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Moment Of A Force

  1. Sep 9, 2005 #1
    hi, i have tried to work this question out it seems so easy but i must be missing something.

    determine the magnitude and directional sense of the moment of the force at A about point O.
    [​IMG]


    these are 3 of my calculations for the question

    F = 520 sin 30(6) - 520 cos 30(0)
    F = 1560N or 1.56kN
    also

    F = 520 sin 120(6) - 0
    F = 2.7kN

    i also tried using vectors

    r = {6i- 0j}m
    F = {520sin30i - 520cos30j}N
    = {260i - 450.33j}
    The moment is therefore

    Mo = r x F = 0i - 0j + [6(-450.33)-(0)(200)]k
    = {-2.7k} kN.m

    the answer in the books telling me 2.88 kN.m can someone point me to where im going wrong i have obviously misunderstood something

    thanks
     
    Last edited: Sep 9, 2005
  2. jcsd
  3. Sep 10, 2005 #2

    Pyrrhus

    User Avatar
    Homework Helper

    Your thinking is right, check the numbers and maybe the signs.

    [tex] \vec{M}_{o} = (6 \vec{i}) \times (-520 \frac{5}{13} \vec{i} + 520 \frac{12}{13} \vec{j}) [/tex]
     
  4. Sep 10, 2005 #3
    thanks Cyclovenom
     
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