College of Engineering

Department of Civil Engineering

Engineering Mechanics I Statics (CEng2103)

1. Scalars and Vectors

By Mubarek Zeyne (MSc)

October, 2024
AASTU, Addis Ababa, Ethiopia

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 Course Learning Outcomes

After completion of this chapter, students

will be able to:
❑ Understand Newton`s Law, and its application in
a real world problems.
❑ Define important concepts such as Statics,
Dynamics, Space, Time, Mass and other physical
quantities
❑ Perform vector operations: dot and cross
product.

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 Introduction
What is mechanics and its application in Engineering science?
♠ Mechanics is the physical science which deals with the effects of forces on
objects. It is divided into three parts: mechanics of rigid bodies, mechanics of
deformable bodies, and mechanics of fluids.
♠ It is the oldest of the physical sciences. The early history of this subject is
synonymous with the very beginnings of engineering.
♠ No other subject plays a greater role in engineering analysis than mechanics.
Although the principles of mechanics are few, they have wide application in
engineering. [1]

♠ The subject of mechanics is logically divided into two parts: statics,

which concerns the equilibrium of bodies under action of forces, and
dynamics, which concerns the motion of bodies. Engineering Mechanics is
divided into these two parts, Vol. 1 Statics and Vol. 2 Dynamics

♠ Statics deals primarily with the calculation of external forces which act on
rigid bodies in equilibrium. Determination of the internal deformations
belongs to the study of the mechanics of deformable bodies or mechanics
of materials.
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 Introduction…
♠ Space is the geometric region occupied by bodies whose positions are
Basic Concepts
described by linear and angular measurements relative to a coordinate system.
For 3D problems, three independent coordinates are needed. For 2D problems,
only two coordinates are required.
♠ Time is the measure of the succession of events and is a basic quantity in
dynamics. Time is not directly involved in the analysis of statics problems.
♠ Mass is a measure of the inertia of a body, which is its resistance to a change of
velocity. Mass can also be thought of as the quantity of matter in a body.

♠ Force is the action of one body on another. A force tends to move a body in the
direction of its action. The action of a force is characterized by its magnitude, by
the direction of its action, and by its point of application. Thus force is a vector
quantity, and its properties are discussed in detail in next session.
♠ A particle is a body of negligible dimensions. In the mathematical sense, a particle is
a body whose dimensions are considered to be near zero so that we may analyze it
as a mass concentrated at a point. We often choose a particle as a differential
element of a body. We may treat a body as a particle when its dimensions are
irrelevant to the description of its position or the action of forces applied to it.
♠ Rigid body. A body is considered rigid when the change in distance between any two
of its points is negligible for the purpose at hand.
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 Introduction…
Fundamental Principles
Newton’s Laws
Law I. A particle remains at rest or continues to move with uniform velocity (in a
straight line with a constant speed) if there is no unbalanced force acting on it.
Law II. The acceleration of a particle is proportional to the vector sum of forces acting
on it, and is in the direction of this vector sum.
Law III. The forces of action and reaction between interacting bodies are equal in
magnitude, opposite in direction, and collinear (they lie on the same line).

♠ The law of gravitation is expressed by the equation

where
F= the mutual force of attraction between two particles
G = a universal constant known as the constant of gravitation,
6.673x10-11 m3/(kg *s2)
m1, m2 = the masses of the two particles
r =the distance between the centers of the particles

The mutual forces F obey the law of action and reaction, since they are
equal and opposite and are directed along the line joining the centers of
the particles.
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 Scalars and Vectors
♠ We use two kinds of quantities in mechanics—scalars and vectors.
✓ Scalars are physical quantities that can be completely described (measured) by their magnitude
alone. These quantities do not need a direction to point out their application (Just a value to
quantify their measurability). They only need the magnitude and the unit of measurement to fully
describe them. Examples of scalar quantities are time(s), area(m2), volume(m3),
mass(kg),density(kg/m3), speed(m/s), and energy(Wat)
♠ Vector quantities, on the other hand, possess direction as well as magnitude with its unit.
E.g.:-displacement(m), velocity(m/s), acceleration(m/s2), force(N, kg.m/s2), moment(N.m), and
momentum(N.s, kg.m/s).
♠ When writing vector equations, always be certain to preserve the mathematical distinction
between vectors and scalars. In handwritten work, use a distinguishing mark for each vector
quantity, such as an underline, V, or an arrow over the symbol, V , to take the place of boldface
type in print.

y Position vector is a vector that locates a given point

in reference to origin(point of interest).
AB B -A
B • Let AB is a vector with initial point (x1,y1) and
V
terminal point B(x2,y2) then its position vector is
A AB=(x2-x1, y2-y1)= (x2-x1,)i + (y2-y1)j
x • Position vector whose initial point is origin(0,0)
(0,0)
and terminal point is (x2-x1, y2-y1) This vector is
V=(x2-x1,)i + (y2-y1)j
• Any vector whose magnitude is unity is called
unit vector. Generally for any vector V its unit
Mubarek Z vector is determined by dividing this vector by
its magnitude. i.e nV= V/(/V/) 6
 Scalars and Vectors…
♠ Vectors representing physical quantities can be classified as free,
sliding, or fixed.
♠ A free vector is one whose action is not confined to or associated
with a unique line in space. we may represent the displacement of
a body by a free vector.
♠ A sliding vector has a unique line of action in space but not a
unique point of application. For example, when an external force
acts on a rigid body, the force can be applied at any point along its
line of action without changing its effect on the body as a whole,*
and thus it is a sliding vector.
♠ A fixed vector is one for which a unique point of application is
specified. The action of a force on a deformable or nonrigid body
must be specified by a fixed vector at the point of application of
the force. In this instance the forces and deformations within the
body depend on the point of application of the force, as well as on
its magnitude and line of action.

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 Scalars and Vectors…
Operation with Vectors
i) Resultant of vectors
♠ Vectors must obey the parallelogram law of combination. This law states that two
vectors V1 and V2, treated as free vectors, Fig. 1/2a, may be replaced by their
equivalent vector V, which is the diagonal of the parallelogram formed by V 1 and V2 as its
two sides, as seen in Fig. 1/2b.This combination is called the vector sum, and is
represented by the vector equation,
The magnitude of
vector V calculated by the law of cosines as follows:
/V/2=/V1/2 + /V2/2 – 2*/V1/*/V2/*cos(180-θ) θ= angle between V1 & V2 when
joined tail to tail as in Fig. 1/2b.
♠ The two vectors V1 and V2, again treated as free vectors, may also be
added head-to-tail by the triangle law, as shown in Fig. 1/2c, to obtain the identical vector
sum V. Knowing θ, α and magnitudes of given vectors,/V1/ and/V2/ then the magnitude of
Vector V, /V/ determined by using the law of sines as follows

(/𝑽𝟏/) (/𝑽𝟐/) (/𝑽/) β

= = θ
𝑺𝒊𝒏 𝜷 𝑺𝒊𝒏 𝜶 𝑺𝒊𝒏 (𝟏𝟖𝟎−𝜽) α

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 Scalars and Vectors …
Operation with Vectors …
♠ The difference V1 - V2 between the two vectors is easily obtained
by adding -V2 to V1 as shown in Fig. 1/3, where either the triangle or
parallelogram procedure may be used. The difference V` between the
two vectors is expressed by the vector equation
where the minus sign denotes vector subtraction.

ii) Decomposition of a vector into its

component for a given coordinate system
Any two or more vectors whose sum equals a
certain vector V are said to be the
components of that vector. Thus, the vectors
V1 and V2 in Fig. 1/4a are the components of
V in the directions 1 and 2, respectively. It is
usually most convenient to deal with vector
components which are mutually
perpendicular; these are called rectangular
components.
Mubarek Z The process of representing a
vector by its component is called resolving 9
a vector.
 Scalars and Vectors…
Operation with Vectors …
♠ vectors Vx and Vy in Fig. 1/4b are the x- and y-components,
respectively, of V. When expressed in rectangular
components, the direction of the vector with respect to,
say, the x-axis is clearly specified by the angle ɵ, where

In many problems, particularly three-dimensional

ones, it is convenient to express the rectangular
components of V, Fig. 1/5, in terms of unit vectors i, j,
and k, which are vectors in the x-, y-, and z-directions,
respectively, with unit magnitudes. Because the
vector V is the vector sum of the components in the
x-, y-, and z-directions, we can express V as follows:

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 Scalars and Vectors…
Vector Multiplication: Dot & Cross
Dot Product

If U and V are non-zero

vectors the angle between
them can be calculated by:

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 Scalars and Vectors…
Vector Multiplication: Dot & Cross
Cross Product
This vector multiplication is important in calculating moment of a force about
an arbitrary point O. Let r is a position vector to the point of application of the
force at A. That means r=OA =rx i+ ry j + rz k
If the force is written in the form of vector i.e F=Fx i+ Fy j + Fz k
Then the moment about O is the cross product of position vector and force
vector, i.e Mo=r X F
✓ Cross product is not commutative i.e r X F=- F X r

𝒊 𝒋 𝒌
𝐌𝒐 = 𝒓𝑿𝑭 = 𝐫𝐱 𝐫𝐲 𝐫𝐳
𝐅𝐱 𝐅𝐲 𝐅𝐳
Mo=(ry*Fz-rz*Fy)i –(rx*Fz - rz*Fx)j + (rx*Fy - ry*Fx)k
Mo =Mx i+ Myj + Mz k

Where Mx , My ,& Mz ,are the scalar component of the

moment. The norm or magnitude of the moment can be
calculated
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by as follows:
/𝑴𝒐/= 𝑴𝟐 𝒙 + 𝑴𝟐 𝒚 + 𝑴𝟐 𝒛 12
 SCALARS AND VECTORS…
Exercise
1. a) Determine the magnitude of the vector sum V=V1 +V2 and the
angle θx which V makes with the positive x-axis.
b) write V as a vector in terms of the unit vectors i and j
c) Determine the unit vector of V.
d) determine the magnitude of the vector difference V`=V2 - V1 and
the angle θx which V` makes with the positive x-axis.

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 SCALARS AND VECTORS…
Solution

1. a) Required: V=V1 +V2 and the angle θx which V makes with +ve x-axis.

Given: y
B Angle analysis
✓ The angle which V1 makes with +ve x-axis
=tan-1(3/4) =36.9ᶱ
✓ The angle b/n V1 and V2 = θ =120 -36.9
=83.1ᶱ AND
✓ β- θ =180-83.1=96.9ᶱ
A
α
θ 83.1ᶱ
60ᶱ 36.9ᶱ
x
O
1. Using Parallelogram law of vector addition see figure above,
The diagonal drawn is the vector sum, V=V1 +V2 .
• Take Triangle OAB. Magnitude of V, obtained by Using law of cosines
/V/2= /V1 /2+/V2/2 - 2/V1/ */V2/cos(96.9ᶱ)
/V/2= 102+122 – 2*10 *12*cos(96.9ᶱ) , /V/=16.52 units…Answer
/𝑽/ /𝑽𝟐/ 𝟏𝟔.𝟓𝟐 𝟏𝟐
Then using law of sine
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= w/c is = 14
𝑺𝒊𝒏(𝟗𝟔.𝟗) 𝑺𝒊𝒏(𝜶) 𝑺𝒊𝒏(𝟗𝟔.𝟗) 𝑺𝒊𝒏(𝜶)
Solving α, α=46.1 Thus θx=46.1+36.9=83ᶱ from positive x-axis …Answer
 SCALARS AND VECTORS…
Solution
1. b) Required: Vector V in terms of i and j
V=16.52*[cos(83) i+ cos(90-83) j]
=2 i + 16.34 j (units)…Ans.

c) unit vector of V, nV= [2 i + 16.34 j]/16.52

= 0.12 i + 0.99 j ….Answer

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 SCALARS AND VECTORS…
Solution
d) Required: V`=V2 - V1 See the diagrammatic representation shown below.

36.9ᶱ
α x
O
83.1ᶱ

B
❖ Take Triangle OAB, Using law of cosines of triangles
i.e /V`/2= /V1 /2+/V2/2 - 2/V1/ */V2/cos(83.1ᶱ)
/V`/2= 102+122 – 2*10 *12*cos(83.1ᶱ) , /V`/=14.67 units…Answer
/𝑉`/ /𝑉2/ 𝟏𝟒.𝟔𝟕 𝟏𝟐
➢ Then using law of sine = → = ; α=54.3ᶱ
𝑆𝑖𝑛(83.1) 𝑆𝑖𝑛(α) 𝑺𝒊𝒏(𝟖𝟑.𝟏) 𝑺𝒊𝒏(𝜶)
Thus
Mubarek Z θx from negative x-axis=54.3 - 36.9=17.4ᶱ

θx from Positive x-axis=180-17.4=162.6ᶱ ….Answer 16