Question

The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of mass = 75.00 and length = 5.000 is supported by two vertical massless strings. String A is attached at a distance = 2.000 from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass = 3000 is supported by the crane at a distance = 4.800 from the left end of the bar. Throughout this problem, positive torque is counterclockwise and use 9.807 for the magnitude of the acceleration due to gravity.

Answer 1

Since the bar is in equilibrium, the net force must be zero and the net torque must be zero.

Net force = 0 = TA -TB - 75 x 9.807 - 3000 x 9.807

This gives us the relationship between the two tensions TA and TB.

Now for the torques. I'll arbitrarily select the left end of the bar as my origin. Now we have

Net torque = 0 = -TA x 2 - TB x 0 - 3000 x9.807 x 4.8 - 75.00 x 9.807 x (5/ 2 )

TA=-71529.81 N

TB=41373.28 N