Buoyant force acting on an inverted glass in water

[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 ...

 

[erobz]

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 ...

What formula?

 

[MatinSAR]

Think about what h=0 physically corresponds to.

Thank you for you rhelp and time.

What formula?

This post :
https://www.physicsforums.com/threa...-inverted-glass-in-water.1047903/post-6831524

 

[erobz]

Thank you for you rhelp and time.

This post :
https://www.physicsforums.com/threa...-inverted-glass-in-water.1047903/post-6831524

I don't understand! I solved it as a quadratic. However, your result indicates that my quadratic must be factorable I believe?

 

[MatinSAR]

I don't understand! I solved it as a quadratic. However, your result indicates that my quadratic must be factorable I believe?

Maybe I made a mistake ...
was my answer correct ??

 

[erobz]

Maybe I made a mistake ...
was my answer correct ??

It plots identically to my solution?

 

[MatinSAR]

It plots identically to my solution?

Let me send you a picture of what I have done ...

 

[MatinSAR]

It plots identically to my solution?

My mistake is here :

So to find delta we have to solve a quadratic equation ...

The question only asked about the increase and decrease of F force, I didn't want to waste your time at all... I'm sorry...
edit :

I have solved that quadratic equation ...

 

[erobz]

My mistake is here :
View attachment 318768

So to find delta we have to solve a quadratic equation ...

The question only asked about the increase and decrease of F force, I didn't want to waste your time at all... I'm sorry...
edit :

View attachment 318770

I have solved that quadratic equation ...

What you did on accident the first time by omitting $\delta$ was actually a really good approximation for $\delta \ll h$. For the parameters I used I couldn’t tell the apart.

 

[erobz]

My mistake is here :
View attachment 318768

So to find delta we have to solve a quadratic equation ...

The question only asked about the increase and decrease of F force, I didn't want to waste your time at all... I'm sorry...
edit :

View attachment 318770

I have solved that quadratic equation ...

I wanted you to explore it. It wasn’t a waste of my time at all!

 

[MatinSAR]

I wanted you to explore it. It wasn’t a waste of my time at all!

Thank you for your help and time ...