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Reduction to a Wrench

  1. Jan 6, 2008 #1
    [SOLVED] Reduction to a Wrench

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    So far I have moved the forces to point O where [itex]\sum F=F_r=-10.0\hat{j}[/itex] I have also re-written each force in Cartesian-vector form where

    [itex]F_1=10[\frac{6i-6j}{\sqrt{72}}=7.071i-7.071j[/itex] [itex]F_2=-10j[/itex] and [itex]F_3=-F_1=-7.071i+7.071j[/itex].

    I am told to decompose each force into its components and then use scalar method to find the sum (which I did above) and to find the sum of the moments about each of the x,y,z axes. Then move the forces to get a wrench.

    I have found each of the moments as follows:
    [tex]\sum M_x=2(10)-2(7.071)=5.858[/tex]
    [tex]\sum M_y=-6(7.071)=-42.426[/tex]
    [tex]\sum M_z=6(7.071)-6(7.071)= 0[/tex]

    Now I am a little confused. I am just not sure where to go from here. I know I need to find M-perp and M-parellel which looks to be what I have just found.

    Are my moments correct? I think they are. And where do I go from here?
     
  2. jcsd
  3. Jan 6, 2008 #2
    I GOT IT! Amidst all my typing I noticed that my moment about y axis is incorrect.

    M_x=5.86=M_perp--->distance from z =5.86/10=.586 ft.
    M_y=-14.1=M_parellel

    YAY!
     
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