# Pascal's law: pressure of fluid at different locations

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.

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:

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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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. –  ABC 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? –  user22761 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 $F/A$ –  ABC 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. –  user22761 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.

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.