1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Force Constant given torque

  1. Jun 30, 2008 #1
    You are designing exercise equiptment to operate as shown in teh figure: Walker.11.80.jpg 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook