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 = 85.00 and length = 6.000 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 = 3500 is supported by the crane at a distance = 5.800 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. Find , the tension in string A

Explanation / Answer

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As bar is in rotational equilibrium,taking torque at left end

Torque due to T(B) =0

Torque due to T(A) =T(A)*1.8 counterclockwise

Torque due to weight of bar = 85*9.81*3=2501.55N clockwise

Torque due to weight of object =3500*9.81*5.8=199143 N clockwise
2501.55N+199143 =201644.55
counterclockwise torque =clockwise torque

T(A)*1.8=201644.55

T(A) =112024.75 N

T(B) + m1g +m2g =T(A)

T(B)=112024.75 -3585*9.81

T(B) =76855.9 N

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