Problem on Pressure due to Surface tension

[phantomvommand]

Homework Statement

Please see the attached image.

Relevant Equations

$\gamma = \frac F L$
$P = \frac F A$

The method to solving this is to equate forces along a portion of the balloon through which $\sigma_L$ acts, and another portion through which $\sigma_t$ acts. The former potion should be a circular cross section of the cylinder, while the latter will be a rectangular cross section. You will thus get the following:

I did exactly the above, except that instead of having $2\pi r \sigma_L$ and $2x\sigma_t$ on the RHS, I had $4\pi r \sigma_L$ and $4x\sigma_t$. Am I right on this? Because I think that in either case, there are 2 surfaces (inner surface of balloon and outer surface of balloon), resulting in double the force exerted by surface tension.

 

[etotheipi]

In the problem they've defined $\sigma_L$ and $\sigma_t$ as forces per unit length of the boundary between two portions in the longitudinal and hoop directions respectively. With that definition there's no factor of 2. Although you're right to be a little skeptical, because there are indeed two surfaces and usually the longitudinal surface tension $\gamma_L$ for instance would be defined such that e.g. $\pi r^2 P' = 4\pi r \gamma_L$, i.e. with the factor of two included.

The reason for the ambiguity is that it's not really a surface tension problem. It's really instead internal stresses in the rubber holding the thing together (although there is a strong analogy). For actual surface tension problems, $\gamma$ is defined thermodynamically as $\gamma = \partial E / \partial A$ where $E$ is the interface energy between two phases and $A$ is the total area of the interface between those two phases, and it's important to account for all of the interfaces.

Here it doesn't really matter how you define $\sigma_L$ and $\sigma_t$ just so long as you're consistent because at the end you're taking a ratio. So I wouldn't worry about it too much!