Engineering Mechanics - Intro to Statics.pdf

1. Engineering Mechanics Is the branch of science, which deals with study of system of forces and their
effects on bodies in motion or at rest.
Thus, the body under consideration will be in motion or at rest. Depending on the
state of the body (in motion or at rest) Engineering Mechanics is further divided
as:
 Statics: Branch of Engineering Mechanics, which deals with the study of
system of forces and their effects on bodies at rest.
 Dynamics: Branch of Engineering Mechanics, which deals with the study of
system of forces and their effects on bodies in motion.
According to Newton’s second law of motion, the motion of a body will depend on
its mass and some external force acting on it. Depending on this, Dynamics is
further divided into:
 Kinematics: a branch of Dynamics, which deals with the study of system of
forces and their effects on bodies in motion, without considering the mass of
the body and the force causing the motion.
 Kinetics: a branch of Dynamics, which deals with the study of system of forces
and their effects on bodies in motion, considering the mass of the body and
the force causing the motion.
 Particle: The point mass (definitely small mass) which possesses negligible
size, shape, mass and volume.
 Rigid Body: The body, which do not undergo any change in its dimensions
even after application of the force.
These are no perfect rigid bodies in the universe. Each body undergoes
negligible deformation, but as the deformation is negligible, the body can be
considered as rigid body.
Force Force is an external agency, which changes or tends to change the state of the
body i.e. force will tend to change the static state of a body to dynamic or vice
versa.
Force is a vector quantity, having ‘N’ (Newton) as S.I. unit.
 Unit Newton force: It is the force, which will produce an acceleration of 1
m/s2
to an object having mass 1 kg.
 Characteristics of a force:
1. Magnitude (unit) 2. Direction (angle)
3. Sense of nature (arrowhead) 4. Point of application.

2. LAW OF TRANSMISSIBILTY OF FORCE: “If a force acts at a point on a rigid body, it
is assumed to act at any other point on the line of action of the force within the
same body”
In fig (a), as per law of transmissibility of forces, the force ‘F’ can be assumed to
act at any other point on the line AB (without changing the magnitude and
direction), but within the limits of the body.
In fig. (b), initially, the forces are trying to create the tension, but after
transmitting these forces, they try to create the compression (fig. (c)). Hence, this
law is not applicable to equal, opposite and collinear forces.
** Limitation of Law of Transmissibility of Force: If two equal, opposite and
collinear forces are acting on the body, and if the forces are transmitted, then
the combined effect is different than the earlier.

3. System of Forces: When two or more forces act on a body, it is the system of forces.
The line, along which the force is acting, is called “the line of action of a force”.
Depending on the nature of line of action of forces, the following are the
different systems of forces
 Coplanar Forces: Lines of action of all forces lie in a single plane.
 Non-Coplanar Forces: Lines of actions of all forces do not lie in a single plane.
 Concurrent forces: Lines of action of all forces are passing through a common
point.
 Non-concurrent forces: Lines of action of all forces are not passing through a
common point.
 Coplanar Concurrent Forces: Lines of action of all forces lie in a single plane
and passes through a common point.
 Coplanar Non-Concurrent Forces: Lines of action of all forces lie in a single
plane but do not pass through a common point.

4.  Non- Coplanar Concurrent Forces: Lines of action of all forces do not lie in a
single plane but passes through a common point.
 Non- Coplanar Non-Concurrent Forces: Lines of action of all forces do not lie
in a single plane and do not pass through a common point.
 Parallel Forces: Lines of action of all forces are parallel to each other.
 Like Parallel Forces: Lines of action of all forces are parallel to each other and
the all forces are acting in same direction.
 Unlike Parallel Forces: Lines of action of all forces are parallel to each other
but all forces are not acting in the same direction.
RESULTANT RESULTANT: Is the single force which will produce the same effect (in
magnitude and direction) as it is produced by number of forces acting together.
Thus, the resultant is a single force, which will give the same result as it is
produced by the number of forces acting on a body. Thus, we can replace all
the forces by the resultant. In other words, we can say that, the resultant is the
combined effect of all the forces acting on the body (in magnitude and
direction).
COMPOSITION
OF FORCES
The process of determining the resultant of number of forces acting
simultaneously on the body is known as the composition of forces.
It is the method of reducing the given force system to its equivalent simplest
system of single force or couple. Combining the forces of any given system is
known as the composition of forces.
HOW TO FIND THE
RESULTANT OF A
GIVEN FORCE SYSTEM
A. The Resultant of TWO coplanar concurrent forces:
 Triangle law of forces.
 Parallelogram law of forces.
B. The Resultant of THREE OR MORE coplanar concurrent forces:
 Polygon law of forces (Graphical method).
 Method of resolution and composition of forces.
C. The Resultant of THREE OR MORE non-concurrent forces:
 Analytical method by using Varignon’s theorem of moments.
 Graphical method.
TRIANGLE LAW OF FORCES: “If two or concurrent forces are represented by two
sides of a triangle taken in order, then the resultant of the two forces shall be
given by the third side of the triangle taken in opposite order”

5. PARALLELOGRAM LAW OF FORCES: “If two coplanar concurrent forces acting at
and away from the point and are represented by the two adjacent sides of a
parallelogram in magnitude and direction, then the resultant is given by the
diagonal of the parallelogram passing through the same common point”
ANALYTICAL SOLUTION FOR PARALLELOGRAM LAW OF FORCES:
Consider two forces ‘P’ and ‘Q’ acting at and away from point ‘A’ as shown in
the Figure below.
Let, the forces P and Q are represented by the two adjacent sides of a
parallelogram AD and AB respectively as shown in fig. Let, θ be the angle
between the force P and Q. Extend line AB and drop perpendicular from point C
on the extended line AB to meet at point E.
Consider Right angle triangle ACE,
AC2
= AE2
+ CE2
=(AB + BE)2
+ CE2
= AB2
+ BE2
+ 2.AB.BE + CE2
= AB2
+ BE2
+ CE2
+ 2.AB.BE ……1
Consider right angle triangle BCE,
BC2
= BE2
+ CE2
and BE = BC.Cos θ ……2
Putting BC2
= BE2
+ CE2
in equation (1), we get
AC2
= AB2
+ BC2
+ 2.AB.BE ……3
Putting BE = BC. Cos θ in equation (3),
AC2
= AB2
+ BC2
+ 2.AB. BC. Cos θ But, AB = P, BC = Q and AC =
R
Consider Right angle triangle ACE,
AC2
= AE2
+ CE2
=(AB + BE)2
+ CE2
= AB2
+ BE2
+ 2.AB.BE + CE2
= AB2
+ BE2
+ CE2
+ 2.AB.BE ……1
θ
θ α 
Q.Cos
Q.Sin

6. Methods for finding
the resultant force
(Cont.)
Consider right angle triangle BCE,
BC2
= BE2
+ CE2
and BE = BC.Cos θ ……2
Putting BC2
= BE2
+ CE2
in equation (1), we get
AC2
= AB2
+ BC2
+ 2.AB.BE ……3
Putting BE = BC. Cos θ in equation (3),
AC2
= AB2
+ BC2
+ 2.AB. BC. Cos θ
But, AB = P, BC = Q and AC = R




 Cos
.
Q
.
P
.
2
Q
P
R 2
2
In triangle ACE,
BE
AB
CE
AE
CE
tan



 But, CE = BC. Sin θ






Cos
.
Q
P
Sin
.
Q
tan
POLYGON LAW OF FORCES: “If ’n ’is the number of coplanar concurrent forces
and these are represented by ‘n’ sides of a polygon taken in order, then the
resultant of the given ‘n’ forces shall be given by the closing (n+1) th
side of the
polygon taken in reverse order”
If several COPLANAR FORCES are acting at a point simultaneously such that each
one of them can be represented in direction and magnitude by a side of a polygon,
taken in order, then the resultant is given by the closing side in the reverse order
Method of Resolution Resolution: The method of splitting a given force into its components, without
changing its effect on the body.
Generally, it is convenient to resolve the force along x and y axis. These
components along x and y axis are known as Orthogonal Components.

7. 









Cos
.
F
Fx
F
Fx
Cos
Sin
.
F
Fy
F
Fy
Sin
A force can also be resolved along the two directions which are not at right
angle to each other.
Let, F1 and F2 be the components of a force ‘F’ along the axes 1 and 2 as shown
in figure at an angles α and β with the direction of force ‘F’.
Complete the parallelogram OACB.
In triangle OAC, applying sine rule, we get
 
 
β
α
180
Sin
F
α
Sin
F
β
Sin
F 2
1




 
β
α
Sin
F
α
Sin
F
β
Sin
F 2
1



   
β
α
Sin
F.Sin
F
β
α
Sin
β
F.Sin
F 2
1





Sign Convention
Along x-axis: Towards right +ve
Towards left - ve
Along y-axis: Upward +ve
Downwards - ve
Steps to find the
Resultant of Concurrent
Forces
1. Resolve all the forces horizontally and find the algebraic sum of all the
horizontal components (Σ Fx). Here, Σ Fx represents the horizontal
component of the resultant.
2. Resolve all the forces vertically and find the algebraic sum of all the vertical
components (Σ Fy). Here, Σ Fy represents the vertical component of the
resultant.
3. Resultant ‘R’ is given by the equation,    2
2
R Fy
Fx 



4.
Let, α be the angle made by the resultant ‘R’ with the horizontal, then
Fx
Fy
tan




β
α
β
α (α+β)
180-(α+β)

8. Analytical Method
Using Varignon’s
Theorem of Moments
Moment of a Force: A force when act on a body produces turning effect is called as
a moment of a force.
Moment (M) of a force is equal to the product of the force (F) and the perpendicular
distance between the line of action of the force and the reference point (x).
x
.
F
M 

If the force is passing through the reference point, then the Moment (M) of a
force is zero.
VARIGNON’S THEOREM OF MOMENTS: “When the number of forces are acting on
a body, then the algebraic sum of moments of all the forces about any point is
equal to the moment of their resultant about the same point”. RA
FA M
M 


Consider, a rectangle ABCD, acted upon by three forces as shown in the figure. Let,
‘R’ be the resultant of the given force system. Let ‘x’ be the perpendicular distance
between the resultant ‘R’ and the point ‘A’.
Now, taking moments of all the forces about point ‘A’, (considering clockwise
moments +ve and anti-clockwise moments –ve) we get
3
3
2
2
1
1 .
.
. x
F
x
F
x
F
M A 



Now, moment of resultant about point ‘A’, x
R
MA .

Now, according to Varignon’s theorem of moments, RA
FA M
M 


R
x
.
F
x
.
F
x
.
F
x
x
.
R
x
.
F
x
.
F
x
.
F
3
3
2
2
1
1
3
3
2
2
1
1








Thus, using Varignon’s theorem of moments, we can find out the position of the

9. resultant w.r.t. to the reference point.
Proof for Varignon’s
Theorem
Consider the forces P1 and P2 acting at point ‘A’ as shown in fig. a. Let, ‘R’ be the
resultant of these forces. Let d,d1 and d2 be the perpendicular distances of R, P1
and P2 respectively from point ‘B’.
Now, according to Varignon’s theorem, we have to prove that,
R.d = P1. d1 + P2.d2
Join AB and consider it as y-axis and draw x-axis perpendicular to it as shown in fig.
b.
Let ‘θ’ be the angle made by the ‘R’ with x-axis, hence angle made by perpendicular
to ‘R’ at point B with AB is also ‘θ’. From figure, we can write
R.d = R. AB Cos θ = AB (R.Cosθ)
R.d = AB.RX --- (1)
Where, RX is the component of R along x-axis
Similarly, if P1X and P2X are the components of P1 and P2 along the x-axis, then:
P1.d1 = AB.P1X --- (2)
P2.d2 = AB.P2X --- (3)
From equations (1) and (2),
P1.d1 + P2.d2 = AB (P1X + P2X) --- (4)
But, the sum of x components of individual forces is
equal to x component of the resultant, we can write,
(P1X + P2X) = RX --- (5)
From equations (4) and (5),
P1.d1 + P2.d2 = AB.RX --- (6)
From equations (1) and (6), R.d = P1.d1 + P2.d2
Steps To Find Resultant
of Non-Concurrent
Forces
1. Resolve all the forces horizontally and find the algebraic sum of all the horizontal
components (Σ Fx). Here, Σ Fx represents the horizontal component of the
resultant.
2. Resolve all the forces vertically and find the algebraic sum of all the vertical

10. components (Σ Fy). Here, Σ Fy represents the vertical component of the resultant.
3. Resultant ‘R’ is given by the equation,    2
2
R Fy
Fx 



4. Let, α be the angle made by the resultant ‘R’ with the horizontal, then
Fx
Fy
tan




5. Select some reference point (say point A) and take moments of all the forces
about that point using correct sign conventions.
6. Find algebraic sum of moments of all forces about the reference point ( A
M
 )
7. Let, ‘x’ be the perpendicular distance between the resultant and the reference
point. Take moment of the resultant about the reference point (MRA).
8. Apply Varignon’s theorem of moments to find the position (i.e. distance x) of the
resultant about the reference point.
Note:
1. If A
M
 is + ve, then the resultant will produce clockwise moment about the
reference point.
2. If A
M
 is - ve, then the resultant will produce anti- clockwise moment about
the reference point.
COUPLE Two equal, opposite and parallel forces acting on a body form a couple.
As the two forces are equal and opposite, their resultant is zero, but they will
produce rotational effect i.e. they will create the moment.
Properties of a Couple:
1. Couple can be replaced by a single force.
2. Couple cannot produce translatory motion (straight line motion) but it can
produce rotation.
3. Moment of a couple = force x lever arm.
4. The moment of a couple is independent of the position of moment centre and it
remains constant.
e.g. from figure, ΣMA = P x a
ΣMO = (P x OB) – ( P x OA) = P ( OB – OA)
ΣMO = P x a
PART - 1: THEORETICAL QUESTIONS
1 Explain the concept of Particle and Rigid body.

11. 2 Write a short note on ‘Systems of Units’.
3 How do you deal with a non-concurrent, non-parallel force system in equilibrium?
4 Write a short note on Analytical and Graphical method
5 State and prove Varignon’s Theorem OR State the Varignon’s Principle.
6 State and explain ‘Law of Transmissibility of Force’
7 Define and explain ‘Moment of force’
8 Differentiate between:
1. Concurrent and non-concurrent force system
2. Coplanar and non-coplanar force system
9 State and prove the Law of Parallelogram of Forces
10 The resultant of a system of parallel forces is zero. What does it signify?
11 Explain the terms ‘Particle’ and ‘Moment’ of a force
12 Write a note on ‘Composition and Resolution of forces’
13 A Force ‘F’ is to be resolved in two components P & Q such that the line of action of P & Q
makes angle α and β respectively with the line of action of ‘F’. Derive the general expressions
for P & Q.
14 State how the resultant of three or more co-planar and non-concurrent forces is calculated
15 State & Explain: Triangle Law of Forces and Polygon Law of Forces
16 Define moment of a couple and prove that moment of couple is always constant in the plane
of couple
17 Define a force system. Name the different force systems
18 Define a ‘Force’. State the characteristics of a force
19 What do you mean by Resultant and Equilibriant force
20 Differentiate between Resolution and Composition of forces
21 Enlist the different types of 2-D force systems with single line characteristics of the system
that can be studied in the system of forces.
22 State and explain the following laws of forces:
1. Law of Parallelogram of forces
2. Law of Polygon of forces