# Fluid statics and pressure

Consider a fluid filled in a cylindrical container, height $$h$$ and area of cross-section $$a$$. The pressure at the bottom will be:

$$p_0+\rho gh$$

but not

$$p_0+\rho gh+\frac{mg}{a}$$ where $$m$$ is the mass of fluid in the container and $$\rho$$ being its density. Why is this so?

We know that, at a depth $$h$$ below the surface of water, pressure will be exerted by the atmosphere and the weight of water present till that level $$h$$, so pressure should be $$P_h = (P_o + \frac{mg}{a})$$

BUT: $$\rho g h =\frac{mg}{a}$$

Proof:

$$m = \rho V$$, where $$\rho$$ is density, $$V$$ is volume

We know that, $$V = ah \implies m = \rho ah$$

Therefore, $$\frac{mg}{a} = \frac{(\rho a h) g}{a} = \rho g h$$

This proves that $$\boxed{P_h = P_o + \rho g h}$$

This is because $$\rho gh$$ is itself the pressure due to the liquid on the bottom of the container.

=>$$\rho gh=\frac{mgh}{v}$$ =>$$\rho gh=\frac{mg}{a}\ \ (since\ h\times a=v)$$

As you can see you counted the term twice.