What is pressure in a fluid?
A fluid is anything that can flow: liquids and gases. A fluid at rest cannot push sideways along a surface; it can only push straight into (normal to) the surface.
Pressure = normal force ÷ area: P = F/A. Unit: pascal (Pa) = N/m². Other units: 1 atm = 1.013 × 10⁵ Pa; 1 bar = 10⁵ Pa; 1 torr = 1 mm of mercury.
Pressure is a scalar. At a point in a still fluid it is the same in all directions. Density ρ = mass ÷ volume (water: 1000 kg/m³).
Pressure increases with depth
Imagine a thin water column of area A and height h. Its weight is mg = ρ(Ah)g. The water below must hold up this weight plus the air pushing from above. So:
P × A = P₀ × A + ρAhg ⇒ P = P₀ + ρgh
- P₀ = atmospheric pressure at the top.
- Gauge pressure = P − P₀ = ρgh (what a tyre gauge reads).
- Absolute pressure = P (includes the air).
This extra ρgh exists only because of gravity. In a spaceship in free fall, the water in a bottle has no weight, so pressure would be the same at the top and the bottom.
Example: 10 m of water gives ρgh = 1000 × 9.8 × 10 ≈ 1 × 10⁵ Pa, about one extra atmosphere.
Same depth, same pressure (hydrostatic paradox)
P = P₀ + ρgh has no area or shape in it. So vessels of any shape, filled to the same height with the same liquid, have the same pressure at the bottom. This surprising fact is called the hydrostatic paradox.
Also, in a connected still liquid, all points at the same level have the same pressure. That is why water stands at one level in the arms of a U-tube, and why a mason's clear water pipe shows a level line on two walls.
Barometer: air pressure holds up a mercury column of 76 cm at sea level, since P₀ = ρgh = 13600 × 9.8 × 0.76 ≈ 1.013 × 10⁵ Pa. An open-tube manometer reads gauge pressure by the height difference of liquid in its two arms.
Pascal's law
Pascal's law: If extra pressure is applied to an enclosed fluid, it is passed on without loss to every part of the fluid and to the walls of the container.
Why? Liquids hardly compress. Push one part and every part must share the push. Pressure at a point is also equal in all directions, because a tiny fluid wedge at rest must have balanced forces on all its faces.
Hydraulic lift and hydraulic brakes
Hydraulic lift: two pistons, small area a and big area A, joined by oil. A force f on the small piston makes pressure f/a. By Pascal's law the same pressure acts on the big piston:
F = (f/a) × A = f × (A/a)
If A/a = 50, a 200 N push lifts 10,000 N. No free energy though: the oil volume moved is the same, so the small piston moves 50 times farther than the big one. Work in = work out.
Hydraulic brakes: the pedal pushes a small piston in the master cylinder. Pressure travels through brake oil in pipes to every wheel. There, wider pistons push the brake pads onto the disc or drum with a bigger force. All wheels get equal pressure, so the car stops evenly.
Try it: the three-hole bottle
Make three small holes one above the other in a plastic bottle (ask an adult to help with a hot needle). Fill it with water over a sink. Predict which jet goes farthest, then look. The lowest jet has the most water above it, so the most pressure. In the 3D (step 6), move the depth slider from 1 m to 10 m and check that gauge pressure grows 10 times.
Key formulas and definitions
- P = F / A (Pa)
- P = P₀ + ρ g h
- Gauge pressure = P − P₀ = ρ g h
- Hydraulic lift: F = f × (A / a)
- Distances: d(small) × a = d(big) × A
- 1 atm = 1.013 × 10⁵ Pa = 76 cm of Hg
Worked examples
1. A 50 kg girl stands on one heel of area 1 cm². Find the pressure on the floor. (g = 10 m/s²)
Step 1: F = mg = 50 × 10 = 500 N. Step 2: A = 1 cm² = 10⁻⁴ m². Step 3: P = F/A = 500/10⁻⁴. Answer: 5 × 10⁶ Pa. That is why pointed heels dent floors.
2. Find the gauge pressure 20 m below the surface of a lake. (ρ = 1000 kg/m³, g = 9.8 m/s²)
Step 1: gauge P = ρgh. Step 2: = 1000 × 9.8 × 20. Answer: 1.96 × 10⁵ Pa (about 2 atm extra).
3. Find the absolute pressure at that 20 m depth. (P₀ = 1.01 × 10⁵ Pa)
Step 1: P = P₀ + ρgh. Step 2: = 1.01 × 10⁵ + 1.96 × 10⁵. Answer: 2.97 × 10⁵ Pa, nearly 3 atm.
4. Why is a mercury barometer about 76 cm tall, but a water barometer would need about 10.3 m?
Step 1: The column must satisfy ρgh = P₀ = 1.013 × 10⁵ Pa. Step 2: For mercury, h = 1.013 × 10⁵ / (13600 × 9.8) ≈ 0.76 m. Step 3: For water, h = 1.013 × 10⁵ / (1000 × 9.8) ≈ 10.3 m. Answer: mercury is 13.6 times denser, so its column is 13.6 times shorter.
5. In a hydraulic lift the pistons have radii 2 cm and 20 cm. What force on the small piston lifts a 1500 kg car? (g = 10 m/s²)
Step 1: Weight F = 1500 × 10 = 15000 N. Step 2: A/a = (20/2)² = 100. Step 3: f = F × a/A = 15000/100. Answer: f = 150 N.
6. In the same lift, the small piston is pushed down by 50 cm. How far does the car rise?
Step 1: Oil volume is conserved: d(small) × a = d(big) × A. Step 2: d(big) = 50 cm × (a/A) = 50/100. Answer: 0.5 cm. The big force comes with a small movement; work is the same (150 × 0.5 = 15000 × 0.005 = 75 J).
Common mistakes
- Thinking a wider vessel gives more pressure at the bottom. Pressure depends only on depth, density and g.
- Mixing up gauge and absolute pressure. Gauge = ρgh; absolute = P₀ + ρgh.
- Using radius ratio instead of area ratio in the hydraulic lift. Area goes as r², so a 10× radius means 100× force.
- Thinking a hydraulic lift makes free energy. The force grows but the distance shrinks by the same factor.