Pascal’s Law

This law states that pressure in a fluid in equilibrium is the same everywhere, if the effect of gravity can be neglected.

Proof:

We can show that pressures at any two points C and D in a liquid in equilibrium state of rest in a vessel are the same.

Let us imagine a right circular cylinder in the liquid with the line CD as axis and points C and D as centers of the flat end faces of the cylinder.

Let us consider the liquid inside the cylinder which is an equilibrium under the action of forces exerted by the liquid outside the cylinder.

These forces are perpendicular to the surfaces of the cylinder everywhere, as shown.

It is clear that the forces F₁ and F₂ on the flat faces at C and D are perpendicular to the forces on the curved surface of the cylinder.

Since the liquid is in equilibrium, the sum of all forces acting on the cylinder must be zero. As such

F₁ = F₂

If p₁ and p₂ be the pressures on the flat faces at C and D, and A be the area of cross section of each face, then

F₁ = p₁A and F₂ = p₂A

.·. p₁ = p₂

Thus, Pascal’s law is proved. A similar argument can be made to show that the pressure at any point does not depend on the direction of the surface on which it acts.

A natural consequence of Pascal’s law is that in an enclosed liquid, if pressure is increased in any part of the liquid, then it is transmitted equally to all parts of the liquid.

Application: (Hydraulic Lift):

It is used to lift heavy loads, such as cars at a service station. 

It is based on the simple principle that when a small force is applied over a piston of small cross-sectional area, then due to transmission of pressure, a large force appears over a piston of large cross-sectional area. 

This force is used to support and lift heavy weights.

Basically, a hydraulic lift consists of two cylinders of cross-sectional areas A₁ and A₂ (A₁ < < A₂) connected by a horizontal pipe. 

Each cylinder is fitted with air-tight friction less piston. The cylinders are filled with an in compressible liquid.

Suppose a force F₁ is applied on the piston of smaller area of cross-section A₁. Then, the pressure exerted over the liquid is

p = F₁/A₁

This pressure is transmitted unchanged to the piston of larger cross-sectional area A₂. Therefore, the upward force F₂ exerted on the larger piston is given by

F₂ = p * A₂ = F₁/A₁ * A₂

Since A₂ > > A₁, therefore F₂ > > F₁. Thus, a heavy load placed on the larger piston is lifted up.