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A force F is exerted on the top right corner of a rectangular plate at

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A force F is exerted on the top right corner of a rectangular plate at an angle 60o
    above the horizontal, as shown below. The magnitude of the torque of the force F
    about point A, the lower left corner is given by

    See picture please : http://tinypic.com/view.php?pic=2njv9xt&s=6


    2. Relevant equations

    Both y and x are being pulled upwards, so that is counterclockwise..
    but answer is Fx sin 60 + Fy cos 60
    I dont understand torque at all.. can someone please help.


    3. The attempt at a solution

    No idea..
     
  2. jcsd
  3. Apr 16, 2013 #2

    CWatters

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    Hint:
     

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  4. Apr 16, 2013 #3
    Thank you!
    but I'm still kinda lost...
     
    Last edited: Apr 16, 2013
  5. Apr 16, 2013 #4
    one question, where do i put the angle in?
     
  6. Apr 16, 2013 #5

    CWatters

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    I'll do a new drawing as I don't think my first was very helpful. Back soon.
     
  7. Apr 16, 2013 #6
    ok thanks, check ur msg too please , thanks
     
  8. Apr 16, 2013 #7

    CWatters

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    Try this drawing. The idea is to split the force F into two component forces. Then multiply each component force by it's distance from the pivot to give two torques (Torque = force * distance). Finally you simply add the two torques.
     

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  9. Apr 16, 2013 #8
    this is the drawing i had, but from this picture, it shows that it is Fy sin 60 and Fx cos 60 , but the answer is opposite.
     
  10. Apr 16, 2013 #9

    CWatters

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    Seems ok to me.

    The two components are

    F*Cos(60)
    F*Sin(60)

    Multiply by their distance from the pivot..

    y*F*Cos(60)
    x*F*Sin(60)

    then add.

    x*F*Sin(60) + y*F*Cos(60)

    Thats' the answer you gave in your first post.
     
  11. Apr 16, 2013 #10
    it still seems to be Fx cos 60 + Fy sin 60 to me :/
     
  12. Apr 16, 2013 #11

    haruspex

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    You need to multiply the x by the component of F that's perpendicular to it, so that's the Y component, F sin(60o).
    However, none of the offered answers is correct. The X and Y components of F act in opposite sense around A, so there should be a minus sign somewhere. E.g. consider the case where y = x tan(60o). There should be no moment.
     
  13. Apr 17, 2013 #12

    CWatters

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    Oops I missed that.

    So answer should be either...

    x*F*Sin(60) - y*F*Cos(60)

    or

    y*F*Cos(60) - x*F*Sin(60)
     

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    Last edited: Apr 17, 2013
  14. Apr 17, 2013 #13
    Thank you! makes sense now
     
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