ENGINEERING SCIENCE

3) How much work is done in lifting a 4) A cart 6 kg load of sand is pulled 5 m
2 kg book onto a shelf 150 cm high? across the ground. The tension in the
rope is 300 N and is directed 30
degrees above the horizontal. How
much work is done pulling the load?

Answer: W = 29.43 J Answer: W = 1299.04 J

5) Determine the amount of work done

a) A 100 N force is applied to move a 15 kg object of a horizontal distance of 5 meters

15 kg 100 N

Answer: W = 500 J

b) A 55 N force at an angle of 300 to the horizontal to move a 10 kg object at a
constant speed for a horizontal distance of 7 meter.
55 N

10 kg 300

Answer: W = 333.42 J

c) An upward force is applied to lift 8kg object to a height of 6 meter of constant speed.
F

8 kg

Answer: W = 470.88 J

Chapter 4–Work, Energy and Power/ Page 144

When No Work Is Done

No work is done when:

a) The object is stationary b) The direction of motion of the object is
perpendicular to that of the applied force

• If the displacement is zero, no work • No work is done if an object is
is done even if a force is applied moving with uniform without the
action of any force
= × where = 0
= 0 = cos 90 ×

= 0

s
s =0

Figure 4.4a: A man standing while Figure 4.4b: A man is carrying a
he is carrying a briefcase briefcase while walking

Video description: Work and energy

Chapter 4–Work, Energy and Power/ Page 145

QUICK CHECK

Quick review: Doing/Not doing work????
1) 2)

A student carrying his bag while waiting at the A man pushes a wall concrete
bus stop Answer:
Answer:

3) 4)

Ali lifting 100 N weigh for 0.16 m high A waiter is carrying a tray of food and
Answer: walking

Answer:

Chapter 4–Work, Energy and Power/ Page 146

4.2 RENEWABLE ENERGY

1. Resources are things we get from the earth. It can be either be renewable or non-renewable
resources.

Concept Renewable Non-renewable
Definition Renewable resources can be replaced and
replenished in our lifetime Non -renewable resources
Sources cannot be replaced in our
Biomass, geothermal, hydropower, solar, lifetime. They do not renew
wind quickly
Coal, natural gas,
petroleum, propane,
uranium.

Advantages • Sustainable • Supplies are abundant so
• Requires little maintenance the cost is cheap
• Produce little waste
• Can produce large
Disadvantages • Difficult to produce large quantities amounts of energy
• The reliability of supply is often
dependent on weather • Not sustainable because
the resources will run out

• The use of them can
produce pollutants.

2. There are examples of renewable energy:

a) Solar This type of renewable energy comes directly from
Figure 4.5a: Solar power plant the capture of solar radiation. Here, the solar radiation
is absorbed by specific sensors and rebroadcasted by
capturing sun rays and directly converting them into
energy through photovoltaic solar panels.
For example: The Kurnool Ultra Solar Park in India.
It has a total generation capacity of 1000MW and over
4 million solar panels installed.

Chapter 4–Work, Energy and Power/ Page 147

b) Wind Wind power is another renewable energy. Here, the
wind’s kinetic energy makes turbines spin and
Figure 4.5b : Wind power plant creates a mechanical movement. Afterward, a
c) Hydro-electric power generator transforms this mechanical energy into
electricity.
Figure 4.5c: Hydropower station For example: The onshore Muppandal Wind Farm
d) Biomass in India with a capacity of 1,500MW and over 3000
turbines.
Figure 4.5d: Wood pellets
e) Geothermal Hydro-electric power consists in the transformation
of the kinetic energy of the water (from rivers,
Figure 4.5e: Geothermal power dams, marine currents or tides) into mechanical
plant energy by turbines.
For example: The 22.5GW Three Gorges
hydroelectric power plant in Yichang, Hubei
province, China, is the world’s biggest hydropower
station.

Biomass is made up of organic materials from
plants or animals that contain stored energy. The
combustion of these natural materials produces
renewable energy.
For example: The Iron Bridge power station in the
UK, with a capacity of 740MW. It uses wood
pellets as the main fuel.

The Earth generates and stores geothermal
energy. In other words, radioactive materials
decaying inside the earth are emitting energy.
For example: The Darajat Power Station in
Indonesia has 3 plants with a total capacity of
259MW.

Chapter 4–Work, Energy and Power/ Page 148

QUICK CHECK

Video description: Renewable energy
Chapter 4–Work, Energy and Power/ Page 149

4.3 CONCEPT OF ENERGY

The definition of energy is capacity to do work. Unit of energy is measured in Joule (J).
Gravitational Potential Energy & Kinetic Energy

Concept Definition and Formula Unit

Gravitational Potential Gravitational potential energy, is the
Energy energy of an object due to its higher position

in the gravitational field

=

m = mass (kg) Joule (J)
h = height (m)

g g= gravitational acceleration (9.81 −2)

Kinetic Energy Kinetic energy, is the energy of an object
due to its motion.

= 1 2
2

m = mass (kg)
v = velocity ( −1)

Video description: Kinetic and potential energy
Chapter 4–Work, Energy and Power/ Page 150

Example 4
A helmet with a mass of 14 kg sits on a shelf 2.5 m off the ground. What is the potential
energy of the helmet?

Solution:

Given: = 14 , ℎ = 2.5 , , ?

= ℎ
= 14 × 9.81 × 2.5
= .

Example 5

A book with mass of 1.5 kg on a table that is 1.2 m high is raised onto a shelf. The shelf is
2 m from the table top.
a) What is the gravitational potential energy of the book relative to the table top?
b) What is the gravitational potential energy of the book relative to the floor?

1.2 m book
shelf

table
2m

Solution:
Given: = 1.5 ℎ ℎ = 1.2 ℎ = 2 ?

a) The zero level is the table top
= ℎ
= 1.5 × 9.81 × 1.2
= .

b) The zero level is the floor.
= ℎ
= 1.5 × 9.81 × 3.2
= .

Chapter 4–Work, Energy and Power/ Page 151

Example 6

A rocket with a mass of 80 kg has an initial velocity of 12 m/s. What is the kinetic energy
of the rocket?

Solution: = 12 ,

Given: = 80 , ?

= 1 2
2

= 1 × 80 × 122
2

=

Example 7

A 50 g flying bullet has the same kinetic energy as a 20 000 kg locomotive which traveling
at 7.2 km/h. Calculate the velocity of the bullet.

Solution:

Given: ( ) = 50 = 0.05 , ( ) = 20 000
1000 1 ℎ

=( )= 7.2 ℎ = 7.2 × 1 × 3600 = 2 /
?

( ) = ( )

1 2 = 1 2
2 2

1 × 0.05 × 2 = 1 × 20 000 × 22
2 2

0.025 2 = 40000

2 = 1600000

= √1600000

= . −

Chapter 4–Work, Energy and Power/ Page 152

QUICK CHECK

1) How much potential energy does a 5000 g ceiling fan have with respect to the
floor when it is 3 meters above it?

3m
floor

Answer: Ep = 147.15 J

2) A 705 kg roller coaster car is moving at a speed of 18.4 −1.
1) Determine its kinetic energy
2) If the roller coaster of moving at twice speed, then what will be its new
kinetic energy

Answer: a) Ek = 119 342.4 J b) Ek = 477369.6 J

Chapter 4–Work, Energy and Power/ Page 153

3) John with the mass 70 kg runs up a set of 10 steps as shown in Figure. What is
his gravitational potential energy at the top of the stairs?
12 cm

Answer: Ep = 824.04 J

4) Calculate the kinetic energy of a car with a mass 2600 kg moving at a speed of:
a) 12 m/s
b) 140 km/hr

Answer: a) Ek = 187.2 k J b) Ek = 1966.16 k
J

Chapter 4–Work, Energy and Power/ Page 154

Principle of Conservation of Energy Video description: Conservation of
energy
The principle of conservative of energy states
that:
• Energy cannot be created and destroyed
• Energy can change from one form to

another form.
• Total of energy is constant

= +

Figure 4.6
Figure 4.6a: Conservation of energy

Ep = max Ep = max
Ek= 0 Ek= 0

h

Ep = kinetic energy Ep = 0
Ek = potential energy Ek=max

Figure 4.6b: Conservation of energy (pendulum)
Chapter 4–Work, Energy and Power/ Page 155

Example 8

The diagram shows part of a roller coaster track. A coach of mass 500kg moves on track
ABCD with negligible friction. At point A, the coach moves with velocity 10 −1.
Assume that gravitational acceleration is 9.81 −2.

a) What is total energy of the coach at point A A C
b) What is the velocity of the coach at point B 5m
c) What is the kinetic energy at point D D
3m

Solution: B

a) At point A

= +

= ℎ + 1 2
2

= (500 × 9.81 × 5) + 1 × 500 × 102)
(2

=

b) At point B ( = 0 )

=

49525 = 1 × 500 × 2
2

= . −

c) At point D
= +
49525 = (500 × 9.81 × 3) +
=

Chapter 4–Work, Energy and Power/ Page 156

QUICK CHECK

1) Figure below shows the transformation of energy as coach moves on track ABCD
with negligible friction. Fill in the blank with values of energy

Ek = 70 kJ Ek =
Ep = 30 kJ Ep = 60 kJ
Type equation here.
C
Ek = 92 kJ
Ep = D Ek = 80 kJ
Ep =
A

B

Answer : B) Ep = 8 kJ C) Ek = 40 kJ D) Ep = 20 kJ

2) A roller coaster with the mass 1200 kg starts at rest from a point A 20 m above the
ground. At point B it is 12 m above the ground

A a) What is the initial potential energy of the car?
20 m b) What is the potential energy at point B?
c) What is the kinetic energy of the car at point
B B?

12 m

ground

Answer: a) Ep = 235 440 J b) Ep = 141 264 J c) Ek = 94 176 J

Chapter 4–Work, Energy and Power/ Page 157

3) At the low point in its swing a pendulum bob with the mass of 0.2 kg has the
velocity of 4 / .
a) What is its kinetic energy at the low point?
b) How high will the bob swing above the low point before reversing
direction?

low point

Answer: a) Ek = 1.6 J b) h = 0.82 m

Chapter 4–Work, Energy and Power/ Page 158

4.4 CONCEPT OF POWER

Definition Power, P is the rate at which work is done, or the amount of work
done per second

Formula = ×
=
Unit F= Force (N)
Type quantity W = work (Joule) v= velocity ( −1)
Other unit t = time (second)

Joule/second (J/s) or Watt, W
Scalar quantity

1 Horse Power (hp)= 746 Watt

Example 9
A crane lifts a heavy bucket to a height of 2.5 meter from the ground in 3.5 seconds.
Calculate the power generated by the crane in lifting the bucket if the mass of the bucket is
840 kg. (Use g = 9.81 −2)

2.5 m

Solution:

Given: = 840 , ℎ = 2.5 , = 3.5 , ?

ℎ
= =

840 × 9.81 × 2.5
= 3.5

=

Chapter 4–Work, Energy and Power/ Page 159

Example 10

A car moves at constant velocity of 20 m/s. Find the power generated by the car if the
force of friction that acts on it is 1500 N.

Solution:

Given: = 1500 , = 20 , ?



= ×

= 1500 × 20

=

QUICK CHECK

1) If you do 100 Joules of work in 4 2) If a weightlifter lifts 2000 Newton to
seconds. How much power is used? a height of 2 meter in 5 seconds, how
powerful is he?

Answer: P = 25 W Answer: P = 800 W

Chapter 4–Work, Energy and Power/ Page 160

3) An elevator must lift 1000 kg load 4) A student has a mass 50 kg climbs a
with constant velocity of 4 −1. flight stair 5.7 m high. If the student
What is the power the elevator in climbs the stairs in 0.2 minutes,
exerts during the trip? what power was required?

Answer: P = 39 240 W Answer: P = 232.99 W

5) What is the time taken by a water pump of 0.67 horse power (hp) to lift 2000
kg of water to a tank, which is at a height of 15 m from the ground?

Answer: t = 588.81 s

Chapter 4–Work, Energy and Power/ Page 161

Efficiency Efficiency, is the ratio between the useful power delivered by the
Definition motor and the power that supply to the engine
Formula
Efficiency of work: Efficiency of power:
Unit = ×
Explanation
= ×

= × = ×



Efficiency has no unit and is usually expressed in %

Input Useful output energy = 80 J
energy = Wasted energy=20 J
100 J

Figure 4.7a: An efficient device

Input Useful output energy = 20 J
energy = Wasted energy=80 J
100 J

Figure 4.7b: An inefficient device

The efficiency of device can be increased in following ways:
a) Reduce friction and heat loss. This can be done by using

quality lubricants. An example is the usage of quality
lubricating oil in car engines.
b) Improve the design of the device to reduce vibration and
sound
c) For moving devices like cars and aeroplanes a streamlined
body will reduce air friction and hence reduce heat loss

Chapter 4–Work, Energy and Power/ Page 162

Example 11

An electric motor has an input power of 120 W. It lifts a 20 kg load to vertical height of
1.5 meters in 5 seconds. What is the efficient of the electric motor?

Solution:
Given: input power, Pinput = 120 W , m = 20 kg , h = 1.5 m , t = 5 s

, ?

= × 100 output = = ℎ



58.8 output = 20 × 9.81 × 1.5
= 120 × 100
5
= %
output = 58.8

Video description: Efficiency

Chapter 4–Work, Energy and Power/ Page 163

QUICK CHECK

1) Table show the input and useful energy of a petrol engine and diesel engine.
a) Complete the table
b) Compare the efficiency of the petrol engine with the diesel engine. Which of
them is more efficient?

Engine Input energy/kJ Useful output Efficiency
energy/kJ %
20
Petrol 80
Diesel 40 14

Answer: = 25%, = 35%

2) Petrol engine has a work output of 78 000 J per minute. What is the power input
if the efficiency of the engine is 30%?

Answer: = 4333.33 W

Chapter 4–Work, Energy and Power/ Page 164

ACTIVITY CHAPTER 4

1) Define the following terms and state its SI unit
a) Work
………………………………………………………………………………….
b) Energy
………………………………………………………………………………….
c) Power
…………………………………………………………………………………..
d) Kinetic energy
…………………………………………………………………………………
e) Gravitational potential energy
…………………………………………………………………………………

2) State the statements below are “TRUE” or “FALSE”.
Work is done when you
a) pick up a box and place it on your shoulder

b) support the box on your shoulder

3) State the principle of Conservation Energy
………………………………………………………………………………………
………………………………………………………………………………………
……………………………………………………………………………………

4) A man of mass 70 kg climbs stairs of vertical height 2.5 m. Calculate the work done
against the force of gravity

Answer: W = 1716.75 J

Chapter 4–Work, Energy and Power/ Page 165

5) A car with mass of 2000 kg moves with constant velocity of 50 ℎ−1. The car
suddenly brakes and stop within 15 m. Calculate the work done to stop the car.

Answer: W = 192 900 J

6) A book of mass 5 kg is placed on a shelf which is 0.2 m above a table. If the table top
is 0.5 m above the floor, find the potential energy of the box with reference to:

0.2 m book
shelf

table

0.5 m

a) The table top

Answer: Ep = 9.81 J

b) The floor

Answer: Ep = 34.34 J

Chapter 4–Work, Energy and Power/ Page 166

7) A cat with mass 2.5 kg is trying to jump off tree which is 3 meters high from the
ground. Calculate the potential energy and kinetic energy based on situation below:

a) While the cat is still on the tree

Answer: = 0 , = 73.58

b) When the cat falls 1 meter from the tree

Answer: = 49.05 , = 24.53

c) When the cat reached the ground

Answer: = 0 , = 73.58

Chapter 4–Work, Energy and Power/ Page 167

8) Figure shows a child playing on swing. The mass of child is 35 kg. The child raised
to point A and then released. She swings downwards towards the equilibrium position
B.

A

B

1.5 m

0.8 m

a) Calculate the loss in gravitational potential energy of the child on reaching
point B

Answer: Ep loss = 240.35 J

b) Assuming that air resistance is negligible, calculate the speed of the child as she
passes through the equilibrium position B

Answer: v = 3.71 m/s

Chapter 4–Work, Energy and Power/ Page 168

9) Figure shows an electric motor on a building site lifts a load of mass 2 kg top of a
house. The house is 15 m high and it takes 7.5 s to lift the load

a) Determine the weight of the load Answer: w = 19.62 N
b) What is the work to lift the load by the motor?

Answer: W = 294.3 J

c) How much gravitational potential energy does the load gain?

Answer: Ep = 294.3 J

d) Compare your answer in (b) and and give the reason. (c)

e) What is the power of the motor?

Answer: P = 39.24 W

Chapter 4–Work, Energy and Power/ Page 169

10) Figure shows a crane lifts a load of 800 kg to a height of 150 m in 20 s. The power
input of the crane is 75 kW

Determine: Answer: =1177.2 kJ
a) the useful energy output

b) the energy input

Answer: =1500 kJ

c) the efficiency of the crane

Answer: = 78.48%

Chapter 4–Work, Energy and Power/ Page 170

CONCEPT MAP

Done

Is Is done when a Causes an
Work
Force Object
The unit
To move in the direction of
Is a the

Scalar Is defined as capacity Is the rate at which
quantity to do
Of

Kinetic Is the
energy
Power

due to its Energy

Motion Stored in the object because Is
of

Is the

Watt (W)

Potential Position
energy

Chapter 4–Work, Energy and Power/ Page 171

5 Solid and
Fluid

5.1 The concept of solid and fluid
5.2 Pascal’s Principle
5.3 Archimedes’s Principle

What led to Archimedes’ discovering his principle?
King Heiron II of Syracuse had a pure gold crown made, but he thought that the crown
maker might have tricked him and used some silver. Heiron asked Archimedes to figure
out whether the crown was pure gold. Archimedes took one mass of gold and one of silver,
both equal in weight to the crown. He filled a vessel to the brim with water, put the silver
in, and found how much water the silver displaced. He refilled the vessel and put the gold
in. The gold displaced less water than the silver. He then put the crown in and found that it
displaced more water than the gold and so was mixed with silver. That Archimedes
discovered his principle when he saw the water in his bathtub rise as he got in and that he
rushed out naked shouting “Eureka!” (“I have found it!”) is believed to be a later
embellishment to the story.

Chapter 5- Solid and Fluid / Page 172

CHAPTER Solid and

5 Fluid

Learning Outcomes

A student should be able to:
❖ Describe the characteristics of solid, liquid and gas.
❖ Define density and its unit.
❖ Define pressure and its unit.
❖ Apply the concept of density and pressure in solving the related problems by using

formula
❖ Explain Pascal's Principle
❖ Apply Pascal's Principle in solving the related problems.
❖ Explain Archimedes’ Principle
❖ Apply Archimedes’ Principle in solving the related problems.

Chapter 5- Solid and Fluid / Page 173

MIND MAP

SOLID AND FLUID

Characteristic Pressure Liquid Gas
s Pressure Pressure

Solid Liquids = = ℎ Atmospheric
pressure

Gas

Archimedes’ Pascal’s
Principle Principle

KEYWORDS Immersed – direndam/ditenggelam
Mass – jisim
Actual weight – berat sebenar Matter – jirim
Apparent weight – berat ketara Pressure – tekanan
Buoyant force – daya apungan Relative density – ketumpatan relatif
Density – ketumpatan Sink – tenggelam
Displaced – tersesar Specific gravity – gravity tentu
Downwards – ke bawah state – fasa,keadaan
Exerted – dikenakan Upward – ke atas
Float – terapung Volume – isipadu
Fluid pressure – tekanan cecair

Chapter 5- Solid and Fluid / Page 174

5.1 CONCEPT OF SOLID AND FLUID

States of Matter

1. Ordinary matter is usually classified into three familiar states : Solids, liquids and gases.

Solids Liquids Gases
Figure 5.1 : Molecules Structures

2. Table 5.1 shows the properties of states of matter for solids, liquids and gases.

Table 5.1 : Properties of states of matter

States Of Matter

Properties Solids Liquids Gases
Molecules are not very Molecules are far apart
Molecules Molecules are very closely held
arrangement closely packed, firmly Molecules can move
and tight with each Molecules can slip out freely and randomly -
Molecules other of position and move molecules are very
movement Molecules are locked in over and around one loosely held by force
place and vibrate and another – molecules are of cohesion
rotational motion about loosely held by force of
their fixed position - cohesion
molecules are firmly
held together by force
of cohesion

Volume Definite Definite Indefinite

Shape Definite Indefinite -takes shape Indefinite -takes shape
Cannot be compressed of container of container
Compression on Easily can be
volume Can be compressed compressed

Chapter 5- Solid and Fluid / Page 175

Video description: States of matter

Density

1. Density of any material is the mass per unit volume. ρ = density (kg m-3 or kg/m3)
m = mass (kilogram, kg)
= v = volume (cube metre, m3)



2. The SI unit of density is kg/m3

3. Figure 5.2 shows 1 meter cubed of water has mass 1000 kilograms when it is put on
scales. Therefore the density of water is 1000 kg/m3

Figure 5.2: The density of water is 1000 kg/m3
Chapter 5- Solid and Fluid / Page 176

Example 1
A liquid has a volume of 74 cm3 and a mass of 94 g.
a) What is its density in g/cm3 ?
b) What is its density in kg/m3 ?

Solution:

Given , = 74 3 , , = 94 , , ?

a) =



94
= 74

= . /

b) = , ( ) ∶ , ( 3) ∶
= 94 74

0.094 = 1000 = 1003
= 0.094 = 7.4 × 10−5 3
7.4 × 10−5

= . /

Example 2

Figure shows two blocks with the same mass 1 kg. Calculate the density of the both block ?
0.08 m

0.06 m

0.10 m

0.05 m

0.05 m

Solution :

Given (cube) : = 0.05 × 0.05 × 0.06 , , ?

= Note that,
, = ℎ


1
= 0.05 × 0.05 × 0.06

= . /

Chapter 5- Solid and Fluid / Page 177

Given (cylinder) : = 0.08 , = 0.04 , ℎ ℎ = 0.10

, = 1 , densi , ? Note that,
, = 2ℎ
=



1
= × 0.042 × 0.10

= . /

Example 3
Beeswax has a density of 960 kg/m3. Calculate the mass of 5 cm3 of beeswax.

Solution :

Given : , = 960 ⁄ 3 , , = 5 3, , ?

, ( 3) ∶
= 5

= 1003
960 = 5 × 10−6 = 5 × 10−6 3
= . @ .

Video description: Density

Chapter 5- Solid and Fluid / Page 178

QUICK CHECK

Complete the following table :

Material Mass, m Volume, V Density,
Aluminum (kg) (m3) (kg m-3)
160 0.060
Lead 11 000
Steel 60 0.032 7 700

Answer: ρ =2 666.67 kg m-3, m = 352 kg, v = 0.008 m3

Chapter 5- Solid and Fluid / Page 179

Relative Density

1. Relative density is a dimensionless ratio of the densities of two materials. The term
specific gravity is similar, except that the reference material is water.

RD = ρobject RD = relative density
ρreference ρobject = density of object ( g cm-3 or kg m-3)
ρreference = density of reference object ( g cm-3 or kg m-3)
or G = specific gravity
ρwater = density of water ( g cm-3 or kg m-3)
G= ρobject ρreference = density of reference object ( g cm-3 or kg m-3)
ρwater

2. A relative density can help quantify the buoyancy between two materials, or determine
the density of one "unknown" material using the "known" density of another material.

3. If a substance's relative density is less than one then it is less dense than the reference;
if greater than 1 then it is denser than the reference. If the relative density is exactly 1
then the densities are equal; that is, equal volumes of the two substances have the same
mass.

4. If the reference material is water then a substance with a relative density (or specific
gravity) less than 1 will float in water. For example, an ice cube, with a relative density
of about 0.91, will float. A substance with a relative density greater than 1 will sink.

Chapter 5- Solid and Fluid / Page 180

Example 4

A rectangular block measures 5 cm x 2 cm x 8 cm. It has a mass of 70 grams.
a) Determine its density
b) Determine its specific gravity and indicate whether it would float or sink in water.

8 cm

5 cm , = 70

Solution :
Given: = 5 × 2 × 8

a) , ? Note that,
= , = ℎ



70
= 5 × 2 × 8

= . −

b) , ? , = 0.875 −3

= Note that,
density of water, ρwater = 1 g cm−3 or 1000 g m−3

0.875
= 1

= . ( < 1, so it will float in water)

Video description: Relative Density
Chapter 5- Solid and Fluid / Page 181

QUICK CHECK

A cylinder of plastic is 0.1 m long, and 0.05 m in diameter. It has a mass of 1 kg.
Determine its density, specific gravity and indicate whether it would float or
sink in water.

Answer: = 5092.96 kg m-3, G = 5.09, sink

Chapter 5- Solid and Fluid / Page 182

Pressure

1. Pressure is defined as the magnitude of force acting perpendicularly to a surface per
unit area of the surface.

P = Pressure (N m-2or Pascal, Pa)
= F = Force (Newton, N)

A = Area (square metre, m2)

2. The SI unit of pressure is N m-2 and often referred to as Pascal (Pa)

3. Tools like knives, axes and saws have sharp cutting
edges because the surface areas of contact are small.
When a force is applied on a cutting tool of small area
of contact with the material, a pressure is exerted to
cut the material.

Figure 5.3: Tools using pressure to cut materials

Example 5

Siti is a girl weighing 500 N. Siti standing on one leg while
dancing on the floor. What is the pressure exerted by Siti’s
shoe on the floor?

Solution : Area of contact 0.002 m2

Given: , = 500 , , = 0.002 2, , ?

F
= A

500
= 0.002

= ⁄

Chapter 5- Solid and Fluid / Page 183

Example 6 0.3 m
0.4 m
A rectangular block with the dimension 0.5 m x 0.4 m x 0.3 m has
a mass of 500 kg. Calculate the pressure exerted by the block on
the ground.

0.5 m

Solution :

Given: mass, m = 500 kg, dimension ∶ 0.5 m × 0.4 m × 0.3 , , ?

F Note that,
= A =
, =
500 × 9.81
= 0.5 × 0.4

4905
= 0.2
= ⁄

Example 7 diameter = 0.5 m

A cylindrical container with a base diameter of 0.5 m in contact
with the floor exerts a pressure of 9 000 Pa on the floor. What
is the mass of the container?

Pressure = 9 000 Pa

Solution :

Given : Pressure, P = 9000 Pa, diameter, d = 0.5 m, radius, r = 0.25 m,
, ?

P=F Note that,
, =
A , = 2

× 9.81
P = π × 0.252

9.81
9 000 = 0.20
1 800 = 9.81

= 183.49 kg

Chapter 5- Solid and Fluid / Page 184

Example 8

A man standing still on reflexology therapy stones. He exerts a
pressure on the stones. The mass of the man is 50 kg and the area
of contact with the stones is 40 cm2. Calculate the pressure exerted
by the man on the stones.

Solution :

Given: mass, m = 50 kg , contact area, A = 40 cm2 , , ?

= F Note that, = , ( 2) ∶
40
A
= 1002
50 × 9.81 = 0.004 2
= 0.004

= ⁄ @

QUICK CHECK

Mimi weighs 450 N. She has three pairs of shoes A, B and C

a) By referring to Figure above, calculate the pressure exerted on the floor by
Mimi for each pair of the shoes.

b) Which pair of shoes is most suitable to be worn by Mimi if she intends to
go to the beach. Explain your answer.

Answer: a) P = 112 500 Pa, 30 000 Pa, 22 500 Pa, b) C because its the pressure
exerted by C on the beach is smallest and hence the least likely to be stuck in the sand

Chapter 5- Solid and Fluid / Page 185

Fluid Pressure

1. The pressure, of a liquid at a particular depth in in the liquid is given by the formula

= P = Pressure (Pascal, Pa)
h = depth (metre, m)

ρ = density of the liquid (kg m-3)

g = gravitational field strength (m s-2 or N kg-
1)

2. In a fluid , pressure acts in all directions. The force produce always the right angle to the

surface under pressure.

Pressure Pressure
Pressure

Example 9

A submarine travelling from the surface of the sea to a depth of 5 000 m. If the density
of sea water is 1025 kg m-3 and by taking g = 9.81 N kg-1, calculate the pressure exerted

by the sea water on the submarine at that depth.

Surface of sea

sea water 5 000 m
pressure

Solution :
Given : depth, h = 5000 m , fluid density, ρ = 1025 kg m−3 , , ?
= hρg
= 5 000 × 1025 × 9.81
= @ .

Chapter 5- Solid and Fluid / Page 186

Example 10
A diver is 1.5 m below the surface of water. If the density of the water is 1000 kg m-3 and
the atmospheric pressure is 1 x 105 Pa, what is:
a) The water pressure exerted on the diver?
b) Total pressure exerted on the diver?

1.5 m

pressure ?

Solution :
Given; depth, h = 1.5 m , fluid density, ρ = 1000 kg m−3
a) , ?
= hρg
= 1.5 × 1000 × 9.81
= @ .
b) , ?
= Pw + Patm
= 14 715 + (1 × 105)
= @ .

Chapter 5- Solid and Fluid / Page 187

Example 11

A scientist has invented a robot to work at seabed. According to his calculation, the amour of
the robot can withstand a maximum pressure of 106 Pa exerted by the sea water. If the density
of sea water is 1025 kg m-3 and g = 9.81 N kg-1,what is the maximum depth of the seabed
that this robot can work at ?

depth, h ?

106 Pa

Solution :
Given: fluid density, ρ = 1025 kg m−3 , maximum pressure on robot, P = 10 6 Pa

g = 9.81 N kg−1 , , ?
P = ρg
106 = × 1025 × 9.81
106 = 10 055.25
= .

Chapter 5- Solid and Fluid / Page 188

Example 12
A deep-sea diver is wearing a watch that can withstand a maximum pressure of 4.5 x 106
Pa due to the sea water.
[Density of the sea water = 1030 kg m-3; g = 9.81 ms-2)
a) Calculate the maximum depth the diver can dive without spoiling the watch.
b) If the surface area of the glass of the watch is 9.0 cm2, calculate the force acting on the

glass of the watch at depth of 100 m.

Solution :
Given: Water Pressure, P = 4.5 × 106 Pa , fluid density, ρ = 1030 kg m−3
a) , ?
P = ρg
4.5 × 106 = × 1030 × 9.81
4.5 × 106 = 10 104.3
= .

Given; fluid density, ρ = 1030 kg m−3 , depth , h = 100 m ,

contact area, A = 9.0 cm2

b) , ? Pressure on the watch :
= ℎ
P = = 100 × 1030 × 9.81
= 1 010 430
A
Contact Area in m2 :
9.0
1 010 430 = 0.0009
= . = 1002
= 0.0009 2

Chapter 5- Solid and Fluid / Page 189

QUICK CHECK

1. Ahmad dives at 5 meter below the surface of water. What is the water
pressure exerted on Ahmad ?

2. A diver swims to a depth of 3.2 m in a freshwater lake. Determine :
a) the water pressure exerted on the diver ?
b) the force pushing in on her eardrum ? ( the area of her eardrum
is 0.6 x 10-5 m2)

Answer: 1) P = 49 050 Pa, 2a) P = 31 392 Pa, 2b) F = 0.19 N

Chapter 5- Solid and Fluid / Page 190

5.2 PASCAL PRINCIPLE

1. In 1651, Pascal wrote
Pressure is exerted equally in all parts of an enclosed static fluid

When pressure is put on some region of a confined liquid, the liquid compresses slightly
and distributes the pressure uniformly everywhere therein. The pascal’s syringe in Figure
5.4 shows the application of Pascal’s principle, as does the piston in Figure 5.5.

pressure Pascal’s syringe is a bottle punctured
with the holes and fitted with a tight
force piston. Pushing down on the piston
increases the pressure on the fluid,
pressure which blows out the holes uniformly in
Figure 5.4: Pascal’s syringe all directions. This phenomenon
suggests that the applied pressure is
distributed equally through the liquid

force

pressure The force divided by the area of
the piston determine the pressure

Figure 5.5: Piston

Chapter 5- Solid and Fluid / Page 191

2. If two chambers fitted with different-sized pistons are connected so that they share a
common working fluid, the pressure generated by one will be transmitted undiminished to
the other as shown in Figure 5.6

Output Force, F2

Input Force, F1

Downwards Area, A2 Upwards
distance, h1 Area, A1 distance, h2

P1 = P2

Figure 5.6 : A hydraulic lift

A hydraulic lift . The downward input force, F1 exerted on input piston creates a input-

pressure, P1 . This pressure, P1 is distributed uniformly and so equals the output-pressure

,P2 which is transmitted to the output piston and produced upward ouput force, F2 which
Therefore,
can lift the car.

=

=


Figure 5.6 shows the small piston will moves downwards , h1 and the larger piston
will moves upwards, h2. In the other words, the volume of liquid displaced at input
side, V1 is equal to the volume of liquid displaced at the output , V2. Accordingly ,

=
=

Chapter 5- Solid and Fluid / Page 192

Example 13 Force = 4.4 N

A nurse applies a force of 4.4 N to the piston of a

syringe.The piston has an area of 5 x 10-5 m2.

What is the pressure increase in the fluid within Fluid pressure
within the syringe?

the syringe ?

Solution :
Given: Force, F = 4.4 N , Contact area, A = 5 × 10−5 m2 , , ?

F
= A

4.4
= 5 × 10−5
= ⁄

Example 14

The figure shows a hydraulic system. How much force is needed to lift a car with a weight of
10 000 N ?

A = 0.01 m2 A = 1 m2

Solution :

Given: (input) ∶ Contact area, A 1 = 0.01 m2 , , ?
(output) ∶ Contact area, A 1 = 0.01 m2, Contact area, A 2 = 1 m2 , F 2 = 10 000 N

P 1 = P2

= F2
A1 A2

= 10 000
0.01 1

= 10 000
0.01

=

Chapter 5- Solid and Fluid / Page 193