INTRODUCTION TO ENGINEERING MECHANICS

INTRODUCTION TO ENGINEERING MECHANICS
ENGINEERING
MECHANICS
KOVVURI VINAY KUMAR
REDDY
Assistant Professor

Contents
2
ENGINEERING MECHANICS KOVVURI VINAY KUMAR REDDY, Assistant Professor
Introduction to Engineering Mechanics
• Composition and resolution of forces
• parallelogram law
• principle of transmissibility
• types of force systems - concurrent and concurrent coplanar forces
• resultant of coplanar force systems couple
• moment of a force Varignon's theorem
• concept of free body diagrams
• concept of equilibrium of coplanar force systems.

ENGINEERING MECHANICS KOVVURI VINAY KUMAR REDDY, Assistant Professor 3
Mechanics is that branch of science which
deals with the behaviour of a body when the
body is at root or in motion.
AUTOMOBILES, ELECTRIC MOTORS, ROBOTS, TELEVISION, MOBILE
PHONE and etc.,

Mechanics
Solid
Mechanics
Rigid Bodies
Static Dynamics
Kinematics Kinetics
Flexible
Bodies
Strength of
Materials
Theory of
Plasticity
Theory of
Elasticity
Fluid
Mechanics
Ideal Fluid
Viscous Fluid
Compressible
Fluid

Static
•Statics deals with the forces on a body at rest.
Dynamics
•Dynamics deals with the forces acting on a body when the body is in motion.
•(a) Kinematics: Deals the motion of a body without considering the forces
causing the motion.
•(b) Kinetics: Deals with the relation between the forces acting on the body
and the resulting motion

Rigid body: The rigid body means the body does not
deform under the action of force.
Note : Engineering Mechanics deals with Rigid body
Dynamics.
Particle: It is an object with its mass concentrated at a
point

IMPORTANCE OF MECHANICS TO ENGINEERING:
1) For designing and manufacturing of various mechanical
tools and equipment's
2) For calculation and estimation of forces of bodies
while they are in use.
3) For designing and construing to dams, roads, sheds,
structure, building etc.
4) For designing a fabrication of rockets.

Force: force is defined as an agency which changes or
tends to change the body at rest or in motion . Force is a
vector quantity. So we have to specify the magnitude,
direction and point of action. The unit of force is Newton.
1 N = 1 kgm
𝑠2

Units and dimensions
The following units are used mostly,
1. Centimetre-Gram Second system of unit.(C.G.S)
2. Metre-kilogram-second system of units. (M.K.S)
3. International system of units. (S.I)

FUNDAMENTAL UNITS
The measurement of physical quantities is one of the most important
operations in engineering. Every quantity is measured in terms of some
arbitrary, but internationally accepted units, called fundamental units.
All the physical quantities, met with in Engineering Mechanics, are
expressed in terms of three fundamental quantities, i.e.
1. length, 2. mass and 3. time.
DERIVED UNITS
Sometimes, the units are also expressed in other units (which are derived
from fundamental units) known as derived units e.g. units of area, velocity,
acceleration, pressure etc.

S.I UNITS
Density 𝑘𝑔/𝑚3 Angular velocity rad/s
Force N Angular acceleration rad/s2
Pressure Pa(pascal) or 𝑁/𝑚2
Work / energy 1𝐽 = 1𝑁 − 𝑚
Power 1W= 1 𝐽
𝑆
Frequency Hz (Hertz)
Acceleration m/s2

Laws of Mechanics
 Newton's laws of mechanics
1. Newton's first law of motion
2. Newton's second law of motion
3. Newton's third law of motion
 Newton's law of gravitational forces
 Law of transmissibility of forces
 Law of superposition of forces
 Parellogram law of forces

1. Newton’s First Law of Motion: It
states, “Every body continues in its state
of rest or of uniform motion in a straight
line, unless acted upon by some external
force”. This is also known as Law of
Inertia.
2. Newton’s Second Law of Motion: It
states, “The rate of change of momentum
is directly proportional to the impressed
force and takes place in the same
direction in which the force acts”.
3. Newton’s Third Law of Motion: It
states, “To every action, there is always
an equal and opposite reaction”.

Law of Gravitation:
It states that two bodies will be attracted towards each other along their
connecting line with a force which is directly proportional to the product of
their masses and inversely proportional to the square of the distance
between the centres.

Principle of transmissibility of forces
It states that “ if a force, acting at a point on a rigid body, is shifted to any
other point which is on the line of action of the force, the external effect of
the force on the body remains unchanged.
For example a force F is acting at point A on
a rigid body along the line of action AB. At
point B, apply two equal and opposite
forces F1 and F2 such that F1 and F2 are
collinear and equal in magnitude with F.
Now, we can transfer F1from B to A such
that F and F1 are equal and opposite and
accordingly they cancel each other. The net
result is force F2 at B. This implies that a
force acting at any point on a body may
also be considered to act at any other point
along its line of action without changing
the equilibrium of the body.

SYSTEM OF FORCES

RESULTANT FORCE
If a number of forces, P, Q, R ... etc. are acting simultaneously on a particle,
then it is possible to find out a single force which could replace them i.e.,
which would produce the same effect as produced by all the given forces.
This single force is called resultant force and the given forces R ... etc. are
called component forces.
COMPOSITION OF FORCES
The process of finding out the resultant force, of a number of given forces,
is called composition of forces or compounding of forces.
METHODS FOR THE RESULTANT FORCE
Though there are many methods for finding out the resultant force of a
number of given forces, yet the following are important from the subject
point of view :
1. Analytical method. 2. Method of resolution.

Characteristics of Force:

RESOLUTION OF A FORCE:
The process of splitting up the given force into a number of components,
without changing its effect on the body is called resolution of a force. A force
is, generally, resolved along two mutually perpendicular directions. In fact,
the resolution of a force is the reverse action of the addition of the
component vectors.
PRINCIPLE OF RESOLUTION
It states, “The algebraic sum of the resolved parts of a no. of forces, in a
given direction, is equal to the resolved part of their resultant in the same
direction.”
Note : In general, the forces are resolved in the vertical and horizontal
directions.

Aditya Engineering College (A)
25
ENGINEERING MECHANICS KOVVURI VINAY KUMAR REDDY, Assistant Professor
PARALLELOGRAM LAW OF FORCES:
It states, “If two forces, acting simultaneously on a particle, be represented in magnitude and direction
by the two adjacent sides of a parallelogram ; their resultant may be represented in magnitude and
direction by the diagonal of the parallelogram, which passes through their point of intersection.”
Mathematically, resultant force,

P and Q= Forces whose resultant is
required to be found out,
ɵ = Angle between the forces F1 and F2,
and
α = Angle which the resultant force makes
with one of the forces P.

Triangle Law of Force:

Lame‟s Theorem:
If three forces acting at a point are in equilibrium each force will be
proportional to the sine of angle between the other two forces.

1. Forces R, S, T, U are collinear. Forces S and T act from left to right. Forces R and U act
from right to left. Magnitudes of the forces R, S, T, U are 40 N, 45 N, 50 N and 55 N
respectively. Find the resultant of R, S,T, U.
PROBLEMS:
Given data:
R=40 N
S=45 N
T=50 N
U=55N Resultant = -R-U+S+T
= -40-55+45+50
= 0

PROBLEMS:
2. Find the resultant of the force system shown in Fig
Given data:
F1=20 KN ; Θ1=60°
F2=26 KN ; Θ2=0°
F3= 6KN ; Θ3=00°
F4=20KN ; Θ4=60°
Solution:
Resolve the given forces horizontally and calculate the algebraic total of all the
horizontal parts (Σ H) =-20cos60°+26cos0°-6cos0°-20cos60°=0
Resolve the given forces vertically and calculate the algebraic total of all the vertical
parts (Σ V)=-20sin60°+20 sin60°=0
R = √( (ΣH)^2+(ΣV)^2 )=0

PROBLEMS:
3. Resolve the 100N force acting a 30° to horizontal into two component one along
horizontal and other along 120° to horizontal.

PROBLEMS:

PROBLEMS:
4. A block of weight 1000 N kept on inclined plane is
pushed by horizontal force 500 N as shown in Fig. Find
the sum of components of forces along x-y axis oriented
parallel and perpendicular to the incline.

PROBLEMS:

PROBLEMS:
5. The following forces act at a point :
(i) 20 N inclined at 30° towards North of East,
(ii) 25 N towards North,
(iii) 30 N towards North West, and
(iv) 35 N inclined at 40° towards South of West.
Find the magnitude and direction of the
resultant force.

forces horizontally
ΣH = 20 cos 30° + 25 cos 90° + 30 cos 135° + 35 cos 220° N
= (20 × 0.866) + (25 × 0) + 30 (– 0.707) + 35 (– 0.766) N
= – 30.7 N
forces vertically
ΣV = 20 sin 30° + 25 sin 90° + 30 sin 135° + 35 sin 220° N
= (20 × 0.5) + (25 × 1.0) + (30 × 0.707) + 35 (– 0.6428) N
= 33.7 N
resultant force
R = √((ΣH)2 + (ΣV)2) = √((–30.7)2 + (33.7)2) = 45.6 N
Tan 𝜃 =
ΣV
ΣH
=
33.7
−30.7
=-1.098 => 𝜃=tan−1(−1.098)=47.7°
Since ΣH is negative and ΣV is positive, therefore resultant lies between 90° and 180°.
Thus actual angle of the resultant = 180° – 47.7° = 132.3°

The Free Body Diagram (F.B.D.) is a sketch of the body showing all active
and reactive forces that acts on it after removing all supports with
consideration of geometrical angles and distance given.

PROBLEMS:
6. Determine the magnitude and direction of the
resultant of forces acting on the hook shown In fig
Given data:
F1=250 N ; Θ1=35°
F2=200 N ; Θ2=20°
F3=110 N
F4=90 N ; Θ4=65°
Solution:
Resolve the given forces horizontally and calculate the algebraic total of all the
horizontal parts Σ H=250cos35°+200cos30°-90cos65°=339.957N
Resolve the given forces vertically and calculate the algebraic total of all the vertical
parts Σ V=250sin35°-200sin20°-110sin90°+90sin65=46.55N
R = √( (ΣH)2+(Σ V)2 )=343.129N
Θ=tan¯( ΣV/ΣH)=7.79°

Static equilibrium
A planar structural system is in a state of static equilibrium when
the resultant of all forces and all moments is equal to zero, i.e.

PROBLEMS:
7. Rope AB shown in fig(1) is 4.5m long and is connected at two points A & B at the
Same level 4m apart. A load of 1500N is suspended from a point C on the rope at 1.5m
from A. What load connected at point D on the rope, 1m from B will be Necessary to
keep the position CD level.
α β
2m
X F B
W=?
1500 N
D
C
E
A
Y
T2
T1
1500 N
C
α T2
T3
W
C β

Given (Longest distance bet A&B) AB = AC+CD+DB
4.5= 1.5+CD+1
⇛ CD = 4.5-2.5 = 2m
Let us take AE=X, DF=CE=Y (to keep the position CD level), CD=EF
(shortest distance bet A&B) AB = AE+EF+FB
4 = X+2+FB
⇛ FB = 4-2-X =2-X
From ∆AEC ⇛ 𝑿𝟐 + 𝒀𝟐 = 𝟏. 𝟓𝟐
⇛ 𝑿𝟐 + 𝒀𝟐 = 𝟐. 𝟐𝟓 ……………….(i)
From ∆BFD ⇛ 𝑭𝑩𝟐
+ 𝑫𝑭𝟐
= 𝟏𝟐
⇛ 𝒀𝟐
= 𝟏 − (𝟐 − 𝑿)𝟐
⇛ 𝒀𝟐
= 𝟏 − 𝟒 + 𝟒𝑿 − 𝑿𝟐
=−𝟑 +𝟒𝑿 − 𝑿𝟐
…………(ii)
SUB (ii) in (i)
⇛ 𝑿𝟐
− 𝟑+𝟒𝑿 − 𝑿𝟐
= 𝟐. 𝟐𝟓
⇛ 4X= 𝟐. 𝟐𝟓 +𝟑
⇛ 4X= 𝟓. 𝟐𝟓
⇛ X= 𝟏. 𝟑𝟏𝟐𝟓𝒎
⇛ 𝒀𝟐 = 𝟏 − (𝟐 − 𝑿)𝟐 ⇛ Y= 𝟎. 𝟕𝟐𝟔𝟐𝒎
⇛FB =2-X=2-1.3125=0.6875m

PROBLEMS:
8. A sphere of mass “M” rests in a V-groove whose sides are inclined at angles α & β to
the horizontal. Another identical sphere of same mass “M” rests on the first sphere and
in contact with the side inclined at an angle α. Find the reaction RB on the lower sphere
at the point B.
C
α β
B
A W=Mg
W=Mg
D

α β
B
RB
A
RA
W=Mg
W=Mg
D
RD
RC
α
α
α β
β
β
α
β
RD
RC
α
β
α
β
RB-RC
RA
W=Mg
W=Mg

MOMENT OF A FORCE:
It is the turning effect produced by a force, on the body, on which it acts. The moment of
a force is equal to the product of the force and the perpendicular distance of the point,
about which the moment is required and the line of action of the force.
M = F × L (N × mm)
F = Force acting on the body, and
L = Perpendicular distance between the point, about which the moment is required
and the line of action of the force.
 clockwise moment as positive
 Anticlockwise moment as negative.

VARIGNON’S PRINCIPLE OF MOMENTS (OR LAW OF MOMENTS)
It states, “If a number of coplanar forces are acting simultaneously on a particle, the
algebraic sum of the moments of all the forces about any point is equal to the moment of
their resultant force about the same point.”

Multiplying both sides by length OA
R x OA sin ɵ = P x OA sin ɵ1 + Q x OA sin ɵ2
But OA sin ɵ = d ,
OA sin ɵ1 = d1 and
OA sin ɵ2 = d2
=> R x d = P x d1 + Q x d2
Hence Proved.
Note: Varignon's theorem is used for determining the position of
resultant of parallel and general force system.

Resultant of Parallel Force System
+𝑣𝑒 ↻/−𝑣𝑒 ↺
+𝑣𝑒 → ⬆/−𝑣𝑒 ← ⬇

(i) ΣF(⬆) and ΣMO −𝑣𝑒 ↺
(ii) ΣF(⬇) and ΣMO −𝑣𝑒 ↺
(iii)ΣF(⬆) and ΣMO +𝑣𝑒 ↻
(iv)ΣF(⬇) and ΣMO(+𝑣𝑒 ↻)

PROBLEMS:
9. Find the resultant of given active forces Fig. (a) w.r.t. point B.
 𝑅 = Σ𝐹 = 100 −150 +200 = 150 N (→)
 Σ𝑀𝐵= 150 + 100 x 5 -150 x 3.5 +200 x 1.5 =
425 N-m (↻)
By Varignon's theorem, we have
 Σ𝑀𝐵= R x h
 ∴h = 2.83 m

PROBLEMS:
10. Replace the system of forces and couple shown in (Fig. a) by a single force couple
system at A.
 ΣFX= −100 cos 36.87° − 75 = −155 N = 155 N ( ← )
 ΣFY=−200 + 50 − 100 sin 36.87°= −210 N = 210 N ( ↓ )
 Magnitude of resultant R
R= (1552 + 2102) = 261N
 Inclination of the resultant ɵ
ɵ = tan−1 210
115
= 53.57°
 ΣMA= -50 x 2 -80 +100 sin 36.87° x 4 = 60 N-m (↻)
 The resultant of forces system is form of single force and
couple is as shown in (Fig.b) at A.

PROBLEMS:
11. A fixed square board EFGH carries two pulley and B which carry load of 20 N and 40 N respectively with
the help of cables fixed at point K and J as shown in (Fig.a). The diameter of each pulley is 400 mm. With
reference to xy-axis the coordinates oi centre of pulleys are A(1, 4) m and B(4, 1) m. Find (1) magnitude of
resultant force on the board and (ii) position r-axis intercept, y-axis intercept of the resultant force.

PROBLEMS:
 ΣFX= −20 cos 60° + 40 cos 30°=24.64 N( → )
 ΣFY=−20− 20 sin 60° − 40+40 sin 30° =−57.32N
=57.32N( ↓ )
 Magnitude of resultant R
R= (24.642 + 57.322) = 62.39 N
 Inclination of the resultant ɵ
ɵ = tan−1 57.32
24.64
= 66.74°
ΣMH= 20 x 1 -20 cos 60° x 4 +20 sin 60° x1 + 40 x 4
+ 40 cos 30° x l - 40 sin 30°x4
= 111.96 N-m (↻)
 By Varignon’s theorem, we have
X =
ΣMA
ΣFy
=
111.96
57.32
= 1.95 m
Y =
ΣMA
ΣFx
=
111.96
24.64
= 4.54 m
 Position of resultant w.r.t. point H (0,0) (Fig.c)

PROBLEMS:
12. A uniform beam AB hinged at A is kept horizontal by
supporting and setting a 50 kN weight with the help of a string
tied at B and passing over a smooth peg at C, as shown in
(Fig.a). The beam weight is 25 kN. Find the reaction at A and C.
 Consider the F.B.D. of BEAM AB
ΣMA= 0
(50 - T)2 +25 x 3 -Tsin 36.87° x 6 = 0
T = 31.25 KN
ΣFX= 0
HA-31.25 cos 36.87° = 0
HA= 25kN (→)
ΣFY= 0
VA-(50- 7) -25 +T sin 36.87° = 0
VA= 25kN (↑)
 Consider the F.B.D. of Block D
ΣFX= 0
R + T − 50 = 0 ⇛ 𝑅 = 50 − 𝑇
 Consider the F.B.D. of Peg C
ΣFX= 0
-HC+ 31.25 cos 36.87°= 0
HC= 25kN (←)
ΣFY= 0
VC-31.25 - 31.25 sin 36.87°= 0
VC= 50kN (↑)

PROBLEMS:
13. A bar AB pinned at A carries a load W= 5 kN. The cord
attached at B is passing through frictionless pulley as
shown in (Fig.a). Find the compression in the bar AB and
also the limiting value of the tension in cord when bar
approaches the vertical position.
ΣFY= 0
𝑅𝐴 sin 𝛳 +T sin 𝛽 - 5000 = 0
T sin 𝛽 = 5000 - 𝑅𝐴 sin 𝛳 .....................(i)
ΣFx= 0
𝑅𝐴 cos 𝛳 -T cos 𝛽 = 0
T cos 𝛽 = 𝑅𝐴 cos 𝛳 ..............................(ii)
Dividing Eq. (i) by Eq. (ii), we get
tan 𝛽 =
5000 − 𝑅𝐴 sin 𝛳
𝑅𝐴 cos 𝛳
.......................(iii)
Consider, ∆DBC
tan 𝛽 =
𝐶𝐷
𝐵𝐷
=
𝐴𝐶−𝐴𝐷
𝐵𝐷
................................(Iv)

PROBLEMS: From Eqs. (iii) and (iv), we get
𝐴𝐶−𝐴𝐷
𝐵𝐷
=
𝑅𝐴 cos 𝛳
⇛
2.5−2𝑠𝑖𝑛 𝛳
2𝑐𝑜𝑠 𝛳
=
𝑅𝐴 cos 𝛳
⇛2.5𝑅𝐴 − 2𝑅𝐴𝑠𝑖𝑛 𝛳 =10000 - 2𝑅𝐴 sin 𝛳
⇛ 𝑅𝐴 = 4000 N
While obtaining RA in the above expression, the
inclination of bar AB is not influencing the
solution. So we concluded that reaction at A
remains 4000 N for all values of 0 as shown in
(Fig.c).
Consider the F.B.D. of Bar AB in vertical position
By equilibrium condition
∑𝐹𝑌 =0 ⇛ 𝑅𝐴+T – 5000 = 0
⇛ T = 5000 - 4000
⇛ T = 1000N

INTRODUCTION TO ENGINEERING MECHANICS

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