Question

A triangular plate is shown in the figure. A force $\overrightarrow{F}=4\hat{i}-3\hat{j}$ is applied at point $P$. The torque acting at point $\text{P}$ with respect to point $\text{O}$ and point $\text{Q}$ respectively are :

• A. $15-20\sqrt{3};15+20\sqrt{3}$
• B. $15+20\sqrt{3};15-20\sqrt{3}$
• C. $-15+20\sqrt{3};15+20\sqrt{3}$
• D. $-15-20\sqrt{3};15-20\sqrt{3}$

Answer 1

The correct option is D $-15-20\sqrt{3};15-20\sqrt{3}$


Given,
$\overrightarrow{F}=4\hat{i}-3\hat{j}$

Position vector of $\overrightarrow{F}$ about point $\text{O}$,

$\overrightarrow{r_1}=10\cos {60}^{\circ }\hat{i}+10\sin {60}^{\circ }\hat{j}$

$\Rightarrow \overrightarrow{r_1}=5\hat{i}+5\sqrt{3}\hat{j}$

Now, torque,
$\overrightarrow{{\tau }_1}=\overrightarrow{r_1}\times \overrightarrow{F}$

$\Rightarrow \overrightarrow{{\tau }_1}=(5\hat{i}+5\sqrt{3}\hat{j})\times (4\hat{i}-3\hat{j})$

$\Rightarrow \overrightarrow{{\tau }_1}=(-15-20\sqrt{3})\hat{k}$

Similarly, position vector of $\overrightarrow{F}$ about point $\text{Q}$,

$\overrightarrow{r_2}=-10\cos {60}^{\circ }\hat{i}+10\sin {60}^{\circ }\hat{j}$

$\Rightarrow \overrightarrow{r_2}=-5\hat{i}+5\sqrt{3}\hat{j}$

Now, torque,
$\overrightarrow{{\tau }_2}=\overrightarrow{r_2}\times \overrightarrow{F}$

$\Rightarrow \overrightarrow{{\tau }_2}=(-5\hat{i}+5\sqrt{3}\hat{j})\times (4\hat{i}-3\hat{j})$

$\Rightarrow \overrightarrow{{\tau }_2}=(15-20\sqrt{3})\hat{k}$