# The shape of the surface of a soap film

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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Can we find the highest L from that equation only?

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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

What

I don't understand that.would you please do it for me?

##
\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

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.

TSny
Homework Helper
Gold Member
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.

Chestermiller
Mentor
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.

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.