# Enclosed liquid

1. Mar 2, 2012

### batballbat

in a container there are some pistons of different areas. If we apply a force at one of the pistons and move it a distance. How can we determine the distance moved by the other pistons?

2. Mar 2, 2012

3. Mar 2, 2012

### batballbat

what if a child asks this question? i thought this rule was for homework section.

4. Mar 2, 2012

### DaveC426913

Yes, this belongs in the homework section. Show your attempt.

5. Mar 2, 2012

### batballbat

it does not depend on the area. Every piston will be moved by an equal distance. Thats my attempt.

6. Mar 2, 2012

It is just conservation of mass. For water, this equates to conservation of volume. If you know how far down one piston moved, you can use that displaced volume to determine how high the rest moved.

7. Mar 2, 2012

### batballbat

everybody knows volume is conserved. so u mean that increase in volume of every piston is same? in which case, the distance can be determined.

8. Mar 2, 2012

No, the pistons will rise the same and the total volume "added" by the rising pistons will equal the volume "lost" by the falling piston(s).

9. Mar 3, 2012

### batballbat

ok. what if the sum of the areas of the other pistons is less than or greater than the area of that piston times the no. of the other pistons?

10. Mar 3, 2012

### batballbat

somebody comment on this problem.

11. Mar 3, 2012

### batballbat

if the other pistons move equal distances. Then the sum of the other areas must equal the no. of the other pistons multiplied by the gien piston.

12. Mar 3, 2012

### batballbat

help?

13. Mar 3, 2012

### DaveC426913

Presumably, only one target piston is allowed to move freely at a time, otherwise, I'm not sure how you could calculate several pistons' movement simultaneously.

This is identical to filling cylindrical glasses of water.

If you start with a glass that is 10cm in diameter, and it is filled with 1L of water, the height of the water will reach X.

If you pour that entire 1L of water (i.e. volume does not change) into another glass of 5cm diameter, what level will the water reach and how do you calculate that?

14. Mar 3, 2012

### batballbat

well imagine a enclosed sphere under no action of gravity. I am guessing there must be some regularity in the movement of the pistons if we push in one of the pistons.

15. Mar 3, 2012

### DaveC426913

I'm not sure how that helps or hinders the problem at-hand. It just seems to kind of restate it.

Did you read my previous post? Do you understand the correlation between the areas and the volume displacements of the pistons?

16. Mar 3, 2012

### batballbat

ok so volume is conserved.
we get, for directed distance
$A_1$ $d_1$ + $A_2$ $d_2$ + ............... = 0
And also the sum of the work done is also 0.
Then?

17. Mar 3, 2012

### DaveC426913

Perhaps you need to rethink what you want from us. If this is a question from a book, perhaps you should write out the entire question.

P1Area x P1Displacement = P2Area x P2Displacement

What more do you want?

18. Mar 4, 2012

forget it