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• JohnnyGui
JohnnyGui
Hello,

I was reading about Archimedes' Principle on the wiki ( https://en.wikipedia.org/wiki/Archimedes'_principle ) and a question popped up regarding the upthrusting force (also called buoyant force) that the liquid exerts on an object that is in the liquid.

I understand that the upthrusting force is equal to the mass of the displaced liquid caused by the object in it. This force is exerted by the liquid right beneath the object. However, I was wondering what exactly causes this upthrusting force.
I first thought that as soon as the object falls in the liquid, it compresses the liquid beneath it and thus increasing the pressure of that liquid beneath which results in a force upwards against the object. In reality however, the object actually pushes the liquid on the sides and the liquid rises in height.
This means that the object isn't actually compressing the liquid beneath it (but just pushing the excess away), thus there shouldn't be an increase in pressure in the liquid beneath the object (the same amount of molecules is there).

If that's so, then where does this upthrusting force come from other than pressure?

Gold Member
Hmmm.. how to explain. First you should avoid thinking in terms of "causing a force" when considering static systems. Rather think more in terms of balancing of forces necessary for the situation to be static. (Cause and effect is better applied to dynamic situations as in what causes a particle to change its direction of motion...)

Now take the boat and ask why doesn't it sink further into the water. The answer is that in general it might but in so doing it will displace water. It also will be deeper into the water where the pressure is greater and the force on the bottom of the boat must increase.

If you are not sure why the pressure deeper down is greater remember that water at a given depth must support the water higher up just like the boat so deeper water supports more weight over the same area and must have higher pressure.

Once you have these facts you have the balance of weights since for the boat to go down the water it displaces must go up and vice versa.

What you say about "compression" isn't quite right as the fluid need not be in any way compressible. I believe you're really thinking "pressible" as in being a fluid and thus able to accept and transfer pressure.

To get into the "cause and effect" of that uou need only think about the atomic substructure of the fluid which is dynamic... a bunch of particles bouncing randomly off each other... the randomness causes the average forces (in a very small volume) to be distributed evenly in all directions in proportion to the areas across which the moving particles may transfer momentum.

Dale
Mentor
JohnnyGui,

Are you comfortable with the idea that the water pressure increases with depth because it is supporting the weight of the water above it?

Gold Member
In a static setting yes.
JohnnyGui,

Are you comfortable with the idea that the water pressure increases with depth because it is supporting the weight of the water above it?
In a static setting yes. Again this is not a "cause and effect" but rather a "solution to a constraint" type of "because".

A fluid by definition cannot sustain a shear force, so the only sustainable force on a level cube of fluid is the normal pressure forces and gravitation. Stack em, add em up, and thar you go.

JohnnyGui
Hmmm.. how to explain. First you should avoid thinking in terms of "causing a force" when considering static systems. Rather think more in terms of balancing of forces necessary for the situation to be static. (Cause and effect is better applied to dynamic situations as in what causes a particle to change its direction of motion...)

Now take the boat and ask why doesn't it sink further into the water. The answer is that in general it might but in so doing it will displace water. It also will be deeper into the water where the pressure is greater and the force on the bottom of the boat must increase.

If you are not sure why the pressure deeper down is greater remember that water at a given depth must support the water higher up just like the boat so deeper water supports more weight over the same area and must have higher pressure.

Once you have these facts you have the balance of weights since for the boat to go down the water it displaces must go up and vice versa.

What you say about "compression" isn't quite right as the fluid need not be in any way compressible. I believe you're really thinking "pressible" as in being a fluid and thus able to accept and transfer pressure.

To get into the "cause and effect" of that uou need only think about the atomic substructure of the fluid which is dynamic... a bunch of particles bouncing randomly off each other... the randomness causes the average forces (in a very small volume) to be distributed evenly in all directions in proportion to the areas across which the moving particles may transfer momentum.

Thanks a lot for your reply. I see that you're partly associating the boat not sinking further with the pressure being higher at depth. Can I say that pressure is the cause lifting the boat up?

I think I'm more focused on understanding the "cause and effect" of all this. Let's say we push a cube into the water. Can I say that the deeper I push the cube, the more particles bounce in a given time against the side of the cube that is pointed downwards? If that's the case, what exactly makes the number of particles that bounce against that side of the cube increase? Because if the cube pushes the excess water to the sides, then the number of particles bounces against the underside of the cube in a given time should stay constant. Unless it's the increasing pressure with depth that's causing this. But if we talk about pressure, doesn't this mean in a sense that water is somewhat compressed down there since its density is increased?

Gold Member
Thanks a lot for your reply. I see that you're partly associating the boat not sinking further with the pressure being higher at depth. Can I say that pressure is the cause lifting the boat up?

I think I'm more focused on understanding the "cause and effect" of all this. Let's say we push a cube into the water. Can I say that the deeper I push the cube, the more particles bounce in a given time against the side of the cube that is pointed downwards? If that's the case, what exactly makes the number of particles that bounce against that side of the cube increase? Because if the cube pushes the excess water to the sides, then the number of particles beneath the cube and thus the number of bounces in a given time should stay constant. Unless it's the increasing pressure with depth that's causing this. But if we talk about pressure, doesn't this mean in a sense that water is somewhat compressed down there since its density is increased?

Well... not necessarily "more particles" but an aggregate of that and the particles moving a bit faster and imparting more momentum when they bounce. And you are right in that there must actually be some level of compression i.e. volume change in practice. But not necessarily in principle one could have the pressure effect rather an increase in temperature keeping the number of particles per unit volume exactly the same, but they'd be moving faster causing more pressure.

Whether it's faster particles or more particles causing the higher pressure doesn't matter in Archimedes' principle. What matters is that things balance out whatever the cause.

JohnnyGui
Well... not necessarily "more particles" but an aggregate of that and the particles moving a bit faster and imparting more momentum when they bounce. And you are right in that there must actually be some level of compression i.e. volume change in practice. But not necessarily in principle one could have the pressure effect rather an increase in temperature keeping the number of particles per unit volume exactly the same, but they'd be moving faster causing more pressure.

Whether it's faster particles or more particles causing the higher pressure doesn't matter in Archimedes' principle. What matters is that things balance out whatever the cause.

I'm having this habit of trying to understand the exact cause of things, which is sometimes tiring :P.

So if I understand correctly, the particles are moving faster beneath the cube the deeper down you push it because the pressure gets higher with depth? And even though there's enough room around the cube for the excess water to "escape" to, there's a net higher weight on the water beneath the cube (weight of cube + "weight of the water above" instead of just the water above) which makes the pressure down there even higher than before?

Mentor
I don't think that viewing a liquid as particles is a very useful mental model. It works well for gasses, where the interactions between particles are limited. For liquids the interactions are strong enough that they simply don't behave like the classical point particle approximation.

If you really want particles for some reason, then you should limit yourself strictly to buoyancy in (ideal) gasses.

Gold Member
I'm having this habit of trying to understand the exact cause of things, which is sometimes tiring :P.

So if I understand correctly, the particles are moving faster beneath the cube the deeper down you push it
faster or there's more of them or both... which is an unimportant detail to understanding the titular principle. Archimedes didn't need an atomic theory of fluids to develop it. (And Dale you're right... my hope was, in bringing it up was to show you don't want to/need to pay attention to this level of detail the "cause and effect" is a distraction) and JonnyGui, the cause and effect is a distraction. Your habit is a bad habit in this case.

because the pressure gets higher with depth? And even though there's enough room around the cube for the excess water to "escape" to, there's a net higher weight on the water beneath the cube (weight of cube + "weight of the water above" instead of just the water above) which makes the pressure down there even higher than before?
eeehhh... this isn't a great way of thinking about it. In an arbitrarily large "sea" the boat may raise the water level an arbitrarily small amount (though of course always by the fixed volume it displaces). Don't focus on the pressure at a given level changing because the surface rises with displacement. (and my apologies if this is not what you meant.)

Focus on the static problem. The pressures on the surface of a volume of water (or boat) must balance the weight of that volume (or boat). At the heart of it is Newton's Law: F=ma. Since acceleration, a=0 in the static case, so too must net force.

When I mentioned increase in pressure I was referring to the fact that bottom of the boat would sink deeper if the buoyancy was less than the weight. (Or I meant to imply that.) There is some equilibrium point (unless the boat as a whole is too heavy and sinks to the bottom). Where that equilibrium point is is the other part.

Replace the volume of the floating boat below its water line with an identically shaped volume of water and the pressures at various points on boundary of these two volumes are identical. They also, given the static situation, must result in a net upward force equal to the weights of the two supported objects. Eureka! The volume of water and the boat weigh the same.
(if the point by point pressures in the two cases were not equal then there would be a different net forces on the remaining body of water. Both could not be in equilibrium. That's a subtle argument and it's easier to reason that pressure of a liquid body is a function of depth only which gives you the same result.)

Gold Member
JohnnyGui,

Are you comfortable with the idea that the water pressure increases with depth because it is supporting the weight of the water above it?
I do see your point... the water could be supporting the weight of the boat above it.. :)

JohnnyGui
Don't focus on the pressure at a given level changing because the surface rises with displacement. (and my apologies if this is not what you meant.)

This is exactly what I'm struggling with. If there's a displacement of water because of the boat, then the net pressure of water beneath the boat should be the same (it only carries the boat's weight), and yet as the boat sinks more in the water the buoyant force increases until there's an equilibrium that compensates for the weight of the boat. Where does this increase in buoyant force come from if the pressure of water beneath the boat doesn't change at all?

I'm sorry if I've missed your explanation on this question.

Mentor
If there's a displacement of water because of the boat, then the net pressure of water beneath the boat should be the same
Be very clear here, please. Exactly which two scenarios are you comparing and claiming are the same?

Gold Member
The pressure in the water increases with depth in any case even if there is no boat present .

Think about a very simple boat consisting of a short vertical axis cylinder with a flat bottom . As this boat sinks further into the water the pressure on the bottom surface increases thus increasing the upward force on the boat .

Chestermiller
JohnnyGui
The pressure in the water increases with depth in any case even if there is no boat present .

Think about a very simple boat consisting of a short vertical axis cylinder with a flat bottom . As this boat sinks further into the water the pressure on the bottom surface increases thus increasing the upward force on the boat .

I was struggling with why the water pressure beneath the bottom surface of the boat should increase with depth because the only thing that this body of water carries is the weight of the boat all the time. Even if the boat sinks further, any extra weight in the form of excess water is pushed to the sides by the boat so there's still a net weight of the boat that is being carried by the water beneath it.

It is as if you're decreasing the volume of the water beneath it (= the boat sinking further), but at the same time you're actually removing water from that volume (the boat pushes water away as it sinks) and thus the pressure beneath the boat would stay the same. Unless ofcourse the increased water height on the sides causes an increase in pressure on the water beneath it and in turn exerts a force against the body of water that is beneath the boat.

@Dale : I've made a simple drawing regarding this theory to explain what I mean (you need to zoom into see the text): https://www.dropbox.com/s/wtx6um71apt7kbe/Archimedes.jpg?dl=0

Could this theory be correct on why the buoyant force increases with depth?

Mentor
In your drawings h is irrelevant and can be made to go to 0 simply by making the water container very wide.

Do you understand that, even without the boat, the water pressure increases with depth? The presence of the boat is not what drives this effect, just the weight of the water itself.

Do you know the formula for calculating the water pressure as a function of depth. (in the fluid itself without any boat)

Chestermiller
JohnnyGui
In your drawings h is irrelevant and can be made to go to 0 simply by making the water container very wide.

Do you understand that, even without the boat, the water pressure increases with depth? The presence of the boat is not what drives this effect, just the weight of the water itself.

Yes, I do understand that. Regarding the h being irrelevant; what I meant is that the increase in height of water in any volume is what eventually increases the pressure of the water beneath the boat as explained in my drawing. Without the boat, the increase in pressure with depth is because of the water above it.

If you make the container very wide such that the water height is 0, this is just according to the previous container. In that wide container, the water would still decrease in height (albeit very little) if you take out the object.

Gold Member

russ_watters and Chestermiller
Mentor
what I meant is that the increase in height of water in any volume is what eventually increases the pressure of the water beneath the boat as explained in my drawing
I understand that is what you meant, but it is wrong. For example, suppose that the container itself expands or contracts in order to maintain an exactly constant height. According to your suggestion, objects would sink in such a situation. Again, h is completely irrelevant to this. You are preventing yourself from understanding by focusing on it.

What is the formula for pressure as a function of depth, in the fluid itself without the boat?

Mentor
Oddly, it seems like you were closer yesterday than you are today. Somehow you now think the weight of the ship matters. But going back to yesterday's's start:
I understand that the upthrusting force is equal to the mass [weight] of the displaced liquid caused by the object in it. This force is exerted by the liquid right beneath the object. However, I was wondering what exactly causes this upthrusting force.

...thus there shouldn't be an increase in pressure in the liquid beneath the object (the same amount of molecules is there).

If that's so, then where does this upthrusting force come from other than pressure?
The upthrusting force comes from the pressure only. There is no increase in pressure when the object is placed in the water: the pressure in the water (at a certain depth) is the same whether there is other water on top of it or an object on top of it. That pressure is where the buoyant force comes from.

JohnnyGui
Oddly, it seems like you were closer yesterday than you are today. Somehow you now think the weight of the ship matters. But going back to yesterday's's start:

The upthrusting force comes from the pressure only. There is no increase in pressure when the object is placed in the water: the pressure in the water (at a certain depth) is the same whether there is other water on top of it or an object on top of it. That pressure is where the buoyant force comes from.

Ah ok, I think I get it now. So the weight of the object that the water beneath the object is carrying, is the same as the weight of the water that was above it before the object, correct? Thus the pressure of the water beneath the object is constant.

Also, thanks @Nidum for drawing this, this helped me understand it better.

This leads me to one last thing though; if the container (where the water is in) is fixed in volume, shouldn't the water displacement, caused by the object, lead to the fact that every other body of water in the container that doesn't carry the object above it, eventually carry that extra displaced volume of water? As a result, this would mean that the water beneath the object is just carrying the weight of the object, while the other bodies of water at the same water level (that is, at the level of water beneath the object) would carry the weight of the object in the form of water plus the weight of the displaced water. Shouldn't this lead to the object rising a bit until an equilibrium is achieved since the other bodies of water are trying to even out that extra displaced body weight?

Mentor
Huh? Can you provide a diagram?

The way I learned it was that the water in the tank surrounding the object does not know that the object isn't water. It still thinks that there is water there. So it exerts the same upward force on the object as if there was water there. This force is just equal to the weight of the water that the object displaced.

Mentor
This leads me to one last thing though; if the container (where the water is in) is fixed in volume, shouldn't the water displacement, caused by the object, lead to the fact that every other body of water in the container that doesn't carry the object above it, eventually carry that extra displaced volume of water? As a result, this would mean that the water beneath the object is just carrying the weight of the object, while the other bodies of water at the same water level (that is, at the level of water beneath the object) would carry the weight of the object in the form of water plus the weight of the displaced water. Shouldn't this lead to the object rising a bit until an equilibrium is achieved since the other bodies of water are trying to even out that extra displaced body weight?
For the third time, what is the formula for the pressure for a fluid in hydrostatic equilibrium? Use that formula consistently to determine the answer to that question.

JohnnyGui
Huh? Can you provide a diagram?

The way I learned it was that the water in the tank surrounding the object does not know that the object isn't water. It still thinks that there is water there. So it exerts the same upward force on the object as if there was water there. This force is just equal to the weight of the water that the object displaced.

Yes I'm indeed aware of this. Here's a diagram of what I mean: https://www.dropbox.com/s/qw06490bqguzpkk/Displacement.jpg?dl=0

For the third time, what is the formula for the pressure for a fluid in hydrostatic equilibrium? Use that formula consistently to determine the answer to that question.

Thinking this through, I myself would conclude that it should be P = (p x V x g) / A in which p is the liquid density, V = volume, g = the gravitational constant and A the surface on which the volume exerts a force on. Implementing depth would give P = (p x g x h) / A. Let me verify this and see how I can use this to answer my above question.

Homework Helper
Gold Member
I looked quickly through this post, but I think some of the confusion may be resulting from a kind of add-on to Archimedes principle that needs to be made to it when it is applied to a floating object in a closed volume. e.g. You can float a 5 lb. boat in a tub using perhaps 1 lb. or less of water because it is the "effective" volume that is displaced is how the law actually works where "effective volume" is the volume of the object below the waterline. The buoyant force is equal to the weight of the "effective" volume of water that is displaced.

russ_watters
Mentor
What is actually happening is that the pressure varies linearly with the depth below the waterline z:
$$P=\rho g z$$
So the deeper you go, the higher the pressure gets. Another characteristic of the pressure is that it acts equally in all directions. So, if you had a small vertical element of surface at the depth z, the horizontal pressure on this surface would also be ##\rho g z##. A third characteristic of the pressure is that it always acts perpendicular to any surface that the fluid contacts. So, if you had a small element of surface at depth z oriented at some angle between the horizontal and the vertical, the pressure force on the surface would have components in both the horizontal and the vertical directions.

Now suppose you had an object in the water having some oddball shape. The object could be partially immersed in the water or it could be fully immersed. If you were to conceptually break the curvy surface of the object into tiny little elements of area, calculate the pressure exerted by the water on each little element of surface area, determine the vertical component of the force on each of the elements of surface, and add up all these vertical components of force, the total upward force exerted by the surrounding water on the object would just add up to the weight of the water that the object displaced below the (present) water line (not the original water line).

Mentor
Ah ok, I think I get it now. So the weight of the object that the water beneath the object is carrying, is the same as the weight of the water that was above it before the object, correct? Thus the pressure of the water beneath the object is constant.
Be careful with that assumption: it is only true for a floating object. The broad description is simply that it's the weight of the water displaced (the water that was there "before the object" got there). If, for example, you dip an object in water but still partly support it from above, the buoyant force doesn't equal the object's weight. And if the object sinks, the buoyant force doesn't equal the object's weight. Again: always the water's weight.
This leads me to one last thing though; if the container (where the water is in) is fixed in volume, shouldn't the water displacement, caused by the object, lead to the fact that every other body of water in the container that doesn't carry the object above it, eventually carry that extra displaced volume of water?
That seems cumbersomely worded, but if I understand, yes, if the volume of water and the volume of the container are fixed, the height of the water in the container goes up when you put another object in it. That's related to Archimedes principle in that you can use it to measure the volume of water displaced and thus the volume of the object.
As a result, this would mean that the water beneath the object is just carrying the weight of the object...
Again: you need to be careful of that because it isn't universally true. It is only true in the case of a freely floating object.
...while the other bodies of water at the same water level (that is, at the level of water beneath the object) would carry the weight of the object in the form of water plus the weight of the displaced water.
I'm not sure what that is supposed to mean. What "other bodies of water"? You mean the water next to your object? Water next to an object can't apply an upward force on the object since it is only next to and not under the object: pressure forces act perpendicular to surfaces in fluids.
Shouldn't this lead to the object rising a bit until an equilibrium is achieved since the other bodies of water are trying to even out that extra displaced body weight?
Yes, floating objects will rise if the water level rises.

Homework Helper
Gold Member
Please look at my post #24. I think it may answer why the OP is somewhat puzzled.

Mentor
Please look at my post #24. I think it may answer why the OP is somewhat puzzled.
I think that's what we've all been trying to say.

Homework Helper
Gold Member
I think that's what we've all been trying to say.
The way Archimedes principle is normally stated, you could not float a 5 lb. boat using only 1 lb. of water. I finally figured out the solution to this dilemma a number of years ago, and subsequently found the same subject (and its solution) addressed in a physics book on all kinds of different topics that I found in a book store. The author described it as one of his favorite puzzles.

Mentor
Please look at my post #24. I think it may answer why the OP is somewhat puzzled.
I looked quickly through this post, but I think some of the confusion may be resulting from a kind of add-on to Archimedes principle that needs to be made to it when it is applied to a floating object in a closed volume. e.g. You can float a 5 lb. boat in a tub using perhaps 1 lb. or less of water because it is the "effective" volume that is displaced is how the law actually works where "effective volume" is the volume of the object below the waterline. The buoyant force is equal to the weight of the "effective" volume of water that is displaced.
Well said.

Mentor
Implementing depth would give P = (p x g x h) / A.
Close. You kept an extra factor of A, so it should be ##P=\rho g h##. Note that this h is not the same as the h in your drawings, it is the depth below the surface.

Are you familiar with calculus? If so then you can more correctly write it ##dP/dh=\rho g##

JohnnyGui
I looked quickly through this post, but I think some of the confusion may be resulting from a kind of add-on to Archimedes principle that needs to be made to it when it is applied to a floating object in a closed volume. e.g. You can float a 5 lb. boat in a tub using perhaps 1 lb. or less of water because it is the "effective" volume that is displaced is how the law actually works where "effective volume" is the volume of the object below the waterline. The buoyant force is equal to the weight of the "effective" volume of water that is displaced.

You actually made me realize what my struggle was. It seems that the whole thing that I was describing (see drawing: https://www.dropbox.com/s/qw06490bqguzpkk/Displacement.jpg?dl=0 ) is actually happening during the Archimedes Principle. It is actually a part of this principle while I thought all this time that it should happening after it.

I was able to conclude a formula for the total hight of the water and the depth of the object beneath the water using the surface area of the object beneath the water, the surface area of the water, the density of water and the volume of the object that is under water. And even after correcting for the problem that I was describing in my posts, the depth of the object stays the same, and the total height of the water rises with the volume of the object that is beneath the water / the surface area of the water. So at the end, the object should rise a bit with the water but the part of it that is under water is the same in depth as before the water displacement.

All this time I thought the depth of the object would have to change after the water displacement because I thought the problem that I discovered happens after the whole Principle (I know it still shouldn't change even if it was after the Principle anyways).Also, keep in mind that I'm talking here about a floating object all along.

@russ_watters : Yes, I was indeed talking about a floating object. Regarding the "other bodies of water", what I meant are indeed the bodies that don't carry the object, but do carry the weight of the object but in the form of water. I think my drawing in post #23 would make it clear what I meant.

@Chestermiller : The present waterline instead of the previous waterline was indeed the keyword here for me.

@Dale : Regarding keeping the A parameter as an extra factor there, isn't pressure related to a force per squared meter? Or is that already covered in the unit of density which makes the A parameter obsolete?

I want to seriously thank you all for the time that you've took to try and explain this to me.

Homework Helper
Gold Member
You actually made me realize what my struggle was. It seems that the whole thing that I was describing (see drawing: https://www.dropbox.com/s/qw06490bqguzpkk/Displacement.jpg?dl=0 ) is actually happening during the Archimedes Principle. It is actually a part of this principle while I thought all this time that it should happening after it.

I was able to conclude a formula for the total hight of the water and the depth of the object beneath the water using the surface area of the object beneath the water, the surface area of the water, the density of water and the volume of the object that is under water. And even after correcting for the problem that I was describing in my posts, the depth of the object stays the same, and the total height of the water rises with the volume of the object that is beneath the water / the surface area of the water. So at the end, the object should rise a bit with the water but the part of it that is under water is the same in depth as before the water displacement.

All this time I thought the depth of the object would have to change after the water displacement because I thought the problem that I discovered happens after the whole Principle (I know it still shouldn't change even if it was after the Principle anyways).Also, keep in mind that I'm talking here about a floating object all along.

@russ_watters : Yes, I was indeed talking about a floating object. Regarding the "other bodies of water", what I meant are indeed the bodies that don't carry the object, but do carry the weight of the object but in the form of water. I think my drawing in post #23 would make it clear what I meant.

@Chestermiller : The present waterline instead of the previous waterline was indeed the keyword here for me.

@Dale : Regarding keeping the A parameter as an extra factor there, isn't pressure related to a force per squared meter? Or is that already covered in the unit of density which makes the A parameter obsolete?

I want to seriously thank you all for the time that you've took to try and explain this to me.
When I first figured this out (about 10-15 years ago), there seemed to be something very problematic with Archimedes principle, but it took a while to pinpoint it. I was helping a high school student design an experiment for a science project about Archimedes principle and the discrepancy arose that the weight of the water that was used to float the object was less than the weight of the object. It took some very careful thinking to figure out just what was the source of the problem. :-) :-)

Mentor
Regarding keeping the A parameter as an extra factor there, isn't pressure related to a force per squared meter? Or is that already covered in the unit of density which makes the A parameter obsolete
Yes, it is already covered. You can check the units to make sure. If you include the factor of A then you wind up with units of ##N/m^4## instead of ##N/m^2##

Homework Helper
Gold Member
For the OP: If your mathematics somewhat advanced, you might find one additional item of interest. You already seem to have a pretty good handle on things, but this last item might explain an item or two. Any pressure variations in a fluid cause a force per unit volume in the fluid given by ## \vec{f}_{vp}=-\nabla P ##. The gravitational force per unit volume on the fluid is given by ## \vec{f}_{vg}=-\rho g \ \hat{k} ## where ## \rho ## is the fluid density (for water ## \rho= 1 \ gm/cm^3 ##) and ## g=9.8 \ m/sec^2 ## (or ##980 \ cm/sec^2 ##) is the gravitational constant.In order for the fluid to be at equilibrium, the net force per unit volume must equal zero so that ##-\nabla P-\rho g \ \hat{k}=0 ##. This is essentially what Dale has shown above for the fluid (here I'm calling upward=+z) with his ## dP/dh=\rho g ## equation. If you are familiar with the gradient (## \nabla ##) operator, this will give you ## -dP/dz=\rho g ##. One additional item to keep in mind is that the atmospheric pressure gets added (as a constant) to the pressure formula as a function of depth ## h \ ##. e.g. ## \ P=\rho g h +P_o \ ## where ## P_o ## is the atmospheric pressure.

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