Mechanical Engineering Principles

21 Hydrostatics

21.1 Pressure

The pressure acting on a surface is defined as the
perpendicular force per unit area of surface. The unit
of pressure is thePascal, Pa, where 1 Pascal is equal
to 1 Newton per square metre. Thus pressure,

p=FAPascal’s

whereF is the force in Newton’s acting at right
angles to a surface of areaAsquare metres.
When a force of 20 N acts uniformly over, and
perpendicular to, an area of 4 m^2 , then the pressure
on the area,p,isgivenby

p=

20 N

4m^2

=5Pa

Problem 1. A table loaded with books has a

force of 250 N acting in each of its legs. If

the contact area between each leg and the

floor is 50 mm^2 , find the pressure each leg

exerts on the floor.

From above, pressure p=

force

area

Hence,

p=

250 N

50 mm^2

=

250 N

50 × 10 −^6 m^2

= 5 × 106 N/m^2 =5MPa

That is,the pressure exerted by each leg on the
floor is 5 MPa.

Problem 2. Calculate the force exerted by

the atmosphere on a pool of water that is

30 m long by 10 m wide, when the

atmospheric pressure is 100 kPa.

From above, pressure=

force

area

, hence,

force=pressure×area.

The area of the pool is 30 m×10 m, i.e. 300 m^2.

Thus, force on pool,F =100 kPa×300 m^2 and

since 1 Pa=1N/m^2 ,

F=( 100 × 103 )

N

m^2

×300 m^2 = 3 × 107 N

= 30 × 106 N=30 MN

That is,the force on the pool of water is 30 MN.

Problem 3. A circular piston exerts a

pressure of 80 kPa on a fluid, when the force

applied to the piston is 0.2 kN. Find the

diameter of the piston.

From above, pressure=

force

area

hence, area=

force

pressure

Force in Newton’s

= 0 .2kN= 0. 2 × 103 N=200 N, and

pressure in Pascal’s is 80 kPa=80 000 Pa

=80 000 N/m^2.

Hence,

area=

200 N

80 000 N/m^2

= 0 .0025 m^2

Since the piston is circular, its area is given by

πd^2 /4, wheredis the diameter of the piston. Hence,

area=

πd^2

4

= 0. 0025

from which, d^2 = 0. 0025 ×

4

π

= 0. 003183

i.e. d=

√

0. 003183

= 0 .0564 m,i.e. 56.4mm

Hence,the diameter of the piston is 56.4 mm.