Buoyant force acting on an inverted glass in water

[MatinSAR]

Homework Statement

Suppose we put a glass upside down in water at a high speed. Therefore, the air inside does not escape. As the glass goes down in the water, the pressure increases and the air inside the glass is compressed (its volume decreases). How to apply the force F so that the object moves downwards at a constant speed? (The depth of the container is infinite.)

Relevant Equations

Newton's second law
Archimedes' principle

My answer : According to the question, the glass and the air inside it entered the water. Let's assume that the net force becomes zero at a moment, that is, the sum of the weight force and F is equal to the buoyancy force. By going down in the water, the gas volume decreases, so the buoyancy force also decreases, as a result, F also decreases. From time to time, the gas volume stops decreasing, as a result, the buoyancy force and F remain constant. So F decreases after equilibrium and remains constant after stopping the decrease in volume.

Can someone tell me Why my answer is wrong ?!

 

[jbriggs444]

Homework Statement:: Suppose we put a glass upside down in water at a high speed. Therefore, the air inside does not escape. As the glass goes down in the water, the pressure increases and the air inside the glass is compressed (its volume decreases). How to apply the force F so that the object moves downwards at a constant speed? (The depth of the container is infinite.)
Relevant Equations:: Newton's second law
Archimedes' principle

My answer : According to the question, the glass and the air inside it entered the water. Let's assume that the net force becomes zero at a moment, that is, the sum of the weight force and F is equal to the buoyancy force. By going down in the water, the gas volume decreases, so the buoyancy force also decreases, as a result, F also decreases. From time to time, the gas volume stops decreasing, as a result, the buoyancy force and F remain constant. So F decreases after equilibrium and remains constant after stopping the decrease in volume.

I am confused both by the question and by your answer.

Why high speed? The air inside an inverted glass will not escape at slow speed either. With high speed comes frictional effects. There would be dynamic effects as well -- compression and rebound as water alternately surges into the glass and bounces back out. We are given no information with which to evaluate those effects.

In your answer you suggest that the gas volume stops decreasing from time to time. Why would this be so?

 

[MatinSAR]

Why high speed? The air inside an inverted glass will not escape at slow speed either. With high speed comes frictional effects. There would be dynamic effects as well -- compression and rebound as water alternately surges into the glass and bounces back out. We are given no information with which to evaluate those effects.

Yes you are right ... This question is at the level of high school courses, so please skip this more specialized information.

In your answer you suggest that the gas volume stops decreasing from time to time. Why would this be so?

I meant that at a certain time the reduction in volume stops.

 

[hutchphd]

The question is: why would it stop "at certain times"?

 

[jbriggs444]

Yes you are right ... This question is at the level of high school courses, so please skip this more specialized information.

I meant that at a certain time the reduction in volume stops.

The question stipulates that "the depth of the container is infinite". This means that the reduction in volume never stops. It may approach a limit asymptotically, but it never stops approaching it.

Since this is high school level, we will ignore the fact that not just air, but also water and glass become more dense with increasing pressure.

 

[Lnewqban]

Air will always be less dense than the water it displaces; therefore, some buoyancy force will always be present.
I don’t see a reason for the volume of air to stop decreasing at a certain value, unless you keep the glass at a constant depth.

Even for the deepest and coldest point in the seas, the air will remain being a gas.
Air is formed mainly by about 80% nitrogen and 20% oxygen (and by small amounts of other gasses).
Both gases have extremely cold condensation temperatures at sea bottom pressure.

Did anybody explain to you why is your reasoning considered to be incorrect?

Please, see what happens to styrofoam when it is submerged:
https://www.sciencefriday.com/educational-resources/high-pressure-in-the-deep-ocean/

 

[MatinSAR]

The question is: why would it stop "at certain times"?

How to apply the force F so that the object moves downwards at a constant speed? (The depth of the container is infinite.)

 

[erobz]

How to apply the force F so that the object moves downwards at a constant speed? (The depth of the container is infinite.)

Write Newtons Second Law for the system. You are only going to be able to talk about it qualitatively (unless there is information you aren't sharing?).

 

[Steve4Physics]

@MatinSAR, in addition to what has already been said, here's an attempt to explain the mistakes (and omissions) in your original answer.

I'm assuming only a qualitative answer was required, not an equation for F as a function of time.

My answer : According to the question, the glass and the air inside it entered the water. Let's assume that the net force becomes zero at a moment,

You have not clearly explained that, because the glass moves at constant velocity, its acceleration is zero. And since $F_{net}=ma$, this means the net force must always be zero while the glass moves downwards through the water.

that is, the sum of the weight force and F is equal to the buoyancy force.

The weight force ($\vec W$) and $\vec F$ both act downwards. Buoyancy ($\vec B$) acts upwards. So $\vec W+\vec F = \vec B$ can never be true. You mean the magnitudes of these forces give $W + F = B$.

By going down in the water, the gas volume decreases,

Correct. Though you haven't explained why the gas volume decreases.

so the buoyancy force also decreases

Correct. Though you haven't explained why the buoyancy force decreases.

, as a result, F also decreases.

OK.

Edit. Do you think it's possible that, at some depth, F would be zero? If so, consider what happens after that point and how F would have to change in order to keep the velocity constant!

From time to time, the gas volume stops decreasing, as a result, the buoyancy force and F remain constant.

That’s just plain wrong, as others have noted. This is a smooth, continuous process. The gas volume will smoothly and continuously decrease . Why would you think otherwise?

So your conclusion about how F and B change is also wrong.

So F decreases after equilibrium

What do you mean 'after equilibrium'? The descending glass is always in equilibrium (zero acceleration, so zero net force). I think you mean 'So F decreases after the glass enters the water.'.

and remains constant after stopping the decrease in volume.

No. F does not remain constant and the volume never stops decreasing. Your conclusion based on incorrect assumptions.

Can someone tell me Why my answer is wrong ?!

See above! Also, you haven't really answered the question about F. Without equations, one way to describe what F does would be to sketch a graph of F's magnitude vs. time. Can you do this?

Edit. minor wording change

 

[MatinSAR]

Write Newtons Second Law for the system. You are only going to be able to talk about it qualitatively (unless there is information you aren't sharing?).

Yes ... It is enough to talk about F decreasing or increasing or remaining constant.

See above! Also, you haven't really answered the question about F. Without equations, one way to describe what F does would be to sketch a graph of F's magnitude vs. time. Can you do this?

Thank you ... I will check your explanation carefully and then I will try to sketch the graph.

 

[haruspex]

You are only going to be able to talk about it qualitatively

Umm… why? Seems to me it is possible to write an equation for the force as a function of depth.

 

[erobz]

Umm… why? Seems to me it is possible to write an equation for the force as a function of depth.

I consider that qualitative in the sense that we can only talk about general characteristics\behavior. I always thought quantitative implied with numerical values.

 

[hutchphd]

I consider that qualitative

There is no reason one cannot assign a volume V to the air and a mass to the glass and figure a good quantitative solution. If you happen to know that 28 ft of depth in water gives one atmosphere you are ahead of the game.
As anyone who has ever worn a wet suite at depth knows it is possible to become negatively buoyant with depth if you rely on bubbles for buoyancy and wear some lead. This rig likely will do the same. Lots of good Physics here.

 

[berkeman]

As anyone who has ever worn a wet suite at depth knows it is possible to become negatively buoyant with depth if you rely on bubbles for buoyancy and wear some lead.

Yep! Free diving off the Northern California coast (which requires a wetsuit), I knew how my buoyancy changed with depth, and where I transitioned from positive to negative buoyancy. I tried not to stay too long below that depth...

We fine tuned our weightbelts to make us neutrally buoyant at the most likely depth for finding those yummy abalones...

 

[haruspex]

I always thought quantitative implied with numerical values.

No, it only implies a solution that would produce a numerical result given the relevant data. A qualitative description is e.g. one that says which way y will change as x increases, but doesn’t say by how much or how fast.

 

[erobz]

@MatinSAR

I've spent some time analyzing this, I think I've worked it out. This isn't a trivial problem IMO ( as far as analysis goes). Once you figure out what is happening with the gas in terms of the work being done by $F$ to compress it, then what is happening with the force becomes obvious as $h$ gets very large. However, try to use the accompanying diagrams to first formulate Newtons 2^nd for the force $F$ required to very slowly (at constant velocity) push the cup down to a depth $l$ such that it is just fully submerged, before moving on to your actual question.

Some good assumptions to make IMO:

1. Isothermal Compression of an Ideal Gas
2. Weight of air is negligible
3. Density of cup ( having mass $M$) is large in comparison to the density of water i.e. $\rho_M \gg \rho_w$
4. Very large reservoir such that its change in height is insignificant from displacing the volume of the cup+ gas

Notes: Take care when dealing with absolute and gage pressures, and the forces acting on the cup due to them.

Spoiler: Hint

I started by equating the absolute hydrostatic pressure at section 1-1 to the absolute pressure of the trapped gas in the cup. The motivation is to determine $\delta(h)$ ( $\delta$ - the change in volume per unit area of the gas as a function of the depth $h$). You will then use that result in formulating Newtons 2^nd Law for the forces acting on the cup.

 

[hutchphd]

This isn't a trivial problem IMO ( as far as analysis goes).

This is not complicated. Achimedes and ideal gas: but I dare not say more .

 

[erobz]

This is not complicated. Achimedes and ideal gas: but I dare not say more .

I said its not "trivial". I didn't say it was un-doable. Are you just commenting to try and hurt my feelings?

 

[hutchphd]

No. I am attempting to not scare away the OP.

 

[erobz]

No. I am attempting to not scare away the OP.

I don't think the OP is scared; they seem quite capable to me. Probably just busy.

 

[hutchphd]

Nontrivial is in the eye of the beholder I guess. No offense intended.

 

[erobz]

Nontrivial is in the eye of the beholder I guess. No offense intended.

Yeah, I have no problem admitting I'm a simpleton.

 

[MatinSAR]

I don't think the OP is scared; they seem quite capable to me. Probably just busy.

No I wasn't busy. I have had problems to login to my account ...
Thank you for your answer.

 

[MatinSAR]

No. F does not remain constant and the volume never stops decreasing. Your conclusion based on incorrect assumptions.

Can't we ignore decrease in gas volume at great depth?

 

[erobz]

Can't we ignore decrease in gas volume at great depth?

No, because it's tied to the pressure pushing back at you from the compressing the gas. The buoyant force from the gas tends to 0, but the compressive force from the gas goes the opposite way.

 

[jbriggs444]

No, because it's tied to the pressure pushing back at you from the compressing the gas. The buoyant force from the gas tends to 0, but the compressive force from the gas goes the opposite way.

Wait, what? "Compressive force from the gas"?

You've drawn your system boundaries in the wrong place. There is no such external force on my closed system consisting of the glass plus the entrained air.

 

[hutchphd]

Assume it behaves as an ideal gas always. But ignore the compressibility of glass and water.

 

[haruspex]

Can't we ignore decrease in gas volume at great depth?

If it gives a different answer to the posed question, you can't ignore it.

 

[erobz]

Wait, what? "Compressive force from the gas"?

You've drawn your system boundaries in the wrong place. There is no such external force on my closed system consisting of the glass plus the entrained air.

The force $F$ is compressing the gas in the container. There is no bottom to the container.

 

[jbriggs444]

The force $F$ is compressing the gas in the container. There is no bottom to the container.

There is a force $F_\text{top}$ acting on the top of the container. There is a force $F_\text{bottom}$ acting on the bottom of the compressed gas.

There is a name for the vector sum of those two forces.

 

[erobz]

There is a force $F_\text{top}$ acting on the top of the container. There is a force $F_\text{bottom}$ acting on the bottom of the compressed gas.

There is a name for the vector sum of those two forces.

That is the buoyant force( that declines as $h \to \infty$ ). There also an unbalanced force on the inside of the container(an internal force) pushing up (there is no bottom of the container for it to cancel). If you are sitting at some depth holding it $F$, and you wish to go deeper by increasing the force $F'$, the force $F'$ must do the amount of work necessary to compress the gas a further distance $\delta$. ( $\delta \neq \Delta h$ )

 

[haruspex]

That is the buoyant force( that declines as $h \to \infty$ ). There also an unbalanced force on the inside of the container pushing up (there is no bottom of the container for it to cancel). If you are sitting at some depth holding it $F$, and you wish to go deeper by increasing the force $F'$, the force $F'$ must do the amount of work necessary to compress the gas a further distance $\delta$. ( $\delta \neq \Delta h$ )

What is your free body here? If the force sum @jbriggs444 mentions is the buoyancy force then the body is container+gas, making the additional force you mention an internal force. If the body is just the container then it is F_top plus your additional force that sums as the buoyancy force.

 

[jbriggs444]

That is the buoyant force. There also an unbalanced force on the inside of the container pushing up (there is no bottom of the container for it to cancel). If you are sitting at some depth holding it $F$, and you wish to go deeper by increasing the force $F'$, the force $F'$ must do the amount of work necessary to compress the gas a further distance $\delta$. ( $\delta \neq \Delta h$ )

I am not following what you are trying to establish here. Let me try to talk it through.

You are at a particular depth. You are applying a force $F$ that is required to hold the container plus entrained air in place at that depth.

Whether the container is currently ascending or descending, the amount of force required to do this is identical. There is no hysteresis. If you increase the downward force that you apply, you will no longer be in an equilibrium situation. The container and the entrained air will be accelerating downward.

But we are told the the container and the entrained air are not accelerating. We are modulating the applied force so that they descend at a uniform pace. It should follow that the applied force is simply the equilibrium force required to balance with weight and buoyancy.

Still, you point out, correctly, that work is being done on the gas as it is compressed. How can this be?

This can be because the glass plus contained air is not descending at a uniform rate. The glass is descending at a uniform rate, but the lower surface of the air (and, hence, the centroid of the volume of air) is not. It is descending at a lower rate.

If we calculate the rate at which work is being done on the system, we can add up the work done by our downward push. That is easy. $F_\text{hand} \cdot v$. But when we calculate the work done by fluid pressure (aka buoyancy), we must be careful not to evaluate $F_\text{buoyancy} \cdot v$. Instead we must evaluate $F_\text{top} \cdot v$ + $F_\text{bottom} \cdot (v - v_{\text{compression}})$.

None of this alters the force balance that is present. It is just reconciling the energy balance.

[I assume that the air is negligibly massive compared to the glass so that we do not have to worry about the acceleration of the air]

 

[erobz]

I'm not sure. I've probably jacked it up as usual. I'll think some more on my confusion later.

 

[MatinSAR]

I'm assuming only a qualitative answer was required, not an equation for F as a function of time.

Yes.

You have not clearly explained that, because the glass moves at constant velocity, its acceleration is zero. And since Fnet=ma, this means the net force must always be zero while the glass moves downwards through the water.

I assumed that we put the object in the water with a force F, which is a big force, and we reduced this force so that the force on the object becomes zero at a moment.

The weight force (W→) and F→ both act downwards. Buoyancy (B→) acts upwards. So W→+F→=B→ can never be true. You mean the magnitudes of these forces give W+F=B.

Yes. Thank you ...
I made a big mistake here ...

Correct. Though you haven't explained why the gas volume decreases.

Because in the depths, the water pressure compresses the gas.

Correct. Though you haven't explained why the buoyancy force decreases.

Because the buoyancy force is directly related to the volume occupied by the object and the volume decreases here.

Edit. Do you think it's possible that, at some depth, F would be zero? If so, consider what happens after that point and how F would have to change in order to keep the velocity constant!

Yes, if the force of buoyancy and the force of weight are equal. In this case, the buoyancy force decreases as it goes down, so F must be increased.

No. F does not remain constant and the volume never stops decreasing. Your conclusion based on incorrect assumptions.

Yes.

See above! Also, you haven't really answered the question about F. Without equations, one way to describe what F does would be to sketch a graph of F's magnitude vs. time. Can you do this?

I think F decreases after entering the water. As F decreases, at moment T, the forces on the object are balanced. From this moment on, since the buoyancy force decreases, F must also decrease so that the forces remain balanced.But I did not understand how to use F=0 in the answer.

 

[MatinSAR]

Thank you for your time and help ... @jbriggs444 @haruspex @hutchphd @berkeman @Steve4Physics @Lnewqban

 

[haruspex]

I think F decreases after entering the water.

Yes, but can you write the equation? I read the question as asking for that, not just whether it increases, decreases or stays the same.

 

[jbriggs444]

But I did not understand how to use F=0 in the answer.

You are distinguishing between $F_\text{net} = 0$ and $F_\text{applied} = 0$, right? I see no particular importance in the depth where $F_\text{applied} = 0$.

It might also be worth thinking of $\vec{F_\text{applied}}$ as a vector. So that the notions of "increasing" or "decreasing" are replaced by "is increasingly downward" or "is increasingly upward".

 

[hutchphd]

I think F decreases after entering the water. As F decreases, at moment T, the forces on the object are balanced. From this moment on, since the buoyancy force decreases, F must also decrease so that the forces remain balanced.But I did not understand how to use F=0 in the answer.

Yes good. (this assumes the glass without air entrapped will sink!). The other place place where the force is zero is when the glass is floating "proud" of the surface at the beginning. If you let F be a signed number then it can be either up or down (a 1D vector) then the equation works out. Can you see that it has some (undetermined) max value (down} when the glass is just at the surface. Sketch the graph of F from way down floating

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

A post with very confusing formatting has been edited, and the thread is back open. Thanks for your patience.

 

[erobz]

Ok, the force that I was thinking was there ( unbalance pressure internal to the cup) was a figment of my carless imagination. I was somehow forgetting about the pressure acting on the top of the cup.

 

[erobz]

Now that I think I'm back on track. Unless I'm not seeing something there isn't a continuous function for the force $F$ from the state where the cup is only partially submerged ( where $F_b \propto h - \delta$):

through where the cup is fully submerged state where $F_b \propto l - \delta$:

$\delta(h)$ is continuous, but the buoyant force changes at $h = l$.

Is that accurate?

 

[jbriggs444]

Now that I think I'm back on track. Unless I'm not seeing something there isn't a continuous function for the force $F$ from the state where the cup is only partially submerged through where the cup is fully submerged state where $F_b \propto l - \delta$:

I disagree. The force $F$ applied to keep the cup in equilibrium is indeed a continuous function of depth. However it is not continuously differentiable. Indeed, at the exact point of submersion, the function is not differentiable at all.

"Continuous" means (intuitively) that you can draw the graph without lifting your pencil from the paper.
"Continuously differential" means (intuitively) that you can't change the direction of the line suddenly either.

I see that you have adopted $h$ as the depth of the cup's bottom. So we can define $F(h)$ as the applied force required to establish an equilibrium at depth $h$. Then $\frac{dF}{dh}$ is the rate of change in F with respect to depth.

The correct claim that I think you want to make is that $\frac{dF}{dh}$ is not continuous (or even defined) at $h=l$

The buoyancy is equal to the volume of displaced water. That volume is roughly the same whether the cup is just barely proud of the water or just barely submerged. But the rate of change of that displaced volume with respect to depth is large while proud of the water and is much smaller when just barely submerged.

 

[erobz]

I disagree. The force $F$ applied to keep the cup in equilibrium is indeed a continuous function of depth. However it is not continuously differentiable. Indeed, at the exact point of submersion, the function is not differentiable at all.

"Continuous" means (intuitively) that you can draw the graph without lifting your pencil from the paper.
"Continuously differential" means (intuitively) that you can't change the direction of the line suddenly either.

I see that you have adopted $h$ as the depth of the cup's bottom. So we can define $F(h)$ as the applied force required to establish an equilibrium at depth $h$. Then $\frac{dF}{dh}$ is the rate of change in F with respect to depth.

The correct claim that I think you want to make is that $\frac{dF}{dh}$ is not continuous (or even defined) at $h=l$

The buoyancy is equal to the volume of displaced water. That volume is roughly the same whether the cup is just barely proud of the water or just barely submerged. But the rate of change of that displaced volume with respect to depth is large while proud of the water and is much smaller when just barely submerged.

I'm thinking my nomenclature was a bit misused(abused).

I agree that $F$ is continuous through the entire domain. What I meant to say was $F$ is piecewise defined for $0 \leq h \leq l$ and $h > l$.

 

[MatinSAR]

Yes, but can you write the equation? I read the question as asking for that, not just whether it increases, decreases or stays the same.

Thank you very much, but specifying How F changes is enough.

I say the equation I found, But I am not sure if you meant this equation or not..

F = -ρ_w Vg-mg

 

[erobz]

Thank you very much, but specifying How F changes is enough.

I say the equation I found, But I am not sure if you meant this equation or not..

F = -ρ_w Vg-mg

The buoyant force acts opposite the force $F$ (you had it correct before the edit - loose the negative sign out front). The next step is to write $V\llap{-}$ as a function of depth using the ideal gas law (and hydrostatic pressure).

 

[MatinSAR]

The buoyant force acts opposite the force $F$ (you had it correct before the edit). The next step is to write $V\llap{-}$ as a function of depth using the ideal gas law (and hydrostatic pressure).

So F= ρ_wVg - mg

P1 : Initial pressure (Pressure, exactly when the glass enters the water)
V1 : Initial volume ( Volume, exactly when the glass enters the water)
P : Final pressure
V : Final volume

P₁V₁=PV ==> V = (P₁/P)V₁
P₁/P=h₁/h

h1 : The height of the glass

 

[erobz]

So F= ρ_wVg - mg

P1 : Initial pressure (Pressure, exactly when the glass enters the water)
V1 : Initial volume ( Volume, exactly when the glass enters the water)
P : Final pressure
V : Final volume

P₁V₁=PV ==> V = (P₁/P)V₁
P₁/P=h₁/h

h1 : The height of the glass

Assume a cylindrical cup with cross-section $A$.

Think about the volume of the gas when the leading edge of the cup is at some depth $h$ w.r.t. its initial volume at atmospheric pressure $P_{atm}$ when it was at the surface.

Try to use the diagram:

Otherwise you are in the ball park.Also, I recommend leaving the expression you develop from the ideal gas law as $P_{atm}V_o\llap{-} = P V\llap{-}$

 

[Frabjous]

There are a lot of assumptions one has to make in order to do the quantified calculations. I think the point of the problem is at entry, there is zero buoyancy force so the initial applied force opposes gravity only. The buoyancy force grows until it reaches a maximum when the glass is fully submerged at which point it begins to decrease and asymptote to a value with a zero volume of air with corresponding changes to the applied force.