Something about Archimedes' Principle

[Dale]

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 height 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.

 

[Charles Link]

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 height 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. :-) :-)

 

[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

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 $N/m^2$

 

[Charles Link]

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.