Question

Two forces $\overrightarrow{F_1}=2\hat{i}-5\hat{j}-6\hat{k}$ and $\overrightarrow{F_2}=-\hat{i}+2\hat{j}-\hat{k}$ are acting on a body at a point $(1,1,0)$ and $(0,1,2)$ respectively. Find the torque acting on the body about the point $(-1,0,1)$.

• A. $-14\hat{i}+10\hat{j}-9\hat{k}$
• B. $-11\hat{i}+10\hat{j}-12\hat{k}$
• C. $-3\hat{i}+3\hat{k}$
• D. $-14\hat{i}-10\hat{j}-9\hat{k}$

Answer 1

The correct option is A $-14\hat{i}+10\hat{j}-9\hat{k}$


From the diagram,
$\overrightarrow{r_1}=(1-(-1\left)\right)\hat{i}+(1-0)\hat{j}+(0-1)\hat{k}$
$\overrightarrow{r_1}=2\hat{i}+\hat{j}-\hat{k}........\left(I\right)$

$\overrightarrow{r_2}=(0-(-1\left)\right)\hat{i}+(1-0)\hat{j}+(2-1)\hat{k}$
$\overrightarrow{r_2}=\hat{i}+\hat{j}+\hat{k}.....\left(II\right)$

As we know, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
$\overrightarrow{{\tau }_1}=\overrightarrow{r_1}\times \overrightarrow{F_1}$
$\overrightarrow{{\tau }_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 2& -5& -6\end{array}\right|$
$\overrightarrow{{\tau }_1}=i(-6-5)-j(-12+2)+\hat{k}(-10-2)$
$=-11\hat{i}+10\hat{j}-12\hat{k}$
And
$\overrightarrow{{\tau }_2}=\overrightarrow{r_2}\times \overrightarrow{F_2}$
$\overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ -1& 2& -1\end{array}\right|$
$\overrightarrow{{\tau }_2}=\hat{i}(-1-2)-\hat{j}(-1+1)+\hat{k}(2+1)$
$\overrightarrow{{\tau }_2}=-3\hat{i}+3\hat{k}$

Therefore, net torque about point $(-1,0,1)$ is
$\overrightarrow{\tau }=\overrightarrow{{\tau }_1}+\overrightarrow{{\tau }_2}$
$\overrightarrow{\tau }=-11\hat{i}+10\hat{j}-12\hat{k}-3\hat{i}+3\hat{k}$
$\overrightarrow{\tau }=-14\hat{i}+10\hat{j}-9\hat{k}$