Reply to: Too good to be true?

[Bob Dalek]

Okay Diver before I let Archimedes rest in peace. Recent thoughts/investigations.

After looking at a dozen or so articles on the web I can see no reference to applications where the bottom surface of an object is removed from the equation (and therefore where Archimedes principle no longer holds). I assume you accept that Archimedes principle *is* due to pressure differentials acting on the top and bottom surfaces rather than because 'it just happens'

Here are some thoughts on experiments where we might make the pressure differential different to what you'd experience in everyday situations.

- An object is sealed to the bottom of a tank of water to prevent water (specifically its pressure) from getting underneath it and creating a net buoyant force. A seal preventing water from getting underneath will have to resist water pressure at that depth though, so this seal would need to be strong enough to resist as much sideways force as the buoyant upward force being eliminated, so without further thought I'm not sure how useful a test this would be.

- Eliminate the underside pressure on an object by making a hole in the bottom of a water filled tank and putting something ordinarily buoyant (less dense than water) in the hole. I imagine an ordinarily buoyant stopper would now not float, but instead would remain stuck in the hole. Or if the stopper is less wide than the hole, would flow out of the hole at the same speed as the water rather than rise.

- A sucker. A ordinary sucker works because when it's pressed against a surface, air is driven out leaving lower pressure air behind in the sucker. The now greater pressure of the atmosphere surrounding the sucker keeps it stuck to a surface. The same principle can be extended to water (I think - I tried it with a sink plug in a water-filled sink). This sink sucker worked underwater, without any air being in the sucker.

It's unsatisfactory that much is written about pressure differentials proving Archimedes correct, where it's applicable, but seemingly so little in situations where the pressure differentials are not as they occur in hum-drum cases :tantrum

You are my old physics teacher and I claim my crisp £5.

[Sysman]

Everywhere is downhill.

Shouldn't that be in one of the many house prices threads?

[TimberWolf]

Okay Diver before I let Archimedes rest in peace. Recent thoughts/investigations.

After looking at a dozen or so articles on the web I can see no reference to applications where the bottom surface of an object is removed from the equation (and therefore where Archimedes principle no longer holds). I assume you accept that Archimedes principle *is* due to pressure differentials acting on the top and bottom surfaces rather than because 'it just happens'

Here are some thoughts on experiments where we might make the pressure differential different to what you'd experience in everyday situations.

- An object is sealed to the bottom of a tank of water to prevent water (specifically its pressure) from getting underneath it and creating a net buoyant force. A seal preventing water from getting underneath will have to resist water pressure at that depth though, so this seal would need to be strong enough to resist as much sideways force as the buoyant upward force being eliminated, so without further thought I'm not sure how useful a test this would be.

- Eliminate the underside pressure on an object by making a hole in the bottom of a water filled tank and putting something ordinarily buoyant (less dense than water) in the hole. I imagine an ordinarily buoyant stopper would now not float, but instead would remain stuck in the hole. Or if the stopper is less wide than the hole, would flow out of the hole at the same speed as the water rather than rise.

- A sucker. A ordinary sucker works because when it's pressed against a surface, air is driven out leaving lower pressure air behind in the sucker. The now greater pressure of the atmosphere surrounding the sucker keeps it stuck to a surface. The same principle can be extended to water (I think - I tried it with a sink plug in a water-filled sink). This sink sucker worked underwater, without any air being in the sucker.

It's unsatisfactory that much is written about pressure differentials proving Archimedes correct, where it's applicable, but seemingly so little in situations where the pressure differentials are not as they occur in hum-drum cases :tantrum

[Diver]

We're doomed.

Seriously...

Totally....

[TimberWolf]

Of course I'm a user. In salvage I have to calculate displacement values that change with depth, tide and wave or swell height for a controlled lift. I also have to calculate in the variance between salt, brackish or fresh water.
In estuaries this can vary considerably with depth as the fresh water rides over the salt.

As for your second point, I covered that in a previous post in this thread; with regard to diving at altitude.

Yeah, employing working practises (being a user) is better than calculating from first principles in most cases. I imagine lifting using deformable bags adds to the trickiness for you too.

I'm beginning to have second thoughts with regard to my recollection about floats sticking together underwater, I may have seen this above the surface. I'll try it next time I'm down the pool. There seems to be little information around about sealing the bottom surface of a submerged object on the Internet - but you needn't concern yourself about such non-practical things

[Diver]

Yeah, you don't want to be calculating anything difficult when 50m under water

It's all very well being a USER , but knowing where those figures come from might come in handy now and then. Pressure = Fluid density * g * h. For instance not only could the density of the fluid change (fresh vs salt water), but the other two variables can change too.

Don't go diving on the Moon Or on Earth, since g varies with latitude amongst other things, as does atmospheric pressure.

Of course I'm a user. In salvage I have to calculate displacement values that change with depth, tide and wave or swell height for a controlled lift. I also have to calculate in the variance between salt, brackish or fresh water.
In estuaries this can vary considerably with depth as the fresh water rides over the salt.

As for your second point, I covered that in a previous post in this thread; with regard to diving at altitude.

[TimberWolf]

0.445 Lbs per Sq Inch per foot for sea water, 0.433 Lbs per Sq inch per foot for Fresh water. <----it's easier

Yeah, you don't want to be calculating anything difficult when 50m under water

It's all very well being a USER , but knowing where those figures come from might come in handy now and then. Pressure = Fluid density * g * h. For instance not only could the density of the fluid change (fresh vs salt water), but the other two variables can change too.

Don't go diving on the Moon Or on Earth, since g varies with latitude amongst other things, as does atmospheric pressure.

[scooterscot]

Bluddy hell, it's as much as I can do to remember Boyle's Law...

Don't eat greasy food?

[cojak]

Bluddy hell, it's as much as I can do to remember Boyle's Law...

[TimberWolf]

Displacement again. Consider the two floats as a whole until separation occurs. Archimedes Principle and the laws of Displacement are still valid.


HTH

What are your 'laws of Displacement'? Last time you used the word displacement it appeared to be instead of volume.

[Diver]

Note also that if the bottom surface isn't subjected to the water pressure, there will be no net buoyant force. Archimedes principle is not valid. For instance if the bottom surface is glued to a flat surface on the bottom. This situation doesn't appear to get discussed much though.

Displacement again. Consider the two floats as a whole until separation occurs. Archimedes Principle and the laws of Displacement are still valid.


HTH

[Diver]

I fast-forwarded through a lot of it, but found it entertaining nevertheless. The proof is actually quite simple. Here's the basics.

Using:

Pressure = Force / Area (P=F/A), therefore F=P*A
Density = Mass / Volume
Let d stand for water density, so the pressure at depth = d * g * depth

Pressure increases (linearly) with depth (P = d * g *depth), so for an object of height h the pressure on the top surface (P1) will be slightly less than on the bottom surface (P2) and so the pressure difference on the object (P2-P1). Using the first equation, this force equals (P2-P1)*A. Horizontal pressures are equal so are ignored.

Using the third formula this force equals (d * g * h) * A, and since volume is height * area, equals d*g*V. And since Mass = density * volume (2nd formula), equals mg.

Mg equals the weight of water displaced, so that's the buoyant force.

In algebraic terms: F= (P2-P1)A = pghA = pgV = mg

I think you have to do things like this yourself before they make sense but that's the basics.

0.445 Lbs per Sq Inch per foot for sea water, 0.433 Lbs per Sq inch per foot for Fresh water. <----it's easier

[TimberWolf]

Note also that if the bottom surface isn't subjected to the water pressure, there will be no net buoyant force. Archimedes principle is not valid. For instance if the bottom surface is glued to a flat surface on the bottom. This situation doesn't appear to get discussed much though.

p.s. I think I've done this when messing about with flat floats in the swimming pool. Put two floats under the water with one flat on top of the other. Just hold down the bottom one, and I think I've seen the top one remain stuck to the bottom one for a while. Archimedes principle not in action. Water works its way between the floats quite quickly though (and how quickly it does so depends on its pressure and therefore the depth), and so after a few seconds Archimedes principle takes over again.

[TimberWolf]

Note also that if the bottom surface isn't subjected to the water pressure, there will be no net buoyant force. Archimedes principle is not valid. For instance if the bottom surface is glued to a flat surface on the bottom. This situation doesn't appear to get discussed much though.

[TimberWolf]

I actually sat right the way through that

The bit where he nips off camera saying "Let me just check"

I fast-forwarded through a lot of it, but found it entertaining nevertheless. The proof is actually quite simple. Here's the basics.

Using:

Pressure = Force / Area (P=F/A), therefore F=P*A
Density = Mass / Volume
Let d stand for water density, so the pressure at depth = d * g * depth

Pressure increases (linearly) with depth (P = d * g *depth), so for an object of height h the pressure on the top surface (P1) will be slightly less than on the bottom surface (P2) and so the pressure difference on the object (P2-P1). Using the first equation, this force equals (P2-P1)*A. Horizontal pressures are equal so are ignored.

Using the third formula this force equals (d * g * h) * A, and since volume is height * area, equals d*g*V. And since Mass = density * volume (2nd formula), equals mg.

Mg equals the weight of water displaced, so that's the buoyant force.

In algebraic terms: F= (P2-P1)A = pghA = pgV = mg

I think you have to do things like this yourself before they make sense but that's the basics.