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Torque and vector product

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data
    find the net torque on the wheel about the axle through O, taking a = 7.00cm and b = 20.00cm. Assume the positive direction is counterclockwise


    2. Relevant equations
    torque = force x distance
    torque = force x (radius x sin(phi))


    3. The attempt at a solution
    10.0 x .2m = -2 (because it's moving clockwise)
    9.00 x .2m = -1.8
    .07 x 12 x sin(30) = .42

    net torque = -3.38 (this is off by 10%-100%)
     

    Attached Files:

  2. jcsd
  3. Apr 25, 2009 #2

    Doc Al

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    Staff: Mentor

    I don't understand what the 30 degree angle represents.
     
  4. Apr 25, 2009 #3
    Me niether, do you think it was put there to confuse?
     
  5. Apr 25, 2009 #4

    Doc Al

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    Staff: Mentor

    Beats me. It looks like the 12 N force is tangential to the circle of radius a. Is that the case?
     
  6. Apr 25, 2009 #5
    It is there to confuse. The net torque turned out to be -2.96 so the 12N force is tangential. Thanks for bringing that up to attention
     
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