Pascals principle and hydraulic lift

In summary, the pressure is distributed evenly around a fluid. However, the principle of work still applies and shows that the two pistons are in equilibrium.
  • #1
zezima1
123
0
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
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
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
hi zezima1! :smile:
zezima1 said:
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
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
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 traveled 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
zezima1 said:
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
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
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
Hello, Studiot! :smile:

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

:biggrin:
 
  • #12
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
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
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?
Studiot said:
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
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
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
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
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 therefore 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
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
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
Studiot said:
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
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
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
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
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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  • #26
Someone called my name?
I don't know if this has been covered yet but...
zezima1 said:
...It is not hard to see, that the force you have to exert one both pistons has to be the same. But unfortunately [...]
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?
attachment.php?attachmentid=43755&stc=1&d=1328941751.png

[Edit: rescaled picture]
The problem with this picture is that the displayed pistons with rod will not be in equilibrium in general, only when the internal and external pressures are equal and at that point there will be no force on the rod.

Sliding this picture back and forth changes the internal volume. Imagine it filled with a gas and the whole device in a vacuum. The gas wants to expand indefinitely. It can do this by the rod and two pistons sliding to the right... and that will happen until the "cork" comes out -Pop!-
 

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  • #27
Studiot said:
I couldn't put it better than this quote from jambaugh's post#16
Let me add...
  • Force is rate of work per distance,
  • Pressure is rate of work per volume (change),
  • Torque is rate of work per radian angle,
  • Voltage is rate of work per coulomb charge,
  • [itex]\vdots[/itex]
 
  • #28
Okay I think I may get what you mean - pressure is something that also connects to statistics, i.e. the force will not concentrate on a single point but rather spread out over the entire volume.
I'm just still not quite sure - can you try to explain what's wrong with my model in post #21?
 
  • #29
zezima1 said:
Okay I think I may get what you mean - pressure is something that also connects to statistics, i.e. the force will not concentrate on a single point but rather spread out over the entire volume.
I'm just still not quite sure - can you try to explain what's wrong with my model in post #21?

Pressure is a force density, force per area. Your areas are changing. In your post #21 remember that the number of particles interacting with the surfaces is proportional to area.

The other thing with that model is that the forces aren't just turned around, they continue to push down when they start pushing down. They also make the fluid want to expand sideways so the walls impart a force but the forces from the left walls cancel with the forces from the right walls. The pressure forces from the floor cancel with the forces from both pistons. Imagine the whole device on a scale. You push down on A1 with F1 and A2 with F2 and the scale will read F1+F2 as it is the SCALE pushing upward on each area of the two pistons through the walls and the fluid.

Try it with the pistons in line, like the ones in your first diagram. The walls push inward and since the area increases some of that inward will be angled toward the bigger area. It geometrically must add up that way and you will get a net force in the direction of the bigger area proportional to the pressure times change in area.

Note this is how a jet or rocket works. The walls push the fluid back and the fluid push the walls forward along with the rest of the rocket.Finally it may help to consider extreme cases. Imagine the area of the smaller piston approaching zero. You then have basically a gun barrel with the "big" piston being your bullet. Try to see from where the force accelerating the bullet originates... why do guns recoil?
 
  • #30
jambaugh said:
Try it with the pistons in line, like the ones in your first diagram. The walls push inward and since the area increases some of that inward will be angled toward the bigger area. It geometrically must add up that way and you will get a net force in the direction of the bigger area proportional to the pressure times change in area.

I really liked this part. Does something similar happen when the pistons are not in line but like in the picture in post #21?
 
  • #31
aaaa202 said:
I really liked this part. Does something similar happen when the pistons are not in line but like in the picture in post #21?

Yes. If you want to calculate it in full glory, you can carry out a surface integral of pressure times inward normal to the surface. This gives the net force on the fluid. I in integrating you leave out the pistons then this total force must equal the forces on the pistons.
 

What is Pascal's principle?

Pascal's principle, also known as the principle of transmission of fluid-pressure, states that a change in pressure applied to an enclosed fluid will be transmitted equally to all parts of the fluid and to the walls of the container.

How does a hydraulic lift work?

A hydraulic lift works by using Pascal's principle. When a small force is applied to a small area of an enclosed fluid, it creates a larger force on a larger area, resulting in a mechanical advantage. This allows for heavy objects to be lifted with less effort.

What are the components of a hydraulic lift?

A hydraulic lift consists of a pump, a reservoir, a valve system, and a piston-cylinder arrangement. The pump creates pressure by forcing fluid into the system, the reservoir stores the fluid, the valve system controls the direction of the fluid, and the piston-cylinder arrangement converts the pressure into mechanical work.

What are the applications of hydraulic lifts?

Hydraulic lifts are commonly used in car lifts, elevators, and heavy machinery such as cranes and forklifts. They are also used in hydraulic presses for metalworking and in aircraft for landing gear and control surfaces.

What are the advantages of using hydraulic lifts?

Hydraulic lifts have several advantages, including their ability to lift heavy objects with minimal effort, their smooth and controlled movements, and their ability to operate in a wide range of temperatures and environments. They also have a long lifespan and require minimal maintenance.

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