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  • Publicado : 26 de septiembre de 2010
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Equilibrium
5-19. A uniform meter stick is balanced at its midpoint with a single support. A 60-N weight is suspended at the 30 cm mark. At what point must a 40-N weight be hung to balance the system? (The 60-N weight is 20 cm from the axis)
 = 0; (60 N)(20 cm) – (40 N)x = 0
40 x = 1200 N cm or x = 30 cm: The weight must be hung at the 80-cm mark.
5-20. Weights of 10 N,20 N, and 30 N are placed on a meterstick at the 20 cm, 40 cm, and 60 cm marks, respectively. The meterstick is balanced by a single support at its midpoint. At what point may a 5-N weight be attached to produce equilibrium.
 = (10 N)(30 cm) + (20 N)(10 cm)

– (30 N)(10 cm) – (5 N) x = 0
5 x = (300 + 200 –300) or x = 40 cm
The 5-N weight must be placed atthe 90-cm mark
5-21. An 8-m board of negligible weight is supported at a point 2 m from the right end where a 50-N weight is attached. What downward force at the must be exerted at the left end to produce equilibrium?
F (6 m) – (50 N)(2 m) = 0
6 F = 100 N m or F = 16.7 N

5-22. A 4-m pole is supported at each end by hunters carrying an 800-Ndeer which is hung at a point 1.5 m from the left end. What are the upward forces required by each hunter?
 = A (0) – (800 N)(1.5 m) + B (4.0 m) = 0
4B = 1200 N or B = 300 N
Fy = A + B – 800 lb = 0; A = 500 N
5-23. Assume that the bar in Fig. 5-16 is of negligible weight. Find the forces F and A provided the system is in equilibrium.
 = (80N)(1.20 m) – F (0.90 m) = 0; F = 107 N
Fy = F – A – 80 N = 0; A = 107 N – 80 N = 26.7 N
F = 107 N, A = 26.7 N
5-24. For equilibrium, what are the forces F1 and F2 in Fig. 5-17. (Neglect weight of bar.)
 = (90 lb)(5 ft) – F2 (4 ft) – (20 lb)(5 ft) = 0;
F2 = 87.5 lb Fy = F1 – F2 – 20 lb – 90 lb = 0
F1 = F2 +110 lb = 87.5 lb +110 lb, F1 = 198 lb
5-25. Consider the light bar supported as shown in Fig. 5-18. What are the forces exerted by the supports A and B?
 = B (11 m) – (60 N)(3 m) – (40 N)( 9 m) = 0;
B = 49.1 N Fy = A + B – 40 N – 60 N = 0
A = 100 N – B = 100 N – 49.1 N; B = 50.9 N
5-26. A V-belt is wrapped around a pulley 16 in. in diameter. If a resultant torqueof 4 lb ft is required, what force must be applied along the belt?
R = ½(16 in.) = 8 in. R = (8/12 ft) = 0.667 ft
F (0.667 ft) = 4 lb ft; F = 6.00 lb
5-27. A bridge whose total weight is 4500 N is 20 m long and supported at each end. Find the forces exerted at each end when a 1600-N tractor is located 8 m from the left end.
 = B (20 m) – (1600 N)(8 m) – (4500 N)(10 m) = 0;
B = 2890 N Fy = A + B – 1600 N – 4500 N = 0
A = 6100 N – B = 6100 N – 2890 N; B = 3210 N
5-28. A 10-ft platform weighing 40 lb is supported at each end by stepladders. A 180-lb painter is located 4 ft from the right end. Find the forces exerted by the supports.
 = B(10 ft) – (40 lb)(5 ft) – (180 lb)( 6 ft) = 0;
B = 128 lbFy = A + B – 40 lb – 180 lb = 0
A = 220 lb – B = 220 lb – 128 lb; A = 92.0 lb
*5-29. A horizontal, 6-m boom weighing 400 N is hinged at the wall as shown in Fig. 5-19. A cable is attached at a point 4.5 m away from the wall, and a 1200-N weight is attached to the right end. What is the tension in the cable?
 = 900 – 370 = 530; Ty = T sin 530
 = (T sin 530)(4.5 m) – (400 N)(3m) – (1200 N)(6 m) = 0;
3.59 T = 1200 N + 7200 N; T = 2340 N
*5-30. What are the horizontal and vertical components of the force exerted by the wall on the boom? What is the magnitude and direction of this force?
Fx = H – Tx = 0; H – T cos 530 = 0; H = (2340 N) cos 530; H = 1408 N
Fy = V + T sin 530 – 400 N – 1200 N = 0; V = 1600 N – (2340 N) sin 530 =...
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