Static lift: How to calculate the volume of displaced fluids?

[physiker278]

Homework Statement

The homework statement is given in the picture.

Relevant Equations

I am not quite sure what the relevant equations are. I have included my approach but I am not getting any further to the value of h1.

 

[Delta2]

I think we can't find the value of $h_1$ unless we are given the total height of the cylinder or its mass density $\rho_z$.

 

[haruspex]

I think we can't find the value of $h_1$ unless we are given the total height of the cylinder or its mass density $\rho_z$.

Seems to me the cylinder is hollow, filled with the same fluid as the tank.

@physiker278, are you saying $h_x=h_2=d=8cm$? That does not make sense.

 

[Delta2]

Seems to me the cylinder is hollow, filled with the same fluid as the tank

So you mean that the cylinder is open from his bottom side? And $m_z$ is the mass of the shell of the cylinder?

 

[Steve4Physics]

There is (at least) one mistake. We are given
$h_X, h_2,d = 8cm$
But from the digram it is obvious that $h_X$ must be smaller than $h_2$.

Edit. @haruspex beat me to it!

 

[Steve4Physics]

Taking the cylinder to be open at its lower end...

We are probably meant to assume that the cylinder walls have negligible thickness. This then means there is negligible upthrust due to the liquid displaced by the actual cylinder walls.

As a consequence, the length of cylinder below the open surface-is irrelevant, (as long as $h_1$ and the cylinder mass remain the same).

As far as I can see $h_X$ and $h_2$.aren't needed are are possibly intended as 'distractors'.

 

[physiker278]

Seems to me the cylinder is hollow, filled with the same fluid as the tank.

@physiker278, are you saying $h_x=h_2=d=8cm$? That does not make sense.

No, I am not saying it. d=8cm and $h_x$ and $h_2$ are constants

 

[physiker278]

So you mean that the cylinder is open from his bottom side? And $m_z$ is the mass of the shell of the cylinder?

Yes this is correct.

 

[Delta2]

Yes this is correct.

Then ignore what I said in post #2. Let me see how we can handle this problem...

 

[Steve4Physics]

@physiker278, there is an upwards force ($F_H=7N$) acting on the top face of the cylinder. What downwards force(s) balance it?

(General hint: make sure you are familiar with the physics of a simple barometer.)

 

[Delta2]

What downwards force(s) balance it?

The only downward force I see is the weight of the cylindrical shell. The force from the fluid must be upwards.

Also i think $F_H$ is the resultant of the weight and the force from the fluid at the top face of the cylinder.

 

[Chestermiller]

In terms of $h_1$, $\rho$, g, and $p_{atm}$, what is the absolute pressure at the top inside face of the cylinder?

 

[Steve4Physics]

The only downward force I see is the weight of the cylindrical shell. The force from the fluid must be upwards.

Also i think $F_H$ is the resultant of the weight and the force from the fluid at the top face of the cylinder.

Are you using g=10m/s²?. What is the weight of the cylinder? How does this compare to the upwards force $F_H$?

There is a second downwards force.

Suppose the liquid were exerting an upwards force on the top face of the cylinder. If you made a small hole in this top face, you would expect the liquid would squirt out upwards. Is this what you would think wiould happen?

 

[Delta2]

In terms of $h_1$, $\rho$, g, and $p_{atm}$, what is the absolute pressure at the top inside face of the cylinder?

is it $p_{atm}+\rho g h_1$?

 

[Delta2]

There is a second downwards force.

Suppose the liquid were exerting an upwards force on the top face of the cylinder. If you made a small hole in this top face, you would expect the liquid would squirt out upwards. Is this what you would think wiould happen?

I can't see what is the second downwards force (really).

And yes i would expect the fluid to squirt upwards...

 

[Delta2]

Ah ok i just saw the second downward force it it is $p_{atm} A$ where A the surface of the top of the cylinder. And the pressure on the inner surface of the top must be $p_{atm}-\rho g h_1$

 

[Chestermiller]

is it $p_{atm}+\rho g h_1$?

I get $p_{atm}-\rho g h_1$, and air would be sucked in through a hole in the top.

 

[Delta2]

Yes ok somehow i reversed gravity in my brain lol...

 

[Steve4Physics]

And yes i would expect the fluid to squirt upwards...

Oh no it wouldn't! If it did, you would have invented a perpetual motion machine! Overall, the liquid would be raised-up and you would have energy created from nothing!

Suppose the top of the cylinder is a disc which can slide up/down the barrel of the cylinder but is being held in position. Which direction would this disc move when released?

 

[Delta2]

Suppose the top of the cylinder is a disc which can slide up/down the barrel of the cylinder but is being held in position. Which direction would this disc move when released?

I think i got it, the disc would move downwards...

 

[Steve4Physics]

I think i got it, the disc would move downwards...

[Edited to make corrections.]

Yes. If it helps, imagine a tiny amount of air at the top of the liquid. There is then a Torricellian vacuum. The lower surface of the cyinder's top then experiences a downwards force.

Even wih no air at the top, a downwards force exists on the cylinder's top because the pressure on the top's upper surface is $p_{atm}$ and the pressure on the top's lower surface is $p_{atm} - \rho g h_1$.

The downwards force (additional to the weight) arises from the pressure difference between the top's upper and lower surfaces. And this force has a very simple physical interpretation.

 

[Steve4Physics]

Yes ok somehow i reversed gravity in my brain lol...

You could have said: "I was using a sign-convention where g is negative".

 

[Delta2]

You could have said: "I was using a sign-convention where g is negative".

No I wasn't doing that, I just was thinking that the weight of the fluid acts towards the up direction.

 

[Steve4Physics]

No I wasn't doing that, I just was thinking that the weight of the fluid acts towards the up direction.

I realized that. My suggestion (to use the sign-convention as an excuse) was meant to be a joke!

 

[Delta2]

E hehe I am kind of sincerely stupid, I admit my mistakes instead of trying to hide...

 

[Steve4Physics]

Hi @physiker278. Are you still reading this thread? Apologies if it's strayed from the original question a bit.
Have you managed to answer the question yet? Let us know if you still need help.