2. Introduction
• This chapter teaches you how to deal with
forces acting on an object
• You will learn how to use several formulae
(inspired by Isaac Newton)
• You will learn how to model situations
involving friction, particles moving on slopes
and when joined over pulleys
• You will also learn laws of momentum and
impulse

4. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
You need to understand all the forces at work in
various situations…
R
The Normal Reaction
The normal reaction acts
perpendicular to the
surface which an object is
resting on
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
mg (mass
x gravity)
It is equal and opposite to
the force exerted on the
surface by the object,
which is determined largely
by gravity and the mass of
the object
The table matches the force from the brick,
which is why the brick remains still on the
table (there of course would be a maximum
possible weight the table could take, but we
will not worry about this for now!
3A

5. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
You need to understand all the forces at work in
various situations…
Frictional
Force
Direction of
motion
Frictional Force
The frictional force opposes
motion between two ‘rough’
surfaces
Although it is a force, friction does not
cause movement in its own direction. It
just reduces the effect of another
force
Surfaces will have a maximum level of
friction where it is unable to completely
prevent movement
3A

6. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
You need to understand all the forces at work in
various situations…
Tension in
string
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
Tension
If an object is being pulled
along (for example by a string),
then the force acting on the
object is called the Tension
Tension = PULLING force
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
3A

7. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
You need to understand all the forces at work in
various situations…
Thrust
Thrust
If an object is being pushed along (for
example by a rod), then the force acting on
the object is called the Thrust (or sometimes
compression)
Tension = PUSHING force
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
3A

8. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
You need to understand all the forces at work in
various situations…
Resistance
Any object moving through air, fluid or a solid
will experience resistance caused by the
particles in the way
Gravity
Gravity is the force between any object and
the earth.
The Force caused by gravity acting on an
object is its weight
Remember Newton’s formula…
 The Force is called the weight
 Mass is just mass!
 The acceleration due to gravity is 9.8ms 2
(or can be left as ‘g’
3A

9. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Before we start looking at question we need to
go through some ‘basics’ that are essential for
you to understand this chapter…
Newton’s second law of motion
“The force needed to accelerate a particle is
equal to the product of the mass of the object
and the acceleration required”
F = ma
You need to understand all the forces at work in
various situations…
Resolving
When there are multiple forces acting on an
object, we ‘resolve’ these forces in different
directions
One direction will usually be the direction of
acceleration
The other will be perpendicular to this
Force is measured in Newtons (N). A Newton is:
“The force that will cause a mass of 1kg to
accelerate at 1ms-2”
3A

10. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the formula F
= ma to solve problems involving forces and
acceleration
Normal
R Reaction
Tension
mg (mass
x gravity)
Frictional
Force
Direction of
motion
Thrust
3A

11. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the weight in Newtons, of a particle
of mass 12kg
The mass is already in kg, and use
acceleration due to gravity
Calculate
As the acceleration was given to
2sf, you should give you answer
to the same accuracy
 Ensure you use the exact
amount in any subsequent
calculations though!
3A

12. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the acceleration when a particle of
mass 1.5kg is acted on by a force of 6N
Sub in F and m
Divide by 1.5
3A

13. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the values of the missing forces
acting on the object in the diagram below
In this example you need to consider the horizontal
forces and vertical forces separately (This is called
resolving)
Resolving Horizontally
Take the direction of acceleration as the positive one
Sub in values. You must subtract any
forces acting in the opposite direction!
2ms-2
Y
Calculate
Add 4
4N
2kg
X
Resolving Vertically
Take the direction of the force Y as positive
2g N
Sub in values. Acceleration is 0 as
there is none in the vertical direction
Calculate
Add 2g
3A

14. Dynamics of a Particle moving in a
Straight Line
You can use Newton’s Laws and the
formula F = ma to solve problems
involving forces and acceleration
Find the values of the missing forces
acting on the object in the diagram below
In this example you need to consider the horizontal
forces and vertical forces separately (This is called
resolving)
Resolving Horizontally
Take the direction of acceleration as the positive one
Sub in values. You must subtract any
forces acting in the opposite direction!
2ms-2
Y
Calculate
20N
80N
4kg
Add X and
Subtract 8
X
Resolving Vertically
Take the direction of the force Y as positive
4g N
Sub in values. Acceleration
is 0 as there is none in the
vertical direction
Calculate
Add 20, add 4g
3A

16. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can solve problems involving forces by
drawing a diagram including all relevant
forces, and then resolving in multiple
directions if necessary
A particle of mass 5kg is pulled along a rough
horizontal table by a force of 20N, with a
frictional force of 4N acting against it. Given
that the particle is initially at rest, find:
a)The acceleration of the particle
b)The distance travelled by the particle in
the first 4 seconds
c)The magnitude of the normal reaction
between the particle and the table
Start by drawing
a diagram
R
4N
5kg
20N
5g N
a)
Resolve horizontally and sub in
values. Take the direction of
acceleration as positive
Calculate a
3B

17. Dynamics of a Particle moving in a
Straight Line
3.2ms-2
You can solve problems involving forces by
drawing a diagram including all relevant
forces, and then resolving in multiple
directions if necessary
A particle of mass 5kg is pulled along a rough
horizontal table by a force of 20N, with a
frictional force of 4N acting against it. Given
that the particle is initially at rest, find:
a)The acceleration of the particle – 3.2ms
b)The distance travelled by the particle in
the first 4 seconds
c)The magnitude of the normal reaction
between the particle and the table
-2
Start by drawing
a diagram
R
4N
5kg
5g N
20N
Use SUVAT
b)
Sub in values
Calculate
3B

18. Dynamics of a Particle moving in a
Straight Line
3.2ms-2
You can solve problems involving forces by
drawing a diagram including all relevant
forces, and then resolving in multiple
directions if necessary
Start by drawing
a diagram
R
A particle of mass 5kg is pulled along a rough
horizontal table by a force of 20N, with a
frictional force of 4N acting against it. Given
that the particle is initially at rest, find:
4N
a)The acceleration of the particle – 3.2ms-2
b)The distance travelled by the particle in
the first 4 seconds – 25.6m
c)The magnitude of the normal reaction
between the particle and the table
c)
5kg
20N
5g N
Resolve vertically, taking R
as the positive direction
Calculate
3B

19. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
T
2ms-2
A small pebble of mass 500g is attached
to the lower end of a light string. Find
the tension in the string when the pebble
is:
a)Moving upwards with an acceleration of
2ms-2
b)Moving downwards with a deceleration
of 4ms-2
Start by drawing
a diagram
0.5kg
0.5g N
a)
Resolve vertically, taking the
direction of the acceleration
as positive
Calculate T
3B

20. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
T
-4ms-2
A small pebble of mass 500g is attached
to the lower end of a light string. Find
the tension in the string when the pebble
is:
a)Moving upwards with an acceleration of
2ms-2 – 5.9N
b)Moving downwards with a deceleration
of 4ms-2
0.5kg
0.5g N
b)
Start by drawing
a diagram
In this case, the pebble is
moving downwards at a
decreasing rate, so you can
put the acceleration on as
negative
Resolve vertically, taking the
direction of movement as
positive
Calculate T
Even though the pebble is moving
downwards, there is more tension in the
string as the pebble is decelerating – the
string is working against gravity!
3B

21. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
A particle of mass 3kg is projected at an
initial speed of 10ms-1 in the horizontal
direction. As it travels, it meets a
constant resistance of magnitude 6N.
Calculate the deceleration of the particle
and the distance travelled by the time it
comes to rest.
a ms-2
R
6N
3kg
3g N
Start by drawing
a diagram
It is important to note that
the initial projection speed
is NOT a force, there are
actually no forces acting in
the positive direction
Deceleration
Take the direction of movement
as positive – remember to include
0 as the positive force!
Calculate a
So the deceleration is 2ms-2
3B

22. Dynamics of a Particle moving in a
Straight Line
You can solve problems involving forces
by drawing a diagram including all
relevant forces, and then resolving in
multiple directions if necessary
A particle of mass 3kg is projected at an
initial speed of 10ms-1 in the horizontal
direction. As it travels, it meets a
constant resistance of magnitude 6N.
Calculate the deceleration of the particle
and the distance travelled by the time it
comes to rest.
a ms-2
R
6N
3kg
3g N
Start by drawing
a diagram
It is important to note that
the initial projection speed
is NOT a force, there are
actually no forces acting in
the positive direction
Distance travelled
Deceleration = 2ms-2
Sub in values
Work through
to calculate s
3B

24. Dynamics of a Particle moving in a
Straight Line
However, a force at an
If a force at applied at an angle to
the direction of motion you can resolve
it to find the component of the force
acting in the direction of motion
A horizontal force has no effect on
the object in the vertical direction
angle will have some effect
in BOTH the horizontal and
vertical directions!
S
O
H
C
A
H
T
O
A
Opp = Sinθ x Hyp Cosθ x Hyp
Adj =
10N
Opp = Sin20Adj = Cos20 x 10
x 10
Hyp
10N
Opp
10sin20
20°
Adj
10N
A vertical force has
no effect on the
object in the
horizontal direction
10cos20
So a force can be split into its horizontal and
vertical components using Trigonometry!
3C

25. Dynamics of a Particle moving in a
Straight Line
y
If a force at applied at an angle to
the direction of motion you can resolve
it to find the component of the force
acting in the direction of motion
9N
9Sin40
40°
Find the component of each force in the
x and y-directions
9Cos40
Force in the x-direction
x
Force in the y-direction
= 9Cos40
= 9Sin40
= 6.89N
= 5.79N
3C

26. Dynamics of a Particle moving in a
Straight Line
y
If a force at applied at an angle to
the direction of motion you can resolve
it to find the component of the force
acting in the direction of motion
12N
12Sin23
Find the component of each force in the
x and y-directions
23°
12Cos23
Force in the x-direction
x
Force in the y-direction
= 12Cos23
= 12Sin23
= 11.05N
= 4.69N
= -11.05N
(This will be negative as it is
the opposite direction to x!)
3C

28. Dynamics of a Particle moving in a
Straight Line
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
It is dependent on two things:
1) The normal reaction between the two surfaces
2) The coefficient of friction between the two
surfaces
The maximum frictional force is calculated as
follows:
If a surface is described as ‘smooth’, the
implication is that the coefficient of friction is 0.
3D

29. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
Draw a diagram
R
F
5kg
10N
5g N
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
a)P = 10N
b)P = 19.6N
c)P = 30N
We need to find the maximum possible frictional force
 To do this we need R, the normal reaction
Resolve vertically
Calculate R
Now we can calculate the maximum possible frictional force
Sub in values
Calculate FMAX
3D

30. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
Draw a diagram
R
10N
F
5kg
10N
5g N
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
The maximum frictional force is 19.6 N
Any force will be opposed by friction up to
this value
For part a), the force is only 10N
Therefore, the frictional force will match
this at 10N, preventing movement
Hence, there is also no acceleration
a)P = 10N
b)P = 19.6N
c)P = 30N
3D

31. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
Draw a diagram
R
19.6N
F
5kg
19.6N
5g N
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
a)P = 10N
b)P = 19.6N
c)P = 30N
The maximum frictional force is 19.6 N
Any force will be opposed by friction up to
this value
For part b), the force is only 19.6N
Therefore, the frictional force will match
this at 19.6N, preventing movement
Hence, there is also no acceleration
This situation is called ‘limiting equilibrium’,
as the object is on the point of movement
3D

32. Dynamics of a Particle moving in a
Straight Line
a ms-2
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
Draw a diagram
R
19.6N
F
5kg
30N
5g N
A block of mass 5kg is lying at rest on rough
horizontal ground. The coefficient of friction
between the block and the ground is 0.4. A rough
horizontal force, P, is applied to the block. Find
the magnitude of the frictional force acting on
the block and its acceleration when:
a)P = 10N
b)P = 19.6N
c)P = 30N
For part c), the force is 30N
The frictional force will oppose 19.6N of this,
but no more.
Hence, the object will accelerate…
Resolve horizontally!
Sub in values and resolve
horizontally
Calculate
Divide by 5
So the acceleration will be 2.08ms-2
3D

33. Dynamics of a Particle moving in a
Straight Line
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
F
Draw a diagram
5kg
P
5g N
Resolve vertically to find the normal reaction
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a)P is applied horizontally
b)P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
Sub in values and
resolve vertically
Calculate
Now find the maximum frictional force
Sub in values
Calculate
So P will have to exceed 24.5N to make the object move!
3D

34. Dynamics of a Particle moving in a
Straight Line
Draw a diagram
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
F
P
5kg
Friction is a force which opposes movement
between two ‘rough’ surfaces.
0.6P
Psinθ
θ
Pcosθ
0.8P
5g N
We need to find the values of Cosθ and Sinθ. The
ratio for Tanθ can be used to find these!
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a)P is applied horizontally – 24.5N
b)P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
S
O
We can find the
hypotenuse using
Pythagoras’
Theorem!
H
C
A
H
T
Hyp
5
O
Tanθ = 3/4
A
So Opp = 3
And Adj = 4
3 Opp
θ
4
Adj
Sinθ = 3/5
Cosθ = 4/5
Sinθ = 0.6
Cosθ = 0.8
3D

35. Dynamics of a Particle moving in a
Straight Line
Draw a diagram
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
F
5kg
P
0.6P
θ
0.8P
5g N
Resolve vertically to find the normal reaction
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a)P is applied horizontally – 24.5N
b)P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
Sub in values and
resolve vertically
We find the normal
reaction in terms of P
Now find the maximum frictional force
Sub in
values
Find Fmax in
terms of P
So, 0.8P will have to exceed this if the box is to
move…
3D

36. Dynamics of a Particle moving in a
Draw a diagram
Straight Line
R
You can calculate the magnitude of a frictional
force using the coefficient of friction
Friction is a force which opposes movement
between two ‘rough’ surfaces.
A 5g box lies at rest on a rough horizontal floor.
The coefficient of friction between the box and
the floor is 0.5. A force P is applied to the box.
Calculate the value of P required to cause the box
to accelerate if:
a)P is applied horizontally – 24.5N
b)P is applied at an angle of θ above the
horizontal, where tanθ = 3/4
24.5 – 0.3P
F
5kg
P
θ
0.6P
0.8P
5g N
We need to find the value for P for which the
box is in ‘limiting equilibrium’ – that is, so the
horizontal forces cancel each other out…
Resolve horizontally…
Sub in values and
resolve horizontally
Careful with the
bracket!
Rearrange and
solve
P must exceed 22N, which is less than when P was horizontal
 The reason is because some of the force is upwards, this
alleviates some of the friction between the surfaces…
3D