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Perpendicular force calculated from torque and point of application?

  1. Feb 11, 2013 #1
    I hope I managed to post my question the right place!

    I have a body consisting of a bunch of mass-points in ℝ[itex]^{3}[/itex], and when torque is applied to this body, I'm interested in finding the force that must have caused the torque based on the point of application and the torque vector, which are given.

    I see the form;
    [itex]\tau[/itex] = r[itex]\times[/itex]F
    quite often, such as it is seen in wikipedia (which offers a nice overview btw).

    I understand that it is not possible to calculate F, but I find it hard to believe that F[itex]_{\bot}[/itex] is impossible to calculate since it should be unique, yet I don't see such an equation anywhere. Any pointers would be greatly appreciated!

  2. jcsd
  3. Feb 11, 2013 #2


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    Staff: Mentor

    Re: Perpendicular force calculated from torque and point of applicatio

    Calculate an arbitrary F which satisfies the equation, subtract the component of ##F_{||}##.
  4. Feb 12, 2013 #3
    Re: Perpendicular force calculated from torque and point of applicatio

    I see, clever :)

    May this arbitrary F (how about we call it F'?) be obtained by;
    F' = r×[itex]\tau[/itex]/r[itex]\bullet[/itex]r ?

    I'm feeding the result I get back into the form
    [itex]\tau[/itex] = r×F
    and as I've realized, the [itex]\tau[/itex] and r I am given are not perfectly perpendicular so my [itex]\tau[/itex]' is similar to [itex]\tau[/itex] at best. But that method seems like it would be correct!

    Thank you very much!
    Last edited: Feb 12, 2013
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