Animated Solution for Physics - Properties of Solids and Liquids: Two liquids of densities ${\rho }_1$ and ${\rho }_2$ (${\rho }_2=2{\rho }_1$) are filled up behind a square wall of side $10\text{ m}$ as shown in figure. Each liquid has a height of $5\text{ m}$. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (assume that the liquids are not mixing)

Unlocking Fluid Forces on Vertical Walls

Imagine standing behind a massive dam. The water at the top barely pushes against the concrete, but at the bottom, the pressure is immense. This problem takes that concept and adds a twist: two different liquids stacked on top of each other. Let's dive into the mechanics of fluid pressure and see how we can elegantly solve this using the concept of average pressure.

We have a square wall of side $10\text{ m}$, divided into two equal vertical halves: MN (upper) and NO (lower). Each half has a height of $h=5\text{ m}$ and an area $A=10\times 5=50{\text{ m}}^2$.

The upper half is exposed to a liquid of density ${\rho }_1$, while the lower half is exposed to a liquid of density ${\rho }_2=2{\rho }_1$.

The force exerted by a fluid on a vertical surface is not uniform because pressure increases with depth. The most efficient way to calculate this force is by using the average pressure:

where $P_{\text{avg}}=\frac{P_{\text{top}}+P_{\text{bottom}}}{2}$.

Let's analyze the upper section MN.

At the top point M, the liquid depth is zero, so the gauge pressure is:

At point N, the depth is $h$, so the pressure due to the first liquid is:

The average pressure on MN is simply the arithmetic mean of the pressures at its boundaries:

Therefore, the force on MN is:

This is where many students make a mistake. The lower section NO is not just experiencing the pressure of the second liquid; it's also bearing the weight of the first liquid above it!

At point N (the top of the lower section), the pressure is already:

At the bottom point O, we must add the pressure contributed by the second liquid of depth $h$:

Now, we find the average pressure on the section NO:

The force on NO is:

We are given a crucial relationship: ${\rho }_2=2{\rho }_1$. Let's substitute this into our equation for $F_{NO}$:

Finally, we calculate the ratio of the forces:

The ${\rho }_{1}ghA$ terms beautifully cancel out, leaving us with:

The force on the lower half is four times greater than the force on the upper half. This perfectly illustrates how rapidly pressure and force escalate with depth, especially when denser fluids are involved!