A triangular plate is shown.

Question:

A triangular plate is shown. A force $\overrightarrow{\mathrm{F}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$ is applied at point $\mathrm{P}$. The torque at point $\mathrm{P}$ with respect to point ' $O^{\prime}$ and ' $Q$ ' are :

  1. $-15-20 \sqrt{3}, 15-20 \sqrt{3}$

  2. $15+20 \sqrt{3}, 15-20 \sqrt{3}$

  3. $15-20 \sqrt{3}, 15+20 \sqrt{3}$

  4. $-15+20 \sqrt{3}, 15+20 \sqrt{3}$


Correct Option: 1

Solution:

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

$\overrightarrow{\mathrm{r}}_{1}=5 \hat{\mathrm{i}}+5 \sqrt{3} \hat{\mathrm{j}} \& \overrightarrow{\mathrm{r}}_{2}=-5 \hat{\mathrm{i}}+5 \sqrt{3} \hat{\mathrm{j}}$

Torque about 'O'

$\vec{\tau}_{\mathrm{O}}=\overrightarrow{\mathrm{r}}_{1} \times \overrightarrow{\mathrm{F}}=(-15-20 \sqrt{3}) \hat{\mathrm{k}}=(15+20 \sqrt{3})(-\hat{\mathrm{k}})$

Torque about 'Q'

$\vec{\tau}_{\mathrm{Q}}=\overrightarrow{\mathrm{r}}_{2} \times \overrightarrow{\mathrm{F}}=(-15+20 \sqrt{3}) \hat{\mathrm{k}}=(15-20 \sqrt{3})(-\hat{\mathrm{k}})$

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