Question

A force $\overrightarrow{F}=4\hat{i}-5\hat{j}+3\hat{k}$ is acting at a point $\overrightarrow{r_1}=\hat{i}-2\hat{j}+3\hat{k}$. Find the torque acting about a point $\overrightarrow{r_2}=3\hat{i}-2\hat{j}-3\hat{k}$.

Answer 1

A force $\overrightarrow{F}=4\hat{i}-5\hat{j}+3\hat{k}$

$\overrightarrow{r_1}=\hat{i}-2\hat{j}+3\hat{k}$

Finding, torque acting about a point

$\overrightarrow{r_2}=3\hat{i}-2\hat{j}-3\hat{k}$

Torque of a force $\overrightarrow{F}$

acting on a point with position vector $\overrightarrow{r}$

Given by

$\tau =\overrightarrow{r_2}\times \overrightarrow{F}$

So, we can fnd torque by finding the cross product of $\overrightarrow{r_2}$ and $\overrightarrow{F}$

We have

$\overrightarrow{r_1}=\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{F}=4\hat{i}-5\hat{j}+3\hat{k}$

$\tau =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -2& -3\\ 4& -5& 3\end{array}\right|$

$\tau =\hat{i}(-6-15)-\hat{j}(9+12)+\hat{k}(-15+8)$

$==-21\hat{i}-21\hat{j}-7\hat{k}$

$=-7(+3\hat{i}+3\hat{j}+\hat{j})=-7(3\hat{i}+3\hat{j}+\hat{j})$