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)

$-17 \hat{ i }+6 \hat{ j }+13 \hat{ k }$ N-m

Answer 2

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, $\vec{F}=2\hat{i}-3\hat{j}+4\,\hat{k},\,$
$r=3\,\hat{i}+2\hat{j}+3\hat{k}$
$\therefore$ $\tau =(2\,\hat{i}-3\hat{j}+4\hat{k})\times (3\,\hat{i}+2\hat{j}+3\hat{k})$
$\tau =\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 2 & -3 & 4 \\\ 3 & 2 & 3 \\\ \end{matrix} \right|$
$\Rightarrow$ $\tau =\hat{i}\,(-9-8)-\hat{j}\,(6-12)+\hat{k}\,(4+9)$
$\Rightarrow$ $\tau =-17\,\hat{i}+6\hat{j}+13\,\hat{k}$ .