Engineering Mechanics
for
First Year B.E. Degree Students
COURSE
CONTENT IN BRIEF
1.
Introduction.
2.
Resultant of concurrent and non

concurrent coplanar forces.
3.
Equilibrium of concurrent and non

concurrent coplanar forces.
4.
Analysis of plane trusses.
5.
Friction.
6.
Centroid and Moment of Inertia.
7.
Resultant and Equilibrium of concurrent non

coplanar forces.
8.
Rectilinear and Projectile motion.
9.
Newton’s second law, D’Alembert’s principle, banking and super
elevation.
10.
Work, Energy, and Power.
11.
Impulse

Momentum principle.
Books
for
Reference
1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.
2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.
3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition
4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.
Definition of Mechanics :
In its broadest sense the term ‘Mechanics’ may be defined
as the ‘Science which describes and predicts the conditions
of rest or motion of bodies under the action of forces’.
CHAPTER
–
I
INTRODUCTION
This Course on Engineering Mechanics comprises of
Mechanics of Rigid bodies and the sub

divisions that come
under it.
1
Engineering
Mechanics
Mechanics of Solids
Mechanics of Fluids
Rigid Bodies
Deformable
Bodies
Statics
Dynamics
Kinematics
Kinetics
Strength of
Materials
Theory of
Elasticity
Theory of
Plasticity
Ideal
Fluids
Viscous
Fluids
Compressible
Fluids
Branches of Mechanics
2
Fundamental Concepts and Axioms
Rigid body :
It
is
defined
as
a
definite
amount
of
matter
the
parts
of
which
are
fixed
in
position
relative
to
one
another
.
Actually
solid
bodies
are
never
rigid
;
they
deform
under
the
action
of
applied
forces
.
In
those
cases
where
this
deformation
is
negligible
compared
to
the
size
of
the
body,
the
body
may
be
considered
to
be
rigid
.
3
Particle
A
body
whose
dimensions
are
negligible
when
compared
to
the
distances
involved
in
the
discussion
of
its
motion
is
called
a
‘Particle’
.
For
example,
while
studying
the
motion
of
sun
and
earth,
they
are
considered
as
particles
since
their
dimensions
are
small
when
compared
with
the
distance
between
them
.
4
Space
The
concept
of
space
is
associated
with
the
notion
of
the
position
of
a
point,
defined
using
a
frame
of
reference,
with
respect
to
which
the
position
of
the
point
is
fixed
through
three
measures
specific
to
the
frame
of
reference
.
These
three
measures
are
known
as
the
co

ordinates
of
the
point,
in
that
particular
frame
of
reference
.
x
y
z
5
Mass
:
It
is
a
measure
of
the
quantity
of
matter
contained
in
a
body
.
It
may
also
be
treated
as
a
measure
of
inertia,
or
resistance
to
change
the
state
of
rest,
or
of
uniform
motion
along
a
straight
line,
of
a
body
.
Two
bodies
of
the
same
mass
will
be
attracted
by
the
earth
in
the
same
manner
.
Continuum
:
A
particle
can
be
divided
into
molecules,
atoms,
etc
.
It
is
not
feasible
to
solve
any
engineering
problem
by
treating
a
body
as
a
conglomeration
of
such
discrete
particles
.
A
body
is
assumed
to
be
made
up
of
a
continuous
distribution
of
matter
.
This
concept
is
called
‘Continuum’
.
6
Force
It
is
that
agent
which
causes
or
tends
to
cause,
changes
or
tends
to
change
the
state
of
rest
or
of
motion
of
a
mass
.
A
force
is
fully
defined
only
when
the
following
four
characteristics
are
known
:
(i)
Magnitude
(ii)
Direction
(iii)
Point
of
application
and
(iv)
Line
of
action
.
7
Scalars and Vectors
A
quantity
is
said
to
be
a
‘scalar’
if
it
is
completely
defined
by
its
magnitude
alone
.
Example
:
Length,
Area,
and
Time
.
Whereas
a
quantity
is
said
to
be
a
‘vector’
if
it
is
completely
defined
only
when
its
magnitude
and
direction
are
specified
.
Example
:
Force,
Velocity,
and
Acceleration
.
8
Classification of force system
Force system
Coplanar Forces
Non

Coplanar Forces
Concurrent
Non

concurrent
Concurrent
Non

concurrent
A force that can replace a set of forces, in a force system,
and cause the same ‘external effect’ is called the
Resultant.
(
More detailed discussion on Resultant will follow in Chapter 2
)
Like
parallel
Unlike parallel
Like
parallel
Unlike parallel
9
Axioms of Mechanics
(1)
Parallelogram
law
of
forces
:
It
is
stated
as
follows
:
‘If
two
forces
acting
at
a
point
are
represented
in
magnitude
and
direction
by
the
two
adjacent
sides
of
a
parallelogram,
then
the
resultant
of
these
two
forces
is
represented
in
magnitude
and
direction
by
the
diagonal
of
the
parallelogram
passing
through
the
same
point
.
’
B
C
A
O
P
2
P
1
R
10
Contd..
In
the
above
figure,
P
1
and
P
2
,
represented
by
the
sides
OA
and
OB
have
R
as
their
resultant
represented
by
the
diagonal
OC
of
the
parallelogram
OACB
.
B
C
A
O
P
2
P
1
R
It can be shown that the magnitude of the resultant is given by:
R =
P
1
2
+ P
2
2
+ 2P
1
P
2
Cos
α
Inclination of the resultant w.r.t. the force P
1
is given by:
= tan

1
[( P
2
Sin
) / ( P
1
+ P
2
Cos
)]
11
Contd..
(
2
)
Principle
of
Transmissibility
:
It
is
stated
as
follows
:
‘The
external
effect
of
a
force
on
a
rigid
body
is
the
same
for
all
points
of
application
along
its
line
of
action’
.
P
A
B
P
For
example,
consider
the
above
figure
.
The
motion
of
the
block
will
be
the
same
if
a
force
of
magnitude
P
is
applied
as
a
push
at
A
or
as
a
pull
at
B
.
The
same
is
true
when
the
force
is
applied
at
a
point
O
.
P
P
O
12
(3) Newton’s Laws of motion:
(i)
First Law :
If
the
resultant
force
acting
on
a
particle
is
zero,
the
particle
will
remain
at
rest
(if
originally
at
rest)
or
will
move
with
constant
speed
in
a
straight
line
(if
originally
in
uniform
motion)
.
(ii) Second Law :
If
the
resultant
force
acting
on
a
particle
is
not
zero,
the
particle
will
have
an
acceleration
proportional
to
the
magnitude
of
the
resultant
and
in
the
direction
of
this
resultant
i
.
e
.
,
F
α
a
,or
F
=
m
.
a
,
where
F,
m,
and
a,
respectively
represent
the
resultant
force,
mass,
and
acceleration
of
the
particle
.
(iii) Third law:
The
forces
of
action
and
reaction
between
bodies
in
contact
have
the
same
magnitude,
same
line
of
action,
and
opposite
sense
.
13
Note :
1
.
‘Axioms’
are
nothing
but
principles
or
postulates
that
are
self
–
evident
facts
which
cannot
be
proved
mathematically
but
can
only
be
verified
experimentally
and/or
demonstrated
to
be
true
.
2
.
The
three
basic
quantities
of
mechanics
are
length,
time,
and
force
.
Throughout
this
Course
we
adopt
SI
units
and
therefore
they
are
expressed
in
meters,
seconds,
and
Newtons,
written
as
m,
s,
and
N
respectively
.
3.
The ‘external effect’ of a force on a body is manifest in a change in
the state of inertia of the body. While the ‘internal effect’ of a force on a
body is in the form of deformation.
14
RESULTANT OF CONCURRENT COPLANAR FORCES
CHAPTER
–
2
Y

Direction
X

Direction
F
3
F
1
R
F
x
F
y
F
2
In the above diagram F
1
, F
2
, F
3
form a system of concurrent
coplanar forces. If R is the resultant of the force system, then its
magnitude and direction are given by:
Composition of forces and Resolution of force
Resultant,
R
:
It
is
defined
as
that
single
force
which
can
replace
a
set
of
forces,
in
a
force
system,
and
cause
the
same
external
effect
.
15
Contd..
(i)
Magnitude, R =
(
F
x
)
2
+
(
F
y
)
2
(ii)
Direction,
θ
=
tan
–
1
(
F
y
/
F
x
)
,
where
:
ΣF
x
=
Algebraic
summation
of
x

components
of
all
individual
forces
.
ΣF
y
=
Algebraic
summation
of
y

components
of
all
individual
forces
.
θ
=
Angle
measured
to
the
resultant
w
.
r
.
t
.
x

direction
.
The
process
of
obtaining
the
resultant
of
a
given
force
system
is
called
‘Composition
of
forces’
.
Note
:
The
orientation
of
x

y
frame
of
reference
is
arbitrary
.
It
may
be
chosen
to
suit
a
particular
problem
.
16
Contd..
Component
of
a
force,
in
simple
terms,
is
the
effect
of
a
force
in
a
certain
direction
.
A
force
can
be
split
into
infinite
number
of
components
along
infinite
directions
.
Usually,
a
force
is
split
into
two
mutually
perpendicular
components,
one
along
the
x

direction
and
the
other
along
y

direction
(generally
horizontal
and
vertical,
respectively)
.
Such
components
that
are
mutually
perpendicular
are
called
‘Rectangular
Components’
.
Component of a force
:
F
y
Fig. 1
F
x
F
F
y
F
F
x
Fig. 2
Fig. 3
F
F
y
F
x
The
process
of
obtaining
the
components
of
a
force
is
called
‘Resolution
of
a
force’
.
17
The
adjacent
diagram
gives
the
sign
convention
for
force
components,
i
.
e
.
,
force
components
that
are
directed
along
positive
x

direction
are
taken
+ve
for
summation
along
the
x

direction
.
Sign Convention for force components:
+ve
+ve
x
x
y
y
Also force components that are directed along +ve y

direction
are taken +ve for summation along the y

direction.
When
the
components
of
a
force
are
not
mutually
perpendicular
they
are
called
‘Oblique
Components’
.
Consider
the
following
case
.
Oblique Components of a force
:
18
Contd..
taken
in
tip
to
tail
order,
the
third
side
of
the
triangle
represents
both
in
magnitude
and
direction
the
resultant
force
F,
the
sense
of
the
same
is
defined
by
its
tail
at
the
tail
of
the
first
force
and
its
tip
at
the
tip
of
the
second
force’
.
F
F
1
F
2
Let
F
1
and
F
2
be
the
oblique
components
of
a
force
F
.
The
components
F
1
and
F
2
can
be
found
using
the
‘triangle
law
of
forces’,
which
states
as
follows
:
‘If
two
forces
acting
at
a
point
can
be
represented
both
in
magnitude
and
direction,
by
the
two
sides
of
a
triangle
F
F
1
F
2
F
1
/ Sin
=
F
2
/ Sin
=
F / Sin(180


)
19
Contd..
Numerical
Problems & Solutions
∑ Fx = + 15 Cos 15
–
75
–
45 Sin 35
+ 60 Cos 40
∑ Fy = + 15 Sin 15 + 105
–
45 Cos 35
–
60 Sin 40
= + 33.453 kN
15
kN
15
0
105
kN
75 kN
45
kN
40
0
60 kN
35
0
Fig.1A
(1A)
+
ve
=

40.359 kN
=
40.359 kN
Obtain the resultant of the concurrent coplanar forces acting
as shown in Fig. 1A.
Solution
:
+ve
20
Contd..
R =
(
∑Fx )
2
+ (
∑Fy)
2
=
(

40.359)
2
+ (33.453)
2
Θ
= tan

1
(
∑Fy/
∑Fx)
Magnitude,R = 52.42 kN
Inclination,
Θ
= 39.69
o
15
kN
15
0
105 kN
75 kN
45 kN
40
0
60 kN
35
0
Fig.1A
(1A)
∑Fx
Θ
Answer
:
(w.r.t. X
–
direction)
∑Fy
R
21
Contd..
(1B)
50 kN
2
3
100 kN
α
26.31
o
75 kN
30
o
1
2
25kN
Obtain the resultant of the
concurrent coplanar forces acting
as shown in Fig. 1B.
∑Fx =

50 Cos 26.31

100 Cos33.69
–
25 Cos 63.43 + 75 Cos 30
+ ve
74.26kN

74.26kN =
75kN
120
2
3
30
1
2
Fig
.
1B
25kN
100kN
50kN
º
º
α
= tan

1
(2/3)=33.69
Solution:
β
º
β
= tan

1
(2/1)
=
63.43
º
22
Contd..
(1B)
50 kN
2
3
100 kN
α
26.31
o
75 kN
30
o
1
2
β
25kN
=

93.17kN
=
93.17kN
23
∑
F
Y
= 50sin26.31

100sin 33.69
–
75sin30
–
25sin63.43
ve
+
Contd..
Contd..
R =
(
∑Fx)
2
+ (∑Fy)
2
= 119.14 kN
Θ
= tan

1
(∑Fy /
∑Fx
)
= 51.44
o
∑Fx
∑
Fy
R
Θ
100
kN
50kN
26.31
o
75 kN
30
o
(1B)
25kN
33.69
º
63.43
º
Answers:
Contd..
24

Assume the fifth force F
5
in the
first quadrant, at an angle
α
, as
shown.

The 150 N force makes an angle
of 20
o
w.r.t. horizontal
(2)
150N
50N
200N
120N
45
°
50
°
110
º
F
5
α
R =250 N
20
º
A
system
of
concurrent
coplanar
forces
has
five
forces
of
which
only
four
are
shown
in
Fig
.
2
.
If
the
resultant
is
a
force
of
magnitude
R
=
250
N
acting
rightwards
along
the
horizontal,
find
the
unknown
fifth
force
.
Fig. 2
120N
150N
50N
200N
45
º
50
°
110
º
Solution:
25
Contd..
200 cos 50
–
150 cos 20
–
50
cos 45 +F
5
cos
α
= 250.
F
5
cos
α
= +297.75 N
+ve
∑
F
X
= R
150N
50N
200N
120N
45
°
50
°
110
º
F
5
α
R =250 N
20
º
26
Contd..
Contd..
∑F
Y
= 0.
F
5
sin
α
+ 200sin 50 + 150 sin 20
–
120 + 50 sin 45 = 0
F
5
sin
α
=

119.87N = 119.87N
α
= 21.90
º
119.87N
297.75N
F
5
= 320.97N
tan
α
= F
5
sin
α
/F
5
cos
α
=0.402
α
= 21.90
º
F
5
= 320.97N
F
5
cos
α
=
F
5
sin
α
=
+ve
Answers
Contd..
27
150N
50N
200N
120N
45
°
50
°
110
º
F
5
α
R =250 N
20
º
(3) A system of concurrent coplanar forces has four forces
of which only three are shown in Fig.3. If the resultant is a
force R = 100N acting as indicated, obtain the unknown
fourth force.
Fig. 3
60
°
45
°
R=100N
50N
25N
70
°
40
°
75N
28
Contd..

Assume the fourth force F
4
in the 1
st
quadrant, making an
angle
α
as shown
α
60
°
45
°
R=100N
50N
25N
70
°
40
°
75N
F
4
F
4
cos
α
+
75cos70
–
50cos45
–
25sin60 =

100cos40
Or, F
4
cos
α
=

45.25N ; or, F
4
cos
α
= 45.25N
Fx =

Rcos40
+ve
29
Contd..
60
°
45
°
R=100N
50N
25N
70
°
40
°
75N
α
F
4
Fy =

Rsin40
F
4
sin
α
+ 75sin70+25cos60+50sin45 =

100sin40
F
4
sin
α
=

182.61N ; or, F
4
sin
α
= 182.61N
+ve
30
Contd..
Contd..
α
=
76.08
º
45.25N
F
4
=188.13N
182.61N
F
4
cos
α
=
F
4
sin
α
=
Answers:
= tan

1
(F
4
sin
/F
4
cos
)
= 76.08
º
& F
4
=188.13N
31
Contd..
.
(4) The resultant of a system of concurrent coplanar forces is a force
acting vertically upwards. Find the magnitude of the resultant, and the
force F
4
acting as shown in Fig. 4.
60
°
30
°
15 kN
5 kN
10 kN
70
°
45
°
F
4
Fig. 4
32
Contd..
F
4
sin70
–
10cos 60
–
15cos 45
–
5cos 30 = 0; or, F
4
sin70 = 19.94
∑Fx = 0
60
°
30
°
15 kN
5 kN
10 kN
70
°
45
°
F
4
Fig. 4
Solution:
+ve
R
33
F
4
= 21.22kN
Contd..
Contd..
F
4
cos70 + 10sin60
–
15sin45 + 5sin30 = +R
+R

0.342F
4
= 0.554
Substituting for F
4
,
R= +7.81kN
∑
Fy = +R
+ve
Solution:
60
°
30
°
15 kN
5 kN
10 kN
70
°
45
°
F
4
Fig. 4
R
Answers:
F
4
= 21.22 kN
R= +7.81kN
34
Contd..
(5) Obtain the magnitudes of the forces P and Q if the resultant of the
system shown in Fig. 5 is zero
.
40
°
60
°
P
50N
Q
70
°
45
°
Fig. 5
100N
35
Contd..
Contd..
40
°
60
°
P
50N
Q
70
°
45
°
Fig
. 5
100N
36
Contd..
For R to be = zero,
∑F
x
= 0 and ∑ F
y
= 0
∑Fx = 0 :

Psin45
–
Qcos40 + 100cos70 + 50cos60 = 0
Or, 0.707P + 0.766Q = 59.2
+ve
40
°
60
°
P
50N
Q
70
°
45
°
Fig
. 5
100N
∑Fy = 0

Pcos45 + Qsin40 + 100sin70
–
50sin60 = 0
or,

0.707P + 0.642Q =

50.67
+ve
Answers:
(b)
Solving (a) & (b)
P = 77.17 N & Q = 6.058N
37
Contd..
30
°
100N
50N
Fig
.
6
(6) Forces of magnitude 50N and 100N are the oblique components of
a force F. Obtain the magnitude and direction of the force F. Refer
Fig.6.
38
Contd..
Rotating the axes to have X parallel to 50N,
∑
Fx = +50 + 100cos30
=
+136.6N
∑
Fy = +100sin30 = +50N
+ve
+ve
30
°
100N
(6)
X

AXIS
Y

AXIS
30
°
100N
50N
Fig
.
6
50N
39
Contd..
Contd..
F = 145.46N
θ
= 20.1
º
w r t X direction (50N force)
50N
F=
(∑Fx)
2
+(
∑
Fy)
2
= tan

1
[(∑Fx)
2
+(∑Fy)
2
]
Fig
.
6
X

AXIS
F
θ
Y

AXIS
30
°
100N
(6)
X

AXIS
Y

AXIS
30
°
100N
50N
50N
θ
40
Contd..
(7)
Resolve the 3kN force along the directions P and Q. Refer Fig. 7.
P
Q
3kN
45
°
60
°
30
°
Fig. 7
41
Contd..
3kN
Fig. 7
Move the force P parallel to itself to complete a triangle. Using
sine rule,
P/sin45 = Q/sin90 = 3/sin45
Answer :
P = 3kN, and Q = 4.243kN
P
45
º
Q
45º
Q
60
°
30
°
X
–
Axis
P
3kN
45
º
42
Contd..
EXERCISE
PROBLEMS
1
.
A
body
of
negligible
weight,
subjected
to
two
forces
F
1
=
1200
N,
and
F
2
=
400
N
acting
along
the
vertical,
and
the
horizontal
respectively,
is
shown
in
Fig
.
1
.
Find
the
component
of
each
force
parallel,
and
perpendicular
to
the
plane
.
Ans : F
1X
=

720 N, F
1Y
=

960N, F
2X
= 320N, F
2Y
=

240N
FIG. 1
=
1200
N
3
4
Y
F
2
F
1
=
400
N
43
2. Determine the X and Y components of each of the forces shown in
(Ans : F
1X
= 259.81 N, F
1Y
=

150 N, F
2X
=

150N, F
2Y
= 360 N,
F
3X
=

306.42 N, F
3Y
=

257.12N )
30
º
40
º
12
5
300 N
390 N
400 N
X
Y
F
1
=
F
2
=
F
3
=
FIG. 2
FIG.2.
44
45
600
N
200
N
800
N
20
º
40
º
30
º
FIG
.
3
3
.
Obtain
the
resultant
of
the
concurrent
coplanar
forces
shown
in
FIG
.
3
(Ans
:
R
=
522
.
67
N,
θ
=
68
.
43
º)
4
.
A
disabled
ship
is
pulled
by
means
of
two
tug
boats
as
shown
in
FIG
.
4
.
If
the
resultant
of
the
two
forces
T
1
and
T
2
exerted
by
the
ropes
is
a
300
N
force
acting
parallel
to
the
X
–
direction,
find
:
(a)
Force
exerted
by
each
of
the
tug
boats
knowing
α
=
30
º
.
(b)
The
value
of
α
such
that
the
force
of
tugboat
2
is
minimum,
while
that
of
1
acts
in
the
same
direction
.
Find
the
corresponding
force
to
be
exerted
by
tugboat
2
.
(
Ans
:
a
.
T
1
=
195
.
81
N,
T
2
=
133
.
94
N
b
.
α
=
70
º
,
T
1
=
281
.
91
N,
T
2
(min)
=
102
.
61
N
)
T
2
R
=
300
N
T
1
α
20
º
FIG
.
4
X

direction
46
5
.
An
automobile
which
is
disabled
is
pulled
by
two
ropes
as
shown
in
FIG
.
5
.
Find
the
force
P
and
resultant
R,
such
that
R
is
directed
as
shown
in
the
figure
.
P
Q
=
5
kN
R
20
º
40
º
FIG
.
5
(Ans
:
P
=
9
.
4
kN
,
R
=
12
.
66
kN)
47
6
.
A
collar,
which
may
slide
on
a
vertical
rod,
is
subjected
to
three
forces
as
shown
in
FIG
.
6
.
The
direction
of
the
force
F
may
be
varied
.
Determine
the
direction
of
the
force
F,
so
that
resultant
of
the
three
forces
is
horizontal,
knowing
that
the
magnitude
of
F
is
equal
to
(a)
2400
N,
(b)
1400
N
(
Ans
:
a
.
θ
=
41
.
81
º
;
b
.
The
resultant
cannot
be
horizontal
.
)
1200 N
800 N
60
º
θ
F
ROD
COLLAR
FIG.6
48
7. Determine the angle
α
and the magnitude of the force Q such that
the resultant of the three forces on the pole is vertically downwards
and of magnitude 12 kN.
Refer Fig. 7.
8kN
5kN
Q
30
º
α
Fig. 7
(Ans:
α
= 10.7
º, Q = 9.479 kN
)
49
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