A rigid uniform horizontal bar of mass m 1 = 95.00kg and length L = 5.600m is su

Question

A rigid uniform horizontal bar of mass m1 = 95.00kg and length L = 5.600m is supported by two vertical massless strings. String A is attached at a distance d = 1.900m 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 = 2000kg is supported by the crane at a distance x = 5.400m 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.

Find TA, the tension in string A.

Find TB, the magnitude of the tension in string B.

Explanation / Answer

Fnet = 0
Ta-Tb+m1*g+m2*g =0

Ta - Tb + 95*9.807 + 2000*9.807 = 0

Ta - Tb = 20545.665 N ---------------------(1)

net torque = 0

-Ta*d + m1*g*(L/2) + m2*g*x = 0

Ta*1.9 = (95*9.807*5.6/2) + (2000*9.807*5.4)

Ta = 57118 N

from equation (1)

Tb = Ta - 20545.665 = 36572.36 N

Answers : TA = 57118 N

        TB = 36572.36 N

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