The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A

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 m1 = 100.0kg and length L = 5.800m is supported by two vertical massless strings. String A is attached at a distance d = 2.000m 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 m2 = 3500kg is supported by the crane at a distance x = 5.600m from the left end of the bar. Throughout this problem, positive torque is counterclockwise and use 9.807m/s2 for the magnitude of the acceleration due to gravity.

Part A

Find TA, the tension in string A

Part B

Find TB, the magnitude of the tension in string B

Explanation / Answer

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

Net force = 0 = TA -TB - ( 100 kg ) g - ( 3500 kg ) ( g )

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 ( 2.000 m ) - TB ( 0 ) - ( 3500 kg ) ( g ) ( 5.600 m ) - ( 100 kg ) ( g ) ( 5.800 m / 2 )

You now have two equations in two unknowns, the tensions TA and TB. Solve for those tensions.

Ta=97461 N

Tb=62181 N

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