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Force Γ=80i+50j60k passes through a point A (6, 2, 6). Compute its moment about a point B(8,1,4).
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Given :- F=80i+50j60krA=6i+2j+6krB=8i+j+4k

To find:-

Moment about B(MB)

Solution:-

Position vector of A w.r.t B is

$\vec{r_{AB} }= \vec{r_A } - \vec{r_B }\\ =(6i+2j+6k)-(8i+j+4k) \\ \vec{r_{AB}} = -2i+j+2k \\ Now, \\ …

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