Moment of a force of magnitude 20 , N acting | vedclass

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(C) The torque (moment of force) is given by the cross product of the position vector and the force vector: $\vec{	au} = \vec{r} 	imes \vec{F}$.Give

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$40$

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(C) The torque (moment of force) is given by the cross product of the position vector and the force vector: $\vec{	au} = \vec{r} 	imes \vec{F}$.Given,the force vector is $\vec{F} = 20 \hat{i} \, N$.The position vector $\vec{r}$ of the point of application $(3, 0, 0)$ with respect to the point $(0, 2, 0)$ is $\vec{r} = (3 - 0) \hat{i} + (0 - 2) \hat{j} + (0 - 0) \hat{k} = (3 \hat{i} - 2 \hat{j}) \, m$.Now,calculating the cross product:$\vec{	au} = (3 \hat{i} - 2 \hat{j}) 	imes (20 \hat{i})$$\vec{	au} = 3 \hat{i} 	imes 20 \hat{i} - 2 \hat{j} 	imes 20 \hat{i}$Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\vec{	au} = 0 - 40 (-\hat{k}) = 40 \hat{k} \, N \cdot m$.The magnitude of the torque is $|\vec{	au}| = 40 \, N \cdot m$.

Moment of a force of magnitude $20 \, N$ acting along the positive $x$-direction at point $(3 \, m, 0, 0)$ about the point $(0, 2, 0)$ (in $N \cdot m$) is ...........

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