1. ENGINEERING MECHANICS
CHAPTER TOPIC
INTRODUCTION AND REVIEW OF MATHEMATICS
    
FORCES AND RESULTANTS
MOMENTS AND COUPLES
EQUILIBRIUM
FRICTION
6     KINEMATICS OF LINEAR & ROTATIONAL MOTION
KINETICS OF LINEAR MOTION
KINETICS OF ROTATIONAL MOTION
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Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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2. CHAPTER 1 :Introduction And Review of Mathematics
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.
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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3. 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 Unit Symbol
1. Length meter m
2. Mass kilogram kg
3. Time second s
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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4. Multiples Submultiples 1 kilogram is 1 kg or 103 g. 1 millimeter is 1
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.
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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5. The following SI derived units are frequently used in this course:
Force – The unit of force is the newton (N) 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.
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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6. Simultaneous equations
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
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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7. 1.2.1 Quadratic Equations
The equation has the standard form as follows
ax2 + bx + c = (1.1)
The standard solution to this equation is
x = – b   ( b2 – 4ac) (1.2) 2a
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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8. 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
= or – 2.56
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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9. 1.2.2 Simultaneous Equation
The equation has two unknowns x and y in the form of ax + by + c = (1.3) px + qy + r = (1.4) Example 2 Find the values of x and y satisfying the given equations: x + 3y + 10 = (1) x + 30y = (2) There are 2 methods of solving these equations
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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10. 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 (3) Substituting (3) into (2) , yielding ( -3y -10) + 30y + 5 = y y + 5 = y – 45 = y = 45 = To find x, substitute the value of y into (3) x = (-3 x ) =
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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11. 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:
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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12. Substitute the value of y into (1) or (2) 4x + 3(3) + 10 = 0
4x + 3y + 10 = (1) 20x + 30y = (2)
Divide (2) by 5
4x + 6y = (3)
Subtract (3) by (1)
3y = 0
y = 3
Substitute the value of y into (1) or (2)
4x + 3(3) + 10 = 0
4x =
x = - 19 4
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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13. 1.2.3 Trigonometry Functions Of A Right-Angle Triangle
 sine  = opposite side = o = cosine  (1.5)
hypotenuse h
cosine  = adjacent side = a = sine  (1.6)
tangent  = opposite side = o (1.7)
adjacent side a
tangent  = sin  cos o  a h 
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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14. 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  (1.8)
b2 = a2 + c2 – 2ac cos 
c2 = a2 + b2 – 2ab cos 
Sine Rule a = b = c
sin  sin  sin 
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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15. If the cosine rule is applied to a right-angle triangle where  = 90 0 , and applying to equation (1.8),
c a
90 0 b
i.e. a2 = b2 + c2 – 2bc cos 90 0
since cos 90 0 = 0
a2 = b2 + c2
(Pythagoras Theorem)
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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16. Example 3 Find the length of the unknown side a and the angle .
Cosine rule : a2 = b2+c2-2bccos
200
6m m 
i.e. a2 = x6x4cos200
= x4xcos200
= 6.9
a = 2.63m
Sine rule : =
sin sin  a
sine  = 6 x sin 200
2.63
= 51.30
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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17. But we know this to be in the second quadrant, Hence  = 180 – 51
Check : = x2.63x4 cos 
cos  = – 62
2x2.63x4 =
 =
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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18. 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
 = ,  = 
 = ,  =     
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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19. Similar triangles ABC and ADE, by proportion
AB = BC = AC
AD DE AE
Hence if AB = 6, AD = 3 and BC = 4,
Then,
6 = 4 DE
DE = (3 x 4) 3
= 2 B D A C E
End of Chapter 1
Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics
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