Force And Pressure

Force And Pressure

PRACTICAL APPLICATIONS OF THE KNOWLEDGE OF PRESSURE
1.paper-pins and nails are made to have pointed ends.Because of their pointed end,pins and nails have very small area .When force is applied over the head of apin or a nail,it will exert a large pressure \( \left[ { = \frac{{force}} {{area}}} \right] \)on the surface and hence will easily penetrate into the surface
2.It is painful to walk on a road covered with pebbles having sharp edges.The weight of our bodyis supported on the pebbles under our feet.In turn ,the pebbles give equal reaction to the feet.As the pebbles have very small area at their sharp edges,the pressure exerted by them on the feet is very large and they prove painful
3.Railway tracks are laid on large sized wooden/concrete sleepers.In the absence of wooden sleepers,the weight of the train acts on the ground through rails and as such pressure would be so large that the rails may get depressed.When wooden sleepers are provided below the rails,the weight acts on greater area and hence pressure is reduced quite appreciably.Due to this,the rails do not get depressed

ATMOSPHERIC PRESSURE
    The earth is surrounded by a gaseous envelope extending upto a few thousand kilometres.This gaseous envelope is made of 78% nitrogen,21% oxygen and a small amount of carbob dioxide,water vapour,etc and this gaseous envelope is called earth’s atmosphere.The density of the atmosphere goes on decreasing as one goes up the surface of earth
    The pressure exerted by the atmosphere is called atmospheric pressure.
Obviously,it will be maximum at the surface of the earth and will decrease as we move up the surface of earth.The atmospheric pressure is about \( 1.013 \times 10^5 Nm^{ - 2} \)
(or Pa)
Thus, at sea level
    1 atm=pressure exerted  by 0.76m of mercury column
        =\( 0.76 \times 13.6 \times 10^3 \times 9.8 = 1.013 \times 10^5 Nm^{ - 2} \)
Thus,     1 atm=\( 1.013 \times 10^5 Nm^{ - 2} = 1.013 \times 10^6 dyncm^{ - 2} \)
The atmospheric pressure is also expressed in terms of the following units:
    1 torr=pressure exerted by 1mm mercury column=133.28Pa
    1 bar=\( 10^5 Pa \)

HEIGHT OF THE ATMOSPHERE
Let us now calculate the height of the air column that would exert a pressure of \( 1.01 \times 10^5 Nm^{ - 2} \) on the surface of earth i.e., just as much that a mercury column of height 0.76m exerts .Since the density of air goes on decreasing as we go up,to determine the value of height of the air column ,the relation \( P = h\rho g \) cannot be applied directly .However taking the density of air as to be uniform(=1.3kg ),the height h of the air column is given by
        \( \begin{gathered} h \times 1.3 \times 9.8 = 0.76 \times \left( {13.6 \times 10^3 } \right) \times 9.8 \hfill \\ (or)h = \frac{{0.76 \times 13.6 \times 10^3 }} {{1.3}} = 7951m \approx 8km \hfill \\ \end{gathered} \)
Following two  assumptions have been made in calculating the height of the atmosphere:
(i)The acceleration due to gravity does not change appreciably upto a height of the order of 8km
(ii)The temperature of the atmosphere is same throughout.