University of Engineering & Technology Peshawar
CE-117: Engineering Mechanics (Spring 2025)
Geometry ,Trigonometry & Vector Algebra used in Engineering
Mechanics
1
Lecture' Objectives
• To review the principles of Geometry ,Trigonometry & Algebra used
in Engineering Mechanics
• Applications of above learned principles to real life problems with
emphasis on Engineering
Applications of Trigonometry
in Real Life Problems
Thales of Miletus (circa 625–547 BC) is known as
the founder of Geometry. The legend is that he
calculated the height of the Great Pyramid of Giza
in Egypt using the theory make use of right angle
triangles for shadows of pyramid.
GPS
Applications of Trigonometry in Civil Engineering Problems
C
B
A
Width of river, AC=?. Distance between
A and B is known
Distance D or height of object H is
required to be calculated
Trigonometry used in Engineering Mechanics
Right angle triangles
For the angle θ , b and h are
base and perpendiculars.
Incase of the other angle β,
the apposite is true
Perpendicular and Base ?
Right Angle Triangle Applications
In designing a house an architect wishes to determine the
amount of overhang of a roof that it shades the entire
south wall at noon during the summer solstice when the
angle elevation of the sun is 81°. Minimally, how much
overhang should be provide for this purpose?
x
Solution
11 ft
θ
81o
Trigonometry used in Engineering Mechanics
Right angle triangle: Applications to real life
With the overhang found, how far will the shadow of the
overhang come down the wall at noon during the winter
solstice when the angle of elevation of the sun is 32°?1.74 ft
θ
x
Ans: 1.07 ft
11 ft
Trigonometry used in Engineering Mechanics
Law of Sines
Law of cosines
Geometry used in Engineering Mechanics
Geometry used in Engineering Mechanics
C
A
α
D
β
γ
B
Relation b/w α & β ?
Angle ACD= γ =180o-90o-α= 90o- α
Angle ABC= γ =180o-90o-α= 90o- α
θ = 90o- γ =90o-(90o-α)
Angle CBD= β= 90o- γ =90o-(90o-α)
 θ= α
 β= α
10
50o
40o
40o
x
x
50o
50o
55o
55o
x
x
30o
55o
θ
30o
Exercise 2.1
x=70o
x=60o
x=70o
x=70o
In order to resolve force F along n- and x-axes, calculate its angle with n and x –
axes.
y
n
θn
x
θx
60o
t
Calculate the angle which the force F makes with n –
axis
θn
30o
10o
30o
O
Exercise 2.2
(a) Calculate angle of force F with n-axis in order to
resolve the force F along n- and t-components, θ=30o,
β=10o
Ans: θn = 70o
(b) Calculate the angle which the force P exerted by
robot arm makes with axes passing through BC and
BC
Ans: θBC= 30o & θAB= 45o
Exercise 2.2
(c) Calculate the angle of cable AB with x and n –
axes
Ans: θx= 33.72o and θn= 29.98o
In order to calculate Moment due to a force about a moment center by direct method,
Perpendicular distance of moment center from line of action of force is required.
Calculate Perpendicular distance of bolt from line of action of the 250 N force applied
on wrench
𝜃
A
𝛼
O
O
15
A′
Where
Calculate Perpendicular distance of point O from line of action
of force F
𝐴𝐵 =
650 + 500 − 2 ∗ 650 ∗ 500 ∗ 𝐶𝑜𝑠60
𝜃
𝐴𝐵 = 589.5 mm
90 − 30 = 60
Using Law of Sines
𝐴𝐵
𝑂𝐴
=
Sin 60
Sin𝜃
589.5
500
=
Sin 60
Sin𝜃
⇒ 𝜃 = 47.3
dO = 650 Sin
650 Sin
o
Exercise 2.3
In order to calculate Moment due to a force about a moment center by direct method,
Perpendicular distance of the moment center from the line of action of force is
required. Calculate these distance in given problems
dA = 2.73 m
do = 4.35 m
dO = 7/8 Sin 50o
Exercise 2.3
θ=25o , β=55o
, OA=50 mm
dO = 6.43 ft
dO = 43.3 mm
dB = 6.33 m
Exercise 2.3
dO = 4.35 m
dO = 93.6 mm
Scalars and Vectors
Scalar. A scalar is physical quantity that can be completely specified by its
magnitude . Examples of scalar quantities include Length, Mass, Time, Volume,
Work, Energy , Power,….
Vector. A vector is any physical quantity that requires both a magnitude and a
direction for its complete description.
 Vector quantities must obey the parallelogram law of addition as described later
in this lecture
 Examples of vectors encountered in statics are force, Moment,.... therefore all the
vectors quantity in remaining part of this lecture could be force, bending moment
,torsion etc
Force as Vector quantity: Proof
 There is no mathematical proof that law of vectors addition can be applied to
forces.
 Applying law of vectors addition to forces is based on results of experiment
performed using Gravesand’s apparatus.
Gravesand’s apparatus
Vectors
 Vector is written as:
• V (computer format)
•
(Hand written format)
V
Magnitude of Vector V is written as:
• |V| or V, (i.e. lightface italic type) - computer format
• V - Hand written format
Classification of Vectors
Vectors representing physical quantities can be classified as Fixed , Free or Sliding
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.
B
B
P
σx
x x
x
𝜎 =
𝑃
𝐴
A
𝜎 =0
x
A
P
In case of deformable bodies, shifting the force
(actually acting A) to any other point, for
instance B in not allowed. what is the reason ??
Classification of Vectors
Sliding vector: A vector which 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.
B
Rigid Body
A
P
B
B
P
C
P
A
A
Free vector: Free vectors refers to a vector which is neither a point nor a line,
and something that can move freely around the space though it has a fixed
magnitude and fixed direction
For example, if a body moves without rotation,
then the movement or displacement of any point
in the body may be taken as a vector.
This vector describes equally well the direction
and magnitude of the displacement of every
point in the body. Thus, we may represent the
displacement of such a body by a free vector.
Parallelogram Law of Vector Addition
Vectors must obey the parallelogram law of combination.
This law states that two vectors V1 and V2, treated as free vectors, Fig. a, may be
replaced by their equivalent vector V, which is the diagonal of the parallelogram
formed by V1 and V2 as its two sides, as shown in Fig. b.
 This combination is called the vector sum, and is represented by the vector
equation
𝟏
𝟐
 The geometry of the parallelogram shows
that V ≠ V1 + V2.
Triangle Law of Vector Addition
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. c, to obtain the identical vector
sum V.
V1
V2
V
(c)
We see from the diagram that the order of addition of the vectors does
not affect their sum, so that V1 +V2 = V2 + V1 (i.e. Vector addition is
commutative)
Components of vector
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 Figure are the components of V in the directions
1 and 2, respectively
Orthogonal (Rectangular) components of a vector
When the components of a vector are mutually perpendicular; these are called
Rectangular components
When expressed in rectangular components, the direction of the vector with
respect to, say, the x-axis is clearly specified by the angle ,where
Unit vector
A vector V may be expressed mathematically by multiplying its magnitude V by a
vector n whose magnitude is one and whose direction coincides with that of V. The
Vy= Vy j
vector n is called a unit vector. Thus, V= V n
Vx= Vx i
Where i and j are vectors in the x- and y-directions, respectively, with unit
magnitudes.
&
For the Force vectors F1 and F2 shown in the figure,
a) Determine the resultant of magnitude R of their vector sum R
= F1+ F2
𝐹 = 4 kN
b) Determine the angle between R and the positive x-axis
c) write R as a vector in terms of the unit vectors i and j and then
write a unit vector n along the vector sum R
d) Determine the vector difference D = F1 - F2
𝐹
C
Solution
𝜃 = 45o+30o = 75o
𝐹 = 3 kN
B
45o
O
A 𝐹
30o
𝛽
B
R
𝜃
O
F2=4 kN
30o
180o − 75o=105o
x
𝜃 = 75o
A
(a)
𝑅=
(b)
𝐹
+𝐹
− 2 ∗ 𝐹 ∗ 𝐹 ∗ Cos 105 =
Using Law of Sines
𝜃 = 𝛽 − 30
3 + 4 − 2 ∗ 3 ∗ 4 ∗ Cos 105
4
𝑆 = 5.59
=
Sin𝛽 Sin 105
⇒ 𝛽 = 43.76
⇒ 𝜃 = 43.76 − 30 = 13.76
⇒ 𝑅 = 5.59 kN
(c)
(d)
𝐅𝟏
𝐹 = 4 kN
−𝐹
−𝐅𝟐
45o
30o
𝐹 = 4 kN
𝟏
𝐹 = 3 kN
𝐹 = 3 kN
𝟐
𝟏
𝟐
Answer: θx = 70.4o
2.4b
Above
prob
Exercise 2.4
2.4a
For the Force vectors V1 and V2 shown in the figure,
a) Determine the magnitude V of their vector sum V = V1+
V2 and the angle between V and the positive x-axis
b) write a unit vector n along the vector sum V
c) Determine the vector difference V′ = V1 - V2 and the
angle between V′ and the positive x-axis
Ans: (a) 16.51 units , 36.9o (b) 0.8i-0.6j (c) 14.67 units ,
162.6o
Problems for Home Assignment
Exercise: 2.2 (a), 2.2 (b), 2.2 (c)
Exercise: 2.3 (b)
Exercise: 2.3 (d)
θ=25o ,
β=55o ,
OA=50
mm
dO = 43.3 mm
Exercise: 2.4
dO = 4.35 m