Engineering Mechanicsfor

First Year B.E. Degree Students

COURSE CONTENT IN BRIEF1. Introduction.

2. Resultant of concurrent and non-concurrent coplanar forces.

3. Equilibrium of concurrent and non-concurrent coplanar forces.

4. Analysis of plane trusses.

5. Friction.

6. Centroid and Moment of Inertia.

7. Resultant and Equilibrium of concurrent non-coplanar forces.

8. Rectilinear and Projectile motion.

9. Newton’s second law, D’Alembert’s principle, banking and super elevation.

10. Work, Energy, and Power.

11. Impulse- Momentum principle.

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.

 

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’.

CHAPTER – I INTRODUCTION

This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.

1

Engineering Mechanics

Mechanics of Solids Mechanics of Fluids

Rigid Bodies DeformableBodies

Statics Dynamics

Kinematics Kinetics

Strength of Materials

Theory of Elasticity

Theory of Plasticity

Ideal Fluids

ViscousFluids

CompressibleFluids

Branches of Mechanics2

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.

3

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.

4

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.

x

y

z

5

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’.

6

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:

(i) Magnitude (ii) Direction (iii) Point of application and(iv) Line of action.

7

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.

8

Classification of force system

Force system

Coplanar Forces Non-Coplanar Forces

Concurrent Non-concurrent Concurrent Non-concurrent

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 )

Like parallel Unlike parallel Like parallel Unlike parallel

9

 Axioms of Mechanics

(1) 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

AO

P2

P1

R

10

Contd..

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.

B C

AO

P2

P1

R

It can be shown that the magnitude of the resultant is given by:R = P1

2 + P22 + 2P1P2Cos α

Inclination of the resultant w.r.t. the force P1 is given by:

= tan-1 [( P2 Sin ) / ( P1 + P2 Cos )]

11

Contd..

(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’.

PA B

P

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.

P P

O

12

(3) Newton’s Laws of motion:

(i) First 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.

13

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.

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RESULTANT OF CONCURRENT COPLANAR FORCES

CHAPTER – 2

Y-Direction

X-DirectionF3

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:

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.

15

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.

16Contd..

  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’. 

Component of a force :

Fy

Fig. 1Fx

FFy

F

FxFig. 2 Fig. 3

FFy

Fx

    The process of obtaining the components of a force is called ‘Resolution of a force’.

17

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.

Sign Convention for force components:

+ve

+vex

xyy

Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.

When the components of a force are not mutually perpendicular they are called ‘Oblique Components’. Consider the following case.

Oblique Components of a force:

18

Contd..

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’.

F

F1

F2Let 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

F

F1

F2

F1 / Sin = F2 / Sin = F / Sin(180 - - )

19

Contd..

Numerical Problems & Solutions

∑ Fx = + 15 Cos 15 – 75 – 45 Sin 35

+ 60 Cos 40

∑ Fy = + 15 Sin 15 + 105 – 45 Cos 35

– 60 Sin 40

= + 33.453 kN

15 kN

150

105 kN

75 kN

45 kN

400

60 kN350

Fig.1A

(1A)

+ve

= - 40.359 kN = 40.359 kN

Obtain the resultant of the concurrent coplanar forces acting as shown in Fig. 1A.

Solution:

+ve

20

Contd..

R = ( ∑Fx )2 + (∑Fy)2 = (- 40.359)2 + (33.453) 2

Θ = tan-1(∑Fy/ ∑Fx)

Magnitude,R = 52.42 kN

Inclination,Θ = 39.69 o

15 kN

150

105 kN

75 kN

45 kN

400

60 kN350

Fig.1A

(1A)

∑Fx

Θ

Answer:

(w.r.t. X – direction)

∑FyR

21

Contd..

(1B)

50 kN

23

100 kN

α26.31o

75 kN

30o1

225kN

Obtain the resultant of the

concurrent coplanar forces acting

as shown in Fig. 1B.

∑Fx = -50 Cos 26.31- 100 Cos33.69 – 25 Cos 63.43 + 75 Cos 30

+ ve 74.26kN -74.26kN =

75kN

1202

3301

2

Fig. 1B25kN

100kN50kN

º

º

α = tan-1(2/3)=33.69

Solution:β

º

β= tan-1(2/1) = 63.43º

22

Contd..

(1B)

50 kN

23

100 kN

α26.31o

75 kN

30o1

2

β

25kN

= -93.17kN = 93.17kN

23

∑FY = 50sin26.31- 100sin 33.69 – 75sin30 – 25sin63.43 ve+

Contd..

Contd..

R = (∑Fx) 2 + (∑Fy) 2 = 119.14 kN

Θ = tan-1(∑Fy / ∑Fx ) = 51.44o

∑Fx

∑FyR

Θ

100 kN50kN

26.31o

75 kN

30o

(1B)

25kN

33.69º

63.43º

Answers:

Contd..

24

- 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

(2)

150N

50N

200N

120N

45°

50°110 º F5

αR =250 N

20º

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. Fig. 2

120N

150N

50N

200N

45º

50°110º

Solution:

25

Contd..

200 cos 50 – 150 cos 20 – 50 cos 45 +F5 cos α = 250.

F5 cos α = +297.75 N

+ve∑FX = R

150N

50N

200N

120N

45°

50°110 º F5

αR =250 N

20º

26

Contd..

Contd..

∑FY = 0.

F5 sin α + 200sin 50 + 150 sin 20

– 120 + 50 sin 45 = 0

F5 sin α= -119.87N = 119.87N

α = 21.90º

119.87N

297.75N

F5 = 320.97N

tan α = F5sin α /F5cos α

=0.402

α = 21.90º

F5= 320.97N

F5cosα =

F5sinα =

+ve

Answers

Contd..

27

150N

50N

200N

120N

45°

50°110 º F5

αR =250 N

20º

(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.

Fig. 3

60°

45°

R=100N

50N

25N

70°

40°

75N

28

Contd..

- Assume the fourth force F4 in the 1st quadrant, making an angle α as shown

α60°

45°

R=100N 50N

25N

70°

40°

75N

F4

F4cosα + 75cos70 – 50cos45 – 25sin60 = -100cos40

Or, F4cosα = - 45.25N ; or, F4cos α = 45.25N

Fx = -Rcos40+ve

29

Contd..

60°

45°

R=100N 50N

25N

70°

40°

75N

α

F4

Fy = -Rsin40

F4sinα + 75sin70+25cos60+50sin45 = - 100sin40

F4sinα = -182.61N ; or, F4sin α = 182.61N

+ve

30Contd..

Contd..

α= 76.08º

45.25N

F4=188.13N

182.61N

F4cosα =F4sinα =

Answers:

= tan-1(F4sin /F4cos)

= 76.08º

& F4 =188.13N

31

Contd..

.  (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.

60°

30°

15 kN

5 kN

10 kN

70°

45°

F4

Fig. 4

32

Contd..

F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94

∑Fx = 0

60°

30°

15 kN

5 kN

10 kN

70°

45°

F4

Fig. 4

Solution:

+ve

R

33

F4 = 21.22kN

Contd..

Contd..

F4cos70 + 10sin60 – 15sin45 + 5sin30 = +R

+R - 0.342F4 = 0.554

Substituting for F4 , R= +7.81kN

∑Fy = +R+ve

Solution:60°

30°

15 kN

5 kN

10 kN

70°

45°

F4

Fig. 4

R

Answers:

F4 = 21.22 kN R= +7.81kN

34

Contd..

(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

35

Contd..

Contd..

40°

60°

P50N

Q

70°

45°

Fig. 5

100N

36

Contd..

For R to be = zero,

∑Fx = 0 and ∑ Fy = 0

∑Fx = 0 :

-Psin45 – Qcos40 + 100cos70 + 50cos60 = 0

Or, 0.707P + 0.766Q = 59.2

+ve

40°

60°

P 50N

Q

70°

45°

Fig. 5

100N

∑Fy = 0

-Pcos45 + Qsin40 + 100sin70 – 50sin60 = 0

or, -0.707P + 0.642Q = -50.67

+ve

Answers:

(b)

Solving (a) & (b)

P = 77.17 N & Q = 6.058N

37

Contd..

30°

100N

50N

Fig. 6

(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.

38

Contd..

Rotating the axes to have X parallel to 50N,

∑Fx = +50 + 100cos30 = +136.6N

∑ Fy = +100sin30 = +50N

+ve

+ve

30°

100N(6)

X - AXIS

Y-AXIS

30°

100N

50N

Fig. 6

50N

39

Contd..

Contd..

F = 145.46N

θ = 20.1º w r t X direction (50N force)

50N

F= (∑Fx)2+(∑Fy)2

= tan-1[(∑Fx)2+(∑Fy)2]

Fig. 6

X - AXISFθ

Y-AXIS

30°

100N(6)

X - AXIS

Y-AXIS

30°

100N

50N50N

θ

40

Contd..

(7) Resolve the 3kN force along the directions P and Q. Refer Fig. 7.

P

Q3kN

45°60°

30°

Fig. 7

41

Contd..

3kN

Fig. 7

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

P

45ºQ

45º

Q

60°

30° X – Axis P

3kN

45º

42

Contd..

EXERCISE PROBLEMS1. 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.

Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N

FIG. 1

= 1200 N

X

34

YF2

F1

= 400 N

43

2. Determine the X and Y components of each of the forces shown in

(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,

F3X = -306.42 N, F3Y= -257.12N )

30º40º

12

5

300 N

390 N

400 N

X

Y

F1 =

F2 =

F3 =

FIG. 2

FIG.2.

44

45

600N

200N

800N

20º 40º

30º

FIG. 3

3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3

(Ans: R = 522.67 N, θ = 68.43º)

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 :

(a) Force exerted by each of the tug boats knowing α = 30º.(b) 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.

( Ans: a. T1= 195.81 N, T2 = 133.94 N b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )

T2

R = 300 N

T1

α

20º

FIG. 4

X - direction

46

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

R20º

40º

FIG. 5

(Ans: P = 9.4 kN , R = 12.66 kN)

47

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

( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)

1200 N

800 N60º

θF

ROD

COLLAR

FIG.6

48

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.

8kN5kN

Q30º

α

Fig. 7(Ans: α = 10.7 º, Q = 9.479 kN )

49