Problem 5: Elongation of Elastic Thread in Soap Film
1. Analysis of Forces
When the film is punctured on one side, the surface tension $\sigma$ from the remaining film pulls the thread. The thread assumes the shape of a circular arc of radius $R$. Since the tension $T$ is very high compared to the surface force, the curvature is small ($R \gg r$).
Consider the equilibrium of the thread. The surface tension force acts perpendicular to the thread along its entire length. For a soap film, there are two surfaces.
Balancing forces on the arc:
$$ T = (2\sigma) R $$Where $2\sigma$ is the force per unit length due to the double-sided film.
2. Geometry and Approximation
Let the angle subtended by the arc at its center be $2\theta$. From geometry:
$$ r = R \sin\theta $$Since $R \gg r$, $\theta$ is small. We use the expansion $\sin\theta \approx \theta – \frac{\theta^3}{6}$.
$$ r \approx R \left( \theta – \frac{\theta^3}{6} \right) \implies \frac{r}{R} \approx \theta – \frac{\theta^3}{6} $$Also, to a first approximation $\theta \approx \frac{r}{R}$.
3. Calculating Elongation ($\Delta l$)
The length of the arc $l_{arc} = 2 R \theta$. The original length was $2r$.
$$ \Delta l = l_{arc} – 2r = 2R\theta – 2r $$Substitute $r = R(\theta – \theta^3/6)$:
$$ \Delta l = 2R\theta – 2R(\theta – \frac{\theta^3}{6}) = 2R \frac{\theta^3}{6} = \frac{R \theta^3}{3} $$Substituting $\theta \approx \frac{r}{R}$:
$$ \Delta l = \frac{R}{3} \left( \frac{r}{R} \right)^3 = \frac{r^3}{3R^2} $$4. Elasticity Relation
From Young’s Modulus formula $Y = \frac{\text{Stress}}{\text{Strain}}$:
$$ Y = \frac{T/S}{\Delta l / (2r)} \implies T = \frac{YS \Delta l}{2r} $$Now we equate the two expressions for Tension $T$:
$$ 2\sigma R = \frac{YS \Delta l}{2r} \implies R = \frac{YS \Delta l}{4\sigma r} $$5. Final Substitution
Substitute this expression for $R$ back into the elongation equation $\Delta l = \frac{r^3}{3R^2}$:
$$ \Delta l = \frac{r^3}{3} \left( \frac{4\sigma r}{YS \Delta l} \right)^2 $$ $$ \Delta l = \frac{r^3}{3} \frac{16 \sigma^2 r^2}{Y^2 S^2 (\Delta l)^2} $$ $$ (\Delta l)^3 = \frac{16 \sigma^2 r^5}{3 Y^2 S^2} $$