Pressure and Pascal's law
A sharp knife cuts easily because the same force acts on a very small area. Link to earlier classes : pressure = force ÷ area of class 8 and 9 is now used for liquids and gases, which are called fluids because they can flow.
Pressure is the normal force per unit area. Pascal's law : a pressure change applied to an enclosed fluid is passed on undiminished to every part of the fluid and to the walls.
Rule 1: Pressure — p = F ÷ A. Unit : pascal (Pa) = N m⁻². 1 atm = 1.013 × 10⁵ Pa. Pressure is a scalar; the force it produces is normal to the surface.
- Easy : F = 500 N on A = 0.25 m² ⇒ p = 2000 Pa.
- Medium : A 50 kg person standing on two feet of total area 0.04 m² exerts p = 50 × 9.8 ÷ 0.04 = 12 250 Pa.
- Exam : Air at 1 atm presses on a window of area 1.5 m² with a force of 1.013 × 10⁵ × 1.5 = 1.5 × 10⁵ N; the same pressure acts inside, so the window does not break.
Rule 2: Hydraulic machines — By Pascal's law F₂ ÷ A₂ = F₁ ÷ A₁, so F₂ = F₁ × (A₂ ÷ A₁). The force is multiplied, but the piston with the greater area moves less, so the work stays the same.
- Easy : A₁ = 10 cm², A₂ = 1000 cm², F₁ = 100 N ⇒ F₂ = 100 × 100 = 10 000 N.
- Medium : To lift a 1500 kg car (weight 14 700 N) with A₂ ÷ A₁ = 100 needs F₁ = 147 N.
- Exam : For the car to rise 10 cm, the small piston moves 100 × 10 cm = 10 m, so the work F₁d₁ = F₂d₂ is conserved.
p = F ÷ A; pressure is transmitted equally in all directions; a hydraulic lift multiplies force, not work.
Thinking the lift gives "free" energy : the distance moved by the small piston is larger by the same factor.
Memory clue : "Pascal passes pressure : same p everywhere, so F follows A."
Spot the difference
| Quantity | Type | Unit |
|---|---|---|
| Force | vector | newton |
| Pressure | scalar | pascal |
Pressure with depth and atmospheric pressure
Your ears feel pressure when you dive deep in a swimming pool, and a dam wall is built thicker at the bottom. Link to earlier classes : density from class 9 now appears in the pressure formula.
The pressure at a depth h in a fluid of density ρ is more than the pressure at the top by the weight of the column above : this extra pressure is the gauge pressure.
Rule 1: Pressure and depth — p = p_a + ρ g h (p_a = pressure at the surface, ρ = density, g = acceleration due to gravity, h = depth). Gauge pressure = p − p_a = ρgh. Pressure is the same at all points at the same depth and does not depend on the shape of the vessel.
- Easy : At h = 10 m in water, ρgh = 1000 × 9.8 × 10 = 9.8 × 10⁴ Pa, about one atmosphere.
- Medium : At 20 m depth the absolute pressure is 1.013 × 10⁵ + 1.96 × 10⁵ = 2.97 × 10⁵ Pa.
- Exam : The average height of the mercury column in a barometer is 76 cm. Then p = ρgh = 13 600 × 9.8 × 0.76 = 1.013 × 10⁵ Pa = 1 atm.
p = p_a + ρgh; pressure depends only on depth, not on the shape of the container.
Using the width of the vessel or the volume of the fluid in the formula.
Memory clue : "Every 10 m of water adds one atmosphere."
Spot the difference
| Pressure | Meaning |
|---|---|
| Absolute | total pressure p_a + ρgh |
| Gauge | only the extra ρgh |
Streamline flow and the equation of continuity
The water from a tap speeds up and becomes narrower as it falls. Link to earlier classes : the conservation of mass is now applied to a flowing liquid.
In streamline (steady) flow every particle passing a point follows the same path with the same velocity. A fluid that cannot be compressed obeys the equation of continuity.
Rule 1: Equation of continuity — A₁ v₁ = A₂ v₂ (A = cross-section area, v = speed). The fluid speeds up where the pipe is narrow.
- Easy : A₁ = 4 cm², v₁ = 2 m s⁻¹, A₂ = 1 cm² ⇒ v₂ = 8 m s⁻¹.
- Medium : If the radius of the pipe halves, the area becomes ¼ and the speed becomes 4 times.
- Exam : A river 20 m wide and 2 m deep flows at 1.5 m s⁻¹; where it narrows to 10 m width and 3 m depth the speed becomes 20 × 2 × 1.5 ÷ (10 × 3) = 2.0 m s⁻¹.
Rule 2: Types of flow — Flow is laminar at low speed (the layers slide smoothly) and turbulent at high speed (eddies form).
- Easy : Smoke rising from an incense stick is smooth at first and swirling higher up.
- Medium : Streamlines never cross.
- Exam : Closely spaced streamlines show a region of high speed.
A v = constant; narrow means fast.
Applying the continuity equation to a compressible gas without allowing for the change in density.
Memory clue : "Small area, Speedy flow : A × v stays put."
Spot the difference
| Flow | Speed | Layers |
|---|---|---|
| Laminar | low | smooth |
| Turbulent | high | eddies |
Bernoulli's principle and Torricelli's law
A passing fast truck seems to pull a nearby light car towards it, and an aeroplane wing gets lifted. Link to earlier classes : the work–energy ideas of chapter 5 are now applied to a flowing fluid.
Bernoulli's principle : for a steady, non-viscous, incompressible flow, the sum of pressure energy, kinetic energy and potential energy per unit volume is the same at every point of a streamline.
Rule 1: Bernoulli's equation — p + ½ρv² + ρgh = constant (p = pressure, ρ = density, v = speed, h = height). In a horizontal pipe, where the speed is high the pressure is low.
- Easy : Water flows at 2 m s⁻¹ in a wide part and at 8 m s⁻¹ in a narrow horizontal part. The pressure difference = ½ × 1000 × (8² − 2²) = 30 000 Pa; the narrow part has the lower pressure.
- Medium : A fast-moving air stream above a roof lowers the pressure above it, so the higher pressure inside can lift the roof.
- Exam : A wing is shaped so that air moves faster over its top than below, which lowers the pressure above and gives an upward lift.
In the figure : In the narrow part the speed is larger and the pressure is smaller.
Rule 2: Torricelli's law (speed of efflux) — A liquid leaves a small hole at a depth h below its surface with speed v = √(2 g h), the same as a body falling freely through h.
- Easy : h = 5 m ⇒ v = √(2 × 9.8 × 5) = 9.9 m s⁻¹.
- Medium : A hole 1.8 m below the surface gives v = √(2 × 9.8 × 1.8) = 5.9 m s⁻¹.
- Exam : A tank 2 m high has a hole at depth h. The horizontal range = 2√(h (H − h)); for H = 2 m the holes at depths 0.5 m and 1.5 m both give 1.73 m and a hole at the middle (1 m) gives the largest range, 2.0 m.
p + ½ρv² + ρgh = constant; high speed means low pressure; v = √(2gh) for efflux.
Using Bernoulli's equation for viscous or turbulent flow.
Memory clue : "Fast flow, Faint pressure."
Spot the difference
| Where | Speed | Pressure |
|---|---|---|
| wide pipe | low | high |
| narrow pipe | high | low |
Viscosity, Stokes' law and Reynolds number
Honey pours slowly while water flows quickly; a raindrop falls at a steady speed. Link to earlier classes : friction between solids now has a counterpart inside a moving fluid.
Viscosity is the internal friction of a fluid that opposes the sliding of one layer over another. The coefficient of viscosity η is measured in Pa s.
Rule 1: Stokes' law — A small sphere of radius r moving at speed v in a fluid of viscosity η feels a drag F = 6 π η r v. A falling sphere reaches a steady terminal speed v_t = 2 r² (ρ − σ) g ÷ (9 η) (ρ = density of the sphere, σ = density of the fluid).
- Easy : Doubling the radius of a falling sphere makes the terminal speed 4 times as large (v_t ∝ r²).
- Medium : A steel ball (ρ = 7800 kg m⁻³, r = 1.0 mm) falls in glycerine (σ = 1260 kg m⁻³, η = 0.85 Pa s). v_t = 2 × (10⁻³)² × 6540 × 9.8 ÷ (9 × 0.85) = 0.017 m s⁻¹.
- Exam : For a sphere lighter than the fluid (ρ < σ) the formula gives a negative value : it rises. Air bubbles rise in water.
Rule 2: Reynolds number — R_e = ρ v d ÷ η (d = diameter of the pipe). Flow is laminar for R_e below about 1000 and turbulent above about 2000.
- Easy : Water (ρ = 1000, η = 1 × 10⁻³) in a 2 cm pipe at 0.5 m s⁻¹ : R_e = 1000 × 0.5 × 0.02 ÷ 10⁻³ = 10 000, which is turbulent.
- Medium : A more viscous fluid has a smaller R_e, so it stays laminar.
- Exam : Doubling the speed doubles R_e.
Viscous drag F = 6πηrv; terminal speed v_t ∝ r²; Reynolds number decides laminar or turbulent flow.
Using Stokes' law for a large or very fast object.
Memory clue : "Terminal speed grows with Radius Squared : double the radius, four times the speed."
Spot the difference
| Quantity | Meaning | Unit |
|---|---|---|
| Viscosity η | internal friction | Pa s |
| Reynolds number | flow type indicator | none |
Surface tension and surface energy
A small needle can float on water, and raindrops are round. Link to earlier classes : the attraction between molecules (cohesion) now explains the behaviour of a liquid surface.
Surface tension S is the force per unit length acting on a line drawn on the liquid surface, or equivalently the surface energy per unit area.
Rule 1: Surface tension — S = F ÷ l = ΔW ÷ ΔA. Unit : N m⁻¹ = J m⁻². For water at room temperature S is about 0.072 N m⁻¹. S decreases as the temperature rises and when soap is added.
- Easy : A soap film is formed on a frame with a slider 5 cm long. With S = 0.030 N m⁻¹ the film pulls with F = 2 S l = 2 × 0.030 × 0.05 = 3 × 10⁻³ N (two surfaces).
- Medium : Enlarging a soap film from nothing to 4 cm × 3 cm (two faces) needs work = S × ΔA = 0.030 × 2 × 0.04 × 0.03 = 7.2 × 10⁻⁵ J.
- Exam : A liquid drop takes a spherical shape because a sphere has the smallest surface area for a given volume, so the surface energy is least.
S = F ÷ l = energy ÷ area; a soap film has two surfaces, so F = 2Sl.
Forgetting the factor 2 for a soap film.
Memory clue : "Soap film, Two faces : F = 2 S l."
Spot the difference
| Liquid | Surface tension (N m⁻¹) |
|---|---|
| Water | about 0.072 |
| Soap solution | about 0.030 |
| Mercury | about 0.465 |
Angle of contact, capillary rise and excess pressure
Water climbs up a thin cloth, and a candle wick brings melted wax upward. Link to earlier classes : the balance between cohesion and adhesion, and the pressure idea of section [2].
The angle of contact θ is the angle between the liquid surface and the solid wall, measured inside the liquid. In a thin tube the liquid rises or falls : this is capillarity.
Rule 1: Capillary rise — h = 2 S cos θ ÷ (ρ g r) (r = radius of the tube). Water has θ near 0° and rises; mercury has θ near 140° and is pushed down. A narrower tube gives a larger change in height.
- Easy : Water (S = 0.072, ρ = 1000) in a tube of r = 0.5 mm : h = 2 × 0.072 ÷ (1000 × 5 × 10⁻⁴ × 9.8) = 0.029 m, about 2.9 cm.
- Medium : Mercury (S = 0.465, ρ = 13 600, θ = 140°) in a tube of r = 1 mm gives h = 2 × 0.465 × cos 140° ÷ (13 600 × 10⁻³ × 9.8) = −5.3 × 10⁻³ m, a fall of about 5.3 mm.
- Exam : If the radius is halved, the rise doubles (h ∝ 1 ÷ r).
Rule 2: Excess pressure — Inside a liquid drop of radius r, the excess pressure is 2S ÷ r; inside a soap bubble (two surfaces) it is 4S ÷ r.
- Easy : A water drop of radius 1 mm has an excess pressure of 2 × 0.072 ÷ 10⁻³ = 144 Pa.
- Medium : A soap bubble of radius 5 mm (S = 0.030) has an excess pressure of 4 × 0.030 ÷ (5 × 10⁻³) = 24 Pa.
- Exam : The smaller bubble has the larger pressure inside, so air flows from a small bubble to a large one when they are connected.
h = 2S cos θ ÷ (ρ g r); excess pressure 2S ÷ r (drop) and 4S ÷ r (bubble).
Using 2S ÷ r for a soap bubble.
Memory clue : "Bubble has Both sides : 4S, drop has one : 2S."
Spot the difference
| Liquid–solid | Angle of contact | Level in the tube |
|---|---|---|
| Water–glass | acute | rises |
| Mercury–glass | obtuse | falls |