Engineering Mechanics Pdf

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INTRODUCTION
In this chapter we will define the most basic concept in mechanics, the 'Force'
and know its various characteristics and effects. We shall further learn to
'break', technically known as 'resolution' of a force and to 'combine'
technically known as 'composition' of forces. We shall be studying the
different 'systems of forces' and finally we shall learn about 'couples' and their
properties.
FORCE
Force is defined as an agency which changes or tends to change the state of
rest or of uniform motion of a body. For example a man trying to push a
cupboard as shown in figure (a) exerts a force 'F'. If the force is sufficient to
cause motion, the cupboard will slide to the right and if not sufficient, the
force F only 'tends' to change the state of rest but not succeeding to cause
motion.
Similarly Fig. shows an arrangement to bring a rotating wheel to a stop by
applying force P to press the brakes.
Force is a vector quantity.
It is characterised by
1) Magnitude
2) Direction
3) Sense
4) Point of application
Coplanar Forces-Res & Comp of Forces

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For the person pushing the cupboard, refer Fig. (a) and (b) the man applies a
force of magnitude F, whose direction is horizontal, sense is to the right and
point of application is at C.
The S.L unit of force is Newton (N) or its multiples i.e. kilo-Newton (kN), where
1 kN = 1000 N
The unit of force has been named after Sir Isaac Newton and one Newton
force is defined as a force required producing an acceleration of 1 m/s2 in a
body of mass 1 kg.
RESOLUTION OF A FORCE
Resolution or resolving a force implies breaking the force into components,
such that the components combined together would have the same effect as
the original force. Fig. (a) Shows a force F acting at an angle 'θ'. This force can
be resolved into components F1 and F2 as shown in Fig. (b) or into
components Fx, and Fy, as shown in Fig. (c).
The components Fx and Fy, of the force F as shown in Fig. (c) are known as
the rectangular or perpendicular components of the force, since the two
components are perpendicular to each other.
Usually we require rectangular components of force and hence let us learn
how to find rectangular components of a force.

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Resolution of a Force into Rectangular Components
Consider a force of magnitude F acting at an angle
0 with the x-axis. Let the force be represented by a
line OA drawn to scale. From A drop a
perpendicular on the x-axis at E and on the y-axis
at P.
Length (OE) represents the magnitude of the
component along x-axis (Fx) with direction from O to E, while length (OP)
represents the magnitude of the component along y-axis ( F ) with direction
from O to P.
From Geometry Fx = F cos θ →
Fy = F sin θ ↑
These are the two rectangular components of force F.
SYSTEM OR FORCES
System of Forces tell us about how the forces are arranged. For example all
the forces may lie on one plane or may lie on different planes. They may meet
at a point, may be parallel to each other or may just neither be parallel, nor
may meet at a point. 'System' defines this arrangement of forces and is
classified as shown.

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Coplanar system
In this arrangement all the forces lie in one plane. Fig. shows a coplanar
system consisting of four forces F1, F2, F3 and F4 all lying in the 'xoy' plane.
A coplanar system is further sub-divided into
a) Concurrent system: In this system all the forces meet at a point.
Examples of concurrent system are,
i) A lamp hanging from two strings. If T1 and T2 are the forces developed
in the strings and W be the weight of the lamp, then we have a
concurrent system at a point A
ii)An electric pole supporting heavy electric cables. If F1, F2, F3 are the
forces in the cable, and W be the weight of the pole, then the forces
form a concurrent system.

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b) Parallel system: In this system the lines of action of the forces are
parallel.
For example;
i) A vegetable vendor weighing the vegetables,
the weight W1 of vegetables, the measured
Weight W2, the force F applied by hand to
hold the weighing scale, all three forces form
a Parallel system.
Refer Fig. (a).
ii)Persons sitting on a bench. Fig. (b) Shows three persons of weights P1,
P2 and P3 sitting on a bench of self-weight W. If R1 and R2 are the
reactions offered by the ground then the six forces form a parallel
system.
c) General system: Also known as 'non-
concurrent and non-parallel' system has
forces which do not meet at a single point,
nor are parallel to each other. For example
the forces acting on the rectangular plate form a general system.
Refer Fig.
Non-coplanar System
When the forces acting in a system do not lie in a single plane, they are
termed as Non-coplanar forces or Space forces. Refer Fig.

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PRINCIPLE OF TRANSMISSIBILITY OF FORCE
It states “A force being a sliding vector continues to act along its line of action
and therefore makes no change if it acts from a different point on its line of
action on a rigid body”. Consider a rigid body as shown in Fig. acted upon by
a force of magnitude F acting at A. The effect on the body would remain
unchanged if it acted from point B, C or any other point on its line of action.
A practical example explaining Principle of Transmissibility, (Refer fig.(b)) is
the case of a locomotive 'L' pulling the wagons „W' to the right by exerting
force F from the front.
This force gets transmitted to all the wagons and they move forward. The
same effect is observed if the locomotive pushes the
wagons from behind.
Again the force F is transmitted to ail the wagons and
they move forward.
Transmissibility of force introduces the concept of the force as a Sliding
Vector when it acts on rigid body. The force can be shown anywhere on its
line of action, causing no change in the force system.
MOMENT OF A FORCE
We all are aware that a force can cause a body to slide, but at the same time
it can cause a body to rotate also. The rotational effect of a force is known as
the moment of the force. When we talk of the rotational effect, it has to be
with respect to a point. The concerned point is known as the moment centre.
The rotational effect of the same force will vary from one moment centre to
another and of course if the point (moment centre) lies on the line of action of
the force, the moment of force about the point would be zero. The rotational

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effect or moment is measured as the product of the force and the
perpendicular distance from the moment centre to the force. This
perpendicular distance is known as the moment arm „d‟.
M= F×d …… 2.1
The tendency to rotate could be either clockwise or anti-clockwise.
Fig. shows a force of magnitude F acting on a rectangular plate ABCD as
shown.
The moment of F
about A = F × d1 (anti-clockwise) = + (F × d1)
about B = F × d2 (clockwise) = - (F × d2)
about C = F × d3 (clockwise) = - (F × d3)
about D = O
Units of moment are N-m or N-mm or kN-m
Sign : we shall take anti-clockwise moments as positive moments, and
Convention: clockwise moments as negative. This shall be indicated as +
ve.
VARIGNON’S THEOREM
Varignon, a French mathematician (1654 - 1722) established that the sum of
the moments of a concurrent system of forces about any point is equal to the
moment of the resultant of the concurrent system about the same point.
Though originally derived for a concurrent system of forces, this theorem can
in fact be applied to any system of forces and is thus stated as "the algebraic
sum of the moments of a system of coplanar forces about and point in the
plane is equal to the moment of the resultant force of the system about the
same point.

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Mathematically it is written as F R
A AM = M .2.2
Sum of moments of all forces = Moment of the resultant about the same
about any point, say point A. point A.
Proof - Let P and O be two concurrent forces
at O, making angle 1 and 2 with the x-axis,
1et R be their resultant making an angle 0
with x-axis.
Let A be a point on the y-axis about which we
shall find the moments of P and Q and also
of the resultant R. Let d1, d2 and d be the
moment arm of P, Q and R from moment
centre A.
Let the x component of forces P, Q and R be Px , Qx and Rx respectively
Now, Moment of P about A = AM P
= P × d1 …..(1)
Moment of Q about A = AM Q
= Q × d2 ……(2)
Moment of R about A = R
AM = R × d
= R (OA cos θ)
= OA (Rx) …..(3)
Adding equations (1) and (2) we have
AM P
+ AM Q
= P d1 + Q d2
Or sum of moments F
AM = (P × OA cos θ1) + ( Q × OA cos θ2)
= OA . Px + OA. Qx since Px = P cos θ1
and Qx = Q cos
θ2
= OA ( Px + Qx )
F
A xM =OA (R ).-----(4)  Px + Qx = Rx since the resultant of
forces in the „x‟ direction is the sum
of components of
Forces in the „x‟ direction
Comparing equation (4) with (3)
F R
A AM =M -----Proved
The above equation can similarly be extended for more than two forces in
the system.

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COMPOSITION OF FORCES (RESULTANT OF FORCES)
Composition means to combine the forces acting in a system into a single
force, which has the same effect as the number of forces acting together.
Such a. single force is known as resultant of the system. Finding the
'resultant' helps to analyse the effect of the forces on the system and may
form an important step in the solution of engineering problems. We shall
learn to find the resultant of
(a) Concurrent system of two forces using law of parallelogram of forces.
(b) Concurrent system of more than two forces using method of resolution.
(c) Parallel system of forces.
(d) General system of forces.
Types of Resultant
Resultant of a force system may either be a (i) a force or (ii) a couple or (iii) a
force couple. Let us understand the three possibilities of a resultant in detail.
1. Resultant- Force
In a given force system consisting of number of forces, if
on composition of these forces it results in a single force,
we call such a resultant as 'Resultant-Force'. Refer Fig. a
2. Resultant – Couple
In a force system if the resultant force is zero but the
resultant moment is not zero, such a system reduces into
a couple. Fig. b shows the 'couple' resultant. Resultant-
Couple is possible in a Parallel system or a General force
system, but a concurrent force system does not result in
a couple. We shall learn in detail about 'couple' later in
this chapter.
3. Resultant-Force Couple
A resultant force when shifted to a new parallel
position without change in its direction and sense,
introduces a couple in the system. Such a resultant
consisting of a single force and a single couple is called
as 'Force Couple'. Fig. c shows a force couple resultant. We shall learn
about shifting of a force by introducing of a couple later in this chapter.

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RESULTANT OF CONCURRENT SYSTEM OF FORCES USING
PARALLELOGRAM LAW OF FORCES
Parallelogram law of forces states "If two forces
acting simultaneously on a body at a point are
represented in magnitude and direction by the two
adjacent sides of a parallelogram, then their
resultant is represented in magnitude and
direction by the diagonal of the parallelogram
which passes through the point of intersection of the two sides representing
the forces".
Let P & Q be the two forces acting at a point and making an angle 'α' with
each other as shown in Fig. (a). The forces are
drawn to scale in figure such that AB and AD
represent forces P & Q. Completing the
parallelogram ABCD, the diagonal AC gives the
magnitude and direction of the resultant R.
Refer Fig. (b).
Mathematically 2 2
R = P + Q + 2PQcos a 2.3 (a)
Q sina
tanθ =
P + Qcosa
2.3 (b)
Where θ is the angle made by resultant R with the force P
Though parallelogram law is for two forces, it can be used for more forces
also, but would require repeated use, taking two forces at a time.

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RESULTANT OF CONCURRENT SYSTEM OF FORCES USING METHOD OF
RESOLUTION
When more than two forces act at a point, the use of method of resolution is
made to avoid tedious repetition of parallelogram law of forces to successive
forces. The following steps are adopted in the solution.
Step 1: Resolve the inclined forces if any along the horizontal x direction and
the Vertically direction.
Step 2: a) Add up the horizontal forces to get ∑Fx. Use sign Convention →+ ve
b) Add up the vertical forces to get ∑Fy . Use sign convention ↑ + ve.
c) The resultant force 2 2
x yR = F + F 
Step 3: The direction of the resultant force is the angle θ made by it with the
x axis. To find angle θ, use
y
x
F
tanθ =
F


Note: while using the above relation take positive values of yF and xF .
The value of θ so obtained will always be less than 90°
Step 4: Decide the quadrant of the resultant, depending on the signs of xF
and yF
EXERCISE 1
1) Resolve the given forces into horizontal and vertical components.
2) Given F1 = 10 N, F2 = 20 N, F3 = 30 N, F4 = 40 N, F5 = 50 N and taking
anticlockwise moment as positive, Find

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a) Moment of force F1 about B and E
b) Moment of force F2 about A and C
c) Moment of force F3 about F
d) Moment of force F4 about B and D
e) Moment of force F5 about A, F and D.
3) Two forces of 18 N and 35 N act away from a point. If
the angle between the forces is 45°, find the
magnitude of the resultant and the angle made by it
with the 18 N force.
4) Four concurrent forces act at a point as shown.
Find their resultant.
5) The striker of carom board lying on the board is being pulled by four
players as shown in the figure. The players are sitting exactly at the
centre of the four sides. Determine the resultant of forces in magnitude
and direction.
6) Find resultant of the force system.

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7) Five concurrent coplanar forces act on a body as shown in figure. Find
the force P and Q such that the resultant of the five forces is zero.
8) Figure shows a concurrent system of four forces. Three of the four forces
are shown. Find the unknown fourth force „P‟ given that the resultant of
the system is a horizontal force of 500 N acting to the right.

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RESULTANT OF PARALLEL SYSTEM OF FORCES
To find the Resultant of Parallel Force System follow the given steps,
Step 1: Since in a parallel system, the forces are directed in one direction
only, they can be simply added up using a sign convention for the
sense of the force. i.e.
R = F
Step 2: Location of the resultant force forms an important step. The point of
application of the resultant force is found out using Varignon's
theorem, discussed earlier in article 2.7. The resultant is initially
assumed to act either to the right or left of the reference point at a 
distance d.
Varignon's theorem F R
A AM = M is used. If a positive value of d is
obtained then the assumption made earlier is true. If a negative
value of d is obtained the resultant lies on the opposite side to what
was assumed.
COUPLE
Couple is a special case of parallel forces. Two parallel forces of equal
magnitude and opposite sense form a couple. The effect of a couple is to
rotate the body on which it acts.
Fig. shows a couple formed by two forces of same magnitude F, separated by
a  distance d known as the arm of the couple.
The magnitude of rotation known as the moment of a couple is M = F × d.
Rotation of a couple could be clockwise Fig. 2.A (a) or could be anti-clockwise
Fig. (b). Couples are represented by curved arrows. Units are N.m, kN.m etc.

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PROPERTIES OF COUPLE
1. Couple tend to cause rotation of the body about an axis I to the plane
containing the two parallel forces.
2. The magnitude of rotation or moment of a couple is equal to the product of
one of the forces and the arm of the couple.
3. Couple is a free vector because of which it can be moved anywhere on the
body on which it acts without causing any change.
4. The resultant force of a couple systems is zero.
5. To balance a system whose resultant is a couple, another couple of the
same magnitude and opposite sense is required to be added.
6. To shift a force to a new parallel position, a couple is required to be added
to the system. For example in Fig. (a), force F is required to be shifted from
its original position at A to a new parallel position B. This is done by
adding two collinear forces of same magnitude F and - F at B Fig. (b). the
two parallel forces F at A and - F at B form a couple. Thus we have a single
force F at B and a couple M = F × d in the system Fig. (c). Such a system is
called as a force couple system.
DM + ve = + (200 × 7) – (300 × 5) – (150 × 3.5)
= - 625 Nm = 625 Nm
Hence, the system of four forces can be replaced by a single force R : 2OO N
t and a couple of 625 Nm ) at D. This is referred to as a force couple system
at D. ........Ans.

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EXERCISE 2
1) Determine the magnitude and position of the resultant with respect to
point A, of the parallel forces shown.
2) Find the magnitude, nature and position of the resultant of the four
parallel forces from B.
3) Determine the resultant V of the coplanar parallel forces shown and locate
it w.r.t. O. Radius of circle is 2m.
4) A system of parallel, non-concurrent forces is acting on a rigid bar. Reduce
this system of forces to
i) A single forces R & its position
ii) A single force R & a couple at B
5) Find the magnitude of two like parallel forces F1
and F2 acting at a distance of 50 cm apart, if
their resultant is 300 N and acts at a distance of
20 cm from one of the force.

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6) Determine the resultant of the parallel forces shown.
7) Determine the resultant of the system shown.
8) Resolve 15 kN force acting at „A‟ into two parallel components at B and C
9) Figure shows a parallel system of four forces and two couples.
1) Replace it by a single force and obtain its location from point A
2) Replace it by a force couple system at point A.
3) Replace it by a force couple system at point D.
4) Replace it by two parallel forces at B and D.
RESULTANT OF GENERAL FORCE SYSTEM
To find the Resultant of General Force System follow the given steps.
Step 1: Follow the same procedure as discussed in article 2.10 and resultant
force R of the system using Method of Resolution.
Step 2: To locate the position of the resultant follow step 2 of article 2.11.

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EXERCISE 3
1) Determine the resultant of the force system shown. The side of each small
square is 1 m. The overall size of the body is 4 m × 4 m.
2) Determine the resultant of the given
force system. Also find out where the
resultant force will meet arm AB.
Take A as the origin.
3) a) A system of forces acting on a bell crank
lever is as shown. Determine the
magnitude, direction and the point of
application of the resultant w.r.t „O‟.
b) Also find the location of the resultant on
the horizontal arm of the lever.
4) Find the resultant of the system of coplanar forces shown in figure.
5) a) A block ABCD of 100 cm
× 40 cm dimensions is
acted upon by four forces
as shown. Calculate the
resultant and then state its
position with reference to A.
b) Also find the location x where the resultant force cuts the base AD.

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6) Determine the resultant of the system of forces shown in figure. Locate the
point where the resultant cuts the base AB.
7) Three force 1 kN, 3 kN and 2.5 kN act on a vertical pole 6 m high.
a. Find the magnitude, direction and position of resultant w.r.t A
b. The position where the resultant cuts the pole from the base
c. Reduce it to a force couple system at A.
8) Resolve the system of forces shown into a force and couple at point 'A'.
9) Replace the force system shown by a single force acting at the origin and
couple.

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10) Replace the system of forces and couples by a single force and locate the
point on the x-axis through which the line of action of the resultant
passes.
11) Replace the loading on the frame by a force and moment at point A.
12) Determine the resultant of the given coplanar system of force and a
couple. Also locate the resultant on the x axis w.r.t. the origin. Reduce the
system to a force couple system at O.
13) A bracket is subjected to a coplanar force system as shown consisting of
three forces and couple. If the resultant force of the system is to pass
through B, determine the value of the couple M which should be applied at
D.

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EXERCISE 4
Theory Questions
1) Define i) Force ii) Moment of a force iii) Resultant.
2) illustrate the classification of system of forces with neat sketch. Give one
example of each.
3) State the difference between Composition and Resolution of Forces.
4) Give example of i) Coplanar, non-concurrent, non-parallel force system
ii) Coplanar, concurrent force system
iii) Non coplanar concurrent force system
5) State Varignon's theorem of moments
6) Prove Varignon's theorem of moments.
7) State "Principle of Transmissibility of Force".
8) What is a couple? State its properties.
UNIVERSITY QUESTIONS
1. Five concurrent coplanar forces act on a body as shown in figure. Find the
forces P and Q such that the resultant of the five forces is zero. (5 marks)
2. Four concurrent forces act at a point as shown. Find their resultant.
(4 Marks)

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3. For given system find resultant and its point of application with respect to
point O on the X – axis (X intercept). Force, along CA = 100 N, along OD =
250N, along ED = 150 N, along OE = 100 N. An clockwise moment of 5000
N-cm is also acting at the point O. (6 Marks)
4. A ring is pulled by three forces as shown in fig.
Find the force F and the angle  if resultant of
these three forces is 100 N acting in vertical
direction. (4 Marks)
5. The resultant of the three forces shown in figure
and other two forces Pand Q acting at „A‟ and ‟B‟ is a couple of magnitude
120 kNm clockwise. Determine the force P and Q. (5 Marks)
6. Two concurrent forces P and Q acts such that their resultant acts along x-
axis. Determine the magnitude of Q and hence the resultant. (4 Marks)

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7. A system of forces acting on a bell crank is as shown. Determine the
magnitude, direction and the point of application of the resultant w.r.t „O‟
(6 Marks)
8. The tension in cable AC is 8 kN. Determine the required tension T in cable
AB such that the net effect of the two cable tensions is a downward force at
point A. Determine the magnitude R of this downward force. (4 Marks)
9. Discuss the resultant of concurrent forces in space. (4 Marks)
10.A ladder of 4 m length weighing 200 N is placed as shown in
0.25 & 0.35B A   Calculate the minimum horizontal force to be applied at
A to prevent slipping. (8 Marks)

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