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

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 ...........

Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$.

The figure shows three forces $\overrightarrow{F}_{1}, \overrightarrow{F}_{2}$ and $\overrightarrow{F}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point $O$ (the center of the triangle) is zero,then the magnitude of $\overrightarrow{F}_{3}$ is:

$A$ wheel of radius $20\, cm$ has forces applied to it as shown in the figure. The net torque produced by the forces $4\, N$ at $A$,$8\, N$ at $B$,$6\, N$ at $C$,and $9\, N$ at $D$ at the angles indicated is:

Let $\vec{F}$ be a force acting on a particle having position vector $\vec{r}$,and $\vec{\tau}$ be the torque of this force about the origin. Then: