The moment of the force,overrightarrow{F} = 4 | vedclass

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(C) The moment of force (torque) is given by the cross product of the position vector relative to the point of rotation and the force vector:$\overrig

Vedclass · Accepted Answer

$-7\hat{i} - 4\hat{j} - 8\hat{k}$

Vedclass · Answer

(C) The moment of force (torque) is given by the cross product of the position vector relative to the point of rotation and the force vector:$\overrightarrow{	au} = (\overrightarrow{r} - \overrightarrow{r_0}) 	imes \overrightarrow{F}$Here,the point of rotation is $\overrightarrow{r_0} = 2\hat{i} - 2\hat{j} - 2\hat{k}$ and the point of application of force is $\overrightarrow{r} = 2\hat{i} + 0\hat{j} - 3\hat{k}$.First,calculate the relative position vector:$\overrightarrow{r} - \overrightarrow{r_0} = (2\hat{i} + 0\hat{j} - 3\hat{k}) - (2\hat{i} - 2\hat{j} - 2\hat{k}) = 0\hat{i} + 2\hat{j} - 1\hat{k}$.Now,calculate the cross product $\overrightarrow{	au} = (0\hat{i} + 2\hat{j} - 1\hat{k}) 	imes (4\hat{i} + 5\hat{j} - 6\hat{k})$:$\overrightarrow{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 & -1 \ 4 & 5 & -6 \end{vmatrix}$$\overrightarrow{	au} = \hat{i}(2(-6) - (-1)(5)) - \hat{j}(0(-6) - (-1)(4)) + \hat{k}(0(5) - 2(4))$$\overrightarrow{	au} = \hat{i}(-12 + 5) - \hat{j}(0 + 4) + \hat{k}(0 - 8)$$\overrightarrow{	au} = -7\hat{i} - 4\hat{j} - 8\hat{k}$.

The moment of the force,$\overrightarrow{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$ at point $(2, 0, -3)$,about the point $(2, -2, -2)$,is given by

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