Question

A force $\overrightarrow{F}=2\hat{i}-3\hat{k}$ acts on a particle at $\overrightarrow{r}=0.5\hat{j}-2\hat{k}$. The torque $\overrightarrow{\Gamma }$ acting on the particle relative to a point with co-ordinates ($2.0$m, $0$; $-3.0$ m) is?

Answer 1

Torque= $\stackrel{⟶}{r}\times \stackrel{⟶}{F}$

$\stackrel{⟶}{\tau }=(0.5\hat{j}-2\hat{k})\times (2\hat{i}-3\hat{k})$

$=\left|\begin{array}{c}\hat{i}\\ 0\\ 2\end{array}\begin{array}{c}\hat{j}\\ 0.5\\ 0\end{array}\begin{array}{c}\hat{k}\\ -2\\ -3\end{array}\right|$

$=\hat{j}(-1.5)-\hat{j}\left(4\right)+\hat{k}(-1)$

$\stackrel{⟶}{\tau }=-1.5\hat{j}-4\hat{j}-\hat{k}$

Therefore, $\stackrel{⟶}{\tau }$ at $(2.0m,0,-3-0m)=-1.5\left(2\right)-4\left(0\right)+3=-3+3$

$\Rightarrow \stackrel{⟶}{\tau }=0$

Therefore, torque acting is $0$ .