Pressure
Pressure is the force acting per unit area at right
angles to a surface.
Pressure can be calculated by using the equation:
𝐹𝑜𝑟𝑐𝑒 (𝑁)
Pressure =
𝐴𝑟𝑒𝑎 (𝑚2 )
SI unit of pressure is Pascal (Pa) or N/m2 and pressure
can also measure in kilopascal (Kpa). 1kpa = 1000pa
Example question:
The base for a statue rests on level ground. It is made
from stone and is 2.0 m long, 2.5 m high and 0.80 m
wide. It has a weight of 96 000 N. Calculate the
pressure.
Ans: Base area hit the ground = l × b
= 2 × 0.80 = 1.6 m2
P = F/A = 96000/1.6
= 60 000 Pa
Pressure act on an object depends on two things
 How much force is applied to the object
If higher force applied on an object the pressure act on
the object will be more.
 How big (or small) the area of the object on which
force applied
If the area of the object on which force applied is more
the pressure act on the object is low and if the area of
the object on which force applied is less the pressure
act on the object is more.
Everyday examples of how pressure varies
with force and area
Increasing pressure by reducing the area
The area under the edge of the
knife’s blade is small. Beneath it,
the pressure is high enough for the
blade to push easily through the
material
The end of the pin is very sharp, it
provide great pressure to the
objects and so easy to put into
objects
Reducing pressure by increasing the area
Skis have a large area to reduce
the pressure on the snow so that
they do not sink in too far
Tractors have wide tyre to reduce
the pressure act on the ground so
that they do not sink too far of the
ground
Pressure in liquids
In a liquid such as water, pressure does not simply act
down wards – it equally in all directions. This is
because the molecules of the liquid move around in all
directions, causing pressure on every surface they
collide with.
Factors affecting the pressure in a liquid
The pressure in a liquid depends on
 Depth of the liquid
The deeper the liquid, the higher pressure
 Density of the liquid
The more dense the liquid, the higher pressure at
particular depth
Pressure of the liquid doesn’t depend shape of the container.
Calculating the Pressure in a Liquid
The pressure on the liquid can be calculated by using
the following equation:
Pressure on the liquid = density of the liquid ×
gravitational field × depth of the liquid
P = ρgh
Example question:
A rectangular storage tank is filled with paraffin to a depth of
0.8m. The density of the paraffin is 800kg/m3. Calculate the
pressure at the bottom of the tank.
Ans: P = ρgh
P = 800 × 10 × 0.8
= 6400 Pa
Hydraulic System
In some machines, the forces are transmitted by
liquids under pressure rather than by levers or
cogs. Machines like this are called hydraulic
machines. In hydraulic machine the following
properties of the liquids is used:
 Liquids are virtually incompressible- they
cannot be squashed.
 If a trapped liquid is put under pressure, the
pressure is transmitted all parts of the liquids.
Hydraulic Jack
Pressure can be transmitted throughout a liquid in
hydraulic presses and hydraulic brakes on vehicles.
The figure below shows a simple hydraulic system.
The hydraulic system shown above when the 15N force
applied to the piston 1, it provides pressure of 1500Pa to
the hydraulic oil. (P = F/A = 15/0.01 = 1500Pa). This
pressure flows through the hydraulic oil and acts on piston
2. As a result the piston 2 produces the larger output force
of 150N. (P = F/A, F = P × A = 1500 × 0.1 = 150N). So
sometimes hydraulic system is also called force multiplier.
Example question:
The system shown in the diagram contains a liquid.
A downward force of 4 N is exerted on piston L.
What will be the upward force exerted by the liquid on
piston K?
Ans: P = F/A = 4/2 = 2Pa
F = P × A = 2 × 40 = 80N
Hydraulic brakes on vehicles
The diagram below shows a braking system of a vehicle.
When the brake pedal is pushed the master piston
provides the pressure on the brake fluid and this pressure
transmit through the brake fluid to slave pistons. As a
result slave pistons pushed outward and cause brake
shoe to rub against the brake drum.
Pressure measurements
The barometer
Simple mercury barometer is used to measure the
atmospheric pressure. The barometer contains liquid
metal mercury. Atmospheric pressure has pushed
mercury up the tube because the space at the top of the
tube has no air in it. It is a vacuum. Normally the height
of the mercury column found to be about 760
millimetres of mercury (760mmHg). Since the value of
the mercury column reflects the value of the atmospheric
pressure, atmospheric pressure can be express in terms
of the height of the column of the mercury.
Standard atmospheric pressure
The pressure that will support a column of mercury 760mm
high is known as standard atmospheric pressure or 1
atmosphere (1atm). Its value in a pascals can be found by
calculating the pressure due to a such column.
Height of the mercury column (h) = 760mm that is 0.76m.
Gravitational field strength (g) = 10N/kg
Density of the mercury (ρ) = 13590kg/m3
So standard atmospheric pressure or 1atm = ρgh
Standard atmospheric pressure or 1atm = 13590 x 10 x 0.76
= 101 300Pa, in calculators for simplicity, you can assume
that 1atm = 100 000Pa
The manometer
Manometer is an instrument used to measures pressure difference.
The gas supplied tube is filled with mercury. The height difference
shows the extra pressure that the gas supply has in addition to
atmospheric pressure. This extra pressure is called excess pressure.
To find the actual pressure of the gas supply, add atmospheric
pressure to this pressure. Standard atmospheric pressure is 1 atm.
How to measure gas pressure?
 First measure the height difference of the liquid column.
(Height difference of the liquid column= 60 – 30 = 30mm
or 0.03m)
 Then calculate the pressure difference of the liquid
column.
(Pressure difference of the liquid column = ρgh = 13590 ×
10 × 0.03 = 4077Pa)
 Pressure of the gas supply = pressure difference of the
liquid column + atmospheric pressure.
(Pressure of the gas supply = 4077 + 100 000 =
104077Pa)
Pressure – volume relationship
The diagram below shows a syringe fill with air and
end of the syringe is connected with pressure gauge.
When the piston of the syringe is pushed inward the
volume of the air trapped inside the syringe decreases
but pressure of the air increases. The pressure increases
because the air molecules hit the wall of the syringe
more frequently.
Boyle’s law
For a fixed mass of a gas at a constant temperature,
pressure of the gas is inversely proportional to the volume
of the gas.
Which means if the temperature of the gas is constant,
when the volume of the gas is decreased the pressure of the
gas increases.
If you plot the graph pressure of the gas against the volume
of the gas you will obtain a smooth curve as shown below.
The following equation is derived from this law:
Initial pressure × initial volume = final pressure × final
volume (P1V1 = P2 V2)
Example question:
The syringe shown below consist the volume of 100cm3 of
gas initially and the pressure of the gas is 1.0 × 105 Pa. The
volume of the gas decreases to 80 cm3. Calculate the final
pressure of the gas.
Ans: P1V1 = P2 V2
1.0 × 105 × 100 = P2 × 80
P2 = 1.0 × 105 × 100 / 80
P2 = 125000 Pa