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Torque of a human arm

  1. Nov 11, 2004 #1
    I can get pretty far in this question, but I can't seem to get the final answer. The problem is:
    Calculate the force required by the "deltoid" muscle, Fm, to hold up an outstretched arm. The total mass of the arm is 3.33kg. On the diagram, the pivot point I chose is the shoulder socket, and the diagram shows the center of mass to be the elbow (with an arrow pointing down labeled 'mg'). By the shoulder socket, there is a drawing of the force Fm we need to find and it's d=.119 m away from the shoulder socket and the force points from .119 meters away upwards at an angle of alpha=14.8deg towards the socket. There is another force Fj (I assume force of the joint?) pointing downwards from the socket at an angle and it appears that the angle it points downwards at would be alpha as well (interior angles) but it is not stated or shown. The center of mass at the elbow is D distance from the socket, D=.238m. I know that we need to find the sum of the torques, but I can't figure out is Fj is supposed to be equal to Fm but opposite sign or what? I have that Sigma Torque= -mgD(torque at elbow)-(Fj)d sin alpha + (Fm) d sin alpha=0 I can't have two variables though, so what am I missing? Any hints are appreciated! Thanks.

    Please please help!
    Last edited: Nov 11, 2004
  2. jcsd
  3. Nov 11, 2004 #2
    I got it! woo hoo...nevermindnow
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