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A light straight wire AB of length 10 cm is free to slide along a vertical frame. A soap film is held between the frame and the wire. What load W must be hung from the wire so that it stays in equilibrium? Ignore friction. Surface tension of the soap solution is 25 dyne/cm and g = 10 m/s².
- 5 * 10⁻³ N
- 2.5 * 10⁻³ N
- 5 * 10⁻⁴ N
- 10 * 10⁻³ N
Correct answer: 5 * 10⁻³ N
Solution
The soap film pulls the wire up with force from both of its surfaces: F = 2*T*L. Convert units: T = 25 dyne/cm = 25*10⁻³ N / 10⁻² m = 2.5*10⁻² N/m? Better keep CGS: T = 25 dyne/cm, L = 10 cm, so F = 2*25*10 = 500 dyne = 500*10⁻⁵ N = 5*10⁻³ N. For equilibrium W = F = 5*10⁻³ N.
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