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Static Equilibrium?

  1. Dec 6, 2005 #1
    Static Equilibrium?????

    Summing the moments up from around the center, it is easy to find the intuitive location for the weight at B which would be the same as the A side. But how are there more locations to make this system balance??

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  3. Dec 6, 2005 #2


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    Staff: Mentor

    Assuming the pans are symmetrically located and of the same size, then one simply needs the sum of the moments to be zero.

    This is essentially a fulcrum problem in which the product of the weight of mass A (mAg) and its moment arm about the central axis (pivot point) must equal the product of weight B (mBg) and its moment. If the weights are identical, then the moment arms must also be identical.
  4. Dec 6, 2005 #3
    Right, so summing up the moments from the center would say you would have to put the weight on Pan B the same distance from the center as the weight on Pan A.

    The questions makes it seem like their are more than one position that the weight can be placed on Pan B to balance the system. Am I missing something?
  5. Dec 7, 2005 #4


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    Staff: Mentor

    I was wondering about that myself. Other than moving the mass B perpendicularly (in or out of the plane of the paper), I don't readily see an alternative. The centers of mass of A and B have to be at the same moment arm, unless I am missing something.
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