Subject: Science
Pressure is defined as force acting per unit area. Its SI unit is Pascal. This note has brief description of pressure.
Pressure is defined as force acting per unit area. Its SI unit is Pascal or N/m2. Its formula is Pressure: \(\frac{Thrust (F)}{Area (A)}\)
The force acting perpendicularly on a surface is called the thrust. It is a vector quantity. The SI unit is Newton (N).
One Pascal pressure is defined as the pressure exerted on a surface of area 1m2by the thrust of 1N.
Pressure depends on following factors:
Let 'F' be the force acting perpendicularly, 'A' be the area and 'P' be the pressure exerted.
Then,
Pressure is directly proportional to the force acting perpendicularly.
Or, P ∝F.............................(i)
Pressure is inversely proportional to the area on which force acts.
P∝\(\frac{1}{A}\) .............................. (ii)
Combining equation (i) and (ii), we get
P∝\(\frac{F}{A}\)
or, P=K\(\frac{F}{A}\) [ where K is a constant ]
If 'F' is 1N, 'A' is 1m2 and the pressure is 1N/m2, then
K=1
so, P=\(\frac{F}{A}\)
Atmospheric Pressure
The earth is surrounded by atmosphere which is estimated to be 9600 km i.e. about 10000 km thick. As we know air is a matter that has weight. So, atmospheric pressure is the weight of air that exerts pressure on the bodies on the earth’s surface. The value of the atmospheric pressure on sea level is one atmosphere which is considered to be standard atmospheric pressure or normal pressure. Thus, the normal pressure is the pressure exerted by 760 mm long column of mercury at the temperature of 0⁰C at the sea level. It is equivalent to 1.01 × 102 Nm-2. Some units used to express atmospheric pressure are atmosphere, torr, and bar. The interrelation between them is shown below:
1 atmosphere = 760 mm of Hg = 1.01 × 105 Pa
1 torr = 1 mm of Hg = 132.322 Pa
1 bar = 106 dynes/ cm2 = 105 Pa (Approx 750 mm of Hg) = 0.987 atm
1 millibar = 100 Pa
The value of atmospheric pressure decreases as the height from the sea level increases. It is difficult for us to survive on a very low atmospheric pressure. As the pressure of air and the blood pressure inside our body is about equal, we do not feel the normal atmospheric pressure on our body. The air pressure is maintained artificially for the passengers, convenience inside airplanes. At very high altitudes, nose, ears, and eyes may bleed due to less atmospheric pressure.
Measuring Atmospheric Pressure
We can measure the atmospheric pressure by using barometers. There are mainly two types of barometer: Mercury Barometer and Aneroid Barometer. We discuss mercury barometer in this unit.
Mercury Barometer
Mercury barometer was invented by an Italian physicist named Evangelista Torricelli. It is also known as a Torricellian barometer. The barometer consists of one long glass tube filled with mercury. The upper end of the tube is closed and has vacuum above the mercury. There is a mark of scale at the side of it.
When the atmospheric pressure increases, the mercury level rises in the tube and when the atmospheric pressure inside decreases, the mercury level falls down. This is the way in which the instrument measures the atmospheric pressure.
Some Instruments Based on Atmospheric Pressure
There are many instruments which are based on the atmospheric pressure. Some of them are: Syringe, Air Pump, Water Pump, etc. Here is the short description on the structure and working methods of these instruments.
A sharp knife has smaller area at its edge than a blunt knife. When a force is applied on the knife, The pressure exerted by the edge of sharp knife is more than the pressure exerted by the blunt knife during cutting. Due to this reason, cutting with a sharp knife is easier than cutting with blunt knife.
When the area is increased by keeping the magnitude of force constant, the pressure exerted is decreased because pressure is inversely proportional to the area of surface that receives the force.
When a man stands with one foot the area of surface where the force is applied is decreased and as a result the pressure is increased because keeping the force constant, the pressure is inversely proportional to the area of surface that receives the force.
P α F……….. (i)
P = K
If 1 N force is applied in 1 square meter area, then the pressure will be 1N/m2 and thus, k=1.
Mass of a man (m) = 60 kg
Acceleration due to gravity (g) = 10 m/s2
Weight of the man (F) = 600 N
Area occupied by one foot (A) = 150 cm2 =0.015 m2
Pressure exerted (P) =?
P =
=
=40,000 N/m2
This pressure exerted by the man on the ground when he stepped with one foot is 40,000 Pa.
Here,
Mass of Anupama (m) = 30 kg
Weight of Anupama (f) = mg = 30 kg × 10 m/s2= 300N
Surface area of a sandal (A) = 150 cm2 =0.015 m2
Pressure exerted (P) =?
P =
=
=20,000 N/m2
This pressure exerted by the sandal on ground when Anupama stands on one foot is 20,000 Pa.
The pressure is defined as “the force acting perpendicularly per unit area”. The pressure is defined as the perpendicularly applied force on a unit area. The SI unit of pressure is N/m2. It is also called Pascal (pa).
Following are the examples show us the extensive use of pressure in our daily life:
The two factors that affect the pressure are as follows:
The pressure developed in 1 meter area when 1N of force is applied is called 1 Pascal of pressure.
1 pa =What is meant by pressure?
Pressure is defined as the prependicular force acting per unit area.
What is lactometer?
The hydrometer which is used to measure the density of milk is called lactometer.
The base of a rectangular vessel measures 10 cm x 18 cm. Water is poured in it to the depth of 4 cm. What are the pressure and force on the base? (Density of water is 1000 kg/\(m^{3}\), take g = 10 m/\(s^{2}\)
Here, Area of the base (A) = 10 cm x 18 cm
= 0.1 m x 0.18 m
= 0.018\(m^{2}\)
Depth of water in the vessel (h) = 4 cm = 0.04 m
Density of water (d) =1000 kg/\(m^{3}\)
∴Force on the base (F) =?
We know that
P = dgh
= 1000 \(\times\) 10\(\times 0.04 = 400Nm^{-2}\)
Again we that
\(P =\frac {F}{A}\)
∴ F =P ×A
= 400 × 0.018
=7.2N
A girl weight 30 kg and surface area of the heel of her shoe is 100 cm2. Find the pressure exerted on the ground if she stands on her single foot(Take g = 10m/s2).
Here, Mass of the girl (m) = 30 kg
Surface area of single shoe (A) = 100 cm2
= 1 ×10-4m2
∴ Weight(F) = mg
= 30 ×10 = 300N
We know that,
p=\(\frac{F}{A}\)
=\(\frac{300}{1 ×10^{-4}}\)
= 3 ×106pa
Hence, pressure exerted on the ground is 3 × 106pa.
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