Force Constant given torque

[baylorbelle]

You are designing exercise equiptment to operate as shown in teh figure:

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?

 

[Irid]

You must plug in the expressions to your torque formula. Use Hook's law to find $$F$$ and geometry of your pictures to find $$r\sin \theta$$.

 

[baylorbelle]

and i use 39 rather than 61 for the theta, because the torque is measured in the arm?

 

[Irid]

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.

 

[RTW69]

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?

 

[baylorbelle]

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