Question

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.

Short answer

(a) Ranking of force for torque about point $P_1$is$F_1>F_2>F_3.$

(b) Ranking of force for torque about point$P_2$ is$F_1=F_2>F_3.$

(c) Ranking of force for torque about point$P_3$ is$F_1=F_3>F_2.$

Step-by-step solution

Step 1: Given information 

The figure and the direction of forces are given.

Understanding the concept of torque

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.

(a) Determining the ranking of force for torque about point.P1

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,$F_1>F_2>F_3.$

(b) Determining the ranking of force for torque about point .P2

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,

(c) Determining theranking of force for torque about point P3.P3

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.