Pascal's law
From the fact that the pressure in a fluid depends on the pressure \(p_0\) at the fluid surface it also follows that any change in the pressure must appear at every other point in the fluid.
Pascal’s law
A change in the pressure applied to a stationary fluid is transmitted undiminished to every point of the fluid and to the walls of the container.
Hydraulic press
One application of Pascal’s law is the hydraulic press. This is essentially a tube filled with an incompressible fluid, terminated by two pistons of different cross-section area at its two ends.
If one applies force \(F_1\) at the smaller end (of cross-section area \(A_1\)), the pressure below the piston increases by
\[p = \frac{F_1}{A_1}.\]
According to Pascal’s law, this is transmitted undiminished to the larger piston, where it causes an upward force
\[F_2 = p \cdot A_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1} > F_1.\]
That is, the force applied at the smaller end is amplified in the ratio of the cross-section area of the greater piston to that of the smaller one.
Supplement: do we gain energy?
We have seen that the hydraulic press can multiply our force input: the force at the larger piston is many times the force we must exert at the smaller piston. Does it mean that the hydraulic press creates energy out of nothing? We shall see that it does not.
The work we invest at the smaller piston is \(W_1 = F_1 \cdot \Delta x_1,\) with \(\Delta x_1\) being the displacement of the smaller piston, whilst the output work at the larger piston is \(W_2 = F_2 \cdot \Delta x_2,\) where \(\Delta x_2\) is the displacement of the greater piston.
How does \(\Delta x_2\) relate to \(\Delta x_1\)? To find out, let us consider that the liquid is incompressible and no volume is added or removed from the system. This means that the volume of the liquid pushed up by the smaller piston must be the same as the volume of liquid pushed up below the larger piston:
\[A_1 \Delta x_1 = A_2 \Delta x_2.\]
This means that
\[\Delta x_2 = \Delta x_1 \frac{A_1}{A_2}.\]
Substituting it into the formula for the output work (the work done at the larger piston):
\[W_2 = F_2 \Delta x_2 = \left(F_1 \frac{A_2}{A_1}\right)\left(\Delta x_1 \frac{A_1}{A_2}\right) = F_1 \Delta x_1 = W_1.\]
This means that there is no energy gain out of nothing: the work done at the output is the same as the work we invested at the smaller piston. Though the hydraulic press can multiply the force we can exert, it does so at the cost of a much longer displacement, and thus the work done is theoretically the same at the input and at the output (in practice, the work at the output is much less due to losses caused by friction).