Surface tension - thread in soap solution

[Saitama]

Homework Statement

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

Homework Equations

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!

 

[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.

 

[Saitama]

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.

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?

 

[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.

 

[Saitama]

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.

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.

 

[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

 

[Saitama]

F = δ2L

Would you please explain how did you get that result?

 

[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.

 

[Saitama]

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.

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?

 

[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.

 

[Saitama]

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.

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

 

[Basic_Physics]

Your result for part a looks right

F_T = 4TR

where

R = L/π

 

[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 F_T due to symmetry.

 

[Saitama]

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

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 $T$.

 

[TSny]

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.

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

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

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?

 

[Saitama]

So, what would be the expression for the force at one endpoint for figure a?

$F'/2=2TR$

Why R/4 rather than R/2?

Yes, sorry about that.

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?

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

Is F' different for both the cases?

 

[TSny]

$F'/2=2TR$

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)?

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

Is F' different for both the cases?

Sorry, I'm not following this at all.

 

[Saitama]

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)?

For case a), the FBD is as follows:

For case b),

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

 

[TSny]

For case a), the FBD is as follows:

For case b),

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

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.

 

[Saitama]

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.

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

 

[TSny]

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

Yes. Think of the unknown force F as broken up into two equal parts: one part to hold the right endpoint of the left semicircle and the other part to hold the left endpoint of the right semicircle. The upward surface tension force of 2TR on the left semicircle is balanced by the two downward forces F'' on the left endpoint and F'' on the right endpoint. Similarly for the right semicircle.

 

[Saitama]

Yes. Think of the unknown force F as broken up into two equal parts: one part to hold the right endpoint of the left semicircle and the other part to hold the left endpoint of the right semicircle. The upward surface tension force of 2TR on the left semicircle is balanced by the two downward forces F' on the left endpoint and F' on the right endpoint. Similarly for the right semicircle.

Thank you!

$$F+2F''=4TR$$
Since $F''=F'/4=TR$
$$F+2TR=4TR$$
Solving, $F=2TR \Rightarrow T=F/2R=\pi F/(2L)$.

This is the right answer. Thanks!

But still, I don't get why the force at endpoints are not equal in both the cases?

 

[TSny]

But still, I don't get why the force at endpoints are not equal in both the cases?

You found by integration that the total surface tension force on a semicircle of radius R is 4TR. This force is balanced by the forces at the two endpoints. Thus, the forces at the endpoints will change if R changes. A smaller semicircle will require less force at the endpoints.

 

[Saitama]

You found by integration that the total surface tension force on a semicircle of radius R is 4TR. This force is balanced by the forces at the two endpoints. Thus, the forces at the endpoints will change if R changes. A smaller semicircle will require less force at the endpoints.

Thanks a lot TSny!

 

[TSny]

You are welcome. Good work.

 

[Chestermiller]

I did this problem two different ways, and got the same result, but the result was different from the answer given. I got T = 2F/L.

Method 1: Draw a dotted vertical line from the top of each semicircle to the top of the frame. Do a force balance on the central region (free body) between the dotted lines. There are no vertical forces acting along the dotted lines. The upward force exerted by the frame on the top of the free body of film is TL/2. The downward force on the bottom of the free body is F. Thus, F = TL/2

Method 2: Focus on the length of string between the bottom of the dotted line and the point where the force F is applied. Let θ be the angle along the arc. The differential arc length is Rdθ, and the surface tension force on the differential arc is normal to the arc, and of magnitude TRdθ. The vertical component of the differential force is TRcosθdθ. If I integrate this vertical component of the force over the entire arc length (θ=0 to θ=π/2), I get TR. There are two arcs that contribute to F, so F = 2TR = TL/2.

I think that the answer that was given in the OP's book was incorrect.

Chet

 

[TSny]

I did this problem two different ways, and got the same result, but the result was different from the answer given. I got T = 2F/L.

I think your two methods will lead to the same answer that Pranav-Arora got.

Note that L represents the length of the string rather than the width of the wire frame.

Also, the film has two surfaces - a front surface and a back surface So, you'll get twice as much surface tension force.

 

[Chestermiller]

I think your two methods will lead to the same answer that Pranav-Arora got.

Note that L represents the length of the string rather than the width of the wire frame.

Also, the film has two surfaces - a front surface and a back surface So, you'll get twice as much surface tension force.

Thanks very much.

Chet

 

[Saitama]

I am interested to learn about the two methods you tried Chet but first let me confirm if I am understanding your reply correctly. Are you talking about the red dotted lines?

 

[Chestermiller]

I am interested to learn about the two methods you tried Chet but first let me confirm if I am understanding your reply correctly. Are you talking about the red dotted lines?

Yes.