A hydraulic press is an ancient machine, but it has retained its significance to the present day. Take a look at Figure 1 depicting a hydraulic press. Two pistons—small and large—can move in a vessel with water. If we press one piston with our hand, the pressure is transmitted to the other piston—it will rise. Just as much water will rise above the initial position of the second piston as the first piston presses down into the vessel.

If the areas of the pistons are \(S_{1}\) and \(S_{2}\) and their displacements are \(l_{1}\) and \(l_{2}\) the equality of the volumes yields \[S_{1} l_{1} = S_{2} l_{2}, \quad \textrm{or} \quad \dfrac{l_{1}}{l_{2}} =\dfrac{S_{2}}{S_{1}}\] We must discover the equilibrium condition for the pistons.

We shall find such a condition without difficulty, starting out from the fact that the work performed by the balancing forces should be equal to zero. Then during the displacement of the pistons the works done by the forces exerted on them should be equal (with opposite signs). Therefore, \[F_{1} l_{1} = F_{2} l_{2}, \quad \textrm{or} \quad \dfrac{F_{2}}{F_{1}} =\dfrac{l_{1}}{l_{2}}\] Comparing this with the preceding equality, we see that \[\dfrac{F_{2}}{F_{1}} =\dfrac{S_{2}}{S_{1}}\] This equation implies the possibility of an enormous multiplication of force. The piston transmitting pressure can have an area which is hundreds or thousands of times smaller. The force acting on the large piston will be just as many times greater compared to the muscular force.

With the aid of a hydraulic press, one can forge and punch metals, press the juice out of grapes and raise weights.

Of course, the gain in force will be accompanied by a loss in path. In order to compress a body by 1 cm with a press, one’s hand would have to cover a path as many times greater as the forces \(F_{2}\) and \(F_{1}\) differ.

Physicists call the ratio of the force to the area, \(F/S\), the pressure (it is denoted by the letter \(p\)). Instead of saying, “One kilogram-force acts on an area of 1 cm\(^{2}\) we shall say more concisely, “The pressure \(p = 1~\mathrm{kgf/cm^2}\).” This pressure is called the technical atmosphere ( \(1~\mathrm{kgf/cm^2} =\) 1 atm).

Instead of the relation \[\dfrac{F_{2}}{F_{1}} =\dfrac{S_{2}}{S_{1}},\] one can now write: \[\dfrac{F_{2}}{S_{2}} =\dfrac{F_{1}}{S_{1}}, \quad \textrm{i.e.} \quad p_{1}=p_{2}\] Thus, the pressures on both the pistons are the same. Our reasoning does not depend on where the pistons are located or whether their surfaces are horizontal, vertical or inclined. And in general, it is not a matter of pistons. One may conceptually choose any two portions of a surface enclosing a liquid, and assert that the pressures on them are identical.

It turns out, therefore, that the pressure within a liquid is the same at all its points and in all the directions. In other words, an identical force is exerted on area elements of a definite size, irrespective of their orientation. This fact is called Pascal’s law.