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Torque problem I think

  • Thread starter marcia888
  • Start date
  • #1
4
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7. Below is a diagram of a baseball/forearm system at rest. It is acted on by four different forces: the weight of the forearm, the weight of the baseball, the bicep force, and a force from the upper arm bone (attached at the elbow).
(Use the following values: L = 14 cm, d = 2 cm, M = 3 kg, and m = 2 kg.)

How large is the force exerted by the bicep?

Diagram:
http://www.webassign.net/userimages/ikoskelo@sfsu/bicep.jpg

I'm thinking that the torque of the forearm + the torque of the baseball combined are going to equal the negative of the torque provided by the bicep. Is this the right way to do this? And if so, I know how to figure out the torque from the baseball because I know it's distance. But how about the torque of the forearm? What is its distance? Can I just use average distance by dividing it by 2? :cry:

Thank you.
 

Answers and Replies

  • #2
4
0
Forget. I got it. 240.1 N

Used 1/2 distance for distance of forarm

mg L/2 + mgL = total torque down

torque up is the same. f=torque / distance , f=240.1

Thanks anyway!!
 

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