Physics notes, chapter 2

Pressure

Textbook Physics Form 5 (DLP), pages 38–89

What this chapter covers

Explain liquid pressure and the factors that affect it; use P = hρg and compare densities with a U-tube (2.1)
Explain atmospheric pressure, how a barometer measures it, and its effects at high altitude and at depth (2.2)
Use a manometer to find gas pressure and convert between Pa, mm Hg, m H₂O and mbar (2.3)
State Pascal's principle and solve problems on hydraulic systems as force multipliers (2.4)
State Archimedes' principle, relate buoyant force to floating and sinking, and solve buoyancy problems (2.5)
State Bernoulli's principle and explain lift force, aerofoils and their applications (2.6)

2.1Pressure in liquids

Pressure is force per unit area, P = F ÷ A. Its S.I. unit is the pascal (Pa), where 1 Pa = 1 N m⁻².

The pressure at the base of a liquid column comes from the weight of the liquid above it. Using weight = ρVg and V = Ah, the pressure at depth h is:

Liquid pressure: P = hρg
h = depth (m), ρ = density of the liquid (kg m⁻³), g = 9.81 m s⁻².

What affects liquid pressure

Factor	Effect
Depth, h	Deeper means higher pressure (a dam wall is thicker at the base)
Density, ρ	A denser liquid gives a higher pressure at the same depth
Gravitational acceleration, g	Larger g gives a higher pressure
Mass, volume, shape or cross-section of the container	No effect

Points at the same level in a static liquid have the same pressure, and pressure at a point acts in all directions.

Comparing densities with a U-tube

Two liquids that do not mix balance in a U-tube. Points A and B at the same level have equal pressure, so h₁ρ₁g = h₂ρ₂g, which gives h₁ρ₁ = h₂ρ₂. Measure both heights from the level of the boundary between the liquids.

U-tube with liquid X and liquid Y, heights h1 and h2 measured from the level of A and B — Two immiscible liquids in a U-tube; A and B are at the same level. (Textbook Figure 2.6, p. 45)

Figure: Physics Form 5 (DLP) textbook, Figure 2.6, p. 45.

Actual pressure

The surface of a liquid is also pushed by the air, so the actual pressure at depth h is P = hρg + P_atm.

Worked example

A fish is 1.5 m deep in an aquarium. ρ = 1 050 kg m⁻³, P_atm = 100 kPa. Find the pressure due to the water and the actual pressure.
Water pressure = 1.5 × 1 050 × 9.81 = 15 451 Pa ≈ 15.5 kPa. Actual pressure = 15.5 + 100 = 115.5 kPa.

Applications

Water tank placed high in a house: the large height difference gives high pressure at the tap.
Intravenous drip bag hung above the patient: pressure from the liquid column pushes liquid into the vein.
Dam: thicker wall at the base, and the penstock to the turbines is low so the water is fast.
Siphon: transfers liquid from a higher container to a lower one; atmospheric pressure pushes liquid up into the tube.

Using the container's volume or width in P = hρg. Only depth, density and g matter. A narrow tube and a wide tank with the same depth give the same pressure.

Find the water pressure at a depth of 24 m in a lake (ρ = 1 000 kg m⁻³).

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P = 24 × 1 000 × 9.81 = 235 440 Pa ≈ 2.35 × 10⁵ Pa.

2.2Atmospheric pressure

Atmospheric pressure is the pressure caused by the weight of the layer of air above a surface on the earth.

It is like water pressure on a diver, except the weight is that of the column of air. It is about 100 kPa at sea level and is lower at high altitude because the air is thinner.

Barometers

In a mercury barometer a tube of mercury is inverted in a bowl of mercury. The column falls until its height is about 760 mm and the space above is a vacuum. The atmospheric pressure on the bowl surface supports the column, so P_atm = 760 mm Hg = 0.76 × 13 600 × 9.81 ≈ 1.01 × 10⁵ Pa.

Figure: Physics Form 5 (DLP) textbook, Figure 2.11, p. 50.

	Fortin barometer	Aneroid barometer
Works by	Height of mercury column	Changes in the shape of a partial-vacuum metal box
Size	Large, not portable	Small, portable
Accuracy	Higher (±0.1 mm Hg)	Lower (±1 mm Hg)
Reading	Slower	Quick, direct reading

Units of pressure

Unit	Use	Conversion
pascal (Pa)	S.I. unit; use in formulae	1 Pa = 1 N m⁻²
mm Hg	Medicine, meteorology, aviation	P = hρg with ρ = 13 600 kg m⁻³
m H₂O	Water in the sea, reservoirs and dams	P = hρg with ρ = 1 000 kg m⁻³
millibar (mbar)	Weather forecasts	1 mbar = 1 hPa = 100 Pa

Worked example

A blood pressure of 160 mm Hg in Pa: h = 0.16 m, P = 0.16 × 13 600 × 9.81 = 2.13 × 10⁴ Pa.

At high altitude and at depth

	High altitude	Extreme depth in the sea
Pressure	Lower (thin air, less oxygen)	Increases by about 1 atmosphere every 10 m
Effects on people	Faster breathing, less oxygen absorbed, dehydration, unclear thinking	Nitrogen dissolves in blood; bubbles form if pressure drops too fast
What is done	Climb slowly, drink water; aircraft cabins are pressurised	Ascend slowly; submarines have strong round hulls and controlled cabin pressure

Remember "10 m of water = 1 atmosphere", so at 30 m depth a diver feels 1 + 3 = 4 atmospheres.

The mercury column in a barometer is 756 mm high. What is the atmospheric pressure in Pa?

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P = 0.756 × 13 600 × 9.81 ≈ 1.01 × 10⁵ Pa (100 863 Pa).

2.3Gas pressure

A manometer is a U-tube of liquid that measures gas pressure. The gas pressure is compared with atmospheric pressure at the same level in the liquid:

P_gas = P_atm + h when the liquid is pushed down on the gas side (and P_gas = P_atm − h when it is pushed up on the gas side). Use the same unit for h and P_atm.

Mercury manometer connected to a flask of gas, height difference 40 cm — A mercury manometer with a height difference of 40 cm. (Textbook Figure 2.19, p. 58)

Figure: Physics Form 5 (DLP) textbook, Figure 2.19, p. 58.

Worked example

With P_atm = 76 cm Hg and h = 40 cm, P_gas = 76 + 40 = 116 cm Hg. In Pa: 1.16 × 13 600 × 9.81 = 1.55 × 10⁵ Pa. A wider tube would not change the answer, because liquid pressure does not depend on cross-section.

Mercury is used for higher gas pressures because it is dense, so the column is short. It is toxic, so digital manometers with pressure transducers are now more common.

Forgetting to add (or subtract) atmospheric pressure. The height h is only the difference from atmospheric pressure, not the gas pressure itself.

A mercury manometer shows a gas pressure of 180 kPa and the atmospheric pressure is 101 kPa. What is the difference in height of the mercury columns?

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Difference = 79 000 Pa. h = 79 000 ÷ (13 600 × 9.81) = 0.592 m ≈ 59 cm.

2.4Pascal's principle

Pascal's principle: pressure applied to an enclosed fluid is transmitted uniformly (equally) in all directions throughout the fluid.

In a hydraulic system the same pressure acts on both pistons: F₁ ÷ A₁ = F₂ ÷ A₂, so F₂ = (A₂ ÷ A₁) × F₁. The ratio A₂ ÷ A₁ is the multiplying factor. A large output piston gives a large output force.

Hydraulic system with a small input piston and a large output piston — Pressure is the same in the fluid, so the larger piston gets the larger force. (Textbook Figure 2.23, p. 62)

Figure: Physics Form 5 (DLP) textbook, Figure 2.23, p. 62.

Hydraulic brake: force on the pedal pushes fluid from the master cylinder to the slave cylinders, which press the brake pads.
Hydraulic jack: small pumping force on the small piston lifts a heavy load on the large piston; valves let the fluid move one way only.
Hydraulic fluid must be incompressible, with a high boiling point and a low freezing point, and should not corrode (water does not suit).

Worked example

An input force of 60 N acts on a cylinder of area 0.80 cm². The brake force needed is 840 N. A₂ = 840 × 0.80 ÷ 60 = 11.2 cm².

Areas can stay in cm² (or any unit) in F₁/A₁ = F₂/A₂ as long as both areas use the same unit. For circular pistons A = πd²/4, so compare diameters squared.

An input force of 4.0 N acts on a piston of area 0.50 cm². What force is produced on a piston of area 6.4 cm²?

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F₂ = (6.4 ÷ 0.50) × 4.0 = 51.2 N.

2.5Archimedes' principle

Archimedes' principle: an object partly or fully immersed in a fluid experiences a buoyant force equal to the weight of the fluid it displaces. F_B = ρVg

Buoyant force arises because pressure at the bottom of an object is greater than pressure at the top. It also equals the apparent loss in weight: F_B = weight in air − weight in liquid. Here ρ is the density of the fluid and V is the volume of fluid displaced (the immersed volume).

Floating and sinking

Condition	Resultant force	What happens
W = F_B	Zero (balanced)	Stationary: floating or suspended
W > F_B	Downwards	Sinks with acceleration
W < F_B	Upwards	Rises with acceleration to the surface

Three beakers: weight equals buoyant force, weight greater than buoyant force, weight less than buoyant force — State of floatation of an object in a fluid. (Textbook Figure 2.31, p. 70)

Figure: Physics Form 5 (DLP) textbook, Figure 2.31, p. 70.

A floating object displaces fluid with the same weight as the object.

Applications

Ship: hollow shape displaces a lot of water; the Plimsoll line shows the safe load for the water density.
Hydrometer: floats lower in a less dense liquid; reads density directly, with the smaller values at the top of the scale.
Submarine: ballast tanks filled with water sink it and emptied with compressed air let it rise.
Hot air balloon: heating the air makes the balloon's weight less than the buoyant force, so it rises; letting hot air out makes it descend.

Worked example

A boat of mass 600 kg floats with 2.8 m³ below the line L. Weight of boat = 600 × 9.81 = 5 886 N. Buoyant force = 1 000 × 2.8 × 9.81 = 27 468 N. Maximum load = 27 468 − 5 886 = 21 582 N.

Thinking a heavy ship sinks because it is heavy. What matters is whether its weight W equals the buoyant force F_B. A hollow ship displaces enough water for F_B to equal W, so it floats.

A small boat displaces 3.8 × 10⁻² m³ of sea water (ρ = 1 050 kg m⁻³). Find the buoyant force.

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F_B = 1 050 × 3.8 × 10⁻² × 9.81 = 391 N.

2.6Bernoulli's principle

Bernoulli's principle: when the velocity of a fluid increases, the pressure in the fluid decreases (and vice versa).

Fast-moving air has low pressure, so the higher pressure of the still air around pushes things towards the fast stream. Blowing across the top of a paper lifts it; blowing between two hanging balloons makes them move together.

In a Venturi tube the water moves fastest in the narrow section, so the pressure there is lowest and the water column in the tube above it is the shortest.

Lift force and the aerofoil

An aeroplane wing (aerofoil) is shaped so air flows faster over the top surface. The top has low pressure and the bottom has higher pressure, so there is a resultant lift force upwards. The tilt of the wing, the angle of attack, deflects air downwards, and by Newton's third law the reaction adds to the lift.

Application	How Bernoulli's principle is used
Bunsen burner	Fast gas at the narrow jet has low pressure and draws air in through the air hole
Racing car spoiler	Inverted aerofoil: air is faster below, so a downforce presses the tyres to the road
Curving football	Spin makes air faster on one side, so pressure differs and the ball curves
Aeroplane wing	Aerofoil shape and angle of attack produce lift

Aeroplane flight uses four forces: lift (up), weight (down), thrust (forward) and drag (backward).

Saying the wing is "sucked" up by low pressure only. It is the pressure difference: higher pressure below pushes up more than the lower pressure above.

Why does a spoiler on a racing car have an inverted aerofoil shape?

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Air flows faster underneath it, so the pressure below is lower than above. The resultant force acts downwards (downforce) and improves grip at high speed.