The figure shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of mass m1 = 80kg and length L = 6.00m is supported by two vertical massless strings. String A is attached at a distance d = 1.4m 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 = 5.8m from the left end of the bar.
Throughout this problem, positive torque is counterclockwise. Use 9.81m/s/s for the magnitude of the acceleration due to gravity.
(1) Find T(A) the tension in string A.
(2) Find T(B) the magnitude of the tension in string B.
Can someone please show how to do this?
Here is the diagram.
1 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)*1.4 counterclockwise
Torque due to weight of bar = 809.813=2354.4 N clockwise
Torque due to weight of object =35009.815.8=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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