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Rotational Cross Product

  1. Apr 14, 2008 #1
    Force F = (9.0 N) i hat + (-4.0 N) k hat acts on a pebble with position vector r = (3.00 m) j hat + (-1.0 m) k hat, relative to the origin.


    (a) What is the resulting torque acting on the pebble about the origin?
    ( N·m) i hat + ( N·m) j hat + ( N·m) k hat

    (b) What is the resulting torque acting on the pebble about a point with coordinates (0.0 m, 0.0 m, 5.0 m)?
    ( N·m) i hat + ( N·m) j hat + ( N·m) k hat



    So all I could come up with was cross product was that torque = r cross F

    Therefore T = fRsin[tex]\theta[/tex]

    I looked up cross product on Wikipedia (not best site I know but w/e) and it came up with:

    i x j = k
    j x k = i
    k x i = j

    I tried those but it doesn't seem to be giving me the right answer. Any help please??



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 14, 2008 #2

    Mentz114

    User Avatar
    Gold Member

    [tex]F = \[ \left[ \begin{array}{c} 9 \\\ 0 \\\ -4 \end{array} \right]\][/tex]

    [tex]R = \[ \left[ \begin{array}{c} 0 \\\ 3 \\\ -1 \end{array} \right]\][/tex]

    [tex]F \times R = \[ \left[ \begin{array}{c} F_2R_3-F_3R_2 \\\ F_3R_1-F_1R_3 \\\ F_1R_2-F_2R_1 \end{array} \right]\][/tex]

    [tex]F \times R = \[ \left[ \begin{array}{c} -12 \\\ -9 \\\ -27 \end{array} \right]\][/tex]

    I did this with software. I haven't hand checked it.
     
    Last edited: Apr 14, 2008
  4. Apr 15, 2008 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Nanuven! :smile:

    Also ii = jj = kk = 0.

    Treat cross products just the same as any product.

    For example:

    (ai + bj)(cj + dk)

    = acij + adik + bcjj + bdjk

    = ack - adj + bdi. :smile:

    (btw, use the B tag to do bold … if that doesn't work, type [noparse] before and after.[/noparse])
     
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