This document provides an overview of Chapter 1 which covers physical quantities and measurements. It discusses basic and derived quantities and their respective SI units. Basic quantities cannot be derived from other quantities, while derived quantities can be derived from basic quantities. It also covers vector and scalar quantities, unit conversion using prefixes, and vector addition and subtraction. Students should be able to define physical quantities and units, convert between units, and perform basic vector operations after completing this chapter.
This is a summary of the topic "Physical quantities, units and measurement" in the GCE O levels subject: Physics. Students taking either the combined science (chemistry/physics) or pure Physics will find this useful. These slides are prepared according to the learning outcomes required by the examinations board.
This is a summary of the topic "Physical quantities, units and measurement" in the GCE O levels subject: Physics. Students taking either the combined science (chemistry/physics) or pure Physics will find this useful. These slides are prepared according to the learning outcomes required by the examinations board.
Physical Quantities--Units and Measurement--Conversion of UnitsKhanSaif2
This presentation covers physical quantities and their types, units and their types, conversion of units and order of magnitude in a very interactive manner. I hope this presentation will be helpful for teachers as well as students.
Math cad vm-001 stress-strain transformationsJulio Banks
This report is intended to clarify that stress P and Q are actually principal stress 1 and 2 but not physical locations
P and Q. Additionally, the strains calculated in Method II
are solved directly with a matrix inversion and not iteratively
as provided in the reference paper. Nonetheless, this is
an excellent VM (Validation Model) recommended for
FEA (Finite Element Analysis) substantiation.
Physical Quantities--Units and Measurement--Conversion of UnitsKhanSaif2
This presentation covers physical quantities and their types, units and their types, conversion of units and order of magnitude in a very interactive manner. I hope this presentation will be helpful for teachers as well as students.
Math cad vm-001 stress-strain transformationsJulio Banks
This report is intended to clarify that stress P and Q are actually principal stress 1 and 2 but not physical locations
P and Q. Additionally, the strains calculated in Method II
are solved directly with a matrix inversion and not iteratively
as provided in the reference paper. Nonetheless, this is
an excellent VM (Validation Model) recommended for
FEA (Finite Element Analysis) substantiation.
This is an introductory lecture on physics. explaining vectors, scalars , The SI unit system and the Dimensions and calculations. explained briefly with examples helps you to understand better concepts of fundamentals
Fisika Bilingual Besaran dan Satuan Kelas 7Frank Nanda
Ini adalah LKS (Lembar Kerja Siswa untuk Bab.Besaran dan Satuan di Kelas 7. Untuk memberi siswa wawasan lebih luas tentang kata-kata pada fisika yang ada di luar negeri.
11 - 3
Experiment 11
Simple Harmonic Motion
Questions
How are swinging pendulums and masses on springs related? Why are these types of
problems so important in Physics? What is a spring’s force constant and how can you measure
it? What is linear regression? How do you use graphs to ascertain physical meaning from
equations? Again, how do you compare two numbers, which have errors?
Note: This week all students must write a very brief lab report during the lab period. It is
due at the end of the period. The explanation of the equations used, the introduction and the
conclusion are not necessary this week. The discussion section can be as little as three sentences
commenting on whether the two measurements of the spring constant are equivalent given the
propagated errors. This mini-lab report will be graded out of 50 points
Concept
When an object (of mass m) is suspended from the end of a spring, the spring will stretch
a distance x and the mass will come to equilibrium when the tension F in the spring balances the
weight of the body, when F = - kx = mg. This is known as Hooke's Law. k is the force constant
of the spring, and its units are Newtons / meter. This is the basis for Part 1.
In Part 2 the object hanging from the spring is allowed to oscillate after being displaced
down from its equilibrium position a distance -x. In this situation, Newton's Second Law gives
for the acceleration of the mass:
Fnet = m a or
The force of gravity can be omitted from this analysis because it only serves to move the
equilibrium position and doesn’t affect the oscillations. Acceleration is the second time-
derivative of x, so this last equation is a differential equation.
To solve: we make an educated guess:
Here A and w are constants yet to be determined. At t = 0 this solution gives x(t=0) = A,
which indicates that A is the initial distance the spring stretches before it oscillates. If friction is
negligible, the mass will continue to oscillate with amplitude A. Now, does this guess actually
solve the (differential) equation? A second time-derivative gives:
Comparing this equation to the original differential equation, the correct solution was
chosen if w2 = k / m. To understand w, consider the first derivative of the solution:
−kx = ma
a = −
k
m
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
x
d 2x
dt 2
= −
k
m
x x(t) = A cos(ωt)
d 2x(t)
dt 2
= −Aω2 cos(ωt) = −ω2x(t)
James Gering
Florida Institute of Technology
11 - 4
Integrating gives
We assume the object completes one oscillation in a certain period of time, T. This helps
set the limits of integration. Initially, we pull the object a distance A from equilibrium and
release it. So at t = 0 and x = A. (one.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
2. Chapter 01Chapter 01PHYSICPHYSIC
SS
2
Overview:
Physical quantitiesPhysical quantities
and measurementsand measurements
Basic andBasic and
derivedderived
quantitiesquantities
Scalar andScalar and
vectorvector
quantitiesquantities
UnitUnit
conversionconversion
VectorsVectors
multiplicationmultiplication
Addition,Addition,
subtractionsubtraction
and vectorand vector
resolutionresolution
Scalar andScalar and
vectorvector
productsproducts
3. Chapter 01Chapter 01PHYSICPHYSIC
SS
3
At the end of this chapter, students should beAt the end of this chapter, students should be
able to:able to:
StateState basic quantities and their respective SIbasic quantities and their respective SI
units: length (m), time (s), mass (kg), electricalunits: length (m), time (s), mass (kg), electrical
current (A), temperature (K), amount ofcurrent (A), temperature (K), amount of
substance (mol) and luminosity (cd).substance (mol) and luminosity (cd).
StateState derived quantities and their respectivederived quantities and their respective
units and symbols: velocity (m sunits and symbols: velocity (m s-1-1
), acceleration), acceleration
(m s(m s-2-2
), work (J), force (N), pressure (Pa), energy), work (J), force (N), pressure (Pa), energy
(J), power (W) and frequency (Hz).(J), power (W) and frequency (Hz).
ConvertConvert units to common SI prefixes.units to common SI prefixes.
Learning Outcome:
1.1 Physical Quantities and units (1 hour)
4. Chapter 01Chapter 01PHYSICPHYSIC
SS
4
Physical quantityPhysical quantity is defined as a quantity which can be measured.quantity which can be measured.
It can be categorised into 2 types
Basic (base) quantityBasic (base) quantity
Derived quantityDerived quantity
Basic quantityBasic quantity is defined as a quantity which cannot be derivedquantity which cannot be derived
from any physical quantities.from any physical quantities.
Table 1.1 shows all the basic (base) quantities.
1.1 Physical Quantities and Units
Quantity Symbol SI Unit Symbol
Length l metre m
Mass m kilogram kg
Time t second s
Temperature T/θ kelvin K
Electric current I ampere A
Amount of substance N mole mol
Luminous Intensity candela cdTable 1.1Table 1.1
5. Chapter 01Chapter 01PHYSICPHYSIC
SS
5
Derived quantityDerived quantity is defined as a quantity which can bequantity which can be
derived from basic quantity.derived from basic quantity.
Table 1.2 shows some examples of derived quantity.
Derived quantity Symbol Formulae SI Unit
Velocity v s/t m s-1
Volume V l × w × t m3
Acceleration a v/t m s-2
Density ρ 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
Power P W/t kg m2
s-3
@ W
Frequency f 1/T s-1
@ Hz
Pressure P F/A kg m-1
s-2
@ Pa
Table 1.2Table 1.2
6. Chapter 01Chapter 01PHYSICPHYSIC
SS
6
UnitUnit is defined as a standard size of measurement ofa standard size of measurement of
physical quantities.physical quantities.
Examples :
1 second1 second is defined as the time required forthe time required for
9,192,631,770 vibrations of radiation emitted by a9,192,631,770 vibrations of radiation emitted by a
caesium-133 atom.caesium-133 atom.
1 kilogram1 kilogram is defined as the mass of a platinum-iridiumthe mass of a platinum-iridium
cylinder kept at International Bureau of Weights andcylinder kept at International Bureau of Weights and
Measures ParisMeasures Paris.
1 meter1 meter is defined as the length of the path travelled bythe length of the path travelled by
light in vacuum during a time interval oflight in vacuum during a time interval of
s
458,792,299
1
second
kilogram
meter
7. Chapter 01Chapter 01PHYSICPHYSIC
SS The unit of basic quantity is called base unit
addition unit for base unit:
unit of plane angle - radian (rd)
The common system of units used today are S.I unit (S.I unit (SystemSystem
International/metric systemInternational/metric system)).
The unit of derived quantity – called derived unit
7
o
o
o
57.296
180
rad1
180rad
==
=
π
π
8. Chapter 01Chapter 01PHYSICPHYSIC
SS
8
It is used for presenting larger and smaller values.for presenting larger and smaller values.
Table 1.3 shows all the unit prefixes.
Examples:
5740000 m = 5740 km = 5.74 Mm
0.00000233 s = 2.33 × 10−6
s = 2.33 µs
Prefix Multiple Symbol
tera × 1012 T
giga × 109 G
mega × 106 M
kilo × 103 k
deci × 10−1 d
centi × 10−2 c
milli × 10−3 m
micro × 10−6
µ
nano × 10−9 n
pico × 10−12 p
1.1.1 Unit Prefixes
Table 1.3Table 1.3
9. Chapter 01Chapter 01PHYSICPHYSIC
SS Note:Note:
Line of prefix: m, s, N, A, g and etc…
T G M k d c m µ n p
1012
109
106
103
100
100
10−1
10−2 10−3
10−6
10−9
10−12
UnitSymbol :
Value :
How to use?How to use?
1 Ts = ? ps
1 Ts =1012−
minus the index - divisionminus the index - division
(−12)
ps
= 1024
ps
9
10. Chapter 01Chapter 01PHYSICPHYSIC
SS
10
Solve the following problems.
a. 45 mm2
= ? m2
b. 37 km h−1
= ? m s−1
c. 30 g cm−3
= ? kg m−3
d. 29 µm = ? m
e. 23 m h−1
= ? m s−1
Solution :Solution :
a.
b. 11stst
method :method :
3
1 37 10 m
37 km h
1 h
− ×
= ÷
Example 1.1 :
3
37 10 m
3600 s
×
= ÷
1 1
37 km h 10.3 m s− −
=
2 2
45 mm 45 1 mm= ×
( )
2
3
45 10 m−
= ×
2 6 2 5 2
45 mm 45 10 m or 4.5 10 m− −
= × ×
1.1.2 Conversion of Unit
11. Chapter 01Chapter 01PHYSICPHYSIC
SS
37 km 1000 m 1 h
1 h 1 km 3600 s
= ÷ ÷ ÷
11
22ndnd
method :method :
c.
1 37 km
37 km h
1 h
−
= ÷
1 1
37 km h 10.3 m s− −
=
( )
3 3
3
33 2 3
30 g 10 kg 1 cm
30 g cm
1 cm 1 g 10 m
−
−
−
÷= ÷ ÷ ÷
3 4 3
30 g cm 3.0 10 kg m− −
= ×
12. Chapter 01Chapter 01PHYSICPHYSIC
SS
12
d.
e.
29 mµ
5
29 m 2.9 10 mµ −
= ×
1 23 m 1 h
23 m h
1 h 3600 s
−
= ÷ ÷
1 3 1
23 m h 6.39 10 m s− − −
= ×
6
29 10 m−
= ×
Note:Note:
Unit conversion is important, but it’s also important to recogniserecognise when
it’s needed.
Always use the SI unitAlways use the SI unit.
13. Chapter 01Chapter 01PHYSICPHYSIC
SS
13
Calculate the volume in SI unit of a wire of length 125 cm and
diameter 0.65 mm.
Solution :Solution :
Given l = 1.25 m ; d =0.65×10−3
m
The radius of the wire is
The volume of the cylindrical wire is given by
3
0.325 10 mr −
= ×
Example 1.2 :
( ) ( )
2
3
0.325 10 1.25π −
= ×
7 3
4.15 10 mV −
= ×
2
V r lπ=
3
0.65 10
2 2
d
r
−
×
= =
14. Chapter 01Chapter 01PHYSICPHYSIC
SS
14
Given 1 (angstrom) Å = 10-10
m
1. Solve all the following problems.
2. Convert the following problems into SI unit.
a. 0.249 mm3
b. 5.87 cm2
c. 12 g km h-1
d. 9.78 g cm-3
e. 10 km h-1
f. 8.5 g cm h-2
Exercise 1.1 :
a. 35 cm =_________m f. 24 mm2
=___________m2
b. 20 km =_________m g. 0.03 m2
=____________ mm2
c. 11 Mm =_________m h. 19 km2
=_____________ m2
d. 23 µF =_________F i. 56 g cm-3
=_________ kg m-3
e. 0.24 Å =_________m j. 560 km h-1
=_________ m s-1
15. Chapter 01Chapter 01PHYSICPHYSIC
SS
15
At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
DefineDefine scalar and vector quantities.scalar and vector quantities.
PerformPerform vector addition and subtraction graphically.vector addition and subtraction graphically.
Resolve vectorResolve vector into two perpendicular components(x andinto two perpendicular components(x and
y axes)y axes)
Ilustrate unit vectors in cartesion coordinate.Ilustrate unit vectors in cartesion coordinate.
State the physical meaning of dot(scalar) product:State the physical meaning of dot(scalar) product:
State the physical meaning of cross(vector) product:State the physical meaning of cross(vector) product:
Direction of cross product is determined by corkscrewDirection of cross product is determined by corkscrew
method or right hand rule.method or right hand rule.
Learning Outcome:
1.2 Scalars and Vectors (2 hours)
( , , )i j k%% %
( cos ) ( cos )A B A B B Aθ θ• = =
r r
( sin ) ( sin )A B A B B Aθ θ× = =
r r
16. Chapter 01Chapter 01PHYSICPHYSIC
SS
16
State the physical meaning of dot(scalar)State the physical meaning of dot(scalar)
product:product:
State the physical meaning of cross(vector)State the physical meaning of cross(vector)
product:product:
Direction of cross product is determined byDirection of cross product is determined by
corkscrew method or right hand rule.corkscrew method or right hand rule.
Learning Outcome:
1.2 Scalars and Vectors (2 hours)
( ) ( )θABθBABA coscos ==•
rr
( ) ( )θABθBABA sinsin ==×
rr
17. Chapter 01Chapter 01PHYSICPHYSIC
SS
17
ScalarScalar quantity is defined as a quantity with magnitudequantity with magnitude only.
e.g. mass, time, temperature, pressure, electric current,
work, energy, power and etc.
Mathematics operational : ordinary algebra
VectorVector quantity is defined as a quantity with both magnitudequantity with both magnitude
& direction.& direction.
e.g. displacement, velocity, acceleration, force, momentum,
electric field, magnetic field and etc.
Mathematics operational : vector algebra
1.2 Scalars and Vectors
18. Chapter 01Chapter 01PHYSICPHYSIC
SS
18
Table 1.4 shows written form (notation) of vectors.
Notation of magnitude of vectors.
1.2.1 Vectors
Vector A
LengthLength of an arrow– magnitudemagnitude of vector A
displacement velocity acceleration
s
r
v
r
a
r
s av
vv =
r
aa =
r
s (bold) v (bold) a (bold)
DirectionDirection of arrow – directiondirection of vector A
Table 1.4Table 1.4
19. Chapter 01Chapter 01PHYSICPHYSIC
SS
19
Two vectorsTwo vectors equal if both magnitude and directionmagnitude and direction are the same.same.
(shown in figure 1.1)
If vector A is multiplied by a scalar quantity k
Then, vector A is
if kk = +ve= +ve, the vector is in the same directionsame direction as vector A.
if kk == --veve, the vector is in the opposite directionopposite direction of vector A.
P
r
Q
r
QP
rr
=
Figure 1.1Figure 1.1
Ak
r
Ak
r
A
r
A
r
−
20. Chapter 01Chapter 01PHYSICPHYSIC
SS
20
Can be represented by using:
a)a) Direction of compassDirection of compass, i.e east, west, north, south, north-east,
north-west, south-east and south-west
b)b) Angle with a reference lineAngle with a reference line
e.g. A boy throws a stone at a velocity of 20 m s-1
, 50° above
horizontal.
1.2.2 Direction of Vectors
50°
v
r
x
y
0
24. Chapter 01Chapter 01PHYSICPHYSIC
SS
24
There are two methods involved in addition of vectors graphically i.e.
ParallelogramParallelogram
TriangleTriangle
For example :
1.2.3 Addition of Vectors
ParallelogramParallelogram TriangleTriangle
B
r
A
r
B
r
A
r
BA
rr
+
O
BA
rr
+
B
r
A
r
BA
rr
+
O
25. Chapter 01Chapter 01PHYSICPHYSIC
SS
25
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
rrrr
+=+ Commutative RuleCommutative Rule
B
r
A
r
AB
rr
+
O
26. Chapter 01Chapter 01PHYSICPHYSIC
SS
26
For example :
1.2.4 Subtraction of Vectors
ParallelogramParallelogram TriangleTriangle
D
rC
r
O
DC
rr
−
O
D
r
−
( )DCDC
rrrr
−+=−
C
r
D
r
−
DC
rr
−
C
r
D
r
−DC
rr
−
27. Chapter 01Chapter 01PHYSICPHYSIC
SS
27
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.
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
rr
+ BA
rr
−
B2A
rr
+ BA2
rr
−
28. Chapter 01Chapter 01PHYSICPHYSIC
SS
28
11stst
methodmethod :
1.2.5 Resolving a Vector
R
r
yR
r
xR
r
θ
0
x
y
θ
R
Rx
cos= cosxR Rθ⇒ =
θ
R
Ry
sin= θRRy sin=⇒
22ndnd
methodmethod :
R
r
yR
r
xR
r
φ
0
x
y
φsin=
R
Rx
φsinRRx =⇒
φcos=
R
Ry
φcosRRy =⇒
φ
AdjacentAdjacent
componentcomponent
OppositeOpposite
componentcomponent
29. Chapter 01Chapter 01PHYSICPHYSIC
SS
29
The magnitude of vectormagnitude of vector R :
Direction of vectorDirection of vector R :
Vector R in terms of unit vectors written as
( ) ( )22
or yx RRRR +=
x
y
R
R
θ =tan or
= −
x
y
R
R
θ 1
tan
jRiRR yx
ˆˆ +=
30. Chapter 01Chapter 01PHYSICPHYSIC
SS
30
A car moves at a velocity of 30 m s-1
in a direction north 60° west.
Calculate the component of the velocity
a) due north. b) due west.
Solution :Solution :
Example 1.3 :
N
EW
S
Nv
v
a)
b)
60cosN vv =
1
N sm51 −
=v
60cos30=
or
30sinN vv =
30sin30=
60sinW vv =
1
W sm26 −
=v
60sin30=
or
30cosW vv =
30cos30=
60°
30°
Wv
31. Chapter 01Chapter 01PHYSICPHYSIC
SS
31
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 :Solution :
Example 1.4 :
Vector x-component y-component
30cosFFx −=
N6.68−=xF
30cos100−=F
30sinFFy −=
N05−=yF
30sin100−=
210°
F
S
x
S
210°
F
x
y
30°
xF
yF
32. Chapter 01Chapter 01PHYSICPHYSIC
SS
32
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 1.5 :
y
O
60o
x
( )N101F
( )N302F
( )N403F
35. Chapter 01Chapter 01PHYSICPHYSIC
SSSolution :Solution :
The magnitude of the resultant force is
and
Its direction is 32.632.6°° from positive x-axis ORfrom positive x-axis OR
35
y
x
O
( ) ( )22
∑∑ += yxr FFF
N.729=rF
( ) ( )22
1625 +=
=
∑
∑−
x
y
F
F
θ 1
tan
6.32
25
16
tan 1
=
= −
θ
∑ yF
rF
32.6
∑ xF
36. Chapter 01Chapter 01PHYSICPHYSIC
SS
36
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, 345ANS. : 5.40 cm, -1.50 cm; 5.60 cm, 345°°; 2.80 cm, -6.00 cm;; 2.80 cm, -6.00 cm;
6.62 cm, 2956.62 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 sANS. : 11.1 m s-1-1
, 77.6, 77.6°°; U think;; U think;
28.5 m s28.5 m s-1-1
, 202, 202°°; 28.5 m s; 28.5 m s-1-1
, 22.2, 22.2°°
Exercise 1.3 :
BA
+
A
BA
+
AB
−
AB
−
B
A
B
BA
+
AB
+
BA
−
AB
−
Figure 1.2Figure 1.2
y
x
0
37.0°
( )-1
sm18.0B
( )-1
sm12.0A
37. Chapter 01Chapter 01PHYSICPHYSIC
SS
37
3. Vector points in the negative x direction. Vector points at an
angle of 30° above the positive x axis. Vector has a magnitude of
15 m and points in a direction 40° below the positive x axis. Given
that , determine the magnitudes of and .
(Walker,pg.78,no. 65)
ANS. : 28 m; 19 mANS. : 28 m; 19 m
4. Given three vectors P, Q and R as shown in Figure 1.3.
Calculate the resultant vector of P, Q and R.
ANS. : 49.4 m sANS. : 49.4 m s−−22
; 70.1; 70.1°° above + x-axisabove + x-axis
Exercise 1.3 :
C
A
B
0=++ CBA
A
B
Figure 1.3Figure 1.3
y
x0
50°
( )2
sm10 −
R
( )2
sm35 −
P
( )2
sm24 −
Q
38. Chapter 01Chapter 01PHYSICPHYSIC
SS
38
notations –
E.g. unit vector a – a vector with a magnitude of 1 unit in thevector with a magnitude of 1 unit in the
direction of vectordirection of vector AA.
Unit vectors have no unitno unit.
Unit vector for 3 dimension axes :
1.2.6 Unit Vectors
A
aˆ
cba ˆ,ˆ,ˆ
1ˆ ==
A
A
a
)(@ˆ⇒- boldjjaxisy 1ˆˆˆ === kji
)(@ˆ⇒- boldiiaxisx
)(@ˆ⇒- boldkkaxisz
39. Chapter 01Chapter 01PHYSICPHYSIC
SS
39
Vector can be written in term of unit vectors as :
Magnitude of vector,
x
z
y
kˆ
jˆ
iˆ
krjrirr zyx
ˆˆˆ ++=
( ) ( ) ( )2
z
2
y
2
x rrrr ++=
41. Chapter 01Chapter 01PHYSICPHYSIC
SS
41
Two vectors are given as:
Calculate
a) the vector and its magnitude,
b) the vector and its magnitude,
c) the vector and its magnitude.
Solution :Solution :
a)
The magnitude,
Example 1.6 :
ab
−
( )mˆ3ˆˆ2 kjia −−=
ba
+
( )mˆ4ˆ2ˆ kjib −+=
( ) ibaba xxx
ˆ312 =+=+=+
( ) jbaba yyy
ˆ21 =+−=+=+
( )mˆ7ˆˆ3 kjiba −+=+
( ) ( ) kbaba zzz
ˆ743 −=−+−=+=+
( ) ( ) ( ) m68.7713
222
=−++=+ ba
ba
+2
43. Chapter 01Chapter 01PHYSICPHYSIC
SS
43
Scalar (dot) productScalar (dot) product
The physical meaning of the scalar productphysical meaning of the scalar product can be explained by
considering two vectors and as shown in Figure 1.4a.
Figure 1.4b shows the projection of vector onto the direction of
vector .
Figure 1.4c shows the projection of vector onto the direction of
vector .
1.2.7 Multiplication of Vectors
A
B
θ
A
B
A
B
Figure 1.4aFigure 1.4a
θ
A
B
A
B
θBcos
Figure 1.4bFigure 1.4b
θ
A
B
θAcos
Figure 1.4cFigure 1.4c
( )ABABA
toparallelofcomponent=•
( )BABBA
toparallelofcomponent=•
44. Chapter 01Chapter 01PHYSICPHYSIC
SS
44
From the Figure 1.4b, the scalar product can be defined as
meanwhile from the Figure 1.4c,
where
The scalar product is a scalar quantityscalar quantity.
The angle θ ranges from 0° to 180 °.
When
The scalar product obeys the commutative law of multiplicationcommutative law of multiplication i.e.
( )θBABA cos=•
vectorsobetween twangle:θ
( )θABAB cos=•
90θ0 << scalar product is positivepositive
180θ09 << scalar product is negativenegative
90θ = scalar product is zerozero
ABBA
•=•
45. Chapter 01Chapter 01PHYSICPHYSIC
SS
45
Example of scalar product is work donework done by a constant force where the
expression is given by
The scalar product of the unit vectors are shown below :
( ) ( )θFsθsFsFW coscos ==•=
x
z
y
kˆ
jˆ
iˆ
( ) ( ) 111cosˆˆ 2
===• o2
0iii
1ˆˆˆˆˆˆ =•=•=• kkjjii
( ) ( ) 111cosˆˆ 2
===• o2
0jjj
( ) ( ) 111cosˆˆ 2
===• o2
0kkk
( )( ) 09cosˆˆ ==• o
011ji
0ˆˆˆˆˆˆ =•=•=• kikjji
( )( ) 09cosˆˆ ==• o
011ki
( )( ) 09cosˆˆ ==• o
011kj
46. Chapter 01Chapter 01PHYSICPHYSIC
SS
46
Calculate the and the angle θ between vectors and for the
following problems.
a) b)
Solution :Solution :
a)
The magnitude of the vectors:
The angle θ ,
Example 1.7 :
A
BA
• B
( )( ) ( )( ) ( )( ) kkjjiiBA ˆˆ33ˆˆ12ˆˆ12 •−+•+•=•
( ) ( ) ( ) 17322
222
=++=A
kjiA ˆ3ˆ2ˆ2 ++=
kjiA ˆˆ3ˆ4 +−=
kjiB ˆ3ˆˆ −+=
kjB ˆ3ˆ2 +=
922 −+=• BA
5−=• BA
( ) ( ) ( ) 11311
222
=−++=B
θABBA cos=•
−
=
•
= −−
1117
5
coscos 11
AB
BA
θ
112=θ
ANS.:ANS.:−−3; 99.43; 99.4°°
47. Chapter 01Chapter 01PHYSICPHYSIC
SS
47
Referring to the vectors in Figure 1.5,
a) determine the scalar product between them.
b) express the resultant vector of C and D in unit vector.
Solution :Solution :
a) The angle between vectors C and D is
Therefore
Example 1.8 :
2
m441.DC −=•
( )
1361925180 =+−=θ
Figure 1.5Figure 1.5
y
x0
( )m1C
( )m2D
19°25°
θCDDC cos=•
( )( )
136cos21=
48. Chapter 01Chapter 01PHYSICPHYSIC
SS
48
b) Vectors C and D in unit vector are
and
Hence
jCiCC yx
ˆˆ +=
( ) ( )ji ˆ25sin1ˆ25cos1
+−=
( )mˆ42.0ˆ910 ji.C +−=
( ) ( ) jiDC ˆ65.042.0ˆ89.191.0 +++−=+
( )mˆ07.1ˆ98.0 ji +=
( ) ( )jiD ˆ19sin2ˆ19cos2
+=
( )mˆ65.0ˆ891 ji.D +=
49. Chapter 01Chapter 01PHYSICPHYSIC
SS
49
Vector (cross) productVector (cross) product
Consider two vectors :
In general, the vector product is defined as
and its magnitudemagnitude is given by
where
The angle θ ranges from 0° to 180 ° so the vector product always
positivepositive value.
Vector product is a vector quantityvector quantity.
The direction of vector is determined by
krjqipB ˆˆˆ ++=
kzjyixA ˆˆˆ ++=
CBA
=×
θABθBACBA sinsin ===×
vectorsobetween twangle:θ
RIGHT-HAND RULERIGHT-HAND RULE
C
Note:Note:
The angle betweenThe angle between
two vectorstwo vectors can only
be determined by
using the scalar (dot)scalar (dot)
productproduct.
50. Chapter 01Chapter 01PHYSICPHYSIC
SS
50
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 perpendicularDirection of the vector product always perpendicular
to the plane containing the vectors andto the plane containing the vectors and .
A
C
B
A
B
C
CBA
=×
CAB
=×
ABBA
×≠× but ( )ABBA
×−=×
B
)(C
A
51. Chapter 01Chapter 01PHYSICPHYSIC
SS
51
The vector product of the unit vectors are shown below :
Example of vector product is a torque (moment of force) on a metrea torque (moment of force) on a metre
rulerule where the expression is given by
Vector form:
Magnitude form:
x
z
y
kˆ
jˆ
iˆ
ijkkj ˆˆˆˆˆ =×−=×
kijji ˆˆˆˆˆ =×−=×
jkiik ˆˆˆˆˆ =×−=×
0ˆˆˆˆˆˆ =×=×=× kkjjii
0inˆˆ ==× o2
0siii
0inˆˆ ==× o2
0sjjj
0inˆˆ ==× o2
0skkk
r Fτ = ×
sinrFθτ =
52. Chapter 01Chapter 01PHYSICPHYSIC
SS
52
Given two vectors :
The angle between both vectors is 155 degree.
Determine
a) the magnitude of and its direction.
b)
Solution :Solution : θ=155°
a) The magnitude of vectors,
Apply:
Example 1.9 :
BA
×
BA
•
jiA ˆˆ3 −−=
kjiB ˆ2ˆ2ˆ4 −+=
sinA B AB θ× =
( ) ( ) 1013
22
=−+−=A
( ) ( ) ( ) 24224
222
=−++=B
( )( )10 24 sin155=
2
6.55 unitA B× =
53. Chapter 01Chapter 01PHYSICPHYSIC
SS
53
Direction: always perpendicular to plane containing vector A
and B.
b) ( ) ( )kjikjiBA ˆ2ˆ2ˆ4ˆ0ˆˆ3 −+•+−−=•
2
14 unitA B• = −
( )( ) ( )( ) ( )( ) kkjjii ˆˆ20ˆˆ21ˆˆ43 •−+•−+•−=
0212 +−−=
54. Chapter 01Chapter 01PHYSICPHYSIC
SS
54
1. If vector and vector , determine
a) b)
ANS. :ANS. :
2. Three vectors are given as follow :
Calculate
a) b)
ANS. :ANS. :
3. If vector and vector ,
determine
a) the direction of
b) the angle between and .
ANS. : U think, 92.8ANS. : U think, 92.8°°
Exercise 1.4 :
26; 46
jia ˆ+ˆ= 53
jib ˆ+ˆ= 42
ba
• ( ) bba
•+
kjickjibkjia ˆˆ2ˆ2andˆ2ˆ4ˆ;ˆ2ˆ3ˆ3 ++=+−−=−+=
( )a b c• −
( )cba
+•
29; 9− −
kjiP ˆˆ2ˆ3 −+=
kjiQ ˆ3ˆ4ˆ2 ++−=
QP
×
P
Q