Can a battleship float in a very tiny amount of water?
up vote
3
down vote
favorite
Okay, we have a battleship.
We construct a tub having exactly the same shape as the hull of the battleship, only 3 cm larger in every direction.
We fill the tub with just enough water to equal the calculated 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? Or does the battleship sink to the bottom of 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? The principle of flotation 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 other physical principles, such as surface tension, surface cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy and not some other physical principle.
buoyancy
asked 1 hour ago
SlowMagic
433
add a comment |
up vote
3
down vote
favorite
Okay, we have a battleship.
We construct a tub having exactly the same shape as the hull of the battleship, only 3 cm larger in every direction.
We fill the tub with just enough water to equal the calculated 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? Or does the battleship sink to the bottom of 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? The principle of flotation 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 other physical principles, such as surface tension, surface cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy and not some other physical principle.
buoyancy
asked 1 hour ago
SlowMagic
433
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.
– Cosmas Zachos
20 mins ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Okay, we have a battleship.
We construct a tub having exactly the same shape as the hull of the battleship, only 3 cm larger in every direction.
We fill the tub with just enough water to equal the calculated 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? Or does the battleship sink to the bottom of 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? The principle of flotation 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 other physical principles, such as surface tension, surface cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy and not some other physical principle.
buoyancy
asked 1 hour ago
SlowMagic
433
Okay, we have a battleship.
We construct a tub having exactly the same shape as the hull of the battleship, only 3 cm larger in every direction.
We fill the tub with just enough water to equal the calculated 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? Or does the battleship sink to the bottom of 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? The principle of flotation 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 other physical principles, such as surface tension, surface cohesion, adhesion, hydroplaning, or whatever. I wanted this to be a question about buoyancy and not some other physical principle.
buoyancy
buoyancy
asked 1 hour ago
SlowMagic
433
asked 1 hour ago
SlowMagic
433
edited 1 hour ago
asked 1 hour ago
SlowMagic
433
asked 1 hour ago
SlowMagic
433
asked 1 hour ago
SlowMagic
433
433
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.
– Cosmas Zachos
20 mins ago
add a comment |
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.
– Cosmas Zachos
20 mins ago
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.
– Cosmas Zachos
20 mins ago
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.
– Cosmas Zachos
20 mins ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
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.
answered 1 hour ago
mike stone
5,9171121
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"...
– SlowMagic
1 hour ago
@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.
– knzhou
19 mins ago
@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.
– Neinstein
19 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
add a comment |
up vote
2
down vote
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 $270times 11.5times 2 approx 6200,rm m^2$ 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.
answered 1 hour ago
Farcher
46.8k33693
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.
– SlowMagic
59 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448673%2fcan-a-battleship-float-in-a-very-tiny-amount-of-water%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
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.
answered 1 hour ago
mike stone
5,9171121
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"...
– SlowMagic
1 hour ago
@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.
– knzhou
19 mins ago
@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.
– Neinstein
19 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
add a comment |
up vote
8
down vote
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.
answered 1 hour ago
mike stone
5,9171121
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"...
– SlowMagic
1 hour ago
@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.
– knzhou
19 mins ago
@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.
– Neinstein
19 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
add a comment |
up vote
8
down vote
up vote
8
down vote
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.
answered 1 hour ago
mike stone
5,9171121
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.
answered 1 hour ago
mike stone
5,9171121
answered 1 hour ago
mike stone
5,9171121
answered 1 hour ago
mike stone
5,9171121
answered 1 hour ago
mike stone
5,9171121
5,9171121
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"...
– SlowMagic
1 hour ago
@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.
– knzhou
19 mins ago
@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.
– Neinstein
19 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
add a comment |
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"...
– SlowMagic
1 hour ago
@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.
– knzhou
19 mins ago
@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.
– Neinstein
19 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
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"...
– SlowMagic
1 hour ago
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"...
– SlowMagic
1 hour ago
@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.
– knzhou
19 mins ago
@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.
– knzhou
19 mins ago
@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.
– Neinstein
19 mins ago
@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.
– Neinstein
19 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
@SlowMagic A better replacement for "water displaced" would be the longer phrase "the volume of the object whose height is less than the highest point of the water", which works here. But this doesn't work in all situations either. The only thing that always works is to turn on your brain and use $F = ma$ from scratch.
– knzhou
17 mins ago
add a comment |
up vote
2
down vote
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 $270times 11.5times 2 approx 6200,rm m^2$ 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.
answered 1 hour ago
Farcher
46.8k33693
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.
– SlowMagic
59 mins ago
add a comment |
up vote
2
down vote
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 $270times 11.5times 2 approx 6200,rm m^2$ 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.
answered 1 hour ago
Farcher
46.8k33693
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.
– SlowMagic
59 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
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 $270times 11.5times 2 approx 6200,rm m^2$ 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.
answered 1 hour ago
Farcher
46.8k33693
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 $270times 11.5times 2 approx 6200,rm m^2$ 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.
answered 1 hour ago
Farcher
46.8k33693
edited 41 mins ago
answered 1 hour ago
Farcher
46.8k33693
answered 1 hour ago
Farcher
46.8k33693
answered 1 hour ago
Farcher
46.8k33693
46.8k33693
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.
– SlowMagic
59 mins ago
add a comment |
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.
– SlowMagic
59 mins ago
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.
– SlowMagic
59 mins ago
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.
– SlowMagic
59 mins ago
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448673%2fcan-a-battleship-float-in-a-very-tiny-amount-of-water%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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.
– Cosmas Zachos
20 mins ago