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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
Example 1 : Solution :
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
Example 1 : Solution :
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
-
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 :
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
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 :
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