Pascals principle and hydraulic lift

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  • #26
jambaugh
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Someone called my name?
I don't know if this has been covered yet but...
...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
jambaugh
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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]
 
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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
jambaugh
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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?
 
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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
jambaugh
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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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