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A force of 1200 KN acts along PQ, P(4,5,-2)m & Q(-3,1,6)m. Calculate its moment about a point A(3,2,0)m
1 Answer
written 8.4 years ago by |
Firstly finding force in vector form
$F = F . rPQ = 1200[(-7i - 4j+8k)/[(\sqrt{7^2+4^2+8^2})] = 739.58 i - 422.62 j - 845.23 k KN $
Now, Finding moment of the force about point $A(3,2,0) $
$MFS = rSP x F .........where\space \space rAP = i + 3 j - 2 k m $
$$MFA= \begin {matrix} i&j&k\\ 1&3&-2 \\ 739.58&-422.62&-845.23 \end{matrix}$$
$MFA = - 3380.93 i - 633.93 j - 2641.36 k KNm $
Moment of force about point is $- 3380.93 i - 633.93 j - 2641.36 k KNm $