Solutionmanual 5a PDF

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© 2010 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. •5–1. Draw the free-...

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© 2010 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.

•5–1. Draw the free-body diagram of the 50-kg paper roll which has a center of mass at G and rests on the smooth blade of the paper hauler. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.)

35 mm G B A 30⬚

5–2. Draw the free-body diagram of member AB, which is supported by a roller at A and a pin at B. Explain the significance of each force on the diagram. (See Fig. 5–7b.)

1950 N 390 lb 13 12

1200 800 lbN⭈· m ft

5

A 8 ftm 2.4 30⬚

0.9 3 ftm

1.2 4 ftm B

1950 N

1200 N · m

0.9 m

2.4 m

1.2 m

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5–3. Draw the free-body diagram of the dumpster D of the truck, which lb and and aa center of gravity which has has aa weight weight of of 5000 25 kN at G. It is supported by a pin at A and a pin-connected hydraulic cylinder BC (short link). Explain the significance of each force on the diagram. (See Fig. 5–7b.)

1.5 m

G D

1m 3m

B

A

C

30⬚

20⬚

25 kN

*5–4. Draw the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A and a cable which wraps around the pulley at D. Explain the significance of each force on the diagram. (See Fig. 5–7b.)

D

5

4 3

A

B

E C

2m

2m

1.5 m

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•5–5. Draw the free-body diagram of the truss that is supported by the cable AB and pin C. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.)

B

30⬚

A

2m C

3 kN 2m

4 kN

2m

5–6. Draw the free-body diagram of the crane boom AB has aa weight weightofof650 3.25lbkN center of gravity G. which has andand center of gravity at G.atThe The boom is supported a pin A and cable load boom is supported by aby pin at Aatand cable BC.BC. TheThe load of of 6.25 is suspended froma acable cableattached attachedat at B. B. Explain 1250 lbkN is suspended from the significance of each force acting on the diagram. (See Fig. 5–7b.)

2m

12 m ft 3.6 B 5.4 18 m ft

13 5 12

C

A

W = 3.25 kN

5.4

3.6

G

30⬚

m

m 6.25 kN

The Significance of Each Force : W is the effect of gravity (weight) on the boom. Ay and Ax are the pin A reactions on the boom. TBC is the cable BC force reactions on the boom. 6.25 kN force is the suspended load reaction on the boom.

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5–7. Draw the free-body diagram of the “spanner Wrench”subjected subjected to to the 20-lb 100-Nforce. force.The Thesupport support at at A A can wrench” be considered a pin, and the surface of contact at B is smooth. Explain the significance of each force on the diagram. (See Fig. 5–7b.)

20 lb 100 N A

125in. mm

B

6 in. 150 mm

100 N

25 mm

175 mm

*5–8. Draw the free-body diagram of member ABC which is supported by a smooth collar at A, roller at B, and short link CD. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.)

D

C

2.5 kN

3m 60⬚ A

4 kN ⭈ m B

45⬚

4m

6m

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•5–9. Draw the free-body diagram of the bar, which has a negligible thickness and smooth points of contact at A, B, and C. Explain the significance of each force on the diagram. (See Fig. 5–7b.)

in. 75 3mm 30⬚ 5 in.mm 125 C

B A 200 mm 8 in.

10 N lb 50 30⬚

75 tan 30° = 43.301 mm 75 mm 125 mm

NA, NB, NC force of wood on bar.

200 mm

50 N force of hand on bar.

75 cos 30° = 86.603 mm 50 N

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5–10. Draw the free-body diagram of the winch, which consists of of aadrum drumofofradius radius 4 mm. in. ItItisispin-connected 100 pin-connectedat at its center C, and at its outer rim is a ratchet gear having a mean radius of in. The serves a two-force member of 6150 mm.pawl TheAB pawl AB as serves as a two-force (short link) andlink) prevents the drumthe from rotating. Explain member (short and prevents drum from rotating. the significance of each forceforce on on thethediagram. Explain the significance of each diagram. (See Fig. 5–7b.)

B

75 mm 3 in. A

502mm in. 6 in.mm 150

C

4 in. 100 mm 500 lb 2500 N

150 mm

Cx, Cy force of pin on drum. FAB force of pawl on drum gear. 2500 N force of cable on drum.

100 mm

2500 N

5–11. Determine the normal reactions at A and B in Prob. 5–1.

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*5–12. Determine the tension in the cord and the horizontal and vertical components of reaction at support A of the beam in Prob. 5–4.

•5–13. Determine the horizontal and vertical components of reaction at C and the tension in the cable AB for the truss in Prob. 5–5.

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5–14. Determine the horizontal and vertical components of reaction at A and the tension in cable BC on the boom in Prob. 5–6. Equations of Equilibrium : The force in cable BC can be obtained directly by summing moments about point A.

哭

+ ΣMA = 0;

= 0;

5.4

Ans

 12  Ax – 55.29   = 0  13 

m

m

6.25 kN

Ax = 51.04 kN +↑ΣFy = 0;

3.6

TBC sin 7.380° (9) – 3.25 cos 30° (5.4) – 6.25 sin 60° (9) = 0 TBC = 55.29 kN

+ → ΣFx

W = 3.25 kN

Ans

 5 Ay – 3.25 – 6.25 – 55.29   = 0  13  Ay = 30.77 kN

Ans

5–15. Determine the horizontal and vertical components of reaction at A and the normal reaction at B on the spanner wrench in Prob. 5–7.

哭

+ ΣMA = 0;

NB (25) – 100 (175) =0 NB = 700 N

+ → ΣFx

= 0;

100 N 150 mm

–Ax + 700 = 0 Ax = 700 N

+↑ΣFy = 0;

Ans

25 mm

Ans

Ay – 100 = 0 Ay = 100 N

Ans

*5–16. Determine the normal reactions at A and B and the force in link CD acting on the member in Prob. 5–8.

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•5–17. Determine the normal reactions at the points of contact at A, B, and C of the bar in Prob. 5–9.

↑

+

ΣFx = 0;

NC sin 60° – 50 sin 30° = 0 NC = 28.87 N

哭

+ ΣMB = 0;

50 cos 30° (325 – 43.30) – NA (125 – 43.30)  75  – 28.87   =0  cos 30°  NA = 118.70 N

↑

+

ΣFy = 0;

Ans 75 tan 30° = 43.30 75 mm 125 mm

Ans

200 mm

NB + 28.87 cos 60° + 50 cos 30° – 118.70 = 0 NB = 60.96 N

50 N

Ans

5–18. Determine the horizontal and vertical components of reaction at pin C and the force in the pawl of the winch in Prob. 5–10.

哭

+ ΣMC = 0;

 3  FAB   150 – 2500 (100) = 0  13  FAB = 2003.1 N = 2.003 kN

+ → ΣFx

= 0;

 3  –Cx + 2003.1   =0  13  Cx = 1666.7 N = 1.667 kN

+↑ΣFy = 0;

Ans

Ans

 2  –2500 + Cy – 2003.1   =0  13  Cy = 3611.1 N = 3.611 kN

150 mm

1000 mm

2500 N

Ans

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5–19. Compare the force exerted on the toe and heel of a 120-lb 600 N woman woman when when she she is is wearing regular shoes and stiletto heels. Assume all her weight is placed on one foot and the reactions occur at points A and B as shown.

600 lb N 120 600 lb N 120

A

B

143.75 mm 5.75 in. 1.25 mm in. 31.25

哭

+ ΣMB = 0;

Ans

Ans 600 N

600 N

Stiletto heal shoe,

哭

600 (93.75) – (NA)s (112.5) = 0 (NA)s = 500 N

+↑ΣFy = 0;

0.75 in. 3.75 93.75in. mm 18.75 mm

(NB)r + 492.9 – 600 = 0 (NB)r = 107.1 N

+ ΣMB = 0;

B

600 (143.75) – (NA)r (175) = 0 (NA)r = 492.9 N

+↑ΣFy = 0;

A

Ans

(NB)s + 500 – 600 = 0 (NB)s = 100 N

Ans

31.25 mm

143.75 mm

18.75 mm 93.75 mm

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*5–20. The 000kN lb and a center Thetrain traincar carhas hasaaweight weightofof24120 of gravity at G. It is suspended from its front and rear on the track by six tires located at A, B, and C. Determine the normal reactions on these tires if the track is assumed to be a smooth surface and an equal portion of the load is supported at both the front and rear tires.

G

C 1.8 6 ftm 4 ftm 1.2 B A 1.5 5 ftm

120 kN

哭

+ ΣMO = 0;

(2 NC) (1.2) – 120 (1.5) = 0 NC = 75 kN

+ → ΣFx

= 0;

1.2 m

2 NA – 2 (75) = 0 NA = 75 kN

+↑ΣFy = 0;

Ans

Ans

2 NB – 120 = 0 NB = 60 kN

1.5 m

Ans

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•5–21. Determine the horizontal and vertical components of reaction at the pin A and the tension developed in cable BC used to support the steel frame.

60 kN 1m

1m

1m

B 30 kN ⭈ m

3m 5

4 3

C

A

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Thearticulated articulatedcrane crane boom a weight of lb 625and N 5–22. The boom hashas a weight of 125 and center of gravity If it supports load 3000 N, center of gravity at G. IfatitG. supports a load ofa600 lb,of determine determine the force thethe pinforce A and thehydraulic force in the force acting at theacting pin Aatand in the the hydraulic cylinder BC when the position boom isshown. in the position cylinder BC when the boom is in the shown.

1.2 4 ftm A

G

1 ft m 0.3 B

2.4 8 ftm 40⬚

1 ft m 0.3

C

哭

+ ΣMA = 0;

FB cos 40° (0.3) + FB sin 40° (0.3) – 625 (1.2) – 3000 (2.7) = 0 FB = 20939.3 N = 20.94 kN

+ → ΣFx

= 0;

Ans

–Ax + 20939.3 cos 40° = 0 Ax = 16040.4 = 16.04 kN

Ans

625 N 1.2 m

+↑ΣFy = 0;

–Ay + 20939.3 sin 40° – 3000 – 625 = 0 Ay = 9.83 kN

0.3 m

Ans

2.4 m 0.3 m

3000 N

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5–23. The airstroke actuator at D is used to apply a force of F = 200 N on the member at B. Determine the horizontal and vertical components of reaction at the pin A and the force of the smooth shaft at C on the member.

C

15⬚ 600 mm

B A

60⬚ D 600 mm 200 mm

F

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*5–24. The airstroke actuator at D is used to apply a force of F on the member at B. The normal reaction of the smooth shaft at C on the member is 300 N. Determine the magnitude of F and the horizontal and vertical components of reaction at pin A.

C

15⬚ 600 mm

B A

60⬚ D 600 mm 200 mm

F

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•5–25. The The300-lb 1500-N electrical transformer with center of •5–25. electrical transformer with center of gravity gravity G is supported byata pin at Aa and a smooth at G is at supported by a pin A and smooth pad atpad B. at B. Determine the horizontal and vertical components of Determine the horizontal and vertical components of reaction reaction and the B on the at the pinat A the and pin the A reaction of reaction the pad Bofonthe thepad transformer. transformer.

0.45 1.5 ftm

A 0.9 3 ftm

G B

Equations of Equilibrium: From the free – body diagram of the transformer, Fig. a, NB and Ay can be obtained by writing the moment equation of equilibrium about point A and the force equation of equilibrium along the y axis.

哭

+ ΣMA = 0;

NB (0.9) – 1500 (0.45) = 0 NB = 750 N

+↑ΣFy = 0;

Ans.

Ay – 1500 = 0 Ay = 1500 N

Ans.

Using the result NB = 750 N and writing the force equation of equilibrium along the x axis, + → ΣFx

= 0;

750 – Ax = 0 Ax = 750 N

Ans.

0.45 m

1500 N

0.9 m

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5–26. A skeletal diagram of a hand holding a load is shown in the upper figure. If the load and the forearm have masses of 2 kg and 1.2 kg, respectively, and their centers of mass are located at G1 and G2, determine the force developed in the biceps CD and the horizontal and vertical components of reaction at the elbow joint B. The forearm supporting system can be modeled as the structural system shown in the lower figure.

D

G1 C B A

G2 D

75⬚

G1

C B A 100 mm

G2 135 mm

65 mm

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5–27. As an airplane’s brakes are applied, the nose wheel exerts two forces on the end of the lan...