ssa915

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Heres a question, and I'm having trouble understanding why the explanation uses 'sin' and not 'cos'

A uniform rod of mass M sticks out from a vertical wall and points toward the floor. If the smaller angle it makes with the wall is theta, and its far end is attached to the ceiling by a string parallel to the wall, find the tension in the supporting string.

Explanation: LTsin(theta) = (L/2)mgsin(theta)

where L = length of rod, T = tension in string, m = mass of rod

I understand that the torques cancel out since we are in equilibrium, but shouldnt we use cosine here, not sine, since cosine would give us the vertical components of the torque due to TENSION and WEIGHT OF ROD respectively?

Thanks!
 

kaml20

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Mar 13, 2005
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your just missing a simple concept ... lever arm

you want to have torque up
(the length from the wall to the end of the rod where force is being applied) * (tension)
L sin theta * T
and set that equal to the torque down
(the length from the wall to the center of mass of the rod) * (force down)
L/2 sin theta * MG

L * sin theta * T = sin theta * .5L * M * G
 
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ssa915

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5+ Year Member
15+ Year Member
Apr 7, 2004
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0
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kaml20 said:
your just missing a simple concept ... lever arm

you want to have torque up
(the length from the wall to the end of the rod where force is being applied) * (tension)
L sin theta * T
and set that equal to the torque down
(the length from the wall to the center of mass of the rod) * (force down)
L/2 sin theta * MG

L * sin theta * T = sin theta * .5L * M * G
Hey thanks so much...that cleared it up, silly mistake on my part
 
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