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Finding moment

  1. May 14, 2013 #1
    I just need help on 39 part a.
    The answer to part a is 84sinθ
    I first converted inches into feet.
    I know how to set up the vector for the wrench which is simply OA= [itex]\frac{3\sqrt{3}}{4}[/itex]i+[itex]\frac{3}{4}[/itex]j
    now as for the force vector I am kind of stuck because I know that Fsinθ is the only component to the force to cause rotation about O. so for F I get 0i and -56sinθ, when I do the cross product i get 42[itex]\sqrt{3}[/itex]sinθ. I know i have the force part wrong I just cant see it.

    [URL=http://s1341.photobucket.com/user/nebula-314/media/20130514_192552_zpsd6a49a82.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20130514_192552_zpsd6a49a82.jpg[/URL][/PLAIN]
    20130514_192628_zps1dc1e90a.jpg "] 20130514_192628_zps1dc1e90a.jpg
     
  2. jcsd
  3. May 14, 2013 #2

    rock.freak667

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    You can evaluate | OA x F| as this is equal to

    |OA x F| = |OA|*|F|*sinθ


    Otherwise, I think you may have made a mistake in the vector for OA.
     
  4. May 14, 2013 #3
    Yup that worked, with my OA.

    I really wanted to know how to set up the cross product on this one, but that worked.


    Thanks,
     
  5. May 15, 2013 #4

    rock.freak667

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    In that case, the reason you did not get the 84sinθ was due to the fact that you resolved your vectors in different directions (meaning that you used different coordinates).

    As the diagram was |OA| = 1.5 ft and your unit vector would have been (sinθi+cosθj). Multiplying these two would have given your vector OA

    Your vector F would have simply been F=-56j
     
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