What is Pascal’s law or Principle? A very common law. In this article, we are going to learn the basic law, along with pascal’s law definition, meaning, formula or equation, examples, applications, etc.

Let’s explore the law!

Let’s start what is Pascal’s law with definition, meaning, statement, etc.

Ever wondered how a hydraulic jack lifts such a huge car when we just rotate the turning rod while applying a small amount of force.

Have you observed that a heavy-duty truck while traveling at a certain speed stops just by a push of a break by his foot? Huge bundles of soft foam, wool, and cotton are compressed and packed with the help of a hydraulic press.

So, what do you think,

- How do these things happen?
- Why a huge heavyweight object is lifted by only a very small pressure?
- How does hydraulic press or hydraulic lift or hydraulic brakes work?

All the above examples use a similar principle though their application is different. Here, Pascal’s principle or law comes into the picture. Let’s get into the statement.

Pascal’s law is discovered in the 1600’s and it is named as per French Mathematician Blaise Pascal.

The main principle of Pascal’s law states that,

The statement can be read also like beow,

The simple meaning of Pascal’s law is that Pressure applied to any region of the confined fluid is equally transferred in all directions through the fluid.

Take a container with two columns which is connected by a horizontal pipe and the container is consisting of water.

- Two columns are Column A and Column B,
- Both the columns were blocked by pistons.
- Column A is blocked by piston A, and Column B is blocked by piston B,

Case-I: If you apply force on piston A, Piston A will go downwards and other Piston B will move upwards or rise.

Case-II: In the same way, if you apply force on piston B, piston A will rise.

This is because of Pascal’s law.

Let us learn Pascal’s law definition.

Pascal’s law also known as the principle of transmission of fluid pressure is used in fluid mechanics for applications involving lifting heavy objects using minimal force with hydraulic pressure.

According to the law when pressure is exerted anywhere in a confined incompressible fluid it is transmitted throughout the fluid equally in all directions in such a way that the pressure ratio remains constant.

This principle can be stated mathematically with the following equation:

**ΔP = ρ. g. Δh**

Where,

- ΔP is the hydrostatic pressure (measured in Pascal in the SI system), or the difference in pressure between two places in a fluid column caused by the fluid’s weight.
- g is the acceleration due to gravity (usually using the sea level acceleration due to Earth’s gravity in meters per second squared), and
- ρ is the density of fluid used in the container (Measured in kilograms per cubic meter in the SI standard).
- ∆h is the height of the fluid above the measurement point or the difference in elevation between two places within the fluid column (in meters in SI unit).
- This formula’s intuitive explanation is that the pressure difference between two elevations is caused by the weight of the fluid between the elevations.

It should be noted that the fluctuation with height is unaffected by any extra forces.

As a result, Pascal’s law may be read as stating that any change in pressure exerted at any particular location of the fluid is communicated equally throughout the fluid.

The most important formula in relation to pressure, force, and the area is given by Pascal’s formula,

**P = F/A**

This expression states that the pressure is equal to force per unit area.

**Considerations**

We can see the power of fluid pressure when it is applied in a U-shaped tube which is smaller in the area on one side with piston A and bigger on the other side with piston B.

Let’s assume the left side of U shaped tube is filled with static fluid.

- The area of the left side column, A1= 1 m2
- The area of the right side column, A2 = 10 m
^{2 }

**Force on Piston A** **& Pressure**

When a force (F_{A}) of 20 N is applied on fluid at the left side column A through a piston, it is transmitted on the other side to a larger piston A2.

The force we get is the multiplication of area and pressure thus we get,

- Force, F
_{A}= pressure x area (A1) - Pressure, P
_{A}= Force/area - Pressure, P
_{A}= 20/10 = 10 N/m^{2}

This pressure will act on the other side.

This happens because when 20 N of force is applied on piston A in area A1, we get 10 N/m^{2} of pressure and this same pressure is when transmitted on the other side of the U tube.

**Force on Piston B**

Pressure will be the same in column B and it is, P_{B} = P_{A}= 10 N/m^{2}.

- Force on piston B, F
_{B}= pressure x area (A2) - Force, F
_{B}= 10 x 10 N - Force, F
_{B}= 200 N

Due to this by applying 20 N of force we can lift 200 N, the larger piston will support 200 N of weight ten times more than that of the smaller piston.

Thus by applying a small amount of force we get substantially large force as the output this principle is used in,

- hydraulic lifts
- hydraulic jacks
- hydraulic pumps
- hydraulic brakes

Hydraulics is used by modern equipment ranging in size from very tiny to extremely large.

Hydraulic pistons, for example, are found in practically all construction machinery that handles huge weights to our local garages for lifting vehicles.

Let’s consider a very small right angled prismatic triangle as per image. Now, we will derive the pascal’s law here.

**Considerations**

- The density of the liquid is ρ,
- This prism is very small and all points are at the same depth,
- Thickness (OQ) is considered as 1
- The Centre of gravity is the same

**Diagram**

This is the simple diagram for the derivation of pascal’s law.

- Px = Pressure on Face OQTR
- Py = Pressure on Face OQSP
- Pz = Pressure on Face SPRT
- Fx = Force on Face OQTR
- Fy = Force on Face OQSP
- Fz = Force on Face SPRT
- θ = Angle QST,
- dx = Fluid element dimension in x direction
- dy = Fluid element dimension in y direction
- dz = Fluid element dimension in angular direction
- ρ = Density of the fluid

**Analysis of Force**

From the definition of force, we can write,

Force = Pressure x the area where pressure is applied.

Pressure Px is applied on the surface OQTR and the corresponding Force is Fx. Hence, we can write,

- Fx = Px x area of OQTR = Px . dy . 1 [Width OQ = 1]
- Fx = Px. dy

In the same way,

- Fy = Py x area of OQSP = Py . dx . 1 [Width OQ = 1]
- Fy = Py. dx

In the same way,

- Fz = Pz x area of SPRT = Pz x dz x 1 [Width OQ = 1]
- Fz = Pz. dz

**Analysis of Weight**

This fluid element will have weight as well. Let’s check the weight.

We know weight, W

- W = volume x density x gravity of acceleration
- W = area x width x density x gravity of acceleration
- W = A x d x ρ x g
- W = ½ x QS x QT x 1 x ρ x g [As, A =½ x QS x QT, d=1]
- W = ½ x dx x dy x 1 x ρ x g
- W = ½ x dx x dy x ρ x g

**Force Balancing & Summary**

Now, let see how the balancing to be done. Here, all force is balanced in the respective direction.

Force balancing in x-direction,

- Px. dy – Pz. dz sinθ = 0
- Px. dy = Pz. dz sinθ
- From the figure QST, dy = dz sinθ
- Px. dy = Pz. dy
- Px = Pz

Force balancing in y-direction,

- Py. dx – Pz. dz cosθ – dy dx/2) x ρ x g = 0
- PY. dx – PZ. ds Sin θ – ½ x dx x dy x ρ x g = 0 [As W =½ x dx x dy x ρ x g]

Weight is neglected, as it is very small. [½ x dx x dy x ρ x g ≃ 0]

Py. dx – Pz. dz cosθ = 0

From the figure QST, dx = dz cosθ,

- Py. dx – Pz dx = 0
- Py = Pz

So, we have seen that

**Px = Py = Pz**

Hence, we have seen the derivation of Pascal’s law and it is proved that pressure at any point in different direction (here, x, y, & z) are same.

Let’s understand the working of the hydraulic Jack, hydraulic braking system,

Hydraulic jack is also known as hydraulic lift works on the principle of Pascal’s law. It is used in automobile workshops for lifting cars, trucks, etc. to change tires or perform repair work.

**Construction**

- Hydraulic jack is constructed using two cylindrical vessels which are connected to each other using a tube or pipe with valves.
- Among the two cylindrical vessels, one is narrower and the other one is comparatively bigger in diameter and size. The small cylindrical vessel piston is attached to a lever which is usually pressed or turned during the lifting operation. The wider cylindrical vessel which is placed on the lifting side has a platform attached to its piston for lifting vehicles.
- The cylinders are filled with fluid, most often water, but in certain circumstances oil. Fluid properties such as viscosity, thermal stability, filterability, hydrolytic stability, and others are used to choose the fluid. A suitable hydraulic fluid will provide optimal performance with self-lubrication, and smooth operation.

**Working**

- The cylinders are filled with the fluid partially and sealed.
- When a little amount of pressure is applied to the smaller cylinder by a piston, the pressure is transmitted evenly to the bigger cylinder via the incompressible fluid.
- There will be a force multiplication in the bigger cylinder, the force exerted on all the points of both the cylinders will be the same and equal.
- The force generated by the bigger cylinder, on the other hand, will be greater and will be directly proportional to the surface area.
- Aside from cylinders, a hydraulic jack will include a pumping mechanism that will force fluid into a cylinder via a one-way valve. The backflow of hydraulic fluid from the cylinder will be restricted by this valve.

When brakes are applied suddenly in a moving vehicle, the vehicle has a good possibility of skidding because the wheels are not slowed equally.

To reduce the risk of skidding while using the brakes, the braking mechanism must be designed such that each wheel is uniformly and simultaneously retarded.

- This is accomplished by the use of a hydraulic braking system.
- It operates on the basis of Pascal’s law.
- Hydraulic brakes are found in automobiles, motorbikes, and trucks.

**Construction**

- It is made up of a master cylinder, four wheel cylinders, and pipelines that transport braking fluid from the master cylinder to the wheel cylinders.
- Brake fluid is used as an operating fluid in hydraulic brakes.
- The brake pedal is connected to the master cylinder piston through a connecting rod.
- The cylinders mounted on the wheel consist of two pistons each that are filled with the brake fluid.
- A cylinder brake drum is used in each wheel brake. This drum is attached to the inner side of the wheel which revolves in tandem with the wheel.
- Brake shoe mounts on the inner side of the drum stay stationary.

**Working**

- When we apply pressure to the brake pedal, a piston in the master cylinder pumps brake fluid through a linkage.
- Due to this, the pressure rises and it is transmitted to all pipes and wheel cylinders connected to the master cylinder in accordance with Pascal’s law.
- As a result of this pressure, both pistons pull out and deliver the braking power to all four wheels giving the breaking effect which in turn reduces the speed of the vehicle.

**Advantages of the hydraulic braking system**

- All four wheels experience the equal braking effect
- As no joints are present there’s minimal or no wear and tear.
- The force required can be increased easily just by changing the size of the cylinders.

**Disadvantages of the hydraulic braking system**

- Leakage or failure of seals can spoil the complete braking system.
- Small air pockets can damage the braking system and eventually result in brake system failure.

There are many other applications of Pascal’s law,

- Hydraulic pumping system
- Lifting equipment
- Water towers
- Air in tires
- Lifting of wheelchairs
- Artesian wells
- Aircraft Hydraulic System

A hydraulic lift has two columns as per the below Fig. 8. Each column is fitted with one piston. The diameter of the small-diameter column is 5cm and the larger column is 70cm.

What will be the value of force will be produced in the larger piston, if 6 N force is applied on the smaller piston?

Input details

Smaller size column

- Diameter, D1 =5cm
- Area, A1 = π D1
^{2}/4= π 5^{2}/4=19.6 cm^{2} - Force, F1 =6 N

Larger size column

- Diameter, D1 =70cm
- Area, A2 = π D2
^{2}/4 = π 70^{2}/4=3846.5 cm^{2} - Force, F2 =?

Pressure at smaller piston, P1= Force/ Area of smaller diameter = F1/A1

Pressure at larger piston P2 = Force, F2/Area of larger diameter=F2/A2

As per Pascal’s Law,

Pressure at larger piston P2 = Pressure at smaller piston, P1

Hence, F1/A1 = F2/A2

- F2 =F1/A1 x A2
- F2 = 6 / 19.6 x 3846.5
- F2 = 1177.5 N

Hence, the force exerted on the larger piston is 1177.5 N if small piston has a force of 6 N.

Hence, we have understood Pascal’s law, definition, examples, equations, etc. Any doubt, please let us know.

Thank you

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