Question

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

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

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

We know that,

The torque is the cross product of the force and distance.

$\tau =F\times r$.....(I)

Now, The distance is

$r=r_2-r_1$

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

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

Now, put the value of F and r in equation (I)

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

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

The torque will be $42\hat{i}+30\hat{j}-6\hat{k}$.

Hence, this is the required solution.