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Homework Help: Simple Statics Equilbrium

  1. Nov 28, 2007 #1
    [SOLVED] Simple Statics Equilbrium

    1. The problem statement, all variables and given/known data

    A diving board of length 3m is supported at a point 1m from the end and a diver weighing 450N stands at the free end. The diving board is of uniform cross section and weighs 285N


    Q) Find the magnitude of the force at the support point.

    2. Relevant equations
    Statics Equilbrium?

    3. The attempt at a solution

    Ok i am kind of confused on this one. I know that there are 2 forces down which is 450N + Force of the push down from supported end + the weight of the board = support force. I hope thats right.

    Im not sure about the exact equation to use. Would it be Torque = 0? But even than i get 2 unknowns (Support Force and force of supported end)

    But how could i do this without given the supported force down?

  2. jcsd
  3. Nov 28, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Unfortunately, the pic link doesn't seem to be working. But don't forget to consider the weight of the board itself.

    Yes, the net torque about any point must be zero. That's one of the conditions for equilibrium.

    What's another condition for equilibrium? What must the sum of the forces equal?

    You can also apply the torque = 0 condition at two different pivot points.
  4. Nov 28, 2007 #3
    I reuploaded image for you:

    Now i am still getting this wrong. I figured Torque in this way:
    T = TF(sp) - TFg - weight of board (set starting point where Fn acts so this is 0)
    This means:
    F(sp) = 3*Fg + weight of board
    F(sp) = 3 * 450 + 285 = 1635 N but this is wrong and i dont understand why.
  5. Nov 28, 2007 #4

    Doc Al

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    Staff: Mentor

    I'm not quite understanding what you are doing. First thing to do is identify all the forces acting on the board and where they act:
    F(sp) = I assume this is the downward support force that acts at the left end of the board?
    Fg = The weight of the girl, which acts at the right end of the board?
    W = weight of the board, which acts at the middle of the board.
    Fn = The upward force located 1 m from the left end.

    OK, If you chose the location of Fn as your pivot, then you must set clockwise and counterclockwise torques equal to each other. And you must measure their distances from the pivot point in order to calculate torque.
    Redo this with the correct distances.
  6. Nov 28, 2007 #5
    Ok still seems to come out to the same thing. If i use the variables you have, i have to set my pivot at left end if i want to find Fn.

    Fn creates counter clockwise direction (+)
    W creates clockwise (-)
    Fg creates clockwise (-) at radius of 3.

    So here's my equation:
    Torque = 1 * Fn - 1 * W -3 * Fg (looking for Fn)
    Fn = W + 3 * Fg
    Fn = 285 + 3 * 450 = 1685 N

    But this seems to be incorrect.

    Thanks for your time!
  7. Nov 28, 2007 #6

    Doc Al

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    Staff: Mentor

    The weight of the board acts at its center of mass, which is not 1 m from the left end. Fix this.
  8. Nov 28, 2007 #7
    Ah! Thank you!
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