Pressure at Depth

Pick a vessel, a depth and a liquid. The page works out the pressure at the base, the force that pressure puts on it, and how that force compares with the weight of the liquid standing above.

Worked example — Pressure at Depth

Density of the liquid	998 kg/m³
Pressure at the base	4.896 kPa
Area of the base	0.0314 m²
Force the liquid puts on the base	153.8 N
What the vessel holds	15.708 L
Weight of that liquid	153.8 N
Force on the base ÷ weight of the liquid	1.000
Force on a vertical wall, per metre of width	1,224.0 N/m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Pressure at Depth

Pressure is force spread over an area, p = F/A, measured in pascals — one pascal is one newton on one square metre. In a liquid that is standing still, the pressure at a depth h below the surface is p = ρgh, where ρ is the density of the liquid and g is 9.81 m/s². Read the formula for what is missing from it: there is no area in it, no volume, and nothing about the container. All three of the page's claims come from that absence. Every pressure here is a gauge pressure, meaning the amount the liquid adds on top of whatever is already pressing on its surface.

What each symbol means

p: pressure, in pascals. 1 Pa = 1 N/m², and a kilopascal is a thousand of them
ρ: density of the liquid, in kg/m³. The only property of the liquid that appears anywhere on this page
h: depth below the surface, in metres. Depth, not height of the vessel, and not the amount of liquid
A: the area being pushed on, in m². Pressure times area is force
g: 9.81 m/s², so that ρgh comes out in pascals

Assumes: That the liquid is still and its density is the same everywhere. Both are good for water in a tank and poor for the atmosphere, which is why this page has liquids in it and no gases. Atmospheric pressure is left out of every figure because it cancels: the air pushes down on the surface of the liquid and up on the underside of the base with the same 101 kPa, so it never appears in the force the base carries. A depth gauge in a pool reads the same way, which is why the figures here are the ones you would measure.
Breaks when: Once the liquid moves. A pipe with flow in it obeys Bernoulli rather than this, and Bernoulli & the Venturi Effect is where the pressure a moving fluid loses to its own speed is worked out. Compressible fluids break the other assumption: air halves in density about every 5.5 km, so ρgh with a constant ρ is wrong for anything more than a few tens of metres of atmosphere. Very small vessels break it too — in a tube a millimetre or so across, surface tension pulls the liquid up or down by a height this page does not model at all. And the wall force here is per metre of a straight vertical wall; a curved dam carries its load partly by arching into the valley sides, which is a different calculation and the reason arch dams can be thin.

Four vessels, forty-two times the water, one reading 🖖

Press Flared, 68 litres and then Tube on a disc, 1.6 litres. Same base, same half metre of water, and the volume changes by a factor of 42. The pressure reads 4.896 kPa both times and the force on the base is 153.8 N both times. Only the depth and the density are in p = ρgh, and neither of them knows anything about the walls. The table below runs all four vessels at once, so you can watch two of its columns stand still while the rest move.

The base carries more than the liquid weighs 🖖

The tube on a disc holds 1.606 litres — about 1.6 kg — and its base is carrying 153.8 N, the weight of 15.7 kg. That is 9.78 times the water standing on it, and the flared vessel is the mirror image at 0.231. Nothing is being created: the sloping walls press down on the liquid in the tapered case and lift it in the flared one, and the difference is exactly what the base sees. The demonstration that carries Pascal's name is this taken to its limit — a few hundred grams of water in a long thin pipe, and the barrel it stands on splits — and the worked problem below is that barrel with the numbers in it.

Twice as deep is four times the load on the wall 🖖

The pressure on a dam wall is not one number, because it grows from nothing at the surface to ρgh at the bottom. Averaged over the wall it is half the maximum, and the wetted area grows with the depth too, so the force per metre of width is ½ρgh² — a square. Take a reservoir from 1 m to 2 m and the load on the wall goes from 4,896 to 19,585 N per metre, four times over. It is also why a dam is thick at the bottom and thin at the top: the resultant acts a third of the way up from the bed, not halfway.

Problem solved in full

Pascal's barrel: how much force does half a litre of water put on the lid? 6 steps
A sealed barrel has a lid 0.60 m across. A thin pipe of radius 5 mm is fitted into the lid and filled with water to a height of 5 m. Work out the force the water puts on the lid, and compare it with the weight of the water in the pipe, before reading on. Press Tube on a disc, 1.6 litres to see the same thing at a tenth of the scale.

Open this problem in the tool
1. The pipe adds pressure through its height and nothing else. Five metres of water:
  
  $p = ρ g h = 998.2 \cdot 9.81 \cdot 5 = 48, 962 Pa$
2. That pressure acts on the whole lid, which is a disc of radius 0.30 m:
  
  $A_{lid} = π R^{2} = π \cdot 0.3 0^{2} = 0.28274 m^{2}$
3. Force is pressure times area. Written as a mass it is easier to feel: the lid is carrying about 1.4 tonnes.
  
  $F = p A_{lid} = 48, 962 \cdot 0.28274 = 13, 844 N (\frac{13 , 844}{9 . 81} = 1, 411 kg)$
4. Now the water that did it. The pipe's bore is 5 mm, so its cross-section is tiny:
  
  $bore radius cross-section π r^{2} volume π r^{2} h mass ρ V weight ρ V g 0.005 m 7.854 \times 1 0^{- 5} m^{2} 3.927 \times 1 0^{- 4} m^{3} = 0.3927 L 0.392 kg 3.85 N$
5. Divide one by the other. Three hundred and ninety grams of water is putting 13,844 N on the lid:
  
  $\frac{F}{W} = \frac{13 , 844}{3 . 85} = 3, 600$
6. And here is the part worth keeping. Write the ratio out in symbols and the height cancels, because it appears in the pressure and again in the volume. So does the density, and so does g. The multiplication is the ratio of the two areas and nothing else — which is exactly how a hydraulic jack lifts a car.
  
  $\frac{F}{W} = \frac{ρ g h A _{lid}}{ρ g A _{pipe} h} = \frac{A _{lid}}{A _{pipe}} = (\frac{0 . 30}{0 . 005})^{2} = 6 0^{2} = 3, 600$
Answer

13,844 N on the lid, from 3.85 N of water: a multiplication of 3,600. That figure is (0.30 ÷ 0.005)² and would be the same for a pipe of any height and a liquid of any density. Pascal did not need a lot of water, he needed a thin pipe — and the barrel does not burst because of what is in the pipe, it bursts because of the ratio between the pipe and the lid.

$13, 844 N from 3.85 N \Rightarrow \frac{A _{lid}}{A _{pipe}} = 3, 600$

References (3)

How the result on this page was arrived at, and by whom. Stevin had the paradox in 1586, half a century before Pascal, and the paper follows what each of them actually added: A. F. Chalmers, "Qualitative novelty in seventeenth-century science: Hydrostatics from Stevin to Pascal." Studies in History and Philosophy of Science Part A 51, 2015, pp. 1-10. DOI 10.1016/j.shpsa.2015.01.001
Stevin's own hydrostatics, where the force on the base was first separated from the weight of the liquid: E. J. Dijksterhuis, "Hydrostatics." In Simon Stevin: Science in the Netherlands around 1600, Springer Netherlands, 1970, pp. 64-69. DOI 10.1007/978-94-010-3207-0_5
The paradox had a name and a readership long before it had a syllabus. This is the 1869 note it carried in Scientific American: "The Hydrostatic Paradox." Scientific American 21, 1869, p. 249. DOI 10.1038/scientificamerican10161869-249

By the 42Math team · Reviewed against published sources. Corrections welcome.