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BvU said:Is the atmosphere really pressing down ?
BvU said:Compare with a concrete ball under water. Is the water pressing down or up ? (Archimedes had something sensible to say about this)
BvU said:Is the atmosphere really pressing down ?
BvU said:Think like Archimedes a little more ...
BvU said:From there it's a small step to the right answer (not c)
I think you are right.Jahnavi said:I don't think so
I think it should be c) .
For simplicity consider a cylinder instead of sphere partially submerged in water . The pressure at the top is PO . Pressure at the bottom is PO+hρg . When we do a force balance the effect of atmospheric pressure cancels leaving us with a net upward buoyant force which depends only on the length of cylinder submerged . This essentially tells us that amount of object submerged in water is independent of air pressure . Even if we increase the pressure inside jar , level of object submerged doesn't change .
@Jahnavi This is an important clue that @BvU has provided. An example: How is it possible that a hot air balloon or a helium balloon "floats"? Does Archimedes principle apply to a helium or hot air balloon? And would a helium balloon float on the moon, which has very little, essentially almost no atmosphere? ## \\ ## One additional hint: Assume the cork is a very low density cork. Work the problem from that perspective.BvU said:Perhaps a thought experiment with a much heavier gas than air ?
Charles Link said:The air by Archimedes principle can in fact be treated as a "liquid" that supplies a buoyant force
You are overlooking that the air pressure will be not quite uniform. It's a subtle point, but it explains why BvU's argument leads to a different answer.Jahnavi said:I don't think so
I think it should be c) .
For simplicity consider a cylinder instead of sphere partially submerged in water . The pressure at the top is PO . Pressure at the bottom is PO+hρg . When we do a force balance the effect of atmospheric pressure cancels leaving us with a net upward buoyant force which depends only on the length of cylinder submerged . This essentially tells us that amount of object submerged in water is independent of air pressure . Even if we increase the pressure inside jar , level of object submerged doesn't change .
Wrong. This exercise is a bit more subtle. The pressure in the air part is being changed -- you can not ignore ##\rho_{\rm air} g h_{\rm air}##Jahnavi said:The pressure at the top is PO . Pressure at the bottom is PO+hρg .
See also the Editing: comment in post 15. In many cases, the buoyant force of the atmosphere is negligible. Archimedes principle is generally quite universal, and also applies to air, even though Archimedes didn't state it that way. ## \\ ## Additional item is Archimedes principle should really more accurately read the "effective volume displaced"=the volume below the water-line. Because it is possible to float a small boat weighing 10 pounds (shaped like the tub but just sightly smaller) in a tub of water using only one pound of water. The "effective" volume of water displaced is 10 pounds for this case, even though you actually have only one pound of water.Jahnavi said:OK . Why doesn't then volume of air displaced by object appear in buoyant force ?
BvU said:Excess noise from too much help. I withdraw.
haruspex said:You are overlooking that the air pressure will be not quite uniform.
The buoyant force from any liquid or gas is indeed a result of a pressure gradient that occurs. In any case, in computing the buoyant force, all that is necessary is to apply Archimedes principle, and to compute the weight of the displaced liquid or gas to get the buoyant force. ## \\ ## The statement of Archimedes principle can be derived by assuming equilibrium between the force per unit volume ## f_p=-\nabla P ## pressure gradient force, and the gravitational force per unit volume ## f_g=\delta g ##. With a vector calculus identity involving the integral of the pressure gradient over a volume (I believe it is a form of Gauss' law), the statement of Archimedes principle follows. ## \\ ## @Jahnavi To respond to your latest post: The details just presented shows a microscopic pressure difference with height in the air necessarily occurs, but all you need to do is consider the air at a fixed pressure and compute the weight of the air displaced to compute the buoyant force from the air.haruspex said:You are overlooking that the air pressure will be not quite uniform. It's a subtle point, but it explains why BvU's argument leads to a different answer.
But I'm not sure what cork ball will do in a vacuum. Lose air while staying the same shape, blow up like a balloon, explode, or...?
Charles Link said:The buoyant force from any liquid or gas is indeed a result of a pressure gradient that occurs.
Yes/No. That means he takes ##\rho_{\rm air} = 0 ## which makes the exercise pointless.Jahnavi said:which means that the question setter is assuming air pressure uniform .Isn't it ?
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". :)Jahnavi said:But to spoil the fun , answer given is c) , which means that the question setter is assuming air pressure uniform .Isn't it ?
BvU said:No. Does the question setter have a good reputation ? I seem to remember not...