# Surface tension of a liquid layer within frame with elastic wire

• peripatein
In summary, a square frame with a thin crust/layer of liquid with surface tension γ is given. An elastic wire of circular shape with radius r_0 is inserted into the plane of the frame and the thin crust/layer of liquid is burst (by pricking). The resulting force of the surface tension on the wire causes it to extend to a new radius, r_1. The tension in the wire is proportional to the increase in length with a constant k. An expression for the surface tension (γ) of the liquid can be found using k, r_1, and r_0, which is γ = πk(r1-r0)/r1. The second part of the question asks for the change in energy
peripatein
Hi,

## Homework Statement

A square frame with a thin crust/layer of liquid with surface tension γ is given. Into the plane of the frame an elastic wire of circular shape with radius r_0 is inserted. The thin crust/layer of liquid is then burst (by pricking) and as a result of the force of the surface tension exerted on the wire, the elastic wire extends and rapidly attains a new radius, r_1.
It is also given that the tension in the wire is proportional to the lengthening of the wire wrt its flaccid length, with a constant k.
I was asked to find an expression for the surface tension (γ) of the liquid, using k, r_1, and r_0.

## The Attempt at a Solution

The question advises to use the diagram in the attachment.
So I wrote down the following equation: 2Tsine(phi) = 2γdl
And T = k(l_1-l_0) = 2pi*k(r_1-r_0)
Hence, pi*k(r_1-r_0) = γ
I am not sure these equations are correct. May you kindly advise?

#### Attachments

• Frame with liquid layer.JPG
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peripatein said:
the tension in the wire is proportional to the lengthening of the wire wrt its flaccid length, with a constant k.
It's not completely clear what that means. Is it proportional to the increase in the length (T=k(r1-r0)) or to the fractional increase in length (T=k(r1/r0-1))?
So I wrote down the following equation: 2Tsine(phi) = 2γdl
That's true, in the limit, for small phi and dl. Breaks down for large angles because the surface tension forces are not all acting in the same direction.
And T = k(l_1-l_0) = 2pi*k(r_1-r_0)
Hence, pi*k(r_1-r_0) = γ
The RHS is wrong. You had 2γdl. What happened to the 2dl?

Hi,
It is proportional to the increase in length.
The LHS should have been 2pi*k(r_1-r_0)dl (while the RHS is still 2*gamma*dl.
Supposing this is correct (is it?), will that indeed be the correct expression for gamma?

peripatein said:
Hi,
It is proportional to the increase in length.
The LHS should have been 2pi*k(r_1-r_0)dl (while the RHS is still 2*gamma*dl.
Supposing this is correct (is it?), will that indeed be the correct expression for gamma?
No, I still think that's wrong. You had, correctly, T = 2πk(r1-r0) and T sin(dθ) = γdl.
Now, what is the relationship between dl and dθ?

Did you mean (dl/r_1) = sin (phi)?

I meant to write (dl/2r_1)=sin(phi). Would you agree?

peripatein said:
I meant to write (dl/2r_1)=sin(phi). Would you agree?
Since the force equation is only valid for small angles (as I mentioned) I'd prefer to make this clear by using dθ for the angle. Also, the equation is really dl = r1 dθ (no sine involved), but of course there's no difference in the limit as dθ→0.
So you have T sin(dθ) = γdl = γ r1 dθ. Since sin(dθ) ≈ dθ, T = γ r1. You already had T = 2πk(r1-r0). So what do you get for γ?

dl is the length of the entire wire segment, so I believe it should be dl=2r_1*d(theta), should it not? Which yields gamma=pi*k(r_1-r_0)/r_1
Is that incorrect?

peripatein said:
dl is the length of the entire wire segment, so I believe it should be dl=2r_1*d(theta), should it not? Which yields gamma=pi*k(r_1-r_0)/r_1
Is that incorrect?
In the OP you had 2Tsine(phi) = 2γdl, implying dl is the half-length. Either way, T = γr1, and T = 2πk(r1-r0).

This is a bubble, i.e. two layers, one from each side, so the surface tension had to be doubled. And the angle in the diagram was indicated as 2phi. Why is dl then the half length? If you consider it to be the full length, gamma will then be equal to pi*k(r_1-r_0)/r_1 and not to 2pi*k(r_1-r_0)/r_1.
So which is correct?

peripatein said:
This is a bubble, i.e. two layers
OK, I missed that. So T = 2γr1, and γ = πk(r1-r0)/r1. But not πk(r1-r0), as you had a few posts back.

The second part of that question asks for the change in energy of the crust between the initial and final states.
As gamma=dE/dS, I wrote down the following equation:
dE=[pi*k(r_1-r_0)/r_1][pi*(r_1)^2-pi*(r_0)^2]
Is that correct?

peripatein said:
The second part of that question asks for the change in energy of the crust between the initial and final states.
As gamma=dE/dS, I wrote down the following equation:
dE=[pi*k(r_1-r_0)/r_1][pi*(r_1)^2-pi*(r_0)^2]
Is that correct?
Almost. dS is the increase in area, right? What would that be?

Why wouldn't that be [pi*(r_1)^2-pi*(r_0)^2]?

peripatein said:
Why wouldn't that be [pi*(r_1)^2-pi*(r_0)^2]?
Two reasons. The area of the film decreases; it's a bubble, remember?

So should it be 2(pi*(r_0)^2-pi*(r_1)^2)?

peripatein said:
So should it be 2(pi*(r_0)^2-pi*(r_1)^2)?
Yes.

Couldn't I have derived that by calculating the change in elastic energy in the wire? Actually, I have tried that but the two expressions are not the same.
For the elastic energy (1/2*kx^2) I get: 1/2k(2pi*r_1-2pi*r_0)^2
Why aren't they the same?

Could it be because of the tension in the wire that the two expressions are not identical?

## 1. What is surface tension?

Surface tension is a physical property of liquids that causes the surface of the liquid to behave like a thin, elastic film. It is the force that allows liquid to resist external forces and maintain its shape.

## 2. How is surface tension measured?

Surface tension can be measured in units of force per unit length, such as newtons per meter (N/m). It can also be measured indirectly through techniques such as capillary rise or drop weight methods.

## 3. What is the role of elastic wire in the surface tension of a liquid layer?

Elastic wire plays a crucial role in the surface tension of a liquid layer within a frame. It provides a boundary for the liquid to adhere to, creating a surface tension force that keeps the liquid in place and maintains its shape.

## 4. How does the thickness of the liquid layer affect surface tension?

The thickness of the liquid layer has a direct impact on surface tension. As the thickness increases, the surface tension decreases, and vice versa. This is because the force of surface tension acts over a smaller area as the liquid layer becomes thinner.

## 5. What are some real-world applications of understanding surface tension of a liquid layer within a frame with elastic wire?

Understanding surface tension in this context has many practical applications, such as in the production of soap bubbles, the functioning of insect wings, and the behavior of droplets on surfaces. It also plays a role in industries like manufacturing, painting, and printing.

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