Q.What is the pressure on a swimmer $10\ \text{m}$ below the surface of a lake?

Hydrostatic Pressure Balance

Imagine you're standing at the bottom of a swimming pool. You feel pressure on your ears — and the deeper you go, the more intense that pressure becomes. Now think about a column of water above you: every kilogram of that water is being pulled down by gravity. That weight has to be supported by the water below it. The deeper you go, the more water is stacked above you, so the greater the weight pressing down.

That's the core intuition: pressure in a fluid at rest increases with depth because the fluid above has to be supported by the fluid below.

The Precise Statement

Hydrostatic pressure balance is the condition that holds for any fluid at rest in a uniform gravitational field. It says:

where:

• $P_{\text{below}}$ is the pressure at a lower point,
• $P_{\text{above}}$ is the pressure at a higher point,
• $\rho$ is the density of the fluid (assumed constant),
• $g$ is the acceleration due to gravity,
• $h$ is the vertical depth between the two points.

Equivalently, the pressure gradient in the vertical direction is:

where $z$ increases upward. The minus sign tells you pressure decreases as you go up.

Why This Makes Sense

Take a thin horizontal slab of fluid of area $A$, thickness $dz$, at some depth. Its weight is $dW = \rho g A \, dz$. For the slab to be in equilibrium (not accelerating), the net upward force from pressure must exactly balance this weight.

The upward force on the slab's bottom face is $P(z) A$, and the downward force on its top face is $P(z+dz) A$. The net upward force is:

Setting this equal to the weight $\rho g A \, dz$ gives:

which is exactly the differential form above.

Key Implications

1. Pressure depends only on depth, not on the shape of the container. A tall thin tube and a wide shallow tank give the same pressure at the same depth — because only the vertical height of fluid above matters.

2. Pressure is the same at all points on the same horizontal level. If you move sideways at constant depth, $\rho g h$ doesn't change, so $P$ doesn't change.

3. Gases are compressible, so $\rho$ is not constant. For air, the density changes with pressure itself, leading to an exponential decrease — but the same principle applies locally.

A Common Mistake …