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Resultant Torque independent of origin?

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Let a system of forces (F1,....Fn) act on a body at points (x1,....xn) respectively. Assume that the resultant or net force vanishes (sum of forces = 0)

    Show that the resultant torque of this system is independent of the choice of origin, i.e. for 2 different origins x0 and x0', we have T = T' where:

    T = [summation](xi-x0) X Fi and T' = [summation](xi-x0') X Fi

    2. Relevant equations

    T = F X d


    3. The attempt at a solution

    I have very little idea as to how to approach this problem, other than the T = Force X distance, and perhaps that the solution may have something to do with a bunch of couple moments that all equate to the same value, regardless of origin. Any help greatly appreciated!
     
  2. jcsd
  3. Jan 24, 2013 #2

    tiny-tim

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    Hi MrLiou168! :smile:
    ok, now subtract one from the other:

    you get … ? :wink:
     
  4. Jan 24, 2013 #3
    Thanks for the reply! But not quite, I'm a bit of a dim light most of the time... So assuming T = T', then T-T' = 0...

    And subtracting the 2 equations nets me x0 - x0'. Therefore, is it sufficient to simply state that x0 - x0' = 0 to prove that the resultant torque is independent of choice of origin?

    Thanks!
     
  5. Jan 24, 2013 #4

    tiny-tim

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    nooo …

    what about all those forces? (∑ Fi) :wink:
     
  6. Jan 24, 2013 #5
    Oh... sorry I didn't realize the F's were summed as well...

    So I have (x0 - x0') X Fi = 0? And since Fi is not zero, the x0 = x0'...?
     
  7. Jan 24, 2013 #6

    tiny-tim

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    (try using the X2 button just above the Reply box :wink:)
    hmm, you're really confused :redface:

    you don't know T - T' = 0, that's what you're trying to prove!

    all you have proved is T - T' = (x0 - x0') X (∑ Fi) …

    now, what do you know about ∑ Fi ? :wink:
     
  8. Jan 24, 2013 #7
    A ha thank you for your patience! OK so one of the givens is that [sum] Fi = 0...

    Therefore the right side of the equation nets zero and we have proven that T-T' = 0 and thus T = T' correct?
     
  9. Jan 24, 2013 #8

    tiny-tim

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    correct!! :smile:
     
  10. Jan 24, 2013 #9
    Tim, thank you very much - you were extremely helpful!
     
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