1. Mechanics
Topic 2.2 Forces and Dynamics

2. Forces and Free-body
Diagrams
To a physicist a force is recognised by
the effect or effects that it produces
A force is something that can cause an
object to
 Deform (i.e. change its shape)
 Speed up
 Slow Down
 Change direction

3. The last three of these can be
summarised by stating that a force
produces a change in velocity
Or an acceleration

4. Free-body Diagrams
A free-body diagram is a diagram in which the
forces acting on the body are represented by
lines with arrows.
The length of the lines represent the relative
magnitude of the forces.
The lines point in the direction of the force.
The forces act from the centre of mass of the
body
The arrows should come from the centre of
mass of the body

5. Example 1
Normal/Contact Force
Weight/Force due to Gravity
A block resting on a worktop

6. Example 2
A car moving with a constant velocity
Normal/Contact Force
Weight/Force due to Gravity
Motor ForceResistance

7. Example 3
A plane accelerating horizontally
Upthrust/Lift
Weight/Force due to Gravity
Motor Force
Air Resistance

8. Resolving Forces
Q. A force of 50N is applied to a block
on a worktop at an angle of 30o
to the
horizontal.
What are the vertical and horizontal
components of this force?

9. Answer
First we need to draw a free-body diagram
30o
50N

10. We can then resolve the force into the 2
components
30o
50NVertical = 50 sin 30o
Horizontal = 50 cos 30o

11. Therefore
 Vertical = 50 sin 30o
= 25N
 Horizontal = 50 cos 30o
= 43.3 = 43N

12. Determining the Resultant
Force
Two forces act on a body P as shown in
the diagram
Find the resultant force on the body.
30o
50N
30N

13. Resolve the forces into the vertical and
horizontal componenets (where
applicable)
Solution
30o
50N
30N
50 sin 30o
50 cos 30o

14. Add horizontal components and add
vertical components.
50 sin 30o
= 25N
50 cos 30o
– 30N= 13.3N

15. Now combine these 2 components
25N
13.3N
R
R2
= 252
+ 13.32
R = 28.3 = 28N

16. Finally to Find the Angle
25N
13.3N
R
θ
tan θ = 25/13.3
θ = 61.987
θ = 62o
The answer is therefore 28N at 62o
upwards from
the horizontal to the right

17. Springs
The extension of a spring which obeys Hooke
´s law is directly proportional to the extending
tension
A mass m attached to the end of a spring
exerts a downward tension mg on it and if it is
stretched by an amount x, then if k is the
tension required to produce unit extension
(called the spring constant and measured
in Nm-1
) the stretching tension is also kx and
so
mg = kx

18. Spring Diagram
x

19. Newton´s Laws
The First Law
Every object continues in a state of
rest or uniform motion in a straight
line unless acted upon by an external
force

20. Examples
Any stationary object!
Difficult to find examples of moving
objects here on the earth due to friction
Possible example could be a puck on
ice where it is a near frictionless surface

21. Equilibrium
If a body is acted upon by a number of
coplanar forces and is in equilibrium ( i.e.
there is rest (static equilibrium) or
unaccelerated motion (dynamic
equilibrium)) then the following condition
must apply
The components of the forces in both of any
two directions (usually taken at right angles)
must balance.

22. Newton´s Laws
The Second Law
There are 2 versions of this law

23. Newton´s Second Law
1st
version
The rate of change of momentum of a
body is proportional to the resultant
force and occurs in the direction of the
force.
F = mv – mu F =∆ρ
t t

24. Newton´s Second Law
2nd
version
The acceleration of a body is
proportional to the resultant force
and occurs in the direction of the
force.
F = ma

25. Linear Momentum
The momentum p of a body of constant mass
m moving with velocity v is, by definition mv
Momentum of a body is defined as the mass
of the body multiplied by its velocity
Momentum = mass x velocity
p = mv
It is a vector quantity
Its units are kg m s-1
or Ns
It is the property of a moving body.

26. Impulse
From Newtons second law
F = mv – mu F =∆ρ
t t
Ft = mv – mu
This quantity Ft is called the impulse of the
force on the body and it is equal to the
change in momentum of a body.
It is a vector quantity
Its units are kg m s-1
or Ns

27. Law of Conservation of Linear
Momentum
The law can be stated thus
When bodies in a system interact the
total momentum remains constant
provided no external force acts on the
system.

28. Deriving This Law
To derive this law we apply Newton´s 2nd
law
to each body and Newton´s 3rd
law to the
system
i.e. Imagine 2 bodies A and B interacting
If A has a mass of mA and B has a mass mB If
A has a velocity change of uA to vA and B has
a velocity change of uB to vB during the time of
the interaction t

29. Then the force on A given by Newton 2 is
FA = mAvA – mAuA
t
And the force on B is
FB = mBvB – mBuB
t
But Newton 3 says that these 2 forces are
equal and opposite in direction

30. Therefore
mAvA – mAuA = -(mBvB – mBuB)
t t
Therefore
mAvA – mAuA = mBuB – mBvB
Rearranging
mAvA + mBvB = mAuA + mBuB
Total Momentum after =
Total Momentum before

31. Newton´s Laws
The Third Law
When two bodies A and B interact,
the force that A exerts on B is equal
and opposite to the force that B
exerts on A.

32. Example of Newton´s 3rd
Q. According to Newton’s third Law
what is the opposite force to your
weight?
A. As your weight is the pull of the Earth
on you, then the opposite is the pull of
you on the Earth!

33. Newton´s 3rd
Law
The law is stating that forces never occur
singularly but always in pairs as a result of
the interaction between two bodies.
For example, when you step forward from
rest, your foot pushes backwards on the
Earth and the Earth exerts an equal and
opposite force forward on you.
Two bodies and two forces are involved.

34. Important
The equal and
opposite forces
do not act on the
same body!