# Surface tension - thread in soap solution

1. May 29, 2013

### Saitama

1. The problem statement, all variables and given/known data

(Ans: $\pi F/(2L)$)

2. Relevant equations

3. The attempt at a solution
I am not too much familiar with these type of questions so I don't really how to begin with this. Also, I don't understand the question here. Do we have to consider it in the vertical or horizontal plane? I think its former because the weight of thread is balanced by force of surface tension. If it is so, here's what I have attempted:

For the first case, consider an element of $Rd\theta$ where $R$ is the radius of semicircle. The force due to surface tension can be resolved into two components. From symmetry, we need to integrate only the sine component. ($dF_T$ represent the force of surface tension on the small part).
$$dF_T=TRd\theta \Rightarrow dF_T\sin\theta=TR\sin\theta d\theta$$
Hence, force due to surface tension is twice the result of the above expression (as there are two sides, I don't know how to put it in words)
$$F_T=4TR$$
As the weight of thread is balanced by force due to surface tension, $mg=4TR$.

In the second case, the force due to surface tension can be separately calculated on each semicircular part by replacing R with R/4 in the above expression for $F_T$.
$$F+mg=4TR/4+4TR/4$$
But solving this gives me a negative value of T.

Any help is appreciated. Thanks!

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2. May 29, 2013

### Basic_Physics

The film attaches itself to the thread and pulls perpendicularly to it. This will cause the thread to be pulled tight by the surface tension as shown in fig. a. There are two sets of forces since there are two surfaces attached to the thread - one for each side of the film.

3. May 29, 2013

### Saitama

I am not sure if I understand your reply correctly but do you mean that there is some force at the point where the thread is connected with wire frame and this force balances the force due to surface tension?

4. May 29, 2013

### Basic_Physics

Yes. Usually a glass slide or wire frame is inserted into the solution or liquid and then it is pulled out with a spring scale. The surface tension is then the force, as measured by the spring scale, needed per unit length to pull it out. The length being twice the width of the slide or wire frame. In this case the endpoints need to supply this force.

5. May 29, 2013

### Saitama

Okay, so I just need to replace mg with the force due to the endpoints. But I will again end up with a negative answer for T.

6. May 29, 2013

### Basic_Physics

The weight of the thread should not be taken into consideration. The total force on the thread due to the surface tension δ in fig. a will be

F = δ2L

7. May 29, 2013

### Saitama

Would you please explain how did you get that result?

8. May 29, 2013

### Basic_Physics

The surface tension, δ, is the force that the film applies on the object it attaches to per unit length. So in this case

δ = F/2L

It is twice the length since there are two surfaces involved. If you look sideways at the thread it will look like a disc. Two surfaces are attached to it so the film pulls on both sides.

9. May 29, 2013

### Saitama

I know the definition of Surface Tension but I don't think in this case you can simply state that force due to surface tension is 2δL. Won't you need to use integration as I have done?

10. May 29, 2013

### Basic_Physics

Yes, you are right. Direction need to be taken into account. The surface tension force will be perpendicular to the thread at all points. So the force is most likely radial from the midpoint of the semi-circle. Another way of looking at it might be that the force on the frame need to balance that on the thread out.

11. May 29, 2013

### Saitama

I still don't know what to do now. Can you have a look at my attempt and tell me where I am wrong?

12. May 29, 2013

### Basic_Physics

Your result for part a looks right

FT = 4TR

where

R = L/π

13. May 29, 2013

### Basic_Physics

Which is exactly what I suspected. The top length of the frame is 2R.

The force at the endpoints of the thread will then be 1/2 of this force FT due to symmetry.

Last edited: May 29, 2013
14. May 29, 2013

### Saitama

The force 4TR is equal to the force exerted by the endpoints i.e 4TR=F', where F' is the total force exerted by the endpoints in the vertical direction.

In case b), $F+F'=4TR/4+4TR/4$ but solving this I get negative value for $T$.

15. May 29, 2013

### TSny

I think that's right. So, what would be the expression for the force at one endpoint for figure a?

I'm not quite following this. Why R/4 rather than R/2? What is the force required to support one endpoint of one of the smaller semi-circles? Why do you say you will get a negative value for T?

16. May 29, 2013

### Saitama

$F'/2=2TR$
$F+F'=4TR/2+4TR/2 \Rightarrow F+4TR=4TR \Rightarrow F=0$

Is F' different for both the cases?

17. May 29, 2013

### TSny

OK. So, that's the external force that must be applied at one endpoint of a semicircle of radius R.

What is the external force that must be applied at one endpoint of one of the smaller semicircles (of radius R/2)?

Sorry, I'm not following this at all.

18. May 29, 2013

### Saitama

For case a), the FBD is as follows:

For case b),

Is $F''$ not equal to $F'/2$?

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19. May 29, 2013

### TSny

I don't think $F''$ equals $F'/2$. According to your picture, $F'/2$ is the force required to hold one endpoint of a semicircle of radius R. But $F''$ according to your picture is the force required to hold one endpoint of a semicircle of radius R/2.

20. May 29, 2013

### Saitama

This is a wild guess but is $F''=F'/4$?