Pascals principle


by zezima1
Tags: pascal's principle
zezima1
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#19
Feb9-12, 12:53 PM
P: 123
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
Studiot
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#20
Feb9-12, 05:33 PM
P: 5,462
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.
zezima1
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#21
Feb9-12, 05:45 PM
P: 123
Quote Quote by Studiot View Post
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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Studiot
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#22
Feb9-12, 05:53 PM
P: 5,462
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.
zezima1
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#23
Feb9-12, 06:47 PM
P: 123
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 :)
Studiot
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#24
Feb10-12, 02:20 AM
P: 5,462
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

http://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
zezima1
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#25
Feb10-12, 05:24 AM
P: 123
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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jambaugh
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#26
Feb11-12, 12:25 AM
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Someone called my name?
I don't know if this has been covered yet but...
Quote Quote by zezima1 View Post
...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?

[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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jambaugh
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#27
Feb11-12, 12:38 AM
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Quote Quote by Studiot View Post
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]
zezima1
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#28
Feb11-12, 05:15 AM
P: 123
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?
jambaugh
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#29
Feb11-12, 03:35 PM
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Quote Quote by zezima1 View Post
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?
aaaa202
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#30
Feb11-12, 03:47 PM
P: 992
Quote Quote by jambaugh View Post
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?
jambaugh
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#31
Feb11-12, 07:49 PM
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Quote Quote by aaaa202 View Post
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.


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