Question

A uniform 30-lb beam 10 ft long is carried by two men A and B, one on each end of the beam. a. If A exerts a force of 25lb, where must the load of 50lb be placed on the beam? b. What force does B exerts

Answer 1

here, F1 = force exerted by man A = 25 lb

F2 = Force exerted by man B

L = length of beam = 10 ft

W = weight of load = 50 lb

Wb = weight of beam = 30 lb

x = distance of load from point A

a.)

By torque balance about point B,

F1*L - W*(L - x) - Wb*(L/2) = 0

F1*L - Wb*(L/2) = W*(L - x)

10 - x = (25*10 - 30*10/2)/50

x = 10 - (25*10 - 30*10/2)/50

x = 8 ft = Position of load from point A

Position of load from point B = 10 - 8 = 2 ft

b.)

By force balance in vertical direction,

F1 + F2 = W + Wb

F2 = 50 + 30 - 25

F2 = 55 lb = Force exerted by man B

"Let me know if you have any query."