A force of -P hat{k} acts on the origin of th | vedclass

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(C) The torque $\vec{	au}$ about a point is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Here,the force $\vec{F} = -P \hat{k}$ acts at the origin $

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$3$

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(C) The torque $\vec{	au}$ about a point is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Here,the force $\vec{F} = -P \hat{k}$ acts at the origin $(0, 0, 0)$.The position vector $\vec{r}$ of the origin with respect to the point $(2, -3)$ is $\vec{r} = (0 - 2)\hat{i} + (0 - (-3))\hat{j} = -2\hat{i} + 3\hat{j}$.Now,calculate the cross product:$\vec{	au} = (-2\hat{i} + 3\hat{j}) 	imes (-P\hat{k})$$\vec{	au} = -P [(-2)(\hat{i} 	imes \hat{k}) + 3(\hat{j} 	imes \hat{k})]$Since $\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$,we get:$\vec{	au} = -P [(-2)(-\hat{j}) + 3(\hat{i})]$$\vec{	au} = -P [2\hat{j} + 3\hat{i}] = P(-3\hat{i} - 2\hat{j})$.Comparing this with $P(a\hat{i} + b\hat{j})$,we get $a = -3$ and $b = -2$.The ratio $\frac{a}{b} = \frac{-3}{-2} = \frac{3}{2}$.Given $\frac{a}{b} = \frac{x}{2}$,we have $\frac{3}{2} = \frac{x}{2}$,which implies $x = 3$.

$A$ force of $-P \hat{k}$ acts on the origin of the coordinate system. The torque about the point $(2, -3)$ is $P(a \hat{i} + b \hat{j})$. If the ratio $\frac{a}{b} = \frac{x}{2}$,find the value of $x$.

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