Question

Moment of a force of magnitude $20N$ acting along positive $x$ direction at point $(3m,0,0)$ about the point $(0,2,0)\left(inNm\right)$ is:

• A. $20$
• B. $60$
• C. $40$
• D. $30$

Answer 1

The correct option is C $40$

According to the definition of torque,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Given that,the force is

$\overrightarrow{F}=20\hat{i}N$

and the arm vector is

$\overrightarrow{r}=(0-3)\hat{i}+(2-0)\hat{j}+(0-0)\hat{k}$

$\overrightarrow{r}=(-3\hat{i}+2\hat{j})m$

Therefore,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

$\overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& 2& 0\\ 20& 0& 0\end{array}\right|=(0-0)\hat{i}-(0-0)\hat{j}+(0-40)\hat{k}$

$|\overrightarrow{\tau }|=40N-m$