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Rotational equilibrium

  1. Feb 5, 2006 #1
    A 10.0-kg solid uniform cube with 0.250-m sides rests on a level surface. What is the minimum amount of work necessary to put the cube in unstable equilibrium?

    My understanding is that the cube has to be on one of its sides with it's center of gravity directly above it. However, I don't know how to find the force required to get the cube into this position. I'm not sure which is the correct distance the force has to travel through...i think it's either d = 0.250 * sin 45[degrees], or it's s = [pi]/4*0.250

    The answer is 5.08 J, but I don't know how to arrive at this answer...
  2. jcsd
  3. Feb 5, 2006 #2


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    The question is not about the force. You are being asked for the amount of energy required. All you have to do is calculate the change in potential energy from the initial orientation to the final orientation.
  4. Feb 6, 2006 #3
    thanks! i got the answer
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