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What is the net torque about the axle?

  1. Feb 14, 2009 #1

    a.a

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    1. The problem statement, all variables and given/known data

    The 19.6 cm diameter disk rotates on an axle through its center. F1=22.8 N, F2=32.5 N, F3=32.5 N, F4=22.8 N, and d=4.60 cm. What is the net torque about the axle?

    2. Relevant equations

    Torque = F r sin (angle)
    F = m g

    3. The attempt at a solution

    I found all the torques acting on the center, disregarded T1 because its sort of passing through the center and isnt really shown on the diagram.

    T1 = 0
    T2= (32.5)(0.196) = 6.37 Nm
    T3= -(32.5)(sin 45)(0.046)= -1.057 Nm (counterclockwise)
    T4= (22.8)(0.046)= -1.0488 (counterclockwise)

    Tnet = 4.26 Nm

    This isn't the right answer, can someone please help and explain where I am going wrong..
     

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    Last edited: Feb 14, 2009
  2. jcsd
  3. Feb 14, 2009 #2

    Hootenanny

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    Re: Torque

    For a standard coordinate axis, counter-clockwise is usually taken as positive.
     
  4. Feb 14, 2009 #3
    Re: Torque

    Re-check the calculation for T2 .
     
  5. Feb 14, 2009 #4

    Hootenanny

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    Re: Torque

    T2 looks fine to me (with the exception of the sign minus of course).
     
  6. Feb 14, 2009 #5
    Re: Torque

    Shouldn't he multiply the force with the radius (as opposed to the diameter)?
     
  7. Feb 14, 2009 #6

    Hootenanny

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    Re: Torque

    Of course he should. And I should pay more attention :redface:
     
  8. Feb 15, 2009 #7

    a.a

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    Re: Torque

    wow.. thanks.. that was a dumb mistake...
     
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