Problem 7: Surface Tension of a Soap Loop
Analysis:
The thread of length $l$ is pulled into a shape consisting of two semicircles by the applied force $F$ and the surface tension of the soap film.
1. Geometry: The thread forms the boundary of the empty space. Since it forms two semicircles, the total perimeter is equal to the circumference of one full circle. $$ 2 \pi r = l \implies r = \frac{l}{2\pi} $$
2. Force Balance: The force $F$ is applied to open the loop against surface tension. The surface tension acts perpendicular to the thread along the diameter where the film was ruptured/pulled.
The total retarding force due to surface tension is: $$ F_{tension} = 2 \times \sigma \times L_{contact} $$ The factor of 2 accounts for the two surfaces of the soap film (front and back). The length of contact $L_{contact}$ corresponds to the total diameter of the shape created (sum of the diameters of the two semicircles). $$ L_{contact} = 2r + 2r = 4r $$
Substituting $r$: $$ F_{tension} = 2\sigma(4r) = 8\sigma r = 8\sigma \left( \frac{l}{2\pi} \right) = \frac{4\sigma l}{\pi} $$
3. Equilibrium: For the system to be in equilibrium, the applied force must balance the tension force: $$ F = \frac{4\sigma l}{\pi} $$
Solving for surface tension $\sigma$: $$ \sigma = \frac{F \pi}{4l} $$
Answer: The surface tension is $\sigma = \frac{F \pi}{4l}$