What is the torque of a force vec{F} = 3hat{i | vedclass

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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 $\vec{F}$:$\vec{	au} = \vec{r} 	imes

Vedclass · Accepted Answer

$\hat{i} - 5\hat{j} + 8\hat{k}$

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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 $\vec{F}$:$\vec{	au} = \vec{r} 	imes \vec{F}$Given $\vec{r} = 2\hat{i} + 2\hat{j} + 1\hat{k}$ and $\vec{F} = 3\hat{i} + 7\hat{j} + 4\hat{k}$.$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 1 \ 3 & 7 & 4 \end{vmatrix}$$\vec{	au} = \hat{i}(2 	imes 4 - 1 	imes 7) - \hat{j}(2 	imes 4 - 1 	imes 3) + \hat{k}(2 	imes 7 - 2 	imes 3)$$\vec{	au} = \hat{i}(8 - 7) - \hat{j}(8 - 3) + \hat{k}(14 - 6)$$\vec{	au} = 1\hat{i} - 5\hat{j} + 8\hat{k}$.

What is the torque of a force $\vec{F} = 3\hat{i} + 7\hat{j} + 4\hat{k}$ about the origin,if the force acts on a particle whose position vector is $\vec{r} = 2\hat{i} + 2\hat{j} + 1\hat{k}$?

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$A$ rod of length $l$ is acted upon by a couple as shown in the figure. The moment of the couple is $\tau \text{ Nm}$. If the force at each end of the rod is $F$,then the magnitude of each force is (given $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$):

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