P6 Matter in bulk: solids and fluids

explain stiffness, strength, pressure, buoyancy and flow.

Before you begin: P2 · Forces and Newton's laws, P3 · Energy and momentum.

Videos coming soon

Five videos (part 1: stiffness and strength; part 2: breaking things; part 3: density and pressure; part 4: floating and surfaces; part 5: flow) and a worked-example video (will it float?) are being made, and will appear here once they're uploaded to YouTube.

233 steps about 159 min

Watch first

Notes available, video soon.

Notes

In this chapter

  • Use stress \(\sigma = F/A\), strain \(\varepsilon = \Delta L/L\) and Young's modulus \(E = \sigma/\varepsilon\) to predict how far a wire stretches, \(\Delta L = FL/(EA)\).
  • Read a stress–strain curve (elastic, yield, plastic, necking, fracture), and tell stiff, strong and tough apart.
  • Explain why holes, corners and cracks weaken parts, how fatigue grows cracks, and why long, thin columns buckle, \(F = \pi^2 EI/L^2\).
  • Use density \(\rho = m/V\) and pressure \(p = F/A\), find the pressure at a depth, \(\rho g h\), and tell gauge from absolute pressure.
  • Explain the barometer and the drinking straw, and use Pascal's principle to multiply force (but not work) in a hydraulic press.
  • Use Archimedes' principle to decide what floats, how much of it is under water, and why a ship stays upright.
  • Use surface tension, \(F = \gamma L\), and capillary rise, \(h = 2\gamma/(\rho g r)\).
  • Use continuity, \(A_1 v_1 = A_2 v_2\), Bernoulli's equation, viscosity and the Reynolds number, and explain how wings really lift.
Part Solids under stress

Stress and strain

FIRST, MAKE A PREDICTION

Two bars carry the same tensile force. One has twice the cross-sectional area. Which has greater stress?

Show the answer

The thinner bar, by a factor of two

Stress is force divided by area, so spreading the same force across twice the area halves the stress.

TRY IT / FOLLOW THE REASONING

Change the bar's cross-sectional area with the load fixed at 1000 N. Compare stress at 5 and 10 mm², then distinguish this from strain, which compares stretch with original length.

Spread the same load

Cross-sectional area: 5 mm²

Tensile stress
200 MPa
Model assumptions

A 1000 N tensile load on a uniform bar. Areas are in mm²; stress is in N/mm², the same as MPa. Stretch is exaggerated in the drawing.

MAKE THE CONNECTION

Key idea

Stress is force per area; strain is stretch per length. Both take the shape out, so you can compare materials.

A spring constant \(k\) (from \(F = kx\)) describes one object, not a material: a wire twice as long stretches twice as far, and two wires side by side (like one with twice the cross-sectional area) stretch half as far. To compare materials, take the shape out. Stress is force shared over the area (like friends sharing a sofa):

Stress and strain

\[ \sigma = \frac{F}{A} \]

with \(\sigma\) (sigma) the stress, \(F\) the force in N and \(A\) the area in m². The unit is the pascal, 1 Pa = 1 N/m² (as for pressure); 1 MPa = \(10^6\) Pa, 1 GPa = \(10^9\) Pa.

Stress and strain

Worked number: a wire

A 10 kg mass on a 1 mm² wire: \(98\ \text{N}/10^{-6}\ \text{m}^2 \approx 98\) MPa. (1 mm² is \(10^{-6}\) m², not \(10^{-3}\).)

Stress and strain

Strain is stretch per length, \(\varepsilon = \Delta L/L\), with \(\Delta L\) the stretch and \(L\) the original length. It has no units.

Stress and strain

Worked number: strain

1 mm on 2 m is 0.0005, or 0.05% (divide a percentage by 100 before using it).

Checkpoint: Stress and strain (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A 500 N load hangs from a rod whose cross-section is 2.5 cm². What is the stress in the rod, in megapascals?

MPa

Show the answer

2 MPa

\(\sigma = F/A = 500/(2.5\times10^{-4}) = 2.0\times10^6\) Pa, which is 2.0 MPa.

Checkpoint: Stress and strain (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A 4.0 m cable stretches by 6.0 mm. What is its strain, as a percentage?

Show the answer

0.15

\(\varepsilon = 6.0/4000 = 0.0015\), which is 0.15%.

Young's modulus and how far it stretches

FIRST, MAKE A PREDICTION

Two wires have the same material and area but one is twice as long. Under the same small elastic load, how do their extensions compare?

Show the answer

The longer wire extends twice as much

The same stress gives the same strain; extension is strain times original length.

TRY IT / FOLLOW THE REASONING

Write ΔL = FL/(AE). Change one of F, L, A and E at a time, then compare a longer wire with a thicker wire without changing the material.

MAKE THE CONNECTION

Key idea

Young's modulus \(E\) is stress ÷ strain: a property of the material, not the object. It measures stiffness.

Below its limit, a material's stress is proportional to its strain. Young's modulus is

Young's modulus and how far it stretches

\[ E = \frac{\sigma}{\varepsilon} \]

in pascals (usually GPa); it belongs to the material, not the object. 100 MPa giving a strain of 0.0005 means 200 GPa: steel.

Young's modulus and how far it stretches

material Young's modulus \(E\)
steel 200 GPa
aluminium, glass about 70 GPa
bone 18 GPa
wood (along the grain) 10 GPa
nylon 2–4 GPa
rubber 0.01 GPa or less

Steel's bonds are stiff springs; rubber's tangled chains just straighten out. Shear (sliding layers) is resisted by the shear modulus, squeezing all round by the bulk modulus.

Combining the definitions gives the stretch:

Young's modulus and how far it stretches

\[ \Delta L = \frac{FL}{EA} \]

Young's modulus and how far it stretches

Worked number: a steel wire

A steel wire 2 m long, 1 mm², carrying 100 N stretches \(200/200\,000 = 1\) mm, storing \(\tfrac12 F\Delta L = 0.05\) J (the same as \(\tfrac12 kx^2\)).

Young's modulus and how far it stretches

Optional maths: the spring constant

\(F = (EA/L)\,\Delta L\) matches \(F = kx\) if \(k = EA/L\): double \(L\) and \(k\) halves; double \(A\) and it doubles.

Checkpoint: Young's modulus and how far it stretches (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A test bar carries a stress of 35 MPa and shows a strain of 0.00050. What is its Young's modulus, in gigapascals?

GPa

Show the answer

70 GPa

\(E = 35\times10^6/0.00050 = 7.0\times10^{10}\) Pa = 70 GPa: aluminium.

Checkpoint: Young's modulus and how far it stretches (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A copper wire 3.0 m long with a cross-section of 0.50 mm² carries 40 N. Taking \(E = 120\) GPa, how far does it stretch, in millimetres?

mm

Show the answer

2 mm

\(\Delta L = 40\times3.0/(120\times10^9\times0.50\times10^{-6}) = 2.0\times10^{-3}\) m = 2.0 mm.

Checkpoint: Young's modulus and how far it stretches (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A wire is replaced by one of the same material and length but twice the cross-sectional area. Under the same load, how far does it stretch compared with before?

Show the answer

Half as far.

\(\Delta L = FL/(EA)\), so doubling \(A\) halves the stretch. Two wires side by side are like one with twice the area.

Size and the square–cube law

FIRST, MAKE A PREDICTION

An animal is scaled to twice every length with the same proportions and material properties. How do weight and load-bearing cross-sectional area scale?

Show the answer

Weight ×8, area ×4

Weight follows volume while supporting cross-section follows area, so the load per area increases.

TRY IT / FOLLOW THE REASONING

Double and then triple the length multiplier. Divide the volume multiplier by the area multiplier to see how supporting stress changes for a geometrically similar object.

Scale the same shape

Length multiplier: 1 ×

Area multiplier
1 ×
Volume multiplier
1 ×
Model assumptions

Similar shapes: every length changes by the same factor. Bars use a common scale.

MAKE THE CONNECTION

Key idea

Scale an object up and its weight grows faster than its strength.

A rope twice as thick has four times the area, so it holds four times the load.

Size and the square–cube law

Worked number: the square–cube law

Double every length of an animal and its bone area grows \(2^2 = 4\) times while its weight grows \(2^3 = 8\) times: the stress in its bones doubles (the square–cube law). Scale lengths by 10 and the stress grows 10 times.

Checkpoint: Size and the square–cube law (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A model of an animal is scaled up so every length is 3 times bigger. By what factor does the stress in its leg bones grow?

Show the answer

3

Weight grows by \(3^3 = 27\) and bone area by \(3^2 = 9\), so the stress grows by \(27/9 = 3\) times.

Checkpoint: Size and the square–cube law (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

An ant can carry many times its own weight, but an elephant cannot. Which explanation fits the square–cube law?

Show the answer

A small animal has more leg area for each kilogram of body weight.

Strength goes as area (length squared) and weight as volume (length cubed), so small animals have relatively strong legs.

Elastic, plastic and breaking

FIRST, MAKE A PREDICTION

A metal strip remains bent after the load is removed. What has happened?

Show the answer

It has undergone plastic deformation

Plastic deformation leaves a permanent shape change; elastic deformation reverses when the load is removed.

TRY IT / FOLLOW THE REASONING

Sketch loading and unloading first within the elastic region and then beyond yield. Mark permanent strain separately from the elastic strain recovered on unloading.

MAKE THE CONNECTION

Key idea

Elastic stretching springs back; plastic stretching is permanent.

In a tensile test, a bar is pulled until it breaks while stress is plotted against strain. For mild steel:

Elastic, plastic and breaking

A tensile test of mild steel, stress in MPa against strain in percent. The stress shoots up the elastic line to the yield strength of 250 MPa at a strain of only 0.125%, stays near it while the steel stretches plastically, climbs to the tensile strength of 400 MPa, then falls as a neck forms before it snaps.51015202550100150200250300350400450yield 250tensile 400strain (%)stress (MPa)
The tensile test for mild steel, from the elastic line to the snap: stress shoots up to a yield strength of 250 MPa and a tensile strength of 400 MPa is reached before a neck forms. The elastic part is too steep to see: it ends at a strain of only 0.125%.

Elastic, plastic and breaking

  1. A straight line of slope \(E\) (elastic: it springs back).
  2. The yield strength, about 250 MPa.
  3. Plastic stretching beyond it, permanent, as layers of atoms slip like a deck of cards (a paperclip bent too far), with strain hardening.
  4. The tensile strength, about 400 MPa, at the peak.
  5. Then a neck forms and it snaps.

Yield comes at a strain of \(250\ \text{MPa}/200\ \text{GPa} = 0.00125\), yet mild steel stretches 20% or more before breaking: almost all plastic. Slow permanent stretching under a long-held load, especially when hot, is creep.

Glass is brittle: a straight line, then a sudden snap. Mild steel is ductile, and sags as a warning.

Elastic, plastic and breaking

Worked number: the piano wire

A 300 kg piano on a 2 mm wire gives about 937 MPa: it would stretch and snap.

Checkpoint: Elastic, plastic and breaking (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

An aluminium alloy yields at 300 MPa and has \(E = 70\) GPa. At what strain does it start to stretch permanently? Give the strain as a plain number.

Show the answer

0.004285714286

\(\varepsilon = 300\times10^6/70\times10^9 = 0.0043\), only 0.43%.

Checkpoint: Elastic, plastic and breaking (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Which of these breaks with a sudden snap after a straight line on its stress–strain graph?

Show the answer

A glass rod.

Glass is brittle: elastic up to the break, with no plastic stretching. Mild steel and copper are ductile.

Stiff, strong and tough

FIRST, MAKE A PREDICTION

A material barely stretches before breaking suddenly. Does being stiff automatically make it tough?

Show the answer

No

Stiffness describes resistance to elastic strain; toughness describes the energy absorbed before fracture, including how much deformation occurs.

TRY IT / FOLLOW THE REASONING

Sketch one steep stress–strain curve that ends early and another that extends much farther. Compare stiffness, maximum stress and energy per volume before failure.

MAKE THE CONNECTION

Key idea

Stiffness, strength and toughness are three different things.

  • Stiffness is how little something stretches (\(E\)).
  • Strength is the stress to yield or break.
  • Toughness is how much energy it absorbs before breaking.

Glass is stiff but not tough; mild steel is all three.

Stiff, strong and tough

Myth

"Spider silk is stronger than steel." Partly true. Dragline silk breaks at roughly 1–1.7 GPa, comparable to a good steel, but it is about six times less dense, so weight for weight it is about five times stronger, and far less stiff. Hanging under its own weight, a thread breaks at length \(L = \sigma/(\rho g)\): about 100 km for silk, 17 km for a high-strength steel, both breaking at 1.3 GPa.

Stiff, strong and tough

Myth

"A thick cable is stiffer and stronger because it's made of tougher stuff." Usually it's the same steel, same \(E\), same yield strength. More area means lower stress.

Checkpoint: Stiff, strong and tough (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A glass rod barely bends under load, but shatters when it is dropped. How would you describe it?

Show the answer

Stiff but not tough.

Glass has a high \(E\) (stiff), but a crack runs through it with little energy absorbed, so it is not tough.

Checkpoint: Stiff, strong and tough (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A fibre breaks at a stress of 2.0 GPa and has a density of 1500 kg/m³. How long a thread of it could hang from the top under its own weight before breaking, in kilometres? Take \(g = 9.81\) m/s².

km

Show the answer

135.9157322 km

\(L = 2.0\times10^9/(1500\times9.81) = 1.36\times10^5\) m, about 136 km.

Stress concentration and cracks

FIRST, MAKE A PREDICTION

Two pieces carry the same nominal stress. One has a sharp crack. Which is generally more at risk of failure near that defect?

Show the answer

The cracked piece

Stress can concentrate strongly at a sharp crack tip, exceeding the average force-per-area value.

TRY IT / FOLLOW THE REASONING

Sketch parallel load paths through a sound bar and around a hole. Compare a rounded hole with a sharp notch and explain why tip geometry matters.

MAKE THE CONNECTION

Key idea

Holes and cracks make stress pile up around them.

Lines of force crowd round a hole, so the stress rises there: stress concentration (like traffic squeezing past a closed lane). A small round hole triples it. For an elliptical hole,

Stress concentration and cracks

\[ \sigma_{\max} = \sigma\left(1 + \frac{2a}{b}\right) \]

with \(\sigma\) the average stress, \(a\) the hole's half-length across the pull and \(b\) its half-width along it; the bracket is the stress concentration factor.

Stress concentration and cracks

Worked numbers: slits and holes

A slit with \(a/b = 10\) turns 20 MPa into 420 MPa; a round hole turns 100 MPa into 300, past steel's 250 MPa yield. Hence rounded corners, and crack-stop holes.

Stress concentration and cracks

A crack is an ellipse with \(b\) almost zero, the worst concentrator; tiny surface flaws are why glass breaks far below the strength of its bonds. Griffith's energy balance: a crack runs (fracture) when the stored energy it releases pays for the new surface, so the bigger the flaw, the weaker the part. Many steels turn brittle in the cold (the ductile-to-brittle transition), which, with welded hulls and square corners, broke wartime ships in two.

Checkpoint: Stress concentration and cracks (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A plate has an average stress of 50 MPa. It has a narrow elliptical slit with \(a/b = 8\). What is the peak stress at the slit, in megapascals?

MPa

Show the answer

850 MPa

\(\sigma_{\max} = 50\times(1 + 2\times8) = 850\) MPa: far above the yield strength of steel.

Checkpoint: Stress concentration and cracks (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Why do engineers round the inside corners of a loaded part?

Show the answer

A sharp corner concentrates stress, and rounding it lowers the peak.

A sharp corner is like a crack: the stress piles up there. A smooth curve spreads it.

Fatigue

FIRST, MAKE A PREDICTION

A metal component repeatedly experiences a load below its single-pull breaking load. Is failure impossible?

Show the answer

No; repeated loading can grow fatigue cracks

A component can fail after many cycles at stresses below its single-test strength.

TRY IT / FOLLOW THE REASONING

Draw repeated load cycles and imagine a tiny crack length increasing after each group of cycles. Identify why inspection history and cycle count belong in the assessment.

MAKE THE CONNECTION

Key idea

Repeated loading, far below the breaking stress, can still break a metal.

Metals fail under repeated stresses well below their yield strength: fatigue. Each load and unload is a cycle, and each grows a tiny crack, usually from a scratch, hole or corner, until the rest snaps. A turning axle goes through a cycle every turn.

An S–N curve plots the stress per cycle, \(S\), against cycles to failure, \(N\) (each step ten times more).

Fatigue

Worked number: an S–N curve

For a steel of tensile strength 600 MPa: about \(10^3\) cycles at 540 MPa, \(10^4\) at 440, \(10^5\) at about 365, and indefinitely below about 300 MPa, the endurance limit (about half the tensile strength for polished lab samples; lower in real parts). Aluminium alloys have none.

Fatigue

An S–N curve for a steel of tensile strength 600 MPa: 540 MPa lasts about 10^3 cycles, 440 MPa lasts about 10^4 cycles, 365 MPa lasts about 10^5 cycles, and below the endurance limit of about 300 MPa it lasts indefinitely.34567250300350400450500550600endurance limit540 MPa440 MPa365 MPacycles to failure (power of ten)stress per cycle (MPa)
An S–N curve for a steel of tensile strength 600 MPa: 540 MPa lasts about a thousand cycles, 440 MPa ten thousand and 365 MPa a hundred thousand. Below the endurance limit of about 300 MPa it lasts indefinitely.

Fatigue

Each pressurised flight is a cycle for a fuselage. The Comet lost near Elba had made about 1290; a short-haul airliner flying 6 times a day for 20 years makes 43 800.

Fatigue

Myth

"The Comets crashed because of their square windows." Partly a myth. Corners did carry high stresses, but the Elba crack began at a rivet hole at the corner of a roof cut-out for a radio aerial, not a passenger window.

Fatigue

Engineers inspect for cracks while they're small, and use damage-tolerant (fail-safe) design with more than one load path.

Checkpoint: Fatigue (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

An axle turns at 600 revolutions per minute and each turn is one stress cycle. How many cycles does it go through in an 8-hour shift?

Show the answer

288000

\(600\times60\times8 = 288\,000\) cycles in a shift, so over a year it adds up to tens of millions.

Checkpoint: Fatigue (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A steel part sees millions of cycles of a stress below its endurance limit. What do you expect?

Show the answer

It should survive indefinitely.

Many steels have an endurance limit, below which they survive indefinitely. Aluminium alloys have none.

Buckling

FIRST, MAKE A PREDICTION

A slender column and a shorter otherwise identical column are compressed. Which generally buckles under a smaller load?

Show the answer

The longer column

For the same material, cross-section and end conditions, Euler's critical load decreases with length squared.

TRY IT / FOLLOW THE REASONING

Sketch a tiny sideways deflection in a compressed column. In Pcrit ∝ 1/L², double the length while keeping the end conditions fixed and calculate the new load ratio.

MAKE THE CONNECTION

Key idea

Slender things fail by bowing sideways, long before they are crushed.

Press on a straw and it bows sideways and folds long before it's crushed: buckling. Below the critical load, stiffness straightens a small bend; above it, the bend feeds on itself. Euler's buckling load, for ends free to tilt, is

Buckling

\[ F = \frac{\pi^2 E I}{L^2} \]

with \(E\) Young's modulus, \(L\) the length and \(I\) the second moment of area, larger the further the material sits from the centre line. - Double the length: a quarter of the load. - Double a solid rod's diameter: 16 times (\(I \propto d^4\)).

Buckling

Worked number: a steel rod

A steel rod 1 cm thick and 1 m long buckles at about 1000 N, a twentieth of its yield load.

Buckling

Optional maths: why hollow tubes win

For a solid rod \(I = \pi d^4/64\): twice the width is four times the material, sitting twice as far out, counted squared, so \(4 \times 4 = 16\). A tube 2.5 cm outside and 1.5 cm inside has the same area as a 2 cm solid rod but about 2.1 times the \(I\): hence hollow bicycle frames, bones and bamboo (though too thin a wall crumples).

Part Fluids at rest

Buckling

Checkpoint: Buckling (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A column is replaced by one of the same material and cross-section but 3 times as long. Its buckling load becomes what fraction of the old one? Give the fraction as a decimal.

Show the answer

0.1111111111

\(F\propto1/L^2\), so the load is \(1/3^2 = 1/9 = 0.111\) of the old one.

Checkpoint: Buckling (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Why is a hollow tube better against buckling than a solid rod of the same mass?

Show the answer

Its material sits further from the centre line, so \(I\) is bigger.

Material far from the centre line counts for much more in \(I\), so a tube resists bowing better than a rod of the same mass.

Density, pressure and depth

FIRST, MAKE A PREDICTION

Two containers of fresh water have different shapes. At the same depth below each surface, how do their gauge pressures compare?

Show the answer

They are equal

For a static fluid, gauge pressure depends on density, gravity and depth, not the overall container shape.

TRY IT / FOLLOW THE REASONING

Move the model's probe down in fresh water and compare pressures at 4, 8 and 16 m. Explain why the readout is extra pressure above the surface pressure.

Go deeper in fresh water

Depth: 10 m

Gauge pressure
98 kPa
Model assumptions

Fresh water density 1000 kg/m³, g = 9.8 m/s². This is gauge pressure: the extra pressure above the surface pressure.

MAKE THE CONNECTION

Key idea

Pressure in a fluid grows with depth: \(p = \rho g h\).

A fluid is anything that flows. Density is mass per volume, \(\rho = m/V\) (rho), in kg/m³. Water is 1000 kg/m³ = 1 kg/L = 1 g/cm³, so the unit must travel with the number. Air about 1.2, steel 7850, mercury 13 600, gold 19 300.

Pressure is \(p = F/A\).

Density, pressure and depth

Worked number: feet and heels

A 60 kg person (589 N) on 150 cm² of sole: about 40 kPa; on a 1 cm² heel: nearly 6 MPa.

Density, pressure and depth

One standard atmosphere is 101 325 Pa (about 101 kPa); a bar is 100 000 Pa. In a fluid, pressure pushes equally every way, at right angles to surfaces, and grows with depth:

Density, pressure and depth

\[ p = \rho g h \]

with \(g = 9.81\) N/kg and \(h\) the vertical depth.

Density, pressure and depth

Optional maths: why \(p = \rho g h\)

A column of base \(A\) has weight \(\rho A h g\); divide by \(A\).

Density, pressure and depth

  • Every 10 m of water adds about one atmosphere (98 kPa).
  • Shape doesn't matter: joined vessels fill to the same level (the hydrostatic paradox).

Absolute pressure = atmospheric + gauge pressure.

Density, pressure and depth

Worked number: a diver

A diver 30 m down in seawater (1025 kg/m³): about 302 kPa gauge, 403 kPa absolute, about 4 atm. Depth is vertical: 20 m along a 30° slope is 10 m down.

Checkpoint: Density, pressure and depth (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A diver is 12 m down in a freshwater lake (density 1000 kg/m³). Take \(g = 9.81\) m/s² and one atmosphere as 101.3 kPa.

The gauge pressure at the diver, in kPa.

kPa

The absolute pressure at the diver, in kPa.

kPa

Show the answer
  1. 117.72 kPa
  2. 219.02 kPa

\(\rho g h = 1000\times9.81\times12 = 117.7\) kPa gauge. Add the atmosphere to get 219 kPa absolute, about 2.2 atm.

Checkpoint: Density, pressure and depth (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Three vessels of different shapes, some wide and some narrow, are filled with water to the same height. How do the pressures at their bases compare?

Show the answer

They are equal: only the depth matters.

The pressure at the base is \(\rho g h\) whatever the shape: the hydrostatic paradox.

The ocean of air

FIRST, MAKE A PREDICTION

When you drink through a straw, what pushes the liquid up the straw after you lower the pressure in your mouth?

Show the answer

The higher external air pressure on the drink

Lower pressure in the straw leaves a pressure difference; outside air pushes the liquid upward.

TRY IT / FOLLOW THE REASONING

Draw the air above the drink and the lower-pressure space in the straw. Mark the resulting pressure difference and compare it with the weight of the raised liquid column.

MAKE THE CONNECTION

Key idea

The weight of the atmosphere pushes on everything. Nothing in a straw "pulls".

In a barometer, the top of the tube is nearly a vacuum, and the air pushes the liquid up until \(\rho g h = p\), so \(h = p/(\rho g)\): 760 mm of mercury, or 10.3 m of water. A straw doesn't pull: you lower the pressure inside and the air pushes the drink up, so even a perfect pump can't raise water more than 10.3 m.

The ocean of air

Myth

"Suction pulls." Empty space can't pull. A sucker is pushed on by the air outside, with almost nothing behind it to push back.

Checkpoint: The ocean of air (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

On a mountain the air pressure is 80 kPa. How tall a column of water could this pressure hold up in a barometer, in metres? Take \(g = 9.81\) m/s².

m

Show the answer

8.154943935 m

\(h = 80\,000/(1000\times9.81) = 8.2\) m, shorter than the 10.3 m at sea level.

Checkpoint: The ocean of air (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

You sip a drink through a straw. What pushes the drink up the straw?

Show the answer

The air pressure on the drink's surface, because you lowered the pressure inside.

You lower the pressure in the straw, and the air pushing on the drink does the lifting.

Pascal's principle and hydraulics

FIRST, MAKE A PREDICTION

An ideal hydraulic lift has an output piston with ten times the input area. If the output force is ten times larger, is energy being created?

Show the answer

No; the output piston moves one tenth as far

Equal pressure gives greater force on greater area, but equal displaced volume trades that force for distance.

TRY IT / FOLLOW THE REASONING

Increase the area ratio in the ideal lift. Compare output force, output movement and work; explain why a force gain comes with a movement reduction.

Trade movement for force

Output ÷ input piston area: 2 ×

Output force
20 N
Output movement
5 cm
Work at either piston
1 J
Model assumptions

Ideal incompressible liquid, neglecting its weight and friction. Input force 10 N; input movement 10 cm downward. The output piston rises. Drawing is schematic; there are no energy losses.

MAKE THE CONNECTION

Key idea

Pressure on an enclosed liquid is passed on everywhere, so a small force on a small piston balances a large force on a large one.

Pascal's principle: pressure applied to an enclosed liquid is passed on undiminished to every part of it and its walls. In a hydraulic press,

Pascal's principle and hydraulics

\[ F_2 = F_1 \times \frac{A_2}{A_1} \]

where \(F_1\) pushes a small piston of area \(A_1\) and \(F_2\) is the force on a big piston of area \(A_2\).

Pascal's principle and hydraulics

A hydraulic press. A force of 20 N on a piston of 10 cm² becomes 1000 N on a piston of 500 cm². The small piston moves 0.5 m and the big one 0.01 m, so the work in and out is the same, 10 J.20 N on 10 cm²1000 N on 500 cm²moves 0.5 mmoves 0.01 m
Force is multiplied by the ratio of the areas, but the big piston moves correspondingly less: with 10 cm² and 500 cm², 20 N becomes 1000 N and the big piston moves 50 times less.

Pascal's principle and hydraulics

Worked number: a hydraulic press

With 10 cm² and 500 cm², 20 N becomes 1000 N. But the same volume moves on both sides, so the big piston moves 50 times less: 20 N × 0.5 m = 10 J in, 1000 N × 0.01 m = 10 J out. Force is multiplied, work is not. (1 cm² = \(10^{-4}\) m².)

Checkpoint: Pascal's principle and hydraulics (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A hydraulic jack has a small piston of area 5.0 cm² and a large one of area 200 cm². You push the small piston with 150 N. What force does the large piston exert, in newtons?

N

Show the answer

6000 N

\(F_2 = 150\times200/5.0 = 6000\) N: the force is multiplied by 40.

Checkpoint: Pascal's principle and hydraulics (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

The jack multiplies your force by 40. Compared with the work you do on the small piston, the work done by the large piston is...

Show the answer

...the same (ignoring friction): the large piston moves 40 times less.

Force is multiplied and distance divided by the same factor, so work is conserved.

Upthrust and Archimedes' principle

FIRST, MAKE A PREDICTION

A fully submerged object is pushed deeper into water of uniform density. If its volume stays fixed, what happens to its buoyant force?

Show the answer

It stays the same

The displaced fluid's weight is unchanged; increasing both top and bottom pressures does not change their net difference for this volume.

TRY IT / FOLLOW THE REASONING

Draw a rectangular block at two depths. Mark the larger pressure on its underside and compute the top-to-bottom difference in each position.

MAKE THE CONNECTION

Key idea

Archimedes' principle: the upthrust equals the weight of the fluid displaced.

The bottom of a submerged object is deeper than its top, so the water pushes up harder than down: the upthrust (buoyancy).

Upthrust and Archimedes' principle

A cube 20 cm on a side with its top 1 m under water. Water pushes down on its top with 392 N and up on its bottom with 471 N, so the net upthrust is 78.5 N, the weight of 8 kg of water.water surfacedown 392 Nup 471 Nnet upthrust 78.5 N= weight of 8 kg of water
Deeper water pushes harder, so the push up on the bottom beats the push down on the top. For a 20 cm cube with its top 1 m down, the net upthrust of 78.5 N is the weight of 8 kg of water.

Upthrust and Archimedes' principle

Worked number: a cube

For a 20 cm cube with its top 1 m down, 471 N up minus 392 N down is 78.5 N, the weight of 8 kg of water.

Upthrust and Archimedes' principle

Optional maths: why the upthrust is \(\rho g V\)

\(\rho g h_2 A - \rho g h_1 A = \rho g (h_2 - h_1) A = \rho g V\), with \(h_1, h_2\) the depths of the top and bottom, \(A\) the face area and \(V\) the volume. The depth drops out.

Upthrust and Archimedes' principle

This holds for any shape (a blob of water of that shape would just hang there).

Upthrust and Archimedes' principle

Worked numbers: a stone and a crown

A 2 kg, 0.8 L stone reads about 11.8 N under water: its apparent weight.

Silver (about 10 500 kg/m³) is less dense than gold (19 300), so a crown with silver in it has more volume. A 1 kg crown that weighs as if 940 g in water displaced 60 cm³: density 16.7 g/cm³, not pure gold.

Checkpoint: Upthrust and Archimedes' principle (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A ball of volume 0.50 L is held completely under water. What is the upthrust on it, in newtons? Take \(g = 9.81\) m/s².

N

Show the answer

4.905 N

It displaces 0.50 kg of water, whose weight is \(0.50\times9.81 = 4.9\) N.

Checkpoint: Upthrust and Archimedes' principle (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A block weighs 30 N in air and reads 22 N when hung under water. What is its volume, in litres? Take \(g = 9.81\) m/s².

L

Show the answer

0.8154943935 L

The upthrust is \(30 - 22 = 8\) N, so \(V = 8/(1000\times9.81) = 8.2\times10^{-4}\) m³, about 0.82 L.

Floating, stability and balloons

FIRST, MAKE A PREDICTION

A uniform object with half the density of water floats at rest. What fraction of its volume is submerged?

Show the answer

One half

The displaced water must weigh as much as the object; the required volume fraction is object density divided by fluid density.

TRY IT / FOLLOW THE REASONING

Write ρobject Vg = ρwater Vsubmerged g and cancel the shared factors. Then distinguish floating at all from returning upright after a small tilt.

MAKE THE CONNECTION

Key idea

Density decides whether something sinks, floats or is suspended.

Denser than the fluid, it sinks; less dense, it rises; equal, it is suspended (a fish's swim bladder, a submarine's ballast tanks). Density decides, not mass.

A floating object displaces its own weight, so

Floating, stability and balloons

\[ \frac{V_{\text{under}}}{V} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}} \]

Floating, stability and balloons

Worked numbers: iceberg and ship

An iceberg (917 kg/m³ in seawater of 1025) floats about 89% under.

A steel ship floats because its average density, air included, is below water's: 50 000 tonnes displaces about 49 000 m³. Floating displaces its own weight; sunk, its own volume.

Floating, stability and balloons

Weight acts at the centre of gravity, G; upthrust at the centre of buoyancy, B, the centre of the displaced water. When a ship tilts, B moves to the low side; if it ends up further out than G, the torque rights the ship. If G is too high, the torque tips it over. Ballast lowers G; cargo moved up onto the deck makes a boat less stable.

A balloon rises if it weighs less than the air it pushes aside.

Floating, stability and balloons

Worked number: a balloon

A cubic metre of air is about 1.2 kg and of helium 0.17 kg, so helium lifts about 1 kg per m³; hot air at about 100 °C, only about a quarter of a kilogram.

Checkpoint: Floating, stability and balloons (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A block of wood of density 700 kg/m³ floats in fresh water. What percentage of its volume is under the water?

Show the answer

70

The fraction submerged is \(700/1000 = 0.70\), so 70% is under water.

Checkpoint: Floating, stability and balloons (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Cargo is stacked high on the deck of a boat. What happens to its stability?

Show the answer

It becomes less stable, because G is higher.

A higher G makes it easier for the torque to tip the boat over instead of righting it. Ballast lowers G.

Checkpoint: Floating, stability and balloons (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

The air outside a hot-air balloon is 1.20 kg/m³, and the hot air inside is 0.95 kg/m³. How much does each cubic metre of hot air lift, in kilograms?

kg

Show the answer

0.25 kg

Each cubic metre displaces 1.20 kg of air but holds only 0.95 kg, so it lifts \(1.20 - 0.95 = 0.25\) kg.

Surface tension and capillary rise

FIRST, MAKE A PREDICTION

For the same wetting liquid in similar clean tubes, which tube produces the greater capillary rise?

Show the answer

The narrower tube

Surface support scales with the tube's circumference while the supported liquid weight scales with area; the resulting rise varies inversely with radius.

TRY IT / FOLLOW THE REASONING

Sketch the upward surface-tension force around a wetting tube's rim and the weight of its liquid column. Halve the radius and compare the predicted height, keeping wetting conditions fixed.

MAKE THE CONNECTION

Key idea

The surface of a liquid behaves like a skin.

A surface molecule has neighbours only beside and below, so it is pulled inwards, and the surface pulls along itself like a skin. Surface tension, \(\gamma\), is the force per length:

Surface tension and capillary rise

\[ F = \gamma L \]

with \(L\) the length of the edge; for water at 20 °C, \(\gamma \approx 0.073\) N/m.

Surface tension and capillary rise

Worked number: a floating needle

A 4 cm needle gets up to \(0.073 \times 0.08 \approx 5.8\) mN, more than its 2.4 mN weight; soap lowers \(\gamma\) to less than half and it sinks.

Surface tension and capillary rise

Water climbs a thin glass tube (capillary action) because adhesion to glass beats cohesion to itself: its meniscus is concave. Mercury doesn't wet glass (low wettability), so it is pushed down. The height it climbs is given by Jurin's law:

Surface tension and capillary rise

\[ h = \frac{2\gamma}{\rho g r} \]

with \(r\) the tube's radius: 3 cm for \(r = 0.5\) mm, 15 cm for 0.1 mm.

Surface tension and capillary rise

Optional maths: where Jurin's law comes from

The rim's pull \(\gamma\,2\pi r\) holds up a column of weight \(\rho g\pi r^2 h\). Setting them equal gives \(h\).

Surface tension and capillary rise

Myth

"Trees lift water by capillary action." In wood's pipes, about 20 micrometres in radius, water rises only about 0.74 m. Instead, evaporation from the leaves pulls up a column held together by cohesion (the cohesion–tension theory, Dixon and Joly, 1890s).

Part Fluids in motion

Surface tension and capillary rise

Checkpoint: Surface tension and capillary rise (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A wire frame with a straight 6.0 cm edge rests on a water surface, which pulls on the edge. With \(\gamma = 0.073\) N/m, what is the pull, in millinewtons?

mN

Show the answer

4.38 mN

\(F = \gamma L = 0.073\times0.060 = 4.4\times10^{-3}\) N, which is 4.4 mN.

Checkpoint: Surface tension and capillary rise (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Why is mercury pushed down in a narrow glass tube, instead of climbing like water?

Show the answer

Mercury does not wet glass: cohesion beats adhesion.

Mercury atoms attract each other more strongly than they attract glass, so its meniscus is domed and the column is pushed down.

Continuity and Bernoulli's equation

FIRST, MAKE A PREDICTION

Steady incompressible flow passes through a pipe that narrows to half its cross-sectional area. What happens to the average flow speed?

Show the answer

It doubles

The same volume per second must pass every cross-section, so Av stays constant.

TRY IT / FOLLOW THE REASONING

Draw equal volumes entering and leaving a narrowing during the same second. Use continuity to find the speed change, then discuss the pressure change for an ideal horizontal flow.

MAKE THE CONNECTION

Key idea

Where a fluid flows faster, its pressure is lower: Bernoulli's equation is conservation of energy.

The flow rate is \(Q = Av\), the volume per second, with \(A\) the cross-section and \(v\) the speed. Water can't be squashed or pile up, so (continuity)

Continuity and Bernoulli's equation

\[ A_1 v_1 = A_2 v_2 \]

Continuity and Bernoulli's equation

A pipe narrowing from 2 cm across to 0.5 cm. The flow at 1 m/s speeds up to 16 m/s, because the area is 16 times smaller.2 cm, 1 m/s0.5 cm, 16 m/sarea ÷ 16, speed × 16
Narrow the pipe and the same flow must speed up; the faster fluid has lower pressure. Here a hose narrows from 2 cm at 1 m/s to 0.5 cm at 16 m/s.

Continuity and Bernoulli's equation

Worked number: a hose

A hose 2 cm across at 1 m/s, squeezed to a 0.5 cm opening (16 times less area), squirts at about 16 m/s (a bit less in practice).

Continuity and Bernoulli's equation

To speed up into a narrowing, a parcel needs higher pressure behind it, so where the flow is fast, the pressure is low. Along a streamline (a parcel's path in steady flow), Bernoulli's equation is

Continuity and Bernoulli's equation

\[ p + \tfrac12 \rho v^2 + \rho g h = \text{constant} \]

with \(p\) the pressure, \(\rho\) the density, \(v\) the speed and \(h\) the height. Each term is an energy per m³ (1 Pa = 1 J/m³): it is conservation of energy.

Continuity and Bernoulli's equation

Worked number: speeding up

Speeding up from 2 to 8 m/s at the same height drops the pressure by \(\tfrac12 \times 1000 \times (64 - 4) = 30\) kPa. (Optional maths: the net work \((p_1 - p_2)V\) on a parcel of volume \(V\) equals its gain in kinetic and potential energy; divide by \(V\).)

Continuity and Bernoulli's equation

From a tank's surface to a hole a depth \(h\) below it, \(\rho g h = \tfrac12\rho v^2\), so \(v = \sqrt{2gh}\) (Torricelli): 3.96 m/s at 0.8 m, twice that at four times the depth. A thumb on a hose likewise turns pressure into speed.

Bernoulli assumes no internal friction, steady flow, an unsqueezed fluid, and one streamline with no pump between. In a long, even hose the pressure falls (friction), and a hairdryer's fast jet is at about room pressure (its fan added energy).

Checkpoint: Continuity and Bernoulli's equation (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A river 20 m wide and 2.0 m deep flows at 0.50 m/s. It narrows to 10 m wide, with the same depth. How fast does it flow there, in metres per second?

m/s

Show the answer

1 m/s

Half the width means half the area, so the speed doubles: \(0.50\times20/10 = 1.0\) m/s.

Checkpoint: Continuity and Bernoulli's equation (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A hole is 1.8 m below the surface of the water in a tank. How fast does water leave it, in metres per second? Take \(g = 9.81\) m/s².

m/s

Show the answer

5.942726647 m/s

\(v = \sqrt{2\times9.81\times1.8} = 5.9\) m/s, the speed of a stone dropped through 1.8 m.

Checkpoint: Continuity and Bernoulli's equation (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

Water flows through a level pipe that narrows. What is true of the narrow section?

Show the answer

The water is faster and its pressure is lower.

By continuity the speed rises; by Bernoulli the pressure falls.

Viscosity and turbulence

FIRST, MAKE A PREDICTION

For the same fluid, speed and shape, doubling a pipe's characteristic diameter does what to its Reynolds number?

Show the answer

It doubles

Re = ρvD/μ grows linearly with the chosen characteristic length when the other quantities are fixed.

TRY IT / FOLLOW THE REASONING

Write the units of ρvD/μ and cancel them. Then vary speed, size and viscosity separately and explain why a Reynolds number compares competing effects rather than directly measuring viscosity.

MAKE THE CONNECTION

Key idea

Viscosity is a fluid's stickiness. The Reynolds number tells you whether the flow is smooth or turbulent.

Viscosity, \(\mu\) (mu), is a fluid's stickiness: a solid pushed sideways leans and stops, but a fluid keeps flowing, in layers that drag on each other and turn flow energy into heat. Sliding a plate of area \(A\) at speed \(v\) over a layer of thickness \(d\) takes \(F = \mu A v/d\). Water is about 0.001 Pa·s, air \(1.8\times10^{-5}\), glycerol 1.5.

At a given pressure, the flow through a pipe goes as the radius to the fourth power (a quarter of the area at a quarter of the average speed).

Viscosity and turbulence

Worked number: an artery

An artery narrowed to 80% of its radius passes about \(0.8^4 \approx 41\%\) of the flow.

Viscosity and turbulence

Flow is laminar (smooth layers) or turbulent (eddies, mixing), depending on the Reynolds number,

Viscosity and turbulence

\[ Re = \frac{\rho v D}{\mu} \]

with \(D\) the pipe's diameter: laminar below about 2000, able to turn between about 2000 and 2300, and turbulent well above. Water at 1 m/s in a 1 cm pipe gives 10 000: turbulent.

Checkpoint: Viscosity and turbulence (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Oil of density 900 kg/m³ and viscosity 0.20 Pa s flows at 0.50 m/s through a pipe 2.0 cm across. What is the Reynolds number?

Show the answer

45

\(Re = 900\times0.50\times0.020/0.20 = 45\), far below 2000: smooth, laminar flow.

Checkpoint: Viscosity and turbulence (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

An artery narrows to half its original radius, at the same pressure difference. What fraction of the flow gets through? Give it as a decimal.

Show the answer

0.0625

\(0.5^4 = 0.0625\): only about 6% of the flow, which is why a small narrowing matters so much.

How wings lift

FIRST, MAKE A PREDICTION

Does air split at a wing's leading edge have to reunite at its trailing edge at the same time?

Show the answer

No

There is no equal-transit-time requirement. Lift involves pressure forces and a downward momentum change of the surrounding air.

TRY IT / FOLLOW THE REASONING

Draw a wing turning a stream of air downward. Apply momentum balance to the air and the reaction force to the wing, then connect that force with the pressure distribution.

MAKE THE CONNECTION

Key idea

A wing lifts by turning a large mass of air downwards.

How wings lift

Myth

"Air over the curved top must meet the air underneath at the back, so it goes faster." There's no such rule: the air over the top arrives first, much faster than the story predicts.

How wings lift

A tilted, curved wing turns a large mass of air downwards, and by Newton's third law the air pushes it up. The same flow has lower pressure above the wing and higher below, and that difference is the lift; Bernoulli is the bookkeeping linking fast air to low pressure, not a separate cause. A 70 000 kg airliner with 120 m² of wing needs only about 5.7 kPa, about 6% of atmospheric pressure.

Checkpoint: How wings lift (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A 400 kg glider has 10 m² of wing. What average pressure difference between the lower and upper surfaces of the wing keeps it in level flight, in pascals? Take \(g = 9.81\) m/s².

Pa

Show the answer

392.4 Pa

The lift must equal the weight, \(400\times9.81 = 3924\) N, spread over 10 m²: \(392\) Pa, a tiny fraction of the atmosphere.

Checkpoint: How wings lift (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Is it true that air going over a wing must meet air going underneath at the back, so the longer path over the top makes it faster?

Show the answer

No: there is no such rule, and the air over the top actually arrives first.

Lift comes from the wing turning air downwards. Bernoulli links the faster, lower-pressure flow, but the equal-transit rule is false.

Key terms

Key terms

Cross-sectional area
The area of the face you'd see if you cut a wire or rod straight across.
Stress (\(\sigma\))
Force per cross-sectional area, \(\sigma = F/A\), in pascals.
Pascal (Pa)
One newton per square metre; 1 kPa = 1000 Pa, 1 MPa = \(10^6\) Pa, 1 GPa = \(10^9\) Pa.
Strain (\(\varepsilon\))
Stretch per original length, \(\varepsilon = \Delta L/L\); it has no units.

Key terms

Young's modulus (\(E\))
Stress over strain, \(E = \sigma/\varepsilon\): the stiffness of a material, in pascals (usually GPa).
Shear
Pushing one face of a block sideways so its layers slide over each other; resisted by the shear modulus.
Square–cube law
Scale something up and its weight grows as length cubed, its strength as length squared.
Tensile test
Pulling a shaped bar harder and harder and plotting stress against strain until it breaks.

Key terms

Elastic
Springing back to the old shape when the load is removed.
Yield strength
The stress at which a material starts to stretch permanently.
Plastic
Stretching or bending that stays after the load is removed.
Creep
Slow, permanent stretching under a steady load held for a long time, especially when hot.

Key terms

Tensile strength
The greatest stress a material can carry before it necks and breaks.
Brittle
Breaking suddenly with little or no plastic stretching, like glass.
Ductile
Stretching plastically a long way before breaking, like mild steel; it can be drawn into wire.
Toughness
How much energy a material soaks up, by bending and stretching, before it breaks.

Key terms

Stress concentration
The rise in stress near a hole, notch, corner or crack.
Stress concentration factor
The peak stress over the average stress: 3 for a round hole, \(1 + 2a/b\) for an ellipse.
Fracture
Breaking apart, when a crack runs through the material.
Ductile-to-brittle transition
The change, below a certain temperature, from ductile to brittle behaviour seen in many steels.

Key terms

Fatigue
Failure caused by a stress applied and removed again and again, growing a crack a little each time.
Cycle
One load and unload.
S–N curve
A graph of the stress in each cycle against the number of cycles to failure; also called a Wöhler curve.
Endurance limit
The stress below which a material (many steels) can survive indefinitely many cycles; also called the fatigue limit.

Key terms

Damage tolerance
Designing so that one crack can't bring a structure down (more than one load path), and inspecting so cracks are found while small; also called fail-safe design.
Buckling
A sudden sideways collapse of a column under a squeeze.
Critical load
The load at which a column buckles, \(F = \pi^2EI/L^2\) for ends free to tilt; also called the buckling load.
Second moment of area (\(I\))
A measure of how far a cross-section's material sits from its centre line; \(\pi d^4/64\) for a solid rod.

Key terms

Fluid
Anything that flows: a liquid or a gas.
Density (\(\rho\))
Mass per volume, \(\rho = m/V\), in kg/m³; water is 1000 kg/m³.
Kilopascal (kPa)
A thousand pascals; air pressure at sea level is about 101 kPa.
Atmosphere (atm)
A unit of pressure, defined as exactly 101 325 Pa, about air pressure at sea level.

Key terms

Hydrostatic paradox
The pressure at a depth depends only on the depth, not on the shape of the container or how much liquid there is.
Absolute pressure
The total pressure: atmospheric pressure plus gauge pressure.
Gauge pressure
The pressure above atmospheric, which most gauges read.
Barometer
A tube of liquid, closed at the top, whose column height measures the air pressure, \(h = p/(\rho g)\).

Key terms

Pascal's principle
Pressure applied to an enclosed liquid is passed on undiminished to every part of the liquid and its walls.
Hydraulic
Worked by liquid under pressure, as in a hydraulic press or car brakes.
Upthrust (buoyancy)
The upward push that a fluid gives anything in it.
Displaced
Pushed out of the way; Archimedes' principle says the upthrust equals the weight of fluid displaced.

Key terms

Apparent weight
What a spring balance reads for an object in a fluid: its weight minus the upthrust.
Suspended
Neither rising nor sinking, because the object's average density equals the fluid's.
Centre of buoyancy
The centre of the displaced fluid, where the upthrust effectively acts.
Surface tension (\(\gamma\))
The pull of a liquid surface along itself, as a force per length, \(F = \gamma L\); about 0.073 N/m for water at 20 °C.

Key terms

Capillary action
A liquid climbing up (or being pushed down) in a narrow tube.
Cohesion
The attraction between molecules of the same substance.
Adhesion
The attraction between a liquid and something else, such as glass.
Meniscus
The curved surface of a liquid in a tube: concave for water in glass, domed for mercury.

Key terms

Wettability
How readily a liquid spreads over a surface.
Flow rate (\(Q\))
The volume of fluid passing a point per second, \(Q = Av\), in m³/s.
Continuity
For an incompressible fluid, the flow rate is the same all along a pipe: \(A_1v_1 = A_2v_2\).
Streamline
The path a small parcel of fluid follows through a steady flow.

Key terms

Viscosity (\(\mu\))
The stickiness of a fluid, its resistance to flowing, in pascal-seconds.
Laminar
Smooth flow in layers that slide past each other without mixing.
Turbulent
Chaotic flow, full of eddies of all sizes, that mixes quickly.
Reynolds number (\(Re\))
\(\rho vD/\mu\), a pure number; in a pipe, flow is laminar below about 2000 and turbulent well above.

History and sources

History

Where a primary text wasn't read, the sources are encyclopedic or institutional pages, and the wording below is cautious.

Hooke's law (1676 and 1678)

Hooke's law (1676 and 1678). In 1676 Robert Hooke published his law first as a scrambled Latin anagram, in A Description of Helioscopes, staking his claim without giving the idea away. In 1678, in his lectures on springs (De Potentia Restitutiva), he gave the answer: ut tensio, sic vis, "as the stretch, so the force". Sources: Hooke's anagram; Hooke quotations (MacTutor).

Cauchy and stress (1822)

Cauchy and stress (1822). Augustin-Louis Cauchy defined stress in 1822 as the force per unit area acting across any surface inside a solid, even when the force is slanted to that surface. A short notice was printed in 1823 and the full account in 1827. Sources: Cauchy and stress (arXiv); Fraser, on the history of stress.

Thomas Young (1807)

Thomas Young (1807). In his Course of Lectures on Natural Philosophy (1807), Young described a measure of a material's stiffness. As historians describe it, his own version used the height of a column of the material itself, and some readers found it hard to follow. Euler and Riccati had pointed the way before him, and the modern form, stress over strain, took shape later in the nineteenth century. Sources: Young's modulus; Thomas Young and the theory of structures, 1807–2007.

Galileo's Two New Sciences (1638)

Galileo's Two New Sciences (1638). After his trial in 1633, Galileo lived under house arrest at Arcetri, near Florence. Two New Sciences was published in 1638 in Leiden, by the Elzevir firm, outside the Inquisition's reach. Its first new science was the strength of materials, and he showed that a beam's strength grows with its cross-section. Sources: Two New Sciences; Christie's, first edition.

Galileo's bones (1638)

Galileo's bones (1638). In the Second Day, Galileo drew a normal bone beside one about three times as long but made far thicker, as it would have to be in a much bigger animal: bones must thicken faster than they lengthen. Sources: Scientific American; OU Libraries: Galileo.

David Kirkaldy's testing works (1860s–1874)

David Kirkaldy's testing works (1860s–1874). From the 1860s the Scottish engineer David Kirkaldy ran a testing business in Southwark, London. His hydraulic machine, about 14.5 m long, pulled iron and steel apart and recorded exactly when they failed. In 1874 he moved into purpose-built works at 99 Southwark Street, with "Facts not Opinions" carved over the door. The machine survives there in the Kirkaldy Testing Museum. Sources: Kirkaldy Testing Museum; Grace's Guide: David Kirkaldy; The Testing Works.

The Tay Bridge (1879–80)

The Tay Bridge (1879–80). The first railway bridge over the Firth of Tay fell on 28 December 1879. In 1880 Kirkaldy tested parts from it, though he was not on the official inquiry; accounts say tie bars failed where they joined the cast-iron columns. Source: Grace's Guide: David Kirkaldy.

Goodyear and vulcanisation (1839)

Goodyear and vulcanisation (1839). In 1839 Charles Goodyear found that heating raw rubber with sulfur made it durable and springy, by linking its long chains together. The story that he found it by accident on a hot stove is told often but contested, and Thomas Hancock patented a similar process in Britain at about the same time, in the 1840s. Sources: Vulcanization; EBSCO: vulcanized rubber.

Kirsch and Inglis (1898, 1913)

Kirsch and Inglis (1898, 1913). In 1898 Gustav Kirsch showed that a small round hole in a wide plate triples the stress at its edge. In 1913 Charles Inglis, writing for the Institution of Naval Architects, solved the problem for an elliptical hole: the longer and sharper the hole across the pull, the higher the peak. Sources: Stress round a hole; Stress round an ellipse.

Griffith (1921)

Griffith (1921). Alan Arnold Griffith, at the Royal Aircraft Establishment, Farnborough, published "The phenomena of rupture and flow in solids" (Phil. Trans. R. Soc. A 221, 1921). Using glass, he argued that a crack grows when the energy it releases pays for the new surface; tests on thin glass fibres, which have few flaws, supported it. Sources: Royal Society blog; Alan Arnold Griffith.

SS Schenectady (16 January 1943)

SS Schenectady (16 January 1943). A new T2 tanker, just back from sea trials, cracked across and broke in two at her dock in Portland, Oregon, in calm, cold weather, with a bang reportedly heard about a mile away. She was not a Liberty ship, but was built the same way, with a welded hull, and was repaired and sailed again. Sources: SS Schenectady; TWI.

Constance Tipper (1940s)

Constance Tipper (1940s). A metallurgist at Cambridge, Tipper tested steel from the broken wartime ships and helped show that the trouble lay chiefly in the steel, not only the welding: below a critical temperature it broke like glass. Her notched test bar, the Tipper test, became a widely used check for brittleness. Sources: Cambridge Engineering: Constance Tipper; Constance Tipper.

The Versailles rail accident (8 May 1842)

The Versailles rail accident (8 May 1842). Near Meudon, on the line between Paris and Versailles, the axle of a train's leading locomotive broke; the carriages piled up and caught fire, with passengers locked in. At least 50 died; the toll is disputed. Afterwards French railways stopped locking passengers in. Source: Versailles rail accident.

Rankine and railway axles (1843)

Rankine and railway axles (1843). William Rankine argued, from broken axles on British railways, that they failed through cracks that spread slowly from sharp shoulders, and recommended smooth, gradual changes of shape (Minutes of Proc. Inst. Civil Engineers 2, 1843). Sources: Rankine's paper (Structurae); Rankine (Encyclopedia.com).

August Wöhler (1852–1870)

August Wöhler (1852–1870). From 1852 Wöhler investigated broken axles for the Prussian state, bending turning axles again and again and counting the turns until they broke. His reports (1858–1870) showed that repeated stresses far below the breaking stress can cause failure, and that below a certain stress a steel axle could go on indefinitely. He mostly worked in tables. Sources: Deutsche Biographie; August Wöhler.

The Comet investigation (1954)

The Comet investigation (1954). Two Comets broke up over the sea off Italy in 1954 (10 January near Elba; 8 April, a South African Airways flight). The Royal Aircraft Establishment pressurised a whole fuselage in a water tank again and again until it split, with cracks from the corners of cut-outs. The Elba break-up was traced to a fatigue crack at a rivet hole at the corner of a roof cut-out for a direction-finding aerial, after about 1290 pressurised flights. Sources: BOAC Flight 781.

Aloha Airlines Flight 243 (28 April 1988)

Aloha Airlines Flight 243 (28 April 1988). At about 24 000 feet over Hawaii a large section of roof tore away; one flight attendant was lost, and the pilots landed on Maui. Investigators found fatigue cracking at the lap joints, in an old aircraft with very many flights, which maintenance had missed. Sources: FAA Lessons Learned: Aloha 243; Airways.

Euler and buckling (1744)

Euler and buckling (1744). In an appendix on elastic curves in his Methodus inveniendi (1744), Leonhard Euler worked out the load at which a slender column suddenly bows, and returned to it in 1757. In 1729 Pieter van Musschenbroek had already reported that long columns are weaker. Sources: Euleriana; Fraser, Euler's elastica.

The Quebec Bridge (29 August 1907)

The Quebec Bridge (29 August 1907). The partly built bridge's south arm collapsed into the river, killing 75 workers, 33 of them Mohawk ironworkers from Kahnawake. Compression chords buckled; a Royal Commission blamed the design of their bracing, too little checking, and the lack of an experienced chief engineer on the owner's side. The bridge was finished in 1917, after a second accident. Sources: CSCE: the Quebec Bridge; Dlubal blog.

The Iron Ring (1922–1925)

The Iron Ring (1922–1925). The Calling of an Engineer was begun by H. E. T. Haultain in 1922, with words written by Rudyard Kipling, and the first ceremonies were held in 1925. The story that the first rings were made from the fallen Quebec Bridge's steel is a legend: they were hand-hammered iron. Sources: Iron Ring; Calling of an Engineer; ironring.ca.

The Gimli Glider (1983)

The Gimli Glider (1983). An Air Canada Boeing 767's fuel was worked out by hand with 1.77, the density in pounds per litre, where the right figure in kilograms per litre was about 0.803. It took off with less than half the fuel it needed, ran dry, and was glided down safely. Sources: Gimli Glider; Code7700 case study.

The pascal (1971)

The pascal (1971). The 14th General Conference on Weights and Measures gave the newton per square metre its own name, the pascal, after Blaise Pascal; the same meeting named the siemens. Source: Pascal (unit).

Stephen Hales's horse (published 1733)

Stephen Hales's horse (published 1733). In the early 1700s the clergyman Stephen Hales joined a tall glass tube to an artery of a living horse, and the blood rose about 8 feet 3 inches (about 2.5 m). He published it in Haemastaticks (1733); it is often called the first measurement of blood pressure. Sources: Wood Library-Museum: Hales; Hales (Encyclopedia.com).

The Trieste (23 January 1960)

The Trieste (23 January 1960). Jacques Piccard and Don Walsh took the bathyscaphe Trieste to the bottom of the Challenger Deep, nearly 11 km down. A window pane in the entrance tunnel cracked on the way down, but the crew sphere held. Sources: US Navy H-Gram 041; IEEE Spectrum: Don Walsh.

Galileo and the suction pump (1638)

Galileo and the suction pump (1638). Pump-makers knew a suction pump would not lift water more than about 18 braccia, about 10 m. In Two New Sciences (First Day) Galileo suggested the column breaks under its own weight, like a rope. Sources: Two New Sciences; Chemistry World: Torricelli's barometer.

Gasparo Berti (around 1640)

Gasparo Berti (around 1640). In Rome, Berti filled a pipe taller than 10 m, with a glass vessel at the top, and opened it at the bottom: the water settled at about 10 m, leaving an empty space above. Sources: Museo Galileo: Gasparo Berti; Berti's experiment (IMSS).

Torricelli's barometer (1643–44)

Torricelli's barometer (1643–44). Evangelista Torricelli, Galileo's successor in Florence, used mercury, about 13.6 times denser than water, and the column fell to about 76 cm. In a letter to Michelangelo Ricci of 11 June 1644 he wrote (in translation) that we live submerged at the bottom of an ocean of air. Sources: Torricelli's letter (Le Moyne); Torricelli and Ricci (IMSS).

Pascal, Périer and the Puy de Dôme (19 September 1648)

Pascal, Périer and the Puy de Dôme (19 September 1648). At Blaise Pascal's request, his brother-in-law Florin Périer measured the mercury near the foot and at the summit of the Puy de Dôme, nearly 1500 m high. At the top it stood about 8 cm lower, a little over three French inches. Sources: Whipple Museum: barometers; Scientific Russia: under atmospheric pressure.

Guericke's hemispheres (1654)

Guericke's hemispheres (1654). Otto von Guericke, mayor of Magdeburg, pumped the air out of two copper hemispheres, and teams of horses could not pull them apart. He showed it at Regensburg in 1654 and described it in his book of 1672; accounts differ on how many horses there were. Sources: Physics Today: Magdeburg hemispheres; Grinnell physics museum.

Pascal's treatise (1663)

Pascal's treatise (1663). Pascal's Treatise on the Equilibrium of Liquids, published in 1663, the year after his death, describes how pressure in an enclosed liquid is passed on, so a small force on a small piston can balance a large force on a large one. Source: Pascal's law.

Bramah and Maudslay (1795)

Bramah and Maudslay (1795). Joseph Bramah patented a hydraulic press in 1795. The hard part was stopping the liquid leaking past the piston; Henry Maudslay, his workshop foreman, is credited with the answer, a leather collar that the pressure itself pressed tighter. Sources: Joseph Bramah; Henry Maudslay.

Pascal's barrel (a later legend)

Pascal's barrel (a later legend). The story that Pascal burst a barrel by pouring water into a long thin tube fixed in its lid is told in many textbooks, but it isn't in his surviving writings and seems to come from later retellings. The physics is right: pressure depends on depth, not volume. Sources: Pascal's law; Pascal's barrel (Observatoire de Paris dictionary).

Archimedes, On Floating Bodies (3rd century BC)

Archimedes, On Floating Bodies (3rd century BC). Archimedes proved that a floating body displaces its own weight of fluid, and treated solids heavier than the fluid too; in Book II he worked out when a floating paraboloid stays upright and when it tips. Sources: On Floating Bodies; Archimedes: floating bodies (NYU).

The king's crown (Vitruvius, 1st century BC)

The king's crown (Vitruvius, 1st century BC). The story of King Hiero's crown comes from Vitruvius, De Architectura, Book IX, about two hundred years after Archimedes, and is not in Archimedes' own writings. The bath, the run through the streets and "Eureka!" are legend. Sources: Vitruvius on the crown (NYU); The golden crown (NYU).

Galileo's little balance (1586)

Galileo's little balance (1586). Aged 22, Galileo wrote La Bilancetta, doubting that Archimedes had used the overflow, and describing instead a balance for weighing in water that could reveal how much of each metal was in a mixture of two. Source: La Bilancetta (NYU).

Samuel Plimsoll and the load line (1870s)

Samuel Plimsoll and the load line (1870s). Plimsoll, a British MP, campaigned against overloaded "coffin ships" and wrote Our Seamen. The Merchant Shipping Act 1876 made a load-line mark compulsory, though at first owners chose where to paint it; its position was fixed by law later. Sources: Royal Museums Greenwich; Lloyd's Register Foundation.

The Vasa (10 August 1628)

The Vasa (10 August 1628). The Swedish warship heeled over and sank on her maiden voyage in Stockholm harbour after about 1300 m. Before sailing, a stability test of 30 men running across her deck had been stopped because she rolled so badly; she was built too high for her keel and ballast. Estimates of the dead range from about 30 to about 50. She was raised in 1961 and is now in a museum in Stockholm. Sources: Britannica: Vasa; The Vasa (John Patrick).

The first balloons (1783)

The first balloons (1783). On 4 June the Montgolfier brothers flew a hot-air balloon in public at Annonay; on 21 November Pilâtre de Rozier and the Marquis d'Arlandes made the first free flight carrying people; on 1 December Jacques Charles and Nicolas-Louis Robert rose in a hydrogen balloon. Sources: V&A: flying high; Centennial of Flight: early balloon flight.

"What good is a newborn baby?" (1783, legend)

"What good is a newborn baby?" (1783, legend). Benjamin Franklin, in France, watched the balloons. The reply often credited to him varies in wording, is tied to different flights, and historians doubt he said it in that form: treat it as legend. Sources: Papers of Benjamin Franklin (Founders Online); American Heritage.

Gao and Jiang on the water strider (2004)

Gao and Jiang on the water strider (2004). A water strider's legs are covered in tiny water-repelling hairs; Gao and Jiang (Nature 432, 2004) found a single leg could push on the surface with about 15 times the insect's body weight before breaking through. Sources: Gao and Jiang, Nature (DOI); BIOKON: water striders.

Franklin and oil on water (1773)

Franklin and oil on water (1773). In a letter to William Brownrigg of 7 November 1773, Franklin described how a little oil calms rough water: on a pond on Clapham Common, about a teaspoon is reported to have spread over roughly half an acre. The Royal Society printed it in 1774. Sources: Founders Online: Franklin to Brownrigg; Royal Society, Making Science.

Agnes Pockels (1891)

Agnes Pockels (1891). With no university education, Pockels measured in her kitchen, with a simple trough, how the surface tension of water changes when the surface is contaminated. Lord Rayleigh forwarded her letter to Nature, published on 12 March 1891; in 1932 the Technical University of Braunschweig gave her an honorary doctorate. Sources: Agnes Pockels; TU Braunschweig.

From Hauksbee to Laplace (1709–1806)

From Hauksbee to Laplace (1709–1806). Francis Hauksbee measured liquids rising in narrow tubes in 1709; around 1718 James Jurin showed the rise is inversely proportional to the tube's width; Thomas Young (1805) and Pierre-Simon Laplace (1806) explained it with surface tension. Sources: Jurin's law; Physics Today: two hundred years of capillarity research.

Leonardo's water studies (about 1508–1513)

Leonardo's water studies (about 1508–1513). Leonardo da Vinci filled notebook pages with moving water: a jet falling into a pool, curling into eddies, swirls made of smaller swirls. The sheets are now in the Royal Collection at Windsor. Sources: Nature (2019); IAS.

Benedetto Castelli (1628)

Benedetto Castelli (1628). Castelli, a pupil of Galileo, is credited with the first clear statement of continuity: in On the Measurement of Running Water he showed that the water passing a section of a river depends on its width, its depth and the speed, so a narrower stretch carries faster water. Sources: Olschki; Lugaresi (SMF sample).

Venturi and Herschel (1797, 1888)

Venturi and Herschel (1797, 1888). Giovanni Battista Venturi published experiments on fluids in narrowing tubes in 1797. Clemens Herschel turned the pressure drop into a water meter, patented in 1888 and named after Venturi. Sources: Giovanni Battista Venturi; Clemens Herschel.

Daniel Bernoulli's Hydrodynamica (1738)

Daniel Bernoulli's Hydrodynamica (1738). Finished in St Petersburg and printed in Strasbourg in 1738, it linked a fluid's speed to its pressure. His father Johann's rival Hydraulica, dated 1732, seems to have appeared later, and priority was disputed; the modern form of the equation is usually credited to Euler. Sources: Hydrodynamica; Linda Hall Library: Daniel Bernoulli.

Torricelli's law (1643)

Torricelli's law (1643). Torricelli found that water leaves a hole at the speed of a free fall through the depth of the water, \(v = \sqrt{2gh}\); it was printed in his Opera geometrica (1644). Source: Britannica: Torricelli's theorem.

Newton and viscosity (1687)

Newton and viscosity (1687). In Principia, Book II, Newton proposed that a fluid resists motion through a "lack of slipperiness" (defectus lubricitatis) that grows with how fast neighbouring parts slide past each other. Fluids like this are still called Newtonian. Sources: arXiv: history of viscosity; Newton's law of viscosity (IIT Delhi).

The Boston molasses flood (15 January 1919)

The Boston molasses flood (15 January 1919). A tank in Boston's North End burst, releasing about 2.3 million US gallons (about 8700 m³) of molasses in a wave several metres high at about 35 mph. 21 people were killed and about 150 injured. Much later, experiments by Harvard researchers suggested the cold made the molasses far more viscous, making escape harder. Sources: Live Science; Popular Science.

Poiseuille and Hagen (1838–1846, 1839)

Poiseuille and Hagen (1838–1846, 1839). Jean Poiseuille, a physician interested in blood in the smallest vessels, measured liquids (apparently water) flowing through very fine glass tubes; Gotthilf Hagen found the same fourth-power law independently in 1839. The old unit of viscosity, the poise, is named after Poiseuille. Sources: Hagen–Poiseuille equation; Poiseuille history (University of Arizona).

The pitch drop experiment (1927 to today)

The pitch drop experiment (1927 to today). Thomas Parnell set up a funnel of pitch at the University of Queensland in 1927, opened in 1930. Drops have fallen about once a decade, the ninth in 2014; its viscosity is about \(2.3\times10^8\) Pa·s. John Mainstone tended it for over 50 years and never saw a drop fall. It won an Ig Nobel Prize in 2005. Sources: University of Queensland; Edgeworth, Dalton and Parnell (1984).

Osborne Reynolds (1883)

Osborne Reynolds (1883). At Owens College, Manchester, Reynolds let a thin stream of dye into water flowing through a glass pipe: slow, it stayed a straight thread; faster, it suddenly broke into eddies (Phil. Trans. R. Soc. 174, 1883). The number was named after him by Arnold Sommerfeld in 1908. Sources: APS News: this month in physics history; Reynolds (DSB, MacTutor).

George Gabriel Stokes (1845, 1851)

George Gabriel Stokes (1845, 1851). In 1845 Stokes put viscosity into the equations of fluid motion, now the Navier–Stokes equations (Navier and others had found them earlier); in 1851 he worked out the drag on a small sphere moving slowly through a sticky fluid. Sources: Stokes (Encyclopedia.com); Nature Physics.

Babinsky and the equal-transit myth (2003)

Babinsky and the equal-transit myth (2003). Holger Babinsky of Cambridge University, in "How do wings work?" (Physics Education 38, 2003), showed the equal-transit story is wrong: the upper air arrives first, and lift comes from curved flow and pressure differences. NASA's Glenn Research Center also lists it as an incorrect theory of lift. Sources: Babinsky, How do wings work? (Cambridge); AVweb: theories of lift.

History

The cloud session that researched these cards could search the web but could not open pages directly (every page fetch was blocked), so the claims above rest on search-result summaries of the cited sources rather than on reading them. The items listed under "Worth checking" in each video's sources.md should be confirmed against the originals before upload.

Numbers used

Numbers used. The reference values are typical textbook values, not exact ones (they vary with grade, temperature and source): \(g = 9.81\) N/kg; one standard atmosphere 101 325 Pa. Young's moduli: steel 200 GPa, aluminium 70, glass about 70, bone about 18, wood along the grain about 10, nylon 2–4, rubber about 0.01. Mild steel: yield strength about 250 MPa, tensile strength about 400 MPa. Densities (kg/m³): water 1000, seawater 1025, ice 917, air 1.2, helium 0.17, steel 7850, mercury 13 600, gold 19 300, silver 10 500, spider silk 1300, glycerol 1260. Surface tension of water at 20 °C: 0.073 N/m. Viscosities: water 0.001 Pa·s, air \(1.8\times10^{-5}\) Pa·s, glycerol about 1.5 Pa·s, pitch about \(2.3\times10^8\) Pa·s. Every worked number is recomputed in each video's numbers_check.py.

Worked example: will it float?

The problem

Problem. A solid steel block, 1 m on each side, has a mass of 7850 kg. The same steel is rolled into plates and welded into an open box 5 m long, 2 m wide and 1 m deep. Which of them floats? How deep does the box sit in fresh water (density 1000 kg/m³) and in sea water (1025 kg/m³), and how much cargo can it carry? Take \(g = 9.81\) m/s².

Solution

  1. The idea. Archimedes' principle: the upthrust equals the weight of the water pushed aside. A floating object pushes aside its own weight of water. Quick test: an object floats if its average density (total mass ÷ the whole volume it takes up, air inside included) is less than the water's.

Solution

  1. The block. Weight \(7850 \times 9.81 \approx 77\,000\) N. The most water it can push aside is 1 m³, weighing \(1000 \times 9.81 = 9810\) N. It sinks (equivalently, \(7850 > 1000\)). On the bottom, a crane feels \(77\,000 - 9810 \approx 67\,200\) N, as if it were 6850 kg: the water takes off exactly one tonne.

Solution

  1. The plates. Bottom \(5\times2 = 10\) m², two sides \(5\times1 = 5\) m² each, two ends \(2\times1 = 2\) m² each: 24 m² in all, so the plate is about \(1/24\ \text{m} \approx 4.2\) cm thick.
  2. Average density. The box encloses \(5\times2\times1 = 10\) m³, so \(\rho_\text{avg} = 7850/10 = 785\) kg/m³ \(< 1000\): it floats.

Solution

  1. Draught in fresh water. It must push aside 7850 kg of water, \(7.85\) m³. With a 10 m² bottom, the draught is \(7.85/10 = 0.785\) m, leaving \(1 - 0.785 = 0.215\) m (21.5 cm) above the water.
  2. In the sea. \(7850/1025 \approx 7.66\) m³, so the draught is \(0.766\) m: about 2 cm higher. That is why load lines have separate fresh and sea marks.

Solution

  1. Cargo. Down to the rim it pushes aside 10 m³ = 10 000 kg, so the cargo can be \(10\,000 - 7850 = 2150\) kg. Keeping 10 cm of side above the water (draught 0.9 m, 9000 kg displaced) leaves \(9000 - 7850 = 1150\) kg.
  2. Checks. Units: kg ÷ m³ is a density and m³ ÷ m² a length ✓. The draught is less than the 1 m depth ✓. Denser water, higher float ✓. Limiting case: a box enclosing less than 7.85 m³ would sink even empty ✓.

Practice

Practice

Use \(g = 9.81\) m/s² unless told otherwise.

★ Warm-up

1.1 A steel wire 1.5 m long, with a cross-section of 0.8 mm², holds up a 10 kg mass. Find the stress and the strain (\(E = 200\) GPa). Then find the extension, in millimetres, and say whether the wire is safely below mild steel's yield strength of about 250 MPa.

mm

Solution

Idea: stress \(\sigma = F/A\), strain \(\varepsilon = \sigma/E\), extension \(\Delta L = \varepsilon L\). \(F = 10\times9.81 = 98.1\) N and \(A = 0.8\ \text{mm}^2 = 0.8\times10^{-6}\ \text{m}^2\), so \(\sigma = 98.1/(0.8\times10^{-6}) \approx 1.23\times10^8\) Pa \(= 123\) MPa. \(\varepsilon = 1.23\times10^8/(200\times10^9) \approx 6.1\times10^{-4}\), and \(\Delta L = 6.1\times10^{-4}\times1.5\ \text{m} \approx 0.92\) mm. 123 MPa is about half of 250 MPa, so it is safely elastic. Check: a strain well under 1%, as expected for steel ✓.

Problem 1.2 (★ Warm-up)

1.2 A block of aluminium measures 25 cm × 10 cm × 5 cm and has a mass of 3.375 kg. What is its density, in kg/m³?

kg/m^3

Solution

Idea: \(\rho = m/V\), with the volume in m³. \(V = 0.25\times0.10\times0.05 = 1.25\times10^{-3}\) m³, so \(\rho = 3.375/1.25\times10^{-3} = 2700\) kg/m³. Check: between water (1000) and steel (7850), as a light metal should be ✓.

Problem 1.3 (★ Warm-up)

1.3 Why your ears hurt at the bottom of a pool. You dive to the bottom of a pool 3.0 m deep. By how much is the water pressure there bigger than at the surface? Your eardrum has an area of about 0.6 cm²; what extra force, in newtons, pushes on it, if the air behind it is still at surface pressure?

N

Solution

Idea: pressure grows with depth as \(\Delta p = \rho g h\), and force = pressure × area. \(\Delta p = 1000\times9.81\times3.0 \approx 29\,400\) Pa, about 0.29 atmospheres. \(A = 0.6\ \text{cm}^2 = 0.6\times10^{-4}\ \text{m}^2\), so \(F = 29\,400\times0.6\times10^{-4} \approx 1.8\) N: like the weight of 180 g pressing on a thin membrane. "Equalising" (swallowing or gently blowing with the nose pinched) lets air into the middle ear until the pressure behind the eardrum matches the water's, and the push disappears. Check: 3 m is less than a third of the 10 m that adds a whole atmosphere ✓.

Problem 1.4 (★ Warm-up)

1.4 In a car's brakes, the pedal pushes a piston of area 1.5 cm² with a force of 200 N. The brake piston at the wheel has an area of 30 cm². What force, in newtons, does it press with? Then find how far it moves for each centimetre the pedal piston moves.

N

Solution

Idea: Pascal's principle: the same pressure everywhere in the liquid, so \(F_2 = F_1 A_2/A_1 = 200\times30/1.5 = 4000\) N. The same volume of liquid moves, so the brake piston moves \(1.5/30 = 0.05\) cm (half a millimetre) per centimetre of pedal piston. Check: force × distance is the same on both sides (200 × 1 = 4000 × 0.05), so no energy is created ✓.

Problem 1.5 (★ Warm-up)

1.5 Water flows at 0.5 m/s in a hose of diameter 4 cm, and leaves through a nozzle of diameter 1 cm. How fast does it come out, in metres per second?

m/s

Solution

Idea: continuity, \(A_1v_1 = A_2v_2\). The area goes as the diameter squared, so the speed goes up by \((4/1)^2 = 16\): \(v_2 = 0.5\times16 = 8\) m/s. Check: a narrower opening means faster flow ✓.

★★ Standard

2.1 Ice has a density of about 917 kg/m³ and sea water about 1025 kg/m³. What percentage of an iceberg is above the water?

Solution

Idea: a floating object's fraction under water is \(\rho_\text{object}/\rho_\text{water} = 917/1025 \approx 0.895\). So about 10.5% is above the surface: "the tip of the iceberg". Check: less than 1, so it floats; close to 1, so most is hidden ✓.

Problem 2.2 (★★ Standard)

2.2 A rock has a mass of 3.0 kg. Hung from a spring balance under water, it reads 20 N. Find its volume, and then its density in kg/m³.

kg/m^3

Solution

Idea: the upthrust is the missing weight, and it equals the weight of the water displaced. Weight \(= 3.0\times9.81 = 29.43\) N, so the upthrust is \(29.43 - 20 = 9.43\) N. Volume \(= 9.43/(1000\times9.81) \approx 9.6\times10^{-4}\) m³ (about 0.96 litres). Density \(= 3.0/9.6\times10^{-4} \approx 3120\) kg/m³. Check: typical rock is 2500–3500 kg/m³ ✓.

Problem 2.3 (★★ Standard)

2.3 A helium balloon must lift 250 kg in all (envelope, basket and passengers). Air has a density of about 1.2 kg/m³ and helium about 0.17 kg/m³. What volume of helium is needed, in cubic metres? Then find the radius of a sphere of that size.

m^3

Solution

Idea: each cubic metre of helium displaces 1.2 kg of air but weighs 0.17 kg itself, so it lifts \(1.2 - 0.17 = 1.03\) kg. \(V = 250/1.03 \approx 243\) m³. A sphere of that volume has radius \(r = (3V/4\pi)^{1/3} \approx 3.9\) m. Check: about 8 m across, a modest crewed gas balloon ✓.

Problem 2.4 (★★ Standard)

2.4 A steel column is replaced by one half as long and twice the diameter, of the same steel. By what factor does its buckling load change?

Solution

Idea: Euler's buckling load \(F = \pi^2EI/L^2\), with \(I\) proportional to the diameter to the fourth power for a solid round rod. Halving \(L\) multiplies the load by \(2^2 = 4\); doubling the diameter multiplies \(I\) by \(2^4 = 16\). Together: \(4\times16 = 64\) times. Check: short, fat columns are very hard to buckle ✓.

Problem 2.5 (★★ Standard)

2.5 How high does water rise, in centimetres, in a clean glass tube of radius 0.25 mm? (Surface tension of water: 0.0728 N/m.)

cm

Solution

Idea: the surface pulls up round the rim, \(2\pi r\gamma\), and holds up a column of weight \(\rho g\pi r^2h\): \(h = 2\gamma/(\rho g r)\). \(h = 2\times0.0728/(1000\times9.81\times2.5\times10^{-4}) \approx 0.059\) m, about 5.9 cm. Check: a thinner tube would give a higher rise (\(h \propto 1/r\)) ✓.

★★★ Stretch

3.1 Bernoulli in a narrowing pipe, and where it fails. Water flows along a level pipe of diameter 5 cm at 1.2 m/s, at a pressure of 200 kPa. The pipe narrows to 2 cm. (a) Find the speed in the narrow part, and the pressure there in kilopascals. (b) Repeat for a narrowing to 0.8 cm. What has gone wrong? (c) Is the flow in the 5 cm pipe smooth or turbulent? (Water's viscosity: 0.001 Pa s.)

Hint

Continuity gives the speed; Bernoulli on a level pipe says \(p + \tfrac12\rho v^2\) is the same in both parts. A pressure can't go below zero.

kPa

Solution

(a) \(v_2 = 1.2\times(5/2)^2 = 7.5\) m/s. \(\Delta p = \tfrac12\rho(v_2^2 - v_1^2) = 500\times(56.25 - 1.44) \approx 27\,400\) Pa, so \(p_2 \approx 172.6\) kPa. (b) \(v_2 = 1.2\times(5/0.8)^2 \approx 46.9\) m/s, and Bernoulli gives \(p_2 \approx 200 - 1098 \approx -898\) kPa: an impossible negative pressure. Long before that, the pressure falls so low that the water boils into vapour bubbles at room temperature (cavitation), and the ideal flow Bernoulli assumes breaks down; friction losses (ignored by Bernoulli) also grow fast at such speeds, so the real speed would be lower. (c) \(Re = \rho vD/\mu = 1000\times1.2\times0.05/0.001 = 60\,000\), far above about 2000: turbulent, so there are real friction losses even in (a), and the narrow-part pressure would be a little lower than the ideal answer. Check: Bernoulli is a good guide for short, smooth contractions, and a bad one when it predicts pressures near or below zero ✓.

Problem 3.2 (★★★ Stretch)

3.2 A hole and fatigue. A steel plate carries a steady-on, steady-off tension of 120 MPa (nominal). It has a small round hole. The steel's tensile strength is 400 MPa, and its endurance limit is about half that. Will it survive millions of load cycles? What is the largest nominal stress that would be safe, in megapascals?

Hint

A round hole multiplies the local stress by about 3 at its edge.

MPa

Solution

The peak stress at the hole's edge is about \(3\times120 = 360\) MPa, well above the endurance limit of about 200 MPa, so a fatigue crack will eventually start there. For a safe design, \(3\sigma \leq 200\) MPa, so \(\sigma \leq 67\) MPa (before any safety factor). Without the hole, 120 MPa would be below the limit. Check: this is the Comet's lesson: the material was strong enough on average, but not at a corner ✓.

Problem 3.3 (★★★ Stretch)

3.3 The window and the cycle. An airliner's cabin is kept at about 75.3 kPa (the pressure at about 2400 m, or 8000 feet). At 11 km, the air outside is at about 22.6 kPa. What net force, in newtons, does the air inside push out with on a window 25 cm × 35 cm? How many times is this load applied in a year if the plane makes six flights a day?

N

Solution

\(\Delta p = 75.3 - 22.6 = 52.7\) kPa. \(A = 0.25\times0.35 = 0.0875\) m². \(F = 52\,700\times0.0875 \approx 4600\) N, like the weight of about 470 kg. Each flight is one load cycle: about \(6\times365 = 2190\) a year, and tens of thousands over the plane's life. This is why pressurised fuselages are designed and inspected for fatigue. Check: half an atmosphere on a small window gives tonnes-scale forces on a whole fuselage panel ✓.

Check yourself

Answer these out loud, from memory. Then try again in a few days: spaced recall is what makes things stick.

Check yourself

  • How do rigid-body models differ from descriptions of deformable matter?

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  • What are stress and strain, and how do they describe deformation?

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  • What do Young's modulus, shear modulus, and bulk modulus measure?

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  • How do tensile, compressive, and shear deformations differ?

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  • What are elastic and plastic deformation?

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  • How can elastic energy be stored in a stretched or compressed material?

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  • How do microscopic interactions produce the macroscopic stiffness of a solid?

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  • What are yield strength, fracture, fatigue, and creep?

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  • Why can structural stability depend on buckling as well as material strength?

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  • How do scale, geometry, and material properties influence the strength of a structure?

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  • What distinguishes a fluid from a solid in its response to shear?

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  • What are density and pressure, and how are they measured?

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  • Why does pressure increase with depth in a fluid at rest?

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  • What do Pascal's principle and Archimedes' principle explain?

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  • What determines whether an object floats, sinks, or remains suspended?

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  • What are surface tension, capillary action, and wettability?

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  • How does conservation of mass produce the continuity equation for flow?

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  • What does Bernoulli's equation say, and what assumptions limit its use?

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  • How do viscosity and the Reynolds number help distinguish flow regimes?

Check yourself

  • How do real fluid losses alter ideal predictions for pipes, pumps, and aerodynamic flow?

Summary

Summary

Your results for each stage appear here when scripting is on. Nothing is graded and nothing is sent anywhere.

Further learning

Free textbooks: OpenStax, College Physics 2e (not yet checked): Chapter 11: Fluid Statics · 11.5 Pascal's Principle · Chapter 12: Fluid Dynamics and Its Biological and Medical Applications. OpenStax, University Physics Volume 1 (not yet checked): 12.3 Stress, Strain, and Elastic Modulus · 12.4 Elasticity and Plasticity.

Maths and physics lessons (found in Khan Academy's search index on 2026-10-14): Pressure at a depth in a fluid.

Simulations (PhET, not yet checked): Under Pressure · Buoyancy · Fluid Pressure and Flow.


Next: M5 · Exponentials, logarithms and waves as functions (coming soon) · Progress