Pascals principle and hydraulic lift

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says that the pressure is distributed evenly around a fluid.

What I'm having a hard time understanding is the hydraulic lift, so I thought I would think of a simple system that could represent a fluid and then use that for intuition.
Regrettably I did not get the same conclusions with that system, which is attached on the picture below.

The idea is that we have a piston with one area A1 connected via a rod to a piston with a larger area A2=3A1, and we want to figure out the force we have to exert on the other piston to keep the system in equilibrium.

It is not hard to see, that the force you have to exert one both pistons has to be the same. But unfortunately that is not the conclusion from using Pascals principle on a fluid, which tells us that the two pressures have to be equal, which in terms of forces would mean:

F2 = F1*A2/A1

I'm really having a hard time getting intuition for this. Why is it that the fluid, on which we uses Pascals principle can't just be replaced with a rigid rod, like in my system?
 

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  • #2
tiny-tim
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hi zezima1! :smile:

if the rod isn't there, and the liquid is in equilibrium, then the forces on the left must equal the forces on the right

but you have to include the forces on the upper and lower part of the left, not just the centre part :wink:

(alternatively, if you consider only the central tube of fluid, the forces are still the same!)
 
  • #3
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Yes okay, I sort of got to that conclusion after thinking about it. But I also think, I now know more clearly what puzzles me:
It is how you can exert a force and get back a force much bigger. I know you can't call force a conserved quantity but it just doesn't fit with my notion of Newtons 3rd law.
 
  • #4
tiny-tim
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hi zezima1! :smile:
It is how you can exert a force and get back a force much bigger.
same as with a lever :wink:
I know you can't call force a conserved quantity but it just doesn't fit with my notion of Newtons 3rd law.
Newtons 3rd law is local

eg, on a lever, it applies between each bit of the lever and the next bit, not from one end to the other

Newtons 3rd law does apply between any two colliding molecules of the fluid :smile:
 
  • #5
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Okay yes, I have seen levers and all such, so I guess you're right.

But could you try to make the lever principle more intuitive, and explain how exactly it applies here? :)
 
  • #6
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Hello zezima,

To understand this in terms of a conservation law (conservation of energy) try this:

In your diagram

The volume of fluid displaced when you push piston1 = Distance travelled by piston x Area = d1A1.

This fluid must be the same volume as 'pushed ' into the second chamber to move the second piston and so = d2A2

Thus d1A1 = d2A2

since in your example A2 = 3A1

d1 = 3d2 or d2 = d1/3 (note the 3 factor is the other way round)

Now work (energy) = force times distance and F2 = 3 F1

So F2d2 = 3F1 * d1/3 = F1d1

That is energy is conserved.

go well
 
  • #7
tiny-tim
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Okay yes, I have seen levers and all such, so I guess you're right.

But could you try to make the lever principle more intuitive, and explain how exactly it applies here? :)
are you familiar with the principle of work done?

W = F.d (work = force . displacement) …

since work in = work out, the product F.d must be the same at both ends

eg, if one end of the lever is displaced twice as much as the other end, then the force at the other end must be half as much :smile:

similarly, when the tube gets narrower, the piston must move further, and so the force must be less :wink:
 
  • #8
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ahh yes, I think I get what you mean :)

But what if the liquid was confined? Then doesn't the principle still hold? And can you apply conservation of energy then?
 
  • #9
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ahh yes, I think I get what you mean
Can you now see that with a fluid each piston travels a different distance,

Whilst with the rod connecting them in your original question each piston travel the same distance?

BTW Hello, Tim
 
  • #10
tiny-tim
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Hello, Studiot! :smile:

Two pistons! :biggrin:
 
  • #11
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Yeah I'm two pistol Pete.

:biggrin:
 
  • #12
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Okay guys I understand it now, in the case, where you can apply conservation of energy =)

But.. What about if I had the system like on the picture:
The pressure is equal so the system must be in equilibrium. But surely you now can't proof that the two are in equilbrium using work and energy since, there is no work done. How can you explain this?
 

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  • #13
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In equilibrium situations we usually find it more convenient to work in terms of the balance of forces (and moments) being zero as the conditions for equilibrium.

Can you do this in this situation?

Hint apply Newton's third law to each of the masses.

I can, however, tell you that there are energy theorems that are not on your syllabus which you could use but it would be a sledgehammer to crack a nut.

So we use the easy way.

:wink:
 
  • #14
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I can try. Since both masses are at rest, the upwards force from the fluid must be mg and 2mg. But since the area is twice as big for the other we actually do get:
P1 = P2
But that is under the assumption, that the two masses are initially at rest, when you put them on them piston simultaneously. And why can you assume that?
I can, however, tell you that there are energy theorems that are not on your syllabus which you could use but it would be a sledgehammer to crack a nut.
:wink:
Tell me the name of these - I know a bit about lagrangian/hamiltonian mechanics..
 
  • #15
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Try this

Let the 2m mass suffer a virtual downward displacement δy so that the m mass suffers an upward virtual displacement βδy, where is a parameter to be determined

Then the total virual work done equals zero in equilibrium

mgβ-(δy) + 2mgδy = 0

β = 2
 
  • #16
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virtual? :eek: I know there's something called virtual work in more advanced mechanics but I don't know what it is sorry. I just meant that I have heard about analytical mechanics, not that I have any idea how to apply it.
Can't you explain it in terms of newtonian? Or was that what you were trying to? :)

But I certainly can see the idea of what you try to say with the virtual work, and while the idea is neat, it to me seems like assuming that out of pure statistical fluctuation the one mass goes down a little bit, which is not really something that would happen in the mechanics I know - though I might be wrong since I don't know what you mean by virtual displacement.

If this can't be rigourosly explained with Newtonian mechanics please tell me.
 
  • #17
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Studiot :)

can you please in terms of forces try to describe how the piston gets acted on by double the force as the other, when initially you push down on the piston with area A. Please don't use energy conservation :(
 
  • #18
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can you please in terms of forces try to describe how the piston gets acted on by double the force as the other, when initially you push down on the piston with area A. Please don't use energy conservation
I'm not quite sure what you mean.

I assume you are referring to the hydraulic lift in post#12.

So the situation is that the piston with mass m and area A exerts a downward force on the fluid =mg.
In exerting this downward force the piston creates a pressure =mg/A in the fluid.

By Newton's third law the fluid therefore exerts an equal upward force on the piston.



Similarly the downward force at the second piston is 2mg and the upward force equals this.

Now let us push down on the smaller piston with and additional force F.

The piston will move down in accordance with Newton's first and second laws. You are now exerting a force mg + 2F on the fluid

So the pressure created in the fluid now equals (mg+2F)/A = mg/A + F/A.

I have split the pressure into the original pressure and the increase (F/A) due to the extra force you are now exerting.

This extra pressure pushes up on the second piston with a extra force equal to (F/A) * (2A) = 2F

The second piston will therfore move upwards in accordance with Newton's first and second laws since it now experiences a net upward force of 2F.

So long as you maintain the downward foce F on the first piston it will continue to move down and the second piston will continue to move up.

Does this help?
 
  • #19
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hmm no I just find it hard to see how you can exert an external force and get a force bigger from that through internal action. I know that it's possible, but I've always found that hard to understand. The most simple example is a pulley in which you have an extra rope. The conservation of energy is still true since you are just pulling over twice the length.
So something equal to that must be applyable here. I want to know how in terms of forces the individual particles in the column with the smallet area pushes on the other particles such that their net force on the bigger area is twice the external force
hope I made myself clear now :) and thank you
 
  • #20
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So something equal to that must be applyable here.
Yes, I thought we has already covered that.

The small piston moves twice as far as the big one.
 
  • #21
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The small piston moves twice as far as the big one.
THAT'S AN ENERGY OBSERVATION! I want to understand it purely in terms of forces. Okay, maybe I'm not making sense so I tried to draw a model of what I think happens in the fluid, and I'll describe how it is to be interpreted. Then you can tell me what it wrong.

So my notion of all this is that when you apply a force at one piston, here at the one with area A, it will try to move down but bump into the top layer of particles which will slow down the piston. This layer will now try to move down but bump into the next layer and so forth until we reach the other piston, which will move upwards if there isn't any force pressing down on it.
See my drawing, I really tried to make it easy to understand what I mean this time. Do note that the force vectors should all be of equal length.
Now, as you can see nowhere in this notion of what happens would a force twice as big on the layer just before signal reaches the other piston appear. So obviously it's wrong - I JUST DONT UNDERSTAND WHY.

Another thing that's wrong about it, is that if you don't apply a force to the other piston it would just keep moving on and on etc....

So yes, I am wrong, but please please try to explain what fundamentally is wrong about my assumptions :( I wouldn't mind a 2 page answer at all, because I'm wrong about the basics of this.

This is so depressing, I can't focus on other things than understanding this, which puts me behind my reading schedule :(
 

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  • #22
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THAT'S AN ENERGY OBSERVATION!
Is it?

I would have said it is a continuity observation.

When I presented the energy argument for your earlier example I said that the volume of fluid moved out of one chamber must equal the volume moved into the second one.

This is the key fact that connects both chambers or both ends of a hydraulic system.

Since the volume does not alter it follows that if the cross section area changes the length or distance must change to compensate.

This principal is known as the equation of continuity and is hidden in many physical processes.
 
  • #23
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Okay that's actually kind of neat.

But still I'd love to have it explained in terms of forces, so that I know it's possible. Will you help me please? - look at my diagram and say where I go wrong :)
 
  • #24
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With regard to your long diagram and sequence;
I will try to think of an explanation that may help,
but meanwhile think about this

I couldn't put it better than this quote from jambaugh's post#16

Meditate on the idea that Force is a Rate of work per distance. Say it out loud, "Force is work per distance!"

As I move a lever I am doing a given amount of work, since the lever transforms distance traveled it "dilutes" or "concentrates" the work per distance i.e. the force.
in this thread

https://www.physicsforums.com/showthread.php?t=571729

There are several laws which appear again and again in Physics in different guises.

Conservation of mass or continuity

(What goes in one end of a hozepipe comes out the other)

Conservation of energy

(You can't get something for nothing)

Geometric compatibility

All systems and changes to systems obey the laws of geometry

But

There is no such thing as a 'law of conservation of force'

Forces can apply all the time like gravity
or they may be turned on and off like someone pushing an autombile
They may be in equilibrium in which case they are balanced
or they may be unbalanced ie have a net resultant in which case something is in motion
 
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  • #25
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Exactly! You finally get my question! :)

But, problem with this is that it is far more complex to understand in forces than just to apply an energy observation - that doesn't however change the fact that I do want to see what happens in terms of forces. And while not so intuitive to me I know, that there is no law of conservation of net force.

To show that I really do understand this, look at the picture. The second mass is being acted on by twice the force, because of the string arrangement. But conservation of energy is still not violated, because when the masses move it will have to travel twice the distance as the other.

I just want the same force intuition for the hydraulic pump.
 

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