Question

A rectagular wire frame with one movable side is convered by a soap film (fig.). What force should be applied to the movable side to counterbalance it ? What work will be done if this side of the frame is moved a distance `S = 2 cm` ? The length of the movable side is `l = 6cm`. The surface tension of the soap film is `alpha = 40 "dyne"//cm`.

Answer 1

Correct Answer - `F = 480 dyn, w = 960 erg`
`P_(excess) = (4T)/(R) = (4 xx 2.5 xx 10^(-2))/(5 xx 10^(-3)) = 20 N//m^(2)`
`p = p_(atm) + hrhog + (2T)/(r) = 1.01 xx 10^(5) + .40 xx 1.2 xx 10^(3) xx 9.8 + (2 xx 2.5 xx 10^(-2))/(5 xx 10^(-3)) = 1.05714 xx 10^(5) N//m^(2)`