$A$ wheel of radius $R$ with an axle of radius $R/2$ is shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces are exerted as shown in the figure. The magnitude of the net torque acting on the system is nearly (in $FR$)

Which component of torque is responsible for rotation about the $Z$-axis?

Three forces $F, 2F$ and $3F$ act on a rod $AB$ which is pivoted at $A$. The anticlockwise moments of forces $F, 2F$ and $3F$ about the pivot $A$ are respectively:

The vector sum of a system of non-collinear forces acting on a rigid body is given to be nonzero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero,does this mean that it is necessarily zero about any arbitrary point?

If $\vec{F} = (4\hat{i} - 10\hat{j})$ and $\vec{r} = (5\hat{i} - 3\hat{j})$,then calculate the torque $\vec{\tau} = \vec{r} \times \vec{F}$. (in $\hat{k}$)

$A$ force $F = 2.0\,N$ acts on a particle $P$ in the $xz-$ plane. The force $F$ is parallel to the $x-$ axis. The particle $P$ (as shown in the figure) is at a distance $3\,m$ from the origin,and the line joining $P$ with the origin makes an angle of $30^\circ$ with the $x-$ axis. The magnitude of the torque on $P$ with respect to the origin $O$ (in $N-m$) is: