The shape of the surface of a soap film

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SUMMARY

The discussion focuses on deriving a differential equation for the shape of a soap film connecting two coaxial rings with a radius of R=10 cm, positioned L distance apart. The participants explore the relationship between surface tension and pressure difference, leading to the conclusion that the solution to the differential equation is of the form r(z) = r0 cosh(z/r0). The critical distance L• at which the soap film breaks is determined by applying boundary conditions, specifically that at z = L/2, the radius must equal 10 cm. The analysis involves understanding the force balance on the soap film and the implications of surface tension.

PREREQUISITES
  • Understanding of differential equations and their applications in physics.
  • Familiarity with concepts of surface tension and pressure in fluid mechanics.
  • Knowledge of cylindrical coordinates and their geometric implications.
  • Basic grasp of hyperbolic functions, specifically cosh(x).
NEXT STEPS
  • Study the derivation of the differential equation for soap films in cylindrical coordinates.
  • Research the application of boundary conditions in solving differential equations.
  • Learn about the physical implications of surface tension in soap films and bubbles.
  • Explore graphical methods for solving transcendental equations, particularly in the context of critical points.
USEFUL FOR

Students and professionals in physics, particularly those focusing on fluid mechanics, materials science, and applied mathematics, will benefit from this discussion.

  • #31
I don't understand that.would you please do it for me?
 
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  • #32
no i can't help you do it but
##
\frac {10}{r_0} = cosh(L/2 r_0)\\
\frac{1}{ r_0} = x\\

10 x = cosh (Lx /2)
##
now you can think of it graphically what is necessary for x to have solution
 
  • #33
Hi. I was trying to solve the same problem you submitted. And I've read Chestermiller's solution to the problem. But I don't understand why in the post #13 there is a sqrt(1+(dr/dz)^2)) ? I would appreciate the help. Thank you.
 
  • #34
LordGfcd said:
Hi. I was trying to solve the same problem you submitted. And I've read Chestermiller's solution to the problem. But I don't understand why in the post #13 there is a sqrt(1+(dr/dz)^2)) ? I would appreciate the help. Thank you.
The square root factor is associated with getting the z-component of the surface tension force.
 
  • #35
LordGfcd said:
Hi. I was trying to solve the same problem you submitted. And I've read Chestermiller's solution to the problem. But I don't understand why in the post #13 there is a sqrt(1+(dr/dz)^2)) ? I would appreciate the help. Thank you.
Let's see your derivation of the force balance on the film.
 
  • #36
Ok thanks very much. It was the project of the tension on the z-axis.
My solution had the same idea as Raihan's - pressure equillibrium.
 

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