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Force Constant given torque

  1. Jun 30, 2008 #1
    You are designing exercise equiptment to operate as shown in teh figure:[​IMG] where a person pulls upward on an elastic stretched length of .31m. If you would like the torque about the elbow joint to be 81 N*m in the position shown, what force constant, k, is required for the cord?

    ::: okay, so since this picture has so many angles, i figured the first thing to do was to break each part down into its x and y components and create two separate pictures:
    : picture 1 is the arm, hypotenuse (length of arm) at 38 cm, x-axis = 38 cos (39)=10.1 cm, and y-axis = 38 sin(39)=36.6 cm.
    : picture 2 is the cord, hypotenuse (length of stretched cord) 44cm, x-axis: 44cos(61)=11.4 cm, y-axis=44sin(61)=42.5cm.

    I think that for the cord, it starts in potential energy of c=.5kx2 and ends in both potential and kinetic (.5mv2) energy.

    I also know that torque= rF sin theta

    However, I dont quite understand how I'm supposed to set it up from here. any advice on getting this one started?
  2. jcsd
  3. Jun 30, 2008 #2
    You must plug in the expressions to your torque formula. Use Hook's law to find [tex]F[/tex] and geometry of your pictures to find [tex]r\sin \theta[/tex].
  4. Jun 30, 2008 #3
    and i use 39 rather than 61 for the theta, because the torque is measured in the arm?
  5. Jul 1, 2008 #4
    Neither of these. You must use the length of arm 39cm for [tex]r[/tex] and the angle between the arm and the string for [tex]\theta[/tex]. That angle is to be determined by you, using the geometry.
  6. Jul 1, 2008 #5
    It looks like there is a third force extending along the arm and away from the hand to provide a balancing force in the positive x direction. Assuming the hand is at the orgin, the force at the hand perpendicular to the arm that causes the torque has a component in the negative x direction. Likewise the force along the elastic band has a component in the negative x direction. There must be a compensating force in the positive x direction or the hand is not static. Do you have an answer so you can check your work?
  7. Jul 1, 2008 #6
    thanks, that really helped a lot : ) i finally figured it out
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