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Surface element meniscus - free body diagram

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opsb
#1
Mar8-10, 04:35 AM
P: 27
Consider the free body diagram of a surface element of a water - glass meniscus in a vacuum. Along the line normal to the surface, the water pressure acts towards the vacuum, and the direction of the surface tension 'curvature force' depends on whether the surface curves like a 'u' or like an 'n'. In the case of water, the curve is a 'u', so the curvature force acts in the same direction as the water pressure force. Which force, acting towards the water along the normal to the surface, balances these?

I was trying to figure out the profile curve of the meniscus, but stumbled quite early on! I can just about convince myself why the curve should be 'u' shaped by thinking about energy minimisation, but when I associate each energy with a force (gravitational p.e. - water pressure; surface energy - surface tension) I can't get over this issue. Unless the meniscus is in fact 'n' shaped in a vacuum, and the atmospheric pressure is the 'mystery' force.
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Studiot
#2
Mar8-10, 04:57 AM
P: 5,462
Liquid pressure increases with depth and is zero at the surface.

Perhaps you would like to draw a diagram of what you mean?
opsb
#3
Mar8-10, 05:15 AM
P: 27
Agreed, but it acts on a surface element in the same direction and the net curvature force due to surface tension - away from the water (towards the vacuum) - along the normal of the element. How are these forces balanced?

Studiot
#4
Mar8-10, 06:30 AM
P: 5,462
Surface element meniscus - free body diagram

Do you really mean the vapour pressure, causing the liquid to (perhaps not so) slowly boil away into the vacuum?
opsb
#5
Mar9-10, 02:27 AM
P: 27
I don't think so. Effect of evaporation and boiling, I think, can be ignored. Consider a water droplet in a vacuum - really it would boil away. However, we can still model a non-boiling stable droplet, where the fluid pressure in the droplet equals the net inwards 'curvature force' due to surface tension, which act oppositely in this instance. I can't see why the meniscus wouldn't also be n - shaped in the vacuum. with the curvature proportional to the depth (due to the proportionality of depth and pressure).
Studiot
#6
Mar9-10, 03:44 AM
P: 5,462
Droplets?
This is the first time you have mentioned droplets; I thought you meant a jug of the stuff.

Please set out the conditions clearly.


Are gravity forces involved?
Droplets are susceptible to electrostatic forces are these involved?
etc
opsb
#7
Mar9-10, 05:29 PM
P: 27
We are considering a jug of the stuff. Apologies - the droplets were a separate thought illustrating a point - that we could forget about boiling.

Here are the conditions:

We are interested in the profile curve of a liquid at a solid - vacuum boundary (say a water meniscus in a test tube with the air pumped out)

We will consider the liquid not to boil (I don't think this is an unreasonable condition, I'll try to justify it if you'd like)

The interface energy between the liquid and the solid is lower that the surface energy of the solid - so the fluid 'creeps up' the solid, forming a 'u-shaped' meniscus (like water in a test-tube).

We assume that the surface energy of the liquid is non-zero

The liquid is in a gravitational field.



Consider, then, a surface element of the fluid, somewhere near the solid. What forces act on it?



Surface tension forces act parallel to the surface of the liquid. Since the meniscus is curved, there is a net force parallel to the normal of the surface element, acting towards the vacuum.

The liquid is in a gravitational field, so there is a pressure acting on the surface element (proportional to the depth below the contact point with the solid (test tube)), also acting towards the vacuum ('pushing' the surface element away from the liquid).

How are these forces balanced?

The only way I could think to resolve the issue would be to assume that the meniscus would be 'n' shaped in a vacuum (in which case the surface tension force would act oppositely to the liquid pressure force), but then the contact angle would be all wrong.

Maybe the problem is in the non-boiling assumption, but I don't think so.

What do you reckon?

Appreciate the input, by the way.
Studiot
#8
Mar9-10, 06:43 PM
P: 5,462
Having complained about the droplet, I am now going to offer you the droplet first, as it is simpler.

There are several ways to analyse the situation, but since you want a force balance here goes.

First the vacuum is a red herring. The pressure inside the droplet is greater than the pressure outside, whether that is atmospheric or non boiling vacuum. I will call the pressure difference p.

The thing to remember is that forces act along a line. Stresses or pressure act over an area.

If you look at the spherical droplet in my sketch you can see that the difference between pressure inside and the pressure on the droplet outside tends to burst the droplet asunder. This tendency is greatest across the greates area ie a diametral disk as shaded.

Opposing this is the hoop force of surface tension. Being tangential at A and B this acts at right angles to the shaded area. I have shown two arrows but, of course this tension acts all the way round the circle which bounds the shaded disk.
The hoop tension acting in the plane of the disk does not participate in the action.

The formula is given in the attachment.

The droplet can exist in free space or air or even stand on a surface.
A meniscus needs the vessel wall to explain it as the difference between the affinity of the liquid molecules for each other and the vessel wall comes into play. I will show this on my next post.
Attached Thumbnails
droplet.jpg  
opsb
#9
Mar10-10, 10:58 AM
P: 27
How do you get pics online?

Nicely explained. Look forward to your next post.
Studiot
#10
Mar10-10, 05:50 PM
P: 5,462
So here is how to work out the forces for a capillary rise.

Can you see how to do a capillary depression?
Also what happens as the walls move apart?

I am simply scanning a sketch in and clicking on the paperclip icon to upload it. I am sticking to greysacale to keep the size down.

Glad you can read them, sorry they are rather hurried hand drawn as it can take me forever to do one on a graphics program.
Attached Thumbnails
meniscus1.jpg  
Andy Resnick
#11
Mar10-10, 09:40 PM
Sci Advisor
P: 5,510
Quote Quote by opsb View Post
Consider the free body diagram of a surface element of a water - glass meniscus in a vacuum. Along the line normal to the surface, the water pressure acts towards the vacuum, and the direction of the surface tension 'curvature force' depends on whether the surface curves like a 'u' or like an 'n'. In the case of water, the curve is a 'u', so the curvature force acts in the same direction as the water pressure force. Which force, acting towards the water along the normal to the surface, balances these?

I was trying to figure out the profile curve of the meniscus, but stumbled quite early on! I can just about convince myself why the curve should be 'u' shaped by thinking about energy minimisation, but when I associate each energy with a force (gravitational p.e. - water pressure; surface energy - surface tension) I can't get over this issue. Unless the meniscus is in fact 'n' shaped in a vacuum, and the atmospheric pressure is the 'mystery' force.
I'm having trouble understanding your question. Laplace's law is a statement saying the curvature of an interface is proportional to the pressure jump across the interface (the interfacial energy is the constant of proportionality). If there was an unbalanced force at the contact line, the contact line may move- the water will wet or de-wet the glass. It's also possible the contact line will not move, but the contact angle will change (canthotaxis). Gravity counteracts capillary rise- in free-fall, water will rise unimpeded in a glass column (assuming the glass is hydrophilic).

Or are you asking something different?
opsb
#12
Mar11-10, 05:01 PM
P: 27
Andy - my problem is in your second sentence. In the water droplet diagram provided by studiot, the curvature clearly counteracts the pressure drop. The water is of a higher pressure than the air around it (or the vacuum in the example given). In the meniscus example, the water surface curves the 'other way'. That is, laplace's law would suggest that the pressure increases as we move across the surface boundary from the water to the vacuum (if the meniscus is u-shaped), but this cannot be the case, clearly. I can't see why the water surface doesn't curve like an 'n'. Although I understand that the water is 'dragged' up the side of the tube by surface forces, I can't see how to resolve the forces (in the stable meniscus sytem) on a surface element of water.

Incidentally (and I don't to get too sidetracked by this, I'm struggling enough already!), even without the presence of a gravitational field, I don't think the water would necessarily rise all the way up the tube, in the same way that a drop of water on a glass surface wouldn't necessarily completely spread out (unless the sum of the surface energies of the water / air and water / glass interfaces was lower than that of the glass / air interface)
opsb
#13
Mar11-10, 05:10 PM
P: 27
Studiot, I'm not conviced by your treatment of the meniscus. In a wide beaker, where the surface curvature is zero in most places, some column of water is clearly not held up by the surface interactions, but rather by the normal force of the bottom of the beaker, I would suggest that this is the case generally. I cannot understand how the depth of the water in the beaker (h) can be proportional to the surface tension (T). Wouldn't this imply that adding washing up liquid to the water (altering its surface tension) would actually change the volume of liquid in the beaker (given by Area x h) I may, however, have misunderstood your diagram (in particular, whatever h denotes).

I'd like to be able to upload a diagram to clarify, but haven't got a scanner and am on a mac which doesn't seem to have any drawing programs. I miss 'paint'.
Studiot
#14
Mar11-10, 05:31 PM
P: 5,462
Perhaps I should have said that in the capillary rise example the pressure of the water in the tube is below atmospheric.
So Andy is correct and there is a pressure rise across the surface.
Try imagining that the surface tension causes a drawing up (I don't like to say suction as suction does not exist) of the water by lowering its pressure.

The issue of the beaker was the point of my second question.

When the walls are a long way apart you get two triangles of water at the walls (supported by the vert component of ST) and a long 'flat' surface in between supporting nothing, which is what you would expect, since the sT is horizontal along that surface so has no vertical component.
opsb
#15
Mar11-10, 06:00 PM
P: 27
In a vacuum, though, the pressure in the water cannot be lower than the pressure outside it. that was what puzzled me.
Studiot
#16
Mar11-10, 06:12 PM
P: 5,462
So there would be no meniscus in a vacuum.

Don't forget the atmospheric pressure also acts on the rest of the fluid surface.
So you could say that this pushes up the water in the tube.
I didn't do this but you could also equate the forces to the pressure differential across the meniscus to calculate this pressure drop.
opsb
#17
Mar11-10, 06:21 PM
P: 27
That's the thing, though, I think there would have to be a meniscus in a vacuum. The water would still 'creep up' the side of the tube to reduce the surface energy. Wouldn't it?
Studiot
#18
Mar11-10, 06:26 PM
P: 5,462
Why does that reduce surface energy?


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