Problem on Pressure due to Surface tension

  • #1
phantomvommand
238
37
Homework Statement:
Please see the attached image.
Relevant Equations:
##\gamma = \frac F L##
##P = \frac F A##
Screenshot 2021-05-11 at 11.26.21 PM.png


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:

Screenshot 2021-05-11 at 11.30.54 PM.png

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.
 

Answers and Replies

  • #2
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! :smile:
 
  • Like
Likes phantomvommand

Suggested for: Problem on Pressure due to Surface tension

  • Last Post
Replies
13
Views
757
Replies
7
Views
384
  • Last Post
Replies
2
Views
401
  • Last Post
Replies
3
Views
335
  • Last Post
Replies
11
Views
1K
Replies
2
Views
214
Replies
2
Views
307
Top