Talk:Buoyancy

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This article was externally reviewed on December 14, 2005 by Nature. It was found to have 2 errors.

This article about buoyancy REALLY ain't cool. Shouldn't there be something about the difference between static and dynamic buoyancy etc? Big B.

I'm troubled by the opening couple of lines.

"If the buoyancy exceeds the weight, then the object will rise; if the weight exceeds the buoyancy, the object will sink. If the buoyancy equals the weight, the body has neutral buoyancy and may remain at its level."

All of this is true only if the buoyancy and weight are the only two vertical forces acting. In practice, there can be many other forces. There is also a problem that rises can convey the idea of constant velocity rather than acceleration. Shouldn't we say something like:

The buoyancy provides an upward force on the object. According to Newton's first law of motion, if the upward forces (including the buoyancy) balance the downward forces (including the weight) the object will remain at rest. Otherwise, it will accelerate upwards or downwards.

In the most common case, where there are no other forces acting on an object, it will float at the level where it displaces a weight of fluid equal to its own weight. If the object is more dense than the fluid, it cannot displace enough weight of fluid to balance its own weight and will not float.

Stuart White 09:15, 1 January 2006 (UTC)

I have restructured the article a bit, and whilst doing so, tidied up the physics in the introduction. I also amended Newtons to newtons, to match SI conventions, and made some minor edits to the phraseology elsewhere.

Stuart White 09:21, 2 January 2006 (UTC)

The introduction states that buoyancy can be found in fluids. as much as i know, buoyancy is also found in air, as, for example, smoke rises or drops down depending on the weight of the air around it. maybe someone can correct either me or the article?

Thanks, --Abdull 15:13, 18 Dec 2004 (UTC)

Air is a fluid. Any liquid or gas is a fluid; a fluid is a substance that flows. Michael Hardy 00:09, 19 Dec 2004 (UTC)

On a little side note below Archimedes and his discovery of buoyancy, I provided the reference to the ancient Chinese child prodigy Cao Chong, and his application of earlier Archimedes' principle of buoyancy. I also edited Cao Chong's article a bit (explaining that no official treatise in ancient China had been written about the buoyancy principle, yet several empirical observations were recorded throughout time). In relation to the story of Hieron's crown, Joseph Needham points to a section of the Zhou Li (the Gao Gong Ji section) which would suggest the Chinese during the Han Dynasty (202 BC - 220 AD) were aware of the same principle. He quotes it, stating: "The workers called Li make measures of capacity (liang). They purify (separately) by successive heatings samples of metal (presumably copper) and tin, until there is no further loss of weight. Then they weigh them."

--PericlesofAthens 19:59, 18 March 2007 (UTC)

Contents 1 Mechanics 2 SI units 3 Citation Question 4 Nature claims errors 5 Errors to fix 6 I removed the "scale" example. 7 about archimedes 8 Someone confused about density section 9 Archimedes discovered buoyancy in his bathtub?? 10 True-ish but not relevant 11 Remove diagram 12 Simple explanation 13 Boyancy force disappears? 14 Provide Proof of Archimedes' Principle 15 dumbing down / folk story 16 Image requests

[edit] Mechanics

It would be nice to have something about the mechanics of buoyancy in here... what exactly is causing the force? It's not very clear.

If an object has a relatively low density, meaning the volume is large and the mass is small -- as compared to the fluid that surrounds the object, then the fluid will attempt to displace the object. This page states clearly, http://theory.uwinnipeg.ca/physics/fluids/node10.html, that the mass of displaced fluid is equivalent to the net pressure acting on the object. --24.131.137.3 15:53, 21 December 2005 (UTC)

That's like saying "because Archimedes said so". Why does the fluid attempt to displace the object? What exactly is happening at microscopic level?

---

I am striking the following: "It is obvious that without taking the displaced fluid element into account, energy would not be conserved during the buoyant motion of an object as it would gain both potential and kinetic energy when rising in the fluid." This statement is incorrect. As a buoyant object ascends in a fluid, it loses potential energy, and has zero potential energy at the surface. Energy is conserved. --24.131.137.3 14:42, 21 December 2005 (UTC)

I think the text was referring to the gain of gravitational potential energy which is in fact gained. It is "compensated" for by considering the loss of gravitational potential energy by the fluid that shifts downward as the object shifts upward. Nonetheless, it think the this "energetic" stuff is probably too fine a point for a general article. So I'm OK with it being removed. I do have to wonder if this is one of the "error" that Nature found. Blimpguy 16:10, 21 December 2005 (UTC)

The text was indeed referring to the gravitational potential energy. I can say this because somebody took this passage (as well as the paragraph about acceleration) literally off my webpage http://www.physicsmyths.org.uk/buoyancy.htm (I changed the latter now to include 'gravitational').

The point is that the total energy of the system is

grav.pot. energy of the body + grav.pot. energy of the fluid + kinetic energy of the body + kinetic energy of the fluid

This means that if the body gains grav.pot. energy as well as kinetic energy by rising, the grav.pot. energy of the fluid must decrease accordingly. Otherwise there can be no energy conservation. So there was actually no error here with the original statement.

Thomas

[edit] SI units

Changed the example's units from pounds to Newtons to maintain the use of SI on Wikipedia.

It is conventional in English to write "3 pounds of water" when one means a quantity of water that weighs 3 pounds. But does one follow that usage with Newtons? Michael Hardy 21:48, 12 Apr 2005 (UTC)

Newtons are a unit of force and therefore are a unit of weight, I believe. I'm sure it's correct, but whether or not it's commonly accepted is a bit dubious.--Ryan Hardy 03:30, 15 Apr 2005 (UTC)

It is not conventional to refer to Newtons of water, but then, one asks, why does one normally measure out peas or potatoes by the pound or kilogram, which can be considered (in violation of current standards) force? The answer is that one is measuring, or ought to be measuring mass; weight is being used as a surrogate. If a spring scale is used, one is achieving a measure that depends on the local acceleration of gravity, but if a balance is used, one measures mass. The bouyancy discussion is really almost independent of the local acceleration of gravity, unless that somehow varies grossly over the experimental region. That never holds in common experience with buoyancy, although the variations of gravity and centrifugal force do affect the geoid. When you speak of such and so many Newtons of water, you refer to a mass that depends on local gravity. It is probably better to couch everything as much as possible in terms of masses, or of densities and volumes.

Pdn 05:39, 15 Apr 2005 (UTC)

But buoyant force is a force, nonetheless, and cannot be measured in kilograms. And, since you have to use subtraction attain that measure of force, the units of measure cannot be changed during the operation(kg-kg ≠ kg∙m/s²). Algebraically, it would make sense to use the term "weight" in the context of force.

If Newtons are unsuitable, then is the kilogram-force a possible option?--Ryan Hardy 21:40, 15 Apr 2005 (UTC)

Newtons are SI units. A Kilogram a unit of mass. A Kilogram-force is a force that a body of 1 kg mass exerts under gravity and is approximately 10*1=10 newtons. But newtons are a better option as they are the Internationally accepted. Kilograms are also used to measure force assuming the gravitational force is the same everywhere om earth.--Jetru 11:59, August 23, 2005 (UTC)

[edit] Citation Question

how do i cite this?

Can you express your question a bit more completely? Cite what exactly? Where?

I think he wants to cite this wikipedia article. --Phinnaeus • T+Σ 11:40, 2 November 2005 (UTC)

There's a page that explains how to cite Wikipedia articles; I'll see if I can find it. Michael Hardy 21:25, 22 December 2005 (UTC)

Just be careful if you're doing an academic citation. That can be problematic, as a lot of academics don't consider Wikipedia reputable. -Phoenixrod 21:21, 19 August 2006 (UTC)

[edit] Nature claims errors

Nature disputes the accuracy of this article; see http://www.nature.com/news/2005/051212/multimedia/438900a_m1.html and Wikipedia:External_peer_review#Nature. We're hoping they will provide a list of the alleged errors soon. —Steven G. Johnson 01:47, 15 December 2005 (UTC)

Indeed. Disputing the accuracy is not useful in this case without specificity. If it said "buoyancy is the opposite of girlancy", I'd know just how to correct the matter. But Steven, don't you have a PhD in physics? And isn't this freshman-level (or secondary school-level) stuff? If so, I'd think you could straighten out this short article in two minutes (maybe). Michael Hardy 01:52, 15 December 2005 (UTC)

[edit] Errors to fix

Archimedes Principle Reviewer: Prof. Timothy J. Pedley, G. I. Taylor Professor of Fluid Dynamics, University of Cambridge, UK.

• In the section on acceleration and energy, which discusses how a body moves when it is not neutrally buoyant, it is rightly stated that the acceleration of a body experiencing a non-zero net force is not the same as in a vacuum, because some of the surrounding fluid has to be accelerated as well. However, it is implied that the mass of fluid that has to be added to that of the body, in using Newton's Law to calculate the acceleration, is equal to the mass of fluid displaced. This is not in general true - for example, the added mass for an immersed sphere is half the mass of fluid displaced.

This is in fact incorrect. You seem to be bringing in the drag of the object here (which obviously would depend on its shape). However, the drag is velocity dependent and has nothing to do with the buoyant gravitational acceleration as such. It merely determines the limiting velocity of the buoyant object. The initial acceleration on the other hand depends only on the mass of the object and the displaced fluid and is given by a=g *(m - m_d)/(m + m_d) (see my webpage http://www.physicsmyths.org.uk/buoyancy.htm from which much of the acceleration paragraph was lifted anyway (not by me)). Regarding the 'drag' issue, you may also be interested in the page http://www.physicsmyths.org.uk/drag.htm ).

Thomas

• The entry is rather imprecise. In line 3, for example, the object is said to "float" if the buoyancy exceeds the weight, so here "float" must mean "rise" and not "stay at the same level", which is probably not what was intended because the word has the other meaning in the second paragraph of the section on "Density".

[edit] I removed the "scale" example.

Buoyancy is about hydrostatics. In the more complex subject of hydrodynamics, you can try to separate out the inertial and viscous forces involved (ignoring turbulence, etc.), but by the time you present the "scale example" (ignoring mass of scale arms, etc., etc.) and then try to explain why the acceleration of the assembly of connected objects that form the "scale system" is F=ma where m is the mass of the object plus one half the mass of the virtual (and presumably spherical) "fluid element", you have lost 90% the readership. You might as well just refer them to the Reynolds number, but probably you should not do either. -- Pinktulip 03:20, 2 January 2006 (UTC)

You shouldn't bring drag into the buoyancy issue which has nothing to do with it. Buoyancy is merely due to the mechanical coupling of the object with the fluid and the fact that both are subject to gravity, and in this sense the 'scale' example is a good analogon to illustrate what goes on when an object rises in a fluid. It was actually (like some other passages in the article) lifted by somebody off my webpage http://www.physicsmyths.org.uk/buoyancy.htm . As shown there, the force equation is actually F= (m+m_d)*a where m is the mass of the object and m_d that of the displaced fluid.

Thomas

[edit] about archimedes

nextime plz.. see the solution of the an exercises..

[edit] Someone confused about density section

User:24.6.14.204 apparently was confused about that section. I don't see how to make it clearer. TimBentley (talk) 17:50, 3 April 2006 (UTC)

[edit] Archimedes discovered buoyancy in his bathtub??

I have doubts about the statement that Archimedes discovered what is called Archimedes' principle in his bathtub. I've always heard that it was in his bathtub that he realized how he could ascertain whether a goldsmith's workmanship included embezzlement of some of the gold. It was suspected that the smith put in some silver in order to bring a crown's weigh up to the weight of the gold with which he had been entrusted, some of which it was suspected that he kept. Archimedes' solution: see whether the crown, when submerged in water, displaced the same amount of water as an amount of gold equal in weight to the crown.

That is not the same as discovering anything about the amount of buoyancy. Michael Hardy 23:19, 29 May 2006 (UTC)

Oh yes it is! Density is weight per unit volume. Weight is easily determined, volume is more difficult. Volume was determined by the bouyancy (reduction in weight) of the object in a liquid of known density (water). The gold was being stolen by its replacement, in part, by silver. Different densities. Same reduction in weight when immersed but lower overall weight! Paul Beardsell 07:04, 30 September 2006 (UTC)

Following on: Archimedes did not tackle the awkward problem of measuring the crown's volume by collecting water displaced. The density of the crown was calculated thus. Paul Beardsell 13:26, 6 October 2006 (UTC)

Are there any other sources beyond De architectura? Because according to De architectura, he was comparing volumes. See here: http://penelope.uchicago.edu/Thayer/L/Roman/Texts/Vitruvius/9*.html#Intro.12 (there is a link to an English translation as well) rado 11:50, 31 October 2006 (UTC)

As an afterthought, he did determine that the crown was not pure gold by comparing volumes, but he found, by calculation, the quantity of silver mixed with the gold so probably both opinions are correct (though, there is a giant mental leap from calculating the ratio of silver - probably as a ratio of displaced water volumes - to realizing there is som sort of force acting upon the body, not even speaking about finding out the formula for the force) rado 11:59, 31 October 2006 (UTC)

If density is determined by d=v/m then both volume and mass must be measured accurately. If density relative to water is determined then only weight need be measured accurately, twice. Density is easier to determine this way as volume need never be measured. I assumed that Archimedes would have realised this and I have asserted that this is the way he measured density of the crown. I have asserted this is what Archimedes did without reference to the original texts. The correct outcome of this is, according to the text quoted, Archimedes did not use buoyancy to determine the density of the crown. This vindicates Michael Hardy's view, earlier, which I contradicted. Paul Beardsell 13:14, 31 October 2006 (UTC)

What remains unresolved (here) is the question of what Archimedes discovered in his bath. Paul Beardsell 13:18, 31 October 2006 (UTC)

The main point must be this: It is not necessary to use Archimedes principle concerning buoyancy to do what Archimedes is said to have done in order to find out whether the crown was adulterated. Michael Hardy 16:10, 31 October 2006 (UTC)

A bit more long-windedly: Archimedes could have weighed the crown, NOT submerged in water, then measured the amount of displaced water to get the volume. That tells him the density, and he can likewise get the density of pure gold. There is no need for him to have known anything about buoyancy to do all this. He did formulate "Archimedes' principle", and he could have done so at the time when he solved the problem of the crown, but we cannot conclude as a matter of logical necessity that he MUST have done that. (I haven't yet dug out my copy of Heath's translations to see the details, though.) Michael Hardy 17:11, 31 October 2006 (UTC)

[edit] True-ish but not relevant

The Acceleration section is just not appropriate. And other bits too. What is said is, when true, true for all objects, not just buoyant or floating ones. And other things are not quite right! The section seems to be written by someone who does not understand that -- the author(s) seems to have an imperfect / incomplete understanding of mechanics. I will attempt a re-write. Paul Beardsell 11:42, 6 October 2006 (UTC)

And with some help from others that is complete. Paul Beardsell 14:56, 30 October 2006 (UTC)

[edit] Remove diagram

I'm keen to remove the diagram if a better one is not forthcoming. It's misleading as there is no spring balance weighing the object. All the diagram is doing is showing the measurement of volume. Paul Beardsell 14:56, 30 October 2006 (UTC)

[edit] Simple explanation

The question how boats stay afloat pops up regularly at the reference desk, but the standard explanations don't quite clarify it very well for people who don't get it. So maybe there should be a 'simple explanation' section, however unencyclopedic that may sound. I will give the answer I gave today as a suggestion:

'Displacement' refers not really to water being moved or something (although in a sense it is) but to the volume of the boat below the waterline. The amount of water that would have been there has to be heavier than the boat itself (plus the air, cargo and crew and what have you). The material of the boat itself may be heavier than water (per volume!), but it forms only the outer layer. Inside it is air, which is negligible in weight. But it does add to the volume. The central term here is density, which is mass ('weight') per volume. The thing to keep in mind is that you have to divide the mass of the boat (plus contents) as a whole by the volume below the waterline. If the boat floats than that is equal to the density of water (1 kg/l). You could say that to the water it is as if there is water there because the average density is the same. If you add weight to the boat, the volume below the waterline will have to increase too, to keep the mass/weight balance equal, so the boat sinks a little to compensate. DirkvdM 09:42, 16 November 2006 (UTC)

I can't see how that is simpler than the article which says everything you say. I reckon if a reader can't understand the article then they won't understand your explanation either. The explanation offered also suffers from not being quite correct in that displacement refers precisely and "really" to the weight of water being moved = displaced. As you will read at displacement, displacement is a weight, not a volume. Paul Beardsell 11:22, 16 November 2006 (UTC)

[edit] Boyancy force disappears?

You write, "The buoyancy force disappears if the fluid is not allowed to flow under the bottom of the object, for example if the object's bottom is fully in contact with the bottom of the container." This needs clarification. What kinds of cases do you have in mind, here? Strictly speaking, the statement is incorrect.

When I take an empty bottle (empty, that is, save the air inside it) with the cap screwed on tight and hold the bottle down at the bottom of a bathtub filled with water, so that the bottom of the bottle is fully in contact with the bottom of the tub, does the buoyancy force disappear? It does not. Rather, the buoyancy is being counteracted by the downward force that I am exerting by holding the bottle down. The net upward force is zero as long as I am holding the bottle down, but the buoyancy force remains. Otherwise, it wouldn't be necessary for me to hold the bottle down. It would stay at the bottom by itself.

So, what would be a real case in which the buoyancy force disappears as the result of an object's bottom being in full contact with the bottom of the container, or as the result of the fluid in some other way not being allowed to flow under the bottom of the object? 66.167.49.46 07:03, 6 December 2006 (UTC)

A real example? A rubber suction cup. Paul Beardsell 08:50, 6 December 2006 (UTC)

Hi, I was the one who had added this sentence. If you place a bucket upside down in an emtpy pool and then fill the pool with water, press the bucket down to prevent water from entering the bucket (no air bubbles seen going up), then you release the bucket, it will stay there by itself. In fact in this case there is a net downward force from water pressure and this will help the remain bucket water-tight. I don't really care that my sentence was deleted, but there is some truth in it so please consider replacing it with a good sentence yourself. Sorry for bad english. Thank you.88.240.9.129 18:54, 9 December 2006 (UTC)

I too have problems with this recent addition to the article. Doubtless the response to your query will be that the bottle is not "fully in contact with the bottom of the container". This does not satisfy me. "Fully in contact with the bottom of the container" must mean there is no space for molecules of the fluid to find their way between object and container. The object thus becomes part of the container. The point is trivial and serves only to confuse. Deleted. Paul Beardsell 08:46, 6 December 2006 (UTC)

In the example of the rubber suction cup, the buoyancy force is being counteracted by the downward pull of the suction, so that the net upward force is again zero. But does this mean that the buoyancy is zero in such a case? It does not. The suction cup still has its buoyancy, but the buoyancy force is being counteracted by the force of the suction. Otherwise, the suction wouldn't be necessary in order to hold the cup in place. For it would then stay on the bottom on its own and under its own weight. 66.167.49.46 11:24, 6 December 2006 (UTC)

I'm not so sure that the ability or inability of molecules of the fluid to get underneath the object bears on the question of whether there is a buoyancy force involved. In the bottle example, it seems to me that even if the bottle were pressed to the bottom of the tub with sufficient force to prevent even a single molecule of water from fitting underneath, thus placing it "fully in contact with the bottom of the container" in the strongest sense of those words, there would still be a buoyancy force in play. One would still be able to feel that force, and to feel one's self counteracting it while pressing down on the bottle to hold it in place. The water molecules would still be pressing differentially on the bottle, with greater pressure toward the bottom and less pressure toward the top. Thus, they would still be "trying" to get underneath, and one would still have to press downward to prevent that from happening. In the event that the bottle became fused to the bottom of the tub as the result of the downward pressure, so that it would now stay on the bottom without being held down, this fusing (or sticking tendency, or structural integrity of the bottle/bottom of the tub, or whatever the right term would be here) would itself represent a force acting counter to the buoyancy that would otherwise push the bottle up. Here again, the buoyancy force would still be present, just counteracted by another force. Just as in the case of the suction cup. No? 66.167.49.46 11:24, 6 December 2006 (UTC)

Well, yes, I agree, I think. But what is the explanation underlying the buoyancy force? I've deleted the one you and I don't like much. Consider if the concrete pool floor were sculpted/moulded to include a hollow concrete bottle just like your glass one. Is there a buoyancy force? Paul Beardsell 21:32, 6 December 2006 (UTC)

In that case, the structure of the pool will include a submerged, hollow projection whose average mass per unit volume (its mean density) will be lower than that of water. So, we can definitely say that if this hollow projection were not attached to the rest of the pool, it would be buoyed up. Is this fact best expressed by saying that there is already a buoyancy force present, but one which is overcome by the strength of the projection's attachment to the rest of the pool? Or, is it best expressed by saying that there is not a buoyancy force in play but that there would be if the projection became detached? I'd say the former. In order for the hollow projection to stay attached indefinitely while immersed, the pool will have to be constructed of a sufficiently strong material. If the pool is made of the sort of stuff that pools normally are, this will not be a problem. But consider a case in which the pool is made of a very weak, thin material. After enough time has passed, the projection might suddenly detach and rise to the surface. This would be due to a buoyancy force acting before the projection detached, as well as after, wouldn't it? (This would of course require that the seam where the projection joins the bottom of the pool break quickly, all the way around, rather than its being breached selectively at one or more weak spots, in which case water would rush in through the breach in the weak areas and fill up the hollow cavity. But given the right material, one could surely construct the pool in such a way that the hollow projection would detach and rise to the surface.) 66.167.49.46 00:00, 7 December 2006 (UTC) Patrick K.

Such a pool costs extra! Consider also a string. One end of the string is "fully in contact with the bottom of the container" - i.e. one end is tied or glued to the container so that no fluid is between it and the container. The other end is tied to a plastic bottle tightly closed and full of air. As by the reasoning now deleted from the article the bottle would not be buoyant and would lie on the floor of the container. But as we know... Paul Beardsell 08:34, 7 December 2006 (UTC)

Well, something similar is the basis of the flush mechanism for a toilet, for example, so there is some basis to it. On the other hand, it's clear that at very least you can't let any water at all displace under the "bucket" or the project is doomed. Gzuckier 20:33, 9 December 2006 (UTC)

You can let some water "displace under the bucket" without the bucket detaching from the bottom. Try it. Not that that proves much of anything. As to toilets I'm sure we're talking sewage. Paul Beardsell 17:46, 16 December 2006 (UTC)

[edit] Provide Proof of Archimedes' Principle

I'd like to add this simple proof to this article, please comment on this proof:

Consider an infinitesimally thin column of water at (x,y), containing a thin longitudinal section of the submerged object. The column of water above the upper surface z=u(x,y) of the object causes a pressure downwards:

In a similar way, the column of water below the low surface z=v(x,y) of the object causes a pressure upwards:

Let the infinitesimal transverse area be ΔA. The infinitesimal resultant force ΔF exerted on the object is therefore:

Integrating all the infinitesimal forces gives the total force on the object:

Assuming ρ and g are constant throughout the fluid, the equation simplifies to:

This proves the simple case of Archimedes' Principle when the object can be represented as the volume between two surfaces. --Freiddy 15:42, 30 January 2007 (UTC)

[edit] dumbing down / folk story

This previously good article has been dumbed down and "augmented" by cod science folk stories. I think I'll just revert wholesale to the version of several months ago. Paul Beardsell 21:16, 4 February 2007 (UTC) == == == ==

Although I, unfortunately, agree, I think there should be a more gentle solution, rather than reverting hundreds of edits and the time spent on them. Perhaps some of the relevant stories may stay etc. -- Syphondu 00:29, 3 March 2007 (UTC)

I think we need some part of the story, at least to explain why it is called "Archimedes' principle" and to say something about the history of the problem. It is not just a "folk" story, as it can be traced specifically to Vitruvius' books on architecture. However it might be appropriate to make the description of the story here even shorter and to point to Eureka (word) for a longer version. —David Eppstein 20:02, 18 March 2007 (UTC)

[edit] Image requests

It is requested that a photograph or photographs be included in this article to improve its quality.

A diagram showing the forces and volumes in question would be very educational. I'm sure there are enough interesting-looking floating things to make a good photo to liven up the page. -- Beland 06:34, 7 April 2007 (UTC)