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Question
A uniform bar of mass M and length L is horizontally suspended from the ceiling by two vertical cables as shown in the above figure. Cable A is connected 1/4th distance from the left end of the bar. Cable B is attached at the far right end of the bar. What is the tension in cable A?
A uniform bar of mass M and length L is horizontally suspended from the ceiling by two vertical cables as shown in the above figure. Cable A is connected 1/4th distance from the left end of the bar. Cable B is attached at the far right end of the bar. What is the tension in cable A?
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Solution
The correct option is B 2/3 Mg
This is a torque problem. While the fulcrum can be placed anywhere, placing it at the far right end of the bar eliminates Cable B from the calculation. There are only two forces acting on the bar; the weight that produces a counterclockwise rotation and the tension in Cable A that produces a clockwise rotation. Since the bar is in equilibrium, these two torques must sum to zero.
∑τ=TA(3/4L)−Mg(1/2L)=0
Therefore, TA=(MgL/2)/(3L/4)=(MgL/2)(4/3L)=2Mg/3
This is a torque problem.
While the fulcrum can be placed anywhere, placing it at the far right end of the bar eliminates Cable B from the calculation. There are only two forces acting on the bar; the weight that produces a counterclockwise rotation and the tension in Cable A that produces a clockwise rotation. Since the bar is in equilibrium, these two torques must sum to zero.
∑τ=TA(3/4L)−Mg(1/2L)=0
Therefore, TA=(MgL/2)/(3L/4)=(MgL/2)(4/3L)=2Mg/3
∑τ=TA(3/4L)−Mg(1/2L)=0
Therefore, TA=(MgL/2)/(3L/4)=(MgL/2)(4/3L)=2Mg/3
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