans: GN>s Gg>N GN>(kg # s) 2

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1–3. Represent each of the following combinations of unit in the correct SI form: (a) Mg>ms, (b) N>mm, (c) mN>(kg # ms).

Solution Mg

(a)   

ms

103 kg 10-3 s

= 106 kg>s = Gg>s

Ans.

N 1N = = 103 N>m = kN>m mm 10-3 m

Ans.

mN 10-3 N = = kN>(kg # s) (kg # ms) 10-6 kg # s

Ans.

(b)   (c)   

=

Ans: Gg>s kN>m kN>(kg # s) 3

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*1–4. Convert: (a) 200 lb # ft to N # m, (b) 350 lb>ft3 to kN>m3, (c) 8 ft>h to mm>s. Express the result to three significant figures. Use an appropriate prefix.

SOLUTION a) (200 lb # ft) ¢

4.4482 N 0.3048 m ≤¢ ≤ = 271 N # m 1 lb 1 ft

Ans.

b) ¢

3 350 lb 1 ft 4.4482 N ≤ ¢ ≤ ¢ ≤ = 55.0 kN>m3 0.3048 m 1 lb 1 ft3

Ans.

c) ¢

8 ft 1h 0.3048 m ≤¢ ≤¢ ≤ = 0.677 mm>s 1h 3600 s 1 ft

Ans.

Ans: 271 N # m 55.0 kN>m3 0.677 mm>s 4

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1–5. Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, and (c) 0.00563 mg.

Solution (a)  45 320 kN = 45.3 MN

Ans.

(b)  568 ( 105 ) mm = 56.8 km

Ans.

(c)  0.00563 mg = 5.63 mg

Ans.

Ans: 45.3 MN 56.8 km 5.63 mg 5

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1–6. Round off the following numbers to three significant figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg.

SOLUTION a) 58.3 km

b) 68.5 s

c) 2.55 kN

Ans.

d) 7.56 Mg

Ans: 58.3 km 68.5 s 2.55 kN 7.56 Mg 6

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1–7.

Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000 431 kg, (b) 35.3(103) N, and (c) 0.005 32 km.

SOLUTION a) 0.000 431 kg = 0.000 431 A 103 B g = 0.431 g

Ans.

b) 35.3 A 103 B N = 35.3 kN

Ans.

c) 0.005 32 km = 0.005 32 A 103 B m = 5.32 m

Ans.

Ans: 0.431 g 35.3 kN 5.32 m 7

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*1–8. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm, (b) mN>ms, (c) mm # Mg.

SOLUTION a) Mg>mm = b) mN>ms =

103 kg -3

10 m 10 - 3 N 10 - 6 s

=

106 kg = Gg>m m

Ans.

=

103 N = kN>s s

Ans.

c) mm # Mg = C 10-6 m D

# C 103 kg D

= (10)-3 m # kg

= mm # kg

Ans.

Ans: Gg>m kN>s mm # kg 8

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1–9.

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m>ms, (b) mkm, (c) ks>mg, and (d) km # mN.

SOLUTION a) m>ms = ¢

11023 m m = ≤ ¢ ≤ = km>s s 1102-3 s

Ans.

b) mkm = 1102-611023 m = 1102-3 m = mm 3

c) ks>mg = d) km # mN =

1102 s 1102

-6

kg

3

m

10

Ans.

9

= 10

1102 s kg -6

= Gs>kg

N = 10

-3

Ans.

m # N = mm # N

Ans.

Ans: km>s mm Gs>kg mm # N 9

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1–10. Represent each of the following combinations of units in the correct SI form: (a) GN # mm, (b) kg>mm, (c) N>ks2, and  (d) KN>ms.

Solution (a)  GN # mm = 109 ( 10-6 ) N # m = kN # m

Ans.

(b)  kg>mm = 103 g>10-6 m = Gg>m 2

6 2

(c)    N>ks = N>10 s = 10

-6

2

Ans. 2

Ans.

N>s = mN>s 

(d)  kN>ms = 103 N>10-6 s = 109 N>s = GN>s

Ans.

Ans: kN # m Gg>m mN>s2 GN>s 10

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1–11. Represent each of the following with SI units having an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg.

SOLUTION a) 8653 ms = 8.653(10)3(10-3) s = 8.653 s

Ans.

b) 8368 N = 8.368 kN

Ans.

c) 0.893 kg = 893(10-3)(103) g = 893 g

Ans.

Ans: 8.653 s 8.368 kN 893 g 11

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*1–12. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) 1684 mm2>143 ms2, (b) 128 ms210.0458 Mm2>1348 mg2, (c) (2.68 mm)(426 Mg).

SOLUTION a) (684 mm)>43 ms =

684(10 -6) m -3

43(10 ) s = 15.9 mm>s

=

15.9(10-3) m s Ans.

b) (28 ms)(0.0458 Mm)>(348 mg) = =

C 28(10-3) s D C 45.8(10-3)(10)6 m D 348(10-3)(10-3) kg

3.69(106) m # s = 3.69 Mm # s>kg kg

Ans.

c) (2.68 mm)(426 Mg) = C 2.68 A 10-3 B m D C 426 A 103 B kg D

= 1.14 A 103 B m # kg = 1.14 km # kg

Ans.

Ans: 15.9 mm>s 3.69 Mm # s>kg 1.14 km # kg 12

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1–13. The density (mass> volume) of aluminum is 5.26 slug>ft3. Determine its density in SI units. Use an appropriate prefix.

SOLUTION 5.26 slug>ft3 = a

5.26 slug 3

ft

ba

3 14.59 kg ft b a b 0.3048 m 1 slug

= 2.71 Mg>m3

Ans.

Ans: 2.71 Mg>m3 13

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1–14. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52800 ms)2, and (c) [548(106)]1>2 ms.

Solution (a)  (212 mN)2 =

3 212(10)-3 N 4 2

(b)  (52 800 ms)2 =

= 0.0449 N2 = 44.9(10)-3 N2

3 52 800(10)-3 4 2 s2

= 2788 s2 = 2.79 ( 103 ) s2

(c)   3 548(10)6 4 ms = (23 409)(10)-3 s = 23.4(10)3(10)-3 s = 23.4 s 1 2

Ans. Ans. Ans.

Ans: 44.9(10)-3 N2 2.79 ( 103 ) s2 23.4 s 14

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1–15.

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

SOLUTION Using Eq. 1–2, F = G N = a

m 1 m2 r2

kg # kg kg # m m3 ba b = 2 2 kg # s m s2

F = G

(Q.E.D.)

m 1 m2 r2

= 66.73 A 10 - 12 B c

200(200) 0.62

d

= 7.41 A 10 - 6 B N = 7.41 mN

Ans.

Ans: 7.41 mN 15

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*1–16. The pascal (Pa) is actually a very small unit of pressure. To show this, convert 1 Pa = 1 N>m2 to lb>ft2. Atmospheric pressure at sea level is 14.7 lb> in2. How many pascals is this?

SOLUTION Using Table 1–2, we have 1 Pa =

1 lb 0.30482 m2 1N a b = 20.9 A 10 - 3 B lb>ft2 2 4.4482 N b a m 1 ft2

1 ATM =

Ans.

144 in2 14.7 lb 4.448 N 1 ft2 ba a ba b 2 2 1 lb in 1 ft 0.30482 m2

= 101.3 A 103 B N>m2 Ans.

= 101 kPa

Ans: 20.9 1 10-3 2 lb>ft 2

101 kPa

16

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1–17. Water has a density of 1.94 slug>ft 3. What is the density expressed in SI units? Express the answer to three significant figures.

SOLUTION Using Table 1–2, we have rw = a

1.94 slug ft

3

ba

14.5938 kg 1 slug

ba

1 ft 3 b 0.30483 m3

= 999.8 kg>m3 = 1.00 Mg>m3

Ans.

Ans: 1.00 Mg>m3 17

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1–18.

Evaluate each of the following to three significant figures and express each answer in Sl units using an appropriate prefix: (a) 354 mg(45 km) > (0.0356 kN), (b) (0.004 53 Mg) (201 ms), and (c) 435 MN> 23.2 mm.

SOLUTION a) (354 mg)(45 km)>(0.0356 kN) =

=

C 354 A 10-3 B g D C 45 A 103 B m D 0.0356 A 103 B N

0.447 A 103 B g # m N

= 0.447 kg # m>N

Ans.

b) (0.00453 Mg)(201 ms) = C 4.53 A 10-3 B A 103 B kg D C 201 A 10-3 B s D = 0.911 kg # s c) 435 MN>23.2 mm =

435 A 106 B N 23.2 A 10-3 B m

=

Ans. 18.75 A 109 B N m

= 18.8 GN>m

Ans.

Ans: 0.447 kg # m>N 0.911 kg # s 18.8 GN>m 18

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1–19. A concrete column has a diameter of 350 mm and a length of 2 m. If the density 1mass>volume2 of concrete is 2.45 Mg>m3, determine the weight of the column in pounds.

SOLUTION 2 3 V = pr2h = p A 0.35 2 m B (2 m) = 0.1924 m

m = rV = ¢

2.45(103)kg m3

≤ A 0.1924 m3 B = 471.44 kg

W = mg = (471.44 kg) A 9.81 m>s2 B = 4.6248 A 103 B N W = C 4.6248 A 103 B N D ¢

1 lb ≤ = 1.04 kip 4.4482 N

Ans.

Ans: 1.04 kip 19

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*1–20. If a man weighs 155 lb on earth, specify (a) his mass in slugs, (b) his mass in kilograms, and (c) his weight in newtons. If the man is on the moon, where the acceleration due to gravity is gm = 5.30 ft>s2, determine (d) his weight in pounds, and (e) his mass in kilograms.

SOLUTION a) m =

155 = 4.81 slug 32.2

b) m = 155 c

Ans.

14.59 kg d = 70.2 kg 32.2

Ans. Ans.

c) W = 15514.44822 = 689 N 5.30 d = 25.5 lb 32.2

Ans.

14.59 kg d = 70.2 kg 32.2

Ans.

d) W = 155c e) m = 155c Also, m = 25.5

14.59 kg 5.30

Ans.

= 70.2 kg

Ans: 4.81 slug 70.2 kg 689 N 25.5 lb 70.2 kg 20

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1–21. Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

SOLUTION F = G

m1 m2 r2

Where G = 66.73 A 10-12 B m3>(kg # s2) F = 66.73 A 10 - 12 B B

8(12) (0.8)2

R = 10.0 A 10 - 9 B N = 10.0 nN

Ans.

W1 = 8(9.81) = 78.5 N

Ans.

W2 = 12(9.81) = 118 N

Ans.

Ans: F = 10.0 nN W1 = 78.5 N W2 = 118 N 21

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2–1. If 60° and 450 N, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

y F

15

x

700 N

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, 7002 497.01 N

4502

2(700)(450) cos 45° Ans.

497 N

This yields sin 700

sin 45° 497.01

Thus, the direction of angle positive axis, is 60°

of F 95.19°

95.19° measured counterclockwise from the 60°

Ans.

155°

Ans: FR = 497 N f = 155° 22

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2–2. y

If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.

F

u 15

x

700 N

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 7002 - 2(500)(700) cos 105° Ans.

= 959.78 N = 960 N Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78

Ans.

u = 45.2°

Ans: F = 960 N u = 45.2° 23

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2–3. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis.

y 250 lb

F1

30

SOLUTION

x 2

2

FR = 2(250) + (375) - 2(250)(375) cos 75° = 393.2 = 393 lb

Ans.

45

393.2 250 = sin 75° sin u u = 37.89° Ans.

f = 360° - 45° + 37.89° = 353°

F2

375 lb

Ans: FR = 393 lb f = 353° 24

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*2–4. The vertical force F acts downward at on the two-membered frame. Determine the magnitudes of the two components of and . Set F directed along the axes of 500 N.

B

SOLUTION

A

Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have sin 60°

F

500 sin 75°

C

Ans.

448 N sin 45°

500 sin 75° Ans.

366 N

Ans: FAB = 448 N FAC = 366 N 25

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2–5. Solve Prob. 2-4 with F = 350 lb.

B

45

SOLUTION

A

Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have

F

FAB 350 = sin 60° sin 75°

30 C

Ans.

FAB = 314 lb FAC 350 = sin 45° sin 75°

Ans.

FAC = 256 lb

Ans: FAB = 314 lb FAC = 256 lb 26

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2–6. v

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.

30 75

F1  4 kN 30 u F2  6 kN

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying Law of cosines by referring to Fig. b,

FR = 242 + 62 - 2(4)(6) cos 105° = 8.026 kN = 8.03 kN

Ans.

Using this result to apply Law of sines, Fig. b, sin u sin 105° = ; 6 8.026

u = 46.22°

Thus, the direction f of FR measured clockwise from the positive u axis is

Ans.

f = 46.22° - 45° = 1.22°

Ans: f = 1.22° 27

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2–7. v

Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 30 75

F1  4 kN 30 u F2  6 kN

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law by referring to Fig. b.

(F1)v sin 45° (F1)u sin 30°

=

4 ; sin 105°

(F1)v = 2.928 kN = 2.93 kN

Ans.

=

4 ; sin 105°

(F1)u = 2.071 kN = 2.07 kN

Ans.

Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 28

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*2–8. v

Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 30 75

F1  4 kN 30 u F2  6 kN

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law of referring to Fig. b,

(F2)u sin 75° (F2)v sin 30°

=

6 ; sin 75°

(F2)u = 6.00 kN

Ans.

=

6 ; sin 75°

(F2)v = 3.106 kN = 3.11 kN

Ans.

Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 29

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2–9. If the resultant force acting on the support is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.

F A u B 60

900 lb

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b,

F = 29002 + 12002 - 2(900)(1200) cos 30° = 615.94 lb = 616 lb

Ans.

Using this result to apply the sines law, Fig. b,

sin u sin 30° = ; 900 615.94

u = 46.94° = 46.9°

Ans.

Ans: F = 616 lb u = 46.9° 30

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2–10. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

800 lb 40

x 35

Solution

500 lb

Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b,

FR = 28002 + 5002 - 2(800)(500) cos 95° = 979.66 lb = 980 lb

Ans.

Using this result to apply the sines law, Fig. b, sin u sin 95° = ; 500 979.66

u = 30.56°

Thus, the direction f of FR measured counterclockwise from the positive x axis is

Ans.

f = 50° - 30.56° = 19.44° = 19.4°

Ans: FR = 980 lb  f = 19.4° 31

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2–11. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

FA u

8 kN

A

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° Ans.

= 10.80 kN = 10.8 kN The angle u can be determined using law of sines (Fig. b).

40 B FB

6 kN

sin 100° sin u = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is Ans.

f = 33.16° - 30° = 3.16°

Ans: FR = 10.8 kN f = 3.16° 32

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*2–12. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?

FA u

8 kN

A

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8

40 B

sin (90° - u) = 0.5745 Ans.

u = 54.93° = 54.9°

FB

6 kN

From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° Ans.

= 10.4 kN

Ans: u = 54.9° FR = 10.4 kN 33

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2–13. The force acting on the gear tooth is F = 20 lb. Resolve this force into two components acting along the lines aa and bb.

b F

a

80 60 a b

SOLUTION Fa 20 = ; sin 40° sin 80°

Fa = 30.6 lb

Ans.

Fb 20 = ; sin 40° sin 60°

Fb = 26.9 lb

Ans.

Ans: Fa = 30.6 lb Fb = 26.9 lb 34

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2–14. The component of force F acting along line aa is required to be 30 lb. Determine the magnitude of F and its component along line bb.

b F

a

80 60 a b

SOLUTION 30 F = ; sin 80° sin 40°

F = 19.6 lb

Ans.

Fb 30 = ; sin 80° sin 60°

Fb = 26.4 lb

Ans.

Ans: F = 19.6 lb Fb = 26.4 lb 35

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2–15. Force F acts on the frame such that its component acting along member AB is 650 lb, directed from B towards A, and the component acting along member BC is 500 lb, directed from B towards C. Determine the magnitude of F and its direction u. Set f = 60°.

B

u F A

f

45

SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 6502 - 2(500)(650) cos 105° Ans.

= 916.91 lb = 917 lb Using this result and applying the law of sines to Fig. b, yields sin u sin 105° = 500 916.91

Ans.

u = 31.8°

Ans: F = 917 lb u = 31.8° 36

C

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*2–16. Force F acts on the frame such that its component acting along member AB is 650 lb, directed from B towards A. Determine the required angle f (0° … f … 45°) and the component acting along member BC. Set F = 850 lb and u = 30°.

B

u F A

f

45

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FBC = 28502 + 6502 - 2(850)(650) cos 30° Ans.

= 433.64 lb = 434 lb Using this result and applying the sine law to Fig. b, yields sin (45° + f) sin 30° = 850 433.64

Ans.

f = 33.5°

Ans: FBC = 434 lb f = 33.5° 37

C

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2–17. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F1 + F2 and then forming FR = F¿ + F3.

y F1

30 N 3

5

F3

4

50 N

20

SOLUTION

F2

20 N

F¿ = 2(20)2 + (30)2 - 2(20)(30) cos 73.13° = 30.85 N 30 30.85 = ; sin 73.13° sin (70° - u¿)

u¿ = 1.47°

FR = 2(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47° = 19.18 = 19.2 N

Ans.

30.85 19.18 = ; sin 1.47° sin u

Ans.

u = 2.37°

Ans: FR = 19.2 N u = 2.37° c 38

x

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2–18. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F2 + F3 and then forming FR = F¿ + F1.

y F1

30 N 3

5

F3

4

50 N

20

SOLUTION ¿

2

F2

20 N

2

F = 2(20) + (50) - 2(20)(50) cos 70° = 47.07 N 20 sin u

¿

=

47.07 ; sin 70°

u¿ = 23.53°

FR = 2(47.07)2 + (30)2 - 2(47.07)(30) cos 13.34° = 19.18 = 19.2 N 30 19.18 = ; sin 13.34° sin f

Ans.

f = 21.15° Ans.

u = 23.53° - 21.15° = 2.37°

Ans: FR = 19.2 N u = 2.37° c 39

x

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2–19. Determine the design angle u (0° … u … 90°) for strut AB so that the 400-lb horizontal force has a component of 500 lb directed from A towards C. What is the component of force acting along member AB? Take f = 40°.

400 lb A u B

f

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a.

C

Trigonometry: Using law of sines (Fig. b), we have sin u sin 40° = 500 400 sin u = 0.8035 Ans.

u = 53.46° = 53.5° Thus, c = 180° - 40° - 53.46° = 86.54° Using law of sines (Fig. b) FAB 400 = sin 86.54° sin 40°

Ans.

FAB = 621 lb

Ans: u = 53.5° FAB = 621 lb 40

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*2–20. Determine the design angle f (0° … f … 90°) between struts AB and AC so that the 400-lb horizontal force has a component of 600 lb which acts up to the left, in the same direction as from B towards A. Take u = 30°.

400 lb A u B

f

SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a.

C

Trigonometry: Using law of cosines (Fig. b), we have FAC = 24002 + 6002 - 2(400)(600) cos 30° = 322.97 lb The angle f can be determined using law of sines (Fig. b). sin f sin 30° = 400 322.97 sin f = 0.6193 Ans.

f = 38.3°

Ans: f = 38.3° 41

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2–21. y

Determine the magnitude and direction of the resultant force, FR measured counterclockwise from the  positive x axis. Solve the problem by first finding the resultant F′ = F1 + F2 and then forming FR = F′ + F3.

F1  400 N 90º

F2  200 N

150º

x

F3  300 N

Solution Parallelogram Law. The parallelogram law of addition for F1 and F2 and then their resultant F′ and F3 are shown in Figs. a and b, respectively. Trigonometry. Referring to Fig. c,

F ′ = 22002 + 4002 = 447.21 N

Thus f′ = 90° - 30° -26.57° = 33.43°

u ′ = tan-1 a

200 b = 26.57° 400

Using these results to apply the law of cosines by referring to Fig. d,   FR = 23002 + 447.212 - 2(300)(447.21) cos 33.43° = 257.05 N = 257 kN Ans. Then, apply the law of sines,

sin u sin 33.43° = ; 300 257.05

u = 40.02°

Thus, the direction f of FR measured counterclockwise from the positive x axis is

Ans.

f = 90° + 33.43° + 40.02° = 163.45° = 163°

Ans: FR = 257 N  f = 163° 42

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2–22. y

Determine the magnitude and direction of the resultant force, measured counterclockwise from the positive x axis. Solve l by first finding the resultant F′ = F2 + F3 and then forming FR = F′ + F1.

F1  400 N 90º

F2  200 N

150º

x

F3  300 N

Solution Parallelogram Law. The parallelogram law of addition for F2 and F3 and then their resultant F′ and F1 are shown in Figs. a and b, respectively. Trigonometry. Applying the law of cosines by referring to Fig. c,

F ′ = 22002 + 3002 - 2(200)(300) cos 30° = 161.48 N

Ans.

Using this result to apply the sines law, Fig. c, sin u′ sin 30° = ; 200 161.48

u′ = 38.26°

Using the results of F′ and u′ to apply the law of cosines by referring to Fig. d,   FR = 2161.482 + 4002 - 2(161.48)(400) cos 21.74° = 257.05 N = 257 N Ans. Then, apply the sines law,

sin u sin 21.74° = ; 161.48 257.05

u = 13.45°

Thus, the direction f of FR measured counterclockwise from the positive x axis is

Ans.

f = 90° + 60° + 13.45° = 163.45° = 163°

Ans:  f = 163° FR = 257 N 43

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2–23. Two forces act on the screw eye. If F1 = 400 N and F2 = 600 N, determine the angle u(0° … u … 180°) between them, so that the resultant force has a magnitude of FR = 800 N.

F1

u

SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. Applying law of cosines to Fig. b, 2

F2

2

800 = 2400 + 600 - 2(400)(600) cos (180° - u°) 8002 = 4002 + 6002 - 480000 cos (180° - u) cos (180° - u) = - 0.25 180° - u = 104.48 Ans.

u = 75.52° = 75.5°

Ans: u = 75.5° 44

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*2–24. Two forces F1 and F2 act on the screw eye. If their lines of action are at an angle u apart and the magnitude of each force is F1 = F2 = F, determine the magnitude of the resultant force FR and the angle between FR and F1.

F1

u

SOLUTION F F = sin f sin (u - f) sin (u - f) = sin f

F2

u - f = f f =

u 2

Ans.

FR = 2(F)2 + (F)2 - 2(F)(F) cos (180° - u) Since cos (180° - u) = -cos u FR = F A 22 B 21 + cos u u 1 + cos u Since cos a b = 2 A 2 Then u FR = 2F cosa b 2

Ans.

Ans: f =

u 2

u FR = 2F cos a b 2 45

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2–25. If F1 = 30 lb and F2 = 40 lb, determine the angles u and f so that the resultant force is directed along the positive x axis and has a magnitude of FR = 60 lb.

y

F1

θ x

φ F2

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosine by referring to Fig. b, 402 = 302 + 602 - 2(30)(60) cos u

Ans.

u = 36.34° = 36.3°

And 302 = 402 + 602 - 2(40)(60) cos f

Ans.

f = 26.38° = 26.4°

Ans: u = 36.3° f = 26.4° 46

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2–26. y

Determine the magnitude and direction u of FA so that the resultant force is directed along the positive x axis and has a magnitude of 1250 N.

FA

θ O

A

x

30° B

SOLUTION

FB = 800 N

+ FR = ΣFx; S x

FRx = FA sin u + 800 cos 30° = 1250

+ c FRy = ΣFy;

FRy = FA cos u - 800 sin 30° = 0 u = 54.3°

Ans.

FA = 686 N

Ans.

Ans: u = 54.3° FA = 686 N 47

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2–27. y

Determine the magnitude and direction, measured counterclockwise from the positive x axis, of the resultant force acting on the ring at O, if FA = 750 N and u = 45°.

FA

θ O

A

x

30° B

SOLUTION

FB = 800 N

Scalar Notation: Suming the force components algebraically, we have + FR = ΣFx; S x

FRx = 750 sin 45° + 800 cos 30° = 1223.15 N S

+ c FRy = ΣFy;

FRy = 750 cos 45° - 800 sin 30° = 130.33 N c

The magnitude of the resultant force FR is FR = 3F 2Rx + F 2Ry = 21223.152 + 130.332 = 1230 N = 1.23 kN

Ans.

The directional angle u measured counterclockwise from positive x axis is u = tan-1

FRy FRx

= tan-1 a

130.33 b = 6.08° 1223.15

Ans.

Ans: FR = 1.23 kN u = 6.08° 48

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*2–28. Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. What is the minimum magnitude of FR?.

8 kN

F 30 6 kN

Solution Parallelogram Law. The parallelogram laws of addition for 6 kN and 8 kN and then their resultant F′ and F are shown in Figs. a and b, respectively. In order for FR to be minimum, it must act perpendicular to F. Trigonometry. Referring to Fig. b, F′ = 262 + 82 = 10.0 kN Referring to Figs. c and d,

8 u = tan - 1 a b = 53.13°. 6



FR = 10.0 sin 83.13° = 9.928 kN = 9.93 kN

Ans.



F = 10.0 cos 83.13° = 1.196 kN = 1.20 kN

Ans.

Ans: FR = 9.93 kN  F = 1.20 kN 49

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2–29. If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u.

y

FA

A

2 kN 30

x

u C

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

FB

B

Applying the law of cosines to Fig. b, FB = 222 + 32 - 2(2)(3)cos 30° Ans.

= 1.615kN = 1.61 kN Using this result and applying the law of sines to Fig. b, yields sin u sin 30° = 2 1.615

Ans.

u = 38.3°

Ans: FB = 1.61 kN u = 38.3° 50

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2–30. If FB = 3 kN and u = 45°, determine the magnitude of the resultant force of the two tugboats and its direction measured clockwise from the positive x axis.

y

FA

A

2 kN 30

x

u C

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

FB

B

Applying the law of cosines to Fig. b, FR = 222 + 32 - 2(2)(3) cos 105° Ans.

= 4.013 kN = 4.01 kN Using this result and applying the law of sines to Fig. b, yields sin 105° sin a = 3 4.013

a = 46.22°

Thus, the direction angle f of FR, measured clockwise from the positive x axis, is Ans.

f = a - 30° = 46.22° - 30° = 16.2°

Ans: FR = 4.01 kN f = 16.2° 51

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2–31. If the resultant force of the two tugboats is required to be directed towards the positive x axis, and FB is to be a minimum, determine the magnitude of FR and FB and the angle u.

y

FA

A

2 kN 30

x

u

SOLUTION

C

FB

For FB to be minimum, it has to be directed perpendicular to FR. Thus, Ans.

u = 90°

B

The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. By applying simple trigonometry to Fig. b, FB = 2 sin 30° = 1 kN

Ans.

FR = 2 cos 30° = 1.73 kN

Ans.

Ans: u = 90° FB = 1 kN FR = 1.73 kN 52

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*2–32. y

Determine the magnitude of the resultant force and  its direction, measured counterclockwise from the positive x axis.

F1  200 N 45 x 30 F2  150 N

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ;  (F ) = 200 sin 45° - 150 cos 30° = 11.518 N S S R x

x

R x

+ c (FR)y = ΣFy;  (FR)y = 200 cos 45° + 150 sin 30° = 216.42 N c Referring to Fig. b, the magnitude of the resultant force FR is

FR = 2(FR)2x + (FR)2y = 211.5182 + 216.422 = 216.73 N = 217 N

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

216.42 b = 86.95° = 87.0° 11.518

Ans.

Ans: FR = 217 N u = 87.0° 53

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2–33. y

Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.

400 N B 30 x 45 800 N

Solution Scalar Notation. Summing the force components along x and y axes by referring to Fig. a, + (F ) = ΣF ;  (F ) = 400 cos 30° + 800 sin 45° = 912.10 N S S R x

x

R x

+ c (FR)y = ΣFy;  (FR)y = 400 sin 30° - 800 cos 45° = - 365.69 N = 365.69 NT Referring to Fig. b, the magnitude of the resultant force is

FR = 2(FR)2x + (FR)2y = 2912.102 + 365.692 = 982.67 N = 983 N

Ans.

And its directional angle u measured clockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

365.69 b = 21.84° = 21.8° 912.10

Ans.

Ans: FR = 983 N u = 21.8° 54

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2–34. y 60

Resolve F1 and F2 into their x and y components.

30 400 N

F1

45

SOLUTION F1 = {400 sin 30°(+ i)+ 400 cos 30°( +j)} N

F2

= {200i +346j} N

x

250 N

Ans.

F2 = {250 cos 45°(+ i)+ 250 sin 45°(-j)} N Ans.

= {177i- 177j} N

55

Ans: F1 = 5200i + 346j 6 N F2 = 5177i - 177j 6 N

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2–35. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y 60 30 400 N

F1

SOLUTION

(F1)x = 400 sin 30° = 200 N

(F1)y = 400 cos 30° = 346.41 N

(F2)x = 250 cos 45° = 176.78 N

(F2)y = 250 sin 45° = 176.78 N

x

45

Rectangular Components: By referring to Fig. a, the x and y components of F1 and F2 can be written as F2

250 N

Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;

(FR)x = 200 + 176.78 = 376.78 N

+ c ©(FR)y = ©Fy;

(FR)y = 346.41 - 176.78 = 169.63 N c

Ans.

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2376.782 + 169.632 = 413 N

Ans.

The direction angle u of FR, Fig. b, measured counterclockwise from the positive axis, is u = tan-1 c

(FR)y 169.63 d = tan-1 a b = 24.2° (FR)x 376.78

Ans.

Ans: FR = 413 N u = 24.2° 56

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*2–36. Resolve each force acting on the gusset plate into its x and y components, and express each force as a Cartesian vector.

y 650 N

F3 3

5

F2

4

750 N

45

F1 = {900(+i)} = {900i} N

Ans.

F2 = {750 cos 45°(+ i) + 750 sin 45°(+j)} N = {530i + 530j} N

Ans.

F3 = e 650a

F1

x 900 N

3 4 b (+i) + 650 a b (-j) f N 5 5

= {520 i - 390j)} N

Ans.

57

Ans: F1 = 5900i6 N F2 = 5530i + 530j 6 N F3 = 5520i - 390j 6 N

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2–37. Determine the magnitude of the resultant force acting on the plate and its direction, measured counterclockwise from the positive x axis.

y 650 N

F3 3

SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = 900 N

5

F2

4

750 N

45 F1

x 900 N

(F1)y = 0

(F2)x = 750 cos 45° = 530.33 N 4 (F3)x = 650 a b = 520 N 5

(F2)y = 750 sin 45° = 530.33 N 3 (F3)y = 650a b = 390 N 5

Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;

(FR)x = 900 + 530.33 + 520 = 1950.33 N :

+ c ©(FR)y = ©Fy;

(FR)y = 530.33 - 390 = 140.33 N c

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 21950.332 + 140.332 = 1955 N = 1.96 kN Ans. The direction angle u of FR, measured clockwise from the positive x axis, is u = tan-1 c

(FR)y 140.33 b = 4.12° d = tan-1 a (FR)x 1950.33

Ans.

Ans: FR = 1.96 kN u = 4.12° 58

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2–38. Express each of the three forces acting on the support in Cartesian vector form and determine the magnitude of the resultant force and its direction, measured clockwise from positive x axis.

y

F1  50 N 5

4

4 3

x F3  30 N

15

F2  80 N

Solution Cartesian Notation. Referring to Fig. a,

3 4 F1 = (F1)x i + (F1)y j = 50 a b i + 50 a b j = {30 i + 40 j} N 5 5

Ans.

F2 = - (F2)x i - (F2)y j = -80 sin 15° i - 80 cos 15° j = { - 20.71 i - 77.27 j} N = { - 20.7 i - 77.3 j} N



Ans.

F3 = (F3)x i = {30 i}



Ans.

Thus, the resultant force is FR = ΣF ;

FR = F1 + F2 + F3 = (30i + 40 j) + ( - 20.71i - 77.27j) + 30i = {39.29 i - 37.27 j} N

Referring to Fig. b, the magnitude of FR is

FR = 239.292 + 37.272 = 54.16 N = 54.2 N

Ans.

And its directional angle u measured clockwise from the positive x axis is

u = tan - 1 a

37.27 b = 43.49° - 43.5° 39.29

Ans.

Ans:  F1 =  F2 =  F3 = FR = u = 59

{30i + 40j} N { - 20.7 i - 77.3 j} N {30 i} 54.2 N 43.5°

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2–39. y

Determine the x and y components of F1 and F2.

45 F1  200 N

30

SOLUTION F1x = 200 sin 45° = 141 N

Ans.

F1y = 200 cos 45° = 141 N

Ans.

F2x = - 150 cos 30° = - 130 N

Ans.

F2y = 150 sin 30° = 75 N

Ans.

F2  150 N x

Ans: F1x = F1y = F2x = F2y = 60

141 N 141 N - 130 N 75 N

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*2–40. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. 45

F1  200 N

30

SOLUTION +R FRx = ©Fx;

FRx = - 150 cos 30° + 200 sin 45° = 11.518 N

Q+FRy = ©Fy;

FRy = 150 sin 30° + 200 cos 45° = 216.421 N

F2  150 N x

FR = 2 (11.518)2 + (216.421)2 = 217 N

Ans.

u = tan - 1 ¢

Ans.

216.421 ≤ = 87.0° 11.518

Ans: FR = 217 N u = 87.0° 61

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2–41. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

F3  8 kN F2  5 kN

60 45

F1  4 kN

x

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ;  (F ) = 4 + 5 cos 45° - 8 sin 15° = 5.465 kN S S R x

x

R x

+ c (FR)y = ΣFy;  (FR)y = 5 sin 45° + 8 cos 15° = 11.263 kN c By referring to Fig. b, the magnitude of the resultant force FR is

FR = 2(FR)2x + (FR)2y = 25.4652 + 11.2632 = 12.52 kN = 12.5 kN

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

11.263 b = 64.12° = 64.1° 5.465

Ans.

Ans: FR = 12.5 kN u = 64.1° 62

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2–42. y

Express F1, F2, and F3 as Cartesian vectors. F3  750 N

45

3

5

x

4

SOLUTION F1 =

30

F1  850 N

F2  625 N

3 4 (850) i - (850) j 5 5 Ans.

= {680 i - 510 j} N F2 = - 625 sin 30° i - 625 cos 30° j

Ans.

= { -312 i - 541 j} N F3 = - 750 sin 45° i + 750 cos 45° j { -530 i + 530 j} N

Ans.

Ans: F1 = {680i - 510j} N F2 = { - 312i - 541j} N F3 = { - 530i + 530j} N 63

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2–43. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y F3  750 N

45

3

5

x

4

SOLUTION

30 F2  625 N

+ F = ©F ; : Rx x

FRx =

+ c FRy = ©Fy ;

3 FRy = - (850) - 625 cos 30° + 750 cos 45° = -520.94 N 5

4 (850) - 625 sin 30° - 750 sin 45° = - 162.83 N 5

FR = 2( - 162.83)2 + ( - 520.94)2 = 546 N f = tan-1 a

F1  850 N

Ans.

520.94 b = 72.64° 162.83 Ans.

u = 180° + 72.64° = 253°

Ans: FR = 546 N u = 253° 64

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*2–44. y

Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.

40 lb 5

4 3

30 lb

Solution

x

5

Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a,

13

12

91 lb

+ (F ) = ΣF ;  (F ) = 40 a 3 b + 91a 5 b + 30 = 89 lb S S R x x R x 5 13

4 12 + c (FR)y = ΣFy;  (FR)y = 40 a b - 91a b = -52 lb = 52 lbT 5 13

By referring to Fig. b, the magnitude of resultant force is

FR = 2(FR)2x + (FR)2y = 2892 + 522 = 103.08 lb = 103 lb

Ans.

And its directional angle u measured clockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

52 b = 30.30° = 30.3° 89

Ans.

Ans: FR = 103 lb u = 30.3° 65

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2–45. Determine the magnitude and direction u of the resultant force FR. Express the result in terms of the magnitudes of the components F1 and F2 and the angle f.

SOLUTION

F1

F 2R

=

F 21

+

F 22

FR

- 2F1F2 cos (180° - f)

Since cos (180° - f) = -cos f,

f u

FR = 2F 21 + F 22 + 2F1F2 cos f

Ans.

F2

From the figure, tan u =

F1 sin f F2 + F1 cos f

u = tan –1 ¢

F1 sin f ≤ F2 + F1 cos f

Ans.

Ans: FR = 2F12 + F22 + 2F1F2 cos f F1 sin f b u = tan-1a F2 + F1 cos f 66

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2–46. y

Determine the magnitude and orientation u of FB so that the resultant force is directed along the positive y axis and has a magnitude of 1500 N.

FB

B

FA  700 N 30

A

u x

SOLUTION Scalar Notation: Suming the force components algebraically, we have + F = ΣF ; S Rz x

0 = 700 sin 30° - FB cos u (1)

FB cos u = 350 + c FRy = ΣFy;

1500 = 700 cos 30° + FB sin u (2)

FB sin u = 893.8 Solving Eq. (1) and (2) yields u = 68.6°

Ans.

FB = 960 N

Ans: u = 68.6° FB = 960 N 67

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2–47. y

Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket, if FB = 600 N and u = 20°.

FB

B

FA  700 N 30

A

u x

SOLUTION Scalar Notation: Suming the force components algebraically, we have + F = ΣF ; S Rx x

FRx = 700 sin 30° - 600 cos 20° = - 213.8 N = 213.8 N d

+ c FRy = ΣFy;

FRy = 700 cos 30° + 600 sin 20° = 811.4 N c

The magnitude of the resultant force FR is FR = 2F 2Rx + F 2Ry = 2213.82 + 811.42 = 839 N

Ans.

The directional angle u measured counterclockwise from positive y axis is u = tan-1

FRx FRy

= tan-1 a

213.8 b = 14.8° 811.4

Ans.

Ans: FR = 839 N u = 14.8° 68

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*2–48. Three forces act on the bracket. Determine the magnitude and direction u of F1 so that the resultant force is directed along the positive x¿ axis and has a magnitude of 800 N.

y F2  200 N

x¿ F3  180 N

SOLUTION

13

60

u

F1

5

+ F = ©F ; : Rx x

12 (180) 800 sin 60° = F1 sin(60° + u) 13

+ c FRy = ©Fy ;

800 cos 60° = F1 cos(60° + u) + 200 +

12

x

5 (180) 13

60° + u = 81.34° u = 21.3°

Ans.

F1 = 869 N

Ans.

Ans: u = 21.3° F1 = 869 N 69

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2–49. If F1 = 300 N and u = 10°, determine the magnitude and direction, measured counterclockwise from the positive x¿ axis, of the resultant force acting on the bracket.

y F2  200 N

x¿ F3  180 N

SOLUTION

13

60

u

F1

5

+ F = ©Fx ; : Rx

FRx

+ c FRy = ©Fy ;

FRy = 300 cos 70° + 200 +

12

12 = 300 sin 70° (180) = 115.8 N 13

x

5 (180) = 371.8 N 13

FR = 2(115.8)2 + (371.8)2 = 389 N f = tan - 1 B

371.8 R = 72.71° 115.8

Ans.

au Ans.

f¿ = 72.71° - 30° = 42.7°

Ans: FR = 389 N f′ = 42.7° 70

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2–50. Express F1, F2, and F3 as Cartesian vectors.

y

F2  26 kN 5

40

F1  15 kN

13 12

SOLUTION

x 30

F1 = { -200 i} lb

Ans. F3  36 kN

F2 = -250 sin 30° i + 250 cos 30° j Ans.

= { -125 i + 217 j} lb F3 = 225 cos 30° i + 225 sin 30° j {195 i

Ans.

112 j} lb

Ans: F1 = { - 200i} lb F2 = { - 125i + 217j} lb F3 = {195i + 112j} lb 71

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2–51. Determine the magnitude of the resultant force and its orientation measured counterclockwise from the positive x axis.

y

F2  26 kN 5

40

F1  15 kN

13 12

SOLUTION

x 30

+ F = ©F ; : Rx x

FRx = 15 sin 40° -

12 (26) + 36 cos 30° = 16.82 kN 13

+ c FRy = ©Fy ;

FRy = 15 cos 40° +

5 (26) - 36 sin 30° = 3.491 kN 13

F3  36 kN

FR = 2(16.82)2 + (3.491)2 = 17.2 kN

Ans.

3.491 b = 11.7° 16.82

Ans.

u = tan-1 a Also,

F1 = {15 sin 40° i + 15 cos 40° j} kN = {9.64i + 11.5j} kN F2 = b -

12 5 (26)i + (26)j r kN = { - 24i + 10j} kN 13 13

F3 = {36 cos 30°i - 36 sin 30°j} kN = {31.2i - 18j} kN FR = F1 + F2 + F3 = {9.64i + 11.5j} + {- 24i + 10j} + {31.2i - 18j}

Ans: FR = 17.2 kN, u = 11.7° 72

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*2–52. y

Determine the x and y components of each force acting on the gusset plate of a bridge truss. Show that the resultant force is zero.

F2  6 kN F1  8 kN 4

5

3

5

4 3

F3  4 kN

F4  6 kN

x

Solution Scalar Notation. Referring to Fig. a, the x and y components of each forces are

4 (F1)x = 8 a b = 6.40 kN S  5

Ans.

3 (F1)y = 8 a b = 4.80 kN T  5

Ans.

3 (F2)x = 6 a b = 3.60 kN S  5

Ans.

4 (F2)y = 6 a b = 4.80 kN c  5

Ans.



(F3)x = 4 kN d 

Ans.



(F3)y = 0

Ans.



(F4)x = 6 kN d 

Ans.



(F4)y = 0

Ans.

Summing these force components along x and y axes algebraically, + (F ) = ΣF ;   (F ) = 6.40 + 3.60 - 4 - 6 = 0 S R x x R x + c (FR)y = ΣFy ;     (FR)y = 4.80 - 4.80 = 0 Thus, FR = 2(FR)2x + (FR)2y = 2O2 + O2 = O

(Q.E.D)

Ans: (F1)x (F1)y (F2)x (F2)y (F3)x (F3)y (F4)x (F4)y 73

= = = = = = = =

6.40 kN S 4.80 kN T 3.60 kN S 4.80 kN c 4 kN d 0 6 kN d 0

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2–53. Express F1 and F2 as Cartesian vectors.

y

F2 = 26 kN 13

12 5

x

SOLUTION F1 = -30 sin 30° i - 30 cos 30° j = 5- 15.0 i - 26.0 j6 kN F2 = =

30°

Ans.

12 5 1262 i + 1262 j 13 13

F1 = 30 kN

Ans.

- 10.0 i + 24.0 j kN

Ans: F1 = { - 15.0i - 26.0j} kN F2 = { - 10.0i + 24.0j} kN 74

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2–54. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y

F2 = 26 kN 13

12 5

x

SOLUTION + F = ©F ; : Rx x + c FRy = ©Fy ;

FRx = - 30 sin 30° FRy

5 1262 = - 25 kN 13

30°

12 = - 30 cos 30° + 1262 = -1.981 kN 13

FR = 21 - 2522 + 1- 1.98122 = 25.1 kN f = tan-1 a

F1 = 30 kN

Ans.

1.981 b = 4.53° 25 Ans.

u = 180° + 4.53° = 185°

Ans: FR = 25.1 kN u = 185° 75

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2–55. Determine the magnitude of force F so that the resultant force of the three forces is as small as possible. What is the magnitude of the resultant force?

F

14 kN 30

45 8 kN

SOLUTION + : FRx = ©Fx ;

FRz = 8 - F cos 45° - 14 cos 30° = - 4.1244 - F cos 45°

+ c FRy = ©Fy ;

FRy = - F sin 45° + 14 sin 30° = 7 - F sin 45° FR2 = (- 4.1244 - F cos 45°)2 + (7 - F sin 45°)2 2FR

From Eq. (1);

(1)

dFR = 2( -4.1244 - F cos 45°)(- cos 45°) + 2(7 - F sin 45°)(-sin 45°) = 0 dF F = 2.03 kN

Ans.

FR = 7.87 kN

Ans.

Also, from the figure require (FR)x¿ = 0 = ©Fx¿;

F + 14 sin 15° - 8 cos 45° = 0 Ans.

F = 2.03 kN (FR)y¿ = ©Fy¿;

FR = 14 cos 15° - 8 sin 45° Ans.

FR = 7.87 kN

Ans: F = 2.03 kN FR = 7.87 kN 76

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*2–56. If the magnitude of the resultant force acting on the bracket is to be 450 N directed along the positive u axis, determine the magnitude of F1 and its direction f.

y u

F1

f 30

SOLUTION

F2

Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, F3, and FR can be written as (F1)x = F1 sin f

(F1)y = F1 cos f

(F2)x = 200 N

( F2)y = 0

(F3)x = 260 ¢

5 ≤ = 100 N 13

(F3)y = 260 ¢

(FR)x = 450 cos 30° = 389.71 N

x 200 N

13

12 5

F3

260 N

12 ≤ = 240 N 13

(FR)y = 450 sin 30° = 225 N

Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

389.71 = F1 sin f + 200 + 100 (1)

F1 sin f = 89.71 + c ©(FR)y = ©Fy;

225 = F1 cos f - 240 (2)

F1 cos f = 465 Solving Eqs. (1) and (2), yields f = 10.9°

Ans.

F1 = 474 N

Ans: f = 10.9° F1 = 474 N 77

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2–57. If the resultant force acting on the bracket is required to be a minimum, determine the magnitudes of F1 and the resultant force. Set f = 30°.

y u

F1

f 30

SOLUTION

F2

Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = F1 sin 30° = 0.5F1

(F1)y = F1 cos 30° = 0.8660F1

(F2)x = 200 N

(F2)y = 0

(F3)x = 260a

5 b = 100 N 13

(F3)y = 260a

x 200 N

13

12 5

F3

260 N

12 b = 240 N 13

Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

(FR)x = 0.5F1 + 200 + 100 = 0.5F1 + 300

+ c ©(FR)y = ©Fy;

(FR)y = 0.8660F1 - 240

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2(0.5F1 + 300)2 + (0.8660F1 - 240)2 = 2F 21 - 115.69F1 + 147 600

(1)

Thus, FR2 = F 21 - 115.69F1 + 147 600

(2)

The first derivative of Eq. (2) is 2FR For FR to be minimum,

dFR = 2F1 - 115.69 dF1

(3)

dFR = 0. Thus, from Eq. (3) dF1

2FR

dFR = 2F1 - 115.69 = 0 dF1 Ans.

F1 = 57.846 N = 57.8 N from Eq. (1), FR = 2(57.846)2 - 115.69(57.846) + 147 600 = 380 N

Ans.

Ans: FR = 380 N F1 = 57.8 N 78

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2–58. y

Three forces act on the bracket. Determine the magnitude and direction u of F so that the resultant force is directed along the positive x′ axis and has a magnitude of 8 kN.

4 kN

F 15

u

x'

30 x 6 kN

Solution Scalar Notation. Equating the force components along the x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ;  8 cos 30° = F sin u + 6 - 4 sin 15° S R x

x



F sin u = 1.9635(1) + c (FR)y = ΣFy ;  8 sin 30° = F cos u + 4 cos 15°



F cos u = 0.1363(2)

Divide Eq (1) by (2)

tan u = 14.406

u = 86.03° = 86.0°

Ans.

Substitute this result into Eq (1) F sin 86.03° = 1.9635

Ans.

F = 1.968 kN = 1.97 kN

Ans:  u = 86.0° F = 1.97 kN 79

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2–59. y

If F = 5 kN and u = 30°, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

4 kN

F 15

u

x'

30 x 6 kN

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; (F ) = 5 sin 30° + 6 - 4 sin 15° = 7.465 kN S S R x x R x + c (FR)y = ΣFy; (FR)y = 4 cos 15° + 5 cos 30° = 8.194 kN c

By referring to Fig. b, the magnitude of the resultant force is FR = 2(FR)2x + (FR)2y = 27.4652 + 8.1942 = 11.08 kN = 11.1 kN Ans.

And its directional angle u measured counterclockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

8.194 b = 47.67° = 47.7° 7.465

Ans.

Ans: FR = 11.1 kN u = 47.7° 80

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*2–60. The force F has a magnitude of 80 lb and acts within the octant shown. Determine the magnitudes of the x, y, z components of F.

z Fz F  80 lb

b  45

SOLUTION 2

2

a  60

2

1 = cos 60° + cos 45° + cos g

Fy

Fx

Solving for the positive root, g = 60°

x

Fx = 80 cos 60° = 40.0 lb

Ans.

Fy = 80 cos 45° = 56.6 lb

Ans.

Fz = 80 cos 60° = 40.0 lb

Ans.

Ans: Fx = 40.0 lb Fy = 56.6 lb Fz = 40.0 lb 81

y

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2–61. The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and a = 60° and g = 45°, determine the magnitudes of its components.

z

Fz g

F Fy

SOLUTION

b y

a Fx

cosb = 21 - cos2 a - cos2g = 21 - cos2 60° - cos2 45°

x

b = 120° Fx = |80 cos 60°| = 40 N

Ans.

Fy = |80 cos 120°| = 40 N

Ans.

Fz = |80 cos 45°| = 56.6 N

Ans.

Ans: Fx = 40 N Fy = 40 N Fz = 56.6 N 82

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2–62. z

Determine the magnitude and coordinate direction angles of the force F acting on the support. The component of F in the x9y plane is 7 kN. F

y

30 40 7 kN x

Solution Coordinate Direction Angles. The unit vector of F is uF = cos 30° cos 40°i - cos 30° sin 40°j + sin 30° k = {0.6634i - 0.5567j + 0.5 k} Thus,

cos a = 0.6634;

a = 48.44° = 48.4°

Ans.



cos b = - 0.5567; b = 123.83° = 124°

Ans.



cos g = 0.5;

Ans.

g = 60°

The magnitude of F can be determined from

F cos 30° = 7;

Ans.

F = 8.083 kN = 8.08 kN

Ans: a = 48.4° b = 124° g = 60° F = 8.08 kN 83

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2–63. Determine the magnitude and coordinate direction angles of the resultant force and sketch this vector on the coordinate system.

z F1

80 lb y

30 40 F2

SOLUTION

130 lb

x

F1 = {80 cos 30° cos 40°i - 80 cos 30° sin 40°j + 80 sin 30°k} lb F1 = {53.1i - 44.5j + 40k} lb F2 = { -130k} lb FR = F1 + F2 FR = {53.1i - 44.5j - 90.0k} lb FR = 2(53.1)2 + (- 44.5)2 + ( -90.0)2 = 114 lb

Ans.

a = cos-1 ¢

53.1 ≤ = 62.1° 113.6

Ans.

b = cos-1 ¢

-44.5 ≤ = 113° 113.6

Ans.

g = cos-1 ¢

-90.0 ≤ = 142° 113.6

Ans.

Ans: FR = 114 lb a = 62.1° b = 113° g = 142° 84

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*2–64. Specify the coordinate direction angles of F1 and F2 and express each force as a Cartesian vector.

z F1

80 lb y

30 40 F2

SOLUTION

x

F1 = {80 cos 30° cos 40°i - 80 cos 30° sin 40°j + 80 sin 30°k} lb F1 = {53.1i - 44.5j + 40k} lb

130 lb

Ans.

a1 = cos-1 ¢

53.1 ≤ = 48.4° 80

Ans.

b 1 = cos-1 ¢

- 44.5 ≤ = 124° 80

Ans.

g1 = cos-1 a

40 b = 60° 80

Ans. Ans.

F2 = {-130k} lb a2 = cos-1 ¢

0 ≤ = 90° 130

Ans.

b 2 = cos-1 ¢

0 ≤ = 90° 130

Ans.

g2 = cos-1 ¢

- 130 ≤ = 180° 130

Ans.

Ans: F1 = a1 = b1 = g1 = F2 = a2 = b2 = g2 = 85

{53.1i - 44.5j + 40k} lb 48.4° 124° 60° { - 130k} lb 90° 90° 180°

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2–65. The screw eye is subjected to the two forces shown. Express each force in Cartesian vector form and then determine the resultant force. Find the magnitude and coordinate direction angles of the resultant force.

z

F1 = 300 N

60° 120°

45° y

SOLUTION 45°

F1 = 300( - cos 60° sin 45° i + cos 60° cos 45° j + sin 60°k) = {-106.07 i + 106.07 j + 259.81 k} N

x

60°

Ans.

= {- 106 i + 106 j + 260 k} N

F2 = 500 N

F2 = 500(cos 60° i + cos 45° j + cos 120° k) = {250.0 i + 353.55 j - 250.0k} N = {250 i + 354 j - 250 k} N

Ans.

FR = F1 + F2 = - 106.07 i + 106.07 j + 259.81 k + 250.0 i + 353.55 j - 250.0 k = 143.93 i + 459.62 j + 9.81k = {144 i + 460 j + 9.81 k} N

Ans.

FR = 2143.932 + 459.622 + 9.812 = 481.73 N = 482 N uFR =

Ans.

143.93i + 459.62j + 9.81k FR = 0.2988i + 0.9541j + 0.02036k = FR 481.73

cos a = 0.2988

a = 72.6°

Ans.

cos b = 0.9541

b = 17.4°

Ans.

cos g

g

88.8°

Ans.

0.02036

Ans: F1 = { - 106i + 106j + 260k} N F2 = {250i + 354j - 250k} N FR = {144i + 460j + 9.81k} N FR = 482 N a = 72.6° b = 17.4° g = 88.8° 86

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2–66. Determine the coordinate direction angles of F1.

z

F1 = 300 N

60° 120°

45° y

SOLUTION 45°

F1 = 300(- cos 60° sin 45° i + cos 60° cos 45° j + sin 60° k) = { - 106.07 i + 106.07 j + 259.81 k} N

x

60°

= { -106 i + 106 j + 260 k} N F2 = 500 N

F1 = - 0.3536 i + 0.3536 j + 0.8660 k u1 = 300 a1 = cos-1 (- 0.3536) = 111°

Ans.

b 1 = cos-1 (0.3536) = 69.3°

Ans.

g1 = cos-1 (0.8660) = 30.0°

Ans.

Ans: a1 = 111° b1 = 69.3° g1 = 30.0° 87

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2–67. Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces acts along the positive y axis and has a magnitude of 600 lb.

z F3

F1  180 lb

SOLUTION

40

FRx = ©Fx ;

0 = - 180 + 300 cos 30° sin 40° + F3 cos a

FRy = ©Fy ;

600 = 300 cos 30° cos 40° + F3 cos b

FRz = ©Fz ;

0 = -300 sin 30° + F3 cos g

x

cos2 a + cos2 b + cos2 g = 1

y

30

F2  300 lb

Solving: F3 = 428 lb

Ans.

a = 88.3°

Ans.

b = 20.6°

Ans.

g = 69.5°

Ans.

Ans: F3 = 428 lb a = 88.3° b = 20.6° g = 69.5° 88

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*2–68. Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero.

z F3

F1  180 lb

SOLUTION

40

FRx = ©Fx;

0 = -180 + 300 cos 30° sin 40° + F3 cos a

FRy = ©Fy;

0 = 300 cos 30° cos 40° + F3 cos b

FRz = ©Fz;

0 = -300 sin 30° + F3 cos g

x

cos2a + cos2 b + cos2g = 1

y

30

F2  300 lb

Solving: F3 = 250 lb

Ans.

a = 87.0°

Ans.

b = 143°

Ans.

g = 53.1°

Ans.

Ans: F3 = 250 lb

a = 87.0° b = 143° g = 53.1°

89

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2–69. Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

z F2  125 N 3

5 4

20

x

y 60 45

60 F1  400 N

Solution Cartesian Vector Notation. For F1 and F2, F1 = 400 (cos 45°i + cos 60°j - cos 60°k) = {282.84i + 200j - 200k} N 4 4 3 F2 = 125 c (cos 20°)i - (sin 20°)j + k d = {93.97i - 34.20j + 75.0k} 5 5 5

Resultant Force.

FR = F1 + F2 = {282.84i + 200j - 200k} + {93.97i - 34.20j + 75.0k} = {376.81i + 165.80j - 125.00k} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2376.812 + 165.802 + ( -125.00)2 = 430.23 N = 430 N

Ans.

The coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

376.81 ; 430.23

a = 28.86° = 28.9°

Ans.

=

165.80 ; 430.23

b = 67.33° = 67.3°

Ans.

=

- 125.00 ; 430.23

g = 106.89° = 107°

Ans.

Ans: FR =  a =  b =  g = 90

430 N 28.9° 67.3° 107°

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2–70. z

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system. F2  525 N

60

120

45 x

y 5

4 3

F1  450 N

Solution Cartesian Vector Notation. For F1 and F2, 3 4 F1 = 450 a j - kb = {270j - 360k} N 5 5

F2 = 525 (cos 45°i + cos 120°j + cos 60°k) = {371.23i - 262.5j + 262.5k} N

Resultant Force. FR = F1 + F2 = {270j - 360k} + {371.23i - 262.5j + 262.5k} = {371.23i + 7.50j - 97.5k} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2371.232 + 7.502 + ( - 97.5)2 = 383.89 N = 384 N

Ans.

The coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

371.23 ; 383.89

a = 14.76° = 14.8°

Ans.

=

7.50 ; 383.89

b = 88.88° = 88.9°

Ans.

=

-97.5 ; 383.89

g = 104.71° = 105°

Ans.

Ans: FR = 384 N 371.23 cos a = ; a = 14.8° 383.89 7.50 cos b = ;  b = 88.9° 383.89 -97.5 cos g = ; g = 105° 383.89 91

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2–71. z

Specify the magnitude and coordinate direction angles a1, b 1, g1 of F1 so that the resultant of the three forces acting on the bracket is FR = 5- 350k6 lb. Note that F3 lies in the x–y plane.

F3 = 400 lb

γ1

30° F2 = 200 lb

SOLUTION

β1 α1

F1 = Fx i + Fy j + Fz k F2 = - 200 j

F1

x

F3 = - 400 sin 30° i + 400 cos 30° j = - 200 i + 346.4 j FR = ©F -350 k = Fx i + Fy j + Fz k - 200 j - 200 i + 346.4 j 0 = Fx - 200 ;

Fx = 200 lb

0 = Fy - 200 + 346.4 ;

Fy = - 146.4 lb

Fz = - 350 lb F1 = 2(200)2 + (-146.4)2 + (- 350)2 Ans.

F1 = 425.9 lb = 429 lb a1 = cos-1 a

200 b = 62.2° 428.9

Ans.

b 1 = cos-1 a

- 146.4 b = 110° 428.9

Ans.

g1 = cos-1

- 350 428.9

Ans.

= 145°

Ans: F1 = a1 = b1 = g1 = 92

429 lb 62.2° 110° 145°

y

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*2–72. z

Two forces F1 and F2 act on the screw eye. If the resultant force FR has a magnitude of 150 lb and the coordinate direction angles shown, determine the magnitude of F2 and its coordinate direction angles.

F2

FR  150 lb

g 130

120

F1  80 lb

y

x

Solution Cartesian Vector Notation. For FR, g can be determined from cos2 a + cos2 b + cos2 g = 1 cos2 120° + cos2 50° + cos2 g = 1 cos g = {0.5804 Here g 6 90°, then g = 54.52° Thus FR = 150(cos 120°i + cos 50°j + cos 54.52°k) = { - 75.0i + 96.42j + 87.05k} lb Also F1 = {80j} lb Resultant Force. FR = F1 + F2 { -75.0i + 96.42j + 87.05k} = {80j} + F2 F2 = { - 75.0i + 16.42j + 87.05k} lb Thus, the magnitude of F2 is



F2 = 2(F2)x + (F2)y + (F2)z = 2( - 75.0)2 + 16.422 + 87.052 = 116.07 lb = 116 lb

Ans.

And its coordinate direction angles are

cos a2 =



cos b2 =



cos g2 =

(F2)x F2 (F2)y F2 (F2)z F2

=

- 75.0 ; 116.07

a2 = 130.25° = 130°

Ans.

=

16.42 ; 116.07

b2 = 81.87° = 81.9°

Ans.

=

87.05 ; 116.07

g2 = 41.41° = 41.4°

Ans.

Ans:     FR  cos a2 cos b 2  cos g2 93

= = = =

116 lb 130° 81.9° 41.4°

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z

2–73. Express each force in Cartesian vector form.

F3  200 N

5

F1  90 N 3

F2  150 N

60

y

4

45

Solution x

Cartesian Vector Notation. For F1, F2 and F3, 4 3 F1 = 90 a i + kb = {72.0i + 54.0k} N 5 5

Ans.

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N Ans.

= {53.0i + 53.0j + 130k} N F3 = {200 k}

Ans.

Ans: F1 = {72.0i + 54.0k} N F2 = {53.0i + 53.0j + 130k} N F3 = {200 k} 94

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z

2–74. Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F3  200 N

5

F1  90 N 3

F2  150 N

60

y

4

45

Solution x

Cartesian Vector Notation. For F1, F2 and F3, 4 3 F1 = 90 a i + kb = {72.0i + 54.0k} N 5 5

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N F3 = {200 k} N Resultant Force. F = F1 + F2 + F3 = (72.0i + 54.0k) + (53.03i + 53.03j + 129.90k) + (200k) = {125.03i + 53.03j + 383.90} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2125.032 + 53.032 + 383.902 = 407.22 N = 407 N



Ans.

And the coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

125.03 ; 407.22

a = 72.12° = 72.1°

Ans.

=

53.03 ; 407.22

b = 82.52° = 82.5°

Ans.

=

383.90 ; 407.22

g = 19.48° = 19.5°

Ans.

Ans: FR =  a =  b =  g = 95

407 N 72.1° 82.5° 19.5°

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2–75. z

The spur gear is subjected to the two forces caused by contact with other gears. Express each force as a Cartesian vector.

F2

SOLUTION F1 =

180 lb

60

135

60

24 7 (50)j (50)k = {14.0j - 48.0k} lb 25 25

Ans.

F2 = 180 cos 60°i + 180 cos 135°j + 180 cos 60°k

y x

25 24 7

Ans.

= {90i - 127j + 90k} lb

F1

50 lb

Ans: F1 = {14.0j - 48.0k} lb F2 = {90i - 127j + 90k} lb 96

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*2–76. z

The spur gear is subjected to the two forces caused by contact with other gears. Determine the resultant of the two forces and express the result as a Cartesian vector.

F2

SOLUTION

180 lb

60

135

60

FRx = 180 cos 60° = 90 FRy =

y x

7 (50) + 180 cos 135° = -113 25

25 24 7

FRz

24 = - (50) + 180 cos 60° = 42 25

F1

FR = {90i - 113j + 42k} lb

50 lb

Ans.

Ans: FRx = 90 FRy = -113 FRz = 42 FR = {90i - 113j + 42k} lb 97

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2–77. z

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F1 = 400 N 60

135 20

60

60

y

x F2  500 N

Solution Cartesian Vector Notation. For F1 and F2, F1 = 400 (sin 60° cos 20°i - sin 60° sin 20°j + cos 60°k) = {325.52i - 118.48j + 200k} N F2 = 500 (cos 60°i + cos 60°j + cos 135°k) = {250i + 250j - 353.55k} N Resultant Force. FR = F1 + F2 = (325.52i - 118.48j + 200k) + (250i + 250j - 353.55k) = {575.52i + 131.52j - 153.55 k} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2575.522 + 131.522 + ( -153.55)2 = 610.00 N = 610 N

 Ans.

The coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

575.52 610.00

a = 19.36° = 19.4°

Ans.

=

131.52 610.00

b = 77.549° = 77.5°

Ans.

=

- 153.55 610.00

g = 104.58° = 105°

Ans.

Ans: FR =  a =  b =  g = 98

610 N 19.4° 77.5° 105°

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2–78. z

The two forces F1 and F2 acting at A have a resultant force of FR = 5 - 100k6 lb. Determine the magnitude and coordinate direction angles of F2 .

B

30°

SOLUTION

A

Cartesian Vector Notation:

x

FR = 5 - 100 k6 lb

F2

y

50°

F1 = 60 lb

F1 = 605 - cos 50° cos 30° i + cos 50° sin 30° j - sin 50° k6 lb = 5 - 33.40 i + 19.28 j - 45.96 k6 lb

F2 = 5F2x i + F2y j + F2z k6 lb

Resultant Force:

FR = F1 + F2 - 100k =

5( F2

x

- 33.40 ) i + ( F2y + 19.28 ) j + ( F2z - 45.96 ) k6

Equating i, j and k components, we have F2x - 33.40 = 0

F2x = 33.40 lb

F2y + 19.28 = 0

F2y = -19.28 lb

F2z - 45.96 = - 100

F2z = -54.04 lb

The magnitude of force F2 is F2 = 2F 22x + F 22y + F 22z = 233.402 + ( -19.28)2 + ( - 54.04)2 Ans.

= 66.39 lb = 66.4 lb The coordinate direction angles for F2 are cos a = cos b = cos g =

F2x F2 F2y F2 F2z F2

=

33.40 66.39

a = 59.8°

Ans.

=

- 19.28 66.39

b = 107°

Ans.

=

-54.04 66.39

g = 144°

Ans.

Ans: F2 = 66.4 lb a = 59.8° b = 107° g = 144° 99

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2–79. z

Determine the coordinate direction angles of the force F1 and indicate them on the figure. B

30°

SOLUTION

A

Unit Vector For Force F1:

x

uF1 = - cos 50° cos 30° i + cos 50° sin 30° j - sin 50° k

F2

y

50°

F1 = 60 lb

= - 0.5567 i + 0.3214 j - 0.7660 k Coordinate Direction Angles: From the unit vector obtained above, we have cos a = - 0.5567

a = 124°

Ans.

cos b = 0.3214

b = 71.3°

Ans.

cos g = -0.7660

g = 140°

Ans.

Ans: a = 124° b = 71.3° g = 140° 100

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*2–80. The bracket is subjected to the two forces shown. Express each force in Cartesian vector form and then determine the resultant force FR. Find the magnitude and coordinate direction angles of the resultant force.

z F2 = 400 N 60°

45° 120°

SOLUTION

y 25°

Cartesian Vector Notation: 35°

F1 = 2505cos 35° sin 25°i + cos 35° cos 25°j - sin 35°k6 N x

= 586.55i + 185.60j - 143.39k6 N

F1 = 250 N

= 586.5i + 186j - 143k6 N

Ans.

F2 = 4005cos 120°i + cos 45°j + cos 60°k6 N = 5-200.0i + 282.84j + 200.0k6 N = 5 -200i + 283j + 200k6 N

Ans.

Resultant Force: FR = F1 + F2 = 5186.55 - 200.02i + 1185.60 + 282.842j + 1- 143.39 + 200.02 k6 = 5-113.45i + 468.44j + 56.61k6 N = 5 -113i + 468j + 56.6k6 N

Ans.

The magnitude of the resultant force is FR = 2F2Rx + F2Ry + F2Rz = 21 -113.4522 + 468.442 + 56.612 Ans.

= 485.30 N = 485 N The coordinate direction angles are cos a = cos b = cos g =

FRx FR F Ry FR FRz FR

=

-113.45 485.30

= =

a = 104°

Ans.

468.44 485.30

b = 15.1°

Ans.

56.61 485.30

g = 83.3°

Ans.

Ans: F1 = {86.5i + 186j - 143k} N F2 = { - 200i + 283j + 200k} N FR = { - 113i + 468j + 56.6k} N FR = 485 N

a = 104° b = 15.1° g = 83.3°

101

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2–81. z

If the coordinate direction angles for F3 are a3 = 120°, b 3 = 60° and g3 = 45°, determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt.

F3  800 lb

5 4 3

SOLUTION Force Vectors: By resolving F1, F2 and F3 into their x, y, and z components, as shown in Figs. a, b, and c, respectively, F1, F2 and F3 can be expressed in Cartesian vector form as F1 = 700 cos 30°(+ i) + 700 sin 30°(+j) = 5606.22i + 350j6 lb

F2  600 lb

30 y

x F1  700 lb

4 3 F2 = 0i + 600 a b ( +j) + 600 a b ( + k) = 5480j + 360k6 lb 5 5 F3 = 800 cos 120°i + 800 cos 60°j + 800 cos 45°k = 3-400i + 400j + 565.69k4 lb Resultant Force: By adding F1, F2 and F3 vectorally, we obtain FR. Thus, FR = F1 + F2 + F3 = (606.22i + 350j) + (480j + 360k) + ( - 400i + 400j + 565.69k) = 3206.22i + 1230j + 925.69k4 lb The magnitude of FR is FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 3(206.22)2 + (1230)2 + (925.69)2 = 1553.16 lb = 1.55 kip

Ans.

The coordinate direction angles of FR are a = cos-1 c

(FR)x 206.22 b = 82.4° d = cos-1 a FR 1553.16

b = cos-1 c

(FR)y

g = cos-1 c

(FR)z

FR FR

Ans.

d = cos-1 a

1230 b = 37.6° 1553.16

Ans.

d = cos-1 a

925.69 b = 53.4° 1553.16

Ans.

Ans: FR = 1.55 kip a = 82.4° b = 37.6° g = 53.4° 102

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2–82. z

If the coordinate direction angles for F3 are a3 = 120°, b 3 = 45° and g3 = 60°, determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt.

F3  800 lb

5 4 3

SOLUTION Force Vectors: By resolving F1, F2 and F3 into their x, y, and z components, as shown in Figs. a, b, and c, respectively, F1, F2, and F3 can be expressed in Cartesian vector form as F1 = 700 cos 30°( + i) + 700 sin 30°(+j) = 5606.22i + 350j6 lb

F2  600 lb

30 y

x F1  700 lb

3 4 F2 = 0i + 600 a b (+ j) + 600 a b( + k) = 5480j + 360k6 lb 5 5 F3 = 800 cos 120°i + 800 cos 45°j + 800 cos 60°k = 5- 400i + 565.69j + 400k6 lb FR = F1 + F2 + F3 = 606.22i + 350j + 480j + 360k - 400i + 565.69j + 400k = 5206.22i + 1395.69j + 760k6 lb FR = 3(206.22)2 + (1395.69)2 + (760)2 Ans.

= 1602.52 lb = 1.60 kip a = cos-1 a

206.22 b = 82.6° 1602.52

Ans.

b = cos-1 a

1395.69 b = 29.4° 1602.52

Ans.

g = cos-1 a

760 b = 61.7° 1602.52

Ans.

Ans: FR = 1.60 kip a = 82.6° b = 29.4° g = 61.7° 103

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2–83. z

If the direction of the resultant force acting on the eyebolt is defined by the unit vector uFR = cos 30°j + sin 30°k, determine the coordinate direction angles of F3 and the magnitude of FR.

F3  800 lb

5 4 3

SOLUTION Force Vectors: By resolving F1, F2 and F3 into their x, y, and z components, as shown in Figs. a, b, and c, respectively, F1, F2, and F3 can be expressed in Cartesian vector form as

F2  600 lb

30 y

x F1  700 lb

F1 = 700 cos 30°( + i) + 700 sin 30°( +j) = 5606.22i + 350j6 lb 3 4 F2 = 0i + 600a b( +j) + 600 a b ( +k) = 5480j + 360k6 lb 5 5 F3 = 800 cos a3i + 800 cos b 3 j + 800 cos g3k Since the direction of FR is defined by uFR = cos 30°j + sin 30°k, it can be written in Cartesian vector form as FR = FRuFR = FR(cos 30°j + sin 30°k) = 0.8660FR j + 0.5FR k Resultant Force: By adding F1, F2, and F3 vectorally, we obtain FR. Thus, FR = F1 + F2 + F3 0.8660FR j + 0.5FR k = (606.22i + 350j) + (480j + 360k) + (800 cos a3i + 800 cos b 3 j + 800 cos g3k) 0.8660FR j + 0.5FR k = (606.22 + 800 cos a3)i + (350 + 480 + 800 cos b 3)j + (360 + 800 cos g3)k Equating the i, j, and k components, we have 0 = 606.22 + 800 cos a3 800 cos a3 = -606.22

(1)

0.8660FR = 350 + 480 + 800 cos b 3 800 cos b 3 = 0.8660FR - 830

(2)

0.5FR = 360 + 800 cos g3 800 cos g3 = 0.5FR - 360

(3)

Squaring and then adding Eqs. (1), (2), and (3), yields 8002 [cos2 a3 + cos2 b 3 + cos2 g3] = FR 2 - 1797.60FR + 1,186,000 2

2

(4)

2

However, cos a3 + cos b 3 + cos g3 = 1. Thus, from Eq. (4) FR 2 - 1797.60FR + 546,000 = 0 Solving the above quadratic equation, we have two positive roots FR = 387.09 N = 387 N

Ans.

FR = 1410.51 N = 1.41 kN

Ans.

From Eq. (1), Ans.

a3 = 139° Substituting FR = 387.09 N into Eqs. (2), and (3), yields b 3 = 128°

Ans.

g3 = 102°

Substituting FR = 1410.51 N into Eqs. (2), and (3), yields b 3 = 60.7°

Ans.

g3 = 64.4° 104

Ans: a3 = 139° b3 = 128°, g3 = 102°, FR1 = 387 N b3 = 60.7°, g3 = 64.4°, FR2 = 1.41 kN

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*2–84. The pole is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 3 kN, b = 30°, and g = 75°, determine the magnitudes of its three components.

z Fz

SOLUTION

F

g b

cos2 a + cos2 b + cos2 g = 1

Fy

a

cos2 a + cos2 30° + cos2 75° = 1

Fx

a = 64.67° Fx = 3 cos 64.67° = 1.28 kN

Ans.

Fy = 3 cos 30° = 2.60 kN

Ans.

Fz = 3 cos 75° = 0.776 kN

Ans.

x

Ans: Fx = 1.28 kN Fy = 2.60 kN Fz = 0.776 kN 105

y

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2–85. The pole is subjected to the force F which has components Fx = 1.5 kN and Fz = 1.25 kN. If b = 75°, determine the magnitudes of F and Fy.

z Fz

SOLUTION

F

g b

cos2 a + cos2 b + cos2 g = 1

Fy

a

1.5 2 1.25 2 a b + cos2 75° + a b = 1 F F

Fx

F = 2.02 kN

Ans.

Fy = 2.02 cos 75° = 0.523 kN

Ans.

x

Ans: F = 2.02 kN Fy = 0.523 kN 106

y

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2–86. y

Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude.

B

300 mm

Solution O

Position Vector. The coordinates of points A and B are A( -150 cos 30°, - 150 sin 30°) mm and B(0, 300) mm respectively. Then

30

rAB = [0 - ( - 150 cos 30°)]i + [300 - ( - 150 sin 30°)]j

x

A 150 mm

= {129.90i + 375j} mm Thus, the magnitude of rAB is

rAB = 2129.902 + 3752 = 396.86 mm = 397 mm

Ans.

Ans: rAB = 397 mm 107

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2–87. Express force F as a Cartesian vector; then determine its coordinate direction angles.

z A

10

ft

F  135 lb

70 30

y

5 ft

Solution

B

rAB = (5 + 10 cos 70° sin 30°)i

7 ft x

+ ( - 7 - 10 cos 70° cos 30°)j - 10 sin 70°k rAB = {6.710i - 9.962j - 9.397k} ft rAB = 2(6.710)2 + ( - 9.962)2 + ( - 9.397)2 = 15.25 uAB =

rAB = (0.4400i - 0.6532j - 0.6162k) rAB

F = 135uAB = (59.40i - 88.18j - 83.18k) Ans.

= {59.4i - 88.2j - 83.2k} lb  a = cos - 1 a b = cos - 1 a g = cos - 1 a

59.40 b = 63.9° 135

Ans.

- 88.18 b = 131° 135

Ans.

- 83.18 b = 128° 135

Ans.

Ans: {59.4i - 88.2j - 83.2k} lb  a = 63.9°  b = 131°  g = 128° 108

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*2–88. Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

z C

13 12 5

O

2.5 ft 4 ft

Solution rAC

x

rAC F1 = 80 lb a b = - 26.20 i - 41.93 j + 62.89 k rAC = { - 26.2 i - 41.9 j + 62.9 k} lb

6 ft

Ans.



B

rAB = {2 i - 4 j - 6 k} ft rAB b = 13.36 i - 26.73 j - 40.09 k rAB

= {13.4 i - 26.7 j - 40.1 k} lb

y A F2  50 lb

12 = e - 2.5 i - 4 j + (2.5) k f ft 5

F2 = 50 lb a

F1  80 lb

2 ft

Ans.



FR = F1 + F2 = - 12.84 i - 68.65 j + 22.80 k = { - 12.8 i - 68.7 j + 22.8 k } lb FR = 2( - 12.84)2 ( - 68.65)2 + (22.80)2 = 73.47 = 73.5 lb

a = cos - 1 a



b = cos - 1 a



g = cos - 1 a

- 12.84 b = 100° 73.47

Ans. Ans.

- 68.65 b = 159° 73.47

Ans.

22.80 b = 71.9° 73.47

Ans.

Ans: F1 = { - 26.2 i - 41.9 j + 62.9 k} lb F2 = {13.4 i - 26.7 j - 40.1 k} lb FR = 73.5 lb  a = 100°  b = 159°  g = 71.9° 109

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2–89. If F = 5350i - 250j - 450k6 N and cable AB is 9 m long, determine the x, y, z coordinates of point A.

z A F

B

SOLUTION Position Vector: The position vector rAB, directed from point A to point B, is given by

x

y

z

x y

rAB = [0 - ( -x)]i + (0 - y)j + (0 - z)k = xi - yj - zk Unit Vector: Knowing the magnitude of rAB is 9 m, the unit vector for rAB is given by uAB =

xi - yj - zk rAB = rAB 9

The unit vector for force F is uF =

F = F

350i - 250j - 450k 3502 + ( - 250)2 + (-450)2

= 0.5623i - 0.4016j - 0.7229k

Since force F is also directed from point A to point B, then uAB = uF xi - yj - zk = 0.5623i - 0.4016j - 0.7229k 9 Equating the i, j, and k components, x = 0.5623 9

x = 5.06 m

Ans.

-y = -0.4016 9

y = 3.61 m

Ans.

-z = 0.7229 9

z = 6.51 m

Ans.

Ans: x = 5.06 m y = 3.61 m z = 6.51 m 110

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2–90. The 8-m-long cable is anchored to the ground at A. If x = 4 m and y = 2 m, determine the coordinate z to the highest point of attachment along the column.

z

B

Solution

z

r = {4i + 2j + zk} m r = 2(4)2 + (2)2 + (z)2 = 8

y

Ans.

z = 6.63 m

x

y

A

x

Ans: 6.63 m 111

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2–91. z

The 8-m-long cable is anchored to the ground at A. If z = 5 m, determine the location + x, + y of point A. Choose a value such that x = y.

B

Solution

z

r = {xi + yj + 5k} m r = 2(x)2 + (y)2 + (5)2 = 8

y

x = y, thus

2x2 = 82 - 52

x

y

A

x

Ans.

x = y = 4.42 m

Ans:  4.42 m 112

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*2–92. z

Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

0.75 m A FAB  250 N FAC  400 N

3m

y 40

2m 1m

Unit Vectors. The coordinates for points A, B and C are (0, - 0.75, 3) m, B(2 cos 40°, 2 sin 40°, 0) m and C(2, -1, 0) m respectively. uAB =

B

C

Solution

2m

x

(2 cos 40° - 0)i + [2 sin 40° - ( -0.75)]j + (0 - 3)k rAB = rAB 2(2 cos 40° - 0)2 + [2 sin 40° - ( - 0.75)]2 + (0 - 3)2 = 0.3893i + 0.5172j - 0.7622k

uAC =

(2 - 0)i + [ - 1 - ( - 0.75)]j + (0 - 3)k rAC = rAC 2(2 - 0)2 + [ - 1 - ( - 0.75)]2 + (0 - 3)2 = 0.5534i - 0.0692j - 0.8301k

Force Vectors FAB = FAB uAB = 250 (0.3893i + 0.5172j - 0.7622k) = {97.32i + 129.30j - 190.56k} N = {97.3i + 129j - 191k} N

Ans.



FAC = FAC uAC = 400 (0.5534i - 0.06917j - 0.8301k) = {221.35i - 27.67j - 332.02k} N = {221i - 27.7j - 332k} N

Ans.



Resultant Force FR = FAB + FAC = {97.32i + 129.30j - 190.56k} + {221.35i - 27.67j - 332.02k} = {318.67i + 101.63j - 522.58 k} N The magnitude of FR is FR = 2(FR)2x + (FR)2y + (FR)2z = 2318.672 + 101.632 + ( -522.58)2 = 620.46 N = 620 N

And its coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

318.67 ; 620.46

101.63 ; = 620.46 =

- 522.58 ; 620.46

Ans.

a = 59.10° = 59.1° b = 80.57° = 80.6°

Ans.

g = 147.38° = 147°

Ans.

113

Ans: FAB FAC  FR cos a cos b cos g

= = = = = =

{97.3i - 129j - 191k} N {221i - 27.7j - 332k} N 620 N 59.1° 80.6° 147°

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2–93. If 560 N and 700 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

6m

A FB FC

SOLUTION Force Vectors: The unit vectors u and u of F and F must be determined first. From Fig. a u u

(2

r

0)i

( 3 2

(2 0) (3 0)i

r

(0

6)k

2

0)2

(2

2 i 7

2

( 3 0) (0 6) (2 0)j (0 6)k

0)2

(3

0)j

6)2

(0

3 j 7

3 i 7

2 j 7

2m B

6 k 7

3m

6 k 7

x

3m

2m C

Thus, the force vectors F and F are given by F

u

560

2 i 7

3 j 7

6 k 7

160i

240j

480k N

F

u

700

3 i 7

2 j 7

6 k 7

300i

200j

600k N

Resultant Force: F

F

F

460i

(160i 40j

240j

480k)

(300i

200j

600k)

1080k N

The magnitude of F is (

)

2

(460)2

(

)

2

( 40)2

(

)

2

( 1080)2

1174.56 N

Ans.

1.17 kN

The coordinate direction angles of F are cos

1

cos

1

cos

1

(

)

(

)

(

)

cos

1

460 1174.56

66.9°

Ans.

cos

1

40 1174.56

92.0°

Ans.

cos

1

1080 1174.56

157°

Ans.

Ans: FR = 1.17 kN a = 66.9° b = 92.0° g = 157° 114

y

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2–94. If 700 N, and 560 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

6m

A FB FC

SOLUTION Force Vectors: The unit vectors u and u of F and F must be determined first. From Fig. a u u

(2

r

0)i 0)2

(2 (3

r

( 3

0)i

(2

0)

(0

0)2

( 3

2

(3

0)j 0)j 0)

6)2

(0

(0 2

(2

6)k 6)k 2

(0

6)

2 i 7

3 j 7

3 i 7

2 j 7

2m B 3m

6 k 7 x

6 k 7

3m

2m C

Thus, the force vectors F and F are given by F

u

700

2 i 7

3 j 7

6 k 7

200i

300j

600k N

F

u

560

3 i 7

2 j 7

6 k 7

240i

160j

480k N

Resultant Force: F

F

F

440i

(200i 140j

300j

600k)

(240i

160j

480k)

1080k N

The magnitude of F is (

)

2

(440)2

(

)

2

( 140)2

(

)

2

( 1080)2

1174.56 N

1.17 kN

Ans.

The coordinate direction angles of F are cos

1

cos

1

cos

1

(

)

(

)

(

)

cos

1

440 1174.56

68.0°

Ans.

cos

1

140 1174.56

96.8°

Ans.

cos

1

1080 1174.56

157°

Ans.

Ans: FR = 1.17 kN a = 68.0° b = 96.8° g = 157° 115

y

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2–95. z

The plate is suspended using the three cables which exert the forces shown. Express each force as a Cartesian vector. A

FCA  500 lb

Solution FBA = 350 a



FCA

= { -109 i + 131 j + 306 k} lb

FDA  400 lb

C

Ans.

B D

3 ft

rCA 3 14 3 = 500 a b = 500 a i + j + kb rCA 14.629 14.629 14.629

3 ft

6 ft

3 ft 3 ft 2 ft

y

x

= {103 i + 103 j + 479 k} lb

FDA = 400 a



rBA 5 6 14 b = 350 a i + j + kb rBA 16.031 16.031 16.031

14 ft

FBA  350 lb

Ans.

rDA 6 14 2 i j + kb b = 400 a rDA 15.362 15.362 15.362

= { -52.1 i - 156 j + 365 k} lb

Ans.

Ans: FBA = { - 109 i + 131 j + 306 k} lb FCA = {103 i + 103 j + 479 k} lb FDA = { - 52.1 i - 156 j + 365 k} lb 116

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*2–96. The three supporting cables exert the forces shown on the sign. Represent each force as a Cartesian vector.

z C

2m E

2m B

FE  350 N 3m

FC  400 N FB  400 N

Solution

D 2m

rC = (0 - 5)i + ( -2 - 0)j + (3 - 0)k = { -5i - 2j + 3k} m

A

rC = 2( - 5)2 + ( - 2)2 + 32 = 238 m

y

3m

x

rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k} m rB = 2( - 5)2 + 22 + 32 = 238 m

rE = (0 - 2)i + (0 - 0)j + (3 - 0)k = { - 2i + 0j + 3k} m rE = 2( - 2)2 + 02 + 32 = 213 m r   F = Fu = F a b r FC = 400 a FB = 400 a FE = 350 a

- 5i - 2j + 3k 138

b = { - 324i - 130j + 195k} N

Ans.

138

b = { - 324i + 130j + 195k} N

Ans.

113

b = { - 194i + 291k} N

Ans.

- 5i + 2j + 3k - 2i + 0j + 3k

Ans: FC = { - 324i - 130j + 195k} N FB = { - 324i + 130j + 195k} N FE = { - 194i + 291k} N 117

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2–97. z

Determine the magnitude and coordinate direction angles of the resultant force of the two forces acting on the sign at point A.

C

2m E

2m B

FE  350 N 3m

FC  400 N FB  400 N

Solution

D 2m

 rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { - 5i - 2j + 3k}

A

 rC = 2( - 5)2 + ( - 2)2 + (3)2 = 238 m

y

3m

x

( - 5i - 2j + 3k) rC FC = 400 a b = 400 a b rC 138

FC = ( -324.4428i - 129.777j + 194.666k) rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k}  rB = 2( - 5)2 + 22 + 32 = 238 m FB = 400 a

( - 5i + 2j + 3k) rB b = 400 a b rB 138

FB = ( -324.443i + 129.777j + 194.666k) FR = FC + FB = ( - 648.89i + 389.33k)

FR = 2( - 648.89)2 + (389.33)2 + 02 = 756.7242

Ans.

FR = 757 N a = cos-1 a b = cos-1 a

g = cos-1 a

- 648.89 b = 149.03 = 149° 756.7242

Ans.

0 b = 90.0° 756.7242

Ans.

389.33 b = 59.036 = 59.0° 756.7242

Ans.

Ans: FR = 757 N a = 149° b = 90.0° g = 59.0° 118

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2–98. The force F has a magnitude of 80 lb and acts at the midpoint C of the thin rod. Express the force as a Cartesian vector.

z B

6 ft C F  80 lb

Solution rAB = ( -3i + 2j + 6k) rCB =

O 3 ft

1 rAB = ( - 1.5i + 1j + 3k) 2

A

rCO = rBO + rCB

= - 6k - 1.5i + 1j + 3k



= - 1.5i + 1j - 3k

y

2 ft x

rCO = 3.5 F = 80 a

rCO b = { - 34.3i + 22.9j - 68.6k} lb rCO

Ans.

Ans: F = { -34.3i + 22.9j - 68.6k} lb 119

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2–99. z

The load at A creates a force of 60 lb in wire AB. Express this force as a Cartesian vector acting on A and directed toward B as shown.

30 5 ft

y B

SOLUTION 10 ft

Unit Vector: First determine the position vector rAB. The coordinates of point B are x

B (5 sin 30°, 5 cos 30°, 0) ft = B (2.50, 4.330, 0) ft

F

Then

60 lb

A

rAB = 5(2.50 - 0)i + (4.330 - 0)j + [0 - ( - 10)]k6 ft = 52.50i + 4.330j + 10k6 ft

rAB = 32.502 + 4.3302 + 10.02 = 11.180 ft uAB =

2.50i + 4.330j + 10k rAB = rAB 11.180 = 0.2236i + 0.3873j + 0.8944k

Force Vector: F = FuAB = 60 50.2236i + 0.3873j + 0.8944k6 lb = 513.4i + 23.2j + 53.7k6 lb

Ans.

Ans: F = {13.4i + 23.2j + 53.7k} lb 120

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*2–100. z

Determine the magnitude and coordinate direction angles of the resultant force acting at point A on the post. FAC 150 N

C

4 5

A

FAB  200 N

3m 3m

3

y O

2m B

4m

Solution Unit Vector. The coordinates for points A, B and C are A(0, 0, 3) m, B(2, 4, 0) m and C( - 3, -4, 0) m respectively

x

rAB = (2 - 0)i + (4 - 0)j + (0 - 3)k = {2i + 4j - 3k} m uAB =

2i + 4j - 3k rAB 2 4 3 = = i + j k rAB 222 + 42 + ( - 3)2 229 229 229

rAC = ( -3 - 0)i + ( -4 - 0)j + (0 - 3)k = { - 3i - 4j - 3k} m uAC =

- 3i - 4j - 3k rAC 3 4 3 = = i j k 2 2 2 rAC 2( - 3) + ( - 4) + ( - 3) 234 234 234

Force Vectors

FAB = FAB uAB = 200 a

2 229

i +

4 229

j -

= {74.28i + 148.56j - 111.42k} N

FAC = FAC uAC = 150 a -

3 234

i -

4 234

= { - 77.17i - 102.90j - 77.17k} N

3 229

j -

kb

3 234

kb

Resultant Force FR = FAB + FAC = {74.28i + 148.56j - 111.42k} + { - 77.17i - 102.90j - 77.17k} = { - 2.896i + 45.66j - 188.59 k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR)2y + (FR)2z = 2( - 2.896)2 + 45.662 + ( - 188.59)2

 Ans.

= 194.06 N = 194 N

And its coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

- 2.896 ; 194.06

a = 90.86° = 90.9°

Ans.

=

45.66 ; 194.06

b = 76.39° = 76.4°

Ans.

=

- 188.59 ; 194.06

g = 166.36° = 166°

Ans.

121

Ans:  FR cos a cos b cos g

= = = =

194 N 90.9° 76.4° 166°

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2–101. The two mooring cables exert forces on the stern of a ship as shown. Represent each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant.

z

C FA = 200 lb

SOLUTION Unit Vector:

10 ft

rCA = 5150 - 02i + 110 - 02j + 1 -30 - 02k6 ft = 550i + 10j - 30k6 ft

x

rCA = 250 + 10 + 1 - 302 = 59.16 ft 2

uCA =

2

2

FB = 150 lb 50 ft

40 ft A

y

30 ft B

50i + 10j - 30k rCA = 0.8452i + 0.1690j - 0.5071k = rCA 59.16

rCB = 5150 - 02i + 150 - 02j + 1 -30 - 02k6 ft = 550i + 50j - 30k6 ft rCB = 2502 + 502 + 1 -3022 = 76.81 ft

uCB =

50i + 50j - 30k rCA = 0.6509i + 0.6509j - 0.3906k = rCA 76.81

Force Vector: FA = FA uCA = 20050.8452i + 0.1690j - 0.5071k6 lb = 5169.03i + 33.81j - 101.42k6 lb = 5169i + 33.8j - 101k6 lb

Ans.

FB = FB uCB = 15050.6509i + 0.6509j - 0.3906k6 lb = 597.64i + 97.64j - 58.59k6 lb = 597.6i + 97.6j - 58.6k6 lb

Ans.

Resultant Force: FR = FA + FB = 51169.03 + 97.642i + 133.81 + 97.642j + 1- 101.42 - 58.592k6 lb = 5266.67i + 131.45j - 160.00k6 lb The magnitude of FR is FR = 2266.672 + 131.452 + 1 -160.0022 Ans.

= 337.63 lb = 338 lb The coordinate direction angles of FR are cos a =

266.67 337.63

a = 37.8°

Ans.

cos b =

131.45 337.63

b = 67.1°

Ans.

g = 118°

Ans.

cos g = -

160.00 337.63

122

Ans: FA = {169i + 33.8j - 101k} lb FB = {97.6i + 97.6j - 58.6k} lb FR = 338 lb a = 37.8° b = 67.1° g = 118°

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2–102. The engine of the lightweight plane is supported by struts that are connected to the space truss that makes up the structure of the plane. The anticipated loading in two of the struts is shown. Express each of these forces as a Cartesian vector.

z 3 ft F2

600 lb 0.5 ft B A F1

SOLUTION F1 = 400 a

C

2.5 ft

D 3 ft

rCD 0.5 0.5 3 i j + kb b = 400a rCD 3.0822 3.0822 3.0822

y x

= {389 i - 64.9 j + 64.9 k} lb F2 = 600 a

400 lb

0.5 ft

Ans.

0.5 ft

rAB 0.5 0.5 3 i j + kb b = 600 a rAB 3.0822 3.0822 3.0822

= {- 584 i + 97.3 j - 97.3k} lb

Ans.

Ans: F1 = {389i - 64.9j + 64.9k} lb F2 = { - 584i + 97.3j - 97.3k} lb 123

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2–103. Determine the magnitude and coordinates on angles of the resultant force.

z A 40 lb 20 lb 4 ft

SOLUTION

2 ft 20

rAC = { -2 sin 20°i + (2 + 2 cos 20°) j - 4 k} ft uAC = a

2 ft B

rAC b = - 0.1218i + 0.6910 j - 0.7125 k rAC

C y

1.5 ft

3 ft x

FAc = 4 lbuAC = { -4.874 i + 27.64 j - 28.50 k} lb rAB = {1.5 i - 1 j - 4 k} ft uAB = a

rAB b = 0.3419 i + 0.2279 j - 0.9117 k rAB

FAB = 20 lb uAB = {6.838 i - 4.558 j - 18.23 k} lb FR = FAB + FAC FR = {1.964 i + 23.08 j - 46.73 k}lb FR = 2(1.964)2 + (23.08)2 + ( -46.73)2 = 52.16 = 52.2 lb

Ans.

a = cos-1 a

1.964 b = 87.8° 52.16

Ans.

b = cos-1 a

23.08 b = 63.7° 52.16

Ans.

g = cos-1 a

- 46.73 b = 154° 52.16

Ans.

Ans: FR = 52.2 lb a = 87.8° b = 63.7° g = 154° 124

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*2–104. If the force in each cable tied to the bin is 70 lb, determine the magnitude and coordinate direction angles of the resultant force.

z

E

FB

Force Vectors: The unit vectors uA, uB, uC, and uD of FA, FB, FC, and FD must be determined first. From Fig. a, uA

FC

FA

SOLUTION

(3 - 0)i + ( -2 - 0)j + (0 - 6)k rA 2 6 3 = = = i - j - k 2 2 2 rA 7 7 7 2(3 - 0) + ( -2 - 0) + (0 - 6)

uB =

(3 - 0)i + (2 - 0)j + (0 - 6)k rB 2 6 3 = = i + j - k 2 2 2 rB 7 7 7 2(3 - 0) + (2 - 0) + (0 - 6)

uC =

( - 3 - 0)i + (2 - 0)j + (0 - 6)k rC 3 2 6 = = - i + j - k 2 2 2 rC 7 7 7 2( -3 - 0) + (2 - 0) + (0 - 6)

uD =

(- 3 - 0)i + ( - 2 - 0)j + (0 - 6)k rD 6 3 2 = = - i- j - k rD 7 7 7 2(- 3 - 0)2 + ( -2 - 0)2 + (0 - 6)2

6 ft

FD D

A x

2 ft

2 ft

C B

3 ft y 3 ft

Thus, the force vectors FA, FB, FC, and FD are given by 2 6 3 FA = FAuA = 70 a i - j - kb = [30i - 20j - 60k] lb 7 7 7 2 6 3 FB = FBuB = 70a i + j - kb = [30i + 20j - 60k] lb 7 7 7 3 2 6 FC = FCuC = 70 a - i + j - kb = [- 30i + 20j - 60k] lb 7 7 7 3 2 6 FD = FDuD = 70 a - i - j - k b = [- 30i - 20j - 60k] lb 7 7 7 Resultant Force: FR = FA + FB + FC + FD = (30i - 20j - 60k) + (30i + 20j - 60k) + (- 30i + 20j - 60k) + ( - 30i - 20j - 60k) = {-240k} N The magnitude of FR is FR = 2(FR)x2 + (FR)y2 + (FR)z2 = 20 + 0 + ( -240)2 = 240 lb

Ans.

The coordinate direction angles of FR are a = cos-1 B b = cos-1 B g = cos-1 B

(FR)x 0 b = 90° R = cos-1 a FR 240 (FR)y FR (FR)z FR

Ans.

R = cos-1 a

0 b = 90° 240

Ans.

R = cos-1 a

-240 b = 180° 240

Ans.

Ans: FR = 240 lb

a = 90° b = 90° g = 180°

125

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2–105. If the resultant of the four forces is FR = 5- 360k6 lb, determine the tension developed in each cable. Due to symmetry, the tension in the four cables is the same.

z

E

FB

Force Vectors: The unit vectors uA, uB, uC, and uD of FA, FB, FC, and FD must be determined first. From Fig. a, uA =

FC

FA

SOLUTION

(3 - 0)i + ( -2 - 0)j + (0 - 6)k rA 2 6 3 = = i - j - k 2 2 2 rA 7 7 7 2(3 - 0) + (-2 - 0) + (0 - 6)

6 ft

FD D

A x

2 ft

2 ft

C B

3 ft y 3 ft

(3 - 0)i + (2 - 0)j + (0 - 6)k rB 3 2 6 uB = = = i + j - k rB 7 7 7 2(3 - 0)2 + (2 - 0)2 + (0 - 6)2 uC =

( -3 - 0)i + (2 - 0)j + (0 - 6)k rC 2 6 3 = = - i + j - k rC 7 7 7 2(-3 - 0)2 + (2 - 0)2 + (0 - 6)2

uD =

( -3 - 0)i + (- 2 - 0)j + (0 - 6)k rD 3 2 6 = = - i- j - k 2 2 2 rD 7 7 7 2(-3 - 0) + (- 2 - 0) + (0 - 6)

Since the magnitudes of FA, FB, FC, and FD are the same and denoted as F, the four vectors or forces can be written as 3 2 6 FA = FAuA = F a i - j - k b 7 7 7 2 6 3 FB = FBuB = F a i + j - kb 7 7 7 2 6 3 FC = FCuC = F a - i + j - kb 7 7 7 2 6 3 FD = FDuD = F a - i - j - kb 7 7 7 Resultant Force: The vector addition of FA, FB, FC, and FD is equal to FR. Thus, FR = FA + FB + FC + FD 2 6 2 6 2 6 2 6 3 3 3 3 {- 360k} = B F a i - j - kb R + B F a i + j - kb R + B Fa - i + j - k b + B F a - i - j - kb R 7 7 7 7 7 7 7 7 7 7 7 7 - 360k = -

24 k 7

Thus, 360 =

24 F 7

Ans.

F = 105 lb

Ans: F = 105 lb 126

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2–106. Express the force F in Cartesian vector form if it acts at the midpoint B of the rod.

z A

B

4m

O

Solution rAB =

4m

F  600 N

C

x

- 3i + 4j - 4k rAC = = - 1.5i + 2j - 2k 2 2

3m

6m

rAD = rAB + rBD

D

4m

y

rBD = rAD - rAB

= (4i + 6j - 4k) - ( - 1.5i + 2j - 2k)



= {5.5i + 4j - 2k} m

rBD = 2(5.5)2 + (4)2 + ( - 2)2 = 7.0887 m        F = 600 a

rBD b = 465.528i + 338.5659j - 169.2829k rBD

F = {466i + 339j - 169k} N

Ans.

Ans: F = {466i + 339j - 169k} N 127

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2–107. z

Express force F in Cartesian vector form if point B is located 3 m along the rod end C.

A

B

4m

O

C

x

Solution

3m

6m

rCA = 3i - 4j + 4k rCA = 6.403124 rCB =

4m

F  600 N

D

4m

y

3 (r ) = 1.4056i - 1.8741j + 1.8741k 6.403124 CA

rOB = rOC + rCB

= -3i + 4j + rCB



= -1.59444i + 2.1259j + 1.874085k

rOD = rOB + rBD rBD = rOD - rOB = (4i + 6j) - rOB

= 5.5944i + 3.8741j - 1.874085k

rBD = 2(5.5914)2 + (3.8741)2 + ( - 1.874085)2 = 7.0582        F = 600 a

rBD b = 475.568i + 329.326j - 159.311k rBD

F = {476i + 329j - 159k} N

Ans.

Ans: F = {476i + 329j - 159k} N 128

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*2–108. The chandelier is supported by three chains which are concurrent at point O. If the force in each chain has a magnitude of 60 lb, express each force as a Cartesian vector and determine the magnitude and coordinate direction angles of the resultant force.

z

O FB

SOLUTION FA = 60

FA

(4 cos 30° i - 4 sin 30° j - 6 k) Ans.

(- 4 cos 30° i - 4 sin 30° j - 6k) 2

2

2( - 4 cos 30°) + (- 4 sin 30°) + ( -6)

A

2(4) + (-6)

120

4 ft

C

120

Ans.

(4 j - 6 k) 2

120

2

= { - 28.8 i - 16.6 j - 49.9 k} lb FC = 60

6 ft

B

2(4 cos 30°)2 + ( -4 sin 30°)2 + (-6)2

= {28.8 i - 16.6 j - 49.9 k} lb FB = 60

FC

x 2

= {33.3 j - 49.9 k} lb

Ans.

FR = FA + FB + FC = { -149.8 k} lb FR = 150 lb

Ans.

a = 90°

Ans.

b = 90°

Ans.

g = 180°

Ans.

Ans: FA = {28.8i - 16.6j - 49.9k} lb FB = { - 28.8i - 16.6j - 49.9k} lb FC = {33.3j - 49.9k} lb FR = 150 lb a = 90° b = 90° g = 180° 129

y

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2–109. The chandelier is supported by three chains which are concurrent at point O. If the resultant force at O has a magnitude of 130 lb and is directed along the negative z axis, determine the force in each chain.

z

O FB

SOLUTION FC = F

FC FA

(4 j - 6 k) 24 2 + ( -6)2

= 0.5547 Fj - 0.8321 Fk

B

FA = FB = FC FRz = ©Fz;

6 ft

120

130 = 3(0.8321F)

A

120

4 ft

C

120

Ans.

F = 52.1 lb

x

Ans: F = 52.1 lb 130

y

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2–110. The window is held open by chain AB. Determine the length of the chain, and express the 50-lb force acting at A along the chain as a Cartesian vector and determine its coordinate direction angles.

z 5 ft

B

F = 50 lb

SOLUTION

12 ft A

Unit Vector: The coordinates of point A are 40˚

A15 cos 40°, 8, 5 sin 40°2 ft = A13.830, 8.00, 3.2142 ft

5 ft 5 ft

Then

8 ft

rAB = 510 - 3.8302i + 15 - 8.002j + 112 - 3.2142k6 ft

x

= 5- 3.830i - 3.00j + 8.786k6 ft

rAB = 21 - 3.83022 + 1 -3.0022 + 8.7862 = 10.043 ft = 10.0 ft uAB =

Ans.

- 3.830i - 3.00j + 8.786k rAB = rAB 10.043 = - 0.3814i - 0.2987j + 0.8748k

Force Vector: F = FuAB = 505- 0.3814i - 0.2987j + 0.8748k6 lb = 5 -19.1i - 14.9j + 43.7k6 lb

Ans.

Coordinate Direction Angles: From the unit vector uAB obtained above, we have cos a = - 0.3814

a = 112°

Ans.

cos b = - 0.2987

b = 107°

Ans.

cos g = 0.8748

g = 29.0°

Ans.

Ans: rAB = 10.0 ft F = { -19.1i - 14.9j + 43.7k} lb a = 112° b = 107° g = 29.0° 131

y

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2–111. The window is held open by cable AB. Determine the length of the cable and express the 30-N force acting at A along the cable as a Cartesian vector.

z 150 mm B

250 mm

SOLUTION

y

rAB = (0- 300 cos 30°)i + (150 - 500)j + (250 + 300 sin 30°)k 30 N

30

300 mm

= -259.81 i -350 j + 400 k rAB = 2(- 259.81)2 + (- 350)2 + (400)2 = 591.61

A

Ans.

= 592 mm F = 30 a

500 mm

x

rAB b = {- 13.2 i - 17.7 j + 20.3k} N rAB

Ans.

Ans: rAB = 592 mm F = { -13.2i - 17.7j + 20.3k} N 132

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*2–112. Given the three vectors A, B, and D, show that A # (B + D) = (A # B) + (A # D).

z A

2m

B 2m

SOLUTION Since the component of (B + D) is equal to the sum of the components of B and D, then A # (B + D) = A # B + A # D

(QED)

y

2m 2m

C

D

F  600 N 3m

Also, A # (B

x

+ D) = (A x i + A y j +

A zk) # [(Bx + Dx)i

E

+ (By + Dy)j + (Bz + Dz)k]

= A x (Bx + Dx) + A y (By + Dy) + A z (Bz + Dz) = (A xBx + A yBy + A zBz) + (A xDx + A yDy + A zDz) = (A # B) + (A # D)

(QED)

133

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2–113. Determine the magnitudes of the components of F = 600 N acting along and perpendicular to segment DE of the pipe assembly.

z A

2m

B 2m

SOLUTION Unit Vectors: The unit vectors uEB and uED must be determined first. From Fig. a, uEB

x

(0 - 4)i + (2 - 5)j + [0 - (- 2)]k rEB = = = - 0.7428i - 0.5571j + 0.3714k rEB 2(0 - 4)2 + (2 - 5)2 + [0 - (- 2)]2

y

2m 2m

C

D

F

600 N

3m

uED = -j

E

Thus, the force vector F is given by F = FuEB = 600 A - 0.7428i - 0.5571j + 0.3714k) = [- 445.66i - 334.25j + 222.83k] N Vector Dot Product: The magnitude of the component of F parallel to segment DE of the pipe assembly is (FED)paral = F # uED =

A -445.66i - 334.25j + 222.83k B # A -j B

= ( - 445.66)(0) + ( -334.25)( -1) + (222.83)(0) Ans.

= 334.25 = 334 N The component of F perpendicular to segment DE of the pipe assembly is (FED)per = 2F 2 - (FED)paral2 = 26002 - 334.252 = 498 N

Ans.

Ans: (FED) = 334 N (FED) # = 498 N 134

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2–114. z

Determine the angle u between the two cables.

C

F2  40 N

A

4m

u F1  70 N 2m

3m 2m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(2, -3, 3) m, B(0, 3, 0) and C( - 2, 3, 4) m respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

3m

B

y

3m x

(0 - 2)i + [3 - ( -3)]j + (0 - 3)k

6 3 2 = - i + j - k 7 7 7 2(0 - 2) + [3 - ( - 3)] + (0 - 3) 2

2

2

( - 2 - 2)i + [3 - ( -3)]j + (4 - 3)k 2

2

2( - 2 - 2) + [3 - ( - 3)] + (4 - 3)

The Angle U Between AB and AC.

2

= -

4

253

i +

6

253

j +

1 253

k

2 6 3 4 6 1 i + j + kb uAB # uAC = a - i + j - kb # a 7 7 7 253 253 253 = a=

2 4 6 6 3 1 ba b + a b + a - ba b 7 7 253 7 253 253

41

7253

Then

u = cos - 1 ( uAB # uAC ) = cos-1a

41 7253

b = 36.43° = 36.4°

Ans.

Ans: u = 36.4° 135

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2–115. Determine the magnitude of the projection of the force F1 along cable AC.

z C

F2  40 N

A

4m

u F1  70 N 2m

3m 2m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(2, -3, 3)m, B(0, 3, 0) and C( -2, 3, 4) m respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

3m

B

y

3m x

(0 - 2)i + [3 - ( - 3)]j + (0 - 3)k

6 3 2 = - i + j - k 7 7 7 2(0 - 2) + [3 - ( - 3)] + (0- 3) 2

2

( - 2 - 2)i + [3 - ( -3)]j + (4 - 3)k

2( - 2 - 2)2 + [3 - ( - 3)]2 + (4 - 3)2

2

4

= -

253

Force Vector, For F1,

i +

6

253

j +

1 253

k

2 6 3 F1 = F1 uAB = 70 a - i + j - kb = { -20i + 60j - 30k} N 7 7 7

Projected Component of F1. Along AC, it is

(F1)AC = F1 # uAC = ( - 20i + 60j - 30k) # a 4

b + 60a



= ( - 20)a -



= 56.32 N = 56.3 N

253

6

253

4 253

b + ( - 30)a

i + 1

253

6 253

j +

b

1 253

kb

 Ans.

The positive sign indicates that this component points in the same direction as uAC.

Ans: (F1)AC = 56.3 N 136

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*2–116. ]

Determine the angle u between the y axis of the pole and the wire AB.

IW

IW [

SOLUTION

e

IW

$

IW

Position Vector: rAC = 5 - 3j6 ft

%

rAB = 512 - 02i + 12 - 32j + 1- 2 - 02k6 ft = 52i - 1j - 2k6 ft

The magnitudes of the position vectors are rAC = 3.00 ft

rAB = 22 2 + 1 - 122 + 1 - 222 = 3.00 ft

The Angles Between Two Vectors U: The dot product of two vectors must be determined first. rAC # rAB = 1 - 3j2 # 12i - 1j - 2k2

= 0122 + 1- 321-12 + 01-22 = 3

Then, u = cos-1

rAO # rAB rAO rAB

= cos-1

3 3.00 3.00

Ans.

= 70.5°

Ans: u = 70.5° 137

\

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2–117. Determine the magnitudes of the projected components of the force F = [60i + 12j - 40k] N along the cables AB and AC.

0.75 m B 1m

z

1m C

1.5 m

u A

Solution

y

F

F = {60 i + 12 j - 40 k} N uAB =

3m

x

- 3 i - 0.75 j + 1 k 1( - 3)2 + ( - 0.75)2 + (1)2

= -0.9231 i - 0.2308 j + 0.3077 k uAC =

- 3 i + 1 j + 1.5 k 1( - 3)2 + (1)2 + (1.5)2

= -0.8571 i + 0.2857 j + 0.4286 k

Proj FAB = F # uAB = (60)( - 0.9231) + (12)( -0.2308) + ( - 40)(0.3077)

= -70.46 N Ans.

|Proj FAB| = 70.5 N

Proj FAC = F # uAC = (60)( - 0.8571) + (12)(0.2857) + ( - 40)(0.4286)

= -65.14 N Ans.

|Proj FAC| = 65.1 N

Ans: |Proj FAB| = 70.5 N |Proj FAC| = 65.1 N 138

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2–118. Determine the angle u between cables AB and AC.

0.75 m B 1m

z

1m C

1.5 m

u A 3m

x

Solution

y

F

rAB = { - 3 i - 0.75 j + 1 k} m rAB = 2( - 3)2 + ( - 0.75)2 + (1)2 = 3.25 m rAC = { - 3 i + 1 j + 1.5 k} m

rAC = 2( - 3)2 + (1)2 + (1.5)2 = 3.50 m

rAB # rAC = ( -3)( -3) + ( - 0.75)(1) + (1)(1.5) = 9.75 u = cos-1 a u = 31.0°

rAB # rAC 9.75 b = cos-1 a b rAB rAC (3.25)(3.50)

Ans.

Ans: u = 31.0° 139

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2–119. A force of F = 5 - 40k6 lb acts at the end of the pipe. Determine the magnitudes of the components F1 and F2 which are directed along the pipe’s axis and perpendicular to it.

z

O

y 3 ft

5 ft

SOLUTION uOA =

3 i + 5 j - 3k 2

2

2

23 + 5 + (- 3)

=

F1 = F # uOA = ( - 40 k) # a

3 ft

x

3i + 5j - 3k

A

243

3i + 5j - 3k 243

b

F2

Ans.

= 18.3 lb

F1 F  {40 k} lb

F2 = 2F2 - F12 F2 = 2402 - 18.32 = 35.6 lb

Ans.

Ans: F1 = 18.3 lb F2 = 35.6 lb 140

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*2–120. z

Two cables exert forces on the pipe. Determine the magnitude of the projected component of F1 along the line of action of F2.

F2

25 lb

60

u

SOLUTION

60

x

Force Vector: uF1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k

30

30

= 0.4330i + 0.75j - 0.5k

F1

y

30 lb

F1 = FRuF1 = 30(0.4330i + 0.75j - 0.5k) lb = {12.990i + 22.5j - 15.0k} lb Unit Vector: One can obtain the angle a = 135° for F2 using Eq. 2–8. cos2 a + cos2 b + cos2 g = 1, with b = 60° and g = 60°. The unit vector along the line of action of F2 is uF2 = cos 135°i + cos 60°j + cos 60°k = - 0.7071i + 0.5j + 0.5k Projected Component of F1 Along the Line of Action of F2: (F1)F2 = F1 # uF2 = (12.990i + 22.5j - 15.0k) # (- 0.7071i + 0.5j + 0.5k) = (12.990)(- 0.7071) + (22.5)(0.5) + (- 15.0)(0.5) = - 5.44 lb Negative sign indicates that the projected component of (F1)F2 acts in the opposite sense of direction to that of uF2. Ans.

The magnitude is (F1)F2 = 5.44 lb

Ans: The magnitude is (F1)F2 = 5.44 lb 141

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2–121. Determine the angle u between the two cables attached to the pipe.

z

F2

25 lb

60

SOLUTION Unit Vectors:

u

60

x

uF1 = cos 30° sin 30°i + cos 30° cos 30°j - sin 30°k = 0.4330i + 0.75j - 0.5k

30

30

y

uF2 = cos 135°i + cos 60°j + cos 60°k F1

= -0.7071i + 0.5j + 0.5k

30 lb

The Angles Between Two Vectors u: uF1 # uF2 = (0.4330i + 0.75j - 0.5k) # ( -0.7071i + 0.5j + 0.5k) = 0.4330(- 0.7071) + 0.75(0.5) + ( -0.5)(0.5) = -0.1812 Then, u = cos - 1 A uF1 # uF2 B = cos - 1( -0.1812) = 100°

Ans.

Ans: u = 100° 142

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2–122. Determine the angle u between the cables AB and AC.

z 1m B

1m

2m C

D F

u

3m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(6, 0, 0) m, x B(0, -1, 2) m and C(0, 1, 3) respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

(0 - 6)i + ( -1 - 0)j + (2 - 0)k 2

2

2(0 - 6) + ( - 1 - 0) + (2 - 0) (0 - 6)i + (1 - 0)j + (3 - 0)k 2

2

2(0 - 6) + (1 - 0) + (3 - 0)

2

= -

2

= -

The Angle u Between AB and AC. uAB # uAC = a -

6 241

= a= Then

6

241

41

i -

1 241

ba -

j +

6

246

2 241

b + a-

kb # a 1

241

ba

6 241 6

246 6 246 1

246

i -

i +

i +

b +

1 241 1

246 1 246 2

j +

j +

j + a

3

A

2 241 3

246 3 246

241 246

6m

y

k

k

kb

b

21886

u = cos - 1 ( UAB # UAC ) = cos-1a

41 21886

b = 19.24998° = 19.2°

Ans.

Ans: u = 19.2° 143

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2–123. z

Determine the magnitude of the projected component of the force F = {400i - 200j + 500k} N acting along the cable BA.

1m B

1m

2m C

D F

u

Solution Unit Vector. Here, the coordinates of points A and B are A(6, 0, 0) m and x B(0, -1, 2) m respectively. Thus the unit vector along BA is uBA =

A

6m

3m

y

(6 - 0)i + [0 - ( - 1)]j + (0 - 2)k rBA 6 1 2 = = i + j k rBA 2(6 - 0)2 + [0 - ( - 1)]2 + (0- 2)2 241 241 241

Projected component of F. Along BA, it is

FBA = F # uBA = (400i - 200j + 500k) # a



= 400 a

6

241

b + ( - 200)a

= 187.41 N = 187 N

6 241 1

241

i +

1 241

j -

b + 500a -

2

2 241

241

b

kb

 Ans.

The positive sign indicates that this component points in the same direction as uBA.

Ans: FBA = 187 N 144

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*2–124. Determine the magnitude of the projected component of the force F = {400i - 200j + 500k} N acting along the cable CA.

z 1m B

1m

2m C

D F

u

3m

Solution Unit Vector. Here, the coordinates of points A and C are A(6, 0, 0) m and C(0, 1, 3) m respectively. Thus, the unit vector along CA is uCA =

x

A

6m

y

(6 - 0)i + (0 - 1)j + (0 - 3)k rCA 6 1 3 = = i j k 2 2 2 rCA 2(6 - 0) + (0 - 1) + (0 - 3) 246 246 246

Projected component of F. Along CA, it is

FCA = F # uCA = (400i - 200j + 500k) # a = 400a

6

246

6 246

b + ( - 200)a -

= 162.19 N = 162 N

i -

1

246

1 246

j -

b + 500 a -

3 246 3

246

b

kb

 Ans.

The positive sign indicates that this component points in the same direction as uCA.

Ans: FCA = 162 N 145

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2–125. z

Determine the magnitude of the projection of force F = 600 N along the u axis.

F

600 N

A 4m

SOLUTION

O

Unit Vectors: The unit vectors uOA and uu must be determined first. From Fig. a, ( - 2 - 0)i + (4 - 0)j + (4 - 0)k rOA 2 2 1 = = - i + j + k uOA = rOA 3 3 3 2 2 2 3( -2 - 0) + (4 - 0) + (4 - 0)

4m 2m 30

x

u

uu = sin30°i + cos30°j Thus, the force vectors F is given by F = F uOA = 600a -

2 2 1 i - j + kb = 5 - 200i + 400j + 400k6 N 3 3 3

Vector Dot Product: The magnitude of the projected component of F along the u axis is Fu = F # uu = ( -200i + 400j + 400k) # (sin30°i + cos 30°j) = ( -200)(sin30°) + 400(cos 30°) + 400(0) Ans.

= 246 N

Ans: Fu = 246 N 146

y

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2–126. z

Determine the magnitude of the projected component of the 100-lb force acting along the axis BC of the pipe.

A B 3 ft u 8 ft x

Solution r gBC = 56i^ + 4j^ r = 100 F =

5- 6i^

6 ft 4 ft D

2 ft C

^ 6 ft - 2k

F

y

100 lb

^6 + 8j^ + 2k

2( - 6)2 + 82 + 22

5- 58.83i^

^ 6 lb + 78.45j^ + 19.61k

r gBC -78.45 m BC = r Fp = r = = - 10.48 F ∙r F∙ r | gBC | 7.483 Ans.

Fp = 10.5 lb

Ans: Fp = 10.5 lb 147

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2–127. z

Determine the angle u between pipe segments BA and BC.

A B 3 ft u 8 ft x

Solution r gBC = 56i^ + 4j^ r gBA =

5- 3i^ 6 ft

D

2 ft C

- 2k^ 6 ft

r gBC # r gBA - 18 b u = cos-1 a r r b = cos-1a | gBC| | gBA| 22.45 u = 143°

6 ft 4 ft F

y

100 lb

Ans.

Ans: u = 142° 148

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*2–128. z

Determine the angle u between BA and BC. A

B

2m

u

5m 4m x

D

F  3 kN

3m C

y

1m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(0, -2, 0) m, B(0, 0, 0) m and C(3, 4, -1) m respectively. Thus, the unit vectors along BA and BC are uBA = -j

uBE =

(3 - 0) i + (4 - 0) j + ( -1 -0) k 2

2

2(3 - 0) + (4 - 0) + ( - 1 - 0)

2

The Angle U Between BA and BC. uBA uBC = ( - j)

Then

= ( - 1) a

#

a

4

3 226

226

u = cos - 1 (uBA # uBC) = cos - 1 a -

i +

b = 4 226

4 226

4

j -

=

3 226 1

226

i +

4 226

j-

1 226

k

kb

226

b = 141.67° = 142°

Ans.

Ans: u = 142° 149

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2–129. z

Determine the magnitude of the projected component of the 3 kN force acting along the axis BC of the pipe. A

B

2m

u

5m 4m x

D

F  3 kN

3m C

y

1m

Solution Unit Vectors. Here, the coordinates of points B, C and D are B (0, 0, 0) m, C(3, 4, - 1) m and D(8, 0, 0). Thus the unit vectors along BC and CD are uBC =

uCD =

(3 - 0) i + (4 - 0) j + ( - 1-0) k 2

2(3 - 0) + (4 - 0)

2

2

2

+ ( - 1 - 0)

2

=

(8 - 3) i + (0 - 4) j + [0 - ( - 1)] k

2(8 - 3) + (0 - 4)

+ [0 - ( - 1)]

2

3 226

5

=

F = FuCD = 3 a = a

5 242

15

242

i -

i -

Projected Component of F. Along BC, it is ` (FBC) ` = ` F # uBC ` = ` a



15 242 15

= `a

242

= `-

21092

6

12

i-

242

ba

3

226

j+

4 242

12

242 3 242

b + a-

1

j +

j +

3

kb # a

12

242

242

226

j +

k

1

242

k

kb kN

3 226

ba

4

1

kb

242

242

j -

226

i -

242

Force Vector. For F,

4

i +

i+

4

226

4 226

b +

j-

3

242

` = ` - 0.1816 kN ` = 0.182 kN

1 226

a-

1

kb `

226

b`

Ans.

The negative signs indicate that this component points in the direction opposite to that of uBC.

Ans: 0.182 kN 150

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2–130. Determine the angles u and f made between the axes OA of the flag pole and AB and AC, respectively, of each cable.

z 1.5 m 2m

B

4m

SOLUTION rA C = {-2i - 4j + 1k} m ;

C

FB

rA C = 4.58 m

rAB = {1.5i - 4j + 3k} m;

rAB = 5.22 m

rA O = { - 4j - 3k} m;

rA O = 5.00 m

6m

u

O

= cos - 1 ¢

rAB # rAO ≤ rAB rAO

f

A

3m

rA B # rA O = (1.5)(0) + ( -4)(-4) + (3)(-3) = 7 u = cos - 1 ¢

40 N

FC

55 N

4m x y

7 ≤ = 74.4° 5.22(5.00)

Ans.

rAC # rAO = (-2)(0) + ( -4)(-4) + (1)(-3) = 13 f = cos - 1 a = cos - 1 a

rAC # rAO b rAC rAO 13 b = 55.4° 4.58(5.00)

Ans.

Ans: u = 74.4° f = 55.4° 151

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2–131. Determine the magnitudes of the components of F acting along and perpendicular to segment BC of the pipe assembly.

z

A

3 ft

4 ft 2 ft

4 ft B

x

y

SOLUTION Unit Vector: The unit vector uCB must be determined first. From Fig. a uCB =

F

{30i

45j

50k} lb 4 ft

(3 - 7)i + (4 - 6)j + [0 - (- 4)]k rCB 2 1 2 = = - i- j + k rCB 3 3 3 2 2 2 3(3 - 7) + (4 - 6) + [0 - ( - 4)]

C

Vector Dot Product: The magnitude of the projected component of F parallel to segment BC of the pipe assembly is (FBC)pa = F # uCB = (30i - 45j + 50k) # ¢-

1 2 2 i - j + k≤ 3 3 3

2 1 2 = (30) ¢ - ≤ + ( -45) ¢ - ≤ + 50 ¢ ≤ 3 3 3 Ans.

= 28.33 lb = 28.3 lb

The magnitude of F is F = 330 2 + ( - 45) 2 + 50 2 = 25425 lb. Thus, the magnitude of the component of F perpendicular to segment BC of the pipe assembly can be determined from (FBC)pr = 3F2 - (FBC)pa2 = 25425 - 28.332 = 68.0 lb

Ans.

Ans: ( FBC )  = 28.3 lb ( FBC ) # = 68.0 lb 152

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*2–132. z

Determine the magnitude of the projected component of F along AC. Express this component as a Cartesian vector.

A

3 ft

4 ft 2 ft

4 ft B

x

SOLUTION Unit Vector: The unit vector uAC must be determined first. From Fig. a uAC =

(7 - 0)i + (6 - 0)j + ( -4 - 0)k 3(7 - 0)2 + (6 - 0)2 + (- 4 - 0)2

F

{30i

45j

50k} lb 4 ft

= 0.6965 i + 0.5970 j - 0.3980 k

C

Vector Dot Product: The magnitude of the projected component of F along line AC is FAC = F # uAC = (30i - 45j + 50k) # (0.6965i + 0.5970j - 0.3980k) = (30)(0.6965) + ( - 45)(0.5970) + 50( -0.3980) Ans.

= 25.87 lb Thus, FAC expressed in Cartesian vector form is FAC = FAC uAC = - 25.87(0.6965i + 0.5970j - 0.3980k)

Ans.

= { - 18.0i - 15.4j + 10.3k} lb

Ans: FAC = 25.87 lb FAC = { - 18.0i - 15.4j + 10.3k} lb 153

y

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2–133. Determine the angle u between the pipe segments BA and BC.

z

A

3 ft

4 ft 2 ft

4 ft B

x

SOLUTION

F

Position Vectors: The position vectors rBA and rBC must be determined first. From Fig. a,

{30i

45j

y

50k} lb 4 ft C

rBA = (0 - 3)i + (0 - 4)j + (0 - 0)k = { - 3i - 4j} ft rBC = (7 - 3)i + (6 - 4)j + (- 4 - 0)k = {4i + 2j - 4k} ft The magnitude of rBA and rBC are rBA = 3(- 3)2 + (- 4)2 = 5 ft rBC = 342 + 22 + (- 4)2 = 6 ft Vector Dot Product: rBA # rBC = (- 3i - 4j) # (4i + 2j - 4k) = (- 3)(4) + ( -4)(2) + 0(- 4) = - 20 ft2 Thus, u = cos-1 a

rBA # rBC -20 b = cos-1 c d = 132° rBA rBC 5(6)

Ans.

Ans: u = 132° 154

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2–134. z¿ z

If the force F = 100 N lies in the plane DBEC, which is parallel to the x–z plane, and makes an angle of 10° with the extended line DB as shown, determine the angle that F makes with the diagonal AB of the crate.

15

F

0.2 m 10u B 30

E 0.2 m



x A 15

F

2

D

6k

N

0.5 m

C y

Solution Use the x, y, z axes. uAB = a

- 0.5i + 0.2j + 0.2k 0.57446

b

= - 0.8704i + 0.3482j + 0.3482k

F = - 100 cos 10°i + 100 sin 10°k u = cos-1a

F # uAB b F uAB

= cos-1 a

-100 (cos 10°)( - 0.8704) + 0 + 100 sin 10° (0.3482)

= cos-1 (0.9176) = 23.4°

100(1)

b

Ans.

Ans: u = 23.4° 155

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2–135. z

Determine the magnitudes of the components of force F = 90 lb acting parallel and perpendicular to diagonal AB of the crate.

F

90 lb 60

B

45

1 ft

SOLUTION

A

Force and Unit Vector: The force vector F and unit vector uAB must be determined x first. From Fig. a

3 ft

1.5 ft

y

C

F = 90(- cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {- 31.82i + 31.82j + 77.94k} lb uAB =

(0 - 1.5)i + (3 - 0)j + (1 - 0)k rAB 6 2 3 = = - i- j + k rAB 7 7 7 3(0 - 1.5)2 + (3 - 0)2 + (1 - 0)2

Vector Dot Product: The magnitude of the projected component of F parallel to the diagonal AB is [(F)AB]pa = F # uAB = (- 31.82i + 31.82j + 77.94k) # ¢-

6 2 3 i + j + k≤ 7 7 7

3 6 2 = (- 31.82) ¢ - ≤ + 31.82 ¢ ≤ + 77.94 ¢ ≤ 7 7 7 Ans.

= 63.18 lb = 63.2 lb The magnitude of the component F perpendicular to the diagonal AB is [(F)AB]pr = 3F2 - [(F)AB]pa2 = 2902 - 63.182 = 64.1 lb

156

Ans.

Ans: 3(F )AB 4  = 63.2 lb 3(F )AB 4 # = 64.1 lb

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*2–136. Determine the magnitudes of the projected components of the force F = 300 N acting along the x and y axes.

z

30

F

A

300 N

30 300 mm

SOLUTION

O

Force Vector: The force vector F must be determined first. From Fig. a, F = -300 sin 30°sin 30°i + 300 cos 30°j + 300 sin 30°cos 30°k

x

300 mm

300 mm y

= [-75i + 259.81j + 129.90k] N Vector Dot Product: The magnitudes of the projected component of F along the x and y axes are Fx = F # i =

A -75i + 259.81j + 129.90k B # i

= - 75(1) + 259.81(0) + 129.90(0) = - 75 N Fy = F # j =

A - 75i + 259.81j + 129.90k B # j

= - 75(0) + 259.81(1) + 129.90(0) = 260 N The negative sign indicates that Fx is directed towards the negative x axis. Thus Fx = 75 N,

Ans.

Fy = 260 N

Ans: Fx = 75 N Fy = 260 N 157

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2–137. Determine the magnitude of the projected component of the force F = 300 N acting along line OA.

30

F

z A

300 N

30 300 mm

O

SOLUTION

x

300 mm

300 mm y

Force and Unit Vector: The force vector F and unit vector uOA must be determined first. From Fig. a F = ( - 300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k) = { - 75i + 259.81j + 129.90k} N uOA =

(-0.45 - 0)i + (0.3 - 0)j + (0.2598 - 0)k rOA = = - 0.75i + 0.5j + 0.4330k rOA 2( - 0.45 - 0)2 + (0.3 - 0)2 + (0.2598 - 0)2

Vector Dot Product: The magnitude of the projected component of F along line OA is FOA = F # uOA =

A - 75i + 259.81j + 129.90k B # A - 0.75i + 0.5j + 0.4330k B

= ( - 75)( -0.75) + 259.81(0.5) + 129.90(0.4330) Ans.

= 242 N

Ans: FOA = 242 N 158

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2–138. z

Determine the angle u between the two cables. 8 ft C

10 ft

B 4 ft

10 ft

Solution u = cos-1 a = cos-1 c

= cos-1 a

u = 82.9°

rAC # rAB b rAC rAB

(2 i - 8 j + 10 k) # ( -6 i + 2 j + 4 k)

122 + ( - 8)2 + 102 1( - 6)2 + 22 + 42

12 b 96.99

FAB  12 lb 6 ft

u

x

8 ft

y

A

d

Ans.

Ans: u = 82.9° 159

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2–139. Determine the projected component of the force F = 12 lb acting in the direction of cable AC. Express the result as a Cartesian vector.

z 8 ft C 10 ft

B 4 ft

10 ft

Solution

rAC = {2 i - 8 j + 10 k} ft



rAB = { - 6 i + 2 j + 4 k} ft



FAB = 12 a



8 ft

x

y

A

rAB 6 2 4 b = 12 a i + j + kb rAB 7.483 7.483 7.483

FAB = { - 9.621 i + 3.207 j + 6.414 k} lb

uAC =

2 8 10 i j + k 12.961 12.961 12.961

Proj FAB = FAB # uAC = - 9.621 a



FAB  12 lb 6 ft

u

= 1.4846

2 8 10 b + 3.207 a b + 6.414 a b 12.961 12.961 12.961

Proj FAB = FAB uAC Proj FAB = (1.4846) c

2 8 10 i j + kd 12.962 12.962 12.962

Proj FAB = {0.229 i - 0.916 j + 1.15 k} lb

Ans.

Ans: Proj FAB = {0.229 i - 0.916 j + 1.15 k} lb 160

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3–1. The members of a truss are pin connected at joint O. Determine the magnitudes of F1 and F2 for equilibrium. Set u = 60°.

y

5 kN

F2

70 30

x O

SOLUTION + ©F = 0; : x

5 4

4 F2 sin 70° + F1 cos 60° - 5 cos 30° - (7) = 0 5

7 kN

u

3

F1

0.9397F2 + 0.5F1 = 9.930 + c ©Fy = 0;

F2 cos 70° + 5 sin 30° - F1 sin 60° -

3 (7) = 0 5

0.3420F2 - 0.8660F1 = 1.7 Solving: F2 = 9.60 kN

Ans.

F1 = 1.83 kN

Ans.

Ans:  F2 = 9.60 kN F1 = 1.83 kN 161

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3–2. The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle u for equilibrium. Set F2 = 6 kN.

y

5 kN

F2

70 30

x O

SOLUTION + ©F = 0; : x

5 4

4 6 sin 70° + F1 cos u - 5 cos 30° - (7) = 0 5

7 kN

u

3

F1

F1 cos u = 4.2920 + c ©Fy = 0;

6 cos 70° + 5 sin 30° - F1 sin u -

3 (7) = 0 5

F1 sin u = 0.3521 Solving: u = 4.69°

Ans.

F1 = 4.31 kN

Ans.

Ans: u = 4.69° F1 = 4.31 kN 162

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3–3. Determine the magnitude and direction u of F so that the particle is in equilibrium.

y 8 kN

30

x 5 kN

60

4 kN

Solution

u

Equations of Equilibrium. Referring to the FBD shown in Fig. a, + ΣFx = 0;   F sin u + 5 - 4 cos 60° - 8 cos 30° = 0    S

(1)

F sin u = 3.9282

F

   +cΣFy = 0;      8 sin 30° - 4 sin 60° - F cos u = 0

(2)

F cos u = 0.5359

Divide Eq (1) by (2), sin u = 7.3301 cos u sin u Realizing that tan u = , then cos u tan u = 7.3301

Ans.

u = 82.23° = 82.2°

Substitute this result into Eq. (1), F sin 82.23° = 3.9282

Ans.

F = 3.9646 kN = 3.96 kN

Ans:  u = 82.2° F = 3.96 kN 163

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*3–4.

The bearing consists of rollers, symmetrically confined within the housing. The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft. Determine the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium.

40°

SOLUTION + c ©Fy = 0;

NB

125 - NC cos 40° = 0 Ans.

NC = 163.176 = 163 N + ©F = 0; : x

NC

C

B A 125 N

NB - 163.176 sin 40° = 0 Ans.

NB = 105 N

Ans: NC = 163 N NB = 105 N 164

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3–5. The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Take u = 90°.

y

9 kN F A 5 3 B 4

SOLUTION 3 f = 90° - tan - 1 a b = 53.13° 4

O

+ ©F = 0; : x

4 T cos 53.13° - F a b = 0 5

+ c ©Fy = 0;

3 9 - T sin 53.13° - Fa b = 0 5

x

u

C T

Solving, T = 7.20 kN

Ans.

F = 5.40 kN

Ans.

Ans: T = 7.20 kN F = 5.40 kN 165

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3–6. The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle u for equilibrium. The forces are concurrent at point O. Take F = 8 kN.

y

9 kN F A 5 3 B 4

SOLUTION + ©F = 0; : x

4 T cos f - 8a b = 0 5

(1)

+ c ©Fy = 0;

3 9 - 8 a b - T sin f = 0 5

(2)

Rearrange then divide Eq. (1) into Eq. (2):

O

x

u

C T

tan f = 0.656, f = 33.27° T = 7.66 kN

Ans.

3 u = f + tan - 1 a b = 70.1° 4

Ans.

Ans: T = 7.66 kN u = 70.1° 166

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3–7. C

The man attempts to pull down the tree using the cable and small pulley arrangement shown. If the tension in AB is 60 lb, determine the tension in cable CAD and the angle u which the cable makes at the pulley.

D

θ 20°

A B 30°

SOLUTION + R©Fx¿ = 0;

60 cos 10° - T - T cos u = 0

+Q©Fy¿ = 0;

T sin u - 60 sin 10° = 0

Thus, T(1 + cos u) = 60 cos 10° u T(2cos2 ) = 60 cos 10° 2 2T sin

(1)

u u cos = 60 sin 10° 2 2

(2)

Divide Eq.(2) by Eq.(1) tan

u = tan 10° 2 Ans.

u = 20° T

Ans.

30.5 lb

Ans: u = 20° T = 30.5 lb 167

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*3–8. The cords ABC and BD can each support a maximum load of 100 lb. Determine the maximum weight of the crate, and the angle u for equilibrium.

D

u B

A

Equations of Equilibrium. Assume that for equilibrium, the tension along the length of rope ABC is constant. Assuming that the tension in cable BD reaches the limit first. Then, TBD = 100 lb. Referring to the FBD shown in Fig. a,

5

C

+ ΣFx = 0;   W a 5 b - 100 cos u = 0 S 13



13

12

Solution

100 cos u =

5W 13

+cΣFy = 0;      100 sin u - W - W a



100 sin u =

25 W 13

(1) 12 b = 0 13 (2)

Divide Eq. (2) by (1),    

sin u = 5 cos u

Realizing that tan u =

sin u , cos u

       tan u = 5 Ans.

     u = 78.69° = 78.7° Substitute this result into Eq. (1),    100 cos 78.69° =

5 W 13

     W = 50.99 lb = 51.0 lb 6 100 lb (O.K)

Ans.

Ans: u = 78.7° W = 51.0 lb 168

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3–9. Determine the maximum force F that can be supported in the position shown if each chain can support a maximum tension of 600 lb before it fails.

B 3 4

5

30 A

C

F

Solution Equations of Equilibrium. Referring to the FBD shown in Fig. a, 4    +cΣFy = 0;   TAB a b - F sin 30° = 0          TAB = 0.625 F 5

+ ΣFx = 0;        TAC + 0.625 F a 3 b - F cos 30° = 0   TAC = 0.4910 F    S 5

Since chain AB is subjected to a higher tension, its tension will reach the limit first. Thus,

TAB = 600;  0.625 F = 600 Ans.

F = 960 lb

Ans: F = 960 lb 169

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3–10. The block has a weight of 20 lb and is being hoisted at uniform velocity. Determine the angle u for equilibrium and the force in cord AB.

B 20

A

u C

Solution

D

F

Equations of Equilibrium. Assume that for equilibrium, the tension along the length of cord CAD is constant. Thus, F = 20 lb. Referring to the FBD shown in Fig. a,

+ ΣFx = 0;   20 sin u - TAB sin 20° = 0 S TAB =



20 sin u sin 20°

(1)

+cΣFy = 0;    TAB cos 20° - 20 cos u - 20 = 0

(2)

Substitute Eq (1) into (2),

20 sin u cos 20° - 20 cos u = 20 sin 20°



sin u cos 20° - cos u sin 20° = sin 20°

Realizing that sin (u - 20°) = sin u cos 20° - cos u sin 20°, then sin (u - 20°) = sin 20° u - 20° = 20°

Ans.

u = 40°

Substitute this result into Eq (1)

TAB =

20 sin 40° = 37.59 lb = 37.6 lb sin 20°

Ans.

Ans: u = 40° TAB = 37.6 lb 170

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3–11. Determine the maximum weight W of the block that can be suspended in the position shown if cords AB and CAD can each support a maximum tension of 80 lb. Also, what is the angle u for equilibrium?

B 20

A

u C

Solution

D

F

Equations of Equilibrium. Assume that for equilibrium, the tension along the length of cord CAD is constant. Thus, F = W. Assuming that the tension in cord AB reaches the limit first, then TAB = 80 lb. Referring to the FBD shown in Fig. a,

+ ΣFx = 0;   W sin u - 80 sin 20° = 0 S W =



80 sin 20° sin u

(1)

+cΣFy = 0;      80 cos 20° - W - W cos u = 0 W =

80 cos 20° 1 + cos u

(2)

Equating Eqs (1) and (2), 80 sin 20° 80 cos 20° = sin u 1 + cos u sin u cos 20° - cos u sin 20° = sin 20° Realizing then sin (u - 20°) = sin u cos 20° - cos u sin 20°, then sin (u - 20°) = sin 20° u - 20° = 20°

Ans.

u = 40°

Substitute this result into Eq (1)   W =          

80 sin 20° = 42.56 lb = 42.6 lb 6 80 lb (O.K) sin 40°

Ans.

Ans: u = 40° W = 42.6 lb 171

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*3–12. The lift sling is used to hoist a container having a mass of 500 kg. Determine the force in each of the cables AB and AC as a function of u. If the maximum tension allowed in each cable is 5 kN, determine the shortest lengths of cables AB and AC that can be used for the lift. The center of gravity of the container is located at G.

F A

SOLUTION

B

Free-Body Diagram: By observation, the force F1 has to support the entire weight of the container. Thus, F1 = 50019.812 = 4905 N.

θ

θ

1.5 m

C

1.5 m

Equations of Equilibrium: + ©F = 0; : x

FAC cos u - FAB cos u = 0

+ c ©Fy = 0;

4905 - 2F sin u = 0

FAC = FAB = F G

F = 52452.5 cos u6 N

Thus, FAC = FAB = F = 52.45 cos u6 kN

Ans.

If the maximum allowable tension in the cable is 5 kN, then 2452.5 cos u = 5000 u = 29.37° From the geometry, l =

1.5 and u = 29.37°. Therefore cos u l =

1.5 = 1.72 m cos 29.37°

Ans.

Ans: FAC = {2.45 cos u} kN l = 1.72 m 172

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3–13. A nuclear-reactor vessel has a weight of 500 ( 103 ) lb. Determine the horizontal compressive force that the spreader bar AB exerts on point A and the force that each cable segment CA and AD exert on this point while the vessel is hoisted upward at constant velocity.

30

C 30

A

B

D

E

Solution At point C : + ΣFx = 0;   FCB cos 30° - FCA cos 30° = 0   S FCB = FCA +cΣFy = 0;      500 ( 103 ) - FCA sin 30° - FCB sin 30° = 0

500 ( 103 ) - 2FCA sin 30° = 0 FCA = 500 ( 103 ) lb Ans.

At point A :

+ ΣFx = 0;          500 ( 103 ) cos 30° - FAB = 0 S FAB = 433 ( 103 ) lb

+cΣFy = 0;      500 ( 10

3



Ans.

) sin 30° - FAD = 0

FAD = 500 ( 103 ) sin 30°

FAD = 250 ( 103 ) lb

Ans.

Ans: FCA = 500 ( 103 ) lb FAB = 433 ( 103 ) lb FAD = 250 ( 103 ) lb 173

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3–14. Determine the stretch in each spring for equlibrium of the 2-kg block. The springs are shown in the equilibrium position.

3m

4m

C

3m

B kAC  20 N/m kAB  30 N/m

SOLUTION FAD = 2(9.81) = xAD(40) + ©F = 0; : x + c ©Fy = 0;

xAD = 0.4905 m

A

Ans.

4 1 FAB a b - FAC a b = 0 5 22 FAC a

kAD  40 N/m

1

3 b + FAB a b - 2(9.81) = 0 5 22

D

FAC = 15.86 N xAC =

15.86 = 0.793 m 20

Ans.

FAB = 14.01 N xAB =

14.01 = 0.467 m 30

Ans.

Ans: xAD = 0.4905 m xAC = 0.793 m xAB = 0.467 m 174

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3–15. The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.

3m

4m

C

3m

B 20 N/m

kAC

kAB

SOLUTION

A

F = kx = 30(5 - 3) = 60 N + ©F = 0; : x

4 Tcos 45° - 60 a b = 0 5 T = 67.88 N

+ c ©Fy = 0;

30 N/m

D

3 -W + 67.88 sin 45° + 60 a b = 0 5 W = 84 N m =

84 = 8.56 kg 9.81

Ans.

Ans: m = 8.56 kg 175

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*3–16. Determine the mass of each of the two cylinders if they cause a sag of s = 0.5 m when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed.

2m

1.5 m

s

1m

2m D

C

k

SOLUTION

100 N/m

k A

100 N/m

B

TAC = 100 N>m (2.828 - 2.5) = 32.84 N + c ©Fy = 0;

32.84 sin 45° - m(9.81) = 0 Ans.

m = 2.37 kg

Ans: m = 2.37 kg 176

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3–17.

Unstretched position

Determine the stiffness kT of the single spring such that the force F will stretch it by the same amount s as the force F stretches the two springs. Express kT in terms of stiffness k1 and k2 of the two springs.

kT F

s k1

k2 F

s

Solution F = ks s = s1 + s2 s =

F F F = + kT k1 k2

1 1 1 = + kT k1 k2

Ans.

Ans: 1 1 1 = + kT k1 k2 177

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3–18. If the spring DB has an unstretched length of 2 m, determine the stiffness of the spring to hold the 40-kg crate in the position shown.

2m

3m

C

B

2m

k

D

Solution

A

Equations of Equilibrium. Referring to the FBD shown in Fig. a,



+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0 S 113 12 +cΣFy = 0;      TBD a

Solving Eqs (1) and (2)

(1)

2 1 b + TCDa b - 40(9.81) = 0 113 12

(2)

TBD = 282.96 N   TCD = 332.96 N The stretched length of the spring is

Then, x = l - l0 =

l = 232 + 22 = 213 m

( 113 - 2 ) m. Thus, Fsp = kx;



282.96 = k ( 113 - 2 )

Ans.

k = 176.24 N>m = 176 N>m

Ans: k = 176 N>m 178

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3–19. Determine the unstretched length of DB to hold the 40-kg crate in the position shown. Take k = 180 N>m.

2m

3m

C

B

2m

k

D

Solution

A

Equations of Equilibrium. Referring to the FBD shown in Fig. a,



+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0 S 113 12 +cΣFy = 0;      TBD a

Solving Eqs (1) and (2)

(1)

2 1 b + TCDa b - 40(9.81) = 0 113 12

(2)

TBD = 282.96 N   TCD = 332.96 N The stretched length of the spring is l = 232 + 22 = 213 m

Then, x = l - l0 = 113 - l0. Thus Fsp = kx;



282.96 = 180 ( 113 - l0 )

Ans.

l0 = 2.034 m = 2.03 m

Ans: l0 = 2.03 m 179

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*3–20. A vertical force P = 10 lb is applied to the ends of the 2-ft cord AB and spring AC. If the spring has an unstretched length of 2 ft, determine the angle u for equilibrium. Take k = 15 lb>ft.

2 ft

B

2 ft

C

u

k A

SOLUTION + ©F = 0; : x

Fs cos f - T cos u = 0

(1)

+ c ©Fy = 0;

T sin u + Fs sin f - 10 = 0

(2) P

s = 2(4)2 + (2)2 - 2(4)(2) cos u - 2 = 2 25 - 4 cos u - 2 Fs = ks = 2k( 25 - 4 cos u - 1) From Eq. (1): T = Fs a

cos f b cos u

T = 2k A 25 - 4 cos u - 1 B ¢

2 - cos u 25 - 4 cos u

≤a

1 b cos u

From Eq. (2): 2k a 25 - 4 cos - 1 b(2 - cos u)

2k a 25 - 4 cos u - 1 b2 sin u tan u +

25 - 4 cos u a 25 - 4 cos u - 1 b 25 - 4 cos u

(2 tan u - sin u + sin u) =

tan u a 25 - 4 cos u - 1 b 25 - 4 cos u

=

= 10

225 - 4 cos u

10 2k

10 4k

Set k = 15 lb>ft Solving for u by trial and error, Ans.

u = 35.0°

Ans: u = 35.0° 180

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3–21. Determine the unstretched length of spring AC if a force P = 80 lb causes the angle u = 60° for equilibrium. Cord AB is 2 ft long. Take k = 50 lb>ft.

2 ft

B

2 ft

C

u

k A

SOLUTION l = 242 + 22 - 2(2)(4) cos 60° l = 212

P

2 212 = sin 60° sin f f = sin - 1 ¢

2 sin 60° 212

≤ = 30°

+ c ©Fy = 0;

T sin 60° + Fs sin 30° - 80 = 0

+ ©F = 0; : x

-T cos 60° + Fs cos 30° = 0

Solving for Fs, Fs = 40 lb Fs = kx 40 = 50(212 - l¿)

l = 212 -

40 = 2.66 ft 50

Ans.

Ans: l = 2.66 ft 181

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3–22. The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m. Determine the horizontal force F applied to the cord which is attached to the small ring B so that the displacement of the ring from the wall is d = 1.5 m.

A k

500 N/m B

6m

F

SOLUTION + ©F = 0; : x

k

1.5 211.25

500 N/m

(T)(2) - F = 0

C d

T = ks = 500(232 + (1.5)2 - 3) = 177.05 N Ans.

F = 158 N

Ans: F = 158 N 182

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3–23. The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m. Determine the displacement d of the cord from the wall when a force F = 175 N is applied to the cord.

A k

500 N/m B

6m

F

SOLUTION + ©F = 0; : x

k

500 N/m

175 = 2T sin u

C

T sin u = 87.5 TC

d 2

23 + d 2

d

S = 87.5

T = ks = 500( 232 + d 2 - 3) d a1 -

3 29 + d2

b = 0.175

By trial and error: Ans.

d = 1.56 m

Ans: d = 1.56 m 183

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*3–24. Determine the distances x and y for equilibrium if F1 = 800 N and F2 = 1000 N.

F1

D

C y

B

F2

2m

Solution Equations of Equilibrium. The tension throughout rope ABCD is constant, that is F1 = 800 N. Referring to the FBD shown in Fig. a,

A x

+cΣFy = 0;   800 sin f - 800 sin u = 0   f = 0

+ ΣFx = 0;        1000 - 2[800 cos u] = 0      u = 51.32° S Referring to the geometry shown in Fig. b,

y = 2 m

Ans.

2 = tan 51.32°;  x = 1.601 m = 1.60 m x

Ans.

and

Ans: y = 2m x = 1.60 m 184

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3–25. Determine the magnitude of F1 and the distance y if x = 1.5 m and F2 = 1000 N.

F1

D

C y

B

F2

2m

Solution Equations of Equilibrium. The tension throughout rope ABCD is constant, that is F1. Referring to the FBD shown in Fig. a,

+cΣFy = 0;      F1a

y 2

2

2y + 1.5 y

2



b - F1a 2

2y + 1.5

=

A x

2 b = 0 2.5

2 2.5

Ans.

y = 2 m

+ ΣFx = 0;         1000 - 2c F1a 1.5 b d = 0 S 2.5

Ans.

F1 = 833.33 N = 833 N  

Ans: y = 2m F1 = 833 N 185

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3–26. The 30-kg pipe is supported at A by a system of five cords. Determine the force in each cord for equilibrium.

5

3

D

4

C B 60°

A

E

H

SOLUTION At H: + c ΣFy = 0;

THA - 30(9.81) = 0 Ans.

THA = 294 N At A: + c ΣFy = 0;

TAB sin 60° - 30(9.81) = 0 Ans.

TAB = 339.83 = 340 N + ΣFx = 0; S

TAE - 339.83 cos 60° = 0 Ans.

TAE = 170 N At B: + c ΣFy = 0;

3 TBD a b - 339.83 sin 60° = 0 5

Ans.

TBD = 490.50 = 490 N + ΣFx = 0; S

4 490.50 a b + 339.83 cos 60° - TBC = 0 5

Ans.

TBC = 562 N

Ans: THA = TAB = TAE = TBD = TBC = 186

294 N 340 N 170 N 490 N 562 N

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3–27. Each cord can sustain a maximum tension of 500 N. Determine the largest mass of pipe that can be supported.

5

3

D

4

C B 60°

SOLUTION

E

A

H

At H: + c ©Fy = 0;

FHA = W

At A: + c ©Fy = 0;

FAB sin 60° - W = 0 FAB = 1.1547 W

+ ©F = 0; : x

FAE - (1.1547 W) cos 60° = 0 FAE = 0.5774 W

At B: + c ©Fy = 0;

3 FBD a b - (1.1547 cos 30°)W = 0 5 FBD = 1.667 W

+ ©F = 0; : x

4 - FBC + 1.667 Wa b + 1.1547 sin 30° = 0 5 FBC = 1.9107 W

By comparison, cord BC carries the largest load. Thus 500 = 1.9107 W W = 261.69 N m =

261.69 = 26.7 kg 9.81

Ans.

Ans: m = 26.7 kg 187

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*3–28. The street-lights at A and B are suspended from the two poles as shown. If each light has a weight of 50 lb, determine the tension in each of the three supporting cables and the required height h of the pole DE so that cable AB is horizontal.

D

A

C

h B

Solution At point B :

E

1 +cΣFy = 0;    FBC - 50 = 0   12

18 ft

24 ft

5 ft

Ans.

FBC = 70.71 = 70.7 lb

10 ft 6 ft

+ ΣFx = 0;         1 (70.71) - FAB = 0 S 12

Ans.

FAB = 50 lb

At point A : + ΣFx = 0;        50 - FAD cos u = 0 S

Ans.

+cΣFy = 0;          FAD sin u - 50 = 0 u = 45° FAD = 70.7 lb

Ans.

h = 18 + 5 = 23 ft

Ans.

Ans: FAB = 50 lb FAD = 70.7 lb h = 23 ft 188

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3–29. A

Determine the tension developed in each cord required for equilibrium of the 20-kg lamp.

E 5

4 3

C

B

SOLUTION

30°

D

45° F

Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. a, we have + ©F = 0; : x

FDE sin 30° - 20(9.81) = 0

FDE = 392.4 N = 392 N

+ c ©Fy = 0;

392.4 cos 30° - FCD = 0

FCD = 339.83 N = 340 N Ans.

Ans.

Using the result FCD = 339.83 N and applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. b, we have + ©F = 0; : x

3 339.83 - FCA a b - FCD cos 45° = 0 5

(1)

+ c ©Fy = 0;

4 FCA a b - FCB sin 45° = 0 5

(2)

Solving Eqs. (1) and (2), yields FCB = 275 N

Ans.

FCA = 243 N

Ans: FDE = FCD = FCB = FCA = 189

392 N 340 N 275 N 243 N

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3–30. A

Determine the maximum mass of the lamp that the cord system can support so that no single cord develops a tension exceeding 400 N.

4

B

SOLUTION

3

E

5

C

30°

D

45° F

Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. a, we have + c ©Fy = 0;

FDE sin 30° - m(9.81) = 0

FDE = 19.62m

+ ©F = 0; : x

19.62m cos 30° - FCD = 0

FCD = 16.99m

Using the result FCD = 16.99m and applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. b, we have + ©F = 0; : x + c ©Fy = 0;

3 16.99m - FCA a b - FCD cos 45° = 0 5 4 FCA a b - FCB sin 45° = 0 5

(1) (2)

Solving Eqs. (1) and (2), yields FCB = 13.73m

FCA = 12.14m

Notice that cord DE is subjected to the greatest tensile force, and so it will achieve the maximum allowable tensile force first. Thus FDE = 400 = 19.62m

m = 20.4 kg

Ans.

Ans: m = 20.4 kg 190

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3–31. Blocks D and E have a mass of 4 kg and 6 kg, respectively. If x = 2 m determine the force F and the sag s for equilibrium.

6m x

C

B

s A

D

Solution

F

E

Equations of Equilibrium. Referring to the geometry shown in Fig. a, cos f = cos u =

s 2

2

2s + 2 s

2

2

2s + 4

   sin u =

Referring to the FBD shown in Fig. b,

+ ΣFx = 0;     6(9.81)a S

2 2

2

2s + 2 3

2



2

2s + 2

2

2s + 22 4

2s + 42

b - 4(9.81)a =

4

2

4 2

2s + 42

b = 0

2

2s + 42

Ans.

s = 3.381 m = 3.38 m

  +cΣFy = 0;    6(9.81)a

2

   sin f =

3.381 2

2

23.381 + 2

b + 4(9.81)a

F = 75.99 N = 76.0 N

3.381 23.3812 + 42

b - F = 0

Ans.

Ans: s = 3.38 m F = 76.0 N 191

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*3–32. Blocks D and E have a mass of 4 kg and 6 kg, respectively. If F = 80 N, determine the sag s and distance x for equilibrium.

6m x

C

B

s A

D

Solution

F

E

Equations of Equilibrium. Referring to the FBD shown in Fig. a,

+ ΣFx = 0;   6(9.81) sin f - 4(9.81) sin u = 0 S sin f =



2 sin u 3

(1)

+cΣFy = 0;      6(9.81) cos f + 4(9.81) cos u - 80 = 0 (2)

3 cos f + 2 cos u = 4.0775 Using Eq (1), the geometry shown in Fig. b can be constructed. Thus cos f =

29 - 4 sin2 u 3

Substitute this result into Eq. (2), 3a

29 - 4 sin2u b + 2 cos u = 4.0775 3

29 - 4 sin2 u = 4.0775 - 2 cos u

9 - 4 sin2 u = 4 cos2 u - 16.310 cos u + 16.6258 16.310 cos u = 4 ( cos2 u + sin2 u ) + 7.6258 Here, cos2 u + sin2 u = 1. Then cos u = 0.7128   u = 44.54° Substitute this result into Eq (1) 2 sin f = sin 44.54°   f = 27.88° 3 From Fig. c,

6 - x x = tan 44.54° and = tan 27.88°. s s

So then, 6 - x x + = tan 44.54° + tan 27.88° s s 6 = 1.5129 s Ans.

s = 3.9659 m = 3.97 m x = 3.9659 tan 27.88°

Ans.

= 2.0978 m = 2.10 m

Ans: s = 3.97 m x = 2.10 m 192

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3–33. The lamp has a weight of 15 lb and is supported by the six cords connected together as shown. Determine the tension in each cord and the angle u for equilibrium. Cord BC is horizontal.

E D u

30 C

B

45

60 A

Solution Equations of Equilibrium. Considering the equilibrium of Joint A by referring to its FBD shown in Fig. a,

+ ΣFx = 0;     TAC cos 45° - TAB cos 60° = 0 S

(1)



+cΣFy = 0;   TAC sin 45° + TAB sin 60° - 15 = 0

(2)

Solving Eqs (1) and (2) yield

Ans.

TAB = 10.98 = 11.0 lb   TAC = 7.764 lb = 7.76 lb

Then, joint B by referring to its FBD shown in Fig. b +cΣFy = 0;  TBE sin 30° - 10.98 sin 60° = 0  TBE = 19.02 lb = 19.0 lb

Ans.

+ ΣFx = 0;  TBC + 10.98 cos 60° - 19.02 cos 30° = 0 S TBC = 10.98 lb = 11.0 lb Finally joint C by referring to its FBD shown in Fig. c

Ans.

+ ΣFx = 0;  TCD cos u - 10.98 - 7.764 cos 45° = 0     S (3)

               TCD cos u = 16.4711     +cΣFy = 0;     TCD sin u - 7.764 sin 45° = 0

(4)

               TCD sin u = 5.4904 Divided Eq (4) by (3)

Ans.

      tan u = 0.3333   u = 18.43° = 18.4° Substitute this result into Eq (3)       TCD cos 18.43° = 16.4711  TCD = 17.36 lb = 17.4 lb

Ans.

Ans: TAB = 11.0 lb TAC = 7.76 lb TBC = 11.0 lb TBE = 19.0 lb TCD = 17.4 lb u = 18.4° 193

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3–34. Each cord can sustain a maximum tension of 20 lb. Determine the largest weight of the lamp that can be supported. Also, determine u of cord DC for equilibrium.

E D u

30 C

B

45

60 A

Solution Equations of Equilibrium. Considering the equilibrium of Joint A by referring to its FBD shown in Fig. a,

+ ΣFx = 0;        TAC cos 45° - TAB cos 60° = 0 S

(1)



+cΣFy = 0;   TAC sin 45° - TAB sin 60° - W = 0

(2)

Solving Eqs (1) and (2) yield

TAB = 0.7321 W   TAC = 0.5176 W

Then, joint B by referring to its FBD shown in Fig. b, +cΣFy = 0;   TBE sin 30° - 0.7321W sin 60° = 0  TBE = 1.2679 W + ΣFx = 0;  TBC + 0.7321 W cos 60° - 1.2679 W cos 30° = 0 S TBC = 0.7321 W Finally, joint C by referring to its FBD shown in Fig. c,

Ans.

+ ΣFx = 0;  TCD cos u - 0.7321 W - 0.5176 W cos 45° = 0     S (3)

               TCD cos u = 1.0981 W     +cΣFy = 0;   TCD sin u - 0.5176 W sin 45° = 0

(4)

               TCD sin u = 0.3660 W Divided Eq (4) by (3)

Ans.

      tan u = 0.3333   u = 18.43° = 18.4° Substitute this result into Eq (3),         TCD cos 18.43° = 1.0981 W  TCD = 1.1575 W

Here cord BE is subjected to the largest tension. Therefore, its tension will reach the limit first, that is TBE = 20 lb. Then          20 = 1.2679 W;

W = 15.77 lb = 15.8 lb

Ans.

Ans: u = 18.4° W = 15.8 lb 194

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3–35. The ring of negligible size is subjected to a vertical force of 200 lb. Determine the required length l of cord AC such that the tension acting in AC is 160 lb. Also, what is the force in cord AB? Hint: Use the equilibrium condition to determine the required angle u for attachment, then determine l using trigonometry applied to triangle ABC.

C

θ

40° l

B

2 ft A

SOLUTION

200 lb

+ ©F = 0; : x

FAB cos 40° - 160 cos u = 0

+ c ©Fy = 0;

160 sin u + FAB sin 40° - 200 = 0

Thus, sin u + 0.8391 cos u = 1.25 Solving by trial and error, u = 33.25° Ans.

FAB = 175 lb l 2 = sin 33.25° sin 40°

Ans.

l = 2.34 ft Also, u = 66.75°

Ans.

FAB = 82.4 lb l 2 = sin 66.75° sin 40° l

Ans.

1.40 ft

Ans: FAB = 175 lb l = 2.34 ft FAB = 82.4 lb l = 1.40 ft 195

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*3–36. 3.5 m x

Cable ABC has a length of 5 m. Determine the position x and the tension developed in ABC required for equilibrium of the 100-kg sack. Neglect the size of the pulley at B.

C

0.75 m A B

SOLUTION Equations of Equilibrium: Since cable ABC passes over the smooth pulley at B, the tension in the cable is constant throughout its entire length. Applying the equation of equilibrium along the y axis to the free-body diagram in Fig. a, we have + c ©Fy = 0;

2T sin f - 100(9.81) = 0

(1)

Geometry: Referring to Fig. b, we can write x 3.5 - x + = 5 cos f cos f f = cos - 1 a

3.5 b = 45.57° 5

Also, x tan 45.57° + 0.75 = (3.5 - x) tan 45.57° Ans.

x = 1.38 m Substituting f = 45.57° into Eq. (1), yields

Ans.

T = 687 N

Ans: x = 1.38 m T = 687 N 196

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3–37. A 4-kg sphere rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown.

y

B

60

SOLUTION Geometry: The angle u which the surface make with the horizontal is to be determined first. tan u `

dy = = 5.0x ` = 2.00 ` dx x = 0.4m x = 0.4 m

x = 0.4 m

A y  2.5x2 0.4 m x 0.4 m

u = 63.43° Free Body Diagram: The tension in the cord is the same throughout the cord and is equal to the weight of block B, WB = mB (9.81). Equations of Equilibrium: + ©F = 0; : x

mB (9.81) cos 60° - Nsin 63.43° = 0 [1]

N = 5.4840mB + c ©Fy = 0;

mB (9.81) sin 60° + Ncos 63.43° - 39.24 = 0 [2]

8.4957mB + 0.4472N = 39.24 Solving Eqs. [1] and [2] yields mB = 3.58 kg

Ans.

N = 19.7 N

Ans: mB = 3.58 kg N = 19.7 N 197

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3–38. Determine the forces in cables AC and AB needed to hold the 20-kg ball D in equilibrium. Take F = 300 N and d = 1 m.

B 1.5 m C

SOLUTION

d A

Equations of Equilibrium: : ©Fx = 0; +

+ c ©Fy = 0;

F

2m

4 2 b - FAC a b = 0 241 25 06247FAB + 0.8944FAC = 300

(1)

5 1 b + FAC a b - 196.2 = 0 241 25 0.7809FAB + 0.4472FAC = 196.2

(2)

300 - FAB a

D

FAB a

Solving Eqs. (1) and (2) yields FAB = 98.6 N

Ans.

FAC = 267 N

Ans: FAB = 98.6 N FAC = 267 N 198

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3–39. The ball D has a mass of 20 kg. If a force of F = 100 N is applied horizontally to the ring at A, determine the largest dimension d so that the force in cable AC is zero.

B 1.5 m C d

SOLUTION

A

Equations of Equilibrium: : ©Fx = 0;

100 - FAB cos u = 0

FAB cos u = 100

(1)

+ c ©Fy = 0;

FAB sin u - 196.2 = 0

FAB sin u = 196.2

(2)

+

F

2m D

Solving Eqs. (1) and (2) yields u = 62.99°

FAB = 220.21 N

From the geometry, d + 1.5 = 2 tan 62.99° Ans.

d = 2.42 m

Ans: d = 2.42 m 199

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*3–40. F

The 200-lb uniform tank is suspended by means of a 6-ftlong cable, which is attached to the sides of the tank and passes over the small pulley located at O. If the cable can be attached at either points A and B, or C and D, determine which attachment produces the least amount of tension in the cable. What is this tension?

O B

1 ft

C D A 2 ft

2 ft

2 ft

SOLUTION Free-Body Diagram: By observation, the force F has to support the entire weight of the tank. Thus, F = 200 lb. The tension in cable AOB or COD is the same throughout the cable. Equations of Equilibrium: + ©F = 0; : x

T cos u - T cos u = 0

+ c ©Fy = 0;

200 - 2T sin u = 0

( Satisfied!) T =

100 sin u

(1)

From the function obtained above, one realizes that in order to produce the least amount of tension in the cable, sin u hence u must be as great as possible. Since the attachment of the cable to point C and D produces a greater u A u = cos - 113 = 70.53° B

as compared to the attachment of the cable to points A and B A u = cos - 1 23 = 48.19° B , the attachment of the cable to point C and D will produce the least amount of tension in the cable.

Ans.

Thus, T =

100 = 106 lb sin 70.53°

Ans.

Ans: T = 106 lb 200

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3–41. 5 ft

The single elastic cord ABC is used to support the 40-lb load. Determine the position x and the tension in the cord that is required for equilibrium. The cord passes through the smooth ring at B and has an unstretched length of 6 ft and stiffness of k = 50 lb>ft.

A

x

1 ft C

B

SOLUTION Equations of Equilibrium: Since elastic cord ABC passes over the smooth ring at B, the tension in the cord is constant throughout its entire length. Applying the equation of equilibrium along the y axis to the free-body diagram in Fig. a, we have + c ©Fy = 0;

(1)

2T sin f - 40 = 0

Geometry: Referring to Fig. (b), the stretched length of cord ABC is lABC =

5 - x 5 x + = cos f cos f cos f

(2)

Also, x tan f + 1 = (5 - x) tan f x =

5 tan f - 1 2 tan f

(3)

Spring Force Formula: Applying the spring force formula using Eq. (2), we obtain Fsp = k(lABC - l0) T = 50 c

5 - 6d cos f

(4)

Substituting Eq. (4) into Eq. (1) yields 5 tan f - 6 sin f = 0.4 Solving the above equation by trial and error f = 40.86° Substituting f = 40.86° into Eqs. (1) and (3) yields T = 30.6 lb

Ans.

x = 1.92 ft

Ans: T = 30.6 lb x = 1.92 ft 201

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3–42. A “scale” is constructed with a 4-ft-long cord and the 10-lb block D. The cord is fixed to a pin at A and passes over two small pulleys at B and C. Determine the weight of the suspended block at B if the system is in equilibrium when s = 1.5 ft.

1 ft

A

C

s

1.5 ft

SOLUTION Free-Body Diagram: The tension force in the cord is the same throughout the cord, that is, 10 lb. From the geometry, u = sin-1 a

D B

0.5 b = 23.58° 1.25

Equations of Equilibrium: + ©F = 0; : x

10 sin 23.58° - 10 sin 23.58° = 0

+ c ©Fy = 0;

2(10) cos 23.58° - WB = 0

(Satisfied!)

Ans.

WB = 18.3 lb

Ans: WB = 18.3 lb 202

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3–43. The three cables are used to support the 40-kg flowerpot. Determine the force developed in each cable for equilibrium.

z

D

1.5 m y A

Solution

ΣFz = 0;   FAD a

B

2m

Equations of Equilibrium. Referring to the FBD shown in Fig. a, 1.5 2

21.5 + 22 + 1.52

x

1.5 m

C

b - 40(9.81) = 0 

FAD = 762.69 N = 763 N  Ans.

Using this result, ΣFx = 0;   FAC - 762.69 a

ΣFy = 0;   FAB - 762.69 a

1.5 2

21.5 + 22 + 1.52 2 2

21.5 + 22 + 1.52

b = 0  

  FAC = 392.4 N = 392 N  Ans.

b = 0

  FAB = 523.2 N = 523 N  Ans.

Ans: FAD = 763 N FAC = 392 N FAB = 523 N 203

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*3–44. Determine the magnitudes of F1, F2, and F3 for equilibrium of the particle.

z F2 4 kN

10 kN 25

30

24

7

y

30 F3

Solution

x

Equations of Equilibrium. Referring to the FBD shown, ΣFy = 0;  10 a

F1

24 b - 4 cos 30° - F2 cos 30° = 0  F2 = 7.085 kN = 7.09 kN  Ans. 25

ΣFx = 0;  F1 - 4 sin 30° - 10 a

7 b = 0 25

F1 = 4.80 kN

Ans.

Using the result of F2 = 7.085 kN, ΣFz = 0;  7.085 sin 30° - F3 = 0

F3 = 3.543 kN = 3.54 kN Ans.

Ans: F2 = 7.09 kN F1 = 4.80 kN F3 = 3.54 kN 204

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3–45. If the bucket and its contents have a total weight of 20 lb, determine the force in the supporting cables DA, DB, and DC.

z

2.5 ft

C

4.5 ft

uDA = {

3 ft

A

SOLUTION

y

3 1.5 3 i j + k} 4.5 4.5 4.5

uDC = {-

3 ft

1 3 1.5 i + j + k} 3.5 3.5 3.5

©Fx = 0;

3 1.5 F F = 0 4.5 DA 3.5 DC

©Fy = 0;

-

©Fz = 0;

3 3 F + F - 20 = 0 4.5 DA 3.5 DC

1.5 ft

B D 1.5 ft

x

1.5 1 F - FDB + F = 0 4.5 DA 3.5 DC

FDA = 10.0 lb

Ans.

FDB = 1.11 lb

Ans.

FDC = 15.6 lb

Ans.

Ans: FDA = 10.0 lb FDB = 1.11 lb FDC = 15.6 lb 205

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3–46. Determine the stretch in each of the two springs required to hold the 20-kg crate in the equilibrium position shown. Each spring has an unstretched length of 2 m and a stiffness of k = 360 N>m.

z C

B A O

12 m

SOLUTION x

Cartesian Vector Notation: FOC = FOC ¢

6i + 4j + 12k 2

2

26 + 4 + 12

FOA = -FOA j

2

4m

6m

≤ = FOCi + FOCj + FOCk 3 7

2 7

6 7

FOB = -FOB i

F = {-196.2k} N Equations of Equilibrium: ©F = 0;

FOC + FOA + FOB + F = 0

3 2 6 a FOC - FOB b i + a FOC - FOA b j + a FOC - 196.2bk = 0 7 7 7 Equating i, j, and k components, we have 3 F - FOB = 0 7 OC

(1)

2 F - FOA = 0 7 OC

(2)

6 FOC - 196.2 = 0 7

(3)

Solving Eqs. (1),(2) and (3) yields FOC = 228.9 N

FOB = 98.1 N

FOA = 65.4 N

Spring Elongation: Using spring formula, Eq. 3–2, the spring elongation is s =

F . k

sOB =

98.1 = 0.327 m = 327 mm 300

Ans.

sOA =

65.4 = 0.218 m = 218 mm 300

Ans.

Ans: sOB = 327 mm sOA = 218 mm 206

y

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3–47. Determine the force in each cable needed to support the 20-kg flowerpot.

z B

6m

4m D

A 3m

Solution Equations of Equilibrium. ΣFz = 0;     FAB a ΣFx = 0;     FAC a

2m C x

2m

y

6 b - 20(9.81) = 0        FAB = 219.36 N = 219 N Ans. 145 2 2 b - FADa b = 0      FAC = FAD = F 120 120

Using the results of FAB = 219.36 N and FAC = FAD = F, ΣFy = 0;     2c F a



4 3 b d - 219.36 a b = 0 120 145

Ans.

FAC = FAD = F = 54.84 N = 54.8 N

Ans: FAB = 219 N FAC = FAD = 54.8 N 207

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*3–48. Determine the tension in the cables in order to support the 100-kg crate in the equilibrium position shown.

z

C

2m

D 1m

A 2.5 m

SOLUTION

B

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

x

2m

2m y

FAB = FAB i FAC = - FAC j FAD = FAD B

(- 2 - 0)i+ (2 - 0)j+ (1 - 0)k 2

2

2

2(-2 - 0) + (2 - 0) + (1 - 0)

R = - FAD i + 2 3

2 1 F j + FAD k 3 AD 3

W = [ -100(9.81)k]N = [ - 981 k]N Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 2 1 FAB i + ( -FAC j) + a - FAD i + FAD j + FAD kb + ( -981k) = 0 3 3 3 a FAB -

2 2 1 F b i + a - FAC + FAD b j + a FAD - 981b k = 0 3 AD 3 3

Equating the i, j, and k components yields FAB - FAC +

2 F = 0 3 AD

(1)

2 F = 0 3 AD

(2)

1 F - 981 = 0 3 AD

(3)

Solving Eqs. (1) through (3) yields FAD = 2943 N = 2.94 kN

Ans.

FAB = FAC = 1962 N = 1.96 kN

Ans.

Ans: FAD = 2.94 kN FAB = 1.96 kN 208

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3–49. Determine the maximum mass of the crate so that the tension developed in any cable does not exceeded 3 kN.

z

C

2m

D 1m

A 2.5 m

SOLUTION

B

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

x

2m

2m y

FAB = FAB i FAC = - FAC j FAD = FAD B

(- 2 - 0)i+ (2 - 0)j+ (1 - 0)k 2(- 2 - 0)2 + (2 - 0)2 + (1 - 0)2

R = - FAD i + 2 3

2 1 F j + FAD 3 AD 3

W = [ -m(9.81)k] Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 2 1 FAB i + ( -FAC j) + a - FAD i + FAD j + FAD kb + [- m(9.81)k] = 0 3 3 3 aFAB -

2 2 1 F b i + a - FAC + FAD b j + a FAD - 9.81mb k = 0 3 AD 3 3

Equating the i, j, and k components yields FAB -

2 F = 0 3 AD

(1)

2 F = 0 3 AD

(2)

1 - 9.81m = 0 F 3 AD

(3)

- FAC +

When cable AD is subjected to maximum tension, FAD = 3000 N. Thus, by substituting this value into Eqs. (1) through (3), we have FAB = FAC = 2000 N Ans.

m = 102 kg

Ans: m = 102 kg 209

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3–50. Determine the force in each cable if F = 500 lb.

z F A

1 ft 6 ft C

Solution

2 ft

2 ft

y

3 ft

B

Equations of Equilibrium. Referring to the FBD shown in Fig. a,

D

1 ft

3 ft x

3 3 2 ΣFx = 0;  FABa b - FAC a b - FADa b = 0 (1) 7 7 146 2 1 3 ΣFy = 0;  - FABa b - FAC a b + FADa b = 0 7 7 146

(2)

6 6 6 ΣFz = 0;  - FABa b - FAC a b - FADa b + 500 = 0 7 7 146

(3)

Solving Eqs (1), (2) and (3)

FAC = 113.04 lb = 113 lb

Ans.



FAB = 256.67 lb = 257 lb

Ans.



FAD = 210 lb

Ans.

Ans: FAC = 113 lb FAB = 257 lb FAD = 210 lb 210

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3–51. Determine the greatest force F that can be applied to the ring if each cable can support a maximum force of 800 lb.

z F A

1 ft 6 ft C

Solution

2 ft

3 3 2 b - FADa b = 0 (1)   ΣFx = 0;  FABa b - FAC a 7 7 146

2 1 3 b + FADa b = 0   ΣFy = 0;  - FABa b - FAC a 7 7 146

2 ft

y

3 ft

B

Equations of Equilibrium. Referring to the FBD shown in Fig. a,

D

1 ft

3 ft x

(2)

6 6 6 b - FADa b + F = 0   ΣFz = 0;  - FABa b - FAC a 7 7 146 Solving Eqs (1), (2) and (3)

(3)

FAC = 0.2261 F  FAB = 0.5133 F  FAD = 0.42 F Since cable AB is subjected to the greatest tension, its tension will reach the limit first that is FAB = 800 lb. Then

800 = 0.5133 F



F = 1558.44 lb = 1558 lb

Ans.

Ans: F = 1558 lb 211

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*3–52. z

Determine the tension developed in cables AB and AC and the force developed along strut AD for equilibrium of the 400-lb crate.

2 ft 2 ft B

C

4 ft

5.5 ft

A

SOLUTION Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as FAB = FAB C FAC = FAC C FAD = FAD C

(- 2 - 0)i + ( -6 - 0)j + (1.5 - 0)k 2( -2 - 0)2 + ( - 6 - 0)2 + (1.5 - 0)2 (2 - 0)i + (- 6 - 0)j + (3 - 0)k 2

2

2

2(2 - 0) + ( - 6 - 0) + (3 - 0)

S = -

S =

2

x

2.5 ft 6 ft y

4 12 3 F i F j + F k 13 AB 13 AB 13 AB

2 6 3 F i - FAC j + FAC k 7 AC 7 7

(0 - 0)i + [0 - ( -6)]j + [0 - (- 2.5)]k 2

D

2

2(0 - 0) + [0 - (- 6)] + (0 - (- 2.5)]

S =

12 5 F j + F k 13 AD 13 AD

W = {- 400k} lb Equations of Equilibrium: Equilibrium requires gF = 0;

FAB + FAC + FAD + W = 0

¢-

2 12 4 12 3 6 3 5 F i F j + F k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + F k ≤ + ( -400 k) = 0 13 AB 13 AB 13 AB 7 7 7 13 13 AD

¢-

4 2 12 6 12 3 5 3 F F ≤ j + ¢ FAB + FAC + F - 400 ≤ k = 0 + FAC ≤ i + ¢ - FAB - FAC + 13 AB 7 13 7 13 AD 13 7 13 AD

Equating the i, j, and k components yields 2 4 F + FAC = 0 13 AB 7 6 12 12 F = 0 - FAB - FAC + 13 7 13 AD 3 3 5 F + FAC + F - 400 = 0 13 AB 7 13 AD

(1)

-

(2) (3)

Solving Eqs. (1) through (3) yields Ans. Ans. Ans.

FAB = 274 lb FAC = 295 lb FAD = 547 lb

Ans: FAB = 274 lb FAC = 295 lb FAD = 547 lb 212

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3–53. z

If the tension developed in each of the cables cannot exceed 300 lb, determine the largest weight of the crate that can be supported. Also, what is the force developed along strut AD?

2 ft 2 ft B

C

4 ft

5.5 ft

A

SOLUTION Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as ( -2 - 0)i + (- 6 - 0)j + (1.5 - 0)k

FAB = FAB C

2

2

2

2( -2 - 0) + (-6 - 0) + (1.5 - 0) (2 - 0)i + ( -6 - 0)j + (3 - 0)k

FAC = FAC C

2(2 - 0)2 + ( -6 - 0)2 + (3 - 0)2

S = -

S =

x

2.5 ft 6 ft y

4 12 3 F i F j + F k 13 AB 13 AB 13 AB

2 6 3 FAC i - FAC j + FAC k 7 7 7

(0 - 0)i + [0 - ( -6)]j + [0 - (- 2.5)]k

FAD = FAD C

D

2(0 - 0)2 + [0 - ( - 6)]2 + [0 - ( -2.5)]2

S =

12 5 FAD j + FAD k 13 13

W = -Wk Equations of Equilibrium: Equilibrium requires gF = 0;

FAB + FAC + FAD + W = 0

¢-

12 3 6 3 5 4 2 12 FAB i FAB j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -Wk) = 0 13 13 13 7 7 7 13 13

¢-

2 12 6 12 3 5 4 3 FAB + FAC ≤ i + ¢- FAB - FAC + FAD ≤ j + ¢ FAB + FAC + FAD - W ≤ k = 0 13 7 13 7 13 13 7 13

Equating the i, j, and k components yields -

2 4 F + FAC = 0 13 AB 7

(1)

-

6 12 12 F - FAC + F = 0 13 AB 7 13 AD

(2)

3 3 5 FAB + FAC + FAD - W = 0 13 7 13

(3)

Let us assume that cable AC achieves maximum tension first. Substituting FAC = 300 lb into Eqs. (1) through (3) and solving, yields FAB = 278.57 lb FAD = 557 lb

Ans.

W = 407 lb

Since FAB = 278.57 lb 6 300 lb, our assumption is correct. Ans: FAD = 557 lb W = 407 lb 213

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3–54. z

Determine the tension developed in each cable for equilibrium of the 300-lb crate. 2 ft

3 ft

B

2 ft

C 3 ft

D 6 ft

x

4 ft

A 3 ft

SOLUTION Force Vectors: We can express each of the forces shown in Fig. a in Cartesian vector form as FAB = FAB C FAC = FAC C

(-3 - 0)i + (-6 - 0)j + (2 - 0)k

3 6 2 S = - FAB i - FAB j + FAB k 7 7 7 2(- 3 - 0) + (-6 - 0) + (2 - 0) 2

2

2

(2 - 0)i + ( - 6 - 0)j + (3 - 0)k 2

2

2

2(2 - 0) + ( -6 - 0) + (3 - 0) (0 - 0)i + (3 - 0)j + (4 - 0)k

FAD = FAD C

2

2

2

2(0 - 0) + (3 - 0) + (4 - 0)

S =

S =

2 6 3 F i - FAC j + FAC k 7 AC 7 7

3 4 F j + FAD k 5 AD 5

W = {-300k} lb Equations of Equilibrium: Equilibrium requires g F = 0;

FAB + FAC + FAD + W = 0

¢ - FAB i -

2 3 6 2 6 3 4 F j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -300k) = 0 7 AB 7 7 7 7 5 5

3 7

Equating the i, j, and k components yields 2 3 - FAB + FAC = 0 7 7 6 3 6 - FAB - FAC + FAD = 0 7 7 5 2 3 4 F + FAC + FAD - 300 = 0 7 AB 7 5

(1) (2) (3)

Solving Eqs. (1) through (3) yields FAB = 79.2 lb

FAC = 119 lb

Ans.

FAD = 283 lb

Ans: FAB = 79.2 lb FAC = 119 lb FAD = 283 lb 214

y

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3–55. z

Determine the maximum weight of the crate that can be suspended from cables AB, AC, and AD so that the tension developed in any one of the cables does not exceed 250 lb. 2 ft

3 ft

B

2 ft

C 3 ft

D 6 ft

x

4 ft

A 3 ft

SOLUTION Force Vectors: We can express each of the forces shown in Fig. a in Cartesian vector form as (- 3 - 0)i + ( -6 - 0)j + (2 - 0)k

FAB = FAB C

3 6 2 S = - FAB i - FAB j + FAB k 7 7 7 2( -3 - 0) + (-6 - 0) + (2 - 0) 2

2

2

(2 - 0)i + ( -6 - 0)j + (3 - 0)k

FAC = FAC C

2

2

2

2(2 - 0) + ( - 6 - 0) + (3 - 0)

FAD = FAD C

(0 - 0)i + (3 - 0)j + (4 - 0)k 2

2

2(0 - 0) + (3 - 0) + (4 - 0)

2

S =

S =

2 6 3 FAC i - FAC j + FAC k 7 7 7

3 4 FAD j + FAD k 5 5

W = -WC k Equations of Equilibrium: Equilibrium requires g F = 0;

FAB + FAC + FAD + W = 0

¢ - FAB i -

6 2 6 3 4 2 3 FAB j + FAB k ≤ + ¢ FAC i - FAC j + FAC k ≤ + ¢ FAD j + FAD k ≤ + ( -WC k) = 0 7 7 7 7 7 5 5

3 7

¢ - FAB + 3 7

2 6 6 3 3 4 2 FAC ≤ i + ¢ - FAB - FAC + FAD ≤ j + ¢ FAB + FAC + FAD - WC ≤ k = 0 7 7 7 5 7 7 5

Equating the i, j, and k components yields 2 3 - FAB + FAC = 0 7 7 6 3 6 - FAB - FAC + FAD = 0 7 7 5 2 3 4 FAB + FAC + FAD - WC = 0 7 7 5

(1) (2) (3)

Assuming that cable AD achieves maximum tension first, substituting FAD = 250 lb into Eqs. (2) and (3), and solving Eqs. (1) through (3) yields FAB = 70 lb WC = 265 lb

FAC = 105 lb

Ans.

Since FAB = 70 lb 6 250 lb and FAC = 105 lb, the above assumption is correct.

Ans: WC = 265 lb 215

y

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*3–56. The 25-kg flowerpot is supported at A by the three cords. Determine the force acting in each cord for equilibrium.

z C

D

B 60

30 45

SOLUTION

30

FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k) = 0.5FADi - 0.75FADj + 0.4330FAD k FAC = FAC (- sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

A x

= -0.5FAC i - 0.75FACj + 0.4330FACk FAB = FAB(sin 45°j + cos 45°k) = 0.7071FAB j + 0.7071FAB k F = -25(9.81)k = {-245.25k} N ©F = 0 ;

FAD + FAB + FAC + F = 0

(0.5FAD i - 0.75FAD j) + 0.4330FAD k + (0.7071FAB j + 0.7071FAB k) + ( -0.5FACi - 0.75FACj + 0.4330FACk) + ( -245.25k) = 0 (0.5FAD - 0.5FAC)i + ( - 0.75FAD + 0.7071FAB - 0.75FAC) j + (0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25) k = 0 Thus, ©Fx = 0;

0.5FAD - 0.5FAC = 0

[1]

©Fy = 0;

-0.75FAD + 0.7071FAB - 0.75FAC = 0

[2]

©Fz = 0;

0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25 = 0

[3]

Solving Eqs. [1], [2], and [3] yields: FAD = FAC = 104 N

Ans.

FAB = 220 N

Ans: FAD = FAC = 104 N FAB = 220 N 216

y

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3–57. If each cord can sustain a maximum tension of 50 N before it fails, determine the greatest weight of the flowerpot the cords can support.

z C

D

B 60

30 45

SOLUTION

30

FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

A

= 0.5FAD i - 0.75FAD j + 0.4330FAD k

y

x

FAC = FAC (- sin 30°i - cos 30° sin 60° j + cos 30° cos 60° k) = -0.5FAC i - 0.75FAC j + 0.4330FAC k FAB = FAB (sin 45° j + cos 45° k) = 0.7071FAB j + 0.7071FAB k W = -Wk ©Fx = 0;

0.5FAD - 0.5FAC = 0 (1)

FAD = FAC ©Fy = 0;

- 0.75FAD + 0.7071FAB - 0.75FAC = 0 (2)

0.7071FAB = 1.5FAC ©Fz = 0;

0.4330FAD + 0.7071FAB + 0.4330FAC - W = 0 0.8660FAC + 1.5FAC - W = 0 2.366FAC = W

Assume FAC = 50 N then FAB =

1.5(50) = 106.07 N 7 50 N (N . G!) 0.7071

Assume FAB = 50 N. Then FAC =

0.7071(50) = 23.57 N 6 50 N (O. K!) 1.5

Thus, Ans.

W = 2.366(23.57) = 55.767 = 55.8 N

Ans: W = 55.8 N 217

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3–58. Determine the tension developed in the three cables required to support the traffic light, which has a mass of 15 kg. Take h = 4 m.

z C 6m D A h

Solution

4m

ΣFx = 0;

ΣFy = 0; ΣFz = 0;

3 6 F - FAC + 5 AB 7 4 3 F - FAC 5 AB 7 2 F - 15(9.81) 7 AC

4m

3m

6 3 2 = e - i - j + kf 7 7 7

4 3 uAD = e i - j f 5 5

B

4m

3 4 uAB = e i + j f 5 5 uAC

3m

x

4m

6m 3m

y

4 F = 0 5 AD 3 F = 0 5 AD = 0



FAB = 441 N

Ans.



FAC = 515 N

Ans.



FAD = 221 N

Ans.

Ans: FAB = 441 N FAC = 515 N FAD = 221 N 218

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3–59.

z

Determine the tension developed in the three cables required to support the traffic light, which has a mass of 20 kg. Take h = 3.5 m.

C 6m D A h

3m B

4m

Solution uAB = uAC = uAD =

4m

3i + 4 j + 0.5k 2

2

23 + 4 + (0.5)

2

3i + 4 j + 0.5k

=

225.25

- 6i - 3 j + 2.5k

2( - 6)2 + ( - 3)2 + 2.52 4i - 3 j + 0.5k

=

242 + ( - 3)2 + 0.52

ΣFx = 0; ΣFy = 0; ΣFz = 0; Solving,

3

225.25 4

225.25 0.5

225.25

FAB FAB FAB +

=

4m

3m x

- 6i - 3 j + 2.5k

4m

6m 3m

y

251.25

4i - 3 j + 0.5k 6

225.25

251.25 3

251.25 2.5

251.25

FAC + FAC FAC +

4 225.25 3

225.25 0.5

225.25

FAD = 0 FAD = 0 FAD - 20(9.81) = 0



FAB = 348 N

Ans.



FAC = 413 N

Ans.



FAD = 174 N

Ans.

Ans: FAB = 348 N FAC = 413 N FAD = 174 N 219

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*3–60. The 800-lb cylinder is supported by three chains as shown. Determine the force in each chain for equilibrium. Take d = 1 ft.

z

D

135

1 ft

B y

135

90 C

SOLUTION FAD = FAD £

FAC = FAC £ FAB = FAB ¢

-1j + 1k 2( -1)2 + 12 1i + 1k 212 + 12

d

≥ = - 0.7071FADj + 0.7071FADk

x

A

≥ = 0.7071FACi + 0.7071FACk

- 0.7071i + 0.7071j + 1k 2( -0.7071)2 + 0.70712 + 12

≤

= -0.5FAB i + 0.5FAB j + 0.7071FAB k F = {-800k} lb ©F = 0;

FAD + FAC + FAB + F = 0

( - 0.7071FADj + 0.7071FADk) + (0.7071FACi + 0.7071FACk) + ( -0.5FAB i + 0.5FAB j + 0.7071FAB k) + (- 800k) = 0 (0.7071FAC - 0.5FAB) i + (- 0 .7071FAD + 0.5FAB)j + (0.7071FAD + 0.7071FAC + 0.7071FAB - 800) k = 0 ©Fx = 0;

0.7071FAC - 0.5FAB = 0

(1)

©Fy = 0;

-0.7071FAD + 0.5FAB = 0

(2)

©Fz = 0;

0.7071FAD + 0.7071FAC + 0.7071FAB - 800 = 0

(3)

Solving Eqs. (1), (2), and (3) yields: FAB = 469 lb

Ans.

FAC = FAD = 331 lb

Ans: FAB = 469 lb FAC = FAD = 331 lb 220

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3–61. Determine the tension in each cable for equilibrium.

z 800 N A

D

3m

5m C

4m

2m O

5m

Solution

4m

  ΣFy = 0;  FABa

4 2 4 b - FAC a b - FADa b = 0 157 138 166

B

(1)

4 3 5 b + FAC a b - FADa b = 0 157 138 166

  ΣFz = 0;  - FABa

y

x

Equations of Equilibrium. Referring to the FBD shown in Fig. a,   ΣFx = 0;  FABa

4m

(2)

5 5 5 b - FAC a b - FADa b + 800 = 0 157 138 166

(3)

Solving Eqs (1), (2) and (3)

FAC = 85.77 N = 85.8 N

Ans.



FAB = 577.73 N = 578 N

Ans.



FAD = 565.15 N = 565 N

Ans.

Ans: FAC = 85.8 N FAB = 578 N FAD = 565 N 221

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3–62. If the maximum force in each rod can not exceed 1500 N, determine the greatest mass of the crate that can be supported.

z

C

B

3m

2m

2m 1m

A

2m

2m O

3m y

1m 3m

Solution Equations of Equilibrium. Referring to the FBD shown in Fig. a,

x

2 3 1 b - FOC a b + FOB a b = 0 (1)   ΣFx = 0;  FOAa 3 114 122   ΣFy = 0;  - FOAa   ΣFz = 0;  FOAa

3 2 2 b + FOC a b + FOB a b = 0 3 114 122

(2)

1 3 2 b + FOC a b - FOB a b - m(9.81) = 0 3 114 122

(3)

Solving Eqs (1), (2) and (3),

FOC = 16.95m  FOA = 15.46m  FOB = 7.745m Since link OC subjected to the greatest force, it will reach the limiting force first, that is FOC = 1500 N. Then

1500 = 16.95 m



m = 88.48 kg = 88.5 kg

Ans.

Ans: m = 88.5 kg 222

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3–63. The crate has a mass of 130 kg. Determine the tension developed in each cable for equilibrium.

z

A x

1m 1m

4m

B

3m C

D 2m

1m

y

Solution Equations of Equilibrium. Referring to the FBD shown in Fig. a,   ΣFx = 0;  FADa

2 2 2 b - FBD a b - FCD a b = 0 3 16 16

  ΣFy = 0;  - FADa   ΣFz = 0;  FADa

(1)

1 1 2 b - FBD a b + FCD a b = 0 3 16 16

(2)

1 1 1 b + FBD a b + FCD a b - 130(9.81) = 0 3 16 16

(3)

Solving Eqs (1), (2) and (3)

FAD = 1561.92 N = 1.56 kN

Ans.



FBD = 520.64 N = 521 N

Ans.



FCD = 1275.3 N = 1.28 kN

Ans.

Ans: FAD = 1.56 kN FBD = 521 N FCD = 1.28 kN 223

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*3–64. If cable AD is tightened by a turnbuckle and develops a tension of 1300 lb, determine the tension developed in cables AB and AC and the force developed along the antenna tower AE at point A.

z A 30 ft

C 10 ft B

E 15 ft

SOLUTION

x

12.5 ft

10 ft

D

15 ft y

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as FAB = FAB C

2(10 - 0) + ( -15 - 0) + ( - 30 - 0)

FAC = FAC C

3 2 6 S = - FAC i - FAC j - FAC k 7 7 7 2( - 15 - 0) + ( -10 - 0) + ( -30 - 0)

FAD = FAD C

(10 - 0)i + ( - 15 - 0)j + ( -30 - 0)k 2

2

2

S =

2 3 6 F i - FAB j - FAB k 7 AB 7 7

(- 15 - 0)i + ( -10 - 0)j + ( -30 - 0)k 2

2

(0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 2(0 - 0)2 + (12.5 - 0)2 + ( -30 - 0)2

2

S = {500j - 1200k} lb

FAE = FAE k Equations of Equilibrium: Equilibrium requires g F = 0;

¢ FAB i 2 7

¢ FAB 2 7

FAB + FAC + FAD + FAE = 0 3 6 3 2 6 FAB j - FAB k ≤ + ¢- FAC i - FAC j - FAC k ≤ + (500j - 1200k) + FAE k = 0 7 7 7 7 7

3 3 2 6 6 F ≤ i + ¢- FAB - FAC + 500 ≤ j + ¢ - FAB - FAC + FAE - 1200 ≤ k = 0 7 AC 7 7 7 7

Equating the i, j, and k components yields 2 3 FAB - FAC = 0 7 7 2 3 - FAB - FAC + 500 = 0 7 7 6 6 - FAB - FAC + FAE - 1200 = 0 7 7

(1) (2) (3)

Solving Eqs. (1) through (3) yields Ans. Ans. Ans.

FAB = 808 lb FAC = 538 lb FAE = 2354 lb = 2.35 kip

Ans: FAB = 808 lb FAC = 538 lb FAE = 2.35 kip 224

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3–65. If the tension developed in either cable AB or AC cannot exceed 1000 lb, determine the maximum tension that can be developed in cable AD when it is tightened by the turnbuckle. Also, what is the force developed along the antenna tower at point A?

z A 30 ft

C 10 ft

SOLUTION

B

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as x FAB = FAB C FAC = FAC C FAD = FC

(10 - 0)i + (- 15 - 0)j + (- 30 - 0)k 2(10 - 0)2 + ( -15 - 0)2 + (- 30 - 0)2

S =

( - 15 - 0)i + ( -10 - 0)j + (-30 - 0)k 2(- 15 - 0)2 + (-10 - 0)2 + ( - 30 - 0)2

(0 - 0)i + (12.5 - 0)j + ( -30 - 0)k 2(0 - 0)2 + (12.5 - 0)2 + (-30 - 0)2

S =

E 15 ft 12.5 ft

10 ft

D

15 ft y

2 3 6 F i - FAB j - FAB k 7 AB 7 7

3 2 6 S = - FAC i - FAC j - FAC k 7 7 7

12 5 Fj Fk 13 13

FAE = FAE k Equations of Equilibrium: Equilibrium requires g F = 0;

¢ FAB i 2 7

¢ FAB 2 7

FAB + FAC + FAD + FAE = 0 12 3 6 3 2 6 5 FAB j - FAB k ≤ + ¢ - FAC i - FAC j - FAC k ≤ + ¢ Fj F k ≤ + FAE k = 0 7 7 7 7 7 13 13

3 3 2 5 6 6 12 F ≤ i + ¢ - FAB - FAC + F ≤ j + ¢ - FAB - FAC F + FAE ≤ k = 0 7 AC 7 7 13 7 7 13

Equating the i, j, and k components yields 3 2 F - FAC = 0 7 AB 7 2 5 3 - FAB - FAC + F = 0 7 7 13 6 12 6 F + FAE = 0 - FAB - FAC 7 7 13

(1) (2) (3)

Let us assume that cable AB achieves maximum tension first. Substituting FAB = 1000 lb into Eqs. (1) through (3) and solving yields FAC = 666.67 lb FAE = 2914 lb = 2.91 kip

Ans.

F = 1610 lb = 1.61 kip

Since FAC = 666.67 lb 6 1000 lb, our assumption is correct. Ans: FAE = 2.91 kip F = 1.61 kip 225

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3–66. Determine the tension developed in cables AB, AC, and AD required for equilibrium of the 300-lb crate.

z

B

C 1 ft

2 ft

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as FAB = FAB C

FAC = FAC C

A

3 ft

(-2 - 0)i + (1 - 0)j + (2 - 0)k 2(-2 - 0)2 + (1 - 0)2 + (2 - 0)2

S = -

2 ft

2 ft

2 ft

SOLUTION

1 ft

y

D x

2 1 2 F i + FAB j + FAB k 3 AB 3 3

(-2 - 0)i + (- 2 - 0)j + (1 - 0)k

2 2 1 S = - FAC i - FAC j + FAC k 3 3 3 2(-2 - 0) + (- 2 - 0) + (1 - 0) 2

2

2

FAD = FAD i W = [- 300k] lb Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

1 2 2 2 1 2 a - FAB i + FAB j + FAB k b + a - FAC i - FAC j + FACkb + FAD i + ( - 300k) = 0 3 3 3 3 3 3 2 2 1 2 2 1 a - FAB - FAC + FAD b i + a FAB - FAC b j + a FAB + FAC - 300b k = 0 3 3 3 3 3 3 Equating the i, j, and k components yields 2 2 - FAB - FAC + FAD = 0 3 3

(1)

2 1 - FAC = 0 F 3 AB 3

(2)

2 1 F + FAC - 300 = 0 3 AB 3

(3)

Solving Eqs. (1) through (3) yields FAB = 360 lb

Ans.

FAC = 180 lb

Ans.

FAD = 360 lb

Ans.

Ans: FAB = 360 lb FAC = 180 lb FAD = 360 lb 226

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3–67. Determine the maximum weight of the crate so that the tension developed in any cable does not exceed 450 lb.

z

B

C 1 ft

2 ft

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as FAB = FAB C

FAC = FAC C

A

3 ft

( - 2 - 0)i + (1 - 0)j + (2 - 0)k 2( -2 - 0)2 + (1 - 0)2 + (2 - 0)2

S = -

2 ft

2 ft

2 ft

SOLUTION

1 ft

y

D x

2 1 2 FAB i + FAB j + FAB k 3 3 3

(- 2 - 0)i + (- 2 - 0)j + (1 - 0)k

2 2 1 S = - FAC i - FAC j + FAC k 3 3 3 2(-2 - 0) + ( -2 - 0) + (1 - 0) 2

2

2

FAD = FAD i W = -Wk Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 1 2 2 2 1 a - FAB i + FAB j + FAB k b + a - FAC i - FAC j + FAC kb + FAD i + (- Wk) = 0 3 3 3 3 3 3 2 1 2 2 1 2 a - FAB - FAC + FAD bi + a FAB - FAC bj + a FAB + FAC - Wbk = 0 3 3 3 3 3 3 Equating the i, j, and k components yields 2 2 - FAB - FAC + FAD = 0 3 3

(1)

2 1 F - FAC = 0 3 AB 3

(2)

1 2 F + FAC - W = 0 3 AB 3

(3)

Let us assume that cable AB achieves maximum tension first. Substituting FAB = 450 lb into Eqs. (1) through (3) and solving, yields FAC = 225 lb

FAD = 450 lb Ans.

W = 375 lb Since FAC = 225 lb 6 450 lb, our assumption is correct.

Ans: W = 375 lb 227

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4–1. If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A : (B + D) = (A : B) + (A : D).

SOLUTION Consider the three vectors; with A vertical. Note obd is perpendicular to A. od = ƒ A * (B + D) ƒ = ƒ A ƒ ƒ B + D ƒ sin u3 ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u1 bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u2 Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross products also form a closed triangle o¿b¿d¿ which is similar to triangle obd. Thus from the figure, A * (B + D) = (A * B) + (A * D)

(QED)

Note also, A = Ax i + Ay j + Az k B = Bx i + By j + Bz k D = Dx i + Dy j + Dz k A * (B + D) = 3

i Ax Bx + Dx

j Ay By + Dy

k Az 3 Bz + Dz

= [A y (Bz + Dz) - A z(By + Dy)]i - [A x(Bz + Dz) - A z(Bx + Dx)]j + [A x(By + Dy) - A y(Bx + Dx)]k = [(A y Bz - A zBy)i - (A x Bz - A z Bx)]j + (A x By - A y Bx)k + [(A y Dz - A z Dy)i - (A x Dz - A z Dx)j + (A x Dy - A y Dx)k i = 3 Ax Bx

j Ay By

k i Az 3 + 3 Ax Bz Dx

j Ay Dy

k Az 3 Dz

= (A * B) + (A * D)

(QED)

228

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4–2. Prove the triple scalar A # (B : C) = (A : B) # C.

product

identity

SOLUTION As shown in the figure Area = B(C sin u) = |B * C| Thus, Volume of parallelepiped is |B * C||h| But, |h| = |A # u(B * C)| = ` A # a

B * C b` |B * C|

Thus, Volume = |A # (B * C)| Since |(A * B) # C| represents this same volume then A # (B : C) = (A : B) # C

(QED)

Also, LHS = A # (B : C) i

= (A x i + A y j +

A z k) # 3 Bx Cx

j By Cy

k Bz 3 Cz

= A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx RHS = (A : B) # C i = 3 Ax Bx

j Ay By

k A z 3 # (Cx i + Cy j + Cz k) Bz

= Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(A xBy - A yBx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx Thus, LHS = RHS A # (B : C) = (A : B) # C

(QED)

229

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4–3. Given the three nonzero vectors A, B, and C, show that if A # (B : C) = 0, the three vectors must lie in the same plane.

SOLUTION Consider, |A # (B * C)| = |A| |B * C | cos u = (|A| cos u)|B * C| = |h| |B * C| = BC |h| sin f = volume of parallelepiped. If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar.

230

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*4–4. Determine the moment about point A of each of the three forces acting on the beam.

F2 = 500 lb

F1 = 375 lb 5

A

4 3

8 ft

6 ft

SOLUTION

B

0.5 ft

5 ft 30˚ F3 = 160 lb

a + 1MF12A = - 375182

= - 3000 lb # ft = 3.00 kip # ft (Clockwise)

Ans.

4 a + 1MF22A = - 500 a b 1142 5

= -5600 lb # ft = 5.60 kip # ft (Clockwise)

Ans.

a + 1MF32A = - 1601cos 30°21192 + 160 sin 30°10.52

= - 2593 lb # ft = 2.59 kip # ft (Clockwise)

Ans.

Ans:

( MF1 ) A = 3.00 kip # ft (Clockwise) ( MF2 ) A = 5.60 kip # ft (Clockwise) ( MF3 ) A = 2.59 kip # ft (Clockwise)

231

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4–5. Determine the moment about point B of each of the three forces acting on the beam.

F2 = 500 lb

F1 = 375 lb 5

A

8 ft

SOLUTION

4 3

6 ft

B

0.5 ft

5 ft 30˚ F3 = 160 lb

a + 1MF12B = 3751112

= 4125 lb # ft = 4.125 kip # ft (Counterclockwise)

Ans.

4 a + 1MF22B = 500 a b 152 5

= 2000 lb # ft = 2.00 kip # ft (Counterclockwise)

Ans.

a + 1MF32B = 160 sin 30°10.52 - 160 cos 30°102 = 40.0 lb # ft (Counterclockwise)

Ans.

Ans: ( MF1 ) B = 4.125 kip # ftd ( MF2 ) B = 2.00 kip # ftd ( MF3 ) B = 40.0 lb # ftd 232

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4–6. The crowbar is subjected to a vertical force of P = 25 lb at the grip, whereas it takes a force of F = 155 lb at the claw to pull the nail out. Find the moment of each force about point A and determine if P is sufficient to pull out the nail. The crowbar contacts the board at point A.

60

F

O 20 3 in. P 14 in.

A 1.5 in.

Solution

a + MP = 25 ( 14 cos 20° + 1.5 sin 20° ) = 341 in # lb (Counterclockwise) c + MF = 155 sin 60°(3) = 403 in # lb (Clockwise)

Since MF 7 MP,  P = 25 lb is not sufficient to pull out the nail.

Ans.

Ans: MP = 341 in. # lbd MF = 403 in. # lbb Not sufficient 233

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4–7. Determine the moment of each of the three forces about point A.

F1

F2

250 N 30

300 N

60

A 2m

3m

4m

SOLUTION The moment arm measured perpendicular to each force from point A is d1 = 2 sin 60° = 1.732 m

B

4

5 3

d2 = 5 sin 60° = 4.330 m

F3

500 N

d3 = 2 sin 53.13° = 1.60 m Using each force where MA = Fd, we have a + 1MF12A = - 25011.7322

= - 433 N # m = 433 N # m (Clockwise)

Ans.

a + 1MF22A = - 30014.3302

= - 1299 N # m = 1.30 kN # m (Clockwise)

Ans.

a + 1MF32A = - 50011.602

= - 800 N # m = 800 N # m (Clockwise)

Ans.

Ans: ( MF1 ) A = 433 N # mb ( MF2 ) A = 1.30 kN # mb ( MF3 ) A = 800 N # mb 234

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*4–8. Determine the moment of each of the three forces about point B.

F1

F2

250 N 30

300 N

60

A 2m

3m

SOLUTION

4m

The forces are resolved into horizontal and vertical component as shown in Fig. a. For F1, a + MB = 250 cos 30°(3) - 250 sin 30°(4) = 149.51 N # m = 150 N # m d

Ans.

B

4

5 3

For F2,

F3

500 N

a + MB = 300 sin 60°(0) + 300 cos 60°(4) = 600 N # m d

Ans.

Since the line of action of F3 passes through B, its moment arm about point B is zero. Thus Ans.

MB = 0

Ans: MB = 150 N # md MB = 600 N # md MB = 0 235

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4–9. Determine the moment of each force about the bolt located at A. Take FB = 40 lb, FC = 50 lb.

0.75 ft B

2.5 ft

30 FC

20

A

C

FB

25

SOLUTION a +MB = 40 cos 25°(2.5) = 90.6 lb # ft d

Ans.

a +MC = 50 cos 30°(3.25) = 141 lb # ftd

Ans.

Ans: MB = 90.6 lb # ftb MC = 141 lb # ftd 236

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4–10. If FB = 30 lb and FC = 45 lb, determine the resultant moment about the bolt located at A.

0.75 ft B

2.5 ft

A

C

30 FC

20 FB

25

SOLUTION a + MA = 30 cos 25°(2.5) + 45 cos 30°(3.25) = 195 lb # ft d

Ans: MA = 195 lb # ftd 237

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4–11. The towline exerts a force of P = 6 kN at the end of the 8-m-long crane boom. If u = 30°, determine the placement x of the hook at B so that this force creates a maximum moment about point O. What is this moment?

A P  6 kN 8m u

O

1m

B x

Solution In order to produce the maximum moment about point O, P must act perpendicular to the boom’s axis OA as shown in Fig. a. Thus

a+ (MO)max = 6 (8) = 48.0 kN # m (counterclockwise)

Ans.

Referring to the geometry of Fig. a,

x = x' + x" =

8 + tan 30° = 9.814 m = 9.81 m cos 30°

Ans.

Ans: (MO)max = 48.0 kN # m d x = 9.81 m 238

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*4–12. A

The towline exerts a force of P = 6 kN at the end of the 8-m-long crane boom. If x = 10 m, determine the position u of the boom so that this force creates a maximum moment about point O. What is this moment?

P  6 kN 8m u

O

1m

B x

Solution In order to produce the maximum moment about point O, P must act perpendicular to the boom’s axis OA as shown in Fig. a. Thus,

a+ (MO)max = 6 (8) = 48.0 kN # m (counterclockwise)

Ans.

Referring to the geometry of Fig. a, 10 =

8 + tan u cos u



10 =

8 sin u + cos u cos u



10 cos u - sin u = 8



x = x' + x";

10 1 8 cos u sin u =  1101 1101 1101



(1)

From the geometry shown in Fig. b,

a = tan-1 a sin a =

Then Eq (1) becomes

1 b = 5.711° 10

1 1101

cos a =

10 1101

cos u cos 5.711° - sin u sin 5.711° =

8 1101

Referring that cos (u + 5.711°) = cos u cos 5.711° - sin u sin 5.711°

cos (u + 5.711°) =

8 1101

u + 5.711° = 37.247° Ans.

u = 31.54° = 31.5°

Ans: (MO)max = 48.0 kN # m (counterclockwise) u = 31.5° 239

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4–13. z

The 20-N horizontal force acts on the handle of the socket wrench. What is the moment of this force about point B. Specify the coordinate direction angles a, b, g of the moment axis.

20 N B

200 mm A

60 10 mm

50 mm O

y

x

Solution Force Vector And Position Vector. Referring to Fig. a, F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rBA = { - 0.01i + 0.2j} m

Moment of Force F about point B. MB = rBA * F  

i = † - 0.01 17.32

j 0.2 - 10

k 0† 0

= { - 3.3641 k} N # m = { - 3.36 k} N # m

Ans.

Here the unit vector for MB is u = - k. Thus, the coordinate direction angles of MB are

a = cos-1 0 = 90°

Ans.



b = cos-1 0 = 90°

Ans.



-1

Ans.

g = cos

( - 1) = 108°

Ans: MB = { - 3.36 k} N # m a = 90° b = 90° g = 180° 240

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4–14. z

The 20-N horizontal force acts on the handle of the socket wrench. Determine the moment of this force about point O. Specify the coordinate direction angles a, b, g of the moment axis.

20 N B

200 mm A

60 10 mm

50 mm O

y

x

Solution Force Vector And Position Vector. Referring to Fig. a, F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rOA = { - 0.01i + 0.2j + 0.05k} m

Moment of F About point O.

MO = rOA * F i = † - 0.01 17.32

j 0.2 - 10

k 0.05 † 0

= {0.5i + 0.8660j - 3.3641k} N # m = {0.5i + 0.866j - 3.36k} N # m

Ans.

The magnitude of MO is MO = 2(MO)2x + (MO)2y + (MO)2z = 20.52 + 0.86602 + ( -3.3641)2 = 3.5096 N # m

Thus, the coordinate direction angles of MO are

a = cos-1 c



b = cos-1 c



g = cos-1 c

(MO)x MO (MO)y MO (MO)z MO

d = cos-1 a d = cos-1 a d = cos-1 a

0.5 b = 81.81° = 81.8° 3.5096 0.8660 b = 75.71° = 75.7° 3.5096

-3.3641 b = 163.45° = 163° 3.5096

Ans. Ans. Ans.

Ans: MO = {0.5i + 0.866j - 3.36k} N # m a = 81.8° b = 75.7° g = 163° 241

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4–15. Two men exert forces of F = 80 lb and P = 50 lb on the ropes. Determine the moment of each force about A. Which way will the pole rotate, clockwise or counterclockwise?

6 ft

P

F

45 B

3

12 ft

5 4

C

SOLUTION

A

4 c + (MA)C = 80 a b (12) = 768 lb # ftb 5

Ans.

a + (MA)B = 50 (cos 45°)(18) = 636 lb # ftd

Ans.

Since (MA)C 7 (MA)B Ans.

Clockwise

Ans: (MA)C = 768 lb # ftb (MA)B = 636 lb # ftd Clockwise 242

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*4–16. If the man at B exerts a force of P = 30 lb on his rope, determine the magnitude of the force F the man at C must exert to prevent the pole from rotating, i.e., so the resultant moment about A of both forces is zero.

6 ft

P

F

45 B

SOLUTION a+

3

12 ft

5 4

C

4 30 (cos 45°)(18) = Fa b(12) = 0 5

A

Ans.

F = 39.8 lb

Ans: F = 39.8 lb 243

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4–17. The torque wrench ABC is used to measure the moment or torque applied to a bolt when the bolt is located at A and a force is applied to the handle at C. The mechanic reads the torque on the scale at B. If an extension AO of length d is used on the wrench, determine the required scale reading if the desired torque on the bolt at O is to be M.

F M O

A d

B

l

C

Solution Moment at A = m = Fl Moment at O = M = (d + l)F

M = (d + l)



m = a

m l

l bM d + l

Ans.

Ans: m = a 244

l bM d + l

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4–18. The tongs are used to grip the ends of the drilling pipe P. Determine the torque (moment) MP that the applied force F = 150 lb exerts on the pipe about point P as a function of u. Plot this moment MP versus u for 0 … u … 90°.

F u

P

SOLUTION

MP

MP = 150 cos u(43) + 150 sin u(6)

43 in.

= (6450 cos u + 900 sin u) lb # in. = (537.5 cos u + 75 sin u) lb # ft dMP = -537.5 sin u + 75 cos u = 0 du

tan u =

75 537.5

6 in.

Ans. u = 7.943°

At u = 7.943° , MP is maximum. (MP)max = 538 cos 7.943° + 75 sin 7.943° = 543 lb # ft Also (MP)max = 150 lb ¢ a

1

43 2 6 2 2 b + a b ≤ = 543 lb # ft 12 12

Ans: MP = (537.5 cos u + 75 sin u) lb # ft 245

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4–19. The tongs are used to grip the ends of the drilling pipe P. If a torque (moment) of MP = 800 lb # ft is needed at P to turn the pipe, determine the cable force F that must be applied to the tongs. Set u = 30°.

SOLUTION MP = F cos 30°(43) + F sin 30°(6)

F u

P

Set MP = 800(12) lb # in.

6 in.

MP 43 in.

800(12) = F cos 30°(43) + F sin 30°(6) Ans.

F = 239 lb

Ans: F = 239 lb 246

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*4–20. The handle of the hammer is subjected to the force of F = 20 lb. Determine the moment of this force about the point A.

F 30

5 in. 18 in.

SOLUTION Resolving the 20-lb force into components parallel and perpendicular to the hammer, Fig. a, and applying the principle of moments,

A B

a +MA = - 20 cos 30°(18) - 20 sin 30°(5) = -361.77 lb # in = 362 lb # in (Clockwise)

Ans.

Ans: MA = 362 lb # in (Clockwise) 247

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4–21. In order to pull out the nail at B, the force F exerted on the handle of the hammer must produce a clockwise moment of 500 lb # in. about point A. Determine the required magnitude of force F.

F 30

5 in. 18 in.

SOLUTION Resolving force F into components parallel and perpendicular to the hammer, Fig. a, and applying the principle of moments,

A B

a + MA = - 500 = -F cos 30°(18) - F sin 30°(5) Ans.

F = 27.6 lb

Ans: F = 27.6 lb 248

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4–22. y

Old clocks were constructed using a fusee B to drive the gears and watch hands. The purpose of the fusee is to increase the leverage developed by the mainspring A as it uncoils and thereby loses some of its tension. The mainspring can develop a torque (moment) Ts = ku, where k = 0.015 N # m>rad is the torsional stiffness and u is the angle of twist of the spring in radians. If the torque Tf developed by the fusee is to remain constant as the mainspring winds down, and x = 10 mm when u = 4 rad, determine the required radius of the fusee when u = 3 rad.

x

A

B y

t

x 12 mm Ts

Tf

Solution When u = 4 rad, r = 10 mm Ts = 0.015(4) = 0.06 N # m F =

0.06 = 5N 0.012

Tf = 5(0.010) = 0.05 N # m (constant) When u = 3 rad,

Ts = 0.015(3) = 0.045 N # m F =

0.045 = 3.75 N 0.012

For the fusee require 0.05 = 3.75 r Ans.

r = 0.0133 m = 13.3 mm

Ans: r = 13.3 mm 249

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4–23. 4m

The tower crane is used to hoist the 2-Mg load upward at constant velocity. The 1.5-Mg jib BD, 0.5-Mg jib BC, and 6-Mg counterweight C have centers of mass at G1 , G2 , and G3 , respectively. Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B.

G2

B

C G3

9.5m

7.5 m

D 12.5 m

G1

23 m

SOLUTION Since the moment arms of the weights and the load measured to points A and B are the same, the resultant moments produced by the load and the weight about points A and B are the same. a + (MR)A = (MR)B = ©Fd;

A

(MR)A = (MR)B = 6000(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5)

- 2000(9.81)(12.5) = 76 027.5 N # m = 76.0 kN # m (Counterclockwise)

Ans.

Ans: (MR)A = (MR)B = 76.0 kN # md 250

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*4–24. The tower crane is used to hoist a 2-Mg load upward at constant velocity. The 1.5-Mg jib BD and 0.5-Mg jib BC have centers of mass at G1 and G2 , respectively. Determine the required mass of the counterweight C so that the resultant moment produced by the load and the weight of the tower crane jibs about point A is zero. The center of mass for the counterweight is located at G3 .

4m G2 C G3

9.5m B

7.5 m

D 12.5 m

G1

23 m

SOLUTION a + (MR)A = ©Fd;

A

0 = MC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) Ans.

MC = 4966.67 kg = 4.97 Mg

Ans: MC = 4.97 Mg 251

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4–25. If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb man have centers of gravity located at points G1 , G2 and G3 , respectively, determine the resultant moment produced by each weight about point A.

G3

B

D

G2 C 2.5 ft 1.75 ft

20 ft

G1

SOLUTION

10 ft 75

Moment of the weight of boom AB about point A:

A

a + (MAB)A = -1500(10 cos 75°) = -3882.29 lb # ft

= 3.88 kip # ft (Clockwise)

Ans.

Moment of the weight of cage BCD about point A: a + (MBCD)A = -200(30 cos 75° + 2.5) = - 2052.91 lb # ft

= 2.05 kip # ft (Clockwise)

Ans.

Moment of the weight of the man about point A: a + (Mman)A = -175(30 cos 75° + 4.25) = - 2102.55 lb # ft

= 2.10 kip # ft (Clockwise)

Ans.

Ans: ( MAB ) A = 3.88 kip # ftb ( MBCD ) A = 2.05 kip # ftb ( Mman ) A = 2.10 kip # ftb 252

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4–26. If the 1500-lb boom AB, the 200-lb cage BCD, and the 175-lb man have centers of gravity located at points G1 , G2 and G3 , respectively, determine the resultant moment produced by all the weights about point A.

G3

B

D

G2 C 2.5 ft 1.75 ft

20 ft

G1

SOLUTION

10 ft

Referring to Fig. a, the resultant moment of the weight about point A is given by a + (MR)A = ©Fd;

75 A

(MR)A = -1500(10 cos 75°) - 200(30 cos 75°+2.5) - 175(30 cos 75° + 4.25) = - 8037.75 lb # ft = 8.04 kip # ft (Clockwise)

Ans.

Ans: (MR)A = 8.04 kip # ftb 253

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4–27. z

Determine the moment of the force F about point O. Express the result as a Cartesian vector.

F  {–6i + 4 j  8k} kN

A

4m P 3m

6m O 1m

y

2m

Solution x

Position Vector. The coordinates of point A are (1, - 2, 6) m. Thus,

rOA = {i - 2j + 6k} m

The moment of F About Point O.

MO = rOA * F



i = † 1 -6



j -2 4

k 6† 8

= { - 40i - 44j - 8k} kN # m

Ans.

Ans: MO = { - 40i - 44j - 8k} kN # m 254

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*4–28. z

Determine the moment of the force F about point P. Express the result as a Cartesian vector.

F  {–6i + 4 j  8k} kN

A

4m P 3m

6m O 1m 2m

Solution Position Vector. The coordinates of points A and P are A (1, - 2, 6) m and P (0, 4, 3) m, respectively. Thus

y

x

rPA = (1 - 0)i + ( - 2 - 4)j + (6 - 3)k = {i - 6j + 3k} m

The moment of F About Point P.

MP = rPA * F



i = † 1 -6



j -6 4

k 3† 8

= { - 60i - 26j - 32k} kN # m

Ans.

Ans: MP = { - 60i - 26j - 32k} kN # m 255

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4–29. z

The force F = {400i - 100j - 700k} lb acts at the end of the beam. Determine the moment of this force about point O.

A F O

B

y

8 ft

1.5 ft

Solution Position Vector. The coordinates of point B are B(8, 0.25, 1.5) ft.

x

0.25 ft

Thus,

rOB = {8i + 0.25j + 1.5k} ft

Moments of F About Point O.

MO = rOB * F



i = † 8 400



j 0.25 - 100

k 1.5 † - 700

= { - 25i + 6200j - 900k} lb # ft

Ans.

Ans: MO = { - 25i + 6200j - 900k} lb # ft 256

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4–30. z

The force F = {400i - 100j - 700k} lb acts at the end of the beam. Determine the moment of this force about point A.

A F O

B

y

8 ft

1.5 ft

Solution Position Vector. The coordinates of points A and B are A (0, 0, 1.5) ft and B (8, 0.25, 1.5) ft, respectively. Thus,

x

0.25 ft

rAB = (8 - 0)i + (0.25 - 0)j + (1.5 - 1.5)k = {8i + 0.25j} ft

Moment of F About Point A.

MA = rAB * F



i = † 8 400



j 0.25 - 100

k 0 † - 700

= { - 175i + 5600j - 900k} lb # ft

Ans.

Ans: MA = { - 175i + 5600j - 900k} lb # ft 257

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4–31. z

Determine the moment of the force F about point P. Express the result as a Cartesian vector.

P 2m

2m 3m

y

O

3m

3m

Solution

x

Position Vector. The coordinates of points A and P are A (3, 3, - 1) m and P ( -2, - 3, 2) m respectively. Thus,

1m A

F  {2i  4j  6k} kN

rPA = [3 - ( - 2)]i + [3 - ( -3)] j + ( - 1 - 2)k = {5i + 6j - 3k} m

Moment of F About Point P.

MP = rAP * F



i = †5 2



j 6 4

k -3 † -6

= { - 24i + 24j + 8k} kN # m

Ans.

Ans: MP = { - 24i + 24j + 8k} kN # m 258

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*4–32. z

The pipe assembly is subjected to the force of F = {600i + 800j - 500k} N. Determine the moment of this force about point A.

A 0.5 m B x

0.3 m

Solution

0.3 m

Position Vector. The coordinates of point C are C (0.5, 0.7, -0.3) m. Thus

C

rAC = {0.5i + 0.7 j - 0.3k} m

Moment of Force F About Point A.

MA = rAC * F



i = † 0.5 600



y

0.4 m

j 0.7 800

F

k - 0.3 † -500

= { - 110i + 70j - 20k} N # m

Ans.

Ans: MA = { - 110i + 70j - 20k} N # m 259

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4–33. z

The pipe assembly is subjected to the force of F = {600i + 800j - 500k} N. Determine the moment of this force about point B.

A 0.5 m B x

0.3 m

Solution Position Vector. The coordinates of points B and C are B (0.5, 0, 0) m and C (0.5, 0.7, -0.3) m, respectively. Thus,

y

0.4 m

0.3 m C

rBC = (0.5 - 0.5)i + (0.7 - 0) j + ( - 0.3 - 0)k = {0.7j - 0.3k} m

F

Moment of Force F About Point B. Applying Eq. 4

MB = rBC * F



i = † 0 600



j 0.7 800

k - 0.3 † -500

= { - 110i - 180j - 420k} N # m

Ans.

Ans: MB = { - 110i - 180j - 420k} N # m 260

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4–34. z

Determine the moment of the force of F = 600 N about point A.

A

45

B

4m

4m

x 6m

F C

Solution

6m

Position Vectors And Force Vector. The coordinates of points A, B and C are A (0, 0, 4) m, B (4 sin 45°, 0, 4 cos 45°) m and C (6, 6, 0) m, respectively. Thus

y

rAB = (4 sin 45° - 0)i + (0 - 0)j + (4 cos 45° - 4)k

= {2.8284i - 1.1716k} m rAC = (6 - 0)i + (6 - 0)j + (0 - 4)k = {6i + 6j - 4k} m rBC = (6 - 4 sin 45°)i + (6 - 0)j + (0 - 4 cos 45°)k = {3.1716i + 6j - 2.8284k} m F = Fa

3.1716i + 6j - 2.8284k rBC b = 600£ ≥ rBC 23.17162 + 62 + ( -2.8284)2

= {258.82i + 489.63j - 230.81k} N

The Moment of Force F About Point A.

MA = rAB * F i = † 2.8284 258.82

j 0 489.63

k - 1.1716 † - 230.81

= {573.64i + 349.62j + 1384.89k} N # m = {574i + 350j + 1385k} N # m

Ans.

OR

MA = rAC * F = †

i 6 258.82

j 6 489.63

k -4 † - 230.81



= {573.64i + 349.62j + 1384.89k} N # m



= {574i + 350j + 1385k}

Ans.

Ans: MA = {574i + 350j + 1385k} N # m 261

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4–35. z

Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 4 m, to fail at the support A. This requires a moment of M = 1500 N # m to be developed at A.

A

45

B

4m

4m

x 6m

F C

Solution

6m

Position Vectors And Force Vector. The coordinates of points A, B and C are A (0, 0, 4) m, B (4 sin 45°, 0, 4 cos 45°) m and C (6, 6, 0) m, respectively.

y

Thus, rAB = (4 sin 45° - 0)i + (0 - 0)j + (4 cos 45° - 4)k

= {2.8284i - 1.1716k} m rAC = (6 - 0)i + (6 - 0)j + (0 - 4)k = {6i + 6j - 4k} m rBC = (6 - 4 sin 45°)i + (6 - 0)j + (0 - 4 cos 45°)k = {3.1716i + 6j - 2.8284k} m F = Fa

3.1716i + 6j - 2.8284k rBC b = F£ ≥ rBC 23.17162 + 62 + ( -2.8284)2

= 0.4314F i + 0.8161Fj - 0.3847F k

The Moment of Force F About Point A.

MA = rAB * F



i = † 2.8284 0.4314F



j 0 0.8161F

k - 1.1716 † - 0.3847F

= 0.9561F i + 0.5827Fj + 2.3081F k

OR

MA = rAC * F = †

i 6 0.4314F

j 6 0.8161F

k -4 † - 0.3847F

= 0.9561F i + 0.5827F j + 2.3081F k

The magnitude of MA is MA = 2(MA)2x + (MA)2y + (MA)2z = 2(0.9561F)2 + (0.5827F)2 + (2.3081F)2

= 2.5654F It is required that MA = 1500 N # m, then

1500 = 2.5654F



F = 584.71 N = 585 N

Ans. Ans: F = 585 N 262

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*4–36. z

Determine the coordinate direction angles a, b, g of force F, so that the moment of F about O is zero. O

y 0.4 m

A 0.5 m

0.3 m

x

Solution

F

Position And Force Vectors. The coordinates of point A are A (0.4, 0.5, -0.3) m. Thus,

rOA = {0.4i + 0.5j - 0.3k} m



u OA =



rAO = { - 0.4i - 0.5j + 0.3k} m



u AO =

0.4i + 0.5j - 0.3k rOA 4 5 3 = = i + j k 2 2 2 rOA 150 150 150 20.4 + 0.5 + ( - 0.3)

- 0.4i - 0.5j + 0.3k rAO 4 5 3 = = i j + k 2 2 2 rAO 150 150 150 2( - 0.4) + ( - 0.5) + 0.3

Moment of F About Point O. To produce zero moment about point O, the line of action of F must pass through point O. Thus, F must directed from O to A (direction defined by uOA). Thus,

cos a = -



cos b = -



cos g =

4 ;  a = 55.56° = 55.6° 150

Ans.

5 ;  b = 45° 150

Ans.

-3 ;  g = 115.10° = 115° 150

Ans.

OR F must directed from A to O (direction defined by uAO). Thus

cos a = -

4 ;  a = 124.44° = 124° 150

Ans.

5 ;  b = 135° 150 3 ;  g = 64.90° = 64.9° cos g = 150 cos b = -

Ans. Ans.

Ans: a = 55.6° b = 45° g = 115° OR a = 124° b = 135° g = 64.9° 263

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4–37. z

Determine the moment of force F about point O. The force has a magnitude of 800 N and coordinate direction angles of a = 60°, b = 120°, g = 45°. Express the result as a Cartesian vector.

O

y 0.4 m

A 0.5 m

0.3 m

x

Solution

F

Position And Force Vectors. The coordinates of point A are A (0.4, 0.5, -0.3) m. Thus

rOA = {0.4i + 0.5j - 0.3k} m



F = FuF = 800 (cos 60°i + cos 120°j + cos 45°k)



= {400i - 400j + 565.69k} N

Moment of F About Point O.

MO = rOA * F



i = † 0.4 400

j 0.5 - 400

k -0.3 † 565.69



= {162.84 i - 346.27j - 360 k} N # m



= {163i - 346j - 360k} N # m

Ans.

Ans: MO = {163i - 346j - 360k} N # m 264

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4–38. z

Determine the moment of the force F about the door hinge at A. Express the result as a Cartesian vector.

4 ft

7 ft

A

F  80 lb

C

1.5 ft

B

45

D

Solution Position Vectors And Force Vector. The coordinates of points A, C and D are A ( -6.5, - 3, 0) ft, C [0, - (3 + 4 cos 45°), 4 sin 45°] ft and D ( - 5, 0, 0) ft, respectively. Thus, rAC = [0 - ( -6.5)]i + [ - (3 + 4 cos 45°) - ( -3)] j + (4 sin 45° - 0)k

1.5 ft 3 ft x

5 ft

y

= {6.5i - 2.8284j + 2.8284k} ft rAD = [ -5 - ( - 6.5)]i + [0 - ( -3)]j + (0 - 0)k = {1.5i + 3j} ft rCD = ( -5 - 0)i + {0 - [ - (3 + 4 cos 45°)]} j + (0 - 4 sin 45°)k = { - 5i + 5.8284j - 2.8284k} ft F = Fa

- 5i + 5.8284j - 2.8284k rCD b = 80 £ ≥ rCD 2( -5)2 + 5.82842 + ( -2.8284)2

= { - 48.88 i + 56.98j - 27.65k} lb Moment of F About Point A.

MA = rAC * F = †

i 6.5 - 48.88

j -2.8284 56.98

k 2.8284 † - 27.65



= { - 82.9496i + 41.47j + 232.10k} lb # ft



= { - 82.9i + 41.5j + 232k} lb # ft

Ans.

OR

MA = rAD * F = †

i 1.5 - 48.88

j 3 56.98

k 0 † - 27.65



= { - 82.9496i + 41.47j + 232.10 k} lb # ft



= { - 82.9i + 41.5j + 232k} lb # ft

Ans.

Ans: MA = { - 82.9i + 41.5j + 232k} lb # ft 265

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4–39. z

Determine the moment of the force F about the door hinge at B. Express the result as a Cartesian vector.

4 ft

7 ft

A

F  80 lb

C

1.5 ft

B

45

D

Solution Position Vectors And Force Vector. The coordinates of points B, C and D are B ( - 1.5, -3, 0) ft, C [0, - (3 + 4 cos 45°), 4 sin 45°] ft and D ( - 5, 0, 0) ft, respectively. Thus, rBC = [0 - ( - 1.5)]i + [ -(3 + 4 cos 45°) - ( - 3)] j + (4 sin 45° - 0)k

1.5 ft 3 ft x

5 ft

y

= {1.5i - 2.8284j + 2.8284k} ft   rBD = [ - 5 - ( - 1.5)]i + [0 - ( - 3)]j + (0 - 0)k = { -3.5i + 3j} ft



rCD = ( - 5- 0)i + {0 - [ -(3 + 4 cos 45°)]} j + (0 - 4 sin 45°)k = { - 5i + 5.8284j - 2.8284k} ft F = Fa

- 5i + 5.8284j - 2.8284k rCD b = 80 £ ≥ rCD 2( -5)2 + 5.82842 + ( -2.8284)2

= { - 48.88 i + 56.98j - 27.65k} lb

Moment of F About Point B.

MB = rBC * F = †

i 1.5 - 48.88

j -2.8284 56.98

k 2.8284 † - 27.65



= { - 82.9496i - 96.77j - 52.78k} lb # ft



= { - 82.9i - 96.8j - 52.8k} lb # ft

Ans.

or

MB = rBD * F



i = † - 3.5 - 48.88

j 3 56.98

k 0 † - 27.65



= { - 82.9496i - 96.77j - 52.78 k} lb # ft



= { - 82.9i - 96.8j - 52.8k} lb # ft

Ans.

Ans: MB = { - 82.9i - 96.8j - 52.8k} lb # ft 266

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*4–40. The curved rod has a radius of 5 ft. If a force of 60 lb acts at its end as shown, determine the moment of this force about point C.

z C

5 ft

60°

A

5 ft

SOLUTION

y 60 lb

Position Vector and Force Vector:

6 ft

B

rCA = 515 sin 60° - 02j + 15 cos 60° - 52k6 m

x

7 ft

= 54.330j - 2.50k6 m

FAB = 60 ¢

16 - 02i + 17 - 5 sin 60°2j + 10 - 5 cos 60°2k

216 - 022 + 17 - 5 sin 60°22 + 10 - 5 cos 60°22

≤ lb

= 551.231i + 22.797j - 21.346k6 lb

Moment of Force FAB About Point C: Applying Eq. 4–7, we have M C = rCA * FAB i 0 = 51.231 =

j 4.330 22.797

k - 2.50 - 21.346

- 35.4i - 128j - 222k lb # ft

Ans.

Ans: MC = { - 35.4i - 128j - 222k} lb # ft 267

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4–41. z

Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 5 ft, to fail at the support C. This requires a moment of M = 80 lb # ft to be developed at C.

C

5 ft

60

A

5 ft

y 60 lb

SOLUTION

6 ft

B

Position Vector and Force Vector:

x

7 ft

rCA = {(5 sin 60° - 0)j + (5 cos 60° - 5)k} m = {4.330j - 2.50 k} m FAB = F a

(6 - 0)i + (7 - 5 sin 60°)j + (0 - 5 cos 60°)k 2(6 - 0)2 + (7 - 5 sin 60°)2 + (0 - 5 cos 60°)2

b lb

= 0.8539Fi + 0.3799Fj - 0.3558Fk Moment of Force FAB About Point C: M C = rCA * FAB = 3

i 0 0.8539F

j 4.330 0.3799F

k - 2.50 3 - 0.3558F

= - 0.5909Fi - 2.135j - 3.697k Require 80 = 2(0.5909)2 + ( - 2.135)2 + ( -3.697)2 F F = 18.6 lb.

Ans.

Ans: F = 18.6 lb 268

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4–42. A 20-N horizontal force is applied perpendicular to the handle of the socket wrench. Determine the magnitude and the coordinate direction angles of the moment created by this force about point O.

z

200 mm

75 mm

A 20 N

SOLUTION

O

rA = 0.2 sin 15°i + 0.2 cos 15°j + 0.075k = 0.05176 i + 0.1932 j + 0.075 k

15

x

F = - 20 cos 15°i + 20 sin 15°j = - 19.32 i + 5.176 j i MO = rA * F = 3 0.05176 -19.32

j 0.1932 5.176

k 0.075 3 0

= { - 0.3882 i - 1.449 j + 4.00 k} N # m MO = 4.272 = 4.27 N # m

Ans.

a = cos -1 a

- 0.3882 b = 95.2° 4.272

Ans.

b = cos -1 a

- 1.449 b = 110° 4.272

Ans.

g = cos -1 a

4 b = 20.6° 4.272

Ans.

Ans: MO = 4.27 N # m a = 95.2° b = 110° g = 20.6° 269

y

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4–43. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A.

z

A 400 mm B x

SOLUTION

300 mm

Position Vector And Force Vector:

200 mm

rAC = {(0.55 - 0)i + (0.4 - 0)j + ( -0.2 - 0)k} m

200 mm

C

250 mm

= {0.55i + 0.4j - 0.2k} m 40

F = 80(cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

30

= (44.53i + 53.07j - 40.0k} N

F

80 N

Moment of Force F About Point A: Applying Eq. 4–7, we have MA = rAC * F i = 3 0.55 44.53

j 0.4 53.07

k - 0.2 3 - 40.0

= {- 5.39i + 13.1j + 11.4k} N # m

Ans.

Ans: MA = {-5.39i + 13.1j + 11.4k} N # m 270

y

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*4–44. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B.

z

A 400 mm B x

SOLUTION

300 mm

Position Vector And Force Vector:

200 mm

rBC = {(0.55 - 0) i + (0.4 - 0.4)j + ( - 0.2 - 0)k} m

200 mm

C

250 mm

= {0.55i - 0.2k} m 40

F = 80 (cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

30

= (44.53i + 53.07j - 40.0k} N

F

80 N

Moment of Force F About Point B: Applying Eq. 4–7, we have MB = rBC * F i = 3 0.55 44.53

j 0 53.07

k - 0.2 3 - 40.0

= {10.6i + 13.1j + 29.2k} N # m

Ans.

Ans: MB = {10.6i + 13.1j + 29.2k} N # m 271

y

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4–45. A force of F = 5 6i - 2j + 1k 6 kN produces a moment of M O = 54i + 5j - 14k6 kN # m about the origin of coordinates, point O. If the force acts at a point having an x coordinate of x = 1 m, determine the y and z coordinates. . Note: The figure shows F and MO in an arbitrary position.

z F

P MO z

d

y

O 1m

SOLUTION y

MO = r * F i 4i + 5j - 14k = 3 1 6

j y -2

x

k z3 1

4 = y + 2z 5 = -1 + 6z -14 = -2 - 6y y = 2m

Ans.

z = 1m

Ans.

Ans: y = 2m z = 1m 272

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4–46. The force F = 56i + 8j + 10k6 N creates a moment about point O of M O = 5 -14i + 8j + 2k6 N # m. If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point.Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F. Note: The figure shows F and MO in an arbitrary position.

z F

P MO z

d

y

O 1m

SOLUTION i -14i + 8j + 2k = 3 1 6

j y 8

y

k z 3 10

x

- 14 = 10y - 8z 8 = -10 + 6z 2 = 8 - 6y y = 1m

Ans.

z = 3m

Ans.

MO = 2( -14)2 + (8)2 + (2)2 = 16.25 N # m F = 2(6)2 + (8)2 + (10)2 = 14.14 N d =

16.25 = 1.15 m 14.14

Ans.

Ans: y = 1m z = 3m d = 1.15 m 273

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4–47. A force F having a magnitude of F = 100 N acts along the diagonal of the parallelepiped. Determine the moment of F about point A, using M A = rB : F and M A = rC : F.

z

F

C

200 mm rC

SOLUTION F = 100 a

400 mm B

- 0.4 i + 0.6 j + 0.2 k b 0.7483

F

F = 5 - 53.5 i + 80.2 j + 26.7 k6 N M A = rB * F = 3

i 0 -53.5

j -0.6 80.2

i - 0.4 -53.5

j 0 80.2

k 0 3 = 5- 16.0 i - 32.1 k6 N # m 26.7

rB 600 mm

A

x

Ans.

Also, M A = rC * F =

k 0.2 = 26.7

- 16.0 i - 32.1 k N # m

Ans.

Ans: MA = { - 16.0i - 32.1k} N # m 274

y

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*4–48. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point A. Express the result as a Cartesian vector.

z 3m

A F

400 N

3m B

SOLUTION

x

C

4m

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF is equal to the unit vector of the cross product, b = rAC * rBC, Fig. a. Here rAC = (0 - 0)i + (4 - 0)j + (0 - 3)k = [4j - 3k] m rBC = (0 - 3)i + (4 - 0)j + (0 - 0)k = [ -3i + 4j] m Thus, i b = rCA * rCB = 3 0 -3

j 4 4

k -3 3 0

= [12i + 9j + 12k] m2 Then, uF =

12i + 9j + 12k b = = 0.6247i + 0.4685j + 0.6247k b 212 2 + 92 + 12 2

And finally F = FuF = 400(0.6247i + 0.4685j + 0.6247k) = [249.88i + 187.41j + 249.88k] N Vector Cross Product: The moment of F about point A is M A = rAC

* F = 3

i 0 249.88

j 4 187.41

k -3 3 249.88

= [1.56i - 0.750j - 1.00k] kN # m

Ans.

Ans: MA = [1.56i - 0.750j - 1.00k] kN # m 275

y

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4–49. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point B. Express the result as a Cartesian vector.

z 3m

A F

400 N

3m B

SOLUTION

x

C

4m

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF is equal to the unit vector of the cross product, b = rAC * rBC, Fig. a. Here rAC = (0 - 0)i + (4 - 0)j + (0 - 3)k = [4j - 3k] m rBC = (0 - 3)i + (4 - 0)j + (0 - 0)k = [- 3k + 4j] m Thus, i b = rCA * rCB = 3 0 -3

j 4 4

k -3 3 = [12i + 9j + 12k] m2 0

Then, uF =

12i + 9j + 12k b = = 0.6247i + 0.4685j + 0.6247k b 2122 + 92 + 122

And finally F = FuF = 400(0.6247i + 0.4685j + 0.6247k) = [249.88i + 187.41j + 249.88k] N Vector Cross Product: The moment of F about point B is MB = rBC

* F = 3

i -3 249.88

j 4 187.41

k 0 3 249.88

= [1.00i + 0.750j - 1.56k] kN # m

Ans.

Ans: MB = {1.00i + 0.750j - 1.56k} kN # m 276

y

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4–50. Strut AB of the 1-m-diameter hatch door exerts a force of 450 N on point B. Determine the moment of this force about point O.

z

B 30°

0.5 m

F = 450 N

O

SOLUTION

y 0.5 m

Position Vector And Force Vector: rOB = 510 - 02i + 11 cos 30° - 02j + 11 sin 30° - 02k6 m

30°

A

x

= 50.8660j + 0.5k6 m

rOA = 510.5 sin 30° - 02i + 10.5 + 0.5 cos 30° - 02j + 10 - 02k6 m = 50.250i + 0.9330j6 m

F = 450 ¢

10 - 0.5 sin 30°2i + 31 cos 30° - 10.5 + 0.5 cos 30°24j + 11 sin 30° - 02k

210 - 0.5 sin 30°22 + 31 cos 30° - 10.5 + 0.5 cos 30°242 + 11 sin 30° - 022

≤N

= 5-199.82i - 53.54j + 399.63k6 N

Moment of Force F About Point O: Applying Eq. 4–7, we have M O = rOB * F = 3

i 0 - 199.82

j 0.8660 - 53.54

k 0.5 3 399.63

= 5373i - 99.9j + 173k6 N # m

Ans.

Or M O = rOA * F =

i 0.250 -199.82

j 0.9330 - 53.54

k 0 399.63

=

373i - 99.9j + 173k N # m

Ans.

Ans: MO = {373i - 99.9j + 173k} N # m 277

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4–51. z

Using a ring collar the 75-N force can act in the vertical plane at various angles u. Determine the magnitude of the moment it produces about point A, plot the result of M (ordinate) versus u (abscissa) for 0° … u … 180°, and specify the angles that give the maximum and minimum moment.

A 2m

1.5 m

SOLUTION i MA = 3 2 0

j 1.5 75 cos u

k 3 0 75 sin u

y

x

75 N

θ

= 112.5 sin u i - 150 sin u j + 150 cos u k MA = 21112.5 sin u22 + 1-150 sin u22 + 1150 cos u22 = 212 656.25 sin2 u + 22 500 dMA 1 1 = 112 656.25 sin2 u + 22 5002- 2 112 656.25212 sin u cos u2 = 0 du 2 sin u cos u = 0;

Ans.

u = 0°, 90°, 180°

umax = 187.5 N # m at u = 90° umin = 150 N # m at u = 0°, 180°

Ans: umax = 90° umin = 0, 180° 278

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*4–52. z

The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force of F = 30 N at A. Determine if this force is adequate, provided 14 N # m of torque about the x axis is initially required to turn the nut. If the 30-N force can be applied at A in any other direction, will it be possible to turn the nut?

B

0.25 m

A

0.3 m

SOLUTION Mx = 30 A 2(0.5) - (0.3) 2

2

B =

12 N #

m 6

14 N #

0.5 m y

m,

No

0.1 m

Ans. x

For (Mx)max , apply force perpendicular to the handle and the x - axis. (Mx)max = 30 (0.5) = 15 N # m 7 14 N # m,

30 N

F

Yes

Ans.

Ans: No Yes 279

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4–53. Solve Prob. 4–52 if the cheater pipe AB is slipped over the handle of the wrench and the 30-N force can be applied at any point and in any direction on the assembly.

z 30 N

F B

0.25 m

A

0.3 m

SOLUTION 4 Mx = 30 (0.75)a b = 18 N # m 7 14 N # m, 5

0.5 m y 0.1 m

Ans.

Yes

x

(Mx)max occurs when force is applied perpendicular to both the handle and the x - axis. (Mx)max = 30(0.75) = 22.5 N # m 7 14N # m,

Ans.

Yes

Ans: Yes Yes 280

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4–54. z

The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the y′ axis passing through points A and B when the frame is in the position shown.

F C A x¿

6 ft x

Solution

15

6 ft

30

y B y¿

Scalar analysis :

My′ = 80 (6 cos 15°) = 464 lb # ft

Vector analysis : uAB = cos 60° i + cos 30° j Coordinates of point C : x = 3 sin 30° - 6 cos 15° cos 30° = - 3.52 ft y = 3 cos 30° + 6 cos 15° sin 30° = 5.50 ft z = 6 sin 15° = 1.55 ft rAC = - 3.52 i + 5.50 j + 1.55 k F = 80 k sin 30° My′ = † - 3.52 0

My′ = 464 lb # ft

cos 30° 5.50 0

0 1.55 † 80 Ans.

Ans: My′ = 464 lb # ft 281

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4–55. z

The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the x axis when the frame is in the position shown.

F C A x¿

6 ft x

Solution

15

6 ft

30

y B y¿

Using x′, y′, z : ux = cos 30° i′ + sin 30° j′ rAC = - 6 cos 15° i′ + 3 j′ + 6 sin 15° k F = 80 k cos 30° Mx = † - 6 cos 15° 0 Mx = 440 lb # ft

sin 30° 3 0

0 6 sin 15° † = 207.85 + 231.82 + 0 80

Also, using x, y, z, Coordinates of point C : x = 3 sin 30° - 6 cos 15° cos 30° = - 3.52 ft y = 3 cos 30° + 6 cos 15° sin 30° = 5.50 ft z = 6 sin 15° = 1.55 ft rAC = - 3.52 i + 5.50 j + 1.55 k F = 80 k 1 Mx = † -3.52 0

0 5.50 0

0 1.55 † = 440 lb # ft 80

Ans.

Ans: Mx = 440 lb # ft 282

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*4–56. Determine the magnitude of the moments of the force F about the x, y, and z axes. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach.

z A

4 ft y

SOLUTION x

a) Vector Analysis

3 ft C

PositionVector: rAB = {(4 - 0) i + (3 - 0)j + ( -2 - 0)k} ft = {4i + 3j - 2k} ft

2 ft

Moment of Force F About x,y, and z Axes: The unit vectors along x, y, and z axes are i, j, and k respectively. Applying Eq. 4–11, we have Mx = i # (rAB * F) 1 = 34 4

0 3 12

B F

{4i

12j

3k} lb

0 -2 3 -3

= 1[3(- 3) - (12)(- 2)] - 0 + 0 = 15.0 lb # ft

Ans.

My = j # (rAB * F) 0 = 34 4

1 3 12

0 -2 3 -3

= 0 - 1[4(- 3) - (4)(-2)] + 0 = 4.00 lb # ft

Ans.

Mz = k # (rAB * F) 0 = 34 4

0 3 12

1 -2 3 -3

= 0 - 0 + 1[4(12) - (4)(3)] = 36.0 lb # ft

Ans.

b) ScalarAnalysis Mx = ©Mx ;

Mx = 12(2) - 3(3) = 15.0 lb # ft

Ans.

My = ©My ;

My = - 4(2) + 3(4) = 4.00 lb # ft

Ans.

Mz = ©Mz ;

Mz = - 4(3) + 12(4) = 36.0 lb # ft

Ans.

Ans: Mx = 15.0 lb # ft My = 4.00 lb # ft Mz = 36.0 lb # ft 283

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4–57. Determine the moment of the force F about an axis extending between A and C. Express the result as a Cartesian vector.

z A

4 ft y

SOLUTION x

PositionVector:

3 ft C

rCB = {- 2k} ft rAB = {(4 - 0)i + (3 - 0)j + ( -2 - 0)k} ft = {4i + 3j - 2k} ft

2 ft

Unit Vector Along AC Axis: uAC =

B

(4 - 0)i + (3 - 0)j 2(4 - 0)2 + (3 - 0)2

F

{4i

12j

3k} lb

= 0.8i + 0.6j

Moment of Force F About AC Axis: With F = {4i + 12j - 3k} lb, applying Eq. 4–7, we have MAC = uAC # (rCB * F) 0.8 = 3 0 4

0.6 0 12

0 -2 3 -3

= 0.8[(0)( -3) - 12( -2)] - 0.6[0(-3) - 4(- 2)] + 0 = 14.4 lb # ft Or MAC = uAC # (rAB * F) 0.8 = 3 4 4

0.6 3 12

0 -2 3 -3

= 0.8[(3)( -3) - 12( - 2)] - 0.6[4(- 3) - 4( -2)] + 0 = 14.4 lb # ft Expressing MAC as a Cartesian vector yields M AC = MAC uAC = 14.4(0.8i + 0.6j) = {11.5i + 8.64j} lb # ft

Ans.

Ans: M AC = {11.5i + 8.64j} lb # ft 284

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4–58. z

The board is used to hold the end of a four-way lug wrench in the position shown when the man applies a force of F = 100 N. Determine the magnitude of the moment produced by this force about the x axis. Force F lies in a vertical plane.

F 60

x

SOLUTION

250 mm y

250 mm

Vector Analysis Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m The force vector F, Fig. a, can be written as F = 100(cos 60°j - sin 60°k) = {50j - 86.60k} N Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = 3 0 0

0 0.25 50

0 0 3 -86.60

= 1[0.25(- 86.60) - 50(0)] + 0 + 0 = - 21.7 N # m

Ans.

The negative sign indicates that Mx is directed towards the negative x axis. Scalar Analysis This problem can be solved by summing the moment about the x axis Mx = ©Mx;

Mx = -100 sin 60°(0.25) + 100 cos 60°(0) = - 21.7 N # m Ans.

Ans: Mx = 21.7 Ν # m 285

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4–59. The board is used to hold the end of a four-way lug wrench in position. If a torque of 30 N # m about the x axis is required to tighten the nut, determine the required magnitude of the force F that the man’s foot must apply on the end of the wrench in order to turn it. Force F lies in a vertical plane.

z

F 60

x

250 mm y

250 mm

SOLUTION Vector Analysis Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m The force vector F, Fig. a, can be written as F = F(cos 60°j - sin 60°k) = 0.5Fj - 0.8660Fk Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = 0 0

0 0.25 0.5F

0  0 -0.8660F

= 1[0.25( -0.8660F) - 0.5F(0)] + 0 + 0 Ans.

= - 0.2165F

The negative sign indicates that Mx is directed towards the negative x axis. The magnitude of F required to produce Mx = 30 N # m can be determined from 30 = 0.2165F Ans.

F = 139 N Scalar Analysis This problem can be solved by summing the moment about the x axis Mx = ©Mx;

- 30 = - F sin 60°(0.25) + F cos 60°(0) Ans.

F = 139 N

Ans: F = 139 Ν 286

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*4–60. z

The A-frame is being hoisted into an upright position by the vertical force of F = 80 lb. Determine the moment of this force about the y axis when the frame is in the position shown.

F C

SOLUTION

A

Using x¿ , y¿ , z:

x¿ 6 ft x

F = 80 k -sin 30° My = 3 - 6 cos 15° 0

cos 30° 3 0

6 ft

30

uy = - sin 30° i¿ + cos 30° j¿ rAC = -6 cos 15°i¿ + 3 j¿ + 6 sin 15° k

15 y B y¿

0 6 sin 15° 3 = -120 + 401.53 + 0 80

My = 282 lb # ft

Ans.

Also, using x, y, z: Coordinates of point C: x = 3 sin 30° - 6 cos 15° cos 30° = -3.52 ft y = 3 cos 30° + 6 cos 15° sin 30° = 5.50 ft z = 6 sin 15° = 1.55 ft rAC = -3.52 i + 5.50 j + 1.55 k F = 80 k 0 My = 3 -3.52 0

1 5.50 0

0 1.55 3 = 282 lb # ft 80

Ans.

Ans: My = 282 lb # ft 287

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4–61. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line AB of the tripod.

z F D 4m

2m 1.5 m

Solution uAB =

C

A

{3.5i + 0.5j}

y

2(3.5)2 + (0.5)2

x

uAB = {0.9899i + 0.1414j} MAB = uAB # ( rAD MAB = 136 N # m

0.9899 * F ) = † 2.5 50

0.1414 0 - 20

2.5 m

2m 0.5 m

0 4 † - 80

1m B

Ans.

Ans: MAB = 136 N # m 288

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4–62. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line BC of the tripod.

z F D 4m

2m 1.5 m

Solution uBC =

C

A

{ -1.5i - 2.5j}

y

2( - 1.5)2 + ( - 2.5)2

x

2m

uBC = { - 0.5145i - 0.8575j} MBC = uBC # ( rCD MBC = 165 N # m

- 0.5145 * F ) = † 0.5 50

- 0.8575 2 - 20

2.5 m

0.5 m

0 4 † - 80

1m B

Ans.

Ans: MBC = 165 N # m 289

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4–63. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line CA of the tripod.

z F D 4m

2m 1.5 m

Solution uCA =

C

A

{ -2i + 2j}

y

2( - 2)2 + (2)2

x

uCA = { -0.707i + 0.707j} MCA = uCA # ( rAD MCA = 226 N # m

- 0.707 * F ) = † 2.5 50

0.707 0 - 20

2.5 m

2m 0.5 m

0 4 † - 80

1m B

Ans.

Ans: MCA = 226 N # m 290

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*4–64. A horizontal force of F = { -50i} N is applied perpendicular to the handle of the pipe wrench. Determine the moment that this force exerts along the axis OA (z axis) of the pipe assembly. Both the wrench and pipe assembly, OABC, lie in the y-z plane. Suggestion: Use a scalar analysis.

z

B

0.8 m

0.2 m C

A 135°

F

0.6 m

Solution

O

Mz = 50(0.8 + 0.2) cos 45° = 35.36 N # m

x

Mz = {35.4 k} N # m

Ans.

y

Ans: Mz = {35.4 k} N # m 291

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4–65. Determine the magnitude of the horizontal force F = -F i acting on the handle of the wrench so that this force produces a component of the moment along the OA  axis (z axis) of the pipe assembly of Mz = {4k} N # m. Both the wrench and the pipe assembly, OABC, lie in the y-z plane. Suggestion: Use a scalar analysis.

z

B

0.8 m

0.2 m C

A 135°

F

0.6 m

Solution

O

Mz = F (0.8 + 0.2) cos 45° = 4

x

Ans.

F = 5.66 N

y

Ans: F = 5.66 N 292

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4–66. The force of F = 30 N acts on the bracket as shown. Determine the moment of the force about the a-a axis of the pipe if a = 60°, b = 60°, and g = 45°. Also, determine the coordinate direction angles of F in order to produce the maximum moment about the a-a axis.What is this moment?

z

F = 30 N y

45•• 60•• 60••

50 mm x

100 mm

100 mm

SOLUTION F = 30 1cos 60° i + cos 60° j + cos 45° k2

a

= 515 i + 15 j + 21.21 k6 N

r = 5 -0.1 i + 0.15 k6 m

a

u = j

0 Ma = 3 -0.1 15

1 0 15

0 0.15 3 = 4.37 N # m 21.21

Ans.

F must be perpendicular to u and r. uF =

0.1 0.15 i + k 0.1803 0.1803

= 0.8321i + 0.5547k a = cos-1 0.8321 = 33.7°

Ans.

b = cos-1 0 = 90°

Ans.

g = cos-1 0.5547 = 56.3°

Ans.

M = 30 0.1803 = 5.41 N # m

Ans.

Ans: Ma = 4.37 N # m a = 33.7° b = 90° g = 56.3° M = 5.41 N # m 293

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4–67. A clockwise couple M = 5 N # m is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces - R and R which act at supports A and B so that the resultant of the two couples is zero.

M

150 mm 60

A

Solution

R

60

B

R

a+MC = -5 + R (2(0.15)>tan 60°) = 0 Ans.

R = 28.9 N

Ans: R = 28.9 N 294

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*4–68. A twist of 4 N # m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle and P exerted on the blade.

–F –P P 5 mm

SOLUTION

4 N·m

30 mm

F

For the handle MC = ©Mx ;

F10.032 = 4 Ans.

F = 133 N For the blade, MC = ©Mx ;

P10.0052 = 4 Ans.

P = 800 N

Ans: F = 133 N P = 800 N 295

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4–69. If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P.

z F C

F 150 N B

SOLUTION

300 mm

BA = 0.5 m The couple created by the 150 - N forces is

x

400 mm

D 500 mm

150 N A

P

600 mm

P

MC1 = 150 (0.5) = 75 N # m Then 3 4 MC1 = 75 a b j + 75 a b k 5 5 = 45 j + 60 k MC2 = - P (0.6) k MC3 = - F (0.6) j Require MC1 + MC2 + MC3 = 0 45 j + 60 k - P (0.6) k - F (0.6) j = 0 Equate the j and k components 45 - F (0.6) = 0 Ans.

F = 75 N 60 - P (0.6) = 0

Ans.

P = 100 N

Ans: F = 75 N P = 100 N 296

y

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4–70. Two couples act on the beam. If F = 125 lb , determine the resultant couple moment.

200 lb

F 30 1.25 ft

1.5 ft

SOLUTION

F

30

200 lb

125 lb couple is resolved in to their horizontal and vertical components as shown in Fig. a.

2 ft

a + (MR)C = 200(1.5) + 125 cos 30° (1.25) = 435.32 lb # ft = 435 lb # ftd

Ans.

Ans: (MR)C = 435 lb # ft d 297

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4–71. Two couples act on the beam. Determine the magnitude of F so that the resultant couple moment is 450 lb # ft, counterclockwise. Where on the beam does the resultant couple moment act?

200 lb

F 30 1.25 ft

1.5 ft

F

30

200 lb 2 ft

SOLUTION a + MR = ©M ;

450 = 200(1.5) + Fcos 30°(1.25) Ans.

F = 139 lb

The resultant couple moment is a free vector. It can act at any point on the beam.

Ans: F = 139 lb 298

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*4–72. Determine the magnitude of the couple force F so that the resultant couple moment on the crank is zero. 30

30

30 5 in.

150 lb 30

SOLUTION

150 lb

–F

5 in.

F

45

45

4 in.

4 in.

By resolving F and the 150-lb couple into components parallel and perpendicular to the lever arm of the crank, Fig. a, and summing the moment of these two force components about point A, we have a + (MC)R = ©MA;

0 = 150 cos 15°(10) - F cos 15°(5) - F sin 15°(4) - 150 sin 15°(8) Ans.

F = 194 lb

Note: Since the line of action of the force component parallel to the lever arm of the crank passes through point A, no moment is produced about this point.

Ans: F = 194 lb 299

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4–73. The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is 400 N # m clockwise.

F

600 N

F

600 N 40

SOLUTION a + MR = ©M;

- 400 = 600 a

40 1m

0.5 0.5 b -F a b -250(1) cos 40° cos 40°

250 N

250 N

Ans.

F = 830 N

Ans: F = 830 N 300

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4–74. The man tries to open the valve by applying the couple forces of F = 75 N to the wheel. Determine the couple moment produced.

150 mm

150 mm

F

SOLUTION a + Mc = ©M;

Mc = - 75(0.15 + 0.15) = - 22.5 N # m = 22.5 N # m b

Ans. F

Ans: MC = 22.5 N # mb 301

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4–75. If the valve can be opened with a couple moment of 25 N # m, determine the required magnitude of each couple force which must be applied to the wheel.

150 mm

150 mm

F

SOLUTION a + Mc = ©M;

- 25 = -F(0.15 + 0.15) F = 83.3 N

Ans. F

Ans: F = 83.3 N 302

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*4–76. Determine the magnitude of F so that the resultant couple moment is 12 kN # m, counterclockwise. Where on the beam does the resultant couple moment act?

F

F 30 30

8 kN

0.3 m

1.2 m 0.4 m

Solution

a + MR = ΣMC;  12 = (F cos 30°)(0.3) + 8(1.2)

8 kN

Ans.

F = 9.238 kN = 9.24 kN

Since the couple moment is a free vector, the resultant couple moment can act at any point on or off the beam.

Ans: F = 9.24 kN 303

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4–77. Two couples act on the beam as shown. If F = 150 lb, determine the resultant couple moment.

–F

5

3 4

200 lb 1.5 ft 200 lb 5

SOLUTION

F

150 lb couple is resolved into their horizontal and vertical components as shown in Fig. a

3

4

4 ft

4 3 a + (MR)c = 150 a b (1.5) + 150 a b (4) - 200(1.5) 5 5 = 240 lb # ftd

Ans.

Ans: (MR)C = 240 lb # ft d 304

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4–78. Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is 300 lb # ft counterclockwise. Where on the beam does the resultant couple act?

–F

5

3 4

200 lb 1.5 ft 200 lb 5

F

SOLUTION a + (MC)R =

3

4

4 ft

4 3 F(4) + F(1.5) - 200(1.5) = 300 5 5 Ans.

F = 167 lb

Ans.

Resultant couple can act anywhere.

Ans: F = 167 lb Resultant couple can act anywhere. 305

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4–79. Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the 80-lb couple forces.

y 2 ft B

3 ft 50 lb

30

1 ft 5

3

d

4

80 lb

SOLUTION a + MC = -50 cos 30°(3) +

5

4 (80)(d) = 0 5

4

80 lb

Ans.

d = 2.03 ft

3

A

30

50 lb

3 ft x

Ans: d = 2.03 ft 306

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*4–80. Two couples act on the frame. If d = 4 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4–13) and (b) summing the moments of all the force components about point A.

y 2 ft B 1 ft 5

3

d

4

80 lb

SOLUTION (a)

5

M C = ©(r * F) = 3

i 3 - 50 sin 30°

4

j 0 - 50 cos 30°

k i 3 3 0 + 0 0 - 45(80)

30

3 ft 50 lb

j 4 - 35(80)

M C = {126k} lb # ft

80 lb

k 03 0

3

A

30

50 lb

3 ft x

Ans.

4 4 (b) a + MC = - (80)(3) + (80)(7) + 50 cos 30°(2) - 50 cos 30°(5) 5 5 MC = 126 lb # ft

Ans.

Ans: MC = 126 lb # ft 307

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4–81. Two couples act on the frame. If d = 4 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 4–13) and (b) summing the moments of all the force components about point B.

y 2 ft B 1 ft 5

3

d

4

80 lb

SOLUTION (a)

5

MC = ©(r * F) = 3

i 3 - 50 sin 30°

4

j 0 - 50 cos 30°

k i 03 + 3 0 4 0 5 (80)

30

3 ft 50 lb

j -4 3 5 (80)

k 03 0

MC = {126k} lb # ft

80 lb

3

A

30

50 lb

3 ft x

Ans.

4 4 (b) a + MC = 50 cos 30°(2) - 50 cos 30°(5)- (80)(1) + (80)(5) 5 5 MC = 126 lb # ft

Ans.

Ans: MC = 126 lb # ftd 308

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4–82. z

Express the moment of the couple acting on the pipe assembly in Cartesian vector form. What is the magnitude of the couple moment?

20 lb x

A

3 ft B 1 ft 1.5 ft 20 lb 2 ft 1 ft

Solution

y

C

rCB = { -3i - 2.5j} ft MC = rCB * F i = † -3 0

j -2.5 0

k 0 † 20

MC = { -50i + 60j} lb # ft

Ans.

MC = 2( - 50) + (60) = 78.1 lb # ft

Ans.

2

2

Ans: MC = 78.1 lb # ft 309

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4–83. If M1 = 180 lb # ft, M2 = 90 lb # ft, and M3 = 120 lb # ft, determine the magnitude and coordinate direction angles of the resultant couple moment.

z 150 lb ft M3 1 ft 2 ft

SOLUTION Since the couple moment is a free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2, M3, and M4 acting on the gear deducer can be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian vector form,

2 ft

45 45

x 2 ft

y 3 ft

M2 M1

M 1 = [180j]lb # ft M 2 = [-90i]lb # ft M 3 = M3u = 120C

(2 - 0)i + ( - 2 - 0)j + (1 + 0)k 2(2 - 0)2 + ( -2 - 0)2 + (1 - 0)2

S = [80i - 80j + 40k]lb # ft

M 4 = 150[cos 45° sin 45°i - cos 45° cos 45°j - sin 45°k] = [75i - 75j - 106.07k]lb # ft The resultant couple moment is given by (M c)R = M 1 + M 2 + M 3 + M 4

(M c)R = ©M;

= 180j - 90i + (80i - 80j + 40k) + (75i - 75j - 106.07k) = [65i + 25j - 66.07k]lb # ft The magnitude of (M c)R is (Mc)R = 2[(Mc)R]x 2 + [(Mc)R]y 2 + [(Mc)R]z 2 = 2(65)2 + (25)2 + (- 66.07)2 = 95.99 lb # ft = 96.0 lb # ft

Ans.

The coordinate angles of (M c)R are a = cos

-1

¢

b = cos

-1

¢

g = cos

-1

¢

[(Mc)R]x 65 b = 47.4° ≤ = cos a (Mc)R 95.99 [(Mc)R]y (Mc)R

[(Mc)R]z (Mc)R

≤ = cos a

≤ = cos a

Ans.

25 b = 74.9° 95.99

Ans.

- 66.07 b = 133° 95.99

Ans.

Ans: MR = 96.0 lb # ft, a = 47.4°, b = 74.9°, g = 133° 310

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*4–84. Determine the magnitudes of couple moments M1, M2, and M3 so that the resultant couple moment is zero.

z 150 lb ft M3 1 ft 2 ft

2 ft

45 45

x

SOLUTION

2 ft

Since the couple moment is a free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2, M3, and M4 acting on the gear deducer can be simplified, as shown in Fig. a. Expressing each couple moment in Cartesian vector form,

y 3 ft

M2 M1

M 1 = M1j M 2 = -M2i M 3 = M3u = M3 C

(2 - 0)i + ( -2 - 0)j + (1 + 0)k 2(2 - 0)2 + ( -2 - 0)2 + (1 - 0)2

S =

2 2 1 M i - M3j + M3k 3 3 3 3

M 4 = 150[cos 45° sin 45°i - cos 45° cos 45°j - sin 45°k] = [75i - 75j - 106.07k]lb # ft The resultant couple moment is required to be zero. Thus, 0 = M1 + M2 + M3 + M4

(M c)R = ©M;

2 2 1 0 = M1j + (-M2i) + a M3i - M3j + M3kb + (75i - 75j - 106.07k) 3 3 3 0 = a -M2 +

Equating the i, j, and k components,

0 = - M2 + 0 = M1 0 =

2 2 1 M + 75b i + aM1 - M3 - 75b j + a M3 - 106.07bk 3 3 3 3

2 M + 75 3 3

(1)

2 M - 75 3 3

(2)

1 M - 106.07 3 3

(3)

Solving Eqs. (1), (2), and (3) yields

M3 = 318 lb # ft

Ans.

M1 = M2 = 287 lb # ft

Ans.

Ans: M3 = 318 lb # ft, M1 = M2 = 287 lb # ft 311

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4–85. The gears are subjected to the couple moments shown. Determine the magnitude and coordinate direction angles of the resultant couple moment.

M1  40 lb  ft

z 20

M2  30 lb  ft

30 15

y

SOLUTION M1 = 40 cos 20° sin 15° i + 40cos 20° cos 15° j - 40 sin 20° k

x

= 9.728 i + 36.307 j - 13.681 k M 2 = -30 sin 30° i + 30 cos 30° j = -15 i + 25.981 j MR = M1 + M2 = - 5.272 i + 62.288 j - 13.681 k MR = 2(- 5.272)2 + (62.288)2 + (- 13.681)2 = 63.990 = 64.0 lb # ft

Ans.

a = cos-1 a

- 5.272 b = 94.7° 63.990

Ans.

b = cos-1 a

62.288 b = 13.2° 63.990

Ans.

g = cos-1 a

- 13.681 b = 102° 63.990

Ans.

Ans: MR = 64.0 lb # ft a = 94.7° b = 13.2° g = 102° 312

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4–86. Determine the required magnitude of the couple moments M2 and M3 so that the resultant couple moment is zero.

M2 45

SOLUTION

M3

Since the couple moment is the free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2, and M3 can be simplified as shown in Fig. a. Since the resultant of M1, M2, and M3 is required to be zero, (MR)y = ©My ;

M1  300 Nm

0 = M2 sin 45° - 300 M2 = 424.26 N # m = 424 N # m

(MR)x = ©Mx ;

Ans.

0 = 424.26 cos 45° - M3 M3 = 300 N # m

Ans.

Ans: M2 = 424 N # m M3 = 300 N # m 313

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4–87. z

Determine the resultant couple moment of the two couples that act on the assembly. Specify its magnitude and coordinate direction angles.

2 in.

60 lb 2 in.

80 lb 30

y

x 4 in.

Solution i MR = † 4 cos 30° 0

j 5 0

k i - 4 sin 30° † + † 4 cos 30° 60 0

j 0 80

k - 4 sin 30° † 0

80 lb

= 300 i - 207.85 j + 160 i + 277.13 k 3 in.

= {460 i - 207.85 j + 277.13 k} lb # in. MR = 2(460)2 + ( - 207.85)2 + (277.13)2 = 575.85 = 576 lb # in.  a = cos-1 a

460 b = 37.0° 575.85

g = cos-1 a

277.13 b = 61.2° 575.85

b = cos-1 a

60 lb

Ans. Ans.

- 207.85 b = 111° 575.85

Ans. Ans.

Ans: MR = 576 lb # in. a = 37.0° b = 111° g = 61.2° 314

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*4–88. Express the moment of the couple acting on the frame in Cartesian vector form. The forces are applied perpendicular to the frame. What is the magnitude of the couple moment? Take F = 50 N.

z

O F

y 3m

30

1.5 m

SOLUTION

x

MC = 80(1.5) = 75 N # m

F

Ans.

MC = - 75(cos 30° i + cos 60° k) = {- 65.0i - 37.5k} N # m

Ans.

Ans: MC = { - 65.0i - 37.5k} N # m 315

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4–89. In order to turn over the frame, a couple moment is applied as shown. If the component of this couple moment along the x axis is Mx = 5 - 20i6 N # m, determine the magnitude F of the couple forces.

z

O F

y 3m

SOLUTION MC = F (1.5)

30

1.5 m

Thus

x

F

20 = F (1.5) cos 30° Ans.

F = 15.4 N

Ans: F = 15.4 N 316

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4–90. Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Take F = 125 N.

z O –F

y

150 mm 600 mm A

SOLUTION

B

MC = rAB * (125 k)

200 mm

x 150 mm

MC = (0.2i + 0.3j) * (125 k)

F

MC = {37.5i - 25j} N # m MC = 2(37.5)2 + ( -25)2 = 45.1 N # m

Ans.

Ans: MC = 45.1 N # m 317

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4–91. If the couple moment acting on the pipe has a magnitude of 300 N # m, determine the magnitude F of the forces applied to the wrenches.

z O –F

y

150 mm 600 mm A

SOLUTION

B

MC = rAB * (F k)

200 mm

x 150 mm

= (0.2i + 0.3j) * (F k)

F

= {0.2Fi - 0.3Fj} N # m MC = F 2(0.2F)2 + ( -0.3F) = 0.3606 F 300 = 0.3606 F Ans.

F = 832 N

Ans: F = 832 N 318

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*4–92. If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x–y plane.

z

F

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m The force vectors F and –F can be written as F = {80 k} N and - F = [-80 k] N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = [40i - 8j] N # m 80

or i Mc = rBA * -F = 3 - 0.1 0

j - 0.5 0

k 0 3 = [40i - 8j] N # m -80

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2402 + ( -8)2 + 02 = 40.79 N # m = 40.8 N # m

Ans.

The coordinate angles of Mc are

¢

a = cos

-1

b = cos

-1

g = cos

-1

¢

¢

Mx 40 ≤ = cos ¢ ≤ = 11.3° M 40.79 My M Mz M

-8 ≤ = 101° 40.79

Ans.

0 ≤ = 90° 40.79

Ans.

≤ = cos ¢

≤ = cos ¢

Ans.

Ans: Mc = 40.8 N # m a = 11.3° b = 101° g = 90° 319

y

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4–93. If the magnitude of the couple moment acting on the pipe assembly is 50 N # m, determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane.

z

F

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of either F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ - 0.1i - 0.5j] m The force vectors F and –F can be written as F = {Fk} N and - F = [-Fk]N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = 0.5Fi - 0.1Fj F

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F Since Mc is required to equal 50 N # m, 50 = 0.5099F Ans.

F = 98.1 N

Ans: F = 98.1 N 320

y

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4–94. z

Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment?

F  { 4i  3j  4k} kN

A 1m x 2m

Position Vector. The coordinates of points A and B are A (0, 0, 1) m and B (3, 2, -1) m, respectively. Thus,

3m

1m

Solution

y

B F  {– 4i + 3j  4k} kN

rAB = (3 - 0)i + (2 - 0)j + ( - 1 - 1)k = {3i + 2j - 2k} m

Couple Moment.

MC = rAB * F



i = † 3 -4



j 2 3

k -2 † -4

= { - 2i + 20j + 17k} kN # m

Ans.

The magnitude of MC is

MC = 2(MC)2x + (MC)2y + (MC)2z = 2( - 2)2 + 202 + 172

= 26.32 kN # m = 26.3 kN # m

Ans.

Ans: MC = { - 2i + 20j + 17k} kN # m MC = 26.3 kN # m

321

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4–95. If F1 = 100 N, F2 = 120 N and F3 = 80 N, determine the magnitude and coordinate direction angles of the resultant couple moment.

z

–F4

[ 150 k] N

0.3 m

0.2 m

0.2 m

SOLUTION Couple Moment: The position vectors r1, r2 , r3 , and r4 , Fig. a, must be determined first. r1 = {0.2i} m

r2 = {0.2j} m

0.3 m

0.2 m F1 0.2 m – F1

x – F2 0.2 m

30

F4

[150 k] N

y

F2

– F3

r3 = {0.2j} m

0.2 m

From the geometry of Figs. b and c, we obtain

F3

r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45°j - 0.3 sin 30°k = {0.1837i + 0.1837j - 0.15k} m The force vectors F1 , F2 , and F3 are given by F1 = {100k} N

F2 = {120k} N

F3 = {80i} N

Thus, M 1 = r1 * F1 = (0.2i) * (100k) = {- 20j} N # m M 2 = r2 * F2 = (0.2j) * (120k) = {24i} N # m

M 3 = r3 * F3 = (0.2j) * (80i) = { -16k} N # m

M 4 = r4 * F4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m

Resultant Moment: The resultant couple moment is given by (M c)R =

M c;

(M c)R = M 1 + M 2 + M 3 + M 4 = (-20j) + (24i) + ( - 16k) + (27.56i -27.56j)

= {51.56i - 47.56j - 16k} N # m The magnitude of the couple moment is (M c)R = 2[(M c)R]x2 + [(M c)R]y 2 + [(M c)R]z 2 = 2(51.56)2 + ( -47.56)2 + ( -16)2 = 71.94 N # m = 71.9 N # m

Ans.

The coordinate angles of (Mc)R are a = cos -1 a

b = cos -1 a

g = cos -1 a

[(Mc)R]x 51.56 b = 44.2° b = cos a (Mc)R 71.94 [(Mc)R]y (Mc)R [(Mc)R]z (Mc)R

b = cos a

b = cos a

Ans.

- 47.56 b = 131° 71.94

Ans.

- 16 b = 103° 71.94

Ans.

322

Ans: (MC)R = 71.9 Ν # m a = 44.2° b = 131° g = 103°

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*4–96. Determine the required magnitude of F1 , F2 , and F3 so that the resultant couple moment is (Mc)R = [50 i - 45 j - 20 k] N # m.

z

–F4

[ 150 k] N

0.3 m

SOLUTION Couple Moment: The position vectors r1 , r2 , r3 , and r4 , Fig. a, must be determined first. r1 = {0.2i} m

0.2 m

0.2 m

r2 = {0.2j} m

0.3 m

0.2 m

F1 0.2 m – F1

x

r3 = {0.2j} m

– F2 0.2 m

From the geometry of Figs. b and c, we obtain

30

F4

[150 k] N

y

F2

– F3

r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45°j - 0.3 sin 30°k

0.2 m

= {0.1837i + 0.1837j - 0.15k} m

F3

The force vectors F1 , F2 , and F3 are given by F1 = F1k

F2 = F2k

F3 = F3i

Thus, M 1 = r1 * F1 = (0.2i) * (F1k) = - 0.2 F1j M 2 = r2 * F2 = (0.2j) * (F2k) = 0.2 F2i M 3 = r3 * F3 = (0.2j) * (F3i) = -0.2 F3k

M 4 = r4 * F4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m

Resultant Moment: The resultant couple moment required to equal (M c)R = {50i - 45j - 20k} N # m. Thus, (M c)R = ©M c;

(M c)R = M 1 + M 2 + M 3 + M 4 50i - 45j - 20k = ( - 0.2F1j) + (0.2F2i) + (- 0.2F3k) + (27.56i - 27.56j) 50i - 45j - 20k = (0.2F2 + 27.56)i + ( -0.2F1 - 27.56)j - 0.2F3k

Equating the i, j, and k components yields 50 = 0.2F2 + 27.56

F2 = 112 N

Ans.

-45 = -0.2F1 - 27.56

F1 = 87.2 N

Ans.

-20 = - 0.2F3

F3 = 100 N

Ans.

Ans: F2 = 112 N F1 = 87.2 N F3 = 100 N 323

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4–97. y

Replace the force system by an equivalent resultant force and couple moment at point O. 455 N

12 13

5

2m

2.5 m O

x

0.75 m 60

Solution Equivalent Resultant Force And Couple Moment At O. 12 + (FR)x = ΣFx;  (FR)x = 600 cos 60° - 455 a b = -120 N = 120 N d S 13 + c (FR)y = ΣFy;  (FR)y = 455 a

As indicated in Fig. a And Also,

(FR)y (FR)x

d = tan-1 a

a+(MR)O = ΣMO;  (MR)O = 455 a



P 1m

600 N

5 b - 600 sin 60° = -344.62 N = 344.62 NT 13

FR = 2 (FR)2x + (FR)2y = 21202 + 344.622 = 364.91 N = 365 N u = tan-1 c

0.75 m

344.62 b = 70.80° = 70.8° d 120

Ans.

Ans.

12 b(2) + 600 cos 60° (0.75) + 600 sin 60° (2.5) 13

= 2364.04 N # m

= 2364 N # m (counterclockwise)

Ans.

Ans: FR = 365 N u = 70.8° d

(MR)O = 2364 N # m (counterclockwise)

324

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4–98. y

Replace the force system by an equivalent resultant force and couple moment at point P. 455 N

12 13

5

2m

2.5 m O

x

0.75 m

0.75 m 60

Solution Equivalent Resultant Force And Couple Moment At P. 12 + (FR)x = ΣFx;  (FR)x = 600 cos 60° - 455 a b = -120 N = 120 N d S 13 5 + c (FR)y = ΣFy;  (FR)y = 455 a b - 600 sin 60° = -344.62 N = 344.62 N T 13

P 1m

600 N

As indicated in Fig. a,

FR = 2 (FR)2x + (FR)2y = 21202 + 344.622 = 364.91 N = 365 N

Ans.

And

u = tan-1 c

Also,



(FR)y (FR)x

d = tan-1 a

344.62 b = 70.80° = 70.8° d 120

a+ (MR)P = ΣMP;  (MR)P = 455 a

Ans.

12 5 b(2.75) - 455 a b(1) + 600 sin 60° (3.5) 13 13

= 2798.65 N # m

= 2799 N # m (counterclockwise)

Ans.

Ans: FR = 365 N u = 70.8° d

(MR)P = 2799 N # m (counterclockwise)

325

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4–99. Replace the force system acting on the beam by an equivalent force and couple moment at point A.

3 kN 2.5 kN 1.5 kN 30 5

4

3

B

A 2m

4m

2m

SOLUTION + F = ©F ; : Rx x

4 FRx = 1.5 sin 30° - 2.5a b 5

= - 1.25 kN = 1.25 kN ;

3 FRy = - 1.5 cos 30° - 2.5 a b - 3 5

+ c FRy = ©Fy ;

= - 5.799 kN = 5.799 kN T Thus, FR = 2F 2Rx + F 2Ry = 21.252 + 5.7992 = 5.93 kN

Ans.

and u = tan

a + MRA = ©MA ;

-1

¢

FRy FRx

≤ = tan

-1

a

5.799 b = 77.8° d 1.25

Ans.

3 MRA = - 2.5 a b (2) - 1.5 cos 30°(6) - 3(8) 5

= -34.8 kN # m = 34.8 kN # m (Clockwise)

Ans.

Ans: FR = 5.93 kN u = 77.8° d MRA = 34.8 kN # m b 326

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*4–100. Replace the force system acting on the beam by an equivalent force and couple moment at point B.

3 kN 2.5 kN 1.5 kN 30 5

4

3

B

A 2m

4m

2m

SOLUTION + F = ©F ; : Rx x

4 FRx = 1.5 sin 30° - 2.5a b 5 = -1.25 kN = 1.25 kN ;

3 FRy = -1.5 cos 30° - 2.5 a b - 3 5

+ c FRy = ©Fy ;

= - 5.799 kN = 5.799 kN T Thus, FR = 2F 2Rx + F 2Ry = 21.252 + 5.7992 = 5.93 kN

Ans.

and u = tan

-1

¢

FRy FRx

a + MRB = ©MRB ;

≤ = tan

-1

a

5.799 b = 77.8° d 1.25

Ans.

3 MB = 1.5cos 30°(2) + 2.5a b(6) 5 = 11.6 kN # m (Counterclockwise)

Ans.

Ans: FR = 5.93 kN u = 77.8° d MB = 11.6 kN # m (Counterclockwise) 327

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4–101.

y 450 N

Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O.

30 200 N  m

0.2 m

x

O 1.5 m

2m

1.5 m 200 N

Solution + d FRx = ΣFx ;

FRx = 450 sin 30° = 225.0

+ T FRy = ΣFy ;

FRy = 450 cos 30° - 200 = 189.7

FR = 2(225)2 + (189.7)2 = 294 N

Ans.

u = tan-1 a

Ans.

189.7 b = 40.1° d 225

c+MRO = ΣMO ;  MRO = 450 cos 30° (1.5) - 450 (sin 30°)(0.2) - 200 (3.5) + 200 MRO = 39.6 N # m b

Ans.

Ans: FR = 294 N u = 40.1° d MRO = 39.6 N # m b 328

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4–102. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point A.

650 N 30

500 N

300 N 1500 N  m

60 B

A

3m

5m

2m

Solution Equivalent Resultant Force And Couple Moment at Point A. + (FR)x = ΣFx ;  (FR)x = 650 sin 30° - 500 cos 60° = 75 N S S + c (FR)y = ΣFy ;  (FR)y = - 650 cos 30° - 300 - 500 sin 60° = - 1295.93 N = 1295.93 NT

As indicated in Fig. a, And Also,

FR = 2(FR)2x + (FR)2y = 2752 + 1295.932 = 1298.10 N = 1.30 kN Ans. u = tan-1 c

(FR)y (FR)x

d = tan-1 a

1295.93 b = 86.69° = 86.7° c 75

Ans.

a+ (MR)A = ΣMA;  (MR)A = 650 cos 30° (3) + 1500 - 500 sin 60° (5)

= 1023.69 N # m

= 1.02 kN # m (counter clockwise)

Ans.

Ans: FR = 1.30 kN u = 86.7° c

(MR)A = 1.02 kN # m (counterclockwise)

329

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4–103. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point B.

650 N 30

500 N

300 N 1500 N  m

60 B

A

3m

5m

2m

Solution Equivalent Resultant Force And Couple Moment At Point B. + (FR)x = ΣFx;  (FR)x = 650 sin 30° - 500 cos 60° = 75 N S S + c (FR)y = ΣFy;  (FR)y = - 650 cos 30° - 300 - 500 sin 60° = - 1295.93 N = 1295.93 NT As indicated in Fig. a, And Also,

FR = 2 (FR)2x + (FR)2y = 2752 + 1295.932 = 1298.10 N = 1.30 kNAns. u = tan-1 c

(FR)y (FR)x

d = tan-1 a

1295.93 b = 86.69° = 86.7° c 75

Ans.

a+ (MR)B = ΣMB;  (MR)B = 650 cos 30° (10) + 300(7) + 500 sin 60°(2) + 1500

= 10,095.19 N # m

= 10.1 kN # m (counterclockwise)

Ans.

Ans: FR = 1.30 kN u = 86.7° c

(MR)B = 1.01 kN # m (counterclockwise)

330

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*4–104. Replace the force system acting on the post by a resultant force and couple moment at point O.

300 lb 30 150 lb 3

2 ft

5 4

SOLUTION Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, we have + ©(F ) = ©F ; : R x x

4 (FR)x = 300 cos 30° - 150 a b + 200 = 339.81 lb : 5

+ c (FR)y = ©Fy;

3 (FR)y = 300 sin 30° + 150a b = 240 lb c 5

2 ft 200 lb 2 ft O

The magnitude of the resultant force FR is given by FR = 2(FR)x2 + (FR)y2 = 2339.812 + 2402 = 416.02 lb = 416 lb

Ans.

The angle u of FR is u = tan-1 c

(FR)y (FR)x

d = tan-1 c

240 d = 35.23° = 35.2° a 339.81

Ans.

Equivalent Resultant Couple Moment: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A, we can write a + (MR)A = ©MA;

4 (MR)A = 150 a b (4) - 200(2) - 300 cos 30°(6) 5

= - 1478.85 lb # ft = 1.48 kip # ft (Clockwise) Ans.

Ans: FR = 416 lb u = 35.2° a ( MR ) A = 1.48 kip # ft (Clockwise) 331

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4–105. Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A.

A 300 N

0.5 m

30

1m

500 N

Solution Equivalent Resultant Force And Couple Moment At A.

0.5 m

+ (FR)x = ΣFx;    (FR)x = 300 cos 30° + 500 = 759.81 N   S

S

  +c (FR)y = ΣFy;    (FR)y = - 300 sin 30° - 400 = -550 N = 550 N

0.3 m 400 N

T

As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 2759.812 + 5502 = 937.98 N = 938 N

Ans.

And

    u = tan-1 c

(FR)y

550 b = 35.90° = 35.9°  c 759.81

Also;

d = tan-1a

a+(MR)A = ΣMA;  

(MR)A = 300 cos 30°(0.5) + 500(1.5) - 400(0.5)

(FR)x

Ans.

= 679.90 N # m

= 680 N # m (counterclockwise)

Ans.

Ans: FR = 938 N u = 35.9° c

(MR)A = 680 N # m (counterclockwise)

332

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4–106. z

The forces F1 = { - 4i + 2j - 3k} kN and F2 = {3i - 4j 2k} kN act on the end of the beam. Replace these forces by an equivalent force and couple moment acting at point O. F1

150 mm F 150 mm 2

O

250 mm y 4m

Solution

x

FR = F1 + F2 = { - 1i - 2j - 5k} kN

Ans.

MRO = r1 * F1 + r2 * F2 i = † 4 -4

j - 0.15 2

k i 0.25 † + † 4 -3 3

j 0.15 -4

k 0.25 † -2

= ( -0.05i + 11j + 7.4k) + (0.7i + 8.75j - 16.45k) = (0.65i + 19.75j - 9.05k)

MRO = {0.650i + 19.75j - 9.05k} kN # m

Ans.



Ans: MRO = {0.650i + 19.75j - 9.05k} kN # m 333

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4–107. A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR = 35 N for the rectus, FO = 45 N for the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erector spinae. These loadings are symmetric with respect to the y–z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form.

z

FR

FR

FO FE

FL

FE FO

FL O

75 mm

15 mm 45 mm

SOLUTION

50 mm

30 mm 40 mm

x

FR = ©Fz ;

FR = {2(35 + 45 + 23 + 32)k } = {270k} N

MROx = ©MOx ;

MR O = [ - 2(35)(0.075) + 2(32)(0.015) + 2(23)(0.045)]i MR O = { - 2.22i} N # m

Ans.

Ans.

334

Ans: FR = 5270k6 N MRO = 5 -2.22i6 N # m

y

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*4–108. z

Replace the force system by an equivalent resultant force  and couple moment at point O. Take F3 = { -200i + 500j - 300k} N.

F1 = 300 N

O 2m

x F3

1.5 m

Solution

y

1.5 m

Position And Force Vectors.

F2 = 200 N

  r1 = {2j} m   r2 = {1.5i + 3.5j}   r3 = {1.5i + 2j} m   F1 = { -300k} N  F2 = {200j} N       F3 = { - 200i + 500j - 300k} N Equivalent Resultant Force And Couple Moment At Point O.   FR = ΣF;       FR = F1 + F2 + F3

= ( - 300k) + 200j + ( - 200i + 500j - 300k)



= { - 200i + 700j - 600k} N

Ans.

  (MR)O = ΣMO;  (MR)O = r1 * F1 + r2 * F2 + r3 * F3

i = 30 0



= ( - 600i) + (300k) + ( -600i + 450j + 1150k)



j 2 0

k i 0 3 + 3 1.5 - 300 0

j 3.5 200

k i 0 3 + 3 1.5 - 200 0

= { - 1200i + 450j + 1450k} N # m

j 2 500

k 0 3 - 300

Ans.

Ans: FR = { - 200i + 700j - 600k} N

(MR)O = { - 1200i + 450j + 1450k} N # m

335

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4–109.

z

Replace the loading by an equivalent resultant force and couple moment at point O.

O

x 0.5 m

y

0.7 m F2 = {–2 i + 5 j – 3 k} kN

0.8 m F1  {8 i – 2 k} kN

Solution Position Vectors. The required position vectors are r1 = {0.8i - 1.2k} m  r2 = { -0.5k} m Equivalent Resultant Force And Couple Moment At Point O.   FR = ΣF;       FR = F1 + F2

= (8i - 2k) + ( - 2i + 5j - 3k)



= {6i + 5j - 5k} kN

Ans.

  (MR)O = ΣMO;  (MR)O = r1 * F1 + r2 * F2 i = 3 0.8 8

j 0 0

k i - 1.2 3 + 3 0 -2 -2

j 0 5

k - 0.5 3 -3

= ( - 8j) + (2.5i + j) = {2.5i - 7j} kN # m

Ans.

Ans: FR = {6i + 5j - 5k} kN

(MR)O = {2.5i - 7j} kN # m

336

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4–110.

z

Replace the force of F = 80 N acting on the pipe assembly by an equivalent resultant force and couple moment at point A.

A 400 mm B [

300 mm

y

200 mm

Solution

200 mm

250 mm

FR = ΣF ;

40

FR = 80 cos 30° sin 40° i + 80 cos 30° cos 40° j - 80 sin 30° k

30

= 44.53 i + 53.07 j - 40 k

F

= {44.5 i + 53.1 j - 40 k} N i MRA = ΣMA ;  MRA = † 0.55 44.53

j 0.4 53.07

80 N

Ans.

 k - 0.2 † -40

= { - 5.39 i + 13.1 j + 11.4 k} N # m

Ans.

Ans: FR = {44.5 i + 53.1 j + 40 k} N MRA = { - 5.39 i + 13.1 j + 11.4 k} N # m 337

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4–111. The belt passing over the pulley is subjected to forces F1 and F2, each having a magnitude of 40 N. F1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Set u = 0° so that F2 acts in the - j direction.

z 

r  80 mm

y

300 mm A

SOLUTION

x

FR = F1 + F2 FR = { -40j - 40 k} N

Ans.

F2

M RA = ©(r * F) i = 3 - 0.3 0

j 0 - 40

F1

k i 0.08 3 + 3 - 0.3 0 0

j 0.08 0

k 0 3 -40

MRA = { -12j + 12k} N # m

Ans.

Ans:

FR = 5 -40j - 40k 6 N 338

MRA = 5 -12j + 12k6 N # m

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*4–112. The belt passing over the pulley is subjected to two forces F1 and F2, each having a magnitude of 40 N. F1 acts in the - k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Take u = 45°.

z r

80 mm

y

300 mm A

SOLUTION

x

FR = F1 + F2 = -40 cos 45°j + ( -40 - 40 sin 45°)k

F2

Ans.

FR = {-28.3j - 68.3k} N

F1

rAF1 = {-0.3i + 0.08j} m rAF2 = -0.3i - 0.08 sin 45°j + 0.08 cos 45°k = {- 0.3i - 0.0566j + 0.0566k} m MRA = (rAF1 * F1) + (rAF2 * F2) i = 3 - 0.3 0

j 0.08 0

i k 0 3 + 3 -0.3 0 - 40

j -0.0566 -40 cos 45°

k 0.0566 3 -40 sin 45°

MRA = {-20.5j + 8.49k} N # m

Ans.

Also, MRAx = ©MAx MRAx = 28.28(0.0566) + 28.28(0.0566) - 40(0.08) MRAx = 0 MRAy = ©MAy MRAy = -28.28(0.3) - 40(0.3) MRAy = -20.5 N # m MRAz = ©MAz MRAz = 28.28(0.3) MRAz = 8.49 N # m MRA = {- 20.5j + 8.49k} N # m

Ans.

Ans:

FR = 5-28.3j - 68.3k6 N 339

MRA = 5- 20.5j + 8.49k6 N # m

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4–113. The weights of the various components of the truck are shown. Replace this system of forces by an equivalent resultant force and specify its location measured from B.

+ c FR = ©Fy;

FR = -1750 - 5500 - 3500

= - 10 750 lb = 10.75 kip T a +MRA = ©MA ;

3500 lb

B

SOLUTION

5500 lb 14 ft

3 ft

A

1750 lb

6 ft 2 ft

Ans.

-10 750d = -3500(3) - 5500(17) - 1750(25) d = 13.7 ft

Ans.

Ans: FR = 10.75 kip T d = 13.7 ft 340

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4–114. The weights of the various components of the truck are shown. Replace this system of forces by an equivalent resultant force and specify its location measured from point A.

5500 lb 14 ft

Equivalent Force: + c FR = ©Fy ;

3500 lb

B

SOLUTION

3 ft

A

1750 lb

6 ft 2 ft

FR = - 1750 - 5500 - 3500 = - 10 750 lb = 10.75 kipT

Ans.

Location of Resultant Force From Point A: a + MRA = ©MA ;

10 750(d) = 3500(20) + 5500(6) - 1750(2) d = 9.26 ft

Ans.

Ans: FR = 10.75 kip T d = 9.26 ft 341

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4–115. Replace the three forces acting on the shaft by a  single resultant force. Specify where the force acts, measured from end A.

5 ft

3 ft

2 ft

4 ft

A

B 5

12

3

4

500 lb

13 5

200 lb

260 lb

SOLUTION + FR = ΣFx; S x + c FRy = ΣFy;

5 4 FRx = -500 a b + 260 a b = - 300 lb = 300 lb d 5 13

12 3 FRy = -500 a b - 200 - 260 a b = -740 lb = 740 lb T 5 13

F = 2( - 300)2 + ( - 740)2 = 798 lb

Ans.

u = tan-1 a

Ans.

740 b = 67.9° d 300

c + MRA = ΣMA;

12 3 740(x) = 500 a b(5) + 200(8) + 260 a b(10) 5 13

740(x) = 5500

Ans.

x = 7.43 ft

Ans: F = 798 lb 67.9° d x = 7.43 ft 342

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*4–116. Replace the three forces acting on the shaft by a  single resultant force. Specify where the force acts, measured from end B.

5 ft

3 ft

2 ft

4 ft

A

B 5

12

3

4

500 lb

13 5

200 lb

260 lb

SOLUTION + ΣFR = ΣFx; S x

+ c FRy = ΣFy;

4 5 FRx = - 500 a b + 260 a b = - 300 lb = 300 lb d 5 13 3 12 FRy = - 500 a b - 200 - 260 a b = - 740 lb = 740 lb T 5 13

F = 2( - 300)2 + ( -740)2 = 798 lb u = tan-1 a

Ans.

740 b = 67.9° d 300

a+ MRB = ΣMB;

Ans.

3 12 740(x) = 500 a b(9) + 200(6) + 260 a b(4) 5 13 Ans.

x = 6.57 ft

Ans: F = 798 lb u = 67.9° d x = 6.57 ft 343

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4–117. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A.

700 N 450 N

30

300 N

60

B

A 2m

4m

3m

1500 N m

SOLUTION ;

+ F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

T

F = 2(-125)2 + ( -1296)2 = 1302 N

Ans.

1296 b = 84.5° 125

Ans.

u = tan-1 a

c + MRA = ©MA ;

d

1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 1500 Ans.

x = 7.36 m

Ans: F = 1302 N u = 84.5° d x = 7.36 m 344

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4–118. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B.

700 N 450 N

30

300 N

60

B

A 2m

4m

3m

1500 N m

SOLUTION ;

+ F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

F = 2(- 125)2 + ( - 1296)2 = 1302 N u = tan-1 a

Ans.

1296 b = 84.5° d 125

c + MRB = ©MB ;

T

Ans.

1296(x) = - 450 sin 60°(4) + 700 cos 30°(3) + 1500 Ans.

x = 1.36 m (to the right)

Ans: F = 1302 N u = 84.5° d x = 1.36 m (to the right) 345

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4–119. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects a vertical line along member AB, measured from A.

400 N

200 N 0.5 m

200 N 0.5 m 600 N

B C

1.5 m

Solution A

Equivalent Resultant Force. Referring to Fig. a, + (FR)x = ΣFx;          S

(FR)x = 600 N S

   + c (FR)y = ΣFy;        (FR)y = - 200 - 400 - 200 = -800 N = 800 NT As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 26002 + 8002 = 1000 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

d = tan-1a

800 b = 53.13° = 53.1°  c 600

Ans.

Location of Resultant Force. Along AB,

  a+ (MR)B = ΣMB;   600(1.5 - d) = -400(0.5) - 200(1)

Ans.

d = 2.1667 m = 2.17 m

Ans: FR = 1000 N u = 53.1° c d = 2.17 m 346

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*4–120. y

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects a vertical line along member AB, measured from A.

1m 600 N

0.5 m B 0.5 m

400 N

3

1.5 m

4

5

900 N

Solution

1m

A

Equivalent Resultant Force. Referring to Fig. a + (FR)x = ΣFx;     (FR)x = 900 a 3 b - 400 a 4 b = 220 N    S 5 5

S

3 4    + c (FR)y = ΣFy;   (FR)y = 600 + 400 a b - 400 - 900 a b 5 5 = - 280 N = 280 N T



As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 22202 + 2802 = 356.09 N = 356 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

d = tan-1a

280 b = 51.84° = 51.8° 220

Ans.

Location of Resultant Force. Referring to Fig. a

3   a+ (MR)A = ΣMA;   280 a - 220 b = 400(1.5) - 600(0.5) - 900 a b(2.5) 5 4 + 400 a b(1) 5



3

(1)

220 b - 280 a = 730

347

5 4

400 N x

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*4–120. Continued

Along AB, a = 0. Then Eq (1) becomes 220 b - 280(0) = 730 b = 3.318 m Thus, the intersection point of line of action of FR on AB measured upward from point A is

Ans.

d = b = 3.32 m

Ans: FR = 356 N u = 51.8° d = b = 3.32 m 348

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4–121. y

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects a horizontal line along member CB, measured from end C.

1m 600 N

0.5 m B 0.5 m

400 N

3

1.5 m

4

5

900 N

Solution

1m

A

Equivalent Resultant Force. Referring to Fig. a + (FR)x = ΣFx;   (FR)x = 900 a 3 b - 400 a 4 b = 220 N    S 5 5

S

3 4    + c (FR)y = ΣFy;         (FR)y = 600 + 400 a b - 400 - 900 a b 5 5 = - 280 N = 280 N T



As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 22202 + 2802 = 356.09 N = 356 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

d = tan-1a

280 b = 51.84° = 51.8° 220

Ans.

Location of Resultant Force. Referring to Fig. a

3    a+ (MR)A = ΣMA;   280 a - 220 b = 400(1.5) - 600(0.5) - 900 a b(2.5) 5 4 + 400 a b(1) 5



3

(1)

220 b - 280 a = 730

349

5 4

400 N x

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4–121. Continued

Along BC, b = 3 m. Then Eq (1) becomes 220(3) - 280 a = 730 a = - 0.25 m Thus, the intersection point of line of action of FR on CB measured to the right of point C is

Ans.

d = 1.5 - ( -0.25) = 1.75 m

Ans: FR = 356 N u = 51.8° d = 1.75 m 350

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4–122. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point A.

0.5 m B

1m

500 N 0.2 m

30

5

3

4

250 N

1m

SOLUTION

300 N

Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, + (F ) = ©F ; : R x x

4 (FR)x = 250 a b - 500 cos 30° - 300 = -533.01 N = 533.01 N ; 5

+ c (FR)y = ©Fy;

3 (FR)y = 500 sin 30° - 250 a b = 100 N c 5

1m A

The magnitude of the resultant force FR is given by FR = 2(FR)x 2 + (FR)y 2 = 2533.012 + 1002 = 542.31 N = 542 N

Ans.

The angle u of FR is u = tan

-1

B

(FR)y (FR)x

R = tan

-1

c

100 d = 10.63° = 10.6° b 533.01

Ans.

Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A, a +(MR)A = ©MA;

4 3 533.01(d) = 500 cos 30°(2) - 500 sin 30°(0.2) - 250 a b(0.5) - 250 a b(3) + 300(1) 5 5 Ans.

d = 0.8274 mm = 827 mm

Ans: FR = 542 N u = 10.6° b d = 0.827 m 351

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4–123. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point B.

0.5 m B 1m

500 N 0.2 m

30

5

3

4

250 N

1m

SOLUTION

300 N

Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

1m A

4 (FR)x = 250 a b - 500 cos 30° - 300 = - 533.01N = 533.01 N ; 5 3 (FR)y = 500 sin 30° - 250 a b = 100 N c 5

+ c (FR)y = ©Fy;

The magnitude of the resultant force FR is given by FR = 2(FR)x 2 + (FR)y 2 = 2533.012 + 1002 = 542.31 N = 542 N

Ans.

The angle u of FR is u = tan

-1

B

(FR)y (FR)x

R = tan

-1

c

100 d = 10.63° = 10.6° b 533.01

Ans.

Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point B, a +(MR)B = ©Mb;

3 -533.01(d) = -500 cos 30°(1) - 500 sin 30°(0.2) - 250 a b(0.5) - 300(2) 5 Ans.

d = 2.17 m

Ans: FR = 542 N u = 10.6° b d = 2.17 m 352

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*4–124. Replace the parallel force system acting on the plate by a resultant force and specify its location on the x–z plane.

z 0.5 m 1m

SOLUTION

2 kN 5 kN

Resultant Force: Summing the forces acting on the plate, (FR)y = ©Fy;

1m

FR = - 5 kN - 2 kN - 3 kN Ans.

= - 10 kN The negative sign indicates that FR acts along the negative y axis.

0.5 m

Resultant Moment: Using the right-hand rule, and equating the moment of FR to the sum of the moments of the force system about the x and z axes, (MR)x = ©Mx;

y 3 kN

x

(10 kN)(z) = (3 kN)(0.5 m) + (5 kN)(1.5 m) + 2 kN(2.5 m) Ans.

z = 1.40 m (MR)z = ©Mz;

1m

-(10 kN)(x) = - (5 kN)(0.5 m) - (2 kN)(1.5 m) - (3 kN)(1.5 m) Ans.

x = 1.00 m

Ans: FR = - 10 kN 353

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4–125. Replace the force and couple system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member AB, measured from A.

A 2 ft

5

3 4

150 lb

4 ft

SOLUTION + F = ©F ; : Rx x + c FRy = ©Fy ;

4 FRx = 150 a b + 50 sin 30° = 145 lb 5

3 FRy = 50 cos 30° + 150 a b = 133.3 lb 5

FR = 2(145)2 + (133.3)2 = 197 lb u = tan

-1

a + MRA = ©MA ;

a

133.3 b = 42.6° 145

500 lb ft B

C 3 ft

Ans.

30 50 lb

Ans.

4 145 d = 150 a b (2) - 50 cos 30° (3) + 50 sin 30° (6) + 500 5 Ans.

d = 5.24 ft

Ans: FR = 197 lb u = 42.6°a d = 5.24 ft 354

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4–126. Replace the force and couple system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member BC, measured from B.

A 2 ft

5

3

4

150 lb

SOLUTION + F = ©F ; : Rx x

4 FRx = 150 a b + 50 sin 30° = 145 lb 5

+ c FRy = ©Fy ;

3 FRy = 50 cos 30° + 150 a b = 133.3 lb 5

2

500 lb  ft

-1

a

3 ft

2

Ans.

133.3 b = 42.6° 145

a + MRA = ©MA ;

B

C

FR = 2(145) + (133.3) = 197 lb u = tan

4 ft

30 50 lb

Ans.

4 145 (6) - 133.3 (d) = 150 a b (2) - 50 cos 30° (3) + 50 sin 30° (6) + 500 5 Ans.

d = 0.824 ft

Ans: FR = 197 lb u = 42.6°a d = 0.824 ft 355

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4–127. If FA = 7 kN and FB = 5 kN, represent the force system acting on the corbels by a resultant force, and specify its location on the x–y plane.

z 150 mm

6 kN

100 mm 650 mm

SOLUTION

x

FB 750 mm O

FA

8kN 700 mm 100 mm

600 mm

150 mm

y

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

- FR = -6 - 5- 7- 8 Ans.

FR = 26 kN

Point of Application: By equating the moment of the forces shown in Fig. a and FR, Fig. b, about the x and y axes, (MR)x = ©Mx;

-26(y) = 6(650) + 5(750) - 7(600) - 8(700) Ans.

y = 82.7 mm (MR)y = ©My;

26(x) = 6(100) + 7(150) - 5(150) - 8(100) Ans.

x = 3.85 mm

Ans: FR = 26 kN y = 82.7 mm x = 3.85 mm 356

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*4–128. Determine the magnitudes of FA and FB so that the resultant force passes through point O of the column.

z 150 mm

6 kN

100 mm 650 mm

SOLUTION

x

FB 750 mm O

FA

8kN 700 mm 100 mm

600 mm

150 mm

y

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

- FR = - FA - FB - 8 - 6 (1)

FR = FA + FB + 14

Point of Application: Since FR is required to pass through point O, the moment of FR about the x and y axes are equal to zero. Thus, (MR)x = ©Mx;

0 = FB (750) + 6(650) - FA (600) - 8(700) (2)

750FB - 600FA - 1700 = 0 (MR)y = ©My;

0 = FA (150) + 6(100) - FB (150) - 8(100) (3)

159FA - 150FB + 200 = 0 Solving Eqs. (1) through (3) yields FA = 18.0 kN

FB = 16.7 kN

FR = 48.7 kN

Ans.

Ans: FA = 18.0 kN FB = 16.7 kN FR = 48.7 kN 357

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4–129. The tube supports the four parallel forces. Determine the magnitudes of forces FC and FD acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube.

FD

z

600 N D FC

A 400 mm

SOLUTION Since the resultant force passes through point O, the resultant moment components about x and y axes are both zero. ©Mx = 0;

500 N C

400 mm z B

200 mm 200 mm y

FD(0.4) + 600(0.4) - FC(0.4) - 500(0.4) = 0 (1)

FC - FD = 100 ©My = 0;

x

O

500(0.2) + 600(0.2) - FC(0.2) - FD(0.2) = 0 (2)

FC + FD = 1100 Solving Eqs. (1) and (2) yields: FC = 600 N

Ans.

FD = 500 N

Ans: FC = 600 N FD = 500 N 358

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4–130. z

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Take F1 = 8 kN and F2 = 9 kN.

12 kN

F1

F2

6 kN

x

8m 16 m

12 m

Solution

6m

y

4m

Equivalent Resultant Force. Sum the forces along z axis by referring to Fig. a      + c (FR)z = ΣFz;    - FR = - 8 - 6 - 12 - 9  FR = 35 kN Ans. Location of the Resultant Force. Sum the moments about the x and y axes by referring to Fig. a, (MR)x = ΣMx;     - 35 y = - 12(8) - 6(20) - 9(20) Ans.

y = 11.31 m = 11.3 m (MR)y = ΣMy;   

  35 x = 12(6) + 8(22) + 6(26) Ans.

       x = 11.54 m = 11.5 m

Ans: FR = 35 kN y = 11.3 m x = 11.5 m 359

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4–131. z

The building slab is subjected to four parallel column loadings. Determine F1 and F2 if the resultant force acts through point (12 m, 10 m).

12 kN

F1

F2

6 kN

x

8m 16 m

12 m

Solution

6m

y

4m

Equivalent Resultant Force. Sum the forces along z axis by referring to Fig. a, + c (FR)z = ΣFz;    - FR = - F1 - F2 - 12 - 6  FR = F1 + F2 + 18 Location of the Resultant Force. Sum the moments about the x and y axes by referring to Fig. a,   (MR)x = ΣMx;    - (F1 + F2 + 18)(10) = - 12(8) - 6(20) - F2(20)

(1)

10F1 - 10F2 = 36

  (MR)y = ΣMy;   (F1 + F2 + 18)(12) = 12(6) + 6(26) + F1(22)

(2)

12F2 - 10F1 = 12

Solving Eqs (1) and (2),

Ans.

F1 = 27.6 kN  F2 = 24.0 kN

Ans: F1 = 27.6 kN F2 = 24.0 kN 360

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*4–132. If FA = 40 kN and FB = 35 kN, determine the magnitude of the resultant force and specify the location of its point of application (x, y) on the slab.

z 30 kN FB

0.75 m 2.5 m

90 kN 20 kN

2.5 m 0.75 m FA

0.75 m

SOLUTION

x

Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

y

3m 3m 0.75 m

- FR = - 30 - 20 - 90 - 35 - 40 Ans.

FR = 215 kN

Point of Application: By equating the moment of the forces and FR, about the x and y axes, (MR)x = ©Mx;

- 215(y) = - 35(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - 40(6.75) Ans.

y = 3.68 m (MR)y = ©My;

215(x) = 30(0.75) + 20(0.75) + 90(3.25) + 35(5.75) + 40(5.75) Ans.

x = 3.54 m

Ans: FR = 215 kN y = 3.68 m x = 3.54 m 361

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4–133. If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings FA and FB and the magnitude of the resultant force.

z 30 kN FB

0.75 m 2.5 m

90 kN 20 kN

2.5 m 0.75 m FA

0.75 m x

SOLUTION

3m

Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force FR, + c FR = ©Fz;

y

3m 0.75 m

- FR = - 30 - 20 - 90 - FA - FB (1)

FR = 140 + FA + FB

Point of Application: By equating the moment of the forces and FR, about the x and y axes, (MR)x = ©Mx;

- FR(3.75) = - FB(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - FA(6.75) (2)

FR = 0.2FB + 1.8FA + 132 (MR)y = ©My;

FR(3.25) = 30(0.75) + 20(0.75) + 90(3.25) + FA(5.75) + FB(5.75) FR = 1.769FA + 1.769FB + 101.54

(3)

Solving Eqs.(1) through (3) yields FA = 30 kN

FB = 20 kN

FR = 190 kN

Ans.

Ans: FA = 30 kN FB = 20 kN FR = 190 kN 362

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4–134. Replace the two wrenches and the force, acting on the pipe assembly, by an equivalent resultant force and couple moment at point O.

100 N · m

z

300 N

C

SOLUTION

O 0.5 m

Force And Moment Vectors:

A 0.6 m

B 0.8 m

100 N

x

F1 = 5300k6 N

F3 = 5100j6 N

45°

F2 = 2005cos 45°i - sin 45°k6 N

200 N 180 N · m

= 5141.42i - 141.42k6 N

M 1 = 5100k6 N # m

M 2 = 1805cos 45°i - sin 45°k6 N # m = 5127.28i - 127.28k6 N # m Equivalent Force and Couple Moment At Point O: FR = ©F;

FR = F1 + F2 + F3 = 141.42i + 100.0j + 1300 - 141.422k

= 5141i + 100j + 159k6 N

Ans.

The position vectors are r1 = 50.5j6 m and r2 = 51.1j6 m. M RO = ©M O ;

M RO = r1 * F1 + r2 * F2 + M 1 + M 2 i = 30 0 +

j 0.5 0 i 0 141.42

k 0 3 300 j 1.1 0

k 0 -141.42

+ 100k + 127.28i - 127.28k =

122i - 183k N # m

Ans.

363

Ans: FR = 5141i + 100j + 159k6 N MRO = 5122i - 183k6 N # m

y

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4–135. z

Replace the force system by a wrench and specify the magnitude of the force and couple moment of the wrench and the point where the wrench intersects the x–z plane.

200 N 400 N 200 N 4

3m

2m x

Resultant Force. Referring to Fig. a 3 4 FR = e c 200 a b - 400 d i - 200j + 200 a bk f 5 5 = { - 280i - 200j + 160k} N

The magnitude of FR is   FR = 2 ( - 280 ) 2 +

( - 200 ) 2 + 1602 = 379.47 N = 379 N

Ans.

The direction of FR is defined by uFR =

- 280i - 200j + 160k FR = - 0.7379i - 0.5270j + 0.4216k = FR 379.47

Resultant Moment. The line of action of MR of the wrench is parallel to that of FR. Also, assume that MR and FR have the same sense. Then uMR = - 0.7379i - 0.5270j + 0.4216k

364

y

5 3

Solution

0.5 m

O

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4–135. Continued

Referring to Fig. a, where the origin of the x′, y′, z′ axes is the point where the wrench intersects the xz plane, (MR)x' = ΣMx'; - 0.7379 MR = - 200(z - 0.5)

 (1)

2 4 (MR)y' = ΣMy'; -0.5270 MR = - 200 a b(z - 0.5) - 200 a b(3 - x) + 400(z - 0.5)(2) 5 5

(MR)z' = ΣMz'; 0.4216 MR = 200 x + 400(2)

 (3)

Solving Eqs (1), (2) and (3) MR = 590.29 N # m = 590 N # m

Ans.

z = 2.6778 m = 2.68 m



Ans.

x = - 2.7556 m = -2.76 m



Ans.

Ans: FR = 379 N

MR = 590 N # m z = 2.68 m x = - 2.76 m

365

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*4–136. Replace the five forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, z) where the wrench intersects the x–z plane.

z

800 N 4m

2m 4m

2m

200 N 400 N

Solution x

Resultant Force. Referring to Fig. a FR = { - 600i - (300 + 200 + 400)j - 800k} N = 5-600i - 900j - 800k6 N

Then the magnitude of FR is

FR = 2( - 600)2 + ( - 900)2 + ( - 800)2 = 1345.36 N = 1.35 kN

Ans.

The direction of FR is defined by uFR =

- 600i - 900j - 800k FR = = -0.4460i - 0.6690j - 0.5946k FR 1345.36

Resultant Moment. The line of action of MR of the wrench is parallel to that of FR. Also, assume that both MR and FR have the same sense. Then uMR = - 0.4460i - 0.6690j - 0.5946k

366

600 N

300 N

y

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*4–136. Continued

Referring to Fig. a, (MR)x′ = ΣMx′;

- 0.4460 MR = - 300z - 200(z - 2) - 400z(1)

(MR)y′ = ΣMy′;

- 0.6690 MR = 800(4 - x) + 600z(2)

(MR)z′ = ΣMz′;

- 0.5946 MR = 200(x - 2) + 400x - 300(4 - x)(3)

Solving Eqs (1), (2) and (3) MR = - 1367.66 N # m = -1.37 kN # m

Ans.

x = 2.681 m = 2.68 m



Ans.

z = - 0.2333 m = -0.233 m



Ans.

The negative sign indicates that the line of action of MR is directed in the opposite sense to that of FR.

Ans: MR = - 1.37 kN # m x = 2.68 m z = -0.233 m 367

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4–137. z

Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where the wrench intersects the plate.

FB  { 300k} N FC  {200j} N C

y

B x

Solution

5m

Resultant Force. Referring to Fig. a, FA  {400i} N

Then, the magnitude of FR is FR = 24002 + 2002 + ( - 300)2 = 538.52 N = 539 N

Ans.

The direction of FR is defined by uFR =

3m A

FR = {400i + 200j - 300k} N



x P

400i + 200j - 300k FR = = 0.7428i + 0.3714j - 0.5571k FR 538.52

Resultant Moment. The line of action of MR of the wrench is parallel to that of FR. Also, assume that both MR and FR have the same sense. Then uMR = 0.7428i + 0.3714j - 0.5571k

368

y

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4–137. Continued

Referring to Fig. a, (MR)x′ = ΣMx′; 0.7428 MR = 300y(1) (MR)y′ = ΣMy′; 0.3714 MR = 300 (3 - x)(2) (MR)z′ = ΣMz; - 0.5571 MR = - 200x - 400 (5 - y)(3) Solving Eqs (1), (2) and (3) MR = 1448.42 N # m = 1.45 kN # m

Ans.

x = 1.2069 m = 1.21 m



Ans.

y = 3.5862 m = 3.59 m



Ans.

Ans: FR = 539 N

MR = 1.45 kN # m x = 1.21 m y = 3.59 m

369

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4–138. Replace the loading by an equivalent resultant force and couple moment acting at point O.

50 lb/ft 9 ft O 9 ft 50 lb/ft

Solution + c FR = ΣF ;

Ans.

FR = 0

a+ MRO = ΣMO ;  MRO = 225 (6) = 1350 lb # ft = 1.35 kip # ft

Ans.

Ans: FR = 0 MRO = 1.35 kip # ft 370

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4–139. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point O.

3 kN/m

O

3m

SOLUTION

1.5 m

Loading: The distributed loading can be divided into two parts as shown in Fig. a. Equations of Equilibrium: Equating the forces along the y axis of Figs. a and b, we have + T FR = ©F;

FR =

1 1 (3)(3) + (3)(1.5) = 6.75 kN T 2 2

Ans.

If we equate the moment of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point O, we have a + (MR)O = ©MO;

1 1 (3)(3)(2) - (3)(1.5)(3.5) 2 2 x = 2.5 m

- 6.75(x) = -

Ans.

Ans: FR = 6.75 kN x = 2.5 m 371

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*4–140. w

Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point A.

5 kN/m 2 kN/m A

x

B 4m

2m

Solution Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a +c (FR)y = ΣFy;

- FR = - 2(6) -

1 (3)(6) 2

Ans. Ans.

FR = 21.0 kNT Location of the Resultant Force. Summing the moments about point A, 1 a + (MR)A = ΣMA; - 21.0(d) = -2(6)(3) - (3)(6)(4) 2 d = 3.429 m = 3.43 m

Ans.

Ans: FR = 21.0 kN d = 3.43 m 372

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4–141. Currently eighty-five percent of all neck injuries are caused by rear-end car collisions. To alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dynamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point A.

A

12 lb/ft

0.5 ft w w  12(1  2x2) lb/ft B 18 lb/ft

x

SOLUTION 0.5

FR =

L

w(x) dx =

L0

12 A 1 + 2 x2 B dx = 12 cx +

0.5

x =

L

x w(x) dx

L

= w(x)dx

L0

x(12) A 1 + 2 x2 B dx 7

12c =

2 3 0.5 x d = 7 lb 3 0

Ans.

x4 0.5 x2 + (2) d 2 4 0 7 Ans.

x = 0.268 ft

Ans: FR = 7 lb x = 0.268 ft 373

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4–142. Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at A.

4 kN/m

2 kN/m

A

Solution

B

3m

3m

Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a, +c (FR)y = ΣFy;

- FR = - 2(6) -

1 (2)(3) 2 Ans.

FR = 15.0 kN T Location of the Resultant Force. Summing the Moments about point A, a + (MR)A = ΣMA;

- 15.0(d) = -2(6)(3) -

1 (2)(3)(5) 2 Ans.

d = 3.40 m

Ans: FR = 15.0 kN d = 3.40 m 374

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4–143. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

6 kN/m 4 kN/m

O 2m

1.5 m

Solution Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a, +c (FR)y = ΣFy;

- FR = - 4(2) -

1 (6)(1.5) 2 Ans.

FR = 12.5 kN Location of the Resultant Force. Summing the Moment about point O, a + (MR)O = ΣMO;

- 12.5(d) = -4(2)(1) -

1 (6)(1.5)(2.5) 2 Ans.

d = 1.54 m

Ans: FR = 12.5 kN d = 1.54 m 375

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*4–144. The distribution of soil loading on the bottom of a building slab is shown. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

O 50 lb/ft

12 ft

SOLUTION + c FR = ©Fy;

300 lb/ft 9 ft

FR = 50(12) + 12 (250)(12) + 12 (200)(9) + 100(9) = 3900 lb = 3.90 kip c

a + MRo = ©MO;

100 lb/ft

Ans.

3900(d) = 50(12)(6) + 12 (250)(12)(8) + 12 (200)(9)(15) + 100(9)(16.5) Ans.

d = 11.3 ft

Ans: FR = 3.90 kip c d = 11.3 ft 376

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4–145. Replace the loading by an equivalent resultant force and couple moment acting at point O.

8 kN/m 5 kN/m O

1.5 m

0.75 m

0.75 m

Solution Equivalent Resultant Force And Couple Moment About Point O. Summing the forces along the y axis by referring to Fig. a, 1 1 +c (FR)y = ΣFy;  FR = - (3)(1.5) - 5(2.25) - (5)(0.75) 2 2 = - 15.375 kN = 15.4 kN T

Ans.

Summing the Moment about point O, 1 a + (MR)O = ΣMO;  (MR)O = - (3)(1.5)(0.5) - 5(2.25)(1.125) 2 1 - (5)(0.75)(2.5) 2

= - 18.46875 kN # m = 18.5 kN # m (clockwise) Ans.

Ans: FR = 15.4 kN

(MR)O = 18.5 kN # m (clockwise)

377

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4–146. 6 kN/m

Replace the distributed loading by an equivalent resultant force and couple moment acting at point A.

6 kN/m 3 kN/m

A

B 3m

3m

Solution Equivalent Resultant Force And Couple Moment About Point A. Summing the forces along the y axis by referring to Fig. a, 1 1 + c (FR)y = ΣFy;  FR = - (3)(3) - 3(6) - (3)(3) 2 2

Ans.

= - 27.0 kN = 27.0 kN T 

Summing the moments about point A, 1 1 a+ (MR)A = ΣMA;  (MR)A = - (3)(3)(1) - 3(6)(3) - (3)(3)(5) 2 2

= - 81.0 kN # m = 81.0 kN # m (clockwise)

Ans.

Ans: FR = 27.0 kN

(MR)A = 81.0 kN # m (clockwise)

378

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4–147. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN # m clockwise.

a

b

4 kN/m

A

2.5 kN/m 9m

Solution + c FR = 0 = ΣFy ;  0 = a+ MRA = ΣMA; 

1 1 (2.5)(9) - (4)(b)  b = 5.625 m 2 2

1 1 2   - 8 = - (2.5)(9)(6) + (4)(5.625) aa + (5.625) b 2 2 3 a = 1.54 m

Ans.

Ans.

Ans: a = 1.54 m 379

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*4–148. The form is used to cast a concrete wall having a width of 5 m. Determine the equivalent resultant force the wet concrete exerts on the form AB if the pressure distribution due to the concrete can be approximated as shown. Specify the location of the resultant force, measured from point B.

B

p

1

p  (4z 2 ) kPa 4m

Solution L

dA =

L0

4

A

1 2

8 kPa

4z dz

4 3 2 = c (4)z2 d 3 0

z

= 21.33 kN>m Ans.

FR = 21.33(5) = 107 kN L

zdA =

L0

4

3

4z2 dz

4 5 2 = c (4)z2 d 5 0

= 51.2 kN z =

51.2 = 2.40 m 21.33

Ans.

Also, from the back of the book, A =

2 2 ab = (8)(4) = 21.33 3 3

FR = 21.33 (5) = 107 kN



Ans.

z = 4 - 1.6 = 2.40 m



Ans.

Ans: FR = 107 kN z = 2.40 m 380

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4–149. If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities w1 and w2 of this distribution needed to support the column loadings.

80 kN

60 kN

1m

50 kN 2.5 m

SOLUTION

3.5 m

1m

w2

Loading: The trapezoidal reactive distributed load can be divided into two parts as shown on the free-body diagram of the footing, Fig. a. The magnitude and location measured from point A of the resultant force of each part are also indicated in Fig. a.

w1

Equations of Equilibrium: Writing the moment equation of equilibrium about point B, we have

a + ©MB = 0; w2(8) ¢ 4 -

8 8 8 8 ≤ + 60 ¢ - 1 ≤ - 80 ¢ 3.5 - ≤ - 50 ¢ 7 - ≤ = 0 3 3 3 3

w2 = 17.1875 kN>m = 17.2 kN>m

Ans.

Using the result of w2 and writing the force equation of equilibrium along the y axis, we obtain + c ©Fy = 0;

1 (w - 17.1875)8 + 17.1875(8) - 60 - 80 - 50 = 0 2 1 w1 = 30.3125 kN>m = 30.3 kN>m

Ans.

Ans: w2 = 17.2 kN>m w1 = 30.3 kN>m 381

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4–150. Replace the loading by an equivalent force and couple moment acting at point O.

6 kN/m

15 kN

500 kN m O

7.5 m

4.5 m

SOLUTION + c FR = ©Fy ;

FR = -22.5 - 13.5 - 15.0 Ans.

= - 51.0 kN = 51.0 kN T a + MRo = ©Mo ;

MRo = - 500 - 22.5(5) - 13.5(9) - 15(12) = - 914 kN # m = 914 kN # m (Clockwise)

Ans.

Ans: FR = 51.0 kN T MRO = 914 kN # m b 382

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4–151. Replace the loading by a single resultant force, and specify the location of the force measured from point O.

6 kN/m

15 kN

500 kN m O

7.5 m

4.5 m

SOLUTION Equivalent Resultant Force: + c FR = ©Fy ;

- FR = - 22.5 - 13.5 - 15 Ans.

FR = 51.0 kN T Location of Equivalent Resultant Force: a + (MR)O = ©MO ;

- 51.0(d) = -500 - 22.5(5) - 13.5(9) - 15(12) Ans.

d = 17.9 m

Ans: FR = 51.0 kN T d = 17.9 m 383

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*4–152. Replace the loading by an equivalent resultant force and couple moment acting at point A.

400 N/m

B

A 3m

3m

Solution Equivalent Resultant Force And Couple Moment At Point A. Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;  FR = - 400(3)

1 (400)(3) 2 Ans.

= - 1800 N = 1.80 kN T 

Summing the moment about point A, a+ (MR)A = ΣMA;  (MR)A = - 400(3)(1.5) -

1 (400)(3)(4) 2

= - 4200 N # m = 4.20 kN # m (clockwise) Ans.

Ans: FR = 1.80 kN

(MR)A = 4.20 kN # m (clockwise)

384

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4–153. Replace the loading by a single resultant force, and specify its location on the beam measured from point A.

400 N/m

B

A 3m

3m

Solution Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;  - FR = - 400(3)

1 (400)(3) 2 Ans.

FR = 1800 N = 1.80 kN T 

Location of Resultant Force. Summing the moment about point A by referring to Fig. a, a+ (MR)A = ΣMA;  - 1800 d = - 400(3)(1.5)

1 (400)(3)(4) 2 Ans.

d = 2.333 m = 2.33 m

Ans: FR = 1.80 kN d = 2.33 m 385

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4–154. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a horizontal line along member AB, measured from A.

3 kN/m B

A 3m

2 kN/m

4m

Solution Equivalent Resultant Force. Summing the forces along the x and y axes by referring to Fig. a, + (FR)x = ΣFx;  (FR)x = - 2(4) = - 8 kN = 8 kN d S + c (FR)y = ΣFy;  (FR)y = - 3(3) = - 9 kN = 9 kN T Then And

FR = 2(FR)2x + (FR)2y = 282 + 92 = 12.04 kN = 12.0 kN u = tan-1 c

(FR)y (FR)x

Ans.

9 d = tan-1 a b = 48.37° = 48.4° dAns. 8

Location of the Resultant Force. Summing the moments about point A, by referring to Fig. a, a+ (MR)A = ΣMA ;   - 8x - 9y = - 3(3)(1.5) - 2(4)(2)

8x + 9y = 29.5

(1)

386

C

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4–154. Continued

Along AB, x = 0.  Then Eq (1) becomes 8(0) + 9y = 29.5 y = 3.278 m Thus, the inter section point of line of action of FR on AB measured to the right from point A is

Ans.

d = y = 3.28 m

Ans: FR = 12.0 kN u = 48.4° d d = 3.28 m 387

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4–155. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a vertical line along member BC, measured from C.

3 kN/m B

A 3m

2 kN/m

4m

Solution Equivalent Resultant Force. Summing the forces along the x and y axes by referring to Fig. a, + (FR)x = ΣFx;  (FR)x = - 2(4) = - 8 kN = 8 kN d S + c (FR)y = ΣFy;  (FR)y = - 3(3) = - 9 kN = 9 kN T Then And

FR = 2(FR)2x + (FR)2y = 282 + 92 = 12.04 kN = 12.0 kN u = tan-1 c

(FR)y (FR)x

Ans.

9 d = tan-1 a b = 48.37° = 48.4° dAns. 8

Location of the Resultant Force. Summing the moments about point A, by referring to Fig. a, a+ (MR)A = ΣMA;   - 8x - 9y = -3(3)(1.5) - 2(4)(2)

8x + 9y = 29.5

(1)

388

C

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4–155. Continued

Along BC, y = 3 m.  Then Eq (1) becomes 8x + 9(3) = 29.5 x = 0.3125 m Thus, the intersection point of line of action of FR on BC measured upward from point C is

d = 4 - x = 4 - 0.3125 = 3.6875 m = 3.69 m

Ans.

Ans: FR = 12.0 kN u = 48.4° d d = 3.69 m 389

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*4–156. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN # m clockwise.

a

b 6 kN/m

A

2 kN/m

4m

Solution Equivalent Resultant Force And Couple Moment At Point A. Summing the forces along the y axis by referring to Fig. a, with the requirement that FR = 0, + c (FR)y = ΣFy;  0 = 2(a + b) -

1 (6)(b) 2

2a - b = 0(1)



Summing the moments about point A, with the requirement that (MR)A = 8 kN # m,



a+ (MR)A = ΣMA;  - 8 = 2(a + b) c4 -

1 1 1 (a + b)d - (6)(b) a4 - bb 2 2 3

- 8 = 8a - 4b - 2ab -a2(2)

Solving Eqs (1) and (2),

a = 1.264 m = 1.26 m

Ans.



b = 2.530 m = 2.53 m 

Ans.

Ans: a = 1.26 m b = 2.53 m 390

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4–157. w

Determine the equivalent resultant force and couple moment at point O.

9 kN/m w  ( 13 x3 ) kN/m

O

x

3m

Solution Equivalent Resultant Force And Couple Moment About Point O. The differential 1 force indicated in Fig. a is dFR = w dx = x3dx. Thus, summing the forces along the 3 y axis, 3m 1 3 x dx + c (FR)y = ΣFy;   FR = - dFR = L L0 3 = -



1 4 3m x 12 L 0 Ans.

= - 6.75 kN = 6.75 kNT 

Summing the moments about point O, a+ (MR)O = ΣMO;  (MR)O =

L L0

3m

L0

3m



=



=



= a



(3 - x)dFR 1 (3 - x) a x3dxb 3 ax3 -

1 4 x b dx 3

1 5 3m x4 x b` 4 15 0

= 4.05 kN # m (counterclockwise)

Ans.

Ans: FR = 6.75 kNT

(MR)O = 4.05 kN # m (counterclockwise)

391

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4–158. w

Determine the magnitude of the equivalent resultant force and its location, measured from point O.

w  (4  2 x ) lb/ft

8.90 lb/ft

4 lb/ft x

O 6 ft

Solution dA = wdx FR =

L

dA =

L0

6

( 4 + 21x ) dx

= c 4x +

4 3 6 x2 d 3 0

   FR = 43.6 lb L

xdF =

x =

L0

6

Ans.

3

( 4x + 2x2 ) dx

= c 2x2 +

4 5 6 x2 d 5 0

= 142.5 lb # ft 142.5 = 3.27 ft 43.6

Ans.

Ans: FR = 43.6 lb x = 3.27 ft 392

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4–159. w

The distributed load acts on the shaft as shown. Determine the magnitude of the equivalent resultant force and specify its location, measured from the support, A.

28 lb/ft w  (2x¤  8x  18) lb/ft

18 lb/ft

10 lb/ft A

B 1 ft

2 ft

x

2 ft

Solution 4

FR = L

L-1

x dF =

x =

2 3

( 2 x2 - 8 x + 18 ) dx = x3 4

L-1

2

4 8x + 18 x ` = 73.33 = 73.3 lb 2 -1

2 4

8 3

( 2 x3 - 8 x2 + 18 x ) dx = x4 - x3 +

89.166 = 1.22 ft 73.3

Ans.

18 2 4 x ` = 89.166 lb # ft 2 -1

Ans.

d = 1 + 1.22 = 2.22 ft

Ans: d = 2.22 ft 393

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*4–160. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A.

w

w  (x2  3x  100) lb/ft

370 lb/ft

100 lb/ft A B

SOLUTION Resultant: The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a. Thus,

15 ft

dFR = w dx = a x2 + 3x + 100b dx Integrating dFR over the entire length of the beam gives the resultant force FR. L

+T

FR =

LL

dFR =

L0

a x2 + 3x + 100b dx = ¢

15 ft 3x2 x3 + + 100x ≤ ` 3 2 0

Ans.

= 2962.5 lb = 2.96 kip

Location: The location of dFR on the beam is xc = x measured from point A. Thus, the location x of FR measured from point A is given by 15 ft

x =

LL

xcdFR

LL

= dFR

L0

xa x2 + 3x + 100bdx 2962.5

=

¢

15 ft x4 + x3 + 50x2 ≤ ` 4 0

2962.5

= 9.21 ft Ans.

Ans: FR = 2.96 kip x = 9.21 ft 394

x

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4–161. w

Replace the loading by an equivalent resultant force and couple moment acting at point O.

p x w  w0 cos ( 2L (

x

O L

Solution Equivalent Resultant Force And Couple Moment About Point O. The differential p xbdx. Thus, summing the force indicated in Fig. a is dFR = w dx = aw0 cos 2L forces along the y axis, + c (FR)y = ΣFy;  FR =

= -



= -

L

dFR = -

L

LO

¢w0 cos

p x≤dx 2L

L 2Lw0 p asin xb ` p 2L O

2Lw0 2Lw0 = T p p

Ans.

Summing the moments about point O, a+ (MR)O = ΣMO;  (MR)O = -

L

xdFR

L

LO

x aw0 cos



= -



= -w0 a



= -a



= a

p xb dx 2L

L 4L2 2L p p x + 2 cos xb ` x sin p 2L 2L p O

2p - 4 bw0L2 p2

2p - 4 bw0L2 (clockwise) p2

Ans.

Ans: FR =

2Lw0 p

(MR)O = a 395

2p - 4 bw0L2 (clockwise) p2

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4–162. Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specify the height h where the bracing strut should be placed so that it lies through the line of action of the resultant force. The wall has a width of 5 m.

p

4m

p

1

(4 z /2) kPa

SOLUTION Equivalent Resultant Force: h

z

+ F = ©F ; : R x

-FR = - LdA = 4m

FR =

L0

L0

8 kPa

wdz

a20z2 b A 103 B dz 1

z

= 106.67 A 103 B N = 107 kN ;

Ans.

Location of Equivalent Resultant Force: z

z =

LA

zdA = dA

LA

zwdz

L0

z

L0

wdz

4m

=

L0

zc A 20z2 B (103) ddz 1

4m

L0 4m

=

L0

A 20z2 B (103)dz 1

c A 20z2 B (10 3) ddz

4m

L0

3

A 20z2 B (103)dz 1

= 2.40 m Thus,

Ans.

h = 4 - z = 4 - 2.40 = 1.60 m

Ans: FR = 107 kN h = 1.60 m 396

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5–10. Determine the components of the support reactions at the fixed support A on the cantilevered beam.

6 kN

30

SOLUTION

30

Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig. a, Ax, Ay, and MA can be obtained by writing the moment equation of equilibrium about point A. + ©F = 0; : x

1.5 m 1.5 m

1.5 m

4 kN

4 cos 30° - A x = 0 Ans.

A x = 3.46 kN + c ©Fy = 0;

A

A y - 6 - 4 sin 30° = 0 Ans.

A y = 8 kN

a+ ©MA = 0;MA - 6(1.5) - 4 cos 30° (1.5 sin 30°) - 4 sin 30°(3 + 1.5 cos 30°) = 0 MA = 20.2 kN # m

Ans.

Ans: Ax = 3.46 kN Ay = 8 kN MA = 20.2 kN # m 397

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5–11. Determine the reactions at the supports.

400 N/m 5

3 4

B A 3m

3m

Solution Equations of Equilibrium. NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the beam’s FBD shown in Fig. a. 4 1 a+ ΣMB = 0;   (400)(6)(3) - NA a b(6) = 0 2 5 NA = 750 N

a+ ΣMA = 0;  By(6) -

Ans.

1 (400)(6)(3) = 0 2 Ans.

By = 600 N

Using the result of NA to write the force equation of equilibrium along the x axis, + ΣFx = 0;  750 a 3 b - Bx = 0 S 5

Ans.

Bx = 450 N

Ans: NA = 750 N By = 600 N Bx = 450 N 398

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*5–12. 4 kN

Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B on the beam.

B

A

30

SOLUTION Equations of Equilibrium: From the free-body diagram of the beam, Fig. a, NB can be obtained by writing the moment equation of equilibrium about point A. a + ©MA = 0;

6m

2m

NB cos 30°(8) - 4(6) = 0 Ans.

NB = 3.464 kN = 3.46 kN

Using this result and writing the force equations of equilibrium along the x and y axes, we have + ©F = 0; : x

A x - 3.464 sin 30° = 0 Ans.

A x = 1.73 kN + c ©Fy = 0;

A y + 3.464 cos 30° - 4 = 0 Ans.

A y = 1.00 kN

Ans: NB = 3.46 kN Ax = 1.73 kN Ay = 1.00 kN 399

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5–13. Determine the reactions at the supports.

900 N/m 600 N/m

B

A 3m

3m

Solution Equations of Equilibrium. NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  600(6)(3) +

1 (300)(3)(5) - NA(6) = 0 2 Ans.

NA = 2175 N = 2.175 kN a+ ΣMA = 0;  By(6) -

1 (300)(3)(1) - 600(6)(3) = 0 2 Ans.

By = 1875 N = 1.875 kN

Also, Bx can be determined directly by writing the force equation of equilibrium along the x axis. + ΣFx = 0;      Bx = 0 S

Ans.

Ans: NA = 2.175 kN By = 1.875 kN Bx = 0 400

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5–14. Determine the reactions at the supports. 800 N/m A

3m

B 1m

3m

Solution Equations of Equilibrium. NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  800(5)(2.5) - NA(3) = 0 Ans.

NA = 3333.33 N = 3.33 kN

Using this result to write the force equations of equilibrium along the x and y axes, + ΣFx = 0;  Bx - 800(5) a 3 b = 0 S 5

Ans.

Bx = 2400 N = 2.40 kN

4 + c ΣFy = 0;  3333.33 - 800 (5)a b - By = 0 5

Ans.

By = 133.33 N = 133 N

Ans: NA = 3.33 kN Bx = 2.40 kN By = 133 N 401

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5–15. Determine the reactions at the supports.

5 kN

2m B

A 6 kN 2m

2m

8 kN 2m

Solution Equations of Equilibrium. Ay and NB can be determined by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the truss shown in Fig. a. a+ ΣMB = 0;  8(2) + 6(4) - 5(2) - Ay(6) = 0 Ans.

Ay = 5.00 kN a+ ΣMA = 0;  NB(6) - 8(4) - 6(2) - 5(2) = 0

Ans.

NB = 9.00 kN

Also, Ax can be determined directly by writing the force equation of equilibrium along x axis. + ΣFx = 0;  5 - Ax = 0  Ax = 5.00 kN S

Ans.

Ans: Ay = 5.00 kN NB = 9.00 kN Ax = 5.00 kN 402

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*5–16. Determine the tension in the cable and the horizontal and vertical components of reaction of the pin A. The pulley at D is frictionless and the cylinder weighs 80 lb.

D 2 1

A

SOLUTION

B

5 ft

C 5 ft

3 ft

Equations of Equilibrium: The tension force developed in the cable is the same throughout the whole cable. The force in the cable can be obtained directly by summing moments about point A. a + ©MA = 0;

+ ©F = 0; : x

T152 + T ¢

≤ 1102 - 801132 = 0 25 T = 74.583 lb = 74.6 lb 2

Ax - 74.583 ¢

1 25

Ans.

≤ = 0 Ans.

Ax = 33.4 lb + c ©Fy = 0;

74.583 + 74.583

2 25

- 80 - By = 0 Ans.

Ay = 61.3 lb

Ans: T = 74.6 lb Ax = 33.4 lb Ay = 61.3 lb 403

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5–17. The man attempts to support the load of boards having a weight W and a center of gravity at G. If he is standing on a smooth floor, determine the smallest angle u at which he can hold them up in the position shown. Neglect his weight.

4 ft

u

SOLUTION a + ©MB = 0;

G

4 ft

- NA (3.5) + W(3 - 4 cos u) = 0

As u becomes smaller, NA : 0 so that,

A

W(3 - 4 cos u) = 0

0.5 ft

B 3 ft

Ans.

u = 41.4°

Ans: u = 41.4° 404

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5–18. Determine the components of reaction at the supports A and B on the rod.

P L –– 2

SOLUTION Equations of Equilibrium: Since the roller at A offers no resistance to vertical movement, the vertical component of reaction at support A is equal to zero. From the free-body diagram, Ax, By, and MA can be obtained by writing the force equations of equilibrium along the x and y axes and the moment equation of equilibrium about point B, respectively. + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

By - P = 0

Pa

B

Ans.

Ans.

By = P a + ©MB = 0;

A

L –– 2

L b - MA = 0 2

MA =

PL 2

Ans.

Ans: Ax = 0 By = P PL MA = 2 405

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5–19. The man has a weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and u.

B

SOLUTION a + ©MB = 0;

L W a cos f b - NA(L cos f ) = 0 2

+ ©F = 0; : x

T cos u -NB sin u = 0

+ c ©Fy = 0;

T sin u +NB cos u +

NA

f

W = 2

L

u

A

(1)

W - W= 0 2

(2)

Solving Eqs. (1) and (2) yields: T= NB =

W sin u 2

Ans.

W cos u 2

Ans: W T= sin u 2 406

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*5–20. A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination u for equilibrium.

u B r A

SOLUTION By observation f = u. Equilibrium: a + ©MA = 0;

NB (2r cos u) - W a

L cos ub = 0 2

+Q ©Fx = 0;

NA cos u - W sin u = 0

+a©Fy = 0;

(W tan u) sin u +

WL 4r

NA = W tan u

WL - W cos u = 0 4r

sin2 u - cos2 u +

L cos u = 0 4r

(1 - cos2 u) - cos2 u + 2 cos2 u cos u =

NB =

L cos u = 0 4r

L cos u - 1 = 0 4r

L ; 2L2 + 128r2 16r

Take the positive root cos u =

L + 2L2 + 128r2 16r

u = cos - 1 ¢

L + 2L2 + 128r2 ≤ 16r

Ans.

Ans: u = cos - 1a 407

L + 2L2 + 12r 2 b 16r

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5–21. The uniform rod AB has a mass of 40 kg. Determine the force in the cable when the rod is in the position shown. There is a smooth collar at A.

A

3m

60 C

Solution

B

Equations of Equilibrium. TBC can be determined by writing the moment equation of equilibrium about point O by referring to the FBD of the rod shown in Fig. a. a+ ΣMO = 0;  40(9.81)(1.5 cos 600°) - TBC (3 sin 60°) = 0 Ans.

TBC = 113.28 N = 113 N

Ans: TBC = 113 N 408

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5–22. If the intensity of the distributed load acting on the beam is w = 3 kN>m, determine the reactions at the roller A and pin B.

A w 30

B

3m 4m

Solution Equations of Equilibrium. NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  3(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0 Ans.

NA = 3.713 kN = 3.71 kN

Using this result to write the force equation of equilibrium along the x and y axes, + ΣFx = 0;  3.713 sin 30° - Bx = 0 S Ans.

Bx = 1.856 kN = 1.86 kN + c ΣFy = 0;  By + 3.713 cos 30° - 3(4) = 0

Ans.

By = 8.7846 kN = 8.78 kN

Ans: NA = 3.71 kN Bx = 1.86 kN By = 8.78 kN 409

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5–23. If the roller at A and the pin at B can support a load up to 4 kN and 8 kN, respectively, determine the maximum intensity of the distributed load w, measured in kN>m, so that failure of the supports does not occur.

A w 30

B

3m 4m

Solution Equations of Equilibrium. NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  w(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0 NA = 1.2376 w Using this result to write the force equation of equilibrium along x and y axes, + ΣFx = 0;  1.2376 w sin 30° - Bx = 0 S

Bx = 0.6188 w

+ c ΣFy = 0;  By + 1.2376 w cos 30° - w(4) = 0

By = 2.9282 w

Thus, FB = 2Bx2 + By2 = 2(0.6188 w)2 + (2.9282 w)2 = 2.9929 w

It is required that FB 6 8 kN;

2.9929 w 6 8

w 6 2.673 kN>m

1.2376 w 6 4

w 6 3.232 kN>m

And NA 6 4 kN;

Thus, the maximum intensity of the distributed load is Ans.

w = 2.673 kN>m = 2.67 kN>m

Ans: w = 2.67 kN>m 410

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*5–24. The relay regulates voltage and current. Determine the force in the spring CD, which has a stiffness of k 120 N m, so that it will allow the armature to make contact at A in figure (a) with a vertical force of 0.4 N. Also, determine the force in the spring when the coil is energized and attracts the armature to E, figure (b), thereby breaking contact at A.

50 mm 50 mm 30 mm 10°

A

B

A

C

E

B

C k

k

D

D

SOLUTION From Fig. (a): a + ©MB = 0;

0.4(100 cos 10°) - Fs (30 cos 10°) = 0

(a)

Ans.

Fs = 1.333 N = 1.33 N Fs = kx;

(b)

1.333 = 120 x x = 0.01111 m = 11.11 mm

From Fig (b), energizing the coil requires the spring to be stretched an additional amount ¢x = 30 sin 10° = 5.209 mm. Thus x¿ = 11.11 + 5.209 = 16.32 mm Ans.

Fs = 120 (0.01632) = 1.96 N

Ans: Fs = 1.33 N Fs = 1.96 N 411

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5–25. Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod.

100 lb 3 ft

3 ft

200 lb  ft

A

2 ft

3 5

4

B

13 12

5

SOLUTION 12 5 b (6) - NB a b (2) = 0 13 13

a + ©MA = 0;

MA - 100 (3) - 200 + NB a

+ ©F = 0; : x

4 5 NA a b - NB a b = 0 5 13

+ c ©Fy = 0;

3 12 NA a b + NB a b - 100 = 0 5 13

Solving, NA = 39.7 lb

Ans.

NB = 82.5 lb

Ans.

MA = 106 lb # ft

Ans.

Ans: NA = 39.7 lb NB = 82.5 lb MA = 106 lb # ft 412

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5–26. The mobile crane is symmetrically supported by two outriggers at A and two at B in order to relieve the suspension of the truck upon which it rests and to provide greater stability. If the crane and truck have a mass of 18 Mg and center of mass at G1, and the boom has a mass of 1.8 Mg and a center of mass at G2, determine the vertical reactions at each of the four outriggers as a function of the boom angle u when the boom is supporting a load having a mass of 1.2 Mg. Plot the results measured from u = 0° to the critical angle where tipping starts to occur.

6.25 m

G2 6m

SOLUTION + ©MB = 0;

θ

- NA (4) + 18 A 103 B (9.81)(1) + 1.8 A 103 B (9.81) (2 - 6 sin u)

G1

+ 1.2 A 103 B (9.81) (2 - 12.25 sin u) = 0 NA = 58 860 - 62 539 sin u

A 2m

Tipping occurs when NA = 0, or

1m 1m

Ans.

u = 70.3° + c ©Fy = 0;

B

NB + 58 860 - 62 539 sin u - (18 + 1.8 + 1.2) A 103 B (9.81) = 0 NB = 147 150 + 62 539 sin u

Since there are two outriggers on each side of the crane, NA ¿ = (29.4 - 31.3 sin u) kN = NA 2 NB¿ =

Ans.

NB = (73.6 + 31.3 sin u) kN 2

Ans.

Ans: u = 70.3° = NA = (29.4 - 31.3 sin u) kN NB= = (73.6 + 31.3 sin u) kN 413

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5–27. Determine the reactions acting on the smooth uniform bar, which has a mass of 20 kg.

B

4m

A

30º

60º

Solution Equations of Equilibrium. NB can be determined directly by writing the moment equation of equilibrium about point A by referring to the FBD of the bar shown in Fig. a. a+ ΣMA = 0;  NB cos 30°(4) - 20(9.81) cos 30°(2) = 0

Ans.

NB = 98.1 N

Using this result to write the force equation of equilibrium along the x and y axes, + ΣFx = 0;  Ax - 98.1 sin 60° = 0 S

Ax = 84.96 N = 85.0 N

Ans.

+ c ΣFy = 0;  Ay + 98.1 cos 60° - 20(9.81) = 0 Ans.

Ay = 147.15 N = 147 N

Ans: NB = 98.1 N Ax = 85.0 N Ay = 147 N 414

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*5–28. A linear torsional spring deforms such that an applied couple moment M is related to the spring’s rotation u in radians by the equation M = (20 u) N # m. If such a spring is attached to the end of a pin-connected uniform 10-kg rod, determine the angle u for equilibrium. The spring is undeformed when u = 0°.

A

u

M  (20 u) N  m 0.5 m

Solution a+ ΣMA = 0;   - 98.1 (0.25 cos u) + 20(u) = 0 Solving for u, Ans.

u = 47.5°

Ans: u = 47.5° 415

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5–29. Determine the force P needed to pull the 50-kg roller over the smooth step. Take u = 30°.

P

u A

50 mm

300 mm

B

Solution Equations of Equilibrium. P can be determined directly by writing the moment equation of Equilibrium about point B, by referring to the FBD of the roller shown in Fig. a. a+ ΣMB = 0;  P cos 30°(0.25) + P sin 30° ( 20.32 - 0.252 2 - 50(9.81) 20.32 - 0.252 = 0 Ans.

P = 271.66 N = 272 N

Ans: P = 272 N 416

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5–30. Determine the magnitude and direction u of the minimum force P needed to pull the 50-kg roller over the smooth step.

P

u A

50 mm

300 mm

B

Solution Equations of Equilibrium. P will be minimum if its orientation produces the greatest moment about point B. This happens when it acts perpendicular to AB as shown in Fig. a. Thus u = f = cos-1 a

0.25 b = 33.56° = 33.6° 0.3

Ans.

Pmin can be determined by writing the moment equation of equilibrium about point B by referring to the FBD of the roller shown in Fig. b. a+ ΣMB = 0;  Pmin (0.3) - 50(9.81)(0.3 sin 33.56°) = 0 Ans.

Pmin = 271.13 N = 271 N

Ans: Pmin = 271 N 417

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5–31. The operation of the fuel pump for an automobile depends on the reciprocating action of the rocker arm ABC, which is pinned at B and is spring loaded at A and D. When the smooth cam C is in the position shown, determine the horizontal and vertical components of force at the pin and the force along the spring DF for equilibrium. The vertical force acting on the rocker arm at A is FA = 60 N, and at C it is FC = 125 N.

E

30° F FC = 125 N

FA = 60 N

B A

C

D

SOLUTION a + ©MB = 0;

- 60(50) - FB cos 30°(10) + 125(30) = 0

50 mm

- Bx + 86.6025 sin 30° = 0 Ans.

Bx = 43.3 N + c ©Fy = 0;

20 mm

Ans.

FB = 86.6025 = 86.6 N + ©F = 0; : x

10 mm

60 - By - 86.6025 cos 30° + 125 = 0 Ans.

By = 110 N

Ans: FB = 86.6 N Bx = 43.3 N By = 110 N 418

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*5–32. Determine the magnitude of force at the pin A and in the cable BC needed to support the 500-lb load. Neglect the weight of the boom AB.

B

8 ft C

22

A

35

SOLUTION Equations of Equilibrium: The force in cable BC can be obtained directly by summing moments about point A. a + ©MA = 0;

FBC sin 13°(8) - 500 cos 35°(8) = 0 Ans.

FBC = 1820.7 lb = 1.82 kip + Q ©Fx = 0;

A x - 1820.7 cos 13° - 500 sin 35° = 0 A x = 2060.9 lb

a + ©Fy = 0;

A y + 1820.7 sin 13° - 500 cos 35° = 0 Ay = 0

Thus,

Ans.

FA = A x = 2060.9 lb = 2.06 kip

Ans: FBC = 1.82 kip FA = 2.06 kip 419

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5–33. The dimensions of a jib crane, which is manufactured by the Basick Co., are given in the figure. If the crane has a mass of 800 kg and a center of mass at G, and the maximum rated force at its end is F 15 kN, determine the reactions at its bearings. The bearing at A is a journal bearing and supports only a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components.

3m A 0.75 m 2m

G F

SOLUTION a + ©MB = 0;

Ans.

Ax = 25.4 kN + c ©Fy = 0;

By - 800 (9.81) - 15 000 = 0 Ans.

By = 22.8 kN + ©F = 0; : x

B

Ax (2) - 800 (9.81) (0.75) - 15 000(3) = 0

Bx - 25.4 = 0 Ans.

Bx = 25.4 kN

Ans: Ax = 25.4 kN By = 22.8 kN Bx = 25.4 kN 420

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5–34. The dimensions of a jib crane, which is manufactured by the Basick Co., are given in the figure. The crane has a mass of 800 kg and a center of mass at G. The bearing at A is a journal bearing and can support a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components. Determine the maximum load F that can be suspended from its end if the selected bearings at A and B can sustain a maximum resultant load of 24 kN and 34 kN, respectively.

3m A 0.75 m 2m

F

SOLUTION a + ©MB = 0;

G

B

Ax (2) - 800 (9.81) (0.75) - F (3) = 0

+ c ©Fy = 0;

By - 800 (9.81) - F = 0

+ ©F = 0; : x

Bx - Ax = 0

Assume Ax = 24 000 N. Solving, Bx = 24 kN By = 21.9 kN Ans.

F = 14.0 kN FB =

(24)2 + (21.9)2 = 32.5 kN 6 34 kN

OK

Ans: F = 14.0 kN 421

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5–35. The smooth pipe rests against the opening at the points of contact A, B, and C. Determine the reactions at these points needed to support the force of 300 N. Neglect the pipe’s thickness in the calculation.

A C

30 300 N

0.5 m

30 0.5 m

0.26 m

B

0.15 m

Solution Equations of Equilibrium. NA can be determined directly by writing the force equation of equilibrium along the x axis by referring to the FBD of the pipe shown in Fig. a. + ΣFx = 0;  NA cos 30° - 300 sin 30° = 0  NA = 173.21 N = 173 N S

Ans.

Using this result to write the moment equations of equilibrium about points B and C, a+ ΣMB = 0;  300 cos 30°(1) - 173.21 cos 30°(0.26) - 173.21 sin 30°(0.15) - NC (0.5) = 0 Ans.

NC = 415.63 N = 416 N

a+ ΣMC = 0;  300 cos 30°(0.5) - 173.21 cos 30°(0.26) - 173.21 sin 30°(0.65) - NB(0.5) = 0 Ans.

NB = 69.22 N = 69.2 N

Ans: NA = 173 N NC = 416 N NB = 69.2 N 422

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*5–36. The beam of negligible weight is supported horizontally by two springs. If the beam is horizontal and the springs are unstretched when the load is removed, determine the angle of tilt of the beam when the load is applied.

B

A 600 N/m

kA = 1 kN/m

kB = 1.5 kN/m D

C 3m

3m

Solution Equations of Equilibrium. FA and FB can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig. a. Assuming that the angle of tilt is small, a+ ΣMA = 0;  FB(6) -

1 (600)(3)(2) = 0 FB = 300 N 2

1 a+ ΣMB = 0;   (600)(3)(4) - FA(6) = 0 FA = 600 N 2 Thus, the stretches of springs A and B can be determined from FA = kAxA;

600 = 1000 xA

xA = 0.6 m

FB = kB xB;

300 = 1500 xB

xB = 0.2 m

From the geometry shown in Fig. b u = sin-1 a

0.4 b = 3.82° 6

Ans.

The assumption of small u is confirmed.

Ans: u = 3.82° 423

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5–37. The cantilevered jib crane is used to support the load of 780 lb. If x = 5 ft, determine the reactions at the supports. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not.

8 ft B T 4 ft

x

SOLUTION

780 lb

Equations of Equilibrium: Referring to the FBD of the jib crane shown in Fig. a, we notice that NA and By can be obtained directly by writing the moment equation of equilibrium about point B and force equation of equilibrium along the y axis, respectively. a+ ©MB = 0;

NA(4) - 780(5) = 0

NA = 975 lb

Ans.

+ c ©Fy = 0;

By - 780 = 0

By = 780

Ans.

A

Using the result of NA to write the force equation of equilibrium along x axis, + : ©Fx = 0;

975 - Bx = 0

Ans.

Bx = 975 lb

Ans: NA = 975 lb Bx = 975 lb By = 780 lb 424

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5–38. The cantilevered jib crane is used to support the load of 780 lb. If the trolley T can be placed anywhere between 1.5 ft … x … 7.5 ft, determine the maximum magnitude of reaction at the supports A and B. Note that the supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, whereas the one at A does not.

8 ft B T 4 ft

x 780 lb

SOLUTION

A

Require x = 7.5 ft a + ©MA = 0;

- 780(7.5) + Bx (4) = 0 Bx = 1462.5 lb

+ ©F = 0; : x

Ax - 1462.5 = 0 Ans.

Ax = 1462.5 = 1462 lb + c ©Fy = 0;

By - 780 = 0 By = 780 lb FB = 2(1462.5)2 + (780)2 Ans.

= 1657.5 lb = 1.66 kip

Ans: Ax = 1.46 kip FB = 1.66 kip 425

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5–39. The bar of negligible weight is supported by two springs, each having a stiffness k = 100 N>m. If the springs are originally unstretched, and the force is vertical as shown, determine the angle u the bar makes with the horizontal, when the 30-N force is applied to the bar.

k 2m

1m C 30 N

B k

Solution Equations of Equilibrium. FA and FB can be determined directly by writing the moment equation of equilibrium about points B and A respectively by referring to the FBD of the bar shown in Fig. a. a+ ΣMB = 0;  30(1) - FA(2) = 0  FA = 15 N

A

a+ ΣMA = 0;  30(3) - FB(2) = 0  FB = 45 N Thus, the stretches of springs A and B can be determined from FA = kxA;

15 = 100 xA

xA = 0.15 m

FB = k xB;

45 = 100 xB

xB = 0.45 m

From the geometry shown in Fig. b, d 2 - d = ; 0.45 0.15

d = 1.5 m

Thus

u = sin-1 a

0.45 b = 17.46° = 17.5° 1.5

Ans.

Note: The moment equations are set up assuming small u, but even with non-small u the reactions come out with the same FA, FB, and then the rest of the solution goes through as before.

Ans: u = 17.5° 426

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*5–40. Determine the stiffness k of each spring so that the  30-N force causes the bar to tip u = 15° when the force is applied. Originally the bar is horizontal and the springs are unstretched. Neglect the weight of the bar.

k 2m

1m C 30 N

B k

Solution Equations of Equilibrium. FA and FB can be determined directly by writing the moment equation of equilibrium about points B and A respectively by referring to the FBD of the bar shown in Fig. a. a+ ΣMB = 0;  30(1) - FA(2) = 0  FA = 15 N

A

a+ ΣMA = 0;  30(3) - FB(2) = 0  FB = 45 N Thus, the stretches of springs A and B can be determined from FA = kxA;

15 = kxA

xA =

15 k

FB = k xB;

45 = kxB

xB =

45 k

From the geometry shown in Fig. b d 2 - d = ; 45>k 15>k

d = 1.5 m

Thus, sin 15° =

45>k 1.5

Ans.

k = 115.91 N>m = 116 N>m

Note: The moment equations are set up assuming small u, but even with non-small u the reactions come out with the same FA, FB, and then the rest of the solution goes through as before.

Ans: k = 116 N>m 427

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5–41. The bulk head AD is subjected to both water and soilbackfill pressures. Assuming AD is “pinned” to the ground at A, determine the horizontal and vertical reactions there and also the required tension in the ground anchor BC necessary for equilibrium. The bulk head has a mass of 800 kg.

D 0.5 m

B

C

F

6m 4m

118 kN/m A

SOLUTION

310 kN/m

Equations of Equilibrium: The force in ground anchor BC can be obtained directly by summing moments about point A. a + ©MA = 0;

1007.512.1672 - 23611.3332 - F162 = 0 Ans.

F = 311.375 kN = 311 kN + ©F = 0; : x

Ax + 311.375 + 236 - 1007.5 = 0 Ans.

Ax = 460 kN + c ©Fy = 0;

Ay - 7.848 = 0

Ans.

Ay = 7.85 kN

Ans: F = 311 kN Ax = 460 kN Ay = 7.85 kN 428

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5–42. The boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Set F1 = 800 N and F2 = 350 N.

C

3

5 4

SOLUTION a + ©MA = 0;

1m

- 800(1.5 cos 30°) - 350(2.5 cos 30°) +

3 4 F (2.5 sin 30°) + FCB(2.5 cos 30°) = 0 5 CB 5

4 Ax - (781.6) = 0 5

Ay - 800 - 350 +

30 A

F1

Ans.

Ax = 625 N + c ©Fy = 0;

D F2

Ans.

FCB = 781.6 = 782 N + ©F = 0; : x

1.5 m

B

3 (781.6) = 0 5 Ans.

Ay = 681 N

Ans: FCB = 782 N Ax = 625 N Ay = 681 N 429

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5–43. The boom is intended to support two vertical loads, F1 and F2. If the cable CB can sustain a maximum load of 1500 N before it fails, determine the critical loads if F1 = 2F2. Also, what is the magnitude of the maximum reaction at pin A?

C

3

5 4

SOLUTION a + ©MA = 0;

1m

- 2F2(1.5 cos 30°) - F2(2.5 cos 30°) +

4 3 (1500)(2.5 sin 30°) + (1500)(2.5 cos 30°) = 0 5 5

1.5 m

F1 = 2F2 = 1448 N

Ax -

30 A

F1

Ans.

F1 = 1.45 kN + ©F = 0; : x

D F2

Ans.

F2 = 724 N

B

4 (1500) = 0 5

Ax = 1200 N + c ©Fy = 0;

Ay - 724 - 1448 +

3 (1500) = 0 5

Ay = 1272 N FA = 2(1200)2 + (1272)2 = 1749 N = 1.75 kN

Ans.

Ans: F2 = 724 N F1 = 1.45 kN FA = 1.75 kN 430

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*5–44. The 10-kg uniform rod is pinned at end A. If it is also subjected to a couple moment of 50 N # m, determine the smallest angle u for equilibrium. The spring is unstretched when u = 0, and has a stiffness of k = 60 N>m.

B k  60 N/m

u

2m

0.5 m

A

Solution

50 N  m

Equations of Equilibrium. Here the spring stretches x = 2 sin u. The force in the spring is Fsp = kx = 60 (2 sin u) = 120 sin u. Write the moment equation of equilibrium about point A by referring to the FBD of the rod shown in Fig. a, a+ ΣMA = 0;  120 sin u cos u (2) - 10(9.81) sin u (1) - 50 = 0

240 sin u cos u - 98.1 sin u - 50 = 0

Solve numerically

Ans.

u = 24.598° = 24.6°

Ans: u = 24.6° 431

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5–45. The man uses the hand truck to move material up the step. If the truck and its contents have a mass of 50 kg with center of gravity at G, determine the normal reaction on both wheels and the magnitude and direction of the minimum force required at the grip B needed to lift the load.

0.4 m B 0.5 m 0.2 m

G

0.4 m

60 0.4 m

Solution

A

0.1 m

Equations of Equilibriums. Py can be determined directly by writing the force equation of equilibrium along y axis by referring to the FBD of the hand truck shown in Fig. a. + c ΣFy = 0;  Py - 50(9.81) = 0

Py = 490.5 N

Using this result to write the moment equation of equilibrium about point A, a+ ΣMA = 0;  Px sin 60°(1.3) - Px cos 60°(0.1) - 490.5 cos 30°(0.1)

- 490.5 sin 30°(1.3) - 50(9.81) sin 60°(0.5)



+ 50(9.81) cos 60°(0.4) = 0



Px = 442.07 N

Thus, the magnitude of minimum force P, Fig. b, is P = 2Px2 + Py2 = 2442.072 + 490.52 = 660.32 N = 660 N

Ans.

and the angle is u = tan-1 a

490.5 b = 47.97° = 48.0° b   442.07

Ans.

Write the force equation of equilibrium along x axis, + ΣFx = 0;  NA - 442.07 = 0  NA = 442.07 N = 442 N S

Ans.

Ans:  P = 660 N NA = 442 N       u = 48.0° b 432

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5–46. Three uniform books, each having a weight W and length a, are stacked as shown. Determine the maximum distance d that the top book can extend out from the bottom one so the stack does not topple over.

a

d

SOLUTION Equilibrium: For top two books, the upper book will topple when the center of gravity of this book is to the right of point A. Therefore, the maximum distance from the right edge of this book to point A is a/2. Equation of Equilibrium: For the entire three books, the top two books will topple about point B. a + ©MB = 0;

a W(a-d) -W ad- b = 0 2 d =

3a 4

Ans.

Ans: d = 433

3a 4

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5–47. Determine the reactions at the pin A and the tension in cord BC. Set F = 40 kN. Neglect the thickness of the beam.

F

26 kN

C 13

12

5

5

3

4

A

B

2m

4m

Solution a+ ΣMA = 0;   - 26 a



12 3 b(2) - 40(6) + FBC(6) = 0 13 5

Ans.

FBC = 80 kN

+ ΣFx = 0;         80 a 4 b - Ax - 26 a 5 b = 0 S 5 13



+ c ΣFy = 0;            Ay - 26 a



Ans.

Ax = 54 kN

12 3 b - 40 + 80 a b = 0 13 5

Ans.

Ay = 16 kN

Ans: FBC = 80 kN Ax = 54 kN Ay = 16 kN 434

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*5–48. If rope BC will fail when the tension becomes  50  kN, determine the greatest vertical load F that can be applied to the beam at B. What is the magnitude of the reaction at A for this loading? Neglect the thickness of the beam.

F

26 kN

C 13

12

5

5

3

4

A

B

2m

4m

Solution a+ ΣMA = 0;   -26 a



12 3 b(2) - F(6) + (50)(6) = 0 13 5

Ans.

F = 22 kN

+ ΣFx = 0;         50 a 4 b - Ax - 26 a 5 b = 0 S 5 13



+ c ΣFy = 0;            Ay - 26 a



Ans.

Ax = 30 kN

12 3 b - 22 + 50 a b = 0 13 5

Ans.

Ay = 16 kN

Ans: F = 22 kN Ax = 30 kN Ay = 16 kN 435

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5–49. The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k = 5 N>m, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C.

50 mm

50 mm

k

B C

A k

10 mm

SOLUTION ©MB = 0;

FA = FC = F

©Fy = 0;

FB = 2F

x 50 - x = yB yA

(1)

kyB 2F = F kyA (2)

2yA = yB Substituting into Eq. (1): 50 - x x = yA 2yA 2x = 50 - x x =

50 = 16.67 mm 3

x 100 - x = yA 10 Set x = 16.67, then yA = 2 mm From Eq. (2), yB = 4 mm Ans.

FC = FA = kyA = (5)(0.002) = 10 mN

Ans: FC = 10 mN 436

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5–50. The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is 0.5 N. Originally the strip is horizontal as shown.

50 mm

50 mm

k

B C

A k

10 mm

SOLUTION ©MB = 0;

FA = FC = F

©Fy = 0;

FB = 2F

x 50 - x = yA yB

(1)

kyB 2F = F kyA (2)

2yA = yB Substituting into Eq. (1): 50 - x x = yA 2yA 2x = 50 - x x =

50 = 16.67 mm 3

x 100 - x = yA 10 Set x = 16.67, then yA = 2 mm From Eq. (2), yB = 4 mm FC = FA = kyA 0.5 = k(0.002) Ans.

k = 250 N/m

Ans: k = 250 N>m 437

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5–51. The cantilever footing is used to support a wall near its edge A so that it causes a uniform soil pressure under the footing. Determine the uniform distribution loads wA and wB, measured in lb ft at pads A and B, necessary to support the wall forces of 8 000 lb and 20 000 lb.

20 000 lb 8000 lb

0.25 ft

1.5 ft B

A

SOLUTION a + ©MA = 0;

- 8000 (10.5) + wB (3)(10.5) + 20 000 (0.75) = 0

8 ft

wB 3 ft

Ans.

wB = 2190.5 lb/ft = 2.19 kip/ft + c ©Fy = 0;

wA 2 ft

2190.5 (3) - 28 000 + wA (2) = 0 Ans.

wA = 10.7 kip/ft

Ans: wB = 2.19 kip>ft wA = 10.7 kip>ft 438

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*5–52. The uniform beam has a weight W and length l and is supported by a pin at A and a cable BC. Determine the horizontal and vertical components of reaction at A and the tension in the cable necessary to hold the beam in the position shown.

C l B A

SOLUTION Equations of Equilibrium: The tension in the cable can be obtained directly by summing moments about point A. a + ©MA = 0;

l T sin 1f - u2l - W cos u a b = 0 2 T =

W cos u 2 sin 1f - u2

Using the result T =

W cos u 2 sin 1f - u2

+ ©F = 0; : x

W cos u b cos f - Ax = 0 2 sin 1f - u2

a

Ax = + c ©Fy = 0;

Ay + a Ay =

Ans.

W cos f cos u 2 sin 1f - u2

Ans.

W cos u b sin f - W = 0 2 sin 1f - u2

W1sin f cos u - 2 cos f sin u2

Ans.

2 sin f - u

Ans: W cos u 2 sin(f - u) Wcos f cos u Ax = 2 sin(f - u) W(sin f cos u - 2 cos f sin u) Ay = 2 sin (f - u) T =

439

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5–53. A boy stands out at the end of the diving board, which is supported by two springs A and B, each having a stiffness of k = 15kN>m. In the position shown the board is horizontal. If the boy has a mass of 40 kg, determine the angle of tilt which the board makes with the horizontal after he jumps off. Neglect the weight of the board and assume it is rigid.

1m A

SOLUTION

3m B

Equations of Equilibrium: The spring force at A and B can be obtained directly by summing moments about points B and A, respectively. a + ©MB = 0;

FA (1) - 392.4(3) = 0

FA = 1177.2 N

a + ©MA = 0;

FB (1) - 392.4(4) = 0

FB = 1569.6 N

Spring Formula: Applying ¢ = ¢A =

F , we have k

1177.2 = 0.07848 m 15(103)

¢B =

1569.6 = 0.10464 m 15(103)

Geometry: The angle of tilt a is a = tan - 1 a

0.10464 + 0.07848 b = 10.4° 1

Ans.

Ans: a = 10.4° 440

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5–54. The 30-N uniform rod has a length of l = 1 m. If s = 1.5 m, determine the distance h of placement at the end A along the smooth wall for equilibrium.

C

SOLUTION

h

Equations of Equilibrium: Referring to the FBD of the rod shown in Fig. a, write the moment equation of equilibrium about point A. a + ©MA = 0;

A

T sin f(1) - 3 sin u(0.5) = 0 T =

s

1.5 sin u sin f

l B

Using this result to write the force equation of equilibrium along y axis, + c ©Fy = 0;

a

15 sin u b cos (u - f) - 3 = 0 sin f (1)

sin u cos (u - f) - 2 sin f = 0

Geometry: Applying the sine law with sin (180° - u) = sin u by referring to Fig. b, sin f sin u = ; h 1.5

sin u = a

h b sin u 1.5

(2)

Substituting Eq. (2) into (1) yields sin u[cos (u - f) -

4 h] = 0 3

since sin u Z 0, then cos (u - f) - (4>3)h

(3)

cos (u - f) = (4>3)h

Again, applying law of cosine by referring to Fig. b, l2 = h2 + 1.52 - 2(h)(1.5) cos (u - f) cos (u - f) =

h2 + 1.25 3h

(4)

Equating Eqs. (3) and (4) yields h2 + 1.25 4 h = 3 3h 3h2 = 1.25 Ans.

h = 0.645 m

Ans: h = 0.645 m 441

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5–55. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Determine the placement h for equilibrium.

C

h s A

SOLUTION Equations of Equilibrium: The tension in the cable can be obtained directly by summing moments about point A.

l B

a + ©MA = 0;

l T sin f(l) - W sin u a b = 0 2 T =

Using the result T = + c ©Fy = 0;

W sin u 2 sin f

W sin u , 2 sin f

W sin u cos (u - f) - W = 0 2 sin f (1)

sin u cos (u - f) - 2 sin f = 0 Geometry: Applying the sine law with sin (180° - u) = sin u, we have sin f sin u = s h

sin f =

h sin u s

(2)

Substituting Eq. (2) into (1) yields cos (u - f) =

2h s

(3)

Using the cosine law, l2 = h2 + s2 - 2hs cos (u - f) cos (u - f) =

h2 + s2 - l2 2hs

(4)

Equating Eqs. (3) and (4) yields h2 + s 2 - l2 2h = s 2hs h =

A

s2 - l2 3

Ans.

Ans: h = 442

s2 - l 2 A 3

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*5–56. The uniform rod of length L and weight W is supported on the smooth planes. Determine its position u for equilibrium. Neglect the thickness of the rod.

L u f

c

SOLUTION a + ©MB = 0;

- Wa

L cos ub + NA cos f (L cos u) + NA sin f (L sin u) = 0 2 NA =

W cos u 2 cos (f - u)

+ ©F = 0; : x

NB sin c - NA sin f = 0

+ c ©Fy = 0;

NB cos c + NA cos f - W = 0 NB =

(1) (2)

W - NA cos f cos c

(3)

Substituting Eqs. (1) and (3) into Eq. (2): aW -

W cos u sin f W cos u cos f b tan c = 0 2 cos (f - u) 2 cos (f - u)

2 cos (f - u) tan c - cos u tan c cos f - cos u sin f = 0 sin u (2 sin f tan c) - cos u (sin f - cos f tan c) = 0 tan u =

sin f - cos f tan c 2 sin f tan c

u = tan - 1 a

1 1 cot c - cot f b 2 2

Ans.

Ans: 1

u = tan - 1 a 2 cot c 443

1 cot f b 2

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5–57. The beam is subjected to the two concentrated loads. Assuming that the foundation exerts a linearly varying load distribution on its bottom, determine the load intensities w1 and w2 for equilibrium if P = 500 lb and L = 12 ft.

P L –– 3

2P L –– 3

L –– 3

w1 w2

SOLUTION Equations of Equilibrium: Referring to the FBD of the beam shown in Fig. a, we notice that W1 can be obtained directly by writing moment equations of equilibrium about point A. a + ©MA = 0;

500(4) - W1(12)(2) = 0 Ans.

W1 = 83.33 lb>ft = 83.3 lb>ft Using this result to write the force equation of equilibrium along y axis, + c ©Fy = 0;

83.33(12) +

1 (W2 - 83.33)(12) - 500 - 1000 = 0 2 Ans.

W2 = 166.67 lb>ft = 167 lb>ft

Ans: w1 = 83.3 lb>ft w2 = 167 lb>ft 444

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5–58. The beam is subjected to the two concentrated loads. Assuming that the foundation exerts a linearly varying load distribution on its bottom, determine the load intensities w1 and w2 for equilibrium in terms of the parameters shown.

P L –– 3

SOLUTION

2P L –– 3

L –– 3

w1

Equations of Equilibrium: The load intensity w1 can be determined directly by summing moments about point A. a + ©MA = 0;

Pa

L L b - w1L a b = 0 3 6 w1 =

+ c ©Fy = 0;

w2

2P L

Ans.

2P 2P 1 a w2 bL + 1L2 - 3P = 0 2 L L w2 =

4P L

Ans.

Ans: w1 = 445

2P 4P ,w = L 2 L

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5–59. The rod supports a weight of 200 lb and is pinned at its end A. If it is also subjected to a couple moment of 100 lb # ft, determine the angle u for equilibrium. The spring has an unstretched length of 2 ft and a stiffness of k = 50 lb/ft.

100 lb ft A

2 ft k

50 lb/ft

u 3 ft

3 ft

B

SOLUTION a + ©MA = 0;

100 + 200 (3 cos u) - Fs (6 cos u) = 0

Fs = kx;

Fs = 50 (6 sin u) 100 + 600 cos u - 1800 sin u cos u = 0 cos u - 1.5 sin 2u + 0.1667 = 0

Solving by trial and error, Ans.

u = 23.2° and u = 85.2°

Ans: u = 23.2° 85.2° 446

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*5–60. Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position u as shown. Neglect the weight of the bar.

d P

u a

SOLUTION + c ©Fy = 0;

R cos u - P = 0

a + ©MA = 0;

- P(d cos u) + R a Rd cos2 u = R a d =

a b = 0 cos u

a b cos u

a cos3 u

Ans.

Also; Require forces to be concurrent at point O. AO = d cos u =

a>cos u cos u

Thus, d =

a cos3 u

Ans.

Ans: d = 447

a cos3 u

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5–61. If d = 1 m, and u = 30°, determine the normal reaction at the smooth supports and the required distance a for the placement of the roller if P = 600 N. Neglect the weight of the bar.

d

SOLUTION

u

Equations of Equilibrium: Referring to the FBD of the rod shown in Fig. a, a+ ©MA = 0;

a b - 600 cos 30°(1) = 0 cos 30° 450 NB = a

P

a

NB = a

a+ ©Fy¿ = 0;

NB - NA sin 30° - 600 cos 30° = 0 NB - 0.5NA = 600 cos 30°

+Q ©F = 0; x¿

NA cos 30° - 600 sin 30° = 0 NA = 346.41 N = 346 N

(1) (2) Ans.

.

Ans.

.

Ans.

.

Substitute this result into Eq (2), NB - 0.5(346.41) = 600 cos 30° NB = 692.82 N = 693 N Substitute this result into Eq (1), 450 a a = 0.6495 m 692.82 =

a = 0.650 m

Ans: NA = 346 N NB = 693 N a = 0.650 m 448

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5–62. The uniform load has a mass of 600 kg and is lifted using a uniform 30-kg strongback beam BAC and four wire ropes as shown. Determine the tension in each segment of rope and the force that must be applied to the sling at A.

F 1.25 m B

SOLUTION

A C

2m

Equations of Equilibrium: Due to symmetry, all wires are subjected to the same tension. This condition statisfies moment equilibrium about the x and y axes and force equilibrium along y axis. ©Fz = 0;

1.25 m

1.5 m 1.5 m

4 4T a b - 5886 = 0 5 Ans.

T = 1839.375 N = 1.84 kN

The force F applied to the sling A must support the weight of the load and strongback beam. Hence ©Fz = 0;

F - 60019.812 - 3019.812 = 0 Ans.

F = 6180.3 N = 6.18 kN

Ans: T = 1.84 kN F = 6.18 kN 449

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5–63. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA = 45 000 lb, WB = 8000 lb, and WC = 6000 lb, determine the normal reactions of the wheels D, E, and F on the ground.

z

D

B

A C E

SOLUTION ©Mx = 0;

8000(6) - RD (14) - 6000(8) + RE (14) = 0

©My = 0;

8000(4) + 45 000(7) + 6000(4) - RF (27) = 0

©Fz = 0;

8 ft

6 ft x

F 8 ft

20 ft 6 ft

4 ft 3 ft

y

RD + RE + RF - 8000 - 6000 - 45 000 = 0

Solving, RD = 22.6 kip

Ans.

RE = 22.6 kip

Ans.

RF = 13.7 kip

Ans.

Ans: RD = 22.6 kip RE = 22.6 kip RF = 13.7 kip 450

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*5–64. z

Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, and z axes, respectively.

600 N

1m 400 N

0.5 m

0.75 m

0.75 m

Solution

A

500 N

x

Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a ΣFx = 0;  Ax - 400 = 0  Ax = 400 N

Ans.

ΣFy = 0;  500 - Ay = 0  Ay = 500 N

Ans.

ΣFz = 0;  Az - 600 = 0  Az = 600 N

Ans.

y

ΣMx = 0;  (MA)x - 500(1.25) - 600(1) = 0 (MA)x = 1225 N # m = 1.225 kN # m

Ans.

ΣMy = 0;  (MA)y - 400(0.75) - 600(0.75) = 0 (MA)y = 750 N # m

Ans.

ΣMz = 0;  (MA)z = 0

Ans.

Ans: Ax = 400 N Ay = 500 N Az = 600 N (MA)x = 1.225 kN # m (MA)y = 750 N # m (MA)z = 0 451

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5–65. z

The 50-lb mulching machine has a center of gravity at G. Determine the vertical reactions at the wheels C and B and the smooth contact point A. G

4 ft

SOLUTION Equations of Equilibrium: From the free-body diagram of the mulching machine, Fig. a, NA can be obtained by writing the moment equation of equilibrium about the y axis.

1.25 ft 1.25 ft

C x

A

B 1.5 ft

©My = 0; 50(2) - NA(1.5 + 2) = 0

2 ft y

Ans.

NA = 28.57 lb = 28.6 lb

Using the above result and writing the moment equation of equilibrium about the x axis and the force equation of equilibrium along the z axis, we have ©Mx = 0; NB(1.25) - NC(1.25) = 0

(1)

©Fz = 0;

(2)

NB + NC + 28.57 - 50 = 0

Solving Eqs. (1) and (2) yields Ans.

NB = NC = 10.71 lb = 10.7 lb

Note: If we write the force equation of equilibrium ©Fx = 0 and ©Fy = 0 and the moment equation of equilibrium ©Mz = 0. This indicates that equilibrium is satisfied.

Ans: NA = 28.6 lb NB = 10.7 lb, NC = 10.7 lb 452

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5–66. z

The smooth uniform rod AB is supported by a ball-and-socket joint at A, the wall at B, and cable BC. Determine the components of reaction at A, the tension in the cable, and the normal reaction at B if the rod has a mass of 20 kg.

C 0.5 m B

2m

Solution A

Force And Position Vectors. The coordinates of points A, B and G are A(1.5, 0, 0) m, B(0, 1, 2) m, C(0, 0, 2.5) m and G(0.75, 0.5, 1) m x

FA = - Axi + Ay j + Azk TBC = TBC a NB = NBi

1m

(0 - 1)j + (2.5 - 2)k rBC 1 0.5 b = TBC c = TBC j + TBC k 2 2 rBC 11.25 11.25 2(0 - 1) + (2.5 - 2)

W = { -20(9.81)k} N rAG = (0.75 - 1.5)i + (0.5 - 0)j + (1 - 0)k = { - 0.75i + 0.5j + k} m rAB = (0 - 1.5)i + (1 - 0)j + (2 - 0)k = { - 1.5i + j + 2k} m Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FA + TBC + NB + W = 0 ( - Ax + NB)i + aAy -

1 0.5 TBC bj + c Az + TBC - 20 (9.81)d k = 0 11.25 11.25

Equating i, j and k components,

(1)

   - Ax + NB = 0   Ay   Az +

1 TBC = 0 11.25

(2)

0.5 TBC - 20(9.81) = 0 11.25

(3)

The moment equation of equilibrium gives ΣMA = 0;  rAG * W + rAB * (TBC + NB) = 0 i † - 0.75 0 a

j 0.5 0

k i 1 † + † - 1.5 - 20(9.81) NB

j 1 -

1 11.25

k 2 TBC

0.5 11.25

TBC

† =0

0.5 2 0.75 1.5 TBC + TBC - 98.1bi + a TBC + 2NB - 147.15bj + a TBC - NB bk = 0 11.25 11.25 11.25 11.25

453

1.5 m

y

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5–66. Continued

Equating i, j and k Components

0.5 2 TBC + TBC - 98.1 = 0 11.25 11.25

(4)

0.75 TBC + 2NB - 147.15 = 0 11.25

(5)

1.5 TBC - NB = 0 11.25

(6)

Solving Eqs. (1) to (6)

TBC = 43.87 N = 43.9 N

Ans.



NB = 58.86 N = 58.9 N

Ans.



Ax = 58.86 N = 58.9 N

Ans.



Ay = 39.24 N = 39.2 N

Ans.



Az = 176.58 N = 177 N

Ans.

Note: One of the equations (4), (5) and (6) is redundant that will be satisfied automatically.

Ans: TBC = NB =  Ax =  Ay =  Az = 454

43.9 N 58.9 N 58.9 N 39.2 N 177 N

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5–67. z

The uniform concrete slab has a mass of 2400 kg. Determine the tension in each of the three parallel supporting cables when the slab is held in the horizontal plane as shown.

TA

TB 15 kN

x

0.5 m

TC A

B

2m C

1m

1m

2m

y

Solution Equations of Equilibrium. Referring to the FBD of the slab shown in Fig. a, we notice that TC can be obtained directly by writing the moment equation of equilibrium about the x axis. ΣMx = 0; TC (2.5) - 2400(9.81)(1.25) - 15 ( 103 ) (0.5) = 0

Ans.

TC = 14,772 N = 14.8 kN

Using this result to write moment equation of equilibrium about y axis and force equation of equilibrium along z axis, ΣMy = 0; TB (2) + 14,772(4) - 2400(9.81)(2) - 15 ( 103 ) (3) = 0 Ans.

TB = 16,500 N = 16.5 kN ΣFz = 0; TA + 16,500 + 14,772 - 2400(9.81) - 15 ( 103 ) = 0

Ans.

TA = 7272 N = 7.27 kN

Ans: TC = 14.8 kN TB = 16.5 kN TA = 7.27 kN 455

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*5–68. z

The 100-lb door has its center of gravity at G. Determine the components of reaction at hinges A and B if hinge B resists only forces in the x and y directions and A resists forces in the x, y, z directions.

18 in.

B 24 in.

SOLUTION Equations of Equilibrium: From the free-body diagram of the door, Fig. a, By, Bx, and Az can be obtained by writing the moment equation of equilibrium about the x¿ and y¿ axes and the force equation of equilibrium along the z axis.

G

A 18 in.

-By(48) - 100(18) = 0

©Mx¿ = 0;

24 in.

By = - 37.5 lb

Ans.

©My¿ = 0;

Bx = 0

Ans.

©Fz = 0;

- 100 + A z = 0;

A z = 100 lb

30 x

y

Ans.

Using the above result and writing the force equations of equilibrium along the x and y axes, we have Ans.

©Fx = 0;

Ax = 0

©Fy = 0;

A y + ( -37.5) = 0

A y = 37.5 lb

Ans.

The negative sign indicates that By acts in the opposite sense to that shown on the free-body diagram. If we write the moment equation of equilibrium ©Mz = 0, it shows that equilibrium is satisfied.

Ans: By = Bx = Az = Ax = Ay = 456

- 37.5 lb 0 100 lb 0 37.5 lb

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5–69. Determine the tension in each cable and the components of reaction at D needed to support the load.

z B 3m

6m

C 2m D

x

A y

Solution

30

Force And Position Vectors. The coordinates of points A, B, and C are A(6, 0, 0) m, B(0, -3, 2) m and C(0, 0, 2) m respectively. FAB

(0 - 6)i + ( - 3 - 0)j + (2 - 0)k rAB 6 3 2 = FAB a b = FAB c d = - FABi - FAB j + FABk rAB 7 7 7 1(0 - 6)2 + ( - 3 - 0)2 + (2 - 0)2

FAC = FAC a

(0 - 6)i + (2 - 0)k rAC 6 2 b = FAC c d = FAC i + FAC k 2 2 rAC 140 140 1(0 - 6) + (2 - 0)

F = 400 (sin 30°j - cos 30°k) = {200j - 346.41k}N FD = Dxi + Dy j + Dzk rDA = {6i} m

Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FAB + FAC + F + FD = 0 6 6 3 a- FAB FAC + Dx bi + a- FAB + Dy + 200bj 7 7 140

2 2 FAC + Dz - 346.41bk = 0 + a FAB + 7 140

457

400 N

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5–69. Continued

Equating i, j and k components, 6 6 - FAB FAC + Dx = 0 7 140

(1)

3 - FAB + Dy + 200 = 0 7

(2)

2 2 F + FAC + Dz - 346.41 = 0 7 AB 140

(3)

Moment equation of equilibrium gives

ΣMD = 0;  rDA * (FAB + FAC + F) = 0

5

i 6

j 0

k 0

6 6 a- FAB F b 140 AC 7

3 a- FAB + 200b 7

2 2 FAC - 346.41b a FAB + 140 7

5 = 0

2 2 3 -6 a FAB + FAC - 346.41bj + 6 a- FAB + 200b k = 0 7 7 140

Equating j and k Components,

2 2 - 6 a FAB + FAC - 346.41b = 0 7 140

(4)

3 6 a - FAB + 200b = 0 7

(5)

Solving Eqs. (1) to (5)

FAB = 466.67 N = 467 N

Ans.

FAC = 673.81 N = 674 N

Ans.

Dx = 1039.23 N = 1.04 kN

Ans.

Dy = 0



Ans.

Dz = 0



Ans.

Ans: FAB = FAC =  Dx =  Dy = Dz = 458

467 N 674 N 1.04 kN 0 0

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5–70. z

The stiff-leg derrick used on ships is supported by a ball-andsocket joint at D and two cables BA and BC. The cables are attached to a smooth collar ring at B, which allows rotation of the derrick about z axis. If the derrick supports a crate having a mass of 200 kg, determine the tension in the cables and the x, y, z components of reaction at D.

B

6m

7.5 m

C 6m D

Solution 2 6 T - TBC 7 BA 9 3 3 ΣFy = 0;      Dy - TBA - TBC 7 9 6 6 ΣFz = 0;      Dz - TBA - TBC 7 9 3 3 ΣMx = 0;   TBA(6) + TBC(6) 7 9 2 6 ΣMy = 0;   TBA(6) - TBC(6) 7 9 ΣFx = 0;      Dx +

1m

2m

y

A

= 0

3m

4m

x

= 0 - 200(9.81) = 0 - 200(9.81)(4) = 0 + 200(9.81)(1) = 0



TBA = 2.00 kN

Ans.



TBC = 1.35 kN

Ans.



Dx = 0.327 kN

Ans.



Dy = 1.31 kN

Ans.



Dz = 4.58 kN

Ans.

Ans: TBA = 2.00 kN TBC = 1.35 kN Dx = 0.327 kN Dy = 1.31 kN Dz = 4.58 kN 459

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5–71. z

Determine the components of reaction at the ball-and-socket joint A and the tension in each cable necessary for equilibrium of the rod.

C

2m 2m D A

x

Solution

B

3m

Force And Position Vectors. The coordinates of points A, B, C, D and E are A(0, 0, 0), B(6, 0, 0), C(0, -2, 3) m, D(0, 2, 3) m and E(3, 0, 0) m respectively. FBC = FBC a

(0 - 6)i + ( - 2 - 0)j + (3 - 0)k rBC 6 2 3 b = FBC £ § = - FBCi - FBC j + FBCk 2 2 2 rBC 7 7 7 2(0 - 6) + ( - 2 - 0) + (3 - 0)

FBD = FBD a

(0 - 6)i + (2 - 0)j + (3 - 0)k rBD 6 2 3 b = FBD £ § = - FBDi + FBD j + FBDk 2 2 2 rBD 7 7 7 2(0 - 6) + (2 - 0) + (3 - 0)

FA = Axi + Ay j + Azk F = { - 600k} N rAB = {6i} m

3m E

rAE = {3i} m

Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FBC + FBD + FA + F = 0 6 6 2 2 3 3 a - FBC - FBD + Ax bi + a FBD - FBC + Ay b j + a FBC + FBD + Az - 600bk = 0 7 7 7 7 7 7

460

600 N

3m y

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5–71. Continued

Equating i, j and k components, 6 6 - FBC - FBD + Ax = 0 7 7

(1)

2 2 F - FBC + Ay = 0 7 BD 7

(2)

3 3 F + FBD + Az - 600 = 0 7 BC 7

(3)

The moment equation of equilibrium gives ΣMA = 0;  rAE * F + rAB * (FBC + FBD ) = 0 i †3 0

j 0 0

k 0 † + 5 -600

c 1800 -

i 6

j 0

k 0

6 - (FBC + FBD) 7

2 (F - FBC) 7 BD

3 (F + FBD) 7 BC

5 = 0

18 12 (FBC + FBD) d j + (F - FBC)k = 0 7 7 BD

Equating j and k components, 1800 -

18 (F + FBD) = 0 7 BC

(4)

12 (F - FBC) = 0 7 BD

(5)

Solving Eqs. (1) to (5), FBD = FBC = 350 N

Ans.

Ax = 600 N

Ans.

Ay = 0

Ans.

Az = 300 N

Ans.

Ans: FBD = Ax = Ay = Az = 461

FBC = 350 N 600 N 0 300 N

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*5–72. z

Determine the components of reaction at the ball-andsocket joint A and the tension in the supporting cables DB and DC.

1.5 m B 1.5 m C D

Solution

1.5 m

Force And Position Vectors. The coordinates of points A, B, C, and D are A(0, 0, 0), B(0, - 1.5, 3) m, C(0, 1.5, 3) m and D(1, 0, 1) m, respectively. x FDC

(0 - 1)i + (1.5 - 0)j + (3 - 1)k rDC = FDC a b = FDC £ § rDC 2(0 - 1)2 + (1.5 - 0)2 + (3 - 1)2 = -

FDB = FDB a

1 1.5 2 FCDi + FDC j + FDC k 17.25 17.25 17.25

(0 - 1)i + ( - 1.5 - 0)j + (3 - 1)k rDB b = FDB £ § rDB 2(0 - 1)2 + ( - 1.5 - 0)2 + (3 - 1)2 = -

1 1.5 2 FDBi + FDB j + FDBk 17.25 17.25 17.25

FA = Axi + Ay j + Azk F = { - 2400k} N

rAD = (1 - 0)i + (1 - 0)k = {i + k} m rF = {4i} m

462

800 N/m

3m

1m

1.5 m

A

3m

1m y

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*5–72. Continued

Equations of Equilibrium. Referring to the FBD of the assembly shown in Fig. a. Force equation of equilibrium gives ΣF = 0;  FDC + FDB + FA + F = 0 a-

1.5 1 1 1.5 FDC FDB + Ax bi + a FDC FDB + Ay bj 17.25 17.25 17.25 17.25 + a

2 2 FDC + FDB + Az - 2400bk = 0 17.25 17.25

Equating i, j and k components, -

1 1 FDC FDB + Ax = 0 17.25 17.25

(1)

1.5 1.5 FDC F + Ay = 0 7.25 DB 17.25

(2)

2 2 FDC + FDB + Az - 2400 = 0 17.25 17.25 Moment equation of equilibrium gives

(3)

ΣMA = 0;  rF * F + rAD * (FDB + FDC) = 0 i †4 0

-

j 0 0

k 0 † + 5 -2400

i 1 -

1 (FDB + FDC) 17.25

j 0

k 1

1.5 (FDC - FDB) 17.25

2 (FDC + FDB) 17.25

5 = 0

3 1.5 1.5 (FDC - FDB)i + c 9600 (FDC + FDB) d j + (FDC + FDB)k = 0 17.25 17.25 17.25

Equating i, j and k Components -

1.5 (FDC - FDB) = 0 17.25

9600 -

(4)

3 (FDC + FDB) = 0 17.25

(5)

1.5 (FDC - FDB) = 0 17.25

(6)

Solving Eqs. (1) to (6)

FDC = FDB = 4308.13 N = 4.31 kN

Ans.

Ax = 3200 N = 3.20 kN

Ans.

Ay = 0

Ans.

Az = - 4000 N = -4 kN

Ans.

Negative sign indicates that Az directed in the sense opposite to that shown in FBD.

463

Ans: FDC = Ax = Ay = Az =

FDB = 4.31 kN 3.20 kN 0 -4 kN

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5–73. z

The bent rod is supported at A, B, and C by smooth journal bearings. Determine the components of reaction at the bearings if the rod is subjected to the force F = 800 N. The bearings are in proper alignment and exert only force reactions on the rod.

C A

2m

2m

x

Solution

B 0.75 m

1m

30 60

Equations of Equilibrium. The x, y and z components of force F are

F

Fx = 800 cos 60° cos 30° = 346.41 N

y

Fy = 800 cos 60° sin 30° = 200 N Fz = 800 sin 60° = 692.82 N Referring to the FBD of the bent rod shown in Fig. a, ΣMx = 0;

- Cy(2) + Bz(2) - 692.82 (2) = 0

(1)

ΣMy = 0;

Bz(1) + Cx(2) = 0

(2)

ΣMz = 0;

- Cy(1.75) - Cx(2) - By(1) - 346.41(2) = 0

(3)

ΣFx = 0;

Ax + Cx + 346.41 = 0

(4)

ΣFy = 0;

200 + By + Cy = 0

(5)

ΣFz = 0;

Az + Bz - 692.82 = 0

(6)

Solving Eqs. (1) to (6) Cy = 800 N Bz = - 107.18 N = 107 N By = 600 N

Ans.

Cx = 53.59 N = 53.6 N Ax = 400 N Az = 800 N

Ans.

The negative signs indicate that Cy, Bz and Az are directed in the senses opposite to those shown in FBD.

Ans: Cy = Bz = By = Cx = Ax = Az = 464

800 N 107 N 600 N 53.6 N 400 N 800 N

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5–74. The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F which will cause the positive x component of reaction at the bearing C to be Cx = 50 N. The bearings are in proper alignment and exert only force reactions on the rod.

z

C A

2m

2m

Solution

x

B 0.75 m

1m

30 60

Equations of Equilibrium. The x, y and z components of force F are

F

Fx = F cos 60° cos 30° = 0.4330 F

y

Fy = F cos 60° sin 30° = 0.25 F Fz = F sin 60° = 0.8660 F Here, it is required that Cx = 50. Thus, by referring to the FBD of the beat rod shown in Fig. a, ΣMx = 0;

(1)

-Cy(2) + Bz(2) - 0.8660 F(2) = 0

ΣMy = 0; ΣMz = 0;

(2)

Bz(1) + 50(2) = 0

(3)

-Cy(1.75) - 50(2) - By(1) - 0.4330 F(2) = 0

ΣFy = 0;

(4)

0.25 F + By + Cy = 0

Solving Eqs. (1) to (4) Ans.

F = 746.41 N = 746 N Cy = -746.41 N Bz = - 100 N By = 559.81 N

Ans: F = 746 N 465

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5–75. z

Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the collar fixed to the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the rod in equilibrium.

1.5 m 400 N

3m

A y C

200 N

Solution x

Force And Position Vectors. The coordinates of points B and C are B(3, 0, - 1) m C(0, 1.5, 0) m, respectively. TBC = TBC a

B

(0 - 3)i + (1.5 - 0)j + [0 - ( - 1)]k rBC b = TBC • ¶ rBC 2(0 - 3)2 + (1.5 - 0)2 + [0 - ( -1)]2

6 3 2 = - TBC i + TBC j + T k 7 7 7 BC

F = {200j - 400k} N FA = Ax i + Ay j MA = (MA)x i + (MA)y j + (MA)z k r1{3 i} m r2 = {1.5 j} m Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a, the force equation of equilibrium gives ΣF = 0; TBC + F + FA = 0 6 3 2 a - TBC + Ax bi + a TBC + 200 + Ayb j + a TBC - 400bk = 0 7 7 7

Equating i, j and k components -

6 T + Ax = 0 7 BC

(1)

3 T + 200 + Ay = 0 7 BC

(2)

2 T - 400 = 0 7 BC

(3)

466

1m

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5–75. Continued

The moment equation of equilibrium gives ΣMA = O;  MA + r1 * F + r2 * TBC = 0 i

( MA ) x i + ( MA ) y j + ( MA ) zk + † 3 0

j 0 200

k 0 † + 5 - 400

i 0 6 - TBC 7

j 1.5 3 TBC 7

k 0 2 TBC 7

3 9 T d i + 3 ( MA ) y + 1200 4 j + c ( MA ) z + TBC + 600 d k = 0 7 BC 7 Equating i, j, and k components, 3 ( MA ) x + TBC = 0 7

5 = 0

c ( MA ) x +

( MA ) y + 1200 = 0

(4) (5)

9 TBC + 600 = 0 7 Solving Eqs. (1) to (6),

( MA ) z +

(6)

Ans.

TBC = 1400 N = 1.40 kN Ay = 800 N

Ans.

Ax = 1200 N = 1.20 kN

Ans.

( MA ) x = 600 N # m

Ans.

( MA ) y = - 1200 N # m = 1.20 kN # m

Ans.

( MA ) z = - 2400 N # m = 2.40 kN # m

Ans.

The negative signs indicate that Ay, ( MA ) x, ( MA ) y and ( MA ) z are directed in sense opposite to those shown in FBD.

Ans: TBC = 1.40 kN Ay = 800 N Ax = 1.20 kN ( MA ) x = 600 N # m ( MA ) y = 1.20 kN # m ( MA ) z = 2.40 kN # m 467

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*5–76. z

The member is supported by a pin at A and cable  BC. Determine the components of reaction at these supports if the cylinder has a mass of 40 kg.

0.5 m B

1m A D

Solution

y 1m

Force And Position Vectors. The coordinates of points B, C and D are B(0, -0.5, 1) m, x C(3, 1, 0) m and D(3, - 1, 0) m, respectively. FCB = FCB a

(0 - 3)i + ( -0.5 - 1)j + (1 - 0)k rCB b = FCB c d rCB 2(0 - 3)2 + ( - 0.5 - 1)2 + (1 - 0)2 6 3 2 = - FCBi - FCBj + FCBk 7 7 7

W = { - 40(9.81)k} N = { -392.4k} N. FA = Ax i + Ay j + Az k MA = ( MA ) x i + ( MA ) z k rAC = {3i + j} m  rAD = {3i - j} m Equations of Equilibrium. Referring to the FBD of the assembly shown in Fig. a. the force equation of equilibrium gives ΣF = 0;  FCB + W + FA = 0; 6 3 2 a - FCB + Ax bi + a - FCB + Ay bj + a FCB + Az - 392.4bk = 0 7 7 7

Equating i, j and k components

6 - FCB + Ax = 0 7 3 - FCB + Ay = 0 7 2 FCB + Az - 392.4 = 0 7 The moment equation of equilibrium gives

(1) (2) (3)

ΣMA = 0;  rAC * FCB + rAD * W + MA = 0 i 3 5 6 - FCB 7

1m

j 1 3 - FCB 7

k i 0 5 + †3 2 0 FCB 7

j -1 0

k 0 † + ( MA ) x i + ( MA ) Z k = 0 - 392.4

2 6 9 6 c FCB + 392.4 + ( MA ) x d i + a - FCB + 1177.2bj + c - FCB + FCB + ( MA ) z d k = 0 7 7 7 7

468

C

3m

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*5–76. Continued

Equating i, j and k components, 2 F + 392.4 + ( MA ) x = 0 7 CB 6 - FCB + 1177.2 = 0 7 9 6 - FCB + FCB + ( MA ) z = 0 7 7 Solving Eqs (1) to (6),

(4) (5) (6)

Ans.

FCB = 1373.4 N = 1.37 kN

( MA ) x = - 784.8 N # m = 785 N # m

Ans.

( MA ) z = 588.6 N # m = 589 N # m

Ans.

Ax = 1177.2 N = 1.18 kN

Ans.

Ay = 588.6 N = 589 N

Ans.

Az = 0

Ans.

Ans: FCB ( MA ) x ( MA ) z Ax Ay Az 469

= = = = = =

1.37 kN 785 N # m 589 N # m 1.18 kN 589 N 0

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5–77. z

The member is supported by a square rod which fits loosely through the smooth square hole of the attached collar at A and by a roller at B. Determine the components of reaction at these supports when the member is subjected to the loading shown.

A

x

B

1m

2m y

2m

Solution

C

Force And Position Vectors. The coordinates of points B and C are B(2,0,0) m and C(3,0,-2) m.

300 N

FA = -Ax i - Ay j F = {300i + 500j - 400k} N NB = NB k MA = - ( MA ) x i + ( MA ) y j - ( MA ) z k rAB = {2i} m  rAC = {3i - 2k} m Equations of Equilibrium. Referring to the FBD of the member shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FA + F + NB = 0

( 300 - Ax ) i + ( 500 - Ay ) j + ( NB - 400 ) k = 0

470

500 N 400 N

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5–77. Continued

Equating i, j and k components, 300 - Ax = 0  Ax = 300 N

Ans.

500 - Ay = 0  Ay = 500 N

Ans.

NB - 400 = 0 

Ans.

     NB = 400 N

The moment equation of equilibrium gives ΣMA = 0;  MA + rAB * NB + rAC * F = 0 i - ( MA ) x i + ( MA ) y j - ( MA ) z k + † 2 0

3 1000

- ( MA ) x 4 i +

3 ( MA ) y

Equating i, j and k components,

j 0 0

- 200 4 j +

k i 0 † + † 3 400 300

3 1500

j 0 500

- ( MA ) z 4 k = 0

1000 - ( MA ) x = 0   ( MA ) x = 1000 N # m = 1.00 kN # m

( MA ) y - 200 = 0   ( MA ) y = 200 N # m 1500 - ( MA ) z = 0   ( MA ) z = 1500 N # m =

k -2 † = 0 - 400

1.50 kN # m

Ans. Ans. Ans.

Ans: Ax = 300 N Ay = 500 N NB = 400 N ( MA ) x = 1.00 kN # m ( MA ) y = 200 N # m ( MA ) z = 1.50 kN # m 471

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5–78. The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z components of reaction at the bearings if the rod is subjected to forces F1 = 300 lb and F2 = 250 lb. F1 lies in the y–z plane. The bearings are in proper alignment and exert only force reactions on the rod.

F1

z

45

1 ft A

C

4 ft

SOLUTION

5 ft

B

F1 = (- 300 cos 45°j - 300 sin 45°k)

2 ft

30

3 ft

= {- 212.1j - 212.1k} lb

y

45

F2 = (250 cos 45° sin 30°i + 250 cos 45° cos 30°j - 250 sin 45°k)

x

F2

= {88.39i + 153.1j - 176.8k} lb ©Fx = 0;

Ax + Bx + 88.39 = 0

©Fy = 0;

Ay + Cy - 212.1 + 153.1 = 0

©Fz = 0;

Bz + Cz - 212.1 - 176.8 = 0

©Mx = 0;

-Bz (3) - Ay (4) + 212.1(5) + 212.1(5) = 0

©My = 0;

Cz (5) + Ax (4) = 0

©Mz = 0;

Ax (5) + Bx (3) - Cy (5) = 0 Ax = 633 lb

Ans.

Ay = - 141 lb

Ans.

Bx = - 721 lb

Ans.

Bz = 895 lb

Ans.

Cy = 200 lb

Ans.

Cz = - 506 lb

Ans.

Ans: Ax = Ay = Bx = Bz = Cy = Cz = 472

633 lb - 141 lb - 721 lb 895 lb 200 lb - 506 lb

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5–79. The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction C y at the bearing C to be equal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Set F1 = 300 lb.

F1

z

45

1 ft A

C

4 ft

SOLUTION

5 ft

B

F1 = ( - 300 cos 45°j - 300 sin 45°k)

2 ft

= { -212.1j - 212.1k} lb

45

F2 = (F2 cos 45° sin 30°i + F2 cos 45° cos 30°j - F2 sin 45°k)

x

= {0.3536F2 i + 0.6124F2 j - 0.7071F2 k} lb ©Fx = 0;

Ax + Bx + 0.3536F2 = 0

©Fy = 0;

Ay + 0.6124F2 - 212.1 = 0

©Fz = 0;

Bz + Cz - 0.7071F2 - 212.1 = 0

©Mx = 0;

- Bz (3) - Ay (4) + 212.1(5) + 212.1(5) = 0

©My = 0;

Cz (5) + Ax (4) = 0

©Mz = 0;

Ax (5) + Bx (3) = 0

30

3 ft

F2

Ax = 357 lb Ay = -200 lb Bx = -595 lb Bz = 974 lb Cz = - 286 lb Ans.

F2 = 674 lb

Ans: F2 = 674 lb 473

y

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*5–80. The bar AB is supported by two smooth collars. At A the connection is with a ball-and-socket joint and at B it is a rigid attachment. If a 50-lb load is applied to the bar, determine the x, y, z components of reaction at A and B.

D B z C

Solution

6 ft

4 ft

6 ft

Ax + Bx = 0 (1) By + 50 = 0

E

50 lb

5 ft x

By = - 50 lb Ans. Az + Bz = 0 (2)

3 ft

A F

y

MBz = 0 Ans. MBx + 50(6) = 0

MBx = - 300 lb # ft Ans. BCD = -9i + 3j BCD = -0.94868i + 0.316228j Require FB # uCD = 0

(Bxi - 50j + Bzk) # ( -0.94868i + 0.316228j) = 0 -0.94868Bx - 50(0.316228) = 0 Ans.

Bx = - 16.667 = - 16.7 lb From Eq. (1);

Ans.

Ax = 16.7 lb Require MB # uCD = 0

( -300i + MByj) # ( - 0.94868i + 0.316228j) = 0 300(0.94868) + MBy(0.316228) = 0

MBy = - 900 lb # ft Ans.

Ans: By = -50 lb MBz = 0 MBx = -300 lb # ft Bx = -16.7 lb Ax = 16.7 lb 474

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5–81. The rod has a weight of 6 lb>ft. If it is supported by a balland-socket joint at C and a journal bearing at D, determine the x, y, z components of reaction at these supports and the moment M that must be applied along the axis of the rod to hold it in the position shown.

z D

60 A 0.5 ft 45

x

Solution

y

ΣFx = 0;      Cx + Dx - 15 sin 45° = 0

(1)

ΣFy = 0;      Cy + Dy = 0

(2)

C

M

B

1 ft

1 ft

ΣFz = 0;      Cz - 15 cos 45° = 0

Ans.

Cz = 10.6 lb

ΣMx = 0;   - 3 cos 45°(0.25 sin 60°) - Dy(2) = 0 Dy = - 0.230 lb

Ans.



Cy = 0.230 lb

Ans.

ΣMy = 0;

- (12 sin 45°)(1) - (3 sin 45°)(1) + (3 cos 45°)(0.25 cos 60°)

From Eq. (2);

+ Dx(2) = 0 Dx = 5.17 lb

Ans.



Cx = 5.44 lb

Ans.

ΣMz = 0;

- M + (3 sin 45°)(0.25 sin 60°) = 0

From Eq. (1);



M = 0.459 lb # ft

Ans.

Ans: Cz = 10.6 lb Dy = - 0.230 lb Cy = 0.230 lb Dx = 5.17 lb Cx = 5.44 lb M = 0.459 lb # ft 475

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5–82. The sign has a mass of 100 kg with center of mass at G. Determine the x, y, z components of reaction at the ball-andsocket joint A and the tension in wires BC and BD.

z 1m

D

2m

C

SOLUTION

1m 2m

Equations of Equilibrium: Expressing the forces indicated on the free-body diagram, Fig. a, in Cartesian vector form, we have

A

FA = A xi + A yj + A zk

x

B

W = {- 100(9.81)k} N = {- 981k} N FBD = FBDuBD = FBD ≥

FBC = FBCuBC = FBC ≥

G

(- 2 - 0)i + (0 - 2)j + (1 - 0)k 2(- 2 - 0)2 + (0 - 2)2 + (1 - 0)2

¥ = a-

2 2 1 FBDi - FBDj + FBDkb 3 3 3

1m

(1 - 0)i + (0 - 2)j + (2 - 0)k

1 2 2 ¥ = a FBCi - FBCj + FBCkb 3 3 3 2(1 - 0) + (0 - 2) + (2 - 0) 2

2

2

Applying the forces equation of equilibrium, we have ©F = 0;

FA + FBD + FBC + W = 0

2 2 1 1 2 2 (A xi + A yj + A zk) + a - FBDi - FBDj + FBDk b + a FBCi - FBCj + FBCkb + ( - 981 k) = 0 3 3 3 3 3 3 a Ax -

2 1 2 2 1 2 FBD + FBC bi + a A y - FBD - FBC b j + aA z + FBD + FBC - 981bk = 0 3 3 3 3 3 3

Equating i, j, and k components, we have Ax -

2 1 F + FBC = 0 3 BD 3

(1)

Ay -

2 2 F - FBC = 0 3 BD 3

(2)

Az +

1 2 FBD + FBC - 981 = 0 3 3

(3)

In order to write the moment equation of equilibrium about point A, the position vectors rAG and rAB must be determined first. rAG = {1j} m rAB = {2j} m

476

1m

y

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5–82. Continued

Thus, ©M A = 0; rAB * (FBC + FBD) + (rAG * W) = 0 2 2 2 2 1 1 (2j) * c a FBC - FBD b i - a FBC + FBD b j + a FBC + FBD bk d + (1j) * ( - 981k) = 0 3 3 3 3 3 3 4 2 4 2 a FBC + FBD - 981 bi + a FBD - FBC b k = 0 3 3 3 3 Equating i, j, and k components we have 2 4 F + FBC - 981 = 0 3 BC 3

(4)

4 2 F - FBC = 0 3 BC 3

(5)

Ans: FBD = 294 N FBC = 589 N Ax = 0 Ay = 589 N Az = 490.5 N 477

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5–83. z

Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B. Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 0°. The bearings are in proper alignment and exert only force reactions on the shaft.

200 mm 250 mm

u

300 mm C

SOLUTION

80 mm A

x

©Fx = 0; ©Fy = 0;

165 + 80210.452 - Cz 10.752 = 0 Ans.

150 + 58.0210.22 - Cy 10.752 = 0 Cy = 28.8 N

Ans.

Dx = 0

Ans.

Dy + 28.8 - 50 - 58.0 = 0 Ans.

Dy = 79.2 N ©Fz = 0;

80 N

Ans.

Cz = 87.0 N ©Mz = 0;

y

65 N

6510.082 - 8010.082 + T10.152 - 5010.152 = 0 T = 58.0 N

©My = 0;

150 mm B T

Equations of Equilibrium: ©Mx = 0;

50 N D

Dz + 87.0 - 80 - 65 = 0 Ans.

Dz = 58.0 N

Ans: T = 58.0 N Cz = 87.0 N Cy = 28.8 N Dx = 0 Dy = 79.2 N Dz = 58.0 N 478

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*5–84. z

Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B. Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 45°. The bearings are in proper alignment and exert only force reactions on the shaft.

200 mm 250 mm

u

300 mm C

SOLUTION

80 mm A

x

©Fx = 0; ©Fy = 0;

165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = 0 Ans.

58.010.22 + 50 cos 45°10.22 - Cy 10.752 = 0 Cy = 24.89 N = 24.9 N

Ans.

Dx = 0

Ans.

Dy + 24.89 - 50 cos 45° - 58.0 = 0 Ans.

Dy = 68.5 N ©Fz = 0;

80 N

Ans.

Cz = 77.57 N = 77.6 N ©Mz = 0;

y

65 N

6510.082 - 8010.082 + T10.152 - 5010.152 = 0 T = 58.0 N

©My = 0;

150 mm B T

Equations of Equilibrium: ©Mx = 0;

50 N D

Dz + 77.57 + 50 sin 45° - 80 - 65 = 0 Ans.

Dz = 32.1 N

Ans: T = Cz = Cy = Dy = Dz = 479

58.0 N 77.6 N 24.9 N 68.5 N 32.1 N

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5–85. Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole at the end joint of the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the 800-lb cylinder in equilibrium.

z C

2 ft

A

x

FBC

3 ft

3 6 2 = FBC a i - j + k b 7 7 7

©Fx = 0;

B

6 ft

SOLUTION

y

3 FBC a b = 0 7 FBC = 0

Ans.

©Fy = 0;

Ay = 0

Ans.

©Fz = 0;

Az = 800 lb

Ans.

©Mx = 0;

(MA)x - 800(6) = 0 (MA)x = 4.80 kip # ft

Ans.

©My = 0;

(MA)y = 0

Ans.

©Mz = 0;

(MA)z = 0

Ans.

Ans: FBC = 0 Ay = 0 Az = 800 lb (MA)x = 4.80 kip # ft (MA)y = 0 (MA)z = 0 480

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6–1. P1

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 20 kN, P2 = 10 kN.

B

C

1.5 m P2 D 2m

Solution

A

Method of Joints. Start at joint C and then proceed to join D. Joint C. Fig. a + ΣFx = 0;     FCB = 0  S + c ΣFy = 0;   

FCD - 20 = 0  

Ans. Ans.

FCD = 20.0 kN (C) 

Joint D. Fig. b 3 + c ΣFy = 0;  FDB a b - 20.0 = 0  FDB = 33.33 kN (T) = 33.3 kN (T) Ans. 5 + ΣFx = 0;    10 + 33.33 a 4 b - FDA = 0 S 5 FDA = 36.67 kN (C) = 36.7 kN (C) Ans.

Ans: FCB = FCD = FDB = FDA = 481

0 20.0 kN (C) 33.3 kN (T) 36.7 kN (C)

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6–2. P1

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 45 kN, P2 = 30 kN.

B

C

1.5 m P2 D 2m

Solution

A

Method of Joints. Start at joint C and then proceed to joint D. Joint C. Fig. a + ΣFx = 0;   FCB = 0  S

Ans.

+ c ΣFy = 0;      FCD - 45 = 0    FCD = 45.0 kN (C) 

Ans.

Joint D. Fig. b 3 + c ΣFy = 0;   FDB a b - 45.0 = 0   FDB = 75.0 kN (T) 5

+ ΣFx = 0;   30 + 75.0 a 4 b - FDA = 0   FDA = 90.0 kN (C) S 5

Ans. Ans.

Ans: FCB = FCD = FDB = FDA = 482

0 45.0 kN (C) 75.0 kN (T) 90.0 kN (C)

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6–3. Determine the force in each member of the truss. State if the members are in tension or compression.

A 13 5

3 ft

B

D

3 ft F

12

4 ft

130 lb

SOLUTION

3 ft

C

4 ft

E

Joint A: + c ©Fy = 0;

4 12 (F ) (130) = 0 5 AC 13 Ans.

FAC = 150 lb (C) + ©F = 0; : x

FAB -

5 3 (150) (130) = 0 5 13 Ans.

FAB = 140 lb (T) Joint B: + ©F = 0; : x

FBD - 140 = 0 Ans.

FBD = 140 lb (T) + c ©Fy = 0;

Ans.

FBC = 0

Joint C: + c ©Fy = 0;

4 4 a b FCD - a b 150 = 0 5 5 Ans.

FCD = 150 lb (T) + ©F = 0; : x

-FCE +

3 3 (150) + (150) = 0 5 5 Ans.

FCE = 180 lb (C) Joint D: + c ©Fy = 0;

FDE -

4 (150) = 0 5 Ans.

FDE = 120 lb (C) + ©F = 0; : x

FDF - 140 -

3 (150) = 0 5 Ans.

FDF = 230 lb (T) Joint E: + ©F = 0; : x

180 -

3 (F ) = 0 5 EF Ans.

FEF = 300 lb (C)

483

Ans: FAC = 150 lb (C) FAB = 140 lb (T) FBD = 140 lb (T) FBC = 0 FCD = 150 lb (T) FCE = 180 lb (C) FDE = 120 lb (C) FDF = 230 lb (T) FEF = 300 lb (C)

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*6–4. Determine the force in each member of the truss and state if the members are in tension or compression.

1.5 kip A

SOLUTION Ay (40) + 1.5(4) + 2(12) - 3(10) - 3(20) = 0

: ©Fx = 0;

1.5 + 2 - Ex = 0

+ c ©Fy = 0;

Ey + 1.5 - 3 - 3 = 0

+

Ay = 1.5 kip

Ex = 3.5 kip Ey = 4.5 kip

Joint A: + c ©Fy = 0;

1.5 - FAl sin 21.80° = 0 FAl = 4.039 kip = 4.04 kip (C)

+ ©F = 0; : x

FAB - 4.039 cos 21.80° = 0

FAB = 3.75 kip (T)

Ans. Ans.

Joint E: + c ©Fy = 0;

4.5 - FEF sin 21.80° = 0 FEF = 12.12 kip = 12.1 kip (C)

+ ©F = 0; : x

Ans.

-FED - 3.5 + 12.12 cos 21.80° = 0 Ans.

FED = 7.75 kip (T) Joint B: + c ©Fy = 0; + ©F = 0; : x

Ans.

FBI = 0 FBC - 3.75 = 0

Ans.

FBC = 3.75 kip (T)

Joint D: + c ©Fy = 0; + ©F = 0; : x

Ans.

FDF = 0 -FDC + 7.75 = 0

Ans.

FDC = 7.75 kip (T)

Joint F: +Q©Fy¿ = 0;

FFC cos 46.40° - 3 cos 21.80° = 0 Ans.

FFC = 4.039 kip = 4.04 kip (C) +R©Fx¿ = 0;

FFG + 3 sin 21.80° + 4.039 sin 46.40° - 12.12 = 0 Ans.

FFG = 8.078 kip = 8.08 kip (C) Joint H: + ©F = 0; : x

+ c ©Fy = 0;

2 - FHG cos 21.80° = 0 FHG = 2.154 kip = 2.15 kip (C)

Ans.

2.154 sin 21.80° - FHI = 0

Ans.

FHI = 0.8 kip (T) 484

I

F B

10 ft

c+ ©ME = 0;

3 kip G

8 ft 4 ft

3 kip

H

2 kip

C 10 ft

E

D 10 ft

10 ft

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*6–4. Continued

Joint C: + ©F = 0; : x

-FCI cos 21.80° - 4.039 cos 21.80° - 3.75 + 7.75 = 0 FCI = 0.2692 kip = 0.269 kip (T)

+ c ©Fy = 0;

Ans.

FCG + 0.2692 sin 21.80° - 4.039 sin 21.80° = 0 FCG = 1.40 kip (T)

Ans.

Joint G: +Q©Fy¿ = 0;

FGI cos 46.40° - 3 cos 21.80° - 1.40 cos 21.80° = 0 Ans.

FGI = 5.924 kip = 5.92 kip (C) + R©Fx¿ = 0;

2.154 + 3 sin 21.80° + 5.924 sin 46.40° + 1.40 sin 21.80° - 8.081 = 0 (Check)

Ans: FAl = 4.04 kip (C) FAB = 3.75 kip (T) FEF = 12.1 kip (C) FED = 7.75 kip (T) FBI = 0 FBC = 3.75 kip (T) FDF = 0 FDC = 7.75 kip (T) FFC = 4.04 kip (C) FFG = 8.08 kip (C) FHG = 2.15 kip (C) FHI = 0.8 kip (T) FCI = 0.269 kip (T) FCG = 1.40 kip (T) FGI = 5.92 kip (C) 485

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6–5. Determine the force in each member of the truss, and state if the members are in tension or compression. Set u = 0°.

D

3 kN

1.5 m A

SOLUTION

B

Support Reactions: Applying the equations of equilibrium to the free-body diagram of the entire truss,Fig.a, we have a + ©MA = 0;

2m

3 - Ax = 0 A x = 3 kN

+ c ©Fy = 0;

A y + 3.125 - 4 = 0 A y = 0.875 kN

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B. Joint C: From the free-body diagram in Fig. b, we can write + c ©Fy = 0;

3 3.125 - FCD a b = 0 5 Ans.

FCD = 5.208 kN = 5.21 kN (C) + ©F = 0; : x

4 5.208 a b - FCB = 0 5 Ans.

FCB = 4.167 kN = 4.17 kN (T) Joint A: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

3 0.875 - FAD a b = 0 5 Ans.

FAD = 1.458 kN = 1.46 kN (C) + : ©Fx = 0;

4 FAB - 3 - 1.458a b = 0 5 Ans.

FAB = 4.167 kN = 4.17 kN (T) Joint B: From the free-body diagram in Fig. d, we can write + c ©Fy = 0;

FBD - 4 = 0 Ans.

FBD = 4 kN (T) + ©F = 0; : x

4.167 - 4.167 = 0

(check!)

Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above. Ans: FCD = FCB = FAD = FAB = FBD = 486

2m 4 kN

NC (2 + 2) - 4(2) - 3(1.5) = 0 NC = 3.125 kN

+ : ©Fx = 0;

C

5.21 kN (C) 4.17 kN (T) 1.46 kN (C) 4.17 kN (T) 4 kN (T)

u

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6–6. Determine the force in each member of the truss, and state if the members are in tension or compression. Set u = 30°.

D

3 kN

1.5 m A

SOLUTION

B

Support Reactions: From the free-body diagram of the truss, Fig. a, and applying the equations of equilibrium, we have a + ©MA = 0;

NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0 NC = 3.608 kN

+ : ©Fx = 0;

3 - 3.608 sin 30° - A x = 0 A x = 1.196 kN

+ c ©Fy = 0;

A y + 3.608 cos 30° - 4 = 0 A y = 0.875 kN

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B. Joint C: From the free-body diagram in Fig. b, we can write + c ©Fy = 0;

3 3.608 cos 30° - FCD a b = 0 5 Ans.

FCD = 5.208 kN = 5.21 kN (C) + ©F = 0; : x

4 5.208 a b - 3.608 sin 30° - FCB = 0 5 Ans.

FCB = 2.362 kN = 2.36 kN (T) Joint A: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

3 0.875 - FAD a b = 0 5 Ans.

FAD = 1.458 kN = 1.46 kN (C) + : ©Fx = 0;

4 FAB - 1.458 a b - 1.196 = 0 5 Ans.

FAB = 2.362 kN = 2.36 kN (T) Joint B: From the free-body diagram in Fig. d, we can write + c ©Fy = 0;

FBD - 4 = 0 Ans.

FBD = 4 kN (T) + : ©Fx = 0;

C

2.362 - 2.362 = 0

(check!)

Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above. Ans: FCD = FCB = FAD = FAB = FBD = 487

5.21 kN (C) 2.36 kN (T) 1.46 kN (C) 2.36 kN (T) 4 kN (T)

2m

2m 4 kN

u

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6–7. Determine the force in each member of the truss and state if the members are in tension or compression.

4 kN 3m

3m B

3m

C

D 3m 5m

A F

SOLUTION

E

Support Reactions: a + ©MD = 0;

4162 + 5192 - Ey 132 = 0

+ c ©Fy = 0;

23.0 - 4 - 5 - D y = 0

+ ©F = 0 : x

5 kN

Ey = 23.0 kN Dy = 14.0 kN

Dx = 0

Method of Joints: Joint D: + c ©Fy = 0;

FDE ¢

5 234

≤ - 14.0 = 0

FDE = 16.33 kN 1C2 = 16.3 kN 1C2 + ©F = 0; : x

16.33 ¢

3 234

Ans.

≤ - FDC = 0

FDC = 8.40 kN 1T2

Ans.

Joint E: + ©F = 0; : x

FEA ¢

3 210

≤ - 16.33 ¢

3 234

≤ = 0

FEA = 8.854 kN 1C2 = 8.85 kN 1C2 + c ©Fy = 0;

23.0 - 16.33 ¢

5 234

≤ - 8.854 ¢

FEC = 6.20 kN 1C2

1 210

Ans.

≤ - FEC = 0 Ans.

Joint C: + c ©Fy = 0; + ©F = 0; : x

6.20 - FCF sin 45° = 0 FCF = 8.768 kN 1T2 = 8.77 kN 1T2

Ans.

8.40 - 8.768 cos 45° - FCB = 0 Ans.

F CB = 2.20 kN T

488

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6–7. Continued

Joint B: + ©F = 0; : x

2.20 - FBA cos 45° = 0 FBA = 3.111 kN 1T2 = 3.11 kN 1T2

+ c ©Fy = 0;

Ans.

FBF - 4 - 3.111 sin 45° = 0 FBF = 6.20 kN 1C2

Ans.

+ c ©Fy = 0;

8.768 sin 45° - 6.20 = 0

(Check!)

+ ©F = 0; : x

8.768 cos 45° - FFA = 0

Joint F:

FFA = 6.20 kN 1T2

Ans.

Ans: FDE = 16.3 kN (C) FDC = 8.40 kN (T) FEA = 8.85 kN (C) FEC = 6.20 kN (C) FCF = 8.77 kN (T) FCB = 2.20 kN (T) FBA = 3.11 kN (T) FBF = 6.20 kN (C) FFA = 6.20 kN (T) 489

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*6–8. Determine the force in each member of the truss, and state if the members are in tension or compression.

600 N

D

4m

SOLUTION

900 N

E

C

Method of Joints: We will begin by analyzing the equilibrium of joint D, and then proceed to analyze joints C and E. Joint D: From the free-body diagram in Fig. a, + : ©Fx = 0;

B

A

3 FDE a b - 600 = 0 5

6m

Ans.

FDE = 1000 N = 1.00 kN (C) + c ©Fy = 0;

4m

4 1000 a b - FDC = 0 5 Ans.

FDC = 800 N (T) Joint C: From the free-body diagram in Fig. b, + ©F = 0; : x

FCE - 900 = 0 Ans.

FCE = 900 N (C) + c ©Fy = 0;

800 - FCB = 0 Ans.

FCB = 800 N (T) Joint E: From the free-body diagram in Fig. c, R+ ©Fx ¿ = 0;

- 900 cos 36.87° + FEB sin 73.74° = 0 Ans.

FEB = 750 N (T) Q+ ©Fy ¿ = 0;

FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0 Ans.

FEA = 1750 N = 1.75 kN (C)

Ans: FDE = FDC = FCE = FCB = FEB = FEA = 490

1.00 kN (C) 800 N (T) 900 N (C) 800 N (T) 750 N (T) 1.75 kN (C)

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6–9. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 3 kN, P2 = 6 kN.

E

6m D

A B

C

4m

4m P1

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, a+ΣMA = 0;   ND(12) - 3(4) - 6(8) = 0   ND = 5.00 kN a+ΣMD = 0;   

6(4) + 3(8) - Ay(12) = 0   Ay = 4.00 kN

+ ΣFx = 0;   Ax = 0 S Method of Joints. We will carry out the analysis of joint equilibrium according to the sequence of joints A, D, B and C. Joint A. Fig. b + c ΣFy = 0;  4.00 - FAEa



1 22

b = 0

Ans.

FAE = 4 22 kN (C) = 5.66 kN (C)

+ ΣFx = 0;    FAB - 4 22 a 1 b = 0  FAB = 4.00 kN (T) S 22

491

Ans.

4m P2

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6–9.   Continued

Joint D. Fig. c + c ΣFy = 0;  5.00 - FDE a

1 22

b = 0  FDE = 5 22 kN (C) = 7.07 kN (C) Ans.

+ ΣFx = 0;    5 22 a 1 b - FDC = 0  FDC = 5.00 kN (T) S 22

Ans.

Joint B. Fig. d

+ c ΣFy = 0;  FBE a

3 210

b - 3 = 0  FBE = 210 kN (T) = 3.16 kN (T) Ans.

+ ΣFx = 0;  FBC + 210 a S

Joint C. Fig. e

+ c ΣFy = 0;  FCE a

3 210

1

210

b - 4.00 = 0  FBC = 3.00 kN (T)

Ans.

b - 6 = 0  FCE = 2210 kN (T) = 6.32 kN (T) Ans.

+ ΣFx = 0;    5.00 - 3.00 - a2210b a S

1

210

b = 0(Check!!)

Ans: FAE = 5.66 kN (C) FAB = 4.00 kN (T) FDE = 7.07 kN (C) FDC = 5.00 kN (T) FBE = 3.16 kN (T) FBC = 3.00 kN (T) FCE = 6.32 kN (T) 492

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6–10. E

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 6 kN, P2 = 9 kN.

6m D

A B

C

4m

4m P1

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, a+ΣMA = 0;   ND(12) - 6(4) - 9(8) = 0   ND = 8.00 kN a+ΣMD = 0;   9(4) + 6(8) - Ay(12) = 0   Ay = 7.00 kN + ΣFx = 0;   Ax = 0 S Method of Joints. We will carry out the analysis of joint equilibrium according to the sequence of joints A, D, B and C. Joint A. Fig. a + c ΣFy = 0;  7.00 - FAE a

1 22

b = 0  FAE = 7 22 kN (C) = 9.90 kN (C)

+ ΣFx = 0;  FAB - 722 a 1 b = 0  FAB = 7.00 kN (T) S 22

Ans. Ans.

Joint D. Fig. c

+ c ΣFy = 0;  8.00 - FDE a

1 22

b = 0  FDE = 8 22 kN (C) = 11.3 kN (C)

+ ΣFx = 0;  822 a 1 b - FDC = 0  FDC = 8.00 kN (T) S 22

Ans. Ans.

493

4m P2

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6–10.   Continued

Joint B. Fig. d + c ΣFy = 0;    FBE a

3 210

b - 6 = 0  FBE = 2 210 kN (T) = 6.32 kN (T) Ans.

+ ΣFx = 0;  FBC - 7.00 + a2210b a 1 b = 0  FBC = 5.00 kN (T) S 210

Ans.

Joint C. Fig. e

+ c ΣFy = 0;    FCE a

3 210

b - 9 = 0  FCE = 3210 kN = 9.49 kN (T) Ans.

+ ΣFx = 0;  8.00 - 5.00 - a3210ba 1 b = 0  (Check!!) S 210

Ans: FAE = 9.90 kN (C) FAB = 7.00 kN (T) FDE = 11.3 kN (C) FDC = 8.00 kN (T) FBE = 6.32 kN (T) FBC = 5.00 kN (T) FCE = 9.49 kN (T) 494

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6–11. Determine the force in each member of the Pratt truss, and state if the members are in tension or compression.

J 2m

SOLUTION

2m

Joint A: + c ©Fy = 0;

2m

20 - FAL sin 45° = 0

I

L

H

A B D C E F 2m 2m 2m 2m 2m 2m

FAL = 28.28 kN (C) + ©F = 0; : x

K

FAB - 28.28 cos 45° = 0

10 kN

FAB = 20 kN (T)

20 kN

G

10 kN

Joint B: + ©F = 0; : x

FBC - 20 = 0 FBC = 20 kN (T)

+ c ©Fy = 0;

FBL = 0

Joint L: R+ ©Fx = 0;

FLC = 0

+Q©Fy = 0;

28.28 - FLK = 0 FLK = 28.28 kN (C)

Joint C: + : ©Fx = 0;

FCD - 20 = 0 FCD = 20 kN (T)

+ c ©Fy = 0;

FCK - 10 = 0 FCK = 10 kN (T)

Joint K: R+ ©Fx - 0;

10 sin 45° - FKD cos (45° - 26.57°) = 0 FKD = 7.454 kN (L)

+Q©Fy = 0;

28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0 FKJ = 23.57 kN (C)

Joint J: + : ©Fx = 0;

23.57 sin 45° - FJI sin 45° = 0 FJI = 23.57 kN (L)

+ c ©Fy = 0;

2 (23.57 cos 45°) - FJD = 0 FJD = 33.3 kN (T)

Ans.

FAL = FGH = FLK = FHI = 28.3 kN (C)

Ans.

Due to Symmetry Ans: Ans. FJD = 33.3 kN (T) Ans. FAL = FGH = FLK = FHI = 28.3 kN (C) FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T) Ans. FBL = FFH = FLC = FHE = 0 F = FEI = 10 kN (T) Ans. CK FKJ = FIJ = 23.6 kN (C) Ans. FKD = FID = 7.45 kN (C)

FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T) FBL = FFH = FLC = FHE = 0 FCK = FEI = 10 kN (T) FKJ = FIJ = 23.6 kN (C) FKD = FID = 7.45 kN (C) 495

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*6–12. Determine the force in each member of the truss and state if the members are in tension or compression.

500 lb 3 ft

500 lb 3 ft

3 ft D

C

6 ft

SOLUTION

B

E

Joint D: + c ©Fy = 0;

6 ft

FDE sin 75.96° - 500 = 0 FDE = 515.39 lb = 515 lb (C)

+ ©F = 0; : x

FCD - 515.39 cos 75.96° = 0

FCD = 125 lb (C)

Ans. Ans.

F

A 9 ft

Joint C: + a©Fy¿ = 0;

FCE cos 39.09° + 125 cos 14.04° - 500 cos 75.96° = 0 Ans.

FCE = 0 +Q©Fx¿ = 0;

FCB - 500 sin 75.96° - 125 sin 14.04° = 0 Ans.

FCB = 515.39 lb = 515 lb (C) Joint E: + Q©Fy¿ = 0;

FEB cos u = 0

+ R©Fx¿ = 0;

515.39 - FEF = 0

Ans.

FEB = 0

Ans.

FEF = 515 lb (C)

Joint B: + a©Fy¿ = 0;

FBF cos u = 0

+Q©Fx¿ = 0;

FBA - 515.39 = 0

Ans.

FBF = 0

Ans.

FBA = 515 lb (C)

Joint B: +a©Fy = 0;

FBF sin 75° - 150 = 0 Ans.

FBF = 155 lb (C) Joint D: +b©Fy = 0;

Ans.

FDF = 0

496

Ans: FDE = 515 lb (C) FCD = 125 lb (C) FCE = 0 FCB = 515 lb (C) FEB = 0 FEF = 515 lb (C) FBF = 0 FBA = 515 lb (C) FBF = 155 lb (C) FDF = 0

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6–13. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

B

3 — a 4

1 — a 4

D A

SOLUTION Joint A: + ©F = 0; : x + c ©Fy = 0;

C a

4 214

(FAD) -

a P

1 22

FAB = 0

1 P 1 (FAB) + (FAD) = 0 2 22 217 FCD = FAD = 0.687 P (T)

Ans.

FCB = FAB = 0.943 P (C)

Ans.

Joint D: + c ©Fy = 0;

FDB - 0.687 P ¢

1 217

≤ -

1 217

(0.687 P) - P = 0 Ans.

FDB = 1.33 P (T)

Ans: FCD = FAD = 0.687P (T) FCB = FAB = 0.943P (C) FDB = 1.33P (T) 497

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6–14. Members AB and BC can each support a maximum compressive force of 800 lb, and members AD, DC, and BD can support a maximum tensile force of 1500 lb. If a = 10 ft , determine the greatest load P the truss can support.

B

3 — a 4

1 — a 4

D A

SOLUTION

C a

Assume FAB = 800 lb (C)

a P

Joint A: + ©F = 0; : x

- 800 ¢

1 22

≤ + FAD ¢

4 217

≤ = 0 OK

FAD = 583.0952 lb 6 1500 lb + c ©Fy = 0;

1 P 1 (800) + (583.0952) = 0 2 22 217 P = 848.5297 lb

Joint D: + c ©Fy = 0;

- 848.5297 - 583.0952(2) ¢

1 217

≤ + FDB = 0 OK

FBD = 1131.3724 lb 6 1500 lb Thus,

Ans.

Pmax = 849 lb

Ans: Pmax = 849 lb 498

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6–15. Members AB and BC can each support a maximum compressive force of 800 lb, and members AD, DC, and BD can support a maximum tensile force of 2000 lb. If a = 6 ft, determine the greatest load P the truss can support.

B

3 — a 4

1 —a 4

D A

C a

Solution

a 3

1) Assume FAB = 800 lb (C) Joint A:



+ ΣFx = 0;    - 800 a 1 b  + FAD a 4 b = 0 S 22 217

OK

FAD = 583.0952 lb 6 2000 lb

+ c ΣFy = 0;   

P 1 1 (800) + (583.0952) = 0 2 22 217

OK

P = 848.5297 lb

Joint D: + c ΣFy = 0;

-848.5297 - 583.0952(2) a

Thus,

Pmax = 849 lb

1 217

b + FDB = 0

OK

FBD = 1131.3724 lb 6 2000 lb 

Ans.

Ans: Pmax = 849 lb 499

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*6–16. Determine the force in each member of the truss. State whether the members are in tension or compression. Set P = 8 kN.

4m

B

A

SOLUTION Method of Joints: In this case, the support reactions are not required for determining the member forces.

60••

C

60••

E

4m

4m P

Joint D: + c ©Fy = 0; + ©F = 0; : x

Joint C: + c ©Fy = 0; + ©F = 0; : x

Joint B: + c ©F y = 0;

FDC sin 60° - 8 = 0 FDC = 9.238 kN 1T2 = 9.24 kN 1T2

Ans.

FDE = 4.619 kN 1C2 = 4.62 kN 1C2

Ans.

FDE - 9.238 cos 60° = 0

FCE sin 60° - 9.238 sin 60° = 0 FCE = 9.238 kN 1C2 = 9.24 kN 1C2

Ans.

FCB = 9.238 kN 1T2 = 9.24 kN 1T2

Ans.

219.238 cos 60°2 - FCB = 0

FBE sin 60° - FBA sin 60° = 0 FBE = FBA = F

+ ©F = 0; : x

9.238 - 2F cos 60° = 0 F = 9.238 kN

Thus,

Joint E:

FBE = 9.24 kN 1C2

Ans.

FBA = 9.24 kN 1T2

+ c ©F y = 0;

Ey - 219.238 sin 60°2 = 0

+ ©F = 0; : x

FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0

Ey = 16.0 kN

Ans.

FEA = 4.62 kN 1C2

Note: The support reactions Ax and Ay can be determinedd by analyzing Joint A using the results obtained above.

500

D

Ans: FDC = 9.24 kN (T) FDE = 4.62 kN (C) FCE = 9.24 kN (C) FCB = 9.24 kN (T) FBA = 9.24 kN (T) FEA = 4.62 kN (C)

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6–17.

If the maximum force that any member can support is 8 kN in tension and 6 kN in compression, determine the maximum force P that can be supported at joint D.

4m B

A

SOLUTION

60°

60°

E

4m

Method of Joints: In this case, the support reactions are not required for determining the member forces.

C

D

4m P

Joint D: FDC = 1.1547P 1T2

+ c ©F y = 0;

FDC sin 60° - P = 0

+ ©F = 0; : x

FDE - 1.1547P cos 60° = 0

FDE = 0.57735P 1C2

Joint C: + c ©Fy = 0; + ©F = 0; : x

FCE sin 60° - 1.1547P sin 60° = 0 FCE = 1.1547P 1C2 211.1547P cos 60°2 - FCB = 0

FCB = 1.1547P 1T2

+ c ©F y = 0;

FBE sin 60° - FBA sin 60° = 0

FBE = FBA = F

+ ©F = 0; : x

1.1547P - 2F cos 60° = 0

Joint B:

F = 1.1547P

Thus, FBE = 1.1547P 1C2

FBA = 1.1547P 1T2

Joint E: + ©F = 0; : x

FEA + 1.1547P cos 60° - 1.1547P cos 60° - 0.57735P = 0 FEA = 0.57735P 1C2

From the above analysis, the maximum compression and tension in the truss member is 1.1547P. For this case, compression controls which requires 1.1547P = 6 Ans.

P = 5.20 kN

Ans: P = 5.20 kN 501

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6–18. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 10 kN, P2 = 8 kN.

G

F

E

2m

B

A

C

1m P1

Solution Support Reactions. Not required. Method of Joints. We will perform the joint equilibrium according to the sequence of joints D, C, E, B and F. Joint D. Fig. a + c ΣFy = 0;    FDE a

2

b - 8 = 0  FDE = 4 25 kN (T) = 8.94 kN (T)Ans.

25

+ ΣFx = 0;  FDC - a4 25ba 1 b = 0  FDC = 4.00 kN (C) S 25

Ans.

Joint C. Fig. b

+ ΣFx = 0;  FCB - 4.00 = 0   FCB = 4.00 kN(C) S

Ans.

+ c ΣFy = 0;    FCE = 0 

Ans.

Joint E. Fig. c + c ΣFy = 0;  FEB a

1 22

b - a425ba

2 25

b =0

FEB = 8 22 kN (C) = 11.3 kN (C)

Ans.

FEF = 12.0 kN (T)

Ans.

+ ΣFx = 0;  a822ba 1 b + a425ba 1 b - FEF = 0  S 22 25



2m

502

D 1m P2

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6–18. Continued

Joint B. Fig. d + ΣFx = 0;  FBA - 4.00 - a8 22ba 1 b = 0  FBA = 12.0 kN (C) Ans. S 22

+ c ΣFy = 0;  FBF - 10 - a8 22ba

Joint F. Fig. e

+ c ΣFy = 0;  FFA a

2 25

1

22

b = 0  FBF = 18.0 kN (T)

Ans.

b - 18.0 = 0  FFA = 9 25 kN (C) = 20.1 kN (C) Ans.

+ ΣFx = 0;  12.0 + a9 25ba 1 b - FFG = 0  FFG = 21.0 kN (T) S 25

Ans.

Ans: FDE = 8.94 kN (T) FDC = 4.00 kN (C) FCB = 4.00 kN (C) FCE = 0 FEB = 11.3 kN (C) FEF = 12.0 kN (T) FBA = 12.0 kN (C) FBF = 18.0 kN (T) FFA = 9 25 kN (C) = 20.1 kN (C) FFG = 21.0 kN (T) 503

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6–19. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 8 kN, P2 = 12 kN.

G

F

E

2m

B

A

C

1m P1

Solution Support Reactions. Not required. Method of Joints. We will perform the joint equilibrium according to the sequence of joints D, C, E, B and F. Joint D. Fig. a + c ΣFy = 0;  FDE a

2 25

b - 12 = 0  FDE = 6 25 kN (T) = 13.4 kN (T)

+ ΣFx = 0;  FDC - a6 25ba 1 b = 0  FDC = 6.00 kN (C) S 25

Ans. Ans.

Joint C. Fig. b

+ ΣFx = 0;  FCB - 6.00 = 0   FCB = 6.00 kN (C) S

Ans.

+ c ΣFy = 0;  FCE = 0 

Ans.

Joint E. Fig. c + c ΣFy = 0;  FEBa





2m

1 22

b - a6 25ba

2 25

b =0

Ans.

FEB = 12 22 kN (C) = 17.0 kN (C)

+ ΣFx = 0;  a12 22ba 1 b + a6 25ba 1 b - FEF = 0 S 22 25

Ans.

FEF = 18.0 kN (T)

504

D 1m P2

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6–19. Continued

Joint B. Fig. d + ΣFx = 0;  FBA - 6.00 - a12 22ba 1 b = 0  FBA = 18.0 kN (C) Ans. S 22

+ c ΣFy = 0;  FBF - 8 - a12 22ba

Joint F. Fig. e

+ c ΣFy = 0;  FFA a

2 25

1

22

b = 0  FBF = 20.0 kN (T)

Ans.

b - 20.0 = 0  FFA = 10 25 kN (C) = 22.4 kN (C) Ans.

+ ΣFx = 0;  a10 25ba 1 b + 18.0 - FFG = 0  FFG = 28.0 kN(T) S 25

Ans.

Ans: FDE = 13.4 kN (T) FDC = 6.00 kN (C) FCB = 6.00 kN (C) FCE = 0 FEB = 17.0 kN (C) FEF = 18.0 kN (T) FBA = 18.0 kN (C) FBF = 20.0 kN (T) FFA = 22.4 kN (C) FFG = 28.0 kN (T) 505

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*6–20. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 9 kN, P2 = 15 kN.

E

F

P1

D

4m

A

Solution

3m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a

a+ΣMC = 0;  15(3) - 9(4) - Ay(6) = 0  Ay = 1.50 kN + ΣFx = 0;      9 - Ax = 0  Ax = 9.00 kN S Method of Joints: By inspecting joints D and F, we notice that members DE, DC and FA are zero force members. Thus Ans.

FDE = FDC = FFA = 0

We will perform the joint equilibrium analysis by following the sequence of joints C, B, A and F. Joint C. Fig. b 4 + c ΣFy = 0;  13.5 - FCE a b = 0   5 FCE = 16.875 kN (C) = 16.9 kN (C)

Ans.

+ ΣFx = 0;  16.875 a 3 b - FCB = 0  FCB = 10.125 kN (T) = 10.1 kN (T) Ans. S 5

Joint B. Fig. c

+ ΣFx = 0;  10.125 - FBA = 0  FBA = 10.125 kN (T) = 10.1 kN (T) Ans. S + c ΣFy = 0;  FBE - 15 = 0  FBE = 15.0 kN (T)

Ans.

506

3m P2

a+ΣMA = 0;  NC(6) - 15(3) - 9(4) = 0  NC = 13.5 kN



C

B

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*6–20. Continued

Joint A. Fig. d 4 + c ΣFy = 0;  1.50 - FAE a b = 0  FAE = 1.875 kN (C) 5

Ans.

+ ΣFx = 0;  10.125 - 1.875a 3 b - 9.00 = 0(Check!!) S 5

Joint F. Fig. e

+ ΣFx = 0;  9 - FFE = 0   FFE = 9.00 kN (C) S

Ans.

Ans: FCE = FCB = FBA = FBE = FAE = FFE = 507

16.9 kN (C) 10.1 kN (T) 10.1 kN (T) 15.0 kN (T) 1.875 kN (C) 9.00 kN (C)

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6–21. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 30 kN, P2 = 15 kN.

E

F

P1

D

4m

A

Solution

C

B 3m

Support Reactions.

3m P2

a+ΣMA = 0;  NC(6) - 15(3) - 30(4) = 0  NC = 27.5 kN Method of Joints: By inspecting joints D and F, we notice that members DE, DC and FA are zero force members. Thus FDE = FDC = FFA = 0 Ans. We will perform the joint equilibrium analysis by following the sequence of joints C, B, F and E. Joint C. Fig. b 4 + c ΣFy = 0;  27.5 - FCEa b = 0  FCE = 34.375 kN (C) = 34.4 kN (C)Ans. 5 + ΣFx = 0;  34.375a 3 b - FCB = 0  FCB = 20.625 kN (T) = 20.6 kN (T)Ans. S 5

Joint B. Fig. c

+ ΣFx = 0;  20.625 - FBA = 0  FBA = 20.625 kN (T) = 20.6 kN (T)Ans. S + c ΣFy = 0;  FBE - 15 = 0  FBE = 15.0 kN (T)Ans.

508

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6–21. Continued

Joint F. Fig. d + ΣFx = 0;  30 - FFE = 0  FFE = 30.0 kN (C)Ans. S Joint E. Fig. e 4 4 + c ΣFy = 0;  34.375a b - 15.0 - FEAa b = 0 5 5



Ans.

FEA = 15.625 kN (T) = 15.6 kN (T)

+ ΣFx = 0;  30.0 - 15.625a 3 b - 34.375a 3 b = 0 (Check!!) S 5 5

Ans: FDE = FDC = FFA = 0 FCE = 34.4 kN (C) FCB = 20.6 kN (T) FBA = 20.6 kN (T) FBE = 15.0 kN (T) FFE = 30.0 kN (C) FEA = 15.6 kN (T) 509

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6–22. B

Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

C

L/3

SOLUTION

L 2L b + Pa b - (Dy)(L) = 0 3 3 Dy = P

c + ©MA = 0; + c ©F y = 0;

E

A

Pa

L/3

Ay = P

F L/3

P

D L/3

P

Joint F: + ©F = 0; : x

FFD - FFE - FFB a

1 22

b = 0

(1)

F FD - FFE = P + ©F = 0; : y

1

FFB a

22

b -P = 0

FFB = 22P = 1.41 P (T) Similarly, FEC = 22P Joint C: + ©F = 0; : x

2 25 +c ©Fy = 0;

2

FCA a

F CA

25

b - 22P a

FCA 1 25

1 22

1 22

b - FCD a

1 22

b = 0

FCD = P

- 22P

1 22

+ FCD

1 22

=0

FCA =

2 25 P = 1.4907P = 1.49P (C) 3

F CD =

22 P = 0.4714P = 0.471P (C) 3

FAE -

1 2 22 2 25 Pa Pa b b = 0 3 3 22 25

FAE =

5 P = 1.67 P (T) 3

Joint A: + ©F = 0; : x

Similarly, FFD=1.67 P (T) From Eq.(1), and Symmetry, FFE = 0.667 P (T)

Ans.

FFD = 1.67 P (T)

Ans.

FAB = 0.471 P (C)

Ans.

FAE = 1.67 P (T)

Ans.

FAC = 1.49 P (C)

Ans.

FBF = 1.41 P (T)

Ans.

FBD = 1.49 P (C)

Ans.

FEC = 1.41 P (T)

Ans.

FCD = 0.471 P (C)

Ans. 510

Ans: FFE = 0.667P (T) FFD = 1.67P (T) FAB = 0.471P (C) FAE = 1.67P (T) FAC = 1.49P (C) FBF = 1.41P (T) FBD = 1.49P (C) FEC = 1.41P (T) FCD = 0.471P (C)

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6–23. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

E

d

D

SOLUTION

d/ 2

Support reactions: a + ©ME = 0;

3 Ax a d b -Pd = 0 2

+ ©F = 0; : x

2 P - Ex = 0 3

+ c ©Fy = 0;

Ey - P = 0

A

C

Ax = 0.6667P

d/ 2 P

Ex = 0.6667P d

B

d

Ey = P

Joint E: + ©F = 0; : x

FEC sin 33.69° - 0.6667P = 0 FEC = 1.202P = 1.20P (T)

+ c ©Fy = 0;

P - FED - 1.202P cos 33.69° = 0

FED = 0

Ans. Ans.

Joint A: + c ©Fy = 0; + ©F = 0; : x

FAB sin 26.57° - FAD sin 26.57° = 0 0.6667P - 2F cos 26.57° = 0

FAB = FAD = F

F = 0.3727P Ans.

FAB = FAD = F = 0.373P (C) Joint D: + ©F = 0; : x

0.3727P cos 26.57° - FDC cos 26.57° = 0 Ans.

FDC = 0.3727P = 0.373P (C) + c ©Fy = 0;

2(0.3727P sin 26.57°) - FDB = 0 Ans.

FDB = 0.3333P = 0.333P (T) Joint B: + ©F = 0; : x

0.3727P cos 26.57° - FBC cos 26.57° = 0 Ans.

FBC = 0.3727P = 0.373P (C) + c ©Fy = 0;

0.3333P - 2(0.3727 sin 26.57°) = 0 (Check)

Ans: FEC = FED = FAB = FDC = FDB = FBC = 511

1.20P (T) 0 FAD = 0.373P (C) 0.373P (C) 0.333P (T) 0.373P (C)

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*6–24. The maximum allowable tensile force in the members of the truss is 1Ft2max = 5 kN, and the maximum allowable compressive force is 1Fc2max = 3 kN. Determine the maximum magnitude of the load P that can be applied to the truss. Take d = 2 m.

E

d

D

SOLUTION Support reactions:

d/ 2

a + ©ME = 0;

3 Ax a db -Pd = 0 2

+ ©F = 0; : x

2 P - Ex = 0 3

+ c ©Fy = 0; Joint E: + ©F = 0; : x

A

C

Ax = 0.6667P d/ 2

Ex = 0.6667P

Ey - P = 0

P d

Ey = P

B

d

FEC sin 33.69° - 0.6667P = 0 FEC = 1.202P = 1.20P (T)

+ c ©Fy = 0;

P - FED - 1.202P cos 33.69° = 0

FED = 0

Joint A: + c ©Fy = 0; + ©F = 0; : x

FAB sin 26.57° - FAD sin 26.57° = 0 0.6667P - 2F cos 26.57° = 0

FAB = FAD = F

F = 0.3727P

FAB = FAD = F = 0.373P (C) Joint D: + ©F = 0; : x

0.3727P cos 26.57° - FDC cos 26.57° = 0 FDC = 0.3727P = 0.373P (C)

+ c ©Fy = 0;

2(0.3727P sin 26.57°) - FDB = 0 FDB = 0.3333P = 0.333P (T)

Joint B: + ©F = 0; : x

0.3727P cos 26.57° - FBC cos 26.57° = 0 FBC = 0.3727P = 0.373P (C)

+ c ©Fy = 0;

0.3333P - 2(0.3727 sin 26.57°) = 0 (Check)

Maximum tension is in member EC. FEC = 1.202 P = 5 P = 4.16 kN Maximum compression is in members AB, AD, DC, and BC. F = 0.3727 P = 3 P = 8.05 kN Thus, the maximum allowable load is Ans.

P = 4.16 kN

Ans: P = 4.16 kN 512

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6–25. P

Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression. Take P = 2 kN.

P 2m

D

C

30 2m

2m

A

Solution

B 2m

Support Reactions. Not required. Method of Joints: We will start the joint equilibrium analysis at joint C, followed by joint D and finally joint A. Joint C. Fig. a + c ΣFy = 0;  FCB sin 60° - 2 = 0        FCB = 2.309 kN (C) = 2.31 kN (C)     Ans. + ΣFx = 0;  FCD - 2.309 cos 60° = 0  FCD = 1.1547 kN (C) = 1.15 kN (C) Ans. S Joint D. Fig. b + ΣFx = 0;  FDB cos 30° - FDA sin 30° - 1.1547 = 0 S + c ΣFy = 0;  FDA cos 30° - FDB sin 30° - 2 = 0

(1) (2)

Solving Eqs. (1) and (2) FDB = 4.00 kN (T)  FDA = 4.6188 kN (C) = 4.62 kN (C)

Ans.

Joint A. Fig. c + ΣFx = 0;  4.6188 sin 30° - FAB = 0  FAB = 2.3094 kN (C) = 2.31 kN (C)Ans. S + c ΣFy = 0;  - 4.6188 cos 30° + NA = 0  NA = 4.00 kN

Ans: FCB = 2.31 kN (C) FCD = 1.15 kN (C) FDB = 4.00 kN (T) FDA = 4.62 kN (C) FAB = 2.31 kN (C) 513

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6–26. P

The maximum allowable tensile force in the members of the truss is (Ft)max = 5 kN, and the maximum allowable compressive force is (Fc)max = 3 kN. Determine the maximum magnitude P of the two loads that can be applied to the truss.

P 2m

D

C

30 2m

2m

A

Solution

B 2m

Support Reactions. Not required. Method of Joints: We will start the joint equilibrium analysis at joint C, followed by joint D and finally joint A. Joint C. Fig. a + c ΣFy = 0;  FCB sin 60° - P = 0  FCB = 1.1547P (C) + ΣFx = 0;  FCD - 1.1547P cos 60° = 0  FCD = 0.5774P (C) S Joint D. Fig. b + ΣFx = 0;  FDB cos 30° - FDA sin 30° - 0.5774P = 0 S

(1)

+ c ΣFy = 0;  FDA cos 30° - FDB sin 30° - P = 0

(2)

Solving Eqs. (1) and (2), FDA = 2.3094P (C)  FDB = 2.00P (T) Joint C. Fig. c + ΣFx = 0;  2.3094 P sin 30° - FAB = 0  FAB = 1.1547P (C) S + c ΣFy = 0;  NA - 2.3094 p cos 30° = 0  NA = 2.00P By observation, members DA and DB are subjected to maximum compression and tension, respectively. Thus, they will reach the limit first. FDA = (FC)max;  2.3094P = 3  P = 1.299 kN = 1.30 kN(Control!) Ans. FDB = (FC)max;  2.00P = 5  P = 2.50 kN

Ans: P = 1.30 kN 514

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6–27. Determine the force in members DC, HC, and HI of the truss, and state if the members are in tension or compression.

50 kN

40 kN 2m

2m

2m D

E

C

F G

SOLUTION

I

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

H

1.5 m 30 kN

1.5 m B 40 kN 1.5 m

A

40(1.5) + 30(3) + 40(2) - Fy(4) = 0 Fy = 57.5 kN

+ ©F = 0; : x

Ax - 30 - 40 = 0;

Ax = 70 kN

+ c ©Fy = 0;

57.5 - 40 - 50 + Ay = 0;

Ay = 32.5 kN

Method of Sections: Using the bottom portion of the free - body diagram, Fig. b. a + ©MC = 0;

70(3) - 32.5(2) - 40(1.5) - FHI(2) = 0 Ans.

FHI = 42.5 kN (T) a + ©MD = 0;

70(4.5) - 40(3) - 30(1.5) - FHC(1.5) = 0 Ans.

FHC = 100 kN (T) + c ©Fy = 0;

3 32.5 + 42.5 - FDC ( ) = 0 5 Ans.

FDC = 125 kN (C)

Ans: FHI = 42.5 kN (T) FHC = 100 kN (T) FDC = 125 kN (C) 515

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*6–28. Determine the force in members ED, EH, and GH of the truss, and state if the members are in tension or compression.

50 kN

40 kN 2m

2m

2m D

E

C

F G

SOLUTION

I

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

H

1.5 m 30 kN

1.5 m B 40 kN 1.5 m

A

40(1.5) + 30(3) + 40(2) - Fy(4) = 0 Fy = 57.5 kN

+ ©F = 0; : x

Ax - 30 - 40 = 0;

Ax = 70 kN

+ c ©Fy = 0;

57.5 - 40 - 50 + Ay = 0;

Ay = 32.5 kN

Method of Sections: Using the left portion of the free - body diagram, Fig. b. a + ©ME = 0;

- 57.5(2) + FGH(1.5) = 0 Ans.

FGH = 76.7 kN (T) a + ©MH = 0;

- 57.5(4) + FED(1.5) + 40(2) = 0 Ans.

FED = 100 kN (C) + c ©Fy = 0;

3 57.5 - FEH ( ) - 40 = 0 5 Ans.

FEH = 29.2 kN (T)

Ans: FGH = 76.7 kN (T) FED = 100 kN (C) FEH = 29.2 kN (T) 516

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6–29. Determine the force in members HG, HE, and DE of the truss, and state if the members are in tension or compression.

J

K

I

H

G

4 ft A B

SOLUTION

3 ft

Method of Sections: The forces in members HG, HE, and DE are exposed by cutting the truss into two portions through section a–a and using the upper portion of the free-body diagram, Fig. a. From this free-body diagram, FHG and FDE can be obtained by writing the moment equations of equilibrium about points E and H, respectively. FHE can be obtained by writing the force equation of equilibrium along the y axis.

C 3 ft

D 3 ft

F

E 3 ft

3 ft

1500 lb 1500 lb 1500 lb 1500 lb 1500 lb

Joint D: From the free-body diagram in Fig. a, a + ©ME = 0;

FHG(4) - 1500(3) = 0 Ans.

FHG = 1125 lb (T) a + ©MH = 0;

FDE(4) - 1500(6) - 1500(3) = 0 Ans.

FDE = 3375 lb (C) + c ©Fy = 0;

4 FHE a b - 1500 - 1500 = 0 5 Ans.

FEH = 3750 lb (T)

Ans: FHG = 1125 lb (T) FDE = 3375 lb (C) FEH = 3750 lb (T) 517

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6–30. Determine the force in members CD, HI, and CJ of the truss, and state if the members are in tension or compression.

J

K

I

H

G

4 ft A B

SOLUTION

3 ft

Method of Sections: The forces in members HI, CH, and CD are exposed by cutting the truss into two portions through section b–b on the right portion of the free-body diagram, Fig. a. From this free-body diagram, FCD and FHI can be obtained by writing the moment equations of equilibrium about points H and C, respectively. FCH can be obtained by writing the force equation of equilibrium along the y axis. a + ©MH = 0;

3 ft

F

E 3 ft

3 ft

1500 lb 1500 lb 1500 lb 1500 lb 1500 lb

Ans.

FHI(4) - 1500(3) - 1500(6) - 1500(9) = 0 Ans.

FHI = 6750 lb (T) + c ©Fy = 0;

3 ft

D

FCD(4) - 1500(6) - 1500(3) = 0 FCD = 3375 lb (C)

a + ©Mc = 0;

C

4 FCH a b - 1500 - 1500 = 0 5 Ans.

FCH = 5625 lb (C)

Ans: FCD = 3375 lb (C) FHI = 6750 lb (T) FCH = 5625 lb (C) 518

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6–31. Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

4000 lb B

A

8000 lb C

5000 lb D

E

F

G 12 ft

SOLUTION a + ©MC = 0;

L 9 ft

Joint D:

I 9 ft

H 9 ft

9 ft

Ans.

-9500(27) + 4000(18) + 8000(9) + FCD(12) = 0 Ans.

FCD = 9375 lb = 9.38 kip (C) + ©F = 0; : x

J 9 ft

- 9500(18) + 4000(9) + FKJ(12) = 0 F KJ = 11 250 lb = 11.2 kip (T)

a + ©MJ = 0;

K 9 ft

- 9375 + 11 250 -

3 FCJ = 0 5

FCJ = 3125 lb = 3.12 kip (C)

Ans.

FDJ = 0

Ans.

Ans: FKJ = 11.25 kip (T) FCD = 9.375 kip (C) FCJ = 3.125 kip (C) FDJ = 0 519

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*6–32.

Determine the force in members EI and JI of the truss which serves to support the deck of a bridge. State if these members are in tension or compression.

4000 lb B

A

8000 lb

5000 lb

C

D

E

F

G 12 ft

SOLUTION a + ©ME = 0;

L 9 ft

J 9 ft

I 9 ft

H 9 ft

9 ft

- 5000(9) + 7500(18) - FJI(12) = 0 Ans.

FJI = 7500 lb = 7.50 kip (T) + c ©Fy = 0;

K 9 ft

7500 - 5000 - FEI = 0 Ans.

FEI = 2500 lb = 2.50 kip (C)

Ans: FJI = 7.50 kip (T) FEI = 2.50 kip (C) 520

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6–33. The Howe truss is subjected to the loading shown. Determine the force in members GF, CD, and GC, and state if the members are in tension or compression.

5 kN G 5 kN 3m

5 kN F

H

2 kN

2 kN

A

SOLUTION

B

a + ©MA = 0;

Ey(8) - 2(8) - 5(6) - 5(4) - 5(2) = 0

a + ©MD = 0;

4 - FGF (1.5) - 2(2) + 9.5(2) = 0 5

Ey = 9.5kN

FGF = 12.5 kN (C) a + ©MG = 0;

2m

C

2m

E

D

2m

2m

Ans.

9.5(4) - 2(4) - 5(2) - FCD(3) = 0 FCD = 6.67 kN (T)

Ans.

Joint C: + c ©Fy = 0;

Ans.

FGC = 0

Ans: FGF = 12.5 kN (C) FCD = 6.67 kN (T) FGC = 0 521

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6–34. The Howe truss is subjected to the loading shown. Determine the force in members GH, BC, and BG of the truss and state if the members are in tension or compression.

5 kN G 5 kN 3m

5 kN F

H

2 kN

2 kN

A

SOLUTION a + ©MB = 0;

B

- 7.5(2) + FGH sin 36.87°(2) = 0

2m

2m

2m

- 5 (2) + FBG sin 56.31°(2) = 0 Ans.

FBG = 6.01 kN (T) a + ©MH = 0;

2m

E

D

Ans.

FGH = 12.5 kN (C) a + ©MA = 0;

C

- 7.5(4) + 5(2) + FBC(3) = 0 Ans.

FBC = 6.67 kN (T)

Ans: FGH = 12.5 kN (C) FBG = 6.01 kN (T) FBC = 6.67 kN (T) 522

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6–35. Determine the force in members EF, CF, and BC, and state if the members are in tension or compression.

4 kN

E

1.5 m

D

2m

8 kN

F C

Solution

2m

Support Reactions. Not required. Method of Sections. FBC and FEF can be determined directly by writing the moment equations of equilibrium about points F and C, respectively, by referring to the FBD of the upper portion of the truss section through a–a shown in Fig. a. a+ΣMF = 0;  FBC(1.5) - 4(2) = 0   FBC = 5.333 kN (C) = 5.33 kN (C)

Ans.

a+ΣMC = 0;  FEF(1.5) - 4(2) = 0   FEF = 5.333 kN (T) = 5.33 kN (T)

Ans.

A

B

Also, by writing the force equation of equilibrium along the x axis, + ΣFx = 0;  4 - FCF = 0  FCF = 4.00 kN (T) S

Ans.

Ans: FBC = 5.33 kN (C) FEF = 5.33 kN (T) FCF = 4.00 kN (T) 523

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*6–36. Determine the force in members AF, BF, and BC, and state if the members are in tension or compression.

4 kN

E

1.5 m

D

2m

8 kN

F C

Solution

2m

Support Reactions. Not required. Method of Sections: Referring to the FBD of the upper portion of the truss section through a–a shown in Fig. a, FAF and FBC can be determined directly by writing the moment equations of equilibrium about points B and F, respectively.

A

B

a+ΣMB = 0;  FAF(1.5) - 8(2) - 4(4) = 0

Ans.

FAF = 21.33 kN (T) = 21.3 kN (T)

a+ΣMF = 0;  FBC(1.5) - 4(2) = 0  

Ans.

FBC = 5.333 kN (C) = 5.33 kN (C)

Also, write the force equation of equilibrium along the x axis, we can obtain FBF directly. + ΣFx = 0;  4 + 8 - FBF a 3 b = 0  FBF = 20.0 kN (C) S 5

Ans.

Ans: FAF = 21.3 kN (T) FBC = 5.33 kN (C) FBF = 20.0 kN (C) 524

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6–37. Determine the force in members EF, BE, BC and BF of the truss and state if these members are in tension or compression. Set P1 = 9 kN, P2 = 12 kN, and P3 = 6 kN.

E

F

P3

3m D

A

B

C 3m

3m P1

3m P2

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  ND(9) - 12(6) - 9(3) - 6(3) = 0  ND = 13.0 kN Method of sections. Referring to the FBD of the right portion of the truss ­sectioned through a–a shown in Fig. b, FEF and FBC can be determined directly by writing the moment equation of equilibrium about points B and E, respectively. a+ΣMB = 0;  13.0(6) - 12(3) - FEF(3) = 0  FEF = 14.0 kN (C)

Ans.

a+ΣME = 0;  13.0(3) - FBC (3) = 0  FBC = 13.0 kN (T)

Ans.

Also, FBE can be determined by writing the force equation of equilibrium along the y axis. + c ΣFy = 0;  13.0 - 12 - FBE a

1 22

b = 0  FBE = 22 kN (T) = 1.41 kN (T)Ans.

Method of Joints. Using the result of FBE, the equilibrium of joint B, Fig. c, requires + c ΣFy = 0;  FBF + a 22ba

1 22

b - 9 = 0  FBF = 8.00 kN (T)

Ans.

Ans: FEF = FBC = FBE = FBF = 525

14.0 kN (C) 13.0 kN (T) 1.41 kN (T) 8.00 kN (T)

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6–38. Determine the force in members BC, BE, and EF of the truss and state if these members are in tension or compression. Set P1 = 6 kN, P2 = 9 kN, and P3 = 12 kN.

E

F

P3

3m D

A

B

C 3m

3m P1

3m P2

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  ND(9) - 12(3) - 6(3) - 9(6) = 0  ND = 12.0 kN Method of Sections. Referring to the FBD of the right portion of the truss s­ ectioned through a–a shown in Fig. b, FEF and FBC can be determined directly by writing the moment equation of equilibrium about points B and E, respectively. a+ΣMB = 0;  12.0(6) - 9(3) - FEF(3) = 0  FEF = 15.0 kN (C)

Ans.

a+ΣME = 0;  12.0(3) - FBC(3) = 0  FBC = 12.0 kN (T)

Ans.

Also, FBE can be obtained directly by writing the force equation of equilibrium along the y axis + c ΣFy = 0;  12.0 - 9 - FBE a



1 22

b = 0

Ans.

FBE = 322 kN = 4.24 kN (T)

Ans: FEF = 15.0 kN (C) FBC = 12.0 kN (T) FBE = 4.24 kN (T) 526

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6–39. Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression.

4 kN

4 kN

B

C

5 kN 3 kN

2 kN A

D

E

3m F

H

2m

G

5m

SOLUTION a+ ΣME = 0;

5m

5m

5m

- Ay(20) + 2(20) + 4(15) + 4(10) + 5(5) = 0 Ay = 8.25 kN

a+ ΣMH = 0;

-8.25(5) + 2(5) + FBC (3) = 0 Ans.

FBC = 10.4 kN (C) a+ ΣMC = 0;

- 8.25(10) + 2(10) + 4(5) +

5 229

FHG (5) = 0 Ans.

FHG = 9.155 = 9.16 kN (T) a+ ΣMO′ = 0;

- 2(2.5) + 8.25(2.5) - 4(7.5) +

3 234

FHC (12.5) = 0 Ans.

FHC = 2.24 kN (T)

Ans: FBC = 10.4 kN (C) FHG = 9.16 kN (T) FHC = 2.24 kN (T) 527

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*6–40. Determine the force in members CD, CF, and CG and state if these members are in tension or compression.

4 kN

4 kN

B

C

5 kN 3 kN

2 kN A

D

E

3m F

H

2m

G

5m

SOLUTION + ΣFx = 0; S a+ ΣMA = 0;

5m

5m

5m

Ex = 0 -4(5) - 4(10) - 5(15) - 3(20) + Ey(20) = 0 Ey = 9.75 kN

a+ ΣMC = 0;

- 5(5) - 3(10) + 9.75(10) -

5 229

FFG(5) = 0

FFG = 9.155 kN (T) a+ ΣMF = 0;

- 3(5) + 9.75(5) - FCD(3) = 0 Ans.

FCD = 11.25 = 11.2 kN (C) a+ ΣMO′ = 0;

- 9.75(2.5) + 5(7.5) + 3(2.5) -

3 234

FCF (12.5) = 0 Ans.

FCF = 3.21 kN (T) Joint G: + ΣFx = 0; S + c ΣFy = 0;

FGH = 9.155 kN (T) 2 229

(9.155)(2) - FCG = 0 Ans.

FCG = 6.80 kN (C)

Ans: FCD = 11.2 kN (C) FCF = 3.21 kN (T) FCG = 6.80 kN (C) 528

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6–41. Determine the force developed in members FE, EB, and BC of the truss and state if these members are in tension or compression.

2m

1.5 m F

E

B

C

2m

2m A

D

11 kN

Solution

22 kN

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  ND(5.5) - 11(2) - 22(3.5) = 0  ND = 18.0 kN Method of Sections. Referring to the FBD of the right portion of the truss ­sectioned through a–a shown in Fig. b, FBC and FFE can be determined directly by writing the moment equations of equilibrium about point E and B, respectively. a+ΣME = 0;  18.0(2) - FBC(2) = 0  FBC = 18.0 kN (T)

Ans.

a+ΣMB = 0;  18.0(3.5) - 22(1.5) - FFE(2) = 0  FFE = 15.0 kN (C) Ans. Also, FEB can be obtained directly by writing force equation of equilibrium along the y axis 4 + c ΣFy = 0;  FEB a b + 18.0 - 22 = 0  FEB = 5.00 kN (C) Ans. 5

Ans: FBC = 18.0 kN (T) FFE = 15.0 kN (C) FEB = 5.00 kN (C) 529

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6–42. Determine the force in members BC, HC, and HG. State if these members are in tension or compression.

12 kN 6 kN

9 kN

G

H

4 kN

6 kN

J 2m

1m

1m

A

E B 1.5 m

Solution

C 1.5 m

D 1.5 m

1.5 m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, NA can be determined directly by writing the moment equation of equilibrium about point E. a+ΣME = 0;  9(1.5) + 12(3) + 6(4.5) + 4(6) - NA(6) = 0  NA = 16.75 kN Method of Sections. Referring to the FBD of the left portion of the truss sectioned through a–a shown in Fig. b, FHG, FHC and FBC can be determined directly by ­writing the moment equations of equilibrium about points C, A, and H, respectively. a+ΣMC = 0;  FHG a

213

b (3) + 6(1.5) + 4(3) - 16.75(3) = 0

FHG = 4.875213 kN (C) = 17.6 kN (C)

a+ΣMA = 0;  FHC a

2

2 213

b (3) - 6(1.5) = 0 

Ans.

Ans.

FHC = 1.5213 kN (C) = 5.41 kN (C)

a+ΣMH = 0;  FBC(1) + 4(1.5) - 16.75(1.5) = 0 

Ans.

FBC = 19.125 kN (T) = 19.1 kN (T)

Ans: FHG = 17.6 kN (C) FHC = 5.41 kN (C) FBC = 19.1 kN (T) 530

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6–43. Determine the force in members CD, CJ, GJ, and CG and state if these members are in tension or compression.

12 kN 6 kN

9 kN

G

H

4 kN

6 kN

J 2m

1m

1m

A

E B 1.5 m

Solution

C 1.5 m

D 1.5 m

1.5 m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, Ey can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  Ey(6) - 6(6) - 9(4.5) - 12(3) - 6(1.5) = 0  Ey = 20.25 kN + ΣFx = 0;  Ex = 0 S Method of Sections. Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FGJ , FCJ and FCD can be determined directly by writing moment equations of equilibrium about point C, E and J, respectively. a+ΣMC = 0;  20.25(3) - 6(3) - 9(1.5) - FGJa



213

b(3) = 0

FGJ = 4.875 213 kN (C) = 17.6 kN (C)

a+ΣME = 0;  9(1.5) - FCJ a



2

2

213

b(3) = 0

Ans.

Ans.

FCJ = 2.25213 kN (C) = 8.11 kN (C)

a+ΣMJ = 0;  20.25(1.5) - 6(1.5) - FCD(1) = 0

Ans.

FCD = 21.375 kN (T) = 21.4 kN (T)

Method of Joints. Using the result of FGJ to consider the equilibrium of joint G, Fig. c, + ΣFx = 0;  FHG a 3 b - (4.875213)a 3 b = 0  FHG = 4.875213 kN (C) S 213 213

2 + c ΣFy = 0;  2 a4.875213ba b - 12 - FCG = 0  FCG = 7.50 kN (T) 213 Ans.

Ans: FGJ = 17.6 kN (C) FCJ = 8.11 kN (C) FCD = 21.4 kN (T) FCG = 7.50 kN (T) 531

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*6–44. Determine the force in members BE, EF, and CB, and state if the members are in tension or compression.

D 5 kN 5 kN

4m C

E

10 kN 4m

SOLUTION + ©F = 0; : x

B

5 + 10 - FBE cos 45° = 0 Ans.

FBE = 21.2 kN (T) a + ©ME = 0;

4m A

- 5 (4) + FCB (4) = 0 Ans.

FCB = 5 kN (T) a + ©MB = 0;

F

10 kN

G 4m

- 5 (8) - 10 (4) - 5 (4) + FEF (4) = 0 Ans.

FEF = 25 kN (C)

Ans: FBE = 21.2 kN (T) FCB = 5 kN (T) FEF = 25 kN (C) 532

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6–45. Determine the force in members BF, BG, and AB, and state if the members are in tension or compression.

D 5 kN 5 kN

4m C

E

10 kN 4m

SOLUTION

B

Joint F: + ©F = 0; : x

Ans.

FBF = 0

G 4m

5 + 10 + 10 - FBG cos 45° = 0 Ans.

FBG = 35.4 kN (C) a + ©MG = 0;

4m A

Section: + ©F = 0; : x

F

10 kN

FAB (4) - 10 (4) - 10 (8) - 5 (12) = 0 Ans.

FAB = 45 kN (T)

Ans: FBF = 0 FBG = 35.4 kN (C) FAB = 45 kN (T) 533

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6–46. Determine the force in members BC, CH, GH, and CG of the truss and state if the members are in tension or compression.

G H

2m

F E

A

B 4m

C 4m

D 4m

4 kN 8 kN

3m

4m 5 kN

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, Ay can be determined directly by writing the moment equation of equilibrium about point E. a+ΣME = 0;  5(4) + 8(8) + 4(12) - Ay(16) = 0  Ay = 8.25 kN + ΣFx = 0;  Ax = 0 S Method of Sections. Referring to the FBD of the left portion of the truss section through a–a shown in Fig. a, FBC, FGH and FCH can be determined directly by writing the moment equations of equilibrium about points H, C, and O, respectively, Ans.

a+ΣMH = 0;  FBC(3) - 8.25(4) = 0  FBC = 11.0 kN (T) a+ΣMC = 0;  FGHa

1 25

b (10) + (4)(4) - 8.25(8) = 0

Ans.

FGH = 525 kN (C) = 11.2 kN (C)

3 a+ΣMO = 0;  FCH a b(10) + (8.25)(2) - 4(6) = 0  FCH = 1.25 kN (C) 5

Ans.

Method of Joints. Using the result of FGH , equilibrium of joint G, Fig. c, requires + ΣFx = 0;  a525ba 2 b - FGF a 2 b = 0  S 25 25

+ c ΣFy = 0;  2a525ba

1

25

b - FCG = 0 

FGF = a525b kN (C)

FCG = 10.0 kN (T)

Ans.

Ans: FBC = 11.0 kN(T) FGH = 11.2 kN(C) FCH = 1.25 kN(C) FCG = 10.0 kN(T) 534

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6–47. 6 kN

Determine the force in members CD, CJ, and KJ and state if these members are in tension or compression.

6 kN

J

6 kN

6 kN I

K

6 kN

L

3m

H G

A B

C

D

E

F

12 m, 6 @ 2 m

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, Ay can be determined directly by writing the moment equation of equilibrium about point G. a+ΣMG = 0;  6(2) + 6(4) + 6(6) + 6(8) + 6(10) - Ay(12) = 0  Ay = 15.0 kN + ΣFx = 0;   Ax = 0   S Method of Sections. Referring to the FBD of the left portion of the truss sectioned through a - a shown in Fig. b, FCD, FCJ and FKJ can be determined directly by writing the moment equations of equilibrium about points J, A and C, respectively. a+ΣMJ = 0;  FCD(3) + 6(2) + 6(4) - 15.0(6) = 0  FCD = 18.0 kN (T)Ans. a+ΣMA = 0;  FCJ a

a+ΣMC = 0;  FKJ a



3 213 1 25

b(4) - 6(4) - 6(2) = 0

FCJ = 3213 kN (T) = 10.8 kN (T)

b(4) + 6(2) - 15.0 (4) = 0

FKJ = 1225 kN (C) = 26.8 kN (C)

Ans.

Ans.

Ans: FCD = 18.0 kN (T) FCJ = 10.8 kN (T) FKJ = 26.8 kN (T) 535

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*6–48. Determine the force in members JK, CJ, and CD of the truss, and state if the members are in tension or compression.

J

K 3m

I H

L

G

A B 2m

2m

4 kN

SOLUTION

C 2m

5 kN

D 2m

F

E 2m

8 kN

2m

6 kN

Method of Joints: Applying the equations of equilibrium to the free - body diagram of the truss, Fig. a, + ©F = 0; : x

Ax = 0

a + ©MG = 0

6(2) + 8(4) + 5(8) + 4(10) - Ay(12) = 0 Ay = 10.33 kN

Method of Sections: Using the left portion of the free - body diagram, Fig. a. a + ©MC = 0;

FJK(3) + 4(2) - 10.33(4) = 0 Ans.

FJK = 11.111 kN = 11.1 kN (C) a + ©MJ = 0;

FCD(3) + 5(2) + 4(4) - 10.33(6) = 0 Ans.

FCD = 12 kN (T) + c ©Fy = 0;

10.33 - 4 - 5 - FCJ sin 56.31° = 0 Ans.

FCJ = 1.602 kN = 1.60 kN (C)

Ans: FJK = 11.1 kN (C) FCD = 12 kN (T) FCJ = 1.60 kN (C) 536

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6–49. Determine the force in members HI, FI, and EF of the truss, and state if the members are in tension or compression.

J

K 3m

I H

L

G

A B 2m

2m

4 kN

SOLUTION

C 2m

5 kN

D 2m

F

E 2m

8 kN

2m

6 kN

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

NG(2) - 4(2) - 5(4) - 8(8) - 6(10) = 0 NG = 12.67 kN

Method of Sections: Using the right portion of the free - body diagram, Fig. b. a + ©MI = 0;

12.67(4) - 6(2) - FEF(3) = 0 Ans.

FEF = 12.89 kN = 12.9 kN (T) a + ©MG = 0;

-FFI sin 56.31°(2) + 6(2) = 0 Ans.

FFI = 7.211 kN = 7.21 kN (T) a + ©MF = 0;

3 12.67(2) - FHI a b (2) = 0 5 Ans.

FHI = 21.11 kN = 21.1 kN (C)

Ans: FEF = 12.9 kN (T) FFI = 7.21 kN (T) FHI = 21.1 kN (C) 537

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6–50.

Determine the force developed in each member of the space truss and state if the members are in tension or compression. The crate has a weight of 150 lb.

z 6 ft

D

B

SOLUTION F CA = FCA B

C

6 ft

6 ft

A

- 1i + 2j + 2 sin 60°k 28

R

6 ft y

x

= - 0.354 F CA i + 0.707 FCA j + 0.612 FCA k FCB = 0.354FCB i + 0.707FCB j + 0.612FCB k F CD = - FCD j W = - 150 k ©F x = 0;

- 0.354FCA + 0.354FCB = 0

©F y = 0;

0.707FCA + 0.707FCB - FCD = 0

©Fz = 0;

0.612FCA + 0.612FCB - 150 = 0

Solving: FCA = FCB = 122.5 lb = 122 lb (C)

Ans.

FCD = 173 lb (T)

Ans.

F BA = FBA i FBD = FBD cos 60°i + FBD sin 60° k F CB = 122.5 ( -0.354i - 0.707 j - 0.612k) = - 43.3i - 86.6j - 75.0k ©F x = 0;

FBA + FBD cos 60° - 43.3 = 0

©Fz = 0;

FBD sin 60° - 75 = 0

Solving: FBD = 86.6 lb (T)

Ans.

FBA = 0

Ans.

FAC = 122.5(0.354 i - 0.707j - 0.612 k) ©Fz = 0;

FDA cos 30° - 0.612(122.5) = 0 Ans.

FDA = 86.6 lb (T)

Ans: FCA = FCD = FBD = FBA = FDA = 538

FCB = 122 lb (C) 173 lb (T) 86.6 lb (T) 0 86.6 lb (T)

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6–51.

Determine the force in each member of the space truss and state if the members are in tension or compression. Hint: The support reaction at E acts along member EB. Why?

z 2m E

SOLUTION

B

5m

3m C

Method of Joints: In this case, the support reactions are not required for determining the member forces.

3m

Joint A: ©Fz = 0;

FAB ¢

5 229

A

≤ - 6 = 0

y

D x

FAB = 6.462 kN 1T2 = 6.46 kN 1T2

4m

6 kN

Ans.

©Fx = 0;

3 3 FAC a b - FAD a b = 0 5 5

©Fy = 0;

4 4 2 FAC a b + FAD a b - 6.462 ¢ ≤ = 0 5 5 229

FAC = FAD

(1)

FAC + FAD = 3.00

(2)

Solving Eqs. (1) and (2) yields FAC = FAD = 1.50 kN 1C2

Ans.

Joint B: ©Fx = 0;

FBC ¢

©Fz = 0;

FBC ¢

3 238 5 238

≤ - FBD ¢ ≤ + FBD ¢

3 238 5 238

≤ = 0

FBC = FBD

≤ - 6.462 ¢

5 229

(1)

≤ = 0

FBC + FBD = 7.397

(2)

Solving Eqs. (1) and (2) yields FBC = FBD = 3.699 kN 1C2 = 3.70 kN 1C2 ©Fy = 0;

2 B 3.699 ¢

2 238

≤ R + 6.462 ¢

2 229

Ans.

≤ - FBE = 0

FBE = 4.80 kN 1T2

Ans.

Note: The support reactions at supports C and D can be determined by analyzing joints C and D, respectively using the results obtained above. Ans: FAB = FAC = FBC = FBE = 539

6.46 kN (T) FAD = 1.50 kN (C) FBD = 3.70 kN (C) 4.80 kN (T)

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*6–52. z

Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at A, B, C, and D.

2m 2m

D 6 kN 4 kN

C G 4m

A B

Solution E

Support Reactions. Not required Method of Joints. Perform the joint equilibrium analysis at joint G first and then proceed to joint E.

y 4m

x

Joint G. Fig. a   ΣFx = 0;  FGDa

2

2

b - FGCa

b = 0  FGD = FGC = F 25 25 1 1   ΣFy = 0;  F a b + Fa b - 4 = 0  F = 225 kN 25 25 Thus,

FGC = 225 kN (T) = 4.47 kN (T)



FGD = 225 kN (C) = 4.47 kN (C)

  ΣFz = 0;  FGE - 6 = 0

Ans. Ans. Ans.

FGE = 6.00 kN (C)

Joint E. Fig. b

2   ΣFz = 0;  FEDa b - 6.00 = 0   FED = 9.00 kN (T)  3   ΣFx = 0;  FEAa

  ΣFy = 0;  FEAa

2

25 1

25

Solving Eqs. (1) and (2)

b - FEBa

Ans.

2 b - 9.00a b = 0 3 25

b + FEBa

2

(1)

1

1 b - 9.00a b = 0 3 25

(2)

FEA = 325 kN (C) = 6.71 kN (C)  FEB = 0

Ans.

Ans:  FGC = 4.47 kN (T)  FGD = 4.47 kN (C) FGE = 6.00 kN (C)  FED = 9.00 kN (T) FEA = 6.71 kN (C)  FEB = 0 540

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6–53. The space truss supports a force F = 5-500i + 600j + 400k6 lb. Determine the force in each member, and state if the members are in tension or compression.

z

C

F

8 ft

6 ft

SOLUTION Method of Joints: In this case, there is no need to compute the support reactions. We will begin by analyzing the equilibrium of joint C, and then that of joints A and D. x

Joint C: From the free - body diagram, Fig. a, 3 ©Fx = 0; FCA a b - 500 = 0 5

©Fz = 0;

B y

A 6 ft

Ans.

FCA = 833.33 lb = 833 lb(T)

©Fy = 0;

6 ft

6 ft D

3 3 FCB a b - FCD a b + 600 = 0 5 5

(1)

4 4 4 400 - 833.33 a b - FCD a b - FCB a b = 0 5 5 5

(2)

Solving Eqs. (1) and (2) yields

FCB = - 666.67 lb = 667 lb(C)

Ans.

FCD = 333.33 lb = 333 lb(T)

Ans.

Joint A: From the free - body diagram, Fig. b, ©Fx = 0; FAD cos 45° - FAB cos 45° = 0 FAD = FAB = F 3 ©Fy = 0; F sin 45° + F sin 45° - 833.33 a b = 0 5 F = 353.55 lb

Ans.

Thus, FAD = FAB = 353.55 lb = 354 lb(C) ©Fz = 0;

4 833.33 a b - Az = 0 5

Az = 666.67 lb

Joint D: From the free - body diagram, Fig. c, ©Fy = 0;

3 FDB + 333.33a b - 353.55 cos 45° = 0 5

Ans.

FDB = 50 lb(T)

©Fx = 0;

Dx - 353.55 sin 45° = 0 Dx = 250 lb

©Fz = 0;

4 333.33 a b - Dz = 0 5

Dz = 266.67 lb

Note: The equilibrium analysis of joint B can be used to determine the components of support reaction of the ball and socket support at B. 541

Ans: FCA = FCB = FCD = FAD = FDB =

833 lb (T) 667 lb (C) 333 lb (T) FAB = 354 lb (C) 50 lb (T)

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6–54. The space truss supports a force F = 5600i + 450j 750k6 lb. Determine the force in each member, and state if the members are in tension or compression.

z

C

F

8 ft

6 ft

SOLUTION Method of Joints: In this case, there is no need to compute the support reactions. We will begin by analyzing the equilibrium of joint C, and then that of joints A and D. x

Joint C: From the free - body diagram, Fig. a, 3 ©Fx = 0; 600 + FCA a b = 0 5

©Fz = 0;

B y

A 6 ft

Ans.

FCA = - 1000 lb = 1000 lb(C)

©Fy = 0;

6 ft

6 ft D

3 3 FCB a b - FCD a b + 450 = 0 5 5

(1)

4 4 4 -FCB a b - FCD a b -(- 1000) a b - 750 = 0 5 5 5

(2)

Solving Eqs. (1) and (2) yields

FCD = 406.25 lb = 406 lb(T)

Ans.

FCB = - 343.75 lb = 344 lb(C)

Ans.

Joint A: From the free - body diagram, Fig. b, ©Fy = 0;

FAB cos 45° - FAD cos 45° = 0 FAB = FAD = F

©Fx = 0;

3 1000 a b - F sin 45° - F sin 45° = 0 5

F = 424.26 lb Ans.

Thus, FAB = FAD = 424.26 lb = 424 lb (T) 4 ©Fz = 0; Az - 1000 a b = 0 5

Az = 800 lb

Joint D: From the free - body diagram, Fig. c, ©Fy = 0;

3 406.25 a b + 406.25 cos 45° - FDB = 0 5

Ans.

FDB = 543.75 lb = 544 lb(C)

©Fx = 0; 424.26 sin 45° - Dx = 0 Dx = 300 lb 4 ©Fz = 0; 406.25 a b - Dz = 0 5

Dz = 325 lb

Note: The equilibrium analysis of joint B can be used to determine the components of support reaction of the ball and socket support at B. 542

Ans: FCA = FCD = FCB = FAB = FDB =

1000 lb (C) 406 lb (T) 344 lb (C) FAD = 424 lb (T) 544 lb (C)

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6–55. z

Determine the force in members EF, AF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and E.

3 kN

4 kN E

2 kN F 3m

3m

C

D

Solution

3m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,

x

  ΣFx = 0;  Dx - Bx - Ex + 2 = 0 

(1)

  ΣFy = 0;  Ay - 3 = 0

(2)

  ΣFz = 0;  Dz + Bz - 4 = 0 

(3)

  ΣMx = 0;  4(1.5) + 3(3 sin 60°) - Dz(3) = 0 

(4)

  ΣMy = 0;  2(3 sin 60°) + 4(5) - Dz (5) - Ex (3 sin 60°) = 0 

(5)

  ΣMz = 0;  Dx (3) + Ay (5) - Ex (1.5) + 2(1.5) - 3(5) = 0 

(6)

Solving Eqs. (1) to (6)

Ay = 3.00 kN   Dz = 4.5981 kN    Bz = -0.5981 kN



Ex = 0.8490 kN  Dx = - 0.5755 kN  Bx = 0.5755 kN

Method of Joints. We will analyse the equilibrium of the joint at joint D first and then proceed to joint F. Joint D. Fig. b + c ΣFz = 0;  4.5981 - FDF sin 60° = 0   FDF = 5.3094 kN (C) = 5.31 kN (C)Ans.

543

B A

5m

y

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6–55. Continued

Joint F. Fig. c   ΣFx = 0;  2 - FEF -

5 234

(7)

FCF = 0

  ΣFy = 0;  FAF cos 60° + 5.3094 cos 60° - 3 - FCF a   ΣFz = 0;  5.3094 sin 60° - 4 - FAF sin 60° - FCF a

Solving Eqs. (7), (8) and (9)

1.5 234

b = 0

3 sin 60° 234

b = 0

(8) (9)

Ans.

FCF = 0  FEF = 2.00 kN (T) FAF = 0.6906 kN (T) = 0.691 kN (T)

Ans.

Ans: FDF = 5.31 kN (C) FEF = 2.00 kN (T) FAF = 0.691 kN (T) 544

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*6–56. The space truss is used to support the forces at joints B and D. Determine the force in each member and state if the members are in tension or compression.

C

2m B

12 kN

90 D

20 kN

3m F A 2.5 m

Solution Support Reactions. Not required Method of Joints. Analysis of joint equilibrium will be in the sequence of joints D, C, B, A and F. Joint D. Fig. a 4   ΣFx = 0;  20 - FDB a b = 0  FDB = 25.0 kN (T) 5

Ans.

3   ΣFy = 0;  25.0 a b - FDC = 0  FDC = 15.0 kN (T) 5

Ans.

  ΣFz = 0;  FDE - 12 = 0     FDE = 12.0 kN (C)

Ans.

Joint C. Fig. b   ΣFx = 0;  FCB = 0   ΣFy = 0;  FCE a

1 25

Ans. b + 15.0 = 0  FCE = - 1525 kN = 33.5 kN (C) Ans.

  ΣFz = 0;   - FCF - a - 1525ba

Joint B. Fig. c

  ΣFy = 0;  FBE a

2

25

b = 0  FCF = 30.0 kN (T)

3 b - 25.0 a b = 0 5 215.25

1.5

FBE = 10215.25 kN (T) = 39.1 kN (T)

4 2 2   ΣFx = 0;  25.0 a b - a10215.25b a b - FBF a b = 0 5 215.25 213 FBF = 0   ΣFz = 0;   - FBA - a10215.25b a

Ans.

3 215.25

b = 0

FBA = - 30.0 kN = 30.0 kN (C)

545

Ans.

Ans.

Ans.

1.5 m E

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*6–56. Continued

Joint A. Fig. d   ΣFy = 0;  FAE = 0

Ans.

  ΣFx = 0;  FAF = 0

Ans.

Joint F. Fig. e   ΣFy = 0;  FFE = 0

Ans.

Ans: FDB = 25.0 kN (C) FDC = 15.0 kN (T) FDE = 12.0 kN (C) FCE = 33.5 kN (C) FCF = 30.0 kN (T) FBE = 39.1 kN (T) FBF = 0 FBA = 30.0 kN (C) FAE = 0 FAF = 0 FFE = 0 546

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6–57. The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Take F1 = 5 -500k6 lb and F2 = 5400j6 lb.

z C D

F 3 ft

SOLUTION ©Mz = 0;

B

- Cy (3) - 400(3) = 0

E x

Cy = - 400 lb Dx = 0

©M y = 0;

Cz = 0

Joint F:

©Fy = 0;

F1

Ans.

FBF = 0

Joint B: ©F z = 0;

FBC = 0

©Fy = 0;

400 -

Ans.

4 F = 0 5 BE Ans.

FBE = 500 lb (T) FAB -

3 (500) = 0 5 Ans.

FAB = 300 lb (C) Joint A: ©Fx = 0;

300 -

3 234

FAC = 0 Ans.

F AC = 583.1 = 583 lb (T) ©Fz = 0;

3 234

(583.1) - 500 +

3 F = 0 5 AD Ans.

FAD = 333 lb (T) ©Fy = 0;

FAE -

A

3 ft y

©Fx = 0;

©Fx = 0;

4 ft

4 4 (583.1) = 0 (333.3) 5 234 Ans.

FAE = 667 lb (C) Joint E: ©Fz = 0;

FDE = 0

©Fx = 0;

FEF -

Ans.

3 (500) = 0 5 Ans.

FEF = 300 lb (C)

547

F2

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6–57. Continued

Joint C:   ΣFx = 0;  

3 234

(583.1) - FCD = 0 Ans.

FCD = 300 lb (C)   ΣFz = 0;  FCF

3 234

(583.1) = 0 Ans.

FCF = 300 lb (C)

  ΣFy = 0;   Joint F:   ΣFx = 0;  

4 234 3 218

(583.1) - 400 = 0Check!

FDF - 300 = 0 Ans.

FDF = 424 lb (T)   ΣFz = 0;  

3 218

(424.3) - 300 = 0Check!

Ans: FBF = 0 FBC = 0 FBE = 500 lb (T) FAB = 300 lb (C) FAC = 583 lb (T) FAD = 333 lb (T) FAE = 667 lb (C) FDE = 0 FEF = 300 lb (C) FCD = 300 lb (C) FCF = 300 lb (C) FDF = 424 lb (T) 548

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6–58. The space truss is supported by a ball-and-socket joint at D and short links at C and E. Determine the force in each member and state if the members are in tension or compression. Take F1 = 5200i + 300j - 500k6 lb and F 2 = 5400j6 lb.

z C D

F 3 ft

SOLUTION ©Fx = 0;

B

Dx + 200 = 0

E x

Dx = - 200 lb ©Mz = 0;

F1

Cz (3) - 200(3) = 0

FBF = 0

Ans.

©F z = 0;

FBC = 0

Ans.

©F y = 0;

400 -

Joint B:

4 F = 0 5 BE Ans.

FBE = 500 lb (T) ©Fx = 0;

FAB -

3 (500) = 0 5 Ans.

F AB = 300 lb (C) Joint A: ©Fx = 0;

300 + 200 -

3 234

FAC = 0 Ans.

F AC = 971.8 = 972 lb (T) ©Fz = 0;

3 234

(971.8) - 500 +

3 FAD = 0 5 Ans.

FAD = 0 ©Fy = 0;

FAE + 300 -

4 234

3 ft y

Cz = 200 lb Joint F:

A

-Cy (3) - 400(3) - 200(4) = 0 Cy = - 666.7 lb

©My = 0;

4 ft

(971.8) = 0

FAE = 367 lb (C)

Ans.

©Fz = 0;

FDE = 0

Ans.

©Fx = 0;

FEF -

Joint E:

3 (500) = 0 5 Ans.

FEF = 300 lb (C)

549

F2

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6–58. Continued

Joint C: ©Fx = 0;

3 234

(971.8) - FCD = 0 Ans.

FCD = 500 lb (C) ©Fz = 0;

FCF -

3 234

(971.8) + 200 = 0

FCF = 300 lb (C) ©F y = 0;

4 234

(971.8) - 666.7 = 0

Ans. Check!

Joint F: ©Fx = 0;

3 218

FDF - 300 = 0 Ans.

FDF = 424 lb (T)

Ans: FBF = 0 FBC = 0 FBE = 500 lb (T) FAB = 300 lb (C) FAC = 972 lb (T) FAD = 0 FAE = 367 lb (C) FDE = 0 FEF = 300 lb (C) FCD = 500 lb (C) FCF = 300 lb (C) FDF = 424 lb (T) 550

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6–59. Determine the force in each member of the space truss and state if the members are in tension or compression. The The truss is supported by ball-and-socket joints at A, B, and E. Set F = 5800j6 N. Hint: The support reaction at E acts along member EC. Why?

z

D F

1m

A

2m C

5m

y

E

x

2m

B 1.5 m

Ans: FAD = FBD = FCD = FBC = FAC = FEC = 551

686 N (T) 0 615 N (C) 229 N (T) 343 N (T) 457 N (C)

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*6–60. Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at A, B, and E. Set F = 5 -200i + 400j6 N. Hint: The support reaction at E acts along member EC. Why?

z

D F

1m

A

2m C

5m E

SOLUTION Joint D:

x

©Fx = 0;

1 5 1 - FAD + FBD + FCD - 200 = 0 3 231.25 27.25

©Fy = 0;

2 1.5 1.5 - FAD + FBD FCD + 400 = 0 3 231.25 27.25

©Fz = 0;

2 2 2 - FAD FBD + FCD = 0 3 231.25 27.25 FAD = 343 N (T)

Ans.

FBD = 186 N (T)

Ans.

FCD = 397.5 = 397 N (C)

Ans.

Joint C: ©Fx = 0;

FBC -

1 27.25

(397.5) = 0 Ans.

FBC = 148 N (T) ©Fy = 0;

1.5 27.25

(397.5) - FAC = 0 Ans.

FAC = 221 N (T) ©Fz = 0;

FEC -

2 27.25

(397.5) = 0 Ans.

FEC = 295 N (C)

552

2m

B 1.5 m

y

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6–61. Determine the force P 100-lb weight in equilibrium.

required

to

hold

the

D

C

SOLUTION Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of pulley A on the free - body diagram, Fig. a, + c ©Fy = 0;

2TA - 100 = 0

B A

P

TA = 50 lb

Applying ©Fy = 0 to the free - body diagram of pulley B, Fig. b, + c ©Fy = 0;

2TB - 50 = 0

TB = 25 lb

From the free - body diagram of pulley C, Fig. c, + c ©Fy = 0;

2P - 25 = 0

P = 12.5 lb

Ans.

Ans: P = 12.5 lb 553

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6–62.

In each case, determine the force P required to maintain equilibrium. The block weighs 100 lb. P P

P

SOLUTION Equations of Equilibrium: a)

+ c ©Fy = 0;

(a)

4P - 100 = 0 P = 25.0 lb

b)

+ c ©Fy = 0;

+ c ©Fy = 0;

(c)

Ans.

3P - 100 = 0 P = 33.3 lb

c)

(b)

Ans.

3P¿ - 100 = 0 P¿ = 33.33 lb

+ c ©Fy = 0;

3P - 33.33 = 0 Ans.

P = 11.1 lb

Ans: P = 25.0 lb P = 33.3 lb

P = 11.1 lb

554

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6–63. Determine the force P required to hold the 50-kg mass in equilibrium.

C B A

SOLUTION

P

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of each pulley. + c ©Fy = 0;

R - 3P = 0;

R = 3P

+ c ©Fy = 0;

T - 3R = 0;

T = 3R = 9P

+ c ©Fy = 0;

2P + 2R + 2T - 50(9.81) = 0

Ans.

Substituting Eqs.(1) and (2) into Eq.(3) and solving for P, 2P + 2(3P) + 2(9P) = 50(9.81) Ans.

P = 18.9 N

Ans: P = 18.9 N 555

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*6–64. Determine the force P required to hold the 150-kg crate in equilibrium. B

SOLUTION

A

C

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of pulley A on the free - body diagram, Fig. a, + c ©Fy = 0;

P

2TA - 150(9.81) = 0 TA = 735.75 N

Using the above result and writing the force equation of equilibrium along the y¿ axis of pulley C on the free - body diagram in Fig. b, ©Fy¿ = 0;

735.75 - 2P = 0

Ans.

P = 367.88 N = 368 N

Ans: P = 368 N 556

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6–65. Determine the horizontal and vertical components of force that pins A and B exert on the frame.

C 2 kN/m

4m

B

A

Solution Free Body Diagram. The frame will be dismembered into members BC and AC. The solution will be very much simplified if one recognizes that member AC is a two force member. The FBDs of member BC and pin A are shown in Figs. a and b, respectively.

3m

Equations of Equilibrium. Consider the equilibrium of member BC, Fig. a, 3 a+ΣMB = 0;  2(4)(2) - FAC a b(4) = 0  FAC = 6.6667 kN 5 a+ΣMC = 0;  Bx (4) - 2(4)(2) = 0  Bx = 4.00 kN

4 + c ΣFy = 0;   6.6667 a b - By = 0  By = 5.333 kN = 5.33 kN 5

Ans. Ans.

Then, the equilibrium of pin A gives

+ ΣFx = 0;  Ax - 6.6667 a 3 b = 0  Ax = 4.00 kN S 5

Ans.

4 + c ΣFy = 0;  Ay - 6.6667 a b = 0  Ay = 5.333 kN = 5.33 kN 5

Ans.

Ans: Bx = By = Ax = Ay = 557

4.00 kN 5.33 kN 4.00 kN 5.33 kN

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6–66. Determine the horizontal and vertical components of force at pins A and D.

D

2m

0.3 m C A

1.5 m

B

1.5 m E

Solution Free Body Diagram. The assembly will be dismembered into member AC, BD and pulley E. The solution will be very much simplified if one recognizes that member BD is a two force member. The FBD of pulley E and member AC are shown in Fig. a and b respectively.

12 kN

Equations of Equilibrium. Consider the equilibrium of pulley E, Fig. a, + c ΣFy = 0;  2T - 12 = 0  T = 6.00 kN Then, the equilibrium of member AC gives 4 a+ΣMA = 0;  FBD a b(1.5) + 6(0.3) - 6(3) - 6(3.3) = 0 5 FBD = 30.0 kN + ΣFx = 0;  Ax - 30.0 a 3 b - 6 = 0   S 5

Ans.

Ax = 24.0 kN

4 + c ΣFy = 0;  30.0 a b - 6 - 6 - Ay = 0  Ay = 12.0 kN 5

Ans.

Thus,

FA = 2A2x + A2y = 224.02 + 12.02 = 26.83 kN = 26.8 kN FB = FBD = 30.0 kN



Dx =

3 (30) = 18.0 kN  5

Ans.



Dy =

4 (30) = 24.0 kN  5

Ans.

Ans: Ax = Ay = Dx = Dy = 558

24.0 kN 12.0 kN 18.0 kN 24.0 kN

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6–67. Determine the force that the smooth roller C exerts on member AB. Also, what are the horizontal and vertical components of reaction at pin A? Neglect the weight of the frame and roller.

60 lb ft

C

D

A

0.5 ft B 3 ft

4 ft

SOLUTION a + ©MA = 0;

-60 + Dx (0.5) = 0 Dx = 120 lb

+ ©F = 0; : x

Ax = 120 lb

Ans.

+ c ©Fy = 0;

Ay = 0

Ans.

a + ©MB = 0;

-NC (4) + 120(0.5) = 0 Ans.

NC = 15.0 lb

Ans: Ax = 120 lb Ay = 0 NC = 15.0 lb 559

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*6–68. The bridge frame consists of three segments which can be considered pinned at A, D, and E, rocker supported at C and F, and roller supported at B. Determine the horizontal and vertical components of reaction at all these supports due to the loading shown.

2 kip/ft A

E 15 ft

C 5 ft

For segment BD: 2(30)(15) - By (30) = 0

+ ©F = 0; : x

Dx = 0

+ c ©Fy = 0;

Dy + 30 - 2(30) = 0

30 ft

D

15 ft

20 ft

SOLUTION a + ©MD = 0;

B

F 5 ft

Ans.

By = 30 kip

Ans. Ans.

Dy = 30 kip

For segment ABC: a + ©MA = 0;

Cy (5) - 2(15)(7.5) - 30(15) = 0

+ ©F = 0; : x + c ©Fy = 0;

Ans.

Cy = 135kip

Ans.

Ax = 0 -Ay + 135 - 2(15) - 30 = 0

Ans.

Ay = 75 kip

For segment DEF: a + ©Mg = 0; + ©F = 0; : x + c ©Fy = 0;

-Fy(5) + 2(15)(7.5) + 30(15) = 0

Ans.

Fy = 135 kip

Ans.

Ex = 0 -Ey + 135 - 2(15) - 30 = 0

Ans.

Ey = 75 kip

Ans: By = 30 kip Dx = 0 Dy = 30 kip Cy = 135 kip Ay = 75 kip Fy = 135 kip Ex = 0 Ey = 75 kip 560

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6–69. Determine the reactions at supports A and B.

700 lb/ ft B D

500 lb/ ft

9 ft C

A 6 ft

SOLUTION

6 ft

6 ft

8 ft

Member DB: a + ©MB = 0;

3 3.15 (6) - FCD a b (9) = 0 5 FCD = 3.50 kip

+ ©F = 0; : x

4 -Bx + 3.50a b = 0 5 Ans.

Bx = 2.80 kip + c ©Fy = 0;

3 By - 3.15 + 3.50 a b = 0 5 Ans.

By = 1.05 kip Member AC: + ©F = 0; : x

4 Ax - 3.50 a b = 0 5 Ans.

Ax = 2.80 kip + c ©Fy = 0;

3 Ay - 3 - 3.50 a b = 0 5 Ans.

Ay = 5.10 kip a + ©MA = 0;

3 MA - 3(6) - 3.50 a b (12) = 0 5 MA = 43.2 kip # ft

Ans.

Ans: Bx = 2.80 kip By = 1.05 kip Ax = 2.80 kip Ay = 5.10 kip MA = 43.2 kip # ft 561

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6–70. Determine the horizontal and vertical components of force at pins B and C. The suspended cylinder has a mass of 75 kg.

0.3 m B

C 1.5 m

Solution

A

Free Body Diagram. The solution will be very much simplified if one realizes that member AB is a two force member. Also, the tension in the cable is equal to the weight of the cylinder and is constant throughout the cable. Equations of Equilibrium. Consider the equilibrium of member BC by referring to its FBD, Fig. a, 3 a+ΣMC = 0;  FAB a b(2) + 75(9.81)(0.3) - 75(9.81)(2.8) = 0 5

2m

0.5 m

FAB = 1532.81 N



a+ΣMB = 0;  Cy (2) + 75(9.81)(0.3) - 75(9.81)(0.8) = 0

Ans.

Cy = 183.94 N = 184 N

+ ΣFx = 0;  1532.81a 4 b - 75(9.81) - Cx = 0 S 5 Cx = 490.5 N

Ans.

Thus,

FB = FAB = 1532.81 N



Bx =

4 (1532.81) = 1226.25 N = 1.23 kN 5

Ans.



By =

3 (1532.81) = 919.69 N = 920 kN 5

Ans.

Ans: Cy = Cx = Bx = By = 562

184 N 490.5 N 1.23 kN 920 kN

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6–71. 12 kN

Determine the reactions at the supports A, C, and E of the compound beam.

3 kN/ m

A 3m

B

C 4m

2m

D

E 6m

3m

Solution Free Body Diagram. The compound beam is being dismembered into members AB, BD and DE of which their respective FBDs are shown in Fig. a, b and c. Equations of Equilibrium. Equilibrium of member DE will be considered first by referring to Fig. c. Ans.

a+ΣMD = 0;  NE(6) - 12(9) = 0  NE = 18.0 kN a+ΣME = 0;  Dy(6) - 12(3) = 0 

Dy = 6.00 kN

+ ΣFx = 0;  Dx = 0 S Next, member BD, Fig. b. a+ΣMC = 0;  6.00(2) + 3(6)(1) - By(4) = 0 

 By = 7.50 kN

a+ΣMB = 0;  NC(4) + 6.00(6) - 3(6)(3) = 0  NC = 4.50 kN

Ans.

+ ΣFx = 0;  Bx = 0 S Finally, member AB, Fig. a + ΣFx = 0;  Ax = 0 S + c ΣFy = 0;  Ay - 7 -50 = 0 

Ans. Ans.

 Ay = 7.50 kN

a+ΣMA = 0;  MA - 7.50(3) = 0  MA = 22.5 kN # m

Ans.

Ans: NE = 18.0 kN NC = 4.50 kN Ax = 0 Ay = 7.50 kN MA = 22.5 kN # m 563

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*6–72. Determine the resultant force at pins A, B, and C on the three-member frame.

2m 800 N C 200 N/ m 2m

A

60 B

Solution Free Body Diagram. The frame is being dismembered into members AC and BC of which their respective FBDs are shown in Fig. a and b. Equations of Equilibrium. Write the moment equation of equilibrium about point A for member AC, Fig. a and point B for member BC, Fig. b. a+ΣMA = 0;  Cy a

2 2 1 b + Cx (2) - 200 a ba b = 0 tan 60° sin 60° sin 60°

a+ΣMB = 0;  Cy(2) - Cx(2) + 800(2) = 0

(1) (2)

Solving Eqs. (1) and (2) Cy = - 338.12 N  Cx = 461.88 N



The negative sign indicates that Cy acts in the sense opposite to that shown in the FBD write the force equation of equilibrium for member AC, Fig. a, 2 b sin 60° - 461.88 = 0   Ax = 61.88 N sin 60° 2 + c ΣFy = 0;  Ay + ( - 338.12) - 200 a b cos 60° = 0  Ay = 569.06 N sin 60° + ΣFx = 0;  Ax + 200 a S

Also, for member BC, Fig. b

+ ΣFx = 0;  Bx + 461.88 - 800 = 0  Bx = 338.12 N S + c ΣFy = 0;   - By - ( - 338.12) = 0   By = 338.12 N Thus,

FC = 2C 2x + C 2y = 2461.88 2 + ( - 338.12)2 = 572.41 N = 572 N

Ans.

FB = 2B2x + B2y = 2338.122 + 338.122 = 478.17 N = 478 N

Ans.

FA = 2A2x + A2y = 261.882 + 569.062 = 572.41 N = 572 N

Ans.

Ans: FC = 572 N FA = 572 N FB = 478 N 564

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6–73. 900 N/m

Determine the reactions at the supports at A, E, and B of the compound beam.

900 N/m B

A 3m

3m

E

D

C 4m

3m

3m

Solution Free Body Diagram. The solution will be very much simplified if one realizes that member CD is a two force member. Equation of Equilibrium. Consider the equilibrium of member BD, Fig. b + ΣFx = 0;  FCD = 0 S 1 a+ΣMB = 0;   (900)(6)(4) - NE (3) = 0  NE = 3600 N = 3.60 kN 2 1 a+ΣME = 0;   (900)(6)(1) - NB (3) = 0  NB = 900 N 2

Ans. Ans.

Then the equilibrium of member AC gives + ΣFx = 0;  Ax = 0 S

Ans.

1 (900)(6) = 0  Ay = 2700 N = 2.70 kN Ans. 2 1 a+ΣMA = 0;  MA - (900)(6)(3) = 0  MA = 8100 N # m = 8.10 kN # m Ans. 2 + c ΣFy = 0;  Ay -

Ans:  NE =  NB = Ax = Ay = MA = 565

3.60 kN 900 N 0 2.70 kN 8.10 kN # m

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6–74. The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W?

D

4 ft 4 ft

4 ft C

SOLUTION A

Pulley E: + c ©Fy = 0;

2T - 700 = 0

E

60

Ans.

T = 350 lb

W

Member ABC: a + ©MA = 0;

B

700 lb

TBD sin 45°(4) - 350 sin 60°(4) # 700 (8) = 0 TBD = 2409 lb

+ c ©Fy = 0; + : ©Fx = 0;

- A y + 2409 sin 45° - 350 sin 60° - 700 = 0 Ans.

A y = 700 lb - A y - 2409 cos 45° - 350 cos 60° + 350 - 350 = 0

Ans.

A x = 1.88 kip At D: Dx = 2409 cos 45° = 1703.1 lb = 1.70 kip

Ans.

Dy = 2409 sin 45° = 1.70 kip

Ans.

Ans: T = 350 lb Ay = 700 lb Ax = 1.88 kip Dx = 1.70 kip Dy = 1.70 kip 566

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6–75. The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable at the winch W? The jib ABC has a weight of 100 lb and member BD has a weight of 40 lb. Each member is uniform and has a center of gravity at its center.

D

4 ft 4 ft

4 ft C

A

SOLUTION

B

Pulley E: + c ©Fy = 0;

60

2T - 700 = 0 Ans.

T = 350 lb

W 700 lb

Member ABC: a + ©MA = 0;

E

By (4) - 700 (8) - 100 (4) - 350 sin 60° (4) = 0 By = 1803.1 lb

+ c ©Fy = 0;

-Ay - 350 sin 60° - 100 - 700 + 1803.1 = 0 Ans.

Ay = 700 lb + ©F = 0; : x

Ax - 350 cos 60° - Bx + 350 - 350 = 0 (1)

Ax = Bx + 175 Member DB: a + ©MD = 0;

- 40 (2) - 1803.1 (4) + Bx (4) = 0 Bx = 1823.1 lb

+ ©F = 0; : x

-Dx + 1823.1 = 0 Ans.

Dx = 1.82 kip + c ©Fy = 0;

Dy - 40 - 1803.1 = 0 Dy = 1843.1 = 1.84 kip

Ans.

Ax = 2.00 kip

Ans.

From Eq. (1)

Ans: T = 350 lb Ay = 700 lb Dx = 1.82 kip Dy = 1.84 kip Ax = 2.00 kip 567

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*6–76. Determine the horizontal and vertical components of force which the pins at A and B exert on the frame.

2m

400 N/m

C

D 1.5 m E

3m

3m

Solution

F

Free Body Diagram. The frame will be dismembered into members AD, EF, CD and BC. The solution will be very much simplified if one realizes that members CD and EF are two force member. Therefore, only the FBD of members AD and BC, Fig. a and b respectively, need to be drawn

1.5 m A

B

Equations of Equilibrium. Write the moment equations of equilibrium about point A for member AD, Fig. a, and point B for member BC, Fig. b. 4 a+ΣMA = 0;  FEF a b(3) - FCD (4.5) - 400 (4.5)(2.25) = 0 5 4 a+ΣMB = 0;   - FEF a b(1.5) + FCD (4.5) = 0 5

(1) (2)

Solving Eqs. (1) and (2)

FEF = 3375 N  FCD = 900 N

Write the force equation of equilibrium for member AD, Fig. a, + ΣFx = 0;  Ax + 400(4.5) + 900 - 3375 a 4 b = 0  Ax = 0 S 5

3 + c ΣFy = 0;  3375 a b - Ay = 0  Ay = 2025 N = 2.025 kN 5

Ans. Ans.

Also, for member BC, Fig. b

+ ΣFx = 0;  3375a 4 b - 900 - Bx = 0  Bx = 1800 N = 1.80 kN S 5 + c ΣFy = 0;  

3 By - 3375a b = 0 5

    By = 2025 N = 2.025 kN

Ans. Ans.

Ans: Ax = Ay = Bx = By = 568

0 2.025 kN 1.80 kN 2.025 kN

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6–77. The two-member structure is connected at C by a pin, which is fixed to BDE and passes through the smooth slot in member AC. Determine the horizontal and vertical components of reaction at the supports.

500 lb B C

D

E

4 ft

SOLUTION

A

Member AC: a + ©MA = 0;

600 lb ft 3 ft

NC (5) - 600 = 0

3 ft

2 ft

NC = 120 lb + ©F = 0; : x

4 Ax - 120a b = 0 5 Ans.

Ax = 96 lb + c ©Fy = 0;

3 - Ay + 120 a b = 0 5 Ans.

Ay = 72 lb Member BDE: a + ©Mg = 0;

3 500 (8) + 120 a b (5) - Dy (2) = 0 5 Ans.

Dy = 2180 lb = 2.18 kip + ©F = 0; : x

4 - Ex + 120 a b = 0 5 Ans.

Ex = 96 lb + c ©Fy = 0;

3 - 500 - 120 a b + 2180 - Ey = 0 5 Ans.

Ey = 1608 lb = 1.61 kip

Ans: Ax = Ay = Dy = Ex = Ey = 569

96 lb 72 lb 2.18 kip 96.0 lb 1.61 kip

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6–78. 2 kN/m

The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports.

A

C D

B 6m

3m

3m

Solution Free Body Diagram. The compound beam will be dismembered into members ABD and CD of which their respective FBD are shown in Fig. a and b. Equations of Equilibrium. First, consider the equilibrium of member CD, Fig. b, a+ΣMD = 0;  NC (3) - 2(3)(1.5) = 0  NC = 3.00 kN

Ans.

a+ΣMC = 0;  2(3)(1.5) - Dy (3) = 0  Dy = 3.00 kN

Ans.

+ ΣFx = 0;  Dx = 0 S Next, the equilibrium of member ABD gives, a+ΣMB = 0;  2(9)(1.5) - 3.00(3) - NA(6) = 0  NA = 3.00 kN

Ans.

a+ΣMA = 0;   By (6) - 2(9)(4.5) - 3.00(9) = 0  By = 18.0 kN

Ans.

+ ΣFx = 0;         Bx = 0  S

Ans.

Ans: NC = NA = By = Bx = 570

3.00 kN 3.00 kN 18.0 kN 0

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6–79. a/2

The toggle clamp is subjected to a force F at the handle. Determine the vertical clamping force acting at E. B A

E

SOLUTION

1.5 a a/2

C 60

D

F

a/2

1.5 a

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CD is a two force member. Equations of Equilibrium: From FBD (a), a + ©MB = 0;

a a FCD cos 30°a b - FCD sin 30°a b - F12a2 = 0 2 2 FCD = 10.93F

+ ©F = 0; : x

Bx - 10.93 sin 30° = 0 Bx = 5.464F

From (b), a + ©MA = 0;

5.464F1a2 - FE 11.5a2 = 0 F E = 3.64F

Ans.

Ans: NC = ND = 2 lb 571

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*6–80. When a force of 2 lb is applied to the handles of the brad squeezer, it pulls in the smooth rod AB. Determine the force P exerted on each of the smooth brads at C and D.

2 lb 0.25 in. P 1.5 in.

Equations of Equilibrium: Applying the moment equation of equilibrium about point E to the free-body diagram of the lower handle in Fig. a, we have + ©ME = 0;

E C

1.5 in. B D P

SOLUTION

2 in.

1 in. A 2 in.

2(2) - FAB(1) = 0

2 lb

FAB = 4 lb Using the result of FAB and considering the free-body diagram in Fig. b, + ©MB = 0; + ©F = 0; : x

NC(1.5) - ND(1.5) = 0 NC = ND

(1)

4 - NC - ND = 0

(2)

Solving Eqs. (1) and (2) yields Ans.

NC = ND = 2 lb

Ans: NC = ND = 2 1b 572

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6–81. The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hydraulic cylinder H.

1m G

F

2m E

2m

SOLUTION

H

Free Body Diagrams: The solution for this problem will be simplified if one realizes that members FB and DB are two-force members.

1226.25(3) - FFB ¢

1938.87 ¢

3 210

B

A

3 210

≤ (2) = 0

2m

1m

Ans.

FFB = 1938.87 N = 1.94 kN + c ©Fy = 0;

1m C

Equations of Equilibrium: For FBD(a), a + ©ME = 0;

D

≤ - 1226.25 - Ey = 0

Ey = 613.125N + ©F = 0; : x

Ex - 1938.87 ¢

1 210

≤ = 0

Ex = 613.125 N From FBD (b), a + ©MC = 0;

613.125(3) - FBD sin 45°(1) = 0 Ans.

FBD = 2601.27 N = 2.60 kN

Ans: FFB = 1.94 kN FBD = 2.60 kN 573

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6–82. A 5-lb force is applied to the handles of the vise grip. Determine the compressive force developed on the smooth bolt shank A at the jaws.

5 lb

0.75 in. A

B D

1 in. E

SOLUTION

20 1.5 in.

From FBD (a) a + ©ME = 0;

5(4) - FCD sin 30.26°(1) = 0

+ ©F = 0; : x

Ex - 39.693 cos 30.26° = 0

C

1 in.

3 in. 5 lb

FCD = 39.693 lb Ex = 34.286 lb

From FBD (b) a + ©MB = 0;

NA sin 20° (0.75) + NA cos 20° (1.5) - 34.286(1.75) = 0 Ans.

NA = 36.0 lb

Ans: NA = 36.0 lb 574

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6–83. Determine the force in members FD and DB of the frame. Also, find the horizontal and vertical components of reaction the pin at C exerts on member ABC and member EDC.

E

G

F

6 kN

2m

D

C 2m

Solution

1m

B

A

1m

Free Body Diagram. The assembly will be dismembered into members GFE, EDC, FD, BD and ABC. The solution will be very much simplified if one recognizes members FD and BD are two force members. The FBDs of members GFE, EDC and ABC are shown in Figs. a, b and c respectively. Equations of Equilibrium. First, consider the equilibrium of member GFE, Fig. a, 2 b(1) = 0  FFD = 925 kN = 20.1 kN 25 a+ΣMF = 0;  6(2) - Ey (1) = 0 Ey = 12.0 kN

a+ΣME = 0;  6(3) - FFD a

+ ΣFx = 0;  Ex - a925ba 1 b = 0 S 25

Ans.

Ex = 9.00 kN

Next, for member EDC, Fig. b,

a+ΣMC = 0;  9.00(3) - a925b a



1 25

b(1) - FBD a

1 22

b(1) = 0

FBD = 1822 kN = 25.5 kN

Ans.

a+ΣMD = 0;  9.00(2) - C′x (1) = 0  C′x = 18.0 kN



+ c ΣFy = 0;  12.0 + a1822ba

1 22

b - a925ba

2

25 C′y = 12.0 kN

Ans.

b - C′y = 0

Ans.

Finally, for member ABC, Fig. c a+ΣMA = 0;  C″y (3) - a1822b a

1 22

b(2) = 0  Cy″ = 12.0 kN

+ ΣFx = 0;  Cx″ - a1822ba 1 b = 0    Cx″ = 18.0 kN S 22

Ans. Ans.

Ans: FFD = 20.1 kN FBD = 25.5 kN Cx″ = 18.0 kN Cy″ = 12.0 kN 575

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*6–84. Determine the force that the smooth 20-kg cylinder exerts on members AB and CDB. Also, what are the horizontal and vertical components of reaction at pin A?

D

C

1m A E 1.5 m

B 2m

Solution Free Body Diagram. The FBDs of the entire assembly, member CDB and the cylinder are shown in Figs. a, b and c, respectively. Equations of Equilibrium. First consider the equilibrium of the entire assembly, Fig. a, a+ΣMA = 0;  ND(1) - 20(9.81)(1.5) = 0  ND = 294.3 N + ΣFx = 0;  Ax - 294.3 = 0      Ax = 294.3 N = 294 N S

Ans.

+ c ΣFy = 0;  Ay - 20(9.81) = 0 

Ans.

Ay = 196.2 N = 196 N

Next, for member CDB, Fig. b a+ΣMB = 0;  294.3(1) - NC (2) = 0  NC = 147.15 N = 147 N

Ans.

Finally for the cylinder, Fig. c + c ΣFy = 0;  NE - 20(9.81) - 147.15 = 0  NE = 343.35 N = 343 N

Ans.

Ans: Ax = 294 N Ay = 196 N NC = 147 N NE = 343 N 576

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6–85.

The three power lines exert the forces shown on the truss joints, which in turn are pin-connected to the poles AH and EG. Determine the force in the guy cable AI and the pin reaction at the support H.

20 ft

20 ft E

A

20 ft B

40 ft

800 lb

C

800 lb

40 ft

D

800 lb 125 ft

SOLUTION AH is a two - force member. I

Joint B: + c ©Fy = 0;

H 50 ft

30 ft

30 ft

F

G 30 ft

30 ft

50 ft

FAB sin 45° - 800 = 0 FAB = 1131.37 lb

Joint C: + c ©Fy = 0;

2FCA sin 18.435° - 800 = 0 FCA = 1264.91 lb

Joint A: + ©F = 0; : x

- TAI sin 21.801° - FH cos 76.504° + 1264.91 cos 18.435° + 1131.37 cos 45° = 0

+ c ©Fy = 0;

- TAI cos 21.801° + FH sin 76.504° - 1131.37 sin 45° - 1264.91 sin 18.435° = 0 TAI(0.3714) + FH(0.2334) = 2000 -TAI(0.9285) + FH(0.97239) = 1200

Solving, TAI = TEF = 2.88 kip

Ans.

FH = FG = 3.99 kip

Ans.

Ans: TAI = 2.88 kip FH = 3.99 kip 577

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6–86.

The pumping unit is used to recover oil. When the walking beam ABC is horizontal, the force acting in the wireline at the well head is 250 lb. Determine the torque M which must be exerted by the motor in order to overcome this load. The horse-head C weighs 60 lb and has a center of gravity at GC. The walking beam ABC has a weight of 130 lb and a center of gravity at GB, and the counterweight has a weight of 200 lb and a center of gravity at GW. The pitman, AD, is pin connected at its ends and has negligible weight.

6 ft

5 ft GB A

C

B

70°

M

1 ft GC

D

Gw

20°

E 250 lb

SOLUTION Free-Body Diagram: The solution for this problem will be simplified if one realizes that the pitman AD is a two force member.

3 ft 2.5 ft

Equations of Equilibrium: From FBD (a), a + ©MB = 0;

FAD sin 70°152 - 60162 - 250172 = 0 FAD = 449.08 lb

From (b), a + ©ME = 0;

449.08132 - 200 cos 20°15.52 - M = 0 M = 314 lb # ft

Ans.

Ans: M = 314 lb # ft 578

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6–87. Determine the force that the jaws J of the metal cutters exert on the smooth cable C if 100-N forces are applied to the handles. The jaws are pinned at E and A, and D and B. There is also a pin at F.

15$ 400 mm

J

15$ A

20 mm

E C B D 30 mm 80 mm

SOLUTION

Free Body Diagram: The solution for this problem will be simplified if one realizes that member ED is a two force member. Equations of Equilibrium: From FBD (b), + ©F = 0; : x

100 N

F

15$ 15$ 20 mm

400 mm 15$

100 N

Ax = 0

From (a), a + ©MF = 0;

Ay sin 15°1202 + 100 sin 15°1202 - 100 cos 15°14002 = 0 Ay = 7364.10 N

From FBD (b), a + ©ME = 0;

7364.101802 - FC1302 = 0 Ans.

FC = 19637.60 N = 19.6 kN

Ans: FC = 19.6 kN 579

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*6–88. The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, which are operated by the hydraulic cylinder H.The toggles push the movable bar G forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P = 12 kN, determine the force F in the hydraulic cylinder when u = 30°.

D

E

200 mm

F 200 mm

F P  12 kN

H G

F

SOLUTION

200 mm

Member EF: a + ©ME = 0;

u  30

A

B u  30

200 mm C

p

- Fy (0.2 cos 30°) + 6 (0.2 sin 30°) = 0 Fy = 3.464 kN

Joint E: + ©F = 0; : x

FDE cos 30° - 6 = 0 FDE = 6.928 kN

+ c ©Fy = 0;

F - 3.464 - 6.928 sin 30° = 0 Ans.

F = 6.93 kN

Ans: F = 6.93 kN 580

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6–89. Determine the horizontal and vertical components of force which pin C exerts on member CDE. The 600-N load is applied to the pin.

2m

2m

E

D

3m A

C

B 1.5 m

Solution

600 N

Free Body Diagram. The solution will be very much simplified if one determines the support reactions first and then considers the equilibrium of two of its three members after they are dismembered. The FBDs of the entire assembly, member DBF and member ABC are shown in Figs. a, b and c, respectively.

F

300 N

Equations of Equilibrium. Consider the equilibrium of the entire assembly, Fig. a, a+ΣME = 0;  NA(3) - 300(4.5) - 600(4) = 0  NA = 1250 N Next, write the moment equation of equilibrium about point D for member DBF, Fig. b. a+ΣMD = 0;  Bx(1.5) - 300(3) = 0  Bx = 600 N Finally, consider the equilibrium of member ABC, Fig. c + ΣFx = 0;  1250 - 600 - Cx = 0  Cx = 650 N S

Ans. Ans.

a+ΣMB = 0;  Cy = 0

Ans: Cx = 650 N Cy = 0 581

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6–90.

The pipe cutter is clamped around the pipe P. If the wheel at A exerts a normal force of FA = 80 N on the pipe, determine the normal forces of wheels B and C on the pipe. The three wheels each have a radius of 7 mm and the pipe has an outer radius of 10 mm.

C 10 mm B

A

10 mm

P

SOLUTION u = sin

-1

a

10 b = 36.03° 17

Equations of Equilibrium: + c ©Fy = 0;

NB sin 36.03° - NC sin 36.03° = 0 N B = NC

+ ©F = 0; : x

80 - NC cos 36.03° - NC cos 36.03° = 0 NB = NC = 49.5 N

Ans.

Ans: NB = NC = 49.5 N 582

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6–91. Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass of 1.25 Mg and a center of gravity at G. All joints are pin connected.

0.25 m

0.25 m

1.5 m

E C

30 H

D 10 60

A

F

2m

0.5 m G

SOLUTION Assembly FHG: a + ©MH = 0;

- 1250(9.81) (0.5) + FEF (1.5 sin 30°) = 0 Ans.

FEF = 8175 N = 8.18 kN (T) Assembly CEFHG: a + ©MC = 0;

FAD cos 40° (0.25) - 1250(9.81) (2 cos 10° + 0.5) = 0 Ans.

FAD = 158 130 N = 158 kN (C)

Ans: FEF = 8.18 kN (T) FAD = 158 kN (C) 583

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*6–92. Determine the horizontal and vertical components of force at pin B and the normal force the pin at C exerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions of force at A. There is a pulley at E.

D

E

4 ft 50 ft

C

SOLUTION

4 ft

BCE: a + ©MB = 0;

B A

- 50(6) - NC(5) + 50(8) = 0 Ans.

NC = 20 lb + ©F = 0; : x

3 ft

4 Bx + 20a b - 50 = 0 5 Ans.

Bx = 34 lb + c ©Fy = 0;

3 ft

3 By - 20a b - 50 = 0 5 Ans.

By = 62 lb ACD: + ©F = 0; : x

4 -Ax - 20a b + 50 = 0 5 Ans.

Ax = 34 lb + c ©Fy = 0;

3 -Ay + 20 a b = 0 5 Ans.

Ay = 12 lb a + ©MA = 0;

4 MA + 20a b(4) - 50(8) = 0 5 MA = 336 lb # ft

Ans.

Ans: Nc = 20 lb Bx = 34 lb By = 62 lb Ax = 34 lb Ay = 12 lb MA = 336 lb ~ ft 584

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6–93. The constant moment of 50 N # m is applied to the crank shaft. Determine the compressive force P that is exerted on the piston for equilibrium as a function of u. Plot the results of P (ordinate) versus u (abscissa) for 0° … u … 90°.

B 0.45 m

u 0.2 m A

C P

50 N  m

SOLUTION Member AB: a + ©MA = 0;

FBC sin f(0.2 sin u) + FBC cos f(0.2 cos f)- 50 = 0 FBC =

250 (sin f sin u + cos f cos u)

Piston: + ©F = 0; : x

FBC cos f -P = 0 P =

250 cos f (sin f sin u + cos f cos u)

(1)

d = 0.2 cos u = 0.45 sin f f = sin-1 a

cos u b 2.25

(2)

Select u, solve for f in Eq. (2), then P in Eq. (1).

P(u) =

=

250 cos c sin-1 a

cos u bd 2.25

sin u cos u cos u + cos csin-1 a b d # cos u 2.25 2.25 250 22.252 - cos2 u

Ans.

sin u cos u + 22.252 - cos2 u # cos u

Ans: P(u) = 585

250 22.252 - cos2 u

sin u cos u + 22.252 - cos2 u # cos u

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6–94. Five coins are stacked in the smooth plastic container shown. If each coin weighs 0.0235 lb, determine the normal reactions of the bottom coin on the container at points A and B.

4

5 3 5 4 3

4

5 3 5

SOLUTION

4

3

All coins: + c ©Fy = 0;

NB - 5 (0.0235) = 0 Ans.

NB = 0.1175 lb

B

Bottom coin: + c ©Fy = 0;

A

4 0.1175 - 0.0235 - N a b = 0 5 N = 0.1175 lb

+ ©F = 0; : x

3 NA - 0.1175 a b = 0 5 Ans.

NA = 0.0705 lb

Ans: NB = 0.1175 lb NA = 0.0705 lb 586

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6–95. The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin connected at B and the surface at D is smooth, determine the normal force on the fingernail when a force of 1 lb is applied to the handles as shown. The pin AC slides through a smooth hole at A and is attached to the bottom member at C.

0.25 in.

1.5 in.

A

B

SOLUTION

D

Handle: a + ©MD = 0;

1 lb

0.25 in.

C 1 lb

FA (0.25) - 1(1.5) = 0 FA = 6 lb

+ c ©Fy = 0;

ND - 1 - 6 = 0 ND = 7 lb

Top blade: a + ©MB = 0;

7(1.5) - FN (2) = 0 Ans.

FN = 5.25 lb Or bottom blade: a + ©MB = 0;

FN (2) - 6 (1.75) = 0 Ans.

FN = 5.25 lb

Ans: FN = 5.25 lb 587

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*6–96. A man having a weight of 175 lb attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform.

A

B A

B

SOLUTION

C

C

(a)

(a)

(b)

Bar: + c ©Fy = 0;

2(F>2) - 2(87.5) = 0 Ans.

F = 175 lb Man: + c ©Fy = 0;

NC - 175 - 2(87.5) = 0 Ans.

NC = 350 lb (b) Bar: + c ©Fy = 0;

2(43.75) - 2(F>2) = 0 Ans.

F = 87.5 lb Man: + c ©Fy = 0;

NC - 175 + 2(43.75) = 0 Ans.

NC = 87.5 lb

Ans: F = 175 lb Nc = 350 lb F = 87.5 lb Nc = 87.5 lb 588

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6–97.

A man having a weight of 175 lb attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has a weight of 30 lb.

A

B A

B

SOLUTION

C

C

(a)

(a)

(b)

Bar: + c ©Fy = 0;

2(F>2) - 102.5 - 102.5 = 0 Ans.

F = 205 lb Man: + c ©Fy = 0;

NC - 175 - 102.5 - 102.5 = 0 Ans.

NC = 380 lb (b) Bar: + c ©Fy = 0;

2(F>2) - 51.25 - 51.25 = 0 Ans.

F = 102 lb Man: + c ©Fy = 0;

NC - 175 + 51.25 + 51.25 = 0 Ans.

NC = 72.5 lb

Ans: F = 205 lb NC = 380 lb F = 102 lb NC = 72.5 lb 589

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6–98. The two-member frame is pin connected at E. The cable is attached to D, passes over the smooth peg at C, and supports the 500-N load. Determine the horizontal and vertical reactions at each pin.

1m

0.5 m 0.5 m

1m

A B

0.5 m E

C D

Solution Free Body Diagram. The frame will be dismembered into members BD and AC, of which their respective FBDs are shown in Figs. a and b.

500 N

Equations of Equilibrium. Write the moment equation of equilibrium about point B for member BD, Fig. a, and about point A for member AC, Fig. b, a+ΣMB = 0;  Ey(0.5) - Ex(0.5) + 500(2) = 0

(1)

a+ΣMA = 0;  Ey (0.5) + Ex (0.5) - 500(2) - 500(2) = 0

(2)

Solving Eqs. (1) and (2),

Ey = 1000 N = 1.00 kN  Ex = 3000 N = 3.00 kN

Ans.

Write the force equations of equilibrium for member BD, Fig. a. + ΣFx = 0;  Bx + 500 - 3000 = 0  Bx = 2500 N = 2.50 kN S

Ans.

+ c ΣFy = 0;  By - 1000 = 0    By = 1000 N = 1.00 kN

Ans.

Also, for member AC, Fig. b + ΣFx = 0;  3000 - 500 - Ax = 0  Ax = 2500 N = 2.50 kN S

Ans.

+ c ΣFy = 0;  1000 - 500 - Ay = 0  Ay = 500 N

Ans.

Ans: Ey = Ex = Bx = By = Ax = Ay = 590

1.00 kN 3.00 kN 2.50 kN 1.00 kN 2.50 kN 500 N

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6–99. If the 300-kg drum has a center of mass at point G, determine the horizontal and vertical components of force acting at pin A and the reactions on the smooth pads C and D. The grip at B on member DAB resists both horizontal and vertical components of force at the rim of the drum.

P

600 mm E

A

SOLUTION

60 mm 60 mm

C

Equations of Equilibrium: From the free - body diagram of segment CAE in Fig. a, a + ©MA = 0;

100 mm

Ans.

D

G

Ax - 12 743.56 = 0 Ans.

Ax = 12 743.56 N = 12.7 kN + c ©Fy = 0;

B

390 mm

300(9.81)(600 cos 30°) - NC(120) = 0 NC = 12 743.56 N = 12.7 kN

+ ©F = 0; : x

30

300(9.81) - Ay = 0 Ans.

Ay = 2943 N = 2.94 kN

Using the results for Ax and Ay obtained above and applying the moment equation of equilibrium about point B on the free - body diagram of segment BAD, Fig. b, a + ©MB = 0;

12 743.56(60) - 2943(100) - ND(450) = 0 Ans.

ND = 1045.14 N = 1.05 kN

Ans: NC = 12.7 kN Ax = 12.7 kN Ay = 2.94 kN ND = 1.05 kN 591

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*6–100. Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the compression in the spring is 20 mm when the valve is open as shown, determine the normal force acting on the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth. The spring has a stiffness of 300 N>m.

40 mm

25 mm

E G

F H

SOLUTION Fs = kx;

Fs = 300 (0.02) = 6 N

+ c ©Fy = 0;

I

- FG + 6 = 0

V

FG = 6 N a + ©MF = 0;

J

- 6(40) + T(25) = 0 Ans.

T = 9.60 N

D C

Ans: T = 9.60 N 592

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6–101. If a clamping force of 300 N is required at A, determine the amount of force F that must be applied to the handle of the toggle clamp.

F

235 mm

70 mm

30 mm C 30 mm A

SOLUTION

30 B

275 mm

E

D 30

Equations of Equilibrium: First, we will consider the free-body diagram of the clamp in Fig. a. Writing the moment equation of equilibrium about point D, a + ©MD = 0;

Cx (60) - 300(235) = 0 Cx = 1175 N

Subsequently, the free - body diagram of the handle in Fig. b will be considered. a + ©MC = 0;

FBE cos 30°(70) - FBE sin 30°(30) - F cos 30°(275 cos 30° + 70) -F sin 30°(275 sin 30°) = 0 (1)

45.62FBE - 335.62F = 0 + ©F = 0; : x

1175 + F sin 30° - FBE sin 30° = 0 (2)

0.5FBE - 0.5F = 1175 Solving Eqs. (1) and (2) yields

Ans.

F = 369.69 N = 370 N FBE = 2719.69N

Ans: F = 370 N 593

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6–102. If a force of F = 350 N is applied to the handle of the toggle clamp, determine the resulting clamping force at A.

F

235 mm

70 mm

30 mm C 30 mm A

SOLUTION

30 B

275 mm

E

D 30

Equations of Equilibrium: First, we will consider the free-body diagram of the handle in Fig. a. a + ©MC = 0;

FBE cos 30°(70) - FBE sin 30°(30) - 350 cos 30°(275 cos 30° + 70) -350 sin 30°(275 sin 30°) = 0 FBE = 2574.81 N

+ ©F = 0; : x

Cx - 2574.81 sin 30° + 350 sin 30° = 0 Cx = 1112.41 N

Subsequently,, the free-body diagram of the clamp in Fig. b will be considered. Using the result of Cx and writing the moment equation of equilibrium about point D, a + ©MD = 0;

1112.41(60)-NA (235) = 0 Ans.

NA = 284.01 N = 284 N

Ans: NA = 284 N 594

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6–103. 4 kN

Determine the horizontal and vertical components of force that the pins at A and B exert on the frame.

2 kN 2m

2m

2m

C B 3m 3 kN

Solution

1m

Free Body Diagram. The assembly will be dismembered into members BC, CD, CE and AD. The solution will be very much simplified if one recognizes member CE is a two force member. The FBDs of members CD, BC and AD are shown in Figs. a, b and c, respectively.

A E

D 3m

3m

Equations of Equilibrium. First, consider the equilibrium of member CD, Fig. a. a+ΣMC = 0;  3(3) - Dx (4) = 0  Dx = 2.25 kN a+ΣMD = 0;  Cx (4) - 3(1) = 0  Cx = 0.75 kN + c ΣFy = 0;  Cy - Dy = 0      Cy = Dy Next write the moment equation of equilibrium about point B for member BC, Fig. b and about point A for member AD, Fig. c, 4 a+ΣMB = 0;  2(2) + 4(4) + Dy (6) - FCE a b(6) = 0 5

(1)

4 a+ΣMA = 0;  FCE a b(3) - Dy(6) = 0 5

(2)

Solving Eqs. (1) and (2)

FCE = 8.3333 kN  Dy = 3.3333 kN

Finally, write the force equation of equilibrium for member BC, Fig. b, 4 + c ΣFy = 0;    By + 8.3333 a b - 3.3333 - 4 - 2 = 0 5 By = 2.6667 kN = 2.67 kN

Ans.

+ ΣFx = 0;  Bx + 0.75 - 8.3333 a 3 b = 0  Bx = 4.25 kN S 5

Ans.

Also, for member AD, Fig. c.

4 + c ΣFy = 0;  Ay + 3.3333 - 8.3333 a b = 0  Ay = 3.3333 kN = 3.33 kN Ans. 5 + ΣFx = 0;  2.25 + 8.3333 a 3 b - Ax = 0  Ax = 7.25 kN S 5

Ans.

Ans: By = 2.67 kN Bx = 4.25 kN Ay = 3.33 kN Ax = 7.25 kN 595

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*6–104. The hydraulic crane is used to lift the 1400-lb load. Determine the force in the hydraulic cylinder AB and the force in links AC and AD when the load is held in the position shown.

1 ft 120

30 C

A

8 ft

1 ft D

70

1 ft

B

SOLUTION a + ©MD = 0;

FCA (sin 60°)(1) - 1400(8) = 0 Ans.

FCA = 12 932.65 lb = 12.9 kip + c ©Fy = 0;

12 932.65 sin 60° - FAB sin 70° = 0 Ans.

F AB = 11 918.79 lb = 11.9 kip + ©F = 0; : x

7 ft

- 11 918.79 cos 70° + 12 932.65 cos 60° - FAD = 0 Ans.

FAD = 2389.86 lb = 2.39 kip

Ans: FCA = 12.9 kip FAB = 11.9 kip FAD = 2.39 kip 596

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6–105. Determine force P on the cable if the spring is compressed 0.025 m when the mechanism is in the position shown. The spring has a stiffness of k = 6 kN>m.

E 150 mm 200 mm

30

k

C

200 mm A

200 mm

Solution

P D

F B

800 mm

Free Body Diagram. The assembly will be dismembered into members ACF, CDE and BD. The solution will be very much simplified if one recognizes member BD is a two force member. Here, the spring force is FSP = kx = 600(0.025) = 150 N. The FBDs of members ACF and CDE are shown in Figs. a and b, respectively. Equations of Equilibrium. Write the moment equation of equilibrium about point A for member ACF, Fig. a, (1)

a+ΣMA = 0;  Cy (0.2) + Cx (0.2) - 150(1) = 0 Next, consider the equilibrium of member CDE

(2)

a+ΣMC = 0;  FBD sin 30°(0.2) - P (0.35) = 0 + ΣFx = 0;  Cx + P - FBD sin 30° = 0 S

(3)

+ c ΣFy = 0;  FBD cos 30° - Cy = 0

(4)

Solving Eqs. (1) to (4),

Cx = 148.77 N  Cy = 601.23 N  FBD = 694.24 N



P = 198.36 N = 198 N

Ans.

Ans: P = 198 N 597

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6–106. If d = 0.75 ft and the spring has an unstretched length of 1 ft, determine the force F required for equilibrium.

B 1 ft

1 ft d

F

SOLUTION

A

Spring Force Formula: The elongation of the spring is x = 2(0.75) - 1 = 0.5 ft. Thus, the force in the spring is given by

F

d

k

1 ft

150 lb/ft 1 ft

C

D

Fsp = kx = 150(0.5) = 75 lb Equations of Equilibrium: First, we will analyze the equilibrium of joint B. From the free-body diagram in Fig. a, + ©F = 0; : x

FAB cos 48.59° - FBC cos 48.59° = 0 FAB = FBC = F¿

+ c ©Fy = 0;

2F¿ sin 48.59° - 75 = 0 F¿ = 50 lb

From the free-body diagram in Fig. b, using the result FBC = F¿ = 50 lb, and analyzing the equilibrium of joint C, we have + c ©Fy = 0;

+ ©F = 0; : x

FCD sin 48.59° - 50 sin 48.59° = 0

FCD = 50 lb

2(50 cos 48.59°) - F = 0 Ans.

F = 66.14 lb = 66.1 lb

Ans: F = 66.1 lb 598

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6–107. If a force of F = 50 lb is applied to the pads at A and C, determine the smallest dimension d required for equilibrium if the spring has an unstretched length of 1 ft.

B

d

F A

SOLUTION

F

d

k

1 ft

150 lb/ft 1 ft

C

D

Geometry: From the geometry shown in Fig. a, we can write sin u = d

1 ft

1 ft

cos u = 21 - d2

Spring Force Formula: The elongation of the spring is x = 2d - 1. Thus, the force in the spring is given by Fsp = kx = 150(2d - 1) Equations of Equilibrium: First, we will analyze the equilibrium of joint B. From the free-body diagram in Fig. b, + ©F = 0; : x

FAB cos u - FBC cos u = 0

+ c ©Fy = 0;

2F¿(d) - 150(2d - 1) = 0

FAB = FBC = F¿ F¿ =

150d - 75 d

From the free-body diagram in Fig. c, using the result FBC = F¿ =

150d - 75 , and d

analyzing the equilibrium of joint C, we have + c ©Fy = 0; + ©F = 0; : x

FCD sin u - a

2c a

150d - 75 b sin u = 0 d

FCD =

150d - 75 d

150d - 75 b a 11 - d2 b d - 50 = 0 d

Ans.

Solving the above equation using a graphing utility, we obtain d = 0.6381 ft = 0.638 ft or d = 0.9334 ft = 0.933 ft

Ans: d = 0.638 ft 599

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*6–108. The skid steer loader has a mass of 1.18 Mg, and in the position shown the center of mass is at G1. If there is a 300-kg stone in the bucket, with center of mass at G2, determine the reactions of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and at the pin E.There is a similar linkage on each side of the loader.

1.25 m

D G2

SOLUTION

E

30

C G1

Entire system: a + ©MA = 0;

0.5 m

300 (9.81)(1.5)-1.18 (103)(9.81)(0.6)+NB (0.75) = 0 NB = 3374.6 N = 3.37 kN

+ c ©Fy = 0;

(Both wheels)

A

0.15 m

Ans. 1.5 m

3374.6 -300 (9.81) -1.18(103)(9.81)+ NA = 0 NA = 11.1 kN

B 0.75 m

Ans.

(Both wheels)

Upper member: a + ©ME = 0;

300(9.81)(2.75)- FCD sin 30° (1.25) = 0 FCD = 12 949 N = 12.9 kN F¿ CD =

+ ©F = 0; : x

FCD 12 949 = = 6.47 kN 2 2

Ans.

Ex - 12 949 cos 30° = 0 Ex = 11 214 N

+ c ©Fy = 0;

-Ey - 300(9.81) + 12 949 sin 30° = 0 Ey = 3532 N

FE = 2(11 214)2 + (3532)2 = 11 757 N Since there are two members, ¿ FE =

FE 11 757 = = 5.88 kN 2 2

Ans.

Ans: NA = 11.1 kN ( Both wheels ) F′CD = 6.47 kN F′E = 5.88 kN 600

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6–109. Determine the force P on the cable if the spring is compressed 0.5 in. when the mechanism is in the position shown. The spring has a stiffness of k = 800 lb>ft.

6 in. 6 in.

A

6 in.

B

4 in.

D

30 C

SOLUTION FE = ks = 800 a

0.5 b = 33.33 lb 12

a + ©MA = 0;

Bx (6) + By (6) - 33.33(30) = 0

k

(1)

Bx + By = 166.67 lb a + ©MD = 0;

P

24 in.

E

By (6) - P(4) = 0 By = 0.6667P

(2)

+ ©F = 0; : x

-Bx + FCD cos 30° = 0

(3)

a + ©MB = 0;

FCD sin 30°(6) - P(10) = 0 FCD = 3.333 P

Thus from Eq. (3) Bx = 2.8867 P Using Eqs. (1) and (2): 2.8867 P + 0.6667 P = 166.67 Ans.

P = 46.9 lb

Ans: P = 46.9 lb 601

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6–110. The spring has an unstretched length of 0.3 m. Determine the angle u for equilibrium if the uniform bars each have a mass of 20 kg.

C

k  150 N/ m

Solution

2m

u u

B

A

Free Body Diagram. The assembly is being dismembered into members AB and BC of which their respective FBD are shown in Fig. b and a. Here, the spring stretches x = 2(2 sin u) - 0.3 = 4 sin u - 0.3. Thus, FSP = kx = 150 (4 sin u - 0.3) = 600 sin u - 45. Equations of Equilibrium. Considered the equilibrium of member BC, Fig. a, a+ΣMC = 0;  By (2 cos u) - Bx (2 sin u) - 20(9.81) cos u = 0

(1)

Also, member AB, Fig. b a+ΣMA = 0;   - By (2 cos u) - Bx (2 sin u) - 20(9.81) cos u = 0 + c ΣFy = 0;  (600 sin u - 45) - 20(9.81) - By = 0

(2) (3)

Solving Eq. (1) and (2) 9.81 cos u sin u Substitute the result of By = 0 into Eq. (3)



By = 0  Bx = -



600 sin u - 45 - 20(9.81) = 0



sin u = 0.402



u = 23.7°

Ans.

Ans: u = 23.7° 602

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6–111. The spring has an unstretched length of 0.3 m. Determine the mass m of each uniform bar if each angle u = 30° for equilibrium.

C

k  150 N/ m

Solution

2m

u u

B

A

Free Body Diagram. The assembly is being dismembered into members AB and BC of which their respective FBD are shown in Fig. b and a. Here, the spring stretches x = 2(2 sin 30°) - 0.3 = 1.7 m. Thus, FSP = kx = 150(1.7) = 255 N. Equations of Equilibrium. Consider the equilibrium of member BC, Fig. a, a+ΣMC = 0;  Bx (2 sin 30°) + By (2 cos 30°) - m(9.81) cos 30° = 0

(1)

Also, member AB, Fig. b a+ΣMA = 0;  Bx (2 sin 30°) - By (2 cos 30°) - m(9.81) cos 30° = 0 + c ΣFy = 0;  255 - m(9.81) - By = 0

(2) (3)

Solving Eqs. (1) and (2)

Bx = 8.4957 m  By = 0

Substitute the result of By = 0 into Eq. (3)

255 - m(9.81) = 0



m = 26.0 kg

Ans.

Ans: m = 26.0 kg 603

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*6–112. The piston C moves vertically between the two smooth walls. If the spring has a stiffness of k = 15 lb>in., and is unstretched when u = 0°, determine the couple M that must be applied to AB to hold the mechanism in equilibrium when u = 30°.

A 8 in.

M u

B

SOLUTION

12 in.

Geometry: sin c sin 30° = 8 12

C

c = 19.47°

k = 15 lb/in.

f = 180° - 30° - 19.47 = 130.53° l¿ AC 12 = sin 130.53° sin 30°

l¿ AC = 18.242 in.

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring spring force is stretches x = lAC - l¿ AC = 20 - 18.242 = 1.758 in. the Fsp = kx = 15 (1.758) = 26.37 lb. Equations of Equilibrium: Using the method of joints, [FBD (a)], + c ©Fy = 0;

FCB cos 19.47° - 26.37 = 0 FCB = 27.97 lb

From FBD (b), a+ ©MA = 0;

27.97 cos 40.53° (8) - M = 0 M = 170.08 lb # in = 14.2 lb # ft

Ans.

Ans: M = 14.2 lb # ft 604

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6–113. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If x = 450 mm, determine the required mass of the counterweight S required to balance a 90-kg load, L.

100 mm

250 mm

150 mm H

E C

F

G

D

150 mm

S

350 mm B

A

x L

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free - body diagram of member AB in Fig. a, a + ©MA = 0;

FBG (500) - 90(9.81)(150) = 0 FBG = 264.87 N

Using the result of FBG and writing the moment equation of equilibrium about point F on the free - body diagram of member EFG in Fig. b, a + ©MF = 0;

FED (250) - 264.87(150) = 0 FED = 158.922 N

Using the result of FED and writing the moment equation of equilibrium about point C on the free - body diagram of member CDI in Fig. c, a + ©MC = 0;

158.922(100) - mS(9.81)(950) = 0 Ans.

mS = 1.705 kg = 1.71 kg

Ans: mS = 1.71 kg 605

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6–114. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If x = 450 mm and, the mass of the counterweight S is 2 kg, determine the mass of the load L required to maintain the balance.

100 mm

250 mm

150 mm H

E C

F

G

D

150 mm

S

350 mm B

A

x L

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free - body diagram of member AB in Fig. a, a + ©MA = 0;

FBG (500) - ML(9.81)(150) = 0 FBG = 2.943 lb

Using the result of FBG and writing the moment equation of equilibrium about point F on the free - body diagram of member EFG in Fig. b, a + ©MF = 0;

FED (250) - 2.943mL(150) = 0 FED = 1.7658mL

Using the result of FED and writing the moment equation of equilibrium about point C on the free - body diagram of member CDI in Fig. c, a + ©MC = 0;

1.7658mL(100) - 2(9.81)(950) = 0 Ans.

mL = 105.56 kg = 106 kg

Ans: mL = 106 kg 606

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6–115. The four-member “A” frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is 800 N, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pin at G only exert force components on the frame.

z

300 mm 300 mm

E 600 mm

x

D

A B

600 mm

600 mm

F C

SOLUTION

G

GF is a two - force member, so the 800 - N force acts along the axis of GF. Using FBD (a), ©Mx = 0;

- P(1.2) + 800 sin 45°(0.6) = 0 - Ay (0.3) + Ey (0.3) = 0

©Fy = 0;

- Ay - Ey + 800 sin 45° = 0

P

Pk

Ans.

P = 283 N ©Mz = 0;

y

Ay = Ey = 283 N ©Mx = 0;

Az (0.6) + Ez (0.6) -283((0.6) = 0

©My = 0;

Az(0.3) -Ez (0.3) = 0 Az = Ez = 118 N

Using FBD (b), ©Fy = 0;

- By - Dy + 800 sin 45° = 0

©Mz = 0;

Dy(0.3) - By (0.3) = 0 Ans.

By = Dy = 283 N ©Fz = 0;

- Bz - Dz + 800 cos 45° = 0

©My = 0;

- Dz (0.3) + Bz (0.3) = 0 Ans.

Bz = Dz = 283 N ©Fx = 0;

- Bx + Dx = 0

Using FBD (c), ©Mz = 0;

- By (0.6) + 283(0.15) - 283(0.3) = 0 Ans.

Bx = Dx = 42.5 N

Ans: P = 283 N Bx = Dx = 42.5 N By = Dy = 283 N Bz = Dz = 283 N 607

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*6–116. The structure is subjected to the loadings shown. Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C.

z 250 N

60 60

45

D A

SOLUTION

4m 800 N  m

From FBD (a) x

©My = 0;

MBy = 0

©Mx = 0;

- MBx + 800 = 0

©Mz = 0;

By (3) -Bx (2) = 0

©Fz = 0;

Az = 0

©Fx = 0;

- Ax + Bx = 0

(2)

©Fy = 0;

- Ay + By = 0

(3)

MBx = 800 N # m

2m 1.5 m

B

3m

C

(1) Ans.

From FBD (b) ©Mg = 0;

By(1.5) + 800 - 250 cos 45°(5.5) = 0

From Eq.(1)

114.85(3) - Bx(2) = 0

From Eq.(2)

Ax = 172 N

Ans.

From Eq.(3)

Ay = 115 N

Ans.

©Fx = 0;

Cx + 250 cos 60° - 172.27 = 0

Cx = 47.3 N

Ans.

©Fy = 0;

250 cos 45° - 114.85 - Cy = 0

Cy = 61.9N

Ans.

©Fz = 0;

250 cos 60° - Cz = 0

©My = 0;

MCy - 172.27(1.5) + 250 cos 60°(5.5) = 0

©Mz = 0;

By = 114.85 N

Bx = 172.27 N

Ans.

Cz = 125 N

MCy = - 429 N # m

Ans.

MCz = 0

Ans.

Negative sign indicates that MCy acts in the opposite sense to that shown on FBD.

Ans: Az = 0 Ax = 172 N Ay = 115 N Cx = 47.3 N Cy = 61.9 N Cz = 125 N MCy = -429 N # m MCz = 0 608

y

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6–117. The structure is subjected to the loading shown. Member AD is supported by a cable AB and roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB.

z B E 0.8 m

SOLUTION ©My = 0;

D

4 - FAB (0.6) + 2.5(0.3) = 0 5

Ans. Ans.

FAB = 1.5625 = 1.56 kN ©Fz = 0;

C

0.5 m

x

0.3 m 0.4 m

A 0.3 m

{ 2.5 } kN

4 (1.5625) - 2.5 + Dz = 0 5 Dz = 1.25 kN

©Fy = 0;

Dy = 0

©Fx = 0;

Dx + Cx -

©Mx = 0;

MDx +

3 (1.5625) = 0 5

(1)

4 (1.5625)(0.4) - 2.5(0.4) = 0 5

MDx = 0.5 kN # m 3 (1.5625)(0.4) - Cx (0.4) = 0 5

(2)

©Mz = 0;

MDz +

©Fz = 0;

Dz¿ = 1.25 kN

©Mx = 0;

MEx = 0.5 kN # m

Ans.

©My = 0;

MEy = 0

Ans.

©Fy = 0;

Ey = 0

Ans.

©Mz = 0;

E x (0.5) - MDz = 0

(3)

Solving Eqs. (1), (2) and (3): Cx = 0.938 kN MDz = 0 Ans.

Ex = 0

Ans: MEx = 0.5 kN # m MEy = 0 Ey = 0 Ex = 0 609

y

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6–118.

The three pin-connected members shown in the top view support a downward force of 60 lb at G. If only vertical forces are supported at the connections B, C, E and pad supports A, D, F, determine the reactions at each pad.

D 6 ft

2 ft A

SOLUTION Equations of Equilibrium : From FBD (a), a + ©MD = 0; + c ©Fy = 0;

2 ft

6 ft

60182 + FC162 - FB1102 = 0

(1)

FB + FD - FC - 60 = 0

(2)

FE162 - FC1102 = 0

(3)

FC + FF - FE = 0

(4)

FE1102 - FB162 = 0

(5)

C G

B

4 ft E

4 ft

6 ft

F

From FBD (b), a + ©MF = 0; + c ©Fy = 0; From FBD (c), a + ©MA = 0; + c ©Fy = 0;

FA + FE - FB = 0

(6)

Solving Eqs. (1), (2), (3), (4), (5) and (6) yields, FE = 36.73 lb FD = 20.8 lb

FC = 22.04 lb FF = 14.7 lb

FB = 61.22 lb FA = 24.5 lb

Ans.

Ans: FD = 20.8 lb FF = 14.7 lb FA = 24.5 lb 610

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7–1.

Determine the shear force and moment at points C and D.

500 lb

300 lb

200 lb B

A

C 6 ft

4 ft

E

D 4 ft

6 ft

2 ft

SOLUTION Support Reactions: FBD (a). a + ©MB = 0;

500(8) - 300(8) -Ay (14) = 0

Ay = 114.29 lb Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have + ©F = 0 : x + c ©Fy = 0;

NC = 0

114.29 - 500 - VC = 0

a + ©MC = 0;

Ans. VC = -386 lb

Ans.

MC + 500(4) - 114.29 (10) = 0 MC = -857 lb # ft

Ans.

Applying the equations of equilibrium to segment ED [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0;

ND = 0

VD - 300 = 0

-MD - 300 (2) = 0

Ans.

VD = 300 lb

Ans.

MD = -600 lb # ft

Ans.

Ans: NC = 0 VC = - 386 lb MC = - 857 lb # ft ND = 0 VD = 300 lb MD = - 600 lb # ft 611

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7–2.

Determine the internal normal force and shear force, and the bending moment in the beam at points C and D. Assume the support at B is a roller. Point C is located just to the right of the 8-kip load.

8 kip 40 kip ft A

C 8 ft

D 8 ft

B 8 ft

SOLUTION Support Reactions: FBD (a). a + ©MA = 0; + c ©Fy = 0;

By (24) + 40 - 8(8) = 0 Ay + 1.00 - 8 = 0

+ ©F = 0 : x

By = 1.00 kip

Ay = 7.00 kip

Ax = 0

Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have + ©F = 0 : x + c ©Fy = 0; a + ©MC = 0;

NC = 0

7.00 - 8 - VC = 0 MC - 7.00(8) = 0

Ans.

VC = -1.00 kip MC = 56.0 kip # ft

Ans. Ans.

Applying the equations of equilibrium to segment BD [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0;

VD + 1.00 = 0

ND = 0

Ans.

VD = -1.00 kip

Ans.

1.00(8) + 40 - MD = 0 MD = 48.0 kip # ft

Ans.

Ans: NC = 0 VC = -1.00 kip MC = 56.0 kip # ft ND = 0 VD = -1.00 kip MD = 48.0 kip # ft 612

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7–3. Two beams are attached to the column such that structural connections transmit the loads shown. Determine the internal normal force, shear force, and moment acting in the column at a section passing horizontally through point A.

30 mm

40 mm 250 mm

16 kN 6 kN 23 kN 185 mm 6 kN

Solution

A

+ ΣFx = 0;     6 - 6 - VA = 0 S Ans.

VA = 0 + c ΣFy = 0;      - NA - 16 - 23 = 0

Ans.

NA = - 39 kN a + ΣMA = 0;

125 mm

- MA + 16(0.155) - 23(0.165) - 6(0.185) = 0

MA = - 2.42 kN # m

Ans.

Ans: VA = 0 NA = -39 kN MA = -2.425 kN # m 613

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*7–4. The beam weighs 280 lb>ft. Determine the internal normal force, shear force, and moment at point C.

B

7 ft 8 ft

3 ft

Solution

A

Entire beam : a + ΣMA = 0;

C

6 ft

- 2.8 (3) + Bx (8) = 0 Bx = 1.05 kip

+ ΣFx = 0;     Ax - 1.05 = 0 S Ax = 1.05 kip + c ΣFy = 0;     Ay - 2.8 = 0 Ay = 2.80 kip Segment AC : + Q ΣFx = 0;         - NC + 1.05 cos 53.13° + 2.80 sin 53.13° - 0.84 sin 53.13° = 0  

Ans.

          NC = 2.20 kip

a + ΣFy = 0;    - VC - 0.84 cos 53.13° + 2.80 cos 53.13° - 1.05 sin 53.13° = 0 Ans.

VC = 0.336 kip

a + ΣMC = 0;   - 2.80 cos 53.13° (3) + 1.05 sin 53.13° (3) + 0.84 cos 53.13° (1.5) + MC = 0

MC = 1.76 kip # ft

Ans.

Ans: NC = 2.20 kip VC = 0.336 kip MC = 1.76 kip # ft 614

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7–5. The pliers are used to grip the tube at B. If a force of 20 lb is applied to the handles,determine the internal shear force and moment at point C. Assume the jaws of the pliers exert only normal forces on the tube.

20 lb 40

10 in.

0.5 in. 1 in. C B

A

SOLUTION + ©MA = 0;

20 lb

-20(10) + RB (1.5) = 0 RB = 133.3 lb

Segment BC: +Q ©Fy = 0;

VC + 133.3 = 0 Ans.

VC = - 133 lb a + ©MC = 0;

- MC + 133.3 (1) = 0 MC = 133 lb # in.

Ans.

Ans: VC = - 133 lb MC = 133 lb # in. 615

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7–6.

Determine the distance a as a fraction of the beam’s length L for locating the roller support so that the moment in the beam at B is zero.

P

P

C

SOLUTION a + ©MA = 0;

a

-P a Cy =

a + ©M = 0;

B

A

L/3 L

2L - a b + Cy1L - a2 + Pa = 0 3 2P A L3 - a B L - a

M =

2P A L3 - a B L - a

2PL a a =

a

L b = 0 3

L - ab = 0 3

L 3

Ans.

Ans: a = 616

L 3

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7–7. Determine the internal shear force and moment acting at point C in the beam.

4 kip/ft

A

B C 6 ft

6 ft

Solution Support Reactions. Referring to the FBD of the entire beam shown in Fig. a,   a+ ΣMA = 0;  By(12) -

1 (4)(6)(4) = 0  By = 4.00 kip 2

Internal Loadings. Referring to the FBD of the right segment of the beam sectioned through C, Fig. b,    + c ΣFy = 0;  VC + 4.00 = 0         VC = - 4.00 kip

Ans.

  a+ ΣMC = 0;     4.00(6) - MC = 0

Ans.

MC = 24.0 kip # ft

Ans: VC = -4.00 kip MC = 24.0 kip # ft 617

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*7–8. Determine the internal shear force and moment acting at point C in the beam.

500 lb/ ft

900 lb  ft

A 3 ft

900 lb  ft

C 6 ft

B 6 ft

3 ft

Solution Support Reactions. Referring to the FBD of the entire beam shown in Fig. a   a+ ΣMB = 0;  500(12)(6) + 900 - 900 - Ay(12) = 0   Ay = 3000 lb + ΣFx = 0;    S

Ax = 0

Internal Loadings. Referring to the FBD of the left segment of beam sectioned through C, Fig. b,    + c ΣFy = 0;     3000 - 500(6) - VC = 0  VC = 0

Ans.

  a+ ΣMC = 0;  MC + 500(6)(3) + 900 - 3000(6) = 0

         MC = 8100 lb # ft = 8.10 kip # ft

Ans.

Ans: VC = 0 MC = 8.10 kip # ft 618

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7–9. Determine the normal force, shear force, and moment at a section passing through point C. Take P = 8 kN.

B

0.1 m

0.5 m C

0.75 m

0.75 m

P

SOLUTION a + ©MA = 0;

0.75 m

A

-T(0.6) + 8(2.25) = 0 T = 30 kN

+ ©F = 0; : x

Ax = 30 kN

+ c ©Fy = 0;

Ay = 8 kN

+ ©F = 0; : x

-NC - 30 = 0 Ans.

NC = - 30 kN + c ©Fy = 0;

VC + 8 = 0 Ans.

VC = - 8 kN a + ©MC = 0;

- MC + 8(0.75) = 0 MC = 6 kN # m

Ans.

Ans: NC = - 30 kN VC = - 8 kN MC = 6 kN # m 619

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7–10. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading.

B

0.1 m

0.5 m C

0.75 m

-2(0.6) + P(2.25) = 0 Ans.

P = 0.533 kN + ©F = 0; : x

Ax = 2 kN

+ c ©Fy = 0;

Ay = 0.533 kN

+ ©F = 0; : x

-NC - 2 = 0 Ans.

NC = - 2 kN + c ©Fy = 0;

VC - 0.533 = 0 Ans.

VC = -0.533 kN a + ©M C = 0;

0.75 m

P

SOLUTION a + ©MA = 0;

0.75 m

A

- MC + 0.533(0.75) = 0 MC = 0.400 kN # m

Ans.

Ans: P = 0.533 kN NC = -2 kN VC = -0.533 kN MC = 0.400 kN # m 620

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7–11. Determine the internal normal force, shear force, and moment at points C and D of the beam.

60 lb /ft

690 lb

40 lb/ ft

13 12 5

A C

B

D

12 ft

SOLUTION

15 ft

5 ft 10 ft

Entire beam: a + ©MA = 0;

- 150 (5) - 600 (7.5) + By (15) -

12 (690) (25) = 0 13

By = 1411.54 lb Segment CBD: + ©F = 0; : x

- NC -

5 (690) = 0 13 Ans.

NC = -265 lb + c ©Fy = 0;

VC - 6 - 120 + 1411.54 - 690 a

12 b = 0 13 Ans.

VC = - 648.62 = - 649 lb a + ©MC = 0;

- 6(1) - 120 (1.5) + 1411.54 (3) -

12 (690) (13) - MC = 0 13

MC = - 4231.38 lb # ft = - 4.23 kip # ft

Ans.

Segment D: + ©F = 0; : x

-ND -

5 (690) = 0 13 Ans.

ND = - 265 lb + c ©Fy = 0;

VD -

12 (690) = 0 13 Ans.

VD = 637 lb a + ©MD = 0;

- MD - 690 a

12 b (5) = 0 13

MD = - 3.18 kip # ft

Ans.

Ans: NC = 265 lb VC = - 649 lb MC = - 4.23 kip # ft ND = - 265 lb VD = 637 lb MD = - 3.18 kip # ft 621

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*7–12. Determine the distance a between the bearings in terms of the shaft’s length L so that the moment in the symmetric shaft is zero at its center.

w

a L

SOLUTION Due to symmetry, Ay = By + c ©Fy = 0;

a + ©M = 0;

Ay + By -

w(L - a) w(L - a) - wa = 0 4 4

Ay = By =

w (L + a) 4

-M -

w(La) a L a w a wa a a b a + - b + (L + a) a b = 0 2 4 4 2 6 6 4 2

Since M = 0; 3a2 + (L - a)(L + 2a) - 3a (L + a) = 0 2a2 + 2a L - L2 = 0 Ans.

a = 0.366 L

Ans: a = 0.366 L 622

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7–13. Determine the internal normal force, shear force, and moment in the beam at sections passing through points D and E. Point D is located just to the left of the 5-kip load.

5 kip 1.5 kip/ft

6 kip  ft

D

A 6 ft

E

B 6 ft

4 ft

C 4 ft

Solution Support Reaction. Referring to the FBD of member AB shown in Fig. a   a+ ΣMB = 0;  5(6) + 6 - Ay(12) = 0  Ay = 3.00 kip   a+ ΣMA = 0;   By(12) - 5(6) + 6 = 0        By = 2.00 kip + ΣFx = 0    S

Bx = 0

Internal Loading. Referring to the left segment of member AB sectioned through D, Fig. b, + ΣFx = 0;    S

ND = 0

   + c ΣFy = 0;

3.00 - VD = 0

Ans. VD = 3.00 kip

  a+ ΣMD = 0;  MD + 6 - 3.00(6) = 0  MD = 12.0 kip # ft

Ans. Ans.

Referring to the left segment of member BC sectioned through E, Fig. c, + ΣFx = 0;    S

Ans.

NE = 0

   + c ΣFy = 0;        - VE - 1.5(4) - 2.00 = 0        VE = -8.00 kip

Ans.

  a+ ΣME = 0;  ME + 1.5(4)(2) + 2.00(4) = 0  ME = -20.0 kip # ft Ans.

Ans: ND = 0 VD = 3.00 kip MD = 12.0 kip # ft NE = 0 VE = - 8.00 kip ME = - 20.0 kip # ft 623

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7–14.

The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the normal force, shear force, and moment at a section passing through (a) point C, which is just to the right of the bearing at A, and (b) point D, which is just to the left of the 3000-lb force.

2500 lb

75 lb/ ft C

3000 lb

D

B

A 6 ft

12 ft

2 ft

SOLUTION a + ©MB = 0;

-Ay (14) + 2500(20) + 900(8) + 3000(2) = 0 Ay = 4514 lb

+ ©F = 0; : x

Bx = 0

+ c ©Fy = 0;

4514 - 2500 - 900 - 3000 + B y = 0 By = 1886 lb

a + ©MC = 0;

2500(6) + MC = 0 MC = -15 000 lb # ft = -15.0 kip # ft

Ans.

+ ©F = 0; : x

NC = 0

Ans.

+ c ©Fy = 0;

- 2500 + 4514 - VC = 0 Ans.

VC = 2014 lb = 2.01 kip a + ©MD = 0;

- MD + 1886(2) = 0 MD = 3771 lb # ft = 3.77 kip # ft

Ans.

+ ©F = 0; : x

ND = 0

Ans.

+ c ©Fy = 0;

VD - 3000 + 1886 = 0 Ans.

VD = 1114 lb = 1.11 kip

Ans: MC = -15.0 kip # ft NC = 0 VC = 2.01 kip MD = 3.77 kip # ft ND = 0 VD = 1.11 kip 624

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7–15. Determine the internal normal force, shear force, and moment at point C.

6 kN/ m

B

A 3m

C

3m

Solution Support Reactions. Referring to the FBD of the entire beam shown in Fig. a,   a+ ΣMA = 0;  By(6) + ΣFx = 0;    S

1 (6)(6)(2) = 0  By = 6.00 kN 2 Bx = 0

Internal Loadings. Referring to the FBD of right segment of the beam sectioned through C, Fig. b + ΣFx = 0;    S

Ans.

NC = 0

1    + c ΣFy = 0; VC + 6.00 - (3)(3) = 0 VC = -1.50 kN 2 1   a+ ΣMC = 0;  6.00(3) - (3)(3)(1) - MC = 0  MC = 13.5 kN # m 2

Ans. Ans.

Ans: NC = 0 VC = - 1.50 kN MC = 13.5 kN # m 625

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*7–16. Determine the internal normal force, shear force, and moment at point C of the beam.

400 N/m 200 N/m A

B

C 3m

3m

SOLUTION Beam: a + ©MB = 0;

600 (2) + 1200 (3) - Ay (6) = 0 Ay = 800 N

+ ©F = 0; : x

Ax = 0

Segment AC: + ©F = 0; : x

NC = 0

+ c ©Fy = 0;

800 - 600 - 150 - VC = 0

Ans.

Ans.

VC = 50 N a + ©MC = 0;

- 800 (3) + 600 (1.5) + 150 (1) + MC = 0 MC = 1350 N # m = 1.35 kN # m

Ans.

Ans: NC = 0 VC = 50 N MC = 1.35 kN # m 626

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7–17. The cantilevered rack is used to support each end of a smooth pipe that has a total weight of 300 lb. Determine the normal force, shear force, and moment that act in the arm at its fixed support A along a vertical section.

C

6 in. B

30

A

SOLUTION Pipe: + c ©Fy = 0;

NB cos 30° - 150 = 0 NB = 173.205 lb

Rack: + ©F = 0; : x

- NA + 173.205 sin 30° = 0 Ans.

NA = 86.6 lb + c ©Fy = 0;

VA - 173.205 cos 30° = 0 Ans.

VA = 150 lb a + ©MA = 0;

MA - 173.205(10.3923) = 0 MA = 1.80 lb # in.

Ans.

Ans: NA = 86.6 lb VA = 150 lb MA = 1.80 kip # in. 627

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7–18.

Determine the internal normal force, shear force, and the moment at points C and D.

A 2m

C 2 kN/m

6m 45˚

B

D

SOLUTION

3m

3m

Support Reactions: FBD (a). a + ©MA = 0;

By 16 + 6 cos 45°2 - 12.013 + 6 cos 45°2 = 0 By = 8.485 kN

+ c ©Fy = 0; + ©F = 0 : x

A y + 8.485 - 12.0 = 0

A y = 3.515 kN

Ax = 0

Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have Q+ ©Fx¿ = 0;

3.515 cos 45° - VC = 0

VC = 2.49 kN

Ans.

a+ ©Fy¿ = 0;

3.515 sin 45° - NC = 0

NC = 2.49 kN

Ans.

a + ©MC = 0;

MC - 3.515 cos 45°122 = 0 MC = 4.97 kN # m

Ans.

Applying the equations of equilibrium to segment BD [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0;

ND = 0 VD + 8.485 - 6.00 = 0

Ans. VD = -2.49 kN

Ans.

8.485132 - 611.52 - MD = 0 MD = 16.5 kN # m

Ans.

Ans: VC = 2.49 kN NC = 2.49 kN MC = 4.97 kN # m ND = 0 VD = -2.49 kN MD = 16.5 kN # m 628

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7–19. Determine the internal normal force, shear force, and moment at point C.

B

0.5 ft 3 ft

A

2 ft C 8 ft

150 lb/ft 4 ft

SOLUTION Entire beam: a + ©MA = 0;

-1.8 (6) + T (2.5) = 0 T = 4.32 kip

+ ©F = 0; : x

Ax - 4.32 = 0 Ax = 4.32 kip

+ c ©Fy = 0;

Ay - 1.8 = 0 Ay = 1.8 kip

Segment AC: + ©F = 0; : x

4.32 + NC = 0 Ans.

NC = - 4.32 kip + c ©Fy = 0;

1.8 - 0.45 - VC = 0 Ans.

VC = 1.35 kip a + ©MC = 0;

- 1.8 (3) + 0.45 (1.5) + MC = 0 MC = 4.72 kip # ft

Ans.

Ans: NC = -4.32 kip VC = 1.35 kip MC = 4.72 kip # ft 629

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*7–20. Rod AB is fixed to a smooth collar D, which slides freely along the vertical guide. Determine the internal normal force, shear force, and moment at point C. which is located just to the left of the 60-lb concentrated load.

60 lb 15 lb/ft A

B

D

C 3 ft

SOLUTION

30

1.5 ft

With reference to Fig. a, we obtain + c ©Fy = 0;

FB cos 30° -

1 1 (15)(3) - 60 - (15)(1.5) = 0 2 2

FB = 108.25 lb

Using this result and referring to Fig. b, we have + ©Fx = 0; :

- NC - 108.25 sin 30° = 0

+ c ©Fy = 0;

VC - 60 -

NC = -54.1 lb

1 (15)(1.5) + 108.25 cos 30° = 0 2 Ans.

VC = - 22.5 lb a + ©MC = 0;

Ans.

108.25 cos 30°(1.5) MC = 135 lb # ft

1 (15)(1.5)(0.5) - MC = 0 2 Ans.

The negative signs indicates that NC and VC act in the opposite sense to that shown on the free-body diagram.

Ans: NC = - 54.1 lb VC = - 22.5 lb MC = 135 lb # ft 630

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7–21. Determine the internal normal force, shear force, and moment at points E and F of the compound beam. Point E is located just to the left of 800 N force.

800 N

1200 N 400 N/m

5

4 3

A E 1.5 m

1m

C

B 2m

1m

D

F 1.5 m

1.5 m

Solution Support Reactions. Referring to the FBD of member BC shown in Fig. a, 4   a+ ΣMB = 0;  Cy(3) - 1200 a b(2) = 0  Cy = 640 N 5 4   a+ ΣMC = 0;  1200 a b(1) - By(3) = 0  By = 320 N 5 3 + ΣFx = 0;    S 1200 a b - Bx = 0 Bx = 720 N 5 Internal Loadings. Referring to the right segment of member AB sectioned through E, Fig. b + ΣFx = 0;    S

720 - NE = 0

NE = 720 N

Ans.

   + c ΣFy = 0;

VE - 800 - 320 = 0

VE = 1120 N = 1.12 kN

Ans.

  a+ ΣME = 0;     - ME - 320(1) = 0     ME = -320 N # m

Ans.

Referring to the left segment of member CD sectioned through F, Fig. c, + ΣFx = 0;     S

 

Ans.

  NF = 0

    + c ΣFy = 0;  - VF - 640 - 400(1.5) = 0  VF = -1240 N = -1.24 kN Ans.   a+ ΣMF = 0;  MF + 400(1.5)(0.75) + 640(1.5) = 0

MF = - 1410 N # m = - 1.41 kN # m

Ans.

Ans: NE = 720 N VE = 1.12 kN ME = - 320 N # m NF = 0 VF = - 1.24 kN MF = - 1.41 kN # m 631

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7–22. Determine the internal normal force, shear force, and moment at points D and E in the overhang beam. Point D is located just to the left of the roller support at B, where the couple moment acts.

2 kN/m

6 kN m

C A

D 3m

B

E

1.5 m

1.5 m

3

5 4

5 kN

SOLUTION The intensity of the triangular distributed load at E can be found using the similar triangles in Fig. b. With reference to Fig. a, a + ©MA = 0;

3 1 By (3) -2(3)(1.5)-6 - (2)(3)(4)-5a b(6) = 0 2 5

By = 15 kN Using this result and referring to Fig. c, 4 + ©F = 0; 5 a b -ND = 0 : x 5 3 1 (2)(3)-5a b = 0 2 5

+ c ©Fy = 0;

VD + 15 -

a + ©MD = 0;

3 1 - MD - 6 - (2)(3)(1)-5 a b (3) = 0 2 5

ND = 4 kN

Ans.

VD = -9 kN

Ans.

MD = -18 kN # m

Ans.

NE = 4 kN

Ans.

VE = 3.75 kN

Ans.

ME = -4.875 kN # m

Ans.

Also, by referring to Fig. d, we can write + ©F = 0; : x

4 5 a b -NE = 0 5

+ c ©Fy = 0;

VE -

a + ©ME = 0;

- ME -

3 1 (1)(1.5)- 5 a b = 0 2 5 3 1 (1)(1.5)(0.5)- 5 a b (1.5) = 0 2 5

The negative sign indicates that VD, MD, and ME act in the opposite sense to that shown on the free-body diagram.

Ans: ND = 4 kN VD = -9 kN MD = -18 kN # m NE = 4 kN  VE = 3.75 kN ME = -4.875 kN # m 632

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7–23. Determine the internal normal force, shear force, and moment at point C.

0.2 m

400 N 1m

A

B

C 1.5 m 3m

SOLUTION

2m

Beam: + ©F = 0; : x

- Ax + 400 = 0 Ax = 400 N

a + ©MB = 0;

A y (5) -400(1.2) = 0 Ay = 96 N

Segment AC: + ©F = 0; : x

NC - 400 = 0 Ans.

NC = 400 N + c ©Fy = 0;

- 96 - VC = 0 Ans.

VC = - 96 N a + ©MC = 0;

MC + 96 (1.5) = 0 MC = -144 N # m

Ans.

Ans: NC = 400 N VC = - 96 N MC = - 144 N # m 633

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*7–24. Determine the ratio of ab for which the shear force will be zero at the midpoint C of the beam.

w

A a

SOLUTION a + ©MB = 0;

-

A

C b/2

C

B b/2

B

a

2 w (2a + b)c (2a + b) - (a + b) d + Ay (b) = 0 2 3

Ay =

w (2a + b)(a - b) 6b

+ ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

-

w w b (2a + b)(a - b) a a + b - VC = 0 6b 4 2

-

1 1 1 (2a + b)(a - b) = (2a + b) a b 6b 4 2

-

1 1 (a - b) = 6b 8

Since V C = 0,

-a+b =

3 b 4

a 1 = b 4

Ans.

Ans: a 1 = b 4 634

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7–25. Determine the normal force, shear force, and moment in the beam at sections passing through points D and E. Point E is just to the right of the 3-kip load.

3 kip 1.5 kip/ ft

A

6 ft

SOLUTION a + ©MB = 0;

D

1 2 (1.5)(12)(4)

E

B 6 ft

4 ft

C

4 ft

- Ay (12) = 0

Ay = 3 kip + ©F = 0; : x

Bx = 0

+ c ©Fy = 0;

By + 3 -

1 2

(1.5)(12) = 0

By = 6 kip + ©F = 0; : x

ND = 0

+ c ©Fy = 0;

3 - 12 (0.75)(6) - VD = 0

Ans.

Ans.

VD = 0.75 kip a ©MD = 0;

MD + 12 (0.75) (6) (2) - 3 (6) = 0 MD = 13.5 kip # ft

Ans.

+ ©F = 0; : x

NE = 0

Ans.

+ c ©Fy = 0;

- VE - 3 - 6 = 0 Ans.

VE = -9 kip ©ME = 0;

ME + 6 (4) = 0 ME = - 24.0 kip # ft

Ans.

Ans: ND = 0 VD = 0.75 kip MD = 13.5 kip # ft NE = 0 VE = - 9 kip ME = - 24.0 kip # ft 635

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7–26.

Determine the internal normal force, shear force, and bending moment at point C.

40 kN 8 kN/m 60° A 3m

C

B 3m

3m 0.3 m

SOLUTION Free body Diagram: The support reactions at A need not be computed. Internal Forces: Applying equations of equilibrium to segment BC, we have + ©F = 0; : x + c ©Fy = 0;

-40 cos 60° - NC = 0

NC = -20.0 kN

VC - 24.0 - 12.0 - 40 sin 60° = 0 Ans.

VC = 70.6 kN a + ©MC = 0;

Ans.

-24.011.52 - 12.0142 - 40 sin 60°16.32 - MC = 0 MC = - 302 kN # m

Ans.

Ans: NC = -20.0 kN VC = 70.6 kN MC = -302 kN # m 636

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7–27. Determine the internal normal force, shear force, and moment at point C.

B

E

200 N 1m

C A

D 1m

1m

2m

800 N  m

Solution Support Reactions. Referring to the FBD of the entire assembly shown in Fig. a,   a+ ΣMA = 0;  FBE (1) - 200(4) - 800 = 0     FBE = 1600 N + ΣFx = 0;    S

Ax - 1600 = 0

Ax = 1600 N

   + c ΣFy = 0;

Ay - 200 = 0

Ay = 200 N

Internal Loading. Referring to the FBD of the left segment of the assembly sectioned through C, Fig. b, + ΣFx = 0;      1600 + NC = 0    S

NC = - 1600 N = - 1.60 kN

Ans.

   + c ΣFy = 0;     200 - VC = 0

VC = 200 N

Ans.

  a+ ΣMA = 0;  MC - 200(1) = 0    MC = 200 N # m

Ans.

Ans: NC = - 1.60 kN VC = 200 N MC = 200 N # m 637

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*7–28. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 10-kN concentrated load.

10 kN

6 kN/m

A

D

C 1.5 m

1.5 m

1.5 m

B

1.5 m

SOLUTION The intensity of the triangular distributed loading at C can be computed using the similar triangles shown in Fig. b, wC 6 = or wC = 3 kN>m 1.5 3 With reference to Fig. a, a + ©MA = 0; a + ©MB = 0; + ©F = 0 : x

1 By(6)- 10(4.5) - (6)(3)(1) = 0 2 1 (6)(3)(5) + 10(1.5) - Ay(6) = 0 2

By = 9 kN Ay = 10 kN

Ax = 0

Using these results and referring to Fig. c, + ©F = 0; : x

NC = 0

+ c ©Fy = 0;

10 -

a + ©MC = 0;

MC + 3(1.5)(0.75) +

Ans.

1 (3)(1.5)-3(1.5) -VC = 0 2

VC = 3.25 kN

1 (3)(1.5)(1) - 10(1.5) = 0 2

Ans.

MC = 9.375 kN # m Ans.

Also, by referring to Fig. d, + ©F = 0; : x

ND = 0

+ c ©Fy = 0;

VD + 9 - 10 = 0

VD = 1 kN

Ans.

a + ©MD = 0;

9(1.5)-MD = 0

MD = 13.5 kN # m

Ans.

Ans.

Ans: NC = 0 VC = 3.25 kN MC = 9.375 kN # m ND = 0 VD = 1 kN MD = 13.5 kN # m 638

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7–29. Determine the normal force, shear force, and moment acting at a section passing through point C.

700 lb

800 lb 3 ft 1.5 ft

2 ft

600 lb 1 ft

D

3 ft

1.5 ft A

C

30

30 B

SOLUTION a + ©M A = 0;

- 800 (3) - 700(6 cos 30°) - 600 cos 30°(6 cos 30° + 3cos 30°) + 600 sin 30°(3 sin 30°) + By (6 cos 30° + 6 cos 30°) = 0 B y = 927.4 lb

+ ©F = 0; : x

800 sin 30° - 600 sin 30° - Ax = 0 Ax = 100 lb

+ c ©Fy = 0;

Ay - 800 cos 30° - 700 - 600 cos 30° + 927.4 = 0 Ay = 985.1 lb

Q + ©Fx = 0;

NC - 100 cos 30° + 985.1 sin 30° = 0 Ans.

N C = - 406 lb + a©Fy = 0;

100 sin 30° + 985.1 cos 30° - VC = 0 Ans.

V C = 903 lb a + ©M C = 0;

-985.1(1.5 cos 30°) - 100(1.5 sin 30°) + MC = 0 MC = 1355 lb # ft = 1.35 kip # ft

Ans.

Ans: NC = - 406 lb VC = 903 lb MC = 1.35 kip # ft 639

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7–30.

Determine the normal force, shear force, and moment acting at a section passing through point D.

700 lb

800 lb 3 ft 1.5 ft

2 ft

600 lb 1 ft

D

3 ft

1.5 ft

SOLUTION a+ ©M A = 0;

A

C

30

30 B

- 800(3) - 700(6 cos 30°) - 600 cos 30°(6 cos 30° + 3 cos 30°) + 600 sin 30°(3 sin 30°) + By (6 cos 30° + 6 cos 30°) = 0 B y = 927.4 lb

+ ©F = 0; : x

800 sin 30° - 600 sin 30° - Ax = 0 Ax = 100 lb

+ c ©Fy = 0;

Ay - 800 cos 30° - 700 - 600 cos 30° + 927.4 = 0 Ay = 985.1 lb

+a©Fx = 0;

ND - 927.4 sin 30° = 0 Ans.

ND = - 464 lb Q + ©Fy = 0;

VD - 600 + 927.4 cos 30° = 0 Ans.

VD = - 203 lb a + ©M D = 0;

- MD - 600(1) + 927.4(4 cos 30°) = 0 MD = 2612 lb # ft = 2.61 kip # ft

Ans.

Ans: ND = -464 lb VD = -203 lb MD = 2.61 kip # ft 640

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7–31. Determine the internal normal force, shear force, and moment acting at points D and E of the frame.

B A

E

2m

1.5 m

D C 4m

900 N . m 600 N

Solution Support Reactions. Notice that member AB is a two force member. Referring to the FBD of member BC,   a+ ΣMC = 0;  FAB (1.5) - 900 - 600(4) = 0  FAB = 2200 N + ΣFx = 0;    S

Cx - 2200 = 0

Cx = 2200 N

   + c ΣFy = 0;

Cy - 600 = 0

Cy = 600 N

Internal Loadings. Referring to the left segment of member AB sectioned through E, Fig. b, + ΣFx = 0;    S

NE - 2200 = 0

NE = 2200 N = 2.20 kN Ans.

   + c ΣFy = 0;

VE = 0

Ans.

  a+ ΣME = 0;

ME = 0

Ans.

Referring to the left segment of member BC sectioned through D, Fig. c + ΣFx = 0;    S

ND + 2200 = 0      ND = - 2200 N = - 2.20 kN

Ans.

   + c ΣFy = 0;

600 - VD = 0       VD = 600 N

Ans.

  a+ ΣMD = 0;  MD - 600(2) = 0    MD = 1200 N # m = 1.20 kN # m Ans.

Ans: NE = 2.20 kN VE = 0 ME = 0 ND = - 2.20 kN VD = 600 N MD = 1.20 kN # m 641

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*7–32. Determine the internal normal force, shear force, and moment at point D.

6 kN E 3m B 1m D 3m

Solution

A

C

Support Reactions. Notice that member BC is a two force member. Referring to the FBD of member ABE shown in Fig. a, 3   a+ ΣMA = 0;  FBC a b(4) - 6(7) = 0  FBC = 17.5 kN 5 3 + ΣFx = 0; Ax = 4.50 kN    S Ax - 17.5 a b + 6 = 0 5 4    + c ΣFy = 0; Ay - 17.5 a b = 0 Ay = 14.0 kN 5 Internal Loadings. Referring to the FBD of the lower segment of member ABE sectioned through D, Fig. b, + ΣFx = 0;    S

4.50 + VD = 0

VD = -4.50 kN

Ans.

   + c ΣFy = 0;

ND + 14.0 = 0

ND = -14.0 kN

Ans.

  a+ ΣMD = 0;

MD + 4.50(3) = 0

MD = -13.5 kN # m

3m

Ans.

Ans: VD = -4.50 kN ND = -14.0 kN MD = -13.5 kN # m 642

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7–33. Determine the internal normal force, shear force, and moment at point D of the two-member frame.

1.5 m

1.5 m B D

1.5 kN/m

1.5 m

E

SOLUTION Member BC: a + ©MC = 0;

1.5 m

4.5 (1.5) - Bx (3) = 0

+ ©F = 0; : x

A

2 kN/m

Bx = 2.25 kN

C

2.25 + Cx - 4.5 = 0 Cx = 2.25 kN

Member AB: a + ©MA = 0;

2.25 (3) - 3 (1) - By (3) = 0 By = 1.25 kN

Segment DB: + ©F = 0; : x

- ND - 2.25 = 0 Ans.

ND = - 2.25 kN + c ©Fy = 0;

VD - 1.25 = 0 Ans.

VD = 1.25 kN a + ©MD = 0;

- MD - 1.25 (1.5) = 0 MD = - 1.88 kN # m

Ans.

Ans: ND = - 2.25 kN VD = 1.25 kN -1.88 kN # m 643

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7–34. Determine the internal normal force, shear force, and moment at point E.

1.5 m

1.5 m B D

1.5 kN/m

1.5 m

E

SOLUTION Member BC: a + ©MC = 0;

1.5 m

4.5 (1.5) - Bx (3) = 0

+ ©F = 0; : x

A

2 kN/m

Bx = 2.25 kN

C

2.25 + Cx - 4.5 = 0 Cx = 2.25 kN

Member AB: a + ©MA = 0;

2.25 (3) - 3 (1) - By (3) = 0 By = 1.25 kN

Segment BE: + c ©Fy = 0;

1.25 - NE = 0 Ans.

NE = 1.25 kN + ©F = 0; : x

VE + 2.25 - 2.25 = 0 Ans.

VE = 0 + ©Mg = 0;

Mg - 2.25 (0.75) = 0 Mg = 1.6875 kN # m = 1.69 kN # m

Ans.

Ans: NE = 1.25 kN VE = 0 MB = 1.69 kN # m 644

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7–35. The strongback or lifting beam is used for materials handling. If the suspended load has a weight of 2 kN and a center of gravity of G, determine the placement d of the padeyes on the top of the beam so that there is no moment developed within the length AB of the beam. The lifting bridle has two legs that are positioned at 45°, as shown.

3m

d

A

3m

45°

45°

0.2 m 0.2 m

SOLUTION

E

B

d

F

Support Reactions: From FBD (a), a + ©ME = 0;

FF162 - 2132 = 0

FE = 1.00 kN

+ c ©Fy = 0;

FF + 1.00 - 2 = 0

FF = 1.00 kN

G

From FBD (b), + ©F = 0; : x + c ©Fy = 0;

FAC cos 45° - FBC cos 45° = 0

FAC = FBC = F

2F sin 45° - 1.00 - 1.00 = 0 FAC = FBC = F = 1.414 kN

Internal Forces: This problem requires MH = 0. Summing moments about point H of segment EH[FBD (c)], we have a + ©MH = 0;

1.001d + x2 - 1.414 sin 45°1x2 - 1.414 cos 45°10.22 = 0 Ans.

d = 0.200 m

Ans: d = 0.200 m 645

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*7–36. Determine the internal normal force, shear force, and moment acting at points B and C on the curved rod.

A B 45 0.5 m

30 C

3

Solution

4

Support Reactions. Not required

5

200 N

Internal Loadings. Referring to the FBD of bottom segment of the curved rod sectioned through C, Fig. a + Q ΣFx = 0; NC - 200 sin (36.87° + 30°) = 0  NC = 183.92 N = 184 N

Ans.

+ aΣFy = 0; - VC - 200 cos (36.87° + 30° ) VC = -78.56 N = - 78.6 N

Ans.

4 3 a+ ΣMC = 0;  200 a b(0.5 sin 30°) - 200 a b[0.5(1 - cos 30°)] + MC = 0 5 5



MC = - 31.96 N # m = - 32.0 N # m

Ans.

Referring to the FBD of bottom segment of the curved rod sectioned through B, Fig. b a + ΣFx = 0;    NB - 200 sin (45° - 36.87°) = 0       NB = 28.28 N = 28.3 N Ans. + bΣFy = 0;    - VB + 200 cos (45° - 36.87°) = 0  VB = 197.99 N = 198 N Ans. 4 3 a+ ΣMB = 0;  MB + 200 a b(0.5 sin 45°) - 200 a b[0.5(1 + cos 45°] = 0 5 5

MB = 45.86 N # m = 45.9 N # m

Ans.

Ans: NC = 184 N VC = - 78.6 N MC = -32.0 N # m NB = 28.3 N VB = 198 N MB = 45.9 N # m 646

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7–37. Determine the internal normal force, shear force, and moment at point D of the two-member frame.

250 N/m B

A

D 2m

1.5 m C

SOLUTION

E

300 N/m

4m

Member AB: a + ©MA = 0;

By (4) - 1000 (2) = 0 By = 500 N

Member BC: a + ©MC = 0;

- 500 (4) + 225 (0.5) + Bx (1.5) = 0 Bx = 1258.33 N

Segment DB: + ©F = 0; : x

- ND + 1258.33 = 0 Ans.

ND = 1.26 kN + c ©Fy = 0;

VD - 500 + 500 = 0 Ans.

VD = 0 a + ©MD = 0;

-MD + 500 (1) = 0 MD = 500 N # m

Ans.

Ans: ND = 1.26 kN VD = 0 MD = 500 N # m 647

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7–38. Determine the internal normal force, shear force, and moment at point E of the two-member frame.

250 N/m B

A

D 2m

1.5 m C

SOLUTION

E

300 N/m

4m

Member AB: a + ©MA = 0;

By (4) - 1000 (2) = 0 By = 500 N

Member BC: a + ©MC = 0;

- 500 (4) + 225 (0.5) + Bx (1.5) = 0 Bx = 1258.33 N

Segment EB: + ©F = 0; : x

- NE - 1258.33 - 225 = 0 Ans.

NE = - 1.48 kN + c ©Fy = 0;

VE - 500 = 0 Ans.

VE = 500 N a + ©ME = 0;

- ME + 225 (0.5) + 1258.33 (1.5) - 500 (2) = 0 ME = 1000 N # m

Ans.

Ans: NE = - 1.48 kN VE = 500 N ME = 1000 N # m 648

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7–39. The distributed loading w = w0 sin u, measured per unit length, acts on the curved rod. Determine the internal normal force, shear force, and moment in the rod at u = 45°.

w = w0 sin θ

r

θ

SOLUTION w = w0 sin u Resultants of distributed loading: u

FRx =

u

w0 sin u(r du) cos u = rw0

L0 u

FRy =

L0

sin u cos u du =

1 r w0 sin2 u 2

u

w0 sin u(r du) sin u = rw0

Q+ ©Fx = 0;

L0

1 l sin2 u du = rw0 c u - sin 2u d 2 4 L0

-V + FRx cos 45° + FRy sin 45° = 0 1 p 1 1 - sin 90°b sin 45° V = a r w0 sin2 45° b cos 45° + w0 a 2 2 4 4 Ans.

V = 0.278 w0r +a©Fy

= 0;

- N - FRy cos 45° + FRx sin 45° = 0 1 p 1 1 N = - r w0 c a b - sin 90° d cos 45° + a r w0 sin2 45°b sin 45° 2 4 4 2 Ans.

N = 0.0759 w0 r a + ©MO = 0;

M - (0.0759 r w0)(r) = 0 M = 0.0759 w0 r2

Ans.

Ans: V = 0.278 w0 r N = 0.0759 w0 r M = 0.0759 w0 r 2 649

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*7–40. Solve Prob. 7–39 for u = 120°.

w = w0 sin θ

r

θ

SOLUTION Resultants of distributed load: L0

u

L0

u

FRy = FRx =

1 2

FRx =

L0

w0 sin u(r du) cos u = rw0

u

sin u cos u =

1 rw sin2 u 2 0

u

w0 sin u(r du) sin u = rw0

r w0 sin2 120° = 0.375 r w0

u 1 1 sin2 u du = rw0 c u sin 2u drw0 (sin u) 2 = r w0 (sin u) 2 4 L0 0

1 120° 1 b - sin 240° d = 1.2637 r w0 FRy = r w0 c (p)a 2 180° 4 +

b©Fx¿ = 0;

N + 0.375 rw0 cos 30° + 1.2637 r w0 sin 30° = 0 Ans.

N = - 0.957 r w0 +a©Fy¿

= 0;

- V + 0.375 rw0 sin 30° - 1.2637 r w0 cos 30° = 0 Ans.

V = - 0.907 rw0 a + ©MO = 0;

- M - 0.957 r w0 (r) = 0 M = - 0.957 r2w0

Ans.

Ans: N = - 0.957 r w0 V = - 0.907 rw0 M = - 0.957 r 2w0 650

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7–41. Determine the x, y, z components of force and moment at point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = 5350i - 400j6 lb and F2 = 5-300j + 150k6 lb.

z B

F1

C

3 ft

SOLUTION x

Internal Forces: Applying the equations of equilibrium to segment BC, we have ©Fx = 0;

NC + 350 = 0

©Fy = 0;

1VC2y - 400 - 300 = 0

©Mx = 0;

©My = 0;

©Mz = 0;

y

2 ft

Free body Diagram: The support reactions need not be computed.

©Fz = 0;

1.5 ft

Ans.

NC = - 350 lb

1VC2z + 150 = 0

1VC2y = 700 lb

Ans.

1VC2z = - 150 lb

Ans.

1MC2x = -1200 lb # ft = - 1.20 kip # ft

Ans.

1MC2y = - 750 lb # ft

Ans.

1MC2x + 400132 = 0

F2

1MC2y + 350132 - 150122 = 0 1MC2z - 300122 - 400122 = 0 MC

z

= 1400 lb # ft = 1.40 kip # ft

Ans.

Ans: NC = -350 lb (VC)y = 700 lb (VC)z = - 150 lb (MC)x = - 1.20 kip # ft (MC)y = - 750 lb # ft (MC)z = 1.40 kip # ft 651

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7–42. Determine the x, y, z components of force and moment at point C in the pipe assembly. Neglect the weight of the pipe. The load acting at (0, 3.5 ft, 3 ft) is F1 = 5-24i - 10k6 lb and M = {-30k} lb # ft and at point (0, 3.5 ft, 0) F2 = {-80i} lb.

z

F1 B M 3 ft

SOLUTION

A

Free body Diagram: The support reactions need not be computed. Internal Forces: Applying the equations of equilibrium to segment BC, we have

1.5 ft x

C

y

2 ft

F2

Ans.

©Fx = 0;

(VC)x - 24- 80 = 0

(VC)x = 104 lb

©Fy = 0;

NC = 0

©Fz = 0;

(VC)z - 10 = 0

(VC)z = 10.0 lb

Ans.

©Mx = 0;

(MC)x - 10(2) = 0

(MC)x = 20.0 lb # ft

Ans.

©My = 0;

(MC)y - 24 (3) = 0

(MC)y = 72.0 lb # ft

Ans.

©Mz = 0;

(MC)z + 24 (2) + 80 (2) - 30 = 0

Ans.

(MC)z = - 178 lb # ft

Ans.

Ans: (VC)x = 104 lb NC = 0 (VC)z = 10 lb (MC)x = 20 lb # ft (MC)y = 72 lb # ft (MC)z = - 178 lb # ft 652

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7–43. Determine the x, y, z components of internal loading at a section passing through point B in the pipe assembly. Neglect the weight of the pipe. Take F1 =  5 200i - 100j - 400k6 N and F2 = 5 300i - 500k 6 N.

z

F2

A B 1m

y

1m x 1.5 m

Solution

F1

Internal Loadings. Referring to the FBD of the free end segment of the pipe assembly sectioned through B, Fig. a,   ΣFx = 0;

Nx + 300 + 200 = 0

Nx = -500 N

Ans.

  ΣFy = 0;

Vy - 100 = 0

Vy = 100 N

Ans.

  ΣFz = 0;

Vz - 500 - 400 = 0

Vz = 900 N

Ans.

  ΣMx = 0;

Mx - 400(1.5) = 0

Mx =

Ans.

  ΣMy = 0;

My + 500(1) + 400(1) = 0

My =

  ΣMz = 0

Mz - 200(1.5) - 100(1) = 0

Mz =

600 N # m -900 N # m 400 N # m

Ans. Ans.

The negative signs indicate that Nx and My act in the opposite sense to those shown in FBD.

Ans: Nx = - 500 N Vy = 100 N Vz = 900 N Mx = 600 N # m My = - 900 N # m Mz = 400 N # m 653

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*7–44. Determine the x, y, z components of internal loading at a section passing through point B in the pipe assembly. Neglect the weight of the pipe. Take F1 = 5100i - 200j - 300k 6  N and F2 = 5 100i + 500j 6 N.

z

F2

A B 1m

y

1m x 1.5 m

Solution

F1

Internal Loadings. Referring to the FBD of the free end segment of the pipe assembly sectioned through B, Fig. a   ΣFx = 0;

Nx + 100 + 100 = 0

Nx = - 200 N

Ans.

  ΣFy = 0;

Vy + 500 - 200 = 0

Vy = - 300 N

Ans.

  ΣFz = 0;

Vz - 300 = 0

Vz = 300 N

Ans.

  ΣMx = 0;

Mx - 300(1.5) = 0

  ΣMy = 0;

My + 300(1) = 0

  ΣMz = 0;

Mz + 500(1) - 100(1.5) - 200(1) = 0



Mx = 450 N # m My = - 300 N # m

Ans. Ans.

Mz = - 150 N # m

Ans.

The negative signs indicates that Nx, Vy, My and Mz act in the senses opposite to those shown in FBD.

Ans: Nx = -200 N Vy = - 300 N Vz = 300 N Mx = 450 N # m My = - 300 N # m Mz = - 150 N # m 654

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7–45. Draw the shear and moment diagrams for the shaft (a) in terms of the parameters shown; (b) set P = 9 kN,a = 2 m, L = 6 m. There is a thrust bearing at A and a journal bearing at B.

P

A

B a L

SOLUTION (a) c + ©MB = 0;

(Ay)(L) - P(L - a) = 0 Ay = a

L - a bP L

Ay = a 1 + c ©Fy = 0;

a bP L

Ay + By - P = 0 By = P - Ay = a

+ ©F = 0; : x

a bP L

Ax = 0

For 0 … x … a + c ©Fy = 0;

a bP - V = 0 L

a1 -

V = a1 + ©F = 0; : x

A = 0

c + ©M = 0;

a1 -

a bP L

Ans.

a b Px - M = 0 L

M = a1 -

a b Px L

Ans.

For a 6 x 6 L + c ©Fy = 0;

a1 -

a bP - P - V = 0 L

V = -a c + ©M = 0;

a1 -

a bP L

Ans.

a b Px - P(x - a) - M = 0 L

M = Px - a

a b Px - Px + Pa L

655

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7–45. Continued

(b) M = Paa c + ©MB = 0;

a xb L

Ans.

Ay (6) - 9(4) = 0 Ay = 6 kN

+ c ©Fy = 0;

By = 3 kN

For 0 … x … 2 m + c ©Fy = 0;

6 - V = 0 Ans.

V = 6 kN c + ©M = 0;

6x - M = 0 M = 6x kN # m

Ans.

For 2 m 6 x … 6 m + c ©Fy = 0;

6 - 9 - V = 0 Ans.

V = - 3 kN c + ©M = 0;

6x - 9(x - 2) - M = 0 M = 18 - 3x kN # m

Ans.

Ans: 0 … x 6 a: V = a1 M = a1 -

a bPx L

a bP L

a a 6 x … L: V = - a bP L a M = Paa - xb L 0 … x 6 2 m: V = 6 kN, M = {6x} kN # m

#

2 m 6 x … 6 m:V = -3 kN M = {18 - 3x} kN # m 656

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7–46.

Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set P = 800 lb, a = 5 ft, L = 12 ft.

P

P

a

a L

SOLUTION (a)

For 0 … x 6 a + c ©Fy = 0;

V = P

Ans.

a + ©M = 0;

M = Px

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

-Px + P(x - a) + M = 0

For a 6 x 6 L- a

M = Pa

Ans.

+ c ©Fy = 0;

V = -P

Ans.

a + ©M = 0;

- M + P(L - x) = 0

For L- a 6 x … L

M = P(L - x)

Ans.

657

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7–46. Continued

(b)

Set P = 800 lb,

a = 5 ft,

L = 12 ft

For 0 … x 6 5 ft + c ©F y = 0;

V = 800 lb

Ans.

a + ©M = 0;

M = 800x lb # ft

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

- 800x + 800(x - 5) + M = 0

For 5 ft 6 x 6 7 ft

M = 4000 lb # ft

Ans.

+ c ©F y = 0;

V = -800 lb

Ans.

a + ©M = 0;

- M + 800(12 - x) = 0

For 7 ft 6 x … 12 ft

M = (9600 - 800x) lb # ft

Ans.

Ans: For 0 … x 6 a, V = P, M = Px For a 6 x 6 L - a, V = 0, M = Pa For L - a 6 x … L, V = - P, M = P(L - x) For 0 … x 6 5 ft, V = 800 lb M = 800x lb # ft For 5 ft 6 x 6 7 ft, V = 0 M = 4000 lb # ft For 7 ft 6 x … 12 ft, V = - 800 lb M = (9600 - 800x) lb # ft 658

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7–47. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set P = 600 lb, a = 5 ft, b = 7 ft.

P

A

B a

SOLUTION (a) For 0 … x 6 a Pb - V = 0 a + b

+ c ©Fy = 0;

V = a + ©M = 0;

M -

Pb a + b

Ans.

Pb x = 0 a + b

M =

Pb x a + b

Ans.

For a 6 x … 1a + b2 Pb - P - V = 0 a + b

+ c ©Fy = 0;

V = a + ©M = 0;

-

Pa a + b

Ans.

Pb x + P1x - a2 + M = 0 a + b M = Pa -

Pa x a + b

Ans.

(b) For P = 600 lb, a = 5 ft, b = 7 ft

659

b

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7–47. Continued

(b) c + ©MB = 0;

A y(12) - 600(7) = 0 A y = 350 lb

+ c ©Fy = 0;

By = 250 lb

For 0 … x … 5 ft + c ©Fy = 0;

350 - V = 0 Ans.

V = 350 lb c + ©M = 0;

350x - M = 0

M = 350x lb # ft

Ans.

For 5 ft 6 x … 12 ft + c ©Fy = 0;

350 - 600 - V = 0 Ans.

V = -250 lb c + ©M = 0;

350x - 600(x - 5) - M = 0

M = {3000 – 250x} lb # ft

Ans.

Ans: Pb Pb ,M = x a + b a + b Pa For a 6 x … a + b, V = a + b Pa M = Pa x a + b For 0 … x 6 5 ft, V = 350 lb M = 350x lb # ft For 5 ft 6 x … 12 ft, V = -250 lb M = 53000 - 250x6 lb # ft

For 0 … x 6 a, V =

660

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*7–48.

Draw the shear and moment diagrams for the cantilevered beam.

100 lb 800 lb ft C

A B 5 ft

5 ft

SOLUTION

Ans: V = 100 lb Mmax = - 1800 lb # ft 661

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7–49.

Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set M0 = 500 N # m, L = 8 m.

M0

L/3

M0

L/3

L/ 3

SOLUTION (a) For 0 … x …

L 3

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = 0

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = M0

Ans.

V = 0

Ans.

M = 0

Ans.

For

For

L 2L 6 x 6 3 3

2L 6 x … L 3

+ c ©Fy = 0; a + ©M = 0; (b)

Set M0 = 500 N # m , L = 8 m For 0 … x 6

8 m 3

+ c ©Fy = 0; c + ©M = 0; For

V = 0

Ans.

M = 0

Ans.

8 16 m 6 x 6 m 3 3

+ c ©F y = 0;

V = 0

Ans.

c + ©M = 0;

M = 500 N # m

Ans.

For

Ans: L : V = 0, M = 0 3 L 2L 6 x 6 : V = 0, M = M0 3 3 2L 6 x … L: V = 0, M = 0 3 8 0 … x 6 m: V = 0, M = 0 3 8 16 m 6 x 6 m: V = 0, M = 500 N # m 3 3 16 m 6 x … 8 m: V = 0, M = 0 3

0 … x 6

16 m 6 x … 8m 3

+ c ©Fy = 0;

V = 0

Ans.

c + ©M = 0;

M = 0

Ans.

662

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7–50.

If L = 9 m, the beam will fail when the maximum shear force is Vmax = 5 kN or the maximum bending moment is Mmax = 2 kN # m. Determine the magnitude M0 of the largest couple moments it will support.

M0

L/3

M0

L/ 3

L/ 3

SOLUTION See solution to Prob. 7–48 a. Mmax = M0 = 2 kN # m

Ans.

Ans: Mmax = 2 kN # m 663

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7–51. Draw the shear and moment diagrams for the beam.

w A

C

B a

a

SOLUTION 0 … x 6 a + c ©Fy = 0;

- V - wx = 0 Ans.

V = - wx a + ©M = 0;

x M + wx a b = 0 2 M = -

w 2 x 2

Ans.

a 6 x … 2a + c ©Fy = 0;

- V + 2 wa - wx = 0 Ans.

V = w (2a - x) a + ©M = 0;

x M + wx a b - 2 wa (x - a) = 0 2 M = 2 wa x - 2 wa2 -

w 2 x 2

Ans.

Ans:

w 2 0 … x 6 a: V = - wx, M = - 2 x a 6 x … 2a: V = w(2a - x) w M = 2wax - 2wa2 - x2 2 664

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*7–52. w

Draw the shear and moment diagrams for the beam. A

B

C

L –– 2 L

SOLUTION Support Reactions: From FBD (a), a + ©MA = 0;

Cy 1L2 -

+ c ©Fy = 0;

Ay +

wL 3L a b = 0 2 4

wL 3wL = 0 8 2

Cy =

Ay =

3wL 8 wL 8

L Shear and Moment Functions: For 0 ◊ x< [FBD (b)], 2 + c ©Fy = 0; a + ©M = 0;

wL - V = 0 8 M -

wL 1x2 = 0 8

V =

wL 8

M =

Ans.

wL x 8

Ans.

L For ft. Determine the maximum and minimum tension in the cable.

50 ft 6 ft

w

SOLUTION From Example 7–12: FH =

250 (50)2 w0 L2 = = 13 021 lb 8h 8 (6)

umax = tan - 1 a Tmax =

250 (50) w0 L b = tan - 1 a b = 25.64° 2FH 2(13 021)

FH 13 021 = = 14.4 kip cos umax cos 25.64°

Ans.

The minimum tension occurs at u = 0°. Ans.

Tmin = FH = 13.0 kip

Ans: Tmax = 14.4 kip Tmin = 13.0 kip 724

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*7–104. The cable AB is subjected to a uniform loading of 200 N/m. If the weight of the cable is neglected and the slope angles at points A and B are 30° and 60°, respectively, determine the curve that defines the cable shape and the maximum tension developed in the cable.

B 60°

y

SOLUTION y =

1 ¢ 200 dx ≤ dx FH L L

A

30° x

1 1100x2 + C1x + C22 y = FH

200 N/m

dy 1 1200x + C12 = dx FH At x = 0,

y = 0;

At x = 0,

dy = tan 30°; dx y =

15 m

C2 = 0 C1 = FH tan 30°

1 1100x2 + FH tan 30°x2 FH dy = tan 60°; dx

At x = 15 m,

FH = 2598 N

y = 138.5x 2 + 577x2110-32 m

Ans.

umax = 60° Tmax =

FH 2598 = = 5196 N cos umax cos 60° Ans.

Tmax = 5.20 kN

Ans: umax = 60° Tmax = 5.20 kN 725

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7–105. If x = 2 ft and the crate weighs 300 lb, which cable segment AB, BC, or CD has the greatest tension? What is this force and what is the sag yB?

2 ft

3 ft

3 ft

yB A

D 3 ft

B

C

SOLUTION The forces FB and FC exerted on joints B and C will be obtained by considering the equilibrium on the free-body diagram, Fig. a. + ©ME = 0;

FC(3) - 300(2) = 0

FC = 200 lb

+ ©MF = 0;

300(1) - FB(3) = 0

FB = 200 lb

x

Referring to Fig. b, we have + ©MA = 0;

TCD sin 45°(8) - 200(5) - 100(2) = 0 TCD = 212.13 lb = 212 lb (max)

Using these results and analyzing the equilibrium of joint C, Fig. c, we obtain + ©F = 0; : x

212.13 cos 45° - TBC cos u = 0

+ c ©Fy = 0;

TBC sin u + 212.13 sin 45° - 200 = 0 Ans.

TAB = TCD = 212 lb (max) Solving, u = 18.43°

TBC = 158.11 lb

Using these results to analyze the equilibrium of joint B, Fig. d, we have + : ©Fx = 0;

158.11 cos 18.43° - TAB cos f = 0

+ c ©Fy = 0;

TAB sin f - 100 - 158.11 sin 18.43° = 0

Solving, f = 45° TAB = 212.13 lb = 212 lb (max) Thus, both cables AB and CD are subjected to maximum tension. The sag yB is given by yB = tan f = tan 45° 2 Ans.

yB = 2 ft

Ans: TAB = TCD = 212 lb (max), yB = 2 ft 725 726

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7–106. If yB = 1.5 ft, determine the largest weight of the crate and its placement x so that neither cable segment AB, BC, or CD is subjected to a tension that exceeds 200 lb.

2 ft

3 ft

yB A

D 3 ft

B

C

SOLUTION The forces FB and FC exerted on joints B and C will be obtained by considering the equilibrium on the free-body diagram, Fig. a. a + ©ME = 0;

FC(3) - w(x) = 0

a + ©MF = 0;

w(3 - x) - FB(3) = 0

3 ft

x

wx 3 w FB = (3 - x) 3 FC =

Since the horizontal component of tensile force developed in each cable is constant, cable CD, which has the greatest angle with the horizontal, will be subjected to the greatest tension. Thus, we will set TCD = 200 lb. First, we will analyze the equilibrium of joint C, Fig. b. + ©Fx = 0; :

200 cos 45° - TBC cos 26.57° = 0

+ c ©Fy = 0;

200 sin 45° + 158.11 sin 26.57° -

TBC = 158.11 lb wx = 0 3 wx = 212.13 3

(1)

Using the result of TBC to analyze the equilibrium of joint B, Fig. c, we have + ©Fx = 0; :

4 158.11 cos 26.57° - TAB a b = 0 5

+ c ©Fy = 0;

3 w 176.78 a b - 158.11 sin 26.57° - (3 - x) = 0 5 3

TAB = 176.78 lb

w (3 - x) = 35.36 3

(2)

Solving Eqs. (1) and (2) x = 2.57 ft

Ans.

W = 247 lb

Ans: x = 2.57 ft W = 247 lb 727 726

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7–107. The cable supports a girder which weighs 850 lb>ft. Determine the tension in the cable at points A, B, and C.

100 ft A 40 ft

C B

SOLUTION y = y =

1 ( w dx) dx FH L L 0

1 (425x2 + C1x + C2) FH dy C1 850 x + = dx FH FH

At x = 0,

dy = 0 C1 = 0 dx

At x = 0 , y = 0

C2 = 0 y =

425 2 x FH

At y = 20 ft, x = x¿ 20 =

425(x¿)2 FH

At y = 40 ft, x = (100- x¿) 40 =

425(100 - x¿)2 FH

2(x¿)2 = (x¿)2 - 200x¿ + 1002 (x¿)2 + 200x¿ - 1002 = 0 x¿ =

- 200 < 22002 + 4(100)2 = 41.42 ft 2 FH = 36 459 lb

At A, dy 2(425)x = tan uA = = 1.366 ` dx FH x = - 58.58 ft uA = 53.79° TA =

FH 36 459 = = 61 714 lb cos uA cos 53.79° Ans.

TA = 61.7 kip

728

20 ft

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7–107. Continued

At B, Ans.

TB = FH = 36.5 kip At C, dy 2(425)x = 0.9657 = tan uC = ` dx FH x = 41.42 ft uC = 44.0° TC =

FH 36 459 = = 50 683 lb cos u C cos 44.0° Ans.

TC = 50.7 kip

Ans: TA = 61.7 kip TB = 36.5 kip TC = 50.7 kip 729 728

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*7–108. The cable is subjected to a uniform loading of w = 200 lb>ft. Determine the maximum and minimum tension in the cable.

100 ft y B

A 20 ft

x

200 lb/ ft

Solution The Equation of The Cable. 1 ( w(x)dx ) dx y = FH L 1 y =

1 wo 2 a x + C1x + C2 b FH 2

(1)

dy 1 = (w x + C1) dx FH o

(2)

Boundary Conditions. At x = 0, y = 0. Then Eq (1) gives 0 = At x = 0,

1 (0 + 0 + C2) FH

dy = 0. Then Eq (2) gives dx 1 0 = (0 + C1) FH

C2 = 0

C1 = 0

Thus, the equation of the cable becomes wo 2 x y = 2FH

(3)

and the slope of the cable is dy wo = x dx FH

(4)

Here, wo = 200 lb>ft. Also, at x = 50 ft, y = 20 ft.Then using Eq (3), 20 =

200 2 (50 ) 2FH

FH = 12,500 lb = 12.5 kip

Thus, Ans.

Tmin = FH = 12.5 kip umax occurs at x = 50 ft.Using Eq. 4 tan umax = Thus, Tmax =

dy 200 ` = a b(50) dx x = 50 ft 12,500

umax = 38.66°

FH 12.5 = = 16.0 kip cos umax cos 38.66°

Ans.

Ans: Tmin = 12.5 kip Tmax = 16.0 kip 729 730

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7–109. If the pipe has a mass per unit length of 1500 kg> m, determine the maximum tension developed in the cable.

30 m A

3m

B

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w0 = 1500(9.81) = 14.715(103) N>m. Using Eq. 7–12, we can write y =

=

1 a w0dxb dx FH L L 1 14.715(103) 2 x + c1x + c2 b a FH 2

dy = 0 at x = 0, results in c1= 0. dx Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus,

Applying the boundary condition

y =

7.3575(103) 2 x FH

Applying the boundary condition y = 3 m at x = 15 m, we have 3 =

7.3575(103) (15)2 FH

FH = 551.81(103) N

Substituting this result into Eq. (1), we have dy = 0.02667x dx The maximum tension occurs at either points at A or B where the cable has the greatest angle with the horizontal. Here, umax = tan - 1 a

dy ` b = tan-1 [0.02667(15)] = 21.80° dx 15 m

Thus, Tmax =

551.8(103) FH = = 594.32(103) N = 594 kN cos umax cos 21.80°

Ans: Tmax = 594 kN 730 731

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7–110. If the pipe has a mass per unit length of 1500 kg> m, determine the minimum tension developed in the cable.

30 m A

3m

B

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w0 = 1500(9.81) = 14.715(103) N>m. Using Eq. 7–12, we can write y =

=

1 a w0dxb dx FH L L 3 1 14.715(10 ) 2 x + c1x + c2 b a FH 2

dy = 0 at x = 0, results in c1= 0. dx Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus, Applying the boundary condition

y =

7.3575(103) 2 x FH

Applying the boundary condition y = 3 m at x = 15 m, we have 3 =

7.3575(103) (15)2 FH

FH = 551.81(103) N

Substituting this result into Eq. (1), we have dy = 0.02667x dx The minimum tension occurs at the lowest point of the cable, where u = 0°. Thus, Tmin = FH = 551.81(103) N = 552 kN

Ans: Tmin = 552 kN 732 731

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7–111. Determine the maximum tension developed in the cable if it is subjected to the triangular distributed load.

y

B

20 ft

15 A

Solution The Equation of The Cable. Here, w(x) = 15x. 1 1 w(x)dx 2 dx y = FH 1 1

y =



y =



y =

300 lb/ft 20 ft

1 1 15x dx 2 dx FH 1 1

1 15 a x2 + C1 bdx 1 FH 2

1 5 a x3 + C1x + C2 b  FH 1 2

(1)

dy 1 15 2 = a x + C1 b  dx FH 2



x

(2)

Boundary Conditions. At x = 0, y = 0. Then Eq (1) gives 1 0 = (0 + 0 + C2)  C2 = 0 FH Also, at x = 0,

dy = tan 15°. Then Eq (2) gives dx 1 (0 + C1)  C1 = FH tan 15° tan 15° = FH

Thus, the equation of the cable becomes 1 5 3 a x + FH tan 15°xb y = FH 2

y =

5 3 x + tan 15° x 2FH

(3)

And the slope of the cable is dy 15 2 = x + tan 15° dx 2FH Also, at x = 20 ft, y = 20 ft.Then using Eq. 3, 20 = a



(4)

5 b(203) + tan 15°(20) 2FH

FH = 1366.03 lb



umax occurs at x = 20 ft. Then Eq (4) gives tan umax = Thus Tmax =

dy 15 ` = c d (202) + tan 15°   umax = 67.91° dx x = 20 ft 2(1366.02)

FH 1366.02 = = 3632.65 lb = 3.63 kip cos umax cos 67.91°

Ans.

733

Ans: Tmax = 3.63 kip

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*7–112. The cable will break when the maximum tension reaches Tmax = 10 kN. Determine the minimum sag h if it supports the uniform distributed load of w = 600 N>m.

25 m h

600 N/m

SOLUTION The Equation of The Cable: y = =

1 ( w(x)dx)dx FH L L 1 w0 2 ¢ x + C1x + C2 ≤ FH 2

[1]

dy 1 (w x + C1) = dx FH 0

[2]

Boundary Conditions: y = 0 at x = 0, then from Eq.[1] 0 =

1 (C ) FH 2

dy 1 = 0 at x = 0, then from Eq.[2] 0 = (C ) dx FH 1

C2 = 0

C1 = 0

Thus, w0 2 x 2FH

[3]

dy w0 x = dx FH

[4]

y =

y = h, at x = 12.5m, then from Eq.[3] h =

w0 (12.52) 2FH

FH =

78.125 w0 h

u = umax at x = 12.5 m and the maximum tension occurs when u = umax. From Eq.[4] tan umax =

Thus, cos umax =

dy 2 = dx x - 12.5m

w0 x 18.125 h w0

= 0.0128h(12.5) = 0.160h

1 20.0256h2 + 1

The maximum tension in the cable is Tmax =

10 =

FH cos umax

18.125 h (0.6)

1 20.0256h2 + 1 Ans.

h = 7.09 m

734

Ans: h = 7.09 m

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7–113. The cable is subjected to the parabolic loading w = 15011 - 1x>50222 lb>ft, where x is in ft. Determine the equation y = f1x2 which defines the cable shape AB and the maximum tension in the cable.

100 ft y A

B 20 ft x

SOLUTION y =

1 ( w(x)dx)dx FH L L

150 lb/ft

x3 1 [150(x ) + C1]dx FH L 3(50)2

y =

y =

1 x4 (75x2 + C1x + C2) FH 200

dy C1 150x 1 = x3 + dx FH 50FH FH At x = 0 ,

dy = 0 C1 = 0 dx

At x = 0 , y = 0

C2 = 0 y =

At x = 50 ft , y = 20 ft

1 x2 b a 75x2 FH 200

FH = 7813 lb y =

x2 x2 a 75 b ft 7813 200

Ans.

dy 1 4x3 2 = a 150x b = tan umax dx 7813 200 x = 50 ft umax = 32.62° Tmax =

FH 7813 = = 9275.9 lb cos umax cos 32.62° Ans.

Tmax = 9.28 kip

Ans:

x2 x2 a75 b 7813 200 Tmax = 9.28 kip y =

735

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7–114. The power transmission cable weighs 10 lb>ft. If the resultant horizontal force on tower BD is required to be zero, determine the sag h of cable BC.

300 ft A

SOLUTION

200 ft B

10 ft

h

C

D

The origin of the x, y coordinate system is set at the lowest point of the cables. Here, w0 = 10 lb>ft. Using Eq. 4 of Example 7–13, y =

FH w0 Bcosh ¢ x ≤ - 1 R w0 FH

y =

FH 10 Bcosh ¢ x ≤ - 1 R ft 10 FH

Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft, 10 =

10(150) (FH)AB Bcosh ¢ ≤ - 1R 10 (FH)AB

Solving by trial and error yields (FH)AB = 11266.63 lb Since the resultant horizontal force at B is required to be zero, (FH)BC = (FH)AB = 11266.62 lb. Applying the boundary condition of cable BC y = h at x = -100 ft to Eq. (1), we obtain

h =

10(- 100) 11266.62 c cosh B R - 1s 10 11266.62 Ans.

= 4.44 ft

Ans: h = 4.44 ft 736

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7–115. The power transmission cable weighs 10 lb>ft. If h = 10 ft, determine the resultant horizontal and vertical forces the cables exert on tower BD.

300 ft A

SOLUTION

200 ft B

10 ft

h

C

D

The origin of the x, y coordinate system is set at the lowest point of the cables. Here, w0 = 10 lb>ft. Using Eq. 4 of Example 7–13, y =

y =

FH w0

Bcosh ¢

w0 x≤ - 1R FH

FH 10 Bcosh ¢ x ≤ - 1 R ft 10 FH

Applying the boundary condition of cable AB, y = 10 ft at x = 150 ft, 10 =

10(150) (FH)AB Bcosh ¢ ≤ - 1R 10 (FH)AB

Solving by trial and error yields (FH)AB = 11266.63 lb Applying the boundary condition of cable BC, y = 10 ft at x = - 100 ft to Eq. (2), we have 10 =

10(100) (FH)BC Bcosh ¢ ≤ - 1R 10 (FH)BC

Solving by trial and error yields (FH)BC = 5016.58 lb Thus, the resultant horizontal force at B is (FH)R = (FH)AB - (FH)BC = 11266.63 - 5016.58 = 6250 lb = 6.25 kip Using Eq. (1), tan (uB)AB = sin h B sin h B

10(150) R = 0.13353 and 11266.63

Ans.

tan (uB)BC =

10( - 100) R = 0.20066. Thus, the vertical force of cables AB and BC acting 5016.58

on point B are (Fv)AB = (FH)AB tan (uB)AB = 11266.63(0.13353) = 1504.44 lb (Fv)BC = (FH)BC tan (uB)BC = 5016.58(0.20066) = 1006.64 lb The resultant vertical force at B is therefore (Fv)R = (Fv)AB + (Fv)BC = 1504.44 + 1006.64 Ans.

= 2511.07 lb = 2.51 kip

737

Ans: (Fh)R = 6.25 kip (Fv)R = 2.51 kip

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*7–116. The man picks up the 52-ft chain and holds it just high enough so it is completely off the ground. The chain has points of attachment A and B that are 50 ft apart. If the chain has a weight of 3 lb/ft, and the man weighs 150 lb, determine the force he exerts on the ground. Also, how high h must he lift the chain? Hint: The slopes at A and B are zero.

h A 25 ft

SOLUTION Deflection Curve of The Cable: x =

L 31 +

ds

2 11>F H 21 1 w0

where w0 = 3 lb>ft

ds2242 1

Performing the integration yields x =

FH 1 b sinh-1 B 13s + C12 R + C2 r 3 FH

(1)

From Eq. 7–14 dy 1 1 13s + C12 w0 ds = = dx FH L FH

(2)

Boundary Conditions: dy = 0 at s = 0. From Eq. (2) dx

0 =

1 10 + C12 FH

C1 = 0

Then, Eq. (2) becomes dy 3s = tan u = dx FH

(3)

s = 0 at x = 0 and use the result C1 = 0. From Eq. (1) x =

FH 1 b sinh-1 B 10 + 02 R + C2 r 3 FH

C2 = 0

Rearranging Eq. (1), we have s =

FH 3 x≤ sinh ¢ 3 FH

(4)

Substituting Eq. (4) into (3) yields dy 3 x≤ = sinh ¢ dx FH Performing the integration y =

FH 3 x cosh 3 FH

y = 0 at x = 0. From Eq. (5) 0 =

B

(5)

+ C3

FH FH cosh 0 + C3 ,thus, C3 = 3 3

738

25 ft

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*7–116. Continued

Then, Eq. (5) becomes 3 26 ft at

cosh

3

(6)

1

25 ft. From Eq. (4) 26

3

sinh

3

25

154.003 lb By trial and error at

25 ft. From Eq. (6) 154.003 3 25 cosh 3 154.003

1

Ans.

6.21 ft

From Eq. (3) tan 26 ft

The vertical force

3 26

26.86°

0.5065

154.003

that each chain exerts on the man is tan

154.003 tan 26.86°

78.00 lb

Equation of Equilibrium: By considering the equilibrium of the man, 0;

150

2 78.00

0

Ans.

306 lb

Ans: h = 6.21 ft Nm = 306 lb 739

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7–117.

The cable has a mass of 0.5 kg>m, and is 25 m long. Determine the vertical and horizontal components of force it exerts on the top of the tower.

A

SOLUTION

B

ds

x =

L

b1 +

30

1

2 1 (w0 ds)2 r F2H

15 m

Performing the integration yields: x =

FH 1 b sinh - 1 c (4.905s + C1) d + C2 r 4.905 FH

(1)

rom Eq. 7-13 dy 1 = w ds dx FH L 0 dy 1 = (4.905s + C1) dx FH At s = 0;

dy = tan 30°. Hence C1 = FH tan 30° dx dy dx

=

4.905s + tan 30° FH

(2)

Applying boundary conditions at x = 0; s = 0 to Eq.(1) and using the result C1 = FH tan 30° yields C2 = - sinh - 1(tan 30°). Hence x =

FH 1 b sinh - 1 c (4.905s+FH tan 30°) d - sinh - 1(tan 30°)r 4.905 FH

(3)

At x = 15 m; s = 25 m. From Eq.(3) 15 =

FH 1 b sinh - 1 c (4.905(25) + FH tan 30°) R - sinh - 1(tan 30°) r 4.905 FH

By trial and error FH = 73.94 N At point A, s = 25 m From Eq.(2) tan uA =

dy 4.905(25) 2 + tan 30° = dx s = 25 m 73.94

uA = 65.90°

(Fv )A = FH tan uA = 73.94 tan 65.90° = 165 N

Ans.

(FH)A = FH = 73.9 N

Ans.

Ans: (Fv) A = 165 N (Fh)A = 73.9 N 740

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7–118.

A 50-ft cable is suspended between two points a distance of 15 ft apart and at the same elevation. If the minimum tension in the cable is 200 lb, determine the total weight of the cable and the maximum tension developed in the cable.

SOLUTION Tmin = FH = 200 lb From Example 7–13: s =

FH w0 x sinh a b w0 FH

w0 15 200 50 sinh a = a bb 2 w0 200 2 Solving, w0 = 79.9 lb>ft Ans.

Total weight = w0 l = 79.9 (50) = 4.00 kip dy w0 s = tan umax = ` dx max FH umax = tan - 1 B

79.9 (25) R = 84.3° 200

Then, Tmax =

FH 200 = 2.01 kip = cos umax cos 84.3°

Ans.

Ans: W = 4.00 kip Tmax = 2.01 kip 741

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7–119. Show that the deflection curve of the cable discussed in Example 7–13 reduces to Eq. 4 in Example 7–12 when the hyperbolic cosine function is expanded in terms of a series and only the first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.)

SOLUTION cosh x = 1 +

x2 + Á 21

Substituting into y =

FH w0 B cosh ¢ x ≤ - 1 R w0 FH

=

FH w 20x2 + Á - 1R B1 + w0 2F 2H

=

w0x2 2FH

Using Eq. (3) in Example 7–12, FH = We get

y =

w0L2 8h

4h 2 x L2

QED

742

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*7–120. A telephone line (cable) stretches between two points which are 150 ft apart and at the same elevation. The line sags 5 ft and the cable has a weight of 0.3 lb> ft. Determine the length of the cable and the maximum tension in the cable.

SOLUTION w = 0.3 lb>ft From Example 7–13, s =

FH w sinh ¢ x≤ w FH

y =

FH w Bcosh ¢ x ≤ - 1 R w FH

At x = 75 ft, y = 5 ft, w = 0.3 lb>ft 5 =

FH 75w Bcosh ¢ ≤ - 1R w FH

FH = 169.0 lb dy w = tan umax = sinh ¢ x≤ ` ` dx max FH x = 75 ft umax = tan-1 csinh ¢ Tmax =

s =

7510.32 169

≤ d = 7.606°

FH 169 = 170 lb = cos umax cos 7.606°

Ans.

169.0 0.3 sinh c 1752 d = 75.22 0.3 169.0 Ans.

L = 2s = 150 ft

Ans: Tmax = 170 lb L = 150 ft 743

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7–121.

A cable has a weight of 2 lb> ft. If it can span 100 ft and has a sag of 12 ft, determine the length of the cable. The ends of the cable are supported from the same elevation.

SOLUTION From Eq. (5) of Example 7–13: h =

12 =

FH w0L Bcosh ¢ ≤ - 1R w0 2FH 211002 FH Bcosh ¢ ≤ - 1R 2 2FH

24 = FH B cosh ¢

100 ≤ - 1R FH

FH = 212.2 lb From Eq. (3) of Example 7–13: s =

FH w0 sinh ¢ x≤ w0 FH

21502 212.2 l = b sinh a 2 2 212.2 Ans.

l = 104 ft

Ans: l = 104 ft 744

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7–122. A cable has a weight of 3 lb ft and is supported at points that are 500 ft apart and at the same elevation. If it has a length of 600 ft, determine the sag.

SOLUTION w0 = 3 lb>ft From Example 7–15, s =

w0 FH sinh a xb w0 FH

At x = 250 ft, 300 =

s = 300 ft

3(250) FH sinh a b 3 FH

FH = 704.3 lb y =

h =

w0 FH c cosh x - 1 d w0 FH 3(250) 704.3 b - 1d c cosh a 3 704.3 Ans.

h = 146 ft

Ans: h = 146 ft 745

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7–123. A cable has a weight of 5 lb/ft. If it can span 300 ft and has a sag of 15 ft, determine the length of the cable. The ends of the cable are supported at the same elevation.

SOLUTION w0 = 5 lb>ft From Example 7–15, y =

w0 FH B cosh ¢ x ≤ - 1 R w0 FH

At x = 150 ft, y = 15 ft 150w0 15w0 = cosh ¢ ≤ - 1 FH FH FH = 3762 lb s =

w0 FH sinh x w0 FH

s = 151.0 ft Ans.

L = 2s = 302 ft

Ans: L = 302 ft 746

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*7–124. The 10 kg m cable is suspended between the supports A and B. If the cable can sustain a maximum tension of 1.5 kN and the maximum sag is 3 m, determine the maximum distance L between the supports

L A

SOLUTION The origin of the x, y coordinate system is set at the lowest point of the cable. Here w0 = 10(9.81) N>m = 98.1 N>m. Using Eq. (4) of Example 7–13, y =

FH w0 Bcosh ¢ x ≤ - 1 R w0 FH

y =

FH 98.1x Bcosh ¢ ≤ - 1R 98.1 FH

Applying the boundary equation y = 3 m at x =

3 =

L , we have 2

FH 49.05L Bcosh ¢ ≤ - 1R 98.1 FH

The maximum tension occurs at either points A or B where the cable makes the greatest angle with the horizontal. From Eq. (1), tan umax = sinh ¢

49.05L ≤ FH

By referring to the geometry shown in Fig. b, we have 1

cos umax = A

1 + sinh2 49.05L ¢ F ≤ H

=

1 cosh ¢

49.05L ≤ FH

hus, Tmax =

FH cos umax

1500 = FH cosh ¢

49.05L ≤ FH

(3)

Solving Eqs. (2) and (3) yields Ans.

L = 16.8 m FH = 1205.7 N

747

3m

B

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8–1. P 2 P 2

Determine the maximum force P the connection can support so that no slipping occurs between the plates. There are four bolts used for the connection and each is tightened so that it is subjected to a tension of 4 kN. The coefficient of static friction between the plates is ms = 0.4.

P

SOLUTION Free-Body Diagram: The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts. Thus, N = 4(4) kN = 16 kN. When the plate is on the verge of slipping, the magnitude of the friction force acting on each contact surface can be computed using the friction formula F = msN = 0.4(16) kN. As indicated on the free-body diagram of the upper plate, F acts to the right since the plate has a tendency to move to the left. Equations of Equilibrium: + ©Fx = 0; :

0.4(16) -

P = 0 2

p = 12.8 kN

Ans.

Ans: P = 12.8 kN 748

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8–2. The tractor exerts a towing force T = 400 lb. Determine the normal reactions at each of the two front and two rear tires and the tractive frictional force F on each rear tire needed to pull the load forward at constant velocity. The tractor has a weight of 7500 lb and a center of gravity located at GT. An additional weight of 600 lb is added to its front having a center of gravity at GA. Take ms = 0.4. The front wheels are free to roll.

GT

C

Equations of Equilibrium:

B 4 ft

3 ft

Ans.

2NC + 2 (2427.78) - 7500 - 600 = 0 Ans.

NC = 1622.22 lb = 1.62 kip + ©F = 0; : x

F 5 ft

2NB (9) + 400(2.5) - 7500(5) - 600(12) = 0 NB = 2427.78 lb = 2.43 kip

+ c ©Fy = 0;

GA

2.5 ft

SOLUTION a + ©MC = 0

A

T

2F - 400 = 0

Ans.

F = 200 lb

Friction: The maximum friction force that can be developed between each of the rear tires and the ground is Fmax = ms NC = 0.4 (1622.22) = 648.89 lb. Since Fmax 7 F = 200 lb, the rear tires will not slip. Hence the tractor is capable of towing the 400 lb load.

Ans: NB = 2.43 kip NC = 1.62 kip F = 200 lb 749

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8–3. The mine car and its contents have a total mass of 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and the tracks is ms = 0.4 when the wheels are locked, find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked. Does the car move?

10 kN

0.9 m

B

SOLUTION

0.15 m

1.5 m

NA 11.52 + 1011.052 - 58.8610.62 = 0 Ans.

NA = 16.544 kN = 16.5 kN + c ©Fy = 0;

A 0.6 m

Equations of Equilibrium: The normal reactions acting on the wheels at (A and B) are independent as to whether the wheels are locked or not. Hence, the normal reactions acting on the wheels are the same for both cases. a + ©MB = 0;

G

NB + 16.544 - 58.86 = 0 Ans.

NB = 42.316 kN = 42.3 kN

When both wheels at A and B are locked, then 1FA2max = msNA = 0.4116.5442 = 6.6176 kN and 1FB2max = msNB = 0.4142.3162 = 16.9264 kN. Since 1FA2max + FB max = 23.544 kN 7 10 kN, the wheels do not slip. Thus, the mine car does not move. Ans.

Ans: NA = 16.5 kN NB = 42.3 kN It does not move. 750

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*8–4. The winch on the truck is used to hoist the garbage bin onto the bed of the truck. If the loaded bin has a weight of 8500 lb and center of gravity at G, determine the force in the cable needed to begin the lift. The coefficients of static friction at A and B are mA = 0.3 and mB = 0.2, respectively. Neglect the height of the support at A.

30

G A

10 ft

12 ft

B

SOLUTION a + ©MB = 0;

8500(12) - NA(22) = 0 NA = 4636.364 lb

+ ©F = 0; : x

T cos 30° - 0.2NB cos 30° - NB sin 30° - 0.3(4636.364) = 0 T(0.86603) - 0.67321 NB = 1390.91

+ c ©Fy = 0;

4636.364 - 8500 + T sin 30° + NB cos 30° - 0.2NB sin 30° = 0 T(0.5) + 0.766025 NB = 3863.636

Solving: Ans.

T = 3666.5 lb = 3.67 kip NB = 2650.6 lb

Ans: T = 3.67 kip 751

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8–5. The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car if the back brakes are locked, and the front wheels are free to roll. Take ms = 0.3.

F G

30

0.6 m

C

0.3 m

0.75 m

A 1m

B

1.50 m

Solution Equations of Equilibrium. Referring to the FBD of the car shown in Fig. a, + ΣFx = 0;  FB - F cos 30° = 0  S

(1)

    + c ΣFy = 0;  NA + NB + F sin 30° - 2000(9.81) = 0

(2)

  a + ΣMA = 0;  F cos 30°(0.3) - F sin 30°(0.75) + (3)

NB (2.5) - 2000(9.81)(1) = 0



Friction. It is required that the rear wheels are on the verge to slip. Thus

(4)

FB = ms NB = 0.3 NB

Solving Eqs. (1) to (4),

Ans.

F = 2,762.72 N = 2.76 kN NB = 7975.30 N  NA = 10, 263.34 N  FB = 2392.59 N

Ans: F = 2.76 kN 752

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8–6. The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car. Both the front and rear brakes are locked. Take ms = 0.3.

F G

30

0.6 m

C

0.3 m

0.75 m

A 1m

B

1.50 m

Solution Equations of Equilibrium. Referring to the FBD of the car shown in Fig. a, + ΣFx = 0;  FA + FB - F cos 30° = 0  S

(1)

 + c ΣFy = 0;  F sin 30° + NA + NB - 2000(9.81) = 0

(2)

  a + ΣMA = 0;  F cos 30°(0.3) - F sin 30°(0.75) + NB (2.5) - 200(9.81)(1) = 0



(3)

Friction. It is required that both the front and rear wheels are on the verge to slip. Thus

FA = ms NA = 0.3 NA

(4)



FB = ms NB = 0.3 NB

(5)

Solving Eqs. (1) to (5),

Ans.

F = 5793.16 N = 5.79 kN NB = 8114.93 N  NA = 8608.49 N  FA = 2582.55 N  FB = 2434.48 N

Ans: F = 5.79 kN 753

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8–7. The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is ms = 0.3, and a torque of 5 N # m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N.

5N m

150 mm 50 mm

SOLUTION

O P

A B 200 mm

400 mm

To hold lever: a + ©MO = 0;

FB (0.15) - 5 = 0;

FB = 33.333 N

Require NB =

33.333 N = 111.1 N 0.3

Lever, a + ©MA = 0;

PReqd. (0.6) - 111.1(0.2) - 33.333(0.05) = 0

PReqd. = 39.8 N a) P = 30 N 6 39.8 N

No

Ans.

b) P = 70 N 7 39.8 N

Yes

Ans.

Ans: No Yes 754

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*8–8. The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is ms = 0.3, and a torque of 5 N # m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N.

5N m

150 mm 50 mm

SOLUTION

O P

A B 200 mm

400 mm

To hold lever: a + ©MO = 0;

- FB(0.15) + 5 = 0;

FB = 33.333 N

Require NB =

33.333 N = 111.1 N 0.3

Lever, a + ©MA = 0;

PReqd. (0.6) - 111.1(0.2) + 33.333(0.05) = 0

PReqd. = 34.26 N a) P = 30 N 6 34.26 N

No

Ans.

b) P = 70 N 7 34.26 N

Yes

Ans.

Ans: No Yes 755

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8–9. The pipe of weight W is to be pulled up the inclined plane of slope a using a force P. If P acts at an angle f, show that for slipping P = W sin(a + u)>cos(f - u), where u is the angle of static friction; u = tan - 1 ms.

P

φ

α

SOLUTION +a©Fy¿ = 0;

N + P sin f - W cos a = 0

+Q©Fx¿ = 0;

P cos f - W sin a - tan u(W cos a - P sin f) = 0 P = =

N = W cos a - P sin f

W(sin a + tan u cos a) cos f + tan u sin f W sin(a + u) W(cos u sin a + sin u cos a) = cos f cos u + sin f sin u cos(f - u)

756

Q.E.D.

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8–10. Determine the angle f at which the applied force P should act on the pipe so that the magnitude of P is as small as possible for pulling the pipe up the incline. What is the corresponding value of P? The pipe weighs W and the slope a is known. Express the answer in terms of the angle of kinetic friction, u = tan - 1 mk.

P

φ

α

SOLUTION +a©Fy¿ = 0;

N + P sin f - W cos a = 0

+Q©Fx¿ = 0;

P cos f - W sin a - tan u (W cos a - P sin f) = 0

P =

N = W cos a - P sin f

W(sin a + tan u cos a) cos f + tan u sin f

=

W(cos u sin a + sin u cos a) cos f cos u + sin f sin u

=

W sin (a + u) cos (f - u)

W sin (a + u) sin (f - u) dP = 0 = df cos2(f - u) W sin (a + u) sin (f - u) = 0

W sin (a + u) = 0

sin (f - u) = 0

f = u

P =

f - u = 0

Ans.

W sin (a + u) = W sin (a + u) cos (u - u)

Ans.

Ans: f = u P = W sin (a + u) 757

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8–11. Determine the maximum weight W the man can lift with constant velocity using the pulley system, without and then with the “leading block” or pulley at A. The man has a weight of 200 lb and the coefficient of static friction between his feet and the ground is ms = 0.6.

B

B 45°

C

A w

w

SOLUTION a) + c ©Fy = 0; + ©F = 0; : x

C

(a)

(b)

W sin 45° + N - 200 = 0 3 -

W cos 45° + 0.6 N = 0 3 Ans.

W = 318 lb b) + c ©Fy = 0;

N = 200 lb

+ ©F = 0; : x

0.612002 =

W 3 Ans.

W = 360 lb

Ans: W = 318 lb W = 360 lb 758

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*8–12. The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M0. If the coefficient of static friction between the wheel and the block is ms, determine the smallest force P that should be applied.

P a b C c

M0

SOLUTION a + ©MC = 0;

Pa - Nb - ms Nc = 0 N =

c + ©MO = 0;

O

r

Pa (b + ms c)

ms Nr - M0 = 0 ms P a P =

a b r = M0 b + ms c

M0 (b + ms c) ms ra

Ans.

Ans: P = 759

M0 (b + ms c) msra

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8–13. If a torque of M = 300 N # m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD to prevent the flywheel from rotating. The coefficient of static friction between the friction pad at B and the flywheel is ms = 0.4.

D

0.6 m

SOLUTION

30

1m B

C

Free-BodyDiagram: First we will consider the equilibrium of the flywheel using the free-body diagram shown in Fig. a. Here, the frictional force FB must act to the left to produce the counterclockwise moment opposing the impending clockwise rotational motion caused by the 300 N # m couple moment. Since the wheel is required to be on the verge of slipping, then FB = msNB = 0.4 NB. Subsequently, the free-body diagram of member ABC shown in Fig. b will be used to determine FCD.

M

0.3 m

60 mm 300 N m

A

O

Equations of Equilibrium: We have a + ©MO = 0;

0.4 NB(0.3) - 300 = 0

NB = 2500 N

Using this result, a + ©MA = 0;

FCD sin 30°(1.6) + 0.4(2500)(0.06) - 2500(1) = 0 Ans.

FCD = 3050 N = 3.05 kN

Ans: FCD = 3.05 kN 760

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8–14. The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is ms = 0.4, determine the greatest slope u the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.

2.5 ft G B 5 ft A

SOLUTION

θ

Tipping: a + ©MA = 0;

- W cos u12.52 + W sin u12.52 = 0 tan u = 1 u = 45°

Slipping: Q + ©Fx = 0;

0.4 N - W sin u = 0

a + ©Fy = 0;

N - W cos u = 0 tan u = 0.4 u = 21.8°

Ans. (car slips before it tips)

Ans: u = 21.8° 761

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8–15. The log has a coefficient of static friction of ms = 0.3 with the ground and a weight of 40 lb/ft. If a man can pull on the rope with a maximum force of 80 lb, determine the greatest length l of log he can drag.

l

80 lb A

B

SOLUTION Equations of Equilibrium: + c ©Fy = 0;

N - 40l = 0

N = 40l

+ ©F = 0; : x

4(80) - F = 0

F = 320 lb

Friction: Since the log slides, F = (F)max = ms N 320 = 0.3 (40l) Ans.

l = 26.7 ft

Ans: l = 26.7 ft 762

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*8–16. The 180-lb man climbs up the ladder and stops at the position shown after he senses that the ladder is on the verge of slipping. Determine the inclination u of the ladder if the coefficient of static friction between the friction pad A and the ground is ms = 0.4.Assume the wall at B is smooth.The center of gravity for the man is at G. Neglect the weight of the ladder.

B

G 10 ft

SOLUTION Free - Body Diagram. Since the weight of the man tends to cause the friction pad A to slide to the right, the frictional force FA must act to the left as indicated on the free - body diagram of the ladder, Fig. a. Here, the ladder is on the verge of slipping. Thus, FA = msNA. 3 ft

Equations of Equilibrium. + c ©Fy = 0; a + ©MB = 0;

u

NA - 180 = 0

A

NA = 180 lb

180(10 cos u°) - 0.4(180)(10 sin u°) - 180(3) = 0 cos u - 0.4 sin u = 0.3 Ans.

u = 52.0°

Ans: u = 52.0° 763

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8–17. The 180-lb man climbs up the ladder and stops at the position shown after he senses that the ladder is on the verge of slipping. Determine the coefficient of static friction between the friction pad at A and ground if the inclination of the ladder is u = 60° and the wall at B is smooth.The center of gravity for the man is at G. Neglect the weight of the ladder.

B

G 10 ft

SOLUTION Free - Body Diagram. Since the weight of the man tends ot cause the friction pad A to slide to the right, the frictional force FA must act to the left as indicated on the free - body diagram of the ladder, Fig. a. Here, the ladder is on the verge of slipping. Thus, FA = msNA.

u 3 ft

Equations of Equilibrium. + c ©Fy = 0;

NA - 180 = 0

a + ©MB = 0;

180(10 cos 60°) - ms(180)(10 sin 60°) - 180(3) = 0

A

NA = 180 lb

180 cos u - 72 sin u = 54 Ans.

ms = 0.231

Ans: ms = 0.231 764

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8–18. The spool of wire having a weight of 300 lb rests on the ground at B and against the wall at A. Determine the force P required to begin pulling the wire horizontally off the spool. The coefficient of static friction between the spool and its points of contact is ms = 0.25.

3 ft O 1 ft

A P

B

Solution Equations of Equilibrium. Referring to the FBD of the spool shown in Fig. a, + ΣFx = 0;  P - NA - FB = 0 S

(1)

    + c ΣFy = 0;  NB - FA - 300 = 0

(2)

a+ ΣMO = 0;  P(1) - FB(3) - FA(3) = 0

(3)

Frictions. It is required that slipping occurs at A and B. Thus,



FA = m NA = 0.25 NA 

(4)



FB = m NB = 0.25 NB

(5)

Solving Eqs. (1) to (5),



Ans.

P = 1350 lb NA = 1200 lb  NB = 600 lb  FA = 300 lb  FB = 150 lb

Ans: P = 1350 lb 765

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8–19. The spool of wire having a weight of 300 lb rests on the ground at B and against the wall at A. Determine the normal force acting on the spool at A if P = 300 lb. The coefficient of static friction between the spool and the ground at B is ms = 0.35. The wall at A is smooth.

3 ft O 1 ft

A P

B

Solution Equations of Equilibrium. Referring to the FBD of the spool shown in Fig. a,  a + ΣMB = 0;  NA(3) - 300(2) = 0  NA = 200 lb

Ans.

a + ΣMO = 0;   300(1)   - FB(3) = 0    FB = 100 lb    

+ c ΣFy = 0;       NB - 300 = 0  NB = 300 lb Friction. Since FB 6 (FB)max = ms NB = 0.35(300) = 105 lb, slipping will not occur at B. Thus, the spool will remain at rest.

Ans: NA = 200 lb 766

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*8–20. The ring has a mass of 0.5 kg and is resting on the surface of the table. In an effort to move the ring a normal force P from the finger is exerted on it. If this force is directed towards the ring’s center O as shown, determine its magnitude when the ring is on the verge of slipping at A. The coefficient of static friction at A is mA = 0.2 and at B, mB = 0.3.

P

B 60

O

75 mm A

Solution

FA = FB



P cos 60° - FB cos 30° - FA = 0



NA - 0.5(9.81) - P sin 60° - FB sin 30° = 0



FA = 0.2 NA



NA = 19.34 N



FA = FB = 3.868 N



P = 14.4 N



(FB)max = 0.3(14.44) = 4.33 N 7 3.868 N(O.K!)

Ans.

Ans: P = 14.4 N 767

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8–21. A man attempts to support a stack of books horizontally by applying a compressive force of F = 120 N to the ends of the stack with his hands. If each book has a mass of 0.95 kg, determine the greatest number of books that can be supported in the stack. The coefficient of static friction between the man’s hands and a book is (ms)h = 0.6 and between any two books (ms)b = 0.4.

F

120 N

F

120 N

SOLUTION Equations of Equilibrium and Friction: Let n¿ be the number of books that are on the verge of sliding together between the two books at the edge. Thus, Fb = (ms)b N = 0.4(120) = 48.0 N. From FBD (a), + c ©Fy = 0;

2(48.0) - n¿(0.95)(9.81) = 0

n¿ = 10.30

Let n be the number of books are on the verge of sliding together in the stack between the hands. Thus, Fk = (ms)k N = 0.6(120) = 72.0 N. From FBD (b), + c ©Fy = 0;

2(72.0) - n(0.95)(9.81) = 0

n = 15.45

Thus, the maximum number of books can be supported in stack is Ans.

n = 10 + 2 = 12

Ans: n = 12 768

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8–22. The tongs are used to lift the 150-kg crate, whose center of mass is at G. Determine the least coefficient of static friction at the pivot blocks so that the crate can be lifted.

P

275 mm E 500 mm

C

30

F H

D

500 mm

SOLUTION Free - Body Diagram. Since the crate is suspended from the tongs, P must be equal to the weight of the crate; i.e., P = 150(9.81)N as indicated on the free - body diagram of joint H shown in Fig.a. Since the crate is required to be on the verge of slipping downward, FA and FB must act upward so that FA = msNA and FB = msNB as indicated on the free - body diagram of the crate shown in Fig. c.

A

G

B

300 mm

Equations of Equilibrium. Referring to Fig. a, + ©F = 0; : x

FHE cos 30° - FHF cos 30° = 0

FHE = FHF = F

+ c ©Fy = 0;

150(9.81) - 2F sin 30° = 0

F = 1471.5 N

Referring to Fig. b, a + ©MC = 0;

1471.5 cos 30°(0.5) + 1471.5 sin 30°(0.275) - NA (0.5) - msNA (0.3) = 0 (1)

0.5NA + 0.3msNA = 839.51 Due to the symmetry of the system and loading, NB = NA. Referring to Fig. c, + c ©Fy = 0;

(2)

2msNA - 150(9.81) = 0

Solving Eqs. (1) and (2), yields NA = 1237.57 N Ans.

ms = 0.595

Ans: ms = 0.595 769

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8–23. The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of static friction is mB = mC = 0.3, determine the largest angle u of the incline so that the roller does not slip for any force P applied to the beam.

P 2m

2m B

A C

u

Solution a+ ΣMO = 0;  FB (15) - FC (15) = 0

(1)

+ ΣFx = 0;   - FB - FC cos u + NC sin u = 0 S

(2)

   + c ΣFy = 0;  NC cos u + FC sin u - NB = 0

(3)

Assume slipping at C so that FC = 0.3 NC



Then from Eqs. (1) and (2),

FB = FC



- 0.3 NC - 0.3 NC cos u + NC sin u = 0



( - 0.3 - 0.3 cos u + sin u ) NC = 0

(4)

The term in parentheses is zero when

Ans.

u = 33.4°

From Eq. (3),    NC (cos 33.4° + 0.3 sin 33.4°) = NB

NC = NB

Since Eq. (4) is satisfied for any value of NC, any value of P can act on the beam. Also, the roller is a “two-force member.”

2(90° - f) + u = 180°



f =



f = tan - 1 a

thus 

 

u 2 mN b = tan - 1 (0.3) = 16.7° N

Ans.

u = 2(16.7°) = 33.4°

Ans: u = 33.4° 770

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*8–24. The uniform thin pole has a weight of 30 lb and a length of 26 ft. If it is placed against the smooth wall and on the rough floor in the position d = 10 ft, will it remain in this position when it is released? The coefficient of static friction is ms = 0.3.

B

26 ft

SOLUTION a + ©MA = 0;

30 (5) - NB (24) = 0 NB = 6.25 lb

+ ©F = 0; : x

A

6.25 - FA = 0 d

FA = 6.25 lb + c ©Fy = 0;

NA - 30 = 0 NA = 30 lb

(FA)max = 0.3 (30) = 9 lb 7 6.25 lb Ans.

Yes, the pole will remain stationary.

Ans: Yes, the pole will remain stationary. 771

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8–25. The uniform pole has a weight of 30 lb and a length of 26 ft. Determine the maximum distance d it can be placed from the smooth wall and not slip. The coefficient of static friction between the floor and the pole is ms = 0.3.

B

26 ft

SOLUTION + c ©Fy = 0;

NA - 30 = 0 A

NA = 30 lb FA = (FA)max = 0.3 (30) = 9 lb + ©F = 0; : x

d

NB - 9 = 0 NB = 9 lb

a + ©MA = 0;

30 (13 cos u) - 9 (26 sin u) = 0 u = 59.04° Ans.

d = 26 cos 59.04° = 13.4 ft

Ans: d = 13.4 ft 772

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8–26. The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M0 = 360 N # m. If the coefficient of static friction between the wheel and the block is ms = 0.6, determine the smallest force P that should be applied.

P

1m 0.4 m C B M0

0.05 m

O 0.3 m

Solution Equations of Equilibrium. Referring to the FBD of the lever arm shown in Fig. a, a + ΣMC = 0;  P(1) + FB (0.05) - NB (0.4) = 0

(1)

Also, the FBD of the wheel, Fig. b, a + ΣMO = 0;  FB (0.3) - 360 = 0  FB = 1200 N Friction. It is required that the wheel is on the verge to rotate thus slip at B. Then

FB = ms NB;  1200 = 0.6 NB  NB = 2000 N

Substitute the result of FB and NB into Eq. (1)

P(1) + 1200(0.05) - 2000(0.4) = 0 Ans.

P = 740 N

Ans: P = 740 N 773

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8–27. Solve Prob. 8–26 if the couple moment M0 is applied counterclockwise.

P

1m 0.4 m C B M0

0.05 m

O 0.3 m

Solution Equations of Equilibrium. Referring to the FBD of the lever arm shown in Fig. a, a+ ΣMC = 0;  P(1) - FB(0.05) - NB(0.4) = 0

(1)

Also, the FBD of the wheel, Fig. b a+ ΣMO = 0;  360 - FB(0.3) = 0  FB = 1200 N Friction. It is required that the wheel is on the verge to rotate thus slip at B. Then FB = ms NB;  1200 = 0.6 NB  NB = 2000 N Substituting the result of FB and NB into Eq. (1),

P(1) - 1200(0.05) - 2000(0.4) = 0 Ans.

P = 860 N

Ans: P = 860 N 774

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*8–28. A worker walks up the sloped roof that is defined by the curve y = (5e0.01x) ft, where x is in feet. Determine how high h he can go without slipping. The coefficient of static friction is ms = 0.6.

y

h 5 ft x

Solution +Q ΣFx = 0;  0.6 N - W sin u = 0 +a ΣFy = 0;  N - W cos u = 0

tan u = 0.6



y = 5 e 0.01 x



dy = tan u = 0.05 e 0.01 x dx



0.6 = 0.05 e 0.01x



ln 12 = ln e 0.01x



2.48 = 0.01 x



x = 248.49 ft



h = 5 e 0.01(248.49)



h = 60.0 ft

Ans.

Ans: h = 60.0 ft 775

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8–29. The friction pawl is pinned at A and rests against the wheel at B. It allows freedom of movement when the wheel is rotating counterclockwise about C. Clockwise rotation is prevented due to friction of the pawl which tends to bind the wheel. If 1ms2B = 0.6, determine the design angle u which will prevent clockwise motion for any value of applied moment M. Hint: Neglect the weight of the pawl so that it becomes a two-force member.

A

θ B

20°

M C

SOLUTION Friction: When the wheel is on the verge of rotating, slipping would have to occur. Hence, FB = mNB = 0.6NB . From the force diagram (FAB is the force developed in the two force member AB) tan120° + u2 =

0.6NB = 0.6 NB Ans.

u = 11.0°

Ans: u = 11.0° 776

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8–30. Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are mA = 0.15 and mB = 0.25. Determine the incline angle u for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2 lb>ft.

k

2 lb/ft

B

A u

SOLUTION Equations of Equilibrium: Using the spring force formula, Fsp = kx = 2x, from FBD (a), +Q©Fx¿ = 0;

2x + FA - 10 sin u = 0

(1)

a + ©Fy¿ = 0;

NA - 10 cos u = 0

(2)

+Q©Fx¿ = 0;

FB - 2x - 6 sin u = 0

(3)

a + ©Fy¿ = 0;

NB - 6 cos u = 0

(4)

From FBD (b),

Friction: If block A and B are on the verge to move, slipping would have to occur at point A and B. Hence. FA = msA NA = 0.15NA and FB = msB NB = 0.25NB. Substituting these values into Eqs. (1), (2),(3) and (4) and solving, we have u = 10.6° NA = 9.829 lb

Ans.

x = 0.184 ft NB = 5.897 lb

Ans: u = 10.6°

x = 0.184 ft

777

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8–31. Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are mA = 0.15 and mB = 0.25. Determine the angle u which will cause motion of one of the blocks. What is the friction force under each of the blocks when this occurs? The spring has a stiffness of k = 2 lb>ft and is originally unstretched.

k

2 lb/ft

B

A u

SOLUTION Equations of Equilibrium: Since neither block A nor block B is moving yet, the spring force Fsp = 0. From FBD (a), +Q©Fx¿ = 0;

FA - 10 sin u = 0

(1)

a+ ©Fy¿ = 0;

NA - 10 cos u = 0

(2)

+Q©Fx¿ = 0;

FB - 6 sin u = 0

(3)

a+ ©Fy¿ = 0;

NB - 6 cos u = 0

(4)

From FBD (b),

Friction: Assuming block A is on the verge of slipping, then (5)

FA = mA NA = 0.15NA Solving Eqs. (1),(2),(3),(4), and (5) yields u = 8.531°

NA = 9.889 lb

FB = 0.8900 lb

FA = 1.483 lb

NB = 5.934 lb

Since (FB)max = mB NB = 0.25(5.934) = 1.483 lb 7 FB, block B does not slip. Therefore, the above assumption is correct. Thus u = 8.53°

FA = 1.48 lb

Ans.

FB = 0.890 lb

Ans: u = 8.53°

FA = 1.48 lb FB = 0.890 lb

778

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*8–32. Determine the smallest force P that must be applied in order to cause the 150-lb uniform crate to move. The coefficent of static friction between the crate and the floor is ms = 0.5.

2 ft P

3 ft

Solution Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0;  F - P = 0 S

(1)

+ c ΣFy = 0;  N - 150 = 0  N = 150 lb a+ ΣMO = 0;  P(3) - 150x = 0

(2)

Friction. Assuming that the crate slides before tipping. Thus

F = m N = 0.5(150) = 75 lb

Substitute this value into Eq. (1)

P = 75 lb

Then Eq. (2) gives

75(3) - 150x = 0  x = 1.5 ft

Since x > 1 ft, the crate tips before sliding. Thus, the assumption was wrong. Substitute x = 1 ft into Eq. (2),

P(3) - 150(1) = 0 Ans.

P = 50 lb

Ans: 50 lb 779

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8–33. The man having a weight of 200 lb pushes horizontally on the crate. If the coefficient of static friction between the 450-lb crate and the floor is ms = 0.3 and between his shoes and the floor is m′s = 0.6, determine if he can move the crate.

2 ft P

3 ft

Solution Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0;  FC - P = 0 S

(1)

+ c ΣFy = 0;  NC - 450 = 0  NC = 450 lb a+ ΣMO = 0;  P(3) - 450(x) = 0

(2)

Also, from the FBD of the man, Fig. b, + ΣFx = 0;  P - Fm = 0  S

(3)

+ c ΣFy = 0;  Nm - 200 = 0  Nm = 200 lb Friction. Assuming that the crate slides before tipping. Thus

FC = ms NC = 0.3(450) = 135 lb

Using this result to solve Eqs. (1), (2) and (3)

Fm = P = 135 lb  x = 0.9 ft

Since x < 1 ft, the crate indeed slides before tipping as assumed. Also, since Fm > (Fm)max = ms ′NC = 0.6(200) = 120 lb, the man slips. Thus he is not able to move the crate.

Ans: No 780

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8–34. The uniform hoop of weight W is subjected to the horizontal force P. Determine the coefficient of static friction between the hoop and the surface of A and B if the hoop is on the verge of rotating.

P

r

B B

A

Solution Equations of Equilibrium. Referring to the FBD of the hoop shown in Fig. a, + ΣFx = 0;  P + FA - NB = 0 S

(1)

+ c ΣFy = 0;  NA + FB - W = 0

(2)

a+ ΣMA = 0;  NB(r) + FB (r) - P(2r) = 0

(3)

Friction. It is required that slipping occurs at point A and B. Thus

FA = ms NA

(4)



FB = ms NB

(5)

Substituting Eq. (5) into (3),

NB r + ms NB r = 2Pr  NB =

2P  1 + ms

(6)

Substituting Eq. (4) into (1) and Eq. (5) into (2), we obtain

NB - ms NA = P

(7)



NA + ms NB = W

(8)

Eliminate NA from Eqs. (7) and (8),

P + msW

NB =

1 + ms2

(9)



Equating Eq. (6) and (9)

P + msW 2P = 1 + ms 1 + m2s



2P(1 + m2s ) = (P + msW)(1 + ms)



2P + 2m2s P = P + Pms + msW + m2s W



(2P - W)m2s - (P + W)ms + P = 0

If P =

1 W, the quadratic term drops out, and then 2 P ms = P + W



=

1 2W 1 W + 2



=

1  3

W Ans.

781

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8–34. Continued

If P ≠

1 W, then 2



ms =

(P + W) { 2[ - (P + W)]2 - 4(2P - W)P



ms =

(P + W) { 2W 2 + 6PW - 7P2



ms =

(P + W) { 2(W + 7P)(W - P)

2(2P - W)

2(2P - W)

2(2P - W)

In order to have a solution, (W + 7P)(W - P) 7 0



Since W + 7P > 0 then W - P > 0  W > P

Also, P > 0. Thus

0 6 P 6 W



Choosing the smaller value of ms,

ms =

(P + W) - 2(W + 7P)(W - P) 2(2P - W)

for 0 6 P 6 W and P ≠

W  Ans. 2

1 1 W and P ≠ W, are continuous. 2 2 Note: Choosing the larger value of ms in the quadratic solution leads to NA, FA < 0, which is nonphysical. Also, (ms)max = 1. For ms > 1, the hoop will tend to climb the wall rather than rotate in place. The two solutions, for P =

Ans: If P = ms =

1 3

If P ≠ ms =

1 W 2 1 W 2

(P + W) - 2(W + 7P)(W - P)

for 0 6 P 6 W 782

2(2P - W)

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8–35. Determine the maximum horizontal force P that can be applied to the 30-lb hoop without causing it to rotate. The coefficient of static friction between the hoop and the surfaces A and B is ms = 0.2. Take r = 300 mm.

P

r

B B

A

Solution Equations of Equilibrium. Referring to the FBD of the hoop shown in Fig. a, + ΣFx = 0;  P + FA - NB = 0 S

(1)

+ c ΣFy = 0;  NA + FB - 30 = 0

(2)

  a+ ΣMA = 0;  FB(0.3) + NB(0.3) - P(0.6) = 0

(3)

Friction. Assuming that the hoop is on the verge to rotate due to the slipping occur at A and B. Then

FA = ms NA = 0.2 NA

(4)



FB = ms NB = 0.2 NB

(5)

Solving Eq. (1) to (5)

NA = 27.27 lb  NB = 13.64 lb  FA = 5.455 lb  FB = 2.727 lb Ans.

P = 8.182 lb = 8.18 lb 

Since NA is positive, the hoop will be in contact with the floor. Thus, the assumption was correct.

Ans: P = 8.18 lb 783

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*8–36. Determine the minimum force P needed to push the tube E up the incline. The force acts parallel to the plane, and the coefficients of static friction at the contacting surfaces are mA = 0.2, mB = 0.3, and mC = 0.4. The 100-kg roller and 40-kg tube each have a radius of 150 mm.

E A

P

B

30

C

Solution Equations of Equilibrium. Referring to the FBD of the roller, Fig. a, + ΣFx = 0;  P - NA cos 30° - FA sin 30° - FC = 0      S

(1)

  + c ΣFy = 0;  NC + FA cos 30° - NA sin 30° - 100(9.81) = 0

(2)

a + ΣMD = 0;  FA(0.15) - FC (0.15) = 0

(3)

Also, for the FBD of the tube, Fig. b,     +QΣFx = 0;  NA - FB - 40(9.81) sin 30° = 0

(4)

+a ΣFy = 0;  NB - FA - 40(9.81) cos 30° = 0

(5)

 a + ΣME = 0;  FA(0.15) - FB(0.15) = 0

(6)

Friction. Assuming that slipping is about to occur at A. Thus

(7)

FA = mA NA = 0.2 NA

Solving Eqs. (1) to (7)

Ans.

P = 285.97 N = 286 N

NA = 245.25 N  NB = 388.88 N  NC = 1061.15 N  FA = FB = FC = 49.05 N Since FB 6 (FB)max = mB NB = 0.3(388.88) = 116.66 N and FC < (FC)max = mC NC = 0.4(1061.15) = 424.46 N, slipping indeed will not occur at B and C. Thus, the assumption was correct.

Ans: 286 N 784

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8–37. The coefficients of static and kinetic friction between the drum and brake bar are ms = 0.4 and mk = 0.3, respectively. If M = 50 N # m and P = 85 N determine the horizontal and vertical components of reaction at the pin O. Neglect the weight and thickness of the brake. The drum has a mass of 25 kg.

300 mm

700 mm

B O

125 mm 500 mm

M P

SOLUTION

A

Equations of Equilibrium: From FBD (b), a + ©MO = 0

50 - FB (0.125) = 0

FB = 400 N

From FBD (a), a + ©MA = 0;

85(1.00) + 400(0.5) - NB (0.7) = 0 NB = 407.14 N

Friction: Since FB 7 (FB)max = msNB = 0.4(407.14) = 162.86 N, the drum slips at point B and rotates. Therefore, the coefficient of kinetic friction should be used. Thus, FB = mkNB = 0.3NB. Equations of Equilibrium: From FBD (b), a + ©MA = 0;

85(1.00) + 0.3NB (0.5) - NB (0.7) = 0 NB = 154.54 N

From FBD (b), + c ©Fy = 0;

Oy - 245.25 - 154.54 = 0

+ ©F = 0; : x

0.3(154.54) - Ox = 0

Ans.

Oy = 400 N

Ans.

Ox = 46.4 N

Ans: Oy = 400 N

Ox = 46.4 N

785

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8–38. The coefficient of static friction between the drum and brake bar is ms = 0.4. If the moment M = 35 N # m, determine the smallest force P that needs to be applied to the brake bar in order to prevent the drum from rotating. Also determine the corresponding horizontal and vertical components of reaction at pin O. Neglect the weight and thickness of the brake bar. The drum has a mass of 25 kg.

300 mm

700 mm

B O

125 mm 500 mm

M P

SOLUTION

A

Equations of Equilibrium: From FBD (b), a + ©MO = 0

35 - FB (0.125) = 0

FB = 280 N

From FBD (a), a + ©MA = 0;

P(1.00) + 280(0.5) - NB (0.7) = 0

Friction: When the drum is on the verge of rotating, FB = msNB 280 = 0.4NB NB = 700 N Substituting NB = 700 N into Eq. [1] yields Ans.

P = 350 N Equations of Equilibrium: From FBD (b), + c ©Fy = 0;

Oy - 245.25 - 700 = 0

+ ©F = 0; : x

280 - Ox = 0

Ans.

Oy = 945 N

Ans.

Ox = 280 N

Ans: P = 350 N Oy = 945 N Ox = 280 N 786

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8–39. Determine the smallest coefficient of static friction at both A and B needed to hold the uniform 100-lb bar in  equilibrium. Neglect the thickness of the bar. Take mA = mB = m.

3 ft

B 13 ft 5 ft

A

Solution Equations of Equilibrium. Referring to the FBD of the bar shown in Fig. a, 12 a + ΣMA = 0;  NB (13) - 100 a b(8) = 0 NB = 56.80 lb 13

+ ΣFx = 0;  FA + FB a 12 b - 56.80 a 5 b = 0(1) S 13 13

5 12 b + 56.80 a b - 100 = 0(2) 13 13 Friction. It is required that slipping occurs at A and B. Thus   + c ΣFy = 0;  NA + FB a



FA = ms NA(3)



FB = ms NB = ms(56.80)(4)

Solving Eqs. (1) to (4)

Ans.

ms = 0.230 NA = 42.54 lb  FA = 9.786 lb  FB = 13.07 lb

Ans: ms = 0.230 787

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*8–40. If u = 30° determine the minimum coefficient of static friction at A and B so that equilibrium of the supporting frame is maintained regardless of the mass of the cylinder C. Neglect the mass of the rods.

C

u

L

u

L

SOLUTION Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of the rod will slip simultaneously. Since end B tends to move to the right, the friction force FB must act to the left as indicated on the free-body diagram shown in Fig. a.

A

B

Equations of Equilibrium: We have + ©Fx = 0; :

FBC sin 30° - FB = 0

FB = 0.5FBC

+ c ©Fy = 0;

NB - FBC cos 30° = 0

NB = 0.8660 FBC

Therefore, to prevent slipping the coefficient of static friction ends A and B must be at least ms =

FB 0 .5FBC = = 0.577 NB 0 .8660FBC

Ans.

Ans: ms = 0.577 788

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8–41. If the coefficient of static friction at A and B is ms = 0.6, determine the maximum angle u so that the frame remains in equilibrium, regardless of the mass of the cylinder. Neglect the mass of the rods.

C

u

L

u

L

SOLUTION Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of the rod will slip simultaneously. Since end B is on the verge of sliding to the right, the friction force FB must act to the left such that FB = msNB = 0.6NB as indicated on the free-body diagram shown in Fig. a.

A

B

Equations of Equilibrium: We have + c ©Fy = 0;

NB - FBC cos u = 0

+ ©Fx = 0; :

FBC sin u - 0.6(FBC cos u) = 0

NB = FBC cos u

tan u = 0 .6 Ans.

u = 31.0°

Ans: u = 31.0° 789

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8–42. The 100-kg disk rests on a surface for which mB = 0.2. Determine the smallest vertical force P that can be applied tangentially to the disk which will cause motion to impend.

P

A 0.5 m B

Solution Equations of Equilibrium. Referring to the FBD of the disk shown in Fig. a,  

 + c ΣFy = 0;    NB - P - 100(9.81) = 0

(1)

a + ΣMA = 0;  P(0.5) - FB(1) = 0

(2)

Friction. It is required that slipping impends at B.Thus,

(3)

FB = mB NB = 0.2 NB

Solving Eqs. (1), (2) and (3)

P = 654 N



NB = 1635 N  FB = 327 N

Ans.

Ans: P = 654 N 790

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8–43. Investigate whether the equilibrium can be maintained. The uniform block has a mass of 500 kg, and the coefficient of static friction is ms = 0.3. 5

4

200 mm

3

A B

600 mm 800 mm

Solution Equations of Equilibrium. The block would move only if it slips at corner O. Referring to the FBD of the block shown in Fig. a, 4 a + ΣMO = 0;  T a b(0.6) - 500(9.81)(0.4) = 0  T = 4087.5 N 5 + ΣFx = 0;  N - 4087.5a 3 b = 0  N = 2452.5 N S 5

4 + c ΣFy = 0;  F + 4087.5a b - 500(9.81) = 0  F = 1635 N 5

Friction. Since F 7 (F)max = ms N = 0.3(2452.5) = 735.75 N, slipping occurs at O. Thus, the block fails to be in equilibrium.

Ans: The block fails to be in equilibrium. 791

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*8–44. The homogenous semicylinder has a mass of 20 kg and mass center at G. If force P is applied at the edge, and r = 300 mm, determine the angle u at which the semicylinder is on the verge of slipping. The coefficient of static friction between the plane and the cylinder is ms = 0.3. Also, what is the corresponding force P for this case?

4r 3p P u r

G

Solution Equations of Equilibrium. Referring to the FBD of the semicylinder shown in Fig. a, + ΣFx = 0;  P sin u - F = 0  S

(1)

 + c ΣFy = 0;  N - P cos u - 20(9.81) = 0

(2)

a+ ΣMA = 0;  P[0.3(1 - sin u)] - 20(9.81) c

P =

4(0.3)

sin u 261.6 a b p 1 - sin u

3p

sin u d = 0 (3)

Friction. Since the semicylinder is required to be on the verge to slip at point A,

(4)

F = ms N = 0.3 N

Substitute Eq. (4) into (1),

(5)

P sin u - 0.3 N = 0

Eliminate N from Eqs. (2) and (5), we obtain

P =

58.86  sin u - 0.3 cos u

(6)

Equating Eq. (3) and (6)

261.6 sin u 58.56 a b = p 1 - sin u sin u - 0.3 cos u

sin u(sin u - 0.3 cos u + 0.225 p) - 0.225 p = 0

Solving by trial and error

Ans.

u = 39.50° = 39.5°

Substitute the result into Eq. 6

P =

58.86 sin 39.50° - 0.3 cos 39.50°



= 145.51 N



= 146 N

Ans.

Ans: 146 N 792

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8–45. The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the minimum force P needed to move the post. The coefficients of static friction at B and C are mB = 0.4 and mC = 0.2, respectively.

800 N/m A

B 2m

P 5

400 mm

4

3

300 mm C

SOLUTION Member AB: a + ©MA = 0;

4 - 800 a b + NB (2) = 0 3 NB = 533.3 N

Post: Assume slipping occurs at C; FC = 0.2 NC a + ©MC = 0;

4 - P(0.3) + FB(0.7) = 0 5

+ ©F = 0; : x

4 P - FB - 0.2NC = 0 5

+ c ©Fy = 0;

3 P + NC - 533.3 - 50(9.81) = 0 5 Ans.

P = 355 N NC = 811.0 N FB = 121.6 N (FB)max = 0.4(533.3) = 213.3 N 7 121.6 N

(O.K.!)

Ans: P = 335 N 793

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8–46. The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P = 150 N, the post slips at both B and C simultaneously.

800 N/m A

B 2m

P 5

400 mm

4

3

300 mm C

SOLUTION Member AB: a + ©MA = 0;

4 -800 a b + NB (2) = 0 3 NB = 533.3 N

Post: + c ©Fy = 0;

3 NC - 533.3 + 150 a b - 50(9.81) = 0 5 NC = 933.83 N

a + ©MC = 0;

4 - (150)(0.3) + FB (0.7) = 0 5 FB = 51.429 N

+ ©F = 0; : x

4 (150) - FC - 51.429 = 0 5 FC = 68.571 N

mC =

FC 68.571 = 0.0734 = NC 933.83

Ans.

mB =

FB 51.429 = 0.0964 = NB 533.3

Ans.

Ans: mC = 0.0734 mB = 0.0964 794

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8–47. Crates A and B weigh 200 lb and 150 lb, respectively. They are connected together with a cable and placed on the inclined plane. If the angle u is gradually increased, determine u when the crates begin to slide. The coefficients of static friction between the crates and the plane are mA = 0.25 and mB = 0.35.

B

D A

C

SOLUTION Free - Body Diagram. Since both crates are required to be on the verge of sliding down the plane, the frictional forces FA and FB must act up the plane so that FA = mANA = 0.25NA and FB = mBNB = 0.35NB as indicated on the free - body diagram of the crates shown in Figs. a and b.

u

Equations of Equilibrium. Referring to Fig. a, a+ ©Fy¿ = 0;

NA - 200 cos u = 0

NA = 200 cos u

+Q©Fx¿ = 0;

FCD + 0.25(200 cos u) - 200 sin u = 0

(1)

Also, by referring to Fig. b, a+ ©Fy¿ = 0;

NB - 150 cos u = 0

NB = 150 cos u

+Q©Fx¿ = 0;

0.35(150 cos u) - FCD - 150 sin u = 0

(2)

Solving Eqs. (1) and (2), yields Ans.

u = 16.3° FCD = 8.23 lb

Ans: u = 16.3° 795

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*8–48. Two blocks A and B, each having a mass of 5 kg, are connected by the linkage shown. If the coefficient of static friction at the contacting surfaces is ms = 0.5, determine the largest force P that can be applied to pin C of the linkage without causing the blocks to move. Neglect the weight of the links.

P 30 B

C 30 30

A

Solution Equations of Equilibrium. Analyze the equilibrium of Joint C Fig. a, + c ΣFy = 0;  FAC sin 30° - P cos 30° = 0  FAC = 23 P

+ ΣFx = 0;  FBC - P sin 30° - ( 23P) cos 30° = 0  FBC = 2 P S

Referring to the FBD of block B, Fig. b

+ a ΣFx = 0;  2 P cos 30° - FB - 5(9.81) sin 30° = 0

(1)

+Q ΣFy = 0;  NB - 2 P sin 30° - 5(9.81) cos 30° = 0

(2)

Also, the FBD of block A, Fig. C + ΣFx = 0;   23P cos 30° - FA = 0 S

(3)

+ c ΣFy = 0;  NA - 23P sin 30° - 5(9.81) = 0

(4)

Friction. Assuming that block A slides first. Then

(5)

FA = ms NA = 0.5 NA

Solving Eqs. (1) to (5)

Ans.

P = 22.99 N = 23.0 N NA = 68.96 N  FA = 34.48 N  FB = 15.29 N  NB = 65.46 N

Since FB 6 (FB)max = ms NB = 0.5(65.46) = 32.73 N, Block B will not slide. Thus, the assumption was correct.

Ans: 23.0 N 796

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8–49. The uniform crate has a mass of 150 kg. If the coefficient of static friction between the crate and the floor is ms = 0.2, determine whether the 85-kg man can move the crate. The coefficient of static friction between his shoes and the floor is m′s = 0.4. Assume the man only exerts a horizontal force on the crate.

2.4 m 1.6 m

1.2 m

Solution Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0;  P - FC = 0 S

(1)

  + c ΣFy = 0;  NC - 150(9.81) = 0  NC = 1471.5 N a + ΣMO = 0;  150(9.81)x - P(1.6) = 0

(2)

Also, from the FBD of the man, Fig. b,  + c ΣFy = 0;  Nm - 85(9.81) = 0  Nm = 833.85 N + ΣFx = 0;  Fm - P = 0 S

(3)

Friction. Assuming that the crate slips before tipping. Then

FC = ms NC = 0.2(1471.5) = 294.3 N

Solving Eqs. (1) to (3) using this result,

Fm = P = 294.3 N  x = 0.32 m

Since x < 0.6 m, the crate indeed slips before tipping as assumed. Also since  Fm 6 (Fm)max = ms′ Nm = 0.4(833.85) = 333.54 N, the man will not slip. Therefore, he is able to move the crate.

Ans: He is able to move the crate. 797

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8–50. The uniform crate has a mass of 150 kg. If the coefficient of static friction between the crate and the floor is ms = 0.2, determine the smallest mass of the man so he can move the crate. The coefficient of static friction between his shoes and the floor is m′s = 0.45. Assume the man exerts only a horizontal force on the crate.

2.4 m 1.6 m

1.2 m

Solution Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0;  P - FC = 0   S

(1)

  + c ΣFy = 0;  NC - 150(9.81) = 0  NC = 1471.5 N a + ΣMO = 0;  150(9.81)x - P(1.6) = 0

(2)

Also, from the FBD of the man, Fig. b,  + c ΣFy = 0;  Nm - m(9.81) = 0  Nm = 9.81 m

(3)

+ ΣFx = 0;  Fm - P = 0   S

(4)

Friction. Assuming that the crate slips before tipping. Then

FC = ms NC = 0.2(1471.5) = 294.3 N

Also, it is required that the man is on the verge of slipping. Then

Fm = ms′ Nm = 0.45 Nm

(5)

Solving Eqs. (1) to (5) using the result of FC,

Fm = P = 294.3 N  x = 0.32 m  Nm = 654 N



m = 66.667 kg = 66.7 kg

Ans.

Since x < 0.6 m, the crate indeed slips before tipping as assumed.

Ans: m = 66.7 kg 798

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8–51. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the minimum force P needed to move the post. The coefficients of static friction at B and C are mB = 0.4 and mC = 0.2, respectively.

A

P

B 1.5 m

5

1.5 m 1m 0.75 m

4

3

C

Solution Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a, a + ΣMA = 0;  NB (3) - 200(9.81)(1.5) = 0  NB = 981 N Then consider the FBD of member BC shown in Fig. b, 3 + c ΣFy = 0;  NC + P a b - 981 - 20(9.81) = 0 5

(1)

4 a + ΣMC = 0;  FB (1.75) - P a b(0.75) = 0 5

(2)

4 a + ΣMB = 0;  P a b(1) - FC (1.75) = 0 5

(3)

Friction. Assuming that slipping occurs at C. Then

(4)

FC = mC NC = 0.2 NC

Solving Eqs. (1) to (4)

Ans.

P = 407.94 N = 408 N NC = 932.44 N  FC = 186.49 N  FB = 139.87

Since FB 6 (FB)max = mB NB = 0.4(981) N = 392.4. Indeed slipping will not occur at B. Thus, the assumption is correct.

Ans: P = 408 N 799

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*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P = 300 N, the post slips at both B and C simultaneously.

A

P

B 1.5 m

5

1.5 m 1m 0.75 m

4

3

C

Solution Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a, a+ ΣMA = 0;  NB(3) - 200(9.81)(1.5) = 0 NB = 981 N Then consider the FBD of member BC shown in Fig. b, 3   + c ΣFy = 0;  NC + 300 a b - 981 - 20(9.81) = 0 NC = 997.2 N 5

4 a + ΣMC = 0;  FB (1.75) - 300 a b(0.75) = 0      FB = 102.86 N 5

4 a + ΣMB = 0;  300 a b(1) - FC (1.75) = 0 5

    

 

FC = 137.14 N

Friction. It is required that slipping occurs at B and simultaneously. Then   mB = 0.1048 = 0.105

Ans.

FC = mC NC;  137.14 = mC (997.2)  mC = 0.1375 = 0.138

Ans.

FB = mB NB;  102.86 = mB (981) 

Ans: mB = 0.105 mC = 0.138 800

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8–53. Determine the smallest couple moment that can be applied to the 150-lb wheel that will cause impending motion. The uniform concrete block has a weight of 300 lb. The coefficients of static friction are mA = 0.2, mB = 0.3, and between the concrete block and the floor, m = 0.4.

1 ft

5 ft

B

M 1.5 ft A

Solution Equations of Equilibrium. Referring to the FBD of the concrete block, Fig. a. + ΣFx = 0;  FC - NB = 0 S

(1)

  + c ΣFy = 0;  NC - FB - 300 = 0

(2)

a + ΣMO = 0;  NB (1.5) - 300x - FB (0.5 + x) = 0

(3)

Also, from the FBD of the wheel, Fig. b. + ΣFx = 0;  NB - FA = 0 S

(4)

  + c ΣFy = 0;  NA - FB - 150 = 0

(5)

a + ΣMA = 0;  M - NB(1.5) - FB(1.5) = 0

(6)

Friction. Assuming that the impending motion is caused by the rotation of wheel due to the slipping at A and B. Thus,

FA = mANA = 0.2NA

(7)



FB = mBNB = 0.3NB

(8)

Solving Eqs. (1) to (8),

NA = 141.51 lb   FA = 28.30 lb   NB = 28.30 lb   FB = 8.491 lb



NC = 308.49 lb   FC = 28.30 lb



  x = 0.1239 ft

M = 55.19 lb # ft = 55.2 lb # ft

Ans.

Since FC 6 (FC)max = mC NC = 0.4(308.49) = 123.40 lb, and x < 0.5 ft, the c­ oncrete block will not slide or tip. Also, NA is positive, so the wheel will be in ­contact with the floor. Thus, the assumption was correct.

Ans: M = 55.2 lb # ft 801

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8–54. Determine the greatest angle X so that the ladder does not slip when it supports the 75-kg man in the position shown. The surface is rather slippery, where the coefficient of static friction at A and B is ms = 0.3.

C

0.25 m

G 2.5 m

u

2.5 m

SOLUTION Free-Body Diagram: The slipping could occur at either end A or B of the ladder. We will assume that slipping occurs at end B. Thus, FB = msNB = 0.3NB . Equations of Equilibrium: Referring to the free-body diagram shown in Fig. b, we have + ©Fx = 0; :

B

FBC sin u>2 - 0.3NB = 0 (1)

FBC sin u>2 = 0.3NB + c ©Fy = 0;

A

NB - FBC cos u>2 = 0 FBC cos u>2 = NB(2)

Dividing Eq. (1) by Eq. (2) yields tan u>2 = 0.3 Ans.

u = 33.40° = 33.4°

Using this result and referring to the free-body diagram of member AC shown in Fig. a, we have a + ©MA = 0;

FBC sin 33.40°(2.5) - 75(9.81)(0.25) = 0

FBC = 133.66 N

+ ©Fx = 0; :

FA - 133.66 sin ¢

FA = 38.40 N

+ c ©Fy = 0;

NA + 133.66 cos ¢

33.40° ≤ = 0 2 33.40° ≤ - 75(9.81) = 0 2

NA = 607.73 N

Since FA 6 (FA) max = msNA = 0.3(607.73) = 182.32 N, end A will not slip. Thus, the above assumption is correct.

Ans: u = 33.4° 802

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8–55. P

The wheel weighs 20 lb and rests on a surface for which mB = 0.2. A cord wrapped around it is attached to the top of the 30-lb homogeneous block. If the coefficient of static friction at D is mD = 0.3, determine the smallest vertical force that can be applied tangentially to the wheel which will cause motion to impend.

1.5 ft

A

C

3 ft

1.5 ft B

D

SOLUTION Cylinder A: Assume slipping at B,

FB = 0.2NB

a+ ΣMA = 0 ;

FB + T = P

+ ΣFx = 0; S

FB = T

+ c ΣFy = 0;

NB = 20 + P NB = 20 + 2(0.2NB) NB = 33.33 lb FB = 6.67 lb T = 6.67 lb Ans.

P = 13.3 lb + ΣFx = 0; S

FD = 6.67 lb

+ c ΣFy = 0;

ND = 30 lb O.K.

(FD)max = 0.3 (30) = 9 lb 7 6.67 lb (No slipping occurs) a+ ΣMD = 0;

- 30(x) + 6.67 (3) = 0 x = 0.667 ft 6

1.5 = 0.75 ft 2

O.K.

(No tipping occurs)

Ans: P = 13.3 lb 803

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*8–56. The disk has a weight W and lies on a plane which has a coefficient of static friction m. Determine the maximum height h to which the plane can be lifted without causing the disk to slip.

z

h a y 2a

SOLUTION

x

Unit Vector: The unit vector perpendicular to the inclined plane can be determined using cross product. A = (0 - 0)i + (0 - a)j + (h - 0)k = - aj + hk B = (2a - 0)i + (0 - a)j + (0 - 0)k = 2ai - aj Then i N = A * B = 3 0 2a n =

j -a -a

k h 3 = ahi + 2ahj + 2a2k 0

ahi + 2ahj + 2a2k N = N a 25h2 + 4a2

Thus cos g =

2a 2

sin g =

hence

2

25h + 4a

25h 25h2 + 4a2

Equations of Equilibrium and Friction: When the disk is on the verge of sliding down the plane, F = mN. ©Fn = 0;

N - W cos g = 0

N = W cos g

(1)

©Ft = 0;

W sin g - mN = 0

N =

W sin g m

(2)

Divide Eq. (2) by (1) yields sin g = 1 m cos g 15h 15h2 + 4a2 2a

ma 2

2

2

5h + 4a

h =

2 25

b

= 1

Ans.

am

Ans: h =

804

2 25

am

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8–57. The man has a weight of 200 lb, and the coefficient of static friction between his shoes and the floor is ms = 0.5. Determine where he should position his center of gravity G at d in order to exert the maximum horizontal force on the door. What is this force? G

SOLUTION 3 ft

Fmax = 0.5 N = 0.5(200) = 100 lb + ©F = 0; : x a + ©MO = 0;

P - 100 = 0;

Ans.

P = 100 lb

d

200(d) - 100(3) = 0 Ans.

d = 1.50 ft

Ans: P = 100 N d = 1.50 ft 805

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8–58. Determine the largest angle X that will cause the wedge to be self-locking regardless of the magnitude of horizontal force P applied to the blocks. The coefficient of static friction between the wedge and the blocks is ms = 0.3. Neglect the weight of the wedge.

P

u

P

SOLUTION Free-Body Diagram: For the wedge to be self-locking, the frictional force F indicated on the free-body diagram of the wedge shown in Fig. a must act downward and its magnitude must be F … msN = 0 .3N. Equations of Equilibrium: Referring to Fig. a, we have + c ©Fy = 0;

2N sin u>2 - 2F cos u>2 = 0 F = N tan u>2

Using the requirement F … 0 .3N, we obtain N tan u>2 … 0 .3N Ans.

u = 33 .4°

Ans: u = 33.4° 806

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8–59. If the beam AD is loaded as shown, determine the horizontal force P which must be applied to the wedge in order to remove it from under the beam.The coefficients of static friction at the wedge’s top and bottom surfaces are mCA = 0.25 and mCB = 0.35, respectively. If P = 0, is the wedge self-locking? Neglect the weight and size of the wedge and the thickness of the beam.

4 kN/m

A

D

10°

C B 3m

4m

SOLUTION Equations of Equilibrium and Friction: If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = ms A NA = 0.25NA and FB = ms B NB = 0.35NB . From FBD (a), a + ©MD = 0;

NA cos 10°172 + 0.25NA sin 10°172 - 6.00122 - 16.0152 = 0 NA = 12.78 kN

From FBD (b), + c ©Fy = 0;

NB - 12.78 sin 80° - 0.25112.782 sin 10° = 0 NB = 13.14 kN

+ ©F = 0; : x

P + 12.78 cos 80° - 0.25112.782 cos 10° - 0.35113.142 = 0 Ans.

P = 5.53 kN

Since a force P 7 0 is required to pull out the wedge, the wedge will be self-locking when P = 0. Ans.

Ans: P = 5.53 kN yes 807

P

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*8–60. The wedge is used to level the member. Determine the horizontal force P that must be applied to begin to push the wedge forward. The coefficient of static friction between the wedge and the two surfaces of contact is ms = 0.2. Neglect the weight of the wedge.

2m 500 N/ m

P

A B

5

1m C

Solution Equations of Equilibrium and Friction. Since the wedge is required to be on the verge to slide to the right, then slipping will have to occur at both of its contact surfaces. Thus, FA = ms NA = 0.2 NA and FB = ms NB. Referring to the FBD ­diagram of member AC shown in Fig. a a+ ΣMC = 0;  500(2)(1) - NA cos 5°(2) - NA sin 5°(1)

- 0.2 NA cos 5°(1) + 0.2 NA sin 5°(2) = 0



NA = 445.65 N

Using this result and the FBD of the wedge, Fig. b,  + c ΣFy = 0;  NB - 445.65 cos 5° + 0.2(445.65) sin 5° = 0

NB = 436.18 N + ΣFx = 0;    P - 0.2(445.65) cos 5° - 445.65 sin 5° - 0.2(436.18) = 0 S



Ans.

P = 214.87 N = 215 N

Ans: 215 N 808

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8–61. The two blocks used in a measuring device have negligible weight. If the spring is compressed 5 in. when in the position shown, determine the smallest axial force P which the adjustment screw must exert on B in order to start the movement of B downward. The end of the screw is smooth and the coefficient of static friction at all other points of contact is ms = 0.3.

k = 20 lb/in. A 60 P B

SOLUTION

45

Note that when block B moves downward, block A will also come downward. Block A: + ©F = 0; : x

N¿ cos 60° + 0.3 N¿ sin 60° - NA = 0

+ c ©Fy = 0;

0.3 NA - 0.3 N¿ cos 60° + N¿ sin 60° - 100 = 0

Block B: + ©F = 0; : x

NB sin 45° - NB sin 45° + P - 0.3N¿ sin 60° - N¿ cos 60° = 0

+ c ©Fy = 0;

NB cos 45° + 0.3 NB cos 45° + 0.3 N¿ cos 60° - N¿ sin 60° = 0

Solving, N¿ = 105.9 lb NB = 82.5 lb NA = 80.5 lb Ans.

P = 39.6 lb

Ans: P = 39.6 lb 809

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8–62. If P = 250 N, determine the required minimum compression in the spring so that the wedge will not move to the right. Neglect the weight of A and B. The coefficient of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.

k  15 kN/m B

SOLUTION

P 3

Free-Body Diagram: The spring force acting on the cylinder is Fsp = kx = 15(10 )x. Since it is required that the wedge is on the verge to slide to the right, the frictional force must act to the left on the top and bottom surfaces of the wedge and their magnitude can be determined using friction formula. (Ff)1 = mN1 = 0.35N1

A

10

(Ff)2 = 0.35N2

Equations of Equilibrium: Referring to the FBD of the cylinder, Fig. a, + c ©Fy = 0;

N1 - 15(103)x = 0

N1 = 15(103)x

Thus, (Ff)1 = 0.35315(103)x4 = 5.25(103)x Referring to the FBD of the wedge shown in Fig. b, + c ©Fy = 0;

N2 cos 10° - 0.35N2 sin 10° - 15(103)x = 0 N2 = 16.233(103)x

+ ©Fx = 0; :

250 - 5.25(103)x - 0.35316.233(103)x4cos 10° - 316.233(103)x4sin 10° = 0 Ans.

x = 0.01830 m = 18.3 mm

Ans: x = 18.3 mm 810

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8–63. Determine the minimum applied force P required to move wedge A to the right. The spring is compressed a distance of 175 mm. Neglect the weight of A and B. The coefficient of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.

k = 15 kN/m B

SOLUTION

P

Equations of Equilibrium and Friction: Using the spring formula, Fsp = kx = 1510.1752 = 2.625 kN. If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = msNA = 0.35NA and FB = msNB = 0.35NB. From FBD (a), + c ©Fy = 0;

NB - 2.625 = 0

A

10°

NB = 2.625 kN

From FBD (b), + c ©Fy = 0;

NA cos 10° - 0.35NA sin 10° - 2.625 = 0 NA = 2.841 kN

+ ©F = 0; : x

P - 0.3512.6252 - 0.3512.8412 cos 10° - 2.841 sin 10° = 0 Ans.

P = 2.39 kN

Ans: P = 2.39 kN 811

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*8–64. If the coefficient of static friction between all the surfaces of contact is ms, determine the force P that must be applied to the wedge in order to lift the block having a weight W.

B

A a

P

C

Solution Equations of Equilibrium and Friction. Since the wedge is required to be on the verge sliding to the left, then slipping will have to occur at both of its contact ­surfaces. Thus, FA = ms NA, FB = ms NB and FC = ms NC. Referring to the FBD of the wedge shown in Fig. a. + ΣFx = 0;  m s NC + m s NA cos a + NA sin a - P = 0 S

(1)

+ c ΣFy = 0;  NC + ms NA sin a - NA cos a = 0

(2)

Also, from the FBD of the block, Fig. b + ΣFx = 0;  NB - NA sin a - ms NA cos a = 0  S

(3)

+ c ΣFy = 0;  NA cos a - ms NA sin a - ms NB - W = 0

(4)

Solving Eqs. (1) to (4) W



NA =



NC = c



cos a ( 1 -

) - 2ms sin a

ms2

cos a + ms sin a

  NB = c

cos a ( 1 - ms2) - 2ms sin a P = c

sin a + ms cos a cos a ( 1 - ms2) - 2ms sin a

d W

dW

2ms cos a + sin a ( 1 - ms2) cos a ( 1 - ms2) - 2ms sin a

Ans.

d W

Ans: P = c 812

2ms cos a + sin a ( 1 - ms2) cos a ( 1 - ms2) - 2ms sin a

dW

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8–65. Determine the smallest force P needed to lift the 3000-lb load. The coefficient of static friction between A and C and between B and D is ms = 0.3, and between A and B ms¿ = 0.4. Neglect the weight of each wedge.

3000 lb

P

SOLUTION

15°

B A

D

C

From FBD (a): + ©F = 0; : x

0.4N cos 15° + N sin 15° - ND = 0

(1)

+ c ©Fy = 0;

N cos 15° - 0.4N sin 15° - 0.3ND - 3000 = 0

(2)

Solving Eqs. (1) and (2) yields: N = 4485.4 lb

ND = 2893.9 lb

From FBD (b): + c ©Fy = 0;

NC + 0.4 (4485.4) sin 15° - 4485.4 cos 15° = 0

+ ©F = 0; : x

P - 0.3(3868.2) - 4485.4 sin 15° - 1794.1 cos 15° = 0

NC = 3868.2 lb

Ans.

P = 4054 lb = 4.05 kip

Ans: P = 4.05 kip 813

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8–66. Determine the reversed horizontal force - P needed to pull out wedge A. The coefficient of static friction between A and C and between B and D is ms = 0.2, and between A and B ms¿ = 0.1. Neglect the weight of each wedge.

3000 lb

B P

SOLUTION

D

15° A C

From FBD (a): + ©F = 0; : x

N sin 15° - 0.1N cos 15° - ND = 0

(1)

+ c ©Fy = 0;

N cos 15° + 0.1N sin 15° + 0.2ND - 3000 = 0

(2)

Solving Eqs. (1) and (2) yields: N = 2929.0 lb

ND = 475.2 lb

From FBD (b): + c ©Fy = 0;

NC - 292.9 sin 15° - 2929.0 cos 15° = 0

+ ©F = 0; : x

0.2(2905.0) + 292.9 cos 15° - 2929.0 sin 15° - P = 0

NC = 2905.0 lb

Ans.

P = 106 lb

Ans: P = 106 lb 814

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8–67. If the clamping force at G is 900 N, determine the horizontal force F that must be applied perpendicular to the handle of the lever at E. The mean diameter and lead of both single square-threaded screws at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.

200 mm G

200 mm A

B

C

D

SOLUTION

E

Referring to the free-body diagram of member GAC shown in Fig. a, we have FCD = 900 N ©MA = 0; FCD(0.2) - 900(0.2) = 0 L b = Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 a 2pr 5 -1 d = 3.643°; tan c 2p(12.5)

125 mm

fs = tan - 1 ms = tan - 1(0.3) = 16.699°; and M = F(0.125). Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)] F(0.125) = 2 [900(0.0125)tan(16.699° + 3.643°)] Ans.

F = 66 .7 N Note: Since fs 7 u, the screw is self-locking.

Ans: F = 66.7 N 815

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*8–68. If a horizontal force of F = 50 N is applied perpendicular to the handle of the lever at E, determine the clamping force developed at G. The mean diameter and lead of the single square-threaded screw at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.

200 mm G

200 mm A

B

C

D

SOLUTION

E

Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 a tan-1 c

5 d = 3.643°; 2p(12.5)

L b = 2pr

125 mm

fs = tan-1ms = tan-1(0.3) = 16.699°; and M = 50(0.125). Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)] 50(0.125) = 2[FCD(0.0125)tan(16.699° + 3.643°)] Ans.

FCD = 674.32 N

Using the result of FCD and referring to the free-body diagram of member GAC shown in Fig. a, we have ©MA = 0; 674.32(0.2) - FG(0.2) = 0 Ans.

FG = 674 N Note: Since fs 7 u, the screws are self-locking.

Ans: FCD = 674.32 N FG = 674 N 816

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8–69. The column is used to support the upper floor. If a force F = 80 N is applied perpendicular to the handle to tighten the screw, determine the compressive force in the column. The square-threaded screw on the jack has a coefficient of static friction of ms = 0.4, mean diameter of 25 mm, and a lead of 3 mm.

0.5 m F

SOLUTION M = W1r2 tan1fs + up2

fs = tan-110.42 = 21.80°

up = tan-1 c

3 d = 2.188° 2p112.52

8010.52 = W10.01252 tan121.80° + 2.188°2 W = 7.19 kN

Ans.

Ans: W = 7.19 kN 817

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8–70. If the force F is removed from the handle of the jack in Prob. 8–69, determine if the screw is self-locking.

0.5 m F

SOLUTION fs = tan-110.42 = 21.80° up = tan-1 c

3 d = 2.188° 2p112.52 Ans.

Since fs 7 up , the screw is self locking.

Ans: The screw is self-locking. 818

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8–71. If couple forces of F = 10 lb are applied perpendicular to the lever of the clamp at A and B, determine the clamping force on the boards. The single square-threaded screw of the clamp has a mean diameter of 1 in. and a lead of 0.25 in. The coefficient of static friction is ms = 0.3.

6 in. A

6 in. B

SOLUTION Since the screw is being tightened, Eq. 8–3 should be used. Here, M = 10(12) = 120 lb # in; u = tan - 1 ¢

L 0.25 ≤ = tan - 1 B R = 4.550°; 2pr 2p(0.5)

fs = tan - 1ms = tan - 1(0.3) = 16.699°. Thus M = Wr tan (fs + u) 120 = P(0.5) tan (16.699° + 4.550°) Ans.

P = 617 lb Note: Since fs 7 u, the screw is self-locking.

Ans: P = 617 lb 819

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*8–72. If the clamping force on the boards is 600 lb, determine the required magnitude of the couple forces that must be applied perpendicular to the lever AB of the clamp at A and B in order to loosen the screw. The single square-threaded screw has a mean diameter of 1 in. and a lead of 0.25 in. The coefficient of static friction is ms = 0.3.

6 in. A

6 in. B

SOLUTION Since the screw is being loosened, Eq. 8–5 should be used. Here, M = F(12); u = tan - 1 ¢

L 0.25 ≤ = tan - 1 B R = 4.550°; 2pr 2p(0.5)

fs = tan - 1ms = tan - 1(0.3) = 16.699°; and W = 600 lb. Thus M = Wr tan (fs - u) F(12) = 600(0.5) tan (16.699° - 4.550°) Ans.

F = 5.38 lb

Ans: F = 5.38 lb 820

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8–73. Prove that the lead l must be less than 2prms for the jack screw shown in Fig. 8–15 to be “self-locking.”

W

M

h

SOLUTION For self–locking, fs 7 uP or tan fs 7 tan up; ms 7

l ; 2p r

Q.E.D.

l 6 2prms

821

r

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8–74. The square-threaded bolt is used to join two plates together. If the bolt has a mean diameter of d = 20 mm and a lead of l = 3 mm, determine the smallest torque M required to loosen the bolt if the tension in the bolt is T = 40 kN. The coefficient of static friction between the threads and the bolt is ms = 0.15. M

SOLUTION f = tan-1 0.15 = 8.531° u = tan-1

d

3 = 2.734° 2p(10)

M = r W tan (f - u) = (0.01)(40 000) tan (8.531° - 2.734°) M = 40.6 N # m

Ans.

Ans: M = 40.6 N # m 822

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8–75. The shaft has a square-threaded screw with a lead of 8 mm and a mean radius of 15 mm. If it is in contact with a plate gear having a mean radius of 30 mm, determine the resisting torque M on the plate gear which can be overcome if a torque of 7 N # m is applied to the shaft. The coefficient of static friction at the screw is mB = 0.2. Neglect friction of the bearings located at A and B.

15 mm

B

M

SOLUTION

30 mm

l 8 b = tan-1 c d = 4.852°, Frictional Forces on Screw: Here, u = tan-1 a 2pr 2p1152 -1 -1 # W = F, M = 7 N m and fs = tan ms = tan 10.22 = 11.310°. Applying Eq. 8–3, we have M = Wr tan 1u + f2

7 = F10.0152 tan 14.852° + 11.310°2

A

7N·m

F = 1610.29 N Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if force F is removed. Equations of Equilibrium: a+©MO = 0;

1610.2910.032 - M = 0 M = 48.3 N # m

Ans.

Ans: M = 48.3 N # m 823

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*8–76. If couple forces of F = 35 N are applied to the handle of the machinist’s vise, determine the compressive force developed in the block. Neglect friction at the bearing A. The guide at B is smooth. The single square-threaded screw has a mean radius of 6 mm and a lead of 8 mm, and the coefficient of static friction is ms = 0.27.

F 125 mm A

B

SOLUTION f = tan - 1 (0.27) = 15.11° u = tan - 1 a

125 mm

8 b = 11.98° 2p(6)

F

M = Wr tan (u + f) 35 (0.250) = P (0.006) tan (11.98° + 15.11°) Ans.

P = 2851 N = 2.85 kN

Ans: P = 2.85 kN 824

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8–77. The square-threaded screw has a mean diameter of 20 mm and a lead of 4 mm. If the weight of the plate A is 5 lb, determine the smallest coefficient of static friction between the screw and the plate so that the plate does not travel down the screw when the plate is suspended as shown.

A

SOLUTION Frictional Forces on Screw: This requires a “self-locking” screw where fs Ú u. l 4 b = tan-1 c d = 3.643°. Here, u = tan-1 a 2pr 2p1102 fs = tan-1ms ms = tan fs

where fs = u = 3.643° Ans.

= 0.0637

Ans: ms = 0.0637 825

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8–78. The device is used to pull the battery cable terminal C from the post of a battery. If the required pulling force is 85 lb, determine the torque M that must be applied to the handle on the screw to tighten it. The screw has square threads, a mean diameter of 0.2 in., a lead of 0.08 in., and the coefficient of static friction is ms = 0.5.

M

SOLUTION

l 0.08 b = tan-1 c d = 7.256°, 2pr 2p(0.1) -1 -1 W = 85 lb and f s = tan ms = tan (0.5) = 26.565°. Applying Eq. 8–3, we have

Frictional Forces on Screw: Here, u = tan-1 a

A

C

M = Wr tan (u + f)

B

= 85(0.1) tan (7.256° + 26.565°) = 5.69 lb # in

Ans.

Note: Since f s 7 u, the screw is self-locking. It will not unscrew even if the moment is removed.

Ans: M = 5.69 lb # in. 826

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8–79. Determine the clamping force on the board A if the screw is tightened with a torque of M = 8 N # m. The squarethreaded screw has a mean radius of 10 mm and a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

M

A

Solution Frictional Forces on Screw. Here u = tan - 1 a

l 3 b = tan - 1 c d = 2.7336°, 2pr 2p (10)

W = F and f s = tan - 1ms = tan - 1(0.35) = 19.2900°.



M = Wr tan (u + f s)



8 = F(0.01) tan (2.7336° + 19.2900°)



F = 1977.72 N = 1.98 kN

Ans.

Note: Since f s > u, the screw is “self-locking”. It will not unscrew even if the torque is removed.

Ans: F = 1.98 kN 827

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*8–80. If the required clamping force at the board A is to be 2 kN, determine the torque M that must be applied to the screw to tighten it down. The square-threaded screw has a mean radius of 10 mm and a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

M

A

Solution Frictional Forces on Screw. Here u = tan - 1 a

l 3 b = tan - 1 c d = 2.7336°, 2pr 2p (10)

W = 2000 N and f s = tan - 1ms = tan - 1(0.35) = 19.2900°.



M = Wr tan (u + f s) = 2000 (0.01) tan (2.7336° + 19.2900°) = 8.09 N # m

Ans.

Ans: 8.09 N # m 828

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8–81. If a horizontal force of P = 100 N is applied perpendicular to the handle of the lever at A, determine the compressive force F exerted on the material. Each single squarethreaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.

A 15

C 15

250 mm

B

SOLUTION Since the screws are being tightened, Eq. 8–3 should be used. Here, L 7 .5 b = tan -1 c d = 5 .455°; u = tan -1 a 2pr 2p(12.5)

fs = tan - 1ms = tan - 1(0.15) = 8.531°; M = 100(0.25) = 25 N # m; and W = T, where T is the tension in the screw shank. Since M must overcome the friction of two screws, M = 2[Wr, tan(fs + u)4 25 = 23T(0.0125) tan (8 .531° + 5 .455°)4 Ans.

T = 4015.09 N = 4.02 kN

Referring to the free-body diagram of wedge B shown in Fig. a using the result of T, we have + ©Fx = 0; : + c ©Fy = 0;

4015 .09 - 0 .2N¿ - 0 .2N cos 15° - N sin 15° = 0

(1)

N¿ + 0 .2N sin 15° - N cos 15° = 0

(2)

Solving, N = 6324.60 N

N¿ = 5781.71 N

Using the result of N and referring to the free-body diagram of wedge C shown in Fig. b, we have + c ©Fy = 0;

2(6324 .60) cos 15° - 230 .2(6324.60) sin 15°4 - F = 0 F = 11563 .42 N = 11 .6 kN

Ans.

Ans: T = 4.02 kN F = 11.6 kN 829

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8–82. Determine the horizontal force P that must be applied perpendicular to the handle of the lever at A in order to develop a compressive force of 12 kN on the material. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.

A 15

C 15

250 mm

B

SOLUTION Referring to the free-body diagram of wedge C shown in Fig. a, we have + c ©Fy = 0;

2N cos 15° - 230 .2N sin 15°4 - 12000 = 0 N = 6563.39 N

Using the result of N and referring to the free-body diagram of wedge B shown in Fig. b, we have + c ©Fy = 0;

N¿ - 6563 .39 cos 15° + 0 .2(6563 .39) sin 15° = 0 N¿ = 6000 N

+ ©Fx = 0; :

T - 6563 .39 sin 15° - 0 .2(6563 .39) cos 15° - 0 .2(6000) = 0 T = 4166 .68 N

Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan - 1 c

L 7 .5 d = tan-1 c d = 5.455°; 2pr 2p(12.5)

fs = tan-1ms = tan-1(0 .15) = 8.531°; M = P(0 .25); and W = T = 4166.68N. Since M must overcome the friction of two screws, M = 23Wr tan (fs + u)4 P(0.25) = 234166 .68(0 .0125) tan (8.531° + 5 .455°)4 P = 104 N

Ans.

Ans: P = 104 N 830

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8–83. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the smallest vertical force F needed to support the load if the cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540°. Take ms = 0.2.

SOLUTION Frictional Force on Flat Belt: Here, T1 = F and T2 = 250(9.81) = 2452.5 N. Applying Eq. 8–6, we have

F

a) If b = 180° = p rad T2 = T1 e mb 2452.5 = Fe 0.2p Ans.

F = 1308.38 N = 1.31 kN b) If b = 540° = 3 p rad T2 = T1 e mb 2452.5 = Fe 0.2(3p)

Ans.

F = 372.38 N = 372 N

Ans: F = 1.31 kN F = 372 N 831

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*8–84. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the largest vertical force F that can be applied to the cord without moving the cylinder. The cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540°. Take ms = 0.2.

SOLUTION

F

Frictional Force on Flat Belt: Here, T1 = 250(9.81) = 2452.5 N and T2 = F. Applying Eq. 8–6, we have a)

If b = 180° = p rad T2 = T1 e mb F = 2452.5e 0.2 p Ans.

F = 4597.10 N = 4.60 kN b)

If b = 540° = 3 p rad T2 = T1e mb F = 2452.5e 0.2(3 p) Ans.

F = 16 152.32 N = 16.2 kN

Ans: F = 4.60 kN F = 16.2 kN 832

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8–85. A 180-lb farmer tries to restrain the cow from escaping by wrapping the rope two turns around the tree trunk as shown. If the cow exerts a force of 250 lb on the rope, determine if the farmer can successfully restrain the cow. The coefficient of static friction between the rope and the tree trunk is ms = 0.15, and between the farmer’s shoes and the ground msœ = 0.3.

SOLUTION Since the cow is on the verge of moving, the force it exerts on the rope is T2 = 250 lb and the force exerted by the man on the rope is T1. Here, b = 2(2p) = 4p rad. Thus, T2 = T1ems b 250 = T1e0.15(4p) T1 = 37.96 lb Using this result and referring to the free - body diagram of the man shown in Fig. a, + c ©Fy = 0;

N - 180 = 0

N = 180 lb

+ ©F = 0; : x

37.96 - F = 0

F = 37.96 lb

Since F 6 Fmax = ms ¿N = 0.3(180) = 54 lb, the man will not slip, and he will successfully restrain the cow.

Ans: He will successfully restrain the cow. 833

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8–86. The 100-lb boy at A is suspended from the cable that passes over the quarter circular cliff rock. Determine if it is possible for the 185-lb woman to hoist him up; and if this is possible, what smallest force must she exert on the horizontal cable? The coefficient of static friction between the cable and the rock is ms = 0.2, and between the shoes of the woman and the ground msœ = 0.8.

SOLUTION b =

p 2

A p 2

T2 = T1 e mb = 100 e 0.2 = 136.9 lb + c ©Fy = 0;

N - 185 = 0 N = 185 lb

+ ©F = 0; : x

136.9 - F = 0 F = 136.9 lb

Fmax = 0.8 (185) = 148 lb 7 136.9 lb Yes, just barely.

Ans.

Ans: Yes, it is possible. F = 137 lb 834

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8–87. The 100-lb boy at A is suspended from the cable that passes over the quarter circular cliff rock. What horizontal force must the woman at A exert on the cable in order to let the boy descend at constant velocity? The coefficients of static and kinetic friction between the cable and the rock are ms = 0.4 and mk = 0.35, respectively.

SOLUTION b =

p 2

T2 = T1 e mb;

A

100 = T1 e0.35

p 2

Ans.

T1 = 57.7 lb

Ans: T1 = 57.7 lb 835

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*8–88.

The uniform concrete pipe has a weight of 800 lb and is unloaded slowly from the truck bed using the rope and skids shown. If the coefficient of kinetic friction between the rope and pipe is mk = 0.3, determine the force the worker must exert on the rope to lower the pipe at constant speed. There is a pulley at B, and the pipe does not slip on the skids. The lower portion of the rope is parallel to the skids.

15 B

30

SOLUTION a + ©MA = 0;

- 800(r sin 30°) + T2 cos 15°(r cos 15° + r cos 30°) + T2 sin 15°(r sin 15° + r sin 15°) = 0

T2 = 203.466 lb b = 180° + 15° = 195° T2 = T1 e mb,

195°

203.466 = T1e(0.3)(180°)(p) Ans.

T1 = 73.3 lb

Ans: T1 = 73.3 lb 836

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8–89. A cable is attached to the 20-kg plate B, passes over a fixed peg at C, and is attached to the block at A. Using the coefficients of static friction shown, determine the smallest mass of block A so that it will prevent sliding motion of B down the plane.

mC  0.3 C

A mA  0.2 mB  0.3 B

SOLUTION

30

Block A: + Q©Fx = 0;

T1 - 0.2 NA - WA sin 30° = 0

(1)

+ a©Fy = 0;

NA - WA cos 30° = 0

(2)

+ Q©Fx = 0;

T2 - 20(9.81) sin 30° + 0.3 NB + 0.2 NA = 0

(3)

+a©Fy = 0;

NB - NA - 20(9.81) cos 30° = 0

(4)

T2 = T1 e 0.3p

(5)

Plate B:

Peg C: T2 = T1 e mb;

Solving Eqs. (1)–(5) yields T1 = 14.68 N;

T2 = 37.68 N;

NA = 18.89 N;

NB = 188.8 N;

WA = 21.81 N

Thus, mA =

21.81 = 2.22 kg 9.81

Ans.

Ans: mA = 2.22 kg 837

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8–90. The smooth beam is being hoisted using a rope which is wrapped around the beam and passes through a ring at A as shown. If the end of the rope is subjected to a tension T and the coefficient of static friction between the rope and ring is ms = 0.3, determine the angle of u for equilibrium.

T

A

θ

SOLUTION Equation of Equilibrium: + c ©Fx = 0;

T - 2T¿ cos

u = 0 2

Frictional Force on Flat Belt: Here, b = T2 = T1 emb, we have

T = 2T¿cos

u 2

(1)

u , T = T and T1 = T¿. Applying Eq. 8–6 2 2

T = T¿e0.31u>22 = T¿e0.15 u

(2)

Substituting Eq. (1) into (2) yields 2T¿cos

u = T¿e0.15 u 2

e0.15 u = 2 cos

u 2

Solving by trial and error Ans.

u = 1.73104 rad = 99.2°

The other solution, which starts with T' = Te 0.3(0/2) based on cinching the ring tight, is 2.4326 rad = 139°. Any angle from 99.2° to 139° is equilibrium.

Ans: u = 92.2° 838

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8–91. The boat has a weight of 500 lb and is held in position off the side of a ship by the spars at A and B. A man having a weight of 130 lb gets in the boat, wraps a rope around an overhead boom at C, and ties it to the end of the boat as shown. If the boat is disconnected from the spars, determine the minimum number of half turns the rope must make around the boom so that the boat can be safely lowered into the water at constant velocity. Also, what is the normal force between the boat and the man? The coefficient of kinetic friction between the rope and the boom is ms = 0.15. Hint: The problem requires that the normal force between the man’s feet and the boat be as small as possible.

C

A B

SOLUTION Frictional Force on Flat Belt: If the normal force between the man and the boat is equal to zero, then, T1 = 130 lb and T2 = 500 lb. Applying Eq. 8–6, we have T2 = T1 e mb 500 = 130e 0.15b b = 8.980 rad The least number of half turns of the rope required is

8.980 = 2.86 turns. Thus p Ans.

Use n = 3 half turns Equations of Equilibrium: From FBD (a), + c ©Fy = 0;

T2 - Nm - 500 = 0

T2 = Nm + 500

T1 + Nm - 130 = 0

T1 = 130 - Nm

From FBD (b), + c ©Fy = 0;

Frictional Force on Flat Belts: Here, b = 3 p rad. Applying Eq. 8–6, we have T2 = T1 e mb Nm + 500 = (130 - Nm) e 0.15 (3 p) Ans.

Nm = 6.74 lb

Ans: n = 3 half turns Nm = 6.74 lb 839

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*8–92. Determine the force P that must be applied to the handle of the lever so that B the wheel is on the verge of turning if M = 300 N # m. The coefficient of static friction between the belt and the wheel is ms = 0.3.

M

P 300 mm 25 mm B C 60 mm

A

D

700 mm

Solution Frictional Force on Flat Belt. Here b = 270° =

TD = TAe ms b



TD = TAe 0.3 ( 2 )



TD = 4.1112 TA

3p rad. 2

3p

(1)

Equations of Equilibrium. Referring to the FBD of the wheel shown in Fig. a, a+ ΣMB = 0;  300 + TA (0.3) - TD (0.3) = 0

(2)

Solving Eqs. (1) and (2),

TA = 321.42 N  TD = 1321.42 N

Subsequently, from the FBD of the lever, Fig. b a+ ΣMC = 0;  1321.42(0.025) - 321.42(0.06) - P(0.7) = 0

Ans.

P = 19.64 N = 19.6 N

Ans: 19.6 N 840

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8–93. If a force of P = 30 N is applied to the handle of the lever, determine the largest couple moment M that can be resisted so that the wheel does not turn. The coefficient of static friction between the belt and the wheel is ms = 0.3.

M

P 300 mm 25 mm B C 60 mm

A

D

700 mm

Solution Frictional Force on Flat Belt. Here b = 270° =

TD = TAe mb



TD = TAe 0.3 ( 2 )



TD = 4.1112 TA

3p rad. 2

3p

(1)

Equations of Equilibrium. Referring to the FBD of the wheel shown in Fig. a, a+ ΣMB = 0;  M + TA (0.3) - TD (0.3) = 0

(2)

Solving Eqs. (1) and (2)

TA = 1.0714 m  TD = 4.4047 m

Subsequently, from the FBD of the lever, Fig. b a+ ΣMC = 0;  4.4047 M(0.025) - 1.0714 M(0.06) - 30(0.7) = 0

M = 458.17 N # m = 458 N # m

Ans.



Ans: M = 458 N # m 841

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8–94. A minimum force of P = 50 lb is required to hold the cylinder from slipping against the belt and the wall. Determine the weight of the cylinder if the coefficient of friction between the belt and cylinder is ms = 0.3 and slipping does not occur at the wall.

30

O

B

SOLUTION

0.1 ft

Equations of Equilibrium: Write the moment equation of equilibrium about point A by referring to the FBD of the cylinder shown in Fig. a, a + ©MA = 0;

50(0.2) + W(0 .1) - T2 cos 30°(0.1 + 0.1 cos 30°)

P

(1)

- T2 sin 30°(0 .1 sin 30°) = 0 Frictional Force on Flat Belt: Here, T1 = 50 lb, b = a

30° p b p = rad. Applying Eq. 8–6 180° 6 T2 = T1emb p

= 50 e0.3 ( 6 ) = 58 .50 lb Substitute this result into Eq. (1), Ans.

W = 9.17 lb

841A

A

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8–95. The cylinder weighs 10 lb and is held in equilibrium by the belt and wall. If slipping does not occur at the wall, determine the minimum vertical force P which must be applied to the belt for equilibrium. The coefficient of static friction between the belt and the cylinder is ms = 0.25.

30°

O

B

SOLUTION

0.1ft

Equations of Equilibrium: a + ©MA = 0;

P10.22 + 1010.12 - T2 cos 30°10.1 + 0.1 cos 30°2 (1)

- T2 sin 30°10.1 sin 30°2 = 0 Frictional Force on Flat Belt: Here, b = 30° =

p rad and T1 = P. Applying Eq. 8–6, 6

T2 = T1 emb, we have T2 = Pe0.251p>62 = 1.140P

(2)

Solving Eqs. (1) and (2) yields Ans.

P = 78.7 lb T2 = 89.76 lb

84 1B

P

A

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*8–96. Determine the maximum and the minimum values of weight W which may be applied without causing the 50-lb block to slip. The coefficient of static friction between the block and the plane is ms = 0.2, and between the rope and the drum D msœ = 0.3.

D

W

45°

SOLUTION Equations of Equilibrium and Friction: Since the block is on the verge of sliding up or down the plane, then, F = msN = 0.2N. If the block is on the verge of sliding up the plane [FBD (a)], a+ ©Fy¿ = 0;

N - 50 cos 45° = 0

Q+ ©Fx¿ = 0;

T1 - 0.2135.362 - 50 sin 45° = 0

N = 35.36 lb T1 = 42.43 lb

If the block is on the verge of sliding down the plane [FBD (b)], a+ ©Fy¿ = 0;

N - 50 cos 45° = 0

Q+ ©Fx¿ = 0;

T2 + 0.2135.362 - 50 sin 45° = 0

N = 35.36 lb T2 = 28.28 lb

3p rad. If the block 4 is on the verge of sliding up the plane, T1 = 42.43 lb and T2 = W.

Frictional Force on Flat Belt: Here, b = 45° + 90° = 135° =

T2 = T1 emb W = 42.43e0.3A 4 B 3p

Ans.

= 86.02 lb = 86.0 lb

If the block is on the verge of sliding down the plane, T1 = W and T2 = 28.28 lb. T2 = T1 emb 28.28 = We0.3A 4 B 3p

Ans.

W = 13.95 lb = 13.9 lb

Ans: W = 86.0 lb W = 13.9 lb 842

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8–97. Granular material, having a density of 1.5 Mg>m3, is transported on a conveyor belt that slides over the fixed surface, having a coefficient of kinetic friction of mk = 0.3. Operation of the belt is provided by a motor that supplies a torque M to wheel A. The wheel at B is free to turn, and the coefficient of static friction between the wheel at A and the belt is mA = 0.4. If the belt is subjected to a pretension of 300 N when no load is on the belt, determine the greatest volume V of material that is permitted on the belt at any time without allowing the belt to stop. What is the torque M required to drive the belt when it is subjected to this maximum load?

mk  0.3 100 mm B

mA  0.4 M

100 mm A

SOLUTION Wheel A: a + ©MA = 0; T2 = T1 emb;

- M - 300 (0.1) + T2(0.1) = 0 T2 = 300e 0.4(p) = 1054.1 N

Thus, M = 75.4 N # m

Ans.

Belt, + ©F = 0; : x

1054.1 - 0.3 (m) (9.81) - 300 = 0 m = 256.2 kg V =

256.2 m = = 0.171 m3 p 1500

Ans.

Ans: M = 75.4 N # m V = 0.171 m3 843

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8–98. Show that the frictional relationship between the belt tensions, the coefficient of friction m, and the angular contacts a and b for the V-belt is T2 = T1emb>sin(a>2).

Impending motion

a b

SOLUTION

T2

FBD of a section of the belt is shown. Proceeding in the general manner: ©Fx = 0;

- (T + dT) cos

du du + T cos + 2 dF = 0 2 2

©Fy = 0;

-(T + dT) sin

du a du - T sin + 2 dN sin = 0 2 2 2

Replace

sin

du du by , 2 2

cos

du by 1, 2

dF = m dN Using this and (dT)(du) : 0, the above relations become dT = 2m dN T du = 2 adN sin

a b 2

Combine dT du = m T sin a2 Integrate from u = 0, T = T1 to u = b, T = T2 we get, mb

T2 = T1 e

¢ sin a ≤

Q.E.D

2

844

T1

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8–99. The wheel is subjected to a torque of M = 50 N # m. If the coefficient of kinetic friction between the band brake and the rim of the wheel is mk = 0.3, determine the smallest horizontal force P that must be applied to the lever to stop the wheel.

P

400 mm

M

SOLUTION

C

Wheel: a + ©MO = 0; T2 = T1 e mb ;

A 150 mm

-T2 (0.150) + T1 (0.150) + 50 = 0

50 mm 25 mm

B

100 mm

b T2 = T1 e 0.3a 3p 2

T1 = 107.14 N Link: a + ©MB = 0;

107.14 (0.05) - F (0.025) = 0 F = 214.28 N

Lever: a + ©MA = 0;

- P (0.4) + 214.28 (0.1) = 0 Ans.

P = 53.6 N

Ans: P = 53.6 N 845

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*8–100. Blocks A and B have a mass of 7 kg and 10 kg, respectively. Using the coefficients of static friction indicated, determine the largest vertical force P which can be applied to the cord without causing motion.

µD = 0.1

300 mm

D

µB = 0.4

B

400 mm P

µC = 0.4 A

SOLUTION

C

µA = 0.3

Frictional Forces on Flat Belts: When the cord pass over peg D, b = 180° = p rad and T2 = P. Applying Eq. 8–6, T2 = T1 e mb, we have P = T1 e 0.1 p

T1 = 0.7304P

When the cord pass over peg C, b = 90° = Applying Eq. 8–6, T2 ′ = T1 ′e mb, we have 0.7304P = T1 ′e 0.4(p>2)

p rad and T2 ′ = T1 = 0.7304P. 2

T1 ′ = 0.3897P

Equations of Equilibrium: From FDB (b), + c ΣFy = 0;

NB - 98.1 = 0

+ ΣFx = 0; S

FB - T = 0

(1)

a+ ΣMO = 0 ;

T(0.4) - 98.1(x) = 0

(2)

NB = 98.1 N

From FDB (b), + c ΣFy = 0;

NA - 98.1 - 68.67 = 0

+ ΣFx = 0; S

0.3897P - FB - FA = 0

NA = 166.77 N (3)

Friction: Assuming the block B is on the verge of tipping, then x = 0.15 m. A1 for motion to occur, block A will have slip. Hence, FA = (ms)ANA = 0.3(166.77) = 50.031 N. Substituting these values into Eqs. (1), (2) and (3) and solving yields Ans.

P = 222.81 N = 223 N FB = T = 36.79 N

Since (FB)max = (ms)B NB = 0.4(98.1) = 39.24 N 7 FB, block B does not slip but tips. Therefore, the above assumption is correct.

Ans: P = 223 N 846

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8–101. The uniform bar AB is supported by a rope that passes over a frictionless pulley at C and a fixed peg at D. If the coefficient of static friction between the rope and the peg is mD = 0.3, determine the smallest distance x from the end of the bar at which a 20-N force may be placed and not cause the bar to move.

C

D

SOLUTION a + ©MA = 0;

- 20 (x) + TB (1) = 0

+ c ©Fy = 0; T2 = T1 e

mb

;

20 N

TA + TB - 20 = 0 TA = TB e

0.3(p2 )

x

= 1.602 TB

A

Solving,

B 1m

TA = 12.3 N TB = 7.69 N Ans.

x = 0.384 m

Ans: x = 0.384 m 847

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8–102. The belt on the portable dryer wraps around the drum D, idler pulley A, and motor pulley B. If the motor can develop a maximum torque of M = 0.80 N # m, determine the smallest spring tension required to hold the belt from slipping. The coefficient of static friction between the belt and the drum and motor pulley is ms = 0.3.

30°

50 mm

M = 0.8 N⋅m

A

SOLUTION

B

a + ©MB = 0;

-T1 10.022 + T2 10.022 - 0.8 = 0

T2 = T1 emb;

T2 = T1 e10.321p2 = 2.5663T1

D

45° C

50 mm

20 mm

T1 = 25.537 N T2 = 65.53 N a + ©MC = 0;

-Fs10.052 + 125.537 + 25.537 sin 30°210.1 cos 45°2 + 25.537 cos 30°10.1 sin 45°2 = 0 Ans.

Fs = 85.4 N

Ans: Fs = 85.4 N 848

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8–103. Blocks A and B weigh 50 lb and 30 lb, respectively. Using the coefficients of static friction indicated, determine the greatest weight of block D without causing motion.

m

B

0.5

mBA

20

0.6

A D

C

mAC

0.4

SOLUTION For block A and B: Assuming block B does not slip + c ΣFy = 0;

NC - (50 + 30) = 0

NC = 80 lb

+ ΣFx = 0; S

0.4(80) - TB = 0

TB = 32 lb

For block B: + c ΣFy = 0;

NB cos 20° + FB sin 20° - 30 = 0

(1)

+ ΣFx = 0; S

FB cos 20° - NB sin 20° - 32 = 0

(2)

Solving Eqs. (1) and (2) yields: FB = 40.32 lb

NB = 17.25 lb

Since FB = 40.32 lb 7 mNB = 0.6(17.25) = 10.35 lb, slipping does occur between A and B. Therefore, the assumption is no good. Since slipping occurs, FB = 0.6 NB. + c ΣFy = 0;

NB cos 20° + 0.6NB sin 20° - 30 = 0

NB = 26.20 lb

+ ΣFx = 0; S

0.6(26.20) cos 20° - 26.20 sin 20° - TB = 0

TB = 5.812 lb

T2 = T1 e mb

Where

T2 = WD, T1 = TB = 5.812 lb, b = 0.5p rad

WD = 5.812e 0.5(0.5p) Ans.

= 12.7 lb

Ans: WD = 12.7 lb 849

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*8–104. The 20-kg motor has a center of gravity at G and is pinconnected at C to maintain a tension in the drive belt. Determine the smallest counterclockwise twist or torque M that must be supplied by the motor to turn the disk B if wheel A locks and causes the belt to slip over the disk. No slipping occurs at A. The coefficient of static friction between the belt and the disk is ms = 0.3.

M A

B

50 mm

G 50 mm

150 mm

C 100 mm

SOLUTION Equations of Equilibrium: From FBD (a), a + ©MC = 0;

T2 11002 + T1 12002 - 196.211002 = 0

(1)

M + T1 10.052 - T2 10.052 = 0

(2)

From FBD (b), a + ©MO = 0;

Frictional Force on Flat Belt: Here, b = 180° = p rad. Applying Eq. 8–6, T2 = T1 emb, we have T2 = T1 e0.3p = 2.566T1

(3)

Solving Eqs. (1), (2), and (3) yields M = 3.37 N # m T1 = 42.97 N

Ans.

T2 = 110.27 N

Ans: M = 3.37 N # m 850

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8–105. A 10-kg cylinder D, which is attached to a small pulley B, is placed on the cord as shown. Determine the largest angle X so that the cord does not slip over the peg at C. The cylinder at E also has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is ms = 0.1.

A

u

u

C

B E

SOLUTION

D

Since pully B is smooth, the tension in the cord between pegs A and C remains constant. Referring to the free-body diagram of the joint B shown in Fig. a, we have + c ©Fy = 0;

T =

2T sin u - 10(9.81) = 0

49.05 sin u

49.05 In the case where cylinder E is on the verge of ascending, T2 = T = and sin u p T1 = 10(9.81) N. Here, + u, Fig. b. Thus, 2 T2 = T1e msb p 49.05 = 10(9.81) e 0.1 a 2 sin u

ln

+ ub

p 0.5 = 0.1 a + u b sin u 2

Solving by trial and error, yields u = 0.4221 rad = 24.2° In the case where cylinder E is on the verge of descending, T2 = 10(9.81) N and 49.05 p . Here, + u. Thus, T1 = sin u 2 T2 = T1e m s b 10(9.81) =

49.05 0.1 a p e 2 sin u

ln (2 sin u) = 0.1 a

+ ub

p + ub 2

Solving by trial and error, yields u = 0.6764 rad = 38.8° Thus, the range of u at which the wire does not slip over peg C is 24.2° 6 u 6 38.8° Ans.

umax = 38.8°

Ans: umax = 38.8° 851

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8–106. A conveyer belt is used to transfer granular material and the frictional resistance on the top of the belt is F = 500 N. Determine the smallest stretch of the spring attached to the moveable axle of the idle pulley B so that the belt does not slip at the drive pulley A when the torque M is applied. What minimum torque M is required to keep the belt moving? The coefficient of static friction between the belt and the wheel at A is ms = 0.2.

0.1 m

M A

F = 500 N

0.1 m B k = 4 kN/m

SOLUTION Frictional Force on Flat Belt: Here, b = 180° = p rad and T2 = 500 + T and T1 = T. Applying Eq. 8–6, we have T2 = T1 emb 500 + T = Te0.2p T = 571.78 N Equations of Equilibrium: From FBD (a), M + 571.7810.12 - 1500 + 578.1210.12 = 0

a + ©MO = 0;

M = 50.0 N # m

Ans.

From FBD (b), + ©F = 0; : x

Fsp - 21578.712 = 0

Fsp = 1143.57 N

Thus, the spring stretch is x =

Fsp k

=

1143.57 = 0.2859 m = 286 mm 4000

Ans.

Ans: M = 50.0 N # m x = 286 mm 852

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8–107. The collar bearing uniformly supports an axial force of P = 5 kN. If the coefficient of static friction is ms = 0.3, determine the smallest torque M required to overcome friction.

P M

150 mm 200 mm

Solution Bearing Friction. With R2 = 0.1 m, R1 = 0.075 m, P = 5 ( 103 ) N, and ms = 0.3, R23 - R13 2 b ms P a 2 3 R2 - R12



M =



=



= 132 N # m

2 0.13 - 0.0753 b (0.3) 3 5 ( 103 )4 a 2 3 0.1 - 0.0752

Ans: M = 132 N # m 853

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*8–108. The collar bearing uniformly supports an axial force of P = 8 kN. If a torque of M = 200 N # m is applied to the shaft and causes it to rotate at constant velocity, determine the coefficient of kinetic friction at the surface of contact.

P M

150 mm 200 mm

Solution

Bearing Friction. With R2 = 0.1 m, R1 = 0.075 m, M = 300 N # m, and P = 8 ( 103 ) N,

M =



200 =



R23 - R13 2 mk P a 2 b 3 R2 - R12

2 0.13 - 0.0753 b mk 3 8 ( 103 )4 a 2 3 0.1 - 0.0752

Ans.

mk = 0.284

Ans: mk = 0.284 854

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8–109. The floor-polishing machine rotates at a constant angular velocity. If it has a weight of 80 lb. determine the couple forces F the operator must apply to the handles to hold the machine stationary. The coefficient of kinetic friction between the floor and brush is mk = 0.3. Assume the brush exerts a uniform pressure on the floor.

1.5 ft

SOLUTION M =

2 mPR 3 2 ft

2 F(1.5) = (0.3) (80)(1) 3 Ans.

F = 10.7 lb

Ans: F = 10.7 lb 855

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8–110. The double-collar bearing is subjected to an axial force P = 4 kN. Assuming that collar A supports 0.75P and collar B supports 0.25P, both with a uniform distribution of pressure, determine the maximum frictional moment M that may be resisted by the bearing. Take ms = 0.2 for both collars.

P M

20 mm B

SOLUTION

A

R32 - R31 2 M = ms P ¢ 2 ≤ 3 R2 - R21 M =

(0.03)3 - (0.01)3 (0.02)3 - (0.01)3 2 (0.75) (4000) + (0.25) (4000) ≤ (0.2) ¢ 5 (0.03)2 - (0.01)2 (0.02)2 - (0.01)2

= 16.1 N # m

10 mm 30 mm

Ans.

Ans: M = 16.1 N # m 856

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8–111. The double-collar bearing is subjected to an axial force P = 16 kN. Assuming that collar A supports 0.75P and collar B supports 0.25P, both with a uniform distribution of pressure, determine the smallest torque M that must be applied to overcome friction. Take ms = 0.2 for both collars.

P 100 mm

M A B

50 mm

75 mm 30 mm

Solution Bearing Friction. Here (RA)2 = 0.1 m, (RA)1 = 0.05 m, PA = 0.75 316 ( 103 ) N4 = 12 ( 103 ) N, (RB)2 = 0.075 m, (RB)1 = 0.05 m and PB = 0.25 316 ( 103 ) N4 = 4 ( 103 ) N. (RA)23 - (RA)13 (RB)23 - (RB)13 2 2 ms PA c d + m P c d 3 3 s B (RB)22 - (RB)12 (RA)22 - (RA)12



m =



=



= 237.33 N # m = 237 N # m

2 2 0.13 - 0.053 0.0753 - 0.053 3 ( ) b + 3 4 10 4 a b (0.2) 312 ( 103 )4 a 2 (0.2) 3 3 0.1 - 0.052 0.0752 - 0.052

Ans.

Ans: M = 237 N # m 857

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*8–112. The pivot bearing is subjected to a pressure distribution at its surface of contact which varies as shown. If the coefficient of static friction is m, determine the torque M required to overcome friction if the shaft supports an axial force P.

P M

R

SOLUTION r

pr b dA dF = m dN = m p0 cos a 2R M =

LA

rm p0 cos a R

= m p0

L0

= m p0 B = mp0 ¢

pr b r dr du 2R

a r2 cos a

2r

r p = p0 cos π 2R

2p

pr du b drb 2R L0

p 2 2 A 2R B r -2

pr cos a b + 2R

p 2 A 2R B

p0

p 3 A 2R B

p 2 16R3 b - 2d c a ≤ 2 p2

sin a

pr R b d 12p2 2R 0

= 0.7577m p0 R3 R

P =

LA

dN =

= p0 B

1

A B

p 2 2R

L0

p0 a cos a

cos a

= 4p0 R 2 a1 -

pr b + 2R

2 b p

2p

pr b rdrb du 2R L0 r

A B p 2R

sin a

pr R b R 12p2 2R 0

= 1.454p0 R2 Thus,

Ans.

M = 0.521 PmR

Ans: M = 0.521 PmR 858

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8–113. The conical bearing is subjected to a constant pressure distribution at its surface of contact. If the coefficient of static friction is ms, determine the torque M required to overcome friction if the shaft supports an axial force P.

P M

R

SOLUTION The differential area (shaded) dA = 2pr ¢ P =

L

p cos u dA =

L

p cos u ¢

2prdr dr ≤ = cos u cos u

P pR2

P = ppR 2

p =

dN = pdA =

2P 2prdr P rdr ¢ ≤ = 2 pR2 cos u R cos u

M =

L

rdF =

L

u

R

2prdr rdr ≤ = 2pp cos u L0

ms rdN = =

2ms P 2

R cos u L0

R

r2 dr

2ms PR 2ms P R 3 = 3 cos u R 2 cos u 3

Ans.

Ans: M = 859

2ms PR 3 cos u

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8–114. The 4-in.-diameter shaft is held in the hole such that the normal pressure acting around the shaft varies linearly with its depth as shown. Determine the frictional torque that must be overcome to rotate the shaft. Take ms = 0.2.

60 lb/in2 M

SOLUTION 6 in.

Express the pressure p as the function of x: P =

60 x = 10x 6

The differential area (shaded) dA = 2p(2)dx = 4pdx dN = pdA = 10x(4pdx) = 40pxdx T = 2

L

dF = 2

L

mdN = 80pm

6

L0

xdx

= 1440pm lb # in. = 1440p(0.2) = 905 lb # in.

Ans.

Ans: T = 905 lb # in. 860

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8–115. The plate clutch consists of a flat plate A that slides over the rotating shaft S. The shaft is fixed to the driving plate gear B. If the gear C, which is in mesh with B, is subjected to a torque of M = 0.8N # m, determine the smallest force P, that must be applied via the control arm, to stop the rotation. The coefficient of static friction between the plates A and D is ms = 0.4. Assume the bearing pressure between A and D to be uniform.

D A F

125 mm P

SOLUTION F =

100 mm S

200 mm

150 mm E

0.8 = 26.667 N 0.03

150 mm

B

30 mm

M = 26.667(0.150) = 4.00 N # m

M C

R32 - R31 2 b M = m P¿ a 2 3 R2 - R21 4.00 =

0.8 N m

(0.125)3 - (0.1)3 2 (0.4) (P¿) a b 3 (0.125)2 - (0.1)2

P¿ = 88.525 N a + ©MF = 0;

88.525(0.2) - P(0.15) = 0 Ans.

P = 118 N

Ans: P = 118 N 861

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*8–116. The collar fits loosely around a fixed shaft that has a radius of 2 in. If the coefficient of kinetic friction between the shaft and the collar is mk = 0.3, determine the force P on the horizontal segment of the belt so that the collar rotates counterclockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt, is 2.25 in.

P 2.25 in. 2 in.

SOLUTION

20 lb

fk = tan-1 mk = tan-1 0.3 = 16.699° rf = 2 sin 16.699°= 0.5747 in. Equilibrium: + c ©Fy = 0;

Ry - 20 = 0

Ry = 20 lb

+ ©F = 0; : x

P - Rx = 0

Rx = P

Hence R = 2R2x + R2y = 2P2 + 202 a + ©MO = 0;

- a 2P2 + 202 b (0.5747) + 20(2.25) - P(2.25) = 0 Ans.

P = 13.8 lb

Ans: P = 13.8 lb 862

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8–117. The collar fits loosely around a fixed shaft that has a radius of 2 in. If the coefficient of kinetic friction between the shaft and the collar is mk = 0.3, determine the force P on the horizontal segment of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt, is 2.25 in.

P 2.25 in. 2 in.

SOLUTION

20 lb

fk = tan-1 mk = tan-1 0.3 = 16.699° rf = 2 sin 16.699° = 0.5747 in. Equilibrium: + c ©Fy = 0;

Ry - 20 = 0

+ ©F = 0; : x

P - Rx = 0

Ry = 20 lb Rx = P

Hence R = 2R2x + R2y = 2P2 + 202 a + ©MO = 0;

a 2P2 + 202 b(0.5747) + 20(2.25) - P(2.25) = 0 Ans.

P = 29.0 lb

Ans: P = 29.0 lb 863

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8–118. The pivot bearing is subjected to a parabolic pressure distribution at its surface of contact. If the coefficient of static friction is ms, determine the torque M required to overcome friction and turn the shaft if it supports an axial force P.

P M

SOLUTION

R

The differential are dA = (rdu)(dr)

r

P =

L

p dA =

L

p0 ¢ 1 -

2

P =

pR p0 2

dN = pdA = M =

L

rdF =

p0 =

2p

R

r2 r2 (rdu)(dr) = p0 du r ¢ 1 - 2 ≤ dr 2≤ R R L0 L0

2P pR2

2

p0

r ) p  p0 (1 –– R2

r2 2P 1 ¢ ≤ (rdu)(dr) pR2 R2 L

ms rdN = =

2ms P

2p

2

pR L0

du

R

L0

r2 ¢ 1 -

r2 ≤ dr R2

8 m PR 15 s

Ans.

Ans: M = 864

8 m PR 15 s

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8–119. A disk having an outer diameter of 120 mm fits loosely over a fixed shaft having a diameter of 30 mm. If the coefficient of static friction between the disk and the shaft is ms = 0.15 and the disk has a mass of 50 kg, determine the smallest vertical force F acting on the rim which must be applied to the disk to cause it to slip over the shaft.

SOLUTION

F

Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.15 = 8.531°. Then the radius of friction circle is rf = r sin fs = 0.015 sin 8.531° = 2.225110 -32 m Equation of Equilibrium: a + ©MP = 0;

490.512.2252110-32 - F30.06 - 12.2252110-324 = 0 Ans.

F = 18.9 N

Ans: F = 18.9 N 865

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*8–120. The 4-lb pulley has a diameter of 1 ft and the axle has a diameter of 1 in. If the coefficient of kinetic friction between the axle and the pulley is mk = 0.20, determine the vertical force P on the rope required to lift the 20-lb block at constant velocity.

6 in.

P

Solution Frictional Force on Journal Bearing. Here f k = tan - 1 mk = tan - 1 0.2 = 11.3099°. Then the radius of the friction circle is

rf = r sin f k = 0.5 sin 11.3099° = 0.09806 in.

Equations of Equilibrium. Referring to the FBD of the pulley shown in Fig. a, a + ΣMP = 0;  P(6 - 0.09806) - 4(0.09806) - 20(6 + 0.09806) = 0

Ans.

P = 20.73 = 20.7 lb

Ans: P = 20.7 lb 866

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8–121. Solve Prob. 8–120 if the force P is applied horizontally to the left.

6 in.

P

Solution Frictional Force on Journal Bearing. Here f k = tan - 1mk = tan - 1 0.2 = 11.3099°. Then the radius of the friction circle is

rf = r sin f k = 0.5 sin 11.3099° = 0.09806 in.

Equations of Equilibrium. Referring to the FBD of the pulley shown in Fig. a. a + ΣMO = 0;  P(6) - 20(6) - R(0.09806) = 0 (1)

R = 61.1882 P - 1223.76



+ ΣFx = 0;  Rx - P = 0    Rx = P S + c ΣFy = 0;  Ry - 4 - 20 = 0  Ry = 24 lb Thus, the magnitude of R is

R = 2Rx2 + Ry2 = 2P2 + 242

(2)

Equating Eqs. (1) and (2)

61.1882 P - 1223.76 = 2P2 + 242

3743.00 P2 - 149,760.00 P + 1,497,024.00 = 0 P2 - 40.01 P + 399.95 = 0

chose the root P 7 20 lb,

Ans.

P = 20.52 lb = 20.5 lb

Ans: P = 20.5 lb 867

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8–122. Determine the tension T in the belt needed to overcome the tension of 200 lb created on the other side. Also, what are the normal and frictional components of force developed on the collar bushing? The coefficient of static friction is ms = 0.21.

2 in. 1.125 in.

SOLUTION Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.21 = 11.86°. Then the radius of friction circle is 200 lb

rf = r sin fk = 1 sin 11.86° = 0.2055 in.

T

Equations of Equilibrium: a + ©MP = 0;

20011.125 + 0.20552 - T11.125 - 0.20552 = 0 Ans.

T = 289.41 lb = 289 lb + c Fy = 0;

R - 200 - 289.41 = 0

R = 489.41 lb

Thus, the normal and friction force are N = R cos fs = 489.41 cos 11.86° = 479 lb

Ans.

F = R sin fs = 489.41 sin 11.86° = 101 lb

Ans.

Ans: T = 289 lb N = 479 lb F = 101 lb 868

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8–123. If a tension force T = 215 lb is required to pull the 200-lb force around the collar bushing, determine the coefficient of static friction at the contacting surface. The belt does not slip on the collar. 2 in. 1.125 in.

SOLUTION Equation of Equilibrium: a + ©MP = 0;

20011.125 + rf2 - 21511.125 - rf2 = 0

200 lb

T

rf = 0.04066 in. Frictional Force on Journal Bearing: The radius of friction circle is rf = r sin fk 0.04066 = 1 sin fk fk = 2.330° and the coefficient of static friction is Ans.

ms = tan fs = tan 2.330° = 0.0407

Ans: ms = 0.0407 869

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*8–124. The uniform disk fits loosely over a fixed shaft having a diameter of 40 mm. If the coefficient of static friction between the disk and the shaft is ms = 0.15, determine the smallest vertical force P, acting on the rim, which must be applied to the disk to cause it to slip on the shaft. The disk has a mass of 20 kg.

150 mm

40 mm

P

Solution Frictional Force on Journal Bearing. Here, f k = tan - 1 ms = tan - 1 0.15 = 8.5308°. Then the radius of the friction circle is

rf = r sin f s = 0.02 sin 8.5308° = 2.9668 ( 10 - 3 ) m

Equations of Equilibrium. Referring to the FBD of the disk shown in Fig. a, a + ΣMP = 0;  20(9.81) 3 2.9668 ( 10-3 ) 4 - P 3 0.075 - 2.9668 ( 10 - 3 ) 4 = 0



P = 8.08 N

Ans.

Ans: 8.08 N 870

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8–125. The 5-kg skateboard rolls down the 5° slope at constant speed. If the coefficient of kinetic friction between the 12.5-mm diameter axles and the wheels is mk = 0.3, determine the radius of the wheels. Neglect rolling resistance of the wheels on the surface. The center of mass for the skateboard is at G.

75 mm G 5 250 mm

300 mm

SOLUTION Referring to the free-body diagram of the skateboard shown in Fig. a, we have ©Fx¿ = 0;

Fs - 5(9.81) sin 5° = 0

Fs = 4.275 N

©Fy¿ = 0;

N - 5(9.81) cos 5° = 0

N = 48.86 N

The effect of the forces acting on the wheels can be represented as if these forces are acting on a single wheel as indicated on the free-body diagram shown in Fig. b.We have ©Fx¿ = 0;

Rx¿ - 4.275 = 0

Rx¿ = 4.275 N

©Fy¿ = 0;

48.86 - Ry¿ = 0

Ry¿ = 48.86 N

Thus, the magnitude of R is R = 2Rx¿ 2 + Ry¿ 2 = 24.2752 + 48.862 = 49.05 N fs = tan-1 ms = tan-1(0.3) = 16.699°. Thus, the moment arm of R from point O is (6.25 sin 16.699°) mm. Using these results and writing the moment equation about point O, Fig. b, we have a + ©MO = 0;

4.275(r) - 49.05(6.25 sin 16.699° = 0) Ans.

r = 20.6 mm

Ans: r = 20.6 mm 871

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8–126. The bell crank fits loosely into a 0.5-in-diameter pin. Determine the required force P which is just sufficient to rotate the bell crank clockwise. The coefficient of static friction between the pin and the bell crank is ms = 0.3.

12 in.

50 lb

45

P

10 in.

SOLUTION + ©F = 0; : x

P cos 45° - Rx = 0

Rx = 0.7071P

+ c ©Fy = 0;

Ry - P sin 45° - 50 = 0

Ry = 0.7071P + 50

Thus, the magnitude of R is R = 2Rx 2 + Ry 2 = 2(0.7071P)2 + (0.7071P + 50)2 = 2P2 + 70.71P + 2500 We find that fs = tan - 1 ms = tan - 1(0.3) = 16.699°. Thus, the moment arm of R from point O is (0.25 sin 16.699°) mm. Using these results and writing the moment equation about point O, Fig. a, a + ©MO = 0;

50(10) + 2P2 + 70.71P + 2500(0.25 sin 16.699°) - P(12) = 0

Choosing the larger root, Ans.

P = 42.2 lb

Ans: P = 42.2 lb 872

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8–127. The bell crank fits loosely into a 0.5-in-diameter pin. If P = 41 lb, the bell crank is then on the verge of rotating counterclockwise. Determine the coefficient of static friction between the pin and the bell crank.

12 in.

50 lb

45

P

10 in.

SOLUTION + ©F = 0; : x

41 cos 45° - Rx = 0

Rx = 28.991 lb

+ c ©Fy = 0;

Ry - 41 sin 45° - 50 = 0

Ry = 78.991 lb

Thus, the magnitude of R is R = 2Rx 2 + Ry 2 = 228.9912 + 78.9912 = 84.144 lb We find that the moment arm of R from point O is 0.25 sin fs. Using these results and writing the moment equation about point O, Fig. a, a + ©MO = 0;

50(10) - 41(12) - 84.144(0.25 sin fs) = 0 fs = 22.35°

Thus, Ans.

ms = tan fs = tan 22.35° = 0.411

Ans: ms = 0.411 873

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*8–128. The vehicle has a weight of 2600 lb and center of gravity at G. Determine the horizontal force P that must be applied to overcome the rolling resistance of the wheels. The coefficient of rolling resistance is 0.5 in. The tires have a diameter of 2.75 ft.

2.5 ft

2 ft

SOLUTION Rolling

Resistance:

W = NA + NB =

Here,

= 2600 lb, a = 0.5 in. and r = a

P

G

5 ft

13000 + 2.5P 5200 - 2.5P + 7 7

2.75 b 1122 = 16.5 in. Applying Eq. 8–11, we have 2

P L L

Wa r 260010.52 16.5 Ans.

L 78.8 lb

Ans: ≈ 78.8 lb 874

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8–129. The tractor has a weight of 16 000 lb and the coefficient of rolling resistance is a = 2 in. Determine the force P needed to overcome rolling resistance at all four wheels and push it forward.

G

P 2 ft

SOLUTION Applying Eq. 8–11 with W = 16 000 lb, a = a

P L

Wa = r

16000 a 2

3 ft

2 b ft and r = 2 ft, we have 12

2 b 12

6 ft

2 ft

Ans.

= 1333 lb

Ans: P = 1333 lb 875

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8–130. The handcart has wheels with a diameter of 6 in. If a crate having a weight of 1500 lb is placed on the cart, determine the force P that must be applied to the handle to overcome the rolling resistance. The coefficient of rolling resistance is 0.04 in. Neglect the weight of the cart.

5

P 3

4

SOLUTION + c ©Fy = 0;

P =

Wa , r

3 N - 1500 - P a b = 0 5 3 c 1500 + P a b d (0.04) 5 4 P = 5 3 2.4 P = 60 + 0.024 P Ans.

P = 25.3 lb

Ans: P = 25.3 lb 876

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8–131. The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for the cylinder’s top and bottom surfaces are aA and aB, respectively, show that a horizontal force having a magnitude of P = [W(aA + aB)]>2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder.

W P A

r

SOLUTION

B

+ ©F = 0; : x

(RA)x - P = 0

(RA)x = P

+ c ©Fy = 0;

(RA)y - W = 0

(RA)y = W

a + ©MB = 0;

(1)

P(r cos fA + r cos fB) - W(aA + aB) = 0

Since fA and fB are very small, cos fA - cos fB = 1. Hence, from Eq. (1) P =

W(aA + aB) 2r

(QED)

877

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*8–132. The 1.4-Mg machine is to be moved over a level surface using a series of rollers for which the coefficient of rolling resistance is 0.5 mm at the ground and 0.2 mm at the bottom surface of the machine. Determine the appropriate diameter of the rollers so that the machine can be pushed forward with a horizontal force of P = 250 N. Hint: Use the result of Prob. 8–131.

P

SOLUTION P =

W(aA + aB) 2r

250 =

1400 (9.81) (0.2 + 0.5) 2r

r = 19.2 mm Ans.

d = 38.5 mm

Ans: d = 38.5 mm 878

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9–1. Locate the center of mass of the homogeneous rod bent into the shape of a circular arc.

y

30 300 mm x

SOLUTION dL = 300 d u 30

' x = 300 cos u ' y = 300 sin u

x =

L

2p 3

' x dL =

L

L-2p3

300 cos u (300du) 2p 3

dL

L-2p3

300d u

(300)2 C sin u D -3 2p3 2p

=

4 300 a p b 3

Ans.

= 124 mm y = 0

Ans.

(By symmetry)

Ans: x = 124 mm y = 0 879

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9–2. y

Determine the location (x, y) of the centroid of the wire.

2 ft

4 ft

SOLUTION Length and Moment Arm: The length of the differential element is dL = 2dx2 + dy2 = ¢

B

1 + a

y = x2

2

dy dy b ≤ dx and its centroid is ∼ = 2x. y = y = x2. Here, dx dx

x

Centroid: Due to symmetry ∼

Ans.

x = 0

Applying Eq. 9–7 and performing the integration, we have 2 ft

∼ ∼

y =

LL

ydL = dL

LL

L-2 ft

x 21 + 4x2 dx

2 ft

L-2 ft =

2

21 + 4x2 dx

16.9423 = 1.82 ft 9.2936

Ans.

Ans: x = 0 y = 1.82 ft 880

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9–3. y

Locate the center of gravity x of the homogeneous rod. If the rod has a weight per unit length of 100 N>m, determine the vertical reaction at A and the x and y components of reaction at the pin B.

1m B

y  x2 A

1m

x

Solution Length

And

Arm. The length of the differential element is dy dy 2 dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~ x = x. Here = 2x. A dx dx Perform the integration L =

Moment

LL

dL =

L0

= 2

1m

L0

= cx

21 + 4x2 dx

1m

A

A

x2 +

x2 +

= 1.4789 m LL

x~ dL =

L0

= 2

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 A 4 0

1m

x21 + 4x2 dx

L0

1m

x

A

x2 +

1 dx 4

2 1 3>2 1 m = c ax2 + b d 3 4 0

= 0.8484 m2 Centroid. x =

~ 0.8484 m2 1L x dL = 0.5736 m = 0.574 m = 1.4789 m 1L dL

Ans.

Equations of Equilibrium. Refering to the FBD of the rod shown in Fig. a + ΣFx = 0; S

Ans.

Bx = 0

a+ΣMB = 0;  100(1.4789) (0.4264) - Ay(1) = 0 Ans.

Ay = 63.06 N = 63.1 N a+ΣMA = 0;  By(1) - 100(1.4789) (0.5736) = 0

Ans.

By = 84.84 N = 84.8 N

Ans: x = 0.574 m Bx = 0 Ay = 63.1 N By = 84.8 N 881

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*9–4. y

Locate the center of gravity y  of the homogeneous rod.

1m B

y  x2 A

1m

x

Solution Length

And

Arm. The length of the differential element is dy dy 2 dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~ y = y. Here = 2x. A dx dx Perform the integration, L =

Moment

LL

L0

dL =

= 2

1m

21 + 4x2 dx

L0

= cx

1m

A

A

x2 +

= 1.4789 m LL

y~ dL =

L0

= 2

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 A 4 0

1m

x2 21 + 4x2 dx

L0

= 2c

x2 +

1m

x

2

A

x2 +

1 dx 4

x 1 1 1 1 1m 1 3 2 x x2 + ln ax + x2 + b d ax + b 4A 32 A 4 128 A 4 0 4

= 0.6063 m2 Centroid.

0.6063 m2 1L y dL y = = 0.40998 m = 0.410 m = 1.4789 m 1L dL ~



Ans.

Ans: y = 0.410 m 882

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9–5. y

Determine the distance y  to the center of gravity of the homogeneous rod.

1m

2m

Solution Length

And

y  2x3

x

Moment

Arm. The length of the differential element is

dy 2 y = y. Here dL = 2dx2 + dy2 = a 1 + a b b dx and its centroid is at ~ A dx dy = 6x2. Evaluate the integral numerically, dx L = LL

~

LL

dL =

y dL =

L0

L0

1m

1m

21 + 36x4 dx = 2.4214 m

3

2x 21 + 36x4 dx = 2.0747 m2

Centroid. Applying Eq. 9–7, y =

~ 2.0747 m2 1L y dL = 0.8568 = 0.857 m = 2.4214 m 1L dL

Ans.

Ans: y = 0.857 m 883

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9–6. Locate the centroid y of the area.

y

y

1 1 – – x2 4

1m x 2m

SOLUTION Area and Moment Arm: The area of the differential element is y 1 1 1 ' = a1 - x 2 b . dA = ydx = a1 - x2 b dx and its centroid is y = 4 2 2 4 Centroid: Due to symmetry Ans.

x = 0 Applying Eq. 9–4 and performing the integration, we have 2m

' ydA

y =

LA

=

1 2 1 1 ¢ 1 - x ≤ ¢ 1 - x2 ≤ dx 2 4 4 L- 2m 2m

dA

LA

L- 2m

=

¢

¢1 -

1 2 x ≤ dx 4

x x3 x 5 2m + ≤` 2 12 160 - 2m x ¢x ≤` 12 - 2m 3

2m

=

2 m 5

Ans.

Ans: y = 884

2 m 5

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9–7. Determine the area and the centroid x of the parabolic area.

y

h

SOLUTION

y

Differential Element:The area element parallel to the x axis shown shaded in Fig. a will be considered. The area of the element is dA = x dy =

a h1>2

h x2 –– a2 x

a

y 1>2 dy

x a ' ' Centroid: The centroid of the element is located at x = = y 1>2 and y = y. 2 2h1>2 Area: Integrating, h

A =

LA

LA

= dA

LA

h

' xdA

x =

dA =

L0

¢

a 1>2

L0 h

y 1>2 dy =

2a 3h

a y1>2 ≤ ¢ 1>2 y 1>2 dy ≤ 1>2 2h h a

2 ah 3

1>2

A y3>2 B 2 = h 0

2 ah 3

Ans.

a2 y2 h a ¢ ≤` y dy 2h 2 0 3 L0 2h = = = a Ans. 2 2 8 ah ah 3 3 h 2

Ans: x = 885

3 a 8

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*9–8. Locate the centroid of the shaded area.

y

y  a cospx L a

L 2

x

L 2

Solution Area And Moment Arm. The area of the differential element shown shaded in y p a p Fig. a is dA = ydx = a cos x dx and its centroid is at y~ = = cos x. 2 2 2 L Centroid. Perform the integration

y =

~ 1A y dA

1A dA

L>2

=

p p a a cos xbaa cos x dxb L L L-L>2 2 L>2

L-L>2

L>2

=

=

L-L>2

p x dx L

a2 2p x + 1b dx acos 4 L L>2

L-L>2

a cos

p x dx L

L>2 a2 L 2p a sin x + xb ` 4 2p L -L>2

a

=

a cos

L>2 aL p sin xb ` p L -L>2

a2 L>4

2aL>p

=

p a 8

Ans.

x = 0

Ans.

Due to Symmetry,

Ans: p a 8 x = 0 y =

886

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9–9. y

Locate the centroid x of the shaded area.

4m y

1 2 x 4 x

4m

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a 1 is dA = x dy and its centroid is at ~ x = x. Here, x = 2y1>2 2 Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

=

L0

4m

1 a2y1>2 ba2y1>2 dyb 2 L0

4m

2y1>2 dy

3 m 2

Ans.

Ans: x = 887

3 m 2

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9–10. y

Locate the centroid y of the shaded area.

4m y

1 2 x 4 x

4m

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a y = y. Here, x = 2y1>2. is dA = x dy and its centroid is at ∼ Centroid. Perform the integration

y =

~ 1A y dA

1A dA

=

=

=

L0

4m

L0

y a2y1>2 dyb

4m

2y1>2 dy

4m 4 a y 5>2 b ` 5 0 4m 4 a y3>2 b ` 3 0

12 m 5

Ans.

Ans: y = 888

12 m 5

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9–11. y

Locate the centroid x of the area.

h y  —2 x2 b

SOLUTION

h

dA = y dx ' x = x

x =

LA

b

' x dA

LA

= dA

h 3 x dx 2 L0 b b

h 2 x dx 2 L0 b

=

B

h 4 x R 4b2 0

h B 2 x3 R 3b 0

b

x

b

b

=

3 b 4

Ans.

Ans: x = 889

3 b 4

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*9–12. y

Locate the centroid y of the shaded area.

h y  —2 x2 b

SOLUTION

h

dA = y dx y ' y = 2

y =

LA

b

' y dA

LA

= dA

2

h 4 x dx 4 L0 2b b

h 2 x dx 2 b L0

=

h2 5 b x R B 10b4 0

B

h 3 x R 3b2 0

b

x

b

=

3 h 10

Ans.

Ans: y = 890

3 h 10

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9–13. y

Locate the centroid x of the shaded area.

SOLUTION

4m

1 dA = 14 - y2dx = a x 2 b dx 16 ' x = x

x =

LA

8

' xdA

LA x = 6m

= dA

L0 L0

8

xa a

y

4

1 2 –– x 16 8m

x2 b dx 16

1 2 x b dx 16 Ans.

Ans: x = 6m 891

x

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9–14. y

Locate the centroid y of the shaded area.

SOLUTION

4m

dA = 14 - y2dx = a y =

y =

1 2 x b dx 16

y

4

1 2 –– x 16 8m

4 + y 2

LA

8

' ydA

LA

= dA

x2 x2 1 b a b dx ¢8 2 L0 16 16 8

L0

a

1 2 x b dx 16 Ans.

y = 2.8 m

Ans: y = 2.8 m 892

x

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9–15. Locate the centroid x of the shaded area. Solve the problem by evaluating the integrals using Simpson’s rule.

y y = 0.5ex2

SOLUTION At x = 1 m 2

y = 0.5e1 = 1.359 m 1

LA

dA =

L0

1

(1.359 - y) dx =

L0

x

2

a 1.359 = 0.5 ex b dx = 0.6278 m2

1m

x = x 1

x dA =

LA

L0

2

x a 1.359 - 0.5 ex b dx

= 0.25 m3

x =

x dA LA LA

dA

=

0.25 = 0.398 m 0.6278

Ans.

Ans: x = 0.398 m 893

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*9–16. Locate the centroid y of the shaded area. Solve the problem by evaluating the integrals using Simpson’s rule.

y y = 0.5ex2

SOLUTION 1

dA =

L0

LA

y =

1

(1.359 - y) dx =

L0

a 1.359 - 0.5ex b dx = 0.6278 m

2

2

1.359 + y 2 1

y dA =

LA

L0

x 1m

a

x2

1.359 + 0.5 e 2 b A 1.359 - 0.5 ex B dx 2

1

=

y =

1 2 a 1.847 - 0.25 e2x b dx = 0.6278 m3 2 L0

y dA LA dA

=

0.6278 = 1.00 m 0.6278

Ans.

LA

Ans: y = 1.00 m 894

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9–17. y

Locate the centroid y of the area.

SOLUTION

y 4 in.

Area: Integrating the area of the differential element gives A =

LA

dA =

8 in.

8 in.

x

L0

2 ––

x3

2>3

3 dx = c x 5>3 d 2 0 5

x

= 19.2 in.2

8 in.

' 1 Centroid: The centroid of the element is located at y = y>2 = x2>3. Applying 2 Eq. 9–4, we have 8 in. '

y =

y dA LA LA c

=

L0

=

dA

8 in.

1 2>3 2>3 1 4>3 x A x B dx x dx L 2 2 0 = 19.2 19.2

8 in.

3 7>3 2 x d 14 0 19.2

Ans.

= 1.43 in.

Ans: y = 1.43 in. 895

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9–18. y

Locate the centroid x of the area.

y

SOLUTION

h

h n — nx a

h

dA = y dx ' x = x

x =

LA

x =

=

L0

dA

h 2

B hx -

h n+1 x ≤ dx an

a

h n x ≤ dx an

¢h -

h1xn + 22

a 1n + 22 n

h1x

2

n+1

a n1n + 12

h h b a2 2 n + 2

ah -

¢ hx -

L0

B x2 -

a

a

' x dA

LA

=

x

a

h ba n + 1

=

R

R

a 0

a 0

a(1 + n) 2(2 + n)

Ans.

Ans: x = 896

a(1 + n) 2(2 + n)

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9–19. y

Locate the centroid y of the area.

y

SOLUTION

h

h n — x an

h

dA = y dx

y =

LA

y =

a

' y dA

LA

=

x

a

y ' y = 2

=

h2 1 h2 ¢ h2 - 2 n xn + 2n x2n ≤ dx 2 L0 a a a

dA L0

¢h -

h n x ≤ dx an

2h21xn + 12 h21x 2n + 12 a 1 2 + 2n Bh x - n R 2 a 1n + 12 a 12n + 12 0

B hx -

h1x n + 12 n

a 1n + 12

2n2 h 21n + 1212n + 12 n n + 1

=

R

a 0

hn 2n + 1

Ans.

Ans: y = 897

hn 2n + 1

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*9–20. Locate the centroid y of the shaded area.

y

y

h xn –– an h

SOLUTION dA = y dx

x a

y y = 2 ' ydA

y =

LA

1 2

= dA

LA

a

L0 L0

h2 a 2n

x 2n dx =

a h an

xn dx

h2(a 2n + 1) 2a 2n(2n + 1) h(a n + 1) a n(n + 1)

=

hn + 1 2(2n + 1)

Ans.

Ans: y = 898

hn + 1 2(2n + 1)

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9–21. y

Locate the centroid x of the shaded area.

1

y  (4  x2 )2

16 ft

4 ft x

4 ft

Solution Area And Moment Arm. The area of the differential element shown shaded in  Fig.  a is dA = y dx = ( 4 - x1>2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is ~ = x. at x Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

=

L0

4 ft

L0 a

a = 1

x(x - 8x1>2 + 16)dx

4 ft

( x - 8x1>2 + 16) dx

4 ft x3 16 5>2 x + 8x2 b ` 3 5 0

4 ft 16 3>2 x2 x + 16xb ` 2 3 0

3 ft 5

Ans.

Ans:

3 x = 1 ft 5

899

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9–22. y

Locate the centroid y of the shaded area.

1

y  (4  x2 )2

16 ft

4 ft x

4 ft

Solution Area And Moment Arm. The area of the differential element shown shaded in 1

Fig.  a is dA = y dx = ( 4 - x 2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is at y 1 = ( 4 - x1>2 ) 2. y~ = 2 2 Centroid. Perform the integration,

y =

~ 1A y dA

1A dA

=

=

=

L0

4 ft

1 ( 4 - x1>2 ) 2 ( x - 8x1>2 + 16 ) dx 2 L0

L0

4 ft

4 ft

(x - 8x1>2 + 16)dx

1 a x2 - 8x3>2 + 48x - 128x1>2 + 128bdx 2 L0

a

= 4

4 ft

( x - 8x1>2 + 16 ) dx

4 ft x3 16 5>2 256 3>2 x + 24x2 x + 128xb ` 6 5 3 0

8 ft 55

a

4 ft 16 3>2 x2 x + 16xb ` 2 3 0



Ans.

Ans: y = 4 900

8 ft 55

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9–23. y

Locate the centroid x of the shaded area.

y   h2 x2h a h

x

a

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a h is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~ x = x. a Centroid. Perform the integration,

x =

~ 1A x dA

1A dA

=

L0

a

L0

x aa

a-

h 2 x + hbdx a2

h

a

2

2

x + hbdx

h 4 h 2 a x + x b` 2 4a2 0 = a h 3 a - 2 x + hxb ` 3a 0 a-

=

3 a 8

Ans.



Ans: x = 901

3 a 8

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*9–24. y

Locate the centroid y of the shaded area.

y   h2 x2h a h

x

a

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a y 1 h2 h is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~ y = = a - x2 + hb. 2 2 a a Centroid. Perform the integration,

y =

a

~ 1A y dA

1 h h a - 2 x2 + hba - 2 x2 + hbdx a a L0 2 = a h dA 2 1A a - 2 x + hbdx L0 a a 1 h2 5 2h2 3 2 a 4x x + h xb ` 2 5a 3a2 0 = a h 3 a - 2 x + hxb ` 3a 0

=

2 h 5



Ans.

Ans: y = 902

2 h 5

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9–25. z

The plate has a thickness of 0.25 ft and a specific weight of g = 180 lb>ft3. Determine the location of its center of gravity. Also, find the tension in each of the cords used to support it.

B

SOLUTION 1 2

Area and Moment Arm: Here, y = x - 8x + 16. The area of the differential 1 ' element is dA = ydx = 1x - 8x2 + 162dx and its centroid is x = x and 1 1 ' y = 1x - 8x2 + 162. Evaluating the integrals, we have 2 16 ft

A =

16 ft

16 ft

LA

dA =

L0

A

1 y2



1 x2

C y

4

x

1x - 8x2 + 162dx 1

16 ft 1 16 3 = a x2 x2 + 16x b ` = 42.67 ft2 2 3 0

LA

' xdA =

16 ft

1

x31x - 8x 2 + 162dx4

L0

16 ft 1 16 5 = a x3 x2 + 8x2 b ` = 136.53 ft3 3 5 0

LA

' ydA =

=

16 ft

L0

1 1 1 1x - 8x2 + 16231x - 8x2 + 162dx4 2

16 ft 1 1 3 32 5 512 3 a x x2 + 48x2 x2 + 256xb ` 2 3 5 3 0

= 136.53 ft3 Centroid: Applying Eq. 9–6, we have

x =

LA

' xdA

LA

y =

LA

=

136.53 = 3.20 ft 42.67

Ans.

=

136.53 = 3.20 ft 42.67

Ans.

dA

' ydA

LA

dA

Equations of Equilibrium: The weight of the plate is W = 42.6710.25211802 = 1920 lb. ©Mx = 0;

192013.202 - TA1162 = 0 TA = 384 lb

Ans.

©My = 0;

TC1162 - 192013.202 = 0 TC = 384 lb

Ans.

©Fz = 0;

TB + 384 + 384 - 1920 = 0 Ans.

TB = 1152 lb = 1.15 kip 903

Ans: x = 3.20 ft y = 3.20 ft TA = 384 lb TC = 384 lb TB = 1.15 kip

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9–26. y

Locate the centroid x of the shaded area.

4 ft

y

1 2 x 4

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a 1 is dA = y dx = x2 dx and its centroid is at ~ x = x. 4 Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

=

L0

4 ft

L0 a a

x 4 ft

1 x a x2 dxb 4

4 ft

1 2 x dx 4

1 4 4 ft x b` 16 0 1 3 4 ft x b` 12 0

= 3 ft

Ans.



Ans: x- = 3 ft 904

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9–27. y

Locate the centroid y of the shaded area.

4 ft

y

1 2 x 4

Solution Area And Moment arm. The area of the differential element shown shaded in Fig. a y 1 1 1 1 is dA = y dx = x2 dx and its centroid is located at ~ y = = a x2 b = x2. 2 2 4 8 4 Centroid. Perform the integration,

y =

~ 1A y dA

1A dA

=

=

L0

4 ft

4 ft

1 2 1 2 x a x dxb 8 4

L0 6 ft 5

x

4 ft

1 2 x dx 4

Ans.



Ans: y = 905

6 ft 5

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*9–28. y

Locate the centroid x of the shaded area.

yx

100 mm

y  1 x2 100 100 mm

x

Solution

1 2 x . Thus the area of the 100 1 2 x ) dx differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx = ( x 100 ~ and its centroid is at x = x. Area And Moment arm. Here, y2 = x and y1 =

Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

=

L0

100 mm

L0 a a

x ax -

100 mm

ax -

1 2 x bdx 100

1 2 x bdx 100

x3 1 4 100 mm x b` 3 400 0 x2 1 3 100 mm x b` 2 300 0

= 50.0 mm

Ans.



Ans: x = 50.0 mm 906

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9–29. Locate the centroid y of the shaded area.

y

yx

100 mm

y  1 x2 100 100 mm

x

Solution Area And Moment arm. Here, x2 = 10y 1>2 and x1 = y. Thus, the area of the differential element shown shaded in Fig. a is dA = ( x2 - x1 ) dy = ( 10y1>2 - y ) dy and its centroid is at ~ y = y. Centroid. Perform the integration,

y =

~ 1A y dA

1A dA

=

=

L0

100 mm

L0

y a10y1>2 - ybdy

100 mm

a4y 5>2 -

a

1>2

a10y

- ybdy

y3 100 mm b` 3 0

y2 100 mm 20 3>2 y - b` 3 2 0

= 40.0 mm

Ans.



Ans: y = 40.0 mm 907

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9–30. y

Locate the centroid x of the shaded area.

a

h

h y  –– ax

y(

h )(xb) ab x

b

Solution

a a - b b y + b. Thus the area y and x2 = a h h a a - b by + b - y d dy = of the differential element is dA = ( x2 - x1 ) dy = c a h h x -x b a b b 1 ( b - y ) dy and its centroid is at ~x = x1 + 2 1 = ( x2 + x1 ) = y - y + . h h 2h 2 2 2

Area And Moment arm. Here x1 =

Centroid. Perform the integration,

x =

h

~ 1A x dA

b b b a a y y + b c ab - ybdy d h 2h 2 h L0 = h b dA 1A ab - ybdy h L0 =

c

h b b b2 (a - b)y2 + (b - 2a)y3 + yd ` 2 2h 2 6h 0 b 2 h aby y b` 2h 0

bh (a + b) 6 = bh 2 1 = (a + b) 3



Ans.

Ans: x = 908

1 (a + b) 3

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9–31. y

Locate the centroid y of the shaded area.

a

h

h y  –– ax

y(

h )(xb) ab x

b

Solution

a - b a by + b. Thus the area y and x2 = a h h a a - b by + b - y d dy = of the differential element is dA = ( x2 - x1 ) dy = c a h h b ab - ybdy and its centroid is at ~ y = y. h Centroid. Perform the integration,

Area And Moment arm. Here, x1 =

y =

~ 1A y dA

1A dA

=

=

L0

h

L0

y ab -

h

ab -

b ybdy h b ybdy h

b b 3 h a y2 y b` 2 3h 0 aby -

b 2 h y b` 2h 0

1 2 bh h 6 = = 1 3 bh 2

Ans.



Ans: y = 909

h 3

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*9–32. y

Locate the centroid x of the shaded area.

y  a sin

SOLUTION

a

Area and Moment Arm: The area of the differential element is x dA = ydx = a sin dx and its centroid are x = x a

x =

LA

x a

pa

' xdA

LA

= dA

L0

x a a sin pa

L0

=

=

c a3 sin

a sin

x

ap

x dx b a x dx a

pa x x - x a a2 cos b d ` a a 0

a -a2 cos

x pa b` a 0

p a 2

Ans.

Ans: x = 910

p a 2

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9–33. y

Locate the centroid y of the shaded area.

y

SOLUTION

a sin

x a

a

Area and Moment Arm: The area of the differential element is y a x x = sin . dA = ydx = a sin dx and its centroid are y = a a 2 2

x

ap

pa 1 2x 1 2 x x a bd` c a a x - a sin sin aa sin dxb a 4 2 a a 0 pa L0 2 y = LA = = = pa pa 8 x x 2 dA a sin dx a -a cos b ` a LA L0 a 0 pa

ydA

Ans.

Ans: y = 911

pa 8

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9–34. y

The steel plate is 0.3 m thick and has a density of 7850 kg>m3. Determine the location of its center of mass. Also compute the reactions at the pin and roller support.

y2  2x 2m

SOLUTION

x

y1 = - x1

A

y 22 = 2x2

2m

dA = 1y2 - y12 dx = ' x = x

A 22x + x B dx

y  x B 2m

y2 + y1 22x - x ' = y = 2 2

x=

LA

=

LA

y =

LA

2

' x dA dA

LA

2

L0

A 22x + x B dx

=

c

2 222 5>2 1 x + x3 d 5 3 0

2 2 22 3>2 1 c x + x2 d 3 2 0

2

' y dA = dA

L0

x A 22x + x B dx

22x - x A 22x + x B dx 2 L0 2

L0

A 22x + x B dx

=

c

= 1.2571 = 1.26 m Ans.

2 x2 1 - x3 d 2 6 0

2 2 22 3>2 1 c x + x2 d 3 2 0

= 0.143 m

Ans.

A = 4.667 m2 W = 785019.81214.667210.32 = 107.81 kN a + ©MA = 0;

-1.25711107.812 + NB A 2 22 B = 0 Ans.

NB = 47.92 = 47.9 kN + : ©Fx = 0;

- A x + 47.92 sin 45° = 0 Ans.

A x = 33.9 kN + c ©Fy = 0;

A y + 47.92 cos 45° - 107.81 = 0 Ans.

A y = 73.9 kN

Ans: x = 1.26 m y = 0.143 m NB = 47.9 kN Ax = 33.9 kN Ay = 73.9 kN 912

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9–35. y

Locate the centroid x of the shaded area.

h y  h  —n xn a h

h yh— a x x

a

Solution

h n h x and y1 = h - x. Thus, the an a area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx Area And Moment Arm. Here, y2 = h -

= ch -

h n h h h x - ah - xb d dx = a x - n xn bdx and its centroid is ~ x = x. an a a a

Centroid. Perform the integration

x =

a

~ 1A x dA

h h x a x - n xn bdx a a L0 = a h h n dA 1A a x - n x bdx a L0 a =

c

c

a h 3 h x - n xn + 2 d ` 3a a (n + 2) 0

a h 2 h x - n xn + 1 b ` 2a a (n + 1) 0

ha2 (n - 1) =

3(n + 2) ha(n - 1) 2(n + 1)

= c

2(n + 1) 3(n + 2)

da

Ans.



Ans: x = c 913

2(n + 1) 3(n + 2)

da

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*9–36. y

Locate the centroid y of the shaded area.

h y  h  —n xn a h

h yh— a x x

a

Solution

h n h x and y1 = h - x. Thus, the an a area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx h h h h = c h - n xn - ah - xb d dx = a x - n xn bdx and its centroid is at a a a a Area And Moment Arm. Here, y2 = h -

∼

y = y1 + a

y2 - y1 1 1 h h 1 h h b = ( y2 + y1 ) = ah - n xn + h - xb = a2h - n xn - xb. 2 2 2 a a 2 a a

Centroid. Perform the integration

y =

a

~ 1A y dA

h 1 h h h a2h - n xn - xba x - n xn bdx 2 a a a a L0 = a h n h dA 1A a x - n x bdx a a L0 =

a 1 h2 2 h2 2h2 h2 x2n + 1 d ` c x - 2 x3 - n xn + 1 + 2n 2 a a (n + 1) 3a a (2n + 1) 0

c

=

h2a c

= c

a h 2 h x - n xn + 1 b ` 2a a (n + 1) 0

(4n + 1) (n - 1)

6(n + 1)(2n + 1) hac

n - 1 d 2(n + 1)

(4n + 1)

3(2n + 1)

d

 Ans.

d h

Ans: y = c 914

(4n + 1) 3(2n + 1)

d h

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9–37. y

Locate the centroid x of the circular sector.

r a

C

x

a x

Solution Area And Moment Arm. The area of the differential element shown in Fig. a is 1 2 dA = r 2 du and its centroid is at x~ = r cos u. 2 3 Centroid. Perform the integration

x =

=

~ 1A x dA

1A dA

L-a a

=

2 1 a r cos u ba r 2 du b 3 2 1 2 r du 2 L-a a

a 1 a r 3 sin u b ` 3 -a a 1 a r2 ub ` 2 -a

2 3 r sin a 3 = r2 a =

2 r sin a a b 3 a

Ans.

Ans: x = 915

2 r sin a a b 3 a

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9–38. Determine the location r of the centroid C for the loop of the lemniscate, r2 = 2a2 cos 2u, ( -45° … u … 45°).

r2 = 2a2 cos 2θ r O

θ C _ r

SOLUTION 1 1 (r) r du = r2 du 2 2

dA =

45°

A = 2

L0

1 (2a2 cos 2u)du = a2 C sin 2u D 45° = a2 0 2 45°

2

x dA

x =

LA

L0

=

A 23 r cos u B A 12r2du B

dA

2 3

=

a2

45°

L0

r3 cos u du 2

a

LA

x dA =

LA

x =

2 3 L0

45°

3

r cos u du =

2 3 L0

45°

A 2a2 B 3/2 cos u (cos 2 u)3/2du = 0.7854 a3

0.7854 a 3 = 0.785 a a2

Ans.

Ans: x = 0.785 a 916

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9–39. Locate the center of gravity of the volume. The material is homogeneous.

z

2m

SOLUTION

y 2 = 2z

2m

Volume and Moment Arm: The volume of the thin disk differential element is ' dV = py2dz = p12z2dz = 2pzdz and its centroid z = z.

y

Centroid: Due to symmetry about z axis - = y -= 0 x

Ans.

Applying Eq. 9–3 and performing the integration, we have

Lv z = Lv

=

z12pzdz2

L0

dV

2m

L0

z3 3

2m

z2 2p 2

2m

2p =

2m

' z dV

0

=

2pzdz

4 m 3

Ans.

0

Ans: x = y = 0 4 z = m 3 917

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*9–40. Locate the centroid y of the paraboloid.

z

z2 = 4y 4m y

SOLUTION Volume and Moment Arm: The volume of the thin disk differential element is ' dV = pz2dy = p14y2dy and its centroid y = y.

4m

Centroid: Applying Eq. 9–3 and performing the integration, we have

y =

LV

4m

' ydV

LV

= dV

L0

y3p14y2dy4 4m

p14y2dy

L0 4p =

y3 3

4m

2

4m

y 4p 2

0

Ans.

= 2.67 m

0

Ans: y = 2.67 m 918

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9–41. Locate the centroid z of the frustum of the right-circular cone.

z r

h R

SOLUTION Volume y =

and

Moment

1r - R2z + Rh h

Arm:

From

the

geometry,

y - r h - z = , R - r h

y

x

. The volume of the thin disk differential element is

dV = py2dz = pc a =

1r - R2z + Rh h

2

b d dz

p c1r - R22z2 + 2Rh1r - R2z + R2h2 ddz h2

and its centroid z = z. Centroid: Applying Eq. 9–5 and performing the integration, we have

z =

LV

h

' z dV

LV

= dV

=

=

L0

zb

p 31r - R22z2 + 2Rh1r - R2z + R2h24dz r h2

h

p 3r - R22z2 + 2Rh1r - R2z + R2h24dz 2 L0 h 4 2 h z3 p 2 z 2 2 z 1r R2 + 2Rh1r R2 + R h B ¢ ≤ ¢ ≤ ¢ ≤ R ` 4 3 2 h2 0 3 p 2 z 1r R2 3 h2

+ 2Rh1r - R2

z2 2

+ R2h21z2 `

R2 + 3r2 + 2rR h 4 R2 + r2 + rR

h 0

Ans.

Ans: z =

919

R2 + 3r 2 + 2rR h 4(R2 + r 2 + rR)

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9–42. z

Determine the centroid y of the solid.

y z  –– (y  1) 6

SOLUTION

1 ft y

Differential Element: The thin disk element shown shaded in Fig. a will be considered. The x volume of the element is 2 y p 4 dV = pz2 dy = pc (y - 1) d dy = (y - 2y3 + y2)dy 6 36

3 ft

Centroid: The centroid of the element is located at yc = y. We have

y =

LV

3 ft

y~ dV

LV

= dV

L0 L0

yc

3 ft

p 4 A y - 2y3 + y2 B dy d 36

p 4 A y - 2y3 + y2 B dy 36

3 ft

=

L0

3 ft

L0

p 5 A y - 2y4 + y3 B dy 36

p 4 A y - 2y3 + y2 B dy 36

=

y4 3 ft p y6 2 c - y5 + d2 36 6 5 4 0 y4 y 3 3 ft p y5 c + d2 36 5 2 3 0 Ans.

= 2.61 ft

Ans: y = 2.61 ft 920

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9–43. Locate the centroid of the quarter-cone.

z

h

SOLUTION

a

' z = z r =

y

a (h - z) h

x

dV =

p 2 p a2 r dz = (h - z)2 dz 4 4 h2

dV =

p a2 2 z3 h p a2 d (h2 - 2hz + z2) dz = c h z - hz2 + 2 2 3 0 4 h L0 4h

h

L

=

pa2 h p a2 h3 a b = 2 3 12 4h 2

h

z3 z4 h p a 2 2 z2 pa ' 2 2 2h + d (h 2hz + z ) z dz = ch z dV = 2 2 3 4 0 4 h2 4 h L0 L =

pa2h2 p a2 h4 b = a 48 4 h2 12

p a2 h2 h 48 L z = = = 4 p a2h dV 12 L ' z dV

h

Ans.

h

pa2 pa 2 4r 4a ' (h - z)2 dz = (h3 - 3h2 z + 3hz2 - z3) dz xdV = 2 2 3p 3 4 h 4 h L0 L0 p h L =

3h4 h4 p a2 4a 4 4 + h b a h 2 4 4 h2 3ph

=

a3 h p a2 a h3 b = a 12 4 h2 3p

a3 h 12 a L = x = y = = p p a2h dV 12 L ' xdV

Ans.

Ans: h 4 a x = y = p

z =

921

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*9–44. z

The hemisphere of radius r is made from a stack of very thin plates such that the density varies with height r = kz, where k is a constant. Determine its mass and the distance to the center of mass G. G

_ z

r

SOLUTION

y

Mass and Moment Arm: The density of the material is r = kz. The mass of the thin disk differential element is dm = rdV = rpy2dz = kz3p(r2 - z2) dz4 and its ' centroid z = z. Evaluating the integrals, we have

x

r

m =

Lm

dm =

L0

kz3p(r2 - z2) dz4

= pk ¢

Lm

' z dm =

r2z2 z4 r pkr4 - ≤` = 2 4 0 4

Ans.

r

L0

z5kz3p(r2 - z2) dz46

= pk ¢

2pkr5 r2z3 z5 r - ≤` = 3 5 0 15

Centroid: Applying Eq. 9–3, we have

z =

Lm

' z dm

Lm

= dm

2pkr5>15 pkr4>4

=

8 r 15

Ans.

Ans:

pkr 4 4 8 z = r 15

m =

922

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9–45. z

Locate the centroid z of the volume.

1m

y2  0.5z 2m

Solution y

Volume And Moment arm. The volume of the thin disk differential element shown shaded in Fig. a is dV = py2 dz = p(0.5z)dz and its centroid is at ~ z = z. Centroid. Perform the integration

z =

~ 1V z dV

1V dV

=

=

=

L0

2m

L0

x

z[p(0.5z)dz]

2m

p(0.5z)dz

0.57 3 2 m z ` 3 0

0.5p 2 2 m z ` 2 0

4 m 3

Ans.

Ans: z = 923

4 m 3

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9–46. Locate the centroid of the ellipsoid of revolution.

z

y2 z2  1 b2 a 2 a x

SOLUTION

b

dV = p z2 dy b

L L

dV =

L0

' ydV =

p a2 a 1 b

L0

y

y2 b

p a2y a1 -

b dy = p a2 c y 2 y b

2

b dy = p a2 c 2

y3

b

3b

0

d = 2

2pa2b 3

2

y4 b y p a2b2 d = 2 2 4 4b 0

pa2b2 3 4 LV = b = y = 8 2pa2b dV 3 LV ' ydV

x = z = 0

Ans.

Ans.

(By symmetry)

Ans: 3 b 8 x = z = 0

y =

924

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9–47. z

Locate the center of gravity z of the solid.

2 ––

z  4y 3

SOLUTION

16 in.

Differential Element: The thin disk element shown shaded in Fig. a will be considered. The volume of the element is y

2 1 p 3 dV = py2 dz = pc z3>2 d dz = z dz 8 64

8 in. x

Centroid: The centroid of the element is located at zc = z. We have

z =

LV

16 in.

~ z dV

LV

= dV

L0 L0

zc

16 in.

p 3 z dz d 64

p 3 z dz 64

16 in.

=

L0 L0

16 in.

p 4 z dz 64 p 3 z dz 64

=

p z5 16 in. a b2 64 5 0 p z4 2 16 in. a b 64 4 0

= 12.8 in. Ans.

Ans: z = 12.8 in. 925

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*9–48. z

Locate the center of gravity y of the volume. The material is homogeneous.

z  1 y2 100 4 in. 1 in.

10 in.

y

10 in.

Solution Volume And Moment Arm. The volume of the thin disk differential element shown p 1 2 2 shaded in Fig. a is dV = pz2 dy = pa y b dy = y4 dy and its centroid is 100 10000 y = y. at ~ Centroid. Perform the integration

y =

~ 1V y dV

1V dV

20 in.

=

=



L10 in.

ya

p y4 dyb 10000

20 in.

p y4 dy L10 in. 10000 a a

20 in. p y6 b ` 60000 10 in. 20 in. p y5 b ` 50000 10 in.

Ans.

= 16.94 in. = 16.9 in.

Ans: y = 16.9 in. 926

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9–49. Locate the centroid z of the spherical segment.

z

z2

a2

y2 1a — 2

C a

SOLUTION

z

2

2

2

dV = py dz = p(a - z ) dz y

z = z

z =

LV

a

' z dV

LV

p =

dV

L

p

=

a 2

La2

x

z (a2 - z2)dz a

(a 2 - z2)dz

p B a2 a

z4 a z2 b - a bR a 2 4 2

z3 a p B a (z) - a b R a 3 2 2

=

pB

a4 a4 a4 a4 + R 2 4 8 64

a3 a3 a3 + pBa R 3 2 24 3

=

pB

9a 4 R 64

pB

5a 3 R 24 Ans.

z = 0.675 a

Ans: z = 0.675a 927

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9–50. Determine the location z of the centroid for the tetrahedron. Hint: Use a triangular “plate” element parallel to the x–y plane and of thickness dz.

z

b a

SOLUTION z = ca1 -

1 1 yb = ca1 - xb a b c

L

dV =

c y

c

1 z abc z 1 (x)(y)dz = a a 1 - b ba 1 - b dz = c c 2 L0 6 L0 2

x

c

z a b c2 z 1 ' z a a 1 - b ba 1 - b dz = z dV = c c 2 L0 24 L a b c2 c 24 L = z = = abc 4 dV 6 L ' z dV

Ans.

Ans: z = 928

c 4

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9–51. The truss is made from five members, each having a length of 4 m and a mass of 7 kg>m. If the mass of the gusset plates at the joints and the thickness of the members can be neglected, determine the distance d to where the hoisting cable must be attached, so that the truss does not tip (rotate) when it is lifted.

y

d B 4m

C

4m 4m

SOLUTION

60

' ©xM = 4(7)(1+ 4 + 2 + 3 + 5) = 420 kg # m

A

©M = 4(7)(5) = 140 kg d = x =

4m

' 420 ©xM = = 3m ©M 140

4m

D

Ans.

Ans: d = 3m 929

x

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*9–52. z

Determine the location (x, y, z) of the centroid of the homogeneous rod.

200 mm x 30 600 mm 100 mm

Solution

y

Centroid. Referring to Fig. a, the length of the segments and the locations of their respective centroids are tabulated below Segment L(mm)

x~(mm)

y~(mm)

z~(mm)

x~ L(mm2)

0

0

100

0

1

200

2

600

300 cos 30° 300 sin 30°

0

3

100

600 cos 30° 600 sin 30°

–50

a

900

y~ L(mm2) z~ L(mm2) 0

155.88 ( 103 ) 90.0 ( 103 )

20.0 ( 103 ) 0

51.96 ( 103 ) 30.0 ( 103 ) - 5.0 ( 103 ) 207.85 ( 103 ) 120.0 ( 103 ) 15.0 ( 103 )

Thus,

x =

207.85 ( 103 ) mm2 Σ~ xL = = 230.94 mm = 231 mm ΣL 900 mm

Ans.



y =

120.0 ( 103 ) mm2 Σ~ yL = = 133.33 mm = 133 mm ΣL 900 mm

Ans.



z =

15.0 ( 103 ) mm2 Σ~ zL = = 16.67 mm = 16.7 mm ΣL 900 mm2

Ans.

Ans: x = 231 mm y = 133 mm z = 16.7 mm 930

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9–53. A rack is made from roll-formed sheet steel and has the cross section shown. Determine the location 1x, y2 of the centroid of the cross section. The dimensions are indicated at the center thickness of each segment.

y 30 mm

80 mm

SOLUTION 50 mm

©L = 15 + 50 + 15 + 30 + 30 + 80 + 15 = 235 mm

' ©xL = 7.5(15) + 0(50) + 7.5(15) + 15(30) + 30(30) + 45(80) + 37.5(15) = 5737.50 mm2 ' ©yL = 0(15) + 25(50) + 50(15) + 65(30) + 80(30) + 40(80) + 0(15) = 9550 mm2 x =

' ©xL 5737.50 = = 24.4 mm ©L 235

Ans.

y =

' ©yL 9550 = = 40.6 mm ©L 235

Ans.

x 15 mm

15 mm

Ans: x = 24.4 mm y = 40.6 mm 931

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9–54. Locate the centroid (x, y) of the metal cross section. Neglect the thickness of the material and slight bends at the corners.

y

50 mm

150 mm

SOLUTION

x

Centroid: The length of each segment and its respective centroid are tabulated below. Segment

L (mm)

' y (mm)

' y L (mm2)

1

50p

168.17

26415.93

2

180.28

75

13520.82

3

400

0

0

4

180.28

75

13520.82

©

917.63

53457.56

- = 0 Due to symmetry about y axis, x

y =

50 mm 100 mm 100 mm 50 mm

Ans.

' ©yL 53457.56 = = 58.26 mm = 58.3 mm ©L 917.63

Ans.

Ans: x = 0 y = 58.3 mm 932

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9–55. Locate the center of gravity 1x, y, z2 of the homogeneous wire.

z

400 mm

SOLUTION 2 (300) p ' a b (300) = 165 000 mm2 ©xL = 150(500) + 0(500) + p 2

y 300 mm

p ©L = 500 + 500 + a b (300) = 1471.24 mm 2 x =

x

' ©xL 165 000 = = 112 mm ©L 1471.24

Ans.

Due to symmetry, Ans.

y = 112 mm p ' © z L = 200(500) + 200(500) + 0 a b (300) = 200 000 mm2 2 z =

' © zL 200 000 = = 136 mm ©L 1471.24

Ans.

Ans: x = 112 mm y = 112 mm z = 136 mm 933

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*9–56. The steel and aluminum plate assembly is bolted together and fastened to the wall. Each plate has a constant width in the z direction of 200 mm and thickness of 20 mm. If the density of A and B is rs = 7.85 Mg>m3, and for C, ral = 2.71 Mg>m3, determine the location x of the center of mass. Neglect the size of the bolts.

y

100 mm 200 mm A

SOLUTION

C

B 300 mm

©m = 2 C 7.85(10)3(0.3)(0.2)(0.02) D + 2.71(10)3(0.3)(0.2)(0.02) = 22.092 kg ' ©xm = 150{2 C 7.85(10)3(0.3)(0.2)(0.02) D }+350 C 2.71(10)3(0.3)(0.2)(0.02) D = 3964.2 kg.mm x =

' 3964.2 ©xm = = 179 mm ©m 22.092

Ans.

Ans: x = 179 mm 934

x

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9–57. Locate the center of gravity G1x, y2 of the streetlight. Neglect the thickness of each segment. The mass per unit length of each segment is as follows: rAB = 12 kg>m, rBC = 8 kg>m, rCD = 5 kg>m, and rDE = 2 kg>m.

1m

y

90 1 m 1m

D

E

1m C

SOLUTION

G (x, y) 3m

2(1) p ' b a b(5) ©xm = 0(4)(12) + 0(3)(8) + 0(1)(5) + a 1 p 2

B

+ 1.5 (1) (5) + 2.75 (1.5) (2) = 18.604 kg # m ©m = 4 (12)+3 (8)+ 1(5)+ x =

1.5 m

p (5) + 1(5) + 1.5 (2) = 92.854 kg 2

4m A

' ©xm 18.604 = = 0.200 m ©m 92.854

x

Ans.

2(1) p ' b a b(5) ©ym = 2 (4) (12) + 5.5 (3)(8) + 7.5(1) (5) + a 8 + p 2 + 9 (1) (5) + 9(1.5) (2) = 405.332 kg # m ' ©ym 405.332 = = 4.37 m y = ©m 92.854

Ans.

Ans: x = 0.200 m y = 4.37 m 935

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9–58. Determine the location y of the centroidal axis x - x of the beam’s cross-sectional area. Neglect the size of the corner welds at A and B for the calculation.

150 mm 15 mm B y 15 mm

150 mm

C

x

SOLUTION

A

' ©yA = 7.5(15) (150) + 90(150) (15) + 215(p) (50)2

50 mm

= 1 907 981.05 mm2 ©A = 15(150) + 150(15) + p(50)2 = 12 353.98 mm2 y =

' ©yA 1 907 981.05 = = 154 mm ©A 12 353.98

Ans.

Ans: y = 154 mm 936

x

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9–59. y

Locate the centroid (x, y) of the shaded area.

6 in.

6 in. x 6 in.

6 in.

Solution Centroid. Referring to Fiq. a, the areas of the segments and the locations of their respective centroids are tabulated below y (in.)

3 x∼A(in. )

∼A(in.3)

0

0

0

0

-4

4

72.0

-72.0

72.0

-72.0

Segment

A(in.2)

x∼(in.)

1

12(12)

2

1 - (6)(6) 2

Σ

126

∼

y

Thus, Σx~A 72.0 in.3 = = 0.5714 in. = 0.571 in. ΣA 126 in.2 Σy~A - 72.0 in.3 = = - 0.5714 in. = - 0.571 in.    y = ΣA 126 in.2

Ans.

   x =

Ans.

Ans: x = 0.571 in. y = - 0.571 in. 937

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*9–60. Locate the centroid y for the beam’s cross-sectional area.

120 mm

240 mm x y

Solution

240 mm

Centroid. The locations of the centroids measuring from the x axis for segments 1  and  2  are indicated in Fig. a. Thus 300(120)(600) + 120(240)(120) Σy~A    y = = ΣA 120(600) + 240(120)

120 mm

240 mm

= 248.57 mm = 249 mm

Ans:  y = 249 mm 938

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9–61. Determine the location y of the centroid C of the beam having the cross-sectional area shown.

150 mm 15 mm B

150 mm

y C

x

15 mm 15 mm

A 100 mm

Solution Centroid. The locations of the centroids measuring from the x axis for segments 1, 2 and 3 are indicated in Fig. a. Thus

y =

7.5(15)(150) + 90(150)(15) + 172.5(15)(100) Σ~ yA = ΣA 15(150) + 150(15) + 15(100) Ans.

= 79.6875 mm = 79.7 mm

Ans: y = 79.7 mm 939

x

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9–62. y

Locate the centroid (x, y) of the shaded area.

6 in. 3 in.

6 in. x

Solution

6 in.

Centroid. Referring to Fig. a, the areas of the segments and the locations of their respective centroids are tabulated below Segment

A(in.2)

x~(in.)

y~(in.)

1

1 (6)(9) 2

2

6

54.0

162.0

2

1 (6)(3) 2

-2

7

-18.0

 63.0

3

6(6)

-3

3

-108.0

108.00

Σ

72.0

-72.0

333.0

x~A(in.3)

y~A(in.3)

Thus, Σx~A - 72.0 in.3 = = - 1.00 in. ΣA 72.0 in.2 Σy~A 333.0 in.3 y = = = 4.625 in. ΣA 72.0 in.2

Ans.

x =

Ans.

Ans: x = - 1.00 in. y = 4.625 in. 940

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9–63. Determine the location y of the centroid of the beam’s crosssectional area. Neglect the size of the corner welds at A and B for the calculation. 35 mm A

SOLUTION

110 mm 2

35 35 ' b = 393 112 mm3 ©yA = p(25)2(25) + 15(110)(50 + 55) + pa b a50 + 110 + 2 2 ©A = p(25)2 + 15(110) + p a

35 2 b = 4575.6 mm2 2

C 15 mm B

y 50 mm

' ©yA 393 112 = = 85.9 mm y = ©A 4575.6

Ans.

Ans: y = 85.9 mm 941

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*9–64. y

Locate the centroid (x, y) of the shaded area.

1 in.

3 in. x

3 in.

3 in.

Solution Centroid. Referring to Fig. a, the areas of the segments and the locations of their respective centroids are tabulated below Segment

A(in.2)

x~(in.)

y~(in.)

1

p 2 (3 ) 4

4 p

4 p

9.00

9.00

2

3(3)

- 1.5

1.5

-13.50

13.50

3

1 (3)(3) 2

-4

1

-18.00

4.50

4

-

0

4 3p

0

-0.67

Σ

18.9978

-22.50

26.33

p 2 (1 ) 2

x~A(in.3)

y~A(in.3)

Thus, Σx~A - 22.50 in.3 = = - 1.1843 in. = -1.18 in. ΣA 18.9978 in.2 Σy~A 26.33 in.3 y = = = 1.3861 in. = 1.39 in. ΣA 18.9978 in.2

x =

Ans. Ans.

Ans: x = - 1.18 in. y = 1.39 in. 942

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9–65. y

Determine the location (x, y) of the centroid C of the area.

1.5 in. 1.5 in.

1.5 in.

1.5 in.

x

Solution

1.5 in.

Centroid. Referring to Fig. a, the areas of the segments and the locations of their respective centroids are tabulated below. Segment

A(in.2)

x~(in.)

y~(in.)

1

3(3)

1.5

1.5

13.5

p (1.52) 4

2 p

2 p

-1.125

1 (1.5)(1.5) 2

2.5

2.5

-2.8125

-2.8125

9.5625

9.5625

-

2 3 Σ

-

6.1079

x~A(in.3)

y~A(in.3) 13.5 -1.125

Thus Σx~A 9.5625 in.3 = = 1.5656 in. = 1.57 in. ΣA 6.1079 in.2 Σy~A 9.5625 in.3 y = = = 1.5656 in. = 1.57 in. ΣA 6.1079 in.2

x =

Ans. Ans.

Ans: x = 1.57 in. y = 1.57 in. 943

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9–66. Determine the location y of the centroid C for a beam having the cross-sectional area shown. The beam is symmetric with respect to the y axis.

y

C y

SOLUTION ' ©yA = 6(4)(2) - 1(1)(0.5) - 3(1)(2.5) = 40 in3

3 in. 1 in.

2 in. 1 in. 2 in. 1 in.

©A = 6(4) - 1(1) - 3(1) = 20 in2 y =

' ©yA 40 = = 2 in. ©A 20

Ans.

Ans: y = 2 in. 944

x

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9–67. Locate the centroid y of the cross-sectional area of the beam constructed from a channel and a plate. Assume all corners are square and neglect the size of the weld at A.

20 mm y 350 mm C A

10 mm 70 mm

SOLUTION

325 mm

325 mm

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (mm2)

' y (mm)

' y A (mm3)

1

350(20)

175

1 225 000

2

630(10)

355

2 236 500

3

70(20)

385

539 000

©

14 700

4 000 500

Thus, ' ©yA 4 000 500 ' y = = = 272.14 mm = 272 mm ©A 14 700

Ans.

Ans: y = 272 mm 945

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*9–68. A triangular plate made of homogeneous material has a constant thickness which is very small. If it is folded over as shown, determine the location y of the plate’s center of gravity G.

z 1 in. 1 in.

SOLUTION

y

1 ©A = (8) (12) = 48 in2 2

G z

6 in.

1 in. 3 in.

1 1 ' ©yA = 2(1) a b (1)(3) + 1.5(6)(3) + 2(2) a b(1)(3) 2 2

3 in. x

= 36 in3 y =

3 in.

' ©yA 36 = = 0.75 in. ©A 48

3 in.

y

1 in.

Ans.

Ans:  y = 0.75 in. 946

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9–69. A triangular plate made of homogeneous material has a constant thickness which is very small. If it is folded over as shown, determine the location z of the plate’s center of gravity G.

z 1 in. 1 in.

SOLUTION ©A =

y

1 (8)(12) = 48 in2 2

G z

6 in.

1 in. 3 in.

1 1 ' © z A = 2(2) a b (2)(6) + 3(6)(2) + 6 a b (2)(3) 2 2

3 in. x

= 78 in3 z =

3 in.

~ 78 ©zA = = 1.625 in. ©A 48

3 in.

y

1 in.

Ans.

Ans: z = 1.625 in. 947

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9–70. Locate the center of mass z of the forked lever, which is made from a homogeneous material and has the dimensions shown.

z

0.5 in.

3 in. 2 in.

SOLUTION

G

1 1 ©A = 2.5(0.5) + c p (2.5)2 - p (2)2 d + 2 [(3)(0.5)] = 7.7843 in2 2 2 4(2.5) 1 2.5 ' © zA = (2.5) (0.5) + a 5 b a p (2.5)2 b 2 3p 2 - a5 z =

2.5 in.

4(2) 1 b a p (2)2 b + 6.5(2)(3)(0.5) = 33.651 in3 3p 2

' 33.651 © zA = = 4.32 in. ©A 7.7843

x

z

0.5 in.

Ans.

Ans: z = 4.32 in. 948

y

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9–71. Determine the location x of the centroid C of the shaded area which is part of a circle having a radius r.

y

r a

SOLUTION

x

2r 1 1 2 ' ©xA = r 2 aa sin a b - (r sin a) (r cos a) a r cos ab 2 3a 2 3 r3 r3 sin a sin a cos2 a 3 3

=

r3 3 sin a 3

©A = =

x

a

Using symmetry, to simplify, consider just the top half:

=

C

1 1 2 r a - (r sin a) (r cos a) 2 2 1 2 sin2a r aa b 2 2

' ©xA = x = ©A

r3 3

1 2

sin3 a

r2 A a -

sin 2 a 2

B

=

2 3

r sin3a

a -

Ans.

sin 2a 2

Ans: x =

949

2 3r

sin3 a

a -

sin 2a 2

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*9–72. A toy skyrocket consists of a solid conical top, rt = 600 kg>m3, a hollow cylinder, rc = 400 kg>m3, and a stick having a circular cross section, rs = 300 kg>m3. Determine the length of the stick, x, so that the center of gravity G of the skyrocket is located along line aa.

a 3 mm

5 mm

100 mm

20 mm

10 mm a G

x

SOLUTION 20 1 x ' ©xm = a b c a b p (5)2 (20) d (600) - 50 C p A 52 - 2.52 B (100) D (400) - C (x) p (1.5)2 D (300) 4 3 2 = - 116.24 A 106 B - x2(1060.29) kg # mm4>m3 1 ©m = c p (5)2 (20) d (600) + p A 52 - 2.52 B (100)(400) + C xp (1.5)2 D (300) 3 = 2.670 A 106 B + 2120.58x kg # mm3/m3 x =

' -116.24(106) - x2(1060.29) ©xm = - 100 = ©m 2.670(106) + 2120.58x

- 116.24 A 106 B - x2 (1060.29) = - 267.0 A 106 B - 212.058 A 103 B x 1060.29x2 - 212.058 A 103 B x - 150.80 A 106 B = 0

Solving for the positive root gives Ans.

x = 490 mm

Ans:  x = 490 mm 950

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9–73. Locate the centroid y for the cross-sectional area of the angle. a

a

–y

C

SOLUTION

t

t

Centroid : The area and the centroid for segments 1 and 2 are A 1 = t1a - t2 t t 22 a - t ' + b cos 45° + = 1a + 2t2 y1 = a 2 2 2cos 45° 4 A 2 = at t t 22 a ' y2 = a - b cos 45° + = 1a + t2 2 2 2cos 45° 4 Listed in a tabular form, we have Segment

A

' y

' yA

1

t1a - t2

22 1a + 2t2 4

22t 2 1a + at - 2t22 4

2

at

22 1a + t2 4

22t 2 1a + at2 4

©

t12a - t2

22t 2 1a + at - t22 2

Thus, ' ©yA y = = ©A =

22t 2 1a + at - t22 2 t12a - t2 22 a2 + at - t2

Ans.

2 2a - t

Ans: y = 951

22 ( a2 + at - t 2 ) 2(2a - t)

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9–74. Determine the location (x, y) of the center of gravity of the three-wheeler. The location of the center of gravity of each component and its weight are tabulated in the figure. If the three-wheeler is symmetrical with respect to the x–y plane, determine the normal reaction each of its wheels exerts on the ground.

1. 2. 3. 4.

y

Rear wheels 18 lb Mechanical components 85 lb Frame 120 lb Front wheel 8 lb

3 2

SOLUTION ' ©xW = 4.51182 + 2.31852 + 3.111202

4

1 ft

= 648.5 lb # ft

1

2 ft

A

B 2.30 ft

©W = 18 + 85 + 120 + 8 = 231 lb x =

1.50 ft

1.30 ft x

1.40 ft 0.80 ft

' 648.5 ©xW = = 2.81 ft ©W 231

Ans.

' ©yW = 1.301182 + 1.51852 + 211202 + 1182 = 398.9 lb # ft y = a + ©MA = 0;

' ©yW 398.9 = = 1.73 ft ©W 231

Ans.

21NB214.52 - 23112.812 = 0 Ans.

NB = 72.1 lb + c ©Fy = 0;

NA + 2172.12 - 231 = 0 Ans.

NA = 86.9 lb

Ans: x = 2.81 ft y = 1.73 ft NB = 72.1 lb NA = 86.9 lb 952

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9–75. Locate the center of mass (x, y, z) of the homogeneous block assembly.

z

250 mm 200 mm

x

SOLUTION

100 mm

150 mm

Centroid: Since the block is made of a homogeneous material, the center of mass of the block coincides with the centroid of its volume. The centroid of each composite segment is shown in Fig. a.

150 mm 150 mm

1 (75)(150)(150)(550) + (225)(150)(150)(200) + (200) a b(150)(150)(100) ' 2.165625(109) 2 ©xV = 120 mm x = = = ©V 1 18(106) (150)(150)(550) + (150)(150)(200) + (150)(150)(100) 2

Ans.

1 (275)(150)(150)(550) + (450)(150)(150)(200) + (50) a b(150)(150)(100) ' ©yV 5.484375(109) 2 = 305 mm y = = = ©V 1 18(106) (150)(150)(550) + (150)(150)(200) + (150)(150)(100) 2

Ans.

1 (75)(150)(150)(550) + (75)(150)(150)(200) + (50) a b(150)(150)(100) ' 1.321875(109) 2 © zV z = = 73.4 mm = = ©V 1 18(106) (150)(150)(550) + (150)(150)(200) + (150)(150)(100) 2

Ans.

y

Ans: x = 120 mm y = 305 mm z = 73.4 mm 953

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*9–76. The sheet metal part has the dimensions shown. Determine the location 1x, y, z2 of its centroid.

z

D

3 in. C

SOLUTION

A

©A = 4(3) +

4 in. B

1 (3)(6) = 21 in2 2

x

1 ' ©xA = - 2(4)(3) + 0 a b (3)(6) = - 24 in3 2

y 6 in.

2 1 ' ©yA = 1.5(4)(3) + (3) a b (3)(6) = 36 in3 3 2 1 1 ' © z A = 0(4)(3) - (6) a b (3)(6) = - 18 in3 3 2 x =

' - 24 ©xA = = - 1.14 in. ©A 21

Ans.

y =

' ©yA 36 = = 1.71 in. ©A 21

Ans.

z =

' - 18 © zA = = - 0.857 in. ©A 21

Ans.

Ans: x = - 1.14 in. y = 1.71 in. z = - 0.857 in. 954

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9–77. The sheet metal part has a weight per unit area of 2 lb>ft2 and is supported by the smooth rod and at C. If the cord is cut, the part will rotate about the y axis until it reaches equilibrium. Determine the equilibrium angle of tilt, measured downward from the negative x axis, that AD makes with the -x axis.

z

D

3 in. C

SOLUTION

A

Since the material is homogeneous, the center of gravity coincides with the centroid. See solution to Prob. 9-74. u = tan

4 in. B

x

y 6 in.

-1

1.14 b = 53.1° a 0.857

Ans.

Ans: u = 53.1° 955

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9–78. The wooden table is made from a square board having a weight of 15 lb. Each of the legs weighs 2 lb and is 3 ft long. Determine how high its center of gravity is from the floor. Also, what is the angle, measured from the horizontal, through which its top surface can be tilted on two of its legs before it begins to overturn? Neglect the thickness of each leg.

4 ft 4 ft

3 ft

SOLUTION z =

' 15(3) + 4(2) (1.5) © zW = = 2.48 ft ©W 15 + 4(2)

u = tan - 1 a

Ans.

2 b = 38.9° 2.48

Ans.

Ans: z = 2.48 ft u = 38.9° 956

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9–79. The buoy is made from two homogeneous cones each having a radius of 1.5 ft. If h = 1.2 ft, find the distance z to the buoy’s center of gravity G.

h 1.5 ft z 4 ft

G

SOLUTION 1 1 1.2 4 ' b + p(1.5)2 (4) a b © z V = p (1.5)2 (1.2) a3 4 3 4 = 8.577 ft4 ©V =

1 1 p (1.5)2 (1.2) + p(1.5)2 (4) 3 3

= 12.25 ft3 ' © zV 8.577 z- = = = 0.70 ft ©V 12.25

Ans.

Ans: z = 0.70 ft 957

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*9–80. The buoy is made from two homogeneous cones each having a radius of 1.5 ft. If it is required that the buoy’s center of gravity G be located at z = 0.5 ft, determine the height h of the top cone.

h 1.5 ft z 4 ft

G

SOLUTION 1 1 h 4 ' © z V = p (1.5)2 (h) a- b + p(1.5)2 (4) a b 3 4 3 4 = -0.5890 h2 + 9.4248 ©V =

1 1 p (1.5)2 (h) + p(1.5)2 (4) 3 3

= 2.3562 h + 9.4248 ' - 0.5890 h2 + 9.4248 © zV ' = = 0.5 z = ©V 2.3562 h + 9.4248 -0.5890 h2 + 9.4248 = 1.1781 h + 4.7124 Ans.

h = 2.00 ft

Ans: h = 2.00 ft 958

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9–81. The assembly is made from a steel hemisphere, rst = 7.80 Mg>m3, and an aluminum cylinder, ral = 2.70 Mg>m3. Determine the mass center of the assembly if the height of the cylinder is h = 200 mm.

z 80 mm

G _ z

SOLUTION © z m = C 0.160 - 38 (0.160) D A 23 B p(0.160)3 (7.80) + A 0.160 +

0.2 2

©m =

3

160 mm y

B p(0.2)(0.08)2(2.70)

= 9.51425(10 - 3) Mg # m 2 3

h

x 2

A B p(0.160) (7.80) + p (0.2)(0.08) (2.70)

= 77.7706(10 - 3) Mg z =

9.51425(10 - 3) © zm = = 0.122 m = 122 mm ©m 77.7706(10 - 3)

Ans.

Ans: z = 122 mm 959

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9–82. The assembly is made from and an rst = 7.80 Mg>m3, ral = 2.70 Mg>m3. Determine the so that the mass center of the z = 160 mm.

a steel hemisphere, aluminum cylinder, height h of the cylinder assembly is located at

z 80 mm

G _ z

SOLUTION © z m = C 0.160 - 38 (0.160) D A 23 B p(0.160)3 (7.80) + A 0.160 +

h 2

©m =

3

160 mm y

B p (h)(0.08)2(2.70)

= 6.691(10 - 3) + 8.686(10 - 3) h + 27.143(10 - 3) h2 2 3

h

x

2

A B p(0.160) (7.80) + p (h)(0.08) (2.70)

= 66.91(10 - 3) + 54.29(10 - 3) h z =

6.691(10 - 3) + 8.686(10 - 3) h + 27.143(10 - 3) h2 © zm = 0.160 = ©m 66.91(10 - 3) + 54.29(10 - 3) h

Solving Ans.

h = 0.385 m = 385 mm

Ans: h = 385 mm 960

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9–83. The car rests on four scales and in this position the scale readings of both the front and rear tires are shown by FA and FB. When the rear wheels are elevated to a height of 3 ft above the front scales, the new readings of the front wheels are also recorded. Use this data to compute the location x and y to the center of gravity G of the car. The tires each have a diameter of 1.98 ft.

G

_ y

B

A _ x

FB

975 lb

SOLUTION In horizontal position

3.0 ft

B

984 lb

9.40 ft FA 1959 lb

1129 lb

1168 lb

G

W = 1959 + 2297 = 4256 lb a + ©MB = 0;

A

2297(9.40) - 4256 x = 0

FA

1269 lb

1307 lb

2576 lb

Ans.

x = 5.0733 = 5.07 ft u = sin - 1 a

2297 lb

3 - 0.990 b = 12.347° 9.40

With rear whells elevated a + ©MB = 0;

2576(9.40 cos 12.347°) - 4256 cos 12.347°(5.0733) - 4256 sin 12.347° y¿ = 0 y¿ = 2.86 ft Ans.

y = 2.815 + 0.990 = 3.80 ft

Ans: x = 5.07 ft y = 3.80 ft 961

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*9–84. z

Determine the distance h to which a 100-mm diameter hole must be bored into the base of the cone so that the center of mass of the resulting shape is located at z = 115 mm. The material has a density of 8 Mg>m3.

1 0.5 2 3 p10.152 10.52 4 1 2 3 p10.152 10.52

A B - p10.05221h2 A h2 B - p10.05221h2

= 0.115

500 mm

0.4313 - 0.2875 h = 0.4688 - 1.25 h2

C

h 50 mm

h2 - 0.230 h - 0.0300 = 0

_ z 150 mm

Choosing the positive root, Ans.

h = 323 mm

x

Ans: h = 323 mm 962

y

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9–85. Determine the distance z to the centroid of the shape which consists of a cone with a hole of height h = 50 mm bored into its base.

z

500 mm

SOLUTION 1 0.5 0.05 ' b - p (0.05)2 (0.05) a b © z V = p (0.15)2 A 0.5 B a 3 4 2 -3

C

h 50 mm

4

= 1.463(10 ) m ©V =

1 p (0.15)2 (0.5) - p (0.05)2 (0.05) 3

z

y

150 mm

x

3

= 0.01139 m z =

' 1.463 (10 - 3) ©z V = = 0.12845 m = 128 mm ©V 0.01139

Ans.

Ans: z = 128 mm 963

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9–86. Locate the center of mass z of the assembly. The cylinder and the cone are made from materials having densities of 5 Mg>m3 and 9 Mg>m3, respectively.

z

0.6 m

0.4 m

SOLUTION Center of mass: The assembly is broken into two composite segments, as shown in Figs. a and b.

z =

=

' © zm = ©m

0.2 m

1 5000(0.4) C p(0.2 2)(0.8) D + 9000(0.8 + 0.15)c p(0.4 2)(0.6) d 3

x

1 5000 C p(0.2 2)(0.8) D + 9000c p(0.4 2)(0.6) d 3

1060.60 = 0.754 m = 754 mm 1407.4

0.8 m

y

Ans.

Ans: z = 754 mm 964

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9–87. Major floor loadings in a shop are caused by the weights of the objects shown. Each force acts through its respective center of gravity G. Locate the center of gravity (x, y) of all these components.

z y 450 lb 1500 lb G2

G1 9 ft 6 ft

SOLUTION Centroid: The floor loadings on the floor and its respective centroid are tabulated below. Loading

W (lb)

x (ft)

y (ft)

xW(lb # ft)

yW(lb # ft)

1 2 3 4

450 1500 600 280

6 18 26 30

7 16 3 8

2700 27000 15600 8400

3150 24000 1800 2240

©

2830

53700

31190

600 lb

7 ft

280 lb

G3

G4

4 ft

5 ft 3 ft

12 ft 8 ft

x

Thus, x =

©xW 53700 = = 18.98 ft = 19.0 ft ©W 2830

Ans.

y =

©yW 31190 = = 11.02 ft = 11.0 ft ©W 2830

Ans.

Ans: x = 19.0 ft y = 11.0 ft 965

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*9–88. The assembly consists of a 20-in. wooden dowel rod and a tight-fitting steel collar. Determine the distance x to its center of gravity if the specific weights of the materials are gw = 150 lb>ft3 and gst = 490 lb>ft3. The radii of the dowel and collar are shown.

5 in.

5 in. 10 in. G

x

x

2 in.

1 in.

SOLUTION ©xW =

E 10p (1)2 (20)(150) + 7.5p (5)(2 2 - 12)(490) F

1 (12)3

= 154.8 lb # in. ©W =

E p (1)2 (20)(150) + p (5)(2 2 - 12)(490) F

1 (12)3

= 18.82 lb x =

©xW 154.8 = = 8.22 in. ©W 18.82

Ans.

Ans: x = 8.22 in. 966

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9–89. The composite plate is made from both steel (A) and brass (B) segments. Determine the mass and location 1x, y, z2 of its mass center G. Take rst = 7.85 Mg>m3 and rbr = 8.74 Mg>m3.

z

A

225 mm

G 150 mm B 150 mm 30 mm

SOLUTION

x

1 1 ©m = ©rV = c 8.74 a (0.15)(0.225)(0.03)b d + c 7.85a (0.15)(0.225)(0.03) b d 2 2 + [7.85(0.15)(0.225)(0.03)] = C 4.4246 A 10 - 3 B D + C 3.9741 A 10 - 3 B D + C 7.9481 A 10 - 3 B D = 16.347 A 10 - 3 B = 16.4 kg

2 1 (0.150) b (4.4246) A 10 - 3 B + a 0.150 + (0.150) b (3.9741) A 10 - 3 B 3 3

©xm = a0.150 + +

Ans.

1 (0.150)(7.9481)(10 -3) = 2.4971(10 -3) kg # m 2

1 2 0.225 b(7.9481) A 10 - 3 B ©zm = a (0.225) b(4.4246) A 10 - 3 B + a (0.225)b (3.9471) A 10 - 3 B + a 3 3 2 = 1.8221 A 10 - 3 B kg # m

x =

2.4971(10 - 3) ©xm = = 0.153 m = 153 mm ©m 16.347(10 - 3)

Ans.

Due to symmetry: Ans.

y = - 15 mm z =

1.8221(10 - 3) ©zm = 0.1115 m = 111 mm = ©m 16.347(10 - 3)

Ans.

Ans: Σm = 16.4 kg x = 153 mm y = - 15 mm z = 111 mm 967

y

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9–90. Determine the volume of the silo which consists of a cylinder and hemispherical cap. Neglect the thickness of the plates.

10 ft 10 ft 10 ft

SOLUTION V = ©u r A = 2p c

80 ft

4(10) 1 a b p (10)2 + 5(80)(10) d 4 3p

= 27.2 A 103 B ft3

Ans.

Ans: V = 27.2 ( 103 ) ft 3 968

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9–91. Determine the outside surface area of the storage tank.

15 ft 4 ft

30 ft

SOLUTION Surface Area: Applying the theorem of Pappus and Guldinus, Eq.9–7. with u = 2p, L1 = 2152 + 4 2 = 2241 ft, L2 = 30 ft, r1 = 7.5 ft and r2 = 15 ft, we have A = u©r~L = 2p C 7.5 A 2241 B + 15(30) D = 3.56 A 103 B ft2

Ans.

Ans: A = 3.56 ( 103 ) ft 2 969

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*9–92. Determine the volume of the storage tank.

15 ft 4 ft

30 ft

SOLUTION Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–8 with u = 2p, 1 r1 = 5 ft, r2 = 7.5 ft, A 1 = (15)(4) = 30.0 ft2 and A 2 = 30(15) = 450 ft2, we have 2 V = u©rA = 2p[5(30.0) + 7.5(450)] = 22.1 A 103 B ft3

Ans.

Ans: V = 22.1 ( 103 ) ft 3 970

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9–93. Determine the surface area of the concrete sea wall, excluding its bottom. 8 ft

SOLUTION 30 ft

Surface Area: Applying Theorem of Pappus and Guldinus, Eq. 9–9 with 5 50 bp = p rad, L1 = 30 ft, L2 = 8 ft, L3 = 272 + 302 = 2949 ft, u = a 180 18 N1 = 75 ft, N2 = 71 ft and N3 = 63.5 ft as indicated in Fig. a, A 1 = u©NL =

60 ft 50

15 ft

5 p [75(30) + 71(8) + 63.5( 2949)] 18

= 4166.25 ft2 The surface area of two sides of the wall is 1 A 2 = 2 c (8 + 15)(30) d = 690 ft2 2 Thus the total surface area is A = A 1 + A 2 = 4166.25 + 690 = 4856.25 ft2 = 4856 ft2

Ans.

Ans: A = 4856 ft2 971

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9–94. A circular sea wall is made of concrete. Determine the total weight of the wall if the concrete has a specific weight of gc = 150 lb>ft3.

8 ft

30 ft

SOLUTION V = ©ur~ A = a

60 ft 50

2 1 50° b p[a 60 + (7)b a b (30) (7) + 71(30)(8)] 180° 3 2

15 ft

= 20 795.6 ft3 W = gV = 150(20 795.6) = 3.12(106) lb

Ans.

Ans: W = 3.12 ( 106 ) lb 972

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9–95. A ring is generated by rotating the quartercircular area about the x axis. Determine its volume. a

SOLUTION

2a

Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–10, with u = 2p, 4a p 6p + 4 r = 2a + = a and A = a2, we have 3p 3p 4 V = urA = 2p

6p + 4 a 3p

p 2 a 4

=

p(6p + 4) 3 a 6

Ans.

x

Ans: V = 973

p(6p + 4) 6

a3

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*9–96. A ring is generated by rotating the quartercircular area about the x axis. Determine its surface area. a

SOLUTION

2a

Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–11, with u = 2p , 2(p + 1) pa 5 , r1 = 2a, r 2 = L1 = L3 = a, L2 = a and r3 = a, we have p 2 2 A = u©rL = 2p 2a(a) +

2(p + 1) pa 5 a a b + a (a) p 2 2

x

= p(2p + 11)a2

Ans.

Ans: A = p(2p + 11)a2 974

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9–97. Determine the volume of concrete needed to construct the curb. 100 mm 150 mm 30

4m

150 mm 150 mm

Solution p p 1 V = Σu A r = a b[(0.15)(0.3)(4.15)] + a b c a b(0.15)(0.1)(4.25)d 6 6 2 V = 0.114 m3



Ans.

Ans: V = 0.114 m3 975

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9–98. Determine the surface area of the curb. Do not include the area of the ends in the calculation. 100 mm 150 mm 30

4m

150 mm 150 mm

Solution A = ΣurL =

p {4(0.15) + 4.075(0.15) + (4.15 + 0.075) ( 20.152 + 0.12 ) 6 + 4.3(0.25) + 4.15(0.3)}

A = 2.25 m2

Ans.

Ans: A = 2.25 m2 976

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9–99. A ring is formed by rotating the area 360° about the x – x axes. Determine its surface area. 80 mm

30 mm

50 mm

30 mm

100 mm x

x

Solution Surface Area. Referring to Fig. a, L1 = 110 mm, L2 = 2302 + 802 = 27300 mm L3 = 50 mm, r1 = 100 mm, r2 = 140 mm and r3 = 180 mm. Applying the theorem of pappus and guldinus, with u = 2p rad,

A = uΣrL



= 2p 3 100(110) + 2 (140) ( 17300 2 + 180 (50) 4



= 276 ( 103 ) mm2

= 275.98 ( 103 ) mm2

Ans.

Ans: A = 276(103) mm2 977

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*9–100. A ring is formed by rotating the area 360° about the x – x axes. Determine its volume. 80 mm

30 mm

50 mm

30 mm

100 mm x

x

Solution

1 (60)(80) = 2400 mm2, A2 = 50(80) = 4000 mm2, 2 r1 = 126.67 mm and r2 = 140 mm. Applying the theorem of pappus and guldinus,

Volume. Referring to Fig. a, A1 = with u = 2p rad,

V = uΣrA = 2p 3 126.67(2400) + 140(4000) 4



= 5.429 ( 106 ) mm3



= 5.43 ( 106 ) mm3

Ans.

Ans: V = 5.43 ( 106 ) mm3 978

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9–101. The water-supply tank has a hemispherical bottom and cylindrical sides. Determine the weight of water in the tank when it is filled to the top at C. Take gw = 62.4 lb>ft3.

6 ft C 8 ft

6 ft

SOLUTION ~ = 2p e 3182162 + V = ©urA

4162 3p

1 a b 1p21622 f 4

V = 1357.17 ft3 Ans.

W = g V = 62.411357.172 = 84.7 kip

Ans: W = 84.7 kip 979

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9–102. Determine the number of gallons of paint needed to paint the outside surface of the water-supply tank, which consists of a hemispherical bottom, cylindrical sides, and conical top. Each gallon of paint can cover 250 ft2.

6 ft C 8 ft

6 ft

SOLUTION ~ = 2p e 3 A 6 22 B + 6182 + A = ©urL

2162 p

a

2162p 4

bf

= 687.73 ft2 Number of gal. =

687.73 ft2 = 2.75 gal. 250 ft2>gal.

Ans.

Ans: Number of gal. = 2.75 gal 980

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9–103. Determine the surface area and the volume of the ring formed by rotating the square about the vertical axis.

b

a

a

SOLUTION ~ = 2c 2p a b A = ©urL

= 4p cba -

45

a a sin 45°b(a) d + 2 c 2p a b + sin 45°b(a) d 2 2

a2 a2 sin 45° + ba + sin 45° d 2 2 Ans.

= 8pba Also A = ©urL = 2p(b)(4a) = 8pba ~ = 2p(b)(a)2 = 2pba 2 V = ©urA

Ans.

Ans: A = 8pba V = 2pba2 981

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*9–104. Determine the surface area of the ring. T he cross section is circular as shown.

8 in. 4 in.

SOLUTION ' A = ur L = 2p (3) 2p (1) = 118 in.2

Ans.

Ans: A = 118 in.2 982

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9–105. The heat exchanger radiates thermal energy at the rate of 2500 kJ h for each square meter of its surface area. Determine how many joules (J) are radiated within a 5-hour period.

0.5 m

0.75 m

0.75 m

SOLUTION A = ©u r L = (2p) B 2 a

0.75 + 0.5 b 2(0.75)2 + (0.25)2 + (0.75)(1.5) + (0.5)(1) R 2

0.75 m

= 16.419 m2 Q = 2500 A 103 B a

1.5 m

1m

h

#

J m2

b A 16.416 m2 B (5 h) = 205 MJ

Ans. 0.5 m

Ans: Q = 205 MJ 983

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9–106. Determine the interior surface area of the brake piston. It consists of a full circular part. Its cross section is shown in the figure.

40 mm 60 mm 80 mm

SOLUTION

20 mm

A = © u r L = 2 p [20(40) + 552(30)2 + (80)2 + 80(20) + 90(60) + 100(20) + 110(40)]

40 mm 30 mm 20 mm

A = 119(103) mm2

Ans.

Ans: A = 119 ( 103 ) mm2 984

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9–107. The suspension bunker is made from plates which are curved to the natural shape which a completely flexible membrane would take if subjected to a full load of coal. This curve may be approximated by a parabola, y = 0.2x2. Determine the weight of coal which the bunker would contain when completely filled. Coal has a specific weight of g = 50 lb>ft3, and assume there is a 20% loss in volume due to air voids. Solve the problem by integration to determine the cross-sectional area of ABC; then use the second theorem of Pappus–Guldinus to find the volume.

y 10 ft C B 20 ft

SOLUTION y  0.2x2

x x = 2

A

x

' y = y dA = x dy 20 y 3 2 dy = y2 2 = 133.3 ft2 0 L0 C 0.2 3 20.2 20

LA

dA =

20

LA

x dA =

x =

L0

y y2 20 2 = 500 ft3 dy = 0.4 0.8 0

x dA LA LA

dA

=

500 = 3.75 ft 133.3

V = u r A = 2p (3.75) (133.3) = 3142 ft3 W = 0.8 g V = 0.8(50)(3142) = 125 664 lb = 126 kip

Ans.

Ans: W = 126 kip 985

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*9–108. Determine the height h to which liquid should be poured into the cup so that it contacts three-fourths the surface area on the inside of the cup. Neglect the cup’s thickness for the calculation.

40 mm

160 mm h

Solution Surface Area. From the geometry shown in Fig. a,

r 40 = ; h 160

r =

1 h 4

1 217 1 2 2 h, Fig. b. Applying the theorem of pappus and h and L = a hb + h = A 4 8 4 Guldinus, with u = 2p rad,

Thus, r =

1 217 p217 2 A = u Σr L = 2p a hba hb = h 8 4 16

For the whole cup, h = 160 mm. Thus Ao = a

It is required that A =

p217 b 1 1602 2 = 1600p217 mm2 16

3 3 A = 1 1600p217 2 = 1200p217 mm2. Thus 4 o 4 1200p217 =



p217 2 h 16

Ans.

h = 138.56 mm = 139 mm

Ans: h = 139 mm 986

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9–109. Determine the surface area of the roof of the structure if it is formed by rotating the parabola about the y axis.

y y

16

(x2/16)

16 m

SOLUTION Centroid: The = ¢

C

1 + a

x

length

of

the

differential

element

is

dL = 2dx2 + dy 2

16 m

dy dy x b ≤ dx and its centroid is x = x. Here, = - . Evaluating the dx dx 8 2

integrals, we nave 16 m

L =

LL

L

dL =

' xdL =

L0 16 m

' x¢

L0

¢

C

C

1 +

1 +

x2 ≤ dx = 23.663 m 64

x2 ≤ dx = 217.181 m2 64

Applying Eq. 9–5, we have

x =

LL

' xdL

LL

= dL

217.181 = 9.178 m 23.663

Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–7, with u = 2p, L = 23.663 m, r = x = 9.178, we have A = urL = 2p(9.178) (23.663) = 1365 m2

Ans.

Ans: A = 1365 m2 987

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9–110. A steel wheel has a diameter of 840 mm and a cross section as shown in the figure. Determine the total mass of the wheel if r = 5 Mg>m3.

100 mm A

30 mm

60 mm 420 mm 250 mm 30 mm 840 mm 80 mm

SOLUTION Volume: Applying the theorem of Pappus and Guldinus, Eq. 9–12, with u = 2p, r1 = 0.095 m, r2 = 0.235 m, r3 = 0.39 m, A 1 = 0.110.032 = 0.003 m2, 2 A 2 = 0.2510.032 = 0.0075 m and A 3 = 10.1210.062 = 0.006 m2, we have

A

Section A–A

V = u©rA = 2p30.09510.0032 + 0.23510.00752 + 0.3910.00624 = 8.775p110-32m3 The mass of the wheel is m = rV = 51103238.775110-32p4 Ans.

= 138 kg

Ans: m = 138 kg 988

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9–111. Half the cross section of the steel housing is shown in the figure. There are six 10-mm-diameter bolt holes around its rim. Determine its mass. The density of steel is 7.85 Mg m3. The housing is a full circular part.

30 mm

20 mm 40 mm

10 mm 30 mm 10 mm

10 mm

10 mm

SOLUTION V = 2p[ (40)(40)(10) + (55)(30)(10) + (75)(30)(10)] - 6 C p (5)2(10) D = 340.9 A 103 B mm3 m = rV = a7850 = 2.68 kg

kg m3

b (340.9) A 103 B A 10-9 B m3 Ans.

Ans: m = 2.68 kg 989

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*9–112. y

The water tank has a paraboloid-shaped roof. If one liter of paint can cover 3 m2 of the tank, determine the number of liters required to coat the roof.

1 (144  x2) y  –– 96 x 2.5 m

SOLUTION Length and Centroid: The length of the differential element shown shaded in Fig. a is dL = 2dx2 + dy2 = where

dy 1 = - x. Thus, dx 48 dL =

B

12 m

LL

dL =

1 + a

dy 2 b dx dx

2 x2 1 1 x b dx = 1 + 2 dx = 2482 + x2 dx 48 B 48 48

1 + a-

Integrating, L =

A

12 m

L0

1 2482 + x2 dx = 12.124 m 48

The centroid x of the line can be obtained by applying Eq. 9–5 with xc = x.

x =

LL

12 m

x~ dL =

L0

xc

1 2482 + x2 dx d 48 73.114 = = 6.031 m 12.124 12.124

dL LL Surface Area: Applying the first theorem of Pappus and Guldinus and using the results obtained above with r = x = 6.031 m, we have A = 2prL = 2p(6.031)(12.124) = 459.39 m2 Thus, the amount of paint required is # of liters =

459.39 = 153 liters 3

Ans.

Ans: 153 liters 990

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9–113. Determine the volume of material needed to make the casting.

2 in.

6 in.

SOLUTION

6 in.

Side View

4 in.

Front View

V = ©uAy 4(6) 4(2) 1 1 b + 2(6)(4) (3) - 2 a pb (2)2 a6 bd = 2 p c2a pb(6)2 a 4 3p 2 3p = 1402.8 in3 V = 1.40(103) in3

Ans.

Ans: V = 1.40 ( 10 3 ) in3 991

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9–114. Determine the height h to which liquid should be poured into the cup so that it contacts half the surface area on the inside of the cup. Neglect the cup’s thickness for the calculation.

30 mm 50 mm h

SOLUTION

10 mm

A = uz ~ rL = 2p{202(20)2 + (50)2 + 5(10)} = 2p(1127.03) mm2 x =

20h 2h = 50 5

2p b 5(10) + a 10 +

2h 2 h 1 b a b + h2 r = (2p)(1127.03) 5 B 5 2

10.77h + 0.2154h2 = 513.5 Ans.

h = 29.9 mm

Ans: h = 29.9 mm 992

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9–115. p

The pressure loading on the plate varies uniformly along each of its edges. Determine the magnitude of the resultant force and the coordinates ( x , y) of the point where the line of action of the force intersects the plate. Hint: The equation defining the boundary of the load has the form p ax by c, where the constants a, b, and c have to be determined.

40 lb/ft

30 lb/ft

20 lb/ft 10 lb/ft

SOLUTION

y 5 ft

10 ft

x

p = ax + by + c At x = 0,

y = 0;

p = 40

40 = 0 + 0 + c; At x = 5,

c = 40

y = 0,

p = 30

30 = a (5) + 0 + 40; At x = 0;

a = -2

y = 10,

p = 20

20 = 0 + b (10) + 40;

b = -2

Thus, p = - 2x - 2y + 40 5

FR =

p(x,y)dA =

LA

10

L0 L0

( -2x - 2y + 40) dy dx

= - 2 A 12(5)2 B (10) - 2 A 12(10)2 B 5 + 40(5)(10) Ans.

= 1250 lb 5

xp(x,y) dA =

LA

10

2

( -2x - 2 yx + 40 x) dy dx

L0 L0

= - 2 A 13(5)2 B (10) - 2 A 12 (10)2 B A 12(5)2 B + 40 A 12(5)2 B (10) = 2916.67 lb # ft

x =

xp(x,y) dA LA

=

p(x,y)dA

2916.67 = 2.33 ft 1250

Ans.

LA

5

yp(x,y) dA =

LA

10

L0 L0

( -2x y - 2y2 + 40y) dy dx

= - 2 A 12(5)2 B A 12(10)2 B - 2 A 13(10)3 B (5) + 40(5) A 12(10)2 B = 5416.67 lb # ft

y =

yp(x,y) dA LA p(x,y) dA LA

=

5416.67 = 4.33 ft 1250

Ans. Ans: FR = 1250 lb x = 2.33 ft y = 4.33 ft 993

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*9–116. p

The load over the plate varies linearly along the sides of the plate such that p = (12 - 6x + 4y) kPa. Determine the magnitude of the resultant force and the coordinates ( x, y ) of the point where the line of action of the force intersects the plate.

12 kPa 18 kPa

x 1.5 m

6 kPa 2m

Solution Centroid. Perform the double integration. FR =

LA

r(x, y)dA = = =

L0

1.5 m

L0

1.5 m

L0

1.5 m

L0

2m

(12 -6x + 4y)dxdy

1 12x

dy

0

(8y + 12)dy

= (4y2 + 12y) `



2m

- 3x2 + 4xy 2 ` 1.5 m 0

Ans.

= 27.0 kN LA

xr(x, y)dA = = = =

L0

1.5 m

L0

1.5 m

L0

1.5 m

L0

2m

1 12x

(6x2 - 2x3 + 2x2y) `

LA

= =

2m

dy

0

(8y + 8)dy

1 4y2

+ 8y 2 `

= 21.0 kN # m yr(x, y)dA =

- 6x2 + 4xy 2 dx dy

L0

1.5 m

L0

1.5 m

L0

1.5 m

L0

2m

1.5 m 0

1 12y

1 12xy

- 6xy + 4y2 2 dx dy

- 3x2y + 4xy2 2 `

(8y2 + 12y)dy

1.5 m 8 = a y3 + 6y2 b ` 3 0

= 22.5 kN # m

994

2m

dy

0

y

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*9–116.  Continued

Thus,



x =

y =

LA

xp(x, y)dA

LA LA

=

21.0 kN # m 7 = m = 0.778 m  27.0 kN 9

Ans.

=

22.5 kN # m = 0.833 m  27.0 kN

Ans.

p(x, y)dA

yp(x, y)dA

LA

p(x, y)dA

Ans: FR = 27.0 kN x = 0.778 m y = 0.833 m 995

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9–117. The load over the plate varies linearly along the sides of the plate such that p = 23[x14 - y2] kPa. Determine the resultant force and its position 1x, y2 on the plate.

p 8 kPa

y 3m

x

4m

SOLUTION Resultant Force and its Location: The volume of the differential element is 2 ' ' dV = d FR = pdxdy = (xdx)[(4 - y)dy] and its centroid is at x = x and y = y. 3 3m

FR =

LFk =

dFR =

LFR

= ' ydFR =

LFR

=

x =

L0

2 2 (x dx) 3 L0

(4 - y)dy Ans.

4m

(4 - y)dy

y2 4 m x3 3 m 2 b 2 R = 48.0 kN # m a 4y Ba b 2 3 3 0 2 0 3m

L0

2 (xdx) 3 L0

4m

y(4 - y)dy

y3 4 m 2 x2 3 m B a b 2 a 2y2 - b 2 R = 32.0 kN # m 3 2 0 3 0 ' xdFR LFR

=

48.0 = 2.00 m 24.0

Ans.

=

32.0 = 1.33 m 24.0

Ans.

dFR LFR ' ydFR

y =

4m

y2 4 m x2 3 m 2 a 4y b 2 R = 24.0 kN Ba b 2 3 2 0 2 0 3m

xdFR =

L0

2 (xdx) 3 L0

LFR

dFR

LFR

Ans: FR = 24.0 kN x = 2.00 m y = 1.33 m 996

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9–118. p

The rectangular plate is subjected to a distributed load over its entire surface. The load is defined by the expression p = p0 sin (px>a) sin (py>b), where p0 represents the pressure acting at the center of the plate. Determine the magnitude and location of the resultant force acting on the plate.

p0

y x

SOLUTION

a b

Resultant Force and its Location: The volume of the differential element is py px dy b . dx b asin dV = dFR = pdxdy = p0 a sin a b a

FR =

dFR = p0

LFR

L0

a sin

= p0 B a =

b py px dx b a sin dy b a b L0

a b px 2 a px 2 b b R cos b a - cos p a p b 0 0

4ab p0 p2

Ans.

Since the loading is symmetric, the location of the resultant force is at the center of the plate. Hence, x =

a 2

y =

b 2

Ans.

Ans: FR =

4ab p0 p2

a 2 b y = 2

x =

997

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9–119. A wind loading creates a positive pressure on one side of the chimney and a negative (suction) pressure on the other side, as shown. If this pressure loading acts uniformly along the chimney’s length, determine the magnitude of the resultant force created by the wind.

p p  p0 cos u

Solution FRx =

p 2l 2p L- 2

u

l

(p0 cos u) cos u r du =

p 2 2rlp0 p L- 2

p = 2rlp0 a b 2

cos2 u du Ans.



FRx = plrp0 FRy = 2l

p 2 p L- 2

(p0 cos u) sin u r du = 0

Thus,

Ans.

FR = plrp0

Ans: p FRx = 2rlp0 a b 2 FR = plrp0 998

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*9–120. When the tide water A subsides, the tide gate automatically swings open to drain the marsh B. For the condition of high tide shown, determine the horizontal reactions developed at the hinge C and stop block D. The length of the gate is 6 m and its height is 4 m. rw = 1.0 Mg/m3.

C 4m A

3m

B

2m

D

SOLUTION Fluid Pressure: The fluid pressure at points D and E can be determined using Eq. 9–13, p = rgz. pD = 1.01103219.812122 = 19 620 N>m2 = 19.62 kN>m2 pE = 1.01103219.812132 = 29 430 N>m2 = 29.43 kN>m2 Thus, wD = 19.62162 = 117.72 kN>m wE = 29.43162 = 176.58 kN>m Resultant Forces: FR1 =

1 1176.582132 = 264.87 kN 2

FR2 =

1 1117.722122 = 117.72 kN 2

Equations of Equilibrium: a + ©MC = 0;

264.87132 - 117.7213.3332 - Dx 142 = 0 Ans.

Dx = 100.55 kN = 101 kN + ©F = 0; : x

264.87 - 117.72 - 100.55 - Cx = 0 Ans.

Cx = 46.6 kN

Ans: Dx = 101 kN Cx = 46.6 kN 999

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9–121. The tank is filled with water to a depth of d = 4 m. Determine the resultant force the water exerts on side A and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so that it creates the same resultant forces? ro = 900 kg>m3 and rw = 1000 kg>m3.

2m

3m

A

B

d

SOLUTION For water At side A: wA = b rw g d = 2(1000)(9.81) (4) = 78 480 N/m FRA =

1 (78 480)(4) = 156 960 N = 157 kN 2

Ans.

At side B: wB = b rw g d = 3(1000)(9.81)(4) = 117 720 N>m FRB =

1 (117 720)(4) = 235 440 N = 235 kN 2

Ans.

For oil At side A: wA = b ro g d = 2(900)(9.81)d = 17 658 d FRA =

1 (17 658 d)(d) = 156 960 N 2 Ans.

d = 4.22 m

Ans: For water: FRA = 157 kN FRB = 235 kN For oil: d = 4.22 m 1000

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9–122. The concrete “gravity” dam is held in place by its own weight. If the density of concrete is rc = 2.5 Mg>m3, and water has a density of rw = 1.0 Mg>m3, determine the smallest dimension d that will prevent the dam from overturning about its end A.

1m

6m

A d–1

d

Solution Loadings. The computation will be based on b = 1 m width of the dam. The pressure at the base of the dam is. P = rgh = 1000(9.81)(6) = 58.86 ( 103 ) pa = 58.86 kPa Thus w = pb = 58.86(1) = 58.86 kN>m The forces that act on the dam and their respective points of application, shown in Fig. a, are W1 = 2500 3 1(6)(1) 4 (9.81) = 147.15 ( 103 ) N = 147.15 kN

1 W2 = 2500c (d - 1)(6)(1) d (9.81) = 73.575 ( d - 1 )( 103 ) = 73.575(d - 1) kN 2 1 2

( FR )v = 1000c (d - 1)(6)(1) d (9.81) = 29.43 ( d - 1 )( 103 ) = 29.43(d - 1) kN ( FR ) h =

1 (58.86)(6) = 176.58 kN 2

x1 = 0.5  x2 = 1 + y =

1 1 2 1 (d - 1) = (d + 2)  x3 = 1 + (d - 1) = (2d + 1) 3 3 3 3

1 (6) = 2 m 3

Equation of Equilibrium. Write the moment equation of equilibrium about A by referring to the FBD of the dam, Fig. a, 1 a+ΣMA = 0;  147.15(0.5) + [73.575(d - 1)]c (d + 2) d 3

1 + [29.43(d - 1)] c (2d + 1) d - 176.58(2) = 0 3

44.145d 2 + 14.715d - 338.445 = 0

Solving and chose the positive root

Ans.

d = 2.607 m = 2.61 m

Ans: d = 2.61 m 1001

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9–123. y

The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment due to the dam’s weight divided by the overturning moment about O due to the water pressure. Determine this factor if the concrete has a density of rconc = 2.5 Mg>m3 and for water rw = 1 Mg>m3.

1m

6m

O

x 4m

Solution Loadings. The computation will be based on b = 1m width of the dam. The pressure at the base of the dam is   P = pwgh = 1000(9.81)(6) = 58.86(103)pa = 58.86 kPa Thus,    w = Pb = 58.86(1) = 58.86 kN>m The forces that act on the dam and their respective points of application, shown in Fig. a, are   W1 = (2500)[(1(6)(1)](9.81) = 147.15(103)] N = 147.15 kN 1   W2 = (2500) c (3)(6)(1) d (9.81) = 220.725(103) N = 220.725 kN 2   FR =

1 (58.86)(6) = 176.58 kN 2

  x1 = 3 +

1 2 1 (1) = 3.5 ft  x2 = (3) = 2m  y = (6) = 2 m 2 3 3

Thus, the overturning moment about O is   MOT = 176.58(2) = 353.16 kN # m And the stabilizing moment about O is   Ms = 147.15(3.5) + 220.725(2) = 956.475 kN # m Thus, the factor of safety is   F.S. =

Ms 956.475 = = 2.7083 = 2.71 MOT 353.16

Ans.

Ans: F.S. = 2.71 1002

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*9–124. The concrete dam in the shape of a quarter circle. Determine the magnitude of the resultant hydrostatic force that acts on the dam per meter of length. The density of water is rw = 1 Mg>m3. 3m

SOLUTION Loading: The hydrostatic force acting on the circular surface of the dam consists of the vertical component Fv and the horizontal component Fh as shown in Fig. a. Resultant Force Component: The vertical component Fv consists of the weight of water contained in the shaded area shown in Fig. a. For a 1-m length of dam, we have Fv = rgAABCb = (1000)(9.81) B (3)(3) -

p 2 (3 ) R (1) = 18947.20 N = 18.95 kN 4

The horizontal component Fh consists of the horizontal hydrostatic pressure. Since the width of the dam is constant (1 m), this loading can be represented by a triangular distributed loading with an intensity of wC = rghCb = 1000(9.81)(3)(1) = 29.43 kN>m at point C, Fig. a. Fh =

1 (29.43)(3) = 44.145 kN 2

Thus, the magnitude of the resultant hydrostatic force acting on the dam is FR = 3F h 2 + F v 2 = 344.1452 + 18.952 = 48.0 kN

Ans.

Ans: FR = 48.0 kN 1003

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9–125. The tank is used to store a liquid having a density of 80 lb>ft3. If it is filled to the top, determine the magnitude of force the liquid exerts on each of its two sides ABDC and BDFE.

C A

D

4 ft

6 ft

B

F 12 ft

3 ft

E

Solution w1 = 80(4)(12) = 3840 lb>ft w2 = 80(10)(12) = 9600 lb>ft ABDC : F1 =

1 (3840)(5) = 9.60 kip 2

Ans.

BDEF : F2 =

1 (9600 - 3840)(6) + 3840(6) = 40.3 kip 2

Ans.

Ans: F1 = 9.60 kip F2 = 40.3 kip 1004

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9–126. y

The parabolic plate is subjected to a fluid pressure that varies linearly from 0 at its top to 100 lb>ft at its bottom B. Determine the magnitude of the resultant force and its location on the plate.

2 ft

2 ft y  x2

4 ft x

Solution FR =

LA

p dA =

L0

= 2

4

(100 - 25y)(2x dy)

L0

4

B

1

(100 - 25y)ay 2 dyb

2 3 2 5 4 = 2c 100 a by 2 - 25a by 2 d = 426.7 lb = 427 lb 3 5 0 FR y =

LA

yp dA;

426.7 y = 2

L0

4

Ans.

1

y(100 - 25y)y 2 dy

2 5 2 7 4 426.7 y = 2c 100 a by 2 - 25a by2 d 5 7 0

426.7 y = 731.4

Ans.

y = 1.71 ft Due to symmetry,

Ans.

x = 0

Ans: FR = 427 lb y = 1.71 ft x = 0 1005

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9–127. The 2-m-wide rectangular gate is pinned at its center A and is prevented from rotating by the block at B. Determine the reactions at these supports due to hydrostatic pressure. rw = 1.0 Mg>m3.

6m 1.5 m A B

SOLUTION

1.5 m

w1 = 1000(9.81)(3)(2) = 58 860 N/m w2 = 1000(9.81)(3)(2) = 58 860 N/m F1 =

1 (3)(58 860) = 88 290 2

F2 = (58 860)(3) = 176 580 a + ©MA = 0;

88 290(0.5) - FB (1.5) = 0 Ans.

FB = 29 430 N = 29.4 kN + ©F = 0; : x

88 290 + 176 580 - 29 430 - FA = 0 Ans.

FA = 235 440 N = 235 kN

Ans: FB = 29.4 kN FA = 235 kN 1006

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*9–128. The tank is filled with a liquid which has a density of 900 kg>m3. Determine the resultant force that it exerts on the elliptical end plate, and the location of the center of pressure, measured from the x axis.

y 1m

1m 4 y2

x2

1

0.5 m x 0.5 m

SOLUTION Fluid Pressure: The fluid pressure at an arbitrary point along y axis can be determined using Eq. 9–13, p = g(0.5 - y) = 900(9.81)(0.5 - y) = 8829(0.5 - y). Resultant Force and its Location: Here, x = 21 - 4y2. The volume of the differential element is dV = dFR = p(2xdy) = 8829(0.5 - y)[221 - 4y 2] dy. Evaluating integrals using Simpson’s rule, we have

the

0.5 m

FR =

LFR

d FR = 17658

L-0.5 m

(0.5 - y)(21 - 4y2)dy Ans.

= 6934.2 N = 6.93 kN 0.5 m

LFR

yd FR = 17658

y(0.5 - y)(21 - 4y2)dy

= - 866.7 N # m ' ydFR

y =

L-0.5 m

LFR

= dFR

- 866.7 = - 0.125 m 6934.2

Ans.

LFR

Ans: FR = 6.93 kN y = - 0.125 m 1007

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9–129. Determine the magnitude of the resultant force acting on the gate ABC due to hydrostatic pressure. The gate has a width of 1.5 m. rw = 1.0 Mg>m3.

1.5 m B

1.25 m C

SOLUTION

2m

w1 = 1000(9.81)(1.5)(1.5) = 22.072 kN>m

60° A

w2 = 1000(9.81)(2)(1.5) = 29.43 kN>m Fx =

1 (29.43)(2) + (22.0725)(2) = 73.58 kN 2

F1 = c (22.072) a 1.25 + F2 =

2 b d = 53.078 kN tan 60°

1 2 (1.5)(2)a b (1000)(9.81) = 16.99 kN 2 tan 60°

Fy = F1 + F2 = 70.069 kN F = 2F2x + F2y = 2(73.58)2 + (70.069)2 = 102 kN

Ans.

Ans: F = 102 kN 1008

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9–130. The semicircular drainage pipe is filled with water. Determine the resultant horizontal and vertical force components that the water exerts on the side AB of the pipe per foot of pipe length; gw = 62.4 lb>ft3.

A 2 ft

B

SOLUTION Fluid Pressure: The fluid pressure at the bottom of the drain can be determined using Eq. 9–13, p = gz. p = 62.4(2) = 124.8 lb>ft2 Thus, w

w = 124.8(1) = 124.8 lb>ft Resultant Forces: The area of the quarter circle is A =

1 2 1 pr = p(2 2) = p ft2. 4 4

Then, the vertical component of the resultant force is Ans.

FR v = gV = 62.4[p(1)] = 196 lb and the horizontal component of the resultant force is FR h =

1 (124.8)(2) = 125 lb 2

Ans.

Ans: FRv = 196 lb FRh = 125 lb 1009

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10–1. y

Determine the moment of inertia about the x axis.

y  bn xn a

b

x a

Solution Differential Element. Here x =

a

1

y n . The area of the differential element parallel b a 1 to the x axis shown shaded in Fig. a is dA = (a - x)dy = aa - 1 y n bdy. bn 1 n

Moment of Inertia. Perform the integration,





Ix =

LA

y2dA = =

L0

b

L0

b

y2aa aay2 -

a 1 n

b

a 1 n

b

1

yn bdy 1

yn + 2 bdy

a a n 3n + 1 = c y3 - a 1 ba by d3 3 3n + 1 n n b



=



=

b

0

1 3 n ab - a bab3 3 3n + 1 ab3 3(3n + 1)

Ans.



Ans: Ix = 1010

ab3 3(3n + 1)

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10–2. y

Determine the moment of inertia about the y axis.

y  bn xn a

b

x a

Solution Differential Element. The area of the differential element parallel to the y axis b shown shaded in Fig. a is dA = ydx = n xndx. a Moment of Inertia. Perform the integration,

Iy =

LA

x2dA =

L0

a

x2 a

b n x dxb an

a



=



=



=

b n+2 dx nx L0 a a b 1 b(xn + 3) ` na a n + 3 0

a3b n + 3

Ans.



Ans: Iy = 1011

a3b n + 3

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10–3. y

Determine the moment of inertia for the shaded area about the x axis.

100 mm

200 mm

y  1 x2 50

Solution 1

Differential Element. Here x = 250y2 . The area of the differential element parallel 1 to the x axis shown shaded in Fig. a is dA = 2x dy = 2250y2dy.

x

Moment of Inertia. Perform the integration,





Ix =

LA

y2dA =

L0

200 mm

= 2250

1

y2 c 2250y2dy d

L0

200 mm

2 7 = 2250 a y2 b 3 7

5

y2dy 200 mm

0



= 457.14(106) mm4



= 457(106) mm4

Ans.



Ans: Ix = 457 ( 106 ) mm4 1012

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*10–4. y

Determine the moment of inertia for the shaded area about the y axis.

100 mm

200 mm

y  1 x2 50

Solution 1

Differential Element. Here x = 250 y2 . The moment of inertia of the differential element parallel to x axis shown in Fig. a about y axis is

x

1 1 2 2 100250 3 (dy)(2x)3 = x3dy = ( 250y2 ) 3dy = y2dy. 12 3 3 3 Moment of Inertia. Perform the integration,

dIy =





Iy =

L

dIy =

L0

200 mm

100150 3 y2dy 3

100250 2 5 3 a y2 b = 3 5



= 53.33(106) mm4



= 53.3(106) mm4

200 mm

0

Ans.



Ans: Iy = 53.3(106) mm4 1013

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10–5. y

Determine the moment of inertia for the shaded area about the x axis.

y  x1/ 2

1m

x 1m

Solution Differential Element. The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = ydx. The moment of inertia of this element about the x axis is dIx = dIx′ + dA ∼ y2

=



=



=

y 2 1 (dx)y3 + ydx a b 12 2 1 3 y dx 3

1 1 3 (x2) dx 3

1 3 x2 dx 3 Moment of Inertia. Perform the integration.



=



Ix =





dIx = L L0

1m

2 53 x2 = 15 =

1 3 x2dx 3 1m

0

2 4 m = 0.133 m4 15

Ans.

Ans: Ix = 0.133 m4 1014

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10–6. y

Determine the moment of inertia for the shaded area about the y axis.

y  x1/ 2

1m

x 1m

Solution Differential Element. The area of the 1differential element parallel to the y axis shown shaded in Fig. a is dA = ydx = x2dx. Moment of Inertia. Perform the integration,





Iy =

LA

x2dA =

L0

1m

2 7 = x2 3 7 =

1

x2 ( x2dx )

1m

0

2 4 m = 0.286 m4 7

Ans.

Ans: Iy = 0.286 m4 1015

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10–7. y

Determine the moment of inertia for the shaded area about the x axis.

y2  1  0.5x 1m

x 2m

Solution Differential Element. Here x = 2(1 - y2). The area of the differential element parallel to the x axis shown shaded in Fig. a is dA = xdy = 2(1 - y2)dy. Moment of Inertia. Perform the integration,



Ix =

LA

y2dA =

L0

= 2

1m

L0

y2[2(1 - y2)dy] 1m

(y2 - y4)dy

y5 y3 = 2a - b 3 3 5

1m

0

4 4 = m = 0.267 m4 15

Ans.



Ans: Ix = 0.267 m4 1016

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*10–8. y

Determine the moment of inertia for the shaded area about the y axis.

y2  1  0.5x 1m

x 2m

Solution Differential Element. Here x = 2(1 - y2). The moment of inertia of the differential element parallel to the x axis shown shaded in Fig. a about the y axis is

∼2 dIy = dIy' + dAx



=



=



=

1 x 2 (dy)x3 + xdya b 12 2 1 3 x dy 3

1 32(1 - y2) 4 3 dy 3

8 ( - y6 + 3y4 - 3y2 + 1)dy 3 Moment of Inertia. Perform the integration,



=

Iy =

L

dIy =



=



=

8 3 L0

1m

( -y6 + 3y4 - 3y2 + 1)dy

1m y7 8 3 a - + y5 - y3 + yb ` 3 7 5 0

128 4 m = 1.22 m4 105

Ans.



Ans: Iy = 1.22 m4 1017

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10–9. Determine the moment of inertia of the area about the x axis. Solve the problem in two ways, using rectangular differential elements: (a) having a thickness dx and (b) having a thickness of dy.

y y = 2.5 – 0.1x2 2.5 ft 5 ft

x

SOLUTION (a) Differential Element: The area of the differential element parallel to y axis is dA = ydx. The moment of inertia of this element about x axis is ' dIx = dIx¿ + dAy 2 =

y 2 1 1dx2y 3 + ydxa b 12 2

=

1 12.5 - 0.1x223 dx 3

=

1 1 - 0.001x6 + 0.075x4 - 1.875x2 + 15.6252 dx 3

Moment of Inertia: Performing the integration, we have 5 ft

Ix =

L

dIx =

1 1 -0.001x6 + 0.075x4 - 1.875x2 + 15.6252 dx 3 L-5 ft

=

5 ft 1 0.075 5 1.875 3 0.001 7 ax + x x + 15.625xb ` 3 7 5 3 -5 ft

= 23.8 ft4

Ans.

(b) Differential Element: Here, x = 225 - 10y. The area of the differential element parallel to x axis is dA = 2xdy = 2 225 - 10y dy. Moment of Inertia: Applying Eq. 10–1 and performing the integration, we have Ix =

LA

y2dA 2.5 ft

= 2

y2 225 - 10ydy

L0

= 2c -

2.5 ft y2 2y 3 3 7 2 125 - 10y22 125 - 10y22 125 - 10y22 d ` 15 375 13125 0

= 23.8 ft4

Ans.

Ans: Ix = 23.8 ft 4 1018

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10–10. Determine the moment of inertia for the shaded area about the x axis.

y y2

h2 —x b h

SOLUTION d Ix = Ix =

x

b

1 3 y dx 3

L

d Ix

=

b 3

b y 1 h2 3>2 3>2 dx = a b x dx L0 3 b L0 3

=

1 h2 3>2 2 5>2 b a b a b x ]0 3 b 5

=

2 bh3 15

Ans.

Also, dA = (b - x) dy = (b Ix =

L

y2 dA h

=

L0

b 2 y ) dy h2

y2 (b -

b 2 y ) dy h2

b b 5 h = c y3 y d 3 5h2 0 =

2 bh3 15

Ans.

Ans: Ix = 1019

2 bh3 15

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10–11. y

Determine the moment of inertia for the shaded area about the x axis.

8m y  1 x3 8 x

Solution

4m 1

Differential Element. Here, x = 2y3. The area of the1 differential element parallel to the x axis shown shaded in Fig. a is dA = xdy = 2y3 dy Moment of Inertia. Perform the integration,

Ix =

LA

2

y dA =

L0

8m

L0



= 2



= 2a



1

2

y (2y3 dy) 8m

7

y3 dy

3 10 8 m y3 b ` 10 0

= 614.4 m4 = 614 m4

Ans.

Ans: Ix = 614 m4 1020

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*10–12. y

Determine the moment of inertia for the shaded area about the y axis.

8m y  1 x3 8 x

Solution

4m

Differential Element. The area of the differential element parallel to the y axis, 1 shown shaded in Fig. a, is dA = (8 - y)d x = a8 - x3 bdx 8

Moment of Inertia. Perform the integration,



Iy =

LA

x2dA = =

L0

4m

L0

4m

x2 a8 a8x2 -

1 3 x b dx 8

1 5 x b dx 8

1 6 3 8 x b = a x3 3 48

4m

0

= 85.33 m4 = 85.3 m4

Ans.

Ans: Iy = 85.3 m4 1021

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10–13. Determine the moment of inertia about the x axis.

y

x2  4y2  4

1m

x 2m

Solution 1 24 - x2. The moment of inertia of the differential 2 element parallel to the y axis shown shaded in Fig. a about x axis is

Differential Element. Here, y =



∼2 dIx = dIx′ + dAy



=



=



=



=

y 2 1 (dx)y3 + ydx a b 12 2 1 3 y dx 3

3 1 1 a 24 - x2 b dx 3 2

1 2(4 - x2)3 dx 24

Moment of Inertia. Perform the integration.

Ix =



=



=

L

dIx =

L0

2m

1 2(4 - x2)3 dx 24

1 x 2m c x2(4 - x2)3 + 6x24 - x2 + 24 sin - 1 d ` 96 2 0 p 4 m 8

Ans.

Ans: Ix = 1022

p 4 m 8

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10–14. y

Determine the moment of inertia about the y axis.

x2  4y2  4

1m

x 2m

Solution

1 24 - x2. The area of the differential element 2 1 parallel to the y axis shown shaded in Fig. a is dA = ydx = 24 - x2dx 2

Differential Element. Here, y =

Moment of Inertia. Perform the integration,

Iy =

LA

x2dA =



=



=



=

L0

2m

1 2 L0

1 x2 c 24 - x2dx d 2

2m

x2 24 - x2dx

2m 1 x 1 x c - 2(4 - x2)3 + ax24 - x2 + 4 sin - 1 b d ` 2 4 2 2 0

p 4 m 2



Ans.

Ans: Iy = 1023

p 4 m 2

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10–15. y

Determine the moment of inertia for the shaded area about the x axis. 4 in.

y2  x x 16 in.

Solution Differential Element. The area of the differential element parallel with the x axis shown shaded in Fig. a is dA = x dy = y2 dy. Moment of Inertia. Perform the integration,

Ix =

LA

y2dA =

L0

4 in.

L0

4 in.

y2(y2dy) y4dy



=



=



= 204.8 in4 = 205 in4

y5 5

`

4 in. 0

Ans.

Ans: Ix = 205 in4 1024

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*10–16. y

Determine the moment of inertia for the shaded area about the y axis. 4 in.

y2  x x 16 in.

Solution Differential Element. The moment of inertia of the differential element parallel to the x axis shown shaded in Fig. a about the y axis is ∼2 dIy = dIy + dAx



1 x 2 (dy)x3 + (xdy) a b 12 2



=

=

1 3 x dy 3



=

1 23 (y ) dy 3



=

1 6 y dy 3

Moment of Inertia. Perform the integration,

Iy =

L

dIy =

L0

4 in.

1 6 y dy 3

1 y7 4 in. a b` 3 7 0



=

 

= 780.19 in4 = 780 in4

Ans.

Ans: Iy = 780 in4 1025

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10–17. y

Determine the moment of inertia for the shaded area about the x axis.

h

y  h x3 b3

x b

Solution Differential Element. The moment of inertia of the differential element parallel to the y axis shown shaded in Fig. a about the x axis is

∼2 dIx = dIx′ + dAy



=



=



=



=

y 2 1 (dx)y3 + ydx a b 12 2 1 3 y dx 3

1 h 3 3 a x b dx 3 b3 h3 9 x dx 3b9

Moment of Inertia. Perform the integration,

Ix =

L

b

dIx =



=



=

h3 9 x dx 9 L0 3b h3 x10 b a b` 3b9 10 0 1 3 bh 30

Ans.



Ans: Ix = 1026

1 bh3 30

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10–18. y

Determine the moment of inertia for the shaded area about the y axis.

h

y  h x3 b3

x b

Solution Differential Element. The area of the differential element parallel to the y axis h shown shaded in Fig. a is dA = ydx = 3 x3dx b Moment of Inertia. Perform the integration, Iy =

LA

x2dA =

L0

b

x2 a b



=



=



=

h 3 x bdx b3

h x5dx b3 L0 h x6 6 a b` b3 6 0

b3h  6

Ans.

Ans: Iy = 1027

b3h 6

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10–19. y

Determine the moment of inertia for the shaded area about the x axis.

y2  1  x 1m x 1m 1m

Solution 1

Differential Element. Here y = (1 - x)2 . The moment of inertia of the differential element parallel to the y axis shown shaded in Fig. a about the x axis is 3 1 1 2 2 2 dIx = (dx)(2y)3 = y3dx = 3(1 - x)2 4 3 dx = (1 - x)2 dx. 12 3 3 3 Moment of Inertia. Perform the integration, Ix =

L

dIx =



=



=

L0

1m

3 2 (1 - x)2 dx 3

1m 5 2 2 c - (1 - x)2 d ` 3 5 0

4 4 m = 0.267 m4 15

Ans.

Ans: Ix = 0.267 m4 1028

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*10–20. y

Determine the moment of inertia for the shaded area about the y axis.

y2  1  x 1m x 1m 1m

Solution Differential Element. Here x = 1 - y2. The moment of inertia of the differential element parallel to the x axis shown shaded in Fig. a about the y axis is ∼2 dIy = dIy′ + dAx

=



=



=

1 x 2 (dy)x3 + xdy a b 12 2 1 3 x dy 3

1 (1 - y2)3dy 3

1 ( -y6 + 3y4 - 3y2 + 1)dy 3 Moment of Inertia. Perform the integration,





=

Iy =

L

1m

dIy =



=



=

1 ( - y6 + 3y4 - 3y2 + 1)dy L-1 m 3 1m 3 1 y7 ( - + y5 - y3 + y) ` 3 7 5 -1 m

32 4 m = 0.305 m4 105

Ans.

Ans: Iy = 0.305 m4 1029

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10–21. y

Determine the moment of inertia for the shaded area about the x axis.

y2  2x yx

2m

2m

x

Solution

1 2 y . The area of the differential element 2 1 parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1)dy = ay - y2 bdy. 2 Moment of Inertia. Perform the integration, Differential Element. Here x2 = y and x1 =

Ix =

LA

y2dA =

L0

L0



=



= a



2m

y2 ay -

2m

ay3 -

1 2 y bdy 2

1 4 y bdy 2

y5 2 m y4 b2 4 10 0

= 0.8 m4

Ans.

Ans: Ix = 0.8 m4 1030

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10–22. Determine the moment of inertia for the shaded area about the y axis.

y y2  2x yx

2m

2m

x

Solution 1

Differential Element. Here, y2 = 22x2 and y1 = x. The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = (y2 - y1) dx 1 = 1 22x2 - x 2 dx.

Moment of Inertia. Perform the integration, Iy =

LA

x2dA =

L0

2m

L0

2m



=



= a



=

1

x2 a 22x2 - xbdx

1 22x

5 2

- x3 2 dx

222 7 x4 2 2 m x2 b 7 4 0

4 4 m = 0.571 m4 7

Ans.

Ans: Iy = 0.571 m4 1031

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10–23. y

Determine the moment of inertia for the shaded area about the x axis.

b2 y2  —x a b x2 y — a2

b

x

a

Solution

a 1 a y2 and x1 = 2 y2. Thus, the area of the 1 b b2 differential element parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1) dy Differential Element. Here x2 =

a a 1 = a 1 y2 - 2 y2 b dy. b 2 b

Moment of Inertia. Perform the integration, Ix =

LA

b

y2dA =

a 1 a y2 a 1 y2 - 2 y2 bdy b 2 L0 b L0



=



= a



=

b

a 5 a a 1 y2 - 2 y4 bdy b 2 b

2a

1 2

7b

7

y2 -

3ab3 35

a 5 2b y b 5b2 0

Ans.

Ans: Ix = 1032

3ab3 35

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*10–24. Determine the moment of inertia for the shaded area about the y axis.

y b2 y2  —x a b x2 y — a2

b

x

a

Solution

b 2 x . Thus, the area of the a2 a differential element parallel to the y axis shown shaded in Fig. a is dA = (y2 - y1)dx b 1 b = a 1 x2 - 2 x2 b dx a a2

Differential Element. Here, y2 =

b

1 2

1

x2 and y1 =

Moment of Inertia. Perform the integration, Iy =

LA

a

x2dA =

b b 1 x2 a 1 x2 - 2 x2 bdx a L0 a2 a

=

b 5 b a 1 x2 - 2 x4 bdx a 2 L0 a

= a

2b

7a

1 2

7

x2 -

b 5 2a x b 5a2 0

=

2 3 1 a b - a3b 7 5

=

3a3b  35

Ans.

Ans: Iy = 1033

3a3b 35

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10–25. Determine the moment of inertia of the composite area about the x axis.

y 3 in.

SOLUTION

3 in.

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

3 in.

6 in.

x

Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel-axis theorem. Thus, -

Ix = Ix¿ + A(dy)2 = B

1 1 1 (3)(33) + (3)(3)(4)2 R + B (3)(33) + 3(3)(1.5)2 R 36 2 12

+ B

1 1 (6)(63) + (6)(6)(2)2 R 36 2

= 209 in4

Ans.

Ans: Ix = 209 in4 1034

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10–26. y

Determine the moment of inertia of the composite area about the y axis.

3 in.

SOLUTION

3 in.

Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated.

3 in.

6 in.

x

Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel-axis theorem. Thus, -

Iy = Iy ¿ + A(dx)2 = B

1 1 1 (3)(33) + (3)(3)(2)2 R + B (3)(33) + 3(3)(1.5)2 R 36 2 12

+ B

1 1 (6)(63) + (6)(6)(5)2 R 36 2

= 533 in4

Ans.

Ans: Iy = 533 in4 1035

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10–27. The polar moment of inertia for the area is JC = 642 (106) mm4, about the z′ axis passing through the centroid C. The moment of inertia about the y′ axis is 264 (106) mm4, and the moment of inertia about the x axis is 938 (106) mm4. Determine the area A.

y¿

C

x¿ 200 mm

Solution Applying the parallel-axis theorem with d y = 200 mm and Ix = 938 ( 106 ) mm4,

x

Ix = Ix′ + Ad 2y   938 ( 106 ) = Ix′ + A ( 2002 )

Ix′ = 938 ( 106 ) - 40 ( 103 ) A

with known polar moment of inertia about C, JC = Ix′ + Iy′

642 ( 10

6

) = 938 ( 106 ) - 40 ( 103 ) A + 264 ( 106 )

  A = 14.0 ( 103 ) mm2

Ans.

Ans: A = 14.0 ( 103 ) mm2 1036

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*10–28. Determine the location y of the centroid of the channel’s cross-sectional area and then calculate the moment of inertia of the area about this axis.

50 mm

50 mm

x

250 mm –y

SOLUTION Centroid: The area of each segment and its respective centroid are tabulated below.

50 mm 350 mm

Segment

A (mm2)

' y (mm)

' y A (mm3)

1

100(250)

125

3.12511062

2

250(50)

25

0.312511062

©

37.511032

3.437511062

Thus, ' 3.437511062 ©yA ' = 91.67 mm = 91.7 mm y = = ©A 37.511032

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ = Ix¿ + Ad2y. Segment

Ai (mm2)

(dy)i (mm)

(Ix¿)i (mm4)

(Ad 2y)i (mm4)

(Ix¿)i (mm4)

1

100(250)

33.33

1 3 12 110021250 2

27.77811062

157.9911062

2

250(50)

66.67

1 3 12 12502150 2

55.55611062

58.1611062

Thus, Ix¿ = © Ix¿

i

= 216.15 106 mm4 = 216 106 mm4

Ans.

Ans: ∼ y = 91.7 mm Ix′ = 216(106) mm4 1037

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10–29. Determine y, which locates the centroidal axis x′ for the cross-sectional area of the T-beam, and then find the moments of inertia Ix′ and Iy′.

y¿

75 mm

75 mm

y

20 mm C x¿

Solution

150 mm

Centroid. Referring to Fig. a, the areas of the segments and their respective centroids are tabulated below.

Thus, y =

y (mm)

∼A(mm3)

150(20)

10

30(103)

2

20(150)

95

285(103)

Σ

6(103)

Segment

A(mm2)

1

∼

y

20 mm

315(103)

∼2 315(103) Σy A = 52.5 mm = ΣA 6(103)

Ans.

Moment of Inertia. The moment of inertia about the x′ axis for each segment can be determined using the parallel axis theorem, Ix′ = Ix′ + Ad 2y. Referring to Fig. b, Segment

Ai(mm2)

(dy)i (mm) (Ix′)i (mm4)

1

150(20)

42.5

2

20(150)

42.5

(Ad 2y)i (mm4)

1 6 (150)(203) 5.41875(10 ) 12 1 6 (20)(1503) 5.41875(10 ) 12

(Ix′)i (mm4) 5.51875(106) 11.04375(106)

Thus     Ix′ = Σ(Ix′)i = 16.5625(106) mm4 = 16.6(106) mm4

Ans.

Since the y' axis passes through the centroids of segments 1 and 2, Iy′ =

1 1 (20)(1503) + (150)(203) 12 12

= 5.725(106) mm4

Ans.

Ans: y = 52.5 mm Ix′ = 16.6 ( 106 ) mm4 Iy′ = 5.725 ( 106 ) mm4 1038

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10–30. y

Determine the moment of inertia for the beam’s crosssectional area about the x axis.

1 in.

1 in. 8 in. 3 in. 1 in.

10 in.

x

Solution Moment of Inertia. The moment of inertia about x axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dy)i (in.)

1

1(8)

4

2

8(1)

0.5

3

1(3)

1.5

(Ix′)i (in4) 1 (1)(83) 12 1 (8)(13) 12 1 (1)(33) 12

(Ad 2y)i (in4)

(Ix)i (in4)

128

170.67

2

  2.67

6.75

  9.00

Thus,

Ix = Σ(Ix)i = 182.33 in4 = 182 in4         Ans.

Ans:  Ix = 182 in4 1039

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10–31. y

Determine the moment of inertia for the beam’s crosssectional area about the y axis.

1 in.

1 in. 8 in. 3 in. 1 in.

Solution

10 in.

x

Moment of Inertia. The moments of inertia about the y axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dx)i (in.)

1

8(1)

9.5

2

1(8)

5

3

3(1)

0.5

Thus,

(Iy')i (in4) 1 (8)(13) 12 1 (1)(83) 12 1 (3)(13) 12

(Adx2)i (in4)

(Iy)i (in4)

722

722.67

200

242.67

0.75

1.00

Iy = Σ(Iy)i = 966.33 in4 = 966 in4          Ans.

Ans: Iy = 966 in4 1040

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*10–32. y

Determine the moment of inertia Ix of the shaded area about the x axis.

100 mm

100 mm

150 mm

150 mm

150 mm

Solution

O

75 mm x

Moment of Inertia. The moment of inertia about the x axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a Segment

Ai(mm2)

(dy)i (mm)

1

200(300)

150

2

1 (150)(300) 2

100

3

- p(752)

150

Thus,

(Ix′)i (mm4)

(Ady)2i (mm4)

(Ix)i (mm4)

1 (200)(3003) 12 1 (150)(3003) 36 p(754) 4

1.35(109)

1.80(109)

0.225(109)

0.3375(109)

- 0.3976(109)

- 0.4225(109)

Ix = Σ(Ix)i = 1.715(109) mm4 = 1.72(109) mm4      Ans.

Ans: Ix = 1.72(109) mm4 1041

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10–33. y

Determine the moment of inertia Ix of the shaded area about the x axis.

100 mm

100 mm

150 mm

150 mm

150 mm

Solution

O

75 mm x

Moment of Inertia. The moment of inertia about the y axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a Segment

Ai(mm2)

(dx)i (mm)

1

200(300)

100

2

1 (150)(300) 2

250

3

- p(752)

100

Iy′(mm4) 1 (300)(2003) 12 1 (300)(1503) 36 p(754) 4

(Ad 2x)i (mm4)

(Iy)i (mm4)

0.6(109)

0.800(109)

1.40625(109) 1.434375(109) - 0.1767(109) - 0.20157(109)

Thus, Iy = Σ(Iy)i = 2.033(109) mm4 = 2.03(109) mm4      Ans.

Ans: Iy = 2.03 ( 109 ) mm4 1042

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10–34. y

Determine the moment of inertia of the beam’s crosssectional area about the y axis.

150 mm 150 mm

50 mm

SOLUTION Moment of Inertia: The dimensions and location of centroid of each segment are shown in Fig. a. Since the y axis passes through the centroid of both segments, the moment of inertia about y axis for each segment is simply (Iy)i = (Iy¿)i. Iy = g (Iy)i =

250 mm

x¿ C x¿

1 1 (50)(3003) + (250)(503) 12 12 6

4

6

_ y 4

Ans.

= 115.10(10 ) mm = 115(10 ) mm

25 mm

25 mm

x

Ans:  Iy = 115 ( 106 ) mm4 1043

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10–35. Determine y, which locates the centroidal axis x¿ for the cross-sectional area of the T-beam, and then find the moment of inertia about the x¿ axis.

y

150 mm 150 mm

50 mm

SOLUTION y =

250 mm

©yA 125(250)(50) + (275)(50)(300) = ©A 250(50) + 50(300)

x¿ C x¿ y

= 206.818 mm Ans.

y = 207 mm Ix¿

1 = c (50)(250)3 + 50(250)(206.818 - 125)2 d 12 +c

25 mm

25 mm

x

1 (300)(50)3 + 50(300)(275 - 206.818)2 d 12

I x¿ = 222(106) mm4

Ans.

Ans: y = 207 mm Ix′ = 222 ( 106 ) mm4 1044

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*10–36. y

Determine the moment of inertia about the x axis. 150 mm

150 mm

20 mm 200 mm x

C 20 mm

Solution Moment of Inertia. Since the x axis passes through the centroids of the two segments, Fig. a, Ix =

200 mm

20 mm

1 1 (300)(4003) (280)(3603) 12 12

= 511.36(106) mm4 = 511(106) mm4

Ans.



Ans: Ix = 511(106) mm4 1045

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10–37. y

Determine the moment of inertia about the y axis. 150 mm

150 mm

20 mm 200 mm x

C 20 mm

Solution Moment of Inertia. Since the y axis passes through the centroid of the two segments, Fig. a, Iy =

200 mm

20 mm

1 1 (360)(203) + (40)(3003) 12 12

= 90.24(106) mm4 = 90.2(106) mm4

Ans.

Ans: Iy = 90.2 ( 106 ) mm4 1046

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10–38. y

Determine the moment of inertia of the shaded area about the x axis. 3 in.

6 in. x 6 in.

6 in.

Solution Moment of Inertia. The moment of inertia about the x axis for each segment can be determined using the parallel-axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dy)i (in.)

1

6(6)

3

2

1 (6)(3) 2

7

3

1 (6)(9) 2

6

Thus,

(Ix′)i (in4) 1 (6)(63) 12 1 (6)(33) 36 1 (6)(93) 36

(Ad 2y)i (in4)

(Ix)i (in4)

324

432.0

441

445.5

972

1093.5

Ix = Σ(Ix)i = 1971 in4            Ans.

Ans: Ix = 1971 in4 1047

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10–39. y

Determine the moment of inertia of the shaded area about the y axis. 3 in.

6 in. x 6 in.

6 in.

Solution Moment of Inertia. The moment of inertia about the y axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a, Segment

Ai(in2)

(dx)i (in.)

1

6(6)

3

2

1 (6)(3) 2

4

3

1 (9)(6) 2

8

(Iy′)i (in4) 1 (6)(63) 12 1 (3)(63) 36 1 (9)(63) 36

(Ad 2x)i (in4)

(Iy)i (in4)

324

432.0

144

162.0

1728

1782.0

Thus, Iy = Σ(Iy)i = 2376 in4            Ans.

Ans: Iy = 2376 in4 1048

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*10–40. Determine the distance y to the centroid of the beam’s cross-sectional area; then find the moment of inertia about the centroidal x′ axis.

y 1 in.

y¿

3 in.

3 in.

1 in.

4 in.

C

x¿

y x

Solution

1 in.

Centroid. Referring to Fig. a, the areas of the segments and their respective centroids are tabulated below.

Thus y =

y (in.)

∼A(in3)

6(1)

0.5

3.00

2(4)

2

16.00

Segment

A(in2)

1 ○ 2 ○ Σ

14.0

∼

y

19.00

Σ∼ yA 19.00 in3 = = 1.357 in. = 1.36 in. ΣA 14.0 in2

Ans.

Moment of Inertia. The moment of inertia about the x′ axis for each segment can be determined using the parallel axis theorem, Ix′ = Ix′ + Ad 2y. Referring to Fig. b, Segment

Ai(in2)

(dy)i (in.)

1

6(1)

0.8571

2

2(4)

0.6429

Thus,

(Ix′)i (in4) 1 (6)(13) 12 1 (2)(43) 12

(Ad 2y)i (in4)

(Ix′)i (in4)

4.4082

4.9082

3.3061

13.9728

Ix′ = Σ(Ix′)i = 18.88 in4 = 18.9 in4         Ans.

Ans: y = 1.36 in. Ix′ = 18.9 in4 1049

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10–41. y

Determine the moment of inertia for the beam’s crosssectional area about the y axis.

1 in.

y¿

3 in.

3 in.

1 in.

4 in.

C

x¿

y x

Solution

1 in.

Moment of Inertia. The moment of inertia about the y axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a, Segment

Ai(in2)

(dx)i (in.)

1

4(1)

0.5

2

1(6)

4

3

4(1)

7.5

Thus,

(Iy′)i (in4) 1 (4)(13) 12 1 (1)(63) 12 1 (4)(13) 12

(Ad 2x)i (in4)

(Iy)i (in4)

1.00

1.3333

96.0

114.00

225.0

225.33

    Iy = Σ(Iy)i = 340.67 in4 = 341 in4           Ans.

Ans: Iy = 341 in4 1050

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10–42. Determine the moment of inertia of the beam’s crosssectional area about the x axis.

y

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

1 (170)(30)3 + 170(30)(15)2 Ix = 12

C y

x¿

30 mm 170 mm

1 + (30)(170)3 + 30(170)(85)2 12 +

x

x

1 (100)(30)3 + 100(30)(185)2 12

Ix = 154(106) mm4

Ans.

Ans: Ix = 154 ( 106 ) mm4 1051

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10–43. y

Determine the moment of inertia of the beam’s crosssectional area about the y axis.

y¿

30 mm 30 mm

SOLUTION Iy =

1 (30)(170)3 + 30(170)(115)2 12 +

140 mm

1 (170)(30)3 + 30(170)(15)2 12

30 mm

1 (30)(100)3 + 30(100)(50)2 + 12

70 mm _ x C _ y

x¿

30 mm 170 mm

Iy = 91.3(106) mm4

x

Ans.

Ans: Iy = 91.3 ( 106 ) mm4 1052

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*10–44. Determine the distance y to the centroid C of the beam’s cross-sectional area and then compute the moment of inertia Ix¿ about the x¿ axis.

y

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

170(30)(15) + 170(30)(85) + 100(30)(185) y = 170(30) + 170(30) + 100(30) Ans.

= 80.68 = 80.7 mm Ix¿ = c

x C y

x¿

30 mm 170 mm x

1 (170)(30)3 + 170(30)(80.68 - 15)2 d 12

+c +

1 (30)(170)3 + 30(170)(85 - 80.68)2 d 12

1 (100)(30)3 + 100(30)(185 - 80.68)2 12

Ix¿ = 67.6(106) mm4

Ans.

Ans: y = 80.7 mm Ix′ = 67.6(106) mm4 1053

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10–45. Determine the distance x to the centroid C of the beam’s cross-sectional area and then compute the moment of inertia Iy¿ about the y¿ axis.

y

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

170(30)(115) + 170(30)(15) + 100(30)(50) x = 170(30) + 170(30) + 100(30) Ans.

= 61.59 = 61.6 mm Iy¿ = c

x C y

x¿

30 mm 170 mm x

1 (30)(170)3 + 170(30)(115 - 61.59)2 d 12

+c +

1 (170)(30)3 + 30(170)(15 - 61.59)2 d 12

1 (30)(100)3 + 100(30)(50 - 61.59)2 12

Iy¿ = 41.2(106) mm4

Ans.

Ans: x = 61.6 mm Iy= = 41.2 ( 106 ) mm4 1054

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10–46. y

Determine the moment of inertia for the shaded area about the x axis.

6 in. 3 in. 3 in. 2 in.

3 in. 3 in.

3 in.

x

Solution Moment of Inertia. The moment of inertia about the x axis for each segment can be determined using the parallel-axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a, Segment

Ai(in2)

(dy)i (in.)

1

6(6)

3

2

1 (6)(3) 2

7

3

1 (6)(9) 2

6

4

- p(22)

3

(Ix′)i (in4) 1 (6)(63) 12 1 (6)(33) 36 1 (6)(93) 36 -

p(24) 4

(Ad 2y)i (in4)

(Ix)i (in4)

324

432.00

441

445.50

972

1093.50

-36p

-40p

Thus,

Ix = Σ(Ix)i = 1845.34 in4 = 1845 in4

Ans.

Ans: Ix = 1845 in4 1055

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10–47. y

Determine the moment of inertia for the shaded area about the y axis.

6 in. 3 in. 3 in. 2 in.

3 in. 3 in.

3 in.

x

Solution Moment of Inertia. The moment of inertia about the x axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a, Segment

Ai(in2)

(dx)i (in.)

1

6(6)

3

2

1 (6)(3) 2

2

3

1 (9)(6) 2

2

4

- p(22)

3

Thus,

(Ix′)i (in4) 1 (6)(63) 12 1 (3)(63) 36 1 (9)(63) 36 -

p(24) 4

(Ad 2x)i (in4)

(Iy)i (in4)

324

432.0

36

54.0

108

162.0

-36p

-40p

   Iy = Σ(Iy)i = 522.34 in4 = 522 in4         Ans.

Ans: Iy = 522 in4 1056

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*10–48. y'

y

Determine the moment of inertia of the parallelogram about the x′ axis, which passes through the centroid C of the area.

C

a

x'

θ

SOLUTION

b

x

h = a sin u Ix′ =

1 1 1 3 bh3 = (b)(a sin u)3 = a b sin3 u 12 12 12

Ans.

Ans: 1 3 Ix′ = a b sin3 u 12 1057

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10–49. y'

y

Determine the moment of inertia of the parallelogram about the y′ axis, which passes through the centroid C of the area.

C

a

x'

θ

SOLUTION x = a cos u + Iy′ = 2 J + =

b

x

1 b - a cos u = (a cos u + b) 2 2

2 a 2 1 1 b (a sin u)(a cos u)3 + (a sin u)(a cos u) a + cos u - a cos u b R 36 2 2 2 3

1 (a sin u)(b - a cos u)3 12

ab sin u 2 ( b + a2 cos2 u ) 12

Ans.

Ans: Iy′ = 1058

ab sin u 2 ( b + a2 cos2 u ) 12

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10–50. Locate the centroid y of the cross section and determine the moment of inertia of the section about the x¿ axis. 0.4 m

x' –y

0.05 m 0.2 m 0.2 m

SOLUTION

0.3 m

0.2 m 0.2 m

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (m2)

y (m)

yA (m3)

1

0.3(0.4)

0.25

0.03

2

1 2 10.4210.42

0.1833

0.014667

3

1.1(0.05)

0.025

0.001375

©

0.255

0.046042

Thus, y =

©yA 0.046042 = = 0.1806 m = 0.181 m ©A 0.255

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ = Ix¿ + Ad2y.

Segment

Ai (m2)

(dy)i (m)

(Ix¿)i (m4)

1

0.3(0.4)

0.06944

2

1 2 10.4210.42

0.002778

3

1.1(0.05)

0.1556

1 3 12 10.3210.4 2 1 3 36 10.4210.4 2 1 3 12 11.1210.05 2

(Ad 2y)i (m4)

(Ix¿)i (m4)

0.5787110-32 2.1787110-32 0.6173110-62 0.7117110-32 1.3309110-32 1.3423110-32

Thus, Ix¿ = © Ix¿

i

= 4.233 10 -3 m4 = 4.23 10-3 m4

Ans.

Ans: y = 0.181 m Ix′ = 4.23(10-3) m4 1059

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10–51. Determine the moment of inertia for the beam’s crosssectional area about the x¿ axis passing through the centroid C of the cross section.

100 mm

100 mm

25 mm 200 mm 200 mm 45

SOLUTION Ix¿ =

45

1 1 1 141.4 2 b d (200)(332.8)3 + 4 c (141.4)(141.4)3 + a (141.4)(141.4) b a 3 12 36 2

C

45 45

x¿

25 mm

1 p 1 - 2 c (200)4 a - sin90° b d 4 4 2 = 520(106) mm4

Ans.

Ans: Ix= = 520 ( 106 ) mm4 1060

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*10–52. Determine the moment of inertia of the area about the x axis.

y 3 in.

3 in.

6 in.

SOLUTION Ix = B

1 1 1 (6)(10)3 + 6(10)(5)2 R - B (3)(6)3 + a b(3)(6)(8)2 R 12 36 2

1 - B p (2)4 + p(2)2(4)2 R = 1.19(103) in4 4

2 in. 4 in.

Ans.

x

Ans: Ix = 1.19(103) in4 1061

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10–53. y

Determine the moment of inertia of the area about the y axis.

3 in.

3 in.

SOLUTION Iy = c

1 1 1 (10)(6)3 + 6(10)(3)2 d - c (6)(3)3 + a b 6(3)(5)2 d 12 36 2

6 in.

1 - c p(2)4 + p(2)2(3)2 d = 365 in4 4

Ans. 2 in. 4 in. x

Ans: Iy = 365 in4 1062

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10–54. y t

Determine the product of inertia of the thin strip of area with respect to the x and y axes. The strip is oriented at an angle u from the x axis. Assume that t V l. l

SOLUTION 1

l

lxy =

=

LA

xydA =

L0

(s cos u)(s sin u)tds = sin u cos ut

L0

u

2

x

s ds

1 3 l t sin 2u 6

Ans.

Ans: Ixy = 1063

1 3 tl sin 2u 3

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10–55. y

Determine the product of inertia of the shaded area with respect to the x and y axes.

SOLUTION

3 in.

Differential Element: The area of the differential element parallel to the y axis 1 shown shaded in Fig. a is dA = y dx = x3 dx. The coordinates of the centroid of 9 y 1 3 = x . Thus, the product of inertia of this this element are xc = x and yc = 2 18 element with respect to the x and y axes is

y  1 x3 9

x 3 in.

'' dIxy = dIx¿y¿ + dAx y 1 1 = 0 + a x3dx b(x) a x3 b 9 18 =

1 7 x dx 162

Product of Inertia: Performing the integration, we have 3 in

Ixy =

L

dIxy =

L0

3 in. 1 7 1 = 5.06 in4 x dx = (x8) 2 162 1296 0

Ans.

Ans: Ixy = 5.06 in4 1064

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*10–56. Determine the product of inertia for the shaded portion of the parabola with respect to the x and y axes.

y 100 mm

200 mm y

SOLUTION

1 2 x 50

1

Differential Element: Here, x = 250y2 . The area of the differential element 1 parallel to the x axis is dA = 2xdy = 2 250y2 dy. The coordinates of the centroid for this element are x = 0, y = y. Then the product of inertia for this element is

x

dIxy = dIx¿y¿ + dAx y = 0 + ¢ 2 250y2 dy ≤ (0)(y) 1

= 0 Product of Inertia: Performing the integration, we have Ixy =

L

Ans.

dIxy = 0

Note: By inspection, Ixy = 0 since the shaded area is symmetrical about the y axis.

Ans: Ixy = 1065

L

dIxy = 0

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10–57. Determine the product of inertia of the shaded area with respect to the x and y axes, and then use the parallel-axis theorem to find the product of inertia of the area with respect to the centroidal x¿ and y¿ axes.

y

y¿

y2

x

2m C

SOLUTION

x

Differential Element: The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = y dx = x 1>2 dx. The coordinates of the centroid of y ' ' 1 1>2 this element are x = x and y = Thus, the product of inertia of this = x 2 2 element with respect to the x and y axes is

4m

~~ dIxy = dIx¿y¿ + dAx y 1 = 0 + A x1>2 dx B (x) a x 1>2 b 2 1 2 x dx 2 Product of Inertia: Performing the integration, we have =

4m

Ixy =

x¿

L

dIxy =

L0

4m 1 1 2 = 10.67 m4 = 10.7 m4 x dx = a x3 b 2 2 6 0

Ans.

Using the information provided on the inside back cover of this book, the location of 2 3 the centroid of the parabolic area is at x = 4 - (4) = 2.4 m and y = (2) = 0.75 m 5 8 2 and its area is given by A = (4)(2) = 5.333 m2. Thus, 3 Ixy = Ix¿y¿ + Adxdy 10.67 = Ix¿y¿ + 5.333(2.4)(0.75) Ix¿y¿ = 1.07 m4

Ans.

Ans: Ixy = 10.7 m4 Ix′y′ = 1.07 m4 1066

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10–58. Determine the product of inertia for the parabolic area with respect to the x and y axes.

y

y

b 1/2 x a1/2

b

x

SOLUTION

a

' x = x y ' y = 2 dA = y dx dIxy = Ixy =

xy2 dx 2

L

d Ixy a

=

1 b2 2 a b x dx L0 2 a

=

a 1 b2 B a b x3 R a 6 0

=

1 2 2 a b 6

Ans.

Ans: Ixy = 1067

1 2 2 ab 6

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10–59. Determine the product of inertia of the shaded area with respect to the x and y axes.

y

1

1

y = (a2 – x 2 )2 a

SOLUTION Differential Element: The area of the differential element parallel to the y axis is 1 1 2 dA = ydx = A a2 - x2 B dx. The coordinates of the centroid for this element are y 1 1 1 ' ' = A a2 - x2 B 2. Then the product of inertia for this element is x = x, y = 2 2

O

x a

'' dIxy = dIx¿y¿ + dAx y 2 1 1 1 1 1 = 0 + c A a2 - x2 B 2 dx d1x2 c a a2 - x2 b d 2

=

3 3 1 5 1 3 A x + a2x + 6ax2 - 4a2x2 - 4a2x2 B dx 2

Product of Inertia: Performing the integration, we have a

Ixy =

L

dIxy =

3 3 1 5 1 ¢ x3 + a2x + 6ax2 - 4a2x2 - 4a2x2 bdx 2 L0

=

8 3 5 8 1 7 a a2 2 1 x4 + x + 2ax3 - a2x2 - a2x2 ` 2 4 2 5 7 0

=

a4 280

Ans.

Ans: Ixy = 1068

a4 280

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*10–60. Determine the product of inertia of the shaded area with respect to the x and y axes.

y

x2 + y2 = 4 2 in.

SOLUTION

x

Differential Element: Here, y = 24 - x2 . The area of the differential element

2 in.

parallel to the y axis is dA = ydx = 24 - x2dx. The coordinates of the centroid y 1 ' ' for this element are x = x, y = = 24 - x2 . Then the product of inertia for this 2 2 element is '' dIxy = dIx¿y¿ + dAx y = 0 + =

A 24 - x2dx B 1x2a 24 - x2 b 1 2

1 14x - x32 dx 2

Product of Inertia: Performing the integration, we have Ixy =

L

dIxy = =

1 2 L0

2in.

14x - x32 dx

1 x4 2x2 2 4

2in. 0

= 2.00 in4

Ans.

Ans: Ixy = 2.00 in4 1069

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10–61. y

Determine the product of inertia of the shaded area with respect to the x and y axes.

1 in. y = 0.25x 2 x 2 in.

SOLUTION 1

Differential Element: Here, x = 2y 2. The area of the differential element parallel 1

to the x axis is dA = xdy = 2y 2 dy. The coordinates of the centroid for this element 1 x are x = = y 2, y = y. Then the product of inertia for this element is 2 dIxy = dIx′y′ + dAx y = 0 + a2y 2 dyb ( y2 ) (y) 1

1

= 2y2 dy Product of Inertia: Performing the integration, we have l in.

Ixy =

L

dIxy =

L0

2y2 dy =

2 3 1 in. y ` = 0.667 in4 3 0

Ans.

Ans: Ixy = 0.667 in4 1070

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10–62. Determine the product of inertia for the beam’s crosssectional area with respect to the x and y axes.

y 1 in.

1 in. 8 in.

SOLUTION

3 in. 1 in. x

Ixy = 0.5(4)(8)(1) + 6(0.5)(10)(1) + 11.5(1.5)(3)(1)

12 in.

4

Ans.

= 97.8 in

Ans: Ixy = 97.8 in4 1071

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10–63. Determine the moments of inertia for the shaded area with respect to the u and v axes.

y 0.5 in. v

u

SOLUTION Moment and Product of Inertia about x and y Axes: Since the shaded area is symmetrical about the x axis, Ixy = 0. Ix = Iy =

1 1 (1)(53) + (4)(13) = 10.75 in4 12 12

0.5 in. x 0.5 in.

30

5 in.

1 in.

4 in.

1 1 (1)(43) + 1(4)(2.5)2 + (5)(13) = 30.75 in4 12 12

Moment of Inertia about the Inclined u and v Axes: Applying Eq. 10-9 with u = 30°, we have Iu = =

I x + Iy 2

+

Ix - I y 2

cos 2u - Ixy sin 2u

10.75 + 30.75 10.75 - 30.75 + cos 60° - 0(sin 60°) 2 2

= 15.75 in4 Iv = =

Ix + I y 2

Ans. -

I x - Iy 2

cos 2u + Ixy sin 2u

10.75 - 30.75 10.75 + 30.75 cos 60° + 0(sin 60°) 2 2

= 25.75 in4

Ans.

Ans: Iu = 15.75 in4 Iu = 25.75 in4 1072

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*10–64. Determine the product of inertia for the beam’s crosssectional area with respect to the u and v axes.

y

v 150 mm

150 mm

u 20

SOLUTION 20 mm

1 1 (300)(400)3 (280)(360)3 = 511.36(10)6 mm4 Ix = 12 12 Iy = 2 c

x

C

Moments of inertia Ix and Iy

200 mm

20 mm

1 1 (20)(300)3 d + (360)(20)3 = 90.24(10)6 mm4 12 12

The section is symmetric about both x and y axes; therefore Ixy = 0. Iuv =

I x - Iy

= a

2

sin 2u + Ixy cos 2u

511.36 - 90.24 sin 40° + 0 cos 40°b 106 2

= 135(10)6 mm4

Ans.

Ans: Iuv = 135(10)6 mm4 1073

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10–65. Determine the product of inertia for the shaded area with respect to the x and y axes.

y 2 in.

2 in.

2 in.

SOLUTION

1 in.

'' Ixy = ©(Ix¿y¿ + x y A) = [0 + 2(3)(4)(6)] - C 0 + 2(4)(p)(1)2 D = 119 in4

4 in.

Ans. x

Ans: Ixy = 119 in4 1074

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10–66. Determine the product of inertia of the cross-sectional area with respect to the x and y axes.

y

100 mm 20 mm

SOLUTION

400 mm

Product of Inertia: The area for each segment, its centroid and product of inertia with respect to x and y axes are tabulated below. Segment

Ai (mm2)

(dx)i (mm)

(dy)i (mm)

(Ixy)i (mm4)

1

100(20)

60

410

49.211062

2

840(20)

0

0

0

3

100(20)

- 60

-410

49.211062

C

x

400 mm

20 mm 100 mm

20 mm

Thus, Ixy = ©1Ixy2i = 98.411062mm4

Ans.

Ans: Ixy = 98.4 ( 106 ) mm4 1075

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10–67. Determine the location (x, y) of the centroid C of the angle’s cross-sectional area, and then compute the product of inertia with respect to the x′ and y′ axes.

y

18 mm

y¿

x

150 mm x¿

C y 150 mm

x 18 mm

SOLUTION Centroid: x =

y =

9(18)(150) + 84(18)(132) ΣxA = = 44.11 mm = 44.1 mm ΣA 18(150) + 18(132) ΣyA ΣA

=

75(18)(150) + 9(18)(132) 18(150) + 18(132)

= 44.11 mm = 44.1 mm

Ans.

Ans.

Product of inertia about x′ and y′ axes: Ix′y′ = 18(150)( - 35.11)(30.89) + 18(132)(39.89)( -35.11) = -6.26(106) mm4

Ans.

Ans: x = y = 44.1 mm Ix=y= = -6.26 ( 106 ) mm4 1076

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*10–68. Determine the distance y to the centroid of the area and then calculate the moments of inertia Iu and Iv for the channel’s cross-sectional area. The u and v axes have their origin at the centroid C. For the calculation, assume all corners to be square.

y v

10 mm

10 mm

C

50 mm

Ans.

1 (300)(10)3 + 300(10)(12.5 - 5)2 d 12

+ 2c

1 (10)(50)3 + 10(50)(35 - 12.5)2 d 12

= 0.9083(106) mm4 Iy =

1 1 (50)(10)3 + 50(10)(150 - 5)2 d (10)(300)3 + 2 c 12 12

= 43.53(106) mm4 Ixy = 0 Iu = =

(By symmetry)

Ix + Iy 2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

0.9083(106) - 43.53(106) 0.9083(106) + 43.53(106) + cos 40° - 0 2 2

= 5.89(106) mm4 Iv = =

Ix + Iy 2

-

y 150 mm

300(10)(5) + 2[(50)(10)(35)] = 12.5 mm y = 300(10) + 2(50)(10) Ix = c

20

10 mm

SOLUTION

Ans.

Ix - Iy 2

cos 2u + Ixy sin 2u

0.9083(106) - 43.53(106) 0.9083(106) + 43.53(106) cos 40°+0 2 2

= 38.5(106) mm4

Ans.

1077

u

150 mm

x

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10–69. Determine the moments of inertia Iu, Iv and the product of inertia Iuv for the beam’s cross-sectional area. Take u = 45°.

y

u

v

16 in. u

SOLUTION Ix =

x

O

1 1 (20)(2)3 + 20(2)(1)2 + (4)(16)3 + 4(16)(8)2 12 12

8 in.

8 in. 2 in. 2 in.

2 in.

= 5.515(103) in4 Iy =

1 1 (2)(20)3 + (16)(4)3 12 12

= 1.419(103) in4 Ixy = 0 Iu = =

Ix + Iy 2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

5.515 + 1.419 3 5.515 - 1.419 3 (10 ) + (10 ) cos 90° - 0 2 2

= 3.47(103) in4

Ans.

Iv = 3.47(103) in4

Ans.

Iuv = =

Ix - Iy 2

sin 2u + Ixy cos 2u

5.515 - 1.419 3 (10 ) sin 90° + 0 2

= 2.05(103) in4

Ans.

Ans: Iu = 3.47 ( 103 ) in4 Iv = 3.47 ( 103 ) in4 Iuv = 2.05 ( 103 ) in4 1078

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10–70. y

Determine the moments of inertia Iu, Iv and the product of inertia Iuv for the rectangular area. The u and v axes pass through the centroid C.

v

u

30 x 30 mm

C

Solution

120 mm

Moment And Product of Inertia About x and y Axes. Since the rectangular area is symmetrical about the x and y axes, Ixy = 0. Ix =

1 1 (120)(303) = 0.270(106) mm4  Iy = (30)(1203) = 4.32(106) mm4 12 12

Moment And Product of Inertia About The Inclined u and v Axes. With u = 30°, Iu =

Ix + Iy 2

= c

+

Ix - Iy 2

cos 2u - Ixy sin 2u

0.270 + 4.32 0.27 - 4.32 + cos 60° - 0 sin 60° d (106) 2 2

= 1.2825(106) mm4 = 1.28(106) mm4

Iv =

Ix + Iy 2

= c

-

Ix - Iy 2

Ans.

cos 2u + Ixy sin 2u

0.27 + 4.32 0.27 - 4.32 cos 60° + 0 sin 60° d (106) 2 2

= 3.3075(106) mm4 = 3.31(106) mm4

Iuv =

Ix - Iy

= c

2

Ans.

sin 2u + Ixy cos 2u

0.270 - 4.32 sin 60° + 0 cos 60° d (106) 2

= - 1.7537(106) mm4 = - 1.75(106) mm4

Ans.

Ans: Iu = 1.28(106) mm4 Iv = 3.31(106) mm4 Iuv = - 1.75(106) mm4 1079

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10–71. y

Solve Prob. 10–70 using Mohr’s circle. Hint: To solve, find the coordinates of the point P(Iu, Iuv) on the circle, measured counterclockwise from the radial line OA. (See Fig. 10–19.) The point Q(Iv, -Iuv) is on the opposite side of the circle.

v

u

30 x 30 mm

C

Solution

120 mm

Moment And Product of Inertia About x And y Axes. Since the rectangular Area is symmetrical about the x and y axes, Ixy = 0. 1 1 (120)(303) = 0.270(106) mm4  Iy = (30)(1203) = 4.32(106) mm4 12 12 Construction of The Circle. The Coordinates of center O of the circle are Ix =

a

Ix + Iy 2

, 0b = a

0.270 + 4.32 , 0b(106) = (2.295, 0)(106) 2

And the reference point A is

(Ix, Ixy) = (0.270, 0)(106) Thus, the radius of the circle is R = OA =

1 2(2.295

- 0.27)2 + 02 2 (106) = 2.025(106) mm4

Using these Results, the circle shown in Fig. a can be constructed. Rotate radial line OA counterclockwise 2u = 60° to coincide with radial line OP where coordinate of point P is 1 Iu, Iuv 2. Then Iu = (2.295 - 2.025 cos 60°)(106) = 1.2825(106) mm4 = 1.28(106) mm4   Ans. Iuv = - 2.025(106) sin 60° = - 1.7537(106) mm4 = - 1.75(106) mm4   Ans.

Iv is represented by the coordinate of point Q. Thus, Iv = (2.295 + 2.025 cos 60°)(106) = 3.3075(106) mm4 = 3.31(106) mm4

Ans.

Ans: Iu = 1.28 ( 106 ) mm4 Iuv = - 1.75 ( 106 ) mm4 Iv = 3.31 ( 106 ) mm4 1080

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*10–72. y

Determine the directions of the principal axes having an origin at point O, and the principal moments of inertia for the triangular area about the axes.

6 in.

Solution

O

x 9 in.

Moment And Product of Inertia About x And y Axes. Using the parallel-axis theorem, Ix = Ix′ + Ad 2y  Ix =

2 1 1 1 (9) ( 63 ) + c (9)(6) d c (6) d = 162 in4 36 2 3

2 1 1 2 (6) ( 93 ) + c (6)(9) d c (9) d = 1093.5 in4 36 2 3 Using the result of Example 10–6,

Iy = Iy′ + Ad 2x  Iy =

b2h2 = 8 Principal Moments of Inertia. Ixy =

I max =

Ix + Iy 2

min

( 92 )( 62 ) 8

{ Aa

= 364.5 in.4

Ix - Iy 2 b + I 2xy 2

162 + 1093.5 162 - 1093.5 2 2 { a = b + 364.5 2 A 2



= 627.75 { 591.42



Imax = 1219.17 in4 = 1219 in4

Ans.

Imin = 36.33 in4 = 36.3 in4

Ans.

The orientation of the principal axes can now be determined - Ixy

tan 2up =

(Ix - Iy)>2

=

- 364.5 = 0.7826 (162 - 1093.5)>2

2uP = 38.04°  And   -141.95° up = 19.02°  And   -70.98° Substitute up = - 70.98° into the equation for Iu, Iu = =

Ix + Iy 2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

162 + 1093.5 162 - 1093.5 + cos ( - 141.95°) - 364.5 sin ( -141.95°) 2 2 = 1219.17 in4

Thus, (uP)1 = - 71.0°

Ans.

(up)2 = 19.0°

Ans: Imax = 1219 in4 Imin = 36.3 in4 (uP)1 = - 71.0° (up)2 = 19.0° 1081

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10–73. Solve Prob. 10–72 using Mohr’s circle.

y

6 in.

Solution Moment And Product of Inertia about x and y Axes. Using the parallel-axis theorem,   Ix = Ix′ + Ad 2y;  Ix =   Iy = Iy′ + Ad 2x;  Iy =

2 1 1 1 (9)(63) + c (9)(6) d c (6) d = 162 in4 36 2 3

2 1 1 2 (6)(93) + c (6)(9) d c (9) d = 1093.5 in4 36 2 3

Using these results of Example 10–6   Ixy =

(92)(62) b2h2 = = 364.5 in4 8 8

Construction of The Circle. The coordinates of center O of the circle are   a

Ix + Iy 2

, 0b = a

162 + 1093.5 , 0b = (627.75, 0) 2

And the reference point A is   (Ix, Ixy) = (162, 364.5) Thus, the radius of the circle is   R = OA = 2(627.75 - 162)2 + (0 - 364.5)2 = 591.42

Using these results the circle shown in Fig. a can be constructed. Here, the coordinates of points B and C represent Imin and Imax respectively. Thus

  Imax = 627.75 + 591.42 = 1219.17 in4 = 1219 in4 4

Ans.

4

Ans.

  Imin = 627.75 - 591.42 = 36.33 in = 36.3 in 

1082

O

x 9 in.

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10–73. Continued

The orientation of principal axes can be determined from the geometry of the shaded triangle on the circle.   tan 2(up)2 =

364.5 = 0.7826 627.75 - 162

2(up)2 = 38.04° Ans.

(up)2 = 19.02° = 19.0° d And 2(up)1 = 180° - 2(up)2 = 180° - 38.04° = 141.95°

Ans.

(up)1 = 70.98° = 71.0° b

Ans: Imax = 1219 in4 Imin = 36.3 in4 (up)2 = 19.0° d (up)1 = 71.0° b 1083

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10–74. y

Determine the orientation of the principal axes having an origin at point C, and the principal moments of inertia of the cross section about these axes.

100 mm

80 mm C 80 mm

x 10 mm

Solution Moment and Product of Inertia About x and y Axes. Using the parallel-axis theorem by referring to Fig. a Ix = Σ ( Ix′ + Ad 2y ) ;  Ix =

10 mm

1 1 (140)(103) + 2 c (10)(1003) + 10(100)(452) d 12 12

= 5.7283 (106) mm4 Iy = Σ ( Iy′ + Ad 2x ) ;  Iy =

1 1 (10)(1403) + 2 c (100)(103) + 100(10)(752) d 12 12

= 13.5533 (106) mm4 Ixy = Σ(Ix′y′ + Ad x d y);  Ixy = 0 +

30

+ 10(100)( - 75)(45) 4

+ [0 + 10(100)(75)( - 45)] = - 6.75(106) mm4 Principal moments of Inertia. I max =

Ix + Iy

min

= £

2

{

C

a

Ix + Iy 2

2

b + I 2xy

5.7283 + 13.5533 5.7283 - 13.5533 2 { a b + ( -6.75)2 § (106) 2 C 2

= (9.6408 { 7.8019)(106)

Imax = 17.44(106) mm4 = 17.4(106) mm4

Ans.

Imin = 1.8389(106) mm4 = 1.84(106) mm4

Ans.

1084

100 mm

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10–74. Continued

The orientation of the principal axes can now be determined tan 2up =

- Ixy (Ix - Iy)>2

=

- ( - 6.75) (5.7283 - 13.5533)>2

= - 1.7252

2up = - 59.90°  and  120.10° up = - 29.95°  and   60.04° Substitute up = 60.05° into the equation for Iu, Iu =

Ix + Iy

= c

2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

5.7283 + 13.5533 5.7283 - 13.5533 + cos 120.10° - ( -6.75) sin 120.10° d (106) 2 2

= 17.44(106) mm4 Thus,

Ans.

(up)1 = 60.0°  (up)2 = - 30.0°

Ans: Imax = 17.4 ( 106 ) mm4 Imin = 1.84 ( 106 ) mm4 (up)1 = 60.0° (up)2 = - 30.0° 1085

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10–75. y

Solve Prob. 10–74 using Mohr’s circle.

100 mm

80 mm C 80 mm

x 10 mm

Solution Moment and Product of Inertia About x and y Axes. Using the parallel-axis theorem by referring to Fig. a Ix = Σ ( Ix′ + Ad 2y ) ;  Ix =

10 mm

1 1 (140)(103) + 2 c (10)(1003) + 10(100)(452) d 12 12

= 5.7283 (106) mm4 Iy = Σ ( Iy′ + Ad 2x ) ;  Iy =

1 1 (10)(1403) + 2 c (100)(103) + 100(10)(752) d 12 12

= 13.5533 (106) mm4 Ixy = Σ ( Ix′y′ + Ad x d y ) ;  Ixy = 0 +

30

+ 10(100)( - 75)(45) 4

+ [0 + 10(100)(75)( - 45)] = - 6.75(106) mm4 Construction of the circle. The coordinates of center O of the circle are a

Ix + Iy 2

, 0b = a

5.7283 + 13.5533 , 0b(106) = (9.6408, 0)(106) 2

And the reference point A is (Ix, Ixy) = (5.7283, - 6.75)(106) Thus, the radius of the circle is R = OA = a 2(9.6408 - 5.7283)2 + ( - 6.75)2 b(106) = 7.8019(106)

Using these results, the circle shown in Fig. b can be constructed. Here, the coordinates of points B and C represent Imin and Imax respectively. Thus   Imax = (9.6408 + 7.8019)(106) = 17.44(106)mm4 = 17.4(106)mm4

Ans.

  Imin = (9.6408 - 7.8019)(106) = 1.8389(106)mm4 = 1.84(106)mm4

Ans.

1086

100 mm

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10–75. Continued

The orientation of the principal axes can be determined from the geometry of the shaded triangle on the circle 6.75   tan 2(up)2 = 9.6408 - 5.7283 2(up)2 = 59.90° (up)2 = 29.95° = 30.0° b

Ans.

And 2(up)1 = 180° - 2(up)1 2(up)1 = 180° - 59.90° = 120.10° Ans.

(up)1 = 60.04° = 60.0° d

Ans: Imax = 17.4 ( 106 ) mm4 Imin = 1.84 ( 106 ) mm4 (up)2 = 30.0° b (up)1 = 60.0° d 1087

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*10–76. y

Determine the orientation of the principal axes having an origin at point O, and the principal moments of inertia for the rectangular area about these axes.

6 in.

3 in.

O

Solution

x

Moment And Product of Inertia About x and y Axes. Using the parallel-axis theorem, 1   Ix = Ix′ + Ad 2y;  Ix = (6)(33) + 6(3)(152) = 54.0 in4 12   Iy = Iy′ + Ad 2x;  Iy =

1 (3)(63) + 3(6)(32) = 216.0 in4 12

  Ixy = Ix′y′ + Ad xd y;  Ixy = 0 + 6(3)(3)(1.5) = 81.0 in4 Principal of Moment of Inertia.   I max =

Ix + Iy 2

min

=

{

C

a

Ix - Iy 2

2

b + I 2xy

54.0 + 216.0 54.0 - 216.0 2 a b + 81.02 { 2 C 2

= 135 { 8122

Imax = 249.55 in4 = 250 in4

Ans.

Imin = 20.44 in4 = 20.4 in4

Ans.

The orientation of principal axes can now be determined.   tan 2up =

- Ixy (Ix - Iy)>2

=

- 81.0 = 1.00 (54.0 - 216.0)>2

2up = 45.0°  And - 135° up = 22.5°  And 67.5° Substitute up = -67.5° into the equation for Iu, Ix + Iy

+

Ix + Iy

cos 2u - Ixy sin 2u 2 2 54.0 + 216.0 54.0 - 216.0 = + cos ( - 135°) - 81.0 sin ( -135°) 2 2

  Iu =

= 249.55 in4 Ans.

thus, (up)1 = - 67.5°  (up)2 = 22.5°

Ans: Imax = 250 in4 Imin = 20.4 in4 (up)1 = - 67.5° (up)2 = 22.5° 1088

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10–77. y

Solve Prob. 10–76 using Mohr’s circle.

6 in.

3 in.

O

Solution Moment And Product of Inertia About x and y Axes. Using the parallel-axis theorem,   Ix = Ix′ + Ad 2y;  Ix =

1 (6)(33) + 6(3)(1.52) = 54.0 in4 12

  Iy = Iy′ + Ad 2x;  Iy =

1 (3)(63) + 3(6)(32) = 216.0 in4 12

  Ixy = Ix′y′ + Ad xd y;  Ixy = 0 + 6(3)(3)(1.5) = 81.0 in4 Construction of the circle. The coordinates of center o of the circle is   a

Ix + Iy 2

, 0b = a

54.0 + 216.0 , 0b = (135, 0) 2

And the reference point A is   (Ix, Ixy) = (54.0, 81.0) Thus, the radius of the circle is R = OA = 2(135 - 54.0)2 + 81.02 = 8122

using these results, the circle shown in Fig. a can be constructed. Here, the coordinated of point B and C represent Imin and Imax respectively. Thus,

1089

x

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10–77. Continued

  Imax = 135 + 8122 = 249.55 in4 = 250 in4 4

Ans.

4

Ans.

  Imin = 135 - 8122 = 20.44 in = 20.4 in 

The orientation of the principal axes can be determined from the geometry of the shaded triangle on the circle.   tan 2(up)2 =

81.0 = 1.00 135 - 54.0

2(up)2 = 45° Ans.

(up)2 = 22.5° d And 2(up)1 = 180° - 2(up)2 2(up)1 = 180° - 45° = 135°

Ans.

(up)1 = 67.5° b

Ans: Imax = 250 in4 Imin = 20.4 in4 (up)2 = 22.5° d (up)1 = 67.5° b 1090

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10–78. The area of the cross section of an airplane wing has the following properties about the x and y axes passing through the centroid C: Ix = 450 in4, Iy = 1730 in4, Ixy = 138 in4. Determine the orientation of the principal axes and the principal moments of inertia.

y

SOLUTION - 2Ixy

tan 2u =

Ix - Iy

- 2(138) 450 - 1730

=

Ans.

u = 6.08° Imax/min = =

x

C

Ix + Iy 2

;

Ix - Iy 2 b + I2xy A 2 a

450 + 1730 450 - 1730 2 ; a b + 1382 2 A 2

Imax = 1.74(103) in4

Ans.

Imin = 435 in4

Ans.

Ans: u = 6.08° Imax = 1.74 ( 103 ) in4 Imin = 435 in4 1091

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10–79. Solve Prob. 10–78 using Mohr’s circle.

y

SOLUTION tan 2u =

- 2Ixy I x - Iy

=

- 2(138) 450 - 1730 Ans.

u = 6.08° Center of circle:

x

C

Ix + Iy 2

=

450 + 1730 = 1090 in4 2

R = 2(1730 - 1090)2 + (138)2 = 654.71 in4 Imax = 1090 + 654.71 = 1.74(103) in4

Ans.

Imin = 1090 - 654.71 = 435 in4

Ans.

Ans: u = 6.08° Imax = 1.74 ( 103 ) in4 Imin = 435 in4 1092

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*10–80. y

Determine the moments and product of inertia for the shaded area with respect to the u and v axes.

u

10 mm

v 60

10 mm x 10 mm

120 mm

Solution Moment And Product of Inertia About x and y Axes. Since the x axis is an axis of symmetry, Ixy = 0. Also, the x axis passes through the centroids of the two segments, Fig. a

Ix =

20 mm

120 mm

1 1 (20)(1203) + (120)(203) = 2.96(106) mm4 12 12

using the parallel-axis theorem,

Iy =

1 1 (120)(203) + c (20)(1203) + 20(120)(702) d = 14.72(106) mm4 12 12

Moment And Product of Inertia About the Inclined u and v Axes. with u = 60°,

Iu =

Ix + Iy 2

= a

Ix - Iy

+

2

cos 2u - Ixy sin 2u

2.96 - 14.72 2.96 + 14.72 + cos 120° - 0 sin 120°b(106) 2 2

= 11.78(106) mm4 = 11.8(106) mm4 Iv =

Ix + Iy 2

-

Ix - Iy 2

cos 2u + Ixy sin u

2.96 + 14.72 2.96 - 14.72 cos 120° + 0 sin 120°b(106) 2 2



= a



Iuv =



=



= - 5.0922(106) mm4 = - 5.09(106) mm4



= 5.90(106) mm4 Ix - Iy 2

sin 2u + Ixy cos 2u

2.96 - 14.72 sin 120° + 0 cos 120° 2

Ans: Iu = 11.8(106) mm4 Iv = 5.90(106) mm4 Iuv = - 5.09(106) mm4 1093

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10–81. y

Solve Prob. 10–80 using Mohr’s circle.

u

10 mm

v 60

10 mm x 10 mm

120 mm

Solution Moment And Product of Inertia About x and y Axes. Since the x axis is an axis of symmetry, Ixy = 0. Also, the x axis passes through the centroids of the two segments, Fig. a

Ix =

20 mm

120 mm

1 1 (20)(1203) + (120)(203) = 2.96(106) mm4 12 12

using the parallel-axis theorem,

Iy =

1 1 (120)(203) + c (20)(1203) + 20(120)(702) d = 14.72(106) mm4 12 12

Construction of the circle. The coordinate of center o of the circle is a

Ix + Iy 2

,0b = a

2.96 + 14.72 , 0b(106) = (8.84, 0)(106) 2

And the reference point A is

(Ix, Ixy) = (2.96, 0)(106) Thus, the radius of the circle is R = OA = a 2(8.84 - 2.96)2 + 02 b(106) = 5.88(106)

using these results, the circle shown in Fig. b can be constructed. Rotate radial line OA counterclockwise 2u = 120° to coincide with radial line OP where the coordinate of point P is ( Iu, Iuv ) . Then

Iu = (8.84 + 5.88 cos 60°)(106) = 11.78(106) mm4 = 11.8(106) mm4

Ans.



Iuv = 5.88(106) sin 60° = - 5.0922(106) mm4 = - 5.09(106) mm4

Ans.

Ir is represented by the coordinate of point Q. Thus

Iv = (8.84 - 5.88 cos 60°)(106) = 5.90(106) mm4

Ans.

Ans: Iu = 11.8 ( 106 ) mm4 Iuv = - 5.09 ( 106 ) mm4 Iv = 5.90 ( 106 ) mm4 1094

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10–82. Determine the directions of the principal axes with origin located at point O, and the principal moments of inertia for the area about these axes.

y

2 in.

2 in. 2 in. 1 in.

SOLUTION Ix = c

4 in.

1 1 (4)(6)3 + (4)(6)(3)2 d - c p(1)4 + p(1)2(4)2 d 12 4

= 236.95 in4 Iy = c

x

O

1 1 (6)(4)3 + (4)(6)(2)2 d - c p(1)4 + p(1)2(2)2 d 12 4

= 114.65 in4 Ixy = [0 + (4)(6)(2)(3)] - [0 + p (1)(2)(4)] = 118.87 in4 tan 2uP =

-Ixy

Ix - Iy

=

2

- 118.87 (236.95 - 114.65) 2

uP = -31.388°;

58.612°

Thus, uP1 = - 31.4°; Imax =

Ix + Iy 2

min

=

Ans.

uP2 = 58.6° ;

C

a

Ix - Iy 2

2

b + I2xy

236.95 + 114.65 236.95 - 114.65 2 ; a b + (118.87)2 2 C 2

Imax = 309 in4

Ans.

Imin = 42.1 in4

Ans.

Ans: up1 = - 31.4° up2 = 58.6° Imax = 309 in4 Imin = 42.1 in4 1095

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10–83. Solve Prob. 10–82 using Mohr’s circle.

y

2 in.

2 in. 2 in. 1 in.

SOLUTION Ix = c

4 in.

1 1 (4)(6)3 + (4)(6)(3)2 d - c p(1)4 + p(1)2(4)2 d 12 4

= 236.95 in4 Iy = c

x

O

1 1 (6)(4)3 + (4)(6)(2)2 d - c p(1)4 + p(1)2(2)2 d 12 4

= 114.65 in4 Ixy = [0 + (4)(6)(2)(3)] - [0 + p (1)(2)(4)] = 118.87 in4 Center of circle:

I x + Iy 2

=

236.95 + 114.65 = 175.8 in4 2

R = 2(236.95 - 175.8)2 + (118.87)2 = 133.68 in4 Imax = (175.8 + 133.68) = 309 in4

Ans.

Imin = (175.8 - 133.68) = 42.1 in4

Ans.

2up1 = tan-1 a

118.87 b = 62.78° (236.95 - 175.8)

up1 = - 31.4°

Ans.

up2 = 90° - 31.4° = 58.6°

Ans.

Ans: Imax = 309 in4 Imin = 42.1 in4 up1 = - 31.4° up2 = 58.6° 1096

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*10–84. y

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

R x

SOLUTION 2p

Iz =

r A(R du)R2 = 2p r A R3

L0 2p

m =

L0

r A R du = 2p r A R

Thus, Iz = m R 2

Ans.

Ans: Iz = m R2 1097

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10–85. Determine the moment of inertia of the ellipsoid with respect to the x axis and express the result in terms of the mass m of the ellipsoid. The material has a constant density r.

y x2 a2

y2 b2

1 b x

SOLUTION a

dm= py2dx L d Ix =

y2dm 2 a

m =

LV

r dV =

L-a

rp b a 1 2

2

x a

2

b dx =

4 2 prab 3

a

Ix =

x2 2 8 1 prab4 rpb4 a 1 - 2 b dx = 2 15 a L-a

Thus, Ix =

2 mb2 5

Ans.

Ans: Ix = 1098

2 mb2 5

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10–86. Determine the radius of gyration kx of the paraboloid. The density of the material is r = 5 Mg>m3.

y y2  50 x 100 mm x

SOLUTION

200 mm

Differential Disk Element: The mass dm = rdV = rpy2 dx = rp(50x) dx. The 1 1 is dIx = dmy2 = [rp(50x) dx](50x) = 2 2

of the differential disk element is mass moment of inertia of this element rp (2500x2) dx. 2

Total Mass: Performing the integration, we have m =

dm =

Lm

200 mm

mm rp(50x)dx = rp(25x2)|200 = 1(106)rp 0

L0

Mass Moment of Inertia: Performing the integration, we have Ix =

L

dIx = =

200 mm

L0

rp (2500x2) dx 2

rp 2500x3 200 mm a b` 2 3 0

= 3.333(109)rp The radius of gyration is kx =

Ix 3.333(109)rp = = 57.7 mm Am A 1(106)rp

Ans.

Ans: kx = 57.7 mm 1099

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10–87. The paraboloid is formed by revolving the shaded area around the x axis. Determine the moment of inertia about the x axis and express the result in terms of the total mass m of the paraboloid. The material has a constant density r.

y 2

y2 = a–h x a x

SOLUTION

h 2

dm = r dV = r (p y dx)

d Ix =

1 1 dm y2 = r p y4 dx 2 2 h

Ix = =

1 a4 r p a 2 b x2 dx h L0 2 1 p ra4 h 6 h

m = = Ix =

1 a2 r p a bx dx h L0 2 1 r p a2 h 2 1 ma2 3

Ans.

Ans: Ix = 1100

1 ma2 3

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*10–88. Determine the moment of inertia of the homogenous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x-y plane having a thickness of dz.

z

z = –h a (x – a)

h

SOLUTION Differential Thin Plate Element: Here, x = a a 1 -

z b . The mass of the h

x

b

a

z differential thin plate element is dm = rdV = rbxdz = rab a1 - b dz. The mass h moment of inertia of this element about y axis is dIy = dIG + dmr2 =

x2 1 dmx2 + dm ¢ + z2 ≤ 12 4

=

1 2 x dm + z2 dm 3

= B =

z 2 a2 z a 1 - b + z2 R B raba 1 - b dz R 3 h h

rab 2 a2 3a2 3a2 3z3 z - 3 z3 + 3z2 b dz ¢ a + 2 z2 3 h h h h

Total Mass: Performing the integration, we have h

m =

Lm

dm =

L0

raba 1 -

z z2 h 1 b dz = rab ¢ z ≤ ` = rabh h 2h 0 2

Mass Moment of Inertia: Performing the integration, we have h

Iy =

L

dIy =

rab 2 3a2 3a2 a2 3z3 z - 3 z3 + 3z2 b dz ¢ a + 2 z2 h h h h L0 3

=

rab 2 a2 3a2 2 a2 4 3z4 h 3 z + z z ¢ a z + 2 z3 ≤` 3 2h 4h 0 h 4h3

=

rabh 2 1a + h22 12

The mass moment of inertia expressed in terms of the total mass is Iy =

1 rabh 6 2

a2 + h2 =

m 2 a + h2 6

Ans.

Ans: Iy = 1101

m 2 (a + h2) 6

y

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10–89. Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density r.

y

x2 y2  1 a2 b2 b x

SOLUTION

a

2 2 Differential Disk Element: Here, y = b a 1 -

x2 b . The mass of the differential disk element is a2

x2 b dx. The mass moment of inertia of this element is a2 2 rp b4 x4 1 1 x x2 2x2 dIx = dmy2 = c rp b2 a 1 - 2 b dx d c b2 a 1 - 2 b d = a 4 - 2 + 1b dx. 2 2 2 a a a a dm = rdV = rp y2 dx = rp b2 a 1 -

Total Mass: Performing the integration, we have a

m =

dm =

Lm

L0

rp b2 a 1 -

x2 x3 a 2 b dx = rp b a x b` a2 3a2 0 =

2 rpab2 3

Mass Moment of Inertia: Performing the integration, we have a

Ix =

L

dIx =

rp b4 x4 2x2 a 4 - 2 + 1b dx a a L0 2

=

a rp b4 x5 2x3 a 4 + x b ` 2 5a 3a2 0

=

4 rp ab4 15

The mass moment of inertia expressed in terms of the total mass is. Ix =

2 2 2 a rp ab2 b b2 = mb2 5 3 5

Ans.

Ans: Ix = 1102

2 mb2 5

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10–90. Determine the radius of gyration kx of the solid formed by revolving the shaded area about x axis. The density of the material is r.

y

n

yn h x a h x

a

Solution Differential Disk Element. The mass of the differential disk element shown h 1 h 1 2 shaded in Fig. a is dm = rdv = rpy2dx. Here y = 1 x n . Thus, dm = rp a 1 xn b an an rph2 2 dx = xndx. The mass moment of Inertia of this element about the x axis is 2 an rxh4 4 1 1 rph2 n2 h 1 2 xn dxb a 2>n x dxb a 2 xn b = a dIx = (dm)y2 = 4 2 2 a an 2an Total Mass. Perform the integration,

m =

Lm

dm =

L0

= a



= a



a

rph2 a

rph2 a

2 n

2 n

2

(x ndx)

ba

n b axn n + 2

+ 2 n b

2

a 0

n brpah2 n + 2

Mass Moment of Inertia. Perform the integration,





Ix =

a rph4 4 dIx = (xn dx) 4 L L0 2an a

rph4

n n+4 bax b3 = a 4>n ba n n + 4 2a = c

0

n d rpah4 2(n + 4)

The radius of gyration is



Ix kx = = Am

c

n d rp ah4 2(n + 4)

n b rp ah2 a c n+4

=

n + 2 h A 2(n + 4)

Ans.

Ans: kx = 1103

n + 2 h A 2(n + 4)

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10–91. y

The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of inertia Iy. The specific weight of concrete is g = 150 lb>ft3.

6 in.

2 2 8 in. y x 9

SOLUTION d Iy = =

1 1 (dm)(10)2 - (dm)x2 2 2 1 1 [pr (10)2 dy](10)2 - prx2 dyx2 2 2 8

Iy =

4 in.

8

1 9 2 pr B (10)4 dy a b y2 dy R 2 L0 2 L0 =

1 2 p (150) 3

32.2(12)

B (10)4(8) - a b a b (8)3 R 9 2

2

1 3

= 324.1 slug # in2 Iy = 2.25 slug # ft2

Ans.

Ans: Iy = 2.25 slug # ft 2 1104

x

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*10–92. Determine the moment of inertia Ix of the sphere and express the result in terms of the total mass m of the sphere. The sphere has a constant density r.

y

x2

y2

r2

r x

SOLUTION d Ix =

y2 dm 2

dm = r dV = r(py2 dx) = rp(r2 - x2)dx d Ix =

1 rp(r2 - x2)2 dx 2 r

Ix = =

1 rp(r2 - x2)2 dx L-r 2 8 prr5 15 r

m = =

L-r

rp(r2 - x2) dx

4 rpr3 3

Thus, Ix =

2 2 mr 5

Ans.

Ans: Ix = 1105

2 mr 2 5

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10–93. The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the cone. The cone has a constant density r.

y y  –hr x r x

SOLUTION dm = r dV = r(p y2 dx)

h

h

1 1 r2 r2 m = r(p) ¢ 2 ≤ x2 dx = rp ¢ 2 ≤ a b h3 = rp r2h 3 3 h h L0 dIx =

1 2 y dm 2

=

1 2 y (rp y2 dx) 2

=

1 r4 r(p)a 4 b x4 dx 2 h h

Ix =

1 1 r4 r(p)a 4 b x4 dx = rp r4 h 2 10 h L0

Thus, Ix =

3 m r2 10

Ans.

Ans: Ix = 1106

3 mr 2 10

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10–94. Determine the mass moment of inertia Iy of the solid formed by revolving the shaded area around the y axis. The total mass of the solid is 1500 kg.

z

4m

z2

1 y3 –– 16

2m y

O

SOLUTION

x

Differential Element: The mass of the disk element shown shaded in 2 1 1 dm = rdV = rpr2dy. Here, r = z = y3>2.Thus, dm = rpa y3>2 b dy = 4 4 The mass moment of inertia of this element about the y rp 4 rp 1 3>2 4 rp 6 1 1 r dy = a y b dy = y dy. dIy = dmr2 = A rpr2dy B r2 = 2 2 2 2 4 512

Fig. a is rp 3 y dy. 16 axis is

Mass: The mass of the solid can be determined by integrating dm. Thus, 4m

m =

L

dm =

L0

4

4m rp y rp 3 = 4 pr y dy = ¢ ≤` 16 16 4 0

The mass of the solid is m = 1500 kg. Thus, 1500 = 4pr

r =

375 kg>m3 p

Mass Moment of Inertia: Integrating dIy, 4m

Iy =

L

Substituting r =

dIy =

L0

rp 6 rp y7 4 m 32p y dy = r = ¢ ≤` 512 512 7 0 7

375 kg>m3 into Iy, p Iy =

32p 375 b = 1.71(103) kg # m2 a p 7

Ans.

Ans: Iy = 1.71 ( 103 ) kg # m2 1107

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10–95. The slender rods have a mass of 4 kg>m. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A.

A

200 mm

Solution

100 mm

100 mm

Mass Moment of Inertia About An Axis Through A. The mass of each segment is mi = (4 kg>m)(0.2 m) = 0.8 kg. The mass moment inertia of each segment shown in Fig. a about an axis through their center of mass can be determined using 1 (IG)i = m l i2. 12 i



IA = Σ c (IG)i + mi d 2i d = c

1 1 (0.8) ( 0.22 ) + 0.8 ( 0.12 ) d + c (0.8) ( 0.22 ) + 0.8 ( 0.22 ) d 12 12



= 0.04533 kg # m2



= 0.0453 kg # m2



Ans.

Ans: IA = 0.0453 kg # m2 1108

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*10–96. The pendulum consists of a 8-kg circular disk A, a 2-kg circular disk B, and a 4-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

A O

0.4 m

1m

0.5 m

B

0.2 m

Solution Mass Moment of Inertia About An Axis Through O. The mass moment of inertia of each rod segment and disk segment shown in Fig. a about an axis passes through 1 1 their center of mass can be determined using (IG)i = m l i2 and (IG)i = mi r i2. 12 i 2 IO = Σ c (IG)i + mi d 2i d   

= c

1 1 (4) ( 1.52 ) + 4 ( 0.252 ) d + c (2) ( 0.12 ) + 2 ( 0.62 ) d 12 2   

  

= 13.41 kg # m2

1 + c (8) ( 0.22 ) + 8 ( 1.22 ) d 2

The total mass is 8 kg + 2 kg + 4 kg = 14 kg The radius of gyration is kO =

13.41 kg # m2 IO = = 0.9787 m = 0.979 m Am C 14 kg

Ans.

Ans: kO = 0.979 m 1109

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10–97. Determine the moment of inertia Iz of the frustum of the cone which has a conical depression. The material has a density of 200 kg>m3.

0.2 m

z

0.8 m 0.6 m

SOLUTION Iz =

3 1 [ p (0.4)2(1.6)(200)](0.4)2 10 3 -

3 1 [ p (0.2)2(0.8)(200)](0.2)2 10 3

-

3 1 [ p (0.4)2(0.6)(200)](0.4)2 10 3

0.4 m

Iz = 1.53 kg # m2

Ans.

Ans: Iz = 1.53 kg # m2 1110

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10–98. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then find the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

O

y

SOLUTION y =

G

©ym 1(3) + 2.25(5) = = 1.781 m = 1.78 m ©m 3 + 5

0.5 m

Ans. 1m

IG = ©IG¿ + md2 =

2m

1 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12

= 4.45 kg # m2

Ans.

Ans: y = 1.78 m IG = 4.45 kg # m2 1111

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10–99. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

O

50 mm

150 mm 50 mm

SOLUTION

150 mm

400 mm

Composite Parts: The plate can be subdivided into the segments shown in Fig. a. Here, the four similar holes of which the perpendicular distances measured from their centers of mass to point C are the same and can be grouped as segment (2). This segment should be considered as a negative part.

400 mm

150 mm 150 mm

Mass Moment of Inertia: The mass of segments (1) and (2) are m1 = (0.4)(0.4)(20) = 3.2 kg and m2 = p(0.052)(20) = 0.05p kg, respectively. The mass moment of inertia of the plate about an axis perpendicular to the page and passing through point C is IC =

1 1 (3.2)(0.4 2 + 0.4 2) - 4 c (0.05p)(0.052) + 0.05p(0.152) d 12 2

= 0.07041 kg # m2

The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O can be determined using the parallel-axis theorem IO = IC + md2, where m = m1 - m2 = 3.2 - 4(0.05p) = 2.5717 kg and d = 0.4 sin 45°m. Thus, IO = 0.07041 + 2.5717(0.4 sin 45°)2 = 0.276 kg # m2

Ans.

Ans: IO = 0.276 kg # m2 1112

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*10–100. The pendulum consists of a plate having a weight of 12 lb and a slender rod having a weight of 4 lb. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

O

1 ft 1 ft

3 ft

2 ft

SOLUTION IO = ©IG + md2 =

1 4 4 1 12 12 a b (5)2 + a b (0.5)2 + a b (12 + 12) + a b(3.5)2 12 32.2 32.2 12 32.2 32.2

= 4.917 slug # ft2 m = a kO =

4 12 b + a b = 0.4969 slug 32.2 32.2

IO 4.917 = = 3.15 ft Am A 0.4969

Ans.

Ans: kO = 3.15 ft 1113

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10–101. If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

4 ft 1 ft O

SOLUTION Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a 3 distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2). 2

A

Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = ¢

100 1 20 20 15 ≤ (42) + 8 B ¢ ≤ (32) + ¢ ≤ (2.52) R + ¢ ≤ (12) 32.2 12 32.2 32.2 32.2

= 84.94 slug # ft2 The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem IA = IO + md2, 100 15 20 where m = + 8¢ = 8.5404 slug and d = 4 ft. Thus, ≤ + 32.2 32.2 32.2 IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2

Ans.

Ans: IA = 222 slug # ft 2 1114

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10–102. z

Determine the mass moment of inertia of the assembly about the z axis. The density of the material is 7.85 Mg> m3.

100 mm

SOLUTION Composite Parts: The assembly can be subdivided into two circular cone segments (1) and (3) and a hemispherical segment (2) as shown in Fig. a. Since segment (3) is a hole, it should be considered as a negative part. From the similar triangles, we obtain z 0.1 = 0.45 + z 0.3

450 mm 300 mm

z = 0.225m

Mass: The mass of each segment is calculated as

300 mm

1 1 m1 = rV1 = r a pr2h b = 7.85(103) c p(0.32)(0.675) d = 158.9625p kg 3 3

x

y

2 2 m2 = rV2 = r a pr3 b = 7.85(103) c p(0.33) d = 141.3p kg 3 3 1 1 m3 = rV3 = r a pr2h b = 7.85(103) c p(0.12)(0.225) d = 5.8875p kg 3 3 Mass Moment of Inertia: Since the z axis is parallel to the axis of the cone and the hemisphere and passes through their center of mass, the mass moment of inertia can be 3 3 2 computed from (Iz)1 = m r12, (Iz)2 = m2r22, and m r32. Thus, 10 1 5 10 3 Iz = ©(Iz)i =

3 2 3 (158.9625p)(0.32) + (141.3p)(0.32) (5.8875p)(0.12) 10 5 10

= 29.4 kg # m2

Ans.

Ans: Iz = 29.4 kg # m2 1115

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10–103. Each of the three slender rods has a mass m. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through the center point O. a

a O

SOLUTION IO = 3 B

1 a sin 60° 2 1 ma2 + m ¢ ≤ R = ma2 12 3 2

Ans.

a

Ans: IO = 1116

1 ma2 2

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*10–104. The thin plate has a mass per unit area of 10 kg>m2. Determine its mass moment of inertia about the y axis.

z 200 mm 100 mm

200 mm

200 mm

SOLUTION

100 mm

Composite Parts: The thin plate can be subdivided into segments as shown in Fig. a. Since the segments labeled (2) are both holes, the y should be considered as negative parts.

200 mm 200 mm x

200 mm

200 mm 200 mm

Mass moment of Inertia: The mass of segments (1) and (2) are m1 = 0.4(0.4)(10) = 1.6 kg and m2 = p(0.12)(10) = 0.1p kg. The perpendicular distances measured from the centroid of each segment to the y axis are indicated in Fig. a. The mass moment of inertia of each segment about the y axis can be determined using the parallel-axis theorem. Iy = © A Iy B G + md2 = 2c

1 1 (1.6)(0.42) + 1.6(0.22) d - 2 c (0.1p)(0.12) + 0.1p(0.22) d 12 4

= 0.144 kg # m2

Ans.

Ans: Iy = 0.144 kg # m2 1117

y

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10–105. The thin plate has a mass per unit area of 10 kg>m2. Determine its mass moment of inertia about the z axis.

z 200 mm 100 mm

200 mm

200 mm

SOLUTION

100 mm

Composite Parts: The thin plate can be subdivided into four segments as shown in Fig. a. Since segments (3) and (4) are both holes, the y should be considered as negative parts.

200 mm 200 mm x

200 mm

200 mm

200 mm

Mass moment of Inertia: Here, the mass for segments (1), (2), (3), and (4) are m1 = m2 = 0.4(0.4)(10) = 1.6 kg and m3 = m4 = p(0.12)(10) = 0.1p kg. The mass moment of inertia of each segment about the z axis can be determined using the parallel-axis theorem. Iz = © A Iz B G + md2 =

1 1 1 1 (1.6)(0.42) + c (1.6)(0.42 + 0.42) + 1.6(0.22) d - (0.1p)(0.12) - c (0.1p)(0.12) + 0.1p(0.22) d 12 12 4 2

= 0.113 kg # m2

Ans.

Ans: Iz = 0.113 kg # m2 1118

y

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10–106. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through the center of mass G. The material has a specific weight of g = 90 lb>ft3.

O

1 ft G 2 ft 0.25 ft

SOLUTION IG =

0.5 ft 1 ft

1 90 1 90 ca b p(2.5)2(1) d (2.5)2 - c a b p(2)2(1) d(2)2 2 32.2 2 32.2 +

1 90 1 90 ca b p(2)2(0.25) d (2)2 - c a b p(1)2(0.25) d(1)2 2 32.2 2 32.2

= 118 slug # ft2

Ans.

Ans: IG = 118 slug # ft 2 1119

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10–107. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through point O. The material has a specific weight of g = 90 lb>ft3.

O

1 ft G 2 ft 0.25 ft

SOLUTION IG =

0.5 ft 1 ft

1 90 1 90 ca b p(2.5)2(1) d (2.5)2 - c a b p(2)2(1) d(2)2 2 32.2 2 32.2 +

1 90 1 90 ca b p(2)2(0.25) d (2)2 - c a b p(1)2(0.25) d(1)2 2 32.2 2 32.2

= 117.72 slug # ft2 IO = IG + md2 m = a

90 90 b p(2 2 - 12)(0.25) + a b p(2.52 - 2 2)(1) = 26.343 slug 32.2 32.2

IO = 117.72 + 26.343(2.5)2 = 282 slug # ft2

Ans.

Ans: IO = 282 slug # ft 2 1120

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*10–108. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg>m. The thin plate has a mass of 12 kg>m2. Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

0.4 m

0.4 m

A

B O

_ y G

1.5 m

SOLUTION y =

1.5(3)(0.75) + p(0.3)2(12)(1.8) - p(0.1)2(12)(1.8)

0.1 m

Ans.

= 0.8878 m = 0.888 m IG =

C

1.5(3) + p(0.3)2(12) - p(0.1)2(12) + 0.8(3)

0.3 m

1 (0.8)(3)(0.8)2 + 0.8(3)(0.8878)2 12 +

1 (1.5)(3)(1.5)2 + 1.5(3)(0.75 - 0.8878)2 12

1 + [p(0.3)2(12)(0.3)2 + [p(0.3)2(12)](1.8 - 0.8878)2 2 -

1 [p(0.1)2(12)(0.1)2 - [p(0.1)2(12)](1.8 - 0.8878)2 2

IG = 5.61 kg # m2

Ans.

Ans: y = 0.888 m IG = 5.61 kg # m2 1121

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10–109. Determine the moment of inertia Iz of the frustrum of the cone which has a conical depression. The material has a density of 200 kg>m3.

z

200 mm

600 mm

SOLUTION

400 mm

z + 1 z = , Mass Moment of Inertia About z Axis: From similar triangles, 0.2 0.8 z = 0.333 m. The mass moment of inertia of each cone about z axis can be 3 determined using Iz = mr2. 10 Iz = ©1Iz2i =

800 mm

3 p c 10.82211.333212002 d10.822 10 3 -

3 p c 10.2 2210.333212002 d10.2 22 10 3

-

3 p c 10.2 2210.6212002 d10.2 22 10 3

= 34.2 kg # m2

Ans.

Ans: Iz = 34.2 kg # m2 1122

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11–1. Use the method of virtual work to determine the tensions in cable AC. The lamp weighs 10 lb.

B C

45°

SOLUTION

A

30°

Free Body Diagram: The tension in cable AC can be determined by releasing cable AC. The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only FAC and the weight of lamp (10 lb force) do work. Virtual Displacement: Force FAC and 10 lb force are located from the fixed point B using position coordinates yA and xA. xA = l cos u

dxA = - l sin udu

(1)

yA = l sin u

dyA = l cos udu

(2)

Virtual–Work Equation: When yA and xA undergo positive virtual displacements dyA and dxA, the 10 lb force and horizontal component of FAC, FAC cos 30°, do positive work while the vertical component of FAC, FAC sin 30°, does negative work. dU = 0;

10dyA - FAC sin 30°dyA + FAC cos 30°, dxA = 0

(3)

Substituting Eqs. (1) and (2) into (3) yields (10 cos u - 0.5FAC cos u - 0.8660FAC sin u)ldu = 0 Since ldu Z 0, then FAC =

10 cos u 0.5 cos u + 0.8660 sin u

At the equilibrium position u = 45°, FAC =

10 cos 45° = 7.32 lb 0.5 cos 45° + 0.8660 sin 45°

Ans.

Ans: FAC = 7.32 lb 1123

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11–2. The scissors jack supports a load P. Determine the axial force in the screw necessary for equilibrium when the jack is in the position u. Each of the four links has a length L and is pin-connected at its center. Points B and D can move horizontally.

P

SOLUTION x = L cos u,

dx = -L sin u du

y = 2L sin u,

dy = 2L cos u du

dU = 0;

- Pdy - Fdx = 0

C

D

A

B u

-P(2L cos u du) - F(- L sin u du) = 0 Ans.

F = 2P cot u

Ans: F = 2P cot u 1124

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11–3. If a force of P = 5 lb is applied to the handle of the mechanism, determine the force the screw exerts on the cork of the bottle. The screw is attached to the pin at A and passes through the collar that is attached to the bottle neck at B.

P  5 lb

D

SOLUTION

A

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the force in the screw Fs and force P do work when the virtual displacements take place.

u  30° 3 in.

B

Virtual Displacement: The position of the points of application for Fs and P are specified by the position coordinates yA and yD, measured from the fixed point B, respectively. yA = 2(3 sin u)

dyA = 6 cos udu

(1)

yD = 6(3 sin u)

dyD = 18 cos udu

(2)

Virtual Work Equation: Since P acts towards the positive sense of its corresponding virtual displacement, it does positive work. However, the work of Fs is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0;

PdyD +

A - FSdyA B = 0

(3)

Substituting P = 5 lb, Eqs. (1) and (2) into Eq. (3), 5(18 cos udu)FS (6 cos udu) = 0 cos udu(90 - 6FS) = 0 Since cos udu Z 0, then 90 - 6FS = 0 Ans.

FS = 15 lb

Ans: FS = 15 lb 1125

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*11–4. The disk has a weight of 10 lb and is subjected to a vertical force P = 8 lb and a couple moment M = 8 lb # ft. Determine the disk’s rotation u if the end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched.

M  8 lb  ft 1.5 ft

k  12 lb/ft

SOLUTION

P  8 lb

dyF = dyP = 1.5du PdyP + Mdu - FdyF = 0

dU = 0;

8(1.5 du) + 8 du - F(1.5 du) = 0 u(20 - 1.5F) = 0 Since du Z 0 20 -1.5F = 0

F = 13.33 lb

F = kx Where x = 1.5u 13.33 = 12(1.5u) Ans.

u = 0.7407 rad = 42.4°

Ans: u = 42.4° 1126

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11–5. The punch press consists of the ram R, connecting rod AB, and a flywheel. If a torque of M = 75 N # m is applied to the flywheel, determine the force F applied at the ram to hold the rod in the position u = 60°.

B

200 mm

600 mm R

u F M

A

Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only force F and couple moment M do work. Virtual Displacement. The position of force F is measured from fixed point O by position coordinate xA. Applying the law of cosines by referring to Fig. b,

0.62 = x2A + 0.22 - 2xA(0.2) cos u

(1)

Differentiating the above expression, 0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA

dxA =

0.4xA sin u du 0.4 cos u - 2xA

(2)

Virtual–Work Equation. When point A undergoes a positive virtual displacement, and the flywheel undergoes positive virtual angular displacement du, both F and M do negative work.

(3)

- FdxA - Mdu = 0

dU = 0;

Substituting Eq. (2) into (3)

Since du ≠ 0, then

ca

0.4 xA sin u bF + M d du = 0 0.4 cos u - 2xA 0.4 xA sin u F + M = 0 0.4 cos u - 2xA



F = -a

0.4 cos u - 2xA bM (4) 0.4 xA sin u

At the equilibrium position u = 60°, Eq. (1) gives

0.62 = xA2 + 0.22 - 2xA(0.2) cos 60° xA = 0.6745 m

Substitute M = 75 N # m, u = 60° and this result into Eq. (4) F = -c

0.4 cos 60° - 2(0.6745) 0.4(0.6745) sin 60°

= 368.81 N = 369 N

d (75)

Ans.

Ans: F = 369 N 1127

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11–6. The flywheel is subjected to a torque of M = 75 N # m. Determine the horizontal compressive force F and plot the result of F (ordinate) versus the equilibrium position u (abscissa) for 0° … u … 180°.

B

200 mm

R

u F M

Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only force F and couple moment M do work. Virtual Displacement. The position of force F is measured from fixed point O by position coordinate xA. Applying the law of cosines by referring to Fig. b, 0.62 = xA2 + 0.22 - 2xA(0.2) cos u



(1)

Differentiating the above expression, 0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA

dxA =

0.4xA sin u 0.4 cos u - 2xA

(2)

Virtual–Work Equation. When point A undergoes a positive virtual displacement and the flywheel undergoes positive virtual angular displacement du, both F and M do negative work.

(3)

- FdxA - Mdu = 0

dU = 0;

Substituting Eq. (2) into (3) ca

Since du ≠ 0, then

0.4 xA sin u bF + M d du = 0 0.4 cos u - 2xA

a

Here, M = 75 N # m. Then

0.4 xA sin u bF + M = 0 0.4 cos u - 2xA F = -M a

0.4 cos u - 2xA b 0.4 xA sin u

F = - 75 a

0.4 cos u - 2xA b (4) 0.4 xA sin u

Using Eq. (1) and (4), the following tabulation can be computed. Subsequently, the graph of F vs u shown in Fig. c can be plotted u(deg.)

0

xA(m)

0.80

F(N)

∞

30

60

90

120

150

0.7648 0.6745 0.5657 0.4745 0.4184 580

369

375

524

1060

600 mm

180 0.4

15

73.0

0.7909 0.6272

∞

1128

1095

356

A

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11–7. When u = 20°, the 50-lb uniform block compresses the two vertical springs 4 in. If the uniform links AB and CD each weigh 10 lb, determine the magnitude of the applied couple moments M needed to maintain equilibrium when u = 20°.

k

2 lb/in.

k

B

D

2 lb/ in. 1 ft

1 ft

SOLUTION

u

Free Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the spring forces Fsp, the weight of the block (50 lb), the weights of the links (10 lb) and the couple moment M do work.

u

M A

1 ft 2 ft M

C

4 ft

Virtual Displacements: The spring forces Fsp, the weight of the block (50 lb) and the weight of the links (10 lb) are located from the fixed point C using position coordinates y3, y2 and y1 respectively. y3 = 1 + 4 cos u y2 = 0.5 + 4 cos u y1 = 2 cos u

(1)

dy3 = -4 sin udu

(2)

dy2 = -4 sin udu

(3)

dy1 = - 2 sin udu

Virtual–Work Equation: When y1, y2 and y3 undergo positive virtual displacements dy1, dy2 and dy3, the spring forces Fsp, the weight of the block (50 lb) and the weights of the links (10 lb) do negative work. The couple moment M does negative work when the links undergo a positive virtual rotation du. dU = 0;

- 2Fspdy3 - 50dy2 - 20dy1 - 2Mdu = 0

(4)

Substituting Eqs. (1), (2) and (3) into (4) yields (8Fsp sin u + 240 sin u - 2M) du = 0 Since du Z 0, then 8Fsp sin u + 240 sin u - 2M = 0 M = sin u(4Fsp + 120) At the equilibrium position u = 20°, Fsp = kx = 2(4) = 8 lb. M = sin 20°[4(8) + 120] = 52.0 lb # ft

Ans.

Ans: M = 52.0 lb # ft 1129

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*11–8. The bar is supported by the spring and smooth collar that allows the spring to be always perpendicular to the bar for any angle u. If the unstretched length of the spring is l0, determine the force P needed to hold the bar in the equilibrium position u. Neglect the weight of the bar.

a A

B u k C l

SOLUTION s = a sin u,

ds = a cos u du

y = l sin u,

dy = l cos u du

Fs = k(a sin u - l0) dU = 0;

P

Pdy - Fsds = 0

Pl cos du - k(a sin u - l0) a cos u du = 0 P =

ka (a sin u - l0) l

Ans.

Ans: P = 1130

ka (a sin u - l0) l

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11–9. The 4-ft members of the mechanism are pin-connected at their centers. If vertical forces P1 = P2 = 30 lb act at C and E as shown, determine the angle u for equilibrium. The spring is unstretched when u = 45°. Neglect the weight of the members.

P1

P2

E

C k = 200 lb/ft 2 ft

2 ft

SOLUTION y = 4 sin u,

x = 4 cos u

dy = 4 cos u du, dU = 0;

B

2 ft

dx = - 4 sin u du A

- Fsdx - 30dy - 30dy = 0

θ

2 ft

D

3 - Fs1 -4 sin u2 - 6014 cos u24du = 0 Fs = 60 a

cos u b sin u

Since Fs = k14 cos u - 4 cos 45°2 = 20014 cos u - 4 cos 45°2 60 cos u = 8001cos u - cos 45°2sin u sin u - 0.707 tan u - 0.075 = 0 Ans.

u = 16.6° and

Ans.

u = 35.8°

Ans: u = 16.6° u = 35.8° 1131

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11–10. The thin rod of weight W rest against the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibrium for a given angle u.

B

SOLUTION

l

Free-Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the weight of the rod W and force P do work. Virtual Displacements: The weight of the rod W and force P are located from the fixed points A and B using position coordinates yC and xA, respectively yC =

1 sin u 2

dyC =

xA = l cos u

1 cos udu 2

dxA = - l sin udu

P

A

θ

(1) (2)

Virtual-Work Equation: When points C and A undergo positive virtual displacements dyC and dxA, the weight of the rod W and force F do negative work. (3)

dU = 0; - WdyC - PdyA = 0 Substituting Eqs. (1) and (2) into (3) yields a Pl sin u -

Wl cos ub du = 0 2

Since du Z 0, then Pl sin u P =

Wl cos u = 0 2

W cot u 2

Ans.

Ans: P = 1132

W cot u 2

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11–11. If each of the three links of the mechanism have a mass of 4 kg, determine the angle u for equilibrium. The spring, which always remains vertical, is unstretched when u = 0°.

A

M  30 N  m u 200 mm

k  3 kN/ m D

200 mm

Solution

B

Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the weights W1 = W2 = W3 = W, couple moment M, and spring force Fsp do work. Virtual Displacement. The positions of the weights W1, W2, W3 and spring force Fsp are measured from fixed point A using position coordinates y1, y2, y3 and y4 respectively

y1 = 0.1 sin u

dy1 = 0.1 cos u du

(1)



y2 = 0.2 sin u + 0.1

dy2 = 0.2 cos u du

(2)



y3 = 0.1 sin u + 0.2

dy3 = 0.1 cos u du

(3)



y4 = 0.5 sin u

dy4 = 0.2 cos u du

(4)

Virtual Work Equation. When all the weights undergo positive virtual displacement, all of them do positive work. However, Fsp does negative work when its undergoes positive virtual displacement. Also, M does positive work when it undergoes positive virtual angular displacement.

dU = 0;

W1dy1 + W2dy2 + W3dy3 - Fspdy4 + Mdu = 0

(5)

Substitute Eqs (1), (2) and (3) into (5), using W1 = W2 = W3 = W. W(0.1 cos u du) + W(0.2 cos u du) + W(0.1 cos u du) - Fsp(0.2 cos u du) + Mdu = 0 (0.4 W cos u - 0.2 Fsp cos u + M)du = 0 Since d ≠ 0, then 0.4 W cos u - 0.2 Fsp cos u + M = 0

Here M = 30 N # m, W = 4(9.81)N = 39.24 N and Fsp = kx = 3000(0.2 sin u) = 600 sin u. Substitute these results into this equation, 0.4(39.24) cos u - 0.2(600 sin u)cos u + 30 = 0 15.696 cos u - 120 sin u cos u + 30 = 0

1133

u 200 mm

C

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11–11. Continued

Solve numerically:

Ans.

u = 23.8354° = 23.8° or



Ans.

u = 72.2895° = 72.3°

Note: u = 23.8° is a stable equilibrium, while u = 72.3° is an unstable one.

Ans: u = 23.8° u = 72.3° 1134

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*11–12. The disk is subjected to a couple moment M. Determine the disk’s rotation u required for equilibrium. The end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched.

k  4 kN/ m

0.5 m

M  300 N  m

Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the spring force Fsp and couple moment M do work. Virtual Work Equation. When the disk undergoes a positive angular displacement du, correspondingly point A undergoes a positive displacement of dxA. As a result couple moment M does positive work whereas spring force Fsp does negative work.

dU = 0;

(1)

Mdu + ( - FspdxA) = 0

Here, Fsp = kxA = 4000(0.50) = 2000u and dxA = 0.5du. Substitute these results into Eq. (1) 300du - 2000u(0.5du) = 0 (300 - 1000u)du = 0 Since du ≠ 0, 300 - 1000u = 0 u = 0.3 rad = 17.19° = 17.2°          Ans.

Ans: u = 17.2° 1135

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11–13. A 5-kg uniform serving table is supported on each side by pairs of two identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg, determine the angle u where the table is in equilibrium. The springs each have a stiffness of k = 200 N>m and are unstretched when u = 90°. Neglect the mass of the links.

250 mm

150 mm

A

E

k

C

250 mm u

u

SOLUTION Free -Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work when the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 200 A 0.25 cos u B = 50 cos u N.

D

B 150 mm

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b

dyGb = 0.25 cos udu

(1)

yGt = 0.25 sin u + a

dyGt = 0.25 cos udu

(2)

xC = 0.25 cos u

dxC = -0.25 sin udu

(3)

Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0;

-WbdyGb +

A - WtdyGt B + A - FspdxC B = 0

(4)

1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 50 cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have - 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 50 cos u( -0.25 sin udu) = 0

du A - 7.3575 cos u + 12.5 sin u cos u B = 0

Since du Z 0, then

- 7.3575 cos u + 12.5 sin u cos u = 0 cos u( -7.3575 + 12.5 sin u) = 0 Solving the above equation, cos u = 0

Ans.

u = 90°

- 7.3575 + 12.5 sin u = 0 Ans.

u = 36.1°

Ans: u = 90° u = 36.1° 1136

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11–14. A 5-kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when u = 45°, determine the stiffness k of each spring. The springs are unstretched when u = 90°. Neglect the mass of the links.

250 mm

150 mm

A

E

k

C

250 mm u

u

SOLUTION Free -Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work when the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = k A 0.25 cos u B = 0.25 k cos u.

D

B 150 mm

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b

dyGb = 0.25 cos udu

(1)

yGt = 0.25 sin u + a

dyGt = 0.25 cos udu

(2)

xC = 0.25 cos u

dxC = -0.25 sin udu

(3)

Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0;

-WbdyGb +

A - WtdyGt B + A - FspdxC B = 0

(4)

1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 0.25k cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have - 4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 0.25k cos u(-0.25 sin udu) = 0 du A -7.3575 cos u + 0.0625k sin u cos u B = 0

Since du Z 0, then

- 7.3575 cos u + 0.0625k sin u cos u = 0 117.72 k = sin u When u = 45°, then 117.72 k = = 166 N>m sin 45°

Ans.

Ans: k = 166 N>m 1137

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11–15. The service window at a fast-food restaurant consists of glass doors that open and close automatically using a motor which supplies a torque M to each door. The far ends, A and B, move along the horizontal guides. If a food tray becomes stuck between the doors as shown, determine the horizontal force the doors exert on the tray at the position u.

a

a

a

u C

M

a u

A

M

B

D

SOLUTION x = 2a cos u, dU = 0;

dx = - 2a sin u du -M du - F dx = 0 -M du + F (2a sin u)du = 0 F =

M 2a sin u

Ans.

Ans: F = 1138

M 2a sin u

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*11–16. The members of the mechanism are pin connected. If a vertical force of 800 N acts at A, determine the angle u for equilibrium. The spring is unstretched when u = 0°. Neglect the mass of the links.

B

k  6 kN/m

1m

1m

1m

Solution

D

u

A

Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only spring force Fsp and force P do work.

800 N

Virtual Displacement. The positions of Fsp and P are measured from fixed point B using position coordinates yc and yA respectively.            yc = 1 sin u      dyc = cos u du

(1)

            yA = 3(1 sin u)  dyA = 3 cos u du

(2)

Virtual Work Equation. When Fsp and P undergo their respective positive virtual displacement, P does positive work whereas Fsp does negative work. (3)

- Fspdyc + PdyA = 0

            dU = 0; Substitute Eqs. (1) and (2) into (3)

- Fsp(cos udu) + P(3 cos udu) = 0 ( -Fsp cos u + 3P cos u)du = 0 Since du ≠ 0, and assuming u 6 90°, then - Fsp cos u + 3P cos u = 0 Fsp = 3P Here Fsp = kx = 6000(1 sin u) = 6000 sin u and P = 800 N, Then 6000 sin u = 3(800)  sin u = 0.4 Ans.

              u = 23.58° = 23.6°

Ans: u = 23.6° 1139

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11–17. When u = 30°, the 25-kg uniform block compresses the two horizontal springs 100 mm. Determine the magnitude of the applied couple moments M needed to maintain equilibrium. Take k = 3 kN>m and neglect the mass of the links.

50 mm

k D

300 mm 100 mm 200 mm

M

C 100 mm

Solution

A

u

k

u M

B

Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only spring force Fsp, the weight of the block W, and couple moment M do work. Virtual Displacement. The positions of Fsp and W are measured from fixed point B using position coordinates x and y respectively.          x = 0.3 cos u + 0.05  dx = -0.3 sin u du

(1)

             y = 0.3 sin u + 0.1      dy = 0.3 cos udu

(2)

Virtual–Work Equation. When Fsp, W and M undergo their respective positive virtual displacement, all of them do negative work. Thus dU = 0;

- 2Fsp dx - Wdy - 2Mdu = 0

(3)

Substitute Eqs. (1) and (2) into (3), - 2Fsp( -0.3 sin udu) - W(0.3 cos udu) - 2Mdu = 0 (0.6Fsp sin u - 0.3 W cos u - 2M)du = 0 Since du ≠ 0, then 0.6 Fsp sin u - 0.3 W cos u - 2M = 0 (4)

M = 0.3 Fsp sin u - 0.15 W cos u

When u = 30°, Fsp = kx = 3000(0.1) = 300 N. Also W = 25(9.81) = 245.25 N. Substitute these values into Eq. 4.   M = 0.3(300) sin 30° - 0.15(245.25) cos 30° = 13.14 N # m = 13.1 N # m Ans.

Ans: M = 13.1 N # m 1140

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11–18. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the angle u for equilibrium. The spring has a stiffness of k = 15 kN>m and is unstretched when u = 15°.

E u

200 mm A k

D

SOLUTION

B 200 mm C

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is a formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 15(103) C 2(0.2 sin u) - 2(0.2 sin 15°) D = 6000(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u

dyA = 0.2 cos udu

(1)

yB = 3(0.2 sin u)

dyB = 0.6 cos udu

(2)

yC = 8(0.2 sin u)

dyB = 1.6 cos udu

(3)

Virtual Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, Fsp dyA + A - FspdyB B + PdyC = 0

dU = 0;

(4)

Substituting Fsp = 6000(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), 6000(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C -2400(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 2400(sin u - 0.2588) + 960 = 0 Ans.

u = 41.2°

Ans: u = 41.2° 1141

P

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11–19. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the stiffness k of the spring for equilibrium when u = 60°. The spring is unstretched when u = 15°.

E u

200 mm A k

D

SOLUTION

B 200 mm C

P

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula. Fsp = kx = k C 2(0.2 sin u) - 2(0.2 sin 15°) D = (0.4)k(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u

dyA = 0.2 cos udu

(1)

yB = 3(0.2 sin u)

dyB = 0.6 cos udu

(2)

yC = 8(0.2 sin u)

dyB = 0.6 cos udu

(3)

Virtual Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0;

Fsp dyA +

A - FspdyB B + PdyC = 0

(4)

Substituting Fsp = k(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), (0.4)k(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C - 0.16k(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 0.16k(sin u - 0.2588) + 960 = 0 k =

6000 sin u - 0.2588

When u = 60°, k =

6000 = 9881N>m = 9.88 kN>m sin 60° - 0.2588

Ans.

Ans: k = 9.88 kN>m 1142

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*11–20. The crankshaft is subjected to a torque of M = 50 N # m. Determine the horizontal compressive force F applied to the piston for equilibrium when u = 60°.

400 mm

100 mm u

F M

SOLUTION (0.4)2 = (0.1)2 + x2 - 2(0.1)(x)(cos u) 0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx dU = 0; For u = 60°,

- 50du - Fdx = 0

x = 0.4405 m Ans.

dx = - 0.09769 du ( -50 + 0.09769F) du = 0

Ans.

F = 512 N

Ans: dx = - 0.09769 du F = 512 N 1143

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11–21. The crankshaft is subjected to a torque of M = 50 N # m. Determine the horizontal compressive force F and plot the result of F (ordinate) versus u (abscissa) for 0° … u … 90°.

400 mm

100 mm u

F M

SOLUTION (0.4)2 = (0.1)2 + x 2 - 2(0.1)(x)(cos u)

(1)

0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx dx = a dU = 0; - 50du - F a F =

0.2x sin u b du 0.2 cos u - 2x -50du - Fdx = 0

0.2x sin u b du = 0, 0.2 cos u - 2x

du Z 0

50(2x - 0.2 cos u) 0.2x sin u

From Eq. (1) x2 - 0.2x cos u - 0.15 = 0 x =

0.2 cos u ; 20.04 cos2 u + 0.6 , 2

x =

0.2 cos u + 20.04 cos2 u + 0.6 2

F =

since20.04 cos2 u + 0.6 7 0.2 cos u

500 20.04 cos2 u + 0.6

Ans.

(0.2 cos u + 20.04 cos2 u + 0.6) sin u

Ans: F =

1144

50020.04 cos2 u + 0.6 (0.2 cos u + 20.04 cos2 u + 0.6) sin u

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11–22. The spring is unstretched when u = 0°. If P = 8 lb, determine the angle u for equilibrium. Due to the roller guide, the spring always remains vertical. Neglect the weight of the links.

k  50 lb/ft 2 ft 2 ft

SOLUTION y1 = 2 sin u,

dy1 = 2 cos u du

y2 = 4 sin u + 4,

dy2 = 4 cos u du

Fs = 50(2 sin u) = 100 sin u dU = 0;

4 ft

u

- Fs dy1 + Pdy2 = 0 4 ft

- 100 sin u(2 cos u du) + 8(4 cos u du) = 0 Assume ® 6 90°, so cos ® Z 0.

P

200 sin u = 32 Ans.

u = 9.21°

Ans: u = 9.21° 1145

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11–23. Determine the weight of block G required to balance the differential lever when the 20-lb load F is placed on the pan. The lever is in balance when the load and block are not on the lever. Take x = 12 in.

4 in.

x

4 in. C

A

G

B E

D 2 in.

SOLUTION Free-Body Diagram: When the lever undergoes a virtual angular displacement of du about point B, the dash line configuration shown in Fig. a is formed. We observe that only the weight WG of block G and the weight WF of load F do work when the virtual displacements take place.

F

Virtual Displacement: Since dyG is very small, the vertical virtual displacement of block G and load F can be approximated as dyG = (12 + 4)du = 16du

(1)

dyF = 2du

(2)

Virtual Work Equation: Since WG acts towards the positive sense of its corresponding virtual displacement, its work is positive. However, force WF does negative work since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0;

WGdyG +

A - WFdyF B = 0

(3)

Substituting WG = 20 lb and Eqs. (1) and (2) into Eq. (3), WG A 16du B - 20(2du) = 0 du(16WG - 40) = 0 Since du Z 0, then 16WG - 40 = 0 Ans.

WG = 2.5 lb

Ans: WG = 2.5 lb 1146

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*11–24. If the load F weighs 20 lb and the block G weighs 2 lb, determine its position x for equilibrium of the differential lever. The lever is in balance when the load and block are not on the lever.

4 in.

x

4 in. C

A

G

B E

D 2 in.

SOLUTION Free -Body Diagram: When the lever undergoes a virtual angular displacement of du about point B, the dash line configuration shown in Fig. a is formed. We observe that only the weight WG of block G and the weight WF of load F do work when the virtual displacements take place.

F

Virtual Displacement: Since dyG is very small, the vertical virtual displacement of block G and load F can be approximated as dyG = (4 + x)du

(1)

dyF = 2du

(2)

Virtual Work Equation: Since WG acts towards the positive sense of its corresponding virtual displacement, its work is positive. However, force WF does negative work since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0;

WGdyG +

A -WFdyF B = 0

(3)

Substituting WF = 20 lb, WG = 2 lb, Eqs.(1) and (2) into Eq. (3), 2(4 + x)du - 20(2du) = 0 du C 2(4 + x) - 40 D = 0 Since du Z 0, then 2(4 + x) - 40 = 0 Ans.

x = 16 in.

Ans: x = 16 in. 1147

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11–25. The dumpster has a weight W and a center of gravity at G. Determine the force in the hydraulic cylinder needed to hold it in the general position u.

b a

SOLUTION

G d

θ

s = 2a2 + c2 - 2a c cos (u + 90°) = 2a2 + c2 + 2a c sin u c

1

ds = (a2 + c2 + 2a c sin u)- 2 ac cos u du y = (a + b) sin u + d cos u dy = (a + b) cos u du - d sin u du dU = 0;

Fds - Wdy = 0 1

F(a2 + c2 + 2a c sin u)- 2 ac cos u du - W(a + b) cos u du + Wd sin u du = 0 F = a

W(a + b - d tan u) b 2a2 + c2 + 2a c sin u ac

Ans.

Ans: F = 1148

W(a + b - d tan u) ac

2a2 + c 2 + 2ac sin u

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11–26. The potential energy of a one-degree-of-freedom system is defined by V = (20x3 - 10x2 - 25x - 10) ft # lb, where x is in ft. Determine the equilibrium positions and investigate the stability for each position.

SOLUTION Equilibrium Configuration: Taking the first derivative of V, we have dV = 60x2 - 20x - 25 dx Equilibrium requires

dV = 0. Thus, dx 60x 2 - 20x - 25 = 0 x =

20 ; 3( -20)2 - 4(60)(- 25) 2(60) Ans.

x = 0.833 ft and - 0.5 ft Stability: The second derivative of V is d2V = 120x - 20 dx2 At x = 0.8333 ft, d2V 2 = 120(0.8333) - 20 = 80 7 0 dx2 x = 0.8333 ft

Ans.

Thus, the equilibrium configuration at x = 0.8333 ft is stable. At x = -0.5ft, d 2V 2 = 120( - 0.5) - 20 = - 80 6 0 dx2 x = -0.5 ft

Ans.

Thus, the equilibrium configuration at x = -0.5 ft is unstable.

Ans: x = - 0.5 ft unstable x = 0.833 ft stable 1149

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11–27. If the potential function for a conservative one-degree-offreedom system is V = (12 sin 2u + 15 cos u) J,where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 12 sin 2u + 15 cos u dV = 0; du

24 cos 2u - 15 sin u = 0

2411 - 2 sin2 u2 - 15 sin u = 0 48 sin2 u + 15 sin u - 24 = 0 Choosing the angle 0° 6 u 6 180° u = 34.6°

Ans.

u = 145°

Ans.

and

d2V = - 48 sin 2u - 15 cos u du2 u = 34.6°,

d 2V = - 57.2 6 0 du2

Unstable

Ans.

u = 145°,

d2V = 57.2 7 0 du2

Stable

Ans.

Ans: Unstable at u = 34.6° stable at u = 145° 1150

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*11–28. If the potential function for a conservative one-degree-offreedom system is V = 18x3 - 2x 2 - 102 J, where x is given in meters, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 8x3 - 2x2 - 10 dV = 24x2 - 4x = 0 dx 124x - 42x = 0 x = 0

Ans.

and x = 0.167 m

d2V = 48x - 4 dx2 x = 0,

d 2V = -4 6 0 dx2

x = 0.167 m,

Ans.

Unstable

d2V = 4 7 0 dx2

Ans.

Stable

Ans: x = 0.167 m d 2V = -4 6 0 Unstable dx2 d 2V = 4 7 0 Stable dx2 1151

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11–29. If the potential function for a conservative one-degree-offreedom system is V = 110 cos 2u + 25 sin u2 J, where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 10 cos 2u + 25 sin u For equilibrium: dV = - 20 sin 2u + 25 cos u = 0 du 1- 40 sin u + 252 cos u = 0 u = sin-1 a

25 b = 38.7° and 141° 40

Ans.

and u = cos-1 0 = 90°

Ans.

Stability: d2V = - 40 cos 2u - 25 sin u du2 u = 38.7°,

d2V = - 24.4 6 0, du2

Unstable

Ans.

u = 141°,

d2V = - 24.4 6 0, du2

Unstable

Ans.

Stable

Ans.

u = 90°,

d 2V = 15 7 0, du2

Ans: u = 38.7° unstable u = 90° stable u = 141° unstable 1152

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11–30. If the potential energy for a conservative one-degree-offreedom system is expressed by the relation V = (4x3 - x 2 - 3x + 10) ft # lb, where x is given in feet, determine the equilibrium positions and investigate the stability at each position.

SOLUTION V = 4x3 - x2 - 3x + 10 Equilibrium Position: dV = 12x 2 - 2x - 3 = 0 dx x =

2 ; 2(- 2)2 - 4(12)(- 3) 24

x = 0.590 ft

and

Ans.

- 0.424 ft

Stability: d 2V = 24x - 2 dx 2 At x = 0.590 ft

d2V = 24(0.590) - 2 = 12.2 7 0 dx 2

Stable

Ans.

At x = -0.424 ft

d 2V = 24( - 0.424) - 2 = -12.2 6 0 dx 2

Unstable

Ans.

Ans: x = - 0.424 ft unstable x = 0.590 ft stable 1153

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11–31. The uniform link AB has a mass of 3 kg and is pin connected at both of its ends. The rod BD, having negligible weight, passes through a swivel block at C. If the spring has a stiffness of k = 100 N>m and is unstretched when u = 0°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. Neglect the size of the swivel block.

400 mm

D k  100 N/m

A

C u

SOLUTION 400 mm

s = 2(0.4)2 + (0.4)2 - 2(0.4)2 cos u

B

= (0.4)22(1 - cos u) V = Vg + Ve = -(0.2)(sin u)3(9.81) +

1 (100) C (0.4)2(2)(1 - cos u) D 2

dV = - (5.886) cos u + 16(sin u) = 0 du

(1) Ans.

u = 20.2° d2V = 5.886 sin u + (16) cos u = 17.0 7 0 du2

stable

Ans.

Ans: u = 20.2° stable 1154

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*11–32. The spring of the scale has an unstretched length of a. Determine the angle u for equilibrium when a weight W is supported on the platform. Neglect the weight of the members. What value W would be required to keep the scale in neutral equilibrium when u = 0°?

W

L

L k

SOLUTION Potential Function: The datum is established at point A. Since the weight W is above the datum, its potential energy is positive. From the geometry, the spring stretches x = 2L sin u and y = 2L cos u. V = Ve + Vg

L

L u

u

a

=

1 2 kx + Wy 2

=

1 (k)(2 L sin u)2 + W(2L cos u) 2

= 2kL2 sin 2 u + 2WL cos u Equilibrium Position: The system is in equilibrium if

dV = 0. du

dV = 4kL2 sin u cos u - 2WL sin u = 0 du dV = 2kL2 sin 2u - 2WLsin u = 0 du Solving, u = 0°

or

u = cos-1 a

Stability: To have neutral equilibrium at u = 0°,

W b 2kL

Ans.

d2V 2 = 0. du2 u - 0°

d 2V = 4kL2 cos 2 u - 2WL cos u du2 d 2V 2 = 4kL2 cos 0° - 2WLcos 0° = 0 du2 u - 0° Ans.

W = 2kL

Ans: u = cos-1a W = 2kL

1155

W b 2kL

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11–33. The uniform bar has a mass of 80 kg. Determine the angle u for equilibrium and investigate the stability of the bar when it is in this position. The spring has an unstretched length when u = 90°. B 4m

k  400 N/ m

Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is above the datum, its potential energy is positive. Here, y = 2 sin u and the spring stretches x = 4(1 - sin u) m. Thus,

A

u

V = Ve + Vg 1 2 kx + Wy 2 1 = (400) 3 4(1 - sin u) 4 2 + 80(9.81)(2 sin u) 2 =

= 3200 sin2 u - 4830.4 sin u + 3200

Equilibrium Position. The bar is in equilibrium of

dV = 0 du

dV = 6400 sin u cos u - 4830.4 cos u = 0 du cos u(6400 sin u - 4830.4) = 0 Solving,

Ans.

u = 90° or u = 49.00° = 49.0°

Using the trigonometry identify sin 2u = 2 sin u cos u, dV = 3200 sin 2u - 4830.4 cos u du d 2V = 6400 cos 2u + 4830.4 sin u du 2 d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 du 2 du 2 d 2V and neutral if = 0. du 2 At u = 90°, d 2V 2 = 6400 cos [2(90°)] + 4830.4 sin 90° = -1569.6 6 0 du 2 u = 90° Thus, the bar is in unstable equilibrium at u = 90° At u = 49.00°, d 2V = 6400 cos [2(49.00°)] + 4830.4 sin 49.00° = 2754.26 7 0 du2 Thus, the bar is in stable equilibrium at u = 49.0°.

Ans: Unstable equilibrium at u = 90° Stable equilibrium at u = 49.0° 1156

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11–34. The uniform bar AD has a mass of 20 kg. If the attached spring is unstretched when u = 90°, determine the angle u for equilibrium. Note that the spring always remains in the vertical position due to the roller guide. Investigate the stability of the bar when it is in the equilibrium position.

C

k  2 kN/m

A

u

B

0.5 m

Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here, y = 0.75 cos u and the spring stretches x = 0.5 cos u. Thus,

1m D

V = Ve + Vg 1 2 kx + ΣWy 2 1 = (2000)(0.5 cos u)2 + 2 =

3 - 20(9.81)(0.75 cos u) 4

= 250 cos2 u - 147.15 cos u

Equilibrium Position. The bar is in equilibrium if

dV = 0. du

dV = - 500 sin u cos u + 147.15 sin u = 0 du sin u(147.15 - 500 cos u) = 0 Solving,

u = 0°

Ans.

   u = 72.88° = 72.9°

Using the trigonometry identity sin 2u = 2 sin u cos u, dV = - 250 sin 2u + 147.15 sin u du d 2V = - 500 cos 2u + 147.15 cos u du 2 d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 2 du du 2 d 2V and neutral if = 0. du 2 d 2V 2 = - 500 cos 0° + 147.15 cos 0° = -352.85 6 0 du 2 u = 0° Thus, the bar is in unstable equilibrium at u = 0°. At u = 72.88°, d 2V 2 = - 500 cos [2(72.88°)] + 147.15 cos 72.88° = 456.69 7 0 du2 u = 72.88° Thus, the bar is in stable equilibrium at u = 72.9°. Ans: Unstable equilibrium at u = 0° Stable equilibrium at u = 72.9° 1157

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11–35.

The two bars each have a weight of 8 lb. Determine the required stiffness k of the spring so that the two bars are in equilibrium when u = 30°. The spring has an unstretched length of 1 ft.

B

2 ft

A

SOLUTION V = 2(8)(1 sin u) +

θ

2 ft

k C

1 k(4 cos u - 1)2 2

dV = 16 cos u + k(4 cos u - 1)( - 4 sin u) du dV = 16 cos u - 4k(4 cos u - 1) sin u du u = 30°,

dV = 0 du

16 cos 30° - 4k(4 cos 30° - 1) sin 30° = 0 Ans.

k = 2.81 lb>ft

Ans: k = 2.81 lb>ft 1158

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*11–36. Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 3 kg and the suspended block D has a mass of 7 kg. Cord DC has a total length of 1 m.

A

SOLUTION

500 mm

500 mm

u

u

C

500 mm

l = 500 mm y1 =

l sin u 2

D

y2 = l + 2l(1 - cos u) = l(3 - 2 cos u) V = 2Wy1 - WDy 2 = Wl sin u - WD l(3 - 2 cos u) dV = l(W cos u - 2WD sin u) = 0 du tan u =

3(9.81) W = 0.2143 = 2WD 14(9.81) Ans.

u = 12.1° d2V = l( -W sin u - 2WD cos u) du2 u = 12.1°,

d 2V = 0.5[- 3(9.81) sin 12.1° - 14(9.81) cos 12.1°] du2 = -70.2 6 0

Ans.

Unstable

Ans: u = 12.1° Unstable 1159

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11–37. Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 10 kg and the spring has an unstretched length of 100 mm.

500 mm

k 1.5 kN/ m u

u

A

500 mm

Solution Potential Function. The Datum the centers of gravity of the energies are positive. Here x = 3 2(0.5 cos u) - 0.5 4 - 0.1 =

is established through point A, Fig. a. Since bars are above the datum, their potential y = 0.25 sin u and the spring stretches (cos u - 0.6) m. Thus

V = Ve + Vg

1 2 kx + ΣWy 2 1 = (1500)(cos u - 0.6)2 + 2[10(9.81)](0.25 sin u) 2 =

= 750 cos2 u - 900 cos u + 49.05 sin u + 270 Equilibrium Position. The system is in equilibrium if

dV = 0. du

dV = - 1500 sin u cos u + 900 sin u + 49.05 cos u = 0 du Solving numerically,

u = 4.713° = 4.71° or u = 51.22° = 51.2°

Ans.

Using the trigonometry identity sin 2u = 2 sin u cos u, dV = - 750 sin 2u + 900 sin u + 49.05 cos u du d 2V = 900 cos u - 1500 cos 2u - 49.05 sin u du 2 d 2V d 2V Stability. The equilibrium configurate is stable if 7 0, unstable if 6 0 and du 2 du 2 d 2V neutral if = 0. du 2

1160

500 mm

C

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11–37. Continued

At u = 51.22°, d 2V 2 = 900 cos 51.22° - 49.05 sin 51.22° - 1500 cos [2(51.22°)] = 848.77 7 0 du 2 u = 51.22° Thus, the system is in stable equilibrium at u = 51.2°. At u = 4.713°, d 2V 2 = 900 cos 4.713° - 49.05 sin 4.713° - 1500 cos [2(4.713°)] = - 586.82 6 0 du2 u = 4.713° Thus, the system is in unstable equilibrium at u = 4.71°.

Ans: Stable equilibrium at u = 51.2° Unstable equilibrium at u = 4.71° 1161

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11–38. The two bars each have a mass of 8 kg. Determine the required stiffness k of the spring so that the two bars are in equilibrium when u = 60°. The spring has an unstretched length of 1 m. Investigate the stability of the system at the equilibrium position.

A

u

1.5 m

B

k

Solution

1.5 m

Potential Function. The Datum is established through point A, Fig. a. Since the centers of gravity of the bars are below the datum, their potential energies are negative. Here, y1 = 0.75 cos u, y2 = 1.5 cos u + 0.75 cos u = 2.25 cos u and the spring stretches x = 2(1.5 cos u) - 1 = (3 cos u - 1) m. Thus, V = Ve + Vg =    =

1 2 kx + ΣWy 2 1 k(3 cos u - 1)2 + [ - 8(9.81)(0.75 cos u)] + [ - 8(9.81)(2.25 cos u)] 2

= 4.5 k cos2 u - 3 k cos u + 0.5 k - 235.44 cos u = 4.5 k cos2 u - (3 k + 235.44) cos u + 0.5 k Equilibrium Position. The system is in equilibrium if

dV = 0 du

dV = - 9 k sin u cos u + (3 k + 235.44) sin u = 0 du sin u( -9 k cos u + 3 k + 235.44) = 0 Since sin u ≠ 0, then

- 9 k cos u + 3 k + 235.44 = 0 k =

When u = 60°,

k =

235.44 9 cos u - 3

235.44 = 156.96 N>m = 157 N>m 9 cos 60° - 3

Using this result, dV = -9(156.96) sin u cos u + [3(156.96) + 235.44] sin u du   = - 1412.64 sin u cos u + 706.32 sin u Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 706.32 sin u - 706.32 sin 2u du d 2V = 706.32 cos u - 1412.64 cos 2u du 2

1162

Ans.

C

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11–38 Continued

d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0, 2 du du 2 d 2V and neutral if = 0. du 2 At u = 60°, d 2V 2 = 706.32 cos 60° - 1412.64 cos [2(60°)] = 1059.16 7 0 du 2 u = 60° Thus, the system is in stable equilibrium at u = 60°.

Ans: k = 157 N>m Stable equilibrium at u = 60° 1163

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11–39. A spring with a torsional stiffness k is attached to the pin at B. It is unstretched when the rod assembly is in the vertical position. Determine the weight W of the block that results in neutral equilibrium. Hint: Establish the potential energy function for a small angle u. i.e., approximate sin u L 0, and cos u L 1 - u2>2.

L 2 A L 2

SOLUTION Potential Function: With reference to the datum, Fig. a, the gravitational potential

B

k

energy of the block is positive since its center of gravity is located above the datum. Here, the rods are tilted with a small angle u. Thus, y =

3 L cos u + L cos u = L cos u. 2 2

u2 .Thus, 2 u2 3WL u2 3 b = a1 - b Vg = Wy = W a L b a1 2 2 2 2

L 2 C

1 -

However, for a small angle u, cos u

The elastic potential energy of the torsional spring can be computed using 1 Ve = kb 2, where b = 2u. Thus, 2 Vg =

1 k(2u)2 = 2ku2 2

The total potential energy of the system is V = Vg + Ve =

u2 3WL a1 - b + 2ku2 2 2

Equilibrium Configuration: Taking the first derivative of V, we have 3WL 3WL dV = u + 4ku = ua + 4k b du 2 2 dV = 0. Thus, du 3WL + 4k b = 0 ua 2

Equilibrium requires

u = 0° Stability: The second derivative of V is 3WL d2V + 4k = 2 du2

To have neutral equilibrium at u = 0°, 3WL + 4k = 0 2 8k W = 3L

d2V 2 = 0. Thus, du2 u = 0°

-

Ans.

Note: The equilibrium configuration of the system at u = 0° is stable if 8k d2V 8k d 2V and is unstable if W > W< > 0 < 0 . 3L du2 3L du2 1164

Ans: W =

8k 3L

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*11–40. A conical hole is drilled into the bottom of the cylinder, and it is then supported on the fulcrum at A. Determine the minimum distance d in order for it to remain in stable equilibrium.

SOLUTION Potential Function: First, we must determine the center of gravity of the cylinder. By referring to Fig. a, d 1 h (rpr2h) - a rpr2d b ©yCm 2 4 3 6h2 - d 2 = y = = 1 ©m 4(3h - d) rpr 2h - rpr2d 3

(1)

With reference to the datum, Fig. a, the gravitational potential energy of the cylinder is positive since its center of gravity is located above the datum. Here, y = (y - d)cos u = B

6h2 - d2 6h2 - 12hd + 3d2 - d R cos u = B R cos u 4(3h - d) 4(3h - d)

Thus, V = Vg = Wy = W B

6h2 - 12hd + 3d2 R cos u 4(3h - d)

Equilibrium Configuration: Taking the first derivative of V, dV 6h2 - 12hd + 3d2 = -W B R sin u du 4(3h - d) Equilibrium requires

dV = 0. Thus, du

-WB

6h2 - 12hd + 3d2 R sin u = 0 4(3h - d)

sin u = 0

u = 0°

Stability: The second derivative of V is 6h2 - 12hd + 3d2 d 2V = -W B R cos u 2 4(3h - d) du To have neutral equilibrium at u = 0°, -WB

d 2V 2 = 0. Thus, d2u u = 0°

6h2 - 12hd + 3d2 R cos 0° = 0 4(3h - d)

6h2 - 12hd + 3d2 = 0 d =

12h ; 2( - 12h)2 - 4(3)(6h2) = 0.5858h = 0.586h 2(3)

Ans.

Note: If we substitute d = 0.5858h into Eq. (1), we notice that the fulcrum must be at the center of gravity for neutral equilibrium. 1165

Ans: d = 0.586h

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11–41. The uniform rod has a mass of 100 kg. If the spring is unstretched when u = 60°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. The spring is always in the horizontal position due to the roller guide at B

A

u

2m k  500 N/m B

2m

Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here y = 2 cos u, and the spring stretches x = 2 sin 60° - 2 sin u = 2(sin 60° - sin u). Thus V = Ve + Vg =

1 2 kx + Wy 2

=

1 (500)[2(sin 60° - sin u)]2 + [ - 100(9.81)(2 cos u)] 2

= 1000 sin2 u - 100013 sin u - 1962 cos u + 750 dV = 0. Equilibrium Position. The bar is in equilibrium if du dV = 2000 sin u cos u - 100013 cos u + 1962 sin u du Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 1000 sin 2u - 100013 cos u + 1962 sin u = 0 du Solved numerically, Ans.

u = 24.62° = 24.6° d 2V = 2000 cos 2u + 100013 sin u + 1962 cos u du 2

d 2V d 2V 7 0, unstable if 6 0 Stability. The equilibrium configuration is stable if 2 du du 2 d 2V and neutral if = 0. du 2 At u = 24.62°, d 2V = 2000 cos[2(24.62°)] + 100013 sin 24.62° + 1962 cos 24.62° = 3811.12 7 0 du 2 Thus, the bar is in stable equilibrium at u = 24.6°.

Ans: Stable equilibrium at u = 24.6° 1166

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11–42. Each bar has a mass per length of m0. Determine the angles u and f at which they are suspended in equilibrium. The contact at A is smooth, and both are pin connected at B.

B

l

u

3l 2

Solution

A l 2

Require G for system to be at its lowest point. 0 x =

f

l l 3 a b - (0.75 l sin 26.565°)a lb 4 2 2 = -0.20937 l l 3 l + + l 2 2

lx 1 3 - a b(l) - l a b - (0.75 l cos 26.565°)a lb 2 2 2 y = = -0.66874 l l 3 l + + l 2 2 f = tan-1a

0.20937 l b = 17.38° = 17.4° 0.66874 l

Ans. Ans.

u = 26.565° - 17.38° = 9.18°

Ans: f = 17.4° u = 9.18° 1167

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11–43. The truck has a mass of 20 Mg and a mass center at G. Determine the steepest grade u along which it can park without overturning and investigate the stability in this position.

G 3.5 m

SOLUTION Potential Function: The datum is established at point A. Since the center of gravity for the truck is above the datum, its potential energy is positive. Here, y = (1.5 sin u + 3.5 cos u) m.

1.5 m u

1.5 m

V = Vg = Wy = W(1.5 sin u + 3.5 cos u) Equilibrium Position: The system is in equilibrium if

dV = 0 du

dV = W(1.5 cos u - 3.5 sin u) = 0 du Since W Z 0, 1.5 cos u - 3.5 sin u = 0 Ans.

u = 23.20° = 23.2° Stability: d2V = W(-1.5 sin u - 3.5 cos u) du2 d2V du2

u = 23.20°

= W( - 1.5 sin 23.20° - 3.5 cos 23.20°) = -3.81W 6 0

Thus, the truck is in unstable equilibrium at u = 23.2°

Ans.

Ans: Unstable equilibrium at u = 23.2° 1168

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*11–44. The small postal scale consists of a counterweight W1, connected to the members having negligible weight. Determine the weight W2 that is on the pan in terms of the angles u and f and the dimensions shown. All members are pin connected.

W2

a

u

SOLUTION

b

y1 = b cos u

a

f

f

W1

y2 = a sin f = a sin (90° - u - g) where g is a constant and f = (90° - u - g) V = -W1y1 + W2y2 = -W1 b cos u + W2 a sin (90° - u - g) dV = W1 b sin u - W2 a cos (90° - u - g) du b sin u W2 = W1 a b a cos f

Ans.

Ans: b sin u W2 = W1a b a cos f 1169

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11–45. A 3-lb weight is attached to the end of rod ABC. If the rod is supported by a smooth slider block at C and rod BD, determine the angle u for equilibrium. Neglect the weight of the rods and the slider.

C

6 in. B

θ

D 4 in.

10 in.

SOLUTION x = 2(6)2 - (4 sin u)2 = 236 - 16 sin2 u

A

16 6 = x (x + 4 cos u) + y x + 4 cos u + y = 2.667x y = - 4 cos u + 1.667 236 - 16 sin2 u 1 1 dy = 4 sin u du + 1.667a b (36 - 16 sin2 u)- 2 ( -32 sin u cos u)du 2

dU = 0;

W dy = 0 W c 4 - 0.8333(36 - 16 sin2 u)- 2 (32 cos u) d sin u du = 0 1

Thus, sin u = 0 Ans.

u = 0° or,

A 36 - 16 sin2u B 2 = 6.667 cos u 1

Ans.

u = 33.0°

Ans: u = 0° u = 33.0° 1170

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11–46. If the uniform rod OA has a mass of 12 kg, determine the mass m that will hold the rod in equilibrium when u = 30°. Point C is coincident with B when OA is horizontal. Neglect the size of the pulley at B.

B

C m 3m

SOLUTION Geometry: Using the law of cosines, 2

A

1m

2

lA¿B = 21 + 3 - 2112132 cos190° - u2 = 210 - 6 sin u

e

O

lAB = 212 + 32 = 210 m l = lAB - lA¿B = 210 - 210 - 6 sin u Potential Function: The datum is established at point O. Since the center of gravity of the rod and the block are above the datum, their potential energy is positive. Here, y1 = 3 - l = 33 - 1210 - 210 - 6 sin u24 m and y2 = 0.5 sin u m. V = Vg = W1y1 + W2y2 = 9.81m33 - 1210 - 210 - 6 sin u24 + 117.7210.5 sin u2 = 29.43m - 9.81m1210 - 210 - 6 sin u2 + 58.86 sin u Equilibrium Position: The system is in equilibrium if dV = 0. ` du u = 30° 1 1 dV = - 9.81m c - 110 - 6 sin u2- 21- 6 cos u2 d + 58.86 cos u du 2

= -

29.43m cos u 210 - 6 sin u

+ 58.86 cos u

At u = 30°, dV 29.43m cos 30° = + 58.86 cos 30° = 0 ` du u = 30° 210 - 6 sin 30° Ans.

m = 5.29 kg

Ans: m = 5.29 kg 1171

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11–47. The cylinder is made of two materials such that it has a mass of m and a center of gravity at point G. Show that when G lies above the centroid C of the cylinder, the equilibrium is unstable.

G C r

SOLUTION Potential Function: The datum is established at point A. Since the center of gravity of the cylinder is above the datum, its potential energy is positive. Here, y = r + d cos u. V = Vg = Wy = mg(r + d cos u) Equilibrium Position: The system is in equilibrium if

dV = 0. du

dV = - mgd sin u = 0 du sin u = 0

u = 0°

Stability: d 2V = - mgd cos u du2 d 2V 2 = - mgd cos 0° = - mgd 6 0 du2 u = 0° Thus, the cylinder is in unstable equilibrium at u = 0°

(Q.E.D.)

1172

a

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*11–48. The bent rod has a weight of 5 lb ft. A pivot of negligible size is attached at its center A and the rod is balanced as shown. Determine the length L of its vertical segments so that it remains in neutral equilibrium. Neglect the thickness of the rod.

8 in. 2 in. L

8 in. A L

SOLUTION To remain in neutral equilibrium, the center of gravity must be located at A. y =

0 =

©yW ©W 2a

L 2L 16 b (5) - a - 2 b a b (5) 12 2 12 a

2L 16 b (5) + a b (5) 12 12

0 = 5L2 - 20L - 160 L =

20 ; 2( - 20)2 - 4(5)(- 160) 10

Take the positive root

Ans.

L = 8 in.

Ans: L = 8 in. 1173

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11–49. The triangular block of weight W rests on the smooth corners which are a distance a apart. If the block has three equal sides of length d, determine the angle u for equilibrium.

d

60

G 60 u

SOLUTION

a

AF = AD sin f = AD sin (60° - u) a AD = sin a sin 60° AD =

a (sin (60° + u)) sin 60°

AF =

a (sin (60° + u)) sin (60° - u) sin 60°

= y=

a (0.75 cos2 u - 0.25 sin2 u) sin 60° d 23

cos u - AF

V = Wy dV a ( -1.5 sin u cos u - 0.5 sin u cos u)d = 0 = Wc( -0.5774 d) sin u du sin 60° Require, sin u = 0 or

- 0.5774 d -

Ans.

u = 0°

a ( -2 cos u) = 0 sin 60° u = cos-1 a

d b 4a

Ans.

Ans: u = 0° u = cos-1a 1174

d b 4a

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12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m>s2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s?

Solution a = 2t - 6 dv = a dt L0

v

dv =

L0

t

(2t - 6) dt

v = t 2 - 6t ds = v dt L0

s

ds =

s =

L0

t

(t2 - 6t) dt

t3 - 3t2 3

When t = 6 s, Ans.

v = 0 When t = 11 s,

Ans.

s = 80.7 m

Ans: s = 80.7 m 1

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12–2. If a particle has an initial velocity of v0 = 12 ft>s to the right, at s0 = 0, determine its position when t = 10 s, if a = 2 ft>s2 to the left.

SOLUTION +2 s = s0 + 1S

v0 t +

1 2 a t 2 c

= 0 + 12(10) +

1 ( - 2)(10)2 2 Ans.

= 20 ft

Ans: s = 20 ft 2

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12–3. A particle travels along a straight line with a velocity v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period.

SOLUTION v = 12 - 3t 2

(1)

dv = - 6t dt

a = s

t=4

= -24 m>s2

t

ds =

L-10

L1

v dt =

t

L1

Ans.

( 12 - 3t 2 ) dt

s + 10 = 12t - t 3 - 11 s = 12t - t 3 - 21 s

t=0

s

t = 10

= - 21 = - 901

∆s = - 901 - ( -21) = -880 m

Ans.

From Eq. (1): v = 0 when t = 2s s

t=2

= 12(2) - (2)3 - 21 = - 5 Ans.

sT = (21 - 5) + (901 - 5) = 912 m

Ans: a = - 24 m>s2 ∆s = - 880 m sT = 912 m 3

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*12–4. A particle travels along a straight line with a constant acceleration. When s = 4 ft, v = 3 ft>s and when s = 10 ft, v = 8 ft>s. Determine the velocity as a function of position.

SOLUTION Velocity: To determine the constant acceleration ac, set s0 = 4 ft, v0 = 3 ft>s, s = 10 ft and v = 8 ft>s and apply Eq. 12–6. + ) (:

v2 = v20 + 2ac (s - s0) 82 = 32 + 2ac (10 - 4) ac = 4.583 ft>s2

Using the result ac = 4.583 ft>s2, the velocity function can be obtained by applying Eq. 12–6. v2 = v20 + 2ac (s - s0) v2 = 32 + 2(4.583) (s - 4)

A

+ ) (:

v = A 29.17s - 27.7 ft>s

Ans.

Ans: v = 4

( 19.17s - 27.7 ) ft>s

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12–5. The velocity of a particle traveling in a straight line is given by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when t  =  0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-s time interval, and what is its average speed?

Solution v = 6t - 3t 2 a =

dv = 6 - 6t dt

At t = 3 s a = - 12 m>s2

Ans.

ds = v dt L0

s

ds =

L0

t

(6t - 3t2)dt

s = 3t 2 - t 3 At t = 3 s Ans.

s = 0 Since v = 0 = 6t - 3t 2, when t = 0 and t = 2 s. when t = 2 s,  s = 3(2)2 - (2)3 = 4 m

Ans.

sT = 4 + 4 = 8 m

( vsp ) avg =

sT 8 = = 2.67 m>s t 3

Ans.

Ans: sT = 8 m vavg = 2.67 m>s 5

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12–6. The position of a particle along a straight line is given by s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

SOLUTION Position: The position of the particle when t = 6 s is s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = - 27.0 ft

Ans.

Total DistanceTraveled: The velocity of the particle can be determined by applying Eq. 12–1. v =

ds = 4.50t2 - 27.0t + 22.5 dt

The times when the particle stops are 4.50t2 - 27.0t + 22.5 = 0 t = 1s

and

t = 5s

The position of the particle at t = 0 s, 1 s and 5 s are s t = 0 s = 1.5(03) - 13.5(02) + 22.5(0) = 0 s t = 1 s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft s t = 5 s = 1.5(53) - 13.5(52) + 22.5(5) = - 37.5 ft From the particle’s path, the total distance is Ans.

stot = 10.5 + 48.0 + 10.5 = 69.0 ft

Ans: s t = 6 s = - 27.0 ft stot = 69.0 ft 6

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12–7. A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s.

Solution s = t2 - 6t + 5 v =

ds = 2t - 6 dt

a =

dv = 2 dt

v = 0 when t = 3 s  t=0 = 5 s  t = 3 = -4 s  t=6 = 5 vavg =

∆s 0 = = 0 ∆t 6

( vsp ) avg =

Ans.

sT 9 + 9 = = 3 m>s ∆t 6

Ans.

a  t = 6 = 2 m>s2

Ans.

Ans: vavg = 0 (vsp)avg = 3 m>s a  t = 6 s = 2 m>s2 7

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*12–8. A particle is moving along a straight line such that its position is defined by s = (10t2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s.

SOLUTION s = 10t2 + 20 (a) s|1 s = 10(1)2 + 20 = 30 mm s|5 s = 10(5)2 + 20 = 270 mm Ans.

¢s = 270 - 30 = 240 mm (b) ¢t = 5 - 1 = 4 s vavg = (c) a =

240 ¢s = = 60 mm>s ¢t 4

d2s = 20 mm s2 dt2

Ans. Ans.

(for all t)

Ans: ∆s = 240 mm vavg = 60 mm>s a = 20 mm>s2 8

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12–9. The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.

Solution a = 2t - 1 dv = a dt L2

v

t

L0

dv =

(2t - 1)dt

v = t2 - t + 2 dx = v dt Lt

s

ds =

s =

L0

t

(t2 - t + 2)dt

1 3 1 t - t 2 + 2t + 1 3 2

When t = 6 s v = 32 m>s

Ans.

s = 67 m

Ans.

Since v ≠ 0 for 0 … t … 6 s, then Ans.

d = 67 - 1 = 66 m

Ans: v = 32 m>s s = 67 m d = 66 m 9

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12–10. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s 5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m . Use a numerical method to evaluate the integral.

SOLUTION a =

5 1 3

5

A 3s + s2 B

a ds = v dv 2

v

5 ds 1 3

L1 A 3s + s 0.8351 =

5 2

B

=

L0

v dv

1 2 v 2 Ans.

v = 1.29 m>s

Ans: v = 1.29 m>s 10

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12–11. A particle travels along a straight-line path such that in 4 s it moves from an initial position sA = - 8 m to a position sB = +3 m. Then in another 5 s it moves from sB to sC = -6 m. Determine the particle’s average velocity and average speed during the 9-s time interval.

SOLUTION Average Velocity: The displacement from A to C is ∆s = sC - SA = -6 - ( - 8) =  2 m. ∆s 2 vavg = = = 0.222 m>s Ans. ∆t 4 + 5 Average Speed: The distances traveled from A to B and B to C are sA S B = 8 + 3 = 11.0 m and sB S C = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled is sTot = sA S B + sB S C = 11.0 + 9.00 = 20.0 m. (vsp)avg =

sTot 20.0 = = 2.22 m>s ∆t 4 + 5

Ans.

Ans: vavg = 0.222 m>s (vsp)avg = 2.22 m>s 11

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*12–12. Traveling with an initial speed of 70 km>h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km>h? Also, through what distance does the car travel during this time?

SOLUTION v = v1 + ac t 120 = 70 + 6000(t) t = 8.33(10 - 3) hr = 30 s

Ans.

v2 = v21 + 2 ac(s - s1) (120)2 = 702 + 2(6000)(s - 0) Ans.

s = 0.792 km = 792 m

Ans: t = 30 s s = 792 m 12

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12–13. Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2, determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don’t drive!

v1

44 ft/s

d

SOLUTION Stopping Distance: For normal driver, the car moves a distance of d¿ = vt = 44(0.75) = 33.0 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 33.0 ft and v = 0. + B A:

v2 = v20 + 2ac (s - s0) 02 = 442 + 2(- 2)(d - 33.0) Ans.

d = 517 ft

For a drunk driver, the car moves a distance of d¿ = vt = 44(3) = 132 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 132 ft and v = 0. + B A:

v2 = v20 + 2ac (s - s0) 02 = 442 + 2(- 2)(d - 132) Ans.

d = 616 ft

Ans: Normal: d = 517 ft drunk: d = 616 ft 13

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12–14. The position of a particle along a straight-line path is defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds. Determine the total distance traveled when t = 10 s. What are the particle’s average velocity, average speed, and the instantaneous velocity and acceleration at this time?

Solution s = t3 - 6t 2 - 15t + 7 v =

ds = 3t2 - 12t - 15 dt

When t = 10 s, Ans.

v = 165 ft>s a =

dv = 6t - 12 dt

When t = 10 s, a = 48 ft>s2

Ans.

When v = 0, 0 = 3t 2 - 12t - 15 The positive root is t = 5s When t = 0,  s = 7 ft When t = 5 s,  s = - 93 ft When t = 10 s,  s = 257 ft Total distance traveled sT = 7 + 93 + 93 + 257 = 450 ft

Ans.

∆s 257 - 7 = = 25.0 ft>s ∆t 10 - 0

Ans.

sT 450 = = 45.0 ft>s ∆t 10

Ans.

vavg =

( vsp ) avg =

Ans: v = 165 ft>s a = 48 ft>s2 sT = 450 ft vavg = 25.0 ft>s (vsp)avg = 45.0 ft>s 14

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12–15. A particle is moving with a velocity of v0 when s = 0 and t = 0. If it is subjected to a deceleration of a = - kv3, where k is a constant, determine its velocity and position as functions of time.

SOLUTION dn = - kn3 dt

a =

t

n

n - 3 dn =

Ln0

L0

- k dt

1 -2 1n - n0- 22 = - kt 2

-

n = a 2kt + a

1

-2 1 b b n20

Ans.

ds = n dt s

L0

s = s =

t

ds =

L0

dt a 2kt + a

2 a 2kt + a 2k

1

2 1 bb v20

1

t 2 1 b b 3 n20

0

1 1 1 B ¢ 2kt + ¢ 2 ≤ ≤ - R n0 k n0 1 2

Ans.

Ans: v = a2kt + s=

15

1 - 1>2 b v20

1 1 1>2 1 c a2kt + 2 b d k v0 v0

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*12–16. A particle is moving along a straight line with an initial velocity of 6 m>s when it is subjected to a deceleration of a = (- 1.5v1>2) m>s2, where v is in m>s. Determine how far it travels before it stops. How much time does this take?

SOLUTION Distance Traveled: The distance traveled by the particle can be determined by applying Eq. 12–3. ds =

vdv a

s

L0

v

ds =

v 1

L6 m>s - 1.5v2

v

s =

dv

1

L6 m>s

- 0.6667 v2 dv 3

= a -0.4444v2 + 6.532 b m When v = 0,

3

Ans.

s = - 0.4444a 0 2 b + 6.532 = 6.53 m

Time: The time required for the particle to stop can be determined by applying Eq. 12–2. dt =

dv a

t

L0

v

dt = 1

t = - 1.333a v2 b When v = 0,

v 6 m>s

dv 1

L6 m>s 1.5v 2 1

= a3.266 - 1.333v 2 b s 1

Ans.

t = 3.266 - 1.333 a0 2 b = 3.27 s

Ans: s = 6.53 m t = 3.27 s 16

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12–17. Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 ft>s when the driver of B suddenly applies the brakes, causing his car to decelerate at 12 ft>s2 . It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 15 ft>s2. Determine the minimum distance d be tween the cars so as to avoid a collision.

A

B d

SOLUTION For B: + ) (:

v = v0 + ac t vB = 60 - 12 t

+ ) (:

s = s0 + v0 t +

1 a t2 2 c

sB = d + 60t -

1 (12) t2 2

(1)

For A: + ) (:

v = v0 + ac t

vA = 60 - 15(t - 0.75), [t 7 0.75] + ) (:

s = s0 + v0 t +

1 a t2 2 c

sA = 60(0.75) + 60(t - 0.75) -

1 (15) (t - 0.75)2, 2

[t 7 0.74]

(2)

Require vA = vB the moment of closest approach. 60 - 12t = 60 - 15(t - 0.75) t = 3.75 s Worst case without collision would occur when sA = sB. At t = 3.75 s, from Eqs. (1) and (2): 60(0.75) + 60(3.75 - 0.75) - 7.5(3.75 - 0.75)2 = d + 60(3.75) - 6(3.75)2 157.5 = d + 140.625 Ans.

d = 16.9 ft

Ans: d = 16.9 ft 17

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12–18. The acceleration of a rocket traveling upward is given by a = 16 + 0.02s2 m>s2, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0.

SOLUTION a ds = n dv

s

s

L0

16 + 0.02 s2 ds =

6 s + 0.01 s2 =

n

L0

n dn

1 2 n 2

n = 212 s + 0.02 s2 ds = n dt 100

L0

ds 212 s + 0.02 s2

1 20.02

t

=

L0

dt

1n B 212s + 0.02s2 + s20.02 +

12 2 20.02

R

100

= t

0

Ans.

t = 5.62 s

Ans: t = 5.62 s 18

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12–19. A train starts from rest at station A and accelerates at 0.5 m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations.

SOLUTION Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus, + B A:

s = s0 + v0t + s1 = 0 + 0 +

+ B A:

1 2 at 2 c

1 (0.5)(602) = 900 m 2

v = v0 + act v1 = 0 + 0.5(60) = 30 m>s

For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus, + B A:

s = s0 + v0t +

1 2 at 2 c

s2 = 900 + 30(900) + 0 = 27 900 m For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = - 1 m>s2. Thus, + B A:

v = v0 + act 0 = 30 + ( -1)t t = 30 s

+ :

s = s0 + v0t +

1 2 at 2 c

s3 = 27 900 + 30(30) +

1 ( - 1)(302) 2 Ans.

= 28 350 m = 28.4 km

Ans: s = 28.4 km 19

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*12–20. The velocity of a particle traveling along a straight line is v = (3t2 - 6t) ft>s, where t is in seconds. If s = 4 ft when t = 0, determine the position of the particle when t = 4 s. What is the total distance traveled during the time interval t = 0 to t = 4 s? Also, what is the acceleration when t = 2 s?

SOLUTION Position: The position of the particle can be determined by integrating the kinematic equation ds = v dt using the initial condition s = 4 ft when t = 0 s. Thus, + B A:

ds = v dt s

t

ds =

L4 ft s2

s

L0

2 A 3t - 6t B dt

= (t 3 - 3t2) 2

4 ft

t 0

3

2

s = A t - 3t + 4 B ft When t = 4 s, s|4 s = 43 - 3(42) + 4 = 20 ft

Ans.

The velocity of the particle changes direction at the instant when it is momentarily brought to rest. Thus, v = 3t2 - 6t = 0 t(3t - 6) = 0 t = 0 and t = 2 s The position of the particle at t = 0 and 2 s is s|0 s = 0 - 3 A 02 B + 4 = 4 ft s|2 s = 23 - 3 A 22 B + 4 = 0 Using the above result, the path of the particle shown in Fig. a is plotted. From this figure, Ans.

sTot = 4 + 20 = 24 ft Acceleration: + B A:

a =

dv d = (3t2 - 6t) dt dt

a = 16t - 62 ft>s2 When t = 2 s, a ƒ t = 2 s = 6122 - 6 = 6 ft>s2 :

Ans. Ans: sTot = 24 ft a  t = 2 s = 6 ft>s2 S 20

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12–21. A freight train travels at v = 6011 - e -t2 ft>s, where t is the elapsed time in seconds. Determine the distance traveled in three seconds, and the acceleration at this time.

s

v

SOLUTION v = 60(1 - e - t) 3

s

L0

ds =

L

v dt =

L0

6011 - e - t2dt

s = 60(t + e - t)|30 Ans.

s = 123 ft a =

dv = 60(e - t) dt

At t = 3 s a = 60e - 3 = 2.99 ft>s2

Ans.

Ans: s = 123 ft a = 2.99 ft>s2 21

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12–22. A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m>s . If the bag is released with the same upward velocity of 6 m>s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant.

SOLUTION (+ T)

s = s0 + v0 t +

1 a t2 2 c

h = 0 + (- 6)(8) +

1 (9.81)(8)2 2

= 265.92 m During t = 8 s, the balloon rises h¿ = vt = 6(8) = 48 m Ans.

Altitude = h + h¿ = 265.92 + 48 = 314 m (+ T)

v = v0 + ac t Ans.

v = - 6 + 9.81(8) = 72.5 m s

Ans: h = 314 m v = 72.5 m>s 22

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12–23. A particle is moving along a straight line such that its acceleration is defined as a = (-2v) m>s2, where v is in meters per second. If v = 20 m>s when s = 0 and t = 0, determine the particle’s position, velocity, and acceleration as functions of time.

Solution a = - 2v dv = - 2v dt v

t dv - 2 dt = L20 v L0

ln

v = - 2t 20

v = ( 20e -2t ) m>s a = L0

dv = dt

Ans.

( - 40e -2t ) m>s2

s

ds = v dt =

L0

Ans.

t

(20e-2t)dt

s = - 10e -2t  t0 = - 10 ( e -2t - 1 ) s = 10 ( 1 - e -2t ) m

Ans.

Ans: v = ( 20e -2t ) m>s a = ( - 40e -2t ) m>s2 s = 10 ( 1 - e -2t ) m 23

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*12–24. The acceleration of a particle traveling along a straight line 1 1>2 is a = s m> s2, where s is in meters. If v = 0, s = 1 m 4 when t = 0, determine the particle’s velocity at s = 2 m.

SOLUTION Velocity: + B A:

v dv = a ds v

L0

s

v dv = v

1 1>2 s ds L1 4

s v2 2 = 1 s3>2 ` 2 0 6 1

v =

1 23

1s3>2 - 121>2 m>s Ans.

When s = 2 m, v = 0.781 m>s.

Ans: v = 0.781 m>s 24

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12–25. If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9.81[1 - v2(10-4)] m>s2, where v is in m>s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as t : q ).

SOLUTION Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2. (+ T)

dt = t

L0

dv a

v

dv 2 L0 9.81[1 - (0.01v) ]

dt = v

t =

v

1 dv dv c + d 9.81 L0 2(1 + 0.01v) L0 2(1 - 0.01v) 9.81t = 50ln a v =

1 + 0.01v b 1 - 0.01v

100(e0.1962t - 1)

(1)

e0.1962t + 1

a) When t = 5 s, then, from Eq. (1) v = b) If t : q ,

e0.1962t - 1 e0.1962t + 1

100[e0.1962(5) - 1] e0.1962(5) + 1

Ans.

= 45.5 m>s

: 1. Then, from Eq. (1) Ans.

vmax = 100 m>s

Ans: (a) v = 45.5 m>s (b) v max = 100 m>s 25

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12–26. The acceleration of a particle along a straight line is defined by a = 12t - 92 m>s2, where t is in seconds. At t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

SOLUTION a = 2t - 9 t

v

L10

dv =

L0

12t - 92 dt

v - 10 = t2 - 9 t v = t2 - 9 t + 10 s

L1

t

ds =

s-1 = s =

L0

1t2 - 9t + 102 dt

13 t - 4.5 t2 + 10 t 3

13 t - 4.5 t2 + 10 t + 1 3

Note when v = t2 - 9 t + 10 = 0: t = 1.298 s and t = 7.701 s When t = 1.298 s,

s = 7.13 m

When t = 7.701 s,

s = - 36.63 m

When t = 9 s,

s = -30.50 m

(a)

s = - 30.5 m

(b)

sTo t = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50)

(c)

Ans.

sTo t = 56.0 m

Ans.

v = 10 m>s

Ans.

Ans: (a) s = - 30.5 m (b) sTot = 56.0 m (c) v = 10 m>s 26

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12–27. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If this variation of the acceleration can be expressed as a = 1g>v2f21v2f - v22, determine the time needed for the velocity to become v = vf>2 . Initially the particle falls from rest.

SOLUTION g dv = a = ¢ 2 ≤ A v2f - v2 B dt vf v

dy

L0 v2f

2¿

- v

=

t

g v2f

L0

dt

vf + v y g 1 ln ¢ ≤` = 2t 2vf vf - v 0 vf t = t =

vf 2g vf 2g

ln ¢

vf + v

ln ¢

vf + vf> 2

t = 0.549 a

vf - v

≤

vf - vf> 2 vf g

≤ Ans.

b

Ans: t = 0.549 a 27

vf g

b

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*12–28. Two particles A and B start from rest at the origin s = 0 and move along a straight line such that a A = (6t - 3) ft>s2 and a B = (12t2 - 8) ft>s2, where t is in seconds. Determine the distance between them when t = 4 s and the total distance each has traveled in t = 4 s.

SOLUTION Velocity: The velocity of particles A and B can be determined using Eq. 12-2. dvA = aAdt vA

t

dvA =

L0

(6t - 3)dt

L0

vA = 3t2 - 3t dvB = aBdt vB

t

dvB =

L0

(12t2 - 8)dt

L0

vB = 4t3 - 8t The times when particle A stops are 3t2 - 3t = 0

t = 0 s and = 1 s

The times when particle B stops are 4t3 - 8t = 0

t = 0 s and t = 22 s

Position:The position of particles A and B can be determined using Eq. 12-1. dsA = vAdt sA

L0

t

dsA =

L0

(3t2 - 3t)dt

sA = t3 -

3 2 t 2

dsB = vBdt sB

L0

t

dsB =

L0

(4t3 - 8t)dt

sB = t4 - 4t2

The positions of particle A at t = 1 s and 4 s are sA |t = 1 s = 13 -

3 2 (1 ) = - 0.500 ft 2

sA |t = 4 s = 43 -

3 2 (4 ) = 40.0 ft 2

Particle A has traveled Ans.

dA = 2(0.5) + 40.0 = 41.0 ft The positions of particle B at t = 22 s and 4 s are sB |t = 12 = (22)4 - 4(22)2 = - 4 ft sB |t = 4 = (4)4 - 4(4)2 = 192 ft Particle B has traveled

Ans.

dB = 2(4) + 192 = 200 ft At t = 4 s the distance beween A and B is

Ans.

¢sAB = 192 - 40 = 152 ft 28

Ans: d A = 41.0 ft d B = 200 ft ∆sAB = 152 ft

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12–29. A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m>s. Determine the height from the ground and the time at which they pass.

Solution Origin at roof: Ball A: 1 2

( + c )   s = s0 + v0t + act 2

- s = 0 + 5t -

1 (9.81)t 2 2

Ball B: 1 2

( + c )   s = s0 + v0t + act 2

- s = - 30 + 20t -

1 (9.81)t 2 2

Solving, Ans.

t = 2 s s = 9.62 m Distance from ground,

Ans.

d = (30 - 9.62) = 20.4 m Also, origin at ground, s = s0 + v0t +

1 2 at 2 c

sA = 30 + 5t +

1 ( -9.81)t 2 2

sB = 0 + 20t +

1 ( -9.81)t 2 2

Require sA = sB 30 + 5t +

1 1 ( - 9.81)t 2 = 20t + ( -9.81)t 2 2 2

t = 2 s

Ans.

sB = 20.4 m

Ans.

29

Ans: h = 20.4 m t = 2s

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12–30. A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = ( -6t) m>s2, where t is in seconds, determine the distance traveled before it stops.

SOLUTION Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have

A+TB

dv = adt v

L27

t

dv =

L0

- 6tdt

v = A 27 - 3t2 B m>s

(1)

At v = 0, from Eq. (1) 0 = 27 - 3t2

t = 3.00 s

Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying Eq. 12–1, we have

A+TB

ds = vdt s

L0

t

ds =

L0

A 27 - 3t2 B dt

s = A 27t - t3 B m

(2)

At t = 3.00 s, from Eq. (2) s = 27(3.00) - 3.003 = 54.0 m

Ans.

Ans: s = 54.0 m 30

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12–31. The velocity of a particle traveling along a straight line is v = v0 - ks, where k is constant. If s = 0 when t = 0, determine the position and acceleration of the particle as a function of time.

SOLUTION Position: + B A:

ds y

dt = t

L0

s

dt =

tt0 = t =

ds v L0 0 - ks

s 1 ln (v0 - ks) 2 k 0

v0 1 ln ¢ ≤ k v0 - ks

ekt =

v0 v0 - ks

s =

v0 A 1 - e - kt B k

v =

d v0 ds = c A 1 - e - kt B d dt dt k

Ans.

Velocity:

v = v0e - kt Acceleration: a =

d dv = A v e - kt B dt dt 0

a = -kv0e - kt

Ans.

Ans: v0 ( 1 - e - kt ) s = k a = - kv0e - kt 31

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*12–32. Ball A is thrown vertically upwards with a velocity of v0. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t 6 2v0>g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

SOLUTION Kinematics: First, we will consider the motion of ball A with (vA)0 = v0, (sA)0 = 0, sA = h, tA = t¿ , and (ac)A = - g.

A+cB

h = 0 + v0t¿ + h = v0t¿ -

A+cB

1 (a ) t 2 2 cA A

sA = (sA)0 + (vA)0tA +

g 2 t¿ 2

1 ( - g)(t¿)2 2

(1)

vA = (vA)0 + (ac)A tA vA = v0 + ( - g)(t¿) (2)

vA = v0 - gt¿

The motion of ball B requires (vB)0 = v0, (sB)0 = 0, sB = h, tB = t¿ - t , and (ac)B = - g.

A+cB

sB = (sB)0 + (vB)0tB + h = 0 + v0(t¿ - t) + h = v0(t¿ - t) -

A+cB

1 (a ) t 2 2 cBB

1 ( - g)(t¿ - t)2 2

g (t¿ - t)2 2

(3)

vB = (vB)0 + (ac)B tB vB = v0 + ( - g)(t¿ - t) (4)

vB = v0 - g(t¿ - t) Solving Eqs. (1) and (3), g 2 g t¿ = v0(t¿ - t) - (t¿ - t)2 2 2 2v0 + gt t¿ = 2g v0t¿ -

Ans.

Substituting this result into Eqs. (2) and (4), vA = v0 - g a = -

2v0 + gt b 2g

1 1 gt = gt T 2 2

Ans.

2v0 + gt 2g 1 vA = gt T 2 1 vB = gt c 2

t= =

2v0 + gt - tb vB = v0 - g a 2g =

Ans:

1 gt c 2

Ans.

32

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12–33. As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = -g0[R2>(R + y)2], where g0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g0 = 9.81 m>s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that v = 0 as y : q.

SOLUTION v dv = a dy 0

Ly

v dv = - g0R

q

2

dy

L0 (R + y)

2

g 0 R2 q v2 2 0 2 = 2 y R + y 0 v = 22g0 R = 22(9.81)(6356)(10)3 Ans.

= 11167 m>s = 11.2 km>s

Ans: v = 11.2 km>s 33

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12–34. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–36), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y0 = 500 km? Use the numerical data in Prob. 12–36.

SOLUTION From Prob. 12–36, (+ c)

a = -g0

R2 (R + y)2

Since a dy = v dv then y

- g0 R 2

dy

2 Ly0 (R + y)

v

=

L0

v dv

g0 R 2 c

y 1 v2 d = R + y y0 2

g0 R2[

1 v2 1 ] = R + y R + y0 2

Thus v = -R

2g0 (y0 - y) A (R + y)(R + y0)

When y0 = 500 km, v = - 6356(103)



Ans.

y = 0,

2(9.81)(500)(103) A 6356(6356 + 500)(106) Ans.

v = -3016 m>s = 3.02 km>s T

Ans: v = -R

2g0 (y0 - y) B (R + y)(R + y0)

vimp = 3.02 km>s 34

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12–35. A freight train starts from rest and travels with a constant acceleration of 0.5 ft>s2. After a time t¿ it maintains a constant speed so that when t = 160 s it has traveled 2000 ft. Determine the time t¿ and draw the v–t graph for the motion.

SOLUTION Total Distance Traveled: The distance for part one of the motion can be related to time t = t¿ by applying Eq. 12–5 with s0 = 0 and v0 = 0. + B A:

s = s0 + v0 t + s1 = 0 + 0 +

1 ac t2 2

1 (0.5)(t¿)2 = 0.25(t¿)2 2

The velocity at time t can be obtained by applying Eq. 12–4 with v0 = 0. + B A:

(1)

v = v0 + act = 0 + 0.5t = 0.5t

The time for the second stage of motion is t2 = 160 - t¿ and the train is traveling at a constant velocity of v = 0.5t¿ (Eq. (1)).Thus, the distance for this part of motion is + B A:

s2 = vt2 = 0.5t¿(160 - t¿) = 80t¿ - 0.5(t¿)2

If the total distance traveled is sTot = 2000, then sTot = s1 + s2 2000 = 0.25(t¿)2 + 80t¿ - 0.5(t¿)2 0.25(t¿)2 - 80t¿ + 2000 = 0 Choose a root that is less than 160 s, then Ans.

t¿ = 27.34 s = 27.3 s

v–t Graph: The equation for the velocity is given by Eq. (1).When t = t¿ = 27.34 s, v = 0.5(27.34) = 13.7 ft>s.

Ans: t′ = 27.3 s. When t = 27.3 s, v = 13.7 ft>s. 35

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*12–36. The s–t graph for a train has been experimentally determined. From the data, construct the v–t and a–t graphs for the motion; 0 … t … 40 s. For 0 … t … 30 s, the curve is s = (0.4t2) m, and then it becomes straight for t Ú 30 s.

s (m)

600

360

Solution 0 … t … 30:   s = 0.4t 2

30

  v =

ds = 0.8t dt

  a =

dv = 0.8 dt

t (s)

40

30 … t … 40:   s - 360 = a

600 - 360 b(t - 30) 40 - 30

      s = 24(t - 30) + 360       v =

ds = 24 dt

      a =

dv = 0 dt

Ans: s = 0.4t 2 ds v = = 0.8t dt dv = 0.8 a = dt s = 24(t - 30) + 360 ds v = = 24 dt dv a = = 0 dt 36

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12–37. Two rockets start from rest at the same elevation. Rocket A accelerates vertically at 20 m>s2 for 12 s and then maintains a constant speed. Rocket B accelerates at 15 m>s2 until reaching a constant speed of 150 m>s. Construct the a–t, v–t, and s–t graphs for each rocket until t = 20 s. What is the distance between the rockets when t = 20 s?

Solution For rocket A For t 6 12 s + c vA = (vA)0 + aA t

vA = 0 + 20 t



vA = 20 t

+ c sA = (sA)0 + (vA)0 t +

sA = 0 + 0 +



sA = 10 t 2

1 a t2 2 A

1 (20) t 2 2

When t = 12 s,

vA = 240 m>s



sA = 1440 m

For t 7 12 s vA = 240 m>s sA = 1440 + 240(t - 12) For rocket B For t 6 10 s + c vB = (vB)0 + aB t

vB = 0 + 15 t



vB = 15 t

+ c sB = (sB)0 + (vB)0 t +

sB = 0 + 0 +



sB = 7.5 t 2

1 aB t 2 2

1 (15) t 2 2

When t = 10 s,

vB = 150 m>s



sB = 750 m

For t 7 10 s vB = 150 m>s sB = 750 + 150(t - 10) When t = 20 s,  sA = 3360 m,  sB = 2250 m ∆s = 1110 m = 1.11 km

Ans. 37

Ans: ∆s = 1.11 km

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12–38. A particle starts from s = 0 and travels along a straight line with a velocity v = (t2 - 4t + 3) m > s, where t is in seconds. Construct the v -t and a - t graphs for the time interval 0 … t … 4 s.

SOLUTION a–t Graph: a =

d 2 dv = 1t - 4t + 32 dt dt

a = (2t - 4) m>s2 Thus, a|t = 0 = 2(0) - 4 = - 4 m>s2 a|t = 2 = 0 a|t = 4 s = 2(4) - 4 = 4 m>s2 The a - t graph is shown in Fig. a. v – t Graph: The slope of the v - t graph is zero when a = a = 2t - 4 = 0

dv = 0. Thus, dt

t = 2s

The velocity of the particle at t = 0 s, 2 s, and 4 s are v|t = 0 s = 02 - 4(0) + 3 = 3 m>s v|t = 2 s = 22 - 4(2) + 3 = - 1 m>s v|t = 4 s = 42 - 4(4) + 3 = 3 m>s The v - t graph is shown in Fig. b.

Ans: a t = 0 = - 4 m>s2 a t = 2 s = 0 a t = 4 s = 4 m>s2 v t = 0 = 3 m>s v t = 2 s = - 1 m>s v t = 4 s = 3 m>s 38

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12–39. If the position of a particle is defined by s = [2 sin (p>5)t + 4] m, where t is in seconds, construct the s -t, v -t, and a- t graphs for 0 … t … 10 s.

SOLUTION

Ans: p s = 2 sin a tb + 4 5 2p p v = cos a tb 5 5 2p2 p a = sin a tb 25 5 39

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*12–40. An airplane starts from rest, travels 5000 ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi>h. It then climbs in a straight line with a uniform acceleration of 3 ft>s2 until it reaches a constant speed of 220 mi>h. Draw the s–t, v–t, and a–t graphs that describe the motion.

SOLUTION v1 = 0 v2 = 162

mi (1h) 5280 ft = 237.6 ft>s h (3600 s)(1 mi)

v22 = v21 + 2 ac(s2 - s1) (237.6)2 = 02 + 2(ac)(5000 - 0) ac = 5.64538 ft>s2 v2 = v1 + act 237.6 = 0 + 5.64538 t t = 42.09 = 42.1 s v3 = 220

mi (1h) 5280 ft = 322.67 ft>s h (3600 s)(1 mi)

v23 = v22 + 2ac(s3 - s2) (322.67)2 = (237.6)2 + 2(3)(s - 5000) s = 12 943.34 ft v3 = v2 + act 322.67 = 237.6 + 3 t t = 28.4 s

Ans: s = 12 943.34 ft v3 = v2 + ac t t = 28.4 s 40

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12–41. The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft>s2 and then decelerate at 2 ft>s2. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a–t, v–t, and s–t graphs for the motion.

40 ft

SOLUTION + c v2 = v1 + act1 vmax = 0 + 5 t1 + c v3 = v2 + ac t 0 = vmax - 2 t2 Thus t1 = 0.4 t2 + c s2 = s1 + v1t1 + h = 0 + 0 +

1 2 a t1 2 c

1 (5)(t21) = 2.5 t21 2

+ c 40 - h = 0 + vmaxt2 -

1 (2) t22 2

+ c v2 = v21 + 2 ac(s - s1) v2max = 0 + 2(5)(h - 0) v2max = 10h 0 = v2max + 2(-2)(40 - h) v2max = 160 - 4h Thus, 10 h = 160 - 4h h = 11.429 ft vmax = 10.69 ft>s t1 = 2.138 s t2 = 5.345 s Ans.

t = t1 + t2 = 7.48 s When t = 2.145, v = vmax = 10.7 ft>s and h = 11.4 ft.

Ans: t = 7.48 s. When t = 2.14 s, v = vmax = 10.7 ft>s h = 11.4 ft 41

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12–42. The velocity of a car is plotted as shown. Determine the total distance the car moves until it stops 1t = 80 s2. Construct the a–t graph.

v (m/s)

10

SOLUTION Distance Traveled: The total distance traveled can be obtained by computing the area under the v - t graph. s = 10(40) +

1 (10)(80 - 40) = 600 m 2

40

80

t (s)

Ans.

dv a – t Graph: The acceleration in terms of time t can be obtained by applying a = . dt For time interval 0 s … t 6 40 s, a =

For time interval 40 s 6 t … 80 s, a =

dv = 0 dt

v - 10 0 - 10 1 , v = a - t + 20 b m>s. = t - 40 80 - 40 4

dv 1 = - = - 0.250 m s2 dt 4

For 0 … t 6 40 s, a = 0. For 40 s 6 t … 80, a = - 0.250 m s2 .

Ans: s = 600 m. For 0 … t 6 40 s, a = 0. For 40 s 6 t … 80 s, a = - 0.250 m>s2 42

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12–43. The motion of a jet plane just after landing on a runway is described by the a–t graph. Determine the time t′ when the jet plane stops. Construct the v–t and s–t graphs for the motion. Here s = 0, and v = 300 ft>s when t = 0.

a (m/s2) 10 10 20

Solution v–t Graph. The v–t function can be determined by integrating dv = a dt. For 0 … t 6 10 s, a = 0. Using the initial condition v = 300 ft>s at t = 0, v



L300 ft>s

L0

dv =

t

0 dt

v - 300 = 0 v = 300 ft>s

For 10 s 6 t 6 20 s,

Ans. a - ( - 20) t - 10

=

- 10 - ( -20) 20 - 10

, a = (t - 30) ft>s2. Using the

initial condition v = 300 ft>s at t = 10 s, t

v



(t - 30) dt L10 s

L300 ft>s

dv =

t 1 v - 300 = a t 2 - 30tb ` 2 10 s

1 v = e t 2 - 30t + 550 f ft>s 2

Ans.

At t = 20 s,

v`

=

t = 20 s

1 ( 202 ) - 30(20) + 550 = 150 ft>s 2

For 20 s 6 t 6 t′, a = -10 ft>s. Using the initial condition v = 150 ft>s at t = 20 s, t

v



L20 s

L150 ft>s

dv =

- 10 dt

v - 150 = ( -10t) `

t 20 s

Ans.

v = ( - 10t + 350) ft>s

It is required that at t = t′, v = 0. Thus 0 = - 10 t′ + 350

Ans.

t′ = 35 s

Using these results, the v9t graph shown in Fig. a can be plotted s-t Graph. The s9t function can be determined by integrating ds = v dt. For 0 … t 6 10 s, the initial condition is s = 0 at t = 0. s



t

300 dt L0 L0 s = {300 t} ft ds =

Ans.

At = 10 s, s 0 t = 10 s = 300(10) = 3000 ft 43

20

t¿

t (s)

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12–43. Continued

For 10 s 6 t 6 20 s, the initial condition is s = 3000 ft at t = 10 s. s t 1 ds = a t 2 - 30t + 550bdt L3000 ft L10 s 2 t 1 s - 3000 = a t 3 - 15t 2 + 550tb ` 6 10 s 1 3 s = e t - 15t 2 + 550t - 1167 f ft 6

Ans.

At t = 20 s,



s =

1 (203) - 15(202) + 550(20) - 1167 = 5167 ft 6

For 20 s 6 t … 35 s, the initial condition is s = 5167 ft at t = 20 s. s



L5167 ft

t

ds =

( -10t + 350) dt L20 s

s - 5167 = ( - 5t 2 + 350t) `

t 20 s

s = 5 - 5t 2 + 350t + 1676 ft

Ans.

At t = 35 s,

s`

t = 35 s

= - 5(352) + 350(35) + 167 = 6292 ft

using these results, the s-t graph shown in Fig. b can be plotted.

Ans: t′ = 35 s For 0 … t 6 10 s, s = {300t} ft v = 300 ft>s For 10 s 6 t 6 20 s, 1 s = e t 3 - 15t 2 + 550t - 1167 f ft 6 1 v = e t 2 - 30t + 550 f ft>s 2 For 20 s 6 t … 35 s, s = 5 -5t 2 + 350t + 1676 ft v = ( - 10t + 350) ft>s 44

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*12–44. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude of 4 m>s2. If the plates are spaced 200 mm apart, determine the maximum velocity vmax and the time t¿ for the particle to travel from one plate to the other. Also draw the s–t graph. When t = t¿>2 the particle is at s = 100 mm.

smax v s

vmax

SOLUTION ac = 4 m/s2

t¿/ 2

t¿

s = 100 mm = 0.1 m 2 v2 = v20 + 2 ac(s - s0) v2max = 0 + 2(4)(0.1 - 0) vmax = 0.89442 m>s

Ans.

= 0.894 m>s

v = v0 + ac t¿ t¿ 0.89442 = 0 + 4( ) 2 t¿ = 0.44721 s s = s0 + v0 t + s = 0 + 0 +

Ans.

= 0.447 s 1 a t2 2 c

1 (4)(t)2 2

s = 2 t2 When t =

0.44721 = 0.2236 = 0.224 s, 2

s = 0.1 m t

v

ds = -

L0.894

4 dt L0.2235

v = -4 t +1.788 s

t

ds =

L0.1

1 -4t +1.7882 dt L0.2235

s = - 2 t2 + 1.788 t - 0.2 When t = 0.447 s, s = 0.2 m

Ans: t = = 0.447 s s = 0.2 m 45

t

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12–45. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where t¿ = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs for the particle. When t = t¿>2 the particle is at s = 0.5 m.

smax v s

vmax

SOLUTION For 0 6 t 6 0.1 s,

t¿/ 2

v = 100 t

t¿

dv = 100 dt

a =

ds = v dt s

L0

t

ds =

L0

100 t dt

s = 50 t 2 When t = 0.1 s, s = 0.5 m For 0.1 s 6 t 6 0.2 s, v = -100 t + 20 a =

dv = - 100 dt

ds = v dt s

L0.5

t

ds =

1 -100t + 202dt L0.1

s - 0.5 = ( - 50 t 2 + 20 t - 1.5) s = - 50 t 2 + 20 t - 1 When t = 0.2 s, s = 1m When t = 0.1 s, s = 0.5 m and a changes from 100 m/s2 to -100 m/s2. When t = 0.2 s, s = 1 m.

Ans: When t = 0.1 s, s = 0.5 m and a changes from 100 m>s2 to - 100 m>s2. When t = 0.2 s, s = 1 m. 46

t

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12–46. The a–s graph for a rocket moving along a straight track has been experimentally determined. If the rocket starts at s = 0 when v = 0, determine its speed when it is at s = 75 ft, and 125 ft, respectively. Use Simpson’s rule with n = 100 to evaluate v at s = 125 ft.

a (ft/s2)

a  5  6(s  10)5/3 5

Solution

s (ft)

100

0 … s 6 100 L0

v

v dv =

L0

s

5 ds

1 2 v = 5s 2 v = 210 s

Ans.

At s = 75 ft,  v = 2750 = 27.4 ft>s At s = 100 ft,  v = 31.623 v dv = ads v

L31.623

v dv =

v

125

L100

35

1 2 v ` = 201.0324 2 31.623

+ 6 ( 2s - 10 ) 5>3 4 ds Ans.

v = 37.4 ft>s

Ans: v` v`

47

s = 75 ft

s = 125 ft

= 27.4 ft>s = 37.4 ft>s

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12–47. A two-stage rocket is fired vertically from rest at s = 0 with the acceleration as shown. After 30 s the first stage, A, burns out and the second stage, B, ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 60 s.

a (m/s2) B 24

A

12

Solution v9t Graph. The v -t function can be determined by integrating dv = a dt. For 0 … t 6 30 s, a =

L0

v

t

12 2 t = a tb m>s2. Using the initial condition v = 0 at t = 0, 30 5

2 t dt L0 5

dv =

1 v = e t 2 f m>s 5

Ans.

At t = 30 s,

v`

t = 30 s

=

1 2 (30 ) = 180 m>s 5

For 30 6 t … 60 s, a = 24 m>s2. Using the initial condition v = 180 m>s at t = 30 s,



L180 m>s v

t

dv =

L30 s

v - 180 = 24 t `

24 dt

t 30 s

Ans.

v = {24t - 540} m>s

At t = 60 s,

v`

t = 60 s

= 24(60) - 540 = 900 m>s

Using these results, v - t graph shown in Fig. a can be plotted. s -t Graph. The s - t function can be determined by integrating ds = v dt. For 0 … t 6 30 s, the initial condition is s = 0 at t = 0.



L0



ds =

s = e

At t = 30 s, s`

t

s

1 2 t dt L0 5

1 3 t f m 15

t = 30 s

=

Ans.

1 (303) = 1800 m 15

48

30

60

t (s)

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12–47. Continued

For 30 s 6 t … 60 s, the initial condition is s = 1800 m at t = 30 s. s





L1800 m

t

ds =

L30 s

(24t - 540)dt

s - 1800 = (12t 2 - 540t) `

t 30 s

s = {12t 2 - 540t + 7200} m

At t = 60 s,

s`

t = 60 s

= 12(602) - 540(60) + 7200 = 18000 m

Using these results, the s -t graph in Fig. b can be plotted.

49

Ans: For 0 … t 6 30 s, 1 v = e t2 f m>s 5 1 s = e t3 f m 15 For 30 … t … 60 s, v = {24t - 540} m>s s = 512t 2 - 540t + 7200} m

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*12–48. v (m/s)

The race car starts from rest and travels along a straight road until it reaches a speed of 26 m>s in 8 s as shown on the v–t graph. The flat part of the graph is caused by shifting gears. Draw the a–t graph and determine the maximum acceleration of the car.

6

26 v  4t  6 14 v  3.5t

Solution For 0 … t 6 4 s a =

4

5

8

t (s)

∆v 14 = = 3.5 m>s2 ∆t 4

For 4 s … t 6 5 s a =

∆v = 0 ∆t

For 5 s … t 6 8 s a =

∆v 26 - 14 = = 4 m>s2 ∆t 8 - 5

a max = 4.00 m>s2

Ans.

Ans: amax = 4.00 m>s2 50

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12–49. The jet car is originally traveling at a velocity of 10 m>s when it is subjected to the acceleration shown. Determine the car’s maximum velocity and the time t′ when it stops. When t = 0, s = 0.

a (m/s2)

6 t¿

Solution v9t Function. The v-t function can be determined by integrating dv = a dt . For 4 0 … t 6 15 s, a = 6 m>s2. Using the initial condition v = 10 m>s at t = 0, v

L0

t (s)

t



L10 m>s



v - 10 = 6t



v = {6t + 10} m>s

dv =

15

6dt

The maximum velocity occurs when t = 15 s. Then

vmax = 6(15) + 10 = 100 m>s

Ans.

For 15 s 6 t … t′, a = -4 m>s, Using the initial condition v = 100 m>s at t = 15 s, v



L100 m>s

dv =

t

L15 s

v - 100 = ( -4t) `

- 4dt t 15 s

v = { -4t + 160} m>s

It is required that v = 0 at t = t′. Then



Ans.

0 = - 4t′ + 160  t′ = 40 s

Ans: vmax = 100 m>s t′ = 40 s 51

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12–50. The car starts from rest at s = 0 and is subjected to an acceleration shown by the a–s graph. Draw the v–s graph and determine the time needed to travel 200 ft.

a (ft/s2)

12

a  0.04s  24

6

Solution

300

For s 6 300 ft

450

s (ft)

a ds = v dv L0

s

12 ds =

12 s =

L0

v

v dv

1 2 v 2

v = 4.90 s1>2 At s = 300 ft,  v = 84.85 ft>s For 300 ft 6 s 6 450 ft a ds = v dv v

s

L300

(24 - 0.04 s) ds =

v dv L84.85

24 s - 0.02 s2 - 5400 = 0.5 v2 - 3600 v = ( - 0.04 s2 + 48 s - 3600)1>2 At s = 450 ft,  v = 99.5 ft>s v = 4.90 s1>2 ds = 4.90 s1>2 dt L0

200

s

2 s1>2 `

-1>2

200 0

ds =

L0

t

4.90 dt

= 4.90 t Ans.

t = 5.77 s

Ans: For 0 … s 6 300 ft, v = 5 4.90 s1>26 m>s. For 300 ft 6 s … 450 ft, v = 5 ( - 0.04s2 + 48s - 3600)1>2 6 m>s. s = 200 ft when t = 5.77 s. 52

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12–51. The v–t graph for a train has been experimentally determined. From the data, construct the s–t and a–t graphs for the motion for 0 … t … 180 s. When t = 0, s = 0.

v (m/s)

10

6

Solution s–t Graph. The s–t function can be determined by integrating ds = v dt. 1 6 For 0 … t 6 60 s, v = t = a t b m>s. Using the initial condition 60 10 s = 0 at t = 0,

L0

s

60

t

ds =

1 a tbdt L0 10

1 2 t f m 20

s = e

Ans.

When t = 60 s,



s  t - 60 s =

1 ( 602 ) = 180 m 20

For 60 s 6 t 6 120 s, v = 6 m>s. Using the initial condition s = 180 m at t = 60 s, s



t

L180 m

ds =

L60 s

s - 180 = 6t `

6 dt

t 60 s

Ans.

s = 56t - 1806 m

At t = 120 s,

s  t - 120 s = 6(120) - 180 = 540 m

For 120 s 6 t … 180 s,

v - 6 10 - 6 1 = ; v = e t - 2 f m>s. Using the initial t - 120 180 - 120 15

condition s = 540 m at t = 120 s, s



L540 m

t

ds =

L120 s

a



s - 540 = a



s = e

At t = 180 s,

1 t - 2b dt 15

t 1 2 t - 2tb ` 30 120 s

1 2 t - 2t + 300 f m 30

s  t = 180 s =

Ans.

1 ( 1802 ) - 2(180) + 300 = 1020 m 30

Using these results, s–t graph shown in Fig. a can be plotted.

53

120

180

t (s)

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12–51. Continued

a–t Graph. The a–t function can be determined using a = For 0 … t 6 60 s,  a =

1 t2 d 1 10

dt

For 60 s 6 t 6 120 s,  a = For 120 s 6 t … 180 s,  a =

= 0.1 m>s2

d(6) dt

dv . dt Ans. Ans.

= 0

1 t - 22 d 1 15

dt

= 0.0667 m>s2

Ans.

Using these results, a–t graph shown in Fig. b can be plotted.

Ans: For 0 … t 6 60 s, 1 s = e t 2 f m, 20 a = 0.1 m>s2. For 60 s 6 t 6 120 s, s = {6t - 180} m, a = 0. For 120 s 6 t … 180 s, 1 s = e t 2 - 2t + 300 f m, 30 a = 0.0667 m>s2.

54

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*12–52. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the total distance the motorcycle travels until it stops when t = 15 s. Also plot the a–t and s–t graphs.

v (m/s)

5

v  1.25t

v5 v  t  15

4

Solution

10

15

t (s)

For t 6 4 s a = L0

dv = 1.25 dt

s

ds =

L0

t

1.25 t dt

s = 0.625 t 2 When t = 4 s,  s = 10 m For 4 s 6 t 6 10 s a =

dv = 0 dt

s

t

L10

ds =

L4

5 dt

s = 5 t - 10 When t = 10 s,  s = 40 m For 10 s 6 t 6 15 s a =

dv = -1 dt

s

L40

ds =

t

(15 - t) dt L10

s = 15 t - 0.5 t 2 - 60 Ans.

When t = 15 s,  s = 52.5 m

Ans: When t = 15 s,  s = 52.5 m 55

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12–53. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the motorcycle’s acceleration and position when t = 8 s and t = 12 s.

v (m/s)

5

v  1.25t

v5 v  t  15

4

Solution

10

15

t (s)

At t = 8 s a =

dv = 0 dt

Ans.

∆ s = 1 v dt s - 0 =

1 (4)(5) + (8 - 4)(5) = 30 2 Ans.

s = 30 m At t = 12 s a =

dv -5 = = - 1 m>s2 dt 5

Ans.

∆ s = 1 v dt s - 0 =

1 1 1 3 3 (4)(5) + (10 - 4)(5) + (15 - 10)(5) - a b(5)a b(5) 2 2 2 5 5

s = 48 m

Ans.

Ans: At t = 8 s, a = 0 and s = 30 m. At t = 12 s, a = - 1 m>s2 and s = 48 m. 56

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12–54. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the s–t and a–t graphs. Also determine the average speed and the distance traveled for the 15-s time interval. When t = 0, s = 0.

v (m/s) 15 v  0.6t 2

Solution s–t Graph. The s–t function can be determined by integrating ds = v dt . For 0 … t 6 5 s, v = 0.6t 2. Using the initial condition s = 0 at t = 0,

L0

s

ds =

L0

t

0.6t 2dt

s = 50.2t 3 6 m

Ans.

At t = 5 s,



s  t = 5 s = 0.2 ( 53 ) = 25 m v - 15 0 - 15 1 = ; v = (45 - 3t). Using the initial condition t - 5 15 - 5 2

For 5 s 6 t … 15 s, s = 25 m at t = 5 s, s

t



L25 m



s - 25 =



ds =

1 (45 - 3t)dt 2 L5 s 45 3 t - t 2 - 93.75 2 4

1 s = e (90t - 3t 2 - 275) f m 4

Ans.

At t = 15 s,

s =

1 3 90(15) - 3 ( 152 ) - 275 4 = 100 m 4

Ans.

Thus the average speed is

vavg =

sT 100 m = = 6.67 m>s t 15 s

Ans.

using these results, the s–t graph shown in Fig. a can be plotted.

57

5

15

t (s)

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12–54. Continued

a–t Graph. The a–t function can be determined using a = For 0 … t 6 5 s,  a =

d(0.6 t 2) dt

dv . dt

= 51.2 t6 m>s2

Ans.

At t = 5 s,  a = 1.2(5) = 6 m>s2 For 5 s 6 t … 15 s,  a =

d 3 12(45 - 3t) 4 dt

Ans.

= - 1.5 m>s2

Ans.

Ans: For 0 … t 6 5 s, s = 5 0.2t 3 6 m a = {1.2t} m>s2 For 5 s 6 t … 15 s, 1 s = e ( 90t - 3t 2 - 275 ) f m 4 a = - 1.5 m>s2 At t = 15 s, s = 100 m vavg = 6.67 m>s 58

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12–55. An airplane lands on the straight runway, originally traveling at 110 ft> s when s = 0. If it is subjected to the decelerations shown, determine the time t¿ needed to stop the plane and construct the s–t graph for the motion.

a (ft/s2) 5

15

20

t'

t (s)

–3

SOLUTION

–8

v0 = 110 ft>s ¢ v = 1 a dt 0 - 110 = - 3(15 - 5) - 8(20 - 15) - 3(t¿ - 20) Ans.

t¿ = 33.3 s = 550 ft

st=

5s

st=

15s

= 1500 ft

st=

20s

= 1800 ft

st=

33.3s

= 2067 ft

Ans: t′ = 33.3 s s  t = 5 s = 550 ft s  t = 15 s = 1500 ft s  t = 20 s = 1800 ft s  t = 33.3 s = 2067 ft 59

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*12–56.  Starting from rest at s = 0, a boat travels in a straight line with the acceleration shown by the a–s graph. Determine the boat’s speed when s = 50 ft, 100 ft, and 150 ft.

a (ft/s2)

8 6

Solution

100

150

s (ft)

v9s Function. The v - s function can be determined by integrating v dv = a ds. a - 8 6 - 8 1 = , a = e - s + 8 f ft>s2. Using the initial For 0 … s 6 100 ft, s - 0 100 - 0 50 condition v - 0 at s = 0, v s 1 v dv = a - s + 8b ds 50 L0 L0 v

s v2 1 2 s + 8 sb ` ` = a2 0 100 0

v2 1 2 = 8s s 2 100 v = e

At s = 50 ft,

1 (800 s - s2) f ft>s A 50

v  s = 50 ft = At s = 100 ft,

1 [800 (50) - 502] = 27.39 ft>s = 27.4 ft>s A 50

Ans.

1 [800 (100) - 1002] = 37.42 ft>s = 37.4 ft>s Ans. A 50 a - 0 6 - 0 3 For 100 ft 6 s … 150 ft, = ; a = e - s + 18 f ft>s2. Using the s - 150 100 - 150 25 initial condition v = 37.42 ft>s at s = 100 ft, v  s = 100 ft =

v

L37.42 ft>s

v dv =

v

s

L100 ft

a-

3 s + 18b ds 25

s 3 v2 ` = a - s2 + 18s b ` 2 37.42 ft>s 50 100 ft

1 v = e 2- 3s2 + 900s - 25000 f ft>s 5

At s = 150 ft v  s = 150 ft =

1 2- 3 ( 1502 ) + 900 (150) - 25000 = 41.23 ft>s = 41.2 ft>s 5

Ans.

Ans: v  s = 50 ft = 27.4 ft>s v  s = 100 ft = 37.4 ft>s v  s = 150 ft = 41.2 ft>s 60

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12–57.  Starting from rest at s = 0, a boat travels in a straight line with the acceleration shown by the a–s graph. Construct the v–s graph.

a (ft/s2)

8 6

Solution

100

150

s (ft)

v9s Graph. The v - s function can be determined by integrating v dv = a ds. For a - 8 6 - 8 1 = , a = e - s + 8 f ft>s2 s - 0 100 - 0 50 condition v = 0 at s = 0, 0 … s 6 100 ft,

L0

v

v dv =

L0

s

a-

using

the

initial

1 s + 8b ds 50

s v2 1 2 ` = as + 8 sb ` 2 0 100 0

v2 1 2 = 8s s 2 100 v = e

1 (800 s - s2) f ft>s A 50

At s = 25 ft, 50 ft, 75 ft and 100 ft v  s = 25 ft = v  s = 50 ft = v  s = 75 ft = v  s = 100 ft =

1 [800 (25) -252] = 19.69 ft>s A 50 1 [800 (50) -502] = 27.39 ft>s A 50 1 [800 (75) -752] = 32.98 ft>s A 50

1 [800 (100) -1002] = 37.42 ft>s A 50

For 100 ft 6 s … 150 ft,

a - 0 6 - 0 3 = ; a = e - s + 18 f ft>s2 using the s - 150 100 - 150 25

initial condition v = 37.42 ft>s at s = 100 ft, v

s

L37.42 ft>s

v dv =

L100 ft

a-

3 s + 18b ds 25

s v2 v 3 ` = a - s2 + 18 s b ` 2 37.42 ft>s 50 100 ft

1 v = e 2- 3s2 + 900s - 25000 f ft>s 5

Ans:

At s = 125 ft and s = 150 ft

1 2- 3 ( 1252 ) + 900 (125) - 25000 = 40.31 ft>s 5 1 = 2- 3 ( 1502 ) + 900 (150) - 25000 = 41.23 ft>s 5

v  s = 125 ft = v  s = 150 ft

61

For 0 … s 6 100 ft, 1 ( 800s - s2 ) f ft>s v = e A 50 For 100 ft 6 s … 150 ft, 1 v = e 2-3s2 + 900s - 25 000 f ft>s 5

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12–58.  A two-stage rocket is fired vertically from rest with the acceleration shown. After 15 s the first stage A burns out and the second stage B ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 40 s.

a (m/s2) B A 20 15

Solution

15

40

t (s)

For 0 … t 6 15 a = t L0

v

dv =

v =

1 2 t 2

L0

t

t dt

v = 112.5 when t = 15 s L0

s

t

ds =

1 2 t dt L0 2

1 3 t 6 s = 562.5 when t = 15 s s =

For 15 6 t 6 40 a = 20 v

L112.5

t

dv =

L1.5

20 dt

v = 20t - 187.5 v = 612.5 when t = 40 s s

L562.5

t

ds =

(20 t - 187.5) dt L15

s = 10 t 2 - 187.5 t + 1125 s = 9625 when t = 40 s

Ans: For 0 … t 6 15 s, 1 v = e t 2 f m>s 2 1 s = e t3 f m 6 For 15 s 6 t … 40 s, v = {20t - 187.5 m>s} s = {10t 2 - 187.5t + 1125} m 62

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12–59. The speed of a train during the first minute has been recorded as follows: t (s) 0 20 40 60 0 16 21 24 v (m>s) Plot the v -t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled.

SOLUTION The total distance traveled is equal to the area under the graph. sT =

1 1 1 (20)(16) + (40 - 20)(16 + 21) + (60 - 40)(21 + 24) = 980 m 2 2 2

Ans.

Ans: sT = 980 m 63

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*12–60. A man riding upward in a freight elevator accidentally drops a package off the elevator when it is 100 ft from the ground. If the elevator maintains a constant upward speed of 4 ft>s, determine how high the elevator is from the ground the instant the package hits the ground. Draw the v–t curve for the package during the time it is in motion. Assume that the package was released with the same upward speed as the elevator.

SOLUTION For package: (+ c )

v2 = v20 + 2ac(s2 - s0) v2 = (4)2 + 2( - 32.2)( 0 - 100) v = 80.35 ft>s T

(+ c )

v = v0 + act

-80.35 = 4 + ( - 32.2)t t = 2.620 s For elevator: (+ c )

s2 = s0 + vt s = 100 + 4(2.620) Ans.

s = 110 ft

Ans: s = 110 ft 64

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12–61. Two cars start from rest side by side and travel along a straight road. Car A accelerates at 4 m>s2 for 10 s and then maintains a constant speed. Car B accelerates at 5 m>s2 until reaching a constant speed of 25 m/s and then maintains this speed. Construct the a–t, v–t, and s–t graphs for each car until t = 15 s. What is the distance between the two cars when t = 15 s?

SOLUTION Car A: v = v0 + ac t vA = 0 + 4t At t = 10 s,

vA = 40 m>s

s = s0 + v0t + sA = 0 + 0 +

1 2 at 2 c 1 (4)t2 = 2t2 2

At t = 10 s,

sA = 200 m

t 7 10 s,

ds = v dt sA

L200

t

ds =

L10

40 dt

sA = 40t - 200 At t = 15 s,

sA = 400 m

Car B: v = v0 + a c t vB = 0 + 5t When vB = 25 m/s,

t =

25 = 5s 5

s = s0 + v0t + sB = 0 + 0 +

1 2 at 2 c

1 (5)t2 = 2.5t2 2

When t = 10 s, vA = (vA)max = 40 m/s and sA = 200 m. When t = 5 s, sB = 62.5 m. When t = 15 s, sA = 400 m and sB = 312.5 m.

65

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12–61.   Continued

At t = 5 s,

sB = 62.5 m

t 7 5 s,

ds = v dt sB

L62.5

t

ds =

L5

25 dt

sB - 62.5 = 25t - 125 sB = 25t - 62.5 When t = 15 s,

sB = 312.5

Distance between the cars is Ans.

¢s = sA - sB = 400 - 312.5 = 87.5 m Car A is ahead of car B.

Ans: When t = 5 s, sB = 62.5 m. When t = 10 s, vA = (vA)max = 40 m>s and sA = 200 m. When t = 15 s, sA = 400 m and sB = 312.5 m. ∆s = sA - sB = 87.5 m 66

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12–62. If the position of a particle is defined as s = 15t - 3t22 ft, where t is in seconds, construct the s–t, v–t, and a–t graphs for 0 … t … 10 s.

SOLUTION

Ans: v = {5 - 6t} ft>s a = - 6 ft>s2 67

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12–63. From experimental data, the motion of a jet plane while traveling along a runway is defined by the n - t graph. Construct the s - t and a - t graphs for the motion. When t = 0, s = 0.

v (m/s) 60

20

SOLUTION s - t Graph: The position in terms of time t can be obtained by applying ds 20 . For time interval 0 s … t 6 5 s, y = y = t = (4t) m>s. dt 5

5

20

30

t (s)

ds = ydt s

L0

t

ds =

L0

4tdt

s = A 2t2 B m s = 2 A 52 B = 50 m,

When t = 5 s, For time interval 5 s 6 t 6 20 s,

ds = ydt s

t

ds =

L50 m

L5 a

20dt

s = (20t - 50) m When t = 20 s,

s = 20(20) - 50 = 350 m

For time interval 20 s 6 t … 30 s,

60 - 20 y - 20 , y = (4t - 60) m>s. = t - 20 30 - 20 ds = ydt

s

t

ds =

L350 m

L20 a

(4t - 60) dt

s = A 2t2 - 60t + 750 B m When t = 30 s,

s = 2 A 302 B - 60(30) + 750 = 750 m

a - t Graph: The acceleration function in terms of time t can be obtained by dy applying a = . For time interval 0 s … t 6 5 s, 5 s 6 t 6 20 s and dt dy dy dy 20 s 6 t … 30 s, a = = 4.00 m s2, a = = 0 and a = = 4.00 m s2, dt dt dt respectively.

68

Ans: For 0 … t 6 5 s, s = 5 2t 2 6 m and a = 4 m>s2. For 5 s 6 t 6 20 s, s = {20t - 50} m and a = 0. For 20 s 6 t … 30 s, s = 5 2t 2 - 60t + 750 6 m and a = 4 m>s2.

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*12–64.  The motion of a train is described by the a–s graph shown. Draw the v–s graph if v = 0 at s = 0.

a (m/s2)

3

300

600

s (m)

Solution v9s Graph. The v - s function can be determined by integrating v dv = a ds. 3 1 For 0 … s 6 300 m, a = a bs = a s b m>s2. Using the initial condition 300 100 v = 0 at s = 0, L0

v

v dv =

L0

1 2 v2 = s 2 200

s

a

1 s b ds 100

1 s f m>s 10

v = e

At s = 300 m,

Ans.

1 (300) = 30 m>s 10 a - 3 0 - 3 1 = ;a = es + 6 f m>s2, using the For 300 m 6 s … 600 m, s - 300 600 - 300 100 initial condition v = 30 m>s at s = 300 m, v  s = 300 m =

v

s

L30 m>s

v dv =

L300 m

2 v

a-

1 s + 6b ds 100

s 1 2 v s + 6 sb ` ` = a2 30 m>s 200 300 m

v2 1 2 - 450 = 6 s s - 1350 2 200

1 2 s - 1800 f m>s 100

v = e

A

v =

12 (600) -

At s = 600 m,

A

12s -

Ans.

1 ( 6002 ) - 1800 = 42.43 m>s 100

Using these results, the v - s graph shown in Fig. a can be plotted.

Ans: v = e 69

1 s f m>s 10

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12–65.  The jet plane starts from rest at s = 0 and is subjected to the acceleration shown. Determine the speed of the plane when it has traveled 1000 ft. Also, how much time is required for it to travel 1000 ft?

a (ft/s2)

75

a  75  0.025s

50

Solution

1000

s (ft)

a - 75 50 - 75 = ; a = {75 - 0.025s} ft>s2. The function v(s) s - 0 1000 - 0 can be determined by integrating v dv = a ds. Using the initial ­condition v = 0 at s = 0, v9s Function. Here,

L0

v

v dv =

L0

s

(75 - 0.025 s) ds

v2 = 75 s - 0.0125 s2 2 v = At s = 1000 ft,

5 2150 s

- 0.025 s2 6 ft>s

v = 2150 (1000) - 0.025 ( 10002 )

Ans.

= 353.55 ft>s = 354 ft>s

ds . Using the initial Time. t as a function of s can be determined by integrating dt = v condition s = 0 at t = 0; L0

t

s

dt =

ds L0 1150 s - 0.025 s2

 

t = c-

 

t =

At s = 1000 ft,  

t =

s 1 150 - 0.05 s bd ` sin-1a 150 10.025 0

1 p 150 - 0.05 s c - sin-1a bd 150 10.025 2

150 - 0.05(1000) 1 p e - sin-1 c df 150 10.025 2

Ans.

   = 5.319 s = 5.32 s

Ans: v = 354 ft>s t = 5.32 s 70

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12–66. The boat travels along a straight line with the speed described by the graph. Construct the s–t and a - s graphs. Also, determine the time required for the boat to travel a distance s = 400 m if s = 0 when t = 0.

v (m/s)

80

SOLUTION s t Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s. + B A:

v

ds v

dt = t

v2 s

ds

dt = L0 2s1>2 L0 1>2 t = s

0.2s

4s

20 s (m) 100

s = A t2 B m

400

When s = 100 m, 100 = t2

t = 10 s

For 100 m 6 s … 400 m, the initial condition is s = 100 m when t = 10 s. + B A:

ds v

dt = t

s

ds 0.2s L10 s L100 m s t - 10 = 5ln 100 s t - 2 = ln 5 100 s et>5 - 2 = 100 s et>5 = 100 e2 dt =

s = A 13.53et>5 B m When s = 400 m, 400 = 13.53et>5 Ans.

t = 16.93 s = 16.9 s The s–t graph is shown in Fig. a. a s Graph: For 0 m … s 6 100 m, a = v

dv = A 2s1>2 B A s - 1>2 B = 2 m>s2 ds

For 100 m 6 s … 400 m, a = v

dv = (0.2s)(0.2) = 0.04s ds Ans: When s = 100 m, t = 10 s. When s = 400 m, t = 16.9 s. a s = 100 m = 4 m>s2 a s = 400 m = 16 m>s2

When s = 100 m and 400 m, a s = 100 m = 0.04(100) = 4 m>s2 a s = 400 m = 0.04(400) = 16 m>s2 The a–s graph is shown in Fig. b. 71

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12–67. The v - s graph of a cyclist traveling along a straight road is shown. Construct the a -s graph.

v (ft/s)

15 v

0.04 s

19

s (ft)

a–s Graph: For 0 … s 6 100 ft, a = v

v

5

5

SOLUTION

+ B A:

0.1s

100

350

dv = A 0.1s + 5 B A 0.1 B = A 0.01s + 0.5 B ft>s2 ds

Thus at s = 0 and 100 ft a ƒ s = 0 = 0.01 A 0 B + 0.5 = 0.5 ft>s2 a ƒ s = 100 ft = 0.01 A 100 B + 0.5 = 1.5 ft>s2 For 100 ft 6 s … 350 ft, + B A:

a = v

dv = ds

A - 0.04s + 19 B A - 0.04 B = A 0.0016s - 0.76 B ft>s2

Thus at s = 100 ft and 350 ft a ƒ s = 100 ft = 0.0016 A 100 B - 0.76 = - 0.6 ft>s2 a ƒ s = 350 ft = 0.0016 A 350 B - 0.76 = - 0.2 ft>s2 The a - s graph is shown in Fig. a.

Thus at s = 0 and 100 ft a ƒ s = 0 = 0.01 A 0 B + 0.5 = 0.5 ft>s2 a ƒ s = 100 ft = 0.01 A 100 B + 0.5 = 1.5 ft>s2 At s = 100 ft, a changes from a max = 1.5 ft>s2 to amin = - 0.6 ft>s2.

Ans: At s = 100 s, a changes from amax = 1.5 ft>s2 to amin = - 0.6 ft>s2. 72

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*12–68. The v–s graph for a test vehicle is shown. Determine its acceleration when s = 100 m and when s = 175 m.

v (m/s)

50

SOLUTION n =

0 … s … 150m:

dn =

1 s, 3

150

200

s (m)

1 ds 3

n dn = a ds 1 1 s a ds b = a ds 3 3 a =

At s = 100 m, 150 … s … 200 m;

a =

1 s 9

1 (100) = 11.1 m>s2 9

Ans.

n = 200 - s, dn = - ds n dn = a ds

(200 - s)( - ds) = a ds a = s - 200 At s = 175 m,

a = 175 - 200 = - 25 m>s2

Ans.

Ans: At s = 100 s,  a = 11.1 m>s2 At s = 175 m,  a = -25 m>s2 73

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12–69.  If the velocity of a particle is defined as v(t) = {0.8t2i + 12t1 > 2j + 5k} m >s, determine the magnitude and coordinate direction angles a, b, g of the particle’s acceleration when t = 2 s.

Solution v(t) = 0.8t 2 i + 12t 1>2 j + 5k a =

dv = 1.6i + 6t 1>2j dt

When t = 2 s,  a = 3.2i + 4.243j a = 2(3.2)2 + (4.243)2 = 5.31 m>s2 uo =

Ans.

a = 0.6022i + 0.7984j a

a = cos-1 (0.6022) = 53.0°

Ans.

b = cos-1 (0.7984) = 37.0°

Ans.

g = cos-1(0) = 90.0°

Ans.

Ans: a = 5.31 m>s2 a = 53.0° b = 37.0° g = 90.0° 74

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12–70. The velocity of a particle is v = 53i + (6 - 2t)j6 m>s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = 1 s to t = 3 s.

SOLUTION Position: The position r of the particle can be determined by integrating the kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the integration limit. Thus, dr = vdt r

L0

t

dr =

L0

C 3i + (6 - 2t)j D dt

r = c3ti + A 6t - t2 B j d m When t = 1 s and 3 s, r t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s r t = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s Thus, the displacement of the particle is ¢r = r t = 3 s - r t = 1 s = (9i + 9j) - (3i + 5j) Ans.

= {6i + 4j} m

75

Ans: ∆r = 5 6i + 4j 6 m

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12–71. A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = 56ti + 12t2k6 ft>s2. Determine the particle’s position (x, y, z) at t = 1 s.

SOLUTION Velocity: The velocity expressed in Cartesian vector form can be obtained by applying Eq. 12–9. dv = adt t

v

L0

dv =

16ti + 12t2k2 dt

L0

v = {3t2i + 4t3k} ft/s Position: The position expressed in Cartesian vector form can be obtained by applying Eq. 12–7. dr = vdt r

Lr1

dr =

t

L0

13t2i + 4t3k2 dt

r - (3i + 2j + 5k) = t3i + t4k r = {(t3 + 3) i + 2j + (t4 + 5)k} ft When t = 1 s, r = (13 + 3)i + 2j + (14 + 5)k = {4i + 2j + 6k} ft. The coordinates of the particle are Ans.

(4 ft, 2 ft, 6 ft)

Ans: (4 ft, 2 ft, 6 ft) 76

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*12–72. The velocity of a particle is given by v = 516t2i + 4t3j + (5t + 2)k6 m>s, where t is in seconds. If the particle is at the origin when t = 0, determine the magnitude of the particle’s acceleration when t = 2 s. Also, what is the x, y, z coordinate position of the particle at this instant?

SOLUTION Acceleration: The acceleration expressed in Cartesian vector form can be obtained by applying Eq. 12–9. a =

dv = {32ti + 12t2j + 5k} m>s2 dt

When t = 2 s, a = 32(2)i + 12 A 22 B j + 5k = {64i + 48j + 5k} m>s2. The magnitude of the acceleration is a = 2a2x + a2y + a2z = 2642 + 482 + 52 = 80.2 m>s2

Ans.

Position: The position expressed in Cartesian vector form can be obtained by applying Eq. 12–7. dr = v dt r

L0

dr = r = c

t

L0

2 3 A 16t i + 4t j + (5t + 2)k B dt

5 16 3 t i + t4j + a t2 + 2tb k d m 3 2

When t = 2 s, r =

5 16 3 A 2 B i + A 24 B j + c A 22 B + 2(2) d k = {42.7i + 16.0j + 14.0k} m. 3 2

Thus, the coordinate of the particle is Ans.

(42.7, 16.0, 14.0) m

Ans: (42.7, 16.0, 14.0) m 77

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12–73. y

The water sprinkler, positioned at the base of a hill, releases a stream of water with a velocity of 15 ft>s as shown. Determine the point B(x, y) where the water strikes the ground on the hill. Assume that the hill is defined by the equation y = (0.05x2) ft and neglect the size of the sprinkler.

y  (0.05x2) ft 15 ft/s B 60 x

Solution vx = 15 cos 60° = 7.5 ft>s  vy = 15 sin 60° = 12.99 ft>s +2 1S

s = v0 t

1 + c2

s = so + vot +





x = 7.5t 1 a t2 2 c 1 y = 0 + 12.99t + ( - 32.2)t 2 2 y = 1.732x - 0.286x2

Since y = 0.05x2, 0.05x2 = 1.732x - 0.286x2 x(0.336x - 1.732) = 0 Ans.

x = 5.15 ft 2

Ans.

y = 0.05(5.15) = 1.33 ft Also, +2 s = vt 1S 0



x = 15 cos 60°t

1 + c2

s = s0 + v0t +



1 a t2 2 c

y = 0 + 15 sin 60°t +

1 ( - 32.2)t 2 2

Since y = 0.05x2 12.99t - 16.1t 2 = 2.8125t 2  t = 0.6869 s So that, x = 15 cos 60° (0.6868) = 5.15 ft

Ans.

y = 0.05(5.15)2 = 1.33 ft

Ans.

Ans: (5.15 ft, 1.33 ft) 78

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12–74.  A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6t i + 12t2 k} ft>s2. Determine the particle’s position (x, y, z) when t = 2 s.

Solution a = 6ti + 12t 2 k L0

v

dv =

L0

t

(6ti + 12t2 k) dt

v = 3t 2i + 4t 3 k r

Lr0

dr =

L0

t

(3t 2 i + 4t 3 k) dt

r - (3i + 2j + 5k) = t 3 i + t 4 k When t = 2 s r = {11i + 2j + 21k} ft

Ans.

Ans: r = {11i + 2j + 21k} ft 79

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12–75. A particle travels along the curve from A to B in 2 s. It takes 4 s for it to go from B to C and then 3 s to go from C to D. Determine its average speed when it goes from A to D.

SOLUTION sT =

vsp =

1 1 (2p)(10)) + 15 + (2p(5)) = 38.56 4 4 sT 38.56 = 4.28 m>s = tt 2 + 4 + 3

Ans.

Ans: (vsp)avg = 4.28 m>s 80

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*12–76.  y

A particle travels along the curve from A to B in 5 s. It takes 8 s for it to go from B to C and then 10 s to go from C to A. Determine its average speed when it goes around the closed path.

B

20 m

Solution

A

30 m

C

x

The total distance traveled is STot = SAB + SBC + SCA p = 20 a b + 2202 + 302 + (30 + 20) 2 = 117.47 m

The total time taken is t Tot = t AB + t BC + t CA = 5 + 8 + 10 = 23 s Thus, the average speed is (vsp)avg =

STot 117.47 m = = 5.107 m>s = 5.11 m>s t Tot 23 s

Ans.

Ans: (vsp)avg = 5.11 m>s 81

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12–77. The position of a crate sliding down a ramp is given by x = (0.25t3) m, y = (1.5t2) m, z = (6 - 0.75t5>2) m, where t is in seconds. Determine the magnitude of the crate’s velocity and acceleration when t = 2 s.

SOLUTION Velocity: By taking the time derivative of x, y, and z, we obtain the x, y, and z components of the crate’s velocity. d # vx = x = A 0.25t3 B = A 0.75t2 B m>s dt d # vy = y = A 1.5t2 B = A 3t B m>s dt d # vz = z = A 6 - 0.75t5>2 B = dt

A - 1.875t3>2 B m>s

When t = 2 s, vx = 0.75 A 22 B = 3 m>s

vz = - 1.875 A 2 B 3>2 = -5.303 m/s

vy = 3(2) = 6 m>s

Thus, the magnitude of the crate’s velocity is v = 2vx 2 + vy 2 + vz 2 = 232 + 62 + ( - 5.303)2 = 8.551 ft>s = 8.55 ft

Ans.

Acceleration: The x, y, and z components of the crate’s acceleration can be obtained by taking the time derivative of the results of vx, vy, and vz, respectively. d # ax = vx = A 0.75t2 B = (1.5t) m>s2 dt d # ay = vy = (3t) = 3 m>s2 dt d # az = vz = A - 1.875t3>2 B = dt

A - 2.815t1>2 B m>s2

When t = 2 s, ax = 1.5(2) = 3 m>s2

ay = 3 m>s2

az = - 2.8125 A 21>2 B = -3.977 m>s2

Thus, the magnitude of the crate’s acceleration is a = 2ax 2 + ay 2 + az 2 = 232 + 32 + ( - 3.977)2 = 5.815 m>s2 = 5.82 m>s

Ans.

Ans: v = 8.55 ft>s a = 5.82 m>s2 82

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12–78. A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y2 = [120(103)x] m. If the 1 x component of acceleration is ax = a t2 b m>s2, where t is 4 in seconds, determine the magnitude of the rocket’s velocity and acceleration when t = 10 s.

SOLUTION Position: The parameter equation of x can be determined by integrating ax twice with respect to t. L

dvx =

L

vx

t

dvx =

L0

vx = a

L

1 2 t dt L0 4

1 3 t b m>s 12

dx = x

L0

axdt

L

vxdt t

dx =

x = a

1 3 t dt 12 L0

1 4 t bm 48

Substituting the result of x into the equation of the path, y2 = 120 A 103 B a

1 4 t b 48

y = A 50t2 B m Velocity: d # vy = y = A 50t2 B = A 100t B m>s dt When t = 10 s, vx =

1 A 103 B = 83.33 m>s 12

vy = 100(10) = 1000 m>s

Thus, the magnitude of the rocket’s velocity is v = 2vx 2 + vy2 = 283.332 + 10002 = 1003 m>s

Ans.

Acceleration: # d ay = vy = (100t) = 100 m>s2 dt When t = 10 s, ax =

1 A 102 B = 25 m>s2 4

Thus, the magnitude of the rocket’s acceleration is a = 2ax 2 + ay2 = 2252 + 1002 = 103 m>s2

Ans. 83

Ans: v = 1003 m>s a = 103 m>s2

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12–79. y

The particle travels along the path defined by the parabola y = 0.5x2. If the component of velocity along the x axis is vx = 15t2 ft>s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.

y

0.5x2

SOLUTION Position: The x position of the particle can be obtained by applying the vx =

dx . dt

x

O

dx = vx dt x

L0

t

dx =

L0

5tdt

x = A 2.50t2 B ft Thus, y = 0.5 A 2.50t2 B 2 = A 3.125t4 B ft. At y = 3.125 A 1

4

t = 1 s, x = 2.5 A 12 B = 2.50 ft

and

B = 3.125 ft. The particle’s distance from the origin at this moment is d = 2(2.50 - 0)2 + (3.125 - 0)2 = 4.00 ft

Ans.

# # Acceleration: Taking the first derivative of the path y = 0.5x2, we have y = xx. The second derivative of the path gives $ # $ y = x2 + xx

(1)

# $ $ However, x = vx, x = ax and y = ay. Thus, Eq. (1) becomes ay = v2x + xax

(2)

dvx = 5 ft>s 2, and x = 2.50 ft . Then, from When t = 1 s, vx = 5(1) = 5 ft>s ax = dt Eq. (2) ay = 52 + 2.50(5) = 37.5 ft>s2 Also, a =

a2x + a2y =

52 + 37.52 = 37.8 ft s2

Ans.

Ans: d = 4.00 ft a = 37.8 ft>s2 84

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*12–80. The motorcycle travels with constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve.

y v0 y

π x) c sin ( –– L x

c L

SOLUTION y = c sin a

c L

p xb L

p p # # ca cos xb x y = L L vy =

p p c vx a cos x b L L

v20 = v2y + v2x v20 = v2x B 1 + a

p 2 p cb cos2 a xb R L L

vx = v0 B 1 + a

-2 p 2 p cb cos2 a xb R L L

1

Ans. 1

vy =

-2 v0 pc p p 2 p a cos xb B 1 + a c b cos2 a x b R L L L L

Ans.

Ans:

1

-2 p 2 p vx = v0 c 1 + a cb cos2 a xb d L L -12 v0 pc p 2 p p vy = acos xb c 1 + a cb cos2 a xb d L L L L

85

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12–81. y

A particle travels along the circular path from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C.

30

C

45 30 m

SOLUTION

B

Position: The coordinates for points B and C are [30 sin 45°, 30 - 30 cos 45°] and [30 sin 75°, 30 - 30 cos 75°]. Thus,

A

rB = (30 sin 45° - 0)i + [(30 - 30 cos 45°) - 30]j = {21.21i - 21.21j} m rC = (30 sin 75° - 0)i + [(30 - 30 cos 75°) - 30]j = {28.98i - 7.765j} m Average Velocity: The displacement from point B to C is ¢rBC = rC - rB = (28.98i - 7.765j) - (21.21i - 21.21j) = {7.765i + 13.45j} m. (vBC)avg =

7.765i + 13.45j ¢rBC = = {3.88i + 6.72j} m>s ¢t 3 - 1

Ans.

Ans: (vBC)avg = {3.88i + 6.72j} m>s 86

x

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12–82.

z

The roller coaster car travels down the helical path at constant speed such that the parametric equations that define its position are x = c sin kt, y = c cos kt, z = h − bt, where c, h, and b are constants. Determine the magnitudes of its velocity and acceleration.

SOLUTION x = c sin kt

# x = ck cos kt

$ x = - ck2 sin kt

y = c cos kt

# y = - ck sin kt

$ y = - ck2 cos kt

z = h - bt

# z = -b

$ z = 0

v =

(ck cos kt)2 + (- ck sin kt)2 + ( - b)2 =

a =

(- ck2 sin kt)2 + (- ck2 cos kt)2 + 0 = ck2

c2k2 + b2

y

Ans.

x

Ans.

Ans: v = 2c 2 k 2 + b2 a = ck 2 87

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12–83. y

Pegs A and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10 m/s, determine the magnitude of the velocity and acceleration of peg A when x = 1 m.

A C

SOLUTION Velocity: The x and y components of the peg’s velocity can be related by taking the first time derivative of the path’s equation. x2 + y2 = 1 4 1 # # (2xx) + 2yy = 0 4 1 # # xx + 2yy = 0 2

B

D

x

v  10 m/s

x2  v2  1 4

or 1 xvx + 2yvy = 0 2

(1)

At x = 1 m, (1)2 + y2 = 1 4

y =

23 m 2

Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1), 1 23 (1)(10) + 2 ¢ ≤ vy = 0 2 2

vy = - 2.887 m>s = 2.887 m>s T

Thus, the magnitude of the peg’s velocity is v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s

Ans.

Acceleration: The x and y components of the peg’s acceleration can be related by taking the second time derivative of the path’s equation. 1 # # ## ## # # (xx + xx) + 2(yy + yy) = 0 2 1 #2 ## ## # A x + xx B + 2 A y 2 + yy B = 0 2 or 1 A vx 2 + xax B + 2 A vy 2 + yay B = 0 2 Since vx is constant, ax = 0. When x = 1 m, y =

(2) 23 m, vx = 10 m>s, and 2

vy = -2.887 m>s. Substituting these values into Eq. (2), 1 23 a d = 0 A 102 + 0 B + 2 c (- 2.887)2 + 2 2 y ay = - 38.49 m>s2 = 38.49 m>s2 T Thus, the magnitude of the peg’s acceleration is a = 2ax 2 + ay 2 = 202 + ( -38.49)2 = 38.5 m>s2

Ans. 88

Ans: v = 10.4 m>s a = 38.5 m>s2

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*12–84. The van travels over the hill described by y = ( -1.5(10–3) x2 + 15) ft. If it has a constant speed of 75 ft>s, determine the x and y components of the van’s velocity and acceleration when x = 50 ft.

y 15 ft

y

( 1.5 (10 3) x2

15) ft

x

SOLUTION

100 ft

Velocity: The x and y components of the van’s velocity can be related by taking the first time derivative of the path’s equation using the chain rule. y = -1.5 A 10 - 3 B x2 + 15 # # y = -3 A 10 - 3 B xx or vy = -3 A 10 - 3 B xvx When x = 50 ft, vy = - 3 A 10 - 3 B (50)vx = - 0.15vx

(1)

The magnitude of the van’s velocity is v = 2vx 2 + vy 2

(2)

Substituting v = 75 ft>s and Eq. (1) into Eq. (2), 75 = 2vx 2 + ( - 0.15vx)2

vx = 74.2 ft>s ;

Ans.

Substituting the result of nx into Eq. (1), we obtain vy = -0.15( - 74.17) = 11.12 ft>s = 11.1 ft>s c

Ans.

Acceleration: The x and y components of the van’s acceleration can be related by taking the second time derivative of the path’s equation using the chain rule. $ # # ## y = - 3 A 10 - 3 B (xx + xx) or ay = - 3 A 10 - 3 B A vx 2 + xax B When x = 50 ft, vx = - 74.17 ft>s. Thus, ay = -3 A 10 - 3 B c ( -74.17)2 + 50ax d (3)

ay = -(16.504 + 0.15ax)

Since the van travels with a constant speed along the path,its acceleration along the tangent of the path is equal to zero. Here, the angle that the tangent makes with the horizontal at dy x = 50 ft is u = tan - 1 ¢ ≤ 2 = tan - 1 c -3 A 10 - 3 B x d 2 = tan - 1( -0.15) = - 8.531°. dx x = 50 ft x = 50 ft Thus, from the diagram shown in Fig. a, (4)

ax cos 8.531° - ay sin 8.531° = 0 Solving Eqs. (3) and (4) yields

ax = -2.42 ft>s = 2.42 ft>s ;

Ans.

ay = -16.1 ft>s = 16.1 ft>s2 T

Ans.

2

89

Ans: vx = vy = ax = ay =

74.2 ft>s d 11.1 ft>s c 2.42 ft>s2 d 16.1 ft>s2 T

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12–85. y

The flight path of the helicopter as it takes off from A is defined by the parametric equations x = 12t22 m and y = 10.04t32 m, where t is the time in seconds. Determine the distance the helicopter is from point A and the magnitudes of its velocity and acceleration when t = 10 s.

SOLUTION 2

A 3

At t = 10 s,

x = 200 m

y = 40 m

d = 3120022 + 14022 = 204 m nx =

dx = 4t dt

ax =

dnx = 4 dt

vy =

dy = 0.12t2 dt

ay =

dny dt

x

y = 0.04t

x = 2t

Ans.

= 0.24t

At t = 10 s, n = 314022 + 11222 = 41.8 m>s

Ans.

a = 31422 + 12.422 = 4.66 m>s2

Ans.

Ans: d = 204 m v = 41.8 m>s a = 4.66 m>s2 90

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12–86. Determine the minimum initial velocity v0 and the corresponding angle u0 at which the ball must be kicked in order for it to just cross over the 3-m high fence.

v0

3m

u0 6m

SOLUTION Coordinate System: The x - y coordinate system will be set so that its origin coincides with the ball’s initial position. x-Motion: Here, (v0)x = v0 cos u, x0 = 0, and x = 6 m. Thus, + B A:

x = x0 + (v0)xt 6 = 0 + (v0 cos u)t t =

6 v0 cos u

(1)

y-Motion: Here, (v0)x = v0 sin u, ay = - g = - 9.81 m > s2, and y0 = 0. Thus,

A+cB

y = y0 + (v0)y t +

1 a t2 2 y

3 = 0 + v0 (sin u) t +

1 ( -9.81)t2 2

3 = v0 (sin u) t - 4.905t2

(2)

Substituting Eq. (1) into Eq. (2) yields v0 =

58.86 B sin 2u - cos2 u

(3)

From Eq. (3), we notice that v0 is minimum when f(u) = sin 2u - cos2 u is df(u) = 0 maximum. This requires du df(u) = 2 cos 2u + sin 2u = 0 du tan 2u = - 2 2u = 116.57° Ans.

u = 58.28° = 58.3° Substituting the result of u into Eq. (2), we have (v0)min =

58.86 = 9.76 m>s B sin 116.57° - cos2 58.28°

Ans.

Ans: u = 58.3° (v0) min = 9.76 m>s 91

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12–87.  The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity vA at which it was launched, the angle of release u, and the height h.

B

h

vA A 3.5 ft

Solution +2 1S



18 ft

s = v0t

v =



0 = vA sin u - 32.2(1.5)

1 + c2

(1)

18 = vA cos u(1.5) 

1 + c2



u

2

v02

+ 2ac (s - s0)

0 = (vA sin u)2 + 2( - 32.2)(h - 3.5) v = v0 + act (2)

To solve, first divide Eq. (2) by Eq. (1) to get u. Then

Ans.

u = 76.0°



vA = 49.8 ft>s

Ans.



h = 39.7 ft

Ans.

Ans: u = 76.0° vA = 49.8 ft>s h = 39.7 ft 92

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*12–88.  Neglecting the size of the ball, determine the magnitude vA of the basketball’s initial velocity and its velocity when it passes through the basket.

B 30 A

vA 3m

2m 10 m

Solution Coordinate System. The origin of the x@y coordinate system will be set to coinside with point A as shown in Fig. a Horizontal Motion. Here (vA)x = vA cos 30° S, (sA)x = 0 and (sB)x = 10 m S .



+2 1S



(sB)x = (sA)x + (vA)x t 10 = 0 + vA cos 30° t 10  vA cos 30°

(1)

(vB)x = (vA)x = vA cos 30°

(2)

t =

Also, +2 1S

Vertical Motion. Here, (vA)y = vA sin 30° c , (sA)y = 0, (sB)y = 3 - 2 = 1 m c and ay = 9.81 m>s2 T

1 + c2



(sB)y = (sA)y + (vA)y t +

1 a t2 2 y

1 ( - 9.81)t 2 2 4.905t 2 - 0.5 vA t + 1 = 0

1 = 0 + vA sin 30° t +

(3)

Also

1 + c2

(vB)y = (vA)y + ay t



(vB)y = vA sin 30° + ( - 9.81)t



(vB)y = 0.5 vA - 9.81t

(4)

Solving Eq. (1) and (3)

vA = 11.705 m>s = 11.7 m>s



t = 0.9865 s

Ans.

Substitute these results into Eq. (2) and (4)

(vB)x = 11.705 cos 30° = 10.14 m>s S



(vB)y = 0.5(11.705) - 9.81(0.9865) = -3.825 m>s = 3.825 m>s T

Thus, the magnitude of vB is

vB = 2(vB)2x + (vB)2y = 210.142 + 3.8252 = 10.83 m>s = 10.8 m>s Ans.

And its direction is defined by

uB = tan-1 c

(vB)y (vB)x

d = tan-1 a

3.825 b = 20.67° = 20.7° 10.14



93

Ans.

Ans: vA = 11.7 m>s vB = 10.8 m>s u = 20.7° c

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12–89.  The girl at A can throw a ball at vA = 10 m>s. Calculate the maximum possible range R = Rmax and the associated angle u at which it should be thrown. Assume the ball is caught at B at the same elevation from which it is thrown.

vA  10 m/s u A

B

R

Solution

+2 1S

s = s0 + v0 t R = 0 + (10 cos u)t

1 + c2

v = v0 + ac t

-10 sin u = 10 sin u - 9.81t t =

20 sin u 9.81

Thus,  R =

R =

200 sin u cos u 9.81

100 sin 2u 9.81

(1)

Require, dR = 0 du 100 cos 2u(2) = 0 9.81 cos 2u = 0 Ans.

u = 45° R =

100 (sin 90°) = 10.2 m 9.81

Ans.

Ans: Rmax = 10.2 m u = 45° 94

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12–90.  Show that the girl at A can throw the ball to the boy at B by launching it at equal angles measured up or down from a 45° inclination. If vA = 10 m>s, determine the range R if this value is 15°, i.e., u1 = 45° − 15° = 30° and u2 = 45° + 15° = 60°. Assume the ball is caught at the same elevation from which it is thrown.

vA  10 m/s u A

B

R

Solution

+2 1S

s = s0 + v0t R = 0 + (10 cos u)t

1 + c2

v = v0 + act

- 10 sin u = 10 sin u - 9.81t t =

20 sin u 9.81

Thus,   R = R =

200 sin u cos u 9.81

100 sin 2u 9.81

(1)

Since the function y = sin 2u is symmetric with respect to u = 45° as indicated, Eq. (1) will be satisfied if | f 1 | = | f 2 | Choosing f = 15° or u1 = 45° - 15° = 30° substituting into Eq. (1) yields

and

u2 = 45° + 15° = 60°,

and Ans.

R = 8.83 m

Ans: R = 8.83 m 95

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12–91.  y

The ball at A is kicked with a speed vA = 80 ft>s and at an angle uA = 30°. Determine the point (x, –y) where it strikes the ground. Assume the ground has the shape of a parabola as shown.

vA A

uA

x y B

Solution

x

y  0.04x2

(vA)x = 80 cos 30° = 69.28 ft>s (vA)y = 80 sin 30° = 40 ft>s



+ 2s 1S



(1)

x = 0 + 69.28t

1 2 at 2 c 1 - y = 0 + 40t + ( - 32.2)t 2 2 y = - 0.04x2

1 + c2



= s0 + v0t

s = s0 + v0t +

(2)

From Eqs. (1) and (2): - y = 0.5774x - 0.003354x2 0.04x2 = 0.5774x - 0.003354x2 0.04335x2 = 0.5774x Ans.

x = 13.3 ft Thus y = - 0.04 (13.3)2 = - 7.09 ft

Ans.

Ans: (13.3 ft, - 7.09 ft) 96

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*12–92.  The ball at A is kicked such that uA = 30°. If it strikes the ground at B having coordinates x = 15 ft, y = -9 ft, determine the speed at which it is kicked and the speed at which it strikes the ground.

y vA A

uA

x y B

Solution +2 1S



y  0.04x2

s = s0 + v0t

15 = 0 + vA cos 30° t 1 1 + c 2 s = s0 + v0t + act 2 2



x

-9 = 0 + vA sin 30° t +

1 ( - 32.2)t 2 2 Ans.

vA = 16.5 ft>s t = 1.047 s + 2 (v ) 1S B x

1 + c2

= 16.54 cos 30° = 14.32 ft>s

v = v0 + act

(vB)y = 16.54 sin 30° + ( -32.2)(1.047)

= -25.45 ft>s

vB = 2(14.32)2 + ( - 25.45)2 = 29.2 ft>s

Ans.

Ans: vA = 16.5 ft>s t = 1.047 s vB = 29.2 ft>s 97

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12–93. A golf ball is struck with a velocity of 80 ft>s as shown. Determine the distance d to where it will land. vA  80 ft/s B A

45

10 d

Solution +2s 1S

= s0 + v0t

d cos 10° = 0 + 80 cos 55° t 1 1 + c 2 s = s0 + v0 t + act 2 2

d sin 10° = 0 + 80 sin 55° t -

1 (32.2) ( t 2 ) 2

Solving t = 3.568 s Ans.

d = 166 ft

Ans: d = 166 ft 98

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12–94.  A golf ball is struck with a velocity of 80 ft>s as shown. Determine the speed at which it strikes the ground at B and the time of flight from A to B.

vA  80 ft/s B A

45

10 d

Solution (vA)x = 80 cos 55° = 44.886 (vA)y = 80 sin 55° = 65.532 +2 1S

s = s0 + v0t

d cos 10° = 0 + 45.886t 1 1+ c 2 s = s0 + v0t + act 2 2

d sin 10° = 0 + 65.532 (t) +

1 ( - 32.2) ( t 2 ) 2

d = 166 ft Ans.

t = 3.568 = 3.57 s (vB)x = (vA)x = 45.886

1 + c2

v = v0 + act

(vB)y = 65.532 - 32.2(3.568) (vB)y = -49.357 vB = 2(45.886)2 + ( - 49.357)2

Ans.

vB = 67.4 ft>s

Ans: t = 3.57 s vB = 67.4 ft>s 99

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12–95. The basketball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Neglecting the size of the ball, determine the magnitude vA of its initial velocity and the height h of the ball when it passes over player B.

C 30 A

vA

B h

7 ft 25 ft

10 f

5 ft

SOLUTION + ) (:

s = s0 + v0t 30 = 0 + vA cos 30° tAC

(+ c )

s = s0 + v0t +

1 2 at 2 c

10 = 7 + vA sin 30° tAC -

1 (32.2)(t2AC) 2

Solving Ans.

vA = 36.73 = 36.7 ft>s tAC = 0.943 s + ) (:

s = s0 + v0t 25 = 0 + 36.73 cos 30° tAB

(+ c )

s = s0 + v0t +

1 2 act 2

h = 7 + 36.73 sin 30° tAB -

1 (32.2)(t2AB) 2

Solving tAB = 0.786 s Ans.

h = 11.5 ft

Ans: vA = 36.7 ft>s h = 11.5 ft 100

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*12–96. It is observed that the skier leaves the ramp A at an angle uA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the time of flight tAB.

vA

uA A 4m

3

SOLUTION + B A:

100 m

s = v0 t 4 100 a b = vA cos 25°tAB 5

A+cB

5 4

s = s0 + v0 t +

B

1 ac t2 2

1 3 -4 - 100a b = 0 + vA sin 25°tAB + ( -9.81)t2AB 5 2 Solving, vA = 19.4 m>s

Ans.

tAB = 4.54 s

Ans.

Ans: vA = 19.4 m>s t AB = 4.54 s 101

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12–97. It is observed that the skier leaves the ramp A at an angle uA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the speed at which he strikes the ground.

vA

uA A 4m

3

5 4

100 m

SOLUTION Coordinate System: x - y coordinate system will be set with its origin to coincide with point A as shown in Fig. a. 4 x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°. 5 + B A:

xB = xA + (vA)xt 80 = 0 + (vA cos 25°)t 80 t = vA cos 25°

(1)

3 y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA sin 25° 5 and ay = -g = -9.81 m>s2.

A+ c B

yB = yA + (vA)y t +

1 a t2 2 y

- 64 = 0 + vA sin 25° t +

1 ( - 9.81)t2 2

4.905t2 - vA sin 25° t = 64

(2)

Substitute Eq. (1) into (2) yieldS 4.905 ¢

¢

2 80 80 ≤ = vA sin 25° ¢ ≤ = 64 vA cos 25° yA cos 25°

2 80 ≤ = 20.65 vA cos 25°

80 = 4.545 vA cos 25° Ans.

vA = 19.42 m>s = 19.4 m>s Substitute this result into Eq. (1), t =

80 = 4.54465 19.42 cos 25°

102

B

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12–97. Continued

Using this result,

A+ c B

(vB)y = (vA)y + ay t = 19.42 sin 25° + ( - 9.81)(4.5446) = - 36.37 m>s = 36.37 m>s T

And + B A:

(vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s :

Thus, vB = 2(vB)2x + (vB)2y = 236.372 + 17.602 Ans.

= 40.4 m>s

Ans: vA = 19.4 m>s vB = 40.4 m>s 103

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12–98. Determine the horizontal velocity vA of a tennis ball at A so that it just clears the net at B. Also, find the distance s where the ball strikes the ground.

vA B

A 7.5 ft

3 ft

C s

21 ft

SOLUTION Vertical Motion: The vertical component of initial velocity is (v0)y = 0. For the ball to travel from A to B, the initial and final vertical positions are (s0)y = 7.5 ft and sy = 3 ft, respectively.

A+cB

sy = (s0)y + (v0)y t + 3 = 7.5 + 0 +

1 (a ) t2 2 cy

1 ( -32.2)t 21 2

t1 = 0.5287 s For the ball to travel from A to C, the initial and final vertical positions are (s0)y = 7.5 ft and sy = 0, respectively.

A+cB

sy = (s0)y + (v0)y t + 0 = 7.5 + 0 +

1 (a ) t2 2 cy

1 ( -32.2)t22 2

t2 = 0.6825 s Horizontal Motion: The horizontal component of velocity is (v0)x = vA. For the ball to travel from A to B, the initial and final horizontal positions are (s0)x = 0 and sx = 21 ft, respectively. The time is t = t1 = 0.5287 s. + B A;

sx = (s0)x + (v0)x t 21 = 0 + vA (0.5287) Ans.

vA = 39.72 ft>s = 39.7 ft>s

For the ball to travel from A to C, the initial and final horizontal positions are (s0)x = 0 and sx = (21 + s) ft, respectively. The time is t = t2 = 0.6825 s. + B A;

sx = (s0)x + (v0)x t 21 + s = 0 + 39.72(0.6825) Ans.

s = 6.11 ft

Ans: vA = 39.7 ft>s s = 6.11 ft 104

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12–99.  The missile at A takes off from rest and rises vertically to B, where its fuel runs out in 8 s. If the acceleration varies with time as shown, determine the missile’s height hB and speed vB. If by internal controls the missile is then suddenly pointed 45° as shown, and allowed to travel in free flight, determine the maximum height attained, hC, and the range R to where it crashes at D.

45

vB

C

B

hC hB

D

A

Solution

R a (m/s2)

40 t = 5t 8 dv = a dt a = L0

v

dv =

L0

40

t

8

5t dt

t (s)

v = 2.5t 2 When t = 8 s,  vB = 2.5(8)2 = 160 m>s

Ans.

ds = v dt L0

s

x =

ds =

L0

t

2.5t 2 dt

2.5 3 t 3

hB =

2.5 (8)3 = 426.67 = 427 m 3

Ans.

(vB)x = 160 sin 45° = 113.14 m>s (vB)y = 160 cos 45° = 113.14 m>s

1 + c2

v2 = v20 + 2ac (s - s0)



02 = (113.14)2 + 2( - 9.81) (sc - 426.67)



hc = 1079.1 m = 1.08 km +2 1S

s = s0 + v0 t

1 + c2

s = s0 + v0t +

Ans.

R = 0 + 113.14t 1 2 at 2 c

1 ( - 9.81)t 2 2 Solving for the positive root, t = 26.36 s

0 = 426.67 + 113.14t +

Then, Ans.

R = 113.14 (26.36) = 2983.0 = 2.98 km 

Ans: vB = 160 m>s hB = 427 m hC = 1.08 km R = 2.98 km 105

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*12–100. The projectile is launched with a velocity v0. Determine the range R, the maximum height h attained, and the time of flight. Express the results in terms of the angle u and v0. The acceleration due to gravity is g.

y

v0 u

h x R

SOLUTION + B A:

s = s0 + v0 t R = 0 + (v0 cos u)t

A+cB

s = s0 + v0 t +

1 a t2 2 c

0 = 0 + (v0 sin u) t + 0 = v0 sin u -

R = t = =

A+cB

1 ( - g)t2 2

1 R (g) ¢ ≤ 2 v0 cos u

v20 sin 2u g

Ans.

v20 (2 sin u cos u) R = v0 cos u v0 g cos u 2v0 sin u g

Ans.

v2 = v20 + 2ac(s - s0) 0 = (v0 sin u)2 + 2(- g)(h - 0) h =

v20 sin2 u 2g

Ans.

Ans: v0 sin 2u g 2v0 t = sin u g v20 sin2 u h = 2g R =

106

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12–101. vA

The drinking fountain is designed such that the nozzle is located from the edge of the basin as shown. Determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C.

40 A 50 mm

B

C

250 mm 100 mm

Solution Horizontal Motion: +2   s = vt 1S 0

R = vA sin 40° t t =

R  vA sin 40°

(1)

Vertical Motion:

1 + c2

  s = s0 + v0 t +

1 2 at 2 c

- 0.05 = 0 + vA cos 40°t +

1 (- 9.81)t2 2

(2)

Substituting Eq.(1) into (2) yields: - 0.05 = vA cos 40° a vA =

2 R 1 R b + ( -9.81) a b vA sin 40° 2 vA sin 40°

4.905R2 A sin 40° (R cos 40° + 0.05 sin 40°)

At point B,  R = 0.1 m. vmin = vA =

4.905 (0.1)2 A sin 40° (0.1 cos 40° + 0.05 sin 40°)

= 0.838 m>s

Ans.

= 1.76 m>s

Ans.

At point C,  R = 0.35 m. vmax = vA =

4.905 (0.35)2 A sin 40° (0.35 cos 40° + 0.05 sin 40°)

Ans: vmin = 0.838 m>s vmax = 1.76 m>s 107

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12–102. If the dart is thrown with a speed of 10 m>s, determine the shortest possible time before it strikes the target. Also, what is the corresponding angle uA at which it should be thrown, and what is the velocity of the dart when it strikes the target?

4m

A vA

uA B

Solution Coordinate System. The origin of the x@y coordinate system will be set to coincide with point A as shown in Fig. a. Horizontal Motion. Here, (vA)x = 10 cos uA S, (sA)x = 0 and (sB)x = 4 m S. + 2 (s ) = (s ) + (v ) t 1S B x

A x

A x



4 = 0 + 10 cos uA t



t =

4  10 cos uA

(1)

Also, +2 1S

(vB)x = (vA)x = 10 cos uA S 

1 + c2



1 ay t 2 2 1 0 = 0 + (10 sin uA) t + ( - 9.81)t 2 2 4.905t 2 - (10 sin uA) t = 0



t (4.905t - 10 sin uA) = 0

(2)

Vertical Motion. Here, (vA)y = 10 sin uA c, (sA)y = (sB)y = 0 and ay = 9.81 m>s2 T



(sB)y = (sA)y + (vA)y t +

Since t ≠ 0, then (3)

4.905t - 10 sin uA = 0 Also

1 + c2

(vB)2y = (vA)2y + 2 ay [(sB)y - (sA)y]



(vB)2y = (10 sin uA)2 + 2 ( -9.81) (0 - 0)



(vB)y = - 10 sin uA = 10 sin uA T 

(4)

Substitute Eq. (1) into (3) 4.905 a

4 b - 10 sin uA = 0 10 cos uA

1.962 - 10 sin uA cos uA = 0

Using the trigonometry identity sin 2uA = 2 sin uA cos uA, this equation becomes 1.962 - 5 sin 2uA = 0 sin 2uA = 0.3924 2uA = 23.10° and 2uA = 156.90° uA = 11.55° and uA = 78.45° 108

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12–102. Continued

Since the shorter time is required, Eq. (1) indicates that smaller uA must be choosen. Thus Ans.

uA = 11.55° = 11.6° and t =

4 = 0.4083 s = 0.408 s 10 cos 11.55°

Ans.

Substitute the result of uA into Eq. (2) and (4) (vB)x = 10 cos 11.55° = 9.7974 m>s S (vB)y = 10 sin 11.55° = 2.0026 m>s T Thus, the magnitude of vB is vB = 2(vB)2x + (vB)2y = 29.79742 + 2.00262 = 10 m>s

Ans.

And its direction is defined by uB = tan-1 c

(vB)y (vB)x

d = tan-1 a

2.0026 b = 11.55° = 11.6° 9.7974

c

Ans.

Ans: uA = 11.6° t = 0.408 s uB = 11.6° c 109

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12–103. If the dart is thrown with a speed of 10 m>s, determine the longest possible time when it strikes the target. Also, what is the corresponding angle uA at which it should be thrown, and what is the velocity of the dart when it strikes the target?

4m

A vA

uA B

Solution Coordinate System. The origin of the x@y coordinate system will be set to coincide with point A as shown in Fig. a. Horizontal Motion. Here, (vA)x = 10 cos uA S , (sA)x = 0 and (sB)x = 4 m S . + 2 (s ) = (s ) + (v ) t 1S B x

A x

A x



4 = 0 + 10 cos uA t



t =

4  10 cos uA

(1)

Also, +2 1S

(vB)x = (vA)x = 10 cos uA S 

1 + c2

(sB)y = (sA)y + (vA)y t +

(2)

Vertical Motion. Here, (vA)y = 10 sin uA c , (sA)y = (sB)y = 0 and ay = - 9.81 m>s2 T. 1 ay t 2 2



1 ( - 9.81) t 2 2 4.905t 2 - (10 sin uA) t = 0



t (4.905t - 10 sin uA) = 0



0 = 0 + (10 sin uA) t +

Since t = 0, then (3)

     4.905 t - 10 sin uA = 0 Also,

(vB)2y = (vA)2y + 2 ay [(sB)y - (sA)y]



(vB)2y = (10 sin uA)2 + 2 ( - 9.81) (0 - 0)



(vB)y = - 10 sin uA = 10 sin uA T 

(4)

Substitute Eq. (1) into (3) 4.905 a

4 b - 10 sin uA = 0 10 cos uA

1.962 - 10 sin uA cos uA = 0

Using the trigonometry identity sin 2uA = 2 sin uA cos uA, this equation becomes 1.962 - 5 sin 2uA = 0 sin 2uA = 0.3924 2uA = 23.10° and 2uA = 156.90° uA = 11.55° and uA = 78.44° 110

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12–103. Continued

Since the longer time is required, Eq. (1) indicates that larger uA must be choosen. Thus, Ans.

uA = 78.44° = 78.4° and t =

4 = 1.9974 s = 2.00 s 10 cos 78.44°

Ans.

Substitute the result of uA into Eq. (2) and (4) (vB)x = 10 cos 78.44° = 2.0026 m>s S (vB)y = 10 sin 78.44° = 9.7974 m>s T Thus, the magnitude of vB is vB = 2(vB)2x + (vB)2y = 22.00262 + 9.79742 = 10 m>s

Ans.

And its direction is defined by uB = tan-1 c

(vB)y (vB)x

d = tan-1 a

9.7974 b = 78.44° = 78.4° 2.0026

c

Ans.

Ans: uA = 78.4° t = 2.00 s uB = 78.4° c 111

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*12–104. The man at A wishes to throw two darts at the target at B so that they arrive at the same time. If each dart is thrown with a speed of 10 m>s, determine the angles uC and uD at which they should be thrown and the time between each throw. Note that the first dart must be thrown at uC 1 7uD2, then the second dart is thrown at uD .

5m uC A

uD

C D B

SOLUTION + ) (:

s = s0 + v0t (1)

5 = 0 + (10 cos u) t (+ c)

v = v0 + act

- 10 sin u = 10 sin u - 9.81 t t =

2(10 sin u) = 2.039 sin u 9.81

From Eq. (1), 5 = 20.39 sin u cos u Since sin 2u = 2 sin u cos u sin 2u = 0.4905 The two roots are uD = 14.7°

Ans.

uC = 75.3°

Ans.

From Eq. (1): tD = 0.517 s tC = 1.97 s Ans.

So that ¢t = tC - tD = 1.45 s

Ans: uD = 14.7° uC = 75.3° ∆t = t C - t D = 1.45 s 112

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12–105. The velocity of the water jet discharging from the orifice can be obtained from v = 22 gh, where h = 2 m is the depth of the orifice from the free water surface. Determine the time for a particle of water leaving the orifice to reach point B and the horizontal distance x where it hits the surface.

2m

A

vA 1.5 m B x

SOLUTION Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. The speed of the water that the jet discharges from A is vA = 22(9.81)(2) = 6.264 m>s x-Motion: Here, (vA)x = vA = 6.264 m>s, xA = 0, xB = x, and t = tA. Thus, + B A:

xB = xA + (vA)xt (1)

x = 0 + 6.264tA

y-Motion: Here, (vA)y = 0, ay = -g = - 9.81 m>s2, yA = 0 m, yB = - 1.5 m, and t = tA. Thus,

A+cB

yB = yA + (vA)yt + -1.5 = 0 + 0 +

1 a t2 2 y

1 ( - 9.81)tA 2 2

tA = 0.553 s

Ans.

x = 0 + 6.264(0.553) = 3.46 m

Ans.

Thus,

Ans: t A = 0.553 s x = 3.46 m 113

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12–106. The snowmobile is traveling at 10 m>s when it leaves the embankment at A. Determine the time of flight from A to B and the range R of the trajectory.

A

40

3

5 4

SOLUTION + ) (S

B

R

sB = sA + vA t R = 0 + 10 cos 40° t

(+ c )

sB = sA + vAt +

1 a t2 2 c

1 3 - R a b = 0 + 10 sin 40°t - (9.81) t 2 4 2 Solving: R = 19.0 m

Ans.

t = 2.48 s

Ans.

Ans: R = 19.0 m t = 2.48 s 114

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12–107. The fireman wishes to direct the flow of water from his hose to the fire at B. Determine two possible angles u1 and u2 at which this can be done. Water flows from the hose at vA = 80 ft>s.

A u vA 20 ft B

SOLUTION + B A:

s = s0 + v0 t

35 ft

35 = 0 + (80)(cos u )t

A+cB

s = s0 + v0 t +

1 2 act 2

- 20 = 0 - 80 (sin u)t +

1 ( -32.2)t 2 2

Thus, 20 = 80 sin u

0.1914 0.4375 t + 16.1 ¢ ≤ cos u cos2 u

20 cos2 u = 17.5 sin 2u + 3.0816 Solving, u1 = 24.9°

(below the horizontal)

Ans.

u2 = 85.2°

(above the horizontal)

Ans.

Ans: u1 = 24.9° c u2 = 85.2° a 115

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*12–108. The baseball player A hits the baseball at vA = 40 ft>s and uA = 60° from the horizontal. When the ball is directly overhead of player B he begins to run under it. Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the ball was hit.

vA = 40 ft/s A

θA

15 ft

B

C

vA

d

SOLUTION Vertical Motion: The vertical component of initial velocity for the football is (v0)y = 40 sin 60° = 34.64 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = 0, respectively. ( + c ) sy = (s0)y + (v0)y t + 0 = 0 + 34.64t +

1 (a ) t2 2 cy

1 ( - 32.2) t2 2

t = 2.152 s Horizontal Motion: The horizontal component of velocity for the baseball is (v0)x = 40 cos 60° = 20.0 ft>s. The initial and final horizontal positions are (s0)x = 0 and sx = R, respectively. + ) (:

sx = (s0)x + (v0)x t R = 0 + 20.0(2.152) = 43.03 ft

The distance for which player B must travel in order to catch the baseball is Ans.

d = R - 15 = 43.03 - 15 = 28.0 ft

Player B is required to run at a same speed as the horizontal component of velocity of the baseball in order to catch it. Ans.

vB = 40 cos 60° = 20.0 ft s

Ans: d = 28.0 ft vB = 20.0 ft>s 116

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12–109. The catapult is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity vA at which it was launched, the angle of release u, and the height h.

B

h

vA A 3.5 ft

SOLUTION + ) (:

s = n0t

18 ft

(1)

18 = nA cos u(1.5) (+ c )

u

n2 = n20 + 2ac(s - s0) 0 = (nA sin u)2 + 2(- 32.2)(h - 3.5)

(+ c )

n = n0 + act (2)

0 = nA sin u - 32.2(1.5) To solve, first divide Eq. (2) by Eq. (1), to get u. Then u = 76.0°

Ans.

nA = 49.8 ft>s

Ans.

h = 39.7 ft

Ans.

Ans: u = 76.0° vA = 49.8 ft>s h = 39.7 ft 117

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12–110. An automobile is traveling on a curve having a radius of 800 ft. If the acceleration of the automobile is 5 ft>s2, determine the constant speed at which the automobile is traveling.

SOLUTION Acceleration: Since the automobile is traveling at a constant speed, at = 0. y2 Thus, an = a = 5 ft>s2. Applying Eq. 12–20, an = , we have r y =

ran =

Ans.

800(5) = 63.2 ft s

Ans: v = 63.2 ft>s 118

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12–111. Determine the maximum constant speed a race car can have if the acceleration of the car cannot exceed 7.5 m>s2 while rounding a track having a radius of curvature of 200 m.

SOLUTION Acceleration: Since the speed of the race car is constant, its tangential component of acceleration is zero, i.e., at = 0. Thus, a = an = 7.5 =

v2 r

v2 200 Ans.

v = 38.7 m>s

Ans: v = 38.7 m>s 119

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*12–112. A boat has an initial speed of 16 ft>s. If it then increases its speed along a circular path of radius r = 80 ft at the rate of # v = (1.5s) ft>s, where s is in feet, determine the time needed for the boat to travel s = 50 ft.

Solution at = 1.5s L0

s

v

1.5s ds =

L16

v dv

0.75 s2 = 0.5 v2 - 128 v = s

ds = 2256 + 1.5 s2 dt

L0 1s2 + 170.7 ds

=

L0

t

1.225 dt s

ln 1 s + 2s2 + 170.7 2 0 0 = 1.225t

ln 1 s + 2s2 + 170.7 2 - 2.570 = 1.225t

At s = 50 ft,

Ans.

t = 1.68 s

Ans: t = 1.68 s 120

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12–113. The position of a particle is defined by r = {4(t - sin t)i + (2t2 - 3)j} m, where t is in seconds and the argument for the sine is in radians. Determine the speed of the particle and its normal and tangential components of acceleration when t = 1 s.

Solution r = 4(t - sin t) i + (2 t 2 - 3)j v =

dr = 4(1 - cos t)i + (4 t)j dt

v 0 t = 1 = 1.83879i + 4j

v = 2(1.83879)2 + (4)2 = 4.40 m>s

u = tan-1 a

Ans.

4 b = 65.312° au 1.83879

a = 4 sin ri + 4j

a  t = 1 = 3.3659i + 4j a = 2(3.3659)2 + (4)2 = 5.22773 m>s2 f = tan-1a

4 b = 49.920° au 3.3659

d = u - f = 15.392°

a2 = 5.22773 cos 15.392° = 5.04 m>s2

Ans.

an = 5.22773 sin 15.392° = 1.39 m>s2

Ans.

Ans: v = 4.40 m>s at = 5.04 m>s2 an = 1.39 m>s2 121

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12–114. The automobile has a speed of 80 ft>s at point A and an acceleration a having a magnitude of 10 ft>s2, acting in the direction shown. Determine the radius of curvature of the path at point A and the tangential component of acceleration.

SOLUTION Acceleration: The tangential acceleration is at = a cos 30° = 10 cos 30° = 8.66 ft>s2

Ans.

and the normal acceleration is an = a sin 30° = 10 sin 30° = 5.00 ft>s2. Applying v2 Eq. 12–20, an = , we have r r =

v2 802 = 1280 ft = an 5.00

Ans.

Ans: at = 8.66 ft>s2 r = 1280 ft 122

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12–115. The automobile is originally at rest at s = 0. If its speed is # increased by v = 10.05t22 ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration when t = 18 s.

300 ft s

240 ft

SOLUTION at = 0.05t2 t

v

dv =

L0

0.05 t2 dt

L0

v = 0.0167 t3 s

L0

t

ds =

L0

0.0167 t3 dt

s = 4.167(10 - 3) t4 When t = 18 s,

s = 437.4 ft

Therefore the car is on a curved path. v = 0.0167(183) = 97.2 ft>s an =

Ans.

(97.2)2 = 39.37 ft>s2 240

at = 0.05(182) = 16.2 ft/s2 a = 2(39.37)2 + (16.2)2 a = 42.6 ft>s2

Ans.

Ans: v = 97.2 ft>s a = 42.6 ft>s2 123

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*12–116. The automobile is originally at rest s = 0. If it then starts to # increase its speed at v = 10.05t22 ft>s2, where t is in seconds, determine the magnitudes of its velocity and acceleration at s = 550 ft.

300 ft s

240 ft

SOLUTION The car is on the curved path. at = 0.05 t2 t

v

dv =

L0

0.05 t2 dt

L0

v = 0.0167 t3 s

L0

t

ds =

L0

0.0167 t3 dt

s = 4.167(10 - 3) t4 550 = 4.167(10-3) t4 t = 19.06 s So that v = 0.0167(19.06)3 = 115.4 Ans.

v = 115 ft>s an =

(115.4)2 = 55.51 ft>s2 240

at = 0.05(19.06)2 = 18.17 ft>s2 a = 2(55.51)2 + (18.17)2 = 58.4 ft>s2

Ans.

Ans: v = 115 ft>s a = 58.4 ft>s2 124

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12–117. vA

The two cars A and B travel along the circular path at constant speeds vA = 80 ft>s and vB = 100 ft>s, respectively. If they are at the positions shown when t = 0, determine the time when the cars are side by side, and the time when they are 90° apart.

A rA  400 ft

rB  390 ft

B

Solution

vB

a) Referring to Fig. a, when cars A and B are side by side, the relation between their angular displacements is (1)

uB = uA + p

s Here, sA = vA t = 80 t and sB = vB t = 100 t. Apply the formula s = ru or u = . Then r uB =

sB 100 t 10 = = t rB 390 39

uA =

sA 80 t 1 = = t rA 400 5

Substitute these results into Eq. (1) 10 1 t = t + p 39 5 Ans.

t = 55.69 s = 55.7 s

(b) Referring to Fig. a, when cars A and B are 90° apart, the relation between their angular displacements is uB +

p = uA + p 2 p uB = uA +  2

(2)

s Here, sA = vA t = 80 t and sB = vB t = 100 t. Applying the formula s = ru or u = . r Then uB =

sB 100 t 10 = = t   rB 390 39

uA =

sA 80 t 1 = = t rA 400 5

Substitute these results into Eq. (2) 10 1 p t = t + 39 5 2 Ans.

t = 27.84 s = 27.8 s

Ans: When cars A and B are side by side, t = 55.7 s. When cars A and B are 90° apart, t = 27.8 s. 125

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12–118. vA

Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 60 ft>s and is increasing its speed at the rate of 15 ft>s2 until it travels for a distance of 100p ft, after which it maintains a constant speed. Car B has a speed of 120 ft>s and is decreasing its speed at 15 ft>s2 until it travels a distance of  65p ft, after which it maintains a constant speed. Determine the time when they come side by side.

A rA  400 ft

rB  390 ft

B vB

Solution Referring to Fig. a, when cars A and B are side by side, the relation between their angular displacements is (1)

uA = uB + p The constant speed achieved by cars A and B can be determined from (vA)2c = (vA)20 + 2(aA)t [sA - (s0)A] (vA)2c = 602 + 2(15) (100p - 0) (vA)c = 114.13 ft>s (vB)2c = (vB)20 + 2(aB)t [sB - (s0)B] (vB)2c = 1202 + 2 ( -15) (65p - 0) (vB)c = 90.96 ft>s The time taken to achieve these constant speeds can be determined from (vA)c = (vA)0 + (aA)t (t A)1 114.13 = 60 + 15(t A)1 (t A)1 = 3.6084 s (vB)c = (vB)0 + (aB)t (t B)1 90.96 = 120 + ( -15) (t B)1 (t B)1 = 1.9359 s

Let t be the time taken for cars A and B to be side by side. Then, the times at which cars A and B travel with constant speed are (t A)2 = t - (t A)1 = t - 3.6084 and (t B)2 = t - (t B)1 = t - 1.9359. Here, (sA)1 = 100p, (sA)2 = (vA)c (t A)2 = 114.13 (t - 3.6084), (sB)1 = 65p and (sB)2 = (vB)c (t B)2 = 90.96 (t - 1.9359). s Using, the formula s = ru or u = , r (sA)1

uA = (uA)1 + (uA)2 =

rA

+

(sA)2 rA

114.13 (t - 3.6084) 100p + 400 400

= 0.2853 t - 0.24414

uB = (uB)1 + (uB)2 =

=

(sB)1 rB

+

(sB)2 rB

=

90.96 (t - 1.9359) 65 p + 390 390

= 0.2332 t + 0.07207 Substitute these results into Eq. (1), 0.2853 t - 0.24414 = 0.2332 t + 0.07207 + p Ans.

t = 66.39 s = 66.4 s

Ans: t = 66.4 s 126

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12–119. The satellite S travels around the earth in a circular path with a constant speed of 20 Mm>h. If the acceleration is 2.5 m>s2, determine the altitude h. Assume the earth’s diameter to be 12 713 km.

S

h

SOLUTION n = 20 Mm>h =

Since at =

dn = 0, then, dt

a = an = 2.5 = r =

20(106) = 5.56(103) m>s 3600

n2 r

(5.56(103))2 = 12.35(106) m 2.5

The radius of the earth is 12 713(103) = 6.36(106) m 2 Hence, h = 12.35(106) - 6.36(106) = 5.99(106) m = 5.99 Mm

Ans.

Ans: h = 5.99 Mm 127

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*12–120. The car travels along the circular path such that its speed is increased by a t = (0.5et) m>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s = 18 m starting from rest. Neglect the size of the car.

s

18 m

SOLUTION v

L0

t

dv =

L0

0.5e t dt ρ

v = 0.5(e t - 1) 18

L0

30 m

t

ds = 0.5

L0

(e t - 1)dt

18 = 0.5(e t - t - 1) Solving, t = 3.7064 s v = 0.5(e 3.7064 - 1) = 19.85 m>s = 19.9 m>s # at = v = 0.5e t ƒ t = 3.7064 s = 20.35 m>s2 an =

Ans.

19.852 v2 = 13.14 m>s2 = r 30

a = 2a2t + a2n = 220.352 + 13.142 = 24.2 m>s2

Ans.

Ans: v = 19.9 m>s a = 24.2 m>s2 128

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12–121. The car passes point A with a speed of 25 m>s after which its speed is defined by v = (25 - 0.15s) m>s. Determine the magnitude of the car’s acceleration when it reaches point B, where s = 51.5 m and x = 50 m.

y y

16 m

16 B

1 2 x 625 s A x

SOLUTION Velocity: The speed of the car at B is vB = C 25 - 0.15 A 51.5 B D = 17.28 m>s Radius of Curvature: y = 16 -

1 2 x 625

dy = -3.2 A 10-3 B x dx dx2

= -3.2 A 10-3 B

B1 + a r =

2

dy 2 b dx

d2y dx2

2

B

d2y

3>2

2 3>2

c 1 + a - 3.2 A 10-3 B x b d =

2 - 3.2 A 10-3 B 2

4

= 324.58 m x = 50 m

Acceleration: vB 2 17.282 = 0.9194 m>s2 = r 324.58 dv = A 25 - 0.15s B A - 0.15 B = A 0.225s - 3.75 B m>s2 at = v ds an =

When the car is at B A s = 51.5 m B a t = C 0.225 A 51.5 B - 3.75 D = - 2.591 m>s2 Thus, the magnitude of the car’s acceleration at B is a = 2a2t + a2n = 2( - 2.591)2 + 0.91942 = 2.75 m>s2

Ans.

Ans: a = 2.75 m>s2 129

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12–122. y

If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of a t = 0.5 m>s2, determine the magnitude of the car’s acceleration when s = 101.68 m and x = 0.

y

16 m

16 B

1 2 x 625 s A x

SOLUTION Velocity: The speed of the car at C is vC 2 = vA 2 + 2a t (sC - sA) vC 2 = 202 + 2(0.5)(100 - 0) vC = 22.361 m>s Radius of Curvature: y = 16 -

1 2 x 625

dy = - 3.2 A 10-3 B x dx d2y dx2

= - 3.2 A 10-3 B

B1 + a r =

2

dy 2 3>2 b R dx

d2y dx2

2

2 3>2

c1 + a -3.2 A 10-3 B xb d =

- 3.2 A 10-3 B

4

= 312.5 m x=0

Acceleration: # a t = v = 0.5 m>s an =

vC 2 22.3612 = = 1.60 m>s2 r 312.5

The magnitude of the car’s acceleration at C is a = 2a2t + a2n = 20.52 + 1.602 = 1.68 m>s2

Ans.

Ans: a = 1.68 m>s2 130

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12–123. y

The motorcycle is traveling at 1 m>s when it is at A. If the # speed is then increased at v = 0.1 m>s2, determine its speed and acceleration at the instant t = 5 s.

y

0.5x2

s

SOLUTION at = n = 0.1 s = s0 + n0 t +

1 2 at 2 c

s = 0 + 1(5) +

1 (0.1)(5)2 = 6.25 m 2

6.25

L0

x

ds =

x

A

L0 A

1 + a

dy 2 b dx dx

y = 0.5x2 dy = x dx 2

dy dx

2

= 1 x

6.25 = 6.25 =

L0

31 + x2 dx

x 1 c x31 + x2 + 1n ax + 31 + x2 b d 2 0

x 31 + x2 + 1n a x + 31 + x2 b = 12.5 Solving, x = 3.184 m 3

c1 + a r =

`

dy 2 2 b d dx

d2y dx2

`

3

=

[1 + x2]2 = 37.17 m ` |1| x = 3.184

n = n0 + act Ans.

= 1 + 0.1(5) = 1.5 m>s an =

(1.5)2 n2 = 0.0605 m>s2 = r 37.17

a = 310.122 + 10.060522 = 0.117 m>s2

Ans.

Ans: v = 1.5 m>s a = 0.117 m>s2 131

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*12–124. y

The box of negligible size is sliding down along a curved path defined by the parabola y = 0.4x2. When it is at A (xA = 2 m, yA = 1.6 m), the speed is v = 8 m >s and the increase in speed is dv >dt = 4 m> s2. Determine the magnitude of the acceleration of the box at this instant.

A y

2

0.4x

x

SOLUTION

2m

y = 0.4 x2 dy 2 = 0.8x 2 = 1.6 dx x = 2 m x=2 m d2y dx2

2

r =

x=2 m

dy 2 3>2 ) D C 1 + (dx

` 2

an =

= 0.8

d 2y 2

dx

`

4

=

C 1 + (1.6)2 D 3>2 |0.8|

= 8.396 m

x=2 m 2

yB 8 = = 7.622 m>s2 r 8.396

a = 2a2t + a2n = 2(4)2 + (7.622)2 = 8.61 m>s2

Ans.

Ans: a = 8.61 m>s2 132

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12–125. y

The car travels around the circular track having a radius of r = 300 m such that when it is at point A it has a velocity of 5 m>s, which is # increasing at the rate of v = (0.06t) m>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration when it has traveled one-third the way around the track.

r A

x

Solution at = v = 0.06t dv = at dt Ls

v

dv =

L0

t

0.06t dt

v = 0.03t 2 + 5 ds = v dt s

t

(0.03t 2 + 5) dt L0 L0 s = 0.01t 3 + 5t ds =

s =

1 (2p(300)) = 628.3185 3

0.01t 3 + 5t - 628.3185 = 0 Solve for the positive root, t = 35.58 s v = 0.03(35.58)2 + 5 = 42.978 m>s = 43.0 m>s an =

Ans.

(42.978)2 v2 = 6.157 m>s2 = r 300

at = 0.06(35.58) = 2.135 m>s2 a = 2(6.157)2 + (2.135)2 = 6.52 m>s2

Ans.

Ans: v = 43.0 m>s a = 6.52 m>s2 133

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12–126. y

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a velocity of 2 ft>s, which is increasing at the rate of # v = (0.002t) ft>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration when it has traveled three-fourths the way around the track.

r A

x

Solution at = 0.002 s at ds = v dv L0

s

0.002s ds =

0.001s2 =

L2

v

v dv

1 2 1 v - (2)2 2 2

v2 = 0.002s2 + 4 s =

3 [2p(500)] = 2356.194 ft 4

v2 = 0.002(2356.194)2 + 4 Ans.

v = 105.39 ft>s = 105 ft>s an =

(105.39)2 v2 = 22.21 ft>s2 = r 500

at = 0.002(2356.194) = 4.712 ft>s2 a = 2(22.21)2 + (4.712)2 = 22.7 ft>s2

Ans.

Ans: v = 105 ft>s a = 22.7 ft>s2 134

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12–127. At a given instant the train engine at E has a speed of 20 m>s and an acceleration of 14 m>s2 acting in the direction shown. Determine the rate of increase in the train’s speed and the radius of curvature r of the path.

v

20 m/s

75 a

SOLUTION

2

14 m/s

E

r

Ans.

at = 14 cos 75° = 3.62 m>s2 an = 14 sin 75° an =

(20)2 r Ans.

r = 29.6 m

Ans: at = 3.62 m>s2 r = 29.6 m 135

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*12–128. The car has an initial speed v0 = 20 m>s. If it increases its speed along the circular track at s = 0, at = (0.8s) m>s2, where s is in meters, determine the time needed for the car to travel s = 25 m. s

Solution The distance traveled by the car along the circular track can be determined by integrating v dv = at ds. Using the initial condition v = 20 m>s at s = 0, v

L20 m>s

v dv =

L0

r  40 m

5

0.8 s ds

v2 v ` = 0.4 s2 2 20 m>s

v = e 20.8 ( s2 + 500 ) f m>s

ds with the initial condition The time can be determined by integrating dt = v s = 0 at t = 0. L0 t =

t

dt =

L0

25 m

ds 10.8 ( s2 + 500 )

25 m 1 3 ln 1 s + 1s2 + 500 24 ` 10.8 0

Ans.

= 1.076 s = 1.08 s

Ans: t = 1.08 s 136

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12–129. The car starts from rest at s = 0 and increases its speed at at = 4 m>s2. Determine the time when the magnitude of acceleration becomes 20 m>s2. At what position s does this occur? s

Solution Acceleration. The normal component of the acceleration can be determined from ur =

r  40 m

v2 v2 ;  ar = r 40

From the magnitude of the acceleration v2 2 b   v = 28.00 m>s B 40 Velocity. Since the car has a constant tangential accelaration of at = 4 m>s2, a = 2a2t + a2n;   20 =

42 + a

v = v0 + at t ;   28.00 = 0 + 4t Ans.

t = 6.999 s = 7.00 s 2

v =

v20

2

2

+ 2at s; 28.00 = 0 + 2(4) s Ans.

s = 97.98 m = 98.0 m

Ans: t = 7.00 s s = 98.0 m 137

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12–130. A boat is traveling along a circular curve having a radius of 100 ft. If its speed at t = 0 is 15 ft/s and is increasing at # v = 10.8t2 ft>s2, determine the magnitude of its acceleration at the instant t = 5 s.

SOLUTION 5

v

L15

dv =

L0

0.8tdt

v = 25 ft>s an =

v2 252 = 6.25 ft>s2 = r 100

At t = 5 s,

# at = v = 0.8(5) = 4 ft>s2 a =

a2t + a2n =

42 + 6.252 = 7.42 ft s2

Ans.

Ans: a = 7.42 ft>s2 138

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12–131. A boat is traveling along a circular path having a radius of 20 m. Determine the magnitude of the boat’s acceleration when the speed is v = 5 m>s and the rate of increase in the # speed is v = 2 m>s2.

SOLUTION at = 2 m>s2 an =

y2 52 = = 1.25 m>s2 r 20

a = 2a2t + a2n = 222 + 1.252 = 2.36 m>s2

Ans.

Ans: a = 2.36 m>s2 139

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*12–132. Starting from rest, a bicyclist travels around a horizontal circular path, r = 10 m, at a speed of v = 10.09t2 + 0.1t2 m>s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m.

SOLUTION s

L0

t

ds =

L0

10.09t2 + 0.1t2dt

s = 0.03t3 + 0.05t2 When s = 3 m,

3 = 0.03t3 + 0.05t2

Solving, t = 4.147 s v =

ds = 0.09t2 + 0.1t dt

v = 0.09(4.147)2 + 0.1(4.147) = 1.96 m>s at =

dv = 0.18t + 0.1 ` = 0.8465 m>s2 dt t = 4.147 s

an =

1.962 v2 = 0.3852 m>s2 = r 10

Ans.

a = 2a2t + a2n = 2(0.8465)2 + (0.3852)2 = 0.930 m>s2

Ans.

Ans: v = 1.96 m>s a = 0.930 m>s2 140

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12–133. A particle travels around a circular path having a radius of 50 m. If it is initially traveling with a speed of 10 m> s and its # speed then increases at a rate of v = 10.05 v2 m>s2, determine the magnitude of the particle’s acceleraton four seconds later.

SOLUTION Velocity: Using the initial condition v = 10 m>s at t = 0 s, dt =

dv a v

t

L0

dt =

t = 20 ln

dv L10 m>s 0.05v v 10

v = (10et>20) m>s When t = 4 s, v = 10e4>20 = 12.214 m>s Acceleration: When v = 12.214 m>s (t = 4 s), at = 0.05(12.214) = 0.6107 m>s2 an =

(12.214)2 v2 = = 2.984 m>s2 r 50

Thus, the magnitude of the particle’s acceleration is a = 2at2 + an2 = 20.61072 + 2.9842 = 3.05 m>s2

Ans.

Ans: a = 3.05 m>s2 141

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12–134. The motorcycle is traveling at a constant speed of 60 km> h. Determine the magnitude of its acceleration when it is at point A.

y

y2  2x A

x 25 m

SOLUTION Radius of Curvature: y = 22x1>2 dy 1 = 22x - 1>2 dx 2 d2y dx2

r =

= -

1 22x - 3>2 4

B1 + a `

dy 2 3>2 b R dx

d2y dx2

`

=

B 1 + ¢ 22x - 1>2 ≤ R 1 2

`-

2

3>2

1 22x - 3>2 ` 4

4

= 364.21 m x = 25 m

Acceleration: The speed of the motorcycle at a is v = ¢ 60 an =

km 1000 m 1h ≤¢ ≤¢ ≤ = 16.67 m>s h 1 km 3600 s

v2 16.672 = 0.7627 m>s2 = r 364.21

Since the motorcycle travels with a constant speed, at = 0. Thus, the magnitude of the motorcycle’s acceleration at A is a = 2at 2 + an2 = 202 + 0.76272 = 0.763 m>s2

Ans.

Ans: a = 0.763 m>s2 142

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12–135. When t = 0, the train has a speed of 8 m>s, which is increasing at 0.5 m>s2. Determine the magnitude of the acceleration of the engine when it reaches point A, at t = 20 s. Here the radius of curvature of the tracks is rA = 400 m.

vt  8 m/s A

Solution Velocity. The velocity of the train along the track can be determined by integrating dv = at dt with initial condition v = 8 m>s at t = 0. v

L8 m>s

dv =

L0

t

0.5 dt

v - 8 = 0.5 t

v = {0.5 t + 8} m>s At t = 20 s, v  t = 20 s = 0.5(20) + 8 = 18 m>s Acceleration. Here, the tangential component is at = 0.5 m>s2. The normal component can be determined from an =

v2 182 = 0.81 m>s2 = r 400

Thus, the magnitude of the acceleration is a = 2a2t + a2n

= 20.52 + 0.812

= 0.9519 m>s2 = 0.952 m>s2

Ans.

Ans: a = 0.952 m>s2 143

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*12–136. At a given instant the jet plane has a speed of 550 m>s and an acceleration of 50 m>s2 acting in the direction shown. Determine the rate of increase in the plane’s speed, and also the radius of curvature r of the path.

550 m/s

70 a  50 m/s2

r

Solution Acceleration. With respect to the n–t coordinate established as shown in Fig. a, the tangential and normal components of the acceleration are at = 50 cos 70° = 17.10 m>s2 = 17.1 m>s2

Ans.

an = 50 sin 70° = 46.98 m>s2 However, an =

v2 5502 ;  46.98 = r r r = 6438.28 m = 6.44 km

Ans.

Ans: at = 17.1 m>s2 an = 46.98 m>s2 r = 6.44 km 144

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12–137. The ball is ejected horizontally from the tube with a speed of 8 m>s. Find the equation of the path, y = f1x2, and then find the ball’s velocity and the normal and tangential components of acceleration when t = 0.25 s.

y vA A

8 m/s

x

SOLUTION vx = 8 m>s

+ ) (:

s = v0t x = 8t

A+cB

s = s0 + v0 t + y = 0 + 0 +

1 2 a t 2 c

1 ( - 9.81)t2 2

y = -4.905t2 x 2 y = -4.905 a b 8 y = -0.0766x2

Ans.

(Parabola)

v = v0 + act vy = 0 - 9.81t When t = 0.25 s, vy = - 2.4525 m>s v = 31822 + 12.452522 = 8.37 m>s u = tan - 1 a ax = 0

Ans.

2.4525 b = 17.04° 8

ay = 9.81 m>s2

an = 9.81 cos 17.04° = 9.38 m>s2

Ans.

at = 9.81 sin 17.04° = 2.88 m>s2

Ans.

Ans: y = -0.0766x2 v = 8.37 m>s an = 9.38 m>s2 at = 2.88 m>s2 145

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12–138. The motorcycle is traveling at 40 m>s when it is at A. If the # speed is then decreased at  v = -  (0.05 s) m>s2, where s is in meters measured from A, determine its speed and acceleration when it reaches B.

60 150 m

150 m B

A

Solution Velocity. The velocity of the motorcycle along the circular track can be determined by integrating vdv = ads with the initial condition v = 40 m>s at s = 0. Here, at = - 0.05s. v

L40 m>s

L0

vdv =

s

- 0.05 sds

v2 v s ` = - 0.025 s2  0 2 40 m>s v =

5 21600

p At B, s = ru = 150a b = 50p m. Thus 3

- 0.05 s2 6 m>s

vB = v s = 50pm = 21600 - 0.05(50p)2 = 19.14 m>s = 19.1 m>s

Ans.

Acceleration. At B, the tangential and normal components are at = 0.05(50p) = 2.5p m>s2 v2B 19.142 an = r = = 2.4420 m>s2 150 Thus, the magnitude of the acceleration is a = 2a2t + a2n = 2(2.5p)2 + 2.44202 = 8.2249 m>s2 = 8.22 m>s

Ans.

And its direction is defined by angle f measured from the negative t-axis, Fig. a. f = tan-1a

an 2.4420 b = tan-1a b at 2.5p

Ans.

= 17.27° = 17.3°

Ans: vB = 19.1 m>s a = 8.22 m>s2 f = 17.3° up from negative - t axis 146

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12–139. y

Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the minimum acceleration experienced by the passengers.

2

x2  y  1 (60)2 (40) 0)2

40 m x 60 m

Solution y2 x2 + = 1 a2 b2 b2x2 + a2y2 = a2b2 b2(2x) + a2(2y)

dy = 0 dx

dy b2x = - 2 dx ay dy - b2x y = dx a2 d 2y dx

2

d 2y dx

y = 2

d 2y dx

y + a

2

y =

d 2y dx

y = 2

d 2y dx

2

y =

d 2y dx

2

r =

=

dy 2 - b2 b = 2 dx a

- b2 - b2x 2 a b a2 a2y

- b2 b4 x2 - a 2 2 ba 2 b 2 a ay a

y2 - b2 b4 a1 b a2 a2y2 b2 - b2 b4 b2 - 2 2 + 2 2 a ay a -b4 a2y3

c1 + a `

- b2x 2 3>2 b d a2y

- b2 ` a2y3

At x = 0, y = h, r =

a2 b

147

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12–139. Continued

Thus at = 0 amin = an =

v2 v2b v2 = 2 = 2 r a a b

Set a = 60 m, b = 40 m, v =

60(10)3 3600

amin =

= 16.67 m>s

(16.67)2(40) (60)2

= 3.09 m>s2

Ans.

Ans: amin = 3.09 m>s2 148

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*12–140. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the maximum acceleration experienced by the passengers.

y 2

x2  y  1 (60)2 (40) 0)2

40 m x 60 m

Solution y2 x2 + = 1 a2 b2 b2x2 + a2y2 = a2b2 b2(2x) + a2(2y)

dy = 0 dx

dy b2x = - 2 dx ay dy - b2x y = dx a2 d 2y dx

2

d 2y dx

y + a

y = 2

d 2y dx

2

=

dy 2 - b2 b = 2 dx a

- b2 - b2x 2 a b a2 a2y

-b4 a2y3

c1 + a

r =

`

-b2x 2 3>2 b d a2y

- b4 ` a2y3

At x = a, y = 0, r =

b2 a

Then at = 0 amax = an =

v2 v2a v2 = 2 = 3 r b b n

Set a = 60 m, b = 40 m, v = amax =

(16.67)2(60) (40)2

60(103) 3600

= 16.67 m>s

= 10.4 m>s2

Ans.

Ans: amax = 10.4 m>s2 149

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12–141. A package is dropped from the plane which is flying with a constant horizontal velocity of vA = 150 ft>s. Determine the normal and tangential components of acceleration and the radius of curvature of the path of motion (a) at the moment the package is released at A, where it has a horizontal velocity of vA = 150 ft>s, and (b) just before it strikes the ground at B.

A

vA

1500 ft

SOLUTION B

Initially (Point A): (an)A = g = 32.2 ft>s2

Ans.

(at)A = 0

Ans.

(an)A =

n2A ; rA

32.2 =

(150)2 rA Ans.

rA = 698.8 ft (nB)x = (nA)x = 150 ft>s n2 = n20 + 2ac(s - s0)

(+ T)

(nB)2y = 0 + 2(32.2)(1500 - 0) (nB)y = 310.8 ft>s nB = 2(150)2 + (310.8)2 = 345.1 ft>s u = tan - 1 a

nBy nBx

b = tan - 1 a

310.8 b = 64.23° 150

(an)B = g cos u = 32.2 cos 64.24° = 14.0 ft>s2

Ans.

(at)B = g sin u = 32.2 sin 64.24° = 29.0 ft>s2

Ans.

(an)B =

n2B ; rB

2

14.0 =

(345.1) rB

rB = 8509.8 ft = 8.51(103) ft

Ans.

Ans: (an)A = g = 32.2 ft>s2 (at)A = 0 rA = 699 ft (an)B = 14.0 ft>s2 (at)B = 29.0 ft>s2 rB = 8.51 ( 103 ) ft 150

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12–142. The race car has an initial speed vA = 15 m>s at A. If it increases its speed along the circular track at the rate a t = 10.4s2 m>s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m. r

SOLUTION

s

n dn at = 0.4s = ds

A

a ds = n dn s

n

0.4s ds =

L0

L15

n dn

n2 n 0.4s2 s ` = ` 2 0 2 15 n2 225 0.4s2 = 2 2 2 n2 = 0.4s2 + 225 n =

ds = 20.4s2 + 225 dt

s

t

ds

L0 20.4s2 + 225 s

ds

L0 2s + 562.5 2

=

L0

dt

= 0.632 456t

1n (s + 2s2 + 562.5) `

s 0

= 0.632 456t

1n (s + 2s2 + 562.5) - 3.166 196 = 0.632 456t At s = 20 m, Ans.

t = 1.21 s

Ans: t = 1.21 s 151

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12–143. The motorcycle travels along the elliptical track at a constant speed v. Determine its greatest acceleration if a 7 b.

y

b

2

y

a x

b

2

a

2

x

2

1

Solution y2 x2 + = 1 a2 b2 b2x2 + a2y2 = a2b2 dy b2x = - 2 dx ay dy - b2x y = dx a2 d 2y dx

2

2

d y dx

y + a

2

d 2y dx2

= =

b =

dx

-b2 a2

-b2 - b2x - a 2 b 2 a ay

-b4 a2y3

c1 + a

r =

2

dy

b2x 2 3>2 b d a2y

- b4 a2y3

At x = a, y = 0, r =

b2 a

Then at = 0 amax = an =

v2 v2a v2 = 2 = 2 r b b a

Ans.

Ans: amax = 152

v2a b2

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*12–144. The motorcycle travels along the elliptical track at a constant speed v. Determine its smallest acceleration if a 7 b.

y

b

2

y

a x

b

2

a

2

x

2

1

Solution y2 x2 + = 1 a2 b2 b2x2 + a2y2 = a2b2 b2(2x) + a2(2y)

dy = 0 dx

dy b2x = - 2 dx ay dy - b2x y = dx a2 d 2y dx

2

d 2y

y + a

dx2 d 2y dx

2

= =

r =

dy 2 - b2 b = 2 dx a

-b2 - b2x a b a2 a2y

-b4 a2y3

c1 + a

b2x 2 3>2 b d a2y

- b4 a2y3

At x = 0, y = b, r =

a2 b

Thus at = 0 amin = an =

v2 v2b v2 = 3 = 2 r a a b

Ans.

Ans: amin = 153

v2b a2

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12–145. Particles A and B are traveling counter-clockwise around a circular track at a constant speed of 8 m>s. If at the instant shown the speed of A begins to increase by (at)A = (0.4sA) m>s2, where sA is in meters, determine the distance measured counterclockwise along the track from B to A when t = 1 s. What is the magnitude of the acceleration of each particle at this instant?

A sA u  120

sB B

SOLUTION

r5m

Distance Traveled: Initially the distance between the particles is d0 = rdu = 5 a

120° b p = 10.47 m 180°

When t = 1 s, B travels a distance of dB = 8(1) = 8 m The distance traveled by particle A is determined as follows: vdv = ads v

L8 m>s

s

vdv =

L0

0.4 sds

v = 0.6325 2s2 + 160

(1)

ds dt = v t

L0

s

dt =

ds

L0 0.6325 2s2 + 160 2s2 + 160 + s 1 S≥ £ In C 0.6325 2160

1 =

s = 8.544 m Thus the distance between the two cyclists after t = 1 s is Ans.

d = 10.47 + 8.544 - 8 = 11.0 m Acceleration: For A, when t = 1 s,

A a t B A = vA = 0.4 A 8.544 B = 3.4176 m>s2 #

vA = 0.6325 28.5442 + 160 = 9.655 m>s (a n)A =

v2A 9.6552 = = 18.64 m>s2 r 5

The magnitude of the A’s acceleration is aA = 23.41762 + 18.642 = 19.0 m>s2

Ans.

For B, when t = 1 s,

A at B B = vA = 0 #

(an)B =

v2B 82 = = 12.80 m>s2 r 5

Ans:

The magnitude of the B’s acceleration is aB = 202 + 12.802 = 12.8 m>s2

Ans. 154

d = 11.0 m aA = 19.0 m>s2 aB = 12.8 m>s2

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12–146. Particles A and B are traveling around a circular track at a speed of 8 m>s at the instant shown. If the speed of B is increasing by (at)B = 4 m>s2, and at the same instant A has an increase in speed of (at)A = 0.8t m>s2, determine how long it takes for a collision to occur. What is the magnitude of the acceleration of each particle just before the collision occurs?

A sA u  120

sB B

SOLUTION Distance Traveled: Initially the distance between the two particles is d0 = ru

r5m

120° p b = 10.47 m. Since particle B travels with a constant acceleration, 180° distance can be obtained by applying equation = 5a

sB = (s0)B + (y0)B t + sB = 0 + 8t +

1 a t2 2 c

1 (4) t2 = A 8 t + 2 t2 B m 2

[1]

The distance traveled by particle A can be obtained as follows. dyA = aA dt t

yA

L8 m>s

dyA =

L0

0.8 tdt

yA = A 0.4 t2 + 8 B m>s

[2]

dsA = yA dt sA

L0

t

dsA =

L0

A 0.4 t2 + 8 B dt

sA = 0.1333t3 + 8 t In order for the collision to occur sA + d0 = sB 0.1333t3 + 8t + 10.47 = 8 t + 2 t2 Solving by trial and error

Ans.

t = 2.5074 s = 2.51 s

240° pb + sB. This equation will result in Note: If particle A strikes B then, sA = 5 a 180° t = 14.6 s 7 2.51 s. Acceleration: The tangential acceleration for particle A and B when t = 2.5074 are (a t)A = 0.8 t = 0.8 (2.5074) = 2.006 m>s2 and (a t)B = 4 m>s2, respectively. When t = 2.5074 s, from Eq. [1], yA = 0.4 A 2.50742 B + 8 = 10.51 m>s and yB = (y0)B + ac t = 8 + 4(2.5074) = 18.03 m>s.To determine the normal acceleration, apply Eq. 12–20. (an)A =

y2A 10.512 = 22.11 m>s2 = r 5

(an)B =

y2B 18.032 = = 65.01 m>s2 r 5

The magnitude of the acceleration for particles A and B just before collision are aA = 2(at)2A + (an)2A = 22.0062 + 22.112 = 22.2 m>s2

Ans.

aB = 2(at)2B + (an)2B = 242 + 65.012 = 65.1 m>s2

Ans. 155

Ans: t = 2.51 s aA = 22.2 m>s2 aB = 65.1 m>s2

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12–147. The jet plane is traveling with a speed of 120 m>s which is decreasing at 40 m>s2 when it reaches point A. Determine the magnitude of its acceleration when it is at this point. Also, specify the direction of flight, measured from the x axis.

y

y  15 lnQ

x R 80

80 m A

x

Solution y = 15 ln a

x b 80

dy 15 ` = 0.1875 = dx x x = 80 m d 2y dx2

r∞

= -

15 ` = - 0.002344 x2 x = 80 m =

c1 + a `

x = 80 m

an =

=

dy dx

d 2y dx2

2 3>2

b d `

∞

x = 80 m

[1 + (0.1875)2]3>2

0 - 0.002344 0

= 449.4 m

(120)2 v2 = 32.04 m>s2 = r 449.4

an = - 40 m>s2 Ans.

a = 2( - 40)2 + (32.04)2 = 51.3 m>s2

Since

dy = tan u = 0.1875 dx Ans.

u = 10.6°

Ans: u = 10.6° 156

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*12–148. The jet plane is traveling with a constant speed of 110 m>s along the curved path. Determine the magnitude of the acceleration of the plane at the instant it reaches point A(y = 0).

y

y  15 lnQ

x R 80

80 m x

A

Solution y = 15 ln a

x b 80

dy 15 ` = 0.1875 = dx x x = 80 m d 2y dx2

r∞

= -

15 † x2 =

= - 0.002344 x = 80 m

c1 + a

x = 80 m

an =

=

31

`

dy 2 3>2 b d dx

d 2y dx2

`

∞

x = 80 m

+ (0.1875)2 4 3>2

0 - 0.002344 0

= 449.4 m

(110)2 v2 = 26.9 m>s2 = r 449.4

Since the plane travels with a constant speed, at = 0. Hence a = an = 26.9 m>s2

Ans.

Ans: a = 26.9 m>s2 157

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12–149. The train passes point B with a speed of 20 m>s which is decreasing at at = – 0.5 m>s2. Determine the magnitude of acceleration of the train at this point.

y x

y  200 e 1000

B

A

SOLUTION x

Radius of Curvature:

400 m

x

y = 200e 1000 x x dy 1 = 200 a b e 1000 = 0.2e 1000 dx 1000

d2y 2

dx

r =

= 0.2 a

x x 1 . b e 1000 = 0.2 A 10-3 B e 1000 1000

c1 + a

2

dy b d dx

d2y dx2

C 1 + ¢ 0.2

2 3>2

2

=

x 2 e 1000 ≤ S

` 0.2 A 10 B -3

x e 1000

`

3>2

6

= 3808.96 m

x = 400 m

Acceleration: # a t = v = -0.5 m>s2 an =

202 v2 = = 0.1050 m>s2 r 3808.96

The magnitude of the train’s acceleration at B is a = 2a2t + a2n = 2 A - 0.5 B 2 + 0.10502 = 0.511 m>s2

Ans.

Ans: a = 0.511 m>s2 158

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12–150. The train passes point A with a speed of 30 m>s and begins to decrease its speed at a constant rate of at = – 0.25 m>s2. Determine the magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m.

y x

y  200 e 1000

B

A

SOLUTION x

Velocity: The speed of the train at B can be determined from

400 m

vB 2 = vA 2 + 2a t (sB - sA) vB 2 = 302 + 2( - 0.25)(412 - 0) vB = 26.34 m>s Radius of Curvature: x

y = 200e1000 dy x = 0.2e1000 dx d2y 2

dx

r =

= 0.2 A 10-3 B e1000 x

B1+ a

2

dy 2 b R dx d2y

dx2

2

2

C 1 + £ 0.2e

3>2

≥ S

x 1000

=

2 0.2 A 10 B e -3

x 1000

2

3>2

6

= 3808.96 m

x = 400 m

Acceleration: # a t = v = - 0.25 m>s2 an =

v2 26.342 = = 0.1822 m>s2 r 3808.96

The magnitude of the train’s acceleration at B is a = 2a2t + a2n = 2 A - 0.5 B 2 + 0.18222 = 0.309 m>s2

Ans.

Ans: a = 0.309 m>s2 159

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12–151. The particle travels with a constant speed of 300 mm>s along the curve. Determine the particle’s acceleration when it is located at point (200 mm, 100 mm) and sketch this vector on the curve.

y (mm)

y

20(103) x

SOLUTION n = 300 mm>s at =

v

dn = 0 dt

P x (mm)

20(103) y = x dy 20(103) ` = = - 0.5 dx x = 200 x2 d2y dx

2

`

x = 200

=

40(103) x3 3

r =

2 2 C 1 + A dy dx B D

an =

2

` ddxy2 `

= 5(10-3) 3

=

[1 + ( - 0.5)2]2

` 5(10 - 3) `

= 279.5 mm

(300)2 n2 = 322 mm>s2 = r 279.5

a = 2a2t + a2n = 2(0)2 + (322)2 = 322 mm>s2 Since

Ans.

dy = -0.5, dx

u = tan - 1(- 0.5) = 26.6° g

Ans.

Ans: a = 322 mm>s2 u = 26.6° g 160

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*12–152. A particle P travels along an elliptical spiral path such that its position vector r is defined by r = 52 cos10.1t2i + 1.5 sin10.1t2j + 12t2k6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. When t = 8 s, determine the coordinate direction angles a, b, and g, which the binormal axis to the osculating plane makes with the x, y, and z axes. Hint: Solve for the velocity vP and acceleration a P of the particle in terms of their i, j, k components. The binormal is parallel to vP * a P . Why?

z

P r

SOLUTION

y

rP = 2 cos (0.1t)i + 1.5 sin (0.1t)j + 2tk # vP = r = - 0.2 sin (0.1t)i + 0.15 cos (0.1t)j + 2k

x

$ aP = r = -0.02 cos 10.1t2i - 0.015 sin 10.1t2j When t = 8 s, vP = -0.2 sin (0.8 rad)i + 0.15 cos (0.8 rad)j + 2k = -0.143 47i + 0.104 51j + 2k aP = -0.02 cos (0.8 rad)i - 0.015 sin (0.8 rad)j = -0.013 934i - 0.010 76 j Since the binormal vector is perpendicular to the plane containing the n–t axis, and ap and vp are in this plane, then by the definition of the cross product, i 3 b = vP * aP = - 0.14 347 - 0.013 934

j k 0.104 51 2 3 = 0.021 52i - 0.027 868j + 0.003k -0.010 76 0

b = 210.0215222 + 1- 0.02786822 + 10.00322 = 0.035 338 ub = 0.608 99i - 0.788 62j + 0.085k a = cos - 1(0.608 99) = 52.5°

Ans.

b = cos - 1(-0.788 62) = 142°

Ans.

g = cos - 1(0.085) = 85.1°

Ans.

Note: The direction of the binormal axis may also be specified by the unit vector ub¿ = - ub, which is obtained from b¿ = ap * vp. Ans.

For this case, a = 128°, b = 37.9°, g = 94.9°

Ans: a b g a 161

= = = =

52.5° 142° 85.1° 128°, b = 37.9°, g = 94.9°

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12–153. The motion of a particle is defined by the equations x = (2t + t2) m and y = (t2) m, where t is in seconds. Determine the normal and tangential components of the particle’s velocity and acceleration when t = 2 s.

SOLUTION Velocity: Here, r =

E A 2 t + t 2 B i + t2 j F m.To determine the velocity v, apply Eq. 12–7. v =

dr = {(2 + 2t) i + 2tj } m>s dt

When t = 2 s, v = [2 + 2(2)]i + 2(2)j = {6i + 4j} m>s. Then v = 262 + 42 = 7.21 m>s. Since the velocity is always directed tangent to the path, vn = 0

and

Ans.

vt = 7.21 m>s

The velocity v makes an angle u = tan-1

4 = 33.69° with the x axis. 6

Acceleration: To determine the acceleration a, apply Eq. 12–9. a =

dv = {2i + 2j} m>s2 dt

Then a = 222 + 22 = 2.828 m>s2 The acceleration a makes an angle f = tan-1 figure, a = 45° - 33.69 = 11.31°. Therefore,

2 = 45.0° with the x axis. From the 2

an = a sin a = 2.828 sin 11.31° = 0.555 m>s2

Ans.

at = a cos a = 2.828 cos 11.31° = 2.77 m>s2

Ans.

Ans: vn = 0 vt = 7.21 m>s an = 0.555 m>s2 at = 2.77 m>s2 162

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12–154. If the speed of the crate at A is 15 ft> s, which is increasing at # a rate v = 3 ft>s2, determine the magnitude of the acceleration of the crate at this instant.

y y  1 x2 16

A

x

SOLUTION

10 ft

Radius of Curvature: y =

1 2 x 16

dy 1 = x dx 8 d2y 1 = 8 dx2 Thus,

r =

B1 + a `

dy 2 3>2 b R dx

d2y dx2

=

B1 + ¢ x≤ R

`

1 8

2

1 ` ` 8

3>2

4

= 32.82 ft x = 10 ft

Acceleration: # at = v = 3ft>s2 an =

v2 152 = = 6.856 ft>s2 r 32.82

The magnitude of the crate’s acceleration at A is a = 2at 2 + an2 = 232 + 6.8562 = 7.48 ft>s2

Ans.

Ans: a = 7.48 ft>s2 163

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12–155. A particle is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by u = cos 2t, where u is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when u = 30°.

SOLUTION When u =

p 6

rad,

p 6

= cos 2t

t = 0.5099 s

# du = - 2 sin 2t 2 u = = - 1.7039 rad>s dt t = 0.5099 s $ d2u = - 2.0944 rad>s2 u = 2 = - 4 cos 2t 2 dt t = 0.5099 s r = 4

# r = 0

$ r = 0

# $ ar = r - ru2 = 0 - 4(- 1.7039)2 = - 11.6135 in.>s2 $ ## au = ru + 2ru = 4(- 2.0944) + 0 = - 8.3776 in.>s2 a =

a2r + a2u =

( -11.6135)2 + ( - 8.3776)2 = 14.3 in. s2

Ans.

Ans: a = 14.3 in.>s2 164

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*12–156. For a short time a rocket travels up and to the right at a constant speed of 800 m>s along the parabolic path y = 600 - 35x2. Determine the radial and transverse components of velocity of the rocket at the instant u = 60°, where u is measured counterclockwise from the x axis.

Solution y = 600 - 35x2 # # y = - 70xx dy = -70x dx y tan 60° = x y = 1.732051x 1.732051x = 600 - 35x2 x2 + 0.049487x - 17.142857 = 0 Solving for the positive root, x = 4.1157 m tan u′ =

dy = - 288.1 dx

u′ = 89.8011° f = 180° - 89.8011° - 60° = 30.1989° vr = 800 cos 30.1989° = 691 m>s

Ans.

vu = 800 sin 30.1989° = 402 m>s

Ans.

Ans: vr = 691 m>s vu = 402 m>s 165

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12–157. A particle moves along a path defined by polar coordinates r = (2et) ft and u = (8t2) rad, where t is in seconds. Determine the components of its velocity and acceleration when t = 1 s.

Solution When t = 1 s, r = 2e t = 5.4366 # r = 2e t = 5.4366 $ r = 2e t = 5.4366 u = 8t 2 # u = 16t = 16 $ u = 16 vr = r = 5.44 ft>s Ans. # vu = ru = 5.4366(16) = 87.0 ft>s Ans. # $ ar = r - r(u)2 = 5.4366 - 5.4366(16)2 = - 1386 ft>s2 Ans. $ ## au = ru + 2ru = 5.4366(16) + 2(5.4366)(16) = 261 ft>s2 Ans.

Ans: vr vu  ar au 166

= = = =

5.44 ft>s 87.0 ft>s - 1386 ft>s2 261 ft>s2

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12–158. An airplane is flying in a straight line with a velocity of 200 mi>h and an acceleration of 3 mi>h2. If the propeller has a diameter of 6 ft and is rotating at an angular rate of 120 rad>s, determine the magnitudes of velocity and acceleration of a particle located on the tip of the propeller.

SOLUTION vPl = ¢

200 mi 5280 ft 1h ≤¢ ≤¢ ≤ = 293.3 ft>s h 1 mi 3600 s

aPl = ¢

3 mi 5280 ft 1h 2 = 0.001 22 ft>s2 2 ≤ ¢ 1 mi ≤ ¢ 3600 s ≤ h

vPr = 120(3) = 360 ft>s v = 2v2Pl + v2Pr = 2(293.3)2 + (360)2 = 464 ft>s aPr =

Ans.

(360)2 v2Pr = = 43 200 ft>s2 r 3

a = 2a2Pl + a2Pr = 2(0.001 22)2 + (43 200)2 = 43.2(103) ft>s2

Ans.

Ans: v = 464 ft>s a = 43.2 ( 103 ) ft>s2 167

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12–159. The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m>s and its acceleration is 7 m>s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical components.

z A

6m O

Solution The position of the washer can be defined using the cylindrical coordinate system (r, u and z) as shown in Fig. a. Since u is constant, there will be no transverse component for v and a. The velocity and acceleration expressed as Cartesian vectors are v = va a = aa -

3m

x

2m

y

(0 - 2)i + (0 - 3)j + (0 - 6)k rAO b = 28c d = { - 8i - 12j - 24k} m>s rAO 2(0 - 2)2 + (0 - 3)2 + (0 - 6)2

(0 - 2)i + (0 - 3)j + (0 - 6)k rAO b = 7c d = { - 2i - 3j - 6k} m2 >s rAO 2(0 - 2)2 + (0 - 3)2 + (0 - 6)2

2i + 3j rOB 2 3 = = i + j 2 2 rOB 22 + 3 213 213 uz = k ur =

Using vector dot product vr = v # ur = ( -8 i - 12 j - 24 k) # a

2 3 2 3 i + j b = -8 a b + c -12 a b d = -14.42 m>s 113 113 113 113

vz = v # uz = ( -8 i - 12 j - 24 k) # (k) = - 24.0 m>s

ar = a # ur = ( -2 i - 3 j - 6 k) # a

2 3 2 3 i + j b = -2 a b + c -3 a b d = -3.606 m>s2 113 113 113 113

az = a # uz = ( -2 i - 3 j - 6 k) # k = - 6.00 m>s2 Thus, in vector form v = { - 14.2 ur - 24.0 uz} m>s

Ans.

2

Ans.

a = { - 3.61 ur - 6.00 uz} m>s

These components can also be determined using trigonometry by first obtain angle f shown in Fig. a. OA = 222 + 32 + 62 = 7 m  OB = 222 + 32 = 113

Thus,

sin f =

6 113 and cos f = .  Then 7 7

vr = - v cos f = -28 a

113 b = - 14.42 m>s 7

6 vz = -v sin f = -28 a b = - 24.0 m>s 7 ar = - a cos f = -7 a

113 b = - 3.606 m>s2 7

6 az = - a sin f = -7 a b = - 6.00 m>s2 7

Ans:

168

v = 5 -14.2ur - 24.0u z 6 m>s a = 5 - 3.61ur - 6.00u z 6 m>s2

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*12–160. A velocity of # radar gun at O rotates with the angular $ 2 u = 0.1 rad>s and angular acceleration of u = 0.025 rad>s , at the instant u = 45°, as it follows the motion of the car traveling along the circular road having a radius of r = 200 m. Determine the magnitudes of velocity and acceleration of the car at this instant. r  200 m u O

SOLUTION Time Derivatives: Since r is constant, # $ r = r = 0 Velocity: # vr = r = 0 # vu = ru = 200(0.1) = 20 m>s Thus, the magnitude of the car’s velocity is v = 2vr2 + vu2 = 202 + 202 = 20 m>s

Ans.

Acceleration: # # ar = r - ru2 = 0 - 200(0.12) = - 2 m>s2 $ # # au = r u + 2ru = 200(0.025) + 0 = 5 m>s2 Thus, the magnitude of the car’s acceleration is a = 2ar2 + au2 = 2(-2)2 + 52 = 5.39 m>s2

Ans.

Ans: v = 20 m>s a = 5.39 m>s2 169

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12–161. If a particle moves along a path such that r = 12 cos t2 ft and u = 1t>22 rad, where t is in seconds, plot the path r = f1u2 and determine the particle’s radial and transverse components of velocity and acceleration.

SOLUTION r = 2 cos t u =

t 2

# r = - 2 sin t # 1 u = 2

$ r = - 2 cos t

$ u = 0

# vr = r = -2 sin t

Ans.

# 1 vu = ru = (2 cos t ) a b = cos t 2

Ans.

#2 1 2 5 $ ar = r - ru = -2 cos t - (2cos t) a b = - cos t 2 2

Ans.

$ 1 ## au = ru + 2ru = 2 cos t102 + 21- 2 sin t2 a b = - 2 sin t 2

Ans.

Ans: vr = - 2 sin t vu = cos t 5 ar = - cos t 2 au = - 2 sin t 170

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12–162. If a particle moves along a path such that r = (eat) m and u = t, where t is in seconds, plot the path r = f(u), and determine the particle’s radial and transverse components of velocity and acceleration.

Solution r = e at u = t

#

# r = ae at # u = 1

$ r = a2e et $ u = 0

vr = r = ae at # va = ru = e at(1) = e at # $ ar = r - r u 2 = a2e at - e at(1)2 = e at(a2 - 1) $ ## au = r u + 2 r u = e at(0) + 2(ae at)(1) = 2ae at

Ans. Ans. Ans. Ans.

Ans: vr vu ar au 171

= = = =

ae at e at e at ( a2 - 1 ) 2ae at

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12–163. The car travels along the circular curve having a radius r = # 400 ft. At the instant shown, its angular rate of rotation is $ u = 0.025 rad>s, which is decreasing at the rate u = -0.008 rad>s2. Determine the radial and transverse components of the car’s velocity and acceleration at this instant and sketch these components on the curve.

r

400 ft

. u

SOLUTION r = 400 # u = 0.025 vr vu ar au

# r = 0

$ r = 0

u = -0.008

# = r = 0 # = ru = 400(0.025) = 10 ft>s # $ = r - r u2 = 0 - 400(0.025)2 = - 0.25 ft>s2 $ # # = r u + 2 r u = 400(- 0.008) + 0 = -3.20 ft>s2

Ans. Ans. Ans. Ans.

Ans: vr = vu = ar = au = 172

0 10 ft>s - 0.25 ft>s2 - 3.20 ft>s2

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*12–164. The car travels along the circular curve of radius r = 400 ft of v = 30 ft>s. Determine the angular with a constant speed # rate of rotation u of the radial line r and the magnitude of the car’s acceleration. r

400 ft

. u

SOLUTION r = 400 ft # vr = r = 0

# r = 0

$ r = 0

# # vu = r u = 400 a u b

# 2 v = 6(0)2 + a 400 u b = 30 # u = 0.075 rad>s $ u = 0 # $ ar = r - r u2 = 0 - 400(0.075)2 = - 2.25 ft>s2 $ # # au = r u + 2 r u = 400(0) + 2(0)(0.075) = 0

Ans.

a = 2( -2.25)2 + (0)2 = 2.25 ft>s2

Ans.

Ans: # u = 0.075 rad>s a = 2.25 ft>s2 173

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12–165. The time rate of change of acceleration is referred to as the jerk, which is often used as a means of measuring passenger # discomfort. Calculate this vector, a, in terms of its cylindrical components, using Eq. 12–32.

SOLUTION $ $ # ## $ a = a r - ru 2 b ur + a ru + 2ru b uu + z uz #$ # # $# $ ### ### $ ## $ #$ $# #$ ## # $# a = a r - ru2 - 2ruu bur + a r - ru2 b u r + a r u + ru + 2ru + 2r u buu + a ru + 2ru buu + z uz + zu z # But, ur = uuu

# # uu = - uur

# uz = 0

Substituting and combining terms yields #$ # $# # ### #$ $# ### # a = a r - 3ru 2 - 3ruu b ur + a 3r u + ru + 3ru - ru3 buu + a z buz

Ans.

Ans: % # # # #$ a = ( r - 3ru 2 - 3ru u )ur % # #$ #$ $ # + ( 3r u + r u + 3r u - r u 3 ) uu + (z )uz 174

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12–166. A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by u = sin 3t, where u is in radians, the argument for the sine are in radians , and t is in seconds. Determine the acceleration of the particle at u = 30°. The particle starts from rest at u = 0°.

SOLUTION r = 6 in.,

# r = 0,

$ r = 0

u = sin 3t # u = 3 cos 3t $ u = -9 sin 3t At u = 30°, 30° p = sin 3t 180° t = 10.525 s Thus, # u = 2.5559 rad>s $ u = -4.7124 rad>s2 # $ ar = r - r u2 = 0 - 6(2.5559)2 = -39.196 $ ## au = r u + 2ru = 6( - 4.7124) + 0 = - 28.274 a = 2(- 39.196)2 + ( - 28.274)2 = 48.3 in.>s2

Ans.

Ans: a = 48.3 in.>s2 175

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12–167. The slotted link is pinned # at O, and as a result of the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = 10.4 u2 m, where u is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant u = p>3 rad.

0.5 m r · u

SOLUTION # u = 3 rad>s

At u =

p , 3

P

r

3 rad/s

0.4 u

u

r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u

O

r = 0.4189 # r = 0.4(3) = 1.20 $ r = 0.4(0) = 0

# v = r = 1.20 m>s # vu = r u = 0.4189(3) = 1.26 m>s # $ ar = r - ru2 = 0 - 0.4189(3)2 = - 3.77 m>s2 $ ## au = r u + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2

Ans. Ans. Ans. Ans.

Ans: vr = 1.20 m>s vu = 1.26 m>s ar = - 3.77 m>s2 au = 7.20 m>s2 176

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*12–168. For a short time the bucket of the backhoe traces the path of the cardioid r = 25(1 − cos u) ft. Determine the magnitudes of the velocity and acceleration of the bucket when # u = 120° if the boom is rotating with an angular velocity of u = 2 rad>s $ and an angular acceleration of u = 0.2 rad>s2 at the instant shown. r

u  120

Solution r = 25(1 - cos u) = 25(1 - cos 120°) = 37.5 ft # # r = 25 sin uu = 25 sin 120°(2) = 43.30 ft>s # $ $ r = 25[cos uu 2 + sin uu ] = 25[cos 120°(2)2 + sin 120°(0.2)] = - 45.67 ft>s2 # vr = r = 43.30 ft>s # vu = ru = 37.5(2) = 75 ft>s v = 2v2r + v2u = 243.302 + 752 = 86.6 ft>s # $ ar = r - ru 2 = - 45.67 - 37.5(2)2 = - 195.67 ft>s2 $ # au = ru + 2ru = 37.5(0.2) + 2(43.30)(2) = 180.71 ft>s2

Ans.

a = 2a2s + a2u = 2( - 195.67)2 + 180.712 = 266 ft>s2

Ans.

Ans: v = 86.6 ft>s a = 266 ft>s2 177

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12–169. The slotted link is pinned # at O, and as a result of the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = 10.4 u2 m, where u is in radians. Determine the velocity and acceleration of the particle at the instant it leaves the slot in the link, i.e., when r = 0.5 m.

0.5 m

P r

· u

SOLUTION

r

3 rad/s

0.4u

u

r = 0.4 u # # r = 0.4 u $ $ r = 0.4 u # u = 3 $ u = 0

O

At r = 0.5 m, u =

0.5 = 1.25 rad 0.4

# r = 1.20 $ r = 0 # vr = r = 1.20 m>s # vu = r u = 0.5(3) = 1.50 m>s # $ ar = r - r(u)2 = 0 - 0.5(3)2 = - 4.50 m>s2 $ ## au = ru + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2

Ans. Ans. Ans. Ans.

Ans: vr = vu = ar = au = 178

1.20 m>s 1.50 m>s - 4.50 m>s2 7.20 m>s2

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12–170. A particle moves in the x–y plane such that its position is defined by r = 52ti + 4t2j6 ft, where t is in seconds. Determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s.

SOLUTION r = 2ti + 4 t2j|t = 2 = 4i + 16j v = 2i + 8 tj|t = 2 = 2i + 16j a = 8j u = tan - 1 a

16 b = 75.964° 4

v = 2(2)2 + (16)2 = 16.1245 ft>s f = tan - 1 a

16 b = 82.875° 2

a = 8 ft>s2 f - u = 6.9112° vr = 16.1245 cos 6.9112° = 16.0 ft>s

Ans.

vu = 16.1245 sin 6.9112° = 1.94 ft>s

Ans.

d = 90° - u = 14.036° ar = 8 cos 14.036° = 7.76 ft>s2

Ans.

au = 8 sin 14.036° = 1.94 ft>s2

Ans.

Ans: vr = vu = ar = au = 179

16.0 ft>s 1.94 ft>s 7.76 ft>s2 1.94 ft>s2

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12–171. At the instant shown, the man # is twirling a hose over his head with an$angular velocity u = 2 rad>s and an angular acceleration u = 3 rad>s2. If it is assumed that the hose lies in a horizontal plane, and water is flowing through it at a constant rate of 3 m>s, determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r = 1.5 m.

· u  2 rad/s ·· u  3 rad/s2 r  1.5 m u

Solution r = 1.5 # r = 3 $ r = 0 # u = 2 $ u = 3 # vr = r = 3 # vu = ru = 1.5(2) = 3 v = 2(3)2 + (3)2 = 4.24 m>s # $ ar = r - r(u)2 = 0 - 1.5(2)2 = 6 $ # # au = r u + 2ru = 1.5(3) + 2(3)(2) = 16.5

Ans.

a = 2(6)2 + (16.5)2 = 17.6 m>s2

Ans.

Ans: v = 4.24 m>s a = 17.6 m>s2 180

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*12–172. The rod OA rotates clockwise with a constant angular velocity of 6 rad>s. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 200(2 − cos u) mm. Determine the speed of the slider blocks at the instant u = 150°.

A B r

6 rad/s u O

400 mm

600 mm 200 mm

Solution Velocity. Using the chain rule, the first and second time derivatives of r can be determined. r = 200(2 - cos u) #

#

#

r = 200 (sin u) u = 5 200 (sin u) u 6 mm>s #

$

$

r = 5 200[(cos u)u 2 + (sin u)u]6 mm>s2

The radial and transverse components of the velocity are #

#

vr = r = 5 200 (sin u)u 6 mm>s #

#

#

vu = ru = 5200(2 - cos u)u 6 mm>s

#

Since u is in the opposite sense to that of positive u, u = - 6 rad>s. Thus, at u = 150°, vr = 200(sin 150°)(- 6) = -600 mm>s vu = 200(2 - cos 150°)( -6) = -3439.23 mm>s Thus, the magnitude of the velocity is v = 2v2r + v2u   2(- 600)2 + (- 3439.23)2 = 3491 mm>s = 3.49 m>s

Ans.

These components are shown in Fig. a

Ans: v = 3.49 m>s 181

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12–173. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–172 when u = 150°.

A B r

6 rad/s u O

400 mm

600 mm 200 mm

Solution Acceleration. Using the chain rule, the first and second time derivatives of r can be determined r = 200(2 - cos u) #

#

#

r = 200 (sin u)u = 5 200 (sin u)u 6 mm>s #

$

#

$

r = 5 200[(cos u)u 2 + (sin u)u] 6 mm>s2 $

#

Here, since u is constant, u = 0. Since u is in the opposite sense to that of positive u, # u = - 6 rad>s. Thus, at u = 150° r = 200(2 - cos 150°) = 573.21 mm #

r = 200(sin 150°)( -6) = - 600 mm>s $

r = 200 3 (cos 150°)( - 6)2 + sin 150°(0) 4 = - 6235.38 mm>s2

The radial and transverse components of the acceleration are #

$

ar = r - ru 2 = - 6235.38 - 573.21 ( - 6)2 = - 26870.77 mm>s2 = - 26.87 m>s2 $

# #

au = ru + 2ru = 573.21(0) + 2( - 600)( - 6) = 7200 mm>s2 = 7.20 m>s2 Thus, the magnitude of the acceleration is a = 2a2r + a2u = 2( - 26.87)2 + 7.202 = 27.82 m>s2 = 27.8 m>s2

Ans.

These components are shown in Fig. a.

Ans: a = 27.8 m>s2 182

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12–174. A double collar C is pin connected together such that one collar slides over a fixed rod and the other slides over a rotating rod. If the geometry of the fixed rod for a short distance can be defined by a lemniscate, r2 = (4 cos 2u) ft2, determine the collar’s radial and transverse components of velocity and acceleration at the instant u = 0° as shown. Rod # OA is rotating at a constant rate of u = 6 rad>s.

r2  4 cos 2 u · u  6 rad/s O

r

A C

Solution r 2 = 4 cos 2u

#

#

rr = - 4 sin 2u u $

$

#

#

rr = r 2 = -4 sin 2u u - 8 cos 2u u 2 $

#

when u = 0, u = 6, u = 0 $

r = 2, r = 0, r = - 144 #

vr = r = 0

Ans.

vu = ru = 2(6) = 12 ft>s

Ans.

#

#

$

ar = r - ru 2 = - 144 - 2(6)2 = - 216 ft>s2 $

Ans.

# #

Ans.

au = ru + 2ru = 2(0) + 2(0)(6) = 0

Ans: vr vu ar au 183

= = = =

0 12 ft>s - 216 ft>s2 0

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12–175. A block moves outward along the slot in the platform with # a speed of r = 14t2 m>s, where t is in seconds. The platform rotates at a constant rate of 6 rad/s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = 1 s.

·

θ = 6 rad/s r

θ

SOLUTION # r = 4t|t = 1 = 4 # $ u = 6 u = 0 1

L0

dr =

$ r = 4

1

L0

4t dt

r = 2t2 D 10 = 2 m

# # v = 3 A r B 2 + A ru B 2 = 2 (4)2 + [2(6)]2 = 12.6 m>s # $ $ ## a = r - ru2 2 + ru + 2 ru 2 = [4 - 2(6)2 ]2 + [0 + 2(4)(6)]2

Ans. Ans.

= 83.2 m s2

Ans: v = 12.6 m>s a = 83.2 m>s2 184

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*12–176. The car travels around the circular track with a constant speed of 20 m>s. Determine the car’s radial and transverse components of velocity and acceleration at the instant u = p>4 rad.

r  (400 cos u) m r

u

Solution v = 20 m>s u =

p = 45° 4

r = 400 cos u #

#

r = -400 sin u u #

$

$

r = - 400(cos u(u)2 + sin u u) #

#

v2 = (r)2 + (ru)2 #

#$

# #

$

0 = rr + ru(ru + ru) Thus r = 282.84 #

#

(20)2 = [ - 400 sin 45° u]2 + [282.84 u]2 #

u = 0.05

#

r = -14.14 $

$

0 = - 14.14[ - 400(cos 45°)(0.05)2 + sin 45° u]  + 282.84(0.05)[- 14.14(0.05) + 282.84u ] $

u = 0 $

r = -0.707 #

Ans.

vr = r = - 14.1 m>s #

Ans.

vu = ru = 282.84(0.05) = 14.1 m>s #

$

ar = r - r (u)2 = - 0.707 - 282.84(0.05)2 = - 1.41 m>s2 $

# #

#

Ans.

2

Ans.

au = ru + 2ru = u + 2( - 14.14)(0.05) = - 1.41 m>s 

Ans: vr vu ar au 185

= = = =

- 14.1 m>s 14.1 m>s - 1.41 m>s2 - 1.41 m>s2

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12–177. The car travels around the circular track such that its transverse component is u = (0.006t2) rad, where t is in seconds. Determine the car’s radial and transverse components of velocity and acceleration at the instant t = 4 s.

r  (400 cos u) m r

u

Solution u = 0.006 t 2 0 t = 4 = 0.096 rad = 5.50° #

u = 0.012 t 0 t = 4 = 0.048 rad>s $

u = 0.012 rad>s2 r = 400 cos u #

#

r = -400 sin u u $

#

$

r = - 400(cos u (u)2 + sinu u) At u = 0.096 rad r = 398.158 m #

r = -1.84037 m>s $

r = - 1.377449 m>s2 #

Ans.

vr = r = - 1.84 m>s #

Ans.

vu = r u = 398.158(0.048) = 19.1 m>s #

$

2

2

2

ar = r - r (u) = - 1.377449 - 398.158(0.048) = - 2.29 m>s 

Ans.

au = r u = 2r u = 398.158 (0.012) + 2( - 1.84037)(0.048) = 4.60 m>s2

Ans.

$

# #

Ans: vr vu ar au 186

= = = =

-1.84 m>s 19.1 m>s - 2.29 m>s2 4.60 m>s2

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12–178. The car travels along a road which for a short distance is defined by r = (200>u) ft, where u is in radians. If it maintains a constant speed of v = 35 ft>s, determine the radial and transverse components of its velocity when u = p>3 rad.

θ r

SOLUTION r =

200 600 ` = ft p u u = p>3 rad

200 # 1800 # # = - 2 u r = - 2 u` u p u = p>3 rad

1800 # # vr = r = - 2 u p

# 600 # vu = ru = u p

v2 = v2r + v2u 352 = a -

600 # 2 1800 # 2 u b + a ub p p2

# u = 0.1325 rad>s vr = -

vu =

1800 (0.1325) = -24.2 ft>s p2

Ans.

600 (0.1325) = 25.3 ft>s p

Ans.

Ans: vr = - 24.2 ft>s vu = 25.3 ft>s 187

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12–179. z

A horse on the merry-go-round moves according to the equations r = 8 ft, u = (0.6t) rad, and z = (1.5 sin u) ft, where t is in seconds. Determine the cylindrical components of the velocity and acceleration of the horse when t = 4 s.

z u r

Solution r = 8  u = 0.6 t #

#

$

$

r = 0  u = 0.6 r = 0  u = 0 z = 1.5 sin u #

#

z = 1.5 cos u u

#

$

$

z = - 1.5 sin u (u)2 + 1.5 cos u u At t = 4 s u = 2.4 #

z = - 0.6637 $

z = - 0.3648 vr = 0

Ans.

vu = 4.80 ft>s

Ans. Ans.

vz = -0.664 ft>s 2

2

ar = 0 - 8(0.6) = - 2.88 ft>s 

Ans.

au = 0 + 0 = 0

Ans. 2

Ans.

az = -0.365 ft>s 

Ans: vr vu vz ar au az 188

= = = = = =

0 4.80 ft>s - 0.664 ft>s - 2.88 ft>s2 0 - 0.365 ft>s2

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*12–180. z

A horse on the merry-go-round moves according to the # equations r = 8 ft, u = 2 rad>s and z = (1.5 sin u) ft, where t is in seconds. Determine the maximum and minimum magnitudes of the velocity and acceleration of the horse during the motion. z u r

Solution r = 8 #

#

$

$

r = 0  u = 2 r = 0  u = 0 z = 1.5 sin u

#

#

z = 1.5 cos u u #

$

$

z = - 1.5 sin u (u)2 + 1.5 cos u u #

vr = r = 0 #

vu = r u = 8(2) = 16 ft>s #

(vz)max = z = 1.5(cos 0°)(2) = 3 ft>s #

(vz)min = z = 1.5(cos 90°)(2) = 0 vmax = 2(16)2 + (3)2 = 16.3 ft>s 2

vmin = 2(16) + (0) = 16 ft>s #

$

2

Ans.

2

2

Ans. 2

ar = r - r(u) = 0 - 8(2) = - 32 ft>s # #

$

au = r u + 2 ru = 0 + 0 = 0 $

(az)max = z = -1.5(sin 90°)(2)2 = - 6 $

(az)min = z = - 1.5(sin 0°)(2)2 = 0 amax = 2( - 32)2 + (0)2 + ( - 6)2 = 32.6 ft>s2

Ans.

amin = 2( - 32)2 + (0)2 + (0)2 = 32 ft>s2

Ans.

Ans: vmax vmin amax amin 189

= = = =

16.3 ft>s 16 ft>s 32.6 ft>s2 32 ft>s2

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12–181. If the slotted arm AB rotates# counterclockwise with a constant angular velocity of u = 2 rad>s, determine the magnitudes of the velocity and acceleration of peg P at u = 30°. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB.

D

P r

(4 sec u) ft

u

A

SOLUTION

C 4 ft

Time Derivatives: r = 4 sec u # u = 2 rad>s $ u = 0

# # r = (4 secu(tanu)u ) ft>s # # # $ $ r = 4 [secu(tanu)u + u (sec u (sec2u)u + tan u secu(tan u)u )] # # = 4[secu(tanu)u + u2(sec3u + tan2u secu)] ft>s2 When u = 30°, r|u = 30° = 4 sec 30° = 4.619 ft # r |u = 30° = (4 sec30° tan30°)(2) = 5.333 ft>s

$ r |u = 30° = 4[0 + 2 2(sec3 30° + tan2 30° sec 30°)] = 30.79 ft>s2 Velocity: # vr = r = 5.333 ft>s

# vu = ru = 4.619(2) = 9.238 ft>s

Thus, the magnitude of the peg’s velocity is v = 2vr2 + vu2 = 25.3332 + 9.2382 = 10.7 ft>s

Ans.

Acceleration: # $ a r = r - ru2 = 30.79 - 4.619(2 2) = 12.32 ft>s2 $ ## au = ru + 2ru = 0 + 2(5.333)(2) = 21.23 ft>s2 Thus, the magnitude of the peg’s acceleration is a = 2ar2 + a u2 = 212.32 2 + 21.232 = 24.6 ft>s2

Ans.

Ans: v = 10.7 ft>s a = 24.6 ft>s2 190

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12–182. D

The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When u = 30°, the angular velocity and angular acceleration of arm AB are $ # u = 2 rad>s and u = 3 rad>s2 , respectively. Determine the magnitudes of the velocity and acceleration of the peg P at this instant.

B P r  (4 sec u) ft

u

A

SOLUTION

C 4 ft

Time Derivatives: r = 4 sec u # # r = (4 secu(tanu)u ) ft>s # # # $ $ r = 4[secu(tanu)u + u (secusec2uu + tanu secu(tanu)u )] # $ = 4[secu(tanu)u + u 2(sec3u° + tan2u°secu°)] ft>s2

# u = 2 rad>s $ u = 3 rad>s2

When u = 30°, r|u = 30° = 4 sec 30° = 4.619 ft # r|u = 30° = (4 sec 30° tan 30°)(2) = 5.333 ft>s $ r|u = 30° = 4[(sec 30° tan 30°)(3) + 22(sec3 30° + tan2 30° sec 30°)] = 38.79 ft>s2 Velocity: # vr = r = 5.333 ft>s

# vu = ru = 4.619(2) = 9.238 ft>s

Thus, the magnitude of the peg’s velocity is v = 2vr2 + vu2 = 25.3332 + 9.2382 = 10.7 ft>s

Ans.

Acceleration: # $ ar = r - ru2 = 38.79 - 4.619(22) = 20.32 ft>s2 $ ## au = ru + 2ru = 4.619(3) + 2(5.333)(2) = 35.19 ft>s2 Thus, the magnitude of the peg’s acceleration is a = 2ar2 + au2 = 220.322 + 35.192 = 40.6 ft>s2

Ans.

Ans: v = 10.7 ft>s a = 40.6 ft>s2 191

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12–183. A truck is traveling along the horizontal circular curve of v = 20 m>s. radius r = 60 m with a constant speed # Determine the angular rate of rotation u of the radial line r and the magnitude of the truck’s acceleration.

· u r  60 m u

SOLUTION r = 60 # r = 0 $ r = 0 n = 20 # nr = r = 0 # # nu = r u = 60 u n = 2(nr)2 + (nu)2 # 20 = 60 u # u = 0.333 rad>s # $ ar = r - r(u)2

Ans.

= 0 - 60(0.333)2 = - 6.67 m>s2 $ # # au = ru + 2r u $ = 60u Since # v = ru $ # ## v = ru + ru $ 0 = 0 + 60 u $ u = 0 Thus, au = 0 a = ` ar ` = 6.67 m>s2

Ans.

Ans:

#

u = 0.333 rad>s a = 6.67 m>s2 192

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*12–184. A truck is traveling along the horizontal circular curve of radius r = 60 m with a speed of 20 m>s which is increasing at 3 m>s2. Determine the truck’s radial and transverse components of acceleration.

· u r  60 m u

SOLUTION r = 60 at = 3 m>s2 an =

(20)2 n2 = 6.67 m>s2 = 60 r

ar = -an = -6.67 m>s2

Ans.

au = at = 3 m>s2

Ans.

Ans: ar = - 6.67 m>s2 au = 3 m>s2 193

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12–185. The rod OA rotates # counterclockwise with a constant angular velocity of u = 5 rad>s. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 100(2 − cos u) mm. Determine the speed of the slider blocks at the instant u = 120°.

· u  5 rad/s

y

A B r u x

O

Solution # u = 5

r = 100(2 - cos u) #

#

r  100 (2  cos u) mm

r = 100 sin uu = 500 sin u #

$

r = 500 cos uu = 2500 cos u At u = 120°, #

vr = r = 500 sin 120° = 433.013 #

vu = ru = 100 (2 - cos 120°)(5) = 1250 v = 2(433.013)2 + (1250)2 = 1322.9 mm>s = 1.32 m>s

Ans.

Ans: v = 1.32 m>s 194

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12–186. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–185 when u = 120°.

· u  5 rad/s

y

A B r u x

O

Solution # u = 5 # u = 0

r = 100(2 - cos u) # # r = 100 sin uu = 500 sin u # $ r = 500 cos uu = 2500 cos u # $ ar = r - ru 2 = 2500 cos u - 100(2 - cos u)(5)2 = 5000(cos 120° - 1) = - 7500 mm>s2 # # as = r u + 2ru = 0 + 2(500 sin u)(5) = 5000 sin 120° = 4330.1 mm>s2 a = 2( - 7500)2 + (4330.1)2 = 8660.3 mm>s2 = 8.66 m>s2

r  100 (2  cos u) mm

Ans.

Ans: a = 8.66 m>s2 195

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12–187. The searchlight on the boat anchored 2000 ft from shore is turned on the automobile, which is traveling along the straight road at a constant speed of 80 ft>s. Determine the angular rate of rotation of the light when the automobile is r = 3000 ft from the boat.

80 ft/s r

u

u

SOLUTION r = 2000 csc u # # r = -2000 csc u ctn u At r = 3000 ft,

2000 ft

u = 41.8103°

# # r = -3354.102 u # # n = 2(r)2 + (r u)2

# (80)2 = [(- 3354.102)2 + (3000)2](u)2 # u = 0.0177778 = 0.0178 rad>s

Ans.

Ans: # u = 0.0178 rad>s 196

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*12–188. If the car in Prob. 12–187 is accelerating at 15 ft >s2 and has a velocity of 80 ft>s at the instant $ r = 3000 ft, determine the required angular acceleration u of the light at this instant.

80 ft/s r

u

u

SOLUTION r = 2000 csc u # # r = - 2000 csc u ctn u u 2000 ft

At r = 3000 ft, u = 41.8103° # # r = - 3354.102 u $ ## au = r u + 2 ru $ au = 3000 u + 2( - 3354.102)(0.0177778)2 Since au = 15 sin 41.8103° = 10 m>s Then, $ u = 0.00404 rad>s2

Ans.

Ans: ## u = 0.00404 rad>s2 197

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12–189. A particle moves along an Archimedean spiral r = (8u) ft, # where u is given in radians. If u = 4 rad>s (constant), determine the radial and transverse components of the particle’s velocity and acceleration at the instant u = p>2 rad. Sketch the curve and show the components on the curve.

y

r

(8 u) ft

r

SOLUTION

u

# $ Time Derivatives: Since u is constant, u = 0. p r = 8u = 8a b = 4p ft 2

# # r = 8 u = 8(4) = 32.0 ft>s

x

$ $ r = 8u = 0

Velocity: Applying Eq. 12–25, we have # v r = r = 32.0 ft>s # vu = ru = 4p (4) = 50.3 ft>s

Ans. Ans.

Acceleration: Applying Eq. 12–29, we have # $ ar = r - ru2 = 0 - 4p A 42 B = - 201 ft>s2 $ # # au = ru + 2r u = 0 + 2(32.0)(4) = 256 ft>s2

Ans. Ans.

Ans: ar = - 201 ft>s2 au = 256 ft>s2 198

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12–190. Solve Prob. $12–189 if the particle has an angular # acceleration u = 5 rad>s2 when u = 4 rad>s at u = p> 2 rad.

y

r

(8 u) ft

r

SOLUTION

u x

Time Derivatives: Here, p r = 8u = 8 a b = 4p ft 2 $ $ r = 8 u = 8(5) = 40 ft>s2

# # r = 8 u = 8(4) = 32.0 ft>s

Velocity: Applying Eq. 12–25, we have # vr = r = 32.0 ft>s # vu = r u = 4p(4) = 50.3 ft>s

Ans. Ans.

Acceleration: Applying Eq. 12–29, we have # $ ar = r - r u2 = 40 - 4p A 42 B = - 161 ft>s2 $ # # au = r u + 2r u = 4p(5) + 2(32.0)(4) = 319 ft>s2

Ans. Ans.

Ans: vr vu ar au 199

= = = =

32.0 ft>s 50.3 ft>s - 161 ft>s2 319 ft>s2

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12–191. The arm of the robot moves so that r = 3 ft is constant, and its grip A moves along the path z = 13 sin 4 u2 ft, where u is in radians. If u = 10.5 t2 rad, where t is in seconds, determine the magnitudes of the grip’s velocity and acceleration when t = 3 s.

A

z r u

SOLUTION u = 0.5 t # u = 0.5 $ u = 0

r = 3

z = 3 sin 2t

# r = 0

# z = 6 cos 2t

$ r = 0

$ z = -12 sin 2t

At t = 3 s, z = -0.8382 # z = 5.761 $ z = 3.353 vr = 0 vu = 3(0.5) = 1.5 vz = 5.761 v = 2(0)2 + (1.5)2 + (5.761)2 = 5.95 ft>s

Ans.

ar = 0 - 3(0.5)2 = - 0.75 au = 0 + 0 = 0 az = 3.353 a = 2(- 0.75)2 + (0)2 + (3.353)2 = 3.44 ft>s2

Ans.

Ans: v = 5.95 ft>s a = 3.44 ft>s2 200

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*12–192. For a short time the arm of the robot is extending at a # constant rate such that r = 1.5 ft>s when r = 3 ft, 2 z = 14t 2 ft, and u = 0.5t rad, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the grip A when t = 3 s.

A

z r u

SOLUTION u = 0.5 t rad # u = 0.5 rad>s $ u = 0

r = 3 ft

z = 4 t2 ft

# r = 1.5 ft>s

# z = 8 t ft>s

$ r = 0

$ z = 8 ft>s2

At t = 3 s, u = 1.5 # u = 0.5 $ u = 0

r = 3

z = 36

# r = 1.5

# z = 24

$ r = 0

$ z = 8

vr = 1.5 vu = 3(0.5) = 1.5 vz = 24 v = 2(1.5)2 + (1.5)2 + (24)2 = 24.1 ft>s

Ans.

ar = 0 - 3(0.5)2 = - 0.75 au = 0 + 2(1.5)(0.5) = 1.5 az = 8 a = 2( -0.75)2 + (1.5)2 + (8)2 = 8.17 ft>s2

Ans.

Ans: v = 24.1 ft>s a = 8.17 ft>s2 201

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12–193. The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the # rotating rod AB. If the angular velocity of AB is given as u = 2 (e0.5 t ) rad>s, where t is in seconds, and the path defined by the fixed rod is r = |(0.4 sin u + 0.2)| m, determine the radial and transverse components of the collar’s velocity and acceleration when t = 1 s. When t = 0, u = 0. Use Simpson’s rule with n = 50 to determine u at t = 1 s.

B

0.6 m

C

r A

u

0.2 m

Solution

0.2 m 0.2 m

# 2 u = e 0.5 t  t = 1 = 1.649 rad>s $ 2 u = e 0.5 t t  t = 1 = 1.649 rad>s2 u =

L0

1

2

e 0.5 t dt = 1.195 rad = 68.47°

r = 0.4 sin u + 0.2 # # r = 0.4 cos u u # $ $ r = - 0.4 sin u u 2 + 0.4 cos u u At t = 1 s, r = 0.5721 # r = 0.2421 $ r = - 0.7697 # vr = r = 0.242 m>s # vu = r u = 0.5721(1.649) = 0.943 m>s # $ ar = r - ru 2 = - 0.7697 - 0.5721(1.649)2

Ans. Ans.

ar = - 2.33 m>s2 $ # # au = ru + 2ru

Ans.

= 0.5721(1.649) + 2(0.2421)(1.649) au = 1.74 m>s2

Ans.

Ans: vr vu ar au 202

= = = =

0.242 m>s 0.943 m>s - 2.33 m>s2 1.74 m>s2

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12–194. The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod AB. If the mechanism is to be designed so that the largest speed given to the collar is # 6 m>s, determine the required constant angular velocity u of rod AB. The path defined by the fixed rod is r = (0.4 sin u + 0.2) m.

B

0.6 m

C

r A

u

0.2 m

Solution

0.2 m 0.2 m

r = 0.4 sin u + 0.2 # # r = 0.4 cos u u # # vr = r = 0.4 cos u u # # vu = r u = (0.4 sin u + 0.2) u v2 = v2r + v3u

# (6)2 = [(0.4 cos u)2 + (0.4 sin u + 0.2)2](u)2 # 36 = [0.2 + 0.16 sin u](u)2 The greatest speed occurs when u = 90°. # u = 10.0 rad>s

Ans.

Ans:

#

u = 10.0 rad>s 203

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12–195. If the end of the cable at A is pulled down with a speed of 2 m>s, determine the speed at which block B rises.

D C 2 m/s

A

SOLUTION Position-Coordinate Equation: Datum is established at fixed pulley D. The position of point A, block B and pulley C with respect to datum are sA, sB, and sC respectively. Since the system consists of two cords, two position-coordinate equations can be derived. (sA - sC) + (sB - sC) + sB = l1

(1)

sB + sC = l2

(2)

B

Eliminating sC from Eqs. (1) and (2) yields sA + 4sB = l1 = 2l2 Time Derivative: Taking the time derivative of the above equation yields (3)

vA + 4vB = 0 Since vA = 2 m>s, from Eq. (3) (+ T)

2 + 4vB = 0 vB = - 0.5 m>s = 0.5 m>s c

Ans.

Ans: vB = 0.5 m>s 204

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*12–196. The motor at C pulls in the cable with an acceleration aC = (3t2) m>s2, where t is in seconds. The motor at D draws in its cable at aD = 5 m>s2. If both motors start at the same instant from rest when d = 3 m, determine (a) the time needed for d = 0, and (b) the velocities of blocks A and B when this occurs.

D B A

C

d=3m

SOLUTION For A: sA + (sA - sC) = l 2vA = vC 2aA = aC = -3t2 aA = -1.5t2 = 1.5t2

:

:

vA = 0.5t3

:

sA = 0.125 t4 For B: aB = 5 m>s2 vB = 5t sB = 2.5t2

;

; ;

Require sA + sB = d 0.125t4 + 2.5t2 = 3 Set u = t2

0.125u2 + 2.5u = 3

The positive root is u = 1.1355. Thus, Ans.

t = 1.0656 = 1.07 s vA = 0.5(1.0656)3 = 0.6050 vB = 5(1.0656) = 5.3281 m>s vA = vB + vA>B 0.6050i = - 5.3281i + vA>B i vA B = 5.93 m s

:

Ans.

Ans: t = 1.07 s vA>B = 5.93 ms>s S 205

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12–197. The pulley arrangement shown is designed for hoisting materials. If BC remains fixed while the plunger P is pushed downward with a speed of 4 ft>s, determine the speed of the load at A. B C P

SOLUTION

4 ft/s A

5 sB + (sB - sA) = l 6 sB - sA = l 6 yB - yA = 0 6(4) = yA Ans.

yA = 24 ft>s

Ans: v = 24 ft>s 206

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12–198. If the end of the cable at A is pulled down with a speed of 5 m>s, determine the speed at which block B rises. A 5 m/s

B

Solution Position Coordinate. The positions of pulley B and point A are specified by position coordinates sB and sA, respectively, as shown in Fig. a. This is a single-cord pulley system. Thus, sB + 2(sB - a) + sA = l (1)

3sB + sA = l + 2a Time Derivative. Taking the time derivative of Eq. (1),

(2)

3vB + vA = 0 Here vA = + 5 m>s, since it is directed toward the positive sense of sA. Thus, 3vB + 5 = 0

vB = - 1.667 m>s = 1.67 m>s c 

Ans.

The negative sign indicates that vB is directed toward the negative sense of sB.

Ans: vB = 1.67 m>s 207

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12–199. Determine the displacement of the log if the truck at C pulls the cable 4 ft to the right.

C

B

SOLUTION 2sB + (sB - sC) = l 3sB - sC = l 3¢sB - ¢sC = 0 Since ¢sC = -4, then 3¢sB = -4 ¢sB = -1.33 ft = 1.33 ft :

Ans.

Ans: ∆sB = 1.33 ft S 208

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*12–200. Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5 s. A

D C B

Solution vB =

6 = 4 m>s c 1.5

sB + (sB - sC) = l1 sC + (sC - sD) = l2 sA + 2 sD = l3 Thus, 2 sB - sC = l1 2 sC - sD = l2 sA + 2 sD = l3 2vA = vC 2vC = vD vA = - 2vD 2(2vB) = vD vA = - 2(4vB) vA = - 8vB Ans.

vA = - 8( - 4) = 32 m>s T

Ans: vA = 32 m>s T 209

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12–201. Starting from rest, the cable can be wound onto the drum of the motor at a rate of vA = (3t2) m>s, where t is in seconds. Determine the time needed to lift the load 7 m. A

D C B

Solution vB =

6 = 4 m>s c 1.5

sB + (sB - sC) = l1 sC + (sC - sD) = l2 sA + 2sD) = l3 Thus, 2sB - sC = l1 2sC - sD = l2 sA + 2sD = l3 2vB = vC 2vC = vD vA = -2vD vA = -8vB 3 t 2 = -8vB vB =

-3 2 t 8 t

L0 8 -1 3 t sB = 8

sB =

-7 =

-3

t 2 dt

-1 3 t 8 Ans.

t = 3.83 s

Ans: t = 3.83 s 210

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12–202. If the end A of the cable is moving at vA = 3 m>s, determine the speed of block B.

C

A

D

vA  3 m/s

B

Solution Position Coordinates. The positions of pulley B, D and point A are specified by position coordinates sB, sD and sA respectively as shown in Fig. a. The pulley system consists of two cords which give (1)

2 sB + sD = l1 and (sA - sD) + (b - sD) = l2

(2)

sA - 2 sD = l2 - b Time Derivative. Taking the time derivatives of Eqs. (1) and (2), we get 2vB + vD = 0

(3)

vA - 2vD = 0

(4)

Eliminate v0 from Eqs. (3) and (4), (5)

vA + 4vB = 0 Here vA = + 3 m>s since it is directed toward the positive sense of sA. Thus 3 + 4vB = 0 vB = - 0.75 m>s = 0.75 m>s d 

Ans.

The negative sign indicates that vD is directed toward the negative sense of s B.

Ans: vB = 0.75 m>s 211

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12–203. Determine the time needed for the load at B to attain a speed of 10 m>s, starting from rest, if the cable is drawn into the motor with an acceleration of 3 m>s2.

A vA C B

Solution Position Coordinates. The position of pulleys B, C and point A are specified by position coordinates sB, sC and sA respectively as shown in Fig. a. The pulley system consists of two cords which gives sB + 2(sB - sC) = l1 3sB - 2sC = l1

(1)

sC + sA = l2

(2)

And

Time Derivative. Taking the time derivative twice of Eqs. (1) and (2), 3aB - 2aC = 0

(3)

aC + aA = 0

(4)

And Eliminate aC from Eqs. (3) and (4) 3aB + 2aA = 0 Here, aA = + 3 m>s2 since it is directed toward the positive sense of sA. Thus, 3aB + 2(3) = 0  aB = - 2 m>s2 = 2 m>s2 c The negative sign indicates that aB is directed toward the negative sense of sB. Applying kinematic equation of constant acceleration, +c

vB = (vB)0 + aBt



10 = 0 + 2 t



t = 5.00 s

Ans.

Ans: t = 5.00 s 212

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*12–204. The cable at A is being drawn toward the motor at vA = 8 m>s. Determine the velocity of the block.

A vA C B

Solution Position Coordinates. The position of pulleys B, C and point A are specified by position coordinates sB, sC and sA respectively as shown in Fig. a. The pulley system consists of two cords which give sB + 2(sB - sC) = l1 3sB - 2sC = l1

(1)

sC + sA = l2

(2)

And Time Derivative. Taking the time derivatives of Eqs. (1) and (2), we get 3vB - 2vC = 0

(3)

vC + vA = 0

(4)

And Eliminate vC from Eqs. (3) and (4), 3vB + 2vA = 0 Here vA = +8 m>s since it is directed toward the positive sense of sA. Thus, 3vB + 2(8) = 0  vB = - 5.33 m>s = 5.33 m>s c

Ans.

The negative sign indicates that vB is directed toward the negative sense of sB.

Ans: vB = 5.33 m>s c 213

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12–205. If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relative velocity of block B with respect to C.

A C

Solution

B

sA + 2 sB + 2 sC = l vA + 2vB + 2vC = 0 6 + 2vB + 2(18) = 0 vB = - 21 ft>s = 21 ft>s c + T vB = vC + vB>C -21 = 18 + vB>C vB>C = - 39 ft>s = 39 ft>s c 

Ans.

Ans: vB>C = 39 ft>sx 214

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12–206. Determine the speed of the block at B.

6 m/s A B

Solution Position Coordinate. The positions of pulley B and point A are specified by position coordinates sB and sA respectively as shown in Fig. a. This is a single cord pulley system. Thus, sB + 2(sB - a - b) + (sB - a) + sA = l (1)

4sB + sA = l + 3a + 2b Time Derivative. Taking the time derivative of Eq. (1),

(2)

4vB + vA = 0 Here, vA = + 6 m>s since it is directed toward the positive sense of sA. Thus, 4vB + 6 = 0 vB = - 1.50 m>s = 1.50 m>s d 

Ans.

The negative sign indicates that vB is directed towards negative sense of sB.

Ans: vB = 1.50 m>s 215

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12–207. Determine the speed of block A if the end of the rope is pulled down with a speed of 4 m>s.

4 m/s B

SOLUTION Position Coordinates: By referring to Fig. a, the length of the cord written in terms of the position coordinates sA and sB is

A

sB + sA + 2(sA - a) = l sB + 3sA = l + 2a Time Derivative: Taking the time derivative of the above equation, (+ T)

vB + 3vA = 0

Here, vB = 4 m>s. Thus, 4 + 3vA = 0

vA = - 133 m>s = 1.33 m>s c

Ans.

Ans: vA = 1.33 m>s 216

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*12–208. The motor draws in the cable at C with a constant velocity of vC = 4 m>s. The motor draws in the cable at D with a constant acceleration of aD = 8 m>s2. If vD = 0 when t = 0, determine (a) the time needed for block A to rise 3 m, and (b) the relative velocity of block A with respect to block B when this occurs.

C

D B

Solution

A

(a) aD = 8 m>s2

vD = 8 t



sD = 4 t 2



sD + 2sA = l



∆sD = -2∆sA



∆sA = - 2 t 2



- 3 = -2 t 2



t = 1.2247 = 1.22 s

(1)

Ans.

(b) vA = sA = - 4 t = - 4(1.2247) = - 4.90 m>s = 4.90 m>s c

sB + (sB - sC) = l



2vB = vC = - 4



vB = - 2 m>s = 2 m>s c

( + T ) vA = vB + vA>B

- 4.90 = - 2 + vA>B



vA>B = - 2.90 m>s = 2.90 m>s c

Ans.

Ans: vA>B = 2.90 m>s c 217

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12–209. The cord is attached to the pin at C and passes over the two pulleys at A and D. The pulley at A is attached to the smooth collar that travels along the vertical rod. Determine the velocity and acceleration of the end of the cord at B if at the instant sA = 4 ft the collar is moving upwards at 5 ft>s, which is decreasing at 2 ft>s2.

3 ft C

D sB

sA

SOLUTION 2 2s2A

3 ft

B

A

2

+ 3 + sB = l

1 1 # # 2a b A s2A + 9 B - 2 a 2sA sA b + s B = 0 2

# 2sA sA # sB = 1 A s2A + 9 B 2

1 1 3 # 1 # ## # = -2sA2 A s2A + 9 B - 2 - a2sA sA b A s2A + 9 B - 2 - a2sA sA b c a - b A s2A + 9 B - 2 a2sA sA b d 2 # $ # 2 A sA + sA sA B 2 A sA sA B 2 $ sB = + 1 3 A s2A + 9 B 2 A s2A + 9 B 2

##

s

B

At sA = 4 ft, 2(4)(-5) # vB = sB = 1 = 8 ft>s T (42 + 9)2 $ aB = sB = -

2 c 1 -522 + 142122 d 14 + 92 2

1 2

+

Ans. 231421- 5242 14 + 92 2

3 2

= - 6.80 ft>s2 = 6.80 ft>s2 c Ans.

Ans: vB = 8 ft>s T aB = 6.80 ft>s2 c 218

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12–210. The 16-ft-long cord is attached to the pin at C and passes over the two pulleys at A and D. The pulley at A is attached to the smooth collar that travels along the vertical rod. When sB = 6 ft, the end of the cord at B is pulled downwards with a velocity of 4 ft>s and is given an acceleration of 3 ft>s2. Determine the velocity and acceleration of the collar at this instant.

3 ft C

D sB

sA

SOLUTION 22s2A

3 ft

B

A

2

+ 3 + sB = l

1 1 # # 2a b (s2A + 9)- 2 a 2sA sA b + sB = 0 2

# sB = -

# 2sA sA 1

(s2A + 9)2

1 1 3 1 $ # $ # # sB = -2s2A(s2A + 9)- 2 - a 2sAsA b (s2A + 9)- 2 - a 2sAsA b c a - b(s2A + 9)- 2 a2sAsA b d 2 # # $ 21sA sA22 21sA2 + sA sA 2 $ + sB = 1 3 1s2A + 922 1s2A + 922

# $ At sB = 6 ft, sB = 4 ft>s, sB =3 ft>s2 22s2A + 32 + 6 = 16

sA = 4 ft 4 = -

# 2(4)(sA) 1

(42 + 9)2

# vA = sA = -2.5 ft/s = 2.5 ft>s c

3 = -

$ 2 C (- 2.5)2 + 4(sA) D 1 2

(42 + 9)

+

Ans.

2[4(- 2.5)]2 3

(42 + 9)2

$ aA = sA = -2.4375 = 2.44 ft>s2 c

Ans.

Ans: vA = 2.5 ft>s c aA = 2.44 ft>s2 c 219

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12–211. The roller at A is moving with a velocity of vA = 4 m>s and has an acceleration of aA = 2 m>s2 when xA = 3 m. Determine the velocity and acceleration of block B at this instant.

vA  4 m/s xA A 4m

Solution Position Coordinates. The position of roller A and block B are specified by position coordinates xA and yB respectively as shown in Fig. a. We can relate these two position coordinates by considering the length of the cable, which is constant 2x2A + 42 + yB = l

yB = l - 2x2A + 16

(1)

Velocity. Taking the time derivative of Eq. (1) using the chain rule, dyB dxA 1 1 = 0 - 1 x2A + 16 2 -2 (2xA) dt 2 dt dyB xA dxA = 2 dt 2xA + 16 dt

However,

dyB dxA = vB and  = vA. Then dt dt vB = -

xA 2x2A

(2)

vA 

+ 16

At xA = 3 m, vA = + 4 m>s since vA is directed toward the positive sense of xA. Then Eq. (2) give vB = -

3 232 + 16

(4) = -2.40 m>s = 2.40 m>s c 

Ans.

The negative sign indicates that vB is directed toward the negative sense of yB. Acceleration. Taking the time derivative of Eq. (2), dvB dxA dxA 1 = - c xA a - b 1 x2A + 16)-3>2 (2xA) + (x2A + 16)-1>2 d dt 2 dt dt dvA dt



vA - xA(x2A + 16)-1>2

However,

dvB dvA dxA = aB, = aA and = vA. Then dt dt dt aB =

xA2 vA2

1

xA2

aB = -

+ 16 2

3>2

-

vA2

1

xA2

+ 16 2 1>2

16 vA2 + aAxA 1 xA2 + 16 2

1 xA2

+ 16 2 3/2

-

xAaA

1 xA2

+ 16 2 1>2

At xA = 3 m, vA = + 4 m>s, aA = + 2 m>s2 since vA and aA are directed toward the positive sense of xA. aB = -

16 ( 42 ) + 2(3) ( 32 + 16 )

( 32 + 16 ) 3>2

= - 3.248 m>s2 = 3.25 m>s2 c  Ans.

The negative sign indicates that aB is directed toward the negative sense of yB. 220

Ans: vB = 2.40 m>s c aB = 3.25 m>s2 c

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*12–212. The girl at C stands near the edge of the pier and pulls in the rope horizontally at a constant speed of 6 ft>s. Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft.

6 ft/s C

xC A

8 ft B xB

SOLUTION The length l of cord is 2(8)2 + x2B + xC = l Taking the time derivative: # # 1 [(8)2 + x2B] - 1/2 2 xBxB + xC = 0 2

(1)

# xC = 6 ft/s When AB = 50 ft, xB = 2(50)2 - (8)2 = 49.356 ft From Eq. (1) # 1 [(8)2 + (49.356)2] - 1/2 2(49.356)(xB) + 6 = 0 2 # xB = - 6.0783 = 6.08 ft>s ;

Ans.

Ans: # xB = 6.08 ft>s d 221

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12–213. If the hydraulic cylinder H draws in rod BC at 2 ft>s, determine the speed of slider A.

A B

C

H

SOLUTION 2sH + sA = l 2vH = -vA 2(2) = -vA vA = -4 ft>s = 4 ft>s ;

Ans.

Ans: vA = 4 ft>s 222

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12–214. At the instant shown, the car at A is traveling at 10 m>s around the curve while increasing its speed at 5 m>s2. The car at B is traveling at 18.5 m>s along the straightaway and increasing its speed at 2 m>s2. Determine the relative velocity and relative acceleration of A with respect to B at this instant.

yB  18.5 m/s B

A 100 m

yA  10 m/s

100 m 45

Solution vA = 10 cos 45°i - 10 sin 45°j = 5 7.071i - 7.071j 6 m>s vB = {18.5i} m>s vA>B = vA - vB   = (7.071i - 7.071j) - 18.5i = 5-11.429i - 7.071j 6 m>s

vA>B = 2(- 11.429)2 + (- 7.071)2 = 13.4 m>s

u = tan-1

Ans.

7.071 = 31.7° d 11.429

Ans.

vA2 102 (aA)t = 5 m>s2 = 1 m>s2 = r 100 = (5 cos 45° - 1 cos 45°)i + ( - 1 sin 45° - 5 sin 45°)j

(aA)n = aA

= {2.828i - 4.243j} m>s2 aB = {2i} m>s2 aA>B = aA - uB   = (2.828i - 4.243j) - 2i = {0.828i - 4.24j} m>s2 aA>B = 20.8282 + ( - 4.243)2 = 4.32 m>s2

u = tan - 1

4.243 = 79.0° 0.828

Ans.

c

Ans.

Ans: vA>B uv = aA>B ua = 223

= 13.4 m>s 31.7° d = 4.32 m>s2 79.0° c

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12–215. The motor draws in the cord at B with an acceleration of aB = 2 m>s2. When sA = 1.5 m, vB = 6 m>s. Determine the velocity and acceleration of the collar at this instant.

B

2m

A

Solution

sA

Position Coordinates. The position of collar A and point B are specified by sA and sC respectively as shown in Fig. a. We can relate these two position coordinates by considering the length of the cable, which is constant. sB + 2sA2 + 22 = l sB = l - 2sA2 + 4

(1)

Velocity. Taking the time derivative of Eq. (1), dsB dsA 1 = 0 - ( sA2 + 4 ) - 1>2 a2sA b dt 2 dt

dsB sA dsA = 2 dt 2sA + 4 dt dsB dsA = vB and = vA. Then this equation becomes However, dt dt sA

vA  (2) 2sA2 + 4 At the instant sA = 1.5 m, vB = + 6 m>s. vB is positive since it is directed toward the positive sense of sB. vB = -

1.5 vA 21.52 + 4 vA = - 10.0 m>s = 10.0 m>s d

6 = -

Ans.

The negative sign indicates that vA is directed toward the negative sense of sA. Acceleration. Taking the time derivative of Eq. (2), dvB dsA dsA 1 = - c sAa - b 1 sA2 + 4 2 - 3>2 a2sA b + ( sA2 + 4 ) - 1>2 d dt 2 dt dt

vA - sA ( sA2 + 4)-1>2 However,

dvA dt

dvB dvA dsA = aB, = aA and  = vA. Then dt dt dt aB =

sA2 vA2

(

sA2

aB = -

+ 4)

3>2

-

vA2

(

sA2

+ 4)

1>2

-

aAsA

(

sA2

+ 4 ) 1>2

4vA2 + aAsA ( sA2 + 4 )

( sA2 + 4 ) 3>2

At the instant sA = 1.5 m, aB = + 2 m>s2. aB is positive since it is directed toward the positive sense of sB. Also, vA = - 10.0 m>s. Then 2 = -c

4( - 10.0)2 + aA(1.5)(1.52 + 4) (1.52 + 4)3>2

aA = -46.0 m>s2 = 46.0 m>s2 d 

d

Ans.

The negative sign indicates that aA is directed toward the negative sense of sA. 224

Ans: vA = 10.0 m>s d aA = 46.0 m>s2 d

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*12–216. If block B is moving down with a velocity vB and has an acceleration a B , determine the velocity and acceleration of block A in terms of the parameters shown.

sA A

h vB, aB B

SOLUTION l = sB + 3s2B + h2 1 # # 0 = sB + (s2A + h2) - 1/2 2sA sA 2 # - sB(s2A + h2)1/2 # vA = sA = sA vA = - vB (1 + a

h 2 1/2 b ) sA

Ans.

h 2 h 2 1 # # # aA = vA = - vB(1 + a b )1/2 - vB a b (1 + a b ) - 1/2(h2)( - 2)(sA) - 3sA sA 2 sA aA = - aB(1 + a

h 2 1/2 vAvBh2 h 2 b ) + (1 + a b ) - 1/2 3 sB sA sA

Ans.

Ans: vA = -vB (1 + a

225

aA = -aB (1 + a

h 2 1>2 b ) sA

h 2 1>2 vAvBh2 h 2 b ) + (1+ a b )-1>2 3 sB SA sA

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12–217. The crate C is being lifted by moving the roller at A downward with a constant speed of vA = 2 m>s along the guide. Determine the velocity and acceleration of the crate at the instant s = 1 m. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hint: Relate the coordinates xC and xA using the problem geometry, then take the first and second time derivatives.

4m B xA xC 4m

SOLUTION xC +

2x2A

A C s

2

+ (4) = l

# # 1 xC + (x2A + 16) - 1/2(2xA)(xA) = 0 2 1 # 2 $ # 2 $ xC - (x2A + 16) - 3/2 (2x2A)(xA ) + (x2A + 16) - 1/2 (xA) + (x2A + 16) - 1/2 (xA)(xA) = 0 2 l = 8 m , and when s = 1 m , xC = 3 m xA = 3 m # vA = xA = 2 m>s $ aA = x A = 0 Thus, vC + [(3)2 + 16] - 1/2 (3)(2) = 0 vC = - 1.2 m>s = 1.2 m>s c

Ans.

aC - [(3)2 + 16] - 3/2 (3)2(2)2 + [(3)2 + 16] - 1/2 (2)2 + 0 = 0 aC = - 0.512 m>s2 = 0.512 m>s2 c

Ans.

Ans: vC = 1.2 m>s c aC = 0.512 m>s2 c 226

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12–218. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 500 km>h and vB = 700 km>h such that the angle between their straight-line courses is u = 60°, determine the velocity of plane B with respect to plane A.

A vA  500 km/h

vB  700 km/h 60

Solution

B

Relative Velocity. Express vA and vB in Cartesian vector form, vA = 5-500 j 6 km>h

vB = 5700 sin 60°i + 700 cos 60°j6 km>h =

Applying the relative velocity equation.

5 35023i + 350j 6 km>h

vB = vA + vB>A 35023i + 350j = - 500j + vB>A vB>A =

5 35023i

Thus, the magnitude of vB>A is

+ 850j 6 km>h

vB>A = 2(35013)2 + 8502 = 1044.03 km>h = 1044 km>h

Ans.

And its direction is defined by angle u, Fig. a.

u = tan-1 a

850 35023

b = 54.50° = 54.5° a

Ans.

Ans: vB>A = 1044 km>h u = 54.5°a 227

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12–219. At the instant shown, cars A and B are traveling at speeds of 55 mi/h and 40 mi/h, respectively. If B is increasing its speed by 1200 mi>h2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.5 mi.

vB = 40 mi/h B 30°

A vA = 55 mi/h

SOLUTION vB = - 40 cos 30°i + 40 sin 30°j = {- 34.64i + 20j} mi>h vA = { - 55i } mi>h vB>A = nB - nA = ( - 34.64i + 20j) - ( -55i) = {20.36i + 20j} mi>h vB>A = 220.362 + 202 = 28.5 mi>h

Ans.

20 = 44.5° 20.36

Ans.

u = tan - 1 (aB)n =

a

v2A 402 = = 3200 mi>h2 r 0.5

(aB)t = 1200 mi>h2

aB = (3200 cos 60° - 1200 cos 30°)i + (3200 sin 60° + 1200 sin 30°)j = {560.77i + 3371.28j}

mi>h2

aA = 0 aB>A = aB - aA = {560.77i + 3371.28j} - 0 = {560.77i + 3371.28j}

mi>h2

aB>A = 2(560.77)2 + (3371.28)2 = 3418 mi>h2 u = tan - 1

3371.28 = 80.6° 560.77

Ans.

a

Ans.

Ans: vB/A uv = aB/A ua = 228

= 28.5 mi>h 44.5° a = 3418 mi>h2 80.6°a

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*12–220. The boat can travel with a speed of 16 km>h in still water. The point of destination is located along the dashed line. If the water is moving at 4 km>h, determine the bearing angle u at which the boat must travel to stay on course.

vW  4 km/h

u 70

Solution vB = vW + vB>W vB cos 70°i + vB sin 70°j = - 4j + 16 sin ui + 16 cos uj + )   v cos 70° = 0 + 16 sin u (S B ( + c )  vB sin 70° = - 4 + 16 cos u



2.748 sin u - cos u + 0.25 = 0

Solving,

Ans.

u = 15.1°

Ans: u = 15.1° 229

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12–221. y

Two boats leave the pier P at the same time and travel in the directions shown. If vA = 40 ft>s and vB = 30 ft>s, determine the velocity of boat A relative to boat B. How long after leaving the pier will the boats be 1500 ft apart?

vA = 40 ft/s A

vB = 30 ft/s B

SOLUTION

30° 45°

Relative Velocity:

x

P

vA = vB + vA>B 40 sin 30°i + 40 cos 30°j = 30 cos 45°i + 30 sin 45°j + vA>B vA>B = {-1.213i + 13.43j} ft>s Thus, the magnitude of the relative velocity vA>B is vA>B = 2(-1.213)2 + 13.432 = 13.48 ft>s = 13.5 ft>s

Ans.

And its direction is u = tan - 1

13.43 = 84.8° 1.213

Ans.

One can obtained the time t required for boats A and B to be 1500 ft apart by noting that boat B is at rest and boat A travels at the relative speed vA>B = 13.48 ft>s for a distance of 1500 ft. Thus t =

1500 1500 = = 111.26 s = 1.85 min vA>B 13.48

Ans.

Ans: vB = 13.5 ft>s u = 84.8° t = 1.85 min 230

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12–222. A car is traveling north along a straight road at 50 km>h. An instrument in the car indicates that the wind is coming from the east. If the car’s speed is 80 km>h, the instrument indicates that the wind is coming from the northeast. Determine the speed and direction of the wind.

SOLUTION Solution I Vector Analysis: For the first case, the velocity of the car and the velocity of the wind relative to the car expressed in Cartesian vector form are vc = [50j] km>h and vW>C = (vW>C)1 i. Applying the relative velocity equation, we have vw = vc + vw>c vw = 50j + (vw>c)1 i vw = (vw>c)1i + 50j

(1)

For the second case, vC = [80j] km>h and vW>C = (vW>C)2 cos 45°i + (vW>C)2 sin 45° j. Applying the relative velocity equation, we have vw = vc + vw>c vw = 80j + (vw>c)2 cos 45°i + (vw>c)2 sin 45° j vw = (vw>c)2 cos 45° i + C 80 + (vw>c)2 sin 45° D j

(2)

Equating Eqs. (1) and (2) and then the i and j components, (vw>c)1 = (vw>c)2 cos 45°

(3)

50 = 80 + (vw>c)2 sin 45°

(4)

Solving Eqs. (3) and (4) yields (vw>c)2 = -42.43 km>h

(vw>c)1 = - 30 km>h

Substituting the result of (vw>c)1 into Eq. (1), vw = [ - 30i + 50j] km>h Thus, the magnitude of vW is vw = 2( -30)2 + 502 = 58.3 km>h

Ans.

and the directional angle u that vW makes with the x axis is u = tan - 1 a

50 b = 59.0° b 30

Ans.

Ans: vw = 58.3 km>h u = 59.0° b 231

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12–223. Two boats leave the shore at the same time and travel in the directions shown. If vA = 10 m>s and vB = 15 m>s, determine the velocity of boat A with respect to boat B. How long after leaving the shore will the boats be 600 m apart?

vA  10 m/s

A

vB  15 m/s

B 30

O

45

Solution Relative Velocity. The velocity triangle shown in Fig. a is drawn based on the relative velocity equation vA = vB + vA>B. Using the cosine law,    vA>B = 2102 + 152 - 2(10)(15) cos 75° = 15.73 m>s = 15.7 m>s

Ans.

Then, the sine law gives

sin f sin 75° =   f = 37.89° 10 15.73 The direction of vA>B is defined by         u = 45° - f = 45° - 37.89° = 7.11°  d

Ans.

Alternatively, we can express vA and vB in Cartesian vector form vA = 5 - 10 sin 30°i + 10 cos 30°j 6 m>s = 5 - 5.00i + 523j 6 m>s

vB = 515 cos 45°i + 15 sin 45°j 6 m>s = 5 7.522i + 7.522j 6 m>s.

Applying the relative velocity equation vA = vB + vA>B

-500i + 523j = 7.522i + 7.522j + vA>B vA>B = 5- 15.61i - 1.946j 6 m>s

Thus the magnitude of vA>B is

     vA>B = 2(- 15.61)2 + (-1.946)2 = 15.73 m>s = 15.7 m>s

Ans.

And its direction is defined by angle u, Fig. b,



u = tan-1a

Here sA>B = 600 m. Thus

t =

sA>B vA>B

1.946 b = 7.1088° = 7.11°  d 15.61

=

600 = 38.15 s = 38.1 s 15.73

Ans.

Ans.

Ans: vA/B = 15.7 m>s u = 7.11° d t = 38.1 s 232

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*12–224. At the instant shown, car A has a speed of 20 km>h, which is being increased at the rate of 300 km>h2 as the car enters an expressway. At the same instant, car B is decelerating at 250 km>h2 while traveling forward at 100 km>h. Determine the velocity and acceleration of A with respect to B. A 100 m

SOLUTION vA = { -20j} km>h

vB = {100j} km>h B

vA>B = vA - vB = (- 20j - 100j) = {- 120j} km>h Ans.

yA>B = 120 km>h T

(aA)n =

y2A 202 = = 4000 km>h2 r 0.1

(aA)t = 300 km>h2

aA = - 4000i + ( -300j) = { -4000i - 300j} km>h2 aB = { -250j} km>h2 aA>B = aA - aB = (- 4000i - 300j) - ( - 250j) = {- 4000i - 50j} km>h2 aA>B = 2(- 4000)2 + (- 50)2 = 4000 km>h2 u = tan - 1

Ans.

50 = 0.716° d 4000

Ans.

Ans: vA>B = 120 km>h T aA>B = 4000 km>h2 u = 0.716° d 233

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12–225. Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 90 ft>s and is increasing its speed at the rate of 15 ft>s2, whereas B has a speed of 105 ft>s and is decreasing its speed at 25 ft>s2. Determine the relative velocity and relative acceleration of car A with respect to car B at this instant.

vA A B rA

300 ft

vB

60 rB

250 ft

SOLUTION vA = vB + vA>B -90i = -105 sin 30° i + 105 cos 30°j + vA>B vA>B = 5- 37.5i - 90.93j6 ft>s vA/B = 2( -37.5)2 + ( - 90.93)2 = 98.4 ft>s

Ans.

90.93 b = 67.6° d 37.5

Ans.

u = tan - 1 a

aA = aB + aA>B -15i -

19022 300

j = 25 cos 60°i - 25 sin 60°j - 44.1 sin 60°i - 44.1 cos 60°j + aA>B

aA>B = {10.69i + 16.70j} ft>s2 aA>B = 2(10.69)2 + (16.70)2 = 19.8 ft>s2

Ans.

16.70 b = 57.4° a 10.69

Ans.

u = tan - 1 a

Ans: vA/B uv = aA>B ua = 234

= 98.4 ft>s 67.6° d = 19.8 ft>s2 57.4°a

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12–226. A man walks at 5 km>h in the direction of a 20-km>h wind. If raindrops fall vertically at 7 km>h in still air, determine the direction in which the drops appear to fall with respect to the man.

SOLUTION Relative Velocity: The velocity of the rain must be determined first. Applying Eq. 12–34 gives vr = vw + vr>w = 20 i + ( - 7 j) = 520 i - 7 j 6 km>h

Thus, the relatives velocity of the rain with respect to the man is vr = vm + vr>m 20 i - 7 j = 5 i + vr>m vr>m = 515 i - 7 j 6 km>h

The magnitude of the relative velocity vr>m is given by r>m

= 2152 + ( - 7)2 = 16.6 km>h

Ans.

And its direction is given by u = tan-1

7 = 25.0° d 15

Ans.

Ans: vr>m = 16.6 km>h u = 25.0° c 235

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12–227. At the instant shown, cars A and B are traveling at velocities of 40 m>s and 30 m>s, respectively. If B is increasing its velocity by 2 m>s2, while A maintains a constant velocity, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is rB = 200 m.

B

A vA  40 m/s

vB  30 m/s 30

Solution Relative velocity. Express vA and vB as Cartesian vectors. vA = 540j 6 m>s  vB = 5 - 30 sin 30°i + 30 cos 30°j6 m>s = 5 - 15i + 1523j 6 m>s

Applying the relative velocity equation, vB = vA + vB>A -15i + 1523j = 40j + vB>A vB>A = 5 -15i - 14.02j 6 m>s

Thus, the magnitude of vB>A is

vB>A = 2(- 15)2 + (- 14.02)2 = 20.53 m>s = 20.5 m>s

Ans.

And its direction is defined by angle u, Fig. a u = tan-1a

14.02 b = 43.06° = 43.1° d 15

Ans.

vB2 302 = 4.50 m>s2 Relative Acceleration. Here, (aB)t = 2 m>s2 and (aB)n = r = 200 and their directions are shown in Fig. b. Then, express aB as a Cartesian vector, aB = ( -2 sin 30° - 4.50 cos 30°)i + (2 cos 30° - 4.50 sin 30°)j   = 5 - 4.8971i - 0.5179j6 m>s2

Applying the relative acceleration equation with aA = 0, aB = aA + aB>A -4.8971i - 0.5179j = 0 + aB>A aB>A = 5 -4.8971i - 0.5179j 6 m>s2

Thus, the magnitude of aB>A is

aB>A = 2(- 4.8971)2 + (-0.5179)2 = 4.9244 m>s2 = 4.92 m>s2

Ans.

And its direction is defined by angle u′, Fig. c, u′ = tan-1a

0.5179 b = 6.038° = 6.04° d 4.8971

Ans.

Ans: vB>A uv = aB>A ua = 236

= 20.5 m>s 43.1° d = 4.92 m>s2 6.04° d

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*12–228. At the instant shown, cars A and B are traveling at velocities of 40 m>s and 30 m>s, respectively. If A is increasing its speed at 4 m>s2, whereas the speed of B is decreasing at 3 m>s2, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is rB = 200 m.

B

A vA  40 m/s

vB  30 m/s 30

Solution Relative velocity. Express vA and vB as Cartesian vector. vA = 540j 6 m>s  vB = 5 - 30 sin 30°i + 30 cos 30°j 6 m>s = 5 - 15i + 1523j 6 m>s

Applying the relative velocity equation, vB = vA + vB>A - 15i + 1523j = 40j + vB>A vB>A = 5 -15i - 14.026 m>s

Thus the magnitude of vB>A is vB>A = 2(- 15)2 + (- 14.02)2 = 20.53 m>s = 20.5 m>s

Ans.

And its direction is defined by angle u, Fig a. u = tan-1 a

14.02 b = 43.06° = 43.1° d 15

Ans.

v2B 302 Relative Acceleration. Here (aB)t = 3 m>s2 and (aB)n = r = = 4.5 m>s2 and 200 their directions are shown in Fig. b. Then express aB as a Cartesian vector, aB = (3 sin 30° - 4.50 cos 30°)i + ( - 3 cos 30° - 4.50 sin 30°)j = { - 2.3971i - 4.8481j} m>s2 Applying the relative acceleration equation with aA = {4j} m>s2, aB = aA + aB>A - 2.3971i - 4.8481j = 4j + aB>A aB>A = { - 2.3971i - 8.8481j} m>s2 Thus, the magnitude of aB>A is aB>A = 2(- 2.3971)2 + (-8.8481)2 = 9.167 m>s2 = 9.17 m>s2

Ans.

And its direction is defined by angle u′, Fig. c u′ = tan-1 a

8.8481 b = 74.84° = 74.8° d 2.3971

Ans.

Ans: vB>A = 20.5 m>s u = 43.1° d aB>A = 9.17 m>s2 u′ = 74.8° d 237

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12–229. A passenger in an automobile observes that raindrops make an angle of 30° with the horizontal as the auto travels forward with a speed of 60 km/h. Compute the terminal (constant) velocity vr of the rain if it is assumed to fall vertically.

vr va = 60 km/h

SOLUTION vr = va + vr>a -vr j = -60i + vr>a cos 30°i - vr>a sin 30°j + ) (:

0 = - 60 + vr>a cos 30°

(+ c)

-vr = 0 - vr>a sin 30° vr>a = 69.3 km>h Ans.

vr = 34.6 km h

Ans: vr = 34.6 km>hT 238

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12–230. A man can swim at 4 ft/s in still water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. If the river flows with a velocity of 2 ft/s, determine the speed of the man and the time needed to make the crossing. Note: While in the water he must not direct himself toward point B to reach this point. Why?

30 ft B vr = 2 ft/s

40 ft

A

SOLUTION Relative Velocity: vm = vr + vm>r 3 4 n i + vm j = 2i + 4 sin ui + 4 cos uj 5 m 5 Equating the i and j components, we have 3 v = 2 + 4 sin u 5 m

(1)

4 v = 4 cos u 5 m

(2)

Solving Eqs. (1) and (2) yields u = 13.29° Ans.

vm = 4.866 ft>s = 4.87 ft>s Thus, the time t required by the boat to travel from points A to B is t =

sAB 2402 + 302 = = 10.3 s vb 4.866

Ans.

In order for the man to reached point B, the man has to direct himself at an angle u = 13.3° with y axis.

Ans: vm = 4.87 ft>s t = 10.3 s 239

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12–231. The ship travels at a constant speed of vs = 20 m>s and the wind is blowing at a speed of vw = 10 m>s, as shown. Determine the magnitude and direction of the horizontal component of velocity of the smoke coming from the smoke stack as it appears to a passenger on the ship.

vs 30 vw

10 m/s

20 m/s 45 y x

SOLUTION Solution I Vector Analysis: The velocity of the smoke as observed from the ship is equal to the velocity of the wind relative to the ship. Here, the velocity of the ship and wind expressed in Cartesian vector form are vs = [20 cos 45° i + 20 sin 45° j] m>s = [14.14i + 14.14j] m>s and vw = [10 cos 30° i - 10 sin 30° j] = [8.660i - 5j] m>s. Applying the relative velocity equation, vw = vs + vw>s 8.660i - 5j = 14.14i + 14.14j + vw>s vw>s = [- 5.482i - 19.14j] m>s Thus, the magnitude of vw/s is given by vw = 2(-5.482)2 + ( -19.14)2 = 19.9 m>s

Ans.

and the direction angle u that vw/s makes with the x axis is u = tan - 1 a

19.14 b = 74.0° d 5.482

Ans.

Solution II Scalar Analysis: Applying the law of cosines by referring to the velocity diagram shown in Fig. a, vw>s = 2202 + 102 - 2(20)(10) cos 75° Ans.

= 19.91 m>s = 19.9 m>s Using the result of vw/s and applying the law of sines, sin f sin 75° = 10 19.91

f = 29.02°

Thus, u = 45° + f = 74.0° d

Ans.

Ans: vw/s = 19.9 m>s u = 74.0° d 240

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*12–232. z

The football player at A throws the ball in the y–z plane at a speed vA = 50 ft>s and an angle uA = 60° with the horizontal. At the instant the ball is thrown, the player is at B and is running with constant speed along the line BC in order to catch it. Determine this speed, vB, so that he makes the catch at the same elevation from which the ball was thrown.

y

C

vA

uA

vB

A

Solution +2 s = s + v t 1S 0 0

B

20 ft

30 ft x

d AC = 0 + (50 cos 60°) t

( + c ) v = v0 + ac t - 50 sin 60° = 50 sin 60° - 32.2 t t = 2.690 s d AC = 67.24 ft d BC = 2(30)2 + (67.24 - 20)2 = 55.96 ft vB =

55.96 = 20.8 ft>s 2.690

Ans.

Ans: vB = 20.8 ft>s 241

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12–233. z

The football player at A throws the ball in the y–z plane with a speed vA = 50 ft>s and an angle uA = 60° with the horizontal. At the instant the ball is thrown, the player is at B and is running at a constant speed of vB = 23 ft>s along the line BC. Determine if he can reach point C, which has the same elevation as A, before the ball gets there.

y

C

vA

uA

vB

A

Solution +2 s = s + v t 1S 0 0

B

20 ft

30 ft x

d AC = 0 + (50 cos 60°) t

( + c ) v = v0 + ac t -50 sin 60° = 50 sin 60° - 32.2 t t = 2.690 s d AC = 67.24 ft d BC = 2(30)2 + (67.24 - 20)2 = 55.96 ft vB =

d BC 55.96 = = 20.8 ft>s t (2.690)

Since vB = 20.8 ft>s 6 (vB)max = 23 ft>s Ans.

Yes, he can catch the ball.

Ans: Yes, he can catch the ball. 242

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12–234. At a given instant the football player at A throws a football C with a velocity of 20 m/s in the direction shown. Determine the constant speed at which the player at B must run so that he can catch the football at the same elevation at which it was thrown. Also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made. Player B is 15 m away from A when A starts to throw the football.

C

20 m/s A

60° B

15 m

SOLUTION Ball: + )s = s + v t (: 0 0 sC = 0 + 20 cos 60° t (+ c )

v = v0 + ac t

- 20 sin 60° = 20 sin 60° - 9.81 t t = 3.53 s sC = 35.31 m Player B: + ) s = s + n t (: B B 0 Require, 35.31 = 15 + vB (3.53) Ans.

vB = 5.75 m>s At the time of the catch (vC)x = 20 cos 60° = 10 m>s : (vC)y = 20 sin 60° = 17.32 m>s T vC = vB + vC>B 10i - 17.32j = 5.751i + (vC>B)x i + (vC>B)y j + ) (:

10 = 5.75 + (vC>B)x

(+ c )

- 17.32 = (vC>B)y (vC>B)x = 4.25 m>s : (vC>B)y = 17.32 m>s T vC>B = 2(4.25)2 + (17.32)2 = 17.8 m>s

Ans.

17.32 b = 76.2° 4.25

Ans.

u = tan - 1 a

c

aC = aB + aC>B - 9.81 j = 0 + aC>B aC B = 9.81 m s2 T

Ans.

243

Ans: vB = 5.75 m>s vC/B = 17.8 m>s u = 76.2° c aC>B = 9.81 m>s2 T

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12–235. At the instant shown, car A travels along the straight portion of the road with a speed of 25 m>s. At this same instant car B travels along the circular portion of the road with a speed of 15 m>s. Determine the velocity of car B relative to car A.

15 r

A

SOLUTION

15

200 m

C 30

Velocity: Referring to Fig. a, the velocity of cars A and B expressed in Cartesian vector form are

B

vA = [25 cos 30° i - 25 sin 30° j] m>s = [21.65i - 12.5j] m>s vB = [15 cos 15° i - 15 sin 15° j] m>s = [14.49i - 3.882j] m>s Applying the relative velocity equation, vB = vA + vB>A 14.49i - 3.882j = 21.65i - 12.5j + vB>A vB>A = [- 7.162i + 8.618j] m>s Thus, the magnitude of vB/A is given by vB>A = 2( -7.162)2 + 8.6182 = 11.2 m>s

Ans.

The direction angle uv of vB/A measured down from the negative x axis, Fig. b is uv = tan - 1 a

8.618 b = 50.3° d 7.162

Ans.

Ans: vB>A = 11.2 m>s u = 50.3° 244

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13–1. The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t2i - 4tj - 1k6 lb, and F3 = 5- 2ti6 lb, where t is in seconds. Determine the distance the ball is from the origin 2 s after being released from rest.

z

F2

y F3

x

F1

SOLUTION ©F = ma;

(2i + 6j - 2tk) + (t2i - 4tj - 1k) - 2ti - 6k = ¢

6 ≤ (axi + ay j + azk) 32.2

Equating components:

¢

6 ≤ a = t2 - 2t + 2 32.2 x

¢

6 ≤ a = - 4t + 6 32.2 y

¢

6 ≤ a = - 2t - 7 32.2 z

Since dv = a dt, integrating from n = 0, t = 0, yields

¢

t3 6 - t2 + 2t ≤ vx = 32.2 3

¢

6 ≤ v = - 2t2 + 6t 32.2 y

¢

6 ≤ v = - t2 - 7t 32.2 z

Since ds = v dt, integrating from s = 0, t = 0 yields

¢

t3 6 t4 + t2 ≤ sx = 32.2 12 3

When t = 2 s then,

¢

6 2t3 + 3t2 ≤ sy = 32.2 3

sx = 14.31 ft,

sy = 35.78 ft

¢

7t2 6 t3 ≤ sz = - 32.2 3 2 sz = -89.44 ft

Thus, s =

(14.31)2 + (35.78)2 + ( - 89.44)2 = 97.4 ft

Ans.

Ans: s = 97.4 ft 245

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13–2. The two boxcars A and B have a weight of 20  000 lb and 30 000 lb, respectively. If they are freely coasting down the incline when the brakes are applied to all the wheels of car A, determine the force in the coupling C between the two cars. The coefficient of kinetic friction between the wheels of A and the tracks is mk = 0.5. The wheels of car B are free to roll. Neglect their mass in the calculation. Suggestion: Solve the problem by representing single resultant normal forces acting on A and B, respectively.

B

A

5° C

SOLUTION Car A: + aΣFy = 0;

NA - 20 000 cos 5° = 0

+ Q ΣFx = max;

0.5(19 923.89) - T - 20 000 sin 5° = a

NA = 19 923.89 lb 20 000 ba 32.2

(1)

Both cars: + Q ΣFx = max;

0.5(19 923.89) - 50 000 sin 5° = a

50 000 ba 32.2

Solving, a = 3.61 ft>s2 Ans.

T = 5.98 kip

Ans: T = 5.98 kip 246

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13–3. P

If the coefficient of kinetic friction between the 50-kg crate and the ground is mk = 0.3, determine the distance the crate travels and its velocity when t = 3 s. The crate starts from rest, and P = 200 N.

30

SOLUTION Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: Here, ay = 0. Thus, + c ©Fy = 0;

N - 50(9.81) + 200 sin 30° = 0 N = 390.5 N

+ ©Fx = max; :

200 cos 30° - 0.3(390.5) = 50a a = 1.121 m>s2

Kinematics: Since the acceleration a of the crate is constant, + B A:

v = v0 + act Ans.

v = 0 + 1.121(3) = 3.36 m>s and + B A:

s = s0 + v0t + s = 0 + 0 +

1 2 at 2 c

1 (1.121) A 32 B = 5.04 m 2

Ans.

Ans: v = 3.36 m>s s = 5.04 m 247

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*13–4. P

If the 50-kg crate starts from rest and achieves a velocity of v = 4 m>s when it travels a distance of 5 m to the right, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is mk = 0.3.

30

SOLUTION Kinematics: The acceleration a of the crate will be determined first since its motion is known. + ) (: v2 = v 2 + 2a (s - s ) 0

2

c

0

2

4 = 0 + 2a(5 - 0) a = 1.60 m>s2 : Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be directed to the left to oppose the motion of the crate which is to the right, Fig. a. Equations of Motion: + c ©Fy = may;

N + P sin 30° - 50(9.81) = 50(0) N = 490.5 - 0.5P

Using the results of N and a, + ©F = ma ; : x x

P cos 30° - 0.3(490.5 - 0.5P) = 50(1.60) Ans.

P = 224 N

Ans: P = 224 N 248

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13–5. If blocks A and B of mass 10 kg and 6 kg, respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.

A

B

SOLUTION

30

Free-Body Diagram: Here, the kinetic friction (Ff)A = mANA = 0.1NA and (Ff)B = mB NB = 0.3NB are required to act up the plane to oppose the motion of the blocks which are down the plane. Since the blocks are connected, they have a common acceleration a. Equations of Motion: By referring to Figs. (a) and (b), +Q©Fy¿ = may¿ ;

NA - 10(9.81) cos 30° = 10(0) NA = 84.96 N

R + ©Fx¿ = max¿ ;

10(9.81) sin 30° - 0.1(84.96) - F = 10a (1)

40.55 - F = 10a and +Q©Fy¿ = may¿ ;

NB - 6(9.81) cos 30° = 6(0) NB = 50.97 N

R + ©Fx¿ = max¿ ;

F + 6(9.81) sin 30° - 0.3(50.97) = 6a (2)

F + 14.14 = 6a Solving Eqs. (1) and (2) yields a = 3.42 m>s2

Ans.

F = 6.37 N

Ans: F = 6.37 N 249

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13–6. The 10-lb block has a speed of 4 ft>s when the force of F = (8t2) lb is applied. Determine the velocity of the block when t = 2 s. The coefficient of kinetic friction at the surface is mk = 0.2.

v  4 ft/s F  (8t2) lb

Solution Equations of Motion. Here the friction is Ff = mk N = 0.2N. Referring to the FBD of the block shown in Fig. a, 10 (0)  N = 10 lb 32.2 10 + ΣFx = max;  8t 2 - 0.2(10) = a S 32.2

+ c ΣFy = may;  N - 10 =

a = 3.22(8t 2 - 2) ft>s2



Kinematics. The velocity of the block as a function of t can be determined by integrating dv = a dt using the initial condition v = 4 ft>s at t = 0. v

L4 ft>s

dv =

L0

t

3.22 (8t 2 - 2)dt

8 v - 4 = 3.22 a t 3 - 2tb 3 When t = 2 s,

v = 5 8.5867t3 - 6.44t + 46ft>s

v = 8.5867(23) - 6.44(2) + 4 = 59.81 ft>s Ans.

= 59.8 ft>s

Ans: v = 59.8 ft>s 250

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13–7. The 10-lb block has a speed of 4 ft>s when the force of F = (8t2) lb is applied. Determine the velocity of the block when it moves s = 30 ft. The coefficient of kinetic friction at the surface is ms = 0.2.

v  4 ft/s F  (8t2) lb

Solution Equations of Motion. Here the friction is Ff = mkN = 0.2N. Referring to the FBD of the block shown in Fig. a, + c ΣFy = may;  N - 10 =

10 (0)  N = 10 lb 32.2

10 + ΣFx = max;  8t 2 - 0.2(10) = a S 32.2 a = 3.22(8t 2 - 2) ft>s2



Kinematics. The velocity of the block as a function of t can be determined by integrating dv = adt using the initial condition v = 4 ft>s at t = 0. v

L4 ft>s

dv =

L0

t

3.22(8t 2 - 2)dt

8 v - 4 = 3.22 a t3 - 2t b 3 v = 58.5867t3 - 6.44t + 46 ft>s

The displacement as a function of t can be determined by integrating ds = vdt using the initial condition s = 0 at t = 0 L0 At s = 30 ft,

s

ds =

L0

t

(8.5867t 3 - 6.44t + 4)dt

s = 5 2.1467t 4 - 3.22t 2 + 4t6 ft

30 = 2.1467t 4 - 3.22t 2 + 4t Solved by numerically, t = 2.0089 s Thus, at s = 30 ft, v = 8.5867(2.00893) - 6.44(2.0089) + 4 = 60.67 ft>s Ans.

= 60.7 ft>s

Ans: v = 60.7 ft>s 251

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*13–8. The speed of the 3500-lb sports car is plotted over the 30-s time period. Plot the variation of the traction force F needed to cause the motion.

v (ft/s) F 80 60

SOLUTION dv 60 Kinematics: For 0 … t 6 10 s. v = , t = {6t} ft>s. Applying equation a = 10 dt we have a =

dv = 6 ft>s2 dt

80 - 60 v - 60 = For 10 6 t … 30 s, , v = {t + 50} ft>s. Applying equation t - 10 30 - 10 dv , we have a = dt a =

10

30

t (s)

v

dv = 1 ft>s2 dt

Equation of Motion: For 0 … t 6 10 s + ; a Fx = max ;

F = ¢

3500 ≤ (6) = 652 lb 32.2

Ans.

F = ¢

3500 ≤ (1) = 109 lb 32.2

Ans.

For 10 6 t … 30 s + ; a Fx = max ;

Ans: + ΣFx = max; F = 652 lb d + ΣFx = max; F = 109 lb d

252

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13–9. B

The conveyor belt is moving at 4 m>s. If the coefficient of static friction between the conveyor and the 10-kg package B is ms = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt.

SOLUTION + ΣFx = max; S

0.2(98.1) = 10 a a = 1.962 m>s2

+ ) v = v0 + ac t (S 4 = 0 + 1.962 t Ans.

t = 2.04 s

Ans: t = 2.04 s 253

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13–10. The conveyor belt is designed to transport packages of various weights. Each 10-kg package has a coefficient of kinetic friction mk = 0.15. If the speed of the conveyor is 5  m>s, and then it suddenly stops, determine the distance the package will slide on the belt before coming to rest.

B

Solution + ΣFx = max;  0.15 m(9.81) = ma S a = 1.4715 m>s2



+ ) v2 = v02 + 2ac(s - s0) (S      

0 = (5)2 + 2( - 1.4715)(s - 0) Ans.

       s = 8.49 m

Ans: s = 8.49 m 254

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13–11. Determine the time needed to pull the cord at B down 4 ft starting from rest when a force of 10 lb is applied to the cord. Block A weighs 20 lb. Neglect the mass of the pulleys and cords.

B

10 lb

Solution + c ΣFy = may;  40 - 20 =

20 a  32.2 A

C

A

Ans.

aA = 32.2 ft>s2

sB + 2sC = l;

aB = - 2aC

2sA - sC = l′;

2aA = aC

aB = - 4aA aB = 128.8 ft>s2 ( + T)

s = s0 + v0t +



4 = 0 + 0 +



t = 0.249 s

1 a t2 2 c

1 (128.8) t 2 2 Ans.

Ans: t = 0.249 s 255

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*13–12. Cylinder B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the block’s vertical position y so that when F is applied the block rises with a constant acceleration aB. Neglect the mass of the cord and pulleys.

d F y

Solution + c ΣFy = may;  2F cos u - mg = maB  where cos u = 2F a

y 2y2 +

F =

1 d2 2 2

b - mg = maB

m(aB + g) 24y2 + d 2 4y

y 2

2y +

aB

1 2

B

d 2 2

Ans.



Ans: F =

256

m(aB + g) 24y2 + d 2 4y

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13–13. Block A has a weight of 8 lb and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic friction is mk = 0.2. If the spring has a stiffness of k = 20 lb>ft, and it is compressed 0.2 ft, determine the acceleration of each block just after they are released.

A

k

B

Solution Block A: 8 + ΣFx = max;  4 - 1.6 = a d 32.2 A aA = 9.66 ft>s2 d 



Ans.

Block B: 6 + ΣFx = max;  4 - 12 = a S 32.2 B

aB = 15.0 ft>s2 S 

Ans.

Ans: aA = 9.66 ft>s2 d aB = 15.0 ft>s2 S 257

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13–14. The 2-Mg truck is traveling at 15 m> s when the brakes on all its wheels are applied, causing it to skid for a distance of 10 m before coming to rest. Determine the constant horizontal force developed in the coupling C, and the frictional force developed between the tires of the truck and the road during this time. The total mass of the boat and trailer is 1 Mg.

C

SOLUTION Kinematics: Since the motion of the truck and trailer is known, their common acceleration a will be determined first. + b a:

v2 = v0 2 + 2ac(s - s0) 0 = 152 + 2a(10 - 0) a = - 11.25 m>s2 = 11.25 m>s2 ;

Free-Body Diagram:The free-body diagram of the truck and trailer are shown in Figs. (a) and (b), respectively. Here, F representes the frictional force developed when the truck skids, while the force developed in coupling C is represented by T. Equations of Motion:Using the result of a and referrning to Fig. (a), + ©F = ma ; : x x

-T = 1000( -11.25) Ans.

T = 11 250 N = 11.25 kN Using the results of a and T and referring to Fig. (b), + c ©Fx = max ;

11 250 - F = 2000( -11.25) Ans.

F = 33 750 N = 33.75 kN

Ans: T = 11.25 kN F = 33.75 kN 258

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13–15. y

The motor lifts the 50-kg crate with an acceleration of 6 m>s2. Determine the components of force reaction and the couple moment at the fixed support A.

4m A

B

x 30

6 m/s2

Solution Equation of Motion. Referring to the FBD of the crate shown in Fig. a, + c ΣFy = may;  T - 50(9.81) = 50(6)  T = 790.5 N Equations of Equilibrium. Since the pulley is smooth, the tension is constant throughout entire cable. Referring to the FBD of the pulley shown in Fig. b, + ΣFx = 0;  790.5 cos 30° - Bx = 0  Bx = 684.59 N S + c ΣFy = 0;  By - 790.5 - 790.5 sin 30° = 0  By = 1185.75 N Consider the FBD of the cantilever beam shown in Fig. c, + ΣFx = 0; S + c ΣFy = 0;

684.59 - Ax = 0

Ans.

Ax = 684.59 N = 685 N

Ay - 1185.75 = 0 Ay = 1185.75 N = 1.19 kN

a+ ΣMA = 0; MA - 1185.75(4)

= 0  MA = 4743 N # m = 4.74 kN # m

Ans. Ans.

Ans: Ax = 685 N Ay = 1.19 kN MA = 4.74 kN # m 259

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*13–16. The 75-kg man pushes on the 150-kg crate with a horizontal force F. If the coefficients of static and kinetic friction between the crate and the surface are ms = 0.3 and mk = 0.2, and the coefficient of static friction between the man’s shoes and the surface is ms = 0.8, show that the man is able to move the crate. What is the greatest acceleration the man can give the crate?

F

Solution Equation of Equilibrium. Assuming that the crate is on the verge of sliding (Ff)C = ms NC = 0.3NC. Referring to the FBD of the crate shown in Fig. a,  + c ΣFy = 0;    NC - 150(9.81) = 0  NC = 1471.5 N + ΣFx = 0;  0.3(1471.5) - F = 0  F = 441.45 N S Referring to the FBD of the man, Fig. b,  + c ΣFy = 0;  Nm - 75(9.81) = 0  NB = 735.75 N + ΣFx = 0;  441.45 - (Ff)m = 0  (Ff)m = 441.45 N S Since (Ff)m 6 m′sNm = 0.8(735.75) = 588.6 N, the man is able to move the crate. Equation of Motion. The greatest acceleration of the crate can be produced when the man is on the verge of slipping. Thus, (Ff)m = m′sNm = 0.8(735.75) = 588.6 N. + ΣFx = 0;  F - 588.6 = 0  F = 588.6 N S Since the crate slides, (Ff)C = mkNC = 0.2(1471.5) = 294.3 N. Thus, + ΣFx = max;  588.6 - 294.3 = 150 a d a = 1.962 m>s2 = 1.96 m>s2

Ans.

Ans: a = 1.96 m>s2 260

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13–17. Determine the acceleration of the blocks when the system is released. The coefficient of kinetic friction is mk, and the mass of each block is m. Neglect the mass of the pulleys and cord.

B

A

Solution Free Body Diagram. Since the pulley is smooth, the tension is constant throughout the entire cord. Since block B is required to slide, Ff = mkN. Also, blocks A and B are attached together with inextensible cord, so aA = aB = a. The FBDs of blocks A and B are shown in Figs. a and b, respectively. Equations of Motion. For block A, Fig. a, + c ΣFy = may;  T - mg = m( - a)

(1)

For block B, Fig. b, + c ΣFy = may;

N - mg = m(0)  N = mg

+ ) ΣFx = max; T - mk mg = ma (d

(2)

Solving Eqs. (1) and (2) a =

1 (1 - mk) g 2

T =

1 (1 + mk) mg 2

Ans.

Ans: a = 261

1 (1 - mk) g 2

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13–18. A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C?

A

20 ft

30 4 ft

SOLUTION + R ©Fx = m ax ;

40 sin 30° =

40 a 32.2

B C R

a = 16.1 ft>s2 (+ R)v2 = v20 + 2 ac(s - s0); v2B = 0 + 2(16.1)(20) vB = 25.38 ft>s ( +R) v = v0 + ac t ; 25.38 = 0 + 16.1 tAB tAB = 1.576 s + )s = (s ) + (v ) t (: x x 0 x 0 R = 0 + 25.38 cos 30°(tBC) ( + T) sy = (sy)0 + (vy)0 t +

1 2 at 2 c

4 = 0 + 25.38 sin 30° tBC +

1 (32.2)(tBC)2 2

tBC = 0.2413 s R = 5.30 ft

Ans.

Total time = tAB + tBC = 1.82 s

Ans.

Ans: R = 5.30 ft t AC = 1.82 s 262

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13–19. Solve Prob. 13–18 if the suitcase has an initial velocity down the ramp of vA = 10 ft>s and the coefficient of kinetic friction along AB is mk = 0.2.

A

20 ft

30

SOLUTION +R©Fx = max;

4 ft

40 40 sin 30° - 6.928 = a 32.2 a = 10.52 ft>s

B C R

2

( +R) v2 = v20 + 2 ac (s - s0); v2B = (10)2 + 2(10.52)(20) vB = 22.82 ft>s ( + R) v = v0 + ac t; 22.82 = 10 + 10.52 tAB tAB = 1.219 s + ) s = (s ) + (v ) t (: x x 0 x 0 R = 0 + 22.82 cos 30° (tBC) 1 a t2 2 c 1 + (32.2)(tBC)2 2

( + T) sy = (sy)0 + (vy)0 t + 4 = 0 + 22.82 sin 30° tBC tBC = 0.2572 s R = 5.08 ft

Ans.

Total time = tAB + tBC = 1.48 s

Ans.

Ans: R = 5.08 ft t AC = 1.48 s 263

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*13–20. The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Take u = 30°.

vA  2.5 m/s A 3m

u

B

Solution Q + ΣFy = may ;

NC - 12(9.81) cos 30° = 0



NC = 101.95 N

+R ΣFx = max;

12(9.81) sin 30° - 0.3(101.95) = 12 aC



aC = 2.356 m>s2

(+R)

v2B = v2A + 2 aC(sB - sA)



v2B = (2.5)2 + 2(2.356)(3 - 0)



vB = 4.5152 = 4.52 m>s

Ans.

Ans: vB = 4.52 m>s 264

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13–21. The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the smallest incline u of the ramp so that the crates will slide off and fall into the cart.

vA  2.5 m/s A 3m

u

B

Solution (+R) v2B = v2A + 2aC(sB - sA)

0 = (2.5)2 + 2(aC)(3 - 0)



aC = 1.0417

Q + ΣFy = may;    NC - 12(9.81) cos u = 0

NC = 117.72 cos u

+R ΣFx = max;   12(9.81) sin u - 0.3(NC) = 12 (1.0417)

117.72 sin u - 35.316 cos u - 12.5 = 0

Solving, Ans.

u = 22.6°

Ans: u = 22.6° 265

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13–22. The 50-kg block A is released from rest. Determine the velocity of the 15-kg block B in 2 s. E

C

D

Solution Kinematics. As shown in Fig. a, the position of block B and point A are specified by sB and sA respectively. Here the pulley system has only one cable which gives

sA + sB + 2(sB - a) = l



sA + 3sB = l + 2a(1)

B

A

Taking the time derivative of Eq. (1) twice, aA + 3aB = 0(2)



Equations of Motion. The FBD of blocks B and A are shown in Fig. b and c. To be consistent to those in Eq. (2), aA and aB are assumed to be directed towards the positive sense of their respective position coordinates sA and sB. For block B, + c ΣFy = may;  3T - 15(9.81) = 15( - aB)(3) For block A, + c ΣFy = may;  T - 50(9.81) = 50( - aA)(4) Solving Eqs. (2), (3) and (4), aB = - 2.848 m>s2 = 2.848 m>s2 c   aA = 8.554 m>s2  T = 63.29 N The negative sign indicates that aB acts in the sense opposite to that shown in FBD. The velocity of block B can be determined using +c

vB = (vA)0 + aBt;  vB = 0 + 2.848(2) vB = 5.696 m>s = 5.70 m>s c 

Ans.

Ans: vB = 5.70 m>s c 266

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13–23. If the supplied force F = 150 N, determine the velocity of the 50-kg block A when it has risen 3 m, starting from rest.

C B F

Solution Equations of Motion. Since the pulleys are smooth, the tension is constant throughout each entire cable. Referring to the FBD of pulley C, Fig. a, of which its mass is negligible.

A

+ c ΣFy = 0;  150 + 150 - T = 0  T = 300 N Subsequently, considered the FBD of block A shown in Fig. b, + c ΣFy = may;  300 + 300 - 50(9.81) = 50a a = 2.19 m>s2 c



Kinematics. Using the result of a, ( + c ) v2 = v20 + 2ac s; v2 = 02 + 2(2.19)(3) Ans.

v = 3.6249 m>s = 3.62 m>s

Ans: v = 3.62 m>s c 267

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*13–24. A 60-kg suitcase slides from rest 5 m down the smooth ramp. Determine the distance R where it strikes the ground at B. How long does it take to go from A to B?

A

5m

C

30

2.5 m

Solution Equation of Motion. Referring to the FBD of the suitcase shown in Fig. a

B R

+b ΣFx′ = max′;   60(9.81) sin 30° = 60a  a = 4.905 m>s2 Kinematics. From A to C, the suitcase moves along the inclined plane (straight line). ( +b) v2 = v20 + 2acs;

v2 = 02 + 2(4.905)(5)



v = 7.0036 m>s

( + b) s = s0 + v0 t +

1 1 a t 2;  5 = 0 + 0 + (4.905)t 2AC 2 c 2



t AC = 1.4278 s

From C to B, the suitcase undergoes projectile motion. Referring to x–y coordinate system with origin at C, Fig. b, the vertical motion gives ( + T)

sy = (s0)y + vy t +

1 a t 2; 2 y 2.5 = 0 + 7.0036 sin 30° t CB +



1 (9.81)t 2CB 2

2 4.905 t CB + 3.5018 t CB - 2.5 = 0 Solve for positive root, t CB = 0.4412 s Then, the horizontal motion gives

+ ) sx = (s0)x + vxt; (d R = 0 + 7.0036 cos 30° (0.4412) = 2.676 m = 2.68 m  The time taken from A to B is t AB = t AC + t CB = 1.4278 + 0.4412 = 1.869 s = 1.87 s

Ans. Ans.

Ans: + ) sx = 2.68 m (d t AB = 1.87 s 268

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13–25. Solve Prob. 13–24 if the suitcase has an initial velocity down the ramp of vA = 2 m>s, and the coefficient of kinetic friction along AC is mk = 0.2.

A

5m

C

30

2.5 m

Solution Equations of Motion. The friction is Ff = mkN = 0.2N. Referring to the FBD of the suitcase shown in Fig. a +

a ΣFy′ = may′;  N - 60(9.81) cos

30 °

B R

= 60(0)

N = 509.74 N

+ b ΣFx′ = max′;  60(9.81) sin 30° - 0.2(509.74) = 60 a a = 3.2059 m>s2 b



Kinematics. From A to C, the suitcase moves along the inclined plane (straight line). (+ b)  v2 = v20 + 2ac s;  v2 = 22 + 2(3.2059)(5) v = 6.0049 m>s b

( + b)

s = s0 + v0 t +

1 2 1 a t ;  5 = 0 + 2t AC + (3.2059)t 2AC 2 c 2 1.6029 t 2AC + 2t AC - 5 = 0

Solve for positive root,

t AC = 1.2492 s



From C to B, the suitcase undergoes projectile motion. Referring to x–y coordinate system with origin at C, Fig. b, the vertical motion gives ( + T)

sy = (s0)y + vyt +

1 2 at ; 2 y

2.5 = 0 + 6.0049 sin 30° t CB +

1 (9.81)t 2CB 2

4.905 t 2CB + 3.0024 t CB - 2.5 = 0 Solve for positive root, t CB = 0.4707 s Then, the horizontal motion gives + ) sx = (s0)x + vxt; (d R = 0 + 6.0049 cos 30° (0.4707) Ans.

= 2.448 m = 2.45 m The time taken from A to B is t AB = t AC + t CB = 1.2492 + 0.4707 = 1.7199 s = 1.72 s

Ans.

Ans: R = 2.45 m t AB = 1.72 s 269

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13–26. The 1.5 Mg sports car has a tractive force of F = 4.5 kN. If it produces the velocity described by v-t graph shown, plot the air resistance R versus t for this time period.

R

F v (m/s) v  (–0.05t 2 + 3t ) m/s 45

Solution Kinematic. For the v9t graph, the acceleration of the car as a function of t is a =

dv = 5- 0.1t + 36m>s2 dt

Equation of Motion. Referring to the FBD of the car shown in Fig. a,

30

t (s)

+ ) ΣFx = max;  4500 - R = 1500( -0.1t + 3) (d

R = 5150t6 N

The plot of R vs t is shown in Fig. b

Ans: R = {150t} N 270

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13–27. The conveyor belt is moving downward at 4 m>s. If  the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

4 m/s B

30

Solution Equations of Motion. It is required that the package is on the verge to slide. Thus, Ff = msN = 0.8N. Referring to the FBD of the package shown in Fig. a, + a ΣFy′ = may′;  N - 15(9.81) cos 30° = 15(0)  N = 127.44 N + Q ΣFx′ = max′;  0.8(127.44) - 15(9.81) sin 30° = 15 a a = 1.8916 m>s2 Q



Kinematic. Since the package is required to stop, v = 0. Here v0 = 4 m>s. (+ b)  v = v0 + a0 t; 0 = 4 + ( - 1.8916) t Ans.

t = 2.1146 s = 2.11 s

Ans: t = 2.11 s 271

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*13–28. At the instant shown the 100-lb block A is moving down the plane at 5 ft>s while being attached to the 50-lb block B. If the coefficient of kinetic friction between the block and the incline is mk = 0.2, determine the acceleration of A and the distance A slides before it stops. Neglect the mass of the pulleys and cables.

A

5 3 4

C D

B

Solution Block A: 3 100 + b ΣFx = max;  - TA - 0.2NA + 100 a b = a ba 5 32.2 A 4 + a ΣFy = may;  NA - 100 a b = 0 5 Thus, TA - 44 = - 3.1056aA(1) Block B: TB - 50 = a

+ c ΣFy = may;

50 baB 32.2

TB - 50 = 1.553aB(2) Pulleys at C and D: + c ΣFy = 0;

2TA - 2TB = 0

TA = TB(3) Kinematics: sA + 2sC = l sD + (sD - sB) = l′ sC + d + sD = d′ Thus, aA = - 2aC 2aD = aB aC = - aD′ so that aA = aB(4) Solving Eqs. (1)–(4): aA = aB = - 1.288 ft>s2 TA = TB = 48.0 lb Thus, aA = 1.29 ft>s2 2

(+ b)  v =

v20

Ans. + 2ac (s - s0) 2



0 = (5) + 2( -1.288)(s - 0)



s = 9.70 ft

Ans.

272

Ans: aA = 1.29 ft>s2 s = 9.70 ft

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13–29. F (lb)

The force exerted by the motor on the cable is shown in the graph. Determine the velocity of the 200-lb crate when t = 2.5 s.

250 lb M 2.5

SOLUTION

t (s)

Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force F must overcome the weight of the crate. Thus, the time required to move the crate is given by + c ©Fy = 0;

100t - 200 = 0

t = 2s

Equation of Motion: For 2 s 6 t 6 2.5 s, F = + c ©Fy = may;

100t - 200 =

A

250 t = (100t) lb. By referring to Fig. a, 2.5

200 a 32.2

a = (16.1t - 32.2) ft>s2 Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 2 s … t 6 2.5 s, v = 0 at t = 2 s will be used as the lower integration limit.Thus, (+ c)

L

dv = v

L0

L

adt t

dv =

L2 s

(16.1t - 32.2)dt

v = A 8.05t2 - 32.2t B 2

t 2s

= A 8.05t2 - 32.2t + 32.2 B ft>s When t = 2.5 s, v = 8.05(2.52) - 32.2(2.5) + 32.2 = 2.01 ft>s

Ans.

Ans: v = 2.01 ft>s 273

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13–30. F (N)

The force of the motor M on the cable is shown in the graph. Determine the velocity of the 400-kg crate A when t = 2 s.

2500 F  625 t 2 M

2

t (s)

SOLUTION Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a. Equilibrium: For the crate to move, force 2F must overcome its weight. Thus, the time required to move the crate is given by + c ©Fy = 0;

2(625t2) - 400(9.81) = 0

A

t = 1.772 s Equations of Motion: F = A 625t2 B N. By referring to Fig. a, + c ©Fy = may;

2 A 625t2 B - 400(9.81) = 400a a = (3.125t2 - 9.81) m>s2

Kinematics: The velocity of the crate can be obtained by integrating the kinematic equation, dv = adt. For 1.772 s … t 6 2 s, v = 0 at t = 1.772 s will be used as the lower integration limit. Thus, (+ c)

L L0

dv = v

dv =

L

adt t

L1.772 s

A 3.125t2 - 9.81 B dt

v = A 1.0417t3 - 9.81t B 2

t 1.772 s

= A 1.0417t - 9.81t + 11.587 B m>s 3

When t = 2 s, v = 1.0417(23) - 9.81(2) + 11.587 = 0.301 m>s

Ans.

Ans: v = 0.301 m>s 274

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13–31. The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to the right at a constant speed of 4 m> s, determine the tension in the rope when sA = 5 m. When sA = 0, sB = 0.

12 m

SOLUTION 12 - sB +

2s2A

sB

B A

2

+ (12) = 24

1 # -sB + A s2A + 144 B - 2 a sAsA b = 0

sA

3 1 1 $ # 2 # $ - sB - A s2A + 144 B - 2 asAsA b + A s2A + 144 B - 2 a s2A b + A s2A + 144 B - 2 asA sA b = 0

$ sB = - C

aB = - C

# s2As2A

-

3

A s2A + 144 B 2 (5)2(4)2 ((5)2 + 144)

+ c ©Fy = may ;

3 2

-

# $ s2A + sAsA 1

A s2A + 144 B 2

S

(4)2 + 0 1

((5)2 + 144)2

S = 1.0487 m>s2

T - 150(9.81) = 150(1.0487) Ans.

T = 1.63 kN

Ans: T = 1.63 kN 275

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*13–32. The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to the right with an acceleration of 3 m>s2 and has a velocity of 4 m> s at the instant sA = 5 m, determine the tension in the rope at this instant. When sA = 0, sB = 0.

12 m

SOLUTION 12 = sB +

2s2A

sB

B A

2

+ (12) = 24 sA

3 1 # # - sB + A s2A + 144 B - 2 a 2sA sA b = 0 2 3 1 1 $ # 2 # $ - sB - A s2A + 144 B - 2 asAsA b + A s2A + 144 B - 2 a s2A b + A s2A + 144 B - 2 asAsA b = 0

$ sB = - C

aB = - C

# s2A s2A

A s2A + 144 B

3 2

-

(5)2(4)2 3

((5)2 + 144)2

+ c ©Fy = may ;

-

# $ s2A + sA sA 1

A s2A + 144 B 2

S

(4)2 + (5)(3) 1

((5)2 + 144)2

S = 2.2025 m>s2

T - 150(9.81) = 150(2.2025) Ans.

T = 1.80 kN

Ans: T = 1.80 kN 276

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13–33. Block A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that it will not slide on A. Also, what is the corresponding acceleration? The coefficient of static friction between A and B is ms. Neglect any friction between A and the horizontal surface.

P

B

A



Solution Equations of Motion. Since block B is required to be on the verge to slide on A, Ff = msNB. Referring to the FBD of block B shown in Fig. a, + c ΣFy = may;  NB cos u - msNB sin u - mg = m(0)

NB =

mg  cos u - ms sin u

(1)

+ ΣFx = max;   P - NB sin u - msNB cos u = ma d (2)

P - NB (sin u + ms cos u) = ma

Substitute Eq. (1) into (2),

P - a



sin u + ms cos u b mg = ma cos u - ms sin u

(3)

Referring to the FBD of blocks A and B shown in Fig. b + ΣFx = max;   P = 2 ma d

(4)

Solving Eqs. (2) into (3), P = 2mg a a = a

sin u + ms cos u b cos u - ms sin u

Ans.

sin u + ms cos u bg cos u - ms sin u

Ans.

Ans: sin u + ms cos u b cos u - ms sin u sin u + ms cos u a = a bg cos u - ms sin u

P = 2mg a

277

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13–34. The 4-kg smooth cylinder is supported by the spring having a stiffness of kAB = 120 N>m. Determine the velocity of the cylinder when it moves downward s = 0.2 m from its equilibrium position, which is caused by the application of the force F = 60 N.

F  60 N

B s kAB  120 N/m

Solution Equation of Motion. At the equilibrium position, realizing that Fsp = kx0 = 120x0 the compression of the spring can be determined from

A

+ c ΣFy = 0;  120x0 - 4(9.81) = 0  x0 = 0.327 m Thus, when 60  N force is applied, the compression of the spring is x = s + x0 = s + 0.327. Thus, Fsp = kx = 120(s + 0.327). Then, referring to the FBD of the collar shown in Fig. a, + c ΣFy = may;  120(s + 0.327) - 60 - 4(9.81) = 4( -a) a = 515 - 30 s 6 m>s2



Kinematics. Using the result of a and integrate 1 vdv = ads with the initial condition v = 0 at s = 0, L0

v

vdv =

L0

s

(15 - 30 s)ds

v2 = 15 s - 15 s2 2 v = At s = 0.2 m,

5 230(s

- s2) 6 m>s

v = 230(0.2 - 0.22) = 2.191 m>s = 2.19 m>s

Ans.

Ans: v = 2.19 m>s 278

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13–35. The coefficient of static friction between the 200-kg crate and the flat bed of the truck is ms = 0.3. Determine the shortest time for the truck to reach a speed of 60 km> h, starting from rest with constant acceleration, so that the crate does not slip.

SOLUTION Free-Body Diagram: When the crate accelerates with the truck, the frictional force Ff develops. Since the crate is required to be on the verge of slipping, Ff = msN = 0.3N. Equations of Motion: Here, ay = 0. By referring to Fig. a, + c ©Fy = may;

N - 200(9.81) = 200(0) N = 1962 N

+ ©F = ma ; : x x

-0.3(1962) = 200(- a) a = 2.943 m>s2 ;

km 1000 m 1h ba ba b = h 1 km 3600 s 16.67 m>s. Since the acceleration of the truck is constant,

Kinematics: The final velocity of the truck is v = a 60

+ ) (;

v = v0 + ac t 16.67 = 0 + 2.943t Ans.

t = 5.66 s

Ans: t = 5.66 s 279

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*13–36. The 2-lb collar C fits loosely on the smooth shaft. If the spring is unstretched when s = 0 and the collar is given a velocity of 15 ft> s, determine the velocity of the collar when s = 1 ft.

15 ft/s

s C

1 ft

SOLUTION Fs = kx;

k

Fs = 4 A 21 + s2 - 1 B -4 A 21 + s2 - 1 B ¢

+ ©F = ma ; : x x 1

-

L0

4 lb/ft

¢ 4s ds -

4s ds 21 + s 1

- C 2s2 - 4 31 + s2 D 0 =

2

≤ =

v

L15

a

s 21 + s2

≤ = a

2 dv b av b 32.2 ds

2 b v dv 32.2

1 A v2 - 152 B 32.2 Ans.

v = 14.6 ft>s

Ans: v = 14.6 ft>s 280

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13–37. The 10-kg block A rests on the 50-kg plate B in the position shown. Neglecting the mass of the rope and pulley, and using the coefficients of kinetic friction indicated, determine the time needed for block A to slide 0.5 m on the plate when the system is released from rest.

0.5 m

Solution

mAB  0.2

Block A:

C

A B

mBC  0.1

30

+ a ΣFy = may;  NA - 10(9.81) cos 30° = 0  NA = 84.96 N + b ΣFx = max;  - T + 0.2(84.96) + 10(9.81) sin 30° = 10aA T - 66.04 = -10aA(1)

Block B: + a ΣFy = may;

NB - 84.96 - 50(9.81) cos 30° = 0



NB = 509.7 N

+ b ΣFx = max;

- 0.2(84.96) - 0.1(509.7) - T + 50(9.81 sin 30°) = 50aB



177.28 - T = 50aB(2)

sA + sB = l ∆sA = - ∆sB aA = - aB(3) Solving Eqs. (1) – (3): aB = 1.854 m>s2 aA = - 1.854 m>s2  T = 84.58 N In order to slide 0.5 m along the plate the block must move 0.25 m. Thus, (+ b)  sB = sA + sB>A - ∆sA = ∆sA + 0.5 ∆sA = - 0.25 m (+ b)  sA = s0 + v0 t + - 0.25 = 0 + 0 +

1 a t2 2 A

1 ( - 1.854)t 2 2 Ans.

t = 0.519 s

Ans: t = 0.519 s 281

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13–38. The 300-kg bar B, originally at rest, is being towed over a series of small rollers. Determine the force in the cable when t = 5 s, if the motor M is drawing in the cable for a short time at a rate of v = (0.4t2) m>s, where t is in seconds (0 … t … 6 s). How far does the bar move in 5 s? Neglect the mass of the cable, pulley, and the rollers.

M

Solution

v

+ ΣFx = max;  T = 300a S

B

v = 0.4t 2 a =

dv = 0.8t dt

When t = 5 s, a = 4 m>s2 Ans.

T = 300(4) = 1200 N = 1.20 kN ds = v dt L0

s

ds =

s = a

L0

5

0.4t 2 ds

0.4 b(5)3 = 16.7 m 3

Ans.

Ans: s = 16.7 m 282

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13–39. y

An electron of mass m is discharged with an initial horizontal velocity of v0. If it is subjected to two fields of force for which Fx = F0 and Fy = 0.3F0, where F0 is constant, determine the equation of the path, and the speed of the electron at any time t.

++++++++++++++

v0

+ ©F = ma ; : x x

F0 = max

+ c ©Fy = may;

0.3 F0 = may

Thus, vx

t

dvx =

Lv0

++++++++++++++

SOLUTION

F0 dt L0 m

F0 t + v0 m vy t 0.3F0 dvy = dt L0 L0 m

vx =

vy =

0.3F0 t m

2 F0 0.3F0 2 t + v0 b + a tb m C m

a

v =

1 21.09F20 t2 + 2F0tmv0 + m2v20 m

= x

t

dx =

L0

L0

a

Ans.

F0 t + v0 b dt m

F0 t2 x = + v0 t 2m y

L0

t

dy =

y =

0.3F0 t dt L0 m

0.3F0 t2 2m

t = a

1 2m by2 B 0.3F0

x =

F0 2m 1 2m a b y + v0 a by2 2m 0.3F0 B 0.3F0

x =

y 1 2m + v0 a by2 0.3 B 0.3F0

Ans.

Ans: 1 21.09F20 t2 + 2F0tmv0 + m2v20 m y 2m x = + v0a by1>2 0.3 A 0.3F0

v =

283

x

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*13–40. The 400-lb cylinder at A is hoisted using the motor and the pulley system shown. If the speed of point B on the cable is increased at a constant rate from zero to vB = 10 ft>s in  t = 5 s, determine the tension in the cable at B to cause the motion.

B

Solution vA

2sA + sB = l

A

2aA = - aB + 2 1S

v = v0 + aBt

10 = 0 + aB(5) aB = 2 ft>s2 aA = - 1 ft>s2 + T ΣFy = may;  400 - 2T = a

400 b( -1) 32.2 Ans.

Thus,  T = 206 lb

Ans: T = 206 lb 284

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13–41. Block A has a mass mA and is attached to a spring having a stiffness k and unstretched length l0. If another block B, having a mass mB, is pressed against A so that the spring deforms a distance d, determine the distance both blocks slide on the smooth surface before they begin to separate. What is their velocity at this instant?

k

A

B

SOLUTION Block A: + ©F = ma ; : x x

-k(x - d) - N = mA aA

Block B: + ©F = ma ; : x x

N = mB aB

Since aA = aB = a, - k(x - d) - mB a = mA a a =

k(d - x) (mA + mB)

N =

N = 0 when d - x = 0,

kmB (d - x) (mA + mB) or

Ans.

x = d

y dy = a dx v

L0

d

y dv =

k(d - x) dx L0 (mA + mB)

d k 1 1 1 2 kd2 y = B (d)x - x2 R = 2 (mA + mB) 2 2 (mA + mB) 0

y =

kd2 C (mA + mB)

Ans.

Ans: x = d v = 285

kd 2 B mA + mB

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13–42. Block A has a mass mA and is attached to a spring having a stiffness k and unstretched length l0. If another block B, having a mass mB, is pressed against A so that the spring deforms a distance d, show that for separation to occur it is necessary that d 7 2mkg(mA + mB)>k, where mk is the coefficient of kinetic friction between the blocks and the ground. Also, what is the distance the blocks slide on the surface before they separate?

k

A

B

SOLUTION Block A: + ©F = ma ; : x x

-k(x - d) - N - mk mA g = mA aA

Block B: + ©F = ma ; : x x

N - mk mB g = mB aB

Since aA = aB = a, k(d - x) k(d - x) - mk g(mA + mB) = - mk g (mA + mB) (mA + mB)

a =

kmB (d - x) (mA + mB)

N =

Ans.

N = 0, then x = d for separation. At the moment of separation: v dv = a dx v

L0

d

v dv =

L0

B

k(d - x) - mk g R dx (mA + mB)

d k 1 1 2 v = B (d)x - x2 - mk g x R 2 (mA + mB) 2 0

v =

kd2 - 2mk g(mA + mB)d B (mA + mB)

Require v 7 0, so that kd2 - 2mk g(mA + mB)d 7 0 Thus, kd 7 2mk g(mA + mB) d 7

2mk g (mA + mB) k

Q.E.D.

Ans: x = d for separation. 286

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13–43. A parachutist having a mass m opens his parachute from an at-rest position at a very high altitude. If the atmospheric drag resistance is FD = kv2, where k is a constant, determine his velocity when he has fallen for a time t. What is his velocity when he lands on the ground? This velocity is referred to as the terminal velocity, which is found by letting the time of fall t S ∞.

FD

v

SOLUTION + T ΣFz = maz;

mg - kv2 = m v

m

t

m dv

( mg - kv2 )

L0

m k L0

dv dt

v

dv mg k

- v2

=

L0

dt

= t

2k + v 1 m ° ¢ln£ mg § = t mg k 22 2k - v k v

mg

0

2k + v mg k t¢2 ≤ = ln mg m A k 2k - v mg

e 2k = mg

2t

mg

A k

2k + v mg

2k - v mg

e 2t 3 k - v e 2t 3 k = mg

mg

mg

A k

+ v

mg e 2t 3 k - 1 £ v = § mg A k 2t 3 k e + 1 mg

When t S ∞

vt =

Ans.

mg A k

Ans.

Ans:

mg>k

mg e 2t 3 - 1 § £ A k mg>k e 2t 3 + 1 mg vt = A k

v =

287

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*13–44. If the motor draws in the cable with an acceleration of 3 m>s2, determine the reactions at the supports A and B. The beam has a uniform mass of 30 kg> m, and the crate has a mass of 200 kg. Neglect the mass of the motor and pulleys.

2.5 m

0.5 m

3m

A

B

3 m/s2

SOLUTION Sc + (Sc - Sp) 2yc = yp

C

2ac = ap 2ac = 3 m>s2 ac = 1.5 m>s2 + c ©Fy = may

2T - 1962 = 200(1.5) T = 1,131 N

a + ©MA = 0;

By (6) - (1765.8 + 1,131)3 - (1,131)(2.5) = 0 Ans.

By = 1,919.65 N = 1.92 kN + c ©Fy = 0; + : ©Fx = 0;

Ay - 1765.8 - (2)(1,131) + 1919.65 = 0 Ay = 2108.15 N = 2.11 kN

Ans.

Ax = 0

Ans.

288

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13–45. If the force exerted on cable AB by the motor is F = (100t3>2) N, where t is in seconds, determine the 50-kg crate’s velocity when t = 5 s. The coefficients of static and kinetic friction between the crate and the ground are ms = 0.4 and mk = 0.3, respectively. Initially the crate is at rest.

A

B

SOLUTION Free-Body Diagram: The frictional force Ff is required to act to the left to oppose the motion of the crate which is to the right. Equations of Motion: Here, ay = 0. Thus, + c ©Fy = may;

N - 50(9.81) = 50(0) N = 490.5 N

Realizing that Ff = mkN = 0.3(490.5) = 147.15 N, + c ©Fx = max;

100t3>2 - 147.15 = 50a a = A 2t3>2 - 2.943 B m>s

Equilibrium: For the crate to move, force F must overcome the static friction of Ff = msN = 0.4(490.5) = 196.2 N. Thus, the time required to cause the crate to be on the verge of moving can be obtained from. + ©F = 0; : x

100t3>2 - 196.2 = 0 t = 1.567 s

Kinematics: Using the result of a and integrating the kinematic equation dv = a dt with the initial condition v = 0 at t = 1.567 as the lower integration limit, + ) (:

L L0

dv = v

dv =

L

adt t

L1.567 s

A 2t3>2 - 2.943 B dt

v = A 0.8t5>2 - 2.943t B 2

t 1.567 s

v = A 0.8t5>2 - 2.943t + 2.152 B m>s When t = 5 s, v = 0.8(5)5>2 - 2.943(5) + 2.152 = 32.16 ft>s = 32.2 ft>s

Ans.

Ans: v = 32.2 ft>s 289

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13–46. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

A

P

u B C

SOLUTION Require aA = aB = a Block A: + c ©Fy = 0; + ©F = ma ; ; x x

N cos u - mg = 0 N sin u = ma a = g tan u

Block B: + ©F = ma ; ; x x

P - N sin u = ma P - mg tan u = mg tan u Ans.

P = 2mg tan u

Ans: P = 2mg tan u 290

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13–47. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip on B. The coefficient of static friction between A and B is ms. Neglect any friction between B and C.

A

P

u B C

SOLUTION Require aA = aB = a Block A: + c ©Fy = 0; + ©F = ma ; ; x x

N cos u - ms N sin u - mg = 0 N sin u + ms N cos u = ma N =

mg cos u - ms sin u

a = ga

sin u + ms cos u b cos u - ms sin u

Block B: + ©F = ma ; ; x x

P - ms N cos u - N sin u = ma P - mg a

sin u + ms cos u sin u + ms cos u b = mga b cos u - ms sin u cos u - ms sin u

P = 2mg a

sin u + ms cos u b cos u - ms sin u

Ans.

Ans: P = 2mga

291

sin u + ms cos u b cos u - ms sin u

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*13–48. The smooth block B of negligible size has a mass m and rests on the horizontal plane. If the board AC pushes on the block at an angle u with a constant acceleration a0, determine the velocity of the block along the board and the distance s the block moves along the board as a function of time t. The block starts from rest when s = 0, t = 0.

a0

C

θ

A

B s

SOLUTION Q+ ©Fx = m ax;

0 = m aB sin f aB = aAC + aB>AC aB = a0 + aB>AC

Q+

aB sin f = - a0 sin u + aB>AC

Thus, 0 = m( - a0 sin u + aB>AC) aB>AC = a0 sin u vB>AC

L0

t

dvB>AC =

L0

a0 sin u dt Ans.

vB>AC = a0 sin u t t

sB>AC = s = s =

L0

a0 sin u t dt

1 a sin u t2 2 0

Ans.

Ans: vB>AC = a0 sin u t s = 292

1 a sin u t 2 2 0

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13–49. If a horizontal force P = 12 lb is applied to block A determine the acceleration of block B. Neglect friction. 15 lb B P

SOLUTION

8 lb A

15°

Block A: + ©F = ma ; : x x

12 - NB sin 15° = ¢

8 ≤a 32.2 A

(1)

NB cos 15° - 15 = ¢

15 ≤a 32.2 B

(2)

Block B: + c ©Fy = may ;

sB = sA tan 15° (3)

aB = aA tan 15° Solving Eqs. (1)–(3) aA = 28.3 ft>s2

NB = 19.2 lb

aB = 7.59 ft s2

Ans.

Ans: aB = 7.59 ft>s2 293

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13–50. A freight elevator, including its load, has a mass of 1 Mg. It is prevented from rotating due to the track and wheels mounted along its sides. If the motor M develops a constant tension T = 4 kN in its attached cable, determine the velocity of the elevator when it has moved upward 6 m starting from rest. Neglect the mass of the pulleys and cables.

M

Solution Equation of Motion. Referring to the FBD of the freight elevator shown in Fig. a, + c ΣFy = may;  3(4000) - 1000(9.81) = 1000a

a = 2.19 m>s2 c

Kinematics. Using the result of a,

(+c )   v2 = v20 + 2as;  v2 = 02 + 2(2.19)(6)

Ans.

v = 5.126 m>s = 5.13 m>s

Ans: v = 5.13 m>s 294

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13–51. The block A has a mass mA and rests on the pan B, which has a mass mB . Both are supported by a spring having a stiffness k that is attached to the bottom of the pan and to the ground. Determine the distance d the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched.

A B y d

k

SOLUTION For Equilibrium + c ©Fy = may;

Fs = (mA + mB)g yeq =

(mA + mB)g Fs = k k

Block: + c ©Fy = may ;

- mA g + N = mA a

Block and pan + c ©Fy = may;

- (mA + mB)g + k(yeq + y) = (mA + mB)a

Thus, - (mA + mB)g + k c a

-mAg + N mA + mB b g + y d = (mA + mB) a b mA k

Require y = d, N = 0 kd = - (mA + mB)g Since d is downward, d =

(mA + mB)g k

Ans.

Ans: d = 295

(mA + mB )g k

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*13–52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5 m from the platform’s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is m = 0.2.

z

5m

SOLUTION Equation of Motion: Since the girl is on the verge of slipping, Ff = msN = 0.2N. Applying Eq. 13–8, we have ©Fb = 0 ; ©Fn = man ;

N - 15(9.81) = 0

N = 147.15 N

0.2(147.15) = 15a

v2 b 5 Ans.

v = 3.13 m>s

Ans: v = 3.13 m>s 296

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13–53. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block is given a speed of v = 10 m>s, determine the radius r of the circular path along which it travels.

r

B v

SOLUTION Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A, i.e., T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the circular path (positive n axis). Equations of Motion: Realizing that an = ©Fn = man;

147.15 = 2 a

A

v2 102 and referring to Fig. (a), = r r

102 b r Ans.

r = 1.36 m

Ans: r = 1.36 m 297

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13–54. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block travels along a circular path of radius r = 1.5 m, determine the speed of the block.

r

B v

SOLUTION Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The tension in the cord is equal to the weight of cylinder A, i.e., T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the circular path (positive n axis). Equations of Motion: Realizing that an = ©Fn = man;

147.15 = 2 a

A

v2 v2 and referring to Fig. (a), = r 1.5

2

v b 1.5 Ans.

v = 10.5 m>s

Ans: v = 10.5 m>s 298

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13–55. Determine the maximum constant speed at which the pilot can travel around the vertical curve having a radius of curvature r = 800 m, so that he experiences a maximum acceleration an = 8g = 78.5 m>s2. If he has a mass of 70 kg, determine the normal force he exerts on the seat of the airplane when the plane is traveling at this speed and is at its lowest point.

r  800 m

Solution an =

v2 v2 ;  78.5 = r 800 Ans.

v = 251 m>s + c ΣFn = man;  N - 70(9.81) = 70(78.5)

Ans.

N = 6.18 kN

Ans: N = 6.18 kN 299

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*13–56. Cartons having a mass of 5 kg are required to move along the assembly line at a constant speed of 8 m/s. Determine the smallest radius of curvature, r, for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are ms = 0.7 and mk = 0.5, respectively.

8 m/s

ρ

SOLUTION + c ©Fb = m ab ;

N - W = 0 N = W Fx = 0.7W

+ ©F = m a ; ; n n

0.7W =

W 82 ( ) 9.81 r Ans.

r = 9.32 m

Ans: r = 9.32 m 300

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13–57. The collar A, having a mass of 0.75 kg, is attached to a spring having a stiffness of k = 200 N>m. When rod BC rotates about the vertical axis, the collar slides outward along the smooth rod DE. If the spring is unstretched when s = 0, determine the constant speed of the collar in order that s = 100 mm. Also, what is the normal force of the rod on the collar? Neglect the size of the collar.

C

k  200 N/m D

SOLUTION ©Fb = 0; ©Fn = man ; ©Ft = mat ;

A

Nb - 0.75(9.81) = 0

Nb = 7.36

E s

n2 b 200(0.1) = 0.75 a 0.10

B

Nt = 0 n = 1.63 m>s

Ans.

N = 2(7.36)2 + (0) = 7.36 N

Ans.

Ans: v = 1.63 m>s N = 7.36 N 301

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13–58. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the minimum constant speed the spool can have so that it does not slip down the rod.

z 5

3

4

S 0.25 m

A

SOLUTION 4 r = 0.25a b = 0.2 m 5 + ©F = m a ; ; n n

v2 3 4 b Ns a b - 0.2Ns a b = 2 a 5 5 0.2

+ c ©Fb = m ab;

4 3 Ns a b + 0.2Ns a b - 2(9.81) = 0 5 5 Ns = 21.3 N Ans.

v = 0.969 m>s

Ans: v = 0.969 m>s 302

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13–59. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the maximum constant speed the spool can have so that it does not slip up the rod.

z 5

3 4

S 0.25 m

A

SOLUTION 4 r = 0.25( ) = 0.2 m 5 + ©F = m a ; ; n n + c ©Fb = m ab ;

v2 3 4 Ns( ) + 0.2Ns( ) = 2( ) 5 5 0.2 4 3 Ns( ) - 0.2Ns( ) - 2(9.81) = 0 5 5 Ns = 28.85 N Ans.

v = 1.48 m s

Ans: v = 1.48 m>s 303

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*13–60. At the instant u = 60°, the boy’s center of mass G has a downward speed vG = 15 ft>s. Determine the rate of increase in his speed and the tension in each of the two supporting cords of the swing at this instant. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

u

10 ft

SOLUTION + R©Ft = mat ; Q+ ©Fn = man ;

G

60 cos 60° =

60 a 32.2 t

2T - 60 sin 60° =

at = 16.1 ft>s2 60 152 a b 32.2 10

Ans.

T = 46.9 lb

Ans.

Ans: at = 16.1 ft>s2 T = 46.9 lb 304

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13–61. At the instant u = 60°, the boy’s center of mass G is momentarily at rest. Determine his speed and the tension in each of the two supporting cords of the swing when u = 90°. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

u

10 ft

SOLUTION

G

60 60 cos u = a 32.2 t

+ R© t = mat;

Q+ ©Fn = man;

2T - 60 sin u =

v dn = a ds

60 v2 a b 32.2 10

(1)

however ds = 10du

90°

v

L0

at = 32.2 cos u

v dn =

L60°

322 cos u du Ans.

v = 9.289 ft>s From Eq. (1) 2T - 60 sin 90° =

60 9.2892 a b 32.2 10

Ans.

T = 38.0 lb

Ans: v = 9.29 ft>s T = 38.0 lb 305

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13–62. A girl having a mass of 25 kg sits at the edge of the merrygo-round so her center of mass G is at a distance of 1.5 m from the axis of rotation. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which she can have before she begins to slip off the merry-go-round. The coefficient of static friction between the girl and the merry-go-round is ms = 0.3.

z

1.5 m

G

SOLUTION + ©F = ma ; : n n

v2 0.3(245.25) = 25( ) 1.5 Ans.

v = 2.10 m>s

Ans: v = 2.10 m>s 306

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13–63. The pendulum bob B has a weight of 5 lb and is released from rest in the position shown, u = 0°. Determine the tension in string BC just after the bob is released, u = 0°, and also at the instant the bob reaches point D, u = 45°. Take r = 3 ft.

B

C

r

SOLUTION Equation of Motion: Since the bob is just being release, y = 0. Applying Eq. 13–8 to FBD(a), we have ©Fn = man ;

T =

02 5 a b = 0 32.2 3

Ans.

Applying Eq. 13–8 to FBD(b), we have ©Ft = mat ; ©Fn = man ;

5 a 32.2 t

5 cos u =

T - 5 sin u =

a t = 32.2 cos u 5 y2 a b 32.2 3

[1]

Kinematics: The speed of the bob at the instant when u = 45° can be determined using ydy = at ds. However, ds = 3du, then ydy = 3a t du. 45°

y

L0

ydy = 3(32.2)

L0

cos udu

y2 = 136.61 ft2>s2 Substitute u = 45° and y2 = 136.61 ft2>s2 into Eq. [1] yields T - 5 sin 45° =

5 136.61 a b 32.2 3 Ans.

T = 10.6 lb

Ans: T = 0 T = 10.6 lb 307

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*13–64. The pendulum bob B has a mass m and is released from rest when u = 0°. Determine the tension in string BC immediately afterwards, and also at the instant the bob reaches the arbitrary position u.

B

C

r

SOLUTION Equation of Motion: Since the bob is just being release, y = 0. Applying Eq. 13–8 to FBD(a), we have ©Fn = man ;

T = ma

02 b = 0 r

Ans.

Applying Eq. 13–8 to FBD(b), we have ©Ft = mat ; ©Fn = man ;

mg cos u = ma t

at = g cos u

T - mg sin u = ma

y2 b r

[1]

Kinematics: The speed of the bob at the arbitrary position u can be detemined using ydy = at ds. However, ds = rdu, then ydy = a t rdu. y

L0

u

ydy = gr

L0

cos udu

y2 = 2gr sin u Substitute y2 = 2gr sin u into Eq. [1] yields T - mg sin u = ma

2gr sin u b r Ans.

T = 3mg sin u

Ans: T = 0 T = 3mg sin u 308

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13–65. Determine the constant speed of the passengers on the amusement-park ride if it is observed that the supporting cables are directed at u = 30° from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what are the components of force in the n, t, and b directions which the chair exerts on a 50-kg passenger during the motion?

4m

b u

SOLUTION

6m t

n

+ ©F = m a ; ; n n

v2 ) T sin 30° = 80( 4 + 6 sin 30°

+ c ©Fb = 0;

T cos 30° - 8019.812 = 0 T = 906.2 N v = 6.30 m>s 16.3022 ) = 283 N Fn = 50( 7

Ans.

©Ft = m at;

Ft = 0

Ans.

©Fb = m ab ;

Fb - 490.5 = 0

©Fn = m an ;

Ans.

Ans.

Fb = 490 N

Ans: v = 6.30 m>s Fn = 283 N Ft = 0 Fb = 490 N 309

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13–66. A motorcyclist in a circus rides his motorcycle within the confines of the hollow sphere. If the coefficient of static friction between the wheels of the motorcycle and the sphere is ms = 0.4, determine the minimum speed at which he must travel if he is to ride along the wall when u = 90°. The mass of the motorcycle and rider is 250 kg, and the radius of curvature to the center of gravity is r = 20 ft. Neglect the size of the motorcycle for the calculation.

u

Solution

v2 + gFn = man;   N = 250 a b S 20 + c gFb = mab;   0.4 N - 250(9.81) = 0

Solving,          v = 22.1 m>s

Ans.

Ans: v = 22.1 m>s 310

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13–67. The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle is traveling at a constant speed of 80 km> h along a circular curved road of radius 100 m, determine the tilt angle u of the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver.

u

SOLUTION Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a). Here, an must be directed towards the center of the circular path (positive n axis). 1h km 1000 m ba ba b h 1 km 3600 s = 22.22 m>s. Thus, the normal component of the passenger’s acceleration is given by 22.222 v2 = 4.938 m>s2. By referring to Fig. (a), = an = r 100

Equations of Motion: The speed of the passenger is v = a80

+ c ©Fb = 0; + ©F = ma ; ; n n

N cos u - m(9.81) = 0

N =

9.81m cos u

9.81m sin u = m(4.938) cos u Ans.

u = 26.7°

Ans: u = 26.7° 311

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*13–68. The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m> s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car.

y

y

20 (1

x2 ) 6400 A

SOLUTION dy d2y = - 0.00625x and 2 = - 0.00625. The slope angle u at point Geometry: Here, dx dx A is given by tan u =

dy 2 = - 0.00625(80) dx x = 80 m

80 m

u = -26.57°

and the radius of curvature at point A is r =

[1 + (dy>dx)2]3>2 |d2y>dx2|

=

[1 + ( -0.00625x)2]3>2 2 = 223.61 m | -0.00625| x = 80 m

Equations of Motion: Here, at = 0. Applying Eq. 13–8 with u = 26.57° and r = 223.61 m, we have ©Ft = mat;

800(9.81) sin 26.57° - Ff = 800(0) Ans.

Ff = 3509.73 N = 3.51 kN ©Fn = man;

800(9.81) cos 26.57° - N = 800 a

92 b 223.61 Ans.

N = 6729.67 N = 6.73 kN

Ans: Ff = 3.51 kN N = 6.73 kN 312

x

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13–69. The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m> s and increasing its speed at 3 m>s2. Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car.

y

y

20 (1

x2 ) 6400 A

SOLUTION dy d2y Geometry: Here, = - 0.00625x and 2 = - 0.00625. The slope angle u at point dx dx A is given by tan u =

dy 2 = - 0.00625(80) dx x = 80 m

80 m

u = -26.57°

and the radius of curvature at point A is r =

C 1 + (dy>dx)2 D 3>2 |d2y>dx2|

=

C 1 + ( - 0.00625x)2 D 3>2 |-0.00625|

2

x = 80 m

= 223.61 m

Equation of Motion: Applying Eq. 13–8 with u = 26.57° and r = 223.61 m, we have ©Ft = mat;

800(9.81) sin 26.57° - Ff = 800(3) Ans.

Ff = 1109.73 N = 1.11 kN ©Fn = man;

800(9.81) cos 26.57° - N = 800 a

92 ≤ 223.61 Ans.

N = 6729.67 N = 6.73 kN

Ans: Ff = 1.11 kN N = 6.73 kN 313

x

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13–70. The package has a weight of 5 lb and slides down the chute. When it reaches the curved portion AB, it is traveling at 8 ft>s 1u = 0°2. If the chute is smooth, determine the speed of the package when it reaches the intermediate point C 1u = 30°2 and when it reaches the horizontal plane 1u = 45°2. Also, find the normal force on the package at C.

45° 8 ft/s

θ = 30° 20 ft A

SOLUTION +b ©Ft = mat ;

45°

B

C

5 a 32.2 t

5 cos f =

at = 32.2 cos f +a©Fn = man ;

N - 5 sin f =

v2 5 ( ) 32.2 20

v dv = at ds v

Lg

f

v dv =

L45°

32.2 cos f (20 df)

1 2 1 v - (8)2 = 644 (sin f - sin 45°) 2 2 At f = 45° + 30° = 75°, vC = 19.933 ft>s = 19.9 ft>s

Ans.

NC = 7.91 lb

Ans.

vB = 21.0 ft s

Ans.

At f = 45° + 45° = 90°

Ans: vC = 19.9 ft>s NC = 7.91 lb vB = 21.0 ft>s 314

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13–71. The 150-lb man lies against the cushion for which the coefficient of static friction is ms = 0.5. Determine the resultant normal and frictional forces the cushion exerts on him if, due to rotation about the z axis, he has a constant speed v = 20 ft>s. Neglect the size of the man.Take u = 60°.

z

8 ft G

SOLUTION + a a Fy = m1an2y ;

N - 150 cos 60° =

150 202 a b sin 60° 32.2 8 Ans.

N = 277 lb + b a Fx = m1an2x ;

- F + 150 sin 60° =

u

150 202 a b cos 60° 32.2 8 Ans.

F = 13.4 lb Note: No slipping occurs Since ms N = 138.4 lb 7 13.4 lb

Ans: N = 277 lb F = 13.4 lb 315

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*13–72. The 150-lb man lies against the cushion for which the coefficient of static friction is ms = 0.5. If he rotates about the z axis with a constant speed v = 30 ft>s, determine the smallest angle u of the cushion at which he will begin to slip off.

z

8 ft G

SOLUTION 2

150 1302 a b 32.2 8

+ ©F = ma ; ; n n

0.5N cos u + N sin u =

+ c ©Fb = 0;

- 150 + N cos u - 0.5 N sin u = 0 N =

u

150 cos u - 0.5 sin u

10.5 cos u + sin u2150 1cos u - 0.5 sin u2

2

=

150 1302 a b 32.2 8

0.5 cos u + sin u = 3.49378 cos u - 1.74689 sin u Ans.

u = 47.5°

Ans: u = 47.5° 316

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13–73. Determine the maximum speed at which the car with mass m can pass over the top point A of the vertical curved road and still maintain contact with the road. If the car maintains this speed, what is the normal reaction the road exerts on the car when it passes the lowest point B on the road?

r

A

r

B r

SOLUTION

r

Free-Body Diagram: The free-body diagram of the car at the top and bottom of the vertical curved road are shown in Figs. (a) and (b), respectively. Here, an must be directed towards the center of curvature of the vertical curved road (positive n axis). Equations of Motion: When the car is on top of the vertical curved road, it is required that its tires are about to lose contact with the road surface. Thus, N = 0. v2 v2 = Realizing that an = and referring to Fig. (a), r r + T ©Fn = man;

mg = m ¢

v2 ≤ r

v = 2gr

Ans.

Using the result of v, the normal component of car acceleration is gr v2 an = = = g when it is at the lowest point on the road. By referring to Fig. (b), r r + c ©Fn = man;

N - mg = mg Ans.

N = 2mg

Ans: v = 2gr N = 2mg 317

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13–74. y

Determine the maximum constant speed at which the 2-Mg car can travel over the crest of the hill at A without leaving the surface of the road. Neglect the size of the car in the calculation.

2

A

y  20 (1  x ) 10 000

x 100 m

Solution Geometry. The radius of curvature of the road at A must be determined first. Here dy 2x = 20 a b = - 0.004x dx 10000 d 2y dx2

= - 0.004

At point A, x = 0. Thus,

r =

c1 + a `

dy ` = 0. Then dx x = 0

dy 2 3>2 b d dx 2

d y dx

` 2

=

( 1 + 02 ) 3>2 0.004

= 250 m

Equation of Motion. Since the car is required to be on the verge to leave the road surface, N = 0. gFn = man;  2000(9.81) = 2000 a

v2 b 250 Ans.

v = 49.52 m>s = 49.5 m>s

Ans: v = 49.5 m>s 318

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13–75. The box has a mass m and slides down the smooth chute having the shape of a parabola. If it has an initial velocity of v0 at the origin, determine its velocity as a function of x. Also, what is the normal force on the box, and the tangential acceleration as a function of x?

y

x x

y = – 0.5 x2

SOLUTION x = -

1 2 x 2

dy = -x dx d 2y dx2

= -1 3

B1 + a r =

2

dy 2 2 b R dx

2

dy dx2

3

=

2

C 1 + x2 D 2 | - 1|

mg ¢

+b©Fn = man ;

mg ¢

+R©Ft = mat ;

1 21 + x

2

x 21 + x2

at = g ¢ v dv = at ds = g ¢

3

= A 1 + x2 B 2

x 21 + x2

≤ - N = m¢

v2 3

(1 + x2)2

≤

(1)

≤ = mat

x 21 + x2

≤

Ans.

≤ ds

1

ds = B 1 + a

dy 2 2 1 b R dx = A 1 + x2 B 2 dx dx

v

Lv0

x

v dv =

L0

gx dx

1 2 1 x2 v - v 20 = g a b 2 2 2 v = 2v20 + gx2

Ans.

From Eq. (1): N =

m 1 + x2

g -

(v20 + gx2)

Ans.

(1 + x2)

Ans: at = g a

x

b 21 + x2 v = 2v20 + gx2 v20 + gx2 m N = cg d 1 + x2 21 + x2

319

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*13–76. Prove that if the block is released from rest at point B of a smooth path of arbitrary shape, the speed it attains when it reaches point A is equal to the speed it attains when it falls freely through a distance h; i.e., v = 22gh.

B

h A

SOLUTION +R©Ft = mat;

mg sin u = mat

v dv = at ds = g sin u ds v

L0

at = g sin u However dy = ds sin u

h

v dv =

L0

g dy

v2 = gh 2 v = 22gh

Q.E.D.

Ans: v = 12gh 320

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12–77. The cylindrical plug has a weight of 2 lb and it is free to move within the confines of the smooth pipe. The spring has a stiffness k = 14 lb>ft and when no motion occurs the distance d = 0.5 ft. Determine the force of the spring on the plug when the plug is at rest with respect to the pipe. The plug is traveling with a constant speed of 15 ft>s, which is caused by the rotation of the pipe about the vertical axis.

3 ft d G

Solution

k  14 lb/ft

2 2 (15) + F = c d = ma ;    ΣF n n s d 32.2 3 - d

Fs = ks;   

  Fs = 14(0.5 - d) 2 2 (15) c d 32.2 3 - d

Thus,

14(0.5 - d) =



(0.5 - d)(3 - d) = 0.9982



1.5 - 3.5d + d 2 = 0.9982



d 2 - 3.5d + 0.5018 = 0

Choosing the root 6 0.5 ft

d = 0.1498 ft



Fs = 14(0.5 - 0.1498) = 4.90 lb

Ans.

Ans: Fs = 4.90 lb 321

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13–78. When crossing an intersection, a motorcyclist encounters the slight bump or crown caused by the intersecting road. If the crest of the bump has a radius of curvature r = 50 ft, determine the maximum constant speed at which he can travel without leaving the surface of the road. Neglect the size of the motorcycle and rider in the calculation. The rider and his motorcycle have a total weight of 450 lb.

Solution

+T gFn = man;   450 - 0 =

450 v2 a b 32.2 50             v = 40.1 ft>s

r  50 ft

Ans.

Ans: v = 40.1 ft>s 322

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13–79. The airplane, traveling at a constant speed of 50 m>s, is executing a horizontal turn. If the plane is banked at u = 15°, when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature r of the turn. Also, what is the normal force of the seat on the pilot if he has a mass of 70 kg.

u

r

SOLUTION + c a Fb = mab;

NP sin 15° - 7019.812 = 0 Ans.

NP = 2.65 kN + ; a Fn = man;

NP cos 15° = 70 a

502 b r Ans.

r = 68.3 m

Ans: NP = 2.65 kN r = 68.3 m 323

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*13–80. The 2-kg pendulum bob moves in the vertical plane with a velocity of 8 m>s when u = 0°. Determine the initial tension in the cord and also at the instant the bob reaches u = 30°. Neglect the size of the bob. 

2m

Solution Equations of Motion. Referring to the FBD of the bob at position u = 0°, Fig. a, 82 b = 64.0 N 2 For the bob at an arbitrary position u, the FBD is shown in Fig. b.

  ΣFn = man;  T = 2 a

Ans.

  ΣFt = mat;   - 2(9.81) cos u = 2at

at = - 9.81 cos u

  ΣFn = man;  T + 2(9.81) sin u = 2 a

v2 b 2

T = v2 - 19.62 sin u (1) Kinematics. The velocity of the bob at the position u = 30° can be determined by integrating vdv = at ds. However, ds = rdu = 2du. Then, 30°

v

L8 m>s

vdv =

L0°

- 9.81 cos u(zdu)

30° v2 v ` = - 19.62 sin u ` 2 8 m>s 0°

82 v2 = - 19.62(sin 30° - 0) 2 2 v2 = 44.38 m2 >s2 Substitute this result and u = 30° into Eq. (1), T = 44.38 - 19.62 sin 30° Ans.

  = 34.57 N = 34.6 N

Ans: T = 64.0 N T = 34.6 N 324

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13–81. The 2-kg pendulum bob moves in the vertical plane with a velocity of 6 m>s when u = 0°. Determine the angle u where the tension in the cord becomes zero.



2m

Solution Equation of Motion. The FBD of the bob at an arbitrary position u is shown in Fig. a. Here, it is required that T = 0. ΣFt = mat;   - 2(9.81) cos u = 2at at = - 9.81 cos u v2 ΣFn = man;  2(9.81) sin u = 2 a b 2 2 v = 19.62 sin u (1) Kinematics. The velocity of the bob at an arbitrary position u can be determined by integrating vdv = at ds. However, ds = rdu = 2du. Then v

u

L6 m>s

vdv =

L0°

- 9.81 cos u(2du)

v

u v2 ` = - 19.62 sin u ` 2 6 m>s 0°

v2 = 36 - 39.24 sin u Equating Eqs. (1) and (2) 19.62 sin u = 36 - 39.24 sin u 58.86 sin u = 36 u = 37.71° = 37.7°

(2)

Ans.

Ans: u = 37.7° 325

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13–82. The 8-kg sack slides down the smooth ramp. If it has a speed of 1.5 m>s when y = 0.2 m, determine the normal reaction the ramp exerts on the sack and the rate of increase in the speed of the sack at this instant.

y

y = 0.2ex

SOLUTION y = 0.2

x

x = 0

y = 0.2ex dy = 0.2ex 2 = 0.2 dx x=0 d 2y dx2

= 0.2ex 2

x=0

= 0.2 3

B1 + a r =

2

dy 2 2 b R dx

d2y dx2

2

3

=

C 1 + (0.2)2 D 2 |0.2|

= 5.303

u = tan - 1 (0.2) = 11.31° +a©Fn = man ;

NB - 8(9.81) cos 11.31° = 8 a

(1.5)2 b 5.303 Ans.

NB = 80.4 N +b©Ft = mat ;

8(9.81) sin 11.31° = 8a t a t = 1.92 m s2

Ans.

Ans: NB = 80.4 N at = 1.92 m>s2 326

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13–83. The ball has a mass m and is attached to the cord of length l. The cord is tied at the top to a swivel and the ball is given a velocity v0. Show that the angle u which the cord makes with the vertical as the ball travels around the circular path must satisfy the equation tan u sin u = v20>gl. Neglect air resistance and the size of the ball.

O

u l

SOLUTION + ©F = ma ; : n n + c ©Fb = 0; Since r = l sin u

T sin u = m a

v20 b r

T cos u - mg = 0 T = a

v0

mv20 l sin2 u

mv20 cos u b a 2 b = mg l sin u

tan u sin u =

v20 gl

Q.E.D.

Ans: tan u sin u = 327

v20 gl

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*13–84. The 2-lb block is released from rest at A and slides down along the smooth cylindrical surface. If the attached spring has a stiffness k = 2 lb>ft, determine its unstretched length so that it does not allow the block to leave the surface until u = 60°.

A

u 2 ft k  2 lb/ft

Solution + bΣFn = map; Fs + 2 cos u =

2 v2 a b 32.2 2

(1)

2 a 32.2 t

+ RΣFt = mat;

2 sin u =



at = 32.2 sin u

v dv = at ds;

L0



1 2 v = 64.4( -cos u + 1) 2

v

v dv =

L0

u

32.2(sin u)2du

When u = 60° v2 = 64.4

From Eq. (1) Fs + 2 cos 60° =

2 64.4 a b 32.2 2

Fs = 1 lb Fs = ks;  l = 2s;  s = 0.5 ft Ans.

l0 = l - s = 2 - 0.5 = 1.5 ft

Ans: l0 = 1.5 ft 328

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13–85. The spring-held follower AB has a weight of 0.75 lb and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.2 ft and z = 10.1 sin u2 ft. If the cam is rotating at a constant rate of 6 rad>s, determine the force at the end A of the follower when u = 90°. In this position the spring is compressed 0.4 ft. Neglect friction at the bearing C.

z  0.1 sin 2u z · u  6 rad/s

0.2 ft

B A

SOLUTION

C

k  12 lb/ft

z = 0.1 sin 2u # # z = 0.2 cos 2uu # $ $ z = -0.4 sin 2uu2 + 0.2 cos 2uu

#

u = 6 rad>s

##

u = 0 $ z = -14.4 sin 2u a Fz = maz;

$ FA - 12(z + 0.3) = mz

FA - 12(0.1 sin 2u + 0.3) =

0.75 ( -14.4 sin 2u) 32.2

For u = 45°, FA - 12(0.4) =

0.75 ( -14.4) 32.2 Ans.

FA = 4.46 lb

Ans: FA = 4.46 lb 329

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13–86. Determine the magnitude of the resultant force acting on a 5-kg particle at the instant t = 2 s, if the particle is moving along a horizontal path defined by the equations r = (2t + 10) m and u = (1.5t2 - 6t) rad, where t is in seconds.

SOLUTION r = 2t + 10|t = 2 s = 14 # r = 2 $ r = 0 u = 1.5t2 - 6t # u = 3t - 6 t = 2 s = 0 $ u = 3 # $ ar = r - ru2 = 0 - 0 = 0 $ ## au = ru + 2ru = 14(3) + 0 = 42 Hence, ©Fr = mar;

Fr = 5(0) = 0

©Fu = mau;

Fu = 5(42) = 210 N

F = 2(Fr)2 + (Fu)2 = 210 N

Ans.

Ans: F = 210 N 330

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13–87. The path of motion of a 5-lb particle in the horizontal plane is described in terms of polar coordinates as r = (2t + 1) ft and u = (0.5t2 - t) rad, where t is in seconds. Determine the magnitude of the unbalanced force acting on the particle when t = 2 s.

SOLUTION # $ r = 2 ft>s r = 0 # = 0 rad u = t - 1|t = 2 s = 1 rad>s

r = 2t + 1|t = 2 s = 5 ft

u = 0.5t2 - t|t = 2 s # $ ar = r - ru2 = 0 - 5(1)2 = - 5 ft>s2 $ ## au = ru + 2ru = 5(1) + 2(2)(1) = 9 ft>s2 ©Fr = mar;

Fr =

5 ( - 5) = - 0.7764 lb 32.2

©Fu = mau;

Fu =

5 (9) = 1.398 lb 32.2

$ u = 1 rad>s2

F = 2F2r + F2u = 2(- 0.7764)2 + (1.398)2 = 1.60 lb

Ans.

Ans: F = 1.60 lb 331

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*13–88. Rod OA rotates # counterclockwise with a constant angular velocity of u = 5 rad>s. The double collar B is pinconnected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation r = 1.512 - cos u2 ft. If both collars weigh 0.75 lb, determine the normal force which the curved rod exerts on one collar at the instant u = 120°. Neglect friction.

A B r

SOLUTION

·

# $ Kinematic: Here, u = 5 rad>s and u = 0. Taking the required time derivatives at u = 120°, we have

O = 5 rad/s

r = 1.5 (2 – cos ) ft.

r = 1.5(2 - cos u)|u = 120° = 3.75 ft # # r = 1.5 sin uu|u = 120° = 6.495 ft>s $ # $ r = 1.5(sin uu + cos uu 2)|u = 120° = - 18.75 ft>s2 Applying Eqs. 12–29, we have # $ ar = r - ru2 = - 18.75 - 3.75(52) = - 112.5 ft>s2 $ # # au = r u + 2r u = 3.75(0) + 2(6.495)(5) = 64.952 ft>s2 Equation of Motion: The angle c must be obtained first. tan c =

1.5(2 - cos u) r 2 = = 2.8867 dr>du 1.5 sin u u = 120°

c = 70.89°

Applying Eq. 13–9, we have a Fr = mar ;

-N cos 19.11° =

0.75 ( - 112.5) 32.2 Ans.

N = 2.773 lb = 2.77 lb a Fu = mau ;

FOA + 2.773 sin 19.11° =

0.75 (64.952) 32.2

FOA = 0.605 lb

Ans: N = 2.77 lb 332

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13–89. z

The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, u = (0.7t) rad, and z = ( - 0.5t) m, where t is in seconds. Determine the components of force Fr, Fu, and Fz which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy.

u

z

Solution r = 1.5 # $ r = r = 0

r  1.5 m

z = - 0.5t # z = - 0.5 $ z = 0



u = 0.7t # u = 0.7 $ u = 0

ΣFr = mar;

Fr = 40( - 0.735) = -29.4 N

Ans.

ΣFu = mau;

Fu = 0

Ans.

ΣFz = maz;

Fz - 40(9.81) = 0



Fz = 392 N

# $ ar = r - r(u)2 = 0 - 1.5(0.7)2 = - 0.735 $ # # au = ru + 2ru = 0 $ az = z = 0

Ans.

Ans: Fr = - 29.4 N Fu = 0 Fz = 392 N 333

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13–90. z

The 40-kg boy is sliding down the smooth spiral slide such that z = - 2 m>s and his speed is 2 m>s. Determine the r, u, z components of force the slide exerts on him at this instant. Neglect the size of the boy.

u

r  1.5 m

z

Solution r = 1.5 m # r = 0 $ r = 0 vu = 2 cos 11.98° = 1.9564 m>s vz = -2 sin 11.98° = - 0.41517 m>s # # vu = ru;  1.9564 = 1.5 u # u = 1.3043 rad>s ΣFr = mar;

- Fr = 40(0 - 1.5 ( 1.3043)2 )



Fr = 102 N

ΣFu = mau;

Nb sin 11.98° = 40(au)

Ans.

ΣFz = maz;

- Nb cos 11.98° + 40(9.81) = 40az az Require tan 11.98° = ,  au = 4.7123az au

Thus,

az = 0.423 m>s2 au = 1.99 m>s2 Nb = 383.85 N Nz = 383.85 cos 11.98° = 375 N

Ans.

Nu = 383.85 sin 11.98° = 79.7 N

Ans.

Ans: Fr = 102 N Fz = 375 N Fu = 79.7 N 334

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13–91. Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 u) m, where u is in radians. If the angular position of the arm is u = (p8 t 2) rad, where t is in seconds, determine the force of the rod on the peg and the normal force of the slot on the peg at the instant t = 2 s. The peg is in contact with only one edge of the rod and slot at any instant.

P r  (0.5) m

r 

Solution

dr = 0.5. The angle c between the du extended radial line and the tangent can be determined from

Equation of Motion. Here, r = 0.5u. Then

tan c =

At the instant t = 25, u =

r 0.5u = = u dr>du 0.5

p 2 p ( 2 ) = rad 8 2 tan c =

p   c = 57.52° 2

The positive sign indicates that c is measured from extended radial line in positive sense of u (counter clockwise) to the tangent. Then the FBD of the peg shown in Fig. a can be drawn.   ΣFr = mar; N sin 57.52° - 0.5(9.81) = 0.5ar

(1)

  ΣFu = mau; F - N cos 57.52° = 0.5au

(2)

Kinematics. Using the chain rule, the first and second derivatives of r and u with respect to t are r = 0.5u = 0.5 a

p p 2 p 2 t b = t    u = t 2 8 16 8

p # r = t 8

# p u = t 4

p $ r = 8

$ p u = 4

When t = 2 s, r =

p 2 p ( 2 ) = m 16 4

u =

p 2 p ( 2 ) = rad 8 2

p p # r = (2) = m>s 8 4

# p p u = (2) = rad>s 4 2

p $ r = m>s2 8

$ p u = rad>s2 4

Thus, # p p p 2 ## ar = r - ru 2 = - a b = - 1.5452 m>s2 8 4 2 $ p p p p # # au = ru + 2ru = a b + 2a ba b = 3.0843 m>s2 4 4 4 2

Substitute these results in Eqs. (1) and (2) N = 4.8987 N = 4.90 N

Ans.

F = 4.173 N = 4.17 N

Ans.

335

Ans: N = 4.90 N F = 4.17 N

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*13–92.

# The arm is rotating at a rate of u = 4 rad>s when $ u = 3 rad>s2 and u = 180°. Determine the force it must exert on the 0.5-kg smooth cylinder if it is confined to move along the slotted path. Motion occurs in the horizontal plane.

r

·

··   4 rad/s,   3 rad/s2

  180

2 r  (—) m 

Solution 2 dr 2 . Then = - 2 . The angle c between the u du u extended radial line and the tangent can be determined from Equation of Motion. Here, r =

tan c =

2>u r = = -u dr>du - 2>u 2

At u = 180° = p rad, tan c = - p  c = - 72.34° The negative sign indicates that c is measured from extended radial line in the negative sense of u (clockwise) to the tangent. Then, the FBD of the peg shown in Fig. a can be drawn. ΣFr = mar;   - N sin 72.34° = 0.5ar

(1)

ΣFu = mau;  F - N cos 72.34° = 0.5au

(2)

Kinematics. Using the chain rule, the first and second time derivatives of r are r = 2u -1 # 2 # # r = -2u -2 u = - a 2 b u u

#

. 2 ( 2u 2 - uu¨ ) u3 # When u = 180° = p rad, u = 4 rad>s and u¨ = 3 rad>s2. Thus

r¨ = - 2 ( - 2u -3 u 2 + u -2 u¨ ) =

r =

#

2 m = 0.6366 m p

r = -a r¨ =

Thus,

2 b(4) = - 0.8106 m>s p2

2 3 2 ( 42 ) - p(3) 4 = 1.4562 m>s2 p3

. ar = r¨ - ru 2 = 1.4562 - 0.6366 ( 42 ) = - 8.7297 m>s2 # . au = ru¨ + 2r u = 0.6366(3) + 2( -0.8106)(4) = -4.5747 m>s2

Substitute these result into Eqs. (1) and (2), N = 4.5807 N Ans.

F = - 0.8980 N = -0.898 N

The negative sign indicates that F acts in the sense opposite to that shown in the FBD.

Ans: F = - 0.898 N 336

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13–93. If arm OA .rotates with a constant clockwise angular velocity of u = 1.5 rad>s. determine the force arm OA exerts on the smooth 4-lb cylinder B when u = 45°. A

B r

SOLUTION

u u

Kinematics: Since the motion of cylinder B is known, ar and au will be determined 4 first. Here, = cos u or r = 4 sec u ft. The value of r and its time derivatives at the r instant u = 45° are

O

4 ft

r = 4 sec u |u = 45° = 4 sec 45° = 5.657 ft # # r = 4 sec u(tan u)u|u = 45° = 4 sec 45° tan 45°(1.5) = 8.485 ft>s $ # # # $ r = 4 C sec u(tan u)u + u A sec u sec2 uu + tan u sec u tan uu B D # $ #2 = 4 C sec u(tan u)u + sec3 uu2 + sec u tan2 uu D 2

u = 45°

= 4 C sec 45° tan 45°(0) + sec3 45°(1.5)2 + sec 45° tan2 45°(1.5)2 D = 38.18 ft>s2 Using the above time derivatives, # $ ar = r - ru 2 = 38.18 - 5.657 A 1.52 B = 25.46 ft>s2 $ ## au = ru - 2ru = 5.657(0) + 2(8.485)(1.5) = 25.46 ft>s2 Equations of Motion: By referring to the free-body diagram of the cylinder shown in Fig. a, ©Fr = mar;

N cos 45° - 4 cos 45° =

4 (25.46) 32.2

N = 8.472 lb ©Fu = mau;

FOA - 8.472 sin 45° - 4 sin 45° =

4 (25.46) 32.2 Ans.

FOA = 12.0 lb

Ans: FOA = 12.0 lb 337

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13–94. Determine the normal and frictional driving forces that the partial spiral track exerts on the 200-kg $motorcycle at # the instant u = 53 p rad, u = 0.4 rad>s, and u = 0.8 rad>s2. Neglect the size of the motorcycle.

r  (5u) m u r

SOLUTION 5 u = a p b = 300° 3

# u = 0.4

$ u = 0.8

5 r = 5u = 5 a p b = 26.18 3 # # r = 5u = 5(0.4) = 2 $ $ r = 5u = 5(0.8) = 4 # $ ar = r - ru2 = 4 - 26.18(0.4)2 = - 0.1888 $ ## au = ru + 2ru = 26.18(0.8) + 2(2)(0.4) = 22.54 r = tan c = dr>du

5 5a pb 3 = 5.236 5

c = 79.19°

+R©Fr = mar;

F sin 10.81° - N cos 10.81° + 200(9.81) cos 30° = 200( -0.1888)

+Q©Fu = mau;

F cos 10.81° - 200(9.81) sin 30° + N sin 10.81° = 200(22.54) F = 5.07 kN

Ans.

N = 2.74 kN

Ans.

Ans: F = 5.07 kN N = 2.74 kN 338

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13–95. A smooth can C, having a mass of 3 kg, is lifted from a feed at A to a ramp at B by a rotating # rod. If the rod maintains a constant angular velocity of u = 0.5 rad>s, determine the force which the rod exerts on the can at the instant u = 30°. Neglect the effects of friction in the calculation and the size of the can so that r = 11.2 cos u2 m. The ramp from A to B is circular, having a radius of 600 mm.

B · u

0.5 rad/s

C r 600 mm

u

A

600 mm

SOLUTION r = 2(0.6 cos u) = 1.2 cos u # # r = -1.2 sin uu $ # $ r = -1.2 cos uu2 - 1.2 sin uu # $ At u = 30°, u = 0.5 rad>s and u = 0 r = 1.2 cos 30° = 1.0392 m # r = -1.2 sin 30°(0.5) = - 0.3 m>s $ r = -1.2 cos 30°(0.5)2 - 1.2 sin 30°(0) = - 0.2598 m>s2 # $ ar = r - ru2 = -0.2598 - 1.0392(0.5)2 = - 0.5196 m>s2 $ ## au = ru + 2ru = 1.0392(0) + 2( -0.3)(0.5) = - 0.3 m>s2 +Q©Fr = mar;

N cos 30° - 3(9.81) sin 30° = 3( -0.5196)

N = 15.19 N

+a©Fu = mau;

F + 15.19 sin 30° - 3(9.81) cos 30° = 3( -0.3) Ans.

F = 17.0 N

Ans: F = 17.0 N 339

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*13–96. The spring-held follower AB has a mass of 0.5 kg and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.15 m and z = (0.02 cos 2u) m. If the cam is rotating at a constant rate of 30 rad>s, determine the force component Fz at the end A of the follower when u = 30°. The spring is uncompressed when u = 90°. Neglect friction at the bearing C.

z  (0.02 cos 2) m z 0.15 m

C

·

  30 rad/s

B A

k  1000 N/m

Solution Kinematics. Using the chain rule, the first and second time derivatives of z are z # z $ z # Here, u $ z

= (0.02 cos 2u) m # # # = 0.02[ - sin 2u(2u)] = [ - 0.04(sin 2u)u] m>s # # $ # $ = - 0.04[cos 2u(2u)u + (sin 2u)u ] = [ - 0.04(2 cos 2u(u)2 + sin 2u(u ))] m>s2 $ = 30 rad>s and u = 0. Then = - 0.04[2 cos 2u(302) + sin 2u(0)] = ( - 72 cos 2u) m>s2

Equation of Motion. When u = 30°, the spring compresses x = 0.02 + 0.02 cos 2(30°) = 0.03 m. Thus, Fsp = kx = 1000(0.03) = 30 N. Also, at this $ position az = z = -72 cos 2(30°) = -36.0 m>s2. Referring to the FBD of the follower, Fig. a, ΣFz = maz;

N - 30 = 0.5( - 36.0) Ans.

N = 12.0 N

Ans: N = 12.0 N 340

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13–97. The spring-held follower AB has a mass of 0.5 kg and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.15 m and z = (0.02 cos 2u) m. If the cam is rotating at a constant rate of 30 rad>s, determine the maximum and minimum force components Fz the follower exerts on the cam if the spring is uncompressed when u = 90°.

z  (0.02 cos 2) m z 0.15 m

C

·

  30 rad/s

B A

k  1000 N/m

Solution Kinematics. Using the chain rule, the first and second time derivatives of z are z = (0.02 cos 2u) m # # # z = 0.02[ - sin 2u(2u)] = ( - 0.04 sin 2uu) m>s # # $ # $ $ z = - 0.04[cos 2u(2u)u + sin 2uu ] = [ - 0.04(2 cos 2u(u)2 + sin 2u(u ))] m>s2 $ # Here u = 30 rad>s and u = 0. Then, $ z = - 0.04[2 cos 2u(302) + sin 2u(0)] = ( - 72 cos 2u) m>s2 Equation of Motion. At any arbitrary u, the spring compresses x = 0.02(1 + cos 2u). Thus, Fsp = kx = 1000[0.02(1 + cos 2u)] = 20 (1 + cos 2u). Referring to the FBD of the follower, Fig. a, ΣFz = maz;

N - 20(1 + cos 2u) = 0.5( - 72 cos 2u)



N = (20 - 16 cos 2u) N

N is maximum when cos 2u = - 1. Then Ans.

(N)max = 36.0 N N is minimum when cos 2u = 1. Then

Ans.

(N)min = 4.00 N

Ans: (N)max = 36.0 N (N)min = 4.00 N 341

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13–98. The particle has a mass of 0.5 kg and is confined to move along the smooth vertical slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when u. = 30°. The rod is rotating with a constant angular velocity u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant.

A

r u O

Solution # 0.5 # = 0.5 sec u,  r = 0.5 sec u tan uu cos u $ # # $ r = 0.5 sec u tan uu + 0.5 sec3 uu 2 + 0.5 sec u tan2 uu 2

u = 2 rad/s 0.5 m

r =

At u = 30°. # u = 2 rad>s $ u = 0 r = 0.5774 m # r = 0.6667 m>s $ r = 3.8490 m>s2 # $ ar = r - ru 2 = 3.8490 - 0.5774(2)2 = 1.5396 m>s2 $ # # au = ru + 2ru = 0 + 2(0.6667)(2) = 2.667 m>s2 + QΣFr = mar;

NP cos 30° - 0.5(9.81)sin 30° = 0.5(1.5396)



NP = 3.7208 = 3.72 N

+ aΣFu = mau;

F - 3.7208 sin 30° - 0.5(9.81) cos 30° = 0.5(2.667)



F = 7.44 N

Ans.

Ans.

Ans: Ns = 3.72 N Fr = 7.44 N 342

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13–99. A car of a roller coaster travels along a track which for a short distance is defined by a conical spiral, r = 34z, u = - 1.5z, where r and# z are in meters and u in radians. If the angular motion u = 1 rad>s is always maintained, determine the r, u, z components of reaction exerted on the car by the track at the instant z = 6 m. The car and passengers have a total mass of 200 kg.

SOLUTION r = 0.75z

# # r = 0.75z # # u = - 1.5z

u = - 1.5z # # u = 1 = - 1.5z

z

$ $ r = 0.75z $ $ u = - 1.5z

# z = - 0.6667 m>s

r

$ z = 0 θ

At z = 6 m, # r = 0.75(6) = 4.5 m r = 0.75( - 0.6667) = - 0.5 m>s # $ at = r - ru2 = 0 - 4.5(1)2 = - 4.5 m>s2 $ ## a u = ru + 2ru = 4.5(0) + 2( -0.5)(1) = - 1 m>s2

$ r = 0.75(0) = 0

$ u = 0

$ az = z = 0 ©Ft = mar;

Fr = 200( - 4.5)

Fr = - 900 N

Ans.

©Fu = mau;

Fu = 200( - 1)

Fu = - 200N

Ans.

©Fz = maz;

Fz - 200(9.81) = 0

Fz = 1962 N = 1.96 kN

Ans.

Ans: Fr = - 900 N Fu = - 200 N Fz = 1.96 kN 343

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*13–100. P

The 0.5-lb ball is guided along the vertical circular path r = 2rc cos# u using the arm OA. If the arm has an angular u = 0.4 rad>s and an angular acceleration velocity $ u = 0.8 rad>s2 at the instant u = 30°, determine the force of the arm on the ball. Neglect friction and the size of the ball. Set rc = 0.4 ft.

r

O

A

rc

u

SOLUTION r = 2(0.4) cos u = 0.8 cos u # # r = -0.8 sin uu $ # $ r = -0.8 cos uu2 - 0.8 sin uu # $ At u = 30°, u = 0.4 rad>s, and u = 0.8 rad>s2 r = 0.8 cos 30° = 0.6928 ft # r = - 0.8 sin 30°(0.4) = -0.16 ft>s $ r = - 0.8 cos 30°(0.4)2 - 0.8 sin 30°(0.8) = - 0.4309 ft>s2 # $ ar = r - ru2 = - 0.4309 - 0.6928(0.4)2 = - 0.5417 ft>s2 $ ## au = ru + 2ru = 0.6928(0.8) + 2( -0.16)(0.4) = 0.4263 ft>s2 +Q©Fr = mar; a+ ©Fu = mau;

N cos 30° - 0.5 sin 30° =

0.5 ( - 0.5417) 32.2

FOA + 0.2790 sin 30° - 0.5 cos 30° =

N = 0.2790 lb

0.5 (0.4263) 32.2 Ans.

FOA = 0.300 lb

Ans: FOA = 0.300 lb 344

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13–101. P

The ball of mass m is guided along the vertical circular path r = 2rc cos u using # the arm OA. If the arm has a constant angular velocity u0, determine the angle u … 45° at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball.

r

O

A

rc

u

SOLUTION r = 2rc cos u # # r = - 2rc sin uu # $ $ r = - 2rc cos uu2 - 2rc sin uu # $ Since u is constant, u = 0. # # # # $ ar = r - ru2 = - 2rc cos uu20 - 2rc cos uu20 = - 4rc cos uu20 +Q©Fr = mar;

# -mg sin u = m(- 4rc cos uu20) # # 4rc u20 4rc u20 -1 u = tan ¢ tan u = ≤ g g

Ans.

Ans: u = tan-1a 345

#

4rcu20 b g

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13–102. Using a forked rod, a smooth cylinder P, having a mass of 0.4 kg, is forced to move along the vertical slotted path r = 10.6u2 m, where u is in radians. If the cylinder has a constant speed of vC = 2 m>s, determine the force of the rod and the normal force of the slot on the cylinder at the instant u = p rad. Assume the cylinder is in contact with only one edge of the rod and slot at any instant. Hint: To obtain the time derivatives necessary to compute the cylinder’s acceleration components ar and au, take the first and second time derivatives of r = 0.6u. Then, for further # information, use Eq. 12–26 to determine u. Also, take the # time derivative of Eq. 12–26, noting that vC = 0, to $ determine u.

P r

r

u

p

0.6u

SOLUTION $ $ # r = 0.6 u r = 0.6 u # # # # vr = r = 0.6u vu = ru = 0.6uu r = 0.6 u

2 # v2 = r 2 + a rub

# 2 # 2 22 = a0.6u b + a 0.6uu b

# u =

2 0.6 21 + u2

#$ # #$ 0 = 0.72u u + 0.36 a 2uu3 + 2u2u u b

At u = p rad,

# u =

2 0.6 21 + p2

$ u = -

# uu2

1 + u2

= 1.011 rad>s

$ (p)(1.011)2 = - 0.2954 rad>s2 u = 1 + p2 r = 0.6(p) = 0.6 p m

# r = 0.6(1.011) = 0.6066 m>s

$ r = 0.6(- 0.2954) = - 0.1772 m>s2 # $ ar = r - ru 2 = - 0.1772 - 0.6 p(1.011)2 = - 2.104 m>s2 $ ## au = ru + 2ru = 0.6p(- 0.2954) + 2(0.6066)(1.011) = 0.6698 m>s2 tan c =

0.6u r = = u = p dr>du 0.6

c = 72.34°

+ ©F = ma ; ; r r

-N cos 17.66° = 0.4(- 2.104)

+ T ©Fu = mau ;

- F + 0.4(9.81) + 0.883 sin 17.66° = 0.4(0.6698)

N = 0.883 N

Ans.

Ans.

F = 3.92 N

Ans: N = 0.883 N F = 3.92 N 346

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13–103. The pilot of the airplane executes a vertical loop which in part follows the path of a cardioid, r = 200(1 + cosu) m, where u is in radians. If his speed at A is a constant vp = 85 m>s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at  A. He has a mass of 80 kg. Hint: To determine the time derivatives necessary to calculate the acceleration components ar and au, take the first and second time derivatives of for further information, use r = 200(1 + cosu). Then, # Eq. 12–26 to determine u.

 r  200 (1  cos  ) m

A

Solution Kinematic. Using the chain rule, the first and second time derivatives of r are r = 200(1 + cos u) # # # r = 200( - sin u)(u) = - 200(sin u)u # $ $ r = - 200[(cos u)(u)2 + (sin u)(u )] When u = 0°, r = 200(1 + cos 0°) = 400 m # # r = - 200(sin 0°) u = 0 # $ # $ r = -200 3 (cos 0°)(u)2 + (sin 0°)(u ) 4 = - 200u 2

Using Eq. 12–26 # # v = 2r2 + (ru)2 # # v2 = r2 + (ru)2 # 852 = 02 + (400u)2 # u = 0.2125 rad>s Thus,

# $ ar = r - ru 2 = - 200 ( 0.21252 ) - 400 ( 0.21252 ) = - 27.09 m>s2

Equation of Motion. Referring to the FBD of the pilot, Fig. a, T + ΣFr = mar;

80(9.81) - N = 80( - 27.09) Ans.

N = 2952.3 N = 2.95 kN

Ans: N = 2.95 N 347

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*13–104. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = 1eu2 m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u =# 45°, if the force F maintains a constant angular motion u = 2 rad>s.

F

r

SOLUTION r = eu

r = eθ

θ

# # r = eu u # # $ r = eu(u)2 + eu u At u = 45° # u = 2 rad>s $ u = 0 r = 2.1933 # r = 4.38656 $ r = 8.7731 # $ ar = r - r(u)2 = 8.7731 - 2.1933(2)2 = 0 $ # # au = r u + 2 r u = 0 + 2(4.38656)(2) = 17.5462 m>s2 tan c =

r = e u>e u = 1 dr a #b du

c = u = 45° Q+ a Fr = mar ; +a a Fu = mau ;

- NC cos 45° + F cos 45° = 2(0) F sin 45° + NC sin 45° = 2(17.5462) N = 24.8 N

Ans.

F = 24.8 N

Ans.

Ans: N = 24.8 N F = 24.8 N 348

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13–105. The particle has a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when u = 30°. The rod is rotating with a constant angular velocity . u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant.

A

· u  2 rad/s

r 0.5 m

u

Solution

O

0.5 = 0.5 sec u cos u # # r = 0.5 sec u tan uu r =

# # # $ $ r = 0.5 5 3 (sec u tan uu)tan u + sec u(sec2 uu) 4 u + sec u tan uu 6 # # $ = 0.5[sec u tan2 uu 2 + sec3 uu 2 + sec u tan uu ] $ # When u = 30°, u = 2 rad>s and u = 0 r = 0.5 sec 30° = 0.5774 m # r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s $ r = 0.5 3 sec 30° tan2 30°(2)2 + sec3 30°(2)2 + sec 30° tan 30°(0) 4 = 3.849 m>s2 # $ ar = r - ru 2 = 3.849 - 0.5774(2)2 = 1.540 m>s2 $ # # au = ru + 2ru = 0.5774(0) + 2(0.6667)(2) = 2.667 m>s2 Q + ΣFr = mar ; + RΣFu = mau;

N cos 30° - 0.5(9.81) cos 30° = 0.5(1.540) Ans.

N = 5.79 N F + 0.5(9.81)sin 30° - 5.79 sin 30° = 0.5(2.667)

Ans.

F = 1.78 N

Ans: Fr = 1.78 N Ns = 5.79 N 349

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13–106. Solve Prob. 13–105 if# the arm has an angular acceleration of $ u = 3 rad>s2 when u = 2 rad>s at u = 30°.

A

· u  2 rad/s

r 0.5 m

u

Solution

O

0.5 = 0.5 sec u cos u # # r = 0.5 sec u tan uu # # # $ $ r = 0.5 5 3 (sec u tan uu)tan u + sec u(sec2 uu) 4 u + sec u tan uu 6 # # $ = 0.5 3 sec u tan2 uu 2 + sec3 uu 2 + sec u tan uu 4 # $ When u = 30°, u = 2 rad>s and u = 3 rad>s2 r =

r = 0.5 sec 30° = 0.5774 m # r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s $ r = 0.5 3 sec 30° tan2 30°(2)2 + sec3 30°(2)2 + sec 30° tan 30°(3) 4 = 4.849 m>s2 # $ ar = r - ru 2 = 4.849 - 0.5774(2)2 = 2.5396 m>s2 $ # # au = ru + 2ru = 0.5774(3) + 2(0.6667)(2) = 4.3987 m>s2 Q + ΣFr = mar;

N cos 30° - 0.5(9.81) cos 30° = 0.5(2.5396) Ans.

  N = 6.3712 = 6.37 N + RΣFu = mau;

F + 0.5(9.81) sin 30° - 6.3712 sin 30° = 0.5(4.3987) Ans.

  F = 2.93 N

Ans: Fr = 2.93 N Ns = 6.37 N 350

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13–107. The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a limaçon, r = (2 + cos u) ft. If u = (0.5t2) rad, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the particle on only one side.

2 ft

r u · u

SOLUTION r = 2 + cos u # r = -sin uu # $ $ r = - cos uu2 - sin uu

3 ft

u = 0.5t2 # u = t $ u = 1 rad>s2

$ At t = 1 s, u = 0.5 rad, u = 1 rad>s, and u = 1 rad>s2 r = 2 + cos 0.5 = 2.8776 ft # r = -sin 0.5(1) = - 0.4974 ft>s2 $ r = -cos 0.5(1)2 - sin 0.5(1) = - 1.357 ft>s2 # $ ar = r - ru2 = -1.375 - 2.8776(1)2 = - 4.2346 ft>s2 $ ## au = ru + 2ru = 2.8776(1) + 2( -0.4794)(1) = 1.9187 ft>s2 tan c =

2 + cos u r 2 = = - 6.002 dr>du -sin u u = 0.5 rad

c = -80.54°

2 ( -4.2346) 32.2

+Q©Fr = mar;

- N cos 9.46° =

+a©Fu = mau;

F - 0.2666 sin 9.46° =

N = 0.2666 lb

2 (1.9187) 32.2 Ans.

F = 0.163 lb

Ans: F = 0.163 lb 351

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*13–108. The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a parabola r = 4>11 - cos u2, where u is in radians and r is #in feet. If the collar’s angular rate is constant and equals u = 4 rad>s, determine the tangential retarding force P needed to cause the motion and the normal force that the collar exerts on the rod at the instant u = 90°.

P

r u

SOLUTION r = # r = $ r =

4 1 - cos u

# -4 sin u u

(1 - cos u)2 $ -4 sin u u

+

(1 - cos u)2

# - 4 cos u (u)2 (1 - cos u)2

# u = 4,

At u = 90°,

+

# 8 sin2 u u2 (1 - cos u)3

$ u = 0 r = 4 # r = - 16 $ r = 128

$ ar = r - r(u)2 = 128 - 4(4)2 = 64 $ ## au = ru + 2 ru = 0 + 2( - 16)(4) = -128 r =

4 1 - cos u

-4 sin u dr = du (1 - cos u)2 tan c =

r dr (du )

=

4 1 - cos u)

2

- 4 sin u ° (1 - cos u)2 u = 90

=

4 = -1 -4

c = - 45° = 135° 3 (64) 32.2

+ c ©Fr = m ar ;

P sin 45° - N cos 45° =

+ © F = ma ; ; u u

- P cos 45° - N sin 45° =

3 ( -128) 32.2

Solving, P = 12.6 lb

Ans.

N = 4.22 lb

Ans.

Ans: P = 12.6 lb N = 4.22 lb 352

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13–109. Rod OA rotates counterclockwise at a constant angular . rate u = 4 rad>s. The double collar B is pin-connected together such that one collar slides over the rotating rod and the other collar slides over the circular rod described by the equation r = (1.6 cos u) m. If both collars have a mass of 0.5 kg, determine the force which the circular rod exerts on one of the collars and the force that OA exerts on the other collar at the instant u = 45°. Motion is in the horizontal plane.

A r  1.6 cos u u = 4 rad/s

B u

O 0.8 m

Solution r = 1.6 cos u # # r = - 1.6 sin uu # $ $ r = - 1.6 cos uu 2 - 1.6 sin uu # $ At u = 45°, u = 4 rad>s and u = 0 r = 1.6 cos 45° = 1.1314 m # r = - 1.6 sin 45°(4) = - 4.5255 m>s $ r = - 1.6 cos 45°(4)2 - 1.6 sin 45°(0) = - 18.1019 m>s2 # $ ar = r - ru 2 = - 18.1019 - 1.1314(4)2 = - 36.20 m>s2 $ # # au = ru + 2ru = 1.1314(0) + 2( - 4.5255)(4) = - 36.20 m>s2 Q + ΣFr = mar ;   - NC cos 45° = 0.5( - 36.20)  NC = 25.6 N

Ans.

+ aΣFu = mau;  FOA - 25.6 sin 45° = 0.5( -36.20)  FOA = 0

Ans.

Ans: Fr = 25.6 N FOA = 0 353

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13–110. Solve Prob. 13–109 if motion is in the vertical plane.

A r  1.6 cos u u = 4 rad/s

B u

O 0.8 m

Solution r = 1.6 cos u # # r = - 1.6 sin uu # $ $ r = -1.6 cos uu 2 - 1.6 sin uu # $ At u = 45°, u = 4 rad>s and u = 0 r = 1.6 cos 45° = 1.1314 m # r = - 1.6 sin 45°(4) = - 4.5255 m>s $ r = -1.6 cos 45°(4)2 - 1.6 sin 45°(0) = -18.1019 m>s2 # $ ar = r - ru 2 = - 18.1019 - 1.1314(4)2 = - 36.20 m>s2 $ # # au = ru + 2ru = 1.1314(0) + 2( - 4.5255)(4) = - 36.20 m>s2 + b ΣFr = mar ;  - NC cos 45° - 4.905 cos 45° = 0.5( -36.204)   + aΣFu = mau;  FOA - NC sin 45° - 4.905 sin 45° = 0.5( -36.204)  NC = 20.7 N

Ans.

FOA = 0

Ans.

Ans: Fr = 20.7 N FOA = 0 354

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13–111. A 0.2-kg spool slides down along a smooth rod. If # the rod has a constant angular rate of rotation u = 2 rad>s in the vertical plane, show that the equations of $ motion for the spool are r - 4r - 9.81 sin u = 0 and # 0.8r + Ns - 1.962 cos u = 0, where Ns is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is # r = C1e -2t + C2e2t - 19.81>82 sin 2t. If r, r, and u are zero when t = 0, evaluate the constants C1 and C2 to determine r at the instant u = p>4 rad.

u

u

2 rad/s

r

SOLUTION

# # Kinematic: Here, u. = 2 rad>s and u = 0. Applying Eqs. 12–29, we have # $ $ $ ar = r - ru2 = r - r(22) = r - 4r $ ## # # au = ru + 2ru = r(0) + 2r(2) = 4r Equation of Motion: Applying Eq. 13–9, we have $ 1.962 sin u = 0.2(r - 4r)

©Fr = mar ;

$ r - 4r - 9.81 sin u = 0

(Q.E.D.)

(1)

(Q.E.D.)

(2)

# 1.962 cos u - Ns = 0.2(4r)

©Fu = mau;

# 0.8r + Ns - 1.962 cos u = 0 # u =

1

u

Since u. = 2 rad>s, then

L0 equation (Eq.(1)) is given by

L0

2dt, u = 2t. The solution of the differential

r = C1 e - 2 t + C2 e2t -

9.81 sin 2t 8

(3)

Thus, 9.81 # cos 2t r = - 2 C1 e - 2t + 2C2 e2t 4

(4)

At t = 0, r = 0 . From Eq.(3) 0 = C1 (1) + C2 (1) - 0

(5)

9.81 # At t = 0, r = 0 . From Eq.(4) 0 = - 2 C1 (1) + 2C2 (1) 4

(6)

Solving Eqs. (5) and (6) yields C1 = -

9.81 16

C2 =

9.81 16

Thus, r = -

9.81 - 2t 9.81 2t 9.81 e e + sin 2t 16 16 8

9.81 -e - 2t + e2t a - sin 2t b 8 2 9.81 (sin h 2t - sin 2t) = 8 =

At u = 2t =

p , 4

r =

p p 9.81 asin h - sin 3 b = 0.198 m 8 4 4

Ans. 355

Ans: r = 0.198 m

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*13–112. The pilot of an airplane executes a vertical loop which in part follows the path of a “four-leaved rose,” r = 1 - 600 cos 2u2 ft, where u is in radians. If his speed at A is a constant vP = 80 ft>s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at A. He weighs 130 lb. Hint: To determine the time derivatives necessary to compute the acceleration components a r and a u , take the first and second time derivatives of r = 40011 + cos u2. # Then, for further information, use Eq. 12–26 to determine u. Also, take # the time derivative of Eq. 12–26, noting that vC = 0, to $ determine u.

SOLUTION r = -600 cos 2u

# # r = 1200 sin 2uu

80 ft/s

A

r

r

600 cos 2u

u

# $ $ r = 1200 A 2 cos 2uu2 + sin 2uu B

At u = 90° # # r = -600 cos 180° = 600 ft r = 1200 sin 180°u = 0 # $ # $ r = 1200 A 2 cos 180°u2 + sin 180°u B = - 2400u2 # # # yr = r = 0 yu = ru = 600u y2p = y2r + y2u # 802 = 02 + A 600 u B 2

# u = 0.1333 rad>s

$ r = -2400(0.1333)2 = - 42.67 ft>s2 # $ a r = r - ru2 = -42.67 - 600(0.1333)2 = -53.33 ft>s2 + c ©Fr = mar;

- N - 130 =

130 (- 53.33) 32.2

N = 85.3 lb

Ans.

Ans: N = 85.3 lb 356

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13–113. The earth has an orbit with eccentricity e = 0.0167 around the sun. Knowing that the earth’s minimum distance from the sun is 146(10 6) km, find the speed at which the earth travels when it is at this distance. Determine the equation in polar coordinates which describes the earth’s orbit about the sun.

SOLUTION e =

Ch2 GMS

e =

GMS 1 ¢1 ≤ (r0v0)2 GMS r0 r0 v20

y0 = =

B

where C =

GMS 1 ¢1 ≤ and h = r0 v0 r0 r0 v20 e = ¢

r0 v20 - 1≤ GMS

r0v20 = e + 1 GMS

GMS (e + 1) r0

66.73(10 - 12)(1.99)(1030)(0.0167 + 1) = 30409 m>s = 30.4 km>s B 146(109)

Ans.

GMS GMS 1 1 = a1 bcos u + 2 2 r r0 r0 v20 r0v0 66.73(10 - 12)(1.99)(1030) 66.73(10 - 12)(1.99)(1030) 1 1 = b cos u + a1 9 9 2 r 146(10 ) 151.3(10 )(30409) C 146(10 9) D 2 (30409)2 1 = 0.348(10 - 12) cos u + 6.74(10 - 12) r

Ans.

Ans: vo = 30.4 km>s 1 = 0.348 ( 10-12 ) cos u + 6.74 ( 10-12 ) r 357

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13–114. A communications satellite is in a circular orbit above the earth such that it always remains directly over a point on the earth’s surface. As a result, the period of the satellite must equal the rotation of the earth, which is approximately 24 hours. Determine the satellite’s altitude h above the earth’s surface and its orbital speed.

SOLUTION The period of the satellite around r0 = h + re = C h + 6.378(106) D m is given by T =

circular

orbit

of

radius

2pr0 vs

24(3600) = vs =

the

2p C h + 6.378(106) D vs

2p C h + 6.378(106)

(1)

86.4(103)

The velocity of the satellite orbiting around the circular orbit of radius r0 = h + re = C h + 6.378(106) D m is given by yS = yS =

GMe

C r0 C

66.73(10-12)(5.976)(1024)

(2)

h + 6.378(106)

Solving Eqs.(1) and (2), h = 35.87(106) m = 35.9 Mm

yS = 3072.32 m>s = 3.07 km>s Ans.

Ans: h = 35.9 mm vs = 3.07 km>s 358

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13–115. The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13–25. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit 800 km from the earth’s surface.

SOLUTION For a 800-km orbit v0 =

66.73(10 - 12)(5.976)(1024) B (800 + 6378)(103) Ans.

= 7453.6 m>s = 7.45 km>s

Ans: v0 = 7.45 km>s 359

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*13–116. The rocket is in circular orbit about the earth at an altitude of 20 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field. 20 Mm

Solution The speed of the rocket in circular orbit is vc =

66.73 ( 10-12 ) 3 5.976 ( 1024 ) 4 GMe = 3888.17 m>s = A A r0 20 ( 106 ) + 6378 ( 103 )

To escape the earth’s gravitational field, the rocket must enter the parabolic trajectory, which require its speed to be ve =

2GMe 2 3 66.73 ( 10-12 ) 4 3 5.976 ( 1024 ) 4 = A = 5498.70 m>s A r0 20 ( 106 ) + 6378 ( 103 )

The required increment in speed is

∆v = ve - vc = 5498.70 - 3888.17 = 1610.53 m>s = 1.61 ( 103 ) m>s

Ans.

Ans:

∆v = 1.61 ( 103 ) m>s

360

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13–117. Prove Kepler’s third law of motion. Hint: Use Eqs. 13–19, 13–28, 13–29, and 13–31.

SOLUTION From Eq. 13–19, GMs 1 = C cos u + r h2 For u = 0° and u = 180°, GMs 1 = C + rp h2 GMs 1 = -C + ra h2 Eliminating C, from Eqs. 13–28 and 13–29, 2GMs 2a = b2 h2 From Eq. 13–31, T =

p (2a)(b) h

b2 =

T2h2 4p2a2

Thus,

GMs 4p2a3 = T2h2 h2 T2 = a

4p2 b a3 GMs

Q.E.D.

Ans: T2 = a 361

4p2 3 ba GMs

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13–118. The satellite is moving in an elliptical orbit with an eccentricity e = 0.25. Determine its speed when it is at its maximum distance A and minimum distance B from the earth. 2 Mm

A

SOLUTION Ch2 GMe

e = where C =

GMe 1 ¢1 ≤ and h = r0 v0. r0 r0v20 e =

GMe 1 ¢1 ≤ (r0 v0)2 GMe r0 r0 v20 e = ¢

r0 v20 = e + 1 GMe

r0 v20 - 1≤ GMe v0 =

B

GMe (e + 1) r0

where r0 = rp = 2 A 106 B + 6378 A 103 B = 8.378 A 106 B m. vB = v0 = ra =

C

66.73(10 - 12)(5.976)(1024)(0.25 + 1) 8.378(106)

= 7713 m>s = 7.71 km>s

Ans.

8.378(106) r0 = = 13.96 A 106 B m 2GMe 2(66.73)(10 - 12)(5.976)(1024) - 1 - 1 r0 v0 8.378(106)(7713)2 vA =

rp ra

vB =

8.378(106) 13.96(106)

(7713) = 4628 m>s = 4.63 km>s

Ans.

Ans: vB = 7.71 km>s vA = 4.63 km>s 362

B

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13–119. The rocket is traveling in free flight along the elliptical orbit. The planet has no atmosphere, and its mass is 0.60 times that of the earth. If the rocket has the orbit shown, determine the rocket’s speed when it is at A and at B.

B

O

18.3 Mm

A

7.60 Mm

Solution Applying Eq. 13–27, ra =

rp

( 2GM>rpv2p ) - 1

rp 2GM - 1 = 2 ra r p vp rp + ra 2GM = 2 ra r p vp vp =

2GM ra A rp(rp + ra)

The elliptical orbit has rp = 7.60(106) m, ra = 18.3(106) m and vp = vA . Then vA =

A

2 3 66.73 ( 10-12 ) 4 3 0.6(5.976) ( 1024 ) 4 3 18.3 ( 106 ) 4 7.60 ( 106 ) 3 7.60 ( 106 ) + 18.3 ( 106 ) 4

= 6669.99 m>s = 6.67 ( 103 ) m>s

Ans.

In this case, h = rp vA = ravB 7.60 ( 106 ) (6669.99) = 18.3 ( 106 ) vB vB = 2770.05 m>s = 2.77 ( 103 ) m>s

Ans.

Ans: vA = 6.67 ( 103 ) m>s vB = 2.77 ( 103 ) m>s 363

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*13–120. Determine the constant speed of satellite S so that it circles the earth with an orbit of radius r = 15 Mm. Hint: Use Eq. 13–1.

r

SOLUTION F = G ms a y =

ms me r2

Also

y2s F = ms a b r

15 Mm

S

Hence

ms me y20 b = G r r2 A

G

me 5.976(1024) = 66.73(10 - 12) a b = 5156 m>s = 5.16 km>s r B 15(106)

Ans.

Ans: v = 5.16 km>s 364

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13–121. The rocket is in free flight along an elliptical trajectory A¿A. The planet has no atmosphere, and its mass is 0.70 times that of the earth. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A.

r A

B

6 Mm

3 Mm

O

A¿

9 Mm

SOLUTION Central-Force Motion: Use ra = M = 0.70Me, we have 9 A 106 B =

r0 (2 GM>r0 y20 B -1

, with r0 = rp = 6 A 106 B m and

6(10)6

¢

2(66.73) (10-12) (0.7) [5.976(1024)] 6(106)y2P

≤ - 1 Ans.

yA = 7471.89 m>s = 7.47 km>s

Ans: vA = 7.47 km>s 365

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13–122. The Viking Explorer approaches the planet Mars on a parabolic trajectory as shown. When it reaches point A its velocity is 10 Mm> h. Determine r0 and the required velocity at A so that it can then maintain a circular orbit as shown. The mass of Mars is 0.1074 times the mass of the earth. A

SOLUTION When the Viking explorer approaches point A on a parabolic trajectory, its velocity at point A is given by vA =

A

2GMM r0

B 10(106)

r0

2(66.73)(10 m 1h R¢ ≤ = h 3600 s D

) C 0.1074(5.976)(1024) D

- 12

r0

r0 = 11.101(106) m = 11.1 Mm

Ans.

When the explorer travels along a circular orbit of r0 = 11.101(106) m, its velocity is vA¿

66.73(10 GMr = = A r0 D

- 12

) C 0.1074(5.976)(1024) D

11.101(106)

= 1964.19 m>s Thus, the required sudden decrease in the explorer’s velocity is ¢vA = vA - vA¿ = 10(106) a

1 b - 1964.19 3600 Ans.

= 814 m>s

Ans: r0 = 11.1 Mm ∆vA = 814 m>s 366

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13–123. The rocket is initially in free-flight circular orbit around the earth. Determine the speed of the rocket at A. What change in the speed at A is required so that it can move in an elliptical orbit to reach point A′?

8 Mm A O

A¿ 19 Mm

Solution The required speed to remain in circular orbit containing point A of which r0 = 8 ( 106 ) + 6378 ( 103 ) = 14.378 ( 106 ) m can be determined from (vA)C = =

GMe A r0 A

366.73 ( 10-12 )4 35.976 ( 1024 )4 14.378 ( 106 )

= 5266.43 m>s = 5.27 ( 103 ) m>s

Ans.

To more from A to A′, the rocket has to follow the elliptical orbit with rp = 8 ( 106 ) + 6378 ( 103 ) = 14.378 ( 106 ) m and ra = 19 ( 106 ) + 6378 ( 103 ) = 25.378 ( 106 ) m. The required speed at A to do so can be determined using Eq. 13–27. ra =

( 2GMe >rpv2p ) - 1

2GMe rpv2p 2GMe rpv2p vp =

rp

- 1 = =

rp ra

rp + ra ra

2GMe ra A rp(rp + ra)

Here, vp = (vA)e. Then (vA)e =

A

2 366.73 ( 10-12 )435.976 ( 1024 )4325.378 ( 106 )4 14.378 ( 106 ) 314.378 ( 106 ) + 25.378 ( 106 )4

= 5950.58 m>s Thus, the required change in speed is

∆v = (vA)e - (vA)c = 5950.58 - 5266.43 = 684.14 m>s = 684 m>s Ans.

Ans: ( vA ) C = 5.27 ( 103 ) m>s ∆v = 684 m>s 367

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*13–124. The rocket is in free-flight circular orbit around the earth. Determine the time needed for the rocket to travel from the inner orbit at A to the outer orbit at A′.

8 Mm A O

A¿ 19 Mm

Solution To move from A to A′, the rocket has to follow the elliptical orbit with   rp = 8 ( 106 ) + 6378 ( 103 ) = 14.378 ( 106 ) m and ra = 19 ( 106 ) + 6378 ( 103 ) 6 ( ) = 25.378 10 m. The required speed at A to do so can be determined using Eq. 13–27. ra =

rp

( 2GMe >rPv2p ) - 1

2GMe rpv2p 2GMe rpv2p vp = Here, vp = vA. Then vA = Then

- 1 = =

rp ra

rp + ra ra

2GMera B rp(rp + ra)

2 3 66.73 ( 10-12 ) 43 5.976 ( 1024 ) 4 3 25.378 ( 106 ) 4

D 14.378 ( 106 ) 314.378 ( 106 ) + 25.378 ( 106 ) 4

= 5950.58 m>s

h = vArp = 5950.58 314.378 ( 106 )4 = 85.5573 ( 109 ) m2 >s



The period of this elliptical orbit can be determined using Eq. 13–31. p (r + ra) 2rpra h p p 314.378 ( 106 ) + 25.378 ( 106 )4 2 314.378 ( 106 )4325.378 ( 106 )4 = 85.5573 ( 109 )

T =

= 27.885 ( 103 ) s

Thus, the time required to travel from A to A′ is

t =

27.885 ( 103 ) T = = 13.94 ( 103 ) s = 3.87 h 2 2

Ans.

Ans: t = 3.87 h 368

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13–125. A satellite is launched with an initial velocity v0 = 2500 mi>h parallel to the surface of the earth. Determine the required altitude (or range of altitudes) above the earth’s surface for launching if the free-flight trajectory is to be (a) circular, (b) parabolic, (c) elliptical, and (d) hyperbolic. Take G = 34.4110-921lb # ft22> slug2, Me = 409110212 slug, the earth’s radius re = 3960 mi, and 1 mi = 5280 ft.

SOLUTION v0 = 2500 mi>h = 3.67(103) ft>s (a)

e = 1 =

C2h = 0 GMe

or C = 0

GMe r0 v20

GMe = 34.4(10 - 9)(409)(1021) = 14.07(1015) r0 =

(b)

GMe v20

=

14.07(1015) [3.67(1013)]2

= 1.046(109) ft

r =

1.047(109) - 3960 = 194(103) mi 5280

e =

C2h = 1 GMe

Ans.

GMe 1 1 (r2 v2) a b ¢ 1 ≤ = 1 GMe 0 0 r0 r0 v20 r0 =

2GMe v20

=

2(14.07)(1015) [3.67(103)]2

= 2.09(109) ft = 396(103) mi

r = 396(103) - 3960 = 392(103) mi

(c)

Ans.

e 6 1 194(103) mi 6 r 6 392(103) mi

(d)

Ans.

e 7 1 r 7 392(103) mi

Ans.

Ans: (a)  r = 194 ( 103 ) mi (b)  r = 392 ( 103 ) mi (c)  194 ( 103 ) mi 6 r 6 392 ( 103 ) mi (d)  r 7 392 ( 103 ) mi 369

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13–126. The rocket is traveling around the earth in free flight along the elliptical orbit. If the rocket has the orbit shown, determine the speed of the rocket when it is at A and at B.

B

A

Solution 30 Mm

Here rp = 20 ( 106 ) m and ra = 30 ( 106 ) m. Applying Eq. 13–27, ra =

rp

( 2GMe >rpv2p ) - 1

2GMe rpv2p 2GMe rpv2p vP = Here vp = vA. Then

vA =

D

20 Mm

- 1 = =

rp ra

rp + ra ra 2GMera

B rp ( rp + ra )

2 3 66.73 ( 10-12 ) 4 3 5.976 ( 1024 )4 3 30 ( 106 )4 20 ( 106 ) 320 ( 106 ) + 30 ( 106 )4

= 4891.49 m>s = 4.89 ( 103 ) m>s



Ans.

For the same orbit h is constant. Thus,

h = rpvp = rava

3 20 ( 10 )4 ( 4891.49 ) 6

=

3 30 ( 106 )4 vB

vB = 3261.00 m>s = 3.26 ( 103 ) m>s

Ans.

Ans: vA = 4.89 ( 103 ) m>s vB = 3.26 ( 103 ) m>s 370

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13–127. An elliptical path of a satellite has an eccentricity e = 0.130. If it has a speed of 15 Mm>h when it is at perigee, P, determine its speed when it arrives at apogee, A. Also, how far is it from the earth’s surface when it is at A?

A

P

SOLUTION e = 0.130 np = n0 = 15 Mm>h = 4.167 km>s e =

GMe r20 v20 1 Ch2 = ¢1 b ≤a 2 r0 GMe GMe r0v0

e = ¢

r0 n20 -1 ≤ GMe

r0 n20 = e +1 GMe r0 = =

(e + 1)GMe n20 1.130166.732110 - 12215.9762110242

C 4.16711032 D 2

= 25.96 Mm GMe

=

r0 n20 rA =

rA = =

1 e + 1

r0 2GMe r0n02

- 1

=

r0

A

2 e + 1

B -1

r01e + 12 1 - e 25.961106211.1302 0.870

= 33.7111062 m = 33.7 Mm nA = =

n0r0 rA 15125.96211062 33.7111062 Ans.

= 11.5 Mm>h d = 33.7111062 - 6.37811062

Ans.

= 27.3 Mm

Ans: vA = 11.5 Mm>h d = 27.3 Mm 371

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*13–128. A rocket is in free-flight elliptical orbit around the planet Venus. Knowing that the periapsis and apoapsis of the orbit are 8 Mm and 26 Mm, respectively, determine (a) the speed of the rocket at point A¿, (b) the required speed it must attain at A just after braking so that it undergoes an 8-Mm free-flight circular orbit around Venus, and (c) the periods of both the circular and elliptical orbits. The mass of Venus is 0.816 times the mass of the earth.

O

A¿

A

8 Mm

SOLUTION 18 Mm

a) Mn = 0.81615.9761102422 = 4.876110242 OA¿ =

OA 2GMn a - 1b OA n2A 811062

2611026 = a 81.3511062 n2A

2166.732110 - 1224.876110242 811062n2A

- 1b

= 1.307

nA = 7887.3 m>s = 7.89 km>s 81106217887.32 OA nA = 2426.9 m>s = 2.43 m>s = OA¿ 2611062

nA =

Ans.

b)

nA¿¿ =

GMn

D OA¿

=

D

66.73110 - 1224.876110242 811062 Ans.

nA¿¿ = 6377.7 m>s = 6.38 km>s c) Circular orbit: Tc =

2p811062 2pOA = 7881.41 s = 2.19 h = nA¿¿ 6377.7

Ans.

Elliptic orbit: Te =

p p 1 8 + 262 A 106 B A 2(8)(26) B A 106 B 1OA + OA¿2 21OA21OA¿2 = OAnA 81106217886.82 Ans.

Te = 24414.2 s = 6.78 h

Ans: vA = 2.43 m>s vA″ = 6.38 km>s Tc = 2.19 h Te = 6.78 h 372

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13–129. The rocket is traveling in a free flight along an elliptical trajectory A′A. The planet has no atmosphere, and its mass is 0.60 times that of the earth. If the rocket has the orbit shown, determine the rocket’s velocity when it is at point A.

r  6 Mm A¿

O

B

A

Solution 100 Mm

Applying Eq. 13–27, ra =

70 Mm

rp

( 2GM>rpv2p ) - 1

rp 2GM - 1 = 2 ra r p vp rp + ra 2GM = 2 ra r p vp vp =

2GMra A rp ( rp + ra )

The rocket is traveling around the elliptical orbit with rp = 70 ( 106 ) m, ra = 100 ( 106 ) m and vp = vA. Then vA =

D

2 366.73 ( 10-12 )430.6 ( 5.976 )( 1024 )4 3100 ( 106 )4 70 ( 106 ) 370 ( 106 ) + 100 ( 106 )4

= 2005.32 m>s = 2.01 ( 103 ) m>s



Ans.

Ans: vA = 2.01 ( 103 ) m>s 373

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13–130. If the rocket is to land on the surface of the planet, determine the required free-flight speed it must have at A′ so that the landing occurs at B. How long does it take for the rocket to land, going from A′ to B? The planet has no atmosphere, and its mass is 0.6 times that of the earth.

r  6 Mm A¿

O

B

A

Solution 100 Mm

Applying Eq. 13–27,

70 Mm

rp



ra =



rp 2GM - 1 = 2 ra r p vp



rp + ra 2GM = ra rpv2p



vp =

( 2GM>rpv2p ) - 1

2GMra B rp ( rp + ra )

To land on B, the rocket has to follow the elliptical orbit A′B with rp = 6 ( 106 ) , ra = 100 ( 106 ) m and vp = vB. vB = In this case

D

2 366.73 ( 10-12 )430.6 ( 5.976 )( 1024 )4 3100 ( 106 )4 6 ( 106 ) 36 ( 106 ) + 100 ( 106 )4



h = rpvB = ravA′



6 ( 106 )( 8674.17 ) = 100 ( 106 ) vA′



vA′ = 520.45 m>s = 521 m>s

= 8674.17 m>s

Ans.

The period of the elliptical orbit can be determined using Eq. 13–31. T = =

p (r + ra) 1rpra h p p 6 ( 106 )( 8674.17 )

= 156.73 ( 103 ) s

36 ( 106 )

+ 100 ( 106 )4 2 36 ( 106 )4 3100 ( 106 )4

Thus, the time required to travel from A′ to B is

t =

T = 78.365 ( 103 ) s = 21.8 h 2

Ans.

Ans: vA′ = 521 m>s t = 21.8 h 374

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13–131. The rocket is traveling around the earth in free flight along an elliptical orbit AC. If the rocket has the orbit shown, determine the rocket’s velocity when it is at point A.

C

A

B

Solution For orbit AC, rp = 10 ( 106 ) m and ra = 16 ( 106 ) m. Applying Eq. 13–27

ra =

rpv2p 2GMe



rpv2p



vp =

Here vp = vA. Then

vA =

D

8 Mm

10 Mm

rp

( 2GMe >rpv2p ) - 1

2GMe



8 Mm

- 1 = =

rp ra

rp + ra ra 2GMe ra

B rp ( rp + ra )

2 366.73 ( 10-12 )4 35.976 ( 1024 )4 316 ( 106 )4 10 ( 106 ) 310 ( 106 ) + 16 ( 106 )4

= 7005.74 m>s = 7.01 ( 103 ) m>s



Ans.

Ans: vA = 7.01 ( 103 ) m>s 375

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*13–132. The rocket is traveling around the earth in free flight along the elliptical orbit AC. Determine its change in speed when it reaches A so that it travels along the elliptical orbit AB.

C

A

B

Solution 8 Mm

Applying Eq. 13–27,

ra =

(2GMe >rpv2p) - 1

rpv2p 2GMe



rav2p



vp =

10 Mm

rp

2GMe



8 Mm

- 1 =

=

rp ra

rp + ra ra

2GMe ra B rp(rp + ra)

For orbit AC, rp = 10 ( 106 ) m, ra = 16 ( 106 ) m and vp = (vA)AC. Then (vA)AC =

D

2 366.73 ( 10-12 )4 35.976 ( 1024 )4 316 ( 106 )4 10 ( 106 ) 310 ( 106 ) + 16 ( 106 )4

= 7005.74 m>s

For orbit AB, rp = 8 ( 106 ) m, ra = 10 ( 106 ) m and vp = vB. Then vB =

D

2 366.73 ( 10-12 )4 35.976 ( 1024 )4 310 ( 106 )4 8 ( 106 ) 38 ( 106 ) + 10 ( 106 )4

= 7442.17 m>s

Since h is constant at any position of the orbit, h = rpvp = rava 8 ( 106 ) (7442.17) = 10 ( 106 ) (vA)AB (vA)AB = 5953.74 m>s Thus, the required change in speed is ∆v = (vA)AB - (vA)AC = 5953.74 - 7005.74 = -1052.01 m>s = - 1.05 km>s

Ans.

The negative sign indicates that the speed must be decreased.

Ans: ∆v = - 1.05 km>s 376

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14–1. The 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N. When s = 15 m, the crate is moving to the right with a speed of 8 m/s. Determine its speed when s = 25 m. The coefficient of kinetic friction between the crate and the ground is mk = 0.25.

F 30°

SOLUTION Equation of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = mkN = 0.25N. Applying Eq. 13–7, we have + c a Fy = may ;

N + 100 sin 30° - 20(9.81) = 20(0) N = 146.2 N

Principle of Work and Energy: The horizontal component of force F which acts in the direction of displacement does positive work, whereas the friction force Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do no work. Applying Eq.14–7, we have T1 + a U1 - 2 = T2 25 m

1 (20)(8 2) + 2 L15 m

100 cos 30° ds

25 m

-

L15 m

36.55 ds =

1 (20)v2 2 Ans.

v = 10.7 m s

Ans: v = 10.7 m>s 377

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14–2. F (lb)

For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is F = (90(103)x1>2) lb, where x is in ft, determine the car’s maximum penetration in the barrier. The car has a weight of 4000 lb and it is traveling with a speed of 75 ft>s just before it hits the barrier.

F  90(10)3 x1/2

x (ft)

SOLUTION Principle of Work and Energy: The speed of the car just before it crashes into the barrier is v1 = 75 ft>s. The maximum penetration occurs when the car is brought to a stop, i.e., v2 = 0. Referring to the free-body diagram of the car, Fig. a, W and N do no work; however, Fb does negative work. T1 + ©U1 - 2 = T2 1 4000 a b (752) + c 2 32.2 L0

xmax

90(103)x1>2dx d = 0 Ans.

xmax = 3.24 ft

Ans: x max = 3.24 ft 378

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14–3. The crate, which has a mass of 100 kg, is subjected to the action of the two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m>s. The coefficient of kinetic friction between the crate and the surface is mk = 0.2.

1000 N

800 N 5

30

3 4

SOLUTION Equations of Motion: Since the crate slides, the friction force developed between the crate and its contact surface is Ff = mk N = 0.2N. Applying Eq. 13–7, we have + c ©Fy = may;

3 N + 1000 a b - 800 sin 30° - 100(9.81) = 100(0) 5 N = 781 N

Principle of Work and Energy: The horizontal components of force 800 N and 1000 N which act in the direction of displacement do positive work, whereas the friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the opposite direction to that of displacement. The normal reaction N, the vertical component of 800 N and 1000 N force and the weight of the crate do not displace, hence they do no work. Since the crate is originally at rest, T1 = 0. Applying Eq. 14–7, we have T1 + a U1-2 = T2 4 1 0 + 800 cos 30°(s) + 1000 a bs - 156.2s = (100) A 62 B 5 2 Ans.

s = 1.35m

Ans: s = 1.35 m 379

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*14–4. The 100-kg crate is subjected to the forces shown. If it is originally at rest, determine the distance it slides in order to attain a speed of v = 8 m>s. The coefficient of kinetic friction between the crate and the surface is mk = 0.2.

500 N

400 N 30

45

Solution Work. Consider the force equilibrium along the y axis by referring to the FBD of the crate, Fig. a, + c ΣFy = 0;

N + 500 sin 45° - 100(9.81) - 400 sin 30° = 0 N = 827.45 N

Thus, the friction is Ff = mkN = 0.2(827.45) = 165.49 N. Here, F1 and F2 do positive work whereas Ff   does negative work. W and N do no work UF1 = 400 cos 30° s = 346.41 s UF2 = 500 cos 45° s = 353.55 s UFf = - 165.49 s Principle of Work And Energy. Applying Eq. 14–7, T1 + ΣU1 - 2 = T2 0 + 346.41 s + 353.55 s + ( -165.49 s) =

1 (100) ( 82 ) 2

s = 5.987 m = 5.99 m

Ans.

Ans: s = 5.99 m 380

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14–5. Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill A it will reach a speed of 100 km>h when it comes to the bottom B. Also, what should be the minimum radius of curvature r for the track at B so that the passengers do not experience a normal force greater than 4mg = (39.24m) N? Neglect the size of the car and passenger.

A

h r B

Solution 100 km>h =

100 ( 103 ) 3600

= 27.778 m>s

T1 + ΣU1 - 2 = T2 0 + m(9.81)h =

1 m(27.778)2 2 Ans.

h = 39.3 m + c ΣFn = man;

(27.778)2 39.24 m - mg = ma b r

Ans.

r = 26.2 m

Ans: h = 39.3 m r = 26.2 m 381

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14–6. When the driver applies the brakes of a light truck traveling 40 km>h, it skids 3 m before stopping. How far will the truck skid if it is traveling 80 km>h when the brakes are applied?

Solution 40 km>h =

40 ( 103 ) 3600

= 11.11 m>s

80 km>h = 22.22 m>s

T1 + ΣU1 - 2 = T2 1 m(11.11)2 - mk mg(3) = 0 2 mk g = 20.576 T1 + ΣU1 - 2 = T2 1 m(22.22)2 - (20.576)m(d) = 0 2 Ans.

d = 12 m

Ans: d = 12 m 382

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14–7. As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block if it has an initial speed of 5 m>s and travels 10 m, both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x¿ axis and moving at a constant velocity of 2 m>s relative to A. Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy.

A B

x x¿

2 m/s 5 m/s 6N

10 m

SOLUTION Observer A: T1 + ©U1 - 2 = T2 1 1 (10)(5)2 + 6(10) = (10)v22 2 2 Ans.

v2 = 6.08 m>s Observer B: F = ma 6 = 10a + B A:

a = 0.6 m>s2 s = s0 + v0t +

1 2 at 2 c

10 = 0 + 5t +

1 (0.6)t2 2

t2 + 16.67t - 33.33 = 0 t = 1.805 s At v = 2 m>s, s¿ = 2(1.805) = 3.609 m Block moves 10 - 3.609 = 6.391 m Thus T1 + ©U1 - 2 = T2 1 1 (10)(3)2 + 6(6.391) = (10)v22 2 2 Ans.

v2 = 4.08 m>s Note that this result is 2 m>s less than that observed by A.

Ans: Observer A: v2 = 6.08 m>s Observer B: v2 = 4.08 m>s 383

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*14–8. A force of F = 250 N is applied to the end at B. Determine the speed of the 10-kg block when it has moved 1.5 m, starting from rest.

Solution Work. with reference to the datum set in Fig. a,

B

SW + 2sF = l (1)

dSW + 2dsF = 0



A

F

Assuming that the block moves upward 1.5 m, then dSW = -1.5 m since it is directed in the negative sense of SW. Substituted this value into Eq. (1), - 1.5 + 2dsF = 0  dsF = 0.75 m Thus, UF = FdSF = 250(0.75) = 187.5 J UW = - WdSW = - 10(9.81)(1.5) = -147.15 J Principle of Work And Energy. Applying Eq. 14–7, T1 + U1 - 2 = T2 0 + 187.5 + ( -147.15) =

1 (10)v2 2 Ans.

v = 2.841 m>s = 2.84 m>s

Ans: v = 2.84 m>s 384

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14–9. The “air spring” A is used to protect the support B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D. The force developed by the air spring as a function of its deflection is shown by the graph. If the block has a mass of 20 kg and is suspended a height d = 0.4 m above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails. Neglect the mass of the pulley and belt.

F (N)

D

1500

C d

0.2

s (m)

A B

Solution Work. Referring to the FBD of the tensioning weight, Fig. a, W does positive work whereas force F does negative work. Here the weight displaces downward SW = 0.4 + xmax where xmax is the maximum compression of the air spring. Thus UW = 20(9.81) ( 0.4 + xmax ) = 196.2 ( 0.4 + xmax ) The work of F is equal to the area under the F-S graph shown shaded in Fig. b, Here F 1500    ; F = 7500xmax. Thus = xmax 0.2 1 UF = - (7500 xmax)(xmax) = - 3750x2max 2 Principle of Work And Energy. Since the block is at rest initially and is required to stop momentarily when the spring is compressed to the maximum, T1 = T2 = 0. Applying Eq. 14–7, T1 + ΣU1 - 2 = T2 0 + 196.2(0.4 + xmax) +

( - 3750x2max ) = 0

3750x2max - 196.2xmax - 78.48 = 0

xmax = 0.1732 m = 0.173 m 6 0.2 m      (O.K!)

Ans.

Ans: xmax = 0.173 m 385

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14–10. v

The force F, acting in a constant direction on the 20-kg block, has a magnitude which varies with the position s of the block. Determine how far the block must slide before its velocity becomes 15 m>s. When s = 0 the block is moving to the right at v = 6 m>s. The coefficient of kinetic friction between the block and surface is mk = 0.3.

F (N)

F

F  50s 1/2

Solution Work. Consider the force equilibrium along y axis, by referring to the FBD of the block, Fig. a,

s (m)

+ c ΣFy = 0  ;   N - 20(9.81) = 0   N = 196.2 N Thus, the friction is Ff = mkN = 0.3(196.2) = 58.86 N. Here, force F does positive work whereas friction Ff does negative work. The weight W and normal reaction N do no work. UF =

L

Fds =

L0

s

1

50s2 ds =

100 3 s2 3

UFf = - 58.86 s Principle of Work And Energy. Applying Eq. 14–7, T1 + ΣU1 - 2 = T2 1 100 3 1 (20)(62) + s2 + ( - 58.86s) = (20)(152) 2 3 2 100 3 s2 - 58.86s - 1890 = 0 3 Solving numerically, Ans.

s = 20.52 m = 20.5 m

Ans: s = 20.5 m 386

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14–11. The force of F = 50 N is applied to the cord when s = 2 m. If the 6-kg collar is orginally at rest, determine its velocity at s = 0. Neglect friction.

F

1.5 m A

Solution

s

Work. Referring to the FBD of the collar, Fig. a, we notice that force F does positive work but W and N do no work. Here, the displacement of F is s = 222 + 1.52 - 1.5 = 1.00 m

UF = 50(1.00) = 50.0 J

Principle of Work And Energy. Applying Eq. 14–7, T1 + ΣU1 - 2 = T2 0 + 50 =

1 (6)v2 2

v = 4.082 m>s = 4.08 m>s

Ans.

Ans: v = 4.08 m>s 387

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*14–12. Design considerations for the bumper B on the 5-Mg train car require use of a nonlinear spring having the loaddeflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to 0.2 m when the car, traveling at 4 m>s, strikes the rigid stop. Neglect the mass of the car wheels.

F (N)

F

ks2

s (m)

SOLUTION B

0.2

1 ks2 ds = 0 (5000)(4)2— 2 L0 40 000 - k

(0.2)3 = 0 3

k = 15.0 MN>m2

Ans.

Ans: k = 15.0 MN>m2 388

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14–13. The 2-lb brick slides down a smooth roof, such that when it is at A it has a velocity of 5 ft>s. Determine the speed of the brick just before it leaves the surface at B, the distance d from the wall to where it strikes the ground, and the speed at which it hits the ground.

y A

5 ft/s

15 ft

5

3 4

B

SOLUTION TA + ©UA-B = TB 30 ft

2 2 1 1 a b(5)2 + 2(15) = a bv2B 2 32.2 2 32.2 Ans.

vB = 31.48 ft>s = 31.5 ft>s + b a:

Ans. d

s = s0 + v0t 4 d = 0 + 31.48 a b t 5

A+TB

s = s0 + v0t -

1 ac t2 2

1 3 30 = 0 + 31.48 a b t + (32.2)t2 5 2 16.1t2 + 18.888t - 30 = 0 Solving for the positive root, t = 0.89916 s 4 d = 31.48a b (0.89916) = 22.6 ft 5

Ans.

TA + ©UA-C = TC 2 2 1 1 a b(5)2 + 2(45) = a bv2C 2 32.2 2 32.2 Ans.

vC = 54.1 ft>s

Ans: vB = 31.5 ft>s d = 22.6 ft vC = 54.1 ft>s 389

x

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14–14. Block A has a weight of 60 lb and block B has a weight of 10 lb. Determine the speed of block A after it moves 5 ft down the plane, starting from rest. Neglect friction and the mass of the cord and pulleys.

B A 5

SOLUTION

3

4

2 sA + sB = l 2 ∆sA + ∆sB = 0 2vA + vB = 0 T1 + ΣU1 - 2 = T2 3 1 10 1 60 0 + 60 a b(5) - 10(10) = a b vA2 + a b(2vA)2 5 2 32.2 2 32.2

Ans.

vA = 7.18 ft>s

Ans: vA = 7.18 ft>s 390

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14–15. The two blocks A and B have weights WA = 60 lb and WB = 10 lb. If the kinetic coefficient of friction between the incline and block A is mk = 0.2, determine the speed of A after it moves 3 ft down the plane starting from rest. Neglect the mass of the cord and pulleys. A

SOLUTION

5

Kinematics: The speed of the block A and B can be related by using position coordinate equation. sA + (sA - sB) = l 2¢sA - ¢sB = 0

B

3 4

2sA - sB = l

¢sB = 2¢sA = 2(3) = 6 ft (1)

2vA - vB = 0 Equation of Motion: Applying Eq. 13–7, we have + ©Fy¿ = may¿ ;

60 4 (0) N - 60 a b = 5 32.2

N = 48.0 lb

Principle of Work and Energy: By considering the whole system, WA which acts in the direction of the displacement does positive work. WB and the friction force Ff = mkN = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite direction to that of displacement Here, WA is being displaced vertically (downward) 3 ¢s and WB is being displaced vertically (upward) ¢sB. Since blocks A and B are 5 A at rest initially, T1 = 0. Applying Eq. 14–7, we have T1 + a U1 - 2 = T2

3 1 1 0 + WA ¢ ¢sA ≤ - Ff ¢sA - WB ¢sB = mAv2A + mB v2B 5 2 2 3 1 60 1 10 60 B (3) R - 9.60(3) - 10(6) = ¢ ≤ v2A + ¢ ≤ v2B 5 2 32.2 2 32.2 1236.48 = 60v2A + 10v2B

(2)

Eqs. (1) and (2) yields Ans.

vA = 3.52 ft>s vB = 7.033 ft s

Ans: vA = 3.52 ft>s 391

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*14–16. A small box of mass m is given a speed of v = 214gr at the top of the smooth half cylinder. Determine the angle u at which the box leaves the cylinder.

A u

r

O

SOLUTION Principle of Work and Energy: By referring to the free-body diagram of the block, Fig. a, notice that N does no work, while W does positive work since it displaces downward though a distance of h = r - r cos u. T1 + ©U1 - 2 = T2 1 1 1 m a gr b + mg(r - r cos u) = mv2 2 4 2 v2 = gr a

9 - 2 cos u b 4

2

v = Equations of Motion: Here, an = r referring to Fig. a, ©Fn = man ;

(1) gr a

mg cos u - N = m c g a N = mg a 3 cos u -

9 - 2 cos u b 4 9 = ga - 2 cos u b . By r 4 9 - 2 cos u b d 4

9 b 4

It is required that the block leave the track. Thus, N = 0. 0 = mg a 3 cos u -

9 b 4

Since mg Z 0, 3 cos u -

9 = 0 4 Ans.

u = 41.41° = 41.4°

Ans: u = 41.4° 392

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14–17. If the cord is subjected to a constant force of F = 30 lb and the smooth 10-lb collar starts from rest at A, determine its speed when it passes point B. Neglect the size of pulley C.

y  1 x2 2

y

C

B

4.5 ft F  30 lb

A

SOLUTION

x 1 ft

3 ft

2 ft

Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a. Principle of Work and Energy: By referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B, W displaces upward through a distance h = 4.5 ft, while force F displaces a distance of s = AC - BC = 262 + 4.52 - 2 = 5.5 ft. The work of F is positive, whereas W does negative work. TA + gUA - B = TB

0 + 30(5.5) + [- 10(4.5)] =

1 10 a bv 2 2 32.2 B Ans.

vB = 27.8 ft>s

Ans: vB = 27.8 ft>s 393

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14–18. When the 12-lb block A is released from rest it lifts the two 15-lb weights B and C. Determine the maximum distance A will fall before its motion is momentarily stopped. Neglect the weight of the cord and the size of the pulleys.

4 ft

4 ft

Solution A

Consider the entire system: t = 2y2 + 42

T1 + ΣU1 - 2 = T2

C

B

(0 + 0 + 0) + 12y - 2(15) ( 2y2 + 42 - 4 ) = (0 + 0 + 0) 0.4y = 2y2 + 16 - 4

(0.4y + 4)2 = y2 + 16

-0.84y2 + 3.20y + 16 = 16 -0.84y + 3.20 = 0 Ans.

y = 3.81 ft

Ans: y = 3.81 ft 394

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14–19. If the cord is subjected to a constant force of F = 300 N and the 15-kg smooth collar starts from rest at A, determine the velocity of the collar when it reaches point B. Neglect the size of the pulley.

200 mm B

C 30 200 mm F  300 N

200 mm

SOLUTION

300 mm

Free-Body Diagram: The free-body diagram of the collar and cord system at an arbitrary position is shown in Fig. a.

A

Principle of Work and Energy: Referring to Fig. a, only N does no work since it always acts perpendicular to the motion. When the collar moves from position A to position B, W displaces vertically upward a distance h = (0.3 + 0.2) m = 0.5 m, while

force 2

F

displaces

a

distance

of

2

s = AC - BC = 20.72 + 0.42 -

20.2 + 0.2 = 0.5234 m . Here, the work of F is positive, whereas W does negative work. TA + gUA - B = TB

0 + 300(0.5234) + [-15(9.81)(0.5)] =

1 (15)vB2 2 Ans.

vB = 3.335 m>s = 3.34 m>s

Ans: vB = 3.34 m>s 395

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*14–20.

Barrier stopping force (kip)

The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having a weight of 4000 lb will penetrate the barrier if it is originally traveling at 55 ft>s when it strikes the first barrel.

SOLUTION T1 + ©U1 - 2 = T2 1 4000 a b (55)2 - Area = 0 2 32.2 Area = 187.89 kip # ft

36 27 18 9 0

2

5

10 15 20 25 Vehicle penetration (ft)

2(9) + (5 - 2)(18) + x(27) = 187.89 (O.K!)

x = 4.29 ft 6 (15 - 5) ft Thus

Ans.

s = 5 ft + 4.29 ft = 9.29 ft

Ans: s = 9.29 ft 396

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14–21. Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is mk = 0.10.

SOLUTION Block A: +a©Fy = may;

A

NA - 60 cos 60° = 0 NA = 30 lb

B 60

30

FA = 0.1(30) = 3 lb Block B: +Q©Fy = may;

NB - 40 cos 30° = 0 NB = 34.64 lb FB = 0.1(34.64) = 3.464 lb

Use the system of both blocks. NA, NB, T, and R do no work. T1 + ©U1-2 = T2 (0 + 0) + 60 sin 60°|¢sA| - 40 sin 30°|¢sB| - 3|¢sA| -3.464|¢sB| =

1 60 1 40 a b v2 + a b v2 2 32.2 A 2 32.2 B

2sA + sB = l 2¢sA = - ¢sB When |¢sB| = 2 ft, |¢sA| = 1 ft Also, 2vA = -vB Substituting and solving, Ans.

vA = 0.771 ft>s vB = -1.54 ft>s

Ans: vA = 0.771 ft>s 397

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14–22. The 25-lb block has an initial speed of v0 = 10 ft>s when it is midway between springs A and B. After striking spring B, it rebounds and slides across the horizontal plane toward spring A, etc. If the coefficient of kinetic friction between the plane and the block is mk = 0.4, determine the total distance traveled by the block before it comes to rest.

2 ft kA = 10 lb/in.

1 ft

kB = 60 lb/in.

v0 = 10 ft/s A

B

SOLUTION Principle of Work and Energy: Here, the friction force Ff = mk N = 0.4(25) = 10.0 lb. Since the friction force is always opposite the motion, it does negative work. When the block strikes spring B and stops momentarily, the spring force does negative work since it acts in the opposite direction to that of displacement. Applying Eq. 14–7, we have Tl + a U1 - 2 = T2

1 1 25 a b (10)2 - 10(1 + s1) - (60)s21 = 0 2 32.2 2 s1 = 0.8275 ft Assume the block bounces back and stops without striking spring A. The spring force does positive work since it acts in the direction of displacement. Applying Eq. 14–7, we have T2 + a U2 - 3 = T3 0 +

1 (60)(0.82752) - 10(0.8275 + s2) = 0 2

s2 = 1.227 ft Since s2 = 1.227 ft 6 2 ft, the block stops before it strikes spring A. Therefore, the above assumption was correct. Thus, the total distance traveled by the block before it stops is Ans.

sTot = 2s1 + s2 + 1 = 2(0.8275) + 1.227 + 1 = 3.88 ft

Ans: sTot = 3.88 ft 398

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14–23. The 8-kg block is moving with an initial speed of 5 m>s. If the coefficient of kinetic friction between the block and plane is mk = 0.25, determine the compression in the spring when the block momentarily stops.

5 m/s B

2m

kA  200 N/m

A

Solution Work. Consider the force equilibrium along y axis by referring to the FBD of the block, Fig. a + c ΣFy = 0;   N - 8(9.81) = 0  N = 78.48 N Thus, the friction is Ff = mkN = 0.25(78.48) = 19.62 N and Fsp = kx = 200 x. Here, the spring force Fsp and Ff both do negative work. The weight W and normal reaction N do no work. UFsp = -

L0

x

200 x dx = - 100 x2

UFf = - 19.62(x + 2) Principle of Work And Energy. It is required that the block stopped momentarily, T2 = 0. Applying Eq. 14–7 T1 + Σ U1 - 2 = T2 1 (8) ( 52 ) + 2

( - 100x2 ) + [ - 19.62(x + 2)] = 0

100x2 + 19.62x - 60.76 = 0 Solved for positive root, Ans.

x = 0.6875 m = 0.688 m

Ans: x = 0.688 m 399

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*14–24. At a given instant the 10-lb block A is moving downward with a speed of 6 ft s. Determine its speed 2 s later. Block B has a weight of 4 lb, and the coefficient of kinetic friction between it and the horizontal plane is mk = 0.2. Neglect the mass of the cord and pulleys.

B

SOLUTION

A

Kinematics: The speed of the block A and B can be related by using the position coordinate equation. sA + (sA - sB) = l 2¢sA - ¢sB = 0

2sA - sB = l [1]

¢sB = 2¢sA

[2]

yB = 2yA Equation of Motion: + ©Fy¿ = may¿ ;

NB - 4 =

4 (0) 32.2

NB = 4.00 lb

Principle of Work and Energy: By considering the whole system, WA, which acts in the direction of the displacement, does positive work. The friction force Ff = mk NB = 0.2(4.00) = 0.800 lb does negative work since it acts in the opposite direction to that of displacement. Here, WA is being displaced vertically (downward) ¢sA. Applying Eq. 14–7, we have T1 + a U1 - 2 = T2 1 1 m A y2A B 0 + mB A y2B B 0 + WA ¢sA - Ff ¢sB 2 A 2 =

1 1 m y2 + mB y2B 2 A A 2

[3] (yA)0 + yA d (2) = 2 (Eq. [2]). Substituting these

From Eq. [1], (yB)0 = 2(yA)0 = 2(6) = 12 ft>s. Also, ¢sA = c (yA)0 + yA = 6 + yA and ¢sB = 2¢sA = 12 + 2yA values into Eq. [3] yields

4 1 1 10 a b A 62 B + a b A 122 B + 10(6 + yA B - 0.800(12 + 2yA) 2 32.2 2 32.2 =

1 10 1 4 a b y2A + a b A 4y2A B 2 32.2 2 32.2 Ans.

yA = 26.8 ft>s

Ans: vA = 26.8 ft>s 400

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14–25. The 5-lb cylinder is falling from A with a speed vA = 10 ft>s onto the platform. Determine the maximum displacement of the platform, caused by the collision. The spring has an unstretched length of 1.75 ft and is originally kept in compression by the 1-ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision.

vA  10 ft/s

Solution

A

3 ft

T1 + Σ U1 - 2 = T2 1 1 5 1 a b ( 102 ) + 5(3 + s) - c (400)(0.75 + s)2 - (400)(0.75)2 d = 0 2 32.2 2 2

200 s2 + 295 s - 22.76 = 0

k  400 lb/ft

(O.K!)

s = 0.0735 ft 6 1 ft

1 ft

Ans.

s = 0.0735 ft

Ans: s = 0.0735 ft 401

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14–26. The catapulting mechanism is used to propel the 10-kg slider A to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the left by means of a piston P. If the piston applies a constant force F = 20 kN to rod BC such that it moves it 0.2 m, determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod BC.

A P F

B

C

SOLUTION 2 sC + sA = l 2 ¢ sC + ¢ sA = 0 2(0.2) = - ¢ sA -0.4 = ¢ sA T1 + ©U1 - 2 = T2 0 + (10 000)(0.4) =

1 (10)(vA)2 2 Ans.

vA = 28.3 m>s

Ans: vA = 28.3 m>s 402

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14–27. The “flying car” is a ride at an amusement park which consists of a car having wheels that roll along a track mounted inside a rotating drum. By design the car cannot fall off the track, however motion of the car is developed by applying the car’s brake, thereby gripping the car to the track and allowing it to move with a constant speed of the track, vt = 3 m>s. If the rider applies the brake when going from B to A and then releases it at the top of the drum, A, so that the car coasts freely down along the track to B 1u = p rad2, determine the speed of the car at B and the normal reaction which the drum exerts on the car at B. Neglect friction during the motion from A to B. The rider and car have a total mass of 250 kg and the center of mass of the car and rider moves along a circular path having a radius of 8 m.

A vt

u

8m

B

SOLUTION TA + ©UA - B = TB 1 1 (250)(3)2 + 250(9.81)(16) = (250)(vB)2 2 2 Ans.

vB = 17.97 = 18.0 m>s + c ©Fn = man

NB - 250(9.81) = 250 a

(17.97)2 b 8 Ans.

NB = 12.5 kN

Ans: vB = 18.0 m>s NB = 12.5 kN 403

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*14–28. y

The 10-lb box falls off the conveyor belt at 5-ft>s. If the coefficient of kinetic friction along AB is mk = 0.2, determine the distance x when the box falls into the cart.

5 ft/s

15 ft

A

3

5 4

B

Solution Work. Consider the force equilibrium along the y axis by referring to Fig. a,

30 ft C

4 + c ΣFy′ = 0;   N - 10 a b = 0   N = 8.00 lb 5

5 ft x

Thus, Ff = mkN = 0.2(8.00) = 1.60 lb. To reach B, W displaces vertically downward 15 ft and the box slides 25 ft down the inclined plane.

x

Uw = 10(15) = 150 ft # lb

UFf = -1.60(25) = -40 ft # lb Principle of Work And Energy. Applying Eq. 14–7 TA + Σ UA - B = TB 1 10 1 10 a b ( 52 ) + 150 + ( - 40) = a b v2B 2 32.2 2 32.2

vB = 27.08 ft>s

Kinematics. Consider the vertical motion with reference to the x-y coordinate system,

( + c ) (SC)y = (SB)y + (vB)yt +

1 a t 2; 2 y

3 1 5 = 30 - 27.08 a bt + ( -32.2)t 2 5 2 16.1t 2 + 16.25t - 25 = 0

Solve for positive root, t = 0.8398 s Then, the horizontal motion gives + S

(Sc)x = (SB)x + (vB)x t ;

4 x = 0 + 27.08 a b(0.8398) = 18.19 ft = 18.2 ft 5

Ans.

Ans: x = 18.2 ft 404

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14–29. The collar has a mass of 20 kg and slides along the smooth rod. Two springs are attached to it and the ends of the rod as shown. If each spring has an uncompressed length of 1 m and the collar has a speed of 2 m>s when s = 0, determine the maximum compression of each spring due to the backand-forth (oscillating) motion of the collar.

s

kA

50 N/m

kB

1m

100 N/m 1m

0.25 m

SOLUTION T1 + ©U1 - 2 = T2 1 1 1 (20)(2)2 - (50)(s)2 - (100)(s)2 = 0 2 2 2 Ans.

s = 0.730 m

Ans: s = 0.730 m 405

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14–30. The 30-lb box A is released from rest and slides down along the smooth ramp and onto the surface of a cart. If the cart is prevented from moving, determine the distance s from the end of the cart to where the box stops. The coefficient of kinetic friction between the cart and the box is mk = 0.6.

A 10 ft 4 ft

s C

B

SOLUTION Principle of Work and Energy: WA which acts in the direction of the vertical displacement does positive work when the block displaces 4 ft vertically. The friction force Ff = mk N = 0.6(30) = 18.0 lb does negative work since it acts in the opposite direction to that of displacement Since the block is at rest initially and is required to stop, TA = TC = 0. Applying Eq. 14–7, we have TA + a UA - C = TC

0 + 30(4) - 18.0s¿ = 0 Thus,

s¿ = 6.667 ft Ans.

s = 10 - s¿ = 3.33 ft

Ans: s = 3.33 ft 406

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14–31. Marbles having a mass of 5 g are dropped from rest at A through the smooth glass tube and accumulate in the can at C. Determine the placement R of the can from the end of the tube and the speed at which the marbles fall into the can. Neglect the size of the can.

A

B 3m 2m

SOLUTION

C

TA + © UA-B = TB 0 + [0.005(9.81)(3 - 2)] =

R

1 (0.005)v2B 2

vB = 4.429 m>s

A+TB

s = s0 + v0t + 2 = 0 + 0 =

1 a t2 2 c

1 (9.81)t2 2

t = 0.6386 s + b a:

s = s0 + v0 t Ans.

R = 0 + 4.429(0.6386) = 2.83 m TA + © UA-C = T1 0 + [0.005(9.81)(3) =

1 (0.005)v2C 2 Ans.

vC = 7.67 m>s

Ans: R = 2.83 m vC = 7.67 m>s 407

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*14–32. The block has a mass of 0.8 kg and moves within the smooth vertical slot. If it starts from rest when the attached spring is in the unstretched position at A, determine the constant vertical force F which must be applied to the cord so that the block attains a speed vB = 2.5 m>s when it reaches B; sB = 0.15 m. Neglect the size and mass of the pulley. Hint: The work of F can be determined by finding the difference ¢l in cord lengths AC and BC and using UF = F ¢l.

0.3 m C

0.4 m B

SOLUTION

sB

lAC = 2(0.3)2 + (0.4)2 = 0.5 m 2

A

2

lBC = 2(0.4 - 0.15) + (0.3) = 0.3905 m

F k

100 N/m

TA + ©UA - B = TB 1 1 0 + F(0.5 - 0.3905) - (100)(0.15)2 - (0.8)(9.81)(0.15) = (0.8)(2.5)2 2 2 Ans.

F = 43.9 N

Ans: F = 43.9 N 408

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14–33. The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at A. If the plane is smooth, determine the distance d, measured from the wall, to where the block strikes the ground. Neglect the size of the block.

B k

100 lb/ft

3 ft

A 4 ft

d

SOLUTION TA + ©UA - B = TB 0 +

1 1 10 (100)(2)2 - (10)(3) = = a b n2B 2 2 32.2

nB = 33.09 ft>s + B A:

s = s0 + n0 t 4 d = 0 + 33.09 a b t 5

A+cB

s = s0 + n0 t +

1 ac t2 2

1 3 -3 = 0 + (33.09) a b t + ( - 32.2)t2 5 2 16.1t2 - 19.853t - 3 = 0 Solving for the positive root, t = 1.369 s 4 d = 33.09a b (1.369) = 36.2 ft 5

Ans.

Ans: d = 36.2 ft 409

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14–34. The spring bumper is used to arrest the motion of the 4-lb block, which is sliding toward it at v = 9 ft>s. As shown, the spring is confined by the plate P and wall using cables so that its length is 1.5 ft. If the stiffness of the spring is k = 50 lb>ft, determine the required unstretched length of the spring so that the plate is not displaced more than 0.2 ft after the block collides into it. Neglect friction, the mass of the plate and spring, and the energy loss between the plate and block during the collision.

k

1.5 ft.

vA

P

5 ft.

A

SOLUTION T1 + ©U1 - 2 = T2 4 1 1 1 a b (9)2 - c (50)(s - 1.3)2 - (50)(s - 1.5)2 d = 0 2 32.2 2 2 0.20124 = s2 - 2.60 s + 1.69 - (s2 - 3.0 s + 2.25) 0.20124 = 0.4 s - 0.560 Ans.

s = 1.90 ft

Ans: s = 1.90 ft 410

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14–35. y

When the 150-lb skier is at point A he has a speed of 5 ft>s. Determine his speed when he reaches point B on the smooth slope. For this distance the slope follows the cosine curve shown. Also, what is the normal force on his skis at B and his rate of increase in speed? Neglect friction and air resistance.

A

B

y

50 cos (

p x) 100

SOLUTION y = 50 cos a

p = 22.70 ft bx ` 100 x = 35

x 35 ft

dy p p p p = tan u = -50 a b sin a bx = - a b sin a bx ` = - 1.3996 dx 100 100 2 100 x = 35 u = -54.45° d2y 2

dx

= -a

p2 p = - 0.02240 b cos a bx ` 200 100 x = 35 3

r =

dy 2 2 c1 + a b d dx

`

d2y dx

` 2

C 1 + ( - 1.3996)2 D 2 3

=

| - 0.02240|

= 227.179

TA + ©UA - B = TB 1 150 2 1 150 a b(5)2 + 150(50 - 22.70) = a b vB 2 32.2 2 32.2 Ans.

vB = 42.227 ft>s = 42.2 ft>s +b©Fn = man;

-N + 150 cos 54.45° = a

2

(42.227) 150 b¢ ≤ 32.2 227.179 Ans.

N = 50.6 lb +R©Ft = mat ;

150 sin 54.45° = a

150 ba 32.2 t

at = 26.2 ft>s2

Ans.

Ans: vB = 42.2 ft>s N = 50.6 lb at = 26.2 ft>s2 411

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*14–36. The spring has a stiffness k = 50 lb>ft and an unstretched length of 2 ft. As shown, it is confined by the plate and wall using cables so that its length is 1.5 ft. A 4-lb block is given a speed vA when it is at A, and it slides down the incline having a coefficient of kinetic friction mk = 0.2. If it strikes the plate and pushes it forward 0.25 ft before stopping, determine its speed at A. Neglect the mass of the plate and spring.

vA 5 4

k  50 lb/ft

3

A 3 ft

1.5 ft

SOLUTION +a©Fy = 0;

4 NB - 4a b = 0 5 NB = 3.20 lb

T1 + ©U1 - 2 = T2 4 3 1 1 1 a b n2A + (3 + 0.25) a b (4) - 0.2(3.20)(3 + 0.25) - c (50)(0.75)2— (50)(0.5)2 d = 0 2 32.2 5 2 2 Ans.

nA = 5.80 ft>s

Ans: vA = 5.80 ft>s 412

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14–37. If the track is to be designed so that the passengers of the roller coaster do not experience a normal force equal to zero or more than 4 times their weight, determine the limiting heights hA and hC so that this does not occur. The roller coaster starts from rest at position A. Neglect friction.

A C hC

rC

20 m rB

15 m

hA

B

SOLUTION Free-Body Diagram:The free-body diagram of the passenger at positions B and C are shown in Figs. a and b, respectively. Equations of Motion: Here, an =

v2 . The requirement at position B is that r

NB = 4mg. By referring to Fig. a, + c ©Fn = man;

4mg - mg = m ¢

vB 2 ≤ 15

vB 2 = 45g At position C, NC is required to be zero. By referring to Fig. b, + T ©Fn = man;

mg - 0 = m ¢

vC 2 ≤ 20

vC 2 = 20g Principle of Work and Energy: The normal reaction N does no work since it always acts perpendicular to the motion. When the rollercoaster moves from position A to B, W displaces vertically downward h = hA and does positive work. We have TA + ©UA-B = TB 0 + mghA =

1 m(45g) 2 Ans.

hA = 22.5 m

When the rollercoaster moves from position A to C, W displaces vertically downward h = hA - hC = (22.5 - hC) m. TA + ©UA-B = TB 0 + mg(22.5 - hC) =

1 m(20g) 2 Ans.

hC = 12.5 m

Ans: hA = 22.5 m hC = 12.5 m 413

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14–38. y

If the 60-kg skier passes point A with a speed of 5 m>s, determine his speed when he reaches point B. Also find the normal force exerted on him by the slope at this point. Neglect friction.

y

(0.025x2

5) m A

B

SOLUTION

15 m

Free-Body Diagram: The free-body diagram of the skier at an arbitrary position is shown in Fig. a.

x

Principle of Work and Energy: By referring to Fig. a, we notice that N does no work since it always acts perpendicular to the motion. When the skier slides down the track from A to B, W displaces vertically downward h = yA - yB = 15 - C 0.025 A 02 B + 5 D = 10 m and does positive work. TA + ©UA - B = TB 1 1 (60)(52) + C 60(9.81)(10) D = (60)vB 2 2 2 Ans.

vB = 14.87 m>s = 14.9 m>s dy>dx = 0.05x d2y>dx2 = 0.05 r =

[1 + 0]3>2 = 20 m 0.5

+ c ©Fn = man ;

N - 60(9.81) = 60 ¢

(14.87)2 ≤ 20 Ans.

N = 1.25 kN

Ans: vB = 14.9 m>s N = 1.25 kN 414

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14–39. If the 75-kg crate starts from rest at A, determine its speed when it reaches point B. The cable is subjected to a constant force of F = 300 N. Neglect friction and the size of the pulley.

C 30 F 6m

B

SOLUTION

A

Free-Body Diagram: The free-body diagram of the crate and cable system at an arbitrary position is shown in Fig. a.

6m

2m

Principle of Work and Energy: By referring to Fig. a, notice that N, W, and R do no work. When the crate moves from A to B, force F displaces through a distance of s = AC - BC = 282 + 62 - 222 + 62 = 3.675 m. Here, the work of F is positive. T1 + ©U1 - 2 = T2 0 + 300(3.675) =

1 (75)vB 2 2 Ans.

vB = 5.42 m>s

Ans: vB = 5.42 m>s 415

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*14–40. If the 75-kg crate starts from rest at A, and its speed is 6 m>s when it passes point B, determine the constant force F exerted on the cable. Neglect friction and the size of the pulley.

C 30 F 6m

B

SOLUTION

A

Free-Body Diagram: The free-body diagram of the crate and cable system at an arbitrary position is shown in Fig. a.

6m

2m

Principle of Work and Energy: By referring to Fig. a, notice that N, W, and R do no work. When the crate moves from A to B, force F displaces through a distance of s = AC - BC = 282 + 62 - 222 + 62 = 3.675 m. Here, the work of F is positive. T1 + ©U1 - 2 = T2 0 + F(3.675) =

1 (75)(62) 2 Ans.

F = 367 N

Ans: F = 367 N 416

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14–41. A 2-lb block rests on the smooth semicylindrical surface. An elastic cord having a stiffness k = 2 lb>ft is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A (u = 0°), determine the unstretched length of the cord so that the block begins to leave the semicylinder at the instant u = 45°. Neglect the size of the block.

k

2 lb/ft B 1.5 ft

C

u

A

SOLUTION +b©Fn = man;

2 sin 45° =

v2 2 a b 32.2 1.5

v = 5.844 ft>s T1 + © U1-2 = T2 0 +

2 2 3p 2 1 1 1 (2) C p(1.5) - l0 D - (2) c (1.5) - l0 d - 2(1.5 sin 45°) = a b(5.844)2 2 2 4 2 32.2

Ans.

l0 = 2.77 ft

Ans: l0 = 2.77 ft 417

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14–42. The jeep has a weight of 2500 lb and an engine which transmits a power of 100 hp to all the wheels. Assuming the wheels do not slip on the ground, determine the angle u of the largest incline the jeep can climb at a constant speed v = 30 ft>s.

θ

SOLUTION P = FJv 100(550) = 2500 sin u(30) Ans.

u = 47.2°

Ans: u = 47.2° 418

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14–43. Determine the power input for a motor necessary to lift 300 lb at a constant rate of 5 ft> s. The efficiency of the motor is P = 0.65.

SOLUTION Power: The power output can be obtained using Eq. 14–10. P = F # v = 300(5) = 1500 ft # lb>s Using Eq. 14–11, the required power input for the motor to provide the above power output is power input = =

power output P 1500 = 2307.7 ft # lb>s = 4.20 hp 0.65

Ans.

Ans: Pi = 4.20 hp 419

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*14–44. An automobile having a mass of 2 Mg travels up a 7° slope at a constant speed of v = 100 km>h. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the automobile has an efficiency P = 0.65.

7

SOLUTION Equation of Motion: The force F which is required to maintain the car’s constant speed up the slope must be determined first. + ©Fx¿ = max¿;

F - 2(103)(9.81) sin 7° = 2(103)(0) F = 2391.08 N

100(103) m 1h b = 27.78 m>s. R * a h 3600 s The power output can be obtained using Eq. 14–10. Power: Here, the speed of the car is y = B

P = F # v = 2391.08(27.78) = 66.418(103) W = 66.418 kW Using Eq. 14–11, the required power input from the engine to provide the above power output is power input = =

power output e 66.418 = 102 kW 0.65

Ans.

Ans: power input = 102 kW 420

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14–45. The Milkin Aircraft Co. manufactures a turbojet engine that is placed in a plane having a weight of 13000 lb. If the engine develops a constant thrust of 5200 lb, determine the power output of the plane when it is just ready to take off with a speed of 600 mi>h.

SOLUTION At 600 ms>h. P = 5200(600)a

88 ft>s 1 b = 8.32 (103) hp 60 m>h 550

Ans.

Ans: P = 8.32 ( 103 ) hp 421

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14–46. To dramatize the loss of energy in an automobile, consider a car having a weight of 5000 lb that is traveling at 35 mi>h. If the car is brought to a stop, determine how long a 100-W light bulb must burn to expend the same amount of energy. 11 mi = 5280 ft.2

SOLUTION Energy: Here, the speed of the car is y = a

5280 ft 1h 35 mi b * a b * a b = h 1 mi 3600 s

51.33 ft>s. Thus, the kinetic energy of the car is U =

1 2 1 5000 my = a b A 51.332 B = 204.59 A 103 B ft # lb 2 2 32.2

The power of the bulb is 73.73 ft # lb>s. Thus, t =

Pbulb = 100 W * a

550 ft # lb>s 1 hp b * a b = 746 W 1 hp

204.59(103) U = 2774.98 s = 46.2 min = Pbulb 73.73

Ans.

Ans: t = 46.2 min 422

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14–47. The escalator steps move with a constant speed of 0.6 m>s. If the steps are 125 mm high and 250 mm in length, determine the power of a motor needed to lift an average mass of 150 kg per step. There are 32 steps.

15 ft

SOLUTION Step height: 0.125 m The number of steps:

4 = 32 0.125

Total load: 32(150)(9.81) = 47 088 N If load is placed at the center height, h =

4 = 2 m, then 2

4 U = 47 088 a b = 94.18 kJ 2 vy = v sin u = 0.6 ¢

t =

4 2 ( 32(0.25) ) 2 + 42

≤ = 0.2683 m>s

h 2 = 7.454 s = vy 0.2683

P =

U 94.18 = = 12.6 kW t 7.454

Ans.

Also, P = F # v = 47 088(0.2683) = 12.6 kW

Ans.

Ans: P = 12.6 kW 423

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*14–48. The man having the weight of 150 lb is able to run up a 15-ft-high flight of stairs in 4 s. Determine the power generated. How long would a 100-W light bulb have to burn to expend the same amount of energy? Conclusion: Please turn off the lights when they are not in use!

15 ft

SOLUTION Power: The work done by the man is U = Wh = 150(15) = 2250 ft # lb Thus, the power generated by the man is given by Pman =

U 2250 = = 562.5 ft # lb>s = 1.02 hp t 4

The power of the bulb is Pbulb = 100 W * a = 73.73 ft # lb>s. Thus,

t =

1 hp 746 W

b * a

550 ft # lb>s 1 hp

2250 U = 30.5 s = Pbulb 73.73

Ans. b Ans.

Ans: Pman = 1.02 hp t = 30.5 s 424

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14–49. The 2-Mg car increases its speed uniformly from rest to 25 m>s in 30 s up the inclined road. Determine the maximum power that must be supplied by the engine, which operates with an efficiency of P = 0.8. Also, find the average power supplied by the engine.

1 10

SOLUTION Kinematics:The constant acceleration of the car can be determined from + B A:

v = v0 + ac t 25 = 0 + ac (30) ac = 0.8333 m>s2

Equations of Motion: By referring to the free-body diagram of the car shown in Fig. a, ©Fx¿ = max¿ ;

F - 2000(9.81) sin 5.711° = 2000(0.8333) F = 3618.93N

Power: The maximum power output of the motor can be determined from (Pout)max = F # vmax = 3618.93(25) = 90 473.24 W Thus, the maximum power input is given by Pin =

Pout 90473.24 = 113 091.55 W = 113 kW = e 0.8

Ans.

The average power output can be determined from (Pout)avg = F # vavg = 3618.93 a

25 b = 45 236.62 W 2

Thus, (Pin)avg =

(Pout)avg e

=

45236.62 = 56 545.78 W = 56.5 kW 0.8

Ans.

Ans: Pmax = 113 kW Pavg = 56.5 kW 425

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14–50. Determine the power output of the draw-works motor M necessary to lift the 600-lb drill pipe upward with a constant speed of 4 ft>s. The cable is tied to the top of the oil rig, wraps around the lower pulley, then around the top pulley, and then to the motor.

SOLUTION 2sP + sM = l

M

2nP = - nM 2( -4) = -nM

4 ft/s

nM = 8 ft>s Po = Fn = a

600 b(8) = 2400 ft # lb>s = 4.36 hp 2

Ans.

Ans: Po = 4.36 hp 426

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14–51. The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, determine the power that must be supplied to the motor at the instant the load has been hoisted s = 15 ft starting from rest. The motor has an efficiency of e = 0.65.

M

Solution Equation of Motion. Referring to the FBD of the elevator, Fig. a, + c ΣFy = may;   3(500) - 1000 =

1000 a 32.2

a = 16.1 ft>s2

When S = 15ft,

+ c v2 = v20 + 2ac(S - S0);  v2 = 02 + 2(16.1)(15) v = 21.98 ft>s



Power. Applying Eq. 14–9, the power output is Pout = F # V = 3(500)(21.98) = 32.97 ( 103 ) lb # ft>s The power input can be determined using Eq. 14–9 Σ =

32.97 ( 103 ) Pout ;  0.65 = Pin Pin

Pin = [50.72 ( 103 ) lb # ft>s] a = 92.21 hp = 92.2 hp

1 hp

550 lb # ft>s

b

Ans.

Ans: P = 92.2 hp 427

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*14–52. The 50-lb crate is given a speed of 10 ft>s in t = 4 s starting from rest. If the acceleration is constant, determine the power that must be supplied to the motor when t = 2 s. The motor has an efficiency e = 0.65. Neglect the mass of the pulley and cable.

Solution + c ΣFy = may;        2T - 50 =

( + c ) v = v0 + act

M

50 a 32.2

10 = 0 + a(4) s

a = 2.5 ft>s2 T = 26.94 lb In t = 2 s

( + c ) v = v0 + ac l v = 0 + 2.5(2) = 5 ft>s sC + (sC - sP) = l 2 vC = vP 2(5) = vP = 10 ft>s P0 = 26.94(10) = 269.4 P1 =

269.4 = 414.5 ft # lb>s 0.65 Ans.

P1 = 0.754 hp

Ans: P1 = 0.754 hp 428

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14–53. The sports car has a mass of 2.3 Mg, and while it is traveling at 28 m/s the driver causes it to accelerate at 5 m>s2. If the drag resistance on the car due to the wind is FD = 10.3v22 N, where v is the velocity in m/s, determine the power supplied to the engine at this instant. The engine has a running efficiency of P = 0.68.

FD

SOLUTION + ©F = m a ; : x x

F - 0.3v2 = 2.3(103)(5) F = 0.3v2 + 11.5(103)

At v = 28 m>s F = 11 735.2 N PO = (11 735.2)(28) = 328.59 kW Pi =

PO 328.59 = 438 kW = e 0.68

Ans.

Ans: Pi = 483 kW 429

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14–54. The sports car has a mass of 2.3 Mg and accelerates at 6 m>s2, starting from rest. If the drag resistance on the car due to the wind is FD = 110v2 N, where v is the velocity in m/s, determine the power supplied to the engine when t = 5 s. The engine has a running efficiency of P = 0.68.

FD

SOLUTION + ©F = m a ; : x x

F - 10v = 2.3(103)(6) F = 13.8(103) + 10 v

+ )v = v + a t (: 0 c v = 0 + 6(5) = 30 m>s PO = F # v = [13.8(103) + 10(30)](30) = 423.0 kW Pi =

PO 423.0 = 622 kW = e 0.68

Ans.

Ans: Pi = 622 kW 430

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14–55. The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of P = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of vE = 4 m>s.

M

C

SOLUTION

vE

Elevator: E

Since a = 0, + c ©Fy = 0;

60(9.81) + 3T - 400(9.81) = 0 T = 1111.8 N

2sE + (sE - sP) = l 3vE = vP Since vE = -4 m>s, Pi =

vP = - 12 m>s

(1111.8)(12) F # vP = = 22.2 kW e 0.6

Ans.

Ans: Pi = 22.2 kW 431

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*14–56. The 10-lb collar starts from rest at A and is lifted by applying a constant vertical force of F = 25 lb to the cord. If the rod is smooth, determine the power developed by the force at the instant u = 60°.

3 ft

4 ft

u F

SOLUTION

A

Work of F U1 - 2 = 25(5 - 3.464) = 38.40 lb # ft T1 + ©U1 - 2 = T2s 0 + 38.40 - 10(4 - 1.732) =

1 10 2 ( )v 2 32.2

v = 10.06 ft>s P = F # v = 25 cos 60°(10.06) = 125.76 ft # lb>s Ans.

P = 0.229 hp

Ans: P = 0.229 hp 432

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14–57. The 10-lb collar starts from rest at A and is lifted with a constant speed of 2 ft>s along the smooth rod. Determine the power developed by the force F at the instant shown.

3 ft

4 ft

u F

SOLUTION + c ©Fy = m ay ;

A

4 F a b - 10 = 0 5 F = 12.5 lb

4 P = F # v = 12.5 a b (2) = 20 lb # ft>s 5 Ans.

= 0.0364 hp

Ans: P = 0.0364 hp 433

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14–58. The 50-lb block rests on the rough surface for which the coefficient of kinetic friction is mk = 0.2. A force F = ( 40 + s2 ) lb, where s is in ft, acts on the block in the direction shown. If the spring is originally unstretched (s = 0) and the block is at rest, determine the power developed by the force the instant the block has moved s = 1.5 ft.

F 30°

k = 20 lb ft

SOLUTION + c ΣFy = 0;

NB - ( 40 + s2 ) sin 30° - 50 = 0 NB = 70 + 0.5s2

T1 + ΣU1 - 2 = T2 1.5

0 +

L0

(40 + s2 ) cos 30° ds -

1 (20)(1.5)2 - 0.2 2 L0

1.5

(70 + 0.5s2 )ds =

1 50 a bv22 2 32.2

0 + 52.936 - 22.5 - 21.1125 = 0.7764v22 v2 = 3.465 ft>s When s = 1.5 ft, F = 40 + (1.5)2 = 42.25 lb P = F # v = (42.25 cos 30°)(3.465) P = 126.79 ft # lb>s = 0.231 hp

Ans.

Ans: P = 0.231 hp 434

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14–59. The escalator steps move with a constant speed of 0.6 m>s. If the steps are 125 mm high and 250 mm in length, determine the power of a motor needed to lift an average mass of 150 kg per step. There are 32 steps.

250 mm 125 mm v

0.6 m/s

4m

SOLUTION Step height: 0.125 m The number of steps:

4 = 32 0.125

Total load: 32(150)(9.81) = 47 088 N If load is placed at the center height, h =

4 = 2 m, then 2

4 U = 47 088 a b = 94.18 kJ 2 ny = n sin u = 0.6 ¢

4 2(32(0.25))2 + 42

t =

h 2 = = 7.454 s ny 0.2683

P =

94.18 U = = 12.6 kW t 7.454

≤ = 0.2683 m>s

Ans.

Also, P = F # v = 47 088(0.2683) = 12.6 kW

Ans.

Ans: P = 12.6 kW 435

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*14–60. If the escalator in Prob. 14–47 is not moving, determine the constant speed at which a man having a mass of 80 kg must walk up the steps to generate 100 W of power—the same amount that is needed to power a standard light bulb.

250 mm 125 mm v

0.6 m/s

4m

SOLUTION P =

(80)(9.81)(4) U1 - 2 = = 100 t t

n =

2(32(0.25))2 + 42 s = = 0.285 m>s t 31.4

t = 31.4 s Ans.

Ans: v = 0.285 m>s 436

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14–61. If the jet on the dragster supplies a constant thrust of T = 20 kN, determine the power generated by the jet as a function of time. Neglect drag and rolling resistance, and the loss of fuel. The dragster has a mass of 1 Mg and starts from rest.

T

SOLUTION Equations of Motion: By referring to the free-body diagram of the dragster shown in Fig. a, + ©F = ma ; : x x

20(103) = 1000(a)

a = 20 m>s2

Kinematics: The velocity of the dragster can be determined from + b a:

v = v0 + ac t v = 0 + 20 t = (20 t) m>s

Power: P = F # v = 20(103)(20 t) = C 400(103)t D W

Ans.

Ans:

437

P = e 400(103)t f W

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14–62. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the power applied as a function of time and the work done in t = 0.3 s.

F (N)

800

0.2

0.3

t (s)

SOLUTION v (m/s)

For 0 … t … 0.2 F = 800 N v =

20

20 t = 66.67t 0.3

P = F # v = 53.3 t kW

0.3

Ans.

t (s)

For 0.2 … t … 0.3 F = 2400 - 8000 t v = 66.67t P = F # v = 1160 t - 533t22 kW

Ans.

0.3

U=

L0

P dt 0.2

U= =

L0

53.3t dt +

0.3

L0.2

1160t - 533t 2 dt 2

160 533 53.3 (0.2)2 + [(0.3)2 - (0.2)2] [(0.3)3 - (0.2)3] 2 2 3 Ans.

= 1.69 kJ

Ans: P = e 160 t - 533t 2 f kW U = 1.69 kJ

438

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14–63. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the maximum power developed during the 0.3-second time period.

F (N)

800

0.2

0.3

t (s)

SOLUTION v (m/s)

See solution to Prob. 14–62. P = 160 t - 533 t2

20

dP = 160 - 1066.6 t = 0 dt 0.3

t = 0.15 s 6 0.2 s

t (s)

Thus maximum occurs at t = 0.2 s Ans.

Pmax = 53.3(0.2) = 10.7 kW

Ans: Pmax = 10.7 kW 439

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*14–64. The block has a weight of 80 lb and rests on the floor for which mk = 0.4. If the motor draws in the cable at a constant rate of 6 ft>s, determine the output of the motor at the instant u = 30°. Neglect the mass of the cable and pulleys.

u

Solution (1)

Time derivative of Eq. (1) yields: # 2SBsB # # # + sP = 0   Where sB = vB and sP = vP 2 2sB + 0 2s2B + 9

3 ft

6 ft/s

2 a2s2B + 32 b + SP = 1

2sBvB

3 ft

u

+ vP = 0   vB =

(2)

2s2B + 9 vp  2sB

(3)

Time derivative of Eq. (2) yields: 1

(

$ $ [2 ( s2B + 9 ) s2B - 2s2Bs2B + 2sB ( s2B + 9 ) sB] + sB = 0

+ 9) $ $ where sp = aP = 0 and sB = aB s2B

3/2

2 ( s2B + 9 ) v2B - 2s2BvB2 + 2sB ( s2B + 9 ) aB = 0 vB =

s2Bv2B - v2B ( s2B ) + 9 sB ( s2B + 9 )

At u = 30°,  sB =

(4)



3 = 5.196 ft tan 30°

From Eq. (3)   vB = -

From Eq. (4)   aB =

25.1962 + 9 (6) = -3.464 ft>s 2(5.196)

( - 3.4642 )( 5.1962 + 9 ) = -0.5773 ft>s2 5.196 ( 5.1962 + 9 )

5.1962( -3.464)2 -

+ S ΣFx = ma;  p - 0.4(80) =

80 ( - 0.5773)   p = 30.57 lb 32.2

F0 = u # v = 30.57(3.464) = 105.9 ft # lb>s = 0.193 hp

Ans.

Also, + S ΣFx = 0    - F + 2T cos 30° = 0

        T =

30.57 = 17.65 lb 2 cos 30°

F0 = T # vp = 17.65(6) = 105.9 ft

#

Ans.

lb>s = 0.193 hp

Ans: F0 = 0.193 hp F0 = 0.193 hp 440

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14–65. The block has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. If a force F = 160t22 N, where t is in seconds, is applied to the cable, determine the power developed by the force when t = 5 s. Hint: First determine the time needed for the force to cause motion.

F

SOLUTION + ©F = 0; : x

2F - 0.5(150)(9.81) = 0 F = 367.875 = 60t2 t = 2.476 s

+ ©F = m a ; : x x

2(60t2) -0.4(150)(9.81) = 150ap ap = 0.8t2 - 3.924

dv = a dt L0

5

v

dv =

v = a

L2.476

A 0.8t2 - 3.924 B dt

5 0.8 3 b t - 3.924t ` = 19.38 m>s 3 2.476

sP + (sP - sF) = l 2vP = vF vF = 2(19.38) = 38.76 m>s F = 60(5)2 = 1500 N P = F # v = 1500(38.76) = 58.1 kW

Ans.

Ans: P = 58.1 kW 441

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14–66. The girl has a mass of 40 kg and center of mass at G. If she is swinging to a maximum height defined by u = 60°, determine the force developed along each of the four supporting posts such as AB at the instant u = 0°. The swing is centrally located between the posts.

A 2m

 G

SOLUTION

30

30

The maximum tension in the cable occurs when u = 0°. B

T1 + V1 = T2 + V2 0 + 40(9.81)(- 2 cos 60°) =

1 (40)v2 + 40(9.81)(- 2) 2

v = 4.429 m>s 4.4292 b 2

+ c ©Fn = ma n;

T - 40(9.81) = (40)a

+ c ©Fy = 0;

2(2F) cos 30° - 784.8 = 0

T = 784.8 N F = 227 N

Ans.

Ans: F = 227 N 442

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14–67. The 30-lb block A is placed on top of two nested springs B and C and then pushed down to the position shown. If it is then released, determine the maximum height h to which it will rise.

A

h

SOLUTION

A

4 in.

kB  200 lb/in.

Conservation of Energy:

B

T1 + V1 = T2 + V2 1 1 mv1 + B a Vg b + A Ve B 1 R = mv2 + B aVg b + A Ve B 2 R 2 2 1 2 0 + 0 +

6 in.

C kC  100 lb/in.

1 1 (200)(4)2 + (100)(6)2 = 0 + h(30) + 0 2 2 Ans.

h = 113 in.

Ans: h = 133 in. 443

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*14–68. The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels down along the smooth guide. Determine the speed of the collar when it reaches point B, which is located just before the end of the curved portion of the rod. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod.

200 mm A

Solution Potential Energy. With reference to the datum set through B the gravitational potential energies of the collar at A and B are

200 mm k  50 N/m

B

(Vg)A = mghA = 5(9.81)(0.2) = 9.81 J (Vg)B = 0 At A and B, the spring stretches xA = 20.22 + 0.22 - 0.1 = 0.1828 m and xB = 0.4 - 0.1 = 0.3 m respectively. Thus, the elastic potential energies in the spring at A and B are (Ve)A =

1 2 1 kx = (50) ( 0.18282 ) = 0.8358 J 2 A 2

(Ve)B =

1 2 1 kxB = (50) ( 0.32 ) = 2.25 J 2 2

Conservation of Energy. TA + VA = TB + VB 1 1 (5) ( 52 ) + 9.81 + 0.8358 = (5)v2B + 0 + 2.25 2 2 Ans.

vB = 5.325 m>s = 5.33 m>s

Equation of Motion. At B, Fsp = kxB = 50(0.3) = 15 N. Referring to the FBD of the collar, Fig. a, ΣFn = man;

  

N + 15 = 5 a

5.3252 b 0.2

Ans.

       N = 693.95 N = 694 N

Ans: vB = 5.33 m>s N = 694 N 444

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14–69. The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its speed when its center reaches point B and the normal force it exerts on the rod at this point. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod.

200 mm A

Solution Potential Energy. With reference to the datum set through B the gravitational potential energies of the collar at A and B are

200 mm k  50 N/m

B

(Vg)A = mghA = 5(9.81)(0.2) = 9.81 J (Vg)B = 0 At A and B, the spring stretches xA = 20.22 + 0.22 - 0.1 = 0.1828 m and xB = 0.4 - 0.1 = 0.3 m respectively. Thus, the elastic potential energies in the spring at A and B are (Ve)A =

1 2 1 kx = (50) ( 0.18282 ) = 0.8358 J 2 A 2

(Ve)B =

1 2 1 kx = (50) ( 0.32 ) = 2.25 J 2 B 2

Conservation of Energy. TA + VA = TB + VB 1 1 (5) ( 52 ) + 9.81 + 0.8358 = (5)v2B + 0 + 2.25 2 2 Ans.

vB = 5.325 m>s = 5.33 m>s

Equation of Motion. At B, Fsp = kxB = 50(0.3) = 15 N. Referring to the FBD of the collar, Fig. a, ΣFn = man;

  

N + 15 = 5 a

5.3252 b 0.2

Ans.

       N = 693.95 N = 694 N

Ans: N = 694 N 445

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14–70. The ball has a weight of 15 lb and is fixed to a rod having a negligible mass. If it is released from rest when u = 0°, determine the angle u at which the compressive force in the rod becomes zero.

θ

3 ft

SOLUTION T1 + V1 = T2 + V2 0 + 0 =

1 15 bv2 - 15(3)(1 - cos u) a 2 32.2

v2 = 193.2(1 - cos u) + bΣFn = man;

15 cos u =

15 193.2(1 - cos u) d c 32.2 3

cos u = 2 - 2 cos u 2 u = cos-1 a b 3

Ans.

u = 48.2°

Ans: u = 48.2° 446

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14–71. The car C and its contents have a weight of 600 lb, whereas block B has a weight of 200 lb. If the car is released from rest, determine its speed when it travels 30 ft down the 20 incline. Suggestion: To measure the gravitational potential energy, establish separate datums at the initial elevations of B and C.

C

30 ft

20°

v

B

SOLUTION 2sB + sC = l 2¢sB = - ¢sC ¢sB = -

30 = - 15 ft 2

2vB = - vC Establish two datums at the initial elevations of the car and the block, respectively. T1 + V1 = T2 + V2 0 + 0 =

- vC 2 1 200 1 600 a b (vC)2 + a ba b + 200(15) - 600 sin 20°(30) 2 32.2 2 32.2 2

vC = 17.7 ft s

Ans.

Ans: vC = 17.7 ft>s 447

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*14–72. The roller coaster car has a mass of 700 kg, including its passenger. If it starts from the top of the hill A with a speed vA = 3 m>s, determine the minimum height h of the hill crest so that the car travels around the inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take rB = 7.5 m and rC = 5 m.

A B C

h

15 m

10 m

Solution Equation of Motion. Referring to the FBD of the roller-coaster car shown in Fig. a, ΣFn = man;   N + 700(9.81) = 700 a

v2 b r

(1)

When the roller-coaster car is about to leave the loop at B and C, N = 0. At B and C, rB = 7.5 m and rC = 5 m. Then Eq. (1) gives

and

0 + 700(9.81) = 700 a 0 + 700(9.81) = 700 a

v2B b   v2B = 73.575 m2 >s2 7.5 v2C b   v2C = 49.05 m2 >s2 5

Judging from the above results, the coster car will not leave the loop at C if it safely passes through B. Thus Ans.

NB = 0

Conservation of Energy. The datum will be set at the ground level. With v2B = 73.575 m2 >s2, TA + VA = TB + VB

1 1 (700) ( 32 ) + 700(9.81)h = (700)(73.575) + 700(9.81)(15) 2 2 Ans.

      h = 18.29 m = 18.3 m And from B to C, TB + VB = TC + VC 1 1 (700)(73.575) + 700(9.81)(15) = (700)v2c + 700(9.81)(10) 2 2 v2c = 171.675 m2 >s2 7 49.05 m2 >s2 

(O.K!)

Substitute this result into Eq. 1 with rC = 5 m, Nc + 700(9.81) = 700 a



171.675 b 5

Nc = 17.17 ( 103 ) N = 17.2 kN

Ans.

Ans: NB = 0 h = 18.3 m Nc = 17.2 kN 448

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14–73. The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, determine the minimum height h of the hill crest so that the car travels around both inside the loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take rB = 7.5 m and rC = 5 m.

A B C

h

15 m

10 m

Solution Equation of Motion. Referring to the FBD of the roller-coaster car shown in Fig. a, ΣFn = man;    N + 700(9.81) = 700 a

v2 b r

(1)

When the roller-coaster car is about to leave the loop at B and C, N = 0. At B and C, rB = 7.5 m and rC = 5 m. Then Eq. (1) gives 0 + 700(9.81) = 700 a

v2B b    v2B = 73.575 m2 >s2 7.5

0 + 700(9.81) = 700 a

v2C b    v2C = 49.05 m2 >s2 5

and

Judging from the above result the coaster car will not leave the loop at C provided it passes through B safely. Thus Ans.

NB = 0



Conservation of Energy. The datum will be set at the ground level. Applying Eq. 14– from A to B with v2B = 73.575 m2 >s2, TA + VA = TB + VB

0 + 700(9.81)h =

1 (700)(73.575) + 700(9.81)(15) 2 Ans.

h = 18.75 m And from B to C, TB + VB = TC + VC 1 1 (700)(73.575) + 700(9.81)(15) = (700)v2C + 700(9.81)(10) 2 2

v2C = 171.675 m2 >s2 > 49.05 m2 >s2

(O.K!)

Substitute this result into Eq. 1 with rC = 5 m,



NC + 700(9.81) = 700 a

171.675 b 5

Nc = 17.17 ( 103 ) N = 17.2 kN

Ans.

Ans: NB = 0 h = 18.75 m NC = 17.2 kN 449

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14–74. The assembly consists of two blocks A and B which have a mass of 20 kg and 30 kg, respectively. Determine the speed of each block when B descends 1.5 m. The blocks are released from rest. Neglect the mass of the pulleys and cords.

Solution 3sA + sB = l 3∆sA = - ∆sB

A

3vA = -vB

B

T1 + V1 = T2 + V2 (0 + 0) + (0 + 0) = vA = 1.54 m>s

1 1 1.5 (20)(vA)2 + (30)( -3vA)2 + 20(9.81)a b - 30(9.81)(1.5) 2 2 3 Ans.

Ans.

vB = 4.62 m>s

Ans: vA = 1.54 m>s vB = 4.62 m>s 450

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14–75. The assembly consists of two blocks A and B, which have a mass of 20 kg and 30 kg, respectively. Determine the distance B must descend in order for A to achieve a speed of 3 m>s starting from rest.

Solution 3sA + sB = l 3∆sA = - ∆sB

A

B

3vA = - vB vB = -9 m>s T1 + V1 = T2 + V2 (0 + 0) + (0 + 0) = sB = 5.70 m

sB 1 1 (20)(3)2 + (30)( - 9)2 + 20(9.81)a b - 30(9.81)(sB) 2 2 3

Ans.

Ans: sB = 5.70 m 451

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*14–76. z

The spring has a stiffness k = 50 N>m and an unstretched length of 0.3 m. If it is attached to the 2-kg smooth collar and the collar is released from rest at A (u = 0°), determine the speed of the collar when u = 60°. The motion occurs in the horizontal plane. Neglect the size of the collar.

A 2m u

Solution

y k  50 N/m

Potential Energy. Since the motion occurs in the horizontal plane, there will be no change in gravitational potential energy when u = 0°, the spring stretches x1 = 4 - 0.3 = 3.7 m. Referring to the geometry shown in Fig. a, the spring stretches x2 = 4 cos 60° - 0.3 = 1.7 m. Thus, the elastic potential energies in the spring when u = 0° and 60° are (Ve)1 =

1 2 1 kx = (50)(3.72) = 342.25 J 2 1 2

(Ve)2 =

1 2 1 kx = (50)(1.72) = 72.25 J 2 2 2

x

Conservation of Energy. Since the collar is released from rest when u = 0°, T1 = 0. T1 + V1 = T2 + V2 0 + 342.25 =

1 (2)v2 + 72.25 2 Ans.

v = 16.43 m>s = 16.4 m>s

Ans: v = 16.4 m>s 452

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14–77. A

The roller coaster car having a mass m is released from rest at point A. If the trackBis to be designed so that the car does 7.5 m not leave it at B, determine the required height h. Also, find the speed of the car when it reaches point C. Neglect h 20 m friction.

A B 7.5 m 20 m

C

h

C

SOLUTION

car is about to leave Equation of Motion: Since it is required that the roller coaster car is about to leave ng to the free-body vB 2 vB 2 = the track at B, NB = 0. Here, an = . By referring to the free-body rB 7.5 diagram of the roller coaster car shown in Fig. a,

s2

©Fn = ma n;

m(9.81) = m ¢

vB 2 ≤ 7.5

vB 2 = 73.575 m2>s2

b, the gravitational Energy: With reference to the datum set in Fig. b, the gravitational A, B, and CPotential are potential energy of the rollercoaster car at positions A, B, and C are 9.81)(20) = 196.2 m, A Vg B A = mghA = m(9.81)h = 9.81mh, A Vg B B = mghB = m(9.81)(20) = 196.2 m, and A Vg B C = mghC = m(9.81)(0) = 0.

ng the motion of the Conservation of Energy: Using the result of vB 2 and considering the motion of the car from position A to B, TA + VA = TB + VB 1 1 mvA 2 + A Vg B A = mvB 2 + A Vg B B 2 2 0 + 9.81mh = Ans.

1 m(73.575) + 196.2m 2 Ans.

h = 23.75 m

Also, considering the motion of the car from position B to C, TB + VB = TC + VC 1 1 mvB 2 + A Vg B B = mvC 2 + A Vg B C 2 2 1 1 m(73.575) + 196.2m = mvC 2 + 0 2 2 Ans.

Ans.

vC = 21.6 m>s

Ans: h = 23.75 m vC = 21.6 m>s 453

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14–78. The spring has a stiffness k = 200 N>m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and the collar is released from rest at A, determine the speed of the collar when it reaches B. Neglect the size of the collar.

A

k  200 N/m 2m

Solution Potential Energy. With reference to the datum set through B, the gravitational potential energies of the collar at A and B are

B

(Vg)A = mghA = 3(9.81)(2) = 58.86 J (Vg)B = 0

1.5 m 2

2

At A and B, the spring stretches xA = 21.5 + 2 - 0.5 = 2.00 m and xB = 1.5 - 0.5 = 1.00 m. Thus, the elastic potential energies in the spring when the collar is at A and B are (Ve)A =

1 2 1 kx = (200) ( 2.002 ) = 400 J 2 A 2

(Ve)B =

1 2 1 kx = (200) ( 1.002 ) = 100 J 2 B 2

Conservation of Energy. Since the collar is released from rest at A, TA = 0. TA + VA = TB + VB 0 + 58.86 + 400 =

1 (3)v2B + 0 + 100 2 Ans.

vB = 15.47 m>s = 15.5 m>s

Ans: vB = 15.5 m>s 454

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14–79. A 2-lb block rests on the smooth semicylindrical surface at A. An elastic cord having a stiffness of k = 2 lb>ft is attached to the block at B and to the base of the semicylinder at C. If the block is released from rest at A, determine the longest unstretched length of the cord so the block begins to leave the cylinder at the instant u = 45°. Neglect the size of the block.

k = 2 lb ft B 1.5 ft C

θ

A

SOLUTION Equation of Motion: It is required that N = 0. Applying Eq. 13–8, we have ΣFn = man;

2 cos 45° =

2 v2 a b 32.2 1.5

v2 = 34.15 m2 >s2

Potential Energy: Datum is set at the base of cylinder. When the block moves to a position 1.5 sin 45° = 1.061 ft above the datum, its gravitational potential energy at this position is 2(1.061) = 2.121 ft # lb. The initial and final elastic potential energy 1 1 are (2) [p (1.5) - l] 2 and (2) [0.75p(1.5) - l]2, respectively. 2 2 Conservation of Energy: ΣT1 + ΣV1 = ΣT2 + ΣV2 0 +

1 1 2 1 b(34.15) + 2.121 + (2)[0.75p(1.5) - l]2 (2) [p(1.5) - l]2 = a 2 2 32.2 2

Ans.

l = 2.77 ft

Ans: l = 2.77 ft 455

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*14–80. When s = 0, the spring on the firing mechanism is unstretched. If the arm is pulled back such that s = 100 mm and released, determine the speed of the 0.3-kg ball and the normal reaction of the circular track on the ball when u = 60°. Assume all surfaces of contact to be smooth. Neglect the mass of the spring and the size of the ball.

Solution

u

Potential Energy. With reference to the datum set through the center of the circular track, the gravitational potential energies of the ball when u = 0° and u = 60° are (Vg)1 = - mgh1 = - 0.3(9.81)(1.5) = - 4.4145 J

s

1.5 m

k  1500 N/m

(Vg)2 = - mgh2 = - 0.3(9.81)(1.5 cos 60°) = - 2.20725 J When u = 0°, the spring compress x1 = 0.1 m and is unstretched when u = 60°. Thus, the elastic potential energies in the spring when u = 0° and 60° are (Ve)1 =

1 2 1 kx = (1500) ( 0.12 ) = 7.50 J 2 1 2

(Ve)2 = 0 Conservation of Energy. Since the ball starts from rest, T1 = 0. T1 + V1 = T2 + V2 0 + ( - 4.4145) + 7.50 =

1 (0.3)v2 + ( - 2.20725) + 0 2

v2 = 35.285 m2 >s2

Ans.

v = 5.94 m>s

Equation of Motion. Referring to the FBD of the ball, Fig. a, ΣFn = man;    N - 0.3(9.81) cos 60° = 0.3 a

35.285 b 1.5

  N = 8.5285 N = 8.53 N

Ans.

Ans: v = 5.94 m>s N = 8.53 N 456

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14–81. When s = 0, the spring on the firing mechanism is unstretched. If the arm is pulled back such that s = 100 mm and released, determine the maximum angle u the ball will travel without leaving the circular track. Assume all surfaces of contact to be smooth. Neglect the mass of the spring and the size of the ball.

Solution

u

Equation of Motion. It is required that the ball leaves the track, and this will occur provided u 7 90°. When this happens, N = 0. Referring to the FBD of the ball, Fig. a ΣFn = man;

0.3(9.81) sin (u - 90°) = 0.3a

v2 b 1.5

s

1.5 m

k  1500 N/m

v2 = 14.715 sin (u - 90°)

(1)

Potential Energy. With reference to the datum set through the center of the circular track Fig. b, the gravitational potential Energies of the ball when u = 0° and u are (Vg)1 = - mgh1 = - 0.3(9.81)(1.5) = -4.4145 J (Vg)2 = mgh2 = 0.3(9.81)[1.5 sin (u - 90°)] = 4.4145 sin (u - 90°) When u = 0°, the spring compresses x1 = 0.1 m and is unstretched when the ball is at u for max height. Thus, the elastic potential energies in the spring when u = 0° and u are (Ve)1 =

1 2 1 kx = (1500) ( 0.12 ) = 7.50 J 2 1 2

(Ve)2 = 0 Conservation of Energy. Since the ball starts from rest, T1 = 0. T1 + V1 = T2 + V2 0 + ( - 4.4145) + 7.50 =

1 (0.3)v2 + 4.4145 sin (u - 90°) + 0 2

v2 = 20.57 - 29.43 sin (u - 90°)

(2)

Equating Eqs. (1) and (2), 14.715 sin (u - 90°) = 20.57 - 29.43 sin (u - 90°) sin (u - 90°) = 0.4660 u - 90° = 27.77° Ans.

u = 117.77° = 118°

Ans: u = 118° 457

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14–82. If the mass of the earth is Me, show that the gravitational potential energy of a body of mass m located a distance r from the center of the earth is Vg = −GMem>r. Recall that the gravitational force acting between the earth and the body is F = G(Mem>r 2), Eq. 13–1. For the calculation, locate the datum at r : q. Also, prove that F is a conservative force.

r2

r

SOLUTION r1

The work is computed by moving F from position r1 to a farther position r2. Vg = - U = -

L

F dr r2

= -G Me m

dr 2 Lr1 r

= - G Me m a

1 1 - b r2 r1

As r1 : q , let r2 = r1, F2 = F1, then Vg :

-G Me m r

To be conservative, require F = - § Vg = =

G Me m 0 ab r 0r

-G Me m

Q.E.D.

r2

Ans: F = 458

- G Me m r2

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14–83. A rocket of mass m is fired vertically from the surface of the earth, i.e., at r = r1 . Assuming no mass is lost as it travels upward, determine the work it must do against gravity to reach a distance r2 . The force of gravity is F = GMem>r 2 (Eq. 13–1), where Me is the mass of the earth and r the distance between the rocket and the center of the earth.

r2

r

SOLUTION F = G F1-2 =

r1

Mem r2 L

r2

F dr = GMem

= GMema

dr

Lr1 r

2

1 1 - b r1 r2

Ans.

Ans: F = GMem a 459

1 1 - b r1 r2

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*14–84. The 4-kg smooth collar has a speed of 3 m>s when it is at s = 0. Determine the maximum distance s it travels before it stops momentarily. The spring has an unstretched length of 1 m.

1.5 m A

3 m/s

s

k  100 N/m

Solution Potential Energy. With reference to the datum set through A the gravitational potential energies of the collar at A and B are

B

(Vg)A = 0    (Vg)B = - mghB = - 4(9.81) Smax = -39.24 Smax At A and B, the spring stretches xA = 1.5 - 1 = 0.5 m and xB = 2S2max + 1.52 - 1. Thus, the elastic potential Energies in the spring when the collar is at A and B are (Ve)A = (Ve)B =

1 2 1 kx = (100) ( 0.52 ) = 12.5 J 2 A 2

1 2 1 kxB = (100) ( 2S2max + 1.52 - 1 ) 2 = 50 ( S2max - 22S2max + 1.52 + 3.25 ) 2 2

Conservation of Energy. Since the collar is required to stop momentarily at B, TB = 0. TA + VA = TB + VB 1 (4) ( 32 ) + 0 + 12.5 = 0 + ( - 39.24 Smax) + 50 ( S2max - 22S2max + 1.52 + 3.25 ) 2 50 S2max - 1002S2max + 1.52 - 39.24 Smax + 132 = 0 Solving numerically, Ans.

Smax = 1.9554 m = 1.96 m

Ans: Smax = 1.96 m 460

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14–85. A 60-kg satellite travels in free flight along an elliptical orbit such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h. What is the speed of the satellite when it reaches point B, where rB = 80 Mm? Hint: See Prob. 14–82, where Me = 5.976(1024) kg and G = 66.73(10-12) m3>(kg # s2).

B vB

80 Mm

rB

rA

20 Mm

vA A

SOLUTION yA = 40 Mm>h = 11 111.1 m>s Since V = -

GMe m r

T1 + V1 = T2 + V2 66.73(10) - 12(5.976)(10)23(60) 66.73(10) - 12(5.976)(10)24(60) 1 1 (60)(11 111.1)2 = (60)v2B 6 2 2 20(10) 80(10)6 Ans.

vB = 9672 m>s = 34.8 Mm>h

Ans: vB = 34.8 Mm>h 461

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14–86. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed vB when he reaches B. Also, compute the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass of 70 kg.

A

vB

50 m B

4m

Solution s

TA + VA = TB + VB

C 30

1 0 + 70(9.81) (46) = (70)v2 + 0 2 Ans.

v = 30.04 m>s = 30.0 m>s ( + T) sy = (sy)0 + (v0)yt +

1 2 at 2 c

4 + s sin 30° = 0 + 0 +

1 (9.81)t 2 2

(1)

+ )s = v t (d x x

(2)

s cos 30° = 30.04t

Ans.

s = 130 m t = 3.75 s

Ans: s = 130 m 462

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14–87. The block has a mass of 20 kg and is released from rest when s 0.5 m. If the mass of the bumpers A and B can be neglected, determine the maximum deformation of each spring due to the collision. kA = 500 N/m

s = 0.5 m

SOLUTION Datum at initial position:

A B kB = 800 N/m

T1 + V1 = T2 + V2 0 + 0 = 0 +

1 1 (500)s2A + (800)s2B + 20(9.81) C - (sA + s B) - 0.5 D 2 2

Also, Fs = 500sA = 800sB

(1) (2)

sA = 1.6sB

Solving Eqs. (1) and (2) yields: sB = 0.638 m

Ans.

sA = 1.02 m

Ans.

Ans: sB = 0.638 m sA = 1.02 m 463

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*14–88. The 2-lb collar has a speed of 5 ft>s at A. The attached spring has an unstretched length of 2 ft and a stiffness of k = 10 lb>ft. If the collar moves over the smooth rod, determine its speed when it reaches point B, the normal force of the rod on the collar, and the rate of decrease in its speed.

y A y  4.5  12 x2

4.5 ft

SOLUTION

k  10 lb/ft

Datum at B: TA + VA = TB + VB

B

2 2 1 1 1 1 a b (5)2 + (10)(4.5 - 2)2 + 2(4.5) = a b (nB)2 + (10)(3 - 2)2 + 0 2 32.2 2 2 32.2 2 Ans.

nB = 34.060 ft>s = 34.1 ft>s y = 4.5 -

3 ft

1 2 x 2

dy = tanu = - x ` = -3 dx x=3 d2y

u = -71.57°

r =

c1 + a

`

3

dy 2 2 b d dx

d2y dx2

= -1

dx2

`

+b©Fn = man ;

3

=

[1 + ( -3)2]2 = 31.623 ft | - 1|

- N + 10 cos 18.43° + 2 cos 71.57° = a

(34.060)2 2 ba b 32.2 31.623 Ans.

N = 7.84 lb + R©Ft = mat ;

2 sin 71.57° - 10 sin 18.43° = a

2 ba 32.2 t

at = -20.4 ft>s2

Ans.

Ans: vB = 34.1 ft>s N = 7.84 lb at = - 20.4 ft>s2 464

x

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14–89. When the 6-kg box reaches point A it has a speed of vA = 2 m>s. Determine the angle u at which it leaves the smooth circular ramp and the distance s to where it falls into the cart. Neglect friction.

vA = 2 m/s

20° A

θ

B

1.2 m

SOLUTION s

At point B: +b©Fn = man;

6(9.81) cos f = 6a

n2B b 1.2

(1)

Datum at bottom of curve: TA + VA = TB + VB 1 1 (6)(2)2 + 6(9.81)(1.2 cos 20°) = (6)(vB)2 + 6(9.81)(1.2 cos f) 2 2 13.062 = 0.5v2B + 11.772 cos f

(2)

Substitute Eq. (1) into Eq. (2), and solving for vB, vB = 2.951 m>s Thus,

f = cos - 1 a

(2.951)2 b = 42.29° 1.2(9.81) Ans.

u = f - 20° = 22.3°

A+cB

s = s0 + v0 t + 12 ac t2 - 1.2 cos 42.29° = 0 - 2.951(sin 42.29°)t +

1 ( -9.81)t2 2

4.905t2 + 1.9857t - 0.8877 = 0 Solving for the positive root: t = 0.2687 s + b a:

s = s0 + v0 t s = 0 + (2.951 cos 42.29°)(0.2687) Ans.

s = 0.587 m

Ans: u = 22.3° s = 0.587 m 465

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14–90. y

When the 5-kg box reaches point A it has a speed vA = 10 m>s. Determine the normal force the box exerts on the surface when it reaches point B. Neglect friction and the size of the box.

yx

9m x 1/ 2  y1/ 2  3

B

Solution

A

Conservation of Energy. At point B, y = x 1

x

9m

1

x2 + x2 = 3 x =

9 m 4

9 m. With reference to the datum set to coincide with the x axis, the 4 gravitational potential energies of the box at points A and B are Then y =

9 (Vg)A = 0    (Vg)B = mghB = 5(9.81)a b = 110.3625 J 4 Applying the energy equation, TA + VA = TB + VB 1 1 (5) ( 102 ) + 0 = (5)v2B + 110.3625 2 2 v2B = 55.855 m2 >s2

dy 1 1 1 = 2 ( 3 - x2 ) a - x - 2 b dx 2 3 9 = . At point B, x = m. Thus, 3 4 2x2 1

Equation of Motion. Here, y = ( 3 - x2 ) 2. Then, 1

=

x2 - 3 x

1 2

= 1 -

3 x

1 2

and

tan uB = d 2y dx

2

`

9 x=4

m

d 2y dx2

=

3 -3 x 2 2

dy ` = 1 dx x = 94 m =

3 3

2 ( 94 ) 2

3 1

( )2 9 4

= - 1  uB = -45° = 45°

= 0.4444

The radius of curvature at B is 3

PB =

[1 + (dy>dx)2]2

0 d 2y>dx2 0

3

=

[1 + ( - 1)2]2 0.4444

= 6.3640 m

Referring to the FBD of the box, Fig. a ΣFn = man;   N - 5(9.81) cos 45° = 5a

55.855 b 6.3640

Ans.

N = 78.57 N = 78.6 N

Ans: N = 78.6 N 466

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14–91. y

When the 5-kg box reaches point A it has a speed vA = 10 m>s. Determine how high the box reaches up the surface before it comes to a stop. Also, what is the resultant normal force on the surface at this point and the acceleration? Neglect friction and the size of the box.

yx

9m x 1/ 2  y1/ 2  3

B

Solution

A

Conservation of Energy. With reference to the datum set coincide with x axis, the gravitational potential energy of the box at A and C (at maximum height) are

x

9m

(Vg)A = 0    (Vg)C = mghc = 5(9.81)(y) = 49.05y

It is required that the box stop at C. Thus, Tc = 0 TA + VA = TC + VC 1 (5) ( 102 ) + 0 = 0 + 49.05y 2 y = 5.0968 m = 5.10 m

Ans.

Then, 1

1

x 2 + 5.0968 2 = 3

x = 0.5511 m 1

Equation of Motion. Here, y = ( 3 - x2 ) 2. Then, 1

=

x2 - 3 x

1 2

3

= 1 -

x

1 2

and

d 2y dx2

=

dy 1 1 1 = 2 ( 3 - x2 ) a - x2 b dx 2

3 -3 3 x2 = At point C, x = 0.5511 m. 3 2 2x 2

Thus

tan uc =

dy 3 ` = 1 = - 3.0410   uC = -71.80° = 71.80° 1 dx x = 0.5511 m 0.55112

Referring to the FBD of the box, Fig. a, ΣFn = man ;

N - 5(9.81) cos 71.80° = 5 a

02 b rC

Ans.

N = 15.32 N = 15.3 N

ΣFt = mat ;

- 5(9.81) sin 71.80° = 5at at = - 9.3191 m>s2 = 9.32 m>s2 R

Since an = 0, Then a = at = 9.32 m>s2 R

Ans.

Ans: y = 5.10 m N = 15.3 N a = 9.32 m>s2 R 467

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*14–92. y

The roller-coaster car has a speed of 15 ft>s when it is at the crest of a vertical parabolic track. Determine the car’s velocity and the normal force it exerts on the track when it reaches point B. Neglect friction and the mass of the wheels. The total weight of the car and the passengers is 350 lb.

15 ft/s

vA

A y

1 200 (40 000

x2)

200 ft

SOLUTION y =

1 (40 000 - x2) 200

dy 1 = x` = - 2, dx 100 x = 200 d2y dx2

= -

B

O 200 ft

u = tan - 1( - 2) = - 63.43°

1 100

Datum at A: TA + VA = TB + VB 1 350 1 350 a b (15)2 + 0 = a b (nB)2 - 350(200) 2 32.2 2 32.2 Ans.

nB = 114.48 = 114 ft>s

r =

c1 + a

`

dy 2 b d dx

d 2y dx

` 2

+b©Fn = man;

3 2 3

=

[1 + (- 2)2]2

`-

1 ` 100

= 1118.0 ft

350 cos 63.43° - NB = a

350 (114.48)2 b 32.2 1118.0 Ans.

NB = 29.1 lb

Ans: vB = 114 ft>s NB = 29.1 lb 468

x

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14–93. The 10-kg sphere C is released from rest when u = 0° and the tension in the spring is 100 N. Determine the speed of the sphere at the instant u = 90°. Neglect the mass of rod AB and the size of the sphere.

E

0.4 m

SOLUTION

k

500 N/m

A

Potential Energy: With reference to the datum set in Fig. a, the gravitational potential

u

44.145 J and A Vg B 2 = mgh2 = 10(9.81)(0) = 0. When the sphere is at position (1), 100 = 0.2 m. Thus, the unstretched length of the spring is the spring stretches s1 = 500

0.3 m

energy of the sphere at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0.45) =

D

B C

0.15 m

l0 = 30.32 + 0.42 - 0.2 = 0.3 m, and the elastic potential energy of the spring is 1 1 A Ve B 1 = ks12 = (500)(0.22) = 10 J. When the sphere is at position (2), the spring 2 2 stretches s2 = 0.7 - 0.3 = 0.4 m, and the elastic potential energy of the spring is 1 1 A Ve B 2 = ks22 = (500)(0.42) = 40 J. 2 2 Conservation of Energy: T1 + V1 = T2 + V2 1 1 m A v B 2 + c A Vg B 1 + A Ve B 1 d = ms A vs B 2 2 + c A Vg B 2 + A Ve B 2 d 2 s s 1 2 0 + (44.145 + 10) =

1 (10)(vs)22 + (0 + 40) 2 Ans.

(vs)2 = 1.68 m>s

Ans: v = 1.68 m>s 469

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14–94. A quarter-circular tube AB of mean radius r contains a smooth chain that has a mass per unit length of m0. If the chain is released from rest from the position shown, determine its speed when it emerges completely from the tube.

A

O r B

SOLUTION Potential Energy: The location of the center of gravity G of the chain at positions (1) and (2) are shown in Fig. a. The mass of the chain is p p 2r p - 2 m = m 0 a r b = m 0r . Thus, the center of mass is at h1 = r = a br . p p 2 2 With reference to the datum set in Fig. a the gravitational potential energy of the chain at positions (1) and (2) are

A Vg B 1 = mgh1 = a m 0rg b a p 2

p - 2 p - 2 br = a b m 0r2g p 2

and

A Vg B 2 = mgh 2 = 0 Conservation of Energy: T1 + V1 = T2 + V2 1 1 mv1 2 + A Vg B 1 = mv2 2 + A Vg B 2 2 2 0+ a

p - 2 1 p b m 0r2g = a m 0r b v 2 2 + 0 2 2 2

v2 =

2 (p - 2)gr Ap

Ans.

Ans: v2 = 470

2 (p - 2)gr Ap

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14–95. The cylinder has a mass of 20 kg and is released from rest when h = 0. Determine its speed when h = 3 m. Each spring has a stiffness k = 40 N>m and an unstretched length of 2 m.

2m

k

2m

k

h

Solution T1 + V1 = T2 + V2 1 1 0 + 0 = 0 + 2c (40) ( 232 + 22 - 22 ) d - 20(9.81)(3) + (20)v2 2 2 v = 6.97 m>s

Ans.

Ans: v = 6.97 m>s 471

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*14–96. If the 20-kg cylinder is released from rest at h = 0, determine the required stiffness k of each spring so that its motion is arrested or stops when h = 0.5 m. Each spring has an unstretched length of 1 m.

2m

k

2m

k

h

Solution T1 + V1 = T2 + V2 1 1 0 + 2c k(2 - 1)2 d = 0 - 20(9.81)(0.5) + 2c k ( 2(2)2 + (0.5)2 - 1 ) 2 d 2 2 k = -98.1 + 1.12689 k

Ans.

k = 773 N>m

Ans: k = 773 N>m 472

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14–97. A pan of negligible mass is attached to two identical springs of stiffness k = 250 N>m. If a 10-kg box is dropped from a height of 0.5 m above the pan, determine the maximum vertical displacement d. Initially each spring has a tension of 50 N.

1m

k

250 N/m

1m

k

250 N/m

0.5 m d

SOLUTION Potential Energy: With reference to the datum set in Fig. a, the gravitational potential energy of the box at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0) = 0 and

A Vg B 2 = mgh2 = 10(9.81) C - A 0.5 + d B D = - 98.1 A 0.5 + d B . Initially, the spring

50 = 0.2 m. Thus, the unstretched length of the spring 250 is l0 = 1 - 0.2 = 0.8 m and the initial elastic potential of each spring is 1 A Ve B 1 = (2) ks1 2 = 2(250 > 2)(0.22) = 10 J . When the box is at position (2), the 2

stretches s1 =

spring stretches s2 = a 2d2 + 12 - 0.8 b m. The elastic potential energy of the springs when the box is at this position is

A Ve B 2 = (2) ks2 2 = 2(250 > 2) c 2d2 + 1 - 0.8 d = 250a d2 - 1.62d2 + 1 + 1.64b . 2

1 2

Conservation of Energy: T1 + V1 + T2 + V2 1 1 mv1 2 + B a Vg b + A Ve B 1 R = mv2 2 + B aVg b + A Ve B 2 R 2 2 1 2 0 + A 0 + 10 B = 0 + B - 98.1 A 0.5 + d B + 250 ¢ d2 - 1.6 2d2 + 1 + 1.64 ≤ R 250d2 - 98.1d - 400 2d2 + 1 + 350.95 = 0 Solving the above equation by trial and error, Ans.

d = 1.34 m

Ans: d = 1.34 m 473

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15–1. A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a distance of 12 m away. Determine the impulse of his foot on the ball at A. Neglect the impulse caused by the ball’s weight while it’s being kicked.

v

60

Solution

A

Kinematics. Consider the vertical motion of the ball where (s0)y = sy = 0, (v0)y = v sin 60° c and ay = 9.81 m>s2 T , 1 2

1 2

( + c )   sy = (s0)y + (v0)yt + ayt2 ;  0 = 0 + v sin 60°t + ( - 9.81)t 2 t(v sin 60° - 4.905t) = 0 Since t ≠ 0, then v sin 60° - 4.905t = 0 (1)

t = 0.1766 v Then, consider the horizontal motion where (v0)x = v cos 60°, and (s0)x = 0, + )   s = (s ) + (v ) t ;  12 = 0 + v cos 60°t (S x 0 x 0 x t =

24  v

(2)

Equating Eqs. (1) and (2) 0.1766 v =

24 v

v = 11.66 m>s Principle of Impulse and Momentum. ( + Q)  mv1 + Σ

Lt1

t2

Fdt = mv2

0 + I = 0.15 (11.66)

I = 1.749 N # s = 1.75 N # s

Ans.

Ans: v = 1.75 N # s 474

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15–2. A 20-lb block slides down a 30° inclined plane with an initial velocity of 2 ft>s. Determine the velocity of the block in 3 s if the coefficient of kinetic friction between the block and the plane is mk = 0.25.

SOLUTION A +a B

m1vvœ2 + ©

t2

Lt1

Fyœdt = m1vyœ22

0 + N(3) - 20 cos 30°(3) = 0

A +b B

N = 17.32 lb

t2

m1vxœ21 + ©

Lt1

Fxœdt = m1vxœ22

20 20 (2) + 20 sin 30°(3) - 0.25(17.32)(3) = v 32.2 32.2 Ans.

v = 29.4 ft>s

Ans: v = 29.4 ft>s 475

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15–3. The uniform beam has a weight of 5000 lb. Determine the average tension in each of the two cables AB and AC if the beam is given an upward speed of 8 ft>s in 1.5 s starting from rest. Neglect the mass of the cables.

P A

4 ft

Solution 25a

B

1000 b = 6.944 m>s 3600

C

3 ft

3 ft

System:

+ )   mv + Σ (S 1

L

F dt = mv2

[0 + 0] + F(35) = (50 + 75) ( 103 ) (6.944) Ans.

F = 24.8 kN Barge: + )   mv + Σ (S 1

L

F dt = mv2

0 + T(35) = (75) ( 103 ) (6.944) Ans.

T = 14.881 = 14.9 kN Also, using this result for T, Tugboat: + )   mv + Σ (S 1

L

F dt = mv2

0 + F(35) - (14.881)(35) = (50) ( 103 ) (6.944) Ans.

F = 24.8 kN

Ans: F = 24.8 kN 476

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*15–4. Each of the cables can sustain a maximum tension of 5000 lb. If the uniform beam has a weight of 5000 lb, determine the shortest time possible to lift the beam with a speed of 10 ft>s starting from rest.

P A

4 ft

Solution

B

4 + c ΣFy = 0 ;  Pmax - 2a b(5000) = 0 5

C

3 ft

3 ft

Pmax = 8000 lb

( + c )   mv1 + Σ

L

F dt = mv2

0 + 8000(t) - 5000(t) =

5000 (10) 32.2 Ans.

t = 0.518 s

Ans: t = 0.518 s 477

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15–5. A hockey puck is traveling to the left with a velocity of v1 = 10 m>s when it is struck by a hockey stick and given a velocity of v2 = 20 m>s as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of 0.2 kg. v2  20 m/s 40 v1  10 m/s

Solution v1 = { -10i} m>s v2 = {20 cos 40°i + 20 sin 40°j} m>s I = m∆v = (0. 2) {[20 cos 40° - ( -10)]i + 20 sin 40°j} = {5.0642i + 2.5712j} kg # m>s I = 2(5.0642)2 + (2.5712)2 = 5.6795 = 5.68 kg # m>s

Ans.

Ans: I = 5.68 N # s 478

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15–6. A train consists of a 50-Mg engine and three cars, each having a mass of 30 Mg. If it takes 80 s for the train to increase its speed uniformly to 40 km>h, starting from rest, determine the force T developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resultant frictional tractive force F which gives the train forward motion, whereas the car wheels roll freely. Also, determine F acting on the engine wheels.

v A

E F

SOLUTION (yx)2 = 40 km>h = 11.11 m>s Entire train: + b a:

m(vx)1 + ©

L

Fx dt = m(vx)2

0 + F(80) = [50 + 3(30)] A 103 B (11.11) Ans.

F = 19.4 kN Three cars: + b a:

m(vx)1 + ©

L

Fx dt = m(vx)2

0 + T(80) = 3(30) A 103 B (11.11)

T = 12.5 kN

Ans.

Ans: F = 19.4 kN T = 12.5 kN 479

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15–7. Crates A and B weigh 100 lb and 50 lb, respectively. If they start from rest, determine their speed when t = 5 s. Also, find the force exerted by crate A on crate B during the motion. The coefficient of kinetic friction between the crates and the ground is mk = 0.25.

A P  50 lb

B

SOLUTION Free-Body Diagram: The free-body diagram of crates A and B are shown in Figs. a and b, respectively. The frictional force acting on each crate is (Ff)A = mkNA = 0.25NA and (Ff)B = mkNB = 0.25NB. Principle of Impulse and Momentum: Referring to Fig. a, (+ c)

t2

m(v1)y + ©

Lt1

Fy dt = m(v2)y

100 100 (0) + NA(5) - 100(5) = (0) 32.2 32.2 NA = 100 lb + ) (:

m(v1)x + ©

t2

Fx dt = m(v2)x

Lt1

100 100 (0) + 50(5) - 0.25(100)(5) - FAB (5) = v 32.2 32.2 (1)

v = 40.25 - 1.61FAB By considering Fig. b, (+ c)

t2

m(v1)y + ©

Lt1

Fy dt = m(v2)y

50 50 (0) + NB(5) - 50(5) = (0) 32.2 32.2 NB = 50 lb + ) (:

m(v1)x + ©

t2

Lt1

Fx dt = m(v2)x

50 50 (0) + FAB(5) - 0.25(50)(5) = v 32.2 32.2 (2)

v = 3.22 FAB - 40.25 Solving Eqs. (1) and (2) yields FAB = 16.67 lb = 16.7 lb

v = 13.42 ft>s = 13.4 ft>s

Ans.

Ans: FAB = 16.7 lb v = 13.4 ft>s 480

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*15–8. The automobile has a weight of 2700 lb and is traveling forward at 4 ft>s when it crashes into the wall. If the impact occurs in 0.06 s, determine the average impulsive force acting on the car. Assume the brakes are not applied. If the coefficient of kinetic friction between the wheels and the pavement is mk = 0.3, calculate the impulsive force on the wall if the brakes were applied during the crash.The brakes are applied to all four wheels so that all the wheels slip.

Solution Impulse is area under curve for hole cavity. I =

L

F dt = 4(3) +

1 1 (8 + 4)(6 - 3) + (8)(10 - 6) 2 2

= 46 lb # s

Ans.

For starred cavity: I =

L

F dt = 6(8) +

1 (6)(10 - 8) 2

= 54 lb # s

Ans.

Ans: I = 46 lb ~ s I = 54 lb ~ s 481

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15–9. The 200-kg crate rests on the ground for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the crate when t = 4 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the crate.

T (N) 800 T  400 t1/2 4

t (s) A T

Solution Equilibrium. The time required to move the crate can be determined by considering the equilibrium of the crate. Since the crate is required to be on the verge of sliding, Ff = msN = 0.5 N. Referring to the FBD of the crate, Fig. a, + c ΣFy = 0 ;  N - 200(9.81) = 0  N = 1962 N + ΣF = 0 ;  2 ( 400t 12 ) - 0.5(1962) = 0  t = 1.5037 s S x Principle of Impulse and Momentum. Since the crate is sliding, Ff = mkN = 0.4(1962) = 784.8 N. Referring to the FBD of the crate, Fig. a + )   m(v ) + Σ (S x 1

0 + 2

4s

t2

Lt1

Fx dt = m(vx)2

400t 2 dt - 784.8(4 - 1.5037) = 200v L1.5037 s 1

Ans.

v = 6.621 m>s = 6.62 m>s

Ans: v = 6.62 m>s 482

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15–10. P

The 50-kg crate is pulled by the constant force P. If the crate starts from rest and achieves a speed of 10 m>s in 5 s, determine the magnitude of P. The coefficient of kinetic friction between the crate and the ground is mk = 0.2.

30

SOLUTION Impulse and Momentum Diagram: The frictional force acting on the crate is Ff = mkN = 0.2N. Principle of Impulse and Momentum: (+ c)

t2

m(v1)y + ©

Fy dt = m(v2)y Lt1 0 + N(5) + P(5) sin 30° - 50(9.81)(5) = 0 (1)

N = 490.5 - 0.5P + ) (:

t2

m(v1)x + ©

Fx dt = m(v2)x Lt1 50(0) + P(5) cos 30° - 0.2N(5) = 50(10) (2)

4.3301P - N = 500 Solving Eqs. (1) and (2), yields N = 387.97 N

Ans.

P = 205 N

Ans: P = 205 N 483

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15–11. During operation the jack hammer strikes the concrete surface with a force which is indicated in the graph. To achieve this the 2-kg spike S is fired into the surface at 90 m>s. Determine the speed of the spike just after rebounding.

F (kN) 1500

S

Solution

0

Principle of Impulse and Momentum. The impulse of the force F is equal to the area under the F–t graph. Referring to the FBD of the spike, Fig. a

( + c )   m(vy)1 + Σ 2( - 90) +

Lt1

0

0.1

0.4

t (ms)

t2

Fy dt = m(vy)2

1 3 0.4 ( 10-3 ) 4 3 1500 ( 103 ) 4 = 2v 2 v = 60.0 m>s c 

Ans.

Ans: v = 60.0 m>s 484

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*15–12. For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is FD = (600t2) N, where t is in seconds. If the van has a speed of 20 km>h when t = 0, determine its speed when t = 5 s.

FD

SOLUTION Principle of Impulse and Momentum: The initial speed of the van is v1 = c20(103) c

m d h

1h d = 5.556 m>s. Referring to the free-body diagram of the van shown in Fig. a, 3600 s

+ ) (:

t2

m(v1)x + ©

Lt1

Fx dt = m(v2)x 5s

2500(5.556) + v2 = 15.6 m>s

L0

2

600 t dt = 2500 v2 Ans.

Ans: v2 = 15.6 m>s 485

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15–13. The 2.5-Mg van is traveling with a speed of 100 km>h when the brakes are applied and all four wheels lock. If the speed decreases to 40 km>h in 5 s, determine the coefficient of kinetic friction between the tires and the road.

SOLUTION Free-Body Diagram: The free-body diagram of the van is shown in Fig. a. The frictional force is Ff = mkN since all the wheels of the van are locked and will cause the van to slide. Principle of Impulse and Momentum: The initial and final speeds of the van are 1h 1h m m v1 = c100(103) d c d = 27.78 m>s and v2 = c40(103) d c d = 11.11 m>s. h 3600 s h 3600 s Referring to Fig. a, (+ c)

t2

m(v1)y + ©

Fy dt = m(v2)y Lt1 2500(0) + N(5) - 2500(9.81)(5) = 2500(0) N = 24 525 N

+ ) (;

t2

m(v1)x + ©

Fx dt = m(v2)x Lt1 2500(27.78) + [ -mk(24525)(5)] = 2500(11.1) Ans.

mk = 0.340

Ans: mk = 0.340 486

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15–14. A tankcar has a mass of 20 Mg and is freely rolling to the right with a speed of 0.75 m>s. If it strikes the barrier, determine the horizontal impulse needed to stop the car if the spring in the bumper B has a stiffness (a) k S ∞ (bumper is rigid), and (b) k = 15 kN>m.

v  0.75 m/s

k B

Solution +

a) b)   ( S )  mv1 + Σ

L

F dt = mv2

  20 ( 103 ) (0.75)  

L

L

F dt = 0

F dt = 15 kN # s

Ans.

The impulse is the same for both cases. For the spring having a stiffness k = 15 kN>m, the impulse is applied over a longer period of time than for k S ∞ .

Ans: I = 15 kN # s in both cases. 487

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15–15. The motor,  M, pulls on the cable with a force F = (10t 2 + 300) N, where t is in seconds. If the 100 kg crate is originally at rest at t = 0, determine its speed when t = 4 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate.

M

Solution Principle of Impulse and Momentum. The crate will only move when 3 ( 10t 2 + 300 ) = 100(9.81). Thus, this instant is t = 1.6432 s. Referring to the FBD of the crate, Fig. a,

( + c )   m(vy)1 + Σ 4s

0 +

Lt1

L1.6432 s

3a

t2

Fy dt = m(vy)2

3 ( 10t 2 + 300 ) dt - 100(9.81)(4 - 1.6432) = 100v

4s 10t 3 + 300tb ` - 2312.05 = 100v 3 1.6432 s

v = 4.047 m>s = 4.05 m>s c 

Ans.

Ans: v = 4.05 m>s 488

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*15–16. The choice of a seating material for moving vehicles depends upon its ability to resist shock and vibration. From the data shown in the graphs, determine the impulses created by a falling weight onto a sample of urethane foam and CONFOR foam.

F (N) 1.2 urethane 0.8

0.5 0.4 0.3

Solution CONFOR foam: Ic =

CONFOR

1 1 1 F dt = c (2)(0.5) + (0.5 + 0.8)(7 - 2) + (0.8)(14 - 7) d ( 10-3 ) 2 2 2 L

= 6.55 N # ms

t (ms) 2

4

7

10

14

Ans.

Urethane foam: Iv =



1 1 1 F dt = c (4)(0.3) + (1.2 + 0.3)(7 - 4) + (1.2 + 0.4)(10 - 7) + 2 2 2 L 1 (14 - 10)(0.4) d ( 10-3 ) 2

= 6.05 N # ms

Ans.

Ans: Ic = 6.55 N # ms Iv = 6.05 N # ms 489

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15–17. The towing force acting on the 400-kg safe varies as shown on the graph. Determine its speed, starting from rest, when t = 8 s. How far has it traveled during this time?

F

F (N) 750

Solution Principle of Impulse and Momentum. The FBD of the safe is shown in Fig. a. 600 For 0 … t 6 5 s, F = t = 120t. 5 + )   m(v ) + Σ (S x 1

0 +

Lt1

L0

t2

Fx dt = m(vx)2 5

t

120t dt = 400v v = 50.15t2 6 m>s

At t = 5 s,

v = 0.15 ( 52 ) = 3.75 m>s For 5 s 6 t … 8 s,

F - 600 750 - 600 = , F = 50t + 350. Here, t - 5 8 - 5

(vx)1 = 3.75 m>s and t 1 = 5 s. + )   m(v ) + Σ (S x 1

Lt1

t2

Fx dt = m(vx)2

t

400(3.75) +

At t = 8 s,

600

(50t + 350) dt = 400v L5 s

v = 50.0625t2 + 0.875t - 2.1875 6 m>s v = 0.0625 ( 82 ) + 0.875(8) - 2.1875 = 8.8125 m>s = 8.81 m>s Ans.

490

8

t (s)

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15–17. Continued

Kinematics. The displacement of the safe can be determined by integrating ds = v dt. For 0 … t 6 5 s, the initial condition is s = 0 at t = 0. L0

s

ds =

L0

t

0.15t 2 dt

s = 50.05t 3 6 m

At t = 5 s,

s = 0.05 ( 53 ) = 6.25 m For 5 s 6 t … 8 s, the initial condition is s = 6.25 m at t = 5 s. s

L6.25 m

t

ds =

L5 s

( 0.0625t 2 + 0.875 t - 2.1875 ) dt

s - 6.25 = ( 0.02083t 3 + 0.4375t 2 - 2.1875t ) ` At t = 8 s,

t 5s

s = 50.02083t 3 + 0.4375t 2 - 2.1875t + 3.64586 m

s = 0.02083 ( 83 ) + 0.4375 ( 82 ) - 2.1875(8) + 3.6458 Ans.

= 24.8125 m = 24.8 m

Ans: v = 8.81 m>s s = 24.8 m 491

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15–18. The motor exerts a force F on the 40-kg crate as shown in the graph. Determine the speed of the crate when t = 3 s and when t = 6 s. When t = 0, the crate is moving downward at 10 m>s.

A F

F (N) 450

B

150 6

t (s)

Solution Principle of Impulse and Momentum. The impulse of force F is equal to the area F - 150 450 - 150 = under the F–t graph. At t = 3 s, F = 300 N. Referring to the 3 - 0 6 - 0 FBD of the crate, Fig. a

( + c )   m(vy)1 + Σ

Lt1

t2

Fy dt = m(vy)2

1 40( - 10) + 2c (150 + 300)(3) d - 40(9.81)(3) = 40v 2

Ans.

v = - 5.68 m>s = 5.68 m>s T 

At t = 6 s,

( + c )   m(vy)1 + Σ

Lt1

t2

Fy dt = m(vy)2

1 40( - 10) + 2c (450 + 150)(6) d - 40(9.81)(6) = 40v 2 v = 21.14 m>s = 21.1 m>s c 

Ans.

Ans: v 0 t = 3 s = 5.68 m>s T 492

v 0 t = 6 s = 21.1 m>s c

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15–19. The 30-kg slider block is moving to the left with a speed of 5  m>s when it is acted upon by the forces F1 and F2. If these loadings vary in the manner shown on the graph, determine the speed of the block at t = 6 s. Neglect friction and the mass of the pulleys and cords.

F2

F1

F (N)

Solution Principle of Impulse and Momentum. The impulses produced by F1 and F2 are equal to the area under the respective F–t graph. Referring to the FBD of the block Fig. a,

(

+ )   m(v ) + Σ S x 1

Lt1

t2

Fx dx = m(vx)2

30 20 F1

10 0

1 - 30(5) + 4c 10(2) + (10 + 30)(4 - 2) + 30(6 - 4) d 2 + c - 40(4) -

F2

40

2

4

6

t (s)

1 (10 + 40)(6 - 4) d = 30v 2

v = 4.00 m>s S 

Ans.

Ans: v = 4.00 m>s 493

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*15–20. The 200-lb cabinet is subjected to the force F = 20(t + 1) lb where t is in seconds. If the cabinet is initially moving to the left with a velocity of 20 ft>s, determine its speed when t = 5 s. Neglect the size of the rollers.

F 30

Solution Principle of Impulse and Momentum. Referring to the FBD of the cabinet, Fig. a + )   m(v ) + Σ (S x 1

Lt1

t2

Fx dt = m(vx)2 5s

200 200 (t + 1) dt = ( - 20) + 20 cos 30° v 32.2 32.2 L0 v = 28.80 ft>s = 28.8 ft>s S 

Ans.

Ans: v = 28.8 ft>s S 494

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15–21. If it takes 35 s for the 50-Mg tugboat to increase its speed uniformly to 25 km>h, starting from rest, determine the force of the rope on the tugboat.The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine F acting on the tugboat. The barge has a mass of 75 Mg.

F

SOLUTION 1000 25a b = 6.944 m/s 3600 System: + 2 1:

my1 + ©

L

F dt = my2

[0 + 0] + F(35) = (50 + 75)(103)(6.944) Ans.

F = 24.8 kN Barge: + ) (:

mv1 + ©

L

F dt = mv2

0 + T(35) = (75)(103)(6.944) Ans.

T = 14.881 = 14.9 kN Also, using this result for T, Tugboat: + 2 1:

mv1 + ©

L

F dt = mv2

0 + F1352 - 114.88121352 = 15021103216.9442 Ans.

F = 24.8 kN

Ans: T = 14.9 kN F = 24.8 kN 495

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15–22. The thrust on the 4-Mg rocket sled is shown in the graph. Determine the sleds maximum velocity and the distance the sled travels when t = 35 s. Neglect friction.

T T (kN) 20 T  4 t1/2

25

35

t (s)

Solution Principle of Impulse And Momentum. The FBD of the rocket sled is shown in Fig. a. For 0 … t 6 25 s, + )   m(v ) + Σ (S x 1 0 +

L0

Lt1

t2

Fx dt = m(vx)2

t

1

4 ( 103 ) t 2dt = 4 ( 103 ) v

2 3 t 4 ( 103 ) a t 2 b ` = 4 ( 103 ) v 3 0 v = e

At t = 25 s, v = For 25 s 6 t 6 35 s,

2 3 t 2 f m>s 3

3 2 (25) 2 = 83.33 m>s 3

20 ( 103 ) - 0 T - 0 = or T = 2 ( 103 ) (35 - t). t - 35 25 - 35

Here, (vx)1 = 83.33 m>s and t 1 = 25 s. + )   m(v ) + Σ (S x 1

Lt1

t2

Fx dt = m(vx)2

4 ( 103 ) (83.33) +

t

2 ( 103 ) (35 - t)dt = 4 ( 103 ) v L25 s

v = 5 - 0.25t2 + 17.5t - 197.91676 m>s

The maximum velocity occurs at t = 35 s, Thus,

vmax = - 0.25 ( 352 ) + 17.5(35) - 197.9167 Ans.

= 108.33 m>s = 108 m>s

Ans: vmax = 108 m>s 496

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15–22. Continued

Kinematics. The displacement of the sled can be determined by integrating ds = vdt. For 0 … t 6 25 s, the initial condition is s = 0 at t = 0. L0

s

t

ds = s

2 2 5 t a bt 2 ` 3 5 0

s` = 0

At t = 25 s,

2 3 t 2 dt L0 3

s = e

4 5 t2f m 15

5 4 (25)2 = 833.33 m 15

s =

For 25 6 t … 35 s, the initial condition is s = 833.33 at t = 25 s. S

t

L833.33 m s2

At t = 35 s,

ds =

( -0.25t 2 + 17.5t - 197.9167) dt L25 s

=

( - 0.08333t 3 + 8.75t 2 - 197.9167t ) 2

S 833.33 m

t 25 s

s = 5 - 0.08333t 3 + 8.75t 2 - 197.9167t + 1614.586 m s = - 0.08333 ( 353 ) + 8.75 ( 352 ) - 197.9167(35) + 1614.58 Ans.

= 1833.33 m = 1833 m

Ans: s = 1.83 km 497

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15–23. The motor pulls on the cable at A with a force F = ( 30 + t 2 ) lb, where t is in seconds. If the 34-lb crate is originally on the ground at t = 0, determine its speed in t = 4 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. A

Solution 30 + t 2 = 34 t = 2 s for crate to start moving

(+c)

mv1 + Σ 0 +

L2

c 30t +

L 4

Fdt = mv2

( 30 + t 2 ) dt - 34(4 - 2) =

34 v 32.2 2

1 3 4 34 t d - 68 = v2 3 2 32.2

Ans.

v2 = 10.1 ft>s

Ans: v = 10.1 ft>s 498

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*15–24. The motor pulls on the cable at A with a force F = (e 2t) lb, where t is in seconds. If the 34-lb crate is originally at rest on the ground at t = 0, determine the crate’s velocity when t = 2 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate. A

Solution F = e 2t = 34 t = 1.7632 s for crate to start moving

(+c)

mv1 + Σ 0#

L

Fdt = mv2

2

L1.7632

e 2tdt - 34(2 - 1.7632) =

34 v 32.2 2

2

1 2t e - 8.0519 = 1.0559 v2 2 L1.7632 Ans.

v2 = 2.13 m>s

Ans: v2 = 2.13 m>s 499

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15–25. The balloon has a total mass of 400 kg including the passengers and ballast. The balloon is rising at a constant velocity of 18 km>h when h = 10 m. If the man drops the 40-kg sand bag, determine the velocity of the balloon when the bag strikes the ground. Neglect air resistance.

vA  18 km/h

A h

Solution Kinematic. When the sand bag is dropped, it will have an upward velocity of km 1000 m 1h v0 = a18 ba ba b = 5 m>s c . When the sand bag strikes the h 1 km 3600 s ground s = 10 m T . The time taken for the sand bag to strike the ground can be determined from 1 2 at ; 2 c

( + c )   s = s0 + v0t +

-10 = 0 + 5t +

1 ( - 9.81t 2 ) 2

4.905t 2 - 5t - 10 = 0 Solve for the positive root, t = 2.0258 s Principle of Impulse and Momentum. The FBD of the ballon when the ballon is rising with the constant velocity of 5 m>s is shown in Fig. a

( + c ) m(vy)1 + Σ

Lt1

t2

Fydt = m(vy)2

400(5) + T(t) - 400(9.81)t = 400(5) T = 3924 N When the sand bag is dropped, the thrust T = 3924 N is still maintained as shown in the FBD, Fig. b.

( + c ) m(vy)1 + Σ

Lt1

t2

Fydt = m(vy)2

360(5) + 3924(2.0258) - 360(9.81)(2.0258) = 360v v = 7.208 m>s = 7.21 m>s c 

Ans.

Ans: v = 7.21 m>s c 500

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15–26. As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which slides along the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block in 4 s if it has an initial speed of 5 m>s measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x¿ axis that moves at a constant velocity of 2 m>s relative to A.

A

x

B

x¿ 2 m/s

5 m/s 6N

SOLUTION Observer A: + 2 1:

m v1 + a

L

F dt = m v2

10(5) + 6(4) = 10v Ans.

v = 7.40 m>s Observer B: + 2 (:

m v1 + a

L

F dt = m v2

10(3) + 6(4) = 10v Ans.

v = 5.40 m>s

Ans: Observer A: v = 7.40 m>s Observer B: v = 5.40 m>s 501

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15–27. The 20-kg crate is lifted by a force of F = (100 + 5t 2) N, where t is in seconds. Determine the speed of the crate when t = 3 s, starting from rest.

F B

Solution Principle of Impulse and Momentum. At t = 0, F = 100 N. Since at this instant, 2F = 200 N 7 W = 20(9.81) = 196.2 N, the crate will move the instant force F is applied. Referring to the FBD of the crate, Fig. a,

(+c)

m(vy)1 + Σ 0 + 2

L0

Lt1 3s

A

t2

Fydt = m(vy)2

( 100 + 5t 2 ) dt - 20(9.81)(3) = 20v

2 a100t +

5 3 3s t b ` - 588.6 = 20v 3 0

Ans.

v = 5.07 m>s

Ans: v = 5.07 m>s 502

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*15–28. The 20-kg crate is lifted by a force of F = ( 100 + 5t 2 ) N, where t is in seconds. Determine how high the crate has moved upward when t = 3 s, starting from rest.

F B

Solution Principle of Impulse and Momentum. At t = 0, F = 100 N. Since at this instant, 2F = 200 N 7 W = 20(9.81) = 196.2 N, the crate will move the instant force F is applied. Referring to the FBD of the crate, Fig. a

(+c)

m(vy)1 + Σ 0 + 2

L0

t

Lt1

A

t2

Fydt = m(vy)2

( 100 + 5t 2 ) dt - 20(9.81)t = 20v

2 a100t +

5 3 t t b ` - 196.2t = 20v 3 0

v = 50.1667t 3 + 0.19t6 m>s

Kinematics. The displacement of the crate can be determined by integrating ds = v dt with the initial condition s = 0 at t = 0. L0

At t = 3 s,

s

ds =

L0

t

( 0.1667t 3 + 0.19t ) dt

s = 5 0.04167t 4 + 0.095t 2 6 m s = 0.04167 ( 34 ) + 0.095 ( 32 ) = 4.23 m

Ans.

Ans: s = 4.23 m 503

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15–29. In case of emergency, the gas actuator is used to move a 75-kg block B by exploding a charge C near a pressurized cylinder of negligible mass. As a result of the explosion, the cylinder fractures and the released gas forces the front part of the cylinder, A, to move B forward, giving it a speed of 200 mm s in 0.4 s. If the coefficient of kinetic friction between B and the floor is mk = 0.5, determine the impulse that the actuator imparts to B.

B A C

v B = 200 mm/s

SOLUTION Principle of Linear Impulse and Momentum: In order for the package to rest on top of the belt, it has to travel at the same speed as the belt. Applying Eq. 15–4, we have m A yy (+ c)

B1 + ©

t2

Lt1

Fy dt = m A yy

B A

B2

6(0) + Nt - 6(9.81) t = 6(0) N = 58.86 N t2

m(yx)1 + © + B A:

Lt1

Fx dt = m(yx)2

6(3) + [ - 0.2(58.86)t] = 6(1) Ans.

t = 1.02 s + B A:

m (vx)1 + © 0 +

L

L

L

Fx dt = m A vx B 2

F dt - (0.5)(9.81)(75)(0.4) = 75(0.2)

F dt = 162 N # s

Ans.

Ans: t = 1.02 s I = 162 N # s 504

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15–30. A jet plane having a mass of 7 Mg takes off from an aircraft carrier such that the engine thrust varies as shown by the graph. If the carrier is traveling forward with a speed of 40 km>h, determine the plane’s airspeed after 5 s.

F (kN) 40 km/h

15

SOLUTION

5

The impulse exerted on the plane is equal to the area under the graph.

0

n1 = 40 km>h = 11.11 m>s + 2 1:

m(vx)1 + ©

L

2

5

t (s)

Fx dt = m1nx22

(7) A 103 B (11.11) -

1 1 (2)(5)(103) + (15 + 5)(5 - 2)(103) = 7(103)n2 2 2 Ans.

n2 = 16.1 m>s

Ans: v = 16.1 m>s 505

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15–31. Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity 1vB21 = 3 ft>s at t = 0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth. Neglect the mass of the pulleys and cords.

A

SOLUTION sA + 2sB = l

(vB)1

vA = - 2vB + 2 1;

mv1 + © -

1+ T2

L

3 ft/s

B

F dt = mv2

10 10 (vA)2 (2)(3) - T(1) = 32.2 32.2

mv1 + ©

L

F dt = mv2

(vA)2 3 3 () (3) + 3(1) - 2T(1) = 32.2 32.2 2 - 32.2T - 10(vA)2 = 60 - 64.4T + 1.5(vA)2 = - 105.6 T = 1.40 lb (vA22 = -10.5 ft>s = 10.5 ft>s :

Ans.

Ans: (vA)2 = 10.5 ft>s S 506

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*15–32. Block A weighs 10 lb and block B weighs 3 lb. If B is moving downward with a velocity 1vB21 = 3 ft>s at t = 0, determine the velocity of A when t = 1 s. The coefficient of kinetic friction between the horizontal plane and block A is mA = 0.15.

A

SOLUTION sA + 2sB = l

(vB)1

vA = -2vB + 2 1;

mv1 + © -

1+ T2

L

3 ft/s

B

F dt = mv2

10 10 (v ) (2)(3) - T(1) + 0.15(10) = 32.2 32.2 A 2

mv1 + ©

L

F dt = mv2

3 3 (vA)2 (3) - 3(1) - 2T(1) = a b 32.2 32.2 2 - 32.2T - 10(vA)2 = 11.70 - 64.4T + 1.51vA22 = - 105.6 T = 1.50 lb (vA)2 = - 6.00 ft>s = 6.00 ft>s :

Ans.

Ans: (vA)2 = 6.00 ft>s S 507

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15–33. The log has a mass of 500 kg and rests on the ground for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the log when t = 5 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log.

T (N) 1800 T

3

t (s) A T

SOLUTION + ©F = 0; : x

200 t2

F - 0.5(500)(9.81) = 0 F = 2452.5 N

Thus, 2T = F 2(200t 2) = 2452.5 t = 2.476 s to start log moving + ) (:

m v1 + ©

L

F dt = m v2

3

0 + 2

L2.476

200t 2 dt + 211800215 - 32 - 0.41500219.81215 - 2.4762 = 500v2

t3 3 400( ) ` + 2247.91 = 500v2 3 2.476 Ans.

v2 = 7.65 m>s

Ans: v = 7.65 m>s 508

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15–34. The 0.15-kg baseball has a speed of v = 30 m>s just before it is struck by the bat. It then travels along the trajectory shown before the outfielder catches it. Determine the magnitude of the average impulsive force imparted to the ball if it is in contact with the bat for 0.75 ms.

v2 15 v1  30 m/s 15 0.75 m

2.5 m

100 m

SOLUTION + ) (:

mA (vA)1 + mB(vB)1 = (mA + mB)v2 3000 7500 4500 (3) (6) = v 32.2 32.2 32.2 2 v2 = -0.600 ft>s = 0.600 ft>s ;

Ans.

Ans: v = 0.6 ft>s d 509

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15–35. vB  20 m/s

The 5-Mg bus B is traveling to the right at 20 m>s. Meanwhile a 2-Mg car A is traveling at 15 m>s to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

B

vA  15 m/s A

Solution Conservation of Linear Momentum. + )         (S mAvA + mBvB = (mA + mB)v

3 5 ( 103 ) 4 (20)

     

+

3 2 ( 103 ) 4 (15)

=

3 5 ( 103 )

+ 2 ( 103 ) 4 v

v = 18.57 m>s = 18.6 m>s S 

Ans.

Ans: v = 18.6 m>s S 510

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*15–36. The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m>s. Determine the distance s the boy reaches up the inclined plane before momentarily coming to rest. Neglect the skateboard’s rolling resistance.

s 30

SOLUTION Free-Body Diagram: The free-body diagram of the boy and skateboard system is shown in Fig. a. Here, Wb,Wsb, and N are nonimpulsive forces. The pair of impulsive forces F resulting from the impact during landing cancel each other out since they are internal to the system. Conservation of Linear Momentum: Since the resultant of the impulsive force along the x axis is zero, the linear momentum of the system is conserved along the x axis. + ) (;

mb(vb)1 + msb(vsb)1 = (mb + msb)v 50(5) + 5(0) = (50 + 5)v v = 4.545 m>s

Conservation of Energy: With reference to the datum set in Fig. b, the gravitational potential energy of the boy and skateboard at positions A and B are

A Vg B A = (mb + msb)ghA = 0 and A Vg B B = (mb + msb)ghB = (50 + 5)(9.81)(s sin 30°) = 269.775s. TA + VA = TB + VB 1 1 (mb + msb)vA 2 + A Vg B A = (mb + msb)vB 2 + A Vg B B 2 2 1 (50 + 5) A 4.5452 B + 0 = 0 + 269.775s 2 Ans.

s = 2.11 m

Ans: s = 2.11 m 511

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15–37. The 2.5-Mg pickup truck is towing the 1.5-Mg car using a cable as shown. If the car is initially at rest and the truck is coasting with a velocity of 30 km>h when the cable is slack, determine the common velocity of the truck and the car just after the cable becomes taut. Also, find the loss of energy.

30 km/h

SOLUTION Free-Body Diagram: The free-body diagram of the truck and car system is shown in Fig. a. Here, Wt,WC, Nt, and NC are nonimpulsive forces. The pair of impulsive forces F generated at the instant the cable becomes taut are internal to the system and thus cancel each other out. Conservation of Linear Momentum: Since the resultant of the impulsive force is zero, the linear momentum of the system is conserved along the x axis. The initial 1h m speed of the truck is A vt B 1 = c30(103) d c d = 8.333 m>s. h 3600 s + ) (;

mt A vt B 1 + mC A vC B 1 = A mt + mC B v2 2500(8.333) + 0 = (2500 + 1500)v2 v2 = 5.208 m>s = 5.21 m>s ;

Ans.

Kinetic Energy: The initial and final kinetic energy of the system is T1 = =

1 1 m (v ) 2 + mC(vC)12 2 t t1 2 1 (2500)(8.3332) + 0 2

= 86 805.56 J and T2 = (mt + mC)v22 =

1 (2500 + 1500)(5.2082) 2

= 54 253.47 Thus, the loss of energy during the impact is ¢T = T1 - T2 = 86 805.56 - 54 253.47 = 32.55(103) J = 32.6 kJ

Ans.

Ans: v = 5.21 m>s d ∆T = -32.6 kJ 512

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15–38. A railroad car having a mass of 15 Mg is coasting at 1.5 m>s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m>s in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred, and explain qualitatively what happened to this energy.

SOLUTION + ) (S

Σmv1 = Σmv2 15 000(1.5) - 12 000(0.75) = 27 000(v2) Ans.

v2 = 0.5 m>s T1 =

1 1 (15 000)(1.5)2 + (12 000)(0.75)2 = 20.25 kJ 2 2

T2 =

1 (27 000)(0.5)2 = 3.375 kJ 2

∆T = T2 - T1 Ans.

= 3.375 - 20.25 = - 16.9 kJ This energy is dissipated as noise, shock, and heat during the coupling.

Ans: v = 0.5 m>s ∆T = - 16.9 kJ 513

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15–39. A ballistic pendulum consists of a 4-kg wooden block originally at rest, u = 0°. When a 2-g bullet strikes and becomes embedded in it, it is observed that the block swings upward to a maximum angle of u = 6°. Estimate the speed of the bullet.

θ

1.25 m

θ

1.25 m

SOLUTION Just after impact: Datum at lowest point. T2 + V2 = T3 + V3 1 (4 + 0.002) (vB)22 + 0 = 0 + (4 + 0.002)(9.81)(1.25)(1 - cos 6°) 2 (vB)2 = 0.3665 m>s For the system of bullet and block: + ) (S

Σmv1 = Σmv2

0.002(vB)1 = (4 + 0.002)(0.3665) Ans.

(vB)1 = 733 m>s

Ans: v = 733 m>s 514

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*15–40. The boy jumps off the flat car at A with a velocity of v = 4 ft>s relative to the car as shown. If he lands on the second flat car B, determine the final speed of both cars after the motion. Each car has a weight of 80 lb. The boy’s weight is 60 lb. Both cars are originally at rest. Neglect the mass of the car’s wheels.

v  4 ft/s 5

13 12

B

A

Solution + )    Σm(v ) = Σm(v ) (d 1 2      0 + 0 = -

80 60 v + (v ) 32.2 A 32.2 b x

     vA = 0.75(vb)x      vb = vA + vb>A + )    (v ) = - v + 4 a (d b x A

     (vb)x = 2.110 ft>s

12 b 13

     vA = 1.58 ft>s S 

Ans.

+ )    Σm(v ) = Σm(v ) (d 1 2     

60 80 60 (2.110) = a + bv 32.2 32.2 32.2

Ans.

     v = 0.904 ft>s

Ans: vA = 1.58 ft>s S v = 0.904 ft>s 515

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15–41. A 0.03-lb bullet traveling at 1300 ft>s strikes the 10-lb wooden block and exits the other side at 50 ft>s as shown. Determine the speed of the block just after the bullet exits the block, and also determine how far the block slides before it stops. The coefficient of kinetic friction between the block and the surface is mk = 0.5.

5

13 12

1300 ft/s

50 ft/s

5 4

SOLUTION + B A:

©m1 n1 = ©m2 n2 a

12 10 4 0.03 0.03 b 113002a b + 0 = a bn + a b1502a b 32.2 13 32.2 B 32.2 5 Ans.

vB = 3.48 ft>s T1 + ©U1 - 2 = T2 1 10 a b 13.4822 - 51d2 = 0 2 32.2

Ans.

d = 0.376 ft

Ans: vB = 3.48 ft>s d = 0.376 ft 516

3

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15–42. A 0.03-lb bullet traveling at 1300 ft>s strikes the 10-lb wooden block and exits the other side at 50 ft>s as shown. Determine the speed of the block just after the bullet exits the block. Also, determine the average normal force on the block if the bullet passes through it in 1 ms, and the time the block slides before it stops. The coefficient of kinetic friction between the block and the surface is mk = 0.5.

5

13 12

1300 ft/s

50 ft/s

5 4

SOLUTION + B A:

©m1 v1 = ©m2 v2 a

12 10 4 0.03 0.03 b 113002a b + 0 = a bv + a b1502a b 32.2 13 32.2 B 32.2 5 Ans.

vB = 3.48 ft>s

A+cB

mv1 + © -a

L

F dt = mv2

5 3 0.03 0.03 b 113002 a b - 10112 A 10 - 3 B + N112 A 10 - 3 B = a b1502a b 32.2 13 32.2 5 Ans.

N = 504 lb + B A:

mv1 + © a

L

F dt = mv2

10 b 13.482 - 51t2 = 0 32.2 Ans.

t = 0.216 s

Ans: vB = 3.48 ft>s Navg = 504 lb t = 0.216 s 517

3

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15–43. 400 m/s

The 20-g bullet is traveling at 400 m>s when it becomes embedded in the 2-kg stationary block. Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the block and the plane is mk = 0.2.

Solution Conservation of Momentum. + )   m v + m v = (m + m )v (S b b B B b B      0.02(400) + 0 = (0.02 + 2)v           v = 3.9604 m>s Principle of Impulse and Momentum. Here, friction Ff = mkN = 0.2 N. Referring to the FBD of the blocks, Fig. a, Lt1

( + c )   m(vy)1 + Σ     

t2

Fydt = m(vy)2

0 + N(t) - 2.02(9.81)(t) = 0

         N = 19.8162 N + )   m(v ) + Σ (S x 1

Lt1

t2

Fxdt = m(vx)2

     2.02(3.9604) + [ - 0.2(19.8162)t] = 2.02 v         v = 5 3.9604 - 1.962 t 6 m>s

Thus, the stopping time can be determined from      0 = 3.9604 - 1.962 t      t = 2.0186 s

Kinematics. The displacement of the block can be determined by integrating ds = v dt with the initial condition s = 0 at t = 0.     

L0

s

ds =

L0

t

(3.9604 - 1.962t) dt

     s = 5 3.9604 t - 0.981t 2 6 m

The block stopped at t = 2.0186 s. Thus          

s = 3.9604(2.0186) - 0.981 ( 2.01862 ) = 3.9971 m = 4.00 m

Ans.

Ans: s = 4.00 m 518

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*15–44. A toboggan having a mass of 10 kg starts from rest at A and carries a girl and boy having a mass of 40 kg and 45 kg, respectively. When the toboggan reaches the bottom of the slope at B, the boy is pushed off from the back with a horizontal velocity of vb>t = 2 m>s, measured relative to the toboggan. Determine the velocity of the toboggan afterwards. Neglect friction in the calculation.

A

3m B

vb/t

vt

SOLUTION Conservation of Energy: The datum is set at the lowest point B. When the toboggan and its rider is at A, their position is 3 m above the datum and their gravitational potential energy is (10 + 40 + 45)(9.81)(3) = 2795.85 N # m. Applying Eq. 14–21, we have T1 + V1 = T2 + V2 0 + 2795.85 =

1 (10 + 40 + 45) y2B + 0 2

yB = 7.672 m>s Relative Velocity: The relative velocity of the falling boy with respect to the toboggan is yb/t = 2 m>s. Thus, the velocity of the boy falling off the toboggan is yb = yt + yb/t + B A;

[1]

yb = yt - 2

Conservation of Linear Momentum: If we consider the tobbogan and the riders as a system, then the impulsive force caused by the push is internal to the system. Therefore, it will cancel out. As the result, the linear momentum is conserved along the x axis. mTyB = mb yb + + B A;

A mt + mg B yt

(10 + 40 + 45)(7.672) = 45yb + (10 + 40) yt

[2]

Solving Eqs. [1] and [2] yields Ans.

yt = 8.62 m>s yb = 6.619 m s

Ans: vt = 8.62 m>s 519

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15–45. The block of mass m is traveling at v1 in the direction u1 shown at the top of the smooth slope. Determine its speed v2 and its direction u2 when it reaches the bottom.

z v1 x

y

u1

h

SOLUTION There are no impulses in the v direction:

v2

mv1 sin u1 = mv2 sin u2

u2

T1 + V1 = T2 + V2 1 1 mv12 + mgh = mv22 + 0 2 2 v2 = 3v21 + 2gh

sin u2 =

Ans.

v1 sin u1 3v21 + 2gh

u2 = sin - 1 £

v1 sin u1 3v21 + 2gh

≥

Ans.

Ans: v2 = 2v21 + 2gh v1sinu u2 = sin-1 a b 2v21 + 2gh 520

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15–46. The two blocks A and B each have a mass of 5 kg and are suspended from parallel cords. A spring, having a stiffness of k = 60 N>m, is attached to B and is compressed 0.3 m against A and B as shown. Determine the maximum angles u and f of the cords when the blocks are released from rest and the spring becomes unstretched.

2m

θ A

2m

φ B

SOLUTION + ) (S

Σmv1 = Σmv2 0 + 0 = -5vA + 5vB vA = vB = v

Just before the blocks begin to rise: T1 + V1 = T2 + V2 (0 + 0) +

1 1 1 (60)(0.3)2 = (5)(v)2 + (5)(v)2 + 0 2 2 2

v = 0.7348 m>s For A or B: Datum at lowest point. T1 + V1 = T2 + V2 1 (5)(0.7348)2 + 0 = 0 + 5(9.81)(2)(1 - cos u) 2 Ans.

u = f = 9.52°

Ans: u = f = 9.52° 521

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15–47. 20 km/h

The 30-Mg freight car A and 15-Mg freight car B are moving towards each other with the velocities shown. Determine the maximum compression of the spring mounted on car A. Neglect rolling resistance.

A

10 km/h k  3 MN/m

B

SOLUTION Conservation of Linear Momentum: Referring to the free-body diagram of the freight cars A and B shown in Fig. a, notice that the linear momentum of the system is conserved along the x axis. The initial speed of freight cars A and B are 1h 1h m m (vA)1 = c 20(103) d a b = 5.556 m>s and (vB)1 = c10(103) d a b h 3600 s h 3600 s = 2.778 m>s. At this instant, the spring is compressed to its maximum, and no relative motion occurs between freight cars A and B and they move with a common speed. + ) (:

mA(vA)1 + mB(vB)1 = (mA + mB)v2 30(103)(5.556) + c - 15(103)(2.778) d = c 30(103) + 15(103) dv2 v2 = 2.778 m>s :

Conservation of Energy: The initial and final elastic potential energy of the spring 1 1 1 is (Ve)1 = ks12 = 0 and (Ve)2 = ks22 = (3)(106)smax2 = 1.5(106)smax2. 2 2 2 ©T1 + ©V1 = ©T2 + ©V2 1 1 1 c mA(vA)12 + mB(vB)12 d + (Ve)1 = (mA + mB)v22 + (Ve)2 2 2 2 1 1 (30) A 103 B (5.5562) + (15) A 103 B A 2.7782 B + 0 2 2 =

1 c 30 A 103 B + 15 A 103 B d A 2.7782 B + 1.5 A 106 B smax2 2 Ans.

smax = 0.4811 m = 481 mm

Ans: smax = 481 mm 522

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*15–48. Blocks A and B have masses of 40 kg and 60 kg, respectively. They are placed on a smooth surface and the spring connected between them is stretched 2 m. If they are released from rest, determine the speeds of both blocks the instant the spring becomes unstretched.

k  180 N/m A

B

SOLUTION + ) (:

©mn1 = ©mn2 0 + 0 = 40 nA - 60 nB T1 + V1 = T2 + V2 0 +

1 1 1 118021222 = 14021nA22 + 16021nB22 2 2 2

vA = 3.29 m>s

Ans.

vB = 2.19 m s

Ans.

Ans: vA = 3.29 m>s vB = 2.19 m>s 523

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15–49. A boy A having a weight of 80 lb and a girl B having a weight of 65 lb stand motionless at the ends of the toboggan, which has a weight of 20 lb. If they exchange positions, A going to B and then B going to A’s original position, determine the final position of the toboggan just after the motion. Neglect friction between the toboggan and the snow.

A

4 ft

B

Solution A goes to B, + )    Σmv = Σmv (S 1 2 0 = mAvA - (mt + mB)vB 0 = mAsA - (mt + mB)sB Assume B moves x to the left, then A moves (4 - x) to the right 0 = mA(4 - x) - (mt + mB)x x = =

4mA mA + mB + mt 4(80) 80 + 65 + 20

= 1.939 ft d

B goes to other end. + )   Σmv = Σmv (S 1 2 0 = - mBvB + (mt + mA)vA 0 = - mB sB + (mt + mA)sA Assume B moves x′ to the right, then A moves (4 - x′) to the left 0 = - mB(4 - x′) + (mt + mA)x′ x′ = =

4mB mA + mB + mt 4(65) 80 + 65 + 20

= 1.576 ft S

Net movement of sled is x = 1.939 - 1.576 = 0.364 ft d 

Ans.

Ans: x = 0.364 ft d 524

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15–50. A boy A having a weight of 80 lb and a girl B having a weight of 65 lb stand motionless at the ends of the toboggan, which has a weight of 20 lb. If A walks to B and stops, and both walk back together to the original position of A, determine the final position of the toboggan just after the motion stops. Neglect friction between the toboggan and the snow.

A

4 ft

B

Solution A goes to B, + )   Σmv = Σmv (S 1 2 0 = mAvA - (mt + mB)vB 0 = mAsA - (mt + mB)sB Assume B moves x to the left, then A moves (4 - x) to the right 0 = mA(4 - x) - (mt + mB)x 4mA mA + mB + mt

x = =

4(80) 80 + 65 + 20

= 1.939 ft d

A and B go to other end. + )   Σmv = Σmv (S 1 2 0 = - mBv - mAv + mt vt 0 = - mBs - mAs + mt st Assume the toboggan moves x′ to the right, then A and B move (4 - x′) to the left 0 = - mB(4 - x′) - mA(4 - x′) + mt x′ x′ = =

4(mB + mA) mA + mB + mt 4(65 + 80) 80 + 65 + 20

= 3.515 ft S

Net movement of sled is + )   x = 3.515 - 1.939 = 1.58 ft S  (S

Ans.

Ans: x = 1.58 ft S 525

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15–51. The 10-Mg barge B supports a 2-Mg automobile A. If someone drives the automobile to the other side of the barge, determine how far the barge moves. Neglect the resistance of the water.

40 m A B

Solution Conservation of Momentum. Assuming that VB is to the left, + )  m v + m v = 0 (d A A B B 2 ( 103 ) vA + 10 ( 103 ) vB = 0      2 vA + 10 vB = 0 Integrate this equation, (1)

2 sA + 10 sB = 0

Kinematics. Here, sA>B = 40 m d, using the relative displacement equation by assuming that sB is to the left, + )  (d

sA = sB + sA>B (2)

sA = sB + 40 Solving Eq. (1) and (2), sB = - 6.6667 m = 6.67 m S 

Ans.

sA = 33.33 m d The negative sign indicates that sB is directed to the right which is opposite to that of the assumed.

Ans: sB = 6.67 m S 526

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*15–52. The free-rolling ramp has a mass of 40 kg. A 10-kg crate is released from rest at A and slides down 3.5 m to point B. If the surface of the ramp is smooth, determine the ramp’s speed when the crate reaches B. Also, what is the velocity of the crate?

3.5 m

A

30

B

SOLUTION Conservation of Energy: The datum is set at lowest point B. When the crate is at point A, it is 3.5 sin 30° = 1.75 m above the datum. Its gravitational potential energy is 1019.81211.752 = 171.675 N # m. Applying Eq. 14–21, we have T1 + V1 = T2 + V2 0 + 171.675 =

1 1 1102v2C + 1402v2R 2 2

171.675 = 5 v2C + 20 v2R

(1)

Relative Velocity: The velocity of the crate is given by vC = vR + vC>R = - vRi + 1vC>R cos 30°i - vC>R sin 30°j2 = 10.8660 vC>R - vR2i - 0.5 vC>Rj

(2)

The magnitude of vC is vC = 2(0.8660 vC>R - vR22 + 1 - 0.5 vC>R22 = 2v2C>R + v2R - 1.732 vR vC>R

(3)

Conservation of Linear Momentum: If we consider the crate and the ramp as a system, from the FBD, one realizes that the normal reaction NC (impulsive force) is internal to the system and will cancel each other. As the result, the linear momentum is conserved along the x axis. 0 = mC 1vC2x + mR vR + ) (:

0 = 1010.8660 vC>R - vR2 + 401-vR2 (4)

0 = 8.660 vC>R - 50 vR Solving Eqs. (1), (3), and (4) yields

Ans.

vR = 1.101 m>s = 1.10 m>s vC = 5.43 m>s vC>R = 6.356 m>s From Eq. (2) vC = 30.866016.3562 - 1.1014i - 0.516.3562j = 54.403i - 3.178j6 m>s Thus, the directional angle f of vC is f = tan - 1

3.178 = 35.8° 4.403

cf

Ans.

527

Ans: vC>R = 6.356 m>s f = 35.8° c

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15–53. Block A has a mass of 5 kg and is placed on the smooth triangular block B having a mass of 30 kg. If the system is released from rest, determine the distance B moves from point O when A reaches the bottom. Neglect the size of block A.

A

B O

Solution

30 0.5 m

+ )   Σmv = Σmv (S 1 2 0 = 30vB - 5(vA)x (vA)x = 6vB vB = vA + vB>A + )  (S

vB = -(vA)x + (vB>A)x vB = -6vB + (vB>A)x (vB>A)x = 7vB Integrate (sB>A)x = 7 sB (sB>A)x = 0.5 m Thus, sB =

0.5 = 0.0714 m = 71.4 mm S  7

Ans.

Ans: sB = 71.4 mm S 528

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15–54. Solve Prob. 15–53 if the coefficient of kinetic friction between A and B is mk = 0.3. Neglect friction between block B and the horizontal plane.

A

B O

Solution

30 0.5 m

+ aΣFy = 0;  NA - 5(9.81) cos 30° = 0     NA = 42.4785 N Q + ΣFx = 0;  FA - 5(9.81) sin 30° = 0     FA = 24.525 N Fmax = mNA = 0.3(42.4785) = 12.74 N 6 24.525 N Block indeed slides. Solution is the same as in Prob. 15–53. Since FA is internal to the system. sB = 71.4 mm S 

Ans.

Ans: sB = 71.4 mm S 529

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15–55. The cart has a mass of 3 kg and rolls freely down the slope. When it reaches the bottom, a spring loaded gun fires a 0.5-kg ball out the back with a horizontal velocity of vb>c = 0.6 m>s, measured relative to the cart. Determine the final velocity of the cart.

A

1.25 m

vc

B

vb/c

SOLUTION Datum at B: TA + VA = TB + VB 0 + (3 + 0.5)(9.81)(1.25) =

1 (3 + 0.5)(nB)22 + 0 2

nB = 4.952 m>s + ) (;

©mn1 = ©mn2 (1)

(3 + 0.5)(4.952) = (3)nc - (0.5)nb + ) (;

nb = nc + nb>c (2)

- nb = nc - 0.6 Solving Eqs. (1) and (2), nc = 5.04 m>s ;

Ans.

nb = - 4.44 m>s = 4.44 m>s ;

Ans: vc = 5.04 m>s d 530

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*15–56. Two boxes A and B, each having a weight of 160 lb, sit on the 500-lb conveyor which is free to roll on the ground. If the belt starts from rest and begins to run with a speed of 3 ft>s, determine the final speed of the conveyor if (a) the boxes are not stacked and A falls off then B falls off, and (b) A is stacked on top of B and both fall off together.

B

A

SOLUTION a) Let vb be the velocity of A and B. + b a:

©mv1 = ©mv2 0 = a

+ b a:

320 500 b (vb) - a b(vc) 32.2 32.2

vb = vc + vb>c vb = -vc + 3

Thus, vb = 1.83 ft>s :

vc = 1.17 ft>s ;

When a box falls off, it exerts no impulse on the conveyor, and so does not alter the momentum of the conveyor. Thus, a) vc = 1.17 ft>s ;

Ans.

b) vc = 1.17 ft>s ;

Ans.

Ans: a) vc = 1.17 ft>s d b) vc = 1.17 ft>s d 531

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15–57. The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the 10-g bullet is traveling at 300 m>s when it becomes embedded in the 10-kg block, determine the distance the block will slide up along the plane before momentarily stopping.

300 m/s

A

30

SOLUTION Conservation of Linear Momentum: If we consider the block and the bullet as a system, then from the FBD, the impulsive force F caused by the impact is internal to the system. Therefore, it will cancel out. Also, the weight of the bullet and the block are nonimpulsive forces. As the result, linear momentum is conserved along the x œ axis. mb(vb)x¿ = (mb + mB) vx œ 0.01(300 cos 30°) = (0.01 + 10) v v = 0.2595 m>s Conservation of Energy: The datum is set at the blocks initial position. When the block and the embedded bullet is at their highest point they are h above the datum. Their gravitational potential energy is (10 + 0.01)(9.81)h = 98.1981h. Applying Eq. 14–21, we have T1 + V1 = T2 + V2 0 +

1 (10 + 0.01) A 0.25952 B = 0 + 98.1981h 2 h = 0.003433 m = 3.43 mm Ans.

d = 3.43 > sin 30° = 6.87 mm

Ans: d = 6.87 mm 532

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15–58. Disk A has a mass of 250 g and is sliding on a smooth horizontal surface with an initial velocity (vA)1 = 2 m>s. It makes a direct collision with disk B, which has a mass of 175 g and is originally at rest. If both disks are of the same size and the collision is perfectly elastic (e = 1), determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same.

Solution + )  (S

(0.250)(2) + 0 = (0.250)(vA)2 + (0.175)(vB)2

+ )  (S

e = 1 =

(vB)2 - (vA)2 2 - 0

Solving (vA)2 = 0.353 m>s

Ans.

(vB)2 = 2.35 m>s

Ans.

T1 =

1 (0.25)(2)2 = 0.5 J 2

T2 =

1 1 (0.25)(0.353)2 + (0.175)(2.35)2 = 0.5 J 2 2

T1 = T2

QED

Ans: (vA)2 = 0.353 m>s (vB)2 = 2.35 m>s 533

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15–59. The 5-Mg truck and 2-Mg car are traveling with the freerolling velocities shown just before they collide. After the collision, the car moves with a velocity of 15 km>h to the right relative to the truck. Determine the coefficient of restitution between the truck and car and the loss of energy due to the collision.

30 km/h 10 km/h

SOLUTION Conservation of Linear Momentum: The linear momentum of the system is conserved along the x axis (line of impact). The initial speeds of the truck and car are (vt)1 = c 30 A 103 B and (vc)1 = c10 A 103 B

1h m da b = 8.333 m>s h 3600 s

m 1h da b = 2.778 m>s. h 3600 s

By referring to Fig. a, + b a:

mt A vt B 1 + mc A vc B 1 = mt A vt B 2 + mc A vc B 2 5000(8.333) + 2000(2.778) = 5000 A vt B 2 + 2000 A vc B 2 5 A vt B 2 + 2 A vc B 2 = 47.22

Coefficient of Restitution: Here, (vc>t) = c15 A 103 B

(1) 1h m da b = 4.167 m>s : . h 3600 s

Applying the relative velocity equation, (vc)2 = (vt)2 + (vc>t)2 + B A:

(vc)2 = (vt)2 + 4.167 (vc)2 - (vt)2 = 4.167

(2)

Applying the coefficient of restitution equation, + B A:

e =

(vc)2 - (vt)2 (vt)1 - (vc)1

e =

(vc)2 - (vt)2 8.333 - 2.778

(3)

534

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15–59. Continued

Substituting Eq. (2) into Eq. (3), e =

4.167 = 0.75 8.333 - 2.778

Ans.

Solving Eqs. (1) and (2) yields (vt)2 = 5.556 m>s (vc)2 = 9.722 m>s Kinetic Energy: The kinetic energy of the system just before and just after the collision are T1 = =

1 1 m (v ) 2 + mc(vc)1 2 2 t t1 2 1 1 (5000)(8.3332) + (2000)(2.7782) 2 2

= 181.33 A 103 B J T2 = =

1 1 m (v ) 2 + mc(vc)2 2 2 t t2 2 1 1 (5000)(5.5562) + (2000)(9.7222) 2 2

= 171.68 A 103 B J Thus, ¢T = T1 - T2 = 181.33 A 103 B - 171.68 A 103 B = 9.645 A 103 B J Ans.

= 9.65 kJ

Ans: e = 0.75 ∆T = -9.65 kJ 535

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*15–60. Disk A has a mass of 2 kg and is sliding forward on the smooth surface with a velocity 1vA21 = 5 m/s when it strikes the 4-kg disk B, which is sliding towards A at 1vB21 = 2 m/s, with direct central impact. If the coefficient of restitution between the disks is e = 0.4, compute the velocities of A and B just after collision.

(vA)1 = 5 m/s

(vB)1 = 2 m/s

A

B

SOLUTION Conservation of Momentum : mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2 + B A:

(1)

2(5) + 4(- 2) = 2(vA)2 + 4(vB)2

Coefficient of Restitution :

+ B A:

e =

(vB)2 - (vA)2 (vA)1 - (vB)1

0.4 =

(vB)2 - (vA)2 5 - ( -2)

(2)

Solving Eqs. (1) and (2) yields (vA)2 = - 1.53 m s = 1.53 m s ;

(vB)2 = 1.27 m s :

Ans.

Ans: (vA)2 = 1.53 m>s d (vB)2 = 1.27 m>s S 536

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15–61. The 15-kg block A slides on the surface for which mk = 0.3. The block has a velocity v = 10 m>s when it is s = 4 m from the 10-kg block B. If the unstretched spring has a stiffness k = 1000 N>m, determine the maximum compression of the spring due to the collision. Take e = 0.6.

10 m/s

k  1000 N/m

A

B

Solution s

Principle of Work and Energy. Referring to the FBD of block A, Fig. a, motion along the y axis gives NA = 15(9.81) = 147.15 N. Thus the friction is Ff = mkNA = 0.3(147.15) = 44.145 N. T1 + ΣU1 - 2 = T2 1 1 2 (15) ( 102 ) + ( - 44.145)(4) = (15)(vA)1 2 2 (vA)1 = 8.7439 m>s d Conservation of Momentum. + )  (d

mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2 15(8.7439) + 0 = 15(vA)2 + 10(vB)2 (1)

3(vA)2 + 2(vB)2 = 26.2317 Coefficient of Restitution. + )  (d

e =

(vB)2 - (vA)2 (vA)1 - (vB)1

;  0.6 =

(vB)2 - (vA)2 8.7439 - 0 (2)

          (vB)2 - (vA)2 = 5.2463 Solving Eqs. (1) and (2) (vB)2 = 8.3942 m>s d (vA)2 = 3.1478 m>s d

Conservation of Energy. When block B stops momentarily, the compression of the spring is maximum. Thus, T2 = 0. T1 + V1 = T2 + V2 1 1 (10) ( 8.39422 ) + 0 = 0 + (1000)x2max 2 2 Ans.

       xmax = 0.8394 m = 0.839 m

Ans: x max = 0.839 m 537

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15–62. The four smooth balls each have the same mass m. If A and B are rolling forward with velocity v and strike C, explain why after collision C and D each move off with velocity v. Why doesn’t D move off with velocity 2v? The collision is elastic, e = 1. Neglect the size of each ball.

v

v

A

B

C D

Solution Collision will occur in the following sequence; B strikes C + ) (S

mv = - mvB + mvC



v = - v B + vC

+ ) (S

e = 1 =



vC = v,  vB = 0

vC + v B v

C strikes D + ) (S

mv = - mvC + mvD vD + v C v

+ ) (S

e = 1 =



vC = 0,  vD = v

Ans.

A strikes B + ) (S

mv = - mvA + mvB vB + v A v

+ ) (S

e = 1 =



vB = v,  vA = 0

Ans.

Finally, B strikes C + ) (S

mv = - mvB + mvC vC + v B v

+ ) (S

e = 1 =



vC = v,  vB = 0

Ans.

Note: If D rolled off with twice the velocity, its kinetic energy would be twice the 1 1 1 energy available from the original two A and B: a mv2 + mv2 ≠ ( 2v ) 2 b 2 2 2

Ans: vC = 0, vD = v vB = v, vA = 0 vC = v, vB = 0 538

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15–63. The four balls each have the same mass m. If A and B are rolling forward with velocity v and strike C, determine the velocity of each ball after the first three collisions. Take e = 0.5 between each ball.

v

v

A

B

C D

Solution Collision will occur in the following sequence; B strikes C + ) (S

mv = mvB + mvc



v = v B + vC

+ ) (S

e = 0.5 =



vC = 0.75v S ,  vB = 0.25v S

vC - v B v

C strikes D + ) (S

m(0.75v) = mvC + mvD

+ ) (S

e = 0.5 =

vD - v C 0.75v



vC = 0.1875v S 

Ans.



vD = 0.5625v S 

Ans.

A strikes B + ) (S

mv + m(0.25v) = mvA + mvB

+ ) (S

e = 0.5 =



vB = 0.8125v S ,  vA = 0.4375v S 

vB - v A (v - 0.25v) Ans.

Ans: vC = vD = vB = vA = 539

0.1875v S 0.5625v S 0.8125v S 0.4375v S

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*15–64. 8 m/s

Ball A has a mass of 3 kg and is moving with a velocity of 8 m>s when it makes a direct collision with ball B, which has a mass of 2 kg and is moving with a velocity of 4 m>s. If e = 0.7, determine the velocity of each ball just after the collision. Neglect the size of the balls.

A

4 m/s B

Solution Conservation of Momentum. The velocity of balls A and B before and after impact are shown in Fig. a + ) (S

mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2 3(8) + 2( -4) = 3vA + 2vB



(1)

3vA + 2vB = 16



Coefficient of Restitution. + ) e = (S

(vB)2 - (vA)2 (vA)1 - (vB)1

; 0.7 =

vB - v A 8 - ( - 4) (2)

vB - vA = 8.4

Solving Eqs. (1) and (2),

vB = 8.24 m>s S 

Ans.



vA = - 0.16 m>s = 0.160 m>s d 

Ans.

Ans: vB = 8.24 m>s S vA = 0.160 m>s d 540

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15–65. A 1-lb ball A is traveling horizontally at 20 ft>s when it strikes a 10-lb block B that is at rest. If the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic friction between the plane and the block is mk = 0.4, determine the time for the block B to stop sliding.

SOLUTION + b a:

©m1 v1 = ©m2 v2 a

1 1 10 b (20) + 0 = a b (vA)2 + a b(vB)2 32.2 32.2 32.2

(vA)2 + 10(vB)2 = 20 + b a:

e =

(vB)2 - (vA)2 (vA)1 - (vB)1

0.6 =

(vB)2 - (vA)2 20 - 0

(vB)2 - (vA)2 = 12 Thus, (vB)2 = 2.909 ft>s : (vA)2 = -9.091 ft>s = 9.091 ft>s ; Block B: + b a:

m v1 + © a

L

F dt = m v2

10 b (2.909) - 4t = 0 32.2 Ans.

t = 0.226 s

Ans: t = 0.226 s 541

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15–66. Block A, having a mass m, is released from rest, falls a distance h and strikes the plate B having a mass 2m. If the coefficient of restitution between A and B is e, determine the velocity of the plate just after collision. The spring has a stiffness k.

A h B k

SOLUTION Just before impact, the velocity of A is T1 + V1 = T2 + V2 0 + 0 =

1 mv2A - mgh 2

vA = 22gh (+ T)

e =

(vB)2 - (vA)2 22gh

e 22gh = (vB)2 - (vA)2 (+ T)

(1)

©mv1 = ©mv2 (2)

m(vA) + 0 = m(vA)2 + 2m(vB)2 Solving Eqs. (1) and (2) for (vB)2 yields; (vB)2 =

1 3

Ans.

2gh(1 + e)

Ans: (vB)2 = 542

1 22gh (1 + e) 3

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15–67. The three balls each weigh 0.5 lb and have a coefficient of restitution of e = 0.85. If ball A is released from rest and strikes ball B and then ball B strikes ball C, determine the velocity of each ball after the second collision has occurred. The balls slide without friction.

A r

3 ft B

C

SOLUTION Ball A: Datum at lowest point. T1 + V1 = T2 + V2 0 + 10.52132 =

1 0.5 1 21vA221 + 0 2 32.2

1vA21 = 13.90 ft>s Balls A and B: + B A:

©mv1 = ©mv 2 1

+ B A:

0.5 0.5 0.5 2113.902 + 0 = 1 21vA22 + 1 21vB22 32.2 32.2 32.2

e = 0.85 =

1vB22 - 1vA222 1vA21 - 1vB21 1vB22 - 1vA22 13.90 - 0

Solving: Ans.

1vA22 = 1.04 ft>s 1vB22 = 12.86 ft>s Balls B and C: + B A:

©mv 2 = ©mv 3 1

+ B A:

0.5 0.5 0.5 2112.862 + 0 = 1 21v 2 + 1 21v 2 32.2 32.2 B 3 32.2 C 3

e = 0.85 =

1vC23 - 1vB23 1vB22 - 1vC22 1vC23 - 1vB23 12.86 - 0

Solving: 1vB23 = 0.964 ft>s

Ans.

1vC23 = 11.9 ft>s

Ans. Ans: (vA)2 = 1.04 ft>s (vB)3 = 0.964 ft>s (vC)3 = 11.9 ft>s 543

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*15–68. A pitching machine throws the 0.5-kg ball toward the wall with an initial velocity vA = 10 m>s as shown. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall if e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C.

B

30 1.5 m

Solution

C

(a)

3m

(vB)x1 = 10 cos 30° = 8.660 m>s S

s

+ ) (S

s = s0 + v0t



3 = 0 + 10 cos 30°t



t = 0.3464 s

( + c )

v = v0 + ac t



(vB)y t = 10 sin 30° - 9.81(0.3464) = 1.602 m>s c

( + c )

s = s0 + v0 t +



h = 1.5 + 10 sin 30°(0.3464) -



vA  10 m/s

A

1 2 at 2 c 1 (9.81)(0.3464)2 2

= 2.643 m (vB)1 = 2(1.602)2 + (8.660)2 = 8.81 m>s

(b) + ) (S

e =

1.602 b = 10.5°  Q  8.660

(vB)2 - (vA)2 (vA)1 - (vB)1

;

0.5 =

(vBx)2 - 0 0 - (8.660)



(vBx)2 = 4.330 m>s d



(vBy)2 = (vBy)1 = 1.602 m>s c



(vB)2 = 2(4.330)2 + (1.602)2 = 4.62 m>s

(c)

( + c )

1.602 b = 20.3°  4.330

s = s0 + vBt +

Ans.

a

Ans.

√

u2 = tan-1a



Ans.

√

u1 = tan-1 a



Ans.

1 a t2 2 c 1 (9.81)(t)2 2



- 2.643 = 0 + 1.602(t) -



t = 0.9153 s

+ ) (d

s = s0 + v0t



s = 0 + 4.330(0.9153) = 3.96 m

Ans.

√

Ans: (vB)1 = 8.81 m>s u1 = 10.5° Q (vB)2 = 4.62 m>s u2 = 20.3° a s = 3.96 m

√

544

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15–69. A 300-g ball is kicked with a velocity of vA = 25 m>s at point A as shown. If the coefficient of restitution between the ball and the field is e = 0.4, determine the magnitude and direction u of the velocity of the rebounding ball at B.

vA

v¿B

25 m>s 30

u B

A

SOLUTION Kinematics: The parabolic trajectory of the football is shown in Fig. a. Due to the symmetrical properties of the trajectory, vB = vA = 25 m>s and f = 30°. Conservation of Linear Momentum: Since no impulsive force acts on the football along the x axis, the linear momentum of the football is conserved along the x axis. + b a;

m A vB B x = m A v Bœ B x 0.3 A 25 cos 30° B = 0.3 Av Bœ B x

A v Bœ B x = 21.65 m>s ; Coefficient of Restitution: Since the ground does not move during the impact, the coefficient of restitution can be written as

A+cB

e =

0 - A v Bœ B y

A vB B y - 0 - A v Bœ B y

0.4 =

- 25 sin 30°

A v Bœ B y = 5 m>s c Thus, the magnitude of vBœ is v Bœ = 2 A v Bœ B x + A vBœ B y = 221.652 + 52 = 22.2 m>s and the angle of vBœ is u = tan - 1 C

A v Bœ B y A v Bœ B x

S = tan - 1 ¢

5 ≤ = 13.0° 21.65

Ans.

Ans.

Ans: vB′ = 22.2 m>s u = 13.0° 545

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15–70. Two smooth spheres A and B each have a mass m. If A is given a velocity of v0, while sphere B is at rest, determine the velocity of B just after it strikes the wall. The coefficient of restitution for any collision is e.

v0

A

B

SOLUTION Impact: The first impact occurs when sphere A strikes sphere B. When this occurs, the linear momentum of the system is conserved along the x axis (line of impact). Referring to Fig. a, + ) (:

mAvA + mBvB = mA(vA)1 + mB(vB)1 mv0 + 0 = m(vA)1 + m(vB)1 (1)

(vA)1 + (vB)1 = v0 + ) (:

e =

(vB)1 - (vA)1 vA - vB

e =

(vB)1 - (vA)1 v0 - 0 (2)

(vB)1 - (vA)1 = ev0 Solving Eqs. (1) and (2) yields (vB)1 = a

1 + e bv0 : 2

(vA)1 = a

1 - e b v0 : 2

The second impact occurs when sphere B strikes the wall, Fig. b. Since the wall does not move during the impact, the coefficient of restitution can be written as + ) (:

e =

0 - C - (vB)2 D (vB)1 - 0 0 + (vB)2

e = c

1 + e d v0 - 0 2

(vB)2 =

e(1 + e) v0 2

Ans.

Ans: (vB)2 = 546

e(1 + e) 2

v0

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15–71. It was observed that a tennis ball when served horizontally 7.5 ft above the ground strikes the smooth ground at B 20 ft away. Determine the initial velocity vA of the ball and the velocity vB (and u) of the ball just after it strikes the court at B. Take e = 0.7.

A

vA

7.5 ft

vB u 20 ft

B

SOLUTION + ) (:

s = s0 + v0t 20 = 0 + vA t

(+ T)

s = s0 + v0 t +

1 2 ac t 2

7.5 = 0 + 0 +

1 (32.2)t 2 2

t = 0.682524 Ans.

vA = 29.303 = 29.3 ft>s vB x 1 = 29.303 ft>s (+ T) v = v0 + ac t vBy 1 = 0 + 32.2(0.68252) = 21.977 ft>s + ) (:

mv1 = mv 2 vB2x = vB 1x = 29.303 ft>s : e =

vBy 2 vBy 1

0.7 =

vBy 2 21.977

,

vBy 2 = 15.384 ft>s c

vB 2 = 2(29.303)2 + (15.384)2 = 33.1 ft>s u = tan - 1

15.384 = 27.7° 29.303

Ans.

a

Ans.

Ans: vA = 29.3 ft>s vB2 = 33.1 ft>s u = 27.7° a 547

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*15–72. The tennis ball is struck with a horizontal velocity vA , strikes the smooth ground at B, and bounces upward at u = 30°. Determine the initial velocity vA , the final velocity vB , and the coefficient of restitution between the ball and the ground.

A

vA

7.5 ft

vB u 20 ft

B

SOLUTION (+ T)

v2 = v20 + 2 ac (s - s0) (vBy)21 = 0 + 2(32.2)(7.5 - 0) vBy1 = 21.9773 m>s

( + T)

v = v0 + ac t 21.9773 = 0 + 32.2 t t = 0.68252 s

( + T)

s = s0 + v0 t 20 = 0 + vA (0.68252) Ans.

vA = 29.303 = 29.3 ft>s + ) (:

mv1 = mv 2 vB x 2 = vB x 1 = vA = 29.303 Ans.

vB2 = 29.303>cos 30° = 33.8 ft>s vBy 2 = 29.303 tan 30° = 16.918 ft>s e =

vBy 2 vBy 1

=

16.918 = 0.770 21.9773

Ans.

Ans: vA = 29.3 ft>s vB2 = 33.8 ft>s e = 0.770 548

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15–73. Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine their final velocities just after collision. The coefficient of restitution is e = 0.75.

y (vA)1

x B

SOLUTION + ) (:

6 m/s

(vB)1

5 4 3

A

4 m/s

©my1 = ©my2 3 0.5(4)( ) - 0.5(6) = 0.5(yB)2x + 0.5(yA)2x 5

+ ) (:

e =

(yA)2 - (yB)2 (yB)1 - (yA)1

0.75 =

(yA)2x - (yB)2x 4 A 35 B - ( - 6)

(yA)2x = 1.35 m>s : (yB)2x = 4.95 m>s ; (+ c)

my1 = my2 4 0.5( )(4) = 0.5(yB)2y 5 (yB)2y = 3.20 m>s c yA = 1.35 m>s :

Ans.

yB = 2(4.59)2 + (3.20)2 = 5.89 m>s

Ans.

u = tan-1

3.20 = 32.9 4.95

b

Ans.

Ans: vA = 1.35 m>s S vB = 5.89 m>s u = 32.9° b 549

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15–74. y

Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision B travels along a line, 30° counterclockwise from the y axis.

(vA)1

6 m/s x

B

SOLUTION

(vB)1

5 4 3

A

4 m/s

©my1 = ©my2 + ) (:

3 0.5(4)( ) - 0.5(6) = -0.5(yB)2x + 0.5(yA)2x 5 -3.60 = -(vB)2x + (yA)2x

(+ c )

4 0.5(4)( ) = 0.5(yB)2y 5 (yB)2y = 3.20 m>s c (yB)2x = 3.20 tan 30° = 1.8475 m>s ; (yA)2x = -1.752 m>s = 1.752 m>s ;

+ ) (:

e = e =

(yA)2 - (yB)2 (yB)1 - (yA)1 - 1.752- ( - 1.8475) 4(35)- ( -6)

Ans.

= 0.0113

Ans: e = 0.0113 550

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15–75. The 0.5-kg ball is fired from the tube at A with a velocity of v = 6 m>s. If the coefficient of restitution between the ball and the surface is e = 0.8, determine the height h after it bounces off the surface.

v  6 m/s

30

A C

2m

h B

Solution Kinematics. Consider the vertical motion from A to B.

( + c ) (vB)2y = (vA)2y + 2ay[(sB)y - (sA)y] ;

(vB)2y = (6 sin 30°)2 + 2( - 9.81)( - 2 - 0)



(vB)y = 6.9455 m>s T

Coefficient of Restitution. The y-component of the rebounding velocity at B is (v′B)y and the ground does not move. Then

( + c ) e =

(vg)2 - (v′B)y (vB)y - (vg)1



;

0.8 =

0 - (v′B)y - 6.9455 - 0

(v′B)y = 5.5564 m>s c

Kinematics. When the ball reach the maximum height h at C, (vc)y = 0.

( + c ) (vc)2y = (v′B)2y + 2ac[(sc)y - (sB)y] ;

02 = 5.55642 + 2( - 9.81)(h - 0) Ans.

h = 1.574 m = 1.57 m

Ans: h = 1.57 m 551

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*15–76. A ball of mass m is dropped vertically from a height h0 above the ground. If it rebounds to a height of h1, determine the coefficient of restitution between the ball and the ground.

h0 h1

SOLUTION Conservation of Energy: First, consider the ball’s fall from position A to position B. Referring to Fig. a, TA + VA = TB + VB 1 1 2 2 mvA + (Vg)A = mvB + (Vg)B 2 2 1 2 0 + mg(h0) = m(vB)1 + 0 2 Subsequently, the ball’s return from position B to position C will be considered. TB + VB = TC + VC 1 1 2 2 mvB + (Vg)B = mvC + (Vg)C 2 2 1 2 m(vB)2 + 0 = 0 + mgh1 2 (vB)2 = 22gh1 c Coefficient of Restitution: Since the ground does not move, (+ c)

e = -

e = -

(vB)2 (vB)1 22gh1 - 22gh0

=

h1 A h0

Ans.

Ans: e = 552

h1 A h0

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15–77. The cue ball A is given an initial velocity (vA)1 = 5 m>s. If it makes a direct collision with ball B (e = 0.8), determine the velocity of B and the angle u just after it rebounds from the cushion at C (e¿ = 0.6). Each ball has a mass of 0.4 kg. Neglect their size.

(vA)1

B

5 m/s A 30

C

SOLUTION

u

Conservation of Momentum: When ball A strikes ball B, we have mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2 (1)

0.4(5) + 0 = 0.4(vA)2 + 0.4(vB)2 Coefficient of Restitution: e = +) (;

(vB)2 - (vA)2 (vA)1 - (vB)1

0.8 =

(vB)2 - (vA)2 5 - 0

(2)

Solving Eqs. (1) and (2) yields (vA)2 = 0.500 m>s

(vB)2 = 4.50 m>s

Conservation of “y” Momentum: When ball B strikes the cushion at C, we have mB(vBy)2 = mB(vBy)3 (+ T)

0.4(4.50 sin 30°) = 0.4(vB)3 sin u (3)

(vB)3 sin u = 2.25 Coefficient of Restitution (x): e = + ) (;

(vC)2 - (vBx)3 (vBx)2 - (vC)1

0.6 =

0 - [- (vB)3 cos u] 4.50 cos 30° - 0

(4)

Solving Eqs. (1) and (2) yields (vB)3 = 3.24 m>s

Ans.

u = 43.9°

Ans: (vB)3 = 3.24 m>s u = 43.9° 553

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15–78. Using a slingshot, the boy fires the 0.2-lb marble at the concrete wall, striking it at B. If the coefficient of restitution between the marble and the wall is e = 0.5, determine the speed of the marble after it rebounds from the wall.

vA

A

45 5 ft

SOLUTION A sB B x = A sA B x + A vA B x t 100 = 0 + 75 cos 45° t t = 1.886 s and a+cb

A sB B y = A sA B y + A vA B y t +

1 a t2 2 y

A sB B y = 0 + 75 sin 45°(1.886) +

1 ( -32.2)(1.8862) 2

= 42.76 ft and a+cb

A vB B y = A vA B y + ay t A vB B y = 75 sin 45° + ( -32.2)(1.886) = -7.684 ft>s = 7.684 ft>s T

Since A vB B x = A vA B x = 75 cos 45° = 53.03 ft>s, the magnitude of vB is vB = 2 A vB B x 2 + A vB B y 2 = 253.032 + 7.6842 = 53.59 ft>s and the direction angle of vB is u = tan - 1 C

A vB B y A vB B x

S = tan - 1 ¢

7.684 ≤ = 8.244° 53.03

Conservation of Linear Momentum: Since no impulsive force acts on the marble along the inclined surface of the concrete wall (x¿ axis) during the impact, the linear momentum of the marble is conserved along the x¿ axis. Referring to Fig. b,

A +QB

C

100 ft

Kinematics: By considering the x and y motion of the marble from A to B, Fig. a, + b a:

B

75 ft>s

mB A v Bœ B x¿ = mB A v Bœ B x¿ 0.2 0.2 A 53.59 sin 21.756° B = A v Bœ cos f B 32.2 32.2 v Bœ cos f = 19.862

(1)

554

60

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15–78. Continued

Coefficient of Restitution: Since the concrete wall does not move during the impact, the coefficient of restitution can be written as

A +a B

e =

0 - A v Bœ B y¿

A v Bœ B y¿ - 0

0.5 =

-v Bœ sin f - 53.59 cos 21.756°

v Bœ sin f = 24.885

(2)

Solving Eqs. (1) and (2) yields v Bœ = 31.8 ft>s

Ans.

Ans: vB′ = 31.8 ft>s 555

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15–79. The two disks A and B have a mass of 3 kg and 5 kg, respectively. If they collide with the initial velocities shown, determine their velocities just after impact. The coefficient of restitution is e = 0.65.

(vB)1  7 m/s

(vA)1  6 m/s

A

B 60

Line of impact

SOLUTION (yAx) = 6 m>s

(yAy)1 = 0

(yBx)1 = -7 cos 60° = - 3.5 m>s + b a:

A yBy B 1 = - 7 cos 60° = -6.062 m>s

mA(yAx)1 + mB(yBx)1 = mA(yAx)2 + mB(yBx)2 3(6) -5(3.5) = 3(yA)x2 + 5(yB)x2

+ b a:

e =

(yBx)2 - (yAx)2 ; (yAx)1 - (yBx)1

0.65 =

(yBx)2 - (yAx)2 6 - (- 3.5)

(yBx)2 - (yAx)2 = 6.175 Solving, (yAx)2 = -3.80 m>s (+ c )

(yBx)2 = 2.378 m>s

mA A yAy B 1 + mA A yAy B 2

A yAy B 2 = 0 (+ c)

mB A yBy B 1 + mB A yAy B 2

A yBy B 2 = - 6.062 m>s (yA)2 = 2(3.80)2 + (0)2 = = 3.80 m>s ;

Ans.

(yB)2 = 2(2.378)2 + ( -6.062)2 = 6.51 m>s

Ans.

(uB)2 = tan - 1 a

6.062 b = 68.6° 2.378

Ans.

Ans: (vA)2 = 3.80 m>s d (vB)2 = 6.51 m>s (uB)2 = 68.6° 556

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*15–80. v0

A ball of negligible size and mass m is given a velocity of v0 on the center of the cart which has a mass M and is originally at rest. If the coefficient of restitution between the ball and walls A and B is e, determine the velocity of the ball and the cart just after the ball strikes A. Also, determine the total time needed for the ball to strike A, rebound, then strike B, and rebound and then return to the center of the cart. Neglect friction.

A

B

d

d

Solution After the first collision; + )  (d

Σmv1 = Σmv2



0 + mv0 = mvb + Mvc

+ )  (d

e =



vc - v b v0

mv0 = mvb +

M v m c

ev0 = vc - vb

v0(1 + e) = a1 +

vc =



vb =

M bv m c

v0(1 + e)m (m + M)

v0(1 + e)m (m + M)

Ans.

 - ev0

= v0 c



= v0 a

m + me - em - eM d m + M



m - eM b m + M

Ans.

The relative velocity on the cart after the first collision is vref e = v0

vref = ev0

Similarly, the relative velocity after the second collision is vref e = ev0

vref = e 2v0

Total time is

t =



=

d 2d d + + 2 v0 ev0 e v0 d 1 2 a1 + b  v0 e

Ans.

Ans: vc =

v0(1 + e)m (m + M)

m - eM b m + M d 1 2 t = a1 + b v0 e vb = v 0 a

557

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15–81. The girl throws the 0.5-kg ball toward the wall with an initial velocity vA = 10 m>s. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C.

B

vA

A

10 m/s

30 1.5 m C

SOLUTION Kinematics: By considering the horizontal motion of the ball before the impact, we have + B A:

sx = (s0)x + vx t 3 = 0 + 10 cos 30°t

t = 0.3464 s

By considering the vertical motion of the ball before the impact, we have (+ c)

vy = (v0)y + (ac)y t = 10 sin 30° + (- 9.81)(0.3464) = 1.602 m>s

The vertical position of point B above the ground is given by (+ c)

sy = (s0)y + (v0)y t +

(sB)y = 1.5 + 10 sin 30°(0.3464) +

1 (a ) t2 2 cy

1 ( - 9.81) A 0.34642 B = 2.643 m 2

Thus, the magnitude of the velocity and its directional angle are (vb)1 = 2(10 cos 30°)2 + 1.6022 = 8.807 m>s = 8.81 m>s u = tan - 1

1.602 = 10.48° = 10.5° 10 cos 30°

Ans. Ans.

Conservation of “y” Momentum: When the ball strikes the wall with a speed of (vb)1 = 8.807 m>s, it rebounds with a speed of (vb)2. mb A vby B 1 = mb A vby B 2 + B A;

mb (1.602) = mb C (vb)2 sin f D (1)

(vb)2 sin f = 1.602 Coefficient of Restitution (x): e = + B A:

0.5 =

(vw)2 - A vbx B 2

A vbx B 1 - (vw)1 0 - C -(vb)2 cos f D

(2)

10 cos 30° - 0

558

3m s

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15–81. Continued

Solving Eqs. (1) and (2) yields f = 20.30° = 20.3°

(vb)2 = 4.617 m>s = 4.62 m>s

Ans.

Kinematics: By considering the vertical motion of the ball after the impact, we have (+ c)

sy = (s0)y + (v0)y t +

1 (a ) t2 2 cy

- 2.643 = 0 + 4.617 sin 20.30°t1 +

1 ( - 9.81)t21 2

t1 = 0.9153 s By considering the horizontal motion of the ball after the impact, we have + B A;

sx = (s0)x + vx t Ans.

s = 0 + 4.617 cos 20.30°(0.9153) = 3.96 m

Ans: (a) (vB)1 = 8.81 m>s, u = 10.5° a (b) (vB)2 = 4.62 m>s, f = 20.3° b (c)  s = 3.96 m 559

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15–82. The 20-lb box slides on the surface for which mk = 0.3. The box has a velocity v = 15 ft>s when it is 2 ft from the plate. If it strikes the smooth plate, which has a weight of 10 lb and is held in position by an unstretched spring of stiffness k = 400 lb>ft, determine the maximum compression imparted to the spring. Take e = 0.8 between the box and the plate. Assume that the plate slides smoothly.

v

15 ft/s k

2 ft

SOLUTION T1 + a U1 - 2 = T2 1 20 1 20 a b (15)2 - (0.3)(20)(2) = a b(v2)2 2 32.2 2 32.2 v2 = 13.65 ft>s + ) (: a

a mv1 = a mv2

20 20 10 b (13.65) = a b vA + vB 32.2 32.2 32.2

e =

(vB)2 - (vA)2 (vA)1 - (vB)1

0.8 =

vP - vA 13.65

Solving, vP = 16.38 ft>s, vA = 5.46 ft>s T1 + V1 = T2 + V2 1 1 10 a b (16.38)2 + 0 = 0 + (400)(s)2 2 32.2 2 Ans.

s = 0.456 ft

Ans: s = 0.456 ft 560

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15–83. The 10-lb collar B is at rest, and when it is in the position shown the spring is unstretched. If another 1-lb collar A strikes it so that B slides 4 ft on the smooth rod before momentarily stopping, determine the velocity of A just after impact, and the average force exerted between A and B during the impact if the impact occurs in 0.002 s. The coefficient of restitution between A and B is e = 0.5.

k  20 lb/ft

SOLUTION

A

3 ft

B

Collar B after impact: T2 + V2 = T3 + V3 1 1 10 a b 1vB222 + 0 = 0 + 120215 - 322 2 2 32.2 (vB)2 = 16.05 ft>s System: + B A:

©m1v1 = ©m1 v2 1 10 1 (v ) + 0 = (v ) + (16.05) 32.2 A 1 32.2 A 2 32.2 1vA21 - 1vA22 = 160.5

+ B A:

e = 0.5 =

1vB22 - 1vA22 1vA21 - 1vB21 16.05 - 1vA22 1vA21 - 0

0.51vA21 + 1vA22 = 16.05 Solving: (vA)1 = 117.7 ft>s = 118 ft/s : (vA)2 = - 42.8 ft>s = 42.8 ft>s ;

Ans.

Collar A: + B A:

mv1 + © a

L

F dt = m v2

1 1 b1117.72 - F10.0022 = a b 1- 42.82 32.2 32.2 Ans.

F = 2492.2 lb = 2.49 kip

Ans: (vA)2 = 42.8 ft>s d F = 2.49 kip 561

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*15–84. A ball is thrown onto a rough floor at an angle u. If it rebounds at an angle f and the coefficient of kinetic friction is m, determine the coefficient of restitution e. Neglect the size of the ball. Hint: Show that during impact, the average impulses in the x and y directions are related by Ix = mIy . Since the time of impact is the same, Fx ¢t = mFy ¢t or Fx = mFy .

y

f

u

x

SOLUTION 0 - 3 - v2 sin f4

( + T)

e =

+ ) (:

m1vx21 +

v1 sin u - 0

e =

v2 sin f v1 sin u

(1)

t2

Fx dx = m1vx22

Lt1

mv1 cos u - Fx ¢t = mv2 cos f Fx = ( + T)

mv1 cos u - mv2 cos f ¢t

m1vy21 +

(2)

t2

Lt1

Fy dx = m1vy22

mv1 sin u - Fy ¢t = - mv2 sin f Fy =

mv1 sin u + mv2 sin f ¢t

(3)

Since Fx = mFy, from Eqs. (2) and (3) m1mv1 sin u + mv2 sin f) mv1 cos u - mv2 cos f = ¢t ¢t cos u - m sin u v2 = v1 m sin f + cos f

(4)

Substituting Eq. (4) into (1) yields: e =

sin f cos u - m sin u a b sin u m sin f + cos f

Ans.

Ans: e = 562

sin f cos u - m sin u a b sin u m sin f + cos f

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15–85. A ball is thrown onto a rough floor at an angle of u = 45°. If it rebounds at the same angle f = 45°, determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is e = 0.6. Hint: Show that during impact, the average impulses in the x and y directions are related by Ix = mIy . Since the time of impact is the same, Fx ¢t = mFy ¢t or Fx = mFy .

y

u

f x

SOLUTION (+ T ) + B A:

e =

0 - [- v2 sin f] v1 sin u - 0

e =

v2 sin f v1 sin u

(1)

t2

m(vx)1 +

Fx dx = m(vx)2

Lt1

mv1 cos u - Fx ¢t = mv2 cos f Fx = (+ c )

mv1 cos u - mv2 cos f ¢t

m A vy B 1 +

(2)

t2

Lt1

Fydx = m A yy B 2

mv1 sin u - Fy ¢t = -mv2 sin f Fy =

mv1 sin u + mv2 sin f ¢t

(3)

Since Fx = mFy, from Eqs. (2) and (3) m(mv1 sin u + mv2 sin f) mv1 cos u - mv2 cos f = ¢t ¢t cos u - m sin u v2 = v1 m sin f + cos f

(4)

Substituting Eq. (4) into (1) yields: e =

sin f cos u - m sin u a b sin u m sin f + cos f

0.6 =

sin 45° cos 45° - m sin 45° a b sin 45° m sin 45° + cos 45°

0.6 =

1-m 1+m

Ans.

m = 0.25

Ans: mk = 0.25 563

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15–86. Two smooth billiard balls A and B each have a mass of 200 g. If A strikes B with a velocity (vA)1 = 1.5 m>s as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.85. Neglect the size of each ball.

y A

(vA)1  1.5 m/s 40 x

B

Solution (vAx)1 = -1.5 cos 40° = -1.1491 m>s

( vAy ) 1 = -1.5 sin 40° = -0.9642 m>s + )  (S

mA ( vAx ) 1 + mB ( vBx ) 1 = mA ( vAx ) 2 + mB ( vBx ) 2



- 0.2 ( 1.1491 ) + 0 = 0.2 ( vAx ) 2 + 0.2 ( vBx ) 2

+ )  (S

e =

( vAx ) 2 - ( vBx ) 2 ( vAx ) 2 - ( vBx ) 2 ;  0.85 = 1.1491 ( vBx ) 1 - ( vAx ) 1

Solving,

( vAx ) 2 = - 0.08618 m>s



( vBx ) 2 = - 1 .0 6 2 9 m>s

For A:

( + T )   mA ( vAy ) 1 = mA ( vAy ) 2

( vAy ) 2 = 0.9642 m>s

For B:

( + c )   mB ( vBy ) 1 = mB ( vBy ) 2

( vBy ) 2 = 0

Hence.

( vB ) 2 = ( vBx ) 2 = 1.06 m>s d 

Ans.



( vA ) 2 = 2 ( - 0.08618 ) 2 + ( 0.9642 ) 2 = 0.968 m>s

Ans.



( uA ) 2 = tan-1a

0.08618 b = 5.11° e 0.9642

Ans.

Ans: (vB)2 = 1.06 m>s d (vA)2 = 0.968 m>s (uA)2 = 5.11° e 564

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15–87. The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has a weight of 47 lb, and the coefficient of restitution between the “stones” is e = 0.8, determine their speeds just after collision. Initially A has a velocity of 8 ft>s and B is at rest. Neglect friction.

3 ft y B A

Solution 30 (vA)1  8 ft/s

Line of impact (x-axis): Σmv1 = Σmv2 ( + a)  0 +

x

47 47 47 (8) cos 30° = (v ) + (v ) 32.2 32.2 B 2x 32.2 A 2x

( + a)  e = 0.8 =

(vB)2x - (vA)2x 8 cos 30° - 0

Solving:

(vA)2x = 0.6928 ft>s



(vB)2x = 6.235 ft>s

Plane of impact (y-axis): Stone A: mv1 = mv2 (Q + )  

47 47 (8) sin 30° = (v ) 32.2 32.2 A 2y

(vA)2y = 4

Stone B: mv1 = mv2 (Q + )  0 =

47 (v ) 32.2 B 2y



(vB)2y = 0



(vA)2 = 2(0.6928)2 + (4)2 = 4.06 ft>s



Ans.

(vB)2 = 2(0)2 + (6.235)2 = 6.235 = 6.24 ft>s

Ans.

Ans: (vA)2 = 4.06 ft>s (vB)2 = 6.24 ft>s 565

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*15–88. The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has a weight of 47 lb, and the coefficient of restitution between the “stone” is e = 0.8, determine the time required just after collision for B to slide off the runway. This requires the horizontal component of displacement to be 3 ft.

3 ft y B A

Solution See solution to Prob. 15–87.

(vB)2 = 6.235 ft>s



s = s0 + v0t



3 = 0 + (6.235 cos 60°)t



t = 0.962 s

30 (vA)1  8 ft/s

x

Ans.

Ans: t = 0.962 s 566

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15–89. Two smooth disks A and B have the initial velocities shown just before they collide. If they have masses mA = 4 kg and mB = 2 kg, determine their speeds just after impact. The coefficient of restitution is e = 0.8.

5

4

3

A

vA  15 m/s

B

Solution Impact. The line of impact is along the line joining the centers of disks A and B represented by y axis in Fig. a. Thus

vB  8 m/s

3 4 [(vA)1]y = 15 a b = 9 m>s b   [(vA)1]x = 15 a b = 12 m>s a 5 5 [(vB)1]y = 8 m>s Q      

[(v  B)1]x = 0

Coefficient of Restitution. Along the line of impact (y axis), ( +Q )

e =

[(vB)2]y - [(vA)2]y [(vA)1]y - [(vB)1]y

;  0.8 =

[(vB)2]y - [(vA)2]y -9 - 8

              [(vA)2]y - [(vB)2]y = 13.6

(1)

Conservation of ‘y’ Momentum. (+Q)

mA[(vA)1]y + mB[(vB)1]y = mA[(vA)2]y + mB[(vB)2]y

    4( -9) + 2(8) = 4[(vA)2]y + 2[(vB)2]y (2)

     2[(vA)2]y + [(vB)2]y = - 10 Solving Eqs. (1) and (2) [(vA)2]y = 1.20 m>s Q    [(vB)2]y = -12.4 m>s = 12.4 m>s b

Conservation of ‘x’ Momentum. Since no impact occurs along the x axis, the component of velocity of each disk remain constant before and after the impact. Thus [(vA)2]x = [(vA)1]x = 12 m>s a   [(vB)2]x = [(vB)1]x = 0 Thus, the magnitude of the velocity of disks A and B just after the impact is (vA)2 = 2[(vA)2]2x + [(vA)2]2y = 2122 + 1.202 = 12.06 m>s = 12.1 m>s (vB)2 =

2[(vB)2]2x

+

[(vB)2]2y

2

2

= 20 + 12.4 = 12.4 m>s 

Ans. Ans.

Ans: (vA)2 = 12.1 m>s (vB)2 = 12.4 m>s 567

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15–90. Before a cranberry can make it to your dinner plate, it must pass a bouncing test which rates its quality. If cranberries having an e Ú 0.8 are to be accepted, determine the dimensions d and h for the barrier so that when a cranberry falls from rest at A it strikes the incline at B and bounces over the barrier at C.

A C

SOLUTION Conservation of Energy: The datum is set at point B. When the cranberry falls from a height of 3.5 ft above the datum, its initial gravitational potential energy is W(3.5) = 3.5 W. Applying Eq. 14–21, we have T1 + V1 = T2 + V2 1 W a b (vc)21 + 0 2 32.2

(vc)1 = 15.01 ft>s Conservation of “x¿ ” Momentum: When the cranberry strikes the plate with a speed of (vc)1 = 15.01 ft>s, it rebounds with a speed of (vc)2. mc A vcx¿ B 1 = mc A vcx¿ B 2 (+ b )

3 mc (15.01)a b = mc C (vc)2 cos f D 5 (1)

(vc)2 cos f = 9.008 Coefficient of Restitution (y¿ ): e = ( a +)

(vP)2 - A vcy¿ B 2

A vcy¿ B 1 - (vP)1

0.8 =

0 - (vc)2 sin f

(2)

4 - 15.01a b - 0 5

Solving Eqs. (1) and (2) yields f = 46.85°

(vc)2 = 13.17 ft>s

Kinematics: By considering the vertical motion of the cranberry after the impact, we have (+ c )

vy = (v0)y + ac t

0 = 13.17 sin 9.978° + (-32.2) t (+ c )

sy = (s0)y + (v0)y t +

t = 0.07087 s

1 (a ) t2 2 cy

= 0 + 13.17 sin 9.978° (0.07087) +

5 4

h

0 + 3.5W =

3.5 ft

1 ( -32.2) A 0.070872 B 2

= 0.080864 ft

568

B

d

3

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15–90. Continued

By considering the horizontal motion of the cranberry after the impact, we have + B A;

sx = (s0)x + vx t 4 d = 0 + 13.17 cos 9.978° (0.07087) 5 Ans.

d = 1.149 ft = 1.15 ft Thus, h = sy +

3 3 d = 0.080864 + (1.149) = 0.770 ft 5 5

Ans.

Ans: d = 1.15 ft h = 0.770 ft 569

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15–91. The 200-g billiard ball is moving with a speed of 2.5 m>s when it strikes the side of the pool table at A. If the coefficient of restitution between the ball and the side of the table is e = 0.6, determine the speed of the ball just after striking the table twice, i.e., at A, then at B. Neglect the size of the ball.

v  2.5 m/s

45

A B

SOLUTION At A: (vA)y 1 = 2.5(sin 45°) = 1.7678 m>s : e =

(vAy)2 (vAy)1

;

0.6 =

(vAy)2 1.7678

(vAy)2 = 1.061 m>s ; (vAx)2 = (vAx)1 = 2.5 cos 45° = 1.7678 m>s T At B: e =

(vBx)3 (vBx)2

;

0.6 =

(vBx)3 1.7678

(vBx)3 = 1.061 m>s

A vBy B 3 = A vAy B 2 = 1.061 m>s Hence, (vB)3 = 2(1.061)2 + (1.061)2 = 1.50 m>s

Ans.

Ans: (vB)3 = 1.50 m>s 570

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*15–92. The two billiard balls A and B are originally in contact with one another when a third ball C strikes each of them at the same time as shown. If ball C remains at rest after the collision, determine the coefficient of restitution.All the balls have the same mass. Neglect the size of each ball.

A v C

B

SOLUTION Conservation of “x” momentum: + b a:

my = 2my œ cos 30° y = 2y œ cos 30°

(1)

Coefficient of restitution: ( +Q)

e =

yœ y cos 30°

(2)

Substituting Eq. (1) into Eq. (2) yields: e =

2y œ

yœ 2 = 3 cos2 30°

Ans.

Ans: e = 571

2 3

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15–93. Disks A and B have a mass of 15 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. The coefficient of restitution between them is e = 0.8.

y 5

4

Line of impact

3

A

10 m/s x

SOLUTION

8 m/s

B

Conservation of Linear Momentum: By referring to the impulse and momentum of the system of disks shown in Fig. a, notice that the linear momentum of the system is conserved along the n axis (line of impact). Thus, ¿ + Q mA A vA B n + mB A vB B n = mA A vA B n + mB A vB¿ B n

3 3 ¿ 15(10)a b - 10(8)a b = 15vA cos fA + 10vB¿ cos fB 5 5 ¿ 15vA cos fA + 10vB¿ cos fB = 42

(1)

Also, we notice that the linear momentum of disks A and B are conserved along the t axis (tangent to? plane of impact). Thus, ¿ + a mA A vA B t = mA A vA Bt

4 ¿ 15(10)a b = 15vA sin fA 5 ¿ vA sin fA = 8

(2)

and +a mB A vB B t = mB A vB¿ B t 4 10(8)a b = 10 vB¿ sin fB 5 vB¿ sin fB = 6.4

(3)

Coefficient of Restitution: The coefficient of restitution equation written along the n axis (line of impact) gives +Q e =

¿ (vB¿ )n - (vA )n (vA)n - (vB)n

0.8 =

¿ cos fA vB¿ cos fB - vA

3 3 10 a b - c -8 a b d 5 5

¿ vB¿ cos fB - vA cos fA = 8.64

(4)

Solving Eqs. (1), (2), (3), and (4), yeilds ¿ = 8.19 m>s vA

Ans.

fA = 102.52° vB¿ = 9.38 m>s

Ans.

fB = 42.99°

Ans: (vA)2 = 8.19 m>s (vB)2 = 9.38 m>s 572

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15–94. Determine the angular momentum HO of the 6-lb particle about point O.

z 6 lb

A

4 ft/s 12 ft P

Solution

8 ft

Position and Velocity Vector. The coordinates of points A and B are A( -8, 8, 12) ft and B(0, 18, 0) ft. Then rOB = 518j 6 ft  rOA = 5 - 8i + 8j + 12k6 ft VA = vAa



= •

B O

10 ft

8 ft

y

x

[0 - ( - 8)]i + (18 - 8)j + (0 - 12)k rAB b = 4• ¶ rAB 2[0 - ( - 8)]2 + (18 - 8)2 + (0 - 12)2 32

2308

i +

40 2308

j -

48 2308

k ¶ ft>s

Angular Momentum about Point O. HO = rOB * mVA

= 5

i 0 6 32 a b 32.2 2308

j 18 40

6 a b 32.2 2308

= 5 - 9.1735i - 6.1156k6 slug # ft 2 >s Also,

k 0

5

6 48 ab 32.2 2308

= 5 - 9.17i - 6.12k6 slug # ft 2 >s

Ans.

HO = rOA * mVA

= 5

i -8 6 32 a b 32.2 2308

j 8

k 12

6 40 a b 32.2 2308

6 48 ab 32.2 2308

= 5 - 9.1735i - 6.1156k6 slug # ft 2 >s

5

= 5 - 9.17i - 6.12k6 slug # ft 2 >s

Ans.

573

Ans: 5 -9.17i - 6.12k6 slug # ft 2 >s

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15–95. Determine the angular momentum Hp of the 6-lb particle about point P.

z 6 lb

A

4 ft/s 12 ft P

Solution

8 ft

Position and Velocity Vector. The coordinates of points A, B and P are A( - 8, 8, 12) ft, B(0, 18, 0) ft and P( - 8, 0, 0). Then rpB = [0 - ( - 8)]i + (18 - 0)j] = 5 8i + 18j 6 ft

B O

10 ft

8 ft

y

x

rpA = [ - 8 - ( - 8)]i + (8 - 0)j + (12 - 0)j] = 58j + 12k6 ft VA = vAa = •

[0 - ( -8)]i + (18 - 8)j + (0 - 12)k rAB b = 4• ¶ rAB [0 - ( - 8)]2 + (18 - 8)2 + (0 - 12)2 32

2308

i +

40 2308

j -

48 2308

k ¶ ft>s

Angular Momentum about Point P. HP = rpA * mVA

= 5

i 0

j 8

k 12

6 32 a b 32.2 2308

6 40 a b 32.2 2308

6 48 ab 32.2 2308

5

= 5 - 9.1735i + 4.0771j - 2.71816 slug # ft 2 >s Also,

= 5 - 9.17i + 4.08j - 2.72k6 slug # ft 2 >s

Ans.

HP = rpB * mVA

= 5

i 8 6 32 a b 32.2 2308

j 18 6 40 a b 32.2 2308

k 0

5

6 48 ab 32.2 2308

= 5 - 9.1735i + 4.0771j - 2.7181k6 slug # ft 2 >s = 5 - 9.17i + 4.08j - 2.72k6 slug # ft 2 >s

Ans.

574

Ans: 5 -9.17i + 4.08j - 2.72k6 slug # ft 2 >s

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*15–96. Determine the angular momentum Ho of each of the two particles about point O.

y 4m

P

5m

1.5 m

4m

Solution O

a+ (HA)O

4 3 = ( -1.5) c 3(8)a b d - (2) c 3(8)a b d = - 57.6 kg # m2 >s 5 5

8 m/s

3 5

4

3 kg A

a+ (HB)O = ( - 1)[4(6 sin 30°)] - (4)[4 (6 cos 30°)] = - 95.14 kg # m2 >s

B 4 kg

2m

1m 30

x

6 m/s

Thus

(HA)O = 5 - 57.6 k6 kg # m2 >s

Ans.

(HB)O = 5 - 95.1 k6 kg # m2 >s

Ans.

575

Ans: (HA)O = 5 - 57.6 k6 kg # m2 >s (HB)O = 5 - 95.1 k6 kg # m2 >s

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15–97. Determine the angular momentum Hp of each of the two particles about point P.

y 4m

P

5m

1.5 m

4m

Solution O

4 3 a+ (HA)p = (2.5) c 3(8)a b d - (7) c 3(8)a b d = - 52.8 kg # m2 >s 5 5

8 m/s

3 5

4

3 kg A

a+ (HB)p = (4)[4(6 sin 30°)] - 8[4 (6 cos 30°)] = - 118.28 kg # m2 >s

B 4 kg

2m

1m 30

x

6 m/s

Thus,

(HA)p = 5 - 52.8k6 kg # m2 >s

Ans.

(HB)p = 5 - 118k6 kg # m2 >s

Ans.

Ans: ( HA ) P = 5 - 52.8k6 kg # m2 >s 576

( HB ) P = 5 - 118k6 kg # m2 >s

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15–98. Determine the angular momentum HO of the 3-kg particle about point O.

z 3 kg

A 6 m/s

P

2m O

2m 1.5 m

Solution

2m

1.5 m 1m

x

Position and Velocity Vectors. The coordinates of points A and B are A(2, - 1.5, 2) m and B(3, 3, 0).

3m B

3m

y

rOB = 53i + 3j 6 m  rOA = 5 2i - 1.5j + 2k6 m VA = vAa

= •

(3 - 2)i + [3 - ( -1.5)]j + (0.2)k rAB b = (6) D T rAB 2(3 - 2)2 + [3 - ( - 1.5)]2 + (0 - 2)2 6

225.25

i +

27 225.25

j -

12 225.25

k ¶ m>s

Angular Momentum about Point O. Applying Eq. 15 HO = rOB * mVA

= 5 3a

i 3 6 225.25

b

3a

j 3 27 225.25

b

3a -

k 0 12 225.25

5 b

= 5 - 21.4928i + 21.4928j + 37.6124k6 kg # m2 >s Also,

= 5 - 21.5i + 21.5j + 37.66 kg # m2 >s

Ans.

HO = rOA * mVA

= 5 3a

i 2 6 225.25

b

3a

j - 1.5 27 225.25

b

3a -

k 2 12 225.25

5 b

= 5 - 21.4928i + 21.4928j + 37.6124k6 kg # m2 >s = 5 - 21.5i + 21.5j + 37.6k6 kg # m2 >s

Ans.

577

Ans: 5 -21.5i + 21.5j + 37.66 kg # m2 >s

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15–99. Determine the angular momentum HP of the 3-kg particle about point P.

z 3 kg

A 6 m/s

P

2m O

2m 1.5 m

Solution

2m

1.5 m 1m

x

Position and Velocity Vectors. The coordinates of points A, B and P are A(2, - 1.5, 2) m, B(3, 3, 0) m and P( -1, 1.5, 2) m.

3m B

3m

y

rPA = [2 - ( - 1)]i + ( - 1.5 - 1.5)j + (2 - 2)k = 53i - 3j 6 m

rPB = [3 - ( - 1)]i + (3 - 1.5)j + (0 - 2)k = 5 4i + 1.5j - 2k6 m VA = vAa

= •

(3 - 2)i + [3 - ( -1.5)]j + (0 - 2)k rAB b = 6D T rAB 2(3 - 2)2 + [3 - ( - 1.5)]2 + (0 - 2)2 6

225.25

i +

27 225.25

j -

12 225.25

k ¶ m>s

Angular Momentum about Point P. Applying Eq. 15 Hp = rpA * mVA

= 5 3a

i 3 6 225.25

b

3a

j -3 27 225.25

b

3a -

k 0 12 225.25

5 b

= 521.4928i + 21.4928j + 59.1052k6 kg # m2 >s Also,

= 521.5i + 21.5j + 59.1k6 kg # m2 >s

Ans.

Hp = rPB * mVA

= 5 3a

i 4 6 225.25

b

3a

j 1.5 27 225.25

b

3a -

k -2 12 225.25

5 b

= 521.4928i + 21.4928j + 59.1052k6 kg # m2 >s = 521.5i + 21.5j + 59.1k6 kg # m2 >s

Ans.

578

Ans: 521.5i + 21.5j + 59.1k6 kg # m2 >s

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*15–100. Each ball has a negligible size and a mass of 10  kg  and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque M = (t 2 + 2) N # m, where t is in seconds, determine the speed of each ball when t = 3 s. Each ball has a speed v = 2 m>s when t = 0.

M  (t 2  2) N  m

v

0.5 m

Solution Principle of Angular Impulse and Momentum. Referring to the FBD of the assembly, Fig. a



(HZ)1 + Σ

2[0.5(10)(2)] +

L0

Lt1

v

t2

MZ dt = (HZ)2

3s

(t 2 + 2)dt = 2[0.5(10v)] Ans.

v = 3.50 m>s

Ans: v = 3.50 m>s 579

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15–101. The 800-lb roller-coaster car starts from rest on the track having the shape of a cylindrical helix. If the helix descends 8 ft for every one revolution, determine the speed of the car when t = 4 s. Also, how far has the car descended in this time? Neglect friction and the size of the car.

r  8 ft 8 ft

Solution u = tan-1a

u b = 9.043° 2p(8)

ΣFy = 0;  N - 800 cos 9.043° = 0 N = 790.1 lb H1 +

0 +

L

L0

M dt = H2

4

8(790.1 sin 9.043°)dt =

800 (8)vt 32.2

vt = 20.0 ft>s v =

20.0 = 20.2 ft>s cos 9.043°

Ans.

T1 + ΣU1 - 2 = T2 0 + 800h = h = 6.36 ft

1 800 a b(20.2)2 2 32.2

Ans.

Ans: v = 20.2 ft>s h = 6.36 ft 580

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15–102. The 800-lb roller-coaster car starts from rest on the track having the shape of a cylindrical helix. If the helix descends 8 ft for every one revolution, determine the time required for the car to attain a speed of 60 ft>s. Neglect friction and the size of the car.

r  8 ft 8 ft

Solution u = tan-1a

8 b = 9.043° 2p(8)

ΣFy = 0;  N - 800 cos 9.043° = 0 N = 790.1 lb vt cos 9.043°

v =

vt cos 9.043°

60 =

vt = 59.254 ft>s H1 + 0 +

L

L0

M dt = Hz

t

8(790.1 sin 9.043°)dt =

800 (8)(59.254) 32.2 Ans.

t = 11.9 s

Ans: t = 11.9 s 581

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15–103. A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine the ball’s speed at the instant r2 = 2 ft. How much work has to be done to pull down the cord? Neglect friction and the size of the ball.

B

r1 = 3 ft (v B )1 = 6 ft>s

SOLUTION H1 = H2

v r = 2 ft>s

4 4 v (2) (6)(3) = 32.2 32.2 u vu = 9 ft>s v2 = 292 + 22 = 9.22 ft>s

Ans.

T1 + ©U1 - 2 = T2 1 4 1 4 ( )(6)2 + ©U1 - 2 = ( )(9.22)2 2 32.2 2 32.2 ©U1 - 2 = 3.04 ft # lb

Ans.

Ans: v2 = 9.22 ft>s ΣU1 - 2 = 3.04 ft # lb 582

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*15–104. A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine how much time is required for the ball to reach a speed of 12 ft/s. How far r2 is the ball from the hole when this occurs? Neglect friction and the size of the ball.

B

r1 = 3 ft (v B )1 = 6 ft>s

SOLUTION v = 2(vu)2 + (2)2

v r = 2 ft>s

12 = 2(vu)2 + (2)2 vu = 11.832 ft>s H1 = H2 4 4 (6)(3) = (11.832)(r2) 32.2 32.2 Ans.

r2 = 1.5213 = 1.52 ft ¢r = v rt (3 - 1.5213) = 2t

Ans.

t = 0.739 s

Ans: r2 = 1.52 ft t = 0.739 s 583

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15–105. The two blocks A and B each have a mass of 400 g. The blocks are fixed to the horizontal rods, and their initial velocity along the circular path is 2 m>s. If a couple moment of M = (0.6) N # m is applied about CD of the frame, determine the speed of the blocks when t = 3 s. The mass of the frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks.

C

A

B 0.3 m

Solution (Ho)1 + Σ

Lt1

M  0.6 N  m

0.3 m

D

t2

Modt = (Ho)2

2[0.3(0.4)(2)] + 0.6(3) = 2[0.3(0.4)v] Ans.

v = 9.50 m>s

Ans: v = 9.50 m>s 584

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15–106. A small particle having a mass m is placed inside the semicircular tube. The particle is placed at the position shown and released. Apply the principle of angular # momentum about point O 1©MO = HO2, and show that the motion $ of the particle is governed by the differential equation u + 1g>R2 sin u = 0.

O

u

R

SOLUTION c + ©MO =

dHO ; dt

-Rmg sin u =

d (mvR) dt

g sin u = -

d 2s dv = - 2 dt dt

But, s = R u $ Thus, g sin u = -Ru $ g or, u + a b sin u = 0 R

Q.E.D.

Ans: ## g u + a b sin u = 0 R 585

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15–107. If the rod of negligible mass is subjected to a couple moment of M = (30t2) N # m and the engine of the car supplies a traction force of F = (15t) N to the wheels, where t is in seconds, determine the speed of the car at the instant t = 5 s. The car starts from rest. The total mass of the car and rider is 150 kg. Neglect the size of the car.

4m

M  (30t 2) Nm

F  15t N

SOLUTION Free-Body Diagram: The free-body diagram of the system is shown in Fig. a. Since the moment reaction MS has no component about the z axis, the force reaction FS acts through the z axis, and the line of action of W and N are parallel to the z axis, they produce no angular impulse about the z axis. Principle of Angular Impulse and Momentum:

A H1 B z + ©

t1

Lt2

Mz dt = A H2 B z

5s

0 +

L0

5s

30 t2 dt +

L0

15t(4)dt = 150v(4) Ans.

v = 3.33 m>s

Ans: v = 3.33 m>s 586

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*15–108. When the 2-kg bob is given a horizontal speed of 1.5 m>s, it begins to rotate around the horizontal circular path A. If the force F on the cord is increased, the bob rises and then rotates around the horizontal circular path B. Determine the speed of the bob around path B. Also, find the work done by force F.

300 mm B

600 mm

A

SOLUTION Equations of Motion: By referring to the free-body diagram of the bob shown in Fig. a, + c ©Fb = 0; ; ©Fn = ma n;

(1)

F cos u - 2(9.81) = 0 F sin u = 2 a

v2 b l sin u

(2)

Eliminating F from Eqs. (1) and (2) yields v2 sin2 u = cos u 9.81l 1 - cos2 u v2 = cos u 9.81l

(3)

When l = 0.6 m, v = v1 = 5 m>s. Using Eq. (3), we obtain 1 - cos2 u1 1.52 = cos u1 9.81(0.6) cos2 u1 + 0.3823 cos u1 - 1 = 0 Solving for the root 6 1, we obtain u1 = 34.21° Conservation of Angular Momentum: By observing the free-body diagram of the system shown in Fig. b, notice that W and F are parallel to the z axis, MS has no z component, and FS acts through the z axis. Thus, they produce no angular impulse about the z axis. As a result, the angular momentum of the system is conserved about the z axis. When u = u1 = 34.21° and u = u2 , r = r1 = 0.6 sin 34.21° = 0.3373 m and r = r2 = 0.3 sin u2 . Thus,

A Hz B 1 = A Hz B 2 r1mv1 = r2mv2 0.3373(2)(1.5) = 0.3 sin u2 (2)v2 (4)

v2 sin u2 = 1.6867

587

F

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*15–108. Continued

Substituting l = 0.3 and u = u2 v = v2 into Eq. (3) yields 1 - cos2 u2 v2 = cos u2 9.81(0.3) 1 - cos2 u2 v2 2 = cos u2 2.943

(5)

Eliminating v2 from Eqs. (4) and (5), sin3 u2 tan u2 - 0.9667 = 0 Solving the above equation by trial and error, we obtain u2 = 57.866° Substituting the result of u2 into Eq. (4), we obtain Ans.

v2 = 1.992 m>s = 1.99 m>s

Principle of Work and Energy: When u changes from u1 to u2, W displaces vertically upward h = 0.6 cos 34.21° - 0.3 cos 57.866° = 0.3366 m. Thus, W does negatives work. T1 + ©U1 - 2 = T2 1 mv1 2 + UF + 2

A - Wh B =

1 mv2 2 2

1 1 (2)(1.52) + UF - 2(9.81)(0.3366) = (2)(1.992)2 2 2 UF = 8.32 N # m

Ans.

Ans: v2 = 1.99 m>s UF = 8.32 N # m 588

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15–109. The elastic cord has an unstretched length l0 = 1.5 ft and a stiffness k = 12 lb>ft. It is attached to a fixed point at A and a block at B, which has a weight of 2 lb. If the block is released from rest from the position shown, determine its speed when it reaches point C after it slides along the smooth guide. After leaving the guide, it is launched onto the smooth horizontal plane. Determine if the cord becomes unstretched. Also, calculate the angular momentum of the block about point A, at any instant after it passes point C.

4 ft B

k  12 lb/ft

C

3 ft

A

Solution TB + VB = TC + VC 0 +

1 1 2 1 (12)(5 - 1.5)2 = a bv2C + (12)(3 - 1.5)2 2 2 32.2 2

Ans.

vC = 43.95 = 44.0 ft>s

There is a central force about A, and angular momentum about A is conserved. HA =

2 (43.95)(3) = 8.19 slug # ft 2 >s 32.2

Ans.

If cord is slack AD = 1.5 ft (HA)1 = (HA)2 8.19 =

2 (v ) (1.5) 32.2 u D

(vu)D = 88 ft>s But TC + VC = TD + VD 1 2 1 1 2 a b(43.95)2 + (12)(3 - 1.5)2 = a b(vD)2 + 0 2 32.2 2 2 32.2

vD = 48.6 ft>s

Ans.

Since vD 6 (vu)D cord will not unstretch.

Ans: vC = 44.0 ft>s

HA = 8.19 slug # ft 2 >s.

The cord will not unstretch. 589

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15–110. The amusement park ride consists of a 200-kg car and passenger that are traveling at 3 m>s along a circular path having a radius of 8 m. If at t = 0, the cable OA is pulled in toward O at 0.5 m>s, determine the speed of the car when t = 4 s. Also, determine the work done to pull in the cable.

O

Solution

r

A

Conservation of Angular Momentum. At t = 4 s, r2 = 8 - 0.5(4) = 6 m. (H0)1 = (H0)2 r1mv1 = r2m(v2)t 8[200(3)] = 6[200(v2)t] (v2)t = 4.00 m>s Here, (v2)t = 0.5 m>s. Thus v2 = 2(v2)2t + (v2)2r = 24.002 + 0.52 = 4.031 m>s = 4.03 m>s

Ans.

Principle of Work and Energy. T1 + ΣU1 - 2 = T2

1 1 (200) ( 32 ) + ΣU1 - 2 = (200)(4.031)2 2 2 ΣU1 - 2 = 725 J

Ans.

Ans: v2 = 4.03 m>s ΣU1 - 2 = 725 J 590

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15–111. A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of 1vA21 = 5 ft>s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft>s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box.

rB rA

1 ft

2 ft vr

4 ft/s B

A

SOLUTION (Hz)A = (Hz)B;

a

8 8 b(2)(5) = a b (1)(vB)tangent 32.2 32.2

(vA)1

5 ft/s

(nB)tangent = 10 ft>s vB = 2(10)2 + (4)2 = 10.77 = 10.8 ft>s a UAB = TB - TA

UAB =

Ans.

1 8 1 8 a b (10.77)2 - a b(5)2 2 32.2 2 32.2

UAB = 11.3 ft # lb

Ans.

Ans: vB = 10.8 ft>s UAB = 11.3 ft # lb 591

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*15–112. z

A toboggan and rider, having a total mass of 150 kg, enter horizontally tangent to a 90° circular curve with a velocity of vA = 70 km>h. If the track is flat and banked at an angle of 60°, determine the speed vB and the angle u of “descent,” measured from the horizontal in a vertical x–z plane, at which the toboggan exists at B. Neglect friction in the calculation.

A

vA

rA

70 km/h B

60 m

60 rB 55 m 90

SOLUTION

57 m 55 m

60

y

u vB

vA = 70 km>h = 19.44 m>s x

(HA)z = (HB)z (1)

150(19.44)(60) = 150(nB) cos u(57) Datum at B: TA + VA = TB + VB 1 1 (150)(19.44)2 + 150(9.81)h = (150)(nB)2 + 0 2 2

(2)

Since h = (rA - rB) tan 60° = (60 - 57) tan 60° = 5.196 Solving Eq. (1) and Eq (2): vB = 21.9 m>s

Ans.

u = 20.9

Ans.

Ans: vB = 21.9 m>s u = 20.9 592

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15–113. An earth satellite of mass 700 kg is launched into a freeflight trajectory about the earth with an initial speed of vA = 10 km>s when the distance from the center of the earth is rA = 15 Mm. If the launch angle at this position is fA = 70°, determine the speed vB of the satellite and its closest distance rB from the center of the earth. The earth has a mass Me = 5.976110242 kg. Hint: Under these conditions, the satellite is subjected only to the earth’s gravitational force, F = GMems>r2, Eq. 13–1. For part of the solution, use the conservation of energy.

vB rB

fA rA vA

SOLUTION (HO)1 = (HO)2 ms (vA sin fA)rA = ms (vB)rB 700[10(103) sin 70°](15)(106) = 700(vB)(rB)

(1)

TA + VA = TB + VB GMe ms GMems 1 1 ms (vA)2 = ms (vB)2 rA rB 2 2 66.73(10-12)(5.976)(1024)(700) 1 1 = (700)(vB)2 (700)[10(103)]2 2 2 [15(106)] -

66.73(10-12)(5.976)(1024)(700) rB

(2)

Solving, vB = 10.2 km>s

Ans.

rB = 13.8 Mm

Ans.

Ans: vB = 10.2 km>s rB = 13.8 Mm 593

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15–114. The fire boat discharges two streams of seawater, each at a flow of 0.25 m3>s and with a nozzle velocity of 50 m> s. Determine the tension developed in the anchor chain needed to secure the boat. The density of seawater is rsw = 1020 kg>m3.

30 45

60

SOLUTION Steady Flow Equation: Here, the mass flow rate of the sea water at nozzles A and dmB dmA = = rsw Q = 1020(0.25) = 225 kg>s. Since the sea water is colB are dt dt lected from the larger reservoir (the sea), the velocity of the sea water entering the control volume can be considered zero. By referring to the free-body diagram of the control volume (the boat),

+ ©F = dmA A v B + dmB A v B ; ; x A x B x dt dt T cos 60° = 225(50 cos 30°) + 225(50 cos 45°) Ans.

T = 40 114.87 N = 40.1 kN

Ans: T = 40.1 kN 594

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15–115. The chute is used to divert the flow of water, Q = 0.6 m3>s. If the water has a cross-sectional area of 0.05 m2, determine the force components at the pin D and roller C necessary for equilibrium. Neglect the weight of the chute and weight of the water on the chute. rw = 1 Mg>m3.

A

0.12 m

C

SOLUTION

2m

Equations of Steady Flow: Here, the flow rate Q = 0.6 m2>s. Then, Q 0.6 dm y = = = 12.0 m>s. Also, = rw Q = 1000 (0.6) = 600 kg>s. Applying A 0.05 dt Eqs. 15–26 and 15–28, we have dm (dDB yB - dDA yA); dt -Cx (2) = 600 [0 - 1.38(12.0)]

dm A youty - yiny B ; dt Dy = 600[0 - ( - 12.0)]

B

1.5 m

a + ©MA =

+ ©F = dm A y - y ); : x Bx Ax dt Dx + 4968 = 600 (12.0 - 0)

D

Cx = 4968 N = 4.97 kN

Dx = 2232N = 2.23 kN

Ans.

Ans.

+ c ©Fy = ©

Ans.

Dy = 7200 N = 7.20 kN

Ans: Cx = 4.97 kN Dx = 2.23 kN Dy = 7.20 kN 595

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*15–116. The 200-kg boat is powered by the fan which develops a slipstream having a diameter of 0.75 m. If the fan ejects air with a speed of 14 m/s, measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest.Assume that air has a constant density of ra = 1.22 kg/m3 and that the entering air is essentially at rest. Neglect the drag resistance of the water.

0.75 m

SOLUTION Equations of Steady Flow: Initially, the boat is at rest hence vB = va>b p dm = 14 m>s. Then, Q = vBA = 14 c A 0.752 B d = 6.185 m3>s and = raQ 4 dt = 1.22(6.185) = 7.546 kg>s. Applying Eq. 15–26, we have ©Fx =

dm (v - vAx); dt Bx

- F = 7.546(- 14 - 0)

F = 105.64 N

Equation of Motion : + ©F = ma ; : x x

105.64 = 200a

a = 0.528 m>s2

Ans.

Ans: a = 0.528 m>s2 596

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15–117. The nozzle discharges water at a constant rate of 2 ft3 >s. The cross-sectional area of the nozzle at A is 4 in2, and at B the cross-sectional area is 12 in2. If the static gauge pressure due to the water at B is 2 lb>in2, determine the magnitude of force which must be applied by the coupling at B to hold the nozzle in place. Neglect the weight of the nozzle and the water within it. gw = 62.4 lb>ft3.

B

Solution 62.4 dm b(2) = 3.876 slug>s = pQ = a dt 32.2 (vBx) =

Q 2 = = 24 ft>s  (vBy) = 0 AB 12>144

(vAy) =

Q 2 = 72 ft>s  (vAx) = 0 = AA 4>144

A

FB = pB AB = 2(12) = 24 lb Equations of steady flow: + dm S ΣFx = (v - vBx);  24 - Fx = 3.876(0 - 24)  Fx = 117.01 lb dt Ax + c ΣFy =

dm (v - vBy);  Fy = 3.876(72 - 0) = 279.06 lb dt Ay

F = 2F 2x + F 2y = 2117.012 + 279.062 = 303 lb

Ans.

Ans: F = 303 lb 597

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15–118. The blade divides the jet of water having a diameter of 4 in. If one-half of the water flows to the right while the other half flows to the left, and the total flow is Q = 1.5 ft3 >s, determine the vertical force exerted on the blade by the jet, gv = 62.4 lb>ft3.

4 in.

Solution dm 62.4 = rwQ = a b(1.5) = 2.9068 slug>s. The dt 32.2 Q 1.5 54 = = ft>s. Referring to the FBD of the velocity of the water jet is vJ = 2 2 p A ) p ( 12

Equation of Steady Flow. Here

control volume shown in Fig. a, + c ΣFy =

dm [(vB)y - (vA)y]; dt

F = 2.9068c 0 - a-

54 b d = 49.96 lb = 50.0 lb p

Ans.

Ans: F = 50.0 lb 598

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15–119. The blade divides the jet of water having a diameter of 3 in. If one-fourth of the water flows downward while the other three-fourths flows upwards, and the total flow is Q = 0.5 ft3>s, determine the horizontal and vertical components of force exerted on the blade by the jet, gw = 62.4 lb>ft3.

3 in.

SOLUTION Equations of Steady Flow: Here, the flow rate Q = 0.5 ft2>s. Then, Q 0.5 dm 62.4 y = = = rw Q = (0.5) = 0.9689 slug>s. = 10.19 ft>s. Also, p 3 2 A dt 32.2 A B 4

12

Applying Eq. 15–25 we have ©Fx = ©

dm A youts - yins B ; - Fx = 0 - 0.9689 (10.19) dt

©Fy = ©

1 dm 3 A youty - yiny B ; Fy = (0.9689)(10.19) + (0.9689)(- 10.19) dt 4 4

Fx = 9.87 lb

Ans.

Ans.

Fy = 4.93 lb

Ans: Fx = 9.87 lb Fy = 4.93 lb 599

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*15–120. The gauge pressure of water at A is 150.5 kPa. Water flows through the pipe at A with a velocity of 18 m>s, and out the pipe at B and C with the same velocity v. Determine the horizontal and vertical components of force exerted on the elbow necessary to hold the pipe assembly in equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of 50 mm at A, and at B and C the diameter is 30 mm. rw = 1000 kg>m3.

B

v

5

v

Solution

4

C 3

A

18 m/s

Continuity. The flow rate at B and C are the same since the pipe have the same diameter there. The flow rate at A is QA = vAAA = (18)[p ( 0.0252 ) ] = 0.01125p m3 >s

Continuity negatives that

QA = QB + QC;  0.01125p = 2Q Thus,

3              Q = 0.005625p m >s

vc = v B = Equation

of

Q 0.005625p = = 25 m>s A p ( 0.0152 )

Steady

Flow. The force due to the pressure at A is dmA P = rAAA = (150.5) ( 10 ) [p ( 0.0252 ) ] = 94.0625p N. = rwQA Here, dt dMc dmA = 1000(0.01125p) = 11.25p kg>s and = = rwQ = 1000(0.005625p) dt dt = 5.625p kg>s. 3

+

d ΣFx =

dmc dmB dmA (vB)x + (vc)x (vA)x; dt dt dt

4 Fx = (5.625p)(25) + (5.625p) c 25 a b d - (11.25p)(0) 5 = 795.22 N = 795 N

+ c ΣFy =

Ans.

dmC dmB dmA (v ) + (v ) (vA)y; dt B y dt C y dt

3 94.0625p - Fy = (5.625p)(0) + (5.625p) c -25a b d - (11.25p)(18) 5

  

   Fy = 1196.75 N = 1.20 kN

Ans.

Ans: Fx = 795 N Fy = 1.20 kN 600

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15–121. 2

n . If water vA = 12 ft>s and vertical w necessary Neglect the of the pipe. A and B the

vA

12 2ft/s

The gauge pressure of water at C is 40 lb >in . If water v B 5 A = 12 ft>s flows out of the pipe at A and B with velocities 4 v 3 and vB = 25 ft>s, determine the horizontal and vertical components of force exerted on the elbow necessary B to hold the pipe assembly in equilibrium. Neglect the A weight of water within the pipe and the weight of the pipe. The pipe has a diameter of 0.75 in. at C, and at A and B the diameter is 0.5 in. gw = 62.4 lb>ft3.

12 ft/s

vA

25 ft/s

4

vB

5

25 ft/s

3

B A

SOLUTION vC dmA 62.4 0.25 2 = b = 0.03171 slug>s (12)(p) a dt 32.2 12 C

vC C

dmB 62.4 0.25 2 = b = 0.06606 slug>s (25)(p) a dt 32.2 12 dm C = 0.03171 + 0.06606 = 0.09777 slug>s dt

25 2 b 2

vC AC = vA AA + vB AB vC(p) a

0.375 2 0.25 2 0.25 2 b = 12(p) a b + 25(p)a b 12 12 12

vC = 16.44 ft>s + ©F = dmB v + dmA v - dm C v : x dt B s dt As dt Cs 0.09777(16.44) 3 40(p)(0.375)2 - Fx = 0 - 0.03171(12)a b - 0.09777(16.44) 5 Ans. Fx = 19.5 lb + c ©Fy =

Ans.

dmB dmA dm C vBy + vAy v dt dt dt Cy

4 Fy = 0.06606(25) + 0.03171 a b (12) - 0 5 Ans. Fy = 1.9559 = 1.96 lb

Ans.

Ans: Fx = 19.5 lb Fy = 1.96 lb 601

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15–122. vA

The fountain shoots water in the direction shown. If the water is discharged at 30° from the horizontal, and the cross-sectional area of the water stream is approximately 2  in2, determine the force it exerts on the concrete wall at B. gw = 62.4 lb>ft3.

A

B

30 3 ft 20 ft

Solution + )    s = s + v t (S 0 0

20 = 0 + vA cos 30°t ( + c)

v = v0 + act

  -(vA sin 30° ) = (vA sin 30° ) - 32.2t Solving, t = 0.8469 s vA = vB = 27.27 ft>s At B: dm 62.4 2 = rvA = a b(27.27)a b = 0.7340 slug>s dt 32.2 144 + RΣF =

dm (v - vB) dt A

-F = 0.7340(0 - 27.27)   F = 20.0 lb

Ans.

Ans: F = 20.0 lb 602

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15–123. A plow located on the front of a locomotive scoops up snow at the rate of 10 ft3>s and stores it in the train. If the locomotive is traveling at a constant speed of 12 ft>s, determine the resistance to motion caused by the shoveling. The specific weight of snow is gs = 6 lb>ft3.

SOLUTION ©Fx = m

dm t dv + vD>t dt dt

F = 0 + (12 - 0) a

10(6) b 32.2 Ans.

F = 22.4 lb

Ans: F = 22.4 lb 603

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*15–124. The boat has a mass of 180 kg and is traveling forward on a river with a constant velocity of 70 km>h, measured relative to the river. The river is flowing in the opposite direction at 5 km>h. If a tube is placed in the water, as shown, and it collects 40 kg of water in the boat in 80 s, determine the horizontal thrust T on the tube that is required to overcome the resistance due to the water collection and yet maintain the constant speed of the boat. rw = 1 Mg>m3.

T

vR

5 km/h

SOLUTION 40 dm = = 0.5 kg>s dt 80 vD>t = (70)a ©Fs = m

1000 b = 19.444 m>s 3600

dm i dv + vD>i dt dt Ans.

T = 0 + 19.444(0.5) = 9.72 N

Ans: T = 9.72 N 604

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15–125. Water is discharged from a nozzle with a velocity of 12 m>s and strikes the blade mounted on the 20-kg cart. Determine the tension developed in the cord, needed to hold the cart stationary, and the normal reaction of the wheels on the cart. The nozzle has a diameter of 50 mm and the density of water is rw = 1000 kg>m3.

45 A

B

SOLUTION Steady Flow Equation: Here, the mass flow rate at sections A and B of the control p dm = rWQ = rWAv = 1000 c (0.052) d(12) = 7.5p kg>s volume is dt 4 Referring to the free-body diagram of the control volume shown in Fig. a, dm + : ©Fx = dt [(vB)x - (vA)x];

- Fx = 7.5p(12 cos 45° - 12) Fx = 82.81 N

dm [(vB)y - (vA)y]; + c ©Fy = dt

Fy = 7.5p(12 sin 45° - 0) Fy = 199.93 N

Equilibrium: Using the results of Fx and Fy and referring to the free-body diagram of the cart shown in Fig. b, + ©F = 0; : x

82.81 - T = 0

T = 82.8 N

Ans.

+ c ©Fy = 0;

N - 20(9.81) - 199.93 = 0

N = 396 N

Ans.

Ans: T = 82.8 N N = 396 N 605

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15–126. A snowblower having a scoop S with a cross-sectional area of As = 0.12 m3 is pushed into snow with a speed of vs = 0.5 m>s. The machine discharges the snow through a tube T that has a cross-sectional area of AT = 0.03 m2 and is directed 60° from the horizontal. If the density of snow is rs = 104 kg>m3, determine the horizontal force P required to push the blower forward, and the resultant frictional force F of the wheels on the ground, necessary to prevent the blower from moving sideways. The wheels roll freely.

P

T

60

Solution dm = rvsAs = (104)(0.5)(0.12) = 6.24 kg>s dt

S vx

F

6.24 dm 1 a b = a b = 2.0 m>s vs = dt rAr 104(0.03) ΣFx =

dm (v - vS2) dt T2

-F = 6.24( -2 cos 60° - 0) F = 6.24 N ΣFy =

Ans.

dm (v - vS2) dt T2

-P = 6.24(0 - 0.5) P = 3.12 N

Ans.

Ans: F = 6.24 N P = 3.12 N 606

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15–127. The fan blows air at 6000 ft3>min. If the fan has a weight of 30 lb and a center of gravity at G, determine the smallest diameter d of its base so that it will not tip over. The specific weight of air is g = 0.076 lb>ft3.

1.5 ft

G

0.5 ft

SOLUTION

4 ft

1 min 6000 ft3 b * a b = 100 ft3>s. Then, Equations of Steady Flow: Here Q = a min 60 s Q dm 100 0.076 y = = p = ra Q = (100) = 0.2360 slug>s. = 56.59 ft>s. Also, 2 A dt 32.2 (1.5 ) 4 Applying Eq. 15–26 we have a + ©MO =

dm d (dOB yB - dOA yA B ; 30 a 0.5 + b = 0.2360 [4(56.59) - 0] dt 2

d

Ans.

d = 2.56 ft

Ans: d = 2.56 ft 607

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*15–128. The nozzle has a diameter of 40 mm. If it discharges water uniformly with a downward velocity of 20 m>s against the fixed blade, determine the vertical force exerted by the water on the blade. rw = 1 Mg>m3. 40 mm

Solution dm = rvA = (1000)(20)(p)(0.02)2 = 25.13 kg>s dt + c ΣFy =

45

45

dm (v - vAy ) dt B[placeholder]

F = (25.13)(20 sin 45° - ( -20)) Ans.

F = 858 N

Ans: F = 858 N 608

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15–129. 250 mm

The water flow enters below the hydrant at C at the rate of 0.75 m3>s. It is then divided equally between the two outlets at A and B. If the gauge pressure at C is 300 kPa, determine the horizontal and vertical force reactions and the moment reaction on the fixed support at C. The diameter of the two outlets at A and B is 75 mm, and the diameter of the inlet pipe at C is 150 mm. The density of water is rw = 1000 kg>m3. Neglect the mass of the contained water and the hydrant.

30 A

B

650 mm

600 mm

SOLUTION Free-Body Diagram: The free-body diagram of the control volume is shown in Fig. a. The force exerted on section A due to the water pressure is FC = pCAC = p 300(103)c A 0.152 B d = 5301.44 N. The mass flow rate at sections A, B, and C, are 4 Q dmC dmB dmA 0.75 = = rW a b = 1000 a b = 375 kg>s = rWQ = and dt dt 2 2 dt 1000(0.75) = 750 kg>s.

C

The speed of the water at sections A, B, and C are vA = vB =

Q>2 0.75>2 = = 84.88 m>s p AA (0.0752) 4

vC =

Q 0.75 = = 42.44 m>s. p AC (0.152) 4

Steady Flow Equation: Writing the force steady flow equations along the x and y axes, + ©F = dmA (v ) + dmB (v ) - dm C (v ) ; : x A x B x C x dt dt dt Cx = - 375(84.88 cos 30°) + 375(84.88) - 0 Ans.

Cx = 4264.54 N = 4.26 kN dmA dmB dmC (vA)y + (vB)y (vC)y; + c ©Fy = dt dt dt - Cy + 5301.44 = 375(84.88 sin 30°) + 0 - 750(42.44)

Ans.

Cy = 21 216.93 N = 2.12 kN Writing the steady flow equation about point C, dmA dmB dmC dvA + dvB dvC; + ©MC = dt dt dt -MC = 375(0.65)(84.88 cos 30°) - 375(0.25)(84.88 sin 30°) + [- 375(0.6)(84.88)] - 0 MC = 5159.28 N # m = 5.16 kN # m

Ans.

Ans: Cx = 4.26 kN Cy = 2.12 kN MC = 5.16 kN # m 609

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15–130. Sand drops onto the 2-Mg empty rail car at 50 kg>s from a conveyor belt. If the car is initially coasting at 4 m>s, determine the speed of the car as a function of time.

4 m/s

Solution Gains Mass System. Here the sand drops vertically onto the rail car. Thus (vi)x = 0. Then VD = Vi + VD>i + ) v = (vi)x + (vD>i)x (S   v = 0 + (vD>i)x (vD>i)x = v Also,

dmi = 50 kg>s and m = 2000 + 50t dt

ΣFx = m

dmi dv + (vD>i)x ; dt dt

0 = (2000 + 50t)

dv + v(50) dt

dv 50 dt = v 2000 + 50t Integrate this equation with initial condition v = 4 m>s at t = 0. t

v

dv dt = - 50 L0 2000 + 50t L4 m>s v ln v `

v 4 m>s

ln

= - ln (2000 + 50t) `

t 0

v 2000 = ln a b 4 2000 + 50t v 2000 = 4 2000 + 50t v = e

8000 f m>s 2000 + 50t

Ans.

Ans: v = e 610

8000 f m>s 2000 + 50t

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15–131. B

Sand is discharged from the silo at A at a rate of 50 kg>s with a vertical velocity of 10 m>s onto the conveyor belt, which is moving with a constant velocity of 1.5 m>s. If the conveyor system and the sand on it have a total mass of 750 kg and center of mass at point G, determine the horizontal and vertical components of reaction at the pin support B roller support A. Neglect the thickness of the conveyor.

1.5 m/s

G

A

SOLUTION

10 m/s

Steady Flow Equation: The moment steady flow equation will be written about point B to eliminate Bx and By. Referring to the free-body diagram of the control volume shown in Fig. a, + ©MB =

dm (dvB - dvA); dt

30

4m

4m

750(9.81)(4) - Ay(8) = 50[0 - 8(5)] Ay = 4178.5 N = 4.18 kN

Ans.

Writing the force steady flow equation along the x and y axes, + ©F = dm [(v ) - (v ) ]; : x B x A x dt

+ c ©Fy =

dm [(vB)y - (vA)y]; dt

-Bx = 50(1.5 cos 30° - 0) Bx = | - 64.95 N| = 65.0 N :

Ans.

By + 4178.5 - 750(9.81) = 50[1.5 sin 30° - ( -10)] By = 3716.25 N = 3.72 kN c

Ans.

Ans: Ay = 4.18 kN Bx = 65.0 N S By = 3.72 kN c 611

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*15–132. Sand is deposited from a chute onto a conveyor belt which is moving at 0.5 m>s. If the sand is assumed to fall vertically onto the belt at A at the rate of 4 kg>s, determine the belt tension FB to the right of A. The belt is free to move over the conveyor rollers and its tension to the left of A is FC = 400 N.

0.5 m/s FC = 400 N

A

FB

Solution + ) ΣFx = (S

dm (v - vAx) dt Bx

FB - 400 = 4(0.5 - 0) Ans.

FB = 2 + 400 = 402 N

Ans: FB = 402 N 612

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15–133. The tractor together with the empty tank has a total mass of 4 Mg. The tank is filled with 2 Mg of water. The water is discharged at a constant rate of 50 kg>s with a constant velocity of 5 m>s, measured relative to the tractor. If the tractor starts from rest, and the rear wheels provide a resultant traction force of 250 N, determine the velocity and acceleration of the tractor at the instant the tank becomes empty.

F

SOLUTION The free-body diagram of the tractor and water jet is shown in Fig. a.The pair of thrust T cancel each other since they are internal to the system. The mass of the tractor and the tank at any instant t is given by m = A 4000 + 2000 B - 50t = A 6000 - 50t B kg. + ©F = m dv - v dm e ; ; s D>e dt dt

250 = A 6000 - 50t B a =

The time taken to empty the tank is t =

dv - 5(50) dt

dv 10 = dt 120 - t

(1)

2000 = 40 s. Substituting the result of t 50

into Eq. (1), 10 = 0.125m>s2 120 - 40

a =

Ans.

Integrating Eq. (1), 40 s

v

L0

dv =

L0

10 dt 120 - t

v = -10 ln A 120 - t B 2

40 s 0

Ans.

= 4.05 m>s

Ans: a = 0.125 m>s2 v = 4.05 m>s 613

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15–134. A rocket has an empty weight of 500 lb and carries 300 lb of fuel. If the fuel is burned at the rate of 15 lb>s and ejected with a relative velocity of 4400 ft>s, determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket. v

Solution + c ΣFs = m

dm e dv - v[placeholder] dt dt

At n time t, m = m0 - ct, where c = 0 = (m0 - ct) L0

v dv =

L0

1

a

dme . In space the weight of the rocket is zero. dt

dv - v[placeholder][placeholder] dt

cv[placeholder] m 0 - ct

v = v[placeholder] lna

bdt

m0 b m0 - ct

(1)

The maximum speed occurs when all the fuel is consumed, that is, when 300 500 + 300 15 t = = 20 s. Here, m0 = = 24.8447 slug, c = = 0.4658 slug>s, 15 32.2 32.2 v[placeholder] = 4400 ft>s. Substitute the numerical values into Eq. (1); vmax = 4400 ln a

24.8447 b 24.8447 - 0.4658(20)

vmax = 2068 ft>s = 2.07 ( 103 ) ft>s

Ans.

Ans: vmax = 2.07 ( 103 ) ft>s 614

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15–135. A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of 6 m>s through an intake unit A, which has a cross-sectional area of A A = 0.25 m2, and then discharging it at the ground, B, where the cross-sectional area is A B = 0.35 m2. If air at A is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of 15 kg with center of mass at G. Assume that air has a constant density of ra = 1.22 kg>m3.

vA A G B

SOLUTION dm = r A A vA = 1.22(0.25)(6) = 1.83 kg>s dt + c ©Fy =

dm ((vB)y - (vA)y) dt

pressure = (0.35) - 15(9.81) = 1.83(0 - ( -6)) Ans.

pressure = 452 Pa

Ans: 452 Pa 615

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*15–136. The rocket car has a mass of 2 Mg (empty) and carries 120 kg of fuel. If the fuel is consumed at a constant rate of 6 kg>s and ejected from the car with a relative velocity of 800 m>s, determine the maximum speed attained by the car starting from rest. The drag resistance due to the atmosphere is FD = ( 6.8v2 ) N, where v is the speed in m>s.

v

Solution ΣFs = m

dv - v[placeholder] [placeholder] dt

At time t 1 the mass of the car is m0 - ct 1 where c =

dmc = 6 kg>s dt

Set F = kv2, then - kv2 = (m0 - ct) t

dv - vD>ec dt

L0 ( cvD>e - kv2 ) dv

t

=

dt (m L0 0 - ct)

cvD>e a

1 22cvD>ek

bln≥

A k cvD>e A k

cvD>e

a

1 22cvD>e

A k bln ± cvD>e k A k

+ v - v + v - v

v t 1 ¥ = - ln(m0 - ct) ` c 0 0

m0 - ct 1 ≤ = - ln a b c m0

Maximum speed occurs at the instant the fuel runs out t =

120 = 20 s 6

Thus, (6)(800) °

1 22(6)(800)(6.8)

Solving,

¢ ln ±

A

6.8 (6)(800)

A

6.8

+ v - v

2120 - 6(20) 1 ≤ = - lna b 6 2120

Ans.

v = 25.0 m>s

Ans: v = 25.0 m>s 616

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15–137. If the chain is lowered at a constant speed v = 4 ft>s, determine the normal reaction exerted on the floor as a function of time. The chain has a weight of 5 lb>ft and a total length of 20 ft.

v  4 ft/s

Solution At time t, the weight of the chain on the floor is W = mg(vt) dv = 0, mi = m(vt) dt

20 ft

dmi = mv dt ΣFs = m

dmi dv + vD>i dt dt

R - mg(vt) = 0 + v(mv) R = m(gvt + v2) R =

5 ( 32.2(4)(t) + (4)2 ) 32.2 Ans.

R = (20t + 2.48) lb

Ans: R = {20t + 2.48} lb 617

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15–138. The second stage of a two-stage rocket weighs 2000 lb (empty) and is launched from the first stage with a velocity of 3000 mi>h. The fuel in the second stage weighs 1000 lb. If it is consumed at the rate of 50 lb>s and ejected with a relative velocity of 8000 ft>s, determine the acceleration of the second stage just after the engine is fired. What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of gravitation.

SOLUTION Initially, ©Fs = m 0 =

dme dv - v D>e a b dt dt

50 3000 a - 8000a b 32.2 32.2

a = 133 ft>s2

Ans.

Finally, 0 =

50 2000 a - 8000 a b 32.2 32.2

a = 200 ft>s2

Ans.

Ans: ai = 133 ft>s2 af = 200 ft>s2 618

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15–139. The missile weighs 40 000 lb. The constant thrust provided by the turbojet engine is T = 15 000 lb. Additional thrust is provided by two rocket boosters B. The propellant in each booster is burned at a constant rate of 150 lb/s, with a relative exhaust velocity of 3000 ft>s. If the mass of the propellant lost by the turbojet engine can be neglected, determine the velocity of the missile after the 4-s burn time of the boosters.The initial velocity of the missile is 300 mi/h.

T B

SOLUTION + ©F = m dv - v dme : s D>e dt dt At a time t, m = m0 - ct, where c = T = (m0 - ct) v

Lv0

t

dv =

v = a

L0

a

dv - vD>e c dt

T + cvD>e m0 - ct

T + cvD>e c

Here, m0 = t = 4 s, v0 =

dme . dt

b1n a

b dt

m0 b + y0 m0 - ct

(1)

40 000 150 = 1242.24 slug, c = 2a b = 9.3168 slug>s, vD>e = 3000 ft>s, 32.2 32.2 300(5280) = 440 ft>s. 3600

Substitute the numerical values into Eq. (1): vmax = a

15 000 + 9.3168(3000) 1242.24 b 1n a b + 440 9.3168 1242.24 - 9.3168(4) Ans.

vmax = 580 ft s

Ans: vmax = 580 ft>s 619

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*15–140. The jet is traveling at a speed of 720 km >h. If the fuel is being spent at 0.8 kg>s, and the engine takes in air at 200  kg >s, whereas the exhaust gas (air and fuel) has a relative speed of 12 000 m>s, determine the acceleration of the plane at this instant. The drag resistance of the air is FD = (55 v2), where the speed is measured in m>s. The jet has a mass of 7 Mg.

720 km/h

Solution Since the mass enters and exits the plane at the same time, we can combine Eqs. 15-29 and 15-30 which resulted in ΣFs = m

dme dm i dv - vD>e + vD>i dt dt dt

dme dv = a, vD>e = 12000 m>s, = 0.8 + 200 = 200.8 kg>s dt dt km 1000 m 1h ba ba b = 200 m>s, vD>i = v = 200 m>s, v = a720 h 1 km 3600 s

Here m = 7000 kg,

dmi = 200 kg>s dt

and FD = 55 ( 2002 ) = 2.2 ( 106 ) N. Referring to the FBD of the jet, Fig. a + ) ΣFs = m (d

dme dmi dv - vD>e + vD>i ; dt dt dt

- 2.2 ( 106 ) = 7000a - 12000(200.8) + 200(200)     a = 24.23 m>s2 = 24.2 m>s2

Ans.

Ans: a = 24.2 m>s2 620

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15–141. The rope has a mass m¿ per unit length. If the end length y = h is draped off the edge of the table, and released, determine the velocity of its end A for any position y, as the rope uncoils and begins to fall.

yh A

SOLUTION + T ©Fs = m

dmi dv + vD>i dt dt

At a time t, m = m¿y and m¿gy = m¿y gy = y

m¿dy dmi dv = = m¿v. Here, vD>i = v, = g. dt dt dt dv + v(m¿v) dt

dv + v2 dt

gy = vy

since v =

dy dy , then dt = dt v

dv + v2 dy

Multiply both sides by 2ydy 2gy2 dy = 2vy2 dv + 2yv2 dy L

2gy2 dy =

2 3 3 gy

L

d A v 2y2 B

+ C = v2y2

v = 0 at y = h 2 3 2 3 3 gy - 3 gh

v =

2 3 3 gh

+ C = 0

C = - 23 gh3

= v 2y 2

2 y3 - h3 b ga C3 y2

Ans.

Ans: v = 621

y3 - h3 2 ga b C3 y2

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15–142. The 12-Mg jet airplane has a constant speed of 950 km>h when it is flying along a horizontal straight line. Air enters the intake scoops S at the rate of 50 m3>s. If the engine burns fuel at the rate of 0.4 kg>s and the gas (air and fuel) is exhausted relative to the plane with a speed of 450 m>s, determine the resultant drag force exerted on the plane by air resistance. Assume that air has a constant density of 1.22 kg>m3. Hint: Since mass both enters and exits the plane, Eqs. 15–28 and 15–29 must be combined to yield dme dmi dv ©Fs = m - vD>e + vD>i . dt dt dt

v

950 km/h

S

SOLUTION ©Fs = m

dm i dm e dv (vD>E) + (v ) dt dt dt D>i

v = 950 km>h = 0.2639 km>s,

(1)

dv = 0 dt

vD>E = 0.45 km>s vD>t = 0.2639 km>s dm t = 50(1.22) = 61.0 kg>s dt dm e = 0.4 + 61.0 = 61.4 kg>s dt Forces T and R are incorporated into Eq. (1) as the last two terms in the equation. + ) - F = 0 - (0.45)(61.4) + (0.2639)(61) (; D Ans.

FD = 11.5 kN

Ans: FD = 11.5 kN 622

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15–143. The jet is traveling at a speed of 500 mi>h, 30° with the horizontal. If the fuel is being spent at 3 lb>s,and the engine takes in air at 400 lb>s, whereas the exhaust gas (air and fuel) has a relative speed of 32 800 ft>s, determine the acceleration of the plane at this instant. The drag resistance 2 of the air is FD = 10.7v 2 lb, where the speed is measured in ft>s.The jet has a weight of 15 000 lb. Hint: See Prob. 15–142.

500 mi/ h

30

SOLUTION dmi 400 = = 12.42 slug>s dt 32.2 dme 403 = = 12.52 slug>s dt 32.2 v = vD>i = 500 mi>h = 733.3 ft>s a + ©Fs = m

dme dmi dv - vD>e + vD>i dt dt dt

- (15 000) sin 30° - 0.7(733.3)2 = a =

15 000 dv - 32 800(12.52) + 733.3(12.42) 32.2 dt

dv = 37.5 ft>s2 dt

Ans.

Ans: a = 37.5 ft>s2 623

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*15–144. A four-engine commercial jumbo jet is cruising at a constant speed of 800 km>h in level flight when all four engines are in operation. Each of the engines is capable of discharging combustion gases with a velocity of 775 m>s relative to the plane. If during a test two of the engines, one on each side of the plane, are shut off, determine the new cruising speed of the jet. Assume that air resistance (drag) is proportional to the square of the speed, that is, FD = cv2, where c is a constant to be determined. Neglect the loss of mass due to fuel consumption.

SOLUTION Steady Flow Equation: Since the air is collected from a large source (the atmosphere), its entrance speed into the engine is negligible. The exit speed of the air from the engine is + b a:

ve + vp + ve>p

When the four engines are in operation, the airplane has a constant speed of m 1h b = 222.22 m>s. Thus, vp = c 800(103) d a h 3600 s + b a:

ve = - 222.22 + 775 = 552.78 m>s :

Referring to the free-body diagram of the airplane shown in Fig. a, + ©F = dm C A v B - A v B D ; : x B x A x dt

C(222.222) = 4

dm (552.78 - 0) dt

C = 0.044775

dm dt

When only two engines are in operation, the exit speed of the air is + b a:

ve = - vp + 775

Using the result for C, + ©F = dm : x dt

C A vB B x - A vA B x D ; ¢ 0.044775

dm dm ≤ A vp 2 B = 2 C -vp + 775 B - 0 D dt dt 0.044775vp 2 + 2vp - 1550 = 0

Solving for the positive root, Ans.

vp = 165.06 m s = 594 km h

Ans: vP = 594 km>h 624

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15–145. a

The 10-Mg helicopter carries a bucket containing 500 kg of water, which is used to fight fires. If it hovers over the land in a fixed position and then releases 50 kg>s of water at 10 m>s, measured relative to thehelicopter, determine the initial upward accelerationthe helicopter experiences as the water is being released.

SOLUTION + c ΣFs = m

dme dv - vD>e dt dt

Initially, the bucket is full of water, hence m = 10 ( 103 ) + 0.5 ( 103 ) = 10.5 ( 103 ) kg 0 = 10.5 ( 103 ) a - (10)(50) a = 0.0476 m>s2

Ans.

Ans: a = 0.0476 m>s2 625

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15–146. A rocket has an empty weight of 500 lb and carries 300 lb of fuel. If the fuel is burned at the rate of 1.5 lb/s and ejected with a velocity of 4400 ft/s relative to the rocket, determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket.

SOLUTION dme dv - vD>e dt dt

+ c ©Fs =

At a time t, m = m0 - ct, where c = 0 = (m0 - ct) v

L0

t

dv =

L0

v = vD> e ln ¢

¢

dme . In space the weight of the rocket is zero. dt

dv - vD>e c dt cvD>e m0 - ct

≤ dt

m0 ≤ m0 - ct

(1)

The maximum speed occurs when all the fuel is consumed, that is, when t = 300 1.5 = 200 s. Here, m0 =

500 + 300 32.2

= 24.8447 slug, c =

1.5 32.2

= 0.04658 slug>s, vD>e = 4400 ft>s.

Substitute the numerical into Eq. (1): vmax = 4400 ln a

24.8447 b 24.8447 - (0.04658(200)) Ans.

vmax = 2068 ft>s

Ans: v max = 2.07 ( 103 ) ft>s 626

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15–147. Determine the magnitude of force F as a function of time, which must be applied to the end of the cord at A to raise the hook H with a constant speed v = 0.4 m>s. Initially the chain is at rest on the ground. Neglect the mass of the cord and the hook. The chain has a mass of 2 kg>m.

A

H v

0.4 m/s

SOLUTION dv = 0, dt

y = vt

mi = my = mvt dmi = mv dt + c ©Fs = m

dm i dv + vD>i ( ) dt dt

F - mgvt = 0 + v(mv) F = m(gvt + v2) = 2[9.81(0.4)t + (0.4)2] Ans.

F = (7.85t + 0.320) N

Ans: F = {7.85t + 0.320} N 627

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*15–148. The truck has a mass of 50 Mg when empty.When it is unloading 5 m3 of sand at a constant rate of 0.8 m3>s, the sand flows out the back at a speed of 7 m> s, measured relative to the truck, in the direction shown. If the truck is free to roll, determine its initial acceleration just as the load begins to empty. Neglect the mass of the wheels and any frictional resistance to motion. The density of sand is rs = 1520 kg>m3.

a

45 7 m/s

SOLUTION A System That Loses Mass: Initially, the total mass of the truck is dme and m = 50(103) + 5(1520) = 57.6(103) kg = 0.8(1520) = 1216 kg>s. dt Applying Eq. 15–29, we have + ©F = m dv - v dme ; : s D>e dt dt

0 = 57.6(103)a - (0.8 cos 45°)(1216) a = 0.104 m>s2

Ans.

Ans: a = 0.104 m>s2 628

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15–149. v

The car has a mass m0 and is used to tow the smooth chain having a total length l and a mass per unit of length m¿. If the chain is originally piled up, determine the tractive force F that must be supplied by the rear wheels of the car, necessary to maintain a constant speed v while the chain is being drawn out. F

SOLUTION + ©F = m dv + v dmi : s D>i dt dt At a time t, m = m0 + ct, where c =

Here, vD>i = v,

dmi m¿dx = = m¿v. dt dt

dv = 0. dt F = (m0 - m¿v )(0) + v(m¿v) = m¿v2

Ans.

Ans: F = m′v2 629

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16–1. The angular velocity of the disk is defined by v = 15t2 + 22 rad>s, where t is in seconds. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = 0.5 s.

A

0.8 m

SOLUTION v = (5 t2 + 2) rad>s a =

dv = 10 t dt

t = 0.5 s v = 3.25 rad>s a = 5 rad>s2 Ans.

vA = vr = 3.25(0.8) = 2.60 m>s a z = ar = 5(0.8) = 4 m>s2 a n = v2r = (3.25)2(0.8) = 8.45 m>s2 a A = 2(4)2 + (8.45)2 = 9.35 m>s2

Ans.

Ans: vA = 2.60 m>s aA = 9.35 m>s2 630

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16–2. The angular acceleration of the disk is defined by a = 3t 2 + 12 rad>s, where t is in seconds. If the disk is originally rotating at v 0 = 12 rad>s, determine the magnitude of the velocity and the n and t components of acceleration of point A on the disk when t = 2 s.

v0  12 rad/s B 0.4 m 0.5 m

A

Solution Angular Motion. The angular velocity of the disk can be determined by integrating dv = a dt with the initial condition v = 12 rad>s at t = 0. v

L12 rad>s

dv =

L0

2s

(3t 2 + 12)dt

v - 12 = (t 3 + 12t) 2

2s 0

v = 44.0 rad>s Motion of Point A. The magnitude of the velocity is

Ans.

vA = vrA = 44.0(0.5) = 22.0 m>s

At t = 2 s, a = 3 ( 22 ) + 12 = 24 rad>s2. Thus, the tangential and normal components of the acceleration are

(aA ) t = arA = 24(0.5) = 12.0 m>s2

Ans.



(aA ) n = v2rA = ( 44.02 ) (0.5) = 968 m>s2

Ans.

Ans: vA = 22.0 m>s ( aA ) t = 12.0 m>s2 ( aA ) n = 968 m>s2 631

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16–3. The disk is originally rotating at v0 = 12 rad>s. If  it is subjected to a constant angular acceleration of a = 20 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point A at the instant t = 2 s.

v0  12 rad/s B 0.4 m 0.5 m

A

Solution Angular Motion. The angular velocity of the disk can be determined using v = v0 + act;

v = 12 + 20(2) = 52 rad>s

Motion of Point A. The magnitude of the velocity is

Ans.

vA = vrA = 52(0.5) = 26.0 m>s

The tangential and normal component of acceleration are (aA)t = ar = 20(0.5) = 10.0 m>s2

Ans.

(aA)n = v2r = ( 522 ) (0.5) = 1352 m>s2

Ans.

Ans: vA = 26.0 m>s ( aA ) t = 10.0 m>s2 ( aA ) n = 1352 m>s2 632

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*16–4. The disk is originally rotating at v0 = 12 rad>s. If it is  subjected to a constant angular acceleration of a = 20 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point B when the disk undergoes 2 revolutions.

v0  12 rad/s B 0.4 m 0.5 m

A

Solution Angular Motion. The angular velocity of the disk can be determined using v2 = v20 + 2ac(u - u0);

v2 = 122 + 2(20)[2(2p) - 0] v = 25.43 rad>s

Motion of Point B. The magnitude of the velocity is

Ans.

vB = vrB = 25.43(0.4) = 10.17 m>s = 10.2 m>s

The tangential and normal components of acceleration are (aB)t = arB = 20(0.4) = 8.00 m>s2

Ans.

(aB)n = v2rB = ( 25.432 ) (0.4) = 258.66 m>s2 = 259 m>s2

Ans.

Ans: vB = 10.2 m>s (aB)t = 8.00 m>s2 (aB)n = 259 m>s2 633

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16–5. The disk is driven by a motor such that the angular position of the disk is defined by u = 120t + 4t22 rad, where t is in seconds. Determine the number of revolutions, the angular velocity, and angular acceleration of the disk when t = 90 s.

0.5 ft

θ

SOLUTION Angular Displacement: At t = 90 s. u = 20(90) + 4 A 902 B = (34200 rad) * ¢

1 rev ≤ = 5443 rev 2p rad

Ans.

Angular Velocity: Applying Eq. 16–1. we have v =

du = 20 + 8t 2 = 740 rad>s dt t = 90 s

Ans.

Angular Acceleration: Applying Eq. 16–2. we have a =

dv = 8 rad s2 dt

Ans.

Ans: u = 5443 rev v = 740 rad>s a = 8 rad>s2 634

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16–6. A wheel has an initial clockwise angular velocity of 10 rad>s and a constant angular acceleration of 3 rad>s2. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity of 15 rad>s. What time is required?

SOLUTION v2 = v20 + 2ac(u - u0) (15)2 = (10)2 + 2(3)(u - 0) u = 20.83 rad = 20.83 ¢

1 ≤ = 3.32 rev. 2p

Ans.

v = v0 + ac t 15 = 10 + 3t Ans.

t = 1.67 s

Ans: u = 3.32 rev t = 1.67 s 635

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16–7. If gear A rotates with a constant angular acceleration of aA = 90 rad>s2, starting from rest, determine the time required for gear D to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear D to attain this angular velocity. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

F

SOLUTION

A

B C

Gear B is in mesh with gear A. Thus, aB rB = aA rA rA 15 aB = a b aA = a b (90) = 27 rad>s2 rB 50 Since gears C and B share the same shaft, aC = aB = 27 rad>s2. Also, gear D is in mesh with gear C. Thus, aD rD = aC rC rC 25 aD = a b aC = a b (27) = 9 rad>s2 rD 75 600 rev 2p rad 1 min ba ba b = min 1 rev 60 s 20p rad>s. Applying the constant acceleration equation,

The final angular velocity of gear D is vD = a

vD = (vD)0 + aD t 20p = 0 + 9t Ans.

t = 6.98 s and vD2 = (vD)02 + 2aD [uD - (uD)0] (20p)2 = 02 + 2(9)(uD - 0) uD = (219.32 rad)a

1 rev b 2p rad Ans.

= 34.9 rev

Ans: t = 6.98 s uD = 34.9 rev 636

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*16–8. If gear A rotates with an angular velocity of vA = (uA + 1) rad>s, where uA is the angular displacement of gear A, measured in radians, determine the angular acceleration of gear D when uA = 3 rad, starting from rest. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

F

SOLUTION

A

B C

Motion of Gear A: aA duA = vA dvA aA duA = (uA + 1) d(uA + 1) aA duA = (uA + 1) duA aA = (uA + 1) At uA = 3 rad, aA = 3 + 1 = 4 rad>s2 Motion of Gear D: Gear A is in mesh with gear B. Thus, aB rB = aA rA rA 15 aB = a b aA = a b (4) = 1.20 rad>s2 rB 50 Since gears C and B share the same shaft aC = aB = 1.20 rad>s2. Also, gear D is in mesh with gear C. Thus, aD rD = aC rC rC 25 aD = a b aC = a b (1.20) = 0.4 rad>s2 rD 75

Ans.

Ans: aD = 0.4 rad>s2 637

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16–9. At the instant vA = 5 rad>s, pulley A is given an angular acceleration a = (0.8u) rad>s2, where u is in radians. Determine the magnitude of acceleration of point B on pulley C when A rotates 3 revolutions. Pulley C has an inner hub which is fixed to its outer one and turns with it.

A

vA aA 50 mm

Solution Angular Motion. The angular velocity of pulley A can be determined by integrating v dv = a du with the initial condition vA = 5 rad>s at uA = 0. vA

L5 rad>s

v dv =

v2 vA ` 2 5 rad>s

40 mm

C

L0

B 60 mm

uA

0.8udu

= ( 0.4u 2 ) `

uA 0

v2A 52 = 0.4u 2A 2 2

vA = e 20.8u 2A + 25 f rad>s

At uA = 3(2p) = 6p rad,

vA = 20.8(6p)2 + 25 = 17.585 rad>s aA = 0.8(6p) = 4.8p rad>s2

Since pulleys A and C are connected by a non-slip belt, vCrC = vArA;

vC(40) = 17.585(50) vC = 21.982 rad>s

aCrC = aArA;

aC(40) = (4.8p)(50) aC = 6p rad>s2



Motion of Point B. The tangential and normal components of acceleration of point B can be determined from (aB)t = aCrB = 6p(0.06) = 1.1310 m>s2 (aB)n = v2CrB = ( 21.9822 ) (0.06) = 28.9917 m>s2 Thus, the magnitude of aB is



aB = 2(aB)2t + (aB)2n = 21.13102 + 28.99172

= 29.01 m>s2 = 29.0 m>s2

Ans.

Ans: aB = 29.0 m>s2 638

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16–10. At the instant vA = 5 rad>s, pulley A is given a constant angular acceleration aA = 6 rad>s2. Determine the magnitude of acceleration of point B on pulley C when A rotates 2 revolutions. Pulley C has an inner hub which is fixed to its outer one and turns with it.

A

vA aA 50 mm

Solution Angular Motion. Since the angular acceleration of pulley A is constant, we can apply

(

B 60 mm

2 vA 0

) + 2aA[uA - ( uA ) 0] ;



v2A



v2A = 52 + 2(6)[2(2p) - 0]



vA = 13.2588 rad>s

=

40 mm

C

Since pulleys A and C are connected by a non-slip belt,

vCrC = vArA  ;   vC(40) = 13.2588(50) vC = 16.5735 rad>s



aCrC = aArA  ;      aC(40) = 6(50) aC = 7.50 rad>s2



Motion of Point B. The tangential and normal component of acceleration of point B can be determined from

( aB ) t = aCrB = 7.50(0.06) = 0.450 m>s2



( aB ) n = v2CrB = ( 16.57352 ) (0.06) = 16.4809 m>s2

Thus, the magnitude of aB is

aB = 2 ( aB ) 2t + ( aB ) 2n = 20.4502 + 16.48092

= 16.4871 m>s2 = 16.5 m>s2

Ans.

Ans: aB = 16.5 m>s2 639

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16–11. The cord, which is wrapped around the disk, is given an acceleration of a = (10t) m>s2, where t is in seconds. Starting from rest, determine the angular displacement, angular velocity, and angular acceleration of the disk when t = 3 s.

a  (10t) m/s2

0.5 m

Solution Motion of Point P. The tangential component of acceleration of a point on the rim is equal to the acceleration of the cord. Thus

( at ) = ar  ;       10t = a(0.5)

When t = 3 s,

a = 520t6 rad>s2



a = 20(3) = 60 rad>s2



Ans.

Angular Motion. The angular velocity of the disk can be determined by integrating dv = a dt with the initial condition v = 0 at t = 0.

L0

v

dv =

L0

t

20t dt

When t = 3 s,

v = 510t 2 6 rad>s



v = 10 ( 32 ) = 90.0 rad>s



Ans.

The angular displacement of the disk can be determined by integrating du = v dt with the initial condition u = 0 at t = 0. When t = 3 s,

L0

u

du =

L0

u = e u =

t

10t 2 dt

10 3 t f rad 3

10 3 ( 3 ) = 90.0 rad 3

Ans.

Ans: a = 60 rad>s2 v = 90.0 rad>s u = 90.0 rad 640

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*16–12. The power of a bus engine is transmitted using the belt-andpulley arrangement shown. If the engine turns pulley A at vA = (20t + 40) rad>s, where t is in seconds, determine the angular velocities of the generator pulley B and the air-conditioning pulley C when t = 3 s.

100 mm 25 mm vB D 75 mm

B vC

Solution

vA

50 mm

A C

When t = 3 s

vA = 20(3) + 40 = 100 rad>s

The speed of a point P on the belt wrapped around A is

vP = vArA = 100(0.075) = 7.5 m>s



vB =

vP 7.5 = = 300 rad>s rD 0.025

Ans.

The speed of a point P′ on the belt wrapped around the outer periphery of B is

v′p = vBrB = 300(0.1) = 30 m>s

Hence,    vC =

v′P 30 = = 600 rad>s rC 0.05

Ans.

Ans: vB = 300 rad>s vC = 600 rad>s 641

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16–13. The power of a bus engine is transmitted using the belt-andpulley arrangement shown. If the engine turns pulley A at vA = 60 rad>s, determine the angular velocities of the generator pulley B and the air-conditioning pulley C. The hub at D is rigidly connected to B and turns with it.

100 mm 25 mm vB D 75 mm

B vC

Solution

vA

The speed of a point P on the belt wrapped around A is

vP = vArA = 60(0.075) = 4.5 m>s



vB =

vP 4.5 = = 180 rad>s rD 0.025

50 mm

A C

Ans.

The speed of a point P′ on the belt wrapped around the outer periphery of B is

v′P = vBrB = 180(0.1) = 18 m>s

Hence,  vC =

v′P 18 = = 360 rad>s rC 0.05

Ans.

Ans: vB = 180 rad>s vC = 360 rad>s 642

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16–14. The disk starts from rest and is given an angular acceleration a = (2t 2) rad>s2, where t is in seconds. Determine the angular velocity of the disk and its angular displacement when t = 4 s.

P 0.4 m

SOLUTION dv = 2 t2 dt

a =

t

v

dv =

L0

v =

2 32t t 3 0

v =

23 t 3

L0

2

2 t dt

When t = 4 s, v =

2 3 (4) = 42.7 rad>s 3

L0 u =

Ans.

t

u

du = 1 4 t 6

2 3 t dt L0 3

When t = 4 s, u =

1 4 (4) = 42.7 rad 6

Ans.

Ans: v = 42.7 rad>s u = 42.7 rad 643

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16–15. The disk starts from rest and is given an angular acceleration a = (5t1>2) rad>s2, where t is in seconds. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when t = 2 s.

P 0.4 m

SOLUTION Motion of the Disk: Here, when t = 0, v = 0. dv = adt

t

v

L0 v2

dv =

v 0

=

v = e

L0

1

5t2dt

10 3 2 t t2 3 0 10 3 t2 f rad>s 3

When t = 2 s, v =

10 3 A 2 2 B = 9.428 rad>s 3

When t = 2 s, 1

a = 5 A 2 2 B = 7.071 rad>s2 Motion of point P: The tangential and normal components of the acceleration of point P when t = 2 s are at = ar = 7.071(0.4) = 2.83 m>s2

Ans.

a n = v2r = 9.4282(0.4) = 35.6 m>s2

Ans.

Ans: at = 2.83 m>s2 an = 35.6 m>s2 644

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*16–16. The disk starts at v0 = 1 rad>s when u = 0, and is given an angular acceleration a = (0.3u) rad>s2, where u is in radians. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when u = 1 rev.

P 0.4 m

SOLUTION a = 0.3u v

L1

vdv =

u

L0

0.3udu

u 1 22v v = 0.15u2 2 2 1 0

v2 - 0.5 = 0.15u2 2 v = 20.3u2 + 1 At u = 1 rev = 2p rad v = 20.3(2p)2 + 1 v = 3.584 rad>s a t = ar = 0.3(2p) rad>s 2 (0.4 m) = 0.7540 m>s2

Ans.

a n = v2r = (3.584 rad>s)2(0.4 m) = 5.137 m>s2

Ans.

a p = 2(0.7540)2 + (5.137)2 = 5.19 m>s2

Ans: at = 0.7540 m>s2 an = 5.137 m>s2 645

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16–17. A motor gives gear A an angular acceleration of aA = (2 + 0.006 u 2) rad>s2, where u is in radians. If this gear is initially turning at vA = 15 rad>s, determine the angular velocity of gear B after A undergoes an angular displacement of 10 rev.

B A

175 mm 100 mm aB

aA vA

Solution Angular Motion. The angular velocity of the gear A can be determined by integrating v dv = a du with initial condition vA = 15 rad>s at uA = 0. vA



L15 rad>s

v dv =

L0

uA

( 2 + 0.006 u 2 ) du

uA v2 vA ` = ( 2u + 0.002 u 3 ) ` 2 15 rad>s 0

v2A 152 = 2uA + 0.002 u 3A 2 2

vA = 20.004 u 3A + 4 u + 225 rad>s

At uA = 10(2p) = 20p rad,

vA = 20.004(20p)3 + 4(20p) + 225 = 38.3214 rad>s

Since gear B is meshed with gear A,

;   vB(175) = 38.3214(100) vBrB = vArA 



vB = 21.8979 rad>s



= 21.9 rad>s d

Ans.

Ans: vB = 21.9 rad>s d 646

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16–18. A motor gives gear A an angular acceleration of aA = (2t 3) rad>s2, where t is in seconds. If this gear is initially turning at vA = 15 rad>s, determine the angular velocity of gear B when t = 3 s.

B A

175 mm 100 mm aB

aA vA

Solution Angular Motion. The angular velocity of gear A can be determined by integrating dv = a dt with initial condition vA = 15 rad>s at t = 0 s. vA



L15 rad>s

dv =

vA - 15 =

At t = 3 s,

L0

t

2t 3 dt

1 4 t t ` 2 0

1 vA = e t 4 + 15 f rad>s 2 vA =

1 4 ( 3 ) + 15 = 55.5 rad>s 2

Since gear B meshed with gear A,

vBrB = vArA ;       vB(175) = 55.5(100)



vB = 31.7143 rad>s



= 31.7 rad>s d

Ans.

Ans: vB = 31.7 rad>s d 647

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16–19. The vacuum cleaner’s armature shaft S rotates with an angular acceleration of a = 4v3>4 rad>s2, where v is in rad>s. Determine the brush’s angular velocity when t = 4 s, starting from v0 = 1 rad>s, at u = 0. The radii of the shaft and the brush are 0.25 in. and 1 in., respectively. Neglect the thickness of the drive belt.

SOLUTION Motion of the Shaft: The angular velocity of the shaft can be determined from dt =

L t

L0 2

t

dt =

A

dvS L aS vs

S

A

S

dvS

L1 4vS 3>4 2

vs

t 0 = vS 1>4 1

t = vS 1>4 – 1 vS = (t+1) 4 When t = 4 s vs = 54 = 625 rad>s Motion of the Beater Brush: Since the brush is connected to the shaft by a non-slip belt, then v B r B = vs r s vB = ¢

rs 0.25 b (625) = 156 rad>s ≤v = a rB s 1

Ans.

Ans: vB = 156 rad>s 648

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*16–20. A motor gives gear A an angular acceleration of aA = (4t 3) rad>s2, where t is in seconds. If this gear is initially turning at (vA)0 = 20 rad>s, determine the angular velocity of gear B when t = 2 s.

(ω A)0 = 20 rad/s

B

A

0.05 m

αA

0.15 m

SOLUTION aA = 4 t 3 dw = a dt wA

L20

t

dwA =

L0

t

aA dt =

L0

4 t 3 dt

wA = t 4 + 20 When t = 2 s, wA = 36 rad>s wA rA = wB rB 36(0.05) = wB (0.15) Ans.

wB = 12 rad>s

Ans: wB = 12 rad>s 649

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16–21. The motor turns the disk with an angular velocity of v = ( 5t 2 + 3t ) rad>s, where t is in seconds. Determine the magnitudes of the velocity and the n and t components of acceleration of the point A on the disk when t = 3 s.

150 mm u

Solution A

Angular Motion. At t = 3 s,

v = 5 ( 32 ) + 3(3) = 54 rad>s

The angular acceleration of the disk can be determined using

a =

At t = 3 s,

dv ;   a = 5 10t + 36 rad>s2 dt

a = 10(3) + 3 = 33 rad>s2

Motion of Point A. The magnitude of the velocity is

Ans.

vA = vrA = 54(0.15) = 8.10 m>s

The tangential and normal component of acceleration are

( aA ) t = arA = 33(0.15) = 4.95 m>s2

Ans.



( aA ) n = v2rA = ( 542 ) (0.15) = 437.4 m>s2 = 437 m>s2

Ans.

Ans: vA = 8.10 m>s ( aA ) t = 4.95 m>s2 ( aA ) n = 437 m>s2 650

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16–22. If the motor turns gear A with an angular acceleration of aA = 2 rad>s2 when the angular velocity is vA = 20 rad>s, determine the angular acceleration and angular velocity of gear D.

B

100 mm C 50 mm

D

SOLUTION Angular Motion: The angular velocity and acceleration of gear B must be determined first. Here, vA rA = vB rB and aA rA = aB rB. Then, vB =

rA 40 b (20) = 8.00 rad>s v = a rB A 100

aB =

rA 40 a = a b (2) = 0.800 rad>s2 rB A 100

Since gear C is attached to gear B, then vC = vB = 8 rad>s aC = aB = 0.8 rad>s2. Realizing that vC rC = vD rD and aC rC = aD rD, then

40 mm 100 mm

and

vD =

rC 50 v = a b (8.00) = 4.00 rad>s rD C 100

Ans.

aD =

rC a = rD C

Ans.

50 (0.800) = 0.400 rad s2 100

A

A

Ans: vD = 4.00 rad>s aD = 0.400 rad>s2 651

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16–23. If the motor turns gear A with an angular acceleration of aA = 3 rad>s2 when the angular velocity is vA = 60 rad>s, determine the angular acceleration and angular velocity of gear D.

B

100 mm C 50 mm

D

SOLUTION Angular Motion: The angular velocity and acceleration of gear B must be determined first. Here, vA rA = vB rB and aA rA = aB rB. Then, vB =

rA 40 b (60) = 24.0 rad>s v = a rB A 100

aB =

rA 40 a = a b (3) = 1.20 rad>s2 rB A 100

A

A

40 mm 100 mm

Since gear C is attached to gear B, then vC = vB = 24.0 rad>s and aC = aB = 1.20 rad>s2. Realizing that vC rC = vD rD and aC rC = aD r D, then vD =

rC 50 v = a b (24.0) = 12.0 rad>s rD C 100

Ans.

aD =

rC a = rD C

Ans.

50 (1.20) = 0.600 rad s2 100

Ans: vD = 12.0 rad>s aD = 0.600 rad>s2 652

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*16–24. The gear A on the drive shaft of the outboard motor has a radius rA = 0.5 in. and the meshed pinion gear B on the propeller shaft has a radius rB = 1.2 in. Determine the angular velocity of the propeller in t = 1.5 s, if the drive shaft rotates with an angular acceleration a = (400t3) rad>s2, where t is in seconds. The propeller is originally at rest and the motor frame does not move.

A 2.20 in.

B

P

SOLUTION Angular Motion: The angular velocity of gear A at t = 1.5 s must be determined first. Applying Eq. 16–2, we have dv = adt 1.5 s

vA

L0

dv =

L0

400t3 dt

s vA = 100t4 |1.5 = 506.25 rad>s 0

However, vA rA = vB rB where vB is the angular velocity of propeller. Then, vB =

rA 0.5 b(506.25) = 211 rad>s v = a rB A 1.2

Ans.

Ans: vB = 211 rad>s 653

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16–25. For the outboard motor in Prob. 16–24, determine the magnitude of the velocity and acceleration of point P located on the tip of the propeller at the instant t = 0.75 s.

A 2.20 in.

B

P

SOLUTION Angular Motion: The angular velocity of gear A at t = 0.75 s must be determined first. Applying Eq. 16–2, we have dv = adt 0.75 s

vA

L0

dv =

L0

400t3 dt

s vA = 100t4 |0.75 = 31.64 rad>s 0

The angular acceleration of gear A at t = 0.75 s is given by aA = 400 A 0.753 B = 168.75 rad>s2 However, vA rA = vB rB and aA rA = aB rB where vB and aB are the angular velocity and acceleration of propeller. Then, vB = aB =

rA 0.5 b (31.64) = 13.18 rad>s v = a rB A 1.2 rA 0.5 a = a b (168.75) = 70.31 rad>s2 rB A 1.2

Motion of P: The magnitude of the velocity of point P can be determined using Eq. 16–8. vP = vB rP = 13.18 a

2.20 b = 2.42 ft>s 12

Ans.

The tangential and normal components of the acceleration of point P can be determined using Eqs. 16–11 and 16–12, respectively. ar = aB rP = 70.31a

2.20 b = 12.89 ft>s2 12

an = v2B rP = A 13.182 B a

2.20 b = 31.86 ft>s2 12

The magnitude of the acceleration of point P is aP = 2a2r + a2n = 212.892 + 31.862 = 34.4 ft>s2

Ans.

Ans: vP = 2.42 ft>s aP = 34.4 ft>s2 654

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16–26. The pinion gear A on the motor shaft is given a constant angular acceleration a = 3 rad>s2. If the gears A and B have the dimensions shown, determine the angular velocity and angular displacement of the output shaft C, when t = 2 s starting from rest. The shaft is fixed to B and turns with it.

B C

125 mm A

35 mm

SOLUTION v = v0 + a c t vA = 0 + 3(2) = 6 rad>s u = u0 + v 0 t + uA = 0 + 0 +

1 ac t2 2

1 (3)(2)2 2

uA = 6 rad vA rA = vB rB 6(35) = vB(125) Ans.

vC = vB = 1.68 rad>s uA rA = uB rB 6(35) = uB (125)

Ans.

uC = uB = 1.68 rad

Ans: vC = 1.68 rad>s uC = 1.68 rad 655

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16–27. The gear A on the drive shaft of the outboard motor has a radius rA = 0.7 in. and the meshed pinion gear B on the propeller shaft has a radius rB = 1.4 in. Determine the angular velocity of the propeller in t = 1.3 s if the drive shaft rotates with an angular acceleration a = 1300 2t2 rad>s2, where t is in seconds. The propeller is originally at rest and the motor frame does not move.

A

SOLUTION

B

2.2 in. P

aArA = aBrB (300 2t)(0.7) = ap(1.4) aP = 150 2t dv = a dt t

v

L0

dv =

L0

150 2t dt

v = 100t3 2|t = 1.3 = 148 rad s

Ans.

Ans: v = 148 rad>s 656

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*16–28. The gear A on the drive shaft of the outboard motor has a radius rA = 0.7 in. and the meshed pinion gear B on the propeller shaft has a radius rB = 1.4 in. Determine the magnitudes of the velocity and acceleration of a point P located on the tip of the propeller at the instant t = 0.75 s. the drive shaft rotates with an angular acceleration a = (300 1t) rad>s2, where t is in seconds. The propeller is originally at rest and the motor frame does not move.

A B

2.2 in. P

Solution Angular Motion: The angular velocity of gear A at t = 0.75 s must be determined first. Applying Eq. 16–2, we have dv = adt L0

vA

dv =

L0

0.75 s

3002t dt

s vA = 200 t 3>2 0 0.75 = 129.9 rad>s 0

The angular acceleration of gear A at t = 0.75 s is given by aA = 30020.75 = 259.81 rad>s2 However, vA rA = vB rB and aA rA = aB rB where vB and aB are the angular velocity and acceleration of propeller. Then, vB = aB =

rA 0.7 v = a b(129.9) = 64.95 rad>s rB A 1.4

rA 0.7 a = a b(259.81) = 129.9 rad>s2 rB A 1.4

Motion of P: The magnitude of the velocity of point P can be determined using Eq. 16–8. vP = vB rP = 64.95 a

2.20 b = 11.9 ft>s 12

The tangential and normal components of the acceleration of point P can be determained using Eqs. 16–11 and 16–12, respectively. at = aB rP = 129.9 a

an = v2B rP = ( 64.95 ) 2 a

2.20 b = 23.82 ft>s2 12

2.20 b = 773.44 ft>s2 12

The magnitude of the acceleration of point P is

aP = 2a2t + a2n = 2 ( 23.82 ) 2 + ( 773.44 ) 2 = 774 ft>s2

Ans.

Ans: aP = 774 ft>s2 657

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16–29. A stamp S, located on the revolving drum, is used to label canisters. If the canisters are centered 200 mm apart on the conveyor, determine the radius rA of the driving wheel A and the radius rB of the conveyor belt drum so that for each revolution of the stamp it marks the top of a canister. How many canisters are marked per minute if the drum at B is rotating at vB = 0.2 rad>s? Note that the driving belt is twisted as it passes between the wheels.

A rA S

SOLUTION l = 2p(rA) rA

rB

200 mm B

200 = = 31.8 mm 2p

Ans.

rB

ω B = 0.2 rad/s

For the drum at B: l = 2p(rB) rB =

200 = 31.8 mm 2p

Ans.

In t = 60 s u = u0 + v0 t u = 0 + 0.2(60) = 12 rad l = urB = 12(31.8) = 382.0 mm Hence, n =

382.0 = 1.91 canisters marked per minute 200

Ans.

Ans: rA = 31.8 mm rB = 31.8 mm 1.91 canisters per minute 658

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16–30. At the instant shown, gear A is rotating with a constant angular velocity of vA = 6 rad>s. Determine the largest angular velocity of gear B and the maximum speed of point C.

C

ωB

100 mm ω A = 6 rad/s

100 mm B

A

100 mm

100 mm

SOLUTION (rB)max = (rA)max = 5022 mm (rB)min = (rA)min = 50 mm When rA is max., rB is min. vB(rB) = vA rA (vB)max = 6 a

rA 5012 b b = 6a rB 50 Ans.

(vB)max = 8.49 rad>s vC = (vB)max rC = 8.49 ( 0.0512 )

Ans.

vC = 0.6 m>s

Ans: (vB)max = 8.49 rad>s (vC)max = 0.6 m>s 659

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16–31. Determine the distance the load W is lifted in t = 5 s using the hoist. The shaft of the motor M turns with an angular velocity v = 100(4 + t) rad>s, where t is in seconds.

300 mm

C

30 mm 225 mm

50 mm

A

40 mm M

D E B

W

SOLUTION Angular Motion: The angular displacement of gear A at t = 5 s must be determined first. Applying Eq. 16–1, we have du = vdt 5s

uA

L0

du =

L0

100(4 + t) dt

uA = 3250 rad Here, rA uA = rB uB. Then, the angular displacement of gear B is given by uB =

rA 40 b (3250) = 577.78 rad u = a rB A 225

Since gear C is attached to the same shaft as gear B, then uC = uB = 577.78 rad. Also, rD uD = rC uC, then, the angular displacement of gear D is given by uD =

rC 30 u = a b (577.78) = 57.78 rad rD C 300

Since shaft E is attached to gear D, uE = uD = 57.78 rad. The distance at which the load W is lifted is Ans.

sW = rE uE = (0.05)(57.78) = 2.89 m

Ans: sW = 2.89 m 660

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*16–32. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If A has a constant angular acceleration of aA = 30 rad>s2, determine the tangential and normal components of acceleration of a point located at the rim of B when t = 3 s, starting from rest.

200 mm vA

B

125 mm

vB A

SOLUTION Motion of Wheel A: Since the angular acceleration of wheel A is constant, its angular velocity can be determined from vA = (vA)0 + aCt = 0 + 30(3) = 90 rad>s Motion of Wheel B: Since wheels A and B are connected by a nonslip belt, then vBrB = vArA vB = a

rA 200 bv = a b (90) = 144 rad>s rB A 125

and aBrB = aArA aB = a

rA 200 ba = a b (30) = 48 rad>s2 rB A 125

Thus, the tangential and normal components of the acceleration of point P located at the rim of wheel B are (ap)t = aBrB = 48(0.125) = 6 m>s2

Ans.

(ap)n = vB 2rB = (144 2)(0.125) = 2592 m>s2

Ans.

Ans: (ap)t = 6 m>s2 (ap)n = 2592 m>s2 661

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16–33. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If the angular displacement of A is uA = (5t3 + 10t2) rad, where t is in seconds, determine the angular velocity and angular acceleration of B when t = 3 s.

200 mm vA

B

125 mm

vB A

SOLUTION Motion of Wheel A: The angular velocity and angular acceleration of wheel A can be determined from vA =

duA = A 15t2 + 20t B rad>s dt

aA =

dvA = A 30t + 20 B rad>s dt

and

When t = 3 s, vA = 15 A 32 B + 20(3) = 195 rad>s aA = 30(3) + 20 = 110 rad>s Motion of Wheel B: Since wheels A and B are connected by a nonslip belt, then vBrB = vArA vB = a

rA 200 bv = a b (195) = 312 rad>s rB A 125

Ans.

aBrB = aArA aB = a

rA 200 ba = a b (110) = 176 rad>s2 rB A 125

Ans.

Ans: vB = 312 rad>s aB = 176 rad>s2 662

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16–34. For a short time a motor of the random-orbit sander drives the gear A with an angular velocity of vA = 401t3 + 6t2 rad>s, where t is in seconds. This gear is connected to gear B, which is fixed connected to the shaft CD. The end of this shaft is connected to the eccentric spindle EF and pad P, which causes the pad to orbit around shaft CD at a radius of 15 mm. Determine the magnitudes of the velocity and the tangential and normal components of acceleration of the spindle EF when t = 2 s after starting from rest.

40 mm 10 mm

B

VA A

C D

15 mm E

SOLUTION vA rA = vB rB

F

vA (10) = vB (40) vB =

P

1 v 4 A

vE = vB rE =

1 1 vA (0.015) = (40) A t3 + 6t B (0.015) 2 4 4 t=2 Ans.

vE = 3 m>s aA =

dvA d = C 40 A t3 + 6t B D = 120t2 + 240 dt dt

aA rA = aB rB aA (10) = aB (40) aB =

1 a 4 A

(aE)t = aB rE =

1 A 120t2 + 240 B (0.015) 2 4 t=2

(aE)t = 2.70 m>s2

Ans.

2 1 (aE)n = v2B rE = c (40) A t3 + 6t B d (0.015) 2 4 t=2

(aE)n = 600 m>s2

Ans.

Ans: vE = 3 m>s (aE)t = 2.70 m>s2 (aE)n = 600 m>s2 663

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16–35. If the shaft and plate rotates with a constant angular velocity of v = 14 rad>s, determine the velocity and acceleration of point C located on the corner of the plate at the instant shown. Express the result in Cartesian vector form.

z

A v

0.6 m

a 0.2 m C

D

SOLUTION

O

0.4 m

We will first express the angular velocity v of the plate in Cartesian vector form. The unit vector that defines the direction of v is uOA =

- 0.3i + 0.2j + 0.6k

3 2 6 = - i + j + k 7 7 7 2(- 0.3) + 0.2 + 0.6 2

2

0.3 m 0.3 m

x

0.4 m B

2

Thus, 2 6 3 v = vuOA = 14 a - i + j + k b = [- 6i + 4j + 12k] rad>s 7 7 7 Since v is constant a = 0 For convenience, rC = [- 0.3i + 0.4j] m is chosen. The velocity and acceleration of point C can be determined from vC = v * rC = ( -6i + 4j + 12k) * (- 0.3i + 0.4j) Ans.

= [ -4.8i - 3.6j - 1.2k] m>s and aC = a * rC + V * (V * rc) = 0 + (- 6i + 4j + 12k) * [( -6i + 4j + 12k) * ( -0.3i + 0.4j)] = [38.4i - 64.8j + 40.8k]m s2

Ans.

664

Ans: vC = 5 -4.8i - 3.6j - 1.2k6 m>s aC = 5 38.4i - 64.8j + 40.8k6 m>s2

y

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*16–36. At the instant shown, the shaft and plate rotates with an angular velocity of v = 14 rad>s and angular acceleration of a = 7 rad>s2. Determine the velocity and acceleration of point D located on the corner of the plate at this instant. Express the result in Cartesian vector form.

z

A v

0.6 m

a 0.2 m C

D

SOLUTION We will first express the angular velocity v of the plate in Cartesian vector form. The unit vector that defines the direction of v and a is uOA =

O

0.4 m

-0.3i + 0.2j + 0.6k

2 6 3 = - i + j + k 7 7 7 2( - 0.3) + 0.2 + 0.6 2

2

0.3 m 0.3 m

x

0.4 m B

2

Thus, 2 6 3 v = vuOA = 14 a - i + j + k b = [- 6i + 4j + 12k] rad>s 7 7 7 3 2 6 a = auOA = 7a - i + j + k b = [- 3i + 2j + 6k] rad>s 7 7 7 For convenience, rD = [- 0.3i + 0.4j] m is chosen. The velocity and acceleration of point D can be determined from vD = v * rD = (-6i + 4j + 12k) * (0.3i - 0.4j) Ans.

= [4.8i + 3.6j + 1.2k] m>s and aD = a * rD - v2 rD

= (- 3i + 2j + 6k) * (- 0.3i + 0.4j) + ( - 6i + 4j + 12k) * [(- 6i + 4j + 12k) * (- 0.3i + 0.4j)] = [-36.0i + 66.6j - 40.2k] m s2

Ans.

Ans: vD = [4.8i + 3.6j + 1.2k] m>s aD = [ - 36.0i + 66.6j - 40.2k] m>s2 665

y

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16–37. The rod assembly is supported by ball-and-socket joints at A and B. At the instant shown it is rotating about the y axis with an angular velocity v = 5 rad>s and has an angular acceleration a = 8 rad>s2. Determine the magnitudes of the velocity and acceleration of point C at this instant. Solve the problem using Cartesian vectors and Eqs. 16–9 and 16–13.

z

C 0.3 m

0.4 m

B A

SOLUTION

A 0.4 m

x

y

V

vC = v * r vC = 5j * (- 0.4i + 0.3k) = {1.5i + 2k} m>s vC = 21.52 + 2 2 = 2.50 m>s

Ans.

a C = a * r - v2r = 8j * (- 0.4i + 0.3k) - 52 ( -0.4i + 0.3k) = {12.4i - 4.3k} m>s2 a C = 212.4 2 + (-4.3)2 = 13.1 m>s2

Ans.

Ans: vC = 2.50 m>s aC = 13.1 m>s2 666

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16–38. The sphere starts from rest at u = 0° and rotates with an angular acceleration of a = (4u + 1) rad>s2, where u is in radians. Determine the magnitudes of the velocity and acceleration of point P on the sphere at the instant u = 6 rad.

P 30 r  8 in.

Solution v dv = a du L0

v

v dv =

L0

u

(4u + 1) du

v = 2u At u = 6 rad, a = 4(6) + 1 = 25 rad>s2,  v = 24(6)2 + 2(6) = 12.49 rad>s

v = ar′ = 12.49(8 cos 30°) = 86.53 in.>s

Ans.

v = 7.21 ft>s ar =

2

2

(86.53) v = = 1080.8 in.>s2 2 (8 cos 30°) r

ar = ar 2 = 25(8 cos 30°) = 173.21 in.>s2 a = 2(1080.8)2 + (173.21)2 = 1094.59 in.>s2 a = 91.2 ft>s2

Ans.

Ans: v = 7.21 ft>s a = 91.2 ft>s2 667

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16–39. The end A of the bar is moving downward along the slotted guide with a constant velocity vA . Determine the angular velocity V and angular acceleration A of the bar as a function of its position y. vA

A

U

y

V, A

SOLUTION

r

Position coordinate equation:

B

sin u =

r y

Time derivatives: # # 2y2 - r2 # r and y = - yA, u = v cos uu = - y# however, cos u = 2 y y

¢

2y2 - r2 r ≤ v = 2 yA y y

v =

ryA

Ans.

y 2y2 - r2

1 3 a = v# = ryA C - y-2y# A y2 - r2 B - 2 + A y-1 B A - 12 B A y2 - r2 B - 2(2yy# ) D

a =

ry2A (2y2 - r2)

Ans.

3

y2(y2 - r2)2

Ans: v = a =

668

rvA y2y2 - r 2 rvA2 ( 2y2 - r 2 ) y2 ( y2 - r 2) 3>2

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*16–40. At the instant u = 60°, the slotted guide rod is moving to the left with an acceleration of 2 m>s2 and a velocity of 5 m>s. Determine the angular acceleration and angular velocity of link AB at this instant.

v  5 m/s a  2 m/s2

A u

Solution Position Coordinate Equation. The rectilinear motion of the guide rod can be related to the angular motion of the crank by relating x and u using the geometry shown in Fig. a, which is

200 mm B

x = 0.2 cos u m Time Derivatives. Using the chain rule, # # x = - 0.2(sin u)u (1) #2 $ $ x = - 0.2[(cos u)u + (sin u)u ] (2) # $ $ # Here x = v, x = a , u = v and u = a when u = 60°. Realizing that the velocity and acceleration of the guide rod are directed toward the negative sense of x, v = - 5 m>s and a = - 2 m>s2. Then Eq (1) gives - s = ( - 0.2(sin 60°)v Ans.

v = 28.87 rad>s = 28.9 rad>s b Subsequently, Eq. (2) gives - 2 = - 0.2[cos 60° ( 28.872 ) + (sin 60°)a] a = -469.57 rad>s2 = 470 rad>s2 d

Ans.

The negative sign indicates that A is directed in the negative sense of u.

Ans: v = 28.9 rad>s b a = 470 rad>s2 d 669

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16–41. At the instant u = 50°, the slotted guide is moving upward with an acceleration of 3 m>s2 and a velocity of 2 m>s. Determine the angular acceleration and angular velocity of link AB at this instant. Note: The upward motion of the guide is in the negative y direction.

B 300 mm V, A

y U

A

SOLUTION y = 0.3 cos u # # y = vy = - 0.3 sin uu $ # ## y = a y = -0.3 A sin uu + cos uu2 B

v a

2 m/s 3 m/s2

# # $ Here vy = - 2 m>s, ay = - 3 m>s2, and u = v, u = v, u = a, u = 50°. - 2 = -0.3 sin 50°(v)

v = 8.70 rad>s

Ans.

- 3 = -0.3[sin 50°(a) + cos 50°(8.70)2]

a = -50.5 rad>s 2

Ans.

Ans: v = 8.70 rad>s a = - 50.5 rad>s2 670

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16–42. At the instant shown, u = 60°, and rod AB is  subjected to a deceleration of 16 m>s2 when the velocity is 10 m>s. Determine the angular velocity and angular acceleration of link CD at this instant.

x

v  10 m/s A

B

D u

u

a  16 m/s2 300 mm

300 mm

Solution C

x = 2(0.3) cos u # x = - 0.6 sin u ( u ) 

(1)

# $ $ x = - 0.6 cos u ( u ) 2 - 0.6 sin u ( u ) 

(2)

$ # Using Eqs. (1) and (2) at u = 60°, x = 10 m>s, x = - 16 m>s2. 10 = - 0.6 sin 60°(v) Ans.

v = -19.245 = - 19.2 rad>s - 16 = -0.6 cos 60°( -19.245)2 - 0.6 sin 60°(a) a = -183 rad>s2

Ans.

Ans: v = - 19.2 rad>s a = - 183 rad>s2 671

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16–43.

C

The crank AB is rotating with a constant angular velocity of 4 rad>s. Determine the angular velocity of the connecting rod CD at the instant u = 30°.

u 600 mm B

300 mm

Solution A

Position Coordinate Equation: From the geometry,

4 rad/s D

[1]

0.3 sin f = (0.6 - 0.3 cos f) tan u Time Derivatives: Taking the time derivative of Eq. [1], we have

0.3 cos f

df df du du = 0.6sec2u - 0.3acos u sec2 u - tan u sin u b dt dt dt dt 0.3(cos f - tan u sin f) df du = c d  dt dt 0.3sec2u(2 - cos f)

[2]

df du = vAB = 4 rad>s. At the instant u = 30°, from Eq. [3], = vBC, dt dt f = 60.0°. Substitute these values into Eq. [2] yields

However,

vBC = c

0.3(cos 60.0° - tan 30° sin 60.0°) 0.3 sec230° (2 - cos 60.0°)

Ans.

d (4) = 0

Ans: vAB = 0 672

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*16–44. Determine the velocity and acceleration of the follower rod CD as a function of u when the contact between the cam and follower is along the straight region AB on the face of the cam. The cam rotates with a constant counterclockwise angular velocity V.

v

A r O

u

D

C B

SOLUTION Position Coordinate: From the geometry shown in Fig. a, xC =

r = r sec u cos u

Time Derivative: Taking the time derivative, # # vCD = xC = r sec u tan uu

(1)

# Here, u = +v since v acts in the positive rotational sense of u. Thus, Eq. (1) gives vCD = rv sec u tan u :

Ans.

The time derivative of Eq. (1) gives $ # # # $ aCD = xC = r{sec u tan uu + u[sec u(sec2uu) + tan u(sec u tan uu)]} $ # aCD = r[sec u tan u u + (sec3 u + sec u tan2 u)u2] # $ Since u = v is constant, u = a = 0. Then, a CD = r[sec u tan u(0) + (sec3 u + sec u tan2 u)v2] = rv2 A sec3 u + sec u tan2 u B :

Ans.

Ans: vCD = rv sec u tan u S aCD = rv2 ( sec3 u + sec u tan2 u ) S 673

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16–45. Determine the velocity of rod R for any angle u of the cam C if the cam rotates with a constant angular velocity V. The pin connection at O does not cause an interference with the motion of A on C.

V

r1

C

r2

SOLUTION

R

u

O

A

Position Coordinate Equation: Using law of cosine.

x

(r1 + r2)2 = x2 + r12 - 2r1x cos u

(1)

Time Derivatives: Taking the time derivative of Eq. (1).we have 0 = 2x

However v =

dx du dx - 2r1 ¢ - x sin u + cos u ≤ dt dt dt

(2)

du dx and v = . From Eq.(2), dt dt 0 = xv - r1(v cos u - xv sin u) v =

r1xv sin u r1 cos u - x

(3)

However, the positive root of Eq.(1) is x = r1 cos u + 2r21 cos 2u + r22 + 2r1r2 Substitute into Eq.(3),we have v = -

r21v sin 2u

2 2r21 cos2u + r22 + 2r# 1r2

Ans.

+ r1v sin u

Note: Negative sign indicates that v is directed in the opposite direction to that of positive x.

Ans: v = -° 674

r 12v sin 2u 22r 12 cos2 u + r 22 + 2r1r2

+ r1v sin u ¢

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16–46. The circular cam rotates about the fixed point O with a constant angular velocity V. Determine the velocity v of the follower rod AB as a function of u.

v R d O

v

u r

Solution

B

A

x = d cos u + 2(R + r)2 - (d sin u)2 # # x = vAB = -d sin uu -

- v = -d sin u(v) -

v = vd °sin u +

d 2 sin 2u 2

2

2

22(R + r) - d sin u d 2 sin 2u

22(R + r)2 - d 2 sin2 u d sin 2u

22(R + r)2 - d 2 sin2 u

# # u  Where u = v and vAB = -v

v

Ans.

¢

Ans: v = vd asin u + 675

d sin 2u 21(R + r)2 - d 2 sin2 u

b

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16–47. V

Determine the velocity of the rod R for any angle u of cam C as the cam rotates with a constant angular velocity V. The pin connection at O does not cause an interference with the motion of plate A on C.

r

u

R

SOLUTION

A

x = r + r cos u

C

O

x

x = - r sin uu Ans.

v = - rv sin u

Ans: v = - rv sin u 676

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*16–48. Determine the velocity and acceleration of the peg A which is confined between the vertical guide and the rotating slotted rod. A v a O

Solution Position Coordinate Equation. The rectilinear motion of peg A can be related to the angular motion of the slotted rod by relating y and u using the geometry shown in Fig. a, which is

u

b

y = b tan u Time Derivatives. Using the chain rule, # # y = b ( sec2 u ) u # # $ $ y = b[2 sec u(sec u tan uu)u + sec2 uu ] # $ $ y = b ( 2 sec2 u tan uu 2 + sec2 uu ) # $ $ y = b sec2 u ( 2 tan uu 2 + u )  # $ # $ Here, y = v, y = a, u = v and u = a. Then Eqs. (1) and (2) become

(1)

(2)

v = vb sec2 u

Ans.

a = b sec2 u ( 2v2 tan u + a ) 

Ans.

Ans: v = vb sec2 u a = b sec2 u ( 2v2 tan u + a ) 677

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16–49. Bar AB rotates uniformly about the fixed pin A with a constant angular velocity V. Determine the velocity and acceleration of block C, at the instant u = 60°.

B

L

V

L A

u

SOLUTION C

L cos u + L cos f = L cos u + cos f = 1 # # sin u u] + sin f f = 0 $ $ # # cos u(u)2 + sin uu + sinf f + cos f ( f)2 = 0

L

(1) (2)

When u = 60°, f = 60°, # # thus, u = -f = v (from Eq. (1)) $ u = 0 $ f = -1.155v2 (from Eq.(2)) Also, sC = L sin f - L sin u # # vC = L cos f f - L cos u u # $ $ # aC = -L sin f (f)2 + L cos f (f) - L cos u(u) + L sin u(u)2 At u = 60°, f = 60° sC = 0 vC = L(cos 60°)(- v) - L cos 60°(v) = - Lv = Lv c

Ans.

aC = -L sin 60°( -v)2 + L cos 60°(-1.155v2) + 0 + L sin 60°(v)2 aC = -0.577 Lv2 = 0.577 Lv2 c

Ans.

Ans: vC = Lv c aC = 0.577 Lv2 c 678

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16–50. The center of the cylinder is moving to the left with a constant velocity v0. Determine the angular velocity V and angular acceleration A of the bar. Neglect the thickness of the bar. V

A u

Solution Position Coordinate Equation. The rectilinear motion of the cylinder can be related to the angular motion of the rod by relating x and u using the geometry shown in Fig. a, which is x =

r = r cot u>2 tan u>2

Time Derivatives. Using the chain rule, 1 # # x = r c ( - csc2 u>2 ) a u b d 2 # r # x = - ( csc2 u>2 ) u 2

(1)

$ r 1 # # $ x = - c 2 csc u>2 ( - csc u>2 cot u>2 ) a u bu + ( csc2 u>2 ) u d 2 2

$ # $ r x = c ( csc2 u>2 cot u>2 ) u 2 - ( csc 2 u>2 ) u d 2 $ # $ r csc2 u>2 x = c ( cot u>2 ) u 2 - u d  2

(2)

# $ Here x = - v0 since v0 is directed toward the negative sense of x and u = v. Then Eq. (1) gives, r - v0 = - (csc2 u>2)v 2 v =

2v0 2 sin u>2 r

Ans.

679

r vO O

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15–50. Continued

$ $ Also, x = 0 since v is constant and u = a. Substitute the results of v into Eq. (2): 0 =

r csc2 u>2 2

c ( cot u>2 ) a

a = ( cot u>2 ) a a = ° a = a =

cos u>2 sin u>2

4v02 r2 2v02 r2

2 2v0 2 sin u>2b - a d r

2 2r0 2 sin u>2b r

¢°

4v02 r2

sin4 u>2¢

( sin3 u>2 )( cos u>2 ) ( 2 sin u>2 cos u>2 )( sin2 u>2 )

Since sin u = 2 sin u>2 cos u>2, then a =

2v02 r2

( sin u )( sin2 u>2 ) 

Ans.

Ans: 2v0 2 sin u>2 r 2 2v0 a = 2 (sin u) ( sin2 u>2 ) r

v =

680

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16–51. The pins at A and B are confined to move in the vertical and horizontal tracks. If the slotted arm is causing A to move downward at vA, determine the velocity of B at the instant shown.

d

θ

y

90°

A

h

vA

SOLUTION

B x

Position coordinate equation: tan u =

d h = x y

h x = ¢ ≤y d Time derivatives: # x = vB =

h # y d h v d A

Ans.

Ans: h vB = ¢ ≤vA d 681

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*16–52. y

The crank AB has a constant angular velocity V . Determine the velocity and acceleration of the slider at C as a function of u. Suggestion: Use the x coordinate to express the motion of C and the f coordinate for CB. x 0 when f = 0°.

B C

l

b

φ

ω

θ A

x

SOLUTION x = l + b - (L cos f + b cos u) b sin u l # # # vC = x = l sin ff + b sin uu

l sin f = b sin u or sin f =

(1)

# # b cos ff = cos uu l

(2)

Since cos f = 21 - sin2 f =

A

b 2 1 - a b sin2 u l

then, # f =

b a b cos uv l

(3)

b 2 1 - a b sin2 u D l

vC = bv D

b a b sin u cos u l b 2 1 - a b sin2 u D l

T + bv sin u

Ans.

From Eq. (1) and (2): ## # # # 2 # aC = vC = lf sin f + lf cos ff + b cos u au b

(4)

# $ # b -sin ff2 + cos ff = - a b sin uu2 l $ f =

# b f2 sin f - v2 sin u l cos f

(5)

Substituting Eqs. (1), (2), (3) and (5) into Eq. (4) and simplifying yields

aC = bv2

b 2 b a b a cos 2u + a b sin4 u b l l 1 -

b l

2

sin2 u

3 2

Ans.

+ cos u

Ans: vC = bv≥

aC = bv2 ≥ 682

b a b sin u cos u l

b 2 2 1 - a b sin u A l

¥ + bv sin u

b b 2 a b acos 2u + a b sin4 u b l l 3

2 b 2 a1 - a b sin2 u b l

+ cos u ¥

x

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16–53. If the wedge moves to the left with a constant velocity v, determine the angular velocity of the rod as a function of u.

L v

u

f

SOLUTION Position Coordinates:Applying the law of sines to the geometry shown in Fig. a, xA L = sin(f - u) sin A 180° - f B xA =

L sin(f - u) sin A 180° - f B

However, sin A 180° - f B = sinf. Therefore, xA =

L sin (f - u) sin f

Time Derivative: Taking the time derivative, # L cos (f - u)(- u) # xA = sin f # L cos (f - u)u # vA = xA = sin f

(1)

Since point A is on the wedge, its velocity is vA = - v. The negative sign indicates that vA is directed towards the negative sense of xA. Thus, Eq. (1) gives # u =

v sin f L cos (f - u)

Ans.

Ans: # u = 683

v sin f L cos (f - u)

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16–54. v

The crate is transported on a platform which rests on rollers, each having a radius r. If the rollers do not slip, determine their angular velocity if the platform moves forward with a velocity v.

r

v

SOLUTION Position coordinate equation: From Example 163, sG = ru. Using similar triangles sA = 2sG = 2ru Time derivatives: # sA = v = 2r u v =

# Where u = v

v 2r

Ans.

Ans: v = 684

v 2r

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16–55. Arm AB has an angular velocity of V and an angular acceleration of A. If no slipping occurs between the disk D and the fixed curved surface, determine the angular velocity and angular acceleration of the disk.

ω ', α ' D

Ar C

SOLUTION

ω ,α B

ds = (R + r) du = r df df du b (R + r)a b = r a dt dt v¿ =

(R + r) v r

Ans.

a¿ =

(R + r) a r

Ans.

R

Ans: v′ = a′ = 685

(R + r)v r (R + r)a r

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*16–56. At the instant shown, the disk is rotating with an angular velocity of V and has an angular acceleration of A. Determine the velocity and acceleration of cylinder B at this instant. Neglect the size of the pulley at C.

A 3 ft u

V, A

C

5 ft

SOLUTION

B

s = 232 + 52 - 2(3)(5) cos u # 1 1 # vB = s = (34 - 30 cos u)- 2(30 sin u)u 2 vB =

15 v sin u

# aB = s =

=

Ans.

1

(34 - 30 cos u) 2 # # 15 v cos uu + 15v sin u 234 - 30 cos u

15 (v2 cos u + a sin u) (34 - 30 cos u)

1 2

-

+

# 1 a - b (15v sin u) a 30 sin uu b 2 3

(34 - 30 cos u) 2 225 v2 sin2 u

Ans.

3

(34 - 30 cos u) 2

Ans: vB = aB =

686

15 v sin u 1

(34 - 30 cos u)2 15 (v2 cos u + a sin u) 1

(34 - 30 cos u)2

-

225 v2 sin2 u 3

(34 - 30 cos u)2

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16–57. At the instant shown the boomerang has an angular velocity v = 4 rad>s, and its mass center G has a velocity vG = 6 in.>s. Determine the velocity of point B at this instant. vG = 6 in./s 30°

G 1.5 in.

SOLUTION

B

ω = 4 rad/s

45°

vB = vG + vB>G vB = 6 + [4(1.5 >sin 45°) = 8.4852] c 30° b + ) (;

5 in.

(vB)x = 6 cos 30° + 0 = 5.196 in.>s

A

( + c )(vB)y = 6 sin 30° + 8.4852 = 11.485 in.>s vB = 2(5.196)2 + (11.485)2 = 12.3 in.>s u = tan - 1

Ans.

11.485 = 65.7° 5.196

Also; vB = vG + v * rB>G (vB)x i + (vB)y j = ( -6 cos 30°i + 6 sin 30°j) + (4k) * (1.5> sin 45°)i (vB)x = - 6 cos 30° = - 5.196 in.>s (vB)y = 6 sin 30° + 8.4853 = 11.485 in.>s vB = 2(5.196)2 + (11.485)2 = 12.6 in.>s u = tan - 1

Ans.

11.485 = 65.7° 5.196

Ans: vB = 12.6 in.>s 65.7° b 687

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16–58. If the block at C is moving downward at 4 ft/s, determine the angular velocity of bar AB at the instant shown.

C 3 ft A

30°

ω AB

SOLUTION

vC = 4 ft/s

B

2 ft

Kinematic Diagram: Since link AB is rotating about fixed point A, then vB is always directed perpendicular to link AB and its magnitude is vB = vAB rAB = 2vAB. At the instant shown, vB is directed towards the negative y axis. Also, block C is moving downward vertically due to the constraint of the guide. Then vc is directed toward negative y axis. Velocity Equation: Here, rC>B = {3 cos 30°i + 3 sin 30°j} ft = {2.598i + 1.50j} ft. Applying Eq. 16–16, we have vC = vB + vBC * rC>B - 4j = - 2vAB j + (vBCk) * (2.598i + 1.50j) - 4j = - 1.50vBCi + (2.598vBC - 2vAB)j Equating i and j components gives 0 = - 1.50vBC - 4 = 2.598(0) - 2vAB

vBC = 0 Ans.

vAB = 2.00 rad s

Ans: vAB = 2.00 rad>s 688

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16–59. The link AB has an angular velocity of 3 rad>s. Determine the velocity of block C and the angular velocity of link BC at the instant u = 45°. Also, sketch the position of link BC when u = 60°, 45°, and 30° to show its general plane motion.

vA B  3 rad/s

B

1.5 m

C

0.5 m

u  45 A

Solution Rotation About Fixed Axis. For link AB, refer to Fig. a. vB = vAB * rAB = (3k) * (0.5 cos 45°i + 0.5 sin 45°j) = { - 1.0607i + 1.0607j} m>s General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, vC = vB + vBC * rC>B - vCi = ( - 1.0607i + 1.0607j) + ( - vBCk) * (1.5i) - vCi = - 1.0607i + (1.0607 - 1.5vBC)j Equating i and j components; - vC = - 1.0607

vC = 1.0607 m>s = 1.06 m>s

Ans.

0 = 1.0607 - 1.5vBC

vBC = 0.7071 rad>s = 0.707 rad>s

Ans.

The general plane motion of link BC is described by its orientation when u = 30°, 45° and 60° shown in Fig. c.

Ans: vC = 1.06 m>s d vBC = 0.707 rad>sd 689

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*16–60. The slider block C moves at 8 m>s down the inclined groove. Determine the angular velocities of links AB and BC, at the instant shown.

2m A

B

45 2m

Solution C

Rotation About Fixed Axis. For link AB, refer to Fig. a. vB = VAB * rAB

vC  8 m/s

vB = ( - vABk) * (2i) = - 2vAB j General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, vB = vC + VBC * rB>C - 2vAB j = (8 sin 45°i - 8 cos 45°j) + (vBCk) * (2j) - 2vAB j = (8 sin 45° - 2vBC)i - 8 cos 45°j Equating i and j components, 0 = 8 sin 45° - 2vBC

vBC = 2.828 rad>s = 2.83 rad>s d

Ans.

- 2vAB = -8 cos 45°

vAB = 2.828 rad>s = 2.83 rad>s b

Ans.

Ans: vBC = 2.83 rad>sd vAB = 2.83 rad>sb 690

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16–61. Determine the angular velocity of links AB and BC at the instant u = 30°. Also, sketch the position of link BC when u = 55°, 45°, and 30° to show its general plane motion.

B u 1 ft A

3 ft C vC  6 ft/s

Solution Rotation About Fixed Axis. For link AB, refer to Fig. a. vB = VAB * rAB vB = (vABk) * j = - vAB i General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, vB = vC + VBC * rB>C - vAB i = 6j + (vBCk) * ( - 3 cos 30°i + 3 sin 30°j) - vABi = - 1.5vBCi + (6 - 2.5981 vBC)j Equating i and j components, 0 = 6 - 2.5981 vBC; - vAB = - 1.5(2.3094);

vBC = 2.3094 rad>s = 2.31 rad>s d

Ans.

vAB = 3.4641 rad>s = 3.46 rad>s d

Ans.

The general plane motion of link BC is described by its orientation when u = 30°, 45° and 55° shown in Fig. c.

Ans: vBC = 2.31 rad>sd vAB = 3.46 rad>sd 691

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16–62. The planetary gear A is pinned at B. Link BC rotates clockwise with an angular velocity of 8 rad/s, while the outer gear rack rotates counterclockwise with an angular velocity of 2 rad/s. Determine the angular velocity of gear A.

C

20 in.

15 in.

ωBC = 8 rad/s

A

B

D

ω = 2 rad/s

SOLUTION Kinematic Diagram: Since link BC is rotating about fixed point C. then vB is always directed perpendicular to link BC and its magnitude is vB = vBC rBC = 8(15) = 120 in. > s. At the instant shown. vB is directed to the left. Also, at the same instant, point E is moving to the right with a speed of vE = vB rCE = 2 (20) = 40 in. >s. Velocity Equation: Here, vB>E = vA rB>E = 5 vA which is directed to the left. Applying Eq. 16–15, we have vB = vE + vB>E + B A:

c 120 d = c 40 d + c 5v d ; : ;A - 120 = 40 - 5vA Ans.

vA = 32.0 rad s

Ans: vA = 32.0 rad>s 692

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16–63. If the angular velocity of link AB is vAB = 3 rad>s, determine the velocity of the block at C and the angular velocity of the connecting link CB at the instant u = 45° and f = 30°.

C θ = 45°

3 ft

ω A B = 3 rad/s

SOLUTION

2 ft φ = 30° B

vC = vB + vC>B

B vC R = C 6 ;

30°c

A

S + DvCB (3) T 45°b

+ ) (:

-vC = 6 sin 30° - vCB (3) cos 45°

(+ c )

0 = - 6 cos 30° + vCB (3) sin 45° vCB = 2.45 rad>s

Ans.

d

vC = 2.20 ft>s ;

Ans.

Also, vC = vB + v * rC>B - vC i = (6 sin 30°i - 6 cos 30°j) + (vCB k) * (3 cos 45°i + 3 sin 45°j) +b a:

-vC = 3 - 2.12vCB

(+ c )

0 = - 5.196 + 2.12vCB

vCB = 2.45 rad s

Ans.

d

vC = 2.20 ft s ;

Ans.

Ans: vCB = 2.45 rad>sd vC = 2.20 ft>s d 693

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*16–64. The pinion gear A rolls on the fixed gear rack B with an angular velocity v = 4 rad>s. Determine the velocity of the gear rack C.

C

A 0.3 ft

v

B

SOLUTION vC = vB + vC>B + ) (;

vC = 0 + 4(0.6) Ans.

vC = 2.40 ft>s Also: vC = vB + v * rC>B -vC i = 0 + (4k) * (0.6j)

Ans.

vC = 2.40 ft>s

Ans: vC = 2.40 ft>s vC = 2.40 ft>s 694

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16–65. The pinion gear rolls on the gear racks. If B is moving to the right at 8 ft>s and C is moving to the left at 4 ft>s, determine the angular velocity of the pinion gear and the velocity of its center A.

C

A 0.3 ft

V

B

SOLUTION vC = vB + vC>B + ) (:

- 4 = 8 - 0.6(v) Ans.

v = 20 rad>s vA = vB + vA>B + ) (:

vA = 8 - 20(0.3) vA = 2 ft>s :

Ans.

Also: vC = vB + v * rC>B - 4i = 8i + (vk) * (0.6j) -4 = 8 - 0.6v Ans.

v = 20 rad>s vA = vB + v * rA>B vAi = 8i + 20k * (0.3j) vA = 2 ft>s :

Ans.

Ans: v = 20 rad>s vA = 2 ft>s S 695

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16–66. Determine the angular velocity of the gear and the velocity of its center O at the instant shown.

A 45 4 ft/s

0.75 ft O 1.50 ft

3 ft/s

SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, vB = vC + v * rB>C 3i = -4i +

A - vk B * A 2.25j B

3i = A 2.25v - 4 B i Equating the i components yields (1)

3 = 2.25v - 4

Ans. (2)

v = 3.111 rad>s For points O and C, vO = vC + v * rO>C = - 4i +

A - 3.111k B * A 1.5j B

= [0.6667i] ft>s Thus, vO = 0.667 ft>s :

Ans.

Ans: v = 3.11 rad>s vO = 0.667 ft>s S 696

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16–67. Determine the velocity of point A on the rim of the gear at the instant shown.

A 45 4 ft/s

0.75 ft O 1.50 ft

3 ft/s

SOLUTION General Plane Motion: Applying the relative velocity equation to points B and C and referring to the kinematic diagram of the gear shown in Fig. a, vB = vC + v * rB>C 3i = - 4i +

A -vk B * A 2.25j B

3i = A 2.25v - 4 B i Equating the i components yields 3 = 2.25v - 4

(1)

v = 3.111 rad>s

(2)

For points A and C, vA = vC + v * rA>C

A vA B x i + A vA B y j = - 4i + A - 3.111k B * A -1.061i + 2.561j B A vA B x i + A vA B y j = 3.9665i + 3.2998j Equating the i and j components yields

A vA B x = 3.9665 ft>s

A vA B y = 3.2998 ft>s

Thus, the magnitude of vA is vA = 2 A vA B x 2 + A vA B y 2 = 23.96652 + 3.29982 = 5.16 ft>s

Ans.

and its direction is u = tan - 1 C

A vA B y A vA B x

S = tan - 1 ¢

3.2998 ≤ = 39.8° 3.9665

Ans.

Ans: vA = 5.16 ft>s u = 39.8° a 697

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*16–68. Knowing that angular velocity of link AB is vAB = 4 rad>s, determine the velocity of the collar at C and the angular velocity of link CB at the instant shown. Link CB is horizontal at this instant.

350 mm

C

45

Solution

vAB  4 rad/s

vB = vAB rAB  

B

= 4(0.5) = 2 m>s

500 mm

60

vB = { -2 cos 30°i + 2 sin 30°j } m/s   vC = - vC cos 45°i - vC sin 45°j A

v = vBCk    rC>B = { - 0.35i } m vC = vB + v * rC>B -vC cos 45°i - vC sin 45°j = ( -2 cos 30°i + 2 sin 30°j ) + (vBCk) * ( -0.35i ) -vC cos 45°i - vC sin 45°j = -2 cos 30°i + (2 sin 30° - 0.35vBC)j Equating the i and j components yields: -vC cos 45° = - 2 cos 30°     vC = 2.45 m>s

Ans.

-2.45 sin 45° = 2 sin 30° - 0.35vBC  vBC = 7.81 rad>s

Ans.

Ans: vC = 2.45 m>s vBC = 7.81 rad>s 698

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16–69. Rod AB is rotating with an angular velocity of vAB = 60 rad>s. Determine the velocity of the slider C at the instant u = 60° and f = 45°. Also, sketch the position of  bar BC when u = 30°, 60° and 90° to show its general plane motion.

300 mm B

600 mm f

u

A vAB  60 rad/s

Solution C

Rotation About Fixed Axis. For link AB, refer to Fig. a. VB = VAB * rAB   = (60k) * ( - 0.3 sin 60°i + 0.3 cos 60°j)   = { - 9i - 923j} m/s General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, VC = VB + VBC * rC>B - vC j = ( - 9i - 923 j ) + (vBCk) * ( - 0.6 sin 45°i - 0.6 cos 45°j) - vC j = ( 0.322vBC - 9 ) i +

( - 0.322vBC - 923 ) j

Equating i components, 0 = 0.322vBC - 9;  vBC = 1522 rad>s = 21.2 rad>s d Then, equating j components, - vC =

( - 0.322)(1522 ) - 923;  vC = 24.59 m>s = 24.6 m>s T  Ans.

The general plane motion of link BC is described by its orientation when u = 30°, 60° and 90° shown in Fig. c.

Ans: vC = 24.6 m>s T 699

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16–70. The angular velocity of link AB is vAB = 5 rad>s. Determine the velocity of block C and the angular velocity of link BC at the instant u = 45° and f = 30°. Also, sketch the position of link CB when u = 45°, 60°, and 75° to show its general plane motion.

A

vAB  5 rad/s 3m u B

Solution f

Rotation About A Fixed Axis. For link AB, refer to Fig. a.

2m

vB = VAB * rAB = (5k) * ( - 3 cos 45°i - 3 sin 45°j ) = •

C

1522 1522 i j ¶ m>s 2 2

General Plane Motion. For link BC, refer to Fig. b. Applying the relative velocity equation, vC = vB + VBC * rC>B vC i = °

1522 1522 i j ¢ + (vBC k) * (2 sin 30° i - 2 cos 30°j) 2 2

vC i = °

1522 1522 + 23 vBC ¢ i + °vBC ¢j 2 2

Equating j components, O = vBC -

1522 1522 rad>s = 10.6 rad>s d ; vBC = 2 2

Ans.

Then, equating i components, vC =

1522 1522 ¢ = 28.98 m>s = 29.0 m>s S  + 23 ° 2 2

Ans.

700

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16–70. Continued

The general plane motion of link BC is described by its orientation when u = 45°, 60° and 75° shown in Fig. c

Ans: vBC = 10.6 rad>s d vC = 29.0 m>s S 701

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16–71. The similar links AB and CD rotate about the fixed pins at A and C. If AB has an angular velocity vAB = 8 rad>s, determine the angular velocity of BDP and the velocity of point P.

B

300 mm

300 mm

D 300 mm

300 mm 60

60

C

A vAB  8 rad/s

700 mm

Solution vD = vB + v * rD>B -vD cos 30°i - vD sin 30°j = - 2.4 cos 30°i + 2.4 sin 30°j + (vk) * (0.6i)

P

-vD cos 30° = -2.4 cos 30° -vD sin 30° = 2.4 sin 30° + 0.6v vD = 2.4 m>s Ans.

v = - 4 rad>s vP = vB + v * rP>B vP = - 2.4 cos 30°i + 2.4 sin 30°j + ( -4k) * (0.3i - 0.7j)

( vP ) x = - 4.88 m>s ( vP ) y = 0 vP = 4.88 m>s d 

Ans.

Ans: vP = 4.88 m>s d 702

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*16–72. If the slider block A is moving downward at vA = 4 m>s, determine the velocities of blocks B and C at the instant shown.

B 4

250 mm 400 mm

E

300 mm 3

5

D

vA  4 m/s

300 mm 30 C

Solution vB = vA + vB>A vB S = 4T + vAB(0.55) 3 5

+ (S )   vB = 0 + vAB(0.55)a b

4 5

( + c)   0 = - 4 + vAB(0.55)a b Solving, vAB = 9.091 rad>s Ans.

vB = 3.00 m>s   vD = vA + vD>A vD = 4 + [(0.3)(9.091) = 2.727]     T

3

   4 Q5

  vC = vD + vC>D vC = 4 + 2.727 + vCE(0.4) S

T  

5 3      h 30° Q 4

3 5

+ (S )   vC = 0 + 2.727 a b - vCE (0.4)(sin 30°)

4 5

( + c )   0 = - 4 + 2.727 a b + vCE(0.4)(cos 30°) vCE = 5.249 rad>s

Ans.

vC = 0.587 m>s Also:   vB = vA + vAB * rB>A   vBi = -4j + ( - vABk) * e

-4 3 (0.55)i + (0.55)j f 5 5

703

A

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*16–72. Continued

vB = vAB(0.33) 0 = - 4 + 0.44vAB vAB = 9.091 rad>s Ans.

vB = 3.00 m>s vD = vA + vAB * rB>A vD = -4j + ( -9.091k) * e

-4 3 (0.3)i + (0.3)j f 5 5

vD = 51.636i - 1.818j 6 m>s vC = vD + vCE * rC>D

vCi = (1.636i - 1.818j) + ( - vCEk) * ( - 0.4 cos 30°i - 0.4 sin 30°j) vC = 1.636 - 0.2vCE 0 = - 1.818 - 0.346vCE vCE = 5.25 rad>s Ans.

vC = 0.587 m>s

Ans: vB = vC = vB = vC = 704

3.00 m>s 0.587 m>s 3.00 m>s 0.587 m>s

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16–73. If the slider block A is moving downward at vA = 4 m>s, determine the velocity of point E at the instant shown.

B 4

250 mm 400 mm

E

300 mm 3

5

D

vA  4 m/s

300 mm 30 C

Solution

A

See solution to Prob. 16–87. vE = vD + vE>D S vE = 4T + 2.727 + (5.249)(0.3) 3

          4

Q5     f 30°

3 5 4 ( + T )   (vE)y = 4 - 2.727 a b + 5.249(0.3)(cos 30°) 5 + (S )   (vE)x = 0 + 2.727 a b + 5.249(0.3)(sin 30°)

(vE)x = 2.424 m>s S (vE)y = 3.182 m>s T

vE = 2(2.424)2 + (3.182)2 = 4.00 m>s

Ans.

u = tan-1 a

Ans.

Also:

3.182 b = 52.7° 2.424

See solution to Prob. 16–87. vE = vD + vCE * rE>D vE = (1.636i - 1.818j) + ( - 5.25k) * {cos 30°(0.3)i - 0.4 sin 30°(0.3)j} vE = 52.424i - 3.182j 6 m>s

vE = 2(2.424)2 + (3.182)2 = 4.00 m>s

Ans.

u = tan-1 a

Ans.

3.182 b = 52.7° 2.424

Ans: vE = 4.00 m>s u = 52.7° c 705

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16–74. The epicyclic gear train consists of the sun gear A which is in mesh with the planet gear B. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the connecting link DE pinned to B and C is rotating at vDE = 18 rad>s about the pin at E, determine the angular velocities of the planet and sun gears.

100 mm 600 mm

A

B C

E

D

SOLUTION

200 mm vD E

vD = rDE vDE = (0.5)(18) = 9 m>s c The velocity of the contact point P with the ring is zero.

18 rad/s 300 mm

R

vD = vP + v * rD>P 9j = 0 + ( -vB k) * ( - 0.1i) vB = 90 rad>s

Ans.

b

Let P¿ be the contact point between A and B. vP¿ = vP + v * rP¿>P vP¿ j = 0 + (- 90k) * ( -0.4i) vP¿ = 36 m>s c vA =

vP¿ 36 = 180 rad>s = rA 0.2

Ans.

d

Ans: vB = 90 rad>s b vA = 180 rad>s d 706

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16–75. If link AB is rotating at vAB = 3 rad>s, determine the angular velocity of link CD at the instant shown.

6 in. B A

vAB

30 8 in.

Solution vB = vAB * rB>A

C

vC = vCD * rC>D

vCD

vC = vB + vBC * rC>B

45

(vCD k) * ( - 4 cos 45°i + 4 sin 45°j) = ( - 3k) * (6i) + (vBCk) * ( -8 sin 30°i - 8 cos 30°j)

D

4 in.

- 2.828vCD = 0 + 6.928vBC - 2.828vCD = - 18 - 4vBC Solving, vBC = - 1.65 rad>s Ans.

vCD = 4.03 rad>s

Ans: vCD = 4.03 rad>s 707

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*16–76. If link CD is rotating at vCD = 5 rad>s, determine the angular velocity of link AB at the instant shown.

6 in. B A

vAB

30 8 in.

Solution vB = vAB * rB>A

C

vC = vCD * rC>D

vCD

vB = vC + vBC * rB>C

45

( - vAB k) * (6i) = (5k) * ( - 4 cos 45°i + 4 sin 45°j) + (vBCk) * (8 sin 30°i + 8 cos 30°j)

D

4 in.

0 = - 14.142 - 6.9282vBC - 6vAB = -14.142 + 4vBC Solving, Ans.

vAB = 3.72 rad>s vBC = - 2.04 rad>s

Ans: vAB = 3.72 rad>s 708

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16–77. The planetary gear system is used in an automatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, vR = 0, and the sun gear S is rotating at vS = 5 rad>s. Determine the angular velocity of each of the planet gears P and shaft A.

40 mm vR P

R

vS

S

SOLUTION

A

80 mm

vA = 5(80) = 400 mm>s ; vB = 0 vB = vA + v * rB>A 0 = -400i + (vp k) * (80j)

40 mm

0 = -400i - 80vp i Ans.

vP = -5 rad>s = 5 rad>s vC = vB + v * rC>B vC = 0 + ( - 5k) * ( - 40j) = -200i vA =

200 = 1.67 rad>s 120

Ans.

Ans: vP = 5 rad>s vA = 1.67 rad>s 709

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16–78. If the ring gear A rotates clockwise with an angular velocity of vA = 30 rad>s, while link BC rotates clockwise with an angular velocity of vBC = 15 rad>s, determine the angular velocity of gear D.

A

vA 30 rad/s D

C vBC  15 rad/s 250 mm

300 mm

Solution

B

Rotation About A Fixed Axis. The magnitudes of the velocity of Point E on the rim and center C of gear D are vE = vArA = 30(0.3) = 9 m>s vC = vBCrBC = 15(0.25) = 3.75 m>s General Plane Motion. Applying the relative velocity equation by referring to Fig. a, vE = vC + VD * rE>C 9i = 3.75i + ( - vDk) * (0.05j) 9i = (3.75 + 0.05vD)i Equating i component, 9 = 3.75 + 0.05vD Ans.

vD = 105 rad>s b

Ans: vD = 105 rad>s b 710

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16–79. The mechanism shown is used in a riveting machine. It consists of a driving piston A, three links, and a riveter which is attached to the slider block D. Determine the velocity of D at the instant shown, when the piston at A is traveling at vA = 20 m>s.

C 150 mm

45° 200 mm

D

SOLUTION

45°

B

60°

300 mm vA = 20 m/s A

30° 45°

Kinematic Diagram: Since link BC is rotating about fixed point B, then vC is always directed perpendicular to link BC. At the instant shown. vC = - vC cos 30°i + vC sin 30°j = - 0.8660 vC i + 0.500 vC j. Also, block D is moving towards the negative y axis due to the constraint of the guide. Then. vD = -vD j. Velocity Equation: Here, vA = { - 20 cos 45°i + 20 sin 45°j}m>s = { -14.14i + 14.14j} m>s and rC>A = { - 0.3 cos 30°i + 0.3 sin 30°j }m = {- 0.2598i + 0.150j } m. Applying Eq. 16–16 to link AC, we have vC = vA + vAC * rC>A -0.8660 vC i + 0.500 vC j = - 14.14i + 14.14j + (vAC k) * (- 0.2598i + 0.150j) - 0.8660 vC i + 0.500 vC j = - (14.14 + 0.150 vAC) i + (14.14 - 0.2598vAC)j Equating i and j components gives - 0.8660 vC = - (14.14 + 0.150 vAC)

[1]

0.500 vC = 14.14 - 0.2598 vAC

[2]

Solving Eqs. [1] and [2] yields vAC = 17.25 rad>s

vC = 19.32 m>s

Thus, vC = { -19.32 cos 30°i + 19.32 sin 30°j} m>s = {- 16.73i + 9.659j} m>s and rD>C = { -0.15 cos 45°i - 0.15 sin 45°j }m = {-0.1061i - 0.1061j } m. Applying Eq. 16–16 to link CD, we have vD = vC + vCD * rD>C - vD j = - 16.73i + 9.659j + (vCD k) * ( -0.1061i - 0.1061j) -vD j = (0.1061vCD - 16.73) i + (9.659 - 0.1061vCD)j Equating i and j components gives 0 = 0.1061vCD - 16.73

[3]

- vD = 9.659 - 0.1061 vCD

[4]

Solving Eqs. [3] and [4] yields vCD = 157.74 rad s Ans.

vD = 7.07 m s

Ans: vD = 7.07 m>s 711

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*16–80. The mechanism is used on a machine for the manufacturing of a wire product. Because of the rotational motion of link AB and the sliding of block F, the segmental gear lever DE undergoes general plane motion. If AB is rotating at vAB = 5 rad>s, determine the velocity of point E at the instant shown.

E

50 mm 45

20 mm F

D

C

20 mm 200 mm

Solution

B

vB = vABrAB = 5(50) = 250 mm>s 45°

45

vC = vB + vC>B

  45°

( + c )   0 = 250 sin 45° - vBC(200) sin 45° + )   vC = 250 cos 45° + vBC(200) cos 45° (d Solving, vC = 353.6 mm>s;  vBC = 1.25 rad>s vp = vC + vp>C vp = 353.6 + [(1.25)(20) = 25] S      d         T vD = vp + vD>p vD = (353.6 + 25) + 20vDE d      d        T     c + )   vD = 353.6 + 0 + 0 (d

( + T )   0 = 0 + (1.25)(20) - vDE(20) Solving, vD = 353.6 mm>s;    vDE = 1.25 rad>s vE = vD + vE>D vE = 353.6 + 1.25(50) f

d   

A

45

vAB  5 rad/s

vC = 250 + vBC(200) d   45°

50 mm

45°

+ )   vE cos f = 353.6 - 1.25(50) cos 45° (d

( + c )   vE sin f = 0 + 1.25(50) sin 45° Solving, vE = 312 mm>s

Ans.

f = 8.13°

Ans.

712

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*16–80. Continued

Also; vB = vAB * rB>A vC = vB + vBC * rC>B - vCi = ( - 5k) * ( - 0.05 cos 45°i - 0.05 sin 45°j) + (vBCk) * ( -0.2 cos 45°i + 0.2 sin 45°j) - vC = -0.1768 - 0.1414vBC 0 = 0.1768 - 0.1414vBC vBC = 1.25 rad>s,  vC = 0.254 m>s vp = vC + vBC * rp>C vD = vp + vDE * rD>p vD = vC + vBC * rp>C + vDE * rD>p vDi = - 0.354i + (1.25k) * ( -0.02i) + (vDEk) * ( -0.02i) vD = - 0.354 0 = -0.025 - vDE(0.02) vD = 0.354 m>s,  vDE = 1.25 rad>s vE = vD + VDE * rE>D (vE)xi + (vE)yj = -0.354i + ( - 1.25k) * ( - 0.05 cos 45°i + 0.05 sin 45°j) (vE)x = -0.354 + 0.0442 = - 0.3098 (vE)y = 0.0442 vE = 2( - 0.3098)2 + (0.0442)2 = 312 mm>s

Ans.

f = tan-1a

Ans.

0.0442 b = 8.13° 0.3098

Ans: vE = 312 mm>s f = 8.13° vE = 312 mm>s f = 8.13° 713

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16–81. In each case show graphically how to locate the instantaneous center of zero velocity of link AB. Assume the geometry is known.

B v

A

A (a) v v

A

SOLUTION a)

B

(c)

b)

c)

714

(b)

C B

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16–82. Determine the angular velocity of link AB at the instant shown if block C is moving upward at 12 in.>s.

A

VAB C 5 in.

45

4 in. 30

B

SOLUTION rIC-B rIC - C 4 = = sin 45° sin 30° sin 105° rIC-C = 5.464 in. rIC-B = 2.828 in. vC = vB C(rIC - C) 12 = vB C(5.464) vB C = 2.1962 rad>s vB = vB C(rIC-B) = 2.1962(2.828) = 6.211 in.>s vB = vAB rAB 6.211 = vAB(5) Ans.

vAB = 1.24 rad>s

Ans: vAB = 1.24 rad>s 715

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16–83. The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the angular velocity of the link CB at the instant shown, if the link AB is rotating at 4 rad>s.

125 mm C 45°

SOLUTION

B

ω AB = 4 rad/s

Kinematic Diagram: Since linke AB is rotating about fixed point A, then vB is always directed perpendicular to link AB and its magnitude is vB = vAB rAB = 4(0.3) = 1.20 m>s. At the instant shown, vB is directed at an angle 30° with the horizontal. Also, block C is moving horizontally due to the constraint of the guide.

300 mm 60°

A

Instantaneous Center: The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC. Using law of sines, we have rB>IC

=

sin 45° rC>IC sin 105°

0.125 sin 30°

=

0.125 sin 30°

rB>IC = 0.1768 m rC>IC = 0.2415 m

The angular velocity of bar BC is given by vBC =

vB 1.20 = 6.79 rad s = rB IC 0.1768

Ans.

Ans: vBC = 6.79 rad>s 716

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*16–84. The conveyor belt is moving to the right at v = 8 ft>s, and at the same instant the cylinder is rolling counterclockwise at v = 2 rad>s without slipping. Determine the velocities of the cylinder’s center C and point B at this instant.

B v

1 ft

C v

A

Solution rA - IC =

8 = 4 ft 2

vC = 2(3) = 6.00 ft>s S

Ans.

vB = 2(2) = 4.00 ft>s S Ans.

Ans: vC = 6.00 ft>s S

vB = 4.00 ft>s S

717

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16–85. The conveyor belt is moving to the right at v = 12 ft>s, and at the same instant the cylinder is rolling counterclockwise at v = 6 rad>s while its center has a velocity of 4 ft>s to the left. Determine the velocities of points A and B on the disk at this instant. Does the cylinder slip on the conveyor?

B v

1 ft

C v

A

Solution rA - IC =

4 = 0.667 ft 6

vA = 6(1 - 0.667) = 2 ft>s S

Ans.

vB = 6(1 + 0.667) = 10 ft>s d Ans. Since vA ≠ 12 ft>s the cylinder slips on the conveyer.

Ans.

Ans: vA = 2 ft>s S vB = 10 ft>s d The cylinder slips. 718

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16–86. As the cord unravels from the wheel’s inner hub, the wheel is rotating at v = 2 rad>s at the instant shown. Determine the velocities of points A and B. ω = 2 rad/s 5 in.

SOLUTION rB>IC = 5 + 2 = 7 in.

B 2

2

rA>IC = 22 + 5 = 229 in.

O 2 in.

yB = v rB>IC = 2(7) = 14 in.>s T

Ans.

yA = v rA>IC = 2 A 229 B = 10.8 in.>s

Ans.

2 ¬ u = tan-1 a b = 21.8° R 5

Ans.

A

Ans: vB = 14 in.>s T vA = 10.8 in.>s u = 21.8° c 719

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16–87. If rod CD is rotating with an angular velocity vCD = 4 rad>s, determine the angular velocities of rods AB and CB at the instant shown.

A

30

0.4 m

1m

C

B 0.5 m vC D  4 rad/s

Solution

D

Rotation About A Fixed Axis. For links AB and CD, the magnitudes of the velocities of C and D are vC = vCDrCD = 4(0.5) = 2.00 m>s

vB = vABrAB = vAB(1)

And their direction are indicated in Fig. a and b. General Plane Motion. With the results of vC and vB, the IC for link BC can be located as shown in Fig. c. From the geometry of this figure, rC>IC = 0.4 tan 30° = 0.2309 m  rB>IC =

0.4 = 0.4619 m cos 30°

Then, the kinematics gives vC = vBCrC>IC;  2.00 = vBC(0.2309) Ans.

         vBC = 8.6603 rad>s = 8.66 rad>s d vB = vBCrB>IC;  vAB(1) = 8.6603(0.4619)

Ans.

          vAB = 4.00 rad>s b

Ans: vBC = 8.66 rad>s d vAB = 4.00 rad>s b 720

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*16–88. If bar AB has an angular velocity vAB = 6 rad>s, determine the velocity of the slider block C at the instant shown.

vAB = 6 rad/s

B 200 mm

A

= 45°

500 mm 30°

C

SOLUTION Kinematic Diagram: Since link AB is rotating about fixed point A, then vB is always directed perpendicular to link AB and its magnitude is yB = vAB rAB = 6(0.2) = 1.20 m>s. At the instant shown. vB is directed with an angle 45° with the horizontal. Also, block C is moving horizontally due to the constraint of the guide. Instantaneous Center: The instantaneous center of zero velocity of bar BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC . Using law of sine, we have rB>IC sin 60° rC>IC sin 75°

=

0.5 sin 45°

rB>IC = 0.6124 m

=

0.5 sin 45°

rC>IC = 0.6830 m

The angular velocity of bar BC is given by vBC =

yB 1.20 = 1.960 rad>s = rB>IC 0.6124

Thus, the velocity of block C is yC = vBC rC IC = 1.960(0.6830) = 1.34 m s ;

Ans.

Ans: vC = 1.34 m>s d 721

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16–89. Show that if the rim of the wheel and its hub maintain contact with the three tracks as the wheel rolls, it is necessary that slipping occurs at the hub A if no slipping occurs at B. Under these conditions, what is the speed at A if the wheel has angular velocity V?

v r2 r1

A

B

Solution IC is at B.

vA = v(r2 - r1) S

Ans.



Ans: vA = v (r2 - r1) 722

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16–90. 10 ft/s B

vB

Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point D at this instant.

D E

45 C A

0.8 ft 30 F vA

5 ft/s

SOLUTION x 1.6 - x = 5 10 5x = 16 - 10x x = 1.06667 ft v =

10 = 9.375 rad>s 1.06667

rIC-D = 2(0.2667)2 + (0.8)2 - 2(0.2667)(0.8) cos 135° = 1.006 ft sin f sin 135° = 0.2667 1.006 f = 10.80° vC = 0.2667(9.375) = 2.50 ft>s :

Ans.

vD = 1.006(9.375) = 9.43 ft>s

Ans.

u = 45° + 10.80° = 55.8° h

Ans: vC = 2.50 ft>s d vD = 9.43 ft>s u = 55.8° h 723

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16–91. Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point E at this instant.

10 ft/s B

vB

D E

45 C A

0.8 ft 30 F vA

5 ft/s

SOLUTION x 1.6 - x = 5 10 5x = 16 - 10x x = 1.06667 ft v =

10 = 9.375 rad>s 1.06667

vC = v(rIC - C) = 9.375(1.06667 - 0.8) = 2.50 ft>s :

Ans.

vE = v(rIC - E) = 9.3752(0.8)2 + (0.26667)2 Ans.

= 7.91 ft>s u = tan–1

) 0.26667 ) = 18.4° 0.8

Ans: vC = 2.50 ft>s d vE = 7.91 ft>s u = 18.4° e 724

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*16–92. Member AB is rotating at vAB = 6 rad>s. Determine the velocity of point D and the angular velocity of members BPD and CD.

B

200 mm

200 mm

D 200 mm

200 mm 60 A

60

250 mm

C

vAB  6 rad/s P

Solution Rotation About A Fixed Axis. For links AB and CD, the magnitudes of the velocities of B and D are vB = vABrAB = 6(0.2) = 1.20 m>s   vD = vCD(0.2) And their directions are indicated in Figs. a and b. General Plane Motion. With the results of vB and vD, the IC for member BPD can be located as show in Fig. c. From the geometry of this figure, rB>IC = rD>IC = 0.4 m Then, the kinematics gives vBPD =

vB 1.20 = = 3.00 rad>s b rB>IC 0.4

Ans. Ans.

vD = vBPDrD>IC = (3.00)(0.4) = 1.20 m>s b Thus, vD = vCD(0.2);  1.2 = vCD(0.2)

Ans.

vCD = 6.00 rad>s d

Ans: vBPD = 3.00 rad>s  b vD = 1.20 m>s b vCD = 6.00 rad>s d 725

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16–93. Member AB is rotating at vAB = 6 rad>s. Determine the velocity of point P, and the angular velocity of member BPD.

B

200 mm

200 mm

D 200 mm

200 mm 60 A

60

250 mm

C

vAB  6 rad/s P

Solution Rotation About A Fixed Axis. For links AB and CD, the magnitudes of the velocities of B and D are vB = vABrAB = 6(0.2) = 1.20 m>s  vD = vCD(0.2) And their direction are indicated in Fig. a and b General Plane Motion. With the results of vB and vD, the IC for member BPD can be located as shown in Fig. c. From the geometry of this figure   rB>IC = 0.4 m  rP>IC = 0.25 + 0.2 tan 60° = 0.5964 m Then the kinematics give vBPD =

vB 1.20 = = 3.00 rad>s b rB>IC 0.4

Ans.

vP = vBPDrP>IC = (3.00)(0.5964) = 1.7892 m>s = 1.79 m>s d Ans.

Ans: vBPD = 3.00 rad>s b vP = 1.79 m>s d 726

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16–94. The cylinder B rolls on the fixed cylinder A without slipping. If connected bar CD is rotating with an angular velocity vCD = 5 rad>s, determine the angular velocity of cylinder B. Point C is a fixed point.

0.3 m

0.1 m D C

A

vCD  5 rad/s

B

Solution vD = 5(0.4) = 2 m>s vB =

2 = 6.67 rad>s 0.3

Ans.

Ans: vB = 6.67 rad>s 727

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16–95. As the car travels forward at 80 ft/s on a wet road, due to slipping, the rear wheels have an angular velocity v = 100 rad>s. Determine the speeds of points A, B, and C caused by the motion.

80 ft/s C B 1.4 ft

A 100 rad/s

SOLUTION r =

80 = 0.8 ft 100

vA = 0.6(100) = 60.0 ft s :

Ans.

vC = 2.2(100) = 220 ft s ;

Ans.

vB = 1.612(100) = 161 ft s

60.3°

b

Ans.

Ans: vA = 60.0 ft>s S vC = 220 ft>s d vB = 161 ft>s u = 60.3° b 728

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*16–96. The pinion gear A rolls on the fixed gear rack B with an angular velocity v = 8 rad>s. Determine the velocity of the gear rack C.

C

A v

150 mm

B

Solution General Plane Motion. The location of IC for the gear is at the bottom of the gear where it meshes with gear rack B as shown in Fig. a. Thus,



vC = vrC>IC = 8(0.3) = 2.40 m>s d

Ans.

Ans: vC = 2.40 m>s d 729

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16–97. If the hub gear H and ring gear R have angular velocities vH = 5 rad>s and vR = 20 rad>s, respectively, determine the angular velocity vS of the spur gear S and the angular velocity of its attached arm OA.

ωR 50 mm

250 mm H

150 mm

SOLUTION

A S

ωS

O

ωH R

5 0.75 = 0.1 - x x x = 0.01304 m vS =

0.75 = 57.5 rad>s d 0.01304

Ans.

vA = 57.5(0.05 - 0.01304) = 2.125 m>s vOA =

2.125 = 10.6 rad>s d 0.2

Ans.

Ans: vS = 57.5 rad>sd vOA = 10.6 rad>sd 730

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16–98. ωR

If the hub gear H has an angular velocity vH = 5 rad>s, determine the angular velocity of the ring gear R so that the arm OA attached to the spur gear S remains stationary (vOA = 0). What is the angular velocity of the spur gear?

50 mm

250 mm H

150 mm

SOLUTION

A S

ωS

O

ωH R

The IC is at A. vS =

0.75 = 15.0 rad>s 0.05

Ans.

vR =

0.75 = 3.00 rad>s 0.250

Ans.

Ans: vS = 15.0 rad>s vR = 3.00 rad>s 731

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16–99. The crankshaft AB rotates at vAB = 50 rad>s about the fixed axis through point A, and the disk at C is held fixed in its support at E. Determine the angular velocity of rod CD at the instant shown.

E

C

75 mm 75 mm

40 mm

SOLUTION

F

D

300 mm

rB>IC =

0.3 = 0.6 m sin 30°

rF>IC =

0.3 = 0.5196 m tan 30°

vBF =

60

vAB

50 rad/s

A

B 100 mm

5 = 8.333 rad>s 0.6

vF = 8.333(0.5196) = 4.330 m>s Thus, vCD =

4.330 = 57.7 rad>sd 0.075

Ans.

Ans: vCD = 57.7 rad>sd 732

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*16–100. Cylinder A rolls on the fixed cylinder B without slipping. If bar CD is rotating with an angular velocity of vCD = 3 rad>s, determine the angular velocity of A.

C 200 mm

A

vCD

Solution

200 mm

Rotation About A Fixed Axis. The magnitude of the velocity of C is



vC = vCDrDC

D B

= 3(0.4) = 1.20 m>s S

General Plane Motion. The IC for cylinder A is located at the bottom of the cylinder where it contacts with cylinder B, since no slipping occurs here, Fig. b. vC = vA rC>IC;  1.20 = vA(0.2)    vA = 6.00 rad>s b

Ans.

Ans: vA = 6.00 rad>s b 733

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16–101. The planet gear A is pin connected to the end of the link BC. If the link rotates about the fixed point B at 4 rad>s, determine the angular velocity of the ring gear R. The sun gear D is fixed from rotating.

R D vBC  4 rad/s B 150 mm

A C 75 mm

vR

Solution Gear A: vC = 4(225) = 900 mm>s vA =

vR 900 = 75 150

vR = 1800 mm>s

Ring gear: vR =

1800 = 4 rad>s 450

Ans.

Ans: vR = 4 rad>s 734

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16–102. Solve Prob. 16–101 if the sun gear D is rotating clockwise at vD = 5 rad>s while link BC rotates counterclockwise at vBC = 4 rad>s.

R D vBC  4 rad/s B 150 mm

A C 75 mm

vR

Solution Gear A: vP = 5(150) = 750 mm>s vC = 4(225) = 900 mm>s x 75 - x = 750 900 x = 34.09 mm v =

750 = 22.0 rad>s 34.09

vR = [75 + (75 - 34.09)](22) = 2550 mm>s

Ring gear: 750 2550 = x x + 450 x = 187.5 mm vR =

750 = 4 rad>s d 187.5

Ans.

Ans: vR = 4 rad>s 735

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16–103. Bar AB has the angular motions shown. Determine the velocity and acceleration of the slider block C at this instant.

B

vAB  4 rad/s

0.5 m

aAB  6 rad/s2

A

Solution

45

1m

60

Rotation About A Fixed Axis. For link AB, refer to Fig. a.

C

vB = vABrAB = 4(0.5) = 2.00 m>s 45°b aB = AAB * rAB - v2AB rAB = 6k * (0.5 cos 45°i + 0.5 sin 45°j) - 42(0.5 cos 45°i + 0.5 sin 45°j) = 5 - 5.522i - 2.522j 6 m>s2

General Plane Motion. The IC of link BC can be located using vB and vC as shown in Fig. b. From the geometry of this figure, rB>IC

=

sin 30° rC>IC sin 105°

=

22 1 m ;  rB>IC = 2 sin 45° 1 ;   rC>IC = 1.3660 m sin 45°

Then the kinematics gives, vB = vBC rB>IC;  2 = vBC a vC = vBCrB>IC;  vC =

22 b   vBC = 222 rad>sb 2

1 222 2 (1.3660)

= 3.864 m>s = 3.86 m>s d  Ans.

Applying the relative acceleration equation by referring to Fig. c, aC = aB + ABC * rC>B - V2BC rC>B - aCi =

( - 5.522i - 2.522j ) + ( - aBCk) * (1 cos 60°i - 1 sin 60°j) - ( 222 ) 2(1 cos 60°i - 1 sin 60°j)

- aCi = a -

23 a - 11.7782bi + (3.3927 - 0.5aBC)j 2 BC

Equating j components, 0 = 3.3927 - 0.5aBC;  aBC = 6.7853 rad>s2b Then, i component gives - aC = -

23 (6.7853) - 11.7782;  aC = 17.65 m>s2 = 17.7 m>s2 d  Ans. 2

Ans: vC = 3.86 m>s d aC = 17.7 m>s2 d 736

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*16–104. At a given instant the bottom A of the ladder has an acceleration aA = 4 ft>s2 and velocity vA = 6 ft>s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant. 16 ft

B

SOLUTION v =

6 = 0.75 rad>s 8

A

30

aB = aA + (aB>A)n + (aB>A)t aB = 4 + (0.75)2 (16) + a(16) ; 30° d 30° f T

+ ) (;

0 = 4 + (0.75)2(16) cos 30° - a(16) sin 30°

(+ T )

aB = 0 + (0.75)2(16) sin 30° + a(16) cos 30°

Solving, a = 1.47 rad>s2

Ans.

aB = 24.9 ft>s2 T

Ans.

Also: aB = aA + a * rB>A - v2rB>A - aBj = - 4i + (ak) * (16 cos 30°i + 16 sin 30°j) - (0.75)2(16 cos 30°i + 16 sin 30°j) 0 = -4 - 8a - 7.794 - aB = 13.856a - 4.5 a = 1.47 rad>s2

Ans.

aB = 24.9 ft>s2 T

Ans.

Ans: a = 1.47 rad>s2 aB = 24.9 ft>s2 T a = 1.47 rad>s2 aB = 24.9 ft>s2 T 737

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16–105. At a given instant the top B of the ladder has an acceleration aB = 2 ft>s2 and a velocity of vB = 4 ft>s, both acting downward. Determine the acceleration of the bottom A of the ladder, and the ladder’s angular acceleration at this instant. 16 ft

B

SOLUTION v =

4 = 0.288675 rad>s 16 cos 30°

A

30

aA = aB + a * rA>B - v2rA>B - aAi = - 2j + (ak) * ( -16 cos 30°i - 16 sin 30°j) - (0.288675)2( -16 cos 30°i - 16 sin 30°j) - aA = 8a + 1.1547 0 = -2 - 13.856 a + 0.6667 a = -0.0962 rad>s2 = 0.0962 rad>s2b

Ans.

aA = -0.385 ft>s2 = 0.385 ft>s2 :

Ans.

Ans: a = 0.0962 rad>s2 b aA = 0.385 ft>s2 S 738

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16–106. Member AB has the angular motions shown. Determine the velocity and acceleration of the slider block C at this instant.

B

2m C

Solution

5

3

A

4 rad/s 5 rad/s2

0.5 m

4

Rotation About A Fixed Axis. For member AB, refer to Fig. a. vB = vABrAB = 4(2) = 8 m>s d aB = AAB * rAB - v2AB rAB = ( - 5k) * (2j) - 42(2j) = 5 10i - 32j 6 m>s2

General Plane Motion. The IC for member BC can be located using vB and vC as shown in Fig. b. From the geometry of this figure

Then

3 f = tan-1a b = 36.87°   u = 90° - f = 53.13° 4 rB>IC - 2 0.5

= tan 53.13;  rB>IC = 2.6667 m

0.5 = cos 53.13;  rC>IC = 0.8333 m rC>IC The kinematics gives vB = vBCrB>IC;  8 = vBC(2.6667)

vBC = 3.00 rad>sd Ans.

vC = vBCrC>IC = 3.00(0.8333) = 2.50 m>s b Applying the relative acceleration equation by referring to Fig. c, aC = aB + ABC * rC>B - v2BC rC>B

4 3 - aC a bi - aC a bj = (10i - 32j) + aBC k * ( - 0.5i - 2j) - ( 3.002 ) ( -0.5i - 2j) 5 5

4 3 - aC i - aC j = (2aBC + 14.5)i + ( - 0.5aBC - 14)j 5 5 Equating i and j components

4 - aC = 2aBC + 14.5(1) 5 3 - aC = - 0.5aBC - 14(2) 5 Solving Eqs. (1) and (2), aC = 12.969 m>s2 = 13.0 m>s2 b 2

aBC = - 12.4375 rad>s

Ans. 2

Ans.

= 12.4 rad>s  b

The negative sign indicates that aBC is directed in the opposite sense from what is shown in Fig. (c). Ans: aC = 13.0 m>s2 b aBC = 12.4 rad>s2 b 739

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16–107. At a given instant the roller A on the bar has the velocity and acceleration shown. Determine the velocity and acceleration of the roller B, and the bar’s angular velocity and angular acceleration at this instant.

4 m/s 6 m/s2

A 30 0.6 m 30

Solution General Plane Motion. The IC of the bar can be located using vA and vB as shown in Fig. a. From the geometry of this figure, rA>IC = rB>IC = 0.6 m

B

Thus, the kinematics give vA = vrA>IC;  4 = v(0.6)

Ans.

v = 6.667 rad>s = 6.67 rad>s d

Ans.

vB = vrB>IC = 6.667(0.6) = 4.00 m>s R Applying the relative acceleration equation, by referring to Fig. b, aB = aA + A * rB>A - v2 rB>A aB cos 30°i - aB sin 30°j = - 6j + (ak) * (0.6 sin 30°i - 0.6 cos 30°j)

- ( 6.6672 ) (0.6 sin 30°i - 0.6 cos 30°j)



23 1 a i - aB j = ( 0.323a - 13.33 ) i + (0.3a + 17.09)j 2 B 2 Equating i and j components, 23 a = 0.323a - 13.33 2 B 1 - aB = 0.3a + 17.09 2

(1) (2)

Solving Eqs. (1) and (2) a = - 15.66 rad>s2 = 15.7 rad>s2 b

Ans.

2

Ans.

2

aB = - 24.79 m>s = 24.8 m>s a

The negative signs indicate that A and aB are directed in the senses that opposite to those shown in Fig. b

Ans: v = 6.67 rad>s d vB = 4.00 m>s R a = 15.7 rad>s2b aB = 24.8 m>s2 a 740

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*16–108. The rod is confined to move along the path due to the pins at its ends. At the instant shown, point A has the motion shown. Determine the velocity and acceleration of point B at this instant.

vA = 6 ft/s aA = 3 ft/s2 A 5f

B

SOLUTION

t

3 ft

vB = vA + v * rB>A vB j = 6i + (- vk) * ( -4i - 3 j) 0 = 6 - 3v,

v = 2 rad>s

vB = 4v = 4(2) = 8 ft>s c

Ans.

aB = aA + a * rB>A - v2 rB>A 21.33i + (aB)t j = - 3i + ak * ( -4i - 3j) - ( - 2)2 ( -4i - 3j) + B A:

21.33 = - 3 + 3a + 16;

A+cB

(aB)t = - (2.778)(4) + 12 = 0.8889 ft>s2

a = 2.778 rad>s2

aB = 2(21.33)2 + (0.8889)2 = 21.4 ft>s2 u = tan - 1 a

Ans.

0.8889 b = 2.39° Q ¬ 21.33

Ans.

Ans: vB = 8 ft>s c aB = 21.4 ft>s2 u = 2.39° 741

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16–109. Member AB has the angular motions shown. Determine the angular velocity and angular acceleration of members CB and DC.

B D

450 mm 200 mm

60

100 mm

vAB  2 rad/s C

aAB  4 rad/s2

A

Solution Rotation About A Fixed Axis. For crank AB, refer to Fig. a. vB = vABrAB = 2(0.2) = 0.4 m>s d  aB = AAB * rAB - v2AB rAB = (4k) * (0.2j) - 22(0.2j) = 5- 0.8i - 0.8j 6 m>s2

For link CD, refer to Fig. b.

vC = vCDrCD = vCD(0.1) aC = aCD * rCD - v2CDrCD = ( - aCD k) * ( - 0.1j) - v2CD( -0.1j) = - 0.1aCD i + 0.1v2CD j General Plane Motion. The IC of link CD can be located using vB and vC of which in this case is at infinity as indicated in Fig. c. Thus, rB>IC = rC>IC = ∞. Thus, kinematics gives vB 0.4 vBC = = = 0 Ans. rB>IC ∞ Then vC = vB;  vCD(0.1) = 0.4  vCD = 4.00 rad>s b

Ans.

Applying the relative acceleration equation by referring to Fig. d, aC = aB + ABC * rC>B - v2BC rC>B -0.1aCD i + 0.1(4.002)j = ( - 0.8i - 0.8j) + (aBC k) * ( -0.45 sin 60°i - 0.45 cos 60°j) - 0 -0.1aCD i + 1.6j = (0.225aBC - 0.8)i + ( - 0.8 - 0.3897aBC)j Equating j components, 1.6 = - 0.8 - 0.3897aBC;  aBC = - 6.1584 rad>s2 = 6.16 rad>s2 b

Ans.

Then i components give - 0.1aCD = 0.225( -6.1584) - 0.8;  aCD = 21.86 rad>s2 = 21.9 rad>s2 bAns.

Ans: vBC = vCD = aBC = aCD = 742

0 4.00 rad>s b 6.16 rad>s2 b 21.9 rad>s2 b

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16–110. The slider block has the motion shown. Determine the angular velocity and angular acceleration of the wheel at this instant.

A

150 mm C 400 mm

B

Solution

vB  4 m/s aB  2 m/s2

Rotation About A Fixed Axis. For wheel C, refer to Fig. a. vA = vC rC = vC (0.15) T aA = AC * rC - v2C rC aA = (aCk) * ( - 0.15i) - v2C ( -0.15i) = 0.15 v2C i - 0.15aC j General Plane Motion. The IC for crank AB can be located using vA and vB as shown in Fig. b. Here rA>IC = 0.3 m  rB>IC = 0.4 m Then the kinematics gives vB = vAB rB>IC;  4 = vAB(0.4)  vAB = 10.0 rad>s d vA = vAB rA>IC;  vC (0.15) = 10.0(0.3)  vC = 20.0 rad>s d

Ans.

Applying the relative acceleration equation by referring to Fig. c, aB = aA + AAB * rB>A - vAB2 rB>A 2i = 0.15 ( 20.02 ) i - 0.15aC j + (aABk) * (0.3i - 0.4j)

-10.02(0.3i - 0.4j)

2i = (0.4aAB + 30)i + (0.3aAB - 0.15aC + 40)j Equating i and j components, 2 = 0.4aAB + 30;  aAB = - 70.0 rad>s2 = 70.0 rad>s2 b 0 = 0.3( - 70.0) + 0.15aC + 40;  aC = - 126.67 rad>s2 = 127 rad>s b Ans. The negative signs indicate that AC and AAB are directed in the sense that those shown in Fig. a and c.

Ans: vC = 20.0 rad>s d aC = 127 rad>s b 743

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16–111. At a given instant the slider block A is moving to the right with the motion shown. Determine the angular acceleration of link AB and the acceleration of point B at this instant.

vA  4 m/s aA  6 m/s2 A

30 B

2m

2m

Solution General Plane Motion. The IC of the link can be located using vA and vB, which in this case is at infinity as shown in Fig. a. Thus rA>IC = rB>IC = ∞ Then the kinematics gives vA = v rA>IC;  4 = v ( ∞ )  v = 0 vB = vA = 4 m>s Since B moves along a circular path, its acceleration will have tangential and normal v2B 42 components. Hence (aB)n = = = 8 m>s2 rB 2 Applying the relative acceleration equation by referring to Fig. b, aB = aA + a * rB>A - v2 rB>A

(aB)ti - 8j = 6i + (ak) * ( - 2 cos 30°i - 2 sin 30°j) - 0



(aB)ti - 8j = (a + 6)i - 23aj

Equating i and j componenets,

823 rad>s2 = 4.62 rad>s2 d 3 823 (aB)t = a + 6;  (aB)t = + 6 = 10.62 m>s2 3 Thus, the magnitude of aB is - 8 = - 23a;  a =

aB = 2(aB)2t + (aB)2n = 210.622 + 82 = 13.30 m>s2 = 13.3 m>s2

And its direction is defined by (aB)n 8 u = tan-1 c d = tan-1a b = 36.99° = 37.0°  c (aB)t 10.62

Ans.

Ans.

Ans.

Ans: aAB = 4.62 rad>s2 d aB = 13.3 m>s2 u = 37.0° c 744

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*16–112. Determine the angular acceleration of link CD if link AB has the angular velocity and angular acceleration shown.

D 0.5 m

0.5 m C

Solution Rotation About A Fixed Axis. For link AB, refer to Fig. a. A

vB = vABrAB = 3(1) = 3.00 m>s T

B

aB = AAB * rAB - vAB2 rAB

1m

= ( - 6k) * (1i) - 32 (1i) = 5 - 9i - 6j 6 m>s

For link CD, refer to Fig. b

vC = vCD rDC = vCD(0.5) S aC = ACD * rDC - vCD2 rDC = (aCDk) * ( - 0.5j) - vCD2( -0.5j) = 0.5aCDi + 0.5v2CD j General Plane Motion. The IC of link BC can be located using vA and vB as shown in Fig. c. Thus rB>IC = 0.5 m  rC>IC = 1 m Then, the kinematics gives vB = vBC rB>IC;  3 = vBC(0.5)  vBC = 6.00 rad>s b vC = vBC rC>IC;  vCD(0.5) = 6.00(1)  vCD = 12.0 rad>s d Applying the relative acceleration equation by referring to Fig. d, aC = aB + ABC * rC>B - vBC2 rC>B 0.5aCDi + 0.5 ( 12.02 ) j = ( - 9i - 6j) + ( - aBCk) * ( - 0.5i + j) -6.002( - 0.5i + j)



1m

aAB  6 rad/s2 vAB  3 rad/s

0.5aCDi + 72j = (aBC + 9)i + (0.5aBC - 42)j

745

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*16–112. Continued

Equating j components, 72 = (0.5aBC - 42);  aBC = 228 rad>s2 b Then i component gives 0.5aCD = 228 + 9;  aCD = 474 rad>s2 d

Ans.

Ans: aCD = 474 rad>s2 d 746

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16–113. The reel of rope has the angular motion shown. Determine the velocity and acceleration of point A at the instant shown.

a  8 rad/s2 v  3 rad/s B

C 100 mm

Solution

A

General Plane Motion. The IC of the reel is located as shown in Fig. a. Here, rA>IC = 20.12 + 0.12 = 0.1414 m

Then, the Kinematics give

vA = vrA>IC = 3(0.1414) = 0.4243 m>s = 0.424 m>s c45°

Ans.

2

Here aC = ar = 8(0.1) = 0.8 m>s T . Applying the relative acceleration equation by referring to Fig. b, aA = aC + A * rA>C - v2 rA>C aA = - 0.8j + (8k) * ( -0.1j) - 32( - 0.1j) = 5 0.8i + 0.1j 6 m>s2

The magnitude of aA is

aA = 20.82 + 0.12 = 0.8062 m>s2 = 0.806 m>s2

Ans.

And its direction is defined by u = tan-1a

0.1 b = 7.125° = 7.13° a 0.8

Ans.

Ans: vA = 0.424 m>s uv = 45° c aA = 0.806 m>s2 ua = 7.13° a 747

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16–114. The reel of rope has the angular motion shown. Determine the velocity and acceleration of point B at the instant shown.

a  8 rad/s2 v  3 rad/s B

C 100 mm

Solution

A

General Plane Motion. The IC of the reel is located as shown in Fig. a. Here, rB>FC = 0.2 m. Then the kinematics gives Ans.

vB = vrB>IC = (3)(0.2) = 0.6 m>s T 

Here, aC = ar = 8(0.1) = 0.8 m>s2 T . Applying the relative acceleration equation, aB = aC + A * rB>C - v2 rB>C aB = - 0.8j + (8k) * ( - 0.1i) - 32( -0.1i) = 50.9i - 1.6j 6 m>s2

The magnitude of aB is

aB = 20.92 + ( - 1.6)2 = 1.8358 m>s2 = 1.84 m>s2

Ans.

And its direction is defined by u = tan-1 a

1.6 b = 60.64° = 60.6° c 0.9

Ans.

Ans: vB = 0.6 m>s T aB = 1.84 m>s2 u = 60.6° c 748

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16–115. A cord is wrapped around the inner spool of the gear. If it is pulled with a constant velocity v, determine the velocities and accelerations of points A and B. The gear rolls on the fixed gear rack.

B 2r A

r

G v

SOLUTION Velocity analysis: v =

v r

vB = vrB>IC =

v (4r) = 4v : r

vA = v rA>IC =

v A 2(2r)2 + (2r)2 B = 2 22v r

Ans. a45°

Ans.

Acceleration equation: From Example 16–3, Since aG = 0, a = 0 rB>G = 2r j

rA>G = - 2r i

aB = aG + a * rB>G - v2rB>G 2v2 v 2 j = 0 + 0 - a b (2rj) = r r aB =

2v2 T r

Ans.

a A = aG + a * rA>G - v2rA>G 2v2 v 2 i = 0 + 0 - a b ( -2ri) = r r aA =

2v2 : r

Ans.

Ans: vB = 4v S vA = 222v u = 45° a 2v2 aB = T r 2 2v S aA = r 749

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*16–116. The disk has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point B.

v  3 rad/s a  8 rad/s2

0.5 m

C

45

45

B

A

Solution General Plane Motion. Since the disk rolls without slipping, aO = ar = 8(0.5) = 4 m>s2 d . Applying the relative acceleration equation by referring to Fig. a, aB = aO + A * rB>O - v2 rB>O aB = ( - 4i) + (8k) * (0.5 sin 45°i - 0.5 cos 45°j) - 32(0.5 sin 45°i - 0.5 cos 45°j)

aB = 5 - 4.354i + 6.010j 6 m>s2

Thus, the magnitude of aB is

aB = 2( - 4.354)2 + 6.0102 = 7.4215 m>s2 = 7.42 m>s2

Ans.

And its direction is given by u = tan-1 a

6.010 b = 54.08° = 54.1° b 4.354

Ans.

Ans: aB = 7.42 m>s2 u = 54.1° b 750

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16–117. The disk has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point C.

v  3 rad/s a  8 rad/s2

0.5 m

C

45

45

B

A

Solution General Plane Motion. Since the disk rolls without slipping, aO = ar = 8(0.5) = 4 m>s2 d . Applying the relative acceleration equation by referring to Fig. a, aC = aO + A * rC>O - v2 rC>O aC = ( - 4i) + (8k) * (0.5 sin 45°i + 0.5 cos 45°j) - 32(0.5 sin 45°i + 0.5 cos 45°j)



aC = 5 - 10.0104i - 0.3536j6 m>s2

Thus, the magnitude of aC is

aC = 2( - 10.0104)2 + ( - 0.3536)2 = 10.017 m>s2 = 10.0 m>s2 Ans.

And its direction is defined by u = tan-1 e

0.3536 f = 2.023° = 2.02° d 10.0104

Ans.

Ans: aC = 10.0 m>s2 u = 2.02° d 751

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16–118. A single pulley having both an inner and outer rim is pinconnected to the block at A. As cord CF unwinds from the inner rim of the pulley with the motion shown, cord DE unwinds from the outer rim. Determine the angular acceleration of the pulley and the acceleration of the block at the instant shown.

D 25 mm

50 mm F

C

E

A

vF = 2 m/s aF = 3 m/s2

SOLUTION Velocity Analysis: The angular velocity of the pulley can be obtained by using the method of instantaneous center of zero velocity. Since the pulley rotates without slipping about point D, i.e: yD = 0, then point D is the location of the instantaneous center. yF = vrC>IC 2 = v(0.075) v = 26.67 rad>s Acceleration Equation: The angular acceleration of the gear can be obtained by analyzing the angular motion points C and D. Applying Eq. 16–18 with rC>D = { -0.075 j} m, we have aC = aD + a * rC>D - v2 rC>D -3i + 17.78 j = - 35.56 j + ( - a k) * (- 0.075 j) - 26.672 ( -0.075 j) - 3i + 17.78 j = - 0.075 a i + 17.78 j Equating i and j components, we have -3 = - 0.075a

a = 40.0 rad>s2

Ans.

17.78 = 17.78 (Check!) The acceleration of point A can be obtained by analyzing the angular motion points A and D. Applying Eq. 16–18 with rA>D = { - 0.05j} m. we have aA = aD + a * rA>D - v2 rA>D = - 35.56 j + ( -40.0k) * ( -0.05 j) - 26.672( -0.05 j) = {-2.00i} m>s2 Thus, aA = 2.00 m s2 ;

Ans.

Ans: a = 40.0 rad>s2 aA = 2.00 m>s2 d 752

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16–119. The wheel rolls without slipping such that at the instant shown it has an angular velocity V and angular acceleration A. Determine the velocity and acceleration of point B on the rod at this instant.

A 2a

O

v, a

a

B

SOLUTION vB = vA + vB/A (Pin) + v = ; B

1 Qv 22aR + 2av¿ a b 2 22

+c O = v¿ =

1

1 22

Qv22aR + 2av¿ a

23 b 2

v 23 Ans.

vB = 1.58 va a A = aO + aA/O (Pin) (a A)x + (aA)y = aa + a(a) + v2a ;

T

;

T

:

(a A)x = aa - v2a (a A)y = aa a B = aA + aB/A (Pin) v 2 23 1 a B = aa - v2a + 2a(a¿) a b - 2a a b 2 2 23 O = -aa + 2aa¿ a

2 23

b + 2a a

2 1 b a b 2 23

v

a¿ = 0.577a - 0.1925v2 a B = 1.58aa - 1.77v2a

Ans.

Ans: vB = 1.58va aB = 1.58 aa - 1.77v2a 753

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*16–120. The collar is moving downward with the motion shown. Determine the angular velocity and angular acceleration of the gear at the instant shown as it rolls along the fixed gear rack.

v  2 m/s a  3 m/s2

A

500 mm

60

Solution General Plane Motion. For gear C, the location of its IC is indicate in Fig. a. Thus vB = vC rB>(IC)1 = vC(0.05) T (1)

O

B

150 mm 200 mm

The IC of link AB can be located using vA and vB, which in this case is at infinity. Thus vAB =

vA 2 = 0 = rA>(IC)2 ∞

Then vB = vA = 2 m>s T Substitute the result of vB into Eq. (1) 2 = vC (0.05) Ans.

vC = 40.0 rad>s d Applying the relative aO = aC rC = aC (0.2) T,

acceleration

equation

to

gear

C, Fig. c, with

aB = aO + AC * rB>O - v2C rB>O aB = - aC (0.2)j + (aCk) * (0.15i) - 40.02(0.15i) = -240i - 0.05aC j For link AB, Fig. d, aB = aA + AAB * rB>A - v2AB rB>A - 240i - 0.05aCj = ( - 3j) + (aABk) * (0.5 sin 60°i - 0.5 cos 60°j) - 0 - 240i - 0.05aC = 0.25aABi + (0.2523aAB - 3)j Equating i and j components - 240 = 0.25aAB;  aAB = - 960 rad>s2 = 960 rad>s2 b - 0.05aC = ( 0.2523 ) ( - 960) - 3;  aC = 8373.84 rad>s2 = 8374 rad>s2 dAns.

Ans: vC = 40.0 rad>s d aC = 8374 rad>s2 d 754

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16–121. The tied crank and gear mechanism gives rocking motion to crank AC, necessary for the operation of a printing press. If link DE has the angular motion shown, determine the respective angular velocities of gear F and crank AC at this instant, and the angular acceleration of crank AC.

C F

100 mm

50 mm

v DE  4 rad/s

75 mm

SOLUTION

100 mm

B

Velocity analysis:

G

yD = vDErD>E = 4(0.1) = 0.4 m>s c 150 mm

vB = vD + vB>D

E D

aDE  20 rad/s2

30

A

yB = 0.4 + (vG)(0.075) a 30°

c

+ ) (:

T

yB cos 30° = 0,

yB = 0

( + c ) vG = 5.33 rad>s Since yB = 0,

yC = 0,

Ans.

vAC = 0

vFrF = vGrG vF = 5.33 a

100 b = 10.7 rad>s 50

Ans.

Acceleration analysis: (aD)n = (4)2(0.1) = 1.6 m>s2 : (aD)t = (20)(0.1) = 2 m>s2 c (aB)n + (aB)t = (aD)n + (aD)t + (aB>D)n + (aB>D)t 0 + (a B)t a 30°

= 1.6 + 2 + (5.33)2(0.075) + aG (0.075) :

c

:

c

(+ c)

(aB)t sin 30° = 0 + 2 + 0 + aG (0.075)

+ ) (:

(aB)t cos 30° = 1.6 + 0 + (5.33)2(0.075) + 0

Solving, (aB)t = 4.31 m>s2,

aG = 2.052 rad>s2b

Hence, aAC =

(a B)t 4.31 = 28.7 rad>s2b = rB>A 0.15

Ans.

Ans: vAC = 0 vF = 10.7 rad>s b aAC = 28.7 rad>s2 b 755

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16–122. If member AB has the angular motion shown, determine the angular velocity and angular acceleration of member CD at the instant shown.

300 mm A

B vAB  3 rad/s aAB  8 rad/s2

500 mm

Solution C

Rotation About A Fixed Axis. For link AB, refer to Fig. a. vB = vABrAB = 3(0.3) = 0.9 m>s T

u  60

200 mm

aB = AAB * rAB - v2AB rAB = ( -8k) * (0.3i) - 32(0.3i)

D

= 5 -2.70i - 2.40 j 6 m>s2

For link CD, refer to Fig. b.

vC = vCDrCD = vCD(0.2) d aC = ACD * rCD - v 2CD rCD aC = (aCD k) * (0.2j) - v2CD(0.2j) = -0.2aCD i - 0.2v2CD j

756

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16–122. Continued

General Plane Motion. The IC of link BC can be located using vB and vC as shown in Fig. c. From the geometry of this figure, rB>IC = 0.5 cos 60° = 0.25 m  rC>IC = 0.5 sin 60° = 0.2523 m Then kinematics gives vB = vBCrB>IC;  0.9 = vBC(0.25)  vBC = 3.60 rad>s b vC = vBCrC>IC;  vCD(0.2) = (3.60) ( 0.2523 ) Ans.

vCD = 7.7942 rad>s = 7.79 rad>s d Applying the relative acceleration equation by referring to Fig. d, aC = aB + ABC * rC>B - v2BC rC>B

- 0.2aCD i - 0.2 ( 7.79422 ) j = ( - 2.70i - 2.40j) + ( - aBC k) * ( - 0.5 cos 60°i - 0.5 sin 60°j) - 3.602( - 0.5 cos 60°i - 0.5 sin 60°j) - 0.2aCD i - 12.15 j = ( 0.54 - 0.2523aBC ) i + (3.2118 + 0.25aBC)j Equating the j components, - 12.15 = 3.2118 + 0.25aBC;  aBC = -61.45 rad>s2 = 61.45 rad>s2 d Then the i component gives - 0.2aCD = 0.54 - 0.2523( - 61.4474);  aCD = -135.74 rad>s2 = 136 rad>s2 b Ans.

Ans: vCD = 7.79 rad>s d aCD = 136 rad>s2 b 757

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16–123. If member AB has the angular motion shown, determine the velocity and acceleration of point C at the instant shown.

300 mm A

B vAB  3 rad/s aAB  8 rad/s2

500 mm

Solution C

Rotation About A Fixed Axis. For link AB, refer to Fig. a. vB = vABrAB = 3(0.3) = 0.9 m>s T

u  60

200 mm

aB = AAB * rAB - v2AB rAB = ( - 8k) * (0.3i) - 32(0.3i)

D

= 5 - 2.70i - 2.40 j 6 m>s2

For link CD, refer to Fig. b. vC = vCDrCD = vCD(0.2) d aC = ACD * rCD - v 2CD rCD aC = (aCD k) * (0.2 j) - v2CD(0.2j) = - 0.2aCD i - 0.2v2CD j

758

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16–123. Continued

General Plane Motion. The IC of link BC can be located using vB and vC as shown in Fig. c. From the geometry of this figure, rB>IC = 0.5 cos 60° = 0.25 m  rC>IC = 0.5 sin 60° = 0.2523 m Then kinematics gives vB = vBCrB>IC;  0.9 = vBC(0.25)  vBC = 3.60 rad>sb vC = vBCrC>IC;  vCD(0.2) = (3.60) ( 0.2523 ) Ans.

vCD = 7.7942 rad>s = 7.79 rad>sd Applying the relative acceleration equation by referring to Fig. d, aC = aB + ABC * rC>B - v2BC rC>B

- 0.2aCD i - 0.2 ( 7.79422 ) j = ( - 2.70i - 2.40j) + ( - aBC k) * ( - 0.5 cos 60°i - 0.5 sin 60°j) - 3.602( - 0.5 cos 60°i - 0.5 sin 60°j) - 0.2aCD i - 12.15j = ( 0.54 - 0.2523aBC ) i + (3.2118 + 0.25aBC)j Equating the j components, - 12.15 = 3.2118 + 0.25aBC;  aBC = - 61.45 rad>s2 = 61.45 rad>s2d Then the i component gives - 0.2aCD = 0.54 - 0.2523( - 61.4474);  aCD = - 135.74 rad>s2 = 136 rad>s2 Ans. From the angular motion of CD,    vC = wCD(0.2) = (7.7942)(0.2) = 1.559 m>s = 1.56 m>s d

Ans.

   aC = - 0.2( - 135.74)i - 12.15j       

= 527.15i - 12.15j6 m>s

The magnitude of aC is

       aC = 227.152 + ( - 12.15)2 = 29.74 m>s2 = 29.7 m>s2

Ans.

And its direction is defined by

          u = tan-1a

12.15 b = 24.11° = 24.1° c 27.15

Ans.

Ans: vC = 1.56 m>s d aC = 29.7 m>s2 u = 24.1° c 759

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*16–124. The disk rolls without slipping such that it has an angular acceleration of a = 4 rad>s2 and angular velocity of v = 2 rad>s at the instant shown. Determine the acceleration of points A and B on the link and the link’s angular acceleration at this instant. Assume point A lies on the periphery of the disk, 150 mm from C.

A v  2 rad/s a  4 rad/s2 C

500 mm

150 mm

B 400 mm

Solution The IC is at ∞, so v = 0. aA = aC + a * rA>C - v2rA>C aA = 0.6i + ( - 4k) * (0.15j) - (2)2(0.15j) aA = (1.20i - 0.6j) m>s2 aA = 2(1.20)2 + ( - 0.6)2 = 1.34 m>s2

Ans.

u = tan-1a

Ans.

0.6 b = 26.6° 1.20



Ans: aA = 1.34 m>s2 u = 26.6° 760

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16–125. The ends of the bar AB are confined to move along the paths shown. At a given instant, A has a velocity of vA = 4 ft>s and an acceleration of aA = 7 ft>s2. Determine the angular velocity and angular acceleration of AB at this instant.

B

2 ft 60

SOLUTION

2 ft

vB = vA + vB>A 30°

vA aA

= 4+ v(4.788) T h 51.21°

vB b

+ ) (:

- vB cos 30° = 0 - v(4.788) sin 51.21°

(+ c )

vB sin 30° = -4 + v(4.788) cos 51.21°

4 ft/s 7 ft/s2

A

30° b

vB = 20.39 ft>s

Ans.

v = 4.73 rad>sd a B = a A + a B>A at

30° b

+

2 07.9 60° d

= 7 + 107.2 + 4.788(a) d 51.21° T h 51.21°

+ ) (;

at cos 30° + 207.9 cos 60° = 0 + 107.2 cos 51.21° + 4.788a(sin 51.21°)

(+ c )

at sin 30° - 207.9 sin 60° = - 7 - 107.2 sin 51.21° + 4.788a(cos 51.21°)

at(0.866) - 3.732a = - 36.78 at (0.5) - 3a = 89.49 at = -607 ft>s2 a = -131 rad>s2 = 131 rad>s2b

Ans.

Also: vB = vA = v * rB>A - vB cos 30°i + vB sin 30°j = - 4j + (vk) * (3i + 3.732j) - vB cos 30° = - v(3.732) vB sin 30° = -4 + v(3) Ans.

v = 4.73 rad>sd vB = 20.39 ft>s a B = a A - v2rB>A + a * rB>A

( - at cos 30°i + at sin 30°j) + ( -207.9 cos 60°i - 207.9 sin 60°j) = -7j - (4.732)2(3i + 3.732j) + (ak) * (3i + 3.732j) - at cos 30° - 207.9 cos 60° = - (4.732)2(3) - a(3.732) at sin 30° - 207.9 sin 60° = - 7 -(4.732)2(3.732) + a(3) at = -607 ft>s2 a = -131 rad>s2 = 131 rad>s2b

Ans. 761

Ans: v = 4.73 rad>s d a = 131 rad>s2 b

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16–126. The mechanism produces intermittent motion of link AB. If the sprocket S is turning with an angular acceleration aS = 2 rad>s2 and has an angular velocity vS = 6 rad>s at the instant shown, determine the angular velocity and angular acceleration of link AB at this instant. The sprocket S is mounted on a shaft which is separate from a collinear shaft attached to AB at A. The pin at C is attached to one of the chain links such that it moves vertically downward.

200 mm B

A

SOLUTION vBC

vS

1.05 = 4.950 rad>s = 0.2121

6 rad/s aS

2 rad/s2

Ans.

a C = aS rS = 2(0.175) = 0.350 m>s2 (aB)n + (a B)t = a C + (aB>C)n + (a B>C)t

D

(7.172)2(0.2)

+ b a:

A+cB

30° d

T + C (aB)t h

30°

S = B 0.350R + C (4.949)2 (0.15) S + D aBC (0.15) T T e 15° e 15°

- (10.29) cos 30° - (aB)t sin 30° = 0 - (4.949)2(0.15) sin 15° - aBC(0.15) cos 15° - (10.29) sin 30° + (aB)t cos 30° = - 0.350 - (4.949)2(0.15) cos 15° + aBC(0.15) sin 15°

aBC = 70.8 rad>s2,

(aB)t = 4.61 m>s2

Hence, aAB =

(aB)t 4.61 = 23.1 rad>s2d = rB>A 0.2

Ans.

Also, vC = vS rS = 6(0.175) = 1.05 m>s T vB = vC + vBC * rB>C vB sin 30°i - vB cos 30°j = -1.05j + ( -vBCk) * (0.15 sin 15°i + 0.15 cos 15°j) + b a:

vB sin 30° = 0 + vBC(0.15) cos 15°

A+ cB

- vB cos 30° = -1.05 - vBC(0.15) sin 15°

vB = 1.434 m>s, vAB =

175 mm

D 50 mm

1.435 = 7.1722 rad>s = 7.17 rad>sb 0.2

vBC = 4.950 rad>s

vB 1.434 = 7.172 = 7.17 rad>sb = rB>A 0.2

Ans. 762

15 150 mm C

S

vB = (4.95)(0.2898) = 1.434 m>s vAB =

30

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16–126. Continued

a B = aC + aBC * rB>C - v2rB>C (aAB k) * (0.2 cos 30°i + 0.2 sin 30°j) - (7.172)2 (0.2 cos 30°i + 0.2 sin 30°j) = -(2)(0.175)j + (aBC k) * (0.15 sin 15°i + 0.15 cos 15°j) - (4.950)2 (0.15 sin 15°i + 0.15 cos 15°j) + b a:

A+cB

- aAB (0.1) - 8.9108 = - 0.1449aBC - 0.9512 aAB (0.1732) - 5.143 = - 0.350 + 0.0388aBC - 3.550

aAB = 23.1 rad>s2d

Ans.

aBC = 70.8 rad>s2

Ans: vAB = 7.17 rad>s b aAB = 23.1 rad>s2 d 763

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16–127. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the angular acceleration of rod AB at the instant shown. 1.5 ft A vB aB

30

5 ft/s 3 ft/s2

2 ft B

SOLUTION Angular Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vAB =

vB rB>IC

=

5 = 5 rad>s 1

Then vA = vAB rA>IC = 5(1.732) = 8.660 ft>s Acceleration and Angular Acceleration: Since point A travels along the circular slot, the normal component of its acceleration has a magnitude of vA 2 8.6602 = 50 ft>s2 and is directed towards the center of the circular (aA)n = = r 1.5 slot. The tangential component is directed along the tangent of the slot. Applying the relative acceleration equation and referring to Fig. b, aA = aB + aAB * rA>B - vAB 2 rA>B 50i - (aA)t j = 3i + (aAB k) * ( - 2 cos 30°i + 2 sin 30°j) - 52( -2 cos 30° i + 2 sin 30°j) 50i - (aA)t j = (46.30 - aAB)i + (1.732aAB + 25)j Equating the i components, 50 = 46.30 - aAB aAB = -3.70 rad>s2 = 3.70 rad>s2 b

Ans.

Ans: aAB = 3.70 rad>s2 b 764

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*16–128. The slider block moves with a velocity of vB = 5 ft>s and an acceleration of aB = 3 ft>s2. Determine the acceleration of A at the instant shown. 1.5 ft A vB aB

30

5 ft/s 3 ft/s2

2 ft B

SOLUTION Angualr Velocity: The velocity of point A is directed along the tangent of the circular slot. Thus, the location of the IC for rod AB is indicated in Fig. a. From the geometry of this figure, rB>IC = 2 sin 30° = 1 ft

rA>IC = 2 cos 30° = 1.732 ft

Thus, vAB =

vB 5 = 5 rad>s = rB>IC 1

Then vA = vAB rA>IC = 5 A 1.732 B = 8.660 ft>s Acceleration and Angular Acceleration: Since point A travels along the circular slot, the normal component of its acceleration has a magnitude of vA 2 8.6602 = 50 ft>s2 and is directed towards the center of the circular = A aA B n = r 1.5 slot. The tangential component is directed along the tangent of the slot. Applying the relative acceleration equation and referring to Fig. b, aA = aB + aAB * rA>B - vAB 2 rA>B 50i - A aA B t j = 3i + A aAB k B *

A - 2cos 30°i + 2 sin 30°j B -52 A -2 cos 30°i + 2 sin 30°j B

50i - A aA B t j = A 46.30 - aAB B i - A 1.732aAB + 25 B j Equating the i and j components, 50 = 46.30-aAB - A aA B t = - A 1.732aAB + 25 B Solving, aAB = - 3.70 rad>s2

A aA B t = 18.59 ft>s2 T Thus, the magnitude of aA is aA = 4A aA B t 2 + A aA B n 2 = 218.592 + 502 = 53.3 ft>s2

Ans.

and its direction is u = tan-1

A aA B t

18.59 C S = tan-1 a b = 20.4° c 50 A aA B n

Ans.

765

Ans: aA = 53.3 ft>s2 u = 20.4° c

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16–129. At the instant shown, ball B is rolling along the slot in the disk with a velocity of 600 mm>s and an acceleration of 150 mm>s2, both measured relative to the disk and directed away from O. If at the same instant the disk has the angular velocity and angular acceleration shown, determine the velocity and acceleration of the ball at this instant.

z v  6 rad/s a  3 rad/s2

0.8 m

x

Solution

B

O 0.4 m

y

Kinematic Equations: vB = vO + Ω * rB>O + (vB>O)xyz(1) aB = aO + Ω * rB>O + Ω * (Ω * rB>O) + 2Ω * (vB>O)xyz + (aB>O)xyz(2) vO = 0 aO = 0 Ω = 56k6 rad>s # Ω = 53k6 rad>s2

rB>O = {0.4i} m

(vB>O)xyz = 50.6i6m>s

(aB>O)xyz = 50.15i6 m>s2

Substitute the date into Eqs. (1) and (2) yields: Ans.

vB = 0 + (6k) * (0.4i) + (0.6i) = 5 0.6i + 2.4j 6 m>s

aB = 0 + (3k) * (0.4i) + (6k) * [(6k) * (0.4i)] + 2(6k) * (0.6i) + (0.15i) = 5 - 14.2i + 8.40j 6m>s2

Ans.

Ans: vB = {0.6i + 2.4j} m>s aB = { - 14.2i + 8.40j} m>s2 766

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16–130. The crane’s telescopic boom rotates with the angular velocity and angular acceleration shown. At the same instant, the boom is extending with a constant speed of 0.5 ft>s, measured relative to the boom. Determine the magnitudes of the velocity and acceleration of point B at this instant.

60 ft B vAB aAB

SOLUTION

30

0.02 rad/s 0.01 rad/s2 A

Reference Frames: The xyz rotating reference frame is attached to boom AB and coincides with the XY fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xy frame with respect to the XY frame is vA = aA = 0

vAB = [- 0.02k] rad>s

# vAB = a = [-0.01k] rad>s2

For the motion of point B with respect to the xyz frame, we have rB>A = [60j] ft

(vrel)xyz = [0.5j] ft>s

(arel)xyz = 0

Velocity: Applying the relative velocity equation, vB = vA + vAB * rB>A + (v rel)xyz = 0 + ( -0.02k) * (60j) + 0.5j = [1.2i + 0.5j] ft > s Thus, the magnitude of vB, Fig. b, is vB = 21.22 + 0.52 = 1.30 ft>s

Ans.

Acceleration: Applying the relative acceleration equation, # aB = aA + vAB * rB>A + vAB * (vAB * rB>A) + 2vAB * (vrel)xyz + (a rel)xyz = 0 + ( - 0.01k) * (60j) + (- 0.02k) * [(-0.02k) * (60j)] + 2( - 0.02k) * (0.5j) + 0 = [0.62i - 0.024 j] ft>s2 Thus, the magnitude of aB, Fig. c, is aB = 20.622 + ( - 0.024)2 = 0.6204 ft>s2

Ans.

Ans: vB = 1.30 ft>s aB = 0.6204 ft>s2 767

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16–131. While the swing bridge is closing with a constant rotation of 0.5 rad>s, a man runs along the roadway at a constant speed of 5 ft>s relative to the roadway. Determine his velocity and acceleration at the instant d = 15 ft.

d

z O

y

x v

0.5 rad/s

SOLUTION Æ = {0.5k} rad>s Æ = 0 rm>o = { -15 j} ft (vm>o)xyz = {- 5j} ft>s (am>o)xyz = 0 vm = vo + Æ * rm>o + (vm>o)xyz vm = 0 + (0.5k) * ( - 15j) - 5j vm = {7.5i - 5j} ft>s

Ans.

am = aO + Æ * rm>O + Æ * (Æ * rm>O) + 2Æ * (vm>O)xyz + (am>O)xyz am = 0 + 0 + (0.5k) * [(0.5k) * ( -15j)] + 2(0.5k) * ( -5j) + 0 am = {5i + 3.75j} ft>s2

Ans.

768

Ans: vm = 57.5i - 5j 6 ft>s am = 55i + 3.75j 6 ft>s2

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*16–132. While the swing bridge is closing with a constant rotation of 0.5 rad>s, a man runs along the roadway such that when d = 10 ft he is running outward from the center at 5 ft>s with an acceleration of 2 ft>s2, both measured relative to the roadway. Determine his velocity and acceleration at this instant.

d

z O

y

x v

0.5 rad/s

SOLUTION Æ = {0.5k} rad>s Æ = 0 rm>o = {- 10 j} ft (vm>O)xyz = {-5j} ft>s (am>O)xyz = {-2j} ft>s2 vm = vo + Æ * rm>o + (vm>o)xyz vm = 0 + (0.5k) * (- 10j) - 5j vm = {5i - 5j} ft>s # am = aO + Æ * rm>O + Æ * (Æ * rm>O) + 2Æ * (vm>O)xyz + (am>O)xyz

Ans.

am = 0 + 0 + (0.5k) * [(0.5k) * ( -10j)] + 2(0.5k) * (-5j) - 2j am = {5i + 0.5j} ft>s2

Ans.

769

Ans: vm = 55i - 5j 6 ft>s am = 5 5i + 0.5j 6 ft>s2

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16–133. y

Water leaves the impeller of the centrifugal pump with a velocity of 25 m>s and acceleration of 30 m>s2, both measured relative to the impeller along the blade line AB. Determine the velocity and acceleration of a water particle at A as it leaves the impeller at the instant shown. The impeller rotates with a constant angular velocity of v = 15 rad>s.

B 30

A

SOLUTION

x

Reference Frame: The xyz rotating reference frame is attached to the impeller and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vO = aO = 0

v 0.3 m

15 rad/s

# v = 0

v = [- 15k] rad > s

The motion of point A with respect to the xyz frame is rA>O = [0.3j] m (vrel)xyz = ( -25 cos 30° i + 25 sin 30° j) = [- 21.65i + 12.5j] m>s (arel)xyz = ( -30 cos 30° i + 30 sin 30° j) = [ - 25.98i + 15j] m>s2 Velocity: Applying the relative velocity equation. vA = vO + v * rA>O + (vrel)xyz = 0 + ( - 15k) * (0.3j) + ( - 21.65i + 12.5j) Ans.

= [- 17.2i + 12.5j] m>s Acceleration: Applying the relative acceleration equation, # aA = aO + v * rA>O + v * (v * rA>O) + 2v * (vrel)xyz + (arel)xyz

= 0 + ( - 15k) * [(- 15k) * (0.3j)] + 2( - 15k) * (- 21.65i + 12.5j) + ( -25.98i + 15j) = [349i + 597j] m>s2

Ans.

Ans: vA = { - 17.2i + 12.5j} m>s aA = {349i + 597j} m>s2 770

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16–134. y

Block A, which is attached to a cord, moves along the slot of a horizontal forked rod. At the instant shown, the cord is pulled down through the hole at O with an acceleration of 4 m>s2 and its velocity is 2 m>s. Determine the acceleration of the block at this instant. The rod rotates about O with a constant angular velocity v = 4 rad>s.

x A

v

O 100 mm

SOLUTION Motion of moving reference. vO = 0 aO = 0 Æ = 4k # Æ = 0 Motion of A with respect to moving reference. rA>O = 0.1i vA>O = - 2i aA>O = - 4i Thus, # aA = a O + Æ * rA>O + Æ * (Æ * rA>O) + 2Æ * (vA>O)xyz + (a A>O)xyz = 0 + 0 + (4k) * (4k * 0.1i) + 2(4k * ( - 2i)) - 4i aA = { -5.60i - 16j} m>s2

Ans.

771

Ans: aA = 5 - 5.60i - 16j 6 m>s2

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16–135. Rod AB rotates counterclockwise with a constant angular velocity v = 3 rad>s. Determine the velocity of point C located on the double collar when u = 30°. The collar consists of two pin-connected slider blocks which are ­constrained to move along the circular path and the rod AB.

C

B

v = 3 rad/s u A 0.4 m

Solution r = 2(0.4 cos 30°) = 0.6928 m rC>A = 0.6928 cos 30°i + 0.6928 sin 30°j = 50.600i + 0.3464j 6 m

vC = -0.866vCi + 0.5vC j

vC = vA + Ω * rC>A + (vC>A)xyz -0.866vCi + 0.5vCj = 0 + (3k) * (0.600i + 0.3464j) + (vC>A cos 30°i + vC>A sin 30°j) -0.866vCi + 0.5vCj = 0 - 1.039i + 1.80j + 0.866vC>Ai + 0.5vC>A j -0.866vC = -1.039 + 0.866vC>A 0.5vC = 1.80 + 0.5vC>A Ans.

vC = 2.40 m>s vC>A = -1.20 m>s

Ans: vC = 2.40 m>s u = 60° b 772

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*16–136. Rod AB rotates counterclockwise with a constant angular velocity v = 3 rad>s. Determine the velocity and acceleration of point C located on the double collar when u = 45°. The collar consists of two pin-connected slider blocks which are constrained to move along the circular path and the rod AB.

C

B

v = 3 rad/s u A 0.4 m

Solution rC>A = 50.400i + 0.400j 6 vC = - vCi

vC = vA + Ω * rC>A + (vC>A)xyz - vCi = 0 + (3k) * (0.400i + 0.400j) + (vC>A cos 45°i + vC>A sin 45°j) - vCi = 0 - 1.20i + 1.20j + 0.707vC>Ai + 0.707vC>A j - vC = - 1.20 + 0.707vC>A 0 = 1.20 + 0.707vC>A Ans.

vC = 2.40 m>s vC>A = -1.697 m>s # aC = aA + Ω * rC>A + Ω * (Ω * rC>A) + 2Ω * (vC>A)xyz + (aC>A)xyz - (aC)ti -

(2.40)2 0.4

j = 0 + 0 + 3k * [3k * (0.4i + 0.4j)] + 2(3k) * [0.707( -1.697)i + 0.707( - 1.697)j] + 0.707aC>Ai + 0.707aC>A j



- (aC)ti - 14.40j = 0 + 0 - 3.60i - 3.60j + 7.20i - 7.20j + 0.707aC>Ai + 0.707aC>A j - (aC)t = -3.60 + 7.20 + 0.707aC>A - 14.40 = - 3.60 - 7.20 + 0.707aC>A aC>A = -5.09 m>s2 (aC)t = 0 Thus, aC = (aC)n =

(2.40)2 0.4

= 14.4 m>s2

aC = 5 -14.4j 6 m>s2

Ans.

773

Ans: vC = 2.40 m>s aC = 5 - 14.4j 6 m>s2

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16–137. Particles B and A move along the parabolic and circular paths, respectively. If B has a velocity of 7 m s in the direction shown and its speed is increasing at 4 m s2, while A has a velocity of 8 m s in the direction shown and its speed is decreasing at 6 m s2, determine the relative velocity and relative acceleration of B with respect to A.

y y = x2

B

yB = 7 m/s 2m

SOLUTION

yA = 8 m/s

8 Æ = { - 8k} rad>s = 8 rad>s2, 1 vB = vA + Æ * rB>A + (vB>A)xyz Æ =

A 1m

7i = - 8i + (8k) * (2 j) + (vB>A)xyz 7i = - 8i - 16i + (vB>A)xyz Ans.

(vB>A)xyz = {31.0i} m>s # 6 = 6 rad>s2, Æ = 1 (aA)n =

# Æ = {- 6k} rad>s2

(8)2 (vA)2 = = 64 m>s2 T 1 1

y = x2 dy = 2x 2 = 0 dx x=0 d2y dx2

= 2 3

r =

dy 2 2 c1 + a b d dx

(aB)n =

2

d2y dx2

2

3

[1 + 0]2 1 = = 2 2

(7)2 = 98 m>s2 c 1 2

# aB = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz 4i + 98 j = 6i - 64 j + (- 6k) * (2 j) + (8k) * (8k * 2 j) + 2(8k) * (31i) + (aB>A)xyz 4i + 98 j = 6i - 64 j + 12i - 128 j + 496 j + (aB>A)xyz (aB A)xyz = { -14.0i - 206j} m s2

Ans.

Ans: ( vB>A ) xyz = {31.0i} m>s ( aB>A ) xyz = { - 14.0i - 206j} m>s2 774

x

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16–138. Collar B moves to the left with a speed of 5 m>s, which is increasing at a constant rate of 1.5 m>s2, relative to the hoop, while the hoop rotates with the angular velocity and angular acceleration shown. Determine the magnitudes of the velocity and acceleration of the collar at this instant.

A

v  6 rad/s a  3 rad/s2

450 mm

SOLUTION Reference Frames: The xyz rotating reference frame is attached to the hoop and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz frame with respect to the XYZ frame is vA = aA = 0

v = [- 6k] rad>s

200 mm B

# v = a = [- 3k] rad>s2

For the motion of collar B with respect to the xyz frame, rB>A = [- 0.45j] m (vrel)xyz = [- 5i] m>s The normal components of (arel)xyz is [(arel)xyz]n =

(vrel)xyz2 r

=

52 = 125 m>s2. Thus, 0.2

(arel)xyz = [- 1.5i + 125j] m>s Velocity: Applying the relative velocity equation, vB = vA + v * rB>A + (vrel)xyz = 0 + (- 6k) * ( -0.45j) + (- 5i) = [- 7.7i] m>s Thus, vB = 7.7 m>s ;

Ans.

Acceleration: Applying the relative acceleration equation,

#

aB = aA + v * rB>A + v * (v * rB>A) + 2v * (vrel)xyz + (arel)xyz = 0 + ( - 3k) * ( - 0.45j) + ( - 6k) * [( -6k) * (- 0.45j)] + 2( -6k) * ( -5i) + ( -1.5i + 125j) = [- 2.85i + 201.2j] m>s2 Thus, the magnitude of aB is therefore aB = 32.852 + 201.22 = 201 m>s2

Ans.

Ans: vB = 7.7 m>s aB = 201 m>s2 775

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16–139. Block D of the mechanism is confined to move within the slot of member CB. If link AD is rotating at a constant rate of vAD = 4 rad>s, determine the angular velocity and angular acceleration of member CB at the instant shown.

B

D

300 mm 200 mm vAD  4 rad/s

30

Solution

C

A

The fixed and rotating X - Y and x - y coordinate system are set to coincide with origin at C as shown in Fig. a. Here the x - y coordinate system is attached to member CB. Thus Motion of moving Reference

 otion of Block D M with respect to moving Reference

vC = 0 aC = 0

rD>C = 5 0.3i6 m

𝛀 = VCB = vCBk

(vD>C)xyz = (vD>C)xyzi

𝛀 = ACB = aCBk

(aD>C)xyz = (aD>C)xyzi

#

The Motions of Block D in the fixed frame are, vD = VA>D * rD>A = (4k) * (0.2 sin 30°i + 0.2 cos 30°j) = 5 -0.423i + 0.4j 6 m>s aD = AAD * rD>A - vAD2(rD>A) = 0 - 42(0.2 sin 30°i + 0.2 cos 30°j) = 5 - 1.6i - 1.623j 6 m>s2



Applying the relative velocity equation,

vD = vC + 𝛀 * rD>C + (vD>C)xyz -0.423i + 0.4j = 0 + (vCBk) * (0.3i) + (vD>C)xyzi - 0.423i + 0.4j = (vD>C)xyzi + 0.3 vCB j Equating i and j components, (vD>C)xyz = - 0.423 m>s Ans.

0.4 = 0.3 vCB;  vCB = 1.3333 rad>s = 1.33 rad>sd Applying the relative acceleration equation, # aD = aC + 𝛀 * rD>C + 𝛀 * (𝛀 * rD>C) + 2𝛀 * (vD>C)xyz + (aD>C)xyz - 1.6i - 1.623j = 0 + (aCDk) * (0.3i) + (1.3333k) * (1.3333k * 0.3i)

+ 2(1.3333k) * ( - 0.423i) + (aD>C)xyzi

1.6i - 1.623j = [(aD>C)xyz - 0.5333]i + (0.3aCD - 1.8475)j Equating i and j components 1.6 = [(aD>C)xyz - 0.5333];  (aD>C)xyz = 2.1333 m>s2 - 1.623 = 0.3 aCD - 1.8475;  aCD = - 3.0792 rad>s2 = 3.08 rad>s2 b

Ans.

Ans: vCB = 1.33 rad>s d aCD = 3.08 rad>s2 b 776

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*16–140. At the instant shown rod AB has an angular velocity vAB = 4 rad>s and an angular acceleration aAB = 2 rad>s2. Determine the angular velocity and angular acceleration of rod CD at this instant.The collar at C is pin connected to CD and slides freely along AB.

vAB aAB

A 60

0.75 m

0.5 m C

D B

SOLUTION Coordinate Axes: The origin of both the fixed and moving frames of reference are located at point A. The x, y, z moving frame is attached to and rotate with rod AB since collar C slides along rod AB. Kinematic Equation: Applying Eqs. 16–24 and 16–27, we have

aC = aA

4 rad/s 2 rad/s2

vC = vA + Æ * rC>A + (vC>A)xyz # + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (v C>A)xyz + (a C>A)xyz

Motion of moving reference vA = 0 aA = 0 Æ = 4k rad>s # Æ = 2k rad>s2

(1) (2)

Motion of C with respect to moving reference rC>A = 50.75i6m (vC>A)xyz = (yC>A)xyz i (a C>A)xyz = (aC>A)xyz i

The velocity and acceleration of collar C can be determined using Eqs. 16–9 and 16–14 with rC>D = {- 0.5 cos 30°i - 0.5 sin 30°j }m = { -0.4330i - 0.250j} m. vC = vCD * rC>D = -vCDk * ( -0.4330i - 0.250j) = -0.250vCDi + 0.4330vCDj aC = a CD * rC>D - v2CD rC>D = -aCD k * (- 0.4330i - 0.250j) - v2CD( -0.4330i - 0.250j) = A 0.4330v2CD - 0.250 aCD B i + A 0.4330aCD + 0.250v2CD B j Substitute the above data into Eq.(1) yields v C = vA + Æ * rC>A + (vC>A)xyz -0.250 vCD i + 0.4330vCDj = 0 + 4k * 0.75i + (yC>A)xyz i -0.250vCD i + 0.4330vCD j = (yC>A)xyz i + 3.00j Equating i and j components and solve, we have (yC>A)xyz = - 1.732 m>s Ans.

vCD = 6.928 rad>s = 6.93 rad>s

777

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*16–140. Continued

Substitute the above data into Eq.(2) yields # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz

C 0.4330 A 6.9282 B - 0.250 aCD D i + C 0.4330aCD + 0.250 A 6.9282 B D j = 0 + 2k * 0.75i + 4k * (4k * 0.75i) + 2 (4k) * ( - 1.732i) + (aC>A)xyz i (20.78 - 0.250aCD)i + (0.4330 aCD + 12)j = C (aC>A)xyz - 12.0 D i - 12.36j Equating i and j components, we have (aC>A)xyz = 46.85 m>s2 aCD = - 56.2 rad>s2 = 56.2 rad>s2

Ans.

d

Ans: vCD = 6.93 rad>s aCD = 56.2 rad>s2 d 778

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16–141. The collar C is pinned to rod CD while it slides on rod AB. If rod AB has an angular velocity of 2 rad>s and an angular acceleration of 8 rad>s2, both acting counterclockwise, determine the angular velocity and the angular acceleration of rod CD at the instant shown.

vAB  2 rad/s aAB  8 rad/s2

A 60

C 1.5 m

B 1m

D

Solution The fixed and rotating X - Y and x - y coordinate systems are set to coincide with origin at A as shown in Fig. a. Here, the x - y coordinate system is attached to link AC. Thus, Motion of moving Reference

 otion of collar C with M respect to moving Reference

vA = 0

rC>A = 51.5i6 m

aA = 0 𝛀 = VAB = 5 2k6 rad>s #

2

𝛀 = AAB = 5 8k6 rad>s

(vC>A)xyz = (vC>A)xyzi (aC>A)xyz = (aC>A)xyzi

The motions of collar C in the fixed system are vC = VCD * rC>D = ( - vCDk) * ( - i) = vCDj aC = ACD * rC>D - vCD2 rC>D = ( - aC>Dk) * ( -i) - v2CD( - i) = v2CDi + aCDj Applying the relative velocity equation, vC = vA + 𝛀 * rC>A + (vC>A)xyz vCDj = 0 + (2k) * (1.5i) = (vC>A)xyzi vCDj = (vC>A)xyzi + 3j Equating i and j components (vC>A)xyz = 0 Ans.

vCD = 3.00 rad>s b Applying the relative acceleration equation, # aC = aA + 𝛀 * rC>A + 𝛀 * (𝛀 * rC>A) + 2Ω * (vC>A)xyz + (aC>A)xyz

3.002i + aCDj = 0 + (8k) * (1.5i) + (2k) * (2k * 1.5i) + 2(2k) * 0 + (aC>A)xyzi 9i + aCDj = 3 (aC>A)xyz - 6 4 i + 12j

Equating i and j components,

9 = (aC>A)xyz - 6;  (aC>A)xyz = 15 m>s2 aCD = 12.0 rad>s2 b

Ans.

Ans: vCD = 3.00 rad>s b aCD = 12.0 rad>s2 b 779

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16–142. At the instant shown, the robotic arm AB is rotating counterclockwise at v = 5 rad>s and has an angular acceleration a = 2 rad>s2. Simultaneously, the grip BC is rotating counterclockwise at v¿ = 6 rad>s and a¿ = 2 rad>s2, both measured relative to a fixed reference. Determine the velocity and acceleration of the object held at the grip C.

y

125 mm 15°

ω ,α

30°

SOLUTION

A

vC = vB + Æ * rC>B + (vC>B)xyz # aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz Motion of moving reference

C

B 300 mm

x

ω, α

(1) (2)

Motion of C with respect to moving reference rC>B = {0.125 cos 15°i + 0.125 sin 15°j} m

Æ = {6k} rad>s # Æ = {2k} rad>s2

(vC>B)xyz = 0 (aC>B)xyz = 0

Motion of B: vB = v * rB>A = (5k) * (0.3 cos 30°i + 0.3 sin 30°j) = {-0.75i + 1.2990j} m>s aB = a * rB>A - v2rB>A = (2k) * (0.3 cos 30°i + 0.3 sin 30°j) - (5)2(0.3 cos 30°i + 0.3 sin 30°j) = { -6.7952i - 3.2304j} m>s2 Substitute the data into Eqs. (1) and (2) yields: vC = ( -0.75i + 1.2990j) + (6k) * (0.125 cos 15°i + 0.125 sin 15°j) + 0 Ans.

= {-0.944i + 2.02j} m>s aC = ( -6.79527i - 3.2304j) + (2k) * (0.125 cos 15°i + 0.125 sin 15°j) + (6k) * [(6k) * (0.125 cos 15°i + 0.125 sin 15°j)] + 0 + 0 = {-11.2i - 4.15j} m s2

Ans.

780

Ans: vC = 5 -0.944i + 2.02j 6 m>s aC = 5 - 11.2i - 4.15j 6 m>s2

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16–143.

Peg B on the gear slides freely along the slot in link AB. If the gear’s center O moves with the velocity and acceleration shown, determine the angular velocity and angular acceleration of the link at this instant.

150 mm

B

vO  3 m/s aO  1.5 m/s2

600 mm O

150 mm

SOLUTION

A

Gear Motion: The IC of the gear is located at the point where the gear and the gear rack mesh, Fig. a. Thus, vO 3 = 20 rad>s = v = rO>IC 0.15 Then,

vB = vrB>IC = 20(0.3) = 6 m>s :

Since the gear rolls on the gear rack, a =

aO 1.5 = = 10 rad>s. By referring to Fig. b, r 0.15

aB = aO + a * rB>O - v2 rB>O (aB)t i - (aB)n j = 1.5i + (-10k) * 0.15j - 202(0.15j) (aB)t i - (aB)n j = 3i - 60j Thus,

(aB)t = 3 m>s2

(aB)n = 60 m>s2

Reference Frame: The x¿y¿z¿ rotating reference frame is attached to link AB and coincides with the XYZ fixed reference frame, Figs. c and d. Thus, vB and aB with respect to the XYZ frame is vB = [6 sin 30°i - 6 cos 30° j] = [3i - 5.196j] m>s aB = (3 sin 30° - 60 cos 30°)i + ( -3 cos 30° - 60 sin 30°)j = [ -50.46i - 32.60j] m>s2 For motion of the x¿y¿z¿ frame with reference to the XYZ reference frame, # vA = aA = 0 vAB = -vABk vAB = -aAB k For the motion of point B with respect to the x¿y¿z¿ frame is rB>A = [0.6j]m

(vrel)x¿y¿z¿ = (vrel)x¿y¿z¿ j

(arel)x¿y¿z¿ = (arel)x¿y¿z¿ j

Velocity: Applying the relative velocity equation, vB = vA + vAB * rB>A + (vrel)x¿y¿z¿ 3i - 5.196j = 0 + ( -vABk) * (0.6j) + (vrel)x¿y¿z¿ j 3i - 5.196j = 0.6vAB i + (vrel)x¿y¿z¿j Equating the i and j components yields vAB = 5 rad>s

3 = 0.6vAB

Ans.

(vrel)x¿y¿z¿ = - 5.196 m>s Acceleration: Applying the relative acceleration equation. # aB = aA + vAB * rB>A + vAB * (vAB * rB>A) + 2vAB * (vrel)x¿y¿z¿ + (a rel)x¿y¿z¿ -50.46i - 32.60j = 0 + (- aABk) * (0.6j) + ( -5k) * [(-5k) * (0.6j)] + 2( -5k) * ( -5.196j) + (arel)x¿y¿z¿j -50.46i - 32.60j = (0.6aAB - 51.96)i + C (arel)x¿y¿z¿ - 15 D j Equating the i components, -50.46 = 0.6a AB - 51.96 aAB = 2.5 rad>s2

Ans. 781

Ans: vAB = 5 rad>s b aAB = 2.5 rad>s2 b

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*16–144. The cars on the amusement-park ride rotate around the axle at A with a constant angular velocity vA>f = 2 rad>s, measured relative to the frame AB. At the same time the frame rotates around the main axle support at B with a constant angular velocity vf = 1 rad>s. Determine the velocity and acceleration of the passenger at C at the instant shown.

y

D

8 ft

SOLUTION vC = vA + Æ * rC>A + (vC>A)xyz # aC = aA + Æ * rC>A + Æ * (Æ * rC>A) + 2Æ * (vC>A)xyz + (aC>A)xyz

(1)

C vA/f

8 ft A

2 rad/s

x 15 ft 30

(2) vf

Motion of moving refernce Æ = {3k} rad>s # Æ = 0

1 rad/s

B

Motion of C with respect to moving reference rC>A = {- 8i} ft (vC>A)xyz = 0 (a C>A)xyz = 0

Motion of A: vA = v * rA>B = (1k) * ( - 15 cos 30°i + 15 sin 30°j) = {- 7.5i - 12.99j} ft>s aA = a * rA>B - v2 rA>B = 0 - (1)2(- 15 cos 30°i + 15 sin 30°j) = {12.99i - 7.5j} ft>s2 Substitute the data into Eqs.(1) and (2) yields: vC = ( -7.5i - 12.99j) + (3k) * (-8i) + 0 Ans.

= {- 7.5i - 37.0j} ft>s aC = (12.99i - 7.5j) + 0 + (3k) * [(3k) * ( - 8i) + 0 + 0] = {85.0i - 7.5j} ft>s2

Ans.

Ans: vC = { - 7.5i - 37.0j} ft>s aC = {85.0i - 7.5j} ft>s2 782

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16–145. A ride in an amusement park consists of a rotating arm AB having a constant angular velocity vAB = 2 rad>s about point A and a car mounted at the end of the arm which has a constant angular velocity V ¿ = 5-0.5k6 rad>s, measured relative to the arm. At the instant shown, determine the velocity and acceleration of the passenger at C.

v¿

0.5 rad/s

B 10 ft

y vAB

SOLUTION

2 rad/s

30

rB>A = (10 cos 30° i + 10 sin 30° j) = {8.66i + 5j} ft

A

2 ft

60 C

x

vB = vAB * rB>A = 2k * (8.66i + 5j) = {-10.0i + 17.32j} ft>s aB = aAB * rB>A - v2AB rB>A = 0 - (2)2 (8.66i + 5j) = { -34.64i - 20j} ft>s2 Æ = (2 - 0.5)k = 1.5k vC = vB + Æ * rC>B + (vC>B)xyz = - 10.0i + 17.32j + 1.5k * ( -2j) + 0 = { -7.00i + 17.3j} ft>s # aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz

Ans.

= -34.64i - 20j + 0 + (1.5k) * (1.5k) * (- 2j) + 0 + 0 = {-34.6i - 15.5j} ft>s2

Ans.

Ans: vC = { - 7.00i + 17.3j} ft>s aC = { - 34.6i - 15.5j} ft>s2 783

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16–146. A ride in an amusement park consists of a rotating arm AB that has an angular acceleration of aAB = 1 rad>s2 when vAB = 2 rad>s at the instant shown. Also at this instant the car mounted at the end of the arm has an angular acceleration of A = 5 -0.6k6 rad>s2 and angular velocity of V ¿ = 5 - 0.5k6 rad>s, measured relative to the arm. Determine the velocity and acceleration of the passenger C at this instant.

v¿

0.5 rad/s

B 10 ft

y vAB

SOLUTION

2 rad/s

30 A

rB>A = (10 cos 30°i + 10 sin 30°j) = {8.66i + 5j} ft

2 ft

60 C

x

vB = vAB * rB>A = 2k * (8.66i + 5j) = { -10.0i + 17.32j} ft>s aB = aAB * rB>A - v2AB rB>A = (1k) * (8.66i + 5j) - (2)2(8.66i + 5j) = {-39.64i - 11.34j} ft>s2 Æ = (2- 0.5)k = 1.5k # Æ = (1 - 0.6)k = 0.4k vC = vB + Æ * rC>B + (vC>B)xyz = - 10.0i + 17.32j + 1.5k * ( -2j) + 0 = {- 7.00i + 17.3j} ft>s # aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz

Ans.

= - 39.64i - 11.34j + (0.4k) * ( -2j) + (1.5k) * (1.5k) * ( -2j) + 0 + 0 = {- 38.8i - 6.84j} ft>s2

Ans.

Ans: vC = { - 7.00i + 17.3j} ft>s aC = { - 38.8i - 6.84j} ft>s2 784

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16–147. If the slider block C is fixed to the disk that has a constant counterclockwise angular velocity of 4 rad>s, determine the angular velocity and angular acceleration of the slotted arm AB at the instant shown.

40 mm C

B

60 mm 30 v  4 rad/s

180 mm

Solution vC = - (4)(60) sin 30°i - 4(60) cos 30°j = - 120i - 207.85j

60

aC = (4)2(60) sin 60°i - (4)2(60) cos 60°j = 831.38i - 480j

A

Thus, vC = vA + Ω * rC>A + (vC>A)xyz - 120i - 207.85j = 0 + ( vABk ) * (180j) - vC>Aj - 120 = -180vAB Ans.

vAB = 0.667 rad>s d - 207.85 = - vC>A vC>A = 207.85 mm>s # aC = aA + 𝛀 * rC>A + 𝛀 *

( 𝛀 * rC>A ) + 2𝛀 * (vC>A)xyz + (aC>A)xyz

831.38i - 480j = 0 + (aABk) * (180j) + (0.667k) * [(0.667k) * (180j)]

+ 2(0.667k) * ( - 207.85j) - aC>A j

831.38i - 480j = - 180 aABi - 80j + 277.13i - aC>Aj 831.38 = - 180aAB + 277.13 aAB = -3.08 Thus, aAB = 3.08 rad>s2 b

Ans.

- 480 = -80 - aC>A aC>A = 400 mm>s2

Ans: vAB = 0.667 rad>s d aAB = 3.08 rad>s2 b 785

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*16–148. At the instant shown, car A travels with a speed of 25 m>s, which is decreasing at a constant rate of 2 m>s2, while car C travels with a speed of 15 m>s, which is increasing at a constant rate of 3 m>s2. Determine the velocity and acceleration of car A with respect to car C.

45

250 m

15 m/s 2 m/s2

C

B

SOLUTION

15 m/s 3 m/s2

200 m

Reference Frame: The xyz rotating reference frame is attached to car C and

A

25 m/s 2 m/s2

coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since car C moves along the circular road, its normal component of acceleration is vC 2 152 = 0.9 m>s2. Thus, the motion of car C with respect to the XYZ = (aC)n = r 250 frame is vC = -15 cos 45°i - 15 sin 45°j = [ - 10.607i - 10.607j] m>s aC = ( -0.9 cos 45° - 3 cos 45°)i + (0.9 sin 45° - 3 sin 45°)j = [-2.758i - 1.485j] m>s2 Also, the angular velocity and angular acceleration of the xyz reference frame is v =

vC 15 = 0.06 rad>s = r 250

v = [ -0.06k] rad>s

(aC)t 3 # = v = = 0.012 rad>s2 r 250

# v = [-0.012k] rad>s2

The velocity and accdeleration of car A with respect to the XYZ frame is vA = [25j] m>s

aA = [-2j] m>s2

From the geometry shown in Fig. a, rA>C = - 250 sin 45°i - (450 - 250 cos 45°)j = [-176.78i - 273.22j] m Velocity: Applying the relative velocity equation, vA = vC + v * rA>C + (v rel)xyz 25j = (- 10.607i - 10.607j) + ( -0.06k) * ( - 176.78i - 273.22j) + (v rel)xyz 25j = - 27i + (vrel)xyz Ans.

(vrel)xyz = [27i + 25j] m>s Acceleration: Applying the relative acceleration equation, # aA = aC + v * rA>C + v * (v * rA>C) + 2v * (v rel)xyz + (a rel)xyz -2j = ( -2.758i - 1.485j) + (-0.012k) * (- 176.78i - 273.22j)

+ ( -0.06k) * [( -0.06k) * ( -176.78i - 273.22j)] + 2( - 0.06k) * (27i + 25j) + (arel)xyz - 2j = - 2.4i - 1.62j + (a rel)xyz (arel)xyz = [2.4i - 0.38j] m>s2

Ans.

786

Ans: (vrel)xyz = [27i + 25j] m>s (arel)xyz = [2.4i - 0.38j] m>s2

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16–149. At the instant shown, car B travels with a speed of 15 m>s, which is increasing at a constant rate of 2 m>s2, while car C travels with a speed of 15 m>s, which is increasing at a constant rate of 3 m>s2. Determine the velocity and acceleration of car B with respect to car C.

45

250 m

15 m/s 2 m/s2

C

B

SOLUTION

15 m/s 3 m/s2

200 m

Reference Frame: The xyz rotating reference frame is attached to C and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Since B and C

A

25 m/s 2 m/s2

move along the circular road, their normal components of acceleration are vB 2 vC 2 152 152 = 0.9 m>s2 and (aC)n = = 0.9 m>s2. Thus, the = = (aB)n = r r 250 250 motion of cars B and C with respect to the XYZ frame are vB = [ - 15i] m>s vC = [ -15 cos 45°i - 15 sin 45°j] = [-10.607i - 10.607j] m>s aB = [ - 2i + 0.9j] m>s2 aC = ( - 0.9 cos 45°- 3 cos 45°)i + (0.9 sin 45°-3 sin 45°)j = [-2.758i - 1.485 j] m>s2 Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are v =

vC 15 = 0.06 rad>s = r 250

(aC)t 3 # = 0.012 rad>s2 = v = r 250

v = [-0.06k] rad>s # v = [- 0.012k] rad>s2

From the geometry shown in Fig. a, rB>C = - 250 sin 45°i - (250 - 250 cos 45°)j = [-176.78i - 73.22 j] m Velocity: Applying the relative velocity equation, vB = vC + v * rB>C + (v rel)xyz -15i = ( - 10.607i - 10.607j) + ( -0.06k) * ( -176.78i - 73.22j) + (vrel)xyz -15i = - 15i + (vrel)xyz Ans.

(vrel)xyz = 0 Acceleration: Applying the relative acceleration equation, # aB = aC + v * rB>C + v * (v * rB>C) + 2v * (vrel)xyz + (a rel)xyz - 2i + 0.9j = ( -2.758i - 1.485j) + (- 0.012k) * (- 176.78i - 73.22j)

+ ( -0.06k) * [(- 0.06k) * ( -176.78i - 73.22j)] + 2( -0.06k) * 0 + (a rel)xyz - 2i + 0.9j = -3i + 0.9j + (arel)xyz 2

Ans.

(a rel)xyz = [1i] m>s

787

Ans: (vrel)xyz = 0 (a rel)xyz = {1i} m>s2

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16–150. The two-link mechanism serves to amplify angular motion. Link AB has a pin at B which is confined to move within the slot of link CD. If at the instant shown, AB (input) has an angular velocity of vAB = 2.5 rad>s, determine the angular velocity of CD (output) at this instant.

B D

150 mm C 30 45

SOLUTION

A

vAB

2.5 rad/s

rBA 0.15 m = sin 120° sin 45° rBA = 0.1837 m vC = 0 aC = 0 Æ = - vDCk # Æ = - aDCk rB>C = { -0.15 i} m (vB>C)xyz = (yB>C)xyzi (aB>C)xyz = (aB>C)xyzi vB = vAB * rB>A = ( -2.5k) * (-0.1837 cos 15°i + 0.1837 sin 15°j) = {0.1189i + 0.4436j} m>s vB = vC + Æ * rB>C + (vB>C)xyz 0.1189i + 0.4436j = 0 + (- vDCk) * (- 0.15i) + (vB>C)xyz i 0.1189i + 0.4436j = (vB>C)xyz i + 0.15vDC j Solving: (vB>C)xyz = 0.1189 m>s Ans.

vDC = 2.96 rad>s b

Ans: vDC = 2.96 rad>s b 788

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16–151. The disk rotates with the angular motion shown. Determine the angular velocity and angular acceleration of the slotted link AC at this instant. The peg at B is fixed to the disk.

A 0.75 m

30

0.3 m

B 30

Solution v  6 rad/s a  10 rad/s2

vB = -6(0.3)i = - 1.8i

C

aB = - 10(0.3)i - (6)2(0.3)j = - 3i - 10.8j vB = vA + 𝛀 * rB>A + (vB>A)xyz - 1.8i = 0 + (vACk) * (0.75i) - (vB>A)xyzi - 1.8i = - (vB>A)xyz (vB>A)xyz = 1.8 m>s 0 = vAC(0.75) Ans.

vAC = 0 #

aB = aA + 𝛀 * rB>A + 𝛀 * (𝛀 * rB>A) + 2𝛀 * (vB>A)xyz + (aB>A)xyz - 3i - 10.8j = 0 + aACk * (0.75i) + 0 + 0 - aA>Bi - 3 = -aA>B aA>B = 3 m>s2 - 10.8 = aA>C(0.75) aA>C = 14.4 rad>s2 b

Ans.

Ans: vAC = 0 aAC = 14.4 rad>s2 b 789

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*16–152. The Geneva mechanism is used in a packaging system to convert constant angular motion into intermittent angular motion. The star wheel A makes one sixth of a revolution for each full revolution of the driving wheel B and the attached guide C. To do this, pin P, which is attached to B, slides into one of the radial slots of A, thereby turning wheel A, and then exits the slot. If B has a constant angular velocity of vB = 4 rad>s, determine V A and AA of wheel A at the instant shown.

vB

B

C P

4 in.

A

SOLUTION

4 rad/s

u

30

The circular path of motion of P has a radius of rP = 4 tan 30° = 2.309 in. Thus, vP = - 4(2.309)j = -9.238j aP = -(4)2(2.309)i = - 36.95i Thus, vP = vA + Æ * rP>A + (vP>A)xyz - 9.238j = 0 + (vA k) * (4j) - vP>A j Solving, Ans.

vA = 0

aP = aA

vP>A = 9.238 in.>s # + Æ * rP>A + Æ * (Æ * rP>A) + 2Æ * (vP>A)xyz + (aP>A)xyz - 36.95i = 0 + (aAk) * (4j) + 0 + 0 - aP>A j

Solving, - 36.95 = -4aA aA = 9.24 rad>s2 d

Ans.

aP>A = 0

Ans: vA = 0 aA = 9.24 rad>s2 d 790

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17–1. z

Determine the moment of inertia Iy for the slender rod. The rod’s density r and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m.

y

l

SOLUTION

A x

Iy =

LM

x 2 dm

l

= =

L0

x 2 (r A dx)

1 r A l3 3

m = rAl Thus, Iy =

1 m l2 3

Ans.

Ans: Iy = 791

1 m l2 3

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17–2. z

The solid cylinder has an outer radius R, height h, and is made from a material having a density that varies from its center as r = k + ar2, where k and a are constants. Determine the mass of the cylinder and its moment of inertia about the z axis.

R

h

SOLUTION Consider a shell element of radius r and mass dm = r dV = r(2p r dr)h R

m =

(k + ar2)(2p r dr)h

L0

m = 2p h(

aR4 kR2 + ) 2 4

m = p h R2(k +

aR2 ) 2

Ans.

dI = r2 dm = r2(r)(2p r dr)h R

Iz =

L0

r2(k + ar2)(2p r dr) h R

Iz = 2ph

L0

Iz = 2ph[ Iz =

(k r3 + a r5) dr

k R4 aR6 + ] 4 6

2 aR2 p h R4 [k + ] 2 3

Ans.

Ans:

aR2 b 2 p h R4 2 aR2 Iz = ck + d 2 3

m = p h R2 ak +

792

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17–3. y

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

R x

SOLUTION 2p

Iz =

L0

r A(R du)R2 = 2p r A R3 2p

m =

L0

r A R du = 2p r A R

Thus, I z = m R2

Ans.

Ans: Iz = mR2 793

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*17–4. y

The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration kx. The density of the material is r = 5 Mg>m3.

y2

50x 100 mm x

SOLUTION 200 mm

dm = r p y2 dx = r p (50x) dx Ix =

1 2 1 y dm = 2 L0 L2

= r pa

200

50 x {p r (50x)} dx

502 1 3 200 bc x d 2 3 0

= rp a

502 b(200)3 6 200

m =

L

dm =

L0

p r (50x) dx

200 1 = r p (50)c x2 d 2 0

= rp a

kx =

50 b (200)2 2

Ix 50 (200) = 57.7 mm = Am A3

Ans.

Ans: kx = 57.7 mm 794

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17–5. y

Determine the radius of gyration kx of the body. The specific weight of the material is g = 380 lb>ft3.

y3 = x 2 in. x

8 in.

SOLUTION dm = r dV = rp y2 dx d Ix =

1 1 (dm) y2 = pry4 dx 2 2 8

Ix =

1 prx4/3 dx = 86.17r L0 2 8

m = kx =

L0

prx2/3 dx = 60.32r

86.17r Ix = = 1.20 in. Am A 60.32r

Ans.

Ans: kx = 1.20 in. 795

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17–6. y

The sphere is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the sphere. The material has a constant density r.

x2 + y2 = r2

x

SOLUTION dIx =

y2 dm 2

dm = r dV = r(py2 dx) = r p(r2 - x2) dx dIx =

1 r p(r2 - x2)2 dx 2 r

Ix = =

1 2 2 2 r p(r - x ) dx 2 L- r 8 pr r5 15 r

m = =

L- r

2

2

Ix =

2 m r2 5

r p(r - x ) dx

4 r p r3 3

Thus, Ans.

Ans: Ix = 796

2 mr2 5

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17–7. y

The frustum is formed by rotating the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the frustum. The frustum has a constant density r.

y

–ba x

b 2b

b

x z

SOLUTION

a

dm = r dV = rpy2 dx = rp A

2b2 b2 2 x + b2 B dx x + a a2

dIx =

1 1 dmy2 = rpy4 dx 2 2

dIx =

1 4 b4 6 b4 4b4 b4 rp A 4 x4 + 3 x3 + 2 x2 + x + b4 B dx a 2 a a a a

Ix =

L

dIx = =

1 4b4 6 b4 4 b4 b4 x + b4 B dx rp A 4 x4 + 3 x3 + 2 x2 + a 2 L0 a a a 31 rpab4 10 a

m = Ix =

Lm

dm = rp

L0

A 2 x2 + b2 a

2b2 7 x + b2 B dx = rpab2 a 3

93 2 mb 70

Ans.

Ans: Ix = 797

93 2 mb 70

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*17–8. y

The hemisphere is formed by rotating the shaded area around the y axis. Determine the moment of inertia Iy and express the result in terms of the total mass m of the hemisphere. The material has a constant density r.

x2

y2

r2

x

SOLUTION r

m =

LV

r dV = r

= rp cr 2 y -

Iy = =

L0

r

p x2 dy = rp

L0

(r2 - y2)dy

1 3 r 2 y d = rp r3 3 3 0

r r r rp 1 (dm) x2 = px4 dy = (r2 - y2)2 dy 2 L0 2 L0 Lm 2

rp 4 y5 r 4rp 5 2 c r y - r2 y3 + d = r 2 3 5 0 15

Thus, Iy =

2 m r2 5

Ans.

Ans: Iy = 798

2 mr 2 5

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17–9. Determine the moment of inertia of the homogeneous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x–y plane and having a thickness dz.

z

–h (x – a) z = –– a

h

SOLUTION dV = bx dz = b(a)(1 -

z ) dz h

x

b

a

y

x dIy = dIy + (dm)[( )2 + z2] 2 =

x2 1 dm(x2) + dm( ) + dmz2 12 4

= dm(

x2 + z2) 3

= [b(a)(1 -

a2 z z )dz](r)[ (1 - )2 + z2] h 3 h

k

Iy = abr

a3 h - z 3 z ( ) + z2(1 - )]dz h h L0 3

= abr[ =

[

3 1 1 1 1 a2 (h4 - h4 + h4 - h4) + ( h4 - h4 )] 2 4 h 3 4 3h3

1 abhr(a2 + h2) 12

m = rV =

1 abhr 2

Thus, Iy =

m 2 (a + h2) 6

Ans.

Ans: Iy = 799

m 2 ( a + h2 ) 6

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17–10. The pendulum consists of a 4-kg circular plate and a 2-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

O

2m

Solution Using the parallel axis theorem by referring to Fig. a, IO = Σ ( IG + md 2 ) = c

1m

1 1 (2) ( 22 ) + 2 ( 12 ) d + c (4) ( 0.52 ) + 4 ( 2.52 ) d 12 2

= 28.17 kg # m2 Thus, the radius of gyration is kO =

28.17 IO = = 2.167 m = 2.17 m Am A4 + 2

Ans.

Ans: kO = 2.17 m 800

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17–11. The assembly is made of the slender rods that have a mass per unit length of 3 kg>m. Determine the mass moment of inertia of the assembly about an axis perpendicular to the page and passing through point O.

O 0.4 m 0.8 m

0.4 m

Solution Using the parallel axis theorem by referring to Fig. a, IO = Σ ( IG + md 2 ) = e

1 33(1.2)4 ( 1.22 ) + 33(1.2)4 ( 0.22 ) f 12 + e

1 33(0.4)4 ( 0.42 ) + 33(0.4)4 ( 0.82 ) f 12

= 1.36 kg # m2

Ans.

Ans: IO = 1.36 kg # m2 801

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*17–12. Determine the moment of inertia of the solid steel assembly about the x axis. Steel has a specific weight of gst = 490 lb>ft3.

0.25 ft 0.5 ft 2 ft

3 ft

SOLUTION Ix =

1 3 3 m (0.5)2 + m (0.5)2 m (0.25)2 2 1 10 2 10 3

1 3 1 3 1 490 a b p(0.5)2 (4)(0.5)2 a bp(0.25)2(2)(0.25)2 d a b = c p(0.5)2(3)(0.5)2 + 2 10 3 10 2 32.2 = 5.64 slug # ft2

Ans.

Ans: Ix = 5.64 slug # ft 2 802

x

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17–13. The wheel consists of a thin ring having a mass of 10 kg and four spokes made from slender rods, each having a mass of 2 kg. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A.

500 mm

SOLUTION A

IA = Io + md3 = c2c

1 (4)(1)2 d + 10(0.5)2 d + 18(0.5)2 12

= 7.67 kg # m2

Ans.

Ans: IA = 7.67 kg # m2 803

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17–14. If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

4 ft 1 ft O

SOLUTION Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a 3 distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2). 2

A

Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = a

100 1 20 20 15 b (4 2) + 8c a b (32) + a b(2.52) d + a b(12) 32.2 12 32.2 32.2 32.2

= 84.94 slug # ft2

The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem 100 20 15 IA = IO + md 2, where m = + 8a b + = 8.5404 slug and d = 4 ft. 32.2 32.2 32.2 Thus, IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2

Ans.

Ans: IA = 222 slug # ft 2 804

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17–15. Determine the moment of inertia about an axis perpendicular to the page and passing through the pin at O. The thin plate has a hole in its center. Its thickness is 50 mm, and the material has a density r = 50 kg>m3.

O

150 mm

SOLUTION IG =

1.40 m

1 1 C 50(1.4)(1.4)(0.05) D C (1.4)2 + (1.4)2 D - C 50(p)(0.15)2(0.05) D (0.15)2 12 2

1.40 m

= 1.5987 kg # m2 IO = IG + md2 m = 50(1.4)(1.4)(0.05) - 50(p)(0.15)2(0.05) = 4.7233 kg IO = 1.5987 + 4.7233(1.4 sin 45°)2 = 6.23 kg # m2

Ans.

Ans: IO = 6.23 kg # m2 805

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*17–16. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

200 mm

O

200 mm

SOLUTION Composite Parts: The plate can be subdivided into two segments as shown in Fig. a. Since segment (2) is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated.

200 mm

Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed as m1 = p(0.2 2)(20) = 0.8p kg and m2 = (0.2)(0.2)(20) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem. IO = ©IG + md2 1 1 = c (0.8p)(0.22) + 0.8p(0.22) d - c (0.8)(0.22 + 0.22) + 0.8(0.22) d 2 12 = 0.113 kg # m2

Ans.

Ans: IO = 0.113 kg # m2 806

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17–17. Determine the location y of the center of mass G of the assembly and then calculate the moment of inertia about an axis perpendicular to the page and passing through G. The  block has a mass of 3 kg and the semicylinder has a mass of 5 kg.

400 mm

300 mm G –y

Solution

200 mm O

Moment inertia of the semicylinder about its center of mass: (IG)cyc =

y =

4R 2 1 mR2 - m a b = 0.3199mR2 2 3p

Σ∼ ym Σm

=

c 0.2 -

4(0.2)

d (5) + 0.35(3) 3p = 0.2032 m = 0.203 m 5 + 3

IG = 0.3199(5)(0.2)2 + 5c 0.2032 - a0.2

4(0.2) 3p

2

bd +

Ans.

1 (3) ( 0.32 + 0.42 ) 12 + 3(0.35 - 0.2032)2

= 0.230 kg # m2

Ans.

Ans: IG = 0.230 kg # m2 807

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17–18. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point O. The block has a mass of 3 kg, and the semicylinder has a mass of 5 kg.

400 mm

300 mm G –y

Solution (IG)cyl =

200 mm O

2

1 4R mR2 - m a b = 0.3199 mR2 2 3p

IO = 0.3199(5)(0.2)2 + 5 a0.2 -

4(0.2) 3p

= 0.560 kg # m2

2

b +

1 (3) ( (0.3)2 + (0.4)2 ) + 3(0.350)2 12 Ans.



Also from the solution to Prob. 17–22, IO = IG + md 2 = 0.230 + 8(0.2032)2 = 0.560 kg # m2



Ans.

Ans: IO = 0.560 kg # m2 808

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17–19. Determine the moment of inertia of the wheel about an axis which is perpendicular to the page and passes through the center of mass G. The material has a specific weight g = 90 lb>ft 3.

0.25 ft

1 ft

G

0.25 ft

2 ft

0.5 ft

Solution IG =

O 1 ft

1 90 1 90 c (p)(2)2(0.25) d (2)2 + c (p)(2.5)2(1) d (2.5)2 2 32.2 2 32.2 -

1 90 1 90 c b(p)(0.25)2(0.25) d (0.25)2 (p)(2)2(1) d (2)2 - 4c a 2 32.2 2 32.2

- 4 ca

90 b(p)(0.25)2(0.25) d (1)2 32.2

= 118.25 = 118 slug # ft 2

Ans.

Ans: IG = 118 slug # ft 2 809

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*17–20. Determine the moment of inertia of the wheel about an axis which is perpendicular to the page and passes through point O. The material has a specific weight g = 90 lb>ft 3. 0.25 ft

1 ft

G

0.25 ft

2 ft

0.5 ft

Solution m =

O 1 ft

90 3p(2)2(0.25) + p 5 (2.5)2(1) - (2)2(1)6 - 4p(0.25)2(0.25)4 = 27.99 slug 32.2

From the solution to Prob. 17–18, IG = 118.25 slug # ft 2

IO = 118.25 + 27.99(2.5)2 = 293 slug # ft 2

Ans.

Ans: IO = 293 slug # ft 2 810

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17–21. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

O

2m

G

SOLUTION y =

y

0.5 m

1(3) + 2.25(5) © ym = = 1.781 m = 1.78 m ©m 3 + 5

Ans. 1m

IG = ©IG + md 2 =

1 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12

= 4.45 kg # m2

Ans.

Ans: y = 1.78 m IG = 4.45 kg # m2 811

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17–22. 20 mm

Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION mc = 7.85(10 ) A (0.05)p(0.01) 3

2

x¿

B = 0.1233 kg

30 mm 20 mm

mr = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg

50 mm

30 mm

1 Ix = 2 c (0.1233)(0.01)2 + (0.1233)(0.06)2 d 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B d 12

= 0.00325 kg # m2 = 3.25 g # m2

Ans.

Ans: Ix = 3.25 g # m2 812

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17–23. 20 mm

Determine the moment of inertia of the overhung crank about the x¿ axis. The material is steel having a destiny of r = 7.85 Mg>m3.

30 mm 90 mm 50 mm x

180 mm

20 mm

SOLUTION

x¿

mc = 7.85 A 103 B A (0.05)p(0.01)2 B = 0.1233 kg

30 mm

mp = 7.85 A 103 B A (0.03)(0.180)(0.02) B = 0.8478 kg

20 mm

50 mm

30 mm

1 1 Ix¿ = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d 2 2 + c

1 (0.8478) A (0.03)2 + (0.180)2 B + (0.8478)(0.06)2 d 12

= 0.00719 kg # m2 = 7.19 g # m2

Ans.

Ans: Ix′ = 7.19 g # m2 813

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*17–24. The door has a weight of 200 lb and a center of gravity at G. Determine how far the door moves in 2 s, starting from rest, if a man pushes on it at C with a horizontal force F = 30 lb. Also, find the vertical reactions at the rollers A and B.

A

F

SOLUTION + ©F = m(a ) ; : x G x

6 ft

G

C

3 ft

200 )aG 30 = ( 32.2

6 ft

B

12 ft 5 ft

aG = 4.83 ft>s2 a+©MA = ©(Mk)A;

NB(12) - 200(6) + 30(9) = (

200 )(4.83)(7) 32.2 Ans.

NB = 95.0 lb + c ©Fy = m(aG)y ;

NA + 95.0 - 200 = 0 Ans.

NA = 105 lb + ) (:

s = s0 + v0t + s = 0 + 0 +

1 2 a t 2 G

1 (4.83)(2)2 = 9.66 ft 2

Ans.

Ans: NB = 95.0 lb NA = 105 lb s = 9.66 ft 814

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17–25. The door has a weight of 200 lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12 ft to the right in 5 s, starting from rest. Also, find the vertical reactions at the rollers A and B.

A

F

SOLUTION

12 = 0 + 0 +

6 ft

G

C

3 ft

+ )s = s + v t + 1 a t2 (: 0 0 G 2

6 ft

B

12 ft 5 ft

1 aG(5)2 2

ac = 0.960 ft>s2 + ©F = m(a ) ; : x G x

F =

200 (0.960) 32.2 Ans.

F = 5.9627 lb = 5.96 lb a + ©MA = ©(Mk)A ;

NB(12) - 200(6) + 5.9627(9) =

200 (0.960)(7) 32.2 Ans.

NB = 99.0 lb + c ©Fy = m(aG)y ;

NA + 99.0 - 200 = 0 Ans.

NA = 101 lb

Ans: F = 5.96 lb NB = 99.0 lb NA = 101 lb 815

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17–26. The jet aircraft has a total mass of 22 Mg and a center of mass at G. Initially at take-off the engines provide a thrust 2T = 4 kN and T¿ = 1.5 kN. Determine the acceleration of the plane and the normal reactions on the nose wheel and each of the two wing wheels located at B. Neglect the mass of the wheels and, due to low velocity, neglect any lift caused by the wings.

T¿ 2.5 m

G

2T 2.3 m

1.2 m B 3m

6m

A

SOLUTION + ©F = ma ; : x x + c ©Fy = 0 ;

1.5 + 4 = 22aG 2By + Ay - 22(9.81) = 0

a + ©MB = ©(MK)B ;

4(2.3) - 1.5(2.5) - 22(9.81)(3) + Ay (9) = - 22aG(1.2) Ay = 72.6 kN

Ans.

By = 71.6 kN

Ans.

aG = 0.250 m>s2

Ans.

Ans: Ay = 72.6 kN By = 71.6 kN aG = 0.250 m>s2 816

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17–27. The sports car has a weight of 4500 lb and center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are ms = 0.5 and mk = 0.3, respectively. Neglect the mass of the wheels.

G 2.5 ft B

4 ft

2 ft

A

SOLUTION a + ©MA = ©(Mk)A ; - 2NB(6) + 4500(2) =

- 4500 a (2.5) 32.2 G

+ ©F = m(a ) ; : x G x

0.3(2NB) =

+ c ©Fy = m(aG)y ;

2NB + 2NA - 4500 = 0

4500 a 32.2 G

Solving, NA = 1393 lb

Ans.

NB = 857 lb

Ans.

aG = 3.68 ft>s2 + ) (:

v = v0 + act 10 = 0 + 3.68 t Ans.

t = 2.72 s

Ans: NA = 1393 lb NB = 857 lb t = 2.72 s 817

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*17–28. The assembly has a mass of 8 Mg and is hoisted using the boom and pulley system. If the winch at B draws in the cable with an acceleration of 2 m>s2, determine the compressive force in the hydraulic cylinder needed to support the boom. The boom has a mass of 2 Mg and mass center at G.

6m

2m

C

4m

Solution

B

sB + 2sL = l 1m

aB = - 2aL 2 = -2aL

G

60 A

D

2m

aL = - 1 m>s2 Assembly: + c ΣFy = may;  2T - 8 ( 103 ) (9.81) = 8 ( 103 ) (1) T = 43.24 kN Boom: FCD(2) - 2 ( 103 ) (9.81)(6 cos 60°) - 2(43.24) ( 103 ) (12 cos 60°) = 0 a+ ΣMA = 0;   Ans.

FCD = 289 kN

Ans: FCD = 289 kN 818

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17–29. The assembly has a mass of 4 Mg and is hoisted using the winch at B. Determine the greatest acceleration of the assembly so that the compressive force in the hydraulic cylinder supporting the boom does not exceed 180 kN. What is the tension in the supporting cable? The boom has a mass of 2 Mg and mass center at G.

6m

2m

B

Boom: a + ΣMA = 0; 

C

4m

Solution   180 ( 103 ) (2) - 2 ( 103 ) (9.81)(6 cos 60°) - 2T(12 cos 60°) = 0

1m

G

60 A

D

Ans.

T = 25 095 N = 25.1 kN

2m

Assembly: + c ΣFy = may ;  2(25 095) - 4 ( 103 ) (9.81) = 4 ( 103 ) a a = 2.74 m>s2

Ans.

Ans: a = 2.74 m>s2 T = 25.1 kN 819

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17–30. The uniform girder AB has a mass of 8 Mg. Determine the internal axial, shear, and bending-moment loadings at the center of the girder if a crane gives it an upward acceleration of 3 m>s2.

3 m/s2 C

Solution Girder: + c ΣFy = may ;  2T sin 60° - 8000(9.81) = 8000(3)

A

T = 59 166.86 N

60

4m

60

B

Segment: + ΣFx = max ;  59 166.86 cos 60° - N = 0 S Ans.

N = 29.6 kN + c ΣFy = may ;  59 166.86 sin 60° - 4000(9.81) + V = 4000(3)

Ans.

V = 0

a+ ΣMC = Σ(Mk)C ;  M + 4000(9.81)(1) - 59 166.86 sin 60°(2) = - 4000(3)(1) M = 51.2 kN # m

Ans.

Ans: N = 29.6 kN V = 0 M = 51.2 kN # m 820

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17–31. z

A car having a weight of 4000 lb begins to skid and turn with the brakes applied to all four wheels. If the coefficient of kinetic friction between the wheels and the road is mk = 0.8, determine the maximum critical height h of the center of gravity G such that the car does not overturn. Tipping will begin to occur after the car rotates 90° from its original direction of motion and, as shown in the figure, undergoes translation while skidding. Hint: Draw a free-body diagram of the car viewed from the front. When tipping occurs, the normal reactions of the wheels on the right side (or passenger side) are zero.

2.5 ft x h 2.5 ft

G

y

Solution NA represents the reaction for both the front and rear wheels on the left side. + ΣF = m(a ) ;  0.8N = 4000 a d A x G x 32.2 G + c ΣFy = m(aG)y ;   NA - 4000 = 0 a + ΣMA = Σ(Mk)A;  4000(2.5) =

4000 (a )(h) 32.2 G

Solving, NA = 4000 lb aG = 25.76 ft>s2 Ans.

h = 3.12 ft

Ans: h = 3.12 ft 821

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*17–32. A force of P = 300 N is applied to the 60-kg cart. Determine the reactions at both the wheels at A and both the wheels at B. Also, what is the acceleration of the cart? The mass center of the cart is at G.

P 30 G 0.4 m 0.3 m B

A

Solution

0.3 m

Equations of Motions. Referring to the FBD of the cart, Fig. a,

0.2 m

0.08 m

+ ΣF = m(a ) ;  300 cos 30° = 60a d x G x a = 4.3301 m>s2 = 4.33 m>s2 d 

Ans.

+c ΣFy = m(aG)y ;  NA + NB + 300 sin 30° - 60(9.81) = 60(0)

(1)

a+ΣMG = 0;  NB(0.2) - NA(0.3) + 300 cos 30°(0.1) - 300 sin 30°(0.38) = 0

(2)

Solving Eqs. (1) and (2), NA = 113.40 N = 113 N

Ans.

NB = 325.20 N = 325 N

Ans.

Ans: a = 4.33 m>s2 d NA = 113 N NB = 325 N 822

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17–33. Determine the largest force P that can be applied to the 60-kg cart, without causing one of the wheel reactions, either at A or at B, to be zero. Also, what is the acceleration of the cart? The mass center of the cart is at G.

P 30 G 0.4 m 0.3 m B

A

Solution Equations of Motions. Since (0.38 m) tan 30° = 0.22 m 7 0.1 m, the line of action of P passes below G. Therefore, P tends to rotate the cart clockwise. The wheels at A will leave the ground before those at B. Then, it is required that NA = 0. Referring, to the FBD of the cart, Fig. a + c ΣFy = m(aG)y;  NB + P sin 30° - 60(9.81) = 60(0)

(1)

a+ ΣMG = 0;  P cos 30°(0.1) - P sin 30°(0.38) + NB(0.2) = 0

(2)

0.3 m

0.2 m

0.08 m

Solving Eqs. (1) and (2) Ans.

P = 578.77 N = 579 N NB = 299.22 N

Ans: P = 579 N 823

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17–34. The trailer with its load has a mass of 150 kg and a center of mass at G. If it is subjected to a horizontal force of P = 600 N, determine the trailer’s acceleration and the normal force on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass.

G P

1.25 m 0.25 m

SOLUTION

0.75 m

Equations of Motion: Writing the force equation of motion along the x axis, + ©F = m(a ) ; : x G x

600 = 150a

a = 4 m>s2 :

B

0.25 m

A

600 N

0.5 m

1.25 m

Ans.

Using this result to write the moment equation about point A, a + ©MA = (Mk)A ;

150(9.81)(1.25) - 600(0.5) - NB(2) = -150(4)(1.25) Ans.

NB = 1144.69 N = 1.14 kN Using this result to write the force equation of motion along the y axis, + c ©Fy = m(a G)y ;

NA + 1144.69 - 150(9.81) = 150(0) Ans.

NA = 326.81 N = 327 N

Ans: a = 4 m>s2 S NB = 1.14 kN NA = 327 N 824

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17–35. The desk has a weight of 75 lb and a center of gravity at G. Determine its initial acceleration if a man pushes on it with a force F = 60 lb. The coefficient of kinetic friction at A and B is mk = 0.2.

F 1 ft

30 G

2 ft A

Solution

+ ΣFx = max;  60 cos 30° - 0.2NA - 0.2NB = 75 aG S 32.2

B 2 ft

2 ft

+ c ΣFy = may;  NA + NB - 75 - 60 sin 30° = 0 a+ΣMG = 0;    60 sin 30°(2) - 60 cos 30°(1) - NA(2) + NB(2) - 0.2NA(2) - 0.2NB(2) = 0 Solving, aG = 13.3 ft>s2

Ans.

NA = 44.0 lb NB = 61.0 lb

Ans: aG = 13.3 ft>s2 825

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*17–36. The desk has a weight of 75 lb and a center of gravity at G. Determine the initial acceleration of a desk when the man applies enough force F to overcome the static friction at A and B. Also, find the vertical reactions on each of the two legs at A and at B. The coefficients of static and kinetic friction at A and B are ms = 0.5 and mk = 0.2, respectively.

F 1 ft

30 G

2 ft A

Solution

B 2 ft

Force required to start desk moving;

2 ft

+ ΣFx = 0;  F cos 30° - 0.5NA - 0.5NB = 0 S + c ΣFy = 0;  NA + NB - F sin 30° - 75 = 0 Solving for F by eliminating NA + NB, F = 60.874 lb



Desk starts to slide. + ΣFx = m(aG)x;  60.874 cos 30° - 0.2NA - 0.2NB = 75 aG S 32.2 + c ΣFy = m(aG)y;  NA + NB - 60.874 sin 30° - 75 = 0 Solving for aG by eliminating NA + NB, aG = 13.58 = 13.6 ft>s2



Ans.

a+ ΣMA = Σ(MK)A;  NB(4) - 75(2) - 60.874 cos 30°(3) =

NB = 61.2 lb

So that

NA = 44.2 lb

-75 (13.58)(2) 32.2

For each leg,

′ NA =

44.2 = 22.1 lb 2

Ans.



NB′ =

61.2 = 30.6 lb 2

Ans.

Ans: aG = 13.6 ft>s2 ′ NA = 22.1 lb NB′ = 30.6 lb 826

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17–37. The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to tip on the cart. Slipping does not occur.

0.5 m

P 1m

Solution Equation of Motion. Tipping will occur about edge A. Referring to the FBD and kinetic diagram of the crate, Fig. a, a+ ΣMA = Σ(MK)A;  150(9.81)(0.25) = (150a)(0.5)

a = 4.905 m>s2

Using the result of a and refer to the FBD of the crate and cart, Fig. b, + ΣF = m(a )   P = (150 + 10)(4.905) = 784.8 N = 785 N d x G x

Ans.

Ans: P = 785 N 827

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17–38. The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to slip or tip on the cart. The coefficient of static friction between the crate and cart is ms = 0.2.

0.5 m

P 1m

Solution Equation of Motion. Assuming that the crate slips before it tips, then Ff = ms N = 0.2 N. Referring to the FBD and kinetic diagram of the crate, Fig. a  + c ΣFy = may;  N - 150 (9.81) = 150 (0)  N = 1471.5 N + ΣF = m(a ) ;         0.2(1471.5) = 150 a  a = 1.962 m>s2 d x G x a+ΣMA = (Mk)A;    150(9.81)(x) = 150(1.962)(0.5) x = 0.1 m Since x = 0.1 m 6 0.25 m, the crate indeed slips before it tips. Using the result of a and refer to the FBD of the crate and cart, Fig. b, + ΣFx = m(aG)x;  P = (150 + 10)(1.962) = 313.92 N = 314 N d

Ans.

Ans: P = 314 N 828

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17–39. The bar has a weight per length w and is supported by the smooth collar. If it is released from rest, determine the internal normal force, shear force, and bending moment in the bar as a function of x. 30

x

Solution Entire bar: ΣFx′ = m(aG)x′;  wl cos 30° =

wl (a ) g G

aG = g cos 30°

Segment: + ΣF = m(a ) ;     N = (wx cos 30°) sin 30° = 0.433wx d x G x

Ans.

+ T ΣFy = m(aG)y;  wx - V = wx cos 30°(cos 30°)

Ans.

V = 0.25wx

x x a+ ΣMS = Σ(Mk)S;  wxa b - M = wx cos 30°(cos 30°)a b 2 2

M = 0.125wx2

Ans.

Ans: N = 0.433wx V = 0.25wx M = 0.125wx2 829

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*17–40. The smooth 180-lb pipe has a length of 20 ft and a negligible diameter. It is carried on a truck as shown. Determine the maximum acceleration which the truck can have without causing the normal reaction at A to be zero. Also determine the horizontal and vertical components of force which the truck exerts on the pipe at B.

20 ft

A 5 ft

B

Solution

12 ft

+ ΣF = ma ;  B = 180 a S x x x 32.2 T + c ΣFy = 0;   By - 180 = 0 c+ΣMB = Σ(Mk)B;  180(10)a Solving,

180 12 5 aT (10)a b b = 13 32.2 13 Ans.

Bx = 432 lb

Ans.

By = 180 lb 2

Ans.

aT = 77.3 ft>s 

Ans: Bx = 432 lb By = 180 lb aT = 77.3 ft>s2 830

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17–41. The smooth 180-lb pipe has a length of 20 ft and a negligible diameter. It is carried on a truck as shown. If the truck accelerates at a = 5 ft>s2, determine the normal reaction at A and the horizontal and vertical components of force which the truck exerts on the pipe at B.

20 ft

A 5 ft

B

Solution

12 ft

+ ΣF = ma ;  B - N a 5 b = 180 (5) S x A x x 13 32.2 + c ΣFy = 0 ;  By - 180 + NAa

12 b = 0 13

a+ΣMB = Σ(Mk)B;  - 180(10)a Solving,

12 180 5 (5)(10)a b b + NA(13) = 13 32.2 13

Bx = 73.9 lb

Ans.

By = 69.7 lb

Ans.

NA = 120 lb

Ans.

Ans: Bx = 73.9 lb By = 69.7 lb NA = 120 lb 831

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17–42. The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and the cart is ms = 0.5.

0.6 m F

1m

SOLUTION 15

Equations of Motion: Assume that the crate slips, then Ff = ms N = 0.5N. a + ©MA = ©(Mk)A ;

50(9.81) cos 15°(x) - 50(9.81) sin 15°(0.5) = 50a cos 15°(0.5) + 50a sin 15°(x)

(1)

+Q©Fy¿ = m(aG)y¿ ;

N - 50(9.81) cos 15° = -50a sin 15°

(2)

R + ©Fx¿ = m(aG)x¿ ;

50(9.81) sin 15° - 0.5N = -50a cos 15°

(3)

Solving Eqs. (1), (2), and (3) yields N = 447.81 N

x = 0.250 m

a = 2.01 m>s2

Ans.

Since x 6 0.3 m , then crate will not tip. Thus, the crate slips.

Ans.

Ans: a = 2.01 m>s2 The crate slips. 832

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17–43. 1 ft

Determine the acceleration of the 150-lb cabinet and the normal reaction under the legs A and B if P = 35 lb. The coefficients of static and kinetic friction between the cabinet and the plane are ms = 0.2 and mk = 0.15, respectively. The cabinet’s center of gravity is located at G. P

G 4 ft

SOLUTION Equations of Equilibrium: The free-body diagram of the cabinet under the static condition is shown in Fig. a, where P is the unknown minimum force needed to move the cabinet. We will assume that the cabinet slides before it tips. Then, FA = msNA = 0.2NA and FB = msNB = 0.2NB. + ©F = 0; : x

P - 0.2NA - 0.2NB = 0

(1)

+ c ©Fy = 0;

NA + NB - 150 = 0

(2)

+ ©MA = 0;

NB(2) - 150(1) - P(4) = 0

(3)

1 ft

3.5 ft

A

B

Solving Eqs. (1), (2), and (3) yields P = 30 lb

NA = 15 lb

NB = 135 lb

Since P 6 35 lb and NA is positive, the cabinet will slide. Equations of Motion: Since the cabinet is in motion, FA = mkNA = 0.15NA and FB = mkNB = 0.15NB. Referring to the free-body diagram of the cabinet shown in Fig. b, + ©F = m(a ) ; : x G x + ©F = m(a ) ; : x G x + ©MG = 0;

35 - 0.15NA - 0.15NB = a

150 ba 32.2

(4) (5)

NA + NB - 150 = 0

NB(1) - 0.15NB(3.5) - 0.15NA(3.5) - NA(1) - 35(0.5) = 0

(6)

Solving Eqs. (4), (5), and (6) yields a = 2.68 ft>s2 NA = 26.9 lb

Ans. Ans.

NB = 123 lb

Ans: a = 2.68 ft>s2 NA = 26.9 lb NB = 123 lb 833

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*17–44. A

The uniform bar of mass m is pin connected to the collar, which slides along the smooth horizontal rod. If the collar is given a constant acceleration of a, determine the bar’s inclination angle u. Neglect the collar’s mass. L

a

u

SOLUTION Equations of Motion: Writing the moment equation of motion about point A, + ©MA = (Mk)A;

mg sin u a

L L b = ma cos u a b 2 2

a u = tan-1 a b g

Ans.

Ans: a u = tan - 1 a b g 834

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17–45. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the tension in the cable when the truck begins to accelerate at 5 m>s2. Also, what are the horizontal and vertical components of reaction at the hinge C?

B

30

$ G

C

1m

1.5 m 45

Solution a+ ΣMC = Σ(Mk)C;  T sin 30°(2.5) - 12 262.5(1.5 cos 45°) = 1250(5)(1.5 sin 45°)

Ans.

 T = 15 708.4 N = 15.7 kN + ΣF = m(a ) ;   - C + 15 708.4 cos 15° = 1250(5) d x x G x

Ans.

  Cx = 8.92 kN + c ΣFy = m(aG)y;  Cy - 12 262.5 - 15 708.4 sin 15° = 0

Ans.

Cy = 16.3 kN

Ans: T = 15.7 kN Cx = 8.92 kN Cy = 16.3 kN 835

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17–46. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the maximum deceleration of the truck so that the gate does not begin to rotate forward. What are the horizontal and vertical components of reaction at the hinge C?

B

30

$ G

C

1m

1.5 m 45

Solution a+ΣMC = Σ(Mk)C;  - 12 262.5(1.5 cos 45°) = - 1250(a)(1.5 sin 45°)   a = 9.81 m>s2

Ans.

+ ΣF = m(a ) ;  C = 1250(9.81) S x x G x

Ans.

Cx = 12.3 kN

 + c ΣFy = m(aG)y;  Cy - 12 262.5 = 0

Ans.

Cy = 12.3 kN

Ans: a = 9.81 m>s2 Cx = 12.3 kN Cy = 12.3 kN 836

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17–47. The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. If the acceleration is a = 20 ft>s2, determine the maximum height h of G2 of the rider so that the snowmobile’s front skid does not lift off the ground. Also, what are the traction (horizontal) force and normal reaction under the rear tracks at A?

0.5 ft

a G2 G1 1 ft

SOLUTION Equations of Motion: Since the front skid is required to be on the verge of lift off, NB = 0. Writing the moment equation about point A and referring to Fig. a, a + ©MA = (Mk)A ;

h

250(1.5) + 150(0.5) =

A

1.5 ft

150 250 (20)(hmax) + (20)(1) 32.2 32.2 Ans.

hmax = 3.163 ft = 3.16 ft Writing the force equations of motion along the x and y axes, + ©F = m(a ) ; ; x G x

FA =

250 150 (20) + (20) 32.2 32.2 Ans.

FA = 248.45 lb = 248 lb + c ©Fy = m(a G)y ;

NA - 250 - 150 = 0 Ans.

NA = 400 lb

Ans: h max = 3.16 ft FA = 248 lb NA = 400 lb 837

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*17–48. The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. If h = 3 ft, determine the snowmobile’s maximum permissible acceleration a so that its front skid does not lift off the ground. Also, find the traction (horizontal) force and the normal reaction under the rear tracks at A.

0.5 ft

a G2 G1 1 ft

SOLUTION Equations of Motion: Since the front skid is required to be on the verge of lift off, NB = 0. Writing the moment equation about point A and referring to Fig. a, a + ©MA = (Mk)A ;

h

250(1.5) + 150(0.5) = a

A

1.5 ft

150 250 a b (3) + a a b(1) 32.2 max 32.2 max

a max = 20.7 ft>s2

Ans.

Writing the force equations of motion along the x and y axes and using this result, we have + ©F = m(a ) ; ; x G x

FA =

150 250 (20.7) + (20.7) 32.2 32.2 Ans.

FA = 257.14 lb = 257 lb + c ©Fy = m(a G)y ;

NA - 150 - 250 = 0 Ans.

NA = 400 lb

Ans: amax = 20.7 ft>s2 FA = 257 lb NA = 400 lb 838

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17–49. A

If the cart’s mass is 30 kg and it is subjected to a horizontal force of P = 90 N, determine the tension in cord AB and the horizontal and vertical components of reaction on end C of the uniform 15-kg rod BC.

30 1m

C

30

B P

SOLUTION Equations of Motion: The acceleration a of the cart and the rod can be determined by considering the free-body diagram of the cart and rod system shown in Fig. a. + ©F = m(a ) ; : x G x

90 = (15 + 30)a

a = 2 m>s2

The force in the cord can be obtained directly by writing the moment equation of motion about point C by referring to Fig. b. + ©MC = (Mk)C;

FAB sin 30°(1) - 15(9.81) cos 30°(0.5) = - 15(2) sin 30°(0.5) Ans.

FAB = 112.44 N = 112 N

Using this result and applying the force equations of motion along the x and y axes, + ©F = m(a ) ; : x G x

-Cx + 112.44 sin 30° = 15(2) Ans.

Cx = 26.22 N = 26.2 N + c ©Fy = m(aG)y;

Cy + 112.44 cos 30° - 15(9.81) = 0 Ans.

Cy = 49.78 N = 49.8 N

Ans: FAB = 112 N Cx = 26.2 N Cy = 49.8 N 839

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17–50. A

If the cart’s mass is 30 kg, determine the horizontal force P that should be applied to the cart so that the cord AB just becomes slack. The uniform rod BC has a mass of 15 kg.

30 1m

C

30

B P

SOLUTION Equations of Motion: Since cord AB is required to be on the verge of becoming slack, FAB = 0. The corresponding acceleration a of the rod can be obtained directly by writing the moment equation of motion about point C. By referring to Fig. a. + ©MC = ©(MC)A;

- 15(9.81) cos 30°(0.5) = - 15a sin 30°(0.5) a = 16.99 m>s2

Using this result and writing the force equation of motion along the x axis and referring to the free-body diagram of the cart and rod system shown in Fig. b, + B ©F = m(a ) ; A: x G x

P = (30 + 15)(16.99) Ans.

= 764.61 N = 765 N

Ans: P = 765 N 840

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17–51. The pipe has a mass of 800 kg and is being towed behind the truck. If the acceleration of the truck is a t = 0.5 m>s2, determine the angle u and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A G 0.4 m

SOLUTION + ©F = ma ; : x x

-0.1NC + T cos 45° = 800(0.5)

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

- 0.1NC(0.4) + T sin f(0.4) = 0

at

45

u

C

NC = 6770.9 N Ans.

T = 1523.24 N = 1.52 kN sin f =

0.1(6770.9) 1523.24

f = 26.39° Ans.

u = 45° - f = 18.6°

Ans: T = 1.52 kN u = 18.6° 841

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*17–52. The pipe has a mass of 800 kg and is being towed behind a truck. If the angle u = 30°, determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

A G 0.4 m

SOLUTION + ©F = ma ; : x x

T cos 45° - 0.1NC = 800a

+ c ©Fy = may ;

NC - 800(9.81) + T sin 45° = 0

a + ©MG = 0;

at

45

u

C

T sin 15°(0.4) - 0.1NC(0.4) = 0

NC = 6161 N T = 2382 N = 2.38 kN

Ans.

a = 1.33 m>s2

Ans.

Ans: T = 2.38 kN a = 1.33 m>s2 842

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17–53. The crate C has a weight of 150 lb and rests on the truck elevator for which the coefficient of static friction is ms = 0.4. Determine the largest initial angular acceleration a, starting from rest, which the parallel links AB and DE can have without causing the crate to slip. No tipping occurs.

B 30 2 ft

a

A

C

E a 2 ft

D

Solution + ΣFx = max ;     0.4NC = 150 (a) cos 30° S 32.2 + c ΣFy = may;      NC - 150 =

150 (a) sin 30° 32.2

NC = 195.0 lb a = 19.34 ft>s2 19.34 = 2a a = 9.67 rad> s2

Ans.

Ans: a = 9.67 rad>s2 843

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17–54. The crate C has a weight of 150 lb and rests on the truck elevator. Determine the initial friction and normal force of the elevator on the crate if the parallel links are given an angular acceleration a = 2 rad>s2 starting from rest.

B 30 2 ft

a

A

C

E a 2 ft

D

Solution a = 2 rad> s2 a = 2a = 4 rad> s2 + ΣFx = max ;   FC = 150 (a) cos 30° S 32.2 + c ΣFy = may;      NC - 150 =

150 (a) sin 30° 32.2

FC = 16.1 lb

Ans.

NC = 159 lb

Ans.

Ans: FC = 16.1 lb NC = 159 lb 844

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17–55. The 100-kg uniform crate C rests on the elevator floor where the coefficient of static friction is ms = 0.4. Determine the largest initial angular acceleration a, starting from rest at u = 90°, without causing the crate to slip. No tipping occurs.

0.6 m

C E

1.5 m D

u

1.2 m

a

B 1.5 m A

u

Solution Equations of Motion. The crate undergoes curvilinear translation. At u = 90°, v = 0. Thus, (aG)n = v2r = 0. However; (aG)t = ar = a(1.5). Assuming that the crate slides before it tips, then, Ff = msN = 0.4 N. ΣFn = m(aG)n;  100(9.81) - N = 100(0)  N = 981 N ΣFt = m(aG)t;  0.4(981) = 100[a(1.5)]  a = 2.616 rad>s2 = 2.62 rad>s2Ans. a+ ΣMG = 0;  0.4(981)(0.6) - 981(x) = 0

x = 0.24 m

Since x 6 0.3 m, the crate indeed slides before it tips, as assumed.

Ans: a = 2.62 rad>s2 845

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*17–56. The two uniform 4-kg bars DC and EF are fixed (welded) together at E. Determine the normal force NE , shear force VE , and moment ME , which DC exerts on EF at E if at the instant u = 60° BC has an angular velocity v = 2 rad>s and an angular acceleration a = 4 rad>s2 as shown.

F

1.5 m E

D

C

2m

Solution Equations of Motion. The rod assembly undergoes curvilinear motion. Thus, (aG)t = ar = 4(2) = 8 m>s2 and (aG)n = v2r = ( 22 ) (2) = 8 m>s2. Referring to the FBD and kinetic diagram of rod EF, Fig. a

u  60 A

2m

a  4 rad/s2 B

v  2 rad/s

+ ΣFx = m(aG)x;  VE = 4(8) cos 30° + 4(8) cos 60° d

Ans.

= 43.71 N = 43.7 N

 + c ΣFy = m(aG)y;  NE - 4(9.81) = 4(8) sin 30° - 4(8) sin 60°

Ans.

NE = 27.53 N = 27.5 N 

a+ΣME = Σ(Mk)E;  ME = 4(8) cos 30°(0.75) + 4(8) cos 60°(0.75 )

= 32.78 N # m = 32.8 N # m

Ans.

Ans: VE = 43.7 N NE = 27.5 N ME = 32.8 N # m 846

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17–57. The 10-kg wheel has a radius of gyration kA = 200 mm. If the wheel is subjected to a moment M = 15t2 N # m, where t is in seconds, determine its angular velocity when t = 3 s starting from rest. Also, compute the reactions which the fixed pin A exerts on the wheel during the motion.

M A

SOLUTION + ©F = m(a ) ; : x G x

Ax = 0

+ c ©Fy = m(aG)y;

Ay - 10(9.81) = 0

c + ©MA = Ia a;

5t = 10(0.2)2a a =

dv = 12.5t dt 3

v =

L0

12.5t dt =

12.5 2 (3) 2 Ans.

v = 56.2 rad>s Ax = 0

Ans.

Ay = 98.1 N

Ans.

Ans: v = 56.2 rad>s Ax = 0 Ay = 98.1 N 847

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17–58. The uniform 24-kg plate is released from rest at the position shown. Determine its initial angular acceleration and the horizontal and vertical reactions at the pin A.

A

0.5 m

Solution

0.5 m

Equations of Motion. The mass moment of inertia of the plate about its center of 1 gravity G is IG = (24) ( 0.52 + 0.52 ) = 1.00 kg # m2. Since the plate is at rest 12 initially v = 0. Thus, (aG)n = v2rG = 0. Here rG = 20.252 + 0.252 = 0.2512 m.

Thus, (aG)t = arG = a ( 0.2512 ) . Referring to the FBD and kinetic diagram of the plate, a+ ΣMA = ( Mk ) A;  -24(9.81)(0.25) = -24 3 a ( 0.2512 )4 ( 0.2512 ) - 1.00 a

a = 14.715 rad>s2 = 14.7 rad>s2

Ans.

Also, the same result can be obtained by applying ΣMA = IAa where 1 IA = (24) ( 0.52 + 0.52 ) + 24 ( 0.2512 ) 2 = 4.00 kg # m2: 12 a+ ΣMA = IA a;  - 24(9.81)(0.25) = -4.00 a

a = 14.715 rad>s2 + ΣF = m(a ) ;  A = 24 314.715 ( 0.2512 ) 4 cos 45° = 88.29 N = 88.3 N Ans. d x G x x

+ c ΣFy = m(aG)y;  Ay - 24(9.81) = - 24 314.715 ( 0.2512 ) 4 sin 45°



Ay = 147.15 N = 147 N

Ans.

Ans: a = 14.7 rad>s2 Ax = 88.3 N Ay = 147 N 848

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17–59. The uniform slender rod has a mass m. If it is released from rest when u = 0°, determine the magnitude of the reactive force exerted on it by pin B when u = 90°.

A

L 3

B

u

2 L 3

SOLUTION C

Equations of Motion: Since the rod rotates about a fixed axis passing through point L L B, (aG)t = a rG = a a b and (aG)n = v2rG = v2 a b . The mass moment of inertia 6 6 1 2 mL . Writing the moment equation of motion about of the rod about its G is IG = 12 point B, - mg cos u a

+ ©MB = ©(Mk)B;

L L 1 L b = -m c a a b d a b - a mL2 ba 6 6 6 12

3g cos u 2L

a =

1 This equation can also be obtained by applying ©MB = IBa, whereIB = mL2 + 12 2 1 L ma b = mL2. Thus, 6 9 - mg cos u a

+ ©MB = IBa;

1 L b = - a mL2 ba 6 9

3g cos u 2L

a =

Using this result and writing the force equation of motion along the n and t axes, mg cos u - Bt = mc a

©Ft = m(aG)t;

3g L cos u b a b d 2L 6

3 mg cos u 4

Bt =

(1)

Bn - mg sin u = m c v2 a

©Fn = m(aG)n;

L bd 6

1 mv2L + mg sin u 6

Bn =

(2)

Kinematics: The angular velocity of the rod can be determined by integrating L

vdv = v

L0

adu u

vdv =

v =

L

3g cos u du L0 2L

3g sin u BL

When u = 90°, v =

3g . Substituting this result and u = 90° into Eqs. (1) and (2), AL

3 mg cos 90° = 0 4 3g 1 3 Bn = m a b(L) + mg sin 90° = mg 6 L 2

Bt =

FA =

3At2

+

An2

Ans:

2 3 3 = 0 + a mg b = mg C 2 2 2

Ans.

849

FA =

3 mg 2

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*17–60. The bent rod has a mass of 2 kg>m. If it is released from rest in the position shown, determine its initial angular acceleration and the horizontal and vertical components of reaction at A.

1.5 m C

A

1.5 m

Solution

B

Equations of Motion. Referring to Fig. a, the location of center of gravity G of the bent rod is at x =

2[0.75(1.5)(2)] + 1.5(2)(1.5) Σx∼m = = 1.00 m Σm 3(1.5)(2)

y =

1.5 = 0.75 m 2

The mass moment of inertia of the bent rod about its center of gravity is 1 1 IG = 2c (3)( 1.52 ) + 3 ( 0.252 + 0.752 )d + c (3)( 1.52 ) + 3 ( 0.52 )d = 6.1875 kg # m2. 12 12 Here, rG = 21.002 + 0.752 = 1.25 m. Since the bent rod is at rest initially, v = 0. Thus, (aG)n = v2rG = 0. Also, (aG)t = arG = a(1.25). Referring to the FBD and kinetic diagram of the plate, a+ ΣMA = (Mk)A;  9(9.81)(1) = 9[a(1.25)](1.25) + 6.1875 a

a = 4.36 rad>s2 d

Ans.

Also, the same result can be obtained by applying ΣMA = IAa where IA =

1 1 (3) ( 1.52 ) + 3 ( 0.752 ) + (3) ( 1.52 ) + 3 ( 1.52 + 0.752 ) 12 12 1 + (3) ( 1.52 ) + 3 ( 1.52 + 0.752 ) = 20.25 kg # m2 : 12

a+ ΣMA = IAa,  9(9.81)(1) = 20.25 a  a = 4.36 rad>s2 + ΣF = m(a ) ;  A = 9[4.36(1.25)] a 3 b = 29.43 N = 29.4 N S x G x x 5

4 + c ΣFy = m(aG)y;  Ay - 9(9.81) = - 9[4.36(1.25)] a b 5



Ans.

Ans.

Ay = 49.05 N = 49.1 N

Ans: a = 4.36 rad>s2 d Ax = 29.4 N Ay = 49.1 N 850

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17–61. If a horizontal force of P = 100 N is applied to the 300-kg reel of cable, determine its initial angular acceleration. The reel rests on rollers at A and B and has a radius of gyration of kO = 0.6 m.

P 0.75 m O 1m 20 A

20 B

Solution Equations of Motions. The mass moment of inertia of the reel about O is IO = Mk 2O = 300 ( 0.62 ) = 108 kg # m2. Referring to the FBD of the reel, Fig. a, a+ ΣMO = IO a;   - 100(0.75) = 108( -a)

a = 0.6944 rad>s2 = 0.694 rad>s2Ans.

Ans: a = 0.694 rad>s2 851

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17–62. The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so that the torque developed is M = 15u2 lb # ft, where u is in radians. If the bar is released from rest when it is vertical at u = 90°, determine its angular velocity at the instant u = 0°.

1 ft u

1 ft O

SOLUTION c + ©MO = IOa; - 5u = [

1 10 ( )(2)2]a 12 32.2

- 48.3 u = a a du = v dv o

-

Lp2

48.3 u du =

v

L0

v dv

1 48.3 p 2 ( ) = v2 2 2 2 Ans.

v = 10.9 rad/s

Ans: v = 10.9 rad>s 852

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17–63. The 10-lb bar is pinned at its center O and connected to a torsional spring. The spring has a stiffness k = 5 lb # ft>rad, so that the torque developed is M = 15u2 lb # ft, where u is in radians. If the bar is released from rest when it is vertical at u = 90°, determine its angular velocity at the instant u = 45°.

1 ft u

1 ft O

SOLUTION c + ©MO = IOa;

5u = [

1 10 ( )(2)2]a 12 32.2

a = - 48.3u a du = v dv -

p 4

Lp2

v

48.3u du =

L0

v dv

p p 1 - 24.15 a( )2 - ( )2 b = v2 4 2 2 Ans.

v = 9.45 rad s

Ans: v = 9.45 rad>s 853

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*17–64. A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = ( ¼ u 2 ) N, where u is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revolutions. The disk has an initial angular velocity of v0 = 1 rad>s.

F 300 mm O

v

Solution Equations of Motion. The mass moment inertia of the disk about O is 1 1 IO = mr 2 = (8)( 0.32 ) = 0.36 kg # m2. Referring to the FBD of the disk, Fig. a, 2 2 1 a+ ΣMO = IO a;  a u 2 b(0.3) = 0.36 a 4

a = ( 0.2083 u 2 ) rad>s2

Kinematics. Using the result of a, integrate vdv = adu with the initial condition v = 0 when u = 0, L1



v

vdv =

L0

5(2p)

0.2083 u 2 du

5(2p) 1 a b ( v2 - 1 ) = 0.06944 u 3 2 2 0

Ans.

v = 65.63 rad>s = 65.6 rad>s

Ans: v = 65.6 rad>s 854

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17–65. Disk A has a weight of 5 lb and disk B has a weight of 10 lb. If no slipping occurs between them, determine the couple moment M which must be applied to disk A to give it an angular acceleration of 4 rad>s2.

a

4 rad/s2 M

0.5 ft A

0.75 ft B

SOLUTION Disk A: c + ©MA = IA aA ;

1 5 b (0.5)2 d(4) M - FD (0.5) = c a 2 32.2

Disk B: + ©MB = IB aB ;

FD (0.75) = c

1 10 a b(0.75)2 daB 2 32.2

rA aA = rB aB 0.5(4) = 0.75aB Solving: aB = 2.67 rad>s2; M = 0.233 lb

#

FD = 0.311 lb Ans.

ft

Ans: M = 0.233 lb # ft 855

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17–66. The kinetic diagram representing the general rotational motion of a rigid body about a fixed axis passing through O is shown in the figure. Show that IGA may be eliminated by moving the vectors m(aG)t and m(aG)n to point P, located a distance rGP = k2G>rOG from the center of mass G of the body. Here kG represents the radius of gyration of the body about an axis passing through G. The point P is called the center of percussion of the body.

a P m(aG)t G m(aG)n

IG a rGP

O

SOLUTION

rOG

m(aG)t rOG + IG a = m(aG)t rOG + A mk2G B a However, k2G = rOG rGP and a =

(aG)t rOG

m(aG)t rOG + IG a = m(aG)t rOG + (mrOG rGP) c = m(aG)t(rOG + rGP)

(aG)t d rOG

Q.E.D.

Ans: m(aG)t rOG + IG a = m(aG)t(rOG + rGP) 856

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17–67. If the cord at B suddenly fails, determine the horizontal and vertical components of the initial reaction at the pin A, and the angular acceleration of the 120-kg beam. Treat the beam as a uniform slender rod.

800 N B A 2m

2m

Solution Equations of Motion. The mass moment of inertia of the beam about A is IA = 1 (120) ( 42 ) + 120 ( 22 ) = 640 kg # m2. Initially, the beam is at rest, v = 0. Thus, 12 (aG)n = v2 r = 0. Also, (aG)t = arG = a(2) = 2a. referring to the FBD of the beam, Fig. a a+ ΣMA = IAa;     800(4) + 120(9.81)(2) = 640 a

a = 8.67875 rad>s2 = 8.68 rad>s2

Ans.

ΣFn = m(aG)n ;     An = 0

Ans.

800 + 120(9.81) + At = 120[2(8.67875)] ΣFt = m(aG)t ;   Ans.

  At = 105.7 N = 106 N

Ans: a = 8.68 rad>s2 An = 0 At = 106 N 857

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*17–68. The device acts as a pop-up barrier to prevent the passage of a vehicle. It consists of a 100-kg steel plate AC and a 200-kg counterweight solid concrete block located as shown. Determine the moment of inertia of the plate and block about the hinged axis through A. Neglect the mass of the supporting arms AB. Also, determine the initial angular acceleration of the assembly when it is released from rest at u = 45°.

0.3 m 0.5 m 0.5 m C

A

B

1.25 m

SOLUTION Mass Moment of Inertia: IA =

1 (100) A 1.252 B + 100 A 0.6252 B 12 +

1 (200) A 0.52 + 0.32 B + 200 A 20.752 + 0.152 B 2 12

= 174.75 kg # m2 = 175 kg # m2

Ans.

Equation of Motion: Applying Eq. 17–16, we have a + ©MA = IA a;

100(9.81)(0.625) + 200(9.81) sin 45°(0.15) - 200(9.81) cos 45°(0.75) = - 174.75a a

1.25 rad s2

Ans.

Ans: IA = 175 kg # m2 a = 1.25 rad>s2 858

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17–69. The 20-kg roll of paper has a radius of gyration kA = 90 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. If the roll rests against a wall for which the coefficient of kinetic friction is μk = 0.2 and a vertical force F = 30 N is applied to the end of the paper, determine the angular acceleration of the roll as the paper unrolls.

B

300 mm

Solution

C

+ S ΣFx = m(aG)x ;  NC - TAB cos 67.38° = 0

A 125 mm

+ c ΣFy = m(aG)y ;  TAB sin 67.38° - 0.2NC - 20(9.81) - 30 = 0 c+ ΣMA = IA a;

- 0.2NC(0.125) + 30(0.125) = 20(0.09)2a F

Solving: NC = 103 N TAB = 267 N a = 7.28 rad>s2

Ans.

Ans: a = 7.28 rad>s2 859

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17–70. The 20-kg roll of paper has a radius of gyration kA = 90 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. If the roll rests against a wall for which the coefficient of kinetic friction is μk = 0.2, determine the constant vertical force F that must be applied to the roll to pull off 1 m of paper in t = 3 s starting from rest. Neglect the mass of paper that is removed.

B

300 mm

Solution

C

1 2

A 125 mm

( + T ) s = s0 + v0 t + aC t 2 1 = 0 + 0 +

1 a (3)2 2 C

F

aC = 0.222 m>s2 a =

aC = 1.778 rad>s2 0.125

+ ΣFx = m(aG x);  NC - TAB cos 67.38° = 0 S + c ΣFy = m(aG)y;        TAB sin 67.38° - 0.2NC - 20(9.81) - F = 0 c+ ΣMA = IAa;        - 0.2NC (0.125) + F(0.125) = 20(0.09)2(1.778) Solving: NC = 99.3 N TAB = 258 N Ans.

F = 22.1 N

Ans: F = 22.1 N 860

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17–71. The reel of cable has a mass of 400 kg and a radius of gyration of kA = 0.75 m. Determine its angular velocity when t = 2 s, starting from rest, if the force P = (20t2 + 80) N, when t is in seconds. Neglect the mass of the unwound cable, and assume it is always at a radius of 0.5 m.

0.5 m

P

1m A

Solution Equations of Motion. The mass moment of inertia of the reel about A is IA = Mk 2A = 400 ( 0.752 ) = 225 kg # m2. Referring to the FBD of the reel, Fig. a a+ ΣMA = IAa ;  - ( 20t 2 + 80 ) (0.5) = 225( - a)   a =

2 2 ( t + 4 ) rad>s2 45

Kinematics. Using the result of a, integrate dv = adt, with the initial condition v = 0 at t = 0, L0

2s

v

dv =

2 2 ( t + 4 ) dt L0 45

v = 0.4741 rad>s = 0.474 rad>s

Ans.

Ans: v = 0.474 rad>s 861

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*17–72. The 30-kg disk is originally spinning at v = 125 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is μC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of force which the member AB exerts on the pin at A during this time? Neglect the mass of AB.

0.5 m

B 0.3 m

0.5 m

C

v  125 rad/s

Solution Equations of Motion. The mass moment of inertia of the disk about B is 1 1 IB = mr 2 = (30) ( 0.32 ) = 1.35 kg # m2. Since it is required to slip at C, 2 2 Ff = mCNC = 0.5 NC. referring to the FBD of the disk, Fig. a,

A

+ 0.5NC - FAB cos 45° = 30(0)(1) S ΣFx = m(aG)x;       + c ΣFy = m(aG)y;   NC - FAB sin 45° - 30(9.81) = 30(0)(2) Solving Eqs. (1) and (2), NC = 588.6 N  FAB = 416.20 N Subsequently, a+ ΣMB = IBa;  0.5(588.6)(0.3) = 1.35a a = 65.4 rad>s2 d Referring to the FBD of pin A, Fig. b, + ΣFx = 0;     416.20 cos 45° - Ax = 0  Ax = 294.3 N = 294 N S

Ans.

+ c ΣFy = 0;  416.20 sin 45° - Ay = 0  Ay = 294.3 N = 294 N

Ans.

Kinematic. Using the result of a, + b v = v0 + at;  0 = 125 + ( - 65.4)t Ans.

t = 1.911 s = 1.91 s

Ans: Ax = 294 N Ay = 294 N t = 1.91 s 862

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17–73. Cable is unwound from a spool supported on small rollers at A and B by exerting a force T = 300 N on the cable. Compute the time needed to unravel 5 m of cable from the spool if the spool and cable have a total mass of 600 kg and a radius of gyration of kO = 1.2 m. For the calculation, neglect the mass of the cable being unwound and the mass of the rollers at A and B. The rollers turn with no friction.

T  300 N

1.5 m 30

0.8 m O

A

Solution

B

1m

IO = mk 2O = 600(1.2)2 = 864 kg # m2

c + ΣMO = IOa;  300(0.8) = 864(a)  a = 0.2778 rad>s2 The angular displacement u = u = u0 + v0 r + 6.25 = 0 + 0 + t = 6.71 s

s 5 = = 6.25 rad. r 0.8

1 2 at 2 c

1 (0.27778)t 2 2 Ans.



Ans: t = 6.71 s 863

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17–74. The 5-kg cylinder is initially at rest when it is placed in contact with the wall B and the rotor at A. If the rotor always maintains a constant clockwise angular velocity v = 6 rad>s, determine the initial angular acceleration of the cylinder. The coefficient of kinetic friction at the contacting surfaces B and C is mk = 0.2. B 125 mm

SOLUTION

v 45

Equations of Motion: The mass moment of inertia of the cylinder about point O is 1 1 given by IO = mr 2 = (5)(0.1252) = 0.0390625 kg # m2. Applying Eq. 17–16, 2 2 we have + ©F = m(a ) ; : x G x

NB + 0.2NA cos 45° - NA sin 45° = 0

(1)

+ c ©Fy = m(a G)y ;

0.2NB + 0.2NA sin 45° + NA cos 45° - 5(9.81) = 0

(2)

a + ©MO = IO a;

0.2NA (0.125) - 0.2NB (0.125) = 0.0390625a

A

C

(3)

Solving Eqs. (1), (2), and (3) yields; NA = 51.01 N

NB = 28.85 N

a = 14.2 rad>s2

Ans.

Ans: a = 14.2 rad>s2 864

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17–75. The wheel has a mass of 25 kg and a radius of gyration kB = 0.15 m. It is originally spinning at v = 40 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is mC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of reaction which the pin at A exerts on AB during this time? Neglect the mass of AB.

0.4 m A 0.3 m B

0.2 m

V

SOLUTION IB =

mk2B

2

= 25(0.15)

A 35 B FAB + NC - 25(9.81) = 0

+ c ©Fy = m(aG)y ; + ©F = m(a ) ; : x G x a + ©MB = IBa;

C

= 0.5625 kg # m2

0.5NC -

(1)

A 45 B FAB = 0

(2) (3)

0.5NC(0.2) = 0.5625( - a)

Solvings Eqs. (1),(2) and (3) yields: FAB = 111.48 N

NC = 178.4 N

a = -31.71 rad>s2 Ax = 45FAB = 0.8(111.48) = 89.2 N

Ans.

Ay = 35 FAB = 0.6(111.48) = 66.9 N

Ans.

v = v0 + ac t 0 = 40 + ( -31.71) t Ans.

t = 1.26 s

Ans: Ax = 89.2 N Ay = 66.9 N t = 1.25 s 865

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*17–76. The 20-kg roll of paper has a radius of gyration kA = 120 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. The roll rests on the floor, for which the coefficient of kinetic friction is μk = 0.2. If a horizontal force F = 60 N is applied to the end of the paper, determine the initial angular acceleration of the roll as the paper unrolls.

F

A 300 mm B C 400 mm

Solution Equations of Motion. The mass moment of inertia of the paper roll about A is IA = mkA2 = 20(0.122) = 0.288 kg # m2. Since it is required to slip at C, the friction is Ff = mkN = 0.2 N. Referring to the FBD of the paper roll, Fig. a 4 + S ΣFx = m(aG)x;  0.2 N - FAB a 5 b + 60 = 20(0)(1)

3 + c ΣFy = m(aG)y;      N - FAB a b - 20(9.81) = 20(0)(2) 5 Solving Eqs. (1) and (2)

FAB = 145.94 N  N = 283.76 N

Subsequently a+ ΣMA = IAa;  0.2(283.76)(0.3) - 60(0.3) = 0.288( -a) a = 3.3824 rad>s2 = 3.38 rad>s2

Ans.

Ans: a = 3.38 rad>s2 866

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17–77. Disk D turns with a constant clockwise angular velocity of 30 rad>s. Disk E has a weight of 60 lb and is initially at rest when it is brought into contact with D. Determine the time required for disk E to attain the same angular velocity as disk D. The coefficient of kinetic friction between the two disks is mk = 0.3. Neglect the weight of bar BC.

2 ft

B E 1 ft

2 ft 1 ft

C

SOLUTION

A

D

v  30 rad/s

Equations of Motion: The mass moment of inertia of disk E about point B is given 1 1 60 by IB = mr2 = a b (12) = 0.9317 slug # ft2. Applying Eq. 17–16, we have 2 2 32.2 + ©F = m(a ) ; : x G x

(1)

0.3N - FBC cos 45° = 0

+ c ©Fy = m(aG)y ;

N - FBC sin 45° - 60 = 0

(2)

0.3N(1) = 0.9317a

(3)

a + ©MO = IO a; Solving Eqs. (1), (2) and (3) yields: FBC = 36.37 lb

N = 85.71 lb

a = 27.60 rad>s2

Kinematics: Applying equation v = v0 + at, we have (a + )

30 = 0 + 27.60t Ans.

t = 1.09 s

Ans: t = 1.09 s 867

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17–78. Two cylinders A and B, having a weight of 10 lb and 5 lb, respectively, are attached to the ends of a cord which passes over a 3-lb pulley (disk). If the cylinders are released from rest, determine their speed in t = 0.5 s. The cord does not slip on the pulley. Neglect the mass of the cord. Suggestion: Analyze the “system” consisting of both the cylinders and the pulley.

0.75 ft O

B

SOLUTION Equation of Motion: The mass moment of inertia of the pulley (disk) about point O 1 1 3 is given by IO = mr2 = a b A 0.752 B = 0.02620 slug # ft2. Here, a = ar or 2 2 32.2 a a . Applying Eq. 17–16, we have = a = r 0.75 a + ©MO = IO a;

5(0.75) - 10(0.75) = - 0.02620 a - ca

A

a b 0.75

5 10 b a d (0.75) - c a ba d(0.75) 32.2 32.2

a = 9.758 ft>s2 Kinematic: Applying equation y = y0 + at, we have Ans.

y = 0 + 9.758 (0.5) = 4.88 ft s

Ans: v = 4.88 ft>s 868

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17–79. The two blocks A and B have a mass of 5 kg and 10 kg, respectively. If the pulley can be treated as a disk of mass 3 kg and radius 0.15 m, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley.

r O

SOLUTION

A

Kinematics: Since the pulley rotates about a fixed axis passes through point O, its angular acceleration is a =

B

a a = 6.6667a = r 0.15

The mass moment of inertia of the pulley about point O is 1 1 Io = Mr2 = (3)(0.152) = 0.03375 kg # m2 2 2 Equation of Motion: Write the moment equation of motion about point O by referring to the free-body and kinetic diagram of the system shown in Fig. a, a + ©Mo = ©(Mk)o;

5(9.81)(0.15) - 10(9.81)(0.15) = -0.03375(6.6667a) - 5a(0.15) - 10a(0.15) a = 2.973 m>s2 = 2.97 m>s2

Ans.

Ans: a = 2.97 m>s2 869

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*17–80. The two blocks A and B have a mass mA and mB , respectively, where mB 7 mA . If the pulley can be treated as a disk of mass M, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. r O

SOLUTION

A

a = ar c + ©MC = © (Mk)C ;

B

1 mB g(r) - mA g(r) = a Mr2 ba + mB r2 a + mA r2 a 2 a =

a =

g(mB - mA) 1 r a M + mB + mA b 2 g(mB - mA) 1 a M + mB + mA b 2

Ans.

Ans: a =

g(mB - mA) 1 2

a M + mB + mA b 870

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17–81. Determine the angular acceleration of the 25-kg diving board and the horizontal and vertical components of reaction at the pin A the instant the man jumps off. Assume that the board is uniform and rigid, and that at the instant he jumps off the spring is compressed a maximum amount of 200 mm, v = 0, and the board is horizontal. Take k = 7 kN>m.

1.5 m

1.5 m

A k

SOLUTION a + a MA = IAa; + c a Ft = m(aG)t ; + ; a Fn = m(aG)n ;

1 1.5(1400 - 245.25) = c (25)(3)2 da 3 1400 - 245.25 - Ay = 25(1.5a) Ax = 0

Solving, Ax = 0

Ans.

A y = 289 N

Ans.

a = 23.1 rad>s2

Ans.

Ans: Ax = 0 Ay = 289 N a = 23.1 rad>s2 871

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17–82. The lightweight turbine consists of a rotor which is powered from a torque applied at its center. At the instant the rotor is horizontal it has an angular velocity of 15 rad/s and a clockwise angular acceleration of 8 rad>s2. Determine the internal normal force, shear force, and moment at a section through A. Assume the rotor is a 50-m-long slender rod, having a mass of 3 kg/m.

10 m A 25 m

SOLUTION + ©F = m(a ) ; ; n G n + T ©Ft = m(aG)t ;

NA = 45(15)2 (17.5) = 177kN

Ans.

VA + 45(9.81) = 45(8)(17.5) Ans.

VA = 5.86 kN c + ©MA = ©(Mk)A ;

MA + 45(9.81)(7.5) = c

1 (45)(15)2 d(8) + [45(8)(17.5)](7.5) 12

MA = 50.7 kN # m

Ans.

Ans: NA = 177 kN VA = 5.86 kN MA = 50.7 kN # m 872

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17–83. The two-bar assembly is released from rest in the position shown. Determine the initial bending moment at the fixed joint B. Each bar has a mass m and length l.

A

l

B

l

SOLUTION

C

Assembly: IA =

1 2 1 l (m)(l)2 + m(l2 + ( )2) ml + 3 12 2

= 1.667 ml2 c + ©MA = IA a;

l mg( ) + mg(l) = (1.667ml2)a 2 a =

0.9 g l

Segment BC: c + ©MB = ©(Mk)B;

M = c M =

l>2 1 l l ml2 d a + m(l 2 + ( )2)1>2 a( )( ) 2 12 2 l + (2l )2 2

0.9g 1 2 1 ml a = ml2 ( ) 3 3 l Ans.

M = 0.3gml

Ans: M = 0.3gml 873

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*17–84. The armature (slender rod) AB has a mass of 0.2 kg and can pivot about the pin at A. Movement is controlled by the electromagnet E, which exerts a horizontal attractive force on the armature at B of FB = 10.2110-32l-22 N, where l in meters is the gap between the armature and the magnet at any instant. If the armature lies in the horizontal plane, and is originally at rest, determine the speed of the contact at B the instant l = 0.01 m. Originally l = 0.02 m.

l

E

B

150 mm

SOLUTION Equation of Motion: The mass moment of inertia of the armature about point A is 1 given by IA = IG + m r2G = (0.2) A 0.152 B + 0.2 A 0.0752 B = 1.50 A 10-3 B kg # m2 12 Applying Eq. 17–16, we have 0.2(10-3)

a + ©MA = IAa;

l2

A

(0.15) = 1.50 A 10-3 B a 0.02 l2

a =

Kinematic: From the geometry, l = 0.02 - 0.15 u. Then dl = -0.15 du or dv dl v . Also, v = hence dv = . Substitute into equation vdv = adu, du = 0.15 0.15 0.15 we have dl dv v ¢ ≤ = a¢ ≤ 0.15 0.15 0.15 vdv = - 0.15 adl 0.01 m

v

L0

vdv =

L0.02 m

- 0.15

0.02 dl l2 Ans.

v = 0.548 m s

Ans: v = 0.548 m>s 874

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17–85. The bar has a weight per length of w. If it is rotating in the vertical plane at a constant rate v about point O, determine the internal normal force, shear force, and moment as a function of x and u.

O v u

L

SOLUTION

x

a = v2 aL -

x bh z u

Forces: wx 2 x v a L - b u = N u + S au + wx T h g z h

(1)

Moments: x Ia = M - S a b 2 O = M -

1 Sx 2

(2)

Solving (1) and (2), N = wx B

x v2 A L - B + cos u R g 2

Ans. Ans.

V = wx sin u M =

1 wx2 sin u 2

Ans.

Ans: N = wxc

v2 x aL - b + cos u d g 2

V = wx sin u M = 875

1 2 wx sin u 2

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17–86. The 4-kg slender rod is initially supported horizontally by a spring at B and pin at A. Determine the angular acceleration of the rod and the acceleration of the rod’s mass center at the instant the 100-N force is applied.

100 N 1.5 m A

1.5 m B k  20 N/m

Solution 1 (4)(32) + 12 4(1.52) = 12.0 kg # m2. Initially, the beam is at rest, v = 0. Thus, (aG)n = v2r = 0. Also,

Equation of Motion. The mass moment of inertia of the rod about A is IA =

(aG)t = arG = a(1.5). The force developed in the spring before the application of the 4(9.81) N 100 N force is Fsp = = 19.62 N. Referring to the FBD of the rod, Fig. a, 2 a+ MA = IAa;  19.62(3) - 100(1.5) - 4(9.81)(1.5) = 12.0( -a) Ans.

a = 12.5 rad>sb

Then

(aG)t = 12.5(1.5) = 18.75 m>s2 T

Since (aG)n = 0. Then

aG = (aG)t = 18.75 m>s2 T 

Ans.

Ans: a = 12.5 rad>sb aG = 18.75 m>s2 T 876

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17–87. The 100-kg pendulum has a center of mass at G and a radius of gyration about G of kG = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant u = 90° when the pendulum is rotating at v = 8 rad>s. Neglect the weight of the beam and the support.

C

u

0.75 m v

1m G

SOLUTION Equations of Motion: Since the pendulum rotates about the fixed axis passing through point C, (aG)t = arG = a(0.75) and (aG)n = v2rG = 82(0.75) = 48 m>s2. Here, the mass moment of inertia of the pendulum about this axis is IC = 100(0.25)2 + 100(0.752) = 62.5 kg # m2. Writing the moment equation of motion about point C and referring to the free-body diagram of the pendulum, Fig. a, we have a + ©MC = ICa;

0 = 62.5a

A

B 0.6 m

0.6 m

a = 0

Using this result to write the force equations of motion along the n and t axes, + ©F = m(a ) ; ; t G t

- Ct = 100[0(0.75)]

+ c ©Fn = m(aG)n;

Cn - 100(9.81) = 100(48)

Ct = 0 Cn = 5781 N

Equilibrium: Writing the moment equation of equilibrium about point A and using the free-body diagram of the beam in Fig. b, we have + ©MA = 0;

NB (1.2) - 5781(0.6) = 0

NB = 2890.5 N = 2.89 kN

Ans.

Using this result to write the force equations of equilibrium along the x and y axes, we have + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

Ay + 2890.5 - 5781 = 0

Ans. Ay = 2890.5 N = 2.89 kN

Ans.

Ans: NB = 2.89 kN Ax = 0 Ay = 2.89 kN 877

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*17–88. The 100-kg pendulum has a center of mass at G and a radius of gyration about G of kG = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant u = 0° when the pendulum is rotating at v = 4 rad>s. Neglect the weight of the beam and the support.

C

u

0.75 m v

1m G

SOLUTION Equations of Motion: Since the pendulum rotates about the fixed axis passing through point C, (aG)t = arG = a(0.75) and (aG)n = v2rG = 4 2(0.75) = 12 m>s2. Here, the mass moment of inertia of the pendulum about this axis is IC = 100(0.252) + 100(0.75)2 = 62.5 kg # m2. Writing the moment equation of motion about point C and referring to the free-body diagram shown in Fig. a, a + ©MC = ICa;

A

B 0.6 m

0.6 m

a = 11.772 rad>s2

- 100(9.81)(0.75) = -62.5a

Using this result to write the force equations of motion along the n and t axes, we have + c ©Ft = m(a G)t ; + ©F = m(a ) ; ; n G n

Ct - 100(9.81) = - 100[11.772(0.75)]

Ct = 98.1 N

Cn = 100(12)

Cn = 1200 N

Equilibrium: Writing the moment equation of equilibrium about point A and using the free-body diagram of the beam in Fig. b, + ©MA = 0;

NB (1.2) - 98.1(0.6) - 1200(1) = 0

NB = 1049.05 N = 1.05 kN Ans.

Using this result to write the force equations of equilibrium along the x and y axes, we have + ©F = 0; : x

1200 - Ax = 0

+ c ©Fy = 0;

1049.05 - 98.1 - A y = 0

Ax = 1200 N = 1.20 kN A y = 950.95 N = 951 N

Ans. Ans.

Ans: NB = 1.05 kN Ax = 1.20 kN Ay = 951 N 878

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17–89. The “Catherine wheel” is a firework that consists of a coiled tube of powder which is pinned at its center. If the powder burns at a constant rate of 20 g>s such as that the exhaust gases always exert a force having a constant magnitude of 0.3 N, directed tangent to the wheel, determine the angular velocity of the wheel when 75% of the mass is burned off. Initially, the wheel is at rest and has a mass of 100 g and a radius of r = 75 mm. For the calculation, consider the wheel to always be a thin disk.

U &

 1

SOLUTION Mass of wheel when 75% of the powder is burned = 0.025 kg Time to burn off 75 % =

0.075 kg = 3.75 s 0.02 kg>s

m(t) = 0.1 - 0.02 t Mass of disk per unit area is r0 =

0.1 kg m = 5.6588 kg>m2 = A p(0.075 m)2

At any time t, 5.6588 = r(t) =

0.1 - 0.02t pr2

0.1 - 0.02t A p(5.6588) + ©MC = ICa;

0.3r = a =

1 2 mr a 2 0.6 = mr

0.6 0.1 - 0.02t A p(5.6588)

(0.1 - 0.02t)

a = 0.6 A 2p(5.6588) B [0.1 - 0.02t]- 2 3

3

a = 2.530[0.1 - 0.02t]- 2 dv = a dt t

v

L0

dv = 2.530

L0

3

[0.1 - 0.02t]- 2 dt

v = 253 C (0.1 - 0.02t)- 2 - 3.162 D 1

For t = 3.75 s, Ans.

v = 800 rad s

Ans: v = 800 rad>s 879

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17–90. If the disk in Fig. 17–21a rolls without slipping, show that when moments are summed about the instantaneous center of zero velocity, IC, it is possible to use the moment equation ©MIC = IICa, where IIC represents the moment of inertia of the disk calculated about the instantaneous axis of zero velocity.

SOLUTION c + ©MlC = ©(MK)lC;

©MlC = IGa + (maG)r

Since there is no slipping, aG = ar Thus, ©MIC = A IG + mr2 B a By the parallel–axis thoerem, the term in parenthesis represents IIC. Thus, Q.E.D.

©MIC = IICa

Ans: ΣMIC = I IC a 880

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17–91. The 20-kg punching bag has a radius of gyration about its center of mass G of kG = 0.4 m. If it is initially at rest and is subjected to a horizontal force F = 30 N, determine the initial angular acceleration of the bag and the tension in the supporting cable AB.

A 1m B 0.3 m

SOLUTION + ©F = m(a ) ; : x G x

30 = 20(aG)x

+ c ©Fy = m(aG)y;

T - 196.2 = 20(aG)y

a + ©MG = IGa;

G

0.6 m F

30(0.6) = 20(0.4)2a a = 5.62 rad>s2

Ans.

(aG)x = 1.5 m>s2 aB = aG + aB>G aB i = (aG)y j + (aG)xi - a(0.3)i (+ c)

(aG)y = 0

Thus, Ans.

T = 196 N

Ans: a = 5.62 rad>s2 T = 196 N 881

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*17–92. FA  100 lb

The uniform 150-lb beam is initially at rest when the forces are applied to the cables. Determine the magnitude of the acceleration of the mass center and the angular acceleration of the beam at this instant.

FB  200 lb

A

B

60

12 ft

SOLUTION Equations of Motion: The mass moment of inertia of the beam about its mass center 1 1 150 ml2 = a b A 122 B = 55.90 slug # ft2. is IG = 12 12 32.2 + ©F = m(a ) ; : x G x

200 cos 60° =

150 (a ) 32.2 G x

(aG)x = 21.47 ft>s2 + c ©Fy = m(aG)y;

100 + 200 sin 60° - 150 =

150 (a ) 32.2 G y

(aG)y = 26.45 ft>s2 + ©MG = IGa;

200 sin 60°(6) - 100(6) = 55.90a a = 7.857 rad>s2 = 7.86 rad>s2

Ans.

Thus, the magnitude of aG is aG = 3(aG)x2 + (aG)y2 = 321.472 + 26.452 = 34.1 ft>s2

Ans.

Ans: a = 7.86 rad>s2 aG = 34.1 ft>s2 882

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17–93. The slender 12-kg bar has a clockwise angular velocity of v = 2 rad>s when it is in the position shown. Determine its angular acceleration and the normal reactions of the smooth surface A and B at this instant.

B

3m

Solution Equations of Motion. The mass moment of inertia of the rod about its center of 1 1 gravity G is IG = ml 2 = (12)(32) = 9.00 kg # m2. Referring to the FBD and 12 12 kinetic diagram of the rod, Fig. a + ΣFx = m(aG)x;  NB = 12(aG)x(1) d + c ΣFy = m(aG)y;        NA - 12(9.81) = - 12(aG)y(2) a+ ΣMO = (Mk)O;  - 12(9.81)(1.5 cos 60°) = -12(aG)x(1.5 sin 60°) - 12(aG)y(1.5 cos 60°) - 9.00a



23(aG)x + (aG)y + a = 9.81(3)

Kinematics. Applying the relative acceleration equation relating aG and aB by referring to Fig. b,

aG = aB + A * rG>B - v2 rG>B



- (aG)xi - (aG)yj = - aB j + ( - ak) * ( - 1.5 cos 60°i - 1.5 sin 60°j)



- 22( - 1.5 cos 60°i - 1.5 sin 60°j) - (aG)x i - (aG)y j = (3 - 0.7523a)i + (0.75a - aB + 323)j

883

60 A

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17–93. Continued

Equating i and j components,

- (aG)x = 3 - 0.7523a

(4)



-(aG)y = 0.75a - aB + 323

(5)

Also, relate aB and aA,

aA = aB + a * rA>B - v2 rA>B



- aAi = -aB j + ( - ak) * ( - 3 cos 60°i - 3 sin 60°j) - 2 2( -3 cos 60°i - 3 sin 60°j)



- aAi = (6 - 1.523a)i + (1.5a - aB + 623)j

Equating j components,

(6)

0 = 1.5a - aB + 623;   aB = 1.5a + 623

Substituting Eq. (6) into (5)

(7)

(aG)y = 0.75a + 323

Substituting Eq. (4) and (7) into (3)

23 1 0.7523a - 3 2 + 0.75a + 323 + a = 9.81

a = 2.4525 rad>s2 = 2.45b rad>s2

Ans.

Substituting this result into Eqs. (4) and (7)

- (aG)x = 3 - (0.7523)(2.4525);  (aG)x = 0.1859 m>s2



 (aG)y = 0.75(2.4525) + 323;    (aG)y = 7.0355 m>s2

Substituting these results into Eqs. (1) and (2)

NB = 12(0.1859);  NB = 2.2307 N = 2.23 N

Ans.



NA - 12(9.81) = -12(7.0355);  NA = 33.2937 N = 33.3 N

Ans.

Ans: a = 2.45 rad>s2 b NB = 2.23 N NA = 33.3 N 884

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17–94. The tire has a weight of 30 lb and a radius of gyration of kG = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are ms = 0.2 and mk = 0.15, determine the tire’s angular acceleration as it rolls down the incline. Set u = 12°.

G 1.25 ft

SOLUTION

u

+b©Fx = m(aG)x ; +a©Fy

= m(aG)y ;

a + ©MG = IG a;

30 sin 12° - F = a

30 ba 32.2 G

N - 30 cos 12° = 0 F(1.25) = c a

30 b (0.6)2 d a 32.2

Assume the wheel does not slip. a G = (1.25)a Solving: F = 1.17 lb N = 29.34 lb a G = 5.44 ft>s2 a = 4.35 rad>s2

Ans. OK

Fmax = 0.2(29.34) = 5.87 lb 7 1.17 lb

Ans: a = 4.32 rad>s2 885

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17–95. The tire has a weight of 30 lb and a radius of gyration of kG = 0.6 ft. If the coefficients of static and kinetic friction between the wheel and the plane are ms = 0.2 and mk = 0.15, determine the maximum angle u of the inclined plane so that the tire rolls without slipping.

G 1.25 ft

SOLUTION

u

Since wheel is on the verge of slipping: +b©Fx = m(aG)x ;

30 sin u - 0.2N = a

+a©Fy = m(a G)y ;

N - 30 cos u = 0

a + ©MC = IG a;

0.2N(1.25) = c a

30 b (1.25a) 32.2

(1) (2)

30 b (0.6)2 d a 32.2

(3)

Substituting Eqs.(2) and (3) into Eq. (1), 30 sin u - 6 cos u = 26.042 cos u 30 sin u = 32.042 cos u tan u = 1.068 Ans.

u = 46.9°

Ans: u = 46.9° 886

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*17–96. The spool has a mass of 100 kg and a radius of gyration of kG = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 50 N.

P 250 mm

G

400 mm

A

SOLUTION + ©F = m(a ) ; : x G x

50 + FA = 100aG

+ c ©Fy = m(aG)y ;

NA - 100(9.81) = 0

c + ©MG = IG a;

50(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4a a = 1.30 rad>s2 a G = 0.520 m>s2

Ans. NA = 981 N

FA = 2.00 N OK

Since (FA)max = 0.2(981) = 196.2 N 7 2.00 N

Ans: a = 1.30 rad>s2 887

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17–97. Solve Prob. 17–96 if the cord and force P = 50 N are directed vertically upwards.

P 250 mm

G

400 mm

A

SOLUTION + ©F = m(a )x; F = 100a : x G A G + c ©Fy = m(aG)y;

NA + 50 - 100(9.81) = 0

c + ©MG = IG a;

50(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4 a a = 0.500 rad>s2 aG = 0.2 m>s2

Ans. NA = 931 N

FA = 20 N OK

Since (FA)max = 0.2(931) = 186.2 N 7 20 N

Ans: a = 0.500 rad>s2 888

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17–98. The spool has a mass of 100 kg and a radius of gyration kG = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 600 N.

P 250 mm

G

400 mm

A

SOLUTION + ©F = m(a ) ; : x G x

600 + FA = 100aG

+ c ©Fy = m(aG)y;

NA - 100(9.81) = 0

c + ©MG = IG a;

600(0.25) - FA(0.4) = [100(0.3)2]a

Assume no slipping: aG = 0.4a a = 15.6 rad>s2 a G = 6.24 m>s2

Ans. NA = 981 N

FA = 24.0 N OK

Since (FA)max = 0.2(981) = 196.2 N 7 24.0 N

Ans: a = 15.6 rad>s2 889

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17–99. The 12-kg uniform bar is supported by a roller at A. If a horizontal force of F = 80 N is applied to the roller, determine the acceleration of the center of the roller at the instant the force is applied. Neglect the weight and the size of the roller.

A

F  80 N

2m

Solution Equations of Motion. The mass moment of inertia of the bar about its center of gravity G 1 1 is IG = ml 2 = (12) ( 22 ) = 4.00 kg # m2. Referring to the FBD and kinetic diagram 12 12 of the bar, Fig. a + S ΣFx = m(aG)x;  80 = 12(aG)x  (aG)x = 6.6667 m>s2 S a + ΣMA = (mk)A;  0 = 12(6.6667)(1) - 4.00 a  a = 20.0 rad>s2 b Kinematic. Since the bar is initially at rest, v = 0. Applying the relative acceleration equation by referring to Fig. b, aG = aA + A * rG>A - v2 rG>A 6.6667i + (aG)y j = aAi + ( - 20.0k) * ( -j) - 0 6.6667i + (aG)y j = (aA - 20)i Equating i and j components, 6.6667 = aA - 20;  aA = 26.67 m>s2 = 26.7 m>s2 S 

Ans.

    (aG)y = 0

Ans: aA = 26.7 m>s2 S 890

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*17–100. A force of F = 10 N is applied to the 10-kg ring as shown. If slipping does not occur, determine the ring’s initial angular acceleration, and the acceleration of its mass center, G. Neglect the thickness of the ring.

F

A G

45

30

0.4 m

Solution

C

Equations of Motion. The mass moment of inertia of the ring about its center of gravity G is IG = mr 2 = 10 ( 0.42 ) = 1.60 kg # m2. Referring to the FBD and kinetic diagram of the ring, Fig. a, a + ΣMC = (mk)C;  (10 sin 45°)(0.4 cos 30°) - (10 cos 45°)[0.4(1 + sin 30°)] = - (10aG)(0.4) - 1.60a 4aG + 1.60a = 1.7932(1) Kinematics. Since the ring rolls without slipping, aG = ar = a(0.4)(2) Solving Eqs. (1) and (2) a = 0.5604 rad>s2 = 0.560 rad>s2 b

Ans.

aG = 0.2241 m>s2 = 0.224 m>s2 S 

Ans.

Ans: a = 0.560 rad>s2 b aG = 0.224 m>s2 S 891

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17–101. If the coefficient of static friction at C is μs = 0.3, determine the largest force F that can be applied to the 5-kg ring, without causing it to slip. Neglect the thickness of the ring.

F

A G

45

30

0.4 m

Solution

C

Equations of Motion: The mass moment of inertia of the ring about its center of gravity G is IG = mr 2 = 10 ( 0.42 ) = 1.60 kg # m2. Here, it is required that the ring is on the verge of slipping at C, Ff = ms N = 0.3 N. Referring to the FBD and kinetic diagram of the ring, Fig. a + c ΣFy = m(aG)y;  F sin 45° + N - 10(9.81) = 10(0)(1) + ΣF = m(a ) ;  F cos 45° - 0.3 N = 10a (2)  S G x G x  a + ΣMG = IGa;  F sin 15°(0.4) - 0.3 N(0.4) = - 1.60a(3) Kinematics. Since the ring rolls without slipping, aG = ar = a(0.4)(4) Solving Eqs. (1) to (4), Ans.

F = 42.34 N = 42.3 N N = 68.16 N  a = 2.373 rad>s2 b  aG = 0.9490 m>s2 S

Ans: F = 42.3 N 892

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17–102. The 25-lb slender rod has a length of 6 ft. Using a collar of negligible mass, its end A is confined to move along the smooth circular bar of radius 3 22 ft. End B rests on the floor, for which the coefficient of kinetic friction is mB = 0.4. If the bar is released from rest when u = 30°, determine the angular acceleration of the bar at this instant.

A

6 ft 3 2 ft B

Solution + ΣF = m(a ) ;  - 0.4 N + N cos 45° = 25 (a )  d B A x G x 32.2 G x - 25 + c ΣFy = m(aG)y;  NB - 25 - NA sin 45° = (a )  32.2 G y c + ΣMG = IG a;  NB(3 cos 30°) - 0.4 NB (3 sin 30°) + NA sin 15°(3) =

  aB = aA + aB>A

u

(1) (2)

1 25 a b(6)2a 12 32.2

(3)

  aB = aA + 6a   d   c45º h 30º ( + c )  0 = - aA sin 45° + 6a(cos 30°) aA = 7.34847a aG = aA + aG>A (aG)x + (aG)y = 7.34847a + 3a d    T    c45º  h 30º +) (d

(aG)x = - 5.196a + 1.5a = - 3.696a

(4)

( + T)

(aG)y = 5.196a - 2.598a = 2.598a

(5)

Solving Eqs. (1)–(5) yields: NB = 9.01 lb NA = - 11.2 lb a = 4.01 rad>s2

Ans.

Ans: a = 4.01 rad>s2 893

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17–103. aA

The 15-lb circular plate is suspended from a pin at A. If the pin is connected to a track which is given an acceleration aA = 5 ft>s2, determine the horizontal and vertical components of reaction at A and the angular acceleration of the plate. The plate is originally at rest.

A G 2 ft

Solution + ΣF = m(a ) ; S x G x + c ΣFy = m(aG)y; c + ΣMG = IGa; aG = aA + aG>A

15 (a ) 32.2 G x 15 Ay - 15 = (a ) 32.2 G y

Ax =

1 15 b(2)2d a Ax(2) = c a 2 32.2

aG = 5i - 2ai

(+ c)

(aG)y = 0

+ ) (S

(aG)x = 5 - 2a

Thus, Ans.

Ay = 15.0 lb

Ans.

Ax = 0.776 lb 2

Ans.

a = 1.67 rad>s  aG = (aG)x = 1.67 ft>s2

Ans: Ay = 15.0 lb Ax = 0.776 lb a = 1.67 rad>s2 894

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*17–104. P

If P = 30 lb, determine the angular acceleration of the 50-lb roller. Assume the roller to be a uniform cylinder and that no slipping occurs.

1.5 ft 30

SOLUTION Equations of Motion: The mass moment of inertia of the roller about its mass center 1 1 50 is IG = mr2 = a b A 1.52 B = 1.7469 slug # ft2. We have 2 2 32.2 + ©F = m(a ) ; : x G x

30 cos 30° - Ff =

+ c ©Fy = m(aG)y;

N - 50 - 30 sin 30° = 0

+ ©MG = IGa;

Ff(1.5) = 1.7469a

50 a 32.2 G

(1) N = 65 lb (2)

Since the roller rolls without slipping, (3)

aG = ar = a(1.5) Solving Eqs. (1) through (3) yields a = 7.436 rad>s2 = 7.44 rad>s2 Ff = 8.660 lb

Ans.

aG = 11.15 ft>s2

Ans: a = 7.44 rad>s2 895

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17–105. P

If the coefficient of static friction between the 50-lb roller and the ground is ms = 0.25, determine the maximum force P that can be applied to the handle, so that roller rolls on the ground without slipping. Also, find the angular acceleration of the roller. Assume the roller to be a uniform cylinder.

1.5 ft 30

SOLUTION Equations of Motion: The mass moment of inertia of the roller about its mass center 1 1 50 b A 1.52 B = 1.7469 slug # ft2. We have is IG = mr2 = a 2 2 32.2 + ©F = m(a ) ; : x G x

P cos 30° - Ff =

+ c ©Fy = m(aG)y;

N - P sin 30° - 50 = 0

(2)

+ ©MG = IGa;

Ff(1.5) = 1.7469a

(3)

50 a 32.2 G

(1)

Since the roller is required to be on the verge of slipping, aG = ar = a(1.5)

(4)

Ff = msN = 0.25N

(5)

Solving Eqs. (1) through (5) yields a = 18.93 rad>s2 = 18.9 rad>s2 N = 88.18 lb

aG = 28.39 ft>s2

P = 76.37 lb = 76.4 lb

Ans.

Ff = 22.05 lb

Ans: a = 18.9 rad>s2 P = 76.4 lb 896

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17–106. The uniform bar of mass m and length L is balanced in the vertical position when the horizontal force P is applied to the roller at A. Determine the bar’s initial angular acceleration and the acceleration of its top point B.

B

L

Solution + ΣFx = m(aG)x;   P = maG d c + ΣMG = IGa; P =

1 mLa 6

a =

6P  mL

aG =

P m

P

A

L 1 P a b = a mL2 ba 2 12

Ans.

aB = aG + aB>G - aB i =

-P L i + ai m 2

+ ) (d

aB =

aB = -

=

P La m 2 P L 6P - a b m 2 mL

2P 2P =  m m

Ans.

Ans: 6P mL 2P aB = m

a =

897

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17–107. Solve Prob. 17–106 if the roller is removed and the coefficient of kinetic friction at the ground is μk.

B

L

Solution + ΣFx = m(aG)x; d

P - mkNA = maG

P

A

L 1 = a mL2 ba 2 12 NA - mg = 0

( P - mkNA )

c + ΣMG = IGa; + c ΣFy = m(aG)y; Solving,

NA = mg



aG =



a =

L a 6 6(P - mkmg) mL

Ans.



aB = aG + aB>G +) a = (S B aB = aB =

L L a + a 6 2

La 3 2(P - mkmg) m

Ans.



Ans: a = aB = 898

6(P - mk mg) mL 2(P - mk mg) m

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*17–108. The semicircular disk having a mass of 10 kg is rotating at v = 4 rad>s at the instant u = 60°. If the coefficient of static friction at A is ms = 0.5, determine if the disk slips at this instant.

v 0.4 m

O

G

4 (0.4) m 3p

u A

SOLUTION Equations of Motion:The mass moment of inertia of the semicircular disk about its center 1 of mass is given by IG = (10) A 0.4 2 B - 10 (0.16982) = 0.5118 kg # m2. From the 2 geometry, rG>A = 20.16982 + 0.4 2 - 2(0.1698) (0.4) cos 60° = 0.3477 m Also, using sin u sin 60° law of sines, , u = 25.01°. Applying Eq. 17–16, we have = 0.1698 0.3477 a + ©MA = ©(Mk)A ;

10(9.81)(0.1698 sin 60°) = 0.5118a + 10(a G)x cos 25.01°(0.3477) + 10(a G)y sin 25.01°(0.3477)

+ ©F = m(a ) ; ; x G x + c Fy = m(a G)y ;

(1) (2)

Ff = 10(aG)x

(3)

N - 10(9.81) = - 10(aG)y

Kinematics:Assume that the semicircular disk does not slip at A, then (a A)x = 0. Here, rG>A = {- 0.3477 sin 25.01°i + 0.3477 cos 25.01°j} m = {- 0.1470i + 0.3151j} m. Applying Eq. 16–18, we have a G = a A + a * rG>A - v2rG>A -(aG)x i - (aG)y j = 6.40j + ak * ( -0.1470i + 0.3151j) - 42( -0.1470i + 0.3151j) -(a G)x i - (aG)y j = (2.3523 - 0.3151 a) i + (1.3581 - 0.1470a)j Equating i and j components, we have (a G)x = 0.3151a - 2.3523

(4)

(a G)y = 0.1470a - 1.3581

(5)

Solving Eqs. (1), (2), (3), (4), and (5) yields: a = 13.85 rad>s2

(aG)x = 2.012 m>s2 Ff = 20.12 N

(aG)y = 0.6779 m>s2

N = 91.32 N

Since Ff 6 (Ff)max = msN = 0.5(91.32) = 45.66 N, then the semicircular disk does not slip.

Ans.

Ans: Since Ff 6 (Ff )max = ms N = 0.5(91.32) = 45.66 N, then the semicircular disk does not slip. 899

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17–109. The 500-kg concrete culvert has a mean radius of 0.5 m. If the truck has an acceleration of 3 m>s2, determine the culvert’s angular acceleration. Assume that the culvert does not slip on the truck bed, and neglect its thickness.

3 m/s2

4m 0.5m

SOLUTION Equations of Motion: The mass moment of inertia of the culvert about its mass center is IG = mr2 = 500 A 0.52 B = 125 kg # m2. Writing the moment equation of motion about point A using Fig. a, a + ©MA = ©(Mk)A ;

(1)

0 = 125a - 500a G(0.5)

Kinematics: Since the culvert does not slip at A, (aA)t = 3 m>s2. Applying the relative acceleration equation and referring to Fig. b, a G = a A + a * rG>A - v2rG>A a Gi = 3i + (a A)n j + (ak * 0.5j) - v2(0.5j) aGi = (3 - 0.5a)i + C (aA)n - 0.5v2 D j Equating the i components, (2)

a G = 3 - 0.5a Solving Eqs. (1) and (2) yields aG = 1.5 m>s2 : a = 3 rad>s2

Ans.

Ans: a = 3 rad>s2 900

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17–110. The 15-lb disk rests on the 5-lb plate. A cord is wrapped around the periphery of the disk and attached to the wall at B. If a torque M = 40 lb # ft is applied to the disk, determine the angular acceleration of the disk and the time needed for the end C of the plate to travel 3 ft and strike the wall. Assume the disk does not slip on the plate and the plate rests on the surface at D having a coefficient of kinetic friction of μk = 0.2. Neglect the mass of the cord.

B

A M  40 lb  ft 1.25 ft D

C

3 ft

Solution Disk: + S ΣFx = m(aG)x;

T - FP =

15 a 32.2 G

Plate:

1 15 b(1.25)2 d a - FP(1.25) + 40 - T(1.25) = c a 2 32.2

+ ΣF = m(a ) ; S x G x

FP - 4 =

a+ ΣMG = IGa;

5 a 32.2 P

aP = aG + aP>G +) (S

aP = aG + a(1.25)



aG = a(1.25)

Thus

aP = 2.5a

Solving,

FP = 9.65 lb



aP = 36.367 ft>s2



a = 14.5 rad>s2



T = 18.1 lb

+) (S

s = s0 + v0 t +



3 = 0 + 0 +



t = 0.406 s

Ans.

1 2 at 2 c

1 (36.367)t 2 2 Ans.

Ans: a = 14.5 rad>s2 t = 0.406 s 901

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17–111. The semicircular disk having a mass of 10 kg is rotating at v = 4 rad>s at the instant u = 60º. If the coefficient of static friction at A is mV = 0.5, determine if the disk slips at this instant.

v 0.4 m

O

G

4 (0.4) m 3p

u A

SOLUTION For roll A. T(0.09) = 12 (8)(0.09)2 aA

c + ©MA = IA a;

(1)

For roll B a + ©MO = ©(Mk)O;

8(9.81)(0.09) = 12 (8)(0.09)2 aB + 8aB (0.09)

(2)

+ c ©Fy = m(aG)y ;

T - 8(9.81) = - 8aB

(3)

Kinematics: aB = aO + (aB>O)t + (aB>O)n c aBd = c aOd + caB (0.09) d + [0] T T T aB = aO + 0.09aB A+TB

(4)

also,

A+TB

(5)

aO = aA (0.09)

Solving Eqs. (1)–(5) yields: aA = 43.6 rad>s2

Ans.

aB = 43.6 rad>s2

Ans.

T = 15.7 N

Ans.

aB = 7.85 m>s2

aO = 3.92 m>s2

Ans: The disk does not slip. 902

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*17–112. v

The circular concrete culvert rolls with an angular velocity of v = 0.5 rad>s when the man is at the position shown. At this instant the center of gravity of the culvert and the man is located at point G, and the radius of gyration about G is kG = 3.5 ft. Determine the angular acceleration of the culvert. The combined weight of the culvert and the man is 500 lb.Assume that the culvert rolls without slipping, and the man does not move within the culvert.

4 ft O

G

0.5 ft

SOLUTIONS Equations of Motion: The mass moment of inertia of the system about its mass 500 (3.52) = 190.22 slug # ft2. Writing the moment equation of center is IG = mkG2= 32.2 motion about point A, Fig. a, + ©MA = ©(Mk)A ; -500(0.5) = -

500 500 (a ) (4) (a ) (0.5) - 190.22a (1) 32.2 G x 32.2 G y

Kinematics: Since the culvert rolls without slipping, a0 = ar = a(4) : Applying the relative acceleration equation and referrring to Fig. b, aG = aO + a * rG>O - v2rG>A (aG)xi - (aG)y j = 4ai + (- ak) * (0.5i) - (0.52)(0.5i) (aG)xi - (aG)y j = (4a - 0.125)i - 0.5aj Equation the i and j components, (aG)x = 4a - 0.125

(2)

(aG)y = 0.5a

(3)

Subtituting Eqs. (2) and (3) into Eq. (1), - 500(0.5) = -

500 500 (4a - 0.125)(4) (0.5a)(0.5) - 190.22a 32.2 32.2

a = 0.582 rad>s2

Ans.

Ans: a = 0.582 rad>s2 903

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17–113. v0

The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the initial angular acceleration of the disk and the acceleration of its mass center. The coefficient of kinetic friction between the disk and the floor is mk.

r

SOLUTION Equations of Motion. Since the disk slips, the frictional force is Ff = mkN. The mass 1 moment of inertia of the disk about its mass center is IG = m r2. We have 2 + c ©Fy = m(aG)y;

N - mg = 0

N = mg

+ ©F = m(a ) ; ; x G x

mk(mg) = maG

aG = mkg ;

Ans.

1 -mk(mg)r = a m r2 b a 2

a =

2mkg b r

Ans.

a+ ©MG = IGa;

Ans: aG = mkg d 2mkg b a = r 904

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17–114. v0

The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the time before it starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is mk.

r

SOLUTION Equations of Motion: Since the disk slips, the frictional force is Ff = mkN. The mass 1 moment of inertia of the disk about its mass center is IG = m r2. 2 + c ©Fy = m(aG)y; N - mg = 0 N = mg + ©F = m(a ) ; ; x G x

mk(mg) = maG

+ ©MG = IGa;

1 -mk(mg)r = - a m r2 b a 2

aG = mkg a =

2mkg r

Kinematics: At the instant when the disk rolls without slipping, vG = vr. Thus, + B A;

vG = (vG)0 + aGt vr = 0 + mkgt t =

vr mkg

(1)

and v = v0 + a t (a +)

v = v0 + a -

2mkg bt r

(2)

Solving Eqs. (1) and (2) yields v =

1 v 3 0

t =

v0r 3mkg

Ans.

Ans: 1 v0 3 v0r t = 3mkg

v =

905

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17–115. A cord is wrapped around each of the two 10-kg disks. If they are released from rest, determine the angular acceleration of each disk and the tension in the cord C. Neglect the mass of the cord.

D

90 mm

A

SOLUTION

C

For A: c + ©MA = IA aA;

1 T(0.09) = c (10)(0.09)2 d aA 2

(1)

1 T(0.09) = c (10)(0.09)2 d aB 2

(2)

90 mm

B

For B: a + ©MB = IBaB; + T ©Fy = m(aB)y;

(3)

10(9.81) - T = 10aB

aB = aP + (aB>P)t + (aB>P)n (4)

( + T) aB = 0.09aA + 0.09aB + 0 Solving, aB = 7.85 m>s2 aA = 43.6 rad>s2

Ans.

aB = 43.6 rad>s2

Ans. Ans.

T = 19.6 N Ay = 10(9.81) + 19.62 = 118 N

Ans: aA = 43.6 rad>s2b aB = 43.6 rad>s2d T = 19.6 N 906

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*17–116. The disk of mass m and radius r rolls without slipping on the circular path. Determine the normal force which the path exerts on the disk and the disk’s angular acceleration if at the instant shown the disk has an angular velocity of V. R

SOLUTION r

Equation of Motion: The mass moment of inertia of the disk about its center of 1 mass is given by IG = mr2. Applying Eq. 17–16, we have 2 a + ©MA = ©(Mk)A ; ©Fn = m(aG)n ;

1 mg sin u(r) = a mr2 b a + m(aG)t (r) 2

[1] [2]

N - mg cos u = m(aG)n

Kinematics: Since the semicircular disk does not slip at A, then yG = vr and (aG)t = ar. Substitute (aG)t = ar into Eq. [1] yields 1 mg sin u(r) = a mr2 b a + m(ar)(r) 2 a =

2g sin u 3r

Ans.

Also, the center of the mass for the disk moves around a circular path having a y2G v2r2 radius of r = R - r. Thus, (aG)n = . Substitute into Eq. [2] yields = r R - r N - mg cos u = ma N = m

v2r2 b R - r

v2r2 + g cos u R - r

Ans.

Ans:

2g sin u 3r v2r 2 N = ma + g cos u b R - r

a =

907

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17–117. The uniform beam has a weight W. If it is originally at rest while being supported at A and B by cables, determine the tension in cable A if cable B suddenly fails. Assume the beam is a slender rod.

A

B

SOLUTION + c ©Fy = m(aG)y; c + ©MA = IAa;

TA - W = Wa

L –– 2

L –– 4

1 W L L W L b = c a b L2 d a + a b aa b g 4 4 12 g 4 1 =

Since aG = a a

L –– 4

W a g G

L 1 L a + ba g 4 3

L b. 4

a =

12 g a b 7 L

TA = W TA =

g W 12 L W L (a) a b = W a ba ba b g g 7 4 L 4

4 W 7

Ans.

Also, + c ©Fy = m(aG)y ; c + ©MG = IG a;

Since a G =

TA - W = TA a

W a g G

L 1 W b = c a b L2 d a 4 12 g

L a 4 TA =

1 W a b La 3 g

W L 1 W a b La - W = - a b a g 4 3 g a =

12 g a b 7 L

TA =

g 12 1 W a bLa b a b 3 g 7 L

TA =

4 W 7

Ans.

Ans: TA = 908

4 W 7

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17–118. The 500-lb beam is supported at A and B when it is subjected to a force of 1000 lb as shown. If the pin support at A suddenly fails, determine the beam’s initial angular acceleration and the force of the roller support on the beam. For the calculation, assume that the beam is a slender rod so that its thickness can be neglected.

1000 lb 5

3

4

B 8 ft

A 2 ft

SOLUTION + ; a Fx = m(aG)x ;

500 4 1000 a b = (a ) 5 32.2 G x

+ T a Fy = m(aG)y ;

3 500 1000 a b + 500 - By = (a ) 5 32.2 G y

a + a MB = a (Mk)B;

3 1 500 500 500(3) + 1000 a b (8) = (a ) (3) + c a b (10)2 d a 5 32.2 G y 12 32.2

aB = aG + aB>G - aBi = - (aG)x i - (aG)yj + a(3)j ( + T)

(aG)y = a(3)

a = 23.4 rad>s2

Ans.

By = 9.62 lb

Ans.

By 7 0 means that the beam stays in contact with the roller support.

Ans: a = 23.4 rad>s2 By = 9.62 lb 909

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17–119. The solid ball of radius r and mass m rolls without slipping down the 60° trough. Determine its angular acceleration.

30°

SOLUTION d = r sin 30° =

30°

r 2

©Ma - a = ©(Mk)a - a;

45°

r r 2 2 mg sin 45°a b = c mr2 + m a b da 2 5 2 a =

10g

Ans.

13 22 r

Ans: a =

910

10g 1322 r

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*17–120. By pressing down with the finger at B, a thin ring having a mass m is given an initial velocity v0 and a backspin V 0 when the finger is released. If the coefficient of kinetic friction between the table and the ring is mk, determine the distance the ring travels forward before backspinning stops.

B

ω0 v0 r

SOLUTION + c ©Fy = 0;

A

NA - mg = 0 NA = mg

+ ©F = m(a ) ; : x G x

mk (mg) = m(aG) aG = mk g

a + ©MG = IG a;

mk (mg)r = mr 2 a a =

(c +)

v = v0 + ac t 0 = v0 - a t =

+ B A;

mk g r

mk g bt r

v0 r mk g 1

s = s0 + v0t + ac t2 2

s = 0 + v0 a s =

v0 r mk g

v0 r v20 r2 1 b - a b (mk g) a 2 2 b mk g 2 mk g v0 -

1 v r 2 0

Ans.

Ans: s = a 911

v0 r 1 b a v0 - v0 r b mk g 2

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18–1. At a given instant the body of mass m has an angular velocity V and its mass center has a velocity vG. Show that its kinetic energy can be represented as T = 12IICv2, where IIC is the moment of inertia of the body determined about the instantaneous axis of zero velocity, located a distance rG>IC from the mass center as shown.

IC rG/IC

G vG

SOLUTION T =

1 1 my2G + IG v2 2 2

=

1 1 m(vrG>IC)2 + IG v2 2 2

=

1 A mr2G>IC + IG B v2 2

=

1 I v2 2 IC

where yG = vrG>IC

However mr2G>IC + IG = IIC Q.E.D.

912

V

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18–2. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, and the wheel is rotated until the torque M = 25 N # m is developed, determine the maximum angular velocity of the wheel if it is released from rest.

0.5 m O M

SOLUTION Kinetic Energy and Work: The mass moment of inertia of the wheel about point O is IO = mRr 2 + 2 ¢

1 m l2 ≤ 12 r

= 5(0.52) + 2 c

1 (2)(12) d 12

= 1.5833 kg # m2 Thus, the kinetic energy of the wheel is T =

1 1 I v2 = (1.5833) v2 = 0.79167 v2 2 O 2

Since the wheel is released from rest, T1 = 0. The torque developed is M = ku = 2u. Here, the angle of rotation needed to develop a torque of M = 25 N # m is 2u = 25

u = 12.5 rad

The wheel achieves its maximum angular velocity when the spacing is unwound that M is when the wheel has rotated u = 12.5 rad. Thus, the work done by q is 12.5 rad

UM =

L

Mdu =

2u du

L0 12.5 rad 2

= u †

= 156.25 J 0

Principle of Work and Energy: T1 + © u 1 - 2 = T2 0 + 156.25 = 0.79167 v2 Ans.

v = 14.0 rad/s

Ans: v = 14.0 rad>s 913

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18–3. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, so that the torque on the center of the wheel is M = 12u2 N # m, where u is in radians, determine the maximum angular velocity of the wheel if it is rotated two revolutions and then released from rest.

0.5 m O M

SOLUTION Io = 2 c

1 (2)(1)2 d + 5(0.5)2 = 1.583 12

T1 + ©U1 - 2 = T2 4p

0 +

L0

2u du =

1 (1.583) v2 2

(4p)2 = 0.7917v2 Ans.

v = 14.1 rad/s

Ans: v = 14.1 rad>s 914

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*18–4. A force of P = 60 N is applied to the cable, which causes the 200-kg reel to turn since it is resting on the two rollers A and B of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the rollers and the mass of the cable. Assume the radius of gyration of the reel about its center axis remains constant at kO = 0.6 m.

P O

0.75 m 1m

A

B 0.6 m

Solution Kinetic Energy. Since the reel is at rest initially, T1 = 0. The mass moment of inertia of the reel about its center O is I0 = mk 20 = 200 ( 0.62 ) = 72.0 kg # m2. Thus, T2 =

1 2 1 I v = (72.0)v2 = 36.0 v2 20 2

Work. Referring to the FBD of the reel, Fig. a, only force P does positive work. When the reel rotates 2 revolution, force P displaces S = ur = 2(2p)(0.75) = 3p m. Thus Up = Ps = 60(3p) = 180p J Principle of Work and Energy. T1 + ΣU1 - 2 = T2   0 + 180p = 36.0 v2 Ans.

v = 3.9633 rad>s = 3.96 rad>s

Ans: v = 3.96 rad>s 915

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18–5. A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn since it is resting on the two rollers A and B of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the rollers and the mass of the cable. The radius of gyration of the reel about its center axis is kG = 0.42 m.

P 30°

250 mm G 500 mm A

B 400 mm

SOLUTION T1 + ΣU1 - 2 = T2 0 + 20(2)(2p)(0.250) = v = 2.02 rad>s

1 3 175(0.42)2 4 v2 2

Ans.

Ans: v = 2.02 rad>s 916

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18–6. A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn without slipping on the two rollers A and B of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the cable. Each roller can be considered as an 18-kg cylinder, having a radius of 0.1 m. The radius of gyration of the reel about its center axis is kG = 0.42 m.

P 30°

250 mm G 500 mm A

B 400 mm

SOLUTION System: T1 + ΣU1 - 2 = T2 [0 + 0 + 0] + 20(2)(2p)(0.250) = v = vr (0.1) = v(0.5)

1 1 3 175(0.42)2 4 v2 + 2c (18)(0.1)2 d v2r 2 2

vr = 5v Solving: Ans.

v = 1.78 rad>s

Ans: v = 1.78 rad>s 917

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18–7. The double pulley consists of two parts that are attached to one another. It has a weight of 50 lb and a radius of gyration about its center of kO = 0.6 ft and is turning with an angular velocity of 20 rad>s clockwise. Determine the kinetic energy of the system. Assume that neither cable slips on the pulley.

v

20 rad/s

0.5ft

1 ft O

SOLUTION T =

1 1 1 I v2O + mA v2A + mB v2B 2 O 2 2

T =

1 50 1 20 1 30 a (0.6)2 b(20)2 + a b C (20)(1) D 2 + a b C (20)(0.5) D 2 2 32.2 2 32.2 2 32.2

= 283 ft # lb

B 30 lb A 20 lb

Ans.

Ans: T = 283 ft # lb 918

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*18–8. The double pulley consists of two parts that are attached to one another. It has a weight of 50 lb and a centroidal radius of gyration of kO = 0.6 ft and is turning with an angular velocity of 20 rad> s clockwise. Determine the angular velocity of the pulley at the instant the 20-lb weight moves 2 ft downward.

v  20 rad/s

0.5 ft

1 ft O

SOLUTION Kinetic Energy and Work: Since the pulley rotates about a fixed axis, vA = vrA = v(1) and vB = vrB = v(0.5). The mass moment of inertia of the pulley about point O is IO = mkO 2 = ¢

kinetic energy of the system is T = =

50 ≤ (0.62) = 0.5590 slug # ft2. Thus, the 32.2

B 30 lb A 20 lb

1 1 1 I v2 + mAvA2 + mBvB2 2 O 2 2 1 1 20 1 30 (0.5590)v2 + ¢ ≤ [v(1)]2 + ¢ ≤ [v(0.5)]2 2 2 32.2 2 32.2

= 0.7065v2 Thus, T1 = 0.7065(202) = 282.61 ft # lb. Referring to the FBD of the system shown in Fig. a, we notice that Ox, Oy, and Wp do no work while WA does positive work and WB does negative work. When A moves 2 ft downward, the pulley rotates u =

SA SB = rA rB

SB 2 = 1 0.5 SB = 2(0.5) = 1 ft c Thus, the work of WA and WB are UWA = WA SA = 20(2) = 40 ft # lb UWB = - WB SB = - 30(1) = -30 ft # lb Principle of Work and Energy: T1 + U1 - 2 = T2 282.61 + [40 + ( - 30)] = 0.7065 v2 Ans.

v = 20.4 rad>s

Ans: v = 20.4 rad>s 919

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18–9. The disk, which has a mass of 20 kg, is subjected to the couple moment of M = (2u + 4) N # m, where u is in radians. If it starts from rest, determine its angular velocity when it has made two revolutions.

300 mm

M

O

Solution Kinetic Energy. Since the disk starts from rest, T1 = 0. The mass moment of inertia 1 1 of the disk about its center O is I0 = mr 2 = ( 20 )( 0.32 ) = 0.9 kg # m2. Thus 2 2 T2 =

1 1 I v2 = (0.9) v2 = 0.45 v2 2 0 2

Work. Referring to the FBD of the disk, Fig. a, only couple moment M does work, which it is positive UM =

L

M du =

L0

Principle of Work and Energy.

2(2p)

(2u + 4)du = u 2 + 4u `

4p 0

= 208.18 J

T1 + ΣU1 - 2 = T2   0 + 208.18 = 0.45 v2 Ans.

  v = 21.51 rad>s = 21.5 rad>s

Ans: v = 21.5 rad>s 920

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18–10. The spool has a mass of 40 kg and a radius of gyration of kO = 0.3 m. If the 10-kg block is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity v = 15 rad>s. Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

300 mm 500 mm O

Solution Kinetic Energy. Since the system is released from rest, T1 = 0. The final velocity of the block is vb = vr = 15(0.3) = 4.50 m>s. The mass moment of inertia of the spool about O is I0 = mk 20 = 40 ( 0.32 ) = 3.60 Kg # m2. Thus T2 = =

1 2 1 I v + mbv2b 20 2 1 1 (3.60) ( 152 ) + (10) ( 4.502 ) 2 2

= 506.25 J For the block, T1 = 0 and T2 =

1 1 m v2 = ( 10 )( 4.502 ) = 101.25 J 2 b b 2

Work. Referring to the FBD of the system Fig. a, only Wb does work when the block displaces s vertically downward, which it is positive. UWb = Wb s = 10(9.81)s = 98.1 s Referring to the FBD of the block, Fig. b. Wb does positive work while T does negative work. UT = - Ts UWb = Wbs = 10(9.81)(s) = 98.1 s Principle of Work and Energy. For the system, T1 + ΣU1 - 2 = T2   0 + 98.1s = 506.25  

Ans.

s = 5.1606 m = 5.16 m

For the block using the result of s, T1 + ΣU1 - 2 = T2 0 + 98.1(5.1606) - T(5.1606) = 101.25      T = 78.48 N = 78.5 N

Ans.

Ans: s = 5.16 m T = 78.5 N 921

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18–11. The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has rotated 4 revolutions starting from rest. The wheel has a radius of gyration of kO = 0.3 m.

0.4 m O

Solution Kinetic Energy. Since the wheel starts from rest, T1 = 0. The mass moment of inertia of the wheel about point O is I0 = mk 20 = 20 ( 0.32 ) = 1.80 kg # m2. Thus, 1 1 I v2 = (1.80) v2 = 0.9 v2 2 0 2 Work. Referring to the FBD of the wheel, Fig. a, only force T does work. This work is positive since T is required to displace vertically downward, sT = ur = 4(2p)(0.4) = 3.2p m. T2 =

T  20 N

UT = TsT = 20(3.2p) = 64p J Principle of Work and Energy.   T1 + ΣU1 - 2 = T2  

0 + 64p = 0.9 v2 Ans.

   v = 14.94 rad>s = 14.9 rad>s

Ans: v = 14.9 rad>s 922

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*18–12. Determine the velocity of the 50-kg cylinder after it has descended a distance of 2 m. Initially, the system is at rest. The reel has a mass of 25 kg and a radius of gyration about its center of mass A of kA = 125 mm.

A

75 mm

SOLUTION T1 + ©U1 - 2 = T2 0 + 50(9.81)(2) =

2 1 v [(25)(0.125)2] ¢ ≤ 2 0.075

+

1 (50) v2 2 Ans.

v = 4.05 m>s

Ans: v = 4.05 m>s 923

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18–13. The 10-kg uniform slender rod is suspended at rest when the force of F = 150 N is applied to its end. Determine the angular velocity of the rod when it has rotated 90° clockwise from the position shown. The force is always perpendicular to the rod.

O

3m

Solution Kinetic Energy. Since the rod starts from rest, T1 = 0. The mass moment of inertia 1 of the rod about O is I0 = (10) ( 32 ) + 10 ( 1.52 ) = 30.0 kg # m2. Thus, 12 F

1 1 T2 = I0 v2 = (30.0) v2 = 15.0 v2 2 2 Work. Referring to the FBD of the rod, Fig. a, when the rod undergoes an angular displacement u, force F does positive work whereas W does negative work. When p 3p m. Thus u = 90°, SW = 1.5 m and SF = ur = a b(3) = 2 2 UF = 150 a

3p b = 225p J 2

UW = - 10(9.81)(1.5) = -147.15 J Principle of Work and Energy. T1 + ΣU1 - 2 = T2 0 + 225p + ( - 147.15) = 15.0 v2   v = 6.1085 rad>s = 6.11 rad>s

Ans.

Ans: v = 6.11 rad>s 924

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18–14. The 10-kg uniform slender rod is suspended at rest when the force of F = 150 N is applied to its end. Determine the angular velocity of the rod when it has rotated 180° clockwise from the position shown. The force is always perpendicular to the rod.

O

3m

Solution Kinetic Energy. Since the rod starts from rest, T1 = 0. The mass moment of inertia 1 of the rod about O is I0 = (10) ( 32 ) + 10 ( 1.52 ) = 30.0 kg # m2. Thus, 12 F

1 1 T2 = I0 v2 = (30.0) v2 = 15.0 v2 2 2 Work. Referring to the FBD of the rod, Fig. a, when the rod undergoes an angular displacement u, force F does positive work whereas W does negative work. When u = 180°, SW = 3 m and SF = ur = p(3) = 3p m. Thus UF = 150(3p) = 450p J UW = - 10(9.81)(3) = - 294.3 J Principle of Work and Energy. Applying Eq. 18, T1 + ΣU1 - 2 = T2 0 + 450p + ( -294.3) = 15.0 v2 v = 8.6387 rad>s = 8.64 rad>s

Ans.

Ans: v = 8.64 rad>s 925

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18–15. The pendulum consists of a 10-kg uniform disk and a 3-kg uniform slender rod. If it is released from rest in the position shown, determine its angular velocity when it rotates clockwise 90°.

A

B 0.8 m

M  30 N  m D

2m

Solution Kinetic Energy. Since the assembly is released from rest, initially, T1 = 0. The mass moment of inertia of the assembly about A is IA = c

1 1 (3) ( 22 ) + 3 ( 12 ) d + c (10) ( 0.42 ) + 10 ( 2.42 ) d = 62.4 kg # m2. Thus, 12 2 T2 =

1 1 I v2 = (62.4) v2 = 31.2 v2 2A 2

Work. Referring to the FBD of the assembly, Fig. a. Both Wr and Wd do positive work, since they displace vertically downward Sr = 1 m and Sd = 2.4 m, respectively. Also, couple moment M does positive work UWr = Wr Sr = 3(9.81)(1) = 29.43 J UWd = WdSd = 10(9.81)(2.4) = 235.44 J p UM = Mu = 30 a b = 15p J 2

Principle of Work and Energy. T1 + ΣU1 - 2 = T2

0 + 29.43 + 235.44 + 15p = 31.2 v2 v = 3.1622 rad>s = 3.16 rad>s

Ans.

Ans: v = 3.16 rad>s 926

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*18–16. A motor supplies a constant torque M = 6 kN # m to the winding drum that operates the elevator. If the elevator has a mass of 900 kg, the counterweight C has a mass of 200 kg, and the winding drum has a mass of 600 kg and radius of gyration about its axis of k = 0.6 m, determine the speed of the elevator after it rises 5 m starting from rest. Neglect the mass of the pulleys.

SOLUTION vE = vC

M

5 s u = = r 0.8

C D

T1 + ©U1 - 2 = T2 0 + 6000(

0.8 m

1 5 1 ) - 900(9.81)(5) + 200(9.81)(5) = (900)(v)2 + (200)(v)2 0.8 2 2 +

1 v [600(0.6)2]( )2 2 0.8 Ans.

v = 2.10 m s

Ans: v = 2.10 m>s 927

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18–17. The center O of the thin ring of mass m is given an angular velocity of v0. If the ring rolls without slipping, determine its angular velocity after it has traveled a distance of s down the plane. Neglect its thickness.

v0 s r O

SOLUTION u

T1 + ©U1-2 = T2 1 1 (mr2 + mr2)v0 2 + mg(s sin u) = (mr2 + mr2)v2 2 2 g v = v0 2 + 2 s sin u A r

Ans.

Ans: v = 928

A

v20 +

g r2

s sin u

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18–18. The wheel has a mass of 100 kg and a radius of gyration of motor supplies a torque kO = 0.2 m. A M = (40u + 900) N # m, where u is in radians, about the drive shaft at O. Determine the speed of the loading car, which has a mass of 300 kg, after it travels s = 4 m. Initially the car is at rest when s = 0 and u = 0°. Neglect the mass of the attached cable and the mass of the car’s wheels.

M s

0.3 m O

Solution s = 0.3u = 4 30

u = 13.33 rad T1 + ΣU1 - 2 = T2 [0 + 0] +

L0

13.33

vC = 7.49 m>s

(40u + 900)du - 300(9.81) sin 30° (4) =

vC 2 1 1 (300)v2C + c 100(0.20)2 d a b 2 2 0.3 Ans.

Ans: vC = 7.49 m>s 929

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18–19. The rotary screen S is used to wash limestone. When empty it has a mass of 800 kg and a radius of gyration of kG = 1.75 m. Rotation is achieved by applying a torque of M = 280 N # m about the drive wheel at A. If no slipping occurs at A and the supporting wheel at B is free to roll, determine the angular velocity of the screen after it has rotated 5 revolutions. Neglect the mass of A and B.

S

2m

Solution

B

A

0.3 m M  280 N  m

TS + ΣU1 - 2 = T2 0 + 280(uA) =

1 [800(1.75)2] v2 2

uS(2) = uA(0.3) 5(2p)(2) = uA(0.3) uA = 209.4 rad Thus Ans.

v = 6.92 rad>s

Ans: v = 6.92 rad>s 930

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*18–20. If P = 200 N and the 15-kg uniform slender rod starts from rest at u = 0°, determine the rod’s angular velocity at the instant just before u = 45°.

600 mm

A 45°

u

P  200 N B

SOLUTION Kinetic Energy and Work: Referring to Fig. a, rA>IC = 0.6 tan 45° = 0.6 m Then rG>IC = 30.32 + 0.62 = 0.6708 m Thus, (vG)2 = v2rG>IC = v2(0.6708) 1 The mass moment of inertia of the rod about its mass center is IG = ml2 12 1 = (15)(0.62) = 0.45 kg # m2. Thus, the final kinetic energy is 12 T2 = =

1 1 m(vG)22 + IG v22 2 2 1 1 (15)[w2(0.6708)]2 + (0.45) v2 2 2 2

= 3.6v22 Since the rod is initially at rest, T1 = 0. Referring to Fig. b, NA and NB do no work, while P does positive work and W does negative work. When u = 45°, P displaces through a horizontal distance sP = 0.6 m and W displaces vertically upwards through a distance of h = 0.3 sin 45°, Fig. c. Thus, the work done by P and W is UP = PsP = 200(0.6) = 120 J UW = - Wh = - 15(9.81)(0.3 sin 45°) = -31.22 J Principle of Work and Energy: T1 + ©U1 - 2 = T2 0 + [120 - 31.22] = 3.6v22 Ans.

v2 = 4.97 rad>s

Ans: v2 = 4.97 rad>s 931

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18–21. A yo-yo has a weight of 0.3 lb and a radius of gyration kO = 0.06 ft. If it is released from rest, determine how far it must descend in order to attain an angular velocity v = 70 rad>s. Neglect the mass of the string and assume that the string is wound around the central peg such that the mean radius at which it unravels is r = 0.02 ft.

SOLUTION r

vG = (0.02)70 = 1.40 ft>s

O

T1 + ©U1 - 2 = T2 0 + (0.3)(s) =

1 0.3 1 0.3 a b (1.40)2 + c (0.06)2 a b d(70)2 2 32.2 2 32.2 Ans.

s = 0.304 ft

Ans: s = 0.304 ft 932

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18–22. If the 50-lb bucket is released from rest, determine its velocity after it has fallen a distance of 10 ft. The windlass A can be considered as a 30-lb cylinder, while the spokes are slender rods, each having a weight of 2 lb. Neglect the pulley’s weight.

B

3 ft

4 ft 0.5 ft

A

0.5 ft

SOLUTION Kinetic Energy and Work: Since the windlass rotates about a fixed axis, vC = vArA vC vC = or vA = = 2vC. The mass moment of inertia of the windlass about its rA 0.5 mass center is IA =

C

2 1 30 1 2 a b A 0.52 B + 4 c a b A 0.52 B + A 0.752 B d = 0.2614 slug # ft2 2 32.2 12 32.2 32.2

Thus, the kinetic energy of the system is T = TA + T C =

1 1 I v2 + mCvC 2 2 A 2

=

1 1 50 (0.2614)(2vC)2 + a bv 2 2 2 32.2 C

= 1.2992vC 2 Since the system is initially at rest, T1 = 0. Referring to Fig. a, WA, Ax, Ay, and RB do no work, while WC does positive work. Thus, the work done by WC, when it displaces vertically downward through a distance of sC = 10 ft, is UWC = WCsC = 50(10) = 500 ft # lb Principle of Work and Energy: T1 + ©U1-2 = T2 0 + 500 = 1.2992vC 2 Ans.

vC = 19.6 ft>s

Ans: vC = 19.6 ft>s 933

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18–23. The coefficient of kinetic friction between the 100-lb disk and the surface of the conveyor belt is mA 0.2. If the conveyor belt is moving with a speed of vC = 6 ft>s when the disk is placed in contact with it, determine the number of revolutions the disk makes before it reaches a constant angular velocity.

0.5 ft v C = 6 ft/s

B

A

SOLUTION Equation of Motion: In order to obtain the friction developed at point A of the disk, the normal reaction NA must be determine first. + c ©Fy = m(aG)y ;

N - 100 = 0

N = 100 lb

Work: The friction Ff = mk N = 0.2(100) = 20.0 lb develops a constant couple moment of M = 20.0(0.5) = 10.0 lb # ft about point O when the disk is brought in contact with the conveyor belt. This couple moment does positive work of U = 10.0(u) when the disk undergoes an angular displacement u. The normal reaction N, force FOB and the weight of the disk do no work since point O does not displace. Principle of Work and Energy: The disk achieves a constant angular velocity when the points on the rim of the disk reach the speed of that of the conveyor i.e; yC = 6 ft>s. This constant angular velocity is given by yC 6 = v = = 12.0 rad>s. The mass moment inertia of the disk about point O is r 0.5

belt,

IO =

1 1 100 mr2 = a b A 0.52 B = 0.3882 slug # ft2. Applying Eq.18–13, we have 2 2 32.2 T1 + a U1 - 2 = T2 0 + U = 0 + 10.0u = u = 2.80 rad *

1 I v2 2 O

1 (0.3882) A 12.02 B 2 1 rev 2p rad

Ans.

= 0.445 rev

Ans: u = 0.445 rev 934

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*18–24. The 30-kg disk is originally at rest, and the spring is unstretched. A couple moment of M = 80 N # m is then applied to the disk as shown. Determine its angular velocity when its mass center G has moved 0.5 m along the plane. The disk rolls without slipping.

M  80 N  m

G

k  200 N/m

A

0.5 m

Solution Kinetic Energy. Since the disk is at rest initially, T1 = 0. The disk rolls without slipping. Thus, vG = vr = v(0.5). The mass moment of inertia of the disk about its 1 1 center of gravity G is IG = mr = (30) ( 0.52 ) = 3.75 kg # m2. Thus, 2 2 T2 = =

1 1 I v2 + Mv2G 2G 2 1 1 (3.75)v2 + (30)[v(0.5)]2 2 2

= 5.625 v2 Work. Since the disk rolls without slipping, the friction Ff does no work. Also when sG 0.5 the center of the disk moves SG = 0.5 m, the disk rotates u = = = 1.00 rad. r 0.5 Here, couple moment M does positive work whereas the spring force does negative work. UM = Mu = 80(1.00) = 80.0 J 1 1 UFsp = - kx2 = - (200) ( 0.52 ) = - 25.0 J 2 2 Principle of Work and Energy. T1 + ΣU1 - 2 = T2 0 + 80 + ( -25.0) = 5.625 v2 Ans.

v = 3.127 rad>s = 3.13 rad>s

Ans: v = 3.13 rad>s 935

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18–25. The 30-kg disk is originally at rest, and the spring is unstretched. A couple moment M = 80 N # m is then applied to the disk as shown. Determine how far the center of mass of the disk travels along the plane before it momentarily stops. The disk rolls without slipping.

M  80 N  m

G

k  200 N/m

A

0.5 m

Solution Kinetic Energy. Since the disk is at rest initially and required to stop finally, T1 = T2 = 0. Work. Since the disk rolls without slipping, the friction Ff does no work. Also, when sG sG = = 2 sG. Here, couple the center of the disk moves sG, the disk rotates u = r 0.5 moment M does positive work whereas the spring force does negative work. UM = Mu = 80(2 sG) = 160 sG 1 1 2 2 UFsp = - kx2 = - (200) sG = - 100 sG 2 2 Principle of Work and Energy. T1 + ΣU1 - 2 = T2 0 + 160 sG +

( - 100 sG2 ) = 0

2 160 sG - 100 sG = 0

sG(160 - 100 sG) = 0 Since sG ≠ 0, then 160 - 100 sG = 0 Ans.

sG = 1.60 m

Ans: sG = 1.60 m 936

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18–26. Two wheels of negligible weight are mounted at corners A and B of the rectangular 75-lb plate. If the plate is released from rest at u = 90°, determine its angular velocity at the instant just before u = 0°.

A 3 ft 1.5 ft

u

SOLUTION

B

Kinetic Energy and Work: Referring Fig. a, (vG)2 = vrA>IC = v a 20.752 + 1.52 b = 1.677v2 The mass moment of inertia of the plate about its mass center is 1 1 75 b(1.52 + 32) = 2.1836 slug # ft2. Thus, the final IG = m(a2 + b2) = a 12 12 32.2 kinetic energy is T2 = =

1 1 m(vG)22 + v22 2 2 1 75 1 b (1.677v2)2 + IG (2.1836)v22 a 2 32.2 2

= 4.3672v2 2 Since the plate is initially at rest, T1 = 0. Referring to Fig. b, NA and NB do no work, while W does positive work. When u = 0°, W displaces vertically through a distance of h = 20.752 + 1.52 = 1.677 ft, Fig. c. Thus, the work done by W is UW = Wh = 75(1.677) = 125.78 ft # lb

Principle of Work and Energy: T1 + ©U1-2 = T2 0 + 125.78 = 4.3672v2 2 Ans.

v2 = 5.37 rad>s

Ans: v2 = 5.37 rad>s 937

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18–27. The link AB is subjected to a couple moment of M = 40 N # m. If the ring gear C is fixed, determine the angular velocity of the 15-kg inner gear when the link has made two revolutions starting from rest. Neglect the mass of the link and assume the inner gear is a disk. Motion occurs in the vertical plane.

C 200 mm

A

Solution Kinetic Energy. The mass moment of inertia of the inner gear about its center 1 1 B is IB = mr 2 = (15) ( 0.152 ) = 0.16875 kg # m2. Referring to the kinematics 2 2 diagram of the gear, the velocity of center B of the gear can be related to the gear’s

M  40 N  m

B 150 mm

angular velocity, which is vB = vrB>IC;  vB = v(0.15) Thus, T = =

1 1 I v2 + Mv2G 2B 2 1 1 (0.16875) v2 + (15)[v(0.15)]2 2 2

= 0.253125 v2 Since the gear starts from rest, T1 = 0. Work. Referring to the FBD of the gear system, we notice that M does positive work whereas W does no work, since the gear returns to its initial position after the link completes two revolutions. UM = Mu = 40[2(2p)] = 160p J Principle of Work and Energy. T1 + ΣU1 - 2 = T2 0 + 160p = 0.253125 v2 Ans.

v = 44.56 rad>s = 44.6 rad>s

Ans: v = 44.6 rad>s 938

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*18–28. The 10-kg rod AB is pin-connected at A and subjected to a couple moment of M 15 N.m. If the rod is released from rest when the spring is unstretched at u 30 , determine the rod’s angular velocity at the instant u 60 . As the rod rotates, the spring always remains horizontal, because of the roller support at C.

C

k = 40 N/m

B

SOLUTION Free Body Diagram: The spring force FsP does negative work since it acts in the opposite direction to that of its displacement ssp, whereas the weight of the cylinder acts in the same direction of its displacement sw and hence does positive work. Also, the couple moment M does positive work as it acts in the same direction of its angular displacement u . The reactions Ax and Ay do no work since point A does not displace. Here, ssp = 0.75 sin 60° - 0.75 sin 30° = 0.2745 m and sW = 0.375 cos 30° - 0.375 cos 60° = 0.1373 m.

0.75 m A

M = 15 N · m

Principle of Work and Energy: The mass moment of inertia of the cylinder about 1 1 point A is IA = ml2 + md 2 = (10) A 0.752 B + 10 A 0.3752 B = 1.875 kg # m2. 12 12 Applying Eq.18–13, we have T1 + a U1-2 = T2 0 + WsW + Mu 0 + 10(9.81)(0.1373) + 15a

1 2 1 ks P = IA v2 2 2

1 p p 1 - b - (40) A 0.27452 B = (1.875) v2 3 6 2 2 Ans.

v = 4.60 rad>s

Ans: v = 4.60 rad>s 939

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18–29. The 10-lb sphere starts from rest at u 0° and rolls without slipping down the cylindrical surface which has a radius of 10 ft. Determine the speed of the sphere’s center of mass at the instant u 45°.

0.5 ft

10 ft

θ

SOLUTION Kinematics: vG = 0.5vG T1 + ©U1 - 2 = T2 0 + 10(10.5)(1 - cos 45°) =

vG 2 1 2 10 1 10 a b v2G + c a b(0.5)2 d a b 2 32.2 2 5 32.2 0.5 Ans.

vG = 11.9 ft>s

Ans: vG = 11.9 ft>s 940

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18–30. Motor M exerts a constant force of P = 750 N on the rope. If the 100-kg post is at rest when u = 0°, determine the angular velocity of the post at the instant u = 60°. Neglect the mass of the pulley and its size, and consider the post as a slender rod.

M P  750 N

C

A

SOLUTION

3m

Kinetic Energy and Work: Since the post rotates about a fixed axis, vG = vrG = v (1.5). The mass moment of inertia of the post about its mass center is 1 IG = (100)(32) = 75 kg # m2. Thus, the kinetic energy of the post is 12 T = =

4m

u B

1 1 mvG2 + IGv2 2 2 1 1 (100)[v(1.5)]2 + (75)v2 2 2

= 150v2 1 This result can also be obtained by applying T = IBv2, where IB = 2 1 (100)(32) + 100 (1.52) = 300 kg # m2. Thus, 12 T =

1 1 I v2 = (300)v2 = 150v2 2 B 2

Since the post is initially at rest, T1 = 0. Referring to Fig. a, Bx, By, and R C do no work, while P does positive work and W does negative work. When u = 60° , P displaces sP = A¿C - AC, where AC = 24 2 + 32 - 2(4)(3) cos 30° = 2.053 m

and A¿C = 24 2 + 32 = 5 m. Thus, sP = 5 - 2.053 = 2.947 m. Also, W displaces vertically upwards through a distance of h = 1.5 sin 60° = 1.299 m. Thus, the work done by P and W is UP = PsP = 750(2.947) = 2210.14 J UW = - Wh = - 100 (9.81)(1.299) = - 1274.36 J Principle of Work and Energy: T1 + ©U1-2 = T2 0 + [2210.14 - 1274.36] = 150v2 Ans.

v = 2.50 rad>s

Ans: v = 2.50 rad>s 941

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18–31. The linkage consists of two 6-kg rods AB and CD and a 20-kg bar BD. When u = 0°, rod AB is rotating with an angular velocity v = 2 rad>s. If rod CD is subjected to a couple moment of M = 30 N # m, determine vAB at the instant u = 90°.

1.5 m B

D u 1m

1m v A

M  30 N  m C

Solution Kinetic Energy. The mass moment of inertia of each link about the axis of rotation 1 is IA = (6) ( 12 ) + 6 ( 0.52 ) = 2.00 kg # m. The velocity of the center of mass of the 12 bar is vG = vr = v(1). Thus, 1 1 T = 2 a IAv2 b + Mbv2G 2 2

1 1 = 2c (2.00)v2 d + (20)[v(1)]2 2 2 = 12.0 v2

Initially, v = 2 rad>s. Then T1 = 12.0 ( 22 ) = 48.0 J Work. Referring to the FBD of the assembly, Fig. a, the weights Wb, Wc and couple moment M do positive work when the links p undergo an angular displacement u. When u = 90° = rad, 2 UWb = Wb sb = 20(9.81)(1) = 196.2 J UWc = Wc sc = 6(9.81)(0.5) = 29.43 J p UM = Mu = 30 a b = 15p J 2

Principle of Work and Energy. T1 + ΣU1 - 2 = T2

48.0 + [196.2 + 2(29.43) + 15p] = 12.0 v2 Ans.

v = 5.4020 rad>s = 5.40 rad>s

Ans: v = 5.40 rad>s 942

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*18–32. The linkage consists of two 6-kg rods AB and CD and a 20-kg bar BD. When u = 0°, rod AB is rotating with an angular velocity v = 2 rad>s. If rod CD is subjected to a couple moment M = 30 N # m, determine v at the instant u = 45°.

1.5 m B

D u 1m

1m v A

M  30 N  m C

Solution Kinetic Energy. The mass moment of inertia of each link about the axis of rotation 1 is IA = (6) ( 12 ) + 6 ( 0.52 ) = 2.00 kg # m2. The velocity of the center of mass of 12 the bar is vG = vr = v(1). Thus, 2 1 1 T = 2 a IAv2A b + mbv2G 2 2

1 1 = 2c (2.00)v2 d + (20)[v(1)]2 2 2 = 12.0 v2

Initially, v = 2 rad>s. Then T1 = 12.0 ( 22 ) = 48.0 J Work. Referring to the FBD of the assembly, Fig. a, the weights Wb, Wc and couple moment M do positive work when the links p undergo an angular displacement u. when u = 45° = rad, 4 UWb = Wb sb = 20(9.81) ( 1 - cos 45° ) = 57.47 J UWc = Wc sc = 6(9.81)[0.5(1 - cos 45°)] = 8.620 J p UM = Mu = 30 a b = 7.5p J 4

Principle of Work and Energy. T1 + ΣU1 - 2 = T2

48.0 + [57.47 + 2(8.620) + 7.5p] = 12.0 v2 Ans.

v = 3.4913 rad>s = 3.49 rad>s 

Ans: v = 3.49 rad>s 943

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18–33. The two 2-kg gears A and B are attached to the ends of a 3-kg slender bar. The gears roll within the fixed ring gear C, which lies in the horizontal plane. If a 10-N # m torque is applied to the center of the bar as shown, determine the number of revolutions the bar must rotate starting from rest in order for it to have an angular velocity of vAB = 20 rad>s. For the calculation, assume the gears can be approximated by thin disks.What is the result if the gears lie in the vertical plane?

C 400 mm

A

SOLUTION

B 150 mm

150 mm M = 10 N · m

Energy equation (where G refers to the center of one of the two gears): Mu = T2 1 1 1 10u = 2a IGv2gear b + 2 a mgear b (0.200vAB)2 + IABv2AB 2 2 2 1 (2)(0.150)2 = 0.0225 kg # m2, 2 1 200 IAB = (3)(0.400)2 = 0.0400 kg # m2 and vgear = v , 12 150 AB 200 2 2 b vAB + 2(0.200)2v2AB + 0.0200v2AB 10u = 0.0225 a 150

Using mgear = 2 kg, IG =

When vAB = 20 rad>s, u = 5.60 rad Ans.

= 0.891 rev, regardless of orientation

Ans: u = 0.891 rev, regardless of orientation 944

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18–34. The linkage consists of two 8-lb rods AB and CD and a 10-lb bar AD. When u = 0°, rod AB is rotating with an angular velocity vAB = 2 rad>s. If rod CD is subjected to a couple moment M = 15 lb # ft and bar AD is subjected to a horizontal force P = 20 lb as shown, determine vAB at the instant u = 90°.

B

vAB

C

M  15 lb · ft

2 ft

2 ft u

A

D

P  20 lb

3 ft

SOLUTION T1 + ΣU1 - 2 = T2 1 1 8 1 10 p 2c e a b(2)2 f(2)2 d + a b (4)2 + c 20(2) + 15 a b - 2(8)(1) - 10(2) d 2 3 32.2 2 32.2 2 1 1 8 1 10 b(2)2 fv2 d + a b(2v)2 = 2c e a 2 3 32.2 2 32.2

Ans.

v = 5.74 rad>s

Ans: v = 5.74 rad>s 945

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18–35. The linkage consists of two 8-lb rods AB and CD and a 10-lb bar AD. When u = 0°, rod AB is rotating with an angular velocity vAB = 2 rad>s. If rod CD is subjected to a couple moment M = 15 lb # ft and bar AD is subjected to a horizontal force P = 20 lb as shown, determine vAB at the instant u = 45°.

AB

B

C

M = 15 lb · ft

2 ft

2 ft

A

D

P = 20 lb

3 ft

SOLUTION T1 + ΣU1 - 2 = T2 1 1 8 1 10 2c e a b(2)2 f(2)2 d + a b(4)2 2 3 32.2 2 32.2

p + c 20(2 sin 45°) + 15a b - 2(8)(1 - cos 45°) - 10(2 - 2 cos 45°) d 4

1 1 8 1 10 b(2)2 fv2 d + a b(2v)2 = 2c e a 2 3 32.2 2 32.2

Ans.

v = 5.92 rad>s

Ans: vAB = 5.92 rad>s 946

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*18–36. The assembly consists of a 3-kg pulley A and 10-kg pulley B. If a 2-kg block is suspended from the cord, determine the block’s speed after it descends 0.5 m starting from rest. Neglect the mass of the cord and treat the pulleys as thin disks. No slipping occurs.

100 mm B 30 mm

A

Solution T1 + V1 = T2 + V2 [0 + 0 + 0] + [0] =

1 1 1 1 1 c (3)(0.03)2 d v2A + c (10)(0.1)2 d v2B + (2)(vC)2 - 2(9.81)(0.5) 2 2 2 2 2

vC = vB (0.1) = 0.03vA Thus, vB = 10vC vA = 33.33vC

Substituting and solving yields, Ans.

vC = 1.52 m>s

Ans: vC = 1.52 m>s 947

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18–37. The assembly consists of a 3-kg pulley A and 10-kg pulley B. If a 2-kg block is suspended from the cord, determine the distance the block must descend, starting from rest, in order to cause B to have an angular velocity of 6 rad>s. Neglect the mass of the cord and treat the pulleys as thin disks. No slipping occurs.

100 mm B 30 mm

A

Solution vC = vB (0.1) = 0.03 vA If vB = 6 rad>s then vA = 20 rad>s vC = 0.6 m>s T1 + V1 = T2 + V2 [0 + 0 + 0] + [0] = sC = 78.0 mm

1 1 1 1 1 c (3)(0.03)2 d (20)2 + c (10)(0.1)2 d (6)2 + (2)(0.6)2 - 2(9.81)sC 2 2 2 2 2 Ans.

Ans: sC = 78.0 mm 948

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18–38. The spool has a mass of 50 kg and a radius of gyration kO = 0.280 m. If the 20-kg block A is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity v = 5 rad>s. Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

0.3 m 0.2 m O

SOLUTION vA = 0.2v = 0.2(5) = 1 m>s System:

A

T1 + V1 = T2 + V2 [0 + 0] + 0 =

1 1 (20)(1)2 + [50(0.280)2](5)2 - 20(9.81) s 2 2 Ans.

s = 0.30071 m = 0.301 m Block: T1 + ©U1 - 2 = T2 0 + 20(9.81)(0.30071) - T(0.30071) =

1 (20)(1)2 2 Ans.

T = 163 N

Ans: s = 0.301 m T = 163 N 949

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18–39. The spool has a mass of 50 kg and a radius of gyration kO = 0.280 m. If the 20-kg block A is released from rest, determine the velocity of the block when it descends 0.5 m.

0.3 m 0.2 m O

SOLUTION Potential Energy: With reference to the datum established in Fig. a, the gravitational potential energy of block A at position 1 and 2 are V1 = (Vg)1 = WAy1 = 20 (9.81)(0) = 0

A

V2 = (Vg)2 = - WAy2 = - 20 (9.81)(0.5) = -98.1 J vA vA = = 5vA. rA 0.2 Here, the mass moment of inertia about the fixed axis passes through point O is

Kinetic Energy: Since the spool rotates about a fixed axis, v =

IO = mkO2 = 50 (0.280)2 = 3.92 kg # m2. Thus, the kinetic energy of the system is T = =

1 1 I v2 + mAvA2 2 O 2 1 1 (3.92)(5vA)2 + (20)vA2 = 59vA2 2 2

Since the system is at rest initially, T1 = 0 Conservation of Energy: T1 + V1 = T2 + V2 0 + 0 = 59vA2 + (- 98.1) vA = 1.289 m>s Ans.

= 1.29 m>s

Ans: vA = 1.29 m>s 950

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*18–40. An automobile tire has a mass of 7 kg and radius of gyration kG = 0.3 m. If it is released from rest at A on the incline, determine its angular velocity when it reaches the horizontal plane. The tire rolls without slipping.

G 0.4 m A 5m 30° 0.4 m B

SOLUTION nG = 0.4v Datum at lowest point. T1 + V1 = T2 + V2 0 + 7(9.81)(5) =

1 1 (7)(0.4v)2 + [7 (0.3)2]v2 + 0 2 2 Ans.

v = 19.8 rad s

Ans: v = 19.8 rad>s 951

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18–41. The spool has a mass of 20 kg and a radius of gyration of kO = 160 mm. If the 15-kg block A is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity v = 8 rad>s. Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

200 mm O

Solution Kinetic Energy. The mass moment of inertia of the spool about its center O is I0 = mk 20 = 20 ( 0.162 ) = 0.512 kg # m2. The velocity of the block is vb = vsrs = vs(0.2). Thus, T =

1 1 I v2 + mb v2b 2 0 2

=

1 1 (0.512)v2s + (15)[vs(0.2)]2 2 2

A

= 0.556 v2s Since the system starts from rest, T1 = 0. When vs = 8 rad>s, T2 = 0.556 ( 82 ) = 35.584 J Potential Energy. With reference to the datum set in Fig. a, the initial and final gravitational potential energy of the block are (Vg)1 = mb g y1 = 0 (Vg)2 = mb g ( -y2) = 15(9.81)( - sb) = - 147.15 sb Conservation of Energy. T1 + V1 = T2 + V2 0 + 0 = 35.584 + ( - 147.15 sb) Ans.

sb = 0.2418 m = 242 mm

Principle of Work and Energy. The final velocity of the block is (vb)2 = (vs)2 rs = 8(0.2) = 1.60 m>s. Referring to the FBD of the block, Fig. b and using the result of Sb, T1 + Σ U1 - 2 = T2 0 + 15(9.81)(0.2418) - T(0.2418) =

1 (15) ( 1.602 ) 2 Ans.

T = 67.75 N = 67.8 N

Ans: sb = 242 mm T = 67.8 N 952

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18–42. The spool has a mass of 20 kg and a radius of gyration of kO = 160 mm. If the 15-kg block A is released from rest, determine the velocity of the block when it descends 600 mm.

200 mm O

Solution Kinetic Energy. The mass moment of inertia of the spool about its center O is I0 = mk 20 = 20 ( 0.162 ) = 0.512 kg # m2. The angular velocity of the spool is vb vb vs = = = 5vb. Thus, rs 0.2 T = =

A

1 1 I0 v2 + mbv2b 2 2 1 1 (0.512)(5vb)2 + (15)v2b 2 2

= 13.9v2b Since the system starts from rest, T1 = 0. Potential Energy. With reference to the datum set in Fig. a, the initial and final gravitational potential energies of the block are (Vg)1 = mb g y1 = 0 (Vg)2 = mb g y2 = 15(9.81)( - 0.6) = -88.29 J Conservation of Energy. T1 + V1 = T2 + V2 0 + 0 = 13.9v2b + ( - 88.29) Ans.

vb = 2.5203 m>s = 2.52 m>s

Ans: vb = 2.52 m>s 953

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18–43. A uniform ladder having a weight of 30 lb is released from rest when it is in the vertical position. If it is allowed to fall freely, determine the angle u at which the bottom end A starts to slide to the right of A. For the calculation, assume the ladder to be a slender rod and neglect friction at A. 10 ft

SOLUTION

u

Potential Energy: Datum is set at point A. When the ladder is at its initial and final position, its center of gravity is located 5 ft and (5 cos u ) ft above the datum. Its initial and final gravitational potential energy are 30(5) = 150 ft # lb and 30(5 cos u ) = 150 cos u ft # lb, respectively. Thus, the initial and final potential energy are V1 = 150 ft # lb

A

V2 = 150 cos u ft # lb

Kinetic Energy: The mass moment inertia of the ladder about point A is 1 30 30 IA = a b (102) + a b (52) = 31.06 slug # ft2. Since the ladder is initially at 12 32.2 32.2 rest, the initial kinetic energy is T1 = 0. The final kinetic energy is given by T2 =

1 1 I v2 = (31.06)v2 = 15.53v2 2 A 2

Conservation of Energy: Applying Eq. 18–18, we have T1 + V1 = T2 + V2 0 + 150 = 15.53v2 + 150 cos u v2 = 9.66(1 - cos u) Equation of Motion: The mass moment inertia of the ladder about its mass center is 1 30 IG = a b (102) = 7.764 slug # ft2. Applying Eq. 17–16, we have 12 32.2 + ©MA = ©(Mk)A;

-30 sin u(5) = - 7.764a - a

30 b[a(5)](5) 32.2

a = 4.83 sin u c

+ ©Fx = m(aG)x;

Ax = -

30 [9.66(1 - cos u)(5)] sin u 32.2 +

Ax = -

30 [4.83 sin u(5)] cos u 32.2

30 (48.3 sin u - 48.3 sin u cos u - 24.15 sin u cos u) 32.2

= 45.0 sin u (1 - 1.5 cos u) = 0 If the ladder begins to slide, then Ax = 0 . Thus, for u>0, 45.0 sin u (1 - 1.5 cos u) = 0 u = 48.2°

Ans.

Ans: u = 48.2° 954

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*18–44. Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10 kg and a radius of gyration of 125 mm about its center of mass. Gear B and drum C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm.

100 mm 150 mm C

A

B 200 mm

SOLUTION Potential Energy: With reference to the datum shown in Fig. a, the gravitational potential energy of block D at position (1) and (2) is V1 = (Vg)1 = WD(yD)1 = 50 (9.81)(0) = 0

D

V2 = (Vg)2 = - WD(yD)2 = - 50(9.81)(2) = -981 J Kinetic Energy: Since gear B rotates about a fixed axis, vB =

vD vD = =10vD. rD 0.1

rB 0.2 b(10vD) = 13.33vD. bv = a rA B 0.15 The mass moment of inertia of gears A and B about their mass centers are IA = mAkA2 = 10(0.1252) = 0.15625 kg # m2 and IB = mBkB2 = 30(0.152) = 0.675 kg # m2.Thus, the kinetic energy of the system is Also, since gear A is in mesh with gear B, vA = a

T = =

1 1 1 I v 2 + IBvB2 + mDvD2 2 A A 2 2 1 1 1 (0.15625)(13.33vD)2 + (0.675)(10vD)2 + (50)vD2 2 2 2

= 72.639vD2 Since the system is initially at rest, T1 = 0. Conservation of Energy: T1 + V1 = T2 + V2 0 + 0 = 72.639vD2 - 981 Ans.

vD = 3.67 m>s

Ans: vD = 3.67 m>s 955

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18–45. The 12-kg slender rod is attached to a spring, which has an unstretched length of 2 m. If the rod is released from rest when u = 30°, determine its angular velocity at the instant u = 90°.

2m

B

A

u 2m

k  40 N/m

Solution

C

Kinetic Energy. The mass moment of inertia of the rod about A is 1 IA = (12) ( 22 ) + 12 ( 12 ) = 16.0 kg # m2. Then 12 T =

1 1 I v2 = (16.0) v2 = 8.00 v2 2 A 2

Since the rod is released from rest, T1 = 0. Potential Energy. With reference to the datum set in Fig. a, the gravitational potential energies of the rod at positions ➀ and ➁ are (Vg)1 = mg( - y1) = 12(9.81)( - 1 sin 30°) = - 58.86 J (Vg)2 = mg( - y2) = 12(9.81)( - 1) = -117.72 J The stretches of the spring when the rod is at positions ➀ and ➁ are x1 = 2(2 sin 75°) - 2 = 1.8637 m x2 = 222 + 22 - 2 = 0.8284 m

Thus, the initial and final elastic potential energies of the spring are 1 2 1 kx = (40) ( 1.86372 ) = 69.47 J 2 1 2 1 2 1 (Ve)2 = kx2 = (40) ( 0.82842 ) = 13.37 J 2 2

(Ve)1 =

Conservation of Energy. T1 + V1 = T2 + V2 0 + ( - 58.86) + 69.47 = 8.00 v2 + ( - 117.72) + 13.73 Ans.

v = 3.7849 rad>s = 3.78 rad>s

Ans: v = 3.78 rad>s 956

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18–46. The 12-kg slender rod is attached to a spring, which has an unstretched length of 2 m. If the rod is released from rest when u = 30°, determine the angular velocity of the rod the instant the spring becomes unstretched.

2m

B

A

u 2m

k  40 N/m

Solution

C

Kinetic Energy. The mass moment of inertia of the rod about A is 1 IA = (12) ( 22 ) + 12 ( 12 ) = 16.0 kg # m3. Then 12 T =

1 1 I v2 = (16.0)v2 = 8.00 v2 2 A 2

Since the rod is release from rest, T1 = 0. Potential Energy. When the spring is unstretched, the rod is at position ➁ shown in Fig. a. with reference to the datum set, the gravitational potential energies of the rod at positions ➀ and ➁ are (Vg)1 = mg( - y1) = 12(9.81)( - 1 sin 30° ) = - 58.86 J (Vg)2 = mg( - y2) = 12(9.81)( - 1 sin 60°) = -101.95 J The stretch of the spring when the rod is at position ➀ is x1 = 2(2 sin 75 ° ) - 2 = 1.8637 m It is required that x2 = 0. Thus, the initial and final elastic potential energy of the spring are (Ve)1 =

1 2 1 kx = (40) ( 1.86372 ) = 69.47 J 2 1 2

(Ve)2 =

1 2 kx = 0 2 2

Conservation of Energy. T1 + V1 = T2 + V2 0 + ( - 58.86) + 69.47 = 8.00 v2 + ( - 101.95) + 0 Ans.

v = 3.7509 rad>s = 3.75 rad>s

Ans: v = 3.75 rad>s 957

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18–47. The 40-kg wheel has a radius of gyration about its center of gravity G of kG = 250 mm. If it rolls without slipping, determine its angular velocity when it has rotated clockwise 90° from the position shown. The spring AB has a stiffness k = 100 N>m and an unstretched length of 500 mm. The wheel is released from rest.

1500 mm 200 mm 200 mm

B G

k  100 N/m A

400 mm

Solution Kinetic Energy. The mass moment of inertia of the wheel about its center of mass G is IG = mk 2G = 40 ( 0.252 ) = 2.50 kg # m2, since the wheel rolls without slipping, vG = vrG = v(0.4). Thus T = =

1 1 I v2 + mv2G 2 G 2

1 1 (2.50)v2 + (40)[v(0.4)]2 = 4.45 v2 2 2

Since the wheel is released from rest, T1 = 0. Potential Energy. When the wheel rotates 90° clockwise from position ➀ p to ➁, Fig. a, its mass center displaces SG = urG = (0.4) = 0.2p m. Then 2 xy = 1.5 - 0.2 - 0.2p = 0.6717 m. The stretches of the spring when the wheel is at positions ➀ and ➁ are x1 = 1.50 - 0.5 = 1.00 m x2 = 20.67172 + 0.22 - 0.5 = 0.2008 m

Thus, the initial and final elastic potential energies are (Ve)1 =

1 2 1 kx = (100) ( 12 ) = 50 J 2 1 2

(Ve)2 =

1 2 1 kx = (100) ( 0.20082 ) = 2.0165 J 2 2 2

Conservation of Energy. T1 + V1 = T2 + V2 0 + 50 = 4.45v2 + 2.0165 Ans.

v = 3.2837 rad>s = 3.28 rad>s

Ans: v = 3.28 rad>s 958

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18–48. The assembly consists of two 10-kg bars which are pin connected. If the bars are released from rest when u = 60°, determine their angular velocities at the instant u = 0°. The 5-kg disk at C has a radius of 0.5 m and rolls without slipping.

B 3m A

u

3m u

C

Solution Kinetic Energy. Since the system is released from rest, T1 = 0. Referring to the kinematics diagram of bar BC at the final position, Fig. a, we found that IC is located at C. Thus, (vc)2 = 0. Also, (vB)2 = (vBC)2 rB>IC;     (vb)2 = (vBC)2(3) (vG)2 = (vBC)2 rG>IC;    (vG)2 = (vBC)2(1.5) Then for bar AB, (vB)2 = (vAB)2 rAB;  (vBC)2(3) = (vAB)2(3) (vAB)2 = (vBC)2 For the disk, since the velocity of its center (vc)2 = 0, then (vd)2 = 0. Thus T2 = =

1 1 1 I (v )22 + IG(vBC)22 + mr(vG)22 2 A AB 2 2 1 1 1 1 1 c (10) ( 32 ) d (vBC)22 + c (10) ( 32 ) d (vBC)22 + (10)[(vBC)2(1.5)]2 2 3 2 12 2

= 30.0(vBC)22

Potential Energy. With reference to the datum set in Fig. b, the initial and final gravitational potential energies of the system are (Vg)1 = 2mgy1 = 2[10(9.81)(1.5 sin 60°)] = 254.87 J (Vg)2 = 2mgy2 = 0 Conservation of Energy. T1 + V1 = T2 + V2 0 + 254.87 = 30.0(vBC)22 + 0 (vBC)2 = 2.9147 rad>s = 2.91 rad>s

Ans.

(vAB)2 = (vBC)2 = 2.91 rad>s

Ans.

Ans: (vBC)2 = 2.91 rad>s (vAB)2 = 2.91 rad>s 959

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18–49. The assembly consists of two 10-kg bars which are pin connected. If the bars are released from rest when u = 60°, determine their angular velocities at the instant u = 30°. The 5-kg disk at C has a radius of 0.5 m and rolls without slipping.

B 3m A

Solution Kinetic Energy. Since the system is released from rest, T1 = 0. Referring to the kinematics diagram of bar BC at final position with IC so located, Fig. a, rB>IC = rC>IC = 3 m   rG>IC = 3 sin 60° = 1.523 m Thus, (vB)2 = (vBC)2 rB>IC ;    (vB)2 = (vBC)2(3) (vC)2 = (vBC)2 rC>IC ;    (vC)2 = (vBC)2(3) (vG)2 = (vBC)2 rG>IC ;   (vG)2 = (vBC)2 ( 1.523 ) Then for rod AB, (vB)2 = (vAB)2 rAB ;    (vBC)2(3) = (vAB)2(3) (vAB)2 = (vBC)2 For the disk, since it rolls without slipping, (vC)2 = (vd)2rd;    (vBC)2(3) = (vd)2(0.5) (vd)2 = 6(vBC)2 Thus, the kinetic energy of the system at final position is T2 = =

1 1 1 1 1 I (v )2 + IG(vBC)22 + mr(vG)22 + IC(vd)22 + md(vC)22 2 A AB 2 2 2 2 2 2 1 1 1 1 1 c (10) ( 32 ) d (vBC)22 + c (10) ( 32 ) d (vBC)22 + (10) c (vBC)2 ( 1.523 ) d 2 3 2 12 2

+ = 86.25(vBC)22

1 1 1 c (5) ( 0.52 ) d [6(vBC)2]2 + (5)[(vBC)2(3)]2 2 2 2

Potential Energy. With reference to the datum set in Fig. b, the initial and final gravitational potential energies of the system are (Vg)1 = 2mgy1 = 2[10(9.81)(1.5 sin 60°)] = 254.87 J (Vg)2 = 2mgy2 = 2[10(9.81)(1.5 sin 30°)] = 147.15 J

960

u

3m u

C

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18–49.  Continued

Conservation of Energy. T1 + V1 = T2 + V2 0 + 254.87 = 86.25(vBC)22 + 147.15 (vBC)2 = 1.1176 rad>s = 1.12 rad>s

Ans.

(vAB)2 = (vBC)2 = 1.12 rad>s

Ans.

Ans: ( vAB ) 2 = ( vBC ) 2 = 1.12 rad>s 961

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18–50. The compound disk pulley consists of a hub and attached outer rim. If it has a mass of 3 kg and a radius of gyration kG = 45 mm, determine the speed of block A after A descends 0.2 m from rest. Blocks A and B each have a mass of 2 kg. Neglect the mass of the cords.

100 mm 30 mm

SOLUTION

B

T1 + V1 = T2 + V2 [0 + 0 + 0] + [0 + 0] = u =

1 1 1 [3(0.045)2]v2 + (2)(0.03v)2 + (2)(0.1v)2 - 2(9.81)sA + 2(9.81)sB 2 2 2

A

sA sB = 0.03 0.1

sB = 0.3 sA Set sA = 0.2 m, sB = 0.06 m Substituting and solving yields, v = 14.04 rad>s vA = 0.1(14.04) = 1.40 m>s

Ans.

Ans: vA = 1.40 m>s 962

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18–51. The uniform garage door has a mass of 150 kg and is guided along smooth tracks at its ends. Lifting is done using the two springs, each of which is attached to the anchor bracket at A and to the counterbalance shaft at B and C. As the door is raised, the springs begin to unwind from the shaft, thereby assisting the lift. If each spring provides a torsional moment of M = (0.7u) N # m, where u is in radians, determine the angle u0 at which both the left-wound and right-wound spring should be attached so that the door is completely balanced by the springs, i.e., when the door is in the vertical position and is given a slight force upwards, the springs will lift the door along the side tracks to the horizontal plane with no final angular velocity. Note: The elastic potential energy of a torsional spring is Ve = 12ku2, where M = ku and in this case k = 0.7 N # m>rad.

C

A B

3m

4m

SOLUTION Datum at initial position. T1 + V1 = T2 + V2 1 0 + 2 c (0.7)u20 d + 0 = 0 + 150(9.81)(1.5) 2 Ans.

u0 = 56.15 rad = 8.94 rev.

Ans: u0 = 8.94 rev 963

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*18–52. The two 12-kg slender rods are pin connected and released from rest at the position u = 60°. If the spring has an unstretched length of 1.5 m, determine the angular velocity of rod BC, when the system is at the position u = 0°. Neglect the mass of the roller at C.

B 2m A

2m

u

C k  20 N/m

Solution Kinetic Energy. Since the system is released from rest, T1 = 0  . Referring to the kinematics diagram of rod BC at the final position, Fig. a we found that IC is located at C. Thus, (vC)2 = 0. Also, (vB)2 = (vBC)2 rB>IC;  (vB)2 = (vBC)2(2) (vG)2 = (vBC)2 rC>IC;  (vG)2 = (vBC)2(1) Then for rod AB, (vB)2 = (vAB)2 rAB;    (vBC)2(2) = (vAB)2(2) (vAB)2 = (vBC)2 Thus, T2 = =

1 1 1 I (v )2 + IG(vBC)22 + mr(vG)22 2 A AB 2 2 2 1 1 1 1 1 c (12) ( 22 ) d (vBC)22 + c (12) ( 22 ) d (vBC)22 + (12)[(vBC)2(1)]2 2 3 2 12 2

= 16.0(vBC)22

Potential Energy. With reference to the datum set in Fig. b, the initial and final gravitational potential energies of the system are (Vg)1 = 2mgy1 = 2[12(9.81)(1 sin 60°)] = 203.90 J (Vg)2 = 2mgy2 = 0 The stretch of the spring when the system is at initial and final position are x1 = 2(2 cos 60°) - 1.5 = 0.5 m x2 = 4 - 1.5 = 2.50 m Thus, the initial and final elastic potential energies of the spring is (Ve)1 =

1 2 1 kx1 = (20) ( 0.52 ) = 2.50 J 2 2

(Ve)2 =

1 2 1 kx2 = (20) ( 2.502 ) = 62.5 J 2 2

964

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18–52.  Continued

Conservation of Energy. T1 + V1 = T2 + V2 0 + (203.90 + 2.50) = 16.0(vBC)22 + (0 + 62.5) Ans.

(vBC)2 = 2.9989 rad>s = 3.00 rad>s

Ans: (vBC)2 = 3.00 rad>s 965

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18–53. The two 12-kg slender rods are pin connected and released from rest at the position u = 60°. If the spring has an unstretched length of 1.5 m, determine the angular velocity of rod BC, when the system is at the position u = 30°.

B 2m

2m A

u

C k  20 N/m

Solution Kinetic Energy. Since the system is released from rest, T1 = 0  . Referring to the kinematics diagram of rod BC at final position with IC so located, Fig. a rB>IC = rC>IC = 2 m   rG>IC = 2 sin 60° = 23 m Thus, (VB)2 = (vBC)2 rB>IC;   (VB)2 = (vBC)2(2) (VC)2 = (vBC)2 rC>IC;   (VC)2 = (vBC)2(2) (VG)2 = (vBC)2 rG>IC;   (VG)2 = (vBC)2 ( 23 )

Then for rod AB,

(VB)2 = (vAB)2 rAB;     (vBC)2(2) = (vAB)2(2) (vAB)2 = (vBC)2 Thus, T2 = =

1 1 1 I (v )2 + IG(vBC)22 + mr(VG)22 2 A AB 2 2 2 1 1 1 1 1 c (12) ( 22 ) d (vBC)22 + c (12) ( 22 ) d (vBC)22 + (12)[(vBC)2 13]2 2 3 2 12 2

= 28.0(vBC)22

Potential Energy. With reference to the datum set in Fig. b, the initial and final gravitational potential energy of the system are (Vg)1 = 2mgy1 = 2[12(9.81)(1 sin 60°)] = 203.90 J (Vg)2 = 2mgy2 = 2[12(9.81)(1 sin 30°)] = 117.72 J The stretch of the spring when the system is at initial and final position are x1 = 2(2 cos 60°) - 1.5 = 0.5 m x2 = 2(2 cos 30°) - 1.5 = 1.9641 m Thus, the initial and final elastic potential energy of the spring are (Ve)1 =

1 2 1 kx1 = (20) ( 0.52 ) = 2.50 J 2 2

(Ve)2 =

1 2 1 kx2 = (20) ( 1.96412 ) = 38.58 J 2 2

966

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18–53.  Continued

Conservation of Energy. T1 + V1 = T2 + V2 0 + (203.90 + 2.50) = 28.0(vBC)22 + (117.72 + 38.58) Ans.

vBC = 1.3376 rad>s = 1.34 rad>s

Ans: vBC = 1.34 rad>s 967

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18–54. If the 250-lb block is released from rest when the spring is unstretched, determine the velocity of the block after it has descended 5 ft. The drum has a weight of 50 lb and a radius of gyration of kO = 0.5 ft about its center of mass O.

0.375 ft k  75 lb/ft 0.75 ft O

SOLUTION Potential Energy: With reference to the datum shown in Fig. a, the gravitational potential energy of the system when the block is at position 1 and 2 is (Vg)1 = W(yG)1 = 250(0) = 0

(Vg)2 = - W(yG)2 = - 250(5) = - 1250 ft # lb When the block descends sb = 5 ft, the drum rotates through an angle of sb 5 u = = = 6.667 rad. Thus, the stretch of the spring is x = s + s0 = rb 0.75 rspu + 0 = 0.375(6.667) = 2.5 ft. The elastic potential energy of the spring is (Ve)2 =

1 2 1 kx = (75)(2.52) = 234.375 ft # lb 2 2

Since the spring is initially unstretched, (Ve)1 = 0. Thus, V1 = (Vg)1 + (Ve)1 = 0 V2 = (Vg)2 + (Ve)2 = - 1250 + 234.375 = - 1015.625 ft # lb Kinetic Energy: Since the drum rotates about a fixed axis passing through point O, vb vb = 1.333vb. The mass moment of inertia of the drum about its mass v = = rb 0.75 50 center is IO = mkO 2 = a 0.52 b = 0.3882 slug # ft2. 32.2

T = =

1 1 I v2 + mbvb2 2 O 2 1 1 250 (0.3882)(1.333vb)2 + a bv 2 2 2 32.2 b

= 4.2271vb2 Since the system is initially at rest, T1 = 0. T1 + V1 = T2 + V2 0 + 0 = 4.2271vb 2 - 1015.625 vb = 15.5 ft>s

Ans.

T

Ans: vb = 15.5 ft>s 968

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18–55. The slender 15-kg bar is initially at rest and standing in the vertical position when the bottom end A is displaced slightly to the right. If the track in which it moves is smooth, determine the speed at which end A strikes the corner D. The bar is constrained to move in the vertical plane. Neglect the mass of the cord BC.

C

5m

Solution 2

B 2

2

x + y = 5

x2 + (7 - y)2 = 42 Thus,

y = 4.1429 m



x = 2.7994 m

4m

(5)2 = (4)2 + (7)2 - 2(4)(7) cos f f = 44.42°

A

h2 = (2)2 + (7)2 - 2(2)(7) cos 44.42°

D 4m

h = 5.745 m T1 + V1 = T2 + V2 0 + 147.15(2) = v = 0.5714 rad>s

1 1 1 7 - 4.1429 c (15)(4)2 d v2 + (15)(5.745v)2 + 147.15 a b 2 12 2 2 Ans.

vA = 0.5714(7) = 4.00 m>s

Ans: vA = 4.00 m>s 969

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*18–56. If the chain is released from rest from the position shown, determine the angular velocity of the pulley after the end B has risen 2 ft. The pulley has a weight of 50 lb and a radius of gyration of 0.375 ft about its axis. The chain weighs 6 lb/ft.

0.5 ft

4 ft

SOLUTION Potential Energy: (yG1)1 = 2 ft, (yG 2)1 = 3 ft, (yG1)2 = 1 ft, and (yG2)2 = 4 ft. With reference to the datum in Fig. a, the gravitational potential energy of the chain at position 1 and 2 is V1 = (Vg)1 = W1(yG1)1 - W2(yG2)1

6 ft B

A

= -6(4)(2) - 6(6)(3) = - 156 ft # lb

V2 = (Vg)2 = - W1(yG1)2 + W2(yG2)2

= -6(2)(1) - 6(8)(4) = - 204 ft # lb

Kinetic Energy: Since the system is initially at rest, T1 = 0. The pulley rotates about a fixed axis, thus, (VG1)2 = (VG2)2 = v2 r = v2(0.5). The mass moment of inertia of 50 the pulley about its axis is IO = mkO 2 = (0.3752) = 0.2184 slug # ft2. Thus, the 32.2 final kinetic energy of the system is T = =

1 1 1 I v 2 + m1(VG1)2 2 + m2 (VG2)2 2 2 O 2 2 2 1 1 6(2) 1 6(8) 1 6(0.5)(p) (0.2184)v2 2 + c d [v2(0.5)]2 + c d[v2(0.5)]2 + c d[v2(0.5)]2 2 2 32.2 2 32.2 2 32.2

= 0.3787v2 2 Conservation of Energy: T1 + V1 = T2 + V2 0 + (- 156) = 0.3787v22 = (- 204) v2 = 11.3 rad >s

Ans.

Ans: v2 = 11.3 rad>s 970

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18–57. If the gear is released from rest, determine its angular velocity after its center of gravity O has descended a distance of 4 ft. The gear has a weight of 100 lb and a radius of gyration about its center of gravity of k = 0.75 ft.

1 ft O

SOLUTION Potential Energy: With reference to the datum in Fig. a, the gravitational potential energy of the gear at position 1 and 2 is V1 = (Vg)1 = W(y0)1 = 100(0) = 0

V2 = (Vg)2 = - W1(y0)2 = - 100(4) = - 400 ft # lb Kinetic Energy: Referring to Fig. b, we obtain vO = vrO / IC = v(1).The mass moment of inertia of the gear about its mass center is IO = mkO 2 = Thus, T = =

100 (0.752) = 1.7469 kg # m2. 32.2

1 1 mvO2 + IOv2 2 2 1 100 1 a b [v (1)]2 + (1.7469)v2 2 32.2 2

= 2.4262v2 Since the gear is initially at rest, T1 = 0. Conservation of Energy: T1 + V1 = T2 + V2 0 + 0 = 2.4262v2 - 400 Ans.

v = 12.8 rad>s

Ans: v = 12.8 rad>s 971

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18–58. The slender 6-kg bar AB is horizontal and at rest and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the bar is momentarily stopped when it has rotated clockwise 90° after being released.

C

k

A

Solution Kinetic Energy. The mass moment of inertia of the bar about A is 1 IA = (6) ( 22 ) + 6 ( 12 ) = 8.00 kg # m2. Then 12 1 1 T = IA v2 = (8.00) v2 = 4.00 v2 2 2

1.5 m

B

2m

Since the bar is at rest initially and required to stop finally, T1 = T2 = 0. Potential Energy. With reference to the datum set in Fig. a, the gravitational potential energies of the bar when it is at positions ➀ and ➁ are (Vg)1 = mgy1 = 0 (Vg)2 = mgy2 = 6(9.81)( - 1) = -58.86 J The stretch of the spring when the bar is at position ➁ is x2 = 222 + 3.52 - 1.5 = 2.5311 m

Thus, the initial and final elastic potential energy of the spring are (Ve)1 =

1 2 kx1 = 0 2

(Ve)2 =

1 k ( 2.53112 ) = 3.2033k 2

Conservation of Energy. T1 + V1 = T2 + V2 0 + (0 + 0) = 0 + ( - 58.86) + 3.2033k k = 18.3748 N>m = 18.4 N>m Ans.

Ans: k = 18.4 N>m 972

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18–59. The slender 6-kg bar AB is horizontal and at rest and the spring is unstretched. Determine the angular velocity of the bar when it has rotated clockwise 45° after being released. The spring has a stiffness of k = 12 N>m.

C

k

A

Solution Kinetic Energy. The mass moment of inertia of the bar about A is 1 IA = (6) ( 22 ) + 6 ( 12 ) = 8.00 kg # m2. Then 12 T =

1.5 m

B

2m

1 1 I v2 = (8.00) v2 = 4.00 v2 2A 2

Since the bar is at rest initially, T1 = 0. Potential Energy. with reference to the datum set in Fig. a, the gravitational potential energies of the bar when it is at positions ① and ② are (Vg)1 = mgy1 = 0 (Vg)2 = mgy2 = 6(9.81)( - 1 sin 45°) = - 41.62 J From the geometry shown in Fig. a, -1 a = 222 + 1.52 = 2.5 m  f = tan a

Then, using cosine law,

1.5 b = 36.87° 2

l = 22.52 + 22 - 2(2.5)(2) cos (45° + 36.87°) = 2.9725 m

Thus, the stretch of the spring when the bar is at position ② is x2 = 2.9725 - 1.5 = 1.4725 m

Thus, the initial and final elastic potential energies of the spring are (Ve)1 =

1 2 kx = 0 2 1

(Ve)2 =

1 (12) ( 1.47252 ) = 13.01 J 2

Conservation of Energy. T1 + V1 = T2 + V2 0 + (0 + 0) = 4.00 v2 + ( - 41.62) + 13.01 v = 2.6744 rad>s = 2.67 rad>s 

Ans.

Ans: v = 2.67 rad>s 973

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*18–60. The pendulum consists of a 6-kg slender rod fixed to a 15-kg disk. If the spring has an unstretched length of 0.2  m, determine the angular velocity of the pendulum when it is released from rest and rotates clockwise 90° from the position shown. The roller at C allows the spring to always remain vertical.

C

0.5 m

k  200 N/m A

B

0.5 m

Solution

0.5 m

D

0.3 m

Kinetic Energy. The mass moment of inertia of the pendulum about B is 1 1 IB = c (6) ( 12 ) + 6 ( 0.52 ) d + c (15) ( 0.32 ) + 15 ( 1.32 ) d = 28.025 kg # m2. Thus 12 2 T =

1 1 I v2 = (28.025) v2 = 14.0125 v2 2B 2

Since the pendulum is released from rest, T1 = 0. Potential Energy. with reference to the datum set in Fig. a, the gravitational potential energies of the pendulum when it is at positions ① and ② are (Vg)1 = mrg(yr)1 + mdg(yd)1 = 0 (Vg)2 = mrg(yr)2 + mdg(yd)2 = 6(9.81)( - 0.5) + 15(9.81)( - 1.3) = -220.725 J The stretch of the spring when the pendulum is at positions ① and ② are x1 = 0.5 - 0.2 = 0.3 m x2 = 1 - 0.2 = 0.8 m Thus, the initial and final elastic potential energies of the spring are (Ve)1 =

1 2 1 kx = (200) ( 0.32 ) = 9.00 J 2 1 2

(Ve)2 =

1 2 1 kx = (200) ( 0.82 ) = 64.0 J 2 2 2

Conservation of Energy. T1 + V1 = T2 + V2 0 + (0 + 9.00) = 14.0125v2 + ( - 220.725) + 64.0 v = 3.4390 rad>s = 3.44 rad>s

Ans.

Ans: v = 3.44 rad>s 974

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18–61. The 500-g rod AB rests along the smooth inner surface of a hemispherical bowl. If the rod is released from rest from the position shown, determine its angular velocity at the instant it swings downward and becomes horizontal.

200 mm B

SOLUTION

A

200 mm

Select datum at the bottom of the bowl. u = sin - 1 a

0.1 b = 30° 0.2

h = 0.1 sin 30° = 0.05 CE = 2(0.2)2 - (0.1)2 = 0.1732 m ED = 0.2 - 0.1732 = 0.02679 T1 + V1 = T2 + V2 0 + (0.5)(9.81)(0.05) =

1 1 1 c (0.5)(0.2)2 d v2AB + (0.5)(vG)2 + (0.5)(9.81)(0.02679) 2 12 2

Since vG = 0.1732vAB Ans.

vAB = 3.70 rad>s

Ans: vAB = 3.70 rad>s 975

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18–62. The 50-lb wheel has a radius of gyration about its center of gravity G of kG = 0.7 ft. If it rolls without slipping, determine its angular velocity when it has rotated clockwise 90° from the position shown. The spring AB has a stiffness k = 1.20 lb/ft and an unstretched length of 0.5 ft. The wheel is released from rest.

3 ft 0.5 ft

B

0.5 ft G

k = 1.20 lb/ft

A

1 ft

SOLUTION T1 + V1 = T2 + V2 0 +

1 1 50 1 50 (1.20)[2(3)2 + (0.5)2 - 0.5]2 = [ (0.7)2]v2 + ( )(1v)2 2 2 32.2 2 32.2 +

1 (1.20)(0.9292 - 0.5)2 2 Ans.

v = 1.80 rad s

Ans: v = 1.80 rad>s 976

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18–63. The system consists of 60-lb and 20-lb blocks A and B, respectively, and 5-lb pulleys C and D that can be treated as thin disks. Determine the speed of block A after block B has risen 5 ft, starting from rest. Assume that the cord does not slip on the pulleys, and neglect the mass of the cord.

C 0.5 ft

A

SOLUTION

0.5 ft

D

Kinematics: The speed of block A and B can be related using the position coordinate equation. (1)

sA + 2sB = l ¢sA + 2¢sB = 0

¢sA + 2(5) = 0

B

¢sA = -10 ft = 10 ftT

Taking time derivative of Eq. (1), we have yA + 2yB = 0

yB = - 0.5yA

Potential Energy: Datum is set at fixed pulley C.When blocks A and B (pulley D) are at their initial position, their centers of gravity are located at sA and sB. Their initial gravitational potential energies are - 60sA, - 20sB, and -5sB. When block B (pulley D) rises 5 ft, block A decends 10 ft. Thus, the final position of blocks A and B (pulley D) are (sA + 10) ft and (sB - 5) ft below datum. Hence, their respective final gravitational potential energy are -60(sA + 10), -20(sB - 5), and -5(sB - 5). Thus, the initial and final potential energy are V1 = - 60sA - 20sB - 5sB = -60sA - 25sB V2 = -60(sA + 10) - 20(sB - 5) - 5(sB - 5) = - 60sA - 25sB - 475 Kinetic Energy: The mass moment inertia of the pulley about its mass center is 1 5 IG = a b A 0.52 B = 0.01941 slug # ft2. Since pulley D rolls without slipping, 2 32.2 yB yB vD = = = 2yB = 2( - 0.5yA) = - yA. Pulley C rotates about the fixed point rD 0.5 yA yA hence vC = = = 2yA. Since the system is at initially rest, the initial kinectic rC 0.5 energy is T1 = 0. The final kinetic energy is given by T2 =

1 1 1 1 1 m y2A + mB y2B + mD y2B + IGv2D + IG v2C 2 A 2 2 2 2

=

1 60 1 20 1 5 a b y2A + a b ( -0.5yA)2 + a b( -0.5yA)2 2 32.2 2 32.2 2 32.2

+

1 1 (0.01941)( -yA)2 + (0.01941)(2yA)2 2 2

= 1.0773y2A Conservation of Energy: Applying Eq. 18–19, we have T1 + V1 = T2 + V2 0 + (-60sA - 25sB) = 1.0773y2A + (- 60sA - 25sB - 475) Ans.

y4 = 21.0 ft>s

977

Ans: vA = 21.0 ft>s

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*18–64. The door is made from one piece, whose ends move along the horizontal and vertical tracks. If the door is in the open position, u = 0°, and then released, determine the speed at which its end A strikes the stop at C. Assume the door is a 180-lb thin plate having a width of 10 ft.

A

C 5 ft

Solution

3 ft

u

B

T1 + V1 = T2 + V2 0 + 0 =

1 1 180 1 180 c a b(8)2 d v2 + a b(1v)2 - 180(4) 2 12 32.2 2 32.2

v = 6.3776 rad>s

vc = v(5) = 6.3776(5) = 31.9 m>s

Ans.

Ans: vc = 31.9 m>s 978

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18–65. The door is made from one piece, whose ends move along the horizontal and vertical tracks. If the door is in the open position, u = 0°, and then released, determine its angular velocity at the instant u = 30°.  Assume the door is a 180-lb thin plate having a width of 10 ft.

A

C 5 ft

Solution

3 ft

u

B

T1 + V1 = T2 + V2 0 + 0 =

1 1 180 1 180 2 c a b(8)2 d v2 + a bv - 180(4 sin 30°) 2 12 32.2 2 32.2 G

(1)

rIC - G = 2(1)2 + (4.3301)2 - 2(1)(4.3301) cos 30°

rIC - G = 3.50 m Thus, vG = 3.50 v

Substitute into Eq. (1) and solving, Ans.

v = 2.71 rad>s

Ans: v = 2.71 rad>s 979

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18–66. The end A of the garage door AB travels along the horizontal track, and the end of member BC is attached to a spring at C. If the spring is originally unstretched, determine the stiffness k so that when the door falls downward from rest in the position shown, it will have zero angular velocity the moment it closes, i.e., when it and BC become vertical. Neglect the mass of member BC and assume the door is a thin plate having a weight of 200 lb and a width and height of 12 ft. There is a similar connection and spring on the other side of the door.

12 ft

A

B

15

7 ft

C 2 ft

SOLUTION

D

6 ft 1 ft

(2)2 = (6)2 + (CD)2 - 2(6)(CD) cos 15° CD2 - 11.591CD + 32 = 0 Selecting the smaller root: CD = 4.5352 ft T1 + V1 = T2 + V2 1 0 + 0 = 0 + 2 c (k)(8 - 4.5352)2 d - 200(6) 2 Ans.

k = 100 lb/ft

Ans: k = 100 lb>ft 980

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18–67. The system consists of a 30-kg disk, 12-kg slender rod BA, and a 5-kg smooth collar A. If the disk rolls without slipping, determine the velocity of the collar at the instant u = 0°. The system is released from rest when u = 45°.

A

2m

Solution Kinetic Energy. Since the system is released from rest, T1 = 0. Referring to the kinematics diagram of the rod at its final position, Fig. a, we found that IC is located at B. Thus, (vB)2 = 0. Also (vA)2 = (vr)2rA>IC;  (vA)2 = (vr)2(2)  (vr)2 =

(vA)2

u

B 30

0.5 m

2

C

Then (vGr)2 = (vr)2(rGr>IC);  (vGr)2 =

(vA)2 2

(1) =

(vA)2 2

For the disk, since the velocity of its center (vB)2 = 0, (vd)2 = 0. Thus, T2 = =

1 1 1 m (v )2 + IGr(vr)22 + mc(vA)22 2 r Gr 2 2 2 (vA)2 2 (vA)2 2 1 1 1 1 (12) c d + c (12) ( 22 ) d c d + (5)(vA)22 2 2 2 12 2 2

= 4.50(vA)22

Potential Energy. Datum is set as shown in Fig. a. Here, SB = 2 - 2 cos 45° = 0.5858 m Then (yd)1 = 0.5858 sin 30° = 0.2929 m (yr)1 = 0.5858 sin 30° + 1 sin 75° = 1.2588 m (yr)2 = 1 sin 30° = 0.5 m (yc)1 = 0.5858 sin 30° + 2 sin 75° = 2.2247 m (yc)2 = 2 sin 30° = 1.00 m Thus, the gravitational potential energies of the disk, rod and collar at the initial and final positions are (Vd)1 = md g(yd)1 = 30(9.81)(0.2929) = 86.20 J (Vd)2 = md g(yd)2 = 0 (Vr)1 = mr g(yr)1 = 12(9.81)(1.2588) = 148.19 J (Vr)2 = mr g(yr)2 = 12(9.81)(0.5) = 58.86 J (Vc)1 = mc g(yc)1 = 5(9.81)(2.2247) = 109.12 J (Vc)2 = mc g(yc)2 = 5(9.81)(1.00) = 49.05 J

981

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18–67.  Continued

Conservation of Energy. T1 + V1 = T2 + V2 0 + (86.20 + 148.19 + 109.12) = 4.50(vA)22 + (0 + 58.86 + 49.05) (vA)2 = 7.2357 m>s = 7.24 m>s

Ans.

Ans: ( vA ) 2 = 7.24 m>s 982

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*18–68. The system consists of a 30-kg disk A, 12-kg slender rod BA, and a 5-kg smooth collar A. If the disk rolls without slipping, determine the velocity of the collar at the instant u = 30°. The system is released from rest when u = 45°.

A

2m

Solution u

Kinetic Energy. Since the system is released from rest, T1 = 0. Referring to the kinematics diagram of the rod at final position with IC so located, Fig. a, rA>IC = 2 cos 30° = 1.7321 m  rB>IC = 2 cos 60° = 1.00 m rGr>IC = 212 + 1.002 - 2(1)(1.00) cos 60° = 1.00 m

B 30

0.5 m

Then

C

(vA)2 = (vr)2(rA>IC);  (vA)2 = (vr)2(1.7321)  (vr)2 = 0.5774(vA)2 (vB)2 = (vr)2(rB>IC);  (vB)2 = [0.5774(vA)2](1.00) = 0.5774(vA)2 (vGr)2 = (vr)2(rGr>IC); 

(vGr)2 = [0.5774(vA)2](1.00) = 0.5774(vA)2

Since the disk rolls without slipping, (vB)2 = vdrd;  0.5774(vA)2 = (vd)2(0.5) (vd)2 = 1.1547(vA)2 Thus, the kinetic energy of the system at final position is T2 = =

1 1 1 1 1 m (v )2 + IGr(vr)22 + md(vB)22 + IB(vd)22 + mc(vA)22 2 r Gr 2 2 2 2 2 1 1 1 (12)[0.5774(vA)2]2 + c (12) ( 22 ) d [0.5774(vA)2]2 2 2 12 +

1 1 1 (3.0)[0.5774(vA)2]2 + c (30) ( 0.52 ) d [1.1547(vA)2]2 2 2 2 +

1 (5)(vA)22 2

= 12.6667(vA)22 Potential Energy. Datum is set as shown in Fig. a. Here, SB = 2 cos 30° - 2 cos 45° = 0.3178 m Then (yd)1 = 0.3178 sin 30° = 0.1589 m (yr)1 = 0.3178 sin 30° + 1 sin 75° = 1.1248 m (yr)2 = 1 sin 60° = 0.8660 m (yc)1 = 0.3178 sin 30° + 2 sin 75° = 2.0908 m (yc)2 = 2 sin 60° = 1.7321 m

983

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*18–68.  Continued

Thus, the gravitational potential energies of the disk, rod and collar at initial and final position are (Vd)1 = mdg(yd)1 = 30(9.81)(0.1589) = 46.77 J (Vd)2 = mdg(yd)2 = 0 (Vr)1 = mrg(yr)1 = 12(9.81)(1.1248) = 132.42 J (Vr)2 = mrg(yr)2 = 12(9.81)(0.8660) = 101.95 J (Vc)1 = mcg(yc)1 = 5(9.81)(2.0908) = 102.55 J (Vc)2 = mcg(yc)2 = 5(9.81)(1.7321) = 84.96 J Conservation of Energy. T1 + V1 = T2 + V2 0 + (46.77 + 132.42 + 102.55) = 12.6667(vA)22 + (0 + 101.95 + 84.96) Ans.

(vA)2 = 2.7362 m>s = 2.74 m>s

Ans: (vA)2 = 2.74 m>s 984

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19–1. mvG

The rigid body (slab) has a mass m and rotates with an angular velocity V about an axis passing through the fixed point O. Show that the momenta of all the particles composing the body can be represented by a single vector having a magnitude mvG and acting through point P, called the center of percussion, which lies at a distance rP>G = k2G>rG>O from the mass center G. Here kG is the radius of gyration of the body, computed about an axis perpendicular to the plane of motion and passing through G.

V vG rG/O O

SOLUTION HO = (rG>O + rP>G) myG = rG>O (myG) + IG v,

where IG = mk2G

rG>O (myG) + rP>G (myG) = rG>O (myG) + (mk2G) v rP>G =

k2G yG>v

However, yG = vrG>O rP>G =

or rG>O =

yG v

k2G rG>O

Q.E.D.

985

rP/G G

P

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19–2. At a given instant, the body has a linear momentum L = mvG and an angular momentum H G = IG V computed about its mass center. Show that the angular momentum of the body computed about the instantaneous center of zero velocity IC can be expressed as H IC = IIC V , where IIC represents the body’s moment of inertia computed about the instantaneous axis of zero velocity. As shown, the IC is located at a distance rG>IC away from the mass center G.

mvG G

rG/IC

SOLUTION HIC = rG>IC (myG) + IG v,

IC

where yG = vrG>IC

= rG>IC (mvrG>IC) + IG v = (IG + mr2G>IC) v Q.E.D.

= IIC v

986

IGV

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19–3. Show that if a slab is rotating about a fixed axis perpendicular to the slab and passing through its mass center G, the angular momentum is the same when computed about any other point P.

V P

G

SOLUTION Since yG = 0, the linear momentum L = myG = 0. Hence the angular momentum about any point P is HP = IG v Since v is a free vector, so is H P.

Q.E.D.

987

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*19–4. The 40-kg disk is rotating at V = 100 rad>s. When the force P is applied to the brake as indicated by the graph. If the coefficient of kinetic friction at B is mk = 0.3, determine the time t needed to stay the disk from rotating. Neglect the thickness of the brake.

P (N) 500

P

t (s)

2

300 mm

300 mm B

Solution Equilibrium. Since slipping occurs at brake pad, Ff = mkN = 0.3 N. Referring to the FBD the brake’s lever, Fig. a,

150 mm

200 mm

V O

A

a+ ΣMA = 0;  N(0.6) - 0.3 N(0.2) - P(0.3) = 0

N = 0.5556 P

Thus, Ff = 0.3(0.5556 P) = 0.1667 P Principle of Impulse and Momentum. The mass moment of inertia of the disk about 1 1 its center O is IO = mr 2 = (40) ( 0.152 ) = 0.45 kg # m2. 2 2 IOv1 + Σ

Lt1

t2

MOdt = IOv2

It is required that v2 = 0. Assuming that t 7 2 s, 0.45(100) + 0.025 L0

L0

L0

t

[ - 0.1667 P(0.15)]dt = 0.45(0)

t

P dt = 45

t

P dt = 1800

1 (500)(2) + 500(t - 2) = 1800 2

Ans.

t = 4.60 s

Since t 7 2 s, the assumption was correct.

Ans: t = 4.60 s 988

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19–5. The impact wrench consists of a slender 1-kg rod AB which is 580 mm long, and cylindrical end weights at A and B that each have a diameter of 20 mm and a mass of 1 kg. This assembly is free to turn about the handle and socket, which are attached to the lug nut on the wheel of a car. If the rod AB is given an angular velocity of 4 rad>s and it strikes the bracket C on the handle without rebounding, determine the angular impulse imparted to the lug nut.

C B

Iaxle = L

300 mm

A

SOLUTION

300 mm

1 1 (1)(0.6 - 0.02)2 + 2 c (1)(0.01)2 + 1(0.3)2 d = 0.2081 kg # m2 12 2

Mdt = Iaxle v = 0.2081(4) = 0.833 kg # m2>s

Ans.

Ans: L 989

M dt = 0.833 kg # m2 >s

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19–6. The airplane is traveling in a straight line with a speed of 300 km> h, when the engines A and B produce a thrust of TA = 40 kN and TB = 20 kN, respectively. Determine the angular velocity of the airplane in t = 5 s. The plane has a mass of 200 Mg, its center of mass is located at G, and its radius of gyration about G is kG = 15 m.

TA  40 kN 8 m A G B TB  20 kN

SOLUTION

8m

Principle of Angular Impulse and Momentum: The mass moment of inertia of the airplane about its mass center is IG = mkG2 = 200 A 103 B A 152 B = 45 A 106 B kg # m2. Applying the angular impulse and momentum equation about point G, t2

Izv1 + ©

Lt1

MGdt = IGv2

0 + 40 A 103 B (5)(8) - 20 A 103 B (5)(8) = 45 A 106 B v Ans.

v = 0.0178 rad>s

Ans: v = 0.0178 rad>s 990

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19–7. The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass of 15 kg and a radius of gyration of kO = 110 mm. If the block at A has a mass of 40 kg, determine the speed of the block in 3 s after a constant force of 2 kN is applied to the rope wrapped around the inner hub of the pulley. The block is originally at rest.

2 kN 200 mm 75 mm O

SOLUTION Principle of Impulse and Momentum: The mass moment inertia of the pulley about

point O is IO = 15 A 0.112 B = 0.1815 kg # m2. The angular velocity of the pulley and yB = 5yB. Applying Eq. 19–15, the velocity of the block can be related by v = 0.2 we have a a syst. angular momentumb

O1

+ a a syst. angular impulseb

O1-2

= a a syst. angular momentumb (a + )

A

O2

0 + [40(9.81)(3)](0.2) - [2000(3)](0.075) = - 40yB(0.2) - 0.1815(5yB) Ans.

yB = 24.1 m>s

Ans: vB = 24.1 m>s 991

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*19–8. The assembly weighs 10 lb and has a radius of gyration kG = 0.6 ft about its center of mass G. The kinetic energy of the assembly is 31 ft # lb when it is in the position shown. If it is rolling counterclockwise on the surface without slipping, determine its linear momentum at this instant.

0.8 ft G

1 ft

1 ft

SOLUTION IG = (0.6)2 a T =

10 b = 0.1118 slug # ft2 32.2

1 1 10 a b v 2 + (0.1118) v2 = 31 2 32.2 G 2

(1)

vG = 1.2 v Substitute into Eq. (1), v = 10.53 rad>s vG = 10.53(1.2) = 12.64 ft>s L = mvG =

10 (12.64) = 3.92 slug # ft>s 32.2

Ans.

Ans: L = 3.92 slug # ft>s 992

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19–9. The disk has a weight of 10 lb and is pinned at its center O. If a vertical force of P = 2 lb is applied to the cord wrapped around its outer rim, determine the angular velocity of the disk in four seconds starting from rest. Neglect the mass of the cord.

0.5 ft O

SOLUTION t2

(c + )

IO v1 + ©

Lt1

P

MO dt = IOv2

0 + 2(0.5)(4) = c

1 10 a b (0.5)2 dv2 2 32.2 Ans.

v2 = 103 rad>s

Ans: v2 = 103 rad>s 993

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19–10. The 30-kg gear A has a radius of gyration about its center of mass O of kO = 125 mm. If the 20-kg gear rack B is subjected to a force of P = 200 N, determine the time required for the gear to obtain an angular velocity of 20 rad>s, starting from rest. The contact surface between the gear rack and the horizontal plane is smooth.

0.15 m O

B

A

P  200 N

SOLUTION Kinematics: Since the gear rotates about the fixed axis, the final velocity of the gear rack is required to be (vB)2 = v2rB = 20(0.15) = 3 m>s : Principle of Impulse and Momentum: Applying the linear impulse and momentum equation along the x axis using the free-body diagram of the gear rack shown in Fig. a, + B A:

t2

m(vB)1 + ©

Fxdt = m(vB)2 Lt1 0 + 200(t) - F(t) = 20(3) (1)

F(t) = 200t - 60

The mass moment of inertia of the gear about its mass center is IO = mkO2 = 30(0.1252) = 0.46875 kg # m2. Writing the angular impulse and momentum equation about point O using the free-body diagram of the gear shown in Fig. b, t2

IOv1 + ©

MOdt = IOv2 Lt1 0 + F(t)(0.15) = 0.46875(20) (2)

F(t) = 62.5 Substituting Eq. (2) into Eq. (1) yields

Ans.

t = 0.6125 s

Ans: t = 0.6125 s 994

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19–11. The pulley has a weight of 8 lb and may be treated as a thin disk. A cord wrapped over its surface is subjected to forces TA = 4 lb and TB = 5 lb. Determine the angular velocity of the pulley when t = 4 s if it starts from rest when t = 0. Neglect the mass of the cord.

0.6 ft

SOLUTION Principle of Impulse and Momentum: The mass moment inertia of the pulley about its 8 1 mass center is Io = a b (0.62) = 0.04472 slug # ft2. Applying Eq. 19–14, we have 2 32.2 t

IOv1 + © 1t12Modt = Iov2 (a +)

TB = 5 lb

TA = 4 lb

0 + [5(4)](0.6) - [4(4)](0.6) = 0.04472v2 Ans.

v2 = 53.7 rad>s

Ans: v2 = 53.7 rad>s 995

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*19–12. The 40-kg roll of paper rests along the wall where the coefficient of kinetic friction is mk = 0.2. If a vertical force of P = 40 N is applied to the paper, determine the angular velocity of the roll when t = 6 s starting from rest. Neglect the mass of the unraveled paper and take the radius of gyration of the spool about the axle O to be kO = 80 mm.

B

P  40 N

13

12 5

A

Solution

O 120 mm

Principle of Impulse and Momentum. The mass moment of inertia of the paper roll about its center is IO = mk 2O = 40 ( 0.082 ) = 0.256 kg # m2. Since the paper roll is required to slip at point of contact, Ff = mkN = 0.2 N. Referring to the FBD of the paper roll, Fig. a, Lt1

+ )  m[(vO)x]1 + Σ (S

(+ c)

t2

Fx dt = m[(vO)x]2

0 + N(6) - FABa FAB =

m[(vO)y]1 + Σ

Lt1

5 b(6) = 0 13

13 N(1) 5

t2

Fy dt = m[(vO)y]2

0 + 0.2 N(6) + FABa

12 b(6) + 40(6) - 40(9.81)(6) = 0 13

0.2 N +



12 F = 352.4(2) 13 AB

Solving Eqs. (1) and (2) N = 135.54 N  FAB = 352.4 N Subsequently a+   IO v1 + Σ MO dt = IO v2 L 0.256(0) + 40(0.12)(6) - 0.2(135.54)(0.12)(6) = 0.256v

Ans.

v = 36.26 rad>s = 36.3 rad>s

Ans: v = 36.3 rad>s 996

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19–13. The slender rod has a mass m and is suspended at its end A by a cord. If the rod receives a horizontal blow giving it an impulse I at its bottom B, determine the location y of the point P about which the rod appears to rotate during the impact.

A

P

SOLUTION

l y

Principle of Impulse and Momentum: t2

IG v1 + ©

(a + )

MG dt = IG v2

Lt1

l 1 0 + I a b = c ml2 d v 2 12 + b a:

I

1 I = mlv 6

B

t2

m(yAx)1 + © 0 +

Lt1

Fx dt = m(yAx)2

1 mlv = mvG 6

yG =

l v 6

Kinematics: Point P is the IC. yB = v y Using similar triangles, vy = y

l v 6 y -

l 2

y =

2 l 3

Ans.

Ans: y = 997

2 l 3

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19–14. The rod of length L and mass m lies on a smooth horizontal surface and is subjected to a force P at its end A as shown. Determine the location d of the point about which the rod begins to turn, i.e, the point that has zero velocity.

P L A

d

Solution L

+ )   m(vGx)1 + Σ Fx dt = m(vGx)2 (S 0 + P(t) = m(vG)x (+ c)

m(vGy)1 + Σ Fy dt = m(vGy)2 L 0 + 0 = m(vG)y

(a+)   (HG)1 + Σ MG dt = (HG)2 L L 1 0 + P(t)a b = mL2v 2 12 vG = yv

L 1 m(vG)x a b = mL2v 2 12 (vG)x =

L v 6

y =

L 6

d =

L L 2 + = L 2 6 3

Ans.

Ans: d = 998

2 l 3

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19–15. z

A 4-kg disk A is mounted on arm BC, which has a negligible mass. If a torque of M = 15e0.5t2 N # m, where t is in seconds, is applied to the arm at C, determine the angular velocity of BC in 2 s starting from rest. Solve the problem assuming that (a) the disk is set in a smooth bearing at B so that it rotates with curvilinear translation, (b) the disk is fixed to the shaft BC, and (c) the disk is given an initial freely spinning angular velocity of V D = 5 -80k6 rad>s prior to application of the torque.

250 mm A 60 mm

B M C

#

(5e0.5 t) N m

SOLUTION a) (Hz)1 + ©

L

Mz dt = (Hz)2

2

0+

L0

5e0.5 tdt = 4(vB)(0.25)

5 0.5 t 2 e 2 = vB 0.5 0 vB = 17.18 m>s Thus, 17.18 = 68.7 rad>s 0.25

vBC =

Ans.

b) (Hz)1 + © 2

0 +

L0

L

Mz dt = (Hz)2

1 5e0.5 tdt = 4(vB)(0.25) + c (4)(0.06)2 d vBC 2

Since vB = 0.25 vBC, then Ans.

vBC = 66.8 rad>s c) (Hz)1 + ©

L

Mz dt = (Hz)2

2

1 1 - c (4)(0.06)2 d(80) + 5e0.5 tdt = 4(vB)(0.25) - c (4)(0.06)2 d(80) 2 2 L0 Since vB = 0.25 vBC, Ans.

vBC = 68.7 rad>s

Ans: (a) vBC = 68.7 rad>s (b) vBC = 66.8 rad>s (c) vBC = 68.7 rad>s 999

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*19–16. The frame of a tandem drum roller has a weight of 4000 lb excluding the two rollers. Each roller has a weight of 1500 lb and a radius of gyration about its axle of 1.25 ft. If a torque of M = 300 lb # ft is supplied to the rear roller A, determine the speed of the drum roller 10 s later, starting from rest. 1.5 ft

SOLUTION

1.5 ft A

Principle of Impulse and Momentum: The mass moments of inertia of the rollers 1500 about their mass centers are IC = ID = A 1.252 B = 72.787 slug # ft2. Since the 32.2 v v = 0.6667v. Using the free-body = rollers roll without slipping, v = r 1.5 diagrams of the rear roller and front roller, Figs. a and b, and the momentum diagram of the rollers, Fig. c,

M

B 300 lb ft

t2

(HA)1 + ©

MAdt = (HA)2

Lt1

0 + 300(10) - Cx(10)(1.5) =

1500 v(1.5) + 72.787(0.6667v) 32.2 (1)

Cx = 200 - 7.893v and t2

(HB)1 + ©

Lt1

MBdt = (HB)2 1500 v(1.5) + 72.787(0.6667v) 32.2

0 + Dx(10)(1.5) =

(2)

Dx = 7.893v Referring to the free-body diagram of the frame shown in Fig. d, + :

m C (vG)x D 1 + ©

t2

Lt1

Fxdt = m C (vG)x D 2

0 + Cx(10) - Dx(10) =

4000 v 32.2

(3)

Substituting Eqs. (1) and (2) into Eq. (3), (200 - 7.893v)(10) - 7.893v(10) =

4000 v 32.2 Ans.

v = 7.09 ft>s

Ans: v = 7.09 ft>s 1000

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19–17. The 100-lb wheel has a radius of gyration of kG = 0.75 ft. If the upper wire is subjected to a tension of T = 50 lb, determine the velocity of the center of the wheel in 3 s, starting from rest.The coefficient of kinetic friction between the wheel and the surface is mk = 0.1.

T 0.5 ft G

1 ft

SOLUTION Principle of Impulse and Momentum: We can eliminate the force F from the analysis if we apply the principle of impulse and momentum about point A. The 100 100 mass moment inertia of the wheel about point A is IA = A 0.752 B + A 0.52 B 32.2 32.2 = 2.523 slug # ft2. Applying Eq. 19–14, we have m A yGy B 1 + © (+ c)

t2

Lt1

Fy dt = m A yGy B 2

0 + N(t) - 100(t) = 0

N = 100 lb

t2

IA v1 + © (a + )

Lt1

MA dt = IA v2 [1]

0 + [50(3)](1) - [0.1(100)(3)](0.5) = 2.523 v2

Kinematics: Since the wheel rolls without slipping at point A, the instantaneous center of zero velocity is located at point A. Thus, yG = v2 rG/IC v2 =

yG yG = 2yG = rG/IC 0.5

[2]

Solving Eqs. [1] and [2] yields Ans.

yG = 26.8 ft>s v2 = 53.50 rad s

Ans: vG = 26.8 ft>s 1001

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19–18. The 4-kg slender rod rests on a smooth floor. If it is kicked so as to receive a horizontal impulse I = 8 N # s at point A as shown, determine its angular velocity and the speed of its mass center. 2m A

1.75 m

SOLUTION + 2 1;

60

m(vGx)1 + ©

L

Fx dt = m(vGx)2 I

8N s

0 + 8 cos 60° = 4(vG)x (vG)x = 1 m>s (+ c)

m(vGy)1 + ©

L

Fy dt = m(vGy)2

0 + 8 sin 60° = 4(vG)y (vG)y = 1.732 m>s vG = 2(1.732)2 + (1)2 = 2 m>s (a + )

(HG)1 + ©

L

Ans.

MG dt = (HG)2

0 + 8 sin 60°(0.75) =

1 (4)(2)2 v 12 Ans.

v = 3.90 rad s

Ans: vG = 2 m>s v = 3.90 rad>s 1002

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19–19. The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass of 15 kg and a radius of gyration kO = 110 mm. If the block at A has a mass of 40 kg, determine the speed of the block in 3 s after a constant force F = 2 kN is applied to the rope wrapped around the inner hub of the pulley. The block is originally at rest. Neglect the mass of the rope.

200 mm

(HO)1 + ©

75 mm

F A

SOLUTION (c + )

O

L

MO dt = (HO)2

0 + 2000(0.075)(3) - 40(9.81)(0.2)(3) = 15(0.110)2v + 40(0.2v) (0.2) v = 120.4 rad>s Ans.

vA = 0.2(120.4) = 24.1 m>s

Ans: vA = 24.1 m>s 1003

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*19–20. The 100-kg spool is resting on the inclined surface for which the coefficient of kinetic friction is mk = 0.1. Determine the angular velocity of the spool when t = 4 s after it is released from rest. The radius of gyration about the mass center is kG = 0.25 m. 0.4 m 0.2 m G

Solution

A

Kinematics. The IC of the spool is located as shown in Fig. a. Thus 30

vG = vrG>IC = v(0.2) Principle of Impulse and Momentum. The mass moment of inertia of the spool about its mass center is IG = mk 2G = 100 ( 0.252 ) = 6.25 kg # m2. Since the spool is required to slip, Ff = mkN = 0.1 N. Referring to the FBD of the spool, Fig. b, t2

a+   m[(vG)y]1 + Σ Fy dt = m[(vG)y]2 Lt1 0 + N(4) - 100(9.81) cos 30°(4) = 0 N = 849.57 N t2

Q   m[(vG)x]1 + Σ Fx dt = m[(vG)x]2 Lt1

+

0 + T(4) + 0.1(849.57)(4) - 100(9.81) sin 30°(4) = 100[ - v(0.2)] (1)

T + 5v = 405.54 t2

a+    IG v1 + Σ MG dt = IG v2 Lt1 0 + 0.1(849.57)(0.4)(4) - T(0.2)(4) = - 6.25 v2 (2)

0.8T - 6.25v2 = 135.93 Solving Eqs. (1) and (2),

Ans.

v = 18.39 rad>s = 18.4 rad>s b T = 313.59 N

Ans: v = 18.4 rad>s b 1004

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19–21. The spool has a weight of 30 lb and a radius of gyration kO = 0.45 ft. A cord is wrapped around its inner hub and the end subjected to a horizontal force P = 5 lb. Determine the spool’s angular velocity in 4 s starting from rest. Assume the spool rolls without slipping.

0.9 ft O

P

5 lb

0.3 ft A

SOLUTION (+ c)

m(vy) + ©

L

Fy dt = m(vy)2

0 + NA(4) - 30(4) = 0 NA = 30 lb + ) (:

m(vx)1 + ©

L

Fxdt = m(vx)2

0 + 5(4) - FA(4) = (c + )

(HG)1 + ©

L

30 vG 32.2

MG dt = (HG)2

0 + FA(4)(0.9) - 5(4)(0.3) =

30 (0.45)2 v 32.2

Since no slipping occurs Set vG = 0.9 v FA = 2.33 lb Ans.

v = 12.7 rad>s Also, (c + )

(HA)1 + ©MA dt = (HA)2 0 + 5(4)(0.6) = c

30 30 (0.45)2 + (0.9)2 d v 32.2 32.2 Ans.

v = 12.7 rad>s

Ans: v = 12.7 rad>s 1005

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19–22. The two gears A and B have weights and radii of gyration of WA = 15 lb, kA = 0.5 ft and WB = 10 lb, kB = 0.35 ft, respectively. If a motor transmits a couple moment to gear B of M = 2(1 - e - 0.5t ) lb # ft, where t is in seconds, determine the angular velocity of gear A in t = 5 s, starting from rest.

0.8 ft A

M

0.5 ft B

Solution vA(0.8) = vB(0.5) vB = 1.6vA Gear B: L

( c + )      (HB)1 + Σ MB dt = ( HB ) 2 0 +

L0

5

2 ( 1 - e -0.5t ) dt -

L

0.5F dt = c a

         6.328 = 0.5 Gear A:

L

F dt + 0.06087vA

0 = 0.8

L (2), and solving for vA.

10 b(0.35)2 d ( 1.6vA ) 32.2

F dt - 0.1165vA L

(2)

L

F dt between Eqs. (1) and

( a+ )   (HA)1 + Σ MA dt = ( HA ) 2  Eliminate

   0 +

L

0.8F dt = c a

(1)

15 b(0.5)2 d vA  vA = 47.3 rad>s 32.2

Ans.

Ans: vA = 47.3 rad>s 1006

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19–23. The hoop (thin ring) has a mass of 5 kg and is released down the inclined plane such that it has a backspin v = 8 rad>s and its center has a velocity vG = 3 m>s as shown. If the coefficient of kinetic friction between the hoop and the plane is mk = 0.6, determine how long the hoop rolls before it stops slipping.

ω = 8 rad/s G vG = 3 m /s

SOLUTION b+

mvx1 + a

L

0.5 m

30°

Fxdt = mvx2

5(3) + 49.05 sin 30° (t) - 25.487t = 5vG a+

(HG)1 + a

L

MG dt = (HG)2

- 5(0.5)2 (8) + 25.487(0.5)(t) = 5(0.5)2 a

vG b 0.5

Solving, vG = 2.75 m>s Ans.

t = 1.32 s

Ans: t = 1.32 s 1007

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*19–24. The 30-kg gear is subjected to a force of P = (20t) N, where t is in seconds. Determine the angular velocity of the gear at t = 4 s, starting from rest. Gear rack B is fixed to the horizontal plane, and the gear’s radius of gyration about its mass center O is kO = 125 mm.

150 mm

P  (20t) N

O B

A

SOLUTION Kinematics: Referring to Fig. a, vO = vrO>IC = v(0.15) Principle of Angular Impulse and Momentum: The mass moment of inertia of the gear about its mass center is IO = mkO2 = 30(0.1252) = 0.46875 kg # m2 . Writing the angular impulse and momentum equation about point A shown in Fig. b, t2

(HA)1 + ©

Lt1

MA dt = (HA)2

4s

0 +

L0

1.5t2 2

4s 0

20t(0.15)dt = 0.46875v + 30 [v(0.15)] (0.15)

= 1.14375v Ans.

v = 21.0 rad>s

Ans: v = 21.0 rad>s 1008

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19–25. The 30-lb flywheel A has a radius of gyration about its center of 4 in. Disk B weighs 50 lb and is coupled to the flywheel by means of a belt which does not slip at its contacting surfaces. If a motor supplies a counterclockwise torque to the flywheel of M = (50t) lb # f t, where t is in seconds, determine the time required for the disk to attain an angular velocity of 60 rad>s starting from rest.

M  (50t) lbft 9 in.

6 in.

A

B

SOLUTION Principle of Impulse and Momentum: The mass moment inertia of the flywheel 4 2 30 about point C is IC = a b = 0.1035 slug # ft2. The angular velocity of the 32.2 12 rB 0.75 v = (60) = 90.0 rad>s. Applying Eq. 19–14 to the flywheel is vA = rA B 0.5 flywheel [FBD(a)], we have t2

IC v1 + © t

(a + )

0 +

L0

50t dt +

C

L

Lt1

MC dt = IC v2

T2 (dt) D (0.5) -

25t2 + 0.5

L

C

L

T1(dt) D (0.5) = 0.1035(90) (1)

(T2 - T1)dt = 9.317

1 50 a b(0.752) 2 32.2 = 0.4367 slug # ft2. Applying Eq. 19–14 to the disk [FBD(b)], we have

The mass moment inertia of the disk about point D is ID =

t2

ID v1 + © (a +)

0 +

C

L

T1 (dt) D (0.75) L

MD dt = ID v2

Lt1

C

L

T2 (dt) D (0.75) = 0.4367(60) (2)

(T2 - T1)dt = - 34.94

Substitute Eq. (2) into Eq. (1) and solving yields Ans.

t = 1.04 s

Ans: t = 1.04 s 1009

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19–26. If the shaft is subjected to a torque of M = (15t2) N # m, where t is in seconds, determine the angular velocity of the assembly when t = 3 s, starting from rest. Rods AB and BC each have a mass of 9 kg.

M

1m

1m

(15t2) N m

A

C

B

SOLUTION Principle of Impulse and Momentum: The mass moment of inertia of the rods 1 1 ml2 = (9) A 12 B = 0.75 kg # m2. Since the about their mass center is IG = 12 12 assembly rotates about the fixed axis, (vG)AB = v(rG)AB = v(0.5) and (vG)BC = v(rG)BC = va 212 + (0.5)2 b = v(1.118). Referring to Fig. a, t2

c+

(Hz)1 + © 3s

0 + 5t3 2

L0 3s 0

Lt1

Mz dt = (Hz)2

15t2dt = 9 C v(0.5) D (0.5) + 0.75v + 9 C v(1.118) D (1.118) + 0.75v

= 15v Ans.

v = 9 rad>s

Ans: v = 9 rad>s 1010

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19–27. The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass of 15 kg and a radius of gyration of kO = 110 mm. If the block at A has a mass of 40 kg and the container at B has a mass of 85 kg, including its contents, determine the speed of the container when t = 3 s after it is released from rest.

200 mm O 75 mm

C

A

Solution The angular velocity of the pulley can be related to the speed of vA container B by v = = 13.333 vB. Also the speed of block A 0.075 vA = v(0.2) = 13.33 vB(0.2) - 2.667 vB,

B

(a+ )( ΣSyst. Ang. Mom. ) O1 + ( Σ Syst. Ang. Imp. ) O(1 - 2) = ( Σ Syst. Ang. Mom. ) O2 0 + 40(9.81)(0.2)(3) - 85(9.81)(0.075)(3) =

3 15(0.110)2 4 (13.333 vB)

+ 85 vB(0.075) + 40(2.667 vB)(0.2) Ans.

vB = 1.59 m>s

Ans: vB = 1.59 m>s 1011

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*19–28. The crate has a mass mc. Determine the constant speed v0 it acquires as it moves down the conveyor. The rollers each have a radius of r, mass m, and are spaced d apart. Note that friction causes each roller to rotate when the crate comes in contact with it.

d A

SOLUTION

30°

The number of rollers per unit length is 1/d. Thus in one second,

v0 rollers are contacted. d

If a roller is brought to full angular speed of v =

v0 in t0 seconds, then the moment r

of inertia that is effected is v0 v0 1 I¿ = I a b(t0) = a m r2ba bt0 d 2 d Since the frictional impluse is F = mc sin u then a + (HG)1 + ©

L

MG dt = (HG)2

v0 v0 1 0 + (mc sin u) r t0 = c a m r2 b a b t0 d a b r 2 d v0 =

A

mc (2 g sin u d) a b m

Ans.

Ans: v0 = 1012

A

(2 g sin u d) a

mc b m

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19–29. The turntable T of a record player has a mass of 0.75 kg and a radius of gyration kz = 125 mm. It is turning freely at vT = 2 rad>s when a 50-g record (thin disk) falls on it. Determine the final angular velocity of the turntable just after the record stops slipping on the turntable.

z 150 mm

T

vT  2 rad/s

Solution (Hz)1 = (Hz)2 0.75(0.125)2(2) = c 0.75(0.125)2 +

v = 1.91 rad>s

1 (0.050)(0.150)2 d v 2

Ans.

Ans: v = 1.91 rad>s 1013

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19–30. The 10-g bullet having a velocity of 800 m>s is fired into the edge of the 5-kg disk as shown. Determine the angular velocity of the disk just after the bullet becomes embedded into its edge. Also, calculate the angle u the disk will swing when it stops. The disk is originally at rest. Neglect the mass of the rod AB.

A

2m

B

v  800 m/s 0.4 m

Solution Conservation of Angular Momentum. The mass moment of inertia of the disk about 1 its mass center is (IG)d = (5) ( 0.42 ) = 0.4 kg # m2. Also, (vb)2 = v2 1 25.92 2 and 2 (vd)2 = v2(2.4). Referring to the momentum diagram with the embedded bullet,

Fig. a,

Σ(HA)1 = Σ(HA)2 Mb(vb)1(rb)1 = (IG)d v2 + Md(vd)2(r2) + Mb(vb)2(rb)2 0.01(800)(2.4) = 0.4v2 + 5 3 v2(2.4) 4 (2.4) + 0.01 3 v2 1 25.92 2 4 1 25.92 2

Ans.

v2 = 0.6562 rad>s = 0.656 rad>s

Kinetic Energy. Since the system is required to stop finally, T3 = 0. Here T2 = =

1 1 1 (I ) v22 + Md(vd)22 + Mb(vb)22 2 Gd 2 2 1 1 1 (0.4) ( 0.65622 ) + (5)[0.6562(2.4)]2 + (0.01) 3 0.6562 1 25.92 2 4 2 2 2 2

= 6.2996 J

Potential Energy. Datum is set as indicated on Fig. b. Here f = tan-1 a

0.4 b = 9.4623. Hence 2.4

y1 = 2.4 cos u  yb = 25.92 cos (u - 9.4623°)

Thus, the gravitational potential energy of the disk and bullet with reference to the datum is (Vg)d = Md g (yd) = 5(9.81)( - 2.4 cos u) = -117.72 cos u (Vg)b = Mbg(yb) = 0.01(9.81) 3 1 - 25.92 cos (u - 9.4623°) 4 = - 0.098125.92 cos (u - 9.4623°)

At u = 0°, [(Vg)d]2 = - 117.72 cos 0° = -117.72 J [(Vg)b]2 = - 0.098125.92 cos (0 - 9.4623°) = - 0.23544 J Conservation of Energy. T2 + V2 = T3 + V3 6.2996 + ( -117.72) + ( -0.23544) = 0 + ( - 117.72 cos u) +

3 - 0.098125.92 cos (u

117.72 cos u + 0.098125.92 cos (u - 9.4623°) = 111.66

- 9.4623°) 4

Solved by numerically,

Ans.

u = 18.83° = 18.8° 1014

Ans: v2 = 0.656 rad>s u = 18.8°

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19–31. The 10-g bullet having a velocity of 800 m>s is fired into the edge of the 5-kg disk as shown. Determine the angular velocity of the disk just after the bullet becomes embedded into its edge. Also, calculate the angle u the disk will swing when it stops. The disk is originally at rest. The rod AB has a mass of 3 kg.

A

2m

B

v  800 m/s 0.4 m

Solution Conservation of Angular Momentum. The mass moments of inertia of the disk and 1 rod about their respective mass centers are (IG)d = (5) ( 0.42 ) = 0.4 kg # m2 and 2 1 (IG)r = (3) ( 22 ) = 1.00 kg # m2. Also, (vb)2 = v2 ( 25.92 ) , (vd)2 = v2(2.4) and 12 ( vr ) 2 = v2 (1). Referring to the momentum diagram with the embedded bullet, Fig. a, Σ(HA)1 = Σ(HA)2 Mb(vb)1(rb)1 = (IG)d v2 + Md(vd)2(rd) + (IG)r v2 + mr(vr)2 (r) + Mb(vb)2(rb)2 0.01(800)(2.4) = 0.4v2 + 5[v2(2.4)](2.4) + 1.00v2 + 3[v2 (1)] (1)



+ 0.01 3 v2 ( 25.92 ) 4 ( 25.92 ) 4

Ans.

v2 = 0.5773 rad>s = 0.577 rad>s

Kinetic Energy. Since the system is required to stop finally, T3 = 0. Here 1 1 1 1 1 (IG)d v22 + Md (vd)22 + ( IG ) r v22 + Mr ( vr ) 22 + Mb ( vb ) 22 2 2 2 2 2 1 1 1 1 1 2 2 = (0.4) ( 0.5773 ) + (5)[0.5773(2.4)] + (1.00)( 0.57732 ) + (3)[0.5773(1)]2 + (0.01) 3 0.5773( 25.92) 4 2 2 2 2 2 2 = 5.5419 J

T2 =

1015

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19–31.  Continued

Potential Energy. Datum is set as indicated on Fig. b. 0.4 b = 9.4623°. Hence 2.4 yd = 2.4 cos u, yr = cos u, yb = 25.92 cos (u - 9.4623°)

Here f = tan-1a

Thus, the gravitational potential energy of the disk, rod and bullet with reference to the datum is (Vg)d = Md g yd = 5(9.81)( - 2.4 cos u) = - 117.72 cos u (Vg)r = Mr g yr = 3(9.81)( - cos u) = - 29.43 cos u (Vg)b = mb g yb = 0.01(9.81) 3 - 25.92 cos (u - 94623°) 4 = - 0.098125.92 cos(u - 9.4623°)

At u = 0°, [(Vg)d]2 = - 117.72 cos 0° = - 117.72 J [(Vg)r]2 = - 29.43 cos 0° = -29.43 J [(Vg)b]2 = - 0.098125.92 cos (0° - 9.4623°) = - 0.23544 J Conservation of Energy. T2 + V2 = T3 + V3 5.5419 + ( -117.72) + ( - 29.43) + ( - 0.23544) = 0 + ( -117.72 cos u) + ( - 29.43 cos u) +

3 - 0.098125.92 cos (u

147.15 cos u + 0.098125.92 cos (u - 9.4623°) = 141.84

- 9.4623°) 4

Solved numerically, Ans.

u = 15.78° = 15.8°

Ans: v2 = 0.577 rad>s u = 15.8° 1016

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*19–32. The circular disk has a mass m and is suspended at A by the wire. If it receives a horizontal impulse I at its edge  B, determine the location y of the point P about which the disk appears to rotate during the impact. A y P a

Solution

I

B

Principle of Impulse and Momentum. The mass moment of inertia of the disk about 1 1 its mass center is IG = mr 2 = ma2 2 2 a+ IGv1 + Σ

MG dt = IG v2 1 0 + Ia = a ma2 bv 2 1 I = ma v(1) 2

+ ) (S

Lt1

t2

m 3 ( vG ) x 4 , + Σ

Lt1

t2

Fx dt = m 3 ( vG ) x 4 2

0 + I = mvG(2)

Equating eqs. (1) and (2), 1 ma v = mvG 2 a vG = v 2 Kinematics. Here, IC is located at P, Fig. b. Thus, vB = v rB>IC = v(2a - y). Using similar triangles, a - y 2a - y a - y 2a - y = ;   = vG vB a v(2a - y) v 2 1 y = a 2

Ans.

Ans: y = 1017

1 a 2

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19–33. 0.20 m 0.65 m

The 80-kg man is holding two dumbbells while standing on a turntable of negligible mass, which turns freely about a vertical axis. When his arms are fully extended, the turntable is rotating with an angular velocity of 0.5 rev>s. Determine the angular velocity of the man when he retracts his arms to the position shown. When his arms are fully extended, approximate each arm as a uniform 6-kg rod having a length of 650 mm, and his body as a 68-kg solid cylinder of 400-mm diameter. With his arms in the retracted position, assume the man as an 80-kg solid cylinder of 450-mm diameter. Each dumbbell consists of two 5-kg spheres of negligible size.

0.3 m

0.3 m

SOLUTION Conservation of Angular Momentum: Since no external angular impulse acts on the system during the motion, angular momentum about the axis of rotation (z axis) is conserved. The mass moment of inertia of the system when the arms are in the fully extended position is (Iz)1 = 2 c 10(0.852) d + 2 c

1 1 (6)(0.652) + 6(0.5252) d + (68)(0.2 2) 12 2 = 19.54 kg # m2

And the mass moment of inertia of the system when the arms are in the retracted position is (Iz)2 = 2 c10(0.32) d +

1 (80)(0.2252) 2

= 3.825 kg # m2 Thus, (Hz)1 = (Hz)2 (Iz)1v1 = (Iz)2v2 19.54(0.5) = 3.825v2 Ans.

v2 = 2.55 rev>s

Ans: v2 = 2.55 rev>s 1018

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19–34. The platform swing consists of a 200-lb flat plate suspended by four rods of negligible weight. When the swing is at rest, the 150-lb man jumps off the platform when his center of gravity G is 10 ft from the pin at A. This is done with a horizontal velocity of 5 ft>s, measured relative to the swing at the level of G. Determine the angular velocity he imparts to the swing just after jumping off.

A

10 ft 11 ft

SOLUTION (a + )

(HA)1 = (HA)2

G

1 200 200 150 0 + 0 = c a (11)2 d v - c a b (4)2 + b (5 - 10v) d (10) 12 32.2 32.2 32.2

4 ft

Ans.

v = 0.190 rad>s

Ans: v = 0.190 rad>s 1019

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19–35. The 2-kg rod ACB supports the two 4-kg disks at its ends. If both disks are given a clockwise angular velocity 1vA21 = 1vB21 = 5 rad>s while the rod is held stationary and then released, determine the angular velocity of the rod after both disks have stopped spinning relative to the rod due to frictional resistance at the pins A and B. Motion is in the horizontal plane. Neglect friction at pin C.

0.75m B (VB)1

0.75m C

0.15 m

A 0.15 m

(VA)1

SOLUTION c+

H1 = H2

1 1 1 2c (4) (0.15)2d(5) = 2c (4)(0.15)2d v + 2[4(0.75 v)(0.75)] + c (2)(1.50)2d v 2 2 12 Ans.

v = 0.0906 rad>s

Ans: v = 0.0906 rad>s 1020

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*19–36. The satellite has a mass of 200 kg and a radius of gyration about z axis of kz = 0.1 m, excluding the two solar panels A and B. Each solar panel has a mass of 15 kg and can be approximated as a thin plate. If the satellite is originally spinning about the z axis at a constant rate vz = 0.5 rad>s when u = 90°, determine the rate of spin if both panels are raised and reach the upward position, u = 0°, at the same instant.

z

vz

B

y

u  90 A 1.5 m

0.2 m

x

0.3 m

Solution Conservation of Angular Momentum. When u = 90°, the mass moment of inertia of the entire satellite is Iz = 200 ( 0.12 ) + 2c

1 (15) ( 0.32 + 1.52 ) + 15 ( 0.952 ) d = 34.925 kg # m2 12

when u = 0°, I′z = 200 ( 0.12 ) + 2c Thus

1 (15) ( 0.32 ) + 15 ( 0.22 ) d = 3.425 kg # m2 12

(Hz)1 = (Hz)2 Iz(vz)1 = I′z(vz)2 34.925(0.5) = 3.425(vz)2 Ans.

(vz)2 = 5.0985 rad>s = 5.10 rad>s

Ans: (vz)2 = 5.10 rad>s 1021

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19–37. Disk A has a weight of 20 lb. An inextensible cable is attached to the 10-lb weight and wrapped around the disk. The weight is dropped 2 ft before the slack is taken up. If the impact is perfectly elastic, i.e., e = 1, determine the angular velocity of the disk just after impact.

0.5 ft

A

SOLUTION For the weight T1 + V1 = T2 + V2 0 + 10(2) =

10 lb

1 10 a b v 22 2 32.2

v2 = 11.35 ft>s (HA)2 = (HA)3 mv2 (0.5) + 0 = m y3 (0.5) + IA v 10 10 20 a 32.2 b (11.35)(0.5) + 0 = a 32.2 b v3 (0.5) + c 12 a 32.2 b (0.5)2 d v

( + T)

e =

0.5 v - v3 v2 - 0

1 =

0.5 v - v3 11.35 - 0

[1]

11.35 = 0.5 v - v3

[2]

Solving Eqs.[1] and [2] yields: Ans.

v = 22.7 rad>s v3 = 0

Ans: v = 22.7 rad>s 1022

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19–38. The plank has a weight of 30 lb, center of gravity at G, and it rests on the two sawhorses at A and B. If the end D is raised 2 ft above the top of the sawhorses and is released from rest, determine how high end C will rise from the top of the sawhorses after the plank falls so that it rotates clockwise about A, strikes and pivots on the sawhorses at B, and rotates clockwise off the sawhorse at A.

2 ft B

A G

C

3 ft

1.5 ft

1.5 ft

D

3 ft

SOLUTION Establishing a datum through AB, the angular velocity of the plank just before striking B is T1 + V1 = T2 + V2 1 1 30 2 30 b(9)2 + (1.5)2 d (vCD)22 + 0 0 + 30c (1.5) d = c a 6 2 12 32.2 32.2 (vCD)2 = 1.8915 rad>s (vG)2 = 1.8915(1.5) = 2.837 m>s (c +) c

(HB)2 = (HB)3

1 30 30 30 1 30 a b(9)2 d(1.8915) (2.837)(1.5) = c a b(9)2 d(vAB)3 + (v ) (1.5) 12 32.2 32.2 2 32.2 32.2 G 3

Since (vG)3 = 1.5(vAB)3 (vAB)3 = 0.9458 rad>s (vG)3 = 1.4186 m>s T3 + V3 = T4 + V4 1 1 30 1 30 c a b(9)2 d (0.9458)2 + a b (1.4186)2 + 0 = 0 + 30hG 2 12 32.2 2 32.2 hG = 0.125 Thus, hC =

6 (0.125) = 0.500 ft 1.5

Ans.

Ans: hC = 0.500 ft 1023

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19–39. The 12-kg rod AB is pinned to the 40-kg disk. If the disk is given an angular velocity vD = 100 rad>s while the rod is held stationary, and the assembly is then released, determine the angular velocity of the rod after the disk has stopped spinning relative to the rod due to frictional resistance at the bearing B. Motion is in the horizontal plane. Neglect friction at the pin A.

2m vD

A

B

0.3 m

Solution Conservation of Angular Momentum Initial: Since the rod is stationary and the disk is not translating, the total angular momentum about A equals the angular momentum of the disk about B, where for 1 1 the disk IB = mDr 2 = (40)(0.3)2 = 1.80 kg # m2. 2 2 (HA)1 = IBv1 = 1.80(100) = 180 kg # m2 >s 1 1 Final: The rod and disk move as a single unit for which IA = mRl 2 + mDr 2 + mDl 2 3 2 1 1 2 2 2 2 # = (12)(2) + (40)(0.3) + (40)(2) = 177.8 kg m . 3 2 (HA)2 = IA v2 = 177.8 v2 Setting (HA)1 = (HA)2 and solving, Ans.

v2 = 1.012 rad>s = 1.01 rad>s

Ans: v2 = 1.01 rad>s 1024

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*19–40. A thin rod of mass m has an angular velocity V 0 while rotating on a smooth surface. Determine its new angular velocity just after its end strikes and hooks onto the peg and the rod starts to rotate about P without rebounding. Solve the problem (a) using the parameters given, (b) setting m = 2 kg, v0 = 4 rad>s, l = 1.5 m.

P

ω0 l

SOLUTION (a) © (HP)0 = ©(HP)1 c

1 1 ml2 d v0 = c ml2 d v 12 3 v =

1 v 4 0

(b) From part (a)

Ans. v =

1 1 v = (4) = 1 rad s 4 0 4

Ans.

Ans:

1 v0 4 v = 1 rad>s v =

1025

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19–41. Tests of impact on the fixed crash dummy are conducted using the 300-lb ram that is released from rest at u = 30°, and allowed to fall and strike the dummy at u = 90°. If the coefficient of restitution between the dummy and the ram is e = 0.4, determine the angle u to which the ram will rebound before momentarily coming to rest.

u 10 ft

10 ft

SOLUTION Datum through pin support at ceiling. T1 + V1 = T2 + V2 0 - 300(10 sin 30°) =

1 300 a b (v)2 - 300(10) 2 32.2

v = 17.944 ft>s + 2 1:

e = 0.4 =

v¿ - 0 0 - (- 17.944) v¿ = 7.178 ft>s

T2 + V2 = T3 + V3 1 300 a b(7.178)2 - 300(10) = 0 - 300(10 sin u) 2 32.2 Ans.

u = 66.9°

Ans: u = 66.9° 1026

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19–42. z

The vertical shaft is rotating with an angular velocity of 3 rad>s when u = 0°. If a force F is applied to the collar so that u = 90°, determine the angular velocity of the shaft. Also, find the work done by force F. Neglect the mass of rods GH and EF and the collars I and J. The rods AB and CD each have a mass of 10 kg.

D

0.3 m

0.3 m 0.3 m

E

I

u

Conservation of Angular Momentum: Referring to the free-body diagram of the assembly shown in Fig. a, the sum of the angular impulses about the z axis is zero. Thus, the angular momentum of the system is conserved about the axis. The mass moments of inertia of the rods about the z axis when u = 0° and 90° are

G

u A

C

SOLUTION

(Iz)1 = 2 c

0.3 m

0.1 m v

F

0.1 m

J

H F

1 (10)(0.62) + 10(0.3 + 0.1)2 d = 3.8 kg # m2 12

(Iz)2 = 2 c 10(0.12) d = 0.2 kg # m2 Thus, (Hz)1 = (Hz)2 3.8(3) = 0.2v2 Ans.

v2 = 57 rad>s

Principle of Work and Energy: As shown on the free-body diagram of the assembly, Fig. b, W does negative work, while F does positive work. The work of W is UW = -Wh = -10(9.81)(0.3) = - 29.43 J. The initial and final kinetic energy of the 1 1 1 assembly is T1 = (Iz)1v1 2 = (3.8)(32) = 17.1 J and T2 = (Iz)2v2 2 = 2 2 2 1 2 (0.2)(57 ) = 324.9 J. Thus, 2 T1 + ©U1 - 2 = T2 17.1 + 2(- 29.43) + UF = 324.9 Ans.

UF = 367 J

Ans: v2 = 57 rad>s UF = 367 J 1027

B

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19–43. The mass center of the 3-lb ball has a velocity of (vG)1 = 6 ft>s when it strikes the end of the smooth 5-lb slender bar which is at rest. Determine the angular velocity of the bar about the z axis just after impact if e = 0.8.

z 2 ft

A

2 ft

B

SOLUTION Conservation of Angular Momentum: Since force F due to the impact is internal to the system consisting of the slender bar and the ball, it will cancel out. Thus, angular momentum is conserved about the z axis. The mass moment of inertia of the slender (yB)2 5 1 a b A 42 B = 0.2070 slug # ft2. Here, v2 = bar about the z axis is Iz = . 12 32.2 2 Applying Eq. 19–17, we have

0.5 ft

O (vG)1

6 ft/s G

r

0.5 ft

(Hz)1 = (Hz)2

C mb (yG)1 D (rb) = Iz v2 + C mb (yG)2 D (rb) a

(yB)2 3 3 b (6)(2) = 0.2070 c d + a b(yG)2(2) 32.2 2 32.2

(1)

Coefficient of Restitution:Applying Eq. 19–20, we have e = 0.8 =

(yB)2 - (yG)2 (yG)1 - (yB)1 (yB)2 - (yG)2 6 - 0

(2)

Solving Eqs. (1) and (2) yields (yG)2 = 2.143 ft>s

(yB)2 = 6.943 ft>s

Thus, the angular velocity of the slender rod is given by v2 =

(yB)2 6.943 = = 3.47 rad>s 2 2

Ans.

Ans: v2 = 3.47 rad>s 1028

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*19–44. The pendulum consists of a slender 2-kg rod AB and 5-kg disk. It is released from rest without rotating. When it falls 0.3 m, the end A strikes the hook S, which provides a permanent connection. Determine the angular velocity of the pendulum after it has rotated 90°. Treat the pendulum’s weight during impact as a nonimpulsive force.

0.5 m A S

B

0.2 m

0.3 m

SOLUTION T0 + V0 = T1 + V1 0 + 2(9.81)(0.3) + 5(9.81)(0.3) =

1 1 (2)(vG)21 + (5)(vG)21 2 2

(vG)1 = 2.4261 m>s ©(Hs)1 = ©(Hs)2 2(2.4261)(0.25) + 5(2.4261)(0.7) = c

1 1 (2)(0.5)2 + 2(0.25)2 + (5)(0.2)2 + 5(0.7)2 dv 12 2

v = 3.572 rad>s T2 + V2 = T3 + V3 1 1 1 c (2)(0.5)2 + 2(0.25) + (5)(0.2)2 + 5(0.7)2 d (3.572)2 + 0 2 12 2 =

1 1 1 c (2)(0.5)2 + 2(0.25)2 + (5)(0.2)2 + 5(0.7)2 dv2 2 12 2

+ 2(9.81)(- 0.25) + 5(9.81)(- 0.7) Ans.

v = 6.45 rad>s

Ans: v = 6.45 rad>s 1029

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19–45. The 10-lb block slides on the smooth surface when the corner D hits a stop block S. Determine the minimum velocity v the block should have which would allow it to tip over on its side and land in the position shown. Neglect the size of S. Hint: During impact consider the weight of the block to be nonimpulsive.

B

v 1 ft

C

A

B

D

S D

C

1 ft A

SOLUTION Conservation of Energy: If the block tips over about point D, it must at least achieve the dash position shown. Datum is set at point D. When the block is at its initial and final position, its center of gravity is located 0.5 ft and 0.7071 ft above the datum. Its initial and final potential energy are and 10(0.5) = 5.00 ft # lb # 10(0.7071) = 7.071 ft lb. The mass moment of inertia of the block about point D is ID =

1 10 10 a b A 12 + 12 B + a b A 20.52 + 0.52 B 2 = 0.2070 slug # ft2 12 32.2 32.2

1 1 The initial kinetic energy of the block (after the impact) is ID v22 = (0.2070) v22. 2 2 Applying Eq. 18–18, we have T2 + V2 = T3 + V3 1 (0.2070) v22 + 5.00 = 0 + 7.071 2 v2 = 4.472 rad>s Conservation of Angular Momentum:Since the weight of the block and the normal reaction N are nonimpulsive forces, the angular momentum is conserves about point D. Applying Eq. 19–17, we have (HD)1 = (HD)2 (myG)(r¿) = ID v2 ca

10 b y d (0.5) = 0.2070(4.472) 32.2 Ans.

y = 5.96 ft>s

Ans: v = 5.96 ft>s 1030

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19–46. Determine the height h at which a billiard ball of mass m must be struck so that no frictional force develops between it and the table at A. Assume that the cue C only exerts a horizontal force P on the ball.

P

C

r h A

SOLUTION For the ball + ) mn + © F dt = mn (; 1 2 1 (1)

0 + P(¢ t) = mn2 a+

(HA)1 + © 1 MA dt = (HA)2 2 0 + (P)¢ t(h) = c m r2 + m r2 d v2 5

(2) (3)

Require n2 = v2 r Solving Eqs. (1)–(3) for h yields h =

7 r 5

Ans.

Ans: h = 1031

7 r 5

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19–47. The pendulum consists of a 15-kg solid ball and 6-kg rod. If it is released from rest when u1 = 90°, determine the angle u2 after the ball strikes the wall, rebounds, and the pendulum swings up to the point of momentary rest. Take e = 0.6.

A

100 mm u 2m

Solution

300 mm

Kinetic Energy. The mass moment of inertia of the pendulum about A is 1 2 IA = (6) ( 22 ) + c (15) ( 0.32 ) + 15 ( 2.32 ) d = 87.89 kg # m2. 3 5 Thus,

T =

1 1 I v2 = (87.89) v2 = 43.945v2 2 A 2

Potential Energy. With reference to datum set in Fig. a, the gravitational potential energy of the pendulum is Vg = mrgyr + msgys = 6(9.81)( - cos u) + 15(9.81)( -2.3 cos u) = - 397.305 cos u Coefficient of Restitution. The velocity of the mass center of the ball is vb = vrG = v(2.3). Thus + )e = (d

( vw ) 2 - ( vb ) 2′ 0 - [ - v2′ (2.3)] ;  0.6 = v2(2.3) - 0 ( vb ) 2 - ( vw ) 1

(1) v2′ = 0.6v2 Conservation of Energy. Consider the pendulum swing from the position u = 90° to u = 0° just before the impact, (Vg)1 = - 397.305 cos 90° = 0 (Vg)2 = - 397.305 cos 0° = - 397.305 J T1 = 0  T2 = 43.945v22 Then T1 + V1 = T2 + V2 0 + 0 = 43.945v22 + ( - 397.305) v2 = 3.0068 rad>s Thus, just after the impact, from Eq. (1) v2′ = 0.6(3.0068) = 1.8041 rad>s Consider the pendulum swing from position u = 90° just after the impact to u, (Vg)2 ′ = (Vg)2 = - 397.305 J (Vg)3 = - 397.305 cos u T2 ′ = 43.945 ( 1.80412 ) = 143.03 J T3 = 0 (required) Then

T2 ′ + V2 ′ = T3 + V3 143.03 + ( - 397.305) = 0 + ( - 397.305 cos u) Ans.

u = 50.21° = 50.2° 1032

Ans: u = 50.2°

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*19–48. The 4-lb rod AB is hanging in the vertical position. A 2-lb block, sliding on a smooth horizontal surface with a velocity of 12 ft/s, strikes the rod at its end B. Determine the velocity of the block immediately after the collision. The coefficient of restitution between the block and the rod at B is e = 0.8.

A

3 ft 12 ft/s B

SOLUTION Conservation of Angular Momentum: Since force F due to the impact is internal to the system consisting of the slender rod and the block, it will cancel out. Thus, angular momentum is conserved about point A. The mass moment of inertia of the 4 1 4 slender rod about point A is IA = a b (32) + (1.52) = 0.3727 slug # ft2. 12 32.2 32.2 (vB)2 Here, v2 = . Applying Eq. 19–17, we have 3 (HA)1 = (HA)2 [mb (vb)1](rb) = IAv2 + [mb (vb)2](rb) a

(vB)2 2 2 b (12)(3) = 0.3727 c d + a b (vb)2 (3) 32.2 3 32.2

[1]

Coefficient of Restitution: Applying Eq. 19–20, we have

+ ) (:

e =

(vB)2 - (vb)2 (vb)1 - (vB)1

0.8 =

(vB)2 - (vb)2 12 - 0

[2]

Solving Eqs. [1] and [2] yields (vb)2 = 3.36 ft s :

Ans.

(vB)2 = 12.96 ft s :

Ans: (vb)2 = 3.36 ft>s S 1033

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19–49. The hammer consists of a 10-kg solid cylinder C and 6-kg uniform slender rod AB. If the hammer is released from rest when u = 90° and strikes the 30-kg block D when u = 0°, determine the velocity of block D and the angular velocity of the hammer immediately after the impact. The coefficient of restitution between the hammer and the block is e = 0.6.

u 100 mm B

SOLUTION

150 mm

Conservation of Energy: With reference to the datum in Fig. a, V1 = (Vg)1 = WAB(yGAB)1 + WC(yGC)1 = 0 and V2 = (Vg)2 = - WAB(yGAB)2 - WC(yGC)2 = - 6(9.81)(0.25) - 10(9.81)(0.55) = - 68.67 J. Initially, T1 = 0. Since the hammer rotates about the fixed axis, and (vGAB)2 = v2rGAB = v2(0.25) (vGC)2 = v2rGC = v2(0.55). The mass moment of inertia of rod AB and cylinder C 1 1 about their mass centers is IGAB = ml2 = (6)(0.52) = 0.125 kg # m2 and 12 12 1 1 m(3r2 + h2) = (10) C 3(0.052) + 0.152 D = 0.025 kg # m2. Thus, IC = 12 12 T2 = =

A

500 mm

1 1 1 1 I v 2 + mAB(vGAB)2 2 + IGC v22 + mC(vGC)22 2 GAB 2 2 2 2 1 1 1 1 (0.125)v22 + (6) C v2(0.25) D 2 + (0.025)v22 + (10) C v2(0.55) D 2 2 2 2 2

= 1.775 v22 Then, T1 + V1 = T2 + V2 0 + 0 = 1.775v2 2 + ( -68.67) v2 = 6.220 rad>s Conservation of Angular Momentum: The angular momentum of the system is conserved point A. Then, (HA)1 = (HA)2 0.125(6.220) + 6[6.220(0.25)](0.25) + 0.025(6.220) + 10[6.220(0.55)](0.55) = 30vD(0.55) - 0.125v3 - 6[v3(0.25)](0.25) - 0.025v3 - 10[v3(0.55)](0.55) (1)

16.5vD - 3.55v3 = 22.08

1034

C 50 mm D

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19–49. Continued

Coefficient of Restitution: Referring to Fig. c, the components of the velocity of the impact point P just before and just after impact along the line of impact are

C (vP)x D 2 = (vGC)2 = v2rGC = 6.220(0.55) = 3.421 m>s :

v3rGC = v3 (0.55) ; . Thus, + :

e =

and

C (vP)x D 3 = (vGC)3 =

(vD)3 - C (vP)x D 3

C (vP)x D 2 - (vD)2

0.6 =

(vD)3 -

C - v3(0.55) D

3.421 - 0 (2)

(vD)3 + 0.55v3 = 2.053 Solving Eqs. (1) and (2), (vD)3 = 1.54 m>s

Ans.

v3 = 0.934 rad>s

Ans: (vD)3 = 1.54 m>s v3 = 0.934 rad>s 1035

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19–50. The 20-kg disk strikes the step without rebounding. Determine the largest angular velocity v1 the disk can have and not lose contact with the step, A.

◊1

200 mm

A

30 mm

Solution Conservation of Angular Momentum. The mass moment of inertia of the disk about 1 1 its mass center is IG = mr 2 = (20) ( 0.22 ) = 0.4 kg # m2. Since no slipping occurs, 2 2 vG = vr = v(0.2). Referring to the impulse and momentum diagram, Fig. a, we notice that angular moment is conserved about point A since W is nonimpulsive. Thus, (HA)1 = (HA)2 20[v1(0.2)](0.17) + 0.4 v1 = 0.4 v2 + 20[v2(0.2)](0.2) (1)

v1 = 1.1111 v2

Equations of Motion. Since the requirement is the disk is about to lose contact with the 0.17 step when it rotates about A, NA ≃ 0. Here u = cos-1 a b = 31.79°. Consider the 0.2 motion along n direction, + R ΣFn = M(aG)n;  20(9.81) cos 31.79° = 203v22(0.2) 4 v2 = 6.4570 rad>s

Substitute this result into Eq. (1) Ans.

v1 = 1.1111(6.4570) = 7.1744 rad>s = 7.17 rad>s

Ans: v1 = 7.17 rad>s 1036

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19–51. The solid ball of mass m is dropped with a velocity v1 onto the edge of the rough step. If it rebounds horizontally off the step with a velocity v2, determine the angle u at which contact occurs. Assume no slipping when the ball strikes the step. The coefficient of restitution is e. u r

v1 v2

SOLUTION Conservation of Angular Momentum: Since the weight of the solid ball is a nonimpulsive force, then angular momentum is conserved about point A. The mass 2 moment of inertia of the solid ball about its mass center is IG = mr2. Here, 5 y2 cos u v2 = . Applying Eq. 19–17, we have r (HA)1 = (HA)2

C mb (yb)1 D (r¿) = IG v2 + C mb (yb)2 D (r–) y2 cos u 2 b + (my2)(r cos u) (my1)(r sin u) = a mr2 b a r 5 y2 5 = tan u y1 7

(1)

Coefficient of Restitution:Applying Eq. 19–20, we have e = e =

0 - (yb)2 (yb)1 - 0 - (y2 sin u) - y1 cos u

y2 e cos u = y1 sin u

(2)

Equating Eqs. (1) and (2) yields 5 e cos u tan u = 7 sin u 7 tan2 u = e 5 u = tan - 1 ¢

7 e≤ A5

Ans.

Ans: u = tan - 1a 1037

7 eb A5

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*19–52. V1

150 mm

The wheel has a mass of 50 kg and a radius of gyration of 125 mm about its center of mass G. Determine the minimum value of the angular velocity V1 of the wheel, so that it strikes the step at A without rebounding and then rolls over it without slipping.

G

A

SOLUTION Conservation of Angular Momentum: Referring to Fig. a, the sum of the angular impulses about point A is zero. Thus, angular momentum of the wheel is conserved about this point. Since the wheel rolls without slipping, (vG)1 = v1r = v1(0.15) and (vG)2 = v2r = v2(0.15). The mass moment of inertia of the wheel about its mass center is IG = mkG2 = 50(0.1252) = 0.78125 kg # m2. Thus, (HA)1 = (HA)2 50[v1(0.15)](0.125) + 0.78125v1 = 50[v2(0.15)](0.15) + 0.78125v2 (1)

v1 = 1.109v2

Conservation of Energy: With reference to the datum in Fig. a, V2 = (Vg)2 = W(yG)2 = 0 and V3 = (Vg)3 = W(yG)3 = 50(9.81)(0.025) = 12.2625 J. Since the wheel is required to be at rest in the final position, T3 = 0. The initial kinetic energy 1 1 1 1 of the wheel is T2 = m(vG)2 2 + IGv2 2 = (50)[v2(0.15)]2 + (0.78125)(v2 2) = 2 2 2 2 0.953125v2 2. Then T2 + V2 = T3 + V3 0.953125v22 + 0 = 0 + 12.2625 v2 = 3.587 rad>s Substituting this result into Eq. (1), we obtain Ans.

v1 = 3.98 rad>s

1038

25 mm

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19–53. The wheel has a mass of 50 kg and a radius of gyration of 125 mm about its center of mass G. If it rolls without slipping with an angular velocity of V1 = 5 rad>s before it strikes the step at A, determine its angular velocity after it rolls over the step. The wheel does not loose contact with the step when it strikes it.

150 mm

V1 G

A

25 mm

SOLUTION Conservation of Angular Momentum: Referring to Fig. a, the sum of the angular impulses about point A is zero. Thus, angular momentum of the wheel is conserved about this point. Since the wheel rolls without slipping, (vG)1 = v1r = (5)(0.15) = 0.75 m>s and v2 = v2r = v2(0.15). The mass moment of inertia of the wheel about its mass center is IG = mkG2 = 50(0.1252) = 0.78125 kg # m2. Thus, (HA)1 = (HA)2 50(0.75)(0.125) + 0.78125(5) = 50[v2(0.15)](0.15) + 0.78125v2 (1)

v2 = 4.508 rad>s

Conservation of Energy: With reference to the datum in Fig. a, V2 = (Vg)2 = W(yG)2 = 0 and V3 = (Vg)3 = W(yG)3 = 50(9.81)(0.025) = 12.2625 J. The initial 1 1 1 kinetic energy of the wheel is T = mvG 2 + IGv2 = (50)[v(0.15)]2 + 2 2 2 1 2 2 2 (0.78125)v = 0.953125v . Thus, T2 = 0.953125v2 = 0.953125(4.5082) = 19.37 J 2 and T3 = 0.953125v32. T2 + V2 = T3 + V3 19.37 + 0 = 0.953125v32 + 12.2625 Ans.

v3 = 2.73 rad>s

Ans: v3 = 2.73 rad>s 1039

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19–54. The rod of mass m and length L is released from rest without rotating. When it falls a distance L, the end A strikes the hook S, which provides a permanent connection. Determine the angular velocity v of the rod after it has rotated 90°. Treat the rod’s weight during impact as a nonimpulsive force.

A L L

S

Solution T1 + V1 = T2 + V2 0 + mgL =

1 2 mv + 0 2 G

vG = 22gL H1 = H2

L 1 m22gL a b = mL2(v2) 2 3 v2 =

3 22gL 2 L

T2 + V2 = T3 + V3 1 1 2 9(2gL) 1 1 L a mL b + 0 = a mL2 bv2 - mga b 2 2 3 2 3 2 4L 3 1 L gL = L2v2 - ga b 4 6 2 v =

A

g 7.5  L

Ans.

Ans: v = 1040

A

7.5

g L

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19–55. B

The 15-lb rod AB is released from rest in the vertical position. If the coefficient of restitution between the floor and the cushion at B is e = 0.7, determine how high the end of the rod rebounds after impact with the floor. 2 ft

SOLUTION

A

T1 = V1 = T2 + V2 0 + 15(1) =

1 1 15 c a b (2)2 d v22 2 3 32.2

v2 = 6.950 rad>s

A+TB

e =

Hence (vB)2 = 6.950(2) = 13.90 rad>s

0 - (vB)3 ; (vB)2 - 0

0.7 =

0 - (vB)3 13.90

(vB)3 = -9.730 ft>s = 9.730 ft>s v3 =

c

(vB)3 9.730 = = 4.865 rad>s 2 2

T3 + V3 = T4 + V4 1 1 15 c a b (2)2 d(4.865)2 = 0 + 15(hG) 2 3 32.2 hG = 0.490 ft Ans.

hB = 2hG = 0.980 ft

Ans: hB = 0.980 ft 1041

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*19–56. A ball having a mass of 8 kg and initial speed of v1 = 0.2 m>s rolls over a 30-mm-long depression. Assuming that the ball rolls off the edges of contact first A, then B, without slipping, determine its final velocity v2 when it reaches the other side.

v2

v1

SOLUTION

B

0.2 = 1.6 rad>s 0.125 15 b = 6.8921° u = sin - 1 a 125 v1 =

y2 v2 = = 8y2 0.125

0.2 m/s

A 125 mm

30 mm

h = 125 - 125 cos 6.8921° = 0.90326 mm T1 + V1 = T2 + V2 1 2 1 (8)(0.2)2 + c (8)(0.125)2 d (1.6)2 + 0 2 2 5 = - (0.90326)(10 - 3)8(9.81) +

1 1 2 (8)v2(0.125)2 + c (8)(0.125)2 d(v)2 2 2 5

v = 1.836 rad>s (HB)2 = (HB)3 2 c (8)(0.125)2 d (1.836) + 8(1.836)(0.125) cos 6.892°(0.125 cos 6.892°) 5 - 8(0.22948 sin 6.892°)(0.125 sin 6.892°) 2 = c (8)(0.125)2 d v3 + 8(0.125)v3 (0.125) 5 v3 = 1.7980 rad>s T3 + V3 = T4 + V4 1 2 1 c (8)(0.125)2 d (1.7980)2 + (8)(1.7980)2(0.125)2 + 0 2 5 2 = 8(9.81)(0.90326(10 - 3)) + +

1 2 c (8)(0.125)2 d(v4)2 2 5

1 (8)(v4)2(0.125)2 2 v4 = 1.56 rad>s

So that Ans.

y2 = 1.56(0.125) = 0.195 m>s

Ans: v2 = 0.195 m>s 1042

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19–57. A solid ball with a mass m is thrown on the ground such that at the instant of contact it has an angular velocity V 1 and velocity components 1vG2x1 and 1vG2y1 as shown. If the ground is rough so no slipping occurs, determine the components of the velocity of its mass center just after impact. The coefficient of restitution is e.

(vG)y1

V1

(vG)x1 G

r

SOLUTION Coefficient of Restitution (y direction):

A+TB

e =

0 - (yG)y 2

(yG)y2 = - e(yG)y1 = e(yG)y1

(yG)y1 - 0

c

Ans.

Conservation of angular momentum about point on the ground: (c + )

(HA)1 = (HA)2

2 2 - mr 2v1 + m(vG)x 1r = mr2v2 + m(vG)x 2 r 5 5 Since no slipping, (vG)x2 = v2 r then, 5 a (vG)x 1 v2 =

2 v rb 5 1

7r

Therefore (yG)x 2 =

5 2 a (yG)x 1 - v1 rb 7 5

Ans.

Ans: (vG)y2 = e(vG)y1 c (vG)x2 = 1043

5 2 a(v ) - v1r b d 7 G x1 5

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19–58. The pendulum consists of a 10-lb solid ball and 4-lb rod. If it is released from rest when u0 = 0°, determine the angle u1 of rebound after the ball strikes the wall and the pendulum swings up to the point of momentary rest. Take e = 0.6.

0.3 ft A

θ 2 ft

SOLUTION IA =

0.3 ft

1 2 10 10 4 a b (2)2 + a b (0.3)2 + a b (2.3)2 = 1.8197 slug # ft2 3 32.2 5 32.2 32.2

Just before impact: T1 + V1 = T2 + V2 0 + 0 +

1 (1.8197)v2 - 4(1) - 10(2.3) 2

v = 5.4475 rad>s vP = 2.3(5.4475) = 12.529 ft>s Since the wall does not move, + b a:

e = 0.6 =

(vP) - 0 0 - ( - 12.529)

(vP) = 7.518 ft>s v¿ =

7.518 = 3.2685 rad>s 2.3

T3 + V3 = T4 + V4 1 (1.8197)(3.2685)2 - 4(1) - 10(2.3) = 0 - 4(1) sin u1 - 10(2.3 sin u1) 2 Ans.

u1 = 39.8°

Ans: u1 = 39.8° 1044

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20–1. The propeller of an airplane is rotating at a constant speed vs i, while the plane is undergoing a turn at a constant rate vt . Determine the angular acceleration of the propeller if (a) the turn is horizontal, i.e., vt k, and (b) the turn is vertical, downward, i.e., vt j.

z

Vs

SOLUTION (a) For vs, Æ = vt k.

y

x

# # (vs)XYZ = (vs)xyz + Æ * vs = 0 + (vt k) * (vs i) = vs vt j For vt, Æ = 0. # # (vt)XYZ = (vt)xyz + Æ * vtk = 0 + 0 = 0 # # # a = v = (vs)XYZ + (vt)XYZ a = vs vt j + 0 = vs vt j

Ans.

(b) For vs, Æ = vt j. # # (vs)XYZ = (vs)xyz + Æ * vs = 0 + (vt j) * (vs i) = - vs vt k For vt, Æ = 0. # # (vt)XYZ = (vt)xyz + Æ * vt = 0 + 0 = 0 # # # a = v = (vs)XYZ + (vt)XYZ a = -vs vt k + 0 = - vs vt k

Ans.

Ans: (a) A = vs vt j (b) A = - vs vt k 1045

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20–2. z

The disk rotates about the z axis at a constant rate vz = 0.5 rad>s without slipping on the horizontal plane. Determine the velocity and the acceleration of point A on the disk.

A

SOLUTION

Vz = 0.5 rad/s

Angular Velocity: The coordinate axes for the fixed frame (X, Y, Z) and rotating frame (x, y, z) at the instant shown are set to be coincident. Thus, the angular velocity of the disk at this instant (with reference to X, Y, Z) can be expressed in terms of i, j, k components. Since the disk rolls without slipping, then its angular velocity v = vs + vz is always directed along the instantaneuos axis of zero x velocity (y axis). Thus,

150 mm

300 mm

y

v = vs + vz - vj = - vs cos 30°j - vs sin 30°k + 0.5k Equating k and j components, we have 0 = - vs sin 30° + 0.5 - v = - 1.00 cos 30°

vs = 1.00 rad>s v = 0.8660 rad>s

Angular Acceleration: The angular acceleration a will be determined by investigating the time rate of change of angular velocity with respect to the fixed XYZ frame. Since v always lies in the fixed X–Y plane, then v = {- 0.8660j} rad>s is observed to have a constant direction from the rotating xyz frame if this frame is # rotating at Æ = vz = {0.5k} rad>s. Applying Eq. 20–6 with (v)xyz = 0, we have # # a = v = (v)xyz + vz * v = 0 + 0.5k * ( -0.8660j) = {0.4330i} rad>s2 Velocity and Acceleration:Applying Eqs. 20–3 and 20–4 with the v and a obtained above and rA = {(0.3 - 0.3 cos 60°)j + 0.3 sin 60°k} m = {0.15j + 0.2598k} m, we have vA = v * rA = ( - 0.8660j) * (0.15j + 0.2598k) = {-0.225i} m>s

Ans.

aA = a * rA + v * (v * rA) = (0.4330i) * (0.15j + 0.2598k) + ( - 0.8660j) * [( -0.8660j) * (0.15j + 0.2598k)] = {- 0.1125j - 0.130k} m s2

Ans.

Ans: vA = { - 0.225i} m>s

aA = { -0.1125j - 0.130k} m>s2

1046

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20–3. The ladder of the fire truck rotates around the z axis with an angular velocity v1 = 0.15 rad>s, which is increasing at 0.8 rad>s2. At the same instant it is rotating upward at a constant rate v2 = 0.6 rad>s. Determine the velocity and acceleration of point A located at the top of the ladder at this instant.

A

z

30

v1 40 ft

SOLUTION v = v1 + v2 = 0.15k + 0.6i = {0.6i + 0.15k} rad>s Angular acceleration: For v1, v = v1 = {0.15k} rad>s. # # (v2)XYZ = (v2)xyz + v * v2

v2

x

= 0 + (0.15k) * (0.6i) = {0.09j} rad>s2 For v1, Æ = 0. # # (v1)XYZ = (v1)xyz + v * v1 = (0.8k) + 0 = {0.8k} rad>s2 # # # a = v = (v1)XYZ + (v2)XYZ a = 0.8k + 0.09j = {0.09j + 0.8k} rad>s2 rA = 40 cos 30°j + 40 sin 30°k = {34.641j + 20k} ft vA = v * rA = (0.6i + 0.15k) * (34.641j + 20k) Ans.

= {-5.20i - 12j + 20.8k} ft>s aA = a * r + v * vA

= (0.09j + 0.8k) * (34.641j + 20k) + (0.6i + 0.15k) * ( -5.20i - 12j + 20.8k) = {- 24.1i - 13.3j - 7.20k} ft>s2

Ans.

1047

Ans: vA = 5-5.20i - 12j + 20.8k6 ft>s aA = 5- 24.1i - 13.3j - 7.20k6 ft>s2

y

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*20–4. The ladder of the fire truck rotates around the z axis with an angular velocity of v1 = 0.15 rad s, which is increasing at 0.2 rad>s2. At the same instant it is rotating upward at v2 = 0.6 rad>s while increasing at 0.4 rad>s2. Determine the velocity and acceleration of point A located at the top of the ladder at this instant.

A

z

30

v1 40 ft

SOLUTION rA/O = 40 cos 30°j + 40 sin 30°k rA/O = {34.641j + 20k} ft Æ = v1 k + v2 i = {0.6i + 0.15k} rad>s

v2

# # # v = v1 k + v2 i + v1 k * v2 i # Æ = 0.2k + 0.4i + 0.15k * 0.6i = {0.4i + 0.09j + 0.2k} rad>s2

x

vA = Æ * rA/O = (0.6i + 0.15k) * (34.641j + 20k) vA = { -5.20i - 12j + 20.8k} ft>s

Ans.

# aA = Æ * (Æ * rA/O) + v * rA/O aA = (0.6i + 0.15k) * [(0.6i + 0.15k) * (34.641j + 20k)] + (0.4i + 0.09j + 0.2k) * (34.641j + 20k) aA = { -3.33i - 21.3j + 6.66k} ft>s2

Ans.

Ans: vA = { - 5.20i - 12j + 20.8k} ft>s

aA = {-3.33i - 21.3j + 6.66k} ft>s2

1048

y

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20–5. y

If the plate gears A and B are rotating with the angular velocities shown, determine the angular velocity of gear C about the shaft DE. What is the angular velocity of DE about the y axis?

vA  5 rad/s

100 mm A

25 mm D

x

E

C

B

Solution The speeds of points P and P′, located at the top and bottom of gear C, are vp = (5)(0.1) = 0.5 m>s

100 mm vB  15 rad/s

vp′ = (15)(0.1) = 1.5 m>s The IC is located as shown. 0.5 1.5 = ;  x = 0.0125 m x (0.05 - x) vp vs = ;  vs = 8 vp 0.1 0.0125 v = vsi - vp j = vsi -

1 v j 8 p

v = v * r 0.5k = avpi i

0.5k = 5 vs - 0.1

1 v j b * ( - 0.1i + 0.025j) 8 s j 1 - vs 8 0.025

0.5 = 40 rad>s 0.0125 1 vp = (40) = 5 rad>s 8 vs =

k

0 5 = 0.0125vs k 0 Ans. (Angular velocity of C about DE) Ans. (Angular velocity of DE about y axis)

Ans: (vC)DE = 40 rad>s (vDE)y = 5 rad>s 1049

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20–6. z

The conical spool rolls on the plane without slipping. If the axle has an angular velocity of v1 = 3 rad>s and an angular acceleration of a1 = 2 rad>s2 at the instant shown, determine the angular velocity and angular acceleration of the spool at this instant.

v1

3 rad/s

a1

2 rad/s2 20

SOLUTION

y B

v1 = 3 rad>s v2 = -

A

3 = -8.7714 rad>s sin 20°

x

v = v1 + v2 = 3k- 8.7714 cos 20°j - 8.7714 sin 20°k Ans.

= {-8.24j} rad>s # A v1 B xyz = 2 rad>s2 #

A v2 B xyz = -

2 = -5.8476 rad>s2 sin 20°

# # # a = v = A v1 B xyz + v1 * v1 + A v2 B xyz + v1 * v2 = 2k + 0 + ( -5.8476 cos 20°j - 5.8476 sin 20°k) + (3k) * ( -8.7714 cos 20°j - 8.7714 sin 20°k) a = {24.7i- 5.49j} rad>s2

Ans.

Ans: V = { - 8.24j} rad>s A = {24.7i - 5.49j} rad>s2 1050

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20–7. At a given instant, the antenna has an angular motion # v1 = 3 rad>s and v1 = 2 rad>s2 about the z axis. At this same instant u = 30°, the angular motion about the x axis is # v2 = 1.5 rad>s, and v2 = 4 rad>s2. Determine the velocity and acceleration of the signal horn A at this instant. The distance from O to A is d = 3 ft.

z v1 v· 1 v2

d

v· 2 O

SOLUTION

A u

30

x

y

rA = 3 cos 30°j + 3 sin 30°k = {2.598j + 1.5k} ft Æ = v1 + v2 = 3k + 1.5i v A = Æ * rA vA = (3k + 1.5i) * (2.598j + 1.5k) = -7.794i + 3.897k - 2.25j Ans.

= { -7.79i - 2.25j + 3.90k} ft>s # # # Æ = v 1 + v2 = (2k + 0) + (4i + 3k * 1.5i) = 4i + 4.5j + 2k # aA = v * rA + Æ * vA

aA = (4i + 4.5j + 2k) * (2.598j + 1.5k) + (3k + 1.5i) * (-7.794i - 2.25j + 3.879k) aA = {8.30i - 35.2j + 7.02k} ft>s2

Ans.

1051

Ans: vA = 5-7.79i - 2.25j + 3.90k6ft>s aA = 58.30i - 35.2j + 7.02k6ft>s2

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*20–8. z

The disk rotates about the shaft S, while the shaft is turning about the z axis at a rate of vz = 4 rad>s, which is increasing at 2 rad>s2. Determine the velocity and acceleration of point A on the disk at the instant shown. No slipping occurs.

2 rad/s2 4 rad/s A

S

B

Solution

x

Angular Velocity. The instantaneous axis of zero velocity (IA) is indicated in Fig. a, Here, the resultant angular velocity is always directed along IA. The fixed XYZ reference frame is set to coincide with the rotating xyz frame. V = V1 + v2 5 226

vi -

1 226

vk = - 4k + v2i

Equating k and i components, -

1 226

5

226

v = -4    v = 4226 rad>s

1 4226 2 5

= v2  v2 = 20 rad>s

( 4226 ) i -

1

( 4226 ) k = 520i - 4k6 rad>s 226 226 Angular Acceleration. The direction of V2 does not change with reference to xyz rotating frame if this frame rotates with 𝛀 = V1 = 5 - 4k6 rad>s. Here # (v2)xyz 5 # # = ;  (v2)xyz = 5(v1)xyz = 5(2) = 10 rad>s2 # (v1)xyz 1 Thus, V =

Therefore # # V2 = (v2)xyz + 𝛀 * V2 = 10i + ( - 4k) * (20i) = 510i - 80j6 rad>s2

Since the direction of V1 will not change that is always along z axis when 𝛀 = V1, then # # V1 = (V1)xyz + V1 * V1 # # V1 = (v1)xyz = 5- 2k6 rad>s2

1052

y 0.1 m 0.5 m

0.1 m

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*20–8. Continued

Finally, # # A = v1 + v2   = -2k + 10i - 80j = 510i - 80j - 2k6 rad>s2

Velocity and Acceleration. Here rA = 50.5i + 0.1k6 m vA = V * rA = (20i - 4k) * (0.5i + 0.1k)



Ans.

= 5- 4.00j6 m>s

aA = A * rA + V * (V * rA)

= (10i - 80j - 2k) * (0.5i + 0.1k) + (20i - 4k) * ( - 4.00j)



= 5- 24i - 2j - 40k6 m>s2

Ans.

1053

Ans: vA = 5-4.00j6m>s aA = 5- 24i - 2j - 40k6 m>s2

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20–9. The disk rotates about the shaft S, while the shaft is turning about the z axis at a rate of vz = 4 rad>s, which is increasing at 2 rad>s2. Determine the velocity and acceleration of point B on the disk at the instant shown. No slipping occurs.

z

2 rad/s2 4 rad/s A

S

B

Solution

x

Angular velocity. The instantaneous axis of zero velocity (IA) is indicated in Fig. a. Here the resultant angular velocity is always directed along IA. The fixed XYZ reference frame is set coincide with the rotating xyz frame. V = V1 + V2 5 226

vi -

1 226

vk = - 4k + v2i

Equating k and i components, -

1 226

5

226

Thus, v =

v = - 4     v = 4226 rad>s

( 4226 ) = v2  v2 = 20 rad>s

5

226

( 4226 ) i -

1 226

( 4226 ) k = 520i - 4k6 rad>s

Angular Acceleration. The direction of V2 does not change with reference to xyz rotating frame if this frame rotates with 𝛀 = V1 = 5- 4k6 rad>s. Here # (v2)xyz 5 # # = ;  (v2)xyz = 5(v1)xyz = 5(2) = 10 rad>s2 # (v1)xyz 1 Therefore, # # v2 = (v2)xyz + 𝛀 * V2

= 10i + ( - 4k) * (20i)



= 510i - 80j6 rad>s2

Since the direction of V1 will not change that is always along z axis when 𝛀 = V1, then # # V1 = (v1)xyz + V1 * v1 # # V1 = (v1)xyz = 5- 2k6 rad>s2

1054

y 0.1 m 0.5 m

0.1 m

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20–9. Continued

Finally, # # a = V1 + V2   = -2k + (10i - 80j) = 510i - 80j - 2k6 rad>s2



Velocity and Acceleration. Here rB = 5 0.5i - 0.1j 6 m vB = V * rB = (20i - 4k) * (0.5i - 0.1j)



Ans.

= 5- 0.4i - 2j - 2k6 m>s

aB = A * rB + V * (V * rB)

= (10i - 80j - 2k) * (0.5i - 0.1j) + (20i - 4k) * ( -0.4i - 2j - 2k) = 5- 8.20i + 40.6j - k6 rad>s2

Ans.

1055

Ans: vB = 5- 0.4i - 2j - 2k6 m>s aB = 5- 8.20i + 40.6j - k6 rad>s2

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20–10. z

The electric fan is mounted on a swivel support such that the fan rotates about the z axis at a constant rate of vz = 1 rad>s and the fan blade is spinning at a constant rate vs = 60 rad>s. If f = 45° for the motion, determine the angular velocity and the angular acceleration of the blade.

Vz Vs f x

\

Solution v = vz + vs   = 1k + 60 cos 45°j + 60 sin 45°k   = 42.426j + 43.426k Ans.

  = 542.4j + 43.4k6 rad>s # # # v = vz + vs   = 0 + 0 + vz * vs   = 1k * 42.426j + 43.426k

Ans.

  = 5- 42.4i6 rad>s2

Ans.

Ans: V = {42.4j + 43.4k} rad>s A = { - 42.4i} rad>s2 1056

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20–11. z

The electric fan is mounted on a swivel support such that the fan rotates about the z axis at a constant rate of vz = 1 rad>s and the fan blade is spinning at# a constant rate vs = 60 rad>s. If  at  the instant f = 45°, f = 2 rad>s for the motion, determine the angular velocity and the angular acceleration of the blade.

Vz Vs f x

\

Solution v = vz + vs + vx = 1k + 60 cos 45°j + 60 sin 45°k + 2i = 2i + 42.426j + 43.426k Ans.

= 52i + 42.4j + 43.4k6 rad>s # # # # v = vz + vs + vx = 0 + (vz + vx) * vs + vz * vx = 0 + (1k + 2i) * (42.426j + 43.426k) + 1k * (2i) = -42.426i + 84.853k - 84.853j + 2j = 5- 42.4i - 82.9j + 84.9k6 rad>s2

Ans.

Ans: V = {2i + 42.4j + 43.4k} rad>s A = { - 42.4i - 82.9j + 84.9k} rad>s2 1057

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*20–12. The drill pipe P turns at a constant angular rate vP = 4 rad>s. Determine the angular velocity and angular acceleration of the conical rock bit, which rolls without slipping. Also, what are the velocity and acceleration of point A?

vP  4 rad/s

P

Solution A

v = v1 + v2 Since v acts along the instantaneous axis of zero velocity, vj = v1k + v2 cos 45°j + v2 sin 45°k

45

Setting v1 = 4 rad>s 50 mm

vj = 4k + 0.707v2 j + 0.707v2k Equating components v = 0.707v2 0 = 4 + 0.707v2 v = - 4 rad>s  v2 = - 5.66 rad>s Thus, Ans.

v = 5- 4.00j6 rad>s

Ω = v1 # # # v = v1 + v2

= 0 + v1 * (v) = 0 + (4k) * ( - 4j) # a = v = 516.016 rad>s2

Ans.

vA = v * rA

= ( - 4j) * [100(0.707)k] = 5- 282.816 mm>s

vA = 5- 0.283i6 m>s

Ans.

aA = a * rA + v * vA = (16i) * (100)(0.707)k + ( - 4j) * ( - 282.8i) = 5- 1131.2j - 1131.2k6 mm>s2

aA = 5- 1.13j - 1.13k6 m>s2

Ans.

Ans: v = 5- 4.00j6 rad>s

1058

a = 516.016 rad>s2 vA = 5- 0.283i6 m>s aA = 5- 1.13j - 1.13k6 m>s2

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20–13. z

The right circular cone rotates about the z axis at a constant rate of v1 = 4 rad>s without slipping on the horizontal plane. Determine the magnitudes of the velocity and acceleration of points B and C.

v1

4 rad/s

C

SOLUTION

50 mm

v = v 1 + v2 A

Since v acts along the instantaneous axis of zero velocity

B

x

vj = 4k + v2 cos 45° j + v2 sin 45°k. Equating components, v = 0.707 v2 0 = 4 + 0.707 v2 v = -4 rad>s,

v2 = - 5.66 rad>s

Thus, v = {-4j} rad>s Æ = v1 # # # v = v1 + v2 = 0 + v1 * v2 = 0 + (4k) * ( -5.66 cos 45°j - 5.66 sin 45°k) # a = v = {16i} rad>s2 vB = v * rB = (- 4j) * (0.1(0.707)j) = 0 Ans.

vB = 0 vC = v * rC = ( -4j) * (0.1(0.707)k) = {-0.2828i} m>s

Ans.

vC = 0.283 m>s aB = a * rB + v * vB = 16i * (0.1)(0.707)j + 0 aB = {1.131k} m>s aB = 1.13 m>s2

Ans.

aC = a * rC + v * vC = 16i * (0.1)(0.707)k + ( -4j) * ( -0.2828i) aC = {-1.131j - 1.131k} m>s2 aC = 1.60 m>s2

Ans.

Ans: vB = vC = aB = aC = 1059

0 0.283 m>s 1.13 m>s2 1.60 m>s2

y

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20–14. The wheel is spinning about shaft AB with an angular velocity of vs = 10 rad>s, which is increasing at a constant # rate of vs = 6 rad>s 2 , while the frame precesses about the z axis with an angular velocity of vp = 12 rad>s, which is # increasing at a constant rate of vp = 3 rad>s 2 . Determine the velocity and acceleration of point C located on the rim of the wheel at this instant.

vs vs

10 rad/s z 6 rad/s2 0.15 m

A

C

B y

x

SOLUTION The XYZ fixed reference frame is set to coincide with the rotating xyz reference frame at the instant considered. Thus, the angular velocity of the wheel at this instant can be obtained by vector addition of vs and vp.

vp vp

12 rad/s 3 rad/s2

v = vs + vp = [10j + 12k] rad>s The angular acceleration of the disk is determined from # # # a = v = v s + vp If we set the xyz rotating frame to have an angular velocity of Æ = vp = [12k] rad>s, the direction of vs will remain unchanged with respect to the xyz rotating frame which is along the y axis. Thus, # # vs = (vs)xyz + vp * vs = 6j + (12k) * (10j) = [- 120i + 6j] rad>s2 Since vp is always directed along the Z axis where Æ = vp, then # # vp = A vp B xyz + vp * vp = 3k + 0 = [3k] rad>s2 Thus, a = 1 -120i + 6j2 + 3k = [-120i + 6j + 3k] rad>s2 Here, rC = [0.15i] m, so that vC = v * rC = (10j + 12k) * (0.15i) = [1.8j - 1.5k] m>s

Ans.

and a C = a * rC + v * (v * rC) = ( - 120i + 6j + 3k) * (0.15i) + (10j + 12k) * [(10j + 12k) * (0.15i)] = [ -36.6i + 0.45j - 0.9k] m>s2

Ans.

1060

Ans: vC = 51.8j - 1.5k6 m>s aC = 5- 36.6i + 0.45j - 0.9k6 m>s2

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20–15. At the instant shown, the tower crane rotates about the z axis with an angular velocity v1 = 0.25 rad>s, which is increasing at 0.6 rad>s2. The boom OA rotates downward with an angular velocity v2 = 0.4 rad>s, which is increasing at 0.8 rad>s2. Determine the velocity and acceleration of point A located at the end of the boom at this instant.

z

v1

0.25 rad/s A

40 ft

SOLUTION

O

v = v1 + v2 = {-0.4 i + 0.25k} rad>s

x v2

30

0.4 rad/s

Æ = {0.25 k} rad>s

y

# # v = (v)xyz + Æ * v = ( - 0.8 i + 0.6k) + (0.25k) * ( -0.4 i + 0.25k) = {- 0.8i - 0.1j + 0.6k} rad>s2 rA = 40 cos 30°j + 40 sin 30°k = {34.64j + 20k} ft vA = v * rA = (1 - 0.4 i + 0.25 k) * (34.64j + 20k) vA = {- 8.66i + 8.00j - 13.9k} ft>s

Ans.

aA = a # rA + v * vA = ( - 0.8i -0.1j + 0.6k) * (34.64j + 20k) + ( -0.4i + 0.25k) * ( -8.66i + 8.00j - 13.9k) aA = {- 24.8i + 8.29j - 30.9k} ft>s2

Ans.

1061

Ans: vA = 5-8.66i + 8.00j - 13.9k6 ft>s aA = 5- 24.8i + 8.29j - 30.9k6 ft>s2

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*20–16. Gear A is fixed while gear B is free to rotate on the shaft S. If the shaft is turning about the z axis at vz = 5 rad>s, while increasing at 2 rad>s2, determine the velocity and acceleration of point P at the instant shown. The face of gear B lies in a vertical plane.

z Vz

P S

A

80 mm

B

80 mm y

160 mm

SOLUTION Æ = {5k - 10j} rad>s # Æ = {50i - 4j + 2k} rad>s2

x

vP = Æ * rP vP

(5k - 10j * (160j + 80k)

vP = {- 1600i} mm>s Ans.

= {-1.60i} m>s # aP = Æ * vP + Æ * rP aP = {50i - 4j + 2k} * (160j + 80k) + ( - 10j + 5k) * ( -1600i) aP = {- 640i - 12000j - 8000k} mm>s2 aP = {- 0.640i - 12.0j - 8.00k} m>s2

Ans.

1062

Ans: vP = 5- 1.60i6 m>s aP = 5- 0.640i - 12.0j - 8.00k6 m>s2

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20–17. z

The truncated double cone rotates about the z axis at vz = 0.4 rad>s without slipping on the horizontal plane. If # at this same instant vz is increasing at vz = 0.5 rad>s2, determine the velocity and acceleration of point A on the cone.

A vz

0.4 rad/s 1.5 ft

0.5 ft 30

SOLUTION u = sin

-1

1 ft

0.5 b = 30° a 1

x

2 ft y

0.4 = 0.8 rad>s vs = sin 30° v = 0.8 cos 30° = 0.6928 rad>s v = {-0.6928j} rad>s Æ = 0.4k # # v = (v)xyz + Æ * v

(1)

= 0.5k + (0.4k) * ( -0.6928j) # v = 0.2771i + 0.5k rA = (3 - 3 sin 30°)j + 3 cos 30°k = (1.5j + 2.598k) ft vA = v * rA = ( -0.6928j) * (1.5j + 2.598k) vA = { -1.80i} ft>s

Ans.

aA = a * rA + v * vA = (0.2771i + 0.5k) * (1.5j + 2.598k) + (- 0.6928j) * (- 1.80i) aA = { -0.750i - 0.720j - 0.831k} ft>s2

Ans.

Ans: vA = { - 1.80i} ft>s aA = { - 0.750i - 0.720j - 0.831k} ft>s2 1063

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20–18. Gear A is fixed to the crankshaft S, while gear C is fixed. Gear B and the propeller are free to rotate. The crankshaft is turning at 80 rad>s about its axis. Determine the magnitudes of the angular velocity of the propeller and the angular acceleration of gear B.

z 0.1 ft B 80 rad/s S

SOLUTION Point P on gear B has a speed of

0.4 ft

vP = 80(0.4) = 32 ft>s C

The IA is located along the points of contant of B and C

A

vs vP = 0.1 0.4 vs = 4vP v = - v P j + vs k = - vP j + 4vP k rP>O = 0.1j = 0.4k vP = -32i vP = v * rP/O i -32i = 3 0 0

j -vP 0.1

k 4vP 3 0.4

-32i = -0.8 vP i vP = 40 rad>s Ans.

vP = { -40j} rad>s vs = 4(40) k = {160k} rad>s Thus, v = vP + vs Let the x,y,z axes have an angular velocity of Æ * vP, then # # # a = v = vP + vs = 0 + vP * (vs + vP) a = (-40j) * (160k - 40j) a = {-6400i} rad>s2

Ans.

1064

Ans: VP = 5-40j6 rad>s AB = 5- 6400i6 rad>s2

y

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20–19. Shaft BD is connected to a ball-and-socket joint at B, and a beveled gear A is attached to its other end. The gear is in mesh with a fixed gear C. If the shaft and gear A are spinning with a constant angular velocity v1 = 8 rad>s, determine the angular velocity and angular acceleration of gear A.

y

300 mm

x

B

v1

SOLUTION g = tan

-1

75 = 14.04° 300

A

b = sin

-1

100 23002 + 752

= 18.87°

75 mm

100 mm D C

The resultant angular velocity v = v1 + v2 is always directed along the instantaneous axis of zero velocity IA. 8 v = sin 147.09° sin 18.87°

v = 13.44 rad>s

v = 13.44 sin 18.87° i + 13.44 cos 18.87° j Ans.

= {4.35i + 12.7j} rad>s v2 8 = sin 14.04° sin 18.87°

v2 = 6.00 rad>s

v2 = {6 j} rad>s v1 = 8 sin 32.91° i + 8 cos 32.91° j = {4.3466i + 6.7162j} rad>s For v1, Æ = v2 = {6j} rad>s (v1)xyz = (v1)xyz + Æ * v1 = 0 + (6j) * (4.3466i + 6.7162j) = {-26.08k} rad>s2 For v2, Æ = 0. # # (v2)XYZ = (v2)xyz + Æ * v2 = 0 + 0 = 0 # # # a = v = (v1)XYZ + (v2)XYZ a = 0 + ( -26.08k) = {-26.1k} rad>s2

Ans.

1065

Ans: V = 54.35i + 12.7j6 rad>s A = 5- 26.1k6 rad>s2

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*20–20. z

Gear B is driven by a motor mounted on turntable C. If gear A is held fixed, and the motor shaft rotates with a constant angular velocity of vy = 30 rad>s, determine the angular velocity and angular acceleration of gear B.

B vy

30 rad/s y

C

0.15 m

A

SOLUTION The angular velocity v of gear B is directed along the instantaneous axis of zero velocity, which is along the line where gears A and B mesh since gear A is held fixed. From Fig. a, the vector addition gives

0.3 m

v = vy + v z 2

vj -

25

1 25

vk = 30j - vzk

Equating the j and k components gives 2 25 -

v = 30

1 25

A 15 25 B = -vz

v = 15 25 rad>s vz = 15 rad>s

Thus, Ans.

v = [30j - 15k] rad>s

Here, we will set the XYZ fixed reference frame to coincide with the xyz rotating frame at the instant considered. If the xyz frame rotates with an angular velocity of Æ = vz = [ -15k] rad>s, then vy will always be directed along the y axis with respect to the xyz frame. Thus, # # vy = A vy B xyz + vz * vy = 0 + (- 15k) * (30j) = [450i] rad>s2 When Æ = vz, vz is always directed along the z axis. Therefore, # # vz = A vz B xyz + vz * vz = 0 + 0 = 0 Thus, # # a = vy + vz = (450i) + 0 = [450i] rad>s2

Ans.

Ans: v = [30j - 15k] rad>s a = [450i] rad>s2 1066

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20–21. Gear B is driven by a motor mounted on turntable C. If gear A and the motor shaft rotate with constant angular speeds of vA = {10k} rad>s and vy = {30j} rad>s, respectively, determine the angular velocity and angular acceleration of gear B.

z B vy  30 rad/s y C

0.15 m

A

SOLUTION If the angular velocity of the turn-table is vz, then the angular velocity of gear B is v = vy + vz = [30j + vzk] rad>s Since gear A rotates about the fixed axis (z axis), the velocity of the contact point P between gears A and B is

0.3 m

vp = vA * rA = (10k) * (0.3j) = [-3i] m>s Since gear B rotates about a fixed point O, the origin of the xyz frame, then rOP = [0.3j - 0.15k] m. vp = v * rOP - 3i = (30j + vzk) * (0.3j - 0.15k) -3i = - (4.5 + 0.3vz)i Thus, -3 = - (4.5 + 0.3vz) vz = -5 rad>s Then, Ans.

v = [30j - 5k] rad>s

Here, we will set the XYZ fixed reference frame to conincide with the xyz rotating frame at the instant considered. If the xyz frame rotates with an angular velocity of Æ = vz = [ -5k] rad>s, then vy will always be directed along the y axis with respect to the xyz frame. Thus, # # vy = A vy B xyz + vz * vy = 0 + (- 5k) * (30j) = [150i] rad>s2 When Æ = vz, vz is always directed along the z axis. Therefore, # # vz = A vz B xyz + vz * vz = 0 + 0 = 0 Thus, # # a = vy + vz = (150i + 0) = [150i] rad>s2

Ans.

1067

Ans: V = 530j - 5k6rad>s A = 5150i6rad>s2

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20–22. The crane boom OA rotates about the z axis with a constant angular velocity of v1 = 0.15 rad>s, while it is rotating downward with a constant angular velocity of v2 = 0.2 rad>s. Determine the velocity and acceleration of point A located at the end of the boom at the instant shown.

A

z

V1 50 ft

SOLUTION v = v1 + v2 = {0.2j + 0.15k} rad>s

110 ft O

Let the x, y, z axes rotate at Æ = v1, then

V2

# # v = (v)xyz + v1 * v2

y

x

# v = 0 + 0.15k * 0.2j = {-0.03i} rad>s2 rA = C 2(110)2 - (50)2 D i + 50k = {97.98i + 50k} ft vA = v * rA = 3

i 0 97.98

j 0.2 0

k 0.15 3 50

vA = {10i + 14.7j - 19.6 k} ft>s i aA = a * rA + v * vA = 3 -0.03 97.98

Ans. j 0 0

i k 0 3 + 3 0 10 50

j 0.2 14.7

k 0.15 3 - 19.6

aA = {-6.12i + 3j - 2k} ft>s2

Ans.

1068

Ans: vA = 510i + 14.7j - 19.6k6ft>s aA = 5- 6.12i + 3j - 2k6ft>s2

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20–23. The differential of an automobile allows the two rear wheels to rotate at different speeds when the automobile travels along a curve. For operation, the rear axles are attached to the wheels at one end and have beveled gears A and B on their other ends. The differential case D is placed over the left axle but can rotate about C independent of the axle. The case supports a pinion gear E on a shaft, which meshes with gears A and B. Finally, a ring gear G is fixed to the differential case so that the case rotates with the ring gear when the latter is driven by the drive pinion H. This gear, like the differential case, is free to rotate about the left wheel axle. If the drive pinion is turning at vH = 100 rad>s and the pinion gear E is spinning about its shaft at vE = 30 rad>s, determine the angular velocity, vA and vB , of each axle.

VH 50 mm

H

From motor z vE

40 mm 180 mm

VA

G

E

60 mm

vB

C To left wheel

O A

B

To right wheel

D

SOLUTION vP = vH rH = 100(50) = 5000 mm>s vG =

5000 = 27.78 rad>s 180

Point O is a fixed point of rotation for gears A, E, and B. Æ = vG + vE = {27.78j + 30k} rad>s vP ¿ = Æ * rP ¿ = (27.78j + 30k) * ( -40j + 60k) = {2866.7i} mm>s vA =

2866.7 = 47.8 rad>s 60

Ans.

vP ¿¿ = Æ * rP ¿¿ = (27.78j + 30k) * (40j + 60k) = {466.7i} mm>s vB =

466.7 = 7.78 rad>s 60

Ans.

Ans: vA = 47.8 rad>s vB = 7.78 rad>s 1069

y

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*20–24. The end C of the plate rests on the horizontal plane, while end points A and B are restricted to move along the grooved slots. If at the instant shown A is moving downward with a constant velocity of vA = 4 ft>s, determine the angular velocity of the plate and the velocities of points B and C.

z 0.4 ft B 0.8 ft A 1 ft vA y

Solution 2 ft

Velocity equation: vA = 5- 4k6 ft>s  vB = - vB j  v = vxi + vyj + vzk

2 ft

rB>A = 50.4j + 0.8k6 ft  rC>A = 52i + 2j - 1k6 ft

C

x

vC = (vC)xi + (vC)yj

vB = vA + v * rB>A i -vB j = ( - 4k) + 3 vx 0

j vy 0.4

k vz 3 0.8

Equating i, j and k components 0.8vy - 0.4vz = 0

(1)

0.8vx = vB

(2)

0.4vx - 4 = 0

(3)

vC = vA + V * rC>A i (vC)xi + (vC)y j = ( - 4k) + 3 vx 1

j vy 2

k vz 3 -1

Equating i, j and k components - vy - 2vz = (vc)x

(4)

2vz + vx = (vC)y

(5)

2vx - 2vy - 4 = 0

(6)

Solving Eqs. [1] to [6] yields: vx = 10 rad>s  vy = 8 rad>s  vz = 16 rad>s  vB = 8 ft>s (vC)x = -40 ft>s  (vC)y = 42 ft>s Then vB = { -8j} ft>s  vC = { - 40i + 42j} ft>s

Ans. Ans.

v = 510i + 8j + 16k6 rad>s

1070

Ans: vB = { - 8j} ft>s  vC = { - 40i + 42j} ft>s v = 510i + 8j + 16k6 rad>s

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20–25. Disk A rotates at a constant angular velocity of 10 rad>s. If rod BC is joined to the disk and a collar by ball-and-socket joints, determine the velocity of collar B at the instant shown. Also, what is the rod’s angular velocity V BC if it is directed perpendicular to the axis of the rod?

z B

D

E

300 mm

SOLUTION vC = {1i} m>s

200 mm C

vB = - vBj

vBC = vx i + vy j + vz k

rB>C = {-0.2i + 0.6j + 0.3k} m

x

100 mm

y v

10 rad/s

A 500 mm

vB = vC + vBC * rB>C i - vB = 1i + 3 vx - 0.2

j vy 0.6

k vz 3 0.3

Equating i, j, and k components 1 - 0.3vy - 0.6vz = 0

(1)

0.3vx + 0.2vz = vB

(2)

0.6vx + 0.2vy = 0

(3)

Since vBC is perpendicular to the axis of the rod, vBC # rB>C = (vx i + vy j + vzk) # ( -0.2i + 0.6j + 0.3k) = 0 (4)

- 0.2vx + 0.6vy + 0.3vz = 0 Solving Eqs. (1) to (4) yields: vx = 0.204 rad>s

vy = -0.612 rad>s

vz = 1.36 rad>s

vB = 0.333 m>s

Then vBC = {0.204i - 0.612j + 1.36k} rad>s

Ans.

vB = {- 0.333j} m>s

Ans.

1071

Ans: VBC = 50.204i - 0.612j + 1.36k6 rad>s vB = 5-0.333j6m>s

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20–26. z

Rod AB is attached to collars at its ends by using­ ball-and-socket joints. If collar A moves along the fixed rod at vA = 5 m>s, determine the angular velocity of the rod and the velocity of collar B at the instant shown. Assume that the rod’s angular velocity is directed perpendicular to the axis of the rod.

vA  5 m/s x

1m 2m

A B 45 y

Solution The velocities of collars A and B are vA = 55i6 m>s  vB = vB sin 45°j + vB cos 45°k =

1 22

vB j +

1 22

vBk

Also, rB>A = (0 - 1)i + (2 - 0)j + (0 - 0)k = 5-1i + 2j6 m and VAB = vxi + vy j + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

1 22 1

22

vB j + vB j +

1 22 1

22

vB k = 5i + (vxi + vy j)vzk * ( -1i + 2j) vBk = (5 - 2vz)i - vz j + (2vx + vy)k

Equating i, j and k components, (1)

0 = 5 - 2 vz  1 22 1

22

vB = - v z 

(2)

vB = 2vx + vy

(3)

Assuming that VAB is directed perpendicular to the axis of rod AB, then VAB # rB>A = 0

(vxi + vy j + vzk) # ( -1i + 2j) = 0 (4)

- vx + 2vy = 0 Solving Eqs. 1 to 4,

vx = - 1.00 rad>s  vy = - 0.500 rad>s  vz = 2.50 rad>s  vB = - 2.5022 m>s Then VAB = 5-1.00i - 0.500j + 2.50k6 rad>s

vB =

1

22

1 - 2.5022 2 j

+

1

22

1 - 2.5022 2 k

Ans. = 5-2.50j - 2.50k6 m>s Ans.

Note: vB can be obtained by solving Eqs. 1 and 2 without knowing the direction of VAB.

Ans: VAB = 5-1.00i - 0.500j + 2.50k6 rad>s 1072

vB = 5-2.50j - 2.50k6 m>s

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20–27. z

Rod AB is attached to collars at its ends by using ball-andsocket joints. If collar A moves along the fixed rod with a velocity of vA = 5 m>s and has an acceleration aA = 2 m>s2 at the instant shown, determine the angular acceleration of the rod and the acceleration of collar B at this instant. Assume that the rod’s angular velocity and angular acceleration are directed perpendicular to the axis of the rod.

vA  5 m/s x

1m 2m

A B 45 y

Solution The velocities of collars A and B are vA = 55i6 m>s  vB = vB sin 45°j + vB cos 45°k =

1 22

vB j +

1 22

vBk

Also, rB>A = (0 - 1)i + (2 - 0)j + (0 - 0)k = 5-1i + 2j6 m and VAB = vxi + vy j + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

1 22 1

22

vB j + vB j +

1 22 1

22

vB k = 5i + (vxi + vy j)vzk * ( -1i + 2j) vBk = (5 - 2vz)i - vz j + (2vx + vy)k

Equating i, j and k components, (1)

0 = 5 - 2 vz  1 22 1

22

vB = - v z 

(2)

vB = 2vx + vy

(3)

Assuming that VAB is directed perpendicular to the axis of rod AB, then VAB # rB>A = 0

(vxi + vy j + vzk) # ( -1i + 2j) = 0 (4)

- vx + 2vy = 0 Solving Eqs. 1 to 4,

vx = - 1.00 rad>s  vy = - 0.500 rad>s  vz = 2.50 rad>s  vB = - 2.5022 m>s Then VAB = 5- 1.00i - 0.500j + 2.50k6 rad>s

vB =

1

22

1 - 2.5022 2 j

+

1

22

1 - 2.5022 2 k

Ans. = 5-2.50j - 2.50k6 m>s Ans.

Note: vB can be obtained by solving Eqs. 1 and 2 without knowing the direction of VAB.

1073

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20–27. Continued

The accelerations of collars A and B are aA = 52i6 m>s2  aB = aB sin 45°j + aB cos 45°k =

Also, aAB = axi + ay j + azk

1 22

aB j +

1 22

aBk

Applying the relative acceleration equation, aB = aA + AAB * rB>A + VAB * ( VAB * rB>A ) 1 22 1 22

aB j +

aB j +

1 22 1 22

aBk = 2i + ( axi + ay j + azk ) * ( -1i + 2j) + ( - 1.00i - 0.500j + 2.50k) * [( -1.00i - 0.500j + 2.50k) * ( - 1i + 2j)] aBk = ( 9.5 - 2az ) i +

( - az - 15 ) j + ( 2ax + ay ) k

Equating i, j and k components 1 22 1

22

0 = 9.5 - 2az

(5)

aB = -az - 15

(6)

aB = zax + ay

(7)

Assuming that AAB is directed perpendicular to the axis of rod AB, then AAB # rB>A = 0

( axi + ay j + azk ) # ( -1i + 2j) = 0 (8)

- ax + 2xy = 0 Solving Eqs. 5 to 8,

ax = - 7.9 rad>s2  ay = - 3.95 rad>s2  az = 4.75 rad>s2  aB = - 19.7522 m>s2 Thus, AAB = 5 - 7.9i - 3.95j + 4.75k6 rad>s2  aB =

1

22

( - 19.7522 ) j +

1

22

Ans.

( - 19.7522 ) j = 5 -19.75j - 19.75k6 m>s2 Ans.

1074

Ans: AAB = 5- 7.9i - 3.95j + 4.75k6 rad>s2 aB = 5- 19.75j - 19.75k6 m>s2

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*20–28. z

If the rod is attached with ball-and-socket joints to smooth collars A and B at its end points, determine the velocity of B at the instant shown if A is moving upward at a constant speed of vA = 5 ft>s. Also, determine the angular velocity of the rod if it is directed perpendicular to the axis of the rod.

vA  5 ft/s A

3 ft x

2 ft

Solution

B

The velocities of collars A and B are

6 ft

vA = 55k6 ft>s  vB = - vB j

y

Also, rB>A = (6 - 0)i + (2 - 0)j + (0 - 3)k = 56i + 2j - 3k6 ft and VAB = vxi+ vy j + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

- vB j = 5k + ( vxi + vy j + vzk ) * (6i + 2j - 3k) -vB j =

( - 3vy - 2vz ) i + ( 3vx + 6vz ) j + ( 2vx - 6vy + 5 ) k

Equating i, j and k components, (1)

0 = - 3vy - 2vz

(2)

- vB = 3vx + 6vz

(3)

0 = 2vx - 6vy + 5 Assuming that VAB is directed perpendicular to the axis of rod AB, then, VAB # rB>A = 0

( vx i + vy j + vzk ) # (6i + 2j - 3k) = 0 (4)

6vx + 2vy - 3vz = 0 Solving Eqs. 1 to 4, vx = -

65 30 rad>s = - 0.6633 rad>s  vy = rad>s = 0.6122 rad>s 98 49

vz = -

45 rad>s = - 0.9183 rad>s  vB = 7.50 ft>s 49

Thus, Ans.

VAB = 5- 0.663i + 0.612j - 0.918k6 rad>s

Ans.

   vB = 5- 7.50j6 ft>s

1075

Ans: VAB = 5- 0.663i + 0.612j - 0.918k6 rad>s vB = 5- 7.50j6 ft>s

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20–29. z

If the collar at A in Prob. 20–28 is moving upward with an acceleration of aA = {-2k} ft>s2, at the instant its speed is vA = 5 ft>s, determine the acceleration of the collar at B at this instant.

vA  5 ft/s A

3 ft x

2 ft

Solution

B

The velocities of collars A and B are

6 ft

vA = 55k6 ft>s  vB = - vB j

y

Also, rB>A = (6 - 0)i + (2 - 0)j + (0 - 3)k = 56i + 2j - 3k6ft and VAB = vxi+ vy j + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

- vB j = 5k + ( vxi + vy j + vzk ) * (6i + 2j - 3k) -vB j =

( - 3vy - 2vz ) i + ( 3vx + 6vz ) j + ( 2vx - 6vy + 5 ) k

Equating i, j and k components, (1)

0 = - 3vy - 2vz

(2)

- vB = 3vx + 6vz

(3)

0 = 2vx - 6vy + 5 Assuming that VAB is directed perpendicular to the axis of rod AB, then, VAB # rB>A = 0

( vx i + vy j + vzk ) # (6i + 2j - 3k) = 0 (4)

6vx + 2vy - 3vz = 0 Solving Eqs. 1 to 4, vx = -

65 30 rad>s = - 0.6633 rad>s  vy = rad>s = 0.6122 rad>s 98 49

vz = -

45 rad>s = - 0.9183 rad>s  vB = 7.50 ft>s 49

Thus, Ans.

VAB = 5-0.663i + 0.612j - 0.918k6 rad>s vB = 5- 7.50j6 ft>s

Ans.

1076

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20–29. Continued

The accelerations of collars A and B are aA = 5- 2k6 ft>s2  aB = aBj

Also, aAB = axi + ay j + azk

Applying the relative acceleration equation, aB = aA + AAB * rB>A + VAB * ( VAB * rB>A ) aBj = - 2k + (axi + ay j + azk) * (6i + 2j - 3k)

+ ( - 0.6633i + 0.6122j - 0.9183k) * [( -0.6633i



+ 0.6122j - 0.9183k) * (6i + 2j - 3k)]

aBj = ( - 3ay - 2az - 9.9490)i + (3ax + 6az - 3.3163)j + (2ax - 6ay + 2.9745)k Equating i, j and k components, (5)

0 = - 3ay - 2az - 9.9490 aB = 3ax + 6az - 3.3163

(6)

0 = 2ax - 6ay + 2.9745

(7)

Eliminate ay from Eqs. 5 and 7 (8)

2ax + 4az = - 22.8724 Multiply Eq. 6 by

2 and rearrange, 3

2ax + 4az =

2 a + 2.2109 3 B

(9)

Equating Eqs. (8) and (9) -22.8724 =

2 a + 2.2109 3 B

aB = - 37.625 ft>s2

Thus, aB = 5-37.6j6 ft>s2

Ans.

Note: There is no need to know the direction of AAB to determine aB.

1077

Ans: aB = 5- 37.6j6 ft>s2

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20–30. z

Rod AB is attached to collars at its ends by ball-andsocket joints. If collar A has a speed vA = 4 m>s, determine the speed of collar B at the instant z = 2 m. Assume the angular velocity of the rod is directed perpendicular to the rod.

vA  4 m/s

2m

A B z

Solution

1.5 m

x

vB = vA + v * rB>A

1.5 m

1m

The velocities of collars A and B are 3 4 3 4 vA = 54k6 m>s  vB = - vB a b j + vB a bk = - vB j + vBk 5 5 5 5

y

3 Also, the coordinates of points A and B are A(1, 0, 2) m and B e 0, c 1.5 - 1.5a b d , 5 4 1.5 a b d = B(0, 0.6, 1.2) m. Thus, rB>A = (0 - 1)i + (0.6 - 0)j + (1.2 - 2)k 5 = 5 - 1i + 0.6j - 0.8k6m. Also vAB = vxi + vy j + vzk. Applying the relative velocity equation

vB = vA + vAB * rB>A 3 4 - vB j + vBk = 4k + ( vxi + vy j + vzk ) * ( -1i + 0.6j - 0.8k ) 5 5 3 4 - vB j + vBk = ( - 0.8vy - 0.6vz)i + (0.8vx - vz)j + (0.6vx + vy + 4)k 5 5 Equating i, j and k components (1)

0 = - 0.8vy - 0.6vz 3 - vB = 0.8vx - vz 5

(2)

4 v = 0.6vx + vy + 4 5 B

(3)

Assuming that VAB is perpendicular to the axis of the rod AB, then VAB # rB>A = 0

( vxi + vyj + vzk ) # ( - 1i + 0.6j - 0.8k) = 0 (4)

- vx + 0.6vy - 0.8vz = 0 Solving Eqs. (1) to (4), vx = - 1.20 rad>s  vy = - 0.720 rad>s  vz = 0.960 rad>s vB = 3.20 m>s 3 4 Then  vB = - (3.20)j - (3.20)k = 5- 1.92j + 2.56k6 m>s 5 5

Ans.

Note: vB can also be obtained by Solving Eqs. (1) to (3) without knowing the direction of VAB.

1078

Ans: vB = 5-1.92j + 2.56k6 m>s

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20–31. z

The rod is attached to smooth collars A and B at its ends using ball-and-socket joints. Determine the speed of B at the instant shown if A is moving at vA = 8 m>s. Also, determine the angular velocity of the rod if it is directed perpendicular to the axis of the rod.

vA  8 m/s

A

1m B y

1.5 m

Solution

2m

vB = vA + rB>A

x

The velocities of collars A and B are 3 4 3 4 vA = 58k6 m>s  vB = vBa bi - vBa bj = vB i - vB j 5 5 5 5

Also, rB>A = (0 - 0)i + (2 - 0)j + (0 - 1)k = 52j - 1k6m and VAB + vxi + vyj + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

3 4 vB i - vB j = 8k + (vxi + vy j + vzk) * (2j - 1k) 5 5 3 4 v i - vB j = ( - vy - 2vz)i + vx j + (2vx + 8)k 5 B 5 Equating i, j and k components, 3 v = -vy - 2vz 5 B

(1)

4 - vB = v x  5

(2)

0 = 2vx + 8

(3)

Assuming that VAB is perpendicular to the axis of rod AB, then VAB # rB>A = 0

(vxi + vy j + vzk) # (2j - 1k) = 0 (4)

2vy - vz = 0 Solving Eq (1) to (4) vx = - 4.00 rad>s  vy = - 0.600 rad>s  vz = -1.20 rad>s vB = 5.00 m>s

Ans.

vAB = 5- 4.00i - 0.600 j - 1.20k6 rad>s

Ans.

Then,

Note. vB can be obtained by solving Eqs (2) and (3) without knowing the direction of VAB.

1079

Ans: vB = 5.00 m>s VAB = 5-4.00i - 0.600j - 1.20k6 rad>s

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*20–32. z

If the collar A in Prob. 20–31 has a deceleration of aA = {-5k} m>s2, at the instant shown, determine the acceleration of collar B at this instant. vA  8 m/s

A

1m B y

1.5 m

Solution

2m

vB = vA + rB>A

x

The velocities of collars A and B are 3 4 3 4 vA = 58k6 m>s  vB = vBa bi - vBa bj = vB i - vB j 5 5 5 5

Also, rB>A = (0 - 0)i + (2 - 0)j + (0 - 1)k = 52j - 1k6 m and VAB + vxi + vyj + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

3 4 vB i - vB j = 8k + (vxi + vy j + vzk) * (2j - 1k) 5 5 3 4 v i - vB j = ( - vy - 2vz)i + vx j + (2vx + 8)k 5 B 5 Equating i, j and k components, 3 v = -vy - 2vz 5 B

(1)

4 - vB = v x  5

(2)

0 = 2vx + 8

(3)

Assuming that VAB is perpendicular to the axis of rod AB, then VAB # rB>A = 0

(vxi + vy j + vzk) # (2j - 1k) = 0 (4)

2vy - vz = 0 Solving Eq (1) to (4) vx = - 4.00 rad>s  vy = - 0.600 rad>s  vz = - 1.20 rad>s

Ans.

vB = 5.00 m>s Then, vAB = 5- 4.00i - 0.600 j - 1.20k6 rad>s

Ans.

Note. vB can be obtained by solving Eqs (2) and (3) without knowing the direction of VAB.

1080

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*20–32. Continued

The accelerations of collars A and B are 3 4 3 4 aA = 5- 5k6 m>s2  aB = - aBa bi + aB a bj = - aB i + aB j 5 5 5 5

Also, aAB = ax i + ay j + azk

Applying the relative acceleration equation, aB = aA + aAB * rB>A + VAB * (VAB * rA>B) 3 4 - aBi + aBj = - 5k + (axi + ayj + azk) * (2j - 1k) 5 5  + ( - 4.00i - 0.600j - 1.20k) * [( -4.00i - 0.600j - 1.20k) * (2j - 1k)] 3 4 - aBi + aBj = ( - ay - 2az)i + (ax - 35.6)j + (2ax + 12.8)k 5 5 Equating i, j and k components, 3 - aB = - ay + 2az 5

(5)

4 a = ax -35.6 5 B

(6)

0 = 2ax + 12.8

(7)

Solving Eqs (6) and (7), ax = - 6.40 rad>s2  aB = - 52.5 m>s2 Then

3 4 aB = - ( -52.5)i + ( -52.5)j 5 5 = 531.5i - 42.0j6 m>s2

Ans.

Note. It is not necessary to know the direction of AAB, if only aB needs to be determined.

1081

Ans: vB = 5.00 m>s vAB = 5-4.00i - 0.600 j - 1.20k6 rad>s aB = 531.5i - 42.0j6 m>s2

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20–33. z

Rod CD is attached to the rotating arms using ball-andsocket joints. If AC has the motion shown, determine the angular velocity of link BD at the instant shown.

vAC  3 rad/s v A C  2 rad/s2 0.4 m

0.8 m A

Solution

C

B

x

vD = vC + VAB * rD>C

D

The velocities of points C and D are

0.6 m

1m y

vC = VAC * rAC = 3k * 0.4 j = 5- 1.2i6 m>s vD = VBD * rBD = vBDj * 0.6i = - 0.6vBDk

Also, rD>C = (0.6 - 0)i + (1.2 - 0.4)j + (0.1)k = 50.6i + 0.8j - k6 m and vCD = vxi + vy j + vzk. Applying the relative velocity equation, vD = vC + VCD * rD>C

-0.6vBDk = -1.2i + (vxi + vy j + vzk) * (0.6i + 0.8j - k) -0.6vBDk = ( - vy - 0.8vz - 1.2)i + (vx + 0.6vz)j + (0.8vx - 0.6vy)k. Equating i, j and k components, - vy - 0.8vz - 1.2 = 0

(1)

vx + 0.6vz = 0

(2)

0.8vx - 0.6vy = - 0.6vBD

(3)

Assuming that VCD is perpendicular to the axis of rod CD, then VCD # rD>C = 0

(vxi + vy j) + vzk) # (0.6i + 0.8 j - k) = 0 0.6vx + 0.8vy - vz = 0(4) Solving Eqs (1) to (4) vx = 0.288 rad>s   vy = - 0.816 rad>s   vz = -0.480 rad>s vBD = - 1.20 rad>s Thus

vBD = 5-1.20 j6 rad>s

Ans.

Note: VBD can be obtained by solving Eqs 1 to 3 without knowing the direction of VAB.

Ans: VBD = { - 1.20j} rad>s 1082

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20–34. z

Rod CD is attached to the rotating arms using balland-socket joints. If AC has the motion shown, determine the angular acceleration of link BD at this instant.

vAC  3 rad/s v A C  2 rad/s2 0.4 m

0.8 m A

Solution

B

x

vD = vC + VAB * rD>C

D

The velocities of points C and D are vC = VAC * rAC = 3k * 0.4 j = 5- 1.2i6 m>s vD = VBD * rBD = vBDj * 0.6i = - 0.6vBDk

Also, rD>C = (0.6 - 0)i + (1.2 - 0.4)j + (0.1)k = 50.6i + 0.8j - k6 m and vCD = vxi + vy j + vzk. Applying the relative velocity equation, vD = vC + VCD * rD>C

- 0.6vBDk = -1.2i + (vxi + vy j + vzk) * (0.6i + 0.8j - k) - 0.6vBDk = ( - vy - 0.8vz - 1.2)i + (vx + 0.6vz)j + (0.8vx - 0.6vy)k. Equating i, j and k components, - vy - 0.8vz - 1.2 = 0

(1)

vx + 0.6vz = 0

(2)

0.8vx - 0.6vy = - 0.6vBD

(3)

Assuming that VCD is perpendicular to the axis of rod CD, then VCD # rD>C = 0

(vxi + vy j) + vzk) # (0.6i + 0.8j - k) = 0 0.6vx + 0.8vy - vz = 0(4) Solving Eqs (1) to (4) vx = 0.288 rad>s   vy = - 0.816 rad>s   vz = -0.480 rad>s vBD = - 1.20 rad>s Thus

C

vBD = 5- 1.20 j6 rad>s

Ans.

Note: VBD can be obtained by solving Eqs 1 to 3 without knowing the direction of VCD.

1083

0.6 m

1m y

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20–34. Continued

The accelerations of points C and D are aC = AAC * rAC - vAC2rAC = (2k * 0.4j) - 32(0.4j) = 5- 0.8i, - 3.6j6 m>s

aD = ABD * rBD - vBD2rBD = (aBD j * 0.6i) - 1.202(0.6i) = - 0.864i - 0.6aBDk

Also, aCD = axi + ayj + azk and VCD = 50.288i - 0.816j - 0.480k6 rad>s

Applying the relative acceleration equation,

aD = aC + ACD * rD>C + VCD * (VCD * rD>C) -0.864i - 0.6aBDk = ( - 0.8i - 3.6j) + (axi + ay j + azk) * (0.6i + 0.8j - k) + (0.288i - 0.816j - 0.480k) * [(0.288i - 0.816j - 0.480k) * (0.6i + 0.8j - k)] -0.864i - 0.6aBDk = ( - ay - 0.8az - 1.38752)i + (ax + 0.6az - 4.38336)j  + (0.8ax - 0.6ay + 0.9792)k Equating i, j and k components, -0.864 = -ay - 0.8az - 1.38752

(5)

0 = ax + 0.6az - 4.38336 

(6)

-0.6aBD = 0.8ax - 0.6ay + 0.9792

(7)

Assuming that ACD is perpendicular to the axis of rod CD, then ACD # rD>C = 0

(axi + ay j + azk) # (0.6i + 0.8j - k) = 0 (8)

0.6ax + 0.8ay - az = 0 Solving Eqs. 5 to 8, ax = 3.72 rad>s2  ay = - 1.408 rad>s2  az = 1.1056 rad>s2 aBD = - 8.00 rad>s2 Thus,

aBD = 5- 8.00j6 rad>s2

Ans.

Note: aBD can be obtained by solving Eqs 5 to 7 without knowing the direction of ACD.

1084

Ans: ABD = 5- 8.00j6 rad>s2

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20–35. n

Solve Prob. 20–28 if the connection at B consists of a pin as shown in the figure below, rather than a ball-andsocket joint. Hint: The constraint allows rotation of the rod both along the bar (j direction) and along the axis of the pin (n  direction). Since there is no rotational component in the u  direction, i.e., perpendicular to n and j where u = j : n, an additional equation for solution can be obtained from V # u = 0. The vector n is in the same direction as rD>B : rC>B.

B j D

Solution The velocities of collars A and B are vA = 55 k6 ft>s  vB = - vB j

Also, rB>A = (6 - 0)i + (2 - 0)j + (0 - 3)k = 56i + 2j - 3k6 ft and vAB = vxi + vy j + vzk. Applying the relative velocity equation, vB = vA + VAB * rB>A

- vB j = 5k + (vxi + vy j + vzk) * (6i + 2j - 3k) - vB j = ( -3vy - 2vz)i + (3vx + 6vz)j + (2vx - 6vy + 5)k Equating i, j and k components, 0 = - 3vy - 2vz - vB = 3vx + 6vz



(1)



(2) (3)

0 = 2 vx - 6vy + 5 Here, rB>A * j = (6i + 2j - 3k) * j = 3i + 6k Then n = Thus

rB>A * j

0 rB>A * j 0

u = j * n = j * ¢ It is required that

=

3i + 6k 2

23 + 6

3 245

i +

2

6 245

=

3 245

k≤ =

i +

6 245

6 245

i -

k

3 245

k

VAB # u = 0 (vxi + vy j + vzk) # a 6

245

vx -

3

245

6 245

vz = 0

i -

3 245

k≤ = 0

(4)

2vx - vz = 0

1085

u

C

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20–35. Continued

Solving Eqs (1) to (4) vx = -0.500 rad>s  vy = 0.6667 rad>s  vz = -1.00 rad>s   vB = 7.50 ft>s Thus, Ans.

VAB = 5- 0.500i + 0.667j - 1.00k6 rad>s

Ans.

vB = 5-7.50j6 ft>s

Ans: VAB = 5-0.500i + 0.667j - 1.00k6 rad>s vB = { - 7.50j} ft>s 1086

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*20–36. Member ABC is pin connected at A and has a ball-andsocket joint at B. If the collar at B is moving along the inclined rod at vB = 8 m>s, determine the velocity of point C at the instant shown. Hint: See Prob. 20–35.

z vB  8 m/s B 30

1.5 m

C

A x

2m

1m

y

Solution Velocities of collars A and B are vA = vA k  vB = 8 cos 30° j - 8 sin 30° k =

5423 j

- 4 k6 m>s

Also, rA>B = 52 i - 1.5 k6 m and vAB = vx i + vy j + vz k. Applying the relative velocity equation, vA = vB + VAB * rA>B vA k =

1 423j

- 4k 2 + (vxi + vy j + vz k) * (2i - 1.5k)

vAk = -1.5vy i +

1 1.5vx

+ 2vz + 423 2 j + ( -2vy - 4)k

Equating i, j and k components, 0 = -1.5 vy  vy = 0

(1)

0 = 1.5 vx + 2vz + 423 vA = -2(0) - 4  vA = - 4 m>s Here n = j. Then u = k * n = k * j = - i. It is required that VAB # u = 0

(vxi + vy j + vzk) # ( -i) = 0 - vx = 0  vx = 0 Substitute this result into Eq (1), 0 = 1.5(0) + 2 vz + 423 vz = - 223 rad>s Thus, vAB = 5- 223k6 rad>s

Here, rC>B = 51j - 1.5k6. Using the result of VAB, vC = vB + VAB * rC>B

vC = (423j - 4k) + ( - 223k) * (1j - 1.5k) = 5 213i + 413j - 4k6 m>s

Ans.

= 53.46i + 6.93j - 4k6 m>s

1087

Ans: vC = 53.46i + 6.93j - 4k6 m>s

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20–37. Solve Example 20.5 such that the x, y, z axes move with curvilinear translation, Æ = 0 in which case the collar appears to have both an angular velocity æxyz = V1 + V2 and radial motion.

SOLUTION Relative to XYZ, let xyz have Æ = 0

# Æ = 0

rB = {- 0.5k} m vB = {2j} m>s aB = {0.75j + 8k} m>s2 Relative to xyz, let x¿ y¿ z¿ be coincident with xyz and be fixed to BD. Then Æ xyz = v1 + v2 = {4i + 5k} rad>s

# # # vxyz = v1 + v2 = {1.5i - 6k} rad>s2

(rC>B)xyz = {0.2j} m # # (vC>B)xyz = (rC>B)xyz = (rC>B)x¿y¿z¿ + (v1 + v2) * (rC>B)xyz = 3j + (4i + 5k) * (0.2j) = {- 1i + 3j + 0.8k} m>s $ $ # (aC>B)xyz = (rC>B)xyz = C (rC>B)x¿y¿z¿ + (v1 + v2) * (rC>B)x¿y¿z¿ D # # # + C (v1 + v2) * (rC>B)xyz D + C (v1 + v2) * (rC>B)xyz D (aC>B)xyz = C 2j + (4i + 5k) * 3j D + C (1.5i - 6k) * 0.2j D + C (4i + 5k) * ( -1i + 3j + 0.8k) D = {- 28.8i - 6.2j + 24.3k} m>s2 vC = vB + Æ * rC>B + (vC>B)xyz = 2j + 0 + ( -1i + 3j + 0.8k) Ans.

= {- 1.00i + 5.00j + 0.800k} m>s aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz = (0.75j + 8k) + 0 + 0 + 0 + ( -28.8i - 6.2j + 24.3k) = { - 28.8i - 5.45j + 32.3k} m>s2

Ans.

1088

Ans: vC = 5-1.00i + 5.00j + 0.800k6 m>s aC = 5- 28.8i - 5.45j + 32.3k6 m>s2

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20–38. Solve Example 20.5 by fixing x, y, z axes to rod BD so that æ = V1 + V2. In this case the collar appears only to move radially outward along BD; hence æxyz = 0.

SOLUTION

# # Relative to XYZ, let x¿ y¿ z¿ be concident with XYZ and have Æ¿ = v1 and Æ ¿ = v1 # # # v = v1 + v2 = {4i + 5k} rad>s # # # # v = v1 + v2 = c a v1 b

x¿y¿z¿

+ v1 * v1 d + c a v2 b

x¿y¿z¿

+ v1 * v2 d

= (1.5i + 0) + C - 6k + (4i) * (5k) D = {1.5i - 20j - 6k} rad>s2 rB = {- 0.5k} m # # vB = rB = a rB b

x¿y¿z¿

# $ a B = rB = c a r B b

+ v1 * rB = 0 + (4i) * ( -0.5k) = {2j} m>s

x¿y¿z¿

# + v1 * arB b

x¿y¿z¿

# # d + v1 * rB + v1 * rB

= 0 + 0 + C (1.5i) * ( -0.5k) D + (4i * 2j) = {0.75j + 8k} m>s2 Relative to x¿y¿z¿ , let xyz have Æ x¿y¿z¿ = 0; a rC>B b

xyz

# Æ x¿y¿z¿ = 0;

= {0.2j} m

(vC>B)xyz = {3j} m>s (aC>B)xyz = {2j} m>s2 vC = vB + Æ * rC>B + (vC>B)xyz = 2j + C (4i + 5k) * (0.2j) D + 3j Ans.

= { -1i + 5j + 0.8k}m>s aC = aB + Æ * rC>B + Æ * (Æ * rC>B) + 2Æ * (vC>B)xyz + (aC>B)xyz

= (0.75j + 8k) + C (1.5i - 20j - 6k) * (0.2j) D + (4i + 5k) * C (4i + 5k) * (0.2j) D + 2 C (4i + 5k) * (3j) D + 2j aC = { -28.2i - 5.45j + 32.3k} m>s2

Ans.

1089

Ans: vC = 5-1i + 5j + 0.8k6 m>s aC = 5- 28.2i - 5.45j + 32.3k6 m>s2

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20–39. At the instant u = 60°, the telescopic boom AB of the construction lift is rotating with a constant angular velocity about the z axis of v1 = 0.5 rad>s and about the pin at A with a constant angular speed of v2 = 0.25 rad>s. Simultaneously, the boom is extending with a velocity of 1.5 ft>s, and it has an acceleration of 0.5 ft>s2, both measured relative to the construction lift. Determine the velocity and acceleration of point B located at the end of the boom at this instant.

z B v1, v1 15 ft

u

SOLUTION

v2, v2

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = 0 Æ = v1 = {0.5k} rad>s

x

2 ft

A

O C

Since point A rotates about a fixed axis (Z axis), its motion can determined from vA = v1 * rOA = (0.5k) * (- 2j) = {1i} ft>s and # aA = v1 * rOA + v1 * (v1 * rOA) = 0 + (0.5k) * (0.5k) * ( - 2j) = {0.5j} ft>s2 In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity of Æ¿ = v2 = {0.25i} rad>s, the direction of rB>A will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of rB>A, # # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * rB>A D = (1.5 cos 60°j + 1.5 sin 60°k) + 0.25i * (15 cos 60°j + 15 sin 60°k) = { -2.4976j + 3.1740k} ft>s Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ = v2 = 0. Taking the time derivative of (rB>A)xyz, ## ## # # # (aB>A)xyz = (r B>A)xyz = C (r>A)x¿y¿z¿ + v2 * (rB>A)x¿y¿z¿ D + v2 * rB>A + v2 * (rB>A)xyz = C (0.5 cos 60°j + 0.5 sin 60°k) + 0.25i * (1.5 cos 60°j + 1.5 sin 60°k) D + 0.25i * ( -2.4976j + 3.1740k) = {-0.8683j - 0.003886k} ft>s2

1090

y

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20–39. Continued

Thus, vB = vA + Æ * rB>A + (vB>A)xyz = (1i) + (0.5k) * (15 cos 60°j + 15 sin 60°k) + ( - 2.4976j + 3.1740k) Ans.

= { - 2.75i - 2.50j + 3.17k} m>s and

# aB = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz = 0.5j + 0 + 0.5k * C (0.5k) * (15 cos 60°j + 15 sin 60°k) D + 2(0.5k) * ( -2.4976j + 3.1740k) + ( - 0.8683j - 0.003886k) = {2.50i - 2.24j - 0.00389k} ft>s2

Ans.

1091

Ans: vB = 5-2.75i - 2.50j + 3.17k6 m>s aB = 52.50i - 2.24j - 0.00389k6 ft>s2

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*20–40. At the instant u = 60°, the construction lift is rotating about the z axis with an angular velocity of v1 = 0.5 rad>s and an # angular acceleration of v1 = 0.25 rad>s2 while the telescopic boom AB rotates about the pin at A with an angular velocity of v2 = 0.25 rad>s and angular # acceleration of v2 = 0.1 rad>s2. Simultaneously, the boom is extending with a velocity of 1.5 ft> s, and it has an acceleration of 0.5 ft> s2, both measured relative to the frame. Determine the velocity and acceleration of point B located at the end of the boom at this instant.

z B v1, v1 15 ft

u v2, v2

SOLUTION The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = {0.25k} rad>s2 Æ = v1 = {0.5k} rad>s Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (0.5k) * (- 2j) = {1i} ft>s # aA = v1 * rOA + v1 * (v1 * rOA) = (0.25k) * (- 2j) + (0 .5k) * [0.5k * ( -2j)] = {0.5i + 0.5j} ft>s2 In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity of Æ¿ = v2 = [0.25i] rad>s, the direction of rB>A will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of rB>A, # # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * rB>A D = (1.5 cos 60°j + 1.5 sin 60°k) + [0.25i * (15 cos 60°j + 15 sin 60°k)] = { - 2.4976j + 3.1740k} ft>s Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ = v2 = [0.1i] rad>s2. Taking the time derivative of (rB>A)xyz, $ $ # # (aB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)x¿y¿z¿ D + v2 * rB>A # + v2 * (rB>A)xyz = (0.5 cos 60°j + 0.5 sin 60°k) + (0.25i) * (1.5 cos 60°j + 1.5 sin 60°k) + (0.1i) * (15 cos 60°j + 15 sin 60°k) + (0.25i) * ( - 2.4976j + 3.1740k) = {-2.1673j + 0.7461k} ft>s2

1092

x

2 ft

A

O C

y

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20–40. Continued

Thus, vB = vA + Æ * rB>A + (vB>A)xyz = [1i] + (0.5k) * (15 cos 60°j + 15 sin 60°k) + ( -2.4976j + 3.1740k) Ans.

= {-2.75i - 2.50j + 3.17k} ft>s and # aB = a A + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz = (0.5i + 0.5j) + (0.25k) * (15 cos 60°j + 15 sin 60°k) + (0.5k) * [(0.5k) * (15 cos 60°j + 15 sin 60°k)] + 2 (0.5k) * ( - 2.4976j + 3.1740k) + (-2.1673j + 0.7461k) = {1.12i - 3.54j + 0.746k} ft>s2

Ans.

1093

Ans: vB = 5- 2.75i - 2.50j + 3.17k6 ft>s aB = 51.12i - 3.54j + 0.746k6 ft>s2

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20–41. At the instant shown, the arm AB is rotating about the fixed pin A with an angular velocity v1 = 4 rad>s and angular acceleration # v 1 = 3 rad>s2. At this same instant, rod BD is rotating relative to rod AB with an angular velocity v2 = 5 rad>s, which is increasing # at v 2 = 7 rad>s2. Also, the collar C is moving along rod BD with a velocity of 3 m>s and an acceleration of 2 m>s2, both measured relative to the rod. Determine the velocity and acceleration of the collar at this instant.

z

v1  4 rad/s v 1  3 rad/s2

A

1.5 m 3 m/s 2 m/s2

Solution

D

Ω = 54i6 rad>s # Ω = 53i6 rad>s2

x

C

B 0.6 m

v2  5 rad/s v 2  7 rad/s2 y

rB = 5-1.5k6 m # vB = (rB)xyz + Ω * rB = 0 + (4i) * ( -1.5k) = 56j6 m>s # $ $ # # aB = rB = 3 (rB)xyz + Ω * (rB)xyz 4 + Ω * rB + Ω * rB = 0 + 0 + (3i) * ( - 1.5k) + (4i) * (6j) = 54.5j + 24k6 m>s2

Ω C>B = 55j6 rad>s # Ω C>B = 57 j6 rad>s2

rC>B = 50.6i6 m # (vC>B)xyz = (rC>B)xyz + Ω C>B * rC>B = (3i) + (5j) * (0.6i)

(aC>B)xyz

= 53i - 3k6 m>s # $ # # = 3 (rC>B)xyz + Ω C>B * (rC>B)xyz 4 + Ω C>B * rC>B + Ω C>B * rC>B = (2i) + (5j) * (3i) + (7j) * (0.6i) + (5j) * (3i - 3k) = 5 -13i - 34.2k6 m>s2

vC = vB + Ω * rC>B + (vC>B)xyz

= (6j) + (4i) * (0.6i) + (3i - 3k) vC = 53i + 6j - 3k6 m>s # aC = aB + Ω * rC>B + Ω * (Ω * rC>B) + 2Ω * (vC>B)xyz + (aC>B)xyz

Ans.

= (4.5j + 24k) + (3i) * (0.6i) + (4i) * [(4i) * (0.6i)]     + 2(4i) * (3i - 3k) + ( - 13i - 34.2k) aC = 5- 13.0i + 28.5j - 10.2k6 m>s2

Ans.

Ans: vC = {3i + 6j - 3k} m>s aC = { - 13.0i + 28.5j - 10.2k} m>s2 1094

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20–42. At the instant u = 30°, the frame of the crane and the boom AB rotate with a constant angular velocity of v1 = 1.5 rad>s and v2 = 0.5 rad>s, respectively. Determine the velocity and acceleration of point B at this instant.

z V1, V1 1.5 m

O

SOLUTION

A

12 m

B

u

y

V2, V2

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. The angular velocity and angular acceleration of this frame with respect to the XYZ frame are # # Æ = v1 = [1.5k] rad>s Æ = v1 = 0 Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (1.5k) * (1.5j) = [- 2.25i] m>s # aA = v1 * rOA + v1 * (v1 * rOA) = 0 + (1.5k) * [(1.5k) * (1.5j)] = [- 3.375j] m>s2 In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity relative to the xyz frame of Æ¿ = v2 = [0.5i] rad>s, the direction of (rB>A)xyz will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of (rB>A)xyz, # # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)xyz D = 0 + (0.5i) * (12 cos 30° j + 12 sin 30°k) = [- 3j + 5.196k] m>s Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ ¿ = v2 = 0. Taking the time derivative of (rA>B)xyz, $ $ # # # (aA>B)xyz = (rA>B)xyz = C (rA>B)x¿y¿z¿ + v2 * (rA>B)x¿y¿z¿ D + v2 * (rA>B)xyz + v2 * (rA>B)xyz = [0 + 0] + 0 + (0.5i) * ( - 3j + 5.196k) = [- 2.598j - 1.5k] m>s2 Thus, vB = vA + Æ * rB>A + (vB>A)xyz = (- 2.25i) + 1.5k * (12 cos 30° j + 12 sin 30° k) + ( - 3j + 5.196k) Ans.

= [- 17.8i - 3j + 5.20k] m>s and

# aB = aA + Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vAB)xyz + (aAB)xyz = ( - 3.375j) + 0 + 1.5k * C (1.5k) * (12 cos 30° j + 12 sin 30° k) D + 2(1.5k) * (-3j + 5.196k) + ( -2.598j - 1.5k) = [9i - 29.4j - 1.5k] m>s2

Ans.

1095

Ans: vB = 5- 17.8i - 3j + 5.20k6 m>s aB = 59i - 29.4j - 1.5k6 m>s2

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20–43. At the instant u = 30°, the frame of the crane is rotating with an angular velocity of v1 = 1.5 rad>s and angular # acceleration of v1 = 0.5 rad>s2, while the boom AB rotates with an angular velocity of v2 = 0.5 rad>s and angular # acceleration of v2 = 0.25 rad>s2. Determine the velocity and acceleration of point B at this instant.

z V1, V1 1.5 m

O

SOLUTION

A

u V2, V2

The xyz rotating frame is set parallel to the fixed XYZ frame with its origin attached to point A, Fig. a. Thus, the angular velocity and angular acceleration of this frame with respect to the XYZ frame are # Æ = v1 = [1.5k] rad>s Æ = [0.5k] rad>s2 Since point A rotates about a fixed axis (Z axis), its motion can be determined from vA = v1 * rOA = (1.5k) * (1.5j) = [ -2.25i] m>s # aA = v1 * rOA + v1 * (v1 * rOA) = (0.5k) * (1.5j) + (1.5k) * C (1.5k) * (1.5j) D = [ -0.75i - 3.375j] m>s2 In order to determine the motion of point B relative to point A, it is necessary to establish a second x¿y¿z¿ rotating frame that coincides with the xyz frame at the instant considered, Fig. a. If we set the x¿y¿z¿ frame to have an angular velocity relative to the xyz frame of Æ¿ = v2 = [0.5i] rad>s, the direction of (rB>A)xyz will remain unchanged with respect to the x¿y¿z¿ frame. Taking the time derivative of (rB>A)xyz, # # (vB>A)xyz = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)xyz D = 0 + (0.5i) * (12 cos 30° j + 12 sin 30° k) = [ -3j + 5.196k] m>s Since Æ ¿ = v2 has a constant direction with respect to the xyz frame, then # # # Æ ¿ = v2 = [0.25i] m>s2. Taking the time derivative of (rB>A)xyz, # $ $ # # (aB>A) = (rB>A)xyz = C (rB>A)x¿y¿z¿ + v2 * (rB>A)x¿y¿z¿ D + Æ 2 * (rB>A)xyz + v2 * (rB>A)xyz = [0 + 0] + (0.25i) * (12 cos 30° j + 12 sin 30°k) + 0.5i * ( -3j + 5.196k) = [- 4.098j + 1.098k] m>s2

1096

12 m

B

y

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20–43. Continued

Thus, vB = vA + Æ * rB>A + (vB>A)xyz = ( -2.25i) + 1.5k * (12 cos 30° j + 12 sin 30°k) + ( -3j + 5.196k) Ans.

= [ - 17.8i - 3j + 5.20k] m>s and

# aB = aA = Æ * rB>A + Æ * (Æ * rB>A) + 2Æ * (vB>A)xyz + (aB>A)xyz = (- 0.75i - 3.375j) + 0.5k * (12 cos 30°j + 12 sin 30° k) + (1.5k) * C (1.5k) * (12 cos 30° j + 12 sin 30° k) D + 2(1.5k) * ( -3j + 5.196k) + (- 4.098j + 1.098k) = [3.05i - 30.9j + 1.10k] m>s2

Ans.

1097

Ans: vB = 5- 17.8i - 3j + 5.20k6 m>s aB = 53.05i - 30.9j + 1.10k6 m>s2

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*20–44. z

At the instant shown, the rod AB is rotating about the z axis with an angular velocity v1 = 4 rad>s and an angular acceleration v# 1 = 3 rad>s2. At this same instant, the circular rod has an angular motion relative to the rod as shown. If the collar C is moving down around the circular rod with a speed of 3 in.>s, which is increasing at 8 in.>s2, both measured relative to the rod, determine the collar’s velocity and acceleration at this instant.

v1  4 rad/s v 1  3 rad/s2 A B 5 in.

Solution Ω = 54k6 rad>s # Ω = 53k6 rad>s2

y

v2  2 rad/s v 2  8 rad/s2

x

C

4 in.

rB = 55j6 in.

vB = (4k) * (5j) = 5- 20i6 in.>s # $ $ # # aB = rB = 3 (rB)xyz + Ω * (rB)xyz 4 + Ω * rB + Ω * rB = (4k) * ( -20i) + (3k) * (5j) = 5- 15i - 80j6 in.>s2

Ω C>B = 52i6 rad>s # Ω C>B = 58i6 rad>s2

rC>B = 54i - 4k6 in. # (vC>B)xyz = (rC>B)xyz + Ω C>B * rC>B = ( -3k) + (2i) * (4i - 4k)

(aC>B)xyz

 

= 58j - 3k6 in.>s # $ # # = 3 (rC>B)xyz + Ω C>B * (rC>B)xyz 4 + Ω C>B * rC>B + Ω C>B * rC>B 32 j - 8kb + (2i) * ( - 3k) + (8i) * (4i - 4k) 4 + (2i) * (8j - 3k)

= a-

= 5- 2.25i + 44j + 8k6 in.>s2

vC = vB + Ω * rC>B + (vC>B)xyz

= ( -20i) + (4k) * (4i - 4k) + (8j - 3k) vC = 5- 20i + 24j - 3k6 in.>s # aC = aB + Ω * rC>B + Ω * (Ω * rC>B) + 2Ω * (vC>B)xyz + (aC>B)xyz = ( - 15i - 80j) + (3k) * (4i - 4k) + (4k) * 3 (4k) * (4i - 4k) 4    + 2(4k) * (8j - 3k) + ( - 2.25i + 44j + 8k) aC = 5- 145i - 24j + 8k6 in.>s2

1098

Ans.

Ans.

Ans: vC = 5-20i + 24j - 3k6 in.>s aC = 5- 145i - 24j + 8k6 in.>s2

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20–45. z

The particle P slides around# the circular hoop with a constant angular velocity of u = 6 rad>s, while the hoop rotates about the x axis at a constant rate of v = 4 rad>s. If at the instant shown the hoop is in the x–y plane and the angle u = 45°, determine the velocity and acceleration of the particle at this instant.

200 mm O θ

y

SOLUTION

P

Relative to XYZ, let xyz have Æ = v = {4i} rad>s, rO = 0;

vO = 0;

# # Æ = v = 0 ( Æ does not change direction relative to XYZ.)

V = 4 rad/s

x

aO = 0

Relative to xyz, let coincident x¿ ‚ y¿ , z¿ ‚ have # Æ xyz = 0 ( Æ xyz does not change direction relative to XYZ.)

Æ xyz = {6k} rad>s,

(rP>O)xyz = 0.2 cos 45° i + 0.2 sin 45°j = {0.1414i + 0.1414j} m (rP>O)xyz ((rP/O)xyz changes direction relative to XYZ.) # (vP>O)xyz = a rP>O b

xyz

# = a rP>O b

x¿y¿z¿

+ Æ xyz * a rP>O b

xyz

= 0 + (6k) * (0.1414i + 0.1414j)

= { -0.8485i + 0.8485j} m>s $ (aP>O)xyz = a rP>O b

xyz

$ = c a rP>O b

x¿y¿z¿

# + Æ xyz * a rP>O b

x¿y¿z¿

d + Æ * arP>O b

xyz

# + Æ * arP>O b

xyz

= C 0 + 0 D + 0 + (6k) * (- 0.8485i + 0.8485j) = {- 5.0912i - 5.0912j} m>s2 Thus, vP = vO + Æ * rP>O + (vP>O)xyz = 0 + (4i) * (0.1414i + 0.1414j) - 0.8485i + 0.8485j Ans.

= { -0.849i + 0.849j + 0.566k} m>s # aP = aO + Æ * rP>O + Æ * (Æ * rP>O) + 2Æ * (vP>O)xyz + (aP>O)xyz = 0 + 0 + (4i) * C (4i) * (0.1414i + 0.1414j) D + 2(4i) * ( -0.8485i + 0.8485j) - 5.0912i - 5.0912j = {- 5.09i - 7.35j + 6.79k} m s2

Ans.

Ans: vP = {-0.849i + 0.849j + 0.566k} m>s aP = {- 5.09i - 7.35j + 6.79k} m>s2 1099

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20–46. z

At the instant shown, the industrial manipulator is rotating about the z axis at v1 = 5 rad>s, and about joint B at v2 = 2 rad>s. Determine the velocity and acceleration of the grip A at this instant, when f = 30°, u = 45°, and r = 1.6 m.

v2 B f 1.2 m

x

Solution

v1 u r

# Ω = 55k6 rad>s  Ω = 0

rB = 1.2 sin 30°j + 1.2 cos 30°k = 50.6j + 1.0392k6 m # # vB = rB = (rB)xyz + Ω * rB = 0 + (5k) * (0.6j + 1.0392k) = 5-3i6 m>s # $ $ # # aB = rB = 3 (rB)xyz + Ω * (rB)xyz 4 + Ω * rB + Ω * rB

A

y

= [0 + 0] + 0 + [(5k) * ( -3i)]

Ω xyz

= 5- 15j6 m>s2 # = 52i6 rad>s  Ω xyz = 0

rA>B = 1.6 cos 45°j - 1.6 sin 45°k = 51.1314j - 1.1314k6 m # # (vA>B)xyz = rA>B = (rA>B)xyz + Ω xyz * rA>B = 0 + (2i) * (1.1314j - 1.1314k)

(aA>B)xyz

= 52.2627j + 2.2627k6 m>s $ $ # = rA>B = 3 ( rA>B)xyz + Ω xyz * (rA>B)xyz 4 +

(aA>B)xyz = [0 + 0] + 0 +

3 (2i)

#

3 Ω xyz

* (2.2627j + 2.2627k) 4

= 5- 4.5255j + 4.5255k6 m>s2

* rA>B 4 +

3 Ω xyz

# * rA>B 4

vA = vB + Ω * rA>B + (vA>B)xyz = ( -3i) +

3 (5k)

* (1.1314j - 1.1314k) 4 + (2.2627j + 2.2627k)

= 5- 8.66i + 2.26j + 2.26k6 m>s # aA = aB + Ω * rA>B + Ω * (Ω * rA>B) + 2Ω * (vA>B)xyz + (aA>B)xyz

Ans.

= ( -15j) + 0 + (5k) * [(5k) * (1.1314j - 1.1314k)] + [2(5k) * (2.2627j + 2.2627k)] + ( -4.5255j + 4.5255k) = 5- 22.6i - 47.8j + 4.53k6 m>s2

Ans.

Ans: vA = { - 8.66i + 2.26j + 2.26k} m>s aA = { - 22.6i - 47.8j + 45.3k} m>s2 1100

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20–47. z

At the instant shown, the industrial manipulator is rotating # about the z axis at v1 = 5 rad>s, and v1 = 2 rad>s2; and # about joint B at v2 = 2 rad>s and v2 = 3 rad>s2. Determine the velocity and acceleration of the grip A at this instant, when f = 30°, u = 45°, and r = 1.6 m.

v2 B f 1.2 m

x

Solution

# Ω = 55k6 rad>s  Ω = {2k} rad>s2

rB = 1.2 sin 30°j + 1.2 cos 30°k = 50.6j + 1.0392k6 m # # vB = rB = (rB)xyz + Ω * rB = 0 + (5k) * (0.6j + 1.0392k) = 5- 3i6 m>s # $ $ # # aB = rB = 3 (rB)xyz + Ω * (rB)xyz 4 + Ω * rB + Ω * rB

v1 u r

A

y

= [0 + 0] + (2k) * (0.6j + 1.0392k) + [(5k) * ( -3i)]

Ω xyz

= 5- 1.2i - 15j6 m>s2 # = 52i6 rad>s  Ω xyz = 53i6 rad>s2

rA>B = 1.6 cos 45°j - 1.6 sin 45°k = 51.1314j - 1.1314k6 m # # (vA>B)xyz = rA>B = (rA>B)xyz + Ω xyz * rA>B = 0 + (2i) * (1.1314j - 1.1314k)

(aA>B)xyz

= 52.2627j + 2.2627k6 m>s $ $ # = rA>B = 3 ( rA>B)xyz + Ω xyz * ( rA>B)xyz 4 +

#

3 Ω xyz

* rA>B 4 +

3 Ω xyz

# * rA>B 4

(aA>B)xyz = [0 + 0] + (3i) * (1.1314j - 1.1314k) + [(2i) * (2.2627j + 2.2627k)] = 5 -1.1313j + 7.9197k6 m>s2

vA = vB + Ω * rA>B + (vA>B)xyz

= ( -3i) + [(5k) * (1.1314j - 1.1314k)] + (2.2627j + 2.2627k) = 5-8.66i + 2.26j + 2.26k6 m>s # aA = aB + Ω * rA>B + Ω * (Ω * rA>B) + 2Ω * (vA>B)xyz + (aA>B)xyz

Ans.

= ( -1.2i - 15j) + (2k) * (1.1314j - 1.1314k) + (5k) * [(5k) * (1.1314j - 1.1314k)]       + [2(5k) * (2.2627j + 2.2627k)] + ( - 1.1313j + 7.9197k) = 5- 26.1i - 44.4j + 7.92k6 m>s2

Ans.

Ans: vA = { - 8.66i + 2.26j + 2.26k} m>s aA = { - 26.1i - 44.4j + 7.92k} m>s2 1101

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*20–48. z

At the given instant, the rod is turning about the z axis with a constant angular velocity v1 = 3 rad>s. At this same instant, the disk is spinning at v2 = 6 rad>s when v# 2 = 4 rad>s2, both measured relative to the rod. Determine the velocity and acceleration of point P on the disk at this instant.

v1  3 rad/s

x

O 2m

v2  6 rad/s   4 rad/s2 v 2

0.5 m P

0.5 m 1.5 m

Solution Motion of point A. Point A is located at the center of the disk. At the instant consider, the fixed XYZ frame and rotating x′ y′ z′ frame are set coincident with origin at Point O. The x′ y′ z′ frame is set rotate with constant angular velocity of 𝛀′ = V1 = 5-3 k6 rad>s of which # the direction does not change relative to XYZ frame. Since 𝛀 ′ is constant 𝛀 ′ = 0. Here rA = { - 0.5i + 2j + 1.5k} m and its direction changes relative to XYZ frame but does not change relative to x′y′z′ frame. Thus, # vA = (rA)x′y′z′ + 𝛀′ * rA = 0 + ( - 3k) * ( - 0.5i + 2j + 1.5k) = {6i + 1.5j} m>s # $ # # aA = 3 (rA)x′y′z′ + 𝛀 ′ * (rA)x′y′z′ 4 + 𝛀 ′ * rA + 𝛀 ′ * rA

y

= [0 + 0] + 0 + ( - 3k) * (6i + 1.5j) = 54.5i - 18j6 m>s2

Motion of P with Respect to A. The xyz and x″ y″ z″ frame are set coincident with origin at A. Here x″ y″ z″ frame is set to rotate at 𝛀 ″ = Vz = { - 6i} rad>s of which its direction does not change relative to xyz frame. Also, Ω″ = 5 -4i6 rad>s2. Here (rP>A)xyz = 50.5j 6 m and its direction changes with respect to xyz frame but does not change relative to x″ y″ z″ frame. # # (vP>A)xyz = (rP>A)xyz = (rP>A)x″y″z″ + 𝛀 * (rP>A)xyz = 0 + ( -6i) * (0.5j) = { -3k} m>s # $ # $ # (aP>A)xyz = (rP>A)xyz = 3(rP>A)x″y″z″ + Ω″ * (rP>A)x″y″z″4 + 𝛀 ″ * (rP>A)xyz + Ω″ * (rP>A)xyz = [0 + 0] + ( - 4i) * (0.5j) + ( - 6i) * ( - 3k) = { -18j - 2k} m>s2

# Motion of P. Here, 𝛀 = V1 = 5- 3 k6 rad>s and 𝛀 = 0. vP = vA + 𝛀 * rP>A + (vP>A)xyz

= (6i + 1.5j) + ( -3k) * (0.5j) + ( - 3k) = 57.50i + 1.50j - 3.00k6m>s # aP = aA + 𝛀 * rP>A + 𝛀 * (𝛀 * rP>A) + 2𝛀 * (vP>A)xyz + (aP>A)xyz

Ans.

= (4.5i - 18j) + 0 + ( -3k) * ( -3k * 0.5j) + 2( -3k) * ( - 3k) + ( -18j - 2k) = {4.50i - 40.5j - 2.00k} m>s2

Ans.

1102

Ans: vP = 57.50i + 1.50j - 3.00k6 m>s aP = 54.50i - 40.5j - 2.00k6 m>s2

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20–49. z

At the instant shown, the backhoe is traveling forward at a constant speed vO = 2 ft>s, and the boom ABC is rotating about the z axis with an angular velocity v1 = 0.8 rad>s and an angular acceleration # v1 = 1.30 rad>s2. At this same instant the boom is # rotating with v2 = 3 rad>s when v2 = 2 rad>s2, both measured relative to the frame. Determine the velocity and acceleration of point P on the bucket at this instant.

v2  3 rad/s v2  2 rad/s2 B P 5 ft

Solution 𝛀 = 50.8k6 rad>s # 𝛀 = 51.3k6 rad>s2

u

A

v1  0.8 rad/s v1  1.30 rad/s2 vO  2 ft/s O

2 ft y

C 4 ft 15 ft 3 ft

x

rA = {2i - 4j} ft # vA = (rA)xyz + 𝛀 * rA = (2i) + (0.8k) * (2i - 4j) = {5.20i + 1.60j} ft>s # $ # # aA = 3 (rA)xyz + 𝛀 * (rA)xyz 4 + 𝛀 * rA + 𝛀 * rA

= 0 + (0.8k) * (2i) + (1.3k) * (2i - 4j) + (0.8k) * (5.20i + 1.60j) = {3.92i + 8.36j} ft>s2

𝛀 P>A = { - 3j} rad>s # 𝛀 P>A = { - 2j} rad>s2 rP>A = {16i + 5k} ft # (vP>A)xyz = (rP>A)xyz + 𝛀 P>A * rP>A = 0 + ( - 3j) * (16i + 5k) = { - 15i + 48k} ft>s # $ # # (aP>A)xyz = 3 (rP>A)xyz + 𝛀 P>A * (rP>A)xyz 4 + 𝛀 P>A * rP>A + 𝛀 P>A * rP>A = 0 + 0 + ( - 2j) * (16i + 5k) + ( -3j) * ( - 15i + 48k) = { - 154i - 13k} ft>s2

vP = vA + 𝛀 * rP>A + (vP>A)xyz = (5.2i + 1.6j) + (0.8k) * (16i + 5k) + ( - 15i + 48k) vP = { -9.80i + 14.4j + 48.0k} ft>s

Ans.

# aP = aA + 𝛀 * rP>A + 𝛀 * (𝛀 * rP>A) + 2𝛀 * (vP>A)xyz + (aP>A)xyz = (3.92i + 8.36j) + (1.3k) * (16i + 5k) + (0.8k) * [(0.8k) * (16i + 5k)]

+ 2(0.8k) * ( - 15i + 48k) + ( - 154i - 13j) aP = { -160i + 5.16j - 13k} ft>s2

Ans.

Ans: vP = {-9.80i + 14.4j + 48.0k} ft>s aP = {- 160i + 5.16j - 13k} ft>s2 1103

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20–50. At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity v1 = 6 rad>s, while at the same instant the arm is rotating upward at a constant rate v2 = 4 rad>s. If the conveyor is # running at a constant rate r = 5 ft>s, determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package.

z

A P v1 = 6 rad/s

SOLUTION Æ = v1 = {6k} rad>s # Æ = 0

r = 6 ft

u = 45

O

rO = vO = aO = 0

x

Æ P/O = {4i} rad>s # Æ P/O = 0

v2 = 4 rad/s

rP/O = {4.243j + 4.243k} ft # (vP/O)xyz = (rP/O)xyz + Æ P/O * rP/O = (5 cos 45°j + 5 sin 45°k) + (4i) * (4.243j + 4.243k) = { -13.44j + 20.51k} ft>s # $ # # (aP/O) = (rP/O)xyz + Æ P/O * (rP/O)xyz + Æ P/O * rP/O + Æ P/O * rP/O = 0 + (4i) * (3.536j + 3.536k) + 0 + (4i) * ( -13.44j + 20.51k) = { -96.18j - 39.60k} ft>s2 vP = vO + Æ * rP/O + (rP/O)xyz = 0 + (6k) * (4.243j + 4.243k) + ( - 13.44j + 20.51k) vP = {-25.5i - 13.4j + 20.5k} ft>s # aP = aO + Æ * rP/O + Æ * (Æ * rP/O) + 2Æ * (vP/O)xyz + (aP/O)xyz

Ans.

= 0 + 0 + (6k) * [(6k) * (4.243j + 4.243k)] + 2(6k) * ( -13.44j + 20.51k) + ( -96.18j - 39.60k) aP = {161i - 249j - 39.6k} ft>s2

Ans.

1104

Ans: vP = 5- 25.5i - 13.4j + 20.5k6 ft>s aP = 5161i - 249j - 39.6k6 ft>s2

y

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20–51. At the instant shown, the arm OA of the conveyor belt is rotating about the z axis with a constant angular velocity v1 = 6 rad>s, while at the same instant the arm is rotating upward at a constant rate v2 = 4 rad>s. If the conveyor is $ # running at a rate r = 5 ft>s, which is increasing at r = 8 ft>s2, determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package.

z

A P v1 = 6 rad/s

SOLUTION Æ = v1 = {6k} rad>s # Æ = 0

r = 6 ft

u = 45

O

rO = vO = aO = 0

x

Æ P/O = {4i} rad>s # Æ P/O = 0

v2 = 4 rad/s

rP/O = {4.243j + 4.243k} ft # (vP/O)xyz = (rP/O)xyz + Æ P/O * rP/O = (5 cos 45°j + 5 sin 45°k) + (4i) * (4.243j + 4.243k) = {-13.44j + 20.51k} ft>s (aP/O)xyz = 8 cos 45j + 8 sin 45°k - 96.18j - 39.60k = { -90.52j - 33.945k} ft>s2 vP = vO + Æ * rP/O + (vP/O)xyz = 0 + (6k) * (4.243j + 4.243k) + ( - 13.44j + 20.51k) vP = {-25.5i - 13.4j + 20.5k} ft>s # aP = aO + Æ * rP/O + Æ * (Æ * rP/O) + 2Æ * (vP/O)xyz + (aP/O)xyz

Ans.

= 0 + 0 + (6k) * [(6k) * (4.243j + 4.243k)] + 2(6k) * ( -13.44j + 20.51k) + ( -90.52j - 33.945k) = - 152.75j + 161.23i - 90.52j - 33.945k aP = {161i - 243j - 33.9k} ft>s2

Ans.

1105

Ans: vP = 5- 25.5i - 13.4j + 20.5k6 ft>s aP = 5161i - 243j - 33.9k6 ft>s2

y

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*20–52. The crane is rotating about the z axis with a constant rate v1 = 0.25 rad>s, while the boom OA is rotating downward with a constant rate v2 = 0.4 rad>s. Compute the velocity and acceleration of point A located at the top of the boom at the instant shown.

z v1  0.25 rad/s A 40 ft

v2  0.4 rad/s

Solution

O

30

x

𝛀 = {0.25k} rad>s # 𝛀 = 0

y

rO = 0 vO = 0 aO = 0 𝛀 A>O = {- 0.4i} rad>s # 𝛀 A>O = 0 rA>O = 4 cos 30°j + 40 sin 30°k = 34.64j + 20k # (vA>O)xyz = (rA>O)xyz + 𝛀 A>O * rA>O = 0 + ( - 0.4i) * (34.64j + 20k)

(aA>O)xyz

= 8j - 13.856k # $ # # = 3 (rA>O)xyz + 𝛀 A>O * (rA>O)xyz 4 + 𝛀 A>O * rA>O + 𝛀 A>O * rA>O

= 0 + 0 + 0 + ( - 4i) * (8j - 13.86k) = -5.542j - 3.2k vA = vO + 𝛀 * rA>O + (vA>O)xyz

= 0 + 0.25k * (34.64j + 20k) + (8j - 13.856k) = { - 8.66i + 8j - 13.9k} ft>s # aA = aO + 𝛀 * rA>O + 𝛀 * (𝛀 * rA>O) + 2𝛀 * (vA>O)xyz + (aA>O)xyz

Ans.

= 0 + 0 + (0.25k) * (0.25k) * (34.64j + 20k) + 2(0.25k) * (8j - 13.856k) - 5.542j - 3.2k aA = { -4i - 7.71j - 3.20k} ft>s2

Ans.

Ans: vA = 5- 8.66i + 8j - 13.9k6 ft>s aA = { - 4i - 7.71j - 3.20k} ft>s2 1106

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20–53. z

Solve Prob. 20–52 if the angular motions are increasing # # at v1 = 0.4 rad>s2 and v2 = 0.8 rad>s2 at the instant shown.

v1  0.25 rad/s A 40 ft

v2  0.4 rad/s

Solution

O

30

x

𝛀 = 50.25k6 rad>s # 𝛀 = 50.4k6 rad>s2

y

rO = 0

vO = 0 aO = 0 𝛀 A>O = 5-0.4i6 rad>s # 𝛀 A>O = 5- 0.8i6 rad>s2

rA>O = 4 cos 30°j + 40 sin 30°k = 34.64j + 20k # (vA>O)xyz = (rA>O)xyz + Ω A>O * rA>O = 0 + ( -0.4i) * (34.64 j + 20k)

(aA>O)xyz

= 8j - 13.856k # $ # # = 3 (rA>O)xyz + 𝛀 A>O * (rA>O)xyz 4 + 𝛀 A>O * rA>O + 𝛀 A>O * rA>O = 0 + 0 + ( - 0.8i) * (34.64j + 20k) + ( -4i) * (8j - 13.86k) = 10.457j - 30.913k

vA = vO + 𝛀 * rA>O + (vA>O)xyz = 0 + 0.25k * (34.64j + 20k) + (8j - 13.856k) = 5- 8.66i + 8j - 13.9k6 ft>s # aA = aO + 𝛀 * rA>O + 𝛀 * (𝛀 * rA>O) + 2𝛀 * (vA>O)xyz + (aA>O)xyz

Ans.

= 0 + (0.4k) * (34.64j + 20k) + (0.25k) * [(0.25k) * (34.64j + 20k)] + 2(0.25k) * (8j - 13.856k) + 10.457j - 30.913k aA = 5- 17.9i + 8.29j - 30.9k6 ft>s2

Ans.

1107

Ans: vA = 5- 8.66i + 8j - 13.9k6 ft>s aA = 5- 17.9i + 8.29j - 30.9k6 ft>s2

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20–54. At the instant shown, the arm AB is rotating about the fixed bearing with an angular velocity v1 = 2 rad>s and angular # acceleration v1 = 6 rad>s2. At the same instant, rod BD is rotating relative to rod AB at v2 = 7 rad>s, which is # increasing at v2 = 1 rad>s2. Also, the collar C is moving # along rod BD with a velocity r = 2 ft>s and a deceleration $ r = - 0.5 ft>s2, both measured relative to the rod. Determine the velocity and acceleration of the collar at this instant.

v2  7 rad/s v2  1 rad/s2

r  1 ft

z

B

C

D

u  30

1.5 ft

v1  2 rad/s v1  6 rad/s2 x

2 ft A

y

Solution 𝛀 = 52k6 rad>s # 𝛀 = 56k6 rad>s2

rB = 5-2i + 1.5k6 ft # # vB = rB = (rB)xyz + 𝛀 * rB = (2k) * ( -2i + 1.5k) = 5- 4j6 ft>s # $ $ # # aB = rB = 3 (rB)xyz + 𝛀 * (rB)xyz 4 + 𝛀 * rB + 𝛀 * rB = (6k) * ( -2i + 1.5k) + (2k) * ( - 4j) = 58i - 12j6 ft>s2

𝛀 C>B = 57i6 rad>s # 𝛀 C>B = 51i6 rad>s2

rC>B = 1 cos 30°j + 1 sin 30°k = 50.866j + 0.5k6 ft # (vC>B)xyz = (rC>B)xyz + 𝛀 C>B * rC>B = (2 cos 30°j + 2 sin 30°k) + (7i) * (0.866j + 0.5k)

(aC>B)xyz

= 5- 1.768j + 7.062k6 ft>s # $ # # = 3 (rC>B)xyz + 𝛀 C>B * (rC>B)xyz 4 + 𝛀 C>B * rC>B + 𝛀 C>B * rC>B

= ( -0.5 cos 30°j - 0.5 sin 30°k) + (7i) * (1.732j + 1k) + (1i) * (0.866j + 0.5k)       + (7i) * ( -1.768j + 7.06k) = 5-57.37j + 0.3640k6 ft>s2

vC = vB + 𝛀 * rC>B + (vC>B)xyz

= ( -4j) + (2k) * (0.866j + 0.5k) + ( - 1.768j + 7.06k) vC = 5- 1.73i - 5.77j + 7.06k6 ft>s # aC = aB + 𝛀 * rC>B + 𝛀 * (𝛀 * rC>B) + 2𝛀 * (vC>B)xyz + (aC>B)xyz

Ans.

= (8i - 12j) + (6k) * (0.866j + 0.5k) + (2k) * 3 (2k) * (0.866j + 0.5k) 4      + 2(2k) * ( - 1.768j + 7.062k) + ( - 57.37j + 0.364k) aC = 5 9.88i - 72.8j + 0.365k6 ft>s2

Ans.

Ans: vC = { - 1.73i - 5.77j + 7.06k} ft>s aC = {9.88i - 72.8j + 0.365k} ft>s2 1108

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21–1. Show that the sum of the moments of inertia of a body, Ixx + Iyy + Izz, is independent of the orientation of the x, y, z axes and thus depends only on the location of its origin.

SOLUTION Ixx + Iyy + Izz =

Lm

= 2

(y2 + z2)dm +

Lm

Lm

(x2 + z2)dm +

Lm

(x2 + y2)dm

(x2 + y2 + z2)dm

However, x2 + y2 + z2 = r2, where r is the distance from the origin O to dm. Since ƒ r ƒ is constant, it does not depend on the orientation of the x, y, z axis. Consequently, Ixx + Iyy + Izz is also indepenent of the orientation of the x, y, z axis. Q.E.D.

1109

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21–2. Determine the moment of inertia of the cone with respect to a vertical y axis passing through the cone’s center of mass. What is the moment of inertia about a parallel axis y¿ that passes through the diameter of the base of the cone? The cone has a mass m.

–y

y

y¿

a x

SOLUTION The mass of the differential element is dm = rdV = r(py2) dx = dIy = = = Iy =

L

rpa2 h2

h

x2dx.

1 dmy2 + dmx2 4 r pa2 a 2 1 rpa2 2 B 2 x dx R a x b + ¢ 2 x2 ≤ x2 dx 4 h h h r pa2

(4h2 + a2) x4 dx

4h4

dIy =

rpa2 4

4h

h

(4h2 + a2)

x4dx =

r pa2h (4h2 + a2) 20

x2 dx =

r pa2h 3

L0

However, m =

Lm

dm =

h

r pa2 2

h

L0

Hence, Iy =

3m (4h2 + a2) 20

Using the parallel axis theorem: Iy = Iy + md2 3h 2 3m (4h2 + a2) = Iy + m a b 20 4 Iy =

3m 2 (h + 4a2) 80

Ans.

Iy' = Iy + md2 =

h 2 3m 2 (h + 4a2) + m a b 80 4

=

m (2h2 + 3a2) 20

Ans.

Ans: 3m 2 ( h + 4a2 ) 80 m ( 2h2 + 3a2 ) Iy′ = 20

Iy =

1110

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21–3. Determine moment of inertia Iy of the solid formed by revolving the shaded area around the x axis. The density of the material is r = 12 slug/ft3.

y

y2 = x 2 ft x

SOLUTION

The mass of the differential element is dm = rdV = r A py2 B dx = rpxdx. dIy =

1 2

dmy2 + dmx2

=

1 4

[rpxdx](x) + (rpxdx)x2

4 ft

1 = rp( x2 + x3) dx 4 4

Iy =

L

dIy = rp

1 ( x2 + x3) dx = 69.33 pr 4 L0

= 69.33(p)(12) = 2614 slug # ft2

Ans.

Ans: Iy = 2614 slug # ft 2 1111

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*21–4. z

Determine the moments of inertia Ix and Iy of the paraboloid of revolution. The mass of the paraboloid is 20 slug.

z2  2y 2 ft y

SOLUTION The mass of the differential element is dm = rdV = r A pz2 B dy = 2rpydy.

x

2

m = 20 =

dm = 2rpydy Lm L0

20 = 4rp dIx =

2 ft

1 4

r =

dmz2 + dm A y2 B

5 p

slug/ft3

= 14 [2rpydy](2y) + [2rpydy]y2

= A 5y2 + 10y3 B dy 2

Ix =

L

dIx = dIy =

L0 1 2

2 A 5y2 + 10y3 B dy = 53.3 slug # ft

Ans.

dmz2 = 2rpy2 dy = 10y2 dy 2

Iy =

L

dIy =

L0

10y2 dy = 26.7 slug # ft2

Ans.

Ans: Ix = 53.3 slug # ft 2 Iy = 26.7 slug # ft 2 1112

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21–5. Determine by direct integration the product of inertia Iyz for the homogeneous prism. The density of the material is r. Express the result in terms of the total mass m of the prism.

z

a

a

h

SOLUTION

x

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy. a

m =

Lm

dm = rh

L0

(a - y)dy =

y

ra2h 2

Using the parallel axis theorem: dIyz = (dIy¿z¿)G + dmyGzG h = 0 + (rhxdy) (y) a b 2

Iyz =

=

rh2 xydy 2

=

rh2 (ay - y2) dy 2

a rh2 ra3h2 1 ra2h m = a b(ah) = ah (ay - y2) dy = 2 L0 12 6 2 6

Ans.

Ans: Iyz = 1113

m ah 6

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21–6. Determine by direct integration the product of inertia Ixy for the homogeneous prism. The density of the material is r. Express the result in terms of the total mass m of the prism.

z

a

a

h

SOLUTION

x

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy. a

m =

Lm

dm = rh

L0

(a - y)dy =

y

ra2h 2

Using the parallel axis theorem: dIxy = (dIx¿y¿)G + dmxGyG x = 0 + (rhxdy)a b(y) 2 =

rh2 2 x ydy 2

=

rh2 3 (y - 2ay2 + a 2 y) dy 2

Ixy =

a rh (y3 - 2ay2 + a 2 y) dy 2 L0

=

ra 4h 1 ra 2h 2 m 2 = ba = a a 24 12 2 12

Ans.

Ans: Ixy = 1114

m 2 a 12

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21–7. Determine the product of inertia Ixy of the object formed by revolving the shaded area about the line x = 5 ft. Express the result in terms of the density of the material, r.

y 3 ft

2 ft

y2  3x

SOLUTION

x

3

L0

3

L0

3

dm = r 2p

L0

3

(5 - x)y dx = r 2p

L0

(5 - x) 23x dx = 38.4rp

3

~

y dm = r2p

y (5 - x)y dx L0 2 3

= rp

(5 - x)(3x) dx L0 = 40.5rp Thus, y =

40.5rp = 1.055 ft 38.4rp

The solid is symmetric about y, thus Ixy¿ = 0 Ixy = Ixy¿ + x ym = 0 + 5(1.055)(38.4rp) Ans.

Ixy = 636r

Ans: Ixy = 636r 1115

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*21–8. Determine the moment of inertia Iy of the object formed by revolving the shaded area about the line x = 5 ft. Express the result in terms of the density of the material, r.

y 3 ft

2 ft

y2  3x

SOLUTION

x

3

Iy¿ =

1 1 dm r2 - (m¿)(2)2 2 L0 2

3

3

1 1 dm r2 = r p(5 - x)4 dy 2 2 L0 L0 3 y2 4 1 rp b dy a5 2 3 L0

=

= 490.29 r p

m¿ = r p (2)2(3) = 12 r p Iy¿ = 490.29 r p -

1 (12 r p)(2)2 = 466.29 r p 2

Mass of body; 3

m =

r p (5 - x)2 dy - m¿

L0

3

r p (5 -

=

L0 = 38.4 r p

y2 2 ) dy - 12 r p 3

Iy = 466.29 r p + (38.4 r p)(5)2 = 1426.29 r p Iy = 4.48(103) r

Ans.

Also, 3

Iy¿ =

r2 dm

L0

3

=

(5 - x)2 r (2p)(5 - x)y dx

L0

3

(5 - x)3 (3x)1/2 dx L0 = 466.29 r p = 2rp

3

m =

L0

dm 3

= 2rp

(5 - x)y dx

L0

3

= 2rp

L0 = 38.4 r p

(5 - x)(3x)1/2 dx

Iy = 466.29 r p + 38.4 r p(5)2 = 4.48(103)r

Ans. 1116

Ans: Iy = 4.48(103) r Iy = 4.48(103) r

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21–9. Determine the moment of inertia of the cone about the z¿ axis. The weight of the cone is 15 lb, the height is h = 1.5 ft, and the radius is r = 0.5 ft.

r h

z z¿

z¿

SOLUTION u = tan-1(

0.5 ) = 18.43° 1.5

Ixx = Iyy = [

3 1.5 m{4(0.5)2 + (1.5)2}] + m[1.5 - ( )]2 80 4

Ixx = Iyy = 1.3875 m Iz =

3 m(0.5)2 = 0.075 m 10

Ixy = Iyz = Izx = 0 Using Eq. 21–5. Iz¿z¿ = u2z¿x Ixx + u2z¿y Iyy + u2z¿z Izz = 0 + [cos(108.43°)]2(1.3875m) + [cos(18.43°)]2(0.075m) = 0.2062m Iz¿z¿ = 0.2062(

15 ) = 0.0961 slug # ft2 32.2

Ans.

Ans: Iz′z′ = 0.0961 slug # ft 2 1117

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21–10. y

Determine the radii of gyration kx and ky for the solid formed by revolving the shaded area about the y axis. The density of the material is r.

0.25 ft

xy  1 4 ft 0.25 ft

SOLUTION For ky: The mass of the differential element is dm = rdV = r(px2) dy = rp dIy = 12 dmx2 = Iy =

L

dIy = 12rp

4

dy y4 L0.25

m =

Hence,

ky =

y2

dy 1 1 dy C rpy 2 D A y 2B = 2rpy 4

+

x

. 4 ft

1 C r(p)(4)2(0.25) D (4)2 2

= 134.03r 4

However,

1 2

dy

dm = rp

Lm

dy y2

L0.25

+ r C p(4)2(0.25) D = 24.35r

Iy 134.03r = = 2.35 ft Am A 24.35r

Ans.

For kx: 0.25 ft 6 y … 4 ft dI¿x = =

1 dmx2 + dmy2 4 1 I dy 2 c rpdy y d a 2 b + a rp y by 4 y 2

2

= rp A 4y1 4 + 1 B dy 4

I¿ x =

L

dI¿ x = rp I¿¿ x =

1 4

L0.25

A 4y1 4 + 1 B dy = 28.53r

C rp(4)2(0.25) D (4)2 + C rp(4)2(0.25) D (0.125)2

= 50.46r Ix = I¿ x + I¿¿ x = 28.53r + 50.46r = 78.99r

Hence,

kx =

Ix = m

78.99r = 1.80 ft 24.35r

Ans.

Ans: ky = 2.35 ft kx = 1.80 ft 1118

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21–11. Determine the moment of inertia of the cylinder with respect to the a–a axis of the cylinder. The cylinder has a mass m.

a

a

SOLUTION

h

The mass of the differential element is dm = rdV = r A pa2 B dy. dIaa = = =

1 2 4 dma 1 4

C r A pa2 B dy D a2 + C r A pa2 B dy D y2

A 14 rpa4 + rpa2y2 B dy h

Iaa =

L

dIaa = =

a

+ dm A y2 B

L0

A 14rpa4 + rpa2y2 B dy

rpa2h (3a2 + 4h2) 12 h

However, Hence,

r A pa2 B dy = rpa2h Lm L0 m = A 3a2 + 4h2 B 12

m = Iaa

dm =

Ans.

Ans: Iaa = 1119

m ( 3a2 + 4h2 ) 12

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*21–12. Determine the moment of inertia Ix of the composite plate assembly. The plates have a specific weight of 6 lb>ft2.

z

0.25 ft 0.5 ft 0.5 ft

SOLUTION

y

0.5 ft

x

Horizontial plate: Ixx =

0.5 ft

1 6(1)(1) ( )(1)2 = 0.0155 12 32.2

Vertical plates: lxx¿ = 0.707,

lxy¿ = 0.707,

lxz¿ = 0

1

Ix¿x¿ =

1 6(4)(122) 1 2 ( )( ) = 0.001372 3 32.2 4

Iy¿y¿ = (

6(14)(122) 32.2

)(

6(14)(1 22) 1 2 1 1 2 )[( ) + (122)2] + ( )( ) 12 4 32.2 8

= 0.01235 Using Eq. 21–5, Ixx = (0.707)2(0.001372) + (0.707)2(0.01235) = 0.00686 Thus, Ixx = 0.0155 + 2(0.00686) = 0.0292 slug # ft2

Ans.

Ans: Ixx = 0.0292 slug # ft 2 1120

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21–13. Determine the product of inertia Iyz of the composite 2 plate assembly. The plates have a weight of 6 lb>ft .

z

0.25 ft 0.5 ft 0.5 ft

SOLUTION

0.5 ft

y

0.5 ft

x

Due to symmetry, Ans.

Iyz y = 0

Ans: Iyz = 0 1121

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21–14. Determine the products of inertia Ixy, Iyz, and Ixz, of the thin plate. The material has a density per unit area of 50 kg>m2.

z

400 mm 200 mm

SOLUTION The masses of segments 1 and 2 shown in Fig. a are m1 = 50(0.4)(0.4) = 8 kg and m2 = 50(0.4)(0.2) = 4 kg. Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for segment 1 and Ix–y– = Iy–z– = Ix–z– = 0 for segment 2 .

x

y

400 mm

Ixy = ©Ix¿y¿ + mxGyG = C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2) D = 0.32 kg # m2

Ans.

Iyz = ©Iy¿z¿ + myGzG = C 0 + 8(0.2)(0) D + C 0 + 4(0.2)(0.1) D = 0.08 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG = C 0 + 8(0.2)(0) D + C 0 + 4(0)(0.1) D Ans.

= 0

Ans: Ixy = 0.32 kg # m2 Iyz = 0.08 kg # m2 Ixz = 0 1122

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21–15. Determine the moment of inertia of both the 1.5-kg rod and 4-kg disk about the z¿ axis. 300 mm

z 100 mm z'

SOLUTION Due to symmetry Ixy = Iyz + Izx = 0 1 1 Iy = Ix = B (4)(0.1)2 + 4(0.3)2 R + (1.5)(0.3)2 4 3 = 0.415 kg # m2 Iz =

1 (4)(0.1)2 = 0.02 kg # m2 2

uz = cos (18.43°) = 0.9487,

uy = cos 90° = 0,

ux = cos (90° + 18.43°) = - 0.3162 Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux = 0.415(- 0.3162)2 + 0 + 0.02(0.9487)2 - 0 - 0 - 0 = 0.0595 kg # m2

Ans.

Ans: Iz′ = 0.0595 kg # m2 1123

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*21–16. z

The bent rod has a mass of 3 kg>m. Determine the moment of inertia of the rod about the O–a axis.

a 0.3 m 1m

0.5 m

Solution

x

The bent rod is subdivided into three segments and the location of center of mass for each segment is indicated in Fig. a. The mass of each segments is m1 = 3(1) = 3 kg, m2 = 3(0.5) = 1.5 kg and m3 = 3(0.3) = 0.9 kg. Ixx = c



1 (3) ( 12 ) + 3 ( 0.52 ) d + 12

= 3.427 kg # m2

Iyy = 0 + c





30

+ 1.5 ( 12 ) 4 + c

1 (0.9) ( 0.32 ) + 0.9 ( 0.152 + 12 ) d 12

1 1 (1.5) ( 0.52 ) + 1.5 ( 0.252 ) d + c (0.9) ( 0.32 ) + 0.9 ( 0.152 + 0.52 ) d 12 12

= 0.377 kg # m2

Izz = c

1 1 (3) ( 12 ) + 3 ( 0.52 ) d + c (1.5) ( 0.52 ) + 1.5 ( 12 + 0.252 ) d + 12 12

= 3.75 kg # m2

30

+ 0.9 ( 12 + 0.52 ) 4

Ixy = [0 + 0] + [0 + 1.5(0.25)( -1)] + [0 + 0.9(0.5)( - 1)] = - 0.825 kg # m2 Iyz = [0 + 0] + [0 + 0] + [0 + 0.9( -1)(0.15)] = - 0.135 kg # m2 Izx = [0 + 0] + [0 + 0] + [0 + 0.9(0.15)(0.5)] = 0.0675 kg # m2 The unit vector that defines the direction of the Oa axis is UOa = Thus,  u x =

O

0.5i - 1j + 0.3k 2

2

2

20.5 + ( - 1) + 0.3 0.5

21.34

  u y = -

=

1

21.34

0.5 21.34

i -

  u z =

1 21.34

0.3

j +

0.3 21.34

21.34

1124

k

y

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*21–16.  Continued

Then IOa = Ixx u 2x + Iyy u 2y + Izz u 2z - 2Ixy u xu y - 2Iyz u yu z - 2Izx u zu x = 3.427a

0.5 21.34

2

b + 0.377a -

- 2( - 0.135)a -

1

21.34

ba

1 21.34 0.3

21.34

= 0.4813 kg # m2 = 0.481 kg # m2

2

b + 3.75a

2

0.3 21.34

b - 2(0.0675)a

b - 2( -0.825)a

0.3

21.34

ba

0.5

21.34

b

0.5 21.34

ba -

1 21.34

b

Ans.

Ans: IOa = 0.481 kg # m2 1125

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21–17. The bent rod has a weight of 1.5 lb>ft. Locate the center of gravity G(x, y) and determine the principal moments of inertia Ix¿ , Iy¿ , and Iz¿ of the rod with respect to the x¿, y¿, z¿ axes.

z¿ z 1 ft

SOLUTION Due to symmetry

x

y = 0.5 ft

Ans.

x =

Ans.

( - 1)(1.5)(1) + 2 C ( - 0.5)(1.5)(1) D ©x W = = - 0.667 ft ©w 3 C 1.5(1) D

Ix¿ = 2 c a

+

y¿

_ x y

Ans.

1 1.5 1.5 1.5 a b (1)2 + a b(0.667 - 0.5)2 d + a b(1 - 0.667)2 12 32.2 32.2 32.2

= 0.0155 slug # ft2 Iz¿ = 2 c

G

1.5 1.5 1 b (0.5)2 d + a b (1)2 32.2 12 32.2

= 0.0272 slug # ft2 Iy¿ = 2 c

1 ft A _ y x¿

Ans.

1 1.5 1.5 a b(1)2 + a b (0.52 + 0.16672) d 12 32.2 32.2 1.5 1 1.5 a b(1)2 + a b (0.3333)2 12 32.2 32.2

= 0.0427 slug # ft2

Ans.

Ans: y = 0.5 ft x = -0.667 ft Ix′ = 0.0272 slug # ft 2 Iy′ = 0.0155 slug # ft 2 Iz′ = 0.0427 slug # ft 2 1126

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21–18. Determine the moment of inertia of the rod-and-disk assembly about the x axis. The disks each have a weight of 12 lb. The two rods each have a weight of 4 lb, and their ends extend to the rims of the disks.

2 ft 1 ft 1 ft x

SOLUTION For a rod: 1 u = tan-1 a b = 45° 1 ux¿ = cos 90° = 0, Ix¿ = Iz¿ = a

uy¿ = cos 45° = 0.7071,

uz¿ = cos (90° + 45°) = -0.7071

4 1 ba b C (2)2 + (2)2 D = 0.08282 slug # ft2 12 32.2

Iy¿ = 0 Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 Ix = 0 + 0 + (0.08282)(- 0.7071)2 = 0.04141 slug # ft2 For a disk: 1 12 Ix = a b a b (1)2 = 0.1863 slug # ft2 2 32.2 Thus. Ix = 2(0.04141) + 2(0.1863) = 0.455 slug # ft2

Ans.

Ans: Ix = 0.455 slug # ft 2 1127

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21–19. Determine the moment of inertia of the composite body about the aa axis. The cylinder weighs 20 lb, and each hemisphere weighs 10 lb.

a

2 ft

a

SOLUTION

2 ft

uaz = 0.707 uax = 0 uay = 0.707 Izz =

1 20 2 10 ( )(1)2 + 2[ ( )(1)2] 2 32.2 5 32.2

= 0.5590 slug # ft2 Ixx = Iyy =

1 20 10 10 11 2 ( )[3(1)2 + (2)2] + 2[0.259( )(1)2 + ( )] 12 32.2 32.2 32.2 8

Ixx = Iyy = 1.6975 slug # ft2 Iaa = 0 + (0.707)2(1.6975) + (0.707)2(0.559) Iaa = 1.13 slug # ft2

Ans.

Ans: Iaa = 1.13 slug # ft 2 1128

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*21–20. Determine the moment of inertia of the disk about the axis of shaft AB. The disk has a mass of 15 kg.

30°

B

A 150 mm

SOLUTION Due to symmetry Ixy = Iyz = Izx = 0 Ix = Iz = Iy =

1 (15)(0.15)2 = 0.084375 kg # m2 4

1 (15)(0.15)2 = 0.16875 kg # m2 2

ux = cos 90° = 0,

uy = cos 30° = 0.8660

uz = cos (30° + 90°) = - 0.5 Iy¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux = 0 + 0.16875(0.8660)2 + 0.084375( - 0.5)2 - 0 - 0 - 0 = 0.148 kg # m2

Ans.

Ans: Iy′ = 0.148 kg # m2 1129

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21–21. z

The thin plate has a weight of 5 lb and each of the four rods weighs 3 lb. Determine the moment of inertia of the assembly about the z axis.

1.5 ft

SOLUTION

0.5 ft 0.5 ft

For the rod: Iz¿ =

2 1 3 a b a 2(0.52 + (0.5)2 b = 0.003882 slug # ft2 12 32.2

0.5 ft x

y

0.5 ft

For the composite assembly of rods and disks: Iz = 4 B 0.003882 + a

3 1 5 20.52 + 0.52 2 b¢ b(12 + 12) ≤ R+ a 32.2 2 12 32.2

= 0.0880 slug # ft2

Ans.

Ans: Iz = 0.0880 slug # ft 2 1130

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21–22. z

If a body contains no planes of symmetry, the principal moments of inertia can be determined mathematically. To show how this is done, consider the rigid body which is spinning with an angular velocity V , directed along one of its principal axes of inertia. If the principal moment of inertia about this axis is I, the angular momentum can be expressed as H = IV = Ivx i + Ivy j + Ivz k. The components of H may also be expressed by Eqs. 21–10, where the inertia tensor is assumed to be known. Equate the i, j, and k components of both expressions for H and consider vx, vy, and vz to be unknown. The solution of these three equations is obtained provided the determinant of the coefficients is zero. Show that this determinant, when expanded, yields the cubic equation

V O

x

I 3 - (Ixx + Iyy + Izz)I 2 + (IxxIyy + IyyIzz + IzzIxx - I2xy - I2yz - I2zx)I - (IxxIyyIzz - 2IxyIyzIzx - IxxI 2yz - IyyI 2zx - IzzI 2xy) = 0 The three positive roots of I, obtained from the solution of this equation, represent the principal moments of inertia Ix, Iy, and Iz.

SOLUTION H = Iv = Ivx i + Ivy j + Ivz k Equating the i, j, k components to the scalar equations (Eq. 21–10) yields (Ixx - I) vx - Ixy vy - Ixz vz = 0 -Ixx vx + (Ixy - I) vy - Iyz vz = 0 - Izx vz - Izy vy + (Izz - I) vz = 0 Solution for vx, vy, and vz requires 3

(Ixx - I) - Iyx - Izx

-Ixy (Iyy - I) - Izy

-Ixz -Iyz 3 = 0 (Izz - I)

Expanding I3 - (Ixx + Iyy + Izz)I2 + A Ixx Iyy + Iyy Izz + Izz Ixx - I2xy - I2yz - I2zx B I

- A Ixx Iyy Izz - 2Ixy Iyz Izx - Ixx I2yz - IyyI2zx - Izz I2xy B = 0 Q.E.D.

Ans: I 3 - (Ixx + Iyy + Izz)I 2 + ( Ixx Iyy + Iyy Izz + Izz Ixx - I 2xy - I 2yz - I 2zx ) I

- ( Ixx Iyy Izz - 2Ixy Iyz Izx - Ixx I 2yz

- Iyy I 2zx - Izz I 2xy ) = 0 Q.E.D. 1131

y

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21–23. Show that if the angular momentum of a body is determined with respect to an arbitrary point A, then H A can be expressed by Eq. 21–9. This requires substituting R A = R G + R G>A into Eq. 21–6 and expanding, noting that 1 R G dm = 0 by definition of the mass center and vG = vA + V : R G>A.

Z

z RG/A G A

RG RA

P y Y

SOLUTION HA = a = a = a

Lm Lm Lm

x

rA dm b * vA +

Lm

rA * (v * rA)dm

(rG + rG>A) dm b * vA +

Lm

X

(rG + rG>A) * C v * rG + rG>A) D dm Lm

rG dmb * vA + (rG>A * vA) + a

Since

Lm

dm + rG * (v * rG) dm Lm Lm

rGdmb * (v * rG>A) + rG>A * av *

rG dm = 0 and from Eq. 21–8 HG =

Lm

Lm

rG dmb + rG>A * (v * rG>A)

Lm

dm

rG * (v * rG)dm

HA = (rG>A * vA)m + HG + rG>A * (v * rG>A)m = rG>A * (vA + (v * rG>A))m + HG Q.E.D.

= (rG>A * mvG) + HG

Ans: HA = (rG>A * mvG) + HG 1132

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*21–24. z

The 15-kg circular disk spins about its axle with a constant angular velocity of v1 = 10 rad>s. Simultaneously, the yoke is rotating with a constant angular velocity of v2 = 5 rad>s. Determine the angular momentum of the disk about its center of mass O, and its kinetic energy.

150 mm O v1  10 rad/s x

SOLUTION

y

The mass moments of inertia of the disk about the x, y, and z axes are Ix = Iz = Iy =

1 1 mr2 = (15)(0.152) = 0.084375 kg # m2 4 4

v2  5 rad/s

1 1 mr2 = (15)(0.152) = 0.16875 kg # m2 2 2

Due to symmetry, Ixy = Iyz = Ixz = 0 Here, the angular velocity of the disk can be determined from the vector addition of v1 and v2. Thus, v = v1 + v2 = [- 10j + 5k] rad>s so that vx = 0

vy = - 10 rad>s

vz = 5 rad>s

Since the disk rotates about a fixed point O, we can apply Hx = Ixvx = 0.084375(0) = 0

Hy = Iyvy = 0.16875(- 10) = - 1.6875 kg # m2>s Hz = Izvz = 0.084375(5) = 0.421875 kg # m2>s

Thus, HO = [ -1.69j + 0.422k] kg # m2>s

Ans.

The kinetic energy of the disk can be determined from 1 1 1 I v 2 + Iyvy2 + Izvz2 2 x x 2 2 1 1 1 = (0.084375)(02) + (0.16875)(- 10)2 + (0.084375)(52) 2 2 2 = 9.49 J

T =

Ans.

Ans: HO = [ - 1.69j + 0.422k] kg # m2 >s T = 9.49 J 1133

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21–25. z

The large gear has a mass of 5 kg and a radius of gyration of kz = 75 mm. Gears B and C each have a mass of 200 g and a radius of gyration about the axis of their connecting shaft of 15 mm. If the gears are in mesh and C has an angular velocity of ω C = {15j} rad>s, determine the total angular momentum for the system of three gears about point A.

40 mm B 45° 100 mm

SOLUTION

40 mm A

vC = {15j} rad/s C

y

x

IA = 5(0.075)2 = 28.125 A 10 - 3 B kg # m2

IB = IC = 0.2(0.015)2 = 45 A 10 - 6 B kg # m2

Kinematics: vC = vB = 15 rad>s v = (0.04)(15) = 0.6 m>s vA = a

0.6 b = 6 rad>s 0.1

HB = IB vB = A 45 A 10 - 6 B B (15) = 675 A 10 - 6 B

HB = - 675 A 10 - 6 B sin 45°i - 675 A 10 - 6 B cos 45° j

HB = - 477.3 A 10 - 6 B i - 477.3 A 10 - 6 B j

HC = IC vC = A 45 A 10 - 6 B B (15) = 675 A 10 - 6 B HC = 675 A 10 - 6 B j

HA = IA vA = 28.125 A 10 - 3 B (6) = 0.16875 HA = 0.16875k The total angular momentum is therefore, H = HB + HC + HA =

- 477 10 - 6 i + 198 10 - 6 j + 0.169k kg # m2 s

1134

Ans.

Ans: H = { - 477(10-6)i + 198(10-6)j + 0.169k} kg # m2 >s

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21–26. The circular disk has a weight of 15 lb and is mounted on the shaft AB at an angle of 45° with the horizontal. Determine the angular velocity of the shaft when t = 3 s if a constant torque M = 2 lb # ft is applied to the shaft. The shaft is originally spinning at v1 = 8 rad>s when the torque is applied.

0.8 ft 45 v1 A

8 rad/s B

M

SOLUTION Due to symmetry Ixy = Iyz = Izx = 0 Iy = Iz = Ix =

A

1 15 2 32.2

A

1 15 2 32.2

B (0.8)2 = 0.07453 slug # ft2

B (0.8)2 = 0.1491 slug # ft2

For x' axis ux = cos 45° = 0.7071

uy = cos 45° = 0.7071

uz = cos 90° = 0 Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux = 0.1491(0.7071)2 + 0.07453(0.7071)2 + 0 - 0 - 0 - 0 = 0.1118 slug # ft2 Principle of impulse and momentum: (Hx¿)1 + ©

L

Mx¿ dt = (Hx¿)2

0.1118(8) + 2(3) = 0.1118 v2 Ans.

v2 = 61.7 rad/s

Ans: v2 = 61.7 rad>s 1135

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21–27. The circular disk has a weight of 15 lb and is mounted on the shaft AB at an angle of 45° with the horizontal. Determine the angular velocity of the shaft when t = 2 s if a torque M = 14e0.1t2 lb # ft, where t is in seconds, is applied to the shaft. The shaft is originally spinning at v1 = 8 rad>s when the torque is applied.

0.8 ft 45 v1 A

8 rad/s B

M

SOLUTION Due to symmetry Ixy = Iyz = Izx = 0 Iy = Iz = Ix =

1 2

1 4

15 A 32.2 B (0.8)2 = 0.07453 slug # ft2

15 A 32.2 B (0.8)2 = 0.1491 slug # ft2

For x' axis ux = cos 45° = 0.7071

uy = cos 45° = 0.7071

uz = cos 90° = 0 Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux = 0.1491(0.7071)2 + 0.07453(0.7071)2 + 0 - 0 - 0 - 0 = 0.1118 slug # ft2 Principle of impulse and momentum: (Hx¿)1 + ©

L

Mx¿ dt = (Hx¿)2 2

0.1118(8) +

L0

4e0.1 t dt = 0.1118v2 Ans.

v2 = 87.2 rad/s

Ans: v2 = 87.2 rad>s 1136

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*21–28. z

The rod assembly is supported at G by a ball-and-socket joint. Each segment has a mass of 0.5 kg/m. If the assembly is originally at rest and an impulse of I = 5- 8k6 N # s is applied at D, determine the angular velocity of the assembly just after the impact.

0.5 m

I = {–8k} N·s

B D

A 1m

G 1m

SOLUTION

x

C

0.5 m y

Moments and products of inertia: Ixx =

1 C 2(0.5) D (2)2 + 2 C 0.5(0.5) D (1)2 = 0.8333 kg # m2 12

Iyy =

1 C 1(0.5) D (1)2 = 0.04166 kg # m2 12

Izz =

1 1 C 2(0.5) D (2)2 + 2c C 0.5(0.5) D (0.5)2 + C 0.5(0.5) D (12 + 0.252) d 12 12

= 0.875 kg # m2 Ixy = C 0.5(0.5) D (- 0.25)(1) + C 0.5(0.5) D (0.25)(- 1) = - 0.125 kg # m2 Iyz = Ixz = 0 From Eq. 21–10 Hx = 0.8333vx + 0.125vy Hy = 0.125vx + 0.04166vy Hz = 0.875vz t2

(HG)1 + ©

Lt1

MGdt = (HG)2

0 + (- 0.5i + 1j) * ( -8k) = (0.8333vx + 0.125vy)i + (0.125vx + 0.04166vy)j + 0.875vzk Equating i, j and k components - 8 = 0.8333vx + 0.125vy

(1)

- 4 = 0.125vx + 0.04166vy

(2)

0 = 0.875vz

(3)

Solving Eqs. (1) to (3) yields: vx = 8.73 rad s Then

-

vy - 122 rad s

vz = 0 Ans.

v = {8.73i - 122j} rad s

Ans: v = {8.73i - 122j} rad>s 1137

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21–29. z

The 4-lb rod AB is attached to the 1-lb collar at A and a 2-lb link BC using ball-and-socket joints. If the rod is released from rest in the position shown, determine the angular velocity of the link after it has rotated 180°.

B 1.3 m 0.5 m C y A

SOLUTION

x

1.2 m

T1 + V1 = T2 + V2 0 + 4(0.25) + 2(0.25) = T2 - 4(0.25) - 2(0.25) T2 = 3 vAB = T2 =

0.5vx = 0.3846vx 1.3

1 1 2 1 1 4 1 4 [ ( )(0.5)2]v2x + [ ( )(1.3)2](0.3846vx)2 + ( )(0.25vx)2 2 3 32.2 2 12 32.2 2 32.2

T2 = 0.007764v2x = 3 Ans.

vx = 19.7 rad/s

Ans: vx = 19.7 rad>s 1138

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21–30. The rod weighs 3 lb>ft and is suspended from parallel cords at A and B. If the rod has an angular velocity of 2 rad>s about the z axis at the instant shown, determine how high the center of the rod rises at the instant the rod momentarily stops swinging.

z 3 ft 3 ft

SOLUTION

v

2 rad/s

A

T1 + V1 = T2 + V2 1 1 W 2 2 c l dv + 0 = 0 + Wh 2 12 g h =

1 l2v2 1 (6)2(2)2 = 24 g 24 (32.2) Ans.

h = 0.1863 ft = 2.24 in.

Ans: h = 2.24 in. 1139

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21–31. The 4-lb rod AB is attached to the rod BC and collar A using ball-and-socket joints. If BC has a constant angular velocity of 2 rad>s, determine the kinetic energy of AB when it is in the position shown. Assume the angular velocity of AB is directed perpendicular to the axis of AB.

z 1 ft

3 ft

A

SOLUTION vA = yAi

vB + {- 2j} ft>s

rB/A = { -3i + 1j + 1k} ft

2 rad/s

B

vAB = vx i + vy j + vz k

1 ft C

x

rG/B = {1.5i - 0.5j - 0.5k}

vB = vA + vAB * rB/A i - 2j = yAi + 3 vx -3

j vy 1

k vz 3 1

Equating i, j and k components vy - vz + yA = 0

(1)

vx + 3vz = 2

(2)

vx + 3vy = 0

(3)

Since vAB is perpendicular to the axis of the rod, vAB

rB/A = (vx i + vy j + vz k)

( -3i + 1j + 1k) = 0 (4)

-3vx + 1vy + 1vz = 0 Solving Eqs. (1) to (4) yields: vx = 0.1818 rad>s

vy = - 0.06061 rad>s

vz = 0.6061 rad>s

yA = 0.6667 ft>s Hence vAB = {0.1818i - 0.06061j + 0.6061k} rad>s

vA = {0.6667i} ft>s

vG = vB + vAB * rG/B i = - 2j + 3 0.1818 1.5

j - 0.06061 - 0.5

k 0.6061 3 - 0.5

= {0.3333i - 1.0j} ft>s v2AB = 0.18182 + ( - 0.06061)2 + 0.60612 = 0.4040 y2G = (0.3333)2 + ( - 1.0)2 = 1.111 T = =

1 1 my2G + IG v2AB 2 2

4 1 1 4 1 a b (1.111) + c a b A 232 + 12 + 12 B 2 d(0.4040) 2 32.2 2 12 32.2

= 0.0920 ft # lb

Ans.

1140

Ans: T = 0.0920 ft # lb

y

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*21–32. The 2-kg thin disk is connected to the slender rod which is fixed to the ball-and-socket joint at A. If it is released from rest in the position shown, determine the spin of the disk about the rod when the disk reaches its lowest position. Neglect the mass of the rod. The disk rolls without slipping.

B

C

0.5 m A

0.1 m

30°

SOLUTION Ix = Iz = Iy =

1 (2)(0.1)2 + 2(0.5)2- = 0.505 kg # m2 4

1 (2)(0.1)2 = 0.01 kg # m2 2

v = vy + vz = - vvj + vz # sin 11.31°j + vz # cos 11.31°k = (0.19612vz - vy)j + (0.98058vz # )k Since vA = vC = 0, then vC = vA + v * rC>A 0 = 0 + C (0.19612vz # - vy)j + (0.98058vz # )k D * (0.5j - 0.1k) 0 = - 0.019612vz + 0.1vy - 0.49029vz # vz = 0.19612vy Thus, v = - 0.96154vyj , 0.19231vyk h1 = 0.5 sin 41.31° = 0.3301 m,

h2 = 0.5 sin 18.69° = 0.1602 m

T1 + V1 = T2 + V2 0 + 2(9.81)(0.3301) = c 0 +

1 1 (0.01)( -0.96154vy)2 + (0.505)(0.19231vy)2 d - 2(9.81)(0.1602 2 2 Ans.

vy = 26.2 rad s

Ans: vy = 26.2 rad>s 1141

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21–33. The 20-kg sphere rotates about the axle with a constant angular velocity of vs = 60 rad>s. If shaft AB is subjected to a torque of M = 50 N # m, causing it to rotate, determine the value of vp after the shaft has turned 90° from the position shown. Initially, vp = 0. Neglect the mass of arm CDE.

z

0.4 m D E

0.3 m

SOLUTION

C

The mass moments of inertia of the sphere about the x¿ , y¿ , and z¿ axes are Ix¿ = Iy¿ = Iz¿ =

2 2 mr2 = (20)(0.12) = 0.08 kg # m2 5 5

x

When the sphere is at position 1 , Fig. a, vp = 0. Thus, the velocity of its mass center is zero and its angular velocity is v1 = [60k] rad>s. Thus, its kinetic energy at this position is T =

A

vs  60 rad/s

0.1 m

B vp

M  50 Nm y

1 1 1 1 m(vG)12 + Ix¿(v1)x¿2 + Iy¿(v1)y¿2 + Iz¿(v1)z¿2 2 2 2 2

=0 + 0 + 0 +

1 (0.08) (602) 2

= 144 J When the sphere is at position 2 , Fig. a, vp = vpi. Then the velocity of its mass center is (vG)2 = vp * vG>C = (vpi) * (- 0.3j + 0.4k) = - 0.4vpj - 0.3vpk. Then (vG)22 = (- 0.4vp)2 + ( - 0.3vp)2 = 0.25vp2. Also, its angular velocity at this position is v2 = vpi - 60j. Thus, its kinetic energy at this position is T = =

1 1 1 1 m(vG)22 + Ix¿(v2)x¿2 + Iy¿(v2)y¿2 + Iz¿(v2)z¿2 2 2 2 2

1 1 1 (20) A 0.25vp2 B + (0.08) A vp 2 B + (0.08)( -60)2 2 2 2

= 2.54vp 2 + 144 When the sphere moves from position 1 to position 2 , its center of gravity raises vertically ¢z = 0.1 m. Thus, its weight W does negative work. UW = -W¢z = - 20(9.81)(0.1) = - 19.62 J Here, the couple moment M does positive work. p UW = Mu = 50 a b = 25pJ 2 Applying the principle of work and energy, T1 + ©U1 - 2 = T2 144 + 25p + ( -19.62) = 2.54vp2 + 144 Ans.

vp = 4.82 rad>s

Ans: vp = 4.82 rad>s 1142

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21–34. The 200-kg satellite has its center of mass at point G. Its radii of gyration about the z¿ , x¿ , y¿ axes are kz¿ = 300 mm, kx¿ = ky¿ = 500 mm, respectively. At the instant shown, the satellite rotates about the x¿, y¿ , and z¿ axes with the angular velocity shown, and its center of mass G has a velocity of vG = 5—250i + 200j + 120k6 m>s. Determine the angular momentum of the satellite about point A at this instant.

z, z¿ Vz¿

G

SOLUTION

600 rad/s

Vx ¿

The mass moments of inertia of the satellite about the x¿ , y¿ , and z¿ axes are

800 mm

x¿

Ix¿ = Iy¿ = 200 A 0.52 B = 50 kg # m2

x

A

1250 rad/s

vG Vy ¿

300 rad/s y¿ y

Iz¿ = 200 A 0.32 B = 18 kg # m2

Due to symmetry, the products of inertia of the satellite with respect to the x¿ , y¿ , and z¿ coordinate system are equal to zero. Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 The angular velocity of the satellite is v = [600i + 300j + 1250k] rad>s Thus, vx¿ = 600 rad>s

vy¿ = - 300 rad>s

vz¿ = 1250 rad>s

Then, the components of the angular momentum of the satellite about its mass center G are (HG)x¿ = Ix¿vx¿ = 50(600) = 30 000 kg # m2>s

(HG)y¿ = Iy¿vy¿ = 50(- 300) = -15 000 kg # m2>s

(HG)z¿ = Iz¿vz¿ = 18(1250) = 22 500 kg # m2>s Thus,

HG = [30 000i - 15 000j + 22 500k] kg # m2>s The angular momentum of the satellite about point A can be determined from HA = rG>A * mvG + HG = (0.8k) * 200( - 250i + 200j + 120k) + (30 000i - 15 000j + 22 500k) = [-2000i - 55 000j + 22 500k] kg # m2/s

Ans.

1143

Ans: HA = { - 2000i - 55 000j + 22 500k6kg # m2 >s

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21–35. The 200-kg satellite has its center of mass at point G. Its radii of gyration about the z¿ , x¿ , y¿ axes are kz¿ = 300 mm, kx¿ = ky¿ = 500 mm, respectively. At the instant shown, the satellite rotates about the x¿ , y¿ , and z¿ axes with the angular velocity shown, and its center of mass G has a velocity of vG = 5—250i + 200j + 120k6 m>s. Determine the kinetic energy of the satellite at this instant.

z, z¿ Vz¿

SOLUTION

G 600 rad/s

Vx ¿

The mass moments of inertia of the satellite about the x¿ , y¿ , and z¿ axes are Ix¿ = Iy¿ = 200 A 0.52 B = 50 kg # m2

800 mm

x¿ x

Iz¿ = 200 A 0.32 B = 18 kg # m2

A

1250 rad/s

vG Vy ¿

300 rad/s y¿ y

Due to symmetry, the products of inertia of the satellite with respect to the x¿ , y¿ , and z coordinate system are equal to zero. ¿ Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 The angular velocity of the satellite is v = [600i - 300j + 1250k] rad>s Thus, vx¿ = 600 rad>s

vy¿ = - 300 rad>s

vz¿ = 1250 rad>s

Since vG 2 = (- 250)2 + 2002 + 1202 = 116 900 m2>s2, the kinetic energy of the satellite can be determined from T = =

1 1 1 1 mvG 2 + Ix¿vx¿ 2 + Iy¿vy¿ 2 + Iz¿vz¿ 2 2 2 2 2 1 1 1 1 (200)(116 900) + (50) A 6002 B + (50)(-300)2 + (18) A 12502 B 2 2 2 2

= 37.0025 A 106 B J = 37.0 MJ

Ans.

Ans: T = 37.0 MJ 1144

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*21–36. The 15-kg rectangular plate is free to rotate about the y axis because of the bearing supports at A and B.When the plate is balanced in the vertical plane, a 3-g bullet is fired into it, perpendicular to its surface, with a velocity v = 5-2000i6 m>s. Compute the angular velocity of the plate at the instant it has rotated 180°. If the bullet strikes corner D with the same velocity v, instead of at C, does the angular velocity remain the same? Why or why not?

z 150 mm 150 mm

D C A

150 mm

SOLUTION

v x

Consider the projectile and plate as an entire system.

B

Angular momentum is conserved about the AB axis. (HAB)1 = -(0.003)(2000)(0.15)j = { -0.9j} (HAB)1 = (HAB)2 -0.9j = Ixvx i + Iyvy j + Izvz k Equating components, vx = 0 vz = 0 vy =

-0.9 = - 8 rad/s 1 c (15)(0.15)2 + 15(0.075)2 d 12

T1 + V1 = T2 + V2 1 1 c (15)(0.15)2 + 15(0.075)2 d(8)2 + 15(9.81)(0.15) 2 12 =

1 1 c (15)(0.15)2 + 15(0.075)2 dv2AB 2 12 Ans.

vAB = 21.4 rad/s

If the projectile strikes the plate at D, the angular velocity is the same, only the impulsive reactions at the bearing supports A and B will be different.

Ans: vAB = 21.4 rad>s 1145

y

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21–37. z

The 5-kg thin plate is suspended at O using a ball-andsocket joint. It is rotating with a constant angular velocity V = 5 2k6 rad>s when the corner A strikes the hook at S, which provides a permanent connection. Determine the angular velocity of the plate immediately after impact.

V {2k} rad/s 300 mm 300 mm O 400 mm

Solution

x y

Angular momentum is conserved about the OA axis. (HO)1 = Izvzk

= c

S A

1 (5)(0.6)2 d (2)k = 0.30k 12

uOA = 50.6j - 0.8k6 (HOA)1 = (HO)1 # uOA

= (0.30)( - 0.8) = -0.24 (1)

v = 0.6vj - 0.8vk (HO)2 = Iyvy j + Izvzk 1 1 (5)(0.4)2vyj + (5)(0.6)2vzk 3 12



=



= 0.2667vyj + 0.150vzk

From Eq. (1), vy = 0.6v vz = - 0.8v (HO)2 = 0.16vj - 0.120vk (HOA)2 = (HO)2 - uOA

= 0.16v(0.6) + (0.12v)(0.8) = 0.192v

Thus, (HOA)1 = (HOA)2 -0.24 = 0.192v v = -1.25 rad>s v = -1.25uOA Ans.

v = 5 -0.750j + 1.00k6rad>s

1146

Ans: V = 5 -0.750j + 1.00k6 rad>s

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21–38. z

Determine the kinetic energy of the 7-kg disk and 1.5-kg rod when the assembly is rotating about the z axis at v = 5 rad>s. B

v  5 rad/s

C 200 mm

Solution Due to symmetry Ixy = Iyz = Izx = 0

D

1 1 Iy = Iz = c (7)(0.1)2 + 7(0.2)2 d + (1.5)(0.2)2 4 3

100 mm

= 0.3175 kg # m2



Iz =

A

1 (7)(0.1)2 = 0.035 kg # m2 2

For z′ axis u z = cos 26.56° = 0.8944  u y = cos 116.57° = -0.4472 u x = cos 90° = 0 Iz′ = Ixu 2x + Iyu 2y + Izu 2z - 2Ixyu xu y - 2Iyzu yu z - 2Izxu zu x

= 0 + 0.3175( -0.4472)2 + 0.035(0.8944)2 - 0 - 0 - 0 = 0.0915 kg # m2

T =

1 1 1 I ′v2x′ + Iy′v2y′ + Iz′v2z′ 2x 2 2

= 0 + 0 +

1 (0.0915)(5)2 2 Ans.

= 1.14 J

Ans: T = 1.14 J 1147

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21–39. z

Determine the angular momentum Hz of the 7-kg disk and 1.5-kg rod when the assembly is rotating about the z axis at v = 5 rad>s. B

v  5 rad/s

C 200 mm

Solution Due to symmetry      Ixy = Iyz = Izx = 0 1 1 Iy = Ix = c (7)(0.1)2 + 7(0.2)2 d + (1.5)(0.2)2 4 3

D

= 0.3175 kg # m2

Iz =

100 mm A

1 (7)(0.1)2 = 0.035 kg # m2 2

For z′ axis u z = cos 26.56° = 0.8944  u y = cos 116.57° = - 0.4472 u x = cos 90° = 0 Iz′ = Ix u 2x + Iy u 2y + Iz u 2z - 2Ixy u x u y - 2Iyz u y u z - 2Izx u z u x = 0 + 0.3175( - 0.4472)2 + 0.035(0.8944)2 - 0 - 0 - 0 = 0.0915 kg # m2 vx = vy = 0 Hz = - Izx vx - Ixy vy + Izz vz = - 0 - 0 + 0.0915(5) = 0.4575 kg # m2 >s

Ans.

1148

Ans: Hz = 0.4575 kg # m2 >s

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*21–40. Derive the scalar form of the rotational equation of motion about the x axis if æ Z V and the moments and products of inertia of the body are not constant with respect to time.

SOLUTION In general d (Hx i + Hy j + Hz k) dt # # # = A Hx i + Hy j + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)

M =

Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields # # M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx b j # + a A Hz B xyz - Æ y Hx + Æ x Hy bk Subsitute Hx, Hy and Hz using Eq. 21–10. For the i component, ©Mx =

d (Ix vx - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx) dt + Æ y (Iz vz - Izx vx - Izy vy)

Ans.

One can obtain y and z components in a similar manner.

Ans: d (Ix vx - Ixy vy - Ixz vz) dt - Ω z (Iy vy - Iyz vz - Iyx vx)

ΣMx =

+ Ω y (Iz vz - Izx vx - Izy vy) 1149

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21–41. Derive the scalar form of the rotational equation of motion about the x axis if æ Z V and the moments and products of inertia of the body are constant with respect to time.

SOLUTION In general d (Hx i + Hy j + Hz k) dt # # # = A Hx i + Hyj + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)

M =

Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields # # M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx bj # + a A Hz B xyz - Æ y Hx + Æ x Hy bk Substitute Hx, Hy and Hz using Eq. 21–10. For the i component ©Mx =

d (I v - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx) dt x x + Æ y (Iz vz - Izx vx - Izy vy)

For constant inertia, expanding the time derivative of the above equation yields # # # ©Mx = (Ix vx - Ixy vy - Ixzvz) - Æ z (Iy vy - Iyz vz - Iyx vx) + Æ y (Iz vz - Izxvx - Izy vy)

Ans.

One can obtain y and z components in a similar manner.

Ans: # # # ΣMx = (Ixvx - Ixyvy - Ixzvz) - Ω z(Iyvy - Iyzvz - Iyxvx) + Ω y(Izvz - Izxvx - Izyvy) Similarly for ΣMy and ΣMz. 1150

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21–42. Derive the Euler equations of motion for æ Z V , i.e., Eqs. 21–26.

SOLUTION In general d (Hx i + Hy j + Hz k) dt # # # = A Hx i + Hyj + Hz k B xyz + Æ * (Hx i + Hyj + Hz k)

M =

Substitute Æ = Æ x i + Æ y j + Æ z k and expanding the cross product yields # # M = a A Hx B xyz - Æ z Hy + Æ yHz b i + a A Hy B xyz - Æ x Hz + Æ zHx bj

# + a A Hz B xyz - Æ y Hx + Æ x Hy bk

Substitute Hx, Hy and Hz using Eq. 21–10. For the i component ©Mx =

d (Ix vx - Ixy vy - Ixz vz) - Æ z (Iy vy - Iyz vz - Iyxvx) dt + Æ y (Iz vz - Izx vx - Izy vy)

Set Ixy = Iyz = Izx = 0 and require Ix, Iy, Iz to be constant. This yields # ©Mx = Ix vx - Iy Æ z vy + Iz Æ yvz

Ans.

One can obtain y and z components in a similar manner.

Ans: . ΣMx = Ix vx - Iy Ω z vy + Iz Ω y vz 1151

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21–43. z

The 4-lb bar rests along the smooth corners of an open box. At the instant shown, the box has a velocity v = 53j6 ft>s and an acceleration a = 5 - 6j6 ft>s2. Determine the x, y, z components of force which the corners exert on the bar.

A

SOLUTION

2 ft

©Fx = m(aG)x ;

Ax + Bx = 0

©Fy = m(aG)y ;

Ay + By = ¢

©Fz = m(aG)z ;

Bz - 4 = 0

[1] 4 ≤ ( -6) 32.2

[2]

B x

1 ft

y

2 ft

Ans.

Bz = 4 lb

# # # Applying Eq. 21–25 with vx = vy = vz = 0 vx = vy = vz = 0 # ©(MG)x = Ixvx - (Iy - Iz)vyvz ;

By (1) - Ay (1) + 4(0.5) = 0

[3]

# ©(MG)y = Iy vy - (Iz - Ix)vz vx ;

Ax (1) - Bx (1) + 4(1) = 0

[4]

Solving Eqs. [1] to [4] yields: Ax = -2.00 lb

Ay = 0.627 lb

Bx = 2.00 lb

By = -1.37 lb

Ans.

# ©(MG)z = Iz vz - (Ix - Iy) vx vy ; (-2.00)(0.5) - (2.00)(0.5) - ( - 1.37)(1) + (0.627)(1) = 0

(O.K!)

Ans: Bz = Ax = Ay = Bx = By = 1152

4 lb - 2.00 lb 0.627 lb 2.00 lb - 1.37 lb

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*21–44. The uniform rectangular plate has a mass of m = 2 kg and is given a rotation of v = 4 rad>s about its bearings at A and B. If a = 0.2 m and c = 0.3 m, determine the vertical reactions at the instant shown. Use the x, y, z axes shown mac c 2 - a 2 and note that Izx = - a b. ba 2 12 c + a2

x

A

V B z c

SOLUTION

y

vx = 0,

vy = 0,

vz = -4

# vx = 0,

# vy = 0,

# vz = 0

a

# # ©My = Iyy vy - (Izz - Ixx)vz vx - Iyz avz - vx vy b # - Izx A v2z - v2x B - Ixy a vx + vy vz b

a 2 c 2 2 a 2 c 2 2 Bx B a b + a b R - Ax B a b + a b R = - Izx (v)2 2 2 2 2 1

Bx - Ax = a

1

mac c2 - a2 2 b£ 3 ≥v 6 C a2 + c2 D 2

©Fx = m(aG)x ;

Ax + Bx - mg = 0

Substitute the data, Bx - Ax =

(0.3)2 - (0.2)2 2(0.2)(0.3) 2 C = 0.34135 3 S (-4) 6 C (0.3)2 + (0.2)2 D 2

Ax + Bx = 2(9.81) Solving:

Ax = 9.64 N

Ans.

Bx = 9.98 N

Ans.

Ans: Ax = 9.64 N Bx = 9.98 N 1153

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21–45. If the shaft AB is rotating with a constant angular velocity of v = 30 rad>s, determine the X, Y, Z components of reaction at the thrust bearing A and journal bearing B at the instant shown. The disk has a weight of 15 lb. Neglect the weight of the shaft AB.

z

1.5 ft A

30

1 ft

x

SOLUTION

0.5 ft

The rotating xyz frame is set with its origin at the plate’s mass center, Fig. a. This frame will be fixed to the disk so that its angular velocity is Æ = v and the x, y, and z axes will always be the principle axes of inertia of the disk. Referring to Fig. b,

B v  30 rad/s

v = [30 cos 30°j- 30 sin 30°k] rad>s = [25.98j-15k] rad>s Thus, vx = 0

vy = 25.98 rad>s

vz = - 15 rad>s

Since v is always directed towards the + Y axis and has a constant magnitude, # # # v = 0. Also, since Æ = v, A vxyz B = v = 0. Thus, # # # vx = vy = vz = 0

The mass moments of intertia of the disk about the x, y, z axes are Ix = Iz = Iy =

1 15 a b A 0.52 B = 0.02911 slug # ft2 4 32.2

1 15 a b A 0.52 B = 0.05823 slug # ft2 2 32.2

Applying the equations of motion, # ©Mx = Ixvx - (Iy - Iz)vyvz; BZ(1) - A Z(1.5) = 0 - (0.05823 - 0.02911)(25.98)( - 15) BZ - 1.5A Z = 11.35 # ©My = Iyvy - (Iz - Ix)vzvx; BX(1 sin 30°) - A X(1.5 sin 30°) = 0 - 0

(1)

BX - 1.5A X = 0 # ©Mz = Izvz - (Ix - Iy)vxvy; BX(1 cos 30°) - A X(1.5 cos 30°) = 0 - 0

(2)

BX - 1.5A X = 0 ©FX = m(aG)X;

A X + BX = 0

©FY = m(aG)Y;

AY = 0

©FZ = m(aG)Z;

A Z + BZ - 15 = 0

(3) Ans. (4)

Solving Eqs. (1) through (4), A Z = 1.461 lb

BZ = 13.54 lb = 13.5 lb

Ans. Ans.

A X = BX = 0

Ans: AZ = 1.46 lb BZ = 13.5 lb AX = AY = BX = 0 1154

y

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21–46. z

The assembly is supported by journal bearings at A and B, which develop only y and z force reactions on the shaft. If the shaft is rotating in the direction shown at V = 52i6 rad>s, determine the reactions at the bearings when the assembly is in the position shown. Also, what is the shaft’s angular acceleration? The mass per unit length of each rod is 5 kg>m.

A

v

B

Solution

1m

1m

y

2m

x

Equations of Motion. The inertia properties of the assembly are Ix =

1 [5(1)] ( 12 ) = 1.6667 kg # m2 3

Iy =

1 [5(3)] ( 32 ) + 3

Iz =

30

+ 5(1) ( 12 ) 4 = 50.0 kg # m2

1 1 [5(3)] ( 32 ) + e [5(1)] ( 12 ) + 5(1) ( 12 + 0.52 ) f = 51.67 kg # m2 3 12

Ixy = 0 + [5(1)](1)(0.5) = 2.50 kg # m2  Iyz = Izx = 0

# # # with vx = 2 rad>s, vy = vz = 0 and vy = vz = 0 by referring to the FBD of the assembly, Fig. a, # # ΣMx = Ix vx;  - [5(1)](9.81)(0.5) = 1.6667 vx # vx = - 14.715 rad>s2 = - 14.7 rad>s2 Ans. # ΣMy = - Ixy vx; [5(1)](9.81)(1) + [5(2)](9.81)(1.5) - Bz(3) = - 2.50( -14.715) Ans.

Bz = 77.6625 N = 77.7 N ΣMz = - Ixy v2x; - By(3) = - 2.50 ( 22 )   By = 3.3333 N = 3.33 N

Ans.

ΣFx = M(aG)x;    Ax = 0

Ans.

ΣFy = M(aG)y; - Ay - 3.333 = [5(1)] 3 - 22(0.5) 4   Ay = 6.667 N = 6.67 N Ans.

ΣFz = M(aG)z; Az + 77.6625 - [5(3)](9.81) - [5(1)](9.81) = [5(1)][ - 14.715(0.5)] Ans.

Az = 81.75 N

Ans: # vx = Bz = By = Ax = Ay = Az = 1155

- 14.7 rad>s2 77.7 N 3.33 N 0 6.67 N 81.75 N

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21–47. z

The assembly is supported by journal bearings at A and B, which develop only y and z force reactions on the shaft. If the shaft A is subjected to a couple moment M = 540i6 N # m, and at the instant shown the shaft has an angular velocity of V = 5 2i6 rad>s, determine the reactions at the bearings of the assembly at this instant. Also, what is the shaft’s angular acceleration? The mass per unit length of each rod is 5 kg>m.

v

B

1m

1m

y

2m

x

Solution

A

Equations of Motions. The inertia properties of the assembly are 1 Ix = [5(1)] ( 12 ) = 1.6667 kg # m2 3 Iy = Iz =

1 [5(3)] ( 32 ) + 3

30

+ [5(1)] ( 12 ) 4 = 50.0 kg # m2

1 1 [5(3)] ( 32 ) + e [5(1)] ( 12 ) + [5(1)] ( 12 + 0.52 ) f = 51.67 kg # m2 3 12

Ixy = 0 + [5(1)](1)(0.5) = 2.50 kg # m2   Iyz = Izx = 0

# # with vx = 2 rad>s, vy = vz = 0 and vy = vz = 0 by referring to the FBD of the assembly, Fig. a, # # ΣMx = Ixvx;  40 - [5(1)](9.81)(0.5) = 1.6667 vx # Ans. vx = 9.285 rad>s2 # ΣMy = - Ixy vx;  [5(1)](9.81)(1) + [5(3)](9.81)(1.5) - Bz(3) = -2.50(9.285) Bz = 97.6625 N = 97.7 N ΣMz =

- Ixy v2x; 

-By(3) = - 2.50 ( 2 )    By = 3.3333 N = 3.33 N 2

ΣFx = M(aG)x;   Ax = 0

Ans. Ans. Ans.

ΣFy = M(aG)y;  -Ay - 3.3333 = [5(1)] 3 - 2 (0.5) 4   Ay = 6.6667 N = 6.67 N Ans. 2

ΣFz = M(aG)z;  Az + 97.6625 - [5(3)](9.81) - [5(1)](9.81) = [5(1)][9.285(0.5)]

Ans.

Az = 121.75 N = 122 N

Ans: # vx = Bz = By = Ax = Ay = Az = 1156

9.285 rad>s2 97.7 N 3.33 N 0 6.67 N 122 N

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*21–48. The man sits on a swivel chair which is rotating with a constant angular velocity of 3 rad>s. He holds the uniform 5-lb rod AB horizontal. He suddenly gives it an angular acceleration of 2 rad>s2, measured relative to him, as shown. Determine the required force and moment components at the grip, A, necessary to do this. Establish axes at the rod’s center of mass G, with + z upward, and + y directed along the axis of the rod towards A.

3 rad/s 3 ft 2 rad/s2 B

SOLUTION I x = Iz =

2 ft

A

1 5 A B (3)2 = 0.1165 ft4 12 32.2

Iy = 0 Æ = v = 3k vx = v y = 0 vz = 3 rad>s # # Æ = (vxyz) + Æ * v = - 2i + 0 # vx = -2 rad>s2 # # vy = vz = 0 (aG)x = 0 (aG)y = (3.5)(3)2 = 31.5 ft>s2 (aG)z = 2(1.5) = 3 ft>s2 ©Fx = m(aG)x;

Ax = 0

©Fy = m(aG)y;

Ay =

©Fz = m(aG)z;

- 5 + Az =

Ans.

5 (31.5) = 4.89 lb 32.2

Ans.

5 (3) 32.2 Ans.

Az = 5.47 lb # ©Mx = Ixvx - (Iy - Iz)vy vz; Mx + 5.47(1.5) = 0.1165(-2) - 0 Mx = - 8.43 lb # ft

Ans.

# ©My = Iyvx - (Iz - Ix)vz vx; 0 + My = 0 - 0 Ans.

My = 0 # ©Mz = Izvz - (Ix - Iy)vx vy; Mz = 0 - 0

Ans.

Mz = 0

1157

Ans: Ax = Ay = Az = Mx = My = Mz =

0 4.89 Ib 5.47 Ib - 8.43 Ib # ft 0 0

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21–49. The rod assembly is supported by a ball-and-socket joint at C and a journal bearing at D, which develops only x and y force reactions. The rods have a mass of 0.75 kg/m. Determine the angular acceleration of the rods and the components of reaction at the supports at the instant v = 8 rad>s as shown.

z D

ω = 8 rad/s

2m

1m

A

B

SOLUTION 2m

Æ = v = 8k vx = vy = 0,

vz = 8 rad>s

# # vx = vy = 0,

# # vz = vz

50 N ⋅ m C y

x

Ixz = Ixy = 0 Iyz = 0.75(1)(2)(0.5) = 0.75 kg # m2 Izz =

1 (0.75)(1)(1)2 = 0.25 kg # m2 3

Eqs. 21–24 become ©Mx = Iyz v2z # ©My = - Iyz vz # ©Mz = Izz vz Thus, - Dy (4) - 7.3575(0.5) = 0.75(8)2 Ans.

Dy = - 12.9 N Dx (4) = - 0.75 vz # 50 = 0.25 vz # vz = 200 rad>s2

Ans.

Dx = -37.5 N

Ans.

©Fx = m(aG)x ;

Cx - 37.5 = - 1(0.75)(200)(0.5) Ans.

Cx = - 37.5 N ©Fy = m(aG)y ;

Cy - 12.9 = - (1)(0.75)(8)2 (0.5) Ans.

Cy = - 11.1 N ©Fz = m(aG)z ;

Cz - 7.3575 - 29.43 = 0 Ans.

Cz = 36.8 N

Ans: # vz = Dy = Dx = Cx = Cy = Cz = 1158

200 rad>s2 - 12.9 N - 37.5 N - 37.5 N - 11.1 N 36.8 N

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21–50. z

The bent uniform rod ACD has a weight of 5 lb>ft and is supported at A by a pin and at B by a cord. If the vertical shaft rotates with a constant angular velocity v = 20 rad>s, determine the x, y, z components of force and moment developed at A and the tension in the cord.

1 ft A

0.5 ft

C B

SOLUTION

1 ft

wx = wy = 0 ω

wz = 20 rad>s

D

# # # wx = wy = wz = 0 The center of mass is located at 1 5(1)a b 2 = 0.25 ft z = 5(2) y = 0.25 ft (symmetry) Iyz =

5 ( - 0.5)(1) = - 0.0776 slug # ft 2 32.2

Izx = 0 Eqs. 21–24 reduce to ΣMx = Iyz(wz)2; - TB(0.5) - 10(0.75) = - 0.0776(20)2 Ans.

TB = 47.1 lb ΣMy = 0;

My = 0

Ans.

ΣMz = 0;

Mz = 0

Ans.

ΣFx = max;

Ax = 0

Ans.

ΣFy = may;

Ay = -a

Ay = -93.2 lb ΣFz = maz;

10 b(20)2(1 - 0.25) 32.2

Ans.

Az - 47.1 - 10 = 0 Ans.

Az = 57.1 lb

Ans: TB = My = Mz = Ax = Ay = Az = 1159

47.1 lb 0 0 0 - 93.2 lb 57.1 lb

y

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21–51. The uniform hatch door, having a mass of 15 kg and a mass center at G, is supported in the horizontal plane by bearings at A and B. If a vertical force F = 300 N is applied to the door as shown, determine the components of reaction at the bearings and the angular acceleration of the door.The bearing at A will resist a component of force in the y direction, whereas the bearing at B will not. For the calculation, assume the door to be a thin plate and neglect the size of each bearing. The door is originally at rest.

z 100 mm 200 mm 200 mm

A

150 mm 150 mm 100 mm B G

x

30 mm

F 30 mm

y

SOLUTION vx = vy = v z = 0 # # v x = vz = 0 Eqs. 21–25 reduce to ©Mx = 0;

300(0.25 - 0.03) + Bz(0.15) - Az(0.15) = 0 (1)

Bz - Az = -440 # ©My = Iyvy;

15(9.81)(0.2) - (300)(0.4 - 0.03) = [

1 # (15)(0.4)2 + 15(0.2)2]vy 12

# vy = -102 rad>s2 ©Mz = 0;

Ans.

-Bx(0.15) + Ax(0.15) = 0

©Fx = m(aG)x;

-Ax + Bx = 0 Ans.

Ax = Bx = 0

Ans.

©Fy = m(aG)y;

Ay = 0

©Fz = m(aG)z;

300 - 15(9.81) + Bz + Az = 15(101.96)(0.2) (2)

Bz + Az = 153.03 Solving Eqs. (1) and (2) yields Az = 297 N

Ans.

Bz = - 143 N

Ans.

Ans: # vy = Ax = Ay = Az = Bz = 1160

-102 rad>s2 Bx = 0 0 297 N -143 N

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*21–52. The 5-kg circular disk is mounted off center on a shaft which is supported by bearings at A and B. If the shaft is rotating at a constant rate of v = 10 rad>s, determine the vertical reactions at the bearings when the disk is in the position shown.

A

100 mm ω

100 mm 20 mm

G

100 mm B

SOLUTION vx = 0,

vy = - 10 rad/s,

# vx = 0,

# vy = 0,

vz = 0

# vz = 0

# a + ©Mx = Ix vx - (Iy - Iz)vy vz - (0.2)(FA) + (0.2)(FB) = 0 FA = FB + T ©Fz = maz ;

FA + FB - 5(9.81) = -5(10)2 (0.02) Ans.

FA = FB = 19.5 N

Ans: FA = FB = 19.5 N 1161

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21–53. Two uniform rods, each having a weight of 10 lb, are pin connected to the edge of a rotating disk. If the disk has a constant angular velocity vD = 4 rad>s, determine the angle u made by each rod during the motion, and the components of the force and moment developed at the pin A. Suggestion: Use the x, y, z axes oriented as shown.

ω D = 4 rad/s z 1.75 ft

A 2 ft

θ

SOLUTION Iy =

10 1 a b(4)2 = 0.4141 slug # ft 2 12 32.2

θ

G

y 2 ft

x

Iz = 0

B

Applying Eq. 21–25 with vy = 4 sin u

vz = 4 cos u

vx = 0

# # # vx = vy = vz = 0 # ΣMx = Ix vx - (Iy - Iz)vy vz;

- Ay(2) = 0 - (0.4141 - 0)(4 sin u)(4 cos u)

# ΣMy = Iy vy - (Iz - Ix)vz vx ;

My + Ax(2) = 0

# ΣMz = Iz vz - (Ix - Iy)vx vy;

Mz = 0

(1) (2) Ans.

Also, ΣFx = m(aG)x; From Eq. (2)

Ans.

Ax = 0

Ans.

My = 0

ΣFy = m(aG)y;

Ay - 10 sin u = - a

ΣFz = m(aG)z;

Az - 10 cos u = a

10 b(1.75 + 2 sin u)(4)2 cos u 32.2

10 b (1.75 + 2 sin u)(4)2 sin u 32.2

(3)

(4)

Solving Eqs. (1), (3) and (4) yields: u = 64.1°

Ay = 1.30 lb

Ans.

Az = 20.2 lb

Ans: Mz = 0 Ax = 0 My = 0 u = 64.1° Ay = 1.30 lb Az = 20.2 lb 1162

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21–54. x

The 10-kg disk turns around the shaft AB, while the shaft rotates about BC at a constant rate of vx = 5 rad>s. If the disk does not slip, determine the normal and frictional force it exerts on the ground. Neglect the mass of shaft AB.

C vx  5 rad/s B

Solution

y

Kinematics. The instantaneous axis of zero velocity (IA) is indicated in Fig. a. Here the resultant angular velocity is always directed along IA. The fixed reference frame is set to coincide with the rotating xyz frame using the similar triangle, vz 2

=

vx 2  ;    vz = (5) = 25.0 rad>s 0.4 0.4

Thus, V = Vx + Vz = 5- 5i + 25.0k6 rad>s # # Here, (vx)xyz = (vz)xyz = 0 since vx is constant. The direction of Vx will not change that always along x axis when Ω = vx. Then # # Vx = (Vx)xyz + Vx * Vx = 0 The direction of Vb does not change with reference to the xyz rotating frame if this frame rotates with Ω = vx = 5- 5i 6 rad>s . Then # # Vz = (vz)xyz + Vx * vz = 0 + ( - 5i) * (25.0k)

Finally

= 5125j 6 rad>s2 # # # V = Vx + Vz = 0 + 125j = 5125j 6 rad>s2

1163

2m

A z 0.4 m

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21–54. Continued

Equations of Motion. The mass moments of inertia of the disk about the x, y and z axes are Ix = Iy = Iz =

1 (10) ( 0.42 ) + 10 ( 22 ) = 40.4 kg # m2 4

1 (10) ( 0.42 ) = 0.800 kg # m2 2

# By referring to the FBD of the disk, Fig. b, ΣMy = Iy vy - (Iz - Ix)vzvx; N(2) - 10(9.81)(2) = 40.4(125) - (0.8 - 40.4)(25.0)( -5) Ans.

N = 148.1 N = 148 N # ΣMz = Izvz - (Ix - Iy)vx vy;    Ff (0.4) = 0 - 0

Ans.

Ff = 0

Ans: N = 148 N Ff = 0 1164

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21–55. z

The 20-kg disk is spinning on its axle at vs = 30 rad>s, while the forked rod is turning at v1 = 6 rad>s. Determine the x and z moment components the axle exerts on the disk during the motion.

200 mm

vs  30 rad/s

x O y

Solution v1  6 rad/s

Solution I Kinematics. The fixed reference XYZ frame is set coincident with the rotating xyz frame. Here, this rotating frame is set to rotate with 𝛀 = Vp = 5-6k6 rad>s. The angular velocity of the disk with respect to the XYZ frame is Then

V = Vp + Vs = 5 30j - 6k6 rad>s

vx = 0  vy = 30 rad>s  vz = - 6 rad>s # Since Vp and Vs does not change with respect to xyz frame, V with respect to this # frame is V = 0. Then # # # vx = 0  vy = 0  vz = 0 Equation of Motion. Although the disk spins about the y axis, but the mass moment of inertia of the disk remain constant with respect to the xyz frame. Ix = Iz =

1 (20) ( 0.22 ) = 0.2 kg # m2 4

Iy =

1 (20) ( 0.22 ) = 0.4 kg # m2 2

1165

A

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21–55. Continued

With 𝛀 ≠ V with Ω x = 0, Ω y = 0 and Ω z = - 6 rad>s # ΣMx = Ixvx - Iy Ω zvy + Iz Ω y vz;

(M0)x = 0 - 0.4( -6)(30) + 0  (M0)x = 72.0 N # m # ΣMy = Iyvy - Iz Ω xvz + Ix Ω zvx

Ans.

0 = 0 (Satisfied!) # ΣMz = Izvz - Ix Ω y vx + Iy Ω x vy; Ans.

(M0)z = 0 - 0 + 0 = 0 Solution II

Here, the xyz frame is set to rotate with 𝛀 = V = 530j - 6k6 rad>s. Setting another x′y′z′ frame coincide with xyz and XYZ frame to have an angular velocity of 𝛀′ = Vp = 5-6k6 rad>s, # # # v = (v)xyz = (v)x′y′z′ + 𝛀 ′ * v = 0 + ( - 6k) * (30j - 6k) = 5180i6 rad>s2

Thus

# # # vx = 180 rad>s  vy = 0  vz = 0 with Ω = v, # ΣMx = Ixvx - (Iy - Iz)vyvz; # ΣMy = Iyvy - (Iz - Ix)vzvx; # ΣMz = Izvz - (Ix - Iy)vxvy

(M0)x = 0.2(180) - (0.4 - 0.2)(30)( - 6) = 72.0 N # m

0 = 0 - 0 (Satisfied) Ans.

(M0)z = 0 - 0 = 0

Ans: (M0)x = 72.0 N # m (M0)z = 0 1166

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*21–56. The 4-kg slender rod AB is pinned at A and held at B by a cord.The axle CD is supported at its ends by ball-and-socket joints and is rotating with a constant angular velocity of 2 rad>s. Determine the tension developed in the cord and the magnitude of force developed at the pin A.

C y

ω B z

40°

SOLUTION Iz =

2m

1 (4)(2)2 = 5.3333 kg # m2 3

A

Iy = 0

D

Applying the third of Eq. 21–25 with vy = 2 cos 40° = 1.5321 rad>s vz = 2 sin 40° = 1.2856 rad>s # ©Mx = Ix vx - (Iy - Iz) vy vz ; T(2 cos 40°) - 4(9.81)(1 sin 40°) = 0 - (0 - 5.3333)(1.5321)(1.2856) Ans.

T = 23.3 N Also, ©Fx¿ = m(aG)x¿ ;

Ax¿ = 0

©Fy¿ = m(aG)y¿ ;

Ay¿ - 23.32 = - 4(2)2 (1 sin 40°)

©Fz¿ = m(aG)z¿ ;

Az¿ - 4(9.81) = 0

FA =

A2x + A2y + A2z =

Ay¿ = 13.03 N

Az¿ = 39.24 N

02 + 13.032 + 39.242 = 41.3 N

Ans.

Ans: T = 23.3 N FA = 41.3 N 1167

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21–57. The blades of a wind turbine spin about the shaft S with a constant angular speed of vs, while the frame precesses about the vertical axis with a constant angular speed of vp. Determine the x, y, and z components of moment that the shaft exerts on the blades as a function of u. Consider each blade as a slender rod of mass m and length l.

z

vp

u

x S

SOLUTION

u

vs

The rotating xyz frame shown in Fig. a will be attached to the blade so that it rotates with an angular velocity of Æ = v, where v = vs + vp. Referring to Fig. b vp = vp sin ui + vp cos uk. Thus, v = vp sin ui + vs j + vp cos uk. Then vx = vp sin u

y

vy = vs vz = vp cos u

# The angular acceleration of the blade v with respect to the XYZ frame can be obtained by setting another x¿y¿z¿ frame having an angular velocity of Æ¿ = vp = vp sin ui + vp cos uk. Thus, # # v = A vx¿y¿z¿ B + Æ ¿ * v

# # = (v1)x¿y¿z¿ + (v2)x¿y¿z¿ + Æ¿ * vS + Æ¿ * vP = 0 + 0 + A vp sin ui + vp cos uk B * (vsj) + 0

= -vsvp cos ui + vsvp sin uk # # Since Æ = v, vx¿y¿z¿ = v. Thus, # vx = - vsvp cos u

# vy = 0

# vz = vsvp sin u

Also, the x, y, and z axes will remain as principle axes of inertia for the blade. Thus, Ix = Iy =

1 2 (2m)(2l)2 = ml2 12 3

Iz = 0

Applying the moment equations of motion and referring to the free-body diagram shown in Fig. a, # ©Mx = Ixvx - A Iy - Iz B vyvz;

Mx =

2 2 2 ml ( -vsvp cos u) - a ml2 - 0b (vs)(vp cos u) 3 3

4 = - ml2 vsvp cos u 3 # ©My = Iyvy - A Iz - Ix B vzvx;

My = 0 - a 0 =

# ©Mz = Izvz - (Ix - Iy)vxvy ;

Ans.

2 2 ml b(vp cos u)(vp sin u) 3

1 2 2 ml vp sin 2u 3

Ans. Ans.

Mz = 0 - 0 = 0

Ans:

4 Mx = - ml 2vsvp cos u 3 1 2 2 My = ml vp sin 2u 3 Mz = 0

1168

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21–58. z

The 15-lb cylinder is rotating about shaft AB with a constant angular speed v = 4 rad>s. If the supporting shaft at C, initially at rest, is given an angular acceleration aC = 12 rad>s2, determine the components of reaction at the bearings A and B. The bearing at A cannot support a force component along the x axis, whereas the bearing at B does.

1 ft 1 ft

ω 0.5 ft

A

G

x

B y

SOLUTION αC

v = { - 4i} rad>s

C

# # v = vxyz + Æ * v = 12k + 0 * ( -4i) = {12k} rad>s2 Hence vx = - 4,

vy = vz = 0,

# # vx = vy = 0,

# vz = 12 rad>s2

# ©Mx = Ix vx - (Iy - Iz) vy vz,

0 = 0

# ©My = Iy vy - (Iz - Ix) vz vx,

Bz (1) - Az (1) = 0

# ©Mz = Iz vz - (Ix - Iy) vx vy,

Ay (1) - By (1) = c

1 15 a b A 3(0.5)2 + (2)2 B d(12) - 0 12 32.2 Ans.

©Fx = m(aG)x ;

Bx = 0

©Fy = m(aG)y ;

Ay + By = - a

©Fz = m(aG)z ;

Az + Bz - 15 = 0

15 b (1)(12) 32.2

Solving, Ay = - 1.69 lb

Ans.

By = - 3.90 lb

Ans.

Az = Bz = 7.5 lb

Ans.

Ans: Bx = By = Ay = Az = 1169

0 - 3.90 lb - 1.69 lb Bz = 7.5 lb

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21–59. The thin rod has a mass of 0.8 kg and a total length of 150 mm. . It is rotating about its midpoint at a constant rate u = 6 rad>s, while the table to which its axle A is fastened is rotating at fastened is rotating at 2 rad> s. Determine the x, y, z moment components which the axle exerts on the rod when the rod is in any position u.

y 2 rad/s

x

A

u

. u

z

SOLUTION The x,y,z axes are fixed as shown. vx = 2 sin u vy = 2 cos u # vz = u = 6 # # vx = 2u cos u = 12 cos u # # vy = - 2u sin u = -12 sin u # vz = 0 Ix = 0 Iy = Iz =

1 (0.8)(0.15)2 = 1.5(10-3) 12

Using Eqs. 21–25: Ans.

©Mx = 0 - 0 = 0 ©My = 1.5(10-3)(- 12 sin u) - [1.5(10-3) -0](6)(2 sin u) ©My = (- 0.036 sin u) N # m

Ans.

©Mz = 0 - [0 - 1.5(10-3)](2 sin u)(2 cos u) ©Mz = 0.006 sin u cos u = (0.003 sin 2 u) N # m

Ans.

Ans: ΣMx = 0 ΣMy = ( -0.036 sin u) N # m ΣMz = (0.003 sin 2u) N # m

1170

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*21–60. Show that the angular velocity of a body, in terms of Euler# angles f , # u , and c , # can be expressed as # v# = (f sin u #sin c + u cos c )i + (f sin u cos c - u sin c ) j + (f cos u + c)k, where i, j, and k are directed along the x, y, z axes as shown in Fig. 21–15d.

SOLUTION

# # From Fig. 21–15b. due to rotation f, the x, y, z components of f are simply f along z axis. # # # From Fig 21–15c,# due to rotation u, the x, y, z #components of f and u are f sin u in the y direction, f cos u in the z direction, and u in the x direction. Lastly, rotation c. Fig. 21–15d, produces the final components which yields # # # # # # v = A f sin u sin c + u cos c B i + A f sin u cos c - u sin c B j + A f cos u + c B k Q.E.D.

Ans: . . v = ( f sin u sin c + u cos c ) i . . + ( f sin u cos c - u sin c ) j . . + ( f cos u + c ) k 1171

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21–61. A thin rod is initially coincident with the Z axis when it is given three rotations defined by the Euler angles f = 30°, u = 45°, and c = 60°. If these rotations are given in the order stated, determine the coordinate direction angles a, b , g of the axis of the rod with respect to the X, Y, and Z axes. Are these directions the same for any order of the rotations? Why?

SOLUTION u = (1 sin 45°) sin 30° i - (1 sin 45°) cos 30°j + 1 cos 45° k u = 0.3536i - 0.6124j + 0.7071k a = cos - 1 0.3536 = 69.3°

Ans.

b = cos - 1(- 0.6124) = 128°

Ans.

g = cos - 1(0.7071) = 45°

Ans.

No, the orientation of the rod will not be the same for any order of rotation, because finite rotations are not vectors.

Ans: a = 69.3° b = 128° g = 45° No, the orientation will not be the same for any order. Finite rotations are not vectors. 1172

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21–62. The gyroscope consists of a uniform 450-g disk D which is attached to the axle AB of negligible mass. The supporting frame has a mass of 180 g and a center of mass at G. If the disk is rotating about the axle at vD = 90 rad>s, determine the constant angular velocity vp at which the frame precesses about the pivot point O. The frame moves in the horizontal plane.

25 mm

80 mm

20 mm

ωp O

35 mm A

SOLUTION

B D

©Mx = Iz Æ y vz (0.450)(9.81)(0.125) + (0.180)(9.81)(0.080) =

25 mm

G

ωD

1 (0.450)(0.035)2 vP (90) 2 Ans.

vP = 27.9 rad>s

Ans: vP = 27.9 rad>s 1173

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21–63. The toy gyroscope consists of a rotor R which is attached to the frame of negligible mass. If it is observed that the frame is precessing about the pivot point O at vp = 2 rad>s, determine the angular velocity vR of the rotor. The stem OA moves in the horizontal plane. The rotor has a mass of 200 g and a radius of gyration kOA = 20 mm about OA.

30 mm

ωR

ωp

A

SOLUTION

R

ΣMx = Iz Ω ywz (0.2)(9.81)(0.03) = vR = 368 rad>s

O

3 0.2(0.02)2 4 (2)(wR)

Ans.

Ans: vR = 368 rad>s 1174

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*21–64. The top consists of a thin disk that has a weight of 8 lb and a radius of 0.3 ft. The rod has a negligible mass and a length of 0.5 ft. If the top is spinning with an angular velocity vs = 300 rad>s, determine the steady-state precessional angular velocity vp of the rod when u = 40°.

0.3 ft

0.5 ft ωp

ωs

θ

SOLUTION

# # # # ΣMx = - I f2 sin u cos u + Izf sin u( f cos u + c)

8 1 8 b(0.3)2 + a b (0.5)2 d v 2p sin 40° cos 40° 8(0.5 sin 40°) = - c a 4 32.2 32.2 1 8 + c a b(0.3)2 d v p sin 40° (v p cos 40° + 300) 2 32.2

0.02783v 2p - 2.1559v p + 2.571 = 0 v p = 1.21 rad>s

Ans.

(Low precession)

v p = 76.3 rad>s

Ans.

(High precession)

Ans: wp = 1.21 rad>s wp = 76.3 rad>s 1175

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21–65. Solve Prob. 21–64 when u = 90°.

0.3 ft

0.5 ft ωp

ωs

θ

SOLUTION ΣMx = Iz Ω y wz 1 8 b (0.3)2 d wp(300) 8(0.5) = c a 2 32.2 Ans.

vp = 1.19 rad>s

Ans: vP = 1.19 rad>s 1176

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21–66. The propeller on a single-engine airplane has a mass of 15 kg and a centroidal radius of gyration of 0.3 m computed about the axis of spin. When viewed from the front of the airplane, the propeller is turning clockwise at 350 rad>s about the spin axis. If the airplane enters a vertical curve having a radius of 80 m and is traveling at 200 km>h, determine the gyroscopic bending moment which the propeller exerts on the bearings of the engine when the airplane is in its lowest position.

r  80 m

SOLUTION vs = 350 rad>s = vz v = 200 km/h = Æy =

200(103) = 55.56 m>s 3600

55.56 = 0.694 rad>s 80

©Mx = Iz Æ y vz Mx = [15(0.3)2](0.694)(350) Mx = 328 N # m

Ans.

Ans: Mx = 328 N # m 1177

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21–67. A wheel of mass m and radius r rolls with constant spin V about a circular path having a radius a. If the angle of inclination is u, determine the rate of precession. Treat the wheel as a thin ring. No slipping occurs.

. f u r

a V

SOLUTION Since no slipping occurs, # # (r) c = (a + r cos u) f or # a + r cos u # c = ( )f r

(1)

Also, # # v = f + c ©Fy¿ = m(aG)y¿ ;

# F = m(a f2)

(2)

©Fz¿ = m(aG)z¿ ;

N - mg = 0

(3)

I x = Iy =

mr2 , 2

Iz = mr2

# # # v = f sin u j + ( -c + f cos u)k Thus, # # # vy = f sin u, vz¿ = - c + f cos u vx = 0, # # # # # v = f * c = - f c sin u # # # # # vx = -f c sin u, vy = vz = 0 Applying # ©Mx = Ix vx + (Iz - Iy)vz vy F r sin u - N r cos u =

# # # # # m r2 mr2 (-f c sin u) + (m r2 )( -c + f cos u)(f sin u) 2 2

# Using Eqs. (1), (2) and (3), and eliminating c, we have # # #2 mr2 -f a # m r2 a + r cos u # ( - f) sin u( ( )f + )f sin u m a f r sin u - m g r cos u = r r 2 2 # #2 m r2 # 2 m r2 - f2a ( (f sin u cos u) m a f sin u r - m g r cos u = ) sin u r 2 2 # # 2 g cos u = a f2 sin u + r f2 sin u cos u # 2 g cot u 1/2 f = ( ) a + r cos u

Ans. Ans: # 2g cos u 1>2 f = a b a + r cos u 1178

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*21–68. The conical top has a mass of 0.8 kg, and the moments of inertia are Ix = Iy = 3.5 ( 10-3 ) kg # m2 and -3 2 # Iz = 0.8 ( 10 ) kg m . If it spins freely in the ball-and socket joint at A with an angular velocity vs = 750 rad>s, compute the precession of the top about the axis of the shaft AB.

B y x

A

Solution

100 mm 30

vs = 750 rad>s

vs

Using Eq. 21 – 30. # # # # ΣMx = - If2 sin u cos u + Iz f sin u (f cos u + f) # # # 0.1(0.8)(9.81) sin 30° = - 3.5 ( 10-3 ) f2 sin 30° cos 30° + 0.8 ( 10-3 ) f sin 30°(f cos 30° + 750)

G

Thus, #

z

#

1.160 ( 10-3 ) f2 - 300 ( 10-3 ) f + 0.3924 = 0 # f = 1.31 rad>s

(low precession)

Ans.

# f = 255 rad>s

(high precession)

Ans.

Ans: # f# = 1.31 rad>s f = 255 rad>s 1179

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21–69. The top has a mass of 90 g, a center of mass at G, and a radius of gyration k = 18 mm about its axis of symmetry. About any transverse axis acting through point O the radius of gyration is kt = 35 mm. If the top is connected to a ball-andsocket joint at O and the precession is vp = 0.5 rad>s, determine the spin V s.

Vp Vs 45

O

SOLUTION

G

60 mm

vp = 0.5 rad>s # # # # ©Mx = -If2 sin u cos u + Izf sin u a f cos u + c b 0.090(9.81)(0.06) sin 45° = -0.090(0.035)2 (0.5)2 (0.7071)2 # + 0.090(0.018)2(0.5)(0.7071) C 0.5(0.7071) + c D vs = c = 3.63 A 103 B rad>s

Ans.

Ans: vs = 3.63 ( 103 ) rad>s 1180

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21–70. The 1-lb top has a center of gravity at point G. If it spins about its axis of symmetry and precesses about the vertical axis at constant rates of vs = 60 rad>s and vp = 10 rad>s, respectively, determine the steady state angle u. The radius of gyration of the top about the z axis is kz = 1 in., and about the x and y axes it is kx = ky = 4 in.

z

u vp

10 rad/s

# # Since c = vs = 60 rad>s and f = vp = - 10 rad>s and u are constant, the top undergoes steady precession. 1 1 b a b = 215.67 A 10 - 6 B slug # ft2 Iz = a 32.2 12

and

= 3.4507 10 - 3 slug # ft2.

A

60 rad/s

G

SOLUTION

2

vs

4 1 ba b I = Ix = Iy = a 32.2 12

3 in. x

2

O y

B

Thus, # # # # ©Mx = -If2 sin u cos u + Izf sin u A f cos + c B

-1 sin u(0.25) = -3.4507 A 10 - 3 B ( - 10)2 sin u cos u + 215.67 A 10 - 6 B ( - 10) sin u[(- 10) cos u + 60] Ans.

u = 68.1°

Ans: u = 68.1° 1181

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21–71. The space capsule has a mass of 2 Mg, center of mass at G, and radii of gyration about its axis of symmetry (z axis) and its transverse axes (x or y axis) of kz = 2.75 m and kx = ky = 5.5 m, respectively. If the capsule has the angular # # velocity shown, determine its precession f and spin c. Indicate whether the precession is regular or retrograde. Also, draw the space cone and body cone for the motion.

y

G x

30 v

SOLUTION The only force acting on the space capsule is its own weight. Thus, it undergoes torque-free motion. Iz = 2000 A 2.752 B = 15 125 kg # m2, I = Ix = Iy = 2000 A 5.52 B = 60 500 kg # m2. Thus,

150 rad/s

z

HG sin u vy = I 150 sin 30° =

HG sin u 60 500 (1)

HG sin u = 4 537 500 vz =

HG cos u Iz

150 cos 30° =

HG cos u 15 125 (2)

HG cos u = 1 964 795.13 Solving Eqs. (1) and (2), HG = 4.9446 A 106 B kg # m2>s

u = 66.59°

Using these results,

4.9446 A 106 B # HG HG f = = = = 81.7 rad>s I 60 500 60 500

Ans.

I - Iz # 60 500 - 15 125 c = HG cos u = B R 4.9446 A 106 B cos 30° IIz 60 500(15125) Ans.

= 212 rad>s

Ans.

Since I 7 Iz , the motion is regular precession.

Ans: # f = 81.7 rad>s # c = 212 rad>s regular precession 1182

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*21–72. The 0.25 kg football is spinning at vz = 15 rad>s as shown. If u = 40°, determine the precession about the z axis. The radius of gyration about the spin axis is kz = 0.042 m, and about a transverse axis is ky = 0.13 m.

Z

z vz  15 rad/s

G

Solution Here, Iz =

# c = vt = 15 rad>s,

mk 2z

= 0.25 ( 0.042

2

I = mk 2y = 0.25 ( 0.132 ) = 0.004225 kg # m2

) = 0.000441 kg # m2.

and

# I - Iz # 0.004225 - 0.000441 # f cos u ;   15 = a bf cos 40° c = Iz 0.000441 # f = 2.282 rad>s2 = 2.28 rad>s2

Ans.

Ans: # f = 2.28 rad>s2 1183

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21–73. The projectile shown is subjected to torque-free motion. The transverse and axial moments of inertia are I and Iz, respectively. If u represents the angle between the precessional axis Z and the axis of symmetry z, and b is the angle between the angular velocity V and the z axis, show that b and u are related by the equation tan u = (I>Iz) tan b .

y Z V z

x

u b G

SOLUTION From Eq. 21–34 vy =

vy Iz HG sin u HG cos u tan u and vz = Hence = I Iz vz I

However, vy = v sin b and vz = v cos b vy vz

= tan b = tan u =

Iz I

tan u

I tan b Iz

Q.E.D.

Ans: tan u = 1184

I tan b Iz

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21–74. The radius of gyration about an axis passing through the axis of symmetry of the 1.6-Mg space capsule is kz = 1.2 m, and about any transverse axis passing through the center of mass G, kt = 1.8 m. If the capsule has a known steady-state precession of two revolutions per hour about the Z axis, determine the rate of spin about the z axis.

Z 20 z

G

SOLUTION I = 1600(1.8)2,

Iz = 1600(1.2)2

Use the result of Prob. 21–75. tan u = a

I b tan b Iz

tan 20° = a

1600(1.8)2 1600(1.2)2

b tan b

b = 9.189° Using the law of sines: sin (20° - 9.189°) sin 9.189° = # 2 c # c = 2.35 rev>h

Ans.

Ans: # c = 2.35 rev>h 1185

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21–75. The rocket has a mass of 4 Mg and radii of gyration kz = 0.85 m and ky = 2.3 m. It is initially spinning about the z axis at vz = 0.05 rad>s when a meteoroid M strikes it at A and creates an impulse I = 5300i6 N # s. Determine the axis of precession after the impact.

y x G

A

ωz z

3m M

SOLUTION The impulse creates an angular momentum about the y axis of Hy = 300(3) = 900 kg # m2>s Since vz = 0.05 rad>s then HG = 900j + [4000(0.85)2](0.05)k = 900j + 144.5k The axis of precession is defined by HG. uHG =

900j + 144.5k = 0.9874j + 0.159k 911.53

Thus, a = cos-1(0) = 90°

Ans.

b = cos-1(0.9874) = 9.12°

Ans.

g = cos-1(0.159) = 80.9°

Ans.

Ans: a = 90° b = 9.12° g = 80.9° 1186

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*21–76. The football has a mass of 450 g and radii of gyration about its axis of symmetry (z axis) and its transverse axes (x or y axis) of k z = 30mm and kx = ky = 50mm, respectively. If the football has an angular momentum of 2 HG = 0.02kg # m >s, determine its precession f and spin c. Also, find the angle b that the angular velocity vector makes with the z axis.

HG

0.02 kg m2/s z V

y 45

B

x

SOLUTION Since the weight is the only force acting on the football, it undergoes torque-free motion.

Iz = 0.45 A 0.032 B = 0.405 A 10 - 3 B kg # m2,

= 1.125 A 10 - 3 B kg # m2, and u = 45°.

G

I = Ix = Iy = 0.45 A 0.052 B

Thus, # HG 0.02 f = = = 17.78 rad>s = 17.8 rad>s I 1.125 A 10 - 3 B

Ans.

1.125 A 10 - 3 B - 0.405 A 10 - 3 B I - Iz # HG cos u = (0.02) cos 45° c = IIz 1.125 A 10 - 3 B (0.405) A 10 - 3 B = 22.35 rad>s = 22.3 rad>s

Ans.

Also, vy = vz =

HG sin u 0.02 sin 45° = = 12.57 rad>s I 1.125 A 10 - 3 B

HG cos u 0.02 cos 45° = = 34.92 rad>s Iz 0.405 A 10 - 3 B

Thus, b = tan - 1 ¢

vy vz

≤ = tan - 1 a

12.57 b = 19.8° 34.92

Ans.

Ans: # f# = 17.8 rad>s c = 22.3 rad>s b = 19.8° 1187

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21–77. z

The satellite has a mass of 1.8 Mg, and about axes passing through the mass center G the axial and transverse radii of gyration are kz = 0.8 m and kt = 1.2 m, respectively. If it is spinning at vs = 6 rad>s when it is launched, determine its angular momentum. Precession occurs about the Z axis.

vs

Z

5

G

Solution

I = 1800(1.2)2 = 2592 kg # m2  Iz = 1800(0.8)2 = 1152 kg # m2

# Applying the third of Eqs. 21–36 with u = 5°  c = 6 rad>s I - Iz # c = HG cos u Hz 0 =

2592 - 1152 H cos 5° 2592(1152) G

HG = 12.5 Mg # m2 >s

Ans.

1188

Ans: HG = 12.5 Mg # m2 >s

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21–78. The radius of gyration about an axis passing through the axis of symmetry of the 1.2 Mg satellite is kz = 1.4 m, and about any transverse axis passing through the center of mass G, k t = 2.20 m. If the satelite has a known spin of 2700 rev>h about the z axis, determine the steady-state precession about the z axis.

Z G

15° z

SOLUTION # 2700(2p) = 1.5p rad>s. Gyroscopic Motion: Here, the spinning angular velocity c = vs = 3600 2 The moment inertia of the satelite about the z axis is Iz = 1200 A 1.4 B = 2352 kg # m2 and the

moment inertia of the satelite about its transverse axis is I = 1200 A 2.202 B = 5808 kg # m2. Applying the third of Eq. 21–36 with u = 15°, we have I - Iz # HG cos u c = I Iz 1.5p = c

5808 - 2352 d HG cos 15° 5808(2352)

HG = 19.28 A 103 B kg # m2>s Applying the second of Eq. 21–36, we have # 19.28(103) HG f = = = 3.32 rad s I 5808

Ans.

Ans: # f = 3.32 rad>s 1189

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22–1. A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m>s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.

SOLUTION $ mg - k(y + yst) = my

+ T ΣFy = may;

where kyst = mg

k $ y + y = 0 m

Hence

p =

=

k Bm

B

Where k =

8(9.81) 0.175

448.46 = 7.487 8 $ y + (7.487)2y = 0

6

= 448.46 N>m

$ y + 56.1y = 0

Ans.

The solution of the above differential equation is of the form: y = A sin pt + B cos pt

(1)

# v = y = Ap cos pt - Bp sin pt

(2)

At t = 0, y = 0.1 m and v = v0 = 1.50 m>s From Eq. (1)

0.1 = A sin 0 + B cos 0

B = 0.1 m v0 1.50 = = 0.2003 m p 7.487

From Eq. (2)

v0 = Ap cos 0 - 0

Hence

y = 0.2003 sin 7.487t + 0.1 cos 7.487t

At t = 0.22 s,

y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]

A =

Ans.

= 0.192 m

1190

Ans: $ y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m

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22–2. A spring has a stiffness of 800 N>m. If a 2-kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation that describes the block’s motion. Assume that positive displacement is downward.

SOLUTION p =

k 800 = = 20 Am A 2

x = A sin pt + B cos pt x = - 0.05 m

when t = 0,

-0.05 = 0 + B;

B = -0.05

v = Ap cos pt - Bp sin pt v = 0 when t = 0, 0 = A(20) - 0;

A = 0

Thus, Ans.

x = - 0.05 cos (20t)

Ans: x = -0.05 cos (20t) 1191

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22–3. A spring is stretched 200 mm by a 15-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 0.75 m>s, determine the equation which describes the motion. What is the phase angle? Assume that positive displacement is downward.

SOLUTION k =

15(9.81) F = 735.75 N>m = y 0.2

vn =

k 735.75 = = 7.00 Am A 15

y = A sin vn t + B cos vn t y = 0.1 m when t = 0, 0.1 = 0 + B;

B = 0.1

v = A vn cos vn t - Bvn sin vn t v = 0.75 m>s when t = 0, 0.75 = A(7.00) A = 0.107 Ans.

y = 0.107 sin (7.00t) + 0.100 cos (7.00t) f = tan - 1 a

B 0.100 b = tan - 1 a b = 43.0° A 0.107

Ans.

Ans: y = 0.107 sin (7.00t) + 0.100 cos (7.00t) f = 43.0° 1192

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*22–4. When a 20-lb weight is suspended from a spring, the spring is stretched a distance of 4 in. Determine the natural frequency and the period of vibration for a 10-lb weight attached to the same spring.

SOLUTION k = vn = t =

20 4 12

= 60 lb>ft

k 60 = = 13.90 rad>s 10 Am A 32.2

Ans.

2p = 0.452 s vn

Ans.

Ans: vn = 13.90 rad>s t = 0.452 s 1193

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22–5. When a 3-kg block is suspended from a spring, the spring is stretched a distance of 60 mm. Determine the natural frequency and the period of vibration for a 0.2-kg block attached to the same spring.

SOLUTION k = vn =

3(9.81) F = = 490.5 N>m ¢x 0.060 k 490.5 = = 49.52 = 49.5 rad>s Am A 0.2

f =

vn 49.52 = = 7.88 Hz 2p 2p

t =

1 1 = = 0.127 s f 7.88

Ans.

Ans.

Ans: vn = 49.5 rad>s t = 0.127 s 1194

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22–6. An 8-kg block is suspended from a spring having a stiffness k = 80 N>m. If the block is given an upward velocity of 0.4 m>s when it is 90 mm above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the block measured from the equilibrium position. Assume that positive displacement is measured downward.

SOLUTION vn =

k 80 = = 3.162 rad>s Am A8

y = - 0.4 m>s,

x = - 0.09 m at t = 0

x = A sin vn t + B cos vn t - 0.09 = 0 + B B = -0.09 y = Avn cos vn t - Bvn sin vn t -0.4 = A(3.162) - 0 A = -0.126 Thus,

x = - 0.126 sin (3.16t) - 0.09 cos (3.16t) m

Ans.

C = 2A2 + B2 = 2( -0.126)2 + ( - 0.09) = 0.155 m

Ans.

Ans: x = 5 - 0.126 sin (3.16t) - 0.09 cos (3.16t) 6 m C = 0.155 m 1195

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22–7. A 2-lb weight is suspended from a spring having a stiffness k = 2 lb>in. If the weight is pushed 1 in. upward from its equilibrium position and then released from rest, determine the equation which describes the motion. What is the amplitude and the natural frequency of the vibration?

SOLUTION k = 2(12) = 24 lb>ft vn =

k 24 = = 19.66 = 19.7 rad>s 2 Am A 32.2

y = -

1 , 12

Ans.

y = 0 at t = 0

From Eqs. 22–3 and 22–4, -

1 = 0 + B 12

B = -0.0833 0 = Avn + 0 A = 0 C = 2A2 + B2 = 0.0833 ft = 1 in.

Ans.

y = (0.0833 cos 19.7t) ft

Ans.

Position equation,

Ans: vn = 19.7 rad>s C = 1 in. y = (0.0833 cos 19.7t) ft 1196

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*22–8. A 6-lb weight is suspended from a spring having a stiffness k = 3 lb>in. If the weight is given an upward velocity of 20 ft>s when it is 2 in. above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.

SOLUTION k = 3(12) = 36 lb>ft vn =

k 36 = = 13.90 rad>s Am A 6 32.2

t = 0,

y = - 20 ft>s,

1 y = - ft 6

From Eq. 22–3, -

1 = 0 + B 6

B = -0.167 From Eq. 22–4, -20 = A(13.90) + 0 A = - 1.44 Thus, Ans.

y = [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ft From Eq. 22–10, C = 2A2 + B2 = 2(1.44)2 + (- 0.167)2 = 1.45 ft

Ans.

Ans: y = [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ft C = 1.45 ft 1197

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22–9. A 3-kg block is suspended from a spring having a stiffness of k = 200 N>m. If the block is pushed 50 mm upward from its equilibrium position and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that positive displacement is downward.

SOLUTION vn =

k 200 = = 8.16 rad>s Am A 3

Ans.

x = A sin vn t + B cos vn t x = -0.05 m when t = 0, - 0.05 = 0 + B;

B = -0.05

v = Ap cos vn t - Bvn sin vn t v = 0 when t = 0, 0 = A(8.165) - 0;

A = 0

Hence, x = - 0.05 cos (8.16t)

Ans.

C = 2A2 + B2 = 2(0)2 + ( - 0.05) = 0.05 m = 50 mm

Ans.

Ans: vn = 8.16 rad>s x = - 0.05 cos (8.16t) C = 50 mm 1198

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22–10. The uniform rod of mass m is supported by a pin at A and a spring at B. If B is given a small sideward displacement and released, determine the natural period of vibration.

A

L

Solution 1 Equation of Motion. The mass moment of inertia of the rod about A is IA = mL2. 3 Referring to the FBD. of the rod, Fig. a, L 1 a + ΣMA = IAa ;  -mg a sin u b - (kx cos u)(L) = a mL2 ba 2 3

B

k

However;   x = L sin u. Then

- mgL 1 sin u - kL2 sin u cos u = mL2a 2 3 Using the trigonometry identity sin 2u = 2 sin u cos u, - mgL KL2 1 sin u sin 2u = mL2a 2 2 3

$ Here since u is small sin u ≃ u and sin 2u ≃ 2u. Also a = u . Then the above equation becomes $ mgL 1 mL2 u + a + kL2 bu = 0 3 2 $ 3mg + 6kL u + u = 0 2mL

Comparing to that of the Standard form, vn = t =

A

3mg + 6kL . Then 2mL

2p 2mL = 2p  vn A 3mg + 6kL

Ans.

Ans: t = 2p 1199

2mL A 3mg + 6kL

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22–11. While standing in an elevator, the man holds a pendulum which consists of an 18-in. cord and a 0.5-lb bob. If the elevator is descending with an acceleration a = 4 ft>s2, determine the natural period of vibration for small amplitudes of swing. a

4 ft/s2

SOLUTION Since the acceleration of the pendulum is (32.2 - 4) = 28.2 ft>s2 Using the result of Example 22–1, We have vn = t =

g 28.2 = = 4.336 rad>s Al A 18>12 2p 2p = 1.45 s = vn 4.336

Ans.

Ans: t = 1.45 s 1200

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*22–12. Determine the natural period of vibration of the uniform bar of mass m when it is displaced downward slightly and released.

O k L — 2

L — 2

Solution Equation of Motion. The mass moment of inertia of the bar about O is I0 =

1 mL2. 12

Referring to the FBD of the rod, Fig. a, L 1 a + ΣM0 = I0a ;  - ky cos ua b = a mL2 ba 2 12 However, y =

L sin u.  Then 2

-ka

1 L L sin u b cos u a b = mL2a 2 2 12

Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain 1 kL2 mL2a + sin 2u = 0 12 8

$ Here since u is small, sin 2u ≃ 2u. Also, a = u .  Then the above equation becomes $ 1 kL2 mL2 u + u = 0 12 4 $ 3k u = 0 u + m Comparing to that of the Standard form, vn = t =

3k .  Then Am

m 2p = 2p  vn A 3k

Ans.

Ans: t = 2p 1201

m A 3k

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22–13. The body of arbitrary shape has a mass m, mass center at G, and a radius of gyration about G of kG. If it is displaced a slight amount u from its equilibrium position and released, determine the natural period of vibration.

O d u G

SOLUTION a + ©MO = IO a;

$ -mgd sin u = C mk2G + md2 D u $ u +

gd k2G

+ d2

sin u = 0

However, for small rotation sin u Lu. Hence $ u +

gd 2 kG

+ d2

u = 0

From the above differential equation, vn = t =

2p = vn

2p gd A k2G + d2

gd B k2G

= 2p

+ d2

.

k2G + d2 C gd

Ans.

Ans: t = 2p 1202

k 2G + d 2 C gd

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22–14. The 20-lb rectangular plate has a natural period of vibration t = 0.3 s, as it oscillates around the axis of rod  AB. Determine the torsional stiffness k, measured in lb # ft>rad, of the rod. Neglect the mass of the rod.

A k

B

Solution T = ku ΣMz = Iza ;  - ku = $ u + k(4.83)u = 0 t =

2p 2k(4.83)

4 ft

$ 1 20 a b(2)2 u 12 32.2 2 ft

= 0.3

k = 90.8 lb # ft>rad

Ans.

Ans: k = 90.8 lb # ft>rad 1203

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22–15. A platform, having an unknown mass, is supported by four springs, each having the same stiffness k. When nothing is on the platform, the period of vertical vibration is measured as 2.35 s; whereas if a 3-kg block is supported on the platform, the period of vertical vibration is 5.23 s. Determine the mass of a block placed on the (empty) platform which causes the platform to vibrate vertically with a period of 5.62 s. What is the stiffness k of each of the springs?

k

k

Solution $ + T ΣFy = may;  mtg - 4k(y + yts) = mty  Where 4k yts = mtg 4k $ y + y = 0 mt Hence             P = t =

4k A mt 2p mt = 2p P A 4k

For empty platform mt = mP, where mP is the mass of the platform.

2.35 = 2p

mP  A 4k

(1)

When 3-kg block is on the platform mt = mP + 3.

5.23 = 2p

mP + 3  A 4k

(2)

When an unknown mass is on the platform mt = mP + mB.

5.62 = 2p

mP + mB  A 4k

(3)

Solving Eqs. (1) to (3) yields : k = 1.36 N>m  mB = 3.58 kg

Ans.

mP = 0.7589 kg

Ans: k = 1.36 N>m mB = 3.58 kg 1204

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*22–16. A block of mass m is suspended from two springs having a stiffness of k1 and k2, arranged a) parallel to each other, and b) as a series. Determine the equivalent stiffness of a single spring with the same oscillation characteristics and the period of oscillation for each case.

k1

k2

k1

k2

SOLUTION (a) When the springs are arranged in parallel, the equivalent spring stiffness is Ans.

keq = k1 + k2 The natural frequency of the system is vn =

keq

Cm

=

(a)

k1 + k2 B m

(b)

Thus, the period of oscillation of the system is t =

2p = vn

2p k1 + k2 B m

= 2p

m

Ans.

C k1 + k2

(b) When the springs are arranged in a series, the equivalent stiffness of the system can be determined by equating the stretch of both spring systems subjected to the same load F. F F F + = k1 k2 keq 1 1 1 + = k1 k2 keq k2 + k1 1 = k1k2 keq keq =

k1k2 k1 + k2

Ans.

The natural frequency of the system is k1k2 b k2 + k1 vn = = m Cm S keq

a

Thus, the period of oscillation of the system is t =

2p = vn

2p k1k2 b a k2 + k1 m S

= 2p

C

m(k1 + k2) k1k2

Ans. Ans: keq = k1 + k2 m t = 2p A k1 + k2 k1k2 keq = k1 + k2 m(k1 + k2) t = 2p A k1k2 1205

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22–17. The 15-kg block is suspended from two springs having a different stiffness and arranged a) parallel to each other, and b) as a series. If the natural periods of oscillation of the parallel system and series system are observed to be 0.5 s and 1.5 s, respectively, determine the spring stiffnesses k1 and k2.

k1

k2

k1

k2

SOLUTION The equivalent spring stiffness of the spring system arranged in parallel is A keq B P = k1 + k2 and the equivalent stiffness of the spring system arranged in a series can be determined by equating the stretch of the system to a single equivalent spring when they are subjected to the same load. F F F + = k1 k2 (keq)S

(a)

(b)

k2 + k1 1 = k1k2 A keq B S

A keq B S =

k1k2 k1 + k2

Thus the natural frequencies of the parallel and series spring system are (vn)P =

(vn)S =

A keq B P

D m

A keq B S

D m

=

=

B

k1 + k2 15

¢

k1k2 ≤ k1 + k2

U

15

=

k1k2

D 15 A k1 + k2 B

Thus, the natural periods of oscillation are tP = tS =

2p 15 = 2p = 0.5 (vn)P B k1 + k2

(1)

15 A k1 + k2 B 2p = 2p = 1.5 (vn)S k1k2 D

(2)

Solving Eqs. (1) and (2), k1 = 2067 N>m or 302 N>m

Ans.

k2 = 302 N>m or 2067 N>m

Ans.

Ans: k1 = 2067 N>m k2 = 302 N>m or vice versa 1206

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22–18. The uniform beam is supported at its ends by two springs A and B, each having the same stiffness k. When nothing is supported on the beam, it has a period of vertical vibration of 0.83 s. If a 50-kg mass is placed at its center, the period of vertical vibration is 1.52 s. Compute the stiffness of each spring and the mass of the beam.

A

B k

k

SOLUTION t = 2p

m Ak

m t2 = k (2p)2 (0.83)2 (2p)2 (1.52)2 (2p)2

=

mB 2k

(1)

=

mB + 50 2k

(2)

Eqs. (1) and (2) become mB = 0.03490k mB + 50 = 0.1170k Ans.

mB = 21.2 kg

Ans.

k = 609 N m

Ans: mB = 21.2 kg k = 609 N>m 1207

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22–19. The slender rod has a mass of 0.2 kg and is supported at O by a pin and at its end A by two springs, each having a stiffness k = 4 N>m. The period of vibration of the rod can be set by fixing the 0.5-kg collar C to the rod at an appropriate location along its length. If the springs are originally unstretched when the rod is vertical, determine the position y of the collar so that the natural period of vibration becomes t = 1 s. Neglect the size of the collar.

O

y C

600 mm

SOLUTION

k A k

Moment of inertia about O: IO =

1 (0.2)(0.6)2 + 0.5y2 = 0.024 + 0.5y2 3

Each spring force Fs = kx = 4x. a + ©MO = IO a;

- 2(4x)(0.6 cos u) - 0.2(9.81)(0.3 sin u) $ - 0.5(9.81)(y sin u) = (0.024 + 0.5y2) u $ - 4.8x cos u - (0.5886 + 4.905y) sin u = (0.024 + 0.5y2) u

However, for small displacement x = 0.6u, sin u L u and cos u = 1. Hence $ 3.4686 + 4.905y u+ u = 0 0.024 + 0.5y2 From the above differential equation, p = t = 1 =

3.4686 + 4.905y B 0.024 + 0.5y2

.

2p p 2p

3.4686 + 4.905y B 0.024 + 0.5y2

19.74y 2 - 4.905y - 2.5211 = 0 Ans.

y = 0.503 m = 503 mm

Ans: y = 503 mm 1208

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*22–20. A uniform board is supported on two wheels which rotate in opposite directions at a constant angular speed. If the coefficient of kinetic friction between the wheels and board is m, determine the frequency of vibration of the board if it is displaced slightly, a distance x from the midpoint between the wheels, and released.

x A

B

d

d

SOLUTION Freebody Diagram: When the board is being displaced x to the right, the restoring force is due to the unbalance friction force at A and B C (Ff)B 7 (Ff)A D . Equation of Motion: a + ©MA = ©(MA)k ;

NB (2d) - mg(d + x) = 0 NB =

+ c ©Fy = m(aG)y ;

NA +

mg(d + x) - mg = 0 2d NA =

+ ©F = m(a ) ; : x G x

mc

Kinematics: Since a =

mg(d - x) 2d

mg(d + x) mg(d-x) d - mc d = ma 2d 2d mg x = 0 a + d

(1)

d2x ## = x, then substitute this value into Eq.(1), we have dt2

##

x+ From Eq.(2), vn 2 =

mg(d + x) 2d

mg x = 0 d

(2)

mg mg , thus, vn = . Applying Eq. 22–4, we have d A d f =

vn 1 = 2p 2p

mg d

Ans.

Ans: f = 1209

mg 1 2p A d

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22–21. If the wire AB is subjected to a tension of 20 lb, determine the equation which describes the motion when the 5-lb weight is displaced 2 in. horizontally and released from rest.

A

6 ft

Solution L′ K L

6 ft

+ ΣF = m a ;  - 2T x = mx$ d x x L

B

2T $ x + x = 0 Lm P =

2(20) 2T = = 6.55 rad>s 5 A Lm A 6 ( 32.2 )

x = A sin pt + B cos pt x =

1 1 ft at t = 0,  Thus B = = 0.167 6 6

v = A p cos pt - B p sin pt v = 0 at t = 0,  Thus A = 0 So that Ans.

x = 0.167 cos 6.55t

Ans: x = 0.167 cos 6.55t 1210

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22–22. The bar has a length l and mass m. It is supported at its ends by rollers of negligible mass. If it is given a small displacement and released, determine the natural frequency of vibration.

R A

B

l

SOLUTION Moment of inertia about point O: IO =

1 2 l2 2 1 ml + ma R2 b = ma R2 - l2 b 12 4 6 B

c + ©MO = IOa;

mg a

B

R2 -

l2 1 $ b u = - ma R2 - l2 bu 4 6 1

$ 3g(4R 2 - l2)2 u + u = 0 6R 2 - l2 1

From the above differential equation, vn =

3g(4R2 - l2)2 D

6R2 - l2

.

Ans.

Ans: vn = 1211

3g(4R2 - l 2)1>2

C

6R2 - l 2

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22–23. The 20-kg disk, is pinned at its mass center O and supports the 4-kg block A. If the belt which passes over the disk is not allowed to slip at its contacting surface, determine the natural period of vibration of the system.

O

Solution

k =300 50 N/m mm

k  200 N/m

Equation of Motion. The mass moment of inertia of the disk about its mass 1 1 center  O is I0 = mr 2 = (20) ( 0.32 ) = 0.9 kg # m2. When the disk undergoes a 2 2 small angular displacement u, the spring stretches further by s = ru = 0.3u. Thus,

A

the total stretch is y = yst + 0.3u. Then Fsp = ky = 200(yst + 0.3u). Referring to the FBD and kinetic diagram of the system, Fig. a, a + ΣM0 = Σ(mk)0;  4(9.81)(0.3) - 200(yst + 0.3u)(0.3) = 0.90a + 4[a(0.3)](0.3) (1)

11.772 - 60yst - 18u = 1.26a When the system is in equilibrium, u = 0°. Then a + ΣM0 = 0;

4(9.81)(0.3) - 200(yst)(0.3) = 0 60yst = 11.772

Substitute this result into Eq. (1), we obtain - 18u = 1.26a a + 14.2857u = 0 $ Since a = u , the above equation becomes $ u + 14.2857u = 0 Comparing to that of standard form, vn = 214.2857 = 3.7796 rad>s. Thus, t =

2p 2p = = 1.6623 s = 1.66 s vn 3.7796

Ans.

Ans: t = 1.66 s 1212

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*22–24. The 10-kg disk is pin connected at its mass center. Determine the natural period of vibration of the disk if the springs have sufficient tension in them to prevent the cord from slipping on the disk as it oscillates. Hint: Assume that the initial stretch in each spring is dO.

150 mm

k  80 N/m O k  80 N/m

Solution Equation of Motion. The mass moment of inertia of the disk about its mass center O 1 1 is I0 = Mr 2 = (10) ( 0.152 ) = 0.1125 kg # m2. When the disk undergoes a small 2 2 angular displacement u, the top spring stretches further but the stretch of the spring is being reduced both by s = ru = 0.15u. Thus, (Fsp)t = Kxt = 80(d0 - 0.15u) and (Fsp)b = 80(d0 - 0.15u). Referring to the FBD of the disk, Fig. a, a + ΣM0 = I0a;  - 80(d0 + 0.15u)(0.15) + 80(d0 - 0.15u)(0.15) = 0.1125a - 3.60u = 0.1125a a + 32u = 0 $ Since a = u , this equation becomes $ u + 32u = 0 Comparing to that of standard form, vn = 232 rad>s. Then t =

2p 2p = = 1.1107 s = 1.11 s vn 232

Ans.

Ans: t = 1.11 s 1213

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22–25. If the disk in Prob. 22–24 has a mass of 10 kg, determine the natural frequency of vibration. Hint: Assume that the initial stretch in each spring is dO.

150 mm

k  80 N/m O k  80 N/m

Solution Equation of Motion. The mass moment of inertia of the disk about its mass center O 1 1 is I0 = mr 2 = (10) ( 0.152 ) = 0.1125 kg # m2 when the disk undergoes a small 2 2 angular displacement u, the top spring stretches but the bottom spring compresses, both by s = ru = 0.15u. Thus, (Fsp)t = (Fsp)b = ks = 80(0.15u) = 12u. Referring to the FBD of the disk, Fig. a, a + ΣM0 = I0a;  - 12u(0.3) = 0.1125a - 3.60u = 0.1125a a + 32u = 0 $ Since a = u , this equation becomes $ u + 32u = 0 Comparing to that of Standard form, vn = 232 rad>s. Then f =

vn 232 = = 0.9003 Hz = 0.900 Hz 2p 2p

Ans: f = 0.900 Hz 1214

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22–26. A flywheel of mass m, which has a radius of gyration about its center of mass of kO, is suspended from a circular shaft that has a torsional resistance of M = Cu. If the flywheel is given a small angular displacement of u and released, determine the natural period of oscillation.

L

SOLUTION Equation of Motion: The mass moment of inertia of the wheel about point O is IO = mkO 2. Referring to Fig. a, $ a+ ©MO = IO a; -Cu = mkO 2u $ u +

O

u

C u = 0 mkO 2

Comparing this equation to the standard equation, the natural circular frequency of the wheel is vn =

1 C C = kO A m A mkO 2

Thus, the natural period of the oscillation is t =

2p m = 2pkO vn AC

Ans.

Ans: t = 2pkO 1215

m AC

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22–27. k = 5 lb/ft

The 6-lb weight is attached to the rods of negligible mass. Determine the natural frequency of vibration of the weight when it is displaced slightly from the equilibrium position and released.

2 ft

SOLUTION

O

TO is the equilibrium force. TO =

3 ft

6(3) = 9 lb 2

Thus, for small u, c + ©MO = IO a;

6(3) - C 9 + 5(2)u D (2) = a

$ 6 b a3u b(3) 32.2

Thus, $ u + 11.926u = 0 vn = 211.926 = 3.45 rad/s

Ans.

Ans: vn = 3.45 rad>s 1216

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*22–28. The platform AB when empty has a mass of 400 kg, center of mass at G1, and natural period of oscillation t1 = 2.38 s. If a car, having a mass of 1.2 Mg and center of mass at G2, is placed on the platform, the natural period of oscillation becomes t2 = 3.16 s. Determine the moment of inertia of the car about an axis passing through G2.

O 1.83 m

2.50 m

G2 G1 A

B

SOLUTION Free-body Diagram: When an object arbitrary shape having a mass m is pinned at O and being displaced by an angular displacement of u, the tangential component of its weight will create the restoring moment about point O. Equation of Motion: Sum moment about point O to eliminate Ox and Oy. (1)

- mg sin u(l) = IOa

a + ©MO = IOa :

$ d2u = u and sin u = u if u is small, then substituting these 2 dt values into Eq. (1), we have Kinematics: Since a =

$ -mglu = IOu From Eq. (2), v2n = t =

or

mgl $ u + u = 0 IO

(2)

mgl mgl , thus, vn = , Applying Eq. 22–12, we have B IO IO

IO 2p = 2p B mgl vn

(3)

When the platform is empty, t = t1 = 2.38 s, m = 400 kg and l = 2.50 m. Substituting these values into Eq. (3), we have 2.38 = 2p

(IO)p

C 400(9.81)(2.50)

(IO)p = 1407.55 kg # m2

When the car is on the platform, t = t2 = 3.16 s, m = 400 kg + 1200 kg = 1600 kg. 2.50(400) + 1.83(1200) and l = = 1.9975 m IO = (IO)C + (IO)p = (IO)C + 1600 1407.55. Substituting these values into Eq. (3), we have 3.16 = 2p

(IO)C + 1407.55 (I ) = 6522.76 kg # m2 D 1600(9.81)(1.9975) O C

Thus, the mass moment inertia of the car about its mass center is (IG)C = (IO)C - mCd2 = 6522.76 - 1200(1.832) = 2.50(103) kg # m2

Ans.

Ans: (IG)C = 2.50(103) kg # m2 1217

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22–29. The plate of mass m is supported by three symmetrically placed cords of length l as shown. If the plate is given a slight rotation about a vertical axis through its center and released, determine the natural period of oscillation. l

l

Solution

120

$ 1 ΣMz = Iza   -3(T sin f)R = mR2 u 2

l 120

R 120

sin f K f $ 6T f = 0 u + Rm ΣFz = 0  3T cos f - mg = 0 f = 0,  T = $ u +

mg R ,  f = u 3 l

6 mg R a ba u b = 0 Rm 3 l

$ 2g u + u = 0 l t =

l 2p = 2p  vn A 2g

Ans.

Ans: t = 2p 1218

l A 2g

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22–30. Determine the differential equation of motion of the 3-kg block when it is displaced slightly and released. The surface is smooth and the springs are originally unstretched.

k

500 N/m

k

500 N/m

3 kg

SOLUTION T + V = const. T =

1 # (3)x2 2

V =

1 1 (500)x 2 + (500)x2 2 2

# T + V = 1.5x 2 + 500x 2 # $ # 1.5(2x) x + 1000xx = 0 $ 3x + 1000x = 0 $ x + 333x = 0

Ans.

Ans: $ x + 333x = 0 1219

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22–31. Determine the natural period of vibration of the pendulum. Consider the two rods to be slender, each having a weight of 8 lb>ft.

O

2 ft

SOLUTION y =

1(8)(2) + 2(8)(2) = 1.5 ft 8(2) + 8(2)

1 ft

1 ft

1 1 IO = c (2)(8)(2)2 + 2(8)(1)2 d 32.2 12 +

1 1 c (2)(8)(2)2 + 2(8)(2)2 d = 2.8157 slug # ft2 32.2 12

h = y (1 - cos u) T + V = const T =

# # 1 (2.8157)(u)2 = 1.4079 u2 2

V = 8(4)(1.5)(1 - cos u) = 48(1 - cos u) # T + V = 1.4079u2 + 48(1 - cos u) # $ # 1.4079 (2u)u + 48(sin u)u = 0 For small u, sin u = u, then $ u + 17.047u = 0 t =

2p = vn

2p 17.047

Ans.

= 1.52 s

Ans: t = 1.52 s 1220

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*22–32. Determine the natural period of vibration of the 10-lb semicircular disk. 0.5 ft

SOLUTION Datum at initial level of center of gravity of disk. ¢ = r(1 - cos u) E = T + V # 1 I (u)2 + Wr(1 - cos u) 2 IC # # $ E = u(IICu + Wr sin u) = 0 =

For small u,

sin u = u $ Wr u = 0 u + IIC r =

4(0.5) = 0.212 ft 3p

IA = IG + mr2 1 10 10 ( )(0.5)2 = IG + (0.212)2 2 32.2 32.2 IG = 0.02483 slug # ft2 IIC = IG + m(r - r)2 = 0.02483 +

10 (0.5 - 0.212)2 32.2

= 0.05056 slug # ft2 t =

IIC 2p 0.05056 = 2p = 2p A Wr A 10(0.212) vn Ans.

t = 0.970 s

Ans: t = 0.970 s 1221

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22–33. If the 20-kg wheel is displaced a small amount and released, determine the natural period of vibration. The radius of gyration of the wheel is kG = 0.36 m. The wheel rolls without slipping.

k  500 N/m

G 0.5 m

Solution Energy Equation. The mass moment of inertia of the wheel about its mass center is IG = mkG = 20(0.361)2 = 2.592 kg # m2. Since the wheel rolls without slipping, vG = vr = v(0.5). Thus,

T =

1 1 I v2 + mv2G 2G 2

1 1 (2.592)v2 + (20)[v10.52]2 2 2 # 2 = 3.796 v = 3.796u 2 =

When the disk undergoes a small angular displacement u, the spring stretches s = u(1) = u, Fig. a. Thus, the elastic potential energy is Ve = Thus, the total energy is

1 2 1 ks = (500)u 2 = 250u 2 2 2

# E = T + V = 3.796u 2 + 250u 2

Time Derivative. Taking the time derivative of the above equation, #$ # 7.592u u + 500uu = 0 # $ u(7.592u + 500u) = 0 # Since u ≠ 0, then $ 7.592u + 500u = 0 $ u + 65.8588u = 0 Comparing to that of standard form, vn = 265.8588 = 8.1153 rad>s. Thus,

t =

2p 2p = = 0.7742 s = 0.774 s vn 8.1153

Ans.

Ans: t = 0.774 s 1222

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22–34. Determine the differential equation of motion of the 3-kg spool. Assume that it does not slip at the surface of contact as it oscillates. The radius of gyration of the spool about its center of mass is kG = 125 mm.

k  400 N/m

200 mm 100 mm G

SOLUTION Kinematics: Since no slipping occurs, sG = 0.1u hence sF =

0.3 S = 0.3u. Also, 0.1 G

# vG = 0.1u. E = T + V # 1 [(3)(0.125)2]u2 + 2 # = 0.03844u2 + 18u2 #$ # 0.076875uu + 36uu = 0 # $ 0.076875u(u + 468.29u) $ u + 468u = 0 E =

1 1 (3)(0.1u)2 + (400)(0.3u)2 = const. 2 2

= 0 Since 0.076875u Z 0 Ans.

Ans: $ u + 468u = 0 1223

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22–35. Determine the natural period of vibration of the 3-kg sphere. Neglect the mass of the rod and the size of the sphere.

k

500 N/m O

300 mm

300 mm

SOLUTION E = T + V # 1 1 (3)(0.3u)2 + (500)(dst + 0.3u)2 - 3(9.81)(0.3u) 2 2 $ # E = u[(3(0.3)2u + 500(dst + 0.3u)(0.3) - 3(9.81)(0.3)] = 0 =

By statics, T(0.3) = 3(9.81)(0.3) T = 3(9.81) N dst =

3(9.81) 500

Thus, $ 3(0.3)2u + 500(0.3)2u = 0 $ u + 166.67u = 0 vn = 2166.67 = 12.91 rad>s t =

2p 2p = 0.487 s = vn 12.91

Ans.

Ans: t = 0.487 s 1224

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*22–36. If the lower end of the 6-kg slender rod is displaced a small amount and released from rest, determine the natural frequency of vibration. Each spring has a stiffness of k = 200 N>m and is unstretched when the rod is hanging vertically.

O

2m

k

Solution

k 2m

Energy Equation. The mass moment of inertia of the rod about O is 1 1 I0 = ml 2 = (6) ( 42 ) = 32 kg # m2. Thus, the Kinetic energy is 3 3 # # 1 1 T = I0v2 = (32)u 2 = 16u 2 2 2 with reference to the datum set in Fig. a, the gravitational potential energy is Vg = mg y = 6(9.81)( - 2 cos u) = -117.72 cos u When the rod undergoes a small angular displacement u the spring deform x = 2 sin Ω. Thus the elastic potential energy is 1 1 Ve = 2a kx2 b = 2c (200)(2 sin u)2 d = 800 sin2 u 2 2

Thus, the total energy is

# E = T + V = 16u 2 + 800 sin2 u - 117.72 cos u

Time Derivative. Taking the first time derivative of the above equation #$ # # 32u u + 1600(sin u cos u)u + 117.72(sin u)u = 0 Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain #$ # # 32u u + 800(sin 2u)u + 117.72(sin u)u = 0 # $ u(32u + 800 sin 2u + 117.72 sin u) = 0 # Since u ≠ 0, $ 32u + 800 sin 2u + 117.72 sin u) = 0 Since u is small, sin 2u ≃ 2u and sin u = u. The above equation becomes $ 32u + 1717.72u = 0 $ u + 53.67875u = 0 Comparing to that of standard form, vn = 253.67875 = 7.3266 rad>s. Thus,

f =

vn 7.3266 = = 1.1661 Hz = 1.17 Hz 2p 2p

Ans.

Ans: f = 1.17 Hz 1225

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22–37. The disk has a weight of 30 lb and rolls without slipping on the horizontal surface as it oscillates about its equilibrium position. If the disk is displaced, by rolling it counterclockwise 0.2 rad, determine the equation which describes its oscillatory motion and the natural period when it is released.

k  80 lb/ft 0.5 ft

Solution Energy Equation. The mass moment of inertia of the disk about its center of gravity 1 1 30 is IG = mr 2 = a b ( 0.52 ) = 0.11646 slug # ft 2. Since the disk rolls without 2 2 32.2 slipping, vG = vr = v(0.5). Thus

T =



1 1 I v2 + mv2G 2G 2

1 30 1 (0.1146) v2 + a b[v(0.5)]2 2 2 32.2 # = 0.17469 v2 = 0.17469u 2 =

When the disk undergoes a small angular displacement u the spring stretches s = ur = u(0.5), Fig. a. Thus, the elastic potential energy is Ve = Thus, the total energy is



1 2 1 ks = (80)[u(0.5)]2 = 10u 2 2 2

# E = T + V = 0.17469u 2 + 10u 2 # E = 0.175u 2 + 10u 2

Ans.

Time Derivative. Taking the time derivative of the above equation, #$ # 0.34938u u + 20uu = 0 # $ u(0.34938u + 20u) = 0 # Since u ≠ 0, then # 0.34938u + 20u = 0 $ u + 57.244u = 0 $ Ans. u = 57.2u = 0 Comparing to that of standard form, vn = 257.2444 = 7.5660 rad>s. Thus

t =

2p 2p = = 0.8304 s = 0.830 s vn 7.5660

Ans.

Ans: # E = 0.175u 2 + 10 u 2 t = 0.830 s 1226

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22–38. The machine has a mass m and is uniformly supported by four springs, each having a stiffness k. Determine the natural period of vertical vibration. G

SOLUTION

k

T + V = const.

V = mgy + T + V =

d — 2

1 # m(y)2 2

T =

k d — 2

1 (4k)(¢s - y)2 2

1 1 # m(y)2 + m g y + (4k)(¢s - y)2 2 2

# $ # # m y y + m g y - 4k(¢s-y )y = 0 $ m y + m g + 4ky - 4k¢s = 0 Since ¢s =

mg 4k

Then $ my + 4ky = 0 y +

4k y = 0 m

vn = t =

4k Cm

2p m = p Ck vn

Ans.

Ans: t = p 1227

m Ak

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22–39. The slender rod has a weight of 4 lb>ft. If it is supported in the horizontal plane by a ball-and-socket joint at A and a cable at B, determine the natural frequency of vibration when the end B is given a small horizontal displacement and then released.

A

0.75 ft B

SOLUTION

1.5 ft

f =

1.5umax 0.75

¢ = 0.75(1 - cos f)  0.75(1 - 1 + = 0.75( ¢G = Tmax = =

f2 ) 2

4u2max ) 2

1 ¢ = 0.75u2max 2 1 I v2max 2 A 1 1 4(1.5) [ ( )(1.5)2]vn2 u2max 2 3 32.2

= 0.0699 vn2 u2max Vmax = W¢ G = 4(1.5)(0.75u2max) Tmax = Vmax 0.0699vn2 u2max = 4.5 u2max vn 2 = 64.40 vn = 8.025 rad>s f =

vn 8.025 = = 1.28 Hz 2p 2p

Ans.

Ans: f = 1.28 Hz 1228

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*22–40. If the slender rod has a weight of 5 lb, determine the natural frequency of vibration. The springs are originally unstretched.

k = 4 lb/ft

1 ft O

SOLUTION Energy Equation: When the rod is being displaced a small angular displacement of u ,

2 ft

the compression of the spring at its ends can be approximated as x1 = 2u and x2 = 1u. Thus, the elastic potential energy when the rod is at this position is 1 1 1 1 Ve = k1 x21 + k2 x2 = (5)(2u)2 + (4)(1u)2 = 12u2. The datum is set at the 2 2 2 2 rod’s mass center when the rod is at its original position. When the rod undergoes a

k = 5 lb/ft

small angular displacement u, its mass center is 0.5(1 - cos u) ft above the datum hence its gravitational potential energy is Vg = 5[0.5(1 - cos u)]. Since u is small, cos u can be approximated by the first two terms of the power series, that is, u2 u2 cos u = 1 - . Thus, Vg = 2.5c 1 - a 1 - b d = 1.25u2 2 2 V = Ve + Vg = 12u2 + 1.25u2 = 13.25u2 5 1 The mass moment inertia of the rod about point O is IO = a b A 32 B 12 32.2 5 + A 0.52 B = 0.1553 slug # ft2. The kinetic energy is 32.2 T =

# # 1 1 IO v2 = (0.1553) u2 = 0.07764u2 2 2

The total energy of the system is # U = T + V = 0.07764u2 + 13.25u2

[1]

Time Derivative: Taking the time derivative of Eq.[1], we have #$ # 0.1553 uu + 26.5uu = 0 # $ u A 0.1553 u + 26.5u B = 0

#

Since u Z 0, then $ 0.1553u + 26.5u = 0 $ u + 170.66u = 0

[2]

From Eq.[2], p2 = 170.66, thus, p = 13.06 rad/s. Applying Eq.22–14, we have f =

p 13.06 = = 2.08 Hz 2p 2p

Ans.

Ans: f = 2.08 Hz 1229

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22–41. If the block-and-spring model is subjected to the periodic force F = F0 cos vt, show that the differential equation of $ motion is x + (k>m)x = (F0>m) cos vt, where x is measured from the equilibrium position of the block. What is the general solution of this equation?

Equilibrium position x k

m

F

F0 cos vt

SOLUTION + ©F = ma ; : x x

$ F0 cos vt - kx = mx F0 k $ x + x = cos vt m m F0 $ cos vt x + p2x = m

(Q.E.D.) Where p =

k

Cm

(1)

The general solution of the above differential equation is of the form of x = xc + xp. The complementary solution: xc = A sin pt + B cos pt The particular solution: sp = .C cos vt

(2)

$ xP = -Cv2 cos v t

(3)

Substitute Eqs. (2) and (3) into (1) yields: -Cv2 cos v t + p2 (C cos vt) = F0 m = C = 2 p - v2

F0>k

1 - a vp b

F0 cos vt m

2

The general solution is therefore s = A sin pt + B cos pt +

F0>k

v 2 1 - a b p

Ans.

cos vt

The constants A and B can be found from the initial conditions.

Ans: x = A sin vnt + B cos vnt + 1230

Fo >k

1 - (v>vn)2

cos vt

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22–42. A block which has a mass m is suspended from a spring having a stiffness k. If an impressed downward vertical force F = FO acts on the weight, determine the equation which describes the position of the block as a function of time.

Solution $ + c ΣFy = may ;   k(yst + y) - mg - F0 = - my $ my + ky + kyst - mg = F0 However, from equilbrium kyst - mg = 0, therefore $ my + ky = F0 F0 k k $ y + y =   where vn = m m Am F0 $ y + v2ny =  m

[1]

The general solution of the above differential equation is of the form of y = yc + yp. yc = A sin vnt + B cos vnt [2]

yP = C $ yP = 0

[3]

Substitute Eqs. [2] and [3] into [1] yields : 0 + v2nC =

F0 F0 F0   C = = 2 m k mp

The general solution is therefore y = A sin vnt + B cos vnt +

F0  k

Ans.

The constants A and B can be found from the initial conditions.

Ans: y = A sin vnt + B cos vnt + 1231

FO k

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22–43. A 4-lb weight is attached to a spring having a stiffness k = 10 lb>ft. The weight is drawn downward a distance of 4 in. and released from rest. If the support moves with a vertical displacement d = 10.5 sin 4t2 in., where t is in seconds, determine the equation which describes the position of the weight as a function of time.

SOLUTION y = A sin vnt + B cos vnt +

d0 1 - a v0 b v

sin v0t

2

n

# v = y = Avn cos vnt - Bvn sin vnt +

d0v0 1 - av b v0

2

cos v0t

n

The initial condition when t = 0, y = y0 , and v = v0 is y0 = 0 + B + 0 v0 = Avn - 0 +

B = y0 d0 v0 1 - a vn b v0

2

A =

v0 vn

d0v0 vn -

v0 2 vn

Thus, d 0 v0 d0 v0 sin v0t y = q 2 r sin vnt + y0 cos vnt + v v 2 0 vn vn 1 - a v0 b vn n vn =

k 10 = = 8.972 Am A 4>32.2

d0 v0 2

1 - Q vn R

=

0.5>12 1 -

2 4 A 8.972 B

= 0.0520

(0.5>12)4 d0 v 0 v0 = 0 = - 0.0232 v02 42 vn vn 8.972 - 8.972 vn

y = (-0.0232 sin 8.97t + 0.333 cos 8.97t + 0.0520 sin 4t) ft

1232

Ans.

Ans: y = 5 -0.0232 sin 8.97 t + 0.333 cos 8.97 t + 0.0520 sin 4 t6 ft

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*22–44. A 4-kg block is suspended from a spring that has a stiffness of k = 600 N>m. The block is drawn downward 50 mm from the equilibrium position and released from rest when t = 0. If the support moves with an impressed displacement of d = 110 sin 4t2 mm, where t is in seconds, determine the equation that describes the vertical motion of the block. Assume positive displacement is downward.

SOLUTION vn =

k 600 = = 12.25 Am A 4

The general solution is defined by Eq. 22–23 with kd0 substituted for F0.

y = A sin vnt + B cos vnt + ±

d0 v 2 c1 - a b d vn

≤ sin vt

d = (0.01 sin 4t)m, hence d0 = 0.01, v = 4, so that y = A sin 12.25t + B cos 12.25t + 0.0112 sin 4t y = 0.05 when t = 0 0.05 = 0 + B + 0;

B = 0.05 m

# y = A(12.25) cos 12.25t - B(12.25) sin 12.25t + 0.0112(4) cos 4t v = y = 0 when t = 0 0 = A(12.25) - 0 + 0.0112(4);

A = - 0.00366 m

Expressing the result in mm, we have Ans.

y = ( - 3.66 sin 12.25t + 50 cos 12.25t + 11.2 sin 4t) mm

Ans: y = ( - 3.66 sin 12.25t + 50 cos 12.25t + 11.2 sin 4t) mm 1233

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22–45. Use a block-and-spring model like that shown in Fig. 22–14a, but suspended from a vertical position and subjected to a periodic support displacement d = d0 sin v0t, determine the equation of motion for the system, and obtain its general solution. Define the displacement y measured from the static equilibrium position of the block when t = 0.

SOLUTION $ k(y - d0 sin v0t + yst) - mg = -my

+ c ©Fx = max;

$ my + ky + kyst - mg = kd0 sin v0t However, from equilibrium kyst - mg = 0, therefore kd0 k $ y = sin v t y + m m

where vn =

k Am

kd0 $ sin v t y + vn 2 y = m

Ans. (1)

The general solution of the above differential equation is of the form of y = yc + yp, where yc = A sin vnt + B cos vnt yp = C sin v0t

(2)

$ yp = -Cv02 sin v0t

(3)

Substitute Eqs. (2) and (3) into (1) yields: -Cv2 sin v0 t + vn2(C sin v0 t) = kd0 m = C = vn 2 - v 0 2

kd0 sin v0t m

d0 v0 2 1 - a b vn

The general solution is therefore y = A sin vnt + B cos vnt +

d0 sin v t v0 2 1 - a b vn

Ans.

The constants A and B can be found from the initial conditions.

Ans: y = A sin vnt + B cos vnt + 1234

d0 1 - (v>vn)2

sin vt

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22–46. A 5-kg block is suspended from a spring having a stiffness of 300 N>m. If the block is acted upon by a vertical force F = 17 sin 8t2 N, where t is in seconds, determine the equation which describes the motion of the block when it is pulled down 100 mm from the equilibrium position and released from rest at t = 0. Assume that positive displacement is downward.

k

300 N/m

SOLUTION The general solution is defined by: F0 k y = A sin vnt + B cos vnt + § ¥ sin v0t v0 2 1 - a b vn

F

7 sin 8t

Since F = 7 sin 8t, vn =

F0 = 7 N,

v0 = 8 rad>s,

k = 300 N>m

k 300 = = 7.746 rad>s Am A 5

Thus, 7 300 ¥ sin 8t y = A sin 7.746t + B cos 7.746t + § 2 8 1 - a b 7.746 y = 0.1 m when t = 0, 0.1 = 0 + B - 0;

B = 0.1 m

# y = A(7.746) cos 7.746t - B(7.746) sin 7.746t - (0.35)(8) cos 8t # y = y = 0 when t = 0, # y = A(7.746) - 2.8 = 0;

A = 0.361

Expressing the results in mm, we have Ans.

y = (361 sin 7.75t + 100 cos 7.75t - 350 sin 8t) mm

Ans: y = (361 sin 7.75t + 100 cos 7.75t) -350 sin 8t) mm 1235

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22–47. The uniform rod has a mass of m. If it is acted upon by a periodic force of F = F0 sin vt, determine the amplitude of the steady-state vibration.

A L 2 k

L 2

SOLUTION Equation of Motion: When the rod rotates through a small angle u, the springs L u. Thus, the force in each spring is compress and stretch s = r AGu = 2 kL Fsp = ks = u. The mass moment of inertia of the rod about point A is 2 1 IA = mL2. Referring to the free-body diagram of the rod shown in Fig. a, 3 + ©MA = IAa;

k

FO sin vt cos u(L) - mg sin u a =

F  F0 sin vt

kL L L b - 2a u bcos u a b 2 2 2

# 1 mL2u 3

Since u is small, sin u  0 and cos u  1. Thus, this equation becomes $ 1 1 mLu + (mg + kL)u = FO sin vt 3 2 $ 3FO 3 g k u + ¢ + ≤u = sin vt m 2 L mL

(1)

The particular solution of this differential equation is assumed to be in the form of (2)

up = C sin vt Taking the time derivative of Eq. (2) twice, $ up = -Cv2 sin vt

(3)

Substituting Eqs. (2) and (3) into Eq. (1), -Cv2 sin vt + CB

3FO 3 g k a + ≤ (C sin vt) = sin vt m 2 L mL

3FO 3 g k a + ≤ - v2 R sin vt = sin vt m 2 L mL

C =

C =

3FO > mL 3 g k a + b - v2 m 2 L 3FO 3 (mg + Lk) - mLv2 2

Ans.

Ans: C = 3

3FO

2 2 (mg + Lk) - mLv

1236

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*22–48. The 30-lb block is attached to two springs having a stiffness of 10 lb>ft. A periodic force F = (8 cos 3t) lb, where t is in seconds, is applied to the block. Determine the maximum speed of the block after frictional forces cause the free vibrations to dampen out.

k  10 lb/ft F  8 cos 3t k  10 lb/ft

SOLUTION Free-body Diagram: When the block is being displaced by amount x to the right, the restoring force that develops in both springs is Fsp = kx = 10x. Equation of Motion: + ©F = 0; : x

- 2(10x) + 8 cos 3t =

30 a 32.2 [1]

a + 21.47x = 8.587 cos 3t Kinematics: Since a =

d2x $ = x, then substituting this value into Eq. [1], we have dt2 $ x + 21.47x = 8.587 cos 3t

[2]

Since the friction will eventually dampen out the free vibration, we are only interested in the particular solution of the above differential equation which is in the form of xp = C cos 3t Taking second time derivative and substituting into Eq. [2], we have - 9C cos 3t + 21.47C cos 3t = 8.587 cos 3t C = 0.6888 ft Thus, [3]

xp = 0.6888 cos 3t Taking the time derivative of Eq. [3], we have # yp = xp = - 2.0663 sin 3t Thus,

A yp B max = 2.07 ft>s

Ans.

Ans: (vp)max = 2.07 ft>s 1237

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22–49. The light elastic rod supports a 4-kg sphere. When an 18-N vertical force is applied to the sphere, the rod deflects 14 mm. If the wall oscillates with harmonic frequency of 2 Hz and has an amplitude of 15 mm, determine the amplitude of vibration for the sphere.

0.75 m

SOLUTION k =

18 F = = 1285.71 N>m ¢y 0.014

v0 = 2 Hz = 2(2p) = 12.57 rad>s d0 = 0.015 m vn =

k 1285.71 = = 17.93 4 Am A

Using Eq. 22–22, the amplitude is (xp)max = 3

d0 0.015 3 = 3 3 12.57 2 v0 2 1 Q 17.93 R 1 - a b vn Ans.

(xp)max = 0.0295 m = 29.5 mm

Ans: (xp)max = 29.5 mm 1238

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22–50. Find the differential equation for small oscillations in terms of u for the uniform rod of mass m. Also show that if c 6 2mk>2, then the system remains underdamped. The rod is in a horizontal position when it is in equilibrium.

2a

a B

A k

C u c

SOLUTION # Equation of Motion: When the rod is in equilibrium, u = 0°, Fc = cyc = 0 and $ u = 0. writing the moment equation of motion about point B by referring to the free-body diagram of the rod, Fig. a, a - FA (a) - mg a b = 0 2

+ ©MB = 0;

FA =

mg 2

mg FA . When the rod rotates about = k 2k point B through a small angle u, the spring stretches further by s1 = au. Thus, the Thus, the initial stretch of the spring is sO =

force in the spring is FA = k(s0 + s1) = k ¢

mg + au ≤ . Also, the velocity of end C 2k # # # # of the rod is vc = yc = 2au. Thus, Fc = cyc = c(2au). The mass moment of inertia of 1 a 2 the rod about B is IB = m(3a)2 + m a b = ma2. Again, referring to Fig. a and 12 2 writing the moment equation of motion about B, ka

©MB = IB a;

# mg a + au b cos u(a) + A 2au B cos u(2a) - mg cos ua b 2k 2

$ = - ma2u $ # 4c k u + cos uu + (cos u)u = 0 m m Since u is small, cos u  1. Thus, this equation becomes $ 4c # k u + u + u = 0 m m

Ans.

Comparing this equation to that of the standard form, vn =

k Am

ceq = 4c

Thus, k = 22mk Am

cc = 2mvn = 2m

For the system to be underdamped, ceq 6 cc 4c 6 22mk c 6

1 2mk 2

Ans.

1239

Ans: $ 4c # k # u + u + u = 0 m m

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22–51. The 40-kg block is attached to a spring having a stiffness of 800 N>m. A force F = (100 cos 2t) N, where t is in seconds is applied to the block. Determine the maximum speed of the block for the steady-state vibration.

k  800 N/m

Solution For the steady-state vibration, the displacement is

yp =

F0 >k

1 - (v0 >vn)2

cos vt

F  (100 cos 2t) N

Here F0 = 100 N, k = 800 N>m, v0 = 2 rad>s and vn =

k 800 = = 220 rad>s. Am A 40

Thus 100>800



yp =



yP = 0.15625 cos 2t

1 - ( 2> 120 ) 2

cos 2t

Taking the time derivative of this equation # vp = yp = -0.3125 sin 2t

(2)

vp is maximum when sin 2t = 1. Thus

Ans.

(vp)max = 0.3125 m>s

Ans: (vp)max = 0.3125 m>s 1240

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*22–52. Use a block-and-spring model like that shown in Fig. 22–14a but suspended from a vertical position and subjected to a periodic support displacement of d = d0 cos v0t, determine the equation of motion for the system, and obtain its general solution. Define the displacement y measured from the static equilibrium position of the block when t = 0.

SOLUTION $ + T ©Fy = may; kd0 cos v0t + W - kdst - ky = my Since W = kdst, kd0 k $ y + y = cos v0t m m

(1)

yC = A sin vny + B cos vny (General sol.) yP = C cos v0t (Particular sol.) Substitute yp into Eq. (1) C(- v02 +

C =

kd0 k ) cos v0t = cos v0t m m

kd0 m Q

k 2 m - v0 R

Thus, y = yC + yP y = A sin vnt + B cos vnt +

Q

kd0 m

k - v02 R m

Ans.

cos v0t

Ans: y = A sin vnt + B cos vnt +

1241

kd0 m k a - v02b m

cos v0t

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22–53. The fan has a mass of 25 kg and is fixed to the end of a horizontal beam that has a negligible mass. The fan blade is mounted eccentrically on the shaft such that it is equivalent to an unbalanced 3.5-kg mass located 100 mm from the axis of rotation. If the static deflection of the beam is 50 mm as a result of the weight of the fan, determine the angular velocity of the fan blade at which resonance will occur. Hint: See the first part of Example 22.8.

V

SOLUTION k = vn =

25(9.81) F = = 4905 N>m ¢y 0.05 k

Cm

=

4905

C 25

= 14.01 rad>s

Resonance occurs when Ans.

v = vn = 14.0 rad>s

Ans: v = 14.0 rad>s 1242

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22–54. In Prob. 22–53, determine the amplitude of steady-state vibration of the fan if its angular velocity is 10 rad>s .

V

SOLUTION k = vn =

25(9.81) F = = 4905 N>m ¢y 0.05 k

Cm

=

4905

C 25

= 14.01 rad>s

The force caused by the unbalanced rotor is F0 = mr v2 = 3.5(0.1)(10)2 = 35 N Using Eq. 22–22, the amplitude is

(xp)max

F0 k 4 = 4 v 2 1 -a b p

(xp)max

35 4905 4 = 0.0146 m = 4 10 2 1 - a b 14.01 Ans.

(xp)max = 14.6 mm

Ans: (xp)max = 14.6 mm 1243

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22–55. What will be the amplitude of steady-state vibration of the fan in Prob. 22–53 if the angular velocity of the fan blade is 18 rad>s? Hint: See the first part of Example 22.8.

V

SOLUTION k = vn =

25(9.81) F = = 4905 N>m ¢y 0.05 k

Cm

=

4905

= 14.01 rad>s

C 25

The force caused by the unbalanced rotor is F0 = mrv2 = 3.5(0.1)(18)2 = 113.4 N Using Eq. 22–22, the amplitude is (xp)max = 4

(xp)max

F0 k v 2 1 - a b p

4

113.4 4905 4 = 0.0355 m = 4 18 2 b 1 - a 14.01 Ans.

(xp)max = 35.5 mm

Ans: (xp)max = 35.5 mm 1244

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*22–56. The small block at A has a mass of 4 kg and is mounted on the bent rod having negligible mass. If the rotor at B causes a harmonic movement dB = (0.1 cos 15t) m, where t is in seconds, determine the steady-state amplitude of vibration of the block.

0.6 m

A

O

1.2 m

SOLUTION $ 4(9.81)(0.6) - Fs (1.2) = 4(0.6)2u

+ ©MO = IO a;

V k

Fs = kx = 15(x + xst - 0.1 cos 15t) xst =

4(9.81)(0.6) 1.2(15)

15 N/m

B

Thus, $ -15(x - 0.1 cos 15t)(1.2) = 4(0.6)2u x = 1.2u u + 15u = 1.25 cos 15t Set xp = C cos 15t -C(15)2 cos 15t + 15(C cos 15t) = 1.25 cos 15t C =

1.25 = -0.00595 m 15 - (15)2

umax = C = 0.00595 rad Ans.

ymax = (0.6 m)(0.00595 rad) = 0.00357 rad

Ans: ymax = 0.00357 rad 1245

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22–57. The electric motor turns an eccentric flywheel which is equivalent to an unbalanced 0.25-lb weight located 10 in. from the axis of rotation. If the static deflection of the beam is 1 in. due to the weight of the motor, determine the angular velocity of the flywheel at which resonance will occur. The motor weights 150 lb. Neglect the mass of the beam.

V

SOLUTION k =

F 150 = = 1800 lb>ft d 1>12

Resonance occurs when

vn =

k 1800 = = 19.66 Am A 150>32.2 Ans.

v = vn = 19.7 rad>s

Ans: v = 19.7 rad>s 1246

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22–58. What will be the amplitude of steady-state vibration of the motor in Prob. 22–57 if the angular velocity of the flywheel is 20 rad>s?

V

SOLUTION The constant value FO of the periodic force is due to the centrifugal force of the unbalanced mass. FO = ma n = mrv2 = a

0.25 10 b a b (20)2 = 2.588 lb 32.2 12

Hence F = 2.588 sin 20t k =

150 F = = 1800 lb>ft d 1>12

vn =

k 1800 = = 19.657 Am A 150>32.2

From Eq. 22–21, the amplitude of the steady state motion is C = 4

F0>k 2.588>1800 4 = 4 4 = 0.04085 ft = 0.490 in. 2 v0 2 20 b 1 - a b 1 - a vn 19.657

Ans.

Ans: C = 0.490 in. 1247

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22–59. Determine the angular velocity of the flywheel in Prob. 22–57 which will produce an amplitude of vibration of 0.25 in.

V

SOLUTION The constant value FO of the periodic force is due to the centrifugal force of the unbalanced mass. FO = man = mrv2 = a

10 0.25 b a bv2 = 0.006470v2 32.2 12

F = 0.006470v2 sin vt k =

150 F = = 1800 lb>ft d 1>12

vn =

k 1800 = = 19.657 Am A 150>32.2

From Eq. 22.21, the amplitude of the steady-state motion is

C = 4

F0>k 4 v 2 1 - a b vn 0.006470 a

v2 b 1800 0.25 4 4 = 2 v 12 b 1 - a 19.657 Ans.

v = 19.0 rad>s

Ans: v = 19.0 rad>s 1248

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*22–60. The 450-kg trailer is pulled with a constant speed over the surface of a bumpy road, which may be approximated by a cosine curve having an amplitude of 50 mm and wave length of 4 m. If the two springs s which support the trailer each have a stiffness of 800 N>m, determine the speed v which will cause the greatest vibration (resonance) of the trailer. Neglect the weight of the wheels.

v s

2m

100 mm

2m

SOLUTION The amplitude is d0 = 50 mm = 0.05 m The wave length is l = 4 m k = 2(800) = 1600 N>m vn = t =

k 1600 = = 1.89 rad>s Am A 450

2p 2p = = 3.33 s vn 1.89

For maximum vibration of the trailer, resonance must occur, i.e., v0 = vn Thus, the trailer must travel l = 4 m, in t = 3.33 s, so that vR =

l 4 = = 1.20 m.s t 3.33

Ans.

Ans: vR = 1.20 m.s 1249

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22–61. Determine the amplitude of vibration of the trailer in Prob. 22–60 if the speed v = 15 km>h.

v s

2m

100 mm

2m

SOLUTION v = 15 km>h =

15(1000) m>s = 4.17 m>s 3600

d0 = 0.05 m As shown in Prob. 22–50, the velocity is inversely proportional to the period. 1 Since = f the the velocity is proportional of f, vn and v0 t Hence, the amplitude of motion is (xp)max = `

1 -

(xp)max = `

2 1 - (4.17 1.20 )

d0

A B

v0 2 vn

` = `

0.05

d0 1 -

A vvR B 2

`

` = 0.00453 m Ans.

(xp)max = 4.53 mm

Ans: (xp)max = 4.53 mm 1250

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22–62. The motor of mass M is supported by a simply supported beam of negligible mass. If block A of mass m is clipped onto the rotor, which is turning at constant angular velocity of v, determine the amplitude of the steady-state vibration. Hint: When the beam is subjected to a concentrated force of P at its mid-span, it deflects d = PL3>48EI at this point. Here E is Young’s modulus of elasticity, a property of the material, and I is the moment of inertia of the beam’s crosssectional area.

r A

L 2

L 2

SOLUTION In this case, P = keqd. Then, keq = frequency of the system is

48EI P P = = . Thus, the natural 3 d PL >48EI L3

48EI 48EI L3 vn = = = Cm R M B ML3 keq

Here, FO = man = m(v2r). Thus, Y =

FO>keq 1 - a

v 2 b vn

m(v2r) 48EI>L3

Y = 1 Y =

v2 48EI>ML3

mrv2L3 48EI - Mv2L3

Ans.

Ans: Y = 1251

mrv2L3 48EI - Mv2L3

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22–63. The spring system is connected to a crosshead that oscillates vertically when the wheel rotates with a constant angular velocity of V . If the amplitude of the steady-state vibration is observed to be 400 mm, and the springs each have a stiffness of k = 2500 N>m, determine the two possible values of V at which the wheel must rotate. The block has a mass of 50 kg.

v

SOLUTION

k

200 mm

k

In this case, keq = 2k = 2(2500) = 5000 N>m Thus, the natural circular frequency of the system is vn =

keq Dm

5000 = 10 rad>s A 50

=

Here, dO = 0.2 m and (YP)max = ; 0.4 m, so that (YP)max =

;0.4 =

dO 1 - ¢

v 2 ≤ vn

0.2 1 - ¢

v 2 ≤ 10

v2 = 1 ; 0.5 100 Thus, v2 = 1.5 100

v = 12.2 rad>s

Ans.

v2 = 0.5 100

v = 7.07 rad>s

Ans.

or

Ans: v = 12.2 rad>s v = 7.07 rad>s 1252

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*22–64. The spring system is connected to a crosshead that oscillates vertically when the wheel rotates with a constant angular velocity of v = 5 rad>s. If the amplitude of the steady-state vibration is observed to be 400 mm, determine the two possible values of the stiffness k of the springs. The block has a mass of 50 kg.

v

SOLUTION

k

200 mm

k

In this case, keq = 2k Thus, the natural circular frequency of the system is vn =

keq Dm

=

2k = 20.04k A 50

Here, dO = 0.2 m and (YP)max = ; 0.4 m, so that (YP)max =

;0.4 =

dO 1 - ¢

v 2 ≤ vn

0.2 1 - ¢

5 20.04k

≤

2

625 = 1 ; 0.5 k Thus, 625 = 1.5 k

k = 417 N>m

Ans.

625 = 0.5 k

k = 1250 N>m

Ans.

or

Ans: k = 417 N>m k = 1250 N>m 1253

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22–65. A 7-lb block is suspended from a spring having a stiffness of k = 75 lb>ft. The support to which the spring is attached is given simple harmonic motion which may be expressed as d = (0.15 sin 2t) ft, where t is in seconds. If the damping factor is c>cc = 0.8, determine the phase angle f of forced vibration.

SOLUTION vn =

k

Cm

=

75 = 18.57 7 b a S 32.2

d = 0.15 sin 2t d0 = 0.15, v = 2 f¿ = tan-1 §

2a

c v ba b cc vn

v 2 1 - a b vn

¥ = tan-1 §

2(0.8)a 1 - a

2 b 18.57

2 2 b 18.57

¥ Ans.

f¿ = 9.89°

Ans: f′ = 9.89° 1254

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22–66. Determine the magnification factor of the block, spring, and dashpot combination in Prob. 22–65.

SOLUTION vn =

k = Am

75

= 18.57

7 Q ¢ 32.2 ≤

d = 0.15 sin 2t d0 = 0.15,

v = 2 1

MF =

C

B1 - a

v v c b R + c2a b a b d cn vn vn 2 2

2

1

=

C

B1 - a

2 2 2 b R + c2(0.8) a bR 18.57 18.57 2 2

Ans.

MF = 0.997

Ans: MF = 0.997 1255

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22–67. A block having a mass of 7 kg is suspended from a spring that has a stiffness k = 600 N>m. If the block is given an upward velocity of 0.6 m>s from its equilibrium position at t = 0, determine its position as a function of time. Assume that positive displacement of the block is downward and that motion takes place in a medium which furnishes a damping force F = 150 ƒ v ƒ 2 N, where v is in m>s.

SOLUTION c = 50 N s>m vn =

k = 600 N>m

m = 7 kg

k 600 = = 9.258 rad>s Am A 7

cc = 2mvn = 2(7)(9.258) = 129.6 N # s>m Since c 6 cz, the system is underdamped, v d = vn

B

1 - a

c 2 50 2 b = 8.542 rad>s b = 9.258 1 - a cc 129.6 B 50 c = = 3.751 2m 2(7)

From Eq. 22-32 y = D C e- A2m Bt sin (vdt + f) S c

c # v = y = D C e- A2m Bt vd cos (vdt + f) +

A-

y = De- A2m Bt C vd cos (vdt + f) c

c c B e- A2m Bt sin (vdt + f) S 2m

c sin (vdt + f) D 2m

Applying the initial condition at t = 0, y = 0 and y = - 0.6 m>s. 0 = D[e - 0 sin (0 + f)] sin f = 0

since D Z 0 f = 0°

- 0.6 = De - 0 [8.542 cos 0° - 0] D = - 0.0702 m y = [ -0.0702 e - 3.57t sin (8.540)] m

Ans.

1256

Ans: y = 5 - 0.0702 e -3.57t sin (8.540)6 m

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*22–68. The 200-lb electric motor is fastened to the midpoint of the simply supported beam. It is found that the beam deflects 2 in. when the motor is not running. The motor turns an eccentric flywheel which is equivalent to an unbalanced weight of 1 lb located 5 in. from the axis of rotation. If the motor is turning at 100 rpm, determine the amplitude of steady-state vibration. The damping factor is c cc 0.20. Neglect the mass of the beam.

SOLUTION d =

2 = 0.167 ft 12

v = 100a k =

2p b = 10.47 rad>s 60

200 = 1200 lb>ft 2 12

FO = mrv2 = a p =

k = Am

5 1 b a b (10.47)2 = 1.419 lb 32.2 12

1200 = 13.90 rad>s 200 Q 32.2 F0 k

C¿ =

C

B1 - ¢ ≤ R + B2¢ ≤ ¢ ≤ R v p

2

2

c cc

v p

2

1.419 1200

=

C

1 -

10.47 13.90

2

2

+

2(0.20)

10.47 13.90

2

= 0.00224 ft C

Ans.

0.0269 in.

Ans: C ′ = 0.0269 in. 1257

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22–69. Two identical dashpots are arranged parallel to each other, as shown. Show that if the damping coefficient c 6 2mk, then the block of mass m will vibrate as an underdamped system.

c

c

k

SOLUTION When the two dash pots are arranged in parallel, the piston of the dashpots have the same velocity. Thus, the force produced is # # # F = cy + cy = 2cy The equivalent damping coefficient ceq of a single dashpot is # 2cy F ceq = # = # = 2c y y For the vibration to occur (underdamped system), ceq 6 cc. However, cc = 2mvn k . Thus, = 2m Am ceq 6 cc k Am

2c 6 2m

c 6 2mk

Ans.

Ans: # F = 2cy k Am c 6 1mk cc = 2m

1258

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22–70. The damping factor, c>cc , may be determined experimentally by measuring the successive amplitudes of vibrating motion of a system. If two of these maximum displacements can be approximated by x1 and x2 , as shown in Fig. 22–16, show that ln x1> x2 = 2p1c>cc2> 21 - 1c> cc22. The quantity ln x1> x2 is called the logarithmic decrement.

SOLUTION Using Eq. 22–32, x = Dc e - A2m B t sin (vdt + f) d c

The maximum displacement is xmax = De - A2m B t c

At t = t1, and t = t2 x1 = De - A2m B t1 c

x2 = De - A2m B t2 c

Hence, x1 De - A2m Bt1 = c c x2 De - A2m Bt2 = e - A2m B(t1 - t2) c

Since vdt2 - vdt1 = 2p then t2 - t1 = so that ln a

2p vd

x1 cp b = x2 mvd

Using Eq. 22–33, cc = 2mvn vd = vn

B

1 - a

cc c 2 c 2 b = 1 - a b cc c 2mA r

So that, x1 ln a b = x2

2p a

c b cc

Q.E.D.

c 2 1- a b cc B

Ans: x1 In a b = x2 1259

2pa A

c b cc

1 - a

c 2 b cc

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22–71. If the amplitude of the 50-lb cylinder’s steady-state vibration is 6 in., determine the wheel’s angular velocity v.

v

9 in.

c

k

200 lb/ft

25 lb s/ft

k

200 lb/ft

SOLUTION In this case, Y = Then

6 9 = 0.5 ft, dO = = 0.75 ft, and keq = 2k = 2(200) = 400 lb>ft. 12 12

vn =

keq

Cm

=

400 = 16.05 rad>s B (50>32.2)

cc = 2mvn = 2 ¢

50 ≤ (16.05) = 49.84 lb # s>ft 32.2

c 25 = 0.5016 = cc 49.84 dO

Y = D

B1 - ¢

2

2(c>cc)v 2 v 2 ≤ R + ¢ ≤ vn vn 0.75

0.5 = D

B1 - ¢

2

2 2(0.5016)v 2 v ≤ R + ¢ ≤ 16.05 16.05

15.07(10 - 6)v4 - 3.858(10 - 3)v2 - 1.25 = 0 Solving for the positive root of this equation, v2 = 443.16 Ans.

v = 21.1 rad>s

Ans: v = 21.1 rad>s 1260

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*22–72. The block, having a weight of 12 lb, is immersed in a liquid such that the damping force acting on the block has a magnitude of F = 10.7 ƒ v ƒ 2 lb, where v is in ft>s. If the block is pulled down 0.62 ft and released from rest, determine the position of the block as a function of time. The spring has a stiffness of k = 53 lb>ft. Assume that positive displacement is downward.

k

SOLUTION c = 0.7 lb # s>ft vn =

k = 53 lb>ft

m =

12 = 0.3727 slug 32.2

k 53 = = 11.925 rad>s Am A 0.3727

cc = 2mvn = 2(0.3727)(11.925) = 8.889 lb # s>ft Since c 6 cc the system is underdamped. vd = vn

B

1 - a

c 2 0.7 2 b = 11.925 1 - a b = 11.888 rad>s cc B 8.889

c 0.7 = = 0.9392 2m 2(0.3727) From Eq. 22–32 y = D c e - A2m B t sin (vdt + f) d c

c c c # b e- A2m B t sin (vdt + f) d y = y = D c e- A2m B tvd cos (vdt + f) + a 2m

y = De- A2m Bt c vd cos (vdt + f) c

c sin (vdt + f) d 2m

Appling the initial condition at t = 0, y = 0.62 ft and y = 0. 0.62 = D C e-0 sin (0 + f) D (1)

D sin f = 0.62 0 = De -0 C 11.888 cos (0 + f) - 0.9392 sin (0 + f) D

since D Z 0 (2)

11.888 cos f - 0.9392 sin u = 0 Solving Eqs. (1) and (2) yields: f = 85.5° = 1.49 rad

D = 0.622 ft

y = 0.622[e-0.939t sin (11.9t + 1.49)]

Ans.

Ans: y = 0.622 [e - 0.939t sin (11.9t + 1.49)] 1261

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22–73. The bar has a weight of 6 lb. If the stiffness of the spring is k = 8 lb>ft and the dashpot has a damping coefficient c = 60 lb # s>ft, determine the differential equation which describes the motion in terms of the angle u of the bar’s rotation. Also, what should be the damping coefficient of the dashpot if the bar is to be critically damped?

A

B

C c

k 2 ft

3 ft

SOLUTION $ 1 6 # 6(2.5) - (60y2)(3) - 8(y1 + yst)(5) = c a b(5)2 du 3 32.2 $ # 1.5528u + 180y2 + 40y1 + 40yst - 15 = 0 [1]

a + ©MA = IA a;

From equilibrium 40yst - 15 = 0. Also, for small u, y1 = 5u and y2 = 3u hence # y2 = 3u. From Eq. [1]

$ # 1.5528u + 180(3u) + 40(5u) = 0 # # 1.55u + 540u + 200u = 0

Ans.

By comparing the above differential equation to Eq. 22-27 m = 1.55

vn =

k = 200 £

A cd # p B c =

9 A cd # p B c 2m

200 = 11.35 rad>s A 1.55

c = 9cd # p

2

≥ -

k = 0 m

2 2 2km = 2200(1.55) = 3.92 lb # s>ft 9 9

Ans.

Ans:$ # 1.55u + 540u + 200u = 0 (cdp)c = 3.92 lb # s>ft 1262

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22–74. A bullet of mass m has a velocity of v0 just before it strikes the target of mass M. If the bullet embeds in the target, and the vibration is to be critically damped, determine the dashpot’s critical damping coefficient, and the springs’ maximum compression. The target is free to move along the two horizontal guides that are “nested” in the springs.

k v0 c

k

SOLUTION Since the springs are arranged in parallel, the equivalent stiffness of the single spring system is keq = 2k. Also, when the bullet becomes embedded in the target, mT = m + M. Thus, the natural frequency of the system is vn =

keq

=

B mT

2k Bm + M

When the system is critically damped c = cc = 2mTvn = 2(m + M)

2k = 28 (m + M)k Bm + M

Ans.

The equation that describes the critically dampened system is x = (A + Bt)e-vn t When t = 0, x = 0. Thus, A = 0 Then, x = Bte-vn t

(1)

Taking the time derivative, # v = x = Be - vn t - Bvn te-vn t v = Be-vn t(1 - vn t)

(2)

Since linear momentum is conserved along the horizontal during the impact, then + B A;

mv0 = (m + M)v v = a

m bv0 m + M

Here, when t = 0, v = a B = a

m b v . Thus, Eq. (2) gives m + M 0

m b v0 m + M

And Eqs. (1) and (2) become x = ca

m bv0 d te-vn t m + M

(3)

v = ca

m bv0 d e-vn t(1 - vn t) m + M

(4) 1263

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22–74. Continued

The maximum compression of the spring occurs when the block stops. Thus, Eq. (4) gives 0 = ca Since a

m b v d (1 - vn t) m + M 0

m b v Z 0, then m + M 0 1 - vn t = 0 t =

1 m + M = vn B 2k

Substituting this result into Eq. (3) xmax = c a = c

m m + M -1 bv d a be m + M 0 A 2k

m 1 dv e A 2k(m + M) 0

Ans.

1264

Ans: cc = 28(m + M)k m 1 xmax = c dv e A 2k(m + M) 0

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22–75. A bullet of mass m has a velocity v0 just before it strikes the target of mass M. If the bullet embeds in the target, and the dashpot’s damping coefficient is 0 6 c V cc, determine the springs’ maximum compression. The target is free to move along the two horizontal guides that are“nested” in the springs.

k v0 c

k

SOLUTION Since the springs are arranged in parallel, the equivalent stiffness of the single spring system is keq = 2k. Also, when the bullet becomes embedded in the target, mT = m + M. Thus, the natural circular frequency of the system vn =

keq

C mT

=

2k Bm + M

The equation that describes the underdamped system is x = Ce - (c>2mT)t sin (vdt + f)

(1)

When t = 0, x = 0. Thus, Eq. (1) gives 0 = C sin f Since C Z 0, sin f = 0. Then f = 0. Thus, Eq. (1) becomes x = Ce - (c>2mT)t sin vd t

(2)

Taking the time derivative of Eq. (2), # v = x = C B vde - (c>2mT)t cos vdt v = Ce - (c>2mT)t B vd cos vdt -

c - (c>2mT)t e sin vdt R 2mT

c sin vdt R 2mT

(3)

Since linear momentum is conserved along the horizontal during the impact, then + B A;

mv0 = (m + M)v v = a

When t = 0, v = a a

m bv m + M 0

m b v . Thus, Eq. (3) gives m + M 0

m b v = Cvd m + M 0

C = a

v0 m b m + M vd

And Eqs. (2) becomes x = ca

v0 m b d e-(c>2mT) t sin vdt m + M vd

(4)

1265

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22–75. Continued

The maximum compression of the spring occurs when sin vdt = 1 p 2

vdt = t =

p 2vd

Substituting this result into Eq. (4), xmax = c a However, vd =

keq

C mT

p v0 m b de -[c>2(m + M)] A 2vd B m + M vd

- a

c2 c 2 1 2k b = = C m + M 4(m + M)2 2(m + M) 2mT

28k(m + M) - c2. Substituting this result into Eq. (5), xmax =

2mv0 2

28k(m + M) - c

e

-c

pc 228k(m + M) - c2)

d

Ans.

Ans: xmax = 1266

2mv0 28k(m + M) - c

e -pc>(228k(m + M) - c ) 2

2

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*22–76. Determine the differential equation of motion for the damped vibratory system shown. What type of motion occurs? Take k = 100 N>m, c = 200 N # s>m, m = 25 kg. k

SOLUTION

k

k

m

Free-body Diagram: When the block is being displaced by an amount y vertically downward, the restoring force is developed by the three springs attached the block. c

Equation of Motion: + c ©Fx = 0;

c

# $ 3ky + mg + 2cy - mg = - my $ # my + 2cy + 3ky = 0

(1)

Here, m = 25 kg, c = 200 N # s>m and k = 100 N>m. Substituting these values into Eq. (1) yields $ # 25y + 400y + 300y = 0 $ # y + 16y + 12y = 0

Ans.

Comparing the above differential equation with Eq. 22–27, we have m = 1kg, k 12 c = 16 N # s>m and k = 12 N>m. Thus, vn = = = 3.464 rad>s. Am A1 cc = 2mvn = 2(1)(3.464) = 6.928 N # s>m Since c 7 cc , the system will not vibrate. Therefore it is overdamped.

Ans.

Ans: # $ y + 16y + 12y = 0 Since c 7 cc, the system will not vibrate. Therefore it is overdamped. 1267

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22–77. Draw the electrical circuit that is equivalent to the mechanical system shown. Determine the differential equation which describes the charge q in the circuit.

k m

F  F0 cos vt

c

SOLUTION For the block, mx + cx + kx = F0 cos vt Using Table 22–1, 1 Lq + Rq + ( )q = E0 cos vt C

Ans.

Ans:

1268

1 Lq + Rq + a bq = E0 cos vt C

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22–78. Draw the electrical circuit that is equivalent to the mechanical system shown. What is the differential equation which describes the charge q in the circuit? c

k

k

SOLUTION m

For the block,

# $ mx + cx + 2k = 0 Using Table 22–1, 2 $ # Lq + Rq + ( )q = 0 C

Ans.

1269

Ans: 2 $ # Lq + Rq + a bq = 0 C

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22–79. Draw the electrical circuit that is equivalent to the mechanical system shown. Determine the differential equation which describes the charge q in the circuit. k

SOLUTION

m

For the block c

$ # my + cy + ky = 0 Using Table 22–1 1 $ # Lq + Rq + q = 0 C

Ans.

Ans: 1 $ # Lq + Rq + q = 0 C 1270