Question

A 25-ft crane supported at its lower end by a pin is elevated by a horizontal cable. A 250-Ib load is suspended through the cable. The center of gravity of the crane is 10 ft from the pin, and the crane weighs 200 lb. Hint: You don't have to convert the unit since they cancel out on both sides. 25. Find the tension in the cable. 26. Find the total force acting on the pin 525016 W(2001)

Answer 1

25] By torque equation about the hinge,

T*L*cos theta - W*L/2*sin theta/2 - 250*L*sin theta = 0

T*25*cos 60 degree - 200*25/2*sin 60 degree - 250*25*sin 60 degree = 0

T = 606.2 lb

26] sum of forces in x direction = 0

force on crane due to pin = T = 606.2 lb i^

Fx on pin = - 606.2 lb i^

Fy on pin = -200 - 250 lb j = -450 lb j^

total force acting =  - 606.2 lb i^ -450 lb j^