Why a Small Soap Bubble Pushes Harder Than a Big One

Honey pours in a slow, thick stream because its layers drag on one another — that internal friction is viscosity. A drop of water on a leaf pulls itself into a ball because its surface behaves like a stretched skin — that is surface tension.

Both decide how raindrops fall, how bubbles form and how a lamp wick draws up kerosene.

This part covers viscosity, Stokes' law and terminal velocity, surface tension and angle of contact, and excess pressure and capillary rise. Take $g=9.8$ m/s${}^2$.

**When fluid layers slide over one another, the viscous force between them is $F=\eta A\frac{dv}{dx}$, where $\eta$ is the coefficient of viscosity, $A$ the area of contact and $\frac{dv}{dx}$ the velocity gradient.**

The SI unit of $\eta$ is the pascal second (Pa s); $1$ poise $=0.1$ Pa s.

Worked example 1. Suppose a $0.1$ m${}^2$ plate slides at $0.05$ m/s over a $1$ mm layer of oil with $\eta =0.8$ Pa s.

$$F=0.8\times 0.1\times \frac{0.05}{0.001}=4\text{ N}$$

**Worked example 2 — finding $\eta$.** A $2$ N force keeps a $0.05$ m${}^2$ plate moving at $0.1$ m/s over a $2$ mm film:

$$\eta =\frac{Fdx}{Adv}=\frac{2\times 0.002}{0.05\times 0.1}=0.8\text{ Pa s}$$

An everyday example. Pouring honey versus pouring water from a spoon shows the difference in viscosity at a glance.

The substance. Viscosity of liquids falls as they warm, but viscosity of gases rises — which is why warm ghee pours easily.

**Stokes' law gives the viscous drag on a small sphere of radius $r$ moving slowly at speed $v$ as $F=6\pi \eta rv$; the sphere reaches terminal velocity when drag plus buoyancy balance its weight, giving $v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$.**

Here $\rho$ is the sphere's density and $\sigma$ the fluid's density.

Derivation. At terminal velocity the net force is zero:

$$\frac{4}{3}\pi r^3\rho g=\frac{4}{3}\pi r^3\sigma g+6\pi \eta r{v}_t\Rightarrow v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }$$

Worked example 1 — a raindrop. Radius $0.1$ mm in air ($\eta =1.8\times 10^{-5}$ Pa s, air density negligible):

$$v_t=\frac{2(10^{-4}{)}^2(1000)(9.8)}{9(1.8\times 10^{-5})}\approx 1.2\text{ m/s}$$

Worked example 2 — a steel ball in oil. Suppose $r=1$ mm, $\rho =7800$ kg/m${}^3$, $\sigma =900$ kg/m${}^3$, $\eta =0.9$ Pa s:

$$v_t=\frac{2(10^{-3}{)}^2(6900)(9.8)}{9\times 0.9}\approx 0.017\text{ m/s}$$

An everyday example. Tiny mist droplets drift in the air for a long time while large raindrops fall fast, because $v_t\propto r^2$.

The substance. **If the fluid is denser than the sphere, $v_t$ is negative** — the body rises, as an air bubble does in water.

**Surface tension $S$ is the force per unit length acting along a liquid surface, and it also equals the extra energy per unit area of surface, so enlarging a surface by $\Delta A$ needs work $W=S\Delta A$; the angle of contact shows whether a liquid wets a solid.

Units.** N/m for surface tension, J/m${}^2$ for surface energy — the same thing.

**Angle of contact $\theta$, measured inside the liquid:

- Acute** ($\theta <90°$): the liquid wets the surface, as water on clean glass
- Obtuse ($\theta >90°$): the liquid does not wet it, as mercury on glass or water on a waxy leaf

Worked example. A soap film ($S=0.03$ N/m) fills a frame with a $5$ cm sliding wire. The film has two surfaces:

$$F=2SL=2\times 0.03\times 0.05=3\times 10^{-3}\text{ N}$$

Pulling the wire $2$ cm adds $2\times 0.05\times 0.02=2\times 10^{-3}$ m${}^2$ of surface:

$$W=S\Delta A=0.03\times 2\times 10^{-3}=6\times 10^{-5}\text{ J}$$

An everyday example. Insects called pond skaters walk on still water without sinking, supported by the surface film.

The substance. Detergents lower surface tension and the angle of contact, so soapy water soaks into cloth and lifts out dirt.

**A curved liquid surface has higher pressure on its concave side: $\Delta P=\frac{2S}{r}$ for a drop or an air bubble in a liquid, $\frac{4S}{r}$ for a soap bubble with two surfaces; and a liquid rises in a capillary tube to $h=\frac{2S\cos \theta }{\rho gr}$.

Worked example 1 — a water drop.** $r=1$ mm, $S=0.072$ N/m:

$$\Delta P=\frac{2\times 0.072}{10^{-3}}=144\text{ Pa}$$

Worked example 2 — a soap bubble. $r=1$ cm, $S=0.03$ N/m:

$$\Delta P=\frac{4\times 0.03}{0.01}=12\text{ Pa}$$

Worked example 3 — capillary rise. Water ($\theta \approx 0$) in a glass tube of radius $0.5$ mm:

$$h=\frac{2\times 0.072\times 1}{1000\times 9.8\times 5\times 10^{-4}}\approx 0.029\text{ m}=2.9\text{ cm}$$

Mercury, with an obtuse angle, is pushed down instead.

Worked example 4 — merging drops. $1000$ drops of radius $0.1$ mm join into one drop of radius $1$ mm. The surface area falls, releasing

$$W=S\cdot 4\pi (1000\times 10^{-8}-10^{-6})=0.072\times 4\pi \times 9\times 10^{-6}\approx 8.1\times 10^{-6}\text{ J}$$

An everyday example. Kerosene climbing up the cotton wick of a lantern is capillary rise through tiny gaps between fibres.

The substance. **A capillary tube shorter than $h$ never overflows** — the liquid reaches the top and its surface simply becomes flatter.