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Old Jan 19th 2010, 08:25 AM   #1
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Red face Apllied Mechanics problem!

Hello! i have an applied mechanics problem. dont know how to start off.. or do any other bit for that matter. Can anybody here help me? This one is beyond me =)


The figure represents the slow flexion and extension movements of the leg and thigh. the length of the leg is L=0,4m, and the weight of the leg is m=4.8Kg, and the thigh m=9,8Kg. We´re also given the distances to the center of mass of the thigh (B) and leg (D) from the extremities. ( AB= CD= 0.43L ).
Determine the force F in function to angle alfa, and the strength of the joint connections of the hip(A), the knee(C) and the ankle(E).

EDIT: figure is attached
Attempt:
AB=CD=0,172m
force apllied at c? m.g.sin -alfa?
F is going to be the opposite force to the angular velocity at C?
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Apllied Mechanics problem!-body.jpg  

Last edited by Corum; Jan 19th 2010 at 09:04 AM.
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Old Jan 19th 2010, 09:35 AM   #2
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Sorry to bump this, but it's a little bit urgent... i would greatly appreciate some assistance
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Old Jan 19th 2010, 08:02 PM   #3
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Originally Posted by Corum View Post
Sorry to bump this, but it's a little bit urgent... i would greatly appreciate some assistance
Try solving it like this; determine the potential energy. U(s), of the legs as a function of the displacement, s, of the feet. Then take the negative of the derivative and that will give you F. I.e. F = -dU/ds.
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Old Jan 20th 2010, 05:03 AM   #4
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hi everybody

I am interested in question but not able to find out answer. Could anybody help me in finding out the solution? I am not able to find out the potential energy of leg (center of mass, D) as a function of alpha.

Please help me!!
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Old Jan 20th 2010, 06:14 AM   #5
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Originally Posted by vyomictor View Post
hi everybody

I am interested in question but not able to find out answer. Could anybody help me in finding out the solution? I am not able to find out the potential energy of leg (center of mass, D) as a function of alpha.

Please help me!!
You need to know how the center of mass of each part of the leg. Then the potential energy of the leg equals the height of that point above the zero reference for potential energy.
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