Floating a cruise ship in a bucket of water

[Rene Dekker]

First of all, a surface is expressed in $m^2$ and not $m^3$ .

Thanks for pointing out that typo.

But I haven’t understood your determination of the wetted surface area (WSA) of the hull .

I took the simplification that the underside of the ship is simply rectangular. We are talking about an imaginary theoretical exercise anyways, so there is no point in discussing all kinds of practical details. We just need an idea of what order of magnitude we are talking about.

Anyway , the point isn’t this. Do you think that the external hull surface is as smooth as a mirror? It isn’t. You cannot speak of a water layer of some micrometers, uniformly distributed on the whole wetted surface. Have you ever examined the hull surface from a distance , say, of 20 cm ? The surface is not uniform at that level you are considering.

We are talking about an imaginary theoretical exercise anyways, so there is no point in discussing all kinds of practical details. And I stated that the shape of the "bucket" will have to follow the surface of the ship (whatever shape it has) to an accuracy of 1 micrometer.

Moreover, the Archimede’s principle , which we can define really a “law of nature” , is a matter of global hydrostatics, doesn’t take into consideration molecular forces of adhesion and cohesion forces of water. This law can be summarised in the vector equation :

$$\ vec\ P + \ vec\ S = 0 $$

nothing more.

Fully agree with that, Archimede's principle does not take into consideration molecular forces. And 2.3 micrometer is definitely a scale where molecular forces must be taken into consideration.

 

[Charles Link]

The vector formula I believe you are referring to above, (the Latex seems to be missing a couple things), with force(s) per unit volume is $-\nabla P-\delta g \hat{z}=0$, for hydrostatic equilibrium. When integrated over a volume, along with a version of Gauss' law for the gradient, the result is Archimedes principle.

 

[gleem]

What keeps a boat afloat? The force of the water pressure on the bottom surface. What is the pressure depend on? The distance from the surface. The height of the surface of the water above the lowest part of the boat. If you come up underneath a floating boat with a container of arbitrary size but large enough to fit the boat and bring the container top edge to the surface of the water the boat remains buoyant and will float in that container. If you continue to raise the container nothing in the container change, the boat is still floating. If the container is made to fit the hull shape to arbitrary close tolerances then the boat can float in a lot less water than it can displace.

A boat float on its waterline no matter how much water is under or around it.

 

[hutchphd]

absolutely

Here's my method:

 

[italicus]

The vector formula I believe you are referring to above, (the Latex seems to be missing a couple things), with force(s) per unit volume is $-\nabla P-\delta g \hat{z}=0$, for hydrostatic equilibrium. When integrated over a volume, along with a version of Gauss' law for the gradient, the result is Archimedes principle.

Thank you Charles, I am not able with Latex ! The letter “S” I wrote is the initial of the Italian word “ Spinta” ( push), which stays for the Archimede’s force that makes equilibrium to weight P .

 

[italicus]

A boat float on its waterline no matter how much water is under or around it.

Absolutely agreed.

 

[Charles Link]

Thank you Charles, I am not able with Latex ! The letter “S” I wrote is the initial of the Italian word “ Spinta” ( push), which stays for the Archimede’s force that makes equilibrium to weight P .

I'm using $P$ for pressure, $\delta$ for the density of water, and $g$ for the gravitational constant.

 

[haruspex]

Archimede’s principle , which we can define really a “law of nature”

the Archimede’s force that makes equilibrium to weight P .

I would caution against such a view of Archimedes' Principle.

The buoyancy comes from the integral of the upward component of the fluid's pressure over the surface in contact with the fluid. Archimedes' insight, I assume, is that if we replace the body by more of the fluid occupying the same displacement it would be in equilibrium, so the buoyancy force on that would be equal to its weight.

But note that there is a hidden assumption in equating that to the buoyancy force on the object: that the fluid is able to reach all parts of the body's surface below the surface level of the fluid. For example, the Principle breaks down for a rubber suction cap stuck to the bottom of a tank.

 

[italicus]

I would caution against such a view of Archimedes' Principle.

The buoyancy comes from the integral of the upward component of the fluid's pressure over the surface in contact with the fluid. Archimedes' insight, I assume, is that if we replace the body by more of the fluid occupying the same displacement it would be in equilibrium, so the buoyancy force on that would be equal to its weight.

Obvious!

But note that there is a hidden assumption in equating that to the buoyancy force on the object: that the fluid is able to reach all parts of the body's surface below the surface level of the fluid. For example, the Principle breaks down for a rubber suction cap stuck to the bottom of a tank.

Could you explain better your example, please? Maybe I haven’t understood it!
As a marine engineer and naval architect, with 40 years of experience in the field, I’ve always found the buoyancy by $\rho g V$ , where V is the immersed volume of a floating body, or even a totally submerged body ( think of a submarine for example) , that displaces water.

One moment...do you mean something like this in the sketch?

but it is obvious , the surface of the cup inside the tank is to be considered “part “ of the tank bottom, no component of water pressure on it is directed towards the surface of tha water!

 

[haruspex]

but it is obvious , the surface of the cup inside the tank is to be considered “part “ of the tank bottom

You can choose to consider it that way, but if you blindly apply Archimedes' Principle to the cap alone it does not work. Hence the need for caution.

 

[italicus]

You can choose to consider it that way, but if you blindly apply Archimedes' Principle to the cap alone it does not work. Hence the need for caution.

Believe me, I have never been so blind...In any case, caution is often useful or even necessary, in all circumstances of science and technology. I agree on that.

 

[Charles Link]

To expound upon what @haruspex is describing, consider a very flat glass plate that is completely submerged, where it will experience a buoyant force by Archimedes equal to the weight of the water of the same volume as the glass plate. Then consider what occurs if you press it against a very flat bottom as it is submerged in a deep tank. It will be stuck to the bottom with a tremendous pressure from the water above, but, as there is no water underneath it to counter, the buoyant force given by Archimedes simply does not hold for this second case.

 

[hutchphd]

as there is no water underneath it to counter

I would say this a little differently: as there is no possibility of water influx (thereby communicating the external pressure)...
Seems a clearer statement to me and some clarity appears to be required.

 

[haruspex]

Believe me, I have never been so blind...In any case, caution is often useful or even necessary, in all circumstances of science and technology. I agree on that.

The difficulty is that Archimedes' Principle is generally expressed as "the buoyant force is equal to the weight of the fluid displaced". I never see it hedged around with any caveat. Your phrasing in post #70 sought to elevate it to a universal law, but in that form it is not.
You might not make the mistake of applying it where it is invalid, but I have seen students do so on this very forum.

 

[Charles Link]

Keeping post 82 in mind, and looking back on post 64, there will be some thickness (very thin layer) of water $d$ that we lack data for, where the air pressure on the boat from above will not be countered or transferred to the water below the boat, and for this case we can ask whether the boat is truly floating. This makes it into a somewhat nebulous problem, and it may be beyond the scope of what @DaveC426913 is looking for.

 

[italicus]

The difficulty is that Archimedes' Principle is generally expressed as "the buoyant force is equal to the weight of the fluid displaced". I never see it hedged around with any caveat. Your phrasing in post #70 sought to elevate it to a universal law, but in that form it is not.
You might not make the mistake of applying it where it is invalid, but I have seen students do so on this very forum.

Dear haruspex
I am not a student! I have sometimes taught this subject to them, and still consider Archimedes’ principle a law of nature, when free floating bodies are concerned. The rubber cup you have proposed isn’t a free floating body, no?
But a more difficult example which I was used to make is the following: consider a heavy rubber ball, (but lighter than an equal volume of water) , kept under water and tied by a rope to the bottom of a tank at rest on Earth What is the tension in the rope?

 

[haruspex]

still consider Archimedes’ principle a law of nature, when free floating bodies are concerned

1. The principle as commonly stated does not specify free-floating
2. In fact, it is not limited to free-floating. It works perfectly well for a body resting on the bottom provided the fluid can reach under it.
3. Ergo, if you wish to teach it well, mention that proviso.

 

[italicus]

1. The principle as commonly stated does not specify free-floating
2. In fact, it is not limited to free-floating. It works perfectly well for a body resting on the bottom provided the fluid can reach under it.
3. Ergo, if you wish to teach it well, mention that proviso.

I have made an example where the ball is tied by a rope to the bottom of a tank, three forces act on the ball, one of which is Archimedes’ force. But what is your position? Mine is that this principle is a law of nature, Archimede simply discovered it, didn’t invent, when playing with the king’s crown in his bathtube, according to the legend…
If a cubic block of steel rests on the bottom of a pool, a thin layer of water underneath is enough to give a hydrostatic force upwards. But you really know nothing about how many contact points are between the two surfaces, which aren’t as smooth as a mirror, and how many small areas are truly supported by water: you cannot really speak of a uniform water layer, so what are we talking about?

You have to “weld” the block to the bottom, or seal accurately the perimeter of contact to prevent water entering underneath, so that Archimedes’ law isn’t applicable.
If a ship is moored to a berth by means of ropes, one should consider also the forces exerted by the ropes, to determine a correct ship’s weight, but this way takes to incorrect evaluation; better, one has to release ropes and read drafts; but the water level isn’t really stable around a draft mark , you can have excursions of 10 cm or more, due to sea and ship’s motion. So you have to use the famous “grain of salt” , which is another way to mean experience. The mean draft of the ship gives you the displacement, its accuracy is as good as possible, everybody in the shipping world knows that.

But I think we have repeated this concept of Archimedes’ law too many times, nothing more needs to be added.

 

[Rene Dekker]

To expound upon what @haruspex is describing, consider a very flat glass plate that is completely submerged, where it will experience a buoyant force by Archimedes equal to the weight of the water of the same volume as the glass plate. Then consider what occurs if you press it against a very flat bottom as it is submerged in a deep tank. It will be stuck to the bottom with a tremendous pressure from the water above, but, as there is no water underneath it to counter, the buoyant force given by Archimedes simply does not hold for this second case.

I would judge that in this case, where the water cannot get beneath the object, there is still the same buoyancy force, equal to the weight of the displaced water. But there is an extra downwards force on the object, due to the pressure difference of the water above the object with the near-vacuum beneath the object.

For example, if we measure the force that the object exerts on the floor, then that is still equal to the weight of the object minus the weight of the displaced water. That is, the object still tries to pull the floor up with the buoyancy force. It is just that the floor is able to counteract that force to keep the object down.

 

[italicus]

I would judge that in this case, where the water cannot get beneath the object, there is still the same buoyancy force, equal to the weight of the displaced water. But there is an extra downwards force on the object, due to the pressure difference of the water above the object with the near-vacuum beneath the object.

Be careful. If the water cannot get beneath the object, because the object is “welded” or accurately sealed around the contact perimeter, there is no buoyancy force on the object! That was the first example by Haruspex, the one of the rubber cup in a hole of the bottom tank. No buoyancy force means that the object’s weight is entirely supported by the bottom tank.

For example, if we measure the force that the object exerts on the floor, then that is still equal to the weight of the object minus the weight of the displaced water. That is, the object still tries to pull the floor up with the buoyancy force. It is just that the floor is able to counteract that force to keep the object down.

No, read above. The floor ( or the tank bottom) doesn’t have to counteract any buoyant force to keep the object down. For clarity, we are speaking of objects heavier than an equal volume of water, which therefore tend to sink when left free.
Instead, consider the example that I have posted in #86 , of a rubber ball tied to the bottom by means of a rope, and evaluate the tension in the rope. There are 3 forces acting on the ball , which are in equilibrium.

 

[Charles Link]

Be careful. If the water cannot get beneath the object, because the object is “welded” or accurately sealed around the contact perimeter, there is no buoyancy force on the object! That was the first example by Haruspex, the one of the rubber cup in a hole of the bottom tank. No buoyancy force means that the object’s weight is entirely supported by the bottom tank.

Yes. In addition, the object can be "plastered" against the bottom by the pressure forces from the above water plus air pressure, acting downward, with no upward force to balance or counter. This happens with a suction cup or even a glass plate, if we pull upwards on it. If we pull the suction cup from the center, (or even the glass plate), we can pull it away from the bottom surface just enough, that the bottom surface is no longer supplying an upward force. There still is no upward buoyancy force, and we find we need to pull very, very hard, before we finally free the suction, and then have an upward buoyancy force. When the upward buoyancy force is absent or reduced, e.g. with the boat, we need to ask whether it is truly floating.

 

[DaveC426913]

the object can be "plastered" against the bottom by the pressure forces from the above water plus air pressure, acting downward, with no upward force to balance or counter. This happens with a suction cup or even a glass plate, if we pull upwards on it. If we pull the suction cup from the center, (or even the glass plate), we can pull it away from the bottom surface just enough, that the bottom surface is no longer supplying an upward force.

Be that as it may, I am not convinced that this scenario is applicable to the proposed scenario. I think it's a faulty analogy.

 

[Charles Link]

Be that as it may, I am not convinced that this scenario is applicable to the proposed scenario. I think it's a faulty analogy.

When you theoretically create a layer of water that is on the order of microns or nanometers in thickness, it is likely, (we seem to lack sufficient data here), that the buoyant forces from underneath will be greatly reduced.

 

[italicus]

Yes. In addition, the object can be "plastered" against the bottom by the pressure forces from the above water plus air pressure, acting downward, with no upward force to balance or counter. This happens with a suction cup or even a glass plate, if we pull upwards on it. If we pull the suction cup from the center, (or even the glass plate), we can pull it away from the bottom surface just enough, that the bottom surface is no longer supplying an upward force. There still is no upward buoyancy force, and we find we need to pull very, very hard, before we finally free the suction, and then have an upward buoyancy force. When the upward buoyancy force is absent or reduced, e.g. with the boat, we need to ask whether it is truly floating.

If air is all around the tank, above and below, you can refer to relative pressure only, because air pressure above the surface is equal to air pressure below, so they compensate.

Here is an exercise taken from Cengel-Cimbala : Fluid mechanics fundamentals ; a car is at the bottom of a lake...

 

[hutchphd]

When you theoretically create a layer of water that is on the order of microns or nanometers in thickness, it is likely, (we seem to lack sufficient data here), that the buoyant forces from underneath will be greatly reduced.

Yes. As I mentioned before there is obviously some scale at which water is no longer a 3D fluid. But the mean free path of a water molecule at STP is few angstroms, so a nanometer is probably the correct scale. This of course assumes that the surface is true to that scale (roughly the precision to which telescope mirrors are ground).
We are discussing angels on a pin. Not a bad question, but I'm out.

 

[DaveC426913]

When you theoretically create a layer of water that is on the order of microns or nanometers in thickness, it is likely, (we seem to lack sufficient data here), that the buoyant forces from underneath will be greatly reduced.

This is simply restating your claim, and it's a red herring. Whether or not it's true, it isn't necessarily germane to the specific scenario at-hand.

An object that is submerged and stuck to the bottom of a tank is not analogous to an object that is partially emmersed* and neutrally buoyant.

* I am sure this is a word. The internet disagrees.

 

[haruspex]

emmersed*

It's one "m", emersed, as with emerge.

 

[DaveC426913]

It's one "m", emersed, as with emerge.

Yeah. That was what I first typed. But PF's spellcheck didn't like that, so I tried variations, which it didn't like either. Tx.

 

[hutchphd]

Never knew the word emerse ! Yay new word. @DaveC426913 Were you looking for "immerse" ?

 

[Charles Link]

* I am sure this is a word.

I think the word is "immersed". :)

 

[DaveC426913]

No. Emersed - as in: some of the boat is above the waterline.

 

[haruspex]

I think the word is "immersed". :)

Different word.
Emerse or emersed: Adjective, rising above the surface (similar to sense of emergent)
Immerge: Rare, verb, transitive or intransitive, plunge below the surface
Immerse: Verb, transitive only, plunge below the surface

It seems there is no verb "to emerse".

So apart from the spelling, @DaveC426913's usage was correct.

 

[DaveC426913]

So apart from the spelling...

I promise I spelled it right at first. (One of my peccadillos is spelling and verbiage.) The spellchecker made me second-guess my convictions.

 

[hutchphd]

It seems there is no verb "to emerse".

I found the same thing...
Which is very weird. Apparently one can immerse an object but not emerse it afterword. That would explain drowning...

 

[Rene Dekker]

For clarity, we are speaking of objects heavier than an equal volume of water, which therefore tend to sink when left free.
Instead, consider the example that I have posted in #86 , of a rubber ball tied to the bottom by means of a rope, and evaluate the tension in the rope. There are 3 forces acting on the ball , which are in equilibrium.

I wasn't, I was speaking of objects that are lighter than water, that are held to the floor through a water-tight seal. Although it does not matter for the principle. Your rubber ball example shows exactly what I was pointing out. There is a buoyancy force on the ball, that is counteracted by the force in the string tying it down to the floor. The same is true for an object that is held to the floor with a water tight seal. There is the same buoyancy force on the object, that is counteracted by a force from the floor. The same is true for an object that is welded to the floor. There is the same buoyancy force on the object, that is counteracted by the force through the weld.
In all cases, the buoyancy force is still there, it does not matter whether there is water under the object. It does not suddenly disappear when you replace the rope with a weld, or with a vacuum.

You can (in principle) measure that counteracting force, if the floor is somewhat movable, and is attached to a force gauge. Setting the force gauge to zero when the object is not there, and then placing the object, and read what the gauge says. It will give the same value for all three cases (assuming all three objects have the same weight).