Animated Solution for Physics - Properties of Solids and Liquids: A light cylindrical vessel is kept on a horizontal surface. Area of base is $A$. A hole of cross-sectional area $a$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is ($a\ll A$).

Thrust, Friction, and Equilibrium

Imagine a cylindrical vessel filled with a liquid, resting peacefully on a horizontal surface. Suddenly, a small hole opens at the bottom, and liquid shoots out. You might expect the vessel to just sit there and drain, but physics tells a different story. As the liquid rushes out, it acts like a miniature rocket engine, pushing the vessel in the opposite direction! Our mission is to find the minimum friction required to keep this "rocket bucket" firmly anchored to the ground.

First, we need to understand how fast the liquid is escaping. This is beautifully described by Torricelli's Law. If the hole is at a depth $h$ below the surface of the liquid, the velocity $v$ of the emerging jet is exactly the same as if a droplet of water had fallen freely from that height.

Mathematically, this is expressed as:

This velocity is the engine driving our problem. The faster the liquid leaves, the harder it pushes back.

Why does the vessel get pushed backwards? It all comes down to Newton's Third Law and the conservation of momentum. The liquid is gaining momentum as it shoots out. To conserve the total momentum of the system, the vessel must receive an equal and opposite rate of change of momentum. This is the thrust force, $F_{\text{thrust}}$.

The thrust force is given by the velocity of the exhaust multiplied by the mass flow rate (how much mass is leaving per second):

The mass flow rate, $\frac{dm}{dt}$, is the density of the liquid $\rho$ multiplied by the volume flowing out per second. The volume per second is simply the cross-sectional area of the hole $a$ times the velocity $v$. Therefore, $\frac{dm}{dt}=\rho av$.

Substituting this back into our thrust equation gives:

Now, let's bring back Torricelli's Law. We know $v=\sqrt{2gh}$, so squaring it gives $v^2=2gh$. Substituting this into our thrust equation reveals the raw force pushing the vessel:

To prevent the vessel from sliding, the static friction $f_s$ between the vessel and the ground must be strong enough to perfectly balance this thrust force. The condition for no sliding is:

We know that the maximum static friction is determined by the coefficient of friction $\mu$ and the normal force $N$, so $f_{s,\text{max}}=\mu N$.

What is the normal force here? The problem states it is a "light" cylindrical vessel, meaning we can ignore the mass of the empty bucket itself. The entire normal force comes from the weight of the liquid inside. The mass of the liquid is its density $\rho$ times its volume (Base Area $A$ times height $h$).

Now, we set up our final epic inequality. The maximum available friction must be greater than or equal to the thrust force:

Look closely at this equation. It is a moment of pure mathematical elegance. The density of the liquid $\rho$, the acceleration due to gravity $g$, and even the height of the liquid $h$ appear on both sides. They perfectly cancel each other out!

Dividing both sides by the base area $A$, we arrive at our final, beautifully simple constraint:

Therefore, the minimum coefficient of friction required to prevent the vessel from sliding is exactly ${\mu }_{\text{min}}=\frac{2a}{A}$. It depends only on the geometric ratio of the hole's area to the vessel's base area. Physics, once again, takes a complex dynamic system and distills it into a profound geometric truth.