# Engineering Mechanics CHAPTER 1 - JeffLamoon.Com

Mechanics

Oct 31, 2013 (4 years and 6 months ago)

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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

EE31102005

Slide 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
.

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 2 of 19

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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

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

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 3 of 19

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

Multiples

Submultiples

1

kilogram

is

1

kg

or

10
3

g
.

1

millimeter

is

1

mm

or

10
-
3

m
.

1

megagram

is

1

Mg

or

10
6

g
.

1

micrometer

is

1

m

-
6
m
.

1

gigagram

is

1

Gg

or

10
9
g
.

1

nanometer

is

1

nm

or

10
-
9
m
.

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 4 of 19

EE31102005

The following SI derived units are frequently
used in this course:

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/s
2

(i.e 1 N = 1 kg.m/s
2
).

Or : Force = mass x acceleration

= kg x m/s
2
= kg m/s
2

Hence a 1 kg mass has a force or weight due to gravity,

equal to (1 kg x 9.81 m/s
2)

= 9.81 N,

where g = 9.81 m/s
2

.

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 5 of 19

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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 :

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 6 of 19

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The equation has the standard form as follows

ax
2

+ bx + c = 0 (1.1)

The

standard

solution

to

this

equation

is

x

=

b

(

b
2

4
ac)

(
1
.
2
)

2
a

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 7 of 19

EE31102005

Example 1

X

=

12

)
2

4
(
5
)(

2
)]

2(5)

=

12

13.56

10

=

+
0
.
156

or

2
.
56

Solve for x in the equation 5x
2

+ 12x

2 = 0.

Comparing equation 1.1 above, and substitute
a=5, b=12 and c=

2 into equation 1.2, the
solution is :

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 8 of 19

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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

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 9 of 19

EE31102005

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)

4

Substituting (3) into (2) , yielding

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

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 10 of 19

EE31102005

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

Slide 11 of 19

EE31102005

4x + 3y + 10 = 0 (1)

20x + 30y + 5 = 0 (2)

Divide (2) by 5

4x + 6y + 1 = 0 (3)

Subtract

(
3
)

by

(
1
)

3
y

-

9

=

0

y

=

3

Substitute

the

value

of

y

into

(
1
)

or

(
2
)

4
x

+

3
(
3
)

+

10

=

0

4
x

=

-

9

-

10

x

=

-

19

4

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 12 of 19

EE31102005

1.2.3

Trigonometry Functions Of A Right
-
Angle Triangle

sine

=

s楤e

=

o

=

c潳楮e

††
(
1
.
5
)
††††††††††††††††††††

hypotenuse

h

cosine

=

s楤e

=

a

=

s楮e

††
(
1
.
6
)

hypotenuse

h

tangent

=

s楤e

=

o
††††††††
(
1
.
7
)

side

a

tangent

=

s楮

cos

h

o

††††††††††

†††††††††††††††

††††††††††††††
a

†††††††††††

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 13 of 19

EE31102005

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

††
s楮

††
s楮

Cosine Rule a
2

= b
2

+ c
2

2bc cos

††
(ㄮ㠩

b
2
= a
2

+ c
2

2ac cos

c
2

= a
2

+ b
2

2ab cos

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 14 of 19

EE31102005

If the cosine rule is applied to a right
-
angle
triangle where

= 90
0
, and applying to
equation (1.8)
,

i
.
e
.

a
2

=

b
2

+

c
2

2
bc

cos

90

0

since

cos

90

0

=

0

a
2

=

b
2

+

c
2

(Pythagoras

Theorem)

c a

90
0

b

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 15 of 19

EE31102005

Example 3

Find the length of the unknown side a and the angle

.

20
0

6m

4m

a

Cosine rule : a
2

= b
2
+c
2
-
2bccos

i.e. a
2
= 6
2
+4
2
-
2x6x4cos20
0

a = 2.63m

Sine rule :
2.63

=
6

sin 20
0

sin

= 6.9

= 36 +16
-
6x4xcos20
0

= 51.3
0

sine

= 6 x
sin 20
0

2.63

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 16 of 19

EE31102005

But we know this to be in the second quadrant,

Hence

= 180

51.4 = 128.6
0

Check

:

6
2

=

2
.
63
2

-

4
2

-

2
x
2
.
63
x
4

cos

cos

=
2.63
2
+ 4
2

6
2

2x2.63x4

=
-

0.634

= 128.6
0

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 17 of 19

EE31102005

1.2.5 Geometry

Some

of

the

basic

rules

are

shown

below
:

Sum of supplementary angles = 180
0

+

㴠180
0

A straight line intersecting
two parallel lines

,

,

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 18 of 19

EE31102005

Similar triangles ABC and ADE, by proportion

B

C

AB

=
BC

=
AC

Hence

if

AB

=

6
,

=

3

and

BC

=

4
,

Then,

6

=

4

3

DE

DE

=

(
3

x

4
)

3

=

2

End of Chapter 1

D

A

E

Engineering Mechanics

Chapter 1: Introduction & Review of Mathematics

Slide 19 of 19

EE31102005