Download NGEE ANN POLYTECHNIC MECHANICAL ENGINEERING DEPARTMENT

ENGINEERING MECHANICS
CHAPTER
TOPIC
1
INTRODUCTION AND REVIEW OF MATHEMATICS
2
FORCES AND RESULTANTS
3
MOMENTS AND COUPLES
4
EQUILIBRIUM
5
FRICTION
6
KINEMATICS OF LINEAR & ROTATIONAL MOTION
7
KINETICS OF LINEAR MOTION
8
KINETICS OF ROTATIONAL MOTION
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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CHAPTER 1 :Introduction And Review of
Mathematics
1.1 Introduction
Basic mechanics involves the study of two principal areas
– statics and dynamics.
Statics is the study of forces on objects or bodies which
are at rest or moving at a constant velocity, and the forces
are in balance, or in static equilibrium.
A ball at rest may have several forces acting on it, such as
gravitational force (weight) and a force opposing that
gravity (reaction). The ball is at rest or static, has forces
in balance or EQUILIBRIUM
Dynamics is the study of forces on moving bodies, and the
forces are in dynamic equilibrium.
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The concept of applied mechanics is useful
when it comes to analyzing stress, designing
of machine structures and hydraulics, etc.
There are only seven basic units in the SI
system but only three are frequently used in
statics and dynamics:
Physical Quantity
1. Length
2. Mass
3. Time
Unit
meter
kilogram
second
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Symbol
m
kg
s
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For large or small figures, multiples or
submultiples are used. For example:
Multiples
Submultiples
1 kilogram is 1 kg or 103 g.
1 millimeter is 1
mm or 10-3 m.
1 megagram is 1 Mg or 106 g.
1 micrometer is 1
m or 10-6m.
1 gigagram is 1 Gg or 109g.
1 nanometer is 1
nm or 10-9m.
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The following SI derived units are frequently
used in this course:
Force – The unit of force is the newton (N) .
1 newton is the force applied to a 1 kg mass to give it an
acceleration of 1 m/s2 (i.e 1 N = 1 kg.m/s2).
Or : Force = mass x acceleration
= kg x m/s2 = kg m/s2
Hence a 1 kg mass has a force or weight due to gravity,
equal to
(1 kg x 9.81 m/s2) = 9.81 N,
where g = 9.81 m/s2 .
Moment – It is the product of a force and its perpendicular
distance, and the unit is newton-meter or N-m.
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1.2 Mathematics Required
The followings are the mathematics skills that are
important for this module :
 Quadratic equations
 Simultaneous equations
 Trigonometry functions of a right-angle triangle
 Sine and cosine rules
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1.2.1 Quadratic Equations
The equation has the standard form as follows
ax2 + bx + c = 0
(1.1)
The standard solution to this equation is
x = – b   ( b2 – 4ac)
2a
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
(1.2)
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Example 1
Solve for x in the equation 5x2 + 12x – 2 = 0.
Comparing equation 1.1 above, and substitute
a=5, b=12 and c= –2 into equation 1.2, the
solution is :
X = – 12   [(12)2 – 4(5)( – 2)]
2(5)
= – 12  13.56
10
= +0.156 or – 2.56
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1.2.2 Simultaneous Equation
The equation has two unknowns x and y in the form of
ax + by + c = 0
(1.3)
px + qy + r = 0
(1.4)
Example 2
Find the values of x and y satisfying the given
equations:
4x + 3y + 10 = 0
(1)
20x + 30y + 5 = 0
(2)
There are 2 methods of solving these equations
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Method of Substitution
We can start by expressing x in terms of y, or y in terms of x.
Let us choose to express x in terms of y, thus from (1)
x = -3y -10
4
Substituting (3) into (2) , yielding
(3)
20( -3y -10) + 30y + 5 = 0
4
-15y -50 + 30y + 5 = 0
15y – 45 = 0
y = 45 = 3
15
To find x, substitute the value of y into (3)
x = (-3 x 3 - 10) = -19
4
4
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Method of Elimination
This method looks for a way to eliminate one of
the unknowns.
This can be done by making the constant
factor of that unknown or variable the same in
both equations by multiplying or dividing one
equation by a selected constant:
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4x + 3y + 10 = 0
20x + 30y + 5 = 0
(1)
(2)
Divide (2) by 5
4x + 6y + 1 = 0
(3)
Subtract (3) by (1)
3y - 9 = 0
y=3
Substitute the value of y into (1) or (2)
4x + 3(3) + 10 = 0
4x = - 9 - 10
x = - 19
4
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1.2.3 Trigonometry Functions Of A Right-Angle Triangle
h

sine  = opposite side = o = cosine 
hypotenuse
h
(1.5)
cosine  = adjacent side = a = sine 
hypotenuse
h
(1.6)
tangent  = opposite side = o
adjacent side
a
(1.7)
o

a
tangent  = sin 
cos
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1.2.4 Sine And Cosine Rules
For triangles that are not right-angle, the following two laws
are important in vector algebra introduced in chapter two
later:
Cosine Rule a2 = b2 + c2 – 2bc cos 
b2 = a2 + c2 – 2ac cos 
c2 = a2 + b2 – 2ab cos 
Sine Rule
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
(1.8)
a = b = c
sin  sin  sin 
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If the cosine rule is applied to a right-angle
triangle where  = 90 0 , and applying to
equation (1.8),
c
a
i.e. a2 = b2 + c2 – 2bc cos 90 0
since cos 90 0 = 0
a2 = b2 + c2
(Pythagoras Theorem)
90 0
b
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Example 3
Find the length of the unknown side a and the angle .
Cosine rule : a2 = b2+c2-2bccos
i.e. a2 = 62+42-2x6x4cos200
200
6m
4m

a
= 36 +16-6x4xcos200
= 6.9
a = 2.63m
Sine rule : 2.63 =
6
sin 200
sin 
sine  = 6 x sin 200
2.63
= 51.30
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But we know this to be in the second quadrant,
Hence  = 180 – 51.4 = 128.6 0
Check :
62 = 2.632 - 42 - 2x2.63x4 cos 
cos  = 2.632 + 42 – 62
2x2.63x4
= - 0.634
 = 128.6 0
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1.2.5 Geometry
Some of the basic rules are shown below:
Sum of supplementary angles = 180 0
 +  = 180 0


A straight line intersecting
two parallel lines




 = ,  = 
 = ,  = 
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Similar triangles ABC and ADE, by proportion
AB = BC = AC
AD
DE
AE
B
D
Hence if AB = 6, AD = 3 and BC = 4,
Then,
6 = 4
3
DE
C
E
A
DE = (3 x 4)
3
=2
End of Chapter 1
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