9.0 Toppling

The phenomena of overbalance or fall of the body after the application of an external force is known as toppling.

Consider two bodies $P$ and $Q$ having smaller and broader base respectively.

An equal horizontal force $F$ is applied on the body as shown in the figure.

Which body has more chances to topple before sliding?

The common understanding suggests that body $P$ will have more chances to topple because the width of the base is small as compared to the body $Q$.

Now, let us understand the physics behind the toppling.

Consider a block of width $W$ and height $H$ as shown in the figure.

Let an external force $F$ is applied at a height $p$ above the base of the block.

The frictional force $f$ starts acting on the block which is sufficient to prevent sliding. $$\begin{equation} \begin{aligned} F = f \\ N = mg \\\end{aligned} \end{equation} $$

So, now the block is in translational equilibrium.

But the body is not in rotational equilibrium due to the torque produced by external force $F$ and frictional force $f$.

This torque has tendency to topple the block about point $D$.

To overcome the effect of toppling, normal reaction $N$ is shifted towards right by a distance $q$.

The normal reaction $(N)$ produces torque in the opposite direction.

$${\overrightarrow \tau _{{E_2}}} = Fp\;\left( { - \widehat k} \right) + mgq\left( {\widehat k} \right)$$ For rotational equilibrium, $${\overrightarrow \tau _{{E_2}}} = 0$$ So, $$\begin{equation} \begin{aligned} Fp\; = mgq \\ q = \frac{{Fp}}{{mg}} \\\end{aligned} \end{equation} $$

Note: $q$ is the distance by which normal reaction $N$ should be shifted in order to prevent toppling.

As the external force $F$ or the height of application of force $p$ increases the torque in the clockwise direction increases.

So, the distance $q$ should also increases in order to balance the torque. But the distance $q$ cannot go beyond the edge of the block. So, in extreme case the normal reaction ($N$) passes through corner $D$ of the block as shown in the figure.

Torque about point $D$,

$${\overrightarrow \tau _D} = Fp\;\left( { - \widehat k} \right) + mg\frac{W}{2}\left( {\widehat k} \right)$$ For rotational equilibrium, $${\overrightarrow \tau _D} = 0$$ So, $$\begin{equation} \begin{aligned} mg\frac{W}{2} = Fp \\ W = \frac{{2Fp}}{{mg}} \\\end{aligned} \end{equation} $$ or $$F = \frac{{Wmg}}{{2p}}$$

Also, $$F \propto W$$

Therefore, the block of larger base requires larger external force for toppling. So, it has less chances of toppling because the normal reaction $(N)$ has more margin to shift as compared to the block of smaller base.

Similarly, in case of circular or spherical objects, normal reaction has zero margin to shift.

So, even of the body is in translational equilibrium, the body can roll very easily.