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Another reduction to Wrench

  1. Jan 7, 2008 #1
    This one is really discouraging me. It looks sooooo easy since the forces are each in one direction

    [​IMG]

    The way I have usually handled these as a procedure is: Find [itex]F_r[/itex] and then find [itex]M_r[/itex] about some point and then decompose [itex]M_r[/itex] into components that run parellel and perpendicular to [itex]F_r[/itex]. Then I can usually find P(x,y)

    If I were to move everything to point A I would have:
    [itex]F_r=500i+300j+800k[/itex]
    And I would also have to find the couple Moments about A:
    [itex]M_x_A=4(800)=3200[/itex]
    [itex]M_y_A=0[/itex]
    [itex]M_z_A=6(300)=1800[/itex]

    I am just unsure where to go from here? Or is this all wrong altogether?
     
    Last edited: Jan 7, 2008
  2. jcsd
  3. Jan 7, 2008 #2
    I have done this one over soooo many times. I am clearly missing a crucial conceptual point. ANY hints or criticism would help.
     
  4. Jan 8, 2008 #3
    Well Good Morning!
     
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