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Equivalent Force couple to single resultant

  1. Sep 14, 2008 #1
    I have reduced the problem involving different forces and a couple in a 2D plane to an equivalent force-couple system. The force F is -27.5lbi + 11.65lbj, and the couple is -66.8inlbk.
    I need the single offset resultant that accounts for the couple.

    The way I thought to solve it was to take the couple and set it equal to r (unknown xi +yj) crossed with the force. But this gives me two unknowns with one equation, and I don't know where to go from here.

    Thanks!
     
  2. jcsd
  3. Sep 14, 2008 #2

    Pyrrhus

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    You could pick a point where one of the components of the force will have zero moment, and equal the product of the lever arm and component with moment respect to said point to the magnitude of the couple, then repeat by reversing the components. You'll end up with an x and y which will give the location of the force vector line of action with respect to the point as it were your origin. If you divide both magnitude of force and couple, you'll get the lever arm from said point to the force line of action.
     
    Last edited: Sep 14, 2008
  4. Sep 14, 2008 #3
    So there is more than one correct solution?
     
  5. Sep 14, 2008 #4

    Pyrrhus

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    What I meant is your resultant at a certain location will account for the couple. It'll be a resultant system (no couple) which is what you are looking for, no?.

    Coplanar resultant-couple systems can always be reduced to a resultant acting on a certain location with respect to the point where you reduced the system. This is because the couple and the resultant will always be normal in this case.

    This means the line of action of the resultant will be in the same place regardless of the point of reference (origin or where you did the moments for the resultant-couple system reduction) you used, so "more than one correct solution" means same place but with different points of reference.
     
    Last edited: Sep 14, 2008
  6. Sep 14, 2008 #5
    I apparently read your first post before you edited it or something, so I was talking about something else. The edited one clarified it for me though.

    Thanks
     
  7. Sep 14, 2008 #6

    Pyrrhus

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    Great!, and Welcome to PF! :smile:
     
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