Can a battleship float in a tiny amount of water?
up vote
3
down vote
favorite
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.
buoyancy
edited 21 mins ago
knzhou
40.1k10113194
asked 2 hours ago
SlowMagic
433
add a comment |
up vote
3
down vote
favorite
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.
buoyancy
edited 21 mins ago
knzhou
40.1k10113194
asked 2 hours 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
48 mins ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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.
buoyancy
edited 21 mins ago
knzhou
40.1k10113194
asked 2 hours ago
SlowMagic
433
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.
buoyancy
buoyancy
edited 21 mins ago
knzhou
40.1k10113194
asked 2 hours ago
SlowMagic
433
edited 21 mins ago
knzhou
40.1k10113194
asked 2 hours ago
SlowMagic
433
edited 21 mins ago
knzhou
40.1k10113194
edited 21 mins ago
knzhou
40.1k10113194
edited 21 mins ago
knzhou
40.1k10113194
40.1k10113194
asked 2 hours ago
SlowMagic
433
asked 2 hours ago
SlowMagic
433
asked 2 hours 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
48 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
48 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
48 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
48 mins ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
9
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 2 hours ago
mike stone
5,9271121
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
47 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
46 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
44 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
1 hour ago
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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active
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active
oldest
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up vote
9
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 2 hours ago
mike stone
5,9271121
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
47 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
46 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
44 mins ago
add a comment |
up vote
9
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 2 hours ago
mike stone
5,9271121
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
47 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
46 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
44 mins ago
add a comment |
up vote
9
down vote
up vote
9
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 2 hours ago
mike stone
5,9271121
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 2 hours ago
mike stone
5,9271121
answered 2 hours ago
mike stone
5,9271121
answered 2 hours ago
mike stone
5,9271121
answered 2 hours ago
mike stone
5,9271121
5,9271121
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
47 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
46 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
44 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
47 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
46 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
44 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
47 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
47 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
46 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
46 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
44 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
44 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
1 hour 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
1 hour 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 1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
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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
48 mins ago