Transcript Document

Engineering Mechanics
for
First Year B.E. Degree Students
COURSE CONTENT IN BRIEF
1. 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.
1
CHAPTER – I
INTRODUCTION
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.
2
Branches of Mechanics
Engineering Mechanics
Mechanics of Solids
Rigid Bodies
Statics Dynamics
Kinematics
Mechanics of Fluids
Deformable
Bodies
Strength of
Materials
Kinetics
Ideal Viscous Compressible
Fluids
Fluids Fluids
Theory of
Elasticity
Theory of
Plasticity
3
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.
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.
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 coordinates of the point, in that particular frame of reference.
z
y
x
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’.
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:
(i)
(ii)
(iii)
(iv)
Magnitude
Direction
Point of application and
Line of action.
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.
Classification of force system
9
Force system
Coplanar Forces
Concurrent Non-concurrent
Non-Coplanar Forces
Concurrent
Like parallel Unlike parallel
Non-concurrent
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 )
10
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
P2
O

C
R

P1
A
Contd..
11
Contd..
B
P2
O

C
R

P1
A
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  )]
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
A
B
P
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.
13
(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.
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.
CHAPTER – 2
15
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
F2
R
Fy

F3
Fx
F1
X-Direction
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..
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.
17
Component of a force :
Fx
Fy
F
F

Fx
Fig. 1

Fx
Fig. 2

Fy
Fy
F
Fig. 3
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’.
18
Sign Convention for force components:
The adjacent diagram gives the
y
x
sign convention for force components,
+ve
i.e., force components that are directed
along positive x-direction are taken +ve
x
+ve
for summation along the x-direction.
Also force components that are directed along +ve y-direction
are taken +ve for summation along the y-direction.
y
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..
19
Let F1 and F2 be the oblique components of a
F
force F. The components F1 and F2 can be
F1 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
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’.
Contd.. F2
F

F1
 F2
F1 / Sin  = F2 / Sin  = F / Sin(180 -  - )
20
Numerical Problems & Solutions
(1A) Obtain the resultant of the concurrent coplanar forces acting
as shown in Fig. 1A.
Solution:
∑ Fx = + 15 Cos 15 – 75 – 45 Sin 35
+ve
15 kN
+ 60 Cos 40
150
= - 40.359 kN
= 40.359 kN
400
105 kN
75 kN
350
45 kN
Fig.1A
60 kN
∑ Fy = + 15 Sin 15 + 105 – 45 Cos 35
+ve
– 60 Sin 40
= + 33.453 kN
Contd..
21
Contd..
(1A)
105 kN
15 kN
150
400
75 kN
350
60 kN
45 kN
Fig.1A
R = ( ∑Fx )2 + (∑Fy)2 = (- 40.359)2 + (33.453) 2
Θ = tan-1(∑Fy/ ∑Fx)
Answer:
Magnitude,R = 52.42 kN
Inclination,Θ = 39.69
(w.r.t. X – direction)
o
R
∑Fy
Θ
∑Fx
22
50kN
100kN
º
120
2
3
30º
75kN
1
2
25kN
Fig. 1B
Solution:
α = tan-1(2/3)=33.69 º
β= tan-1(2/1)=63.43º
(1B) Obtain the resultant of the
concurrent coplanar forces acting
as shown in Fig. 1B.
50 kN
100 kN
2
o
26.31
α 3
30o
1 β
2
25kN
75 kN
∑Fx = -50 Cos 26.31- 100 Cos33.69 – 25 Cos 63.43 + 75 Cos 30
Contd..
-74.26kN =
74.26kN
+ ve
23
Contd..
50 kN
(1B)
26.31o
1
β
α 3
100 kN
2
30o
2
25kN
75 kN
∑FY = 50sin26.31- 100sin 33.69 – 75sin30 – 25sin63.43
+ ve
= -93.17kN = 93.17kN
Contd..
24
Contd..
50kN
(1B)
100 kN
26.31o
33.69º
30o
63.43º
75 kN
25kN
∑Fx
Answers:
Θ
∑Fy
R
R = (∑Fx) 2 + (∑Fy) 2 = 119.14 kN
Θ = tan-1(∑Fy / ∑Fx ) = 51.44o
25
(2)
150N
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
Fig. 2
horizontal, find the unknown fifth force.
200N
110º
50°
45º
120N
Solution:
- 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
150N
20º
50N
200N
F5
110 º
50°
α
45° R =250 N
120N
50N
Contd..
26
Contd..
150N
20º
∑FX = R
+ve
200N
F5
110 º
50°
α
45° R =250 N
120N
50N
200 cos 50 – 150 cos 20 – 50 cos 45 +F5 cos α = 250.
 F5 cos α = +297.75 N
Contd..
27
Contd..
150N
20º
∑FY = 0.+ve
F5 sin α + 200sin 50 + 150 sin 20
– 120 + 50 sin 45 = 0
F5 sin α= -119.87N = 119.87N
F5cosα = 297.75N
α = 21.90º
F5 = 320.97N
F5sinα = 119.87N
200N
F5
110 º
50°
α
45° R =250 N
120N
50N
tan α = F5sin α /F5cos α
Answers
α = 21.90º
F5= 320.97N
=0.402
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°
40°
45°
Fig. 3
50N
R=100N
Contd..
29
Contd..
- Assume the fourth force F4 in the 1st quadrant, making an
angle α as shown
75N
25N
F4
60°
70°
40°
Fx = -Rcos40
α
45°
R=100N
50N
+ve
F4cosα + 75cos70 – 50cos45 – 25sin60 = -100cos40
Or, F4cosα = - 45.25N ; or, F4cos α = 45.25N
30
Contd..
75N
25N
F4
60°
70°
40°
α
45°
R=100N
50N
Fy = -Rsin40
+ve
F4sinα + 75sin70+25cos60+50sin45 = - 100sin40
 F4sinα = -182.61N ;
or, F4sin α = 182.61N
Contd..
31
Contd..
Answers:
F4cosα = 45.25N
F4sinα =
α= 76.08º
182.61N
 = tan-1(F4sin /F4cos)
= 76.08º
& F4 =188.13N
F4=188.13N
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..
33
Contd..
R
Solution:
F4
10 kN
70°
60°
45°
30°
5 kN
15 kN
∑Fx = 0
Fig. 4
+ve
F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94
F4 = 21.22kN
Contd..
34
Contd..
10 kN
R
Solution:
F4
70°
60°
30°
45°
5 kN
∑Fy = +R
15 kN
+ve
Fig. 4
F4cos70 + 10sin60 – 15sin45 + 5sin30 = +R
 +R - 0.342F4 = 0.554
Substituting for F4 ,
R= +7.81kN
Answers:
F4 = 21.22 kN
R= +7.81kN
35
(5) Obtain the magnitudes of the forces P and Q if the resultant of the
system shown in Fig. 5 is zero .
100N
Q
40°
70°
60°
45°
50N
P
Fig. 5
Contd..
36
Contd..
100N
Q
40°
70°
45°
60°
50N
P
For R to be = zero,
Fig. 5
∑Fx = 0 and ∑ Fy = 0
∑Fx = 0 :
+ve
-Psin45 – Qcos40 + 100cos70 + 50cos60 = 0
Or, 0.707P + 0.766Q = 59.2
Contd..
37
Contd..
100N
Q
40°
70°
45°
Solving (a) & (b)
Answers:
P = 77.17 N & Q = 6.058N
50N
P
∑Fy = 0
60°
Fig. 5
+ve
-Pcos45 + Qsin40 + 100sin70 – 50sin60 = 0
or, -0.707P + 0.642Q = -50.67
(b)
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..
39
Contd..
(6)
100N
100N
Y-AXIS
50N
30°
X - AXIS
30°
50N
Fig. 6
Rotating the axes to have X parallel to 50N,
∑Fx = +50 + 100cos30 = +136.6N
+ve
∑ Fy = +100sin30 = +50N
+ve
Contd..
40
Contd..
100N
(6)
100N
Y-AXIS
X - AXIS
50N
30°
30°
50N
Fig. 6
F=
(∑Fx)2+(∑Fy)2
θ = tan-1[(∑Fx)2+(∑Fy)2]
F = 145.46N
θ = 20.1º w r t X direction (50N force)
Y-AXIS
F
X - AXIS
θ
50N
41
(7) Resolve the 3kN force along the directions P and Q. Refer Fig. 7.
Q
3kN
45°
60°
30°
P
Fig. 7
Contd..
42
Contd..
Q
3kN
45º P
Fig. 7
45º
60°
30° X – Axis
P
Q
45º
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
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.
F1 = 1200 N
Y
F2 = 400 N
4
3
FIG. 1
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
44
2. Determine the X and Y components of each of the forces shown in
FIG.2.
F2 = 390 N
Y
12
X
5
30º
40º
F1 = 300 N
F3 =400 N
FIG. 2
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,
F3X = -306.42 N, F3Y= -257.12N )
45
3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3
600N
800N
40º
20º
30º
FIG. 3
200N
(Ans: R = 522.67 N, θ = 68.43º)
46
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.
T2
α
20º
R = 300 N
X - direction
FIG. 4
T1
( Ans: a. T1= 195.81 N, T2 = 133.94 N
b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )
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
20º
R
40º
Q = 5 kN
FIG. 5
(Ans: P = 9.4 kN , R = 12.66 kN)
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
FIG.6
θ
F
COLLAR
ROD
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
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.
5kN
8kN
α
30º
Q
(Ans: α = 10.7 º, Q = 9.479 kN )
Fig. 7