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Torque? Rotation it has the Word Pivot

  1. Oct 31, 2006 #1
    Torque? Rotation it has the Word Pivot....

    A forearm can be modeled as a 2.50 kg, 32-cm-long "beam" that pivots at the elbow and is supported by the biceps.
    http://session.masteringphysics.com/problemAsset/1001007/10/knight_Figure_13_58.jpg
    How much force must the biceps exert to hold a 430 g ball with the forearm parallel to the floor?
    Okay So I have no clue how to aproach. i'm thinking torque=rFsin theta and solve for F......but I don't knwo how to do that

    Anyone want to give me a small clue to start this?

    Oh yeah when you look at the picture is r=2.5 cm......?

    Thanks
     
  2. jcsd
  3. Oct 31, 2006 #2

    berkeman

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    Staff: Mentor

    Is 2.5cm the distance to where the bicep muscle attaches to the horizontal forearm beam?

    If so, draw the free body diagram, with the tension vector up the bicep muscle broken down into its vertical and horizontal components.
     
  4. Oct 31, 2006 #3

    Astronuc

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    Staff: Mentor

    The picture makes it clear that the biceps are attached at 2.5 cm from the elbow (pivot).

    This is a matter of determining the net moment. Also realize that the forearm "can be modeled as a 2.50 kg, 32-cm-long "beam" that pivots at the elbow and is supported by the biceps."

    One must know how to take the moment of the forearm, which is a distributed load. Hint: think center of mass.
     
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