Animated Solution for Physics - Rotational Motion: A cubical block of side $L$ rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force $F$ is applied on the block as shown. If the coefficient of friction is sufficiently high, so that the block does not slide before toppling, the minimum force required to topple the block is

Toppling vs. Sliding

Imagine you are trying to push a heavy refrigerator across the floor. If you push it right at the bottom, it might slide. But if you push it at the very top, it might tip over and fall on its face. This tipping over is what physicists call toppling.

In this classic JEE problem, we are given a cubical block of side $L$ and mass $m$. A horizontal force $F$ is applied at the top edge. The problem explicitly states that the friction is high enough to prevent sliding. This means our block is locked in a battle of torques, and we need to find the exact moment when the overturning torque defeats the restoring torque.

To understand toppling, we must first understand the behavior of the normal reaction. When the block is just resting on the ground without any horizontal force, the normal reaction $N$ acts exactly through the center of mass, perfectly balancing gravity $mg$.

However, the moment you apply a horizontal force $F$ at the top, you create a torque that tries to rotate the block clockwise. To counter this and maintain equilibrium, the normal reaction $N$ must shift to the right. It shifts just enough to create an equal and opposite counter-clockwise torque.

But there is a limit! The normal reaction cannot shift beyond the physical boundary of the block. When the force $F$ is increased to a critical value, the normal reaction shifts entirely to the extreme bottom-right edge (let's call it point $P$). If you increase $F$ even slightly beyond this point, the normal reaction has nowhere else to go, equilibrium breaks, and the block topples over point $P$.

At the critical moment just before toppling, the block is on the verge of rotating about the pivot point $P$. To find the required force, we take the torque about this exact point $P$. Why point $P$? Because both the normal reaction $N$ and the static friction $f$ pass directly through $P$. Their perpendicular distances from the pivot are zero, meaning their torques are zero. This brilliantly simplifies our equation!

We are left with only two torques battling it out: 1. The Overturning Torque: Created by the applied force $F$. Its perpendicular distance from point $P$ is the full height of the block, $L$. So, ${\tau }_F=F\times L$. 2. The Restoring Torque: Created by gravity $mg$. Since gravity acts through the center of mass, its line of action is exactly halfway across the block. Its perpendicular distance from point $P$ is $L/2$. So, ${\tau }_{mg}=mg\times \frac{L}{2}$.

For the block to topple, the overturning torque must be greater than or equal to the restoring torque:

The $L$ beautifully cancels out from both sides, leaving us with the final condition:

Thus, the minimum force required to topple the block is exactly half of its weight, $\frac{mg}{2}$. It is a beautifully elegant result that showcases the power of choosing the right pivot point in rotational mechanics!