The shape of the surface of a soap film

  • #1
1. Two coaxial rings of radius R=10 cm are placed to a distance L from each other.There is a soap film connecting the two rings(that looks like a cylinder which have different radii with z coordinate. (The rings lie in xy plane)).Derive a differential equation describing the shape r(z) of the film where r is the radial distance from symmetry axis,as a function of distance z along the axis.
Show that cosh x is one of its solutions.
When the distance between rings is slowly increased ,at a certain critical distance L• ,the soap film breaks.Find L•



2.surface tension T=F/l and
Excess pressure inside a soap bubble is ∆P=4T/r


3. I write the force balance equation for a divided part of soap film.
T.4Πr=∆P.Πr^2
and then
∆p=4T/r but I can't move ahead .Long hints is requested.
Thanks in advance.
 
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  • #3
Delta2
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I suppose the soap film surface . which is kind of the collateral surface of the cylinder with height L and radius R=10cm, can be described (in cylindrical coordinates) by ##r=r(z,\phi)## and I suppose it has symmetry with respect to ##\phi## so its just ##r(z)## as the exercise claims. I also suppose the soap film in the top and bottom surface of the cylinder can be taken to be flat or I don't know if we want for soap film to exist there. Still I cant be much of help for this problem never studied the laws of surface tension...
 
  • #4
Draw a catenoid type line (a string hanging loosely between two fixed points) connecting two points of the rings .it is the surface.
 
  • #5
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So the axis of the rings is vertical and they share a common axis? You have a pressure difference in your equations. What is forcing this pressure difference? Is gravity a factor? Is the thickness of the film assumed uniform, or does it vary with z?
 
  • #6
They share a common axis .but as mass is not included i dont think gravity is a factor.and the question only says that r=r(z). To Illustrate the fig. I told you that the rings is in xy plane.
 

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  • #7
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OK. I’ll help you analyze this a little later when i’m back at my computer.
 
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  • #8
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OK. We're going to be deriving a differential force balance on a small window shaped panel of film situated between ##\theta## and ##\theta + \Delta \theta## in the circumferential direction and between z and ##z+\Delta z## in the axial direction. This differential force balance will involve the unit vectors in the r, ##\theta##, and z directions and the derivatives of r with respect to z. For the window shaped panel of film that I have identified, what are

1. The lengths of the edges of the panel
2. The unit vectors within the plane of the film normal to the edges of the panel
 
  • #9
Can you show your work.
 
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  • #10
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Can you sbow your work.
No. I'm expecting a response from you on my questions. If you don't understand what I'm asking, I'll be glad to help. But I need to see some effort on your part. That's how we do things on Physics Forums.

Chet
 
  • #11
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There is a simpler way of doing this than the method I originally had in mind. Suppose we mathematically cut the film by hypothetical planes at z = 0 (the axial location of minimum radius ##r_0##, half way between the two rings) and the other at arbitrary z. The axial tensile force on the film at z = 0 will be in the negative z direction, and of magnitude ##2T(2\pi r_0)##, where T is the surface tension. What is the magnitude and direction of the axial tensile force on the film at the plane of arbitrary z, in terms of T, r, and the slope dr/dz?
 
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  • #12
T(z)=T(2πr)/[(1+r'²)½]; In the negative z direction.
Is it?
 
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  • #13
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T(z)=T(2πr)/[(1+r'²)½]; In the negative z direction.
Is it?
Very close. There would be a factor of 2 out front to account for the surface tension at both the inner and outer interfaces of the film with the air; and, for the portion of the film between 0 and z, the tension exerted by the film at ##z^+## on the film at ##z^-## would be in the positive z direction.

So the overall force balance on the portion of the film between 0 and z would be:
$$\frac{2T(2\pi r)}{\sqrt{1+(\frac{dr}{dz})^2}}-2T(2\pi r_0)=0$$or$$\sqrt{1+\left(\frac{dr}{dz}\right)^2}=\frac{r}{r_0}$$
 
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  • #14
I got it.But it doesn't yield a cosh x like solution.is it?
 
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  • #15
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I got it.But it doesn't yield a cosh x like solution.is it?
Have you tried to solve the differential equation for r as a function of z. Does ##r=r_0 \cosh(z/r_0)## satisfy the differential equation?
 
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  • #16
Now what to do?
 

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  • #18
No. cosh x is not the solution of avove equation.
 
  • #21
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See if the solution I posted in post #15 satisfies the equation.
 
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  • #22
OK.then how we will find L• ?
 
  • #23
I think this is correct.Now we have to find L•
 

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  • #24
i just saw this thread
to find l you have to apply boundary conditions
 
  • #25
But we have only one boundary value
 

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