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Torque on a Pulley with two weights

  • Thread starter Quincy
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  • #1
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Homework Statement


Two weights (one is 125 N, other is 75 N) are connected on each side by a very light flexible cord that passes over a 40.0 N frictionless pulley of radius 0.300 m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the ceiling exert on the hook?


Homework Equations





The Attempt at a Solution


I'm not really sure about what the question is asking... wouldn't the the force on the hook be the tension in the hook, which would be equal to the weight of the pulley plus the 125 & 75 N weights?
 

Answers and Replies

  • #2
Doc Al
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I'm not really sure about what the question is asking... wouldn't the the force on the hook be the tension in the hook, which would be equal to the weight of the pulley plus the 125 & 75 N weights?
The weights of those hanging masses do not directly act on the pulley. What does act on the pulley is the tension in the cord. You'll need to solve for it.
 
  • #3
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The weights of those hanging masses do not directly act on the pulley. What does act on the pulley is the tension in the cord. You'll need to solve for it.
F1 = m1a = 125 N - T
F2 = m2a = -75 + T

-- If I solve for T, it gives 93.75 N, added that to 40 N of the pulley, gives 133.75 n, which isn't correct... what am i doing wrong?

Edit: I found the torque on the pulley from the tension to be 28.125 N*m... I'm guessing that has something to do with finding the right answer?
 
  • #4
Doc Al
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F1 = m1a = 125 N - T
F2 = m2a = -75 + T
Careful. Since the pulley has mass the tension in the cord will be different on each side of the pulley. You need three equations to solve for the tensions. (Write a torque equation for the pulley.)
 
  • #5
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Thanks, got it now.

Last one: A block with mass m = 5.00 kg slides down a surface inclined 36.9o to the horizontal. The coefficient of kinetic friction is 0.23. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 29.0 kg and moment of inertia 0.500 kg*m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.400 m from that axis. What is the acceleration of the block down the plane?

Torque = I*(alpha) = I*(a/r) = F*r = (0.5)(a/.1313) = (.1313)(mgsin(36.9) - 0.23mgcos(36.9))

--> a = 0.704, which is not correct; what am i doing wrong? Isn't the tangential acceleration of the pulley the same as the acceleration of the block?
 
  • #6
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bump, it's due pretty soon :S
 
  • #7
Doc Al
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Torque = I*(alpha) = I*(a/r) = F*r = (0.5)(a/.1313) = (.1313)(mgsin(36.9) - 0.23mgcos(36.9))
I don't quite understand where that equation came from.

Show me two equations. One for the mass; one for the flywheel. Then combine them to solve for the acceleration.

Isn't the tangential acceleration of the pulley the same as the acceleration of the block?
Sure.
 

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