Properties of Solids and Liquids

Why does a liquid at rest push only perpendicular to any surface, never along it?

Because a parallel push is a shear — and a liquid cannot resist shear. It would simply flow instead.

That one sentence generates most of this unit:

A liquid does resist shape change in exactly two places — at its surface (surface tension) and between layers sliding past each other (viscosity). Sections 5 and 6 are entirely those two exceptions.

Second organising fact. The recurring currency here is per unit area. Stress, pressure, surface tension, heat current density — all of them. When a formula looks unfamiliar, ask what is being divided by area.

1. Stress, strain and the three moduli

Stress — internal restoring force per unit area, in Pa. Strain — fractional deformation, dimensionless.

$Y=\frac{FL}{A\Delta L},K=-\frac{\Delta P}{\Delta V/V}$

A modulus is a stiffness, not a strength. Steel's $Y$ is ~100× rubber's, meaning it resists stretching far more — it says nothing about the load at which either breaks.

• $Y$ and $\eta$ exist for solids only. A liquid has no definite length to stretch and no rigidity at all.
• Water's $K\approx 2.2\times 10^9$ Pa, so one atmosphere compresses it by 0.005% — which is why liquids are treated as incompressible in all of fluid mechanics.
• For a given solid, $Y>\eta$ by a factor of two or three: stretching bonds is harder than bending them.

2. The stress–strain curve

• Proportional limit — Hooke's law stops, but the material still returns fully.
• Elastic limit / yield point — permanent deformation begins; what remains after unloading is the permanent set.
• Ultimate tensile strength — the maximum stress carried. After it the sample necks.
• Fracture — it breaks.

Ductile (copper): long plastic region, so it can be drawn into wire. Brittle (glass): fractures almost at the elastic limit. Elastomer (rubber): no straight-line region at all, yet returns from enormous strains — elastic without obeying Hooke's law.

$U=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$

Energy density is half stress times strain, which is exactly the area under the curve.

Illustration 1

A steel wire of length $2\text{m}$ and diameter $1\text{mm}$ carries a $50\text{kg}$ load. Take $Y=2\times 10^{11}$ Pa, $g=10$. Find the stress, the extension and the elastic energy stored.

$A=\pi r^2=\pi (0.5\times 10^{-3}{)}^2=7.85\times 10^{-7}{\text{m}}^2$

$\text{stress}=\frac{500}{7.85\times 10^{-7}}=6.37\times 10^8\text{Pa}$

$\Delta L=\frac{\text{stress}\times L}{Y}=\frac{6.37\times 10^8(2)}{2\times 10^{11}}=6.4\text{mm}$

Energy, as the area under a straight-line graph — so half the product, not the whole:

$U=\frac{1}{2}F\Delta L=\frac{1}{2}(500)(6.37\times 10^{-3})=1.59\text{J}$

Check by the density route: $\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$ with strain $=3.18\times 10^{-3}$ and volume $=1.57\times 10^{-6}{\text{m}}^3$ gives $1.59$ J. The two agree.

Worth noticing: the stress is already about half a typical steel yield stress. Doubling the load would take this wire past its elastic limit, and the neat linear formulas would stop applying.

3. Fluid statics

$P=P_0+\rho gh$

Hydrostatic paradox. Pressure depends on depth alone — never on shape or amount. A narrow tube and a wide tank filled to the same height press equally hard on their bases.

Pascal's law — pressure applied to an enclosed fluid is transmitted undiminished everywhere. The hydraulic lift follows: force multiplies in the ratio of areas. But it multiplies force, not energy — the big piston moves proportionally less, so the work is equal on both sides.

Archimedes — buoyant force = weight of fluid displaced. Not a separate law of nature: it is just the pressure on the bottom exceeding that on the top.

The commonest flotation error. A floating body displaces its own weight of fluid. A fully submerged body displaces its own volume. Confusing the two ruins the problem.

Illustration 2

A hydraulic lift has pistons of diameter $5\text{cm}$ and $50\text{cm}$. What force on the small piston will raise a $1500\text{kg}$ car? Through what distance must the small piston move to lift the car by $10\text{cm}$? $g=10$.

Pascal's law makes the pressure equal on both sides, so the forces go as the areas — and area goes as the diameter squared:

$\frac{A_2}{A_1}={\left(\frac{50}{5}\right)}^2=100$

$F_1=\frac{15,000}{100}=150\text{N}$

The fluid is incompressible, so the two swept volumes must match:

$A_1{d}_1=A_2{d}_2\Rightarrow d_1=100(0.1)=10\text{m}$

Where the "free" force went: the work is identical on both sides — $150(10)=1500$ J and $15,000(0.1)=1500$ J. A hydraulic lift multiplies force by exactly the factor it divides distance, which is why it is a machine and not a violation.

Illustration 3

What fraction of an iceberg floats above water? ${\rho }_{ice}=917$, ${\rho }_{water}=1030{\text{kg/m}}^3$.

Floating $\Rightarrow$ weight = buoyancy: ${\rho }_{ice}Vg={\rho }_w{V}_{sub}g$

$\frac{V_{sub}}{V}=\frac{917}{1030}=0.89$

About 11% above water. Note the answer is a pure density ratio — the size of the iceberg never enters.

4. Surface tension

A molecule in the bulk is pulled equally in all directions; one at the surface has neighbours only below, so it is pulled inward. The surface behaves like a stretched membrane.

$T$ = force per unit length in the surface = energy per unit area of new surface. Same units either way: N m⁻¹ = J m⁻².

This is why drops are spherical: a sphere has least area for a given volume, hence least surface energy.

$T$ falls with temperature and vanishes at the critical temperature — which is why hot water cleans better, wetting and penetrating fabric more easily.

Excess pressure

A soap bubble has an inner and an outer surface, so its excess pressure is double a drop's. This is tested in almost every paper.

Smaller bubble $\Rightarrow$ higher internal pressure. So when two soap bubbles are connected, the small one empties into the large one rather than equalising.

Illustration 4

A water drop of radius $2\text{mm}$ is broken into $1000$ identical droplets. Find the work required. $T=0.073\text{N/m}$.

Volume is conserved, so $1000\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$, giving

$r=\frac{R}{1000^{1/3}}=\frac{R}{10}=0.2\text{mm}$

Surface tension is energy per unit area, so the work is $T$ times the increase in area:

$A_i=4\pi R^2=5.03\times 10^{-5}{\text{m}}^2$

$A_f=1000(4\pi r^2)=1000(5.03\times 10^{-7})=5.03\times 10^{-4}{\text{m}}^2$

$W=T\Delta A=0.073(4.52\times 10^{-4})=3.3\times 10^{-5}\text{J}$

The general result is worth keeping: $W=4\pi R^{2}T\left(n^{1/3}-1\right)$. Splitting into $n$ pieces multiplies the area by $n^{1/3}$, so a thousandfold subdivision only costs nine times the original surface energy — and it is why fine sprays take real energy to produce, and why droplets spontaneously coalesce back.

Capillarity

The angle of contact $\theta$ is measured inside the liquid.

• Adhesion beats cohesion $\Rightarrow$ $\theta$ acute, concave meniscus, liquid rises. Water in glass, $\theta \approx 0$.
• Cohesion beats adhesion $\Rightarrow$ $\theta$ obtuse, convex meniscus, liquid is depressed. Mercury in glass, $\theta \approx 140°$.

$h=\frac{2T\cos \theta }{\rho gr}$

Rise is inversely proportional to radius — narrower tube, higher lift. This is what draws water through soil and up a paper towel.

If the tube is shorter than the calculated rise, the liquid does not overflow. The meniscus flattens to a larger radius of curvature, and the reduced height is still supported.

Illustration 5

Water rises $5\text{cm}$ in a clean glass capillary. Find the tube's radius. Then predict what happens if the same tube is cut to a length of $3\text{cm}$ above the water surface. Take $T=0.073$, $\theta =0$, $\rho =1000$, $g=9.8$.

$r=\frac{2T\cos \theta }{\rho gh}=\frac{2(0.073)}{1000(9.8)(0.05)}=\frac{0.146}{490}=3.0\times 10^{-4}\text{m}$, i.e. $0.3$ mm.

Cut to 3 cm. The water rises to the top and stops. It does not spill.

The height a meniscus can support is set by its radius of curvature $R$, through $\rho gh=2T/R$. Fix $h$ at $3$ cm instead and solve for $R$:

$R=\frac{2T}{\rho gh}=\frac{0.146}{294}=4.97\times 10^{-4}\text{m}$

Since $R=r/\cos \theta$, the contact angle simply adjusts:

$\cos {\theta }^{\prime }=\frac{r}{R}=\frac{3.0}{4.97}=0.60\Rightarrow {\theta }^{\prime }=53°$

The meniscus flattens from fully concave to a $53°$ contact angle, which is exactly enough to hold the shorter column. Nature adjusts the one variable it can.

5. Viscosity and terminal velocity

$F=-\eta A\frac{dv}{dy},F_{Stokes}=6\pi \eta rv$

Viscosity is internal friction between layers — the one place a liquid resists shear, and only while it is actually shearing.

Liquids and gases behave oppositely. Liquid viscosity falls with temperature (heating weakens intermolecular attraction); gas viscosity rises (heating increases momentum transfer between layers).

At terminal velocity, drag + buoyancy = weight:

$v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$

$v_t\propto r^2$ — which is why fine mist hangs in the air almost indefinitely while raindrops fall fast.

Streamline flow, turbulence and critical velocity

Below a certain speed, fluid moves in orderly layers that never cross — streamline or laminar flow, in which every particle passing a given point follows the same path. Above it the layers break up into eddies and the flow becomes turbulent.

The changeover is governed by a single dimensionless number comparing inertia with viscosity:

$R_e=\frac{\rho vD}{\eta }$

Setting $R_e$ to its critical value gives the critical velocity:

$v_c=\frac{R_e\eta }{\rho D}$

Being dimensionless is what makes $R_e$ powerful: two flows with the same Reynolds number behave alike whatever their actual size. That is why a scale model in a wind tunnel predicts the behaviour of a full-size aircraft.

Turbulence matters practically because it raises drag sharply, and because Bernoulli's equation — derived for streamline flow — stops applying.

Illustration 6

Find the critical speed above which water flow in a $2\text{cm}$ pipe turns turbulent. Take $\rho =1000$, $\eta =1.0\times 10^{-3}$ Pa s, $R_e=2000$.

$v_c=\frac{R_e\eta }{\rho D}=\frac{2000(1.0\times 10^{-3})}{1000(0.02)}=\frac{2}{20}=0.1\text{m/s}$

Read it back, because the number is surprising. Ten centimetres per second is a barely visible trickle. Water from an ordinary household tap moves at over a metre per second, so domestic plumbing is almost always turbulent — and the tidy streamline picture used in Bernoulli problems is the exception in real life, not the rule.

Illustration 7

A fog droplet has radius $10\mu \text{m}$; a raindrop has radius $1\text{mm}$. How long does each take to fall 1 km? Take $\eta =1.8\times 10^{-5}$ Pa s, $\rho =1000$, $\sigma \approx 0$, $g=10$.

$v_t=\frac{2r^2\rho g}{9\eta }$, so the two speeds are in the ratio ${\left(\frac{10^{-3}}{10^{-5}}\right)}^2=10^4$.

Raindrop: $v_t=\frac{2(10^{-6})(10^4)}{1.62\times 10^{-4}}\approx 123\text{m/s}$, so about 8 seconds.

Fog droplet: $v_t\approx 1.2\times 10^{-2}\text{m/s}$, so about 23 hours.

That single factor of $r^2$ is the entire difference between rain and a cloud that simply hangs there. (Real raindrops are slower than 123 m/s because Stokes' law breaks down once the flow turns turbulent.)

6. Fluid flow

$A_1{v}_1=A_2{v}_2\text{(continuity — conservation of mass)}$

$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}\text{(Bernoulli — energy per unit volume)}$

All three terms are in pascals — a useful dimensional check.

Faster fluid has lower pressure. At constant height, raising the kinetic term must lower the pressure term.

Assumes non-viscous, incompressible, steady streamline flow, along a single streamline. Real pipes lose pressure to viscosity, which is why long pipelines need booster pumps.

Torricelli: for a hole at depth $h$, efflux speed $v=\sqrt{2gh}$ — exactly free-fall speed through that height.

Illustration 8

Water flows through a horizontal pipe that narrows from $8{\text{cm}}^2$ to $4{\text{cm}}^2$. The pressure drop across the constriction is $700\text{Pa}$. Find the flow rate. $\rho =1000$.

Continuity first — halving the area doubles the speed:

$A_1{v}_1=A_2{v}_2\Rightarrow v_2=2v_1$

The pipe is horizontal, so the $\rho gh$ terms cancel and Bernoulli reduces to

$P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)=\frac{1}{2}(1000)\left(4v_1^2-v_1^2\right)=1500v_1^2$

$700=1500v_1^2\Rightarrow v_1=0.68\text{m/s},v_2=1.37\text{m/s}$

$Q=A_1{v}_1=8\times 10^{-4}(0.68)=5.5\times 10^{-4}{\text{m}}^3\text{/s}$, about $0.55$ litres per second.

This is a Venturi meter, and it is how flow rate is actually measured in a pipe you cannot see into: read a pressure difference, get a speed. Note the pressure fell where the pipe got narrower — the opposite of most people's intuition, and the whole content of Bernoulli's principle.

Aerofoil lift, the Magnus effect on a spinning ball, and a fast train pulling people toward the track are all the same reduced-pressure phenomenon.

7. Thermal expansion and calorimetry

$\Delta L=\alpha L\Delta T,\beta =2\alpha ,\gamma =3\alpha$

A hole expands as if it were filled with the same material. Heating makes the hole bigger, not smaller. This is the most frequently mishandled fact in the topic.

Water is anomalous between 0 and 4 °C — it contracts as it warms, with maximum density at 4 °C. That is why ice floats and lakes freeze from the top down, leaving liquid beneath for fish.

Illustration 9

A steel measuring tape is correct at $20°\text{C}$. On a hot day at $40°\text{C}$ it reads a distance as $50.00\text{m}$. What is the true distance? Take $\alpha =1.2\times 10^{-5}°{\text{C}}^{-1}$.

Think about what expanded. The tape's own markings have spread apart, so each interval labelled "1 m" is now slightly longer than a metre:

$\text{true length of one division}=1\left(1+\alpha \Delta T\right)=1+1.2\times 10^{-5}(20)=1.00024\text{m}$

$\text{True distance}=50.00(1.00024)=50.012\text{m}$

The tape under-reads by 12 mm.

The trap is the direction. The instinct is that heat makes things longer so the reading must be too big. But the tape is the ruler, not the object — a stretched ruler fits fewer of its own divisions into a fixed distance, so it reports a smaller number than the truth.

$Q=mc\Delta T\text{(temperature change)},Q=mL\text{(phase change)}$

Water's $c\approx 4186$ J kg⁻¹ K⁻¹ is unusually large, which is why coastal climates are milder than inland ones. $L_{fusion}\approx 3.34\times 10^5$, $L_{vaporisation}\approx 2.26\times 10^6$ J kg⁻¹.

Temperature stays constant during a phase change — the energy breaks bonds rather than raising kinetic energy. Omitting the latent heat term is the standard calorimetry error.

8. Heat transfer

$\frac{dQ}{dt}=\frac{kA(T_1-T_2)}{L}$

$L/kA$ is thermal resistance. Slabs in series add resistances; slabs in parallel add conductances — exactly like electrical circuits.

Illustration 10

Two slabs of equal thickness and equal area, with conductivities $k$ and $2k$, are placed in series between $100°\text{C}$ and $0°\text{C}$. Find the interface temperature.

In the steady state the same heat current passes through both — nothing accumulates in between. That single statement is the whole solution.

$R_1=\frac{L}{kA},R_2=\frac{L}{2kA}=\frac{R_1}{2}$

$\frac{100-T}{R_1}=\frac{T-0}{R_2}\Rightarrow \frac{100-T}{2}=T$

$100-T=2T\Rightarrow T=33.3°\text{C}$

Read it back: the poorer conductor takes the larger temperature drop — two thirds of it here — because it needs a steeper gradient to push the same current through. It is exactly the voltage divider, with temperature for potential and heat current for current.

Convection needs bulk fluid motion, so it cannot happen in solids. Hence heating elements at the bottom of a kettle and cooling coils at the top of a fridge.

Radiation needs no medium:

$P=\sigma Ae{T}^4$

The fourth power makes radiation dominate at high temperature — double the absolute temperature and the power goes up sixteen-fold.

Wien: ${\lambda }_{m}T=2.898\times 10^{-3}$ m K. Hotter bodies radiate shorter — which is why heated iron goes red, then orange, then white.

A good absorber is a good emitter, so $e=a$ at equilibrium; a black body has both equal to 1.

Newton's law of cooling — rate of cooling $\propto$ excess temperature. It is the small-difference approximation to Stefan–Boltzmann, not an independent law.

Illustration 11

How much heat converts 100 g of ice at $-10°\text{C}$ to water at $20°\text{C}$? Take $c_{ice}=2100$, $c_{water}=4186$, $L_f=3.34\times 10^5$ J/kg.

Three stages, and the middle one is the trap:

Total $\approx 43900\text{J}$

The phase change alone is 76% of the total, at constant temperature throughout. Omit it and the answer is out by a factor of four.

Summary

• A solid resists shape change; a liquid does not. Hence $Y$ and $\eta$ for solids only, $K$ for both.
• A modulus is stiffness, not strength. Liquids are incompressible in practice ($K$ huge).
• Stress–strain: proportional limit, elastic limit, ultimate strength, fracture. Area under the curve is stored energy per unit volume.
• $P=P_0+\rho gh$ — depth only, never shape. Hydraulic lifts multiply force, not energy.
• Floating displaces its own weight; submerged displaces its own volume.
• Soap bubble $4T/R$ (two surfaces) versus drop $2T/R$. Small bubbles empty into large ones.
• $h=2T\cos \theta /\rho gr$ — narrower tube, higher rise. A short tube does not overflow.
• $v_t\propto r^2$. Liquid viscosity falls with temperature; gas viscosity rises.
• $R_e=\rho vD/\eta$ decides streamline versus turbulent. Critical speed in a 2 cm water pipe is only 0.1 m/s, so household plumbing is normally turbulent.
• Splitting a drop into $n$ pieces costs $4\pi R^{2}T\left(n^{1/3}-1\right)$ — area goes as $n^{1/3}$, not $n$.
• Bernoulli: faster means lower pressure. Torricelli $v=\sqrt{2gh}$.
• A heated hole gets bigger. Water is anomalous below 4 °C.
• Never omit $mL$ in calorimetry — temperature is constant during a phase change.
• $P=\sigma Ae{T}^4$: double $T$, sixteen times the power. Thermal resistances add in series.
• A heated ruler under-reads: its own divisions have stretched, so fewer of them span a fixed distance.