# Surface tension and capillary tube

1. Dec 17, 2013

### Rugile

1. The problem statement, all variables and given/known data
A capillary tube of radius r = 0.3mm is filled with water. A water droplet is hanging on the bottom of the tube, as shown in the picture. The water level is h = 5.2cm. Estimate the radius of curvature of the droplet R. The coefficient of surface tension of water is σ = 7*10-2 N/m.

2. Relevant equations
$F = \delta L$
$p = \frac{F}{S} = \rho gh$

3. The attempt at a solution

If i understand correctly, the surface tension will be $F = \delta L = \delta \frac{2\pi R}{2}=\delta\pi R$. Now I'm not sure what does the tension has to be equal to? I assumed it might be the force of the pressure of liquid: $F=p*S=\rho ghS=\rho gh\pi r^2$. Though I'm not quite sure about this solution - I don't really get what is the surface tension equal to, in a general case?

2. Dec 17, 2013

### haruspex

By δ you mean σ, right?
Since you didn't post the diagram, there's something that's not entirely clear. Is the top of the liquid level or is there a meniscus there too? I'll assume it's level, since it says "filled".
Remember that force is a vector. The radius of the drop is greater than the radius of the tube. At some point around the edge of the drop (i.e. where it meets the tube), what is the direction of that force? When you integrate around the drop edge, the direction of the force changes, so there will be some cancellation. What component of the force is relevant?

3. Dec 18, 2013

### Rugile

Oops, sorry...

Seems like I'm too absent minded for physics these days..

Yes:)

Wait a sec, I'm not sure - what force are we talking about? The force of pressure?

Last edited: Dec 18, 2013
4. Dec 18, 2013

### haruspex

No, sorry, I mean the force due to surface tension.
There are two menisci. The one at the top is easy. Since the contact angle of water is zero, the surface tension force around that meniscus acts straight upwards, so it's just the surface tension value (a force per unit length) multiplied by the length of the meniscus (interior circumference of tube).
The force from the lower meniscus is more complicated. The meniscus forms a spherical cap. The surface tension force where it meets the end of the tube will be tangential to that cap.
Can you figure out what the net vertical force from it will be?

5. Dec 18, 2013

### Rugile

I'm not getting it - what do you mean by 'contact angle of water is zero'? Since we multiply by the interior circumference of tube, the force of surface tension is only where the tube has contact with the meniscus?

How do you know the direction of the force?

6. Dec 18, 2013

### haruspex

See http://en.wikipedia.org/wiki/Contact_angle and http://en.wikipedia.org/wiki/Wetting.
I can't find it stated in either of those places, but when I learnt this stuff we were taught to assume that the contact angle for water on clean glass is zero ('perfect wetting').
The direction of the force is tangential to the surface of the liquid. Where the contact angle is zero, that means it will be parallel to the surface, and you can use that at the upper meniscus.
The lower meniscus is actually rather complicated. Since the contact angle should again be zero, the water surface at the edge of the meniscus will be horizontal, but as you come away from the edge it will dip down to form the suspended drop. (You can see this with water dripping slowly from a tap.) But it's a good enough approximation to treat the water surface as a spherical cap having a nonzero contact angle with the base of the glass tube.
So, suppose the water at the edge of the lower meniscus is at angle theta to the horizontal. That means the surface tension force will also be at that angle. For the purpose of supporting the weight of water in the tube, what component of that is relevant?
From that and the given information you can deduce theta.
What is the relationship between theta, the radius of the tube, and the radius of curvature of the cap?

7. Dec 18, 2013

### Rugile

So we need the vertical component of lowers meniscus' surface tension force $sin \theta = \frac{F_y}{F}; F_y = Fsin \theta$? And the force itself will be (just like the top meniscus') $F = \sigma L = \sigma 2 \pi r$?
Also, $Fsin \theta + F = mg$ and then $\sigma 2 \pi r sin \theta + \sigma 2 \pi r = \sigma 2 \pi r(sin \theta + 1) = mg => sin \theta = \frac{mg}{\sigma 2 \pi r} - 1$. And I also figured that $sin \theta = \frac{r}{R}$.
Is any of this true?

8. Dec 18, 2013

### haruspex

That all looks right to me.

9. Dec 18, 2013

!! Thanks :)