Practicing Success
The moment of the force, F = 4 i + 5 j - 6 k at (2, 0, –3), about the point (2, –2, –2), is given by : |
- 8 i - 4 j - 7 k - 4 i - j - 8 k - 7 i - 4 j - 8 k - 7 i - 8 j - 4 k |
- 7 i - 4 j - 8 k |
$\vec{\tau} = (\vec{r} - \vec{r_0}) x \vec{F}$ (r - ro) = (2 i - 3 k) - (2 i - 2 j - 2 k) = 2 j - k $\vec{\tau}$ = (2 j - k) $\times$ (4 i + 5 j - 6 k) = - 7 i - 4 j - 8 k |