Ball floating on the surface of water

[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.