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1 CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT 1.1 BASE/FUNDAMENTAL QUANTITIES Definition: A base or fundamental quantity is a quantity that stand on its own and is not a product from the combination of other quantities. Name of Base/Fundamental Quantity Symbol SI Unit 1. Length L/s metre(m) 2. Mass m kilogram(kg) 3. Time t saat(s) 4. Temperature T kelvin(K) 5. Electric current I ampere(A) 6. Amount of substance mol(mol) 7. Luminous intensity I v candela(cd) 1.2 DERIVED QUANTITIES Definition: A derived quantity is a quantity that is a product from the combination of base quantities. Examples: Name of Derived Quantity Symbol Formula SI Unit 1. Area A A = P x L m 2 , cm 2 , mm 2 2. Volume V V = P x L x T m 3 , cm 3 , mm 3 3. Velocity v v = s / t ms -1 , kmj -1 4. Acceleration a a = ∆v / t ms -2 5. Density ρ ρ = m / V kgm -3 , gcm -3 6. Pressure p p = F / A Nm -2 , pascal(Pa) 7. Work W W = F x s Nm, joule(J) 8. Power P P = W / t Js -1 , watt(W) 9. Energy E E = mc 2 J BB101 – ENGINEERING SCIENCE

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1CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

1.1 BASE/FUNDAMENTAL QUANTITIES

Definition: A base or fundamental quantity is a quantity that stand on its own and is not a product from the combination of other quantities.

Name of Base/Fundamental Quantity Symbol SI Unit

1. Length L/s metre(m)

2. Mass m kilogram(kg)

3. Time t saat(s)

4. Temperature T kelvin(K)

5. Electric current I ampere(A)

6. Amount of substance mol(mol)

7. Luminous intensity Iv candela(cd)

1.2 DERIVED QUANTITIES

Definition: A derived quantity is a quantity that is a product from the combination ofbase quantities.

Examples:

Name of Derived Quantity Symbol Formula SI Unit1. Area A A = P x L m2 , cm2, mm2

2. Volume V V = P x L x T m3, cm3, mm3

3. Velocity v v = s / t ms-1, kmj-1

4. Acceleration a a = ∆v / t ms-2

5. Density ρ ρ = m / V kgm-3, gcm-3

6. Pressure p p = F / A Nm-2, pascal(Pa)

7. Work W W = F x s Nm, joule(J)

8. Power P P = W / t Js-1, watt(W)

9. Energy E E = mc2 J

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2CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

1.3 UNIT CONVERSION OF BASE AND DERIVED QUANTITIES

Prefix of multiplication factor is used for a very large or small value of quantity so that it could be written briefly and easily understood.

Prefix(symbol) Multiplication Factor Numerical Value Examples

tera ( T ) 1012 1 000 000 000 000 2 TN (teranewton)

giga ( G ) 109 1 000 000 000 5 Gm (gigameter)

mega ( M ) 106 1 000 000 13 MW (megawatt)

kilo ( k ) ↑ 103 1 000 18 kg (kilogram)

centi ( c ) ↓ 10-2 0.01 55 cm (centimeter)

mili ( m ) 10-3 0.001 24 mm (milimeter)

micro ( µ ) 10-6 0.000 001 89 µm (micrometer)

nano ( n ) 10-9 0.000 000 001 17 nm (nanometer)

pico ( p ) 10-12 0.000 000 000 001 21 pm (picometer)

1.3.1 EXAMPLES OF UNIT CONVERSION FOR BASE QUANTITIES

a) LENGTH

Example 1:

The distance of Kota Bharu city to Kuala lumpur city is 474 km. Find the distance in the units of meter and centimetre.

Solution:

1 km = 1000 m(103m), 1 m = 100 cm(102 cm)

474 km = 474 km x 1000 m = 474 000 m = 4.74 x 10 5 m 1 km

474 km = 4.74 x 105 m x 10 2 cm = 4.74 x 10 7 cm 1 m

Example 2:

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3CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

A student is 160 cm tall. Determine his height in kilometre and millimetre.

Solution:

1 km = 1000 m(103 m), 1 m = 100 cm(102 cm), 1cm = 10 mm

160 cm = 160 cm x 1 m_ _ x 1 km = 1.6 x 102 x 10-5 = 1.6 x 10 -3 km 102 cm 103m

160 cm = 160 cm x 10 mm = 1600 mm 1 cm

Example 3:

The distance from planet Earth to Sun is 1.5 x 108 km. Find the distance in megameter(Mm)?

Solution:

1 Mm = 106 m, 1 km = 103 m

1.5 x 108 km = 1.5 x 108 km x 10 3 m x 1 Mm = 1.5 x 10 5 Mm 1 km 106 m

b) MASS

Example 1:

A bag of cement has a mass of 50 kg. Convert the unit to gram(g)?

Solution:

1 kg = 1000 g(103 g)

50 kg = 50 kg x 1000 g = 50 000 g = 5.0 x 10 4 g 1 kg

Example 2:

A wooden block has a mass of 380 g. Convert the unit to kilogram(kg)?

Solution:

380 g = 380 g x 1 kg__ = 0.38 kg 1000 g

Example 3:

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4CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

The mass of a lorry is 2 tons. Determine its mass in gram(g)?

Solution:

1 ton = 1000 kg(103 kg), 1 kg = 1000 g(103 g)

2 ton = 2 ton x 10 3 kg x _10 3 g_ = 2 x 10 6 g 1 ton 1 kg

c) TIME

Example 1:

Convert 8 hours to minute?

Solution:

1 h = 60 minutes, 1minute = 60 s

8 hr = 8 hr x 60 minutes = 480 minutes 1 hr

8 hr = 480 min x 60 s = 28 800 s 1 min

Example 2:

Convert 300 seconds to hour?

Solution:

300 s = 300 s x 1 hr = 0.083 hr 3600 s

Example 3:

An aeroplane took off from Kuala Lumpur airport at 1400 hours and arrived at New Delhi airport at 1830 hours. Calculate the duration of the journey in minute and second?

Solution:

Duration, t = 1830 – 1400 = 4 hr 30 min

t = (4 x 60 minutes) + 30 minutes = 240 + 30 = 270 minutes

and, t = 270 min x 60 s = 16 200 s 1 min

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5CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

1.3.2 EXAMPLES OF UNIT CONVERSION FOR DERIVED QUANTITIES

a) AREA

Example 1:

The floor area of a science laboratory is 40 m2. Find the area in cm2 and mm2?

Solution:

1 m = 102 cm (100 cm)

1 m2 = (102 cm)2 = 104 cm2

40 m2 = 40 m2 x 10 4 cm 2 = 40 x 104 cm2 = 4 x 10 5 cm 2 1 m2

and 1 cm = 10 mm

1 cm2 = (10 mm)2 = 102 mm2

40 m2 = 4 x 105 cm2 x 10 2 mm 2 = 4 x 10 7 mm 2 1 cm2

Example 2:

The measurements of A4 size paper is 300 mm long and 210 mm wide. Find the area of the paper in m2?

Solution:

1 m = 100 cm, 1 cm = 10 mm, because of that,

1m = 100 x 10 mm = 1000 mm (103 mm)

L = 300 mm = 300 mm x 1 m = 0.30 m 1000 mm

W = 210 mm = 210 mm x 1m = 0.21 m 1000 mm

Area of paper, A = L x W = 0.30 m x 0.21 m = 0.063 m 2

Example 3:

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6CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Convert the area of A4 paper in example 2 to cm2?

Solution:

P = 300 mm = 300/10 = 30 cm, L = 210 mm = 210/10 = 21 cm

A = P x L = 30 cm x 21 cm = 630 cm 2

b) VOLUME

Example 1:

A steel water tank could be filled with 25 m3 of water. What is its volume in cm3?

Solution:

1 m = 102 cm (100 cm), oleh itu,

1 m3 = (102 cm)3 = 106 cm3 atau 1 m3 = 100 cm x 100 cm x 100 cm = 106 cm3

Tank volume, V = 25 m3 = 25 m3 x 10 6 cm 3 = 25 x 10 6 cm 3 1 m3

Example 2:

A beaker has a volume of 7.68 x 104 mm3. Determine its volume in cm3?

Solution:

1 cm = 10 mm, so that,

1 cm3 = (10 mm)3 = 103 mm3 or,

1 cm3 = 10 mm x 10 mm x 10 mm = 103 mm

V = 7.68 x 104 mm3 = 7.68 x 104 mm3 x 1 cm 3 = 76.8 cm 3 103 mm3

Example 3:

Find the volume of a beaker in example 2 in m3?

Solution:

76.8 cm3 = 76.8 cm3 x 1 m 3 = 7.68 x 10 -5 m 3 106 cm3

c) VELOCITY

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7CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Example 1:

A van was driven at a speed of 90 km/h along a straight road. Convert the unit to m/s?

Solution:

1 km = 1000 m, 1 h = 3600 s

90 km/h = 90 km x 1000 m x 1 h = 25 m/s h 1 km 3600 s

Example 2:

The speed of sound in the air is 330 ms-1. Convert the unit to cm/min?

Solution:

1 m = 100 cm, 1 min = 60 s

330 ms-1 = 330 m x 100 cm x 60 s = 1.98 x 10 6 cm/min s 1 m 1 min

Example 3:

According to scientific studies, Tsunami waves move at a speed of 700 km/h. Are the waves faster or slower than the speed of sound?

Solution:

Speed of sound = 330 m/s, convert the unit to km/h,

330 m/s = 330 m x 1 km x 3600 s = 1080 km/h s 1000 m 1 h

Tsunami waves are slower than the sound.

Exercise 1:

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8CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

The speed of light in vacuum is 3 x 108 ms-1. What is its value in km/h?

Exercise 2:

An aeroplane flies at a constant speed of 700 km/h. Determine its speed in m/min. How far can it goes in 30 minutes? (Give your answer in metre)

Exercise 3:

A cheetah can run at a speed of 130 kph. Find its speed in ms-1?

d) ACCELERATION

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9CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Example 1:

The free fall acceleration of an object under the influence of Earth’s gravity is 9.81 ms-2. What is its value in km/min2?

Solution:

1 km = 1000 m, 1 min2 = (60 s)2 = 3600 s2

9.81 ms-2 = 9.81 m x 1 km x 3600 s 2 = 35.316 km/min 2 s2 1000 m 1 min2

Example 2:

The free fall acceleration of an object under the influence of Sun’s gravity is 274 ms-2. What is its value in cmh-2?

Solution:

1 m = 102 cm, 1 h2 = (3600s)2 = 1.296 x 107 s2

274 ms-2 = 274 m x 10 2 cm x 1.296 x 10 7 s 2 = 3.55 x 10 9 cmh -2 s2 1m 1 h2

Example 3:

A car increases its velocity at a rate of 28.8 km/min2. Determine its acceleration in m/s2?

Solution:

28.8 km/min2 = 28.8 km x 1000 m x 1 min 2 = 8 m/s 2 min2 1 km 3600 s2

Exercise 1:

The surface gravitational acceleration of a moon is 1.62 ms-2. Convert its unit to cm/min2?

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10CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Exercise 2:

Along a straight road, a car tries to over take a lorry by accelerating at a rate of 4 ms-2. Determine the acceleration in km/h2?

Exercise 3:

The acceleration of an electric car is 8 ms-2. Convert the unit to mm/min2?

e) DENSITY

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11CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Example 1:

The density of sea water is 1024 kgm-3. Find the density in gcm-3?

Solution:

1 kg = 103 g, 1 m3 = (102 cm)3 = 106 cm3

1024 kgm-3 = 1024 kg x 10 3 g x 1 m 3 = 1.024 gcm -3 m3 1 kg 106 cm3

Example 2:

The density of Mercury is 13.6 g/cm3. Convert its unit to Mg/mm3?

Solution:

1 Mg = 106 g, 1 cm3 = (10 mm)3 = 103 mm3

13.6 g/cm3 = 13.6 g x 1 Mg x 1 cm 3 = 13.6 x 10 -9 Mg/mm 3 cm3 106 g 103 mm3

Example 3:

If the density of mercury is 13 600 kgm-3, what is its density in kg/cm3?

Solution:

13 600 kgm-3 = 13 600 kg x 1 m 3 = 0.0136 kg/cm 3 m3 106 cm3

Exercise 1:

The mean density of Earth is 5.52 x 10 kgm-3. Determine its value in g/mm3?

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12CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Exercise 2:

If the density of palm oil is 0.86 g/cm3, what is its value in kg/m3?

Exercise 3:

Convert the density of palm oil in exercise 2 to kg/cm3?

1.4 MEASURING INSTRUMENTS

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13CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

The following are three types of instruments to measure the length of an object:

1. Metre Ruler

The length of a metre ruler is 1 metre and it is divided to 100 sections(100 cm). Each section is 1 cm long and it is further divided into 10 sections(10 mm). The scales of meter ruler are as follows:

1 m = 100 cm, 1 cm = 10 mm, 1 m = 1000 mm

The shortest length that can be measured with a metre ruler is 1 mm or 0.1 cm. Due to that, each measurement taken with a metre ruler must be recorded accurate to one decimal point or 0.1 cm. For example, the length of a section of a writing table is measured as 72 cm. So, its reading must be recorded as 72.0 cm or (72 ± 0.1) cm.

Picture #1: Samples of wooden Metre Rulers

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14CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

2. Vernier Calliper

A vernier calliper has two scales – consisting of a main scale and a vernier scale. The accuracy of the reading measured with a vernier calliper is up to 0.1 mm or 0.01 cm. Any reading measured with a vernier calliper must be recorded to two decimal point. For example, the diameter of a marble measured with a vernier calliper is found to be 2 cm sharp. Its measurement must be recorded as 2.00 cm.

The following examples illustrate the method of using those instruments:

Example 1: ( Figure 2.1)

The reading is 3.7 mm or 0.37 cm

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15CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Example 2: (Figure 2.2)

The reading is 15.8 mm or 1.58 cm

Example 3: (Figure 2.3)

The reading is 34.60 mm or 3.46 cm

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16CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Exercise 1:

Determine the measurements taken by a vernier calliper below:

1)

2)

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The reading is ………….. mm or …………. cm

The reading is ………….. mm or …………. cm

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17CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

3. Micrometer Screw Gauge

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18CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

A micrometer screw gauge consists of a main scale and a thimble scale. The main scale is callibrated in milimeter(mm) and each shortest section is 0.5 mm. Thimble scale has 50 small sections of equal length. Micrometer screw gauge reading is accurate up to 0.01 mm atau 0.001 cm. Its reading must be recorded up to three decimal points. For example, the diameter of a marble is measured to be 3 cm sharp, so the reading must be recorded as 3.000 cm.

The following examples illustrate the use of Micrometer Screw Gauge:

[Figure 3.1(a)]

Example 1: [Figure 3.1(b)]

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19CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Recorded measurement: 18.91 mm or 1.891 cm

Example 2:

Recorded measurement: 7.5 + 0.38 = 7.88 mm or 0.788 cm

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20CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Example 3:

Recorded measurement: 7.5 + 0.22 = 7.72 mm or 0.772 cm

Example 4:

Recorded measurement: 3.5 + 0.46 = 3.96 mm or 0.396 cm

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21CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Exercise 1:Fill in the blanks for the readings of a micrometer screw gauge below:

1)

2)

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Recorded measurement: ………. + ………. = ………. mm or ……….. cm

Recorded measurement: ………. + ………. = ………. mm or ……….. cm

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22CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

3)

1.4.1 RECORDING THE MEASUREMENT

The correct way of recording the measurement of the above instruments:

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Recorded measurement: ………. + ………. = ………. mm or ……….. cm

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23CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

END OF CHAPTER ( Questions )

Question 1

a) Explain the differences of base quantities and derived quantities? b) Name the five base quantities and their SI units. c) State the unit of force in terms of base SI units.

Question 2

Convert the units of the following quantities:

(a) A length of 15 meters to feet(b) A month with 30 days to seconds(c) A speed of 50 mph to m/s

[ Hint: 1 m = 3.28 ft, 1 mile = 1069 m ]

Question 3

Convert the units of velocities below:

i. 90 km/h to unit m/min

ii. 268 m/s to unit km/h

Question 4

Convert the units of accelerations below:

i. 10 m/s2 to unit cm/min2

ii. 20 cm/min2 to unit mm/s2

Question 5

Convert the units of pressures below:

i. 20 MN/m2 to unit N/cm2

ii. 40 kN/cm2 to unit N/mm2

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24CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT

Question 6

Convert the units of densities below:

i. 2400 kg/m3 to unit g/cm3

ii. 5 g/cm3 to unit kg/mm3

iii. 900 g/cm3 to unit kg/m3

Question 7

Two students measure the lengths of adjacent sides of their dorm room. One reports 15ft 8in. and the other reports 4.25 m. What is the area of the room in square meter?

[ 20.3 m2 ]

Question 8

A solid sphere has a radius of 12 cm. What is its surface are in (a) square centimetres, (b) square meters? (c) If it has a mass of 9 kg, what is its density in kg/m3?

[ Hint: Asphere = 4ᴫr2 , Vsphere = 4 ᴫr3 ]3

[ (a) 1.8 × 103 cm2, (b) 0.18 m2, (c) 1.2 × 103 kgm-3 ]

Question 9

Identify the measuring instruments used for the measurement of lengths below:

Measuring instrument Length

12.0 cm

8.75 cm

0.633 cm

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