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Net moment in a circular disc

  1. Sep 23, 2009 #1
    I understand that to find moment of force, we look at direction of force and perpendicular distance. In the diagram, O is the centre and pivot of the circular disc.

    Would the anticlockwise moment at A be (30N x 4m)? I'm not sure if this is true, cos the distance 4m is only AB. It does not include O which is the pivot.
    Should we resort to resolving the 30N force to 2 components at 5m (OA)?
    (Clockwise moment is straightforward here, isn't it?)
     

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  2. jcsd
  3. Sep 23, 2009 #2

    Doc Al

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    Staff: Mentor

    Yes.
    What counts is the perpendicular distance from the line of the force to the pivot, which equals AB.
    That's a perfectly fine thing to do. Try it and you'll see that you get the same answer for the moment, since OA*F*sinθ = AB*F.
    Sure.
     
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