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Statics in 3D, there must a be faster way to do this.

  1. Mar 20, 2016 #1

    Given this, is there a faster way to compute this or must I do 3 separate 3x3 determinants?

    I can probably use cross product of each term, for example , (2rk) X (Axi) = (-2rAx j) (not forgetting the negative sign for j)

    next, (2rk) X (Ayj) = (2rAyi)

    and so on.... but that feels too slow as well and can get very messy.
    Last edited: Mar 20, 2016
  2. jcsd
  3. Mar 20, 2016 #2


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    Break up each of the forces into their x y and z components and sum them separately equal to 0. Then for the sum of moments equals 0 for each force component about an axis, it's

    [itex] M_x = F_yz + F_zy[/itex]

    M_y = F_xz + F_zx

    [itex]M_z = F_xy + F_yx [/itex]

    where x y and z are the perpendicular distances from the line of action of the component force to the axis about which moments are being summed.
    I don't know if that's easier for you, but it avoids determinants and vector math equations .
  4. Apr 21, 2016 #3
    Divide your 3D problems into 2D problems. Just remember, the forces and moments in a plane must be in equilibrium.
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