How come there is a buoyant force?

[albertrichardf]

Hello.
Suppose you have an incompressible fluid in a column of infinite height and area and that is in a vacuum. The formula for the net pressure at any point is hpg, where p is the density and h the height.
The formula comes from considering that at h = 0, the net pressure is zero, but that at h = h the pressure there must not only balance the exerted by the random collisions of the molecules, but also that of the weight of all the water above. The pressure from random collisions is balanced because throughout the fluids the molecules move in the same way, but what force balances the pressure from gravity?
I would think it to be a reaction force to gravity, like the normal force when you place an object on a table. In the same sense, the weight of the fluid above is balanced by the normal force of the fluid below, and thus there is a pressure difference due to the normal force exerted by the fluid, but I'm not sure if this is correct. If it is not, then from where does the difference in pressures arise from? I know the force is there because the fluid is stationary, but what causes this force?

Assuming the above is correct, then I have problems seeing the buoyant force.
In most derivations that I have seen, the buoyant force arises due to the difference in pressure in the fluid below and above the object submerged. If the object has volume V, then the buoyant force is pVg, where p is the density of the fluid.
But these derivations seem to neglect the reaction force to the weight of the object. The fluid below the submerged object should feel a force due to the pressure of the fluid above it, and the weight of the object submerged. But in that case, it would exert a force by Newton's third law which would be equal to the weight of the object plus the weight of the fluid above it. This is akin to how I suppose the fluid exerts a reaction force to the weight of the fluid above to balance it. In the derivations, the reaction force to the fluid does not change despite the change in weight on the fluid below. So I must be missing something, but what?

Thanks for answers

 

[BvU]

the force is there because the fluid is stationary, but what causes this force?

Precisely that object that causes the fluid to be stationary: the container (walls/bottom).

it would exert a force by Newton's third law which would be equal to the weight of the object plus the weight of the fluid above it

it certainly does.

despite the change in weight on the fluid below

if there is a change, things start to move (another Newton law): objects lighter than the displaced liquid will be pushed up, and an object heavier than the displaced liquid will start falling down

 

[Buckleymanor]

Density of the object.
If the density is less than the fluid it moves up if it's more it goes down.
A battle ship can float in a bucket of water weight or how heavy an object is in relation to the fluid it is put in does not stop it floating it is the difference in density that does.

 

[albertrichardf]

Precisely that object that causes the fluid to be stationary: the container (walls/bottom).

So the fluid from above exerts a force on the fluid below, and the fluid below transfers this force to the container's walls and the container's walls exert a reaction force?

it certainly does.

If it does exert such a force on the object, doesn't the pressure increase?
Suppose that the pressure at the top of the object is P. At the bottom it would be P + (mg/A) = P + phg, where p is the density of the object and h its height. But that is exactly equal to the force exerted by the liquid above the object plus the weight of the object. Then there would never be any net force on the object which can't be true. And I have never seen any derivation take this reaction force into account.

if there is a change, things start to move (another Newton law): objects lighter than the displaced liquid will be pushed up, and an object heavier than the displaced liquid will start falling down

I know that there are changes in forces, but in the end, would they not all sum up to zero based on what I said above? If the process of submerging the object was done very slowly, such that the forces had time to reach their final values, then the object would not sink according the these points. But that does not make any difference.

My main question is why does the pressure exerted by the fluid below not change? The force exerted by the liquid below changes, since it exerts a reaction force to the weight of the object submerged. So why does that not affect the pressure?

 

[BvU]

So the fluid from above exerts a force on the fluid below, and the fluid below transfers this force to the container's walls and the container's walls exert a reaction force?

Yes. (vertical walls exert only horizontal forces).

If it does exert such a force on the object, doesn't the pressure increase?

fluids have a tendency to equate pressure through flowing. If 'object + fluid above' heavier than before, sinking object results. If lighter, it goes up.

At the bottom it would be P + (mg/A) = P + phg, where p is the density of the object and h its height. But that is exactly equal to the force pressure exerted by the liquid above the object plus the weight / area of the object

Correct. But the liquid at that height was used to $P + \rho g h$ where now $\rho$ is the density of the fluid. So there remains a net force ($\Delta \rho gh$) that results in acceleration.

And I have never seen any derivation take this reaction force into account

Yes you have: if you step on a boat you will expect it to go a bit deeper in the water

I know that there are changes in forces, but in the end, would they not all sum up to zero based on what I said above?

no they don't if $\Delta \rho \ne 0$.

If the process of submerging the object was done very slowly, such that the forces had time to reach their final values, then the object would not sink according the these points.

Let's pick an object with a density lower than that of the fluid as an example. First of all you have to push it to go under. Then you have to push it to its final position. To keep it there, you have to keep pushing push with a force $A \Delta \rho gh$.
Note that the liquid level has gone up (with a certain $\Delta h_{\rm liq}$, so the pressures at top and bottom of object are somewhat bigger, but both increase with the same amount.

But that does not make any difference.

Do you mean 'that does not make any sense ?

 

[russ_watters]

But these derivations seem to neglect the reaction force to the weight of the object. The fluid below the submerged object should feel a force due to the pressure of the fluid above it, and the weight of the object submerged.

I think this is simply a naming convention issue. Scientists have simply chosen to refer to the upward force caused by the fluid's weight "buoyant force" and the resulting force on the object (including its weight) something else, like "resultant force". It doesn't mean there isn't also a resultant force on the object that must include its weight. This was probably done for simplicity reasons, but it is very common for people to say "buoyancy" when they really mean "resultant force". For example, people often refer to a floating object as "buoyant", but wrongly say a sinking one "isn't buoyant".

To make the resultant force even more complicated, for a static case where the object has a different density than the fluid, There needs to be some sort of tether. And note that in the static case, the resultant force must be zero.

 

[albertrichardf]

Thanks for correcting the force to pressure.

Correct. But the liquid at that height was used to $P + \rho g h$ where now $\rho$ is the density of the fluid. So there remains a net force ($\Delta \rho gh$) that results in acceleration.

That is the part I don't get. The pressure at that height should be the force there divided by the area. Without an object the forces there are the weight of the fluid and the pressure of the fluid. Summing up the two gives the formula you just gave, with the fluid density. But the pressure at that depth is a result of the reaction force of the liquid to the weight of the fluid above it.
But if were to place an object in the fluid, then doesn't the force change because of that? Say there is some object with height ∆h. At h1, the fluid exerts a pressure ph1g. At h2, the fluid would exert a force ph2g. But there is no reason for it to do so. According to what was said earlier, this pressure is a result of the reaction force to the weight of all the fluid above. But instead of fluid, there is the object. The object's weight is different from that of the fluid. Why does this not affect the pressure at h2? Unless the fluid that flows out is the reaction to the weight of the object, and the fluid that is below that is what pushes the fluid.

Let's pick an object with a density lower than that of the fluid as an example. First of all you have to push it to go under. Then you have to push it to its final position. To keep it there, you have to keep pushing push with a force $A \Delta \rho gh$.
Note that the liquid level has gone up (with a certain $\Delta h_{\rm liq}$, so the pressures at top and bottom of object are somewhat bigger, but both increase with the same amount.

So you mean to say that the reason the pressure at h2 does not change is because it has not had time to change? As the fluid feels the object coming it flows out, so that the fluid below it does not feel the weight of the object yet, and thus does not exert a reaction force in regards to the weight, so that the pressure at h2 does not change? It would be like some fluid "cushion", which would essentially counteract the weight of the object by Newton's third law, but that would also flow out of the way of the object at the same time because it is incompressible?

Yes you have: if you step on a boat you will expect it to go a bit deeper in the water

I meant that in the derivation of the equation on paper, they go through the same derivation as you do, saying that the pressure at the bottom of the object is ph2g.

 

[Buzz Bloom]

That is the part I don't get. The pressure at that height should be the force there divided by the area.

Hi Albertrichardf:

It may help If you think of an example where the object that is experiencing buoyancy is a cube. The net downward force on the cube is the gravitation force on the cube plus the fluid pressure times area on the top of the cube minus the fluid pressure times area on the bottom of the cube. If this sum is positive the cube will sink. If it is negative the cube will rise.

Hope this helps.

Regards,
Buzz

 

[A.T.]

According to what was said earlier, this pressure is a result of the reaction force to the weight of all the fluid above.

It doesn't have to be directly above a point, just higher than the point, but eventually to the side. Pressure acts in all directions equally.

The object's weight is different from that of the fluid.

Then you don't have a static situation, as the object sinks or rises. So the condition for hydro-static pressure is not fullfilled.

 

[BvU]

Perhaps this helps:

On the right (the fluid can't get past the object) you get equilibrium with the same level at the top only if $\rho_{\rm liq} = \rho_{\rm obj}$.
When $\rho_{\rm liq} < \rho_{\rm obj}$ you need a higher level in the right leg and
When $\rho_{\rm liq} > \rho_{\rm obj}$ you need a higher level in the left leg of the U-tube.

On the left the fluid can get past the object and the equilibrium condition
$$P + \rho_{\rm\ liq} g h_{\rm tot} = P + \rho_{\rm liq} g h_1 + \rho_{\rm liq} g h_{\rm obj} $$again requires $\rho_{\rm liq} = \rho_{\rm obj}$. If this is not the case, there is no equilibrium and acceleration occurs if there is no external force keeping the object in place.

 

[albertrichardf]

So then the pressure does not change because the fluid flows?
Supposing you place an object, under its weight, the fluid will flow, and in the liquid column on the left, it would flow upwards because it is incompressible. Then its weight is still on the liquid below it, and the pressure difference will p∆hg, where p is the density of the fluid, and ∆h the height of the submerged object. Then this is what would cause the buoyancy force.
So thank you for answering my question, and thank you for bearing with me. However I just have three more questions in regards to this.

If the object had density equal to the fluid, and it was suspended somewhere in it, let's say in the middle of the column, and at the bottom of the column, I made a tiny hole so that the liquid could flow out. Then liquid below the object would start flowing, so it will exert less pressure, so the object would sink as well? If it was lighter it would sink as well?

Now, if the fluid was actually compressible, would the same thing hold? If I understood well, the buoyancy came from the liquid flowing to adjust for the pressure difference. But if the fluid was compressible, then it wouldn't flow, would it? It would be compressed. And the density of the fluid would change till the pressure of the fluid exactly matched that due to the weight of the object. Is that correct?

And finally, what would happen if the liquid column had the same area as the object, like in the U-tube picture? And the liquid was incompressible. When I place the object the fluid can't flow past it. Would the container explode because of all the pressure? And if the liquid was not incompressible, would it compress under that pressure?

Thanks for taking time to answer my question. I appreciate it.

 

[Stephen Tashi]

In the derivations, the reaction force to the fluid does not change despite the change in weight on the fluid below. So I must be missing something, but what?

That is an interesting question.

If we consider a stack of metal washers, a washer in the stack is acted by the force due to the weight of the washers above it and by the reaction force of the washer below it. These forces balance. If we replace one of the washers in the stack by a wooden washer, which is less dense than the metal washers, the wooden washer is still in equilibrium. The stack of washers has the property that there is more pressure on washers that are deeper in the stack - at least in terms of pressure on their upper and lower faces.

If we have table covered with stacks of metal washers and somewhere near the bottom of one of the stacks there is a wooden washer, we can't formulate any Newtonian argument that proves the wooden washer is somehow going to float to the top in this "sea" of washers.

So a proof that there is a "buoyant" force on a object is going to have to introduce some properties that distinguish the properties of a fluid from the properties of stacks of washers in order to explain why there can be a net force on a submerged object.

One property that distinguishes a fluid from a stack of washer is that a column of fluid would not stand upright without something to contain it. Another property of a fluid is that two (un-seperated) columns of fluid exert a sideways pressure on each other. So the concept of "pressure" implies "a force in all directions", so to speak. Another thought is that the molecular motion in the fluid causes movement that is not present in the stacks of washers. . Perhaps if the stacks of washers were in a box and we jiggled it around violently, the wooden washer would rise to the top.

Let's compare the free body diagrams of a wooden washer in a stack of washers with the free body diagram of a wooden washer submerged in water.

In the stack of washers, the weight of the metal washers above the wooden washer exerts a force on the wooden washer. The weight of the wooden washer itself exerts a force on it. The force of the metal washer below the wooden washer exerts a "reaction" force. Reaction forces have the property that they "know" how much force to exert in order to cancel out other forces.

In the case of wooden washer submerged in water, I don't know if it is literally correct to say that the force on the top of the wooden washer is "due to to column of water above it". But I'm willing to accept that the pressure on top of the wooden washer can be calculated as if that is the case. The weight of the wooden washer acts on the wooden washer. However, below the wooden washer, we do not have a "reaction" force. The way that a fluid acts is that it exerts a force due to its pressure - this need not be a force that "knows" how to cancel out other forces.

The pressure coming from below depends on the depth of the water. (We can't say the pressure results from the "column" of stuff above the bottom of the washer since we aren't assuming there is an actual tube containing that column - a column that would include the washer itself.) Once we accept that the pressure on the bottom of the washer is not a reaction force, then we can accept the idea that the wooden washer is not in equilibrium. There is some net force acting on it.

Fluid mechanics arguments that show pictures of parts of a fluid and analyze them as free bodies are going to be different that similar arguments with solids due to the different behavior of pressure forces vs reaction forces. Such arguments in fluid dynamics don't prove the properties of pressure forces. Instead they assume them. I don't know of a simple way to take a model of a fluid as a collection of Newtonian particles and derive the properties of pressures. The "Kinetic theory of gases" accomplishes those derivations - but not just by drawing free body diagrams.

 

[Buckleymanor]

Perhaps if the stacks of washers were in a box and we jiggled it around violently, the wooden washer would rise to the top.

Yes it would the same way a gold or lead weight would sink in a box of jiggled sand .The sand would rise to the top like the wooden washer.