Force required to keep arm in position

AI Thread Summary
To determine the force required for Lawrence's elbow flexor muscles to hold a 9kg dumbbell at a 90-degree angle, the mass must be converted to Newtons, resulting in a force of approximately 88.29N. The torque calculation involves the distances from the elbow to the dumbbell's center of mass and the elbow flexor tendons, which should be consistently measured in the same units. The relevance of the 90-degree angle is acknowledged, as it affects torque calculations, but the specific angle does not need to be converted to radians for this problem. It's emphasized that demonstrating understanding through explanation is more critical than arriving at the exact numerical answer. The discussion also raises questions about how the force would change if the arm were positioned at different angles, highlighting the importance of torque in various positions.
DrowningKthnx
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Homework Statement


Lawrence is holding a 9kg dumbbell in his hand with his elbow flexed 90degrees. The horizontal distance from the center of mass of the dumbbell to the center of Lawrence's elbow is 33cm. The horizontal distance from the insertion of his elbow flexor tendons to his elbow is 3cm. If we ignore the mass of Lawrence's arm, how much Force do his elbow flexor muscles have to produce to hold his arm in this position?

Homework Equations


T = F x r ?

The Attempt at a Solution


I drew a diagram and labeled the known values. I am unsure if I should convert the 9kg dumbbell to Newtons (or if that is even right to do). The professor has previously converted cm to m, but I found nothing about that in our book. I converted 33cm to 0.33m and 3cm to 0.03m, but also don't know if that is right to do. Also, converted 90degrees to radians: 1.570rad, but once again unsure. I just need some guidance on if I'm doing Anything correctly and what step to take next.
 
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Forces are measured in Newton so yes, you have to convert the mass of the dumbell to Newtons (noting of course that it is in a gravitational field).

The distances you can give in cm, m, yards, or light years - as long as you use the same units for both (resulting in the same unitd for torque, i.e., Ncm, Nm, Nyards, or Nly).
 
Show us your diagram.
There was no need to convert degrees to radians..

What can you tell us about the sum of torques about the elbow?
 
Is the 90degrees relevant to the equation?
I looked over another, similar problem and then did this:
T = 9kg x 33cm = 88.29N x 0.33m = 29.13Nm
Then
29.13Nm / 0.03m = 971.19N

But the other problem didn't have a value for degrees.
 
An explanation that demonstrates you understand the problem is more important than getting the correct numerical answer. Suppose you made a minor error in the calculation? If you only give an incorrect numerical answer you will score zero. If you explain your working you might still get most of the marks.

If I was to be picky I would point out that..

9kg x 33cm = 88.29N x 0.33m

..is incorrect because the units don't balance.
 
DrowningKthnx said:
Is the 90degrees relevant to the equation?
Yes.
Could you answer the question if the arm was at 80 degrees?
If the arm was at 180 degrees what would the answer be? You should be able to answer that without doing any calculations.
 
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