Challenge Questions

Sign???

No cutting needed - but the crevasse could do with filling

You give (correctly):
m_s - m_sa = m_p - m_pa
and
m_sa < m_pa

Adding the two gives
m_s - m_sa + m_sa < m_p - m_pa + m_pa
and simplifying this gives:
m_s < m_p

Right. Silly me .

It always happened long back in my exams (so many marks lost at last step) But then fortunately we had long questions, not like todays objective type, tick the correct answer. The examiners were considerate and for methods they used to ginve marks.

So as the hairs turn grey the grey matter inside don't change..

BTW are the challenges so easy now ?

This one, the earlier one (car brake), the current carrying conductor.

Though I am not fully convinced about the atomic explosion. Since there we are in quantum and

a) released Energy I am not sure has a mass (at least weighable- remember photon is a mass less particle)

b) I am not sure the energy released is from the nuclear binding energy - if the binding energy is released, then the mass may in fact increase - since the energy requirement within nucleus to keep the protons together has reduced.

The problem is it is such a nanoscopic mass (is it a term for far beyond microscopic ? ) I don't simply want to add the new nuclii, alpha and beta particles and see.

Archimedes' principle:

"Any object, wholly or partly immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object."

The problem can be looked at from Archimedes' principle to clarify my answer.

Instead of plastic let us consider a small iron ball and a larger Aluminium ball. Instead of air let it be water. Immerse the system in water. If the scale balances is the Al ball not heavier?

Although deducing the correct answer, as given above, is relatively simple. Knowing the principle behind the result, Archimedes principle, as elucidated by Dedaelus is not necessarily obvious to all people even engineers.

More interesting to me is the challenge of explaining how Archimedes's principle acts upon the two masses which formerly appeared identical and now appear different!

According to both Newton and Einstein the acceleration acting on them has not changed but the weight has???? How can this be??? If you believe their is a difference in the top air pressure and the bottom air pressure you will find that there is not enough difference to explain the change in apparent weight. In spite of this the balls act as if they are "aware" of the presence or absence of the weight of the column of air above and below them. It is very difficult indeed to explain how this can be and yet there it is....

Sincerely,

Mr. Gee

All based on a false premise - because the difference in pressure is absolutely (and exactly) sufficient to account for the upwards force. However, calculation via pressure is an overly complex way of reaching the solution unless the top and the base happen to be entirely horizontal and the sides entirely vertical (not possible for a ball of standard shape, of course).
If this was not the case the net force on air occupying equivalent vertical space would be upwards, so that air would rise. (But you can also check the veracity by calculating the buoyancy effects on a cylinder)

(A linguistic note: the weight of course does not change at all - only the net force on the supporting system (weight of ball minus weight of displaced fluid)

I am relatively certain that it is your assumption that there is a difference in pressure between the top and bottom of the ball that accounts for the buoyant effect which is false.

The air is a fluid and exerts a uniform pressure on both balls and all surfaces of each ball. There is an infinitesimally small pressure difference due to the height of the volume of air displaced by the volume of the ball but that "pressure" difference is approximately equal to zero within five orders of magnitude.

If the problem is reversed so that the air flows into the chamber containing the balance then, while the air is raising the plastic ball to the level of the steel one there will be a column of air of lower density directly above the plastic ball and a flow of air around the ball working to raise the mass.

Once the chamber is full of air and the two balls are in static balance no more work is being done and the flow and the column of rarefied air above the plastic ball disappears. Similarly, there is also appears to be a column of higher density fluid below the ball although this column is more difficult to observe.

In air these effects are very subtle and difficult to observe or measure. In water the refractive index of the fluid is dependent upon the density of the fluid and the rarefied column is readily apparent as it has a lower refractive index than the surrounding water.

The following illustration shows the termination of a bubble bursting at the surface of a fluid and the small diameter high density fluid columns momentum is revealed as a small but macroscopic water jet.

http://www.islandnet.com/%7Esee/weather/graphics/photos/bubble1.jpg

The following link shows a computation fluid model of meander in a rising bubble column.

http://www.fluent.com/about/news/newsletters/03v12i2/imgs/a7.gif

Typically the bubble train will rotate about a central core of rising fluid as they raise giving rise to a spiral formation. With the fluid flowing in the core more rapidly than the bubbles themselves rise.

http://hal.archives-ouvertes.fr/docs/00/06/88/92/PDF/Bubble_current.pdf

This paper discusses measurement of a single bubbles turbulent path as is rises in a long column. The paper does not mention the rarefied fluid column Per Se, as this feature of the fluid dynamics of buoyancy is little known and even less understood.

I have seen and used to have on file a photo of a single bubble train spiraling about a central fluid column in which the rarefaction of the central column was visible due to the interaction of the lighting, the background and the round cylinder but I have had a computer failure or two in the intervening time and do not have this reference, nor could I find it after searching online.

Best wishes,

Mr. Gee

The pressure difference is small*, not infinitesimal, and so is the weight of the displaced air.
However, it has become apparent that you are neither going to believe me, nor check whether your statements make sense by doing the sums for yourself, so:

For simplicity of calculation, consider the buoyancy effect on a cube of side 10-cm (i.e. volume 1-litre). Assume sea level, 20-degree atmosphere. The air density is about 1.2gm/m³. Now we'll do the calculation both ways:

First, via the displaced air: the volume is 10^-3-m³, so the displaced air weighs 1.2-mg, which is the expected buoyancy.
Now via the applied pressure: the pressure differential for the 0.1-m height (= density x height) will be 0.12-gm/m². The area of the base and top of the cube are equal at (0.1-m)², or 0.01-m². So the difference between the forces applied by the pressure at top and bottom (= area x pressure_differential) is 0.12 x 0.01 gm, or 1.2-mg.

Same answer both ways. No mystery, no magic, just plain simple old physics. OK?

*You say "equal to zero within five orders of magnitude". Actually, if you are comparing the pressure differential with the atmospheric pressure the difference is about eight orders below the absolute pressure for the 10-cm height I've chosen - but that sort of difference is not negligible if the effect of the much larger reference is by its nature self-cancelling. If you are comparing the buoyancy with the weight of a typical plastic ball it is about a part in 1500; now that may be negligible or not - depending on how accurately you need to know the exact mass of the ball.

I was skeptical of the buoyancy factor until I ran some numbers. I decided to compare two solid balls, one of steel and the other PVC. I chose steel because most metal balls are steel. I chose PVC because it seems to be so common.

Using common values given for the density of these materials, I found the ratio of PVC to steel is almost 1:12.

Then I assigned the steel ball a diameter of 1". I calculated the volume to be 2.19 cu. in., and the mass to be .62 lbs. (Yes I know about the mass/weight thing, but the effect of gravity on both balls is the same, so I'm using weight as a substitute for mass. Besides, I didn't want to do the extra math. Anyway, if I'd converted to slugs, people would have thought I was talking about their brother-in-law, or people of the opposing political persuasion.)

Using this value for mass, I calculated the PVC ball's diameter to be 3.68 in. Because of the larger diameter of the PVC ball, the buoyancy for it must be larger than the buoyancy of the steel ball.

However, the Challenge Question as stated still has problems in the real world, such as getting balls to remain still in a scale's pans. The question would have been better with cylinders instead of spheres.

That's the problem with asking engineering types questions that are primarily theoretical. Engineers think in terms of the real world, and those real world considerations distract them. But then, if this question had been presented with a real world scenario of a keg of beer on one side of the scale, and a case of longnecks on the other, then most CR4ers would have been too distracted by their thirst, and their efforts to satisfy it, to even offer any answers!

Presumably the pans on this balance had curved sides (designed for comparing powder weights?). Or perhaps this was the reason for needing a draught-free evacuation (scatology intended)?

But I don't know where you got the density ratio from. PVC ranges from about 1.37 to 1.42 gm/cm³, steels from perhaps 7.8 to 8 gm/cm³. I think that gives a maximum of 5.8:1.

The other numbers can be handled more simply as well. The ratio of diameters will be the cube-root of the ratio of densities (because everything else cancels). So using your 12:1 density ratio would give the diameter of the pvc ball to be about 2.29". But you don't really need to go into diameters - only the volumes are important, and these depend directly on the densities. [Of course plastic:steel density ratio is the same as steel:plastic volume ratio]. Then you can use the ratio to air density to calculate each buoyancy.

I find I owe Physicist and the group an apology. I stated that your premise was in error and it was not. It was I that was mistaken.

The point I was attempting to make has to do with the way that Avogadro's principle is applied in the real world and not with methods of calculation of the net effect. I have spent quite a bit of time on the physics of buoyancy and there really is more to be learned about the subject than the straightforward calculations you give shows. However, the difference only becomes clear however when you consider the time rate of change, not the net effect.

In this example the balance would have a strong tendency to return to level faster than simple Newtonian physics would allow as there truly is a fluid dynamics air flow component to the problem.

If you accept that there is a pressure differential in the air due to the presence of the plastic ball as the jar is either emptied or filled then there is obviously going to be an air flow caused by this differential.

The structure of the flow is along the line I indicated and the pressure densities and core flow rates are much larger than the simple calculations above would tend to indicate.

In evacuated balances, these differences are, I assume completely unimportant. In calculating flow rates in marine seeps, they are not. The flows are anomalously fast. In fact there is a marine phenomena called a water spout in which gas bubbles forming in deep stratified layers of sea water saturated with dissolved gas can progressively accelerate a water column to speeds of 60 miles per hour or so.

These water columns can and do punch holes right through vessels and the gas clouds released are very likely to have killed entire crews. Giving rise to what are commonly reported as "ghost ships" in which everything aboard ship seems fine, but the entire crew is either missing or dead.

In the rising water column problems I have worked upon it is necessary to assume that the water column is a semi-rigid column which undergoes a buoyant acceleration until the viscous drag on the column is equal to the acceleration due to buoyancy and an upward terminal velocity is reached.

These same principles apply only very very slightly to this challenge question and it was a far stretch to attempt to use this scenario as a "teaching moment" additionally, I was both rude and wrong. My apologize.

Sincerely,

Mr. Gee

Apology appreciated and accepted (at least by me).

Depending on the velocities, dynamic situations move through multiple regions. But the challenge was quite explicit - the flow rates have to be too slow to be significant. If they were not, we would need additional information about the enclosure, as the direction of force would depend on the positions of the extraction vents. But of course the arms of precision balances are supported until the weighing conditions are established, so even that is not too relevant.

I'm at a loss as to how a vessel that had been punctured could self-repair so that it appears as a ghost ship, but I suspect I'm misreading your intent. The only natural phenomenon I know that would on its own cause a populated ghost ship is poisoning due to a release of previously trapped suboceanic gas. The equivalent effect is well documented in several lakes. (CO2 is possibly the most insidious, because it leaves no chemical trace, but the non-lethal effects of methane will also dissipate in a reasonable time.) Of course, ships can also be abandoned by their crews for good precautionary reasons.

You clearly do not understand Avogadro's hypothesis.

Your babbling dissertation on bubbles is totally irrelevant.

In fact, it could be said that "Eureka sophistry"

The answer has been given. It couldn't be clearer. No moving air. The removal of the unequal buoyancy forces on the two bodies causes an imbalance.

Yes, physicist is right.

You put very nice explanation, but alas, did small mistake in mathematics.

So conclusion is Plastic ball will go down. (Which physicist did not put in words)

m_s<m_p

Wow this is a gross misrepresentation of actual physics. Mass or weight have nothing to do with air and you should be ashamed of that GA you got. Mass is an independent property of a material. Weight is a force that is dependent on gravity. You threw in a bunch of garbage equations. You get an F on your home work.

A balance measures mass which is an intrinsic property of an object and has no dependence on atmosphere. Where is everyone getting this other garbage.

Just try the thought experiment of tying a helium balloon to one side of the balance. Are you saying that it won't pull upwards just because it is tied to a balance? Are you saying that a helium balloon has negative mass?

Where did you get the idea that a balance measures mass directly? It relies on gravity, so it is always comparing forces. The balance will compare the differences between the mass of the balls and the air that each displaces. Because the plastic ball is larger, it will displace more air - so it needs to be heavier in the first instance.

N.B. even if you "know" you are right, please refrain from insults in future.

No, a balance measures relative weights and is dependant on a gravitational field to function. (Just try using a balance on zero gee!) It is not possible to measure mass directly. And yes, weight is dependant on buoyancy which does exist in atmosphere. If you doubt any of this, just try using a balance to measure the mass of a helium filled balloon!

I might mention that even if you were correct (which, of course you are not), insulting others does not help to make your point. To be honest, it erodes your point, because much as we would like to say that we judge only on the facts (in which case your explanations here come up short) we also cannot help but judge on the basis of politeness and kindness (where your "explanation" also comes up short).

SB only made a small mistake in math.

I am sorry that I am not at all ashamed of the GA i got (though I don't know who gave it). Though i ought to be ashamed of the mistake on last line, but as I said in an earlier post, I am habituated of doing it right till the last leg.

Mass has nothing to do with air. Accepted , weight has nothing to do with air accepted. But you should try to study the equations, and read the explanations before jumping to conclusions.

Well Physicist being what he is has got it as have a few others. If you didn't then it is your luck.

Please study the force balance and remove common factors and then come back and check the logic.

How many times did You proof read this to Yourself before You determined You had the proper # of reversals?

Your post reminded Me of Daffy Duck, and Bugs Bunny arguing which Hunting Season it is, but unlike the Looney Tunes characters debating with each other, You seem to carry both halves of the conversation internally, or You have a Split-Personality...

Sufferin' Sucotash! That was a Big Mouse.

Nice work concerning their masses but they are balanced thus the folcrum has not moved. Therefore regardless if the air is removed creating a vacuum there would be no effect, that is if the plastic ball is solid. Then, the vacuum would collapse the hollow plastic ball creating the imbalance. The question does not address the issue.

Concider

How does the vacuum collapse a hollow plastic ball? Are you assuming the hollow ball to be sealed? (In which case there would be residual pressure inside the ball.) Or does the ball leak? (In which case the internal pressure would equalize with the vacuum.) Please explain your reasoning.

Look at it in another way.

The amount of air supporting the plastic ball is more than the iron ball. There need be no air in the balls and they could be solid. Remove the air and the plastic ball will tip the scale and go down and is heavier. The supporting air which kept up the heavier Plastic ball is no more

Look at it from Galileo experiment. If a large plastic ball and a small steel ball of equal weight were dropped from the tower of Pisa, with air around, the plastic ball will hit the ground after the iron ball due to larger surface area and air resistance. In vacuum they would hit at the same time.

Look at it in another way. Make a heavier plastic ball such that the iron ball and plastic hit the ground at the same time with air all around. Create a vacuum and the plastic ball will hit the ground faster as it is heavier

You write as if there is a connection between the resistance of air to a moving object and the mass of air it displaces. In a sense that is true (particularly once the motion is turbulent), but unlike buoyancy the air resistance depends on the shape of the object as well as its volume. So I would say this was at best an unhelpful analogy.

Unfortunately, your last paragraph is incorrect as written - perhaps you intended something different:
"Make a heavier plastic ball such that the iron ball and plastic hit the ground at the same time with air all around"
Yes I can do that - as air resistance rises according to surface area and mass rises as volume there will be a size of plastic ball at which the ratio of weight to air_resistance is the same as for the smaller steel ball.
"Create a vacuum and the plastic ball will hit the ground faster as it is heavier"
This will not happen if the local gravity and drop-heights are the same - in vacuum all objects will have equal acceleration.

You need not go to Pisa. You can disprove yourself even in Mumbai.

Air resistance is only applicable to very light (effective) objects having very large surface areas, as compared to their weight, like feather or parachutes.

In your experiment in last para, the plastic ball and the iron ball will hit the ground simultaneously in presence and absence of air.

Nowhere can you assume there is air in the balls. Not given in the problem. It says "BALL"

When you remove the air, it stops supporting* the balls.

How do you figure that the surrounding air is supporting the plastic ball? Since you have to put the air under at least some pressure to fill the plastic ball, the air within is already denser than the surrounding air. There is no buoyancy to begin with.

Since the plastic ball is a closed system, the mass of air inside does not change. Without the buoyancy factor, the mass of the 2 will still balance out.

You have made an assumption that the plastic ball is hollow. The problem does not specifically state, but I believe the proper assumption is that both the steel and plastic balls are solid & homogeneous.

Actually, it doesn't matter if the plastic ball is hollow or not. The buoyancy depends on the amount of air displaced.

Let's say the metal ball has a diameter of 10 cm and the plastic ball has a diameter of 20 cm. Then the mass difference in the displaced air is

4/3*pi*(r1³ - r2³) * 1.29kg/m³, where r1= 0.1 m (the plastic ball) and r2 = 0.05 m (the metal ball), and 1.29 is the density of air at STP. The weight difference is this times g, or 0.046 N.

Think of a Navy diver. He can walk on the ocean floor in his heavy diving suit despite weighing hundreds of pounds on land. The weight of the water he displaces gives him some bouyancy.

If either of the two balls is hollow, it had better survive the experience of having a vacuum pulled on it. Otherwise, at bang, the scales will certainly move.

Thin walled pressure vessel.

Perhaps a balloon would expand and burst but not a plastic ball. Again over analyzing the question.

OK. Got it. The fourth link was the one that made sense to me (true mass vs conventional mass).

It's the connotation of buoyancy (your floating in air), that tripped me.

Floating, meaning completely supported by the surrounding medium: No. The only time I have floated in/on air was when I had 1 beer too many! And that is a subjective observation.

Buoyant, meaning to have a measurable effect on the "apparent" weight; even if the overall buoyancy is negative: Yes.

Thanks for the lesson.

Thanks Kilowatt.

I do not understand why other people do not accept the things ... even in this thread.. as you accepted.

Very nice of you.

The surrounding air is supporting both balls to a small degree.

Forget about the inside of the plastic ball it could be solid, hollow, doesn't matter.

what does matter is the plastic ball is larger, and therefore displaces more air. Remove that air and the scale tips down on the side of the plastic ball.

Haha, interest.

this is a question from primary school. the same problem is

One kg iron, One kg cotton. which is weighter in vacuum? or which has more mass?

Not quite - the two initially appear the same when supported by air, and are therefore not the same mass. (The same reason people can float, except the effect in air is nearly 1000 times smaller)

The question is, will the two balls remain balanced. If the balls are solid and homogeneous and there is no entrapped air as a result of porosity, the balls will remain balanced. If, however, there is ambient (pre-vacuum) air entrapped in either ball greater than that of the other, that ball will descend.

No, the larger ball is heavier.

Obviously the plastic ball will descend.

And it's final.

Formulas to follow.

Yes

Yes. Since the form in question is a ball, any influence air pressure could have on it's weight on the scale is negligible.

Why does the shape make a difference?

I think should have said since both are balls, meaning of the same shape. Thus the ratio of buoyancy would be same regardless of air pressure. However, thinking this through again, I am not so certain: If removing the air from the bell glass creates a complete vacumn.....

A complete vacuum does not have to be created.

If the air's density is reduced by creating a partial vacuum, then the buoyancy effect will change. Therefore, assuming the the plastic ball is of greater volume than the iron ball, (in other words, we are not assuming that the plastic ball is made of lead-filled epoxy) if we reduce the air density by half, the plastic ball will descend.

To prove this effect to yourself, you can go for a hot air balloon ride. If you prefer, use a helium balloon.

Lets suppose that in air the metal ball's weight = plastic ball's weight

Now this is not the real weights of the balls because each is actually lighter with the weight of the air they replace.

Therefore without air the plastic ball will weigh more than the metal ball because the plastic ball replaced more air because of its bigger volume.

The scale will therefore tip in favour of the plastic ball.

The plastic ball will go down - it was bouyed up more by the air it displaces (more than the metallic ball)

Of course they'll stay balanced. Weight is a result of "atraction of the earth". Air or no air doesn't matter in this case. Even if you fill the jar with water or no matter what substance, the weight will always stay the same, so they both will keep putting the same weight on each arm of the balance. Yes ?

that was one of the dumbest replys ive ever read

i seriously think i got dumber reading that.

Weight will stay the same....

Take for example, two submarines of exactly the same "weight" (mass too.) First, we place them on a huge scale under water and find that they do indeed weigh the same. One we will leave in the form of a submarine. The other we will crumple up into a block as small as we can (OK, all the hatches remain closed and there are no ruptures as we crumple so the air inside is compressed, nothing is lost). Now we will weigh them on the scale again. One submarine has more bouyancy, this is the one that was not crumpled. The other as you could imagine would not have much bouyancy. The mass has not changed and weight have not changed, and assuming that we place them carefully so that their CG are equidistant from the fulcrum, the scale will still read that they are equivalent, right?

This is ridiculous of course, as the principle of bouyancy relates to the mass of a displaced fluid. The crumpled sub with the same mass displaces much less fluid than the uncrumpled one. Does it make sense now why they will not be balanced in a vacuum?

A helium-filled balloon could pull one side of the balance upwards when it is in the air. Are you suggesting it would still do that when you have pumped the air out?

Lets suppose that the plastic ball has small air bubbles trapped inside. Then as the air in the bell gets less the pressure inside the plastic ball builds up until it explodes driving down the arm at the plastic ball's side

Then as the air in the bell gets less the pressure inside the plastic ball builds up

And what might be causing the pressure to build up?

For those still in doubt that the plastic ball will go down, what do you think would happen if the scale and balls were initially balanced in water instead of air?

The Plastic Ball would float, and not remain on the scale.

NO, it will not remain balanced.

Come on people. This either proves or disproves your very basic understanding of the physical universe around us. MASS DOES NOT CHANGE. And do go quantum on me (you guys know who you are). Air bubbles inside the plastic ball have no bearing on this. Asking about what if these balls were in water is a completely different problem.

Humour me with a thought experiment...then see if you are so confident.
Imagine the plastic ball has extremely thin walls and is full of helium so that in air it just lifts off the ground, now imagine it's weight is inceased by adding a few weights such that it balances with the steel ball.
Now imagine whta happens when the air is pumped out.
Or to put it another way....why do you think water is sooo different to air?
I do sympathis as my knee jerk reaction was...'it doesn't make ant different', but I have dealt with Fyz before and realise he's not as daft as I look .
Del

You have dealt with me before too. tell me how I look ? It will not make me vain I promise <blush blush blush>

BTW: you can not correct the incorrigible.

Whoever has heard that the scales balance mass ? .

They balance force (or to be correct the torque/ moment of force = force x distance)

F₁d₁ = F₂d₂

Nah - I'm far dafter than that

Asking about what if these balls were in water is a completely different problem.

Actually, it is essentially the same problem. Air, like any other fluid, has mass and volume, and therefore density. Helium is thought of as very low density. Therefore, helium balloons float in air. Water is relatively high density, so even "heavier" things float in water. Mercury is higher density yet, so even metals will float in mercury.

As should be obvious, (I hope) when you ask the clerk at the grocery store to fill your helium balloon, its mass increases as it is filled. Yet it begins to float. If you tied it to a balance, (and placed an identical unfilled balloon on the other side of the balance) you would find that as you increase the mass of the balloon as it fills, the balance will tip toward the unfilled balloon -- the one with less mass. To someone unaware of physical principals, it would appear that the balance is working backwards, or that the unfilled balloon has more mass.

We like to imagine that our balances are accurately measuring only the mass of the objects placed upon them. But obviously there is more to it than that. Clearly a filled helium balloon has more mass than an unfilled one. Certainly you accept that. Then why does it appear to have less mass than an unfilled one when the two are placed on a balance?

This either proves or disproves your very basic understanding of the physical universe around us.

You are probably right.

GA from me Ken

Good example, which hope all who have not understood the buoyancy of air, will understand.

Thank you, kind sir.

Humm, if I remember correctly gravity is 9.81 m/s per sec or 32.2 ft/sec per sec. (regardless). A feather and a rock will fall at the same speed, void of air so resistance of air holds the lighter one back. The question states that the plastic ball is larger, keeping in mind they weigh the same the only difference is volume. So in a vacuum the only variable left is the non presence of air which exerts a force of 14.6 lbs pressure (again can't remember but again regardless). This air pressure does exert a little more pressure on the larger object (helps hold it up)... So when this pressure is taken away the larger object will weigh slightly more therefore the larger plastic ball will go down. (even if both balls were metal the larger one would drop due to the loss of upwards force of the air pressure.)

I hope this clarifies the answer in a simple explanation.

Josh.

There is no weight in this question. A balance measures mass. You guys have to get your physics straight and ignore the air. Mass does not change if you are in a vacuum. Acceleration of gravity and atmospheric pressure have no bearing on mass. You guys are killing me with all this extra stuff. Keep in mind that a balance measures mass and functions the same here on Earth as it does anywhere else. If we took these 2 balls and did this on the moon with the same balance you would have the same result.

A balance measures mass.

No it doesn't!
Without gravity the masses would not exert any downward force to 'ballance'....The act of balancing assumes that gravity is acting uniformly on either end of the ballance arm, which of course it generally is, but what if one pan of the balance was above the earth and the other above the moon...ok it would need to be a very long balance arm and I don't know where the fulcrum would be mounted (on a space ship)!
I suggest you just slow up and carefully read some of the explanations.
Play devils advocate with yourself....try to prove yourself wrong....

Does a helium filled balloon fall to earth if there was no air around it? Yes it does...its mass doesn't change nor arguably does it's weight, but it will still fall to earth.

Del

And also will a scale unbalance a mass if no gravity ? say I put a 5o Kgs on one side of scale and keep the other pan empty in a zero gravity situation, will the 50 Kgs pan go down ?

The answer is, of course, that without gravity, there is no "down".

That is what we have been trying desparately to tell.

The scales do not measure the mass (the masses are unbalanced 50 Kgs vs 0 ) but the moment of the force about the fulcrum.

Yes, Mr. B seems unable to grasp that even a balance does not measure mass directly. He has not replied to any of the posts suggesting a thought experiment with a helium balloon. It is okay to be wrong, or even to misapply a scientific principle. Every one of us has learned that to humbly admit our mistakes does not make us any less of a person.

You are right, it is not the weight or density in question. It is the volume. You have to keep in mind this test is started in one atmosphere and then changed to another (that being vacuum). What you are missing is that fact of the change of atmosphere. The variable of atmospheric pressure will float the larger object, although balanced in one atmosphere not in the other. Doesn't a barometer work on this principle?

Analogy~ (Ones body will float or be more buoyant in salt water than in fresh water)

Either I am right or I have convinced myself that I am right! LOL

You are right, it is not the weight or density in question. It is the volume.

My own observations have indicated that things with volume possess mass, and things with mass have a weight. Given that the question is about buoyancy, density could reasonably be expected to relate to the question, I suppose: we often think that things of high density will sink in a lower-density fluid.

I suppose humans can have their "blinders on" when observing the world around them. We don't realize that we are, to some extent, floating in the fluid surrounding us. If we were much less dense (and I am using that term with some consideration) we would float around just as helium balloons do. Then, kicking and screaming to gain traction, we would start to perceive things as they really are.

To help you regain perspective on how things work, imagine yourself to be a fish for a while. You would, no doubt, see all sorts of things that seem to be suspended in your environment. You'd see a few things that appear to rest on top of your environment. You'd also see many things that seem to rest at the very bottom of your environment. What might you conclude, assuming you are a reasonably smart and observant fish?

1. Suppose, as a smart and observant fish, you stop by the local fish market to pick up a water balloon (of 1kg mass) and a steel cue ball, also of 1kg mass. The store keeper places each on opposite sides of a balance. What happens?

2. Now, the store keeper places a different amount of steel on one side of the balance to get the balance to "balance". Would that amount of steel be less or more than 1kg?

3. Imagine, now, that some mean-spirited human decides to remove the water from your pond, by pumping it out. Your balance was balanced in question 2. Does it remain balanced after the water is removed?

4. How might all this relate to the challenge question?

Do you experience fear when boating?

If you find yourself disoriented by all this, then perhaps reading this will help.

A little exercise in reverse logic. Suppose we had a scale with a steel ball and a wooden block balanced on a scale, place it in a empty fish tank and slowly fill it with water. As the water level rises the wooden block due to displacement will rise, and the steel block will fall. this is due to the density of the environment. So there is the density of water, the density of air, and the lack of any density in a vacuum, so the larger displacement will become heavier in a vacuum if balanced in a more dense environment, or it will become lighter if the environment density is increased.

Regards JD.

The cause is not the absolute pressure (14.6 psi) that air applies, but the difference between the forces on the tops and bottoms of the spheres. More pressure further down means the pressure on the bottoms of the spheres (directed upwards) is greater than the pressure on the tops of the spheres. The larger sphere will experience both a larger pressure difference and have a larger area - both of which will result in more upwards force.

But for calculation it is simpler to consider what would happen if you replaced the ball with air - the air would be in equilibrium, so its weight must exactly equal the total upwards force due to pressure differences.

People, get a job, the question has been answered!! Let go!

The question was answered correctly by Physicist in #2. Some posts since then contain confusing (wrong) comments. This is how I see it

Archimedes Principle -"When a body is wholly or partially placed in a fluid, it experiences an up-thrust equal to the weight of fluid displaced by the body" *. There is no restriction on the body. It can be any shape and in-homogeneous as long as it displaces fluid. Also note that ther is no restriction placed on the fluid which can be compressible.

A simple way of looking at this is to imagine a bath of fluid (choose water): Mentally take out a body shaped portion (a sphere is easy to imagine); This removed sphere of water has a weight; Before you removed it, the weight was being supported by the water configuration left in the bath; Now fill the spherical cavity in the bath water with a solid steel sphere which exactly fits; The water in the bath doesn't "know' that you've substituted steel for water an continues to supply an upward force to balance the weight of the water which it thinks is there; The up-thrust in this case is not sufficient to support the steel ball and so it sinks. If we now consider replacing the solid sphere with a hollow one having the same external diameter, we can see that the up-thrust remains unchanged and the weight of the body can be made to be less than the up-thrust, and so a hollow steel ball may float in water.

Once Archimedes principle has been thoroughly understood it becomes clear that on a balance beam (where forces are being compared) weighing in a fluid (air), the up-thrusts on the bodies are not equal since we are told that the plastic ball is larger. The plastic ball therefore experiences a larger up-thrust (regardless of whether it is hollow or not; regardless if it contains air or even helium or not and similarly regardless of the internal composition of the steel ball).

As soon as any vacuum is applied, the up-thrusts will diminish and because the plastic ball is displacing more fluid the up-thrust on the plastic ball is reduced by more than the up-thrust on the steel ball and the plastic ball balance arm goes down. Note that it is not necessary to remove all the air from the container to witness the unbalance.

* I have deliberately used the term up-thrust rather than buoyancy since I find that buoyancy is sometimes understood to be a characteristic of the material being immersed, whereas up-thrust describes perfectly what force is generated. Also up-thrust is how I first learned Archimedes Principle many years ago

Right on the money, SlideRuler. Seems a lot of misconceptions are "floating" around out there. What I want to know is what difference it makes whether there is any turbulence as the air is evacuated? When the final steady state is reached, the same effect will be observed in either case. Or was this a red herring to hint that the solution involved "upthrust" (I like that term)?

Aside from going back to physics class to perform this experiment, I think the guest's responses in #36 and #43 are bang on! Ha Ha Yeah, it's me I just registered! .

I think I am right and if not I can't wait for the correct answer!

Josh.