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Fluids: pressure, upthrust and viscosity

Pressure is force per unit area, and in a fluid at rest it grows with depth as p = ρgh, whatever the shape of the container holding it. The deeper face of a submerged object is pushed harder, giving an upthrust equal to the weight of fluid displaced. Stokes' law gives the viscous drag on a small sphere in laminar flow.

Pressure, upthrust and viscosity

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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.

Floating depends on density, not on being light. An object floats when the upthrust, equal to the weight of fluid displaced, can equal the object's weight; that needs its average density below the fluid's.

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 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 restrict stiletto heels for that reason. A snowshoe does the reverse, spreading the same weight over a much larger area and reducing the pressure on the snow.

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 the pressure results from molecular collisions with the surface. Pressure acting in all directions 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 above the stress any timber can carry. The force is the same in both cases; only the area differs.

Pressure grows with depth

A fluid pushes on any surface inside it, and the pressure increases with depth. The result follows from the definitions of pressure and density alone.

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. 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. The shape and width of the container never enter, so 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

Three vessels of different shape filled to the same depth: the pressure at the base is the same in all three, though the volumes are wildly different — FIG. 1Three vessels of very different shape standing side by side, a narrow chimney, a cone that widens downwards and a flask that bulges and then necks, each filled with water to the same depth of one metre. Under every base the same 9810 Pa is printed, worked out as ρgh from the depth drawn, while the volume of each drawn shape is printed beneath it and the largest holds nine times the smallest. A dashed line across all three at half the depth carries half the pressure. Beside them stands the column of area A and depth Δh the result comes from, its weight ρAgΔh divided by the area A it rests on. The area cancels, so depth and density fix the pressure and the shape of the container does not enter it.

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 the pressure depends on the depth and not on the 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:

A grey block submerged below a water surface, with short pressure arrows pressing down on its top face and longer amber arrows pressing up on its deeper bottom face: the imbalance is the upthrust. — FIG. 2The 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. 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. 3A 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 viscous liquid, so the flow around it stays laminar and Stokes' law was the right model. Whether it 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. The motion is slow and steady enough to time, which is why the falling-sphere method is used to measure viscosity.

Viscosity falls steeply as a liquid warms. Cold engines are harder on their oil pumps for that reason, and a falling-sphere measurement is only meaningful if the temperature is recorded.

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 A lists the solid case at 3.2.4, so solid-contact examples such as a heel, a snowshoe or a drawing pin may be set.
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 where a question describes turbulence, or an object that is large or fast, state that the law does not apply.
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.

Stokes' law applies to a small sphere in laminar flow only.

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.