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

Click For Summary

Homework Help Overview

The original poster is designing exercise equipment that involves calculating the force constant needed for an elastic cord to achieve a specific torque about the elbow joint. The problem involves breaking down the geometry of the arm and cord to understand the forces at play.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss breaking down the problem into x and y components and using torque formulas. There are questions about which angles to use in the calculations and how to apply Hook's law to find the force.

Discussion Status

Some participants have provided guidance on using the correct lengths and angles in the torque formula. There is an ongoing exploration of the forces involved, with some participants questioning the assumptions made about the angles and components of the forces.

Contextual Notes

There is a mention of a potential third force that may need to be considered for balance, indicating that the setup may involve more complexity than initially assumed. The discussion reflects a collaborative effort to clarify the problem without reaching a definitive conclusion.

baylorbelle
Messages
14
Reaction score
0
You are designing exercise equiptment to operate as shown in the 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?
 
Physics news on Phys.org
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].
 
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 [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.
 
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 [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.

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

Similar threads

Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
7K
Replies
2
Views
1K