I can my self move any megalithic stone on hundreds of tons with physics

[J.C.MacSwell]

Mistermack is correct. You don't need to "dis-place" the same mass if the container is just a little larger than the object, you just have "displace" it in terms of taking up the space with the right fluid head in place around it.

10 hours ago, swansont said:

That violates Archimedes principle: an object will displace its own weight of the fluid. If it is less dense it will float, BUT there needs to be enough of the fluid present for this to happen. Otherwise it will bottom out. So 500 kg of mercury (a little less than 100L) will not float anything over 500 kg. exchemist notes this above.

Might want to rethink that.

Think of putting a block of hardwood in a tub that's dimensioned slightly larger than the block. Add a bit of water and you will float it.

If the tub was full of water from the start you would displace the weight in water as the water overflows but you get the same final result.

Edited June 30, 2022 by J.C.MacSwell

[swansont]

4 hours ago, sethoflagos said:

I see this as a false dichotomy. Hydraulic head is hydraulic head whether it is generated statically by an elevated reservoir, or dynamically by some form of pump. The body of fluid physically engaged in the lift doesn't see any difference in the two. 

The OP employs a statically generated head via the depth of mercury in the canal. But this is by the by. Somewhere along the line mercury had to be raised to the level of the canal surface in order to fill it. There's always a pump in there somewhere to provide the initial input energy. 

So where is this pump, in the OP? How do you lift the rock? All there is is the static pressure, as far as I can tell.

4 hours ago, mistermack said:

The pressure is supplied by the head of mercury, which only has to be one fifth of the way up the side of the rock, because it's five times as dense. It just boils down the the pressure exerted on the bottom surface of the rock, and that is purely down to the depth of mercury. It doesn't matter if it's open or closed, the pressure on the bottom is just a result in the depth of mercury. 

And if it’s not that deep?

1 hour ago, J.C.MacSwell said:

Mistermack is correct. You don't need to "dis-place" the same mass if the container is just a little larger than the object, you just have "displace" it in terms of taking up the space with the right fluid head in place around it.

Might want to rethink that.

Think of putting a block of hardwood in a tub that's dimensioned slightly larger than the block. Add a bit of water and you will float it.

If the tub was full of water from the start you would displace the weight in water as the water overflows but you get the same final result.

Just tried it. Unsurprisingly it didn’t work. The wood bottomed out.

Why do you think the wood would displace less than its mass?

[sethoflagos]

1 hour ago, swansont said:

How do you lift the rock? All there is is the static pressure, as far as I can tell.

Static pressure is all you need.

A 1 metre thick slab of rock exerts of the order 3,000 kgf per m^2 on the surface supporting it.

Mercury could provide that supporti at a depth of 3,000/13,600 = 0.221 m

If the mercury surface level is >0.221 m above the bottom of the channel, the rock floats.

Increase the mercury level, the rock slab rises accordingly.

1 hour ago, swansont said:

So where is this pump, in the OP?

The pump wasn't introduced in response to the OP, it was introduced in response to

7 hours ago, swansont said:

Except that in hydraulics you exert a force via a source of external pressure which is lacking here, so it's really not.  

Hence my suggestion that if a rock is being raised, an external source of energy input is required. The OP seems to imply it may be via a bloke with a bucket (which I tend to view as a low technology pump anyway). 

Edited July 1, 2022 by sethoflagos
Concatenation

[J.C.MacSwell]

1 hour ago, swansont said:

 

Just tried it. Unsurprisingly it didn’t work. The wood bottomed out.

Why do you think the wood would displace less than its mass?

Try wood with a density lass than 1.0. I assure you it will float.

Are you sure your tub was bigger in each dimension?

Here's an experiment we've surely all done:

Try filling a tub with water, freezing it, then letting it melt a bit. The ice will float well before half the ice is melted, therefore requiring less mass of water than the ice to make it float. (Or does the ice stay at the bottom where you live, until it knows the by the size of the container that it's allowed up?)

2 hours ago, swansont said:

 

Why do you think the wood would displace less than its mass?

Displace in this context doesn't mean you actually need the fluid to be present.

14 hours ago, swansont said:

That violates Archimedes principle: an object will displace its own weight of the fluid. If it is less dense it will float, BUT there needs to be enough of the fluid present for this to happen. Otherwise it will bottom out. So 500 kg of mercury (a little less than 100L) will not float anything over 500 kg. exchemist notes this above.

There certainly needs to be enough, but enough can be significantly less than the mass of the object it makes buoyant.

Edited July 1, 2022 by J.C.MacSwell

[mistermack]

2 hours ago, swansont said:

And if it’s not that deep?

I'll try a different approach. Imagine the trench is 20 feet long and 2 feet deep. You fill it with mercury to a depth of 1 foot. Your rock slab is just a little narrower than the trench, and it's four foot thick and four feet long. You place it in the trench, and it will float, because it will only sink to a fifth of it's height, as the mercury is five times as dense. So it sinks less than a foot, and doesn't touch the bottom. 

So there you have your block, floating in the trench. Now, if you block off the trench in front of it, and behind it, do you claim that that will make it sink? Of course it doesn't, it will continue to float. So there you have the block floating in very small amount of mercury, just as in the OP. 

Edited July 1, 2022 by mistermack

[TheVat]

I don't understand the resistance to Archimedes principle here.

Pressure increases with depth below the surface of a liquid. Any object with a non-zero vertical depth will have different pressures on its top and bottom, with the pressure on the bottom being greater. This difference in pressure causes the upward buoyancy force.  You need a lot of mercury to get that pressure differential to be sufficient.  An object will displace its own weight of mercury, which means you need a ton of it.  Listen to the guy who did an experiment at home.

[mistermack]

1 minute ago, TheVat said:

An object will displace its own weight of mercury, which means you need a ton of it.

No, that's where the mistake is being made. The rock IS displacing a ton of mercury, because if you were to remove the rock, you would need a ton of mercury to fill the void that it left behind. Imagine the mercury was in grain form, like sand. You remove the rock, you need a ton of mercury to fill the hole that it left. Thats how, in the close fitting container, Archimedes principle still applies. Your rock IS displacing it's own weight of mercury.

[TheVat]

 Say Hg has an SG of 12, and the rock 2.0.  You will need a ton of Hg to float a ton of rock, so you need at least one-sixth the volume of the rock in mercury, your flotation fluid.  That is your minimum required volume of a fluid that's six times denser than rock.  If your fluid had a SG of 4, then you would need fluid with half the volume of the rock, at minimum.   All the close fitting container does is make it difficult to add enough fluid volume to float the rock.  Your rock will bottom out.

[J.C.MacSwell]

Archimedes is turning in his grave...

A 20 cm x 20 cm x 10 cm high  (approx 3.6 kg )block of ice will float in your bath tub..due to the pressure difference on the bottom from that of the top, which will be about 1 cm above the surface of the water.

Now make a smaller tank 21 cm x 21 cm x 11 cm high.  Add 1 litre of water (approx 1 kg). Water level approx 2.27 cm from bottom.

Now carefully place into it the above 3.6 kg block, into the tank with the 1 kg of water. Water level will rise.

The water level will be the same as for 4.6 litres of water, about 10.43 cm from the bottom, and the block will float about 1 cm above that, and with the top surface above the top of the tank. 

1. It will float

2. It won't go to the bottom.

3. Only 1 litre of liquid water present

4. Archimedes would describe this as displacing 3.6 kg of water

5. Don't make me get him up

 

 

 

Edited July 1, 2022 by J.C.MacSwell

[sethoflagos]

14 minutes ago, TheVat said:

 Say Hg has an SG of 12, and the rock 2.0.  You will need a ton of Hg to float a ton of rock, so you need at least one-sixth the volume of the rock in mercury, your flotation fluid.  That is your minimum required volume of a fluid that's six times denser than rock.  If your fluid had a SG of 4, then you would need fluid with half the volume of the rock, at minimum.   All the close fitting container does is make it difficult to add enough fluid volume to float the rock.  Your rock will bottom out.

No. Hydraulic pressure is determined by the depth of the liquid column, irrespective of its cross-section.

So a 2" depth of mercury will float a 12" depth of rock (using your figures).  Even if the mercury connects to the atmosphere via a pinhole.

However, the rise of the rock vs the fall of the mercury level is in proportion to their respective cross-sectional areas. This ratio is also called the 'mechanical advantage' which can permit say 1 kg of mercury to lift say 1 tonne of rock.  

[J.C.MacSwell]

2 hours ago, TheVat said:

I don't understand the resistance to Archimedes principle here.

 

There is none.

2 hours ago, TheVat said:

Pressure increases with depth below the surface of a liquid. Any object with a non-zero vertical depth will have different pressures on its top and bottom, with the pressure on the bottom being greater. This difference in pressure causes the upward buoyancy force.

Good so far...

 

2 hours ago, TheVat said:

You need a lot of mercury to get that pressure differential to be sufficient. 

You need sufficient mercury to get that pressure differential to be sufficient, and the smaller the container the less required.

2 hours ago, TheVat said:

An object will displace its own weight of mercury, which means you need a ton of it. 

Wrong. That (the bold) is not what that means.

 

2 hours ago, TheVat said:

Listen to the guy who did an experiment at home.

Do you really think he tried it?

[exchemist]

4 hours ago, TheVat said:

 Say Hg has an SG of 12, and the rock 2.0.  You will need a ton of Hg to float a ton of rock, so you need at least one-sixth the volume of the rock in mercury, your flotation fluid.  That is your minimum required volume of a fluid that's six times denser than rock.  If your fluid had a SG of 4, then you would need fluid with half the volume of the rock, at minimum.   All the close fitting container does is make it difficult to add enough fluid volume to float the rock.  Your rock will bottom out.

I initially thought this, but I now think it is wrong. One has to consider carefully what "displace" means in this context. It means the object to be floated has to occupy a volume below the surface of the fluid which, if it were filled with the fluid, would have a weight equal to its own weight. It does not mean you have to have that volume of fluid present initially and then physically move it out of the way.

The upthrust required to float the object results from the pressure exerted at depth. So it results from the height of the column of fluid exerting the pressure, integrated over the dimensions of the object.

(Hg has a density of 13.6, by the way - a number familiar to anyone who has worked with manometers 😉.)   

Edited July 1, 2022 by exchemist

[swansont]

11 hours ago, sethoflagos said:

Static pressure is all you need.

A 1 metre thick slab of rock exerts of the order 3,000 kgf per m^2 on the surface supporting it.

Mercury could provide that supporti at a depth of 3,000/13,600 = 0.221 m

If the mercury surface level is >0.221 m above the bottom of the channel, the rock floats.

Increase the mercury level, the rock slab rises accordingly.

"Add more mercury" is a new item The OP had a fixed amount. There is no more mercury to add.

9 hours ago, mistermack said:

I'll try a different approach. Imagine the trench is 20 feet long and 2 feet deep. You fill it with mercury to a depth of 1 foot. Your rock slab is just a little narrower than the trench, and it's four foot thick and four feet long. You place it in the trench, and it will float, because it will only sink to a fifth of it's height, as the mercury is five times as dense. So it sinks less than a foot, and doesn't touch the bottom. 

It sinks less than a foot because the weight of the rock is less than the mercury it displaces. So if the rock is heavier, this is not the case.

So you are giving a specific case where this works, and my objection is that there are conditions where it won't, specifically the scenario where  "it would be theoretically possible to RAISE a rock, in a tight fitting container, with less than it's own weight of mercury, using the head of mercury to exert the required hydraulic pressure on the base of the rock" the problem being that this won't work under the parameters of the OP, since a tight fitting container is not the described scenario, and as such, you would violate Archimedes principle. As I stated.

IOW, your counterexample is not one to which I was voicing an objection.

3 hours ago, exchemist said:

Hg has a density of 13.6, by the way

Yeah, my bad on this in a previous post - I googled it and saw a different number. Doesn't change the overall issue, though, just the numbers.

11 hours ago, J.C.MacSwell said:

Try wood with a density lass than 1.0. I assure you it will float.

Let's say I have a 100cm tall pole with a 1 cm^2 cross section and I put it (long side up) in water, 1 cm deep. The mass of the water is, at most, 1 gram. The mass of the wood is, say 50 grams. You are assuring me it will float. Why does Archimedes principle fail in this case?

If your assertion were true, ships/boats would never run aground.

[J.C.MacSwell]

1 hour ago, swansont said:

Let's say I have a 100cm tall pole with a 1 cm^2 cross section and I put it (long side up) in water, 1 cm deep. The mass of the water is, at most, 1 gram. The mass of the wood is, say 50 grams. You are assuring me it will float. Why does Archimedes principle fail in this case?

If your assertion were true, ships/boats would never run aground.

Ships generally don't run aground in water deeper than their maximum depth

14 hours ago, J.C.MacSwell said:

Might want to rethink that.

Think of putting a block of hardwood in a tub that's dimensioned slightly larger than the block. Add a bit of water and you will float it.

 

11 hours ago, J.C.MacSwell said:

 

Are you sure your tub was bigger in each dimension?

 

I really thought I had given you enough to recognize your assertion was wrong.

On 6/30/2022 at 8:03 AM, swansont said:

That violates Archimedes principle: an object will displace its own weight of the fluid. If it is less dense it will float, BUT there needs to be enough of the fluid present for this to happen. Otherwise it will bottom out. So 500 kg of mercury (a little less than 100L) will not float anything over 500 kg. exchemist notes this above.

Have you recognized it yet?

There is no theoretical minimum of liquid require to float an object...you just need the right container.

[swansont]

33 minutes ago, J.C.MacSwell said:

I really thought I had given you enough to recognize your assertion was wrong.

Will making the walls taller do anything in my problem? Please explain how that makes any difference.

[J.C.MacSwell]

Now, we can consider scenarios of stability, density, shape...lack of gravity...or just recognize your assertion was not correct and let Archimedes rest in peace. He knew the displaced liquid did not need to be present. He doesn't need to wake up and rethink it.

10 minutes ago, swansont said:

Will making the walls taller do anything in my problem? Please explain how that makes any difference.

It will allow a greater pressure head to act on the bottom of the object, after sufficient liquid, which can be well under that of the mass of the object, is added.

Edited July 1, 2022 by J.C.MacSwell

[swansont]

Just now, J.C.MacSwell said:

Now, you can consider scenarios of stability, density, shape...lack of gravity...or just recognize your assertion was not correct and let Archimedes rest in peace. He knew the displaced liquid did not need to be present. He doesn't need to wake up and rethink it.

Stability was not a criterion. You keep adding caveats.

I'm pretty sure Archimedes needed a fluid. His tub was not empty when he shouted, "Eureka!"

Explain to my why taller walls will not make the wood bottom out under the scenario I described. Where's the physics?

3 minutes ago, J.C.MacSwell said:

 

It will allow a greater pressure head to act on the bottom of the object, after sufficient liquid, which can be well under the mass of the object, is added.

Sufficient liquid? You said "Add a bit of water" - there was no threshold.

 

47 minutes ago, J.C.MacSwell said:

Ships generally don't run aground in water deeper than their maximum depth

No, and that's irrelevant. But they do run aground in shallow water, which is my point. If your assertion of "add a bit of water" was true, then they would not run aground.

[J.C.MacSwell]

8 minutes ago, swansont said:

 

Sufficient liquid? You said "Add a bit of water" - there was no threshold.

Your claim was that you needed at least as much as the mass of the object.

My bad for assuming you would immediately recognize that was not the case , after a brief thought. 

15 hours ago, J.C.MacSwell said:

 

Think of putting a block of hardwood in a tub that's dimensioned slightly larger than the block. Add a bit of water and you will float it.

 

 

Apologies if I made you pull out a block of wood, stick it in a container, and add a bit of water to no avail.

30 minutes ago, swansont said:

I'm pretty sure Archimedes needed a fluid. His tub was not empty when he shouted, "Eureka!"

We're supposed to stand on the shoulders of giants to get further...not to drown them in the tub.

[TheVat]

5 hours ago, exchemist said:

I initially thought this, but I now think it is wrong. One has to consider carefully what "displace" means in this context. It means the object to be floated has to occupy a volume below the surface of the fluid which, if it were filled with the fluid, would have a weight equal to its own weight. It does not mean you have to have that volume of fluid present initially and then physically move it out of the way.

The upthrust required to float the object results from the pressure exerted at depth. So it results from the height of the column of fluid exerting the pressure, integrated over the dimensions of the object.

(Hg has a density of 13.6, by the way - a number familiar to anyone who has worked with manometers 😉.)   

I just made something up on the density, for simple math.  I will think about your explanation - am having minor surgery (basal cell carcinoma lopped off) in about an hour, so hope to have a clearer head after that. 

Not sure I entirely follow @sethoflagos equivalency between buoyant force and hydraulic pressure, either.   Will also try to revisit that. 

Please, everyone, wear hats, slather sunblock, follow the "10 and 4" rule if you can.  

[studiot]

I really can't see the interest in this idea. Cladking's hydraulic system at least made mechanical sense and could have been implemented with the technology of the time in ancient Egypt.

If you are going to make a containing channel strong enough to contain the pressures involved in supporting a 100 (remember the OP actually said hundreds) tons block you have to ask how would neolithic Man have constructed it ?
It would surely have been a more onerous task then making the block itself.
Having consrtucted the channel how would the stone have been lifted in and out ?

And of course how would they have made 180 miles of channel ?

[mistermack]

1 hour ago, swansont said:

Where's the physics?

The physics is remarkably simple. It's only clouded by the meaning of the word "displace". 

The rock will float purely and simply when the pressure on it's lower surface reaches a certain level P. The pressure in the liquid is purely down to the depth D. If the rock is placed in an open pond, it will float when the bottom is at depth D. 

Explain how and why the pressure on the bottom is not exactly the same, at depth D, in a close fitting container as described. 

[J.C.MacSwell]

5 minutes ago, mistermack said:

The physics is remarkably simple. It's only clouded by the meaning of the word "displace". 

Exactly. As per my first post:

17 hours ago, J.C.MacSwell said:

Mistermack is correct. You don't need to "dis-place" the same mass if the container is just a little larger than the object, you just have "displace" it in terms of taking up the space with the right fluid head in place around it.

 

I'm sure I could have stated it better, but I really thought that would be sufficient for most here.

[exchemist]

1 hour ago, TheVat said:

I just made something up on the density, for simple math.  I will think about your explanation - am having minor surgery (basal cell carcinoma lopped off) in about an hour, so hope to have a clearer head after that. 

Not sure I entirely follow @sethoflagos equivalency between buoyant force and hydraulic pressure, either.   Will also try to revisit that. 

Please, everyone, wear hats, slather sunblock, follow the "10 and 4" rule if you can.  

Oh, my commiserations. Hope it goes well and they get it all out.

I dimly remember a derivation of Archimedes' principle at school, based on calculating the upthrust, from pressure at a point at given depth, due to a column of liquid above, multiplied by the area in cross-section, parallel to the surface, of the object to be floated. This gives you a force equal to the weight of fluid occupied by the object.  So I think all you need is a column of fluid of sufficient depth to make a "hole" that it big enough. It's the hole that does it, rather than the fluid that would be in it.  

 

[swansont]

1 hour ago, mistermack said:

Explain how and why the pressure on the bottom is not exactly the same, at depth D, in a close fitting container as described. 

I am talking about the scenario of the OP. I am not, and have not been, addressing any of the variants with new variables that others have introduced.

Quote

 

The physics is remarkably simple. It's only clouded by the meaning of the word "displace". 

The rock will float purely and simply when the pressure on it's lower surface reaches a certain level P. The pressure in the liquid is purely down to the depth D. If the rock is placed in an open pond, it will float when the bottom is at depth D.

 

 You need to define your variables. What is D?

The pressure in the fluid (above atmosphere) will be pgh, where p is the density and h is the depth of the fluid. 

In my example h is 10^-2 m.

The pressure will be about 100 Pa. But your area is 10^-4 m^2. How are you lifting 5 N with that?

 

2 hours ago, J.C.MacSwell said:

Your claim was that you needed at least as much as the mass of the object.

My bad for assuming you would immediately recognize that was not the case , after a brief thought. 

Here's my brief thought: Archimedes principle says that the buoyancy force is equal to the weight of the liquid displaced. If you have a smaller mass displaced, the weight will be greater than the buoyancy force, and the object will sink.

Now it's your turn: back up your claim. Where does the force come from to float the object?

[jlivingstonsg]

-------------------------

Please learn some physics.

 

https://www.quora.com/Is-it-possible-to-float-a-100-ton-boat-in-just-1-ton-of-water/answer/Kim-Aaron?ch=10&oid=303591043&share=33e3ccd4&target_type=answer

 

Master of Science in Engineering Physics