Ball floating on the surface of water

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

When air is evacuated , atmospheric pressure pushing down the cork is eliminated , cork should rise a little i.e option b) . This is incorrect .

What am I missing ?

 

[BvU]

Is the atmosphere really pressing down ?

 

[Jahnavi]

Is the atmosphere really pressing down ?

Suppose 1/4th part of the ball is out of water , then the net force exerted by the air surrounding the ball will be downwards . No ?

 

[BvU]

Compare with a concrete ball under water. Is the water pressing down or up ? (Archimedes had something sensible to say about this)

 

[Jahnavi]

Compare with a concrete ball under water. Is the water pressing down or up ? (Archimedes had something sensible to say about this)

If the ball is completely under water ( water surrounding it from all sides ) then net force exerted by water on ball will be upwards . But in this case only a part of the ball is exposed to air .If a small part is exposed to air , net force exerted by air will be downwards . Although net force exerted by water will always be upwards .

 

[BvU]

Think like Archimedes a little more ... (*)

After all, you already know that b) is not the right answer. Process of elimination of unlikely answers leads you to ... the right answer.

But then: how to explain that ?

(*)
Perhaps a thought experiment with a much heavier gas than air ?

 

[Jahnavi]

Is the atmosphere really pressing down ?

I mean that force exerted by air will be radially inwards ( towards the center of the ball) at each part, but if we calculate the net force then that force will be vertically downwards . Is this incorrect ?

 

[BvU]

Yes.

I repeat: think Archmedes.

 

[Jahnavi]

Think like Archimedes a little more ...

OK . Let me wear my Archimedes hat .

Since air is eliminated effect of atmospheric pressure is eliminated from all sides . The net upthrust is the weight of volume of water displaced . This remains same . Should that be option c) remains unchanged ?

 

[BvU]

Nice hat

The incarnation I was phishing for has to do with

The weight of the displaced portion of the fluid is equivalent to the magnitude of the buoyant force.​

(I think that should be: 'is equal to')

From there it's a small step to the right answer (not c)

 

[Jahnavi]

From there it's a small step to the right answer (not c)

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 P_O . Pressure at the bottom is P_O+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 .

 

[ehild]

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 P_O . Pressure at the bottom is P_O+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 .

I think you are right.
"net upward buoyant force which depends only on the length volume of cylinder submerged"
Edit: I am wrong, I forgot the buoyant force of air

 

[Charles Link]

Perhaps a thought experiment with a much heavier gas than air ?

@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.

 

[Jahnavi]

Hi Charles ,

Here is the key issue .

Do you agree that if we consider a cylinder partially submerged in water , the net buoyant force does not include weight of volume of air displaced by the cylinder AND instead if we consider a cylinder completely submerged in a container containing two different liquids such that part of cylinder lies in one liquid and part in other , buoyant force includes volume of both the liquids displaced by the cylinder ?

 

[Charles Link]

No. The air, with Archimedes principle, can in fact be treated as a "liquid" that supplies a buoyant force. See e.g. https://www.physicsforums.com/threads/lifting-a-man-with-100m-3-of-helium.943829/#post-5972396 and the "links" of post 3 of that thread. $\\$ Editing: Consider also the cork to be a ping-pong (table tennis) ball, that will float very high up on the surface because its density is so low. What happens when you remove the air in the beaker?

 

[Jahnavi]

The air by Archimedes principle can in fact be treated as a "liquid" that supplies a buoyant force

OK . Why doesn't then volume of air displaced by object appear in buoyant force ?

 

[BvU]

Weight of volume of air displaced = $\rho\; g V$ with $V$ volume of object and $\rho$ density of fluid.

 

[haruspex]

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 P_O . Pressure at the bottom is P_O+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 .

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...?

 

[BvU]

The pressure at the top is P_O . Pressure at the bottom is P_O+hρg .

Wrong. This exercise is a bit more subtle. The pressure in the air part is being changed -- you can not ignore $\rho_{\rm air} g h_{\rm air}$

 

[Charles Link]

OK . Why doesn't then volume of air displaced by object appear in buoyant force ?

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.

 

[BvU]

Excess noise from too much help. I withdraw.

 

[Jahnavi]

Excess noise from too much help. I withdraw.

It might seem noise to you , but it is actually helping me understand things . I am thinking about your suggestions .

Don't withdraw !

 

[Jahnavi]

You are overlooking that the air pressure will be not quite uniform.

Excellent point

We generally assume atmospheric pressure is uniform .Don't we ?

Assuming air pressure is uniform ,

Consider a cylinder partially submerged in water with length 'l' . The pressure at the top circular surface is P_O . Force is P_OA . Pressure at the bottom is P_O+lρg .

Net Buoyant force = [P_O+lρg]A -
P_OA = lρgA = Vρg

V is volume of water displaced by the cylinder . So , volume of air doesn't come in picture . This is due to the assumption that atmospheric pressure is constant .

This leads to the question - Why can't we assume atmospheric pressure constant ?

@BvU , @haruspex , @Charles Link please tell what do you think ?

Note : I agree that if we consider air pressure non uniform , volume of air does play a role in buoyant force .

 

[Charles Link]

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...?

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.

 

[BvU]

all that is necessary is to apply Archimedes principle

 

[Jahnavi]

OK . Back to the original question

Assuming air pressure is non uniform ,

Initially , buoyant force = V_air ρ_airg + V_water ρ_waterg .

When air is removed , buoyant force = V_water ρ_waterg .

Since this buoyant force has to balance the weight of the ball , volume of water displaced by ball in latter case is more . Hence the ball sinks a little i.e option a) .

 

[BvU]

Note: Archi principle takes all this nonuniformity into account. Pretty neat !

 

[Jahnavi]

I am glad everyone is happy especially @BvU .

 

[BvU]

Let's not turn PF into a mutual admiration society...

 

[Jahnavi]

But to spoil the fun , answer given is c) , which means that the question setter is assuming air pressure uniform .Isn't it ?

 

[Jahnavi]

The buoyant force from any liquid or gas is indeed a result of a pressure gradient that occurs.

Yes it is , but I believe we need to neglect it in this problem .

 

[BvU]

which means that the question setter is assuming air pressure uniform .Isn't it ?

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

Does the question setter have a good reputation ? I seem to remember not...

 

[Charles Link]

But to spoil the fun , answer given is c) , 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]

No. Does the question setter have a good reputation ? I seem to remember not...

Yes . It is from the test paper of a top institute .

The previous case you are remembering was from a reference book .

 

[BvU]

My bad. I'm lost. What top level institute, what level test ?

 

[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 ?