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= 85.00 kg and length = 6.000 m is supported by two vertical massless strings. String A is attached at a distance d = 1.800 m 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 = 2500 kg is supported by the crane at a distance x = 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. Part A Find the tension in string A. Part B Find the magnitude of the tension in string B.

Explanation / Answer

As bar is in rotational equilibrium,taking torque at left end Torque due to T(B) =0 Torque due to T(A) =T(A)*1.4 counterclockwise Torque due to weight of bar = 80*9.81*3=2354.4 N clockwise Torque due to weight of object =3500*9.81*5.8=199143 N clockwise counterclockwise torque =clockwise torque T(A)*1.4 =201497.4 T(A) =143926.71 N T(B) + m1g +m2g =T(A) T(B)=143926.71 -3580*9.81 T(B) =108806.91 N (1) T(A) the tension in string A.is 143926.71 N (2) T(B) the magnitude of the tension in string B is 108806.91 N ______________________________

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