A pully with radius .2 m is fixed on a shaft and on the other end there's a gear with radius .01 m. If a force of .01 N pulls down on the pully, what's the force on the tip of the small gear?(adsbygoogle = window.adsbygoogle || []).push({});

Here's my work:

The force produces a torque about the shaft T = Fr = (.01 N)(.2 m)= .002 Nm

The small gear is subjected to the same torque so T = Fr again,

.002 Nm = (F2) (.01 m)

F2 = .2 N

I know this is a simple problem, but I have to make sure I did it right because it's part of a bigger problem. So, did I do it right?

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# Homework Help: Torque calculations

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