Big picture
A hydraulic lift raises a vehicle, a straw draws liquid and a small insect can stand on a water surface. These effects involve different aspects of fluid mechanics: pressure, flow, viscosity and surface tension. Liquids and gases are fluids because they cannot sustain a static shear stress without continuing deformation.
Choose the governing model before using an equation. Hydrostatic formulas apply to fluids at rest, while Bernoulli's equation requires ideal-flow assumptions. A capillary problem depends on surface forces and wetting, not simply on the pressure at the bottom of a wide tank.
Learning goals
Calculate fluid pressure and apply Pascal's law and buoyancy.
Use continuity and Bernoulli's equation with stated assumptions.
Explain viscosity, Stokes drag and terminal speed.
Calculate surface-energy changes and excess pressure.
Relate capillary action to contact angle and tube radius.
Core ideas
1. Pressure in a fluid at rest
Pressure is normal force per unit area, P = F/A, measured in pascals. Unlike force, pressure is a scalar; its local action on a surface is normal to that surface. At depth h in a liquid of uniform density ρ, P = P₀ + ρgh. Gauge pressure is ρgh, while absolute pressure includes surface pressure P₀.
For water at depth 1.8 m, taking ρ = 1,000 kg m⁻³ and g = 10 m s⁻², gauge pressure = 1,000 × 10 × 1.8 = 18,000 Pa. If atmospheric pressure is 100,000 Pa, absolute pressure is 118,000 Pa. Pressure at a depth does not depend directly on container shape, explaining the hydrostatic paradox. A mercury barometer balances atmospheric pressure with ρgh; a manometer compares pressures through a liquid-column difference.
2. Pascal's law and buoyancy
Pascal's law states that an externally applied pressure change is transmitted undiminished throughout an enclosed fluid. In an ideal hydraulic system, F₁/A₁ = F₂/A₂. If areas are 4 cm² and 160 cm² and the small piston receives 75 N, the large-piston force is 75 × 160/4 = 3,000 N.
The lift does not create energy: volume conservation means the smaller piston moves farther, A₁x₁ = A₂x₂. Archimedes' principle states that buoyant force equals the weight of displaced fluid. A submerged volume of 0.0025 m³ in water experiences upthrust 1,000 × 10 × 0.0025 = 25 N. A floating object displaces its own weight of fluid; submerged volume depends on the density ratio. An object need not be entirely underwater to experience buoyancy.
3. Streamlines and continuity
In steady flow, the velocity at a given point does not change with time. A streamline is tangent to local velocity, and streamlines cannot cross in a well-defined steady velocity field. Laminar flow occurs in ordered layers, while turbulent flow involves irregular fluctuations and mixing.
For steady incompressible flow through a tube, conservation of volume gives A₁v₁ = A₂v₂. If diameter halves, area becomes one-quarter, so speed becomes four times larger. With initial speed 1.5 m s⁻¹, the narrowed section has 6.0 m s⁻¹. This relation conserves volume flux, not speed. For compressible flow, use mass conservation ρAv constant instead of automatically taking Av constant.
4. Bernoulli's equation
For steady, incompressible, nonviscous flow along a streamline, P + ½ρv² + ρgh = constant. The terms represent pressure energy, kinetic energy and gravitational potential energy per unit volume. Every term has pressure units. In a horizontal pipe, increased speed accompanies reduced static pressure under these assumptions.
For water speeding from 2 to 5 m s⁻¹ at equal height, P₁ − P₂ = ½ρ(v₂² − v₁²) = 0.5 × 1,000 × (25 − 4) = 10,500 Pa. Do not infer that higher speed always means lower pressure in every real flow: pumps, friction and height differences can change the relationship.
Torricelli's law, v = √(2gh), follows for a small opening below a large tank's free surface with atmospheric pressure at both locations. At h = 0.80 m and g = 10, v = √16 = 4 m s⁻¹. A Venturi meter uses pressure differences associated with speed changes to infer flow. Dynamic lift and atomiser operation are related applications, but a complete account of aerodynamic lift also involves momentum changes in the surrounding fluid.
5. Viscosity and flow resistance
Viscosity describes resistance to relative motion between fluid layers. For a Newtonian fluid, shear stress is proportional to velocity gradient: F/A = η(dv/dy). Dynamic viscosity η has SI unit Pa s and dimensions M L⁻¹ T⁻¹. A larger η requires more force for the same layer geometry and speed difference.
For parallel layers with area 0.020 m², separation 0.001 m, relative speed 0.10 m s⁻¹ and η = 0.50 Pa s, F = ηAv/d = 0.50 × 0.020 × 0.10/0.001 = 1.0 N. Liquid viscosity usually decreases as temperature rises, whereas gas viscosity usually increases over ordinary ranges. Reynolds number, Re = ρvd/η, compares inertial and viscous effects; it is dimensionless. Transition values depend on geometry and disturbances, so do not present a single threshold as universal.
6. Stokes drag and terminal velocity
For a small sphere moving slowly through a viscous fluid at low Reynolds number, Stokes' law gives drag Fd = 6πηrv. Drag opposes relative motion. A falling sphere also experiences weight downward and buoyancy upward. At terminal speed, net force is zero:
(4/3)πr³ρs g = (4/3)πr³ρf g + 6πηrvₜ.
Therefore vₜ = 2r²(ρs − ρf)g/(9η) for a denser sphere. For r = 0.001 m, density difference 1,000 kg m⁻³, g = 10 and η = 1.0 Pa s, vₜ = 2 × 10⁻⁶ × 1,000 × 10/9 = 0.00222 m s⁻¹. Doubling radius would quadruple this speed if the low-Reynolds-number approximation remained valid. Terminal speed does not mean gravity disappears; the forces balance.
7. Surface tension and surface energy
Surface tension, S, is force per unit length acting tangentially to a liquid surface and perpendicular to an imaginary line on it. Its unit is N m⁻¹, equivalent to J m⁻². Surface molecules have a different neighbour environment from interior molecules, so increasing surface area requires work. For one interface, ΔW = SΔA when S is constant.
A soap film on a movable wire has two surfaces, giving F = 2Sl. With S = 0.030 N m⁻¹ and l = 0.080 m, F = 0.0048 N. If the wire moves 0.020 m, total new area is 2 × 0.080 × 0.020 = 0.0032 m² and work is 0.000096 J. The factor two comes from two interfaces, not from two ends of the wire.
8. Drops, bubbles and capillarity
The excess pressure inside a liquid drop is ΔP = 2S/r. A soap bubble has inner and outer interfaces, giving ΔP = 4S/r. With r = 0.002 m and S = 0.040 N m⁻¹, drop excess pressure is 40 Pa and soap-bubble excess pressure 80 Pa. Smaller bubbles have greater excess pressure, so two connected unequal soap bubbles tend to transfer air from the smaller to the larger.
The contact angle, θ, is measured through the liquid between surface tangent and solid wall. A wetting liquid with θ less than 90° rises in a fine tube; a nonwetting liquid can be depressed. Balancing upward surface force with column weight gives h = 2S cosθ/(ρgr). For water with S = 0.072 N m⁻¹, θ = 0°, r = 0.00050 m and g = 10, h = 0.144/5 = 0.0288 m, or 2.88 cm. A narrower tube increases the rise. Mention the idealised meniscus and uniform-radius assumptions when explaining the derivation.
9. Experiments and diagrams
A useful hydrostatic demonstration connects side holes at different depths in a container. Lower holes initially produce stronger jets, consistent with larger pressure head; compare holes while accounting for the falling water level. A capillarity diagram should label tube radius, meniscus, contact angle and rise relative to the outer liquid level.
In a falling-sphere viscosity experiment, measure speed only after the sphere reaches approximately uniform motion. Use a sphere small enough for Stokes' assumptions and a container wide enough to reduce wall effects. Observation of constant speed supports balanced forces; it does not imply absence of fluid resistance.
A reliable approach to questions
Decide whether the fluid is static or flowing and list model assumptions.
Distinguish absolute pressure from gauge or excess pressure.
Use continuity before Bernoulli when areas change.
Draw all forces before finding buoyancy or terminal speed.
Count interfaces before using surface-energy or bubble formulas.
Connections that matter
Newton's laws explain hydrostatic force balance and terminal motion.
Energy conservation underlies Bernoulli's equation and hydraulic work.
Molecular interactions explain viscosity, wetting and surface energy.
Common mistakes
Using diameter as radius causes errors in capillary and bubble formulas.
Omitting atmospheric pressure confuses absolute and gauge values.
Applying ideal Bernoulli across large friction losses ignores dissipation.
Giving a soap bubble the drop formula omits its second surface.
Revision checklist
I can solve hydrostatic and hydraulic problems with units.
I can combine continuity and Bernoulli appropriately.
I can derive terminal speed from a force diagram.
I can distinguish drops, bubbles and films.
I can explain capillary rise through wetting and force balance.