Question

Moment of the force, $\overrightarrow{F}=(4\hat{i}+5\hat{j}-6\hat{k})\text{N}$ at $P(2,0,-3)\text{m}$ about the point $Q(2,-2,-2)\text{m}$ is given by

• A. $(-7\hat{i}-8\hat{j}-4\hat{k})\text{N.m}$
• B. $(-4\hat{i}-\hat{j}-8\hat{k})\text{N.m}$
• C. $(-8\hat{i}-4\hat{j}-7\hat{k})\text{N.m}$
• D. $(-7\hat{i}-4\hat{j}-8\hat{k})\text{N.m}$

Answer 1

The correct option is D $(-7\hat{i}-4\hat{j}-8\hat{k})\text{N.m}$

Moment of the force $\overrightarrow{F}$ is torque produced due to force given as,
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}...\left(i\right)$


$\overrightarrow{F}=4\hat{i}+5\hat{j}-6\hat{k}$
Position vector$\left(\overrightarrow{r}\right)$ of $P$ with respect to reference point $Q$ is,
$\overrightarrow{r}=(2\hat{i}+0\hat{j}-3\hat{k})-(2\hat{i}-2\hat{j}-2\hat{k})$
$\therefore \overrightarrow{r}=0\hat{i}+2\hat{j}-1\hat{k}$
Substituting in Eq $\left(i\right)$,
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& -1\\ 4& 5& -6\end{array}\right|$
$\overrightarrow{\tau }=\hat{i}(-12-(-5\left)\right)-\hat{j}(0-(-4\left)\right)+\hat{k}(0-8)$
$\therefore \overrightarrow{\tau }=-7\hat{i}-4\hat{j}-8\hat{k}$
The moment of the force $\overrightarrow{F}$ about point $Q$ at point $P$ is $(-7\hat{i}-4\hat{j}-8\hat{k})\text{N.m}$