# How do hydraulic machines work

## Hydraulic machines

In some machines, the forces are transmitted by fluids under pressure instead of levers or gears. Machines like this are called hydraulic machines. They use the following properties of liquids:

• Liquids are practically incompressible - they cannot be compressed.
• When a liquid is pressurized in a closed container, the pressure is transferred to all objects in the liquid.

### Hydraulic brakes

The brakes work hydraulically. The above diagram shows the principle. When the brake pedal is depressed, a piston pushes the brake fluid from one cylinder along a connecting pipe to another cylinder. There the liquid presses on another piston. This in turn presses the brake lining against the brake disc, which is attached to the rotating wheel of the vehicle. The resulting friction slows the wheel down.

In today's automotive braking systems, there are lines to all four wheels, brake pads on both sides of each brake disc, and usually "brake boosters" as well.

### Hydraulic jack

When changing tires, it is easier to lift the load of the car if you use a jack. The diagram below shows a simple hydraulic load lifter. A downward force on the input piston releases pressure on the oil. The pressure is transmitted through the oil and creates a greater upward force on the output piston.

 The force acts on a small area, which causes high pressure The great upward force is caused by the high pressure acting on a large area

If the input force and the piston area are known, the output force can be calculated:

In the input cylinder:

An input force of 12 N acts on an area of ​​0.01 m2.

So:

\$ \ mathrm {pressure \ on \ the \ oil \ = \ \ tfrac {force} {area} \ = \ \ tfrac {12 \ N} {0.01 \ m ^ 2} \ = \ 1200 \ Pa} \$

In the connection pipe:

The pressure of 1200 Pa is transmitted through the oil.

In the exit cylinder:

The pressure of 1200 Pa acts on a piston with an area of ​​0.1 m2.

So:

\$ \ mathrm {output force \ = \ pressure \ \ cdot \ area \ = \ 1200 \ Pa \ \ cdot \ 0.1 \ m ^ 2 \ = \ 120 \ N} \$

### A power booster

With the jack on top you only apply a force of 12 N, but a force of 120 N is applied. So the jack is a force booster. In this case it multiplies the input force by a factor of 10. But there is a price for the gain in force: The output piston is only raised by \$ \ tfrac {1} {10} \$ the way the input piston is pushed down.

The output force calculation above assumes that the jack works without friction. In a real jack there is friction, which reduces the output force.