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Net torque

  1. Jan 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the net torque about the axle of the wheel shown in https://jigsaw.vitalsource.com/books/9781269392464/content/id/ch08fig39 [Broken]. Assume that a friction torque of 0.40 m · N opposes the motion. Where F=18N (in photo).

    The picture is here: http://www.webassign.net/giancoli/8-39alt.gif

    2. Relevant equations
    τ (tau) = rF

    3. The attempt at a solution
    Direction of the positive torque: counterclockwise direction

    T1 = 28N*24cm = 6.72 N-m

    T2 = -35N*12cm = -4.2 N-m

    T3 = -18N*24cm = -4.32 N-m

    Now you add all these torques

    6.72 N-m -4.2 N-m -4.32 N-m = -1.8 N-m

    Since I know the direction of rotation is clockwise, I added the friction torque to that result in the opposite direction:

    -1.8 N-m +0.40 N-m = -1.4 N-m (clockwise)

    This is the only way I was able to get the right answer. However, my first try was using -35Nsin135. Why do I not have to take the component of F in this example?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 18, 2016 #2

    Suraj M

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    Aren't all the forces in the diagram, tangential?
    And how did you arrive at -35Sin135 as the component?
     
  4. Jan 18, 2016 #3
    That's where I was confused. I assumed it wouldn't be tangential because it was on the inside of the wheel.
     
  5. Jan 18, 2016 #4

    Suraj M

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    Inside of the wheel? I don't really think that matters, it still has to be tangential, even if it's not, again, how did you arrive at -35 sin 135?
     
  6. Jan 18, 2016 #5
    I was thinking that the F of 35N would have to be resolved into components and the only one being effective in producing rotational motion would be F sin θ. I'm pretty sure I'm just mixing up the theory behind torque.
     
  7. Jan 18, 2016 #6

    Suraj M

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    That ##\theta## in ## \sin (\theta) ## is the angle which the force makes with the radius, in this case I don't thing 135 is that angle, hence I feel we should take it as a tangential force.
     
  8. Jan 18, 2016 #7
    I definitely agree with that. I just have to revisit some of the theory so I'm not confusing concepts.
     
  9. Jan 18, 2016 #8

    haruspex

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