Question

$Aforce\underset{F}{\overset{}{\to }}=4\hat{i}-5\hat{j}+3\hat{k}isactingapoint\underset{r\_1}{\overset{}{\to }}=\hat{i}+2\hat{j}+3\hat{k}$. The torque acting about a point $\underset{r\_1}{\overset{}{\to }}=3\hat{i}-2\hat{j}-3\hat{k}$ is

• A. zero
• B. $42\hat{i}-30\hat{j}+6\hat{k}$
• C. $42\hat{i}+30\hat{j}+6\hat{k}$
• D. $42\hat{i}+30\hat{j}-6\hat{k}$

Answer 1

The correct option is D $42\hat{i}+30\hat{j}-6\hat{k}$

Given that,

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

Position vector ${\overrightarrow{r}}_1=\hat{i}+2\hat{j}+3\hat{k}$

Position vector ${\overrightarrow{r}}_2=3\hat{i}-2\hat{j}-3\hat{k}$

Now, the torque is

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

$\overrightarrow{r}={\overrightarrow{r}}_1-{\overrightarrow{r}}_2$

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

$\overrightarrow{r}=-2\hat{i}+4\hat{j}+6\hat{k}$

$\overrightarrow{\tau }=(-2\hat{i}+4\hat{j}+6\hat{k})\times (4\hat{i}-5\hat{j}+3\hat{k})$

$\overrightarrow{\tau }=42\hat{i}+30\hat{j}-6\hat{k}$

Hence, the torque is $42\hat{i}+30\hat{j}-6\hat{k}$