Pascal's Law And Its Application, Practice Problems, FAQs

Ever imagined how technicians in garages lift heavy cars or bikes which cannot be humanly possible? They use hydraulic lifts, which can lift objects of several thousand kilograms that we can’t do with our bare hands. Hydraulic brakes work on the same principle as that of a hydraulic lift; when a pressure is applied on a braking fluid with a piston, it gets equally transmitted in all directions, reaching the brake shoe, stopping the wheel. Ever seen a hydraulic press used to punch holes or deform metal objects in steel factories? It consists of a piston, with a fluid like oil, to uniformly transmit pressure onto a die which compresses the metal placed in contact. The hydraulic lift ,hydraulic brake and hydraulic press all work on the principle of Pascal's law. 

Table of contents

• What is Pascal’s law?
• Applications of Pascal’s law
• Practice Problems
• FAQs

What is Pascal’s law?

Pascal’s law states that the pressure applied to an incompressible, enclosed fluid is transmitted undiminished to every portion of the fluid, as well as to the walls of the container. Take the following example of a liquid filled in a container upto a height h upon which a lid of weight W and area of cross section A. A small hole is made at the top of the container. The liquid has a density . Now the liquid is acted upon by three different pressures

1. Atmospheric pressure P0. One can imagine this to be the pressure exerted by the gas molecules of the atmosphere. The container experiences atmospheric pressure due to the hole on the top.
2. Pressure exerted by the lid W/A
3. Pressure exerted by the liquid column ,gh. Here acceleration due to gravity is denoted by g.

Fig shows three pressures experienced by the liquid

Now according to Pascal’s law, the pressure exerted at the bottom of the container can be written as

Pbottom=P0+gh+WA

Applications of Pascal’s Law

Hydraulic Lift

The hydraulic lift works on the principle of Pascal’s law. It consists of two pistons with different areas of cross sections A1 and A2. The heavy object to be lifted is placed on the piston with bigger cross sectional area A2. A force F1 is exerted on the piston having area A1. Now,the pressure exerted on the first piston is ratio of the force to its cross sectional area .i.e.,P1= F1A1 . 

This pressure is equally transmitted in all directions towards the second piston where the car to be lifted is placed on. If P2 is the pressure on the second piston, then P2=F2A2 

By pascal’s law, P1=P2F1A1=F2A2; F2=A2A1F1 .Since A2>>A1; F2>>F1

What does this mean? The hydraulic lift is a force multiplier.Whatever force we apply at the piston of the shorter cross section area gets multiplied; this enables us to lift heavy objects like trucks placed on the bigger piston.

Fig shows hydraulic lift having two pistons of areas of cross section A1 and A2 (A2>>A1)

Hydraulic Brakes

Hydraulic brakes also work on the principle of Pascal’s law. It consists of a master cylinder fitted with a pipeline that contains the fluid. When the driver manning the vehicle depresses the pedal, the oil flows into the second cylinder which presses the brake pads (also known as brake calipers) against the wheel.

Fig shows working of a hydraulic brake system. The master cylinder transmits pressure evenly so that it reaches the caliper

Practice Problems

Q1. A hydraulic machine has two pistons with diameters of 2 cm and 20 cm. On the smaller piston, a 50 kg wt weight is placed. What is the force applied to the larger piston? 
Answer. Area of the smaller piston A1=r12=110-22=110-4m2

Area of the larger piston A2=r22=1010-22=110-2m2

According to Pascal’s law;

F2=A2A1F1=110-2110-450 10=1025010=5104 N

Q2. Two pistons of cross sectional areas 5 cm2 and 2 cm2 are fitted to both ends of a hydraulic lift. When a force of 10 N is applied on the smaller piston, how much force is experienced by the larger piston? 
Answer. Here A2=5 cm2

A1=2 cm2 ; F1=10 N

F2=A2A1F15210=25 N

Q3. Water is filled in airtight container upto a height of 10 cm. What is the force experienced by the base of the container?(Take g=10 ms-2)
Answer. Total pressure=hg=0.1 1000 10=1000 Pa

Q4. A cylindrical container is filled with water upto height of 10 m upon which a lid of mass 750 Kg is kept. Find the force on the bottom surface if area of cross section is 2 m2?
Answer. Pbottom=P0+hg+mgA

Pbottom=1105+10100010+750102

Pbottom=2105+3750

Pbottom=203.75 KPa

Force=PressureArea

Force=203.751032

Force=4.075105 N

FAQs

Question 1. Why is Pascal's law not applicable for compressible fluid?
Answer. Density of compressible fluid varies because of applied pressure, so Pascal’s law is not valid for compressible fluid.

Question 2. Is Pascal’s law valid for accelerated fluid?
Answer. No, Pascal’s law is not applicable for accelerated fluid.

Question 3. A U-tube manometer has a liquid filled with density upto a height h. What is the pressure exerted by the liquid on the walls of the manometer?
Answer. P=hg

Question 4. Will pressure at the same level always be the same?
Answer. Not always, pressure at the same level will be only if fluid is incompressible and static.