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A Bar Suspended by Two Wires

  1. Aug 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A nonuniform horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle (phi) with the horizontal, and the right wire makes an angle . The bar has length L.

    What is the position of the center of mass of the bar, measured as distance from the bar's left end?

    x=?

    2. Relevant equations

    -F₁ · sin(φ₁) + F₂ · sin(φ₂) = 0
    ___________________________________
    F₁x = -F₁ · sin(φ₁)
    F₁y = F₁ · cos(φ₁)
    F₂x = F₂ · sin(φ₂)
    F₂y = F₂ · cos(φ₂)


    forces in terms of magnitude and angle
    x = L / ( (F₁/F₂)·(cos(φ₁)/cos(φ₂)) + 1)


    3. The attempt at a solution


    x = L / ( (tan(φ₂)/tan(φ₁) + 1)

    this seems right, but I'm repeatedly getting it wrong no matter how I input the answer.
     
  2. jcsd
  3. Aug 26, 2008 #2

    tiny-tim

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    take moments …

    Hi PSEYE! :smile:

    Sorry, but I've no idea how you got that result. :confused:

    You have three unknown forces, F1 F2 and W, which you don't want in the final equation.

    The only ways I know to eliminate unknown forces are:

    i] take components perpendicular to them … which isn't going to work in this case … or

    ii] take moments about a point through their line of action.

    Try ii]! :smile:
     
  4. Aug 26, 2008 #3

    PhanthomJay

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    If phi_1 and phi_2 are the angles with the horizontal, you've got your sines and cosines mixed up.
     
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