Buoyant force acting on an inverted glass in water

[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.

 

[erobz]

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

Clarification on the above statement:

The absolute pressure $P$ is a function of $h,\delta$, and the gas volume is a function of $\delta$.

$$P_{atm}V_o\llap{-} = P( h,\delta) V\llap{-}( \delta)$$

The objective is to solve that for $\delta(h)$.

 

[MatinSAR]

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.

Thank you.

Clarification on the above statement:

The absolute pressure $P$ is a function of $h,\delta$, and the gas volume is a function of $\delta$.

$$P_{atm}V_o\llap{-} = P( h,\delta) V\llap{-}( \delta)$$

The objective is to solve that for $\delta(h)$.

Thank you. I will try again today ...

 

[MatinSAR]

Clarification on the above statement:

The absolute pressure $P$ is a function of $h,\delta$, and the gas volume is a function of $\delta$.

$$P_{atm}V_o\llap{-} = P( h,\delta) V\llap{-}( \delta)$$

The objective is to solve that for $\delta(h)$.

If we assume that the temperature of the gas is constant, the pressure at depth h is equal to:
P=(P_atmL) / (L-𝛿)

Is it true ?!

 

[Frabjous]

If we assume that the temperature of the gas is constant, the pressure at depth h is equal to:
P=(P_atmL) / (L-𝛿)

Is it true ?!

The pressure at depth h, P(h), is P_atm plus the weight of the water column, ρ_watergh.
The pressure of the air in the glass is P_air=P(h-δ)
You can then use the EOS to determine V_air.
You can then use the shape of the glass to determine δ.

 

[erobz]

If we assume that the temperature of the gas is constant, the pressure at depth h is equal to:
P=(P_atmL) / (L-𝛿)

Is it true ?!

You are on the right track. $P$ is the pressure of the gas in the cup when the leading edge of the cup is at depth $h$. So what is left to do is to write $P$ ( LHS ) as a function of the depth $h$ and total change in volume per unit area $\delta$.

Remember: $P$ is an absolute pressure.

 

[MatinSAR]

The pressure of the air in the glass is Pair=P(h-δ)

Can you please tell why ?

You are on the right track. $P$ is the pressure of the gas in the cup when the leading edge of the cup is at depth $h$. So what is left to do is to write $P$ ( LHS ) as a function of the depth $h$ and total change in volume per unit area $\delta$.

Remember: $P$ is an absolute pressure.

Thank you ...

 

[erobz]

I think (hope) they mean that the pressure of the trapped gas is a function of $h- \delta$. I can see how that could be confusing.

So far you are doing good. You just need to get the pressure of the trapped gas as a function of $h$ and $\delta$. That is what I meant by $P(h,\delta)$.

Spoiler: Hint

Find the pressure of the gas at the gas-liquid interface by using the hydrostatic relationship.

 

[Frabjous]

Sorry for the confusion. You are on track. You have written the pressure of the air in the glass (assuming a constant glass shape and an EOS). You need to set it equal to the pressure in the water, P(h-δ) to solve for delta. It is at h-δ because that where the air-water interface is located.

 

[MatinSAR]

Find the pressure of the gas at the gas-liquid interface by using the hydrostatic relationship.

Is that equal to P_atm+ρg(h-δ) ?

Sorry for the confusion. You are on track. You have written the pressure of the air in the glass (assuming a constant glass shape and an EOS). You need to set it equal to the pressure in the water, P(h-δ) to solve for delta. It is at h-δ because that where the air-water interface is located.

Thank you.

 

[erobz]

Is that equal to P_atm+ρg(h-δ) ?

Good... plug that in, solve for $\delta(h)$. I would suggest evaluating that solution at $h = 0$ to ensure you have chosen the correct form.