chapter 1 note

1CHAPTER 1:PHYSICAL QUANTITIES AND MEASUREMENTS

(3 Hours)

2At the end of this chapter, students should be able to:

a) State basic quantities and their respective SI units: length (m), time (s), mass (kg), electrical current (A), temperature (K), amount of substance (mol) and luminosity (cd).

b) State derived quantities and their respective units and symbols: velocity (m s-1), acceleration (m s-2), work (J), force (N), pressure (Pa), energy (J), power (W) and frequency (Hz).

c) State and convert units with common SI prefixes.

Learning Outcome:

1.1 Physical Quantities and Units (1 hours)

1.1 Physical Quantities and

Units

Physical quantity is defined as a quantity which can be measured.

It can be categorized into 2 types

Basic (base) quantity

Derived quantity

3

4 Basic quantity is defined as a

quantity which

cannot be

derived from any

physical

quantities.

Table 1.1 shows all the basic

(base) quantities.

Quantity Symbol SI Unit Symbol

Length l metre m

Mass mkilogra

m kg

Time t second s

Temperature T/ kelvin K

Electric

current I ampere A

Amount of

substance N mole mol

Luminous

Intensity candela cd

Table 1.1

5 Derived quantityis defined as a

quantity which

can be

expressed in

term of base

quantity.

Table 1.2 shows some examples

of derived

quantity.

Derived

quantity

Symbol Formulae Unit

Velocity v s/t m s-1

Volume V l w t M3

Acceleration a v/t m s-2

Density m/V kg m-3

Momentum p m v kg m s-1

Force F m a kg m s-2 @

N

Work W F s kg m2 s-2

@ J

Table 1.2

6 Unit is defined as a standard size of measurement of physical quantities.

Examples :

1 second is defined as the time required for 9,192,631,770 vibrations of

radiation emitted by a caesium-133

atom.

7 1 kilogram is defined as the mass of a platinum-iridium cylinder kept at

International Bureau of Weights and

Measures Paris.

1 meter is defined as the length of the path travelled by light in vacuum

during a time interval ofs

458,792,299

1

8oo

o

57.296180 rad 1

180 rad

The unit of basic quantity is called base unit

addition unit for base unit:

unit of plane angle radian (rd)

unit of solid angle-steradian (sr)

9Notes:

oThe common system of units

used today are S.I unit

(System International/metric

system) and cgs unit - UK.

o The unit of derived quantity called derived unit.

10

It is used for

presenting larger

and smaller

values.

Table 1.3 shows all

the unit prefixes.

Prefix Value Symbol

tera 1012 T

giga 109 G

mega 106 M

Kilo 103 k

Deci 101 d

centi 102 c

milli 103 m

micro 106

nano 109 n

pico 1012 p

1.1.1 Unit Prefixes

Table 1.3

Examples:

(i) 5740000 m = 5740 km

= 5.74 Mm

(ii)0.00000233 s = 2.33 106 s

= 2.33 s

11

12

Table 1.4 shows the conversion factors between SI

and British units for length and mass only.

1.1.2 Conversion of Unit

Length Mass

1 m = 39.37 in = 3.281 ft 1 kg = 103 g

1 in = 2.54 cm 1 slug = 14.59 kg

1 km = 0.621 mi 1 lb = 0.453 592 kg

1 mi = 5280 ft = 1.609 km 1 kg = 0.0685 slug

1 angstrom () = 1010 m

Table 1.4

13

Solve the following

problems of unit conversion.

a. 30 mm2 = ? m2

b. 865 km h1 = ? m s1

c. 300 g cm3 = ? kg m3

d. 17 cm = ? in

e. 24 mi h1 = ? km s1

Example 1 :

232 m10mm 1 262 m 10mm 1

25

262

m 103.0

m 1030mm 30

Solution :

a. 30 mm2 = ? m2

14

Solve the following

problems of unit

conversion.

a. 30 mm2 = ? m2

b. 865 km h1 = ? ms1

c. 300 g cm3 = ? kgm3

d. 17 cm = ? in

e. 24 mi h1 = ? kms1

Example 1 : Solution :

b. 865 km h-1 = ? m s-1

1st method :

h 1

m10865h km 865

31

s 3600

m10865h km 865

31

11 s m 240h km 865

15

Solve the following

problems of unit

conversion.

a. 30 mm2 = ? m2

b. 865 km h1 = ? ms1

c. 300 g cm3 = ? kgm3

d. 17 cm = ? in

e. 24 mi h1 = ? kms1


b. 865 km h-1 = ? m s-1

2nd method :

s 3600

h 1

km 1

m 1000

h 1

km 865h km 865 1

s 3600

h 1

km 1

m 1000

h 1

km 865h km 865 1

11 s m 240h km 865

16

Solve the following

problems of unit

conversion.

a. 30 mm2 = ? m2

b. 865 km h1 = ? ms1

c. 300 g cm3 = ? kgm3

d. 17 cm = ? in

e. 24 mi h1 = ? km s1


c. 300 g cm-3 = ? kg m-3

332-

33-

3

3-

m 10

cm 1

g 1

kg 10

cm 1

g 300cm g 300

-353 m kg 103.0cm g 300

17

Solve the following

problems of unit

conversion.

a. 30 mm2 = ? m2

b. 865 km h1 = ? ms1

c. 300 g cm3 = ? kgm3

d. 17 cm = ? in

e. 24 mi h1 = ? kms1

Example 1 :

d. 17 cm = ? in

e. 24 mi h-1 = ? km s-1

Solution :

cm 1

in cm 17cm 17 2.54

1

in 6.69cm 17

s 3600

h 1

mi 1

km 1.609

h 1

mi 24h mi 24 1-

-1-21 s km 101.07h mi 24

18

At the end of this chapter, students should be able to:

a) Define scalar and vector quantities.

b) Perform vector addition and subtraction

operations graphically.

c) Resolve vector into two perpendicular

component (2-D):

i) components in the x and y axes

ii) component in the unit vectors in Cartesian

coordinate.

Learning Outcome:

1.2 Scalars and Vectors (2 hours)

19

At the end of this chapter, students should be able to:

d) Define and use dot (scalar) product;

c) Define and use cross (vector) product;

Direction of cross product is determined by corkscrew method or right hand rule.

Learning Outcome:

1.2 Scalars and Vectors (2 hours)

ABBABA coscos

ABBABA sinsin

20

Scalar quantity is defined as a quantity with magnitude only.

e.g. mass, time, temperature, pressure, electric current, work, energy and etc.

Mathematics operational : ordinary algebra

Vector quantity is defined as a quantity with both magnitude & direction.

e.g. displacement, velocity, acceleration, force, momentum, electric field, magnetic field

and etc.

Mathematics operational : vector algebra

1.2 Scalars and Vectors

21

Table 1.6 shows written form (notation) of vectors.

Notation of magnitude of vectors.

1.2.1 Vectors

Vector ALength of an arrow magnitude of vector A

displacement velocity acceleration

vv

aa

s

v

a

s avs (bold) v (bold) a (bold)

Direction of arrow direction of vector A

22

Two vectors equal if both magnitude and direction are the same.

(shown in figure 1.1)

If vector A is multiplied by a scalar quantity k

Then, vector A is

if k = +ve, the vector is in the same direction as vector A.

if k = - ve, the vector is in the opposite direction of vector A.

Ak

Figure 1.1

P Q

QP

Ak

A

A

23

Can be represented by using:

a) Direction of compass east, west,

north, south, north-east, north-west,

south-east and south-west.

b)Angle with a reference line

e.g. A man throws a stone with a

velocity of 10 m s-1, 30 above

horizontal.

c) Cartesian coordinates

2-Dimension (2-D)

1.2.2 Direction of Vectors

30v

x

y

0

m) 4 m, 2(),( yxs

y/m

x/m

4

20

s

24

1.2.2 Direction of Vectors

c)Cartesian coordinates

3-Dimension (3-D)

m 2) 3, 4,(),,( zyxs

y/m

x/m

z/m

0

s

4

2

3

25

d)Polar coordinates

e)Denotes with + or signs.

N,120 50F

F

120

+

+-

-

26

There are two methods involved in addition of vectors graphically;

Parallelogram

Triangle

1.2.3 Addition of Vectors

Parallelogram

Triangle

B

A

BA

O

B

A

BA

O

A

B

For example : BA

27

28

Triangle of vectors method:

a)Use a suitable scale to

draw vector A.

b)From the head of vector A

draw a line to represent the

vector B.

c)Complete the triangle.

Draw a line from the tail of

vector A to the head of

vector B to represent the

vector A + B.

ABBA

Commutative Rule

B

A

AB

O

29

If there are more than 2 vectors therefore

Use vector polygon and associative rule.

E.g. RQP

RQPRQP

Associative Rule

P

R

Q

RQP

Q

P

R

30

Distributive Rule :

a.

b.

For example :

i Proof of case a: let = 2

BABA

AAA

number real are ,

BABA

2

A

B

BA

O BA

2

31

A

2O

B

2

BA

22

BABA

222

BABA

22 ii

32

Proof of case b: let = 2 and = 1

AAA

312

A

A

3

A

3

AAAA

12

A

2 A

AAA

1212

Parallelogram

Triangle

C

D

D

O

D

DC

O

D

DC

1.2.4 Subtraction of Vectors

33

34

Vectors subtraction can be used

to determine the velocity of one object relative to another object i.e. to

determine the relative velocity.

to determine the change in velocity of a moving object.

35

1.Vector A has a magnitude of 8.00 units and 45

above the positive x axis. Vector B also has a

magnitude of 8.00 units and is directed along the

negative x axis. Using graphical methods and

suitable scale to determine

a) b)

c) d)

Hint : use 1 cm = 2.00 units)

Exercise 1.2 :

BA

BA

B2A

BA2

1.2.5 Resolving a Vector

o1st method : 2nd method :

0x

y

D

yD

xD

D

xD

yD

0x

y

D

Dx cos D Dx cos

D

Dysin D Dy sin

sinD

Dx sin D Dx

cosD

Dy cos D Dy

36

37

The magnitude of vector D :

Direction of vector D :

Vector D in terms of unit vectors written as

2y2x DDDD or

x

y

D

D tan or

x

y

D

D 1tan

jDiDD yx

38

A car moves at a

velocity of 50 m s-1 in

a direction north 30

east. Calculate the

component of the

velocity

a) due north.

b) due east.

Example 6 :N

EW

S

Nv

Ev

v30

60

a)

b)

30vvN cos

1s m 43.3 Nv

3050vN cosor

60vvN sin6050vN sin

30vvE sin

1s m 25 Ev

3050vE sin

or60vvE cos6050vE cos

Solution :

39

A particle S experienced a

force of 100 N as shown in

figure above. Determine the

x-component and the y-

component of the force.

Example 7 :

120

F

Sx

Sx

y

12060

F

yF

xF

Solution :

40

A particle S experienced a force of 100 N as shown in

figure above. Determine the x-component and the y-

component of the force.

Solution :

Example 7 :120

F

Sx

Vector x-component y-component

60FFx cos

N 50xF

60100Fx cos

orF

120FFx cos

N 50xF

120100Fx cos

60FFy sin

N 86.6yF

60100Fy sin

or120FFy sin

N 86.6yF

120100Fy sin

41

The figure above shows three forces F1, F2 and F3 acted on a particle O. Calculate the magnitude and

direction of the resultant force on particle O.

Example 8 : y

45o

O

)( N30F2

)( N10F1

30ox

)( N40F3

20

42

Solution :

321r FFFFF

yxr FFF

x3x2x1x FFFF

y3y2y1y FFFF

O

y

x

3F

45o30o

20

1F

y1F

2F

y2F

x1F

y3F

x3F

x2F

43

Solution :

Vector x-component y-component

20FF 1x1 cos

1F

3F

2F

2010F x1 cosN 9.40x1F

20FF 1y1 sin2010F y1 sin

N 3.42y1F4530F x2 cos

N 21.2x2F

4530F y2 sin

N 21.2y2F3040F x3 cos

N 34.6x3F

3040F y3 sinN 20.0y3F

Vector

sum

34.621.29.40 xFN 4.00 xF

20.021.23.42 yFN 37.8 yF

44

Solution :

The magnitude of the resultant force is

and its direction is

22 yxr FFF

N 38.0rF

22 37.84.00 rF

x

y

F

F 1tan

iseanticlockwaxis - xpositive from 264or 84.0

4.00

37.8tan 1

y

xO

rF

yF

xF

84.0

264

45

1. Vector has components Ax = 1.30 cm, Ay = 2.25 cm; vector has

components Bx = 4.10 cm, By = -3.75 cm. Determine

a) the components of the vector sum ,

b) the magnitude and direction of ,

c) the components of the vector ,

d) the magnitude and direction of .

( Young & freedman,pg.35,no.1.42)

ANS. : 5.40 cm, -1.50 cm; 5.60 cm, 345; 2.80 cm, -6.00 cm;

6.62 cm, 295

2. For the vectors and in figure 1.2, use the method of vector resolution to

determine the magnitude and direction of

a) the vector sum ,

b) the vector sum ,

c) the vector difference ,

d) the vector difference .

(Young & freedman,pg.35,no.1.39)

ANS. : 11.1 m s-1, 77.6; U think;

28.5 m s-1, 202; 28.5 m s-1, 22.2

Exercise 1.3 :

BA

A

BA

AB

AB

B

A

B

BA

AB

BA

AB

Figure 1.2

y

x0

37.0

-1s m 18.0B

-1s m 12.0A

46

notations

E.g. unit vector a : a vector with a magnitude of 1 unit in the direction of vector A.

Unit vector for 3 dimension axes :

1.2.6 Unit Vectors

A

a

cba , ,

1 A

Aa

)(@- boldjjaxisy 1 kji

)(@- boldiiaxisx

)(@- boldkkaxisz

47

Vector can be written in term of unit vectors as :

Magnitude of vector,

x

z

y

k

j

i

krjrirr zyx

2z2y2x rrrr

48

E.g. : m 234 kjis

m 5.39234 222 s

j3

x/m

y/m

z/m

0

s

i4k2

49

Two vectors are given as:

Calculate

a)the vector and its

magnitude,

b)the vector and its

magnitude,

c) the vector and its

magnitude.

Example 9 :

m 542 kjia

m 87 kjib

ab

ba

ba

2

Solution :

a)

The magnitude,

2i 7i 5x xx

a b a b i

4j 8j 4y yy

a b a b j

m 645 kjiba

5k 1k 6z zz

a b a b k

2 2 2

5 4 6a b

8.78 m


Calculate

a)the vector and its

magnitude,

b)the vector and its

magnitude,


magnitude.

Example 9 :

m 542 kjia

m 87 kjib

ab

ba

ba

2

Solution :

b) 7i 2i 9x xx

b a b a i

8j 4j 12y yy

b a b a j

1k 5k 4z zz

b a b a k

m 4129 kjiab

The magnitude,

2 2 2

9 12 4b a

15.5 m

50

51


Calculate

a) the vector and its

magnitude,

b) the vector and its

magnitude,


magnitude.

Example 9 :

m 542 kjia

m 87 kjib

ab

ba

ba

2

Solution :

c)

The magnitude,

m 1132 kiba

m 11.41132 22 ba

2 2 x xx

a b a b

2 2 7i 3i i

2 2 y yy

a b a b

2 4 8j 0j j

2 2 z zz

a b a b

2 5 1k 11k k

52

Scalar (dot) product

The physical meaning of the scalar product can be

explained by considering two

vectors and as shown in

figure 1.3a.

1.2.7 Multiplication of Vectors

A

B

A

B

Figure 1.3a

53

Scalar (dot) product

Figure 1.3b shows the projection of vector onto the direction of vector

.

Figure 1.3c shows the projection of vector onto the direction of vector

.

1.2.7 Multiplication of Vectors

A

B

A

B

A

B

Bcos

Figure 1.3b

A

B

AcosFigure 1.3c

BABBA

toparallel ofcomponent

ABABA

toparallel ofcomponent

54

From the figure 1.3b, the scalar product can be defined as

meanwhile from the figure 1.3c,

where

The scalar product is a scalar quantity.

The angle ranges from 0 to 180 .

When

BABA cos

vectorsobetween tw angle :

ABAB cos

ABBA

55

The scalar product obeys the commutative law of multiplication i.e.

Example of scalar product is work done by a constant force where the expression is

given by

900 scalar product is positive

18009 scalar product is negative

90 scalar product is zero

FssFsFW coscos

The scalar product of the unit vectors are shown below :

x

z

y

k

j

i

111 cos 2 o2 0iii

1 kkjjii

111 cos 2 o2 0jjj 111 cos 2 o2 0kkk

09 cos o011ji0 kikjji

09 cos o011ki 09 cos o011kj

57

Vector (cross) product

Consider two vectors :

In general, the vector product is defined as

and its magnitude is given by

where

The angle ranges from 0 to 180 so the vector product always positive value.

krjqipB

kzjyixA

CBA

ABBACBA sinsin

vectorsobetween tw angle :

58

Vector product is a vector quantity.

The direction of vector is determined by CORKSCREW

METHOD or RIGHT HAND RULE

C

59

For example:

How to use right hand rule :

Point the 4 fingers to the direction of the 1st vector.

Swept the 4 fingers from the 1st

vector towards the 2nd vector.

The thumb shows the direction of the vector product.

Direction of the vector product always perpendicular to the

plane containing the vectors

and .

A

C

B

A

B

C

CBA

CAB

ABBA

but

ABBA

B

)(C

A

60

The vector product of the unit vectors are shown below :

x

z

y

k

j

i

ijkkj

kijji

jkiik

0 kkjjii

0in o2 0siii

0in o2 0sjjj

0in o2 0skkk

61

Example of vector product is a magnetic force on the straight conductor carrying

current places in magnetic field where the

expression is given by ;

The vector product can also be expressed in determinant form as

BlIF

IlBF sin

rqp

zyx

kji

BA

62

1st method :

2nd method :

Note :

The angle between two vectors can only be determined by using the scalar (dot)

product.

kypxqjzpxrizqyrBA

kypxqjxrzpizqyrBA

63

Given two vectors :

Determine

a)

b) the angle between

vectors and .

Example 11 :

BA

kjiA 425

kjiB 5

A

B

Solution :

a)

kjikjiBA 5425

kkjjiiBA 541215

2025 BA

23BA

64

Given two vectors :

Determine

a)

b) the angle between

vectors and .

Example 11 :

BA

kjiA 425

kjiB 5

A

B

Solution :

b)The magnitude of vectors,

Using the scalar (dot) product

formula,

45425 222 A

27511 222 B

ABBA cos

2745

23coscos 11

AB

BA

7.48

65

1. If vector and vector , determine

a) , b) , c) .

ANS. :

2. Three vectors are given as follow :

Calculate

a) , b) , c) .

ANS. :

Exercise 1.4 :

46 ;26 ;2k

jia += 53

jib += 42

ba

ba

bba

kjickjibkjia 22 and 24; 233

cba cba

cba

kji 9115 ;9 ;21

66

3. If vector and vector

. Determine

a)

b) the angle between and .

ANS. :

Exercise 1.4 :

kjiP 23

kjiQ 342

QP

P

Q

92.8 ;16710 kji

67

THE ENDNext

ChapterCHAPTER 2

:

Kinematics

of

Linear

chapter 1 note

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