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This question already has an answer here:

Given a battleship, suppose we construct a tub with exactly the same shape as the hull of the battleship, but 3 cm larger. We fill the tub with just enough water to equal the volume of space between the hull and the tub. Now, we very carefully lower the battleship into the tub.

Does the battleship float in the tub?

I tried it with two large glass bowls, and the inner bowl seemed to float. But if the battleship floats, doesn't this contradict what we learned in school? Archimedes' principle says "Any floating object displaces its own weight of fluid." Surely the battleship weighs far more than the small amount of water that it would displace in the tub.

Note: I originally specified the tub to be just 1 mm larger in every direction, but I figured you would probably tell me when a layer of fluid becomes so thin, buoyancy is overtaken by surface tension, cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy alone.

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marked as duplicate by John Rennie, Jon Custer, Qmechanic Dec 17 '18 at 17:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Sink the ship an extra 0.5mm off equilibrium. This requires displacing Ax0.5mm of water with an equal volume of ship of smaller density. Buoyancy is opposing it. $\endgroup$ – Cosmas Zachos Dec 16 '18 at 15:37
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    $\begingroup$ If you've only started with just enough water to equal the volume it would displace, you've already done the 'displacement'. Fill the tub, then put it in; now either go find a mop or run around naked screaming. $\endgroup$ – Mazura Dec 16 '18 at 20:08
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    $\begingroup$ Possible duplicate of Can a ship float in a (big) bathtub? $\endgroup$ – Kamil Maciorowski Dec 17 '18 at 5:59
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    $\begingroup$ Similar to an idea my father (a construction engineer) told me when I was a kid, and I could not believe it: If the Hoover Dam had just a vertical sheet of water behind it, a few inches thick (and then a rock face or whatever), but the same water height, it would have to be exactly as strong as it is now, holding all of Lake Mead. $\endgroup$ – Peter A. Schneider Dec 17 '18 at 11:51
  • $\begingroup$ You can search for 'tight dry dock' on google images for examples; liners and tankers are closer in shape to the dock so may only have a metre or less around them. $\endgroup$ – Pete Kirkham Dec 17 '18 at 13:40
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The question you ask is a good one, and a common one. The phrasing of Achimedes' principle you use is most apropos in situations where you have a large body of water compared to the size of the vessel, so that you don't noticeably raise the water level.

The full mathematical phrasing of the principle holds true in all of these situations.

If your boat meaningfully increases the water level, like it must in your example, you have to play some mental gymnastics with the definition of "displace" to arrive at the correct answer. That's not to say that Archimedes' Principle is wrong, just that the informal phrasing gets a bit tricky to apply. The formal version, involving moving water upwards against the pull of gravity, does not run into any such complications.

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Yes it floats. And it has displaced its "own weight of water" in the sense that if you had filled the container with water and only then lowered the ship into the container, nearly all that water would have been dispaced and is now sloshing around on the floor.

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    $\begingroup$ Thanks, I understand what you're saying: If we filled the tub all the way up to the top and then lowered in the battleship, then a full battleship of water would have to be displaced, which would have spilled over the top. But in the scenario I describe in my question, that's not what happens. Very little water is actually displaced, and yet the battleship still ends up floating. Maybe I'm being too pedantic or literal about the word "displaced"... $\endgroup$ – SlowMagic Dec 16 '18 at 14:57
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    $\begingroup$ @SlowMagic Fundamentally the buoyant force occurs because the pressure of the water pushing up the ship on the bottom is higher than the pressure of the air pushing it down at the top. The "water displaced" thing is just a mnemonic for the result of this calculation, not the law itself. $\endgroup$ – knzhou Dec 16 '18 at 15:38
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    $\begingroup$ @SlowMagic that "Any floating object displaces its own weight of fluid" is more an analogy than an anything distantly resembling an accurate definition. Physics tells you that if you submerge something with $V$ volume in water, then a force of $F = \rho_{w} g h$ will push it upwards ($\rho_{w}$ being the density of water, $g$ the gravitational acceleration). Which, by chance, is the same force Earth would exert on $V$ volume of water. The displacement does not have to happen at all. $\endgroup$ – Neinstein Dec 16 '18 at 15:38
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    $\begingroup$ @SlowMagic Another way to think about it: fill the tub, letting water slosh onto the floor so the battleship is floating in the tub. Remove the battleship. Note how much water is in the tub (and clean up the rest on the floor). At some interminably point in the hypothetical futurepast, reconstruct the bathtub with this much water in it. Add battleship. No water will slosh over the edge, but the ship will still float even though no water was "displaced outside the tub." The overslosh is just that our container was too small to hold the displacement, not that water had to slosh over the edge. $\endgroup$ – Draco18s Dec 16 '18 at 18:12
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    $\begingroup$ @Neinstein I wouldn't say it is by chance this happens. The buoyant force is directly related to how much water is displaced. You can see this by thinking about the buoyant force on a "stationary" section of water in a larger body of water. Since there is no net force acting on this water, it must be that the buoyant force is equal to this water weight. $\endgroup$ – Aaron Stevens Dec 16 '18 at 20:01
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Note that the water does not need to have been present - this calculation gives just the way to calculate the non water volume occupied by the floating object (that hence is unavailable for water).

So in your example, assume you put the ship into the container. If you filled it up to the same water level without the ship being present - that is the displacement caused by the ship.

 BBB
wBBBw     wwwww     w   w
wBBBw vs. wwwww NOT w   w
wwwww     wwwww     wwwww

In this sketch, 6 w have been "displaced".

Hence your example works even with a tiny gap. You can argue that hydraulic machines work with 0 gaps. The piston is the boat in the cylinder.

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  • $\begingroup$ I like the ASCII graphic... However, the hydraulic machines are a bad analogy imho: They work with dynamic pressure supplied by a pump rather than relying on the hydrostatic pressure that makes a ship float. They also rely on the impossibility of the liquid of escaping, which allows them to operate overhead. Don't try that with a boat! $\endgroup$ – cmaster Dec 16 '18 at 21:38
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The USS Missouri $5.8 \times 10^7\,\rm kg, \, 270\,\rm m$ long with a fully laden draft of $11.5\,\rm m$ has an underwater surface area in excess of $270\times 11.5\times 2 \approx 6200\,\rm m^2$ and needs to "displace" $5.8 \times 10^7\,\rm kg$ of salt water (density $\approx 1020 \,\rm kg \, m^{-3}$) to float.

Assume a custom made tank so that an even thickness of water (total volume $1 \,\rm litre = 0.001 \,m^{-3}$) surrounds the USS Missouri below its waterline.
This thickness of the water layer would be smaller than $\frac{0.001}{6200} \approx 1.6 \times 10^{-7} \rm m$.

So in theory possible but in practice very highly unlikely.


The OP has changed the title from "1 litre" to "a small amount of water".
All that needs to be done is to choose a volume of water such that it is practically possible to float the USS Missouri in a suitably shaped dock and the OP's layer of water 3 cm thick might be possible in practice?

The picture in this answer gives a flavour of an "apparent lack of water" being able to float a ship.

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  • $\begingroup$ I edited the question title so as not to get fixated on a specific amount of water. I just mean, a very small amount of water, much less than the weight of the battleship. $\endgroup$ – SlowMagic Dec 16 '18 at 14:58
  • $\begingroup$ Also, the image isn't fair, because a huge amount of water is displaced, it's just that it's almost entirely dispersed along the entirety of the water for quite a ways in both directions, rather than just directly around the boat. $\endgroup$ – Mooing Duck Dec 17 '18 at 5:54
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The issue is just in your "definition" of displaced. When we say "the buoyant force is equal to the weight of the displaced fluid" (which is more true than it seems people are saying it is), displaced does not mean "how much fluid overflows out of our container" (unless we started with a full container).

The displaced fluid really just means how much fluid is pushed out of the way. What this leads to is that whatever volume of the object is submerged under the fluid surface, this is the volume of the fluid displaced. If we were to calculate the weight of this volume of water, we would find that it is equal to the buoyant force exerted on the object.

Therefore, in your example, if the volume of the boat that is submerged would give a volume of water that weighs the same as the boat, then the boat will float. How you get to this final configuration is irrelevant.

As a counter example to using the idea of water spilling out of a container, just imagine a boat in the ocean, where no water is spilling out of a container, yet the boat still floats.

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Like other answers have said, the issue comes from an incorrect understanding of "displaced".

Imagine you lie in a bathtub. Now you fill up the bath until it covers your legs and body (not your head). Your body is very slightly less dense than water, so you verrrry slightly float - maybe only a few mm, not much, and probably still touching the sides and bottom of the tub.

There was never any water where your body is. The water filled the tub around you.

Then, the "displaced water" is the water that would have been in the tub, if your body vanished and enough water was added to keep the surface level of the water at the same level.

What Archimedes is saying us that, if the tub was so big that you floated (without moving) and you weren't touching the sides or bottom, then your body's level in the water is set at the level which displaces the same mass of water. That "missing" water doesn't need to have been there originally (see definition above).

Going in the other direction, imagine I trim the bathtub until it's1/2 an inch from your body in all directions, and you're still floating. The amount of water is tiny. What matters is (as above) the among of "missing" water where your body is, that would be below the surface level of the water. Not the remaining water around you in the bath.

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  • $\begingroup$ On the "trim the bathtub" bit: Imagine an idealised cubic battleship floating in a shallow dock, 3cm from the bottom. Then, slide walls in 3cm from each edge, and weld/seal them to be water tight. Is the Battleship still floating? Next, remove the Battleship - how much water is left in your new box? Then, gently replace the Battleship - does it resume floating? $\endgroup$ – Chronocidal Dec 17 '18 at 10:19
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For the interested, here is a little illustration I made once, to defend someone called 'Marilyn vos Savant'.
I didn't believe what people wrote in there against her statement, about a battleship floating in a bucket of water.
I am very happy to see, that everybody in here knows what is right.
Here is the illustration:

Battleship

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  • $\begingroup$ Nice illustration! $\endgroup$ – SlowMagic Dec 23 '18 at 17:53
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As the battleship is placed in the giant tub, the water along the sides rises to a height, h. This produce a pressure = water density x 9.8 x h in the thin layer of water beneath the ship. This pressure x the area of the bottom of the ship equals the weight of the ship, so it floats. If, instead of a giant tub, we place the ship in the ocean, h is achieved not by the water rising, but the ship descending to an equilibrium depth (hopefully less than the distance between the ship’s bottom and the deck).

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