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(A) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.Given $\vec{r} = (1\hat{i} +

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(A) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.Given $\vec{r} = (1\hat{i} + 1\hat{j} + 1\hat{k}) \ m$ and $\vec{F} = (1\hat{i} - 1\hat{j} + 1\hat{k}) \ N$.$\vec{	au} = \vec{r} 	imes \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -1 & 1 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i}(1 - (-1)) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$$\vec{	au} = \hat{i}(2) - \hat{j}(0) + \hat{k}(-2)$$\vec{	au} = 2\hat{i} - 2\hat{k} \ N \cdot m$The torque in the $z$-direction is the component associated with the $\hat{k}$ unit vector,which is $-2 \ N \cdot m$.The magnitude of this component is $|-2| = 2 \ N \cdot m$.

The coordinates of a particle with respect to origin in a given reference frame is $(1, 1, 1) \ m$. If a force of $\vec{F} = \hat{i} - \hat{j} + \hat{k} \ N$ acts on the particle,then the magnitude of the torque (with respect to origin) in the $z$-direction is . . . . . . .

Similar Questions

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$A$ couple produces:

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