Question

The torque of force $\vec{F}=(2\hat{i}-3\hat{j}+4\hat{k})N$ acting at a point $\vec{r}=(3\hat{i}+2\hat{j}+3\hat{k})m$ about origin is

• A. $6\hat{i}-6\hat{j}+12\hat{k}N-m$
• B. $-6\hat{i}+6\hat{j}-12\hat{k}N-m$
• C. $17\hat{i}-6\hat{j}-13\hat{k}N-m$
• D. $-17\hat{i}+6\hat{j}+13\hat{k}$ N-m

Answer 1

Answer: (D)

The moment of force or torque about an axis is equal to the vector product of force $(F)$ and perpendicular distance of line of action of force from the axis of rotation $(r)$
$\therefore$ $\tau =F\times r$
Given, $\vec{F}=2\hat{i}-3\hat{j}+4\hat{k},$ $r=3\hat{i}+2\hat{j}+3\hat{k}$
$\therefore$ $\tau =(2\hat{i}-3\hat{j}+4\hat{k})\times (3\hat{i}+2\hat{j}+3\hat{k})$
$\tau =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 4\\ 3& 2& 3\end{array}\right|$
$\Rightarrow$ $\tau =\hat{i}(-9-8)-\hat{j}(6-12)+\hat{k}(4+9)$
$\Rightarrow$ $\tau =-17\hat{i}+6\hat{j}+13\hat{k}$ .