$A$ force $\vec{F} = 5\hat{i} + 2\hat{j} - 5\hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$. What is the torque about the origin?

$A$ 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 $FR$)

$A$ force $F_1 = A \hat{j}$ is applied to a point whose radius vector is $r_1 = a \hat{i}$, while a force $F_2 = B \hat{i}$ is applied to the point whose radius vector is $r_2 = b \hat{j}$. Both the radius vectors are determined relative to the origin of the coordinate axes $O$. The moment of the force relative to $O$ is

If force $\vec{F} = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$ acts on a particle having position vector $\vec{r} = 2 \hat{i} + \hat{j} + 2 \hat{k}$,then the torque about the origin will be:

The center of an equilateral triangle $ABC$ is $O$. Three forces $F_1$,$F_2$,and $F_3$ are applied along the sides $AB$,$BC$,and $AC$ respectively. If the total torque about $O$ is zero,what is the relationship between $F_1$,$F_2$,and $F_3$?

$A$ force of $-P \hat{k}$ acts on the origin of the coordinate system. The torque about the point $(2, -3)$ is $P(a \hat{i} + b \hat{j})$. If the ratio $\frac{a}{b} = \frac{x}{2}$,find the value of $x$.