Pressure is the ratio of normal force to area. Although both force and area are vectors, pressure is a scalar quantity and has no direction. Pressure is a way to describe force in a region of a continuous system.
CONCEPT OF PRESSURE
Definition,
Pressure is defined as the force per unit area. OR Pressure is the force acting normally (perpendicularly) per unit surface area.
Pressure is defined as the force per unit area. OR Pressure is the force acting normally (perpendicularly) per unit surface area.
It is calculated by the formula:
Pressure = Force (f)
Area (A)
Area (A)
P = F/A
Where P – Pressure
F – Force
A – Area
The SI unit of Pressure is Newton per square metre (N/M2). This unit is usually referred to as the Pascal (Pa).
1Pa = 1 N/M2
1Pa = 1 N/M2
The other units of pressure are atmosphere, torr bar and mmHg
1 atmosphere = 780mmHg
1 atmosphere = 1 105 N/M2 = 1bar (used by meteorologists)
Note: for a given amount of force, the smaller the area of application the greater the pressure exerted.
When a man lifts a bucket of water by its handle that is made with a thin metal, he would experience some discomfort but if the bucket was made with a thicker handle the discomfort will be much less if any.
This is because the area over which the force is applied is larger.
PRESSURE DUE TO SOLID:
The pressure in solid depends on the surface area of contact. A force (F) applied onto a small area exerts a higher pressure as compared to when it is applied onto a large surface.
Pressure in solid = Force applied
Area of contact.
Area of contact.
Example 1;
A block of wood that weighs 30N and measures 5m by 10m by 4m. If it was placed on a table with the largest possible area (5mx10m) in contact with table, exerts less pressure than it would when placed with its smallest possible area (5mx4m) in contact with table.
Solution
Data:
Force = 30N
Largest base area = ?
Solution
Data:
Force = 30N
Largest base area = ?
AL = 10mx5m = 50m2
P = F/A = 30N = 0.6N = 0.6N/m2
30m2
P = F/A = 30N = 0.6N = 0.6N/m2
30m2
- Pressure = 0.6N/m2
(b) Force (F) = 30N.
Small base are = ?
As = 5mx4m
As = 20m2
P= F/A = 30N/20M2 = 1.5 N/M2
- Pressure = 1.5 N/M2
Example 2:
A tip of needle has a cross- sectional area of 1x106m2. If doctor applies a force of 2N to a syringe that is connected to the needle, what is pressure is exerted at the tip of the needle?
Soln:
Data:
Area (A) = 1x106m2
Data:
Area (A) = 1x106m2
Force (F) = 20N
Pressure =?
P = F/A
P = 20 = 2.0x107 N/M2
- The pressure extended by the needle lip is 2.0x107N/m2
Example 3:
A rectangle metal block with sides 105m by 1.0m by 1.2m rests on a horizontal surface. If the density of the metal is 7000kg/m3. Calculate the maximum and minimum pressure that the block can exerts on the surface.
(Take the weight of 1kg mass to be 10N)
Solution:
Data:
Dimension = 1.2m,1.5m,1.0m.
Density = 7000kg/m3
Maximum pressure (Pmx) = ?
Minimum pressure (Pmn) = ?
Volume = 1.2x1.5x1=18m3
Density = 7000kg/m3
Maximum pressure (Pmx) = ?
Minimum pressure (Pmn) = ?
Volume = 1.2x1.5x1=18m3
Density = mass
Volume
Volume
7000kg/m3 = mass
18
18
Mass = 7000x18 = 126000Kg
1kg = 10N
126000Kg = ?
F = 1260000 = 1.26 x 106
Force of metal = 1.26 x 106
1kg = 10N
126000Kg = ?
F = 1260000 = 1.26 x 106
Force of metal = 1.26 x 106
A1 = 1.2 x 1.5= 18m2
A2= 1.2m2= 1.2mx1m=1.2m2
A3= 1.5 x 1m= 1.5m2
Amx = 18m2
Amin = 102m2
PMX = F/AMX
PMX = 1.26X106N = 1.05x106N/m2
12m2
12m2
Pmin = large area
Pmn = F/Amx =1.26 x 106N = 1.26X106
18M2 1.8 X101
18M2 1.8 X101
- Maximum pressure = 1.05x106N/m2
maxmum pressure= 0.7x105N/m2
Application of pressure in solid:
It is used to make different objects like screw, nails, pins, spears and arrows. This item is given sharp points to increase their penetrating power.
It helps some of living organisms for self-defence. A fish uses its sharp fins to protect itself.
When one walks on shoes with sharp pointed heels they exert greater pressure on the ground than when wearing flat shoes.
It helps in construction of railways. During the constructions of railways, wide wooden or concrete sleepers are placed below the railways tracks. This provides a larger surface area over which the weight of the train acts. This is a safety measure to train tracks.
Buildings are constructed with wide foundations to ensure that the weight of the building acts over the layer area.
PRESSURE IN LIQUIDS
A liquid will exert pressure on an immersed object as well as on the walls of the container holding it. Note that the pressure exerted by liquids is due to the weight of the liquid. Also increase in water level in liquid cause increase in pressure.
H= Height of the liquid column
A= Area of the base
Characteristic of pressure liquids
Note: Pressure in liquids =Force
Area
But force= hApg
A
Area
But force= hApg
A
=Hsg
Hence pressure in liquids is given as hsg.
The pressure at any point in a liquid at rest then depends
The pressure at any point in a liquid at rest then depends
- Depth
- Density of liquid.
Note: It does not defend on the area
Pressure in liquids is characterized using the following parameters
Pressure in liquids is characterized using the following parameters
- Pressure in a liquid increase with depth
- Pressure in a liquid acts equally in all directors
- Pressure in a liquid increases with increase in density of the liquid.
Mercury exerts more pressure than an equal volume of water. This because mercury is denser than water.
Variation of pressure with depth.
Variation of pressure with depth.
a. Demonstration of water spurting from holes at different heights.
As pressure at point A is low due to small height of water above it but at point B the height increase and the distance of water increase while at point C the height is greatest therefore the level of water distance will be large distance.
As pressure at point A is low due to small height of water above it but at point B the height increase and the distance of water increase while at point C the height is greatest therefore the level of water distance will be large distance.
b. Demonstration of pressure in a communicating vessel
The shape of vessel found in point A, B, C and D is different but the pressure is the same due to the equal height (L) of the liquid above the points.
The shape of vessel found in point A, B, C and D is different but the pressure is the same due to the equal height (L) of the liquid above the points.
Examples: 1
A cube of side 2cm is completely submerged in water so that the bottom of the cube is at a depth of 10cm. use g=10m/s2 and s=1000kg/m3
A cube of side 2cm is completely submerged in water so that the bottom of the cube is at a depth of 10cm. use g=10m/s2 and s=1000kg/m3
- What is different between the pressure on the bottom of the cable and the pressure on its tap?
- Determine the difference in the force on the top and bottom.
- What is the weight of the water displaced by the cube?
Solution;
Data.
Depth = 10cm = 0.1m
s = 1000kg/m3
g = 10m/s
h= 8cm= 0.08
Data.
Depth = 10cm = 0.1m
s = 1000kg/m3
g = 10m/s
h= 8cm= 0.08
formular
p= sgh
p=1000kg/m3x10m/s2x0.1m=1000N/m2
At the h= 0.08m p= 1000x10x0.08= 800N/m2
Change in pressure = 1000N/m =200N/m2
So that different in pressure between the bottom the top is 200N/m2
p= sgh
p=1000kg/m3x10m/s2x0.1m=1000N/m2
At the h= 0.08m p= 1000x10x0.08= 800N/m2
Change in pressure = 1000N/m =200N/m2
So that different in pressure between the bottom the top is 200N/m2
(b) Pressure = Force = P= F
Area A
Area A
Area =L2= (10.02)2= 0.0004m2
Distance in water 200N/m2x0.0004m2
This is the upthrust acting on the cube
( c) water displaced by the cube in weight= volume of the cube
Volume = l3 (2cm) 3 = 8cm3
For water, 1cm3= 1g 50.8cm3 = 8g
Weight of displaced water
For water, 1cm3= 1g 50.8cm3 = 8g
Weight of displaced water
= mass of displaced water=8g = 0.08N
100
100
Example 2:
Calculate the pressure at the bottom of tank of water 15m deep due to the water above it is (s=1000kg/m3).
Sln:
Data:
Height =15m
S=1000kg/m3
g= 10m/s2
Formula
Pressure in liquid sgh
= 1000x10x15=150000N/M2
Paschal’s principle of the hydraulic press:
It state that “Any external pressure applied to the surface of an enclosed liquid will be transmitted equally throughout liquid.”
Illustration of Pascal’s Principle.
Note: in all side of vessel will experience equal pressure.
Calculate the pressure at the bottom of tank of water 15m deep due to the water above it is (s=1000kg/m3).
Sln:
Data:
Height =15m
S=1000kg/m3
g= 10m/s2
Formula
Pressure in liquid sgh
= 1000x10x15=150000N/M2
Paschal’s principle of the hydraulic press:
It state that “Any external pressure applied to the surface of an enclosed liquid will be transmitted equally throughout liquid.”
Illustration of Pascal’s Principle.
Note: in all side of vessel will experience equal pressure.
The hydraulic press
The hydraulic press works on Pascal’s principle, when using this principle a small force converted into a large force and vice versa.
From Pascal’s Principle
From Pascal’s Principle
P1=P2
F1=F2
A1 A2
F1=F2
A1 A2
F2 =F1, A2, but F1=P1
A1
F2=P1,A2
A1
F2=P1,A2
OR
F2= F1,A2
A1
A1
Note: The distance moved (d) is inversely proportional to the cross sectional area.
Therefore, d1 =A1
d1 A2
d1 A2
therefore d2 = A1d1
A2
A2
Example,
The pistons of a hydraulic press have their areas given as 3.0x10-4m2 and 2x102m2 respectively. If the smaller piston is pushed down with a force of 12N, what is the force required to push the larger piston?
The pistons of a hydraulic press have their areas given as 3.0x10-4m2 and 2x102m2 respectively. If the smaller piston is pushed down with a force of 12N, what is the force required to push the larger piston?
Soln:
Data:
F1 = 120N, A2= 2X102m2
A2 = 3X104m2
Data:
F1 = 120N, A2= 2X102m2
A2 = 3X104m2
Therefore, P1=F1=120N = 120 = 4X105N/m2
A1 3.0X10-4 3x0.0003
Force on the larger piston
F2 = P2xA2
A1 3.0X10-4 3x0.0003
Force on the larger piston
F2 = P2xA2
= 4.0x105N/M2x 2.0x102m = 8.0x4x103N
APPLICATION OF THE HYDRAULIC PRESS
- Hydraulic press is used in industries to express bulk items.
- Hydraulic brake system
When pressure is applied to the brake pedal, it pushes the piston in the master cylinder forward creating a pressure in the brake fluid. This pressure is transferred to the slave cylinders where it is multiplied and pushes the brake shoes against the brake dram that is attached to the wheel of the vehicle. - It is used in industries in the forming of metals.
- Since the hydraulic press act like a lift, it is used for lifting heavy loads.
Measuring the pressure of liquids using a manometer. Manometer is the device for measuring pressure (commonly gas).
ATMOSPHERIC PRESSURE:
Atmospheric pressure is as a result of the weight of this layer of gases, the act at atmospheric pressure is surrounded the layer of air consisting of a mixture of gases.
Note. The atmospheric pressure on the earth’s surface and objects on the earth is a approximately 1.01x105N/m2.
The experiment which demonstrate the existence of atmospheric pressure.
- Tumbler experiment
When you fill a glass timber with cord and gentle turned upside down. It will be seen that water will not poured down. - Plunger
Pulling the plunger is not easy. This is so because all the air is squeezed out from the table when the surrounding pressure being high the plunger sticks. - Crushed bottle
When putting hot water in, the bottle is cooled under cold water the steam condenses; leaving partial vacuum inside the bottle consequently the greater atmospheric pressure outside the bottle crushes it in wards.Atmospheric pressure is measured by using barometer;
Types of barometer;-
- The simple barometer
- Fortin barometer
- Aneroid barometer
A simple barometer is the most fundamental of the other types of barometer. The barometric liquid used is mercury.
A simple barometer consists of a hard glass tube closed at one end.
A simple barometer consists of a hard glass tube closed at one end.
Fortin barometer
A Fortin barometer is a modified simple barometer. It consists of an inverted tube closed at its upper end with the lower open end immersed in a reserve of mercury. The atmospheric pressure is measured in terms of the height of the column of mercury.
Disadvantages of fortin barometer;
Disadvantages of fortin barometer;
- Mercury is very expensive to use and is very toxic. For these reasons, an aneroid barometer is usually preferred.
- It is not portable as it is generally big and contains liquid.
- It must be mounted in a vertical position.
Application of atmospheric pressure.
(a) A siphon
A siphon is applied in areas and devices that we use everyday
A siphon is applied in areas and devices that we use everyday
- It is used in the toilet flushing cisterns (chain and ball tank). The flush is triggered by a handle that operates a simple diaphragm like piston pump that lifts enough water.
- It is used in special rain gauges called siphon rain gauge which are able to automatically drain out excess water.
- A siphon cup is a reservoir attached to a gum.
- It is used is some drainage systems to drain water to one point.
(b)The lift pump
A lift pump is used to raise water from underground sources. This is a pump that is used to lift the liquid, rather than force liquid up.
A lift pump is used to raise water from underground sources. This is a pump that is used to lift the liquid, rather than force liquid up.
(c) A syringe is a simple piston pump that lifts a tube. The plunger that lifts can be pulled and pushed through inside a cylindrical tube or barrel. This enables the syringe to take in or expel fluid through the opening (nozzle) at the end of the tube.
Uses of syringe:
- They can be fitted with hypodermic needles and used to administer injections.
- They are used to measure liquids and gases in a laboratory.
- They are used to apply certain compounds such as a glue or lubricant.
(c) Bicycle pump
This is a type of force pump that consists of a hollow metal cylinder and a movable piston.
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