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Discover Physics for GCE ‘O’ Level
Science
Unit 3: Forces and Pressure
3.1 Forces
Learning Outcomes
At the end of this section, you’ll be able to:
• understand what a force is
• describe some types of forces
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3.1 Forces
What is a Force?
• A Force is a push or pull that one object exerts on
another.
• It produces or tends to produce motion, and stops and
tends to stop motion:
Boy exerts a push on the boat
Boy exerts a pull on the boat
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3.1 Forces
What types of forces are there?
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3.1 Forces
Quantifying Forces
• The SI unit of force is newton (N).
• A force of 1 N is roughly the amount of force the Earth’s
gravity pulls on a 100 g mass.
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3.1 Forces
Adding Forces
• A force always acts in a particular direction.
• Two or more forces may act on an object at the same
time.
• These forces can be added using vector diagrams.
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3.1 Forces
Key Ideas
• A force is a push or pull that one object exerts on
another. It produces or tends to produce motion, and
stops or tends to stop motion.
• There are many types of forces, eg. Friction, weight,
tension, magnetic force and electric force.
• The SI unit for force is the newton (N).
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3.1 Forces
Test Yourself 3.1
1. Name the types of forces in our daily lives.
Answer:
• Pushing or pulling force
• Friction force
• Repulsive/attractive force between poles of magnets
• Gravitational force, weight
• Contact force between you and the floor
• Repulsive/attractive force between electric charges
• Air Resistance
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3.2 Scalars and Vectors
Learning Outcomes
At the end of this section, you’ll be able to:
• understand and distinguish between scalar and vector
quantities
• add two vectors using a graphical method
• solve problems involving three vectors acting on a
static body
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3.2 Scalars and Vectors
What are scalar and vector quantities?
• Scalar quantities are physical quantities that have
magnitude only.
• Vector quantities are physical quantities that possess
both magnitude and direction.
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3.2 Scalars and Vectors
When referring to a scalar quantity, we only need to
consider its magnitude.
• Example:
– The mass of an object is 2.0 kg
– The volume of the box is 5 m3
– The distance traveled by the car is 800 m
Scalar quantities are added by summing the magnitude
• Example:
A mass of 100 g added to 200 g gives a total of
mass 300 g.
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3.2 Scalars and Vectors
When referring to a vector quantity, we must consider
both its magnitude and direction.
Example:
The car travels with a velocity of 20 ms-1 in the direction of
45o North of East
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3.2 Scalars and Vectors
How do we add vectors?
• When adding vectors, both the magnitude and direction
of the vectors must be considered.
• Addition of two or more vectors together gives a single
vector called the RESULTANT VECTOR
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3.2 Scalars and Vectors
Addition of parallel vectors (Example 1)
Resultant Force = 3 N + 5 N
=8 N
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3.2 Scalars and Vectors
Addition of Parallel Vectors (Example 2)
Resultant Force = (-3 N) + 5 N
=2N
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3.2 Scalars and Vectors
Addition of Parallel Vectors (Example 3)
Resultant Force = (-3 N) + 3 N
= 0N
The object is in a state of equilibrium
i.e. it remains stationary or continues
moving in a straight line.
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3.2 Scalars and Vectors
Addition of non-parallel vectors using the
parallelogram method
In most cases, vectors such as forces act at an angle to
each other, such as in the diagram below.
We can use the parallelogram
method (on the next slide) to
find its resultant.
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3.2 Scalars and Vectors
Parallelogram Method
Addition of vectors: Using the parallelogram method
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3.2 Scalars and Vectors
Tip-To-Tail Method
Another method of adding non-parallel vectors is the
tip-to-tail method. We can use the tip-to-tail method
to find the resultant of the diagram below.
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3.2 Scalars and Vectors
Tip-To-Tail Method
Addition of vectors: Using the tip-to-tail method
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3.2 Scalars and Vectors
Worked Example 3.1
A weight W (6.0 N) hangs on the
end of a string, which is pulled
sideways by a force F. The string
makes an angle of 30 with the
vertical, as show in the diagram.
The string supports the weight by
exerting a pull known as tension T
of 7.0 N.
Determine the force F by using the
(a) parallelogram method,
(b) tip-to-tail method.
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3.2 Scalars and Vectors
Solution to Worked Example 3.1(a)
(a) Parallelogram Method: For the
weight to be stationary, the resultant
force must be zero. Therefore, force F
must balance out the resultant of
weight W and tension T. Hence, we will
first find the resultant of W and T, then
determine F.
From the force parallelogram (as shown
in the diagram), drawn with a scale of
1 cm:2 N, the diagonal, which is the
resultant of T and W, has a length of
1.75 cm. In order to balance this
resultant, F must be 1.75 cm also. This
means force F should be 3.5 N.
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3.2 Scalars and Vectors
Solution to Worked Example 3.1(b)
(b) Tip-to-tail Method: Using an
appropriate scale, e.g. 1 cm:2 N, let W
be the first force vector drawn, followed
by T as the second vector. Force F is
found by joining the end point of T to
the start point of W to form a closed
triangle.
In doing so, the resultant force will be
zero, and the system is in equilibrium
(i.e. stationary in this case). By measuring
the vector, F has a length of 1.75 cm.
Hence, force F = 3.5 N. The answer is the
same using the parallelogram method.
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3.2 Scalars and Vectors
Key Ideas
• A scalar quantity has magnitude only.
• A vector quantity has magnitude and direction.
• When there are two or more forces acting on an object,
the resultant can be found by adding the forces
together.
> For parallel forces, the resultant force is found by
taking one direction as positive and the opposite as
negative, and then adding up the forces.
> For non-parallel forces, the resultant force is found
by adding the vectors using the parallelogram
method or the tip-to-tail method.
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3.2 Scalars and Vectors
Test Yourself 3.2
1. A man can row a boat in still water at a speed of 1.0 m s-1.
The man sets out to row the boat in a river from A to B.
The water in the river flows at 0.5 m s-1 in the direction B
to A. Find the velocity of the boat through the water.
Answer: Since the velocity of water is acting opposite to
that of the boat, the velocity of the boat through water,
Vbw, is
Vbw = Vb - Vw = 1.0 - 0.5 = 0.5 m s-1
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3.2 Scalars and Vectors
Test Yourself 3.2
2. A lorry, which has been stuck
in muddy ground, is being
pulled by two jeeps. Each jeep
exerts a force of 3000 N at an
angle of 20o to the horizontal
in the direction shown. Find,
using a scale diagram, the
resultant force pulling the
lorry forward.
Answer: Using an appropriate
scale diagram, the resultant
force F = 5600 N
3000 N
20o
20o
5600 N
3000 N
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3.3 Forces and Motion
Learning Outcomes
In this section, you’ll be able to:
• describe how a force changes the motion of a body
• describe the effects of balanced and unbalanced forces
on a body
• identify forces acting on an object
• state Newton’s three laws of motion
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3.3 Forces and Motion
What is the effect of a force on motion?
Forces can affect the movement of objects, as can be
observed from our daily activities.
•
A force can cause a stationary object to start moving e.g.
a football player kicking a football.
•
A force can cause a moving object to increase speed e.g. a
person rollerblading gives a push to move even faster
•
A force can cause a moving object to decrease speed e.g.
when cycling down a slope, a cyclist applies the brakes to
slow down the bicycle.
•
A force can cause a moving object to changes its direction
of motion e.g. during a game of badminton, a player
intercepts and hits the shuttlecock to change its direction
and motion.
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3.3 Forces and Motion
• In each of the four examples given, there is a change in
the motion of the object.
• The velocity of an object changes with time i.e. there is
acceleration.
• A force can cause an object to accelerate or decelerate.
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3.3 Forces and Motion
Forces and Zero Acceleration
• Zero acceleration implies that object could be stationary
or moving with constant velocity.
• It does not mean there are no forces acting on it.
A car at rest has zero acceleration.
An ice skater gliding at
constant velocity has zero
acceleration.
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3.3 Forces and Motion
Forces and Zero Acceleration
For an object with zero acceleration, the different forces
acting on it are balanced or add up to zero – i.e. the
resultant or net force is zero.
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3.3 Forces and Motion
Newton’s First Law of Motion
Every object will continue in its state of rest or uniform
motion in a straight line unless a resultant force acts on
it to change its state.
Forces acting on a book resting
on a table show that F = W
Book sliding along a frictionless
surface at constant velocity.
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3.3 Forces and Motion
Newton’s First Law of Motion
Newton’s first Law of Motion is also know as the Law of
Inertia. You will learn more about inertia in Unit 4: Mass,
Weight and Density.
Newton’s First Law
is summarised in
this simple flow chart.
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3.3 Forces and Motion
Unbalanced Forces and Newton’s Second Law
If the resultant force acting on an object is not zero, we
say the forces are unbalanced.
Forces cause the book to accelerate or decelerate.
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3.3 Forces and Motion
Unbalanced Forces and Newton’s Second Law
• Unbalanced forces causes an object to accelerate,
decelerate or change direction.
• The direction of acceleration is in the direction of
the resultant force.
Unbalanced forces
and their relation
to Newton’s Second
Law
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3.3 Forces and Motion
Unbalanced Forces and Newton’s Second Law
Newton’s Second Law of Motion: When a resultant force
acts on an object of constant mass, the object will accelerate
and move in the direction of the resultant force. The product
of the mass and acceleration of the object is equal to the
resultant force.
F = ma
where
F = resultant force (in N)
m = mass of object (in kg)
a = acceleration of object (in m s-2)
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3.3 Forces and Motion
Unbalanced forces and Newton’s Second Law
The SI unit of force is newton (N). One newton (1 N) is
defined as the force that will produce an acceleration of
1 m s-2 on a mass of 1 kg.
If m = 1 kg and a = 1 m s-2, then using the equation
F = ma
= (1 kg)(1 m s-2)
=1N
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3.3 Forces and Motion
Worked Example 3.2
A boy pushes a box of mass 20 kg with a force of 50 N.
What is the acceleration of the box? (Assume no friction.)
50 N
20 kg
Solution: Given that mass m = 20 kg and force F = 50 N
From Newton’s Second Law, F = ma. Then,
a=
50
F
= 2.5 m s-2
=
m
20
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3.3 Forces and Motion
Worked Example 3.3
A car of mass 1000 kg accelerates from rest to 20 m s-1
in 5 s. Calculate the forward thrust of the car (assume no
friction).
Solution: Given mass m = 1000 kg, initial speed u = 0 m s-1,
final speed v = 20 m s-1 and time t = 5 s
From Newton’s Second Law,
forward thrust F = ma where a = acceleration produced
But a =
20 - 0
v-u
= 4 m s-2
=
t
5
 F = ma = (1000)(4) = 4000 N
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3.3 Forces and Motion
Weight is a force
Weight is an ever-present force exerted on every object
due to gravity. The acceleration of free fall ‘g’ due to
Earth’s gravity is 10 m s-2.
Weight = Force
W= F
F = ma
F = mg
W = mg
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3.3 Forces and Motion
Newton’s Third Law (Optional)
Newton’s Third Law of Motion states that:
For every action, there is an equal and opposite reaction,
and these forces act on mutually opposite bodies.
Newton’s Third Law of Motion tells us four characteristics:
•
Forces always occur in pairs.
•
Action and reaction forces are equal in magnitude.
•
Action and reaction forces act in opposite directions.
•
Action and reaction forces act on different bodies.
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3.3 Forces and Motion
Newton’s Third Law (Optional)
An example of action and reaction force is shown in the diagram above.
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3.3 Forces and Motion
Key Ideas
• The change in motion of an object is caused by
unbalanced forces acting on it.
• Forces acting on an object are balanced if the resultant
force is zero.
• Forces acting on an object are unbalanced if the
resultant force is not zero.
• Newton’s First Law states that every object will continue
in its state of rest or uniform motion in a straight line
unless a resultant force acts on it to change its state.
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3.3 Forces and Motion
Key Ideas (continued)
• Newton’s Second Law of Motion states that when a
resultant force acts on an object of constant mass, the
object will accelerate. The product of the mass and
acceleration of the object is equal to the resultant force.
In equation form, this is represened as F = ma
• A resultant force is 1 N if the acceleration it produces
on a mass of 1 kg is 1 m s-2.
• Newton’s Third Law of Motion states that for every
action, there is an equal and opposite reaction, and
these forces act on mutually opposite bodies.
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3.3 Forces and Motion
Test Yourself 3.3
1. What can you deduce about the resultant force acting
on an object that is
(a) moving with constant speed in a straight line?
(b) accelerating?
Answer:
(a) When an object is moving at constant speed in a
straight line, then the resultant force F = 0.
(b) If the object is accelerating, then there must be a
resultant force. This resultant force F is given by
F = ma.
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3.3 Forces and Motion
Test Yourself 3.3
2. An unloaded van has an acceleration of 5 m s-2. A fully
loaded van weighs twice as much as the unloaded van.
What is the acceleration of the fully loaded van if the
forward thrust remains the same?
Solution: Let a = acceleration of unloaded van
a’ = acceleration of the fully loaded van
m = mass of unloaded van
Then, F = ma
- - - - (1)
F = 2ma’ - - - - (2)
Take equation (2)  (1)
1=
2ma'
ma
 a' = 1 a = 1  5 = 2.5 m s-2
2
2
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3.3 Forces and Motion
Test Yourself 3.3
3. The Republic of Singapore Air Force uses the AIM-9
Sidewinder air-to-air missile, which has a mass of 86.5
kg. If the missile can accelerate from 300 m s-1 to 700 m
s-1 in 6 s, what is the average force exerted on the
missile?
Answer:
v - u = 700 - 300
a =
t
6
Acceleration of missile,
= 66.7 m s-2
By Newton’s Second Law, F = ma
= 86.5  66.7
= 5770 N
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3.4 Friction and Its Effects
Learning Outcomes
In this section, you’ll be able to:
• explain the effects of friction on the motion of a body
• identify the forces acting on a body and draw free body
diagrams
• apply the relationship F = ma to solve related problems
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3.4 Friction and Its Effects
How does friction affect motion?
Friction always opposes motion between two surfaces in
contact.
The diagram below shows a microscopic view of two
surfaces in contact.
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3.4 Friction and Its Effects
Effects of friction
Friction has both positive and negative effects in our lives.
Negative effects of friction
• Friction reduces efficiency of cars by up to 20%.
• Causes wear and tear of moving parts in engines, motors
and machines.
Positive effects of friction
• Needed for walking or holding a pair of chopsticks
• Used in braking pads to slow down cars
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3.4 Friction and Its Effects
Friction as a negative force
Friction causes wear and tear of moving parts in machinery.
Ways to reduce friction include:
• Wheels
• Ball Bearings
• Lubricants and polishing surfaces
• Air cushion e.g. hovercraft and magnetic levitation in trains
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3.4 Friction and Its Effects
Friction as a useful force
Friction can also be useful for our daily lives.
Examples include:
• Car tyres
• Parachutes
• Rock climbing
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3.4 Friction and Its Effects
Identifying forces and free body diagrams
• Block diagrams with arrows used to represent forces
acting on a body are called free body diagrams.
• These diagrams help you to solve problems on forces
in Physics.
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3.4 Friction and Its Effects
An example of a free body diagram is shown below.
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3.4 Friction and Its Effects
Key Ideas
• Friction opposes motion between two surfaces in contact.
• A non-zero resultant frictional force will cause a moving
object to slow down to a complete stop.
• There are both positive and negative effects of friction in
our daily lives.
• Free body diagrams are drawn to help us identify forces
acting on a body or system of bodies, in order to solve
problems.
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3.4 Friction and Its Effects
Test Yourself 3.4
1. Can you lean against a wall if frictional forces are absent?
Answer: No. When you lean against the wall, the
frictional force between your body and the wall, as well as
the frictional force between your feet and the floor, keep
you from slipping.
2. How can you reduce the effect of friction on the motion
of a body?
Answer: Using wheels, ball bearings and lubrication.
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3.4 Friction and Its Effects
Test Yourself 3.4
3. A force F of 50 N is needed to keep a trolley of mass 60 kg
moving at a uniform velocity of 2 m s-1. What is the
frictional force f on the trolley?
a=0
F = 50 N
friction f = ?
Answer:
Since the trolley is moving at uniform velocity, this means
that acceleration a = 0. The resultant force is thus zero.
Hence the pushing force F is equal to the friction force f.
 f = F = 50 N
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3.4 Friction and Its Effects
Test Yourself 3.4
4. A feather and a stone are released simultaneously from the same
height. Explain why the feather will fall more slowly to the ground
than the stone even though the acceleration due to free fall is the
same for both the feather and the stone.
fair
stone
W
Answer: The air resistance fair
experienced by the stone is
relatively much smaller compared
to its weight W. Hence the stone
has an acceleration close to g,
the acceleration due to gravity.
fair
W
Answer: The feather experienced
an air resistance fair that is quite
comparable to its own weight W.
Hence the feather’s acceleration
is very low.
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3.5 Pressure
Learning Outcomes
In this section, you’ll be able to:
•
Define the term pressure in terms of force and area
•
Recall and apply the relationship pressure =
force
area
to new situations or to solve related problems
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3.5 Pressure
What is pressure?
Pressure is defined as the force acting per unit area.
Pressure = force
area
In symbols, p =
where
F
A
p = pressure
F = force (in N) and
A = area (in m2)
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3.5 Pressure
The SI unit of pressure is newton per square metre (N m-2)
or pascal (Pa).
You may have carried plastic
bags filled with groceries
before. If so, your fingers
must have hurt!
Padding the handles of heavy
plastic bags helps to reduce
the pressure acting on your
fingers.
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3.5 Pressure
Figure 7.3(a)
Figure 7.3(b)
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3.5 Pressure
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