Zero Torque and Static Equilibrium pulley problem

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SUMMARY

The discussion focuses on calculating the frictional torque required for a pulley system to achieve static equilibrium with a 0.301 kg mass on one side and a 0.635 kg mass on the other. The pulley has a radius of 9.50 cm and experiences friction in its axle. It is established that the tensions in the string will not be equal due to the differing weights, and the total torque must equal zero. The relevant equation for torque is identified as torque = Fr sin(theta), where F represents the force applied at the radius of the pulley.

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just.karl
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A string that passes over a pulley has a .301kg mass attached to one end and a 0.635 kg mass attached to the other end. The pulley, which is a disk of radius 9.50cm, has friction in its axle. What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium?

I know I have to figure out the torque required by the axle to get both tensions equal. But I don't know what equations you use to figure out the torque with two masses on each side.


Thanks
 
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just.karl said:
I know I have to figure out the torque required by the axle to get both tensions equal.

No, the tensions will not be equal. total torque will be zero. Two different tensions will be applied to the wheel at the given radius. The angles are assumed to be 90 degrees, since both masses will hang down (safe assumption).

torque = Fr sin(theta)
 

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