In Fig. 11-26, three forces of the same magnitude are applied to a particle at the origin (${\overrightarrow{F}}_{\mathbf{1}}$acts directly into the plane of the figure). Rank the forces according to the magnitudes of the torques they create about (a) point ,$\mathbf{P}\mathbf{1}$(b) point, $\mathbf{P}\mathbf{2}$and (c) point,$\mathbf{P}\mathbf{3}$retest first.

The figure and the direction of forces are given.

The torque acting on the rotating object is equal to the moment of force. The magnitude of the torque is equal to the product of the magnitude of the radius vector, magnitude of the force, and sin of the angle between the radius vector and force.

Use the concept of torque to find the forces which create torques about a given point.

Formulae are as follows:

$\begin{array}{c}\left|\tau \right|=\left|\overrightarrow{r}\right|\times \left|\overrightarrow{F}\right|\\ =rF\text{sin}\theta \end{array}$

Here, r is the radius, F is force and τ is torque.

Now, calculate for the magnitude of torques about:$P_1$

Magnitude if torque due to,$F_1\text{\hspace{0.17em}aboutpoint}{P}_1$

${\tau }_1=r_1{F}_1\text{sin}\left({90}^0\right)$

The magnitude of torque due to$F_2\text{point}{P}_1$,

${\tau }_2=r_2{F}_2\text{sin}({90}^0+\theta )$

Magnitude of torque due to$F_3$ is ${\tau }_3=0$because position vector and force are in the same direction.

Therefore,$F_1>F_2>F_3.$

Now, calculate for magnitude of torques about:$P_2$

Similarly,

${\tau }_1=r_1{F}_1\text{sin}\left({90}^0\right)$

${\tau }_2=r_2{F}_2\text{sin}\left({90}^0\right)$

${\tau }_3=r_3{F}_2\text{sin}({90}^0+\theta )$

Therefore,

Now, calculatefor magnitude of torques about:

${\tau }_1=r_1{F}_1\text{sin}\left({90}^0\right)$

${\tau }_2=r_2{F}_2\text{sin}({90}^0+\theta )$

${\tau }_3=r_3{F}_2\text{sin}\left({90}^0\right)$

Therefore,$F_1=F_3>F_2$

Therefore, we can rank the forces which create torques by using the formula for torque.