Pascals Principle Formula

When the question arises how to lift a heavy substance by just applying a smaller force with the help of Fluid, Blaise Pascal The French Mathematician introduced Pascals Law to answer this by using the property of liquid and pressure of a fluid. Most aircraft use the Pressure and Property of liquid in the braking systems and landing gear.

Pascal’s Law 

This law is given by Blaise Pascal who states that A pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. Consider the given figure, 

Initial setup

Here, P₁ and P₂ are the pressure at a particular point inside the fluid neglecting the atmospheric pressure.

Piston applying force P

Then, according to Pascal, this pressure P will be transmitted throughout the fluid, and pressure at every point will be increased by P i.e.; P₁ + P and P₂ + P.

Some Points to remember

• Pascal law is applicable only in a closed vessel i.e.; there should be no leakage in the vessel.
• The piston is a device that prevents leakage from the vessel from the open ends
• All the hydraulic things work on the principle of pascals
• JCB Machine, Injection works on this principle

Pascal’s Formula

Consider a closed vessel where two-piston is used, A₁ and A₂ are the areas of a cross-section of their respective sections.

Initial setup.

When a force F is applied on the narrower side the piston goes down and from the border side, the piston goes up. D₁ and D₂ are the displacements from the initial position of the piston.

After applying pressure

Since the volume remain the same after displacement of fluid, therefore,

A₁D₁ = A₂D₂ ⇢ (Equation 1)

From the law of conservation of energy, Work done in both the cross-section will be equal.

F₁D₁ = F₂D₂ ⇢  (Equation 2)

Dividing equation 2 by equation 1,

F₁/A₁ = F₂/A₂

We know F(Force)/A(Area) = P(Pressure).

Mechanical Advantage

Mechanical advantage is Calculated by,

(Output force)/(Input Force) = F₂/F₁.

Sample Problems

Question 1: A Downward force of 100N is applied to the small piston with a diameter of 50cm in the hydraulic lift system. What is the upward force exerted by the large piston with a diameter of 2m?

Solution:

From Pascal’s law, we know force exerted on one side will be equal to force experienced on another side.

⇒ F₁/A₁ = F₂/A₂

Given, F₁ = 100N , Diameter of small piston = 50cm = 0.5m, Area of cross section (A₁) = π × 0.5 × 0.5.

Diameter of large piston = 2m, Area of cross section (A₂) = π × 2 × 2, F₂ = ?

⇒ 100/(π × 0.5 × 0.5) = F₂/(π × 2 × 2)

⇒ 100/(π × 0.5 × 0.5) × (π × 2 × 2) = F₂

⇒ F₂ = 1600N.

Question 2: Calculate the Mechanical Advantage of the above question?

Solution:

Mechanical Advantage is ⇒ F₂/F₁ = 1600/100= 16, which means, applying 1 unit of force on the smaller side will result in 16 times of force on the larger side.

Question 3: If the input force of 100N pushes the small piston down by 2m, of diameter 50 cm find how high the large piston will rise of diameter 2m?

Solution: 

We Know Volume of fluid inside the vessel will be constant

⇒ A₁D₁ = A₂D₂

Given, A₁ = π × 0.5 × 0.5, D₁ = 2m, A₂ = π × 2 × 2, D₂ = ?

⇒ (π × 0.5 × 0.5) × 2 = (π × 2 × 2) × D₂

⇒ D₂ =  (π × 0.5 × 0.5) × 2 ÷ (π × 2 × 2)

⇒ D₂ = 0.125m

This shows larger piston will move less than the smaller piston when force is applied to the smaller piston.

Question 4: In a car lift compressed air exerts a focus F₁ on the small piston having a radius of 0.5cm. This pressure is transmitted to the second piston of a radius of 15cm. If the mass of the car to be lifted is 1350kg, calculate F₁. What is the pressure necessary to accomplish this task? (g = 9.8 m.s² )

Solution: 

Given,

F₁ (small piston), r₁ = 5cm, r₂ = 15cm, m₂ = 1350.

⇒ F₂ = m₂ × g = 1350 × 10 = 13500N

⇒ F₁/A₁ = F₂/A₂

⇒ F₁ = (F₂/A₂) × A₁

⇒ F₁ = (13500 × π × 5 × 5)/(π × 15 × 15)

⇒ F₁ = 2250N

Pressure(P₁) = F₁/A₁ = 2250/π × (5)²

⇒ P₁ = 28.64N/cm².

Question 5: The large piston in a hydraulic lift has a radius of 20cm. What force must be applied to the small Piston of radius 2.0cm to raise a car of mass 1500kg? ( g=10m/s² )

Solution:

Radius of large piston (R₂) = 20 cm = 0.2 m

Radius of small piston (R₁) = 2 cm = 0.02m

Mass of car (M₂) = 1500kg 

Force due to car on large piston (F₂) = M₂ × g = 1500 × 10 = 15000N

Force on small piston (F₁) = ?

We know, 

F₁/A₁ = F₂/A₂

⇒ A₁ = π × 0.02 × 0.02 m²

⇒ A₂ = π × 0.2 × 0.2 m²

⇒ F₁/(π × 0.02 × 0.02) = 15000/(π × 0.2 × 0.2)

⇒ F₁ = 15000/(π × 0.2 × 0.2) × (π × 0.02 × 0.02)

⇒ F₁ = 150N

Question 6: Find the work done in the above question for lifting a car to a height of 0.3m?

Solution:

We know,

Work Done (W) = Force(F₂) × height(displacement)

⇒ W = 15000 × 0.3

⇒ W = 4500N