In the overhead view of Fig. 10 - 24, five forces of the same magnitude act on a strange merry-go-round; it is a square that can rotate about point P, at mid-length along one of the edges. Rank the forces according to the magnitude of the torque they create about point P, greatest first.

Figure 10-24 – overhead view of the square with different forces acting on it at different angles.

From the given figure and directions of forces, predict the angles between the radius of force and the magnitude of radius. The forces producing torque can be easily ranked.

Formulae are as follows:

$$\begin{gathered} \mathrm{\tau }=\mathrm{r}\times \mathrm{F} \\ =\mathrm{rFsin\varphi } \end{gathered}$$

Where, r is radius, F is force and is torque.

As for the formula of torque, it is clear that the torque is dependent on the radius and the angle between force and radius.

If we draw a line from point P to the vertices of each side and see it, then by the 1^st sight, it is clear that the radius of force${\mathrm{F}}_4\&\hspace{0.33em}{\mathrm{F}}_5$is greater than others.

Now,

Case (a)

Force 1 – The angle between radius and force isslightly less than$180°\&\sin \left(180\right)=0$

So,$\tau >0$

Case (b)

Force 2 – The angle between radius and force is around$135°\&\sin \left(135\right)=0.707$

So,$\tau$is slightly greater than force1.

Case (c)

Force 3 – The angle between radius and force is around$0°\&\sin \left(0\right)=0.$

So,$\mathrm{\tau }=0$

Case (d)

Force 4 – The angle between radius and force is around$150°\&\hspace{0.33em}\sin \left(150\right)=0.5$

Case (e)

Force 5 – The angle between radius and force is around$120°\&\sin \left(120\right)=0.866$

From the above, the ranking of forces of greater torque is clear as

${\mathrm{F}}_5>{\mathrm{F}}_4>{\mathrm{F}}_2>{\mathrm{F}}_1>{\mathrm{F}}_3$

Therefore, Using the given diagram of forces and point of rotation, we found the ranking of forces for higher torque.