I A proof of Archimedes' Principle

AI Thread Summary
Archimedes' Principle can be proven through the application of Pascal's principle and the understanding of pressure in a fluid at equilibrium. The buoyant force arises from the difference in pressure exerted on the top and bottom surfaces of an object submerged in a fluid, leading to the conclusion that this force equals the weight of the displaced fluid. The proof is general and applies to any solid shape, relying on the isotropic nature of fluid stress as described by Pascal's Law. The discussion raises the question of why it is referred to as a principle rather than a law or theorem, attributing this to historical context. Overall, the proof demonstrates that Archimedes' Principle is fundamentally rooted in established physical laws.
DaTario
Messages
1,092
Reaction score
46
TL;DR Summary
Hi All, is there a proof to Arquimedes Principle and the expression for the buoyancy force?
Hi All, is there a proof to Arquimedes Principle and the expression for the buoyancy force? In case there is a proof, may we refer to this as Arquimedes theorem? (Buoyancy force = density of the fluid x acceleration of gravity x submerged volume)

Best Regards,
DaTario
 
Physics news on Phys.org
DaTario said:
Hi All, is there a proof to Arquimedes Principle and the expression for the buoyancy force? In case there is a proof, may we refer to this as Arquimedes theorem? (Buoyancy force = density of the fluid x acceleration of gravity x submerged volume)
It follow more or less automatically from Pascal's principle that the pressure in a stationary fluid at equilibrium and under no net force is the same everywhere.

You layer onto this the idea that a uniform gravity field on a fluid of uniform density results in a fixed pressure at any given depth given by ##P=\rho g h##.

Then you realize that "buoyancy" is nothing more than the difference in the net force along the bottom of an object compared to the force along its top.

From there you realize that the net force difference on any given column within the object is equal to the volume of that column times ##\rho## times ##g##. Add up all the columns and you get that ##\sum F=\rho g V##.

If one is going to be careful about the proof, one needs that the object is completely surrounded by the fluid and that the fluid is all connected -- free to flow so as to equalize pressure.
 
It's fairly straightforward to transform the buoyant force integral ##-\int_{\partial V} p d\mathbf{S}## into a volume integral over ##V## by pure vector calculus.

A more intuitive explanation is that said buoyant force integral depends only on the shape of the boundary ##\partial V##. Therefore, for the purpose of determining the buoyant force, one can replace the body with fluid. The parcel of fluid is obviously in equilibrium along with the rest of the fluid, which implies that the buoyant force is nothing but minus the weight of the parcel of fluid.
 
  • Like
Likes hutchphd, vanhees71 and jbriggs444
Just to work that out: For a static fluid the total force density must be 0, i.e.,
$$-\vec{\nabla} P +\rho \vec{g}=0.$$
This implies that the buoyancy force of a body with volume ##V## and boundary surface ##\partial V## is
$$\vec{F}_{\text{buo}}=-\int_{\partial V} \mathrm{d}^2 \vec{f} P=\int_V \mathrm{d}^3 x (-\vec{\nabla} P)=-\vec{g} \int_V \mathrm{d}^3 x \rho=-\vec{g} m_V,$$
where ##m_V## is the mass of the displaced fluid, which is Archimedes's principle.
 
  • Like
Likes hutchphd and DaTario
vanhees71 said:
Just to work that out: For a static fluid the total force density must be 0, i.e.,
$$-\vec{\nabla} P +\rho \vec{g}=0.$$
This implies that the buoyancy force of a body with volume ##V## and boundary surface ##\partial V## is
$$\vec{F}_{\text{buo}}=-\int_{\partial V} \mathrm{d}^2 \vec{f} P=\int_V \mathrm{d}^3 x (-\vec{\nabla} P)=-\vec{g} \int_V \mathrm{d}^3 x \rho=-\vec{g} m_V,$$
where ##m_V## is the mass of the displaced fluid, which is Archimedes's principle.
So we use Gauss Law, basically. Is there any obstacle to including every possible solid shape in this proof? My guess is that this is the reason why Arquimedes Principle is not named as a law or a theorem (as in Stevin's).
 
Yes, it's Gauss's law, and the proof shows that it's completely general, i.e., independent of the shape of the solid. Of course it rests on Pascal's Law that the static stress tensor for a fluid is isotrophic, i.e., ##\sigma_{jk}=-P\delta_{jk}##. It's a good question, why it's named Archimendes's Principle and not Archimendes's Law or Theorem. I guess it's simply for historical reasons.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...

Similar threads

Replies
6
Views
2K
Replies
32
Views
8K
Replies
48
Views
4K
Replies
6
Views
452
Replies
35
Views
4K
Replies
3
Views
1K
Replies
14
Views
6K
Back
Top