lecture note 1_1

| Chapter 1 | PHYSICAL QUANTITIES & MEASUREMENTS |

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TABLE : Contents of SF 017 Physics Semester 1

Chapter Date Topic Check List [ √ ]

1 …………. Physical Quantities & Measurements. [ ]

2 …………. Kinematics of Linear Motion. [ ]

3 …………. Force, Momentum & Impulse. [ ]

4 …………. Work, Energy & Power. [ ]

5 …………. Static. [ ]

6 …………. Circular Motion. [ ]

7 …………. Rotational Motion. [ ]

8 …………. Gravitation. [ ]

9 …………. Simple Harmonics Motion (SHM). [ ]

10 …………. Mechanical Waves. [ ]

11 …………. Sound Wave. [ ]

12 …………. The Mechanical Properties of Matter. [ ]

13 …………. Fluid Mechanics. [ ]

14 …………. Temperature & Heat Transfer. [ ]

15 …………. Kinetic Theory of Gases. [ ]

16 …………. Thermodynamics. [ ]

Retention rate from various methods of learning :

Method Percentage

Lecture …..5%

Reading …..10%

Audio visual presentation …..15%

Demonstration …..30%

Discuss in group …..50%

Practice by doing …..75%

Teaching others …..90%


FLOW CHART : Recommended Student Learning Process.

Designed By : KMM Physics Lecturer.

Read and Try to Understand the Content of the Topic Before Coming into the Lecture Class.

Pay Attention to the Lecture and Take Useful Notes / Information To Get Better Understanding.

Retry all the examples discussed During Lecture Session.

Try to Solve all the Exercises Provided or From Any Other Reference Book / Website.

Do Self-revision by Making Short Notes / Answering all the Questions Given in the Conceptual Map.

Do Discussion in a Small Study Group for Any Misunderstanding.

Answer the Tutorial Questions as well as the Past Years Final Examination Questions.

See Any Physics Lecturer if Having Unsolvable Problems.

Get Ready For The Next Topic (Don’t Study Last Minute).


Do the information complete to be substituted into

the main equation?

FLOW CHART : Strategy On Solving Problem In Physics

Designed By : KMM Physics Lecturer.

Think the related subsidiary formula (any equation that relates to

the sub-target)

Identify the sub-target (unknown useful

quantity) of the question

Read & understand the question

Relate the question to the related physics concept

Identify the target (quantity) of the question

Sketch an appropriate diagram to describe the situation

Determine the main physics formula (any equation that

relates to the target)

List down all the given information

Find out all the hidden information

Do calculation in standard SI unit

Get the answer within the correct unit

Explain the physical meaning

No

Yes


METHODS USED

RESOLVED INTO AXES

UNIT VECTOR

PARALLE-LOGRAM

TIP TO TAIL

| CONCEPTUAL MAP |

CHAPTER 1 : PHYSICAL QUANTITIES & MEASUREMENTS Answer all the following questions and store permanently in your memory.

1. What is physical quantity? 2. What is scientific notation? 3. What is the difference between basic quantity and derived quantity? 4. State all the basic quantities with their respective symbol and unit. 5. What is dimensions? 6. State 3 uses of dimensions. 7. What is the difference between scalar quantity and vector quantity? 8. “A vector can be represented graphically as an arrow”. Explain this statement. 9. Explain the magnitude and direction of an arrow which representing a resultant vector. 10. State the 3 rules of vector additional operation. 11. State the formula to resolve a vector to its component. 12. State the formula of the magnitude and the direction of a resultant vector from its resolved vector. 13. What is unit vector? 14. State the formula of additional / subtraction of vectors through the unit vector method. 15. What is the difference between scalar product and vector product? 16. State the formula the formula of scalar and vector product. 17. State and explain the way to determine the direction of vector product. 18. What is the difference between resolved angle and angle between two vectores?


CHAPTER 1 : PHYSICAL QUANTITES & MEASUREMENTS

(5 Hours)

Terminology :

1. Physical Quantity : Kuantiti Fizik 2. Basic Quantity : Kuantiti Asas 3. Derived Quantity : Kuantiti Terbitan 4. Dimensions : Dimensi 5. Parallelogram Method : Kaedah Segiempat Selari 6. Triangle Method : Kaedah Segitiga 7. Product : Hasil Darab

Introduction :

1. The word ‘physics’ comes from the Greek word which means ‘nature’. Physics is the study of the physical objects and the natural phenomena. It involves the

process of doing experiments to observe how the nature is related to the laws and principles.

2. The study of physics ranges from fundamental particle of size 10-16 m to galactic objects

larger than 1024 m.

Quantities & Units :

1. In order to describe the phenomena quantitatively, we introduce the so-called physical

quantities which have to be observed, measured and described by numbers. 2. Physical quantities are quantities that are measurable with instruments or derived from

other quantities. Each quantity is represented with a symbol and uses a standard size called the ‘unit’.

© Physics Teaching Courseware

microscopic-scale normal-scale macroscopic-scale


3. Scientific notation : = A way of presenting numbers with a value between 0 to 10 multiplied by ten raised to a certain power.

Normally, the value of a physical quantity is represented by the Unit Prefixes :

Factor Prefixa Symbol Factor Prefix

a Symbol

1024

1021

1018

1015

1012

109

106

103

102

101

yotta- zeta- exa- peta- tera- giga- mega- kilo-

hecto- deka-

Y Z E P T G M k

h da

10-1

10-2

10-3

10-6

10-9

10-12

10-15

10-18

10-21

10-24

deci- centi- milli- micro- nano- pico- femto- atto- zepto- yocto-

d c m

µµµµ n p

f a z y

The most commonly used prefixes are shown in bold type.

Example : 0.001 m = 1.0x10-3 m = 1.0 mm 1000 m = 1.0x103 m = 1.0 km

4. Unit Conversion : It is sometimes necessary or more useful

to change from one set of units to another :

Mass Length

1 kg = 1000 g = 6.02x1026 u 1 slug = 14.6 kg 1 u = 1.66x10-27 kg

Volume

1 m3 = 1000 L = 35.3 ft3 = 264 gal

1 m = 100 cm = 39.4 in = 3.281 ft 1 mi = 1.609 km = 5280 ft 1 in = 2.54 cm 1 nm = 10 Å 1 light-year = 9.46x1015 m

Time Angular Measure

1 day = 86,400 s 1 year = 365.25 days π rad = 180o = 0.5 rev




Force Pressure

1 N = 105 dyne = 0.2248 lb 1 ton = 2000 lb

1 Pa = 1 N m-2 1 atm = 1.013x105 Pa = 76 cm Hg = 1.013 bar 1 mmHg = 1 torr

Energy Power

1 hp = 746 W = 550 ft lb s-1 Magnetism

1 J = 107 erg = 0.239 cal 1 kW.h = 3.6x106 J 1 eV = 1.6x10-19 J 1 T = 104 gauss = 1 Wb m-2

Basic quantity and Derived quantity :

1. Physical quantities are classified in to 2 types : Basic quantity and Derived quantity. 3. Basic Quantity :

A basic quantity (also known as base quantity or fundamental quantity) is a quantity which cannot be derived from any other quantities. Table below shows all the basic quantities :

Physical Quantity SI unit Symbol

Length Mass Time

Temperature Electric Current

Amount of Substance Luminous Intensity

meter, m kilogram, kg second, s kelvin, K ampere, A

mol candela, cd

l

m t T I n I

4. Derived Quantity :

A derived quantity is a quantity that is derived from the combination of several basic quantities by multiplication, division, differentiation or integration (according to defining equation).

Table below shows some examples of derived quantities :

Derived Quantity

Defining Equation

Unit Name Unit

Symbol Unit Expressed by Basic Units

Velocity v = s / t meter per second

m s-1 m s-1

Acceleration a = v / t Meter per second

persecond m s-2 m s-2

Work W = Fs joule J kg m2 s-2

Force F = ma newton N kg m s-2

Pressure p = F / A pascal Pa kg m-1 s-2

Energy E = Fs joule J kg m2 s-2

Power P = E / t watt W kg m2 s-3

Frequency f = 1 / T hertz Hz s-1

Electric Charge Q = It coulomb C A s

Voltage V = E / Q volt V kg m2 s-3 A-1

Resistance R = V / I ohm Ω kg m2 s-3 A-2


Note : Some ranges in physical studies.


Dimensions :

1. A new quantity can be derived from 2 or more basic quantities. These basic quantities

are combined in some specified way such as addition, subtraction, multiplication or division.

2. The dimensions of a physical quantity indicate the way it is related to the basic

quantities. Usually, the dimension of a physical quantity is shown as [X].

Basic Quantity Dimensions Symbol

Length Mass Time

Temperature Electric Current

[ l ] [ m] [ t ] [ T ] [ I ]

L M T θ I

3. In mechanics, a physical quantity is normally involved with only mass, length and time.

Hence, its dimensions can be expressed in the following way : [X] = MxLyTz where x, y and z are dimensionless constant. 4. Some physical quantities may have no dimensions (dimensionless quantity). 5. Applications (uses) of Dimensions :

a) To check the homogeneity of an equation (an equation is homogeneous if the dimension on both sides of the equation and all parts at right side are equal).

b) To construct an equation with the given quantities. c) To determined the dimensions as well as the SI unit for certain unknown

constants.

Technique 1 : Checking The Homogeneity of an Equation

(Zahidi, 2008)

Step Technique Explanation

1 Part division of right side. Divide the right of the equation into several parts if it involves the additional and/or subtraction operations (N operations → N + 1 parts).

2 Dimensional check up for every

part. Check the dimensions of every part at the right side of the equation separately.

3 Dimensional check up for the

left side.

The left side of the equation is normally a quantity and called the main quantity. Check the dimensions of this main quantity

4 Dimensional Comparison. Compare the dimensions of the main quantity (left side) with the dimensions of all parts of the right sides of the equation (compare step 2 with step 3).

5 Conclusion (determination). The equation is homogeneous if the dimension on both sides of the equation and all parts at right side are the same and vise versa.


Technique 2 : Constructing a Physical Equation

(Zahidi, 2003)


1 Factors identification. Identify all physical quantities (factors) that might be considered to be related to the main physical quantity.

2 Proportional relationship. Write a directly proportional relationship between the main quantity and the product of all the factors.

3 Power factor. Insert the power factor (dimensionless constant) to all the factors (such as x, y, z ect.)

4 Conversion into mathematical relationship (raw equation).

Convert the proportional relationship to the form of mathematical equation (∝ → =) by inserting a constant, k (normally dimensionless). This is actually the form of equation needed to achieve.

5 Dimensional determination of main quantity and all factors.

Determine the dimension of the main quantity (left side) as well as all the factors involved (right side). If the number of dimensions is unbalanced (normally for the main quantity), insert new dimensions with power factor of 0 (mathematical concept).

6 Distribution Distribute the equation into smaller parts involving identical dimensions to determine the value of the power factor.

7 Substitution into the raw

equation. Substitute all the values of power factor to its factors respectively.

8 Simplification of the equation. Rewrite (reform) the equation into the simplest form.


Contextual example : Constructing equation of the period, T of oscillation of a simple pendulum.

The period of oscillation, T of a simple pendulum is said to be influenced by mass, m, length of string, L and the acceleration due to gravity, g. 1. Main quantity : T Factors : m, L, g

2. T ∝ m L g

3. T ∝ mx Ly gz

4. T = k mx Ly gz

5. [ T ] = [ k ] [ m ] x [L ]y [ g ]z (T)1 = (1) (M)x (L)y (LT-2)z (T)1 = (M)x (L)y (L)z (T-2z) (T)1 = (M)x (Ly +z) (T-2z)

(T)1 (M)0 (L)0 = (M)x (Ly +z) (T-2z)

6. Dimensions of time : T1 = T-2z → z = -1/2

Dimensions of mass : M0 = Mx → x = 0 This means mass does not affected the period.

Dimensions of length : L0 = Ly +z → y + z = 0

y + (-1/2) = 0 y = ½

7. T = k L1/2

g-1/2

8. The equation is T = kg

L

Note : The value of k just only be determined through

experimental way. The actual equation of the period of oscillation of a

simple pendulum is :

T = 2πg

L



Technique 3 : Determination of Dimensions and SI unit of a Constant

(Zahidi, 2003)


1 Identification of the constant. Identify the constant of unidentified dimensions.

2 Reformation of equation. Reform the involved equation. Let the unknown quantity to be the main quantity (left side equation).

3 Dimension the equation.

Dimensions the both sides of the equation. Determine the dimensions of the right equation. If the equation consists of additional or subtraction operation of several parts, don’t add or subtract the dimensions of all the parts, whereas the dimensions of all the parts are equal!. However, dimensions can be multiplied and divided.

4 Determination of dimensions. The dimensions of the quantity is now equal to the dimensions of the right side of the equation.

5 Determination of SI unit. The SI unit of the unknown quantity can be derived based on the dimensions.

Contextual example : Determining the dimensions and SI unit of the coefficient of viscosity.

According to Stoke’s law (discussed in chapter 13 later), the speed of a sphere through a liquid is given by :

v = 9η

ρ)g(σ2r 2 −

where r = radius of the sphere. σ = density of the sphere.

ρ = density of the liquid. η = coefficient of viscosity (a constant). g = acceleration due to gravity.

1. The unknown constant = η.

2. η = 9v

ρg2r

9v

σg2r 22

−

3. [ η ] =

=

9v

ρg2r

9v

σg2r 22

Then, it can be simplified as [ η ] =

9v

σg2r 2

or [ η ] =

9v

ρg2r 2

, either one.

Then, [ η ] = [9][v]

[σσ][g[2][r] 2

= )LT)(1(

)LT)(ML()L)(1(1

232

−

−−

4. Then, the dimension of η is M L-1 T-1 5. Then, the SI unit of η is kg m-1 s-1.


Scalar and Vector Quantity :

1. A scalar quantity is a quantity which has magnitude only and has no direction. 2. A vector quantity is a quantity which has both magnitude and direction.

Table below shows some examples of scalar and vector quantities :

Scalar quantities Vector quantities

Distance, Length Speed Mass,

Inertia, Moment of Inertia

Area, Volume Density

Time, Period, Frequency

Temperature Heat, Energy, Photon

Intensity Work Power

Charge Electric Current Voltage, e.m.f Resistance Capacitance

Capacitive Reactance Inductance

Inductive Reactance Impedance All constant

Displacement Velocity

Weight, Tension, Force Acceleration Momentum Impulse Pressure

Gravitational Field Electric Field Magnetic Filed

3. A vector quantity can be labeled by writing :

a) a bold-faced type letter (X), or

b) a normal letter with an overhead arrow ( Xr

) The magnitude of a vector can be labeled by writing :

a) a plain italic letter, X or b) | X |

A vector quantity can be represented graphically as an arrow (directed line). The length OP = Magnitude of vector A.

The angle θ = Angle that indicates the direction of vector A with x-axis.

O

P

θ

A


Vector Operations

1. There are three (3) mathematical operations involving 2 or more vectors : a) Additional. b) Subtraction. c) Products.

2. Every operation can be solved by using 2 general methods : a) Graphical method (scaled diagram). b) Mathematical method (vector resolution).

Addition and Subtraction of Vectors :

1. Consider two vectors, A and B as follow :

Note : • Negative vector means the vector of the same magnitude (same length of the

arrow) but directed opposite to that of its original vector. 2. Graphical Method :

Vector A and B can be added or subtracted graphically by using the following methods :

a) Parallelogram Method : The resultant vector is represented by the diagonal of the parallelogram (with proper scale).

R1 = A + B R2 = A – B = A + (-B) Note :

• Make sure the tip of vector A and B (and all individual vectors) is concentrated at the same point (origin).

b) Tip to Tail (poligon) Method :

The resultant vector is represented by an arrow which is connecting the tip of the first vector to the tail of the last vector.

R1 = A + B R2 = A – B = A + (-B)

Note :

• Make sure the tail of the 1st vector is connected to the tip of the 2nd vector and so on.

A B

A

B

- A

- B

A

-B

R2

R1

A B

A

-B

R2 R1


Note : Example of ‘Tip to Tail’ Method. The rules of vector addition operation :

Rules Formula

Commutative A + B = B + A

Associative (A + B) + C = A + (B + C)

Distributive n(A + B) = nA + nB

Example : Commutative Rule

B

A A + B

B

A B + A


Example : Associative Rule

Example : Distributive Rule

3. Mathematical Method : Vector A and B also can be added or subtracted by using vector resolution. Basically, a vector can be resolved into its components in two ways :

a) Components in the x, y, z axes.

Resolving Vector Into 2 Perpendicular Components : A = Ax + Ay

where mathematically the magnitude of the resolved vector : | Ax | = | A | cosθ

| Ay | = | A | sinθ with θ is called resolved angle

upon x-axis. and mathematically the magnitude of the resultant vector :

| A | = 2| |y

2

x A||A +

with θ = tan-1

|A|

|A|

x

y


B

(A + B) C (A + B)

(A + B) + C

B (B + C)

C

A

A

A + (B + C)

(B + C)

B 1.5 (A + B)

A

1.5B

(1.5A + 1.5B)

1.5A

(A + B)


Resolution of vectors in each quadrant :

Quadrant x-component y-component

I Ax = A cosθ1 Ay = A sinθ1

II Bx = -B cosθ2 = B cos(180o - θ2) By = B sinθ2 = B sin(180

o - θ2)

III Cx = -C cosθ3 = C cos(180o + θ3) Cy = -C sinθ3 = C sin(180

o + θ3)

IV Dx = D cosθ4 = C cos(360o - θ4) Dy = -D sinθ4 = C sin(360

o - θ4)

b) Components in the i, j, k unit vectors. A vector quantity A can also represented by an identical vector as follows :

A = Axi + Ayj …( 2D ) Or A = Axi + Ayj + Azk …( 3D )

where i, j dan k are called as the unit vector.

Note : • All unit vectors have magnitude equal to 1 and direction as follows :

a) i in the positive x direction. b) j in the positive y direction. c) k in the positive z direction.

• Relationship between vector components and identical vectors :

Ax = Axi Ay = Ayj Az = Azk

x

y

θ1 θ2

θ4 θ3

A B

C D

i

j

Ax

Ay


Two (2) vector are easier to be added as well as to be subtracted by expressing in their unit vector as follows :

If A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk

then :

A + B = (Axi + Ayj + Azk) + (Bxi + Byj + Bzk) = (Ax + Bx)i + (Ay + By)j + (Az + Bz)k A - B = (Axi + Ayj + Azk) - (Bxi + Byj + Bzk) = (Ax - Bx)i + (Ay - By)j + (Az - Bz)k

Vector Products

1. There are two (2) types of vector product :

a) Scalar product. b) Vector product.

2. Scalar ( or Dot ) Product : = Multiplication of a vector by the same or other vector, as a result of which produces a scalar.

A • B = (Axi + Ayj + Azk) • (Bxi + Byj + Bzk)

= (AxBx) + (AyBy) + (AzBz)

The product also can be written as :

A • B = | A || B | cosθAB ; where θAB = angle between A and B

Note : The scalar product is commutative : A • B = B • A

Contextual Examples : Work, W = F • s

3. Vector ( or Cross ) Product : = Multiplication of a vector by the same or other vector, as a result of which produces a new vector.

A x B =

zyx

zyx

BBB

AAA

kji

= (AyBz – ByAz)i – (AxBz – BxAz)j + (AxBy – BxAy)k

The product also can be written as :

A x B = | A || B | sinθAB ; where θAB = angle between A and B

The direction of the vector product can be determined by two ways :

a) Right Hand Rule. b) Corkscrew Rule.


A

(1st Vector)

B (2

nd Vector)

A x B

(Vector Product)

θ

Right Hand Rule :

A (First Vector) Direction of First Finger

B (Second Vector) Direction of Second Finger

Vector Product Direction of Thumb

Corkscrew Rule :

Note : The vector product is not commutative : A x B = -B x A

Contextual Examples : Torque, τ = r X F Magnetic Force, F = qv X B


| Examples | Chapter 1 | Physical Quantities and Measurements |

Unit Conversion :

Example 1 : What is the difference between 1 m s-1 and 1 ms-1

Answer : a) 1 m s-1 is 1 meter per second.

↓ meter

b) 1 ms-1 is 1 per milisecond. ↓ mili

Example 2 :

What is the difference between m and m

Answer : a) m represents meter (unit of length) b) m represents mass (symbol of mass) Example 3 :

By using SI unit, explain the physical meaning of the following statements : a) A motor works at rate 1.61 hp. b) A man has weight of 143 lbs. c) An F1 car has maximum speed of 218 miles per hour (mph).

Answer :

a) P = 1.61(746 W) = 1200 W.

b) W = N2248.0

143

= 636.12 N (contributed by mass, m = 65 kg)

c) v = 218 mi per hour = hour1

mi218 =

s3600

)m10x609.1(218 3 = 97.43 m s-1.

Example 4 :

Complete the following unit conversion : a) 5 cm → km b) 0.5 mm2 → cm2 c) 50 µm3 → mm3


Answer :

a) 5 cm =

km1

km1xcm5 =

−

m10x1

km1xm10x5

32 = 5 x10-5 km

b) 0.5mm2 = (0.5 mm)(1 mm) = (0.5 x10-3 m) (1 x10-3 m)

= 5 x10-7 m2

= 5 x10-7 m2 x

2

2

cm1

cm1

= 5 x10-7 m2 x

)cm1)(cm1(

cm1 2

= 5 x10-7 m2 x

−− )m10x1)(m10x1(

cm122

2 = 5 x10-3 cm2

c) 50µm3 = (50 µm)(1µm)(1µm) = (50 x10-6 m)(1 x10-6 m) (1 x10-6 m) = 5 x10-17 m3

= 5 x10-17 m3 x

3

3

mm1

mm1

= 5 x10-17 m3 x

)mm1)(mm1)(mm1(

mm1 3

= 5 x10-17 m3 x

−−− )m10x1)(m10x1)(m10x1(

mm1333

3 = 5 x10-8 mm3

Dimension :

Example 1 :

Which of the following equations is dimensionally homogenous? a) s = ut + ½ at b) s = ut + ½ at

2

c) s = 2ut + at2

Answer : a) The equation is dimensionally homogenous if : [ s ] = [ ut ] = [ ½ at ] Then : [ s ] = L [ ut ] = [ u ][ t ] = (LT-1)(T) = L [ ½ at ] = [ ½ ][ a ][ t ] = (1)(LT-2)(T) = LT-1 Since [ s ] = [ ut ] ≠ [ ½ at ] → The equation is not homogenous (physically invalid).


b) The equation is dimensionally homogenous if : [ s ] = [ ut ] = [ ½ at

2 ] Then : [ s ] = L [ ut ] = [ u ][ t ] = (LT-1)(T) = L [ ½ at

2 ] = [ ½ ][ a ][ t ]2 = (1)(LT-2)(T)2 = L Since [ s ] = [ ut ] = [ ½ at

2 ] → The equation is homogenous (and physically valid). c) The equation is dimensionally homogenous if : [ s ] = [ 2ut ] = [ at

2 ] Then : [ s ] = L [ 2ut ] = [ 2 ][ u ][ t ] = (1)(LT-1)(T) = L [ at

2 ] = [ a ][ t ]2 = (LT-2)(T)2 = L Since [ s ] = [ ut ] = [ ½ at

2 ] → The equation is homogenous (but physically invalid). Note : - Any equation which is not homogenous definitely invalid physically. - A homogenous equation however is not necessary physically valid. - The validity of a homogenous equation can only be determine experimentally. Example 2 :

Using dimensional analysis, derive an expression for the gravitational force, F of an object which shows how F is related to the mass, m of the object, the mass, M of the earth and the distance, r between the centre of both masses.

Example 3 : The rate of heat flow along a 1D metal rod is given by :

t

Q= - )T(T

x

kAif −

where A = area of cross sectional of the metal rod. Tf = temperature at the cold end. Ti = temperature at the hot end. x = length of the rod. k = thermal conductivity of the metal.

a) Determine the dimensional expression of k. b) Determine the SI unit of k.


Vector Resolution : Example 1 : Two forces act upon a body at a point O as shown in FIGURE 1.

FIGURE 1

a) Resolve these forces along the y-axis and x-axis respectively. b) Determine : i) the sum of both components. ii) the resultant force (including magnitude and direction) of the system. Answer :

a) x-component : Ax = +A cosθ1 = 10 cos(90 – 25)

o = 4.23 N Bx = -B cosθ2 = 30 cos(15

o) = -28.98 N y-component : Ay = +A sinθ1 = 10 sin(90 – 25)

o = 9.06 N By = +B sinθ2 = 30 sin(15

o) = 7.76 N b) i) Sum of x-component, ∑Fx = 4.23 + (-28.98) = -24.75 N Sum of y-component, ∑Fy = 9.06 + 7.76 = 16.82 N Note : - Question (a) and (b) actually can be simplified using a tabling method as follow :

Vector x-component y-component

A B

Ax = +A cosθ1 = 10 cos(90 – 25)o = 4.23 N

Bx = -B cosθ2 = -30 cos(15o) = -28.98 N

Ay = +A sinθ1 = 10 sin(90 – 25)o = 9.06 N

By = +B sinθ2 = 30 sin(15o) = 7.76 N

∑Fx = -24.75 N ∑Fy = 16.82 N

ii) Resultant force, | F | = ( ) ( )2y2

x FF ∑∑ + = ( ) ( )22 82.1675.24 +− = 29.92 N

Angle, θ = tan-1

∑∑

x

y

F

F = tan-1

− 75.24

82.16 = -34.20o (quadrant II)

y

x

A = 10N

B = 30N 250

150

O


Or graphically presented as follow :

Example 2 :

Two vectors A and B respectively can be represented as follow :

A = 5i - 6j + 7k B = 2i + 3j - 4k

Calculate :

a) A + B b) A - B c) m(A + B) ; where m = 3 d) (A - B) ; where m = 3 m e) A • B and what do you think if B • A f) A X B and what do you think if B X A g) angle θ between A and B

Answer : a) A + B = ( 5i - 6j + 7k ) b) A - B = A + (-B) = ( 5i - 6j + 7k ) + ( 2i + 3j - 4k ) + ( -2i - 3j + 4k )

7i - 3j + 3k 3i - 9j + 11k

c) Based in question (a) : 3(A + B) = 3 ( 7i - 3j + 3k ) = 21i - 9j + 9k

d) Based in question (a) : ( )

3

BA + = )BA(

3

1+ =

3

1( 7i - 3j + 3k )

= 3

7i - j + k

e) A • B = ( 5i - 6j + 7k ) B • A = ( 2i + 3j - 4k ) •( 2i + 3j - 4k ) • ( 5i - 6j + 7k ) = [10( i•i)] + [-18( j•j)] + [-28(k•k)] = [10( i•i)] + [-18( j•j)] + [-28(k•k)] = [10(1)] + [-18(1)] + [-28(1)] = [10(1)] + [-18(1)] + [-28(1)] = -36 = -36

It shows that A • B = B • A

y

x

F = 29.92 N

34.20

O

∑Fy = 16.82 N

∑Fx = 24.75 N


f) A X B =432

765

kji

−

−

= [(-6)(-4) – (3)(7)] i - [(5)(-4) – (2)(7)] j + [(5)(3) – (2)(-6)] k = [(-24) – (21)] i - [(-20) – (14)] j + [(15) – (-12)] k = 3i - 34j + 33k

B X A =765

432

kji

−

−

= [(3)(7) – (-6)(-4)] i - [(2)(7) – (5)(-4)] j + [(2)(-6) – (5)(3)] k = [(21) – (24)] i - [(14) – (-20)] j + [(-12) – (15)] k = -3i + 34j - 33k It shows that A X B = -( B X A) g) Based on equation | A • B | = |A||B|cosθ

θ = cos-1 BA

|BA| •

= cos-1

−++

+−+

−

222222 )4()3()2()7()6()5(

36

= cos-1 )39.5)(49.10(

36− = = cos-1 (0.637)

= 129.57o

Example 3 : Three vectors A, B and C respectively can be represented as follow :

A = 5i - 6j + 7k B = 2i + 3j - 4k

C = 3i - 2k Show that : a) A + B = B + A … to obey the ‘comutative rule’. b) (A + B) + C = A + (B + C) … to obey the ‘associative rule’. c) 2(A + B) = 2A + 2B … to obey the ‘distributive’ rule. Answer : a) A + B = ( 5i - 6j + 7k ) B + A = ( 2i + 3j - 4k ) + ( 2i + 3j - 4k ) + ( 5i - 6j + 7k )

7i - 3j + 3k 7i - 3j + 3k

It shows that the ‘comutative rule’ is obeyed.


b) Based on equation (a) : A + B = 7i - 3j + 3k Then, (A + B) + C = ( 7i - 3j + 3k ) + ( 3i + 0j - 2k ) = 10i - 3j + k B + C = ( 2i + 3j - 4k ) + ( 3i + 0j - 2k ) = 5i + 3j - 6k Then, A + ( B + C ) = (5i - 6j + 7k) + (5i + 3j - 6k) = 10i - 3j + k It shows that the ‘associative rule’ is obeyed. c) Based on question (a) : A + B = 7i - 3j + 3k Then, 2(A + B) = 2(7i - 3j + 3k) = 14i - 6j + 6k 2A + 2B = 2( 5i - 6j + 7k ) + 2( 2i + 3j - 4k) = ( 10i - 12j + 14k) + ( 4i + 6j - 8k) = 14i - 6j + 6k

It shows that the ‘distributive rule’ is obeyed.

Example 4 : Two vectors A and B respectively can be represented as follow :

A = 20 ‘unit’ (directed eastwards) B = 15 ‘unit’ (directed north-westwards)

Determine (by using resolved vectors into axes and unit vector) : a) A + B

b) A • B f) A X B Answer : The vectors can be presented graphically as follow. a) Method 1 : Resolved into axes :

Vector x-component y-component

A B

Ax = +A cosθ1 = 20 cos0o = 20.0 unit

Bx = -B cosθ2 = -15 cos(45o) = -10.61 unit

Ay = +A sinθ1 = 20 sin0o = 0

By = +B sinθ2 = 15 sin(45o) = 10.61 unit

Sum 9.39 unit 10.61 unit

Hence, A + B = Resultant vector = ( ) ( )22 61.1039.9 + = 14.17 unit

Angle, θ = tan-1

39.9

61.10 = 48.49o (quadrant I)

x A = 20 unit

B = 15 unit

450

O

y


Method 2 : Unit vector : From the table, the vectors can be written as :

A = 20i + 0j + 0k B = -10.61i + 10.61j + 0k

Hence, A + B = ( 20i + 0j + 0k) + ( -10.61i + 10.61j + 0k ) = 9.39i + 10.61j + 0k

with the magnitude, A + B = ( ) ( )22 61.1039.9 + = 14.17 unit

and angle, θ = tan-1

39.9

61.10 = 48.49o (quadrant I)

b) Method 1 : Resolved into axes : Angle between A and B = 180o – 45o = 135o. Hence, A • B = A B cosθ = (20)(15)cos(135º) = -212.13 unit

Method 2 : Unit vector :

A • B = ( 20i + 0j + 0k ) • ( -10.16i + 10.61j + 0k ) = [-101.6( i•i)] + [0( j•j)] + [0(k•k)]

= -212.13 unit c) Method 1 : Resolved into axes : Angle between A and B = 180o – 45o = 135o. Hence, A X B = A B sinθ = (20)(15)sin(135º) = 212.13 unit

Direction : Based on above figure, and by using right hand rule or corkscrew rule outwards the plane. positive z axis.

Method 2 : Unit vector :

A X B =061.1061.10

0020

kji

−

= [0] i - [0] j + [(20)(10.61) – (0)] k = 0 i - 0 j + 212.13 k = 212.13k (magnitude = 212.13 unit, direction = positive z axis)


MEASUREMENT & UNCERTAINTY

(This topic is being discussed during practical session)

1.0 Introduction :

Measurements are trials to determine the actual value of a particular physical quantity. The difference between the actual value of a quantity and the value obtained in measurement is called error. To measure is to make an acceptable estimate. A suitable and even the best measuring instrument (apparatus) must be determined to be used. 2.0 Properties of Measurement :

A measurement depends on following properties :

1. Accuracy of the apparatus.

= An ability of the apparatus to give readings close (or almost equal) to the actual value of a quantity.

2. Sensitivity of the apparatus.

= An ability of the apparatus to respond (or to detect) a small change in the value of a measurement.

3. Consistency of the apparatus.

= An ability of the apparatus to register the same (almost the same) reading when a measurement is repeated.

High consistency → Small deviation from the mean value. Example :

Equipment Range Sensitivity

Metre ruler (0.0 - 100.0)cm 1mm Vernier caliper (0.0 – 150.0)mm 0.1mm Micrometer screw gauge (0.00 – 25.00)mm 0.01mm Spherometer (0.00 – 7.00)mm 0.01mm

Length

Traveling microscope (0.00 – 220.00)mm 0.01mm Stop watch (analog) (0.0 – 900.0)s 0.1s

Time Timer (0.000000 – 100.000000)s 0.000001s

The use of Range of a measuring apparatus : To determine the suitable apparatus can be used to measure a reading. An apparatus is suitable to measure a reading if the range is grater than the sample. Example : The suitable apparatus to measure the thickness of a reference book are metre ruler, vernier calipers and micrometer screw gauge. The use of Sensitivity of a measuring apparatus : To determine the best apparatus can be used to measure a reading. An apparatus is the best apparatus if it has the smallest division of the scale. Example : The best apparatus to measure the thickness of a reference book is micrometer screw gauge.


0 5 10 15 20

mm 0

45

0.01mm

1

2

0 5 10 15 20

mm 25

20 0.01mm

1

2

2.1 Taking reading skill :

1. Vernier caliper : Example : Main scale = 1.1 cm Vernier scale = 0.04 cm (i.e. 4 x 0.01 cm) Actual reading = 1.14 cm 2. Micrometer screw gauge : Example : Main scale = 11.0 mm Vernier scale = 0.48 mm (i.e. 48 x 0.01 mm) Actual reading = 11.48 mm Example : Main scale = 7.5 mm Vernier scale = 0.24 mm (i.e. 24 x 0.01 mm) Actual reading = 7.74 mm

0 1 2

cm

3

0.01cm

1

2


7

0

-7

2

1

mm

0.01mm 30

0 10

20

40 50

60

70

80

90

7

0

-7

1

mm

0.01mm 20

0

10

30 40

50

60

70

80 90

2

3. Spherometer : Example : Main scale = 3.0 mm (upwards) Vernier scale = 0.27 mm (i.e. 27 x 0.01 mm) Actual reading = 3.27 mm (height) Example : Main scale = - 4.0 mm (downwards) Vernier scale = - 0.87 mm (opposite direction) Actual reading = - 4.87 = 4.87 mm ( -ve = depth)


4. Traveling microscope : Example : Main scale = 7.1 cm = 71.0 mm Vernier scale = 0.41 mm Actual reading = 71.41 mm Example : Main scale = 7.25 cm = 72.5 mm Vernier scale = 0.21 mm (lower scale) Actual reading = 72.71 mm OR Main scale = 7.20 cm = 72.0 mm Vernier scale = 0.71 mm (upper scale) Actual reading = 72.71 mm

↓ .1 .2 .3 .4 .5

.6 .7 .8 .9 1.0

0.01mm

7 8 9

cm

1

2

↓ .1 .2 .3 .4 .5

.6 .7 .8 .9 1.0

0.01mm

7 8 9

cm

1

2


3.0 Error :

There is no such thing as a perfect measurement! All measurements have errors and uncertainties, no matter how hard we might try to minimize them. Error and uncertainty in a measurement can arise from three possible origins: the measuring device, the environment and the observer. Basically, there are two different types of errors : SYSTEMATIC ERROR & RANDOM ERROR

Systematic Error

Characteristics The reading is always bias in one specific direction

(greater / less than the actual value).

Sources

Device 1. Zero error. 2. Fault in the device (metal expands when the temperature increases).

Environment 1. Gravity is not a constant. 2. Uniform air resistance

Observer 1. Reaction varies from one person to another : - reaction time. - short / long-signed.

Precaution Use perfect device.

1. Use correct value of gravity. 2. Create a ‘closed system’.

1. Reaction almost constant for a particular person (carry out carefully).

Random Error

Characteristics The reading is unbiased

(can be greater / can be less than the actual value).

Sources

Device 1. Quivering pointer (ammeter, voltmeter).

Environment 1. Changes in the surrounding during experiment).

Observer 1. Parallax error. 2. Wrong count (number of oscillation).

Precaution Taking several precise readings and then calculate the mean.

The way to overcome the zero error :

- Actual reading = (Final reading – zero error) - Value of zero error = difference value between 0 and the initial reading. - Sign of zero error : → +ve (if initial reading is greater than 0). → -ve (if initial reading is less than 0). Example : Zero error = – 0.02mm Zero error = + 0.04mm Then, actual reading : Then, actual reading : = (Final reading – ( – 0.02))mm = (Final reading – ( + 0.04))mm = (Final reading + 0.02))mm = (Final reading – 0.04))mm

0 5 10 15 20 mm

0

45

0.01mm

0 5 10 15 20 mm

5

0

0.01mm


4.0 Significant Figure / Digit :

The number of significant figures in the measurement of a physical quantity : = the number of digits before the estimated digit + the estimated digit itself. Example : By using vernier caliper, the diameter of a marker pen is recorded as, d = ( 2.74 + 0.01) cm.

The degree of precision that a measured value possesses in a number of significant figures carries by the value. The most significant figures a value carries, the higher is the degree of precision. There are several rules which detail how many significant figures a number has :

1. All non-zero digits are significant. (e.g. 3.12 has 3 s.f., 45.229 has 5 s.f.)

2. Zeros between non-zero digits are significant. (e.g. 3.012 has 4 s.f., 45.0009 has 6 s.f.)

3. Zeros beyond the decimal point at the end of a number are significant. (e.g. 3.340 has 4 s.f.)

4. Zeros preceding the first non-zero digit are not significant. There are merely placeholders. (e.g., 0.0034 has only 2 s.f.)

5. Digits in the exponent in exponential notation are not significant. (e.g. 1.34 x107 has 3 s.f. (1.34) not 4)

Processing significant figures :

(a) Additional and subtraction : When two or more measured values are added and/ or subtracted, the final calculated value must have the same number of decimal places as that measured value which has the least number of decimal places.

Example : 2.345 cm + 1.25 cm = 3.595 cm → 3.60 cm. (b) Multiplication and division :

When two or more measured values are multiplied/ divided, the final calculated value must have as many significant figures as that measured value which has the least number of significant figures.

Example : 2.345 cm X 1.25 cm = 2.93125 cm2 → 2.93 cm2

Note : Sometimes, the final answer may be obtained only after performing several intermediate

calculations. In this case results produced in intermediate calculations need not be rounded off. Round off only the final answer. Rounded off values too early may result in greater cumulative errors.

Two digit before the estimated digit

One estimated digit (i.e. digit 4)

Number of significant figures = (2+1) = 3


5.0 Uncertainty :

In an experiment, measuring a physical quantity is important but not all measured values are exactly same as the actual values. It is because of errors we made and the apparatus we used may not in perfect conditions. Therefore, the uncertainty of a measurement must been taken, so that the information about the accuracy of a measurement can be obtained and it has to be recorded together with the result of the experiment. For a quantity (best value), x with the uncertainty, ∆x, its measurement is recorded as below:

- relative uncertainty = x

x∆

- percentage uncertainty = x

x∆x100%.

The result should be written as (x + ∆x) ‘unit’.

Basically, there are four different types of uncertainty:

1. Uncertainty in a single reading. 2. Uncertainty in repeated readings. 3. Uncertainty in a straight line graph. 4. Uncertainty in a function.

a) Single reading :

o If the reading is taken from a single point or at the end of the scale we used : ∆x = ½ x (the smallest division from the scale).

o If the reading are taken from two points on the scale :

∆x = 2 x ½ x (the smallest division from the scale).

o If the apparatus is using the vernier scale : ∆x = 1 x (the smallest division from the scale).

Equipment / apparatus

Type of reading

Smallest scale

Example of reading

Thermometer Metre ruler Stopwatch

Micrometer screw gauge Vernier caliper

Timer

Single point Two points Two points Vernier scale Vernier scale Vernier scale

0.1oC 0.1cm 0.1s

0.01mm 0.1mm

0.000001s

(37.50 + 0.05)oC (24.0 + 0.1)cm (23.9 + 0.1)s

(2.43 + 0.01)mm (2.4 + 0.1)mm

(23.972659 + 0.000001)s

b) Repeated reading :

For a set of n repeated measurements, the best value is their average value.

<x> = n

xn

1ii∑

= , ∆x = n

|xx|n

1ii∑

=

−><


Example :

No Time (+ 0.1s) 1 2 3

10.3 10.1 10.4

Best value (mean), <t> = 3

4.101.103.10 ++= 10.26667s = 10.3s

Uncertainty ∆t = 3

|4.103.10||1.103.10||3.103.10I −+−+−= 0.1s

The result, t = (10.3 + 0.1)s

c) Straight line graph :

When plotting a straight line graph, the line always does not pass through all the points. Therefore it is important to determine the uncertainties ∆m and ∆c for the gradient of the graph, m and the y-interception respectively. Method to determine ∆m and ∆c : - Consider the data obtained are :

X x1 x2 x3 x4 x5 x1 x1 xn Y y1 y2 y3 y4 y5 x1 x1 yn

- Find the centroid, ( )y,x :

x = n

xn

1ii∑

= , y = n

yn

1ii∑

=

- Draw the best straight line through all the point and the centroid. - Draw the second and third straight lines to obtain the maximum and minimum gradients. Hence :

∆m = 2

mm minmax − and ∆c = 2

cc minmax −

…………. ………….

y

x

centroid

m

mmax

mmin

cmax

cmin c


d) Functions : Basically, the uncertainty for any function is represented by a formula as follow :

If the function is r = f(x,y,z,…) then the uncertainty :

∆r = zdz

dry

dy

drx

dx

dr∆+∆+∆

Function Best Value Uncertainty

Addition / subtraction r = x + y + z ∆r = ∆x + ∆y + ∆z

Multiplication with constant k r = kx ∆r = k∆x

Multiplication / division r = z

xy ∆r = r

z

z

y

y

x

x

∆+

∆+

∆

Index r = xn ∆r = n rx

x

∆


Example :

If the function is g = 2mt2

1

Then, the uncertainty : g = f(m,t)

∴ ∆g = tdt

dgm

dm

dg∆+∆ …..(i)

dm

dg = )g(

dm

d

dt

dg = )g(

dt

d

=

2mt2

1

dm

d =

2mt2

1

dt

d

= )m(dm

d

t2

1 12

−

= )t(

dt

d

m2

1 2−

= )m(t2

1 22

−−

= )t2(

m2

1 3−−

= -

22mt2

1 …..(ii) = -

mt

13

…..(iii)

Substitute (ii) and (iii) into (i), we obtain :

∴ ∆g = mmt2

122∆

+ t

mt

13

∆

Or ∆g = m

m

mt2

12

∆

+

t

t

mt

1

2

22

∆

∆g =

∆+

∆

mt2

1

t

t2

m

m2

∴ ∆g = gt

t2

m

m

∆+

∆#

lecture note 1_1

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