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Need help with 3D system with applied forces and moment

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    The system is a rectangular box with dimensions of 3 ft x 4 ft x 2 ft. There's an applied couple moment of 600 lb-ft on the x-z plane rotating positively. The moment is 2 ft above the x-y plane in the positive z direction.

    Applied force 1 is 450 lbs in the negative x direction with a position vector of (3,4,2). Applied force 2 is 600 lb in the positive y direction with a position vector of (0,4,2). Applied force 3 is 300 lb in the positive z direction with a position vector of (3,4,0).

    I'm being asked to find the magnitude of the resultant force and couple moment of a wrench and where the line of action intersects the x-y plane.

    2. Relevant equations

    M = F × r

    3. The attempt at a solution

    ƩF = (-450, 600, 300) lb
    |F| = (4522+6002+3002)1/2 = 808 lb

    directions are based on RHR
    ƩMx = 300 lb * 4 ft - 600 lb * 2 ft + 600 lb-ft = 600 lb-ft
    ƩMy = -450 lb * 2 ft - 300 lb * 3 ft = -1800 lb-ft
    ƩMz = 450lb * 4ft = 1800 lb-ft

    I know to get where the line of action intersects the x-y plane I need to cross (x,y,0) with the Fr vector and set that equal to Mr then just solve for x and y but my math is all wrong with it.

    Fr×r=Mr
    (300y)i - (300x)j + (600x-(-450y))k = 600i - 1800j + 1800k

    that gives me 300y = 600 so y = 2, -300x = -1800 so x = 6, but 600 * 6 + 450 * 2 ≠ 1800

    I'm guessing I'm screwing up on the moments part but even if I separate the couple moment into two equal and opposite vectors that would cause a rotation there I still get all these numbers. I even tried simplifying all vectors and moments to one point but still I'm getting these numbers.

    Not looking for the answer just some help seeing the forest through the trees.
     
  2. jcsd
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