ENGINEERING MECHANICSCHAPTER 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 EE31102005Slide 1 of 19

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.

Statics & Dynamics.exe Bridges.exe

equilibrium.exe

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The concept of applied mechanics is useful when it comes to analyzing stress, designing of machine structures and hydraulics, etc.

Physical Quantity Unit Symbol

1. Length meter m

2. Mass kilogram kg

3. Time second s

Beam under stress.exe Hydranlics.exe

There are only seven basic units in the SI system but only three are frequently used in statics and dynamics:

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For large or small figures, multiples or submultiples are used. For example:

Multiples Submultiples1 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:

Forces.exe

Moment – It is the product of a force and its perpendicular distance, and the unit is newton-meter or N-m.

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 .

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Quadratic equations Simultaneous equations Trigonometry functions of a right-angle triangle

Sine and cosine rules

1.2 Mathematics Required

The followings are the mathematics skills that are important for this module :

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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) (1.2) 2a

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Example 1

X = – 12 [(12)2 – 4(5)( – 2)]

2(5) = – 12 13.56 10 = +0.156 or – 2.56

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 :

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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 (3) 4Substituting (3) into (2) , yielding 20( -3y -10) + 30y + 5 = 0 4 -15y -50 + 30y + 5 = 0 15y – 45 = 0 y = 45 = 3 15To 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 (1)20x + 30y + 5 = 0 (2)

Divide (2) by 5 4x + 6y + 1 = 0 (3)Subtract (3) by (1) 3y - 9 = 0 y = 3Substitute 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

 sine = opposite side = o = cosine (1.5)

hypotenuse h cosine = adjacent side = a = sine (1.6) hypotenuse h tangent = opposite side = o (1.7) adjacent side a

tangent = sin

cos

h

o

a

Right-Angle Triangle.exe

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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:

Sine Rule a = b = c sin sin sin

Cosine Rule a2 = b2 + c2 – 2bc cos (1.8) b2 = a2 + c2 – 2ac cos c2 = a2 + b2 – 2ab cos

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If the cosine rule is applied to a right-angle triangle where = 90 0 , and applying to equation (1.8),

i.e. a2 = b2 + c2 – 2bc cos 90 0

since cos 90 0 = 0

a2 = b2 + c2

(Pythagoras Theorem) c a

90 0

b

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Example 3Find the length of the unknown side a and the angle .

200

6m 4m

a

Cosine rule : a2 = b2+c2-2bccosi.e. a2 = 62+42-2x6x4cos200

a = 2.63m

Sine rule : 2.63 = 6 sin 200 sin

= 6.9

= 36 +16-6x4xcos200

= 51.30

sine = 6 x sin 200

2.63

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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

B

C

AB = BC = AC AD DE AE

Hence if AB = 6, AD = 3 and BC = 4,Then,

6 = 4 3 DE

DE = (3 x 4) 3 = 2

End of Chapter 1

D

A

E

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