What Force Constant Is Needed for Exercise Equipment to Achieve Specific Torque?

AI Thread Summary
To determine the required force constant (k) for the exercise equipment, the torque about the elbow joint must equal 81 N*m. The discussion involves breaking down the arm and cord into their x and y components to analyze the forces involved. Torque is calculated using the formula torque = rF sin(theta), where r is the arm length and theta is the angle between the arm and the cord. Participants emphasize the importance of using the correct angles and lengths derived from geometry to set up the equations properly. The conversation concludes with one participant successfully resolving the problem after clarifying these concepts.
baylorbelle
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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 don't quite understand how I'm supposed to set it up from here. any advice on getting this one started?
 
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You must plug in the expressions to your torque formula. Use Hook's law to find F and geometry of your pictures to find r\sin \theta.
 
and i use 39 rather than 61 for the theta, because the torque is measured in the arm?
 
Neither of these. You must use the length of arm 39cm for r and the angle between the arm and the string for \theta. That angle is to be determined by you, using the geometry.
 
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?
 
Irid said:
Neither of these. You must use the length of arm 39cm for r and the angle between the arm and the string for \theta. That angle is to be determined by you, using the geometry.

thanks, that really helped a lot : ) i finally figured it out
 
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