Be mounted on a truck. A rigid uniform horizontal bar of mass m_1 = 100.0 kg and

Question

Be mounted on a truck. A rigid uniform horizontal bar of mass m_1 = 100.0 kg and length L = 5.500 m is supported by two vertical massless strings. String A is attached at a distance d = 2.000 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 m_2 = 3500 kg is supported by the crane at a distance x = 5.300 m from the left end of the bar. Throughout this problem, positive torque is counterclockwise and use 9.807 m/s^2 for the magnitude of the acceleration due to gravity. Find T_A, the tension in string A. Express your answer in newtons using four significant figures. Find T_B, the magnitude of the tension in string B. Express your answer in newtons using four significant figures.

Explanation / Answer

Given :

m1 = 100 kg
L= 5.5m
d = 2.0m
m2 = 3500kg
x = 5.30m
g = 9.807 m/s/s

part A

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

Torque due to T(B) =0

Torque due to T(A) = T(A)*d = T(A)*2.0 counterclockwise

Torque due to weight of bar =m1*g * L/2 = 100kg*9.807*2.75m=2696.92 N clockwise

Torque due to weight of object =m2*g*X=3500*9.807*5.3m=181919.85 N clockwise

counterclockwise torque =clockwise torque

T(A)*2.0m =184616.77 N

T(A) = 92308.385 N

Part B

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

T(B) = 92308.385 -3600*9.807

T(B) = 57003.2 N

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