Hydraulic Systems

Hydraulic systems use a liquid (usually oil) in an enclosed system to transmit pressure and multiply force.

The key principle

Pressure in an enclosed liquid is transmitted equally in all directions (Pascal’s principle).

If a force $F_1$ is applied to a small piston of area $A_1$, the pressure in the liquid is:

$p = \frac{F_1}{A_1}$

This pressure acts on a larger piston of area $A_2$, producing a larger output force:

$F_2 = p \times A_2 = \frac{F_1}{A_1} \times A_2$

Force multiplication

$\frac{F_2}{F_1} = \frac{A_2}{A_1}$

If the output piston has 10 times the area of the input piston, the output force is 10 times larger.

Example: A hydraulic jack has pistons of area 5 cm$^2$ and 200 cm$^2$. A force of 50 N is applied to the small piston.

$F_2 = 50 \times \frac{200}{5} = 2000 \text{ N}$

Distance trade-off

The force is multiplied, but the distance is reduced by the same factor. The small piston moves a large distance; the large piston moves a small distance.

Work input = work output (ideal): $F_1 d_1 = F_2 d_2$

Applications

Hydraulic brakes, hydraulic car jacks, hydraulic presses, digger arms.

Common errors and how to correct them

• Thinking the hydraulic system creates energy. It multiplies force at the expense of distance (energy is conserved).
• Using inconsistent units for area (both areas must be in the same unit).

How to apply this in an exam

Calculate the pressure using $p = F/A$ at the input piston. Apply this pressure to the output piston using $F = pA$. Show the force multiplication ratio.