Calculate Force on Small Gear from Torque with Pully and Shaft

[Physics_wiz]

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?

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?

 

[christinono]

Yeah, i think so

 

[thepatientmental]

Yes, you have correctly calculated the force on the tip of the small gear. Your method of using the torque equation (T=Fr) to find the force is correct. Since the small gear has a smaller radius than the pulley, it will experience a larger force to produce the same torque. This is because torque is the product of force and distance, and since the distance (radius) is smaller for the small gear, the force must be larger to achieve the same torque. Overall, your calculation is correct and you can use this method for similar problems in the future.