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I know that's stupid question, but I'm really confused what my teachers says, so I need to check that theory.

Here are just two ordinary connected containers, which are full of water.

Image NO 1

On grounds of theory of hydrostatics we can say that :

  • p3 is greater than p1
  • p4 is greater than p2
  • p1 equals p2
  • p3 equals p4

Let's say:

  • p1 = 2Pa
  • p2 = 2Pa
  • p3 = 5Pa
  • p4 = 5Pa

But what if piston 1 goes down, like here:

Image NO 2

What pressure will be at places p1, p2, p3, and p4?

Will they increase by the same number with maintaining amounts based on hydrostatic theory, or they will be same?

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3 Answers 3

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enter image description here

Hoping you know the elementary law:The points at same horizontal heights have same pressures.

Let's see , this also holds for a system with closed top. Now we can see when a force F is applied on the piston 1 , it depresses by some height h.where $\rho g h A=F$ and thus is the pressure on the $p1$ ie. $F/A$ is equal to pressure of p2 ie. $\rho gh$. And P3=P1+$\rho g d$,P4=P2+$\rho gd$ where d is depth of p3 point. So, they also have same pressures.

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is the initial difference between heights same as final differences? and are initially pistons relaxed. –  Mr.ØØ7 Apr 4 '13 at 17:29
Case 1: After pushing piston will be all values same. (E.g. p1=p2=p3=p4=8Pa) | Case 2: After pushing piston all values will increase by same number, but they itself will be different (E.g. Before pushing piston - p1&p2 equals 3, p3&p4 equals 8. After pushing piston, p1&p2 equals 5, p3&p4 equals 10) Which case will happen? –  Dundee Apr 4 '13 at 17:34
if the assumptions of mine in comment above are correct then case2. and all increase by $\rho g h$ or $F/A$ –  Mr.ØØ7 Apr 4 '13 at 17:36
thank you very much, that's what I wanted to see. They'll increase BY some number, not TO some number. –  Dundee Apr 4 '13 at 17:38
Your reasonning was a bit fast and not quite correct (see my answer). The pressure increase is $F/(A_1+A_2)=h_2 \rho g$ where $A_1$ and $A_2$ are the surfaces of the two pistons, and $h_2$ is the height piston 2 rises. –  babou Sep 5 '13 at 21:09

Simplifying the problem I would make piston 2 open and piston 1 fixed. So in case 2 the open water column over p2 and p4 is higher and the pressure should increase accordingly. p1=p2 and p3=p4 still holds. If you then calculate the pressure at piston1 you check the required force in case it is not fixed. Simply said: you must add the pressure of the water column height difference.

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Analysis of the effect of force $F$ on the system

I assume that $A_i$ is the surface area of piston $i$, for $i=1,2$.

If the force $F$ pushes piston 1 down by some height $h_1$, there is a pression increase on $p_1$ due to the force $F$ applied to surface $A_1$, and a decrease due to the loss of the water level above $p_1$ of height $h_1$, which is $\rho g h_1$. Thus the pressure in $p_1$ is increased by $F/A_1-\rho g h_1$.

Whatever happens, the pressure will be the same in $p_1$ and $p_2$ since they are at the same height. The same is true for $p_3$ and $p_4$.

(Note that what matters is height, not depth, if you can have different water levels, or different pressures at the top)

Thus the same pressure increase occurs in $p_2$. But the pressure in $p_2$ is balanced by an increase $h_2$ in water level, i.e. a raising of piston 2, since there is no force on piston 2. This pressure increase is $\rho g h_2$.

Hence we have $F/A_1-\rho g h_1=\rho g h_2$ , i.e., $F/(\rho g)=A_1 h_1 + A_1 h_2$

(Remember that $h_1$ and $h_2$ are measured in opposite directions.)

Now, since the amount of water stays the same, the volume lost on one side equal the volume gained on the other side: $h_1 A_1=h_2 A_2$.

Hence $F/(\rho g)=h_2 A_2 + A_1 h_2$ , i.e., $h_2=F/(\rho g (A_1+A_2))$.

and $h_1=h_2 A_2/A_1= F A_2/(A_1 \rho g (A_1+A_2))$

The pressure change

The pressure under piston 2 was initially the atmospheric pressure. As the water level rises by height $h_2$, the pressure at this initial level is increased accordingly by $h_2 \rho g$, which is equal to $F/(A_1+A_2)$ (see computation of $h_2$ above).

Then, according to Pascal's law, the increase in pressure is the same everywhere in the liquid, so that everywhere the pressure is increased by $F/(A_1+A_2)$.

This is true in particular for the 4 points under consideration. But the pressure differential between any two points stays of course the same.

A minor additional remark

This reasonning, including the use of Pascal's law, assume that water is not compressible, which is not absolutely true, though it can generally be neglected, except for extremely high pressures.

The pressure increase of $F/(A_1+A_2)$ does cause a very minute compression of the fluid, resulting in a minute increase of specific mass (density) $\rho$. Hence the pressure difference between $p_1$ and $p_3$ is actually very slightly increased.

As I said, this is negligible for water in most situations, but it could be more significant for a more compressible fluid. If more significant, it should also be accounted for in writing the equations above, particularly regarding volume of fluid which would be reduced.

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