# Pascal’s principle

How is the working principle of a hydraulic lift used to lift a car? What is the working principle of hydraulic brakes when used to reduce the car’s speed? Please learn Pascal’s Principle to understand this.

Every fluid is always exerting pressure on all objects that in contact with it. The water we put in the glass will exert pressure on the glass wall. Likewise, if we take a bath in a swimming pool or sea water, the pool water or sea water also exerting pressure on our entire body.

The total pressure of water at a certain depth, for example, sea water pressure at a depth of 200 meters is the amount of atmospheric pressure and measured pressure at a depth of 200 meters. So besides the water pressure

there is also the atmosphere that suppresses the surface of the sea water.

Atmospheric pressure works on all fluid surfaces, and the pressure is distributed to all parts of the fluid. Therefore the total pressure of the fluid at a certain depth other than caused by the pressure of the fluid layer at the top is also affected by atmospheric pressure.

To further understand this explanation, let’s review the liquid in a container.

The liquid pressure at the bottom of the container is, of course, higher than the pressure of the liquid in the upper. The more downward, the higher the pressure of the liquid, on the contrary, the closer to the top surface of the container, the smaller the pressure of the liquid.

The amount of pressure is proportional to ρ g h (ρ = density, g = gravity acceleration and h = height or depth). At each point at the same depth, the amount of pressure is the same. This applies to all liquids in any container and does not depend on the shape of the container.

If we add outside pressure, for example by pressing the surface of the liquid, the increase in pressure in the liquid is the same everywhere. So if external pressure is given, each part of the liquid gets the same amount of pressure. Therefore the pressure is always the same at every point at the same depth. This is Pascal’s Principle, initiated and named according to the name of Blaise Pascal (1623-1662). Pascal is a French philosopher and scientist.

**Pascal’s principle states that the pressure applied to a liquid in a closed place will be forwarded to each part of the fluid and the walls of the container.**

Application of Pascal’s Principle

Based on Pascal’s principle, humans have produced several tools, both simple and sophisticated to help make life easier. Some of them are Hydraulic Lifts, Hydraulic Brakes, etc.

**Hydraulic Lift**

The hydraulic lift consists of a vessel that has two surfaces. On both surfaces, there is a piston, where the piston surface area on the left is smaller than the piston surface area on the right. The piston surface area is adjusted to the surface area of the vessel. Vessels are filled with liquids, such as lubricants.

If the piston with a small surface area is pressed down, each part of the liquid also

Pressed down. The amount of pressure given by a piston with a small surface is passed to all parts of the liquid. As a result, the liquid presses the piston with a larger surface area until the piston is pushed up. The surface area of the piston pressed is small, so that the force needed to suppress the liquid is also small. But because the pressure is continued on all parts of the liquid, the small force changes when the liquid presses the piston on the right with a large surface area. Rarely do people provide an entry force on a piston with a large surface area, because it is not profitable? At the top of the piston that has a large surface area is usually placed objects or things that want to be lifted (for example a car).

Do not be surprised if a car with a very large mass is easily lifted just by pressing one of the pistons. The surface area of the piston is so small that the force we give is also small. However, the small input force can change to a very large output force if the output surface area is very large.

Example problem 1 :

__Known :__

A_{1} = 100 cm^{2}

A_{2} = 250 cm^{2}

F_{1} = 200 N

__Wanted :__ F_{2 }

__Solution :__

Example problem 2 :

__Known :__

A_{1} = 100 cm^{2 }= 100 x 10^{-4} m^{2} = 0.01 m^{2}

A_{2} = 250 cm^{2 }= 250 x 10^{-4} m^{2} = 0.025 m^{2}

Mass of load = 200 kg

Density of oil (ρ) = 780 kg/m^{3}

Height of oil column (h) = 2 m

Acceleration of gravity (g) = 10 m/s^{2}

__Wanted :__

What is the minimum input force (F) so that the load is in a balanced state (the load is not moving)?

__Solution :__