1. Engineering Mechanics First Year B.E. Degree Students
for
First Year B.E. Degree Students

2. COURSE CONTENT IN BRIEF
Introduction.
Resultant of concurrent and non-concurrent coplanar forces.
Equilibrium of concurrent and non-concurrent coplanar forces.
Analysis of plane trusses.
Friction.
Centroid and Moment of Inertia.
Resultant and Equilibrium of concurrent non-coplanar forces.
Rectilinear and Projectile motion.
Newton’s second law, D’Alembert’s principle, banking and super elevation.
Work, Energy, and Power.
Impulse- Momentum principle.

3. 1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.
Books for Reference
1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.
2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.
3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition
4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.

4. Definition of Mechanics :
CHAPTER – I
INTRODUCTION 1
Definition of Mechanics :
In its broadest sense the term ‘Mechanics’ may be defined as the ‘Science which describes and predicts the conditions of rest or motion of bodies under the action of forces’.
This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.

5. Engineering Mechanics2
Branches of Mechanics
Engineering Mechanics
Mechanics of Solids
Mechanics of Fluids
Rigid Bodies
Deformable
Bodies
Ideal
Fluids
Viscous
Fluids
Compressible
Fluids
Statics
Dynamics
Theory of Elasticity
Strength of Materials
Theory of Plasticity
Kinematics
Kinetics

6. Fundamental Concepts and Axioms3
Fundamental Concepts and Axioms
Rigid body :
It is defined as a definite amount of matter the parts of which are fixed in position relative to one another. Actually solid bodies are never rigid; they deform under the action of applied forces. In those cases where this deformation is negligible compared to the size of the body, the body may be considered to be rigid.

7. 4
Particle
A body whose dimensions are negligible when compared to the distances involved in the discussion of its motion is called a ‘Particle’. For example, while studying the motion of sun and earth, they are considered as particles since their dimensions are small when compared with the distance between them.

8. 5
Space
The concept of space is associated with the notion of the position of a point, defined using a frame of reference, with respect to which the position of the point is fixed through three measures specific to the frame of reference. These three measures are known as the co-ordinates of the point, in that particular frame of reference. z y x

9. 6
Mass :
It is a measure of the quantity of matter contained in a body. It may also be treated as a measure of inertia, or resistance to change the state of rest, or of uniform motion along a straight line, of a body. Two bodies of the same mass will be attracted by the earth in the same manner.
Continuum :
A particle can be divided into molecules, atoms, etc. It is not feasible to solve any engineering problem by treating a body as a conglomeration of such discrete particles. A body is assumed to be made up of a continuous distribution of matter. This concept is called ‘Continuum’.

10. 7
Force
It is that agent which causes or tends to cause, changes or tends to change the state of rest or of motion of a mass. A force is fully defined only when the following four characteristics are known:
Magnitude
Direction
Point of application and
Line of action.

11. 8
Scalars and Vectors
A quantity is said to be a ‘scalar’ if it is completely defined by its magnitude alone.
Example : Length, Area, and Time.
Whereas a quantity is said to be a ‘vector’ if it is completely defined only when its magnitude and direction are specified.
Example : Force, Velocity, and Acceleration.

12. Classification of force system 9
Coplanar Forces
Non-Coplanar Forces
Concurrent
Non-concurrent
Concurrent
Non-concurrent
Like parallel
Unlike parallel
Like parallel
Unlike parallel
A force that can replace a set of forces, in a force system,
and cause the same ‘external effect’ is called the Resultant.
( More detailed discussion on Resultant will follow in Chapter 2 )

13. 10
 Axioms of Mechanics
Parallelogram law of forces : It is stated as follows : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forces is represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’ B C A O P2 P1 R  
Contd..

14. 11
Contd.. B C A O P2 P1 R  
In the above figure, P1 and P2, represented by the sides OA and OB have R as their resultant represented by the diagonal OC of the parallelogram OACB.
It can be shown that the magnitude of the resultant is given by:
R = P12 + P22 + 2P1P2Cos α
Inclination of the resultant w.r.t. the force P1 is given by:
 = tan-1 [( P2 Sin ) / ( P1 + P2 Cos  )]

15. 12
(2)   Principle of Transmissibility :  It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’. P P A B P P O
For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B.
The same is true when the force is applied at a point O.

16. 13 (3) Newton’s Laws of motion: (i) First Law : (ii) Second Law :
If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in uniform motion). 
(ii) Second Law :
If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant i.e., F α a ,or F = m.a , where F, m, and a, respectively represent the resultant force, mass, and acceleration of the particle. 
(iii) Third law:
  The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

17. 14
Note :  
1. ‘Axioms’ are nothing but principles or postulates that are self – evident facts which cannot be proved mathematically but can only be verified experimentally and/or demonstrated to be true.
2. The three basic quantities of mechanics are length, time, and force. Throughout this Course we adopt SI units and therefore they are expressed in meters, seconds, and Newtons, written as m, s, and N respectively. 
  3. The ‘external effect’ of a force on a body is manifest in a change in the state of inertia of the body. While the ‘internal effect’ of a force on a body is in the form of deformation.

18. 15
CHAPTER – 2
RESULTANT OF CONCURRENT COPLANAR FORCES
Composition of forces and Resolution of force
Resultant, R : It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect.
Y-Direction
X-Direction F3 F1 R
Fx
Fy  F2
In the above diagram F1, F2, F3 form a system of concurrent coplanar forces. If R is the resultant of the force system, then its magnitude and direction are given by:
Contd..

19. 16 Contd.. (i) Magnitude, R =  (Fx)2 + (Fy)2
   (ii) Direction, θ = tan –1(Fy / Fx) , where:
 ΣFx = Algebraic summation of x-components of all individual forces. 
 ΣFy = Algebraic summation of y-components of all individual forces. 
θ = Angle measured to the resultant w.r.t. x-direction. 
The process of obtaining the resultant of a given force system is called ‘Composition of forces’.
Note: The orientation of x-y frame of reference is arbitrary. It may be chosen to suit a particular problem.

20. 17
Component of a force : Fx  F F Fy Fy Fy F   Fx Fx
Fig. 3
Fig. 1
Fig. 2
  Component of a force, in simple terms, is the effect of a force in a certain direction. A force can be split into infinite number of components along infinite directions. Usually, a force is split into two mutually perpendicular components, one along the x-direction and the other along y-direction (generally horizontal and vertical, respectively). Such components that are mutually perpendicular are called ‘Rectangular Components’. 
    The process of obtaining the components of a force is called ‘Resolution of a force’.

21. Sign Convention for force components:18
Sign Convention for force components:
The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction. y y x
+ve x
+ve
Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.
Oblique Components of a force:
When the components of a force are not mutually perpendicular they are called ‘Oblique Components’. Consider the following case.
Contd..

22. 19
Let F1 and F2 be the oblique components of a force F. The components F1 and F2 can be found using the ‘triangle law of forces’, which states as follows: ‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle
Contd.. F2 F F1  
taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
F1 / Sin  = F2 / Sin  = F / Sin(180 -  - ) F  F2  F1

23. Numerical Problems & Solutions20
Numerical Problems & Solutions
(1A)
Obtain the resultant of the concurrent coplanar forces acting
as shown in Fig. 1A.
Solution:
105 kN
∑ Fx = + 15 Cos 15 – 75 – 45 Sin 35
+ 60 Cos 40
∑ Fy = + 15 Sin – 45 Cos 35
– 60 Sin 40
= kN
+ve
15 kN
150
= kN
75 kN
400
= kN
350
60 kN
45 kN
Fig.1A
+ve
Contd..

24. 21 105 kN Contd.. (1A) 15 kN 150 75 kN 400 350 60 kN 45 kN Fig.1A
R = ( ∑Fx )2 + (∑Fy)2 = ( )2 + (33.453) 2
Θ = tan-1(∑Fy/ ∑Fx) R
Answer:
∑Fy
Magnitude,R = kN
Inclination,Θ = o Θ
(w.r.t. X – direction)
∑Fx

25. ∑Fx = -50 Cos 26.31- 100 Cos33.69 – 25 Cos 63.43 + 75 Cos 3022
75kN
120 2 3 30 1
Fig. 1B
25kN
100kN
50kN
Obtain the resultant of the
concurrent coplanar forces acting
as shown in Fig. 1B.
(1B)
50 kN
100 kN 2
26.31o 3 α
Solution: 1
30o β
α = tan-1(2/3)=33.69
75 kN 2
β= tan-1(2/1) =
63.43º
25kN
∑Fx = -50 Cos Cos33.69 – 25 Cos Cos 30
Contd..
+ ve
-74.26kN =
74.26kN

26. 23 Contd.. 50 kN 100 kN (1B) 2 26.31o 3 α β 30o 1 75 kN 2 25kN
∑FY = 50sin sin – 75sin30 – 25sin63.43 + ve
= kN
= 93.17kN
Contd..

27. 24 Contd.. (1B) 50kN 100 kN 26.31o 33.69º 30o 63.43º 75 kN 25kN ∑Fx R
∑Fy R Θ
Answers:
R = (∑Fx) 2 + (∑Fy) 2 = kN
Θ = tan-1(∑Fy / ∑Fx ) = 51.44o

28. 25 (2) 150N 200N A system of concurrent coplanar forces has
five forces of which only four are shown in
Fig.2. If the resultant is a force of magnitude
R = 250 N acting rightwards along the
horizontal, find the unknown fifth force.
110º
50°
45º
Fig. 2
50N
120N
Solution:
150N
50N
200N
120N
45°
50°
110 º F5 α
R =250 N
20º
- Assume the fifth force F5 in the first quadrant, at an angle α, as shown.
- The 150 N force makes an angle of 20o w.r.t. horizontal
Contd..

29. 26 Contd.. 200N 150N F5 110 º 50° 20º α R =250 N 45° ∑FX = R +ve 50N
200 cos 50 – 150 cos 20 – 50 cos 45 +F5 cos α = 250.
 F5 cos α = N
Contd..

30. 27 Contd.. 150N 50N 200N 120N 45° 50° 110 º F5 α R =250 N 20º ∑FY = 0.
F5 sin α + 200sin sin 20
– sin 45 = 0
F5 sin α= N = N
+ve
tan α = F5sin α /F5cos α
=0.402
α = 21.90º
F5= N
F5cosα =
297.75N
Answers
α = 21.90º
F5 = N
F5sinα =
119.87N

31. 28
(3) A system of concurrent coplanar forces has four forces of which only three are shown in Fig.3. If the resultant is a force R = 100N acting as indicated, obtain the unknown fourth force.
75N
25N
60°
70°
45°
40°
Fig. 3
50N
R=100N
Contd..

32. 29
Contd..
- Assume the fourth force F4 in the 1st quadrant, making an angle α as shown
75N
25N F4
60°
70° α
45°
40°
R=100N
50N
Fx = -Rcos40
+ve
F4cosα + 75cos70 – 50cos45 – 25sin60 = -100cos40
Or, F4cosα = N ; or, F4cos α = 45.25N

33. 30 Contd.. 75N 25N 60° 70° 45° 40° R=100N 50N Fy = -Rsin40α
70°
45°
40°
R=100N
50N
Fy = -Rsin40
F4sinα + 75sin70+25cos60+50sin45 = - 100sin40
 F4sinα = N ; or, F4sin α = N
+ve
Contd..

34. 31 Contd.. Answers:  = tan-1(F4sin /F4cos) F4cosα = 45.25N = 76.08º
& F4 =188.13N
α= 76.08º
45.25N
F4=188.13N
182.61N
F4cosα =
F4sinα =

35. 32
.  (4) The resultant of a system of concurrent coplanar forces is a force acting vertically upwards. Find the magnitude of the resultant, and the force F4 acting as shown in Fig. 4.
10 kN F4
70°
60°
30°
45°
5 kN
15 kN
Fig. 4
Contd..

36. 33
Contd.. R
Solution: F4
10 kN
70°
60°
30°
45°
5 kN
15 kN
F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94
∑Fx = 0
Fig. 4
+ve
F4 = 21.22kN
Contd..

37. Substituting for F4 , R= +7.81kN F4 = 21.22 kN34
Contd..
10 kN R F4
Solution:
70°
60°
30°
45°
5 kN
15 kN
∑Fy = +R
+ve
Fig. 4
Answers:
F4cos sin60 – 15sin45 + 5sin30 = +R
 +R F4 = 0.554
Substituting for F4 , R= +7.81kN
F4 = kN
R= +7.81kN

38. 35
(5) Obtain the magnitudes of the forces P and Q if the resultant of the system shown in Fig. 5 is zero .
40°
60° P
50N Q
70°
45°
Fig. 5
100N
Contd..

39. 36
Contd..
100N Q
40°
70°
60°
50N
45° P
For R to be = zero,
∑Fx = 0 and ∑ Fy = 0
∑Fx = 0 :
-Psin45 – Qcos cos cos60 = 0
Or, 0.707P Q = 59.2
Fig. 5
+ve
Contd..

40. 37 Contd.. 100N Q 40° 70° 60° 45° Solving (a) & (b) 50N P Answers:
Fig. 5
100N
Solving (a) & (b)
Answers:
∑Fy = 0
-Pcos45 + Qsin sin70 – 50sin60 = 0
or, P Q =
P = N & Q = 6.058N
+ve
(b)

41. 38
(6) Forces of magnitude 50N and 100N are the oblique components of a force F. Obtain the magnitude and direction of the force F. Refer Fig.6.
100N
50N
30°
Fig. 6
Contd..

42. Rotating the axes to have X parallel to 50N, 39
Contd..
(6)
100N
100N
Y-AXIS
X - AXIS
50N
30°
30°
50N
Fig. 6
Rotating the axes to have X parallel to 50N,
∑Fx = cos30 = N
∑ Fy = +100sin30 = +50N
+ve
Contd..
+ve

43. 40
Contd..
100N
(6)
100N
Y-AXIS
X - AXIS
50N
30°
30°
50N
Fig. 6
F = N
θ = 20.1º w r t X direction (50N force)
F= (∑Fx)2+(∑Fy)2
Y-AXIS θ
= tan-1[(∑Fx)2+(∑Fy)2] F
X - AXIS θ
50N

44. 41
(7) Resolve the 3kN force along the directions P and Q. Refer Fig. 7. Q
3kN
45°
60°
30° P
Contd..
Fig. 7

45. 42
Contd.. Q
60°
30°
X – Axis P
3kN
45º P
45º Q
Fig. 7
3kN
Move the force P parallel to itself to complete a triangle. Using sine rule,
P/sin45 = Q/sin90 = 3/sin45
Answer :
P = 3kN, and Q = 4.243kN

46. 43
EXERCISE PROBLEMS
1. A body of negligible weight, subjected to two forces F1= 1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in Fig.1. Find the component of each force parallel, and perpendicular to the plane.
FIG. 1
= 1200 N X 3 4 Y F2 F1
= 400 N
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N

47. 44
2. Determine the X and Y components of each of the forces shown in
FIG.2.
30º
40º 12 5
300 N
390 N
400 N X Y
F1 =
F2 =
F3 =
FIG. 2
(Ans : F1X = N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,
F3X = N, F3Y= N )

48. 45
3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3
600N
200N
800N
20º
40º
30º
FIG. 3
(Ans: R = N, θ = 68.43º)

49. 46 T2 R = 300 N T1 α 20º FIG. 4 X - direction
4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find :
Force exerted by each of the tug boats knowing α = 30º.
The value of α such that the force of tugboat 2 is minimum,
while that of 1 acts in the same direction.
Find the corresponding force to be exerted by tugboat 2. T2
R = 300 N T1 α
20º
FIG. 4
X - direction
( Ans: a. T1= N, T2 = N
b. α = 70º, T1 = N, T2(min) = N )

50. 47
5. An automobile which is disabled is pulled by two ropes as shown in FIG. 5. Find the force P and resultant R, such that R is directed as shown in the figure. P
Q = 5 kN R
20º
40º
FIG. 5
(Ans: P = 9.4 kN , R = kN)

51. ( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)48
6. A collar, which may slide on a vertical rod, is subjected to three forces
as shown in FIG.6. The direction of the force F may be varied .
Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to
(a) 2400 N, (b)1400N
1200 N
800 N
60º θ F
ROD
COLLAR
FIG.6
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)

52. 49
7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kN.
Refer Fig. 7.
8kN
5kN Q
30º α
Fig. 7
(Ans: α = 10.7 º, Q = kN )