Buoyant force acting on an inverted glass in water

[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.

 

[MatinSAR]

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.

Ok I will but why should I find this ?

 

[erobz]

Ok I will but why should I find this ?

Because you to eliminate $\delta$ as an independent variable and get everything in terms of the depth $h$.

 

[MatinSAR]

Because you to eliminate $\delta$ as an independent variable and get everything in terms of the depth $h$.

If we consider h=0 :

 

[erobz]

If we consider h=0 :
View attachment 318757

Thats isn't what I meant.

Did you solve for $\delta$ as a function of $h$ yet?

Remember, $\delta$ is part of both the pressure and the volume of the gas. When I say solve for $\delta$ I mean solve the following for $\delta$:

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

 

[Frabjous]

If we consider h=0 :
View attachment 318757

Think about what h=0 physically corresponds to.

 

[MatinSAR]

Thats isn't what I meant.

Did you solve for $\delta$ as a function of $h$ yet?

Remember, $\delta$ is part of both the pressure and the volume of the gas. When I say solve for $\delta$ I mean solve the following for $\delta$:

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

Can I find δ using :

P_atmL/(L -δ) = P_atm+ρg(h-δ)

Think about what h=0 physically corresponds to.

The moment that the glass enters the water.

 

[erobz]

Can I find δ using :

P_atmL/(L -δ) = P_atm+ρg(h-δ)

Give it a try.

 

[MatinSAR]

Give it a try.

At h=0 delta should be zero. The above equation confirms this.
And we can see that if h goes to ∞ the delta goes to L.

 

[erobz]

View attachment 318764
At h=0 delta should be zero. The above equation confirms this.
And we can see that if h goes to ∞ the delta goes to L.

Seems good. I had to check whether that quadratic simplified, I didn't take it that far! Good Job!

 

[MatinSAR]

Seems good. I had to check whether that quadratic simplified, I didn't take it that far! Good Job!

I appreciate your help. I didn't use P_atmL/(L -δ) = P_atm+ρg(h-δ) because
2nd degree equation was obtained ...

I have used the formula you suggested ...