How a Roadside Jack Lets One Person Lift a Whole Car

Dive into a swimming pool and your ears feel the water pressing from all sides — harder the deeper you go. A fluid at rest pushes perpendicular to every surface it touches, and that push per unit area is pressure.

Pressure explains why dams are thicker at the bottom, how a small jack lifts a car, and how a column of mercury measures the weight of the atmosphere.

This part covers pressure at a depth, Pascal's law, gauge and absolute pressure, and barometers and manometers. Take $g=9.8$ m/s${}^2$.

**Pressure is normal force per unit area, $P=\frac{F}{A}$, measured in pascal; the extra pressure at depth $h$ in a fluid of density $\rho$ is $\rho gh$.

Why $\rho gh$.** A column of fluid of height $h$ and base area $A$ has weight $\rho (Ah)g$, which pressed on area $A$ gives

$$P=\frac{\rho Ahg}{A}=\rho gh$$

Worked example 1 — force over area. A $600$ N person standing on feet of total area $0.04$ m${}^2$ exerts

$$P=\frac{600}{0.04}=1.5\times 10^4\text{ Pa}$$

Worked example 2 — under water. At $10$ m depth in water:

$$\rho gh=1000\times 9.8\times 10=9.8\times 10^4\text{ Pa}$$

An everyday example. A sharp knife cuts vegetables easily because the same push acts on a tiny area, giving a huge pressure.

The substance. Pressure at a point in a fluid at rest is the same in all directions — it is a scalar, not a vector.

**Pascal's law states that a pressure change applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls, so a small force on a small piston produces a large force on a large piston: $F_2=F_1\frac{A_2}{A_1}$.

Worked example — a hydraulic lift.** A small piston of area $0.01$ m${}^2$ is pushed with $200$ N; the large piston has area $0.5$ m${}^2$.

$$P=\frac{200}{0.01}=2\times 10^4\text{ Pa}{F}_2=2\times 10^4\times 0.5=10^4\text{ N}$$

That lifts about $1000$ kg. But to raise the car $2$ cm, the small piston must move $50\times 2=100$ cm, because the fluid volume moved is the same:

$$W_{in}=200\times 1=200\text{ J}{W}_{out}=10^4\times 0.02=200\text{ J}$$

Hydraulic brakes. Pressing the brake pedal pushes a small piston in the master cylinder; the pressure travels through brake oil to larger pistons at each wheel, which press the brake pads with a much larger force — and equally on all wheels.

An everyday example. A hydraulic jack at a roadside tyre shop lets one mechanic lift a car with a few strokes of a handle.

The substance. A hydraulic machine multiplies force, not energy — the small piston must travel farther.

**Because of gravity, pressure increases with depth and is the same at all points on one horizontal level of a connected fluid at rest; absolute pressure is $P=P_a+\rho gh$, and gauge pressure is the excess over atmospheric, $P-P_a=\rho gh$.**

Atmospheric pressure at sea level is $P_a\approx 1.013\times 10^5$ Pa.

Worked example 1 — a diver. At $5$ m in fresh water:

$$P_{gauge}=1000\times 9.8\times 5=4.9\times 10^4\text{ Pa}{P}_{abs}\approx 1.013\times 10^5+0.49\times 10^5\approx 1.50\times 10^5\text{ Pa}$$

Worked example 2 — a terrace tank. Water stands $12$ m above a ground-floor tap and $3$ m above a top-floor tap.

$$\text{ground floor: }1000\times 9.8\times 12\approx 1.18\times 10^5\text{ Pa}\text{top floor: }\approx 2.9\times 10^4\text{ Pa (gauge)}$$

The hydrostatic paradox. Vessels of different shapes filled to the same height have the same pressure at the base, whatever the amount of water.

An everyday example. Taps on the ground floor gush harder than those on the top floor of a building fed by a terrace tank.

The substance. A tyre gauge reads gauge pressure — a flat tyre reads zero but still contains air at atmospheric pressure.

**A mercury barometer balances the atmosphere with a mercury column, $P_a={\rho }_{Hg}gh$; an open-tube manometer reads gauge pressure from the height difference, $\Delta P=\rho g\Delta h$; and in layered liquids the pressures of each layer simply add.

Worked example 1 — barometer.** A column of $76$ cm of mercury ($\rho =13600$ kg/m${}^3$):

$$P_a=13600\times 9.8\times 0.76\approx 1.013\times 10^5\text{ Pa}$$

A water barometer would need $\frac{1.013\times 10^5}{1000\times 9.8}\approx 10.3$ m.

Worked example 2 — manometer. The mercury levels differ by $20$ cm:

$$\Delta P=13600\times 9.8\times 0.20\approx 2.67\times 10^4\text{ Pa}$$

Worked example 3 — two liquids in a U-tube. A $12.5$ cm oil column balances $10$ cm of water:

$${\rho }_{oil}(0.125)=1000(0.10)\Rightarrow {\rho }_{oil}=800{\text{ kg/m}}^3$$

Worked example 4 — layers. A tank holds $2$ m of that oil above $3$ m of water. Gauge pressure at the bottom:

$$800\times 9.8\times 2+1000\times 9.8\times 3=15680+29400=45080\text{ Pa}$$

An everyday example. A doctor's blood-pressure instrument reports readings in millimetres of mercury, a manometer unit.

The substance. Only the vertical height of the column matters — tilting or widening a barometer tube does not change it.