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Rotational problems

  1. Mar 13, 2009 #1
    1. The problem statement, all variables and given/known data

    1) A force F = 1.5 i + 2.4 j N is applied at the point x = 3.0 m, y = 0 m. Find the torque about the point x = -1.3 m, y = 2.4 m.

    I believe the lever arm should be the distance between the two points. but then the force that acts perpendicular to this is where i cannot get the right answer. How do i do this problem?
     
  2. jcsd
  3. Mar 13, 2009 #2

    Doc Al

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    Use the cross product definition of torque:

    [tex]\tau = \vec{r} \times \vec{F}[/tex]
     
  4. Mar 13, 2009 #3
    ok so F = 1.5 i + 2.4 j and x = -1.3 m, y = 2.4
    i get -3.6 K and -1.95K so is it.. -5.55 k?
     
  5. Mar 14, 2009 #4

    Doc Al

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    OK.
    No, that's just the point about which you are finding the torque. [itex]\vec{r}[/itex] is the displacement from that point: r = (x2 - x1) i + (y2 - y1) j
     
  6. Mar 14, 2009 #5
    F = 1.5 i + 2.4 j N is applied at the point x = 3.0 m, y = 0 m.
    x = -1.3 m, y = 2.4

    so the r displacement would be 3-(-1.3) i + 0-2.4 j
    so r would be sqrt( 4.3^2 + 2.4^2) ?
     
  7. Mar 14, 2009 #6

    Doc Al

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    Right.
    That would be the magnitude of r.
     
  8. Mar 14, 2009 #7
    ok so since i have 4.3 i - 2.4 j how do i begin to find the torque and such?
     
  9. Mar 14, 2009 #8

    Doc Al

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    Take the cross-product, as I indicated in post #2.
     
  10. Mar 15, 2009 #9
    so i get 13.92 k is that right?
     
  11. Mar 15, 2009 #10

    Doc Al

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    Sounds good to me. (Don't neglect the units.)
     
  12. Mar 15, 2009 #11
    ok thanks a lot!

    i ask if it was right because i am unable to check my answer.
     
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