A rigid uniform horizontal bar of mass = 85.00 and length = 5.600 is supported b

Question

A rigid uniform horizontal bar of mass = 85.00 and length = 5.600 is supported by two vertical massless strings. String A is attached at a distance = 1.800 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 = 2000 is supported by the crane at a distance = 5.400 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. Find the tension in string A. Find the magnitude of 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 - ( 85.00 kg ) g - ( 2000 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 ) - ( 75.00 kg ) ( g ) ( 5.800 m / 2 )

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