# Understanding buoyancy

1. Jan 27, 2012

### TheLil'Turkey

I think that buoyancy is caused by the increase of density with depth (the deeper you go, the more molecules there are per unit volume). Therefore an object in a fluid will be hit by more of the fluid molecules from below than from above (even if the difference is only a tiny fraction of 1%). Is this correct?

More thorough explanation of my hypothesis (same as post 13): If a solid object is in water, the pressure on it is caused by the impacts of the water molecules. If the pressure is twice as high on the bottom of the object than on the top, then twice as many water molecules hit the bottom per unit time (assuming no macroscopic movement and everything's at the same temperature). Obviously the water's density at the bottom of the object isn't twice as high as it is at the top, but I'm pretty sure that the average time that a water molecule travels between impacts (and therefore also the average intermolecular distance, or average distance from the "edge" of one molecule to the "edge" of another) decreases extremely quickly with increasing density. I crudely visualize this as water molecules being huge (and in constant motion) with tiny spaces between them. This would mean that a tiny percentage increase in density would lead to an enormous percentage decrease in the average intermolecular distance. If the average intermolecular distance halves, the pressure doubles.

Last edited: Jan 27, 2012
2. Jan 27, 2012

### DaveC426913

No. Buoyancy works perfectly fine even in fluids that have a very low density gradient over their depth.

It may be better to think - not about buoyancy of the object - but about the weight of the surrounding heavier objects.

i.e. A 1in3 cork released near the bottom of a bucket of water will have water all around that that wants to be in the 1in3 space occupied by the cork. The water has enough weight (since 1in3 of water is significantly heavier), pushing down from above, to shove the cork out of the way. So does the water above that, all the way up to the surface.

Last edited: Jan 27, 2012
3. Jan 27, 2012

### Staff: Mentor

No. What matters is the increasing pressure with depth.

4. Jan 27, 2012

### TheLil'Turkey

Of course, Al. But I want to understand what causes the increase of pressure with depth in terms of the molecules of the fluid.
I'd really appreciate a response from someone knowledgable.

5. Jan 27, 2012

### Staff: Mentor

It's certainly true that a fluid under pressure will be squashed somewhat, even in nominally 'incompressible' fluids. But that will be a trivial increase in density. Is that what you're looking for?

6. Jan 27, 2012

### TheLil'Turkey

I want to be sure that I understand buoyancy in terms of the molecules of the fluid and the forces they exert on a submerged solid.

Last edited: Jan 27, 2012
7. Jan 27, 2012

### sophiecentaur

Not sure about that as being the crucial issue. Is it not just a matter of relative density and. as Archimedes put it, volume of 'displaced fluid'.

If you take a cuboidal stick of uniform density, will it float higher out of the water (as a percentage of its volume) if it lays horizontally or if you force it to an upright position using frictionless rails or some such arrangement? Having it cuboidal (rectangular faces) makes for the simplest model as the upthrust will only be due to the horizontal, lowest face.
When it lays down, it will get upthrust from a large area with low pressure but, when pointing upright, the upthrust will be over a small area at a greater depth. In both cases, the hydrostatic pressure times the area of the lowest face is the same (I think?) - namely, the volume times ρg. Hence I think it will float with the same fraction immersed.
It is true to say that the horizontal position is the position with least potential - but is that relevant?

8. Jan 27, 2012

### Staff: Mentor

Clarification: it is the pressure gradient across an object. An object does not gain buoyancy as it sinks.

9. Jan 27, 2012

### sophiecentaur

Yep. Ok.

10. Jan 27, 2012

### Staff: Mentor

Right. It's the pressure gradient (increasing pressure with depth) in the fluid and thus across an object that gives rise to the buoyant force. Archimedes' principle follows from that.

11. Jan 27, 2012

### Naty1

As others have implied, that is NOT the way to understand it.

As directly noted above, that is [virtually] irrelevant to describing buoyancy.

But such thinking..such ideas and maybe different perspectives ARE good.....

12. Jan 27, 2012

Perhaps the simplest way to look at it is that the deeper you go in a fluid, the more of that fluid you have above it pushing down on it, so that fluid has to support all of the weight above it. Of course, the way a fluid exerts force is through molecular collisions, the integrated effect of which is called pressure. So, the deeper you go, the more molecular collisions must occur in a given area to supply a pressure high enough to hold the water above it up and the higher the pressure gets.

13. Jan 27, 2012

### TheLil'Turkey

I'll explain myself in more detail now.

If a solid object is in water, the pressure on it is caused by the impacts of the water molecules. If the pressure is twice as high on the bottom of the object than on the top, then twice as many water molecules hit the bottom per unit time (assuming no macroscopic movement and everything's at the same temperature). Obviously the water's density at the bottom of the object isn't twice as high as it is at the top, but I'm pretty sure that the average time that a water molecule travels between impacts (and therefore also the average intermolecular distance, or average distance from the "edge" of one molecule to the "edge" of another) decreases extremely quickly with increasing density. I crudely visualize this as water molecules being huge (and in constant motion) with tiny spaces between them. This would mean that a tiny percentage increase in density would lead to an enormous percentage decrease in the average intermolecular distance. If the average intermolecular distance halves, the pressure doubles.

Is the General Physics Forum the correct one for this question? Should I have posted this in a more advanced forum?

Last edited: Jan 27, 2012
14. Jan 27, 2012

Did you just skip over my response or just decide not to acknowledge it? The density need not change for the pressure to go up. That is how water still exerts a buoyant force on an object, because it is about as close to incompressible as you can get.

15. Jan 27, 2012

### TheLil'Turkey

You don't understand my question in this thread. Of course the bolded is true, but how does a molecule deeper down "know" it has to collide more often? My hypothesis is that it's simply due to increased density. You haven't offered an alternative hypothesis.
If my hypothesis is correct then the bolded is completely wrong. And there are substances far less compressible than water, so I don't know where you got that idea.

16. Jan 27, 2012

But that is just it. Your hypothesis is 100% incorrect. There are many, many counterexamples. Take for instance, anything floating in water. Water is so incompressible that it can effectively be treated as such. You could measure the density at the top of a lake, say Loch Ness since it is so deep, and get right around $1000 kg/m^3$. Then you could go down to the bottom of the lake and get, to within 1% or so, that exact same reading.

Now, if we are talking air, then yes, as you get "deeper", the air gets more dense. This is precisely what happens in the atmosphere.

Now, how does it "know" it needs to exert more pressure? Newton's third law. The weight of the entire column of water above a given point is exerting pressure on that point. To remain in equilibrium, the water at that point must exert and equal pressure to that. Pascal's law, which is easy to derive, shows that pressure acts in all directions at once, so at a given point, the pressure exerted on a surface, including on an object, is equal to the pressure as a result of the hydrostatic pressure as described above. Density is not required to change.

17. Jan 27, 2012

### DaveC426913

I see where he's going.

Take one cubic inch of water and examine the molecules in it. They bounce around inside that one cubic inch and, when they get near the boundary, they are repelled by other atoms in adjacent cubes.

Now, move that system one mile downward in the water column. The volume of water may not have compressed very much, but what exactly is happening to cause the molecules to be under pressure?

It seems to me, that, while the density has changed very little, it is enough to be the cause of the pressure.

If not, I don't see the flaw in the argument.

I can imagine replacing the water with something even less compressible. Say a cubic inch of steel. The atoms in this cube are super dense to start, crammed right up against each other and crammed right up against the boundary - no free room. Put that under a mile of steel, and you've got cubes pressing on the surface of our target cube with zero tolerance. It takes a vanishingly small amount of compression before the atoms from from cube are pushing on the target cube.

18. Jan 27, 2012

### TheLil'Turkey

Thanks for your detailed reply, Dave. Since I'm still not 100% sure I'm right about this, I'd appreciate it if the forum mentors (and others) could chime in on this and hopefully even provide some proof. And sorry for being a little rude in this thread; I was frustrated with the way some people dismissed my hypothesis without offering any alternative, but I think this was in large part due to how poorly I initially explained my hypothesis.

Last edited: Jan 27, 2012
19. Jan 27, 2012

### DaveC426913

I am not sure either, so don't go on me.

But I believe I got your argument, and tried to advocate for it to get you some definitive answers from the experts.

20. Jan 28, 2012

### maimonides

In the kinetic theory of gases we consider the molecules to move freely and to interact only in collisions, or in other words, we have kinetic energy only and neglect long range forces and their potential.
You can´t do that for a fluid or a solid. If you work it out, there will be something like a mean potential energy which depends on the mean distance of the molecules.
So yes, you need compression for bouyancy - but you might not need very much of it.