Hydrostatics — pressure = ρgh, derivation: Question bank

Before we start, one reminder of the symbols so nothing here is a surprise:

$P$ = pressure (force per area, in pascals $Pa = N/m^{2}$ ).
$ρ$ = fluid density (mass per volume, $kg/m^{3}$ ), see Density and Specific Gravity.
$g$ = gravitational field strength ( $\approx 9.8 m/s^{2}$ ).
$h$ = vertical depth below the free surface.
$P_{0}$ = pressure at the surface (usually atmospheric).

True or false — justify

Each item is a claim. Verdict + reason on the reveal.

A dam wall must be built thicker at the bottom than at the top.

True. Pressure grows linearly with depth (

P = P_{0} + ρ g h

), so the water pushes hardest near the base; the wall needs the most material where the load is largest.

Two containers of water with the same depth but very different widths have the same pressure at the bottom.

True. Bottom pressure depends only on depth

h

, not on how much water sits beside the column — the classic hydrostatic paradox.

Doubling the depth doubles the gauge pressure of the fluid.

True for gauge pressure (

ρ g h

is proportional to

h

), but false for absolute pressure

P_{0} + ρ g h

, since

P_{0}

does not double.

If you tilt a straight tube of water, the pressure at its lowest end depends on the tube's length.

False. It depends on the vertical drop from the surface, not the slanted length; a longer tilted tube with the same vertical depth gives the same pressure.

Pressure at a point in a fluid acts only downward.

False. In a fluid at rest pressure acts equally in all directions at a point; that is exactly why the sideways forces on our imaginary column cancel and why a diver feels squeezed from every side.

Oil floating on water, both at depth

h

, produce the same bottom pressure as pure water of depth

h

False. Oil is less dense, so the same depth of oil contributes less

ρ g h

; the fluid's density matters, not just the height.

Atmospheric pressure adds nothing to the pressure a fish feels 5 m down.

False. The atmosphere still presses on the water surface and that push is transmitted downward (Pascal), so the fish feels

P_{0} + ρ g h

; only gauge readings hide

P_{0}

At the free surface of an open lake the gauge pressure is zero.

True. At

h = 0

the fluid column above vanishes, so

ρ g h = 0

; the absolute pressure there is just

P_{0}

Spot the error

Each item is a flawed argument. The reveal names the specific broken step.

"The pool holds 50,000 L and the bucket holds 5 L, so the pool floor feels far more pressure."

The error is confusing total weight with pressure. Only the column directly above a point presses on it; equal depth gives equal pressure regardless of total volume.

h

is the distance from where I'm standing to the edge of the tank, about 2 m."

h

is the vertical depth below the free surface, measured straight down — never a horizontal or diagonal distance to a wall.

"The question asks for absolute pressure, so the answer is

ρ g h

ρ g h

is gauge pressure. Absolute pressure requires adding

P_{0}

P = P_{0} + ρ g h

"Water is

1 g/cm^{3}

, so I plug

ρ = 1

into

P = ρ g h

Units mismatch — SI needs

ρ = 1000 kg/m^{3}

. Using

1

underestimates pressure by a factor of 1000.

"In the derivation I must include the force from the tank walls on the column."

The walls' forces are horizontal, so they contribute nothing to the vertical balance and correctly drop out. Including them as vertical forces is the mistake.

"A wider column of the same depth has more weight, so it should raise the bottom pressure."

A wider column has more weight but also more area (

W = ρ A h g

P = W / A

); the

A

cancels, leaving

ρ g h

independent of width.

"Since a barometer tube is thin, mercury only rises a little; use a fat tube to read a bigger pressure."

Barometer height depends on depth balance (

P_{0} = ρ g h

), not tube cross-section — a fat tube gives the same 760 mm. Area cancels here too.

Why questions

Why do the horizontal forces on the fluid column cancel in the derivation?

By symmetry, every sideways push on the curved wall is matched by an equal, opposite push across the column, and none of them point vertically — so they play no role in the up–down balance.

Why does pressure depend on depth but not on the shape of the container?

Only the fluid directly above a point contributes to its pressure; the extra water in a wide or oddly shaped vessel is supported by the base and walls, not stacked over that single point.

Why is pressure quoted in "mm of Hg"?

Because a mercury barometer converts pressure into a column height via

P = ρ g h

; stating the height (with

ρ_{H g}

and

g

known) fully specifies the pressure — see Atmospheric Pressure & Barometer.

Why does a submarine experience a net upward buoyant force even though pressure squeezes it from all sides?

Pressure is larger on the deeper bottom face than the shallower top face (because

P

grows with depth); that pressure difference is the buoyant force — see Buoyancy & Archimedes' Principle.

Why can we use Newton's first law (not the second) in this derivation?

The fluid is hydrostatic — at rest with zero acceleration — so net force on any parcel is zero; that force balance is exactly what gives

P = P_{0} + ρ g h

Why does a manometer read pressure as a difference in height between two liquid surfaces?

Because each side obeys

P = ρ g h

; equal pressures demand equal

ρ g h

, so a pressure difference shows up as a height difference

Δ h

— see Manometers.

Edge cases

What is the pressure exactly at the free surface (

h = 0

Gauge pressure is

0

and absolute pressure equals

P_{0}

; the formula

ρ g h

correctly gives zero fluid contribution there.

What happens to

ρ g h

in "zero gravity" (a freely falling or orbiting tank)?

With effective

g = 0

the hydrostatic pressure difference vanishes — the fluid presses equally everywhere and no depth dependence remains, since the weight that caused it is gone.

In a very deep ocean, is

P = ρ g h

with constant

ρ

exactly correct?

Only approximately — water is slightly compressible, so

ρ

rises with depth and true pressure grows a little faster than the constant-density formula predicts. For most problems the constant-

ρ

answer is fine.

If two immiscible liquids stack (oil over water), how do you find the bottom pressure?

Add each layer's contribution:

P = P_{0} + ρ_{oil} g h_{oil} + ρ_{water} g h_{water}

; each layer uses its own density and its own vertical thickness.

At the top of the vacuum in a mercury barometer, what is the pressure?

Essentially zero — there is (nearly) nothing above the mercury to press down, which is why the atmosphere's

P_{0}

can balance a full

ρ g h

column below.

For a static fluid, does pressure change as you move horizontally at the same depth?

No — at equal depth in a connected fluid the pressure is the same everywhere; only vertical movement changes

h

and hence

P

Recall One-line summary of every trap here

Pressure cares about vertical depth and density only ( $P_{0} + ρ g h$ ); it acts in all directions, ignores container shape and total volume, hides $P_{0}$ in "gauge" readings, and vanishes when either $h$ or $g$ goes to zero.

Connections

Parent: the full derivation
Pascal's Law — why $P_{0}$ is felt at depth
Buoyancy & Archimedes' Principle — pressure difference with depth
Manometers — pressure difference as height difference
Atmospheric Pressure & Barometer — mm-of-Hg reasoning
Density and Specific Gravity — the $ρ$ that must be in SI units