T-Level Unit 5 Session 13 & 14

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

5. Essential science for engineering and
manufacturing

5.6

Forces and motion in engineering

PowerPoint 9: Vector representation of forces and
coplanar forces

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Objectives

By the end of this session, learners should be able to:

•Carry out calculations using forces represented by vectors.

•State the conditions for equilibrium.

•Carry out calculations for coplanar forces.

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

A force has magnitude and direction and can be represented graphically
in two dimensions as a vector. The length of the line represents the
magnitude, with the angle indicating the direction.

y

x

F

θ

Vector representation of force

The force F acts at an angle θ to the
horizontal and as such, can be split into a
vertical and a horizontal component.

The vertical component
𝑽= F sin θ

The horizontal component
𝑯= F cos θ

Resultant magnitude R =
𝒗𝟐

𝑯
𝟐

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Resultant angle θ

•The angle in which the resultant force
is applied can be either measured from
the force diagram using a protractor or
calculated from the equation below

•Resultant angle

𝑽

𝑯

y

x

F

θ

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Sign convention

M +

M -

1. Upward forces are positive (+) and downward forces are negative(-).

2. Horizontal forces acting to the right are positive(+) and to the left are

negative(-).

3. Clockwise acting moments / couples are positive(+) and

anticlockwise ones are negative(-).

+

+

-

-

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Calculate the resultant of the two forces shown.

F1 = 7N

F2 = 5N

30°

40°

Force vector example

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

FV = 7 sin 70° + 5 sin 30° = 9.08N

Resultant force (R) =
�
�

�
� = √(9.082+6.722) = 11.3N

Resultant angle (θ) =

Resolving vertical components:

��

�

�

= 53.5°

Force vector example

Resolving horizontal components:
FH = 7 cos 70° + 5 cos 30° = 6.72N

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Equilibrium

Equilibrium means that all forces acting on an object or system are
arranged in a way that there is an overall resultant force of zero. This
means that there can be no acceleration and so the object must either be
static or moving at a constant velocity.

There are several conditions that need to be met in order to bring an
object or system into equilibrium:

•The sum of the forces in the horizontal direction is zero.

•The sum of the forces in the vertical direction is zero.

•The sum of all moments is zero.

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Conditions for static equilibrium

•An object is in equilibrium if the net forces on it are zero.

•This does not mean there are no forces on it – it means that the
forces on the object balance out.

•Mathematically this is represented as:

•

Ʃ FV = 0

•

Ʃ FH = 0

•

Ʃ M = 0

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Equilibrant force

5N@30°

Resultant = 11.3N

7N@70°

53.5o

Equilibrant =
11.3N @ 233.5o

53.5o

This is the force that would need to be present in order to bring the
system into equilibrium. It would need to be equal in magnitude and
opposite in direction to the resultant as shown below.

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Equilibrium conditions

•For a system with two forces, it is in equilibrium when:

F1 + F2 = 0, i.e. F1 = - F2.

•For a system with 3 forces, it is in equilibrium when:

F1 + F2 + F3 = 0, i.e. F1 = - (F2 + F3).

Trigonometry, Pythagoras’ theorem and vector addition may be needed
to calculate the balancing forces.

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Trigonometry and Pythagoras’ Theorem

• Trigonometry may be required to calculate

the angles between force vectors

• Tan θ = O / A

• Rearranging, θ = tan-1(O / A)

• Pythagoras’ Theorem (a2+ b2= c2) may be

needed to calculate the magnitude
O

H

A

θ

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Example calculation

A motion-damping system in a machine is at equilibrium when three
forces are acting on it:

R N, (5i – 8j) N and (3i + 4j) N.

a)

Express R in the form ai + bj

Answer: Rearranging R + F1 + F2 = 0, R = - (F1 + F2)

therefore R = -((5i – 8j) + (3i + 4j)) = (-8i + 4j) N

b)

Calculate the magnitude of R

Answer: Magnitude = √ (i2+ j2) = √ (-82+ 42) = √ (64 + 16) = 8.94 N

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Example calculation

c) Calculate the angle that R makes to the i direction.

Answer: θ = tan-1(4 / -8) = tan-1-0.5 = 26.6o

However, the i value of the vector is negative.

Hence, relative to the i direction the angle of the cable, is

(180 + 26.6) o= 206.6o(or -153.4o)

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Coplanar forces

Forces that act within a two dimensional plane are referred to as
coplanar forces.

If all of the lines of action of these forces pass through the same point,
known as the point of concurrence, then we have a concurrent
coplanar force system.

F4

F1

F3

F2
Point of

concurrence

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

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‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Four co-planar forces act at a point O. The values and direction of the
forces are:

Calculate the resultant force and describe the magnitude and direction of
the force required to bring the system into equilibrium.

10N

Coplanar forces example

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Force (N)

Horizontal Component (N)

Vertical Component (N)

50

50 cos 0o= 50 x 1 = 50N

50 sin 0o= 50 x 0 = 0N

10

10 cos 90o= 10 x 0 = 0N

-10 sin 90o= 10 x 1 = -10N

20

-20 cos 30o= -20 x 0.866 = -17.32N

-20 sin 30o= -20 x 0.5 = -10N

30

-30 cos 60o= -30 x 0.5 = -15N

30 sin 60o= 30 x 0.866 = 25.98N

Resultant

50 – 17.32 – 15 = 17.68N

-10 -10 +25.98 = 5.98N

Resultant force R
OB2= OA2+ AB2
R2= (17.68)2+ (5.98)2
R = √312.58 + 35.76
R = 18.66N

��

𝐨

Coplanar Forces Example

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

The following force is required to bring the system into equilibrium.

18.69o

198.69o

Resultant =
18.66N @ 18.69o

Equilibrant =
18.66N @ 198.69o

Coplanar Forces Example

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Non-concurrent coplanar force systems

Another type of coplanar system exists where the lines of action of the
forces do not pass through the same point. These type of systems are
called NON-CONCURRENT COPLANAR FORCE SYSTEMS.

In this type of system, there is not only a tendency for the force to
move the body in a certain linear direction, but also to make it rotate.

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

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‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Forces in the same plane that
do not meet at 1 point

Resultant force can be
calculated as concurrent forces.

However, a turning force occurs
as well.

This can be calculated using
moments.

F2
F1

F3
F4

Non-concurrent coplanar force systems

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

O

Non-concurrent coplanar force system example

The members in this structure
are acted upon by the four
forces shown.

Determine the magnitude and
direction of the resultant force
and the perpendicular
distance of its line of action
from O.

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T Level Technical Qualification in
Engineering and Manufacturing (Level 3)
300 Engineering Common Core Content

© 2022 City and Guilds of London Institute. All rights reserved.

‘T-LEVELS’ is a registered trade mark of the Department for Education.
‘T Level’ is a registered trade mark of the Institute for Apprenticeships and Technical Education

Force

Angle

FH
FV
MH
MV

4kN

40°

4cos40 =
3.06kN

4sin40 =
2.57kN

0

2.57 x 0.25 =

-0.64kNm

7kN

180°

7cos180 =

-7kN

7sin180 =

0kN

0

0

8kN

135°

8cos135 =

-5.66kN

8sin135 =

5.66kN

5.66 x 0.4

= -

2.26kNm

0

Total

-9.60kN

8.23kN

kNm

Anti-clockwise

�
�

�
�

��

�

�