Consider an ideal, incompressible fluid of density $\rho$ in a uniform gravitational field, in a rigid container and in contact with two massless pistons of differing area $A_1$ and $A_2$. At the same level in the fluid as the pistons is a pressure gauge reading zero absolute pressure. There is no pressure outside the container. See the below drawing: multi piston pressure

Two masses $m_1$ and $m_2$ are placed on their respective pistons. We may find the new pressure reading at the gauge by considering the new rest positions of the pistons and masses. Let $y_1$ and $y_2$ denote both positive (upward) and negative (downward) displacement of the pistons. Then to conserve volume one requires:

$$ y_1 A_1 + y_2 A_2 = 0 $$

Using Pascal's law, one can also relate the difference in pressure between the two piston's positions as:

$$ \frac{ m_1 g }{ A_1 } = \frac{ m_2 g }{A_2} + (y_2 - y_1 )\rho g $$

These equations can be solved for the positions of the pistons, and ​then choose either piston to sum up the pressure change to the $y=0$ gauge. Choosing piston 2:

$$ P = \frac{m_2 g}{A_2} + y_2 \rho g $$

When one perform the subsitution the values for $y$ and $\rho$ cancel out, leaving:

$$P = \frac{ m_1 + m_2 }{ A_1 + A_2 } g $$


For N number of pistons and masses arranged as above, the final equilibrium pressure at the $y=0$ pressure gauge will be:

$$ P = \frac{ \Sigma m_k }{ \Sigma A_k }g $$


Can the above conjecture be shown from a physical argument, without resorting to solving the system of linear equations? I.e., why would an arbitrary number of pistons act like a single piston of the same total area and mass?


2 Answers 2


The force in any given pipe is $PA_{i} = gm_{i} + y_{i}\rho g A_{i}$, intuitively it is carrying the mass and the water difference.

The total force is then $P\sum{A_{i}} = g\sum{m_{i}} + \rho g \sum{y_{i}A_{i}}$.

Argue from conservation that $\sum{y_{i}A_{i}} = 0$.

Then $P = \frac{\sum{m_{i}}}{\sum{A_{i}}}g$

In words, the total force (pressure at y=0 * total area) must balance the total mass (water + masses) against gravity, and the total effect of the water is 0 by definition of the water level, leaving only the weight of the masses.


If we started with one large mass and a large area $$P = \frac{Mg}{A}$$

Then split this into two masses over different areas, we could separate one of the desired masses/areas $m_1$, $A_1$ , leaving the extra mass over the extra area

$$P = \frac{ m_1 + m_{ex} }{ A_1 + A_{ex} } g$$

and from the analysis in the question the measured pressure would be the same.

We could then split up the extra mass into two masses and areas and the measured pressure would be unchanged, provided that the height of $m_1$ didn't change.

Since splitting up a larger mass over one piston into two masses over two pistons didn't change the pressure, (again from the analysis in the question), we can presume that the height of $m_1$ doesn't change.

In a similar way, this splitting up could continue until the original mass was divided into any number of smaller masses over different areas, giving the result

$$P = \frac{\sum{m_{i}}}{\sum{A_{i}}}g$$

  • $\begingroup$ Why exactly can we presume the height of m1 doesn’t change? $\endgroup$
    – cms
    Commented Sep 28, 2021 at 10:41
  • 1
    $\begingroup$ @cms because the splitting up doesn't change the pressure (according to the analysis in the question) and so if the original pressure at $y=0$ was $P$ and $P=\frac{m_1g}{A_1} +h_1\rho g$ and no other quantities in that equation are changing, then h_1 must be constant $\endgroup$ Commented Sep 28, 2021 at 11:16

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