Have You Ever Photographed Something You Can't Explain?

P = rho * g * h

Pressure = density * gravity * height where height is the height of the fluid above the object.

So, for example, a cube 1m on a side with the top 1m below the surface of the water:

Top pressure = 1000kg/m^3 * 9.8m/s^2 * 1m = 9,800 Pascals
Bottom pressure = 1000kg/m^3 * 9.8m/s^2 * 2m = 19,200 Pascals
(not including barometric pressure, which is added on).

I explained above why there's more pressure below - because the pressure is the weight of the fluid above you. More fluid above you = more pressure.

Good luck with the explanation with which I fully agree ... but David keeps talking about "forces" ... and telling him the formula for hydrostatic pressure has yet to satisfy his years-long curiosity in the matter.

Perhaps if we figured the forces in Newtons - the one on the top of your cube pushing down and the one on the bottom pushing up, he might agree but I am rather doubtful.

David, as somebody wrote above, hydrostatic prassure is actually created by the weight of the fluid above given level. So it's depth x density x g, OK.? Note, that depth (liquid column height in other words) x density x any bottom surface is just the mass of fluid column. Now mass x g = the weight of this column (the force pointed down). Now pressure is by definition force / surface (in this case bottom surface of the column).
Now according to Pascal law hydrostatic pressure in liquids/gases can create forces on surrouding object in any direction.Not like a solid object, which weight acts only down (because inside it there are internal forces maintaining its shape). This is due to fact, that liquids/gases are not solid (do not maintain their shape) and, so any force within liquid/gas is transferred multidirectionaly by chaotic forces between particles.
So now You understand, that hydrostatic pressure is dependant on depth. If You have any object submerged in liquid/gas, than its higher parts (top) are highier (shallower) than the lower ones (at the bottom of the object) and this is the very cause of pressure difference that causes buoyancy. You really claim You can't understand this? I really hope at this point, that You're not just trolling.
Now about hot air balloon. This is also nothing hard to understand, really. As we described before, one of the effecting/practical principles of the basic cause of buoyancy is that buoyant object should have total density equal to the density of surrounding liquid (full submersion case, like balloon in air or submarine underwater).
That means, that if constructed from and carrying as a payload heavy/dense material, it must be empty or filled with material that is really buoyant, having density much lesser than surrounding fluid.
Submarines have some air inside, right? Deep sea vehicles can use also gasoline tanks or special hard foam as flotations tanks. The balloon in general have to compensate its own weight and that of gondole, so it must be filled with gas which is less densier than air surrounding the baloon. Why we can use not only different, light gas like hydrogen or helium (and many more btw)? Because the density of any gas depends on 3 things:
1) molecular weight of the gas (that's why we can use hydrogen/helium)
2) pressure (no meaning for buoyancy, becasue inside soft balloon at the given height it's equal to the outside, unless we use the ballon with hard shell)
3) the temperature
No 3 is the key. When You heat the air in hot air balloon, its density decreases (because it expands) compared to cold air outside. So the hot air is buoyant material compared to colder air outside. For the same reason vertical currnets exists in athmosphere or even in oceans (liquids do change density with temperature, though not so much like gases): colder fluids go down, warmer go up. Simple.
Regards
-J.

OK, let me try and explain my thoughts, step by step, starting with a basic scenario and building it up to the real world situation then someone can jump in and say "Stop! That point there is where you are going wrong". Please bear with me and be patient as I build up the scenario, it's the only way to put a finger on any mis-apprehensions on my part.

Initial scenario

1. If we consider a large volume of air somehow trapped in outer space (by a force field, say), where there is no strong gravity to consider.
2. In the middle of this volume of air is a hollow sphere made of a thin super strong skin.
3. This sphere is therefore entirely immersed in a fluid of air.
4. In this scenario, there are forces at play: the sphere is being impinged all around its surface by randomly moving air molecules
5. Collisions with these random kinetic movements will apply a tiny force somewhere on the sphere every time each molecule collides with the sphere
6. Statistically, these little forces will balance out all around the sphere and there will be no net force applied to the sphere and therefore:
7. Apart from little random jiggles, the sphere will just sit there motionless with respect to the volume of air.

Is this toy model correct so far?

OK, let me try and explain my thoughts, step by step, starting with a basic scenario and building it up to the real world situation then someone can jump in and say "Stop! That point there is where you are going wrong". Please bear with me and be patient as I build up the scenario, it's the only way to put a finger on any mis-apprehensions on my part.

Initial toy model scenario

1. Consider a large volume of air somehow trapped in outer space (by a force field, say), where there is no strong gravity.
2. In the middle of this volume of air is a hollow sphere made of a thin super strong skin.
3. This sphere is therefore entirely immersed in a fluid of air.
4. In this scenario, there are forces at play: the sphere is being impinged all around its surface by randomly moving air molecules
5. Collisions with these random kinetic movements will apply a tiny force somewhere on the sphere every time each molecule collides with the sphere
6. Statistically, these little forces will balance out all around the sphere and there will be no net force applied to the sphere and therefore:
7. Apart from little random jiggles, the sphere will just sit there motionless with respect to the volume of air.

Is this toy model correct so far?

Scenario 2

Now we complicate the model by adding a flat surface with a Star Trek like artificial gravity device outside the volume of air
This device creates an "up" and a "down" and a force of gravity that now acts on the air molecules and the sphere

At this point, sorting out what changes starts to get tricky for me... I think what happens is this:

1. The air molecules are attracted to the ground by gravity and fall towards it
2. The first molecules that arrive at the surface form an initial layer lying on top of the surface
3. Subsequent layers then arrive and the air starts to pile up on the surface

4. At this point, each air molecule in the layers lying on the ground are feeling:

5. a downwards force caused by its own gravity

6. further downward forces caused by being impacted from above by all the remaining air molecules dropping down on top of it

7. Each molecule must also be feeling upwards forces from the surface and all the molecules already below it. However, I'm completely muddled as to exactly what the forces are. I guess there must be reaction forces (Newton 3rd) and there must be further little random kinetic jiggles from each molecules temperature, but how all these little forces add up, I'm just not sure.

8. But what I'm imagining is that we now a new scenario in which all the air is stacked on the surface with a density gradient with the densest most crushed air at the bottom of the stack and the thinnest least crushed air at the top of the stack.

9. And this density gradient must mean that the cumulative kinetic forces applied at the bottom of the stack are stronger than those at the top simply because there are more molecules at the bottom than at the top applying said forces

10. Now we return to our sphere. In scenario 1 in the absence of gravity, our sphere was surrounded by equal forces in all directions and and subject to no net force. And thus it stayed exactly where it was. Now things are completely different.

11. it now feels a downwards force from its own gravity

12. it now feels additional downwards forces from all the air molecules above it that are now not applying random kinetic forces but directed downwards forces because they are falling themselves and colliding with the top of the sphere as they do so.

13. in summary the downwards acting forces are now adding up to cause a considerable downwards motion to the sphere

14. However, there are also forces operating upwards to counteract this downwards force:

15. there are denser layers of air below it that contain many more molecules than the number of molecules above it, so the number of molecular impacts operating in an upwards direction now exceed the number of molecular impacts operating in the downwards direction. This provides some lift.

16. From this point, it follows that to make our sphere into an effective balloon we need to maximise the discrepancy between the forces at the top and the bottom to create sufficient lift to overcome the sphere's self gravity. And the way to do this, is to increase the surface area of the sphere so it feels relatively more impacts from the air molecules and thus magnifies the relative difference between the upward and downward forces.

17. Once the balloon is big enough so that the relative difference between forces operating top and bottom are large enough to create a net force that exceeds the downward force it feel because of its own self gravity (its weight), it will start to rise.

18. To make this possible, we need to construct the sphere from materials that are light enough such that as its size increases, it weight increases more slowly so the lift dominates gravity. So, a super thin lightweight skin made of infinitely rigid material that allows a total vacuum to exist inside the sphere would be ideal. Then we could make it as large as it needs to be to generate sufficient lift.

Conclusion
To my mind, this is a plausible toy model that explains where the buoyancy force comes from in terms of net newtonian forces. I've deliberately avoided talking about more complex concepts such as pressure, Archimedes and so on because I want a model that operates using the simplest principles of mechanics: forces and net forces only, no other more sophisticated concepts wanted.

Now, I'm fairly sure that my little model is plain wrong, that it doesn't work, that I have missed out something vital or I have mis-described a scenario somewhere along the line and that my model simply doesn't apply.

However, I can't figure out where my errors lie.

Ok, understood. But I do not want to introduce more complicated concepts such as pressure. Let's keep it basic and talk only in terms of all the individual tiny forces at play and their effects. I can imagine little forces caused by molecules colliding but my imagination starts to get confused when we abstract this into concepts like pressure. We are talking about different levels of explanation here - let's keep it at my explanation level 😄

So no useful gradient. That wipes out my entire model at source. 😄😄😄 And means I'm left with nothing to rescue from it. Got to start completely from scratch and forget everything I have dreamt up so far. Oh well. That's science for you, no respect for opinion, you just have to take it on the chin 😄

Thank you, anyway, for the correction. While disappointing, it is helpful to understand what is the fundamental flaw in your assumptions underpinning the house of cards.

The problem for me now, is exactly how to reset and compile a completely different model where things work differently from how I always supposed. I'm still looking for a ball bearing forces model rather than a higher level abstracted model. Ideally it would work so the correspondence between the ball bearings and Archimedes is obvious and both explained the other formulation.

Where to start...

David, ArvoJ is right. Don't mix the density change/gradient into the problem. Forget temporarily about the gas/air. Try to model the whole thing first in non-compresible medium, like liquid. You will see there's no need for any density difference to create buoyancy force, just the pressure difference. Which is caused only by the height/depth difference. If You compare 2 parts of submerged object (of any shape), You will see that that deeper one has more fluid above it, which weight creates more hydrostatic pressure, that's all. Going into particle's level there's a lot more particles of fluid above it that create hydrostatic pressure by their weight. The "shallower" part of submerged object has above it the lesser crowd of fluid particles trying to squeeze it. So there must be some difference in forces here. The fluid may have (as in liquid) the same density around submerged object. That's difference in the layer thickness (and so its weight) that creates pressure difference here.
"Above" means location in gravity field. No gravity, no hydrostatic pressure.
Regards,
-J.

Ok, let's consider a scenario of cube of solid metal of size 1m * 1m * 1m that masses 1kg submerged in an incompressible liquid in a gravity field set conveniently to g=10ms^-2.

1. If the cube were floating above the surface, the downwards force on the cube must be 1kg * 10ms^-2 = 1 Newton

2. But because it is submerged, we must also consider the contribution of the weight of the water on top of it when working out the downwards force.

How do we do that?

(Let's ignore upwards forces for now).

I'm guessing we need to work out the mass of the liquid above contained within the square section prism volume above the cube in order to calculate the downward force acting on that liquid, then add it to the 1N of the cube. And of course, the mass of the liquid will depend on the length of the prism it occupies and that length is the depth of the cube below the surface (assuming the liquid has the same density throughout)?

Sorry, I initially missed out the 0, corrected it, but forgot to correct the calculation

EDIT (following Arvo's reply)

So...

1. Corrected downwards force on the cube when above surface = 10N

2. Downwards force when submerged at say 2m depth is:

10N + force acting on the volume of water above the cube

1. Force acting on the volume of water above the cube at 2m depth is: mass of 1m^2 * 2m length of liquid * 10ms*-2 (acceleration due to gravity)

2. I looked up how much 2 cubic m of water masses and conveniently it is 2000kg.

3. So, downwards force applied to the cube @ 2m depth is 10N (weight of cube) + 2000kg * 10ms^-2 = 20,000N (weight of the water)

4. So total downwards force acting on the cube @ 2m depth is 20,010N

Is water really that heavy? Have I screwed up the calc?

Part 2 Upwards force

This is where I get hazy about how things work...

So we currently have a downwards force of 20010N pressing down on the water underneath the cube. That will create an upwards Newtons 3rd law reaction force of equal and opposite direction. So that will be a 20010N upward force completely balancing the downwards force.

My conclusion therefore is that it is impossible for any object to sink..... wait a minute, something wrong here.

In order to sink, the cube has to move the water under it out of the way so it has a space to occupy. So maybe some of the downwards force is redirected sideways shifting the water out of the way? This reduces the downward force acting on the cube. The difference between this reduced downwards force and the original reaction force creates lift.

Something like that.......??

Somewhere I read a quote by Cartier-Bresson, something like: If I see something I cannot perceive right now, I take a photograph.

This is getting far too complicated.

Take a 1 litre plastic bowl to a beach, & pushing it down into the sand produces resistance, but once pushed down, it stays there.
That’s because the sand particles have no interacting force between them.

Liquids, such as water, have intermolecular forces holding the molecules together. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapour at 100°C.

Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!

Imagine pushing the bowl down onto a trampoline made of sheet rubber. The rubber molecules stretch but want to return to normal, thus providing a reactionary opposite force.

But water molecules, unlike rubber, do not stretch or compress.

Since the water does not get compressed/stretched downwards or sideways, it simply rises upwards against the force of gravity, as the displaced water has absolutely nowhere else to go. This means gravity is the reactionary opposite force acting on any object placed in water.

When removing the object, the intermolecular forces of the water will normalise back to the previous level.

This was Archimedes observation & resulting principle of buoyancy.

Unlike metal, except for mercury, its difficult to make a dent in liquid water.