Physics_wiz
Apr17-05, 07:01 PM
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?
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?