EM025 CHAPTER 2 ENGINEERING MECHANICS

M AT R I C U L AT I O N ENGINEERING
MECHANIC

KEJURUTERAAN MEKANIKAL EM025

2.1 THREE DIMENSIONAL FORCE SYSTEM
(3D)

a) Define vector using Cartesian coordinate system.
b) Resolve forces into components in 3D by applying vector
operations.
c) Solve 3D resultant forces by applying vector operations.
d) Solve multiplication of 3D forces (Dot product and cross
product).

Cartesian coordinate system

• Simple vector quantities can be expressed geometrically.
• As the applications become more complex, or involve 3D,

express vectors in Cartesian coordinates (x, y, and z) is needed.
• Basic vector problems using Cartesian coordinate system.

- Cartesian Vector
- Position Vector
- Dot Product
- Cross Product

3

INTRODUCTION 4
• A coordinate system that specifies each point uniquely in a

plane by a set of numerical coordinates, which are the signed
distances to the point from perpendicular directed lines,
measured in the same unit of length.
• Sense (or arrowhead) of these vectors are described by a plus
or minus sign (depending on pointing towards the positive
or negative axes).
CONCEPT
Right-Handed Coordinate System:

Thumb of right hand points
in the direction of the
positive z axis when the
right-hand fingers are curled
about this axis and directed
from the positive x towards
the positive y axis.

UNIT VECTOR

• Vectors have both a magnitude and a direction and are labeled
with an arrow ( → ). Example :

• A unit vector is a vector that has a magnitude of 1. They are
labeled with a (^). Example :

• Any vector can become a unit vector by dividing it by the
vector's magnitude.

• Vectors are often written in xyz coordinates. Example :
OR

• The magnitude of a vector is
• So :
• In bracket format:

5

RECTANGULAR COMPONENTS OF A VECTOR

o-corAimevnpetocanttoieornntA.s amloanyghtahveexo, ynea,ntdwoz oaxretsh, rdeeepreencdtiannggounlar
- By two successive application of the parallelogram law

A = A’ + Az
A’ = Ax + Ay
- Combining the equations, A can be
expressed as
A = Ax + Ay + Az

6

Addition and Subtraction of Cartesian Vectors

Given: A = BAxxii + BAyyjj + ABZZkk
B = + +

Vector Addition
Resultant R = A + B
= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) k

Vector Subtraction
Resultant R = A - B
= (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k

Concurrent Force Systems
Force resultant is the vector sum of all the forces in the system

FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk

7

CARTESIAN VECTOR REPRESENTATIONS

Magnitude of a Cartesian Vector :
From the shaded triangle,

A'= Ax2 + A2y
From the colored triangle,

A = A'2 + Az2

Combining equations gives magnitude A,

A = Ax2 + Ay2 + Az2

Three components of A act in the positive i, j and k directions, 8
as expressed in Cartesian vector form

A = AuA
= Acosαi + Acosβj + Acosγk
= Axi + Ayj + Azk

DIRECTION OF A CARTESIAN VECTOR

• Direction of A can be specified using a unit vector
• Unit vector has a magnitude of 1
• Orientation of A is defined as the coordinate direction angles

α, β and γ measured between the tail of A and the positive x, y
and z axes
• 0° ≤ α, β and γ ≤ 180 °

cos2  + cos2  + cos2  = 1

9

Direction of a Cartesian Vector

- For angles α, β and γ :

cos = Ax cos  = Ay cos = Az 10

A A A

RESOLVING & RESULTANT FORCES IN 3D
Example 1 :
Express the forces FA and FB shown as a Cartesian vector
(FA=140N, FB=100N)

11

RESOLVING & RESULTANT FORCES IN 3D

12

RESOLVING & RESULTANT FORCES IN 3D
Example 2 :
Express the force F as Cartesian vectors.

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RESOLVING & RESULTANT FORCES IN 3D 14
Solution :
Since two angles are specified, the third angle is found by :

cos2  + cos2  + cos2  = 1

cos2  + cos2 60 + cos2 45 = 1

cos = 1− (0.5)2 − (0.707)2 = 0.5

Two possibilities exit, namely :

 = cos−1(0.5) = 60 OR  = cos−1(− 0.5) = 120

* By inspection, α = 60° since Fx is in the + x direction

Given F = 200N
F = F cos α i + Fc os β j + Fc os γ k
= (200cos60°N)i + (200cos60°N)j + (200cos45°N)k
= {100.0i + 100.0j + 141.4k}N

RESOLVING & RESULTANT FORCES IN 3D

Example 3:
Two forces act on the hook. Specify the coordinate direction
angles of aFx2i,ssaontdhhatasthaemreasgunlittaundtefoofrc8e00FNR .acts along the
positive y

15

RESOLVING & RESULTANT FORCES IN 3D

Solution :
Cartesian vector form
FR = F1 + F2

F1 = F1cosα1i + F1cosβ1j + F1cosγ1k
= (300cos45°N)i + (300cos60°N)j + (300cos120°N)k
= {212.1i + 150j - 150k}N

F2 = F2xi + F2yj + F2zk

Since FR has a magnitude of 800N and acts in the +j direction

FR = F1 + F2
800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk
= (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)k
16

RESOLVING & RESULTANT FORCES IN 3D cos = Ax

Hence, A
0 = 212.1 + F2x
F2x = -212.1N cos = Ay

800 = 150 + F2y A
F2y = 650N cos = Az

0 = -150 + F2z A
F2z = 150N
17
Since magnitude of F2 and its components are known :
α1 = cos-1(-212.1/700) = 108°
β1 = cos-1(650/700) = 21.8°
γ1 = cos-1(150/700) = 77.6°

Concept

• A position vector represents the position of a point in space in
relation to reference origin O in x, y, z coordinates.

• Position vector r is defined as a fixed vector which locates a
point in space relative to another point.

• Example :
If r extends from the origin, O to point P (x, y, z) then, in Cartesian
vector form r = xi + yj + zk .
The head to tail vector addition of the three components.
Start at origin O, one travels x in the +i direction, y in the + j
direction and z in the +k direction, arriving at point P (x, y, z).

18

Concept

Position vector maybe directed from point A to point B.
Designated by r or rAB

Vector addition gives 19

rA + r = rB
Solving

rAB = rB – rA = (xB – xA) i + (yB – yA) j + (zB –zA) k

RESOLVING & RESULTANT FORCES IN 3D
Example 1:
Express the position vector r in Cartesian Vector form. Determine
its magnitude and coordinates direction angles.

A

B

20

RESOLVING & RESULTANT FORCES IN 3D

Solution :
Coordinate A(3,4,0) and B(0,-8,4)
rAB = rB-rA

=[(0-3) i +(-8-4) j +(4-0) k]
= -3 i -12 j +4 k

= −32 + −122 + 42 = 13

Assume angles directed from r to x-axis =α , y-axis=β , z-axis=γ

= cos− − = ᵒ



= cos− − = . ᵒ


= cos− = ᵒ. 21


RESOLVING & RESULTANT FORCES IN 3D
Example 2:
Determine the magnitude and coordinates direction angles of the
resultant force.

22

RESOLVING & RESULTANT FORCES IN 3D 23

Solution :
Coordinate A(0,2,4) and B(0,0,0)
rAB = rB-rA

= 0 i -2 j -4 k

= 02 + −22 + −42 = 4.47
UAB = (0/4.47) i –(2/4.47) j – (4/4.47) k

= 0i – 0.45j – 0.89k
FAB = 600 (0i – 0.45j – 0.89k)

= [0i – 270j – 534k] N

Coordinate A(0,2,4) and C(4,8,0)
rAC = rC-rA

= 4 i +6 j -4 k

= 42 + 62 + −42 = 8.25
UAC = (4/8.25) i +(6/8.25) j – (4/8.25) k

= 0.48i + 0.73j – 0.48k
FAC = 500 (0.48i + 0.73j – 0.48k)

= [240i + 365j – 240k] N

RESOLVING & RESULTANT FORCES IN 3D

Solution :
FR = FAB + FAC

= [0i – 270j – 534k] + [240i + 365j – 240k]
= 240i + 95j – 774k
= 2402 + 952 + −7742
= 815 N

Coordinated direction:

= cos− = ᵒ


= cos− = ᵒ



= cos− − = ᵒ 24


CONCEPT

• Dot product of vectors A and B is written as A·B
• Define the magnitudes of A and B and the angle between

their tails
A·B = │A││B│ cos θ (0°≤ θ ≤180°)

25

Cartesian Vector Formulation 26

• Dot product of Cartesian unit vectors
Eg: i·i = (1)(1)cos0° = 1
i·j = (1)(1)cos90° = 0

• Similarly
i·i = 1 j·j = 1 k·k = 1
i·j = 0 i·k = 0 j·k = 0

• Dot product of 2 vectors A and B
A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)
= AxBAx(yiB·iz)(j+·kA)x+ByA(ziB·jx)(+k·Ai)xB+z(Aiz·Bky)(+kA·jy)B+x(Aj·ziB) z+(kA·ykB)y(j·j) +
A·B = AxBx + AyBy + AzBz

RESOLVING & RESULTANT FORCES IN 3D
Example 1 :
Suppose that the coordinates of point B are changed to
(6, 4, 4) m. What is the angle of Ɵ?

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RESOLVING & RESULTANT FORCES IN 3D
Solution :
Using the new coordinates we have

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RESOLVING & RESULTANT FORCES IN 3D
Example 2 :
a) Determine the component of a unit vector that is parallel to the

forestay AB and points from A towards B
b) Determine the component of a unit vector that is parallel to the

backstay BC and points from C towards B

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RESOLVING & RESULTANT FORCES IN 3D

a)

b)

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CONCEPT

Cross product of two vectors A and B written as: A X B = C

Magnitude of C is defined as:
• The product of the magnitudes of A and B ;
• The sine of the angle θ between their tails.

Therefore: (0° ≤ θ ≤ 180)
C = │A││B│ sinθ

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In determinant form,

A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk)

= AxBx (i X i) + AxBy (i X j) + AxBz (i X k) + AyBx (j X i)
+ AyBy (j X j) + AyBz (j X k) + AzBx (k X i) +AzBy (k X j)
+AzBz (k X k)

= (AyBz – AzBy)i – (AxBz – AzBx)j + (AxBy – AyBx)k

OR

In matrix form,   
 i j k
AXB = Ax Ay Az
Bz
Bx By

= (AyBz – AzBy)i – (AxBz – AzBx)j + (AxBy – AyBx)k 32

RESOLVING & RESULTANT FORCES IN 3D
Example 3:
If A = (0, –1, 3) and B = (2, 0, –1), find A X B.
Solution:
AXB=

=[(-1)(-1) – (0)(3)]i – [(0)(-1) – (2)(3)]j + [(0)(0) – (2)(-1)]k
=( 1 - 0)i – (0 – 6)j + (0+2)k
=1i + 6j + 2k

33

RESOLVING & RESULTANT FORCES IN 3D
Example 1:
Determine the resultant moment produced by the two forces about
point O. Express the result as a Cartesian vector.

1m N
2m
3m 34

2m
N

RESOLVING & RESULTANT FORCES IN 3D
Solution:
Coordinate O = (0, 0, 0)
Coordinate A = (3, 3, –2)

rOA =

= [3i + 3j – 2k]N

MOA = (rOA X F1) + (rOA X F2)

=

= N.m

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RESOLVING & RESULTANT FORCES IN 3D
Example 2:
The rope exerts a force of magnitude F=2000N on the top of the
pole at B. Determine the magnitude of the moment of F about A.

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RESOLVING & RESULTANT FORCES IN 3D

37

2.2 EQUILIBRIUM OF PARTICLES IN TWO
DIMENSION (2D)

a) Identify force system using equilibrium particle concept.
b) Solve the force system using the equilibrium particle.

PARTICLE

• A particle is a body whose rotational aspects are not of interest
at the moment.

• This is the case of course when the body itself is extremely
small so that we think of it as a single point at which all its mass
is concentrated.

• Often we take a large body and shift all its mass to its center of
mass and just look at the motion of the center of mass.

EQUILIBRIUM 39

• Equilibrium can be thought of as an unchanging or stable
condition.

• All the bodies that are at rest are in equilibrium.

EQUILIBRIUM OF PARTICLE

• Equilibrium condition for Particle:
➢ at rest
➢ moving at a constant velocity

• Newton’s first law of motion:
∑F = 0

*sum of all the forces acting on the particle equal to zero.

• From Newton’s second law of motion:

∑F = ma *F = force , m = mass, a = acceleration

*Particle is moving in constant velocity or at rest and these fulfill

the equilibrium condition.

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PROCEDURE FOR ANALYSIS:
COPLANAR METHOD

i. Free-Body Diagram 41

i. Establish the x, y axes in any suitable orientation.

ii. Label all the unknown and known forces magnitudes
and directions.

iii. Sense of the unknown forces can be assumed

ii. 2. Equations of Equilibrium

i. apply the equations of equilibrium:

∑Fx = 0 , ∑Fy = 0
i. components are positive if they are directed along the

positive axis and negative if directed along the negative axis.

Example 1:
Determine the tension in cables AB and AD for equilibrium of
the 250kg engine.

42

Solution: 43

FBD at Point A
- Initially, two forces acting,
forces of cables AB and AD.

- Engine Weight,
= (250kg)(9.81m/s2) = 2.452kN
[supported by cable CA]

- Finally, three forces acting,
forces TB and TD and engine
weight on cable CA.

Solution:

+→ ∑Fx = 0;
TBcos30° - TD = 0

+↑ ∑Fy = 0;
TBsin30° - 2.452kN = 0

solving,

TB = 4.90kN
TD = 4.25kN

*Note: 44
Neglect the weights of the cables since they are too small compared to the
weight of the engine.

Example 2:
If the sack at A has a weight of 20N, determine the weight of
the sack at B and the force in each cord needed to hold the
system in the equilibrium position shown.

45

Explanation :
To solve this problem, two FBD need to be sketched. At point E
and C

46

Solution:

i) FBD at Point E
- Three forces acting, forces of cables EG and EC and the

weight of the sack on cable EA

47

Solution :

+→ ∑Fx = 0;
TEGsin30° - TEC cos45° = 0

+↑ ∑Fy = 0;
TEGcos30° - TEC sin45° - 20N = 0

Solving,

TEC = 38.6kN
TEG = 54.6kN

* Note: 48
Uasnedequilibrium at the ring to determine tension in CD
weight of B with TEC known

Solution :

ii) FBD at Point C
- Three forces acting, forces by cable CD and EC (known)
and weight of sack B on cable CB

49

Solution :

+→ ∑Fx = 0;
38.6cos45° - (4/5)TCD = 0

+↑ ∑Fy = 0;
(3/5)TCD + 38.6sin45°N – WB = 0

Solving,

TCD = 34.1kN
WB = 47.8kN

*Note: 50
Components of TCD are proportional to the slope of the cord
by the 3-4-5 triangle