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Torque using i, j & k-hat

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A wheel of diameter 34.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force vector F = (-30.0 i-hat + 39.6 j-hat) N acts at a point on the edge of the wheel that lies exactly on the x axis at a particular instant. What is the torque about the rotation axis at this instant?


    2. Relevant equations
    Torque=R*F


    3. The attempt at a solution
    Tx= 0
    Ty= 0
    Tz= 6.732
    *EDIT* SOLVED *EDIT*
     
    Last edited: Mar 8, 2009
  2. jcsd
  3. Mar 8, 2009 #2
    Hi, Stryder_SW!

    What component of the force creates the torque? I can tell you that it's not the vector given.
     
    Last edited: Mar 8, 2009
  4. Mar 8, 2009 #3

    tiny-tim

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    Hi Stryder_SW! :smile:
    Nooo! :cry:

    torque = r x F :smile:
     
  5. Mar 8, 2009 #4
    You have written Torque = R*F but it should be T = R x F. This is important.
     
  6. Mar 8, 2009 #5
    Um...what part of that is important?
     
  7. Mar 8, 2009 #6

    tiny-tim

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    Dr.D :smile: and I may be wrong, but we think you're trying to multiply F by the number r,

    instead of cross-producting it with the vector r :wink:
     
  8. Mar 8, 2009 #7
    The x is important.
     
  9. Mar 8, 2009 #8
    No, you're right. by cross-producting it do you mean like dot product? and even if you did this how is the torque in the Tz/k-hat direction not zero when there is no force applied in that direction
     
  10. Mar 8, 2009 #9
    Most specifically the cross product is not a dot product. A cross product is a vector, perpendicular to the two original vectors, in the direction given by rotating the first vector to the second vector and applying the right hand rule. The magnitude of the cross product is mag(r x F) = r*F*sin(r,F).
     
  11. Mar 8, 2009 #10

    tiny-tim

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    still can't tell whether you've got it :redface:

    dot product is almost the opposite

    have a look at the PF library entry on cross product :wink:

    EDIT: oooh, the doctor beat me this time! :biggrin:
     
  12. Mar 8, 2009 #11
    Well I've never seen cross-product before, and the PF library explanation...kinda confuses me. So I looked it up in my physics book, apparently its not discussed until the chapter AFTER the one that this homework is in. I'm gonna take a look at the book now and hopefully I'll get somewhere.
     
  13. Mar 8, 2009 #12
    Awesome, the explanation in my book was surprisingly good. Got the problem using the cross-product. And I think I get why that works, anyways thank you very much for your help.
     
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