10 min readConcept sheetSection A MCQ (1 mark) and objective (2 marks, two parts); Section B very short answer (2 marks): derive P = Patm + ρgh or explain why dam walls are thick at the base

Pressure in a liquid, atmospheric pressure and buoyancy

Liquid pressure grows with depth: P = Patm + ρgh. A body in a fluid feels an upward buoyant force equal to the weight of fluid it displaces.

PressureP = thrust ÷ area; SI unit N m⁻², called the pascal (Pa)

Pressure at depth hP = Patm + ρgh, where ρ is the density of the liquid and g = 9.8 m s⁻²

Atmospheric pressure76 cm of mercury = 1.01 × 10⁵ Pa at sea level

Archimedes' principleBuoyant force on a body = weight of the fluid it displaces

Thrust and pressure

Thrust is a force that acts at right angles (normal) to a surface. Pressure is thrust per unit area: P = F/A. The same force gives more pressure on a smaller area. That is why a sharp pin goes into a board easily, and a blunt one does not. A fluid (a liquid or a gas) at rest pushes at right angles on every surface it touches. The pressure of a fluid at rest is called hydrostatic pressure. Its SI unit is N m⁻², also called the pascal (Pa).

Derivation: pressure at a depth h

Imagine a small upright cylinder of the liquid itself, of cross-section area A and height h, inside a liquid of density ρ.

1
Let the pressure on the bottom face be P₁ and on the top face P₂. The liquid pushes up on the bottom with force P₁A and down on the top with force P₂A.
2
Mass of liquid in the cylinder = density × volume = ρAh. Its weight = ρAhg, acting downward.
3
The cylinder is at rest, so the net force on it is zero: P₁A − P₂A − ρghA = 0.
4
Divide by A: P₁ − P₂ = ρgh.
5
Now let the top face be at the open surface. Then P₂ = Patm, the atmospheric pressure, and P₁ = P, the pressure at depth h.
6★
So, the pressure at depth h is P = Patm + ρgh. It grows in a straight line with depth. The area A has cancelled, so the pressure does not depend on the shape of the vessel.

★ key step

Atmospheric pressure: how to draw Torricelli's barometer

The air around the Earth, up to a height of about 200 km, presses on everything. This is **atmospheric pressure**. Torricelli measured it with a tube of mercury (density 13600 kg m⁻³).

1
Draw a dish of mercury. Stand a glass tube about 1 m long upside down in it.
2
Shade the mercury in the tube up to a height marked h = 76 cm above the level in the dish.
3
Leave the space above the mercury empty and label it vacuum (almost zero pressure).
4
Draw small arrows pressing down on the open mercury surface in the dish. Label them atmospheric pressure.
5★
Working: the air pressure on the dish holds up the column. Patm = hρg = 0.76 m × 13600 kg m⁻³ × 9.8 m s⁻² = 1.01 × 10⁵ Pa.

★ key step

Buoyancy and Archimedes' principle

A bucket feels lighter while it is still under the water in a well. The water pushes it up. This upward push of a fluid on a body is the buoyant force (also called upthrust). Archimedes' principle: a body dipped in a fluid, fully or partly, feels a buoyant force equal to the weight of the fluid it displaces. To displace means to push aside. Compare the buoyant force B with the weight W. If B equals W, the body floats at rest. A body less dense than the fluid gets more push than its weight, so it rises. A denser body sinks. So a floating body displaces fluid equal to its own weight. A hot air balloon rises because hot air is less dense than the cold air around it.

Worked example: pressure on a diver

Problem

A diver works 20 m below the surface of the sea. Find the total pressure on her. (Density of sea water = 1030 kg m⁻³, Patm = 1.01 × 10⁵ Pa, g = 9.8 m s⁻²)

Working

Given: h = 20 m, ρ = 1030 kg m⁻³, Patm = 1.01 × 10⁵ Pa.
Formula: P = Patm + ρgh.
ρgh = 1030 kg m⁻³ × 9.8 m s⁻² × 20 m = 201880 Pa ≈ 2.02 × 10⁵ Pa.
P = 1.01 × 10⁵ Pa + 2.02 × 10⁵ Pa = 3.03 × 10⁵ Pa.

Answer

About 3.0 × 10⁵ Pa, roughly three times the atmospheric pressure

Worked example: a floating block of wood

Problem

A wooden block of volume 2.0 × 10⁻³ m³ floats in water with 60% of its volume under the surface. Find (a) the buoyant force on it, (b) the density of the wood. (Density of water = 1000 kg m⁻³, g = 9.8 m s⁻²)

Working

Volume under water = 0.60 × 2.0 × 10⁻³ m³ = 1.2 × 10⁻³ m³.
(a) Buoyant force = weight of water displaced = ρVg = 1000 × 1.2 × 10⁻³ × 9.8 N = 11.76 N.
(b) The block floats, so its weight = buoyant force = 11.76 N. Its mass = 11.76 ÷ 9.8 = 1.2 kg.
Density of wood = mass ÷ volume = 1.2 kg ÷ 2.0 × 10⁻³ m³ = 600 kg m⁻³.

Answer

(a) About 11.8 N; (b) 600 kg m⁻³. Shortcut: for a floating body, its density ÷ the liquid's density = the fraction of it under the surface

In real life

Big river dams such as the Bhakra dam on the Sutlej have walls far thicker at the base, because water pressure grows with depth. A steel ship floats because its hollow hull displaces water equal to its own weight.

Exam angle

2 marks: “Derive an expression for the pressure at a depth h below the free surface of a liquid of density ρ.” Take a liquid cylinder of area A and height h. Upward force P₁A, downward force P₂A, weight ρAhg. At rest: P₁A − P₂A = ρghA, so P₁ − P₂ = ρgh. With the top at the surface, P = Patm + ρgh.

2 marks: “Use the formula for liquid pressure to explain why a dam wall is built thickest at its foot.” The water pressure at depth h is ρgh, so it grows with depth. The foot of the wall faces the greatest pressure, so it is made thickest to withstand it.

1 marks: “State Archimedes' principle.” A body dipped fully or partly in a fluid feels an upward buoyant force equal to the weight of the fluid it displaces.

Remember it

Deeper means a bigger push: P = Patm + ρgh. Archimedes: the fluid you push aside pushes you up.