Find the torque of a force $\overrightarrow F = 2\hat i + \hat j + 4\hat k$ acting at the point $\overrightarrow r = 7\hat i + 3\hat j + \hat k$.

$A$ 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 $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$):

$A$ couple produces

Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$.

The torque of a force $\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $\vec{\tau}$. If the force acts on a particle whose position vector is $\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$,then the value of $\vec{\tau}$ will be:

$A$ slab is subjected to two forces $\vec F_1$ and $\vec F_2$ of the same magnitude $F$ as shown in the figure. Force $\vec F_2$ is in the $XY$-plane,while force $\vec F_1$ acts along the $z$-axis at the point $(2\hat i + 3\hat j)$. The moment of these forces about point $O$ is: