Engineering Mechanics Statics 13E - Chapter 05 [Solutions] (Hibbeler) PDF

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5–1. Draw the free-body diagram of the dumpster D of the truck, which has a weight of 5000 lb and a center of gravity 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 A

B 20

30

C

SOLUTION The Significance of Each Force: W is the effect of gravity (weight) on the dumpster. Ay and Ax are the pin A reactions on the dumpster. FBC is the hydraulic cylinder BC reaction on the dumpster.

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5–2. Draw the free-body diagram of member ABC which is supported by a smooth collar at A, rocker at B, and short link CD. Explain the significance of each force acting on the diagram. (See Fig. 5–7b.)

SOLUTION

3m

60 A

4 kN m B

45 4m

D

C

2.5 kN

6m

The Significance of Each Force: NA is the smooth collar reaction on member ABC. NB is the rocker support B reaction on member ABC. FCD is the short link reaction on member ABC. 2.5 kN is the effect of external applied force on member ABC. 4 kN # m is the effect of external applied couple moment on member ABC.

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5–3. 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

E

B

SOLUTION T force of cable on beam.

C 2m

2m

1.5 m

Ax, Ay force of pin on beam. 80(9.81)N force of load on beam.

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*5–4. Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B.

F ⫽ 8 lb

1.5 ft

SOLUTION

A B

0.2 ft

2 ft

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5–5. Draw the free-body diagram of the uniform bar, which has a mass of 100 kg and a center of mass at G. The supports A, B, and C are smooth.

0.5 m

0.2 m

1.25 m C G

1.75 m

SOLUTION

A

0.1 m 30⬚

B

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5–6. Draw the free-body diagram of the beam,which is pin supported at A and rests on the smooth incline at B.

800 lb

800 lb 600 lb

600 lb

400 lb 3 ft

3 ft

3 ft

3 ft

0.6 ft 1.2 ft 0.6 ft

SOLUTION

A

B 5

3

4

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5–7. Draw the free-body diagram of the beam, which is pin connected at A and rocker-supported at B.

500 N

800 N⭈m

SOLUTION B

5m

A 8m

4m

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*5–8. 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.)

3 in. 30 C

5 in. B A 8 in.

SOLUTION NA, NB, NC force of wood on bar.

10 lb 30

10 lb force of hand on bar.

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5–9. Draw the free-body diagram of the jib crane AB, which is pin connected at A and supported by member (link) BC.

C

5

3 4

SOLUTION

B

0.4 m A

3m

4m 8 kN

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5–10. 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 6m

2m

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;

NB cos 30°(8) - 4(6) = 0 NB = 3.464 kN = 3.46 kN

Ans.

Using this result and writing the force equations of equilibrium along the x and y axes, we have + ©Fx = 0; :

A x - 3.464 sin 30° = 0 A x = 1.73 kN

+ c ©Fy = 0;

Ans.

A y + 3.464 cos 30° - 4 = 0 A y = 1.00 kN

Ans.

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5–11. Determi ne th e magnitud e of th e reacti ons on th e b eam at A and B. Neglect the thickness of the beam.

600 N

15⬚ 400 N

5

3 4

B

A 4m

8m

SOLUTION a + ©MA = 0;

By (12) - (400 cos 15°)(12) - 600(4) = 0 By = 586.37 = 586 N

+ ©F = 0; : x

Ans.

Ax - 400 sin 15° = 0 Ax = 103.528 N

+ c ©Fy = 0;

Ay - 600 - 400 cos 15° + 586.37 = 0 Ay = 400 N FA = 2(103.528)2 + (400)2 = 413 N

Ans.

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*5–12. Determine the components of the support reactions at the fixed support A on the cantilevered beam.

6 kN

30⬚

SOLUTION 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. + ©Fx = 0; :

A 1.5 m

4 kN

1.5 m

4 cos 30° - A x = 0 A x = 3.46 kN

+ c ©Fy = 0;

30⬚ 1.5 m

Ans.

A y - 6 - 4 sin 30° = 0 A y = 8 kN

Ans.

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.

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5–13. The 75-kg gate has a center of mass located at G. If A supports only a horizontal force and B can be assumed as a pin, determine the components of reaction at these supports.

A

1.25 m G

1m B

SOLUTION Equations of Equilibrium: From the free-body diagram of the gate, Fig. a, By and Ax can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about point B. + c ©Fy = 0;

By - 75(9.81) = 0 By = 735.75 N = 736 N

a+©MB = 0;

Ans.

A x(1) - 75(9.81)(1.25) = 0 A x = 919.69 N = 920 N

Ans.

Using the result Ax = 919.69 N and writing the force equation of equilibrium along the x axis, we have + ©F = 0; : x

Bx - 919.69 = 0 Bx = 919.69 N = 920 N

Ans.

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5–14. The overhanging beam is supported by a pin at A and the two-force strut BC. Determine the horizontal and vertical components of reaction at A and the reaction at B on the beam.

1m

A

SOLUTION

800 N 2m

B

900 N⭈m

1.5 m

Equations of Equilibrium: Since line BC is a two-force member, it will exert a force FBC directed along its axis on the beam as shown on the free-body diagram, Fig. a. From the free-body diagram, FBC can be obtained by writing the moment equation of equilibrium about point A. a+ ©MA = 0;

600 N 1m

3 FBC a b (2) - 600(1) - 800(4) - 900 = 0 5

FBC = 3916.67 N = 3.92 kN

C

Ans.

Using this result and writing the force equations of equilibrium along the x and y axes, we have + : ©Fx = 0;

+ c ©Fy = 0;

4 3916.67a b - A x = 0 5

A x = 3133.33 N = 3.13 kN

Ans.

A y = 950 N

Ans.

3 -A y - 600 - 800 + 3916.67a b = 0 5

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5–15. Determine the horizontal and vertical components of reaction at the pin at A and the reaction of the roller at B on the lever.

14 in. 30⬚ F ⫽ 50 lb

A

SOLUTION Equations of Equilibrium: From the free-body diagram, FB and A x can be obtained by writing the moment equation of equilibrium about point A and the force equation of equilibrium along the x axis, respectively. a+ ©MA = 0;

18 in.

50 cos 30°(20) + 50 sin 30°(14) - FB(18) = 0 FB = 67.56 lb = 67.6 lb

+ ©F = 0; : x

20 in.

B

Ans.

A x - 50 sin 30° = 0 A x = 25 lb

Ans.

Using the result FB = 67.56 lb and writing the force equation of equilibrium along the y axis, we have + c ©Fy = 0;

A y - 50 cos 30° - 67.56 = 0 A y = 110.86 lb = 111 lb

Ans.

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*5–16. Determine the components of reaction at the supports A and B on the rod.

P L –– 2

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 + c ©Fy = 0;

Ax = 0

B

Ans.

By - P = 0 By = P

a + ©MB = 0;

A

L Pa b - MA = 0 2

Ans.

MA =

Ans.

PL 2

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5–17. If the wheelbarrow and its contents have a mass of 60 kg and center of mass at G, determine the magnitude of the resultant force which the man must exert on each of the two handles in order to hold the wheelbarrow in equilibrium. B

G 0.6 m

0.5 m A

SOLUTION a +©MB = 0;

0.5 m

0.9 m

-Ay (1.4) + 60(9.81)(0.9) = 0 Ay = 378.39 N

+ c ©Fy = 0;

378.39 - 60(9.81) + 2By = 0 By = 105.11 N

+ ©F = 0; : x

Bx = 0 FB = 105 N

Ans.

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5–18. 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

B C

SOLUTION 5 ft

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. 2 a + ©MA = 0; T152 + T ¢ ≤ 1102 - 801132 = 0 25 Ans. T = 74.583 lb = 74.6 lb + ©F = 0; : x

Ax - 74.583 ¢

1

25

≤ = 0

Ax = 33.4 lb + c ©Fy = 0;

74.583 + 74.583

2

25

Ans.

- 80 - By = 0

Ay = 61.3 lb

Ans.

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5–19. The shelf supports the electric motor which has a mass of 15 kg and mass center at Gm. The platform upon which it rests has a mass of 4 kg and mass center at Gp. Assuming that a single bolt B holds the shelf up and the bracket bears against the smooth wall at A, determine this normal force at A and the horizontal and vertical components of reaction of the bolt on the bracket.

150 mm

Gm

50 mm 40 mm B

Gp

60 mm

SOLUTION a + ©MA = 0;

200 mm

Bx (60) - 4(9.81)(200) - 15(9.81)(350) = 0

A

Bx = 989.18 = 989 N

Ans.

+ ©F = 0; : x

Ax = 989.18 = 989 N

Ans.

+ c ©Fy = 0;

By = 4(9.81) + 15(9.81) By = 186.39 = 186 N

Ans.

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*5–20. The pad footing is used to support the load of 12 000 lb. Determine the intensities w1 and w2 of the distributed loading acting on the base of the footing for the equilibrium.

12 000 lb

5 in.

9 in.

9 in.

w2 w1

SOLUTION

35 in.

Equations of Equilibrium: The load intensity w2 can be determined directly by summing moments about point A. a+ ©MA = 0;

w2 a

35 b 117.5 - 11.672 - 12114 - 11.672 = 0 12

w2 = 1.646 kip>ft = 1.65 kip> ft + c ©Fy = 0;

Ans.

35 35 1 1w1 - 1.6462 a b + 1.646a b - 12 = 0 12 12 2 w1 = 6.58 kip> ft

Ans.

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