Animated Solution for Physics - Rotational Motion: Consider a uniform cubical box of side $a$ on a rough floor that is to be moved by applying minimum possible force $F$ at a point $b$ above its centre of mass (see figure). If the coefficient of friction is $\mu =0.4$, the maximum possible value of $100\times \frac{b}{a}$ for box not to topple before moving is ……… .

The Delicate Dance of Sliding and Toppling

Imagine you are trying to push a heavy, uniform cubical box across a rough floor. If you push too low, the box simply slides. But if you push too high, the box tips over before it even starts moving! This problem explores the exact threshold where the box is on the verge of doing both simultaneously.

We are applying a horizontal force $F$ at a height $b$ above the center of mass of a cubical box of side $a$. The center of mass itself is at a height of $a/2$ from the floor. Therefore, the total height at which the force is applied from the ground is $(b+a/2)$.

For the box to just start sliding, the applied force must overcome the maximum static friction.

Since there is no vertical motion, the normal force $N$ simply balances the weight of the box, so $N=mg$. This gives us our sliding condition:

Now, let's consider the rotational aspect. As you push higher and harder, the box tends to rotate forward. To prevent this, the floor pushes back harder on the front edge. At the exact moment the box is about to topple, the back edge lifts off the ground, and the entire normal force $N$ shifts to the front bottom corner (let's call it point $A$).

To find the critical height, we can balance the torques about this pivot point $A$. - The applied force $F$ tries to rotate the box clockwise. Its lever arm from point $A$ is the vertical height $(b+a/2)$. - Gravity $mg$ acts downwards through the center of mass, trying to pull the box back counter-clockwise. Its lever arm from point $A$ is the horizontal distance $a/2$. - The normal force $N$ and friction $f$ both pass directly through point $A$, so they create zero torque!

Equating the clockwise and counter-clockwise torques gives us our master equation:

We know that at this critical threshold, the box is also on the verge of sliding, so we can substitute $F=\mu mg$ into our torque equation:

Notice how beautifully the mass $m$ and gravity $g$ cancel out from both sides! The condition for toppling depends entirely on the geometry of the box and the coefficient of friction, not on how heavy the box is.

We are given the coefficient of friction $\mu =0.4$. Let's plug this in and solve for the ratio $b/a$:

The question asks for the value of $100\times \frac{b}{a}$: