Ball floating on the surface of water

[Jahnavi]

I think the consensus would be, that that answer (c) is incorrect. The effect that we computed could be significant enough that it is more than microscopic. When one of their answers says "it will sink a little", they didn't define "little", but the small effect we computed could certainly mean a "little". It didn't say "a lot", and we didn't compute " a lot", but we did compute " a little". :)

Very nice !

 

[Jahnavi]

The same question is given in a book with correct option given as a) .

What do you think of this explanation ?

 

[Jahnavi]

Yes/No. That means he takes $\rho_{\rm air} = 0$ which makes the exercise pointless.

Ok . I think this point also needs to be cleared .

@BvU you are saying that if density of air is assumed 0 , then level doesn't change .

I am getting the impression that you believe that if air pressure is assumed uniform , then too, the ball sinks . Is that so ?

Please clarify what option would you choose if we assume uniform air pressure .

 

[Charles Link]

The same question is given in a book with correct option given as a) .

What do you think of this explanation ?

View attachment 223921

Accurate (editing: But not accurate... see post 41 at the bottom=this is inaccurate in the logic they used), but really getting overly detailed. What temperature is the water at, so that the vapor pressure of the water can be included as part of the problem? Was there water vapor, and what was its pressure in the initial case? $\\$ Then, it could even get more detailed: Does the cork expand when the vacuum is made above the cork? $\\$ Also, in regards to the water vapor, do we continually pump the water vapor out, so that its pressure is negligible? $\\$ I think we all came up with a pretty good answer considering the question didn't contain all of these minute details.

 

[Jahnavi]

Accurate,

Really ?

Upthrust remains same . Upthrust is always equal to the weight of the ball .

 

[Charles Link]

Really ?

Upthrust remains same . Upthrust is always equal to the weight of the ball .

Upthrust is another word for buoyant force. They are claiming water vapor will create a buoyant force after the beaker is evacuated, but that it will be less than than the buoyant force from the air pressure. That same water vapor pressure would have been present in the beaker before it was evacuated. $\\$ They got the correct answer, but really are over-thinking the problem as it was intended. It should not be necessary to include water vapor considerations in this problem, and I don't think the people who designed the problem needed to specify that. $\\$Editing: In addition, the water molecule is lighter than $N_2$ or $O_2$, (they got that part right), but the water vapor pressure will not be anywhere near 1 atmosphere. Even though they presented this extra detail, they overlooked the biggest reason why the water vapor can be ignored=the vapor pressure of water, (except near boiling temperature), is much less than 1 atmosphere.

 

[Jahnavi]

OK . Thank you .

Even though post#38 is directed to @BvU , please tell what do you think .

 

[BvU]

The same question is given in a book with correct option given as a) .
What do you think of this explanation ?
View attachment 223921

That a book from a top institute too ?

I agree with Charles. Fortunately it doesn't work against a) as the correct answer.

 

[Charles Link]

@Jahnavi See also my edited comments of post 41.

 

[Jahnavi]

That a book from a top institute too ?

No . This explanation is from a reference book .

The same question has appeared in different places . Reference books are going with option a) but the institute's answer key of the test paper states c) .

@BvU Please reply to post#38 .

 

[BvU]

I am getting the impression that you believe that if air pressure is assumed uniform , then too, the ball sinks . Is that so ?
Please clarify what option would you choose if we assume uniform air pressure

The only way the air pressure can be uniform is if $\rho\; g = 0$. We can assume $g\ne 0$ (no space ships or anything). If indeed $\rho_{\rm air} = 0$ then pumping it away does not change anything. In fact, then the evaporation of the water will make b) the better answer.

Please do not let the cork be expandable

Over 40 posts for a simple Archimedes exercise

 

[Jahnavi]

@Charles Link ,

If air pressure is assumed uniform , what option would you choose ?

 

[Charles Link]

@Charles Link ,

If air pressure is assumed uniform , what option would you choose ?

That is an assumption that you can not make. This item came up several years ago when I was teaching another student about the buoyancy forces on a helium balloon. The pressure can be assumed to be 1.0 atmosphere, but if the pressure is exactly 1.000000000 atmospheres all around the balloon, there would be no buoyant force.

 

[BvU]

In fact, then the evaporation of the water will make b) the better answer.

Now I've sunk to quoting myself. Not so sure anymore: if $\rho_{\rm air} = 0$ there is no atmospheric pressure and the partial pressure from the water will be the same before and after 'removing' the air, so: answer c). But the 'evacuation' evacuates water vapor too -- back to answer a) until the pumping stops; then it's c) again.

Can you give a link to this 'top institute', as I asked in #35 ?
And while you're at it, what book in #37 ?



All this roadrunning is in vain

 

[Jahnavi]

That is an assumption that you can not make.

Fair point .

But do you agree with BvU's inference that if air pressure is uniform implies density of air is zero ?

@Charles Link we usually make these assumptions in intro Physics numericals . We invariable assume value of atmospheric pressure constant on surface of Earth . Don't we ?

 

[Charles Link]

Fair point .

But do you agree with BvU's inference that if air pressure is uniform implies density of air is zero ?

@Charles Link we usually make these assumptions in intro Physics numericals . We invariable assume value of atmospheric pressure constant on surface of Earth . Don't we ?

In a gravitational field, even in a closed container, there will be a gravitational force per unit volume on the air molecules of $f_g=- \delta_{air} g \hat{z}$. The air in the container does not accelerate from this force because there is a force per unit volume $f_p =-\nabla P$ as a result of pressure gradients that necessarily must occur that balances/counters the gravitational force. (There is no mechanism for any other forces that can balance the gravitational force=it must come from a pressure gradient that necessarily must occur). A little advanced mathematics shows these pressure gradient forces, when integrated over a volume are equivalent to a buoyant force of $mg$. Basically, the result is $\int \nabla P \, d^3x=\int P \, \hat{n} dA=\int \delta _{air} g \, d^3x=mg$. (The alternative is the container and its contents are accelerating in a free fall. Then there is no balancing force from pressure gradients and the air molecules are, in fact, accelerating). $\\$ And I agree with the statement by @BvU . $\\$ To compute the buoyant force by calculating the weight of the air, you can assume atmospheric pressure of $P=1.00$ atm to calculate the density of the air. There still must necessarily be some pressure gradient in the $\hat{z}$ direction for there to be equilibrium of air molecules in a gravitational field. (Pressure changes with height will be small in most cases=they may be 1 part in 1000 or less over the object of interest, but they are non-zero). The result is the buoyant force given by Archimedes principle.$\\$ If you assume no pressure gradient, yes, then you get no buoyant force, but you then must also assume no gravity. The alternative, as correctly stated by @BvU is that $\delta_{air}=0$.$\\$ Editing: Archimedes principle works extremely well in determining buoyant forces in both liquids and gases. There is perhaps one major item with it that needs a qualifier and that is described in the last paragraph of post 20. $\\$ Additional item: If you @Jahnavi want to see an example on how to apply Archimedes principle quantitatively to gases, in particular, using the ideal gas law $PV=nRT$, a suggestion would be to read through the thread of the "link" in post 15. Qualitative problems have answers like "a little" or "a lot", but there is more precision in quantitative calculations that actually compute things such as "given a density of cork=... how much does the cork sink?", or "given the man weighs 150 pounds, and the payload is another 50 pounds, what helium balloon volume do we need to lift the man off the ground?" Anyway, you might find the "link" of post 15 of interest.

 

[haruspex]

Please do not let the cork be expandable

The most likely is that it would expand somewhat, and some air would leak out, both causing it to rise!

 

[Charles Link]

@Jahnavi Please see the edited additions to post 51 above.

 

[haruspex]

Archimedes principle works extremely well in determining buoyant forces in both liquids and gases. There is perhaps one major item with it that needs a qualifier and that is described in the last paragraph of post 20. \\

There is one other situation I am aware of.
An object sitting on the floor of the vessel and making a watertight seal with it lacks the upward pressure on the underside. In this case the buoyant force is (V-Ah)ρg, where V is the volume of the object, A its base area and h the depth of water at its base. Note this will often be negative.

 

[Jahnavi]

In a gravitational field, even in a closed container, there will be a gravitational force per unit volume on the air molecules of $f_g=- \delta_{air} g \hat{z}$. The air in the container does not accelerate from this force because there is a force per unit volume $f_p =-\nabla P$ as a result of pressure gradients that necessarily must occur that balances/counters the gravitational force. (There is no mechanism for any other forces that can balance the gravitational force=it must come from a pressure gradient that necessarily must occur). A little advanced mathematics shows these pressure gradient forces, when integrated over a volume are equivalent to a buoyant force of $mg$. Basically, the result is $\int \nabla P \, d^3x=\int P \, \hat{n} dA=\int \delta _{air} g \, d^3x=mg$. (The alternative is the container and its contents are accelerating in a free fall. Then there is no balancing force from pressure gradients and the air molecules are, in fact, accelerating). $\\$ And I agree with the statement by @BvU . $\\$ To compute the buoyant force by calculating the weight of the air, you can assume atmospheric pressure of $P=1.00$ atm to calculate the density of the air. There still must necessarily be some pressure gradient in the $\hat{z}$ direction for there to be equilibrium of air molecules in a gravitational field. (Pressure changes with height will be small in most cases=they may be 1 part in 1000 or less over the object of interest, but they are non-zero). The result is the buoyant force given by Archimedes principle.$\\$ If you assume no pressure gradient, yes, then you get no buoyant force, but you then must also assume no gravity. The alternative, as correctly stated by @BvU is that $\delta_{air}=0$.$\\$ Editing: Archimedes principle works extremely well in determining buoyant forces in both liquids and gases. There is perhaps one major item with it that needs a qualifier and that is described in the last paragraph of post 20. $\\$ Additional item: If you @Jahnavi want to see an example on how to apply Archimedes principle quantitatively to gases, in particular, using the ideal gas law $PV=nRT$, a suggestion would be to read through the thread of the "link" in post 15. Qualitative problems have answers like "a little" or "a lot", but there is more precision in quantitative calculations that actually compute things such as "given a density of cork=... how much does the cork sink?", or "given the man weighs 150 pounds, and the payload is another 50 pounds, what helium balloon volume do we need to lift the man off the ground?" Anyway, you might find the "link" of post 15 of interest.

I am really grateful for your wonderfully detailed responses . Thanks !

This might interest you

https://physics.stackexchange.com/q...-exerted-by-atmospheric-pressure-in-mechanics

 

[BvU]

Goes to show that science doesn't benefit from one man one vote. What a mess.

 

[Charles Link]

@Jahnavi I think you might be starting to get somewhat of a handle on Archimedes principle with the details of this homework problem. There are two items that I think are worth an additional closer look though, and one is the last paragraph of my post 20, and the other is that of @haruspex post 54. Before he mentioned it here, I did know that you can stick together two optically flat glass surfaces with something that is referred to as an "optical contact bond" and it can be very hard to get them apart, but only with his post did I see that this is a result of what is a pressure force that is actually the exception to Archimedes principle that he is describing. And thank you @haruspex . :) $\\$ Additional edit: Wikipedia is not what could be called an authoritative source, and they seem to overlook the effects of atmospheric pressure for what holds two optically flat surfaces together in a google of the topic of "optical contact bonding". Perhaps their discussion is about such surfaces in contact in a vacuum... I don't want to take the discussion off of the Archimedes principle topic.