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Fluids: pressure, upthrust and viscosity

Liquids and gases exert pressure on everything within them, and the pressure grows with depth. That explains flotation through Archimedes' principle, and a second property, viscosity, explains why objects fall slowly through a fluid. Between them they determine what floats, what sinks, and how fast a sphere falls through a liquid.

Year 12CIE 4.3.2, 4.3.3, 4.3.4, 4.3.5, 4.3.6 OCR A 3.2.4(b), 3.2.4(c)EDEXCEL 50, 51

Builds on Density and Hooke's law and Drag and terminal speed.

IN THIS TOPIC

Use p = F/A for a solid, a liquid or a gas, and say what changes between the three.
Derive Δp = ρgΔh from the definitions of pressure and density, and use p = ρgh for the pressure due to a column of fluid.
Explain upthrust as the weight of fluid displaced, and state the condition for floating.
Use Stokes' law for the viscous drag on a small sphere, and find its terminal velocity.

COMMON MISCONCEPTION

Things float because they are light.

Pressure: a force spread over an area

Pressure is the force acting at right angles to a surface, divided by the area of that surface.

p = \frac{F}{A}

The unit is the pascal, one newton per square metre, or kg m⁻¹ s⁻² in base units. Only a force and an area go in, so the same definition serves solids, liquids and gases alike. What differs between them is not the arithmetic but the directions the push is free to act in.

For a solid resting on something, the force is usually the weight and the area is the contact patch, so the geometry sets the pressure. A 60 kg woman standing flat-footed spreads about 600 N over some 0.03 m² of sole and presses the floor at roughly 20 kPa. Put her in stiletto heels and, for the instant her whole weight lands on one heel tip of half a square centimetre, the pressure passes 10 MPa. Sprung dance floors and aircraft cabins have rules about stiletto heels for exactly that reason. Snowshoes run the trick the other way, spreading one unchanged weight over a large area so the wearer stays on top of the snow. In every case the force never moved. Only the area did.

A liquid or a gas is not confined to one contact patch. A fluid presses at right angles on every surface it touches, and at any point inside it the pressure acts equally in all directions, sideways and upwards as much as down. In a gas that push is the drumming of molecules on the wall. That is what lets a fluid carry a force round a corner, as the brake line of a car does, where a solid rod could not.

WORKED EXAMPLE

One force, two pressures

A drawing pin is pushed into a board with a force of 20 N. Its flat head has an area of 1.0 cm² and its point an area of 0.010 mm². Find the pressure under each.

Convert both areas before anything else. 1.0 cm² = 1.0 × 10⁻⁴ m², and 0.010 mm² = 0.010 × 10⁻⁶ = 1.0 × 10⁻⁸ m².

Under the thumb, p = F/A = 20/(1.0 × 10⁻⁴) = 2.0 × 10⁵ Pa, about two atmospheres, and perfectly comfortable.

Under the point, p = 20/(1.0 × 10⁻⁸) = 2.0 × 10⁹ Pa, ten thousand times larger and far past the stress any timber can carry. Same hand, same force. The pin does nothing except change the area.

Pressure grows with depth

A fluid pushes on any surface inside it, and the deeper you go the harder the push. Where the extra pressure comes from is worth deriving rather than memorising, because the derivation uses nothing beyond the two definitions you already have.

Picture a column of the fluid standing on the surface in question, of cross-sectional area A, with that surface a depth Δh below the top of the column. Density gives the mass of the column, $m = \rho V = \rho A \Delta h$ , so its weight is $\rho A g \Delta h$ , and the whole of that weight rests on the area A. Pressure is force over area, so divide by A.

\Delta p = \frac{\rho A g \Delta h}{A} = \rho g \Delta h

The area cancels, and that cancellation is the entire result. The derivation assumes a fluid at rest, of uniform density, with g effectively constant over the depth; grant those and depth and density set the pressure and nothing else does, so the shape and width of the container never enter, and ten metres down feels identical in a lake and in a flooded mine shaft the width of a chimney. Measuring the depth h from the surface itself, the same equation is usually written

p = \rho g h

WORKED EXAMPLE

Pressure on a diver

Find the extra pressure on a diver 10 m below the surface of fresh water (ρ = 1000 kg m⁻³).

p = ρgh = 1000 × 9.81 × 10 = 9.8 × 10⁴ Pa.

Almost exactly one extra atmosphere, on top of the atmosphere already pressing on the surface: p = ρgh is the pressure difference from the surface, and the absolute pressure adds whatever acts there. Every ten metres of water adds another atmosphere, so divers count depth and never distance swum.

Upthrust: why anything floats

Because pressure grows with depth, the bottom of a submerged object is pushed up harder than its top is pushed down. The imbalance is a net upward force, the upthrust, and its size follows from p = ρgh applied to both faces:

FIG. 1The pressure arrows on a submerged block: longer from below than from above, because the bottom face is deeper. The difference is the upthrust, equal to the weight of the fluid the block displaces.

the upthrust equals the weight of fluid displaced by the object. An object floats when it manages to displace its own weight of fluid before it goes fully under. It sinks when the fluid it displaces, fully submerged, still weighs less than the object does. Lightness has nothing to do with it. A steel bar sinks and a steel ship floats, because the ship's shape shoves aside a hull-sized volume of water and the bar shoves aside a bar-sized one.

Viscosity and Stokes' law

Real fluids resist flowing, and the resistance is measured by the viscosity η. Smooth, layered flow is laminar. Push past a certain speed and it breaks up into chaotic turbulent flow, at which point the drag laws change completely. For a small sphere of radius r moving slowly enough to keep the flow around it laminar, Stokes' law gives the viscous drag.

F = 6\pi\eta r v

Drag proportional to speed means a falling sphere cannot accelerate for ever. It reaches terminal velocity once its weight is balanced by upthrust plus viscous drag.

FIG. 2A small sphere falling through a viscous fluid at terminal velocity: weight down, balanced by upthrust and Stokes' drag up. The three forces sum to zero and the speed is steady.

GUIDED PRACTICE

A ball bearing in glycerol

A steel sphere of radius 1.0 mm and weight 3.2 × 10⁻⁴ N falls through glycerol of viscosity 0.85 Pa s. The upthrust on it is 5.2 × 10⁻⁵ N. Find its terminal velocity.

Show the working

At terminal velocity, drag = weight − upthrust = 3.2 × 10⁻⁴ − 5.2 × 10⁻⁵ = 2.68 × 10⁻⁴ N.

Rearranging Stokes' law, v = F/6πηr = 2.68 × 10⁻⁴ / (6π × 0.85 × 1.0 × 10⁻³) = 0.017 m s⁻¹.

Under two centimetres per second for a small sphere in a thick oil, so the flow around it stays laminar and the law we used was the right model; whether Stokes' law applies depends on the sphere's size and the fluid as well as the speed, which is why the method uses small spheres in viscous liquids, well away from any container wall. Slow, steady and easy to time, so the falling-sphere method is how viscosity actually gets measured.

Viscosity falls steeply as a liquid warms. Cold engines are hard on their oil pumps for that reason, and a falling-sphere experiment that fails to record its temperature has measured nothing.

ASSESSMENT FOCUS

p = F/A is the definition, and it holds for a solid standing on a bench as much as for a fluid. The area wanted is the contact area, and halving it doubles the pressure with the force untouched. OCR names the solid case in its specification, so expect a heel, a snowshoe or a drawing pin to turn up.
Asked to derive Δp = ρgΔh, set the chain out in full: volume AΔh, mass ρAΔh, weight ρAgΔh, then divide by A. The step that matters is the cancellation of A, because that is what makes the result independent of the container.
p = ρgh gives the pressure due to the fluid column alone. Where a question wants the total pressure at depth, add atmospheric pressure on top of it.
Upthrust arguments score by naming the mechanism. Pressure on the lower face exceeds pressure on the upper face, and the resultant equals the weight of fluid displaced.
Terminal velocity working starts from the force balance, weight = upthrust + drag, written out before a single number is substituted. Stokes' law holds only for small spheres in laminar flow, so if a question flags turbulence or something large and fast, say the law does not apply and take the mark for it.
Board coverage varies here. Edexcel examines all of it, viscosity and Stokes' law included, while CIE stops at pressure and upthrust and OCR A at density and pressure. AQA keeps only the qualitative drag ideas in the mechanics unit, although its Turning Points option does print Stokes' law in the booklet.

CHECK YOURSELF

A wooden block of density 600 kg m⁻³ floats in water of density 1000 kg m⁻³. What fraction of the block is submerged?

Show a hint

Floating means the displaced water weighs exactly what the block weighs.

Show the answer

Floating means the block's weight equals the weight of the water it displaces, so ρ_blockV_wholeg = ρ_waterV_subg.

V_sub/V_whole = 600/1000 = 0.6. Three fifths of the block sits below the waterline, whatever its size or shape.

Pressure is force per unit area, for a solid as much as a fluid.

Pressure in a fluid is depth times density times g.

Upthrust is the weight of fluid displaced.

Drag on a slow small sphere is Stokes' law, and nothing faster.

WORKBOOK

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Use p = F/A for a solid, a liquid or a gas, and say what changes between the three.
Derive Δp = ρgΔh from the definitions of pressure and density, and use p = ρgh for the pressure due to a column of fluid.
Explain upthrust as the weight of fluid displaced, and state the condition for floating.
Use Stokes' law for the viscous drag on a small sphere, and find its terminal velocity.

Open the full revision checklist to track your progress across the whole unit.