# Pressure on sides of container by two immiscible liquids

Today after my physics class a question just strikes in my brain. I have been pondering over it since an hour but couldn't arrive at a particular result. We were taught to find the average pressure on the sides of a container by a fluid which simply becomes $\frac {\rho gh}{2}$. But what if two immiscible liquids like water and oil with different height columns are kept in a container and we have to find the $\mathbf {force}$ on the side walls.

For example consider a cuboidal container having length $b$ breadth $a$ and total height $H$. Now I fill the container with two immiscible liquids with different densities say $\rho_1$ and $\rho_2$. Let these two liquids occupy a height columns $H_1$ and $H_2$ respectively such that $$H_1+H_2=H$$. Now if I need to calculate the force on the side wall with shaded part how must I proceed. A general way could be integration $\int PdA$ where $A$ is the area of surface and I have tried it but I am getting different answers some or the other time. Can somebody please shed some light over this topic.

• A very useful result is that the force will be equal to the pressure at the center of mass of the fluid, multiplied by the surface area of the container. Try deducing why. – aditya_stack Dec 30 '19 at 14:42

The pressure at depth $h$ (measured from the top) is proprtional to the wight of the fluid on top of it

$$P(y)= g\int_0^y \rho(\tilde y) d\tilde y$$

and therefore the average pressure is

$$\langle P \rangle= \frac{g}{h} \int_0^h \left[\int_0^y \rho(\tilde y) d\tilde y\right] dy$$

For constant $\rho$ this becomes

$$\langle P \rangle = \frac{\rho g}{h} \int_0^h y dy \\ =\frac{1}{2}\rho g h$$

I am hoping that when you said "we were taught to find average pressure" this is what you mean. If so then the rest is straightforwards. Otherwise, understand the above first.

For the case you want, the force on the side wall (for the fron wall replace $b \to a$) we have

$$d F(y)= P(y) b dy \\ =g b \left[\int_0^y \rho(\tilde y) d\tilde y \right] dy \\ = \begin{cases} gb \rho_1 y dy & y \le H_1 \\ gb (\rho_1 H_1 + \rho_2 (y-H_1)) & y > H_1 \end{cases}$$

Which is very intuitive if you think about it. The force at any depth is proportional to the weight of the fluid above it.

If we integrate this we get the result

$$F = g b \left[ \rho_1 \frac{H_1^2}{2} + \rho_1 H_1 H_2 + \rho_2 \frac{H_2^2}{2} \right]$$

To make this more illuminating take $H_1 = \eta H$ and $H_2 = (1-\eta) H$. Then the above expression becomes

$$F = g b H^2 \left[ \frac{(\rho_2 - \rho_1) \eta^2}{2} + \rho_1 \eta + \rho_2 (\frac{1}{2}- \eta) \right]$$

You can then take various limits $\eta \to 0,1$ and $\rho_1 \to \rho_2$ to make sure the result is sensible and gain intuition.