Question

Force $\overrightarrow{F_1}=(2\hat{i}–3\hat{j})\text{N}$ and $\overrightarrow{F_2}=(-5\hat{i}+2\hat{j})\text{N}$ act through the points whose position vectors are $\hat{k}$ and $3\hat{k}$ respectively. The vector sum of the torque due to these forces is

• A. $3\hat{i}+13\hat{j}\text{Nm}$
• B. $-3\hat{i}-13\hat{j}\text{Nm}$
• C. $-13\hat{i}-3\hat{j}\text{Nm}$
• D. $-3\hat{i}+13\hat{j}\text{Nm}$

Answer 1

The correct option is B $-3\hat{i}-13\hat{j}\text{Nm}$

Given forces,
$\overrightarrow{F_1}=(2\hat{i}–3\hat{j})\text{N}$ and $\overrightarrow{F_2}=(-5\hat{i}+2\hat{j})\text{N}$
Position vector are $\overrightarrow{r_1}=\hat{k}\text{m}$ and $\overrightarrow{r_2}=3\hat{k}\text{m}$
$\therefore$ Vector sum of the torque is given by
$\overrightarrow{\tau }={\overrightarrow{r}}_1\times \overrightarrow{F_1}+{\overrightarrow{r}}_2\times \overrightarrow{F_2}$
$\Rightarrow \overrightarrow{\tau }=\hat{k}\times (2\hat{i}–3\hat{j})+3\hat{k}\times (-5\hat{i}+2\hat{j})$
$\therefore \overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& 1\\ 2& -3& 0\end{array}\right|+\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& 3\\ -5& 2& 0\end{array}\right|$
$\Rightarrow \overrightarrow{\tau }=(3\hat{i}+2\hat{j})+(-6\hat{i}-15\hat{j})$
$\Rightarrow \overrightarrow{\tau }=(-3\hat{i}-13\hat{j})\text{Nm}$