Visual walkthrough — Hydrostatics — pressure = ρgh, derivation

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Step 0 — What do the words even mean?

Before any physics, let's pin down the two ideas that carry the whole story: pressure and depth. A smart 12-year-old who has never seen these should be fine.

Look at the picture: the same hand pushes on a wide board and on a thin nail. Same force $F$, but the nail's tiny area makes a huge pressure.

WHY start here? Every later equation is really the sentence "force per area." If "pressure" stays fuzzy, the algebra is meaningless. Now it isn't.

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Step 1 — Draw an imaginary column of water

We can't do force-balance on a whole ocean at once. So we cut out a shape we can reason about: a straight vertical cylinder of fluid.

WHAT: Imagine a tube-shaped chunk of the fluid — cross-section area $A$, standing from the free surface (the top, open to air) straight down to depth $h$.

WHY this shape: A vertical cylinder has only two flat faces that point up/down (top and bottom). Everything vertical happens on just those two faces. Its side wall is curved and vertical, so pushes on it are sideways — they will cancel and never bother us. Simple geometry = simple physics.

PICTURE: The red cylinder is our chosen "body." It is still made of ordinary water; we've only drawn an imaginary boundary around it.

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Step 2 — Weigh the column

Pressure comes from weight, and weight needs mass. Let's get the mass of our red cylinder.

WHAT: Its volume is height $\times$ area, and mass is density $\times$ volume:

$$\underset{\text{volume }({\text{m}}^3)}{\underbrace{V}}=\underset{\text{area of face}}{\underbrace{A}}\times \underset{\text{height of column}}{\underbrace{h}}$$ $$\underset{\text{mass }(\text{kg})}{\underbrace{m}}=\underset{\begin{array}{c}\text{density}\\ {\text{kg/m}}^3\end{array}}{\underbrace{\rho }}\times \underset{Ah}{\underbrace{V}}=\rho Ah$$

WHY: Density $\rho$ answers "how many kilograms per cubic metre?" Multiply by how many cubic metres we have ($Ah$) and out pops the kilograms. (Density comes from Density and Specific Gravity.)

PICTURE: The column split into little unit cubes — count the cubes ($V$), each cube weighs $\rho$ per unit volume, total mass stacks up as $\rho Ah$.

Now turn mass into weight (the downward gravity force):

$$\underset{\text{weight }(\text{N})}{\underbrace{W}}=\underset{\rho Ah}{\underbrace{m}}\underset{\begin{array}{c}\text{gravity}\\ 9.8{\text{m/s}}^2\end{array}}{\underbrace{g}}=\rho Ahg$$

The symbol $g$ is doing one job: converting kilograms into newtons of downward pull.

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Step 3 — Name every vertical force on the column

Now list all the up/down forces. There are exactly three, and each is a pressure $\times$ area.

WHAT: three forces —

$$\begin{array}{ll}\text{Weight, down:}& W=\rho Ahg\\ \text{Air on the top, down:}& F_{\text{top}}=\underset{\begin{array}{c}\text{pressure at}\\ \text{surface}\end{array}}{\underbrace{P_0}}\underset{\text{top face}}{\underbrace{A}}\\ \text{Fluid below, up:}& F_{\text{bot}}=\underset{\begin{array}{c}\text{pressure at}\\ \text{depth }h\end{array}}{\underbrace{P}}\underset{\text{bottom face}}{\underbrace{A}}\end{array}$$

WHY these three, and no more: The top face feels the atmosphere pressing down ($P_0$ is the pressure right at the surface). The bottom face feels the fluid underneath pressing up ($P$ — this is the mystery we're solving for). Gravity pulls the whole mass down. Nothing else points vertically.

PICTURE: Three black arrows for the vertical forces, plus little grey sideways arrows on the wall — drawn only to show they cancel (each left push is matched by an equal right push at the same depth). This is why we never need them.

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Step 4 — Balance the forces (the physics happens here)

The column is at rest (that's what "hydrostatic" means). Newton's first law: at rest ⇒ net force $=0$ ⇒ up-forces equal down-forces.

WHAT:

$$\underset{\text{only up-force}}{\underbrace{F_{\text{bot}}}}=\underset{\text{air down}}{\underbrace{F_{\text{top}}}}+\underset{\text{weight down}}{\underbrace{W}}$$

Substitute each one from Step 3:

$$\underset{\text{fluid pushes up}}{\underbrace{PA}}=\underset{\text{air pushes down}}{\underbrace{P_{0}A}}+\underset{\text{weight of column}}{\underbrace{\rho Ahg}}$$

WHY: If the up-push were bigger, the column would shoot upward; if smaller, it would sink. It does neither — it just sits — so the two sides are exactly equal. This single equality carries all the answer.

PICTURE: A see-saw / balance beam. On the left pan sits $PA$ (up); on the right pan sit $P_{0}A$ and $\rho Ahg$ (down). The beam is level — perfectly balanced.

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Step 5 — Cancel the area, read off the law

Every term has an $A$ in it. The column's thickness shouldn't decide the pressure at a point, so let's remove it.

WHAT: Divide the whole equation by $A$:

$$\frac{PA}{A}=\frac{P_{0}A}{A}+\frac{\rho Ahg}{A}⟹\boxed{P=P_0+\rho gh}$$

WHY dividing by $A$ is allowed and meaningful: $A$ was just the size of our imaginary tube — we invented it, so it must vanish from the physical answer. Its disappearance is a promise: pressure does not depend on how fat you drew the column. (That's the hydrostatic paradox, coming in Step 7.)

Term by term, the final law reads:

$$\underset{\begin{array}{c}\text{total pressure}\\ \text{at depth }h\end{array}}{\underbrace{P}}=\underset{\begin{array}{c}\text{pressure at}\\ \text{the surface}\end{array}}{\underbrace{P_0}}+\underset{\text{how heavy}}{\underbrace{\rho }}\underset{\text{gravity}}{\underbrace{g}}\underset{\begin{array}{c}\text{how deep}\end{array}}{\underbrace{h}}$$

PICTURE: A vertical depth-axis with a ruler. At $h=0$ the pressure bar has height $P_0$. As you go down, a red wedge $\rho gh$ is added on top — the bar grows in a perfectly straight line with depth.

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Step 6 — The edge cases (never leave a scenario undrawn)

A law you trust must survive its extremes. Let's check all of them on one figure.

• At the surface, $h=0$: $P=P_0+\rho g(0)=P_0$. Correct — no fluid above you, only air. The red wedge has zero height.
• Zero gravity, $g=0$: $P=P_0$ everywhere. In free-fall or deep space, water has no weight, so depth stops mattering — pressure is uniform. The wedge collapses flat.
• Very light fluid, $\rho \to 0$ (like thin gas): $\rho gh$ becomes tiny; pressure barely changes with depth. This is why we usually ignore the "depth" of the air in a room.
• Going up instead of down ($h$ negative): above the surface $\rho gh$ is negative — pressure drops. Same straight line, extended upward.

PICTURE: One graph of pressure vs depth showing four lines: the normal fluid (steep red line), the light gas (nearly flat), the $g=0$ case (vertical — no change), and the surface point where all meet at $P_0$.

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Step 7 — Why the container's shape is invisible

The area $A$ cancelled in Step 5. Let's see what that promise means physically.

WHAT: Three containers — a thin straw, a fat barrel, a slanted funnel — all filled to the same depth $h$. Bottom pressure is identical in all three, even though the barrel holds a thousand times more water.

WHY: Only the fluid directly above a point presses on it. In the fat barrel, the extra width of water off to the sides isn't stacked over that point — its weight is carried by the bottom and the walls, not by the point below the middle. So volume is irrelevant; only vertical depth counts.

PICTURE: Three differently shaped vessels side by side, water level dead-flat across all, a red dot at each bottom labelled with the same pressure $P_0+\rho gh$.

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The one-picture summary

Here is the entire derivation compressed into a single frame: the red column, its three vertical forces, the balance, and the resulting straight pressure line — all at once.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a pool. I imagine drawing a skinny invisible tube of water, from the top surface straight down to my feet at depth $h$. That tube of water isn't moving, so it must be perfectly balanced — whatever holds it up equals whatever pulls it down. Pulling down: the air pressing on its top ($P_{0}A$) plus its own weight ($\rho Ahg$ — density times how tall times gravity). Pushing up: the water underneath, pressing on its bottom face ($PA$). Set up equal to down: $PA=P_{0}A+\rho Ahg$. The area $A$ was just the fatness of my imaginary tube — it can't be part of the real answer, so I divide it away and get $P=P_0+\rho gh$. That's it. The $\rho gh$ piece is the extra squeeze from the water column; go twice as deep and it doubles. And because $A$ cancelled, it never mattered how wide or weirdly-shaped the container was — only how far down you go. At the surface ($h=0$) there's no water above you, so you feel just the air $P_0$. In zero gravity the water is weightless and depth stops mattering at all. Everything checks out.

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Connections

• ← Parent: full derivation & examples
• Pascal's Law — the cancelling-$A$ idea, pushed further
• Atmospheric Pressure & Barometer — where $P_0$ comes from
• Buoyancy & Archimedes' Principle — pressure difference between top and bottom faces
• Manometers — reading $\rho gh$ as a height
• Bernoulli's Equation — add motion; static limit returns this law
• Density and Specific Gravity — supplies $\rho$