Class 11 Physics Mechanical Properties of Fluids Application : Hydraulic Brakes

Hydraulic Brakes

  • Hydraulic brakes work on the principle of Pascal’s law.
  • According to this law whenever pressure is applied on a fluid it travels uniformly in all the directions.
  • Therefore when we apply force on a small piston, pressure gets created which is transmitted through the fluid to a larger piston. As a result of this larger force,uniformbrakingis applied on all four wheels.
  • As braking force is generateddue to hydraulic pressure,theyare known as hydraulic brakes.
  • Liquids are used instead of gas as liquids are incompressible.

Construction

  • The fluid in the hydraulic brake is known as brake fluid.
  • It consists of a master cylinder, four wheel cylinders and pipes carrying brake fluid from master cylinder to wheel cylinders.
  • Master cylinder consists of a piston which is connected to pedal through connecting rod. 
  • The wheel cylinders consist of two pistons between which fluid is filled.
  • Each wheel brake consists of a cylinder brake drum. This drum is mounted on the inner side of wheel. The drum revolves with the wheel.
  • Two brake shoes whichare mounted inside the drum remain stationary.

Working

  • When we press the brake pedal, piston in the master cylinder forces the brake fluid through a linkage.
  • As a result pressure increases and gets transmitted to all the pipes and to all the wheel cylinders according to Pascal’s law.
  • Because of this pressure,both the pistons move outand transmit the braking force on all the wheels.

Advantages:-

  • Equal braking effort to all the four wheels.
  • Less rate of wear due to absence of joints.
  • By just changing the size of one piston and cylinder, force can be increased or decreased.

Disadvantages:-

  • Leakage of brake fluid spoils the brake shoes.
  • Even the slightest presence of air pockets can spoil the whole system.

 

Inside of the cylinder

Problem:- Two syringes of differentcross sections (without needles) filled withwater are connected with a tightly fittedrubber tube filled with water. Diametersof the smaller piston and larger piston are1.0 cm and 3.0 cm respectively. (a) Findthe force exerted on the larger piston whena force of 10 N is applied to the smallerpiston. (b) If the smaller piston is pushed in through 6.0 cm, how much does thelarger piston move out?

Answer:

  • Since pressure is transmitted undiminished throughout the fluid,

F2 = (A2/A1) F1= (3/2 10-2m2/1/2 10m-2 m2) 10N

=90N.

(b) Water is considered to be perfectlyincompressible. Volume covered by themovement of smaller piston inwards is equal tovolume moved outwards due to the larger piston.

L1 A1 = L2 A2

= 0.67 × 10-2 m = 0.67 cm

Note, atmospheric pressure is common to bothpistons and has been ignored.

Problem:- In a car lift compressed airexerts a force F1 on a small piston havinga radius of 5.0 cm. This pressure istransmitted to a second piston of radius 15 cm. If the mass of the car tobe lifted is 1350 kg, calculate F1. What isthe pressure necessary to accomplish thistask? (g = 9.8 ms-2).

 Answer:-

Since pressure is transmittedundiminished throughout the fluid,

F1=A1/A2 F2

= (5x10-2 m2/15x10-2m2) 1350N x 9.8ms-2

=1470N = 1.5x103N

The air pressure that will produce thisforce is

P=F1/A1 = (1.5x103N/5x10-2m2)1.9x105 Pa

This is almost double the atmosphericpressure.

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