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Complicated Centre of Mass Question

  1. Jun 16, 2004 #1
    :cry: Hey all... I'm having trouble deciding where the heck I should even start with this question. Soooo.... here it is (I've attached a picture of the diagram because the question wouldn't make any sense without it).

    A non-uniform bar of weight W is suspended at rest in a horizontal position by two light ropes as shown in the following diagram. The angle one rope makes with the vertical is theta=36.9 degrees, the other makes the angle phi=53.1 degrees with the vertical. If the length L of the bar is 6.1 m, compute the distance x from the left hand end to the centre of gravity.

    (see attachment)

    Where I was headed, was to assign a weight to this bar, and calculate the tensions in the string. But, even if this was correct, I'm not sure where to go from there. Maybe I'm taking too much from a simple drawing, but would the fact that where the strings are suspended from on either side are at different heights have any effect on the outcome? Please advise...


    Out of Equilibrium in Calgary

    Attached Files:

  2. jcsd
  3. Jun 16, 2004 #2

    Doc Al

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    Staff: Mentor

    Treat this like any other equilibrium problem: Apply the conditions for equilibrium. The net force on the bar must be zero and the net torque (about any axis) must be zero. You can get three equations (x-components; y-components; torque) which will allow you to solve for the tension in each rope and "x" (the location of the center of mass).
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