Manipur Technical univerisity

1. The general principle states that the effect of a force acting on a rigid body does not
change if the force is moved along its line of action to another point on the body.
In clear, the point of application of a force can be moved anywhere along its line of
action without changing the external reaction forces on a rigid body.
Any force that has the same magnitude and direction, and which has a point of
application somewhere along the same line of action will cause the same acceleration
and will result in the same moment.
Therefore, the points of application of forces may be moved along the line of action to
simplify the analysis of rigid bodies.
Principle of transmissibility of a force

2. The main thing to be noted is when analyzing
the internal forces (stress) in a rigid body, the
exact point of application does matter.
This difference in stresses may also result in
changes in geometry which will may in turn
affect reaction forces. For this reason, the
principle of transmissibility should only be
used when examining external forces on
bodies that are assumed to be rigid.
Limitation of principle of transmissibility: Principle of transmissibility can be used
only for rigid bodies and cannot be used for deformable bodies.

3. The Principle Of Superposition states that when two waves of the
same kind meet at a point in space, the resultant displacement at that point is the
vector sum of the displacements that the two waves would separately produce at
that point.
Net effect of forces applied in any sequence on a body is given by the algebraic sum
of effect of individual forces on the body

4. Assumptions made in Engineering Mechanics
i) All bodies are RIGID.
ii) PARTICLE concept can be used wherever applicable.
iii) PRINCIPLE OF PHYSICAL INDEPENDENCE of forces is valid.
iv) PRINCIPLE OF SUPERPOSITION of forces is valid.
v) PRINCIPLE OF TRANSMISSIBILITY of forces is valid.

5. SYSTEM OF FORCES
A group or set of forces is called system of forces.
Coplanar force system: If all the forces in the
system lie in a single plane, it is called
coplanar force system.
Concurrent force system: If line of action of all
the forces in a system passes through a single
point it is called concurrent force system.
Collinear force system: In a system, all the
forces parallel to each other, if line of action
of all forces lie along a single line then it is called a
collinear force system.

6. Concurrent force system: If the line of action of
forces forming the system pass through a common
point (point of concurrence) then the system is
said to be concurrent.
Non-concurrent force system : f the lines of
action of forces forming the system do not pass
through a common point, then the system is said
to be non-concurrent.
Parallel force system :It is a particular case of non-
concurrent force system in which the line of
action of forces forming the system are parallel.

7. COMPOSITION OF FORCES
Composition of Forces or compounding is the method to find resultant force of
a number of forces acting simultaneously on a body.
Methods for finding resultant force(FR) are
• Parallelogram Law
• Triangle Law
• Polygon law
• Resolution of forces or Method of Resolution

8. GRAPHICAL METHOD
In Graphical method, forces are drawn on graph with some suitable scale and their
resultant is measured by Parallelogram law, Triangle Law or Polygon Law.
While Parallelogram law is also used as analytical method for resolution of forces.
Let’s study the laws used for graphical method for composition of forces.
• Parallelogram Law
• Triangle Law
• Polygon law

9. PARALLELOGRAM LAW OF FORCES
“If two consecutive sides of a parallelogram represents the magnitude
and direction of two coplaner forces acting on a particle, then its
diagonal represents the Resultant Force(FR) in magnitude and
direction.”
Consider two vectors P⃗ and Q⃗
as shown in the figure. Vector Q⃗ is displaced parallel to itself till the tail end of both the vectors touch at
a point O.
The parallelogram is completed as shown in the figure. Applying the law of triangle of vectors, to the
triangle OAC, we have,
OA + AC = OC
or P⃗ + Q⃗ = R⃗
→ → →

10. Two forces F1 and F2 of magnitudes 80KN and 50KN respectively acting simultaneously on a particle such that
the angle between the two forces is 60o. Considering bigger force as F1.
Taking some suitable scale and drawing a
parallelogram ABCD with Sides AB and AD
representing the forces F1 and F2 and having an
angle (θ) between them.
F1 as AB = 80 mm = 8 cm,
F2 as AD = 50 mm = 5 cm
θ = 60o
According to law of parallelogram diagonal AC is representing
the resultant force (FR). Length of the diagonal AC is the
magnitude of the resultant force and the angle (α), which
diagonal AC is making with the side AB, represents the
direction of resultant force (FR).
By Measurement
Thus the Resultant FR is equal to 113.5 KN acting at an angle (α) 22.5o with F1.

11. TRIANGLE LAW OF FORCES
“If two sides of a triangle, taken in order, represents the magnitude and direction of
two coplaner forces acting on a partical, then its third side, taken in opposite order
represents the Resultant Force(FR) in magnitude and direction.”
Taking some suitable scale (1KN = 1 mm)
drawing a triangle ABC with sides AB and BC
representing the forces F1 and F2 and having an
angle (θ) between BC and extended side AB
F1 as AB = 80 mm = 8 cm,
F2 as BC = 50 mm = 5 cm
θ = 60o
According to triangle law of forces side AC is representing the resultant force (FR).
Length of side AC is the magnitude of the resultant force and the angle (α), which
side AC is making with side AB represents the direction of resultant force (FR).
By Measurement

12. Triangle and parallelogram law are best suited for system of forces
having two forces acting simultaneously on a body. For a system of
forces having more than two forces, Polygon law of forces is applied.
Thus the Resultant FR is equal to 113.5 KN
acting at an angle (α) 22.5o with F1.

13. POLYGON LAW OF FORCE
“If sides of a polygon, taken in order, represents the magnitude and direction of more
than two coplaner forces, acting on a partical, then its closing side, taken in opposite
order, represents the Resultant Force(FR) in magnitude and direction.”

14. Let us consider a system of forces F1, F2, F3 & F4. Having magnitude as 25 KN, 35 KN, 30
KN & 20 KN, making an angle of 30o, 45o, 40o and 80o with horizontal
Taking some suitable scale (1KN = 1 mm) drawing the sides of
polygon ABCDE with Sides AB, BC, CD & DE representing the
forces F1, F2, F3 & F4, having respective angle between them.
Joining point A and E to draw the closing side AE.

15. Joining point A and E to draw the closing side AE.
F1 as AB = 25 mm = 2.5 cm,
F2 as BC = 35 mm = 3.5 cm
F3 as CD = 30 mm = 3 cm,
F4 as DE = 20 mm = 2 cm
Angle between F1 and F2 is 105o
Angle between F2 and F3 is 95o
A
B
C
D
E

16. F1 as AB = 25 mm = 2.5 cm,
F2 as BC = 35 mm = 3.5 cm
F3 as CD = 30 mm = 3 cm,
F4 as DE = 20 mm = 2 cm
A B
B C
C D
D E

17. Rotate 330Original

20. According to polygon law of forces, side AE is representing the resultant force (FR).
Length of side AE is the magnitude of the resultant force and the angle (α) which
side AE is making with side AB represents the direction of resultant force (FR).
Thus the Resultant FR is equal to 22.6 KN acting at an angle (α)
154.4o with F1.

21. ANALYTICAL METHOD
Analytical method is different from graphical method. Let us study
analytical method for finding the resultant force for number of forces
acting simultaneously on a particle. There are two analytical methods
for composition of forces, Parallelogram law and Method of
Resolution.
PARALLELOGRAM LAW OF FORCES

22. “If two consecutive sides of a parallelogram represents the magnitude
and direction of two coplaner forces acting on a partical,
Then the Resultant Force (FR) and angle α is obtained by

23. Solving Illustration 1 by parallelogram law
Resultant Force FR will be
The angle (α) which resultant FR is making with force F1

24. RESOLUTION OF FORCES FOR RESULTANT FORCE
Before understanding the method of finding resultant force by method
of Resolution let us understand what resolution of forces is.
Resolution of Forces
Resolution of forces is the process of breaking a force into two components, basically horizontal and
vertical components.
Suppose force (F) acting on a particle such that, it is making an angle (θ) with horizontal axis (AB).

25. Horizontal Component
It is the product of magnitude of force and the cosine
of the angle, it is making with horizontal axis.
Direction of the component is considered same as
the direction of horizontal axis, with which it is
making the angle θ.
Vertical Component
It is the product of magnitude of force and the sine of
the angle, it is making with horizontal axis. Direction
of the component is considered same as the direction
of vertical axis, normal to the horizontal axis with
which it is making the angle θ.

26. PRINCIPLE OF FINDING RESULTANT FORCE BY THE METHOD OF
RESOLUTION OF FORCES
“The resolved parts of the forces in a particular direction will be equal to the
resolved parts of resultant force in same direction.”
Method of resolution of forces starts with summation of Horizontal Components in
a system of forces is represented By ∑H while summation of Vertical Components is
represented by ∑V.
∑H = Sum of Horizontal Components in a system of forces.
= F1cos θ1 + F2cos θ2 + F3cos θ3 +_ _ _+ Fn cos θn
∑V = Sum of Vertical Components in a system of forces.
= F1sin θ1 + F2sin θ2 + F3sin θ3 +_ _ _+ Fn sin θn

27. Direction of resultant in terms of angle(α) is obtained by
Magnitude of resultant force is obtained by

28. Solving Illustration 2 with method of resolution
Evaluating ∑H and ∑V (Using sign convention)
∑H = 25 cos 30o + 20 cos 80o + (-30 cos 40o) + (-35 cos 45o)
= 25 cos 30o + 20 cos 80o – 30 cos 40o – 35 cos 45o
= – 22.60
∑H = – 22.60 (-ve)
Similarly
∑V = 25 sin 30o + 35 sin 45o + (- 30 sin 40o) + (- 20 sin 80o)
= 25 sin 30o + 35 sin 45o – 30 sin 40o – 20 sin 80o
= – 1.73
∑V = – 1.73 (-ve)
Resultant Force FR

29. Resultant Force FR
The angle (α), which resultant (FR) making with horizontal axis.

30. In Solution of illustration 2 by polygon law, α is the angle which resultant FR is making
with force F1. Here α is the angle, which resultant FR is making with horizontal axis in
negative direction.
Thus angle between FR and F1 will be
Thus sum of both horizontal and vertical
Component of forces is negative, the
resultant will lie in 3rd quadrant.

31. Two forces 5 N and 20 N are acting at an angle of 120
degree between them . Find the resultant force in
magnitude and direction.
When vectors act as maximum and minimum?
Solution : It is to be noted that the magnitude of the resultant of two vectors is maximum , when the vectors
act in the same direction and is minimum when they act in opposite directions.
Two forces 10 N and 14 N are acting upon the a body . What
can be the maximum and minimum resultant force on the body
?

32. Two forces 5 N and 20 N are acting at an angle of 120
degree between them . Find the resultant force in
magnitude and direction.
When vectors act as maximum and minimum?
Solution : It is to be noted that the magnitude of the resultant of two vectors is maximum , when the vectors
act in the same direction and is minimum when they act in opposite directions.
Two forces 10 N and 14 N are acting upon the a body . What
can be the maximum and minimum resultant force on the body
?

33. Two forces 5 N and 20 N are acting at an angle of 120
degree between them . Find the resultant force in
magnitude and direction.
Two forces 10 N and 14 N are acting upon the a body . What
can be the maximum and minimum resultant force on the body
?

34. Two forces 5 N and 20 N are acting at an angle of 120
degree between them . Find the resultant force in
magnitude and direction.
Two forces 10 N and 14 N are acting upon the a body . What
can be the maximum and minimum resultant force on the body
?

35. Drawing Free-Body Diagrams
Free-body diagrams are diagrams used to show the relative
magnitude and direction of all forces acting upon an object in a
given situation.
1. The size of the arrow in a free-body diagram reflects the
magnitude of the force.
2. The direction of the arrow shows the direction that the force is
acting.
3. Each force arrow in the diagram is labelled to indicate the exact
type of force.
It is generally customary in a free-body diagram to represent the
object by a box and to draw the force arrow from the centre of the
box outward in the direction that the force is acting

37. Force Symbol Magnitude Direction
Gravity (weight) F(G) = mobject*g (9.8 m/s2) Downward
Normal (surface) F(N) any (up to breaking load) perpendicular to surface
Tension F(T) any (up to breaking load) along string/rope/chain
Friction F(fr)
not slipping: between zero
and ms*F(N)
slipping: = mk*F(N)
Direction opposing relative
motion.

38. Horizontal Component
It is the product of magnitude of force and the cosine
of the angle, it is making with horizontal axis.
Direction of the component is considered same as
the direction of horizontal axis, with which it is
making the angle θ.
Vertical Component
It is the product of magnitude of force and the sine of
the angle, it is making with horizontal axis. Direction
of the component is considered same as the direction
of vertical axis, normal to the horizontal axis with
which it is making the angle θ.

39. •A book is at rest on a tabletop. Diagram the forces
acting on the book.
•A girl is suspended motionless from the ceiling by
two ropes. Diagram the forces acting on the
combination of girl and bar..

40. A book on a table
In this example, there are two forces acting on a book at rest on a table:
1) The weight W exerted by the earth on the book
2) The normal force N exerted by the table on the book.
A suspended block
In this example, there are two forces acting on the suspended block at rest:
1) The weight W exerted by the earth on the block
2) The tension force T exerted by the string (or rope) on the block.

41. A block on a floor with an acting force Fa to pull the block
In this example, the block is being pulled and therefore a force
of friction acts on the block. So there are four forces acting on
the block.
1) The weight W exerted by the earth on the block
2) The normal force N exerted by floor on the block.
3) The acting force Fa to pull the block.
4) The force of friction Ff exerted by floor on the block in the
direction opposite the motion due to Fa.

42. A falling object
In this example, there is only one force action on the falling object.
1) The weight W exerted by the earth on the falling object.
A falling object with Air Friction
here only two force action on the falling object.
1) The weight W exerted by the earth on the falling object.
2) The Friction FAir

43. An box on an incline plane (with no frictions)
In this example, we assume that the inclined plane is frictionless;
two forces act on the box:
1) The weight W exerted by the earth on the box.
2) The normal force N exerted by the inclined plane on the box.
N is normal to the inclined plane.

44. An box on an incline plane with an acting force
and friction considered.
In this example, a force Fa pulls the box upward
and frictions are not negligible. Four forces act on
the box:
1) The weight W exerted by the earth on the box.
2) The normal force N exerted by the inclined plane
on the box. N is normal to the inclined plane.
3) The acting force Fa.
4) The force of friction Fs exerted by the inclined
plane on the box in the direction opposite the
motion due to Fa.

50. A block suspended to the ceiling using three strings.
A) free body diagram for the block; two forces (lower part
of figure below)
1) The weight W exerted by the earth on the box.
2) The tension force T '3 exerted by the string on the block.
B) free body diagram of point P; three forces (upper part of
figure below)
1) Tension T1
2) Tension T2
3) Tension T3

54. A system with two blocks, an inclined plane and a pulley.
A) free body diagram for block m1 (left of figure below)
1) The weight W1 exerted by the earth on the box.
2) The normal force N
3) The force of friction Fk
4) The tension force T exerted by the string on the block m1.
B) free body diagram of block m2 (right of figure below)
1) The weight of the block W2
2) Tension T '.

55. Determining the Net Force
In the statement of Newton's first law, the unbalanced force refers to that force that
does not become completely balanced (or canceled) by the other individual forces. If
either all the vertical forces (up and down) do not cancel each other and/or all
horizontal forces do not cancel each other, then an unbalanced force exists. The
existence of an unbalanced force for a given situation can be quickly realized by looking
at the free-body diagram for that situation. Free-body diagrams for three situations are
shown below. Note that the actual magnitudes of the individual forces are indicated on
the diagram.

60. Free-body diagrams for four situations are shown below. The net force is known for each situation.
However, the magnitudes of a few of the individual forces are not known. Analyze each situation
individually and determine the magnitude of the unknown forces.

61. The force F is producing, simultaneous
x displacement and y-displacement.
The part of the force F which is
producing x displacement is called x
component or horizontal component of
the force F(Fx). The part of the force F
which produces y – displacement is
called y component of the force or
vertical component of force F(Fy).
The technique of finding a component of a force along any direction is called
resolution of force. The component of a force along any direction is called the
resolved component. The components of a force determined along two mutually
perpendicular direction are called rectangular components.
RESOLUTION OF A FORCE

62. In the diagram the force F is resolved into two perpendicular
component forces that is the Fy and Fx components (using
parallelogram method).
To calculate the magnitude of the vertical (Fy) and
horizontal (Fx) forces, we can use simple trigonometry.
Fx = F cos θ , Since cos θ = (Fx/F)
Fy = F sin θ, Since sin θ = (Fy/ F)

63. Horizontal Component
It is the product of magnitude of force and the cosine
of the angle, it is making with horizontal axis.
Direction of the component is considered same as
the direction of horizontal axis, with which it is
making the angle θ.
Vertical Component
It is the product of magnitude of force and the sine of
the angle, it is making with horizontal axis. Direction
of the component is considered same as the direction
of vertical axis, normal to the horizontal axis with
which it is making the angle θ.

64. To calculate the magnitude of the vertical
(Fy) and horizontal (Fx) forces, we can use
simple trigonometry.
Fx = F cos θ , Since cos θ = (Fx/F)
Fy = F sin θ, Since sin θ = (Fy/ F)

66. Note:
1. Sign convention for the direction of components.
The horizontal component or ‘X’ component of a force acting
along x direction is the force itself. Whereas, its vertical
component or y-component is zero.
2.

67. ‘x’ component of a force acting along Y direction is
zero. Whereas, its ‘y’ component is equal to itself.
4. If a force is inclined at 45o
w.r.t. x axis or y axis then its x component will
be equal to y component (Fx= Fy).