Ocean liner in bucket full of water

[McQueen]

Doc,
I presume that there would have to be a certain volume of water to be dispaced in the first place, otherwise nothing makes sense.

 

[Doc Al]

Doc,
I presume that there would have to be a certain volume of water to be dispaced in the first place, otherwise nothing makes sense.

You assume wrong. The amount of water 'displaced' is a theoretical construct. That water need never have existed.

Example: Imagine a perfect bucket and something with exactly the same shape as the inside of the bucket just a tad smaller. If that something has a density just a bit less than water, how much water do you need to float it? A bucketful? Hardly. All you is a volume of water equal to the difference between the two volumes. Much less than a bucket. Yet, in the sense used here, the amount of water displaced will almost equal the volume of the bucket. Similarly, with the ocean liner: you won't need an amount of water equal to an ocean liner in order for it to displace that much water.

Sure, if you start with a full bucket, the water displaced will end up on the floor. But there's no reason to start with a full bucket. If you started with half a bucketful, you'd still 'displace' the same amount of water. When discussing things like Archimedes' principle, 'displaced' has a specific meaning.

 

[chingel]

Why does pressure depend only on the height of the column?

If I have a cylindrical tank that gets narrower at the top and replace it with a cylindrical tank that is of uniform diameter of the same width that the first tank it at the bottom, and it is also as tall, it would take more water in it, but why doesn't the extra mass cause extra force at the bottom of the tank?

The height is the same, the area at the bottom is the same, but why doesn't the extra mass cause extra pressure?

 

[cmb]

why doesn't the extra mass cause extra force at the bottom of the tank?

Because whatever restriction is in the way, limiting the aperture, it unloads the burden of the pressure above it. [wheras an aperture below it (if it expands again) re-applies the pressure back down on the water. The upper and lower parts of this aperture, that closes up then opens out again, must be in mechanical load-bearing contact to do this - else it'd leak!]

 

[chingel]

Because whatever restriction is in the way, limiting the aperture, it unloads the burden of the pressure above it. [wheras an aperture below it (if it expands again) re-applies the pressure back down on the water. The upper and lower parts of this aperture, that closes up then opens out again, must be in mechanical load-bearing contact to do this - else it'd leak!]

But what if the aperture doesn't increase again, but rather is increasing all the way, as in the case of a cylinder getting narrower at the top?

 

[Doc Al]

Why does pressure depend only on the height of the column?

If I have a cylindrical tank that gets narrower at the top and replace it with a cylindrical tank that is of uniform diameter of the same width that the first tank it at the bottom, and it is also as tall, it would take more water in it, but why doesn't the extra mass cause extra force at the bottom of the tank?

The height is the same, the area at the bottom is the same, but why doesn't the extra mass cause extra pressure?

In the case of the uniform diameter cylinder, the walls exert no downward force. Thus you can imagine that the pressure on the bottom is just the weight of the water divided by the area. But in the case of the tank that gets narrower at the top, the walls do exert a downward force. This adds to the weight of the water--just enough to give you the same pressure at the bottom, independent of how much water is in the tank.

 

[chingel]

In the case of the uniform diameter cylinder, the walls exert no downward force. Thus you can imagine that the pressure on the bottom is just the weight of the water divided by the area. But in the case of the tank that gets narrower at the top, the walls do exert a downward force. This adds to the weight of the water--just enough to give you the same pressure at the bottom, independent of how much water is in the tank.

How do the walls exert a downward force?

 

[Doc Al]

How do the walls exert a downward force?

Walls exert force perpendicular to their surface. If the diameter of the tank varies, that means the walls must curve in, and thus will exert a component of force downward.

 

[Q_Goest]

How do the walls exert a downward force?

Hi chingel. Maybe I could help explain what Doc Al is trying to say (not to put words in his mouth) but take a look at the picture OmCheeto posted:

These three containers have the same pressure on the bottom where the red dot is because the height of the water in the containers is the same. In C, the pressure on the walls of the container results in a horizontal force, so the walls don't pull up or push down on the bottom of the container (other than the weight of the walls of course). However, in A, the pressure on the walls is again perpendicular to the wall and it has a vertical component of force. That force is downwards and is transmitted through the wall and into the bottom of the container. This is the downwards force Doc Al is referring to.

For the sake of better understanding, let's say A and C have the same area on the bottom of the container and let's neglect the weight of the container and just look at the force downwards due to water pressure. In C, it's simply the pressure on the bottom times the area of the bottom. The horizontal component of force on the walls can't contribute to the downward force. On A however, you could break it up into the force down on the bottom of the container (pressure times area of bottom) plus the additional downwards component of force due to pressure on the walls. The pressure is integrated over the area of the wall (PdA) to get the force, and that force is acting on the wall at some angle. The downward component of that force (the force being a vector) puts stress in the wall and that stress in the wall is what is transmitting the pressure force down to the bottom of the container so that it weighs more than C.

In B, there is also a force on the walls due to pressure, but this time it's actually pushing UP on the walls. That upwards force is transmitted through the walls and down to the bottom of the container. So although we might think the container force downwards is the pressure at the red dot times the area of the bottom of the container, there is also a force UP on the bottom of the container due to the pressure exerted on the walls. That upwards force due to the pressure on the walls (which is transmitted through the walls as stress) results in a force up on the bottom of the container so that the total force due to the walls and bottom of the container is exactly equal to the weight of the water (plus container).

 

[DaveC426913]

Excellent diagram!

And if there's any doubt that the walls of B are applying downward pressure on the water near the edges, imagine what would happen if we removed that pressure that the wall is exerting!

 

[russ_watters]

Excellent diagram!

And if there's any doubt that the walls of B are applying downward pressure on the water near the edges, imagine what would happen if we removed that pressure that the wall is exerting!

See: Artesian well: http://www.google.com/search?sourceid=chrome&ie=UTF-8&q=artesian+well

 

[russ_watters]

You assume wrong. The amount of water 'displaced' is a theoretical construct. That water need never have existed.

Example:

I think a more direct example is a drydock. When you tow/sail a ship into a drydock, you can literally watch the water flow out of the drydock. But if you start with an empty drydock with a ship in it and open the gates, you fill it up to the same level to float the ship. The amount of water in the drydock when filled and with a ship in it is the same either way, regardless of if it started from empty or started with water and no ship.

 

[AlephZero]

A fluid supporting an aircraft carrrier and a piston is that different? In a general sense?

It is different. The constant part of the force (i.e. the dead weight of the moving parts) is supported by the oil being under pressure. That's why engines have oil pumps. The dynamic part of the force is resisted because of the geometry of the situation. Suppose the bearing is 1 inch diameter and the oil film is 0.001 inches thick. To move the shaft 0.001 inches so it contacts the metal, you would have to "squidge" the oil through the 0.001 inch wide gap around half the circumference of the bearing (i.e a distance of about 1.5 inches).

When the engine is not running, the forces in the oil are too small to support the rotor. That's why, unless there is a failure (e.g. an oil leak, or the oil pump fails) most of the wear in engine bearings happens when the engine is started, before the oil film is doing its job, not when it is running.

 

[DaveC426913]

I've encountered a blip in my engineering idea.

Say our boat is a cube-esque block length 1m, width 1m, height 1.1m.
It weighs 1T, and so will float with .1m out of the water.

As a control, we float it in a calm pool and mark off where the waterline is, approximately .1m from the top.

And to prove it is indeed floating (this will become important later), we drop a 10kg cinder block on top. This causes the boat to sink into the pool by an additional 1cm, i.e. it is now floating a mere 9cm out of the water, our mark has sunk 1cm below the waterline. (The water level in the pool does not rise noticeably with the addition of 10kg.)

Agreed so far?

Now we want to demonstrate the exact same thing in our boat-hugging container. The container is 1.004m on a side, leaving a 2mm gap all around. It is as tall as we need it to be.

We drop the boat in and it floats to the exact same mark as above. So far so good.

Now for the pièce de resistance, we must prove that the boat is indeed floating above the bottom of the container (that it will rise and sink freely with only a change in weight).

We drop the 10kg cinder block on top and the boat sinks into the water by 1cm, just as before. In doing so, it displaces 10L of water.

But wait - the water level in the container does not rise a mere 1cm, as before, it rockets out of the gap and climbs much higher because the 10L of water that the boat is now displacing must squeeze into the 2mm gap all around. I get a ridiculous figure when I try to calculate how high the water level in the tank must rise in the tank to accommodate 10L of water in a 2mm gap.

10L of water, distributed in a 4m x 2mm area works out to a ridiculous height of 125cm. That is obviously stupid because it means a 10kg addition makes the boat-and-water-level rise up in the container by 1.25m.

Where is the flaw in my logic?

 

[Gordianus]

I've encountered a blip in my engineering idea.

Say our boat is a cube-esque block length 1m, width 1m, height 1.1m.
It weighs 1T, and so will float with .1m out of the water.

As a control, we float it in a calm pool and mark off where the waterline is, approximately .1m from the top.

And to prove it is indeed floating (this will become important later), we drop a 10kg cinder block on top. This causes the boat to sink into the pool by an additional 1cm, i.e. it is now floating a mere 9cm out of the water, our mark has sunk 1cm below the waterline. (The water level in the pool does not rise noticeably with the addition of 10kg.)

Agreed so far?

Now we want to demonstrate the exact same thing in our boat-hugging container. The container is 1.004m on a side, leaving a 2mm gap all around. It is as tall as we need it to be.

We drop the boat in and it floats to the exact same mark as above. So far so good.

Now for the pièce de resistance, we must prove that the boat is indeed floating above the bottom of the container (that it will rise and sink freely with only a change in weight).

We drop the 10kg cinder block on top and the boat sinks into the water by 1cm, just as before.

But wait - the water level in the container does not rise a mere 1cm, as before, it rockets out of the gap and climbs much higher because the 10L of water that the boat is now displacing must squeeze into the 2mm gap all around. It get a ridiculous figure when I try to calculate how high the water level in the tank must rise in the tank to accommodate 10L of water in a 2mm gap.

10L of water, distributed in a 4m x 2mm area works out to a ridiculous height of 125cm. That is obviously stupid because it means a 10kg addition makes the boat-and-water-level rise up in the container by 1.25m.

Where is the flaw in my logic?

I think the block sinks 1 cm when it floats in veeeeeeery large container (a lake). In your example when the water level rises aprox 1 cm, the pressure grows to the amount we need to keep the 1ton+10 kG object floating.

 

[Borek]

Where is the flaw in my logic?

Is there one? If I understand you correctly you are surprised by the fact that 1L volume can occupy 10x10x10 cube or hundred times longer 1x1x1000 cuboid.

 

[DaveC426913]

Is there one?

Yes. If my logic is not flawed, the inevitable conclusion is that the act of dropping a 10kg cinder block on top of the boat would cause the boat-and-water-level-together to rise in the container by 125cm.

That is a ridiculous outcome.

Question: when the cinder block pushes down on the boat, the boat displaces an additional 10L of water. Where does that displaced 10L of water go? If it simply seeps up the 4m x 2mm gap, it would have to seep upward by 125cm.

 

[AlephZero]

There is nothing wrong with the experiment, but your logic is wrong.

When you add the 10kg weight in the lake, then measured relative the the bottom of the lake, the boat sinks (1-x) cm and the water rises x cm, where x is very small.

In the tank, measured relative the bottom of the tank, the boat sinks x cm and the water rises (1-x) cm where x is very small.

The level of the water relative to the boat changes by the same amount (1cm) in both situations.

 

[DaveC426913]

The level of the water relative to the boat changes by the same amount in both situations.

Absolutely. Which I too state, and have no qualms with. Regardless of the water level in the tank, the boat will float 9cm above the water and 101cm below it.

My issue is with the water level in the tank.

measured relative the the bottom of the lake, the boat sinks (1-x) cm and the water rises x cm, where x is very small.

The displaced 10L is distributed throughout the lake. Lake water level rises a tiny amount. No prob.

In the tank, measured relative the bottom of the tank, the boat sinks x cm and the water rises (1-x) cm where x is very small.

The displaced 10L is distributed into only a tiny gap, resulting in a calculated water level rise in the tank of 125cm. (Gap is 4.016m x 2mm = 80cm^2. A 10L volume on an 80cm area must be 125cm in height.) The inevitable conclusion is that the water level in the tank rises 125cm.This is not just intuitively ridiculous, it actually results in a paradox. (How can the water level in the tank rise 125cm? That would mean the boat (which is happily remaining floating 9cm out of the water and 101cm into the water at all times) is now physically sitting 125cm higher above the bottom of the container than it was before - meaning there's room for 1,250L of water underneath it - meaning it can't sit that high.)

I know it's wrong, I just can't spot it.

 

[DaveC426913]

Reexaming what you said instead of my own flawed logic. I know you're right (because I know that's the way it it really does work), I just haven't made the leap of logic as to why yet.

When you add the 10kg weight in the lake, then measured relative the the bottom of the lake, the boat sinks (1-x) cm and the water rises x cm, where x is very small.

In the tank, measured relative the bottom of the tank, the boat sinks x cm and the water rises (1-x) cm where x is very small.

Why do you treat these scenarios differently? Why does the same math not apply?

Is 10L of water not displaced from under the boat? That's a specific volume that has to go somewhere.

 

[maimonides]

Displacement (as has been mentioned) does not mean actual water transported to some place, but refers to the volume of the floating body below the water line. In the experiment discussed a very small amount of water makes a big change in displacement.

 

[AlephZero]

Why do you treat these scenarios differently? Why does the same math not apply?

It is the same math. I just renamed the variables, so "x" was the small quantity in both cases. If you prefer, call the movement of the boat down and water surface up b and w, both measured relative to the bottom, and where where b+w = 1cm. The relative size of b and w depends on the relative surface areas of the boat and the water.

Is 10L of water not displaced from under the boat? That's a specific volume that has to go somewhere.

I think that is the key point of your mistake. "The displacement increases by 10L" doesn't say anything directly about where the water goes. What it means is "The volume of the boat which is below the water level increases by 10L". In the lake, you need to move very nearly 10L of water to make that happen. In the tank, you only need to move a small amount of water to get the same effect.

 

[Per Oni]

This causes the boat to sink into the pool by an additional 1cm,

Where is the flaw in my logic?

To add to what other are saying:

The water will rise by that amount. The boat doesn’t sink that amount.

2mm gap x 4 lengths of 1 meter (approx. container size ) is a surface area of 8x10^-3 m^2. The volume of the rising water is that amount times the height of rising water (1 cm) which gives 8x10^-5 m^3. This is the volume of water the boat has to displace. Therefore it needs to sink 8x10^-5m^3 / 1m^2 (bottom area) = 8x10^-5m. Your 2mm gap all round should be enough.

I hope this makes sense.

 

[DaveC426913]

I think that is the key point of your mistake. "The displacement increases by 10L" doesn't say anything directly about where the water goes. What it means is "The volume of the boat which is below the water level increases by 10L". In the lake, you need to move very nearly 10L of water to make that happen. In the tank, you only need to move a small amount of water to get the same effect.

Hm. I concede, though I do not quite yet get the logic.

Counter-intuitively the movement of a mere 4m x 2mm x 1cm of (80mL/80g) of water is enough to support a 10kg load.

 

[nasu]

This is maybe because the image of "water displaced" from Archimedes' law is very strong in our minds. It is not the displacement or the movement of water that matters. Water displaced is just an easy to remember calculation tool.
The increased pressure is what matters. To support the 10 kg (100 N) the pressure over the 1 m^2 area needs to increase by 100 Pa or 1 cm of water. The block goes down in water until the level of the water relative to the bottom of the block increases by 1 cm. In the given conditions this will not displace 10 L of water (the original block did not have to displace/move 1000 L of water to float).

 

[DaveC426913]

(the original block did not have to displace/move 1000 L of water to float).

You're right!

The solution to the paradox is ... the experiment itself!

Part A of the experiment just demonstrated that a paltry bucketful of water is all that needs to be displaced in order to float a 1T mass. Why did I suddenly forget that when it came to part B!

 

[DaveC426913]

So back to the engineering component.

Trying to figure out:

- best shape
- - rectangle? cylinder? cone?
- - - which is easiest to make (with appropriate precision)
- - - which is least likely to present technical problems during the experiment (eg. cylinder, rectangle could tilt, bind, cone is self-calibrating)

- best manufacturing procedure
- - make inner shell, cover in wax, paint on outer shell, remove wax (somehow)
- - make outer shell, cover in wax, paint on inner shell, remove wax (somehow)

- test procedure
- - put water in first then inner shell
- - put shells together then pour water into gap

- scale, cost
- - 1 cubic metre / 1 tonne mass (too small? too big?)
- - 1mm gap / 5mm gap (impossible precision versus doesn't prove the point)
- - cost (assuming I did this myself)
- - - I've got a winch and a barn, a good start

Mostly, I'm trying figure out what to build the inner shell out of so that it's easy to make but does not flex even 1 mm.

 

[chingel]

Thank you all for the answers.

Another problem I have is that if I apply a 1kg per cm2 pressure through a small piston to a closed tank, why does the same pressure get applied to every cm2 of the tank? Isn't the only way that the force can transfer because the water molecules bump into each other and apply the force. But if one water molecule pushes two water molecules, shouldn't the force be divided by the two of them?

In the case of a solid block, the force gets divided by the area of contact at the other end. What is different with a liquid? The water molecules themselves are solid, and shouldn't the force transfer like in the case of solid objects? I can understand why the force goes in all directions in a liquid, because water molecules aren't exactly on top of each other and they are allowed to roll out to the sides if something pushes them at a slight angle.

 

[DaveC426913]

Another problem I have

chingel, please start a new thread.

 

[AlephZero]

Hm. I concede, though I do not quite yet get the logic.

Counter-intuitively the movement of a mere 4m x 2mm x 1cm of (80mL/80g) of water is enough to support a 10kg load.

I agree, Pascal's law of hydrostatics (pressure varies with depth of fluid, independent of the shape of the container) is counterintuitive till you "get used to it".

Here's another scenario that might help. Suppose the boat is the lake and resting on the bottom, but just at the point of floating off. Now add 10kg to the weight of the boat. To bring it back to be just on the point of floating, you have to add enough water to raise the level of the whole lake by 1 cm.

In the tank, the physics of situation is exactly the same, but you need a much smaller amount of water to raise the level by 1 cm.

I think you could make a first version of the experiment with something as simple as a glass aquarium tank and a suitable sized watertight box (e.g. make a wooden box and cover the outside with plastic film). Weight the box with sand so it floats at a suitable depth. With a glass tank, you can (literally) see what is happening.

 

[Per Oni]

Counter-intuitively the movement of a mere 4m x 2mm x 1cm of (80mL/80g) of water is enough to support a 10kg load.

That’s the principle of a hydraulic ram.

In the end it all comes down to energy. For example a miserly 1 joule of energy can in principle lift a vast amount of mass. Not very high but it will lift. Of course pretty soon we get into practical troubles but in theory it can be done.

 

[nasu]

That’s the principle of a hydraulic ram.

Not really. The buoyancy is not involved in the functioning of the hydraulic press (if this is what you have in mind).

 

[DaveC426913]

I agree, Pascal's law of hydrostatics (pressure varies with depth of fluid, independent of the shape of the container) is counterintuitive till you "get used to it".

No, as an owner of swimming pools and fish tanks, I'm comfortable with the concept that pressure varies with depth, not with shape or area.

Here's another scenario that might help.

I got it with nasu's explanation. When the water level in the tank rises 1cm yet the boat does not, that means the boat is now displacing 10 more litres of water. As he points out, it is not essential for 10L of displaced water to actually go anywhere.

We just proved that in the part A of the experiment, where we started with naught but a bucket of water in the bottom of a giant vat, and dropped a cubic metre vessel into it. There was no cubic metre of water to move in the first place, so it's not like we had to actually move a cubic metre of water to get the boat to float. The whole point of the entire experiment is that we only need to move a tiny amount of water to effect* buoyancy. I had just forgotten that, by the time I got to Part B is all.
*No, I did not use the wrong word.

 

[DaveC426913]

I think you could make a first version of the experiment with something as simple as a glass aquarium tank and a suitable sized watertight box (e.g. make a wooden box and cover the outside with plastic film). Weight the box with sand so it floats at a suitable depth. With a glass tank, you can (literally) see what is happening.

This is what I want to do, except most ideas I have for the inner vessel will experience deformation under pressure. We're talking < mm tolerances here for the experiment to work.

 

[AlephZero]

If you want to everything to stay "rigid", you have two basic choices. Either you do a proper stress analysis, or you just overdesign everything.

For a small scale experiment, I'm assuming a glass aquarium tank isn't going to deform much under the water pressure, otherwise the glass would crack. You could check that assumption with a measuring rod cut to fit accurately inside the tank when it is empty, and check the amount of clearance when it is full.

I don't think an aquarium-sized wooden box made from say 1/2'' or 1'' thick wood will deform much either, so long as you keep the wood dry.

Designing experiments is often a matter of trial and error. Starting with something simple and cheap to find out what is really important in practice is often a good strategy. Often the biggest problems are caused by something that nobody had thought about.