Properties of Solids and Liquids

A sealed tank is completely filled with water and carries one small air bubble. The tank sits on a trolley that accelerates forward at $a$. Which way does the bubble drift?

A pendulum bob hanging inside the same trolley swings backwards. The bubble does the opposite — it drifts forwards, and the harder the trolley accelerates the more sharply it goes.

Buoyancy has no independent existence. It is the resultant of pressure forces, and pressure is set by gravity. Work in the trolley's frame and add the pseudo-force $-m\vec{a}$ to every element of fluid. Gravity and pseudo-force merge into a single vector,

${\vec{g}}_{eff}=\vec{g}-\vec{a}$

which points down and backwards. Every hydrostatic statement now refers to this vector instead of to $\vec{g}$. The free surface sets itself perpendicular to ${\vec{g}}_{eff}$, pressure grows along it, and buoyancy acts opposite to it. Down-and-backwards, reversed, is up-and-forwards.

The bubble is lighter than the fluid it displaces, so it follows buoyancy and moves forward. The surface tilts through

$\tan \theta =\frac{a}{g}$

and a submerged cork tethered to the floor leans forward through the same angle.

That is the pattern of the whole chapter. Replace $g$ by ${\vec{g}}_{eff}$, replace forces by integrals, and replace the remembered formula by the energy or momentum statement it came from. Main asks you to substitute into $P=\rho gh$, $F=6\pi \eta rv$ and $2T/R$. Advanced asks what happens when the depth, the radius or the frame refuses to stay constant.

1. Fluids in accelerated and rotating frames

For a fluid at rest in an accelerating frame, the pressure gradient is set by ${\vec{g}}_{eff}$ component by component:

$\frac{\partial P}{\partial x}=-\rho a,\frac{\partial P}{\partial y}=-\rho g$

taking $x$ along the acceleration and $y$ upward. Integrating between two points $1$ and $2$,

$P_2-P_1=-\rho a\Delta x-\rho g\Delta y$

Two consequences do most of the work. Any surface of constant pressure — the free surface included — is perpendicular to ${\vec{g}}_{eff}$. And in a completely filled sealed tank there is no free surface to tilt, so the tilt shows up entirely as a pressure difference across the tank instead.

Illustration 1

A closed tank of length $L=2$ m and height $H=1$ m is completely filled with water and accelerated horizontally at $a=4$ m s${}^{-2}$. Find the difference in pressure between the bottom rear corner and the top front corner.

Going from the top front corner to the bottom rear corner, $\Delta x=-L$ and $\Delta y=-H$:

$P_{rear,bot}-P_{front,top}=\rho aL+\rho gH$

$=1000\left(4\times 2+10\times 1\right)=1000\times 18=1.8\times 10^4$ Pa

Read the structure: the $\rho aL$ term is exactly what a column of height $aL/g=0.8$ m of water would produce. The acceleration behaves like an extra $0.8$ m of depth measured backwards along the tank.

A rotating vessel is the same idea with a radial pseudo-force. In the frame of a vessel spinning at $\omega$, an element at radius $r$ feels $\rho {\omega }^{2}r$ outward, so

$\frac{dP}{dr}=\rho {\omega }^{2}r\Rightarrow P=P_0+\frac{1}{2}\rho {\omega }^2{r}^2-\rho gz$

Setting $P$ constant on the free surface gives

$z=\frac{{\omega }^2{r}^2}{2g}$

a paraboloid. Because a paraboloid encloses exactly half the volume of its bounding cylinder, the liquid at the rim rises by ${\omega }^2{R}^2/4g$ while the centre falls by the same amount.

Illustration 2

A cylindrical vessel of radius $R=0.1$ m is filled with water to a depth $h=0.2$ m and spun about its axis. At what $\omega$ does the water just expose the centre of the base?

The centre must fall by $h$, and the fall of the centre is ${\omega }^2{R}^2/4g$:

$\frac{{\omega }^2{R}^2}{4g}=h\Rightarrow \omega =\frac{2}{R}\sqrt{gh}$

$\omega =\frac{2}{0.1}\sqrt{10\times 0.2}=20\times 1.414=28.3$ rad s${}^{-1}$

The centrifuge follows from the same equation. Pressure now rises outward, so buoyancy points inward. Anything denser than the fluid is driven to the rim and anything lighter collects on the axis — which is why a bubble in a spinning tube runs to the centre.

2. Hydrostatic thrust and where it acts

Pressure on a submerged wall grows linearly with depth, so the total force needs an integral and the point of application is not the centre of the wall. For a vertical rectangular wall of width $b$ holding water of depth $h$,

$F={\int }_0^h\rho gy(bdy)=\frac{1}{2}\rho gb{h}^2$

which equals the average pressure $\rho gh/2$ times the area. The torque about the base, however, needs the distribution:

$\tau ={\int }_0^h\rho gy(bdy)(h-y)=\frac{1}{6}\rho gb{h}^3$

Dividing torque by force puts the resultant at height $h/3$ above the base — that is, at depth $2h/3$. This is the centre of pressure, and it is what decides whether a dam topples.

Curved surfaces never need a surface integral in an exam. Take the horizontal component as the thrust on the vertical projection of the surface, and the vertical component as the weight of the fluid column standing above it — real fluid or imagined, depending on which side the fluid lies.

Illustration 3

A dam of width $b$ holds water to depth $h$. Its own weight per unit width is $W$, acting through a point $d$ from the downstream toe. Find the condition for it not to topple about the toe.

Restoring torque about the toe is $Wd$ per unit width; overturning torque is $\left(\frac{1}{2}\rho g{h}^2\right)\left(\frac{h}{3}\right)=\frac{1}{6}\rho g{h}^3$ per unit width.

$Wd\ge \frac{1}{6}\rho g{h}^3$

The cube is the point. Doubling the water depth multiplies the thrust by four but the overturning torque by eight, which is why dams thicken so sharply towards the base rather than uniformly.

Illustration 4

A hemispherical bulge of radius $R$ projects outward from the vertical wall of a tank, its centre at depth $H$ below the surface. Find the horizontal force on it.

The vertical projection of the hemisphere is a circle of radius $R$ centred at depth $H$. By symmetry the pressure above the centre is deficient by exactly what the pressure below is in excess, so the average pressure over the projection is $\rho gH$:

$F_H=\rho gH\left(\pi R^2\right)$

The bulge's own shape is irrelevant. Any surface with the same vertical projection and the same centroid depth takes the same horizontal thrust — a cone, a hemisphere or a flat plate alike.

3. Floating bodies, and how they oscillate

A floating body displaces its own weight, so the submerged fraction is ${\rho }_b/{\rho }_l$. Two Advanced extensions matter.

In an accelerated frame the fraction does not change. Both the weight and the buoyant force are proportional to $g_{eff}$, so it cancels out of the equilibrium condition. A block floating in a beaker inside a lift floats exactly as deep whether the lift accelerates up, down or not at all.

Push a floating body down and it oscillates. Depress a uniform cylinder of base area $A$ by $x$ and the extra buoyancy is ${\rho }_{l}gAx$, always restoring. With mass $m={\rho }_{b}Ah$,

${\omega }^2=\frac{{\rho }_{l}gA}{{\rho }_{b}Ah}=\frac{g}{h_{sub}}\Rightarrow T=2\pi \sqrt{\frac{h_{sub}}{g}}$

using ${\rho }_{b}h={\rho }_l{h}_{sub}$. The period depends only on the depth to which it floats — a simple pendulum of that length.

Illustration 5

A wooden cylinder of density $600$ kg m${}^{-3}$ and length $0.3$ m floats upright in water. Find its period of small vertical oscillation.

$h_{sub}=\frac{600}{1000}\times 0.3=0.18$ m

$T=2\pi \sqrt{\frac{0.18}{10}}=2\pi \left(0.134\right)=0.84$ s

Neither the radius nor the mass appears. Both the restoring force and the inertia scale with the cross-section, so the area cancels — the same reason a sphere and a plank of the same draught bob at the same rate.

Illustration 6

A beaker of water stands on a balance. A steel sphere of volume $V$, hanging from a spring balance, is lowered until fully submerged without touching the bottom. What happens to each reading?

The spring balance falls by the buoyant force ${\rho }_{l}Vg$. The beaker's balance rises by exactly the same amount, because the sphere pushes down on the water with an equal and opposite reaction.

The system is the check. Nothing has been added to or removed from the beaker-plus-sphere-plus-spring system, so the two changes must cancel — a one-line way to catch a sign error.

4. Elasticity where the stress is not uniform

The elastic energy stored per unit volume is

$u=\frac{1}{2}(\text{stress})(\text{strain})=\frac{{\sigma }^2}{2Y}$

which is the elastic analogue of $\frac{1}{2}k{x}^2$ — and indeed a wire of length $L$ and area $A$ is a spring of stiffness $k=YA/L$. Wires joined end to end share the load and add extensions, like springs in series; wires side by side share the extension and add loads, like springs in parallel.

Advanced problems make the stress vary along the body, which turns the extension into an integral.

A rod hanging under its own weight. At a distance $x$ from the bottom the tension is $\rho Agx$, so

$\Delta L={\int }_0^L\frac{\rho Agx}{AY}dx=\frac{\rho g{L}^2}{2Y}=\frac{MgL}{2AY}$

exactly half the extension the same total weight would cause if hung at the end.

A rod spun about one end in a horizontal plane. The tension at radius $r$ must supply the centripetal force for everything beyond it, giving $T(r)=\frac{1}{2}\rho A{\omega }^2\left(L^2-r^2\right)$ and

$\Delta L=\frac{\rho {\omega }^2{L}^3}{3Y}$

A clamped rod that is heated cannot expand, so the thermal strain $\alpha \Delta \theta$ is cancelled by an equal elastic strain:

$\sigma =Y\alpha \Delta \theta ,F=YA\alpha \Delta \theta$

Notice that the length has vanished — a long rail and a short one develop the same thermal stress.

Illustration 7

A steel wire of length $2$ m and area $1$ mm${}^2$ is clamped at both ends at $40^{\circ }$C and cooled to $20^{\circ }$C. Find the tension. Take $Y=2\times 10^{11}$ Pa, $\alpha =1.2\times 10^{-5}$ K${}^{-1}$.

$F=YA\alpha \Delta \theta =\left(2\times 10^{11}\right)\left(10^{-6}\right)\left(1.2\times 10^{-5}\right)(20)$

$F=48$ N

This is why rails are laid with gaps and bridges sit on rollers. A $48$ N tension in a $1$ mm${}^2$ wire is a stress of $48$ MPa for a mere $20$ K, and the result is independent of how long the span is.

Illustration 8

Two wires of the same material and length, of areas $A$ and $2A$, are joined end to end and a load $F$ is hung from the free end. Compare the extensions and the energies stored.

The same $F$ passes through both, so $\Delta L\propto 1/A$: the thin wire extends twice as much.

Energy $=\frac{1}{2}F\Delta L$, so the thin wire also stores twice the energy.

Check by energy density: $u={\sigma }^2/2Y$ is four times larger in the thin wire, but its volume is half — giving the factor of two, as it must.

5. Surface tension read as energy

Surface tension is more usefully a surface energy: creating area $\Delta A$ costs $T\Delta A$. Blowing a soap bubble of radius $R$ costs $T\left(8\pi R^2\right)$, because a film has two surfaces.

The excess pressure follows from that alone. Expand a drop by $dR$; the work done by the excess pressure equals the energy stored in the new area:

$P_{ex}\left(4\pi R^{2}dR\right)=Td\left(4\pi R^2\right)=8\pi RTdR\Rightarrow P_{ex}=\frac{2T}{R}$

and $4T/R$ for a soap bubble with its two surfaces. Smaller means higher pressure, which drives several standard results.

Coalescence. If $n$ drops of radius $r$ merge into one of radius $R=n^{1/3}r$, volume is conserved but area is not. The energy released is

$\Delta E=4\pi T\left(n{r}^2-R^2\right)=4\pi R^{2}T\left(n^{1/3}-1\right)$

which appears as a temperature rise of the liquid.

Two bubbles connected by a tube. The smaller one has the larger excess pressure, so it empties into the larger — the opposite of most students' first guess. The film across the junction bulges into the bigger bubble with radius $r=\frac{r_1{r}_2}{r_2-r_1}$.

A capillary tube shorter than the rise. The liquid does not spurt out of the top. Instead the meniscus flattens until the pressure balance is satisfied at the available height. Since $h\propto 1/R$, the new radius of curvature satisfies

$R^{\prime }=\frac{hR}{L}$

where $L<h$ is the tube's length. The contact angle adjusts; the liquid stays put.

Illustration 9

Eight mercury drops of radius $1$ mm each coalesce into a single drop. Find the energy released, taking $T=0.465$ N m${}^{-1}$.

$R=8^{1/3}\times 1=2$ mm

$\Delta E=4\pi T\left(n{r}^2-R^2\right)=4\pi (0.465)\left(8\times 10^{-6}-4\times 10^{-6}\right)$

$\Delta E=4\pi (0.465)\left(4\times 10^{-6}\right)=2.34\times 10^{-5}$ J

Where it goes: into internal energy, warming the mercury. The surface area has halved while the volume stayed fixed, which is the general reason droplets always merge spontaneously and never split without help.

Illustration 10

Water rises to $6$ cm in a capillary tube. The tube is cut down to $4$ cm and held vertically with its lower end in the water. Does water flow out of the top?

No. The rise height is $h=2T\cos \theta /\rho gr$, so $hR$ is fixed for the given liquid. With the column limited to $4$ cm:

$R^{\prime }=\frac{hR}{L}=\frac{6}{4}R=1.5R$

Physically: the meniscus simply becomes less curved, which reduces the pressure drop across it to exactly what a $4$ cm column needs. A capillary tube is never a perpetual fountain.

6. Bernoulli's theorem, used properly

Bernoulli's relation is the work-energy theorem per unit volume along a streamline, valid for steady, incompressible, non-viscous flow:

$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$

Applied to a hole of area $a$ at depth $h$ in a tank of area $A\gg a$, it gives Torricelli's result $v=\sqrt{2gh}$. Advanced questions then push past that in three directions.

The range on the ground. A hole at depth $h$ in a tank filled to height $H$ launches a horizontal jet that falls $H-h$:

$x=v\sqrt{\frac{2(H-h)}{g}}=2\sqrt{h(H-h)}$

which is maximum at $h=H/2$, where $x_{max}=H$. Because the expression is symmetric under $h\leftrightarrow H-h$, two holes equidistant from the top and the bottom throw water the same distance.

Time to empty. Now the level falls, so $v$ is not constant and the problem becomes a separable differential equation:

$A\frac{dH}{dt}=-a\sqrt{2gH}\Rightarrow t=\frac{A}{a}\sqrt{\frac{2}{g}}\left(\sqrt{H_1}-\sqrt{H_2}\right)$

The reaction on the tank. Water leaves at rate $\rho av$ carrying momentum, so the thrust on the tank is

$F=\rho a{v}^2=2\rho gha$

which is twice the naive $\rho gh\times a$ that "pressure times area" suggests, and it points backwards.

Illustration 11

A tank of cross-section $1$ m${}^2$ filled to $H=1.8$ m drains through a hole of area $5$ cm${}^2$ at its base. Find the time to empty and the initial thrust on the tank.

$t=\frac{1}{5\times 10^{-4}}\sqrt{\frac{2}{10}}\left(\sqrt{1.8}-0\right)=2000\times 0.447\times 1.342=1200$ s

$F=2\rho gha=2(1000)(10)(1.8)\left(5\times 10^{-4}\right)=18$ N

Why the factor of two survives: the fluid does not merely feel the static pressure, it is accelerated from rest to $\sqrt{2gh}$ on its way out, and the extra momentum flux is what doubles the reaction.

Illustration 12

A Venturi meter of throat area $a$ and pipe area $A$ carries a liquid, and the two vertical tubes show a level difference $\Delta h$. Find the volume flow rate.

Continuity gives $v_{1}A=v_{2}a$; Bernoulli gives $P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)=\rho g\Delta h$.

$Q=aA\sqrt{\frac{2g\Delta h}{A^2-a^2}}$

Read the limit: for $a\ll A$ this reduces to $a\sqrt{2g\Delta h}$, which is Torricelli again — the throat behaves like a hole draining a column of height $\Delta h$.

7. Viscous flow and its resistance

For steady laminar flow through a pipe, Poiseuille's law gives

$Q=\frac{\pi \Delta P{r}^4}{8\eta L},R_{fl}=\frac{8\eta L}{\pi r^4}$

The fourth power is the whole story: halving the radius cuts the flow to a sixteenth. Written as $\Delta P=Q{R}_{fl}$ this is Ohm's law, and pipe networks combine exactly like resistors — in series the resistances add, in parallel the reciprocals do.

A sphere falling through a viscous fluid reaches

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

and if $\sigma >\rho$ the sign flips, which is a bubble rising. The approach to that speed is exponential: with $m\dot{v}=m^{\prime }g-kv$,

$v=v_T\left(1-e^{-t/\tau }\right),\tau =\frac{m}{k}=\frac{v_T}{g_{eff}}$

Flow stays laminar only while the Reynolds number $Re=\rho vD/\eta$ stays below about $2000$; beyond roughly $3000$ it is turbulent and none of these results apply.

Illustration 13

Two capillaries of the same length, radii $r$ and $2r$, are connected in parallel across the same pressure difference. What fraction of the total flow passes through the wider one?

$Q\propto r^4$, so the flows are in the ratio $1:16$.

The wider tube carries $\frac{16}{17}=94\%$ of the total.

The design consequence is severe. Adding a second narrow tube alongside a wide one barely helps, which is why arterial narrowing is so dangerous — a $20\%$ loss of radius costs about $60\%$ of the flow.

Summary

• In any accelerated frame, replace $\vec{g}$ by ${\vec{g}}_{eff}=\vec{g}-\vec{a}$. The free surface sets itself perpendicular to it and buoyancy acts opposite to it.
• A bubble in an accelerating liquid drifts forwards, opposite to a pendulum bob, because buoyancy tilts with ${\vec{g}}_{eff}$.
• Surface tilt $\tan \theta =a/g$; in a sealed full tank the tilt appears as a pressure difference $\rho aL$ instead.
• Rotating vessel: $P=P_0+\frac{1}{2}\rho {\omega }^2{r}^2-\rho gz$, free surface $z={\omega }^2{r}^2/2g$, rim up and centre down by ${\omega }^2{R}^2/4g$ each.
• Thrust on a vertical wall is $\frac{1}{2}\rho gb{h}^2$ but acts at depth $2h/3$; overturning torque goes as $h^3$.
• Curved surfaces: horizontal component from the vertical projection, vertical component from the fluid column above.
• Submerged fraction is unchanged by any frame acceleration, since $g_{eff}$ cancels from both weight and buoyancy.
• A floating body bobs with $T=2\pi \sqrt{h_{sub}/g}$ — area and mass both cancel.
• Elastic energy density ${\sigma }^2/2Y$; a wire is a spring of stiffness $YA/L$, so wires combine in series and parallel like springs.
• Rod under its own weight: $\Delta L=\rho g{L}^2/2Y$, half the end-loaded value. Clamped and heated: $\sigma =Y\alpha \Delta \theta$, independent of length.
• Excess pressure follows from surface energy: $2T/R$ for a drop, $4T/R$ for a bubble. Coalescing $n$ drops releases $4\pi R^{2}T\left(n^{1/3}-1\right)$.
• The smaller bubble empties into the larger; a capillary shorter than the rise flattens its meniscus to $R^{\prime }=hR/L$ and never overflows.
• Efflux range $x=2\sqrt{h(H-h)}$ is maximum at mid-depth and equal for holes symmetric about it; emptying takes $\frac{A}{a}\sqrt{2/g}\left(\sqrt{H_1}-\sqrt{H_2}\right)$; the reaction is $2\rho gha$.
• Poiseuille resistance $8\eta L/\pi r^4$ combines like electrical resistance, and the $r^4$ makes narrow branches almost irrelevant.