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Prepared By
Prof. V. V. Nalawade
ENGINEERING MECHANICS
Prof. V.V.Nalawade 1
Module 1: Force System 8hrs
Module 2 : Equilibrium 7hrs
Module 3: Center of Gravity and Moment of Inertia 7 hrs.
Module 4: Friction 6hrs
Module 5 : Kinematics 6hrs
Module 6 : Kinetics 6hrs
C
u
r
r
i
c
u
l
u
m
2
Prof. V.V.Nalawade 3
Engineering Mechanics
• Dr. R.K. Bansal
• N.H. Dubey
• Beer & Johnston
• K. L. Kumar
• R. V. Kulkarni
Applied Mechanics
• R. S. Khurmi
• Sunil Deo
• R. K. Singer
Reference Books
Prof. V.V.Nalawade 4
CASIO
Fx-991MS
Prerequisite
Basic Concepts
Resolution &
composition of forces
Contents of the presentation
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Prof. V.V.Nalawade 6
Coordinate Geometry
Prof. V.V.Nalawade 7
II
Quadrant
(-,+)
I
Quadrant
(+,+)
III
Quadrant
(-,-)
IV
Quadrant
(+,-)
Slope of line
Prof. V.V.Nalawade 8
θ
(X2-X1)
(Y2-Y1)
P
Q (X2,Y2)
(X1,Y1)
X Axis
Y Axis
m = tan θ =
Few more to Recall……
Trigonometry
Pythagoras Theorem
Cosine & Sine rule
Sign conversion
Unit conversion
Prof. V.V.Nalawade 9
1.
Force
System
Prof. V.V.Nalawade 1
0
Prof. V.V.Nalawade 11
Sr.
No.
Topic Learning Objective
(TLO)
CO BL
CA
Code
1
To recall the basic principles of
mechanics
CO1 L1 1.2, 1.3
2
To describe the concepts on
mechanics and its practical
implementation
CO1 L2 1.1, 1.2
3
To identify the force system and
calculate the resultant of it
CO1 L3
1.2, 2.1,
12.2
4
To analyze the numerical of
different cases
CO1 L3 2.1, 12.2
Learning Outcome:
At the end of the topic the student should be able to:
Branches of Mechanics
Engineering
Mechanics
Statics
(Rest)
Dynamics
(Motion)
Kinetics Kinematics
EM is the branch of physics which deals with the study of forces and their
effect on body when body is at rest or in motion.
With Reference to the
Cause of motion
Without Reference to the
Cause of motion
Prof. V.V.Nalawade 1
2
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3
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4
Prof. V.V.Nalawade 15
We learn
ENGINEERING
MECHANICS
Prof. V.V.Nalawade 16
automobiles, aircrafts, electric motors, robots,
television, mobile , satellite, projectile of missiles,
launching of rockets, radar communication, trusses,
lifting machines like crane, hoist, screw jack,
elevator, conveyor belt, cargo ship, submarine, etc.
Prof. V.V.Nalawade 17
Laws of Mechanics
The following are the fundamental laws of mechanics:
(i) Newton’s first law
(ii) Newton’s second law
(iii) Newton’s third law
(iv) Newton’s gravitational law
(v) Law of transmissibility of forces
(vi) Parallelogram law of forces
Prof. V.V.Nalawade 1
8
Prof. V.V.Nalawade 1
9
Image source: https://issuu.com/daennagonzalez/docs/newton___s_laws_of_motion
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Because the
amount of
acceleration of a
body is
proportional to
the acting force
and inversely
proportional to
the mass of the
body
Prof. V.V.Nalawade 25
Prof. V.V.Nalawade 26
Balloon
goes up
Air goes
down
27
Prof. V.V.Nalawade
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1.What happens according to Newton if you let an
untied balloon go????
Prof. V.V.Nalawade 30
1.What happens according to Newton if you let an
untied balloon go????
Balloon
goes up
Air goes
down
3 rd Law
Air will rush out of the balloon
forcing the balloon to move
through the air in the opposite
direction, but equal in force.
Prof. V.V.Nalawade 31
2. Describe what happens if you are riding a skateboard
and hit something (like a curb) with the front wheels???
Prof. V.V.Nalawade 32
2. Describe what happens if you are riding a skateboard
and hit something (like a curb) with the front wheels???
1 st Law
Your body will keep moving forward and fly off your
skateboard since the curb only stops the board, not
yourself.
Prof. V.V.Nalawade 33
3. Describe why you hold your gun next to your
shoulder while deer hunting????
Prof. V.V.Nalawade 34
3. Describe why you hold your gun next to your
shoulder while deer hunting????
3 rd Law
When you pull the gun’s trigger, it
forces the bullet out of the gun, but
at the same time, the gun is forced
in the opposite direction of the
bullet (towards you). Your shoulder
is a new force that is introduced in
order to keep your gun from flying
away from you.
Prof. V.V.Nalawade 35
4. Why should we wear seatbelts – use one of Newton’s
Laws in your answer?
Prof. V.V.Nalawade 36
4. Why should we wear seatbelts – use one of Newton’s
Laws in your answer?
We should wear seatbelts so if we are in an accident our body
doesn’t keep moving at the same speed and in the same direction
that the car was going. A new force would be introduced to our
bodies (the seatbelt) in order to keep our bodies in place.
Prof. V.V.Nalawade 37
Prof. V.V.Nalawade 38
Newton’s third law would tell us that when the rocket
pushes out fire with a specific amount of force, the rocket
will move in the opposite direction, but with the same
amount of force. This is what causes the rocket to shoot
up into the air.
Prof. V.V.Nalawade 39
6. Explain how each of Newton’s laws affects a game of
Tug of War.
Prof. V.V.Nalawade 40
6. Explain how each of Newton’s laws affects a game of
Tug of War.
•First Law: The rope will stay in the same place until the tugging starts
(a new force is introduced)
•Second Law: We could measure a team’s force that they can pull the
rope with based on their body masses and the acceleration that they
are causing the rope to move at.
•Third Law: 1 team pulls the rope towards themselves with a certain
amount of force and the opposing team is also putting force on the rope.
The same amount of force is applied from the ground to the people as
they are putting on the ground.
41
Concept of force and its measurements
Prof. V.V.Nalawade
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43
1. Magnitude: 2. Direction :
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3. Point of application : 4. Sense or
Nature :
Prof. V.V.Nalawade
Prof. V.V.Nalawade 45
Questions:
1. Define Mechanics. What are the different branches of
mechanics?
2. What are the characteristics of force?
Prof. V.V.Nalawade 46
System of forces
System of Forces
Several forces acting simultaneously upon a body
Force System
Coplanar
Concurrent Collinear
Parallel
Like Unlike
Non Concurrent
& Non Parallel
(General)
Non-coplanar
Concurrent Collinear
Parallel
Like Unlike
Non Concurrent
& Non Parallel
(General)
Prof. V.V.Nalawade 4
7
Coplanar System of Forces 2D
4
8
Prof. V.V.Nalawade
Non-Coplanar System of Forces
4
9
Prof. V.V.Nalawade
Composition of forces
 Forces added to obtain a single force which produces
the same effect as the original system of forces.
 This single force is known as Resultant force.
 The process of finding the resultant force is called
composition of forces.
Prof. V.V.Nalawade 50
Composition of forces
 There are two methods of finding resultant
1. Analytical method
2. Graphical method
 Analytical methods are
 Parallelogram law &
 Method of Resolution
Prof. V.V.Nalawade 51
Prof. V.V.Nalawade 5
2
Type I: Problems on Composition of Forces by
Parallelogram and Triangle Law
Sin Sin Sin
R P Q
  
 
Where,
R = Resultant of force P & Q
θ = Angle Between P & R
β = Angle Between P & Q
α = Angle Between Q & R
1.1 Law Of Parallelogram:- 1.2 Triangle Law :-
Prof. V.V.Nalawade 5
3
Ex.1. Find the resultant of the following forces
• Solution : Method i) By Parallelogram Law
Prof. V.V.Nalawade 54
3 N
4N
3 N
4N
R
α
R = 5 N
Continue……..
Prof. V.V.Nalawade 55
• Solution : Mathod ii) By Triangle Law
3 N
4N
R
α
By Cosine rule
R = 5 N
By Sine rule
Prof. V.V.Nalawade 56
70 N
50 N
60ᵒ
70 N
50 N
α
R
α
70 N
50 N
120ᵒ 60ᵒ
R
Prof. V.V.Nalawade 57
Prof. V.V.Nalawade 58
500 N
50
300 N
500 N
300 N
50
Prof. V.V.Nalawade 59
Resolution of forces
Definition
Problems
Resolution of forces
• The way of representing a single force into number of
forces without changing the effect of the force on
the body is called as resolution of forces.
Prof. V.V.Nalawade 60
R
Fx
Fy
Prof. V.V.Nalawade 61
Fx
Fy
R = 10
= 90
Ex. 1. Two Forces act at an angle of 120°. The bigger force is of 40N and the resultant
is perpendicular to the smaller one. Find the smaller force.
Prof. V.V.Nalawade 6
2
F2 = 20 N
Ex. 2. Resolve the 100 N force acting a 30° to horizontal into two component one along
horizontal and other along 120° to horizontal.
Prof. V.V.Nalawade 6
3
Resolution of a force into two mutually perpendicular
components (Rectangular Components)
• Let a force F be inclined at an angle as shown in fig.
We have to resolve it into two mutually perpendicular
components Fx along X- Axis and Fy along Y- Axis.
Prof. V.V.Nalawade 64
A
B
O
• From point A on the line of action of
a force, draw perpendicular AB on
X-Axis.
• Now we have to calculate the
lengths OB & AB.
• Length OB represents the
magnitude of X component i.e. (Fx)
• & Similarly AB represents (Fy)
Prof. V.V.Nalawade 65
• In ∆ AOB,
OB = OA cos θ
But OA = F
Therefore, OB = F cos θ
Lets say OB = Fx, as it is the magnitude of
x-component
Hence, Fx = F cos θ
A
B
O
Fy
Fx
Prof. V.V.Nalawade 66
• In ∆ OBA,
AB = OA sin θ
But OA = F
Therefore, AB = F sin θ
Lets say AB = Fy, as it is the magnitude of y-
component
Hence, Fy = F sin θ
A
B
O
Fy
Fx
Resolution of a force into two non perpendicular
components (Oblique Components)
• A force can also be resolved along the two
directions which are not at right angles to
each other.
• In ∆OAC, Applying sine rule, we get
Prof. V.V.Nalawade 67
α
β
(α+β)
F1
F2
F1
F2
O
B C
A
F
F2
F1
α
β
Resolution of Force By Perpendicular component
cos
sin
x
y
F F
F F




1st Quad = Fx (+ve) & Fy (-ve)
2nd Quad = Fx (-ve) & Fy (+ve)
3rd Quad = Fx (-ve) & Fy (-ve)
4th Quad = Fx (+ve) & Fy (-ve)
Where,
Fx = Horizontal component of
Force
Fy = Vertical Component of
force
Prof. V.V.Nalawade 6
8
Q. 2 Find the Component of force 100 N passing
through the points (0,2) & (-1,2)
Prof. V.V.Nalawade 69
-1
1
-2
2
(-1,2) (0,2)
F = 100 N
-1
1
-2
2
(-1,2) (0,2)
F = 100 N
Θ = 0
cos
sin
x
y
F F
F F




Fx = 100N
Fy = 0
Θ = 180
Fx = -100N
Fy = 0
Resolution of Force By Non-Perpendicular component
1st Quad = Fx (+ve) & Fy (-ve)
2nd Quad = Fx (-ve) & Fy (+ve)
3rd Quad = Fx (-ve) & Fy (-ve)
4th Quad = Fx (+ve) & Fy (-ve)
Prof. V.V.Nalawade 7
0
F
F2
F1
α
β
Prof. V.V.Nalawade 71
60°
105°
330°
X- axis
2000 N
F1
F2
Prof. V.V.Nalawade 72
X- axis
F1
F2
F
30°
60°
Method of resolution
 STEPWISE PROCEDURE OF METHOD OF
RESOLUTION:
i. Resolve all forces horizontally and find the algebraic
sum of all the horizontal components (i.e., ΣFx)
ii. Resolve all forces vertically and find the algebraic sum
of all the vertical components (i.e., ΣFy).
iii. The resultant R of the given forces will be given by the
equation:
iv. The resultant force will be inclined at an angle θ, with
the horizontal, such that
v. Position of the resultant
Prof. V.V.Nalawade 73
1.4 Design steps of resolution of concurrent force system :-
Case I :- When magnitude and direction
of all forces in the force system is given
& resultant is to be determined
Step 1:- Find ΣFx (Horizontal
Component)
Step 2:- Find ΣFy (Vertical Component)
Step 3:- Find Magnitude of Resultant
Step 4:- Find Direction of Resultant
Step 5:- Find Position of Resultant
2 2
x y
R F F
  
1
tan y
x
F
F
 



Case II :- When resultant is horizontal
Σ Fx = R and Σ Fy = 0
Case III :- When resultant is Vertical
Σ Fx = 0 and Σ Fy = R
Case IV :- When resultant is Zero
Σ Fx = 0 and Σ Fy = 0
Case V :- When magnitude & Direction
of resultant is given & magnitude &
direction of any one force among the
force system is to be determined
Σ Fx = R cos θ and Σ Fy = R sin θ
θ is measured w.r.t X-axis
Prof. V.V.Nalawade 7
4
Prof. V.V.Nalawade 75
Prof. V.V.Nalawade 76
X-Axis
Y-Axis
155.8 N
76.6ᵒ
Prof. V.V.Nalawade 77
132.3ᵒ
45.6 N
X-Axis
Y-Axis
47.7ᵒ
Prof. V.V.Nalawade 78
X-Axis
Y-Axis
29.09 N
45.89ᵒ
Prof. V.V.Nalawade 79
132.3ᵒ
45.6 N
X-Axis
Y-Axis
47.7ᵒ
132.3ᵒ
45.6 N
X-Axis
Y-Axis
47.7ᵒ
X-Axis
Y-Axis
29.09 N
45.89ᵒ
Prof. V.V.Nalawade 80
X-Axis
Y-Axis
29.09 N
45.89ᵒ
Prof. V.V.Nalawade 81
Moment of force
Law of moments
Varignon’s Theorem
Couples
Problems
Moment of forces
• The rotational effect produced by force is
known as moment of force.
• It is equal to the magnitude of force
multiplied by the perpendicular distance of
the point from the line of action of the force.
• M = F x d
• Unit N.m , KN.m , N.mm etc.,
Prof. V.V.Nalawade 82
Sign Convention
Prof. V.V.Nalawade 83
Clockwise (+VE) Anti-Clockwise (-VE)
F
d
O
F
d
O
F
d O
F
d
O
M@O = F x d
Law of Moments
• It states that, “ In equilibrium when no of
coplanar forces act on a body, the sum of the
clockwise moments@ any point in their plane is
equal to the sum of the anticlockwise
moments @ the same point.
• Algebraic sum Clockwise moments = Algebraic
sum Anti-Clockwise moments @ same point
Prof. V.V.Nalawade 84
• Moment about pivot
• 500*2 = 1000*1
• 1000 = 1000
• Algebraic sum Clockwise moments = Algebraic
sum Anti-Clockwise moments @ same point
Prof. V.V.Nalawade 85
 To find beam
reaction
 To find forces in
frames
Prof. V.V.Nalawade 86
Use of Law of Moments
Varignon’s theorem of moments
• It status that, “ The algebraic sum of moments of all
forces about any point is equal to the moments of
their resultant about the same point.”
• Let ƩMFA = Algebraic sum of moments of all forces
about any point A
• ƩMRA = Moment of resultant force about same point A
• Then ƩMFA = ƩMRA
• i.e. F1.x1 + F2.x2 + F3.x3+……..+Fn.Xn = R.x
Prof. V.V.Nalawade 87
Use of Varignon’s theorem of moments
Prof. V.V.Nalawade 88
• This theorem is very useful in locating the
position of the resultant of non- concurrent
forces.
Examples of Moment of forces
• Rotation of door
• Tightening of nut by spanner
• Compass etc.,
Prof. V.V.Nalawade 89
Couple
• Two non-colinear, equal, unlike, parallel forces
forms a couple.
• As the forces are equal & opposite their
resultant is ZERO.
• Hence couple produces only rotary motion
without producing linear motion.
Prof. V.V.Nalawade 90
Prof. V.V.Nalawade 91
Examples of Couple
Rotation of Steering Wheel, key, tap etc.,
Lever Arm OR Arm of the Couple
• The distance between two forces of a couple
is known as lever arm.
• SI unit of couple is same as moment i.e. N.m,
N.mm, KN.m, KN.mm etc.,
Prof. V.V.Nalawade 92
a
P
P
Sign Convention
Prof. V.V.Nalawade 93
a
P
P
a
P
P
Clockwise (+VE) Anti-Clockwise (-VE)
Properties of Couple
• The resultant of the force of a couple is always
ZERO. i.e. R = P – P = 0
• The moment of couple is equal to the product
of one of the force and lever arm.
i.e. M = P x a
Prof. V.V.Nalawade 94
a
P
P
Properties of Couple
• The moment of couple about any point is
constant.
• Moment of couple = P x a
• Moment of couple @ C = P*AC – P*BC = P*a
• Moment of couple @ D = - P*AD + P*BD = P*a
Prof. V.V.Nalawade 95
a
P
P
D C
B
A
Properties of Couple
• A couple can be balanced only by another
couple of equal and opposite moment.
• Two or more couples are said to be equal
when they have same sense or moment
• Moment = 100 *1=100 N. m= 10*10=100N.m
= 50*2=100 N.m Prof. V.V.Nalawade 96
2 m
50 N
50 N
10 m
10 N
10 N
1 m
100 N
100 N
Properties of Couple
• Couple can only rotate the body but cannot
translate the body.
• A couple does not have moment centre, like
moment of force.
Prof. V.V.Nalawade 97
Prof. V.V.Nalawade 9
8
Prof. V.V.Nalawade 9
9
1.5 Moment of force:-
M= F x d
Unit = N.m or KN.m
Clockwise = +ve
Anticlockwise = -ve
Couple:-
M= F x d
Unit = N.m or KN.m
Clockwise = +ve
Anticlockwise = -ve
Varignon’s Theorem:-
Moment of resultant about any point A = Σ of moments
of all the forces about same point A.
R x d = Σ M
Prof. V.V.Nalawade 1
0
1.6 Resultant of parallel force
system:-
Step 1:- Find R = ΣF
Step 2:- Find ΣM @ O
Step 3:- Apply Varignon’s theorem
ΣM @ O = R. d
Step 4:- Find position of resultant
w.r.t. O
i) ΣF (upward) = Σmo (+ve)
ii) ΣF (downward) =Σmo (-ve)
1.7 Resultant of general force
system:-
Step 1:- Find ΣFx (Horizontal Component)
Step 2:- Find ΣFy (Vertical Component)
Step 3:- Find Magnitude of Resultant
Step 4:- Find Direction of Resultant
Step 5:- Find ΣM @O
Step 6:-:- Apply Varignon’s theorem
ΣM @ O = R. d
Step 4:- Find position of resultant w.r.t. O
2 2
x y
R F F
  
1
tan y
x
F
F
 



Prof. V.V.Nalawade 1
0
Method of approach to solve Coplanar (2D) problems
Problem
2-D
Concurrent
Not Equilibrium
(Resultant)
1. Parallelogram Law
2. Triangle law
3. Polygon law
4. Method of projections
Equilibrium
(Unknowns)
F =0
x
Fy=0
Non-concurrent
Not Equilibrium
(Resultant)
1.Choose a reference Point
2.Shift all the forces to a point
3.Find the resultant force and
couple at that point
4.Reduce the force-couple
system to a single force
Equilibrium
(Unknowns)
ΣFx=0
ΣFy=0
ΣMz=0
Prof. V.V.Nalawade 1
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Prof. V.V.Nalawade 1
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Problem – 2D- Concurrent - Resultant
Ex.1. Determine the resultant of the following figure
Prof. V.V.Nalawade 1
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Prof. V.V.Nalawade 1
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Ex.2. The resultant of the four concurrent forces as shown in
Fig acts along Y-axis and is equal to 300N.Determine the
forces P and Q.
Problem – 2D- Concurrent - Resultant
 F x
 0
9
F y
 R  300N
Prof. V.V.Nalawade 1
0
F x
 800380QSin45 Psin50  0
F y
 QCos45 PCos50  R  300
Solve quadratic equation and find
P = 511 N
Q = - 40.3N
Problem solution
 F x  0
Fy
 R  300N
Prof. V.V.Nalawade 1
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Prof. V.V.Nalawade 1
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Determine the resultant of the following figure
Problem – 2D- Non Concurrent - Resultant
Problem
Prof. V.V.Nalawade 1
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Resultant of General forces in a plane –
Coplanar non-concurrent
Step 2: Shift all the forces to a point
Step 3: Find the resultant force Step 4: Reduce resultant force
Step 1: Choose a reference point
Prof. V.V.Nalawade 1
1
Resultant – Non-concurrent general forces in a plane
Step:1: Choose A as reference Point Step:2: Shift all forces to point A
Step:4: Reduce it to a single force
Step 3: Find resultant force and couple
x = 1880/600
x = 3.13m
Problem solution:
Prof. V.V.Nalawade 1
1
Determine the resultant of the following figure
Problem – 2D- Non Concurrent - Resultant
Problem
Prof. V.V.Nalawade 1
1
Example:
Resultant – Non-concurrent general forces in a plane
Determine the resultant force of the non-concurrent forces as shown in
plate and distance of the resultant force from point O͛.
Step:2
Step:4
Step:3
Step:1
Problem solution
Prof. V.V.Nalawade 1
1
Continued……
Prof. V.V.Nalawade 1
1
Prof. V.V.Nalawade 1
1
Prof. V.V.Nalawade 1
1
Prof. V.V.Nalawade 1
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Exercise 2
Prof. V.V.Nalawade 1
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Prof. V.V.Nalawade 11
9
References
• Engg. Mechanics ,Timoshenko & Young.
• 2.Engg. Mechanics, R.K. Bansal , Laxmi publications
• 3.Engineering Mechanics,Fedinand.L.Singer , Harper – Collins.
• 4. Engineering Mechanics statics and dynamics, A Nelson, Mc Gra Hill
publications
• 5. Engg. Mechanics Umesh Regl, Tayal.
• 6. Engineering Mechanics by N H Dubey
• 7. Engineering Mechanics , statics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd.
• 8. Engineering Mechanics , dynamics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd.
• 9. Mechanics For Engineers , statics - F.P.Beer & E.R.Johnston 5th Edn Mc Graw
Hill Publ.
• 10.Mechanics For Engineers, dynamics - F.P.Beer & E.R.Johnston – 5th Edn Mc
Graw Hill Publ.
• 11. www.google.com
• 12. http://nptel.iitm.ac.in/
Prof. V.V.Nalawade 12
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Thank YOU

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Engineering Mechanics Ch 1 Force System.ppt

  • 1. Prepared By Prof. V. V. Nalawade ENGINEERING MECHANICS Prof. V.V.Nalawade 1
  • 2. Module 1: Force System 8hrs Module 2 : Equilibrium 7hrs Module 3: Center of Gravity and Moment of Inertia 7 hrs. Module 4: Friction 6hrs Module 5 : Kinematics 6hrs Module 6 : Kinetics 6hrs C u r r i c u l u m 2
  • 3. Prof. V.V.Nalawade 3 Engineering Mechanics • Dr. R.K. Bansal • N.H. Dubey • Beer & Johnston • K. L. Kumar • R. V. Kulkarni Applied Mechanics • R. S. Khurmi • Sunil Deo • R. K. Singer Reference Books
  • 5. Prerequisite Basic Concepts Resolution & composition of forces Contents of the presentation Prof. V.V.Nalawade 5
  • 7. Coordinate Geometry Prof. V.V.Nalawade 7 II Quadrant (-,+) I Quadrant (+,+) III Quadrant (-,-) IV Quadrant (+,-)
  • 8. Slope of line Prof. V.V.Nalawade 8 θ (X2-X1) (Y2-Y1) P Q (X2,Y2) (X1,Y1) X Axis Y Axis m = tan θ =
  • 9. Few more to Recall…… Trigonometry Pythagoras Theorem Cosine & Sine rule Sign conversion Unit conversion Prof. V.V.Nalawade 9
  • 11. Prof. V.V.Nalawade 11 Sr. No. Topic Learning Objective (TLO) CO BL CA Code 1 To recall the basic principles of mechanics CO1 L1 1.2, 1.3 2 To describe the concepts on mechanics and its practical implementation CO1 L2 1.1, 1.2 3 To identify the force system and calculate the resultant of it CO1 L3 1.2, 2.1, 12.2 4 To analyze the numerical of different cases CO1 L3 2.1, 12.2 Learning Outcome: At the end of the topic the student should be able to:
  • 12. Branches of Mechanics Engineering Mechanics Statics (Rest) Dynamics (Motion) Kinetics Kinematics EM is the branch of physics which deals with the study of forces and their effect on body when body is at rest or in motion. With Reference to the Cause of motion Without Reference to the Cause of motion Prof. V.V.Nalawade 1 2
  • 15. Prof. V.V.Nalawade 15 We learn ENGINEERING MECHANICS
  • 17. automobiles, aircrafts, electric motors, robots, television, mobile , satellite, projectile of missiles, launching of rockets, radar communication, trusses, lifting machines like crane, hoist, screw jack, elevator, conveyor belt, cargo ship, submarine, etc. Prof. V.V.Nalawade 17
  • 18. Laws of Mechanics The following are the fundamental laws of mechanics: (i) Newton’s first law (ii) Newton’s second law (iii) Newton’s third law (iv) Newton’s gravitational law (v) Law of transmissibility of forces (vi) Parallelogram law of forces Prof. V.V.Nalawade 1 8
  • 19. Prof. V.V.Nalawade 1 9 Image source: https://issuu.com/daennagonzalez/docs/newton___s_laws_of_motion
  • 24. Prof. V.V.Nalawade 24 Because the amount of acceleration of a body is proportional to the acting force and inversely proportional to the mass of the body
  • 29. Prof. V.V.Nalawade 29 1.What happens according to Newton if you let an untied balloon go????
  • 30. Prof. V.V.Nalawade 30 1.What happens according to Newton if you let an untied balloon go???? Balloon goes up Air goes down 3 rd Law Air will rush out of the balloon forcing the balloon to move through the air in the opposite direction, but equal in force.
  • 31. Prof. V.V.Nalawade 31 2. Describe what happens if you are riding a skateboard and hit something (like a curb) with the front wheels???
  • 32. Prof. V.V.Nalawade 32 2. Describe what happens if you are riding a skateboard and hit something (like a curb) with the front wheels??? 1 st Law Your body will keep moving forward and fly off your skateboard since the curb only stops the board, not yourself.
  • 33. Prof. V.V.Nalawade 33 3. Describe why you hold your gun next to your shoulder while deer hunting????
  • 34. Prof. V.V.Nalawade 34 3. Describe why you hold your gun next to your shoulder while deer hunting???? 3 rd Law When you pull the gun’s trigger, it forces the bullet out of the gun, but at the same time, the gun is forced in the opposite direction of the bullet (towards you). Your shoulder is a new force that is introduced in order to keep your gun from flying away from you.
  • 35. Prof. V.V.Nalawade 35 4. Why should we wear seatbelts – use one of Newton’s Laws in your answer?
  • 36. Prof. V.V.Nalawade 36 4. Why should we wear seatbelts – use one of Newton’s Laws in your answer? We should wear seatbelts so if we are in an accident our body doesn’t keep moving at the same speed and in the same direction that the car was going. A new force would be introduced to our bodies (the seatbelt) in order to keep our bodies in place.
  • 38. Prof. V.V.Nalawade 38 Newton’s third law would tell us that when the rocket pushes out fire with a specific amount of force, the rocket will move in the opposite direction, but with the same amount of force. This is what causes the rocket to shoot up into the air.
  • 39. Prof. V.V.Nalawade 39 6. Explain how each of Newton’s laws affects a game of Tug of War.
  • 40. Prof. V.V.Nalawade 40 6. Explain how each of Newton’s laws affects a game of Tug of War. •First Law: The rope will stay in the same place until the tugging starts (a new force is introduced) •Second Law: We could measure a team’s force that they can pull the rope with based on their body masses and the acceleration that they are causing the rope to move at. •Third Law: 1 team pulls the rope towards themselves with a certain amount of force and the opposing team is also putting force on the rope. The same amount of force is applied from the ground to the people as they are putting on the ground.
  • 41. 41 Concept of force and its measurements Prof. V.V.Nalawade
  • 43. 43 1. Magnitude: 2. Direction : Prof. V.V.Nalawade
  • 44. 44 3. Point of application : 4. Sense or Nature : Prof. V.V.Nalawade
  • 45. Prof. V.V.Nalawade 45 Questions: 1. Define Mechanics. What are the different branches of mechanics? 2. What are the characteristics of force?
  • 47. System of Forces Several forces acting simultaneously upon a body Force System Coplanar Concurrent Collinear Parallel Like Unlike Non Concurrent & Non Parallel (General) Non-coplanar Concurrent Collinear Parallel Like Unlike Non Concurrent & Non Parallel (General) Prof. V.V.Nalawade 4 7
  • 48. Coplanar System of Forces 2D 4 8 Prof. V.V.Nalawade
  • 49. Non-Coplanar System of Forces 4 9 Prof. V.V.Nalawade
  • 50. Composition of forces  Forces added to obtain a single force which produces the same effect as the original system of forces.  This single force is known as Resultant force.  The process of finding the resultant force is called composition of forces. Prof. V.V.Nalawade 50
  • 51. Composition of forces  There are two methods of finding resultant 1. Analytical method 2. Graphical method  Analytical methods are  Parallelogram law &  Method of Resolution Prof. V.V.Nalawade 51
  • 53. Type I: Problems on Composition of Forces by Parallelogram and Triangle Law Sin Sin Sin R P Q      Where, R = Resultant of force P & Q θ = Angle Between P & R β = Angle Between P & Q α = Angle Between Q & R 1.1 Law Of Parallelogram:- 1.2 Triangle Law :- Prof. V.V.Nalawade 5 3
  • 54. Ex.1. Find the resultant of the following forces • Solution : Method i) By Parallelogram Law Prof. V.V.Nalawade 54 3 N 4N 3 N 4N R α R = 5 N
  • 55. Continue…….. Prof. V.V.Nalawade 55 • Solution : Mathod ii) By Triangle Law 3 N 4N R α By Cosine rule R = 5 N By Sine rule
  • 56. Prof. V.V.Nalawade 56 70 N 50 N 60ᵒ 70 N 50 N α R α 70 N 50 N 120ᵒ 60ᵒ R
  • 58. Prof. V.V.Nalawade 58 500 N 50 300 N 500 N 300 N 50
  • 59. Prof. V.V.Nalawade 59 Resolution of forces Definition Problems
  • 60. Resolution of forces • The way of representing a single force into number of forces without changing the effect of the force on the body is called as resolution of forces. Prof. V.V.Nalawade 60 R Fx Fy
  • 62. Ex. 1. Two Forces act at an angle of 120°. The bigger force is of 40N and the resultant is perpendicular to the smaller one. Find the smaller force. Prof. V.V.Nalawade 6 2 F2 = 20 N
  • 63. Ex. 2. Resolve the 100 N force acting a 30° to horizontal into two component one along horizontal and other along 120° to horizontal. Prof. V.V.Nalawade 6 3
  • 64. Resolution of a force into two mutually perpendicular components (Rectangular Components) • Let a force F be inclined at an angle as shown in fig. We have to resolve it into two mutually perpendicular components Fx along X- Axis and Fy along Y- Axis. Prof. V.V.Nalawade 64 A B O • From point A on the line of action of a force, draw perpendicular AB on X-Axis. • Now we have to calculate the lengths OB & AB. • Length OB represents the magnitude of X component i.e. (Fx) • & Similarly AB represents (Fy)
  • 65. Prof. V.V.Nalawade 65 • In ∆ AOB, OB = OA cos θ But OA = F Therefore, OB = F cos θ Lets say OB = Fx, as it is the magnitude of x-component Hence, Fx = F cos θ A B O Fy Fx
  • 66. Prof. V.V.Nalawade 66 • In ∆ OBA, AB = OA sin θ But OA = F Therefore, AB = F sin θ Lets say AB = Fy, as it is the magnitude of y- component Hence, Fy = F sin θ A B O Fy Fx
  • 67. Resolution of a force into two non perpendicular components (Oblique Components) • A force can also be resolved along the two directions which are not at right angles to each other. • In ∆OAC, Applying sine rule, we get Prof. V.V.Nalawade 67 α β (α+β) F1 F2 F1 F2 O B C A F F2 F1 α β
  • 68. Resolution of Force By Perpendicular component cos sin x y F F F F     1st Quad = Fx (+ve) & Fy (-ve) 2nd Quad = Fx (-ve) & Fy (+ve) 3rd Quad = Fx (-ve) & Fy (-ve) 4th Quad = Fx (+ve) & Fy (-ve) Where, Fx = Horizontal component of Force Fy = Vertical Component of force Prof. V.V.Nalawade 6 8
  • 69. Q. 2 Find the Component of force 100 N passing through the points (0,2) & (-1,2) Prof. V.V.Nalawade 69 -1 1 -2 2 (-1,2) (0,2) F = 100 N -1 1 -2 2 (-1,2) (0,2) F = 100 N Θ = 0 cos sin x y F F F F     Fx = 100N Fy = 0 Θ = 180 Fx = -100N Fy = 0
  • 70. Resolution of Force By Non-Perpendicular component 1st Quad = Fx (+ve) & Fy (-ve) 2nd Quad = Fx (-ve) & Fy (+ve) 3rd Quad = Fx (-ve) & Fy (-ve) 4th Quad = Fx (+ve) & Fy (-ve) Prof. V.V.Nalawade 7 0 F F2 F1 α β
  • 72. Prof. V.V.Nalawade 72 X- axis F1 F2 F 30° 60°
  • 73. Method of resolution  STEPWISE PROCEDURE OF METHOD OF RESOLUTION: i. Resolve all forces horizontally and find the algebraic sum of all the horizontal components (i.e., ΣFx) ii. Resolve all forces vertically and find the algebraic sum of all the vertical components (i.e., ΣFy). iii. The resultant R of the given forces will be given by the equation: iv. The resultant force will be inclined at an angle θ, with the horizontal, such that v. Position of the resultant Prof. V.V.Nalawade 73
  • 74. 1.4 Design steps of resolution of concurrent force system :- Case I :- When magnitude and direction of all forces in the force system is given & resultant is to be determined Step 1:- Find ΣFx (Horizontal Component) Step 2:- Find ΣFy (Vertical Component) Step 3:- Find Magnitude of Resultant Step 4:- Find Direction of Resultant Step 5:- Find Position of Resultant 2 2 x y R F F    1 tan y x F F      Case II :- When resultant is horizontal Σ Fx = R and Σ Fy = 0 Case III :- When resultant is Vertical Σ Fx = 0 and Σ Fy = R Case IV :- When resultant is Zero Σ Fx = 0 and Σ Fy = 0 Case V :- When magnitude & Direction of resultant is given & magnitude & direction of any one force among the force system is to be determined Σ Fx = R cos θ and Σ Fy = R sin θ θ is measured w.r.t X-axis Prof. V.V.Nalawade 7 4
  • 77. Prof. V.V.Nalawade 77 132.3ᵒ 45.6 N X-Axis Y-Axis 47.7ᵒ
  • 79. Prof. V.V.Nalawade 79 132.3ᵒ 45.6 N X-Axis Y-Axis 47.7ᵒ 132.3ᵒ 45.6 N X-Axis Y-Axis 47.7ᵒ X-Axis Y-Axis 29.09 N 45.89ᵒ
  • 81. Prof. V.V.Nalawade 81 Moment of force Law of moments Varignon’s Theorem Couples Problems
  • 82. Moment of forces • The rotational effect produced by force is known as moment of force. • It is equal to the magnitude of force multiplied by the perpendicular distance of the point from the line of action of the force. • M = F x d • Unit N.m , KN.m , N.mm etc., Prof. V.V.Nalawade 82
  • 83. Sign Convention Prof. V.V.Nalawade 83 Clockwise (+VE) Anti-Clockwise (-VE) F d O F d O F d O F d O M@O = F x d
  • 84. Law of Moments • It states that, “ In equilibrium when no of coplanar forces act on a body, the sum of the clockwise moments@ any point in their plane is equal to the sum of the anticlockwise moments @ the same point. • Algebraic sum Clockwise moments = Algebraic sum Anti-Clockwise moments @ same point Prof. V.V.Nalawade 84
  • 85. • Moment about pivot • 500*2 = 1000*1 • 1000 = 1000 • Algebraic sum Clockwise moments = Algebraic sum Anti-Clockwise moments @ same point Prof. V.V.Nalawade 85
  • 86.  To find beam reaction  To find forces in frames Prof. V.V.Nalawade 86 Use of Law of Moments
  • 87. Varignon’s theorem of moments • It status that, “ The algebraic sum of moments of all forces about any point is equal to the moments of their resultant about the same point.” • Let ƩMFA = Algebraic sum of moments of all forces about any point A • ƩMRA = Moment of resultant force about same point A • Then ƩMFA = ƩMRA • i.e. F1.x1 + F2.x2 + F3.x3+……..+Fn.Xn = R.x Prof. V.V.Nalawade 87
  • 88. Use of Varignon’s theorem of moments Prof. V.V.Nalawade 88 • This theorem is very useful in locating the position of the resultant of non- concurrent forces.
  • 89. Examples of Moment of forces • Rotation of door • Tightening of nut by spanner • Compass etc., Prof. V.V.Nalawade 89
  • 90. Couple • Two non-colinear, equal, unlike, parallel forces forms a couple. • As the forces are equal & opposite their resultant is ZERO. • Hence couple produces only rotary motion without producing linear motion. Prof. V.V.Nalawade 90
  • 91. Prof. V.V.Nalawade 91 Examples of Couple Rotation of Steering Wheel, key, tap etc.,
  • 92. Lever Arm OR Arm of the Couple • The distance between two forces of a couple is known as lever arm. • SI unit of couple is same as moment i.e. N.m, N.mm, KN.m, KN.mm etc., Prof. V.V.Nalawade 92 a P P
  • 93. Sign Convention Prof. V.V.Nalawade 93 a P P a P P Clockwise (+VE) Anti-Clockwise (-VE)
  • 94. Properties of Couple • The resultant of the force of a couple is always ZERO. i.e. R = P – P = 0 • The moment of couple is equal to the product of one of the force and lever arm. i.e. M = P x a Prof. V.V.Nalawade 94 a P P
  • 95. Properties of Couple • The moment of couple about any point is constant. • Moment of couple = P x a • Moment of couple @ C = P*AC – P*BC = P*a • Moment of couple @ D = - P*AD + P*BD = P*a Prof. V.V.Nalawade 95 a P P D C B A
  • 96. Properties of Couple • A couple can be balanced only by another couple of equal and opposite moment. • Two or more couples are said to be equal when they have same sense or moment • Moment = 100 *1=100 N. m= 10*10=100N.m = 50*2=100 N.m Prof. V.V.Nalawade 96 2 m 50 N 50 N 10 m 10 N 10 N 1 m 100 N 100 N
  • 97. Properties of Couple • Couple can only rotate the body but cannot translate the body. • A couple does not have moment centre, like moment of force. Prof. V.V.Nalawade 97
  • 100. 1.5 Moment of force:- M= F x d Unit = N.m or KN.m Clockwise = +ve Anticlockwise = -ve Couple:- M= F x d Unit = N.m or KN.m Clockwise = +ve Anticlockwise = -ve Varignon’s Theorem:- Moment of resultant about any point A = Σ of moments of all the forces about same point A. R x d = Σ M Prof. V.V.Nalawade 1 0
  • 101. 1.6 Resultant of parallel force system:- Step 1:- Find R = ΣF Step 2:- Find ΣM @ O Step 3:- Apply Varignon’s theorem ΣM @ O = R. d Step 4:- Find position of resultant w.r.t. O i) ΣF (upward) = Σmo (+ve) ii) ΣF (downward) =Σmo (-ve) 1.7 Resultant of general force system:- Step 1:- Find ΣFx (Horizontal Component) Step 2:- Find ΣFy (Vertical Component) Step 3:- Find Magnitude of Resultant Step 4:- Find Direction of Resultant Step 5:- Find ΣM @O Step 6:-:- Apply Varignon’s theorem ΣM @ O = R. d Step 4:- Find position of resultant w.r.t. O 2 2 x y R F F    1 tan y x F F      Prof. V.V.Nalawade 1 0
  • 102. Method of approach to solve Coplanar (2D) problems Problem 2-D Concurrent Not Equilibrium (Resultant) 1. Parallelogram Law 2. Triangle law 3. Polygon law 4. Method of projections Equilibrium (Unknowns) F =0 x Fy=0 Non-concurrent Not Equilibrium (Resultant) 1.Choose a reference Point 2.Shift all the forces to a point 3.Find the resultant force and couple at that point 4.Reduce the force-couple system to a single force Equilibrium (Unknowns) ΣFx=0 ΣFy=0 ΣMz=0 Prof. V.V.Nalawade 1 0
  • 104. Problem – 2D- Concurrent - Resultant Ex.1. Determine the resultant of the following figure Prof. V.V.Nalawade 1 0
  • 106. Ex.2. The resultant of the four concurrent forces as shown in Fig acts along Y-axis and is equal to 300N.Determine the forces P and Q. Problem – 2D- Concurrent - Resultant  F x  0 9 F y  R  300N Prof. V.V.Nalawade 1 0
  • 107. F x  800380QSin45 Psin50  0 F y  QCos45 PCos50  R  300 Solve quadratic equation and find P = 511 N Q = - 40.3N Problem solution  F x  0 Fy  R  300N Prof. V.V.Nalawade 1 0
  • 109. Determine the resultant of the following figure Problem – 2D- Non Concurrent - Resultant Problem Prof. V.V.Nalawade 1 0
  • 110. Resultant of General forces in a plane – Coplanar non-concurrent Step 2: Shift all the forces to a point Step 3: Find the resultant force Step 4: Reduce resultant force Step 1: Choose a reference point Prof. V.V.Nalawade 1 1
  • 111. Resultant – Non-concurrent general forces in a plane Step:1: Choose A as reference Point Step:2: Shift all forces to point A Step:4: Reduce it to a single force Step 3: Find resultant force and couple x = 1880/600 x = 3.13m Problem solution: Prof. V.V.Nalawade 1 1
  • 112. Determine the resultant of the following figure Problem – 2D- Non Concurrent - Resultant Problem Prof. V.V.Nalawade 1 1
  • 113. Example: Resultant – Non-concurrent general forces in a plane Determine the resultant force of the non-concurrent forces as shown in plate and distance of the resultant force from point O͛. Step:2 Step:4 Step:3 Step:1 Problem solution Prof. V.V.Nalawade 1 1
  • 119. Prof. V.V.Nalawade 11 9 References • Engg. Mechanics ,Timoshenko & Young. • 2.Engg. Mechanics, R.K. Bansal , Laxmi publications • 3.Engineering Mechanics,Fedinand.L.Singer , Harper – Collins. • 4. Engineering Mechanics statics and dynamics, A Nelson, Mc Gra Hill publications • 5. Engg. Mechanics Umesh Regl, Tayal. • 6. Engineering Mechanics by N H Dubey • 7. Engineering Mechanics , statics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd. • 8. Engineering Mechanics , dynamics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd. • 9. Mechanics For Engineers , statics - F.P.Beer & E.R.Johnston 5th Edn Mc Graw Hill Publ. • 10.Mechanics For Engineers, dynamics - F.P.Beer & E.R.Johnston – 5th Edn Mc Graw Hill Publ. • 11. www.google.com • 12. http://nptel.iitm.ac.in/