A triangular plate is shown in the figure. A | vedclass

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(D) The coordinates of point $P$ are $(5, 5 \sqrt{3})$ cm. The coordinates of point $O$ are $(0, 0)$ and point $Q$ are $(10, 0)$.Given force $\overrig

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

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(D) The coordinates of point $P$ are $(5, 5 \sqrt{3})$ cm. The coordinates of point $O$ are $(0, 0)$ and point $Q$ are $(10, 0)$.Given force $\overrightarrow{F} = 4 \hat{i} - 3 \hat{j}$.Position vector of $P$ with respect to $O$ is $\overrightarrow{r}_1 = 5 \hat{i} + 5 \sqrt{3} \hat{j}$.Torque about $O$ is $\vec{	au}_O = \overrightarrow{r}_1 	imes \overrightarrow{F} = (5 \hat{i} + 5 \sqrt{3} \hat{j}) 	imes (4 \hat{i} - 3 \hat{j}) = (-15 \hat{k} - 20 \sqrt{3} \hat{k}) = (-15 - 20 \sqrt{3}) \hat{k}$.Position vector of $P$ with respect to $Q$ is $\overrightarrow{r}_2 = (5-10) \hat{i} + 5 \sqrt{3} \hat{j} = -5 \hat{i} + 5 \sqrt{3} \hat{j}$.Torque about $Q$ is $\vec{	au}_Q = \overrightarrow{r}_2 	imes \overrightarrow{F} = (-5 \hat{i} + 5 \sqrt{3} \hat{j}) 	imes (4 \hat{i} - 3 \hat{j}) = (15 \hat{k} + 20 \sqrt{3} \hat{k}) = (15 + 20 \sqrt{3}) \hat{k}$.Thus,the values are $(-15 - 20 \sqrt{3})$ and $(15 + 20 \sqrt{3})$.

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

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