Let a force vec{F} = -Fhat{k} act on the orig | vedclass

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(A) The point $P$ is given as $(1, -1)$,so its position vector is $\vec{r}_P = \hat{i} - \hat{j}$.Since the force acts at the origin $O(0, 0)$,the pos

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$-F(\hat{i} + \hat{j})$

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(A) The point $P$ is given as $(1, -1)$,so its position vector is $\vec{r}_P = \hat{i} - \hat{j}$.Since the force acts at the origin $O(0, 0)$,the position vector of the point of application relative to point $P$ is $\vec{r} = \vec{r}_O - \vec{r}_P = 0 - (\hat{i} - \hat{j}) = -\hat{i} + \hat{j}$.The torque (moment of force) about point $P$ is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Substituting the values: $\vec{	au} = (-\hat{i} + \hat{j}) 	imes (-F\hat{k})$.Using the distributive property of cross product: $\vec{	au} = F[(\hat{i} 	imes \hat{k}) - (\hat{j} 	imes \hat{k})]$.Using the cross product relations $\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$:$\vec{	au} = F[-\hat{j} - \hat{i}] = -F(\hat{i} + \hat{j})$.

Let a force $\vec{F} = -F\hat{k}$ act on the origin of a Cartesian frame of reference. The moment of force about a point $(1, -1)$ will be

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