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Torques in rotational equilibrium

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Why, for any system that is in rotational equilibrium, the torque about 1) any point on the object or 2) any point in space, must be zero.


    2. Relevant equations

    N/A


    3. The attempt at a solution

    What I do not understand is, why the torque about ANY point on the object is zero? Isn't it only about the point of rotation/center of gravity? Say, if I have a seesaw that is balanced (therefore it's in rotational equilibrium?) on both side. If I move the pivot point away from the middle, the equilibrium will not exist anymore, right? How can this be for torque about ANY point?

    Also, as 2) stated in the original question, about ANY POINT IN SPACE, too? I.e. A balanced seesaw with torque about a random position on the moon??

    There gotta be something important here I missed or didn't understand. Can someone clarify these for me? Thank you!
     
  2. jcsd
  3. Aug 10, 2011 #2

    Dick

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    Homework Helper

    If you have a balanced seesaw and you look at torques around the pivot point then the force at the pivot point doesn't contribute any torque because it's at the pivot point. And it doesn't rotate because the torques sum to zero. If you compute the torques around some other point, like an endpoint of the seesaw, then the force at the end of the seesaw now contributes no torque, but now the force at the pivot does. But they still have to sum to zero because it still doesn't rotate. Write down a concrete example and try it.
     
  4. Aug 10, 2011 #3
    Thank you so much Dr.! Now it makes sense! I think the reason I couldn't understand this before was I always assumes that the actual pivot point "moves" with different ways of torque calculation! After all, the physical pivot for the seesaw is still at the same location but the imaginary rotation center used for torque calculation changes for different ways torque combination! :)
     
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