Fluid statics in Year 2: pressure, Stevin, Archimedes

End of September 2026: in Year 2 of the science high school (liceo scientifico), physics restarts with fluid statics — pressure, Stevin’s law, Pascal’s principle, Archimedes’ buoyant force, a nod to Torricelli. With 2 hours a week the first test typically arrives between late October and early November, so five to seven weeks from now. It is a topic that looks like all direct formulas, and indeed it is: what loses marks is not the reasoning, it is keeping the units in order while the formulas are combined one inside the other.

Pressure: force over area, and four units that coexist

Pressure is defined as force divided by the area on which it acts:

The International System unit is the pascal ($\text{Pa}$, $1\text{Pa}=1\text{N}/{\text{m}}^2$), but in fluid statics problems you regularly meet three others, and no test gives them to you already made uniform:

• atmosphere ($\text{atm}$): $1\text{atm}\approx 1.013\times 10^5\text{Pa}$, the average atmospheric pressure at sea level;
• bar: $1\text{bar}=10^5\text{Pa}$ exactly, so $1\text{atm}\approx 1.013\text{bar}$;
• millimetres of mercury ($\text{mmHg}$): $1\text{mmHg}\approx 133.3\text{Pa}$, the historical unit of Torricelli’s barometer, where $760\text{mmHg}=1\text{atm}$.

For the general method of converting between units — powers of ten, conversion factors, significant figures — what I have already written for measurements and significant figures in Year 3 applies: I don’t explain it again here, I concentrate on the specific conversions that, in fluid statics, trip up almost everyone.

Stevin’s law and communicating vessels

In a fluid at rest, pressure grows linearly with depth. Stevin’s law writes it like this:

where $p_0$ is the pressure at the surface of the fluid (usually atmospheric), $\rho$ is the density of the fluid, $g$ the acceleration due to gravity and $h$ the depth measured from the surface. The term $\rho gh$ is the pressure due to the weight of the fluid above alone: at the same depth, a denser fluid generates more pressure.

A direct consequence is the principle of communicating vessels: in two (or more) containers connected at the base and filled with the same fluid, the level of the liquid is the same in all the branches, regardless of the shape or cross-section of the container. If it were not so, there would be a pressure difference between two points at the same depth in the same fluid, and the fluid would move until that difference disappeared.

Pascal’s principle and the hydraulic press

Pascal’s principle says that a change in pressure applied at one point of an enclosed incompressible fluid is transmitted unchanged to every other point of the fluid and to the walls of the container. It is the principle behind the hydraulic press: two pistons of different area, connected by the same fluid, for which

A small force applied to the small piston generates, on the large piston, a much greater force — not because energy is “created” from nothing, but because the large piston moves much less than the small one: the work done stays the same.

Archimedes’ buoyant force and floating

A body immersed (partly or completely) in a fluid receives an upward force equal to the weight of the fluid displaced:

where $V_{\text{immersed}}$ is the volume of the part of the body actually below the surface of the fluid, not the total volume of the body. When a body floats in equilibrium, the buoyant force exactly equals the weight of the body:

from which you get the fraction immersed, which depends only on the ratio of the two densities and nothing else:

If ${\rho }_{\text{body}}<{\rho }_{\text{fluid}}$ the body floats (the fraction immersed is less than 1); if ${\rho }_{\text{body}}>{\rho }_{\text{fluid}}$ the body sinks, because the maximum possible buoyant force (body fully immersed) is not enough to support the weight anyway.

A nod to Torricelli and atmospheric pressure

Torricelli’s experiment — a glass tube closed at one end, filled with mercury and turned upside down in a basin of the same liquid — measures atmospheric pressure from the height of the column of mercury that stays suspended: at sea level, about $760\text{mm}$. It is the same Stevin’s law applied in reverse: the column of mercury stops at the height where the pressure it exerts at the base equals the atmospheric pressure pushing on the basin. This is where the mmHg unit you saw above comes from.

The typical mistake: pressure units, not physics

The point on which most marks are lost in a test is not understanding Stevin or Archimedes — the formulas, once seen, are linear and direct — it is keeping the units consistent as they pass from one piece of data to another. Three recurring traps:

1. Density in g/cm³ instead of kg/m³. Textbooks and lab data often give density in ${\text{g/cm}}^3$ (water is $1{\text{g/cm}}^3$), but in Stevin’s formula, with $g$ in ${\text{m/s}}^2$, you need it in ${\text{kg/m}}^3$. The conversion is to multiply by $1000$: $1{\text{g/cm}}^3=1000{\text{kg/m}}^3$. An error of a factor of 1000 in the density produces an error of a factor of 1000 in the calculated pressure, and often goes unnoticed because the number “looks” reasonable.

2. Areas in cm² left as they are in the hydraulic press. The ratio $F_1/S_1=F_2/S_2$ works with any unit of area, provided it is the same on both sides — but if a datum of the problem is in ${\text{cm}}^2$ and you also have to calculate the pressure in pascals, you need the area in ${\text{m}}^2$: you divide by $10^4$, not by $100$ (a common mistake, because people forget that area has two dimensions: $1{\text{m}}^2=100\text{cm}\times 100\text{cm}=10^4{\text{cm}}^2$).

3. Mixing atm, bar and Pa in the same equation. Stevin’s law requires $p_0$ and $\rho gh$ to be in the same unit before adding them: adding $1\text{atm}$ (without converting) to a $\rho gh$ calculated in pascals gives a number with no physical meaning. The safest rule: convert everything to pascals at the start of the problem, do the calculations, convert the final result to the required unit only at the end.

Three worked exercises

Exercise 1 — pressure at 10 metres under water.

A diver is at a depth of $h=10\text{m}$ in sea water, which we approximate with $\rho =1000{\text{kg/m}}^3$. The atmospheric pressure at the surface is $p_0=1.013\times 10^5\text{Pa}$. Calculate the total pressure at that depth, with $g=9.8{\text{m/s}}^2$.

I apply Stevin’s law directly:

Converting to atmospheres to get a concrete idea: $p\approx 1.99\times 10^5/1.013\times 10^5\approx 1.97\text{atm}$. At 10 metres, the pressure is almost double that at the surface: almost all the addition comes from the layer of water above the diver alone.

Before moving on to the next exercise, if you want to check that you have understood Stevin on other depth values, you will find the simulated test on fluid statics: the correct answer is free, the step-by-step guided solution is unlocked with your first lesson.

Exercise 2 — the hydraulic press.

A hydraulic press has a small piston of area $S_1=5{\text{cm}}^2$ and a large one of area $S_2=200{\text{cm}}^2$. A force $F_1=50\text{N}$ is applied to the small piston. What is the force $F_2$ transmitted by the large piston?

I convert both areas to square metres, dividing by $10^4$: $S_1=5\times 10^{-4}{\text{m}}^2$, $S_2=200\times 10^{-4}=2\times 10^{-2}{\text{m}}^2$. By Pascal, the pressure is the same on both pistons:

With a force of only $50\text{N}$ on the small piston, the press generates $2000\text{N}$ on the large one — 40 times as much, exactly the ratio between the two areas ($S_2/S_1=200/5=40$).

Exercise 3 — a wooden cube that floats.

A wooden cube has side $L=20\text{cm}=0.2\text{m}$ and density ${\rho }_{\text{wood}}=600{\text{kg/m}}^3$. It floats in water (${\rho }_{\text{water}}=1000{\text{kg/m}}^3$). Calculate the fraction of volume immersed and the immersed volume in cubic metres.

The fraction immersed depends only on the ratio of the densities:

60% of the cube stays under water, 40% emerges. For the volume in cubic metres I first calculate the total volume of the cube:

Check with the balance of forces: the weight of the cube is $P={\rho }_{\text{wood}}{V}_{\text{cube}}g=600\times 0.008\times 9.8\approx 47.0\text{N}$, and the buoyant force on the immersed volume found is $F_A={\rho }_{\text{water}}{V}_{\text{immersed}}g=1000\times 4.8\times 10^{-3}\times 9.8\approx 47.0\text{N}$: they match, the cube is in equilibrium.

Before the test

The quickest way to get stuck on these problems is not not knowing the formulas — there are four, and you will find them all above — it is arriving at the numerical calculation with a wrong unit in the middle: a density in g/cm³ not converted, an area in cm² left as it is. It is worth redoing the exercises in this post changing only the numbers, keeping the units in view on every line.

If this is the first block of Year 2 and you also want to review systems and first-degree equations before the year speeds up, there is the maths formula sheet for the start of Year 2. For dynamics and the inclined plane, the topic that follows fluid statics in the syllabus of many Year 2 classes, you will find the step-by-step method in this post on the inclined plane — it is written for the technical track, but the resolution of forces is identical.

The chapter that usually follows, in December, is thermology: specific heat and thermal equilibrium.

The overall picture of the subject is on the physics page, and the practice test on fluid statics is there for one last run before the test. If you want a hand, start from the free introductory call.