Fluid Pressure and Pascal's Principle

Beginner

Pressure tells you how hard a fluid pushes on each piece of a surface. Water and air are made of countless tiny molecules that never stop moving. They bump into everything they touch: the sides of a glass, your skin, your eardrums. Each bump is a tiny push, and all of them together, spread over an area, make the pressure. That is also why a fluid pushes on a surface whichever way the surface faces.

Step 1: pressure grows with depth. The deeper you go in a still liquid, the more liquid sits above you, and its weight presses on everything below. That is why your ears ache at the bottom of a pool, and why a dam is built much thicker at its base than at its top. A denser liquid, such as seawater, presses a little harder at the same depth.

Step 2: only the depth counts. A thin tube and a wide lake, filled to the same level, have the same pressure at the same depth; the amount of water makes no difference.

Check yourself: in the animation, which of the three holes squirts fastest, and why? Answer: the lowest one, because the water there is deepest and presses hardest.

Step 3: a trapped liquid passes a push along. Press on it in one place and the extra pressure reaches every point of the liquid equally (Pascal's principle). In the animation a push on the thin piston lifts a car standing on the wide piston. The wide piston has four times the area, so it receives four times the force.

Step 4: nothing comes for free. To raise the car, the thin piston must travel four times as far as the car rises.

Intermediate

What pressure measures

The air around us presses on every square metre with the weight of some 10 tonnes, yet our bodies are not crushed. Pressure is the force that a fluid exerts on each square metre of a surface it touches:
$p=\frac{F}{A}$
Its unit is the pascal, $1\mathrm{Pa}=1N/{\mathrm{m}}^2$. The air presses with about 101 kPa, and the fluids inside our bodies push back just as hard.

Pressure is not a force and has no direction of its own. At a point in a fluid at rest it has the same value whichever way a small surface faces, and the force it produces, $F=pA$, is always perpendicular to that surface. The absolute pressure is counted from vacuum, the gauge pressure from the atmosphere's 101 kPa: a tyre gauge reading 220 kPa means about 321 kPa absolute.

Why pressure grows with depth

Inside a liquid of density $\rho$ at rest, picture the vertical column of liquid above a small area $A$ at depth $h$. Its weight is $\rho Ahg$, and since the pressure on its sides is horizontal, only the liquid below can carry it; dividing by the area gives the gauge pressure at depth $h$:
$p=\rho gh$
Adding the pressure $p_0$ on the free surface, usually the atmosphere's, gives the absolute pressure:
$p=p_0+\rho gh$

In water the pressure rises by 9.81 kPa per metre, so 10 m add about 98 kPa, almost one atmosphere: a diver at 10 m bears twice the surface pressure. Mercury is 13.6 times denser, so in a barometer a column of only $101325/(13600\cdot 9.81)\approx 0.76\mathrm{m}$ balances the atmosphere; a water column would need 10.3 m, the limit for any suction pump.

Only the depth counts, not the shape of the vessel or the amount of liquid: a thin pipe and a wide lake filled to the same level have the same pressure at the same depth. This is the hydrostatic paradox. All points at one level of a connected, still liquid thus share one pressure, and connected vessels fill to the same height.

Pascal's principle and the hydraulic press

Pascal's principle states that a change of pressure applied to an enclosed liquid is transmitted undiminished to every point of the liquid and to the walls. The liquid's weight fixes only the differences of pressure, through $\rho gh$, so a push on the boundary must raise every pressure by the same amount.

A hydraulic press is built on this principle. With its two pistons, of areas $A_1$ and $A_2$, at about the same height, the pressures under them are equal, $F_1/A_1=F_2/A_2$, so
$F_2=\frac{F_1{A}_2}{A_1}$
For round pistons the area goes as the square of the diameter: doubling the diameter quadruples the force.

The price is distance. A liquid hardly compresses, so the volume pushed out of the narrow cylinder must enter the wide one:
$A_1{d}_1=A_2{d}_2$
Combined with the ratio of the forces, this gives
$F_1{d}_1=F_2{d}_2$
The work put in equals the work taken out: like a lever, the press multiplies force but divides distance, never creating energy.

Worked example: the tank and press in the animation

The tank is kept full, with holes 15, 35 and 55 cm below the surface. The purple labels give $\rho gh$ at each hole; for the lowest one
$p=1000\cdot 9.81\cdot 0.55\approx 5400\mathrm{Pa}=5.4\mathrm{kPa}$
and likewise 1.5 and 3.4 kPa for the upper two. Only this gauge part drives the jets, because the atmosphere also presses on the outside of each hole. The push becomes speed, $v=\sqrt{2gh}$ (see Bernoulli's Principle): about 1.7, 2.6 and 3.3 m/s, so the lowest jet is the fastest.

On the right, the wide piston is twice as wide as the thin one, so $A_2=4A_1$. To hold up the 1000 kg car it must push with $F_2=1000\cdot 9.81=9810\mathrm{N}$, so the hand needs only
$F_1=\frac{F_2{A}_1}{A_2}=\frac{9810}{4}\approx 2450\mathrm{N}$
These are the 2.45 kN and 9.81 kN of the label, and the green arrow under the wide piston is four times as long as the red one above the handle.

As the handle goes down, the readouts show $d_1$ reaching 48 cm while $d_2$ reaches only 12 cm, a quarter as much. The works agree: $F_1{d}_1=2452.5\cdot 0.48\approx 1177\mathrm{J}$ and $F_2{d}_2=9810\cdot 0.12\approx 1177\mathrm{J}$, the energy that raises the car 12 cm.

Worked example: the front glass of an aquarium

An aquarium is filled 60 cm deep, and its front pane is 1.5 m wide. The gauge pressure on the glass grows from zero at the surface to $1000\cdot 9.81\cdot 0.60\approx 5890\mathrm{Pa}$ at the bottom, so its average $\bar{p}$ over the pane is the value halfway down, about 2940 Pa. With a wetted area of $1.5\cdot 0.60=0.90{\mathrm{m}}^2$,
$F=\bar{p}A\approx 2943\cdot 0.90\approx 2650\mathrm{N}$
the weight of a 270 kg mass. The atmosphere presses on both faces of the glass and cancels.

The push does not depend on how far the tank extends behind the glass: a tank only 5 cm from front to back pushes just as hard. Doubling the depth to 1.20 m doubles both the average pressure and the area, so the force grows fourfold, to about 10.6 kN. Since most of it acts low down, tanks and dams are built strongest at the bottom.

Reading a pressure–depth diagram

Plotted against depth, the pressure is a straight line of slope $\rho g$: 9.81 kPa per metre in water but only 12 Pa per metre in air. The gauge-pressure line starts at zero; the absolute-pressure line is the same line lifted by $p_0\approx 101\mathrm{kPa}$.

Arrows drawn on a wall, one per depth and as long as the pressure there, outline a triangle whose area times the width of the wall is the total force: hence the aquarium's average of half the bottom value. Around a submerged object the arrows are longer underneath than on top, and their sum is the upward buoyant force.

Common mistakes

• More water does not mean more pressure: a narrow pipe and a lake, both 10 m deep, give the same 98 kPa at the bottom.
• Pressure is not a force: the 5.4 kPa at the lowest hole pushes with 0.54 N on a 1 cm² plug but with 5.4 N on a 10 cm² one.
• Keep gauge and absolute pressure apart: 20 m down in a lake the water adds 196 kPa, but the absolute pressure is 297 kPa.
• Use the ratio of areas, not of diameters: pistons 3 cm and 12 cm across multiply a force by $(12/3{)}^2=16$, not 4.

Where you meet it

Car lifts, jacks, hydraulic brakes and excavators run on Pascal's principle, usually with oil at several megapascals. Water towers stand high because each metre of height adds 9.81 kPa at the taps below, and when you stand, the blood pressure in your feet is about 100 mmHg (13 kPa) higher than at your heart. Next, Buoyancy and Archimedes' Principle turns the extra push from below into the force that floats ships, and Bernoulli's Principle follows pressure into moving fluids.

Advanced

Why pressure has no direction

A fluid is a material that cannot resist a shear: push sideways on a layer of water, however gently, and it flows. A viscous fluid such as honey resists the rate of shearing, but once it is at rest it carries no shear stress at all. So inside a fluid at rest the force across any small surface is at right angles to that surface, and the force per unit area is called the pressure. This section shows that the pressure at a point is a single number, the same whichever way the surface faces; everything later in the lesson rests on that fact.

Legend

$\hat{n}$	Unit vector at right angles to the slanted face, pointing out of the wedge.
$p_x$	Pressure on the face normal to $x$.	$\mathrm{Pa}$
$p_z$	Pressure on the face normal to $z$.	$\mathrm{Pa}$
$p_n$	Pressure on the slanted face.	$\mathrm{Pa}$
$dA$	Area of the slanted face.	${\mathrm{m}}^2$
$n_{x}dA$	Area of the face normal to $x$: the slanted face projected onto it, with $n_x$ the $x$-component of $\hat{n}$.	${\mathrm{m}}^2$
$n_{z}dA$	Area of the face normal to $z$: the slanted face projected onto it, with $n_z$ the $z$-component of $\hat{n}$.	${\mathrm{m}}^2$
$x$	Horizontal coordinate axis.	$\mathrm{m}$
$z$	Vertical coordinate axis, pointing up.	$\mathrm{m}$

Figure 1 shows a small wedge of fluid seen from the side. Look at its three faces. The left face is vertical, so its normal points along $x$; the bottom face is horizontal, so its normal points along $z$; the slanted face has area $dA$ and an outward unit normal $\hat{n}$ with components $n_x$ and $n_z$. The left and bottom faces are the slanted face projected onto planes at right angles to $x$ and to $z$, so their areas are $n_{x}dA$ and $n_{z}dA$. In the drawing $n_x=0.6$ and $n_z=0.8$, so the three faces have areas in the ratio $0.6:0.8:1$ and carry 3, 4 and 5 equal arrows. The two triangular ends of the wedge, parallel to the page, push only along $y$, so they play no part in the balance along $x$ or $z$.

Call the pressures on the three faces $p_x$, $p_z$ and $p_n$, without assuming that they are equal. The fluid outside pushes each face inward. Along $x$, the left face is pushed toward $+x$ with the force $p_x{n}_{x}dA$. The slanted face is pushed along $-\hat{n}$ with the force $p_{n}dA$, and the $x$-part of that push is $p_n{n}_{x}dA$, toward $-x$. The bottom face pushes only along $z$. The wedge is at rest, so the forces along $x$ cancel:
$p_x{n}_{x}dA=p_n{n}_{x}dA$
and therefore $p_x=p_n$.

Along $z$ the same count gives $p_z{n}_{z}dA$ up and $p_n{n}_{z}dA$ down, but the wedge's weight also enters. The weight is proportional to the wedge's volume, which shrinks as the cube of its size, while the surface forces shrink only as the square. Divide the balance by $dA$ and let the wedge shrink to a point: the weight term goes to zero, and $p_z=p_n$.

So $p_x=p_z=p_n$, and since the slant of the face was arbitrary, the pressure at a point is the same on every surface through it. Pressure is a scalar field $p(\vec{r})$: one number at each point, with no direction of its own. The direction of a pressure force comes from the surface. On a surface element of area $dA$ with outward unit normal $\hat{n}$, the fluid pushes with $d\vec{F}=-p\hat{n}dA$, the minus sign meaning inward.

The equation of hydrostatics

Pressure changes from point to point, and that change is what holds a fluid up against its own weight. This section turns the idea into one equation that gives the pressure everywhere in a fluid at rest.

Legend

$dx$	Width of the block along $x$.	$\mathrm{m}$
$dz$	Height of the block.	$\mathrm{m}$
$p(z)$	Pressure on the bottom face, at height $z$; it pushes up.	$\mathrm{Pa}$
$p(z+dz)$	Pressure on the top face, at height $z+dz$; it pushes down and is the smaller of the two.	$\mathrm{Pa}$
$p$	Pressure on each side face; the same on both, since they are at the same height.	$\mathrm{Pa}$
$\rho gdV$	Weight of the block: the density $\rho$ times $g$ times its volume $dV=dxdydz$.	$\mathrm{N}$
$x$	Horizontal coordinate axis.	$\mathrm{m}$
$z$	Height, measured upward.	$\mathrm{m}$

Figure 2 shows a small block of fluid at rest, $dx$ wide, $dz$ high and $dy$ deep into the page, so its volume is $dV=dxdydz$. Compare the arrows in opposite pairs. A uniform pressure squeezes the block equally from opposite sides and gives no net force; only a difference between opposite faces can push it.

Start with the sides. The two side faces are at the same height and carry the same pressure $p$, so their pushes are equal and opposite, and they cancel.

Now the top and the bottom. The bottom face, at height $z$, is pushed up with $p(z)dxdy$; the top face, at height $z+dz$, is pushed down with $p(z+dz)dxdy$. Over the small height $dz$ the pressure changes by $p(z+dz)-p(z)=(\partial p/\partial z)dz$, so the net upward push is
$[p(z)-p(z+dz)]dxdy=-\frac{\partial p}{\partial z}dV$
The pressure is larger below, so the arrow $p(z)$ is longer than the arrow $p(z+dz)$ and the net push is upward. The green arrow $\rho gdV$ is the block's weight, with $\rho$ the density and $g$ the gravitational acceleration. At rest the net push and the weight must be equal, and the figure draws the weight exactly as long as the difference between the two pressure arrows.

The same count works for every pair of faces. The faces normal to $x$ leave a net push $-(\partial p/\partial x)dV$ along $x$, which is zero for the two equal side arrows of the figure, and the faces normal to $y$ leave $-(\partial p/\partial y)dV$. The three pairs together give a pressure force per unit volume of $-\nabla p$. In equilibrium it cancels the weight per unit volume, $\rho \vec{g}$, so
$\nabla p=\rho \vec{g}$
This is the equation of hydrostatics. With the gravitational potential energy per unit mass $\varphi =gz$, for which $\vec{g}=-\nabla \varphi$, it reads $\nabla p=-\rho \nabla \varphi$. With $z$ measured upward, $\vec{g}$ has only a $z$-component, equal to $-g$, so the vertical part of the equation is
$\frac{dp}{dz}=-\rho g$
and the horizontal parts, $\partial p/\partial x=\partial p/\partial y=0$, say that the pressure is the same all along a level.

For a liquid the density is nearly constant, and the equation integrates at once to $p+\rho gz=\mathrm{const}$. The boundary fixes the constant: at the free surface, a height $h$ above the point considered, the pressure is the atmospheric pressure $p_0$, so
$p=p_0+\rho gh$
Here $h$ is the depth, measured down from the surface. The pressure grows with $h$ even though $dp/dz$ is negative; the two statements agree, because $z$ points up and $h$ points down.

In the animation's expert layer a sinking probe draws this law. It plots the gauge pressure $p-p_0$, the pressure above atmospheric, against the depth $h$: a straight line through the origin with slope $\rho g\approx 9.81kPa/m$, reaching about 7.0 kPa at the probe's deepest point, 71 cm down. The absolute pressure $p$ would give the same line lifted by $p_0\approx 101\mathrm{kPa}$. Both lines have the same slope, and only the slope, the gradient of $p$, exerts a net force on the fluid.

Isobars, levels and convection

The hydrostatic equation says more than how pressure grows with depth. It fixes the shape of the surfaces of equal pressure, and it tells when a fluid cannot stay at rest at all.

Since $\nabla p=-\rho \nabla \varphi$, the gradient of $p$ is parallel to the gradient of $\varphi$ at every point. A gradient is perpendicular to the surfaces on which its quantity is constant, so the surfaces of equal pressure, the isobars, are also surfaces of constant $\varphi$: the equipotentials of gravity. In a laboratory these are horizontal planes; on the scale of the planet they follow the curved shape of a calm sea. That is why connected vessels fill to one level: the free surface in each of them is part of the same isobar, $p=p_0$.

The same equation shows when no state of rest is possible. The curl of a gradient is always zero, so the curl of $\nabla p$ vanishes, and so must the curl of $\rho \nabla \varphi$:
$\nabla \times (\rho \nabla \varphi )=\nabla \rho \times \nabla \varphi =0$
A cross product vanishes only when its two vectors are parallel, so the density gradient must be vertical: $\rho$ must be the same all along each level. Warm one side of a room, or one side of a pan of water, and the density changes along a level. No pressure field can then balance gravity, and the fluid has to move. That motion is convection; it drives sea breezes and spreads the heat of a radiator round a room.

Pascal's principle and the work of the press

A hydraulic press lifts a car with the push of a hand. Two short arguments explain how: one from the pressure field, one from energy.

First the field. The equation $\nabla p=\rho \vec{g}$ involves only the gradient of $p$, so it fixes the pressure only up to an added constant, which the boundary sets. Raise the pressure at the boundary of an incompressible liquid by $\Delta p$ and the whole field rises by $\Delta p$, while its gradient stays $\rho \vec{g}$. Every point of the liquid feels the same increase: that is Pascal's principle. In a real liquid the increase does not arrive everywhere at once. It spreads as a pressure wave at the speed of sound, about 1480 m/s in water, so it crosses the 1.5 m wide press of the animation in about a millisecond.

Now energy, which gives the force ratio without mentioning pressure at all. Push the small piston, of area $A_1$, down by a small distance $\delta d_1$. The liquid does not compress, so the volume $A_1\delta d_1$ that leaves the small cylinder fills the large one, of area $A_2$, whose piston rises by $\delta d_2=(A_1/A_2)\delta d_1$. In an ideal press, with no friction, the work put in equals the work taken out, $F_1\delta d_1=F_2\delta d_2$, and so
$F_2=F_1\frac{A_2}{A_1}$
This is the result Pascal's principle gives, with $F_1=\Delta p{A}_1$ and $F_2=\Delta p{A}_2$. The argument is the principle of virtual work, the same one that explains a lever: the press trades distance for force.

Over a whole stroke each piston sweeps the same volume, $V=A_1{d}_1=A_2{d}_2$, so both works equal $\Delta pV$. In the animation the small piston is 26 cm across and is pushed with 2452.5 N, so the pressure under it is $\Delta p=2452.5/(\pi \cdot 0.13^2)\approx 46.2\mathrm{kPa}$. Its area is $0.0531{\mathrm{m}}^2$ and its stroke 0.48 m, so it sweeps $V\approx 0.0531\cdot 0.48\approx 0.0255{\mathrm{m}}^3$, and $W_1=W_2=\Delta pV\approx 1177\mathrm{J}$. That is the largest value both work readouts reach, at the bottom of the stroke.

Gases and the exponential atmosphere

In a liquid the density hardly changes with pressure; in a gas it is nearly proportional to it. To see what that does to the hydrostatic equation, we first need the pressure of a gas in terms of its molecules.

Gas pressure is the momentum that molecules deliver to a wall per unit time and per unit area. Take a wall at right angles to $x$. A molecule of mass $m$ that hits it with velocity component $v_x$ bounces back with $-v_x$, so it hands over the momentum $2m{v}_x$. The molecules with a given $v_x$ that strike an area $dA$ in a time $dt$ are those within a distance $v_{x}dt$ of it, a number proportional to $v_x$, so each group contributes in proportion to $v_x^2$. Only the half of the molecules moving toward the wall arrive, which cancels the factor 2, and averaging over all the molecules gives
$p=nm\overline{v_x^2}$
where $n$ is the number of molecules per unit volume and $\overline{v_x^2}$ is the mean of $v_x^2$. In thermal equilibrium each molecule has on average the kinetic energy $\frac{1}{2}kT$ for its motion along $x$, with $k$ Boltzmann's constant and $T$ the absolute temperature. So $m\overline{v_x^2}=kT$, and
$p=nkT$
For air at 101 kPa and 288 K this gives $n\approx 2.55\times 10^{25}{\mathrm{m}}^{-3}$.

Now the atmosphere, with $m$ the mean mass of an air molecule. The density is $\rho =nm$, so at a uniform temperature $\rho =mp/(kT)$: proportional to the pressure. The hydrostatic equation becomes
$\frac{dp}{dz}=-\frac{mg}{kT}p$
The pressure falls at a rate proportional to itself, so it falls exponentially with height. With $p_0$ now the pressure at sea level, $z=0$:
$p=p_0{e}^{-z/H},H=\frac{kT}{mg}\approx 8.4\mathrm{km}$
for air at 288 K. $H$ is the scale height, the climb over which the pressure falls by a factor $e$. The factor $e^{-mgz/(kT)}$ is the Boltzmann factor, which compares the potential energy $mgz$ of a molecule with the thermal energy $kT$; here it appears from the mechanical side.

The law agrees well with measurement. On the summit of Everest, 8.85 km up, it predicts about 35 kPa, close to the 34 kPa measured there; the real air is colder than 288 K, which lowers the pressure a little further. In 1648 a mercury barometer carried up the Puy de Dôme at Pascal's request showed the column falling by about 8 cm, because less air was pressing down from above.

Accelerating fluids and buoyancy

The hydrostatic equation also holds in a container that accelerates, provided we work in the container's own frame. It then explains why the surface of a sloshing tank tilts, and, with one more step, where buoyancy comes from.

Legend

$\vec{a}$	Acceleration of the lorry and its tank; also drawn reversed, as $-\vec{a}$, to build $\vec{g}-\vec{a}$.	$m/{\mathrm{s}}^2$
$\vec{g}$	Gravitational acceleration, pointing down.	$m/{\mathrm{s}}^2$
$\vec{g}-\vec{a}$	Effective gravity in the tank's frame; the pressure grows along it.	$m/{\mathrm{s}}^2$
$\theta$	Tilt of the free surface from the horizontal, equal to the angle between $\vec{g}$ and $\vec{g}-\vec{a}$; $\tan \theta =a/g$.	$\mathrm{rad}$

In a frame that accelerates with $\vec{a}$, every bit of fluid feels an inertial force $-\rho \vec{a}$ per unit volume besides its weight, so the equation becomes
$\nabla p=\rho (\vec{g}-\vec{a})$
The fluid behaves as if gravity were $\vec{g}-\vec{a}$, the effective gravity. Figure 3 shows a tank on a lorry that accelerates to the right with $\vec{a}$. On the right of the figure, $\vec{g}-\vec{a}$ is built by adding $\vec{g}$, straight down, and $-\vec{a}$, pointing backwards: the result points down and backwards. The pressure grows along $\vec{g}-\vec{a}$, so the isobars, drawn dashed, are perpendicular to it. The free surface is one of these isobars, $p=p_0$, so it tilts with them, and the water piles up at the back of the tank.

To find the tilt, look at the angle $\theta$ between the surface and the horizontal. The surface is perpendicular to $\vec{g}-\vec{a}$ and the horizontal is perpendicular to $\vec{g}$, so $\theta$ is also the angle between $\vec{g}$ and $\vec{g}-\vec{a}$. In the vector triangle on the right, the side opposite $\theta$ is $a$ and the side next to it is $g$, so
$\tan \theta =a/g$
The figure uses $a=g/2$ to make the tilt easy to see. Water in a tank on a lorry accelerating at 3 m/s² slopes at an angle $\theta$ with $\tan \theta =3/9.81\approx 0.31$, about 17°.

Buoyancy comes from the same field. The total pressure force on a body under the surface is the sum of $-p\hat{n}dA$ over its closed surface. A theorem of vector calculus turns this surface integral into the volume integral of $\nabla p$ over the space the body occupies, so for a fluid at rest
$-\oint p\hat{n}dA=-\int \nabla pdV=-\rho \vec{g}V$
with $V$ the body's volume. This force points up and equals the weight of the fluid the body displaces: Archimedes' principle. The pressure on the surface does not depend on what fills it, so the result holds for any body. In the accelerating tank the same integral gives $-\rho (\vec{g}-\vec{a})V$, which has a forward part; that is why a helium balloon in a car leans forward as the car speeds up.

Worked problem: the weight of the liquid in the press

The analysis of the press above ignored the weight of the liquid itself. This problem puts it back, to show how large the correction is and where the energy goes.

Legend

$A_1$	Area of the small piston.	${\mathrm{m}}^2$
$A_2$	Area of the large piston, four times $A_1$.	${\mathrm{m}}^2$
$F_1$	Force the hand exerts on the small piston.	$\mathrm{N}$
$F_2$	Force of the liquid on the large piston, equal to the car's weight.	$\mathrm{N}$
$p_1$	Pressure of the liquid under the small piston.	$\mathrm{Pa}$
$p_2$	Pressure of the liquid under the large piston, $F_2/A_2$.	$\mathrm{Pa}$
$z_1$	Height of the bottom of the small piston.	$\mathrm{m}$
$z_2$	Height of the bottom of the large piston.	$\mathrm{m}$
$z_1-z_2$	How much higher the small piston's bottom is; the pressure under it is lower by $\rho g(z_1-z_2)$.	$\mathrm{m}$

Figure 4 shows the animation's press filled with water, at the start of the stroke. Look first at the two piston bottoms. The small piston's, at height $z_1$, is 40 cm above the large piston's, at height $z_2$. During the stroke the small piston moves 48 cm down and the large one 12 cm up, so at the end the small piston's bottom is 20 cm below the other's.

The large piston carries the car, so the liquid must push it up with the car's weight, $F_2$. Over the area $A_2$ this needs the pressure $p_2=F_2/A_2\approx 46.19\mathrm{kPa}$, all through the stroke. The pressure falls with height, as the hydrostatic equation says, so under the small piston, higher by $z_1-z_2$, it is lower:
$p_1=p_2-\rho g(z_1-z_2)$
That is the shorter arrow $p_1$ in the figure. The hand must then push with $F_1=p_1{A}_1$.

At the start $\rho g(z_1-z_2)=9810\cdot 0.40\approx 3.92\mathrm{kPa}$; at the end the height difference is $-0.20\mathrm{m}$ and the term is $-1.96\mathrm{kPa}$. With $A_1=0.0531{\mathrm{m}}^2$ the hand must push with
$F_1=(46.19-3.92)\mathrm{kPa}\cdot 0.0531{\mathrm{m}}^2\approx 2.24\mathrm{kN}$
at the start and
$F_1=(46.19+1.96)\mathrm{kPa}\cdot 0.0531{\mathrm{m}}^2\approx 2.56\mathrm{kN}$
at the end, instead of the steady 2.45 kN found without the liquid's weight.

The height difference changes linearly along the stroke, from $0.40$ to $-0.20\mathrm{m}$, and so does the pressure under the small piston. The work of the hand is therefore the swept volume times the average pressure. The average height difference is 0.10 m, so the hand does
$W_1=V(p_2-\rho g\cdot 0.10\mathrm{m})=0.02548{\mathrm{m}}^3\cdot 45.21\mathrm{kPa}\approx 1152\mathrm{J}$

That is 25 J less than the 1177 J the car gains, and the difference comes from the water itself. The 25.5 kg of water pushed across leaves a column at the top of the small cylinder, centred 24 cm below the small piston's starting level, and ends as a layer under the large piston, centred 6 cm above that piston's starting level, which was 40 cm lower. Its centre of mass therefore drops by $0.40-0.24-0.06=0.10\mathrm{m}$ and releases $25.5\cdot 9.81\cdot 0.10\approx 25\mathrm{J}$: energy is conserved once the liquid is counted. Here the correction reaches 8.5 % of the force; real jacks work at megapascals, where a few kilopascals of $\rho g\Delta z$ are negligible.

Limits of validity

Every result above rests on a few assumptions, and each of them fails somewhere.

The first is the absence of shear, which holds only for a fluid at rest or moving as a rigid body, like the water in the lorry. In a flowing fluid the pressure also depends on the speed, and viscosity brings shear stresses back.

The second is constant density. It suits liquids very well: water compresses by about 0.005 % per atmosphere, and even at the bottom of the deepest ocean trench, nearly 11 km down under some 110 MPa, it is only a few per cent denser. For a gas it holds only over heights small compared with the scale height $H$: across a 10 m room the air pressure changes by 0.12 %, but across a mountain range the exponential law is needed.

The third is that surfaces are flat. Where a liquid surface is curved, surface tension adds a jump in pressure across it. In a water-filled glass tube of radius $r=0.5\mathrm{mm}$, the surface tension $\gamma \approx 0.072N/m$ lowers the pressure just under the curved surface by $2\gamma /r$, about 0.3 kPa, enough to pull the water some 3 cm up the tube.

The last is the continuum itself. A smooth field $p(\vec{r})$ needs fluid elements small compared with the apparatus yet holding many molecules. An air molecule at sea level travels about 70 nm between collisions, so that is easy there; above about 120 km it travels more than a metre, and the continuum picture fails.

Where this leads

Pressure in a fluid at rest is the starting point of several other lessons. Integrating it over a submerged body is the subject of Buoyancy and Archimedes' Principle. Bernoulli's Principle extends $p+\rho gz=\mathrm{const}$ to moving fluids by adding $\frac{1}{2}\rho v^2$, which is where the jet speed $\sqrt{2gh}$ comes from, and Drag and Terminal Velocity brings back the shear that a fluid at rest lacks. The molecular origin of $p=nkT$ is worked out in Kinetic Theory of Gases, convection returns in Conduction, Convection and Radiation, and the press's trade of force for distance is the one met in Simple Machines: Levers, Pulleys and Gears. Sound, finally, is a small and rapid ripple of pressure, the subject of Sound and the Doppler Effect.