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

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SUMMARY

The discussion focuses on calculating the required force constant (k) for an elastic cord in exercise equipment to achieve a torque of 81 N*m about the elbow joint. The user breaks down the problem into x and y components, using lengths of 38 cm for the arm and 44 cm for the cord, with angles of 39 degrees and 61 degrees respectively. Key formulas utilized include torque (τ = rF sin θ) and potential energy (PE = 0.5kx²). The correct angle for torque calculations is determined to be the angle between the arm and the string, emphasizing the importance of geometry in the setup.

PREREQUISITES
  • Understanding of torque calculations using τ = rF sin θ
  • Familiarity with Hook's Law for elastic materials (F = kx)
  • Basic knowledge of potential and kinetic energy equations
  • Ability to decompose vectors into x and y components
NEXT STEPS
  • Research the application of Hook's Law in mechanical systems
  • Learn about torque and its applications in biomechanics
  • Explore vector decomposition techniques in physics
  • Study the principles of energy conservation in mechanical systems
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Engineers, biomechanics researchers, and fitness equipment designers who are involved in the design and analysis of exercise equipment and torque applications.

baylorbelle
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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?
 
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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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