Class 10 | Pressure (Pascal's Law, Density and Relative Density, Archimedes' Principle, Principle of Flotation)

PRESSURE

---

Thrust

A force can be applied on a surface in any direction. When the force is applied on a surface in a direction normal (or perpendicular) to the surface, it is called the thrust. A body, when placed on a surface, exerts a thrust on the surface which is equal to the weight of the body. Thus, thrust is defined as:

The force acting normally on a surface is called the thrust. Thrust is a vector quantity.

Unit of thrust: It is measured in the units of force. The SI unit of thrust is Newton (N) and CGS unit is dyne, where;

1N =10^5 dyne.

Pascal's Law

Blaise Pascal 's was a noted French physicist who discovered that a closed container of fluid could be used to transfer force from one place to another or to multiply forces by its transmission through a fluid. It should be noted that Pascal's law applies to fluids-both gas and liquid.

Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Application of Pascal's Law

The figure shows two cylindrical vessels P and Q connected by a horizontal tube R. The vessels contain a liquid (or water) and they are provided with water-tight pistons P1 and Q1. The vessel P is smaller diameter than the vessel Q. Let are of cross section of vessel P be A1 and that of vessel Q be A2. A force F1 is applied on the piston P. Therefore, the pressure applied on the piston P is:

P1 = F1/ A1 

According to Pascal's law, the pressure exerted on piston P is transmitted through the liquid to piston Q. Thus, the upward pressure P2 exerted on Q is:

P2 = P1 = F1/A2

Hence the upward force exerted on piston Q is F2 - Pressure on piston P x Area (A2)

or, F2 = (F1/A1) X A2

or, F2/F1 = A2/A1

Since A2>A1, therefore F2>F1.

Thus, a small force F1 applied on the smaller piston P1 can be used to produce a large force F2 on the bigger piston Q1. This is the principle of a hydraulic machine. The formula used in above hydraulic press is:

Pressure in large piston = Pressure in small piston

F2/F1 = A2/A1


Example: Calculate the cross sectional area of the small piston from the given figure of the hydraulic press.

Solution :
          Force on large piston (F2)=12000N
          Force on small piston (F1)=200N
          Cross sectional area of large                 piston (A2)=2.5 sq.m
          Cross sectional area of small                 piston (A1)= ?
          We have; using Pascal's Law,
          F2/F1=A2/A1
      or, 12000/200=2.5/A1
      or, A1=2.5/60
       ∴  A1=0.0417 sq.m
Hence, the cross sectional area of small piston is 0.0417 sq.m.

Uses of hydraulic press
A hydraulic press is used mainly for the following purposes:
i.      For pressing the cotton bales and goods, quilts, books etc.
ii.     For extracting the juice of sugarcane, sugar beet, etc.
iii.    For squeezing the oil out of linseed and cotton seeds.
iv.    For engraving the monogram on goods. 

Density and Relative Density

Density
The density of substance is defined as its mass per unit volume.

Density of substance (d)= Mass of the substance (m)/ Volume of substance (v)
d=m/v

Unit of density
Unit of density = Unit of mass (kg)/ Unit of volume (m^3)

The SI unit of mass is kilogram (kg) and the SI unit of volume is cubic meter (m^3). So, the unit of density is kg/m^3. In CGS system. the unit of mass is gram (g) and the SI unit of volume is cubic centimeter (cm^3). So, in CGS system the unit of density is g/cm^3.

Relation between the SI and C.G.S unit of density
     1kg/m^3=1kg/(1m x 1m x 1m)
                   =1000g/(100cm x 100cm x 100cm)
     1kg/m^3=1g/1000cm^3
or, 1000kg/m^3=1g/m^3
Thus, 1g/cm^3= 1000 kg/m^3

---