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Torque calculations

  1. Apr 17, 2005 #1
    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?
     
  2. jcsd
  3. Apr 17, 2005 #2
    Yeah, i think so
     
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