The shape of the surface of a soap film

  • #26
The boundary condition is that at z = L/2 the radius must be 10cm with the centre of the soapbfilm having some ro after certain l there is no solution for ro and hence no film
 
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  • #27
Can we find the highest L from that equation only?
 

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  • #28
Yes but the form you behave written in seems a bit complicated let 1/r be X if 1/r has to have solution then X have to have solution
 
  • #31
I don't understand that.would you please do it for me?
 
  • #32
no i cant 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
11
1
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
TSny
Homework Helper
Gold Member
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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
21,394
4,807
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
11
1
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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