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Homework Help: Torque on a Massful Beam with Load

  1. Apr 22, 2007 #1
    This is the problem I have with the question with a beam with mass.

    1. The problem statement, all variables and given/known data
    A 2.90 -m-long, 390 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 71.0 kg construction worker stands at the far end of the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

    2. Relevant equations
    The equation that I used:
    Torque = mgl
    m = mass
    g = 9.8
    l = length

    As provided by my professor.

    3. The attempt at a solution
    Since the beam is bolted at the edge, and is horizontal, m=390, and l=2.9
    In addition, there is a load applied to the edge of the beam so:
    (390 + 71)9.8 * 2.9 = torque

    Torque is 13101.62

    However, that is wrong, can someone tell me why and what I am doing wrong? Thanks
  2. jcsd
  3. Apr 22, 2007 #2
    How do you express your equation as sum of all the torque? Also is the length value same for the beam and the person if you draw a freebody diagram? Another words, is the man standing in the middle or the end? Did you take consideration center of gravity? You need to learn to draw things out, put the arrows pointing down at the right places, it will help you visulize where the force is acting so you can correctly express it in your Net torque equations
    Last edited: Apr 22, 2007
  4. Apr 22, 2007 #3
    The sum of the torque is T1+T2, and the person is standing at the edge of the beam. I do know the torque of the person standing at the edge, which is 2.9 * 71 * 9.8 = 2017.82 however, I do not know how to find the torque of the beam.
  5. Apr 22, 2007 #4
    Your stealbeam assuming that it is uniform in mass, it should have its downward force eactly in the middle, think of the stealbeam as a point particle, and all the mass is centered at the center :) that will give you its torque. And the person is standing at the edge at that distance, there is another downward force there. So sum up all the Torque.
    Last edited: Apr 22, 2007
  6. Apr 22, 2007 #5
    Ok, thanks. The torque of the beam is mgl/2. Thats what I forgot.
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