A triangular plate is shown. A force F→=4i^−3j^ is applied at point P. The torque at point P with respect to point ‘O’ and ‘Q’ are:

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

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

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

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

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

answer is B.

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Detailed Solution

Torque of force $\vec{F} = 4 \hat{i} - 3 \hat{j}$ w.r.t O

$\vec{τ_{O}} = \vec{r_{OP}} \times \vec{F} = (10 \cos 60^{0} \hat{i} + 10 \sin 60^{0} \hat{j}) \times (4 \hat{i} - 3 \hat{j})$
$\Rightarrow \vec{τ_{O}} = (5 \hat{i} + 5 \sqrt{3} \hat{j}) \times (4 \hat{i} - 3 \hat{j}) = |\begin{array}{l} \hat{i} \hat{j} \hat{k} \\ 5 5 \sqrt{3} 0 \\ 4 - 3 0 \end{array}|$

$\Rightarrow \vec{τ_{O}} = \hat{k} (- 15 - 20 \sqrt{3})$

Torque of force $\vec{F} = 4 \hat{i} - 3 \hat{j}$ w.r.t Q
${\vec{τ}}_{Q} = {\overset{}{\vec{r}}}_{PQ} \times \overset{}{\vec{F}}$
$\begin{array}{l} \overset{}{{\vec{r}}_{PQ}} = {\overset{}{\vec{r}}}_{P} - {\overset{}{\vec{r}}}_{Q} = (5 \hat{i} + 5 \sqrt{3} \hat{j}) - 10 \hat{i} = (- 5 \hat{i} + 5 \sqrt{3} \hat{j}) \\ {\overset{}{\vec{τ}}}_{Q} = (- 5 \hat{i} + 5 \sqrt{3} \hat{j}) \times (4 \hat{i} - 3 \hat{j}) \end{array}$

$= |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 5 \sqrt{3} & 0 \\ 4 & - 3 & 0 \end{array}| = (15 - 20 \sqrt{3}) \hat{k}$

So the answer for given question is $- 15 - 20 \sqrt{3}, 15 - 20 \sqrt{3}$