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Math Help - Physics easy torque question

  1. #1
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    Physics easy torque question

    A beam is held in place by two wires, one at each end. A monkey with a mass of 50kg is hanging somewhere on the beam. There is a force of 180 N on one of the wires. Draw the picture, find the force on the 2nd wire and where the monkey is hanging if the beam is 2m long.

    Any insight? T = F(d) will be used I am guessing and so far I have a line drawn labeled 2m with 2 vertical lines on the ends, and an arrow on the left vertical line to show the 180 N of force. Thanks in advance.
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  2. #2
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    Well assuming that beam is in equilibrium, there must be zero moments, and also zero total force in all directions. In this case there is only forces in 1 direction, the vertical.

    So, yes. You have a beam which is 2m long. Let's imagine that the far left is A, and the far right is B. So length AB = 2m.

    Imagine that the wire at B is the one with a force of 180N.

    Then the wire at A will have an upwards force,  F_A which creates a torque  T = -F_{A}(2) about point B. This torque will be clockwise, and that's why it's negative.

    Then there's the monkey. The monkey has mass 50kg, and hence it has a weight  F_{monkey} = 9.81(50)  .

    The weight of the monkey will cause an anticlockwise torque about B, given by  T_{monkey} = 9.81(50) \times d where d is the distance between the monkey and B.

    Now we know that the sum of the moments about B must be zero, hence:

     \Sigma T = -2F_{A} + 9.81(50)d = 0

    And we also know that the sum of the forces is zero, hence:

     \Sigma F_{y} = 180 + F_{A} - 50(9.81) = 0

    Using the last equation, you can get a value for  F_{A} , and if you substitute that value into the 2nd to last equation, you can get a value for  d , and that's everything!

    One major assumption here is that the beam itself has zero/negligible weight.
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  3. #3
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    Excellent, thanks.
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