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Calculate the net torque about the axle of the wheel

  1. Oct 29, 2005 #1
    This problem is due on my webassign homework in 3 hours and I can't get it right!!

    Calculate the net torque about the axle of the wheel shown in Fig. 8-39. Assume that a friction torque of 0.43 mN opposes the motion and that F = 26.

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

    What I did:

    The 35 N force will move the wheel clockwise so the torque will be -

    28 N force will move it counterclockwise so torque will be +

    26 N force will move it clockwise so force will be -

    They forces are all perpendicular to the radius so its just T = r X F

    35 N force T = 35 N X .12 m = -4.2 (moves cw)

    28 N force T = 28 X .24 m = 6.72 (moves ccw)

    26 N force T = 26 X .24 m = -6.24 (moves cw)

    So then including the friction torque which opposes the motion of .43 I put

    6.72 - 4.2 - 6.24 - .43 = -4.15 is the net torque and because its - it movies in the clockwise direction

    Web assign says -4.15 is WRONG! What am I doing wrong, please help!
     
  2. jcsd
  3. Oct 29, 2005 #2

    lightgrav

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    You haven't said yet which direction (cw , ccw) the disk is rotating...
    if it was already rotating ccw, the friction torque would be cw.
    have you tried that?
     
  4. Oct 29, 2005 #3
    It says 0.43 mN, which isnt the same as 0.43N
     
  5. Oct 29, 2005 #4
    Ohhhh...so if the motion is going clockwise than the friction will be counter clockwise so it would be +.43 instead of -.43?

    6.72 - 4.2 - 6.24 + .43 = -3.29

    is that right?? ( I only have one guess left on my webassign and I dont want to screw it up, lol)
     
  6. Oct 29, 2005 #5

    lightgrav

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    Well, I don't know the whole scenario of the problem, but ...
    the 3 forces in the diagram MIGHT be slowing the spin of the disk.
    There should be some indication of the spin, in the problem text.

    I expect your .43 mN means meter x Newton (not milliNewton)
    since it is a torque - unless it is 43 milliN applied at the .12m hub!
     
  7. Oct 29, 2005 #6
    Seems right.
     
  8. Oct 29, 2005 #7
    Ok I just guessed and -3.29 is right. That was stupid of me not to catch that the friction is positive. Thanks for the help!
     
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